CHEMISTRY Raymond Chang Williams College Kenneth A. Goldsby Florida State University CHEMISTRY, TWELFTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright 2016 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2013, 2010, and 2007. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 ISBN 978–0–07–802151–0 MHID 0–07–802151–0 Senior Vice President, Products & Markets: Kurt L. 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Library of Congress Cataloging-in-Publication Data Chang, Raymond. Chemistry.—Twelfth edition / Raymond Chang, Williams College, Kenneth A. Goldsby, Florida State University. pages cm Includes index. ISBN 978-0-07-802151-0 (0-07-802151-0 : alk. paper) 1. Chemistry—Textbooks. I. Goldsby, Kenneth A. II. Title. QD31.3.C38 2016 540—dc23 2014024893 The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites. www.mhhe.com  About the Authors Raymond Chang was born in Hong Kong and grew up in Shanghai and Hong Kong. He received his B.Sc. degree in chemistry from London University, and his Ph.D. in chemistry from Yale University. After doing postdoctoral research at Washington University and teaching for a year at Hunter College of the City University of New York, he joined the chemistry department at Williams College. Professor Chang has served on the American Chemical Society Examination Committee, the National Chemistry Olympiad Examination, and the Graduate Record Examination (GRE) Committee. He has written books on physical chemistry, industrial chemistry, and physical science. He has also coauthored books on the Chinese language, children’s pic- ture books, and a novel for young readers. For relaxation, Professor Chang does gardening, plays the harmon- ica, and practices the piano. Ken Goldsby was born and raised in Pensacola, Florida. He received his B.A. in chemistry and mathematical science from Rice University. After obtaining his Ph.D. in chemistry from the University of North Carolina at Chapel Hill, Ken carried out postdoctoral research at Ohio State University. Since joining the Department of Chemistry and Biochemistry at Florida State University in 1986, Ken has received several teaching and advising awards, including the Cottrell Family Professorship for Teaching in Chemistry. In 1998 he was selected as the Florida State University Distinguished Teaching Professor. Ken also works with students in his laboratory on a project to initiate collaborations between science depart- ments and technical arts programs. When he is not working, Ken enjoys hanging out with his family. They especially like spending time together at the coast. iii Contents in Brief 1 Chemistry: The Study of Change 1 2 Atoms, Molecules, and Ions 38 3 Mass Relationships in Chemical Reactions 75 4 Reactions in Aqueous Solutions 118 5 Gases 172 6 Thermochemistry 230 7 Quantum Theory and the Electronic Structure of Atoms 274 8 Periodic Relationships Among the Elements 326 9 Chemical Bonding I: Basic Concepts 368 10 Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals 412 11 Intermolecular Forces and Liquids and Solids 465 12 Physical Properties of Solutions 518 13 Chemical Kinetics 562 14 Chemical Equilibrium 621 15 Acids and Bases 666 16 Acid-Base Equilibria and Solubility Equilibria 720 17 Entropy, Free Energy, and Equilibrium 776 18 Electrochemistry 812 19 Nuclear Chemistry 862 20 Chemistry in the Atmosphere 900 21 Metallurgy and the Chemistry of Metals 930 22 Nonmetallic Elements and Their Compounds 956 23 Transition Metals Chemistry and Coordination Compounds 994 24 Organic Chemistry 1025 25 Synthetic and Natural Organic Polymers 1058 Appendix 1 Derivation of the Names of Elements A-1 Appendix 2 Units for the Gas Constant A-7 Appendix 3 Thermodynamic Data at 1 atm and 25°C A-8 Appendix 4 Mathematical Operations A-13 iv  Contents List of Applications xix List of Animations xx Preface xxi Setting the Stage for Learning xxix A Note to the Student xxxii CHAPTER 1 Chemistry: Chem The Study of Change 1 1.1 Chemistry: A Science for the Twenty-First Century 2 1.2 The Study of Chemistry 2 1.3 The Scientific Method 4 CHEMISTRY in Action The Search for the Higgs Boson 6 1.4 Classifications of Matter 6 1.5 The Three States of Matter 9 1.6 Physical and Chemical Properties of Matter 10 1.7 Measurement 11 CHEMISTRY in Action The Importance of Units 17 1.8 Handling Numbers 18 1.9 Dimensional Analysis in Solving Problems 23 1.10 Real-World Problem Solving: Information, Assumptions, and Simplifications 27 Key Equations 28 Summary of Facts & Concepts 29 Key Words 29 Questions & Problems 29 CHEMICAL M YS TERY The Disappearance of the Dinosaurs 36 v vi Contents CHAPTER 2 Atoms, A tom Molecules, and Ions 38 2.1 The Atomic Theory 39 2.2 The Structure of the Atom 40 2.3 Atomic Number, Mass Number, and Isotopes 46 2.4 The Periodic Table 48 CHEMISTRY in Action Distribution of Elements on Earth and in Living Systems 49 2.5 Molecules and Ions 50 2.6 Chemical Formulas 52 2.7 Naming Compounds 56 2.8 Introduction to Organic Compounds 65 Key Equation 67 Summary of Facts & Concepts 67 Key Words 67 Questions & Problems 68 CHAPTER 3 Mass M as Relationships in Chemical Reactions 75 3.1 Atomic Mass 76 3.2 Avogadro’s Number and the Molar Mass of an Element 77 3.3 Molecular Mass 81 3.4 The Mass Spectrometer 83 3.5 Percent Composition of Compounds 85 3.6 Experimental Determination of Empirical Formulas 88 3.7 Chemical Reactions and Chemical Equations 90 3.8 Amounts of Reactants and Products 95 3.9 Limiting Reagents 99 3.10 Reaction Yield 103 CHEMISTRY in Action Chemical Fertilizers 105 Key Equations 106 Summary of Facts & Concepts 106 Key Words 106 Questions & Problems 106  Contents vii CHAPTER 4 Reactions in Aqueous Solutions 118 4.1 General Properties of Aqueous Solutions 119 4.2 Precipitation Reactions 121 CHEMISTRY in Action An Undesirable Precipitation Reaction 126 4.3 Acid-Base Reactions 126 4.4 Oxidation-Reduction Reactions 132 CHEMISTRY in Action Breathalyzer 144 4.5 Concentration of Solutions 145 4.6 Gravimetric Analysis 149 4.7 Acid-Base Titrations 151 4.8 Redox Titrations 155 CHEMISTRY in Action Metal from the Sea 156 Key Equations 157 Summary of Facts & Concepts 158 Key Words 158 Questions & Problems 158 CHEMICAL M YS TERY Who Killed Napoleon? 170 CHAPTER 5 Gases 172 5.1 Substances That Exist as Gases 173 5.2 Pressure of a Gas 174 5.3 The Gas Laws 178 5.4 The Ideal Gas Equation 184 5.5 Gas Stoichiometry 193 5.6 Dalton’s Law of Partial Pressures 195 CHEMISTRY in Action Scuba Diving and the Gas Laws 200 5.7 The Kinetic Molecular Theory of Gases 202 CHEMISTRY in Action Super Cold Atoms 208 5.8 Deviation from Ideal Behavior 210 Key Equations 213 Summary of Facts & Concepts 214 Key Words 214 Questions & Problems 215 CHEMICAL M YS TERY Out of Oxygen 228 viii Contents CHAPTER 6 Thermochemistry Ther 230 6.1 The Nature of Energy and Types of Energy 231 6.2 Energy Changes in Chemical Reactions 232 6.3 Introduction to Thermodynamics 234 CHEMISTRY in Action Making Snow and Inflating a Bicycle Tire 240 6.4 Enthalpy of Chemical Reactions 240 6.5 Calorimetry 246 CHEMISTRY in Action White Fat Cells, Brown Fat Cells, and a Potential Cure for Obesity 250 6.6 Standard Enthalpy of Formation and Reaction 253 CHEMISTRY in Action How a Bombardier Beetle Defends Itself 256 6.7 Heat of Solution and Dilution 258 Key Equations 261 Summary of Facts & Concepts 261 Key Words 262 Questions & Problems 262 CHEMICAL M YS TERY The Exploding Tire 272 Quantum Q uan an Theory and the CHAPTER 7 Electronic Elec Structure of Atoms 274 7.1 From Classical Physics to Quantum Theory 275 7.2 The Photoelectric Effect 279 7.3 Bohr’s Theory of the Hydrogen Atom 282 7.4 The Dual Nature of the Electron 287 CHEMISTRY in Action Laser—The Splendid Light 288 7.5 Quantum Mechanics 291 CHEMISTRY in Action Electron Microscopy 292 7.6 Quantum Numbers 295 7.7 Atomic Orbitals 297 7.8 Electron Configuration 301  Contents ix 7.9 The Building-Up Principle 308 CHEMISTRY in Action Quantum Dots 312 Key Equations 313 Summary of Facts & Concepts 314 Key Words 315 Questions & Problems 315 CHEMICAL M YS TERY Discovery of Helium and the Rise and Fall of Coronium 324 Periodic Relationships CHAPTER 8 Among the Elements 326 8.1 Development of the Periodic Table 327 8.2 Periodic Classification of the Elements 329 8.3 Periodic Variation in Physical Properties 333 8.4 Ionization Energy 340 CHEMISTRY in Action The Third Liquid Element? 341 8.5 Electron Affinity 345 8.6 Variation in Chemical Properties of the Representative Elements 347 CHEMISTRY in Action Discovery of the Noble Gases 358 Key Equation 359 Summary of Facts & Concepts 359 Key Words 360 Questions & Problems 360 CHAPTER 9 Chemical Bonding I: Basic Concepts 368 9.1 Lewis Dot Symbols 369 9.2 The Ionic Bond 370 9.3 Lattice Energy of Ionic Compounds 372 CHEMISTRY in Action Sodium Chloride—A Common and Important Ionic Compound 376 9.4 The Covalent Bond 377 9.5 Electronegativity 380 9.6 Writing Lewis Structures 384 9.7 Formal Charge and Lewis Structure 387 x Contents 9.8 The Concept of Resonance 390 9.9 Exceptions to the Octet Rule 392 CHEMISTRY in Action Just Say NO 397 9.10 Bond Enthalpy 398 Key Equation 403 Summary of Facts & Concepts 403 Key Words 403 Questions & Problems 403 Chemical C hem Bonding II: Molecular Geometry CHAPTER 10 aand nd Hybridization of Atomic Orbitals 412 10.1 Molecular Geometry 413 10.2 Dipole Moments 423 CHEMISTRY in Action Microwave Ovens—Dipole Moments at Work 426 10.3 Valance Bond Theory 429 10.4 Hybridization of Atomic Orbitals 431 10.5 Hybridization in Molecules Containing Double and Triple Bonds 440 10.6 Molecular Orbital Theory 443 10.7 Molecular Orbital Configurations 446 10.8 Delocalized Molecular Orbitals 452 CHEMISTRY in Action Buckyball, Anyone? 454 Key Equations 456 Summary of Facts & Concepts 456 Key Words 456 Questions & Problems 457 IIntermolecular nter Forces and Liquids CHAPTER 11 aand nd Solids 465 11.1 The Kinetic Molecular Theory of Liquids and Solids 466 11.2 Intermolecular Forces 467 11.3 Properties of Liquids 473 CHEMISTRY in Action A Very Slow Pitch 475 11.4 Crystal Structure 477 CHEMISTRY in Action Why Do Lakes Freeze from the Top Down? 478 11.5 X-Ray Diffraction by Crystals 483  Contents xi 11.6 Types of Crystals 486 CHEMISTRY in Action High-Temperature Superconductors 488 CHEMISTRY in Action And All for the Want of a Button 492 11.7 Amorphous Solids 492 11.8 Phase Changes 493 11.9 Phase Diagrams 503 CHEMISTRY in Action Hard-Boiling an Egg on a Mountaintop, Pressure Cookers, and Ice Skating 505 CHEMISTRY in Action Liquid Crystals 506 Key Equations 508 Summary of Facts & Concepts 508 Key Words 509 Questions & Problems 509 CHAPTER 12 Physical Phys Properties of Solutions 518 12.1 Types of Solutions 519 12.2 A Molecular View of the Solution Process 520 12.3 Concentration Units 522 12.4 The Effect of Temperature on Solubility 527 12.5 The Effect of Pressure on the Solubility of Gases 529 CHEMISTRY in Action The Killer Lake 531 12.6 Colligative Properties of Nonelectrolyte Solutions 532 12.7 Colligative Properties of Electrolyte Solutions 544 CHEMISTRY in Action Dialysis 546 12.8 Colloids 546 Key Equations 549 Summary of Facts & Concepts 549 Key Words 550 Questions & Problems 550 CHEMICAL M YS TERY The Wrong Knife 560 xii Contents CHAPTER 13 Chemical C hem Kinetics 562 13.1 The Rate of a Reaction 563 13.2 The Rate Law 571 13.3 The Relation Between Reactant Concentration and Time 575 CHEMISTRY in Action Radiocarbon Dating 586 13.4 Activation Energy and Temperature Dependence of Rate Constants 588 13.5 Reaction Mechanisms 594 13.6 Catalysis 599 CHEMISTRY in Action Pharmacokinetics 606 Key Equations 608 Summary of Facts & Concepts 608 Key Words 609 Questions & Problems 609 CHAPTER 14 Chemical C hem Equilibrium 621 14.1 The Concept of Equilibrium and the Equilibrium Constant 622 14.2 Writing Equilibrium Constant Expressions 625 14.3 The Relationship Between Chemical Kinetics and Chemical Equilibrium 637 14.4 What Does the Equilibrium Constant Tell Us? 638 14.5 Factors That Affect Chemical Equilibrium 644 CHEMISTRY in Action Life at High Altitudes and Hemoglobin Production 651 CHEMISTRY in Action The Haber Process 652 Key Equations 654 Summary of Facts & Concepts 654 Key Words 655 Questions & Problems 655 CHAPTER 15 Acids A cid and Bases 666 15.1 Brønsted Acids and Bases 667 15.2 The Acid-Base Properties of Water 668 15.3 pH—A Measure of Acidity 670 15.4 Strength of Acids and Bases 673 15.5 Weak Acids and Acid Ionization Constants 677 15.6 Weak Bases and Base Ionization Constants 685 15.7 The Relationship Between the Ionization Constants of Acids and Their Conjugate Bases 687  Contents xiii 15.8 Diprotic and Polyprotic Acids 688 15.9 Molecular Structure and the Strength of Acids 692 15.10 Acid-Base Properties of Salts 696 15.11 Acid-Base Properties of Oxides and Hydroxides 702 15.12 Lewis Acids and Bases 704 CHEMISTRY in Action Antacids and the pH Balance in Your Stomach 706 Key Equations 708 Summary of Facts & Concepts 709 Key Words 709 Questions & Problems 709 CHEMICAL M YS TERY Decaying Papers 718 Acid-Base A cid Equilibria and Solubility CHAPTER 16 Equilibria Equi 720 16.1 Homogeneous versus Heterogeneous Solution Equilibria 721 16.2 The Common Ion Effect 721 16.3 Buffer Solutions 724 16.4 Acid-Base Titrations 730 CHEMISTRY in Action Maintaining the pH of Blood 732 16.5 Acid-Base Indicators 739 16.6 Solubility Equilibria 742 16.7 Separation of Ions by Fractional Precipitation 749 16.8 The Common Ion Effect and Solubility 751 16.9 pH and Solubility 753 16.10 Complex Ion Equilibria and Solubility 756 CHEMISTRY in Action How an Eggshell Is Formed 760 16.11 Application of the Solubility Product Principle to Qualitative Analysis 761 Key Equations 763 Summary of Facts & Concepts 764 Key Words 764 Questions & Problems 764 CHEMICAL M YS TERY A Hard-Boiled Snack 774 xiv Contents CHAPTER 17 Entropy, Entr Free Energy, and Equilibrium 776 17.1 The Three Laws of Thermodynamics 777 17.2 Spontaneous Processes 777 17.3 Entropy 778 17.4 The Second Law of Thermodynamics 783 17.5 Gibbs Free Energy 789 CHEMISTRY in Action The Efficiency of Heat Engines 790 17.6 Free Energy and Chemical Equilibrium 796 17.7 Thermodynamics in Living Systems 800 CHEMISTRY in Action The Thermodynamics of a Rubber Band 801 Key Equations 803 Summary of Facts & Concepts 803 Key Words 803 Questions & Problems 804 CHAPTER 18 Electrochemistry Elec 812 18.1 Redox Reactions 813 18.2 Galvanic Cells 816 18.3 Standard Reduction Potentials 818 18.4 Thermodynamics of Redox Reactions 824 18.5 The Effect of Concentration of Cell Emf 827 18.6 Batteries 832 CHEMISTRY in Action Bacteria Power 837 18.7 Corrosion 838 18.8 Electrolysis 841 CHEMISTRY in Action Dental Filling Discomfort 846 Key Equations 848 Summary of Facts & Concepts 848 Key Words 849 Questions & Problems 849 CHEMICAL M YS TERY Tainted Water 860  Contents xv CHAPTER 19 Nuclear N ucl Chemistry 862 19.1 The Nature of Nuclear Reactions 863 19.2 Nuclear Stability 865 19.3 Natural Radioactivity 870 19.4 Nuclear Transmutation 874 19.5 Nuclear Fission 877 CHEMISTRY in Action Nature’s Own Fission Reactor 882 19.6 Nuclear Fusion 883 19.7 Uses of Isotopes 886 19.8 Biological Effects of Radiation 888 CHEMISTRY in Action Food Irradiation 890 Key Equations 890 CHEMISTRY in Action Boron Neutron Capture Therapy 891 Summary of Facts & Concepts 891 Key Words 892 Questions & Problems 892 CHEMICAL M YS TERY The Art Forgery of the Twentieth Century 898 CHAPTER 20 Chemistry C hem in the Atmosphere 900 20.1 Earth’s Atmosphere 901 20.2 Phenomena in the Outer Layers of the Atmosphere 905 20.3 Depletion of Ozone in the Stratosphere 907 20.4 Volcanoes 911 20.5 The Greenhouse Effect 912 20.6 Acid Rain 916 20.7 Photochemical Smog 919 20.8 Indoor Pollution 921 Summary of Facts & Concepts 924 Key Words 924 Questions & Problems 925 xvi Contents CHAPTER 21 Metallurgy M eta and the Chemistry of Metals 930 21.1 Occurrence of Metals 931 21.2 Metallurgical Processes 932 21.3 Band Theory of Electrical Conductivity 939 21.4 Periodic Trends in Metallic Properties 941 21.5 The Alkali Metals 942 21.6 The Alkaline Earth Metals 946 21.7 Aluminum 948 CHEMISTRY in Action Recycling Aluminum 950 Summary of Facts & Concepts 952 Key Words 952 Questions & Problems 952 Nonmetallic N on Elements CHAPTER 22 aand nd Their Compounds 956 22.1 General Properties of Nonmetals 957 22.2 Hydrogen 958 CHEMISTRY in Action Metallic Hydrogen 962 22.3 Carbon 963 CHEMISTRY in Action Synthetic Gas from Coal 966 22.4 Nitrogen and Phosphorus 967 CHEMISTRY in Action Ammonium Nitrate—The Explosive Fertilizer 974 22.5 Oxygen and Sulfur 975 22.6 The Halogens 982 Summary of Facts & Concepts 989 Key Words 989 Questions & Problems 990  Contents xvii Transition Tran Metals Chemistry and CHAPTER 23 Coordination C oor Compounds 994 23.1 Properties of the Transition Metals 995 23.2 Chemistry of Iron and Copper 998 23.3 Coordination Compounds 1000 23.4 Structure of Coordination Compounds 1005 23.5 Bonding in Coordination Compounds: Crystal Field Theory 1009 23.6 Reactions of Coordination Compounds 1015 CHEMISTRY in Action Coordination Compounds in Living Systems 1016 23.7 Applications of Coordination Compounds 1016 CHEMISTRY in Action Cisplatin—The Anticancer Drug 1018 Key Equation 1020 Summary of Facts & Concepts 1020 Key Words 1020 Questions & Problems 1021 CHAPTER 24 Organic O rga Chemistry 1025 24.1 Classes of Organic Compounds 1026 24.2 Aliphatic Hydrocarbons 1026 CHEMISTRY in Action Ice That Burns 1038 24.3 Aromatic Hydrocarbons 1039 24.4 Chemistry of the Functional Groups 1042 CHEMISTRY in Action The Petroleum Industry 1048 Summary of Facts & Concepts 1050 Key Words 1051 Questions & Problems 1051 CHEMICAL M YS TERY The Disappearing Fingerprints 1056 xviii Contents Synthetic Synt and Natural Organic CHAPTER 25 Polymers Poly 1058 25.1 Properties of Polymers 1059 25.2 Synthetic Organic Polymers 1059 25.3 Proteins 1065 CHEMISTRY in Action Sickle Cell Anemia—A Molecular Disease 1072 25.4 Nucleic Acids 1073 CHEMISTRY in Action DNA Fingerprinting 1076 Summary of Facts & Concepts 1077 Key Words 1077 Questions & Problems 1077 CHEMICAL M YS TERY A Story That Will Curl Your Hair 1082 Appendix 1 Derivation of the Names of Elements A-1 Appendix 2 Units for the Gas Constant A-7 Appendix 3 Thermodynamic Data at 1 atm and 25°C A-8 Appendix 4 Mathematical Operations A-13 Glossary G-1 Answers to Even-Numbered Problems AP-1 Credits C-1 Index I-1  List of Applications The opening sentence of this text is, “Chemistry Dialysis 546 is an active, evolving science that has vital Radiocarbon Dating 586 importance to our world, in both the realm of nature and Pharmacokinetics 606 the realm of society.” Throughout the text, Chemistry Life at High Altitudes and Hemoglobin Production 651 in Action boxes and Chemical Mysteries give specific The Haber Process 652 examples of chemistry as active and evolving in all facets Antacids and the pH Balance in Your Stomach 706 of our lives. Maintaining the pH of Blood 732 How an Eggshell Is Formed 760 Chemistry in Action The Efficiency of Heat Engines 790 The Thermodynamics of a Rubber Band 801 The Search for the Higgs Boson 6 Bacteria Power 837 The Importance of Units 17 Dental Filling Discomfort 846 Distribution of Elements on Earth Nature’s Own Fission Reactor 882 and in Living Systems 49 Food Irradiation 890 Chemical Fertilizers 105 Boron Neutron Capture Therapy 891 An Undesirable Precipitation Reaction 126 Recycling Aluminum 950 Breathalyzer 144 Metallic Hydrogen 962 Metal from the Sea 156 Synthetic Gas from Coal 966 Scuba Diving and the Gas Laws 200 Ammonium Nitrate—The Explosive Fertilizer 974 Super Cold Atoms 208 Coordination Compounds in Living Systems 1016 Making Snow and Inflating a Bicycle Tire 240 Cisplatin—The Anticancer Drug 1018 White Fat Cells, Brown Fat Cells, and a Potential Cure Ice That Burns 1038 for Obesity 250 The Petroleum Industry 1048 How a Bombardier Beetle Defends Itself 256 Sickle Cell Anemia—A Molecular Disease 1072 Laser—The Splendid Light 288 DNA Fingerprinting 1076 Electron Microscopy 292 Quantum Dots 312 The Third Liquid Element? 341 Chemical Mystery Discovery of the Noble Gases 358 The Disappearance of the Dinosaurs 36 Sodium Chloride—A Common and Important Who Killed Napoleon? 170 Ionic Compound 376 Out of Oxygen 228 Just Say NO 397 The Exploding Tire 272 Microwave Ovens—Dipole Moments at Work 426 Discovery of Helium and the Rise Buckyball, Anyone? 454 and Fall of Coronium 324 A Very Slow Pitch 475 The Wrong Knife 560 Why Do Lakes Freeze from the Top Down? 478 Decaying Papers 718 High-Temperature Superconductors 488 A Hard-Boiled Snack 774 And All for the Want of a Button 492 Tainted Water 860 Hard-Boiling an Egg on a Mountaintop, Pressure The Art Forgery of the Twentieth Century 898 Cookers, and Ice Skating 505 The Disappearing Fingerprints 1056 Liquid Crystals 506 A Story That Will Curl Your Hair 1081 The Killer Lake 531 xix List of Animations The animations below are correlated to Chemistry. Neutralization Reactions (4.3) Within the chapter are icons letting the student and Orientation of Collision (13.4) instructor know that an animation is available for a spe- Osmosis (12.6) cific topic. Animations can be found online in the Chang Oxidation-Reduction Reactions (4.4) Connect site. Packing Spheres (11.4) Polarity of Molecules (10.2) Chang Animations Precipitation Reactions (4.2) Preparing a Solution by Dilution (4.5) Absorption of Color (23.5) Radioactive Decay (19.3) Acid-Base Titrations (16.4) Resonance (9.8) Acid Ionization (15.5) Sigma and Pi Bonds (10.5) Activation Energy (13.4) Strong Electrolytes, Weak Electrolytes, Alpha, Beta, and Gamma Rays (2.2) and Nonelectrolytes (4.1) α-Particle Scattering (2.2) VSEPR (10.1) Atomic and Ionic Radius (8.3) Base Ionization (15.6) Buffer Solutions (16.3) More McGraw-Hill Education Catalysis (13.6) Animations Cathode Ray Tube (2.2) Aluminum Production (21.7) Chemical Equilibrium (14.1) Atomic Line Spectra (7.3) Chirality (23.4, 24.2) Cubic Unit Cells and Their Origins (11.4) Collecting a Gas over Water (5.6) Cu/Zn Voltaic Cell (18.2) Diffusion of Gases (5.7) Current Generation from a Voltaic Cell (18.2) Dissolution of an Ionic and a Covalent Compound (12.2) Dissociation of Strong and Weak Acids (15.4) Electron Configurations (7.8) Emission Spectra (7.3) Equilibrium Vapor Pressure (11.8) Formation of Ag2S by Oxidation-Reduction (4.4) Galvanic Cells (18.2) Formation of an Ionic Compound (2.7) The Gas Laws (5.3) Formation of a Covalent Bond (9.4) Heat Flow (6.2) Influence of Shape on Polarity (10.2) Hybridization (10.4) Ionic and Covalent Bonding (9.4) Hydration (4.1) Molecular Shape and Orbital Hybridization (10.4) Ionic vs. Covalent Bonding (9.4) Operation of a Voltaic Cell (18.2) Le Chátelier’s Principle (14.5) Phase Diagrams and the States of Matter (11.9) Limiting Reagent (3.9) Properties of Buffers (16.3) Line Spectra (7.3) Reaction of Cu with AgNO3 (4.4) Making a Solution (4.5) Reaction of Magnesium and Oxygen (4.4, 9.2) Millikan Oil Drop (2.2) Rutherford’s Experiment (2.2) Nuclear Fission (19.5) VSEPR Theory (10.1) xx  Preface T he twelfth edition continues the tradition by pro- • Solution is the process of solving a problem given in viding a firm foundation in chemical concepts and a stepwise manner. principles and to instill in students an appreciation • Check enables the student to compare and verify of the vital part chemistry plays in our daily life. It is the with the source information to make sure the answer responsibility of the textbook authors to assist both in- is reasonable. structors and their students in their pursuit of this objec- • Practice Exercise provides the opportunity to solve tive by presenting a broad range of topics in a logical a similar problem in order to become proficient in manner. We try to strike a balance between theory and this problem type. The Practice Exercises are avail- application and to illustrate basic principles with every- able in the Connect electronic homework system. day examples whenever possible. The margin note lists additional similar problems to As in previous editions, our goal is to create a text work in the end-of-chapter problem section. that is clear in explaining abstract concepts, concise so that it does not overburden students with unnecessary ex- End-of-Chapter Problems are organized in various traneous information, yet comprehensive enough so that ways. Each section under a topic heading begins with it prepares students to move on to the next level of learn- Review Questions followed by Problems. The Additional ing. The encouraging feedback we have received from Problems section provides more problems not organized instructors and students has convinced us that this by section, followed by the new problem type of approach is effective. Interpreting, Modeling & Estimating. The art program has been extensively revised in this Many of the examples and end-of-chapter prob- edition. Many of the laboratory apparatuses and scientific lems present extra tidbits of knowledge and enable the instruments were redrawn to enhance the realism of the student to solve a chemical problem that a chemist components. Several of the drawings were updated to re- would solve. The examples and problems show stu- flect advances in the science and applications described dents the real world of chemistry and applications to in the text; see, for example, the lithium-ion battery de- everyday life situations. picted in Figure 18.10. Molecular structures were created using ChemDraw, the gold standard in chemical drawing software. Not only do these structures introduce students Visualization to the convention used to represent chemical structures in Graphs and Flow Charts are important in science. In three dimensions that they will see in further coursework, Chemistry, flow charts show the thought process of a con- they also provide better continuity with the ChemDraw cept and graphs present data to comprehend the concept. application they will use in Connect, the online home- A significant number of Problems and Review of work and practice system for our text. Concepts, including many new to this edition, include In addition to revising the art program, over 100 new graphical data. photographs are added in this edition. These photos pro- Molecular art appears in various formats to serve vide a striking look at processes that can be understood different needs. Molecular models help to visualize the by studying the underlying chemistry (see, for example, three-dimensional arrangement of atoms in a molecule. Figure 19.15, which shows the latest attempt of using Electrostatic potential maps illustrate the electron density lasers to induce nuclear fusion). distribution in molecules. Finally, there is the macro- scopic to microscopic art helping students understand Problem Solving processes at the molecular level. Photos are used to help students become familiar The development of problem-solving skills has always with chemicals and understand how chemical reactions been a major objective of this text. The two major catego- appear in reality. ries of learning are shown next. Figures of apparatus enable the student to visualize Worked examples follow a proven step-by-step the practical arrangement in a chemistry laboratory. strategy and solution. • Problem statement is the reporting of the facts needed to solve the problem based on the question Study Aids posed. Setting the Stage • Strategy is a carefully thought-out plan or method to Each chapter starts with the Chapter Outline and A Look serve as an important function of learning. Ahead. xxi xxii Preface Chapter Outline enables the student to see at a glance Solution Manual and online in the accompanying the big picture and focus on the main ideas of the Connect Chemistry companion website. chapter. End-of-Chapter Problems enable the student to A Look Ahead provides the student with an overview of practice critical thinking and problem-solving skills. concepts that will be presented in the chapter. The problems are broken into various types: • By chapter section. Starting with Review Quest- Tools to Use for Studying ions to test basic conceptual understanding, fol- Useful aids for studying are plentiful in Chemistry and lowed by Problems to test the student’s skill in should be used constantly to reinforce the comprehension solving problems for that particular section of the of chemical concepts. chapter. Marginal Notes are used to provide hints and feedback • Additional Problems uses knowledge gained from to enhance the knowledge base for the student. the various sections and/or previous chapters to Worked Examples along with the accompanying solve the problem. Practice Exercises are very important tools for • Interpreting, Modeling & Estimating problems learning and mastering chemistry. The problem- teach students the art of formulating models and solving steps guide the student through the critical estimating ballpark answers based on appropriate thinking necessary for succeeding in chemistry. assumptions. Using sketches helps student understand the inner workings of a problem. (See Example 6.1 on page 238.) A margin note lists similar problems in Real-Life Relevance the end-of-chapter problems section, enabling the Interesting examples of how chemistry applies to life are student to apply new skill to other problems of used throughout the text. Analogies are used where the same type. Answers to the Practice Exercises appropriate to help foster understanding of abstract are listed at the end of the chapter problems. chemical concepts. Review of Concepts enables the student to evaluate End-of-Chapter Problems pose many relevant if they understand the concept presented in the questions for the student to solve. Examples section. include Why do swimming coaches sometimes Key Equations are highlighted within the chapter, place a drop of alcohol in a swimmer’s ear to drawing the student’s eye to material that needs to draw out water? How does one estimate the be understood and retained. The key equations are pressure in a carbonated soft drink bottle before also presented in the chapter summary materials for removing the cap? easy access in review and study. Chemistry in Action boxes appear in every chapter on Summary of Facts and Concepts provides a quick a variety of topics, each with its own story of how review of concepts presented and discussed in detail chemistry can affect a part of life. The student can within the chapter. learn about the science of scuba diving and nuclear Key Words are a list of all important terms to help the medicine, among many other interesting cases. student understand the language of chemistry. Chemical Mystery poses a mystery case to the student. A series of chemical questions provide Testing Your Knowledge clues as to how the mystery could possibly be solved. Chemical Mystery will foster a high Review of Concepts lets students pause and check to level of critical thinking using the basic problem- see if they understand the concept presented and solving steps built up throughout the text. discussed in the section occurred. Answers to the Review of Concepts can be found in the Student  Digital Resources McGraw-Hill Education offers various tools and tech- assignments. As an instructor, you can edit existing ques- nology products to support Chemistry, 12th edition. tions and author entirely new problems. Track individual student performance—by question, assignment, or in rela- tion to the class overall—with detailed grade reports. Integrate grade reports easily with Learning Management chemistry Systems (LMS), such as WebCT and Blackboard—and much more. ConnectPlus Chemistry offers 24/7 online McGraw-Hill ConnectPlus Chemistry provides online pre- access to an eBook. This media-rich version of the book sentation, assignment, and assessment solutions. It con- allows seamless integration of text, media, and assessment. nects your students with the tools and resources they’ll To learn more visit connect.mheducation.com need to achieve success. With ConnectPlus Chemistry, you can deliver assignments, quizzes, and tests online. A robust set of questions, problems, and interactives are presented and aligned with the textbook’s learning goals. The integra- SmartBook is the first and only adaptive reading expe- tion of ChemDraw by PerkinElmer, the industry standard rience designed to change the way students read and in chemical drawing software, allows students to create learn. It creates a personalized reading experience by accurate chemical structures in their online homework highlighting the most impactful concepts a student needs to learn at that moment in time. As a student engages with SmartBook, the reading experience continuously adapts by highlighting content based on what the student knows and doesn’t know. This ensures that the focus is on the content he or she needs to learn, while simultaneously promoting long-term retention of ma- terial. Use SmartBook’s real-time re- ports to quickly identify the concepts that require more attention from indi- vidual students—or the entire class. The end result? Students are more engaged with course content, can better prioritize their time, and come to class ready to participate. Many questions within Connect Chemistry will allow students a chemical drawing experience that can be assessed directly inside of their homework. iii xxiii xxiv Digital Resources content through a series of adaptive questions. It pin- points concepts the student does not understand and maps McGraw-Hill LearnSmart is available as a standalone out a personalized study plan for success. This innovative product or as an integrated feature of McGraw-Hill study tool also has features that allow instructors to see Connect® Chemistry. It is an adaptive learning system exactly what students have accomplished and a built-in designed to help students learn faster, study more effi- assessment tool for graded assignments. Visit the follow- ciently, and retain more knowledge for greater success. ing site for a demonstration. www.mhlearnsmart.com LearnSmart assesses a student’s knowledge of course Adaptive Probes A student’s knowledge is intelligently probed by ask- ing a series of questions. These questions dynamically change both in the level of difficulty and in content based on the student’s weak and strong areas. Each practice session is based on the previous performance, and LearnSmart uses sophisticated models for predict- ing what the student will forget and how to reinforce that material typically forgotten. This saves students study time and ensures that they have actual mastery of the concepts. Immediate Feedback When a student incorrectly answers a probe, the correct answer is provided, along with feedback. Time Out When LearnSmart has identified a specific subject area where the student is struggling, he or she is given a “time out” and directed to the textbook section or learning objective for remediation. With ConnectPlus, students are provided with a link to the specific page of the eBook where they can study the material immediately. Reporting Dynamically generated reports document student prog- ress and areas for additional reinforcement, offering at-a-glance views of their strengths and weaknesses. Reports Include: • Most challenging learning objectives • Current learning status • Missed questions • Tree of wisdom • Metacognitive skills • Learning plan • Test results  Digital Resources xxv LearnSmart Labs for General Chemistry™ problem solving skills. And with ALEKS 360, your student also has access to this text’s eBook. Learn more at www.aleks.com/highered/ science THE Virtual Lab Experience. LearnSmart Labs is a must-see, outcomes-based lab simulation. It assesses a student’s knowledge and adap- tively corrects deficiencies, allowing the student to learn McGraw-Hill Create™ is a self-service website that faster and retain more knowledge with greater success. allows you to create customized course materials us- First, a student’s knowledge is adaptively leveled on ing McGraw-Hill Education’s comprehensive, cross- core learning outcomes: Questioning reveals knowledge disciplinary content and digital products. You can even deficiencies that are corrected by the delivery of content access third party content such as readings, articles, that is conditional on a student’s response. Then, a simu- cases, videos, and more. Arrange the content you’ve lated lab experience requires the student to think and act selected to match the scope and sequence of your like a scientist: Recording, interpreting, and analyzing course. Personalize your book with a cover design and data using simulated equipment found in labs and clinics. choose the best format for your students–eBook, color The student is allowed to make mistakes—a powerful print, or black-and-white print. And, when you are part of the learning experience! A virtual coach provides done, you’ll receive a PDF review copy in just minutes! subtle hints when needed; asks questions about the stu- www.mcgrawhillcreate.com dent’s choices; and allows the student to reflect upon and correct those mistakes. Whether your need is to overcome the logistical challenges of a traditional lab, provide bet- ter lab prep, improve student performance, or make your ® online experience one that rivals the real world, LearnSmart Labs accomplishes it all. Learn more at Tegrity Campus is a fully automated lecture capture solu- www.mhlearnsmart.com tion used in traditional, hybrid, “flipped classes” and on- line courses to record lesson, lectures, and skills. Its personalized learning features make study time incredi- bly efficient and its ability to affordably scale brings this benefit to every student on campus. Patented search tech- nology and real-time LMS integrations make Tegrity the LearnSmart Prep is an adaptive tool that prepares stu- market-leading solution and service. Tegrity is available dents for the course they are about to take. It identifies the as an integrated feature of McGraw-Hill Connect® prerequisite knowledge each student doesn’t know or Chemistry and as a standalone. fully understand and provides learning resources to teach essential concepts so he or she enters the classroom pre- pared to succeed. Presentation Tools Build instructional materials wherever, whenever, and however you want! Access instructor tools from your text’s Connect website to find photo’s, artwork, animations, and other media that can be used to create customized lectures, ALEKS (Assessment and LEarning in Knowledge visually enhanced tests and quizzes, compelling course Spaces) is a web-based system for individualized as- websites, or attractive printed support materials. All assets sessment and learning available 24/7 over the Internet. are copyrighted by McGraw-Hill Higher Education, but ALEKS uses artificial intelligence to accurately deter- can be used by instructors for classroom purposes. The mine a student’s knowledge and then guides her to the visual resources in this collection include: material that she is most ready to learn. ALEKS offers • Art Full-color digital files of all illustrations in the immediate feedback and access to ALEKSPedia—an book. interactive text that contains concise entries on chemis- • Photos The photo collection contains digital files of try topics. ALEKS is also a full-featured course man- photographs from the text. agement system with rich reporting features that allow instructors to monitor individual and class performance, • Tables Every table that appears in the text is avail- set student goals, assign/grade online quizzes, and able electronically. more. ALEKS allows instructors to spend more time on • Animations Numerous full-color animations illus- concepts while ALEKS teaches students practical trating important processes are also provided. xxvi Digital Resources • PowerPoint Lecture Outlines Ready made presen- and a summary of the corresponding text. Following tations for each chapter of the text. the summary are sample problems with detailed solu- • PowerPoint Slides All illustrations, photos, and tions. Each chapter has true–false questions and a tables are pre-inserted by chapter into blank self-test, with all answers provided at the end of the PowerPoint slides. chapter. Computerized Test Bank Online Animations for MP3/iPod A comprehensive bank of questions is provided within a A number of animations are available for download to computerized test bank, enabling professors to prepare your MP3/iPod through the textbook’s Connect website. and access tests or quizzes. Instructors can create or edit questions, or drag-and-drop questions to prepare tests quickly and easily. Tests can be published to their online Acknowledgments course, or printed for paper-based assignments. We would like to thank the following reviewers and sym- posium participants, whose comments were of great help Instructor’s Solution’s Manual to us in preparing this revision: The Instructor’s Solution Manual, written by Raymond Kathryn S. Asala, University of North Carolina, Chang and Ken Goldsby, provides the solutions to most Charlotte end-of-chapter problems. The manual also provides the Mohd Asim Ansari, Fullerton College difficulty level and category type for each problem. This Keith Baessler, Suffolk County Community College manual is available to instructors online in the text’s Christian S. Bahn, Montana State University Connect library tab. Mary Fran Barber, Wayne State University H. Laine Berghout, Weber State University Instructor’s Manual Feri Billiot, Texas A&M University Corpus Christi The Instructor’s Manual provides a brief summary of the John Blaha, Columbus State Community College contents of each chapter, along with the learning goals, references to background concepts in earlier chapters, Marco Bonizzoni, University of Alabama–Tuscaloosa and teaching tips. This manual can be found online for Christopher Bowers, Ohio Northern University instructors on the text’s Connect library tab. Bryan Breyfogle, Missouri State University Steve Burns, St. Thomas Aquinas College For the Student Mark L. Campbell, United States Naval Academy Tara Carpenter, University of Maryland Students can order supplemental study materials by con- David Carter, Angelo State tacting their campus bookstore, calling 1-800-262-4729, or online at http://shop.mheducation.com Daesung Chong, Ball State University Elzbieta Cook, Louisiana State University Student Solutions Manual Robert L. Cook, Louisiana State University ISBN 1-25-928622-3 Colleen Craig, University of Washington The Student Solutions Manual is written by Raymond Brandon Cruickshank, Northern Arizona Chang and Ken Goldsby. This supplement contains de- University–Flagstaff tailed solutions and explanations for even-numbered Elizabeth A. Clizbe, SUNY Buffalo problems in the main text. The manual also includes a Mohammed Daoudi, University of Central Florida detailed discussion of different types of problems and ap- Jay Deiner, New York City College of Technology proaches to solving chemical problems and tutorial solu- Dawn Del Carlo, University of Northern Iowa tions for many of the end-of-chapter problems in the text, along with strategies for solving them. Note that solutions Milagros Delgado, Florida International University, to the problems listed under Interpreting, Modeling & Biscayne Bay Campus Estimating are not provided in the manual. Michael Denniston, Georgia Perimeter College Stephanie R. Dillon, Florida State University Student Study Guide ISBN 1-25-928623-1 Anne Distler, Cuyahoga Community College This valuable ancillary contains material to help the Bill Donovan, University of Akron student practice problem-solving skills. For each sec- Mathilda D. Doorley, Southwest Tennessee Community tion of a chapter, the author provides study objectives College  Digital Resources xxvii Jack Eichler, University of California–Riverside Ramin Radfar, Wofford College Bradley D. Fahlman, Central Michigan University Betsy B. Ratcliff, West Virginia University Lee Friedman, University of Maryland–College Park Mike Rennekamp, Columbus State Community College Tiffany Gierasch, University of Maryland– Thomas G. Richmond, University of Utah Baltimore County Steven Rowley, Middlesex County College Cameon Geyer, Olympic College Joel W. Russell, Oakland University John Gorden, Auburn University Raymond Sadeghi, University of Texas– Tracy Hamilton, University of Alabama–Birmingham San Antonio Tony Hascall, Northern Arizona University Richard Schwenz, University of Northern Colorado Lindsay M. Hinkle, Harvard University Allison S. Soult, University of Kentucky Rebecca Hoenigman, Community College of Aurora Anne M. Spuches, East Carolina University T. Keith Hollis, Mississippi State University John Stubbs, University of New England Byron Howell, Tyler Junior College Katherine Stumpo, University of Tennessee–Martin Michael R. Ivanov, Northeast Iowa Community College Jerry Suits, University of Northern Colorado David W. Johnson, University of Dayton Charles Taylor, Florence Darlington Steve Johnson, University of New England Technical College Mohammad Karim, Tennessee State University– Mark Thomson, Ferris State University Nashville Eric M. Todd, University of Wisconsin–Stevens Point Jeremy Karr, College of Saint Mary Yijun Tang, University of Wisconsin–Oshkosh Vance Kennedy, Eastern Michigan University Steve Theberge, Merrimack College Katrina Kline, University of Missouri Lori Van Der Sluys, Penn State University An-Phong Lee, Florida Southern College Lindsay B. Wheeler, University of Virginia Debbie Leedy, Glendale Community College Gary D. White, Middle Tennessee State University Willem R. Leenstra, University of Vermont Stan Whittingham, Binghamton University Barbara S. Lewis, Clemson University Troy Wolfskill, Stony Brook University Scott Luaders, Quincy University Anthony Wren, Butte College Vicky Lykourinou, University of South Florida Fadi Zaher, Gateway Technical College Yinfa Ma, Missouri University of Science and Connect: Chemistry has been greatly enhanced by the Technology efforts of Yasmin Patell, Kansas State University; Sara-Kaye Madsen, South Dakota State University MaryKay Orgill, University of Nevada–Las Vegas; Sharyl Majorski, Central Michigan University Mirela Krichten, The College of New Jersey; who did a Roy McClean, United States Naval Academy masterful job of authoring hints and feedback to augment Helene Maire-Afeli, University of South Carolina–Union all of the system’s homework problems. The following individuals helped write and review Tracy McGill, Emory University learning goal–oriented content for LearnSmart for David M. McGinnis, University of Arkansas–Fort Smith General Chemistry: Thomas McGrath, Baylor University Margaret Ruth Leslie, Kent State University Deb Mlsna, Mississippi State University David G. Jones, Vistamar School Patricia Muisener, University of South Florida Erin Whitteck Kim Myung, Georgia Perimeter College Margaret Asirvatham, University of Anne-Marie Nickel, Milwaukee School of Engineering Colorado–Boulder Krista Noren-Santmyer, Hillsborough Community Alexander J. Seed, Kent State University College–Brandon Benjamin Martin, Texas State University–San Marcos Greg Oswald, North Dakota State University Claire Cohen, University of Toledo John W. Overcash, University of Illinois– Manoj Patil, Western Iowa Tech Community College Urbana–Champaign Adam I. Keller, Columbus State Community College Shadrick Paris, Ohio University Peter de Lijser, California State University–Fullerton Manoj Patil, Western Iowa Technical College Lisa Smith, North Hennepin Community College John Pollard, University of Arizona xxviii Digital Resources We have benefited much from discussions with our specialist, and Tami Hodge, the marketing manager, for colleagues at Williams College and Florida State, and their suggestions and encouragement. We also thank our correspondence with instructors here and abroad. sponsoring editor, David Spurgeon, for his advice and as- It is a pleasure to acknowledge the support given us sistance. Our special thanks go to Jodi Rhomberg, the by the following members of McGraw-Hill Education’s product developer, for her supervision of the project at College Division: Tammy Ben, Thomas Timp, Marty Lange, every stage of the writing of this edition. and Kurt Strand. In particular, we would like to mention Finally, we would like to acknowledge Toni Michaels, Sandy Wille for supervising the production, David Hash the photo researcher, for her resourcefulness in acquiring for the book design, and John Leland, the content licensing the new images under a very tight schedule.  Setting the Stage for Learning Real-Life Relevance Interesting examples of how chemistry applies to life are used throughout the text. Analogies are used where appropriate to help foster understanding of abstract chemical concepts. cha21510_ch04_118-171.indd Page 144 31/05/14 1:11 PM f-w-196 /203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles Chemistry in Action boxes appear in every chapter on a variety of topics, each with its own story of how chemistry can CHEMISTRY in Action affect a part of life. The student can learn Breathalyzer about the science of scuba diving and nuclear medicine, among many other cha21510_ch03_075-117.indd Pagevery 99 year Ein the United 7/19/14 11:02States AMabout f-49625,000 people are killed and 500,000 more are injured as a result of drunk driving. In spite of efforts to educate the public about the dangers of driving /203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles interesting cases. while intoxicated and stiffer penalties for drunk driving offenses, law enforcement agencies still have to devote a great deal of work to removing drunk drivers from America’s roads. The police often use a device called a breathalyzer to Chemical Mystery poses a mystery case test drivers suspected of being drunk. The chemical basis of this device is a redox reaction. A sample of the driver’s breath is drawn into the breathalyzer, where it is treated with to the student. A series of chemical ques- an acidic solution of potassium dichromate. The alcohol (ethanol) in the breath is converted to acetic acid as shown in tions provide clues as to how the mystery the following equation: could possibly be solved. Chemical 3CH3CH2OH 1 2K2Cr2O7 1 8H2SO4 ¡ ethanol potassium dichromate sulfuric acid A driver being tested for blood alcohol content with a handheld breathalyzer. Mystery will foster a high level of critical (orange yellow) 3CH3COOH 1 2Cr2(SO4)3 1 2K2SO4 1 11H2O thinking using the basic problem-solving to the green chromium(III) ion (see Figure 4.22). The driver’s acetic acid chromium(III) sulfate (green) potassium sulfate blood alcohol level can be determined readily by measuring the degree of this color change (read from a calibrated meter on the steps built up throughout the text. In this reaction, the ethanol is oxidized to acetic acid and the instrument). The current legal limit of blood alcohol content is chromium(VI) in the orange-yellow dichromate ion is reduced 0.08 percent by mass. Anything higher constitutes intoxication. Breath Schematic diagram of a breathalyzer. The alcohol in the driver’s breath is reacted with a potassium dichromate solution. The change in the absorption of light due to the formation of chromium(III) sulfate is registered by the detector and shown on a meter, which directly displays the alcohol Meter content in blood. The filter selects only one wavelength of light for Light Filter Photocell measurement. source detector K2Cr2O7 solution 510_ch05_172-229.indd Page 211 31/05/14 1:15 PM f-w-196 /203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles Before reaction has started Visualization CH4 2.0 H2 Molecular art NH3 Graphs and PV 1.0 Ideal gas Flow Charts RT 0 200 400 600 800 1000 1200 P (atm) After reaction is complete H2 CO CH3OH xxix xxx Setting the Stage for Learning Key Equations cha21510_ch02_038-074.indd Page 67 27/05/14 10:37 AM f-w-196 /203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles DU 5 q 1 w (6.1) Mathematical statement of the first law of thermodynamics. w 5 2PDV cha21510_ch02_038-074.indd Page 67(6.3) 27/05/14 10:37 AM f-w-196 Calculating work done in gas expansion or gas compression. /203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles H 5 U 1 PV (6.6) Definition of enthalpy. DH 5 DU 1 PDV (6.8) Calculating enthalpy (or energy) change for a constant-pressure process. C 5 ms (6.11) Definition of heat capacity. q 5 msDt (6.12) Calculating heat change in terms of specific heat. q 5 CDt (6.13) Calculating heat change in terms of heat capacity. ¢H°rxn 5 on¢H°f (products) 2 om¢H°f (reactants) (6.18) Calculating standard enthalpy of reaction. DHsoln 5 U 1 DHhydr (6.20) Lattice energy and hydration contributions to heat of solution. Summary of Facts & Concepts 1. Modern chemistry began with Dalton’s atomic theory, determines the identity of an element. The mass which states that all matter is composed of tiny, indi- number is the sum of the number of protons and the visible particles called atoms; that all atoms of the number of neutrons in the nucleus. same element are identical; that compounds contain 6. Isotopes are atoms of the same element with the same atoms of different elements combined in whole- number of protons but different numbers of neutrons. Study Aids number ratios; and that atoms are neither created nor 7. Chemical formulas combine the symbols for the con- destroyed in chemical reactions (the law of conserva- stituent elements with whole-number subscripts to Key Equations—material to tion of mass). show the type and number of atoms contained in the retain 2. Atoms of constituent elements in a particular compound smallest unit of a compound. are always combined in the same proportions by mass 8. The molecular formula conveys the specific number (law of definite proportions). When two elements can and type of atoms combined in each molecule of a com- Summary of Facts & combine to form more than one type of compound, the pound. The empirical formula shows the simplest ratios masses of one element that combine with a fixed mass of the atoms combined in a molecule. Concepts—quick review of of the other element are in a ratio of small whole num- 9. Chemical compounds are either molecular compounds important concepts bers (law of multiple proportions). (in which the smallest units are discrete, individual mol- 3. An atom consists of a very dense central nucleus ecules) or ionic compounds, which are made of cations containing protons and neutrons, with electrons and anions. Key Words—important terms moving about the nucleus at a relatively large dis- 10. The names of many inorganic compounds can be tance from it. to understand 4. Protons are positively charged, neutrons have no charge, deduced from a set of simple rules. The formulas can be written from the names of the compounds. and electrons are negatively charged. Protons and neu- 11. Organic compounds contain carbon and elements like trons have roughly the same mass, which is about 1840 hydrogen, oxygen, and nitrogen. Hydrocarbon is the times greater than the mass of an electron. simplest type of organic compound. 5. The atomic number of an element is the number of protons in the nucleus of an atom of the element; it Key Words Acid, p. 62 Chemical formula, p. 52 Law of conservation of Nonmetal, p. 48 Alkali metals, p. 50 Diatomic molecule, p. 50 mass, p. 40 Nucleus, p. 44 Alkaline earth metals, p. 50 Electron, p. 41 Law of definite Organic compound, p. 56 Allotrope, p. 52 Empirical formula, p. 53 proportions, p. 40 Oxoacid, p. 62 Alpha (α) particles, p. 43 Families, p. 48 Law of multiple Oxoanion, p. 63 Alpha (α) rays, p. 43 Gamma (γ) rays, p. 43 proportions, p. 40 Periods, p. 48 Anion, p. 51 Groups, p. 48 Mass number (A), p. 46 Periodic table, p. 48 Atom, p. 40 Halogens, p. 50 Metal, p. 48 Polyatomic ion, p. 51 Atomic number (Z), p. 46 Hydrate, p. 64 Metalloid, p. 48 Polyatomic molecule, p. 50 Base, p. 64 Inorganic Molecular formula, p. 52 Proton, p. 44 Beta (β) particles, p. 43 compounds, p. 56 Molecule, p. 50 Radiation, p. 41 Beta (β) rays, p. 43 Ion, p. 50 Monatomic ion, p. 51 Radioactivity, p. 43 Binary compound, p. 56 Ionic compound, p. 51 Neutron, p. 45 Structural formula, p. 53 Cation, p. 51 Isotope, p. 46 Noble gases, p. 50 Ternary compound, p. 57  Setting the Stage for Learning xxxi Chang Learning System Review of Concepts The diagrams here show three compounds AB2 (a), AC2 (b), and AD2 (c) dissolved Review the section content by using this quick test for in water. Which is the strongest electrolyte and which is the weakest? (For simplicity, water molecules are not shown.) acquired knowledge. (a) (b) (c) Example 4.6 Classify the following redox reactions and indicate changes in the oxidation numbers of the elements: (a) 2N2O(g) ¡ 2N2(g) 1 O2(g) (b) 6Li(s) 1 N2(g) ¡ 2Li3N(s) (c) Ni(s) 1 Pb(NO3)2(aq) ¡ Pb(s) 1 Ni(NO3)2(aq) (d) 2NO2(g) 1 H2O(l) ¡ HNO2(aq) 1 HNO3(aq) Strategy Review the definitions of combination reactions, decomposition reactions, displacement reactions, and disproportionation reactions. Solution Learn a problem-solving process of (a) This is a decomposition reaction because one reactant is converted to two different strategizing, solving, and checking products. The oxidation number of N changes from 11 to 0, while that of O changes from 22 to 0. your way to a solution. (b) This is a combination reaction (two reactants form a single product). The oxidation number of Li changes from 0 to 11 while that of N changes from 0 to 23. (c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb21 ion. The oxidation number of Ni increases from 0 to 12 while that of Pb decreases from 12 to 0. (d) The oxidation number of N is 14 in NO2 and it is 13 in HNO2 and 15 in HNO3. Because the oxidation number of the same element both increases and decreases, this is a disproportionation reaction. Practice Exercise Identify the following redox reactions by type: (a) Fe 1 H2SO4 ¡ FeSO4 1 H2 (b) S 1 3F2 ¡ SF6 (c) 2CuCl ¡ Cu 1 CuCl2 (d) 2Ag 1 PtCl2 ¡ 2AgCl 1 Pt Use the problem-solving approach on real-world problems. Interpreting, Modeling & Estimating problems provide students the opportunity to solve problems like a chemist. 4.172 Potassium superoxide (KO2), a useful source of oxygen employed in breathing equipment, reacts with water to form potassium hydroxide, hydro- gen peroxide, and oxygen. Furthermore, potas- • sium superoxide also reacts with carbon dioxide to form potassium carbonate and oxygen. (a) Write equations for these two reactions and comment on the effectiveness of potassium superoxide in this application. (b) Focusing only on the reaction be- tween KO2 and CO2, estimate the amount of KO2 needed to sustain a worker in a polluted environ- ment for 30 min. See Problem 1.69 for useful information. A Note to the Student G eneral chemistry is commonly perceived to be • Definitions of the key words can be studied in con- more difficult than most other subjects. There text on the pages cited in the end-of-chapter list or in is some justification for this perception. For the glossary at the back of the book. one thing, chemistry has a very specialized vocabulary. • Careful study of the worked-out examples in the At first, studying chemistry is like learning a new lan- body of each chapter will improve your ability to guage. Furthermore, some of the concepts are abstract. analyze problems and correctly carry out the calcula- Nevertheless, with diligence you can complete this tions needed to solve them. Also take the time to course successfully, and you might even enjoy it. Here work through the practice exercise that follows each are some suggestions to help you form good study example to be sure you understand how to solve the habits and master the material in this text. type of problem illustrated in the example. The • Attend classes regularly and take careful notes. answers to the practice exercises appear at the end of • If possible, always review the topics discussed in the chapter, following the questions and problems. class the same day they are covered in class. Use this For additional practice, you can turn to similar prob- book to supplement your notes. lems referred to in the margin next to the example. • Think critically. Ask yourself if you really under- • The questions and problems at the end of the chapter stand the meaning of a term or the use of an equation. are organized by section. A good way to test your understanding is to explain • The back inside cover shows a list of important a concept to a classmate or some other person. figures and tables with page references. This index • Do not hesitate to ask your instructor or your teach- makes it convenient to quickly look up information ing assistant for help. when you are solving problems or studying related subjects in different chapters. The twelfth edition tools for Chemistry are designed to enable you to do well in your general chemistry course. If you follow these suggestions and stay up-to-date The following guide explains how to take full advantage with your assignments, you should find that chemistry is of the text, technology, and other tools. challenging, but less difficult and much more interesting than you expected. • Before delving into the chapter, read the chapter out- line and the chapter introduction to get a sense of the important topics. Use the outline to organize your —Raymond Chang and Ken Goldsby note taking in class. • At the end of each chapter you will find a summary of facts and concepts, the key equations, and a list of key words, all of which will help you review for exams. xxxii  CHAPTER 1 Chemistry The Study of Change By applying electric fields to push DNA molecules through pores created in graphene, scientists have developed a technique that someday can be used for fast sequencing the four chemical bases according to their unique electrical properties. CHAPTER OUTLINE A LOOK AHEAD 1.1 Chemistry: A Science for  We begin with a brief introduction to the study of chemistry and describe its the Twenty-First Century role in our modern society. (1.1 and 1.2) 1.2 The Study of Chemistry  Next, we become familiar with the scientific method, which is a systematic approach to research in all scientific disciplines. (1.3) 1.3 The Scientific Method  We define matter and note that a pure substance can either be an element or 1.4 Classifications of Matter a compound. We distinguish between a homogeneous mixture and a hetero- 1.5 The Three States of Matter geneous mixture. We also learn that, in principle, all matter can exist in one of three states: solid, liquid, and gas. (1.4 and 1.5) 1.6 Physical and Chemical  To characterize a substance, we need to know its physical properties, which Properties of Matter can be observed without changing its identity and chemical properties, 1.7 Measurement which can be demonstrated only by chemical changes. (1.6) 1.8 Handling Numbers  Being an experimental science, chemistry involves measurements. We learn the basic SI units and use the SI-derived units for quantities like volume and 1.9 Dimensional Analysis density. We also become familiar with the three temperature scales: Celsius, in Solving Problems Fahrenheit, and Kelvin. (1.7) 1.10 Real-World Problem Solving:  Chemical calculations often involve very large or very small numbers and a Information, Assumptions, convenient way to deal with these numbers is the scientific notation. In cal- and Simplifications culations or measurements, every quantity must show the proper number of significant figures, which are the meaningful digits. (1.8)  We learn that dimensional analysis is useful in chemical calculations. By carrying the units through the entire sequence of calculations, all the units will cancel except the desired one. (1.9)  Solving real-world problems frequently involves making assumptions and simplifications. (1.10) 1 2 Chapter 1 ■ Chemistry: The Study of Change C hemistry is an active, evolving science that has vital importance to our world, in both the realm of nature and the realm of society. Its roots are ancient, but as we will see, chem- istry is every bit a modern science. We will begin our study of chemistry at the macroscopic level, where we can see and measure the materials of which our world is made. In this chapter, we will discuss the scientific method, which provides the framework for research not only in chemistry but in all other sciences as well. Next we will discover how scientists define and characterize matter. Then we will spend some time learning how to handle numerical results of chemical measurements and solve numerical problems. In Chapter 2, we will begin to explore the microscopic world of atoms and molecules. 1.1 Chemistry: A Science for the Twenty-First Century Chemistry is the study of matter and the changes it undergoes. Chemistry is often called the central science, because a basic knowledge of chemistry is essential for students of biology, physics, geology, ecology, and many other subjects. Indeed, it is central to our way of life; without it, we would be living shorter lives in what we would consider primitive conditions, without automobiles, electricity, computers, CDs, and many other everyday conveniences. Although chemistry is an ancient science, its modern foundation was laid in the The Chinese characters for chem- nineteenth century, when intellectual and technological advances enabled scientists to istry mean “The study of change.” break down substances into ever smaller components and consequently to explain many of their physical and chemical characteristics. The rapid development of increas- ingly sophisticated technology throughout the twentieth century has given us even greater means to study things that cannot be seen with the naked eye. Using comput- ers and special microscopes, for example, chemists can analyze the structure of atoms and molecules—the fundamental units on which the study of chemistry is based—and design new substances with specific properties, such as drugs and environmentally friendly consumer products. It is fitting to ask what part the central science will have in the twenty-first century. Almost certainly, chemistry will continue to play a pivotal role in all areas of science and technology. Before plunging into the study of matter and its transfor- mation, let us consider some of the frontiers that chemists are currently exploring (Figure 1.1). Whatever your reasons for taking general chemistry, a good knowledge of the subject will better enable you to appreciate its impact on society and on you as an individual. 1.2 The Study of Chemistry Compared with other subjects, chemistry is commonly believed to be more difficult, at least at the introductory level. There is some justification for this perception; for one thing, chemistry has a very specialized vocabulary. However, even if this is your first course in chemistry, you already have more familiarity with the subject than you may realize. In everyday conversations we hear words that have a chemical connec- tion, although they may not be used in the scientifically correct sense. Examples are “electronic,” “quantum leap,” “equilibrium,” “catalyst,” “chain reaction,” and “critical mass.” Moreover, if you cook, then you are a practicing chemist! From experience gained in the kitchen, you know that oil and water do not mix and that boiling water left on the stove will evaporate. You apply chemical and physical principles when you use baking soda to leaven bread, choose a pressure cooker to shorten the time it takes to prepare soup, add meat tenderizer to a pot roast, squeeze lemon juice over sliced  1.2 The Study of Chemistry 3 (a) (b) (c) (d) Figure 1.1 (a) The output from an automated DNA sequencing machine. Each lane displays the sequence (indicated by different colors) obtained with a separate DNA sample. (b) A graphene supercapacitor. These materials provide some of the highest known energy-to-volume ratios and response times. (c) Production of photovoltaic cells, used to convert light into electrical current. (d) The leaf on the left was taken from a tobacco plant that was not genetically engineered but was exposed to tobacco horn worms. The leaf on the right was genetically engineered and is barely attacked by the worms. The same technique can be applied to protect the leaves of other types of plants. pears to prevent them from turning brown or over fish to minimize its odor, and add vinegar to the water in which you are going to poach eggs. Every day we observe such changes without thinking about their chemical nature. The purpose of this course is to make you think like a chemist, to look at the macroscopic world—the things we can see, touch, and measure directly—and visualize the particles and events of the microscopic world that we cannot experience without modern technology and our imaginations. At first some students find it confusing that their chemistry instructor and textbook seem to be continually shifting back and forth between the macroscopic and micro- scopic worlds. Just keep in mind that the data for chemical investigations most often come from observations of large-scale phenomena, but the explanations frequently lie in the unseen and partially imagined microscopic world of atoms and molecules. In other words, chemists often see one thing (in the macroscopic world) and think another (in the microscopic world). Looking at the rusted nails in Figure 1.2, for example, a chemist might think about the basic properties of individual atoms of iron and how these units interact with other atoms and molecules to produce the observed change. 4 Chapter 1 ■ Chemistry: The Study of Change O2 88n Fe2O3 Fe Figure 1.2 A simplified molecular view of rust (Fe2O3) formation from iron (Fe) atoms and oxygen molecules (O2). In reality, the process requires water and rust also contains water molecules. 1.3 The Scientific Method All sciences, including the social sciences, employ variations of what is called the scientific method, a systematic approach to research. For example, a psychologist who wants to know how noise affects people’s ability to learn chemistry and a chem- ist interested in measuring the heat given off when hydrogen gas burns in air would follow roughly the same procedure in carrying out their investigations. The first step is to carefully define the problem. The next step includes performing experiments, making careful observations, and recording information, or data, about the system— the part of the universe that is under investigation. (In the examples just discussed, the systems are the group of people the psychologist will study and a mixture of hydrogen and air.) The data obtained in a research study may be both qualitative, consisting of general observations about the system, and quantitative, comprising numbers obtained by various measurements of the system. Chemists generally use standardized symbols and equations in recording their measurements and observations. This form of repre- sentation not only simplifies the process of keeping records, but also provides a com- mon basis for communication with other chemists. When the experiments have been completed and the data have been recorded, the next step in the scientific method is interpretation, meaning that the scientist attempts to explain the observed phenomenon. Based on the data that were gathered, the researcher formulates a hypothesis, a tentative explanation for a set of observations. Further experiments are devised to test the validity of the hypothesis in as many ways as possible, and the process begins anew. Figure 1.3 summarizes the main steps of the research process. After a large amount of data has been collected, it is often desirable to sum- marize the information in a concise way, as a law. In science, a law is a concise verbal or mathematical statement of a relationship between phenomena that is always the same under the same conditions. For example, Sir Isaac Newton’s sec- ond law of motion, which you may remember from high school science, says that force equals mass times acceleration (F 5 ma). What this law means is that an  1.3 The Scientific Method 5 Observation Representation Interpretation Figure 1.3 The three levels of studying chemistry and their relationships. Observation deals with events in the macroscopic world; atoms and molecules constitute the microscopic world. Representation is a scientific shorthand for describing an experiment in symbols and chemical equations. Chemists use their knowledge of atoms and molecules to explain an observed phenomenon. increase in the mass or in the acceleration of an object will always increase its force proportionally, and a decrease in mass or acceleration will always decrease the force. Hypotheses that survive many experimental tests of their validity may evolve into theories. A theory is a unifying principle that explains a body of facts and/or those laws that are based on them. Theories, too, are constantly being tested. If a theory is disproved by experiment, then it must be discarded or modified so that it becomes consistent with experimental observations. Proving or disproving a theory can take years, even centuries, in part because the necessary technology may not be available. Atomic theory, which we will study in Chapter 2, is a case in point. It took more than 2000 years to work out this fundamental principle of chemistry proposed by Dem- ocritus, an ancient Greek philosopher. A more contemporary example is the search for the Higgs boson discussed on page 6. Scientific progress is seldom, if ever, made in a rigid, step-by-step fashion. Some- times a law precedes a theory; sometimes it is the other way around. Two scientists may start working on a project with exactly the same objective, but will end up tak- ing drastically different approaches. Scientists are, after all, human beings, and their modes of thinking and working are very much influenced by their background, train- ing, and personalities. The development of science has been irregular and sometimes even illogical. Great discoveries are usually the result of the cumulative contributions and expe- rience of many workers, even though the credit for formulating a theory or a law is usually given to only one individual. There is, of course, an element of luck involved in scientific discoveries, but it has been said that “chance favors the prepared mind.” It takes an alert and well-trained person to recognize the signifi- cance of an accidental discovery and to take full advantage of it. More often than not, the public learns only of spectacular scientific breakthroughs. For every suc- cess story, however, there are hundreds of cases in which scientists have spent years working on projects that ultimately led to a dead end, and in which positive achievements came only after many wrong turns and at such a slow pace that they went unheralded. Yet even the dead ends contribute something to the continually growing body of knowledge about the physical universe. It is the love of the search that keeps many scientists in the laboratory. Review of Concepts Which of the following statements is true? (a) A hypothesis always leads to the formulation of a law. (b) The scientific method is a rigid sequence of steps in solving problems. (c) A law summarizes a series of experimental observations; a theory provides an explanation for the observations. CHEMISTRY in Action The Search for the Higgs Boson I n this chapter, we identify mass as a fundamental property of matter, but have you ever wondered: Why does matter even have mass? It might seem obvious that “everything” has mass, but is that a requirement of nature? We will see later in our stud- ies that light is composed of particles that do not have mass when at rest, and physics tells us under different circumstances the universe might not contain anything with mass. Yet we know that our universe is made up of an uncountable number of par- ticles with mass, and these building blocks are necessary to form the elements that make up the people to ask such ques- tions. The search for the answer to this question illustrates nicely the process we call the scientific method. Current theoretical models tell us that everything in the uni- Illustration of the data obtained from decay of the Higgs boson into other verse is based on two types of elementary particles: bosons and particles following an 8-TeV collision at the Large Hadron Collider at CERN. fermions. We can distinguish the roles of these particles by consid- ering the building blocks of matter to be constructed from fermi- ons, while bosons are particles responsible for the force that holds On July 4, 2012, scientists at CERN announced the discov- the fermions together. In 1964, three different research teams inde- ery of the Higgs boson. It takes about 1 trillion proton-proton pendently proposed mechanisms in which a field of energy perme- collisions to produce one Higgs boson event, so it requires a ates the universe, and the interaction of matter with this field is due tremendous amount of data obtained from two independent sets to a specific boson associated with the field. The greater the number of experiments to confirm the findings. In science, the quest for of these bosons, the greater the interaction will be with the field. answers is never completely done. Our understanding can always This interaction is the property we call mass, and the field and the be improved or refined, and sometimes entire tenets of accepted associated boson came to be named for Peter Higgs, one of the science are replaced by another theory that does a better job ex- original physicists to propose this mechanism. plaining the observations. For example, scientists are not sure if This theory ignited a frantic search for the “Higgs boson” the Higgs boson is the only particle that confers mass to matter, or that became one of the most heralded quests in modern science. if it is only one of several such bosons predicted by other theories. The Large Hadron Collider at CERN in Geneva, Switzerland But over the long run, the scientific method has proven to (described on p. 875) was constructed to carry out experiments be our best way of understanding the physical world. It took designed to find evidence for the Higgs boson. In these experi- 50 years for experimental science to validate the existence of the ments, protons are accelerated to nearly the speed of light in Higgs boson. This discovery was greeted with great fanfare and opposite directions in a circular 17-mile tunnel, and then al- recognized the following year with a 2013 Nobel Prize in lowed to collide, generating even more fundamental particles at Physics for Peter Higgs and François Englert, another one of the very high energies. The data are examined for evidence of an six original scientists who first proposed the existence of a uni- excess of particles at an energy consistent with theoretical pre- versal field that gives particles their mass. It is impossible to dictions for the Higgs boson. The ongoing process of theory imagine where science will take our understanding of the uni- suggesting experiments that give results used to evaluate and verse in the next 50 years, but we can be fairly certain that many ultimately refine the theory, and so on, is the essence of the of the theories and experiments driving this scientific discovery scientific method. will be very different than the ones we use today. 1.4 Classifications of Matter We defined chemistry in Section 1.1 as the study of matter and the changes it under- goes. Matter is anything that occupies space and has mass. Matter includes things we can see and touch (such as water, earth, and trees), as well as things we cannot (such as air). Thus, everything in the universe has a “chemical” connection. 6  1.4 Classifications of Matter 7 Chemists distinguish among several subcategories of matter based on compo- sition and properties. The classifications of matter include substances, mixtures, elements, and compounds, as well as atoms and molecules, which we will consider in Chapter 2. Substances and Mixtures A substance is a form of matter that has a definite (constant) composition and distinct properties. Examples are water, ammonia, table sugar (sucrose), gold, and oxygen. Substances differ from one another in composition and can be identified by their appearance, smell, taste, and other properties. A mixture is a combination of two or more substances in which the substances retain their distinct identities. Some familiar examples are air, soft drinks, milk, and cement. Mixtures do not have constant composition. Therefore, samples of air col- lected in different cities would probably differ in composition because of differences in altitude, pollution, and so on. Mixtures are either homogeneous or heterogeneous. When a spoonful of sugar dissolves in water we obtain a homogeneous mixture in which the composition of the mixture is the same throughout. If sand is mixed with iron filings, however, the sand grains and the iron filings remain separate (Figure 1.4). This type of mixture is called a heterogeneous mixture because the composition is not uniform. Any mixture, whether homogeneous or heterogeneous, can be created and then separated by physical means into pure components without changing the identities of the components. Thus, sugar can be recovered from a water solution by heating the solution and evaporating it to dryness. Condensing the vapor will give us back the water component. To separate the iron-sand mixture, we can use a magnet to remove the iron filings from the sand, because sand is not attracted to the magnet [see Figure 1.4(b)]. After separation, the components of the mixture will have the same composition and properties as they did to start with. Elements and Compounds Substances can be either elements or compounds. An element is a substance that cannot be separated into simpler substances by chemical means. To date, 118 ele- ments have been positively identified. Most of them occur naturally on Earth. The others have been created by scientists via nuclear processes, which are the subject of Chapter 19 of this text. Figure 1.4 (a) The mixture contains iron filings and sand. (b) A magnet separates the iron filings from the mixture. The same technique is used on a larger scale to separate iron and steel from nonmagnetic objects such as aluminum, glass, and plastics. (a) (b) 8 Chapter 1 ■ Chemistry: The Study of Change Table 1.1 Some Common Elements and Their Symbols Name Symbol Name Symbol Name Symbol Aluminum Al Fluorine F Oxygen O Arsenic As Gold Au Phosphorus P Barium Ba Hydrogen H Platinum Pt Bismuth Bi Iodine I Potassium K Bromine Br Iron Fe Silicon Si Calcium Ca Lead Pb Silver Ag Carbon C Magnesium Mg Sodium Na Chlorine Cl Manganese Mn Sulfur S Chromium Cr Mercury Hg Tin Sn Cobalt Co Nickel Ni Tungsten W Copper Cu Nitrogen N Zinc Zn For convenience, chemists use symbols of one or two letters to represent the ele- ments. The first letter of a symbol is always capitalized, but any following letters are not. For example, Co is the symbol for the element cobalt, whereas CO is the formula for the carbon monoxide molecule. Table 1.1 shows the names and symbols of some of the more common elements; a complete list of the elements and their symbols appears inside the front cover of this book. The symbols of some elements are derived from their Latin names—for example, Au from aurum (gold), Fe from ferrum (iron), and Na from natrium (sodium)—whereas most of them come from their English names. Appendix 1 gives the origin of the names and lists the discoverers of most of the elements. Atoms of most elements can interact with one another to form compounds. Hydrogen gas, for example, burns in oxygen gas to form water, which has proper- ties that are distinctly different from those of the starting materials. Water is made up of two parts hydrogen and one part oxygen. This composition does not change, regardless of whether the water comes from a faucet in the United States, a lake in Outer Mongolia, or the ice caps on Mars. Thus, water is a compound, a substance composed of atoms of two or more elements chemically united in fixed proportions. Unlike mixtures, compounds can be separated only by chemical means into their pure components. The relationships among elements, compounds, and other categories of matter are summarized in Figure 1.5. Review of Concepts Which of the following diagrams represent elements and which represent compounds? Each color sphere (or truncated sphere) represents an atom. Different colored atoms indicate different elements. (a) (b) (c) (d)  1.5 The Three States of Matter 9 Matter Separation by Mixtures Substances physical methods Homogeneous Heterogeneous Separation by Compounds Elements mixtures mixtures chemical methods Figure 1.5 Classification of matter. 1.5 The Three States of Matter All substances, at least in principle, can exist in three states: solid, liquid, and gas. As Figure 1.6 shows, gases differ from liquids and solids in the distances between the molecules. In a solid, molecules are held close together in an orderly fashion with little freedom of motion. Molecules in a liquid are close together but are not held so rigidly in position and can move past one another. In a gas, the molecules are separated by distances that are large compared with the size of the molecules. The three states of matter can be interconverted without changing the composition of the substance. Upon heating, a solid (for example, ice) will melt to form a liquid (water). (The temperature at which this transition occurs is called the melting point.) Further heating will convert the liquid into a gas. (This conversion takes place at the boiling point of the liquid.) On the other hand, cooling a gas will cause it to condense into a liquid. When the liquid is cooled further, it will freeze into the solid form. Figure 1.6 Microscopic views of a solid, a liquid, and a gas. Solid Liquid Gas 10 Chapter 1 ■ Chemistry: The Study of Change Figure 1.7 The three states of matter. A hot poker changes ice into water and steam. Figure 1.7 shows the three states of water. Note that the properties of water are unique among common substances in that the molecules in the liquid state are more closely packed than those in the solid state. Review of Concepts An ice cube is placed in a closed container. On heating, the ice cube first melts and the water then boils to form steam. Which of the following statements is true? (a) The physical appearance of the water is different at every stage of change. (b) The mass of water is greatest for the ice cube and least for the steam. 1.6 Physical and Chemical Properties of Matter Substances are identified by their properties as well as by their composition. Color, melting point, and boiling point are physical properties. A physical property can be measured and observed without changing the composition or identity of a substance. For example, we can measure the melting point of ice by heating a block of ice and recording the temperature at which the ice is converted to water. Water differs from ice only in appearance, not in composition, so this is a physical change; we can freeze  1.7 Measurement 11 the water to recover the original ice. Therefore, the melting point of a substance is a physical property. Similarly, when we say that helium gas is lighter than air, we are referring to a physical property. On the other hand, the statement “Hydrogen gas burns in oxygen gas to form water” describes a chemical property of hydrogen, because to observe this property we must carry out a chemical change, in this case burning. After the change, the original chemical substance, the hydrogen gas, will have vanished, and all that will be left is a different chemical substance—water. We cannot recover the hydrogen from the water by means of a physical change, such as boiling or freezing. Every time we hard-boil an egg, we bring about a chemical change. When sub- jected to a temperature of about 1008C, the yolk and the egg white undergo changes that alter not only their physical appearance but their chemical makeup as well. When eaten, the egg is changed again, by substances in our bodies called enzymes. This digestive action is another example of a chemical change. What happens during diges- tion depends on the chemical properties of both the enzymes and the food. All measurable properties of matter fall into one of two additional categories: extensive properties and intensive properties. The measured value of an extensive property depends on how much matter is being considered. Mass, which is the quan- Hydrogen burning in air to form tity of matter in a given sample of a substance, is an extensive property. More matter water. means more mass. Values of the same extensive property can be added together. For example, two copper pennies will have a combined mass that is the sum of the masses of each penny, and the length of two tennis courts is the sum of the lengths of each tennis court. Volume, defined as length cubed, is another extensive property. The value of an extensive quantity depends on the amount of matter. The measured value of an intensive property does not depend on how much mat- ter is being considered. Density, defined as the mass of an object divided by its volume, is an intensive property. So is temperature. Suppose that we have two beakers of water at the same temperature. If we combine them to make a single quantity of water in a larger beaker, the temperature of the larger quantity of water will be the same as it was in two separate beakers. Unlike mass, length, and volume, temperature and other intensive properties are not additive. Review of Concepts The diagram in (a) shows a compound made up of atoms of two elements (represented by the green and red spheres) in the liquid state. Which of the diagrams in (b)–(d) represents a physical change and which diagrams represent a chemical change? (a) (b) (c) (d) 1.7 Measurement The measurements chemists make are often used in calculations to obtain other related quantities. Different instruments enable us to measure a substance’s properties: The meterstick measures length or scale; the buret, the pipet, the graduated cylinder, and 12 Chapter 1 ■ Chemistry: The Study of Change the volumetric flask measure volume (Figure 1.8); the balance measures mass; the thermometer measures temperature. These instruments provide measurements of macroscopic properties, which can be determined directly. Microscopic properties, on the atomic or molecular scale, must be determined by an indirect method, as we will see in Chapter 2. A measured quantity is usually written as a number with an appropri- ate unit. To say that the distance between New York and San Francisco by car along a certain route is 5166 is meaningless. We must specify that the distance is 5166 kilometers. The same is true in chemistry; units are essential to stating measure- ments correctly. SI Units For many years, scientists recorded measure- ments in metric units, which are related deci- mally, that is, by powers of 10. In 1960, however, the General Conference of Weights and Mea- sures, the international authority on units, pro- posed a revised metric system called the International System of Units (abbreviated SI, from the French Système Internationale d’Unites). Table 1.2 shows the seven SI base units. All other units of measurement can be derived from these base units. Like metric units, SI units are modi- fied in decimal fashion by a series of prefixes, as shown in Table 1.3. We will use both metric and SI units in this book. Figure 1.8 Some common measuring devices found in a chemistry laboratory. These devices are not drawn to scale relative to one another. We will discuss the uses of these measuring devices in Chapter 4. Volumetric flask Graduated cylinder Pipet Buret  1.7 Measurement 13 Note that a metric prefix simply represents Table 1.2 SI Base Units a number: 1 mm 5 1 3 1023 m Base Quantity Name of Unit Symbol Length meter m Mass kilogram kg Time second s Electrical current ampere A Temperature kelvin K Amount of substance mole mol Luminous intensity candela cd An astronaut jumping on the surface of the moon. Table 1.3 Prefixes Used with SI Units Prefix Symbol Meaning Example tera- T 1,000,000,000,000, or 1012 1 terameter (Tm) 5 1 3 1012 m giga- G 1,000,000,000, or 109 1 gigameter (Gm) 5 1 3 109 m mega- M 1,000,000, or 106 1 megameter (Mm) 5 1 3 106 m kilo- k 1,000, or 103 1 kilometer (km) 5 1 3 103 m deci- d 1/10, or 1021 1 decimeter (dm) 5 0.1 m centi- c 1/100, or 1022 1 centimeter (cm) 5 0.01 m milli- m 1/1,000, or 1023 1 millimeter (mm) 5 0.001 m micro- μ 1/1,000,000, or 1026 1 micrometer (μm) 5 1 3 1026 m nano- n 1/1,000,000,000, or 1029 1 nanometer (nm) 5 1 3 1029 m pico- p 1/1,000,000,000,000, or 10212 1 picometer (pm) 5 1 3 10212 m femto- f 1/1,000,000,000,000,000, or 10215 1 femtometer (fm) 5 1 3 10215 m atto- a 1/1,000,000,000,000,000,000 or 10218 1 attometer (am) 5 1 3 10218 m Measurements that we will utilize frequently in our study of chemistry include time, mass, volume, density, and temperature. Mass and Weight The terms “mass” and “weight” are often used interchangeably, although, strictly speaking, they are different quantities. Whereas mass is a measure of the amount of matter in an object, weight, technically speaking, is the force that gravity exerts on an object. An apple that falls from a tree is pulled downward by Earth’s gravity. The mass of the apple is constant and does not depend on its location, but its weight does. For example, on the surface of the moon the apple would weigh only one-sixth what it does on Earth, because the moon’s gravity is only one-sixth that of Earth. The moon’s smaller gravity enabled astronauts to jump about rather freely on its surface despite their bulky suits and equipment. Chemists are interested primarily in mass, which can be determined readily with a balance; the process of measuring mass, oddly, is called weighing. Figure 1.9 The prototype The SI unit of mass is the kilogram (kg). Unlike the units of length and time, kilogram is made of a platinum- which are based on natural processes that can be repeated by scientists anywhere, the iridium alloy. It is kept in a vault kilogram is defined in terms of a particular object (Figure 1.9). In chemistry, however, at the International Bureau of Weights and Measures in Sèvres, the smaller gram (g) is more convenient: France. In 2007 it was discovered that the alloy has mysteriously lost 1 kg 5 1000 g 5 1 3 103 g about 50 μg! 14 Chapter 1 ■ Chemistry: The Study of Change Volume: 1000 cm3; 1000 mL; Volume 1 dm3; The SI unit of length is the meter (m), and the SI-derived unit for volume is the 1L cubic meter (m3). Generally, however, chemists work with much smaller volumes, such as the cubic centimeter (cm3) and the cubic decimeter (dm3): 1 cm3 5 (1 3 1022 m) 3 5 1 3 1026 m3 1 dm3 5 (1 3 1021 m) 3 5 1 3 1023 m3 Another common unit of volume is the liter (L). A liter is the volume occupied by one cubic decimeter. One liter of volume is equal to 1000 milliliters (mL) or 1000 cm3: 1 cm 1 L 5 1000 mL 10 cm = 1 dm 5 1000 cm3 Volume: 1 cm3; 5 1 dm3 1 mL 1 cm and one milliliter is equal to one cubic centimeter: Figure 1.10 Comparison of two volumes, 1 mL and 1000 mL. 1 mL 5 1 cm3 Figure 1.10 compares the relative sizes of two volumes. Even though the liter is not an SI unit, volumes are usually expressed in liters and milliliters. Density The equation for density is mass density 5 volume Table 1.4 Densities of Some or Substances at 25°C Density m Substance (g/cm3) d5 (1.1) V Air* 0.001 Ethanol 0.79 where d, m, and V denote density, mass, and volume, respectively. Because density is Water 1.00 an intensive property and does not depend on the quantity of mass present, for a given Graphite 2.2 substance the ratio of mass to volume always remains the same; in other words, V Table salt 2.2 increases as m does. Density usually decreases with temperature. Aluminum 2.70 The SI-derived unit for density is the kilogram per cubic meter (kg/m3). This unit Diamond 3.5 is awkwardly large for most chemical applications. Therefore, grams per cubic centi- Iron 7.9 meter (g/cm3) and its equivalent, grams per milliliter (g/mL), are more com- Lead 11.3 monly used for solid and liquid densities. Because gas densities are often very low, Mercury 13.6 we express them in units of grams per liter (g/L): Gold 19.3 Osmium† 22.6 1 g/cm3 5 1 g/mL 5 1000 kg/m3 1 g/L 5 0.001 g/mL *Measured at 1 atmosphere. † Osmium (Os) is the densest element known. Table 1.4 lists the densities of several substances.  1.7 Measurement 15 Examples 1.1 and 1.2 show density calculations. Example 1.1 Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold. Solution We are given the mass and volume and asked to calculate the density. Therefore, from Equation (1.1), we write m d5 V 301 g 5 15.6 cm3 5 19.3 g/cm3 Practice Exercise A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass? Example 1.2 Gold bars and the solid-state The density of mercury, the only metal that is a liquid at room temperature, is 13.6 g/mL. arrangement of the gold atoms. Calculate the mass of 5.50 mL of the liquid. Similar problems: 1.21, 1.22. Solution We are given the density and volume of a liquid and asked to calculate the mass of the liquid. We rearrange Equation (1.1) to give m5d3V g 5 13.6 3 5.50 mL mL 5 74.8 g Practice Exercise The density of sulfuric acid in a certain car battery is 1.41 g/mL. Calculate the mass of 242 mL of the liquid. Temperature Scales Mercury. Three temperature scales are currently in use. Their units are 8F (degrees Fahren- heit), 8C (degrees Celsius), and K (kelvin). The Fahrenheit scale, which is the most Similar problems: 1.21, 1.22. commonly used scale in the United States outside the laboratory, defines the normal freezing and boiling points of water to be exactly 328F and 2128F, respectively. The Celsius scale divides the range between the freezing point (08C) and boiling point (1008C) of water into 100 degrees. As Table 1.2 shows, the kelvin is the SI Note that the Kelvin scale does not have the degree sign. Also, temperatures base unit of temperature: It is the absolute temperature scale. By absolute we mean expressed in kelvins can never be that the zero on the Kelvin scale, denoted by 0 K, is the lowest temperature that negative. can be attained theoretically. On the other hand, 08F and 08C are based on the behavior of an arbitrarily chosen substance, water. Figure 1.11 compares the three temperature scales. The size of a degree on the Fahrenheit scale is only 100/180, or 5/9, of a degree on the Celsius scale. To convert degrees Fahrenheit to degrees Celsius, we write 5°C ?°C 5 (°F 2 32°F) 3 (1.2) 9°F 16 Chapter 1 ■ Chemistry: The Study of Change Figure 1.11 Comparison of the three temperature scales: Celsius, 373 K 100°C Boiling point 212°F and Fahrenheit, and the absolute of water (Kelvin) scales. Note that there are 100 divisions, or 100 degrees, between the freezing point and the boiling point of water on the Celsius scale, and there are 180 divisions, or 180 degrees, between the same two temperature limits Body on the Fahrenheit scale. The 310 K 37°C temperature 98.6°F Celsius scale was formerly called 298 K 25°C Room 77°F the centigrade scale. temperature 273 K 0°C Freezing point 32°F of water Kelvin Celsius Fahrenheit The following equation is used to convert degrees Celsius to degrees Fahrenheit: 9°F ?°F 5 3 (°C) 1 32°F (1.3) 5°C Both the Celsius and the Kelvin scales have units of equal magnitude; that is, one degree Celsius is equivalent to one kelvin. Experimental studies have shown that absolute zero on the Kelvin scale is equivalent to 2273.158C on the Celsius scale. Thus, we can use the following equation to convert degrees Celsius to kelvin: 1K ? K 5 (°C 1 273.15°C) (1.4) 1°C We will frequently find it necessary to convert between degrees Celsius and degrees Fahrenheit and between degrees Celsius and kelvin. Example 1.3 illustrates these conversions. The Chemistry in Action essay on page 17 shows why we must be careful with units in scientific work. Example 1.3 (a)  Below the transition temperature of 21418C, a certain substance becomes a superconductor; that is, it can conduct electricity with no resistance. What is the temperature in degrees Fahrenheit? (b) Helium has the lowest boiling point of all the elements at 24528F. Convert this temperature to degrees Celsius. (c) Mercury, the only metal that exists as a liquid at room temperature, melts at 238.98C. Convert its melting point to kelvins. Magnet suspended above ­superconductor cooled below Solution  These three parts require that we carry out temperature conversions, so we its transition temperature by need Equations (1.2), (1.3), and (1.4). Keep in mind that the lowest temperature on the liquid nitrogen. Kelvin scale is zero (0 K); therefore, it can never be negative. (a) This conversion is carried out by writing 9°F 3 (2141°C) 1 32°F 5 2222°F 5°C (Continued) CHEMISTRY in Action The Importance of Units I n December 1998, NASA launched the 125-million dollar Mars Climate Orbiter, intended as the red planet’s first weather satellite. After a 416-million mi journey, the spacecraft was sup- scientist said: “This is going to be the cautionary tale that will be embedded into introduction to the metric system in elemen- tary school, high school, and college science courses till the end posed to go into Mars’ orbit on September 23, 1999. Instead, it of time.” entered Mars’ atmosphere about 100 km (62 mi) lower than planned and was destroyed by heat. The mission controllers said the loss of the spacecraft was due to the failure to convert English measurement units into metric units in the navigation software. Engineers at Lockheed Martin Corporation who built the spacecraft specified its thrust in pounds, which is an English unit. Scientists at NASA’s Jet Propulsion Laboratory, on the other hand, had assumed that thrust data they received were expressed in metric units, as newtons. Normally, pound is the unit for mass. Expressed as a unit for force, however, 1 lb is the force due to gravitational attraction on an object of that mass. To carry out the conversion between pound and newton, we start with 1 lb 5 0.4536 kg and from Newton’s second law of motion, force 5 mass 3 acceleration 5 0.4536 kg 3 9.81 m/s2 5 4.45 kg m/s2 5 4.45 N because 1 newton (N) 5 1 kg m/s2. Therefore, instead of converting 1 lb of force to 4.45 N, the scientists treated it as 1 N. The considerably smaller engine thrust expressed in new- tons resulted in a lower orbit and the ultimate destruction of the spacecraft. Commenting on the failure of the Mars mission, one Artist’s conception of the Martian Climate Orbiter. (b) Here we have 5°C (2452°F 2 32°F) 3 5 2269°C 9°F (c) The melting point of mercury in kelvins is given by 1K (238.9°C 1 273.15°C) 3 5 234.3 K Similar problems: 1.24, 1.25, 1.26. 1°C Practice Exercise Convert (a) 327.58C (the melting point of lead) to degrees Fahrenheit; (b) 172.98F (the boiling point of ethanol) to degrees Celsius; and (c) 77 K, the boiling point of liquid nitrogen, to degrees Celsius. 17 18 Chapter 1 ■ Chemistry: The Study of Change Review of Concepts The density of copper is 8.94 g/cm3 at 208C and 8.91 g/cm3 at 608C. This density decrease is the result of which of the following? (a) The metal expands. (b) The metal contracts. (c) The mass of the metal increases. (d) The mass of the metal decreases. 1.8 Handling Numbers Having surveyed some of the units used in chemistry, we now turn to techniques for han- dling numbers associated with measurements: scientific notation and significant figures. Scientific Notation Chemists often deal with numbers that are either extremely large or extremely small. For example, in 1 g of the element hydrogen there are roughly 602,200,000,000,000,000,000,000 hydrogen atoms. Each hydrogen atom has a mass of only 0.00000000000000000000000166 g These numbers are cumbersome to handle, and it is easy to make mistakes when using them in arithmetic computations. Consider the following multiplication: 0.0000000056 3 0.00000000048 5 0.000000000000000002688 It would be easy for us to miss one zero or add one more zero after the decimal point. Consequently, when working with very large and very small numbers, we use a system called scientific notation. Regardless of their magnitude, all numbers can be expressed in the form N 3 10n where N is a number between 1 and 10 and n, the exponent, is a positive or negative integer (whole number). Any number expressed in this way is said to be written in scientific notation. Suppose that we are given a certain number and asked to express it in scientific notation. Basically, this assignment calls for us to find n. We count the number of places that the decimal point must be moved to give the number N (which is between 1 and 10). If the decimal point has to be moved to the left, then n is a positive inte- ger; if it has to be moved to the right, n is a negative integer. The following examples illustrate the use of scientific notation: (1) Express 568.762 in scientific notation: 568.762 5 5.68762 3 102 Note that the decimal point is moved to the left by two places and n 5 2. (2) Express 0.00000772 in scientific notation: 0.00000772 5 7.72 3 1026 Here the decimal point is moved to the right by six places and n 5 26.  1.8 Handling Numbers 19 Keep in mind the following two points. First, n 5 0 is used for numbers that are Any number raised to the power zero is not expressed in scientific notation. For example, 74.6 3 100 (n 5 0) is equivalent equal to one. to 74.6. Second, the usual practice is to omit the superscript when n 5 1. Thus, the scientific notation for 74.6 is 7.46 3 10 and not 7.46 3 101. Next, we consider how scientific notation is handled in arithmetic operations. Addition and Subtraction To add or subtract using scientific notation, we first write each quantity—say, N1 and N2—with the same exponent n. Then we combine N1 and N2; the exponents remain the same. Consider the following examples: (7.4 3 103 ) 1 (2.1 3 103 ) 5 9.5 3 103 (4.31 3 104 ) 1 (3.9 3 103 ) 5 (4.31 3 104 ) 1 (0.39 3 104 ) 5 4.70 3 104 (2.22 3 10 ) 2 (4.10 3 10 ) 5 (2.22 3 1022 ) 2 (0.41 3 1022 ) 22 23 5 1.81 3 1022 Multiplication and Division To multiply numbers expressed in scientific notation, we multiply N1 and N2 in the usual way, but add the exponents together. To divide using scientific notation, we divide N1 and N2 as usual and subtract the exponents. The following examples show how these operations are performed: (8.0 3 104 ) 3 (5.0 3 102 ) 5 (8.0 3 5.0)(10412 ) 5 40 3 106 5 4.0 3 107 (4.0 3 1025 ) 3 (7.0 3 103 ) 5 (4.0 3 7.0)(102513 ) 5 28 3 1022 5 2.8 3 1021 7 6.9 3 10 6.9 25 5 3 1072 (25) 3.0 3 10 3.0 5 2.3 3 1012 4 8.5 3 10 8.5 9 5 3 10429 5.0 3 10 5.0 5 1.7 3 1025 Significant Figures Except when all the numbers involved are integers (for example, in counting the number of students in a class), it is often impossible to obtain the exact value of the quantity under investigation. For this reason, it is important to indicate the margin of error in a measurement by clearly indicating the number of significant figures, which are the meaningful digits in a measured or calculated quantity. When significant figures are used, the last digit is understood to be uncertain. For example, we might measure the volume of a given amount of liquid using a graduated cyl- inder with a scale that gives an uncertainty of 1 mL in the measurement. If the volume is found to be 6 mL, then the actual volume is in the range of 5 mL to 7  mL. We represent the volume of the liquid as (6 6 1) mL. In this case, there is only one significant figure (the digit 6) that is uncertain by either plus or minus 1  mL. For greater accuracy, we might use a graduated cylinder that has finer divi- sions, so that the volume we measure is now uncertain by only 0.1 mL. If the volume of the liquid is now found to be 6.0 mL, we may express the quantity as (6.0 6 0.1) mL, and the actual value is somewhere between 5.9 mL and 6.1 mL. 20 Chapter 1 ■ Chemistry: The Study of Change We can further improve the measuring device and obtain more significant figures, but in every case, the last digit is always uncertain; the amount of this uncertainty depends on the particular measuring device we use. Figure 1.12 shows a modern balance. Balances such as this one are available in many general chemistry laboratories; they readily measure the mass of objects to four decimal places. Therefore, the measured mass typically will have four significant figures (for example, 0.8642 g) or more (for example, 3.9745 g). Keeping track of the number of significant figures in a measurement such as mass ensures that calculations involving the data will reflect the precision of the measurement. Guidelines for Using Significant Figures We must always be careful in scientific work to write the proper number of significant figures. In general, it is fairly easy to determine how many significant figures a num- ber has by following these rules: Figure 1.12 A Fisher Scientific 1. Any digit that is not zero is significant. Thus, 845 cm has three significant figures, A-200DS Digital Recorder 1.234 kg has four significant figures, and so on. Precision Balance. 2. Zeros between nonzero digits are significant. Thus, 606 m contains three sig- nificant figures, 40,501 kg contains five significant figures, and so on. 3. Zeros to the left of the first nonzero digit are not significant. Their purpose is to indicate the placement of the decimal point. For example, 0.08 L contains one significant figure, 0.0000349 g contains three significant figures, and so on. 4. If a number is greater than 1, then all the zeros written to the right of the decimal point count as significant figures. Thus, 2.0 mg has two significant figures, 40.062 mL has five significant figures, and 3.040 dm has four signifi- cant figures. If a number is less than 1, then only the zeros that are at the end of the number and the zeros that are between nonzero digits are significant. This means that 0.090 kg has two significant figures, 0.3005 L has four significant figures, 0.00420 min has three significant figures, and so on. 5. For numbers that do not contain decimal points, the trailing zeros (that is, zeros after the last nonzero digit) may or may not be significant. Thus, 400 cm may have one significant figure (the digit 4), two significant figures (40), or three significant figures (400). We cannot know which is correct without more information. By using scientific notation, however, we avoid this ambiguity. In this particular case, we can express the number 400 as 4 3 102 for one significant figure, 4.0 3 102 for two significant figures, or 4.00 3 102 for three significant figures. Example 1.4 shows the determination of significant figures. Example 1.4 Determine the number of significant figures in the following measurements: (a) 394 cm, (b) 5.03 g, (c) 0.714 m, (d) 0.052 kg, (e) 2.720 3 1022 atoms, (f ) 3000 mL. Solution (a) Three , because each digit is a nonzero digit. (b) Three , because zeros between nonzero digits are significant. (c) Three , because zeros to the left of the first nonzero digit do not count as significant figures. (d) Two . Same reason as in (c). (e) Four . Because the number is greater than one, all the zeros written to the right of the decimal point count as significant figures. (f) This is an ambiguous case. The number of significant figures may be four (3.000 3 103), three (3.00 3 103), two (Continued)  1.8 Handling Numbers 21 (3.0 3 103), or one (3 3 103). This example illustrates why scientific notation must be used to show the proper number of significant figures. Similar problems: 1.33, 1.34. Practice Exercise Determine the number of significant figures in each of the following measurements: (a) 35 mL, (b) 2008 g, (c) 0.0580 m3, (d) 7.2 3 104 molecules, (e) 830 kg. A second set of rules specifies how to handle significant figures in calculations. 1. In addition and subtraction, the answer cannot have more digits to the right of the decimal point than either of the original numbers. Consider these examples: 89.332 1 1.1 ←— one digit after the decimal point 90.432 ←— round off to 90.4 2.097 2 0.12 ←— two digits after the decimal point 1.977 ←— round off to 1.98 The rounding-off procedure is as follows. To round off a number at a certain point we simply drop the digits that follow if the first of them is less than 5. Thus, 8.724 rounds off to 8.72 if we want only two digits after the decimal point. If the first digit following the point of rounding off is equal to or greater than 5, we add 1 to the preceding digit. Thus, 8.727 rounds off to 8.73, and 0.425 rounds off to 0.43. 2. In multiplication and division, the number of significant figures in the final prod- uct or quotient is determined by the original number that has the smallest number of significant figures. The following examples illustrate this rule: 2.8 3 4.5039 5 12.61092 — round off to 13 6.85 5 0.0611388789 — round off to 0.0611 112.04 3. Keep in mind that exact numbers obtained from definitions or by counting num- bers of objects can be considered to have an infinite number of significant figures. For example, the inch is defined to be exactly 2.54 centimeters; that is, 1 in 5 2.54 cm Thus, the “2.54” in the equation should not be interpreted as a measured number with three significant figures. In calculations involving conversion between “in” and “cm,” we treat both “1” and “2.54” as having an infinite number of significant figures. Similarly, if an object has a mass of 5.0 g, then the mass of nine such objects is 5.0 g 3 9 5 45 g The answer has two significant figures because 5.0 g has two significant figures. The number 9 is exact and does not determine the number of significant figures. Example 1.5 shows how significant figures are handled in arithmetic operations. Example 1.5 Carry out the following arithmetic operations to the correct number of significant figures: (a) 12,343.2 g 1 0.1893 g, (b) 55.67 L 2 2.386 L, (c) 7.52 m 3 6.9232, (d) 0.0239 kg 4 46.5 mL, (e) 5.21 3 103 cm 1 2.92 3 102 cm. (Continued) 22 Chapter 1 ■ Chemistry: The Study of Change Solution In addition and subtraction, the number of decimal places in the answer is determined by the number having the lowest number of decimal places. In multiplication and division, the significant number of the answer is determined by the number having the smallest number of significant figures. (a) 12,343.2 g 1 0.1893 g 12,343.3893 g ←— round off to 12,343.4 g (b) 55.67 L 2 2.386 L 53.284 L ←— round off to 53.28 L (c) 7.52 m 3 6.9232 5 52.06246 m ←— round off to 52.1 m 0.0239 kg (d) 5 0.0005139784946 kg/mL ←— round off to 0.000514 kg/mL 46.5 mL or 5.14 3 1024 kg/mL (e) First we change 2.92 3 10 cm to 0.292 3 103 cm and then carry out the addition 2 (5.21 cm 1 0.292 cm) 3 103. Following the procedure in (a), we find the answer Similar problems: 1.35, 1.36. is 5.50 3 103 cm. Practice Exercise Carry out the following arithmetic operations and round off the answers to the appropriate number of significant figures: (a) 26.5862 L 1 0.17 L, (b) 9.1 g 2 4.682 g, (c) 7.1 3 104 dm 3 2.2654 3 102 dm, (d) 6.54 g 4 86.5542 mL, (e) (7.55 3 104 m) 2 (8.62 3 103 m). The preceding rounding-off procedure applies to one-step calculations. In chain calculations, that is, calculations involving more than one step, we can get a different answer depending on how we round off. Consider the following two- step calculations: First step:   A3B5C Second step:    C 3 D 5 E Let’s suppose that A 5 3.66, B 5 8.45, and D 5 2.11. Depending on whether we round off C to three or four significant figures, we obtain a different number for E: Method 1 Method 2 3.66 3 8.45 5 30.9 3.66 3 8.45 5 30.93 30.9 3 2.11 5 65.2 30.93 3 2.11 5 65.3 However, if we had carried out the calculation as 3.66 3 8.45 3 2.11 on a calcula- tor without rounding off the intermediate answer, we would have obtained 65.3 as the answer for E. Although retaining an additional digit past the number of sig- nificant figures for intermediate steps helps to eliminate errors from rounding, this procedure is not necessary for most calculations because the difference between the answers is usually quite small. Therefore, for most examples and end-of-chapter problems where intermediate answers are reported, all answers, intermediate and final, will be rounded. Accuracy and Precision In discussing measurements and significant figures, it is useful to distinguish between accuracy and precision. Accuracy tells us how close a measurement is to the true value of the quantity that was measured. Precision refers to how closely two or more measurements of the same quantity agree with one another (Figure 1.13).  1.9 Dimensional Analysis in Solving Problems 23 Figure 1.13 The distribution of holes formed by darts on a dart board shows the difference between precise and accurate. (a) Good accuracy and good precision. (b) Poor accuracy and good precision. (c) Poor accuracy and poor precision. (a) (b) (c) The difference between accuracy and precision is a subtle but important one. Suppose, for example, that three students are asked to determine the mass of a piece of copper wire. The results of two successive weighings by each student are Student A Student B Student C 1.964 g 1.972 g 2.000 g 1.978 g 1.968 g 2.002 g Average value 1.971 g 1.970 g 2.001 g The true mass of the wire is 2.000 g. Therefore, Student B’s results are more precise than those of Student A (1.972 g and 1.968 g deviate less from 1.970 g than 1.964 g and 1.978  g from 1.971 g), but neither set of results is very accurate. Student C’s results are not only the most precise, but also the most accurate, because the average value is closest to the true value. Highly accurate measurements are usually precise too. On the other hand, highly precise measurements do not necessarily guarantee accurate results. For example, an improperly calibrated meterstick or a faulty balance may give precise readings that are in error. Review of Concepts Give the length of the pencil with proper significant figures according to which ruler you use for the measurement. 1.9 Dimensional Analysis in Solving Problems Careful measurements and the proper use of significant figures, along with correct calculations, will yield accurate numerical results. But to be meaningful, the answers also must be expressed in the desired units. The procedure we use to convert between units in solving chemistry problems is called dimensional analysis (also called the factor-label method). A simple technique requiring little memorization, dimensional analysis is based on the relationship between different units that express the same 24 Chapter 1 ■ Chemistry: The Study of Change physical quantity. For example, by definition 1 in 5 2.54 cm (exactly). This equiva- lence enables us to write a conversion factor as follows: 1 in 2.54 cm Because both the numerator and the denominator express the same length, this fraction is equal to 1. Similarly, we can write the conversion factor as 2.54 cm 1 in which is also equal to 1. Conversion factors are useful for changing units. Thus, if we wish to convert a length expressed in inches to centimeters, we multiply the length by the appropriate conversion factor. 2.54 cm 12.00 in 3 5 30.48 cm 1 in We choose the conversion factor that cancels the unit inches and produces the desired unit, centimeters. Note that the result is expressed in four significant figures because 2.54 is an exact number. Next let us consider the conversion of 57.8 meters to centimeters. This problem can be expressed as ? cm 5 57.8 m By definition, 1 cm 5 1 3 10 22 m Because we are converting “m” to “cm,” we choose the conversion factor that has meters in the denominator, 1 cm 1 3 1022 m and write the conversion as 1 cm ? cm 5 57.8 m 3 1 3 1022 m 5 5780 cm 5 5.78 3 103 cm Note that scientific notation is used to indicate that the answer has three significant figures. Again, the conversion factor 1 cm/1 3 1022 m contains exact numbers; there- fore, it does not affect the number of significant figures. In general, to apply dimensional analysis we use the relationship given quantity 3 conversion factor 5 desired quantity and the units cancel as follows: desired unit Remember that the unit we want appears given unit 3 5 desired unit in the numerator and the unit we want to given unit cancel appears in the denominator. In dimensional analysis, the units are carried through the entire sequence of calcula- tions. Therefore, if the equation is set up correctly, then all the units will cancel except the desired one. If this is not the case, then an error must have been made somewhere, and it can usually be spotted by reviewing the solution.  1.9 Dimensional Analysis in Solving Problems 25 A Note on Problem Solving At this point you have been introduced to scientific notation, significant figures, and dimensional analysis, which will help you in solving numerical problems. Chemistry is an experimental science and many of the problems are quantitative in nature. The key to success in problem solving is practice. Just as a marathon runner cannot prepare for a race by simply reading books on running and a pianist cannot give a successful concert by only memorizing the musical score, you cannot be sure of your understand- ing of chemistry without solving problems. The following steps will help to improve your skill at solving numerical problems. 1. Read the question carefully. Understand the information that is given and what you are asked to solve. Frequently it is helpful to make a sketch that will help you to visualize the situation. 2. Find the appropriate equation that relates the given information and the unknown quantity. Sometimes solving a problem will involve more than one step, and you may be expected to look up quantities in tables that are not provided in the problem. Dimensional analysis is often needed to carry out conversions. 3. Check your answer for the correct sign, units, and significant figures. 4. A very important part of problem solving is being able to judge whether the answer is reasonable. It is relatively easy to spot a wrong sign or incorrect units. But if a number (say, 9) is incorrectly placed in the denominator instead of in the numerator, the answer would be too small even if the sign and units of the calculated quantity were correct. 5. One quick way to check the answer is to round off the numbers in the calculation in such a way so as to simplify the arithmetic. The answer you get will not be exact, but it will be close to the correct one. Example 1.6 A person’s average daily intake of glucose (a form of sugar) is 0.0833 pound (lb). What is this mass in milligrams (mg)? (1 lb 5 453.6 g.) Strategy The problem can be stated as ? mg 5 0.0833 lb Glucose tablets can provide The relationship between pounds and grams is given in the problem. This relationship diabetics with a quick method for raising their blood sugar levels. will enable conversion from pounds to grams. A metric conversion is then needed to convert grams to milligrams (1 mg 5 1 3 1023 g). Arrange the appropriate conversion factors so that pounds and grams cancel and the unit milligrams is obtained in your answer. Solution The sequence of conversions is pounds ¡ grams ¡ milligrams Conversion factors for some of the English system units commonly used in the United States for nonscientific Using the following conversion factors measurements (for example, pounds and inches) are provided inside the back cover of this book. 453.6 g 1 mg   and   1 lb 1 3 1023 g (Continued) 26 Chapter 1 ■ Chemistry: The Study of Change we obtain the answer in one step: 453.6 g 1 mg ? mg 5 0.0833 lb 3 3 5 3.78 3 104 mg 1 lb 1 3 1023 g Check As an estimate, we note that 1 lb is roughly 500 g and that 1 g 5 1000 mg. Therefore, 1 lb is roughly 5 3 105 mg. Rounding off 0.0833 lb to 0.1 lb, we get Similar problem: 1.45. 5 3 104 mg, which is close to the preceding quantity. Practice Exercise A roll of aluminum foil has a mass of 1.07 kg. What is its mass in pounds? As Examples 1.7 and 1.8 illustrate, conversion factors can be squared or cubed in dimensional analysis. Example 1.7 A liquid helium storage tank has a volume of 275 L. What is the volume in m3? Strategy The problem can be stated as ? m3 5 275 L How many conversion factors are needed for this problem? Recall that 1 L 5 1000 cm3 and 1 cm 5 1 3 1022 m. Solution We need two conversion factors here: one to convert liters to cm3 and one to convert centimeters to meters: 1000 cm3 1 3 1022 m   and   1L 1 cm Because the second conversion deals with length (cm and m) and we want volume here, it must therefore be cubed to give 1 3 1022 m 1 3 1022 m 1 3 1022 m 1 3 1022 m 3 A cryogenic storage tank for 3 3 5a b 1 cm 1 cm 1 cm 1 cm liquid helium. This means that 1 cm3 5 1 3 1026 m3. Now we can write Remember that when a unit is raised to a 1000 cm3 1 3 1022 m 3 power, any conversion factor you use ? m3 5 275 L 3 3a b 5 0.275 m3 must also be raised to that power. 1L 1 cm Check From the preceding conversion factors you can show that 1 L 5 1 3 1023 m3. Therefore, a 275-L storage tank would be equal to 275 3 1023 m3 or 0.275 m3, which Similar problem: 1.50(d). is the answer. Practice Exercise The volume of a room is 1.08 3 108 dm3. What is the volume in m3? Example 1.8 Liquid nitrogen is obtained from liquefied air and is used to prepare frozen goods and in low-temperature research. The density of the liquid at its boiling point (21968C or 77 K) is 0.808 g/cm3. Convert the density to units of kg/m3. (Continued)  1.10 Real-World Problem Solving: Information, Assumptions, and Simplifications 27 Strategy The problem can be stated as ? kg/m3 5 0.808 g/cm3 Two separate conversions are required for this problem: g ¡ kg and cm3 ¡ m3. Recall that 1 kg 5 1000 g and 1 cm 5 1 3 1022 m. Solution In Example 1.7 we saw that 1 cm3 5 1 3 1026 m3. The conversion factors are 1 kg 1 cm3   and   1000 g 1 3 1026 m3 Finally, Liquid nitrogen is used for frozen 3 0.808 g 1 kg 1 cm3 foods and low-temperature ? kg/m 5 3 3 5 808 kg/m3 research. 1 cm3 1000 g 1 3 1026 m3 Check Because 1 m3 5 1 3 106 cm3, we would expect much more mass in 1 m3 than in 1 cm3. Therefore, the answer is reasonable. Similar problem: 1.51. 2 3 Practice Exercise The density of the lightest metal, lithium (Li), is 5.34 3 10 kg/m . Convert the density to g/cm3. Review of Concepts The Food and Drug Administration recommends no more than 65 g of daily intake of fat. What is this mass in pounds? (1 lb 5 453.6 g.) 1.10 Real-World Problem Solving: Information, Assumptions, and Simplifications In chemistry, as in other scientific disciplines, it is not always possible to solve a numerical problem exactly. There are many reasons why this is the case. For example, our understanding of a situation is not complete or data are not fully available. In these cases, we must learn to make an intelligent guess. This approach is sometimes called “ball-park estimates,” which are simple, quick calculations that can be done on the “back of an envelope.” As you can imagine, in many cases the answers are only order-of-magnitude estimates.† In most of the example problems that you have seen so far, as well as the questions given at the end of this and subsequent chapters, the necessary information is provided; however, in order to solve important real-world problems such as those related to med- icine, energy, and agriculture, you must be able to determine what information is needed and where to find it. Much of the information you might need can be found in the various tables located throughout the text, and a list of tables and important figures is given on the inside back cover. In many cases, however, you will need to go to outside sources to find the information you need. Although the Internet is a fast way to find information, you must take care that the source is reliable and well referenced. One excellent source is the National Institute of Standards and Technology (NIST). In order to know what information you need, you will first have to formulate a plan for solving the problem. In addition to the limitations of the theories used in science, typically assumptions are made in setting up and solving the problems based on those theories. These assumptions come at a price, however, as the accuracy of the answer is reduced with increasing simplifications of the problem, as illustrated in Example 1.9. † An order of magnitude is a factor of 10. 28 Chapter 1 ■ Chemistry: The Study of Change Example 1.9 A modern pencil “lead” is actually composed primarily of graphite, a form of carbon. Estimate the mass of the graphite core in a standard No. 2 pencil before it is sharpened. Strategy Assume that the pencil lead can be approximated as a cylinder. Measurement of a typical unsharpened pencil gives a length of about 18 cm (subtracting the length of the eraser head) and a diameter of roughly 2 mm for the lead. The volume of a cylinder V is given by V 5 πr2l, where r is the radius and l is the length. Assuming that the lead is pure graphite, you can calculate the mass of the lead from the volume using the density of graphite given in Table 1.4. Solution Converting the diameter of the lead to units of cm gives 1 cm 2 mm 3 5 0.2 cm 10 mm which, along with the length of the lead, gives 0.2 cm 2 V5πa b 3 18 cm 2 5 0.57 cm3 Rearranging Equation (1.1) gives m5d3V g 5 2.2 3 0.57 cm3 cm3 5 1g Check Rounding off the values used to calculate the volume of the lead gives 3 3 (0.1 cm)2 3 20 cm 5 0.6 cm3. Multiplying that volume by roughly 2 g/cm3 gives Similar problems: 1.105, 1.106, 1.114. around 1 g, which agrees with the value just calculated. Practice Exercise Estimate the mass of air in a ping pong ball. Considering Example 1.9, even if the dimensions of the pencil lead were mea- sured with greater precision, the accuracy of the final answer would be limited by the assumptions made in modeling this problem. The pencil lead is actually a mixture of graphite and clay, where the relative amounts of the two materials determine the soft- ness of the lead, so the density of the material is likely to be different than 2.2 g/cm3. You could probably find a better value for the density of the mixture used to make No. 2 pencils, but it is not worth the effort in this case. Key Equations m d5 (1.1) Equation for density V 5°C ?°C 5 (°F 2 32°F) 3 (1.2) Converting °F to °C 9°F 9°F ?°F 5 3 (°C) 1 32°F (1.3) Converting °C to °F 5°C 1K ? K 5 (°C 1 273.15°C) (1.4) Converting °C to K 1°C  Questions & Problems 29 Summary of Facts & Concepts 1. The study of chemistry involves three basic steps: ob- substances. Mixtures, whether homogeneous or hetero- servation, representation, and interpretation. Observa- geneous, can be separated into pure components by tion refers to measurements in the macroscopic world; physical means. representation involves the use of shorthand notation 4. The simplest substances in chemistry are elements. symbols and equations for communication; interpreta- Compounds are formed by the chemical combination of tions are based on atoms and molecules, which belong atoms of different elements in fixed proportions. to the microscopic world. 5. All substances, in principle, can exist in three states: 2. The scientific method is a systematic approach to re- solid, liquid, and gas. The interconversion between search that begins with the gathering of information these states can be effected by changing the tempera- through observation and measurements. In the process, ture. hypotheses, laws, and theories are devised and tested. 6. SI units are used to express physical quantities in all 3. Chemists study matter and the changes it undergoes. sciences, including chemistry. The substances that make up matter have unique physi- 7. Numbers expressed in scientific notation have the form cal properties that can be observed without changing N 3 10n, where N is between 1 and 10, and n is a posi- their identity and unique chemical properties that, when tive or negative integer. Scientific notation helps us they are demonstrated, do change the identity of the handle very large and very small quantities. Key Words Accuracy, p. 22 Homogeneous mixture, p. 7 Macroscopic property, p. 12 Quantitative, p. 4 Chemical property, p. 11 Hypothesis, p. 4 Mass, p. 11 Scientific method, p. 4 Chemistry, p. 2 Intensive property, p. 11 Matter, p. 6 Significant figures, p. 19 Compound, p. 8 International System of Units Microscopic property, p. 12 Substance, p. 7 Density, p. 11 (SI), p. 12 Mixture, p. 7 Theory, p. 5 Element, p. 7 Kelvin, p. 15 Physical property, p. 10 Volume, p. 11 Extensive property, p. 11 Law, p. 4 Precision, p. 22 Weight, p. 13 Heterogeneous mixture, p. 7 Liter, p. 14 Qualitative, p. 4 Questions & Problems • Problems available in Connect Plus to music would have been much greater if he had Red numbered problems solved in Student Solutions Manual married. (b) An autumn leaf gravitates toward the ground because there is an attractive force between The Scientific Method the leaf and Earth. (c)  All matter is composed of Review Questions very small particles called atoms. 1.1 Explain what is meant by the scientific method. Classification and Properties of Matter 1.2 What is the difference between qualitative data and quantitative data? Review Questions 1.5 Give an example for each of the following terms: Problems (a) matter, (b) substance, (c) mixture. 1.3 Classify the following as qualitative or quantitative • 1.6 Give an example of a homogeneous mixture and an statements, giving your reasons. (a) The sun is ap- example of a heterogeneous mixture. proximately 93 million mi from Earth. (b) Leonardo 1.7 Using examples, explain the difference between a da Vinci was a better painter than Michelangelo. (c) physical property and a chemical property. Ice is less dense than water. (d) Butter tastes better 1.8 How does an intensive property differ from an ex- than margarine. (e) A stitch in time saves nine. tensive property? Which of the following properties • 1.4 Classify each of the following statements as a hypoth- are intensive and which are extensive? (a) length, esis, a law, or a theory. (a) Beethoven’s contribution (b) volume, (c) temperature, (d) mass.  30 Chapter 1 ■ Chemistry: The Study of Change 1.9 Give an example of an element and a compound. Problems How do elements and compounds differ? 1.21 Bromine is a reddish-brown liquid. Calculate its density 1.10 What is the number of known elements? (in g/mL) if 586 g of the substance occupies 188 mL. • 1.22 The density of methanol, a colorless organic liquid Problems used as solvent, is 0.7918 g/mL. Calculate the mass • 1.11 Do the following statements describe chemical or of 89.9 mL of the liquid. physical properties? (a) Oxygen gas supports • 1.23 Convert the following temperatures to degrees combustion. (b) Fertilizers help to increase agricul- Celsius or Fahrenheit: (a) 958F, the temperature on a tural production. (c) Water boils below 1008C on top hot summer day; (b) 128F, the temperature on a cold of a mountain. (d) Lead is denser than aluminum. winter day; (c) a 1028F fever; (d) a furnace operating (e) Uranium is a radioactive element. at 18528F; (e) 2273.158C (theoretically the lowest • 1.12 Does each of the following describe a physical attainable temperature). change or a chemical change? (a) The helium gas • 1.24 (a) Normally the human body can endure a tempera- inside a balloon tends to leak out after a few hours. ture of 1058F for only short periods of time without (b) A flashlight beam slowly gets dimmer and fi- permanent damage to the brain and other vital or- nally goes out. (c) Frozen orange juice is reconsti- gans. What is this temperature in degrees Celsius? tuted by adding water to it. (d) The growth of plants (b) Ethylene glycol is a liquid organic compound depends on the sun’s energy in a process called pho- that is used as an antifreeze in car radiators. It tosynthesis. (e) A spoonful of table salt dissolves in freezes at 211.58C. Calculate its freezing tempera- a bowl of soup. ture in degrees Fahrenheit. (c) The temperature on • 1.13 Give the names of the elements represented by the surface of the sun is about 63008C. What is this the chemical symbols Li, F, P, Cu, As, Zn, Cl, Pt, temperature in degrees Fahrenheit? (d) The ignition Mg, U, Al, Si, Ne. (See Table 1.1 and the inside temperature of paper is 4518F. What is the tempera- front cover.) ture in degrees Celsius? 1.14 Give the chemical symbols for the following elements: • 1.25 Convert the following temperatures to kelvin: (a) cesium, (b) germanium, (c) gallium, (d) strontium, (a) 1138C, the melting point of sulfur, (b) 378C, the (e) uranium, (f) selenium, (g) neon, (h) cadmium. (See normal body temperature, (c) 3578C, the boiling Table 1.1 and the inside front cover.) point of mercury. 1.15 Classify each of the following substances as an ele- 1.26 Convert the following temperatures to degrees Cel- ment or a compound: (a) hydrogen, (b) water, (c) gold, sius: (a) 77 K, the boiling point of liquid nitrogen, (d) sugar. (b) 4.2 K, the boiling point of liquid helium, (c) 601 K, the melting point of lead. • 1.16 Classify each of the following as an element, a compound, a homogeneous mixture, or a heteroge- neous mixture: (a) water from a well, (b) argon gas, Handling Numbers (c) sucrose, (d) a bottle of red wine, (e) chicken Review Questions noodle soup, (f ) blood flowing in a capillary, (g) ozone. 1.27 What is the advantage of using scientific notation over decimal notation? 1.28 Define significant figure. Discuss the importance of Measurement using the proper number of significant figures in Review Questions measurements and calculations. 1.17 Name the SI base units that are important in chemis- try. Give the SI units for expressing the following: Problems (a) length, (b) volume, (c) mass, (d) time, (e) energy, • 1.29 Express the following numbers in scientific notation: (f ) temperature. (a) 0.000000027, (b) 356, (c) 47,764, (d) 0.096. • 1.18 Write the numbers represented by the following pre- 1.30 Express the following numbers as decimals: fixes: (a) mega-, (b) kilo-, (c) deci-, (d) centi-, (e) milli-, (a) 1.52 3 1022, (b) 7.78 3 1028. (f) micro-, (g) nano-, (h) pico-. 1.19 What units do chemists normally use for density of • 1.31 Express the answers to the following calculations in scientific notation: liquids and solids? For gas density? Explain the differences. (a) 145.75 1 (2.3 3 1021) 1.20 Describe the three temperature scales used in the (b) 79,500 4 (2.5 3 102) laboratory and in everyday life: the Fahrenheit scale, (c) (7.0 3 1023) 2 (8.0 3 1024) the Celsius scale, and the Kelvin scale. (d) (1.0 3 104) 3 (9.9 3 106)  Questions & Problems 31 1.32 Express the answers to the following calculations in Dimensional Analysis scientific notation: Problems (a) 0.0095 1 (8.5 3 1023) (b) 653 4 (5.75 3 1028) • 1.39 Carry out the following conversions: (a) 22.6 m to decimeters, (b) 25.4 mg to kilograms, (c) 556 mL to (c) 850,000 2 (9.0 3 105) liters, (d) 10.6 kg/m3 to g/cm3. (d) (3.6 3 1024) 3 (3.6 3 106) 1.40 Carry out the following conversions: (a) 242 lb to • 1.33 What is the number of significant figures in each of milligrams, (b) 68.3 cm3 to cubic meters, (c) 7.2 m3 the following measurements? to liters, (d) 28.3 μg to pounds. (a) 4867 mi • 1.41 The average speed of helium at 258C is 1255 m/s. (b) 56 mL Convert this speed to miles per hour (mph). (c) 60,104 tons 1.42 How many seconds are there in a solar year (365.24 days)? (d) 2900 g 1.43 How many minutes does it take light from the (e) 40.2 g/cm3 sun to reach Earth? (The distance from the sun (f) 0.0000003 cm to Earth is 93 million mi; the speed of light 5 (g) 0.7 min 3.00 3 10 8 m/s.) (h) 4.6 3 1019 atoms • 1.44 A jogger runs a mile in 8.92 min. Calculate the speed 1.34 How many significant figures are there in each of in (a) in/s, (b) m/min, (c) km/h. (1 mi 5 1609 m; the following? (a) 0.006 L, (b) 0.0605 dm, 1 in 5 2.54 cm.) (c) 60.5 mg, (d) 605.5 cm2, (e) 960 3 1023 g, • 1.45 A 6.0-ft person weighs 168 lb. Express this person’s (f) 6 kg, (g) 60 m. height in meters and weight in kilograms. (1 lb 5 • 1.35 Carry out the following operations as if they were 453.6 g; 1 m 5 3.28 ft.) calculations of experimental results, and express 1.46 The speed limit on parts of the German autobahn each answer in the correct units with the correct was once set at 286 kilometers per hour (km/h). Cal- number of significant figures: culate the speed limit in miles per hour (mph). (a) 5.6792 m 1 0.6 m 1 4.33 m 1.47 For a fighter jet to take off from the deck of an air- (b) 3.70 g 2 2.9133 g craft carrier, it must reach a speed of 62 m/s. Calcu- (c) 4.51 cm 3 3.6666 cm late the speed in miles per hour (mph). (d) (3 3 104 g 1 6.827 g)/(0.043 cm3 2 0.021 cm3) • 1.48 The “normal” lead content in human blood is about 0.40 part per million (that is, 0.40 g of lead per mil- • 1.36 Carry out the following operations as if they were lion grams of blood). A value of 0.80 part per mil- calculations of experimental results, and express each answer in the correct units with the correct lion (ppm) is considered to be dangerous. How number of significant figures: many grams of lead are contained in 6.0 3 103 g of blood (the amount in an average adult) if the lead (a) 7.310 km 4 5.70 km content is 0.62 ppm? (b) (3.26 3 1023 mg) 2 (7.88 3 1025 mg) • 1.49 Carry out the following conversions: (a) 1.42 light- (c) (4.02 3 106 dm) 1 (7.74 3 107 dm) years to miles (a light-year is an astronomical measure (d) (7.8 m 2 0.34 m)/(1.15 s 1 0.82 s) of distance—the distance traveled by light in a year, or 365 days; the speed of light is 3.00 3 108 m/s). 1.37 Three students (A, B, and C) are asked to deter- (b) 32.4 yd to centimeters. (c) 3.0 3 1010 cm/s to ft/s. mine the volume of a sample of ethanol. Each stu- dent measures the volume three times with a 1.50 Carry out the following conversions: (a) 70 kg, the graduated cylinder. The results in milliliters are: average weight of a male adult, to pounds. (b) 14 bil- A (87.1, 88.2, 87.6); B (86.9, 87.1, 87.2); C (87.6, lion years (roughly the age of the universe) to sec- 87.8, 87.9). The true volume is 87.0 mL. Com- onds. (Assume there are 365 days in a year.) (c) 7 ft ment on the precision and the accuracy of each 6 in, the height of the basketball player Yao Ming, to student’s results. meters. (d) 88.6 m3 to liters. • 1.38 Three apprentice tailors (X, Y, and Z) are assigned • 1.51 Aluminum is a lightweight metal (density 5 2.70 g/ the task of measuring the seam of a pair of trousers. cm3) used in aircraft construction, high-voltage Each one makes three measurements. The results in transmission lines, beverage cans, and foils. What is inches are X (31.5, 31.6, 31.4); Y (32.8, 32.3, 32.7); its density in kg/m3? Z (31.9, 32.2, 32.1). The true length is 32.0 in. Com- • 1.52 Ammonia gas is used as a refrigerant in large-scale cool- ment on the precision and the accuracy of each tai- ing systems. The density of ammonia gas under certain lor’s measurements. conditions is 0.625 g/L. Calculate its density in g/cm3.  32 Chapter 1 ■ Chemistry: The Study of Change Additional Problems • 1.64 Lithium is the least dense metal known (density: 1.53 Give one qualitative and one quantitative statement 0.53 g/cm3). What is the volume occupied by 1.20 3 about each of the following: (a) water, (b) carbon, 103 g of lithium? (c) iron, (d) hydrogen gas, (e) sucrose (cane sugar), • 1.65 The medicinal thermometer commonly used in (f) table salt (sodium chloride), (g) mercury, homes can be read 60.18F, whereas those in the (h) gold, (i) air. doctor’s office may be accurate to 60.18C. In de- • 1.54 Which of the following statements describe physi- grees Celsius, express the percent error expected cal properties and which describe chemical proper- from each of these thermometers in measuring a ties? (a) Iron has a tendency to rust. (b) Rainwater person’s body temperature of 38.98C. in industrialized regions tends to be acidic. (c) He- • 1.66 Vanillin (used to flavor vanilla ice cream and other moglobin molecules have a red color. (d) When a foods) is the substance whose aroma the human glass of water is left out in the sun, the water gradu- nose detects in the smallest amount. The threshold ally disappears. (e) Carbon dioxide in air is con- limit is 2.0 3 10211 g per liter of air. If the current verted to more complex molecules by plants during price of 50 g of vanillin is $112, determine the cost photosynthesis. to supply enough vanillin so that the aroma could be 1.55 In 2008, about 95.0 billion lb of sulfuric acid were detected in a large aircraft hangar with a volume of produced in the United States. Convert this quantity 5.0 3 107 ft3. to tons. • 1.67 At what temperature does the numerical reading on a Celsius thermometer equal that on a Fahrenheit • 1.56 In determining the density of a rectangular metal thermometer? bar, a student made the following measurements: length, 8.53 cm; width, 2.4 cm; height, 1.0 cm; • 1.68 Suppose that a new temperature scale has been devised mass, 52.7064 g. Calculate the density of the metal on which the melting point of ethanol (2117.38C) and to the correct number of significant figures. the boiling point of ethanol (78.38C) are taken as 08S and 1008S, respectively, where S is the symbol for the • 1.57 Calculate the mass of each of the following: (a) a new temperature scale. Derive an equation relating a sphere of gold with a radius of 10.0 cm [the volume of a sphere with a radius r is V 5 (4/3)πr3; the den- reading on this scale to a reading on the Celsius scale. sity of gold 5 19.3 g/cm3], (b) a cube of platinum of What would this thermometer read at 258C? edge length 0.040 mm (the density of platinum 5 • 1.69 A resting adult requires about 240 mL of pure 21.4 g/cm3), (c) 50.0 mL of ethanol (the density of oxygen/min and breathes about 12 times every min- ethanol 5 0.798 g/mL). ute. If inhaled air contains 20 percent oxygen by volume and exhaled air 16 percent, what is the vol- • 1.58 A cylindrical glass bottle 21.5 cm in length is filled ume of air per breath? (Assume that the volume of with cooking oil of density 0.953 g/mL. If the mass inhaled air is equal to that of exhaled air.) of the oil needed to fill the bottle is 1360 g, calculate the inner diameter of the bottle. • 1.70 (a) Referring to Problem 1.69, calculate the total volume (in liters) of air an adult breathes in a day. • 1.59 The following procedure was used to determine the (b) In a city with heavy traffic, the air contains 2.1 3 volume of a flask. The flask was weighed dry and 1026 L of carbon monoxide (a poisonous gas) per then filled with water. If the masses of the empty liter. Calculate the average daily intake of carbon flask and filled flask were 56.12 g and 87.39 g, re- monoxide in liters by a person. spectively, and the density of water is 0.9976 g/cm3, calculate the volume of the flask in cm3. • 1.71 Three different 25.0-g samples of solid pellets are added to 20.0 mL of water in three different measur- 1.60 The speed of sound in air at room temperature is ing cylinders. The results are shown here. Given the about 343 m/s. Calculate this speed in miles per densities of the three metals used, identify the cylin- hour. (1 mi 5 1609 m.) der that contains each sample of solid pellets: A • 1.61 A piece of silver (Ag) metal weighing 194.3 g is (2.9 g/cm3), B (8.3 g/cm3), and C (3.3 g/cm3). placed in a graduated cylinder containing 242.0 mL of water. The volume of water now reads 260.5 mL. From these data calculate the density 30 30 30 of silver. 1.62 The experiment described in Problem 1.61 is a crude but convenient way to determine the density of some solids. Describe a similar experiment that would en- able you to measure the density of ice. Specifically, 20 20 20 what would be the requirements for the liquid used in your experiment? • 1.63 A lead sphere of diameter 48.6 cm has a mass of 6.852 3 105 g. Calculate the density of lead. (a) (b) (c)  Questions & Problems 33 1.72 The circumference of an NBA-approved basketball • 1.81 Chalcopyrite, the principal ore of copper (Cu), is 29.6 in. Given that the radius of Earth is about contains 34.63 percent Cu by mass. How many 6400 km, how many basketballs would it take to grams of Cu can be obtained from 5.11 3 103 kg of circle around the equator with the basketballs touch- the ore? ing one another? Round off your answer to an inte- 1.82 It has been estimated that 8.0 3 104 tons of gold ger with three significant figures. (Au) have been mined. Assume gold costs $948 per • 1.73 A student is given a crucible and asked to prove ounce. What is the total worth of this quantity of whether it is made of pure platinum. She first weighs gold? the crucible in air and then weighs it suspended in • 1.83 A 1.0-mL volume of seawater contains about 4.0 3 water (density 5 0.9986 g/mL). The readings are 10212 g of gold. The total volume of ocean water is 860.2 g and 820.2 g, respectively. Based on these 1.5 3 1021 L. Calculate the total amount of gold (in measurements and given that the density of plati- grams) that is present in seawater, and the worth of num is 21.45 g/cm3, what should her conclusion be? the gold in dollars (see Problem 1.82). With so much (Hint: An object suspended in a fluid is buoyed up gold out there, why hasn’t someone become rich by by the mass of the fluid displaced by the object. Ne- mining gold from the ocean? glect the buoyance of air.) 1.84 Measurements show that 1.0 g of iron (Fe) contains • 1.74 The surface area and average depth of the Pacific 1.1 3 1022 Fe atoms. How many Fe atoms are in 4.9 g Ocean are 1.8 3 108 km2 and 3.9 3 103 m, respec- of Fe, which is the total amount of iron in the body tively. Calculate the volume of water in the ocean in of an average adult? liters. • 1.85 The thin outer layer of Earth, called the crust, • 1.75 The unit “troy ounce” is often used for precious contains only 0.50 percent of Earth’s total mass metals such as gold (Au) and platinum (Pt). (1 troy and yet is the source of almost all the elements ounce 5 31.103 g.) (a) A gold coin weighs 2.41 (the atmosphere provides elements such as oxy- troy ounces. Calculate its mass in grams. (b) Is a gen, nitrogen, and a few other gases). Silicon troy ounce heavier or lighter than an ounce? (1 lb 5 (Si) is the second most abundant element in 16 oz; 1 lb 5 453.6 g.) Earth’s crust (27.2 percent by mass). Calculate • 1.76 Osmium (Os) is the densest element known (density 5 the mass of silicon in kilograms in Earth’s crust. 22.57 g/cm3). Calculate the mass in pounds and in (The mass of Earth is 5.9 3 1021 tons. 1 ton 5 kilograms of an Os sphere 15 cm in diameter (about 2000 lb; 1 lb 5 453.6 g.) the size of a grapefruit). See Problem 1.57 for vol- • 1.86 The radius of a copper (Cu) atom is roughly 1.3 3 ume of a sphere. 10210 m. How many times can you divide evenly a • 1.77 Percent error is often expressed as the absolute value piece of 10-cm copper wire until it is reduced to two of the difference between the true value and the ex- separate copper atoms? (Assume there are appropri- perimental value, divided by the true value: ate tools for this procedure and that copper atoms Ztrue value 2 experimental valueZ are lined up in a straight line, in contact with each percent error 5 3 100% other. Round off your answer to an integer.) Ztrue valueZ • 1.87 One gallon of gasoline in an automobile’s engine The vertical lines indicate absolute value. Calculate produces on the average 9.5 kg of carbon dioxide, the percent error for the following measurements: which is a greenhouse gas, that is, it promotes the (a) The density of alcohol (ethanol) is found to be warming of Earth’s atmosphere. Calculate the an- 0.802 g/mL. (True value: 0.798 g/mL.) (b) The mass nual production of carbon dioxide in kilograms if of gold in an earring is analyzed to be 0.837 g. (True there are 250 million cars in the United States and value: 0.864 g.) each car covers a distance of 5000 mi at a consump- • 1.78 The natural abundances of elements in the human tion rate of 20 miles per gallon. body, expressed as percent by mass, are: oxygen • 1.88 A sheet of aluminum (Al) foil has a total area (O), 65 percent; carbon (C), 18 percent; hydro- of 1.000 ft2 and a mass of 3.636 g. What is the gen (H), 10 percent; nitrogen (N), 3 percent; cal- thickness of the foil in millimeters? (Density of cium (Ca), 1.6  percent; phosphorus (P), 1.2 Al 5 2.699 g/cm3.) percent; all other elements, 1.2 percent. Calcu- • 1.89 Comment on whether each of the following is a late the mass in grams of each element in the homogeneous mixture or a heterogeneous mix- body of a 62-kg person. ture: (a) air in a closed bottle and (b) air over New • 1.79 The men’s world record for running a mile outdoors York City. (as of 1999) is 3 min 43.13 s. At this rate, how long • 1.90 Chlorine is used to disinfect swimming pools. The would it take to run a 1500-m race? (1 mi 5 1609 m.) accepted concentration for this purpose is 1 ppm • 1.80 Venus, the second closest planet to the sun, has a chlorine, or 1 g of chlorine per million grams of surface temperature of 7.3 3 102 K. Convert this water. Calculate the volume of a chlorine solution temperature to 8C and 8F. (in milliliters) a homeowner should add to her  34 Chapter 1 ■ Chemistry: The Study of Change swimming pool if the solution contains 6.0 percent • 1.95 A human brain weighs about 1 kg and contains chlorine by mass and there are 2.0 3 104 gallons of about 1011 cells. Assuming that each cell is com- water in the pool. (1 gallon 5 3.79 L; density of pletely filled with water (density 5 1 g/mL), calcu- liquids 5 1.0 g/mL.) late the length of one side of such a cell if it were a • 1.91 An aluminum cylinder is 10.0 cm in length and has cube. If the cells are spread out in a thin layer that a radius of 0.25 cm. If the mass of a single Al atom is a single cell thick, what is the surface area in is 4.48 3 10223g, calculate the number of Al atoms square meters? present in the cylinder. The density of aluminum is • 1.96 (a) Carbon monoxide (CO) is a poisonous gas be- 2.70 g/cm3. cause it binds very strongly to the oxygen carrier • 1.92 A pycnometer is a device for measuring the density hemoglobin in blood. A concentration of 8.00 3 102 of liquids. It is a glass flask with a close-fitting ppm by volume of carbon monoxide is considered ground glass stopper having a capillary hole lethal to humans. Calculate the volume in liters oc- through it. (a) The volume of the pycnometer is cupied by carbon monoxide in a room that measures determined by using distilled water at 208C with a 17.6 m long, 8.80 m wide, and 2.64 m high at this known density of 0.99820 g/mL. First, the water is concentration. (b) Prolonged exposure to mercury filled to the rim. With the stopper in place, the fine (Hg) vapor can cause neurological disorders and re- hole allows the excess liquid to escape. The pyc- spiratory problems. For safe air quality control, the nometer is then carefully dried with filter paper. concentration of mercury vapor must be under 0.050 Given that the masses of the empty pycnometer mg/m3. Convert this number to g/L. (c) The general and the same one filled with water are 32.0764 g test for type II diabetes is that the blood sugar (glu- and 43.1195 g, respectively, calculate the volume cose) level should be below 120 mg per deciliter of the pycnometer. (b) If the mass of the pycnom- (mg/dL). Convert this number to micrograms per eter filled with ethanol at 208C is 40.8051 g, calcu- milliliter (μg/mL). late the density of ethanol. (c) Pycnometers can 1.97 A bank teller is asked to assemble “one-dollar” sets also be used to measure the density of solids. First, of coins for his clients. Each set is made of three small zinc granules weighing 22.8476 g are placed quarters, one nickel, and two dimes. The masses of in the pycnometer, which is then filled with water. the coins are: quarter: 5.645 g; nickel: 4.967 g; If the combined mass of the pycnometer plus the dime: 2.316 g. What is the maximum number of zinc granules and water is 62.7728 g, what is the sets that can be assembled from 33.871 kg of quar- density of zinc? ters, 10.432 kg of nickels, and 7.990 kg of dimes? What is the total mass (in g) of the assembled sets of coins? • 1.98 A graduated cylinder is filled to the 40.00-mL mark with a mineral oil. The masses of the cylinder before and after the addition of the mineral oil are 124.966 g and 159.446 g, respectively. In a separate experi- ment, a metal ball bearing of mass 18.713 g is placed in the cylinder and the cylinder is again filled to the 40.00-mL mark with the mineral oil. The combined mass of the ball bearing and mineral oil is 50.952 g. Calculate the density and radius of the ball bearing. • 1.93 In 1849 a gold prospector in California collected a [The volume of a sphere of radius r is (4/3)πr3.] bag of gold nuggets plus sand. Given that the density of 1.99 A chemist in the nineteenth century prepared an un- gold and sand are 19.3 g/cm3 and 2.95 g/cm3, known substance. In general, do you think it would respectively, and that the density of the mixture is be more difficult to prove that it is an element or a 4.17 g/cm3, calculate the percent by mass of gold in compound? Explain. the mixture. • 1.100 Bronze is an alloy made of copper (Cu) and tin (Sn) • 1.94 The average time it takes for a molecule to diffuse a used in applications that require low metal-on-metal distance of x cm is given by friction. Calculate the mass of a bronze cylinder of radius 6.44 cm and length 44.37 cm. The composi- x2 t5 tion of the bronze is 79.42 percent Cu and 20.58 per- 2D cent Sn and the densities of Cu and Sn are 8.94 g/cm3 where t is the time in seconds and D is the diffusion and 7.31 g/cm3, respectively. What assumption coefficient. Given that the diffusion coefficient of should you make in this calculation? glucose is 5.7 3 1027 cm2/s, calculate the time it 1.101 You are given a liquid. Briefly describe steps you would take for a glucose molecule to diffuse 10 μm, would take to show whether it is a pure substance or which is roughly the size of a cell. a homogeneous mixture.  Answers to Practice Exercises 35 1.102 A chemist mixes two liquids A and B to form a ho- the gastric juice (hydrochloric acid) in the stomach mogeneous mixture. The densities of the liquids are to give off carbon dioxide gas. When a 1.328-g tab- 2.0514 g/mL for A and 2.6678 g/mL for B. When let reacted with 40.00 mL of hydrochloric acid (den- she drops a small object into the mixture, she finds sity: 1.140 g/mL), carbon dioxide gas was given off that the object becomes suspended in the liquid; that and the resulting solution weighed 46.699 g. Calcu- is, it neither sinks nor floats. If the mixture is made late the number of liters of carbon dioxide gas re- of 41.37 percent A and 58.63 percent B by volume, leased if its density is 1.81 g/L. what is the density of the metal? Can this procedure 1.104 A 250-mL glass bottle was filled with 242 mL of be used in general to determine the densities of sol- water at 208C and tightly capped. It was then left ids? What assumptions must be made in applying outdoors overnight, where the average temperature this method? was 258C. Predict what would happen. The density • 1.103 Tums is a popular remedy for acid indigestion. A of water at 208C is 0.998 g/cm3 and that of ice at typical Tums tablet contains calcium carbonate plus 258C is 0.916 g/cm3. some inert substances. When ingested, it reacts with Interpreting, Modeling & Estimating 1.105 What is the mass of one mole of ants? (Useful infor- 1.110 Estimate the annual consumption of gasoline by mation: A mole is the unit used for atomic and sub- passenger cars in the United States. atomic particles. It is approximately 6 3 1023. A 1.111 Estimate the total amount of ocean water in liters. 1-cm-long ant weighs about 3 mg.) 1.112 Estimate the volume of blood in an adult in liters. 1.106 How much time (in years) does an 80-year-old per- 1.113 How far (in feet) does light travel in one nanosecond? son spend sleeping during his or her life span? 1.114 Estimate the distance (in miles) covered by an NBA 1.107 Estimate the daily amount of water (in gallons) used player in a professional basketball game. indoors by a family of four in the United States. 1.115 In water conservation, chemists spread a thin film of 1.108 Public bowling alleys generally stock bowling balls a certain inert material over the surface of water to from 8 to 16 lb, where the mass is given in whole cut down on the rate of evaporation of water in res- numbers. Given that regulation bowling balls have a ervoirs. This technique was pioneered by Benjamin diameter of 8.6 in, which (if any) of these bowling Franklin three centuries ago. Franklin found that balls would you expect to float in water? 0.10 mL of oil could spread over the surface of wa- 1.109 Fusing “nanofibers” with ter about 40 m2 in area. Assuming that the oil forms diameters of 100–300 nm a monolayer, that is, a layer that is only one mole- gives junctures with very cule thick, estimate the length of each oil molecule small volumes that would in nanometers. (1 nm 5 1 3 1029 m.) potentially allow the study of reactions involving 1 μm only a few molecules. Es- timate the volume in liters of the junction formed between two such fibers with internal diameters of 200 nm. The scale reads 1 μm. Answers to Practice Exercises 1.1 96.5 g. 1.2 341 g. 1.3 (a) 621.58F, (b) 78.38C, (d) 0.0756 g/mL, (e) 6.69 3 104 m. 1.6 2.36 lb. 1.7 1.08 3 (c) 21968C. 1.4 (a) Two, (b) four, (c) three, (d) two, 105 m3. 1.8 0.534 g/cm3. 1.9 Roughly 0.03 g. (e) three or two. 1.5 (a) 26.76 L, (b) 4.4 g, (c) 1.6 3 107 dm2, CHEMICAL M YS TERY The Disappearance of the Dinosaurs D inosaurs dominated life on Earth for millions of years and then disappeared very suddenly. To solve the mystery, paleontologists studied fossils and skeletons found in rocks in various layers of Earth’s crust. Their findings enabled them to map out which species existed on Earth during specific geologic periods. They also revealed no dinosaur skeletons in rocks formed immediately after the Cretaceous period, which dates back some 36 65 million years. It is therefore assumed that the dinosaurs became extinct about 65 mil- lion years ago. Among the many hypotheses put forward to account for their disappearance were disruptions of the food chain and a dramatic change in climate caused by violent volcanic eruptions. However, there was no convincing evidence for any one hypothesis until 1977. It was then that a group of paleontologists working in Italy obtained some very puzzling data at a site near Gubbio. The chemical analysis of a layer of clay deposited above sediments formed during the Cretaceous period (and therefore a layer that records events occurring after the Cretaceous period) showed a surprisingly high content of the element iridium (Ir). Iridium is very rare in Earth’s crust but is comparatively abundant in asteroids. This investigation led to the hypothesis that the extinction of dinosaurs occurred as follows. To account for the quantity of iridium found, scientists suggested that a large asteroid several miles in diameter hit Earth about the time the dinosaurs disappeared. The impact of the asteroid on Earth’s surface must have been so tremendous that it literally vaporized a large quantity of surrounding rocks, soils, and other objects. The resulting dust and debris floated through the air and blocked the sunlight for months or perhaps years. Without ample sunlight most plants could not grow, and the fossil record confirms that many types of plants did indeed die out at this time. Consequently, of course, many plant- eating animals perished, and then, in turn, meat-eating animals began to starve. Dwindling food sources would obviously affect large animals needing great amounts of food more quickly and more severely than small animals. Therefore, the huge dinosaurs, the largest of which might have weighed as much as 30 tons, vanished due to lack of food. Chemical Clues 1. How does the study of dinosaur extinction illustrate the scientific method? 2. Suggest two ways that would enable you to test the asteroid collision hypothesis. 3. In your opinion, is it justifiable to refer to the asteroid explanation as the theory of dinosaur extinction? 4. Available evidence suggests that about 20 percent of the asteroid’s mass turned to dust and spread uniformly over Earth after settling out of the upper atmosphere. This dust amounted to about 0.02 g/cm2 of Earth’s surface. The asteroid very likely had a density of about 2 g/cm3. Calculate the mass (in kilograms and tons) of the asteroid and its radius in meters, assuming that it was a sphere. (The area of Earth is 5.1 3 1014 m2; 1 lb 5 453.6 g.) (Source: Consider a Spherical Cow—A Course in Environ- mental Problem Solving by J. Harte, University Science Books, Mill Valley, CA 1988. Used with permission.) 37  CHAPTER 2 Atoms, Molecules, and Ions Illustration depicting Marie and Pierre Curie at work in their laboratory. The Curies studied and identified many radioactive elements. CHAPTER OUTLINE A LOOK AHEAD 2.1 The Atomic Theory  We begin with a historical perspective of the search for the fundamental units of matter. The modern version of atomic theory was laid by John Dalton 2.2 The Structure of the Atom in the nineteenth century, who postulated that elements are composed of 2.3 Atomic Number, Mass extremely small particles, called atoms. All atoms of a given element are Number, and Isotopes identical, but they are different from atoms of all other elements. (2.1) 2.4 The Periodic Table  We note that, through experimentation, scientists have learned that an atom is composed of three elementary particles: proton, electron, and neutron. 2.5 Molecules and Ions The proton has a positive charge, the electron has a negative charge, and the 2.6 Chemical Formulas neutron has no charge. Protons and neutrons are located in a small region at the center of the atom, called the nucleus, while electrons are spread out 2.7 Naming Compounds about the nucleus at some distance from it. (2.2) 2.8 Introduction to Organic  We will learn the following ways to identify atoms. Atomic number is the Compounds number of protons in a nucleus; atoms of different elements have different atomic numbers. Isotopes are atoms of the same element having a different number of neutrons. Mass number is the sum of the number of protons and neutrons in an atom. Because an atom is electrically neutral, the number of protons is equal to the number of electrons in it. (2.3)  Next we will see how elements can be grouped together according to their chemical and physical properties in a chart called the periodic table. The periodic table enables us to classify elements (as metals, metalloids, and nonmetals) and correlate their properties in a systematic way. (2.4)  We will see that atoms of most elements interact to form compounds, which are classified as molecules or ionic compounds made of positive (cations) and negative (anions) ions. (2.5)  We learn to use chemical formulas (molecular and empirical) to represent molecules and ionic compounds and models to represent molecules. (2.6)  We learn a set of rules that help us name the inorganic compounds. (2.7)  Finally, we will briefly explore the organic world to which we will return in a later chapter. (2.8) 38  2.1 The Atomic Theory 39 S ince ancient times humans have pondered the nature of matter. Our modern ideas of the structure of matter began to take shape in the early nineteenth century with Dalton’s atomic theory. We now know that all matter is made of atoms, molecules, and ions. All of chemistry is concerned in one way or another with these species. 2.1 The Atomic Theory In the fifth century b.c. the Greek philosopher Democritus expressed the belief that all matter consists of very small, indivisible particles, which he named atomos (meaning uncuttable or indivisible). Although Democritus’ idea was not accepted by many of his contemporaries (notably Plato and Aristotle), somehow it endured. Experimental evidence from early scientific investigations provided support for the notion of “atomism” and gradually gave rise to the modern definitions of elements and compounds. In 1808 an English scientist and school teacher, John Dalton,† formulated a precise definition of the indivisible building blocks of matter that we call atoms. Dalton’s work marked the beginning of the modern era of chemistry. The hypoth- eses about the nature of matter on which Dalton’s atomic theory is based can be summarized as follows: 1. Elements are composed of extremely small particles called atoms. 2. All atoms of a given element are identical, having the same size, mass, and chemical properties. The atoms of one element are different from the atoms of all other elements. 3. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction. 4. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction. Figure 2.1 is a schematic representation of the last three hypotheses. Dalton’s concept of an atom was far more detailed and specific than Democritus’. The second hypothesis states that atoms of one element are different from atoms of all other elements. Dalton made no attempt to describe the structure or composition of atoms—he had no idea what an atom is really like. But he did realize that the † John Dalton (1766–1844). English chemist, mathematician, and philosopher. In addition to the atomic theory, he also formulated several gas laws and gave the first detailed description of color blindness, from which he suffered. Dalton was described as an indifferent experimenter, and singularly wanting in the language and power of illustration. His only recreation was lawn bowling on Thursday afternoons. Perhaps it was the sight of those wooden balls that provided him with the idea of the atomic theory. Figure 2.1 (a) According to Dalton’s atomic theory, atoms of the same element are identical, but atoms of one element are different from atoms of other elements. (b) Compound formed from atoms of elements X and Y. In this case, the ratio of the atoms of element X to the atoms of element Y is 2:1. Note that a chemical reaction results only in the rearrangement of atoms, not Atoms of element X Atoms of element Y Compounds of elements X and Y in their destruction or creation. (a) (b) 40 Chapter 2 ■ Atoms, Molecules, and Ions Carbon monoxide different properties shown by elements such as hydrogen and oxygen can be explained by assuming that hydrogen atoms are not the same as oxygen atoms. O 1 ± 5 ±±± 5 ± The third hypothesis suggests that, to form a certain compound, we need not only C 1 atoms of the right kinds of elements, but specific numbers of these atoms as well. This idea is an extension of a law published in 1799 by Joseph Proust,† a French chemist. Proust’s law of definite proportions states that different samples of the same Carbon dioxide compound always contain its constituent elements in the same proportion by mass. O 2 Thus, if we were to analyze samples of carbon dioxide gas obtained from different ± 5 ±±±±±±± 5 ± sources, we would find in each sample the same ratio by mass of carbon to oxygen. C 1 It stands to reason, then, that if the ratio of the masses of different elements in a given compound is fixed, the ratio of the atoms of these elements in the compound also Ratio of oxygen in must be constant. carbon monoxide to Dalton’s third hypothesis supports another important law, the law of multiple oxygen in carbon dioxide: 1:2 proportions. According to the law, if two elements can combine to form more than Figure 2.2 An illustration of the one compound, the masses of one element that combine with a fixed mass of the law of multiple proportions. other element are in ratios of small whole numbers. Dalton’s theory explains the law of multiple proportions quite simply: Different compounds made up of the same elements differ in the number of atoms of each kind that combine. For example, carbon forms two stable compounds with oxygen, namely, carbon monoxide and carbon dioxide. Modern measurement techniques indicate that one atom of carbon combines with one atom of oxygen in carbon monoxide and with two atoms of oxygen in carbon dioxide. Thus, the ratio of oxygen in carbon monoxide to oxygen in carbon dioxide is 1:2. This result is consistent with the law of multiple propor- tions (Figure 2.2). Dalton’s fourth hypothesis is another way of stating the law of conservation of mass,‡ which is that matter can be neither created nor destroyed. Because matter is made of atoms that are unchanged in a chemical reaction, it follows that mass must be conserved as well. Dalton’s brilliant insight into the nature of matter was the main stimulus for the rapid progress of chemistry during the nineteenth century. Review of Concepts The atoms of elements A (blue) and B (orange) form two compounds shown here. Do these compounds obey the law of multiple proportions? 2.2 The Structure of the Atom On the basis of Dalton’s atomic theory, we can define an atom as the basic unit of an element that can enter into chemical combination. Dalton imagined an atom that was both extremely small and indivisible. However, a series of investigations that began in the 1850s and extended into the twentieth century clearly demonstrated that atoms actually possess internal structure; that is, they are made up of even smaller particles, which are called subatomic particles. This research led to the discovery of three such particles—electrons, protons, and neutrons. † Joseph Louis Proust (1754–1826). French chemist. Proust was the first person to isolate sugar from grapes. ‡ According to Albert Einstein, mass and energy are alternate aspects of a single entity called mass-energy. Chemical reactions usually involve a gain or loss of heat and other forms of energy. Thus, when energy is lost in a reaction, for example, mass is also lost. Except for nuclear reactions (see Chapter 19), however, changes of mass in chemical reactions are too small to detect. Therefore, for all practical purposes mass is conserved.  2.2 The Structure of the Atom 41 Figure 2.3 A cathode ray tube with an electric field perpendicular (−) A to the direction of the cathode rays and an external magnetic field. The symbols N and S denote Cathode ray the north and south poles of the Anode magnet. The cathode rays will (+) B strike the end of the tube at A in Cathode the presence of a magnetic field, (−) at C in the presence of an electric field, and at B when there are no external fields present or when C the effects of the electric field and magnetic field cancel each other. Evacuated tube (+) Fluorescent screen Magnet The Electron In the 1890s, many scientists became caught up in the study of radiation, the emission and transmission of energy through space in the form of waves. Information gained from this research contributed greatly to our understanding of atomic structure. One Animation device used to investigate this phenomenon was a cathode ray tube, the forerunner of Cathode Ray Tube the television tube (Figure 2.3). It is a glass tube from which most of the air has been evacuated. When the two metal plates are connected to a high-voltage source, the negatively charged plate, called the cathode, emits an invisible ray. The cathode ray is drawn to the positively charged plate, called the anode, where it passes through a hole and continues traveling to the other end of the tube. When the ray strikes the specially coated surface, it produces a strong fluorescence, or bright light. In some experiments, two electrically charged plates and a magnet were added to the outside of the cathode ray tube (see Figure 2.3). When the magnetic field is on and the electric field is off, the cathode ray strikes point A. When only the electric field is on, the ray strikes point C. When both the magnetic and the electric fields are off or when they are both on but balanced so that they cancel each other’s influence, the ray strikes point B. According to electromagnetic theory, a moving charged body behaves like a magnet and can interact with electric and magnetic fields through which it passes. Because the cathode ray is attracted by the plate bearing positive charges and repelled by the plate bearing negative Electrons are normally associated with charges, it must consist of negatively charged particles. We know these negatively charged atoms. However, they can also be studied particles as electrons. Figure 2.4 shows the effect of a bar magnet on the cathode ray. individually. (a) (b) (c) Figure 2.4 (a) A cathode ray produced in a discharge tube traveling from the cathode (left) to the anode (right). The ray itself is invisible, but the fluorescence of a zinc sulfide coating on the glass causes it to appear green. (b) The cathode ray is bent downward when a bar magnet is brought toward it. (c) When the polarity of the magnet is reversed, the ray bends in the opposite direction. 42 Chapter 2 ■ Atoms, Molecules, and Ions Figure 2.5 Schematic diagram of Millikan’s oil drop experiment. Atomizer Fine mist of oil particles Electrically charged plates (+) X ray source to produce charge on oil droplet Viewing microscope (–) An English physicist, J. J. Thomson,† used a cathode ray tube and his knowledge of electromagnetic theory to determine the ratio of electric charge to the mass of an individual electron. The number he came up with was 21.76 3 108 C/g, where C stands for coulomb, which is the unit of electric charge. Thereafter, in a series of Animation experiments carried out between 1908 and 1917, R. A. Millikan‡ succeeded in measur- Millikan Oil Drop ing the charge of the electron with great precision. His work proved that the charge on each electron was exactly the same. In his experiment, Millikan examined the motion of single tiny drops of oil that picked up static charge from ions in the air. He suspended the charged drops in air by applying an electric field and followed their motions through a microscope (Figure 2.5). Using his knowledge of electrostatics, Millikan found the charge of an electron to be 21.6022 3 10219 C. From these data he calculated the mass of an electron: charge mass of an electron 5 charge/mass 21.6022 3 10219 C 5 21.76 3 108 C/g 5 9.10 3 10228 g This is an exceedingly small mass. Radioactivity In 1895 the German physicist Wilhelm Röntgen§ noticed that cathode rays caused glass and metals to emit very unusual rays. This highly energetic radiation penetrated matter, darkened covered photographic plates, and caused a variety of substances to † Joseph John Thomson (1856–1940). British physicist who received the Nobel Prize in Physics in 1906 for discovering the electron. ‡ Robert Andrews Millikan (1868–1953). American physicist who was awarded the Nobel Prize in Physics in 1923 for determining the charge of the electron. § Wilhelm Konrad Röntgen (1845–1923). German physicist who received the Nobel Prize in Physics in 1901 for the discovery of X rays.  2.2 The Structure of the Atom 43 Figure 2.6 Three types of rays emitted by radioactive elements. β rays consist of negatively charged particles (electrons) and are therefore attracted by the positively charged plate. The opposite holds true for α rays— – they are positively charged and α are drawn to the negatively Lead block charged plate. Because γ rays have no charges, their path is unaffected by an external γ electric field. β + Radioactive substance fluoresce. Because these rays could not be deflected by a magnet, they could not contain charged particles as cathode rays do. Röntgen called them X rays because their nature was not known. Not long after Röntgen’s discovery, Antoine Becquerel,† a professor of physics in Paris, began to study the fluorescent properties of substances. Purely by accident, he found that exposing thickly wrapped photographic plates to a certain uranium compound caused them to darken, even without the stimulation of cathode rays. Like X rays, the rays from the uranium compound were highly energetic and could not be deflected by a magnet, but they differed from X rays because they arose spontane- ously. One of Becquerel’s students, Marie Curie,‡ suggested the name radioactivity to describe this spontaneous emission of particles and/or radiation. Since then, any element that spontaneously emits radiation is said to be radioactive. Three types of rays are produced by the decay, or breakdown, of radioactive substances such as uranium. Two of the three are deflected by oppositely charged metal plates (Figure 2.6). Alpha (α) rays consist of positively charged particles, called Animation Alpha, Beta, and Gamma Rays α particles, and therefore are deflected by the positively charged plate. Beta (β) rays, or β particles, are electrons and are deflected by the negatively charged plate. The third type of radioactive radiation consists of high-energy rays called gamma (γ) rays. Like X rays, γ rays have no charge and are not affected by an external field. Positive charge spread over the entire sphere The Proton and the Nucleus By the early 1900s, two features of atoms had become clear: They contain electrons, and they are electrically neutral. To maintain electric neutrality, an atom must contain – an equal number of positive and negative charges. Therefore, Thomson proposed that – – an atom could be thought of as a uniform, positive sphere of matter in which electrons are embedded like raisins in a cake (Figure 2.7). This so-called “plum-pudding” model – was the accepted theory for a number of years. – – † Antoine Henri Becquerel (1852–1908). French physicist who was awarded the Nobel Prize in Physics in – 1903 for discovering radioactivity in uranium. ‡ Marie (Marya Sklodowska) Curie (1867–1934). Polish-born chemist and physicist. In 1903 she and her Figure 2.7 Thomson’s model of French husband, Pierre Curie, were awarded the Nobel Prize in Physics for their work on radioactivity. In the atom, sometimes described as 1911, she again received the Nobel prize, this time in chemistry, for her work on the radioactive elements the “plum-pudding” model, after a radium and polonium. She is one of only three people to have received two Nobel prizes in science. Despite traditional English dessert her great contribution to science, her nomination to the French Academy of Sciences in 1911 was rejected containing raisins. The electrons by one vote because she was a woman! Her daughter Irene, and son-in-law Frederic Joliot-Curie, shared are embedded in a uniform, the Nobel Prize in Chemistry in 1935. positively charged sphere. 44 Chapter 2 ■ Atoms, Molecules, and Ions Figure 2.8 (a) Rutherford’s Gold foil experimental design for measuring the scattering of α a –Particle particles by a piece of gold foil. emitter Most of the α particles passed through the gold foil with little or no deflection. A few were deflected at wide angles. Occasionally an α particle was turned back. (b) Magnified view of α particles passing through and Detecting screen Slit being deflected by nuclei. (a) (b) Animation In 1910 the New Zealand physicist Ernest Rutherford,† who had studied with α-Particle Scattering Thomson at Cambridge University, decided to use α particles to probe the structure of atoms. Together with his associate Hans Geiger‡ and an undergraduate named Animation Ernest Marsden,§ Rutherford carried out a series of experiments using very thin foils Rutherford’s Experiment of gold and other metals as targets for α particles from a radioactive source (Figure 2.8). They observed that the majority of particles penetrated the foil either unde- flected or with only a slight deflection. But every now and then an α particle was scattered (or deflected) at a large angle. In some instances, an α particle actually bounced back in the direction from which it had come! This was a most surprising finding, for in Thomson’s model the positive charge of the atom was so diffuse that the positive α particles should have passed through the foil with very little deflection. To quote Rutherford’s initial reaction when told of this discovery: “It was as incred- ible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Rutherford was later able to explain the results of the α-scattering experiment in terms of a new model for the atom. According to Rutherford, most of the atom must be empty space. This explains why the majority of α particles passed through the gold foil with little or no deflection. The atom’s positive charges, Rutherford proposed, are all concentrated in the nucleus, which is a dense central core within the atom. When- ever an α particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an α particle trav- eling directly toward a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the nucleus are called protons. In separate experiments, it was found that each proton carries the same quantity of charge as an electron and has a mass of 1.67262 3 10224 g—about 1840 times the mass of the oppositely charged electron. At this stage of investigation, scientists perceived the atom as follows: The mass of a nucleus constitutes most of the mass of the entire atom, but the nucleus occupies A common non-SI unit for atomic length only about 1/1013 of the volume of the atom. We express atomic (and molecular) is the angstrom (Å; 1 Å 5 100 pm). dimensions in terms of the SI unit called the picometer (pm), where 1 pm 5 1 3 10212 m † Ernest Rutherford (1871–1937). New Zealand physicist. Rutherford did most of his work in England (Manchester and Cambridge Universities). He received the Nobel Prize in Chemistry in 1908 for his inves- tigations into the structure of the atomic nucleus. His often-quoted comment to his students was that “all science is either physics or stamp-collecting.” ‡ Johannes Hans Wilhelm Geiger (1882–1945). German physicist. Geiger’s work focused on the structure of the atomic nucleus and on radioactivity. He invented a device for measuring radiation that is now com- monly called the Geiger counter. § Ernest Marsden (1889–1970). English physicist. It is gratifying to know that at times an undergraduate can assist in winning a Nobel prize. Marsden went on to contribute significantly to the development of science in New Zealand.  2.2 The Structure of the Atom 45 A typical atomic radius is about 100 pm, whereas the radius of an atomic nucleus is only about 5 3 1023 pm. You can appreciate the relative sizes of an atom and its nucleus by imagining that if an atom were the size of a sports stadium, the volume of its nucleus would be comparable to that of a small marble. Although the protons are confined to the nucleus of the atom, the electrons are conceived of as being spread out about the nucleus at some distance from it. The concept of atomic radius is useful experimentally, but we should not infer that atoms have well-defined boundaries or surfaces. We will learn later that the outer regions of atoms are relatively “fuzzy.” If the size of an atom were expanded to that of this sports stadium, the size of the nucleus The Neutron would be that of a marble. Rutherford’s model of atomic structure left one major problem unsolved. It was known that hydrogen, the simplest atom, contains only one proton and that the helium atom contains two protons. Therefore, the ratio of the mass of a helium atom to that of a hydrogen atom should be 2:1. (Because electrons are much lighter than protons, their contribution to atomic mass can be ignored.) In reality, however, the ratio is 4:1. Rutherford and others postulated that there must be another type of subatomic particle in the atomic nucleus; the proof was provided by another English physicist, James Chadwick,† in 1932. When Chadwick bom- barded a thin sheet of beryllium with α particles, a very high-energy radiation similar to γ rays was emitted by the metal. Later experiments showed that the rays actually consisted of a third type of subatomic particles, which Chadwick named neutrons, because they proved to be electrically neutral particles having a mass slightly greater than that of protons. The mystery of the mass ratio could now be explained. In the helium nucleus there are two protons and two neutrons, but in the hydrogen nucleus there is only one proton and no neutrons; therefore, the ratio is 4:1. Figure 2.9 shows the location of the elementary particles (protons, neutrons, and electrons) in an atom. There are other subatomic particles, but the electron, the † James Chadwick (1891–1972). British physicist. In 1935 he received the Nobel Prize in Physics for proving the existence of neutrons. Figure 2.9 The protons and neutrons of an atom are packed in an extremely small nucleus. Electrons are shown as “clouds” around the nucleus. Proton Neutron 46 Chapter 2 ■ Atoms, Molecules, and Ions Table 2.1 Mass and Charge of Subatomic Particles Charge Particle Mass (g) Coulomb Charge Unit 228 219 Electron* 9.10938 3 10 21.6022 3 10 21 Proton 1.67262 3 10224 11.6022 3 10219 11 Neutron 1.67493 3 10224 0 0 *More refined measurements have given us a more accurate value of an electron’s mass than Millikan’s. proton, and the neutron are the three fundamental components of the atom that are important in chemistry. Table 2.1 shows the masses and charges of these three elementary particles. 2.3 Atomic Number, Mass Number, and Isotopes All atoms can be identified by the number of protons and neutrons they contain. The atomic number (Z) is the number of protons in the nucleus of each atom of an ele- ment. In a neutral atom the number of protons is equal to the number of electrons, so the atomic number also indicates the number of electrons present in the atom. The chemical identity of an atom can be determined solely from its atomic number. For example, the atomic number of fluorine is 9. This means that each fluorine atom has 9 protons and 9 electrons. Or, viewed another way, every atom in the universe that contains 9 protons is correctly named “fluorine.” The mass number (A) is the total number of neutrons and protons present in the Protons and neutrons are collectively nucleus of an atom of an element. Except for the most common form of hydrogen, called nucleons. which has one proton and no neutrons, all atomic nuclei contain both protons and neutrons. In general, the mass number is given by mass number 5 number of protons 1 number of neutrons (2.1) 5 atomic number 1 number of neutrons The number of neutrons in an atom is equal to the difference between the mass number and the atomic number, or (A 2 Z). For example, if the mass number of a particular boron atom is 12 and the atomic number is 5 (indicating 5 protons in the nucleus), then the number of neutrons is 12 2 5 5 7. Note that all three quantities (atomic number, number of neutrons, and mass number) must be positive integers, or whole numbers. Atoms of a given element do not all have the same mass. Most elements have two or more isotopes, atoms that have the same atomic number but different mass numbers. For example, there are three isotopes of hydrogen. One, simply known as hydrogen, has one proton and no neutrons. The deuterium isotope contains one proton and one neutron, and tritium has one proton and two neutrons. The accepted way to denote the atomic number and mass number of an atom of an element (X) is as follows: mass number 8n A ZX atomic number 8n Thus, for the isotopes of hydrogen, we write 1 2 3 1H 1H 1H 1 2 3 1H 1H 1H hydrogen deuterium tritium  2.3 Atomic Number, Mass Number, and Isotopes 47 As another example, consider two common isotopes of uranium with mass numbers of 235 and 238, respectively: 235 238 92U 92U The first isotope is used in nuclear reactors and atomic bombs, whereas the second isotope lacks the properties necessary for these applications. With the exception of hydrogen, which has different names for each of its isotopes, isotopes of elements are identified by their mass numbers. Thus, the preceding two isotopes are called uranium-235 (pronounced “uranium two thirty-five”) and uranium-238 (pronounced “uranium two thirty-eight”). The chemical properties of an element are determined primarily by the protons and electrons in its atoms; neutrons do not take part in chemical changes under nor- mal conditions. Therefore, isotopes of the same element have similar chemistries, forming the same types of compounds and displaying similar reactivities. Example 2.1 shows how to calculate the number of protons, neutrons, and elec- trons using atomic numbers and mass numbers. Example 2.1 Give the number of protons, neutrons, and electrons in each of the following species: (a) 20 22 11Na, (b) 11Na, (c) 17 O, and (d) carbon-14. Strategy Recall that the superscript denotes the mass number (A) and the subscript denotes the atomic number (Z). Mass number is always greater than atomic number. (The only exception is 11H, where the mass number is equal to the atomic number.) In a  case where no subscript is shown, as in parts (c) and (d), the atomic number can be deduced from the element symbol or name. To determine the number of electrons, remember that because atoms are electrically neutral, the number of electrons is equal to the number of protons. Solution (a) The atomic number is 11, so there are 11 protons. The mass number is 20, so the number of neutrons is 20 2 11 5 9. The number of electrons is the same as the number of protons; that is, 11. (b) The atomic number is the same as that in (a), or 11. The mass number is 22, so the number of neutrons is 22 2 11 5 11. The number of electrons is 11. Note that the species in (a) and (b) are chemically similar isotopes of sodium. (c) The atomic number of O (oxygen) is 8, so there are 8 protons. The mass number is 17, so there are 17 2 8 5 9 neutrons. There are 8 electrons. (d) Carbon-14 can also be represented as 14C. The atomic number of carbon is 6, so there are 14 2 6 5 8 neutrons. The number of electrons is 6. Similar problems: 2.15, 2.16. Practice Exercise How many protons, neutrons, and electrons are in the following 63 isotope of copper: Cu? Review of Concepts (a) What is the atomic number of an element if one of its isotopes has 117 neutrons and a mass number of 195? (b) Which of the following two symbols provides more information? 17 O or 8O. 48 Chapter 2 ■ Atoms, Molecules, and Ions 2.4 The Periodic Table More than half of the elements known today were discovered between 1800 and 1900. During this period, chemists noted that many elements show strong similarities to one another. Recognition of periodic regularities in physical and chemical behavior and the need to organize the large volume of available information about the structure and properties of elemental substances led to the development of the periodic table, a chart in which elements having similar chemical and physical properties are grouped together. Figure 2.10 shows the modern periodic table in which the elements are arranged by atomic number (shown above the element symbol) in horizontal rows called periods and in vertical columns known as groups or families, according to similarities in their chemical properties. Note that elements 113–118 have recently been synthesized, although they have not yet been named. The elements can be divided into three categories—metals, nonmetals, and metalloids. A metal is a good conductor of heat and electricity while a non- metal is usually a poor conductor of heat and electricity. A metalloid has proper- ties that are intermediate between those of metals and nonmetals. Figure 2.10 shows that the majority of known elements are metals; only 17 elements are nonmetals, and 8 elements are metalloids. From left to right across any period, the physical and chemical properties of the elements change gradually from metal- lic to nonmetallic. 1 18 1A 8A 1 2 2 13 14 15 16 17 H 2A 3A 4A 5A 6A 7A He 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Metals Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Metalloids Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Nonmetals Figure 2.10 The modern periodic table. The elements are arranged according to the atomic numbers above their symbols. With the exception of hydrogen (H), nonmetals appear at the far right of the table. The two rows of metals beneath the main body of the table are conventionally set apart to keep the table from being too wide. Actually, cerium (Ce) should follow lanthanum (La), and thorium (Th) should come right after actinium (Ac). The 1–18 group designation has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) but is not yet in wide use. In this text, we use the standard U.S. notation for group numbers (1A–8A and 1B–8B). No names have yet been assigned to elements 113, 115, 117, and 118. CHEMISTRY in Action Distribution of Elements on Earth and in Living Systems T he majority of elements are naturally occurring. How are these elements distributed on Earth, and which are essential to living systems? elements, we should keep in mind that (1) the elements are not evenly distributed throughout Earth’s crust, and (2) most ele- ments occur in combined forms. These facts provide the basis Earth’s crust extends from the surface to a depth of about for most methods of obtaining pure elements from their com- 40 km (about 25 mi). Because of technical difficulties, scientists pounds, as we will see in later chapters. have not been able to study the inner portions of Earth as easily The accompanying table lists the essential elements in the as the crust. Nevertheless, it is believed that there is a solid core human body. Of special interest are the trace elements, such as consisting mostly of iron at the center of Earth. Surrounding the iron (Fe), copper (Cu), zinc (Zn), iodine (I), and cobalt (Co), core is a layer called the mantle, which consists of hot fluid con- which together make up about 0.1 percent of the body’s mass. taining iron, carbon, silicon, and sulfur. These elements are necessary for biological functions such as Of the 83 elements that are found in nature, 12 make up growth, transport of oxygen for metabolism, and defense 99.7 percent of Earth’s crust by mass. They are, in decreasing against disease. There is a delicate balance in the amounts of order of natural abundance, oxygen (O), silicon (Si), aluminum these elements in our bodies. Too much or too little over an (Al), iron (Fe), calcium (Ca), magnesium (Mg), sodium (Na), extended period of time can lead to serious illness, retardation, potassium (K), titanium (Ti), hydrogen (H), phosphorus (P), or even death. and manganese (Mn). In discussing the natural abundance of the Mantle Essential Elements in the Human Body Crust Element Percent by Mass* Element Percent by Mass* Oxygen 65 Sodium 0.1 Carbon 18 Magnesium 0.05 Core Hydrogen 10 Iron ,0.05 Nitrogen 3 Cobalt ,0.05 Calcium 1.6 Copper ,0.05 Phosphorus 1.2 Zinc ,0.05 Potassium 0.2 Iodine ,0.05 Sulfur 0.2 Selenium ,0.01 2900 km 3480 km Chlorine 0.2 Fluorine ,0.01 Structure of Earth’s interior. *Percent by mass gives the mass of the element in grams present in a 100-g sample. All others 5.3% Magnesium 2.8% Calcium 4.7% Oxygen All others 1.2% 45.5% Oxygen 65% Phosphorus 1.2% Iron 6.2% Calcium 1.6% Nitrogen 3% Silicon Aluminum 8.3% Carbon 27.2% Hydrogen 10% 18% (a) (b) (a) Natural abundance of the elements in percent by mass. For example, oxygen’s abundance is 45.5 percent. This means that in a 100-g sample of Earth’s crust there are, on the average, 45.5 g of the element oxygen. (b) Abundance of elements in the human body in percent by mass. 49 50 Chapter 2 ■ Atoms, Molecules, and Ions Elements are often referred to collectively by their periodic table group number (Group 1A, Group 2A, and so on). However, for convenience, some element groups have been given special names. The Group 1A elements (Li, Na, K, Rb, Cs, and Fr) are called alkali metals, and the Group 2A elements (Be, Mg, Ca, Sr, Ba, and Ra) are called alkaline earth metals. Elements in Group 7A (F, Cl, Br, I, and At) are known as halogens, and elements in Group 8A (He, Ne, Ar, Kr, Xe, and Rn) are called noble gases, or rare gases. The periodic table is a handy tool that correlates the properties of the elements in a systematic way and helps us to make predictions about chemical behavior. We will take a closer look at this keystone of chemistry in Chapter 8. The Chemistry in Action essay on p. 49 describes the distribution of the elements on Earth and in the human body. Review of Concepts In viewing the periodic table, do chemical properties change more markedly across a period or down a group? 2.5 Molecules and Ions Of all the elements, only the six noble gases in Group 8A of the periodic table (He, Ne, Ar, Kr, Xe, and Rn) exist in nature as single atoms. For this reason, they are called monatomic (meaning a single atom) gases. Most matter is composed of mol- ecules or ions formed by atoms. Molecules We will discuss the nature of chemical A molecule is an aggregate of at least two atoms in a definite arrangement held bonds in Chapters 9 and 10. together by chemical forces (also called chemical bonds). A molecule may contain atoms of the same element or atoms of two or more elements joined in a fixed ratio, in accordance with the law of definite proportions stated in Section 2.1. Thus, a mol- ecule is not necessarily a compound, which, by definition, is made up of two or more elements (see Section 1.4). Hydrogen gas, for example, is a pure element, but it consists of molecules made up of two H atoms each. Water, on the other hand, is a molecular compound that contains hydrogen and oxygen in a ratio of two H atoms and one O atom. Like atoms, molecules are electrically neutral. 1A 8A H 2A 3A 4A 5A 6A 7A The hydrogen molecule, symbolized as H2, is called a diatomic molecule because N O F Cl it contains only two atoms. Other elements that normally exist as diatomic molecules Br are nitrogen (N2) and oxygen (O2), as well as the Group 7A elements—fluorine (F2), I chlorine (Cl2), bromine (Br2), and iodine (I2). Of course, a diatomic molecule can contain atoms of different elements. Examples are hydrogen chloride (HCl) and car- Elements that exist as diatomic bon monoxide (CO). molecules. The vast majority of molecules contain more than two atoms. They can be atoms of the same element, as in ozone (O3), which is made up of three atoms of oxygen, or they can be combinations of two or more different elements. Molecules containing more than two atoms are called polyatomic molecules. Like ozone, water (H2O) and ammonia (NH3) are polyatomic molecules. Ions An ion is an atom or a group of atoms that has a net positive or negative charge. The number of positively charged protons in the nucleus of an atom remains the same during ordinary chemical changes (called chemical reactions), but negatively charged  2.5 Molecules and Ions 51 electrons may be lost or gained. The loss of one or more electrons from a neutral atom results in a cation, an ion with a net positive charge. For example, a sodium In Chapter 8 we will see why atoms of different elements gain (or lose) a atom (Na) can readily lose an electron to become a sodium cation, which is repre- specific number of electrons. sented by Na1: Na Atom Na1 Ion 11 protons 11 protons 11 electrons 10 electrons On the other hand, an anion is an ion whose net charge is negative due to an increase in the number of electrons. A chlorine atom (Cl), for instance, can gain an electron to become the chloride ion Cl2: Cl Atom Cl2 Ion 17 protons 17 protons 17 electrons 18 electrons Sodium chloride (NaCl), ordinary table salt, is called an ionic compound because it is formed from cations and anions. An atom can lose or gain more than one electron. Examples of ions formed by the loss or gain of more than one electron are Mg21, Fe31, S22, and N32. These ions, as well as Na1 and Cl2, are called monatomic ions because they contain only one atom. Figure 2.11 shows the charges of a number of monatomic ions. With very few exceptions, metals tend to form cations and nonmetals form anions. In addition, two or more atoms can combine to form an ion that has a net positive or net negative charge. Polyatomic ions such as OH2 (hydroxide ion), CN2 (cyanide ion), and NH14 (ammonium ion) are ions containing more than one atom. Review of Concepts (a) What does S8 signify? How does it differ from 8S? (b) Determine the number of protons and electrons for the following ions: (a) P32 and (b) Ti41. 1 18 1A 8A 2 13 14 15 16 17 2A 3A 4A 5A 6A 7A Li+ C4– N3– O2– F– Na+ Mg2+ 3 4 5 6 7 8 9 10 11 12 Al3+ P3– S2– Cl– 3B 4B 5B 6B 7B 8B 1B 2B Cr 2+ Mn2+ Fe2+ Co2+ Ni2+ Cu+ K+ Ca2+ Zn2+ Se2– Br– Cr 3+ Mn3+ Fe3+ Co3+ Ni3+ Cu2+ Sn2+ Rb+ Sr2+ Ag+ Cd2+ Te2– I– Sn4+ Au+ Hg2+ 2 Pb2+ Cs+ Ba2+ Au3+ Hg2+ Pb4+ Figure 2.11 Common monatomic ions arranged according to their positions in the periodic table. Note that the Hg221 ion contains two atoms. 52 Chapter 2 ■ Atoms, Molecules, and Ions 2.6 Chemical Formulas Chemists use chemical formulas to express the composition of molecules and ionic compounds in terms of chemical symbols. By composition we mean not only the elements present but also the ratios in which the atoms are combined. Here we are concerned with two types of formulas: molecular formulas and empirical formulas. Molecular Formulas A molecular formula shows the exact number of atoms of each element in the small- est unit of a substance. In our discussion of molecules, each example was given with its molecular formula in parentheses. Thus, H2 is the molecular formula for hydrogen, O2 is oxygen, O3 is ozone, and H2O is water. The subscript numeral indicates the number of atoms of an element present. There is no subscript for O in H2O because there is only one atom of oxygen in a molecule of water, and so the number “one” is omitted from the formula. Note that oxygen (O2) and ozone (O3) are allotropes of oxygen. An allotrope is one of two or more distinct forms of an element. Two allo- tropic forms of the element carbon—diamond and graphite—are dramatically different not only in properties but also in their relative cost. Molecular Models Molecules are too small for us to observe directly. An effective means of visualizing them is by the use of molecular models. Two standard types of molecular models are currently in use: ball-and-stick models and space-filling models (Figure 2.12). In ball- and-stick model kits, the atoms are wooden or plastic balls with holes in them. Sticks or springs are used to represent chemical bonds. The angles they form between atoms approximate the bond angles in actual molecules. With the exception of the H atom, See back endpaper for color codes the balls are all the same size and each type of atom is represented by a specific color. for atoms. In space-filling models, atoms are represented by truncated balls held together by snap Hydrogen Water Ammonia Methane Molecular H2 H2O NH3 CH4 formula H W Structural H±H H±O±H H±N±H H±C±H formula W W H H Ball-and-stick model Space-filling model Figure 2.12 Molecular and structural formulas and molecular models of four common molecules.  2.6 Chemical Formulas 53 fasteners, so that the bonds are not visible. The balls are proportional in size to atoms. The first step toward building a molecular model is writing the structural formula, which shows how atoms are bonded to one another in a molecule. For example, it is known that each of the two H atoms is bonded to an O atom in the water molecule. Therefore, the structural formula of water is H¬O¬H. A line connecting the two atomic symbols represents a chemical bond. Ball-and-stick models show the three-dimensional arrangement of atoms clearly, and they are fairly easy to construct. However, the balls are not proportional to the size of atoms. Furthermore, the sticks greatly exaggerate the space between atoms in a molecule. Space-filling models are more accurate because they show the variation in atomic size. Their drawbacks are that they are time-consuming to put together and they do not show the three-dimensional positions of atoms very well. Molecular mod- eling software can also be used to create ball-and-stick and space-filling models. We will use both models extensively in this text. Empirical Formulas The molecular formula of hydrogen peroxide, a substance used as an antiseptic and as a bleaching agent for textiles and hair, is H2O2. This formula indicates that each hydrogen peroxide molecule consists of two hydrogen atoms and two oxygen atoms. The ratio of hydrogen to oxygen atoms in this molecule is 2:2 or 1:1. The empirical formula of hydrogen peroxide is HO. Thus, the empirical formula tells us which ele- ments are present and the simplest whole-number ratio of their atoms, but not neces- sarily the actual number of atoms in a given molecule. As another example, consider the compound hydrazine (N2H4), which is used as a rocket fuel. The empirical for- H2O2 mula of hydrazine is NH2. Although the ratio of nitrogen to hydrogen is 1:2 in both the molecular formula (N2H4) and the empirical formula (NH2), only the molecular formula tells us the actual number of N atoms (two) and H atoms (four) present in a hydrazine molecule. Empirical formulas are the simplest chemical formulas; they are written by reduc- The word “empirical” means “derived from experiment.” As we will see in Chapter 3, ing the subscripts in the molecular formulas to the smallest possible whole numbers. empirical formulas are determined Molecular formulas are the true formulas of molecules. If we know the molecular experimentally. formula, we also know the empirical formula, but the reverse is not true. Why, then, do chemists bother with empirical formulas? As we will see in Chapter 3, when chem- ists analyze an unknown compound, the first step is usually the determination of the compound’s empirical formula. With additional information, it is possible to deduce the molecular formula. For many molecules, the molecular formula and the empirical formula are one C and the same. Some examples are water (H2O), ammonia (NH3), carbon dioxide N (CO2), and methane (CH4). Examples 2.2 and 2.3 deal with writing molecular formulas from molecular mod- H els and writing empirical formulas from molecular formulas. Methylamine Cl Example 2.2 H Write the molecular formula of methylamine, a colorless gas used in the production of C pharmaceuticals and pesticides, from its ball-and-stick model, shown in the margin. Solution Refer to the labels (also see back end papers). There are five H atoms, one C atom, and one N atom. Therefore, the molecular formula is CH5N. However, the standard way of writing the molecular formula for methylamine is CH3NH2 because it shows how the atoms are joined in the molecule. Chloroform Practice Exercise Write the molecular formula of chloroform, which is used as a Similar problems: 2.47, 2.48. solvent and a cleaning agent. The ball-and-stick model of chloroform is shown in the margin. 54 Chapter 2 ■ Atoms, Molecules, and Ions Example 2.3 Write the empirical formulas for the following molecules: (a) diborane (B2H6), used in rocket propellants; (b) dimethyl fumarate (C8H12O4), a substance used to treat psoriasis, a skin disease; and (c) vanillin (C8H8O3), a flavoring agent used in foods and beverages. Strategy Recall that to write the empirical formula, the subscripts in the molecular formula must be converted to the smallest possible whole numbers. Solution (a) There are two boron atoms and six hydrogen atoms in diborane. Dividing the subscripts by 2, we obtain the empirical formula BH3. (b) In dimethyl fumarate there are 8 carbon atoms, 12 hydrogen atoms, and 4 oxygen atoms. Dividing the subscripts by 4, we obtain the empirical formula C2H3O. Note that if we had divided the subscripts by 2, we would have obtained the formula C4H6O2. Although the ratio of carbon to hydrogen to oxygen atoms in C4H6O2 is the same as that in C2H3O (2:3:1), C4H6O2 is not the simplest formula because its subscripts are not in the smallest whole-number ratio. (c) Because the subscripts in C8H8O3 are already the smallest possible whole numbers, Similar problems: 2.45, 2.46. the empirical formula for vanillin is the same as its molecular formula. Practice Exercise Write the empirical formula for caffeine (C8H10N4O2), a stimulant found in tea and coffee. Formula of Ionic Compounds The formulas of ionic compounds are usually the same as their empirical formulas because ionic compounds do not consist of discrete molecular units. For example, a solid sample of sodium chloride (NaCl) consists of equal numbers of Na1 and Cl2 ions arranged in a three-dimensional network (Figure 2.13). In such a compound there is a 1:1 ratio of cations to anions so that the compound is electrically neutral. As you can see in Figure 2.13, no Na1 ion in NaCl is associated with just one par- ticular Cl2 ion. In fact, each Na1 ion is equally held by six surrounding Cl2 ions and vice versa. Thus, NaCl is the empirical formula for sodium chloride. In other ionic compounds, the actual structure may be different, but the arrangement of cations and anions is such that the compounds are all electrically neutral. Note that the charges on the cation and anion are not shown in the formula for an ionic compound. Sodium metal reacting with chlorine For ionic compounds to be electrically neutral, the sum of the charges on the gas to form sodium chloride. cation and anion in each formula unit must be zero. If the charges on the cation and anion are numerically different, we apply the following rule to make the formula (a) (b) (c) Figure 2.13 (a) Structure of solid NaCl. (b) In reality, the cations are in contact with the anions. In both (a) and (b), the smaller spheres represent Na1 ions and the larger spheres, Cl2 ions. (c) Crystals of NaCl.  2.6 Chemical Formulas 55 electrically neutral: The subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges are numerically equal, then no subscripts are necessary. This rule follows from the fact that because the formulas of ionic compounds are usually empirical formulas, the subscripts must always be reduced to the smallest ratios. Let us consider some examples. • Potassium Bromide. The potassium cation K1 and the bromine anion Br2 com- Refer to Figure 2.11 for charges of cations and anions. bine to form the ionic compound potassium bromide. The sum of the charges is 11 1 (21) 5 0, so no subscripts are necessary. The formula is KBr. • Zinc Iodide. The zinc cation Zn21 and the iodine anion I2 combine to form zinc iodide. The sum of the charges of one Zn21 ion and one I2 ion is 12 1 (21) 5 11. To make the charges add up to zero we multiply the 21 charge of the anion by 2 and add the subscript “2” to the symbol for iodine. Therefore the formula for zinc iodide is ZnI2. • Aluminum Oxide. The cation is Al31 and the oxygen anion is O22. The follow- ing diagram helps us determine the subscripts for the compound formed by the cation and the anion: Al 3 1 O22 Al2 O3 The sum of the charges is 2(13) 1 3(22) 5 0. Thus, the formula for aluminum Note that in each of the three examples, the subscripts are in the smallest ratios. oxide is Al2O3. Example 2.4 Magnesium nitride is used to prepare Borazon, a very hard compound employed in cutting tools and machine parts. Write the formula of magnesium nitride, containing the Mg21 and N32 ions. Strategy Our guide for writing formulas for ionic compounds is electrical neutrality; that is, the total charge on the cation(s) must be equal to the total charge on the anion(s). Because the charges on the Mg21 and N32 ions are not equal, we know the formula cannot be MgN. Instead, we write the formula as MgxNy, where x and y are subscripts to be determined. When magnesium burns in air, it Solution To satisfy electrical neutrality, the following relationship must hold: forms both magnesium oxide and magnesium nitride. (12)x 1 (23)y 5 0 Solving, we obtain xyy 5 3y2. Setting x 5 3 and y 5 2, we write Mg 2 1 N 3 2 Mg3 N2 Check The subscripts are reduced to the smallest whole-number ratio of the atoms because the chemical formula of an ionic compound is usually its empirical formula. Similar problems: 2.43, 2.44. Practice Exercise Write the formulas of the following ionic compounds: (a) chromium sulfate (containing the Cr31 and SO 422 ions) and (b) titanium oxide (containing the Ti41 and O22 ions). 56 Chapter 2 ■ Atoms, Molecules, and Ions Review of Concepts Match each of the diagrams shown here with the following ionic compounds: Al2O3, LiH, Na2S, Mg(NO3)2. (Green spheres represent cations and red spheres represent anions.) (a) (b) (c) (d) 2.7 Naming Compounds When chemistry was a young science and the number of known compounds was small, it was possible to memorize their names. Many of the names were derived from their physical appearance, properties, origin, or application—for example, milk of magnesia, laughing gas, limestone, caustic soda, lye, washing soda, and baking soda. Today the number of known compounds is well over 66 million. Fortunately, it is not necessary to memorize their names. Over the years chemists have devised a clear system for naming chemical substances. The rules are accepted worldwide, facilitating communication among chemists and providing a useful way of labeling an overwhelming variety of substances. Mastering these rules now will prove benefi- cial almost immediately as we proceed with our study of chemistry. To begin our discussion of chemical nomenclature, the naming of chemical com- pounds, we must first distinguish between inorganic and organic compounds. Organic compounds contain carbon, usually in combination with elements such as hydrogen, oxygen, nitrogen, and sulfur. All other compounds are classified as inorganic com- pounds. For convenience, some carbon-containing compounds, such as carbon mon- oxide (CO), carbon dioxide (CO2), carbon disulfide (CS2), compounds containing the cyanide group (CN2), and carbonate (CO322) and bicarbonate (HCO32) groups are considered to be inorganic compounds. Section 2.8 gives a brief introduction to organic compounds. For names and symbols of the elements, To organize and simplify our venture into naming compounds, we can divide see front endpapers. inorganic compounds into four categories: ionic compounds, molecular compounds, acids and bases, and hydrates. Ionic Compounds 1A 8A 2A 3A 4A 5A 6A 7A In Section 2.5 we learned that ionic compounds are made up of cations (positive ions) Li Na Mg Al N O F S Cl and anions (negative ions). With the important exception of the ammonium ion, NH1 4, K Ca Rb Sr Br I all cations of interest to us are derived from metal atoms. Metal cations take their Cs Ba names from the elements. For example, The most reactive metals (green) Element Name of Cation and the most reactive nonmetals 1 (blue) combine to form ionic Na sodium Na sodium ion (or sodium cation) compounds. K potassium K1 potassium ion (or potassium cation) Mg magnesium Mg21 magnesium ion (or magnesium cation) Al aluminum Al31 aluminum ion (or aluminum cation) Animation Many ionic compounds are binary compounds, or compounds formed from just Formation of an Ionic Compound two elements. For binary compounds, the first element named is the metal cation, followed by the nonmetallic anion. Thus, NaCl is sodium chloride. The anion is named by taking the first part of the element name (chlorine) and adding “-ide.”  2.7 Naming Compounds 57 The “-ide” Nomenclature of Some Common Monatomic Anions Table 2.2 According to Their Positions in the Periodic Table Group 4A Group 5A Group 6A Group 7A 42 32 22 C carbide (C )* N nitride (N ) O oxide (O ) F fluoride (F2) Si silicide (Si42) P phosphide (P32) S sulfide (S22) Cl chloride (Cl2) Se selenide (Se22) Br bromide (Br2) Te telluride (Te22) I iodide (I2) *The word “carbide” is also used for the anion C22 2 . Potassium bromide (KBr), zinc iodide (ZnI2), and aluminum oxide (Al2O3) are also binary compounds. Table 2.2 shows the “-ide” nomenclature of some common mon- atomic anions according to their positions in the periodic table. The “-ide” ending is also used for certain anion groups containing different ele- ments, such as hydroxide (OH2) and cyanide (CN2). Thus, the compounds LiOH and KCN are named lithium hydroxide and potassium cyanide, respectively. These and a number of other such ionic substances are called ternary compounds, meaning com- pounds consisting of three elements. Table 2.3 lists alphabetically the names of a number of common cations and anions. Certain metals, especially the transition metals, can form more than one type of cation. Take iron as an example. Iron can form two cations: Fe21 and Fe31. An older 3B 4B 5B 6B 7B 8B 1B 2B nomenclature system that is still in limited use assigns the ending “-ous” to the cation with fewer positive charges and the ending “-ic” to the cation with more positive charges: The transition metals are the elements in Groups 1B and 3B–8B Fe21 ferrous ion (see Figure 2.10). Fe31 ferric ion The names of the compounds that these iron ions form with chlorine would thus be FeCl2 ferrous chloride FeCl3 ferric chloride This method of naming ions has some distinct limitations. First, the “-ous” and “-ic” suffixes do not provide information regarding the actual charges of the two cations involved. Thus, the ferric ion is Fe31, but the cation of copper named cupric has the formula Cu21. In addition, the “-ous” and “-ic” designations provide names for only two different elemental cations. Some metallic elements can assume three or more different positive charges in compounds. Therefore, it has become increasingly com- mon to designate different cations with Roman numerals. This is called the Stock† system. In this system, the Roman numeral I indicates one positive charge, II means two positive charges, and so on. For example, manganese (Mn) atoms can assume FeCl2 (left) and FeCl3 (right). several different positive charges: Mn21: MnO manganese(II) oxide Keep in mind that the Roman numerals Mn31: Mn2O3 manganese(III) oxide refer to the charges on the metal cations. Mn41: MnO2 manganese(IV) oxide These names are pronounced “manganese-two oxide,” “manganese-three oxide,” and “manganese-four oxide.” Using the Stock system, we denote the ferrous ion † Alfred E. Stock (1876–1946). German chemist. Stock did most of his research in the synthesis and char- acterization of boron, beryllium, and silicon compounds. He was the first scientist to explore the dangers of mercury poisoning. 58 Chapter 2 ■ Atoms, Molecules, and Ions Names and Formulas of Some Common Inorganic Cations Table 2.3 and Anions Cation Anion 31 aluminum (Al ) bromide (Br2) ammonium (NH14) carbonate (CO22 3 ) barium (Ba21) chlorate (ClO2 3) cadmium (Cd21) chloride (Cl2) calcium (Ca21) chromate (CrO22 4 ) cesium (Cs1) cyanide (CN ) 2 chromium(III) or chromic (Cr31) dichromate (Cr2O22 7 ) cobalt(II) or cobaltous (Co21) dihydrogen phosphate (H2PO2 4) copper(I) or cuprous (Cu1) fluoride (F2) copper(II) or cupric (Cu21) hydride (H2) hydrogen (H1) hydrogen carbonate or bicarbonate (HCO2 3) iron(II) or ferrous (Fe21) hydrogen phosphate (HPO422) iron(III) or ferric (Fe31) hydrogen sulfate or bisulfate (HSO2 4) lead(II) or plumbous (Pb21) hydroxide (OH ) 2 lithium (Li1) iodide (I2) magnesium (Mg21) nitrate (NO2 3) manganese(II) or manganous (Mn21) nitride (N32) mercury(I) or mercurous (Hg221)* nitrite (NO22) mercury(II) or mercuric (Hg21) oxide (O22) potassium (K1) permanganate (MnO2 4) rubidium (Rb1) peroxide (O222) silver (Ag1) phosphate (PO432) sodium (Na1) sulfate (SO224 ) strontium (Sr21) sulfide (S22) tin(II) or stannous (Sn21) sulfite (SO22 3 ) zinc (Zn21) thiocyanate (SCN2) *Mercury(I) exists as a pair as shown. Nontransition metals such as tin (Sn) and and the ferric ion as iron(II) and iron(III), respectively; ferrous chloride becomes lead (Pb) can also form more than one type of cations. iron(II) chloride, and ferric chloride is called iron(III) chloride. In keeping with modern practice, we will favor the Stock system of naming compounds in this textbook. Examples 2.5 and 2.6 illustrate how to name ionic compounds and write formu- las for ionic compounds based on the information given in Figure 2.11 and Tables 2.2 and 2.3. Example 2.5 Name the following compounds: (a) Fe(NO3)2, (b) Na2HPO4, and (c) (NH4)2SO3. Strategy Our reference for the names of cations and anions is Table 2.3. Keep in mind that if a metal can form cations of different charges (see Figure 2.11), we need to use the Stock system. (Continued)  2.7 Naming Compounds 59 Solution (a) The nitrate ion (NO23 ) bears one negative charge, so the iron ion must have two positive charges. Because iron forms both Fe21 and Fe31 ions, we need to use the Stock system and call the compound iron(II) nitrate. (b) The cation is Na1 and the anion is HPO22 4 (hydrogen phosphate). Because sodium only forms one type of ion (Na1), there is no need to use sodium(I) in the name. The compound is sodium hydrogen phosphate. (c) The cation is NH1 22 4 (ammonium ion) and the anion is SO3 (sulfite ion). The compound is ammonium sulfite. Similar problems: 2.57(b), (e), (f). Practice Exercise Name the following compounds: (a) PbO and (b) LiClO3. Example 2.6 Write chemical formulas for the following compounds: (a) mercury(I) nitrate, (b) cesium oxide, and (c) strontium nitride. Strategy We refer to Table 2.3 for the formulas of cations and anions. Recall that the Roman numerals in the Stock system provide useful information about the charges of the cation. Solution (a) The Roman numeral shows that the mercury ion bears a 11 charge. According to Note that the subscripts of this ionic Table 2.3, however, the mercury(I) ion is diatomic (that is, Hg221) and the nitrate ion compound are not reduced to the smallest ratio because the Hg(I) ion is NO23 . Therefore, the formula is Hg2(NO3)2. exists as a pair or dimer. (b) Each oxide ion bears two negative charges, and each cesium ion bears one positive charge (cesium is in Group 1A, as is sodium). Therefore, the formula is Cs2O. (c) Each strontium ion (Sr21) bears two positive charges, and each nitride ion (N32) bears three negative charges. To make the sum of the charges equal zero, we must adjust the numbers of cations and anions: 3(12) 1 2(23) 5 0 Thus, the formula is Sr3N2. Similar problems: 2.59(a), (b), (d), (h), (i). Practice Exercise Write formulas for the following ionic compounds: (a) rubidium sulfate and (b) barium hydride. Molecular Compounds Unlike ionic compounds, molecular compounds contain discrete molecular units. They are usually composed of nonmetallic elements (see Figure 2.10). Many molecular compounds are binary compounds. Naming binary molecular compounds is similar to naming binary ionic compounds. We place the name of the first element in the for- mula first, and the second element is named by adding -ide to the root of the element name. Some examples are HCl hydrogen chloride HBr hydrogen bromide SiC silicon carbide 60 Chapter 2 ■ Atoms, Molecules, and Ions Table 2.4 It is quite common for one pair of elements to form several different compounds. In these cases, confusion in naming the compounds is avoided by the use of Greek Greek Prefixes Used prefixes to denote the number of atoms of each element present (Table 2.4). Consider in Naming Molecular Compounds the following examples: Prefix Meaning CO carbon monoxide mono- 1 CO2 carbon dioxide di- 2 SO2 sulfur dioxide tri- 3 SO3 sulfur trioxide tetra- 4 NO2 nitrogen dioxide penta- 5 N2O4 dinitrogen tetroxide hexa- 6 The following guidelines are helpful in naming compounds with prefixes: hepta- 7 • The prefix “mono-” may be omitted for the first element. For example, PCl3 is octa- 8 named phosphorus trichloride, not monophosphorus trichloride. Thus, the absence nona- 9 of a prefix for the first element usually means there is only one atom of that deca- 10 element present in the molecule. • For oxides, the ending “a” in the prefix is sometimes omitted. For example, N2O4 may be called dinitrogen tetroxide rather than dinitrogen tetraoxide. Exceptions to the use of Greek prefixes are molecular compounds containing hydrogen. Traditionally, many of these compounds are called either by their common, nonsystematic names or by names that do not specifically indicate the number of H atoms present: Binary compounds containing carbon B2H6 diborane and hydrogen are organic compounds; they do not follow the same naming CH4 methane conventions. We will discuss the naming SiH4 silane of organic compounds in Chapter 24. NH3 ammonia PH3 phosphine H2O water H2S hydrogen sulfide Note that even the order of writing the elements in the formulas for hydrogen com- pounds is irregular. In water and hydrogen sulfide, H is written first, whereas it appears last in the other compounds. Writing formulas for molecular compounds is usually straightforward. Thus, the name arsenic trifluoride means that there are three F atoms and one As atom in each molecule, and the molecular formula is AsF3. Note that the order of elements in the formula is the same as in its name. Example 2.7 Name the following molecular compounds: (a) PBr5 and (b) As2O5. Strategy We refer to Table 2.4 for the prefixes used in naming molecular compounds. Solution (a) Because there are five bromine atoms present, the compound is phosphorus pentabromide. (b) There are two arsenic atoms and five oxygen atoms present, so the compound is Similar problems: 2.57(c), (i), ( j). diarsenic pentoxide. Note that the “a” is omitted in “penta.” Practice Exercise Name the following molecular compounds: (a) NF3 and (b) Cl2O7.  2.7 Naming Compounds 61 Example 2.8 Write chemical formulas for the following molecular compounds: (a) bromine trifluoride and (b) diboron trioxide. Strategy We refer to Table 2.4 for the prefixes used in naming molecular compounds. Solution (a) Because there are three fluorine atoms and one bromine atom present, the formula is BrF3. (b) There are two boron atoms and three oxygen atoms present, so the formula is B2O3. Similar problems: 2.59(g), ( j). Practice Exercise Write chemical formulas for the following molecular compounds: (a) sulfur tetrafluoride and (b) dinitrogen pentoxide. Figure 2.14 summarizes the steps for naming ionic and binary molecular compounds. Review of Concepts Why is it that the name for SeCl2, selenium dichloride, contains a prefix, but the name for SrCl2, strontium chloride, does not? Compound Ionic Molecular Cation: metal or NH+4 • Binary compounds Anion: monatomic or of nonmetals polyatomic Naming Cation has Cation has more • Use prefixes for only one charge than one charge both elements present (Prefix “mono–” • Alkali metal cations • Other metal cations usually omitted for • Alkaline earth metal cations the first element) • Ag+, Al3+, Cd2+, Zn2+ • Add “–ide” to the root of the second Naming element Naming • Name metal first • Specify charge of • Name metal first metal cation with • If monatomic anion, Roman numeral add “–ide” to the in parentheses root of the element • If monatomic anion, name add “–ide” to the • If polyatomic anion, root of the element use name of anion name (see Table 2.3) • If polyatomic anion, use name of anion (see Table 2.3) Figure 2.14 Steps for naming ionic and binary molecular compounds. 62 Chapter 2 ■ Atoms, Molecules, and Ions Acids and Bases HCl Naming Acids An acid can be described as a substance that yields hydrogen ions (H1) when dissolved in water. (H1 is equivalent to one proton, and is often referred to that way.) Formulas for acids contain one or more hydrogen atoms as well as an anionic group. Anions whose names end in “-ide” form acids with a “hydro-” prefix and an “-ic” ending, as shown in Table 2.5. In some cases two different names seem to be assigned to the same chemical formula. H3O+ HCl hydrogen chloride HCl hydrochloric acid Cl– The name assigned to the compound depends on its physical state. In the gaseous or pure liquid state, HCl is a molecular compound called hydrogen chloride. When it is When dissolved in water, the HCl molecule is converted to the dissolved in water, the molecules break up into H1 and Cl2 ions; in this state, the H1 and Cl– ions. The H1 ion is substance is called hydrochloric acid. associated with one or more Oxoacids are acids that contain hydrogen, oxygen, and another element (the cen- water molecules, and is usually tral element). The formulas of oxoacids are usually written with the H first, followed represented as H3O1. by the central element and then O. We use the following five common acids as our references in naming oxoacids: H H2CO3 carbonic acid HClO3 chloric acid O HNO3 nitric acid C H3PO4 phosphoric acid H2SO4 sulfuric acid Often two or more oxoacids have the same central atom but a different number of O atoms. Starting with our reference oxoacids whose names all end with “-ic,” we use H2CO3 the following rules to name these compounds. 1. Addition of one O atom to the “-ic” acid: The acid is called “per . . . -ic” acid. H Thus, adding an O atom to HClO3 changes chloric acid to perchloric acid, O HClO4. 2. Removal of one O atom from the “-ic” acid: The acid is called “-ous” acid. Thus, nitric acid, HNO3, becomes nitrous acid, HNO2. N 3. Removal of two O atoms from the “-ic” acid: The acid is called “hypo . . . -ous” acid. Thus, when HBrO3 is converted to HBrO, the acid is called hypobromous acid. HNO3 Table 2.5 Some Simple Acids Acid Corresponding Anion Note that these acids all exist as molecular HF (hydrofluoric acid) F2 (fluoride) compounds in the gas phase. HCl (hydrochloric acid) Cl2 (chloride) HBr (hydrobromic acid) Br2 (bromide) HI (hydroiodic acid) I2 (iodide) HCN (hydrocyanic acid) CN2 (cyanide) H2S (hydrosulfuric acid) S22 (sulfide)  2.7 Naming Compounds 63 Removal of Figure 2.15 Naming oxoacids Oxoacid Oxoanion all H+ ions and oxoanions. per– –ic acid per– –ate +[O] Reference “–ic” acid –ate –[O] “–ous” acid –ite –[O] hypo– –ous acid hypo– –ite The rules for naming oxoanions, anions of oxoacids, are as follows: 1. When all the H ions are removed from the “-ic” acid, the anion’s name ends with “-ate.” For example, the anion CO322 derived from H2CO3 is called carbonate. 2. When all the H ions are removed from the “-ous” acid, the anion’s name ends with “-ite.” Thus, the anion ClO2 2 derived from HClO2 is called chlorite. 3. The names of anions in which one or more but not all the hydrogen ions have been removed must indicate the number of H ions present. For example, consider the anions derived from phosphoric acid: O H3PO4 phosphoric acid H2PO42 dihydrogen phosphate H HPO422 hydrogen phosphate P PO432 phosphate Note that we usually omit the prefix “mono-” when there is only one H in the anion. Figure 2.15 summarizes the nomenclature for the oxoacids and oxoanions, and Table 2.6 H3PO4 gives the names of the oxoacids and oxoanions that contain chlorine. Example 2.9 deals with the nomenclature for an oxoacid and an oxoanion. Table 2.6 Names of Oxoacids and Oxoanions That Contain Chlorine Acid Corresponding Anion HClO4 (perchloric acid) ClO2 4 (perchlorate) HClO3 (chloric acid) ClO2 3 (chlorate) HClO2 (chlorous acid) ClO2 2 (chlorite) HClO (hypochlorous acid) ClO2 (hypochlorite) 64 Chapter 2 ■ Atoms, Molecules, and Ions Example 2.9 Name the following oxoacid and oxoanions: (a) H2SO3, a very unstable acid formed when SO2(g) reacts with water, (b) H2AsO2 4, once used to control ticks and lice on livestock, and (c) SeO232, used to manufacture colorless glass. H3AsO4 is arsenic acid, and H2SeO4 is selenic acid. Strategy We refer to Figure 2.15 and Table 2.6 for the conventions used in naming oxoacids and oxoanions. Solution (a) We start with our reference acid, sulfuric acid (H2SO4). Because H2SO3 has one fewer O atom, it is called sulfurous acid. (b) Because H3AsO4 is arsenic acid, the AsO432 ion is named arsenate. The H2AsO2 4 anion is formed by adding two H1 ions to AsO432, so H2AsO2 4 is called dihydrogen arsenate. (c) The parent acid is H2SeO3. Because the acid has one fewer O atom than selenic acid (H2SeO4), it is called selenous acid. Therefore, the SeO322 anion derived from Similar problems: 2.58(f). H2SeO3 is called selenite. Practice Exercise Name the following oxoacid and oxoanion: (a) HBrO and (b) HSO24 . Naming Bases A base can be described as a substance that yields hydroxide ions (OH2) when dis- solved in water. Some examples are NaOH sodium hydroxide KOH potassium hydroxide Ba(OH)2 barium hydroxide Ammonia (NH3), a molecular compound in the gaseous or pure liquid state, is also classified as a common base. At first glance this may seem to be an exception to the definition of a base. But note that as long as a substance yields hydroxide ions when dissolved in water, it need not contain hydroxide ions in its structure to be considered a base. In fact, when ammonia dissolves in water, NH3 reacts partially with water to yield NH41 and OH2 ions. Thus, it is properly classified as a base. Review of Concepts Why is the following question ambiguous: What is the name of HF? What additional information is needed to answer the question? Hydrates Hydrates are compounds that have a specific number of water molecules attached to them. For example, in its normal state, each unit of copper(II) sulfate has five water molecules associated with it. The systematic name for this compound is copper(II) sulfate pentahydrate, and its formula is written as CuSO4 ? 5H2O. The water molecules can be driven off by heating. When this occurs, the resulting compound is CuSO4, which is sometimes called anhydrous copper(II) sulfate; “anhydrous” means that the compound no longer has water molecules associated with it (Figure 2.16). Some other hydrates are BaCl2 ? 2H2O barium chloride dihydrate LiCl ? H2O lithium chloride monohydrate MgSO4 ? 7H2O magnesium sulfate heptahydrate Sr(NO3)2 ? 4H2O strontium nitrate tetrahydrate  2.8 Introduction to Organic Compounds 65 Figure 2.16 CuSO4 ? 5H2O (left) is blue; CuSO4 (right) is white. Table 2.7 Common and Systematic Names of Some Compounds Formula Common Name Systematic Name H2O Water Dihydrogen monoxide NH3 Ammonia Trihydrogen nitride CO2 Dry ice Solid carbon dioxide NaCl Table salt Sodium chloride N2O Laughing gas Dinitrogen monoxide CaCO3 Marble, chalk, limestone Calcium carbonate CaO Quicklime Calcium oxide Ca(OH)2 Slaked lime Calcium hydroxide NaHCO3 Baking soda Sodium hydrogen carbonate Na2CO3 ? 10H2O Washing soda Sodium carbonate decahydrate MgSO4 ? 7H2O Epsom salt Magnesium sulfate heptahydrate Mg(OH)2 Milk of magnesia Magnesium hydroxide CaSO4 ? 2H2O Gypsum Calcium sulfate dihydrate Familiar Inorganic Compounds Some compounds are better known by their common names than by their systematic chemical names. Familiar examples are listed in Table 2.7. CH3OH 2.8 Introduction to Organic Compounds The simplest type of organic compounds is the hydrocarbons, which contain only carbon and hydrogen atoms. The hydrocarbons are used as fuels for domestic and industrial heating, for generating electricity and powering internal combustion engines, and as starting materials for the chemical industry. One class of hydrocarbons is called the alkanes. Table 2.8 shows the names, formulas, and molecular models of the first 10 straight-chain alkanes, in which the carbon chains have no branches. Note that all CH3NH2 the names end with -ane. Starting with C5H12, we use the Greek prefixes in Table 2.4 to indicate the number of carbon atoms present. The chemistry of organic compounds is largely determined by the functional groups, which consist of one or a few atoms bonded in a specific way. For example, when an H atom in methane is replaced by a hydroxyl group (¬OH), an amino group (¬NH2), and a carboxyl group (¬COOH), the following molecules are generated: H H H O H C OH H C NH2 H C C OH H H H Methanol Methylamine Acetic acid CH3COOH 66 Chapter 2 ■ Atoms, Molecules, and Ions Table 2.8 The First Ten Straight-Chain Alkanes Name Formula Molecular Model Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12 Hexane C6H14 Heptane C7H16 Octane C8H18 Nonane C9H20 Decane C10H22 The chemical properties of these molecules can be predicted based on the reactivity of the functional groups. Although the nomenclature of the major classes of organic compounds and their properties in terms of the functional groups will not be discussed until Chapter 24, we will frequently use organic compounds as examples to illustrate chemical bonding, acid-base reactions, and other properties throughout the book. Review of Concepts How many different molecules can you generate by replacing one H atom with a hydroxyl group (¬OH) in butane (see Table 2.8)?  Key Words 67 Key Equation mass number 5 number of protons 1 number of neutrons 5 atomic number 1 number of neutrons  (2.1) Summary of Facts & Concepts 1. Modern chemistry began with Dalton’s atomic theory, determines the identity of an element. The mass which states that all matter is composed of tiny, indi- number is the sum of the number of protons and the visible particles called atoms; that all atoms of the number of neutrons in the nucleus. same element are identical; that compounds contain 6. Isotopes are atoms of the same element with the same atoms of different elements combined in whole- number of protons but different numbers of neutrons. number ratios; and that atoms are neither created nor 7. Chemical formulas combine the symbols for the con- destroyed in chemical reactions (the law of conserva- stituent elements with whole-number subscripts to tion of mass). show the type and number of atoms contained in the 2. Atoms of constituent elements in a particular compound smallest unit of a compound. are always combined in the same proportions by mass 8. The molecular formula conveys the specific number (law of definite proportions). When two elements can and type of atoms combined in each molecule of a com- combine to form more than one type of compound, the pound. The empirical formula shows the simplest ratios masses of one element that combine with a fixed mass of the atoms combined in a molecule. of the other element are in a ratio of small whole num- 9. Chemical compounds are either molecular compounds bers (law of multiple proportions). (in which the smallest units are discrete, individual mol- 3. An atom consists of a very dense central nucleus ecules) or ionic compounds, which are made of cations containing protons and neutrons, with electrons and anions. moving about the nucleus at a relatively large dis- 10. The names of many inorganic compounds can be tance from it. deduced from a set of simple rules. The formulas can be 4. Protons are positively charged, neutrons have no charge, written from the names of the compounds. and electrons are negatively charged. Protons and neu- 11. Organic compounds contain carbon and elements like trons have roughly the same mass, which is about 1840 hydrogen, oxygen, and nitrogen. Hydrocarbon is the times greater than the mass of an electron. simplest type of organic compound. 5. The atomic number of an element is the number of protons in the nucleus of an atom of the element; it Key Words Acid, p. 62 Chemical formula, p. 52 Law of conservation of Nonmetal, p. 48 Alkali metals, p. 50 Diatomic molecule, p. 50 mass, p. 40 Nucleus, p. 44 Alkaline earth metals, p. 50 Electron, p. 41 Law of definite Organic compound, p. 56 Allotrope, p. 52 Empirical formula, p. 53 proportions, p. 40 Oxoacid, p. 62 Alpha (α) particles, p. 43 Families, p. 48 Law of multiple Oxoanion, p. 63 Alpha (α) rays, p. 43 Gamma (γ) rays, p. 43 proportions, p. 40 Periods, p. 48 Anion, p. 51 Groups, p. 48 Mass number (A), p. 46 Periodic table, p. 48 Atom, p. 40 Halogens, p. 50 Metal, p. 48 Polyatomic ion, p. 51 Atomic number (Z), p. 46 Hydrate, p. 64 Metalloid, p. 48 Polyatomic molecule, p. 50 Base, p. 64 Inorganic Molecular formula, p. 52 Proton, p. 44 Beta (β) particles, p. 43 compounds, p. 56 Molecule, p. 50 Radiation, p. 41 Beta (β) rays, p. 43 Ion, p. 50 Monatomic ion, p. 51 Radioactivity, p. 43 Binary compound, p. 56 Ionic compound, p. 51 Neutron, p. 45 Structural formula, p. 53 Cation, p. 51 Isotope, p. 46 Noble gases, p. 50 Ternary compound, p. 57  68 Chapter 2 ■ Atoms, Molecules, and Ions Questions & Problems • Problems available in Connect Plus • 2.16 Indicate the number of protons, neutrons, and elec- Red numbered problems solved in Student Solutions Manual trons in each of the following species: 15 33 63 84 130 186 202 7N, 16S, 29Cu, 38Sr, 56Ba, 74 W, 80Hg Structure of the Atom • 2.17 Write the appropriate symbol for each of the fol- Review Questions lowing isotopes: (a) Z 5 11, A 5 23; (b) Z 5 28, 2.1 Define the following terms: (a) α particle, (b) β par- A 5 64. ticle, (c) γ ray, (d) X ray. • 2.18 Write the appropriate symbol for each of the fol- 2.2 Name the types of radiation known to be emitted by lowing isotopes: (a) Z 5 74, A 5 186; (b) Z 5 80, radioactive elements. A 5 201. 2.3 Compare the properties of the following: α parti- cles, cathode rays, protons, neutrons, electrons. The Periodic Table 2.4 What is meant by the term “fundamental particle”? Review Questions 2.5 Describe the contributions of the following scientists 2.19 What is the periodic table, and what is its signifi- to our knowledge of atomic structure: J. J. Thomson, cance in the study of chemistry? R. A. Millikan, Ernest Rutherford, James Chadwick. 2.20 State two differences between a metal and a 2.6 Describe the experimental basis for believing that nonmetal. the nucleus occupies a very small fraction of the 2.21 Write the names and symbols for four elements volume of the atom. in each of the following categories: (a) nonmetal, Problems (b) metal, (c) metalloid. • 2.22 Define, with two examples, the following terms: • 2.7 The diameter of a helium atom is about 1 3 102 pm. (a) alkali metals, (b) alkaline earth metals, (c) halo- Suppose that we could line up helium atoms side by gens, (d) noble gases. side in contact with one another. Approximately how many atoms would it take to make the distance from end to end 1 cm? Problems 2.8 Roughly speaking, the radius of an atom is about 2.23 Elements whose names end with -ium are usually 10,000 times greater than that of its nucleus. If an metals; sodium is one example. Identify a nonmetal atom were magnified so that the radius of its nucleus whose name also ends with -ium. became 2.0 cm, about the size of a marble, what would 2.24 Describe the changes in properties (from metals to be the radius of the atom in miles? (1 mi 5 1609 m.) nonmetals or from nonmetals to metals) as we move (a) down a periodic group and (b) across the peri- Atomic Number, Mass Number, and Isotopes odic table from left to right. Review Questions 2.25 Consult a handbook of chemical and physical data 2.9 Use the helium-4 isotope to define atomic number (ask your instructor where you can locate a copy and mass number. Why does a knowledge of atomic of the handbook) to find (a) two metals less dense number enable us to deduce the number of electrons than water, (b) two metals more dense than mer- present in an atom? cury, (c) the densest known solid metallic ele- ment, (d) the densest known solid nonmetallic 2.10 Why do all atoms of an element have the same element. atomic number, although they may have different mass numbers? • 2.26 Group the following elements in pairs that you would expect to show similar chemical properties: 2.11 What do we call atoms of the same elements with K, F, P, Na, Cl, and N. different mass numbers? 2.12 Explain the meaning of each term in the symbol ZAX. Molecules and Ions Problems Review Questions • 2.13 What is the mass number of an iron atom that has 28 2.27 What is the difference between an atom and a neutrons? molecule? • 2.14 Calculate the number of neutrons in 239Pu. 2.28 What are allotropes? Give an example. How are • 2.15 For each of the following species, determine the number allotropes different from isotopes? of protons and the number of neutrons in the nucleus: 2.29 Describe the two commonly used molecular 3 4 24 25 48 79 195 2He, 2He, 12Mg, 12Mg, 22Ti, 35Br, 78Pt models.  Questions & Problems 69 2.30 Give an example of each of the following: (a) a mona- Chemical Formulas tomic cation, (b) a monatomic anion, (c) a polyatomic Review Questions cation, (d) a polyatomic anion. 2.37 What does a chemical formula represent? What is the ratio of the atoms in the following molecular Problems formulas? (a) NO, (b) NCl3, (c) N2O4, (d) P4O6 • 2.31 Which of the following diagrams represent diatomic 2.38 Define molecular formula and empirical formula. molecules, polyatomic molecules, molecules that What are the similarities and differences between are not compounds, molecules that are compounds, the empirical formula and molecular formula of a or an elemental form of the substance? compound? 2.39 Give an example of a case in which two molecules have different molecular formulas but the same em- pirical formula. 2.40 What does P4 signify? How does it differ from 4P? 2.41 What is an ionic compound? How is electrical neu- trality maintained in an ionic compound? 2.42 Explain why the chemical formulas of ionic com- pounds are usually the same as their empirical formulas. Problems (a) (b) (c) • 2.43 Write the formulas for the following ionic com- pounds: (a) sodium oxide, (b) iron sulfide (contain- • 2.32 Which of the following diagrams represent diatomic ing the Fe21 ion), (c) cobalt sulfate (containing the molecules, polyatomic molecules, molecules that Co31 and SO422 ions), and (d) barium fluoride. are not compounds, molecules that are compounds, (Hint: See Figure 2.11.) or an elemental form of the substance? • 2.44 Write the formulas for the following ionic com- pounds: (a) copper bromide (containing the Cu1 ion), (b) manganese oxide (containing the Mn31 ion), (c) mercury iodide (containing the Hg221 ion), and (d) magnesium phosphate (containing the PO432 ion). (Hint: See Figure 2.11.) • 2.45 What are the empirical formulas of the following compounds? (a) C2N2, (b) C6H6, (c) C9H20, (d) P4O10, (e) B2H6 • 2.46 What are the empirical formulas of the following compounds? (a) Al2Br6, (b) Na2S2O4, (c) N2O5, (d) K2Cr2O7 (a) (b) (c) • 2.47 Write the molecular formula of glycine, an amino acid present in proteins. The color codes are: black (carbon), blue (nitrogen), red (oxygen), and gray • 2.33 Identify the following as elements or compounds: (hydrogen). NH3, N2, S8, NO, CO, CO2, H2, SO2. • 2.34 Give two examples of each of the following: (a) a diatomic molecule containing atoms of the same element, (b) a diatomic molecule containing at- oms of different elements, (c) a polyatomic mol- H O ecule containing atoms of the same element, (d) a polyatomic molecule containing atoms of differ- ent elements. C • 2.35 Give the number of protons and electrons in each of the following common ions: Na1, Ca21, Al31, Fe21, I2, F2, S22, O22, and N32. N • 2.36 Give the number of protons and electrons in each of the following common ions: K1, Mg21, Fe31, Br2, Mn21, C42, Cu21.  70 Chapter 2 ■ Atoms, Molecules, and Ions • 2.48 Write the molecular formula of ethanol. The color 2.61 Sulfur (S) and fluorine (F) form several different codes are: black (carbon), red (oxygen), and gray compounds. One of them, SF6, contains 3.55 g of F (hydrogen). for every gram of S. Use the law of multiple propor- tions to determine n, which represents the number of F atoms in SFn, given that it contains 2.37 g of F for H every gram of S. O 2.62 Name the following compounds. C O Br F Al N B • 2.49 Which of the following compounds are likely to be ionic? Which are likely to be molecular? SiCl4, LiF, BaCl2, B2H6, KCl, C2H4 • 2.50 Which of the following compounds are likely to be 2.63 Pair the following species that contain the same ionic? Which are likely to be molecular? CH4, NaBr, number of electrons: Ar, Sn41, F2, Fe31, P32, V, BaF2, CCl4, ICl, CsCl, NF3 Ag1, N32. 2.64 Write the correct symbols for the atoms that contain: Naming Inorganic Compounds (a) 25 protons, 25 electrons, and 27 neutrons; (b) 10 Review Questions protons, 10 electrons, and 12 neutrons; (c) 47 pro- 2.51 What is the difference between inorganic com- tons, 47 electrons, and 60 neutrons; (d) 53 protons, pounds and organic compounds? 53 electrons, and 74 neutrons; (e) 94 protons, 94 electrons, and 145 neutrons. 2.52 What are the four major categories of inorganic compounds? Additional Problems 2.53 Give an example each for a binary compound and a 2.65 A sample of a uranium compound is found to be los- ternary compound. ing mass gradually. Explain what is happening to 2.54 What is the Stock system? What are its advantages the sample. over the older system of naming cations? 2.55 Explain why the formula HCl can represent two dif- • 2.66 In which one of the following pairs do the two spe- cies resemble each other most closely in chemical ferent chemical systems. properties? Explain. (a) 11H and 11H1, (b) 147N and 2.56 Define the following terms: acids, bases, oxoacids, 14 32 12 13 7N , (c) 6C and 6C. oxoanions, and hydrates. • 2.67 One isotope of a metallic element has mass number 65 and 35 neutrons in the nucleus. The cation de- Problems rived from the isotope has 28 electrons. Write the • 2.57 Name these compounds: (a) Na2CrO4, (b) K2HPO4, symbol for this cation. (c) HBr (gas), (d) HBr (in water), (e) Li2CO3, 2.68 One isotope of a nonmetallic element has mass (f) K2Cr2O7, (g) NH4NO2, (h) PF3, (i) PF5, (j) P4O6, number 127 and 74 neutrons in the nucleus. The (k) CdI2, (l) SrSO4, (m) Al(OH)3, (n) Na2CO3 ? 10H2O. anion derived from the isotope has 54 electrons. • 2.58 Name these compounds: (a) KClO, (b) Ag2CO3, Write the symbol for this anion. (c) FeCl2, (d) KMnO4, (e) CsClO3, (f) HIO, (g) FeO, • 2.69 Determine the molecular and empirical formulas of (h) Fe2O3, (i) TiCl4, ( j) NaH, (k) Li3N, (l) Na2O, the compounds shown here. (Black spheres are (m) Na2O2, (n) FeCl3 ? 6H2O. carbon and gray spheres are hydrogen.) • 2.59 Write the formulas for the following compounds: (a) rubidium nitrite, (b) potassium sulfide, (c) sodium hydrogen sulfide, (d) magnesium phosphate, (e) cal- cium hydrogen phosphate, (f) potassium dihydrogen phosphate, (g) iodine heptafluoride, (h) ammonium sulfate, (i) silver perchlorate, (j) boron trichloride. • 2.60 Write the formulas for the following compounds: (a) copper(I) cyanide, (b) strontium chlorite, (c) perbro- (a) (b) (c) (d) mic acid, (d) hydroiodic acid, (e) disodium ammonium phosphate, (f ) lead(II) carbonate, (g) tin(II) fluoride, (h) tetraphosphorus decasulfide, (i) mercury(II) oxide, 2.70 What is wrong with or ambiguous about the phrase ( j) mercury(I) iodide, (k) selenium hexafluoride. “four molecules of NaCl”?  Questions & Problems 71 2.71 The following phosphorus sulfides are known: P4S3, 2.79 Caffeine, shown here, is a psychoactive stimulant P4S7, and P4S10. Do these compounds obey the law drug. Write the molecular formula and empirical of multiple proportions? formula of the compound. 2.72 Which of the following are elements, which are molecules but not compounds, which are com- pounds but not molecules, and which are both H compounds and molecules? (a) SO 2, (b) S 8, (c) Cs, (d) N2O5, (e) O, (f) O2, (g) O3, (h) CH4, (i) KBr, (j) S, (k) P4, (l) LiF O N • 2.73 The following table gives numbers of electrons, protons, and neutrons in atoms or ions of a num- ber of elements. Answer the following: (a) Which C of the species are neutral? (b) Which are nega- tively charged? (c) Which are positively charged? (d) What are the conventional symbols for all the species? Atom or Ion 2.80 Acetaminophen, shown here, is the active ingredient of Element A B C D E F G in Tylenol. Write the molecular formula and empiri- Number of electrons 5 10 18 28 36 5 9 cal formula of the compound. Number of protons 5 7 19 30 35 5 9 Number of neutrons 5 7 20 36 46 6 10 H O 2.74 Identify the elements represented by the following N symbols and give the number of protons and neu- trons in each case: (a) 20 63 107 182 10X, (b) 29X, (c) 47X, (d) 74X, C (e) 203 84 X, (f) 234 94X. 2.75 Each of the following pairs of elements will react to form an ionic compound. Write the formulas and name these compounds: (a) barium and oxygen, 2.81 What is wrong with the chemical formula for each (b) calcium and phosphorus, (c) aluminum and sul- of the following compounds: (a) magnesium fur, (d) lithium and nitrogen. iodate [Mg(IO4)2], (b) phosphoric acid (H3PO3), 2.76 Match the descriptions [(a)–(h)] with each of the (c) barium sulfite (BaS), (d) ammonium bicarbon- following elements: P, Cu, Kr, Sb, Cs, Al, Sr, Cl. ate (NH3HCO3)? (a) A transition metal, (b) a nonmetal that forms a 2.82 What is wrong with the names (in parentheses) for 23 ion, (c) a noble gas, (d) an alkali metal, (e) a each of the following compounds: SnCl4 (tin chlo- metal that forms a 13 ion, (f) a metalloid, (g) an ride), (b) Cu2O [copper(II) oxide], (c) Co(NO3)2 element that exists as a diatomic gas molecule, (cobalt nitrate), (d) Na2Cr2O7 (sodium chromate)? (h) an alkaline earth metal. • 2.83 Fill in the blanks in the following table. 2.77 Explain why anions are always larger than the atoms from which they are derived, whereas cations are always smaller than the atoms from which they are Symbol 54 26Fe 21 derived. (Hint: Consider the electrostatic attraction Protons 5 79 86 between protons and electrons.) Neutrons 6 16 117 136 2.78 (a) Describe Rutherford’s experiment and how it led to the structure of the atom. How was he able to Electrons 5 18 79 estimate the number of protons in a nucleus from the Net charge 23 0 scattering of the α particles? (b) Consider the 23Na atom. Given that the radius and mass of the nucleus are 3.04 3 10215 m and 3.82 3 10223 g, respectively, 2.84 (a) Which elements are most likely to form ionic calculate the density of the nucleus in g/cm3. The compounds? (b) Which metallic elements are radius of a 23Na atom is 186 pm. Calculate the den- most likely to form cations with different sity of the space occupied by the electrons in the charges? sodium atom. Do your results support Rutherford’s • 2.85 Write the formula of the common ion derived from model of an atom? [The volume of a sphere of each of the following: (a) Li, (b) S, (c) I, (d) N, radius r is (4/3)πr3.] (e) Al, (f) Cs, (g) Mg  72 Chapter 2 ■ Atoms, Molecules, and Ions 2.86 Which of the following symbols provides more in- element whose anion contains 36 electrons, (d) an formation about the atom: 23Na or 11Na? Explain. alkali metal cation that contains 36 electrons, (e) a • 2.87 Write the chemical formulas and names of binary ac- Group 4A cation that contains 80 electrons. ids and oxoacids that contain Group 7A elements. Do • 2.100 Write the molecular formulas for and names of the the same for elements in Groups 3A, 4A, 5A, and 6A. following compounds. 2.88 Of the 118 elements known, only two are liquids at room temperature (25°C). What are they? (Hint: One element is a familiar metal and the other ele- ment is in Group 7A.) • 2.89 For the noble gases (the Group 8A elements), 42He, 20 40 84 132 S 10Ne, 18Ar, 36Kr, and 54Xe, (a) determine the num- N ber of protons and neutrons in the nucleus of each F Br P atom, and (b) determine the ratio of neutrons to pro- Cl tons in the nucleus of each atom. Describe any gen- eral trend you discover in the way this ratio changes with increasing atomic number. • 2.90 List the elements that exist as gases at room tem- perature. (Hint: Most of these elements can be found 2.101 Show the locations of (a) alkali metals, (b) alkaline in Groups 5A, 6A, 7A, and 8A.) earth metals, (c) the halogens, and (d) the noble 2.91 The Group 1B metals, Cu, Ag, and Au, are called gases in the following outline of a periodic table. coinage metals. What chemical properties make Also draw dividing lines between metals and metal- them specially suitable for making coins and loids and between metalloids and nonmetals. jewelry? 2.92 The elements in Group 8A of the periodic table are 1A 8A called noble gases. Can you suggest what “noble” means in this context? 2A 3A 4A 5A 6A 7A • 2.93 The formula for calcium oxide is CaO. What are the formulas for magnesium oxide and strontium oxide? • 2.94 A common mineral of barium is barytes, or bar- ium sulfate (BaSO4). Because elements in the same periodic group have similar chemical prop- erties, we might expect to find some radium sul- fate (RaSO4) mixed with barytes since radium is the last member of Group 2A. However, the only • 2.102 Fill the blanks in the following table. source of radium compounds in nature is in ura- nium minerals. Why? 2.95 List five elements each that are (a) named after Cation Anion Formula Name places, (b) named after people, (c) named after a Magnesium bicarbonate color. (Hint: See Appendix 1.) SrCl2 2.96 One isotope of a nonmetallic element has mass Fe31 NO2 2 number 77 and 43 neutrons in the nucleus. The anion derived from the isotope has 36 electrons. Manganese(II) chlorate Write the symbol for this anion. SnBr4 2.97 Fluorine reacts with hydrogen (H) and deuterium Co21 PO32 4 (D) to form hydrogen fluoride (HF) and deuterium Hg221 I2 fluoride (DF), where deuterium (21H) is an isotope Cu2CO3 of hydrogen. Would a given amount of fluorine re- Lithium nitride act with different masses of the two hydrogen iso- 31 22 topes? Does this violate the law of definite Al S proportion? Explain. • 2.98 Predict the formula and name of a binary compound • 2.103 Some compounds are better known by their com- formed from the following elements: (a) Na and H, mon names than by their systematic chemical (b) B and O, (c) Na and S, (d) Al and F, (e) F and O, names. Give the chemical formulas of the following (f ) Sr and Cl. substances: (a) dry ice, (b) table salt, (c) laughing • 2.99 Identify each of the following elements: (a) a halo- gas, (d) marble (chalk, limestone), (e) quicklime, gen whose anion contains 36 electrons, (b) a radio- (f) slaked lime, (g) baking soda, (h) washing soda, active noble gas with 86 protons, (c) a Group 6A (i) gypsum, (j) milk of magnesia.  Questions & Problems 73 • 2.104 On p. 40 it was pointed out that mass and energy are Pt is 21.45 g/cm3 and the mass of a single Pt atom alternate aspects of a single entity called mass- is 3.240 3 10222 g. [The volume of a sphere of energy. The relationship between these two physical radius r is (4/3)πr3.] quantities is Einstein’s famous equation, E 5 mc2, • 2.110 A monatomic ion has a charge of 12. The nucleus where E is energy, m is mass, and c is the speed of of the parent atom has a mass number of 55. If the light. In a combustion experiment, it was found that number of neutrons in the nucleus is 1.2 times that 12.096 g of hydrogen molecules combined with of the number of protons, what is the name and sym- 96.000 g of oxygen molecules to form water and bol of the element? released 1.715 3 103 kJ of heat. Calculate the cor- responding mass change in this process and com- • 2.111 In the following 2 3 2 crossword, each letter must be correct four ways: horizontally, vertically, diago- ment on whether the law of conservation of mass nally, and by itself. When the puzzle is complete, the holds for ordinary chemical processes. (Hint: The four spaces will contain the overlapping symbols of Einstein equation can be used to calculate the 10 elements. Use capital letters for each square. change in mass as a result of the change in energy. There is only one correct solution.* 1 J 5 1 kg m2/s2 and c 5 3.00 3 108 m/s.) • 2.105 Draw all possible structural formulas of the following 1 2 hydrocarbons: CH4, C2H6, C3H8, C4H10, and C5H12. 2.106 (a) Assuming nuclei are spherical in shape, show that its radius r is proportional to the cube root of 3 4 mass number (A). (b) In general, the radius of a nucleus is given by r 5 r0 A1/3, where r0 is a propor- tionality constant given by 1.2 3 10215 m. Calculate the volume of the 73Li nucleus. (c) Given that the Horizontal radius of a Li atom is 152 pm, calculate the fraction of the atom’s volume occupied by the nucleus. Does 1–2: Two-letter symbol for a metal used in ancient times your result support Rutherford’s model of an atom? 3–4: Two-letter symbol for a metal that burns in air and is • 2.107 Draw two different structural formulas based on the found in Group 5A molecular formula C2H6O. Is the fact that you can have more than one compound with the same molecular for- Vertical mula consistent with Dalton’s atomic theory? 1–3: Two-letter symbol for a metalloid • 2.108 Ethane and acetylene are two gaseous hydrocar- 2–4: Two-letter symbol for a metal used in U.S. coins bons. Chemical analyses show that in one sample of ethane, 2.65 g of carbon are combined with 0.665 g Single Squares of hydrogen, and in one sample of acetylene, 4.56 g of carbon are combined with 0.383 g of hydrogen. 1: A colorful nonmetal (a) Are these results consistent with the law of mul- 2: A colorless gaseous nonmetal tiple proportions? (b) Write reasonable molecular 3: An element that makes fireworks green formulas for these compounds. 4: An element that has medicinal uses • 2.109 A cube made of platinum (Pt) has an edge length of 1.0 cm. (a) Calculate the number of Pt atoms in the Diagonal cube. (b) Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all of the avail- 1–4: Two-letter symbol for an element used in electronics able space. If only 74 percent of the space inside 2–3: Two-letter symbol for a metal used with Zr to make the cube is taken up by Pt atoms, calculate the wires for superconducting magnets radius in picometers of a Pt atom. The density of • 2.112 Name the following acids. H N S Cl C O *Reproduced with permission of S. J. Cyvin of the University of Trondheim (Norway). This puzzle appeared in Chemical & Engineering News, December 14, 1987 (p. 86) and in Chem Matters, October 1988. 74 Chapter 2 ■ Atoms, Molecules, and Ions 2.113 Calculate the density of the nucleus of a 56 26 Fe atom, 2.115 Methane, ethane, and propane are shown in Table 2.8. given that the nuclear mass is 9.229 3 10223 g. From Show that the following data are consistent with the your result, comment on the fact that any nucleus law of multiple proportions. containing more than one proton must have neutrons present as well. (Hint: See Problem 2.106.) Mass of Carbon Mass of Hydrogen in 1 g Sample in 1 g Sample 2.114 Element X reacts with element Y to form an ionic compound containing X41 and Y22 ions. Write a Methane 0.749 g 0.251 g formula for the compound and suggest in which Ethane 0.799 g 0.201 g periodic groups these elements are likely to be found. Name a representative compound. Propane 0.817 g 0.183 g Interpreting, Modeling & Estimating 2.116 In the Rutherford scattering experiment, an α parti- 2.121 Sodium and potassium are roughly equal in natural cle is heading directly toward a gold nucleus. The abundance in Earth’s crust and most of their com- particle will come to a halt when its kinetic energy is pounds are soluble. However, the composition of completely converted to electrical potential energy. seawater is much higher in sodium than potassium. When this happens, how close will the α particle Explain. with a kinetic energy of 6.0 3 10214 J be from the 2.122 One technique proposed for recycling plastic gro- nucleus? [According to Coulomb’s law, the electri- cery bags is to heat them at 700°C and high pressure cal potential energy between two charged particles to form carbon microspheres that can be used in a is E 5 kQ1Q2/r, where Q1 and Q2 are the charges (in number of applications. Electron microscopy shows coulombs) of the α particle and the gold nucleus, r some representative carbon microspheres obtained is the distance of separation in meters, and k is a in this manner, where the scale is given in the bot- constant equal to 9.0 3 109 kg ? m3/s2 ? C2. Joule (J) tom right corner of the figure. Determine the num- is the unit of energy where 1 J 5 1 kg ? m2/s2.] ber of carbon atoms in a typical carbon microsphere. 2.117 Estimate the relative sizes of the following species: Li, Li1, Li2. 2.118 Compare the atomic size of the following two mag- nesium isotopes: 24Mg and 26Mg. 2.119 Using visible light, we humans cannot see any object smaller than 2 3 1025 cm with an unaided eye. Roughly how many silver atoms must be lined up for us to see the atoms? 2.120 If the size of the nucleus of an atom were that of a 5 μm pea, how far would the electrons be (on average) from the nucleus in meters? Answers to Practice Exercises 2.1 29 protons, 34 neutrons, and 29 electrons. 2.2 CHCl3. 2.7 (a) Nitrogen trifluoride, (b) dichlorine heptoxide. 2.3 C4H5N2O. 2.4 (a) Cr2(SO4)3, (b) TiO2. 2.5 (a) Lead(II) 2.8 (a) SF4, (b) N2O5. 2.9 (a) Hypobromous acid, oxide, (b) lithium chlorate. 2.6 (a) Rb2SO4, (b) BaH2. (b) hydrogen sulfate ion.  CHAPTER 3 Mass Relationships in Chemical Reactions Fireworks are chemical reactions noted for the spectacular colors rather than the energy or useful substances they produce. CHAPTER OUTLINE A LOOK AHEAD 3.1 Atomic Mass  We begin by studying the mass of an atom, which is based on the carbon-12 isotope scale. An atom of the carbon-12 isotope is assigned a mass of exactly 3.2 Avogadro’s Number and the 12 atomic mass units (amu). To work with the more convenient scale of Molar Mass of an Element grams, we use the molar mass. The molar mass of carbon-12 has a mass 3.3 Molecular Mass of exactly 12 grams and contains an Avogadro’s number (6.022 3 1023) of atoms. The molar masses of other elements are also expressed in grams and 3.4 The Mass Spectrometer contain the same number of atoms. (3.1 and 3.2) 3.5 Percent Composition  Our discussion of atomic mass leads to molecular mass, which is the sum of of Compounds the masses of the constituent atoms present. We learn that the most direct 3.6 Experimental Determination way to determine atomic and molecular mass is by the use of a mass spec- trometer. (3.3 and 3.4) of Empirical Formulas  To continue our study of molecules and ionic compounds, we learn how to 3.7 Chemical Reactions and calculate the percent composition of these species from their chemical Chemical Equations formulas. (3.5) 3.8 Amounts of Reactants  We will see how the empirical and molecular formulas of a compound are and Products determined by experiment. (3.6) 3.9 Limiting Reagents  Next, we learn how to write a chemical equation to describe the outcome of a chemical reaction. A chemical equation must be balanced so that we have 3.10 Reaction Yield the same number and type of atoms for the reactants, the starting materials, and the products, the substances formed at the end of the reaction. (3.7)  Building on our knowledge of chemical equations, we then proceed to study the mass relationships of chemical reactions. A chemical equation enables us to use the mole method to predict the amount of product(s) formed, knowing how much the reactant(s) was used. We will see that a reaction’s yield depends on the amount of limiting reagent (a reactant that is used up first) present. (3.8 and 3.9)  We will learn that the actual yield of a reaction is almost always less than that predicted from the equation, called the theoretical yield, because of various complications. (3.10) 75 76 Chapter 3 ■ Mass Relationships in Chemical Reactions I n this chapter, we will consider the masses of atoms and molecules and what happens to them when chemical changes occur. Our guide for this discussion will be the law of conser- vation of mass. 3.1 Atomic Mass In this chapter, we will use what we have learned about chemical structure and for- mulas in studying the mass relationships of atoms and molecules. These relationships in turn will help us to explain the composition of compounds and the ways in which composition changes. Section 3.4 describes a method for The mass of an atom depends on the number of electrons, protons, and neutrons determining atomic mass. it contains. Knowledge of an atom’s mass is important in laboratory work. But atoms are extremely small particles—even the smallest speck of dust that our unaided eyes can detect contains as many as 1 3 1016 atoms! Clearly we cannot weigh a single atom, but it is possible to determine the mass of one atom relative to another exper- imentally. The first step is to assign a value to the mass of one atom of a given element so that it can be used as a standard. One atomic mass unit is also called By international agreement, atomic mass (sometimes called atomic weight) is the one dalton. mass of the atom in atomic mass units (amu). One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. Carbon-12 is the carbon isotope that has six protons and six neutrons. Setting the atomic mass of carbon-12 at 12 amu provides the standard for measuring the atomic mass of the other elements. For example, experiments have shown that, on average, a hydrogen atom is only 8.400 per- cent as massive as the carbon-12 atom. Thus, if the mass of one carbon-12 atom is exactly 12 amu, the atomic mass of hydrogen must be 0.08400 3 12 amu or 1.008 amu. Simi- lar calculations show that the atomic mass of oxygen is 16.00 amu and that of iron is 55.85 amu. Thus, although we do not know just how much an average iron atom’s mass is, we know that it is approximately 56 times as massive as a hydrogen atom. Average Atomic Mass When you look up the atomic mass of carbon in a table such as the one on the inside Atomic front cover of this book, you will find that its value is not 12.00 amu but 12.01 amu. 6 number The reason for the difference is that most naturally occurring elements (including carbon) have more than one isotope. This means that when we measure the atomic C mass of an element, we must generally settle for the average mass of the naturally Atomic 12.01 mass occurring mixture of isotopes. For example, the natural abundances of carbon-12 and carbon-13 are 98.90 percent and 1.10 percent, respectively. The atomic mass of carbon-13 has been determined to be 13.00335 amu. Thus, the average atomic mass 13 C of carbon can be calculated as follows: 12 C 1.10% 98.90% average atomic mass of natural carbon 5 (0.9890)(12 amu) 1 (0.0110)(13.00335 amu) 5 12.01 amu Note that in calculations involving percentages, we need to convert percentages to fractions. For example, 98.90 percent becomes 98.90/100, or 0.9890. Because there are many more carbon-12 atoms than carbon-13 atoms in naturally occurring carbon, the average atomic mass is much closer to 12 amu than to 13 amu. It is important to understand that when we say that the atomic mass of carbon is 12.01 amu, we are referring to the average value. If carbon atoms could be examined individually, we would find either an atom of atomic mass exactly 12 amu or one of Natural abundances of C-12 and 13.00335 amu, but never one of 12.01 amu. Example 3.1 shows how to calculate the C-13 isotopes. average atomic mass of an element.  3.2 Avogadro’s Number and the Molar Mass of an Element 77 Example 3.1 Boron is used in the manufacture of ceramics and polymers such as fiberglass. The atomic masses of its two stable isotopes, 105B (19.80 percent) and 115B (80.20 percent), are 10.0129 amu and 11.0093 amu, respectively. The boron-10 isotope is also important as a neutron-capturing agent in nuclear reactors. Calculate the average atomic mass of boron. Strategy Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. Solution First the percent abundances are converted to fractions: 19.80 percent to 19.80/100 or 0.1980 and 80.20 percent to 80.20/100 or 0.8020. We find the contribution to the average atomic mass for each isotope, and then add the contributions together to obtain the average atomic mass. (0.1980) (10.0129 amu) 1 (0.8020) (11.0093 amu) 5 10.8129 amu Check The average atomic mass should be between the two isotopic masses; therefore, the answer is reasonable. Note that because there are more 115B isotopes than 105B isotopes, the average atomic mass is closer to 11.0093 amu than to 10.0129 amu. 63 Practice Exercise The atomic masses of the two stable isotopes of copper, 29 Cu (69.17 percent) and 65 29Cu (30.83 percent), are 62.9296 amu and 64.9278 amu, respectively. Calculate the average atomic mass of copper. The atomic masses of many elements have been accurately determined to five or six significant figures. However, for our purposes we will normally use atomic masses accurate only to four significant figures (see table of atomic masses inside the front Boron and the solid-state structure cover). For simplicity, we will omit the word “average” when we discuss the atomic of boron. masses of the elements. Review of Concepts There are two stable isotopes of iridium: 191Ir (190.96 amu) and 193Ir (192.96 amu). Similar problems: 3.5, 3.6. If you were to randomly pick an iridium atom from a large collection of iridium atoms, which isotope are you more likely to select? 3.2 Avogadro’s Number and the Molar Mass of an Element Atomic mass units provide a relative scale for the masses of the elements. But because atoms have such small masses, no usable scale can be devised to weigh them in calibrated units of atomic mass units. In any real situation, we deal with macroscopic samples containing enormous numbers of atoms. Therefore, it is convenient to have a special unit to describe a very large number of atoms. The idea of a unit to denote a particular number of objects is not new. For example, the pair (2 items), the dozen (12 items), and the gross (144 items) are all familiar units. Chemists measure atoms and molecules in moles. In the SI system the mole (mol) is the amount of a substance that contains as The adjective formed from the noun “mole” is “molar.” many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in 12 g of carbon-12 is determined experimentally. This number is called Avogadro’s 78 Chapter 3 ■ Mass Relationships in Chemical Reactions Figure 3.1 One mole each of several common elements. Carbon (black charcoal powder), sulfur (yellow powder), iron (as nails), copper wires, and mercury (shiny liquid metal). number (NA), in honor of the Italian scientist Amedeo Avogadro.† The currently accepted value is NA 5 6.0221413 3 1023 Generally, we round Avogadro’s number to 6.022 3 1023. Thus, just as 1 dozen oranges contains 12 oranges, 1 mole of hydrogen atoms contains 6.022 3 1023 H atoms. Figure 3.1 shows samples containing 1 mole each of several common elements. The enormity of Avogadro’s number is difficult to imagine. For example, spreading 6.022 3 1023 oranges over the entire surface of Earth would produce a layer 9 mi into space! Because atoms (and molecules) are so tiny, we need a huge number to study them in manageable quantities. In calculations, the units of molar mass We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and are g/mol or kg/mol. contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass (m), defined as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance. Note that the molar mass of carbon-12 (in grams) is numerically equal to its atomic mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu and its molar mass is 22.99 g; the atomic mass of phosphorus is 30.97 amu and its molar mass is 30.97 g; and so on. If we know the atomic mass of an element, we also know its molar mass. The molar masses of the elements are Knowing the molar mass and Avogadro’s number, we can calculate the mass of given on the inside front cover of the book. a single atom in grams. For example, we know the molar mass of carbon-12 is 12 g and there are 6.022 3 1023 carbon-12 atoms in 1 mole of the substance; therefore, the mass of one carbon-12 atom is given by 12 g carbon-12 atoms 5 1.993 3 10223 g 6.022 3 1023 carbon-12 atoms † Lorenzo Romano Amedeo Carlo Avogadro di Quaregua e di Cerreto (1776–1856). Italian mathematical physicist. He practiced law for many years before he became interested in science. His most famous work, now known as Avogadro’s law (see Chapter 5), was largely ignored during his lifetime, although it became the basis for determining atomic masses in the late nineteenth century.  3.2 Avogadro’s Number and the Molar Mass of an Element 79 Mass of m /} Number of moles nNA Number of atoms element (m) n} of element (n) N/NA of element (N) Figure 3.2 The relationships between mass (m in grams) of an element and number of moles of an element (n) and between number of moles of an element and number of atoms (N) of an element. m is the molar mass (g/mol) of the element and NA is Avogadro’s number. We can use the preceding result to determine the relationship between atomic mass units and grams. Because the mass of every carbon-12 atom is exactly 12 amu, the number of atomic mass units equivalent to 1 gram is amu 12 amu 1 carbon-12 atom 5 3 gram 1 carbon-12 atom 1.993 3 10223 g 5 6.022 3 1023 amu/g Thus, 1 g 5 6.022 3 1023 amu and 1 amu 5 1.661 3 10224 g This example shows that Avogadro’s number can be used to convert from the atomic mass units to mass in grams and vice versa. The notions of Avogadro’s number and molar mass enable us to carry out conver- sions between mass and moles of atoms and between moles and number of atoms (Figure 3.2). We will employ the following conversion factors in the calculations: 1 mol X 1 mol X After some practice, you can use the   and   equations in Figure 3.2 in calculations: molar mass of X 6.022 3 1023 X atoms n 5 m/m and N 5 nNA. where X represents the symbol of an element. Using the proper conversion factors we can convert one quantity to another, as Examples 3.2–3.4 show. Example 3.2 Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks, and balloons. How many moles of He atoms are in 6.46 g of He? Strategy We are given grams of helium and asked to solve for moles of helium. What conversion factor do we need to convert between grams and moles? Arrange the appropriate conversion factor so that grams cancel and the unit moles is obtained for your answer. Solution The conversion factor needed to convert between grams and moles is the molar mass. In the periodic table (see inside front cover) we see that the molar mass of He is 4.003 g. This can be expressed as 1 mol He 5 4.003 g He From this equality, we can write two conversion factors 4.003 g He A scientific research helium balloon. 1 mol He   and   4.003 g He 1 mol He (Continued) 80 Chapter 3 ■ Mass Relationships in Chemical Reactions The conversion factor on the left is the correct one. Grams will cancel, leaving the unit mol for the answer, that is, 1 mol He 6.46 g He 3 5 1.61 mol He 4.003 g He Thus, there are 1.61 moles of He atoms in 6.46 g of He. Check Because the given mass (6.46 g) is larger than the molar mass of He, we expect Similar problem: 3.15. to have more than 1 mole of He. Practice Exercise How many moles of magnesium (Mg) are there in 87.3 g of Mg? Example 3.3 Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating iron to prevent corrosion. How many grams of Zn are in 0.356 mole of Zn? Strategy We are trying to solve for grams of zinc. What conversion factor do we need to convert between moles and grams? Arrange the appropriate conversion factor so that moles cancel and the unit grams are obtained for your answer. Solution The conversion factor needed to convert between moles and grams is the molar mass. In the periodic table (see inside front cover) we see the molar mass of Zn is 65.39 g. This can be expressed as Zinc. 1 mol Zn 5 65.39 g Zn From this equality, we can write two conversion factors 1 mol Zn 65.39 g Zn   and   65.39 g Zn 1 mol Zn The conversion factor on the right is the correct one. Moles will cancel, leaving unit of grams for the answer. The number of grams of Zn is 65.39 g Zn 0.356 mol Zn 3 5 23.3 g Zn 1 mol Zn Thus, there are 23.3 g of Zn in 0.356 mole of Zn. Check Does a mass of 23.3 g for 0.356 mole of Zn seem reasonable? What is the Similar problem: 3.16. mass of 1 mole of Zn? Practice Exercise Calculate the number of grams of lead (Pb) in 12.4 moles of lead. Example 3.4 The C60 molecule is called buckminsterfullerene because its shape resembles the geodesic domes designed by the visionary architect R. Buckminster Fuller. What is the mass (in grams) of one C60 molecule? Strategy The question asks for the mass of one C60 molecule. Determine the moles of C atoms in one C60 molecule, and then use the molar mass of C to calculate the mass of one molecule in grams. Buckminsterfullerene (C60) or (Continued) “buckyball.”  3.3 Molecular Mass 81 Solution Because one C60 molecule contains 60 C atoms, and 1 mole of C contains 6.022 3 1023 C atoms and has a mass of 12.011 g, we can calculate the mass of one C60 molecule as follows: 60 C atoms 1 mol C 12.01 g 1 C60 molecule 3 3 3 5 1.197 3 10221 g 1 C60 molecule 6.022 3 1023 C atoms 1 mol C Check Because 6.022 3 1023 atoms of C have a mass 12.01 g, a molecule containing only 60 carbon atoms should have a significantly smaller mass. Similar problems: 3.20, 3.21. Practice Exercise Gold atoms form small clusters containing a fixed number of atoms. What is the mass (in grams) of one Au31 cluster? Review of Concepts Referring to the periodic table in the inside front cover and Figure 3.2, determine which of the following contains the largest number of atoms: (a) 7.68 g of He, (b) 112 g of Fe, and (c) 389 g of Hg. 3.3 Molecular Mass If we know the atomic masses of the component atoms, we can calculate the mass of a molecule. The molecular mass (sometimes called molecular weight) is the sum of the atomic masses (in amu) in the molecule. For example, the molecular mass of H2O is 2(atomic mass of H) 1 atomic mass of O or 2(1.008 amu) 1 16.00 amu 5 18.02 amu In general, we need to multiply the atomic mass of each element by the number of atoms of that element present in the molecule and sum over all the elements. Exam- ple 3.5 illustrates this approach. Example 3.5 Calculate the molecular masses (in amu) of the following compounds: (a) sulfur dioxide (SO2), a gas that is responsible for acid rain, and (b) caffeine (C8H10N4O2), a stimulant present in tea, coffee, and cola beverages. Strategy How do atomic masses of different elements combine to give the molecular mass of a compound? SO2 Solution To calculate molecular mass, we need to sum all the atomic masses in the molecule. For each element, we multiply the atomic mass of the element by the number of atoms of that element in the molecule. We find atomic masses in the periodic table (inside front cover). (a) There are two O atoms and one S atom in SO2, so that molecular mass of SO2 5 32.07 amu 1 2(16.00 amu) 5 64.07 amu (Continued) 82 Chapter 3 ■ Mass Relationships in Chemical Reactions (b) There are eight C atoms, ten H atoms, four N atoms, and two O atoms in caffeine, so the molecular mass of C8H10N4O2 is given by Similar problems: 3.23, 3.24. 8(12.01 amu) 1 10(1.008 amu) 1 4(14.01 amu) 1 2(16.00 amu) 5 194.20 amu Practice Exercise What is the molecular mass of methanol (CH4O)? From the molecular mass we can determine the molar mass of a molecule or compound. The molar mass of a compound (in grams) is numerically equal to its molecular mass (in amu). For example, the molecular mass of water is 18.02 amu, so its molar mass is 18.02 g. Note that 1 mole of water weighs 18.02 g and contains 6.022 3 1023 H2O molecules, just as 1 mole of elemental carbon contains 6.022 3 1023 carbon atoms. As Examples 3.6 and 3.7 show, a knowledge of the molar mass enables us to calculate the numbers of moles and individual atoms in a given quantity of a compound. Example 3.6 Methane (CH4) is the principal component of natural gas. How many moles of CH4 are present in 6.07 g of CH4? Strategy We are given grams of CH4 and asked to solve for moles of CH4. What conversion factor do we need to convert between grams and moles? Arrange the CH4 appropriate conversion factor so that grams cancel and the unit moles are obtained for your answer. Solution The conversion factor needed to convert between grams and moles is the molar mass. First we need to calculate the molar mass of CH4, following the procedure in Example 3.5: molar mass of CH4 5 12.01 g 1 4(1.008 g) 5 16.04 g Methane gas burning on a cooking Because range. 1 mol CH4 5 16.04 g CH4 the conversion factor we need should have grams in the denominator so that the unit g will cancel, leaving the unit mol in the numerator: 1 mol CH4 16.04 g CH4 We now write 1 mol CH4 6.07 g CH4 3 5 0.378 mol CH4 16.04 g CH4 Thus, there is 0.378 mole of CH4 in 6.07 g of CH4. Check Should 6.07 g of CH4 equal less than 1 mole of CH4? What is the mass of Similar problem: 3.26. 1  mole of CH4? Practice Exercise Calculate the number of moles of chloroform (CHCl3) in 198 g of chloroform.  3.4 The Mass Spectrometer 83 Example 3.7 How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g. Strategy We are asked to solve for atoms of hydrogen in 25.6 g of urea. We cannot convert directly from grams of urea to atoms of hydrogen. How should molar mass and Avogadro’s number be used in this calculation? How many moles of H are in 1 mole of urea? Solution To calculate the number of H atoms, we first must convert grams of urea to moles of urea using the molar mass of urea. This part is similar to Example 3.2. The Urea. molecular formula of urea shows there are four moles of H atoms in one mole of urea molecule, so the mole ratio is 4:1. Finally, knowing the number of moles of H atoms, we can calculate the number of H atoms using Avogadro’s number. We need two conversion factors: molar mass and Avogadro’s number. We can combine these conversions grams of urea ¡ moles of urea ¡ moles of H ¡ atoms of H into one step: 1 mol (NH2 ) 2CO 4 mol H 6.022 3 1023 H atoms 25.6 g (NH2 ) 2CO 3 3 3 60.06 g (NH2 ) 2CO 1 mol (NH2 ) 2CO 1 mol H 5 1.03 3 1024 H atoms Check Does the answer look reasonable? How many atoms of H would 60.06 g of urea contain? Similar problems: 3.27, 3.28. Practice Exercise How many H atoms are in 72.5 g of isopropanol (rubbing alcohol), C3H8O? Finally, note that for ionic compounds like NaCl and MgO that do not contain discrete molecular units, we use the term formula mass instead. The formula unit of NaCl consists of one Na1 ion and one Cl2 ion. Thus, the formula mass of NaCl is the mass of one formula unit: formula mass of NaCl 5 22.99 amu 1 35.45 amu Note that the combined mass of a Na1 ion and a Cl2 ion is equal to the combined 5 58.44 amu mass of a Na atom and a Cl atom. and its molar mass is 58.44 g. Review of Concepts Determine the molecular mass and the molar mass of citric acid, H3C6H5O7. 3.4 The Mass Spectrometer The most direct and most accurate method for determining atomic and molecular masses is mass spectrometry, which is depicted in Figure 3.3. In one type of a mass spectrometer, a gaseous sample is bombarded by a stream of high-energy electrons. Collisions between the electrons and the gaseous atoms (or molecules) produce positive ions by dislodging an electron from each atom or molecule. These positive 84 Chapter 3 ■ Mass Relationships in Chemical Reactions Detecting screen Accelerating plates Electron beam Magnet Ion beam Sample gas Filament Figure 3.3 Schematic diagram of one type of mass spectrometer. ions (of mass m and charge e) are accelerated by two oppositely charged plates as they pass through the plates. The emerging ions are deflected into a circular path by a magnet. The radius of the path depends on the charge-to-mass ratio (that is, e/m). Ions of smaller e/m ratio trace a wider curve than those having a larger e/m ratio, so that ions with equal charges but different masses are separated from one another. The mass of each ion (and hence its parent atom or molecule) is determined from the magnitude of its deflection. Eventually the ions arrive at the detector, which registers a current for each type of ion. The amount of current generated is directly proportional to the number of ions, so it enables us to determine the relative abundance of isotopes. The first mass spectrometer, developed in the 1920s by the English physicist F. W. Aston,† was crude by today’s standards. Nevertheless, it provided indisputable evidence of the existence of isotopes—neon-20 (atomic mass 19.9924 amu and natu- ral abundance 90.92 percent) and neon-22 (atomic mass 21.9914 amu and natural abundance 8.82 percent). When more sophisticated and sensitive mass spectrometers became available, scientists were surprised to discover that neon has a third stable isotope with an atomic mass of 20.9940 amu and natural abundance 0.257 percent (Figure 3.4). This example illustrates how very important experimental accuracy is to Note that it is possible to determine the a quantitative science like chemistry. Early experiments failed to detect neon-21 molar mass of a compound without knowing its chemical formula. because its natural abundance is just 0.257 percent. In other words, only 26 in 10,000 Ne atoms are neon-21. The masses of molecules can be determined in a similar man- ner by the mass spectrometer. Review of Concepts Explain how the mass spectrometer enables chemists to determine the average atomic mass of chlorine, which has two stable isotopes (35Cl and 37Cl). † Francis William Aston (1877–1945). English chemist and physicist. He was awarded the Nobel Prize in Chemistry in 1922 for developing the mass spectrometer.  3.5 Percent Composition of Compounds 85 20 10 Ne(90.92%) Intensity of peaks 21 22 10 Ne(0.26%) 10 Ne(8.82%) 19 20 21 22 23 Atomic mass (amu) Figure 3.4 The mass spectrum of the three isotopes of neon. 3.5 Percent Composition of Compounds As we have seen, the formula of a compound tells us the numbers of atoms of each element in a unit of the compound. However, suppose we needed to verify the purity of a compound for use in a laboratory experiment. From the formula we could cal- culate what percent of the total mass of the compound is contributed by each element. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample. The percent composition by mass is the percent by mass of each element in a compound. Percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent. Mathematically, the percent composition of an element in a compound is expressed as n 3 molar mass of element percent composition of an element 5 3 100% (3.1) molar mass of compound where n is the number of moles of the element in 1 mole of the compound. For example, in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H2O2, H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H2O2 is calculated as follows: 2 3 1.008 g H %H 5 3 100% 5 5.926% 34.02 g H2O2 H2O2 2 3 16.00 g O %O 5 3 100% 5 94.06% 34.02 g H2O2 The sum of the percentages is 5.926% 1 94.06% 5 99.99%. The small discrep- ancy from 100 percent is due to the way we rounded off the molar masses of the elements. If we had used the empirical formula HO for the calculation, we 86 Chapter 3 ■ Mass Relationships in Chemical Reactions would have obtained the same percentages. This is so because both the molecular formula and empirical formula tell us the percent composition by mass of the compound. Example 3.8 Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound. Strategy Recall the procedure for calculating a percentage. Assume that we have 1  mole of H3PO4. The percent by mass of each element (H, P, and O) is given by the combined molar mass of the atoms of the element in 1 mole of H3PO4 divided by the H3PO4 molar mass of H3PO4, then multiplied by 100 percent. Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows: 3(1.008 g) H %H 5 3 100% 5 3.086% 97.99 g H3PO4 30.97 g P %P 5 3 100% 5 31.61% 97.99 g H3PO4 4(16.00 g) O %O 5 3 100% 5 65.31% 97.99 g H3PO4 Check Do the percentages add to 100 percent? The sum of the percentages is (3.086% 1 31.61% 1 65.31%) 5 100.01%. The small discrepancy from 100 percent Similar problem: 3.40. is due to the way we rounded off. Practice Exercise Calculate the percent composition by mass of each of the elements in sulfuric acid (H2SO4). The procedure used in the example can be reversed if necessary. Given the percent Mass percent composition by mass of a compound, we can determine the empirical formula of the compound (Figure 3.5). Because we are dealing with percentages and the sum of all Convert to grams and the percentages is 100 percent, it is convenient to assume that we started with 100 g divide by molar mass of a compound, as Example 3.9 shows. Moles of each element Example 3.9 Divide by the smallest Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C), number of moles 4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula. Mole ratios of elements Strategy In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form one mole of the compound. How can we Change to convert from mass percent to moles? If we assume an exactly 100-g sample of the integer subscripts compound, do we know the mass of each element in the compound? How do we then convert from grams to moles? Empirical formula Solution If we have 100 g of ascorbic acid, then each percentage can be converted directly to grams. In this sample, there will be 40.92 g of C, 4.58 g of H, and 54.50 g of O. Because the subscripts in the formula represent a mole ratio, we need to Figure 3.5 Procedure for calculating the empirical formula convert the grams of each element to moles. The conversion factor needed is the of a compound from its percent (Continued) compositions.  3.5 Percent Composition of Compounds 87 molar mass of each element. Let n represent the number of moles of each element so that 1 mol C nC 5 40.92 g C 3 5 3.407 mol C 12.01 g C 1 mol H nH 5 4.58 g H 3 5 4.54 mol H 1.008 g H 1 mol O nO 5 54.50 g O 3 5 3.406 mol O 16.00 g O Thus, we arrive at the formula C3.407H4.54O3.406, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to The molecular formula of ascorbic acid is C6H8O6. convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406): 3.407 4.54 3.406 C: < 1  H: 5 1.33  O: 51 3.406 3.406 3.406 where the < sign means “approximately equal to.” This gives CH1.33O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer. This can be done by a trial-and-error procedure: 1.33 3 1 5 1.33 1.33 3 2 5 2.66 1.33 3 3 5 3.99 < 4 Because 1.33 3 3 gives us an integer (4), we multiply all the subscripts by 3 and obtain C3H4O3 as the empirical formula for ascorbic acid. Check Are the subscripts in C3H4O3 reduced to the smallest whole numbers? Similar problems: 3.49, 3.50. Practice Exercise Determine the empirical formula of a compound having the following percent composition by mass: K: 24.75 percent; Mn: 34.77 percent; O: 40.51 percent. Chemists often want to know the actual mass of an element in a certain mass of a compound. For example, in the mining industry, this information will tell the sci- entists about the quality of the ore. Because the percent composition by mass of the elements in the substance can be readily calculated, such a problem can be solved in a rather direct way. Example 3.10 Chalcopyrite (CuFeS2) is a principal mineral of copper. Calculate the number of kilograms of Cu in 3.71 3 103 kg of chalcopyrite. Strategy Chalcopyrite is composed of Cu, Fe, and S. The mass due to Cu is based on its percentage by mass in the compound. How do we calculate mass percent of an element? Solution The molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g, respectively. Chalcopyrite. The mass percent of Cu is therefore molar mass of Cu %Cu 5 3 100% molar mass of CuFeS2 63.55 g 5 3 100% 5 34.63% 183.5 g To calculate the mass of Cu in a 3.71 3 103 kg sample of CuFeS2, we need to convert the percentage to a fraction (that is, convert 34.63 percent to 34.63/100, or 0.3463) and write mass of Cu in CuFeS2 5 0.3463 3 (3.71 3 103 kg) 5 1.28 3 103 kg (Continued) 88 Chapter 3 ■ Mass Relationships in Chemical Reactions Check As a ball-park estimate, note that the mass percent of Cu is roughly 33 percent, so that a third of the mass should be Cu; that is, 13 3 3.71 3 103 kg < 1.24 3 103 kg. Similar problem: 3.45. This quantity is quite close to the answer. Practice Exercise Calculate the number of grams of Al in 371 g of Al2O3. Review of Concepts Without doing detailed calculations, estimate whether the percent composition by mass of Sr is greater than or smaller than that of O in strontium nitrate [Sr(NO3)2]. 3.6 Experimental Determination of Empirical Formulas The fact that we can determine the empirical formula of a compound if we know the percent composition enables us to identify compounds experimentally. The procedure is as follows. First, chemical analysis tells us the number of grams of each element present in a given amount of a compound. Then, we convert the quantities in grams to number of moles of each element. Finally, using the method given in Example 3.9, we find the empirical formula of the compound. As a specific example, let us consider the compound ethanol. When ethanol is burned in an apparatus such as that shown in Figure 3.6, carbon dioxide (CO2) and water (H2O) are given off. Because neither carbon nor hydrogen was in the inlet gas, we can conclude that both carbon (C) and hydrogen (H) were present in etha- nol and that oxygen (O) may also be present. (Molecular oxygen was added in the combustion process, but some of the oxygen may also have come from the original ethanol sample.) The masses of CO2 and of H2O produced can be determined by measuring the increase in mass of the CO2 and H2O absorbers, respectively. Suppose that in one experiment the combustion of 11.5 g of ethanol produced 22.0 g of CO2 and 13.5 g of H2O. We can calculate the mass of carbon and hydrogen in the original 11.5-g sample of ethanol as follows: 1 mol CO2 1 mol C 12.01 g C mass of C 5 22.0 g CO2 3 3 3 44.01 g CO2 1 mol CO2 1 mol C 5 6.00 g C 1 mol H2O 2 mol H 1.008 g H mass of H 5 13.5 g H2O 3 3 3 18.02 g H2O 1 mol H2O 1 mol H 5 1.51 g H Figure 3.6 Apparatus for determining the empirical formula of ethanol. The absorbers are substances that can retain water and carbon dioxide, respectively. CuO is used to ensure complete combustion of all carbon to CO2. Sample CO2 absorber H2O O2 absorber CuO Furnace  3.6 Experimental Determination of Empirical Formulas 89 Thus, 11.5 g of ethanol contains 6.00 g of carbon and 1.51 g of hydrogen. The remainder must be oxygen, whose mass is mass of O 5 mass of sample 2 (mass of C 1 mass of H) 5 11.5 g 2 (6.00 g 1 1.51 g) 5 4.0 g The number of moles of each element present in 11.5 g of ethanol is 1 mol C moles of C 5 6.00 g C 3 5 0.500 mol C 12.01 g C 1 mol H moles of H 5 1.51 g H 3 5 1.50 mol H 1.008 g H 1 mol O moles of O 5 4.0 g O 3 5 0.25 mol O 16.00 g O The formula of ethanol is therefore C0.50H1.5O0.25 (we round off the number of moles to two significant figures). Because the number of atoms must be an integer, we divide the subscripts by 0.25, the smallest subscript, and obtain for the empirical formula C2H6O. Now we can better understand the word “empirical,” which literally means “based only on observation and measurement.” The empirical formula of ethanol is deter- mined from analysis of the compound in terms of its component elements. No knowl- It happens that the molecular edge of how the atoms are linked together in the compound is required. formula of ethanol is the same as its empirical formula. Determination of Molecular Formulas The formula calculated from percent composition by mass is always the empirical formula because the subscripts in the formula are always reduced to the smallest whole numbers. To calculate the actual, molecular formula we must know the approx- imate molar mass of the compound in addition to its empirical formula. Knowing that the molar mass of a compound must be an integral multiple of the molar mass of its empirical formula, we can use the molar mass to find the molecular formula, as Example 3.11 demonstrates. Example 3.11 A sample of a compound contains 30.46 percent nitrogen and 69.54 percent oxygen by mass, as determined by a mass spectrometer. In a separate experiment, the molar mass of the compound is found to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound. Strategy To determine the molecular formula, we first need to determine the empirical formula. Comparing the empirical molar mass to the experimentally determined molar mass will reveal the relationship between the empirical formula and molecular formula. Solution We start by assuming that there are 100 g of the compound. Then each percentage can be converted directly to grams; that is, 30.46 g of N and 69.54 g of O. Let n represent the number of moles of each element so that 1 mol N nN 5 30.46 g N 3 5 2.174 mol N 14.01 g N 1 mol O nO 5 69.54 g O 3 5 4.346 mol O 16.00 g O (Continued) 90 Chapter 3 ■ Mass Relationships in Chemical Reactions Thus, we arrive at the formula N2.174O4.346, which gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing the subscripts by the smaller subscript (2.174). After rounding off, we obtain NO2 as the empirical formula. The molecular formula might be the same as the empirical formula or some integral multiple of it (for example, two, three, four, or more times the empirical formula). Comparing the ratio of the molar mass to the molar mass of the empirical formula will show the integral relationship between the empirical and molecular formulas. The molar mass of the empirical formula NO2 is empirical molar mass 5 14.01 g 1 2(16.00 g) 5 46.01 g Next, we determine the ratio between the molar mass and the empirical molar mass molar mass 90 g 5 <2 empirical molar mass 46.01 g N 2O4 The molar mass is twice the empirical molar mass. This means that there are two NO2 units in each molecule of the compound, and the molecular formula is (NO2)2 or N2O4. The actual molar mass of the compound is two times the empirical molar mass, that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g. Check Note that in determining the molecular formula from the empirical formula, we need only know the approximate molar mass of the compound. The reason is that the true molar mass is an integral multiple (13, 23, 33, . . .) of the empirical molar mass. Therefore, the ratio (molar mass/empirical molar mass) will always be close to Similar problems: 3.52, 3.53, 3.54. an integer. Practice Exercise A sample of a compound containing boron (B) and hydrogen (H) contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g. What is its molecular formula? Review of Concepts What is the molecular formula of a compound containing only carbon and hydrogen if combustion of 1.05 g of the compound produces 3.30 g CO2 and 1.35 g H2O and its molar mass is about 70 g? 3.7 Chemical Reactions and Chemical Equations Having discussed the masses of atoms and molecules, we turn next to what happens to atoms and molecules in a chemical reaction, a process in which a substance (or substances) is changed into one or more new substances. To communicate with one another about chemical reactions, chemists have devised a standard way to represent them using chemical equations. A chemical equation uses chemical symbols to show what happens during a chemical reaction. In this section, we will learn how to write chemical equations and balance them. Writing Chemical Equations Consider what happens when hydrogen gas (H2) burns in air (which contains oxygen, O2) to form water (H2O). This reaction can be represented by the chem- ical equation H2 1 O2 ¡ H2 O (3.2)  3.7 Chemical Reactions and Chemical Equations 91 + Two hydrogen molecules + One oxygen molecule Two water molecules 2H2 + O2 2H2O 2 moles H2 + 1 mole O2 2 moles H2O 2(2.02 g) = 4.04 g H2 + 32.00 g O2 2(18.02 g) = 36.04 g H2O 36.04 g reactants 36.04 g product Figure 3.7 Three ways of representing the combustion of hydrogen. In accordance with the law of conservation of mass, the number of each type of atom must be the same on both sides of the equation. where the “plus” sign means “reacts with” and the arrow means “to yield.” Thus, this symbolic expression can be read: “Molecular hydrogen reacts with molecular oxygen to yield water.” The reaction is assumed to proceed from left to right as the arrow indicates. Equation (3.2) is not complete, however, because there are twice as many oxygen We use the law of conservation of mass as our guide in balancing chemical atoms on the left side of the arrow (two) as on the right side (one). To conform with equations. the law of conservation of mass, there must be the same number of each type of atom on both sides of the arrow; that is, we must have as many atoms after the reaction ends as we did before it started. We can balance Equation (3.2) by placing the appro- priate coefficient (2 in this case) in front of H2 and H2O: 2H2 1 O2 ¡ 2H2O When the coefficient is 1, as in the case of O2, it is not shown. This balanced chemical equation shows that “two hydrogen molecules can combine or react with one oxygen molecule to form two water molecules” (Figure 3.7). Because the ratio of the number of molecules is equal to the ratio of the number of moles, the equation can also be read as “2 moles of hydrogen molecules react with 1 mole of oxygen molecules to produce 2 moles of water molecules.” We know the mass of a mole of each of these substances, so we can also interpret the equation as “4.04 g of H2 react with 32.00 g of O2 to give 36.04 g of H2O.” These three ways of reading the equation are summarized in Figure 3.7. We refer to H2 and O2 in Equation (3.2) as reactants, which are the starting materials in a chemical reaction. Water is the product, which is the substance formed as a result of a chemical reaction. A chemical equation, then, is just the chemist’s shorthand description of a reaction. In a chemical equation, the reactants are conven- tionally written on the left and the products on the right of the arrow: reactants ¡ products To provide additional information, chemists often indicate the physical states of the reactants and products by using the letters g, l, and s to denote gas, liquid, and solid, respectively. For example, 2CO(g) 1 O2 (g) ¡ 2CO2 (g) The procedure for balancing chemical equations is shown on p. 92. 2HgO(s) ¡ 2Hg(l) 1 O2 (g) To represent what happens when sodium chloride (NaCl) is added to water, we write H2O NaCl(s) ¡ NaCl(aq) 92 Chapter 3 ■ Mass Relationships in Chemical Reactions where aq denotes the aqueous (that is, water) environment. Writing H2O above the arrow symbolizes the physical process of dissolving a substance in water, although it is sometimes left out for simplicity. Knowing the states of the reactants and products is especially useful in the labo- ratory. For example, when potassium bromide (KBr) and silver nitrate (AgNO3) react in an aqueous environment, a solid, silver bromide (AgBr), is formed. This reaction can be represented by the equation: KBr(aq) 1 AgNO3 (aq) ¡ KNO3 (aq) 1 AgBr(s) If the physical states of reactants and products are not given, an uninformed person might try to bring about the reaction by mixing solid KBr with solid AgNO3. These solids would react very slowly or not at all. Imagining the process on the microscopic level, we can understand that for a product like silver bromide to form, the Ag1 and Br2 ions would have to come in contact with each other. However, these ions are locked in place in their solid compounds and have little mobility. (Here is an example of how we explain a phenomenon by thinking about what happens at the molecular level, as discussed in Section 1.2.) Balancing Chemical Equations Suppose we want to write an equation to describe a chemical reaction that we have just carried out in the laboratory. How should we go about doing it? Because we know the identities of the reactants, we can write their chemical formulas. The iden- tities of products are more difficult to establish. For simple reactions it is often pos- sible to guess the product(s). For more complicated reactions involving three or more products, chemists may need to perform further tests to establish the presence of specific compounds. Once we have identified all the reactants and products and have written the cor- rect formulas for them, we assemble them in the conventional sequence—reactants on the left separated by an arrow from products on the right. The equation written at this point is likely to be unbalanced; that is, the number of each type of atom on one side of the arrow differs from the number on the other side. In general, we can balance a chemical equation by the following steps: 1. Identify all reactants and products and write their correct formulas on the left side and right side of the equation, respectively. 2. Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of the equation. We can change the coefficients (the numbers preceding the formulas) but not the subscripts (the numbers within formulas). Changing the subscripts would change the identity of the substance. For example, 2NO2 means “two molecules of nitrogen dioxide,” but if we double the subscripts, we have N2O4, which is the formula of dinitrogen tetroxide, a completely different compound. 3. First, look for elements that appear only once on each side of the equation with the same number of atoms on each side: The formulas containing these elements must have the same coefficient. Therefore, there is no need to adjust the coeffi- cients of these elements at this point. Next, look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. Finally, balance elements that appear in two or more formulas on the same side of the equation. 4. Check your balanced equation to be sure that you have the same total number of each type of atoms on both sides of the equation arrow.  3.7 Chemical Reactions and Chemical Equations 93 Let’s consider a specific example. In the laboratory, small amounts of oxygen gas can be prepared by heating potassium chlorate (KClO3). The products are oxygen gas (O2) and potassium chloride (KCl). From this information, we write KClO3 ¡ KCl 1 O2 (For simplicity, we omit the physical states of reactants and products.) All three ele- ments (K, Cl, and O) appear only once on each side of the equation, but only for K and Cl do we have equal numbers of atoms on both sides. Thus, KClO3 and KCl must have the same coefficient. The next step is to make the number of O atoms the same on both sides of the equation. Because there are three O atoms on the left and two O atoms on the right of the equation, we can balance the O atoms by placing a 2 in Heating potassium chlorate front of KClO3 and a 3 in front of O2. produces oxygen, which supports the combustion of a wood splint. 2KClO3 ¡ KCl 1 3O2 Finally, we balance the K and Cl atoms by placing a 2 in front of KCl: 2KClO3 ¡ 2KCl 1 3O2 (3.3) As a final check, we can draw up a balance sheet for the reactants and products where the number in parentheses indicates the number of atoms of each element: Reactants Products K (2) K (2) Cl (2) Cl (2) O (6) O (6) Note that this equation could also be balanced with coefficients that are multiples of 2 (for KClO3), 2 (for KCl), and 3 (for O2); for example, 4KClO3 ¡ 4KCl 1 6O2 However, it is common practice to use the simplest possible set of whole-number coefficients to balance the equation. Equation (3.3) conforms to this convention. Now let us consider the combustion (that is, burning) of the natural gas compo- nent ethane (C2H6) in oxygen or air, which yields carbon dioxide (CO2) and water. The unbalanced equation is C2H6 1 O2 ¡ CO2 1 H2O We see that the number of atoms is not the same on both sides of the equation for any of the elements (C, H, and O). In addition, C and H appear only once on each side of the equation; O appears in two compounds on the right side (CO2 and H2O). C2H6 To balance the C atoms, we place a 2 in front of CO2: C2H6 1 O2 ¡ 2CO2 1 H2O To balance the H atoms, we place a 3 in front of H2O: C2H6 1 O2 ¡ 2CO2 1 3H2O At this stage, the C and H atoms are balanced, but the O atoms are not because there are seven O atoms on the right-hand side and only two O atoms on the left-hand side 94 Chapter 3 ■ Mass Relationships in Chemical Reactions 7 of the equation. This inequality of O atoms can be eliminated by writing 2 in front of the O2 on the left-hand side: C2H6 1 72 O2 ¡ 2CO2 1 3H2O The “logic” for using 72 as a coefficient is that there were seven oxygen atoms on the right-hand side of the equation, but only a pair of oxygen atoms (O2) on the left. To balance them we ask how many pairs of oxygen atoms are needed to equal seven oxygen atoms. Just as 3.5 pairs of shoes equal seven shoes, 72 O2 molecules equal seven O atoms. As the following tally shows, the equation is now balanced: Reactants Products C (2) C (2) H (6) H (6) O (7) O (7) However, we normally prefer to express the coefficients as whole numbers rather than as fractions. Therefore, we multiply the entire equation by 2 to convert 72 to 7: 2C2H6 1 7O2 ¡ 4CO2 1 6H2O The final tally is Reactants Products C (4) C (4) H (12) H (12) O (14) O (14) Note that the coefficients used in balancing the last equation are the smallest possible set of whole numbers. In Example 3.12 we will continue to practice our equation-balancing skills. Example 3.12 When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between aluminum and oxygen, and it is the reason that aluminum beverage cans do not corrode. [In the case of iron, the rust, or iron(III) oxide, that forms is too porous to protect the iron metal underneath, so rusting continues.] Write a balanced equation for the formation of Al2O3. Strategy Remember that the formula of an element or compound cannot be changed when balancing a chemical equation. The equation is balanced by placing the appropriate coefficients in front of the formulas. Follow the procedure described on p. 92. Solution The unbalanced equation is Al 1 O2 ¡ Al2O3 In a balanced equation, the number and types of atoms on each side of the equation must be the same. We see that there is one Al atom on the reactants side and there are two Al atoms on the product side. We can balance the Al atoms by placing a coefficient of 2 in front of Al on the reactants side. 2Al 1 O2 ¡ Al2O3 There are two O atoms on the reactants side, and three O atoms on the product side of the equation. We can balance the O atoms by placing a coefficient of 32 in front of O2 on the reactants side. 2Al 1 32 O2 ¡ Al2O3 (Continued)  3.8 Amounts of Reactants and Products 95 This is a balanced equation. However, equations are normally balanced with the smallest set of whole-number coefficients. Multiplying both sides of the equation by 2 gives whole-number coefficients. 2(2Al 1 32 O2 ¡ Al2O3 ) or 4Al 1 3O2 ¡ 2Al2O3 Check For an equation to be balanced, the number and types of atoms on each side of the equation must be the same. The final tally is Reactants Products Al (4) Al (4) O (6) O (6) The equation is balanced. Also, the coefficients are reduced to the simplest set of Similar problems: 3.59, 3.60. whole numbers. Practice Exercise Balance the equation representing the reaction between iron(III) oxide, Fe2O3, and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO2). Review of Concepts Which parts of the equation shown here are essential for a balanced equation and which parts are helpful if we want to carry out the reaction in the laboratory? BaH2 (s) 1 2H2O(l) ¡ Ba(OH) 2 (aq) 1 2H2 (g) 3.8 Amounts of Reactants and Products A basic question raised in the chemical laboratory is “How much product will be formed from specific amounts of starting materials (reactants)?” Or in some cases, we might ask the reverse question: “How much starting material must be used to obtain a specific amount of product?” To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mole concept. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. Whether the units given for reactants (or products) are moles, grams, liters (for gases), or some other units, we use moles to calculate the amount of product formed in a reaction. This approach is called the mole method, which means simply that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. For example, industrially ammonia is synthesized from hydrogen and nitrogen as follows: N2 (g) 1 3H2 (g) ¡ 2NH3 (g) The stoichiometric coefficients show that one molecule of N2 reacts with three mol- ecules of H2 to form two molecules of NH3. It follows that the relative numbers of moles are the same as the relative number of molecules: N2(g) 1 3H2(g) ¡ 2NH3(g) 1 molecule 3 molecules 2 molecules 6.022 3 1023 molecules 3(6.022 3 1023 molecules) 2(6.022 3 1023 molecules) The synthesis of NH3 from H2 1 mol 3 mol 2 mol and N2. 96 Chapter 3 ■ Mass Relationships in Chemical Reactions Thus, this equation can also be read as “1 mole of N2 gas combines with 3 moles of H2 gas to form 2 moles of NH3 gas.” In stoichiometric calculations, we say that three moles of H2 are equivalent to two moles of NH3, that is, 3 mol H2 ∞ 2 mol NH3 where the symbol ∞ means “stoichiometrically equivalent to” or simply “equivalent to.” This relationship enables us to write the conversion factors 3 mol H2 2 mol NH3   and   2 mol NH3 3 mol H2 Similarly, we have 1 mol N2 ∞ 2 mol NH3 and 1 mol N2 ∞ 3 mol H2. Let’s consider a simple example in which 6.0 moles of H2 react completely with N2 to form NH3. To calculate the amount of NH3 produced in moles, we use the conversion factor that has H2 in the denominator and write 2 mol NH3 moles of NH3 produced 5 6.0 mol H2 3 3 mol H2 5 4.0 mol NH3 Now suppose 16.0 g of H2 react completely with N2 to form NH3. How many grams of NH3 will be formed? To do this calculation, we note that the link between H2 and NH3 is the mole ratio from the balanced equation. So we need to first convert grams of H2 to moles of H2, then to moles of NH3, and finally to grams of NH3. The conversion steps are grams of H2 ¡ moles of H2 ¡ moles of NH3 ¡ grams of NH3 First, we convert 16.0 g of H2 to number of moles of H2, using the molar mass of H2 as the conversion factor: 1 mol H2 moles of H2 5 16.0 g H2 3 2.016 g H2 5 7.94 mol H2 Next, we calculate the number of moles of NH3 produced. 2 mol NH3 moles of NH3 5 7.94 mol H2 3 3 mol H2 5 5.29 mol NH3 Finally, we calculate the mass of NH3 produced in grams using the molar mass of NH3 as the conversion factor 17.03 g NH3 grams of NH3 5 5.29 mol NH3 3 1 mol NH3 5 90.1 g NH3 These three separate calculations can be combined in a single step as follows: 1 mol H2 2 mol NH3 17.03 g NH3 grams of NH3 5 16.0 g H2 3 3 3 2.016 g H2 3 mol H2 1 mol NH3 5 90.1 g NH3  3.8 Amounts of Reactants and Products 97 Figure 3.8 The procedure for Mass (g) Mass (g) calculating the amounts of of compound A of compound B reactants or products in a reaction using the mole method. Use molar Use molar mass (g/mol) mass (g/mol) of compound A of compound B Use mole ratio Moles of of A and B Moles of compound A from balanced compound B equation Similarly, we can calculate the mass in grams of N2 consumed in this reaction. The conversion steps are grams of H2 ¡ moles of H2 ¡ moles of N2 ¡ grams of N2 By using the relationship 1 mol N2 ∞ 3 mol H2, we write 1 mol H2 1 mol N2 28.02 g N2 grams of N2 5 16.0 g H2 3 3 3 2.016 g H2 3 mol H2 1 mol N2 5 74.1 g N2 The general approach for solving stoichiometry problems is summarized next. 1. Write a balanced equation for the reaction. 2. Convert the given amount of the reactant (in grams or other units) to number of moles. 3. Use the mole ratio from the balanced equation to calculate the number of moles of product formed. 4. Convert the moles of product to grams (or other units) of product. Figure 3.8 shows these steps. Sometimes we may be asked to calculate the amount of a reactant needed to form a specific amount of product. In those cases, we can reverse the steps shown in Figure 3.8. Examples 3.13 and 3.14 illustrate the application of this approach. Example 3.13 The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O): C6H12O6 1 6O2 ¡ 6CO2 1 6H2O If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced? C6H12O6 Strategy Looking at the balanced equation, how do we compare the amounts of C6H12O6 and CO2? We can compare them based on the mole ratio from the balanced equation. Starting with grams of C6H12O6, how do we convert to moles of C6H12O6? Once moles of CO2 are determined using the mole ratio from the balanced equation, how do we convert to grams of CO2? Solution We follow the preceding steps and Figure 3.8. (Continued) 98 Chapter 3 ■ Mass Relationships in Chemical Reactions Step 1: The balanced equation is given in the problem. Step 2: To convert grams of C6H12O6 to moles of C6H12O6, we write 1 mol C6H12O6 856 g C6H12O6 3 5 4.750 mol C6H12O6 180.2 g C6H12O6 Step 3: From the mole ratio, we see that 1 mol C6H12O6 ∞ 6 mol CO2. Therefore, the number of moles of CO2 formed is 6 mol CO2 4.750 mol C6H12O6 3 5 28.50 mol CO2 1 mol C6H12O6 Step 4: Finally, the number of grams of CO2 formed is given by 44.01 g CO2 28.50 mol CO2 3 5 1.25 3 103 g CO2 1 mol CO2 After some practice, we can combine the conversion steps grams of C6H12O6 ¡ moles of C6H12O6 ¡ moles of CO2 ¡ grams of CO2 into one equation: 1 mol C6H12O6 6 mol CO2 44.01 g CO2 mass of CO2 5 856 g C6H12O6 3 3 3 180.2 g C6H12 O6 1 mol C6H12O6 1 mol CO2 5 1.25 3 103 g CO2 Check Does the answer seem reasonable? Should the mass of CO2 produced be larger than the mass of C6H12O6 reacted, even though the molar mass of CO2 is considerably less than the molar mass of C6H12O6? What is the mole ratio between Similar problem: 3.72. CO2 and C6H12O6? Practice Exercise Methanol (CH3OH) burns in air according to the equation 2CH3OH 1 3O2 ¡ 2CO2 1 4H2O If 209 g of methanol are used up in a combustion process, what is the mass of H2O produced? Example 3.14 All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water: 2Li(s) 1 2H2O(l) ¡ 2LiOH(aq) 1 H2 (g) How many grams of Li are needed to produce 9.89 g of H2? Lithium reacting with water to produce hydrogen gas. Strategy The question asks for number of grams of reactant (Li) to form a specific amount of product (H2). Therefore, we need to reverse the steps shown in Figure 3.8. From the equation we see that 2 mol Li ∞ 1 mol H2. Solution The conversion steps are grams of H2 ¡ moles of H2 ¡ moles of Li ¡ grams of Li (Continued)  3.9 Limiting Reagents 99 Combining these steps into one equation, we write 1 mol H2 2 mol Li 6.941 g Li 9.89 g H2 3 3 3 5 68.1 g Li 2.016 g H2 1 mol H2 1 mol Li Check There are roughly 5 moles of H2 in 9.89 g H2, so we need 10 moles of Li. From the approximate molar mass of Li (7 g), does the answer seem reasonable? Similar problem: 3.66. Practice Exercise The reaction between nitric oxide (NO) and oxygen to form nitrogen dioxide (NO2) is a key step in photochemical smog formation: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) How many grams of O2 are needed to produce 2.21 g of NO2? Animation Review of Concepts Limiting Reagent Which of the following statements is correct for the equation shown here? 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) Before reaction has started (a) 6 g of H2O are produced for every 4 g of NH3 reacted. (b) 1 mole of NO is produced per mole of NH3 reacted. (c) 2 moles of NO are produced for every 3 moles of O2 reacted. 3.9 Limiting Reagents When a chemist carries out a reaction, the reactants are usually not present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation. Because the goal of a reaction is to produce the maximum quantity of a useful com- pound from the starting materials, frequently a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted to the desired product. Consequently, some reactant will be left over at the end of the reaction. The reactant used up first in a reaction is called the limiting reagent, because the maxi- mum amount of product formed depends on how much of this reactant was originally present. When this reactant is used up, no more product can be formed. Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent. The concept of the limiting reagent is analogous to the relationship between men and women in a dance contest at a club. If there are 14 men and only 9 women, then After reaction is complete only 9 female/male pairs can compete. Five men will be left without partners. The number of women thus limits the number of men that can dance in the contest, and H2 CO CH3OH there is an excess of men. Figure 3.9 At the start of the Consider the industrial synthesis of methanol (CH3OH) from carbon monoxide reaction, there were six H2 and hydrogen at high temperatures: molecules and four CO molecules. At the end, all the H2 molecules are gone and only one CO CO(g) 1 2H2 (g) ¡ CH3OH(g) molecule is left. Therefore, H2 molecule is the limiting reagent Suppose initially we have 4 moles of CO and 6 moles of H2 (Figure 3.9). One way and CO is the excess reagent. Each molecule can also be to determine which of two reactants is the limiting reagent is to calculate the number treated as one mole of the of moles of CH3OH obtained based on the initial quantities of CO and H2. From the substance in this reaction. 100 Chapter 3 ■ Mass Relationships in Chemical Reactions preceding definition, we see that only the limiting reagent will yield the smaller amount of the product. Starting with 4 moles of CO, we find the number of moles of CH3OH produced is 1 mol CH3OH 4 mol CO 3 5 4 mol CH3OH 1 mol CO and starting with 6 moles of H2, the number of moles of CH3OH formed is 1 mol CH3OH 6 mol H2 3 5 3 mol CH3OH 2 mol H2 Because H2 results in a smaller amount of CH3OH, it must be the limiting reagent. Therefore, CO is the excess reagent. In stoichiometric calculations involving limiting reagents, the first step is to decide which reactant is the limiting reagent. After the limiting reagent has been identified, the rest of the problem can be solved as outlined in Section 3.8. Example 3.15 illus- trates this approach. Example 3.15 The synthesis of urea, [(NH2)2CO], is considered to be the first recognized example of preparing a biological compound from nonbiological reactants, challenging the notion that biological processes involved a “vital force” present only in living systems. Today urea is produced industrially by reacting ammonia with carbon dioxide: 2NH3 (g) 1 CO2 (g) ¡ (NH2 ) 2CO(aq) 1 H2O(l) In one process, 637.2 g of NH3 are treated with 1142 g of CO2. (a) Which of the two (NH2)2CO reactants is the limiting reagent? (b) Calculate the mass of (NH2)2CO formed. (c) How much excess reagent (in grams) is left at the end of the reaction? (a) Strategy The reactant that produces fewer moles of product is the limiting reagent because it limits the amount of product that can be formed. How do we convert from the amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, (NH2)2CO, formed by the given amounts of NH3 and CO2 to determine which reactant is the limiting reagent. Solution We carry out two separate calculations. First, starting with 637.2 g of NH3, we calculate the number of moles of (NH2)2CO that could be produced if all the NH3 reacted according to the following conversions: grams of NH3 ¡ moles of NH3 ¡ moles of (NH2 ) 2CO Combining these conversions in one step, we write 1 mol NH3 1 mol (NH2 ) 2CO moles of (NH2 ) 2CO 5 637.2 g NH3 3 3 17.03 g NH3 2 mol NH3 5 18.71 mol (NH2 ) 2CO Second, for 1142 g of CO2, the conversions are grams of CO2 ¡ moles of CO2 ¡ moles of (NH2 ) 2CO The number of moles of (NH2)2CO that could be produced if all the CO2 reacted is 1 mol CO2 1 mol (NH2 ) 2CO moles of (NH2 ) 2CO 5 1142 g CO2 3 3 44.01 g CO2 1 mol CO2 5 25.95 mol (NH2 ) 2CO (Continued)  3.9 Limiting Reagents 101 It follows, therefore, that NH3 must be the limiting reagent because it produces a smaller amount of (NH2)2CO. (b) Strategy We determined the moles of (NH2)2CO produced in part (a), using NH3 as the limiting reagent. How do we convert from moles to grams? Solution The molar mass of (NH2)2CO is 60.06 g. We use this as a conversion factor to convert from moles of (NH2)2CO to grams of (NH2)2CO: 60.06 g (NH2 ) 2CO mass of (NH2 ) 2CO 5 18.71 mol (NH2 ) 2CO 3 1 mol (NH2 ) 2CO 5 1124 g (NH2 ) 2CO Check Does your answer seem reasonable? 18.71 moles of product are formed. What is the mass of 1 mole of (NH2)2CO? (c) Strategy Working backward, we can determine the amount of CO2 that reacted to produce 18.71 moles of (NH2)2CO. The amount of CO2 left over is the difference between the initial amount and the amount reacted. Solution Starting with 18.71 moles of (NH2)2CO, we can determine the mass of CO2 that reacted using the mole ratio from the balanced equation and the molar mass of CO2. The conversion steps are moles of (NH2 ) 2CO ¡ moles of CO2 ¡ grams of CO2 so that 1 mol CO2 44.01 g CO2 mass of CO2 reacted 5 18.71 mol (NH2 ) 2CO 3 3 1 mol (NH2 ) 2CO 1 mol CO2 5 823.4 g CO2 The amount of CO2 remaining (in excess) is the difference between the initial amount (1142 g) and the amount reacted (823.4 g): mass of CO2 remaining 5 1142 g 2 823.4 g 5 319 g Similar problem: 3.86. Practice Exercise The reaction between aluminum and iron(III) oxide can generate temperatures approaching 30008C and is used in welding metals: 2Al 1 Fe2O3 ¡ Al2O3 1 2Fe In one process, 124 g of Al are reacted with 601 g of Fe2O3. (a) Calculate the mass (in grams) of Al2O3 formed. (b) How much of the excess reagent is left at the end of the reaction? Example 3.15 brings out an important point. In practice, chemists usually choose the more expensive chemical as the limiting reagent so that all or most of it will be converted to products in the reaction. In the synthesis of urea, NH3 is invariably the limiting reagent because it is more expensive than CO2. At other times, an excess of one reagent is used to help drive the reaction to completion, or to compensate for a side reaction that consumes that reagent. Synthetic chemists often have to calculate the amount of reagents to use based on this need to have one or more components in excess, as Example 3.16 shows. 102 Chapter 3 ■ Mass Relationships in Chemical Reactions Example 3.16 The reaction between alcohols and halogen compounds to form ethers is important in organic chemistry, as illustrated here for the reaction between methanol (CH3OH) and methyl bromide (CH3Br) to form dimethylether (CH3OCH3), which is a useful precursor to other organic compounds and an aerosol propellant. CH3OH 1 CH3Br 1 LiC4H9 ¡ CH3OCH3 1 LiBr 1 C4H10 This reaction is carried out in a dry (water-free) organic solvent, and the butyl lithium (LiC 4H9) serves to remove a hydrogen ion from CH3OH. Butyl lithium will also react with any residual water in the solvent, so the reaction is typically carried out with 2.5  molar equivalents of that reagent. How many grams of CH3Br and LiC4H9 will be needed to carry out the preceding reaction with 10.0 g of CH 3OH? Solution We start with the knowledge that CH3OH and CH3Br are present in stoichiometric amounts and that LiC4H9 is the excess reagent. To calculate the quantities of CH3Br and LiC4H9 needed, we proceed as shown in Example 3.14. 1 mol CH3OH 1 mol CH3Br 94.93 g CH3Br grams of CH3Br 5 10.0 g CH3OH 3 3 3 32.04 g CH3OH 1 mol CH3OH 1 mol CH3Br 5 29.6 g CH3Br 1 mol CH3OH 2.5 mol LiC4H9 64.05 g LiC4H9 grams of LiC4H9 5 10.0 g CH3OH 3 3 3 32.04 g CH3OH 1 mol CH3OH 1 mol LiC4H9 Similar problems: 3.137, 3.138. 5 50.0 g LiC4H9 Practice Exercise The reaction between benzoic acid (C6H5COOH) and octanol (C8H17OH) to yield octyl benzoate (C6H5COOC8H17) and water C6H5COOH 1 C8H17OH ¡ C6H5COOC8H17 1 H2O is carried out with an excess of C8H17OH to help drive the reaction to completion and maximize the yield of product. If an organic chemist wants to use 1.5 molar equivalents of C8H17OH, how many grams of C8H17OH would be required to carry out the reaction with 15.7 g of C6H5COOH? Review of Concepts Consider the following reaction: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) Starting with the reactants shown in (a), which of the diagrams shown in (b)–(d) best represents the situation in which the limiting reagent has completely reacted? NO O2 NO2 (a) (b) (c) (d)  3.10 Reaction Yield 103 3.10 Reaction Yield The amount of limiting reagent present at the start of a reaction determines the Keep in mind that the theoretical yield is the yield that you calculate using the theoretical yield of the reaction, that is, the amount of product that would result balanced equation. The actual yield is if all the limiting reagent reacted. The theoretical yield, then, is the maximum the yield obtained by carrying out the reaction. obtainable yield, predicted by the balanced equation. In practice, the actual yield, or the amount of product actually obtained from a reaction, is almost always less than the theoretical yield. There are many reasons for the difference between actual and theoretical yields. For instance, many reactions are reversible, and so they do not proceed 100 percent from left to right. Even when a reaction is 100 percent complete, it may be difficult to recover all of the product from the reaction medium (say, from an aqueous solution). Some reactions are complex in the sense that the products formed may react further among themselves or with the reactants to form still other products. These additional reactions will reduce the yield of the first reaction. To determine how efficient a given reaction is, chemists often figure the percent yield, which describes the proportion of the actual yield to the theoretical yield. It is calculated as follows: actual yield %yield 5 3 100% (3.4) theoretical yield Percent yields may range from a fraction of 1 percent to 100 percent. Chemists strive to maximize the percent yield in a reaction. Factors that can affect the percent yield include temperature and pressure. We will study these effects later. In Example 3.17 we will calculate the yield of an industrial process. Example 3.17 Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 9508C and 11508C: TiCl4 (g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2 (l) In a certain industrial operation 3.54 3 107 g of TiCl4 are reacted with 1.13 3 107 g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 3 106 g of Ti are actually obtained. (a) Strategy Because there are two reactants, this is likely to be a limiting reagent problem. The reactant that produces fewer moles of product is the limiting reagent. How do we convert from amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, Ti, formed. Solution Carry out two separate calculations to see which of the two reactants is the limiting reagent. First, starting with 3.54 3 107 g of TiCl4, calculate the number of moles of Ti that could be produced if all the TiCl4 reacted. The conversions are grams of TiCl4 ¡ moles of TiCl4 ¡ moles of Ti (Continued) 104 Chapter 3 ■ Mass Relationships in Chemical Reactions so that 1 mol TiCl4 1 mol Ti moles of Ti 5 3.54 3 107 g TiCl4 3 3 189.7 g TiCl4 1 mol TiCl4 5 1.87 3 105 mol Ti Next, we calculate the number of moles of Ti formed from 1.13 3 107 g of Mg. The conversion steps are grams of Mg ¡ moles of Mg ¡ moles of Ti and we write 1 mol Mg 1 mol Ti moles of Ti 5 1.13 3 107 g Mg 3 3 24.31 g Mg 2 mol Mg 5 2.32 3 105 mol Ti Therefore, TiCl4 is the limiting reagent because it produces a smaller amount of Ti. The mass of Ti formed is 47.88 g Ti 1.87 3 105 mol Ti 3 5 8.95 3 106 g Ti 1 mol Ti (b) Strategy The mass of Ti determined in part (a) is the theoretical yield. The amount given in part (b) is the actual yield of the reaction. Solution The percent yield is given by An artificial hip joint made of titanium and the structure of actual yield solid titanium. %yield 5 3 100% theoretical yield 6 7.91 3 10 g 5 3 100% 8.95 3 106 g 5 88.4% Similar problems: 3.89, 3.90. Check Should the percent yield be less than 100 percent? Practice Exercise Industrially, vanadium metal, which is used in steel alloys, can be obtained by reacting vanadium(V) oxide with calcium at high temperatures: 5Ca 1 V2O5 ¡ 5CaO 1 2V In one process, 1.54 3 103 g of V2O5 react with 1.96 3 103 g of Ca. (a) Calculate the theoretical yield of V. (b) Calculate the percent yield if 803 g of V are obtained. Industrial processes usually involve huge quantities (thousands to millions of tons) of products. Thus, even a slight improvement in the yield can significantly reduce the cost of production. A case in point is the manufacture of chemical fertil- izers, discussed in the Chemistry in Action essay on p. 105. Review of Concepts Can the percent yield ever exceed the theoretical yield of a reaction? CHEMISTRY in Action Chemical Fertilizers F eeding the world’s rapidly increasing population requires that farmers produce ever-larger and healthier crops. Every year they add hundreds of millions of tons of chemical fertiliz- Another method of preparing ammonium sulfate requires two steps: ers to the soil to increase crop quality and yield. In addition to 2NH3 (aq) 1 CO2 (aq) 1 H2O(l) ¡ (NH4 ) 2CO3 (aq) (1) carbon dioxide and water, plants need at least six elements for (NH4 ) 2CO3 (aq) 1 CaSO4 (aq) ¡ satisfactory growth. They are N, P, K, Ca, S, and Mg. The (NH4 ) 2SO4 (aq) 1 CaCO3 (s) (2) preparation and properties of several nitrogen- and phosphorus- containing fertilizers illustrate some of the principles introduced This approach is desirable because the starting materials— in this chapter. carbon dioxide and calcium sulfate—are less costly than sulfu- Nitrogen fertilizers contain nitrate (NO23) salts, ammonium ric acid. To increase the yield, ammonia is made the limiting 1 (NH4 ) salts, and other compounds. Plants can absorb nitrogen reagent in Reaction (1) and ammonium carbonate is made the in the form of nitrate directly, but ammonium salts and ammonia limiting reagent in Reaction (2). (NH3) must first be converted to nitrates by the action of soil The table lists the percent composition by mass of nitrogen bacteria. The principal raw material of nitrogen fertilizers is in some common fertilizers. The preparation of urea was dis- ammonia, prepared by the reaction between hydrogen and cussed in Example 3.15. nitrogen: Percent Composition by Mass of Nitrogen in Five Common Fertilizers 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) Fertilizer % N by Mass (This reaction will be discussed in detail in Chapters 13 and 14.) In its liquid form, ammonia can be injected directly into the soil. NH3 82.4 Alternatively, ammonia can be converted to ammonium NH4NO3 35.0 nitrate, NH4NO3, ammonium sulfate, (NH4)2SO4, or ammonium (NH4)2SO4 21.2 hydrogen phosphate, (NH4)2HPO4, in the following acid-base (NH4)2HPO4 21.2 reactions: (NH2)2CO 46.7 NH3 (aq) 1 HNO3 (aq) ¡ NH4NO3 (aq) 2NH3 (aq) 1 H2SO4 (aq) ¡ (NH4 ) 2SO4 (aq) Several factors influence the choice of one fertilizer over an- 2NH3 (aq) 1 H3PO4 (aq) ¡ (NH4 ) 2HPO4 (aq) other: (1) cost of the raw materials needed to prepare the fertilizer; (2) ease of storage, transportation, and utilization; (3) percent composition by mass of the desired element; and (4) suitability of the compound, that is, whether the compound is soluble in water and whether it can be readily taken up by plants. Considering all these factors together, we find that NH4NO3 is the most important nitrogen-containing fertilizer in the world, even though ammonia has the highest percentage by mass of nitrogen. Phosphorus fertilizers are derived from phosphate rock, called fluorapatite, Ca5(PO4)3F. Fluorapatite is insoluble in water, so it must first be converted to water-soluble calcium dihydrogen phosphate [Ca(H2PO4)2]: 2Ca5 (PO4 ) 3F(s) 1 7H2SO4 (aq) ¡ 3Ca(H2PO4 ) 2 (aq) 1 7CaSO4 (aq) 1 2HF(g) For maximum yield, fluorapatite is made the limiting reagent in this reaction. The reactions we have discussed for the preparation of fer- tilizers all appear relatively simple, yet much effort has been expended to improve the yields by changing conditions such as temperature, pressure, and so on. Industrial chemists usually run promising reactions first in the laboratory and then test them Liquid ammonia being applied to the soil before planting. in a pilot facility before putting them into mass production. 105 106 Chapter 3 ■ Mass Relationships in Chemical Reactions Key Equations percent composition of an element in a compound 5 n 3 molar mass of element 3 100% (3.1) molar mass of compound actual yield %yield 5 3 100% (3.4) theoretical yield Summary of Facts & Concepts 1. Atomic masses are measured in atomic mass units (amu), 4. Chemical changes, called chemical reactions, are repre- a relative unit based on a value of exactly 12 for the C-12 sented by chemical equations. Substances that undergo isotope. The atomic mass given for the atoms of a particu- change—the reactants—are written on the left and the lar element is the average of the naturally occurring iso- substances formed—the products—appear to the right tope distribution of that element. The molecular mass of a of the arrow. Chemical equations must be balanced, in molecule is the sum of the atomic masses of the atoms in accordance with the law of conservation of mass. The the molecule. Both atomic mass and molecular mass can number of atoms of each element in the reactants must be accurately determined with a mass spectrometer. equal the number in the products. 2. A mole is Avogadro’s number (6.022 3 1023) of atoms, 5. Stoichiometry is the quantitative study of products and molecules, or other particles. The molar mass (in grams) reactants in chemical reactions. Stoichiometric calcu- of an element or a compound is numerically equal to its lations are best done by expressing both the known mass in atomic mass units (amu) and contains Avogadro’s and unknown quantities in terms of moles and then number of atoms (in the case of elements), molecules converting to other units if necessary. A limiting (in the case of molecular substances), or simplest for- reagent is the reactant that is present in the smallest mula units (in the case of ionic compounds). stoichiometric amount. It limits the amount of product 3. The percent composition by mass of a compound is the that can be formed. The amount of product obtained in percent by mass of each element present. If we know a reaction (the actual yield) may be less than the max- the percent composition by mass of a compound, we imum possible amount (the theoretical yield). The can deduce the empirical formula of the compound and ratio of the two multiplied by 100 percent is expressed also the molecular formula of the compound if the as the percent yield. approximate molar mass is known. Key Words Actual yield, p. 103 Chemical reaction, p. 90 Mole method, p. 95 Product, p. 91 Atomic mass, p. 76 Excess reagent, p. 99 Molecular mass, p. 81 Reactant, p. 91 Atomic mass unit (amu), p. 76 Limiting reagent, p. 99 Percent composition by Stoichiometric amount, p. 99 Avogadro’s number (NA), p. 77 Molar mass (m), p. 78 mass, p. 85 Stoichiometry, p. 95 Chemical equation, p. 90 Mole (mol), p. 77 Percent yield, p. 103 Theoretical yield, p. 103 Questions & Problems • Problems available in Connect Plus 3.2 What is the mass (in amu) of a carbon-12 atom? Red numbered problems solved in Student Solutions Manual Why is the atomic mass of carbon listed as 12.01 amu in the table on the inside front cover of this book? Atomic Mass 3.3 Explain clearly what is meant by the statement “The Review Questions atomic mass of gold is 197.0 amu.” 3.1 What is an atomic mass unit? Why is it necessary to introduce such a unit?  Questions & Problems 107 3.4 What information would you need to calculate the • 3.21 Which of the following has more atoms: 1.10 g of average atomic mass of an element? hydrogen atoms or 14.7 g of chromium atoms? • 3.22 Which of the following has a greater mass: 2 atoms Problems of lead or 5.1 3 10223 mole of helium. • 3.5 The atomic masses of 35 37 17Cl (75.53 percent) and 17Cl (24.47 percent) are 34.968 amu and 36.956 amu, Molecular Mass respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the Problems relative abundances. • 3.23 Calculate the molecular mass or formula mass (in 3.6 The atomic masses of 63Li and 73Li are 6.0151 amu amu) of each of the following substances: (a) CH4, and 7.0160 amu, respectively. Calculate the natural (b) NO2, (c) SO3, (d) C6H6, (e) NaI, (f) K2SO4, abundances of these two isotopes. The average (g) Ca3(PO4)2. atomic mass of Li is 6.941 amu. 3.24 Calculate the molar mass of the following substances: • 3.7 What is the mass in grams of 13.2 amu? (a) Li2CO3, (b) CS2, (c) CHCl3 (chloroform), • 3.8 How many amu are there in 8.4 g? (d) C6H8O6 (ascorbic acid, or vitamin C), (e) KNO3, (f) Mg3N2. • 3.25 Calculate the molar mass of a compound if 0.372 mole Avogadro’s Number and Molar Mass of it has a mass of 152 g. Review Questions • 3.26 How many molecules of ethane (C2H6) are present 3.9 Define the term “mole.” What is the unit for mole in in 0.334 g of C2H6? calculations? What does the mole have in common • 3.27 Calculate the number of C, H, and O atoms in 1.50 g with the pair, the dozen, and the gross? What does of glucose (C6H12O6), a sugar. Avogadro’s number represent? • 3.28 Dimethyl sulfoxide [(CH 3) 2SO], also called 3.10 What is the molar mass of an atom? What are the DMSO, is an important solvent that penetrates commonly used units for molar mass? the skin, enabling it to be used as a topical drug-delivery agent. Calculate the number of C, S, H, and O atoms in 7.14 3 103 g of dimethyl Problems sulfoxide. • 3.11 Earth’s population is about 7.2 billion. Suppose that • 3.29 Pheromones are a special type of compound every person on Earth participates in a process of secreted by the females of many insect species to counting identical particles at the rate of two parti- attract the males for mating. One pheromone has cles per second. How many years would it take to the molecular formula C19H38O. Normally, the count 6.0 3 1023 particles? Assume that there are amount of this pheromone secreted by a female in- 365 days in a year. sect is about 1.0 3 10212 g. How many molecules 3.12 The thickness of a piece of paper is 0.0036 in. Suppose are there in this quantity? a certain book has an Avogadro’s number of pages; • 3.30 The density of water is 1.00 g/mL at 48C. How many calculate the thickness of the book in light-years. (Hint: water molecules are present in 2.56 mL of water at See Problem 1.49 for the definition of light-year.) this temperature? • 3.13 How many atoms are there in 5.10 moles of sulfur (S)? • 3.14 How many moles of cobalt (Co) atoms are there in Mass Spectrometry 6.00 3 109 (6 billion) Co atoms? Review Questions • 3.15 How many moles of calcium (Ca) atoms are in 77.4 g 3.31 Describe the operation of a mass spectrometer. of Ca? 3.32 Describe how you would determine the isotopic • 3.16 How many grams of gold (Au) are there in 15.3 moles abundance of an element from its mass spectrum. of Au? • 3.17 What is the mass in grams of a single atom of each of the following elements? (a) Hg, (b) Ne. Problems • 3.18 What is the mass in grams of a single atom of each • 3.33 Carbon has two stable isotopes, 126C and 136C, and of the following elements? (a) As, (b) Ni. fluorine has only one stable isotope, 199F. How many • 3.19 What is the mass in grams of 1.00 3 1012 lead (Pb) peaks would you observe in the mass spectrum of atoms? the positive ion of CF14? Assume that the ion does • 3.20 A modern penny weighs 2.5 g but contains only not break up into smaller fragments. 0.063 g of copper (Cu). How many copper atoms are 3.34 Hydrogen has two stable isotopes, 11H and 21H, and present in a modern penny? sulfur has four stable isotopes, 32 33 34 36 16S, 16S, 16S, and 16S.  108 Chapter 3 ■ Mass Relationships in Chemical Reactions How many peaks would you observe in the mass molecular formula given that its molar mass is spectrum of the positive ion of hydrogen sulfide, about 120 g? H2S1? Assume no decomposition of the ion into • 3.45 The formula for rust can be represented by Fe2O3. smaller fragments. How many moles of Fe are present in 24.6 g of the compound? Percent Composition and Chemical Formulas • 3.46 How many grams of sulfur (S) are needed to react completely with 246 g of mercury (Hg) to form Review Questions HgS? 3.35 Use ammonia (NH3) to explain what is meant by the • 3.47 Calculate the mass in grams of iodine (I2) that will percent composition by mass of a compound. react completely with 20.4 g of aluminum (Al) to 3.36 Describe how the knowledge of the percent compo- form aluminum iodide (AlI3). sition by mass of an unknown compound can help us • 3.48 Tin(II) fluoride (SnF2) is often added to toothpaste identify the compound. as an ingredient to prevent tooth decay. What is the 3.37 What does the word “empirical” in empirical for- mass of F in grams in 24.6 g of the compound? mula mean? • 3.49 What are the empirical formulas of the compounds 3.38 If we know the empirical formula of a compound, with the following compositions? (a) 2.1 percent H, what additional information do we need to deter- 65.3 percent O, 32.6 percent S, (b) 20.2 percent Al, mine its molecular formula? 79.8 percent Cl. • 3.50 What are the empirical formulas of the compounds Problems with the following compositions? (a) 40.1 percent C, 6.6 percent H, 53.3 percent O, (b) 18.4 percent C, • 3.39 Tin (Sn) exists in Earth’s crust as SnO2. Calculate 21.5 percent N, 60.1 percent K. the percent composition by mass of Sn and O in • 3.51 The anticaking agent added to Morton salt is cal- SnO2. cium silicate, CaSiO3. This compound can absorb • 3.40 For many years chloroform (CHCl3) was used as an up to 2.5 times its mass of water and still remains a inhalation anesthetic in spite of the fact that it is also free-flowing powder. Calculate the percent compo- a toxic substance that may cause severe liver, kidney, sition of CaSiO3. and heart damage. Calculate the percent composi- • 3.52 The empirical formula of a compound is CH. If the tion by mass of this compound. molar mass of this compound is about 78 g, what is • 3.41 Cinnamic alcohol is used mainly in perfumery, its molecular formula? particularly in soaps and cosmetics. Its molecular • 3.53 The molar mass of caffeine is 194.19 g. Is the formula is C9H10O. (a) Calculate the percent compo- molecular formula of caffeine C 4H 5N 2O or sition by mass of C, H, and O in cinnamic alcohol. C8H10N4O2? (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g? • 3.54 Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for “Chinese restaurant 3.42 All of the substances listed here are fertilizers that syndrome,” the symptoms of which are headaches contribute nitrogen to the soil. Which of these is and chest pains. MSG has the following composi- the richest source of nitrogen on a mass percentage tion by mass: 35.51 percent C, 4.77 percent H, basis? 37.85 percent O, 8.29 percent N, and 13.60 percent (a) Urea, (NH2)2CO Na. What is its molecular formula if its molar mass (b) Ammonium nitrate, NH4NO3 is about 169 g? (c) Guanidine, HNC(NH2)2 (d) Ammonia, NH3 • 3.43 Allicin is the compound responsible for the charac- Chemical Reactions and Chemical Equations teristic smell of garlic. An analysis of the compound Review Questions gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; 3.55 Use the formation of water from hydrogen and oxy- O: 9.86 percent. Calculate its empirical formula. gen to explain the following terms: chemical reac- What is its molecular formula given that its molar tion, reactant, product. mass is about 162 g? 3.56 What is the difference between a chemical reaction 3.44 Peroxyacylnitrate (PAN) is one of the compo- and a chemical equation? nents of smog. It is a compound of C, H, N, and 3.57 Why must a chemical equation be balanced? What O. Determine the percent composition of oxygen law is obeyed by a balanced chemical equation? and the empirical formula from the following 3.58 Write the symbols used to represent gas, liquid, percent composition by mass: 19.8 percent C, solid, and the aqueous phase in chemical 2.50 percent H, 11.6 percent N. What is its equations.  Questions & Problems 109 Problems A • 3.59 Balance the following equations using the method outlined in Section 3.7: B (a) C 1 O2 ¡ CO 8n C (b) CO 1 O2 ¡ CO2 (c) H2 1 Br2 ¡ HBr D (d) K 1 H2O ¡ KOH 1 H2 (e) Mg 1 O2 ¡ MgO 3.64 Which of the following equations best represents the (f) O3 ¡ O2 reaction shown in the diagram? (g) H2O2 ¡ H2O 1 O2 (a) A 1 B ¡ C 1 D (h) N2 1 H2 ¡ NH3 (b) 6A 1 4B ¡ C 1 D (i) Zn 1 AgCl ¡ ZnCl2 1 Ag (c) A 1 2B ¡ 2C 1 D (j) S8 1 O2 ¡ SO2 (d) 3A 1 2B ¡ 2C 1 D (k) NaOH 1 H2SO4 ¡ Na2SO4 1 H2O (e) 3A 1 2B ¡ 4C 1 2D (l) Cl2 1 NaI ¡ NaCl 1 I2 (m) KOH 1 H3PO4 ¡ K3PO4 1 H2O (n) CH4 1 Br2 ¡ CBr4 1 HBr A 3.60 Balance the following equations using the method B outlined in Section 3.7: 8n (a) N2O5 ¡ N2O4 1 O2 C (b) KNO3 ¡ KNO2 1 O2 (c) NH4NO3 ¡ N2O 1 H2O D (d) NH4NO2 ¡ N2 1 H2O (e) NaHCO3 ¡ Na2CO3 1 H2O 1 CO2 • 3.65 Consider the combustion of carbon monoxide (CO) (f) P4O10 1 H2O ¡ H3PO4 in oxygen gas: (g) HCl 1 CaCO3 ¡ CaCl2 1 H2O 1 CO2 2CO(g) 1 O2 (g) ¡ 2CO2 (g) (h) Al 1 H2SO4 ¡ Al2 (SO4 ) 3 1 H2 (i) CO2 1 KOH ¡ K2CO3 1 H2O Starting with 3.60 moles of CO, calculate the num- (j) CH4 1 O2 ¡ CO2 1 H2O ber of moles of CO2 produced if there is enough oxygen gas to react with all of the CO. (k) Be2C 1 H2O ¡ Be(OH) 2 1 CH4 (l) Cu 1 HNO3 ¡ Cu(NO3 ) 2 1 NO 1 H2O • 3.66 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: (m) S 1 HNO3 ¡ H2SO4 1 NO2 1 H2O (n) NH3 1 CuO ¡ Cu 1 N2 1 H2O Si(s) 1 2Cl2 (g) ¡ SiCl4 (l) In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used Amounts of Reactants and Products in the reaction? Review Questions • 3.67 Ammonia is a principal nitrogen fertilizer. It is 3.61 On what law is stoichiometry based? Why is it prepared by the reaction between hydrogen and essential to use balanced equations in solving stoi- nitrogen. chiometric problems? 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) 3.62 Describe the steps involved in the mole method. In a particular reaction, 6.0 moles of NH3 were pro- duced. How many moles of H2 and how many moles Problems of N2 were reacted to produce this amount of NH3? • 3.63 Which of the following equations best represents the • 3.68 Certain race cars use methanol (CH3OH, also called reaction shown in the diagram? wood alcohol) as a fuel. The combustion of metha- (a) 8A 1 4B ¡ C 1 D nol occurs according to the following equation: (b) 4A 1 8B ¡ 4C 1 4D 2CH3OH(l) 1 3O2 (g) ¡ 2CO2 (g) 1 4H2O(l) (c) 2A 1 B ¡ C 1 D In a particular reaction, 9.8 moles of CH3OH are (d) 4A 1 2B ¡ 4C 1 4D reacted with an excess of O2. Calculate the number (e) 2A 1 4B ¡ C 1 D of moles of H2O formed.  110 Chapter 3 ■ Mass Relationships in Chemical Reactions • 3.69 The annual production of sulfur dioxide from burn- • 3.76 Nitrous oxide (N2O) is also called “laughing gas.” It ing coal and fossil fuels, auto exhaust, and other can be prepared by the thermal decomposition of sources is about 26 million tons. The equation for ammonium nitrate (NH4NO3). The other product is the reaction is H2O. (a) Write a balanced equation for this reaction. (b) How many grams of N2O are formed if 0.46 S(s) 1 O2 (g) ¡ SO2 (g) mole of NH4NO3 is used in the reaction? How much sulfur (in tons), present in the original • 3.77 The fertilizer ammonium sulfate [(NH4)2SO4] is pre- materials, would result in that quantity of SO2? pared by the reaction between ammonia (NH3) and • 3.70 When baking soda (sodium bicarbonate or sodium sulfuric acid: hydrogen carbonate, NaHCO3) is heated, it releases 2NH3 (g) 1 H2SO4 (aq) ¡ (NH4 ) 2SO4 (aq) carbon dioxide gas, which is responsible for the ris- ing of cookies, donuts, and bread. (a) Write a bal- How many kilograms of NH3 are needed to produce anced equation for the decomposition of the 1.00 3 105 kg of (NH4)2SO4? compound (one of the products is Na2CO3). (b) Cal- • 3.78 A common laboratory preparation of oxygen gas is culate the mass of NaHCO3 required to produce the thermal decomposition of potassium chlorate 20.5 g of CO2. (KClO3). Assuming complete decomposition, calcu- • 3.71 If chlorine bleach is mixed with other cleaning prod- late the number of grams of O2 gas that can be ucts containing ammonia, the toxic gas NCl3(g) can obtained from 46.0 g of KClO3. (The products are form according to the equation: KCl and O2.) 3NaClO(aq) 1 NH3 (aq) ¡ 3NaOH(aq) 1 NCl3 (g) Limiting Reagents When 2.94 g of NH3 reacts with an excess of NaClO according to the preceding reaction, how many Review Questions grams of NCl3 are formed? 3.79 Define limiting reagent and excess reagent. What is • 3.72 Fermentation is a complex chemical process of wine the significance of the limiting reagent in predicting making in which glucose is converted into ethanol the amount of the product obtained in a reaction? and carbon dioxide: Can there be a limiting reagent if only one reactant is present? C6H12O6 ¡ 2C2H5OH 1 2CO2 3.80 Give an everyday example that illustrates the limit- glucose ethanol ing reagent concept. Starting with 500.4 g of glucose, what is the maxi- mum amount of ethanol in grams and in liters that can Problems be obtained by this process? (Density of ethanol 5 0.789 g/mL.) • 3.81 Consider the reaction • 3.73 Each copper(II) sulfate unit is associated with five 2A 1 B ¡ C water molecules in crystalline copper(II) sulfate pentahydrate (CuSO4 ? 5H2O). When this compound (a) In the diagram here that represents the reaction, is heated in air above 1008C, it loses the water mol- which reactant, A or B, is the limiting reagent? ecules and also its blue color: (b) Assuming complete reaction, draw a molecular- model representation of the amounts of reactants CuSO4 ? 5H2O ¡ CuSO4 1 5H2O and products left after the reaction. The atomic arrangement in C is ABA. If 9.60 g of CuSO4 are left after heating 15.01 g of the blue compound, calculate the number of moles of H2O originally present in the compound. • 3.74 For many years the recovery of gold—that is, the A separation of gold from other materials—involved the use of potassium cyanide: B 4Au 1 8KCN 1 O2 1 2H2O ¡ 4KAu(CN) 2 1 4KOH What is the minimum amount of KCN in moles needed to extract 29.0 g (about an ounce) of gold? • 3.75 Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how 3.82 Consider the reaction many grams of quicklime can be produced from 1.0 kg of limestone. N2 1 3H2 ¡ 2NH3  Questions & Problems 111 Assuming each model represents 1 mole of the Problems substance, show the number of moles of the prod- uct and the excess reagent left after the complete • 3.89 Hydrogen fluoride is used in the manufacture of reaction. Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is pre- pared by the reaction CaF2 1 H2SO4 ¡ CaSO4 1 2HF H2 In one process, 6.00 kg of CaF2 are treated with an excess of H2SO4 and yield 2.86 kg of HF. Calculate N2 the percent yield of HF. • 3.90 Nitroglycerin (C3H5N3O9) is a powerful explosive. Its decomposition may be represented by NH3 4C3H5N3O9 ¡ 6N2 1 12CO2 1 10H2O 1 O2 This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, • 3.83 Nitric oxide (NO) reacts with oxygen gas to form that produces the explosion. (a) What is the maxi- nitrogen dioxide (NO2), a dark-brown gas: mum amount of O2 in grams that can be obtained from 2.00 3 102 g of nitroglycerin? (b) Calculate 2NO(g) 1 O2 (g) ¡ 2NO2 (g) the percent yield in this reaction if the amount of O2 generated is found to be 6.55 g. In one experiment 0.886 mole of NO is mixed with 0.503 mole of O2. Calculate which of the two reac- • 3.91 Titanium(IV) oxide (TiO2) is a white substance pro- tants is the limiting reagent. Calculate also the num- duced by the action of sulfuric acid on the mineral ber of moles of NO2 produced. ilmenite (FeTiO3): • 3.84 Ammonia and sulfuric acid react to form ammo- FeTiO3 1 H2SO4 ¡ TiO2 1 FeSO4 1 H2O nium sulfate. (a) Write an equation for the reac- tion. (b) Determine the starting mass (in g) of Its opaque and nontoxic properties make it suitable each reactant if 20.3 g of ammonium sulfate is as a pigment in plastics and paints. In one process, produced and 5.89 g of sulfuric acid remains 8.00 3 103 kg of FeTiO3 yielded 3.67 3 103 kg of unreacted. TiO2. What is the percent yield of the reaction? • 3.85 Propane (C3H8) is a component of natural gas and is 3.92 Ethylene (C2H4), an important industrial organic used in domestic cooking and heating. (a) Balance chemical, can be prepared by heating hexane (C6H14) the following equation representing the combustion at 8008C: of propane in air: C6H14 ¡ C2H4 1 other products C3H8 1 O2 ¡ CO2 1 H2O If the yield of ethylene production is 42.5 percent, what (b) How many grams of carbon dioxide can be mass of hexane must be reacted to produce 481 g of produced by burning 3.65 moles of propane? ethylene? Assume that oxygen is the excess reagent in this • 3.93 When heated, lithium reacts with nitrogen to form reaction. lithium nitride: • 3.86 Consider the reaction 6Li(s) 1 N2 (g) ¡ 2Li3N(s) MnO2 1 4HCl ¡ MnCl2 1 Cl2 1 2H2O What is the theoretical yield of Li3N in grams when 12.3 g of Li are heated with 33.6 g of N2? If the If 0.86 mole of MnO2 and 48.2 g of HCl react, which actual yield of Li3N is 5.89 g, what is the percent reagent will be used up first? How many grams of yield of the reaction? Cl2 will be produced? 3.94 Disulfide dichloride (S2Cl2) is used in the vulcanization of rubber, a process that prevents the slippage of rubber Reaction Yield molecules past one another when stretched. It is pre- pared by heating sulfur in an atmosphere of chlorine: Review Questions 3.87 Why is the theoretical yield of a reaction determined S8 (l) 1 4Cl2 (g) ¡ 4S2Cl2 (l) only by the amount of the limiting reagent? What is the theoretical yield of S2Cl2 in grams when 3.88 Why is the actual yield of a reaction almost always 4.06 g of S8 are heated with 6.24 g of Cl2? If the actual smaller than the theoretical yield? yield of S2Cl2 is 6.55 g, what is the percent yield?  112 Chapter 3 ■ Mass Relationships in Chemical Reactions Additional Problems 3.99 Ethylene reacts with hydrogen chloride to form 3.95 Gallium is an important element in the production of ethyl chloride: semiconductors. The average atomic mass of 69 31 Ga C2H4 (g) 1 HCl(g) ¡ C2H5Cl(g) (68.9256 amu) and 7131 Ga (70.9247 amu) is 69.72 amu. Calculate the natural abundances of the gallium Calculate the mass of ethyl chloride formed if 4.66 g isotopes. of ethylene reacts with an 89.4 percent yield. 3.96 Rubidium is used in “atomic clocks” and other pre- 3.100 Write balanced equations for the following reactions cise electronic equipment. The average atomic mass described in words. of 85 87 37 Rb (84.912 amu) and 37 Rb (86.909 amu) is (a) Pentane burns in oxygen to form carbon 85.47 amu. Calculate the natural abundances of the dioxide and water. rubidium isotopes. (b) Sodium bicarbonate reacts with hydrochloric • 3.97 The following diagram represents the products acid to form carbon dioxide, sodium chloride, (CO2 and H2O) formed after the combustion of a and water. hydrocarbon (a compound containing only C and H (c) When heated in an atmosphere of nitrogen, atoms). Write an equation for the reaction. (Hint: lithium forms lithium nitride. The molar mass of the hydrocarbon is about 30 g.) (d) Phosphorus trichloride reacts with water to form phosphorus acid and hydrogen chloride. CO2 (e) Copper(II) oxide heated with ammonia will form copper, nitrogen gas, and water. H2O • 3.101 Industrially, nitric acid is produced by the Ostwald process represented by the following equations: 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(l) 2NO(g) 1 O2 (g) ¡ 2NO2 (g) 2NO2 (g) 1 H2O(l) ¡ HNO3 (aq) 1 HNO2 (aq) What mass of NH3 (in g) must be used to produce • 3.98 Consider the reaction of hydrogen gas with oxygen gas: 1.00 ton of HNO3 by the above procedure, assuming 2H2 (g) 1 O2 (g) ¡ 2H2O(g) an 80 percent yield in each step? (1 ton 5 2000 lb; 1 lb 5 453.6 g.) 3.102 A sample of a compound of Cl and O reacts with an H2 excess of H2 to give 0.233 g of HCl and 0.403 g of H2O. Determine the empirical formula of the compound. 3.103 How many grams of H2O will be produced from the O2 complete combustion of 26.7 g of butane (C4H10)? 3.104 A 26.2-g sample of oxalic acid hydrate (H2C2O4 ? H2O 2H2O) is heated in an oven until all the water is driven off. How much of the anhydrous acid is left? • 3.105 The atomic mass of element X is 33.42 amu. A 27.22-g sample of X combines with 84.10 g of another element Y to form a compound XY. Calcu- Assuming complete reaction, which of the diagrams late the atomic mass of Y. shown next represents the amounts of reactants and products left after the reaction? • 3.106 How many moles of O are needed to combine with 0.212 mole of C to form (a) CO and (b) CO2? 3.107 A research chemist used a mass spectrometer to study the two isotopes of an element. Over time, she recorded a number of mass spectra of these isotopes. On analy- sis, she noticed that the ratio of the taller peak (the more abundant isotope) to the shorter peak (the less abundant isotope) gradually increased with time. As- suming that the mass spectrometer was functioning normally, what do you think was causing this change? 3.108 The aluminum sulfate hydrate [Al2(SO4)3 ? xH2O] contains 8.10 percent Al by mass. Calculate x, that is, the number of water molecules associated with (a) (b) (c) (d) each Al2(SO4)3 unit.  Questions & Problems 113 3.109 The explosive nitroglycerin (C3H5N3O9) has also been 3.119 Hemoglobin (C2952H4664N812O832S8Fe4) is the oxygen used as a drug to treat heart patients to relieve pain carrier in blood. (a) Calculate its molar mass. (b) An (angina pectoris). We now know that nitroglycerin pro- average adult has about 5.0 L of blood. Every milliliter duces nitric oxide (NO), which causes muscles to relax of blood has approximately 5.0 3 109 erythrocytes, or and allows the arteries to dilate. If each nitroglycerin red blood cells, and every red blood cell has about molecule releases one NO per atom of N, calculate the 2.8 3 108 hemoglobin molecules. Calculate the mass mass percent of NO available from nitroglycerin. of hemoglobin molecules in grams in an average adult. 3.110 The carat is the unit of mass used by jewelers. One 3.120 Myoglobin stores oxygen for metabolic processes in carat is exactly 200 mg. How many carbon atoms muscle. Chemical analysis shows that it contains are present in a 24-carat diamond? 0.34 percent Fe by mass. What is the molar mass of • 3.111 An iron bar weighed 664 g. After the bar had been myoglobin? (There is one Fe atom per molecule.) standing in moist air for a month, exactly one-eighth • 3.121 Calculate the number of cations and anions in each of the iron turned to rust (Fe2O3). Calculate the final of the following compounds: (a) 0.764 g of CsI, mass of the iron bar and rust. (b) 72.8 g of K2Cr2O7, (c) 6.54 g of Hg2(NO3)2. • 3.112 A certain metal oxide has the formula MO where M 3.122 A mixture of NaBr and Na2SO4 contains 29.96 per- denotes the metal. A 39.46-g sample of the com- cent Na by mass. Calculate the percent by mass of pound is strongly heated in an atmosphere of hydro- each compound in the mixture. gen to remove oxygen as water molecules. At the 3.123 Consider the reaction 3A 1 2B S 3C. A student end, 31.70 g of the metal is left over. If O has an mixed 4.0 moles of A with 4.0 moles of B and atomic mass of 16.00 amu, calculate the atomic obtained 2.8 moles of C. What is the percent yield of mass of M and identify the element. the reaction? • 3.113 An impure sample of zinc (Zn) is treated with an 3.124 Balance the following equation shown in molecular excess of sulfuric acid (H2SO4) to form zinc sulfate models. (ZnSO4) and molecular hydrogen (H2). (a) Write a balanced equation for the reaction. (b) If 0.0764 g of H2 is obtained from 3.86 g of the sample, calculate the percent purity of the sample. (c) What assump- 1 8n 1 tions must you make in (b)? • 3.114 One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is • 3.125 Aspirin or acetyl salicylic acid is synthesized by Fe2O3 1 3CO ¡ 2Fe 1 3CO2 reacting salicylic acid with acetic anhydride: Suppose that 1.64 3 103 kg of Fe are obtained from C7H6O3   1   C4H6O3    ¡ C9H8O4 1 C2H4O2 a 2.62 3 103-kg sample of Fe2O3. Assuming that the salicylic acid acetic anhydride aspirin acetic acid reaction goes to completion, what is the percent pu- rity of Fe2O3 in the original sample? (a) How much salicylic acid is required to produce • 3.115 Carbon dioxide (CO2) is the gas that is mainly 0.400 g of aspirin (about the content in a tablet), as- responsible for global warming (the greenhouse suming acetic anhydride is present in excess? effect). The burning of fossil fuels is a major cause of (b) Calculate the amount of salicylic acid needed if the increased concentration of CO2 in the atmosphere. only 74.9 percent of salicylic acid is converted to Carbon dioxide is also the end product of metabolism aspirin. (c) In one experiment, 9.26 g of salicylic (see Example 3.13). Using glucose as an example acid is reacted with 8.54 g of acetic anhydride. Cal- of  food, calculate the annual human production of culate the theoretical yield of aspirin and the percent CO2 in grams, assuming that each person consumes yield if only 10.9 g of aspirin is produced. 5.0 3 102 g of glucose per day. The world’s popula- • 3.126 Calculate the percent composition by mass of all the tion is 7.2 billion, and there are 365 days in a year. elements in calcium phosphate [Ca3(PO4)2], a major 3.116 Carbohydrates are compounds containing carbon, component of bone. hydrogen, and oxygen in which the hydrogen to 3.127 Lysine, an essential amino acid in the human body, oxygen ratio is 2:1. A certain carbohydrate contains contains C, H, O, and N. In one experiment, the 40.0 percent carbon by mass. Calculate the empiri- complete combustion of 2.175 g of lysine gave 3.94 g cal and molecular formulas of the compound if the CO2 and 1.89 g H2O. In a separate experiment, approximate molar mass is 178 g. 1.873 g of lysine gave 0.436 g NH3. (a) Calculate the 3.117 Which of the following has the greater mass: 0.72 g of empirical formula of lysine. (b) The approximate O2 or 0.0011 mole of chlorophyll (C55H72MgN4O5)? molar mass of lysine is 150 g. What is the molecular • 3.118 Analysis of a metal chloride XCl3 shows that it con- formula of the compound? tains 67.2 percent Cl by mass. Calculate the molar 3.128 Does 1 g of hydrogen molecules contain as many mass of X and identify the element. H atoms as 1 g of hydrogen atoms?  114 Chapter 3 ■ Mass Relationships in Chemical Reactions 3.129 Avogadro’s number has sometimes been described because of their role in systems that convert solar as a conversion factor between amu and grams. Use energy to electricity. The compound [Ru(C10H8N2)3] the fluorine atom (19.00 amu) as an example to Cl2 is synthesized by reacting RuCl3 ? 3H2O(s) show the relation between the atomic mass unit and with three molar equivalents of C10H8N2(s), along the gram. with an excess of triethylamine, N(C2H5)3(l), to 3.130 The natural abundances of the two stable isotopes of convert ruthenium(III) to ruthenium(II). The den- hydrogen (hydrogen and deuterium) are 11H: 99.985 sity of triethylamine is 0.73 g/mL, and typically percent and 21H: 0.015 percent. Assume that water eight molar equivalents are used in the synthesis. exists as either H2O or D2O. Calculate the number (a) Assuming that you start with 6.5 g of RuCl3 ? of D2O molecules in exactly 400 mL of water. 3H2O, how many grams of C10H8N2 and what vol- (Density 5 1.00 g/mL.) ume of N(C2H5)3 should be used in the reaction? 3.131 A compound containing only C, H, and Cl was ex- (b) Given that the yield of this reaction is 91 per- amined in a mass spectrometer. The highest mass cent, how many grams of [Ru(C10H8N2)3]Cl2 will peak seen corresponds to an ion mass of 52 amu. be obtained? The most abundant mass peak seen corresponds to • 3.139 Heating 2.40 g of the oxide of metal X (molar an ion mass of 50 amu and is about three times as mass of X 5 55.9 g/mol) in carbon monoxide intense as the peak at 52 amu. Deduce a reasonable (CO) yields the pure metal and carbon dioxide. molecular formula for the compound and explain The mass of the metal product is 1.68 g. From the the positions and intensities of the mass peaks data given, show that the simplest formula of the mentioned. (Hint: Chlorine is the only element that oxide is X2O3 and write a balanced equation for has isotopes in comparable abundances: 35 17Cl: 75.5 the reaction. percent; 35 1 17Cl: 24.5 percent. For H, use 1H; for C, 3.140 A compound X contains 63.3 percent manganese use 121C.) (Mn) and 36.7 percent O by mass. When X is heated, 3.132 In the formation of carbon monoxide, CO, it is oxygen gas is evolved and a new compound Y con- found that 2.445 g of carbon combine with 3.257 g taining 72.0 percent Mn and 28.0 percent O is of oxygen. What is the atomic mass of oxygen if the formed. (a) Determine the empirical formulas of X atomic mass of carbon is 12.01 amu? and Y. (b) Write a balanced equation for the conver- 3.133 What mole ratio of molecular chlorine (Cl2) to sion of X to Y. molecular oxygen (O2) would result from the • 3.141 The formula of a hydrate of barium chloride is breakup of the compound Cl2O7 into its constituent BaCl2 ? xH2O. If 1.936 g of the compound gives elements? 1.864 g of anhydrous BaSO4 upon treatment with sulfuric acid, calculate the value of x. 3.134 Which of the following substances contains the greatest mass of chlorine? (a) 5.0 g Cl2, (b) 60.0 g 3.142 It is estimated that the day Mt. St. Helens erupted NaClO3, (c) 0.10 mol KCl, (d) 30.0 g MgCl2, (May 18, 1980), about 4.0 3 105 tons of SO2 were (e) 0.50 mol Cl2. released into the atmosphere. If all the SO2 were 3.135 A compound made up of C, H, and Cl contains 55.0 eventually converted to sulfuric acid, how many tons percent Cl by mass. If 9.00 g of the compound con- of H2SO4 were produced? tain 4.19 3 1023 H atoms, what is the empirical for- 3.143 Cysteine, shown here, is one of the 20 amino acids mula of the compound? found in proteins in humans. Write the molecular formula and calculate its percent composition • 3.136 Platinum forms two different compounds with by mass. chlorine. One contains 26.7 percent Cl by mass, and the other contains 42.1 percent Cl by mass. Determine the empirical formulas of the two compounds. 3.137 The following reaction is stoichiometric as written H C4H9Cl 1 NaOC2H5 ¡ C4H8 1 C2H5OH 1 NaCl but it is often carried out with an excess of S NaOC2H5 to react with any water present in the reaction mixture that might reduce the yield. If the O reaction shown was carried out with 6.83 g of C4H9Cl, how many grams of NaOC2H5 would be C needed to have a 50 percent molar excess of that reactant? 3.138 Compounds containing ruthenium(II) and bipyr- idine, C10H8N2, have received considerable interest  Questions & Problems 115 • 3.144 Isoflurane, shown here, is a common inhalation 1.74 g/cm3 and the volume of a sphere of radius r anesthetic. Write its molecular formula and calcu- is 43πr3.) late its percent composition by mass. • 3.151 A certain sample of coal contains 1.6 percent sul- fur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollu- Cl tion, this sulfur dioxide is treated with calcium oxide (CaO) to form calcium sulfite (CaSO3). Calculate the daily mass (in kilograms) of CaO C O needed by a power plant that uses 6.60 3 106 kg of coal per day. • 3.152 Air is a mixture of many gases. However, in calcu- lating its “molar mass” we need consider only the H three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is F made up of 78.08 percent nitrogen, 20.95 percent oxygen, and 0.97 percent argon, what is the molar mass of air? • 3.145 A mixture of CuSO4 ? 5H2O and MgSO4 ? 7H2O is heated until all the water is lost. If 5.020 g of the 3.153 (a) Determine the mass of calcium metal that con- mixture gives 2.988 g of the anhydrous salts, what tains the same number of moles as 89.6 g of zinc is the percent by mass of CuSO4 ? 5H2O in the metal. (b) Calculate the number of moles of molecu- mixture? lar fluorine that has the same mass as 36.9 moles of argon. (c) What is the mass of sulfuric acid that con- • 3.146 When 0.273 g of Mg is heated strongly in a nitrogen tains 0.56 mole of oxygen atoms? (d) Determine the (N2) atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Calculate number of moles of phosphoric acid that contains the empirical formula of the compound containing 2.12 g of hydrogen atoms. Mg and N. Name the compound. 3.154 A major industrial use of hydrochloric acid is in metal pickling. This process involves the removal • 3.147 A mixture of methane (CH4) and ethane (C2H6) of of metal oxide layers from metal surfaces to pre- mass 13.43 g is completely burned in oxygen. If the total mass of CO2 and H2O produced is 64.84 g, pare them for coating. (a) Write an equation calculate the fraction of CH4 in the mixture. between iron(III) oxide, which represents the rust layer over iron, and HCl to form iron(III) chloride • 3.148 Leaded gasoline contains an additive to prevent and water. (b) If 1.22 moles of Fe2O3 and 289.2 g engine “knocking.” On analysis, the additive com- of HCl react, how many grams of FeCl3 will be pound is found to contain carbon, hydrogen, and produced? lead (Pb) (hence, “leaded gasoline”). When 51.36 g of this compound are burned in an apparatus such as 3.155 Octane (C8H18) is a component of gasoline. Com- that shown in Figure 3.6, 55.90 g of CO2 and 28.61 g plete combustion of octane yields H2O and CO2. In- of H2O are produced. Determine the empirical for- complete combustion produces H2O and CO, which mula of the gasoline additive. not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gal 3.149 Because of its detrimental effect on the environment, of octane is burned in an engine. The total mass of the lead compound described in Problem 3.148 has CO, CO2, and H2O produced is 11.53 kg. Calculate been replaced by methyl tert-butyl ether (a compound the efficiency of the process; that is, calculate the of C, H, and O) to enhance the performance of gaso- fraction of octane converted to CO2. The density of line. (This compound is also being phased out because octane is 2.650 kg/gal. of its contamination of drinking water.) When 12.1 g of the compound are burned in an apparatus like the • 3.156 Industrially, hydrogen gas can be prepared by re- one shown in Figure 3.6, 30.2 g of CO2 and 14.8 g of acting propane gas (C3H8) with steam at about H2O are formed. What is the empirical formula of the 4008C. The products are carbon monoxide (CO) compound? and hydrogen gas (H2). (a) Write a balanced equa- tion for the reaction. (b) How many kilograms • 3.150 Suppose you are given a cube made of magnesium of H2 can be obtained from 2.84 3 103 kg of (Mg) metal of edge length 1.0 cm. (a) Calculate propane? the number of Mg atoms in the cube. (b) Atoms are spherical in shape. Therefore, the Mg atoms in • 3.157 In a natural product synthesis, a chemist prepares a the cube cannot fill all of the available space. If complex biological molecule entirely from nonbio- only 74  percent of the space inside the cube is logical starting materials. The target molecules are taken up by Mg atoms, calculate the radius in pi- often known to have some promise as therapeutic cometers of a Mg atom. (The density of Mg is agents, and the organic reactions that are developed  116 Chapter 3 ■ Mass Relationships in Chemical Reactions along the way benefit all chemists. The overall syn- the United States goes into fertilizer. The major thesis, however, requires many steps, so it is impor- sources of potash are potassium chloride (KCl) and tant to have the best possible percent yields at each potassium sulfate (K2SO4). Potash production is step. What is the overall percent yield for such a often reported as the potassium oxide (K2O) equiv- synthesis that has 24 steps with an 80 percent yield alent or the amount of K2O that could be made at each step? from a given mineral. (a) If KCl costs $0.55 per kg, 3.158 What is wrong or ambiguous with each of the state- for what price (dollar per kg) must K2SO4 be sold ments here? to supply the same amount of potassium on a per (a) NH4NO2 is the limiting reagent in the reaction dollar basis? (b) What mass (in kg) of K2O contains the same number of moles of K atoms as 1.00 kg NH4NO2 (s) ¡ N2 (g) 1 2H2O(l) of KCl? (b) The limiting reagents for the reaction shown • 3.161 A 21.496-g sample of magnesium is burned in air to here are NH3 and NaCl. form magnesium oxide and magnesium nitride. When the products are treated with water, 2.813 g of NH3 (aq) 1 NaCl(aq) 1 H2CO3 (aq) ¡ gaseous ammonia are generated. Calculate the NaHCO3 (aq) 1 NH4Cl(aq) amounts of magnesium nitride and magnesium 3.159 (a) For molecules having small molecular masses, oxide formed. mass spectrometry can be used to identify their for- • 3.162 A certain metal M forms a bromide containing 53.79 mulas. To illustrate this point, identify the mole- percent Br by mass. What is the chemical formula of cule that most likely accounts for the observation the compound? of a peak in a mass spectrum at: 16 amu, 17 amu, 3.163 A sample of iron weighing 15.0 g was heated with 18 amu, and 64 amu. (b) Note that there are (among potassium chlorate (KClO3) in an evacuated con- others) two likely molecules that would give rise to tainer. The oxygen generated from the decomposi- a peak at 44 amu, namely, C3H8 and CO2. In such tion of KClO3 converted some of the Fe to Fe2O3. If cases, a chemist might try to look for other peaks the combined mass of Fe and Fe2O3 was 17.9 g, cal- generated when some of the molecules break apart culate the mass of Fe2O3 formed and the mass of in the spectrometer. For example, if a chemist sees KClO3 decomposed. a peak at 44 amu and also one at 15 amu, which • 3.164 A sample containing NaCl, Na2SO4, and NaNO3 molecule is producing the 44-amu peak? Why? gives the following elemental analysis: Na: 32.08 per- (c) Using the following precise atomic masses— cent; O: 36.01 percent; Cl: 19.51 percent. Calculate 1 H (1.00797 amu), 12C (12.00000 amu), and 16O the mass percent of each compound in the sample. (15.99491 amu)—how precisely must the masses 3.165 A sample of 10.00 g of sodium reacts with oxygen of C3H8 and CO2 be measured to distinguish to form 13.83 g of sodium oxide (Na2O) and sodium between them? peroxide (Na2O2). Calculate the percent composi- • 3.160 Potash is any potassium mineral that is used for its tion of the mixture. potassium content. Most of the potash produced in Interpreting, Modeling & Estimating 3.166 While most isotopes of light elements such as oxy- 3.167 Without doing any detailed calculations, arrange the gen and phosphorus contain relatively equal num- following substances in the increasing order of num- bers of protons and neutrons, recent results indicate ber of moles: 20.0 g Cl, 35.0 g Br, and 94.0 g I. that a new class of isotopes called neutron-rich iso- 3.168 Without doing any detailed calculations, estimate topes can be prepared. These neutron-rich isotopes which element has the highest percent composition push the limits of nuclear stability as the large num- by mass in each of the following compounds: ber of neutrons approach the “neutron drip line.” (a) Hg(NO3)2 They may play a critical role in the nuclear reactions (b) NF3 of stars. An unusually heavy isotope of aluminum (43 13Al) has been reported. How many more neutrons (c) K2Cr2O7 does this atom contain compared to an average (d) C2952H4664N812O832S8Fe4 aluminum atom?  Answers to Practice Exercises 117 3.169 Consider the reaction number using stearic acid (C18H36O2) shown here. When stearic acid is added to water, its molecules 6Li(s) 1 N2 (g) ¡ 2Li3N(s) collect at the surface and form a monolayer; that is, Without doing any detailed calculations, choose one the layer is only one molecule thick. The cross- of the following combinations in which nitrogen is sectional area of each stearic acid molecule has been the limiting reagent: measured to be 0.21 nm2. In one experiment it is (a) 44 g Li and 38 g N2 found that 1.4 3 1024 g of stearic acid is needed to form a monolayer over water in a dish of diameter (b) 1380 g Li and 842 g N2 20 cm. Based on these measurements, what is (c) 1.1 g Li and 0.81 g N2 Avogadro’s number? 3.170 Estimate how high in miles you can stack up an Avogadro’s number of oranges covering the entire H3C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 OH Earth. CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C 3.171 The following is a crude but effective method for estimating the order of magnitude of Avogadro’s O Answers to Practice Exercises 3.1 63.55 amu. 3.2 3.59 moles. 3.3 2.57 3 103 g. 3.10 196 g. 3.11 B2H6. 3.12 Fe2O3 1 3CO S 2Fe 1 3CO2. 3.4 1.0 3 10220 g. 3.5 32.04 amu. 3.6 1.66 moles. 3.13 235 g. 3.14 0.769 g. 3.15 (a) 234 g, (b) 234 g. 3.7 5.81 3 1024 H atoms. 3.8 H: 2.055%; S: 32.69%; 3.16 25.1 g. 3.17 (a) 863 g, (b) 93.0%. O: 65.25%. 3.9 KMnO4 (potassium permanganate).  CHAPTER 4 Reactions in Aqueous Solutions Black smokers form when superheated water, rich in minerals, flows out onto the ocean floor through the lava from an ocean volcano. The hydrogen sulfide present converts the metal ions to insoluble metal sulfides. CHAPTER OUTLINE A LOOK AHEAD 4.1 General Properties of  We begin by studying the properties of solutions prepared by dissolving Aqueous Solutions substances in water, called aqueous solutions. Aqueous solutions can be classified as nonelectrolyte or electrolyte, depending on their ability to con- 4.2 Precipitation Reactions duct electricity. (4.1) 4.3 Acid-Base Reactions  We will see that precipitation reactions are those in which the product is an 4.4 Oxidation-Reduction insoluble compound. We learn to represent these reactions using ionic equa- Reactions tions and net ionic equations. (4.2)  Next, we learn acid-base reactions, which involve the transfer of proton 4.5 Concentration of Solutions (H1) from an acid to a base. (4.3) 4.6 Gravimetric Analysis  We then learn oxidation-reduction (redox) reactions in which electrons are 4.7 Acid-Base Titrations transferred between reactants. We will see that there are several types of redox reactions. (4.4) 4.8 Redox Titrations  To carry out quantitative studies of solutions, we learn how to express the concentration of a solution in molarity. (4.5)  Finally, we will apply our knowledge of the mole method from Chapter 3 to the three types of reactions studied here. We will see how gravimetric analy- sis is used to study precipitation reactions, and the titration technique is used to study acid-base and redox reactions. (4.6, 4.7, and 4.8) 118  4.1 General Properties of Aqueous Solutions 119 M any chemical reactions and virtually all biological processes take place in water. In this chapter, we will discuss three major categories of reactions that occur in aqueous solu- tions: precipitation reactions, acid-base reactions, and redox reactions. In later chapters, we will study the structural characteristics and properties of water—the so-called universal solvent— and its solutions. 4.1 General Properties of Aqueous Solutions A solution is a homogeneous mixture of two or more substances. The solute is the substance present in a smaller amount, and the solvent is the substance present in a larger amount. A solution may be gaseous (such as air), solid (such as an alloy), or liquid (seawater, for example). In this section we will discuss only aqueous solutions, in which the solute initially is a liquid or a solid and the solvent is water. Electrolytic Properties All solutes that dissolve in water fit into one of two categories: electrolytes and nonelectrolytes. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte does not conduct electric- ity when dissolved in water. Figure 4.1 shows an easy and straightforward method of distinguishing between electrolytes and nonelectrolytes. A pair of inert electrodes (copper or platinum) is immersed in a beaker of water. To light the bulb, electric current must flow from one electrode to the other, thus completing the circuit. Pure water is a very poor conductor of electricity. However, if we add a small amount of sodium chloride (NaCl), the bulb will glow as soon as the salt dissolves in the water. Solid NaCl, an ionic compound, breaks up into Na1 and Cl2 ions when it dissolves in water. The Na1 ions are attracted to the negative electrode, and the Cl2 ions to the positive electrode. This movement sets up an electric current that is equivalent to the flow of electrons along a metal wire. Because the NaCl solution conducts Tap water does conduct electricity because it contains many dissolved ions. electricity, we say that NaCl is an electrolyte. Pure water contains very few ions, so it cannot conduct electricity. Comparing the lightbulb’s brightness for the same molar amounts of dissolved Animation Strong Electrolytes, Weak Electrolytes, substances helps us distinguish between strong and weak electrolytes. A character- and Nonelectrolytes istic of strong electrolytes is that the solute is assumed to be 100 percent dissoci- ated into ions in solution. (By dissociation we mean the breaking up of the Figure 4.1 An arrangement for distinguishing between electrolytes and nonelectrolytes. A solution's ability to conduct electricity depends on the number of ions it contains. (a) A nonelectrolyte solution does not contain ions, and the lightbulb is not lit. (b) A weak electrolyte solution contains a small number of ions, and the lightbulb is dimly lit. (c) A strong electrolyte solution contains a large number of ions, and the lightbulb is brightly lit. The molar amounts of the dissolved solutes are equal in all three cases. (a) (b) (c) 120 Chapter 4 ■ Reactions in Aqueous Solutions Table 4.1 Classification of Solutes in Aqueous Solution Strong Electrolyte Weak Electrolyte Nonelectrolyte HCl CH3COOH (NH2)2CO (urea) HNO3 HF CH3OH (methanol) HClO4 HNO2 C2H5OH (ethanol) H2SO4* NH3 C6H12O6 (glucose) NaOH H2O† C12H22O11 (sucrose) Ba(OH)2 Ionic compounds *H2SO4 has two ionizable H1 ions, but only one of the H1 ions is totally ionized. † Pure water is an extremely weak electrolyte. compound into cations and anions.) Thus, we can represent sodium chloride dis- solving in water as H2O NaCl(s) ¡ Na1 (aq) 1 Cl2 (aq) This equation says that all sodium chloride that enters the solution ends up as Na1 and Cl2 ions; there are no undissociated NaCl units in solution. Table 4.1 lists examples of strong electrolytes, weak electrolytes, and nonelectro- lytes. Ionic compounds, such as sodium chloride, potassium iodide (KI), and calcium nitrate [Ca(NO3)2], are strong electrolytes. It is interesting to note that human body fluids contain many strong and weak electrolytes. Water is a very effective solvent for ionic compounds. Although water is an electrically neutral molecule, it has a positive region (the H atoms) and a negative region (the O atom), or positive and negative “poles”; for this reason it is a polar solvent. When an ionic compound such as sodium chloride dissolves in water, the three-dimensional network of ions in the solid is destroyed. The Na1 and Cl2 ions Animation are separated from each other and undergo hydration, the process in which an ion is Hydration surrounded by water molecules arranged in a specific manner. Each Na1 ion is sur- rounded by a number of water molecules orienting their negative poles toward the cation. Similarly, each Cl2 ion is surrounded by water molecules with their positive poles oriented toward the anion (Figure 4.2). Hydration helps to stabilize ions in solu- tion and prevents cations from combining with anions. Acids and bases are also electrolytes. Some acids, including hydrochloric acid (HCl) and nitric acid (HNO3), are strong electrolytes. These acids are assumed to ionize completely in water; for example, when hydrogen chloride gas dissolves in water, it forms hydrated H1 and Cl2 ions: H2O HCl(g) ¡ H1 (aq) 1 Cl2 (aq) In other words, all the dissolved HCl molecules separate into hydrated H1 and Cl2 ions. Thus, when we write HCl(aq), it is understood that it is a solution of only H1(aq) Figure 4.2 Hydration of Na1 and Cl2 ions. 1 2  4.2 Precipitation Reactions 121 and Cl2(aq) ions and that there are no hydrated HCl molecules present. On the other hand, certain acids, such as acetic acid (CH3COOH), which gives vinegar its tart flavor, do not ionize completely and are weak electrolytes. We represent the ionization of acetic acid as CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq) where CH3COO2 is called the acetate ion. We use the term ionization to describe the separation of acids and bases into ions. By writing the formula of acetic acid as CH3COOH, we indicate that the ionizable proton is in the COOH group. CH3COOH The ionization of acetic acid is written with a double arrow to show that it is a reversible reaction; that is, the reaction can occur in both directions. Initially, a number of CH3COOH molecules break up into CH3COO2 and H1 ions. As time goes on, some of the CH3COO2 and H1 ions recombine into CH3COOH mole- cules. Eventually, a state is reached in which the acid molecules ionize as fast as the ions recombine. Such a chemical state, in which no net change can be observed (although activity is continuous on the molecular level), is called chemical equi- There are different types of chemical equilibrium. We will return to this very librium. Acetic acid, then, is a weak electrolyte because its ionization in water is important topic in Chapter 14. incomplete. By contrast, in a hydrochloric acid solution the H1 and Cl2 ions have no tendency to recombine and form molecular HCl. We use a single arrow to represent complete ionizations. Review of Concepts The diagrams here show three compounds AB2 (a), AC2 (b), and AD2 (c) dissolved in water. Which is the strongest electrolyte and which is the weakest? (For simplicity, water molecules are not shown.) (a) (b) (c) 4.2 Precipitation Reactions One common type of reaction that occurs in aqueous solution is the precipitation Animation Precipitation Reactions reaction, which results in the formation of an insoluble product, or precipitate. A precipitate is an insoluble solid that separates from the solution. Precipitation reac- tions usually involve ionic compounds. For example, when an aqueous solution of lead(II) nitrate [Pb(NO3)2] is added to an aqueous solution of potassium iodide (KI), a yellow precipitate of lead(II) iodide (PbI2) is formed: Pb(NO3 ) 2 (aq) 1 2KI(aq) ¡ PbI2 (s) 1 2KNO3 (aq) Potassium nitrate remains in solution. Figure 4.3 shows this reaction in progress. The preceding reaction is an example of a metathesis reaction (also called a double-displacement reaction), a reaction that involves the exchange of parts between the two compounds. (In this case, the cations in the two compounds exchange anions, so Pb21 ends up with I2 as PbI2 and K1 ends up with NO2 3 as 122 Chapter 4 ■ Reactions in Aqueous Solutions K1 Pb21 K1 NO2 3 88n I2 Pb21 I2 NO2 3 Figure 4.3 Formation of yellow PbI2 precipitate as a solution of Pb(NO3)2 is added to a solution of KI. KNO3.) As we will see, the precipitation reactions discussed in this chapter are examples of metathesis reactions. Solubility How can we predict whether a precipitate will form when a compound is added to a solution or when two solutions are mixed? It depends on the solubility of the solute, which is defined as the maximum amount of solute that will dissolve in a given quan- tity of solvent at a specific temperature. Chemists refer to substances as soluble, slightly soluble, or insoluble in a qualitative sense. A substance is said to be soluble if a fair amount of it visibly dissolves when added to water. If not, the substance is described as slightly soluble or insoluble. All ionic compounds are strong electrolytes, but they are not equally soluble. Table 4.2 classifies a number of common ionic compounds as soluble or insolu- ble. Keep in mind, however, that even insoluble compounds dissolve to a certain extent. Figure 4.4 shows several precipitates. Table 4.2 Solubility Rules for Common Ionic Compounds in Water at 258C Soluble Compounds Insoluble Exceptions Compounds containing alkali metal ions (Li1, Na1, K1, Rb1, Cs1) and the ammonium ion (NH14 ) Nitrates (NO23 ), acetates (CH3COO2), bicarbonates (HCO23 ), chlorates (ClO23 ), and perchlorates (ClO24 ) Halides (Cl2, Br2, I2) Halides of Ag1, Hg221, and Pb21 22 Sulfates (SO4 ) Sulfates of Ag1, Ca21, Sr21, Ba21, Hg221, and Pb21 Insoluble Compounds Soluble Exceptions Carbonates (CO322), phosphates Compounds containing alkali metal ions (PO432), chromates (CrO422), and the ammonium ion sulfides (S22) Hydroxides (OH2) Compounds containing alkali metal ions and the Ba21 ion  4.2 Precipitation Reactions 123 Figure 4.4 Appearance of several precipitates. From left to right: CdS, PbS, Ni(OH)2, and Al(OH)3. Example 4.1 applies the solubility rules in Table 4.2. Example 4.1 Classify the following ionic compounds as soluble or insoluble: (a) silver sulfate (Ag2SO4), (b) calcium carbonate (CaCO3), (c) sodium phosphate (Na3PO4). Strategy Although it is not necessary to memorize the solubilities of compounds, you should keep in mind the following useful rules: All ionic compounds containing alkali metal cations; the ammonium ion; and the nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, we need to refer to Table 4.2. Solution (a) According to Table 4.2, Ag2SO4 is insoluble. (b) This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO3 is insoluble. (c) Sodium is an alkali metal (Group 1A) so Na3PO4 is soluble. Similar problems: 4.19, 4.20. Practice Exercise Classify the following ionic compounds as soluble or insoluble: (a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2. Molecular Equations, Ionic Equations, and Net Ionic Equations The equation describing the precipitation of lead(II) iodide on page 121 is called a molecular equation because the formulas of the compounds are written as though all species existed as molecules or whole units. A molecular equation is useful because it identifies the reagents [that is, lead(II) nitrate and potassium iodide]. If we wanted to bring about this reaction in the laboratory, we would use the molecular equation. However, a molecular equation does not describe in detail what actually is happening in solution. As pointed out earlier, when ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. Therefore, returning to the reaction between potassium iodide and lead(II) nitrate, we would write Pb21 (aq) 1 2NO23 (aq) 1 2K1 (aq) 1 2I2 (aq) ¡ PbI2 (s) 1 2K 1 (aq) 1 2NO23 (aq) 124 Chapter 4 ■ Reactions in Aqueous Solutions The preceding equation is an example of an ionic equation, which shows dis- solved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds; that is, PbI2 and KNO3. Referring to Table 4.2, we see that PbI2 is an insoluble compound and KNO3 is soluble. Therefore, the dissolved KNO3 remains in solution as separate K1 and NO32 ions, which are called spectator ions, or ions that are not involved in the over- all reaction. Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation Pb21 (aq) 1 2NO23 (aq) 1 2K1 (aq) 1 2I2 (aq) ¡ PbI2 (s) 1 2K1 (aq) 1 2NO23 (aq) Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction: Figure 4.5 Formation of BaSO4 precipitate. Pb21 (aq) 1 2I2 (aq) ¡ PbI2 (s) Looking at another example, we find that when an aqueous solution of barium chloride (BaCl2) is added to an aqueous solution of sodium sulfate (Na2SO4), a white precipitate is formed (Figure 4.5). Treating this as a metathesis reaction, the products are BaSO4 and NaCl. From Table 4.2 we see that only BaSO4 is insoluble. Therefore, we write the molecular equation as BaCl2 (aq) 1 Na2SO4 (aq) ¡ BaSO4 (s) 1 2NaCl(aq) The ionic equation for the reaction is Ba21 (aq) 1 2Cl2 (aq) 1 2Na1 (aq) 1 SO22 4 (aq) ¡ BaSO4 (s) 1 2Na1 (aq) 1 2Cl2 (aq) Canceling the spectator ions (Na1 and Cl2) on both sides of the equation gives us the net ionic equation Ba21 (aq) 1 SO22 4 (aq) ¡ BaSO4 (s) The following four steps summarize the procedure for writing ionic and net ionic equations: 1. Write a balanced molecular equation for the reaction, using the correct formulas for the reactant and product ionic compounds. Refer to Table 4.2 to decide which of the products is insoluble and therefore will appear as a precipitate. 2. Write the ionic equation for the reaction. The compound that does not appear as the precipitate should be shown as free ions. 3. Identify and cancel the spectator ions on both sides of the equation. Write the net ionic equation for the reaction. 4. Check that the charges and number of atoms balance in the net ionic equation. These steps are applied in Example 4.2. Example 4.2 Predict what happens when a potassium phosphate (K3PO4) solution is mixed with a calcium nitrate [Ca(NO3)2] solution. Write a net ionic equation for the reaction. Precipitate formed by the reaction between K3PO4 (aq) and (Continued) Ca(NO3)2(aq).  4.2 Precipitation Reactions 125 Strategy From the given information, it is useful to first write the unbalanced equation K3PO4 (aq) 1 Ca(NO3 ) 2 (aq) ¡ ? What happens when ionic compounds dissolve in water? What ions are formed from the dissociation of K3PO4 and Ca(NO3)2? What happens when the cations encounter the anions in solution? Solution In solution, K3PO4 dissociates into K1 and PO432 ions and Ca(NO3)2 dissociates into Ca21 and NO23 ions. According to Table 4.2, calcium ions (Ca21) and phosphate ions (PO432) will form an insoluble compound, calcium phosphate [Ca3(PO4)2], while the other product, KNO3, is soluble and remains in solution. Therefore, this is a precipitation reaction. We follow the stepwise procedure just outlined. Step 1: The balanced molecular equation for this reaction is 2K3PO4 (aq) 1 3Ca(NO3 ) 2 (aq) ¡ Ca3 (PO4 ) 2 (s) 1 6KNO3 (aq) Step 2: To write the ionic equation, the soluble compounds are shown as dissociated ions: 6K1 (aq) 1 2PO32 21 2 4 (aq) 1 3Ca (aq) 1 6NO3 (aq) ¡ 6K (aq) 1 6NO23 (aq) 1 Ca3 (PO4 ) 2 (s) 1 Step 3: Canceling the spectator ions (K1 and NO32) on each side of the equation, we obtain the net ionic equation: 3Ca21 (aq) 1 2PO32 4 (aq) ¡ Ca3 (PO4 ) 2 (s) Step 4: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to the number of atoms on each side and the number of positive (16) and negative (26) charges on the left-hand side is the same. Similar problems: 4.21, 4.22. Practice Exercise Predict the precipitate produced by mixing an Al(NO3)3 solution with a NaOH solution. Write the net ionic equation for the reaction. Review of Concepts Which of the diagrams here accurately describes the reaction between Ca(NO3)2(aq) and Na2CO3(aq)? For simplicity, only the Ca21 (yellow) and CO322 (blue) ions are shown. (a) (b) (c) The Chemistry in Action essay on p. 126 discusses some practical problems associated with precipitation reactions. CHEMISTRY in Action An Undesirable Precipitation Reaction L imestone (CaCO3) and dolomite (CaCO3 ? MgCO3), which are widespread on Earth’s surface, often enter the water sup- ply. According to Table 4.2, calcium carbonate is insoluble in water. However, in the presence of dissolved carbon dioxide (from the atmosphere), calcium carbonate is converted to solu- ble calcium bicarbonate [Ca(HCO3)2]: FPO CaCO3 (s) 1 CO2 (aq) 1 H2O(l) ¡ Ca21 (aq) 1 2HCO23 (aq) where HCO2 3 is the bicarbonate ion. Water containing Ca21 and/or Mg21 ions is called hard wa- ter, and water that is mostly free of these ions is called soft water. Hard water is unsuitable for some household and industrial uses. When water containing Ca21 and HCO2 3 ions is heated or boiled, the solution reaction is reversed to produce the CaCO3 precipitate Boiler scale almost fills this hot-water pipe. The deposits consist mostly of Ca21 (aq) 1 2HCO23 (aq) ¡ CaCO3 with some MgCO3. CaCO3 (s) 1 CO2 (aq) 1 H2O(l) and gaseous carbon dioxide is driven off: totally block the flow of water. A simple method used by plumbers to remove scale deposits is to introduce a small CO2 (aq) ¡ CO2 (g) amount of hydrochloric acid, which reacts with (and therefore dissolves) CaCO3: Solid calcium carbonate formed in this way is the main compo- nent of the scale that accumulates in boilers, water heaters, CaCO3 (s) 1 2HCl(aq) ¡ pipes, and teakettles. A thick layer of scale reduces heat trans- CaCl2 (aq) 1 H2O(l) 1 CO2 (g) fer and decreases the efficiency and durability of boilers, pipes, and appliances. In household hot-water pipes it can restrict or In this way, CaCO3 is converted to soluble CaCl2. 4.3 Acid-Base Reactions Acids and bases are as familiar as aspirin and milk of magnesia although many people do not know their chemical names—acetylsalicylic acid (aspirin) and magne- sium hydroxide (milk of magnesia). In addition to being the basis of many medicinal and household products, acid-base chemistry is important in industrial processes and essential in sustaining biological systems. Before we can discuss acid-base reactions, we need to know more about acids and bases themselves. General Properties of Acids and Bases In Section 2.7 we defined acids as substances that ionize in water to produce H1 ions and bases as substances that ionize in water to produce OH2 ions. These definitions were formulated in the late nineteenth century by the Swedish chemist 126  4.3 Acid-Base Reactions 127 Svante Arrhenius† to classify substances whose properties in aqueous solutions were well known. Acids • Acids have a sour taste; for example, vinegar owes its sourness to acetic acid, and lemons and other citrus fruits contain citric acid. • Acids cause color changes in plant dyes; for example, they change the color of litmus from blue to red. • Acids react with certain metals, such as zinc, magnesium, and iron, to produce hydrogen gas. A typical reaction is that between hydrochloric acid and magnesium: 2HCl(aq) 1 Mg(s) ¡ MgCl2 (aq) 1 H2 (g) Figure 4.6 A piece of blackboard chalk, which is mostly CaCO3, • Acids react with carbonates and bicarbonates, such as Na2CO3, CaCO3, and reacts with hydrochloric acid. NaHCO3, to produce carbon dioxide gas (Figure 4.6). For example, 2HCl(aq) 1 CaCO3 (s) ¡ CaCl2 (aq) 1 H2O(l) 1 CO2 (g) HCl(aq) 1 NaHCO3 (s) ¡ NaCl(aq) 1 H2O(l) 1 CO2 (g) • Aqueous acid solutions conduct electricity. Bases • Bases have a bitter taste. • Bases feel slippery; for example, soaps, which contain bases, exhibit this property. • Bases cause color changes in plant dyes; for example, they change the color of litmus from red to blue. • Aqueous base solutions conduct electricity. Brønsted Acids and Bases Arrhenius’ definitions of acids and bases are limited in that they apply only to aque- ous solutions. Broader definitions were proposed by the Danish chemist Johannes Brønsted‡ in 1932; a Brønsted acid is a proton donor, and a Brønsted base is a proton acceptor. Note that Brønsted’s definitions do not require acids and bases to be in aqueous solution. Hydrochloric acid is a Brønsted acid because it donates a proton in water: HCl(aq) ¡ H1 (aq) 1 Cl2 (aq) Note that the H1 ion is a hydrogen atom that has lost its electron; that is, it is just a bare proton. The size of a proton is about 10215 m, compared to a diameter of 10210 m for an average atom or ion. Such an exceedingly small charged particle cannot exist as a separate entity in aqueous solution owing to its strong attraction † Svante August Arrhenius (1859–1927). Swedish chemist. Arrhenius made important contributions in the study of chemical kinetics and electrolyte solutions. He also speculated that life had come to Earth from other planets, a theory now known as panspermia. Arrhenius was awarded the Nobel Prize in Chemistry in 1903. ‡ Johannes Nicolaus Brønsted (1879–1947). Danish chemist. In addition to his theory of acids and bases, Brønsted worked on thermodynamics and the separation of mercury isotopes. In some texts, Brønsted acids and bases are called Brønsted-Lowry acids and bases. Thomas Martin Lowry (1874–1936). English chemist. Brønsted and Lowry developed essentially the same acid-base theory independently in 1923. 128 Chapter 4 ■ Reactions in Aqueous Solutions Figure 4.7 Ionization of HCl in water to form the hydronium ion and the chloride ion. 1 8n 1 HCl 1 H2O 8n H3O1 1 Cl2 for the negative pole (the O atom) in H2O. Consequently, the proton exists in the hydrated form, as shown in Figure 4.7. Therefore, the ionization of hydrochloric acid should be written as HCl(aq) 1 H2O(l) ¡ H3O1 (aq) 1 Cl2 (aq) The hydrated proton, H3O1, is called the hydronium ion. This equation shows a reac- tion in which a Brønsted acid (HCl) donates a proton to a Brønsted base (H2O). Experiments show that the hydronium ion is further hydrated so that the proton may have several water molecules associated with it. Because the acidic properties of the proton are unaffected by the degree of hydration, in this text we will generally use H1(aq) to represent the hydrated proton. This notation is for convenience, but H3O1 is closer to reality. Keep in mind that both notations represent the same species in aqueous solution. Acids commonly used in the laboratory include hydrochloric acid (HCl), nitric Electrostatic potential map of the acid (HNO3), acetic acid (CH3COOH), sulfuric acid (H2SO4), and phosphoric acid H3O1 ion. In the rainbow color spectrum representation, the most (H3PO4). The first three are monoprotic acids; that is, each unit of the acid yields one electron-rich region is red and the hydrogen ion upon ionization: most electron-poor region is blue. HCl(aq) ¡ H1 (aq) 1 Cl2 (aq) In most cases, acids start with H in the HNO3 (aq) ¡ H1 (aq) 1 NO23 (aq) formula or have a COOH group. CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq) As mentioned earlier, because the ionization of acetic acid is incomplete (note the double arrows), it is a weak electrolyte. For this reason it is called a weak acid (see Table 4.3 Table 4.1). On the other hand, HCl and HNO3 are strong acids because they are strong Some Common Strong electrolytes, so they are completely ionized in solution (note the use of single arrows). and Weak Acids Sulfuric acid (H2SO4) is a diprotic acid because each unit of the acid gives up two H1ions, in two separate steps: Strong Acids Hydrochloric HCl H2SO4 (aq) ¡ H1 (aq) 1 HSO24 (aq) acid HSO24 (aq) Δ H1 (aq) 1 SO22 4 (aq) Hydrobromic HBr acid H2SO4 is a strong electrolyte or strong acid (the first step of ionization is complete), Hydroiodic HI but HSO2 4 is a weak acid or weak electrolyte, and we need a double arrow to repre- acid sent its incomplete ionization. Nitric acid HNO3 Triprotic acids, which yield three H1ions, are relatively few in number. The best Sulfuric acid H2SO4 known triprotic acid is phosphoric acid, whose ionizations are Perchloric acid HClO4 H3PO4 (aq) Δ H1 (aq) 1 H2PO24 (aq) Weak Acids H2PO24 (aq) Δ H1 (aq) 1 HPO22 4 (aq) Hydrofluoric HF HPO22 4 (aq) Δ H 1 (aq) 1 PO 32 4 (aq) acid Nitrous acid HNO2 All three species (H3PO4, H2PO2 22 4 , and HPO4 ) in this case are weak acids, and we Phosphoric acid H3PO4 use the double arrows to represent each ionization step. Anions such as H2PO2 4 and Acetic acid CH3COOH HPO224 are found in aqueous solutions of phosphates such as NaH2PO4 and Na2HPO4. Table 4.3 lists several common strong and weak acids.  4.3 Acid-Base Reactions 129 1 34 1 NH3 1 H2O 34 NH14 1 OH2 Figure 4.8 Ionization of ammonia in water to form the ammonium ion and the hydroxide ion. Review of Concepts Which of the following diagrams best represents a weak acid? Which represents a very weak acid? Which represents a strong acid? The proton exists in water as the hydronium ion. All acids are monoprotic. (For simplicity, water molecules are not shown.) (a) (b) (c) Table 4.1 shows that sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2] are strong electrolytes. This means that they are completely ionized in solution: H2O NaOH(s) ¡ Na1(aq) 1 OH2 (aq) H2O Ba(OH) 2 (s) ¡ Ba21(aq) 1 2OH2 (aq) The OH2 ion can accept a proton as follows: H1 (aq) 1 OH2 (aq) ¡ H2O(l) Thus, OH2 is a Brønsted base. Ammonia (NH3) is classified as a Brønsted base because it can accept a H1 ion (Figure 4.8): NH3 (aq) 1 H2O(l) Δ NH 14 (aq) 1 OH2 (aq) Ammonia is a weak electrolyte (and therefore a weak base) because only a small fraction of dissolved NH3 molecules react with water to form NH14 and OH2 ions. The most commonly used strong base in the laboratory is sodium hydroxide. It is cheap and soluble. (In fact, all of the alkali metal hydroxides are soluble.) The most commonly used weak base is aqueous ammonia solution, which is sometimes erroneously called ammonium hydroxide. There is no evidence that the species NH4OH actually exists other than the NH14 and OH2 ions in solution. All of the Group 2A elements form hydroxides of the type M(OH)2, where M Note that this bottle of aqueous denotes an alkaline earth metal. Of these hydroxides, only Ba(OH)2 is soluble. ammonia is erroneously labeled. 130 Chapter 4 ■ Reactions in Aqueous Solutions Magnesium and calcium hydroxides are used in medicine and industry. Hydrox- ides of other metals, such as Al(OH)3 and Zn(OH)2 are insoluble and are not used as bases. Example 4.3 classifies substances as Brønsted acids or Brønsted bases. Example 4.3 Classify each of the following species in aqueous solution as a Brønsted acid or base: (a) HBr, (b) NO22 , (c) HCO23 . Strategy What are the characteristics of a Brønsted acid? Does it contain at least an H atom? With the exception of ammonia, most Brønsted bases that you will encounter at this stage are anions. Solution (a) We know that HCl is an acid. Because Br and Cl are both halogens (Group 7A), we expect HBr, like HCl, to ionize in water as follows: HBr(aq) ¡ H1 (aq) 1 Br2 (aq) Therefore HBr is a Brønsted acid. (b) In solution the nitrite ion can accept a proton from water to form nitrous acid: NO22 (aq) 1 H1 (aq) ¡ HNO2 (aq) This property makes NO22 a Brønsted base. (c) The bicarbonate ion is a Brønsted acid because it ionizes in solution as follows: HCO23 (aq) Δ H1 (aq) 1 CO22 3 (aq) It is also a Brønsted base because it can accept a proton to form carbonic acid: HCO23 (aq) 1 H1 (aq) Δ H2CO3 (aq) Comment The HCO23 species is said to be amphoteric because it possesses both acidic Similar problems: 4.31, 4.32. and basic properties. The double arrows show that this is a reversible reaction. Practice Exercise Classify each of the following species as a Brønsted acid or base: (a) SO22 4 , (b) HI. Acid-Base Neutralization Animation A neutralization reaction is a reaction between an acid and a base. Generally, aque- Neutralization Reactions ous acid-base reactions produce water and a salt, which is an ionic compound made up of a cation other than H1 and an anion other than OH2 or O22: acid 1 base ¡ salt 1 water The substance we know as table salt, NaCl, is a product of the acid-base reaction Acid-base reactions generally go to HCl(aq) 1 NaOH(aq) ¡ NaCl(aq) 1 H2O(l) completion. However, because both the acid and the base are strong electrolytes, they are com- pletely ionized in solution. The ionic equation is H1 (aq) 1 Cl2 (aq) 1 Na1 (aq) 1 OH2 (aq) ¡ Na1 (aq) 1 Cl1 (aq) 1 H2O(l)  4.3 Acid-Base Reactions 131 Therefore, the reaction can be represented by the net ionic equation H1 (aq) 1 OH2 (aq) ¡ H2O(l) Both Na1 and Cl2 are spectator ions. If we had started the preceding reaction with equal molar amounts of the acid and the base, at the end of the reaction we would have only a salt and no leftover acid or base. This is a characteristic of acid-base neutralization reactions. A reaction between a weak acid such as hydrocyanic acid (HCN) and a strong base is HCN(aq) 1 NaOH(aq) ¡ NaCN(aq) 1 H2O(l) Because HCN is a weak acid, it does not ionize appreciably in solution. Thus, the ionic equation is written as HCN(aq) 1 Na1 (aq) 1 OH2 (aq) ¡ Na1 (aq) 1 CN2 (aq) 1 H2O(l) and the net ionic equation is HCN(aq) 1 OH2 (aq) ¡ CN2 (aq) 1 H2O(l) Note that only Na1 is a spectator ion; OH2 and CN2 are not. The following are also examples of acid-base neutralization reactions, represented by molecular equations: HF(aq) 1 KOH(aq) ¡ KF(aq) 1 H2O(l) H2SO4 (aq) 1 2NaOH(aq) ¡ Na2SO4 (aq) 1 2H2O(l) HNO3 (aq) 1 NH3 (aq) ¡ NH4NO3 (aq) The last equation looks different because it does not show water as a product. How- ever, if we express NH3(aq) as NH14 (aq) and OH2(aq), as discussed earlier, then the equation becomes HNO3 (aq) 1 NH1 2 4 (aq) 1 OH (aq) ¡ NH4NO3 (aq) 1 H2O(l) Example 4.4 Write molecular, ionic, and net ionic equations for each of the following acid-base reactions: (a) hydrobromic acid(aq) 1 barium hydroxide(aq) ¡ (b) sulfuric acid(aq) 1 potassium hydroxide(aq) ¡ Strategy The first step is to identify the acids and bases as strong or weak. We see that HBr is a strong acid and H2SO4 is a strong acid for the first step ionization and a weak acid for the second step ionization. Both Ba(OH)2 and KOH are strong bases. Solution (a) Molecular equation: 2HBr(aq) 1 Ba(OH) 2 (aq) ¡ BaBr2 (aq) 1 2H2O(l) Ionic equation: 2H1 (aq) 1 2Br2 (aq) 1 Ba21 (aq) 1 2OH2 (aq) ¡ Ba21 (aq) 1 2Br2 (aq) 1 2H2O(l) (Continued) 132 Chapter 4 ■ Reactions in Aqueous Solutions Net ionic equation: 2H 1 (aq) 1 2OH 2 (aq) ¡ 2H2O(l) or H 1 (aq) 1 OH 2 (aq) ¡ H2O(l) Both Ba21 and Br2 are spectator ions. (b) Molecular equation: H2SO4 (aq) 1 2KOH(aq) ¡ K2SO4 (aq) 1 2H2O(l) Ionic equation: H 1 (aq) 1 HSO24 (aq) 1 2K 1 (aq) 1 2OH 2 (aq) ¡ 2K 1 (aq) 1 SO22 4 (aq) 1 2H2O(l) Net ionic equation: H 1 (aq) 1 HSO24 (aq) 1 2OH 2 (aq) ¡ SO22 4 (aq) 1 2H2O(l) Note that because HSO24 is a weak acid and does not ionize appreciably in water, Similar problem: 4.33(b). the only spectator ion is K1. Practice Exercise Write a molecular equation, an ionic equation, and a net ionic equation for the reaction between aqueous solutions of phosphoric acid and sodium hydroxide. Acid-Base Reactions Leading to Gas Formation Certain salts like carbonates (containing the CO322 ion), bicarbonates (containing the HCO23 ion), sulfites (containing the SO322 ion), and sulfides (containing the S22 ion) react with acids to form gaseous products. For example, the molecular equation for the reaction between sodium carbonate (Na2CO3) and HCl(aq) is (see Figure 4.6) Na2CO3 (aq) 1 2HCl(aq) ¡ 2NaCl(aq) 1 H2CO3 (aq) Carbonic acid is unstable and if present in solution in sufficient concentrations decom- poses as follows: H2CO3 (aq) ¡ H2O(l) 1 CO2 (g) Similar reactions involving other mentioned salts are NaHCO3 (aq) 1 HCl(aq) ¡ NaCl(aq) 1 H2O(l) 1 CO2 (g) Na2SO3 (aq) 1 2HCl(aq) ¡ 2NaCl(aq) 1 H2O(l) 1 SO2 (g) K2S(aq) 1 2HCl(aq) ¡ 2KCl(aq) 1 H2S(g) 4.4 Oxidation-Reduction Reactions Animation Whereas acid-base reactions can be characterized as proton-transfer processes, the Oxidation-Reduction Reactions class of reactions called oxidation-reduction, or redox, reactions are considered electron- transfer reactions. Oxidation-reduction reactions are very much a part of the world around us. They range from the burning of fossil fuels to the action of household  4.4 Oxidation-Reduction Reactions 133 Mg 1 88n Mg21 O22 O2 Figure 4.9 Magnesium burns in oxygen to form magnesium oxide. bleach. Additionally, most metallic and nonmetallic elements are obtained from their ores by the process of oxidation or reduction. Many important redox reactions take place in water, but not all redox reactions Animation Reaction of Magnesium and Oxygen occur in aqueous solution. We begin our discussion with a reaction in which two elements combine to form a compound. Consider the formation of magnesium oxide (MgO) from magnesium and oxygen (Figure 4.9): 2Mg(s) 1 O2 (g) ¡ 2MgO(s) Animation Formation of Ag2S by Oxidation- Reduction Magnesium oxide (MgO) is an ionic compound made up of Mg21 and O22 ions. In this reaction, two Mg atoms give up or transfer four electrons to two O atoms (in O2). For convenience, we can think of this process as two separate steps, one involving the loss of four electrons by the two Mg atoms and the other being the gain of four electrons by an O2 molecule: 2Mg ¡ 2Mg21 1 4e 2 Note that in an oxidation half-reaction, O2 1 4e 2 ¡ 2O22 electrons appear as the product; in a reduction half-reaction, electrons appear as the reactant. Each of these steps is called a half-reaction, which explicitly shows the electrons involved in a redox reaction. The sum of the half-reactions gives the overall reaction: 2Mg 1 O2 1 4e 2 ¡ 2Mg21 1 2O22 1 4e 2 or, if we cancel the electrons that appear on both sides of the equation, 2Mg 1 O2 ¡ 2Mg21 1 2O22 Finally, the Mg21 and O22 ions combine to form MgO: 2Mg21 1 2O22 ¡ 2MgO The term oxidation reaction refers to the half-reaction that involves loss of elec- A useful mnemonic for redox is OILRIG: Oxidation Is Loss (of electrons) and trons. Chemists originally used “oxidation” to denote the combination of elements Reduction Is Gain (of electrons). with oxygen. However, it now has a broader meaning that includes reactions not involving oxygen. A reduction reaction is a half-reaction that involves gain of elec- trons. In the formation of magnesium oxide, magnesium is oxidized. It is said to act 134 Chapter 4 ■ Reactions in Aqueous Solutions Oxidizing agents are always reduced and as a reducing agent because it donates electrons to oxygen and causes oxygen to be reducing agents are always oxidized. This statement may be somewhat confusing, reduced. Oxygen is reduced and acts as an oxidizing agent because it accepts elec- but it is simply a consequence of the trons from magnesium, causing magnesium to be oxidized. Note that the extent of definitions of the two processes. oxidation in a redox reaction must be equal to the extent of reduction; that is, the number of electrons lost by a reducing agent must be equal to the number of electrons gained by an oxidizing agent. The occurrence of electron transfer is more apparent in some redox reactions than others. When metallic zinc is added to a solution containing copper(II) sulfate (CuSO4), zinc reduces Cu21 by donating two electrons to it: Zn(s) 1 CuSO4 (aq) ¡ ZnSO4 (aq) 1 Cu(s) In the process, the solution loses the blue color that characterizes the presence of hydrated Cu21 ions (Figure 4.10): Zn(s) 1 Cu21 (aq) ¡ Zn21 (aq) 1 Cu(s) The Zn bar is in aqueous Cu2+ solution of CuSO4 2e– Zn2+ Cu Ag+ Zn Cu2+ Zn 2e– Zn2+ Cu Ag Cu2+ Cu When a piece of copper wire is Cu2+ ions are placed in an aqueous AgNO3 solution Ag converted to Cu atoms. Cu atoms enter the solution as Cu2+ ions, Zn atoms enter the and Ag+ ions are converted to solid Ag. solution as Zn2+ ions. (a) (b) Figure 4.10 Metal displacement reactions in solution. (a) First beaker: A zinc strip is placed in a blue CuSO4 solution. Immediately Cu21 ions are reduced to metallic Cu in the form of a dark layer. Second beaker: In time, most of the Cu21 ions are reduced and the solution becomes colorless. (b) First beaker: A piece of Cu wire is placed in a colorless AgNO3 solution. Ag1 ions are reduced to metallic Ag. Second beaker: As time progresses, most of the Ag1 ions are reduced and the solution acquires the characteristic blue color due to the presence of hydrated Cu21 ions.  4.4 Oxidation-Reduction Reactions 135 The oxidation and reduction half-reactions are Zn ¡ Zn21 1 2e 2 21 Cu 1 2e 2 ¡ Cu Similarly, metallic copper reduces silver ions in a solution of silver nitrate (AgNO3): Animation Reaction of Cu with AgNO3 Cu(s) 1 2AgNO3 (aq) ¡ Cu(NO3 ) 2 (aq) 1 2Ag(s) or Cu(s) 1 2Ag 1 (aq) ¡ Cu21 (aq) 1 2Ag(s) Oxidation Number The definitions of oxidation and reduction in terms of loss and gain of electrons apply to the formation of ionic compounds such as MgO and the reduction of Cu21 ions by Zn. However, these definitions do not accurately characterize the formation of hydrogen chloride (HCl) and sulfur dioxide (SO2): H2 (g) 1 Cl2 (g) ¡ 2HCl(g) S(s) 1 O2 (g) ¡ SO2 (g) Because HCl and SO2 are not ionic but molecular compounds, no electrons are actu- ally transferred in the formation of these compounds, as they are in the case of MgO. Nevertheless, chemists find it convenient to treat these reactions as redox reactions because experimental measurements show that there is a partial transfer of electrons (from H to Cl in HCl and from S to O in SO2). To keep track of electrons in redox reactions, it is useful to assign oxidation numbers to the reactants and products. An atom’s oxidation number, also called oxidation state, signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. For example, we can rewrite the previous equations for the formation of HCl and SO2 as follows: 0 0 1121 H2(g) 1 Cl2(g) 88n 2HCl(g) 0 0 14 22 S(s) 1 O2(g) 88n SO2(g) The numbers above the element symbols are the oxidation numbers. In both of the reactions shown, there is no charge on the atoms in the reactant molecules. Thus, their oxidation number is zero. For the product molecules, however, it is assumed that complete electron transfer has taken place and that atoms have gained or lost electrons. The oxidation numbers reflect the number of electrons “transferred.” Oxidation numbers enable us to identify elements that are oxidized and reduced at a glance. The elements that show an increase in oxidation number—hydrogen and sulfur in the preceding examples—are oxidized. Chlorine and oxygen are reduced, so their oxidation numbers show a decrease from their initial values. Note that the sum of the oxidation numbers of H and Cl in HCl (11 and 21) is zero. Likewise, if we add the oxidation numbers of S (14) and two atoms of O [2 3 (22)], the 136 Chapter 4 ■ Reactions in Aqueous Solutions total is zero. The reason is that the HCl and SO2 molecules are neutral, so the charges must cancel. We use the following rules to assign oxidation numbers: 1. In free elements (that is, in the uncombined state), each atom has an oxidation number of zero. Thus, each atom in H2, Br2, Na, Be, K, O2, and P4 has the same oxidation number: zero. 2. For ions composed of only one atom (that is, monatomic ions), the oxidation number is equal to the charge on the ion. Thus, Li1 ion has an oxidation number of 11; Ba21 ion, 12; Fe31 ion, 13; I2 ion, 21; O22 ion, 22; and so on. All alkali metals have an oxidation number of 11 and all alkaline earth metals have an oxidation number of 12 in their compounds. Aluminum has an oxidation number of 13 in all its compounds. 3. The oxidation number of oxygen in most compounds (for example, MgO and H2O) is 22, but in hydrogen peroxide (H2O2) and peroxide ion (O22 2 ), it is 21. 4. The oxidation number of hydrogen is 11, except when it is bonded to metals in binary compounds. In these cases (for example, LiH, NaH, CaH2), its oxidation number is 21. 5. Fluorine has an oxidation number of 21 in all its compounds. Other halogens (Cl, Br, and I) have negative oxidation numbers when they occur as halide ions in their compounds. When combined with oxygen—for example in oxoacids and oxoanions (see Section 2.7)—they have positive oxidation numbers. 6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion. For example, in the ammo- nium ion, NH1 4 , the oxidation number of N is 23 and that of H is 11. Thus, the sum of the oxidation numbers is 23 1 4(11) 5 11, which is equal to the net charge of the ion. 7. Oxidation numbers do not have to be integers. For example, the oxidation number of O in the superoxide ion, O22 , is 212. We apply the preceding rules to assign oxidation numbers in Example 4.5. Example 4.5 Assign oxidation numbers to all the elements in the following compounds and ion: (a) Li2O, (b) HNO3, (c) Cr2O722. Strategy In general, we follow the rules just listed for assigning oxidation numbers. Remember that all alkali metals have an oxidation number of 11, and in most cases hydrogen has an oxidation number of 11 and oxygen has an oxidation number of 22 in their compounds. Solution (a) By rule 2 we see that lithium has an oxidation number of 11 (Li1) and oxygen’s oxidation number is 22 (O22). (b) This is the formula for nitric acid, which yields a H1 ion and a NO23 ion in solution. From rule 4 we see that H has an oxidation number of 11. Thus the other group (the nitrate ion) must have a net oxidation number of 21. Oxygen has an (Continued)  4.4 Oxidation-Reduction Reactions 137 oxidation number of 22, and if we use x to represent the oxidation number of nitrogen, then the nitrate ion can be written as 3N(x)O3(22) 4 2 so that x 1 3(22) 5 21 or x 5 15 (c) From rule 6 we see that the sum of the oxidation numbers in the dichromate ion Cr2O722 must be 22. We know that the oxidation number of O is 22, so all that remains is to determine the oxidation number of Cr, which we call y. The dichromate ion can be written as 3Cr2( y)O7(22) 4 22 so that 2(y) 1 7(22) 5 22 or y 5 16 Check In each case, does the sum of the oxidation numbers of all the atoms equal the net charge on the species? Similar problems: 4.47, 4.49. Practice Exercise Assign oxidation numbers to all the elements in the following compound and ion: (a) PF3, (b) MnO24 . Figure 4.11 shows the known oxidation numbers of the familiar elements, arranged according to their positions in the periodic table. We can summarize the content of this figure as follows: • Metallic elements have only positive oxidation numbers, whereas nonmetallic elements may have either positive or negative oxidation numbers. • The highest oxidation number an element in Groups 1A–7A can have is its group number. For example, the halogens are in Group 7A, so their highest possible oxidation number is 17. • The transition metals (Groups 1B, 3B–8B) usually have several possible oxidation numbers. Types of Redox Reactions Among the most common oxidation-reduction reactions are combination, decomposi- tion, combustion, and displacement reactions. A more involved type is called dispro- portionation reactions, which will also be discussed in this section. Combination Reactions A combination reaction is a reaction in which two or more substances combine to Not all combination reactions are redox in nature. The same holds for decomposition form a single product. Figure 4.12 shows some combination reactions. For example, reactions. 0 0 14 22 S(s) 1 O2(g) 88n SO2(g) 0 0 13 21 2Al(s) 1 3Br2(l) 88n 2AlBr3(s) 138 Chapter 4 ■ Reactions in Aqueous Solutions 1 18 1A 8A 1 2 H He +1 –1 2 13 14 15 16 17 2A 3A 4A 5A 6A 7A 3 4 5 6 7 8 9 10 Li Be B C N O F Ne +1 +2 +3 +4 +5 +2 –1 +2 +4 – 12 –4 +3 +2 –1 +1 –2 –3 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar +1 +2 +3 +4 +5 +6 +7 –4 +3 +4 +6 –3 +2 +5 –2 +4 3 4 5 6 7 8 9 10 11 12 +3 +1 3B 4B 5B 6B 7B 8B 1B 2B –1 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr +1 +2 +3 +4 +5 +6 +7 +3 +3 +2 +2 +2 +3 +4 +5 +6 +5 +4 +3 +4 +5 +6 +2 +2 +1 –4 +3 +4 +3 +2 +2 +3 +4 +4 –3 –2 +1 +2 +3 +3 –1 +2 +2 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe +1 +2 +3 +4 +5 +6 +7 +8 +4 +4 +1 +2 +3 +4 +5 +6 +7 +6 +4 +4 +6 +6 +3 +2 +2 +3 +4 +5 +4 +3 +4 +4 +2 –3 –2 +1 +2 +3 –1 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn +1 +2 +3 +4 +5 +6 +7 +8 +4 +4 +3 +2 +3 +4 +5 +2 –1 +4 +6 +4 +3 +2 +1 +1 +1 +2 +3 +4 Figure 4.11 The oxidation numbers of elements in their compounds. The more common oxidation numbers are in color. (a) (b) (c) Figure 4.12 Some simple combination redox reactions. (a) Sulfur burning in air to form sulfur dioxide. (b) Sodium burning in chlorine to form sodium chloride. (c) Aluminum reacting with bromine to form aluminum bromide.  4.4 Oxidation-Reduction Reactions 139 Figure 4.13 (a) On heating, mercury(II) oxide (HgO) decomposes to form mercury and oxygen. (b) Heating potassium chlorate (KClO3) produces oxygen, which supports the combustion of the wood splint. (a) (b) Decomposition Reactions Decomposition reactions are the opposite of combination reactions. Specifically, a We show oxidation numbers only for elements that are oxidized or reduced. decomposition reaction is the breakdown of a compound into two or more components (Figure 4.13). For example, All combustion reactions are redox processes. 12 22 0 0 2HgO(s) 88n 2Hg(l) 1 O2(g) 15 22 21 0 2KClO3(s) 88n 2KCl(s) 1 3O2(g) 1121 0 0 2NaH(s) 88n 2Na(s) 1 H2(g) Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the release of heat and light to produce a flame. The reactions between magne- sium and sulfur with oxygen described earlier are combustion reactions. Another example is the burning of propane (C3H8), a component of natural gas that is used for domestic heating and cooking: (a) C3H8 (g) 1 5O2 (g) ¡ 3CO2 (g) 1 4H2O(l) Assigning an oxidation number to C atoms in organic compounds is more involved. Here, we focus only on the oxidation number of O atoms, which changes from 0 to 22. Displacement Reactions In a displacement reaction, an ion (or atom) in a compound is replaced by an ion (or atom) of another element: Most displacement reactions fit into one of three subcategories: hydrogen displacement, metal displacement, or halogen displacement. 1. Hydrogen Displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba), which are the most reactive of the metallic elements, will displace hydro- gen from cold water (Figure 4.14): (b) 0 11 11 11 0 Figure 4.14 Reactions of 2Na(s) 1 2H2O(l) 88n 2NaOH(aq) 1 H2(g) (a) sodium (Na) and (b) calcium (Ca) with cold water. Note that the 0 11 12 11 0 reaction is more vigorous with Na Ca(s) 1 2H2O(l) 88n Ca(OH)2(s) 1 H2(g) than with Ca. 140 Chapter 4 ■ Reactions in Aqueous Solutions (a) (b) (c) Figure 4.15 Reactions of (a) iron (Fe), (b) zinc (Zn), and (c) magnesium (Mg) with hydrochloric acid to form hydrogen gas and the metal chlorides (FeCl2, ZnCl2, MgCl2). The reactivity of these metals is reflected in the rate of hydrogen gas evolution, which is slowest for the least reactive metal, Fe, and fastest for the most reactive metal, Mg. Many metals, including those that do not react with water, are capable of displac- ing hydrogen from acids. For example, zinc (Zn) and magnesium (Mg) do not react with cold water but do react with hydrochloric acid, as follows: 0 11 12 0 Zn(s) 1 2HCl(aq) 88n ZnCl2(aq) 1 H2(g) 0 11 12 0 Mg(s) 1 2HCl(aq) 88n MgCl2(aq) 1 H2(g) Figure 4.15 shows the reactions between hydrochloric acid (HCl) and iron (Fe), zinc (Zn), and magnesium (Mg). These reactions are used to prepare hydrogen gas in the laboratory. 2. Metal Displacement. A metal in a compound can be displaced by another metal in the elemental state. We have already seen examples of zinc replacing copper ions and copper replacing silver ions (see p. 134). Reversing the roles of the metals would result in no reaction. Thus, copper metal will not displace zinc ions from zinc sulfate, and silver metal will not displace copper ions from copper nitrate. An easy way to predict whether a metal or hydrogen displacement reaction will actually occur is to refer to an activity series (sometimes called the electrochemical series), shown in Figure 4.16. Basically, an activity series is a convenient summary of the results of many possible displacement reactions similar to the ones already discussed. According to this series, any metal above hydrogen will displace it from water or from an acid, but metals below hydrogen will not react with either water or an acid. In fact, any metal listed in the series will react with any metal (in a com- pound) below it. For example, Zn is above Cu, so zinc metal will displace copper ions from copper sulfate.  4.4 Oxidation-Reduction Reactions 141 Li n Li1 1 e2 K n K1 1 e2 React with cold Ba n Ba21 1 2e2 water to produce H2 Ca n Ca21 1 2e2 Na n Na1 1 e2 Mg n Mg21 1 2e2 Reducing strength increases Al n Al31 1 3e2 Zn n Zn21 1 2e2 React with steam Cr n Cr31 1 3e2 to produce H2 Fe n Fe21 1 2e2 Cd n Cd21 1 2e2 Co n Co21 1 2e2 Ni n Ni21 1 2e2 React with acids Sn n Sn21 1 2e2 to produce H2 Pb n Pb21 1 2e2 H2 n 2H1 1 2e2 Cu n Cu21 1 2e2 Ag n Ag1 1 e2 Do not react with water Hg n Hg21 1 2e2 or acids to produce H2 Pt n Pt21 1 2e2 Au n Au31 1 3e2 Figure 4.16 The activity series for metals. The metals are arranged according to their ability to displace hydrogen from an acid or water. Li (lithium) is the most reactive metal, and Au (gold) is the least reactive. Metal displacement reactions find many applications in metallurgical processes, the goal of which is to separate pure metals from their ores. For example, vanadium is obtained by treating vanadium(V) oxide with metallic calcium: V2O5 (s) 1 5Ca(l) ¡ 2V(l) 1 5CaO(s) Similarly, titanium is obtained from titanium(IV) chloride according to the reaction TiCl4 (g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2 (l) In each case, the metal that acts as the reducing agent lies above the metal that is reduced (that is, Ca is above V and Mg is above Ti) in the activity series. We will see more examples of this type of reaction in Chapter 18. 3. Halogen Displacement. Another activity series summarizes the halogens’ behavior 1A 8A 2A 3A 4A 5A 6A 7A in halogen displacement reactions: F Cl Br F2 . Cl2 . Br2 . I2 I The power of these elements as oxidizing agents decreases as we move down Group The halogens. 7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, and iodide ions in solution. In fact, molecular fluorine is so reactive that it also attacks water; thus these reactions cannot be carried out in aqueous solutions. On the other hand, molecular chlorine can displace bromide and iodide ions in aqueous solution. The displacement equations are 0 21 21 0 Cl2(g) 1 2KBr(aq) 88n 2KCl(aq) 1 Br2(l) 0 21 21 0 Cl2(g) 1 2NaI(aq) 88n 2NaCl(aq) 1 I2(s) 142 Chapter 4 ■ Reactions in Aqueous Solutions The ionic equations are 0 21 21 0 Cl2(g) 1 2Br(aq) 88n 2Cl2(aq) 1 Br2(l) 0 21 21 0 Cl2(g) 1 2I2(aq) 88n 2Cl2(aq) 1 I2(s) Molecular bromine, in turn, can displace iodide ion in solution: 0 21 21 0 Br2(l) 1 2I2(aq) 88n 2Br2(aq) 1 I2(s) Reversing the roles of the halogens produces no reaction. Thus, bromine cannot dis- place chloride ions, and iodine cannot displace bromide and chloride ions. The halogen displacement reactions have a direct industrial application. The hal- ogens as a group are the most reactive of the nonmetallic elements. They are all strong oxidizing agents. As a result, they are found in nature in the combined state (with metals) as halides and never as free elements. Of these four elements, chlorine is by far the most important industrial chemical. In 2010 the amount of chlorine produced in the United States was about 25 billion pounds, making chlorine the tenth-ranking industrial chemical. The annual production of bromine is only one-hundredth that of chlorine, while the amounts of fluorine and iodine produced are even less. Recovering the halogens from their halides requires an oxidation process, which is represented by 2X 2 ¡ X2 1 2e 2 Bromine is a fuming red liquid. where X denotes a halogen element. Seawater and natural brine (for example, under- ground water in contact with salt deposits) are rich sources of Cl2, Br2, and I2 ions. Minerals such as fluorite (CaF2) and cryolite (Na3AlF6) are used to prepare fluorine. Because fluorine is the strongest oxidizing agent known, there is no way to convert F2 ions to F2 by chemical means. The only way to carry out the oxidation is by electrolytic means, the details of which will be discussed in Chapter 18. Industrially, chlorine, like fluorine, is produced electrolytically. Bromine is prepared industrially by oxidizing Br2 ions with chlorine, which is a strong enough oxidizing agent to oxidize Br2 ions but not water: 2Br 2 (aq) ¡ Br2 (l) 1 2e 2 One of the richest sources of Br2 ions is the Dead Sea—about 4000 parts per million (ppm) by mass of all dissolved substances in the Dead Sea is Br. Follow- Figure 4.17 The industrial ing the oxidation of Br2 ions, bromine is removed from the solution by blowing manufacture of liquid bromine by air over the solution, and the air-bromine mixture is then cooled to condense the oxidizing an aqueous solution bromine (Figure 4.17). containing Br2 ions with chlorine gas. Iodine is also prepared from seawater and natural brine by the oxidation of I2 ions with chlorine. Because Br2 and I2 ions are invariably present in the same source, they are both oxidized by chlorine. However, it is relatively easy to separate Br2 from I2 because iodine is a solid that is sparingly soluble in water. The air-blowing proce- dure will remove most of the bromine formed but will not affect the iodine present. 5A 6A 7A N O 7B 1B 2B P S Cl Disproportionation Reaction Mn Cu Br I A special type of redox reaction is the disproportionation reaction. In a dispropor- Au Hg tionation reaction, an element in one oxidation state is simultaneously oxidized and reduced. One reactant in a disproportionation reaction always contains an element that Elements that are most likely to undergo disproportionation can have at least three oxidation states. The element itself is in an intermediate reactions. oxidation state; that is, both higher and lower oxidation states exist for that element  4.4 Oxidation-Reduction Reactions 143 in the products. The decomposition of hydrogen peroxide is an example of a dispro- portionation reaction: 21 22 0 2H2O2(aq) 88n 2H2O(l) 1 O2(g) Note that the oxidation number of H remains unchanged at 11. Here the oxidation number of oxygen in the reactant (21) both increases to zero in O2 and decreases to 22 in H2O. Another example is the reaction between molecular chlorine and NaOH solution: 0 11 21 Cl2(g) 1 2OH2(aq) 88n ClO2(aq) 1 Cl2(aq) 1 H2O(l) This reaction describes the formation of household bleaching agents, for it is the hypochlorite ion (ClO2) that oxidizes the color-bearing substances in stains, convert- ing them to colorless compounds. Finally, it is interesting to compare redox reactions and acid-base reactions. They are analogous in that acid-base reactions involve the transfer of protons while redox reactions involve the transfer of electrons. However, while acid-base reactions are quite easy to recognize (because they always involve an acid and a base), there is no simple procedure for identifying a redox process. The only sure way is to compare the oxidation numbers of all the elements in the reactants and products. Any change in oxidation number guarantees that the reaction is redox in nature. The classification of different types of redox reactions is illustrated in Example 4.6. Example 4.6 Classify the following redox reactions and indicate changes in the oxidation numbers of the elements: (a) 2N2O(g) ¡ 2N2(g) 1 O2(g) (b) 6Li(s) 1 N2(g) ¡ 2Li3N(s) (c) Ni(s) 1 Pb(NO3)2(aq) ¡ Pb(s) 1 Ni(NO3)2(aq) (d) 2NO2(g) 1 H2O(l) ¡ HNO2(aq) 1 HNO3(aq) Strategy Review the definitions of combination reactions, decomposition reactions, displacement reactions, and disproportionation reactions. Solution (a) This is a decomposition reaction because one reactant is converted to two different products. The oxidation number of N changes from 11 to 0, while that of O changes from 22 to 0. (b) This is a combination reaction (two reactants form a single product). The oxidation number of Li changes from 0 to 11 while that of N changes from 0 to 23. (c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb21 ion. The oxidation number of Ni increases from 0 to 12 while that of Pb decreases from 12 to 0. (d) The oxidation number of N is 14 in NO2 and it is 13 in HNO2 and 15 in HNO3. Because the oxidation number of the same element both increases and decreases, this is a disproportionation reaction. Similar problems: 4.55, 4.56. Practice Exercise Identify the following redox reactions by type: (a) Fe 1 H2SO4 ¡ FeSO4 1 H2 (b) S 1 3F2 ¡ SF6 (c) 2CuCl ¡ Cu 1 CuCl2 (d) 2Ag 1 PtCl2 ¡ 2AgCl 1 Pt CHEMISTRY in Action Breathalyzer E very year in the United States about 25,000 people are killed and 500,000 more are injured as a result of drunk driving. In spite of efforts to educate the public about the dangers of driving while intoxicated and stiffer penalties for drunk driving offenses, law enforcement agencies still have to devote a great deal of work to removing drunk drivers from America’s roads. The police often use a device called a breathalyzer to test drivers suspected of being drunk. The chemical basis of this device is a redox reaction. A sample of the driver’s breath is drawn into the breathalyzer, where it is treated with an acidic solution of potassium dichromate. The alcohol (ethanol) in the breath is converted to acetic acid as shown in the following equation: 3CH3CH2OH 1 2K2Cr2O7 1 8H2SO4 ¡ ethanol potassium sulfuric dichromate acid A driver being tested for blood alcohol content with a handheld breathalyzer. (orange yellow) 3CH3COOH 1 2Cr2(SO4)3 1 2K2SO4 1 11H2O to the green chromium(III) ion (see Figure 4.22). The driver’s acetic acid chromium(III) potassium sulfate (green) sulfate blood alcohol level can be determined readily by measuring the degree of this color change (read from a calibrated meter on the In this reaction, the ethanol is oxidized to acetic acid and the instrument). The current legal limit of blood alcohol content is chromium(VI) in the orange-yellow dichromate ion is reduced 0.08 percent by mass. Anything higher constitutes intoxication. Breath Schematic diagram of a breathalyzer. The alcohol in the driver’s breath is reacted with a potassium dichromate solution. The change in the absorption of light due to the formation of chromium(III) sulfate is registered by the detector and shown on a meter, which directly displays the alcohol Meter content in blood. The filter selects only one wavelength of light for Light Filter Photocell measurement. source detector K2Cr2O7 solution Review of Concepts Which of the following combination reactions is not a redox reaction? (a) 2Mg(s) 1 O2(g) ¡ 2MgO(s) (b) H2(g) 1 F2(g) ¡ 2HF(g) (c) NH3(g) 1 HCl(g) ¡ NH4Cl(s) (d) 2Na(s) 1 S(s) ¡ Na2S(s) The above Chemistry in Action essay describes how law enforcement makes use of a redox reaction to apprehend drunk drivers. 144  4.5 Concentration of Solutions 145 4.5 Concentration of Solutions To study solution stoichiometry, we must know how much of the reactants are present in a solution and also how to control the amounts of reactants used to bring about a reaction in aqueous solution. The concentration of a solution is the amount of solute present in a given amount of solvent, or a given amount of solution. (For this discussion, we will assume the solute is a liquid or a solid and the solvent is a liquid.) The concentration of a solu- tion can be expressed in many different ways, as we will see in Chapter 12. Here we will consider one of the most commonly used units in chemistry, molarity (M), or molar concentration, which is the number of moles of solute per liter of solution. Molarity is defined as moles of solute molarity 5 (4.1) liters of solution Equation (4.1) can also be expressed algebraically as n M5 (4.2) Keep in mind that volume (V) is liters of V solution, not liters of solvent. Also, the molarity of a solution depends on temperature. where n denotes the number of moles of solute and V is the volume of the solution in liters. A 1.46 molar glucose (C6H12O6) solution, written as 1.46 M C6H12O6, contains 1.46 moles of the solute (C6H12O6) in 1 L of the solution. Of course, we do not always work with solution volumes of 1 L. Thus, a 500-mL solution containing 0.730 mole of C6H12O6 also has a concentration of 1.46 M: 0.730 mol C6H12O6 1000 mL soln molarity 5 3 5 1.46 M C6H12O6 500 mL soln 1 L soln Note that concentration, like density, is an intensive property, so its value does not depend on how much of the solution is present. It is important to keep in mind that molarity refers only to the amount of solute originally dissolved in water and does not take into account any subsequent processes, such as the dissociation of a salt or the ionization of an acid. Consider what happens when a sample of potassium chloride (KCl) is dissolved in enough water to make a 1 M solution: H2O KCl(s) ¡ K1(aq) 1 Cl2(aq) Because KCl is a strong electrolyte, it undergoes complete dissociation in solu- tion. Thus, a 1 M KCl solution contains 1 mole of K1 ions and 1 mole of Cl2 ions, and no KCl units are present. The concentrations of the ions can be expressed as [K1] 5 1 M and [Cl2] 5 1 M, where the square brackets [ ] indicate that the concentration is expressed in molarity. Similarly, in a 1 M barium nitrate [Ba(NO3)2] solution H2O Ba(NO3 ) 2 (s) ¡ Ba21 (aq) 1 2NO23 (aq) we have [Ba21] 5 1 M and [NO23 ] 5 2 M and no Ba(NO3)2 units at all. The procedure for preparing a solution of known molarity is as follows. First, the Animation Making a Solution solute is accurately weighed and transferred to a volumetric flask through a funnel 146 Chapter 4 ■ Reactions in Aqueous Solutions Figure 4.18 Preparing a solution of known molarity. (a) A known amount of a solid solute is transferred into the volumetric flask; then water is added through a funnel. (b) The solid is slowly dissolved by gently swirling the flask. (c) After the solid has completely dissolved, more water is added to bring the level of solution to the mark. Knowing the volume of the solution and the Meniscus amount of solute dissolved in it, Marker showing we can calculate the molarity of known volume the prepared solution. of solution (a) (b) (c) (Figure 4.18). Next, water is added to the flask, which is carefully swirled to dissolve the solid. After all the solid has dissolved, more water is added slowly to bring the level of solution exactly to the volume mark. Knowing the volume of the solution in the flask and the quantity of compound (the number of moles) dissolved, we can calculate the molarity of the solution using Equation (4.1). Note that this procedure does not require knowing the amount of water added, as long as the volume of the final solution is known. Examples 4.7 and 4.8 illustrate the applications of Equations (4.1) and (4.2). Example 4.7 How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250-mL solution whose concentration is 2.16 M? Strategy How many moles of K2Cr2O7 does a 1-L (or 1000 mL) 2.16 M K2Cr2O7 solution contain? A 250-mL solution? How would you convert moles to grams? Solution The first step is to determine the number of moles of K2Cr2O7 in 250 mL or 0.250 L of a 2.16 M solution. Rearranging Equation (4.1) gives moles of solute 5 molarity 3 L soln A K2Cr2O7 solution. Thus, 2.16 mol K2Cr2O7 moles of K2Cr2O7 5 3 0.250 L soln 1 L soln 5 0.540 mol K2Cr2O7 The molar mass of K2Cr2O7 is 294.2 g, so we write 294.2 g K2Cr2O7 grams of K2Cr2O7 needed 5 0.540 mol K2Cr2O7 3 1 mol K2Cr2O7 5 159 g K2Cr2O7 (Continued)  4.5 Concentration of Solutions 147 Check As a ball-park estimate, the mass should be given by [molarity (mol/L) 3 volume (L) 3 molar mass (g/mol)] or [2 mol/L 3 0.25 L 3 300 g/mol] 5 150 g. So the answer is reasonable. Similar problems: 4.65, 4.68. Practice Exercise What is the molarity of an 85.0-mL ethanol (C2H5OH) solution containing 1.77 g of ethanol? Example 4.8 A chemist needs to add 3.81 g of glucose to a reaction mixture. Calculate the volume in milliliters of a 2.53 M glucose solution she should use for the addition. Strategy We must first determine the number of moles contained in 3.81 g of glucose and then use Equation (4.2) to calculate the volume. Solution From the molar mass of glucose, we write 1 mol C6H12O6 Note that we have carried an additional 3.81 g C6H12O6 3 5 2.114 3 1022 mol C6H12O6 digit past the number of significant figures 180.2 g C6H12O6 for the intermediate step. Next, we calculate the volume of the solution that contains 2.114 3 1022 mole of the solute. Rearranging Equation (4.2) gives n V5 M 2.114 3 1022 mol C6H12O6 1000 mL soln 5 3 2.53 mol C6H12O6/L soln 1 L soln 5 8.36 mL soln Check One liter of the solution contains 2.53 moles of C6H12O6. Therefore, the number of moles in 8.36 mL or 8.36 3 1023 L is (2.53 mol 3 8.36 3 1023) or 2.12 3 1022 mol. The small difference is due to the different ways of rounding off. Similar problem: 4.67. Practice Exercise What volume (in milliliters) of a 0.315 M NaOH solution contains 6.22 g of NaOH? Dilution of Solutions Concentrated solutions are often stored in the laboratory stockroom for use as needed. Animation Preparing a Solution by Dilution Frequently we dilute these “stock” solutions before working with them. Dilution is the procedure for preparing a less concentrated solution from a more concentrated one. Suppose that we want to prepare 1 L of a 0.400 M KMnO4 solution from a solu- tion of 1.00 M KMnO4. For this purpose we need 0.400 mole of KMnO4. Because there is 1.00 mole of KMnO4 in 1 L of a 1.00 M KMnO4 solution, there is 0.400 mole of KMnO4 in 0.400 L of the same solution: 1.00 mol 0.400 mol 5 1 L soln 0.400 L soln Therefore, we must withdraw 400 mL from the 1.00 M KMnO4 solution and dilute it to 1000 mL by adding water (in a 1-L volumetric flask). This method gives us 1 L of the desired solution of 0.400 M KMnO4. In carrying out a dilution process, it is useful to remember that adding more Two KMnO4 solutions of different solvent to a given amount of the stock solution changes (decreases) the concentration concentrations. 148 Chapter 4 ■ Reactions in Aqueous Solutions Figure 4.19 The dilution of a more concentrated solution (a) to a less concentrated one (b) does not change the total number of solute particles (18). (a) (b) of the solution without changing the number of moles of solute present in the solution (Figure 4.19). In other words, moles of solute before dilution 5 moles of solute after dilution Molarity is defined as moles of solute in 1 liter of solution, so the number of moles of solute is given by [see Equation (4.2)] moles of solute 3 volume of soln (in liters) 5 moles of solute liters of soln M V n or MV 5 n Because all the solute comes from the original stock solution, we can conclude that n remains the same; that is, MiVi 5 MfVf moles of solute moles of solute (4.3) before dilution after dilution where Mi and Mf are the initial and final concentrations of the solution in molarity and Vi and Vf are the initial and final volumes of the solution, respectively. Of course, the units of Vi and Vf must be the same (mL or L) for the calculation to work. To check the reasonableness of your results, be sure that Mi . Mf and Vf . Vi. We apply Equation (4.3) in Example 4.9. Example 4.9 Describe how you would prepare 5.00 3 102 mL of a 1.75 M H2SO4 solution, starting with an 8.61 M stock solution of H2SO4. Strategy Because the concentration of the final solution is less than that of the original one, this is a dilution process. Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same. Solution We prepare for the calculation by tabulating our data: Mi 5 8.61 M    Mf 5 1.75 M Vi 5 ?    Vf 5 5.00 3 102 mL (Continued)  4.6 Gravimetric Analysis 149 Substituting in Equation (4.3), (8.61 M) (Vi ) 5 (1.75 M) (5.00 3 102 mL) (1.75 M) (5.00 3 102 mL) Vi 5 8.61 M 5 102 mL Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with sufficient water to give a final volume of 5.00 3 102 mL in a 500-mL volumetric flask to obtain the desired concentration. Check The initial volume is less than the final volume, so the answer is reasonable. Similar problems: 4.75, 4.76. Practice Exercise How would you prepare 2.00 3 102 mL of a 0.866 M NaOH solution, starting with a 5.07 M stock solution? Review of Concepts What is the final concentration of a 0.6 M NaCl solution if its volume is doubled and the number of moles of solute is tripled? Now that we have discussed the concentration and dilution of solutions, we can examine the quantitative aspects of reactions in aqueous solution, or solution stoichi- ometry. Sections 4.6–4.8 focus on two techniques for studying solution stoichiometry: gravimetric analysis and titration. These techniques are important tools of quantitative analysis, which is the determination of the amount or concentration of a substance in a sample. 4.6 Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of mass. One type of gravimetric analysis experiment involves the formation, isolation, and mass determination of a precipitate. Generally, this procedure is applied to ionic com- pounds. First, a sample substance of unknown composition is dissolved in water and allowed to react with another substance to form a precipitate. Then the precipitate is filtered off, dried, and weighed. Knowing the mass and chemical formula of the pre- cipitate formed, we can calculate the mass of a particular chemical component (that is, the anion or cation) of the original sample. Finally, from the mass of the component and the mass of the original sample, we can determine the percent composition by mass of the component in the original compound. A reaction that is often studied in gravimetric analysis, because the reactants can be obtained in pure form, is AgNO3 (aq) 1 NaCl(aq) ¡ NaNO3 (aq) 1 AgCl(s) The net ionic equation is Ag 1 (aq) 1 Cl 2 (aq) ¡ AgCl(s) The precipitate is silver chloride (see Table 4.2). As an example, let us say that we This procedure would enable us to determine the purity of the NaCl sample. wanted to determine experimentally the percent by mass of Cl in NaCl. First, we would accurately weigh out a sample of NaCl and dissolve it in water. Next, we would add enough AgNO3 solution to the NaCl solution to cause the precipitation of all 150 Chapter 4 ■ Reactions in Aqueous Solutions (a) (b) (c) Figure 4.20 Basic steps for gravimetric analysis. (a) A solution containing a known amount of NaCl in a beaker. (b) The precipitation of AgCl upon the addition of AgNO3 solution from a measuring cylinder. In this reaction, AgNO3 is the excess reagent and NaCl is the limiting reagent. (c) The solution containing the AgCl precipitate is filtered through a preweighed sintered-disk crucible, which allows the liquid (but not the precipitate) to pass through. The crucible is then removed from the apparatus, dried in an oven, and weighed again. The difference between this mass and that of the empty crucible gives the mass of the AgCl precipitate. the Cl2 ions present in solution as AgCl. In this procedure, NaCl is the limiting reagent and AgNO3 the excess reagent. The AgCl precipitate is separated from the solution by filtration, dried, and weighed. From the measured mass of AgCl, we can calculate the mass of Cl using the percent by mass of Cl in AgCl. Because this same amount of Cl was present in the original NaCl sample, we can calcu- late the percent by mass of Cl in NaCl. Figure 4.20 shows how this procedure is performed. Gravimetric analysis is a highly accurate technique, because the mass of a sample can be measured accurately. However, this procedure is applicable only to reactions that go to completion, or have nearly 100 percent yield. Thus, if AgCl were slightly soluble instead of being insoluble, it would not be possible to remove all the Cl2 ions from the NaCl solution and the subsequent calculation would be in error. Example 4.10 shows the calculations involved in a gravimetric experiment. Example 4.10 A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? Strategy We are asked to calculate the percent by mass of Cl in the unknown sample, which is mass of Cl %Cl 5 3 100% 0.5662 g sample The only source of Cl2 ions is the original compound. These chloride ions eventually end up in the AgCl precipitate. Can we calculate the mass of the Cl2 ions if we know the percent by mass of Cl in AgCl? (Continued)  4.7 Acid-Base Titrations 151 Solution The molar masses of Cl and AgCl are 35.45 g and 143.4 g, respectively. Therefore, the percent by mass of Cl in AgCl is given by 35.45 g Cl %Cl 5 3 100% 143.4 g AgCl 5 24.72% Next, we calculate the mass of Cl in 1.0882 g of AgCl. To do so we convert 24.72 percent to 0.2472 and write mass of Cl 5 0.2472 3 1.0882 g 5 0.2690 g Because the original compound also contained this amount of Cl2 ions, the percent by mass of Cl in the compound is 0.2690 g %Cl 5 3 100% 0.5662 g 5 47.51% Check AgCl is about 25 percent chloride by mass, so the roughly 1 g of AgCl precipitate that formed corresponds to about 0.25 g of chloride, which is a little less than half of the mass of the original sample. Therefore, the calculated percent chloride of 47.51 percent is reasonable. Similar problem: 4.82. Practice Exercise A sample of 0.3220 g of an ionic compound containing the bromide ion (Br2) is dissolved in water and treated with an excess of AgNO3. If the mass of the AgBr precipitate that forms is 0.6964 g, what is the percent by mass of Br in the original compound? Note that gravimetric analysis does not establish the whole identity of the unknown. Thus, in Example 4.10 we still do not know what the cation is. However, knowing the percent by mass of Cl greatly helps us to narrow the possibilities. Because no two compounds containing the same anion (or cation) have the same percent com- position by mass, comparison of the percent by mass obtained from gravimetric anal- ysis with that calculated from a series of known compounds would reveal the identity of the unknown. Review of Concepts Calculate the mass of AgBr formed if a solution containing 6.00 g of KBr is treated with an excess of AgNO3. 4.7 Acid-Base Titrations Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In titration, a solution of accurately known concen- tration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete. If we know the volumes of the standard and unknown solutions used in the titration, along with the concentration of the standard solution, we can calculate the concentration of the unknown solution. 152 Chapter 4 ■ Reactions in Aqueous Solutions Sodium hydroxide is one of the bases commonly used in the laboratory. However, it is difficult to obtain solid sodium hydroxide in a pure form because it has a tendency to absorb water from air, and its solution reacts with carbon dioxide. For these reasons, a solution of sodium hydroxide must be standardized before it can be used in accurate analytical work. We can standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is a monoprotic acid called potassium hydrogen phthalate (KHP), for which the molecular formula is KHC8H4O4 (molar mass 5 204.2 g). KHP is a white, solu- ble solid that is commercially available in highly pure form. The reaction between KHP and sodium hydroxide is Potassium hydrogen KHC8H4O4 (aq) 1 NaOH(aq) ¡ KNaC8H4O4 (aq) 1 H2O(l) phthalate (KHP). KHP is a weak acid. and the net ionic equation is HC8H4O2 2 22 4 (aq) 1 OH (aq) ¡ C8H4O4 (aq) 1 H2O(l) The procedure for the titration is shown in Figure 4.21. First, a known amount of KHP is transferred to an Erlenmeyer flask and some distilled water is added to make up a solution. Next, NaOH solution is carefully added to the KHP solu- tion from a buret until we reach the equivalence point, that is, the point at which the acid has completely reacted with or been neutralized by the base. The equiv- alence point is usually signaled by a sharp change in the color of an indicator in the acid solution. In acid-base titrations, indicators are substances that have distinctly different colors in acidic and basic media. One commonly used indica- tor is phenolphthalein, which is colorless in acidic and neutral solutions but reddish pink in basic solutions. At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic. Example 4.11 illustrates such a titration. Figure 4.21 (a) Apparatus for acid-base titration. A NaOH solution is added from the buret to a KHP solution in an Erlenmeyer flask. (b) A reddish-pink color appears when the equivalence point is reached. The color here has been intensified for visual display. (a) (b)  4.7 Acid-Base Titrations 153 Example 4.11 In a titration experiment, a student finds that 23.48 mL of a NaOH solution are needed to neutralize 0.5468 g of KHP. What is the concentration (in molarity) of the NaOH solution? Strategy We want to determine the molarity of the NaOH solution. What is the definition of molarity? need to find 8n mol NaOH molarity of NaOH 5 ]]]]]]]]] L soln 8n n want to calculate given 8 The volume of NaOH solution is given in the problem. Therefore, we need to find the number of moles of NaOH to solve for molarity. From the preceding equation for the reaction between KHP and NaOH shown in the text we see that 1 mole of KHP neutralizes 1 mole of NaOH. How many moles of KHP are contained in 0.5468 g of KHP? Solution First we calculate the number of moles of KHP consumed in the titration: 1 mol KHP moles of KHP 5 0.5468 g KHP 3 Recall that KHP is KHC8H4O4. 204.2 g KHP 5 2.678 3 1023 mol KHP Because 1 mol KHP ∞ 1 mol NaOH, there must be 2.678 3 1023 mole of NaOH in 23.48 mL of NaOH solution. Finally, we calculate the number of moles of NaOH in 1 L of the solution or the molarity as follows: 2.678 3 1023 mol NaOH 1000 mL soln molarity of NaOH soln 5 3 23.48 mL soln 1 L soln 5 0.1141 mol NaOH/1 L soln 5 0.1141 M Similar problems: 4.89, 4.90. Practice Exercise How many grams of KHP are needed to neutralize 18.64 mL of a 0.1004 M NaOH solution? The neutralization reaction between NaOH and KHP is one of the simplest types of acid-base neutralization known. Suppose, though, that instead of KHP, we wanted to use a diprotic acid such as H2SO4 for the titration. The reaction is rep- resented by 2NaOH(aq) 1 H2SO4 (aq) ¡ Na2SO4 (aq) 1 2H2O(l) Because 2 mol NaOH ∞ 1 mol H2SO4, we need twice as much NaOH to react com- pletely with a H2SO4 solution of the same molar concentration and volume as a H2SO4 has two ionizable protons. monoprotic acid like HCl. On the other hand, we would need twice the amount of HCl to neutralize a Ba(OH)2 solution compared to a NaOH solution having the same concentration and volume because 1 mole of Ba(OH)2 yields 2 moles of OH2 ions: 2HCl(aq) 1 Ba(OH) 2 (aq) ¡ BaCl2 (aq) 1 2H2O(l) In calculations involving acid-base titrations, regardless of the acid or base that takes place in the reaction, keep in mind that the total number of moles of H1 ions that have reacted at the equivalence point must be equal to the total number of moles of OH2 ions that have reacted. Example 4.12 shows the titration of a NaOH solution with a diprotic acid. 154 Chapter 4 ■ Reactions in Aqueous Solutions Example 4.12 The sodium hydroxide solution standardized in Example 4.11 is used to titrate 25.00 mL of a sulfuric acid solution. The titration requires 43.79 mL of the 0.1172 M NaOH solution to completely neutralize the acid. What is the concentration of the H2SO4 solution? Strategy We want to calculate the concentration of the H2SO4 solution. Starting with the volume of NaOH solution required to neutralize the acid, we calculate the moles of NaOH. want to find h mol NaOH L soln 3 5 mol NaOH L soln h h measured given From the equation for the neutralization reaction just shown, we see that 2 moles of NaOH neutralize 1 mole of H2SO4. How many moles of NaOH are contained in 43.79 mL of a 0.1172 M NaOH solution? How many moles of H2SO4 would this quantity of NaOH neutralize? What would be the concentration of the H2SO4 solution? Solution First, we calculate the number of moles of NaOH contained in 43.79 mL of solution: 1 L soln 0.1172 mol NaOH 43.79 mL 3 3 5 5.132 3 1023 mol NaOH 1000 mL soln L soln From the stoichiometry we see that 1 mol H2SO4 ∞ 2 mol NaOH. Therefore, the number of moles of H2SO4 reacted must be 1 mol H2SO4 5.132 3 1023 mol NaOH 3 5 2.566 3 1023 mol H2SO4 2 mol NaOH From the definition of molarity [see Equation (4.1)], we have moles of solute molarity 5 liters of soln So the molarity of the H2SO4 solution is 2.566 3 1023 mol H2SO4 Similar problem: 4.91(b), (c). 5 0.1026 M H2SO4 25 mL 3 (1 L/1000 mL) Practice Exercise If 60.2 mL of 0.427 M KOH solution are required to neutralize 10.1 mL of H2SO4 solution, what is the concentration of the H2SO4 solution in molarity? Review of Concepts A NaOH solution is initially mixed with an acid solution shown in (a). Which of the diagrams shown in (b)–(d) corresponds to one of the following acids: HCl, H2SO4, H3PO4? Color codes: Blue spheres (OH2 ions); red spheres (acid molecules); green spheres (anions of the acids). Assume all the acid-base neutralization reactions go to completion. (a) (b) (c) (d)  4.8 Redox Titrations 155 Figure 4.22 Left to right: Solutions containing the MnO42, Mn21, Cr2O722, and Cr 31 ions. 4.8 Redox Titrations As mentioned earlier, redox reactions involve the transfer of electrons, and acid-base reactions involve the transfer of protons. Just as an acid can be titrated against a base, we can titrate an oxidizing agent against a reducing agent, using a similar procedure. We can, for example, carefully add a solution containing an oxidizing agent to a solu- tion containing a reducing agent. The equivalence point is reached when the reducing agent is completely oxidized by the oxidizing agent. Like acid-base titrations, redox titrations normally require an indicator that clearly There are not as many redox indicators as there are acid-base indicators. changes color. In the presence of large amounts of reducing agent, the color of the indicator is characteristic of its reduced form. The indicator assumes the color of its oxidized form when it is present in an oxidizing medium. At or near the equivalence point, a sharp change in the indicator’s color will occur as it changes from one form to the other, so the equivalence point can be readily identified. Two common oxidizing agents are potassium permanganate (KMnO4) and potas- sium dichromate (K2Cr2O7). As Figure 4.22 shows, the colors of the permanganate and dichromate anions are distinctly different from those of the reduced species: MnO24 ¡ Mn21 purple light pink Cr2O22 7 ¡ Cr31 orange green yellow Thus, these oxidizing agents can themselves be used as an internal indicator in a redox titration because they have distinctly different colors in the oxidized and reduced forms. Redox titrations require the same type of calculations (based on the mole method) as acid-base neutralizations. The difference is that the equations and the stoichiom- etry tend to be more complex for redox reactions. The following is an example of a redox titration. Example 4.13 A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution in molarity? The net ionic equation is 5Fe21 1 MnO42 1 8H 1 ¡ Mn21 1 5Fe31 1 4H2O (Continued) Addition of a KMnO4 solution from a buret to a FeSO4 solution. CHEMISTRY in Action Metal from the Sea M agnesium is a valuable, lightweight metal used as a structural material as well as in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth’s crust, it is cheaper to “mine” the metal from sea- water. Magnesium forms the second most abundant cation in the sea (after sodium); there are about 1.3 g of magne- sium in a kilogram of seawater. The process for obtaining magnesium from seawater employs all three types of reac- tions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone (CaCO3) is heated at high temperatures to produce quicklime, or calcium oxide (CaO): CaCO3 (s) ¡ CaO(s) 1 CO2 (g) Magnesium hydroxide was precipitated from processed seawater in When calcium oxide is treated with seawater, it forms calcium settling ponds at the Dow Chemical Company once operated in hydroxide [Ca(OH)2], which is slightly soluble and ionizes to Freeport, Texas. give Ca21 and OH2 ions: CaO(s) 1 H2O(l) ¡ Ca21 (aq) 1 2OH2 (aq) both Mg21 and Cl2 ions. In a process called electrolysis, an electric current is passed through the cell to reduce the Mg21 The surplus hydroxide ions cause the much less soluble magne- ions and oxidize the Cl2 ions. The half-reactions are sium hydroxide to precipitate: Mg21 1 2e 2 ¡ Mg 21 2 Mg (aq) 1 2OH (aq) ¡ Mg(OH) 2 (s) 2Cl 2 ¡ Cl2 1 2e 2 The solid magnesium hydroxide is filtered and reacted with The overall reaction is hydrochloric acid to form magnesium chloride (MgCl2): MgCl2 (l) ¡ Mg(l) 1 Cl2 (g) Mg(OH) 2 (s) 1 2HCl(aq) ¡ MgCl2 (aq) 1 2H2O(l) This is how magnesium metal is produced. The chlorine gas After the water is evaporated, the solid magnesium chloride is generated can be converted to hydrochloric acid and recycled melted in a steel cell. The molten magnesium chloride contains through the process. Strategy We want to calculate the molarity of the FeSO4 solution. From the definition of molarity need to find 8n mol FeSO molarity of FeSO4 5 ]]]]]]]]]4 L soln 8n n want to calculate given 8 (Continued) 156  Key Equations 157 The volume of the FeSO4 solution is given in the problem. Therefore, we need to find the number of moles of FeSO4 to solve for the molarity. From the net ionic equation, what is the stoichiometric equivalence between Fe21 and MnO42? How many moles of KMnO4 are contained in 16.42 mL of 0.1327 M KMnO4 solution? Solution The number of moles of KMnO4 in 16.42 mL of the solution is 0.1327 mol KMnO4 moles of KMnO4 5 3 16.42 mL 1000 mL soln 23 5 2.179 3 10 mol KMnO4 From the net ionic equation we see that 5 mol Fe21 ∞ 1 mol MnO2 4 . Therefore, the number of moles of FeSO4 oxidized is 5 mol FeSO4 moles FeSO4 5 2.179 3 1023 mol KMnO4 3 1 mol KMnO4 5 1.090 3 1022 mol FeSO4 The concentration of the FeSO4 solution in moles of FeSO4 per liter of solution is mol FeSO4 molarity of FeSO4 5 L soln 1.090 3 1022 mol FeSO4 1000 mL soln 5 3 25.00 mL soln 1 L soln 5 0.4360 M Similar problems: 4.95, 4.96. Practice Exercise How many milliliters of a 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according to the following equation: 10HI 1 2KMnO4 1 3H2SO4 ¡ 5I2 1 2MnSO4 1 K2SO4 1 8H2O The Chemistry in Action essay on p. 156 describes an industrial process that involves the types of reactions discussed in this chapter. Review of Concepts If a solution of a reducing agent is titrated with a solution of an oxidizing agent, and the initial concentrations of the two solutions are the same, does that mean that the equivalence point will be reached when an equal volume of oxidizing has been added? Explain. Key Equations moles of solute molarity 5 (4.1) Calculating molarity liters of solution n M5 (4.2) Calculating molarity V MiVi 5 MfVf (4.3) Dilution of solution 158 Chapter 4 ■ Reactions in Aqueous Solutions Summary of Facts & Concepts 1. Aqueous solutions are electrically conducting if the sol- 8. Many redox reactions can be subclassified as combina- utes are electrolytes. If the solutes are nonelectrolytes, tion, decomposition, combustion, displacement, or dis- the solutions do not conduct electricity. proportionation reactions. 2. Three major categories of chemical reactions that 9. The concentration of a solution is the amount of solute take place in aqueous solution are precipitation reac- present in a given amount of solution. Molarity ex- tions, acid-base reactions, and oxidation-reduction presses concentration as the number of moles of solute reactions. in 1 L of solution. 3. From general rules about solubilities of ionic com- 10. Adding a solvent to a solution, a process known as pounds, we can predict whether a precipitate will form dilution, decreases the concentration (molarity) of the in a reaction. solution without changing the total number of moles of 4. Arrhenius acids ionize in water to give H1 ions, and solute present in the solution. Arrhenius bases ionize in water to give OH2 ions. 11. Gravimetric analysis is a technique for determining the Brønsted acids donate protons, and Brønsted bases ac- identity of a compound and/or the concentration of a cept protons. solution by measuring mass. Gravimetric experiments 5. The reaction of an acid and a base is called neutralization. often involve precipitation reactions. 6. In redox reactions, oxidation and reduction always 12. In acid-base titration, a solution of known concentration occur simultaneously. Oxidation is characterized by (say, a base) is added gradually to a solution of un- the loss of electrons, reduction by the gain of known concentration (say, an acid) with the goal of de- electrons. termining the unknown concentration. The point at 7. Oxidation numbers help us keep track of charge distri- which the reaction in the titration is complete, as shown bution and are assigned to all atoms in a compound by the change in the indicator’s color, is called the or ion according to specific rules. Oxidation can be equivalence point. defined as an increase in oxidation number; reduc- 13. Redox titrations are similar to acid-base titrations. The tion can be defined as a decrease in oxidation point at which the oxidation-reduction reaction is com- number. plete is called the equivalence point. Key Words Activity series, p. 140 Electrolyte, p. 119 Net ionic equation, p. 124 Reducing agent, p. 134 Aqueous solution, p. 119 Equivalence point, p. 152 Neutralization reaction, p. 130 Reduction reaction, p. 133 Brønsted acid, p. 127 Gravimetric analysis, p. 149 Nonelectrolyte, p. 119 Reversible reaction, p. 121 Brønsted base, p. 127 Half-reaction, p. 133 Oxidation number, p. 135 Salt, p. 130 Combination reaction, p. 137 Hydration, p. 120 Oxidation reaction, p. 133 Solubility, p. 122 Combustion reaction, p. 139 Hydronium ion, p. 128 Oxidation-reduction Solute, p. 119 Concentration of a Indicator, p. 152 reaction, p. 132 Solution, p. 119 solution, p. 145 Ionic equation, p. 124 Oxidation state, p. 135 Solvent, p. 119 Decomposition reaction, p. 139 Metathesis reaction, p. 121 Oxidizing agent, p. 134 Spectator ion, p. 124 Dilution, p. 147 Molar concentration, p. 145 Precipitate, p. 121 Standard solution, p. 151 Diprotic acid, p. 128 Molarity (M), p. 145 Precipitation reaction, p. 121 Titration, p. 151 Displacement reaction, p. 139 Molecular equation, p. 123 Quantitative analysis, p. 149 Triprotic acid, p. 128 Disproportionation Monoprotic acid, p. 128 Redox reaction, p. 132 reaction, p. 142 Questions & Problems • Problems available in Connect Plus 4.2 What is the difference between a nonelectrolyte and Red numbered problems solved in Student Solutions Manual an electrolyte? Between a weak electrolyte and a strong electrolyte? Properties of Aqueous Solutions 4.3 Describe hydration. What properties of water enable Review Questions its molecules to interact with ions in solution? 4.1 Define solute, solvent, and solution by describing 4.4 What is the difference between the following sym- the process of dissolving a solid in a liquid. bols in chemical equations: ¡ and Δ ?  Questions & Problems 159 4.5 Water is an extremely weak electrolyte and there- electrolyte or a nonelectrolyte. If it is an electrolyte, fore cannot conduct electricity. Why are we often how would you determine whether it is strong cautioned not to operate electrical appliances when or weak? our hands are wet? 4.14 Explain why a solution of HCl in benzene does not • 4.6 Sodium sulfate (Na2SO4) is a strong electrolyte. conduct electricity but in water it does. What species are present in Na2SO4(aq)? Problems Precipitation Reactions Review Questions • 4.7 The aqueous solutions of three compounds are shown in the diagram. Identify each compound as a 4.15 What is the difference between an ionic equation nonelectrolyte, a weak electrolyte, and a strong and a molecular equation? electrolyte. 4.16 What is the advantage of writing net ionic equations? Problems • 4.17 Two aqueous solutions of AgNO3 and NaCl are mixed. Which of the following diagrams best rep- resents the mixture? For simplicity, water mole- cules are not shown. (Color codes are: Ag1 5 gray, (a) (b) (c) Cl2 5 orange, Na1 5 green, NO23 5 blue.) • 4.8 Which of the following diagrams best represents the hydration of NaCl when dissolved in water? The Cl2 ion is larger in size than the Na1 ion. (a) (b) (c) (d) • 4.18 Two aqueous solutions of KOH and MgCl2 are mixed. Which of the following diagrams best repre- sents the mixture? For simplicity, water molecules are not shown. (Color codes are: K1 5 purple, (a) (b) (c) OH2 5 red, Mg21 5 green, Cl2 5 orange.) • 4.9 Identify each of the following substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: (a) H2O, (b) KCl, (c) HNO3, (d) CH3COOH, (e) C12H22O11. • 4.10 Identify each of the following substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: (a) Ba(NO3)2, (b) Ne, (c) NH3, (d) NaOH. • 4.11 The passage of electricity through an electrolyte so- lution is caused by the movement of (a) electrons (a) (b) (c) (d) only, (b) cations only, (c) anions only, (d) both cat- ions and anions. 4.12 Predict and explain which of the following sys- • 4.19 Characterize the following compounds as soluble or tems are electrically conducting: (a) solid NaCl, insoluble in water: (a) Ca3(PO4)2, (b) Mn(OH)2, (b) molten NaCl, (c) an aqueous solution of (c) AgClO3, (d) K2S. NaCl. • 4.20 Characterize the following compounds as soluble 4.13 You are given a water-soluble compound X. or insoluble in water: (a) CaCO3, (b) ZnSO4, Describe how you would determine whether it is an (c) Hg(NO3)2, (d) HgSO4, (e) NH4ClO4.  160 Chapter 4 ■ Reactions in Aqueous Solutions • 4.21 Write ionic and net ionic equations for the following • 4.34 Balance the following equations and write the reactions: corresponding ionic and net ionic equations (if (a) AgNO3 (aq) 1 Na2SO4 (aq) ¡ appropriate): (b) BaCl2 (aq) 1 ZnSO4 (aq) ¡ (a) CH3COOH(aq) 1 KOH(aq) ¡ (c) (NH4 ) 2CO3 (aq) 1 CaCl2 (aq) ¡ (b) H2CO3 (aq) 1 NaOH(aq) ¡ • 4.22 Write ionic and net ionic equations for the following (c) HNO3 (aq) 1 Ba(OH) 2 (aq) ¡ reactions: (a) Na2S(aq) 1 ZnCl2 (aq) ¡ Oxidation-Reduction Reactions (b) K3PO4 (aq) 1 3Sr(NO3 ) 2 (aq) ¡ Review Questions (c) Mg(NO3 ) 2 (aq) 1 2NaOH(aq) ¡ 4.35 Give an example of a combination redox reaction, a • 4.23 Which of the following processes will likely result decomposition redox reaction, and a displacement in a precipitation reaction? (a) Mixing a NaNO3 so- redox reaction. lution with a CuSO4 solution. (b) Mixing a BaCl2 4.36 All combustion reactions are redox reactions. True solution with a K2SO4 solution. Write a net ionic or false? Explain. equation for the precipitation reaction. 4.37 What is an oxidation number? How is it used to 4.24 With reference to Table 4.2, suggest one method by identify redox reactions? Explain why, except for which you might separate (a) K1 from Ag1, (b) Ba21 ionic compounds, oxidation number does not have from Pb21, (c) NH1 21 4 from Ca , (d) Ba 21 from Cu21. any physical significance. All cations are assumed to be in aqueous solution, 4.38 (a) Without referring to Figure 4.11, give the oxi- and the common anion is the nitrate ion. dation numbers of the alkali and alkaline earth metals in their compounds. (b) Give the highest oxidation numbers that the Groups 3A–7A ele- Acid-Base Reactions ments can have. Review Questions 4.39 How is the activity series organized? How is it used 4.25 List the general properties of acids and bases. in the study of redox reactions? 4.26 Give Arrhenius’ and Brønsted’s definitions of an 4.40 Use the following reaction to define redox reaction, acid and a base. Why are Brønsted’s definitions half-reaction, oxidizing agent, reducing agent: more useful in describing acid-base properties? 4Na(s) 1 O2 (g) ¡ 2Na2O(s) 4.27 Give an example of a monoprotic acid, a diprotic 4.41 Is it possible to have a reaction in which oxidation acid, and a triprotic acid. occurs and reduction does not? Explain. 4.28 What are the characteristics of an acid-base neutral- 4.42 What is the requirement for an element to undergo ization reaction? disproportionation reactions? Name five common • 4.29 What factors qualify a compound as a salt? Specify elements that are likely to take part in such which of the following compounds are salts: CH4, reactions. NaF, NaOH, CaO, BaSO4, HNO3, NH3, KBr? • 4.30 Identify the following as a weak or strong acid or base: (a) NH3, (b) H3PO4, (c) LiOH, (d) HCOOH Problems (formic acid), (e) H2SO4, (f) HF, (g) Ba(OH)2. • 4.43 For the complete redox reactions given here, (i) break down each reaction into its half-reactions; Problems (ii) identify the oxidizing agent; (iii) identify the reducing agent. • 4.31 Identify each of the following species as a Brønsted (a) 2Sr 1 O2 ¡ 2SrO acid, base, or both: (a) HI, (b) CH3COO2, (b) 2Li 1 H2 ¡ 2LiH (c) H2PO2 2 4 , (d) HSO 4. (c) 2Cs 1 Br2 ¡ 2CsBr • 4.32 Identify each of the following species as a Brønsted (d) 3Mg 1 N2 ¡ Mg3N2 acid, base, or both: PO432, (b) ClO22, (c) NH14 , (d) HCO23 . • 4.44 For the complete redox reactions given here, write • 4.33 Balance the following equations and write the the half-reactions and identify the oxidizing and re- corresponding ionic and net ionic equations (if ducing agents. appropriate): (a) 4Fe 1 3O2 ¡ 2Fe2O3 (a) HBr(aq) 1 NH3 (aq) ¡ (b) Cl2 1 2NaBr ¡ 2NaCl 1 Br2 (b) Ba(OH) 2 (aq) 1 H3PO4 (aq) ¡ (c) Si 1 2F2 ¡ SiF4 (c) HClO4 (aq) 1 Mg(OH) 2 (s) ¡ (d) H2 1 Cl2 ¡ 2HCl  Questions & Problems 161 • 4.45 Arrange the following species in order of increas- 4.57 Which of the following are redox processes? ing oxidation number of the sulfur atom: (a) H2S, (a) CO2 ¡ CO22 3 (b) S8, (c) H2SO4, (d) S22, (e) HS2, (f) SO2, (b) VO3 ¡ VO2 (g) SO3. (c) SO3 ¡ SO22 4 • 4.46 Phosphorus forms many oxoacids. Indicate the (d) NO22 ¡ NO23 oxidation number of phosphorus in each of the fol- lowing acids: (a) HPO3, (b) H3PO2, (c) H3PO3, (e) Cr31 ¡ CrO22 4 (d) H3PO4, (e) H4P2O7, (f) H5P3O10. 4.58 Of the following, which is most likely to be the strongest oxidizing agent? O2, O12 , O22 , O22 2 . • 4.47 Give the oxidation number of the underlined atoms in the following molecules and ions: (a) ClF, (b) IF7, (c) CH4, (d) C2H2, (e) C2H4, (f) K2CrO4, (g) K2Cr2O7, Concentration of Solutions (h) KMnO4, (i) NaHCO3, (j) Li2, (k) NaIO3, (l) KO2, Review Questions (m) PF2 6 , (n) KAuCl4. 4.59 Write the equation for calculating molarity. Why is • 4.48 Give the oxidation number for the following spe- molarity a convenient concentration unit in cies: H2, Se8, P4, O, U, As4, B12. chemistry? • 4.49 Give oxidation numbers for the underlined atoms 4.60 Describe the steps involved in preparing a solution in the following molecules and ions: (a) Cs2O, of known molar concentration using a volumetric (b) CaI 2, (c) Al2O3, (d) H3AsO3, (e) TiO2, flask. (f) MoO22 22 22 4 , (g) PtCl4 , (h) PtCl6 , (i) SnF2, (j) ClF3, 2 (k) SbF6 . Problems • 4.50 Give the oxidation numbers of the underlined atoms • 4.61 Calculate the mass of KI in grams required to pre- in the following molecules and ions: (a) Mg3N2, pare 5.00 3 102 mL of a 2.80 M solution. (b) CsO2, (c) CaC2, (d) CO22 22 22 3 , (e) C2O4 , (f) ZnO2 , 4.62 Describe how you would prepare 250 mL of a 22 (g) NaBH4, (h) WO4 . 0.707 M NaNO3 solution. • 4.51 Nitric acid is a strong oxidizing agent. State which • 4.63 How many moles of MgCl2 are present in 60.0 mL of the following species is least likely to be pro- of 0.100 M MgCl2 solution? duced when nitric acid reacts with a strong reducing agent such as zinc metal, and explain why: N2O, • 4.64 How many grams of KOH are present in 35.0 mL of a 5.50 M solution? NO, NO2, N2O4, N2O5, NH1 4. • 4.52 Which of the following metals can react with water? • 4.65 Calculate the molarity of each of the following solu- tions: (a) 29.0 g of ethanol (C2H5OH) in 545 mL of (a) Au, (b) Li, (c) Hg, (d) Ca, (e) Pt. solution, (b) 15.4 g of sucrose (C12H22O11) in 74.0 mL • 4.53 On the basis of oxidation number considerations, of solution, (c) 9.00 g of sodium chloride (NaCl) in one of the following oxides would not react with 86.4 mL of solution. molecular oxygen: NO, N2O, SO2, SO3, P4O6. Which one is it? Why? • 4.66 Calculate the molarity of each of the following solutions: (a) 6.57 g of methanol (CH 3OH) in • 4.54 Predict the outcome of the reactions represented by 1.50 3 102 mL of solution, (b) 10.4 g of calcium the following equations by using the activity series, chloride (CaCl2) in 2.20 3 102 mL of solution, and balance the equations. (c) 7.82 g of naphthalene (C10H8) in 85.2 mL of (a) Cu(s) 1 HCl(aq) ¡ benzene solution. (b) I2 (s) 1 NaBr(aq) ¡ • 4.67 Calculate the volume in mL of a solution required to (c) Mg(s) 1 CuSO4 (aq) ¡ provide the following: (a) 2.14 g of sodium chloride from a 0.270 M solution, (b) 4.30 g of ethanol from (d) Cl2 (g) 1 KBr(aq) ¡ a 1.50 M solution, (c) 0.85 g of acetic acid • 4.55 Classify the following redox reactions: (CH3COOH) from a 0.30 M solution. (a) 2H2O2 ¡ 2H2O 1 O2 • 4.68 Determine how many grams of each of the follow- (b) Mg 1 2AgNO3 ¡ Mg(NO3 ) 2 1 2Ag ing solutes would be needed to make 2.50 3 102 mL (c) NH4NO2 ¡ N2 1 2H2O of a 0.100 M solution: (a) cesium iodide (CsI), (d) H2 1 Br2 ¡ 2HBr (b) sulfuric acid (H2SO4), (c) sodium carbonate (Na2CO3), (d) potassium dichromate (K2Cr2O7), • 4.56 Classify the following redox reactions: (e) potassium permanganate (KMnO4). (a) P4 1 10Cl2 ¡ 4PCl5 4.69 What volume of 0.416 M Mg(NO3)2 should be (b) 2NO ¡ N2 1 O2 added to 255 mL of 0.102 M KNO3 to produce a (c) Cl2 1 2KI ¡ 2KCl 1 I2 solution with a concentration of 0.278 M NO23 ions? (d) 3HNO2 ¡ HNO3 1 H2O 1 2NO Assume volumes are additive.  162 Chapter 4 ■ Reactions in Aqueous Solutions 4.70 Barium hydroxide, often used to titrate weak organic the percent by mass of Ba in the original unknown acids, is obtained as the octahydrate, Ba(OH)2 ? 8H2O. compound? What mass of Ba(OH)2 ? 8H2O would be required to • 4.83 How many grams of NaCl are required to precipitate make 500.0 mL of a solution that is 0.1500 M in most of the Ag1 ions from 2.50 3 102 mL of 0.0113 M hydroxide ions? AgNO3 solution? Write the net ionic equation for the reaction. Dilution of Solutions • 4.84 The concentration of sulfate in water can be deter- mined by adding a solution of barium chloride to Review Questions precipitate the sulfate ion. Write the net ionic equa- 4.71 Describe the basic steps involved in diluting a solu- tion for this reaction. Treating a 145-mL sample of tion of known concentration. water with excess BaCl2(aq) precipitated 0.330 g of 4.72 Write the equation that enables us to calculate the BaSO4. Determine the concentration of sulfate in concentration of a diluted solution. Give units for all the original water sample. the terms. Problems Acid-Base Titrations Review Questions • 4.73 Describe how to prepare 1.00 L of 0.646 M HCl solution, starting with a 2.00 M HCl solution. 4.85 Describe the basic steps involved in an acid-base • 4.74 Water is added to 25.0 mL of a 0.866 M KNO3 titration. Why is this technique of great practical solution until the volume of the solution is exactly value? 500 mL. What is the concentration of the final 4.86 How does an acid-base indicator work? solution? 4.87 A student carried out two titrations using a NaOH • 4.75 How would you prepare 60.0 mL of 0.200 M HNO3 solution of unknown concentration in the buret. from a stock solution of 4.00 M HNO3? In one titration she weighed out 0.2458 g of KHP (see p. 152) and transferred it to an Erlenmeyer • 4.76 You have 505 mL of a 0.125 M HCl solution and flask. She then added 20.00 mL of distilled water you want to dilute it to exactly 0.100 M. How much water should you add? Assume volumes are to dissolve the acid. In the other titration she additive. weighed out 0.2507 g of KHP but added 40.00 mL of distilled water to dissolve the acid. Assuming • 4.77 A 35.2-mL, 1.66 M KMnO4 solution is mixed with no experimental error, would she obtain the 16.7 mL of 0.892 M KMnO4 solution. Calculate the same result for the concentration of the NaOH concentration of the final solution. solution? • 4.78 A 46.2-mL, 0.568 M calcium nitrate [Ca(NO3)2] so- 4.88 Would the volume of a 0.10 M NaOH solution lution is mixed with 80.5 mL of 1.396 M calcium needed to titrate 25.0 mL of a 0.10 M HNO2 (a nitrate solution. Calculate the concentration of the weak acid) solution be different from that needed final solution. to titrate 25.0 mL of a 0.10 M HCl (a strong acid) solution? Gravimetric Analysis Review Questions Problems 4.79 Describe the basic steps involved in gravimetric • 4.89 A quantity of 18.68 mL of a KOH solution is needed analysis. How does this procedure help us determine to neutralize 0.4218 g of KHP. What is the concen- the identity of a compound or the purity of a com- tration (in molarity) of the KOH solution? pound if its formula is known? • 4.90 Calculate the concentration (in molarity) of a NaOH 4.80 Distilled water must be used in the gravimetric anal- solution if 25.0 mL of the solution are needed to ysis of chlorides. Why? neutralize 17.4 mL of a 0.312 M HCl solution. • 4.91 Calculate the volume in mL of a 1.420 M NaOH Problems solution required to titrate the following solutions: (a) 25.00 mL of a 2.430 M HCl solution • 4.81 If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of (b) 25.00 mL of a 4.500 M H2SO4 solution 0.100 M AgNO3, what is the mass in grams of AgCl precipitate? (c) 25.00 mL of a 1.500 M H3PO4 solution • 4.82 A sample of 0.6760 g of an unknown compound • 4.92 What volume of a 0.500 M HCl solution is needed to containing barium ions (Ba21) is dissolved in water neutralize each of the following: and treated with an excess of Na2SO4. If the mass of (a) 10.0 mL of a 0.300 M NaOH solution the BaSO4 precipitate formed is 0.4105 g, what is (b) 10.0 mL of a 0.200 M Ba(OH)2 solution  Questions & Problems 163 Redox Titrations 4.100 A 15.0-mL sample of an oxalic acid solution re- Review Questions quires 25.2 mL of 0.149 M NaOH for neutraliza- tion. Calculate the volume of a 0.122 M KMnO4 4.93 What are the similarities and differences between solution needed to react with a second 15.0-mL acid-base titrations and redox titrations? sample of the oxalic acid solution. (Hint: Oxalic 4.94 Explain why potassium permanganate (KMnO4) acid is a diprotic acid. See Problem 4.99 for redox and potassium dichromate (K2Cr2O7) can serve as equation.) internal indicators in redox titrations. • 4.101 Iodate ion, IO23 , oxidizes SO322 in acidic solution. The half-reaction for the oxidation is Problems SO22 22 1 2 3 1 H2O ¡ SO4 1 2H 1 2e • 4.95 Iron(II) can be oxidized by an acidic K2Cr2O7 solu- A 100.0-mL sample of solution containing 1.390 g tion according to the net ionic equation: of KIO3 reacts with 32.5 mL of 0.500 M Na2SO3. Cr2O22 7 1 6Fe 21 1 14H 1 ¡ What is the final oxidation state of the iodine after 2Cr31 1 6Fe31 1 7H2O the reaction has occurred? 4.102 Calcium oxalate (CaC2O4), the main component of If it takes 26.0 mL of 0.0250 M K2Cr2O7 to titrate kidney stones, is insoluble in water. For this reason 25.0 mL of a solution containing Fe21, what is the it can be used to determine the amount of Ca21 ions molar concentration of Fe21? in fluids such as blood. The calcium oxalate isolated 4.96 The SO2 present in air is mainly responsible for the from blood is dissolved in acid and titrated against a acid rain phenomenon. Its concentration can be de- standardized KMnO4 solution, as shown in Problem termined by titrating against a standard permanga- 4.99. In one test it is found that the calcium oxalate nate solution as follows: isolated from a 10.0-mL sample of blood requires 24.2 mL of 9.56 3 1024 M KMnO4 for titration. 5SO2 1 2MnO24 1 2H2O ¡ Calculate the number of milligrams of calcium per 5SO22 4 1 2Mn 21 1 4H 1 milliliter of blood. Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800 M KMnO4 solution are Additional Problems required for the titration. 4.103 Classify the following reactions according to the 4.97 A sample of iron ore (containing only Fe21 ions) types discussed in the chapter: weighing 0.2792 g was dissolved in dilute acid solu- tion, and all the Fe(II) was converted to Fe(III) ions. (a) Cl2 1 2OH 2 ¡ Cl 2 1 ClO 2 1 H2O The solution required 23.30 mL of 0.0194 M (b) Ca21 1 CO22 3 ¡ CaCO3 K2Cr2O7 for titration. Calculate the percent by mass (c) NH3 1 H 1 ¡ NH41 of iron in the ore. (Hint: See Problem 4.95 for the (d) 2CCl4 1 CrO22 4 ¡ balanced equation.) 2COCl2 1 CrO2Cl2 1 2Cl2 4.98 The concentration of a hydrogen peroxide solution (e) Ca 1 F2 ¡ CaF2 can be conveniently determined by titration against (f) 2Li 1 H2 ¡ 2LiH a standardized potassium permanganate solution in (g) Ba(NO3 ) 2 1 Na2SO4 ¡ 2NaNO3 1 BaSO4 an acidic medium according to the following equation: (h) CuO 1 H2 ¡ Cu 1 H2O (i) Zn 1 2HCl ¡ ZnCl2 1 H2 2MnO24 1 5H2O2 1 6H1 ¡ (j) 2FeCl2 1 Cl2 ¡ 2FeCl3 5O2 1 2Mn21 1 8H2O (k) LiOH 1 HNO3 ¡ LiNO3 1 H2O If 36.44 mL of a 0.01652 M KMnO4 solution are 4.104 Oxygen (O2) and carbon dioxide (CO2) are color- required to oxidize 25.00 mL of a H2O2 solution, less and odorless gases. Suggest two chemical tests calculate the molarity of the H2O2 solution. that would enable you to distinguish between these • 4.99 Oxalic acid (H2C2O4) is present in many plants two gases. and vegetables. If 24.0 mL of 0.0100 M KMnO4 • 4.105 Which of the following aqueous solutions would solution is needed to titrate 1.00 g of a sample of you expect to be the best conductor of electricity at H2C2O4 to the equivalence point, what is the per- 258C? Explain your answer. cent by mass of H2C2O4 in the sample? The net (a) 0.20 M NaCl ionic equation is (b) 0.60 M CH3COOH 2MnO24 1 16H1 1 5C2O22 4 ¡ (c) 0.25 M HCl 2Mn21 1 10CO2 1 8H2O (d) 0.20 M Mg(NO3)2  164 Chapter 4 ■ Reactions in Aqueous Solutions • 4.106 A 5.00 3 102-mL sample of 2.00 M HCl solution is • 4.112 Acetic acid (CH3COOH) is an important ingredient treated with 4.47 g of magnesium. Calculate the of vinegar. A sample of 50.0 mL of a commercial concentration of the acid solution after all the metal vinegar is titrated against a 1.00 M NaOH solution. has reacted. Assume that the volume remains un- What is the concentration (in M) of acetic acid pres- changed. ent in the vinegar if 5.75 mL of the base are needed 4.107 Shown here are two aqueous solutions containing for the titration? various ions. The volume of each solution is 200 mL. • 4.113 A 15.00-mL solution of potassium nitrate (KNO3) (a) Calculate the mass of the precipitate (in g) after was diluted to 125.0 mL, and 25.00 mL of this solu- the solutions are mixed. (b) What are the concentra- tion were then diluted to 1.000 3 103 mL. The con- tions (in M) of the ions in the final solution? Treat centration of the final solution is 0.00383 M. each sphere as 0.100 mol. Assume the volumes Calculate the concentration of the original solution. are additive. 4.114 When 2.50 g of a zinc strip were placed in a AgNO3 solution, silver metal formed on the surface of the strip. After some time had passed, the strip was re- moved from the solution, dried, and weighed. If the mass of the strip was 3.37 g, calculate the mass of Ag and Zn metals present. Ba21 • 4.115 Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L Cl2 of 0.0664 M Na2SO4. Na1 • 4.116 Calculate the concentration of the acid (or base) re- maining in solution when 10.7 mL of 0.211 M HNO3 SO22 4 are added to 16.3 mL of 0.258 M NaOH. 4.117 (a) Describe a preparation for magnesium hydroxide [Mg(OH)2] and predict its solubility. (b) Milk of 4.108 Shown here are two aqueous solutions containing magnesia contains mostly Mg(OH)2 and is effective various ions. The volume of each solution is 200 mL. in treating acid (mostly hydrochloric acid) indiges- (a) Calculate the mass of the precipitate (in g) after tion. Calculate the volume of a 0.035 M HCl solu- the solutions are mixed. (b) What are the concentra- tion (a typical acid concentration in an upset tions (in M) of the ions in the final solution? Treat stomach) needed to react with two spoonfuls (ap- each sphere as 0.100 mol. Assume the volumes are proximately 10 mL) of milk of magnesia [at 0.080 g additive. Mg(OH)2/mL]. • 4.118 A 1.00-g sample of a metal X (that is known to form X21 ions) was added to 0.100 L of 0.500 M H2SO4. After all the metal had reacted, the remaining acid required 0.0334 L of 0.500 M NaOH solution for neutralization. Calculate the molar mass of the metal and identify the element. Al31 4.119 Carbon dioxide in air can be removed by an aqueous NO2 3 metal hydroxide solution such as LiOH and K1 Ba(OH)2. (a) Write equations for the reactions. (Carbon dioxide reacts with water to form carbonic OH2 acid.) (b) Calculate the mass of CO2 that can be re- moved by 5.00 3 102 mL of a 0.800 M LiOH and a 0.800 M Ba(OH)2 solution. (c) Which solution would you choose for use in a space capsule and • 4.109 Calculate the volume of a 0.156 M CuSO4 solution which for use in a submarine? that would react with 7.89 g of zinc. 4.120 The molecular formula of malonic acid is C3H4O4. • 4.110 Sodium carbonate (Na2CO3) is available in very If a solution containing 0.762 g of the acid requires pure form and can be used to standardize acid solu- 12.44 mL of 1.174 M NaOH for neutralization, tions. What is the molarity of a HCl solution if 28.3 mL how many ionizable H atoms are present in the of the solution are required to react with 0.256 g of molecule? Na2CO3? 4.121 A quantitative definition of solubility is the maxi- • 4.111 A 3.664-g sample of a monoprotic acid was dis- mum number of grams of a solute that will dissolve solved in water. It took 20.27 mL of a 0.1578 M in a given volume of water at a particular tempera- NaOH solution to neutralize the acid. Calculate the ture. Describe an experiment that would enable you molar mass of the acid. to determine the solubility of a soluble compound.  Questions & Problems 165 • 4.122 A 60.0-mL 0.513 M glucose (C6H12O6) solution is (Na2SO4) to exactly 500 mL of the water. (a) Write mixed with 120.0 mL of 2.33 M glucose solution. the molecular and net ionic equations for the What is the concentration of the final solution? As- reaction. (b) Calculate the molar concentration of sume the volumes are additive. Pb21 if 0.00450 g of Na2SO4 was needed for the 4.123 An ionic compound X is only slightly soluble in complete precipitation of Pb21 ions as PbSO4. water. What test would you employ to show that the 4.132 Hydrochloric acid is not an oxidizing agent in the compound does indeed dissolve in water to a cer- sense that sulfuric acid and nitric acid are. Explain tain extent? why the chloride ion is not a strong oxidizing agent 4.124 A student is given an unknown that is either iron(II) like SO422 and NO32. sulfate or iron(III) sulfate. Suggest a chemical pro- 4.133 Explain how you would prepare potassium iodide cedure for determining its identity. (Both iron com- (KI) by means of (a) an acid-base reaction and pounds are water soluble.) (b) a reaction between an acid and a carbonate 4.125 You are given a colorless liquid. Describe three compound. chemical tests you would perform on the liquid to 4.134 Sodium reacts with water to yield hydrogen gas. show that it is water. Why is this reaction not used in the laboratory prep- 4.126 Using the apparatus shown in Figure 4.1, a stu- aration of hydrogen? dent found that a sulfuric acid solution caused 4.135 Describe how you would prepare the following the lightbulb to glow brightly. However, after the compounds: (a) Mg(OH)2, (b) AgI, (c) Ba3(PO4)2. addition of a certain amount of a barium hydrox- 4.136 Someone spilled concentrated sulfuric acid on the ide [Ba(OH)2] solution, the light began to dim floor of a chemistry laboratory. To neutralize the even though Ba(OH) 2 is also a strong electrolyte. acid, would it be preferable to pour concentrated Explain. sodium hydroxide solution or spray solid sodium 4.127 The molar mass of a certain metal carbonate, MCO3, bicarbonate over the acid? Explain your choice and can be determined by adding an excess of HCl acid the chemical basis for the action. to react with all the carbonate and then “back titrat- 4.137 Describe in each case how you would separate ing” the remaining acid with a NaOH solution. the cations or anions in an aqueous solution of: (a) Write an equation for these reactions. (b) In a (a) NaNO3 and Ba(NO3)2, (b) Mg(NO3)2 and KNO3, certain experiment, 20.00 mL of 0.0800 M HCl were (c) KBr and KNO3, (d) K3PO4 and KNO3, (e) Na2CO3 added to a 0.1022-g sample of MCO3. The excess and NaNO3. HCl required 5.64 mL of 0.1000 M NaOH for neu- 4.138 The following are common household com- tralization. Calculate the molar mass of the carbon- pounds: table salt (NaCl), table sugar (sucrose), ate and identify M. vinegar (contains acetic acid), baking soda 4.128 A 5.012-g sample of an iron chloride hydrate was (NaHCO3), washing soda (Na2CO3 ? 10H2O), bo- dried in an oven. The mass of the anhydrous com- ric acid (H3BO3, used in eyewash), epsom salt pound was 3.195 g. The compound was then (MgSO4 ? 7H2O), sodium hydroxide (used in dissolved in water and reacted with an excess of drain openers), ammonia, milk of magnesia AgNO 3. The AgCl precipitate formed weighed [Mg(OH)2], and calcium carbonate. Based on 7.225 g. What is the formula of the original what you have learned in this chapter, describe compound? test(s) that would enable you to identify each of 4.129 You are given a soluble compound of unknown these compounds. molecular formula. (a) Describe three tests that 4.139 Sulfites (compounds containing the SO22 3 ions) are would show that the compound is an acid. used as preservatives in dried fruit and vegetables (b) Once you have established that the compound and in wine making. In an experiment to test the is an acid, describe how you would determine its presence of sulfite in fruit, a student first soaked sev- molar mass using a NaOH solution of known con- eral dried apricots in water overnight and then fil- centration. (Assume the acid is monoprotic.) tered the solution to remove all solid particles. She (c) How would you find out whether the acid is then treated the solution with hydrogen peroxide weak or strong? You are provided with a sample of (H2O2) to oxidize the sulfite ions to sulfate ions. NaCl and an apparatus like that shown in Figure 4.1 Finally, the sulfate ions were precipitated by treating for comparison. the solution with a few drops of a barium chloride 4.130 You are given two colorless solutions, one contain- (BaCl2) solution. Write a balanced equation for each ing NaCl and the other sucrose (C12H22O11). Suggest of the preceding steps. a chemical and a physical test that would allow you • 4.140 A 0.8870-g sample of a mixture of NaCl and KCl is to distinguish between these two solutions. dissolved in water, and the solution is then treated • 4.131 The concentration of lead ions (Pb21) in a sample of with an excess of AgNO3 to yield 1.913 g of AgCl. polluted water that also contains nitrate ions (NO23 ) Calculate the percent by mass of each compound in is determined by adding solid sodium sulfate the mixture.  166 Chapter 4 ■ Reactions in Aqueous Solutions 4.141 Based on oxidation number consideration, explain • 4.148 A 325-mL sample of solution contains 25.3 g of why carbon monoxide (CO) is flammable but car- CaCl2. (a) Calculate the molar concentration of Cl2 bon dioxide (CO2) is not. in this solution. (b) How many grams of Cl2 are in 4.142 Which of the diagrams shown here corresponds to 0.100 L of this solution? the reaction between AgOH(s) and HNO3(aq)? • 4.149 Phosphoric acid (H3PO4) is an important indus- Write a balanced equation for the reaction. The trial chemical used in fertilizers, in detergents, green spheres represent the Ag1 ions and the red and in the food industry. It is produced by two spheres represent the NO23 ions. different methods. In the electric furnace method, elemental phosphorus (P 4) is burned in air to form P4O10, which is then reacted with water to give H3PO4. In the wet process, the mineral phos- phate rock fluorapatite [Ca5(PO4) 3F] is reacted with sulfuric acid to give H 3PO 4 (and HF and CaSO4). Write equations for these processes and classify each step as precipitation, acid-base, or redox reaction. • 4.150 Ammonium nitrate (NH4NO3) is one of the most important nitrogen-containing fertilizers. Its pu- (a) (b) (c) rity can be analyzed by titrating a solution of NH4NO3 with a standard NaOH solution. In one • 4.143 Chlorine forms a number of oxides with the follow- experiment a 0.2041-g sample of industrially pre- ing oxidation numbers: 11, 13, 14, 16, and 17. pared NH4NO3 required 24.42 mL of 0.1023 M Write a formula for each of these compounds. NaOH for neutralization. • 4.144 A useful application of oxalic acid is the removal of (a) Write a net ionic equation for the reaction. rust (Fe2O3) from, say, bathtub rings according to (b) What is the percent purity of the sample? the reaction 4.151 Is the following reaction a redox reaction? Explain. Fe2O3 (s) 1 6H2C2O4 (aq) ¡ 3O2 (g) ¡ 2O3 (g) 2Fe(C2O4 ) 32 1 3 (aq) 1 3H2O 1 6H (aq) Calculate the number of grams of rust that can be 4.152 What is the oxidation number of O in HFO? removed by 5.00 3 102 mL of a 0.100 M solution of 4.153 Use molecular models like those in Figures 4.7 and oxalic acid. 4.8 to represent the following acid-base reactions: • 4.145 Acetylsalicylic acid (C9H8O4) is a monoprotic (a) OH 2 1 H3O 1 ¡ 2H2O acid commonly known as “aspirin.” A typical as- (b) NH41 1 NH22 ¡ 2NH3 pirin tablet, however, contains only a small Identify the Brønsted acid and base in each case. amount of the acid. In an experiment to determine 4.154 The alcohol content in a 10.0-g sample of blood its composition, an aspirin tablet was crushed and from a driver required 4.23 mL of 0.07654 M dissolved in water. It took 12.25 mL of 0.1466 M K2Cr2O7 for titration. Should the police prosecute NaOH to neutralize the solution. Calculate the the individual for drunken driving? (Hint: See the number of grains of aspirin in the tablet. (One Chemistry in Action essay on p. 144.) grain 5 0.0648 g.) 4.155 On standing, a concentrated nitric acid gradually 4.146 A 0.9157-g mixture of CaBr2 and NaBr is dissolved turns yellow in color. Explain. (Hint: Nitric in water, and AgNO3 is added to the solution to form acid slowly decomposes. Nitrogen dioxide is a AgBr precipitate. If the mass of the precipitate is colored gas.) 1.6930 g, what is the percent by mass of NaBr in the original mixture? 4.156 Describe the laboratory preparation for the follow- ing gases: (a) hydrogen, (b) oxygen, (c) carbon di- 4.147 Hydrogen halides (HF, HCl, HBr, HI) are highly reac- oxide, and (d) nitrogen. Indicate the physical states tive compounds that have many industrial and labora- of the reactants and products in each case. [Hint: tory uses. (a) In the laboratory, HF and HCl can be Nitrogen can be obtained by heating ammonium ni- generated by reacting CaF2 and NaCl with concen- trite (NH4NO2).] trated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) 4.157 Referring to Figure 4.18, explain why one must first (b) Why is it that HBr and HI cannot be prepared dissolve the solid completely before making up the similarly, that is, by reacting NaBr and NaI with con- solution to the correct volume. centrated sulfuric acid? (Hint: H2SO4 is a stronger 4.158 Can the following decomposition reaction be char- oxidizing agent than both Br2 and I2.) (c) HBr can be acterized as an acid-base reaction? Explain. prepared by reacting phosphorus tribromide (PBr3) with water. Write an equation for this reaction. NH4Cl(s) ¡ NH3 (g) 1 HCl(g)  Questions & Problems 167 4.159 Give a chemical explanation for each of the fol- 0.0200 M KMnO4 (in dilute sulfuric acid). As a re- lowing: (a) When calcium metal is added to a sul- sult, all of the Fe21 ions are oxidized to Fe31 ions. furic acid solution, hydrogen gas is generated. Next, the solution is treated with Zn metal to convert After a few minutes, the reaction slows down and all of the Fe31 ions to Fe21 ions. Finally, the solution eventually stops even though none of the reactants containing only the Fe21 ions requires 40.0 mL of is used up. Explain. (b) In the activity series, alu- the same KMnO4 solution for oxidation to Fe31. minum is above hydrogen, yet the metal appears to Calculate the molar concentrations of Fe21 and Fe31 be unreactive toward steam and hydrochloric acid. in the original solution. The net ionic equation is Why? (c) Sodium and potassium lie above copper in the activity series. Explain why Cu21 ions in a MnO42 1 5Fe21 1 8H 1 ¡ CuSO4 solution are not converted to metallic cop- Mn21 1 5Fe31 1 4H2O per upon the addition of these metals. (d) A metal 4.163 Use the periodic table framework shown to show the M reacts slowly with steam. There is no visible names and positions of two metals that can change when it is placed in a pale green iron(II) (a) displace hydrogen from cold water, (b) displace sulfate solution. Where should we place M in the hydrogen from steam, and (c) displace hydrogen activity series? (e) Before aluminum metal was from acid. Also show two metals that can react nei- obtained by electrolysis, it was produced by re- ther with water nor acid. ducing its chloride (AlCl3) with an active metal. What metals would you use to produce aluminum in that way? • 4.160 The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of a stock solution. Instead, it is done by a series of dilutions. A sample of 0.8214 g of KMnO4 was dissolved in water and made up to the volume in a 500-mL volumetric flask. A 2.000-mL sample of this solution was transferred to a 1000-mL volumetric flask and di- 4.164 Referring to the Chemistry in Action essay on page luted to the mark with water. Next, 10.00 mL of the 156, answer the following questions: (a) Identify the diluted solution were transferred to a 250-mL flask precipitation, acid-base, and redox processes. (b) In- and diluted to the mark with water. (a) Calculate the stead of calcium oxide, why don’t we simply add concentration (in molarity) of the final solution. sodium hydroxide to seawater to precipitate magne- (b) Calculate the mass of KMnO4 needed to directly sium hydroxide? (c) Sometimes a mineral called prepare the final solution. dolomite (a mixture of CaCO3 and MgCO3) is sub- • 4.161 The following “cycle of copper” experiment is per- stituted for limestone to bring about the precipita- formed in some general chemistry laboratories. The tion of magnesium hydroxide. What is the advantage series of reactions starts with copper and ends with of using dolomite? metallic copper. The steps are as follows: (1) A piece • 4.165 A 22.02-mL solution containing 1.615 g Mg(NO3)2 of copper wire of known mass is allowed to react is mixed with a 28.64-mL solution containing 1.073 g with concentrated nitric acid [the products are NaOH. Calculate the concentrations of the ions copper(II) nitrate, nitrogen dioxide, and water]. remaining in solution after the reaction is complete. (2) The copper(II) nitrate is treated with a sodium Assume volumes are additive. hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide de- • 4.166 Chemical tests of four metals A, B, C, and D show the following results. composes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfu- (a) Only B and C react with 0.5 M HCl to give ric acid to yield copper(II) sulfate. (5) Copper(II) H2 gas. sulfate is treated with an excess of zinc metal to (b) When B is added to a solution containing the form metallic copper. (6) The remaining zinc metal ions of the other metals, metallic A, C, and D is removed by treatment with hydrochloric acid, and are formed. metallic copper is filtered, dried, and weighed. (c) A reacts with 6 M HNO3 but D does not. (a) Write a balanced equation for each step and clas- Arrange the metals in the increasing order as reducing sify the reactions. (b) Assuming that a student started agents. Suggest four metals that fit these descriptions. with 65.6 g of copper, calculate the theoretical yield 4.167 The antibiotic gramicidin A can transport Na1 ions at each step. (c) Considering the nature of the steps, into a certain cell at the rate of 5.0 3 107 Na1 ions comment on why it is possible to recover most of the s21. Calculate the time in seconds to transport copper used at the start. enough Na1 ions to increase its concentration by 4.162 A quantity of 25.0 mL of a solution containing both 8.0 3 1023 M in a cell whose intracellular volume is Fe21 and Fe31 ions is titrated with 23.0 mL of 2.0 3 10210 mL. 168 Chapter 4 ■ Reactions in Aqueous Solutions 4.168 Shown here are two aqueous solutions containing various ions. The volume of each solution is 600 mL. (a) Write a net ionic equation for the reac- tion after the solutions are mixed. (b) Calculate the mass of the precipitates formed and the concentra- Cu21 tions of the ions in the mixed solution. Treat each SO22 4 sphere as 0.0500 mol. Ba21 OH2 Interpreting, Modeling & Estimating 4.169 Many proteins contain metal ions for structural and/or redox functions. Which of the following metals fit into one or both categories: Ca, Cu, Fe, Mg, Mn, Ni, Zn? 4.170 The fastest way to introduce therapeutic agents into the bloodstream is by direct delivery into a vein (in- travenous therapy, or IV therapy). A clinical re- searcher wishes to establish an initial concentration of 6 3 1024 mmol/L in the bloodstream of an adult male participating in a trial study of a new drug. The drug serum is prepared in the hospital’s pharmacy at a concentration of 1.2 3 1023 mol/L. How much of the serum should be introduced intravenously in or- der to achieve the desired initial blood concentration of the drug? 4.171 Public water supplies are often “fluoridated” by the addition of compounds such as NaF, H2SiF6, and Na2SiF6. It is well established that fluoride helps prevent tooth decay; however, care must be taken not to exceed safe levels of fluoride, which can stain or etch tooth enamel (dental fluorosis). A safe and effective concentration of fluoride in 4.173 Muriatic acid, a commercial-grade hydrochloric drinking water is generally considered to be acid used for cleaning masonry surfaces, is typically around 1 mg/L. How much fluoride would a per- around 10 percent HCl by mass and has a density of son consume by drinking fluoridated water in 1.2 g/cm3. A 0.5-in layer of boiler scale has accumu- 1  year? What would be the equivalent mass as lated on a 6.0-ft section of hot water pipe with an sodium fluoride? internal diameter of 2.0 in (see the Chemistry in Ac- tion essay on p. 126). What is the minimum volume 4.172 Potassium superoxide (KO2), a useful source of of muriatic acid in gallons that would be needed to oxygen employed in breathing equipment, reacts remove the boiler scale? with water to form potassium hydroxide, hydro- gen peroxide, and oxygen. Furthermore, potas- • 4.174 Because acid-base and precipitation reactions dis- sium superoxide also reacts with carbon dioxide to cussed in this chapter all involve ionic species, their form potassium carbonate and oxygen. (a) Write progress can be monitored by measuring the electri- equations for these two reactions and comment on cal conductance of the solution. Match the following the effectiveness of potassium superoxide in this reactions with the diagrams shown here. The electri- application. (b) Focusing only on the reaction be- cal conductance is shown in arbitrary units. tween KO2 and CO2, estimate the amount of KO2 (1) A 1.0 M KOH solution is added to 1.0 L of 1.0 M needed to sustain a worker in a polluted environ- CH3COOH. ment for 30 min. See Problem 1.69 for useful (2) A 1.0 M NaOH solution is added to 1.0 L of information. 1.0 M HCl.  Answers to Practice Exercises 169 (3) A 1.0 M BaCl2 solution is added to 1.0 L of (5) A 1.0 M CH3COOH solution is added to 1.0 L of 1.0 M K2SO4. 1.0 M NH3. (4) A 1.0 M NaCl solution is added to 1.0 L of 1.0 M AgNO3. 4 Electrical conductance 3 2 1 1.0 2.0 1.0 2.0 1.0 2.0 1.0 2.0 Volume (L) Volume (L) Volume (L) Volume (L ) (a) (b) (c) (d) Answers to Practice Exercises 4.1 (a) Insoluble, (b) insoluble, (c) soluble. 4.2 Al31(aq) 1 (b) Mn: 17, O: 22. 4.6 (a) Hydrogen displacement 3OH2(aq) ¡ Al(OH)3(s). 4.3 (a) Brønsted base, reaction, (b) combination reaction, (c) disproportionation (b) Brønsted acid. 4.4 Molecular equation: H3PO4(aq) 1 reaction, (d) metal displacement reaction. 4.7 0.452 M. 3NaOH(aq) ¡ Na3PO4(aq) 1 3H2O(l); ionic equation: 4.8 494 mL. 4.9 Dilute 34.2 mL of the stock solution to H3PO4(aq) 1 3Na1(aq) 1 3OH2(aq) ¡ 3Na1(aq) 1 200 mL. 4.10 92.02%. 4.11 0.3822 g. 4.12 1.27 M. PO432(aq) 1 3H2O(l); net ionic equation: H3PO4(aq) 1 4.13 204 mL. 3OH2(aq) ¡ PO432(aq) 1 3H2O(l). 4.5 (a) P: 13, F: 21; CHEMICAL M YS TERY Who Killed Napoleon? A fter his defeat at Waterloo in 1815, Napoleon was exiled to St. Helena, a small island in the Atlantic Ocean, where he spent the last 6 years of his life. In the 1960s, samples of his hair were analyzed and found to contain a high level of arsenic, suggesting that he might have been poisoned. The prime suspects are the governor of St. Helena, with whom Napoleon did not get along, and the French royal family, who wanted to prevent his return to France. Elemental arsenic is not that harmful. The commonly used poison is actually arsenic(III) oxide, As2O3, a white compound that dissolves in water, is tasteless, and if administered over a period of time, is hard to detect. It was once known as the “inheritance powder” because it could be added to grandfather’s wine to hasten his demise so that his grandson could inherit the estate! In 1832 the English chemist James Marsh devised a procedure for detecting arsenic. This test, which now bears Marsh’s name, combines hydrogen formed by the reaction between zinc and sulfuric acid with a sample of the suspected poison. If As2O3 is present, it reacts with hydrogen to form a toxic gas, arsine (AsH3). When arsine gas is heated, it decomposes to form arsenic, which is recognized by its metallic luster. The Marsh test is an effective deterrent to murder by As2O3, but it was invented too late to do Napoleon any good, if, in fact, he was a victim of deliberate arsenic poisoning. Apparatus for Marsh’s test. Sulfuric acid is added to zinc metal and a solution containing arsenic(III) oxide. The H2SO4 Hydrogen flame hydrogen produced reacts with As2O3 to yield arsine (AsH3 ). On heating, arsine decomposes to elemental arsenic, which has a metallic appearance, and hydrogen gas. Shiny metallic ring As2O3 solution Zinc granules 170  Doubts about the conspiracy theory of Napoleon’s death developed in the early 1990s, when a sample of the wallpaper from his drawing room was found to contain copper arsenate (CuHAsO4), a green pigment that was commonly used at the time Napoleon lived. It has been suggested that the damp climate on St. Helena promoted the growth of molds on the wallpaper. To rid themselves of arsenic, the molds could have converted it to tri- methyl arsine [(CH3)3As], which is a volatile and highly poisonous compound. Prolonged exposure to these vapors would have ruined Napoleon’s health and would also account for the presence of arsenic in his body, though it may not have been the primary cause of his death. This provocative theory is supported by the fact that Napoleon’s regular guests suf- fered from gastrointestinal disturbances and other symptoms of arsenic poisoning and that their health all seemed to improve whenever they spent hours working outdoors in the garden, their main hobby on the island. We will probably never know whether Napoleon died from arsenic poisoning, inten- tional or accidental, but this exercise in historical sleuthing provides a fascinating example of the use of chemical analysis. Not only is chemical analysis used in forensic science, but it also plays an essential part in endeavors ranging from pure research to practical applications, such as quality control of commercial products and medical diagnosis. Chemical Clues 1. The arsenic in Napoleon’s hair was detected using a technique called neutron activa- A lock of Napoleon’s hair. tion. When As-75 is bombarded with high-energy neutrons, it is converted to the radioactive As-76 isotope. The energy of the g rays emitted by the radioactive isotope is characteristic of arsenic, and the intensity of the rays establishes how much arsenic is present in a sample. With this technique, as little as 5 ng (5 3 1029 g) of arsenic can be detected in 1 g of material. (a) Write symbols for the two isotopes of As, showing mass number and atomic number. (b) Name two advantages of analyzing the arsenic content by neutron activation instead of a chemical analysis. 2. Arsenic is not an essential element for the human body. (a) Based on its position in the periodic table, suggest a reason for its toxicity. (b) In addition to hair, where else might one look for the accumulation of the element if arsenic poison- ing is suspected? 3. The Marsh test for arsenic involves the following steps: (a) The generation of hydrogen gas when sulfuric acid is added to zinc. (b) The reaction of hydrogen with As(III) oxide to produce arsine. (c) Conversion of arsine to arsenic by heat- ing. Write equations representing these steps and identify the type of the reaction in each step. 171  CHAPTER 5 Gases Water vapor and methane have recently been detected in significant amounts in the Martian atmosphere. (The concentration increases from purple to red.) The methane could be released by geothermal activity, or it could be produced by bacteria, fueling speculation that there may be life on Mars. CHAPTER OUTLINE A LOOK AHEAD 5.1 Substances That Exist  We begin by examining the substances that exist as gases and their general as Gases properties. (5.1) 5.2 Pressure of a Gas  We learn units for expressing gas pressure and the characteristics of atmo- spheric pressure. (5.2) 5.3 The Gas Laws  Next, we study the relationship among pressure, volume, temperature, and 5.4 The Ideal Gas Equation amount of a gas in terms of various gas laws. We will see that these laws can 5.5 Gas Stoichiometry be summarized by the ideal gas equation, which can be used to calculate the density or molar mass of a gas. (5.3 and 5.4) 5.6 Dalton’s Law of Partial  We will see that the ideal gas equation can be used to study the stoichiometry Pressures involving gases. (5.5) 5.7 The Kinetic Molecular Theory  We learn that the behavior of a mixture of gases can be understood by of Gases Dalton’s law of partial pressures, which is an extension of the ideal gas 5.8 Deviation from Ideal Behavior equation. (5.6)  We will see how the kinetic molecular theory of gases, which is based on the properties of individual molecules, can be used to describe macroscopic properties such as the pressure and temperature of a gas. We learn that this theory enables us to obtain an expression for the speed of molecules at a given temperature, and understand phenomena such as gas diffusion and effusion. (5.7)  Finally, we will study the correction for the nonideal behavior of gases using the van der Waals equation. (5.8) 172  5.1 Substances That Exist as Gases 173 U nder certain conditions of pressure and temperature, most substances can exist in any one of three states of matter: solid, liquid, or gas. Water, for example, can be solid ice, liquid water, steam, or water vapor. The physical properties of a substance often depend on its state. Gases, the subject of this chapter, are simpler than liquids and solids in many ways. Molecular motion in gases is totally random, and the forces of attraction between gas molecules are so small that each molecule moves freely and essentially independently of other molecules. Subjected to changes in temperature and pressure, it is easier to predict the behavior of gases. The laws that govern this behavior have played an important role in the development of the atomic theory of matter and the kinetic molecular theory of gases. 5.1 Substances That Exist as Gases We live at the bottom of an ocean of air whose composition by volume is roughly 78  percent N2, 21 percent O2, and 1 percent other gases, including CO2. Today, the chemistry of this vital mixture of gases has become a source of great interest because of the detrimental effects of environmental pollution. The chemistry of the atmosphere and polluting gases is discussed in Chapter 20. Here we will focus generally on the behavior of substances that exist as gases under normal atmospheric conditions, which are defined as 25°C and 1 atmosphere (atm) pressure. Figure 5.1 shows the elements that are gases under normal atmospheric condi- tions. Note that hydrogen, nitrogen, oxygen, fluorine, and chlorine exist as gaseous diatomic molecules: H2, N2, O2, F2, and Cl2. An allotrope of oxygen, ozone (O3), is also a gas at room temperature. All the elements in Group 8A, the noble gases, are monatomic gases: He, Ne, Ar, Kr, Xe, and Rn. Ionic compounds do not exist as gases at 25°C and 1 atm, because cations and anions in an ionic solid are held together by very strong electrostatic forces; that is, forces between positive and negative charges. To overcome these attractions we must apply a large amount of energy, which in practice means strongly heating the solid. Under normal conditions, all we can do is melt the solid; for example, NaCl melts at the rather high temperature of 801°C. In order to boil it, we would have to raise the temperature to well above 1000°C. 1A 8A H He 2A 3A 4A 5A 6A 7A Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3B 4B 5B 6B 7B 8B 1B 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv Figure 5.1 Elements that exist as gases at 25°C and 1 atm. The noble gases (the Group 8A elements) are monatomic species; the other elements exist as diatomic molecules. Ozone (O3 ) is also a gas. 174 Chapter 5 ■ Gases Table 5.1 Some Substances Found as Gases at 1 atm and 25°C Elements Although “gas” and “vapor” are often used H2 (molecular hydrogen) N2 (molecular nitrogen) interchangeably, there is a difference. A gas is a substance that is normally in the O2 (molecular oxygen) O3 (ozone) gaseous state at ordinary temperatures F2 (molecular fluorine) Cl2 (molecular chlorine) and pressures; a vapor is the gaseous form of any substance that is a liquid or a solid He (helium) Ne (neon) at normal temperatures and pressures. Thus, at 258C and 1 atm pressure, we Ar (argon) Kr (krypton) speak of water vapor and oxygen gas. Xe (xenon) Rn (radon) Compounds HF (hydrogen fluoride) HCl (hydrogen chloride) HBr (hydrogen bromide) HI (hydrogen iodide) CO (carbon monoxide) CO2 (carbon dioxide) CH4 (methane) C2H2 (acetylene) NH3 (ammonia) NO (nitric oxide) NO2 (nitrogen dioxide) N2O (nitrous oxide) SO2 (sulfur dioxide) SF6 (sulfur hexafluoride) H2S (hydrogen sulfide) HCN (hydrogen cyanide)* *The boiling point of HCN is 26°C, but it is close enough to qualify as a gas at ordinary atmospheric conditions. The behavior of molecular compounds is more varied. Some—for example, CO, CO2, HCl, NH3, and CH4 (methane)—are gases, but the majority of molecular com- pounds are liquids or solids at room temperature. However, on heating they are con- verted to gases much more easily than ionic compounds. In other words, molecular compounds usually boil at much lower temperatures than ionic compounds do. There is no simple rule to help us determine whether a certain molecular compound is a gas under normal atmospheric conditions. To make such a determination we need to understand the nature and magnitude of the attractive forces among the molecules, called intermolecular forces (discussed in Chapter 11). In general, the stronger these attractions, the less likely a compound can exist as a gas at ordinary temperatures. Of the gases listed in Table 5.1, only O2 is essential for our survival. Hydrogen sulfide (H2S) and hydrogen cyanide (HCN) are deadly poisons. Several others, such as CO, NO2, O3, and SO2, are somewhat less toxic. The gases He, Ne, and Ar are chemically inert; that is, they do not react with any other substance. Most gases are colorless. Exceptions are F2, Cl2, and NO2. The dark-brown color of NO2 is some- times visible in polluted air. All gases have the following physical characteristics: • Gases assume the volume and shape of their containers. • Gases are the most compressible of the states of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids. 5.2 Pressure of a Gas NO2 gas. Gases exert pressure on any surface with which they come in contact, because gas molecules are constantly in motion. We humans have adapted so well physiologically to the pressure of the air around us that we are usually unaware of it, perhaps as fish are not conscious of the water’s pressure on them. It is easy to demonstrate atmospheric pressure. One everyday example is the ability to drink a liquid through a straw. Sucking air out of the straw reduces the pressure inside the straw. The greater atmospheric pressure on the liquid pushes it up into the straw to replace the air that has been sucked out.  5.2 Pressure of a Gas 175 SI Units of Pressure Pressure is one of the most readily measurable properties of a gas. In order to under- stand how we measure the pressure of a gas, it is helpful to know how the units of measurement are derived. We begin with velocity and acceleration. Velocity is defined as the change in distance with elapsed time; that is, distance moved velocity 5 elapsed time The SI unit for velocity is m/s, although we also use cm/s. Acceleration is the change in velocity with time, or change in velocity acceleration 5 elapsed time Acceleration is measured in m/s2 (or cm/s2). The second law of motion, formulated by Sir Isaac Newton† in the late seven- teenth century, defines another term, from which the units of pressure are derived, namely, force. According to this law, force 5 mass 3 acceleration In this context, the SI unit of force is the newton (N), where 1 N is roughly equivalent to the force exerted by Earth’s gravity on an apple. 1 N 5 1 kg m/s2 Finally, we define pressure as force applied per unit area: force pressure 5 area The SI unit of pressure is the pascal (Pa),‡ defined as one newton per square meter: 1 Pa 5 1 N/m2 Atmospheric Pressure The atoms and molecules of the gases in the atmosphere, like those of all other matter, are subject to Earth’s gravitational pull. As a consequence, the atmosphere is much denser near the surface of Earth than at high altitudes. (The air outside the pressurized cabin of an airplane at 9 km is too thin to breathe.) In fact, the density of air decreases very rapidly with increasing distance from Earth. Measurements show that about 50 percent of the atmosphere lies within 6.4 km of Earth’s surface, 90 percent within 16 km, and 99 percent within 32 km. Not surprisingly, the denser the air is, the greater the pressure it exerts. The force experienced by any area exposed to Earth’s atmosphere is equal to the weight of the column of air above it. Atmospheric pressure is the pressure exerted by Earth’s atmosphere (Figure 5.2). The actual value of atmospheric pressure depends on location, temperature, and Column of air weather conditions. † Sir Isaac Newton (1642–1726). English mathematician, physicist, and astronomer. Newton is regarded by many as one of the two greatest physicists the world has known (the other is Albert Einstein). There was hardly a branch of physics to which Newton did not make a significant contribution. His book Principia, published in 1687, marks a milestone in the history of science. ‡ Blaise Pascal (1623–1662). French mathematician and physicist. Pascal’s work ranged widely in mathemat- Figure 5.2 A column of air ics and physics, but his specialty was in the area of hydrodynamics (the study of the motion of fluids). He extending from sea level to the also invented a calculating machine. upper atmosphere. 176 Chapter 5 ■ Gases Does atmospheric pressure act only downward, as you might infer from its defi- nition? Imagine what would happen, then, if you were to hold a piece of paper tight (with both hands) above your head. You might expect the paper to bend due to the pressure of air acting on it, but this does not happen. The reason is that air, like water, is a fluid. The pressure exerted on an object in a fluid comes from all directions— downward and upward, as well as from the left and from the right. At the molecular level, air pressure results from collisions between the air molecules and any surface 76 cm with which they come in contact. The magnitude of pressure depends on how often and how strongly the molecules impact the surface. It turns out that there are just as Atmospheric pressure many molecules hitting the paper from the top as there are from underneath, so the paper stays flat. How is atmospheric pressure measured? The barometer is probably the most familiar instrument for measuring atmospheric pressure. A simple barometer con- sists of a long glass tube, closed at one end and filled with mercury. If the tube is carefully inverted in a dish of mercury so that no air enters the tube, some mercury will flow out of the tube into the dish, creating a vacuum at the top Figure 5.3 A barometer for (Figure 5.3). The weight of the mercury remaining in the tube is supported by measuring atmospheric pressure. Above the mercury in the tube is a atmospheric pressure acting on the surface of the mercury in the dish. Standard vacuum. The column of mercury is atmospheric pressure (1 atm) is equal to the pressure that supports a column of supported by the atmospheric mercury exactly 760 mm (or 76 cm) high at 0°C at sea level. In other words, the pressure. standard atmosphere equals a pressure of 760 mmHg, where mmHg represents the pressure exerted by a column of mercury 1 mm high. The mmHg unit is also called the torr, after the Italian scientist Evangelista Torricelli,† who invented the barometer. Thus, 1 torr 5 1 mmHg and 1 atm 5 760 mmHg   (exactly) The relation between atmospheres and pascals (see Appendix 2) is 1 atm 5 101,325 Pa 5 1.01325 3 105 Pa and because 1000 Pa 5 1 kPa (kilopascal) 1 atm 5 1.01325 3 102 kPa Examples 5.1 and 5.2 show the conversion from mmHg to atm and kPa. Example 5.1 The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized to protect the passengers. What is the pressure in atmospheres in the cabin if the barometer reading is 688 mmHg? Strategy Because 1 atm 5 760 mmHg, the following conversion factor is needed to obtain the pressure in atmospheres: 1 atm 760 mmHg (Continued) † Evangelista Torricelli (1608–1674). Italian mathematician. Torricelli was supposedly the first person to recognize the existence of atmospheric pressure.  5.2 Pressure of a Gas 177 Solution The pressure in the cabin is given by 1 atm pressure 5 688 mmHg 3 760 mmHg 5 0.905 atm Similar problem: 5.13. Practice Exercise Convert 749 mmHg to atmospheres. Example 5.2 Hurricane Sandy (“Superstorm Sandy”) was one the most destructive hurricanes in recent years, affecting the Caribbean, Cuba, the Bahamas, and 24 states along U.S. east coast. The lowest pressure recorded for Hurricane Sandy was 705 mmHg. What was the pressure in kPa? Strategy Here we are asked to convert mmHg to kPa. Because 1 atm 5 1.01325 3 105 Pa 5 760 mmHg the conversion factor we need is 1.01325 3 105 Pa 760 mmHg Solution The pressure in kPa is 1.01325 3 105 Pa pressure 5 705 mmHg 3 760 mmHg 5 9.40 3 104 Pa 5 94.0 kPa Similar problem: 5.14. Practice Exercise Convert 295 mmHg to kilopascals. Review of Concepts Rank the following pressures from lowest to highest: (a) 736 mmHg, (b) 0.928 atm, (c) 728 torr, (d) 1.12 3 105 Pa. A manometer is a device used to measure the pressure of gases other than the atmosphere. The principle of operation of a manometer is similar to that of a barom- eter. There are two types of manometers, shown in Figure 5.4. The closed-tube manometer is normally used to measure pressures below atmospheric pressure [Fig- ure 5.4(a)], whereas the open-tube manometer is better suited for measuring pressures equal to or greater than atmospheric pressure [Figure 5.4(b)]. Nearly all barometers and many manometers use mercury as the working fluid, despite the fact that it is a toxic substance with a harmful vapor. The reason is that mercury has a very high density (13.6 g/mL) compared with most other liquids. Because the height of the liquid in a column is inversely proportional to the liquid’s density, this property enables the construction of manageably small barometers and  manometers. Review of Concepts Would it be easier to drink water with a straw on top or at the foot of Mt. Everest? 178 Chapter 5 ■ Gases Figure 5.4 Two types of manometers used to measure gas pressures. (a) Gas pressure Vacuum may be less or greater than atmospheric pressure. (b) Gas pressure is greater than atmospheric pressure. h h Gas Gas Mercury Pgas = Ph Pgas = Ph + Patm (a) (b) 5.3 The Gas Laws The gas laws we will study in this chapter are the product of countless experiments on the physical properties of gases that were carried out over several centuries. Each of these generalizations regarding the macroscopic behavior of gaseous substances represents a milestone in the history of science. Together they have played a major role in the development of many ideas in chemistry. The Pressure-Volume Relationship: Boyle’s Law Animation In the seventeenth century, Robert Boyle† studied the behavior of gases systemati- The Gas Laws cally and quantitatively. In one series of studies, Boyle investigated the pressure- volume relationship of a gas sample. Typical data collected by Boyle are shown in Table 5.2. Note that as the pressure (P) is increased at constant temperature, the volume (V ) occupied by a given amount of gas decreases. Compare the first data point with a pressure of 724 mmHg and a volume of 1.50 (in arbitrary unit) to the last data point with a pressure of 2250 mmHg and a volume of 0.58. Clearly there is an inverse relationship between pressure and volume of a gas at constant tem- perature. As the pressure is increased, the volume occupied by the gas decreases. Conversely, if the applied pressure is decreased, the volume the gas occupies increases. This relationship is now known as Boyle’s law, which states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas. † Robert Boyle (1627–1691). British chemist and natural philosopher. Although Boyle is commonly associ- ated with the gas law that bears his name, he made many other significant contributions in chemistry and physics. Despite the fact that Boyle was often at odds with scientists of his generation, his book The Skeptical Chymist (1661) influenced generations of chemists. Table 5.2 Typical Pressure-Volume Relationship Obtained by Boyle P (mmHg) 724 869 951 998 1230 1893 2250 V (arbitrary units) 1.50 1.33 1.22 1.18 0.94 0.61 0.58 PV 1.09 3 103 1.16 3 103 1.16 3 103 1.18 3 103 1.2 3 103 1.2 3 103 1.3 3 103  5.3 The Gas Laws 179 Figure 5.5 Apparatus for studying the relationship between pressure and volume of a gas. (a) The levels of mercury are equal and the pressure of the gas is equal to the atmospheric pressure (760 mmHg). The gas volume is 100 mL. (b) Doubling the pressure by adding more mercury reduces the gas volume 1520 mmHg to 50 mL. (c) Tripling the pressure decreases the gas volume to one-third of the original value. The temperature and amount of gas are kept constant. 760 mmHg Gas 33 mL 50 mL 100 mL Hg (a) (b) (c) The apparatus used by Boyle in this experiment was very simple (Figure 5.5). The pressure applied to a gas is equal to the gas pressure. In Figure 5.5(a), the pressure exerted on the gas is equal to atmospheric pressure and the volume of the gas is 100 mL. (Note that the tube is open at the top and is therefore exposed to atmospheric pressure.) In Figure 5.5(b), more mercury has been added to double the pressure on the gas, and the gas volume decreases to 50 mL. Tripling the pressure on the gas decreases its volume to a third of the original value [Figure 5.5(c)]. We can write a mathematical expression showing the inverse relationship between pressure and volume: 1 Pr V where the symbol r means proportional to. We can change r to an equals sign and write 1 P 5 k1 3 (5.1a) V where k1 is a constant called the proportionality constant. Equation (5.1a) is the  mathematical expression of Boyle’s law. We can rearrange Equation (5.1a) and  obtain PV 5 k1 (5.1b) This form of Boyle’s law says that the product of the pressure and volume of a gas at constant temperature and amount of gas is a constant. The top diagram in Figure 5.6 is a schematic representation of Boyle’s law. The quantity n is the number of moles of the gas and R is a constant to be defined in Section 5.4. We will see in Section 5.4 that the proportionality constant k1 in Equations (5.1) is equal to nRT. The concept of one quantity being proportional to another and the use of a proportionality constant can be clarified through the following analogy. The daily income of a movie theater depends on both the price of the tickets (in dollars per 180 Chapter 5 ■ Gases Increasing or decreasing the volume of a gas at a constant temperature P P P Volume decreases Volume increases Boyle’s Law (Pressure increases) (Pressure decreases) Boyle’s Law P = (nRT) 1 nRT is constant V Heating or cooling a gas at constant pressure P P P Lower temperature Higher temperature (Volume decreases) (Volume increases) Charles’ Law V = (nR) T nR is constant P P Charles’ Law Heating or cooling a gas at constant volume P P P Lower temperature Higher temperature (Pressure decreases) (Pressure increases) Charles’ Law P = (nR nR V ) T V is constant Dependence of volume on amount of gas at constant temperature and pressure P P P Gas cylinder Remove gas Add gas molecules (Volume decreases) (Volume increases) Valve Avogadro’s Law V = (RT ) n RT is constant P P Figure 5.6 Schematic illustrations of Boyle’s law, Charles’ law, and Avogadro’s law.  5.3 The Gas Laws 181 Figure 5.7 Graphs showing variation of the volume of a gas with the pressure exerted on the gas, at constant temperature. P P (a) P versus V. Note that the volume of the gas doubles as the 0.6 atm pressure is halved. (b) P versus 1yV. The slope of the line is equal to k1. 0.3 atm 2L 4L V 1 –– V (a) (b) ticket) and the number of tickets sold. Assuming that the theater charges one price for all tickets, we write income 5 (dollar/ticket) 3 number of tickets sold Because the number of tickets sold varies from day to day, the income on a given day is said to be proportional to the number of tickets sold: income r number of tickets sold 5 C 3 number of tickets sold where C, the proportionality constant, is the price per ticket. Figure 5.7 shows two conventional ways of expressing Boyle’s findings graphi- cally. Figure 5.7(a) is a graph of the equation PV 5 k1; Figure 5.7(b) is a graph of the equivalent equation P 5 k1 3 1yV. Note that the latter is a linear equation of the form y 5 mx 1 b, where b 5 0 and m 5 k1. Although the individual values of pressure and volume can vary greatly for a given sample of gas, as long as the temperature is held constant and the amount of the gas does not change, P times V is always equal to the same constant. Therefore, for a given sample of gas under two different sets of conditions at constant tem- perature, we have P1V1 5 k1 5 P2V2 or P1V1 5 P2V2 (5.2) where V1 and V2 are the volumes at pressures P1 and P2, respectively. The Temperature-Volume Relationship: Charles’ and Gay-Lussac’s Law Boyle’s law depends on the temperature of the system remaining constant. But sup- pose the temperature changes: How does a change in temperature affect the volume and pressure of a gas? Let us first look at the effect of temperature on the volume of a gas. The earliest investigators of this relationship were French scientists, Jacques Charles† and Joseph Gay-Lussac.‡ Their studies showed that, at constant pressure, the † Jacques Alexandre Cesar Charles (1746–1823). French physicist. He was a gifted lecturer, an inventor of scientific apparatus, and the first person to use hydrogen to inflate balloons. ‡ Joseph Louis Gay-Lussac (1778–1850). French chemist and physicist. Like Charles, Gay-Lussac was a balloon enthusiast. Once he ascended to an altitude of 20,000 ft to collect air samples for analysis. 182 Chapter 5 ■ Gases volume of a gas sample expands when heated and contracts when cooled (Figure 5.8). The quantitative relations involved in changes in gas temperature and volume turn out Capillary to be remarkably consistent. For example, we observe an interesting phenomenon tubing when we study the temperature-volume relationship at various pressures. At any given pressure, the plot of volume versus temperature yields a straight line. By extending Mercury the line to zero volume, we find the intercept on the temperature axis to be 2273.15°C. At any other pressure, we obtain a different straight line for the volume-temperature Gas plot, but we get the same zero-volume temperature intercept at 2273.15°C (Figure 5.9). (In practice, we can measure the volume of a gas over only a limited temperature Low High range, because all gases condense at low temperatures to form liquids.) temperature temperature In 1848 Lord Kelvin† realized the significance of this phenomenon. He identified Figure 5.8 Variation of the 2273.15°C as absolute zero, theoretically the lowest attainable temperature. Then he volume of a gas sample with set up an absolute temperature scale, now called the Kelvin temperature scale, with temperature, at constant absolute zero as the starting point (see Section 1.7). On the Kelvin scale, one kelvin pressure. The pressure exerted on the gas is the sum of the (K) is equal in magnitude to one degree Celsius. The only difference between the atmospheric pressure and the absolute temperature scale and the Celsius scale is that the zero position is shifted. pressure due to the weight of Important points on the two scales match up as follows: the mercury. Kelvin Scale Celsius Scale Under special experimental conditions, Absolute zero 0K 2273.15°C scientists have succeeded in approaching absolute zero to within a small fraction Freezing point of water 273.15 K 0°C of a kelvin. Boiling point of water 373.15 K 100°C The conversion between °C and K is given on p. 16. In most calculations we will use 273 instead of 273.15 as the term relating K and °C. By convention, we use T to denote absolute (kelvin) temperature and t to indicate temperature on the Celsius scale. The dependence of the volume of a gas on temperature is given by VrT V 5 k2T Remember that temperature must be in V kelvins in gas law calculations. or 5 k2 (5.3) T where k2 is the proportionality constant. Equation (5.3) is known as Charles’ and Gay-Lussac’s law, or simply Charles’ law, which states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute † William Thomson, Lord Kelvin (1824–1907). Scottish mathematician and physicist. Kelvin did important work in many branches of physics. Figure 5.9 Variation of the volume of a gas sample with temperature, at constant pressure. Each line 50 P1 represents the variation at a certain pressure. The pressures increase 40 P2 from P1 to P4. All gases ultimately V (mL) condense (become liquids) if they 30 are cooled to sufficiently low P3 temperatures; the solid portions of the lines represent the temperature 20 –273.15°C region above the condensation P4 point. When these lines are 10 extrapolated, or extended (the dashed portions), they all intersect at 0 the point representing zero volume –300 –200 –100 0 100 200 300 400 and a temperature of 2273.15°C. t (°C)  5.3 The Gas Laws 183 temperature of the gas. Charles’ law is also illustrated in Figure 5.6. We see that the proportionality constant k2 in Equation (5.3) is equal to nRyP. Just as we did for pressure-volume relationships at constant temperature, we can compare two sets of volume-temperature conditions for a given sample of gas at constant pressure. From Equation (5.3) we can write V1 V2 5 k2 5 T1 T2 V1 V2 or 5 (5.4) T1 T2 where V1 and V2 are the volumes of the gas at temperatures T1 and T2 (both in kelvins), respectively. Another form of Charles’ law shows that at constant amount of gas and volume, the pressure of a gas is proportional to temperature PrT P 5 k3T P or 5 k3 (5.5) T From Figure 5.6 we see that k3 5 nRyV. Starting with Equation (5.5), we have P1 P2 5 k3 5 T1 T2 P1 P2 or 5 (5.6) T1 T2 where P1 and P2 are the pressures of the gas at temperatures T1 and T2, respectively. Review of Concepts Compare the changes in volume when the temperature of a gas is increased at constant pressure from (a) 200 K to 400 K and (b) 200°C to 400°C. The Volume-Amount Relationship: Avogadro’s Law The work of the Italian scientist Amedeo Avogadro complemented the studies of Avogadro’s name first appeared in Section 3.2. Boyle, Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of moles of molecules present; that is, Vrn V 5 k4n (5.7) where n represents the number of moles and k4 is the proportionality constant. Equa- tion (5.7) is the mathematical expression of Avogadro’s law, which states that at constant 184 Chapter 5 ■ Gases Figure 5.10 Volume relationship of gases in a chemical reaction. The ratio of the volumes of molecular hydrogen to molecular nitrogen is 3:1, and that of + ammonia (the product) to molecular hydrogen and molecular nitrogen combined (the reactants) is 2:4, or 1:2. 3H2(g) + N2(g) 2NH3(g) 3 molecules + 1 molecule 2 molecules 3 moles + 1 mole 2 moles 3 volumes + 1 volume 2 volumes pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. From Figure 5.6 we see that k4 5 RTyP. According to Avogadro’s law we see that when two gases react with each other, their reacting volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac). For example, consider the synthesis of ammonia from molec- ular hydrogen and molecular nitrogen: 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) 3 mol 1 mol 2 mol Because, at the same temperature and pressure, the volumes of gases are directly proportional to the number of moles of the gases present, we can now write 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) 3 volumes 1 volume 2 volumes The volume ratio of molecular hydrogen to molecular nitrogen is 3:1, and that of ammonia (the product) to the sum of the volumes of molecular hydrogen and molec- ular nitrogen (the reactants) is 2:4 or 1:2 (Figure 5.10). Worked examples illustrating the gas laws are presented in Section 5.4. 5.4 The Ideal Gas Equation Let us summarize the gas laws we have discussed so far: 1 Boyle’s law: V r     (at constant n and T )   P Charles’ law: V r T    (at constant n and P) Avogadro’s law: V r n    (at constant P and T )   We can combine all three expressions to form a single master equation for the behav- ior of gases: nT Vr P nT V5R P or PV 5 nRT (5.8) Keep in mind that the ideal gas equation, unlike the gas laws discussed in Section 5.3, applies to systems that do not undergo changes in pressure, volume, where R, the proportionality constant, is called the gas constant. Equation (5.8), temperature, and amount of a gas. which is called the ideal gas equation, describes the relationship among the four  5.4 The Ideal Gas Equation 185 Figure 5.11 A comparison of the molar volume at STP (which is approximately 22.4 L) with a basketball. variables P, V, T, and n. An ideal gas is a hypothetical gas whose pressure-volume- temperature behavior can be completely accounted for by the ideal gas equation. The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container. Although there is no such thing in nature as an ideal gas, the ideal gas approximation works rather well for most reasonable temperature and pressure ranges. Thus, we can safely use the ideal gas equation to solve many gas problems. Before we can apply the ideal gas equation to a real system, we must evalu- ate the gas constant R. At 0°C (273.15 K) and 1 atm pressure, many real gases behave like an ideal gas. Experiments show that under these conditions, 1 mole of an ideal gas occupies 22.414 L, which is somewhat greater than the volume of a basketball, as shown in Figure 5.11. The conditions 0°C and 1 atm are called standard temperature and pressure, often abbreviated STP. From Equation (5.8) we can write PV The gas constant can be expressed in R5 different units (see Appendix 2). nT (1 atm)(22.414 L) 5 (1 mol)(273.15 K) L ? atm 5 0.082057 K ? mol 5 0.082057 L ? atmyK ? mol The dots between L and atm and between K and mol remind us that both L and atm are in the numerator and both K and mol are in the denominator. For most calculations, we will round off the value of R to three significant figures (0.0821 L ? atmyK ? mol) and use 22.41 L for the molar volume of a gas at STP. Example 5.3 shows that if we know the quantity, volume, and temperature of a gas, we can calculate its pressure using the ideal gas equation. Unless otherwise stated, we assume that the temperatures given in °C in calculations are exact so that they do not affect the number of significant figures. Example 5.3 Sulfur hexafluoride (SF6) is a colorless and odorless gas. Due to its lack of chemical reactivity, it is used as an insulator in electronic equipment. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C. (Continued) SF6 186 Chapter 5 ■ Gases Strategy The problem gives the amount of the gas and its volume and temperature. Is the gas undergoing a change in any of its properties? What equation should we use to solve for the pressure? What temperature unit should we use? Solution Because no changes in gas properties occur, we can use the ideal gas equation to calculate the pressure. Rearranging Equation (5.8), we write nRT P5 V (1.82 mol) (0.0821 L ? atmyK ? mol) (69.5 1 273) K 5 5.43 L Similar problem: 5.32. 5 9.42 atm Practice Exercise Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54 atm and 76°C. By using the fact that the molar volume of a gas occupies 22.41 L at STP, we can calculate the volume of a gas at STP without using the ideal gas equation. Example 5.4 Ammonia gas is used as a refrigerant in food processing and storage industries. Calculate the volume (in liters) occupied by 7.40 g of NH3 at STP. NH3 Strategy What is the volume of one mole of an ideal gas at STP? How many moles are there in 7.40 g of NH3? Solution Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP and using the molar mass of NH3 (17.03 g), we write the sequence of conversions as grams of NH3 ¡ moles of NH3 ¡ liters of NH3 at STP so the volume of NH3 is given by 1 mol NH3 22.41 L V 5 7.40 g NH3 3 3 17.03 g NH3 1 mol NH3 5 9.74 L It is often true in chemistry, particularly in gas-law calculations, that a problem can be solved in more than one way. Here the problem can also be solved by first converting 7.40 g of NH3 to number of moles of NH3, and then applying the ideal gas equation (V 5 nRTyP). Try it. Industrial ammonia refrigeration system. Check Because 7.40 g of NH3 is smaller than its molar mass, its volume at STP Similar problem: 5.40. should be smaller than 22.41 L. Therefore, the answer is reasonable. Practice Exercise What is the volume (in liters) occupied by 49.8 g of HCl at STP? Review of Concepts Assuming ideal behavior, which of the following gases will have the greatest volume at STP? (a) 0.82 mole of He. (b) 24 g of N2. (c) 5.0 3 1023 molecules of Cl2. Which gas will have the greatest density?  5.4 The Ideal Gas Equation 187 The ideal gas equation is useful for problems that do not involve changes in P, V, T, and n for a gas sample. Thus, if we know any three of the variables we can calculate the fourth one using the equation. At times, however, we need to deal with changes in pressure, volume, and temperature, or even in the amount of gas. When conditions change, we must employ a modified form of the ideal gas equation that takes into account the initial and final conditions. We derive the modified equation as follows. From Equation (5.8), P1V1 P2V2 The subscripts 1 and 2 denote the initial R5 (before change) and R5 (after change) and final states of the gas, respectively. n1T1 n2T2 Therefore, P1V1 P2V2 5 (5.9) n1T1 n2T2 It is interesting to note that all the gas laws discussed in Section 5.3 can be derived from Equation (5.9). If n1 5 n2, as is usually the case because the amount of gas normally does not change, the equation then becomes P1V1 P2V2 5 (5.10) T1 T2 Applications of Equation (5.9) are shown in Examples 5.5, 5.6, and 5.7. Example 5.5 An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? Strategy The amount of gas inside the balloon and its temperature remain constant, but both the pressure and the volume change. What gas law do you need? Solution We start with Equation (5.9) P1V1 P2V2 5 n1T1 n2T2 Because n1 5 n2 and T1 5 T2, P1V1 5 P2V2 A scientific research helium which is Boyle’s law [see Equation (5.2)]. The given information is tabulated: balloon. Initial Conditions Final Conditions P1 5 1.0 atm P2 5 0.40 atm V1 5 0.55 L V2 5 ? Therefore, P1 V 2 5 V1 3 P2 1.0 atm 5 0.55 L 3 0.40 atm 5 1.4 L (Continued) 188 Chapter 5 ■ Gases Check When pressure applied on the balloon is reduced (at constant temperature), the helium gas expands and the balloon’s volume increases. The final volume is greater Similar problem: 5.19. than the initial volume, so the answer is reasonable. Practice Exercise A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. Calculate the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL. Example 5.6 Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure (in atm). Strategy The temperature and pressure of argon change but the amount and volume of gas remain the same. What equation would you use to solve for the final pressure? What temperature unit should you use? Solution Because n1 5 n2 and V1 5 V2, Equation (5.9) becomes P1 P2 Electric lightbulbs are usually filled 5 with argon. T1 T2 which is Charles’ law [see Equation (5.6)]. Next we write Initial Conditions Final Conditions Remember to convert °C to K when P1 5 1.20 atm P2 5 ? solving gas-law problems. T1 5 (18 1 273) K 5 291 K T2 5 (85 1 273) K 5 358 K The final pressure is given by T2 One practical consequence of this P2 5 P1 3 relationship is that automobile tire T1 pressures should be checked only when 358 K the tires are at normal temperatures. 5 1.20 atm 3 After a long drive (especially in the 291 K summer), tires become quite hot, and 5 1.48 atm the air pressure inside them rises. Check At constant volume, the pressure of a given amount of gas is directly proportional Similar problem: 5.36. to its absolute temperature. Therefore the increase in pressure is reasonable. Practice Exercise A sample of oxygen gas initially at 0.97 atm is cooled from 21°C to 268°C at constant volume. What is its final pressure (in atm)? Example 5.7 A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressure is 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL. (Continued)  5.4 The Ideal Gas Equation 189 Strategy In solving this kind of problem, where a lot of information is given, it is sometimes helpful to make a sketch of the situation, as shown here: What temperature unit should be used in the calculation? Solution According to Equation (5.9) P1V1 P2V2 5 n1T1 n2T2 We assume that the amount of air in the bubble remains constant, that is, n1 5 n2 so that P1V1 P2V2 5 T1 T2 which is Equation (5.10). The given information is summarized: Initial Conditions Final Conditions P1 5 6.4 atm P2 5 1.0 atm We can use any appropriate units for V1 5 2.1 mL V2 5 ? volume (or pressure) as long as we use the same units on both sides of the equation. T1 5 (8 1 273) K 5 281 K T2 5 (25 1 273) K 5 298 K Rearranging Equation (5.10) gives P1 T2 V2 5 V1 3 3 P2 T1 6.4 atm 298 K 5 2.1 mL 3 3 1.0 atm 281 K 5 14 mL Check We see that the final volume involves multiplying the initial volume by a ratio of pressures (P1yP2) and a ratio of temperatures (T2yT1). Recall that volume is inversely proportional to pressure, and volume is directly proportional to temperature. Because the pressure decreases and temperature increases as the bubble rises, we expect the bubble’s volume to increase. In fact, here the change in pressure plays a greater role in the volume change. Similar problem: 5.35. Practice Exercise A gas initially at 4.0 L, 1.2 atm, and 66°C undergoes a change so that its final volume and temperature are 1.7 L and 42°C. What is its final pressure? Assume the number of moles remains unchanged. Density Calculations If we rearrange the ideal gas equation, we can calculate the density of a gas: n P 5 V RT 190 Chapter 5 ■ Gases The number of moles of the gas, n, is given by m n5 m where m is the mass of the gas in grams and m is its molar mass. Therefore m P 5 mV RT Because density, d, is mass per unit volume, we can write m Pm d5 5 (5.11) V RT Unlike molecules in condensed matter (that is, in liquids and solids), gaseous mole- cules are separated by distances that are large compared with their size. Consequently, the density of gases is very low under atmospheric conditions. For this reason, gas densities are usually expressed in grams per liter (g/L) rather than grams per milliliter (g/mL), as Example 5.8 shows. Example 5.8 CO2 Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C. Strategy We need Equation (5.11) to calculate gas density. Is sufficient information provided in the problem? What temperature unit should be used? Solution To use Equation (5.11), we convert temperature to kelvins (T 5 273 1 55 5 328 K) and use 44.01 g for the molar mass of CO2: Pm d5 RT (0.990 atm) (44.01 g/mol) 5 5 1.62 g/L (0.0821 L ? atmyK ? mol) (328 K) Alternatively, we can solve for the density by writing mass density 5 volume Being an intensive property, density is Assuming that we have 1 mole of CO2, the mass is 44.01 g. The volume of the gas can independent of the amount of substance. be obtained from the ideal gas equation Therefore, we can use any convenient amount to help us solve the problem. nRT V5 P (1 mol) (0.0821 L ? atmyK ? mol) (328 K) 5 0.990 atm 5 27.2 L Therefore, the density of CO2 is given by 44.01 g d5 5 1.62 g/L 27.2 L (Continued)  5.4 The Ideal Gas Equation 191 Comment ln units of grams per milliliter, the gas density is 1.62 3 1023 g/mL, which is a very small number. In comparison, the density of water is 1.0 g/mL and that of gold is 19.3 g/cm3. Similar problem: 5.48. Practice Exercise What is the density (in g/L) of uranium hexafluoride (UF6) at 779  mmHg and 62°C? The Molar Mass of a Gaseous Substance From what we have seen so far, you may have the impression that the molar mass of a substance is found by examining its formula and summing the molar masses of its component atoms. However, this procedure works only if the actual formula of the substance is known. In practice, chemists often deal with substances of unknown or only partially defined composition. If the unknown substance is gaseous, its molar mass can nevertheless be found thanks to the ideal gas equation. All that is needed is an experimentally determined density value (or mass and volume data) for the gas at a known temperature and pressure. By rearranging Equation (5.11) we get dRT m5 (5.12) P Figure 5.12 An apparatus for measuring the density of a gas. In a typical experiment, a bulb of known volume is filled with the gaseous sub- A bulb of known volume is filled stance under study. The temperature and pressure of the gas sample are recorded, and with the gas under study at a certain temperature and pressure. the total mass of the bulb plus gas sample is determined (Figure 5.12). The bulb is First the bulb is weighed, and then evacuated (emptied) and weighed again. The difference in mass is the mass of then it is emptied (evacuated) and the gas. The density of the gas is equal to its mass divided by the volume of the bulb. weighed again. The difference in masses gives the mass of the Once we know the density of a gas, we can calculate the molar mass of the substance gas. Knowing the volume of the using Equation (5.12). bulb, we can calculate the density The mass spectrometer has become the dominant instrument for determining of the gas. Under atmospheric conditions, 100 mL of air weigh molar mass, but the determination of molar mass by the density method is still useful, about 0.12 g, an easily measured as illustrated by Example 5.9. quantity. Example 5.9 A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36°C and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula. Strategy Because Equations (5.11) and (5.12) are rearrangements of each other, we can calculate the molar mass of a gas if we know its density, temperature, and pressure. The molecular formula of the compound must be consistent with its molar mass. What temperature unit should we use? Solution From Equation (5.12) dRT m5 P (7.71 g/L) (0.0821 L ? atmyK ? mol) (36 1 273) K Note that we can determine the molar 5 mass of a gaseous compound by this 2.88 atm procedure without knowing its chemical 5 67.9 g/mol formula. (Continued) 192 Chapter 5 ■ Gases Alternatively, we can solve for the molar mass by writing mass of compound molar mass of compound 5 moles of compound From the given density we know there are 7.71 g of the gas in 1 L. The number of moles of the gas in this volume can be obtained from the ideal gas equation PV n5 RT (2.88 atm) (1.00 L) 5 (0.0821 L ? atmyK ? mol) (309 K) 5 0.1135 mol Therefore, the molar mass is given by mass 7.71 g m5 5 5 67.9 g/mol number of moles 0.1135 mol ClO2 We can determine the molecular formula of the compound by trial and error, using only the knowledge of the molar masses of chlorine (35.45 g) and oxygen (16.00 g). We know that a compound containing one Cl atom and one O atom would have a molar mass of 51.45 g, which is too low, while the molar mass of a compound made up of two Cl atoms and one O atom is 86.90 g, which is too high. Thus, the compound must contain one Cl atom and two O atoms and have the formula ClO2, which has a molar Similar problems: 5.43, 5.47. mass of 67.45 g. Practice Exercise The density of a gaseous organic compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass? Because Equation (5.12) is derived from the ideal gas equation, we can also calculate the molar mass of a gaseous substance using the ideal gas equation, as shown in Example 5.10. Example 5.10 Chemical analysis of a gaseous compound showed that it contained 33.0 percent silicon (Si) and 67.0 percent fluorine (F) by mass. At 35°C, 0.210 L of the compound exerted a pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g, calculate the molecular formula of the compound. Strategy This problem can be divided into two parts. First, it asks for the empirical formula of the compound from the percent by mass of Si and F. Second, the information Si2F6 provided enables us to calculate the molar mass of the compound and hence determine its molecular formula. What is the relationship between empirical molar mass and molar mass calculated from the molecular formula? Solution We follow the procedure in Example 3.9 (p. 86) to calculate the empirical formula by assuming that we have 100 g of the compound, so the percentages are converted to grams. The number of moles of Si and F are given by 1 mol Si nSi 5 33.0 g Si 3 5 1.17 mol Si 28.09 g Si 1 mol F nF 5 67.0 g F 3 5 3.53 mol F 19.00 g F (Continued)  5.5  Gas Stoichiometry 193 Therefore, the empirical formula is Si1.17F3.53, or, dividing by the smaller subscript (1.17), we obtain SiF3. To calculate the molar mass of the compound, we need first to calculate the ­number of moles contained in 2.38 g of the compound. From the ideal gas equation PV n5 RT (1.70 atm) (0.210 L) 5 5 0.0141 mol (0.0821 L ? atmyK ? mol) (308 K) Because there are 2.38 g in 0.0141 mole of the compound, the mass in 1 mole, or the molar mass, is given by 2.38 g m5 5 169 g/mol 0.0141 mol The molar mass of the empirical formula SiF3 is 85.09 g. Recall that the ratio (molar mass/empirical molar mass) is always an integer (169y85.09 < 2). Therefore, the molecular formula of the compound must be (SiF3)2 or Si2F6. Similar problem: 5.49. Practice Exercise  A gaseous compound is 78.14 percent boron and 21.86 percent hydrogen. At 27°C, 74.3 mL of the gas exerted a pressure of 1.12 atm. If the mass of the gas was 0.0934 g, what is its molecular formula? 5.5 Gas Stoichiometry In Chapter 3 we used relationships between amounts (in moles) and masses (in grams) The key to solving stoichiometry problems is mole ratio, regardless of of reactants and products to solve stoichiometry problems. When the reactants and/or the physical state of the reactants and products are gases, we can also use the relationships between amounts (moles, n) and products. volume (V) to solve such problems (Figure 5.13). Examples 5.11, 5.12, and 5.13 show how the gas laws are used in these calculations. Example 5.11 The combustion of acetylene with pure oxygen produces a very high-temperature flame used for welding and cutting metals. Calculate the volume of O2 (in liters) required for the complete combustion of 7.64 L of acetylene (C2H2) measured at the same temperature and pressure. 2C2H2 (g) 1 5O2 (g) ¡ 4CO2 (g) 1 2H2O(l) Strategy  Note that the temperature and pressure of O2 and C2H2 are the same. Which gas law do we need to relate the volume of the gases to the moles of gases? Solution  According to Avogadro’s law, at the same temperature and pressure, the number of moles of gases are directly related to their volumes. From the equation, we have 5 mol O2 ∞ 2 mol C2H2; therefore, we can also write 5 L O2 ∞ 2 L C2H2. The volume of O2 that will react with 7.64 L C2H2 is given by 5 L O2 volume of O2 5 7.64 L C2H2 3 The reaction of calcium carbide 2 L C2H2 (CaC2) with water produces 5 19.1 L ­acetylene (C2H2), a flammable gas. Practice Exercise  Assuming no change in temperature and pressure, calculate the Similar problem: 5.26. volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10): 2C4H10 (g) 1 13O2 (g) ¡ 8CO2 (g) 1 10H2O(l) 194 Chapter 5 ■ Gases Figure 5.13 Stoichiometric calculations involving gases. Amount of Amount of Moles of Moles of reactant (grams product (grams reactant product or volume) or volume) Example 5.12 Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows: 2NaN3 (s) ¡ 2Na(s) 1 3N2 (g) The nitrogen gas produced quickly inflates the bag between the driver and the windshield An air bag can protect the driver and dashboard. Calculate the volume of N2 generated at 80°C and 823 mmHg by the in an automobile collision. decomposition of 60.0 g of NaN3. Strategy From the balanced equation we see that 2 mol NaN3 ∞ 3 mol N2 so the conversion factor between NaN3 and N2 is 3 mol N2 2 mol NaN3 Because the mass of NaN3 is given, we can calculate the number of moles of NaN3 and hence the number of moles of N2 produced. Finally, we can calculate the volume of N2 using the ideal gas equation. Solution First we calculate number of moles of N2 produced by 60.0 g NaN3 using the following sequence of conversions grams of NaN3 ¡ moles of NaN3 ¡ moles of N2 so that 1 mol NaN3 3 mol N2 moles of N2 5 60.0 g NaN3 3 3 65.02 g NaN3 2 mol NaN3 5 1.38 mol N2 The volume of 1.38 moles of N2 can be obtained by using the ideal gas equation: nRT (1.38 mol) (0.0821 L ? atmyK ? mol) (80 1 273 K) V5 5 P (823y760) atm Similar problem: 5.62. 5 36.9 L Practice Exercise The equation for the metabolic breakdown of glucose (C6H12O6) is the same as the equation for the combustion of glucose in air: C6H12O6 (s) 1 6O2 (g) ¡ 6CO2 (g) 1 6H2O(l) Calculate the volume of CO2 produced at 37°C and 1.00 atm when 5.60 g of glucose is used up in the reaction. Example 5.13 Aqueous lithium hydroxide solution is used to purify air in spacecrafts and submarines because it absorbs carbon dioxide, which is an end product of metabolism, according to the equation 2LiOH(aq) 1 CO2 (g) ¡ Li2CO3 (aq) 1 H2O(l) (Continued)  5.6 Dalton’s Law of Partial Pressures 195 The pressure of carbon dioxide inside the cabin of a submarine having a volume of 2.4 3 105 L is 7.9 3 1023 atm at 312 K. A solution of lithium hydroxide (LiOH) of negligible volume is introduced into the cabin. Eventually the pressure of CO2 falls to 1.2 3 1024 atm. How many grams of lithium carbonate are formed by this process? Strategy How do we calculate the number of moles of CO2 reacted from the drop in CO2 pressure? From the ideal gas equation we write V The air in submerged submarines n5P3a b and space vehicles needs to be RT purified continuously. At constant T and V, the change in pressure of CO2, ≤P, corresponds to the change in the number of moles of CO2, ≤n. Thus, V ¢n 5 ¢P 3 a b RT What is the conversion factor between CO2 and Li2CO3? Solution The drop in CO2 pressure is (7.9 3 1023 atm) 2 (1.2 3 1024 atm) or 7.8 3 1023 atm. Therefore, the number of moles of CO2 reacted is given by 2.4 3 105 L ¢n 5 7.8 3 10 23 atm 3 (0.0821 L ? atmyK ? mol) (312 K) 5 73 mol From the chemical equation we see that 1 mol CO2 ∞ 1 mol Li2CO3, so the amount of Li2CO3 formed is also 73 moles. Then, with the molar mass of Li2CO3 (73.89 g), we calculate its mass: 73.89 g Li2CO3 mass of Li2CO3 formed 5 73 mol Li2CO3 3 1 mol Li2CO3 5 5.4 3 103 g Li2CO3 Similar problem: 5.100. Practice Exercise A 2.14-L sample of hydrogen chloride (HCl) gas at 2.61 atm and 28°C is completely dissolved in 668 mL of water to form hydrochloric acid solution. Calculate the molarity of the acid solution. Assume no change in volume. Review of Concepts Alkanes (CnH2n12) are discussed in Section 2.8. For which alkanes, if any, does the number of moles of gas remain constant as the gas-phase combustion reaction alkane(g) 1 oxygen(g) S carbon dioxide(g) 1 water vapor(g) proceeds from reactants to products? 5.6 Dalton’s Law of Partial Pressures Thus far we have concentrated on the behavior of pure gaseous substances, but experimental studies very often involve mixtures of gases. For example, for a study of air pollution, we may be interested in the pressure-volume-temperature 196 Chapter 5 ■ Gases Volume and temperature are constant Combining 1 the gases P1 P2 PT 5 P1 1 P2 Figure 5.14 Schematic illustration of Dalton’s law of partial pressures. relationship of a sample of air, which contains several gases. In this case, and all cases involving mixtures of gases, the total gas pressure is related to partial pressures, that is, the pressures of individual gas components in the mixture. In 1801 Dalton formulated a law, now known as Dalton’s law of partial pressures, which states that the total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. Figure 5.14 illus- trates Dalton’s law. Consider a case in which two gases, A and B, are in a container of volume V. The pressure exerted by gas A, according to the ideal gas equation, is nART PA 5 V where nA is the number of moles of A present. Similarly, the pressure exerted by gas B is nBRT PB 5 V In a mixture of gases A and B, the total pressure PT is the result of the collisions of both types of molecules, A and B, with the walls of the container. Thus, according to Dalton’s law, PT 5 PA 1 PB nART nBRT 5 1 V V RT 5 (nA 1 nB ) V nRT 5 V where n, the total number of moles of gases present, is given by n 5 nA 1 nB, and PA and PB are the partial pressures of gases A and B, respectively. For a mixture of gases, then, PT depends only on the total number of moles of gas present, not on the nature of the gas molecules.  5.6 Dalton’s Law of Partial Pressures 197 In general, the total pressure of a mixture of gases is given by PT 5 P1 1 P2 1 P3 1 . . . where P1, P2, P3, . . . are the partial pressures of components 1, 2, 3, . . . . To see how each partial pressure is related to the total pressure, consider again the case of a mixture of two gases A and B. Dividing PA by PT, we obtain PA nARTyV 5 PT (nA 1 nB )RTyV nA 5 nA 1 nB 5 XA where XA is called the mole fraction of A. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. In general, the mole fraction of compo- nent i in a mixture is given by ni Xi 5 (5.13) nT where ni and nT are the number of moles of component i and the total number of moles present, respectively. The mole fraction is always smaller than 1. We can now express the partial pressure of A as PA 5 XAPT Similarly, PB 5 XBPT Note that the sum of the mole fractions for a mixture of gases must be unity. If only two components are present, then nA nB XA 1 XB 5 1 51 nA 1 nB nA 1 nB If a system contains more than two gases, then the partial pressure of the ith compo- For gas mixtures, the sum of partial pressures must equal the total pressure nent is related to the total pressure by and the sum of mole fractions must equal 1. Pi 5 XiPT (5.14) How are partial pressures determined? A manometer can measure only the total pressure of a gaseous mixture. To obtain the partial pressures, we need to know the mole fractions of the components, which would involve elaborate chem- ical analyses. The most direct method of measuring partial pressures is using a mass spectrometer. The relative intensities of the peaks in a mass spectrum are directly proportional to the amounts, and hence to the mole fractions, of the gases present. From mole fractions and total pressure, we can calculate the partial pressures of individual components, as Example 5.14 shows. A direct application of Dalton’s law of partial pressures to scuba diving is discussed in the Chemistry in Action essay on p. 200. 198 Chapter 5 ■ Gases Example 5.14 A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a certain temperature. Strategy What is the relationship between the partial pressure of a gas and the total gas pressure? How do we calculate the mole fraction of a gas? Solution According to Equation (5.14), the partial pressure of Ne (PNe) is equal to the product of its mole fraction (XNe) and the total pressure (PT) need to find o PNe 5 XNePT p rgiven want to calculate Using Equation (5.13), we calculate the mole fraction of Ne as follows: nNe 4.46 mol XNe 5 5 nNe 1 nAr 1 nXe 4.46 mol 1 0.74 mol 1 2.15 mol 5 0.607 Therefore, PNe 5 XNePT 5 0.607 3 2.00 atm 5 1.21 atm Similarly, PAr 5 XArPT 5 0.10 3 2.00 atm 5 0.20 atm and PXe 5 XXePT 5 0.293 3 2.00 atm 5 0.586 atm Check Make sure that the sum of the partial pressures is equal to the given total Similar problem: 5.67. pressure; that is, (1.21 1 0.20 1 0.586) atm 5 2.00 atm. Practice Exercise A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole of ethane (C2H6), and 0.116 mole of propane (C3H8). If the total pressure of the gases is 1.37 atm, what are the partial pressures of the gases? Animation An important application of Dalton’s law of partial pressures involves calculating Collecting a Gas over Water the amount of a gas collected over water. Gases that are commonly used in the labo- ratory are generally obtained from pressurized gas cylinders, but if there is an occa- sional need for a small amount of a certain gas, it may be more convenient to prepare it chemically. For example, when potassium chlorate (KClO3) is heated, it decomposes to KCl and O2: 2KClO3 (s) ¡ 2KCl(s) 1 3O2 (g)  5.6 Dalton’s Law of Partial Pressures 199 Bottle being filled with oxygen gas KClO3 and MnO2 Bottle filled with water Bottle full of oxygen gas ready to be placed in plus water vapor the plastic basin Figure 5.15 An apparatus for collecting gas over water. The oxygen generated by heating potassium chlorate (KClO3 ) in the presence of a small amount of manganese dioxide (MnO2 ), which speeds up the reaction, is bubbled through water and collected in a bottle as shown. When collecting a gas over water, the Water originally present in the bottle is pushed into the trough by the oxygen gas. total pressure (gas plus water vapor) is equal to the atmospheric pressure. The oxygen gas can be collected over water, as shown in Figure 5.15. Initially, the Table 5.3 inverted bottle is completely filled with water. As oxygen gas is generated, the gas Pressure of Water Vapor at bubbles rise to the top and displace water from the bottle. This method of collecting Various Temperatures a gas is based on the assumptions that the gas does not react with water and that it Water is not appreciably soluble in it. These assumptions are valid for oxygen gas, but not Vapor for gases such as NH3, which dissolves readily in water. The oxygen gas collected in Temperature Pressure this way is not pure, however, because water vapor is also present in the bottle. The (°C) (mmHg) total gas pressure is equal to the sum of the pressures exerted by the oxygen gas and 0 4.58 the water vapor: 5 6.54 PT 5 PO2 1 PH2O 10 9.21 15 12.79 Consequently, we must allow for the pressure caused by the presence of water vapor 20 17.54 when we calculate the amount of O2 generated. Table 5.3 shows the pressure of water 25 23.76 vapor at various temperatures. These data are plotted in Figure 5.16. 30 31.82 Example 5.15 shows how to use Dalton’s law to calculate the amount of a gas 35 42.18 collected over water. 40 55.32 45 71.88 50 92.51 Example 5.15 55 118.04 60 149.38 Oxygen gas generated by the decomposition of potassium chlorate is collected as shown 65 187.54 in Figure 5.15. The volume of oxygen collected at 24°C and atmospheric pressure of 70 233.7 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The 75 289.1 pressure of the water vapor at 24°C is 22.4 mmHg. 80 355.1 Strategy To solve for the mass of O2 generated, we must first calculate the partial 85 433.6 pressure of O2 in the mixture. What gas law do we need? How do we convert pressure 90 525.76 of O2 gas to mass of O2 in grams? 95 633.90 (Continued) 100 760.00  CHEMISTRY in Action Scuba Diving and the Gas Laws S cuba diving is an exhilarating sport, and, thanks in part to the gas laws, it is also a safe activity for trained individuals who are in good health. (“Scuba” is an acronym for self- of air bubbles in these vessels can block normal blood flow to the brain. As a result, the diver might lose consciousness before reaching the surface. The only cure for an air embo- contained underwater breathing apparatus.) Two applications of lism is recompression. For this painful process, the victim is the gas laws to this popular pastime are the development of placed in a chamber filled with compressed air. Here bubbles guidelines for returning safely to the surface after a dive and the in the blood are slowly squeezed down to harmless size over determination of the proper mix of gases to prevent a potentially the course of several hours to a day. To avoid these unpleasant fatal condition during a dive. complications, divers know they must ascend slowly, pausing A typical dive might be 40 to 65 ft, but dives to 90 ft are not at certain points to give their bodies time to adjust to the fall- uncommon. Because seawater has a slightly higher density than ing pressure. fresh water—about 1.03 g/mL, compared with 1.00 g/mL—the Our second example is a direct application of Dalton’s law. pressure exerted by a column of 33 ft of seawater is equivalent Oxygen gas is essential for our survival, so it is hard to believe to 1 atm pressure. Pressure increases with increasing depth, so that an excess of oxygen could be harmful. Nevertheless, the at a depth of 66 ft the pressure of the water will be 2 atm, toxicity of too much oxygen is well established. For example, and so on. newborn infants placed in oxygen tents often sustain damage to What would happen if a diver rose to the surface from a the retinal tissue, which can cause partial or total blindness. depth of, say, 20 ft rather quickly without breathing? The total Our bodies function best when oxygen gas has a partial decrease in pressure for this change in depth would be pressure of about 0.20 atm, as it does in the air we breathe. The (20 fty33 ft) 3 1 atm, or 0.6 atm. When the diver reached the oxygen partial pressure is given by surface, the volume of air trapped in the lungs would have increased by a factor of (1 1 0.6) atmyl atm, or 1.6 times. nO2 PO2 5 XO2PT 5 PT This sudden expansion of air can fatally rupture the mem- nO2 1 nN2 branes of the lungs. Another serious possibility is that an air embolism might develop. As air expands in the lungs, it is where PT is the total pressure. However, because volume is forced into tiny blood vessels called capillaries. The presence directly proportional to the number of moles of gas present (at 800 760 Solution From Dalton’s law of partial pressures we know that 600 P (mmHg) PT 5 PO2 1 PH2O 400 Therefore, 200 PO2 5 PT 2 PH2O 5 762 mmHg 2 22.4 mmHg 0 20 40 60 80 100 5 740 mmHg t (°C) Figure 5.16 The pressure of From the ideal gas equation we write water vapor as a function of temperature. Note that at the m PV 5 nRT 5 RT boiling point of water (100°C) the m pressure is 760 mmHg, which is exactly equal to 1 atm. (Continued) 200 constant temperature and pressure), we can now write VO2 PO2 5 PT VO2 1 VN2 Thus, the composition of air is 20 percent oxygen gas and 80 percent nitrogen gas by volume. When a diver is submerged, the pressure of the water on the diver is greater than atmospheric pressure. The air pressure inside the body cavities (for example, lungs, sinuses) must be the same as the pressure of the sur- rounding water; otherwise they would collapse. A special valve automatically adjusts the pressure of the air breathed from a scuba tank to ensure that the air pressure equals the water pres- sure at all times. For example, at a depth where the total pres- sure is 2.0 atm, the oxygen content in air should be reduced to 10 percent by volume to maintain the same partial pressure of 0.20 atm; that is, VO2 PO2 5 0.20 atm 5 3 2.0 atm Scuba divers. VO2 1 VN2 VO2 0.20 atm 5 5 0.10 or 10% VO2 1 VN2 2.0 atm those associated with alcohol intoxication. Divers suffering Although nitrogen gas may seem to be the obvious choice from nitrogen narcosis have been known to do strange things, to mix with oxygen gas, there is a serious problem with it. such as dancing on the seafloor and chasing sharks. For this When the partial pressure of nitrogen gas exceeds 1 atm, reason, helium is often used to dilute oxygen gas. An inert gas, enough of the gas dissolves in the blood to cause a condition helium is much less soluble in blood than nitrogen and pro- known as nitrogen narcosis. The effects on the diver resemble duces no narcotic effects. where m and m are the mass of O2 collected and the molar mass of O2, respectively. Rearranging the equation we obtain PVm (740y760)atm(0.128 L) (32.00 g/mol) m5 5 RT (0.0821 L ? atmyK ? mol) (273 1 24) K 5 0.164 g Check The density of the oxygen gas is (0.164 gy0.128 L), or 1.28 g/L, which is a reasonable value for gases under atmospheric conditions (see Example 5.8). Similar problem: 5.72. Practice Exercise Hydrogen gas generated when calcium metal reacts with water is collected as shown in Figure 5.15. The volume of gas collected at 30°C and pressure of 988 mmHg is 641 mL. What is the mass (in grams) of the hydrogen gas obtained? The pressure of water vapor at 30°C is 31.82 mmHg. 201 202 Chapter 5 ■ Gases Review of Concepts Each of the color spheres represents a different gas molecule. Calculate the partial pressures of the gases if the total pressure is 2.6 atm. 5.7 The Kinetic Molecular Theory of Gases The gas laws help us to predict the behavior of gases, but they do not explain what happens at the molecular level to cause the changes we observe in the macroscopic world. For example, why does a gas expand on heating? In the nineteenth century, a number of physicists, notably Ludwig Boltzmann† and James Clerk Maxwell,‡ found that the physical properties of gases can be explained in terms of the motion of individual molecules. This molecular movement is a form of energy, which we define as the capacity to do work or to produce change. In mechanics, work is defined as force times distance. Because energy can be measured as work, we can write energy 5 work done 5 force 3 distance The joule (J)§ is the SI unit of energy 1 J 5 1 kg m2/s2 51Nm Alternatively, energy can be expressed in kilojoules (kJ): 1 kJ 5 1000 J As we will see in Chapter 6, there are many different kinds of energy. Kinetic energy (KE) is the type of energy expended by a moving object, or energy of motion. The findings of Maxwell, Boltzmann, and others resulted in a number of gener- alizations about gas behavior that have since been known as the kinetic molecular † Ludwig Eduard Boltzmann (1844–1906). Austrian physicist. Although Boltzmann was one of the greatest theoretical physicists of all time, his work was not recognized by other scientists in his own lifetime. Suffering from poor health and great depression, he committed suicide in 1906. ‡ James Clerk Maxwell (1831–1879). Scottish physicist. Maxwell was one of the great theoretical physicists of the nineteenth century; his work covered many areas in physics, including kinetic theory of gases, thermodynamics, and electricity and magnetism. § James Prescott Joule (1818–1889). English physicist. As a young man, Joule was tutored by John Dalton. He is most famous for determining the mechanical equivalent of heat, the conversion between mechanical energy and thermal energy.  5.7 The Kinetic Molecular Theory of Gases 203 theory of gases, or simply the kinetic theory of gases. Central to the kinetic theory are the following assumptions: 1. A gas is composed of molecules that are separated from each other by distances The kinetic theory of gases treats molecules as hard spheres without far greater than their own dimensions. The molecules can be considered to be internal structure. “points”; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. In other words, energy can be transferred from one molecule to another as a result of a collision. Nevertheless, the total energy of all the molecules in a system remains the same. 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same aver- age kinetic energy. The average kinetic energy of a molecule is given by KE 5 12mu2 where m is the mass of the molecule and u is its speed. The horizontal bar denotes an average value. The quantity u2 is called mean square speed; it is the average of the square of the speeds of all the molecules: u21 1 u22 1 . . . 1 u2N u2 5 N where N is the number of molecules. Assumption 4 enables us to write KE r T 1 2 2 mu rT Hence, KE 5 12mu2 5 CT (5.15) where C is the proportionality constant and T is the absolute temperature. According to the kinetic molecular theory, gas pressure is the result of collisions between molecules and the walls of their container. It depends on the frequency of collision per unit area and on how “hard” the molecules strike the wall. The theory also provides a molecular interpretation of temperature. According to Equation (5.15), the absolute temperature of a gas is a measure of the average kinetic energy of the molecules. In other words, the absolute temperature is an indication of the random motion of the molecules—the higher the temperature, the more energetic the mole- cules. Because it is related to the temperature of the gas sample, random molecular motion is sometimes referred to as thermal motion. Application to the Gas Laws Although the kinetic theory of gases is based on a rather simple model, the mathe- matical details involved are very complex. However, on a qualitative basis, it is possible to use the theory to account for the general properties of substances in the gaseous state. The following examples illustrate the range of its utility. • Compressibility of Gases. Because molecules in the gas phase are separated by large distances (assumption 1), gases can be compressed easily to occupy less volume. 204 Chapter 5 ■ Gases • Boyle’s Law. The pressure exerted by a gas results from the impact of its mol- ecules on the walls of the container. The collision rate, or the number of molecular collisions with the walls per second, is proportional to the number density (that is, number of molecules per unit volume) of the gas. Decreasing the volume of a given amount of gas increases its number density and hence its collision rate. For this reason, the pressure of a gas is inversely propor- tional to the volume it occupies; as volume decreases, pressure increases and vice versa. • Charles’ Law. Because the average kinetic energy of gas molecules is propor- tional to the sample’s absolute temperature (assumption 4), raising the tem- perature increases the average kinetic energy. Consequently, molecules will collide with the walls of the container more frequently and with greater impact if the gas is heated, and thus the pressure increases. The volume of gas will expand until the gas pressure is balanced by the constant external pressure (see Figure 5.8). Another way of stating Avogadro’s law is • Avogadro’s Law. We have shown that the pressure of a gas is directly propor- that at the same pressure and temperature, equal volumes of gases, whether they are tional to both the density and the temperature of the gas. Because the mass of the same or different gases, contain equal the gas is directly proportional to the number of moles (n) of the gas, we can numbers of molecules. represent density by nyV. Therefore n Pr T V For two gases, 1 and 2, we write n1T1 n1T1 P1 r 5C V1 V1 n2T2 n2T2 P2 r 5C V2 V2 where C is the proportionality constant. Thus, for two gases under the same conditions of pressure, volume, and temperature (that is, when P1 5 P2, T1 5 T2, and V1 5 V2), it follows that n1 5 n2, which is a mathematical expression of Avogadro’s law. • Dalton’s Law of Partial Pressures. If molecules do not attract or repel one another (assumption 3), then the pressure exerted by one type of molecule is unaffected by the presence of another gas. Consequently, the total pressure is given by the sum of individual gas pressures. Distribution of Molecular Speeds The kinetic theory of gases enables us to investigate molecular motion in more detail. Suppose we have a large number of gas molecules, say, 1 mole, in a container. As long as we hold the temperature constant, the average kinetic energy and the mean- square speed will remain unchanged as time passes. As you might expect, the motion of the molecules is totally random and unpredictable. At a given instant, how many molecules are moving at a particular speed? To answer this question Maxwell ana- lyzed the behavior of gas molecules at different temperatures. Figure 5.17(a) shows typical Maxwell speed distribution curves for nitrogen gas at three different temperatures. At a given temperature, the distribution curve tells us the number of molecules moving at a certain speed. The peak of each curve represents the most probable speed, that is, the speed of the largest number of molecules. Note that the most probable speed increases as temperature increases  5.7 The Kinetic Molecular Theory of Gases 205 N2 (28.02 g/mol) T 5 300 K 100 K Cl2 (70.90 g/mol) Number of molecules Number of molecules 300 K N2 (28.02 g/mol) 700 K He (4.003 g/mol) 500 1000 1500 500 1000 1500 2000 2500 Molecular speed (m/s) Molecular speed (m/s) (a) (b) Figure 5.17 (a) The distribution of speeds for nitrogen gas at three different temperatures. At the higher temperatures, more molecules are moving at faster speeds. (b) The distribution of speeds for three gases at 300 K. At a given temperature, the lighter molecules are moving faster, on the average. (the peak shifts toward the right). Furthermore, the curve also begins to flatten out with increasing temperature, indicating that larger numbers of molecules are moving at greater speed. Figure 5.17(b) shows the speed distributions of three gases at the same temperature. The difference in the curves can be explained by noting that lighter molecules move faster, on average, than heavier ones. The distribution of molecular speeds can be demonstrated with the apparatus shown in Figure 5.18. A beam of atoms (or molecules) exits from an oven at a known temperature and passes through a pinhole (to collimate the beam). Two circular plates mounted on the same shaft are rotated by a motor. The first plate is called the “chop- per” and the second is the detector. The purpose of the chopper is to allow small bursts of atoms (or molecules) to pass through it whenever the slit is aligned with the beam. Within each burst, the faster-moving molecules will reach the detector earlier than the slower-moving ones. Eventually, a layer of deposit will accumulate on the detector. Because the two plates are rotating at the same speed, molecules in the next burst will hit the detector plate at approximately the same place as molecules from To vacuum pump Figure 5.18 (a) Apparatus for studying molecular speed Motor distribution at a certain temperature. The vacuum pump Slow causes the molecules to travel molecules from left to right as shown. (b) The Oven Fast spread of the deposit on the molecules detector gives the range of molecular speeds, and the density of the deposit is proportional to the fraction of molecules moving at different speeds. Detector Chopper with rotating slit Average molecules Detector (a) (b) 206 Chapter 5 ■ Gases the previous burst having the same speed. In time, the molecular deposition will become visible. The density of the deposition indicates the distribution of molecular speeds at that particular temperature. Root-Mean-Square Speed How fast does a molecule move, on the average, at any temperature T ? One way to estimate molecular speed is to calculate the root-mean-square (rms) speed There are comparable ways to estimate (urms), which is an average molecular speed. One of the results of the kinetic the “average” speed of molecules, of which root-mean-square speed is one. theory of gases is that the total kinetic energy of a mole of any gas equals 32RT . Earlier we saw that the average kinetic energy of one molecule is 12mu2 and so we can write KE 5 32RT NA ( 12mu2 ) 5 32RT where NA is Avogadro’s number and m is the mass of a single molecule. Because NAm 5 m, the above equation can be rearranged to give 3RT u2 5 m Taking the square root of both sides gives 3RT 2u2 5 urms 5 (5.16) B m Equation (5.16) shows that the root-mean-square speed of a gas increases with the square root of its temperature (in kelvins). Because m appears in the denominator, it follows that the heavier the gas, the more slowly its molecules move. If we substitute 8.314 J/K ? mol for R (see Appendix 2) and convert the molar mass to kg/mol, then urms will be calculated in meters per second (m/s). This procedure is illustrated in Example 5.16. Example 5.16 Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in m/s at 25°C. Strategy To calculate the root-mean-square speed we need Equation (5.16). What units should we use for R and m so that urms will be expressed in m/s? Solution To calculate urms, the units of R should be 8.314 J/K ? mol and, because 1 J 5 1 kg m2/s2, the molar mass must be in kg/mol. The molar mass of He is 4.003 g/mol, or 4.003 3 1023 kg/mol. From Equation (5.16), 3RT urms 5 B m 3(8.314 J/K ? mol) (298 K) 5 B 4.003 3 10 23 kg/mol 5 21.86 3 106 J/kg (Continued)  5.7 The Kinetic Molecular Theory of Gases 207 Using the conversion factor 1 J 5 1 kg m2/s2 we get urms 5 2 1.86 3 106 kg m2ykg ? s2 5 2 1.86 3 106 m2/s2 5 1.36 3 103 m/s The procedure is the same for N2, the molar mass of which is 28.02 g/mol, or Figure 5.19 The path traveled by a single gas molecule. Each 2.802 3 1022 kg/mol so that we write change in direction represents a collision with another molecule. 3(8.314 JyK ? mole) (298 K) urms 5 B 2.802 3 10 22 kg/mol 5 2 2.65 3 105 m2/s2 5 515 m/s Check Because He is a lighter gas, we expect it to move faster, on average, than N2. A quick way to check the answers is to note that the ratio of the two urms values (1.36 3 103y515 < 2.6) should be equal to the square root of the ratios of the molar masses of N2 to He, that is, 2 28y4 < 2.6. Similar problems: 5.81, 5.82. Practice Exercise Calculate the root-mean-square speed of molecular chlorine in m/s at 20°C. The calculation in Example 5.16 has an interesting relationship to the composition of Earth’s atmosphere. Unlike Jupiter, Earth does not have appreciable amounts of hydrogen or helium in its atmosphere. Why is this the case? A smaller planet than Jupiter, Earth has a weaker gravitational attraction for these lighter molecules. A fairly straightforward calculation shows that to escape Earth’s gravitational field, a molecule must possess an escape velocity equal to or greater than 1.1 3 104 m/s. Because the average speed of helium is considerably greater than that of molecular nitrogen or molecular oxygen, more helium atoms escape from Earth’s atmosphere into outer space. Consequently, only a trace amount of helium is present in our atmosphere. On the other hand, Jupiter, with a mass about 320 times greater than that of Earth, retains both heavy and light gases in its atmosphere. The Chemistry in Action essay on p. 208 describes a fascinating phenomenon Jupiter. The interior of this massive involving gases at extremely low temperatures. planet consists mainly of hydrogen. Gas Diffusion and Effusion We will now discuss two phenomena based on gaseous motion. Gas Diffusion A direct demonstration of gaseous random motion is provided by diffusion, the grad- Diffusion always proceeds from a region of higher concentration to one where the ual mixing of molecules of one gas with molecules of another by virtue of their kinetic concentration is lower. properties. Despite the fact that molecular speeds are very great, the diffusion process takes a relatively long time to complete. For example, when a bottle of concentrated Animation Diffusion of Gases ammonia solution is opened at one end of a lab bench, it takes some time before a person at the other end of the bench can smell it. The reason is that a molecule expe- riences numerous collisions while moving from one end of the bench to the other, as shown in Figure 5.19. Thus, diffusion of gases always happens gradually, and not instantly as molecular speeds seem to suggest. Furthermore, because the root-mean- square speed of a light gas is greater than that of a heavier gas (see Example 5.16), CHEMISTRY in Action Super Cold Atoms W hat happens to a gas when cooled to nearly absolute zero? More than 85 years ago, Albert Einstein, extending work by the Indian physicist Satyendra Nath Bose, predicted that at ex- predicted. Although this BEC was invisible to the naked eye (it measured only 5 3 1023 cm across), the scientists were able to capture its image on a computer screen by focusing another laser tremely low temperatures gaseous atoms of certain elements would beam on it. The laser caused the BEC to break up after about 15 “merge” or “condense” to form a single entity and a new form of seconds, but that was long enough to record its existence. matter. Unlike ordinary gases, liquids, and solids, this supercooled The figure shows the Maxwell velocity distribution† of the substance, which was named the Bose-Einstein condensate (BEC), Rb atoms at this temperature. The colors indicate the number of would contain no individual atoms because the original atoms atoms having velocity specified by the two horizontal axes. The would overlap one another, leaving no space in between. blue and white portions represent atoms that have merged to Einstein’s hypothesis inspired an international effort to pro- form the BEC. duce the BEC. But, as sometimes happens in science, the neces- Within weeks of the Colorado team’s discovery, a group of sary technology was not available until fairly recently, and so early scientists at Rice University, using similar techniques, succeeded investigations were fruitless. Lasers, which use a process based on in producing a BEC with lithium atoms and in 1998 scientists at another of Einstein’s ideas, were not designed specifically for the Massachusetts Institute of Technology were able to produce BEC research, but they became a critical tool for this work. a BEC with hydrogen atoms. Since then, many advances have Finally, in 1995, physicists found the evidence they had been made in understanding the properties of the BEC in general sought for so long. A team at the University of Colorado was the and experiments are being extended to molecular systems. It is first to report success. They created a BEC by cooling a sample expected that studies of the BEC will shed light on atomic prop- of gaseous rubidium (Rb) atoms to about 5.0 3 1028 K using a erties that are still not fully understood (see Chapter 7) and on technique called “laser cooling,” a process in which a laser light the mechanism of superconductivity (see the Chemistry in is directed at a beam of atoms, hitting them head on and dra- Action essay on this topic in Chapter 11). An additional benefit matically slowing them down. The Rb atoms were further cooled might be the development of better lasers. Other applications in an “optical molasses” produced by the intersection of six la- will depend on further study of the BEC itself. Nevertheless, the sers. The slowest, coolest atoms were trapped in a magnetic field discovery of a new form of matter has to be one of the foremost while the faster-moving, “hotter” atoms escaped, thereby remov- scientific achievements of the twentieth century. ing more energy from the gas. Under these conditions, the ki- netic energy of the trapped atoms was virtually zero, which † Velocity distribution differs from speed distribution in that velocity has both accounts for the extremely low temperature of the gas. At this magnitude and direction. Thus, velocity can have both positive and negative values point the Rb atoms formed the condensate, just as Einstein had but speed can have only zero or positive values. Maxwell velocity distribution of Rb atoms at three different temperatures during the formation of Bose-Einstein condensate. In each case, the velocity increases from the center (zero) outward along the two axes. The red color represents the lower number of Rb atoms and the white color the highest. 208  5.7 The Kinetic Molecular Theory of Gases 209 Figure 5.20 A demonstration of gas diffusion. NH3 gas (from a bottle containing aqueous ammonia) combines with HCl gas (from a bottle containing hydrochloric acid) to form solid NH4Cl. Because NH3 is lighter and therefore diffuses faster, solid NH4Cl first appears nearer the HCl bottle (on the right). a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Figure 5.20 illustrates gaseous diffusion. In 1832 the Scottish chemist Thomas Graham† found that under the same conditions of temperature and pressure, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. This statement, now known as Graham’s law of diffusion, is expressed mathematically as r1 m2 5 (5.17) r2 B m1 where r1 and r2 are the diffusion rates of gases 1 and 2, and m1 and m2 are their molar masses, respectively. Gas Effusion Whereas diffusion is a process by which one gas gradually mixes with another, effu- sion is the process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening. Figure 5.21 shows the effusion of a gas into a vacuum. Although effusion differs from diffusion in nature, the rate of effusion of a gas has the same form as Graham’s law of diffusion [see Equation (5.17)]. A helium-filled rubber balloon deflates faster than an air-filled one because the rate of effusion through the pores of the rubber is faster for the lighter helium atoms than for the air molecules. Industrially, gas effusion is used to separate uranium isotopes in the forms of gaseous 235UF6 and 238UF6. By subjecting the gases to many stages of effusion, scientists were able to obtain highly enriched 235U isotope, which was used in the construction of atomic bombs during World War II. Example 5.17 shows an application of Graham’s law. Gas Vacuum Example 5.17 A flammable gas made up only of carbon and hydrogen is found to effuse through a porous barrier in 1.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be. (Continued) Figure 5.21 Gas effusion. Gas molecules move from a high- † Thomas Graham (1805–1869). Scottish chemist. Graham did important work on osmosis and characterized pressure region (left) to a low- a number of phosphoric acids. pressure one through a pinhole. 210 Chapter 5 ■ Gases Strategy The rate of diffusion is the number of molecules passing through a porous barrier in a given time. The longer the time it takes, the slower is the rate. Therefore, the rate is inversely proportional to the time required for diffusion. Equation (5.17) can now be written as r1yr2 5 t2yt1 5 2m2ym1, where t1 and t2 are the times for effusion for gases 1 and 2, respectively. Solution From the molar mass of Br2, we write 1.50 min m 5 4.73 min B 159.8 g/mol where m is the molar mass of the unknown gas. Solving for m, we obtain 1.50 min 2 m5a b 3 159.8 g/mol 4.73 min 5 16.1 g/mol Because the molar mass of carbon is 12.01 g and that of hydrogen is 1.008 g, the gas is Similar problems: 5.87, 5.88. methane (CH4). Practice Exercise It takes 192 s for an unknown gas to effuse through a porous wall and 84 s for the same volume of N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas? Review of Concepts If one mole each of He and Cl2 gases are compared at STP, which of the following quantities will be equal to each other? (a) Root-mean-square speed, (b) effusion rate, (c) average kinetic energy, (d) volume. 5.8 Deviation from Ideal Behavior The gas laws and the kinetic molecular theory assume that molecules in the gaseous state do not exert any force, either attractive or repulsive, on one another. The other assumption is that the volume of the molecules is negligibly small compared with that of the container. A gas that satisfies these two conditions is said to exhibit ideal behavior. Although we can assume that real gases behave like an ideal gas, we cannot expect them to do so under all conditions. For example, without intermolecular forces, gases could not condense to form liquids. The important question is: Under what conditions will gases most likely exhibit nonideal behavior? Figure 5.22 shows PVyRT plotted against P for three real gases and an ideal gas at a given temperature. This graph provides a test of ideal gas behavior. Accord- ing to the ideal gas equation (for 1 mole of gas), PVyRT equals 1, regardless of the actual gas pressure. (When n 5 1, PV 5 nRT becomes PV 5 RT, or PVyRT 5 1.) For real gases, this is true only at moderately low pressures (# 5 atm); significant deviations occur as pressure increases. Attractive forces operate among molecules at relatively short distances. At atmospheric pressure, the molecules in a gas are far apart and the attractive forces are negligible. At high pressures, the density of the gas increases; the molecules are much closer to one another. Intermolecular forces can then be significant enough to affect the motion of the molecules, and the gas will not behave ideally.  5.8 Deviation from Ideal Behavior 211 Figure 5.22 Plot of PV/RT versus CH4 P of 1 mole of a gas at 0°C. For 1 mole of an ideal gas, PV/RT is 2.0 H2 equal to 1, no matter what the pressure of the gas is. For real gases, we observe various NH3 deviations from ideality at high pressures. At very low pressures, PV all gases exhibit ideal behavior; 1.0 Ideal gas that is, their PV/RT values all RT converge to 1 as P approaches zero. 0 200 400 600 800 1000 1200 P (atm) Another way to observe the nonideal behavior of gases is to lower the tempera- ture. Cooling a gas decreases the molecules’ average kinetic energy, which in a sense deprives molecules of the drive they need to break from their mutual attraction. To study real gases accurately, then, we need to modify the ideal gas equation, taking into account intermolecular forces and finite molecular volumes. Such an anal- ysis was first made by the Dutch physicist J. D. van der Waals† in 1873. Besides being mathematically simple, van der Waals’ treatment provides us with an interpretation of real gas behavior at the molecular level. Consider the approach of a particular molecule toward the wall of a container (Figure 5.23). The intermolecular attractions exerted by its neighbors tend to soften the impact made by this molecule against the wall. The overall effect is a lower gas pressure than we would expect for an ideal gas. Van der Waals suggested that the pressure exerted by an ideal gas, Pideal, is related to the experimentally measured pres- sure, Preal, by the equation an2 Pideal 5 Preal 1 V2 h h observed correction pressure term where a is a constant and n and V are the number of moles and volume of the container, respectively. The correction term for pressure (an2yV2) can be understood as follows. The intermolecular interaction that gives rise to nonideal behavior depends on how frequently any two molecules approach each other closely. The frequency of such “encounters” increases with the square of the number of mole- cules per unit volume (n2yV2), because the probability of finding each of the two Figure 5.23 Effect of molecules in a particular region is proportional to nyV. Thus, a is just a proportion- intermolecular forces on the ality constant. pressure exerted by a gas. The speed of a molecule that is Another correction concerns the volume occupied by the gas molecules. In moving toward the container wall the ideal gas equation, V represents the volume of the container. However, each (red sphere) is reduced by the molecule does occupy a finite, although small, intrinsic volume, so the effective attractive forces exerted by its neighbors (gray spheres). volume of the gas becomes (V 2 nb), where n is the number of moles of the gas Consequently, the impact this and b is a constant. The term nb represents the volume occupied by n moles of molecule makes with the wall is the gas. not as great as it would be if no intermolecular forces were present. In general, the measured gas pressure is lower than the † Johannes Diderck van der Waals (1837–1923). Dutch physicist. Van der Waals received the Nobel Prize pressure the gas would exert if it in Physics in 1910 for his work on the properties of gases and liquids. behaved ideally. 212 Chapter 5 ■ Gases Having taken into account the corrections for pressure and volume, we can rewrite the ideal gas equation as follows: an2 Keep in mind that in Equation (5.18), P 1 P 1 V 2 (V 2 nb) 5 nRT 2 (5.18) is the experimentally measured gas pressure and V is the volume of the corrected corrected gas container. pressure volume Equation (5.18), relating P, V, T, and n for a nonideal gas, is known as the van der Waals equation. The van der Waals constants a and b are selected to give the best possible agreement between Equation (5.18) and observed behavior of a par- ticular gas. Table 5.4 lists the values of a and b for a number of gases. The value of a indi- cates how strongly molecules of a given type of gas attract one another. We see that helium atoms have the weakest attraction for one another, because helium has the smallest a value. There is also a rough correlation between molecular size and b. Generally, the larger the molecule (or atom), the greater b is, but the relationship between b and molecular (or atomic) size is not a simple one. Example 5.18 compares the pressure of a gas calculated using the ideal gas equa- tion and the van der Waals equation. Example 5.18 Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation. Strategy To calculate the pressure of NH3 using the ideal gas equation, we proceed as in Example 5.3. What corrections are made to the pressure and volume terms in the van Table 5.4 der Waals equation? van der Waals Constants of Some Common Gases Solution (a) We have the following data: a b V 5 5.20 L atm ? L2 L T 5 (47 1 273) K 5 320 K Gas a b a b n 5 3.50 mol mol2 mol R 5 0.0821 L ? atmyK ? mol He 0.034 0.0237 Ne 0.211 0.0171 Substituting these values in the ideal gas equation, we write Ar 1.34 0.0322 Kr 2.32 0.0398 nRT P5 Xe 4.19 0.0266 V (3.50 mol) (0.0821 L ? atmyK ? mol) (320 K) H2 0.244 0.0266 5 5.20 L N2 1.39 0.0391 5 17.7 atm O2 1.36 0.0318 Cl2 6.49 0.0562 (b) We need Equation (5.18). It is convenient to first calculate the correction terms in CO2 3.59 0.0427 Equation (5.18) separately. From Table 5.4, we have CH4 2.25 0.0428 CCl4 20.4 0.138 a 5 4.17 atm ? L2/mol2 NH3 4.17 0.0371 b 5 0.0371 L/mol H2O 5.46 0.0305 (Continued)  Key Equations 213 so that the correction terms for pressure and volume are an2 (4.17 atm ? L2/mol2 ) (3.50 mol) 2 5 5 1.89 atm V2 (5.20 L) 2 nb 5 (3.50 mol) (0.0371 L/mol) 5 0.130 L Finally, substituting these values in the van der Waals equation, we have (P 1 1.89 atm) (5.20 L 2 0.130 L) 5 (3.50 mol) (0.0821 L ? atmyK ? mol) (320 K) P 5 16.2 atm Check Based on your understanding of nonideal gas behavior, is it reasonable that the pressure calculated using the van der Waals equation should be smaller than that using the ideal gas equation? Why? Similar problem: 5.93. Practice Exercise Using the data shown in Table 5.4, calculate the pressure exerted by 4.37 moles of molecular chlorine confined in a volume of 2.45 L at 38°C. Compare the pressure with that calculated using the ideal gas equation. Review of Concepts What pressure and temperature conditions cause the most deviation from ideal gas behavior? Key Equations P1V1 5 P2V2 (5.2) Boyle’s law. For calculating pressure or volume changes. V1 V2 5 (5.4) Charles’ law. For calculating temperature T1 T2 or volume changes. P1 P2 5 (5.6) Charles’ law. For calculating temperature T1 T2 or pressure changes. V 5 k4n (5.7) Avogadro’s law. Constant P and T. PV 5 nRT (5.8) Ideal gas equation. P 1V 1 P 2V 2 5 (5.9) For calculating changes in pressure, temperature, n1T1 n2T2 volume, or amount of gas. P 1V 1 P 2V 2 5 (5.10) For calculating changes in pressure, temperature, T1 T2 or volume when n is constant. Pm d5 (5.11) For calculating density or molar mass. RT ni Xi 5 (5.13) Definition of mole fraction. nT Pi 5 XiPT (5.14) Dalton’s law of partial pressures. For calculating partial pressures. KE 5 12mu2 5 CT (5.15) Relating the average kinetic energy of a gas to its absolute temperature. 214 Chapter 5 ■ Gases 3RT urms 5 (5.16) For calculating the root-mean-square speed of gas B m molecules. r1 m2 5 (5.17) Graham’s law of diffusion and effusion. r2 B m1 an2 aP 1 b1V 2 nb2 5 nRT (5.18) van der Waals equation. For calculating the pressure of a V2 nonideal gas. Summary of Facts & Concepts 1. At 25°C and 1 atm, a number of elements and molecular 8. Dalton’s law of partial pressures states that each gas in compounds exist as gases. Ionic compounds are solids a mixture of gases exerts the same pressure that it would rather than gases under atmospheric conditions. if it were alone and occupied the same volume. 2. Gases exert pressure because their molecules move 9. The kinetic molecular theory, a mathematical way of freely and collide with any surface with which they describing the behavior of gas molecules, is based on make contact. Units of gas pressure include millimeters the following assumptions: Gas molecules are sepa- of mercury (mmHg), torr, pascals, and atmospheres. rated by distances far greater than their own dimen- One atmosphere equals 760 mmHg, or 760 torr. sions, they possess mass but have negligible volume, 3. The pressure-volume relationships of ideal gases are they are in constant motion, and they frequently collide governed by Boyle’s law: Volume is inversely propor- with one another. The molecules neither attract nor tional to pressure (at constant T and n). repel one another. 4. The temperature-volume relationships of ideal gases 10. A Maxwell speed distribution curve shows how many are described by Charles’ and Gay-Lussac’s law: Vol- gas molecules are moving at various speeds at a given ume is directly proportional to temperature (at constant temperature. As temperature increases, more molecules P and n). move at greater speeds. 5. Absolute zero (2273.15°C) is the lowest theoretically 11. In diffusion, two gases gradually mix with each other. attainable temperature. The Kelvin temperature scale In effusion, gas molecules move through a small open- takes 0 K as absolute zero. In all gas law calculations, ing under pressure. Both processes are governed by the temperature must be expressed in kelvins. same mathematical law—Graham’s law of diffusion 6. The amount-volume relationships of ideal gases are and effusion. described by Avogadro’s law: Equal volumes of gases 12. The van der Waals equation is a modification of the contain equal numbers of molecules (at the same T ideal gas equation that takes into account the nonideal and P). behavior of real gases. It corrects for the fact that real 7. The ideal gas equation, PV 5 nRT, combines the laws gas molecules do exert forces on each other and that of Boyle, Charles, and Avogadro. This equation de- they do have volume. The van der Waals constants are scribes the behavior of an ideal gas. determined experimentally for each gas. Key Words Absolute temperature Dalton’s law of partial Kelvin temperature Pressure, p. 175 scale, p. 182 pressures, p. 196 scale, p. 182 Root-mean-square (rms) speed Absolute zero, p. 182 Diffusion, p. 207 Kinetic energy (KE), p. 202 (urms), p. 206 Atmospheric pressure, p. 175 Effusion, p. 209 Kinetic molecular theory of Standard atmospheric pressure Avogadro’s law, p. 183 Gas constant (R), p. 184 gases, p. 202 (1 atm), p. 176 Barometer, p. 176 Graham’s law of Manometer, p. 177 Standard temperature and Boyle’s law, p. 178 diffusion, p. 209 Mole fraction, p. 197 pressure (STP), p. 185 Charles’ and Gay-Lussac’s Ideal gas, p. 185 Newton (N), p. 175 van der Waals equation, p. 212 law, p. 182 Ideal gas equation, p. 184 Partial pressure, p. 196 Charles’ law, p. 182 Joule (J), p. 202 Pascal (Pa), p. 175  Questions & Problems 215 Questions & Problems • Problems available in Connect Plus 5.11 Why is it that if the barometer reading falls in one Red numbered problems solved in Student Solutions Manual part of the world, it must rise somewhere else? 5.12 Why do astronauts have to wear protective suits Substances That Exist as Gases when they are on the surface of the moon? Review Questions Problems • 5.1 Name five elements and five compounds that exist as gases at room temperature. 5.13 Convert 562 mmHg to atm. 5.2 List the physical characteristics of gases. • 5.14 The atmospheric pressure at the summit of Mt. McKinley is 606 mmHg on a certain day. What is Pressure of a Gas the pressure in atm and in kPa? Review Questions 5.3 Define pressure and give the common units for The Gas Laws pressure. Review Questions 5.4 When you are in a plane flying at high altitudes, your ears often experience pain. This discomfort can 5.15 State the following gas laws in words and also in the be temporarily relieved by yawning or swallowing form of an equation: Boyle’s law, Charles’ law, some water. Explain. Avogadro’s law. In each case, indicate the condi- tions under which the law is applicable, and give the 5.5 Why is mercury a more suitable substance to use in units for each quantity in the equation. a barometer than water? 5.16 A certain amount of gas is contained in a closed 5.6 Explain why the height of mercury in a barometer is mercury manometer as shown here. Assuming no independent of the cross-sectional area of the tube. other parameters change, would h increase, Would the barometer still work if the tubing were decrease, or remain the same if (a) the amount of tilted at an angle, say 15° (see Figure 5.3)? the gas were increased; (b) the molar mass of the 5.7 Explain how a unit of length (mmHg) can be used as gas were doubled; (c) the temperature of the gas a unit for pressure. was increased; (d) the atmospheric pressure in 5.8 Describe what would happen to the column of the room was increased; (e) the mercury in the mercury in the following manometers when the tube were replaced with a less dense fluid; stopcock is opened. (f) some gas was added to the vacuum at the top of the right-side tube; (g) a hole was drilled in the top of the right-side tube? Vacuum Vacuum h h h (a) (b) 5.9 What is the difference between a gas and a vapor? At 25°C, which of the following substances in the gas phase should be properly called a gas and which should be called a vapor: molecular nitrogen (N2), mercury? Problems 5.10 If the maximum distance that water may be brought up a well by a suction pump is 34 ft (10.3 m), how is • 5.17 A gaseous sample of a substance is cooled at con- it possible to obtain water and oil from hundreds of stant pressure. Which of the following diagrams best feet below the surface of Earth? represents the situation if the final temperature is  216 Chapter 5 ■ Gases (a) above the boiling point of the substance and • 5.22 A sample of air occupies 3.8 L when the pressure (b) below the boiling point but above the freezing is 1.2 atm. (a) What volume does it occupy at point of the substance? 6.6 atm? (b) What pressure is required in order to compress it to 0.075 L? (The temperature is kept constant.) • 5.23 A 36.4-L volume of methane gas is heated from 25°C to 88°C at constant pressure. What is the final volume of the gas? • 5.24 Under constant-pressure conditions a sample of (a) (b) (c) (d) hydrogen gas initially at 88°C and 9.6 L is cooled until its final volume is 3.4 L. What is its final temperature? • 5.18 Consider the following gaseous sample in a cylinder • 5.25 Ammonia burns in oxygen gas to form nitric oxide fitted with a movable piston. Initially there are n (NO) and water vapor. How many volumes of NO moles of the gas at temperature T, pressure P, and are obtained from one volume of ammonia at the volume V. same temperature and pressure? • 5.26 Molecular chlorine and molecular fluorine combine to form a gaseous product. Under the same condi- tions of temperature and pressure it is found that one volume of Cl2 reacts with three volumes of F2 to yield two volumes of the product. What is the for- mula of the product? The Ideal Gas Equation Review Questions Choose the cylinder that correctly represents the gas 5.27 List the characteristics of an ideal gas. Write the after each of the following changes. (1) The pressure ideal gas equation and also state it in words. Give on the piston is tripled at constant n and T. (2) The the units for each term in the equation. temperature is doubled at constant n and P. (3) n moles 5.28 Use Equation (5.9) to derive all the gas laws. of another gas are added at constant T and P. (4) T is 5.29 What are standard temperature and pressure (STP)? halved and pressure on the piston is reduced to a What is the significance of STP in relation to the quarter of its original value. volume of 1 mole of an ideal gas? 5.30 Why is the density of a gas much lower than that of a liquid or solid under atmospheric conditions? What units are normally used to express the density of gases? Problems • 5.31 A sample of nitrogen gas kept in a container of vol- (a) (b) (c) ume 2.3 L and at a temperature of 32°C exerts a pressure of 4.7 atm. Calculate the number of moles of gas present. • 5.19 A gas occupying a volume of 725 mL at a pressure • 5.32 Given that 6.9 moles of carbon monoxide gas are of 0.970 atm is allowed to expand at constant tem- present in a container of volume 30.4 L, what is the perature until its pressure reaches 0.541 atm. What pressure of the gas (in atm) if the temperature is its final volume? is 62°C? • 5.20 At 46°C a sample of ammonia gas exerts a pressure • 5.33 What volume will 5.6 moles of sulfur hexafluoride of 5.3 atm. What is the pressure when the volume of (SF6) gas occupy if the temperature and pressure of the gas is reduced to one-tenth (0.10) of the original the gas are 128°C and 9.4 atm? value at the same temperature? • 5.34 A certain amount of gas at 25°C and at a pressure of • 5.21 The volume of a gas is 5.80 L, measured at 1.00 atm. 0.800 atm is contained in a glass vessel. Suppose What is the pressure of the gas in mmHg if the vol- that the vessel can withstand a pressure of 2.00 atm. ume is changed to 9.65 L? (The temperature remains How high can you raise the temperature of the gas constant.) without bursting the vessel?  Questions & Problems 217 • 5.35 A gas-filled balloon having a volume of 2.50 L at 5.50 A compound has the empirical formula SF4. At 20°C, 1.2 atm and 25°C is allowed to rise to the strato- 0.100 g of the gaseous compound occupies a volume sphere (about 30 km above the surface of Earth), of 22.1 mL and exerts a pressure of 1.02 atm. What is where the temperature and pressure are 223°C and the molecular formula of the gas? 3.00 3 1023 atm, respectively. Calculate the final 5.51 What pressure will be required for neon at 30°C volume of the balloon. to have the same density as nitrogen at 20°C and 5.36 The temperature of 2.5 L of a gas initially at STP is 1.0 atm? raised to 250°C at constant volume. Calculate the 5.52 The density of a mixture of fluorine and chlorine final pressure of the gas in atm. gases is 1.77 g/L at 14°C and 0.893 atm. Calculate • 5.37 The pressure of 6.0 L of an ideal gas in a flexible the mass percent of the gases. container is decreased to one-third of its original pressure, and its absolute temperature is de- creased by one-half. What is the final volume of Gas Stoichiometry the gas? Problems • 5.38 A gas evolved during the fermentation of glucose • 5.53 Consider the formation of nitrogen dioxide from (wine making) has a volume of 0.78 L at 20.1°C and nitric oxide and oxygen: 1.00 atm. What was the volume of this gas at the fermentation temperature of 36.5°C and 1.00 atm 2NO(g) 1 O2 (g) ¡ 2NO2 (g) pressure? If 9.0 L of NO are reacted with excess O2 at STP, • 5.39 An ideal gas originally at 0.85 atm and 66°C was what is the volume in liters of the NO2 produced? allowed to expand until its final volume, pressure, and temperature were 94 mL, 0.60 atm, and 45°C, • 5.54 Methane, the principal component of natural gas, respectively. What was its initial volume? is used for heating and cooking. The combustion process is 5.40 Calculate its volume (in liters) of 88.4 g of CO2 at STP. CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l) • 5.41 A gas at 772 mmHg and 35.0°C occupies a volume If 15.0 moles of CH4 are reacted, what is the volume of 6.85 L. Calculate its volume at STP. of CO2 (in liters) produced at 23.0°C and 0.985 atm? • 5.42 Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at • 5.55 When coal is burned, the sulfur present in coal is 30°C. Calculate the pressure inside the vessel after converted to sulfur dioxide (SO2), which is respon- all the dry ice has been converted to CO2 gas. sible for the acid rain phenomenon. • 5.43 At STP, 0.280 L of a gas weighs 0.400 g. Calculate S(s) 1 O2 (g) ¡ SO2 (g) the molar mass of the gas. 5.44 At 741 torr and 44°C, 7.10 g of a gas occupy a vol- If 2.54 kg of S are reacted with oxygen, calculate the ume of 5.40 L. What is the molar mass of the gas? volume of SO2 gas (in mL) formed at 30.5°C and 1.12 atm. • 5.45 Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. Typically, the • 5.56 In alcohol fermentation, yeast converts glucose to temperature and pressure of ozone in the strato- ethanol and carbon dioxide: sphere are 250 K and 1.0 3 1023 atm, respectively. How many ozone molecules are present in 1.0 L of C6H12O6 (s) ¡ 2C2H5OH(l) 1 2CO2 (g) air under these conditions? If 5.97 g of glucose are reacted and 1.44 L of CO2 5.46 Assuming that air contains 78 percent N2, 21 percent gas are collected at 293 K and 0.984 atm, what is the O2, and 1 percent Ar, all by volume, how many mol- percent yield of the reaction? ecules of each type of gas are present in 1.0 L of air 5.57 A compound of P and F was analyzed as follows: at STP? Heating 0.2324 g of the compound in a 378-cm3 • 5.47 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm container turned all of it to gas, which had a pressure and 27.0°C. (a) Calculate the density of the gas of 97.3 mmHg at 77°C. Then the gas was mixed in grams per liter. (b) What is the molar mass of with calcium chloride solution, which turned all of the gas? the F to 0.2631 g of CaF2. Determine the molecular • 5.48 Calculate the density of hydrogen bromide (HBr) formula of the compound. gas in grams per liter at 733 mmHg and 46°C. • 5.58 A quantity of 0.225 g of a metal M (molar mass 5 • 5.49 A certain anesthetic contains 64.9 percent C, 13.5 per- 27.0 g/mol) liberated 0.303 L of molecular hydrogen cent H, and 21.6 percent O by mass. At 120°C and (measured at 17°C and 741 mmHg) from an excess 750 mmHg, 1.00 L of the gaseous compound of hydrochloric acid. Deduce from these data the weighs 2.30 g. What is the molecular formula of corresponding equation and write formulas for the the compound? oxide and sulfate of M.  218 Chapter 5 ■ Gases • 5.59 What is the mass of the solid NH4Cl formed when pressure of the mixture. (b) Calculate the volume in 73.0 g of NH3 are mixed with an equal mass of liters at STP occupied by He and Ne if the N2 is HCl? What is the volume of the gas remaining, removed selectively. measured at 14.0°C and 752 mmHg? What gas • 5.69 Dry air near sea level has the following composition is it? by volume: N2, 78.08 percent; O2, 20.94 percent; Ar, • 5.60 Dissolving 3.00 g of an impure sample of cal- 0.93 percent; CO2, 0.05 percent. The atmospheric cium carbonate in hydrochloric acid produced pressure is 1.00 atm. Calculate (a) the partial pressure 0.656 L of carbon dioxide (measured at 20.0°C of each gas in atm and (b) the concentration of each and 792 mmHg). Calculate the percent by mass gas in moles per liter at 0°C. (Hint: Because volume of calcium carbonate in the sample. State any as- is proportional to the number of moles present, mole sumptions. fractions of gases can be expressed as ratios of vol- • 5.61 Calculate the mass in grams of hydrogen chloride umes at the same temperature and pressure.) produced when 5.6 L of molecular hydrogen mea- • 5.70 A mixture of helium and neon gases is collected sured at STP react with an excess of molecular chlo- over water at 28.0°C and 745 mmHg. If the partial rine gas. pressure of helium is 368 mmHg, what is the par- 5.62 Ethanol (C2H5OH) burns in air: tial pressure of neon? (Vapor pressure of water at 28°C 5 28.3 mmHg.) C2H5OH(l) 1 O2 (g) ¡ CO2 (g) 1 H2O(l) • 5.71 A piece of sodium metal reacts completely with Balance the equation and determine the volume of water as follows: air in liters at 35.0°C and 790 mmHg required to 2Na(s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2 (g) burn 227 g of ethanol. Assume that air is 21.0 per- cent O2 by volume. The hydrogen gas generated is collected over water 5.63 (a) What volumes (in liters) of ammonia and oxy- at 25.0°C. The volume of the gas is 246 mL mea- gen must react to form 12.8 L of nitric oxide sured at 1.00 atm. Calculate the number of grams of according to the equation at the same temperature sodium used in the reaction. (Vapor pressure of and pressure? water at 25°C 5 0.0313 atm.) 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) • 5.72 A sample of zinc metal reacts completely with an excess of hydrochloric acid: (b) What volumes (in liters) of propane and water Zn(s) 1 2HCl(aq) ¡ ZnCl2 (aq) 1 H2 (g) vapor must react to form 8.96 L of hydrogen accord- ing to the equation at the same temperature and The hydrogen gas produced is collected over water at pressure? 25.0°C using an arrangement similar to that shown in Figure 5.15. The volume of the gas is 7.80 L, and the C3H8 (g) 1 3H2O(g) ¡ 3CO(g) 1 7H2 (g) pressure is 0.980 atm. Calculate the amount of zinc 5.64 A 4.00-g sample of FeS containing nonsulfide im- metal in grams consumed in the reaction. (Vapor purities reacted with HCl to give 896 mL of H2S at pressure of water at 25°C 5 23.8 mmHg.) 14°C and 782 mmHg. Calculate mass percent purity • 5.73 Helium is mixed with oxygen gas for deep-sea divers. of the sample. Calculate the percent by volume of oxygen gas in the mixture if the diver has to submerge to a depth where Dalton’s Law of Partial Pressures the total pressure is 4.2 atm. The partial pressure of Review Questions oxygen is maintained at 0.20 atm at this depth. 5.65 State Dalton’s law of partial pressures and explain • 5.74 A sample of ammonia (NH3) gas is completely de- composed to nitrogen and hydrogen gases over what mole fraction is. Does mole fraction have heated iron wool. If the total pressure is 866 mmHg, units? calculate the partial pressures of N2 and H2. 5.66 A sample of air contains only nitrogen and oxygen gases whose partial pressures are 0.80 atm and • 5.75 Consider the three gas containers shown here. All of them have the same volume and are at the same 0.20 atm, respectively. Calculate the total pressure temperature. (a) Which container has the smallest and the mole fractions of the gases. mole fraction of gas A (blue sphere)? (b) Which container has the highest partial pressure of gas B Problems (green sphere)? • 5.67 A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressures of the gases. • 5.68 A 2.5-L flask at 15°C contains a mixture of N2, He, and Ne at partial pressures of 0.32 atm for N2, 0.15 atm for He, and 0.42 atm for Ne. (a) Calculate the total (i) (ii) (iii)  Questions & Problems 219 5.76 The volume of the box on the right is twice that of only 0.72 percent. To separate it from the more the box on the left. The boxes contain helium at- abundant 238U isotope, uranium is first converted oms (red) and hydrogen molecules (green) at the to UF6, which is easily vaporized above room same temperature. (a) Which box has a higher total temperature. The mixture of the 235UF6 and 238UF6 pressure? (b) Which box has a lower partial pres- gases is then subjected to many stages of effusion. sure of helium? Calculate the separation factor, that is, the enrich- ment of 235U relative to 238U after one stage of effusion. 5.87 A gas evolved from the fermentation of glucose is found to effuse through a porous barrier in 15.0 min. Under the same conditions of temperature and pressure, it takes an equal volume of N2 12.0 min to effuse through the same barrier. Calculate the molar mass of the gas and suggest what the gas Kinetic Molecular Theory of Gases might be. Review Questions 5.88 Nickel forms a gaseous compound of the formula Ni(CO)x. What is the value of x given the fact that 5.77 What are the basic assumptions of the kinetic mo- under the same conditions of temperature and pres- lecular theory of gases? How does the kinetic sure, methane (CH4) effuses 3.3 times faster than the molecular theory explain Boyle’s law, Charles’ compound? law, Avogadro’s law, and Dalton’s law of partial pressures? 5.78 What does the Maxwell speed distribution curve tell Deviation from Ideal Behavior us? Does Maxwell’s theory work for a sample of Review Questions 200 molecules? Explain. 5.89 Cite two pieces of evidence to show that gases do • 5.79 Which of the following statements is correct? not behave ideally under all conditions. (a) Heat is produced by the collision of gas mol- ecules against one another. (b) When a gas is 5.90 Under what set of conditions would a gas be ex- heated, the molecules collide with one another pected to behave most ideally? (a) High temperature more often. and low pressure, (b) high temperature and high pressure, (c) low temperature and high pressure, 5.80 What is the difference between gas diffusion and (d) low temperature and low pressure. effusion? State Graham’s law and define the terms in Equation (5.17). 5.91 Shown here are plots of PVyRT against P for one mole of a nonideal gas at two different temperatures. Problems Which curve is at the higher temperature? 5.81 Compare the root-mean-square speeds of O2 and UF6 at 65°C. • 5.82 The temperature in the stratosphere is 223°C. Cal- culate the root-mean-square speeds of N2, O2, and O3 molecules in this region. 5.83 The average distance traveled by a molecule between successive collisions is called mean free path. For a PV given amount of a gas, how does the mean free path RT 1.0 P of a gas depend on (a) density, (b) temperature at constant volume, (c) pressure at constant tempera- ture, (d) volume at constant temperature, and (e) size of the atoms? 5.84 At a certain temperature the speeds of six gaseous molecules in a container are 2.0 m/s, 2.2 m/s, 2.6 m/s, 2.7 m/s, 3.3 m/s, and 3.5 m/s. Calculate the root- mean-square speed and the average speed of the mol- ecules. These two average values are close to each 5.92 (a) A real gas is introduced into a flask of volume other, but the root-mean-square value is always the V. Is the corrected volume of the gas greater or less larger of the two. Why? than V? (b) Ammonia has a larger a value than 5.85 Based on your knowledge of the kinetic theory of neon does (see Table 5.4). What can you conclude gases, derive Graham’s law [Equation (5.17)]. about the relative strength of the attractive forces • 5.86 The 235U isotope undergoes fission when bombarded between molecules of ammonia and between atoms with neutrons. However, its natural abundance is of neon?  220 Chapter 5 ■ Gases Problems completion.) (b) At temperatures above 43°C, the pressure of the gas is observed to increase much • 5.93 Using the data shown in Table 5.4, calculate the more rapidly than predicted by the ideal gas equa- pressure exerted by 2.50 moles of CO2 confined in a tion. Explain. volume of 5.00 L at 450 K. Compare the pressure with that predicted by the ideal gas equation. 5.103 The partial pressure of carbon dioxide varies with seasons. Would you expect the partial pressure in the 5.94 At 27°C, 10.0 moles of a gas in a 1.50-L container Northern Hemisphere to be higher in the summer or exert a pressure of 130 atm. Is this an ideal gas? winter? Explain. Additional Problems 5.104 A healthy adult exhales about 5.0 3 102 mL of a 5.95 Discuss the following phenomena in terms of the gaseous mixture with each breath. Calculate the gas laws: (a) the pressure increase in an automobile number of molecules present in this volume at 37°C tire on a hot day, (b) the “popping” of a paper bag, and 1.1 atm. List the major components of this gas- (c) the expansion of a weather balloon as it rises in eous mixture. the air, (d) the loud noise heard when a lightbulb • 5.105 Sodium bicarbonate (NaHCO3) is called baking shatters. soda because when heated, it releases carbon diox- 5.96 Under the same conditions of temperature and pres- ide gas, which is responsible for the rising of cook- sure, which of the following gases would behave ies, doughnuts, and bread. (a) Calculate the volume most ideally: Ne, N2, or CH4? Explain. (in liters) of CO2 produced by heating 5.0 g of NaHCO3 at 180°C and 1.3 atm. (b) Ammonium bi- • 5.97 Nitroglycerin, an explosive compound, decomposes carbonate (NH4HCO3) has also been used for the according to the equation same purpose. Suggest one advantage and one dis- 4C3H5 (NO3 ) 3 (s) ¡ advantage of using NH4HCO3 instead of NaHCO3 12CO2 (g) 1 10H2O(g) 1 6N2 (g) 1 O2 (g) for baking. 5.106 A barometer having a cross-sectional area of 1.00 Calculate the total volume of gases when collected cm2 at sea level measures a pressure of 76.0 cm of at 1.2 atm and 25°C from 2.6 3 102 g of nitroglyc- mercury. The pressure exerted by this column of erin. What are the partial pressures of the gases mercury is equal to the pressure exerted by all the under these conditions? air on 1 cm2 of Earth’s surface. Given that the den- • 5.98 The empirical formula of a compound is CH. At sity of mercury is 13.6 g/mL and the average radius 200°C, 0.145 g of this compound occupies 97.2 mL of Earth is 6371 km, calculate the total mass of at a pressure of 0.74 atm. What is the molecular Earth’s atmosphere in kilograms. (Hint: The sur- formula of the compound? face area of a sphere is 4πr2 where r is the radius of • 5.99 When ammonium nitrite (NH4NO2) is heated, it the sphere.) decomposes to give nitrogen gas. This property is • 5.107 Some commercial drain cleaners contain a mixture used to inflate some tennis balls. (a) Write a bal- of sodium hydroxide and aluminum powder. When anced equation for the reaction. (b) Calculate the the mixture is poured down a clogged drain, the fol- quantity (in grams) of NH 4NO2 needed to inflate lowing reaction occurs: a tennis ball to a volume of 86.2 mL at 1.20 atm and 22°C. 2NaOH(aq) 1 2Al(s) 1 6H2O(l) ¡ 2NaAl(OH) 4 (aq) 1 3H2 (g) • 5.100 The percent by mass of bicarbonate (HCO2 3 ) in a certain Alka-Seltzer product is 32.5 percent. Calcu- The heat generated in this reaction helps melt away late the volume of CO2 generated (in mL) at 37°C obstructions such as grease, and the hydrogen gas and 1.00 atm when a person ingests a 3.29-g tablet. released stirs up the solids clogging the drain. (Hint: The reaction is between HCO2 3 and HCl acid Calculate the volume of H2 formed at 23°C and in the stomach.) 1.00 atm if 3.12 g of Al are treated with an excess of 5.101 The boiling point of liquid nitrogen is 2196°C. On NaOH. the basis of this information alone, do you think nitrogen is an ideal gas? • 5.108 The volume of a sample of pure HCl gas was 189 mL at 25°C and 108 mmHg. It was completely dissolved • 5.102 In the metallurgical process of refining nickel, in about 60 mL of water and titrated with an NaOH the metal is first combined with carbon monox- solution; 15.7 mL of the NaOH solution were re- ide to form tetracarbonylnickel, which is a gas at quired to neutralize the HCl. Calculate the molarity 43°C: of the NaOH solution. Ni(s) 1 4CO(g) ¡ Ni(CO) 4 (g) • 5.109 Propane (C 3H 8) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a This reaction separates nickel from other solid balanced equation for this reaction. (b) Calculate impurities. (a) Starting with 86.4 g of Ni, calcu- the number of liters of carbon dioxide measured late the pressure of Ni(CO)4 in a container of vol- at STP that could be produced from 7.45 g of ume 4.00 L. (Assume the above reaction goes to propane.  Questions & Problems 221 5.110 Consider the following apparatus. Calculate the mixture of gases of the following composition: partial pressures of helium and neon after the stop- (a) CO2 and H2, (b) He and N2. cock is open. The temperature remains constant • 5.114 A certain hydrate has the formula MgSO4 ? xH2O. A at 16°C. quantity of 54.2 g of the compound is heated in an oven to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.00-L container at 120°C, calculate x. He Ne 5.115 A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a volume of 1.67 L at 1.24 atm and 26°C. From these data, calculate the 1.2 L 3.4 L 0.63 atm 2.8 atm percent composition by mass of Na2CO3 in the mixture. • 5.116 The following apparatus can be used to measure • 5.111 Nitric oxide (NO) reacts with molecular oxygen as atomic and molecular speed. Suppose that a beam follows: of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows 2NO(g) 1 O2 (g) ¡ 2NO2 (g) the atoms to strike a target area. Because the cylin- der is rotating, atoms traveling at different speeds Initially NO and O2 are separated as shown here. will strike the target at different positions. In time, When the valve is opened, the reaction quickly a layer of the metal will deposit on the target area, goes to completion. Determine what gases remain and the variation in its thickness is found to cor- at the end and calculate their partial pressures. respond to Maxwell’s speed distribution. In one Assume that the temperature remains constant experiment it is found that at 850°C some bismuth at 25°C. (Bi) atoms struck the target at a point 2.80 cm from the spot directly opposite the slit. The diam- eter of the cylinder is 15.0 cm and it is rotating at 130 revolutions per second. (a) Calculate the speed (m/s) at which the target is moving. (Hint: The NO O2 circumference of a circle is given by 2πr, where r is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel 2.80 cm. (c) Deter- 4.00 L at 2.00 L at mine the speed of the Bi atoms. Compare your re- 0.500 atm 1.00 atm sult in (c) with the urms of Bi at 850°C. Comment on the difference. 5.112 Consider the apparatus shown here. When a small amount of water is introduced into the flask by Rotating cylinder squeezing the bulb of the medicine dropper, water is squirted upward out of the long glass tubing. Explain Target this observation. (Hint: Hydrogen chloride gas is Bi atoms soluble in water.) Slit HCl gas H2O • 5.117 If 10.00 g of water are introduced into an evacu- ated flask of volume 2.500 L at 65°C, calculate the mass of water vaporized. (Hint: Assume that the volume of the remaining liquid water is neg- Rubber ligible; the vapor pressure of water at 65°C is bulb 187.5 mmHg.) H2O 5.118 Commercially, compressed oxygen is sold in metal cylinders. If a 120-L cylinder is filled with oxygen to a pressure of 132 atm at 22°C, what is the mass (in grams) of O2 present? How many liters of O2 gas 5.113 Describe how you would measure, by either chemi- at 1.00 atm and 22°C could the cylinder produce? cal or physical means, the partial pressures of a (Assume ideal behavior.)  222 Chapter 5 ■ Gases • 5.119 The shells of hard-boiled eggs sometimes crack due the plunger) is 5.58 mL and the atmospheric pres- to the rapid thermal expansion of the shells at high sure is 760 mmHg. Given that the compound’s em- temperatures. Suggest another reason why the shells pirical formula is CH2, determine the molar mass of may crack. the compound. 5.120 Ethylene gas (C2H4) is emitted by fruits and is known to be responsible for their ripening. Based on this information, explain why a bunch of Rubber tip 5 4 3 2 1 bananas ripens faster in a closed paper bag than in a bowl. • 5.121 About 8.0 3 106 tons of urea [(NH2)2CO] are used annually as a fertilizer. The urea is prepared at 200°C and under high-pressure conditions from carbon dioxide and ammonia (the products are 5.128 In 1995 a man suffocated as he walked by an aban- urea and steam). Calculate the volume of ammonia doned mine in England. At that moment there was a (in liters) measured at 150 atm needed to prepare sharp drop in atmospheric pressure due to a change 1.0 ton of urea. in the weather. Suggest what might have caused the 5.122 Some ballpoint pens have a small hole in the main man’s death. body of the pen. What is the purpose of this hole? • 5.129 Acidic oxides such as carbon dioxide react with 5.123 The gas laws are vitally important to scuba divers. basic oxides like calcium oxide (CaO) and barium The pressure exerted by 33 ft of seawater is equiv- oxide (BaO) to form salts (metal carbonates). alent to 1 atm pressure. (a) A diver ascends (a) Write equations representing these two quickly to the surface of the water from a depth of reactions. (b) A student placed a mixture of BaO 36 ft without exhaling gas from his lungs. By and CaO of combined mass 4.88 g in a 1.46-L what factor will the volume of his lungs increase flask containing carbon dioxide gas at 35°C and by the time he reaches the surface? Assume that 746 mmHg. After the reactions were complete, the temperature is constant. (b) The partial pres- she found that the CO2 pressure had dropped to sure of oxygen in air is about 0.20 atm. (Air is 20 252 mmHg. Calculate the percent composition by percent oxygen by volume.) In deep-sea diving, mass of the mixture. Assume volumes of the solids the composition of air the diver breathes must be are negligible. changed to maintain this partial pressure. What 5.130 Identify the Maxwell speed distribution curves must the oxygen content (in percent by volume) shown here with the following gases: Br2, CH4, be when the total pressure exerted on the diver is N2, SO3. 4.0 atm? (At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gases.) (Hint: See the Chemistry Number of molecules in Action essay on p. 200.) 5.124 Nitrous oxide (N2O) can be obtained by the ther- mal decomposition of ammonium nitrate (NH4NO3). (a) Write a balanced equation for the reaction. (b) In a certain experiment, a student ob- tains 0.340 L of the gas at 718 mmHg and 24°C. If the gas weighs 0.580 g, calculate the value of the gas constant. 0 500 1000 1500 • 5.125 Two vessels are labeled A and B. Vessel A contains Molecular speed (m/s) NH3 gas at 70°C, and vessel B contains Ne gas at the same temperature. If the average kinetic energy of • 5.131 The running engine of an automobile produces car- NH3 is 7.1 3 10221 J/molecule, calculate the mean- bon monoxide (CO), a toxic gas, at the rate of about square speed of Ne atoms in m2/s2. 188 g CO per hour. A car is left idling in a poorly 5.126 Which of the following molecules has the largest a ventilated garage that is 6.0 m long, 4.0 m wide, and value: CH4, F2, C6H6, Ne? 2.2 m high at 20°C. (a) Calculate the rate of CO 5.127 The following procedure is a simple though some- production in moles per minute. (b) How long would what crude way to measure the molar mass of a gas. it take to build up a lethal concentration of CO of A liquid of mass 0.0184 g is introduced into a sy- 1000 ppmv (parts per million by volume)? ringe like the one shown here by injection through 5.132 Interstellar space contains mostly hydrogen atoms at the rubber tip using a hypodermic needle. The sy- a concentration of about 1 atom/cm3. (a) Calculate ringe is then transferred to a temperature bath heated the pressure of the H atoms. (b) Calculate the vol- to 45°C, and the liquid vaporizes. The final volume ume (in liters) that contains 1.0 g of H atoms. The of the vapor (measured by the outward movement of temperature is 3 K.  Questions & Problems 223 5.133 Atop Mt. Everest, the atmospheric pressure is of NO2 and N2O4. At 25°C and 0.98 atm, the density 210 mmHg and the air density is 0.426 kg/m3. of this gas mixture is 2.7 g/L. What is the partial (a) Calculate the air temperature, given that the pressure of each gas? molar mass of air is 29.0 g/mol. (b) Assuming no 5.140 The Chemistry in Action essay on p. 208 de- change in air composition, calculate the percent scribes the cooling of rubidium vapor to 5.0 3 decrease in oxygen gas from sea level to the top 10 28 K. Calculate the root-mean-square speed of Mt. Everest. and average kinetic energy of a Rb atom at this 5.134 Relative humidity is defined as the ratio (expressed temperature. as a percentage) of the partial pressure of water va- • 5.141 Lithium hydride reacts with water as follows: por in the air to the equilibrium vapor pressure (see Table 5.3) at a given temperature. On a certain sum- LiH(s) 1 H2O(l) ¡ LiOH(aq) 1 H2 (g) mer day in North Carolina the partial pressure of water vapor in the air is 3.9 3 103 Pa at 30°C. During World War II, U.S. pilots carried LiH tablets. Calculate the relative humidity. In the event of a crash landing at sea, the LiH would 5.135 Under the same conditions of temperature and pres- react with the seawater and fill their life belts and sure, why does one liter of moist air weigh less than lifeboats with hydrogen gas. How many grams of one liter of dry air? In weather forecasts, an oncom- LiH are needed to fill a 4.1-L life belt at 0.97 atm ing low-pressure front usually means imminent rain- and 12°C? fall. Explain. 5.142 The atmosphere on Mars is composed mainly of car- • 5.136 Air entering the lungs ends up in tiny sacs called bon dioxide. The surface temperature is 220 K and alveoli. It is from the alveoli that oxygen diffuses the atmospheric pressure is about 6.0 mmHg. Tak- into the blood. The average radius of the alveoli is ing these values as Martian “STP,” calculate the mo- 0.0050 cm and the air inside contains 14 percent lar volume in liters of an ideal gas on Mars. oxygen. Assuming that the pressure in the alveoli 5.143 Venus’s atmosphere is composed of 96.5 percent is 1.0 atm and the temperature is 37°C, calculate CO2, 3.5 percent N2, and 0.015 percent SO2 by vol- the number of oxygen molecules in one of the al- ume. Its standard atmospheric pressure is 9.0 3 106 veoli. (Hint: The volume of a sphere of radius r Pa. Calculate the partial pressures of the gases in is 43πr3 .) pascals. 5.137 A student breaks a thermometer and spills most of 5.144 A student tries to determine the volume of a bulb the mercury (Hg) onto the floor of a laboratory like the one shown on p. 191. These are her results: that measures 15.2 m long, 6.6 m wide, and 2.4 m Mass of the bulb filled with dry air at 23°C and high. (a) Calculate the mass of mercury vapor (in 744 mmHg 5 91.6843 g; mass of evacuated bulb 5 grams) in the room at 20°C. The vapor pressure of 91.4715 g. Assume the composition of air is 78 per- mercury at 20°C is 1.7 3 1026 atm. (b) Does the cent N2, 21 percent O2, and 1 percent argon. What is concentration of mercury vapor exceed the air the volume (in milliliters) of the bulb? (Hint: First quality regulation of 0.050 mg Hg/m3 of air? calculate the average molar mass of air, as shown in (c) One way to treat small quantities of spilled Problem 3.152.) mercury is to spray sulfur powder over the metal. 5.145 Apply your knowledge of the kinetic theory of Suggest a physical and a chemical reason for gases to the following situations. (a) Two flasks of this action. volumes V1 and V2 (V2 . V1) contain the same 5.138 Consider two bulbs containing argon (left) and number of helium atoms at the same temperature. oxygen (right) gases. After the stopcock is opened, (i) Compare the root-mean-square (rms) speeds the pressure of the combined gases is 1.08 atm. and average kinetic energies of the helium (He) Calculate the volume of the right bulb. The tem- atoms in the flasks. (ii) Compare the frequency perature is kept at 20°C. Assume ideal behavior. and the force with which the He atoms collide with the walls of their containers. (b) Equal num- bers of He atoms are placed in two flasks of the same volume at temperatures T1 and T2 (T2 . T1). (i) Compare the rms speeds of the atoms in the two flasks. (ii) Compare the frequency and the force Ar O2 with which the He atoms collide with the walls of their containers. (c) Equal numbers of He and neon (Ne) atoms are placed in two flasks of the same volume, and the temperature of both gases is n 5 0.227 mol n 5 0.144 mol 74°C. Comment on the validity of the following V 5 3.60 L V5? statements: (i) The rms speed of He is equal to that of Ne. (ii) The average kinetic energies of the two • 5.139 Nitrogen dioxide (NO2) cannot be obtained in a pure gases are equal. (iii) The rms speed of each He form in the gas phase because it exists as a mixture atom is 1.47 3 103 m/s.  224 Chapter 5 ■ Gases • 5.146 It has been said that every breath we take, on aver- 5.153 A mixture of calcium carbonate (CaCO3) and mag- age, contains molecules that were once exhaled by nesium carbonate (MgCO3) of mass 6.26 g reacts Wolfgang Amadeus Mozart (1756–1791). The fol- completely with hydrochloric acid (HCl) to gener- lowing calculations demonstrate the validity of this ate 1.73 L of CO2 at 48°C and 1.12 atm. Calculate statement. (a) Calculate the total number of mole- the mass percentages of CaCO3 and MgCO3 in the cules in the atmosphere. (Hint: Use the result in mixture. Problem 5.106 and 29.0 g/mol as the molar mass of 5.154 A 6.11-g sample of a Cu-Zn alloy reacts with HCl air.) (b) Assuming the volume of every breath (in- acid to produce hydrogen gas. If the hydrogen gas hale or exhale) is 500 mL, calculate the number of has a volume of 1.26 L at 22°C and 728 mmHg, molecules exhaled in each breath at 37°C, which is what is the percent of Zn in the alloy? (Hint: Cu the body temperature. (c) If Mozart’s life span was does not react with HCl.) exactly 35 years, what is the number of molecules he exhaled in that period? (Given that an average • 5.155 A stockroom supervisor measured the contents of a partially filled 25.0-gallon acetone drum on a day person breathes 12 times per minute.) (d) Calculate when the temperature was 18.0°C and atmospheric the fraction of molecules in the atmosphere that pressure was 750 mmHg, and found that 15.4 gal- was exhaled by Mozart. How many of Mozart’s lons of the solvent remained. After tightly sealing molecules do we breathe in with every inhalation the drum, an assistant dropped the drum while car- of air? Round off your answer to one significant rying it upstairs to the organic laboratory. The figure. (e) List three important assumptions in drum was dented and its internal volume was de- these calculations. creased to 20.4 gallons. What is the total pressure 5.147 At what temperature will He atoms have the same inside the drum after the accident? The vapor pres- urms value as N2 molecules at 25°C? sure of acetone at 18.0°C is 400 mmHg. (Hint: At 5.148 Estimate the distance (in nanometers) between the time the drum was sealed, the pressure inside molecules of water vapor at 100°C and 1.0 atm. the drum, which is equal to the sum of the pres- Assume ideal behavior. Repeat the calculation sures of air and acetone, was equal to the atmo- for liquid water at 100°C, given that the density spheric pressure.) of water is 0.96 g/cm3 at that temperature. Com- • 5.156 In 2.00 min, 29.7 mL of He effuse through a small ment on your results. (Assume water molecule to hole. Under the same conditions of pressure and be a sphere with a diameter of 0.3 nm.) (Hint: temperature, 10.0 mL of a mixture of CO and CO2 First calculate the number density of water mol- effuse through the hole in the same amount of time. ecules. Next, convert the number density to linear Calculate the percent composition by volume of the density, that is, number of molecules in one mixture. direction.) 5.157 Referring to Figure 5.22, explain the following: 5.149 Which of the noble gases would not behave ideally (a) Why do the curves dip below the horizontal line under any circumstance? Why? labeled ideal gas at low pressures and then why do • 5.150 A relation known as the barometric formula is use- they arise above the horizontal line at high pres- ful for estimating the change in atmospheric pres- sures? (b) Why do the curves all converge to 1 at sure with altitude. The formula is given by very low pressures? (c) Each curve intercepts the P 5 P0e2gmhyRT , where P and P0 are the pressures at horizontal line labeled ideal gas. Does it mean that height h and sea level, respectively, g is the accelera- at that point the gas behaves ideally? tion due to gravity (9.8 m/s2), m is the average mo- • 5.158 A mixture of methane (CH4) and ethane (C2H6) is lar mass of air (29.0 g/mol), and R is the gas constant. stored in a container at 294 mmHg. The gases are Calculate the atmospheric pressure in atm at a height burned in air to form CO2 and H2O. If the pressure of 5.0 km, assuming the temperature is constant at of CO2 is 356 mmHg measured at the same tempera- 5°C and P0 5 1.0 atm. ture and volume as the original mixture, calculate • 5.151 A 5.72-g sample of graphite was heated with 68.4 g the mole fractions of the gases. of O2 in a 8.00-L flask. The reaction that took 5.159 Use the kinetic theory of gases to explain why hot place was air rises. 5.160 One way to gain a physical understanding of b in the C(graphite) 1 O2 (g) ¡ CO2 (g) van der Waals equation is to calculate the “excluded After the reaction was complete, the temperature in volume.” Assume that the distance of closest ap- the flask was 182°C. What was the total pressure proach between two similar atoms is the sum of their inside the flask? radii (2r). (a) Calculate the volume around each atom into which the center of another atom cannot 5.152 An equimolar mixture of H2 and D2 effuses through penetrate. (b) From your result in (a), calculate the an orifice (small hole) at a certain temperature. Cal- excluded volume for 1 mole of the atoms, which is culate the composition (in mole fractions) of the the constant b. How does this volume compare with gases that pass through the orifice. The molar mass the sum of the volumes of 1 mole of the atoms? of D2 is 2.014 g/mol.  Questions & Problems 225 5.161 Use the van der Waals constants in Table 5.4 to esti- 5.167 A gaseous hydrocarbon (containing C and H atoms) mate the radius of argon in picometers. (Hint: See in a container of volume 20.2 L at 350 K and Problem 5.160.) 6.63 atm reacts with an excess of oxygen to form 5.162 Identify the gas whose root-mean-square speed is 205.1 g of CO2 and 168.0 g of H2O. What is the 2.82 times that of hydrogen iodide (HI) at the same molecular formula of the hydrocarbon? temperature. 5.168 Three flasks containing gases A (red) and B (green) 5.163 A 5.00-mole sample of NH 3 gas is kept in a are shown here. (i) If the pressure in (a) is 4.0 atm, 1.92-L container at 300 K. If the van der Waals what are the pressures in (b) and (c)? (ii) Calculate equation is assumed to give the correct answer the total pressure and partial pressure of each gas for the pressure of the gas, calculate the percent after the valves are opened. The volumes of (a) and error made in using the ideal gas equation to cal- (c) are 4.0 L each and that of (b) is 2.0 L. The tem- culate the pressure. perature is the same throughout. 5.164 The root-mean-square speed of a certain gaseous oxide is 493 m/s at 20°C. What is the molecular for- mula of the compound? 5.165 Referring to Figure 5.17, we see that the maximum of each speed distribution plot is called the most probable speed (ump) because it is the speed pos- sessed by the largest number of molecules. It is given by ump 5 12RTym. (a) Compare ump with (a) (b) (c) urms for nitrogen at 25°C. (b) The following diagram shows the Maxwell speed distribution curves for an ideal gas at two different temperatures T1 and T2. 5.169 (a) Show that the pressure exerted by a fluid P (in Calculate the value of T2. pascals) is given by P 5 hdg, where h is the column of the fluid in meters, d is the density in kg/m3, and g is the acceleration due to gravity (9.81 m/s2). (Hint: See Appendix 2.) (b) The volume of an air T1 5 300 K bubble that starts at the bottom of a lake at 5.24°C increases by a factor of 6 as it rises to the surface of Number of molecules water where the temperature is 18.73°C and the air pressure is 0.973 atm. The density of the lake water is 1.02 g/cm3. Use the equation in (a) to determine T2 5 ? the depth of the lake in meters. 5.170 A student first measured the total pressure of a mix- ture of gases methane (CH4), ethane (C2H6), and propane (C3H8) at a certain temperature, which turned out to be 4.50 atm. She then recorded the 0 500 1000 1500 2000 Molecular speed (m/s) mass spectra of the gases shown here. Calculate the partial pressure of the gases. 5.166 A gaseous reaction takes place at constant volume and constant pressure in a cylinder shown here. 5 Which of the following equations best describes the 4 Intensity of peaks reaction? The initial temperature (T1) is twice that of (arbitrary units) the final temperature (T2). 3 (a) A 1 B ¡ C 2 (b) AB ¡ C 1 D (c) A 1 B ¡ C 1 D 1 (d) A 1 B ¡ 2C 1 D 0 0 10 20 30 40 50 Molecular mass (amu) 88n 5.171 In 2012, Felix Baumgartner jumped from a bal- loon roughly 24 mi above Earth, breaking the re- T1 T2 cord for the highest skydive. He reached speeds of more than 700 miles per hour and became the first skydiver to exceed the speed of sound during free 226 Chapter 5 ■ Gases fall. The helium-filled plastic balloon used to at the time of the jump (8.5 3 108 L, 267.8°C, carry Baumgartner to the edge of space was 0.027 mmHg). (b) Determine the volume of the designed to expand to 8.5 3 108 L in order to helium in the balloon just before it was released, accommodate the low pressures at the altitude assuming a pressure of 1.0 atm and a temperature required to break the record. (a) Calculate the of 23°C. mass of helium in the balloon from the conditions Interpreting, Modeling & Estimating 5.172 Which of the following has a greater mass: a sample emits visible (white) light. Estimate the mass of Hg of air of volume V at a certain temperature T and vapor present in the type of long, thin fluorescent pressure P or a sample of air plus water vapor hav- tubes used in offices. (b) Ordinary tungsten incan- ing the same volume and at the same temperature descent lightbulbs used in households are filled and pressure? with argon gas at about 0.5 atm to retard the subli- 5.173 A flask with a volume of 14.5 L contains 1.25 moles mation of the tungsten filament. Estimate the num- of helium gas. Estimate the average distance be- ber of moles of Ar in a typical lightbulb. tween He atoms in nanometers. 5.176 (a) Estimate the volume of air at 1.0 atm and 22°C 5.174 Hyperbaric oxygen therapy (HBOT) is very effec- needed to fill a bicycle tire to a pressure of 5.0 atm tive in treating burns, crush injuries that impede at the same temperature. (Note that the 5.0 atm is the blood flow, and tissue-damaging infections, as gauge pressure, which is the difference between the well as carbon monoxide poisoning. However, it pressure in the tire and atmospheric pressure.) has generated some controversy in its application (b) The tire is pumped by filling the cylinder of a to other maladies (for example, autism, multiple hand pump with air at 1.0 atm and then, by com- sclerosis). A typical oxygen hyperbaric chamber pressing the gas in the cylinder, adding all the air in is shown here. HBOT can be administered using the pump to the air in the tire. If the volume of the pressure up to six atmospheres, but lower pres- pump is 33 percent of the tire’s volume, what is the sures are more common. (a) If this chamber was gauge pressure in the tire after three full strokes of pressurized to 3.0 atm with pure oxygen, how the pump? many moles of O2 would be contained in an empty 5.177 On October 15, 2009, a homemade helium balloon chamber? (b) Given that a full tank of oxygen was released, and for a while authorities were led contains about 2500 moles of the gas, how many to believe that a 6-year-old boy had been carried times could the chamber be filled with a single away in the balloon. (The incident was later re- tank of oxygen? vealed to be a hoax.) The balloon traveled more than 50 mi and reached a height of 7000 ft. The shape and span of the balloon are shown in the fig- ure. How much weight could this balloon lift? (A helium balloon can lift a mass equal to the differ- ence in the mass of air and the mass of helium that would be contained in the balloon.) Could it actu- ally lift a 6-year-old boy? 20 ft 5.175 (a) Fluorescent lightbulbs contain a small amount of mercury, giving a mercury vapor pressure of around 1 3 1025 atm. When excited electrically, the Hg atoms emit UV light, which excites the phosphor coating of the inner tube, which then  Answers to Practice Exercises 227 Answers to Practice Exercises 5.1 0.986 atm. 5.2 39.3 kPa. 5.3 9.29 L. 5.4 30.6 L. 0.0657 atm; C3H8: 0.0181 atm. 5.15 0.0653 g. 5.5 4.46 3 103 mmHg. 5.6 0.68 atm. 5.7 2.6 atm. 5.16 321 m/s. 5.17 146 g/mol. 5.18 30.0 atm; 5.8 13.1 g/L. 5.9 44.1 g/mol. 5.10 B2H6. 5.11 96.9 L. 45.5 atm using the ideal gas equation. 5.12 4.75 L. 5.13 0.338 M. 5.14 CH4: 1.29 atm; C2H6: CHEMICAL M YS TERY Out of Oxygen† I n September 1991 four men and four women entered the world’s largest glass bubble, known as Biosphere II, to test the idea that humans could design and build a totally self-contained ecosystem, a model for some future colony on another planet. Biosphere II (Earth is considered Biosphere I) was a 3-acre mini-world, complete with a tropical rain forest, savanna, marsh, desert, and working farm that was intended to be fully self- sufficient. This unique experiment was to continue for 2 to 3 years, but almost immediately there were signs that the project could be in jeopardy. Soon after the bubble had been sealed, sensors inside the facility showed that the concentration of oxygen in Biosphere II’s atmosphere had fallen from its initial level of 21 percent (by volume), while the amount of carbon dioxide had risen from a level of 0.035 percent (by volume), or 350 ppm (parts per million). Alarmingly, the oxygen level contin- ued to fall at a rate of about 0.5 percent a month and the level of carbon dioxide kept rising, forcing the crew to turn on electrically powered chemical scrubbers, similar to those on submarines, to remove some of the excess CO2. Gradually the CO2 level stabilized around 4000 ppm, which is high but not dangerous. The loss of oxygen did not stop, though. By January 1993—16 months into the experiment—the oxygen concentration had dropped to 14 percent, which is equivalent to the O2 concentration in air at an elevation of 4360 m (14,300 ft). The crew began having trouble performing normal tasks. For their safety it was necessary to pump pure oxygen into Biosphere II. With all the plants present in Biosphere II, the production of oxygen should have been greater as a consequence of photosynthesis. Why had the oxygen concentration declined to such a low level? A small part of the loss was blamed on unusually cloudy weather, which had slowed down plant growth. The possibility that iron in the soil was reacting with oxygen to form iron(III) oxide or rust was ruled out along with several other explanations for lack of evidence. The most plausible hypothesis was that microbes (micro- organisms) were using oxygen to metabolize the excess organic matter that had been added to the soils to promote plant growth. This turned out to be the case. Identifying the cause of oxygen depletion raised another question. Metabolism pro- duces carbon dioxide. Based on the amount of oxygen consumed by the microbes, the CO2 level should have been at 40,000 ppm, 10 times what was measured. What happened to the excess gas? After ruling out leakage to the outside world and reactions between CO2 with compounds in the soils and in water, scientists found that the concrete inside Biosphere II was consuming large amounts of CO2! Concrete is a mixture of sand and gravel held together by a binding agent that is a mixture of calcium silicate hydrates and calcium hydroxide. The calcium hydroxide is the key ingredient in the CO2 mystery. Carbon dioxide diffuses into the porous structure of concrete, then reacts with calcium hydroxide to form calcium carbonate and water: Ca(OH) 2 (s) 1 CO2 (g) ¡ CaCO3 (s) 1 H2O(l) Under normal conditions, this reaction goes on slowly. But CO2 concentrations in Biosphere II were much higher than normal, so the reaction proceeded much faster. In † Adapted with permission from “Biosphere II: Out of Oxygen,” by Joe Alper, CHEM MATTERS, February, 1995, p. 8. Copyright 1995 American Chemical Society. 228  Vegetations in Biosphere II. fact, in just over 2 years, CaCO3 had accumulated to a depth of more than 2 cm in Biosphere II’s concrete. Some 10,000 m2 of exposed concrete was hiding 500,000 to 1,500,000 moles of CO2. The water produced in the reaction between Ca(OH)2 and CO2 created another prob- lem: CO2 also reacts with water to form carbonic acid (H2CO3), and hydrogen ions pro- duced by the acid promote the corrosion of the reinforcing iron bars in the concrete, thereby weakening its structure. This situation was dealt with effectively by painting all concrete surfaces with an impermeable coating. In the meantime, the decline in oxygen (and hence also the rise in carbon dioxide) slowed, perhaps because there was now less organic matter in the soils and also because new lights in the agricultural areas may have boosted photosynthesis. The project was terminated prematurely and in 1996, the facility was transformed into a science educa- tion and research center. As of 2011, the Biosphere is under the management of the University of Arizona. The Biosphere II experiment is an interesting project from which we can learn a lot about Earth and its inhabitants. If nothing else, it has shown us how complex Earth’s ecosystems are and how difficult it is to mimic nature, even on a small scale. Chemical Clues 1. What solution would you use in a chemical scrubber to remove carbon dioxide? 2. Photosynthesis converts carbon dioxide and water to carbohydrates and oxygen gas, while metabolism is the process by which carbohydrates react with oxygen to form carbon dioxide and water. Using glucose (C6H12O6) to represent carbohydrates, write equations for these two processes. 3. Why was diffusion of O2 from Biosphere II to the outside world not considered a possible cause for the depletion in oxygen? 4. Carbonic acid is a diprotic acid. Write equations for the stepwise ionization of the acid in water. 5. What are the factors to consider in choosing a planet on which to build a structure like Biosphere II? 229  CHAPTER 6 Thermochemistry The analysis of particles formed from burning methane in a flame is performed with a visible laser. CHAPTER OUTLINE A LOOK AHEAD 6.1 The Nature of Energy  We begin by studying the nature and different types of energy, which, in and Types of Energy principle, are interconvertible. (6.1) 6.2 Energy Changes in Chemical  Next, we build up our vocabulary in learning thermochemistry, which is the Reactions study of heat change in chemical reactions. We see that the vast majority of reactions are either endothermic (absorbing heat) or exothermic (releasing 6.3 Introduction to heat). (6.2) Thermodynamics  We learn that thermochemistry is part of a broader subject called the first 6.4 Enthalpy of Chemical law of thermodynamics, which is based on the law of conservation of Reactions energy. We see that the change in internal energy can be expressed in terms of the changes in heat and work done of a system. (6.3) 6.5 Calorimetry  We then become acquainted with a new term for energy, called enthalpy, 6.6 Standard Enthalpy of whose change applies to processes carried out under constant-pressure con- Formation and Reaction ditions. (6.4) 6.7 Heat of Solution and Dilution  We learn ways to measure the heats of reaction or calorimetry, and the meaning of specific heat and heat capacity, quantities used in experimental work. (6.5)  Knowing the standard enthalpies of formation of reactants and products enables us to calculate the enthalpy of a reaction. We will discuss ways to determine these quantities either by the direct method from the elements or by the indirect method, which is based on Hess’s law of heat summation. (6.6)  Finally, we will study the heat changes when a solute dissolves in a solvent (heat of solution) and when a solution is diluted (heat of dilution). (6.7) 230  6.1 The Nature of Energy and Types of Energy 231 E very chemical reaction obeys two fundamental laws: the law of conservation of mass and the law of conservation of energy. We discussed the mass relationship between reactants and products in Chapter 3; here we will look at the energy changes that accompany chemical reactions. 6.1 The Nature of Energy and Types of Energy “Energy” is a much-used term that represents a rather abstract concept. For instance, when we feel tired, we might say we haven’t any energy; and we read about the need to find alternatives to nonrenewable energy sources. Unlike matter, energy is known and recognized by its effects. It cannot be seen, touched, smelled, or weighed. Energy is usually defined as the capacity to do work. In Chapter 5 we defined work as “force 3 distance,” but we will soon see that there are other kinds of work. All forms of energy are capable of doing work (that is, of exerting a force over a distance), but not all of them are equally relevant to chemistry. The energy contained in tidal waves, for example, can be harnessed to perform useful work, but the rela- tionship between tidal waves and chemistry is minimal. Chemists define work as directed energy change resulting from a process. Kinetic energy—the energy pro- duced by a moving object—is one form of energy that is of particular interest to Kinetic energy was introduced in Chapter 5 (p. 202). chemists. Others include radiant energy, thermal energy, chemical energy, and poten- tial energy. Radiant energy, or solar energy, comes from the sun and is Earth’s primary energy source. Solar energy heats the atmosphere and Earth’s surface, stimulates the growth of vegetation through the process known as photosynthesis, and influences global climate patterns. Thermal energy is the energy associated with the random motion of atoms and molecules. In general, thermal energy can be calculated from temperature measurements. The more vigorous the motion of the atoms and molecules in a sample of matter, the hotter the sample is and the greater its thermal energy. However, we need to distinguish carefully between thermal energy and temperature. A cup of coffee at 708C has a higher temperature than a bathtub filled with warm water at 408C, but much more thermal energy is stored in the bathtub water because it has a much larger volume and greater mass than the coffee and therefore more water molecules and more molecular motion. Chemical energy is stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of constituent atoms. When sub- stances participate in chemical reactions, chemical energy is released, stored, or con- verted to other forms of energy. Potential energy is energy available by virtue of an object’s position. For instance, because of its altitude, a rock at the top of a cliff has more potential energy and will make a bigger splash if it falls into the water below than a similar rock located part- way down the cliff. Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance. All forms of energy can be converted (at least in principle) from one form to another. We feel warm when we stand in sunlight because radiant energy is As the water falls over the dam, its converted to thermal energy on our skin. When we exercise, chemical energy potential energy is converted to stored in our bodies is used to produce kinetic energy. When a ball starts to roll kinetic energy. Use of this energy downhill, its potential energy is converted to kinetic energy. You can undoubtedly to generate electricity is called think of many other examples. Although energy can assume many different forms hydroelectric power. that are interconvertible, scientists have concluded that energy can be neither destroyed nor created. When one form of energy disappears, some other form of energy (of equal magnitude) must appear, and vice versa. This principle is sum- marized by the law of conservation of energy: the total quantity of energy in the universe is assumed constant. 232 Chapter 6 ■ Thermochemistry 6.2 Energy Changes in Chemical Reactions Often the energy changes that take place during chemical reactions are of as much practical interest as the mass relationships we discussed in Chapter 3. For example, combustion reactions involving fuels such as natural gas and oil are carried out in daily life more for the thermal energy they release than for their products, which are water and carbon dioxide. Almost all chemical reactions absorb or produce (release) energy, generally in This infrared photo shows where the form of heat. It is important to understand the distinction between thermal energy energy (heat) leaks through the and heat. Heat is the transfer of thermal energy between two bodies that are at dif- house. The more red the color, the more energy is lost to the outside. ferent temperatures. Thus, we often speak of the “heat flow” from a hot object to a cold one. Although the term “heat” by itself implies the transfer of energy, we cus- Animation Heat Flow tomarily talk of “heat absorbed” or “heat released” when describing the energy changes that occur during a process. Thermochemistry is the study of heat change in chemical reactions. When heat is absorbed or released To analyze energy changes associated with chemical reactions we must first define during a process, energy is conserved, but it is transferred between system and the system, or the specific part of the universe that is of interest to us. For chemists, surroundings. systems usually include substances involved in chemical and physical changes. For example, in an acid-base neutralization experiment, the system may be a beaker con- taining 50 mL of HCl to which 50 mL of NaOH is added. The surroundings are the rest of the universe outside the system. There are three types of systems. An open system can exchange mass and energy, usually in the form of heat with its surroundings. For example, an open system may consist of a quantity of water in an open container, as shown in Figure 6.1(a). If we close the flask, as in Figure 6.1(b), so that no water vapor can escape from or condense into the container, we create a closed system, which allows the transfer of energy (heat) but not mass. By placing the water in a totally insulated container, we can construct an isolated system, which does not allow the transfer of either mass or energy, as shown in Figure 6.1(c). The combustion of hydrogen gas in oxygen is one of many chemical reactions that release considerable quantities of energy (Figure 6.2): 2H2 (g) 1 O2 (g) ¡ 2H2O(l) 1 energy In this case, we label the reacting mixture (hydrogen, oxygen, and water molecules) the system and the rest of the universe the surroundings. Because energy cannot be Water vapor Heat Heat (a) (b) (c) Figure 6.1 Three systems represented by water in a flask: (a) an open system, which allows the exchange of both energy and mass with surroundings; (b) a closed system, which allows the exchange of energy but not mass; and (c) an isolated system, which allows neither energy nor mass to be exchanged (here the flask is enclosed by a vacuum jacket).  6.2 Energy Changes in Chemical Reactions 233 Figure 6.2 The Hindenburg disaster. The Hindenburg, a German airship filled with hydrogen gas, was destroyed in a spectacular fire at Lakehurst, New Jersey, in 1937. created or destroyed, any energy lost by the system must be gained by the sur- roundings. Thus, the heat generated by the combustion process is transferred from the system to its surroundings. This reaction is an example of an exothermic pro- Exo- comes from the Greek word meaning “outside”; endo- means “within.” cess, which is any process that gives off heat—that is, transfers thermal energy to the surroundings. Figure 6.3(a) shows the energy change for the combustion of hydrogen gas. Now consider another reaction, the decomposition of mercury(II) oxide (HgO) at high temperatures: energy 1 2HgO(s) ¡ 2Hg(l) 1 O2 (g) This reaction is an endothermic process, in which heat has to be supplied to the system (that is, to HgO) by the surroundings [Figure 6.3(b)]. From Figure 6.3 you can see that in exothermic reactions, the total energy of the products is less than the total energy of the reactants. The difference is the heat sup- plied by the system to the surroundings. Just the opposite happens in endothermic reactions. Here, the difference between the energy of the products and the energy of On heating, HgO decomposes to the reactants is equal to the heat supplied to the system by the surroundings. give Hg and O2. Figure 6.3 (a) An exothermic 2H2(g) + O2(g) 2Hg(l) + O2(g) process. (b) An endothermic process. Parts (a) and (b) are not drawn to the same scale; that is, the heat released in the formation Exothermic: Endothermic: of H2O from H2 and O2 is not heat given off heat absorbed Energy Energy equal to the heat absorbed in the by the system by the system decomposition of HgO. to the surroundings from the surroundings 2H2O(l) 2HgO(s) (a) (b) 234 Chapter 6 ■ Thermochemistry Review of Concepts Classify each of the following as an open system, a closed system, or an isolated system. (a) Milk kept in a closed thermo flask. (b) A student reading in her dorm room. (c) Air inside a tennis ball. 6.3 Introduction to Thermodynamics Thermochemistry is part of a broader subject called thermodynamics, which is the scientific study of the interconversion of heat and other kinds of energy. The laws of thermodynamics provide useful guidelines for understanding the energetics and direc- tions of processes. In this section we will concentrate on the first law of thermody- namics, which is particularly relevant to the study of thermochemistry. We will continue our discussion of thermodynamics in Chapter 17. In thermodynamics, we study changes in the state of a system, which is defined by the values of all relevant macroscopic properties, for example, composition, energy, temperature, pressure, and volume. Energy, pressure, volume, and temperature are said Changes in state functions do not depend to be state functions—properties that are determined by the state of the system, on the pathway, but only on the initial and final state. regardless of how that condition was achieved. In other words, when the state of a system changes, the magnitude of change in any state function depends only on the initial and final states of the system and not on how the change is accomplished. The state of a given amount of a gas is specified by its volume, pressure, and temperature. Consider a gas at 2 atm, 300 K, and 1 L (the initial state). Suppose a process is carried out at constant temperature such that the gas pressure decreases to 1 atm. According to Boyle’s law, its volume must increase to 2 L. The final state then corresponds to 1 atm, 300 K, and 2 L. The change in volume (DV) is The Greek letter delta, D, symbolizes ¢V 5 Vf 2 Vi change. We use D in this text to mean final 2 initial; that is, final “minus” initial. 52L21L 51L where Vi and Vf denote the initial and final volume, respectively. No matter how we arrive at the final state (for example, the pressure of the gas can be increased first and then decreased to 1 atm), the change in volume is always 1 L. Thus, the volume of a gas is a state function. In a similar manner, we can show that pressure and tem- perature are also state functions. Recall that an object possesses potential Energy is another state function. Using potential energy as an example, we find energy by virtue of its position or chemical composition. that the net increase in gravitational potential energy when we go from the same starting point to the top of a mountain is always the same, regardless of how we get there (Figure 6.4). The First Law of Thermodynamics The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed.† How do we know this is so? It would be impossible to prove the valid- ity of the first law of thermodynamics if we had to determine the total energy content † See footnote on p. 40 (Chapter 2) for a discussion of mass and energy relationship in chemical reactions.  6.3 Introduction to Thermodynamics 235 Figure 6.4 The gain in gravitational potential energy that occurs when a person climbs from the base to the top of a mountain is independent of the path taken. of the universe. Even determining the total energy content of 1 g of iron, say, would be extremely difficult. Fortunately, we can test the validity of the first law by measur- ing only the change in the internal energy of a system between its initial state and its final state in a process. The change in internal energy DU is given by ¢U 5 Uf 2 Ui where Ui and Uf are the internal energies of the system in the initial and final states, respectively. The internal energy of a system has two components: kinetic energy and potential energy. The kinetic energy component consists of various types of molecular motion and the movement of electrons within molecules. Potential energy is determined by the attractive interactions between electrons and nuclei and by repulsive interactions between electrons and between nuclei in individual molecules, as well as by interac- tion between molecules. It is impossible to measure all these contributions accurately, so we cannot calculate the total energy of a system with any certainty. Changes in energy, on the other hand, can be determined experimentally. Consider the reaction between 1 mole of sulfur and 1 mole of oxygen gas to produce 1 mole of sulfur dioxide: S(s) 1 O2 (g) ¡ SO2 (g) In this case, our system is composed of the reactant molecules S and O2 (the initial state) and the product molecules SO2 (the final state). We do not know the internal energy content of either the reactant molecules or the product molecules, but we can accurately measure the change in energy content, DU, given by ¢U 5 U(product) 2 U(reactants) 5 energy content of 1 mol SO2 (g) 2 energy content of [1 mol S(s) 11 mol O2 (g)] We find that this reaction gives off heat. Therefore, the energy of the product is less than that of the reactants, and DU is negative. Sulfur burning in air to form SO2. Interpreting the release of heat in this reaction to mean that some of the chemical energy contained in the molecules has been converted to thermal energy, we conclude that the transfer of energy from the system to the surroundings does not change the total energy of the universe. That is, the sum of the energy changes must be zero: ¢Usys 1 ¢Usurr 5 0 or ¢Usys 5 2¢Usurr where the subscripts “sys” and “surr” denote system and surroundings, respectively. Thus, if one system undergoes an energy change DUsys, the rest of the universe, or 236 Chapter 6 ■ Thermochemistry the surroundings, must undergo a change in energy that is equal in magnitude but opposite in sign (2DUsurr); energy gained in one place must have been lost somewhere else. Furthermore, because energy can be changed from one form to another, the energy lost by one system can be gained by another system in a different form. For example, the energy lost by burning oil in a power plant may ultimately turn up in our homes as electrical energy, heat, light, and so on. In chemistry, we are normally interested in the energy changes associated with the system (which may be a flask containing reactants and products), not with its surroundings. Therefore, a more useful form of the first law is We use lowercase letters (such as w and q) to represent thermodynamic quantities that ¢U 5 q 1 w (6.1) are not state functions. (We drop the subscript “sys” for simplicity.) Equation (6.1) says that the change in For convenience, we sometimes omit the the internal energy, DU, of a system is the sum of the heat exchange q between the word “internal” when discussing the energy of a system. system and the surroundings and the work done w on (or by) the system. The sign conventions for q and w are as follows: q is positive for an endothermic process and negative for an exothermic process and w is positive for work done on the system by the surroundings and negative for work done by the system on the surroundings. We can think of the first law of thermodynamics as an energy balance sheet, much like a money balance sheet kept in a bank that does currency exchange. You can withdraw or deposit money in either of two different currencies (like energy change due to heat exchange and work done). However, the value of your bank account depends only on the net amount of money left in it after these transactions, not on which currency you used. Equation (6.1) may seem abstract, but it is actually quite logical. If a system loses heat to the surroundings or does work on the surroundings, we would expect its internal energy to decrease because those are energy-depleting processes. For this reason, both q and w are negative. Conversely, if heat is added to the system or if work is done on the system, then the internal energy of the system would increase. In this case, both q and w are positive. Table 6.1 summarizes the sign conventions for q and w. Work and Heat We will now look at the nature of work and heat in more detail. Work We have seen that work can be defined as force F multiplied by distance d: w5F3d (6.2) In thermodynamics, work has a broader meaning that includes mechanical work (for example, a crane lifting a steel beam), electrical work (a battery supplying electrons Table 6.1 Sign Conventions for Work and Heat Process Sign Work done by the system on the surroundings 2 Work done on the system by the surroundings 1 Heat absorbed by the system from the surroundings (endothermic process) 1 Heat absorbed by the surroundings from the system (exothermic process) 2  6.3 Introduction to Thermodynamics 237 Figure 6.5 The expansion of a P gas against a constant external pressure (such as atmospheric pressure). The gas is in a cylinder P fitted with a weightless movable piston. The work done is given by 2PDV. Because DV . 0, the ΔV work done is a negative quantity. to light the bulb of a flashlight), and surface work (blowing up a soap bubble). In this section we will concentrate on mechanical work; in Chapter 18 we will discuss the nature of electrical work. One way to illustrate mechanical work is to study the expansion or compression of a gas. Many chemical and biological processes involve gas volume changes. Breath- ing and exhaling air involves the expansion and contraction of the tiny sacs called alveoli in the lungs. Another example is the internal combustion engine of the auto- mobile. The successive expansion and compression of the cylinders due to the com- bustion of the gasoline-air mixture provide power to the vehicle. Figure 6.5 shows a gas in a cylinder fitted with a weightless, frictionless movable piston at a certain temperature, pressure, and volume. As it expands, the gas pushes the piston upward against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is w 5 2P¢V (6.3) where DV, the change in volume, is given by Vf 2 Vi. The minus sign in Equation (6.3) takes care of the sign convention for w. For gas expansion (work done by the system), DV . 0, so 2PDV is a negative quantity. For gas compression (work done on the system), DV , 0, and 2PDV is a positive quantity. Equation (6.3) derives from the fact that pressure 3 volume can be expressed as (force/area) 3 volume; that is, F P3V5     3  d3   5 F 3 d 5 w d2 pressure volume where F is the opposing force and d has the dimension of length, d2 has the dimen- sions of area, and d3 has the dimensions of volume. Thus, the product of pressure and volume is equal to force times distance, or work. You can see that for a given increase in volume (that is, for a certain value of DV), the work done depends on the magni- tude of the external, opposing pressure P. If P is zero (that is, if the gas is expanding against a vacuum), the work done must also be zero. If P is some positive, nonzero value, then the work done is given by 2PDV. According to Equation (6.3), the units for work done by or on a gas are liter atmospheres. To express the work done in the more familiar unit of joules, we use the conversion factor (see Appendix 2). 1 L ? atm 5 101.3 J 238 Chapter 6 ■ Thermochemistry Example 6.1 A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm. Strategy A simple sketch of the situation is helpful here: The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. What is the conversion factor between L ? atm and J? Solution (a) Because the external pressure is zero, no work is done in the expansion. w 5 2P¢V 5 2(0) (6.0 2 2.0) L 5 0 (b) The external, opposing pressure is 1.2 atm, so w 5 2P¢V 5 2(1.2 atm) (6.0 2 2.0) L 5 24.8 L ? atm To convert the answer to joules, we write 101.3 J w 5 24.8 L ? atm 3 1 L ? atm 5 24.9 3 102 J Check Because this is gas expansion (work is done by the system on the surroundings), Similar problems: 6.15, 6.16. the work done has a negative sign. Practice Exercise A gas expands from 264 mL to 971 mL at constant temperature. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm. Because temperature is kept constant, Example 6.1 shows that work is not a state function. Although the initial and final you can use Boyle’s law to show that the final pressure is the same in (a) and (b). states are the same in (a) and (b), the amount of work done is different because the external, opposing pressures are different. We cannot write Dw 5 wf 2 wi for a change. Work done depends not only on the initial state and final state, but also on how the process is carried out, that is, on the path.  6.3 Introduction to Thermodynamics 239 Heat The other component of internal energy is heat, q. Like work, heat is not a state func- tion. For example, it takes 4184 J of energy to raise the temperature of 100 g of water from 208C to 308C. This energy can be gained (a) directly as heat energy from a Bunsen burner, without doing any work on the water; (b) by doing work on the water without adding heat energy (for example, by stirring the water with a magnetic stir bar); or (c) by some combination of the procedures described in (a) and (b). This simple illustration shows that heat associated with a given process, like work, depends on how the process is carried out. It is important to note that regardless of which procedure is taken, the change in internal energy of the system, DU, depends on the sum of (q 1 w). If changing the path from the initial state to the final state increases the value of q, then it will decrease the value of w by the same amount and vice versa, so that DU remains unchanged. In summary, heat and work are not state functions because they are not properties of a system. They manifest themselves only during a process (during a change). Thus, their values depend on the path of the process and vary accordingly. Example 6.2 The work done when a gas is compressed in a cylinder like that shown in Figure 6.5 is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process. Strategy Compression is work done on the gas, so what is the sign for w? Heat is released by the gas to the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution To calculate the energy change of the gas, we need Equation (6.1). Work of compression is positive and because heat is released by the gas, q is negative. Therefore, we have ¢U 5 q 1 w 5 2128 J 1 462 J 5 334 J As a result, the energy of the gas increases by 334 J. Similar problems: 6.17, 6.18. Practice Exercise A gas expands and does P-V work on the surroundings equal to 279 J. At the same time, it absorbs 216 J of heat from the surroundings. What is the change in energy of the system? Review of Concepts Two ideal gases at the same temperature and pressure are placed in two equal- volume containers. One container has a fixed volume, while the other is a cylinder fitted with a weightless movable piston like that shown in Figure 6.5. Initially, the gas pressures are equal to the external atmospheric pressure. The gases are then heated with a Bunsen burner. What are the signs of q and w for the gases under these conditions? CHEMISTRY in Action Making Snow and Inflating a Bicycle Tire M any phenomena in everyday life can be explained by the first law of thermodynamics. Here we will discuss two examples of interest to lovers of the outdoors. cools on expansion, helps to lower the temperature of the water vapor. Inflating a Bicycle Tire Making Snow If you have ever pumped air into a bicycle tire, you probably If you are an avid downhill skier, you have probably skied on noticed a warming effect at the valve stem. This phenome- artificial snow. How is this stuff made in quantities large enough non, too, can be explained by the first law of thermodynam- to meet the needs of skiers on snowless days? The secret of ics. The action of the pump compresses the air inside the snowmaking is in the equation DU 5 q 1 w. A snowmaking pump and the tire. The process is rapid enough to be treated machine contains a mixture of compressed air and water vapor as approximately adiabatic, so that q 5 0 and DU 5 w. at about 20 atm. Because of the large difference in pressure be- Because work is done on the gas in this case (it is being com- tween the tank and the outside atmosphere, when the mixture is pressed), w is positive, and there is an increase in energy. sprayed into the atmosphere it expands so rapidly that, as a good Hence, the temperature of the system increases also, accord- approximation, no heat exchange occurs between the system ing to the equation (air and water) and its surroundings; that is, q 5 0. (In thermo- ¢U 5 C¢T dynamics, such a process is called an adiabatic process.) Thus, we write ¢U 5 q 1 w 5 w Because the system does work on the surroundings, w is a nega- tive quantity, and there is a decrease in the system’s energy. Kinetic energy is part of the total energy of the system. In Section 5.7 we saw that the average kinetic energy of a gas is directly proportional to the absolute temperature [Equation (5.15)]. It follows, therefore, that the change in energy DU is given by ¢U 5 C¢T where C is the proportionality constant. Because DU is nega- tive, DT must also be negative, and it is this cooling effect (or the decrease in the kinetic energy of the water molecules) that is responsible for the formation of snow. Although we need only water to form snow, the presence of air, which also A snowmaking machine in operation. 6.4 Enthalpy of Chemical Reactions Our next step is to see how the first law of thermodynamics can be applied to pro- cesses carried out under different conditions. Specifically, we will consider two situ- ations most commonly encountered in the laboratory; one in which the volume of the system is kept constant and one in which the pressure applied on the system is kept constant. 240  6.4 Enthalpy of Chemical Reactions 241 If a chemical reaction is run at constant volume, then DV 5 0 and no P-V work Recall that w 5 2PDV. will result from this change. From Equation (6.1) it follows that ¢U 5 q 2 P¢V 5 qv (6.4) We add the subscript “v” to remind us that this is a constant-volume process. This equality may seem strange at first, for we showed earlier that q is not a state function. The process is carried out under constant-volume conditions, however, so that the heat change can have only a specific value, which is equal to DU. Enthalpy Constant-volume conditions are often inconvenient and sometimes impossible to achieve. Most reactions occur under conditions of constant pressure (usually atmo- spheric pressure). If such a reaction results in a net increase in the number of moles of a gas, then the system does work on the surroundings (expansion). This result follows from the fact that for the gas formed to enter the atmosphere, it must push the surrounding air back. Conversely, if more gas molecules are consumed than are produced, work is done on the system by the surroundings (compression). Finally, no work is done if there is no net change in the number of moles of gases from reactants to products. In general, for a constant-pressure process we write ¢U 5 q 1 w 5 qp 2 P¢V or qp 5 ¢U 1 P¢V (6.5) where the subscript “p” denotes constant-pressure condition. We now introduce a new thermodynamic function of a system called enthalpy (H), which is defined by the equation H 5 U 1 PV (6.6) where U is the internal energy of the system and P and V are the pressure and volume of the system, respectively. Because U and PV have energy units, enthalpy also has energy units. Furthermore, U, P, and V are all state functions, that is, the changes in (U 1 PV) depend only on the initial and final states. It follows, therefore, that the change in H, or DH, also depends only on the initial and final states. Thus, H is a state function. For any process, the change in enthalpy according to Equation (6.6) is given by ¢H 5 ¢U 1 ¢(PV) (6.7) If the pressure is held constant, then ¢H 5 ¢U 1 P¢V (6.8) Comparing Equation (6.8) with Equation (6.5), we see that for a constant-pressure process, qp 5 DH. Again, although q is not a state function, the heat change at constant pressure is equal to DH because the “path” is defined and therefore it can have only a specific value. 242 Chapter 6 ■ Thermochemistry In Section 6.5 we will discuss ways to We now have two quantities—DU and DH—that can be associated with a reaction. measure heat changes at constant volume and constant pressure. If the reaction occurs under constant-volume conditions, then the heat change, qv, is equal to DU. On the other hand, when the reaction is carried out at constant pressure, the heat change, qp, is equal to DH. Enthalpy of Reactions Because most reactions are constant-pressure processes, we can equate the heat change in these cases to the change in enthalpy. For any reaction of the type reactants ¡ products we define the change in enthalpy, called the enthalpy of reaction, DHrxn, as the difference between the enthalpies of the products and the enthalpies of the reactants: We often omit the subscript “rxn” and ¢H 5 H(products) 2 H(reactants) (6.9) simply write DH for enthalpy changes of reactions. The enthalpy of reaction can be positive or negative, depending on the process. For an endothermic process (heat absorbed by the system from the surroundings), DH is positive (that is, DH . 0). For an exothermic process (heat released by the system to the surroundings), DH is negative (that is, DH , 0). This analogy assumes that you will not An analogy for enthalpy change is a change in the balance in your bank account. overdraw your bank account. The enthalpy of a substance cannot be negative. Suppose your initial balance is $100. After a transaction (deposit or withdrawal), the change in your bank balance, DX, is given by ¢X 5 Xfinal 2 Xinitial where X represents the bank balance. If you deposit $80 into your account, then DX 5 $180 2 $100 5 $80. This corresponds to an endothermic reaction. (The balance increases and so does the enthalpy of the system.) On the other hand, a withdrawal of $60 means DX 5 $40 2 $100 5 2$60. The negative sign of DX means your balance has decreased. Similarly, a negative value of DH reflects a decrease in enthalpy of the system as a result of an exothermic process. The difference between this analogy and Equation (6.9) is that while you always know your exact bank balance, there is no way to know the enthalpies of individual products and reactants. In practice, we can only measure the difference in their values. Now let us apply the idea of enthalpy changes to two common processes, the first involving a physical change, the second a chemical change. Thermochemical Equations At 08C and a pressure of 1 atm, ice melts to form liquid water. Measurements show that for every mole of ice converted to liquid water under these conditions, 6.01 kilojoules (kJ) of heat energy are absorbed by the system (ice). Because the pressure is constant, the heat change is equal to the enthalpy change, DH. Further- more, this is an endothermic process, as expected for the energy-absorbing change of melting ice [Figure 6.6(a)]. Therefore, DH is a positive quantity. The equation for this physical change is H2O(s) ¡ H2O(l) ¢H 5 6.01 kJ/mol For simplicity, we use “per mole” rather The “per mole” in the unit for DH means that this is the enthalpy change per mole than “per mole of reaction” for DH in thermochemical equations. of the reaction (or process) as it is written; that is, when 1 mole of ice is converted to 1 mole of liquid water.  6.4 Enthalpy of Chemical Reactions 243 Figure 6.6 (a) Melting 1 mole H2O(l) CH4(g) + 2O2(g) of ice at 08C (an endothermic process) results in an enthalpy increase in the system of 6.01 kJ. (b) Burning 1 mole of methane Heat absorbed Heat given off in oxygen gas (an exothermic by the system by the system Enthalpy Enthalpy process) results in an enthalpy from the surroundings to the surroundings decrease in the system of 890.4 kJ. DH 5 6.01 kJ/mol DH 5 2890.4 kJ/mol Parts (a) and (b) are not drawn to the same scale. H2O(s) CO2(g) + 2H2O(l) (a) (b) As another example, consider the combustion of methane (CH4), the principal component of natural gas: CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l) ¢H 5 2890.4 kJ/mol From experience we know that burning natural gas releases heat to the surroundings, so it is an exothermic process. Under constant-pressure condition this heat change is equal to enthalpy change and DH must have a negative sign [Figure 6.6(b)]. Again, the per mole of reaction unit for DH means that when 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of liquid H2O, 890.4 kJ of heat energy are released to the surroundings. It is important to keep in mind that the DH value does not refer to a particular reactant or product. It simply means that the quoted DH value refers to all the reacting species in molar quantities. Thus, the following conversion factors can be created: 2890.4 kJ 2890.4 kJ 2890.4 kJ 2890.4 kJ 1 mol CH4 2 mol O2 1 mol CO2 2 mol H2O Expressing DH in units of kJ/mol (rather than just kJ) conforms to the standard con- vention; its merit will become apparent when we continue our study of thermodynam- ics in Chapter 17. The equations for the melting of ice and the combustion of methane are examples of thermochemical equations, which show the enthalpy changes as well as the mass relationships. It is essential to specify a balanced equation when quoting the enthalpy change of a reaction. The following guidelines are helpful in writing and interpreting thermochemical equations. 1. When writing thermochemical equations, we must always specify the physical states of all reactants and products, because they help determine the actual enthalpy changes. For example, in the equation for the combustion of methane, if we show water vapor rather than liquid water as a product, CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(g) ¢H 5 2802.4 kJ/mol the enthalpy change is 2802.4 kJ rather than 2890.4 kJ because 88.0 kJ are needed to convert 2 moles of liquid water to water vapor; that is, Methane gas burning from a 2H2O(l) ¡ 2H2O(g) ¢H 5 88.0 kJ/mol Bunsen burner. 244 Chapter 6 ■ Thermochemistry 2. If we multiply both sides of a thermochemical equation by a factor n, then DH must also change by the same factor. Returning to the melting of ice H2O(s) ¡ H2O(l) ¢H 5 6.01 kJ/mol If we multiply the equation throughout by 2; that is, if we set n 5 2, then 2H2O(s) ¡ 2H2O(l) ¢H 5 2(6.01 kJ/mol) 5 12.0 kJ/mol 3. When we reverse an equation, we change the roles of reactants and products. Consequently, the magnitude of DH for the equation remains the same, but its sign changes. For example, if a reaction consumes thermal energy from its surroundings (that is, if it is endothermic), then the reverse reaction must release thermal energy back to its surroundings (that is, it must be exothermic) and the enthalpy change expression must also change its sign. Thus, reversing the melting of ice and the combustion of methane, the thermochemical equa- tions become H2O(l) ¡ H2O(s) ¢H 5 26.01 kJ/mol CO2 (g) 1 2H2O(l) ¡ CH4 (g) 1 2O2 (g) ¢H 5 890.4 kJ/mol and what was an endothermic process becomes exothermic, and vice versa. Example 6.3 Given the thermochemical equation 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g) ¢H 5 2198.2 kJ/mol calculate the heat evolved when 87.9 g of SO2 (molar mass 5 64.07 g/mol) is converted to SO3. Strategy The thermochemical equation shows that for every 2 moles of SO2 reacted, 198.2 kJ of heat are given off (note the negative sign). Therefore, the conversion factor is 2198.2 kJ 2 mol SO2 How many moles of SO2 are in 87.9 g of SO2? What is the conversion factor between grams and moles? Solution We need to first calculate the number of moles of SO2 in 87.9 g of the compound and then find the number of kilojoules produced from the exothermic reaction. The sequence of conversions is as follows: grams of SO2 ¡ moles of SO2 ¡ kilojoules of heat generated Therefore, the enthalpy change for this reaction is given by 1 mol SO2 2198.2 kJ Keep in mind that the DH for a reaction ¢H 5 87.9 g SO2 3 3 5 2136 kJ can be positive or negative, but the heat 64.07 g SO2 2 mol SO2 released or absorbed is always a positive quantity. The words “released” and and the heat released to the surroundings is 136 kJ. “absorbed” give the direction of heat transfer, so no sign is needed. (Continued)  6.4 Enthalpy of Chemical Reactions 245 Check Because 87.9 g is less than twice the molar mass of SO2 (2 3 64.07 g) as shown in the preceding thermochemical equation, we expect the heat released to be smaller than 198.2 kJ. Similar problem: 6.26. Practice Exercise Calculate the heat evolved when 266 g of white phosphorus (P4) burns in air according to the equation P4 (s) 1 5O2 (g) ¡ P4O10 (s) ¢H 5 23013 kJ/mol A Comparison of DH and DU What is the relationship between DH and DU for a process? To find out, let us consider the reaction between sodium metal and water: 2Na(s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2 (g) ¢H 5 2367.5 kJ/mol This thermochemical equation says that when two moles of sodium react with an excess of water, 367.5 kJ of heat are given off. Note that one of the products is hydrogen gas, which must push back air to enter the atmosphere. Consequently, some of the energy produced by the reaction is used to do work of pushing back a volume of air (DV ) against atmospheric pressure (P) (Figure 6.7). To calculate the change in internal energy, we rearrange Equation (6.8) as follows: ¢U 5 ¢H 2 P¢V If we assume the temperature to be 258C and ignore the small change in the volume Sodium reacting with water to form hydrogen gas. of the solution, we can show that the volume of 1 mole of H2 gas at 1.0 atm and 298 K is 24.5 L, so that 2PDV 5 224.5 L ? atm or 22.5 kJ. Finally, Recall that 1 L ? atm 5 101.3 J. ¢U 5 2367.5 kJ/mol 2 2.5 kJ/mol 5 2370.0 kJ/mol This calculation shows that DU and DH are approximately the same. The reason DH For reactions that do not result in a change in the number of moles of gases is smaller than DU in magnitude is that some of the internal energy released is used from reactants to products [for example, to do gas expansion work, so less heat is evolved. For reactions that do not involve H2(g) 1 F2(g) ¡ 2HF(g)], DU 5 DH. gases, DV is usually very small and so DU is practically the same as DH. Another way to calculate the internal energy change of a gaseous reaction is to assume ideal gas behavior and constant temperature. In this case, ¢U 5 ¢H 2 ¢(PV ) 5 ¢H 2 ¢(nRT ) 5 ¢H 2 RT¢n (6.10) P Figure 6.7 (a) A beaker of water inside a cylinder fitted with a movable piston. The pressure P inside is equal to the atmospheric pressure. (b) As the sodium Air + water vapor + metal reacts with water, the H2 gas hydrogen gas generated pushes Air + water vapor the piston upward (doing work on the surroundings) until the pressure inside is again equal to that of outside. (a) (b) 246 Chapter 6 ■ Thermochemistry where Dn is defined as ¢n 5 number of moles of product gases 2 number of moles of reactant gases Example 6.4 Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 258C: 2CO(g) 1 O2 (g) ¡ 2CO2 (g) ¢H 5 2566.0 kJ/mol Strategy We are given the enthalpy change, DH, for the reaction and are asked to calculate the change in internal energy, DU. Therefore, we need Equation (6.10). What is the change in the number of moles of gases? DH is given in kilojoules, so what units should we use for R? Solution From the chemical equation we see that 3 moles of gases are converted to 2 moles of gases so that ¢n 5 number of moles of product gas 2 number of moles of reactant gases 5223 5 21 Using 8.314 J/K ? mol for R and T 5 298 K in Equation (6.10), we write ¢U 5 ¢H 2 RT¢n 1 kJ 5 2566.0 kJ/mol 2 (8.314 J/K ? mol)a b(298 K) (21) 1000 J 5 2563.5 kJ/mol Check Knowing that the reacting gaseous system undergoes a compression (3 moles to 2 moles), is it reasonable to have DH . DU in magnitude? Carbon monoxide burns in air to Practice Exercise What is DU for the formation of 1 mole of CO at 1 atm and 258C? form carbon dioxide. Similar problem: 6.27. C(graphite) 1 12O2 (g) ¡ CO(g) ¢H 5 2110.5 kJ/mol Review of Concepts Which of the constant-pressure processes shown here has the smallest difference between DU and DH? (a) water ¡ water vapor (b) water ¡ ice (c) ice ¡ water vapor 6.5 Calorimetry In the laboratory, heat changes in physical and chemical processes are measured with a calorimeter, a closed container designed specifically for this purpose. Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity, so let us consider them first. Specific Heat and Heat Capacity The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. It has the units J/g ? 8C. The heat capacity (C) of a substance is the amount of heat required to  6.5 Calorimetry 247 raise the temperature of a given quantity of the substance by one degree Celsius. Table 6.2 Its units are J/8C. Specific heat is an intensive property, whereas heat capacity is an The Specific Heats extensive property. The relationship between the heat capacity and specific heat of of Some Common a substance is Substances Specific C 5 ms (6.11) Heat Substance (J/g ? 8C) where m is the mass of the substance in grams. For example, the specific heat of water is 4.184 J/g ? 8C, and the heat capacity of 60.0 g of water is Al 0.900 Au 0.129 (60.0 g)(4.184 J/g ? °C) 5 251 J/°C C (graphite) 0.720 Table 6.2 shows the specific heat of some common substances. C (diamond) 0.502 If we know the specific heat and the amount of a substance, then the change in Cu 0.385 the sample’s temperature (Dt) will tell us the amount of heat (q) that has been absorbed Fe 0.444 or released in a particular process. The equations for calculating the heat change are Hg 0.139 given by Pb 0.158 H2O 4.184 q 5 ms¢t (6.12) C2H5OH (ethanol) 2.46 q 5 C¢t (6.13) where Dt is the temperature change: ¢t 5 tfinal 2 tinitial The sign convention for q is the same as that for enthalpy change; q is positive for endothermic processes and negative for exothermic processes. Example 6.5 A 466-g sample of water is heated from 8.508C to 74.608C. Calculate the amount of heat absorbed (in kilojoules) by the water. Strategy We know the quantity of water and the specific heat of water. With this information and the temperature rise, we can calculate the amount of heat absorbed (q). Solution Using Equation (6.12), we write q 5 ms¢t 5 (466 g) (4.184 J/g ? °C) (74.60°C 2 8.50°C) 1 kJ 5 1.29 3 105 J 3 1000 J 5 129 kJ Check The units g and 8C cancel, and we are left with the desired unit kJ. Because heat is absorbed by the water from the surroundings, it has a positive sign. Similar problem: 6.33. Practice Exercise An iron bar of mass 869 g cools from 948C to 58C. Calculate the heat released (in kilojoules) by the metal. “Constant volume” refers to the volume of the container, which does not change Constant-Volume Calorimetry during the reaction. Note that the container remains intact after the measurement. The Heat of combustion is usually measured by placing a known mass of a compound in term “bomb calorimeter” connotes the explosive nature of the reaction (on a a steel container called a constant-volume bomb calorimeter, which is filled with small scale) in the presence of excess oxygen at about 30 atm of pressure. The closed bomb is immersed in a known amount oxygen gas. 248 Chapter 6 ■ Thermochemistry Figure 6.8 A constant-volume Thermometer Stirrer bomb calorimeter. The calorimeter is filled with oxygen gas before it is placed in the bucket. The Ignition wires sample is ignited electrically, and the heat produced by the reaction can be accurately determined by measuring the temperature increase in the known amount of Calorimeter bucket surrounding water. Insulated jacket Water O2 inlet Bomb Sample cup of water, as shown in Figure 6.8. The sample is ignited electrically, and the heat produced by the combustion reaction can be calculated accurately by recording the rise in temperature of the water. The heat given off by the sample is absorbed by the water and the bomb. The special design of the calorimeter enables us to assume that no heat (or mass) is lost to the surroundings during the time it takes to make measure- ments. Therefore, we can call the bomb and the water in which it is submerged an isolated system. Because no heat enters or leaves the system throughout the process, the heat change of the system (qsystem) must be zero and we can write qsystem 5 qcal 1 qrxn 50 (6.14) where qcal and qrxn are the heat changes for the calorimeter and the reaction, respec- tively. Thus, qrxn 5 2qcal (6.15) Note that Ccal comprises both the bomb To calculate qcal, we need to know the heat capacity of the calorimeter (Ccal) and the and the surrounding water. temperature rise, that is, qcal 5 Ccal ¢t (6.16) The quantity Ccal is calibrated by burning a substance with an accurately known heat of combustion. For example, it is known that the combustion of 1 g of benzoic acid (C6H5COOH) releases 26.42 kJ of heat. If the temperature rise is 4.6738C, then the heat capacity of the calorimeter is given by Note that although the combustion qcal reaction is exothermic, qcal is a positive Ccal 5 quantity because it represents the heat ¢t absorbed by the calorimeter. 26.42 kJ 5 5 5.654 kJ/°C 4.673°C Once Ccal has been determined, the calorimeter can be used to measure the heat of combustion of other substances. Note that because reactions in a bomb calorimeter occur under constant- volume rather than constant-pressure conditions, the heat changes correspond to ≤U,  6.5 Calorimetry 249 not the enthalpy change ≤H (see Section 6.4). Equation (6.10) can be used to correct the measured heat changes so that they correspond to ≤H values, but the corrections usually are quite small so we will not concern ourselves with the details here. Finally, it is interesting to note that the energy contents of food and fuel (usually expressed in calories where 1 cal 5 4.184 J) are measured with constant- volume calorimeters. Example 6.6 A quantity of 1.435 g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.288C to 25.958C. If the heat capacity of the bomb plus water was 10.17 kJ/8C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion. Strategy Knowing the heat capacity and the temperature rise, how do we calculate the C10H8 heat absorbed by the calorimeter? What is the heat generated by the combustion of 1.435 g of naphthalene? What is the conversion factor between grams and moles of naphthalene? Solution The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. From Equation (6.16), assuming no heat is lost to the surroundings, we write qcal 5 Ccal ¢t 5 (10.17 kJ/°C) (25.95°C 2 20.28°C) 5 57.66 kJ Because qsys 5 qcal 1 qrxn 5 0, qcal 5 2qrxn. The heat change of the reaction is 257.66 kJ. This is the heat released by the combustion of 1.435 g of C10H8; therefore, we can write the conversion factor as 257.66 kJ 1.435 g C10H8 The molar mass of naphthalene is 128.2 g, so the heat of combustion of 1 mole of naphthalene is 257.66 kJ 128.2 g C10H8 molar heat of combustion 5 3 1.435 g C10H8 1 mol C10H8 5 25.151 3 103 kJ/mol Check Knowing that the combustion reaction is exothermic and that the molar mass of naphthalene is much greater than 1.4 g, is the answer reasonable? Under the reaction conditions, can the heat change (257.66 kJ) be equated to the enthalpy change of the reaction? Similar problem: 6.37. Practice Exercise A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by 4.208C. If the heat capacity of the bomb plus water was 10.4 kJ/8C, calculate the molar heat of combustion of methanol. Constant-Pressure Calorimetry A simpler device than the constant-volume calorimeter is the constant-pressure calo- rimeter, which is used to determine the heat changes for noncombustion reactions. A crude constant-pressure calorimeter can be constructed from two Styrofoam coffee CHEMISTRY in Action White Fat Cells, Brown Fat Cells, and a Potential Cure for Obesity T he food we eat is broken down, or metabolized, in stages by a group of complex biological molecules called en- zymes. Most of the energy released at each stage is captured internal organs and they cushion and insulate the body. Obese people have a high content of WFC in their bodies. BFC, on the other hand, contain a high concentration of mitochondria, for function and growth. One interesting aspect of metabolism which are specialized subunits within a cell. The main role of is that the overall change in energy is the same as it is in com- BFC is to burn fat molecules and generate heat. Its name is bustion. For example, the total enthalpy change for the conver- derived from the fact that mitochondria contain iron, giving sion of glucose (C6H12O6) to carbon dioxide and water is the tissue a reddish brown color. In general, women have the same whether we burn the substance in air or digest it in more BFC than men. our bodies: We lose our brown fat as we age, but several studies carried out in 2009 showed that adults possess metabolically active C6H12O6 (s) 1 6CO2 (g) ¡ 6CO2 (g) 1 6H2O(l) BFC. In one experiment, PET/CT (positron emission tomogra- ¢H 5 22801 kJ/mol phy and computerized tomography) scans of 24 men exposed to cold and room temperature show that the chilly temperature The energy content of food is generally measured in calories. The activates the BFC as they burn off fat molecules to generate heat calorie (cal) is a non-SI unit of energy that is equivalent to 4.184 J: (see figure). Furthermore, it was found that lean people have more active BFC than obese people. 1 cal 5 4.184 J Mice have the same type of fat cells as humans. In 2013 it was demonstrated by genetically labeling the fat cells of mice In the context of nutrition, however, the calorie we speak of that WFC could be converted into BFC by exposure to cold (sometimes called a “big calorie”) is actually equal to a kilo- (8°C) for one week. Unfortunately, BFC were converted back to calorie; that is, WFC a few weeks after the mice were returned to normal room temperature. A separate study suggested that a different type of 1 Cal 5 1000 cal 5 4184 J BFC can be formed from WFC by exercise. The bomb calorimeter described in Section 6.5 is ideally suited Obesity is a major health hazard in the United States. for measuring the energy content, or “fuel values,” of foods Treatments for obesity so far are focused on diet to lower the (see table). amount of energy consumed, or exercise to increase the The excess energy from food is stored in the body in the amount of energy the body needs. Most current antiobesity form of fats. Fats are a group of organic compounds (triesters drugs work on the diet half of treatment. If scientists can find of glycerol and fatty acids) that are soluble in organic sol- a way to convert WFC to BFC by biological means, and signs vents but insoluble in water. There are two types of fat cells are encouraging, drugs will one day be developed that would called the white fat cells (WFC) and brown fat cells (BFC). fight obesity based on energy expenditure rather than appetite. The WFC are designed to store energy for use in time of need And one can accomplish this goal without having to exercise for body function. They accumulate under the skin and around in a cold environment. Fuel Values of Foods Substance DHcombustion (kJ/g) Apple 22 Beef 28 Beer 21.5 Bread 211 Butter 234 Cheese 218 Eggs 26 Milk 23 Potatoes 23 PET/CT scans of a person exposed to cold temperature (A) and room temperature (B). 250  6.5 Calorimetry 251 Thermometer Table 6.3 Heats of Some Typical Reactions Measured at Constant Pressure Type of ≤H Stirrer Styrofoam cups Reaction Example (kJ/mol) Heat of neutralization HCl(aq) 1 NaOH(aq) ¡ NaCl(aq) 1 H2O(l) 256.2 Heat of ionization H2O(l) ¡ H1 (aq) 1 OH2 (aq) 56.2 Heat of fusion H2O(s) ¡ H2O(l) 6.01 Heat of vaporization H2O(l) ¡ H2O(g) 44.0* Heat of reaction MgCl2 (s) 1 2Na(l) ¡ 2NaCl(s) 1 Mg1s2 2180.2 Reaction *Measured at 258C. At 1008C, the value is 40.79 kJ. mixture cups, as shown in Figure 6.9. This device measures the heat effects of a variety of reactions, such as acid-base neutralization, as well as the heat of solution and heat of dilution. Because the pressure is constant, the heat change for the process (qrxn) is equal to the enthalpy change (DH). As in the case of a constant-volume calorimeter, we treat the calorimeter as an isolated system. Furthermore, we neglect the small heat capacity of the coffee cups in our calculations. Table 6.3 lists some reactions that have Figure 6.9 A constant-pressure calorimeter made of two been studied with the constant-pressure calorimeter. Styrofoam coffee cups. The outer cup helps to insulate the reacting mixture from the surroundings. Two solutions of known volume Example 6.7 containing the reactants at the same temperature are carefully mixed in the calorimeter. The A lead (Pb) pellet having a mass of 26.47 g at 89.988C was placed in a constant-pressure heat produced or absorbed by calorimeter of negligible heat capacity containing 100.0 mL of water. The water the reaction can be determined temperature rose from 22.508C to 23.178C. What is the specific heat of the lead pellet? by measuring the temperature change. Strategy A sketch of the initial and final situation is as follows: We know the masses of water and the lead pellet as well as the initial and final temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost by the lead pellet to the heat gained by the water. Knowing the specific heat of water, we can then calculate the specific heat of lead. Solution Treating the calorimeter as an isolated system (no heat lost to the surroundings), we write qPb 1 qH2O 5 0 or qPb 5 2qH2O The heat gained by the water is given by qH2O 5 ms¢t (Continued) 252 Chapter 6 ■ Thermochemistry where m and s are the mass and specific heat and Dt 5 tfinal 2 tinitial. Therefore, qH2O 5 (100.0 g) (4.184 J/g ? °C) (23.17°C 2 22.50°C) 5 280.3 J Because the heat lost by the lead pellet is equal to the heat gained by the water, qPb 5 2280.3 J. Solving for the specific heat of Pb, we write qPb 5 ms¢t 2280.3 J 5 (26.47 g) (s) (23.17°C 2 89.98°C) Similar problem: 6.88. s 5 0.158 J/g ? °C Practice Exercise A 30.14-g stainless steel ball bearing at 117.828C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.448C. If the specific heat of the ball bearing is 0.474 J/g ? 8C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity. Example 6.8 A quantity of 1.00 3 102 mL of 0.500 M HCl was mixed with 1.00 3 102 mL of 0.500 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 22.508C, and the final temperature of the mixed solution was 25.868C. Calculate the heat change for the neutralization reaction on a molar basis: NaOH(aq) 1 HCl(aq) ¡ NaCl(aq) 1 H2O(l) Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g ? 8C, respectively). Strategy Because the temperature rose, the neutralization reaction is exothermic. How do we calculate the heat absorbed by the combined solution? What is the heat of the reaction? What is the conversion factor for expressing the heat of reaction on a molar basis? Solution Assuming no heat is lost to the surroundings, qsys 5 qsoln 1 qrxn 5 0, so qrxn 5 2qsoln, where qsoln is the heat absorbed by the combined solution. Because the density of the solution is 1.00 g/mL, the mass of a 100-mL solution is 100 g. Thus, qsoln 5 ms¢t 5 (1.00 3 102 g 1 1.00 3 102 g) (4.184 J/g ? °C) (25.86°C 2 22.50°C) 5 2.81 3 103 J 5 2.81 kJ Because qrxn 5 2qsoln, qrxn 5 22.81 kJ. From the molarities given, the number of moles of both HCl and NaOH in 1.00 3 102 mL solution is 0.500 mol 3 0.100 L 5 0.0500 mol 1L Therefore, the heat of neutralization when 1.00 mole of HCl reacts with 1.00 mole of NaOH is 22.81 kJ heat of neutralization 5 5 256.2 kJ/mol 0.0500 mol Check Is the sign consistent with the nature of the reaction? Under the reaction Similar problem: 6.38. condition, can the heat change be equated to the enthalpy change? Practice Exercise A quantity of 4.00 3 102 mL of 0.600 M HNO3 is mixed with 4.00 3 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.468C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation.)  6.6 Standard Enthalpy of Formation and Reaction 253 Review of Concepts A 1-g sample of Al and a 1-g sample of Fe are heated from 408C to 1008C. Which metal has absorbed a greater amount of heat? 6.6 Standard Enthalpy of Formation and Reaction So far we have learned that we can determine the enthalpy change that accompanies a reaction by measuring the heat absorbed or released (at constant pressure). From Equation (6.9) we see that DH can also be calculated if we know the actual enthalpies of all reactants and products. However, as mentioned earlier, there is no way to measure the absolute value of the enthalpy of a substance. Only values relative to an arbitrary reference can be determined. This problem is similar to the one geographers face in expressing the elevations of specific mountains or valleys. Rather than trying to devise some type of “absolute” elevation scale (perhaps based on distance from the center of Earth?), by common agreement all geographic heights and depths are expressed relative to sea level, an arbitrary reference with a defined elevation of “zero” meters or feet. Similarly, chemists have agreed on an arbitrary reference point for enthalpy. The “sea level” reference point for all enthalpy expressions is called the standard enthalpy of formation (DH8f ). Substances are said to be in the standard state at 1 atm, † hence the term “standard enthalpy.” The superscript “°” represents standard-state con- ditions (1 atm), and the subscript “f” stands for formation. By convention, the standard enthalpy of formation of any element in its most stable form is zero. Take the element oxygen as an example. Molecular oxygen (O2) is more stable than the other allotropic form of oxygen, ozone (O3), at 1 atm and 258C. Thus, we can write DH8f (O2) 5 0, but DH8f (O3) 5 142.2 kJ/mol. Similarly, graphite is a more stable allotropic form of carbon than diamond at 1 atm and 258C, so we have DH8f (C, graphite) 5 0 and DH8f (C, diamond) 5 1.90 kJ/mol. Based on this reference for elements, we can now define the standard enthalpy of formation of a compound as the heat change that results when 1 mole of the compound is formed from its elements at a pressure of 1 atm. Table 6.4 lists the standard enthalpies of formation for a number of elements and compounds. (For a more complete list of DH8f values, see Appendix 3.) Note that although the standard state does not specify a temperature, we will always use DH8f values measured at 258C for our discussion because most of the thermodynamic data are collected at this temperature. Graphite (top) and diamond (bottom). Review of Concepts Which of the following does not have DH8f 5 0 at 258C? N2(g) Cu(s) Kr(g) Hg(s) H2(g) I2(s) The importance of the standard enthalpies of formation is that once we know their values, we can readily calculate the standard enthalpy of reaction, DH8rxn, defined as the enthalpy of a reaction carried out at 1 atm. For example, consider the hypothetical reaction aA 1 bB ¡ cC 1 dD where a, b, c, and d are stoichiometric coefficients. For this reaction, DH8rxn is given by ¢H°rxn 5 [c¢H°f (C) 1 d¢H°f (D)] 2 [a¢H°f (A) 1 b¢H°f (B)] (6.17) † In thermodynamics, the standard pressure is defined as 1 bar, where 1 bar 5 105 Pa 5 0.987 atm. Because 1 bar differs from 1 atm by only 1.3 percent, we will continue to use 1 atm as the standard pressure. Note that the normal melting point and boiling point of a substance are defined in terms of 1 atm. 254 Chapter 6 ■ Thermochemistry Table 6.4 Standard Enthalpies of Formation of Some Inorganic Substances at 25°C Substance ≤H8f (kJ/mol) Substance ≤H8f (kJ/mol) Ag(s) 0 H2O2(l) 2187.6 AgCl(s) 2127.0 Hg(l) 0 Al(s) 0 I2(s) 0 Al2O3(s) 21669.8 HI(g) 25.9 Br2(l) 0 Mg(s) 0 HBr(g) 236.2 MgO(s) 2601.8 C(graphite) 0 MgCO3(s) 21112.9 C(diamond) 1.90 N2(g) 0 CO(g) 2110.5 NH3(g) 246.3 CO2(g) 2393.5 NO(g) 90.4 Ca(s) 0 NO2(g) 33.85 CaO(s) 2635.6 N2O(g) 81.56 CaCO3(s) 21206.9 N2O4(g) 9.66 Cl2(g) 0 O(g) 249.4 HCl(g) 292.3 O2(g) 0 Cu(s) 0 O3(g) 142.2 CuO(s) 2155.2 S(rhombic) 0 F2(g) 0 S(monoclinic) 0.30 HF(g) 2271.6 SO2(g) 2296.1 H(g) 218.2 SO3(g) 2395.2 H2(g) 0 H2S(g) 220.15 H2O(g) 2241.8 Zn(s) 0 H2O(l) 2285.8 ZnO(s) 2348.0 We can generalize Equation (6.17) as ¢H°rxn 5 on¢H°f (products) 2 om¢H°f (reactants) (6.18) where m and n denote the stoichiometric coefficients for the reactants and products, and o (sigma) means “the sum of.” Note that in calculations, the stoichiometric coef- ficients are just numbers without units. To use Equation (6.18) to calculate DH8rxn, we must know the DH8f values of the compounds that take part in the reaction. These values can be determined by applying the direct method or the indirect method. The Direct Method This method of measuring DH8f works for compounds that can be readily synthesized from their elements. Suppose we want to know the enthalpy of formation of carbon dioxide. We must measure the enthalpy of the reaction when carbon (graphite) and molecular oxygen in their standard states are converted to carbon dioxide in its standard state: C(graphite) 1 O2 (g) ¡ CO2 (g) ¢H°rxn 5 2393.5 kJ/mol We know from experience that this combustion easily goes to completion. Thus, from Equation (6.18) we can write ¢H°rxn 5 ¢H°f (CO2, g) 2 [¢H°f (C, graphite) 1 ¢H°f (O2, g)] 5 2393.5 kJ/mol  6.6 Standard Enthalpy of Formation and Reaction 255 Because both graphite and O2 are stable allotropic forms of the elements, it follows that DH8f (C, graphite) and DH8f (O2, g) are zero. Therefore, ¢H°rxn 5 ¢H°f (CO2, g) 5 2393.5 kJ/mol or ¢H°f (CO2, g) 5 2393.5 kJ/mol Note that arbitrarily assigning zero DH8f for each element in its most stable form at the standard state does not affect our calculations in any way. Remember, in thermochem- istry we are interested only in enthalpy changes because they can be determined experi- mentally whereas the absolute enthalpy values cannot. The choice of a zero “reference P4 level” for enthalpy makes calculations easier to handle. Again referring to the terrestrial altitude analogy, we find that Mt. Everest is 8708 ft higher than Mt. McKinley. This dif- ference in altitude is unaffected by the decision to set sea level at 0 ft or at 1000 ft. Other compounds that can be studied by the direct method are SF6, P4O10, and CS2. The equations representing their syntheses are S(rhombic) 1 3F2 (g) ¡ SF6 (g) P4 (white) 1 5O2 (g) ¡ P4O10 (s) C(graphite) 1 2S(rhombic) ¡ CS2 (l) Note that S(rhombic) and P(white) are the most stable allotropes of sulfur and phos- phorus, respectively, at 1 atm and 258C, so their DH8f values are zero. The Indirect Method Many compounds cannot be directly synthesized from their elements. In some cases, White phosphorus burns in air to the reaction proceeds too slowly, or side reactions produce substances other than the form P4O10. desired compound. In these cases, DH8f can be determined by an indirect approach, which is based on Hess’s law of heat summation, or simply Hess’s law, named after the Swiss-Russian chemist Germain Hess.† Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. In other words, if we can break down the reaction of interest into a series of reactions for which DH8rxn can be measured, we can calculate DH8rxn for the overall reaction. Hess’s law is based on the fact that because H is a state function, DH depends only on the initial and final state (that is, only on the nature of reactants and products). The enthalpy change would be the same whether the overall reaction takes place in one step or many steps. An analogy for Hess’s law is as follows. Suppose you go from the first floor to the sixth floor of a building by elevator. The gain in your gravitational potential energy (which corresponds to the enthalpy change for the overall process) is the same whether you go directly there or stop at each floor on your way up (breaking the trip into a series of steps). Let’s say we are interested in the standard enthalpy of formation of carbon mon- oxide (CO). We might represent the reaction as C(graphite) 1 12 O2 (g) ¡ CO(g) However, burning graphite also produces some carbon dioxide (CO2), so we cannot measure the enthalpy change for CO directly as shown. Instead, we must employ an indirect route, based on Hess’s law. It is possible to carry out the following two separate reactions, which do go to completion: (a) C(graphite) 1 O2 (g) ¡ CO2 (g)  ¢H°rxn 5 2393.5 kJ/mol (b) CO(g) 1 12 O2 (g) ¡ CO2 (g)  ¢H°rxn 5 2283.0 kJ/mol † Germain Henri Hess (1802–1850). Swiss-Russian chemist. Hess was born in Switzerland but spent most of his life in Russia. For formulating Hess’s law, he is called the father of thermochemistry.  CHEMISTRY in Action How a Bombardier Beetle Defends Itself S urvival techniques of insects and small animals in a fiercely competitive environment take many forms. For example, chameleons have developed the ability to change color to match their surroundings and the butterfly Limenitis has evolved into a form that mimics the poisonous and unpleasant-tasting monarch butterfly (Danaus). A less passive defense mechanism is em- ployed by bombardier beetles (Brachinus), which repel preda- tors with a “chemical spray.” The bombardier beetle has a pair of glands at the tip of its abdomen. Each gland consists of two compartments. The inner compartment contains an aqueous solution of hydroquinone and hydrogen peroxide, and the outer compartment holds a mixture of enzymes. (Enzymes are biological molecules that can speed up a reaction.) When threatened, the beetle squeezes some fluid from the inner compartment into the outer compartment, where, in the presence of the enzymes, an exothermic reaction takes place: A bombardier beetle discharging a chemical spray. (a) C6H4 (OH) 2 (aq) 1 H2O2 (aq) ¡ hydroquinone C6H4O2 (aq) 1 2H2O(l) Therefore, we write quinone ¢H°a 5 ¢H°b 1 ¢H°c 1 ¢H°d To estimate the heat of reaction, let us consider the following 5 (177 2 94.6 2 286) kJ/mol steps: 5 2204 kJ/mol (b) C6H4 (OH) 2 (aq) ¡ C6H4O2 (aq) 1 H2 (g) The large amount of heat generated is sufficient to bring the ¢H° 5 177 kJ/mol mixture to its boiling point. By rotating the tip of its abdomen, (c) H2O2 (aq) ¡ H2O(l) 1 12 O2 (g) the beetle can quickly discharge the vapor in the form of a fine ¢H° 5 294.6 kJ/mol mist toward an unsuspecting predator. In addition to the thermal (d) H2 (g) 1 12 O2 (g) ¡ H2O(l) ¢H° 5 2286 kJ/mol effect, the quinones also act as a repellent to other insects and animals. One bombardier beetle carries enough reagents to pro- Recalling Hess’s law, we find that the heat of reaction for (a) is duce 20 to 30 discharges in quick succession, each with an au- simply the sum of those for (b), (c), and (d). dible detonation. Remember to reverse the sign of DH First, we reverse Equation (b) to get when you reverse a chemical equation. (c) CO2 (g) ¡ CO(g) 1 12 O2 (g)  ¢H°rxn 5 1283.0 kJ/mol Because chemical equations can be added and subtracted just like algebraic equations, we carry out the operation (a) 1 (c) and obtain (a) C(graphite) 1 O2 (g) ¡ CO2 (g)  ¢H°rxn 5 2393.5 kJ/mol (c) CO2 (g) ¡ CO(g) 1 12 O2 (g)   ¢H°rxn 5 1283.0 kJ/mol (d) C(graphite) 1 12 O2 (g) ¡ CO(g)  ¢H°rxn 5 2110.5 kJ/mol 256  6.6 Standard Enthalpy of Formation and Reaction 257 Thus, DH8f (CO) 5 2110.5 kJ/mol. Looking back, we see that the overall reaction is C(graphite) + O2(g) the formation of CO2 [Equation (a)], which can be broken down into two parts [Equa- tions (d) and (b)]. Figure 6.10 shows the overall scheme of our procedure. The general rule in applying Hess’s law is to arrange a series of chemical equa- ΔH8 5 –110.5 kJ tions (corresponding to a series of steps) in such a way that, when added together, all species will cancel except for the reactants and products that appear in the overall CO(g) + ]12 O2(g) reaction. This means that we want the elements on the left and the compound of Enthalpy interest on the right of the arrow. Further, we often need to multiply some or all of ΔH 8 5 the equations representing the individual steps by the appropriate coefficients. –393.5 kJ ΔH8 5 –283.0 kJ Example 6.9 CO2(g) Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C(graphite) 1 H2 (g) ¡ C2H2 (g) Figure 6.10 The enthalpy change for the formation of 1 mole of CO2 from graphite and O2 can The equations for each step and the corresponding enthalpy changes are be broken down into two steps according to Hess’s law. (a) C(graphite) 1 O2 (g) ¡ CO2 (g)  ¢H°rxn 5 2393.5 kJ/mol (b) H2 (g) 1 12 O2 (g) ¡ H2O(l)  ¢H°rxn 5 2285.8 kJ/mol (c) 2C2H2 (g) 1 5O2 (g) ¡ 4CO2 (g) 1 2H2O(l)   ¢H°rxn 5 22598.8 kJ/mol Strategy Our goal here is to calculate the enthalpy change for the formation of C2H2 from its elements C and H2. The reaction does not occur directly, however, so we must use an indirect route using the information given by Equations (a), C2H2 (b), and (c). Solution Looking at the synthesis of C2H2, we need 2 moles of graphite as reactant. So we multiply Equation (a) by 2 to get (d) 2C(graphite) 1 2O2 (g) ¡ 2CO2 (g)  ¢H°rxn 5 2(2393.5 kJ/mol) 5 2787.0 kJ/mol Next, we need 1 mole of H2 as a reactant and this is provided by Equation (b). Last, we need 1 mole of C2H2 as a product. Equation (c) has 2 moles of C2H2 as a reactant so we need to reverse the equation and divide it by 2: (e) 2CO2 (g) 1 H2O(l) ¡ C2H2 (g) 1 52 O2 (g)  ¢H°rxn 5 12 (2598.8 kJ/mol) 5 1299.4 kJ/mol Adding Equations (d), (b), and (e) together, we get 2C(graphite) 1 2O2 (g) ¡ 2CO2 (g)   ¢H°rxn 5 2787.0 kJ/mol H2 (g) 1 12 O2 (g) ¡ H2O(l)  ¢H°rxn 5 2285.8 kJ/mol 2CO2 (g) 1 H2O(l) ¡ C2H2 (g) 1 52 O2 (g)   ¢H°rxn 5 1299.4 kJ/mol 2C(graphite) 1 H2 (g) ¡ C2H2 (g)  ¢H°rxn 5 226.6 kJ/mol An oxyacetylene torch has a high Therefore, DH8f 5 DH8rxn 5 226.6 kJ/mol. The DH8f value means that when 1 mole of C2H2 flame temperature (30008C) and is is synthesized from 2 moles of C(graphite) and 1 mole of H2, 226.6 kJ of heat are absorbed used to weld metals. by the reacting system from the surroundings. Thus, this is an endothermic process. Similar problems: 6.62, 6.63. Practice Exercise Calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, given that C(graphite) 1 O2 (g) ¡ CO2 (g)  ¢H°rxn 5 2393.5 kJ/mol S(rhombic) 1 O2 (g) ¡ SO2 (g)  ¢H°rxn 5 2296.4 kJ/mol CS2 (l) 1 3O2 (g) ¡ CO2 (g) 1 2SO2 (g)   ¢H°rxn 5 21073.6 kJ/mol We can calculate the enthalpy of reactions from the values of DH8f, as shown in Example 6.10. 258 Chapter 6 ■ Thermochemistry Example 6.10 The thermite reaction involves aluminum and iron(III) oxide 2Al(s) 1 Fe2O3 (s) ¡ Al2O3 (s) 1 2Fe(l) This reaction is highly exothermic and the liquid iron formed is used to weld metals. Calculate the heat released in kilojoules per gram of Al reacted with Fe2O3. The DH8f for Fe(l ) is 12.40 kJ/mol. Strategy The enthalpy of a reaction is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. The enthalpy of each species (reactant or product) is given by its stoichiometric coefficient times the standard The molten iron formed in a enthalpy of formation of the species. thermite reaction is run down into a mold between the ends of two Solution Using the given DH8f value for Fe(l) and other DH8f values in Appendix 3 and railroad rails. On cooling, the rails Equation (6.18), we write are welded together. ¢H°rxn 5 [¢H°f (Al2O3 ) 1 2¢H°f (Fe)] 2 [2¢H°f (Al) 1 ¢H°f (Fe2O3 )] 5 [(21669.8 kJ/mol) 1 2(12.40 kJ/mol)] 2 [2(0) 1 (2822.2 kJ/mol)] 5 2822.8 kJ/mol This is the amount of heat released for two moles of Al reacted. We use the following ratio 2822.8 kJ 2 mol Al to convert to kJ/g Al. The molar mass of Al is 26.98 g, so 2822.8 kJ 1 mol Al heat released per gram of Al 5 3 2 mol Al 26.98 g Al 5 215.25 kJ/g Check Is the negative sign consistent with the exothermic nature of the reaction? As a quick check, we see that 2 moles of Al weigh about 54 g and give off about 823 kJ of heat when reacted with Fe2O3. Therefore, the heat given off per gram of Al reacted is Similar problems: 6.54, 6.57. approximately 2830 kJ/54 g or 215.4 kJ/g. Practice Exercise Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. Calculate the heat released (in kilojoules) per gram of the compound reacted with oxygen. The standard enthalpy of formation of benzene is 49.04 kJ/mol. Review of Concepts Explain why reactions involving reactant compounds with positive DH8f values are generally more exothermic than those with negative DH8f values. 6.7 Heat of Solution and Dilution Although we have focused so far on the thermal energy effects resulting from chem- ical reactions, many physical processes, such as the melting of ice or the condensa- tion of a vapor, also involve the absorption or release of heat. Enthalpy changes occur as well when a solute dissolves in a solvent or when a solution is diluted. Let us look at these two related physical processes, involving heat of solution and heat of dilution.  6.7 Heat of Solution and Dilution 259 Heat of Solution In the vast majority of cases, dissolving a solute in a solvent produces measurable heat change. At constant pressure, the heat change is equal to the enthalpy change. The heat of solution, or enthalpy of solution, DHsoln , is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. The quantity DHsoln represents the difference between the enthalpy of the final solution and the enthalpies of its original components (that is, solute and solvent) before they are mixed. Thus, ¢Hsoln 5 Hsoln 2 Hcomponents (6.19) Neither Hsoln nor Hcomponents can be measured, but their difference, DHsoln, can be read- ily determined in a constant-pressure calorimeter. Like other enthalpy changes, DHsoln is positive for endothermic (heat-absorbing) processes and negative for exothermic (heat-generating) processes. Consider the heat of solution of a process in which an ionic compound is the solute and water is the solvent. For example, what happens when solid NaCl dis- solves in water? In solid NaCl, the Na1 and Cl2 ions are held together by strong positive-negative (electrostatic) forces, but when a small crystal of NaCl dissolves in water, the three-dimensional network of ions breaks into its individual units. (The structure of solid NaCl is shown in Figure 2.13.) The separated Na1 and Cl2 ions are stabilized in solution by their interaction with water molecules (see Fig- ure 4.2). These ions are said to be hydrated. In this case water plays a role similar to that of a good electrical insulator. Water molecules shield the ions (Na1 and Cl2) from each other and effectively reduce the electrostatic attraction that held them together in the solid state. The heat of solution is defined by the fol- lowing process: H2O NaCl(s) ¡ Na1 (aq) 1 Cl2 (aq) ¢Hsoln 5 ? Dissolving an ionic compound such as NaCl in water involves complex interactions among the solute and solvent species. However, for the sake of analysis we can imagine that the solution process takes place in two separate steps, illustrated in Figure 6.11. First, the Na1 and Cl2 ions in the solid crystal are separated from each other and converted to the gaseous state: energy 1 NaCl(s) ¡ Na1 (g) 1 Cl 2 (g) The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U). The lattice energy of NaCl is 788 kJ/mol. The word “lattice” describes arrangement in space of isolated points (occupied by In other words, we would need to supply 788 kJ of energy to break 1 mole of solid ions) in a regular pattern. Lattice energy NaCl into 1 mole of Na1 ions and 1 mole of Cl2 ions. is a positive quantity. Beware that lattice Next, the “gaseous” Na1 and Cl2 ions enter the water and become hydrated: energy and internal energy share the same symbol. H2O Na1 (g) 1 Cl2 (g) ¡ Na1 (aq) 1 Cl2 (aq) 1 energy The enthalpy change associated with the hydration process is called the heat of hydration, DHhydr (heat of hydration is a negative quantity for cations and anions). Applying Hess’s law, it is possible to consider DHsoln as the sum of two related quantities, lattice energy (U) and heat of hydration (DHhydr), as shown in Figure 6.11: ¢Hsoln 5 U 1 ¢Hhydr (6.20) 260 Chapter 6 ■ Thermochemistry – + – + + + – – – + + – Na+ and Cl– ions in the gaseous state H D ea = Step 2 Hh t o –7 – f h 84 yd r yd kJ ra tio mol n / + kJ rgy Step 1 ol e 78 e en /m c tti 8 + La + – = U – – + – – + – + Heat of solution + – + – D Hsoln = 4 kJ/mol – + – + + + – Na+ and Cl– ions in the solid state Hydrated Na+ and Cl– ions Figure 6.11 The solution process for NaCl. The process can be considered to occur in two separate steps: (1) separation of ions from the crystal state to the gaseous state and (2) hydration of the gaseous ions. The heat of solution is equal to the energy changes for these two steps, DHsoln 5 U 1 DHhydr. Therefore, NaCl(s) ¡ Na1 (g) 1 Cl2 (g) U 5 788 kJ/mol H2O Na (g) 1 Cl2 (g) ¡ Na1 (aq) 1 Cl2 (aq) 1 ¢Hhydr 5 2784 kJ/mol H2O NaCl(s) ¡ Na1 (aq) 1 Cl2 (aq) ¢Hsoln 5 4 kJ/mol Thus, when 1 mole of NaCl dissolves in water, 4 kJ of heat will be absorbed from the surroundings. We would observe this effect by noting that the beaker containing the solution becomes slightly colder. Table 6.5 lists the DHsoln of several ionic com- pounds. Depending on the nature of the cation and anion involved, DHsoln for an ionic compound may be either negative (exothermic) or positive (endothermic). Table 6.5 Heats of Solution of Some Ionic Compounds Review of Concepts ≤Hsoln Use the data in Appendix 3 to calculate the heat of solution for the following Compound (kJ/mol) process: KNO3 (s) ¡ K1 (aq) 1 NO2 3 (aq) LiCl 237.1 CaCl2 282.8 NaCl 4.0 KCl 17.2 Heat of Dilution NH4Cl 15.2 When a previously prepared solution is diluted, that is, when more solvent is added NH4NO3 26.6 to lower the overall concentration of the solute, additional heat is usually given off or absorbed. The heat of dilution is the heat change associated with the dilution  Summary of Facts & Concepts 261 process. If a certain solution process is endothermic and the solution is subsequently diluted, more heat will be absorbed by the same solution from the surroundings. The converse holds true for an exothermic solution process—more heat will be liberated if additional solvent is added to dilute the solution. Therefore, always be cautious when working on a dilution procedure in the laboratory. Because of its highly exo- thermic heat of dilution, concentrated sulfuric acid (H2SO4) poses a particularly haz- ardous problem if its concentration must be reduced by mixing it with additional water. Concentrated H2SO4 is composed of 98 percent acid and 2 percent water by mass. Diluting it with water releases considerable amount of heat to the surroundings. Generations of chemistry students have been reminded of the safe procedure for This process is so exothermic that you must never attempt to dilute the concentrated diluting acids by the venerable saying, acid by adding water to it. The heat generated could cause the acid solution to boil “Do as you oughter, add acid to water.” and splatter. The recommended procedure is to add the concentrated acid slowly to the water (while constantly stirring). Key Equations DU 5 q 1 w (6.1) Mathematical statement of the first law of thermodynamics. w 5 2PDV (6.3) Calculating work done in gas expansion or gas compression. H 5 U 1 PV (6.6) Definition of enthalpy. DH 5 DU 1 PDV (6.8) Calculating enthalpy (or energy) change for a constant-pressure process. C 5 ms (6.11) Definition of heat capacity. q 5 msDt (6.12) Calculating heat change in terms of specific heat. q 5 CDt (6.13) Calculating heat change in terms of heat capacity. ¢H°rxn 5 on¢H°f (products) 2 om¢H°f (reactants) (6.18) Calculating standard enthalpy of reaction. DHsoln 5 U 1 DHhydr (6.20) Lattice energy and hydration contributions to heat of solution. Summary of Facts & Concepts 1. Energy is the capacity to do work. There are many 6. Enthalpy is a state function. A change in enthalpy DH is forms of energy and they are interconvertible. The law equal to DU 1 PDV for a constant-pressure process. of conservation of energy states that the total amount of 7. The change in enthalpy (DH, usually given in kilo- energy in the universe is constant. joules) is a measure of the heat of reaction (or any other 2. A process that gives off heat to the surroundings is exo- process) at constant pressure. thermic; a process that absorbs heat from the surround- 8. Constant-volume and constant-pressure calorimeters ings is endothermic. are used to measure heat changes that occur in physical 3. The state of a system is defined by properties such as and chemical processes. composition, volume, temperature, and pressure. These 9. Hess’s law states that the overall enthalpy change in a properties are called state functions. reaction is equal to the sum of enthalpy changes for in- 4. The change in a state function for a system depends dividual steps in the overall reaction. only on the initial and final states of the system, and not 10. The standard enthalpy of a reaction can be calculated on the path by which the change is accomplished. En- from the standard enthalpies of formation of reactants ergy is a state function; work and heat are not. and products. 5. Energy can be converted from one form to another, but 11. The heat of solution of an ionic compound in water is the it cannot be created or destroyed (first law of thermo- sum of the lattice energy of the compound and the heat dynamics). In chemistry we are concerned mainly of hydration. The relative magnitudes of these two quan- with thermal energy, electrical energy, and mechanical tities determine whether the solution process is endo- energy, which is usually associated with pressure- thermic or exothermic. The heat of dilution is the heat volume work. absorbed or evolved when a solution is diluted.  262 Chapter 6 ■ Thermochemistry Key Words Calorimetry, p. 246 First law of Lattice energy (U), p. 259 Standard state, p. 253 Chemical energy, p. 231 thermodynamics, p. 234 Law of conservation of State function, p. 234 Closed system, p. 232 Heat, p. 232 energy, p. 231 State of a system, p. 234 Endothermic process, p. 233 Heat capacity (C), p. 246 Open system, p. 232 Surroundings, p. 232 Energy, p. 231 Heat of dilution, p. 260 Potential energy, p. 231 System, p. 232 Enthalpy (H), p. 241 Heat of hydration Radiant energy, p. 231 Thermal energy, p. 231 Enthalpy of reaction (DHhydr), p. 259 Specific heat (s), p. 246 Thermochemical (DHrxn), p. 242 Heat of solution Standard enthalpy of equation, p. 243 Enthalpy of solution (DHsoln), p. 259 formation (DH8f ), p. 253 Thermochemistry, p. 232 (DHsoln), p. 259 Hess’s law, p. 255 Standard enthalpy of reaction Thermodynamics, p. 234 Exothermic process, p. 233 Isolated system, p. 232 (DH8rxn), p. 253 Work, p. 231 Questions & Problems • Problems available in Connect Plus First Law of Thermodynamics Red numbered problems solved in Student Solutions Manual Review Questions Definitions 6.11 On what law is the first law of thermodynamics based? Explain the sign conventions in the equation Review Questions DU 5 q 1 w. • 6.1 Define these terms: system, surroundings, open sys- 6.12 Explain what is meant by a state function. Give two tem, closed system, isolated system, thermal energy, examples of quantities that are state functions and chemical energy, potential energy, kinetic energy, two that are not. law of conservation of energy. 6.13 The internal energy of an ideal gas depends only 6.2 What is heat? How does heat differ from thermal on its temperature. Do a first-law analysis of this energy? Under what condition is heat transferred process. A sample of an ideal gas is allowed to ex- from one system to another? pand at constant temperature against atmospheric 6.3 What are the units for energy commonly employed pressure. (a) Does the gas do work on its surround- in chemistry? ings? (b) Is there heat exchange between the sys- 6.4 A truck initially traveling at 60 km per hour is brought tem and the surroundings? If so, in which direction? to a complete stop at a traffic light. Does this change (c) What is DU for the gas for this process? violate the law of conservation of energy? Explain. • 6.14 Consider these changes. 6.5 These are various forms of energy: chemical, heat, (a) Hg(l) ¡ Hg(g) light, mechanical, and electrical. Suggest ways of (b) 3O2 (g) ¡ 2O3 (g) interconverting these forms of energy. (c) CuSO4 ? 5H2O(s) ¡ CuSO4 (s) 1 5H2O(g) 6.6 Describe the interconversions of forms of energy oc- (d) H2 (g) 1 F2 (g) ¡ 2HF(g) curring in these processes: (a) You throw a softball up At constant pressure, in which of the reactions is into the air and catch it. (b) You switch on a flashlight. work done by the system on the surroundings? By (c) You ride the ski lift to the top of the hill and then the surroundings on the system? In which of them is ski down. (d) You strike a match and let it burn down. no work done? Energy Changes in Chemical Reactions Problems Review Questions • 6.7 Define these terms: thermochemistry, exothermic • 6.15 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. Calculate the process, endothermic process. work done in joules if the gas expands (a) against a 6.8 Stoichiometry is based on the law of conservation of vacuum, (b) against a constant pressure of 0.80 atm, mass. On what law is thermochemistry based? and (c) against a constant pressure of 3.7 atm. 6.9 Describe two exothermic processes and two endo- • 6.16 A gas expands in volume from 26.7 mL to 89.3 mL thermic processes. at constant temperature. Calculate the work done (in 6.10 Decomposition reactions are usually endothermic, joules) if the gas expands (a) against a vacuum, whereas combination reactions are usually exother- (b) against a constant pressure of 1.5 atm, and mic. Give a qualitative explanation for these trends. (c) against a constant pressure of 2.8 atm.  Questions & Problems 263 • 6.17 A gas expands and does P-V work on the surround- If 2.0 moles of H2O(g) are converted to H2(g) and ings equal to 325 J. At the same time, it absorbs 127 J O2(g) against a pressure of 1.0 atm at 1258C, what is of heat from the surroundings. Calculate the change DU for this reaction? in energy of the gas. • 6.28 Consider the reaction • 6.18 The work done to compress a gas is 74 J. As a result, H2 (g) 1 Cl2 (g) ¡ 2HCl(g) 26 J of heat is given off to the surroundings. Calcu- ¢H 5 2184.6 kJ/mol late the change in energy of the gas. • 6.19 Calculate the work done when 50.0 g of tin dis- If 3 moles of H2 react with 3 moles of Cl2 to form solves in excess acid at 1.00 atm and 258C: HCl, calculate the work done (in joules) against a pressure of 1.0 atm at 258C. What is DU for this re- Sn(s) 1 2H1 (aq) ¡ Sn21 (aq) 1 H2 (g) action? Assume the reaction goes to completion. Assume ideal gas behavior. 6.20 Calculate the work done in joules when 1.0 mole of Calorimetry water vaporizes at 1.0 atm and 1008C. Assume that Review Questions the volume of liquid water is negligible compared with that of steam at 1008C, and ideal gas behavior. 6.29 What is the difference between specific heat and heat capacity? What are the units for these two quantities? Which is the intensive property and Enthalpy of Chemical Reactions which is the extensive property? Review Questions 6.30 Define calorimetry and describe two commonly 6.21 Define these terms: enthalpy, enthalpy of reaction. used calorimeters. In a calorimetric measurement, Under what condition is the heat of a reaction equal why is it important that we know the heat capacity to the enthalpy change of the same reaction? of the calorimeter? How is this value determined? 6.22 In writing thermochemical equations, why is it im- Problems portant to indicate the physical state (that is, gas- eous, liquid, solid, or aqueous) of each substance? • 6.31 Consider the following data: 6.23 Explain the meaning of this thermochemical equation: Metal Al Cu 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) ¢H 5 2904 kJ/mol Mass (g) 10 30 Specific heat (J/g ? 8C) 0.900 0.385 6.24 Consider this reaction: Temperature (8C) 40 60 2CH3OH(l) 1 3O2 (g) ¡ 4H2O(l) 1 2CO2 (g) ¢H 5 21452.8 kJ/mol When these two metals are placed in contact, which What is the value of DH if (a) the equation is multi- of the following will take place? plied throughout by 2, (b) the direction of the reac- (a) Heat will flow from Al to Cu because Al has tion is reversed so that the products become the a larger specific heat. reactants and vice versa, (c) water vapor instead of (b) Heat will flow from Cu to Al because Cu has liquid water is formed as the product? a larger mass. (c) Heat will flow from Cu to Al because Cu has Problems a larger heat capacity. • 6.25 The first step in the industrial recovery of zinc from (d) Heat will flow from Cu to Al because Cu is at a the zinc sulfide ore is roasting, that is, the conver- higher temperature. sion of ZnS to ZnO by heating: (e) No heat will flow in either direction. 2ZnS(s) 1 3O2 (g) ¡ 2ZnO(s) 1 2SO2 (g) • 6.32 A piece of silver of mass 362 g has a heat capacity ¢H 5 2879 kJ/mol of 85.7 J/8C. What is the specific heat of silver? Calculate the heat evolved (in kJ) per gram of ZnS • 6.33 A 6.22-kg piece of copper metal is heated from 20.58C to 324.38C. Calculate the heat absorbed (in roasted. kJ) by the metal. • 6.26 Determine the amount of heat (in kJ) given off when • 6.34 Calculate the amount of heat liberated (in kJ) from 1.26 3 104 g of NO2 are produced according to the 366 g of mercury when it cools from 77.08C to 12.08C. equation • 6.35 A sheet of gold weighing 10.0 g and at a temperature 2NO(g) 1 O2 (g) ¡ 2NO2 (g) of 18.08C is placed flat on a sheet of iron weighing ¢H 5 2114.6 kJ/mol 20.0 g and at a temperature of 55.68C. What is the final temperature of the combined metals? Assume that no • 6.27 Consider the reaction heat is lost to the surroundings. (Hint: The heat gained 2H2O(g) ¡ 2H2 (g) 1 O2 (g) by the gold must be equal to the heat lost by the iron. ¢H 5 483.6 kJ/mol The specific heats of the metals are given in Table 6.2.)  264 Chapter 6 ■ Thermochemistry 6.36 To a sample of water at 23.48C in a constant-pressure • 6.51 Calculate the heat of decomposition for this process calorimeter of negligible heat capacity is added a at constant pressure and 258C: 12.1-g piece of aluminum whose temperature is 81.78C. If the final temperature of water is 24.98C, CaCO3 (s) ¡ CaO(s) 1 CO2 (g) calculate the mass of the water in the calorimeter. (Look up the standard enthalpy of formation of the (Hint: See Table 6.2.) reactant and products in Table 6.4.) • 6.37 A 0.1375-g sample of solid magnesium is burned in • 6.52 The standard enthalpies of formation of ions in a constant-volume bomb calorimeter that has a heat aqueous solutions are obtained by arbitrarily as- capacity of 3024 J/8C. The temperature increases by signing a value of zero to H1 ions; that is, 1.1268C. Calculate the heat given off by the burning ¢H°f [H 1 (aq)] 5 0. Mg, in kJ/g and in kJ/mol. (a) For the following reaction • 6.38 A quantity of 85.0 mL of 0.900 M HCl is mixed with H2O 85.0 mL of 0.900 M KOH in a constant-pressure HCl(g) ¡ H1 (aq) 1 Cl2 (aq) calorimeter that has a heat capacity of 325 J/8C. If ¢H° 5 274.9 kJ/mol the initial temperatures of both solutions are the same at 18.248C, what is the final temperature of calculate DH8f for the Cl2 ions. the mixed solution? The heat of neutralization is (b) Given that DH8f for OH2 ions is 2229.6 kJ/mol, 256.2 kJ/mol. Assume the density and specific heat calculate the enthalpy of neutralization when of the solutions are the same as those for water. 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 258C. Standard Enthalpy of Formation and Reaction • 6.53 Calculate the heats of combustion for the following Review Questions reactions from the standard enthalpies of formation 6.39 What is meant by the standard-state condition? listed in Appendix 3: 6.40 How are the standard enthalpies of an element and (a) 2H2 (g) 1 O2 (g) ¡ 2H2O(l) of a compound determined? (b) 2C2H2 (g) 1 5O2 (g) ¡ 4CO2 (g) 1 2H2O(l) 6.41 What is meant by the standard enthalpy of a reaction? • 6.54 Calculate the heats of combustion for the following 6.42 Write the equation for calculating the enthalpy of a reactions from the standard enthalpies of formation reaction. Define all the terms. listed in Appendix 3: 6.43 State Hess’s law. Explain, with one example, the (a) C2H4 (g) 1 3O2 (g) ¡ 2CO2 (g) 1 2H2O(l) usefulness of Hess’s law in thermochemistry. (b) 2H2S(g) 1 3O2 (g) ¡ 2H2O(l) 1 2SO2 (g) 6.44 Describe how chemists use Hess’s law to determine • 6.55 Methanol, ethanol, and n-propanol are three com- the DH8f of a compound by measuring its heat (en- mon alcohols. When 1.00 g of each of these alcohols thalpy) of combustion. is burned in air, heat is liberated as shown by the following data: (a) methanol (CH3OH), 222.6 kJ; (b) ethanol (C2H5OH), 229.7 kJ; (c) n-propanol Problems (C3H7OH), 233.4 kJ. Calculate the heats of com- • 6.45 Which of the following standard enthalpy of forma- bustion of these alcohols in kJ/mol. tion values is not zero at 258C? Na(s), Ne(g), CH4(g), • 6.56 The standard enthalpy change for the following re- S8(s), Hg(l), H(g). action is 436.4 kJ/mol: • 6.46 The DH8f values of the two allotropes of oxygen, O2 H2 (g) ¡ H(g) 1 H(g) and O3, are 0 and 142.2 kJ/mol, respectively, at 258C. Which is the more stable form at this Calculate the standard enthalpy of formation of temperature? atomic hydrogen (H). • 6.47 Which is the more negative quantity at 258C: DH8f • 6.57 From the standard enthalpies of formation, calculate for H2O(l) or DH8f for H2O(g)? DH8rxn for the reaction • 6.48 Predict the value of DH8f (greater than, less than, or equal to zero) for these elements at 258C: (a) Br2(g); C6H12 (l) 1 9O2 (g) ¡ 6CO2 (g) 1 6H2O(l) Br2(l). (b) I2(g); I2(s). For C6H12 (l), ¢H°f 5 2151.9 kJ/mol. 6.49 In general, compounds with negative DH8f values are more stable than those with positive DH8f values. • 6.58 Pentaborane-9, B5H9, is a colorless, highly reactive H2O2(l) has a negative DH8f (see Table 6.4). Why, liquid that will burst into flame when exposed to then, does H2O2(l) have a tendency to decompose to oxygen. The reaction is H2O(l) and O2(g)? 2B5H9 (l) 1 12O2 (g) ¡ 5B2O3 (s) 1 9H2O(l) 6.50 Suggest ways (with appropriate equations) that would enable you to measure the DH8f values of Calculate the kilojoules of heat released per gram of Ag2O(s) and CaCl2(s) from their elements. No cal- the compound reacted with oxygen. The standard culations are necessary. enthalpy of formation of B5H9 is 73.2 kJ/mol.  Questions & Problems 265 • 6.59 Determine the amount of heat (in kJ) given off when Heat of Solution and Dilution 1.26 3 104 g of ammonia are produced according to Review Questions the equation 6.65 Define the following terms: enthalpy of solution, N2 (g) 1 3H2 (g) ¡ 2NH3 (g) heat of hydration, lattice energy, heat of dilution. ¢H°rxn 5 292.6 kJ/mol 6.66 Why is the lattice energy of a solid always a positive quantity? Why is the hydration of ions always a neg- Assume that the reaction takes place under standard- ative quantity? state conditions at 258C. 6.67 Consider two ionic compounds A and B. A has a • 6.60 At 8508C, CaCO3 undergoes substantial decomposi- larger lattice energy than B. Which of the two com- tion to yield CaO and CO2. Assuming that the DH8f pounds is more stable? values of the reactant and products are the same at 8508C as they are at 258C, calculate the enthalpy 6.68 Mg21 is a smaller cation than Na1 and also carries change (in kJ) if 66.8 g of CO2 are produced in one more positive charge. Which of the two species has reaction. a larger hydration energy (in kJ/mol)? Explain. • 6.61 From these data, 6.69 Consider the dissolution of an ionic compound such as potassium fluoride in water. Break the process S(rhombic) 1 O2 (g) ¡ SO2 (g) into the following steps: separation of the cations ¢H°rxn 5 2296.06 kJ/mol and anions in the vapor phase and the hydration of S(monoclinic) 1 O2 (g) ¡ SO2 (g) the ions in the aqueous medium. Discuss the energy ¢H°rxn 5 2296.36 kJ/mol changes associated with each step. How does the heat of solution of KF depend on the relative magni- calculate the enthalpy change for the transformation tudes of these two quantities? On what law is the relationship based? S(rhombic) ¡ S(monoclinic) 6.70 Why is it dangerous to add water to a concentrated (Monoclinic and rhombic are different allotropic acid such as sulfuric acid in a dilution process? forms of elemental sulfur.) • 6.62 From the following data, Additional Problems 6.71 Which of the following does not have DH8f 5 0 at C(graphite) 1 O2 (g) ¡ CO2 (g) 258C? ¢H°rxn 5 2393.5 kJ/mol He(g) Fe(s) Cl(g) S8(s) O2(g) Br2(l) H2 (g) 1 12O2 (g) ¡ H2O(l) ¢H°rxn 5 2285.8 kJ/mol 6.72 Calculate the expansion work done when 3.70 moles 2C2H6 (g) 1 7O2 (g) ¡ 4CO2 (g) 1 6H2O(l) of ethanol are converted to vapor at its boiling point ¢H°rxn 5 23119.6 kJ/mol (78.38C) and 1.0 atm. 6.73 The convention of arbitrarily assigning a zero en- calculate the enthalpy change for the reaction thalpy value for the most stable form of each element 2C(graphite) 1 3H2 (g) ¡ C2H6 (g) in the standard state at 258C is a convenient way of dealing with enthalpies of reactions. Explain why this • 6.63 From the following heats of combustion, convention cannot be applied to nuclear reactions. CH3OH(l) 1 32 O2 (g) ¡ CO2 (g) 1 2H2O(l) 6.74 Given the thermochemical equations: ¢H°rxn 5 2726.4 kJ/mol Br2 (l) 1 F2 (g) ¡ 2BrF(g) C(graphite) 1 O2 (g) ¡ CO2 (g) ¢H° 5 2188 kJ/mol ¢H°rxn 5 2393.5 kJ/mol Br2 (l) 1 3F2 (g) ¡ 2BrF3 (g) H2 (g) 1 1 2 O2 (g) ¡ H2O(l) ¢H° 5 2768 kJ/mol ¢H°rxn 5 2285.8 kJ/mol calculate the DH8rxn for the reaction calculate the enthalpy of formation of methanol BrF(g) 1 F2 (g) ¡ BrF3 (g) (CH3OH) from its elements: • 6.75 The standard enthalpy change DH8 for the thermal C(graphite) 1 2H2 (g) 1 12 O2 (g) ¡ CH3OH(l) decomposition of silver nitrate according to the fol- lowing equation is 178.67 kJ: • 6.64 Calculate the standard enthalpy change for the reaction AgNO3 (s) ¡ AgNO2 (s) 1 12 O2 (g) 2Al(s) 1 Fe2O3 (s) ¡ 2Fe(s) 1 Al2O3 (s) The standard enthalpy of formation of AgNO3(s) is given that 2123.02 kJ/mol. Calculate the standard enthalpy of 2Al(s) 1 32 O2 (g) ¡ Al2O3 (s) formation of AgNO2(s). ¢H°rxn 5 21669.8 kJ/mol 6.76 Hydrazine, N2H4, decomposes according to the fol- 3 lowing reaction: 2Fe(s) 1 2 O2 (g) ¡ Fe2O3 (s) ¢H°rxn 5 2822.2 kJ/mol 3N2H4 (l) ¡ 4NH3 (g) 1 N2 (g)  266 Chapter 6 ■ Thermochemistry (a) Given that the standard enthalpy of formation of Calculate DH8 for the reaction hydrazine is 50.42 kJ/mol, calculate DH8 for its de- H(g) 1 Br(g) ¡ HBr(g) composition. (b) Both hydrazine and ammonia burn in oxygen to produce H2O(l) and N2(g). Write bal- 6.85 A gaseous mixture consists of 28.4 mole percent of anced equations for each of these processes and cal- hydrogen and 71.6 mole percent of methane. A 15.6-L culate DH8 for each of them. On a mass basis (per gas sample, measured at 19.48C and 2.23 atm, is kg), would hydrazine or ammonia be the better fuel? burned in air. Calculate the heat released. 6.77 A quantity of 2.00 3 102 mL of 0.862 M HCl is 6.86 When 2.740 g of Ba reacts with O2 at 298 K and mixed with an equal volume of 0.431 M Ba(OH)2 in 1 atm to form BaO, 11.14 kJ of heat are released. a constant-pressure calorimeter of negligible heat What is DH8f for BaO? capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.488C, For the process • 6.87 Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From the H1 (aq) 1 OH2 (aq) ¡ H2O(l) following data, calculate the standard enthalpy of formation of methanol: the heat of neutralization is 256.2 kJ/mol. What is 2CH3OH(l) 1 3O2 (g) ¡ 2CO2 (g) 1 4H2O(l) the final temperature of the mixed solution? ¢H°rxn 5 21452.8 kJ/mol 6.78 A 3.53-g sample of ammonium nitrate (NH4NO3) was added to 80.0 mL of water in a constant- • 6.88 A 44.0-g sample of an unknown metal at 99.08C was pressure calorimeter of negligible heat capacity. As placed in a constant-pressure calorimeter containing a result, the temperature of the water decreased 80.0 g of water at 24.08C. The final temperature of from 21.68C to 18.18C. Calculate the heat of solution the system was found to be 28.48C. Calculate the (DHsoln) of ammonium nitrate. specific heat of the metal. (The heat capacity of the • 6.79 Consider the reaction calorimeter is 12.4 J/8C.) • 6.89 Using the data in Appendix 3, calculate the enthalpy N2 (g) 1 3H2 (g) ¡ 2NH3 (g) change for the gaseous reaction shown here. (Hint: ¢H°rxn 5 292.6 kJ/mol First determine the limiting reagent.) If 2.0 moles of N2 react with 6.0 moles of H2 to form NH3, calculate the work done (in joules) against a pressure of 1.0 atm at 258C. What is DU for this re- action? Assume the reaction goes to completion. • 6.80 Calculate the heat released when 2.00 L of Cl2(g) with a density of 1.88 g/L react with an excess of sodium metal at 258C and 1 atm to form sodium chloride. • 6.81 Photosynthesis produces glucose, C6H12O6, and oxygen from carbon dioxide and water: CO NO CO2 N2 6CO2 1 6H2O ¡ C6H12O6 1 6O2 (a) How would you determine experimentally the 6.90 Producer gas (carbon monoxide) is prepared by DH8rxn value for this reaction? (b) Solar radiation passing air over red-hot coke: produces about 7.0 3 1014 kg glucose a year on C(s) 1 12 O2 (g) ¡ CO(g) Earth. What is the corresponding DH8 change? 6.82 A 2.10-mole sample of crystalline acetic acid, ini- Water gas (mixture of carbon monoxide and hydro- tially at 17.08C, is allowed to melt at 17.08C and is gen) is prepared by passing steam over red-hot coke: then heated to 118.18C (its normal boiling point) at C(s) 1 H2O(g) ¡ CO(g) 1 H2 (g) 1.00 atm. The sample is allowed to vaporize at 118.18C and is then rapidly quenched to 17.08C, so For many years, both producer gas and water gas that it recrystallizes. Calculate DH8 for the total pro- were used as fuels in industry and for domestic cess as described. cooking. The large-scale preparation of these gases • 6.83 Calculate the work done in joules by the reaction was carried out alternately, that is, first producer gas, then water gas, and so on. Using thermochem- 2Na(s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2 (g) ical reasoning, explain why this procedure was when 0.34 g of Na reacts with water to form hydro- chosen. gen gas at 08C and 1.0 atm. • 6.91 Compare the heat produced by the complete com- bustion of 1 mole of methane (CH4) with a mole of • 6.84 You are given the following data: water gas (0.50 mole H2 and 0.50 mole CO) under H2 (g) ¡ 2H(g)    ¢H° 5 436.4 kJ/mol the same conditions. On the basis of your answer, Br2 (g) ¡ 2Br(g)   ¢H° 5 192.5 kJ/mol would you prefer methane over water gas as a fuel? H2 (g) 1 Br2 (g) ¡ 2HBr(g) Can you suggest two other reasons why methane is ¢H° 5 272.4 kJ/mol preferable to water gas as a fuel?  Questions & Problems 267 • 6.92 The so-called hydrogen economy is based on hydro- • 6.101 Calculate the standard enthalpy of formation for dia- gen produced from water using solar energy. The mond, given that gas may be burned as a fuel: C(graphite) 1 O2 (g) ¡ CO2 (g) 2H2 (g) 1 O2 (g) ¡ 2H2O(l) ¢H° 5 2393.5 kJ/mol C(diamond) 1 O2 (g) ¡ CO2 (g) A primary advantage of hydrogen as a fuel is that it ¢H° 5 2395.4 kJ/mol is nonpolluting. A major disadvantage is that it is a gas and therefore is harder to store than liquids or 6.102 (a) For most efficient use, refrigerator freezer com- solids. Calculate the volume of hydrogen gas at partments should be fully packed with food. What is 258C and 1.00 atm required to produce an amount of the thermochemical basis for this recommendation? energy equivalent to that produced by the combus- (b) Starting at the same temperature, tea and coffee tion of a gallon of octane (C8H18). The density of remain hot longer in a thermal flask than chicken octane is 2.66 kg/gal, and its standard enthalpy of noodle soup. Explain. formation is 2249.9 kJ/mol. 6.103 Calculate the standard enthalpy change for the fer- • 6.93 Ethanol (C2H5OH) and gasoline (assumed to be all mentation process. (See Problem 3.72.) octane, C8H18) are both used as automobile fuel. If gasoline is selling for $4.50/gal, what would the • 6.104 Portable hot packs are available for skiers and peo- ple engaged in other outdoor activities in a cold cli- price of ethanol have to be in order to provide the mate. The air-permeable paper packet contains a same amount of heat per dollar? The density and mixture of powdered iron, sodium chloride, and DH8f of octane are 0.7025 g/mL and 2249.9 kJ/mol other components, all moistened by a little water. and of ethanol are 0.7894 g/mL and 2277.0 kJ/mol, The exothermic reaction that produces the heat is a respectively. 1 gal 5 3.785 L. very common one—the rusting of iron: • 6.94 The combustion of what volume of ethane (C2H6), measured at 23.08C and 752 mmHg, would be 4Fe(s) 1 3O2 (g) ¡ 2Fe2O3 (s) required to heat 855 g of water from 25.08C to 98.08C? When the outside plastic envelope is removed, O2 6.95 If energy is conserved, how can there be an energy molecules penetrate the paper, causing the reaction crisis? to begin. A typical packet contains 250 g of iron to • 6.96 The heat of vaporization of a liquid (DHvap) is the warm your hands or feet for up to 4 hours. How energy required to vaporize 1.00 g of the liquid at its much heat (in kJ) is produced by this reaction? boiling point. In one experiment, 60.0 g of liquid (Hint: See Appendix 3 for DH8f values.) nitrogen (boiling point 21968C) are poured into a • 6.105 A person ate 0.50 pound of cheese (an energy intake Styrofoam cup containing 2.00 3 102 g of water at of 4000 kJ). Suppose that none of the energy was 55.38C. Calculate the molar heat of vaporization of stored in his body. What mass (in grams) of water liquid nitrogen if the final temperature of the water would he need to perspire in order to maintain his is 41.08C. original temperature? (It takes 44.0 kJ to vaporize 6.97 Explain the cooling effect experienced when ethanol 1 mole of water.) is rubbed on your skin, given that • 6.106 The total volume of the Pacific Ocean is estimated to be 7.2 3 108 km3. A medium-sized atomic bomb C2H5OH(l) ¡ C2H5OH(g) ¢H° 5 42.2 kJ/mol produces 1.0 3 1015 J of energy upon explosion. Calculate the number of atomic bombs needed to • 6.98 For which of the following reactions does DH8rxn 5 release enough energy to raise the temperature of DH8f ? the water in the Pacific Ocean by 18C. (a) H2 (g) 1 S(rhombic) ¡ H2S(g) (b) C(diamond) 1 O2 (g) ¡ CO2 (g) • 6.107 A 19.2-g quantity of dry ice (solid carbon diox- ide) is allowed to sublime (evaporate) in an appa- (c) H2 (g) 1 CuO(s) ¡ H2O(l) 1 Cu(s) ratus like the one shown in Figure 6.5. Calculate (d) O(g) 1 O2 (g) ¡ O3 (g) the expansion work done against a constant exter- 6.99 Calculate the work done (in joules) when 1.0 mole nal pressure of 0.995 atm and at a constant tem- of water is frozen at 08C and 1.0 atm. The volumes perature of 228C. Assume that the initial volume of one mole of water and ice at 08C are 0.0180 L and of dry ice is negligible and that CO2 behaves like 0.0196 L, respectively. an ideal gas. • 6.100 A quantity of 0.020 mole of a gas initially at 0.050 L • 6.108 The enthalpy of combustion of benzoic acid and 208C undergoes a constant-temperature expan- (C6H5COOH) is commonly used as the standard for sion until its volume is 0.50 L. Calculate the work calibrating constant-volume bomb calorimeters; done (in joules) by the gas if it expands (a) against its value has been accurately determined to be a vacuum and (b) against a constant pressure of 23226.7 kJ/mol. When 1.9862 g of benzoic acid are 0.20  atm. (c) If the gas in (b) is allowed to expand burned in a calorimeter, the temperature rises from unchecked until its pressure is equal to the external 21.848C to 25.678C. What is the heat capacity of the pressure, what would its final volume be before it bomb? (Assume that the quantity of water surround- stopped expanding, and what would be the work done? ing the bomb is exactly 2000 g.)  268 Chapter 6 ■ Thermochemistry • 6.109 The combustion of a 25.0-g gaseous mixture of H2 the temperature rises from 19.258C to 22.178C. If and CH4 releases 2354 kJ of heat. Calculate the the heat capacity of the calorimeter is 98.6 J/8C, cal- amounts of the gases in grams. culate the enthalpy change for the above reaction on • 6.110 Calcium oxide (CaO) is used to remove sulfur diox- a molar basis. Assume that the density and specific ide generated by coal-burning power stations: heat of the solution are the same as those for water, and ignore the specific heats of the metals. 2CaO(s) 1 2SO2 (g) 1 O2 (g) ¡ 2CaSO4 (s) 6.116 (a) A person drinks four glasses of cold water Calculate the enthalpy change for this process if (3.08C) every day. The volume of each glass is 6.6 3 105 g of SO2 are removed by this process 2.5 3 102 mL. How much heat (in kJ) does the body every day. have to supply to raise the temperature of the water • 6.111 Glauber’s salt, sodium sulfate decahydrate (Na2SO4 ? to 378C, the body temperature? (b) How much heat 10H2O), undergoes a phase transition (that is, would your body lose if you were to ingest 8.0 3 melting or freezing) at a convenient temperature of 102 g of snow at 08C to quench thirst? (The amount about 328C: of heat necessary to melt snow is 6.01 kJ/mol.) 6.117 A driver’s manual states that the stopping distance Na2SO4 ? 10H2O(s) ¡ Na2SO4 ? 10H2O(l) quadruples as the speed doubles; that is, if it takes 30 ¢H° 5 74.4 kJ/mol ft to stop a car moving at 25 mph then it would take As a result, this compound is used to regulate the 120 ft to stop a car moving at 50 mph. Justify this temperature in homes. It is placed in plastic bags in statement by using mechanics and the first law of ther- the ceiling of a room. During the day, the endother- modynamics. [Assume that when a car is stopped, its mic melting process absorbs heat from the surround- kinetic energy ( 12 mu2 ) is totally converted to heat.] ings, cooling the room. At night, it gives off heat as • 6.118 At 258C, the standard enthalpy of formation of it freezes. Calculate the mass of Glauber’s salt in HF(aq) is given by 2320.1 kJ/mol; of OH2(aq), it is kilograms needed to lower the temperature of air in 2229.6 kJ/mol; of F2(aq), it is 2329.1 kJ/mol; and a room by 8.28C at 1.0 atm. The dimensions of the of H2O(l), it is 2285.8 kJ/mol. room are 2.80 m 3 10.6 m 3 17.2 m, the specific (a) Calculate the standard enthalpy of neutralization heat of air is 1.2 J/g ? 8C, and the molar mass of air of HF(aq): may be taken as 29.0 g/mol. HF(aq) 1 OH2 (aq) ¡ F2 (aq) 1 H2O(l) • 6.112 A balloon 16 m in diameter is inflated with helium at 188C. (a) Calculate the mass of He in the balloon, (b) Using the value of 256.2 kJ as the standard assuming ideal behavior. (b) Calculate the work enthalpy change for the reaction done (in joules) during the inflation process if the atmospheric pressure is 98.7 kPa. H 1 (aq) 1 OH2 (aq) ¡ H2O(l) 6.113 Acetylene (C2H2) can be hydrogenated (reacting calculate the standard enthalpy change for the with hydrogen) first to ethylene (C2H4) and then to reaction ethane (C2H6). Starting with one mole of C2H2, label the diagram shown here analogous to Figure 6.10. HF(aq) ¡ H 1 (aq) 1 F2 (aq) Use the data in Appendix 3. 6.119 Why are cold, damp air and hot, humid air more un- comfortable than dry air at the same temperatures? (The specific heats of water vapor and air are approx- imately 1.9 J/g ? 8C and 1.0 J/g ? 8C, respectively.) 6.120 From the enthalpy of formation for CO2 and the fol- lowing information, calculate the standard enthalpy Enthalpy of formation for carbon monoxide (CO). CO(g) 1 12O2 (g) ¡ CO2 (g) ¢H° 5 2283.0 kJ/mol Why can’t we obtain it directly by measuring the enthalpy of the following reaction? C(graphite) 1 12 O2 (g) ¡ CO(g) 6.114 Calculate the DH8 for the reaction Fe31 (aq) 1 3OH2 (aq) ¡ Fe(OH) 3 (s) • 6.121 A 46-kg person drinks 500 g of milk, which has a “caloric” value of approximately 3.0 kJ/g. If only 6.115 An excess of zinc metal is added to 50.0 mL of a 17 percent of the energy in milk is converted to me- 0.100 M AgNO3 solution in a constant-pressure cal- chanical work, how high (in meters) can the person orimeter like the one pictured in Figure 6.9. As a climb based on this energy intake? [Hint: The work result of the reaction done in ascending is given by mgh, where m is the mass (in kilograms), g the gravitational accelera- Zn(s) 1 2Ag1 (aq) ¡ Zn21 (aq) 1 2Ag(s) tion (9.8 m/s2), and h the height (in meters).]  Questions & Problems 269 • 6.122 The height of Niagara Falls on the American side • 6.129 A gas company in Massachusetts charges $1.30 for is 51 m. (a) Calculate the potential energy of 1.0 g 15 ft3 of natural gas (CH4) measured at 208C and of water at the top of the falls relative to the 1.0 atm. Calculate the cost of heating 200 mL of ground level. (b) What is the speed of the falling water (enough to make a cup of coffee or tea) from water if all of the potential energy is converted to 208C to 1008C. Assume that only 50 percent of the kinetic energy? (c) What would be the increase in heat generated by the combustion is used to heat the temperature of the water if all the kinetic energy water; the rest of the heat is lost to the surroundings. were converted to heat? (See Problem 6.121 for 6.130 Calculate the internal energy of a Goodyear blimp suggestions.) filled with helium gas at 1.2 3 105 Pa. The volume • 6.123 In the nineteenth century two scientists named of the blimp is 5.5 3 103 m3. If all the energy were Dulong and Petit noticed that for a solid element, the used to heat 10.0 tons of copper at 218C, calculate product of its molar mass and its specific heat is the final temperature of the metal. (Hint: See Sec- approximately 25 J/8C. This observation, now called tion 5.7 for help in calculating the internal energy of Dulong and Petit’s law, was used to estimate the spe- a gas. 1 ton 5 9.072 3 105 g.) cific heat of metals. Verify the law for the metals 6.131 Decomposition reactions are usually endothermic, listed in Table 6.2. The law does not apply to one of whereas combination reactions are usually exother- the metals. Which one is it? Why? mic. Give a qualitative explanation for these trends. 6.124 Determine the standard enthalpy of formation of 6.132 Acetylene (C2H2) can be made by reacting cal- ethanol (C2H5OH) from its standard enthalpy of cium carbide (CaC2) with water. (a) Write an combustion (21367.4 kJ/mol). equation for the reaction. (b) What is the maxi- • 6.125 Acetylene (C2H2) and benzene (C6H6) have the same mum amount of heat (in joules) that can be ob- empirical formula. In fact, benzene can be made tained from the combustion of acetylene, starting from acetylene as follows: with 74.6 g of CaC2? 3C2H2 (g) ¡ C6H6 (l) 6.133 The average temperature in deserts is high during the day but quite cool at night, whereas that in regions The enthalpies of combustion for C2H2 and C6H6 are along the coastline is more moderate. Explain. 21299.4 kJ/mol and 23267.4 kJ/mol, respectively. • 6.134 When 1.034 g of naphthalene (C10H8) are burned Calculate the standard enthalpies of formation of in a constant-volume bomb calorimeter at 298 K, C2H2 and C6H6 and hence the enthalpy change for 41.56 kJ of heat are evolved. Calculate DU and DH the formation of C6H6 from C2H2. for the reaction on a molar basis. • 6.126 Ice at 08C is placed in a Styrofoam cup containing 6.135 From a thermochemical point of view, explain why 361 g of a soft drink at 238C. The specific heat of a carbon dioxide fire extinguisher or water should the drink is about the same as that of water. Some not be used on a magnesium fire. ice remains after the ice and soft drink reach an 6.136 Calculate the DU for the following reaction at 298 K: equilibrium temperature of 08C. Determine the mass of ice that has melted. Ignore the heat capac- 2H2 (g) 1 O2 (g) ¡ 2H2O(l) ity of the cup. (Hint: It takes 334 J to melt 1 g of ice at 08C.) • 6.137 Lime is a term that includes calcium oxide (CaO, also called quicklime) and calcium hydroxide 6.127 After a dinner party, the host performed the following [Ca(OH)2, also called slaked lime]. It is used in the trick. First, he blew out one of the burning candles. steel industry to remove acidic impurities, in air- He then quickly brought a lighted match to about 1 in pollution control to remove acidic oxides such as above the wick. To everyone’s surprise, the candle SO2, and in water treatment. Quicklime is made in- was relighted. Explain how the host was able to ac- dustrially by heating limestone (CaCO3) above complish the task without touching the wick. 20008C: CaCO3 (s) ¡ CaO(s) 1 CO2 (g) ¢H° 5 177.8 kJ/mol Slaked lime is produced by treating quicklime with water: CaO(s) 1 H2O(l) ¡ Ca(OH) 2 (s) ¢H° 5 265.2 kJ/mol The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime (0.946 J/g ? 8C) and slaked lime (1.20 J/g ? 8C) make it hazardous to store and transport lime in vessels made 6.128 How much heat is required to decompose 89.7 g of of wood. Wooden sailing ships carrying lime would NH4Cl? (Hint: You may use the enthalpy of forma- occasionally catch fire when water leaked into the tion values at 258C for the calculation.) hold. (a) If a 500-g sample of water reacts with an 270 Chapter 6 ■ Thermochemistry equimolar amount of CaO (both at an initial tempera- ture of 258C), what is the final temperature of the prod- uct, Ca(OH)2? Assume that the product absorbs all of 2 B C the heat released in the reaction. (b) Given that the stan- dard enthalpies of formation of CaO and H2O are P (atm) 2635.6 kJ/mol and 2285.8 kJ/mol, respectively, cal- culate the standard enthalpy of formation of Ca(OH)2. 1 6.138 A 4.117-g impure sample of glucose (C6H12O6) was A D burned in a constant-volume calorimeter having a heat capacity of 19.65 kJ/8C. If the rise in temperature is 3.1348C, calculate the percent by mass of the glu- 1 2 cose in the sample. Assume that the impurities are V (L) unaffected by the combustion process. See Appendix 3 for thermodynamic data. • 6.145 For reactions in condensed phases (liquids and 6.139 Construct a table with the headings q, w, DU, and DH. solids), the difference between DH and DU is usu- For each of the following processes, deduce whether ally quite small. This statement holds for reactions each of the quantities listed is positive (1), negative carried out under atmospheric conditions. For cer- (2), or zero (0). (a) Freezing of benzene. (b) Compres- tain geochemical processes, however, the external sion of an ideal gas at constant temperature. (c) Reac- pressure may be so great that DH and DU can dif- tion of sodium with water. (d) Boiling liquid ammonia. fer by a significant amount. A well-known exam- (e) Heating a gas at constant volume. (f) Melting of ice. ple is the slow conversion of graphite to diamond 6.140 The combustion of 0.4196 g of a hydrocarbon re- under Earth’s surface. Calculate (DH 2 DU) for leases 17.55 kJ of heat. The masses of the products the conversion of 1 mole of graphite to 1 mole of are CO2 5 1.419 g and H2O 5 0.290 g. (a) What is diamond at a pressure of 50,000 atm. The densities of the empirical formula of the compound? (b) If the graphite and diamond are 2.25 g/cm3 and 3.52 g/cm3, approximate molar mass of the compound is 76 g, respectively. calculate its standard enthalpy of formation. 6.146 The diagrams shown here represent various physical 6.141 Metabolic activity in the human body releases and chemical processes. (a) 2A(g) ¡ A2(g). approximately 1.0 3 10 4 kJ of heat per day. (b) MX(s) ¡ M1(aq) 1 X2(aq). (c) AB(g) 1 Assuming the body is 50 kg of water, how much C(g) ¡ AC(g) 1 B(g). (d) B(l) ¡ B(g). Pre- would the body temperature rise if it were an iso- dict whether the situations shown are endothermic lated system? How much water must the body or exothermic. Explain why in some cases no clear eliminate as perspiration to maintain the normal conclusions can be made. body temperature (98.68F)? Comment on your re- sults. The heat of vaporization of water may be taken as 2.41 kJ/g. 6.142 Give an example for each of the following situations: (a) Adding heat to a system raises its temperature, (b) adding heat to a system does not change (raise) its temperature, and (c) a system’s temperature is (a) (b) changed even though no heat is added or removed from it. 6.143 From the following data, calculate the heat of solu- tion for KI: NaCl NaI KCl KI Lattice energy 788 686 699 632 (c) (d) (kJ/mol) Heat of solution 4.0 25.1 17.2 ? 6.147 A 20.3-g sample of an unknown metal and a 28.5-g (kJ/mol) sample of copper, both at 80.68C, are added to 100 g of water at 11.28C in a constant-pressure calorimeter 6.144 Starting at A, an ideal gas undergoes a cyclic pro- of negligible heat capacity. If the final temperature of cess involving expansion and compression, as the metals and water is 13.78C, determine the spe- shown here. Calculate the total work done. Does cific heat of the unknown metal. your result support the notion that work is not a state function?  Answers to Practice Exercises 271 Interpreting, Modeling & Estimating 6.148 For most biological processes, DH < DU. Explain. would the hydrogen gas need to be kept for the tank to contain an equivalent amount of chemical energy 6.149 Estimate the potential energy expended by an average as a tank of gasoline? adult male in going from the ground to the top floor of the Empire State Building using the staircase. 6.155 A press release announcing a new fuel-cell car to the public stated that hydrogen is “relatively cheap” and 6.150 The fastest serve in tennis is about 150 mph. Can the “some stations in California sell hydrogen for $5 a kinetic energy of a tennis ball traveling at this speed kilogram. A kg has the same energy as a gallon of be sufficient to heat 1 mL of water by 308C? gasoline, so it’s like paying $5 a gallon. But you go 6.151 Can the total energy output of the sun in one second two to three times as far on the hydrogen.” Analyze be sufficient to heat all of the ocean water on Earth this claim. to its boiling point? 6.156 We hear a lot about how the burning of hydrocar- 6.152 It has been estimated that 3 trillion standard cubic bons produces the greenhouse gas CO2, but what feet of methane is released into the atmosphere every about the effect of increasing energy consumption year. Capturing that methane would provide a source on the amount of oxygen in the atmosphere required of energy, and it would also remove a potent green- to sustain life. The figure shows past and projected house gas from the atmosphere (methane is 25 times energy world consumption. (a) How many moles of more effective at trapping heat than an equal number oxygen would be required to generate the addi- of molecules of carbon dioxide). Standard cubic feet tional energy expenditure for the next decade? is measured at 608F and 1 atm. Determine the amount (b) What would be the resulting decrease in atmo- of energy that could be obtained by combustion of spheric oxygen? the methane that escapes each year. 6.153 Biomass plants generate electricity from waste ma- 800 World energy consumption terial such as wood chips. Some of these plants con- vert the feedstock to ethanol (C2H5OH) for later use 700 as a fuel. (a) How many grams of ethanol can be (1015 kJ) produced from 1.0 ton of wood chips, if 85 percent 600 of the carbon is converted to C2H5OH? (b) How much energy would be released by burning the etha- 500 nol obtained from 1.0 ton of wood chips? (Hint: Treat the wood chips as cellulose.) 400 6.154 Suppose an automobile carried hydrogen gas in its 2005 2010 2015 2020 2025 2030 fuel tank instead of gasoline. At what pressure Year Answers to Practice Exercises 6.1 (a) 0, (b) 2286 J. 6.2 263 J. 6.3 26.47 3 103 kJ. 6.7 21.198C. 6.8 22.498C. 6.9 87.3 kJ/mol. 6.4 2111.7 kJ/mol. 6.5 234.3 kJ. 6.6 2728 kJ/mol. 6.10 241.83 kJ/g. CHEMICAL M YS TERY The Exploding Tire† I t was supposed to be a routine job: Fix the flat tire on Harvey Smith’s car. The owner of Tom’s Garage, Tom Lee, gave the tire to Jerry to work on, while he went outside to pump gas. A few minutes later, Tom heard a loud bang. He rushed inside to find the tire blown to pieces, a wall collapsed, equipment damaged, and Jerry lying on the floor, unconscious and bleeding. Luckily Jerry’s injury was not serious. As he lay in the hospital recovering, the mystery of the exploding tire unfolded. The tire had gone flat when Harvey drove over a nail. Being a cautious driver, Harvey carried a can of instant tire repair in the car, so he was able to reinflate the tire and drive safely home. The can of tire repair Harvey used contained latex (natural rubber) dissolved in a liquid propellant, which is a mixture of propane (C3H8) and butane (C4H10). Propane and butane are gases under atmospheric conditions but exist as liquids under compression in the can. When the valve on the top of the can is pressed, it opens, releasing the pressure inside. The mixture boils, forming a latex foam which is propelled by the gases into the tire to seal the puncture while the gas reinflates the tire. The pressure in a flat tire is approximately one atmosphere, or roughly 15 pounds per square inch (psi). Using the aerosol tire repair, Harvey reinflated his damaged tire to a pressure of 35 psi. This is called the gauge pressure, which is the pressure of the tire above the atmospheric pressure. Thus, the total pressure in the tire was actually (15 1 35) psi, or 50 psi. One problem with using natural gases like propane and butane as propellants is that they are highly flammable. In fact, these gases can react explo- sively when mixed with air at a concentration of 2 percent to 9 percent by volume. Jerry was aware of the hazards of repairing Harvey’s tire and took precautions to avoid an accident. First he let out the excess gas in the tire. Next he reinflated the tire to 35 psi with air. And he repeated the procedure once. Clearly, this is a dilution process intended to gradually decrease the concentrations of propane and butane. The fact that the tire exploded means that Jerry had not diluted the gases enough. But what was the source of ignition? When Jerry found the nail hole in the tire, he used a tire reamer, a metal file-like instrument, to clean dirt and loose rubber from the hole before applying a rubber plug and liquid sealant. The last thing Jerry remembered was pulling the reamer out of the hole. The next thing he knew he was lying in the hospital, hurting all over. To solve this mystery, make use of the following clues. † Adapted from “The Exploding Tire,” by Jay A. Young, CHEM MATTERS, April, 1988, p. 12. Copyright 1995 American Chemical Society. 272 Chemical Clues 1. Write balanced equations for the combustion of propane and butane. The products are carbon dioxide and water. 2. When Harvey inflated his flat tire to 35 psi, the composition by volume of the propane and butane gases is given by (35 psiy50 psi) 3 100%, or 70 percent. When Jerry deflated the tire the first time, the pressure fell to 15 psi but the composition remained at 70 percent. Based on these facts, calculate the percent composition of propane and butane at the end of two deflation-inflation steps. Does it fall within the explosive range? 3. Given that Harvey’s flat tire is a steel-belted tire, explain how the ignition of the gas mixture might have been triggered. (A steel-belted tire has two belts of steel wire for outer reinforcement and two belts of polyester cord for inner reinforcement.) Instant flat tire repair. 273  CHAPTER 7 Quantum Theory and the Electronic Structure of Atoms “Neon light” is a generic term for atomic emission involving various noble gases, mercury, and phosphor. The UV light from excited mercury atoms causes phosphor-coated tubes to fluoresce white light and other colors. CHAPTER OUTLINE A LOOK AHEAD 7.1 From Classical Physics  We begin by discussing the transition from classical physics to quantum to Quantum Theory theory. In particular, we become familiar with properties of waves and elec- tromagnetic radiation and Planck’s formulation of the quantum theory. (7.1) 7.2 The Photoelectric Effect  Einstein’s explanation of the photoelectric effect is another step toward the 7.3 Bohr’s Theory of the development of the quantum theory. To explain experimental observations, Hydrogen Atom Einstein suggested that light behaves like a bundle of particles called 7.4 The Dual Nature photons. (7.2) of the Electron  We then study Bohr’s theory for the emission spectrum of the hydrogen atom. In particular, Bohr postulated that the energies of an electron in the 7.5 Quantum Mechanics atom are quantized and transitions from higher levels to lower ones account 7.6 Quantum Numbers for the emission lines. (7.3) 7.7 Atomic Orbitals  Some of the mysteries of Bohr’s theory are explained by de Broglie who suggested that electrons can behave like waves. (7.4) 7.8 Electron Configuration  We see that the early ideas of quantum theory led to a new era in physics 7.9 The Building-Up Principle called quantum mechanics. The Heisenberg uncertainty principle sets the lim- its for measurement of quantum mechanical systems. The Schrödinger wave equation describes the behavior of electrons in atoms and molecules. (7.5)  We learn that there are four quantum numbers to describe an electron in an atom and the characteristics of orbitals in which the electrons reside. (7.6 and 7.7)  Electron configuration enables us to keep track of the distribution of elec- trons in an atom and understand its magnetic properties. (7.8)  Finally, we apply the rules in writing electron configurations to the entire periodic table. In particular, we group elements according to their valence electron configurations. (7.9) 274  7.1 From Classical Physics to Quantum Theory 275 Q uantum theory enables us to predict and understand the critical role that electrons play in chemistry. In one sense, studying atoms amounts to asking the following questions: 1. How many electrons are present in a particular atom? 2. What energies do individual electrons possess? 3. Where in the atom can electrons be found? The answers to these questions have a direct relationship to the behavior of all substances in chemical reactions, and the story of the search for answers provides a fascinating backdrop for our discussion. 7.1 From Classical Physics to Quantum Theory Early attempts by nineteenth-century physicists to understand atoms and molecules met with only limited success. By assuming that molecules behave like rebounding balls, physicists were able to predict and explain some macroscopic phenomena, such as the pressure exerted by a gas. However, this model did not account for the stability of molecules; that is, it could not explain the forces that hold atoms together. It took a long time to realize—and an even longer time to accept—that the properties of atoms and molecules are not governed by the same physical laws as larger objects. The new era in physics started in 1900 with a young German physicist named Max Planck.† While analyzing the data on radiation emitted by solids heated to vari- ous temperatures, Planck discovered that atoms and molecules emit energy only in certain discrete quantities, or quanta. Physicists had always assumed that energy is continuous and that any amount of energy could be released in a radiation process. Planck’s quantum theory turned physics upside down. Indeed, the flurry of research that ensued altered our concept of nature forever. Properties of Waves To understand Planck’s quantum theory, we must first know something about the nature of waves. A wave can be thought of as a vibrating disturbance by which energy is transmitted. The fundamental properties of a wave are illustrated by a familiar type—water waves (Figure 7.1). The regular variation of the peaks and troughs enable us to sense the propagation of the waves. Waves are characterized by their length and height and by the number of waves that pass through a certain point in one second (Figure 7.2). Wavelength λ (lambda) is the distance between identical points on successive waves. The frequency ν (nu) is the Figure 7.1 Ocean water waves. number of waves that pass through a particular point in 1 second. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Another important property of waves is their speed, which depends on the type of wave and the nature of the medium through which the wave is traveling (for example, air, water, or a vacuum). The speed (u) of a wave is the product of its wavelength and its frequency: u 5 λn (7.1) The inherent “sensibility” of Equation (7.1) becomes apparent if we analyze the phys- ical dimensions involved in the three terms. The wavelength (λ) expresses the length † Max Karl Ernst Ludwig Planck (1858–1947). German physicist. Planck received the Nobel Prize in Physics in 1918 for his quantum theory. He also made significant contributions in thermodynamics and other areas of physics. 276 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms Wavelength Wavelength Amplitude Amplitude Direction of wave Wavelength propagation Amplitude (a) (b) Figure 7.2 (a) Wavelength and amplitude. (b) Two waves having different wavelengths and frequencies. The wavelength of the top wave is three times that of the lower wave, but its frequency is only one-third that of the lower wave. Both waves have the same speed and amplitude. of a wave, or distance/wave. The frequency (n) indicates the number of these waves that pass any reference point per unit of time, or waves/time. Thus, the product of these terms results in dimensions of distance/time, which is speed: distance distance waves 5 3 time wave time Wavelength is usually expressed in units of meters, centimeters, or nanometers, and frequency is measured in hertz (Hz), where 1 Hz 5 1 cycle/s The word “cycle” may be left out and the frequency expressed as, for example, 25/s or 25 s21 (read as “25 per second”). Review of Concepts Which of the waves shown here has (a) the highest frequency, (b) the longest wavelength, (c) the greatest amplitude? (a) (b) (c) Electromagnetic Radiation There are many kinds of waves, such as water waves, sound waves, and light waves. Sound waves and water waves are not electromagnetic waves, but X rays and In 1873 James Clerk Maxwell proposed that visible light consists of electromag- radio waves are. netic waves. According to Maxwell’s theory, an electromagnetic wave has an electric  7.1 From Classical Physics to Quantum Theory 277 field component and a magnetic field component. These two components have the same wavelength and frequency, and hence the same speed, but they travel in mutu- ally perpendicular planes (Figure 7.3). The significance of Maxwell’s theory is that it provides a mathematical description of the general behavior of light. In particu- lar, his model accurately describes how energy in the form of radiation can be propagated through space as vibrating electric and magnetic fields. Electromag- netic radiation is the emission and transmission of energy in the form of electro- magnetic waves. Electromagnetic waves travel 3.00 3 108 meters per second (rounded off), or A more accurate value for the speed of light is given on the inside back cover of 186,000 miles per second in a vacuum. This speed differs from one medium to the book. another, but not enough to distort our calculations significantly. By convention, we use the symbol c for the speed of electromagnetic waves, or as it is more commonly z Electric field component called, the speed of light. The wavelength of electromagnetic waves is usually given y in nanometers (nm). x Example 7.1 Magnetic field component The wavelength of the green light from a traffic signal is centered at 522 nm. What is Figure 7.3 The electric field the frequency of this radiation? and magnetic field components of an electromagnetic wave. Strategy We are given the wavelength of an electromagnetic wave and asked to calculate These two components have the its frequency. Rearranging Equation (7.1) and replacing u with c (the speed of light) gives same wavelength, frequency, and amplitude, but they oscillate in two c mutually perpendicular planes. n5 λ Solution Because the speed of light is given in meters per second, it is convenient to first convert wavelength to meters. Recall that 1 nm 5 1 3 1029 m (see Table 1.3). We write 1 3 1029 m λ 5 522 nm 3 5 522 3 1029 m 1 nm 5 5.22 3 1027 m Substituting in the wavelength and the speed of light (3.00 3 108 m/s), the frequency is 3.00 3 108 m/s n5 5.22 3 1027 m 5 5.75 3 1014/s, or 5.75 3 1014 Hz Check The answer shows that 5.75 3 1014 waves pass a fixed point every second. This very high frequency is in accordance with the very high speed of light. Similar problem: 7.7. Practice Exercise What is the wavelength (in meters) of an electromagnetic wave whose frequency is 3.64 3 107 Hz? Figure 7.4 shows various types of electromagnetic radiation, which differ from one another in wavelength and frequency. The long radio waves are emitted by large antennas, such as those used by broadcasting stations. The shorter, visible light waves are produced by the motions of electrons within atoms and molecules. The shortest waves, which also have the highest frequency, are associated with γ (gamma) rays, which result from changes within the nucleus of the atom (see Chapter 2). As we will see shortly, the higher the frequency, the more energetic the radiation. Thus, ultravio- let radiation, X rays, and γ rays are high-energy radiation. 278 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms 10–3 10–1 10 103 105 107 109 1011 1013 Wavelength (nm) 1020 1018 1016 1014 1012 1010 108 106 104 Frequency (Hz) Visible Gamma X rays Ultra- Infrared Microwave Radio waves rays violet Type of radiation X ray Sun lamps Heat lamps Microwave ovens, UHF TV, FM radio, VHF TV AM radio police radar, cellular satellite stations telephones (a) 400 nm 500 600 700 (b) Figure 7.4 (a) Types of electromagnetic radiation. Gamma rays have the shortest wavelength and highest frequency; radio waves have the longest wavelength and the lowest frequency. Each type of radiation is spread over a specific range of wavelengths (and frequencies). (b) Visible light ranges from a wavelength of 400 nm (violet) to 700 nm (red). Planck’s Quantum Theory When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. The dull red glow of an electric heater and the bright white light of a tungsten lightbulb are examples of radiation from heated solids. Measurements taken in the latter part of the nineteenth century showed that the amount of radiant energy emitted by an object at a certain temperature depends on its wavelength. Attempts to account for this dependence in terms of established wave theory and thermodynamic laws were only partially successful. One theory explained short-wavelength dependence but failed to account for the longer wavelengths. The failure in the short wavelength region Another theory accounted for the longer wavelengths but failed for short wave- is called the ultraviolet catastrophe. lengths. It seemed that something fundamental was missing from the laws of clas- sical physics. Planck solved the problem with an assumption that departed drastically from accepted concepts. Classical physics assumed that atoms and molecules could emit (or absorb) any arbitrary amount of radiant energy. Planck said that atoms and mol- ecules could emit (or absorb) energy only in discrete quantities, like small packages or bundles. Planck gave the name quantum to the smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation. The energy E of a single quantum of energy is given by E 5 hn (7.2)  7.2 The Photoelectric Effect 279 where h is called Planck’s constant and n is the frequency of radiation. The value of Incident light Planck’s constant is 6.63 3 10234 J ? s. Because n 5 cyλ, Equation (7.2) can also be expressed as c E5h (7.3) λ e– According to quantum theory, energy is always emitted in integral multiples of Metal hn; for example, hn, 2 hn, 3 hn, . . . , but never, for example, 1.67 hn or 4.98 hn. At + – the time Planck presented his theory, he could not explain why energies should be fixed or quantized in this manner. Starting with this hypothesis, however, he had no trouble correlating the experimental data for emission by solids over the entire range of wavelengths; they all supported the quantum theory. The idea that energy should be quantized or “bundled” may seem strange, but the concept of quantization has many analogies. For example, an electric charge is also quantized; there can be only whole-number multiples of e, the charge of one electron. Matter itself is quantized, for the numbers of electrons, protons, and neutrons and the numbers of atoms in a sample of matter must also be integers. Our money system is based on a “quantum” of value called a penny. Even processes in living systems involve quantized phenomena. The eggs laid by hens are quantized, and a pregnant cat gives birth to an integral number of kittens, not to one-half or three- quarters of a kitten. Review of Concepts Voltage Why is radiation only in the UV but not the visible or infrared region responsible source Meter for sun tanning? Figure 7.5 An apparatus for studying the photoelectric effect. Light of a certain frequency falls on a clean metal surface. Ejected electrons are attracted toward 7.2 The Photoelectric Effect the positive electrode. The flow of electrons is registered by a In 1905, only five years after Planck presented his quantum theory, Albert Einstein† detecting meter. Light meters used in cameras are based on the used the theory to solve another mystery in physics, the photoelectric effect, a phe- photoelectric effect. nomenon in which electrons are ejected from the surface of certain metals exposed to light of at least a certain minimum frequency, called the threshold frequency (Figure 7.5). The number of electrons ejected was proportional to the intensity (or brightness) of the light, but the energies of the ejected electrons were not. Below the threshold frequency no electrons were ejected no matter how intense the light. The photoelectric effect could not be explained by the wave theory of light. Einstein, however, made an extraordinary assumption. He suggested that a beam of light is really a stream of particles. These particles of light are now called photons. Using Planck’s quantum theory of radiation as a starting point, Einstein deduced that each photon must possess energy E, given by the equation This equation has the same form as Equation (7.2) because, as we will see shortly, electromagnetic radiation is E 5 hn emitted as well as absorbed in the form of photons. where n is the frequency of light. † Albert Einstein (1879–1955). German-born American physicist. Regarded by many as one of the two greatest physicists the world has known (the other is Isaac Newton). The three papers (on special relativity, Brownian motion, and the photoelectric effect) that he published in 1905 while employed as a technical assistant in the Swiss patent office in Berne have profoundly influenced the development of physics. He received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect. 280 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms Example 7.2 Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 3 104 nm (infrared region) and (b) a photon with a wavelength of 5.00 3 1022 nm (X-ray region). Strategy In both (a) and (b) we are given the wavelength of a photon and asked to calculate its energy. We need to use Equation (7.3) to calculate the energy. Planck’s constant is given in the text and also on the back inside cover. Solution (a) From Equation (7.3), c E5h λ (6.63 3 10234 J ? s) (3.00 3 108 m/s) 5 1 3 1029 m (5.00 3 104 nm) 1 nm 5 3.98 3 10221 J This is the energy of a single photon with a 5.00 3 104 nm wavelength. (b) Following the same procedure as in (a), we can show that the energy of the photon that has a wavelength of 5.00 3 1022 nm is 3.98 3 10215 J . Check Because the energy of a photon increases with decreasing wavelength, we see that Similar problem: 7.15. an “X-ray” photon is 1 3 106, or a million times, more energetic than an “infrared” photon. Practice Exercise The energy of a photon is 5.87 3 10220 J. What is its wavelength (in nanometers)? Electrons are held in a metal by attractive forces, and so removing them from the metal requires light of a sufficiently high frequency (which corresponds to sufficiently high energy) to break them free. Shining a beam of light onto a metal surface can be thought of as shooting a beam of particles—photons—at the metal atoms. If the fre- quency of photons is such that hn is exactly equal to the energy that binds the electrons in the metal, then the light will have just enough energy to knock the electrons loose. If we use light of a higher frequency, then not only will the electrons be knocked loose, but they will also acquire some kinetic energy. This situation is summarized by the equation hn 5 KE 1 W (7.4) where KE is the kinetic energy of the ejected electron and W is the work function, which is a measure of how strongly the electrons are held in the metal. Rewriting Equation (7.4) as KE 5 hn 2 W shows that the more energetic the photon (that is, the higher the frequency), the greater the kinetic energy of the ejected electron. Now consider two beams of light having the same frequency (which is greater than the threshold frequency) but different intensities. The more intense beam of light consists of a larger number of photons; consequently, it ejects more electrons from the metal’s surface than the weaker beam of light. Thus, the more intense the light, the greater the number of electrons emitted by the target metal; the higher the fre- quency of the light, the greater the kinetic energy of the ejected electrons.  7.2 The Photoelectric Effect 281 Example 7.3 The work function of cesium metal is 3.42 3 10219 J. (a) Calculate the minimum frequency of light required to release electrons from the metal. (b) Calculate the kinetic energy of the ejected electron if light of frequency 1.00 3 1015 s21 is used for irradiating the metal. Strategy (a) The relationship between the work function of an element and the frequency of light is given by Equation (7.4). The minimum frequency of light needed to dislodge an electron is the point where the kinetic energy of the ejected electron is zero. (b) Knowing both the work function and the frequency of light, we can solve for the kinetic energy of the ejected electron. Solution (a) Setting KE 5 0 in Equation (7.4), we write hn 5 W Thus, W 3.42 3 10219 J n5 5 h 6.63 3 10234 J ? s 5 5.16 3 1014 s21 (b) Rearranging Equation (7.4) gives KE 5 hn 2 W 5 (6.63 3 10234 J ? s) (1.00 3 1015 s21 ) 2 3.42 3 10219 J 5 3.21 3 10219 J Check The kinetic energy of the ejected electron (3.21 3 10219 J) is smaller than the energy of the photon (6.63 3 10219 J). Therefore, the answer is reasonable. Similar problems: 7.21, 7.22. 219 Practice Exercise The work function of titanium metal is 6.93 3 10 J. Calculate the kinetic energy of the ejected electrons if light of frequency 2.50 3 1015 s21 is used to irradiate the metal. Einstein’s theory of light posed a dilemma for scientists. On the one hand, it explains the photoelectric effect satisfactorily. On the other hand, the particle theory of light is not consistent with the known wave behavior of light. The only way to resolve the dilemma is to accept the idea that light possesses both particle- like and wavelike properties. Depending on the experiment, light behaves either as a wave or as a stream of particles. This concept, called particle-wave duality, was totally alien to the way physicists had thought about matter and radiation, and it took a long time for them to accept it. We will see in Section 7.4 that a dual nature (particles and waves) is not unique to light but is also characteristic of all matter, including electrons. Review of Concepts A clean metal surface is irradiated with light of three different wavelengths λ1, λ2, and λ3. The kinetic energies of the ejected electrons are as follows: λ1: 2.9 3 10220 J; λ2: approximately zero; λ3: 4.2 3 10219 J. Which light has the shortest wavelength and which has the longest wavelength? 282 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms 7.3 Bohr’s Theory of the Hydrogen Atom Einstein’s work paved the way for the solution of yet another nineteenth-century “mystery” in physics: the emission spectra of atoms. Emission Spectra Animation Ever since the seventeenth century, when Newton showed that sunlight is composed Emission Spectra of various color components that can be recombined to produce white light, chemists and physicists have studied the characteristics of emission spectra, that is, either continuous or line spectra of radiation emitted by substances. The emission spectrum of a substance can be seen by energizing a sample of material either with thermal Animation energy or with some other form of energy (such as a high-voltage electrical dis- Line Spectra charge). A “red-hot” or “white-hot” iron bar freshly removed from a high-temperature source produces a characteristic glow. This visible glow is the portion of its emission spectrum that is sensed by eye. The warmth of the same iron bar represents another portion of its emission spectrum—the infrared region. A feature common to the emission spectra of the sun and of a heated solid is that both are continuous; that is, all wavelengths of visible light are represented in the spectra (see the visible region in Figure 7.4). The emission spectra of atoms in the gas phase, on the other hand, do not show a continuous spread of wavelengths from red to violet; rather, the atoms produce bright lines in different parts of the visible spectrum. These line spectra are the light emis- sion only at specific wavelengths. Figure 7.6 is a schematic diagram of a discharge tube that is used to study emission spectra, and Figure 7.7 shows the color emitted by hydrogen atoms in a discharge tube. Every element has a unique emission spectrum. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms, much as fin- When a high voltage is applied gerprints are used to identify people. When the lines of the emission spectrum of a between the forks, some of the known element exactly match the lines of the emission spectrum of an unknown sodium ions in the pickle are sample, the identity of the sample is established. Although the utility of this proce- converted to sodium atoms in an excited state. These atoms emit dure was recognized some time ago in chemical analysis, the origin of these lines the characteristic yellow light as was unknown until early in the twentieth century. Figure 7.8 shows the emission they relax to the ground state. spectra of several elements. Emission Spectrum of the Hydrogen Atom Animation In 1913, not too long after Planck’s and Einstein’s discoveries, a theoretical explana- Atomic Line Spectra tion of the emission spectrum of the hydrogen atom was presented by the Danish physicist Niels Bohr.† Bohr’s treatment is very complex and is no longer considered to be correct in all its details. Thus, we will concentrate only on his important assump- tions and final results, which do account for the spectral lines. When Bohr first tackled this problem, physicists already knew that the atom contains electrons and protons. They thought of an atom as an entity in which elec- trons whirled around the nucleus in circular orbits at high velocities. This was an appealing model because it resembled the motions of the planets around the sun. In the hydrogen atom, it was believed that the electrostatic attraction between the positive “solar” proton and the negative “planetary” electron pulls the electron inward and that this force is balanced exactly by the outward acceleration due to the circular motion of the electron. † Niels Henrik David Bohr (1885–1962). Danish physicist. One of the founders of modern physics, he received the Nobel Prize in Physics in 1922 for his theory explaining the spectrum of the hydrogen atom.  Photographic plate Slit High voltage Line Prism spectrum Discharge tube Light separated into various components (a) Figure 7.7 Color emitted by hydrogen atoms in a discharge tube. The color observed results from the combination of the colors (b) emitted in the visible spectrum. Figure 7.6 (a) An experimental arrangement for studying the emission spectra of atoms and molecules. The gas under study is in a discharge tube containing two electrodes. As electrons flow from the negative electrode to the positive electrode, they collide with the gas. This collision process eventually leads to the emission of light by the atoms (or molecules). The emitted light is separated into its components by a prism. Each component color is focused at a definite position, according to its wavelength, and forms a colored image of the slit on the photographic plate. The colored images are called spectral lines. (b) The line emission spectrum of hydrogen atoms. Figure 7.8 The emission spectra of various elements. The units for Mercury the spectral lines are angstroms (Å), where 1 Å 5 1 3 10210 m. Helium Lithium Thallium Cadmium Strontium Barium Calcium Hydrogen Sodium 283 284 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms According to the laws of classical physics, however, an electron moving in an Photon orbit of a hydrogen atom would experience an acceleration toward the nucleus by radiating away energy in the form of electromagnetic waves. Thus, such an electron would quickly spiral into the nucleus and annihilate itself with the proton. To explain why this does not happen, Bohr postulated that the electron is allowed to occupy only n=1 certain orbits of specific energies. In other words, the energies of the electron are n=2 quantized. An electron in any of the allowed orbits will not spiral into the nucleus and therefore will not radiate energy. Bohr attributed the emission of radiation by an energized hydrogen atom to the electron dropping from a higher-energy allowed orbit n=3 to a lower one and emitting a quantum of energy (a photon) in the form of light (Figure 7.9). Bohr showed that the energies that an electron in a hydrogen atom can Figure 7.9 The emission process in an excited hydrogen atom, occupy are given by according to Bohr’s theory. An electron originally in a higher- energy orbit (n 5 3) falls back to 1 a lower-energy orbit (n 5 2). As En 5 2RH a b (7.5) a result, a photon with energy hn n2 is given off. The value of hn is equal to the difference in energies of the two orbits occupied by the where RH, the Rydberg† constant for the hydrogen atom, has the value 2.18 3 10218 J. electron in the emission process. The number n is an integer called the principal quantum number; it has the values For simplicity, only three orbits are shown. n 5 1, 2, 3, . . . . The negative sign in Equation (7.5) is an arbitrary convention, signifying that the energy of the electron in the atom is lower than the energy of a free electron, which is an electron that is infinitely far from the nucleus. The energy of a free electron is arbitrarily assigned a value of zero. Mathematically, this corresponds to setting n equal to infinity in Equation (7.5), so that E` 5 0. As the electron gets closer to the nucleus (as n decreases), En becomes larger in absolute value, but also more negative. The most negative value, then, is reached when n 5 1, which corresponds to the most stable energy state. We call this the ground state, or the ground level, which refers to the lowest energy state of a system (which is an atom in our discussion). The sta- bility of the electron diminishes for n 5 2, 3, . . . . Each of these levels is called an excited state, or excited level, which is higher in energy than the ground state. A hydrogen electron for which n is greater than 1 is said to be in an excited state. The radius of each circular orbit in Bohr’s model depends on n2. Thus, as n increases from 1 to 2 to 3, the orbit radius increases very rapidly. The higher the excited state, the farther away the electron is from the nucleus (and the less tightly it is held by the nucleus). Bohr’s theory enables us to explain the line spectrum of the hydrogen atom. Radiant energy absorbed by the atom causes the electron to move from a lower- energy state (characterized by a smaller n value) to a higher-energy state (character- ized by a larger n value). Conversely, radiant energy (in the form of a photon) is emitted when the electron moves from a higher-energy state to a lower-energy state. The quantized movement of the electron from one energy state to another is analo- gous to the movement of a tennis ball either up or down a set of stairs (Figure 7.10). The ball can be on any of several steps but never between steps. The journey from a lower step to a higher one is an energy-requiring process, whereas movement from a higher step to a lower step is an energy-releasing process. The quantity of energy involved in either type of change is determined by the distance between the begin- ning and ending steps. Similarly, the amount of energy needed to move an electron in the Bohr atom depends on the difference in energy levels between the initial and final states. Figure 7.10 A mechanical analogy for the emission † processes. The ball can rest on Johannes Robert Rydberg (1854–1919). Swedish physicist. Rydberg’s major contribution to physics was any step but not between steps. his study of the line spectra of many elements.  7.3 Bohr’s Theory of the Hydrogen Atom 285 To apply Equation (7.5) to the emission process in a hydrogen atom, let us sup- pose that the electron is initially in an excited state characterized by the principal quantum number ni. During emission, the electron drops to a lower energy state char- acterized by the principal quantum number nf (the subscripts i and f denote the initial and final states, respectively). This lower energy state may be either a less excited state or the ground state. The difference between the energies of the initial and final states is ¢E 5 Ef 2 Ei From Equation (7.5), 1 Ef 5 2RH a b n2f 1 and Ei 5 2RH a b n2i 2RH 2RH Therefore, ¢E 5 a b2a b n2f n2i 1 1 5 RH a 2 2b n2i nf Because this transition results in the emission of a photon of frequency n and energy hn, we can write 1 1 ¢E 5 hn 5 RH a 2 2 2b (7.6) ni nf When a photon is emitted, ni . nf. Consequently the term in parentheses is negative and DE is negative (energy is lost to the surroundings). When energy is absorbed, ni , nf and the term in parentheses is positive, so DE is positive. Each spectral line in the emission spectrum corresponds to a particular transition in a hydrogen atom. When we study a large number of hydrogen atoms, we observe all possible transitions and hence the corresponding spectral lines. The brightness of a spectral line depends on how many photons of the same wavelength are emitted. The emission spectrum of hydrogen includes a wide range of wavelengths from the infrared to the ultraviolet. Table 7.1 lists the series of transitions in the hydrogen Table 7.1 The Various Series in Atomic Hydrogen Emission Spectrum Series nf ni Spectrum Region Lyman 1 2, 3, 4, . . . Ultraviolet Balmer 2 3, 4, 5, . . . Visible and ultraviolet Paschen 3 4, 5, 6, . . . Infrared Brackett 4 5, 6, 7, . . . Infrared 286 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms ∞ 7 6 5 4 Brackett series 3 Paschen series Energy 2 Balmer series n=1 Lyman series Figure 7.11 The energy levels in the hydrogen atom and the various emission series. Each energy level corresponds to the energy associated with an allowed energy state for an orbit, as postulated by Bohr and shown in Figure 7.9. The emission lines are labeled according to the scheme in Table 7.1. spectrum; they are named after their discoverers. The Balmer series was particularly easy to study because a number of its lines fall in the visible range. Figure 7.9 shows a single transition. However, it is more informative to express transitions as shown in Figure 7.11. Each horizontal line represents an allowed energy level for the electron in a hydrogen atom. The energy levels are labeled with their principal quantum numbers. Example 7.4 illustrates the use of Equation (7.6). Example 7.4 What is the wavelength of a photon (in nanometers) emitted during a transition from the ni 5 5 state to the nf 5 2 state in the hydrogen atom? Strategy We are given the initial and final states in the emission process. We can calculate the energy of the emitted photon using Equation (7.6). Then from Equations (7.2) and (7.1) we can solve for the wavelength of the photon. The value of Rydberg’s constant is given in the text. Solution From Equation (7.6) we write 1 1 ¢E 5 RH a 2 2 2b ni nf 1 1 5 2.18 3 10218 J a 2 2 2b 5 2 5 24.58 3 10219 J (Continued)  7.4 The Dual Nature of the Electron 287 The negative sign indicates that this is energy associated with an emission process. To The negative sign is in accord with our calculate the wavelength, we will omit the minus sign for DE because the wavelength convention that energy is given off to the surroundings. of the photon must be positive. Because ¢E 5 hn or n 5 ¢E/h, we can calculate the wavelength of the photon by writing c λ5 n ch 5 ¢E (3.00 3 108 m/s) (6.63 3 10234 J ? s) 5 4.58 3 10219 J 27 5 4.34 3 10 m 1 nm 5 4.34 3 1027 m 3 a b 5 434 nm 1 3 1029 m Check The wavelength is in the visible region of the electromagnetic region (see Figure 7.4). This is consistent with the fact that because nf 5 2, this transition gives Similar problems: 7.31, 7.32. rise to a spectral line in the Balmer series (see Figure 7.6). Practice Exercise What is the wavelength (in nanometers) of a photon emitted during a transition from ni 5 6 to nf 5 4 state in the H atom? Review of Concepts Which transition in the hydrogen atom would emit light of a shorter wavelength? (a) ni 5 5 ¡ nf 5 3 or (b) ni 5 4 ¡ nf 5 2. The Chemistry in Action essay on p. 288 discusses a special type of atomic emission—lasers. 7.4 The Dual Nature of the Electron Physicists were both mystified and intrigued by Bohr’s theory. They questioned why the energies of the hydrogen electron are quantized. Or, phrasing the question in a more concrete way, Why is the electron in a Bohr atom restricted to orbiting the nucleus at certain fixed distances? For a decade no one, not even Bohr himself, had a logical explanation. In 1924 Louis de Broglie† provided a solution to this puzzle. De Broglie reasoned that if light waves can behave like a stream of particles (photons), then perhaps particles such as electrons can possess wave properties. According to de Broglie, an electron bound to the nucleus behaves like a standing wave. Standing waves can be generated by plucking, say, a guitar string (Figure 7.12). The waves are described as standing, or stationary, because they do not travel along the string. Some points on the string, called nodes, do not move at all; that is, the amplitude † Louis Victor Pierre Raymond Duc de Broglie (1892–1977). French physicist. Member of an old and noble family in France, he held the title of a prince. In his doctoral dissertation, he proposed that matter and radiation have the properties of both wave and particle. For this work, de Broglie was awarded the Nobel Prize in Physics in 1929.  CHEMISTRY in Action Laser—The Splendid Light L aser is an acronym for light amplification by stimulated emission of radiation. It is a special type of emission that involves either atoms or molecules. Since the discovery of la- radiation with wavelengths ranging from infrared through visi- ble and ultraviolet. The advent of laser has truly revolutionized science, medicine, and technology. ser in 1960, it has been used in numerous systems designed to Ruby laser was the first known laser. Ruby is a deep-red operate in the gas, liquid, and solid states. These systems emit mineral containing corundum, Al2O3, in which some of the Al31 Totally reflecting mirror Flash lamp The emission of laser light from a ruby laser. Laser beam λ 5 694.3 nm Ruby rod Partially reflecting mirror The stimulated emission of one photon by another photon in a cascade event that leads to the emission of laser light. The synchronization of the light waves produces an intensely penetrating laser beam. of the wave at these points is zero. There is a node at each end, and there may be nodes between the ends. The greater the frequency of vibration, the shorter the wavelength of the standing wave and the greater the number of nodes. As Fig- ure 7.12 shows, there can be only certain wavelengths in any of the allowed motions of the string. Figure 7.12 The standing waves generated by plucking a guitar string. Each dot represents a node. The length of the string (l) must be equal to a whole λ number times one-half the l= – 2 l = 2 λ–2 l = 3 λ–2 wavelength (λ/2). 288 ions have been replaced by Cr31 ions. A flashlamp is used to Laser light is characterized by three properties: It is in- excite the chromium atoms to a higher energy level. The excited tense, it has precisely known wavelength and hence energy, and atoms are unstable, so at a given instant some of them will return it is coherent. By coherent we mean that the light waves are all to the ground state by emitting a photon in the red region of the in phase. The applications of lasers are quite numerous. Their spectrum. The photon bounces back and forth many times be- high intensity and ease of focus make them suitable for doing tween mirrors at opposite ends of the laser tube. This photon can eye surgery, for drilling holes in metals and welding, and for stimulate the emission of photons of exactly the same wave- carrying out nuclear fusion. The fact that they are highly direc- length from other excited chromium atoms; these photons in turn tional and have precisely known wavelengths makes them very can stimulate the emission of more photons, and so on. Because useful for telecommunications. Lasers are also used in isotope the light waves are in phase—that is, their maxima and minima separation, in holography (three-dimensional photography), in coincide—the photons enhance one another, increasing their compact disc players, and in supermarket scanners. Lasers power with each passage between the mirrors. One of the mirrors have played an important role in the spectroscopic investiga- is only partially reflecting, so that when the light reaches a cer- tion of molecular properties and of many chemical and bio- tain intensity it emerges from the mirror as a laser beam. De- logical processes. pending on the mode of operation, the laser light may be emitted in pulses (as in the ruby laser case) or in continuous waves. State-of-the-art lasers used in the research laboratory of Dr. A. H. Zewail at the California Institute of Technology. De Broglie argued that if an electron does behave like a standing wave in the hydrogen atom, the length of the wave must fit the circumference of the orbit exactly (Figure 7.13). Otherwise the wave would partially cancel itself on each successive orbit. Eventually the amplitude of the wave would be reduced to zero, and the wave would not exist. The relation between the circumference of an allowed orbit (2πr) and the wave- length (λ) of the electron is given by 2πr 5 nλ (7.7) where r is the radius of the orbit, λ is the wavelength of the electron wave, and n 5 1, 2, 3, . . . . Because n is an integer, it follows that r can have only certain values as n increases from 1 to 2 to 3 and so on. And because the energy of the electron depends on the size of the orbit (or the value of r), its value must be quantized. 289 290 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms De Broglie’s reasoning led to the conclusion that waves can behave like particles and particles can exhibit wavelike properties. De Broglie deduced that the particle and wave properties are related by the expression In using Equation (7.8), m must be in h kilograms and u must be in m/s. λ5 (7.8) mu where λ, m, and u are the wavelengths associated with a moving particle, its mass, and its velocity, respectively. Equation (7.8) implies that a particle in motion can be treated as a wave, and a wave can exhibit the properties of a particle. Note that the left side of Equation (7.8) involves the wavelike property of wavelength, whereas the right side makes references to mass, a distinctly particlelike property. Example 7.5 (a) Calculate the wavelength of the “particle” in the following two cases: (a) The fastest serve in tennis is about 150 miles per hour, or 68 m/s. Calculate the wavelength associated with a 6.0 3 1022-kg tennis ball traveling at this speed. (b) Calculate the wavelength associated with an electron (9.1094 3 10231 kg) moving at 68 m/s. Strategy We are given the mass and the speed of the particle in (a) and (b) and asked to calculate the wavelength so we need Equation (7.8). Note that because the units of Planck’s constants are J ? s, m and u must be in kg and m/s (1 J 5 1 kg m2/s2), respectively. Solution (a) Using Equation (7.8) we write (b) h λ5 Figure 7.13 (a) The circum- mu ference of the orbit is equal to an 6.63 3 10234 J ? s integral number of wavelengths. 5 (6.0 3 1022 kg) 3 68 m/s This is an allowed orbit. (b) The circumference of the orbit is not 5 1.6 3 10234 m equal to an integral number of wavelengths. As a result, the Comment This is an exceedingly small wavelength considering that the size of an electron wave does not close in on atom itself is on the order of 1 3 10210 m. For this reason, the wave properties of a itself. This is a nonallowed orbit. tennis ball cannot be detected by any existing measuring device. (b) In this case, h λ5 mu 6.63 3 10234 J ? s 5 (9.1094 3 10231 kg) 3 68 m/s 5 1.1 3 1025 m Comment This wavelength (1.1 3 1025 m or 1.1 3 104 nm) is in the infrared region. Similar problems: 7.40, 7.41. This calculation shows that only electrons (and other submicroscopic particles) have measurable wavelengths. Practice Exercise Calculate the wavelength (in nanometers) of a H atom (mass 5 1.674 3 10227 kg) moving at 7.00 3 102 cm/s.  7.5 Quantum Mechanics 291 Figure 7.14 (a) X-ray diffraction pattern of aluminum foil. (b) Electron diffraction of aluminum foil. The similarity of these two patterns shows that electrons can behave like X rays and display wave properties. (a) (b) Review of Concepts Which quantity in Equation (7.8) is responsible for the fact that macroscopic objects do not show observable wave properties? Example 7.5 shows that although de Broglie’s equation can be applied to diverse systems, the wave properties become observable only for submicroscopic objects. Shortly after de Broglie introduced his equation, Clinton Davisson† and Lester Germer‡ in the United States and G. P. Thomson§ in England demonstrated that elec- trons do indeed possess wavelike properties. By directing a beam of electrons through a thin piece of gold foil, Thomson obtained a set of concentric rings on a screen, similar to the pattern observed when X rays (which are waves) were used. Figure 7.14 shows such a pattern for aluminum. The Chemistry in Action essay on p. 292 describes electron microscopy. 7.5 Quantum Mechanics The spectacular success of Bohr’s theory was followed by a series of disappointments. In reality, Bohr’s theory accounted for the observed emission spectra of He1 and Bohr’s approach did not account for the emission spectra of atoms containing more Li21 ions, as well as that of hydrogen. than one electron, such as atoms of helium and lithium. Nor did it explain why extra However, all three systems have one feature in common—each contains a lines appear in the hydrogen emission spectrum when a magnetic field is applied. single electron. Thus, the Bohr model Another problem arose with the discovery that electrons are wavelike: How can the worked successfully only for the hydrogen atom and for “hydrogenlike ions.” “position” of a wave be specified? We cannot define the precise location of a wave because a wave extends in space. † Clinton Joseph Davisson (1881–1958). American physicist. He and G. P. Thomson shared the Nobel Prize in Physics in 1937 for demonstrating wave properties of electrons. ‡ Lester Halbert Germer (1896–1972). American physicist. Discoverer (with Davisson) of the wave proper- ties of electrons. § George Paget Thomson (1892–1975). English physicist. Son of J. J. Thomson, he received the Nobel Prize in Physics in 1937, along with Clinton Davisson, for demonstrating wave properties of electrons.  CHEMISTRY in Action Electron Microscopy T he electron microscope is an extremely valuable application of the wavelike properties of electrons because it produces images of objects that cannot be seen with the naked eye or with light microscopes. According to the laws of optics, it is impos- sible to form an image of an object that is smaller than half the wavelength of the light used for the observation. Because the range of visible light wavelengths starts at around 400 nm, or 4 3 1025 cm, we cannot see anything smaller than 2 3 1025 cm. In principle, we can see objects on the atomic and molecular scale by using X rays, whose wavelengths range from about 0.01 nm to 10 nm. However, X rays cannot be focused, so they An electron micrograph showing a normal red blood cell and a sickled red do not produce well-formed images. Electrons, on the other blood cell from the same person. hand, are charged particles, which can be focused in the same way the image on a TV screen is focused, that is, by applying an electric field or a magnetic field. According to Equation (7.8), maintained between the needle and the surface of the sample the wavelength of an electron is inversely proportional to its to induce electrons to tunnel through space to the sample. As velocity. By accelerating electrons to very high velocities, we the needle moves over the sample, at a distance of a few can obtain wavelengths as short as 0.004 nm. atomic diameters from the surface, the tunneling current is A different type of electron microscope, called the scan- measured. This current decreases with increasing distance ning tunneling microscope (STM), makes use of another from the sample. By using a feedback loop, the vertical posi- quantum mechanical property of the electron to produce an tion of the tip can be adjusted to a constant distance from the image of the atoms on the surface of a sample. Because of its surface. The extent of these adjustments, which profile the extremely small mass, an electron is able to move or “tunnel” sample, is recorded and displayed as a three-dimensional through an energy barrier (instead of going over it). The false-colored image. STM consists of a tungsten metal needle with a very fine Both the electron microscope and the STM are among the point, the source of the tunneling electrons. A voltage is most powerful tools in chemical and biological research. To describe the problem of trying to locate a subatomic particle that behaves like a wave, Werner Heisenberg† formulated what is now known as the Heisenberg uncertainty principle: it is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certainty. Stated mathematically, The $ sign means that the product DxDp can be greater than or equal to h/4π, but h it can never be smaller than h/4π. Also, ¢x¢p $ (7.9) in using Equation (7.9), m must be in 4π kilograms and u must be in m/s. where Dx and Dp are the uncertainties in measuring the position and momentum of the particle, respectively. The $ signs have the following meaning. If the measured uncer- tainties of position and momentum are large (say, in a crude experiment), their product can be substantially greater than hy4π (hence the . sign). The significance of Equation 7.9 is that even in the most favorable conditions for measuring position and momentum, † Werner Karl Heisenberg (1901–1976). German physicist. One of the founders of modern quantum theory, Heisenberg received the Nobel Prize in Physics in 1932. 292  7.5 Quantum Mechanics 293 the product of the uncertainties can never be less than hy4π (hence the 5 sign). Thus, making measurement of the momentum of a particle more precise (that is, making Dp a small quantity) means that the position must become correspondingly less precise (that is, Dx will become larger). Similarly, if the position of the particle is known more precisely, its momentum measurement must become less precise. Applying the Heisenberg uncertainty principle to the hydrogen atom, we see that in reality the electron does not orbit the nucleus in a well-defined path, as Bohr thought. If it did, we could determine precisely both the position of the electron (from its location on a particular orbit) and its momentum (from its kinetic energy) at the same time, a violation of the uncertainty principle. Example 7.6 (a) An electron is moving at a speed of 8.0 3 106 m/s. If the uncertainty in measuring the speed is 1.0 percent of the speed, calculate the uncertainty in the electron’s position. The mass of the electron is 9.1094 3 10231 kg. (b) A baseball of mass 0.15 kg thrown at 100 mph has a momentum of 6.7 kg ? m/s. If the uncertainty in measuring this momentum is 1.0 3 1027 of the momentum, calculate the uncertainty in the baseball’s position. Strategy To calculate the minimum uncertainty in both (a) and (b), we use an equal sign in Equation (7.9). Solution (a) The uncertainty in the electron’s speed u is ¢u 5 0.010 3 8.0 3 106 m/s 5 8.0 3 104 m/s Momentum ( p) is p 5 mu, so that ¢p 5 m¢u 5 9.1094 3 10231 kg 3 8.0 3 104 m/s 5 7.3 3 10226 kg ? m/s From Equation (7.9), the uncertainty in the electron’s position is h ¢x 5 4π¢p 6.63 3 10234 J ? s 5 4π(7.3 3 10226 kg ? m/s) 5 7.2 3 10210 m This uncertainty corresponds to about 4 atomic diameters. (b) The uncertainty in the position of the baseball is h ¢x 5 4π¢p 6.63 3 10234 J ? s 5 4π 3 1.0 3 1027 3 6.7 kg ? m/s 5 7.9 3 10229 m This is such a small number as to be of no consequence; that is, there is practically no uncertainty in determining the position of the baseball in the macroscopic world. Similar problems: 7.128, 7.146. Practice Exercise Estimate the uncertainty in the speed of an oxygen molecule if its position is known to be 63 nm. The mass of an oxygen molecule is 5.31 3 10226 kg. 294 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms To be sure, Bohr made a significant contribution to our understanding of atoms, and his suggestion that the energy of an electron in an atom is quantized remains unchallenged. But his theory did not provide a complete description of electronic behavior in atoms. In 1926 the Austrian physicist Erwin Schrödinger,† using a com- plicated mathematical technique, formulated an equation that describes the behavior and energies of submicroscopic particles in general, an equation analogous to Newton’s laws of motion for macroscopic objects. The Schrödinger equation requires advanced calculus to solve, and we will not discuss it here. It is important to know, however, that the equation incorporates both particle behavior, in terms of mass m, and wave behavior, in terms of a wave function Ψ (psi), which depends on the location in space of the system (such as an electron in an atom). The wave function itself has no direct physical meaning. However, the probabil- ity of finding the electron in a certain region in space is proportional to the square of the wave function, Ψ2. The idea of relating Ψ2 to probability stemmed from a wave theory analogy. According to wave theory, the intensity of light is proportional to the square of the amplitude of the wave, or Ψ2. The most likely place to find a photon is where the intensity is greatest, that is, where the value of Ψ2 is greatest. A similar argument associates Ψ2 with the likelihood of finding an electron in regions surround- ing the nucleus. Schrödinger’s equation began a new era in physics and chemistry, for it launched a new field, quantum mechanics (also called wave mechanics). We now refer to the developments in quantum theory from 1913—the time Bohr presented his analysis for the hydrogen atom—to 1926 as “old quantum theory.” The Quantum Mechanical Description of the Hydrogen Atom The Schrödinger equation specifies the possible energy states the electron can occupy in a hydrogen atom and identifies the corresponding wave functions (Ψ). These energy states and wave functions are characterized by a set of quantum numbers (to be discussed shortly), with which we can construct a comprehensive model of the hydrogen atom. Although quantum mechanics tells us that we cannot pinpoint an electron in an atom, it does define the region where the electron might be at a given time. The concept of electron density gives the probability that an electron will be found in a particular region of an atom. The square of the wave function, Ψ2, defines the distribution of electron density in three-dimensional space around the nucleus. Regions of high electron density represent a high probability of locating the electron, whereas the opposite holds for regions of low electron density (Figure 7.15). To distinguish the quantum mechanical description of an atom from Bohr’s model, we speak of an atomic orbital, rather than an orbit. An atomic orbital can Figure 7.15 A representation of the electron density distribution be thought of as the wave function of an electron in an atom. When we say that surrounding the nucleus in the an electron is in a certain orbital, we mean that the distribution of the electron hydrogen atom. It shows a high density or the probability of locating the electron in space is described by the probability of finding the electron closer to the nucleus. square of the wave function associated with that orbital. An atomic orbital, there- fore, has a characteristic energy, as well as a characteristic distribution of elec- tron density. The Schrödinger equation works nicely for the simple hydrogen atom with its one proton and one electron, but it turns out that it cannot be solved exactly for any atom containing more than one electron! Fortunately, chemists and physicists have † Erwin Schrödinger (1887–1961). Austrian physicist. Schrödinger formulated wave mechanics, which laid the foundation for modern quantum theory. He received the Nobel Prize in Physics in 1933.  7.6 Quantum Numbers 295 learned to get around this kind of difficulty by approximation. For example, although the behavior of electrons in many-electron atoms (that is, atoms containing two or Although the helium atom has only two electrons, in quantum mechanics it is more electrons) is not the same as in the hydrogen atom, we assume that the differ- regarded as a many-electron atom. ence is probably not too great. Thus, we can use the energies and wave functions obtained from the hydrogen atom as good approximations of the behavior of electrons in more complex atoms. In fact, this approach provides fairly reliable descriptions of electronic behavior in many-electron atoms. Review of Concepts What is the difference between Ψ and Ψ2 for the electron in a hydrogen atom? 7.6 Quantum Numbers In quantum mechanics, three quantum numbers are required to describe the distribu- tion of electrons in hydrogen and other atoms. These numbers are derived from the mathematical solution of the Schrödinger equation for the hydrogen atom. They are called the principal quantum number, the angular momentum quantum number, and the magnetic quantum number. These quantum numbers will be used to describe atomic orbitals and to label electrons that reside in them. A fourth quantum number— the spin quantum number—describes the behavior of a specific electron and completes the description of electrons in atoms. The Principal Quantum Number (n) The principal quantum number (n) can have integral values 1, 2, 3, and so forth; it Equation (7.5) holds only for the hydrogen atom. corresponds to the quantum number in Equation (7.5). In a hydrogen atom, the value of n determines the energy of an orbital. As we will see shortly, this is not the case for a many-electron atom. The principal quantum number also relates to the average distance of the electron from the nucleus in a particular orbital. The larger n is, the greater the average distance of an electron in the orbital from the nucleus and there- fore the larger the orbital. The Angular Momentum Quantum Number (<) The angular momentum quantum number (/) tells us the “shape” of the orbitals The value of / is fixed based on the type of the orbital. (see Section 7.7). The values of / depend on the value of the principal quantum number, n. For a given value of n, / has possible integral values from 0 to (n 2 1). If n 5 1, there is only one possible value of /; that is, / 5 n 2 1 5 1 2 1 5 0. If n 5 2, there are two values of /, given by 0 and 1. If n 5 3, there are three values of /, given by 0, 1, and 2. The value of / is generally designated by the letters s, p, d, . . . as follows: / 0 1 2 3 4 5 Name of orbital s p d f g h Thus, if / 5 0, we have an s orbital; if / 5 1, we have a p orbital; and so on. The unusual sequence of letters (s, p, and d) has a historical origin. Physicists who studied atomic emission spectra tried to correlate the observed spectral lines with the particular energy states involved in the transitions. They noted that some of the lines were sharp; some were rather spread out, or diffuse; and some were 296 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms very strong and hence referred to as principal lines. Subsequently, the initial letters of each adjective were assigned to those energy states. However, after the letter d and starting with the letter f (for fundamental), the orbital designations follow alpha- betical order. A collection of orbitals with the same value of n is frequently called a shell. One or more orbitals with the same n and / values are referred to as a subshell. For Remember that the “2” in 2s refers to the example, the shell with n 5 2 is composed of two subshells, / 5 0 and 1 (the allowed value of n and the “s” symbolizes the value of /. values for n 5 2). These subshells are called the 2s and 2p subshells where 2 denotes the value of n, and s and p denote the values of /. The Magnetic Quantum Number (m<) The magnetic quantum number (m/ ) describes the orientation of the orbital in space (to be discussed in Section 7.7). Within a subshell, the value of m/ depends on the value of the angular momentum quantum number, /. For a certain value of /, there are (2/ 1 1) integral values of m/ as follows: 2/, (2/ 1 1), . . . 0, . . . (1/ 2 1), 1/ If / 5 0, then m/ 5 0. If / 5 1, then there are [(2 3 1) 1 1], or three values of m/ , namely, 21, 0, and 1. If / 5 2, there are [(2 3 2) 1 1], or five values of m/ , namely, 22, 21, 0, 1, and 2. The number of m/ values indicates the number of orbitals in a subshell with a particular / value. N S To conclude our discussion of these three quantum numbers, let us consider a situation in which n 5 2 and / 5 1. The values of n and / indicate that we have a 2p subshell, and in this subshell we have three 2p orbitals (because there are three S N values of m/ , given by 21, 0, and 1). The Electron Spin Quantum Number (ms) (a) (b) Experiments on the emission spectra of hydrogen and sodium atoms indicated that Figure 7.16 The (a) clockwise lines in the emission spectra could be split by the application of an external mag- and (b) counterclockwise spins of an electron. The magnetic fields netic field. The only way physicists could explain these results was to assume that generated by these two spinning electrons act like tiny magnets. If electrons are thought of as spinning on their own motions are analogous to those from the two magnets. The upward axes, as Earth does, their magnetic properties can be accounted for. According to and downward arrows are used to electromagnetic theory, a spinning charge generates a magnetic field, and it is this denote the direction of spin. motion that causes an electron to behave like a magnet. Figure 7.16 shows the two possible spinning motions of an electron, one clockwise and the other counterclock- wise. To take the electron spin into account, it is necessary to introduce a fourth quantum number, called the electron spin quantum number (ms), which has a value of 112 or 212. In their experiment, Stern and Gerlach Conclusive proof of electron spin was provided by Otto Stern† and Walther used silver atoms, which contain just one unpaired electron. To illustrate the Gerlach‡ in 1924. Figure 7.17 shows the basic experimental arrangement. A beam principle, we can assume that hydrogen of gaseous atoms generated in a hot furnace passes through a nonhomogeneous atoms are used in the study. magnetic field. The interaction between an electron and the magnetic field causes the atom to be deflected from its straight-line path. Because the spinning motion is completely random, the electrons in half of the atoms will be spinning in one direc- tion, and those atoms will be deflected in one way; the electrons in the other half † Otto Stern (1888–1969). German physicist. He made important contributions to the study of magnetic properties of atoms and the kinetic theory of gases. Stern was awarded the Nobel Prize in Physics in 1943. ‡ Walther Gerlach (1889–1979). German physicist. Gerlach’s main area of research was in quantum theory.  7.7 Atomic Orbitals 297 ms 5 22 1 2 Oven Atom beam ms 5 12 1 2 Detecting screen Magnet Slit screen Figure 7.17 Experimental arrangement for demonstrating the spinning motion of electrons. A beam of atoms is directed through a magnetic field. For example, when a hydrogen atom with a single electron passes through the field, it is deflected in one direction or the other, depending on the direction of the spin. In a stream consisting of many atoms, there will be equal distributions of the two kinds of spins, so that two spots of equal intensity are detected on the screen. of the atoms will be spinning in the opposite direction, and those atoms will be deflected in the other direction. Thus, two spots of equal intensity are observed on the detecting screen. Review of Concepts Give the four quantum numbers for each of the two electrons in a 6s orbital. 7.7 Atomic Orbitals Table 7.2 shows the relation between quantum numbers and atomic orbitals. We see that when / 5 0, 12/ 1 12 5 1 and there is only one value of m/ , thus we have an s orbital. When / 5 1, 12/ 1 12 5 3, so there are three values of m/ or three p orbitals, labeled px, py, and pz. When / 5 2, 12/ 1 12 5 5 and there are five values of m/ , and the corresponding five d orbitals are labeled with more elaborate subscripts. In the following sections we will consider the s, p, and d orbitals separately. Table 7.2 Relation Between Quantum Numbers and Atomic Orbitals Number Atomic n < m< of Orbitals Orbital Designations 1 0 0 1 1s An s subshell has one orbital, a p subshell has three orbitals, and a d subshell has 2 0 0 1 2s five orbitals. 1 21, 0, 1 3 2px, 2py, 2pz 3 0 0 1 3s 1 21, 0, 1 3 3px, 3py, 3pz 2 22, 21, 0, 1, 2 5 3dxy, 3dyz, 3dxz, 3dx2 2 y2, 3dz2 . . . . . . . . . . . . . . . 298 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms (a) (b) probability Radial Distance from nucleus (c) Figure 7.18 (a) Plot of electron density in the hydrogen 1s orbital as a function of the distance from the nucleus. The electron density falls off rapidly as the distance from the nucleus increases. (b) Boundary surface diagram of the hydrogen 1s orbital. (c) A more realistic way of viewing electron density distribution is to divide the 1s orbital into successive spherical thin shells. A plot of the probability of finding the electron in each shell, called radial probability, as a function of distance shows a maximum at 52.9 pm from the nucleus. Interestingly, this is equal to the radius of the innermost orbit in the Bohr model. That the wave function for an orbital s Orbitals. One of the important questions we ask when studying the properties of theoretically has no outer limit as one moves outward from the nucleus raises atomic orbitals is, What are the shapes of the orbitals? Strictly speaking, an orbital interesting philosophical questions does not have a well-defined shape because the wave function characterizing the regarding the sizes of atoms. Chemists have agreed on an operational definition orbital extends from the nucleus to infinity. In that sense, it is difficult to say what of atomic size, as we will see in later an orbital looks like. On the other hand, it is certainly convenient to think of orbitals chapters. as having specific shapes, particularly in discussing the formation of chemical bonds between atoms, as we will do in Chapters 9 and 10. Although in principle an electron can be found anywhere, we know that most of the time it is quite close to the nucleus. Figure 7.18(a) shows the distribution of electron density in a hydrogen 1s orbital moving outward from the nucleus. As you can see, the electron density falls off rapidly as the distance from the nucleus increases. Roughly speaking, there is about a 90 percent probability of finding the electron within a sphere of radius 100 pm (1 pm 5 1 3 10212 m) surrounding the nucleus. Thus, we can represent the 1s orbital by drawing a boundary surface diagram that encloses about 90 percent of the total electron density in an orbital, as shown in Figure 7.18(b). A 1s orbital represented in this manner is merely a sphere. Figure 7.19 shows boundary surface diagrams for the 1s, 2s, and 3s hydrogen 1s 2s atomic orbitals. All s orbitals are spherical in shape but differ in size, which increases 3s as the principal quantum number increases. Although the details of electron density variation within each boundary surface are lost, there is no serious disadvantage. For Figure 7.19 Boundary surface diagrams of the hydrogen 1s, 2s, us the most important features of atomic orbitals are their shapes and relative sizes, and 3s orbitals. Each sphere which are adequately represented by boundary surface diagrams. contains about 90 percent of the total electron density. All s orbitals p Orbitals. It should be clear that the p orbitals start with the principal quantum number are spherical. Roughly speaking, n 5 2. If n 5 1, then the angular momentum quantum number / can assume only the the size of an orbital is proportional to n2, where n is the value of zero; therefore, there is only a 1s orbital. As we saw earlier, when / 5 1, the principal quantum number. magnetic quantum number m/ can have values of 21, 0, 1. Starting with n 5 2 and  7.7 Atomic Orbitals 299 z z z Figure 7.20 The boundary surface diagrams of the three 2p orbitals. These orbitals are identical in shape and energy, but their orientations are different. The p orbitals of higher principal quantum x y x y x y numbers have similar shapes. 2px 2py 2pz / 5 1, we therefore have three 2p orbitals: 2px, 2py, and 2pz (Figure 7.20). The letter subscripts indicate the axes along which the orbitals are oriented. These three p orbitals Orbitals that have the same energy are said to be degenerate orbitals. are identical in size, shape, and energy; they differ from one another only in orientation. Note, however, that there is no simple relation between the values of m/ and the x, y, and z directions. For our purpose, you need only remember that because there are three possible values of m/ , there are three p orbitals with different orientations. The boundary surface diagrams of p orbitals in Figure 7.20 show that each p orbital can be thought of as two lobes on opposite sides of the nucleus. Like s orbit- als, p orbitals increase in size from 2p to 3p to 4p orbital and so on. d Orbitals and Other Higher-Energy Orbitals. When / 5 2, there are five values of m/ , which correspond to five d orbitals. The lowest value of n for a d orbital is 3. Because / can never be greater than n 2 1, when n 5 3 and / 5 2, we have five 3d orbitals (3dxy, 3dyz, 3dxz, 3dx2 2y2 , and 3dz2 ), shown in Figure 7.21. As in the case of the p orbitals, the different orientations of the d orbitals cor- respond to the different values of m/ , but again there is no direct correspondence between a given orientation and a particular m/ value. All the 3d orbitals in an atom are identical in energy. The d orbitals for which n is greater than 3 (4d, 5d, . . .) have similar shapes. Orbitals having higher energy than d orbitals are labeled f, g, . . . and so on. The f orbitals are important in accounting for the behavior of elements with atomic numbers greater than 57, but their shapes are difficult to represent. In general chemistry, we are not concerned with orbitals having / values greater than 3 (the g orbitals and beyond). Examples 7.7 and 7.8 illustrate the labeling of orbitals with quantum numbers and the calculation of total number of orbitals associated with a given principal quan- tum number. Example 7.7 List the values of n, /, and m/ for orbitals in the 4d subshell. Strategy What are the relationships among n, /, and m/ ? What do “4” and “d” represent in 4d? Solution As we saw earlier, the number given in the designation of the subshell is the principal quantum number, so in this case n 5 4. The letter designates the type of orbital. Because we are dealing with d orbitals, / 5 2. The values of m/ can vary from 2/ to /. Therefore, m/ can be 22, 21, 0, 1, or 2. Check The values of n and / are fixed for 4d, but m/ can have any one of the five values, which correspond to the five d orbitals. Similar problem: 7.57. Practice Exercise Give the values of the quantum numbers associated with the orbitals in the 3p subshell. 300 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms z z z z z x y x y x y x y x y 3dx2 – y2 3dz2 3dxy 3dxz 3dyz Figure 7.21 Boundary surface diagrams of the five 3d orbitals. Although the 3dz2 orbital looks different, it is equivalent to the other four orbitals in all other respects. The d orbitals of higher principal quantum numbers have similar shapes. Example 7.8 What is the total number of orbitals associated with the principal quantum number n 5 3? Strategy To calculate the total number of orbitals for a given n value, we need to first write the possible values of /. We then determine how many m/ values are associated with each value of /. The total number of orbitals is equal to the sum of all the m/ values. Solution For n 5 3, the possible values of / are 0, 1, and 2. Thus, there is one 3s orbital (n 5 3, / 5 0, and m/ 5 0); there are three 3p orbitals (n 5 3, / 5 1, and m/ 5 21, 0, 1); there are five 3d orbitals (n 5 3, / 5 2, and m/ 5 22, 21, 0, 1, 2). The total number of orbitals is 1 1 3 1 5 5 9. Check The total number of orbitals for a given value of n is n2. So here we have 32 5 9. Similar problem: 7.62. Can you prove the validity of this relationship? Practice Exercise What is the total number of orbitals associated with the principal quantum number n 5 4? Review of Concepts Why is it not possible to have a 2d orbital but a 3d orbital is allowed? The Energies of Orbitals Now that we have some understanding of the shapes and sizes of atomic orbitals, we are ready to inquire into their relative energies and look at how energy levels affect the actual arrangement of electrons in atoms. According to Equation (7.5), the energy of an electron in a hydrogen atom is determined solely by its principal quantum number. Thus, the energies of hydrogen orbitals increase as follows (Figure 7.22): 1s , 2s 5 2p , 3s 5 3p 5 3d , 4s 5 4p 5 4d 5 4f , . . . Although the electron density distributions are different in the 2s and 2p orbitals, hydrogen’s electron has the same energy whether it is in the 2s orbital or a 2p orbital. The 1s orbital in a hydrogen atom corresponds to the most stable condition, the ground state. An electron residing in this orbital is most strongly held by the nucleus because it is closest to the nucleus. An electron in the 2s, 2p, or higher orbitals in a hydrogen atom is in an excited state. The energy picture is more complex for many-electron atoms than for hydrogen. The energy of an electron in such an atom depends on its angular momentum quantum  7.8 Electron Configuration 301 Figure 7.22 Orbital energy levels in the hydrogen atom. Each 4s 4p 4d 4f short horizontal line represents one orbital. Orbitals with the 3s 3p 3d same principal quantum number (n) all have the same energy. Energy 2s 2p 1s 4d 5s 4p 3d 4s 3p 3s Energy 2p 2s 1s Figure 7.23 Orbital energy levels in a many-electron atom. Note that the energy level depends on both n and / values. number as well as on its principal quantum number (Figure 7.23). For many-electron atoms, the 3d energy level is very close to the 4s energy level. The total energy of an atom, however, depends not only on the sum of the orbital energies but also on the energy of repulsion between the electrons in these orbitals (each orbital can 1s accommodate up to two electrons, as we will see in Section 7.8). It turns out that the 2s 2p total energy of an atom is lower when the 4s subshell is filled before a 3d subshell. 3s 3p 3d Figure 7.24 depicts the order in which atomic orbitals are filled in a many-electron atom. We will consider specific examples in Section 7.8. 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7.8 Electron Configuration 7s 7p The four quantum numbers n, /, m/ , and ms enable us to label completely an electron in any orbital in any atom. In a sense, we can regard the set of four quantum numbers Figure 7.24 The order in which atomic subshells are filled in a as the “address” of an electron in an atom, somewhat in the same way that a street many-electron atom. Start with the address, city, state, and postal ZIP code specify the address of an individual. For exam- 1s orbital and move downward, ple, the four quantum numbers for a 2s orbital electron are n 5 2, / 5 0, m/ 5 0, and following the direction of the arrows. Thus, the order goes as ms 5 112 or 212. It is inconvenient to write out all the individual quantum numbers, and follows: 1s , 2s , 2p , 3s , so we use the simplified notation (n, /, m/ , ms). For the preceding example, the quantum 3p , 4s , 3d , . . . . 302 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms numbers are either (2, 0, 0, 112 ) or (2, 0, 0, 212 ). The value of ms has no effect on the energy, size, shape, or orientation of an orbital, but it determines how electrons are arranged in an orbital. Example 7.9 shows how quantum numbers of an electron in an orbital are assigned. Example 7.9 Write the four quantum numbers for an electron in a 3p orbital. Strategy What do the “3” and “p” designate in 3p? How many orbitals (values of m/ ) are there in a 3p subshell? What are the possible values of electron spin quantum number? Solution To start with, we know that the principal quantum number n is 3 and the angular momentum quantum number / must be 1 (because we are dealing with a p orbital). For / 5 1, there are three values of m/ given by 21, 0, and 1. Because the electron spin quantum number ms can be either 112 or 212, we conclude that there are six possible ways to designate the electron using the (n, /, m/ , ms) notation: (3, 1, 21, 112 ) (3, 1, 21, 212 ) (3, 1, 0, 112 ) (3, 1, 0, 212 ) (3, 1, 1, 112 ) (3, 1, 1, 212 ) Check In these six designations we see that the values of n and / are constant, but the Similar problem: 7.58. values of m/ and ms can vary. Practice Exercise Write the four quantum numbers for an electron in a 4d orbital. Animation The hydrogen atom is a particularly simple system because it contains only one Electron Configurations electron. The electron may reside in the 1s orbital (the ground state), or it may be found in some higher-energy orbital (an excited state). For many-electron atoms, how- ever, we must know the electron configuration of the atom, that is, how the electrons are distributed among the various atomic orbitals, in order to understand electronic 1A 8A H 2A 3A 4A 5A 6A 7A He behavior. We will use the first 10 elements (hydrogen to neon) to illustrate the rules Li Be B C N O F Ne for writing electron configurations for atoms in the ground state. (Section 7.9 will describe how these rules can be applied to the remainder of the elements in the peri- odic table.) For this discussion, recall that the number of electrons in an atom is equal to its atomic number Z. Figure 7.22 indicates that the electron in a ground-state hydrogen atom must be in the 1s orbital, so its electron configuration is 1s1: denotes the number of electrons 8 in the orbital or subshell 8 1s1m m 8n 88 denotes the principal 8 denotes the angular momentum quantum number n quantum number , The electron configuration can also be represented by an orbital diagram that shows the spin of the electron (see Figure 7.16): H h 1s1  7.8 Electron Configuration 303 The upward arrow denotes one of the two possible spinning motions of the electron. Remember that the direction of electron spin has no effect on the energy of the (Alternatively, we could have represented the electron with a downward arrow.) The electron. box represents an atomic orbital. The Pauli Exclusion Principle For many-electron atoms we use the Pauli† exclusion principle to determine electron configurations. This principle states that no two electrons in an atom can have the same set of four quantum numbers. If two electrons in an atom should have the same n, /, and m/ values (that is, these two electrons are in the same atomic orbital), then they must have different values of ms. In other words, only two electrons may occupy the same atomic orbital, and these electrons must have opposite spins. Consider the helium atom, which has two electrons. The three possible ways of placing two elec- trons in the 1s orbital are as follows: He hh gg hg 2 2 1s 1s 1s2 (a) (b) (c) Diagrams (a) and (b) are ruled out by the Pauli exclusion principle. In (a), both elec- trons have the same upward spin and would have the quantum numbers (1, 0, 0, 112 ); in (b), both electrons have downward spins and would have the quantum numbers (1, 0, 0, 212 ). Only the configuration in (c) is physically acceptable, because one elec- tron has the quantum numbers (1, 0, 0, 112 ) and the other has (1, 0, 0, 212 ). Thus, the helium atom has the following configuration: He hg Electrons that have opposite spins are said to be paired. In helium, ms 5 1 21 for one electron; ms 5 2 21 for the other. 1s2 Note that 1s2 is read “one s two,” not “one s squared.” Diamagnetism and Paramagnetism The Pauli exclusion principle is one of the fundamental principles of quantum mechan- ics. It can be tested by a simple observation. If the two electrons in the 1s orbital of a helium atom had the same, or parallel, spins (↑↑ or ↓↓), their net magnetic fields would reinforce each other [Figure 7.25(a)]. Such an arrangement would make the † Wolfgang Pauli (1900–1958). Austrian physicist. One of the founders of quantum mechanics, Pauli was awarded the Nobel Prize in Physics in 1945. Figure 7.25 The (a) parallel and (b) antiparallel spins of two electrons. In (a) the two magnetic fields reinforce each other. In (b) the two magnetic fields cancel N N N S each other. S S S N (a) (b) 304 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms helium gas paramagnetic. Paramagnetic substances are those that contain net unpaired spins and are attracted by a magnet. On the other hand, if the electron spins are paired, or antiparallel to each other (↑↓ or ↓↑), the magnetic effects cancel out [Figure 7.25(b)]. Diamagnetic substances do not contain net unpaired spins and are slightly repelled by a magnet. Measurements of magnetic properties provide the most direct evidence for specific electron configurations of elements. Advances in instrument design during the last Paramagnetic 30 years or so enable us to determine the number of unpaired electrons in an atom substance (Figure 7.26). By experiment we find that the helium atom in its ground state has no net magnetic field. Therefore, the two electrons in the 1s orbital must be paired in accord Electromagnet with the Pauli exclusion principle and the helium gas is diamagnetic. A useful rule to keep in mind is that any atom with an odd number of electrons will always contain one or more unpaired spins because we need an even number of electrons for complete Figure 7.26 Initially the pairing. On the other hand, atoms containing an even number of electrons may or may paramagnetic substance was weighed on a balance in the not contain unpaired spins. We will see the reason for this behavior shortly. absence of a magnetic field. As another example, consider the lithium atom (Z 5 3), which has three elec- When the electromagnet is turned trons. The third electron cannot go into the 1s orbital because it would inevitably on, the balance is offset because the sample tube is drawn into have the same set of four quantum numbers as one of the first two electrons. There- the magnetic field. Knowing the fore, this electron “enters” the next (energetically) higher orbital, which is the 2s concentration and the additional orbital (see Figure 7.23). The electron configuration of lithium is 1s22s1, and its mass needed to reestablish balance, it is possible to calculate orbital diagram is the number of unpaired electrons in the sample. hg h Li 1s2 2s1 The lithium atom contains one unpaired electron and the lithium metal is therefore paramagnetic. The Shielding Effect in Many-Electron Atoms Experimentally we find that the 2s orbital lies at a lower energy level than the 2p orbital in a many-electron atom. Why? In comparing the electron configurations of 1s22s1 and 1s22p1, we note that, in both cases, the 1s orbital is filled with two elec- trons. Figure 7.27 shows the radial probability plots for the 1s, 2s, and 2p orbitals. Because the 2s and 2p orbitals are larger than the 1s orbital, an electron in either of these orbitals will spend more time away from the nucleus than an electron in the 1s 1s orbital. Thus, we can speak of a 2s or 2p electron being partly “shielded” from the attractive force of the nucleus by the 1s electrons. The important consequence of the Radial probability shielding effect is that it reduces the electrostatic attraction between the protons in the nucleus and the electron in the 2s or 2p orbital. The manner in which the electron density varies as we move from the nucleus 2p outward depends on the type of orbital. Although a 2s electron spends most of its 2s time (on average) slightly farther from the nucleus than a 2p electron, the electron density near the nucleus is actually greater for the 2s electron (see the small maximum for the 2s orbital in Figure 7.27). For this reason, the 2s orbital is said to be more “penetrating” than the 2p orbital. Therefore, a 2s electron is less shielded by the 1s electrons and is more strongly held by the nucleus. In fact, for the same principal Distance from nucleus quantum number n, the penetrating power decreases as the angular momentum quan- Figure 7.27 Radial probability tum number / increases, or plots (see Figure 7.18) for the 1s, 2s, and 2p orbitals. The 1s electrons s . p . d . f . ### effectively shield both the 2s and 2p electrons from the nucleus. The 2s orbital is more penetrating than Because the stability of an electron is determined by the strength of its attraction to the 2p orbital. the nucleus, it follows that a 2s electron will be lower in energy than a 2p electron.  7.8 Electron Configuration 305 To put it another way, less energy is required to remove a 2p electron than a 2s elec- tron because a 2p electron is not held quite as strongly by the nucleus. The hydrogen atom has only one electron and, therefore, is without such a shielding effect. Continuing our discussion of atoms of the first 10 elements, we go next to beryl- lium (Z 5 4). The ground-state electron configuration of beryllium is 1s22s2, or Be hg hg 2 1s 2s2 Beryllium is diamagnetic, as we would expect. The electron configuration of boron (Z 5 5) is 1s22s22p1, or B hg hg h 2 2 1s 2s 2p1 Note that the unpaired electron can be in the 2px, 2py, or 2pz orbital. The choice is completely arbitrary because the three p orbitals are equivalent in energy. As the diagram shows, boron is paramagnetic. Hund’s Rule The electron configuration of carbon (Z 5 6) is 1s22s22p2. The following are different ways of distributing two electrons among three p orbitals: hg h g h h 2px 2py 2pz 2px 2py 2pz 2px 2py 2pz (a) (b) (c) None of the three arrangements violates the Pauli exclusion principle, so we must determine which one will give the greatest stability. The answer is provided by Hund’s rule,† which states that the most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. The arrangement shown in (c) satisfies this condition. In both (a) and (b) the two spins cancel each other. Thus, the orbital diagram for carbon is C hg hg h h 1s2 2s2 2p2 Qualitatively, we can understand why (c) is preferred to (a). In (a), the two elec- trons are in the same 2px orbital, and their proximity results in a greater mutual repulsion than when they occupy two separate orbitals, say 2px and 2py. The choice of (c) over (b) is more subtle but can be justified on theoretical grounds. The fact that carbon atoms contain two unpaired electrons is in accord with Hund’s rule. The electron configuration of nitrogen (Z 5 7) is 1s22s22p3: N hg hg h h h 1s2 2s2 2p3 Again, Hund’s rule dictates that all three 2p electrons have spins parallel to one another; the nitrogen atom contains three unpaired electrons. † Frederick Hund (1896–1997). German physicist. Hund’s work was mainly in quantum mechanics. He also helped to develop the molecular orbital theory of chemical bonding. 306 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms The electron configuration of oxygen (Z 5 8) is 1s22s22p4. An oxygen atom has two unpaired electrons: O hg hg hg h h 1s2 2s2 2p4 The electron configuration of fluorine (Z 5 9) is 1s22s22p5. The nine electrons are arranged as follows: F hg hg hg hg h 1s2 2s2 2p5 The fluorine atom has one unpaired electron. In neon (Z 5 10), the 2p subshell is completely filled. The electron configuration of neon is 1s22s22p6, and all the electrons are paired, as follows: Ne hg hg hg hg hg 1s2 2s2 2p6 The neon gas should be diamagnetic, and experimental observation bears out this prediction. General Rules for Assigning Electrons to Atomic Orbitals Based on the preceding examples we can formulate some general rules for determin- ing the maximum number of electrons that can be assigned to the various subshells and orbitals for a given value of n: 1. Each shell or principal level of quantum number n contains n subshells. For example, if n 5 2, then there are two subshells (two values of /) of angular momentum quantum numbers 0 and 1. 2. Each subshell of quantum number / contains (2/ 1 1) orbitals. For example, if / 5 1, then there are three p orbitals. 3. No more than two electrons can be placed in each orbital. Therefore, the maxi- mum number of electrons is simply twice the number of orbitals that are employed. 4. A quick way to determine the maximum number of electrons that an atom can have in a principal level n is to use the formula 2n2. Examples 7.10 and 7.11 illustrate the procedure for calculating the number of electrons in orbitals and labeling electrons with the four quantum numbers. Example 7.10 What is the maximum number of electrons that can be present in the principal level for which n 5 3? Strategy We are given the principal quantum number (n) so we can determine all the possible values of the angular momentum quantum number (/). The preceding rule shows that the number of orbitals for each value of / is (2/ 1 1). Thus, we can determine the total number of orbitals. How many electrons can each orbital accommodate? (Continued)  7.8 Electron Configuration 307 Solution When n 5 3, / 5 0, 1, and 2. The number of orbitals for each value of / is given by Number of Orbitals Value of < (2< 1 1) 0 1 1 3 2 5 The total number of orbitals is nine. Because each orbital can accommodate two electrons, the maximum number of electrons that can reside in the orbitals is 2 3 9, or 18. Check If we use the formula (n2) in Example 7.8, we find that the total number of orbitals is 32 and the total number of electrons is 2(32) or 18. In general, the number of electrons in a given principal energy level n is 2n2. Similar problems: 7.64, 7.65. Practice Exercise Calculate the total number of electrons that can be present in the principal level for which n 5 4. Example 7.11 An oxygen atom has a total of eight electrons. Write the four quantum numbers for each of the eight electrons in the ground state. Strategy We start with n 5 1 and proceed to fill orbitals in the order shown in Figure 7.24. For each value of n we determine the possible values of /. For each value of /, we assign the possible values of m/ . We can place electrons in the orbitals according to the Pauli exclusion principle and Hund’s rule. Solution We start with n 5 1, so / 5 0, a subshell corresponding to the 1s orbital. This orbital can accommodate a total of two electrons. Next, n 5 2, and / may be either 0 or 1. The / 5 0 subshell contains one 2s orbital, which can accommodate two electrons. The remaining four electrons are placed in the / 5 1 subshell, which contains three 2p orbitals. The orbital diagram is O hg hg hg h h 2 2 1s 2s 2p4 The results are summarized in the following table: Electron n < m< ms Orbital 1 1 0 0 112 1s 2 1 0 0 212 3 2 0 0 112 2s 4 2 0 0 212 5 2 1 21 112 6 2 1 0 112 2px, 2py, 2pz 7 2 1 1 112 8 2 1 1 212 Of course, the placement of the eighth electron in the orbital labeled m/ 5 1 is completely arbitrary. It would be equally correct to assign it to m/ 5 0 or m/ 5 21. Similar problem: 7.91. Practice Exercise Write a complete set of quantum numbers for each of the electrons in boron (B). 308 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms At this point let’s summarize what our examination of the first 10 elements has revealed about ground-state electron configurations and the properties of electrons in atoms: 1. No two electrons in the same atom can have the same four quantum numbers. This is the Pauli exclusion principle. 2. Each orbital can be occupied by a maximum of two electrons. They must have opposite spins, or different electron spin quantum numbers. 3. The most stable arrangement of electrons in a subshell is the one that has the greatest number of parallel spins. This is Hund’s rule. 4. Atoms in which one or more electrons are unpaired are paramagnetic. Atoms in which all the electron spins are paired are diamagnetic. 5. In a hydrogen atom, the energy of the electron depends only on its principal quantum number n. In a many-electron atom, the energy of an electron depends on both n and its angular momentum quantum number /. 6. In a many-electron atom the subshells are filled in the order shown in Figure 7.24. 7. For electrons of the same principal quantum number, their penetrating power, or proximity to the nucleus, decreases in the order s . p . d . f. This means that, for example, more energy is required to separate a 3s electron from a many- electron atom than is required to remove a 3p electron. Review of Concepts The ground-state electron configuration of an atom is 1s22s22p63s23p3. Which of the four quantum numbers would be the same for the three 3p electrons? 7.9 The Building-Up Principle Here we will extend the rules used in writing electron configurations for the first 10 elements to the rest of the elements. This process is based on the Aufbau principle. The German word “Aufbau” means The Aufbau principle dictates that as protons are added one by one to the nucleus “building up.” to build up the elements, electrons are similarly added to the atomic orbitals. Through this process we gain a detailed knowledge of the ground-state electron configurations of the elements. As we will see later, knowledge of electron configurations helps us to understand and predict the properties of the elements; it also explains why the periodic table works so well. 1A 8A 2A 3A 4A 5A 6A 7A He Table 7.3 gives the ground-state electron configurations of elements from H Ne (Z 5 1) through the named elements up to Lv (Z 5 116). The electron configurations Ar Kr of all elements except hydrogen and helium are represented by a noble gas core, Xe Rn which shows in brackets the noble gas element that most nearly precedes the element being considered, followed by the symbol for the highest filled subshells in the out- The noble gases. ermost shells. Notice that the electron configurations of the highest filled subshells in the outermost shells for the elements sodium (Z 5 11) through argon (Z 5 18) follow a pattern similar to those of lithium (Z 5 3) through neon (Z 5 10). As mentioned in Section 7.7, the 4s subshell is filled before the 3d subshell in a many-electron atom (see Figure 7.24). Thus, the electron configuration of potas- sium (Z 5 19) is 1s22s22p63s23p64s1. Because 1s22s22p63s23p6 is the electron con- figuration of argon, we can simplify the electron configuration of potassium by writing [Ar]4s1, where [Ar] denotes the “argon core.” Similarly, we can write the electron configuration of calcium (Z 5 20) as [Ar]4s2. The placement of the outermost  7.9 The Building-Up Principle 309 Table 7.3 The Ground-State Electron Configurations of the Elements* Atomic Electron Atomic Electron Atomic Electron Number Symbol Configuration Number Symbol Configuration Number Symbol Configuration 1 H 1s1 39 Y [Kr]5s24d 1 77 Ir [Xe]6s24f 145d 7 2 He 1s2 40 Zr [Kr]5s24d 2 78 Pt [Xe]6s14f 145d 9 3 Li [He]2s1 41 Nb [Kr]5s14d 4 79 Au [Xe]6s14f 145d 10 4 Be [He]2s2 42 Mo [Kr]5s14d 5 80 Hg [Xe]6s24f 145d10 5 B [He]2s22p1 43 Tc [Kr]5s24d 5 81 Tl [Xe]6s24f 145d 106p1 6 C [He]2s22p2 44 Ru [Kr]5s14d 7 82 Pb [Xe]6s24f 145d 106p2 7 N [He]2s22p3 45 Rh [Kr]5s14d 8 83 Bi [Xe]6s24f 145d 106p3 8 O [He]2s22p4 46 Pd [Kr]4d 10 84 Po [Xe]6s24f 145d 106p4 9 F [He]2s22p5 47 Ag [Kr]5s14d 10 85 At [Xe]6s24f 145d 106p5 10 Ne [He]2s22p6 48 Cd [Kr]5s24d 10 86 Rn [Xe]6s24f 145d 106p6 11 Na [Ne]3s1 49 In [Kr]5s24d 105p1 87 Fr [Rn]7s1 12 Mg [Ne]3s2 50 Sn [Kr]5s24d 105p2 88 Ra [Rn]7s2 13 Al [Ne]3s23p1 51 Sb [Kr]5s24d 105p3 89 Ac [Rn]7s26d 1 14 Si [Ne]3s23p2 52 Te [Kr]5s24d 105p4 90 Th [Rn]7s26d 2 15 P [Ne]3s23p3 53 I [Kr]5s24d105p5 91 Pa [Rn]7s25f 26d 1 16 S [Ne]3s23p4 54 Xe [Kr]5s24d 105p6 92 U [Rn]7s25f 36d 1 17 Cl [Ne]3s23p5 55 Cs [Xe]6s1 93 Np [Rn]7s25f 46d 1 18 Ar [Ne]3s23p6 56 Ba [Xe]6s2 94 Pu [Rn]7s25f 6 19 K [Ar]4s1 57 La [Xe]6s25d 1 95 Am [Rn]7s25f 7 20 Ca [Ar]4s2 58 Ce [Xe]6s24f 15d 1 96 Cm [Rn]7s25f 76d 1 21 Sc [Ar]4s23d 1 59 Pr [Xe]6s24f 3 97 Bk [Rn]7s25f 9 22 Ti [Ar]4s23d 2 60 Nd [Xe]6s24f 4 98 Cf [Rn]7s25f 10 23 V [Ar]4s23d 3 61 Pm [Xe]6s24f 5 99 Es [Rn]7s25f 11 24 Cr [Ar]4s13d 5 62 Sm [Xe]6s24f 6 100 Fm [Rn]7s25f 12 25 Mn [Ar]4s23d 5 63 Eu [Xe]6s24f 7 101 Md [Rn]7s25f 13 26 Fe [Ar]4s23d 6 64 Gd [Xe]6s24f 75d 1 102 No [Rn]7s25f 14 27 Co [Ar]4s23d 7 65 Tb [Xe]6s24f 9 103 Lr [Rn]7s25f 146d 1 28 Ni [Ar]4s23d 8 66 Dy [Xe]6s24f 10 104 Rf [Rn]7s25f 146d 2 29 Cu [Ar]4s13d 10 67 Ho [Xe]6s24f 11 105 Db [Rn]7s25f 146d 3 30 Zn [Ar]4s23d 10 68 Er [Xe]6s24f 12 106 Sg [Rn]7s25f 146d 4 31 Ga [Ar]4s23d 104p1 69 Tm [Xe]6s24f 13 107 Bh [Rn]7s25f 146d 5 32 Ge [Ar]4s23d 104p2 70 Yb [Xe]6s24f 14 108 Hs [Rn]7s25f 146d 6 33 As [Ar]4s23d 104p3 71 Lu [Xe]6s24f 145d 1 109 Mt [Rn]7s25f 146d 7 34 Se [Ar]4s23d 104p4 72 Hf [Xe]6s24f 145d 2 110 Ds [Rn]7s25f 146d 8 35 Br [Ar]4s23d 104p5 73 Ta [Xe]6s24f 145d 3 111 Rg [Rn]7s25f 146d 9 36 Kr [Ar]4s23d 104p6 74 W [Xe]6s24f 145d 4 112 Cn [Rn]7s25f 146d 10 37 Rb [Kr]5s1 75 Re [Xe]6s24f 145d 5 114 Fl [Rn]7s25f 146d107p2 38 Sr [Kr]5s2 76 Os [Xe]6s24f 145d 6 116 Lv [Rn]7s25f 146d107p4 *The symbol [He] is called the helium core and represents 1s2. [Ne] is called the neon core and represents 1s22s22p6. [Ar] is called the argon core and represents [Ne]3s23p6. [Kr] is called the krypton core and represents [Ar]4s23d 104p6. [Xe] is called the xenon core and represents [Kr]5s24d 105p6. [Rn] is called the radon core and represents [Xe]6s24f 145d 106p6. 310 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms electron in the 4s orbital (rather than in the 3d orbital) of potassium is strongly sup- ported by experimental evidence. The following comparison also suggests that this is the correct configuration. The chemistry of potassium is very similar to that of lithium and sodium, the first two alkali metals. The outermost electron of both lith- ium and sodium is in an s orbital (there is no ambiguity in assigning their electron configurations); therefore, we expect the last electron in potassium to occupy the 4s rather than the 3d orbital. The elements from scandium (Z 5 21) to copper (Z 5 29) are transition metals. 3B 4B 5B 6B 7B 8B 1B 2B Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells. Consider the first transition metal series, from scandium through copper. In this series additional electrons are placed in the 3d orbitals, according to Hund’s rule. However, there are two irregularities. The The transition metals. electron configuration of chromium (Z 5 24) is [Ar]4s13d 5 and not [Ar]4s23d 4, as we might expect. A similar break in the pattern is observed for copper, whose electron configuration is [Ar]4s13d10 rather than [Ar]4s23d 9. The reason for these irregularities is that a slightly greater stability is associated with the half-filled (3d 5) and completely filled (3d10) subshells. Electrons in the same subshell (in this case, the d orbitals) have equal energy but different spatial distributions. Consequently, their shielding of one another is relatively small, and the electrons are more strongly attracted by the nucleus when they have the 3d 5 configuration. According to Hund’s rule, the orbital diagram for Cr is Cr [Ar] h h h h h h 1 4s 3d 5 Thus, Cr has a total of six unpaired electrons. The orbital diagram for copper is Cu [Ar] h hg hg hg hg hg 1 4s 3d10 Again, extra stability is gained in this case by having the 3d subshell completely filled. In general, half-filled and completely filled subshells have extra stability. For elements Zn (Z 5 30) through Kr (Z 5 36), the 4s and 4p subshells fill in a straightforward manner. With rubidium (Z 5 37), electrons begin to enter the n 5 5 energy level. The electron configurations in the second transition metal series [yttrium (Z 5 39) to silver (Z 5 47)] are also irregular, but we will not be concerned with the details here. The sixth period of the periodic table begins with cesium (Z 5 55) and barium (Z 5 56), whose electron configurations are [Xe]6s1 and [Xe]6s2, respectively. Next we come to lanthanum (Z 5 57). From Figure 7.24 we would expect that after filling the 6s orbital we would place the additional electrons in 4f orbitals. In reality, the energies of the 5d and 4f orbitals are very close; in fact, for lanthanum 4f is slightly higher in energy than 5d. Thus, lanthanum’s electron configuration is [Xe]6s25d1 and not [Xe]6s24f 1. Following lanthanum are the 14 elements known as the lanthanides, or rare earth series [cerium (Z 5 58) to lutetium (Z 5 71)]. The rare earth metals have incompletely filled 4f subshells or readily give rise to cations that have incompletely filled 4f subshells. In this series, the added electrons are placed in 4f orbitals. After the 4f subshell is completely filled, the next electron enters the 5d subshell of lute- tium. Note that the electron configuration of gadolinium (Z 5 64) is [Xe]6s24f 75d 1 rather than [Xe]6s24f 8. Like chromium, gadolinium gains extra stability by having a half-filled subshell (4f 7).  7.9 The Building-Up Principle 311 Figure 7.28 Classification of 1s 1s groups of elements in the periodic table according to 2s 2p the type of subshell being filled with electrons. 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 7p 4f 5f The third transition metal series, including lanthanum and hafnium (Z 5 72) and extending through gold (Z 5 79), is characterized by the filling of the 5d subshell. With Hg (Z 5 80), both the 6s and 5d orbitals are now filled. The 6p subshell is filled next, which takes us to radon (Z 5 86). The last row of elements is the actinide series, which starts at thorium (Z 5 90). Most of these elements are not found in nature but have been synthesized. With few exceptions, you should be able to write the electron configuration of any element, using Figure 7.24 as a guide. Elements that require particular care are the transition metals, the lanthanides, and the actinides. As we noted earlier, at larger values of the principal quantum number n, the order of subshell filling may reverse from one element to the next. Figure 7.28 groups the elements according to the type of subshell in which the outermost electrons are placed. 1A 8A Example 7.12 2A 3A 4A 5A 6A 7A 3B 4B 5B 6B 7B 8B 1B 2B S Write the ground-state electron configurations for (a) sulfur (S) and (b) palladium (Pd), Pd which is diamagnetic. (a) Strategy How many electrons are in the S (Z 5 16) atom? We start with n 5 1 and proceed to fill orbitals in the order shown in Figure 7.24. For each value of /, we assign the possible values of m/. We can place electrons in the orbitals according to the Pauli exclusion principle and Hund’s rule and then write the electron configuration. The task is simplified if we use the noble gas core preceding S for the inner electrons. Solution Sulfur has 16 electrons. The noble gas core in this case is [Ne]. (Ne is the noble gas in the period preceding sulfur.) [Ne] represents 1s22s22p6. This leaves us 6 electrons to fill the 3s subshell and partially fill the 3p subshell. Thus, the electron configuration of S is 1s22s22p63s23p4 or [Ne]3s23p4 . (b) Strategy We use the same approach as that in (a). What does it mean to say that Pd is a diamagnetic element? (Continued)  CHEMISTRY in Action Quantum Dots W e normally consider the color of a chemical substance to be an intensive property (p. 11) because the color does not depend on the amount of that substance that is being considered. or a semiconductor (see Section 21.3 for a description of semiconductors). By confining the electrons to such a small volume, the allowed energies of these electrons are quan- As we are learning in this chapter, however, the “normal” be- tized. Therefore, if quantum dots are excited to higher en- havior of matter is much harder to define as we enter the quan- ergy, only certain wavelengths of light are emitted when the tum world of the very small. electrons go back to their ground states, just as in the case of Quantum dots are tiny pieces of matter, typically on the the emission spectra of atoms. But unlike atoms, the energy order of a few nanometers in diameter, composed of a metal of light omitted from a quantum dot can be “tuned” by Emission from dispersed solutions of CdSe quantum dots arranged from left to right in order of increasing diameter (2 nm to 7 nm). Solution Palladium has 46 electrons. The noble gas core in this case is [Kr]. (Kr is the noble gas in the period preceding palladium.) [Kr] represents 1s22s22p63s23p64s23d104p6 The remaining 10 electrons are distributed among the 4d and 5s orbitals. The three choices are (1) 4d10, (2) 4d 95s1, and (3) 4d 85s2. Because palladium is diamagnetic, all the electrons are paired and its electron configuration must be 1s22s22p63s23p64s23d104p64d10 or simply [Kr]4d10 . The configurations in (2) and (3) both represent paramagnetic elements. Similar problems: 7.87, 7.88. Check To confirm the answer, write the orbital diagrams for (1), (2), and (3). Practice Exercise Write the ground-state electron configuration for phosphorus (P). 312 varying the size of the quantum dot because that changes the volume within which the electrons are confined. This phe- nomenon is due to the wavelike behavior of electrons and is analogous to changing the pitch (frequency) of the sound made by plucking a guitar string (see Figure 7.12) by press- ing against the neck of the instrument, effectively shortening the string. The ability to regulate the energy of light emitted by a quantum dot is quite remarkable, enabling one to gener- ate the visible spectrum using a single chemical substance by simply varying the diameter of the quantum dots over a range of a few nanometers. Besides illustrating the quantum behavior of matter and enabling that behavior to be studied on the nanometer scale (as opposed to on a picometer scale at the atomic level), quantum dots offer great promise for yielding important applications in the fields of technology and medicine. Like bulk semiconduct- ing materials, quantum dots can be made to function as LEDs A light micrograph showing the fluorescence of quantum dots in a proto- (light emitting diodes), but unlike these bulk materials, quantum zoan, used to study the movement of nanoparticles through the food chain and their potential for accumulation by way of bioconcentration. dots emit light symmetrically in very narrow ranges. By com- bining three quantum dots that emit light of appropriate colors, it is possible to create devices that produce white light at much lower energy costs than required for incandescent bulbs or even fluorescent bulbs, which carry an additional environmen- tal concern because they contain mercury. Quantum dots can added potential to act therapeutically, either by being incorpo- also be used to label biological tissue. Besides offering the rated into the more permeable cancer cells and destroying advantage of greater stability over traditional biological dyes, those cells, or by attaching a known antitumor agent to the the surface of quantum dots can be chemically modified to quantum dot. Other potential applications for quantum dots target certain cells such as cancer cells. In addition to enabling include quantum computing and photovoltaic cells for har- tumors to be imaged, these modified quantum dots have the vesting solar energy. Review of Concepts Identify the atom that has the following ground-state electron configuration: [Ar]4s 23d 6 Key Equations u 5 λn (7.1) Relating speed of a wave to its wavelength and frequency. E 5 hn (7.2) Relating energy of a quantum (and of a photon) to the frequency. c E5h (7.3) Relating energy of a quantum (and of a photon) to the wavelength. λ hn 5 KE 1 W (7.4) The photoelectric effect. 313 314 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms 1 En 5 2RH a b (7.5) Energy of an electron in the nth state in a hydrogen atom. n2 1 1 ¢E 5 hn 5 RH a 2 2 2 b (7.6) Energy of a photon absorbed or emitted as the electron undergoes a ni nf transition from the ni level to the nf level. h λ5 (7.8) Relating wavelength of a particle to its mass m and velocity u. mu h ¢x¢p $ (7.9) Calculating the uncertainty in the position or in the momentum of a particle. 4π Summary of Facts & Concepts 1. The quantum theory developed by Planck successfully 7. The Schrödinger equation tells us the possible energy explains the emission of radiation by heated solids. The states of the electron in a hydrogen atom and the quantum theory states that radiant energy is emitted probability of its location in a particular region sur- by atoms and molecules in small discrete amounts rounding the nucleus. These results can be applied with (quanta), rather than over a continuous range. This be- reasonable accuracy to many-electron atoms. havior is governed by the relationship E 5 hn, where E 8. An atomic orbital is a function (Ψ) that defines the dis- is the energy of the radiation, h is Planck’s constant, tribution of electron density (Ψ2) in space. Orbitals are and n is the frequency of the radiation. Energy is al- represented by electron density diagrams or boundary ways emitted in whole-number multiples of hn (1 hn, surface diagrams. 2 hn, 3 hn, . . . ). 9. Four quantum numbers characterize each electron in an 2. Using quantum theory, Einstein solved another mys- atom: the principal quantum number n identifies the tery of physics—the photoelectric effect. Einstein pro- main energy level, or shell, of the orbital; the angular posed that light can behave like a stream of particles momentum quantum number / indicates the shape of (photons). the orbital; the magnetic quantum number m/ specifies 3. The line spectrum of hydrogen, yet another mystery the orientation of the orbital in space; and the electron to nineteenth-century physicists, was also explained spin quantum number ms indicates the direction of the by applying the quantum theory. Bohr developed a electron’s spin on its own axis. model of the hydrogen atom in which the energy of its 10. The single s orbital for each energy level is spherical single electron is quantized—limited to certain en- and centered on the nucleus. The three p orbitals present ergy values determined by an integer, the principal at n 5 2 and higher; each has two lobes, and the pairs quantum number. of lobes are arranged at right angles to one another. 4. An electron in its most stable energy state is said to Starting with n 5 3, there are five d orbitals, with more be in the ground state, and an electron at an energy complex shapes and orientations. level higher than its most stable state is said to be in 11. The energy of the electron in a hydrogen atom is deter- an excited state. In the Bohr model, an electron mined solely by its principal quantum number. In many- emits a photon when it drops from a higher-energy electron atoms, the principal quantum number and the state (an excited state) to a lower-energy state (the angular momentum quantum number together deter- ground state or another, less excited state). The re- mine the energy of an electron. lease of specific amounts of energy in the form of 12. No two electrons in the same atom can have the same photons accounts for the lines in the hydrogen emis- four quantum numbers (the Pauli exclusion principle). sion spectrum. 13. The most stable arrangement of electrons in a subshell 5. De Broglie extended Einstein’s wave-particle descrip- is the one that has the greatest number of parallel spins tion of light to all matter in motion. The wavelength of (Hund’s rule). Atoms with one or more unpaired elec- a moving particle of mass m and velocity u is given by tron spins are paramagnetic. Atoms in which all elec- the de Broglie equation λ 5 h/mu. trons are paired are diamagnetic. 6. The Schrödinger equation describes the motions 14. The Aufbau principle provides the guideline for build- and energies of submicroscopic particles. This ing up the elements. The periodic table classifies the equation launched quantum mechanics and a new elements according to their atomic numbers and thus era in physics. also by the electronic configurations of their atoms.  Questions & Problems 315 Key Words Actinide series, p. 311 Electron Heisenberg uncertainty Pauli exclusion Amplitude, p. 275 configuration, p. 302 principle, p. 292 principle, p. 303 Atomic orbital, p. 294 Electron density, p. 294 Hund’s rule, p. 305 Photoelectric effect, p. 279 Aufbau principle, p. 308 Emission spectra, p. 282 Lanthanide (rare earth) Photon, p. 279 Boundary surface Excited level (or series, p. 310 Quantum, p. 278 diagram, p. 298 state), p. 284 Line spectra, p. 282 Quantum numbers, p. 295 Diamagnetic, p. 304 Frequency (n), p. 275 Many-electron atom, p. 295 Rare earth series, p. 310 Electromagnetic Ground level (or Noble gas core, p. 308 Transition metals, p. 310 radiation, p. 277 state), p. 284 Node, p. 287 Wave, p. 275 Electromagnetic wave, p. 276 Ground state, p. 284 Paramagnetic, p. 304 Wavelength (λ), p. 275 Questions & Problems • Problems available in Connect Plus 7.10 How many minutes would it take a radio wave to Red numbered problems solved in Student Solutions Manual travel from the planet Venus to Earth? (Average dis- tance from Venus to Earth is 28 million miles.) Quantum Theory and • 7.11 The SI unit of time is the second, which is de- Electromagnetic Radiation fined as 9,192,631,770 cycles of radiation associ- Review Questions ated with a certain emission process in the cesium atom. Calculate the wavelength of this radiation 7.1 What is a wave? Explain the following terms associ- (to three significant figures). In which region of ated with waves: wavelength, frequency, amplitude. the electromagnetic spectrum is this wavelength 7.2 What are the units for wavelength and frequency of found? electromagnetic waves? What is the speed of light in 7.12 The SI unit of length is the meter, which is defined meters per second and miles per hour? as the length equal to 1,650,763.73 wavelengths of 7.3 List the types of electromagnetic radiation, starting with the light emitted by a particular energy transition in the radiation having the longest wavelength and ending krypton atoms. Calculate the frequency of the light with the radiation having the shortest wavelength. to three significant figures. 7.4 Give the high and low wavelength values that define the visible region of the electromagnetic spectrum. The Photoelectric Effect 7.5 Briefly explain Planck’s quantum theory and explain Review Questions what a quantum is. What are the units for Planck’s 7.13 What are photons? What role did Einstein’s expla- constant? nation of the photoelectric effect play in the devel- 7.6 Give two everyday examples that illustrate the con- opment of the particle-wave interpretation of the cept of quantization. nature of electromagnetic radiation? 7.14 Consider the plots shown here for the photoelectric Problems effect of two different metals A (green line) and • 7.7 (a) What is the wavelength (in nanometers) of light hav- B (red line). (a) Which metal has a greater work ing a frequency of 8.6 3 1013 Hz? (b) What is the fre- function? (b) What does the slope of the lines tell us? quency (in Hz) of light having a wavelength of 566 nm? • 7.8 (a) What is the frequency of light having a wave- length of 456 nm? (b) What is the wavelength (in nanometers) of radiation having a frequency of Kinetic energy 2.45 3 109 Hz? (This is the type of radiation used in microwave ovens.) • 7.9 The average distance between Mars and Earth is about 1.3 3 108 miles. How long would it take TV pictures transmitted from the Viking space vehicle on Mars’ surface to reach Earth? (1 mile 5 1.61 km.) ␯  316 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms Problems 7.26 Some copper compounds emit green light when they are heated in a flame. How would you determine • 7.15 A photon has a wavelength of 624 nm. Calculate the whether the light is of one wavelength or a mixture energy of the photon in joules. of two or more wavelengths? • 7.16 The blue color of the sky results from the scattering of 7.27 Is it possible for a fluorescent material to emit radia- sunlight by air molecules. The blue light has a fre- tion in the ultraviolet region after absorbing visible quency of about 7.5 3 1014 Hz. (a) Calculate the light? Explain your answer. wavelength, in nm, associated with this radiation, and (b) calculate the energy, in joules, of a single photon 7.28 Explain how astronomers are able to tell which ele- associated with this frequency. ments are present in distant stars by analyzing the electromagnetic radiation emitted by the stars. • 7.17 A photon has a frequency of 6.0 3 104 Hz. (a) Con- • 7.29 Consider the following energy levels of a hypotheti- vert this frequency into wavelength (nm). Does this frequency fall in the visible region? (b) Calculate the cal atom: energy (in joules) of this photon. (c) Calculate the E4 __________ 21.0 3 10219 J energy (in joules) of 1 mole of photons all with this E3 __________ 25.0 3 10219 J frequency. E2 __________ 210 3 10219 J 7.18 What is the wavelength, in nm, of radiation that has E1 __________ 215 3 10219 J an energy content of 1.0 3 103 kJ/mol? In which (a) What is the wavelength of the photon needed to region of the electromagnetic spectrum is this radia- excite an electron from E1 to E4? (b) What is the en- tion found? ergy (in joules) a photon must have in order to excite • 7.19 When copper is bombarded with high-energy elec- an electron from E2 to E3? (c) When an electron trons, X rays are emitted. Calculate the energy (in drops from the E3 level to the E1 level, the atom is joules) associated with the photons if the wavelength said to undergo emission. Calculate the wavelength of the X rays is 0.154 nm. of the photon emitted in this process. • 7.20 A particular form of electromagnetic radiation has a 7.30 The first line of the Balmer series occurs at a wave- frequency of 8.11 3 1014 Hz. (a) What is its wave- length of 656.3 nm. What is the energy difference length in nanometers? In meters? (b) To what region between the two energy levels involved in the emis- of the electromagnetic spectrum would you assign sion that results in this spectral line? it? (c) What is the energy (in joules) of one quantum • 7.31 Calculate the wavelength (in nanometers) of a photon of this radiation? emitted by a hydrogen atom when its electron drops • 7.21 The work function of potassium is 3.68 3 10219 J. from the n 5 5 state to the n 5 3 state. (a) What is the minimum frequency of light needed to • 7.32 Calculate the frequency (Hz) and wavelength (nm) eject electrons from the metal? (b) Calculate the ki- of the emitted photon when an electron drops from netic energy of the ejected electrons when light of fre- the n 5 4 to the n 5 2 level in a hydrogen atom. quency equal to 8.62 3 1014 s21 is used for irradiation. • 7.33 Careful spectral analysis shows that the familiar • 7.22 When light of frequency equal to 2.11 3 1015 s21 yellow light of sodium lamps (such as street shines on the surface of gold metal, the kinetic energy lamps) is made up of photons of two wavelengths, of ejected electrons is found to be 5.83 3 10219 J. 589.0 nm and 589.6 nm. What is the difference in What is the work function of gold? energy (in joules) between photons with these wavelengths? Bohr’s Theory of the Hydrogen Atom • 7.34 An electron in the hydrogen atom makes a transition Review Questions from an energy state of principal quantum numbers ni to the n 5 2 state. If the photon emitted has a 7.23 (a) What is an energy level? Explain the difference wavelength of 434 nm, what is the value of ni? between ground state and excited state. (b) What are emission spectra? How do line spectra differ from continuous spectra? Particle-Wave Duality 7.24 (a) Briefly describe Bohr’s theory of the hydrogen Review Questions atom and how it explains the appearance of an emis- 7.35 Explain the statement, Matter and radiation have a sion spectrum. How does Bohr’s theory differ from “dual nature.” concepts of classical physics? (b) Explain the mean- ing of the negative sign in Equation (7.5). 7.36 How does de Broglie’s hypothesis account for the fact that the energies of the electron in a hydrogen atom are quantized? Problems 7.37 Why is Equation (7.8) meaningful only for submi- 7.25 Explain why elements produce their own character- croscopic particles, such as electrons and atoms, and istic colors when they emit photons? not for macroscopic objects?  Questions & Problems 317 7.38 (a) If a H atom and a He atom are traveling at the 7.53 Which quantum number defines a shell? Which quan- same speed, what will be the relative wavelengths of tum numbers define a subshell? the two atoms? (b) If a H atom and a He atom have • 7.54 Which of the four quantum numbers (n, /, m/, ms) the same kinetic energy, what will be the relative determine (a) the energy of an electron in a hydrogen wavelengths of the two atoms? atom and in a many-electron atom, (b) the size of an orbital, (c) the shape of an orbital, (d) the orientation Problems of an orbital in space? • 7.39 Thermal neutrons are neutrons that move at speeds comparable to those of air molecules at room tem- Problems perature. These neutrons are most effective in initi- ating a nuclear chain reaction among 235U isotopes. • 7.55 An electron in a certain atom is in the n 5 2 quan- Calculate the wavelength (in nm) associated with a tum level. List the possible values of / and m/ that it beam of neutrons moving at 7.00 3 102 m/s. (Mass can have. of a neutron 5 1.675 3 10227 kg.) • 7.56 An electron in an atom is in the n 5 3 quantum • 7.40 Protons can be accelerated to speeds near that of light level. List the possible values of / and m/ that it in particle accelerators. Estimate the wavelength can have. (in nm) of such a proton moving at 2.90 3 108 m/s. • 7.57 Give the values of the quantum numbers associated (Mass of a proton 5 1.673 3 10227 kg.) with the following orbitals: (a) 2p, (b) 3s, (c) 5d. • 7.41 What is the de Broglie wavelength, in cm, of a 12.4-g • 7.58 Give the values of the four quantum numbers of hummingbird flying at 1.20 3 102 mph? (1 mile 5 an electron in the following orbitals: (a) 3s, (b) 4p, 1.61 km.) (c) 3d. • 7.42 What is the de Broglie wavelength (in nm) associ- 7.59 Discuss the similarities and differences between a 1s ated with a 2.5-g Ping-Pong ball traveling and a 2s orbital. 35 mph? 7.60 What is the difference between a 2px and a 2py orbital? Quantum Mechanics • 7.61 List all the possible subshells and orbitals associated with the principal quantum number n, if n 5 5. Review Questions • 7.62 List all the possible subshells and orbitals associated 7.43 What are the inadequacies of Bohr’s theory? with the principal quantum number n, if n 5 6. 7.44 What is the Heisenberg uncertainty principle? What • 7.63 Calculate the total number of electrons that can oc- is the Schrödinger equation? cupy (a) one s orbital, (b) three p orbitals, (c) five d 7.45 What is the physical significance of the wave orbitals, (d) seven f orbitals. function? 7.64 What is the total number of electrons that can be 7.46 How is the concept of electron density used to held in all orbitals having the same principal quan- describe the position of an electron in the quantum tum number n? mechanical treatment of an atom? • 7.65 Determine the maximum number of electrons that can be found in each of the following subshells: 3s, 3d, 4p, 4f, 5f. Atomic Orbitals Review Questions • 7.66 Indicate the total number of (a) p electrons in N (Z 5 7); (b) s electrons in Si (Z 5 14); and (c) 3d 7.47 What is an atomic orbital? How does an atomic electrons in S (Z 5 16). orbital differ from an orbit? 7.67 Make a chart of all allowable orbitals in the first four 7.48 Describe the shapes of s, p, and d orbitals. How are principal energy levels of the hydrogen atom. Desig- these orbitals related to the quantum numbers n, /, nate each by type (for example, s, p) and indicate how and m/? many orbitals of each type there are. 7.49 List the hydrogen orbitals in increasing order of 7.68 Why do the 3s, 3p, and 3d orbitals have the same energy. energy in a hydrogen atom but different energies in 7.50 Describe the characteristics of an s orbital, a p or- a many-electron atom? bital, and a d orbital. Which of the following orbitals • 7.69 For each of the following pairs of hydrogen orbitals, do not exist: 1p, 2s, 2d, 3p, 3d, 3f, 4g? indicate which is higher in energy: (a) 1s, 2s; (b) 2p, 7.51 Why is a boundary surface diagram useful in repre- 3p; (c) 3dxy, 3dyz; (d) 3s, 3d; (e) 4f, 5s. senting an atomic orbital? • 7.70 Which orbital in each of the following pairs is lower 7.52 Describe the four quantum numbers used to charac- in energy in a many-electron atom? (a) 2s, 2p; (b) 3p, terize an electron in an atom. 3d; (c) 3s, 4s; (d) 4d, 5f.  318 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms Electron Configuration Problems Review Questions • 7.87 Use the Aufbau principle to obtain the ground-state 7.71 What is electron configuration? Describe the roles electron configuration of selenium. that the Pauli exclusion principle and Hund’s rule 7.88 Use the Aufbau principle to obtain the ground-state play in writing the electron configuration of elements. electron configuration of technetium. • 7.72 Explain the meaning of the symbol 4d 6. • 7.89 Write the ground-state electron configurations for 7.73 Explain the meaning of diamagnetic and para- the following elements: B, V, Ni, As, I, Au. magnetic. Give an example of an element that is • 7.90 Write the ground-state electron configurations for diamagnetic and one that is paramagnetic. What the following elements: Ge, Fe, Zn, Ni, W, Tl. does it mean when we say that electrons are • 7.91 The electron configuration of a neutral atom is paired? 1s 22s 22p 63s 2. Write a complete set of quantum 7.74 What is meant by the term “shielding of electrons” in numbers for each of the electrons. Name the an atom? Using the Li atom as an example, describe element. the effect of shielding on the energy of electrons in • 7.92 Which of the following species has the most an atom. unpaired electrons? S1, S, or S2. Explain how you arrive at your answer. Problems • 7.75 Indicate which of the following sets of quantum num- Additional Problems bers in an atom are unacceptable and explain why: (a) (1, 0, 12 , 12 ), (b) (3, 0, 0, 1 12 ), (c) (2, 2, 1, 1 12 ), 7.93 A sample tube consisted of atomic hydrogens in (d) (4, 3, 22, 1 12 ), (e) (3, 2, 1, 1). their ground state. A student illuminated the at- oms with monochromatic light, that is, light of a 7.76 The ground-state electron configurations listed here single wavelength. If only two spectral emission are incorrect. Explain what mistakes have been made lines in the visible region are observed, what is in each and write the correct electron configurations. the wavelength (or wavelengths) of the incident Al: 1s22s22p43s23p3 radiation? B: 1s22s22p5 7.94 A laser produces a beam of light with a wave- F: 1s22s22p6 length of 532 nm. If the power output is 25.0 mW, • 7.77 The atomic number of an element is 73. Is this ele- how many photons does the laser emit per second? ment diamagnetic or paramagnetic? (1 W 5 1 J/s.) • 7.78 Indicate the number of unpaired electrons present in • 7.95 When a compound containing cesium ion is heated each of the following atoms: B, Ne, P, Sc, Mn, Se, in a Bunsen burner flame, photons with an energy Kr, Fe, Cd, I, Pb. of 4.30 3 10219 J are emitted. What color is the cesium flame? 7.96 Discuss the current view of the correctness of the The Building-Up Principle following statements. (a) The electron in the hydro- Review Questions gen atom is in an orbit that never brings it closer than 100 pm to the nucleus. (b) Atomic absorption 7.79 State the Aufbau principle and explain the role it plays spectra result from transitions of electrons from in classifying the elements in the periodic table. lower to higher energy levels. (c) A many-electron 7.80 Describe the characteristics of the following groups atom behaves somewhat like a solar system that has of elements: transition metals, lanthanides, actinides. a number of planets. 7.81 What is the noble gas core? How does it simplify the 7.97 What is the basis for thinking that atoms are spherical writing of electron configurations? in shape even though the atomic orbitals p, d, . . . have • 7.82 What are the group and period of the element distinctly nonspherical shapes? osmium? • 7.98 What is the maximum number of electrons in an 7.83 Define the following terms and give an example of atom that can have the following quantum numbers? each: transition metals, lanthanides, actinides. Specify the orbitals in which the electrons would be 7.84 Explain why the ground-state electron configurations found. (a) n 5 2, ms 5 1 12; (b) n 5 4, m/ 5 11; of Cr and Cu are different from what we might expect. (c) n 5 3, / 5 2; (d) n 5 2, / 5 0, ms 5 212; 7.85 Explain what is meant by a noble gas core. Write the (e) n 5 4, / 5 3, m/ 5 22. electron configuration of a xenon core. • 7.99 Identify the following individuals and their contribu- 7.86 Comment on the correctness of the following tions to the development of quantum theory: Bohr, de statement: The probability of finding two elec- Broglie, Einstein, Planck, Heisenberg, Schrödinger. trons with the same four quantum numbers in an 7.100 What properties of electrons are used in the opera- atom is zero. tion of an electron microscope?  Questions & Problems 319 • 7.101 In a photoelectric experiment a student uses a feasible to use sunlight as a source of energy for this light source whose frequency is greater than that process? needed to eject electrons from a certain metal. 7.109 Spectral lines of the Lyman and Balmer series do However, after continuously shining the light on not overlap. Verify this statement by calculating the the same area of the metal for a long period of longest wavelength associated with the Lyman time the student notices that the maximum ki- series and the shortest wavelength associated with netic energy of ejected electrons begins to de- the Balmer series (in nm). crease, even though the frequency of the light is held constant. How would you account for this • 7.110 An atom moving at its root-mean-square speed at 208C has a wavelength of 3.28 3 10211 m. Identify behavior? the atom. • 7.102 A certain pitcher’s fastballs have been clocked at • 7.111 Certain sunglasses have small crystals of silver about 100 mph. (a) Calculate the wavelength of a chloride (AgCl) incorporated in the lenses. When 0.141-kg baseball (in nm) at this speed. (b) What is the lenses are exposed to light of the appropriate the wavelength of a hydrogen atom at the same wavelength, the following reaction occurs: speed? (1 mile 5 1609 m.) 7.103 A student carried out a photoelectric experiment by AgCl ¡ Ag 1 Cl shining visible light on a clean piece of cesium The Ag atoms formed produce a uniform gray metal. The table here shows the kinetic energies color that reduces the glare. If DH for the preced- (KE) of the ejected electrons as a function of wave- ing reaction is 248 kJ/mol, calculate the maxi- lengths (λ). Determine graphically the work func- mum wavelength of light that can induce this tion and the Planck constant. process. 7.112 The He1 ion contains only one electron and is there- fore a hydrogenlike ion. Calculate the wavelengths, λ (nm) 405 435.8 480 520 577.7 in increasing order, of the first four transitions in the 2.360 3 2.029 3 1.643 3 1.417 3 1.067 3 Balmer series of the He1 ion. Compare these wave- KE (J) 10219 10219 10219 10219 10219 lengths with the same transitions in a H atom. Com- ment on the differences. (The Rydberg constant for He1 is 8.72 3 10218 J.) 7.104 (a) What is the lowest possible value of the principal 7.113 Ozone (O3) in the stratosphere absorbs the harmful quantum number (n) when the angular momentum radiation from the sun by undergoing decomposi- quantum number (/) is 1? (b) What are the possible tion: O3 ¡ O 1 O2 . (a) Referring to Table 6.4, values of the angular momentum quantum number calculate the DH8 for this process. (b) Calculate the (/) when the magnetic quantum number (m/) is 0, maximum wavelength of photons (in nm) that pos- given than n # 4? sess this energy to cause the decomposition of ozone 7.105 Considering only the ground-state electron configu- photochemically. ration, are there more diamagnetic or paramagnetic • 7.114 The retina of a human eye can detect light when elements? Explain. radiant energy incident on it is at least 4.0 3 10217 J. • 7.106 A ruby laser produces radiation of wavelength 633 nm For light of 600-nm wavelength, how many photons in pulses whose duration is 1.00 3 1029 s. (a) If the does this correspond to? laser produces 0.376 J of energy per pulse, how many 7.115 A helium atom and a xenon atom have the same photons are produced in each pulse? (b) Calculate kinetic energy. Calculate the ratio of the de Broglie the power (in watts) delivered by the laser per pulse. wavelength of the helium atom to that of the xenon (1 W 5 1 J/s.) atom. • 7.107 A 368-g sample of water absorbs infrared radiation 7.116 A laser is used in treating retina detachment. The at 1.06 3 104 nm from a carbon dioxide laser. Sup- wavelength of the laser beam is 514 nm and the pose all the absorbed radiation is converted to heat. power is 1.6 W. If the laser is turned on for 0.060 s Calculate the number of photons at this wavelength during surgery, calculate the number of photons required to raise the temperature of the water by emitted by the laser. (1 W 5 1 J/s.) 5.008C. 7.117 An electron in an excited state in a hydrogen atom • 7.108 Photodissociation of water can return to the ground state in two different ways: (a) via a direct transition in which a photon of wave- H2O(l) 1 hn ¡ H2 (g) 1 12O2 (g) length λ1 is emitted and (b) via an intermediate ex- has been suggested as a source of hydrogen. The cited state reached by the emission of a photon of ¢H°rxn for the reaction, calculated from thermochem- wavelength λ2. This intermediate excited state then ical data, is 285.8 kJ per mole of water decomposed. decays to the ground state by emitting another pho- Calculate the maximum wavelength (in nm) that ton of wavelength λ3. Derive an equation that relates would provide the necessary energy. In principle, is it λ1 to λ2 and λ3.  320 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms • 7.118 A photoelectric experiment was performed by correctness of the following statements (true separately shining a laser at 450 nm (blue light) or false). and a laser at 560 nm (yellow light) on a clean (a) n 5 4 is the first excited state. metal surface and measuring the number and ki- (b) It takes more energy to ionize (remove) the netic energy of the ejected electrons. Which light electron from n 5 4 than from the ground state. would generate more electrons? Which light (c) The electron is farther from the nucleus (on would eject electrons with greater kinetic energy? average) in n 5 4 than in the ground state. Assume that the same amount of energy is deliv- ered to the metal surface by each laser and that the (d) The wavelength of light emitted when the frequencies of the laser lights exceed the thresh- electron drops from n 5 4 to n 5 1 is longer old frequency. than that from n 5 4 to n 5 2. 7.119 Draw the shapes (boundary surfaces) of the following (e) The wavelength the atom absorbs in going orbitals: (a) 2py, (b) 3dz2, (c) 3dx2 2y2. (Show coordi- from n 5 1 to n 5 4 is the same as that emitted nate axes in your sketches.) as it goes from n 5 4 to n 5 1. • 7.120 The electron configurations described in this chapter • 7.127 The ionization energy of a certain element is 412 all refer to gaseous atoms in their ground states. An kJ/mol (see Problem 7.125). However, when the atoms atom may absorb a quantum of energy and promote of this element are in the first excited state, the ion- one of its electrons to a higher-energy orbital. When ization energy is only 126 kJ/mol. Based on this in- this happens, we say that the atom is in an excited formation, calculate the wavelength of light emitted state. The electron configurations of some excited in a transition from the first excited state to the atoms are given. Identify these atoms and write their ground state. ground-state configurations: • 7.128 Alveoli are the tiny sacs of air in the lungs (see Prob- (a) 1s12s1 lem 5.136) whose average diameter is 5.0 3 1025 m. Consider an oxygen molecule (5.3 3 10226 kg) trapped (b) 1s22s22p23d 1 within a sac. Calculate the uncertainty in the velocity of (c) 1s22s22p64s1 the oxygen molecule. (Hint: The maximum uncer- (d) [Ar]4s13d 104p4 tainty in the position of the molecule is given by the (e) [Ne]3s23p43d1 diameter of the sac.) • 7.121 Draw orbital diagrams for atoms with the following • 7.129 How many photons at 660 nm must be absorbed to electron configurations: melt 5.0 3 102 g of ice? On average, how many (a) 1s22s22p5 H2O molecules does one photon convert from ice (b) 1s22s22p63s23p3 to water? (Hint: It takes 334 J to melt 1 g of ice at 08C.) (c) 1s22s22p63s23p64s23d 7 7.122 If Rutherford and his coworkers had used electrons • 7.130 Shown are portions of orbital diagrams representing the ground-state electron configurations of certain instead of alpha particles to probe the structure of elements. Which of them violate the Pauli exclusion the nucleus as described in Section 2.2, what might principle? Hund’s rule? they have discovered? • 7.123 Scientists have found interstellar hydrogen atoms with quantum number n in the hundreds. Calculate h h hh h hg g h hg h the wavelength of light emitted when a hydrogen atom undergoes a transition from n 5 236 to n 5 235. (a) (b) (c) In what region of the electromagnetic spectrum does this wavelength fall? hg h h h h h h g hg 7.124 Calculate the wavelength of a helium atom whose speed is equal to the root-mean-square speed at (d) (e) 208C. hg hg gg hg hg 7.125 Ionization energy is the minimum energy required to remove an electron from an atom. It is usually (f) expressed in units of kJ/mol, that is, the energy in kilojoules required to remove one mole of electrons from one mole of atoms. (a) Calculate the ionization • 7.131 The UV light that is responsible for tanning the skin energy for the hydrogen atom. (b) Repeat the calcu- falls in the 320- to 400-nm region. Calculate the total lation, assuming in this second case that the elec- energy (in joules) absorbed by a person exposed to this trons are removed from the n 5 2 state. radiation for 2.0 h, given that there are 2.0 3 1016 pho- • 7.126 An electron in a hydrogen atom is excited from the tons hitting Earth’s surface per square centimeter ground state to the n 5 4 state. Comment on the per second over a 80-nm (320 nm to 400 nm) range  Questions & Problems 321 and that the exposed body area is 0.45 m2. Assume • 7.137 A microwave oven operating at 1.22 3 108 nm is that only half of the radiation is absorbed and the used to heat 150 mL of water (roughly the volume other half is reflected by the body. (Hint: Use an of a tea cup) from 208C to 1008C. Calculate the average wavelength of 360 nm in calculating the en- number of photons needed if 92.0 percent of ergy of a photon.) microwave energy is converted to the thermal en- • 7.132 The sun is surrounded by a white circle of gaseous ergy of water. material called the corona, which becomes visible • 7.138 The radioactive Co-60 isotope is used in nuclear during a total eclipse of the sun. The temperature medicine to treat certain types of cancer. Calcu- of the corona is in the millions of degrees Celsius, late the wavelength and frequency of an emitted which is high enough to break up molecules and gamma photon having the energy of 1.29 3 remove some or all of the electrons from atoms. 1011 J/mol. One way astronomers have been able to estimate • 7.139 (a) An electron in the ground state of the hydrogen the temperature of the corona is by studying the atom moves at an average speed of 5 3 106 m/s. If emission lines of ions of certain elements. For the speed is known to an uncertainty of 1 percent, example, the emission spectrum of Fe141 ions has what is the uncertainty in knowing its position? been recorded and analyzed. Knowing that it Given that the radius of the hydrogen atom in the takes 3.5 3 104 kJ/mol to convert Fe131 to Fe141, ground state is 5.29 3 10211 m, comment on your estimate the temperature of the sun’s corona. result. The mass of an electron is 9.1094 3 10231 kg. (Hint: The average kinetic energy of one mole of (b) A 3.2-g Ping-Pong ball moving at 50 mph has a a gas is 32RT .) momentum of 0.073 kg ? m/s. If the uncertainty in 7.133 In 1996 physicists created an anti-atom of hydrogen. measuring the momentum is 1.0 3 1027 of the mo- In such an atom, which is the antimatter equivalent mentum, calculate the uncertainty in the Ping-Pong of an ordinary atom, the electrical charges of all the ball’s position. component particles are reversed. Thus, the nucleus 7.140 One wavelength in the hydrogen emission spectrum of an anti-atom is made of an anti-proton, which has is 1280 nm. What are the initial and final states of the same mass as a proton but bears a negative the transition responsible for this emission? charge, while the electron is replaced by an anti- 7.141 Owls have good night vision because their eyes can electron (also called positron) with the same mass as detect a light intensity as low as 5.0 3 10213 W/m2. an electron, but bearing a positive charge. Would Calculate the number of photons per second that you expect the energy levels, emission spectra, and an owl’s eye can detect if its pupil has a diameter of atomic orbitals of an antihydrogen atom to be differ- 9.0 mm and the light has a wavelength of 500 nm. ent from those of a hydrogen atom? What would (1 W 5 1 J/s.) happen if an anti-atom of hydrogen collided with a hydrogen atom? • 7.142 For hydrogenlike ions, that is, ions containing only one electron, Equation (7.5) is modified as 7.134 Use Equation (5.16) to calculate the de Broglie follows: En 5 2RHZ2(1yn2), where Z is the atomic wavelength of a N2 molecule at 300 K. number of the parent atom. The figure here repre- 7.135 When an electron makes a transition between energy sents the emission spectrum of such a hydrogen- levels of a hydrogen atom, there are no restrictions on like ion in the gas phase. All the lines result from the initial and final values of the principal quantum the electronic transitions from the excited states to number n. However, there is a quantum mechanical the n 5 2 state. (a) What electronic transitions cor- rule that restricts the initial and final values of the or- respond to lines B and C? (b) If the wavelength of bital angular momentum /. This is the selection rule, line C is 27.1 nm, calculate the wavelengths of which states that ¢/ 5 61; that is, in a transition, lines A and B. (c) Calculate the energy needed the value of / can only increase or decrease by one. to remove the electron from the ion in the n 5 4 According to this rule, which of the following transi- state. (d) What is the physical significance of the tions are allowed: (a) 2s ¡ 1s, (b) 3p ¡ 1s, continuum? (c) 3d ¡ 4f , (d) 4d ¡ 3s? In view of this se- lection rule, explain why it is possible to observe the various emission series shown in Figure 7.11. Continuum C B A • 7.136 In an electron microscope, electrons are accelerated by passing them through a voltage difference. The kinetic energy thus acquired by the electrons is equal to the voltage times the charge on the electron. Thus, a voltage difference of 1 V imparts a kinetic energy of 1.602 3 10219 C 3 V or 1.602 3 10219 J. Calculate the wavelength associated with electrons accelerated by 5.00 3 103 V. λ  322 Chapter 7 ■ Quantum Theory and the Electronic Structure of Atoms 7.143 When two atoms collide, some of their kinetic energy 7.149 In the beginning of the twentieth century, some scien- may be converted into electronic energy in one or both tists thought that a nucleus may contain both electrons atoms. If the average kinetic energy is about equal to and protons. Use the Heisenberg uncertainty principle the energy for some allowed electronic transition, an to show that an electron cannot be confined within a appreciable number of atoms can absorb enough nucleus. Repeat the calculation for a proton. Comment energy through an inelastic collision to be raised to an on your results. Assume the radius of a nucleus to be excited electronic state. (a) Calculate the average 1.0 3 10215 m. The masses of an electron and a proton kinetic energy per atom in a gas sample at 298 K. are 9.109 3 10231 kg and 1.673 3 10227 kg, respec- (b) Calculate the energy difference between the n 5 1 tively. (Hint: Treat the diameter of the nucleus as the and n 5 2 levels in hydrogen. (c) At what temperature uncertainty in position.) is it possible to excite a hydrogen atom from the n 5 1 7.150 Blackbody radiation is the term used to describe the level to n 5 2 level by collision? [The average kinetic dependence of the radiation energy emitted by an energy of 1 mole of an ideal gas is ( 32 )RT.] object on wavelength at a certain temperature. 7.144 Calculate the energies needed to remove an elec- Planck proposed the quantum theory to account for tron from the n 5 1 state and the n 5 5 state in this dependence. Shown in the figure is a plot of the the Li 21 ion. What is the wavelength (in nm) of radiation energy emitted by our sun versus wave- the emitted photon in a transition from n 5 5 to length. This curve is characteristic of the tempera- n 5 1? The Rydberg constant for hydrogenlike ture at the surface of the sun. At a higher temperature, ions is (2.18 3 10 218 J)Z2, where Z is the atomic the curve has a similar shape but the maximum will number. shift to a shorter wavelength. What does this curve 7.145 The de Broglie wavelength of an accelerating proton reveal about two consequences of great biological in the Large Hadron Collider is 2.5 3 10214 m. What significance on Earth? is the kinetic energy (in joules) of the proton? 7.146 The minimum uncertainty in the position of a cer- tain moving particle is equal to its de Broglie wave- length. If the speed of the particle is 1.2 3 105 m/s, what is the minimum uncertainty in its speed? Solar radiation energy • 7.147 According to Einstein’s special theory of relativity, the mass of a moving particle, mmoving, is related to its mass at rest, mrest, by the following equation mrest mmoving 5 u 2 12a b B c 0 500 1000 λ (nm) where u and c are the speeds of the particle and light, respectively. (a) In particle accelerators, protons, electrons, and other charged particles are often 7.151 All molecules undergo vibrational motions. Quan- accelerated to speeds close to the speed of light. Cal- tum mechanical treatment shows that the vibra- culate the wavelength (in nm) of a proton moving at tional energy, Evib, of a diatomic molecule like HCl 50.0 percent the speed of light. The mass of a proton is given by is 1.673 3 10227 kg. (b) Calculate the mass of a 6.0 1 3 1022 kg tennis ball moving at 63 m/s. Comment Evib 5 an 1 b hn on your results. 2 • 7.148 The mathematical equation for studying the photo- where n is a quantum number given by n 5 0, 1, 2, electric effect is 3, . . . and n is the fundamental frequency of vibra- hn 5 W 1 12meu2 tion. (a) Sketch the first three vibrational energy lev- els for HCl. (b) Calculate the energy required to where n is the frequency of light shining on the excite a HCl molecule from the ground level to the metal, W is the work function, and me and u are the first excited level. The fundamental frequency of mass and speed of the ejected electron. In an experi- vibration for HCl is 8.66 3 1013 s21. (c) The fact that ment, a student found that a maximum wavelength the lowest vibrational energy in the ground level is of 351 nm is needed to just dislodge electrons from not zero but equal to 12hn means that molecules will a zinc metal surface. Calculate the speed (in m/s) of vibrate at all temperatures, including the absolute an ejected electron when she employed light with a zero. Use the Heisenberg uncertainty principle to wavelength of 313 nm. justify this prediction. (Hint: Consider a nonvibrating  Answers to Practice Exercises 323 molecule and predict the uncertainty in the momen- there were evenly spaced dents (due to melting) about tum and hence the uncertainty in the position.) 6 cm apart. Based on her observations, calculate the • 7.152 The wave function for the 2s orbital in the hydrogen speed of light given that the microwave frequency is atom is 2.45 GHz. (Hint: The energy of a wave is propor- tional to the square of its amplitude.) 1 ρ Ψ 2s 5 a1 2 be2ρy2 7.154 The wave properties of matter can generally be ig- 22a30 2 nored for macroscopic objects such as tennis balls; however, wave properties have been measured at the where a0 is the value of the radius of the first Bohr fringe of detection for some very large molecules. orbit, equal to 0.529 nm, ρ is Z(r/a0), and r is the For example, wave patterns were detected for distance from the nucleus in meters. Calculate the C60(C12F25)8 molecules moving at a velocity of location of the node of the 2s wave function from 63 m/s. (a) Calculate the wavelength of a C60(C12F25)8 the nucleus. molecule moving at this velocity. (b) How does the 7.153 A student placed a large unwrapped chocolate bar in wavelength compare to the size of the molecule a microwave oven without a rotating glass plate. After given that its diameter is roughly 3000 pm? turning the oven on for less than a minute, she noticed Interpreting, Modeling & Estimating 7.155 Atoms of an element have only two accessible ex- visible light. Roughly how many photons are emitted cited states. In an emission experiment, however, by the lightbulb per second? (1 W 5 1 J/s.) three spectral lines were observed. Explain. Write 7.158 Photosynthesis makes use of photons of visible light an equation relating the shortest wavelength to the to bring about chemical changes. Explain why heat other two wavelengths. energy in the form of infrared photons is ineffective 7.156 According to Wien’s law, the wavelength of maxi- for photosynthesis. (Hint: Typical chemical bond mum intensity in blackbody radiation, λmax, is energies are 200 kJ/mol or greater.) given by 7.159 A typical red laser pointer has a power of 5 mW. b How long would it take a red laser pointer to emit λmax 5 the same number of photons emitted by a 1-W blue T laser in 1 s? (1 W 5 1 J/s.) where b is a constant (2.898 3 106 nm ? K) and T 7.160 Referring to the Chemistry in Action essay on is the temperature of the radiating body in kelvins. p. 312, estimate the wavelength of light that would (a) Estimate the temperature at the surface of the be emitted by a cadmium selenide (CdSe) quan- sun. (b) How are astronomers able to determine the tum dot with a diameter of 10 nm. Would the emit- temperature of stars in general? (See Problem 7.150 ted light be visible to the human eye? The diameter for a definition of blackbody radiation.) and emission wavelength for a series of quantum 7.157 Only a fraction of the electrical energy supplied to an dots are given here. incandescent-tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared Diameter (nm) 2.2 2.5 3.3 4.2 4.9 6.3 radiation (that is, heat). A 60-W lightbulb converts about 15.0 percent of the energy supplied to it into Wavelength (nm) 462 503 528 560 583 626 Answers to Practice Exercises 7.1 8.24 m. 7.2 3.39 3 103 nm. 7.3 9.65 3 10219 J. (4, 2, 2, 212 ). 7.10 32. 7.11 (1, 0, 0, 1 12 ), (1, 0, 0, 212 ), 7.4 2.63 3 103 nm. 7.5 56.6 nm. 7.6 0.2 m/s. 7.7 n 5 3, (2, 0, 0, 1 12 ), (2, 0, 0, 212 ), (2, 1, 21, 212 ). There are five / 5 1, m/ 5 21, 0, 1. 7.8 16. 7.9 (4, 2, 22, 1 12 ), other acceptable ways to write the quantum numbers for the (4, 2, 21, 1 12 ), (4, 2, 0, 1 12 ), (4, 2, 1, 1 12 ), (4, 2, 2, 1 12 ), last electron (in the 2p orbital). 7.12 [Ne]3s23p3. (4, 2, 22, 212 ), (4, 2, 21, 212 ), (4, 2, 0, 212 ), (4, 2, 1, 212 ), CHEMICAL M YS TERY Discovery of Helium and the Rise and Fall of Coronium S cientists know that our sun and other stars contain certain elements. How was this information obtained? In the early nineteenth century, the German physicist Josef Fraunhofer studied the emission spectrum of the sun and noticed certain dark lines at specific wave- lengths. We interpret the appearance of these lines by supposing that originally a continuous band of color was radiated and that, as the emitted light moves outward from the sun, some of the radiation is reabsorbed at those wavelengths by the atoms in space. These dark lines are therefore absorption lines. For atoms, the emission and absorption of light occur at the same wavelengths. By matching the absorption lines in the emission spectra of a star with the emission spectra of known elements in the laboratory, scientists have been able to deduce the types of elements present in the star. Another way to study the sun spectroscopically is during its eclipse. In 1868 the French physicist Pierre Janssen observed a bright yellow line (see Figure 7.8) in the emission spectrum of the sun’s corona during the totality of the eclipse. (The corona is the pearly white crown of light visible around the sun during a total eclipse.) This line did not match the emission lines of known elements, but did match one of the dark lines in the spectrum sketched by Fraunhofer. The name helium (from Helios, the sun god in Fraunhofer’s original drawing, in 1814, showing the dark absorption lines in the sun’s emission spectrum. The top of the diagram shows the overall bright- ness of the sun at different colors. 324 Greek mythology) was given to the element responsible for the emission line. Twenty- seven years later, helium was discovered on Earth by the British chemist William Ram- say in a mineral of uranium. On Earth, the only source of helium is through radioactive decay processes—a particles emitted during nuclear decay are eventually converted to helium atoms. The search for new elements from the sun did not end with helium. Around the time of Janssen’s work, scientists also detected a bright green line in the spectrum from the corona. They did not know the identity of the element giving rise to the line, so they called it coronium because it was only found in the corona. Over the following years, additional mystery coronal emission lines were found. The coronium problem proved During the total eclipse of the sun, much harder to solve than the helium case because no matchings were found with the which lasts for only a few minutes, the emission lines of known elements. It was not until the late 1930s that the Swedish phys- corona becomes visible. icist Bengt Edlén identified these lines as coming from partially ionized atoms of iron, calcium, and nickel. At very high temperatures (over a million degrees Celsius), many atoms become ionized by losing one or more electrons. Therefore, the mystery emission lines come from the resulting ions of the metals and not from a new element. So, after some 80 years the coronium problem was finally solved. There is no such element as coronium after all! Chemical Clues 1. Sketch a two-energy-level system (E1 and E2) to illustrate the absorption and emission processes. 2. Explain why the sun’s spectrum provides only absorption lines (the dark lines), whereas the corona spectrum provides only emission lines. 3. Why is it difficult to detect helium on Earth? 4. How are scientists able to determine the abundances of elements in stars? 5. Knowing the identity of an ion of an element giving rise to a coronal emission line, describe in qualitative terms how you can estimate the temperature of the corona. 325  CHAPTER 8 Periodic Relationships Among the Elements While the recurring or “periodic” trends in the properties of elements are most commonly illustrated in tabular form, alternative geometric arrangements are possible. CHAPTER OUTLINE A LOOK AHEAD 8.1 Development of the  We start with the development of the periodic table and the contributions Periodic Table made by nineteenth-century scientists, in particular by Mendeleev. (8.1) 8.2 Periodic Classification  We see that electron configuration is the logical way to build up the periodic of the Elements table, which explains some of the early anomalies. We also learn the rules for writing the electron configurations of cations and anions. (8.2) 8.3 Periodic Variation in Physical  Next, we examine the periodic trends in physical properties such as the size Properties of atoms and ions in terms of effective nuclear charge. (8.3) 8.4 Ionization Energy  We continue our study of periodic trends by examining chemical properties 8.5 Electron Affinity like ionization energy and electron affinity. (8.4 and 8.5) 8.6 Variation in Chemical  We then apply the knowledge acquired in the chapter to systematically Properties of the study the properties of the representative elements as individual groups and also across a given period. (8.6) Representative Elements 326  8.1 Development of the Periodic Table 327 M any of the chemical properties of the elements can be understood in terms of their elec- tron configurations. Because electrons fill atomic orbitals in a fairly regular fashion, it is not surprising that elements with similar electron configurations, such as sodium and potas- sium, behave similarly in many respects and that, in general, the properties of the elements exhibit observable trends. Chemists in the nineteenth century recognized periodic trends in the physical and chemical properties of the elements, long before quantum theory came onto the scene. Although these chemists were not aware of the existence of electrons and protons, their efforts to systematize the chemistry of the elements were remarkably successful. Their main sources of information were the atomic masses of the elements and other known physical and chemical properties. 8.1 Development of the Periodic Table In the nineteenth century, when chemists had only a vague idea of atoms and mole- cules and did not know of the existence of electrons and protons, they devised the periodic table using their knowledge of atomic masses. Accurate measurements of the atomic masses of many elements had already been made. Arranging elements accord- ing to their atomic masses in a periodic table seemed logical to those chemists, who felt that chemical behavior should somehow be related to atomic mass. In 1864 the English chemist John Newlands† noticed that when the elements were arranged in order of atomic mass, every eighth element had similar properties. Newlands referred to this peculiar relationship as the law of octaves. However, this “law” turned out to be inadequate for elements beyond calcium, and Newlands’s work was not accepted by the scientific community. In 1869 the Russian chemist Dmitri Mendeleev‡ and the German chemist Lothar Meyer§ independently proposed a much more extensive tabulation of the elements based on the regular, periodic recurrence of properties. Mendeleev’s classification system was a great improvement over Newlands’s for two reasons. First, it grouped the elements together more accurately, according to their properties. Equally impor- tant, it made possible the prediction of the properties of several elements that had not yet been discovered. For example, Mendeleev proposed the existence of an unknown element that he called eka-aluminum and predicted a number of its prop- erties. (Eka is a Sanskrit word meaning “first”; thus eka-aluminum would be the first element under aluminum in the same group.) When gallium was discovered four years later, its properties matched the predicted properties of eka-aluminum remarkably well: Eka-Aluminum (Ea) Gallium (Ga) Atomic mass 68 amu 69.9 amu Gallium melts in a person’s hand Melting point Low 29.78°C (body temperature is about 37°C). Density 5.9 g/cm3 5.94 g/cm3 Formula of oxide Ea2O3 Ga2O3 Mendeleev’s periodic table included 66 known elements. By 1900, some 30 more had Appendix 1 explains the names and symbols of the elements. been added to the list, filling in some of the empty spaces. Figure 8.1 charts the discovery of the elements chronologically. † John Alexander Reina Newlands (1838–1898). English chemist. Newlands’s work was a step in the right direction in the classification of the elements. Unfortunately, because of its shortcomings, he was subjected to much criticism, and even ridicule. At one meeting he was asked if he had ever examined the elements according to the order of their initial letters! Nevertheless, in 1887 Newlands was honored by the Royal Society of London for his contribution. ‡ Dmitri Ivanovich Mendeleev (1836–1907). Russian chemist. His work on the periodic classification of elements is regarded by many as the most significant achievement in chemistry in the nineteenth century. § Julius Lothar Meyer (1830–1895). German chemist. In addition to his contribution to the periodic table, Meyer also discovered the chemical affinity of hemoglobin for oxygen. 328 Chapter 8 ■ Periodic Relationships Among the Elements Ancient times 1735–1843 1894–1918 Middle Ages–1700 1843–1886 1923–1961 1965– 1 2 H He 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Figure 8.1 A chronological chart of the discovery of the elements. To date, 118 elements have been identified. Although this periodic table was a celebrated success, the early versions had some glaring inconsistencies. For example, the atomic mass of argon (39.95 amu) is greater than that of potassium (39.10 amu). If elements were arranged solely according to increasing atomic mass, argon would appear in the position occupied by potassium in our modern periodic table (see the inside front cover). But no chemist would place argon, an inert gas, in the same group as lithium and sodium, two very reactive metals. This and other discrepancies suggested that some funda- mental property other than atomic mass must be the basis of periodicity. This prop- erty turned out to be associated with atomic number, a concept unknown to Mendeleev and his contemporaries. Using data from α-particle scattering experiments (see Section 2.2), Rutherford estimated the number of positive charges in the nucleus of a few elements, but the significance of these numbers was overlooked for several more years. In 1913 a young English physicist, Henry Moseley,† discovered a correlation between what he called atomic number and the frequency of X rays generated by bombarding an element with high-energy electrons. Moseley noticed that the frequencies of X rays emitted from the elements could be correlated by the equation 1n 5 a(Z 2 b) (8.1) † Henry Gwyn-Jeffreys Moseley (1887–1915). English physicist. Moseley discovered the relationship between X-ray spectra and atomic number. A lieutenant in the Royal Engineers, he was killed in action at the age of 28 during the British campaign in Gallipoli, Turkey.  8.2 Periodic Classification of the Elements 329 where v is the frequency of the emitted X rays and a and b are constants that are the same for all the elements. Thus, from the square root of the measured frequency of the X rays emitted, we can determine the atomic number of the element. With a few exceptions, Moseley found that atomic number increases in the same order as atomic mass. For example, calcium is the twentieth element in order of increasing atomic mass, and it has an atomic number of 20. The dis- crepancies that had puzzled earlier scientists now made sense. The atomic num- ber of argon is 18 and that of potassium is 19, so potassium should follow argon in the periodic table. A modern periodic table usually shows the atomic number along with the ele- ment symbol. As you already know, the atomic number also indicates the number of electrons in the atoms of an element. Electron configurations of elements help to explain the recurrence of physical and chemical properties. The importance and usefulness of the periodic table lie in the fact that we can use our understanding of the general properties and trends within a group or a period to predict with consid- erable accuracy the properties of any element, even though that element may be unfamiliar to us. 8.2 Periodic Classification of the Elements Figure 8.2 shows the periodic table together with the outermost ground-state electron configurations of the elements. (The electron configurations of the elements are also given in Table 7.3.) Starting with hydrogen, we see that subshells are filled in the order shown in Figure 7.24. According to the type of subshell being filled, the ele- ments can be divided into categories—the representative elements, the noble gases, 1 18 1A 8A 1 2 1 H 2 13 14 15 16 17 He 1s1 2A 3A 4A 5A 6A 7A 1s2 3 4 5 6 7 8 9 10 2 Li Be B C N O F Ne 2s1 2s2 2s22p1 2s22p2 2s22p3 2s22p4 2s22p5 2s22p6 11 12 13 14 15 16 17 18 3 Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar 3s1 3s2 3B 4B 5B 6B 7B 8B 1B 2B 3s23p1 3s23p2 3s23p3 3s23p4 3s23p5 3s23p6 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4s1 4s2 4s23d1 4s23d2 4s23d3 4s13d5 4s23d5 4s23d6 4s23d7 4s23d8 4s13d10 4s23d10 4s24p1 4s24p2 4s24p3 4s24p4 4s24p5 4s24p6 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 5s1 5s2 5s24d1 5s24d2 5s14d4 5s14d5 5s24d5 5s14d7 5s14d8 4d10 5s14d10 5s24d10 5s25p1 5s25p2 5s25p3 5s25p4 5s25p5 5s25p6 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 6s1 6s2 6s25d1 6s25d2 6s25d3 6s25d4 6s25d5 6s25d6 6s25d7 6s15d9 6s15d10 6s25d10 6s26p1 6s26p2 6s26p3 6s26p4 6s26p5 6s26p6 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv 7s1 7s2 7s26d1 7s26d2 7s26d3 7s26d4 7s26d5 7s26d6 7s26d7 7s26d8 7s26d9 7s26d10 7s27p1 7s27p2 7s27p3 7s27p4 7s27p5 7s27p6 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 6s24f15d1 6s24f3 6s24f4 6s24f5 6s24f6 6s24f7 6s24f75d1 6s24f9 6s24f10 6s24f11 6s24f12 6s24f13 6s24f14 6s24f145d1 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 7s26d2 7s25f26d1 7s25f36d1 7s25f46d1 7s25f6 7s25f7 7s25f76d1 7s25f9 7s25f10 7s25f11 7s25f12 7s25f13 7s25f14 7s25f146d1 Figure 8.2 The ground-state electron configurations of the elements. For simplicity, only the configurations of the outer electrons are shown. 330 Chapter 8 ■ Periodic Relationships Among the Elements the transition elements (or transition metals), the lanthanides, and the actinides. The representative elements (also called main group elements) are the elements in Groups 1A through 7A, all of which have incompletely filled s or p subshells of the highest principal quantum number. With the exception of helium, the noble gases (the Group 8A elements) all have a completely filled p subshell. (The electron configurations are 1s2 for helium and ns2np6 for the other noble gases, where n is the principal quantum number for the outermost shell.) The transition metals are the elements in Groups 1B and 3B through 8B, which have incompletely filled d subshells, or readily produce cations with incompletely filled d subshells. (These metals are sometimes referred to as the d-block transition elements.) The nonsequential numbering of the transition metals in the periodic table (that is, 3B–8B, followed by 1B–2B) acknowledges a correspondence between the outer electron configurations of these elements and those of the representative elements. For example, scandium and gallium both have three outer electrons. However, because they are in different types of atomic orbitals, they are placed in different groups (3B and 3A). The metals iron (Fe), cobalt (Co), and nickel (Ni) do not fit this classification and are all placed in Group 8B. The Group 2B ele- ments, Zn, Cd, and Hg, are neither representative elements nor transition metals. There is no special name for this group of metals. It should be noted that the designation of A and B groups is not universal. In Europe the practice is to use B for representative elements and A for transition metals, which is just the opposite of the American convention. The International Union of Pure and Applied Chem- istry (IUPAC) has recommended numbering the columns sequentially with Arabic numerals 1 through 18 (see Figure 8.2). The proposal has sparked much contro- versy in the international chemistry community, and its merits and drawbacks will be deliberated for some time to come. In this text we will adhere to the American designation. The lanthanides and actinides are sometimes called f-block transition elements because they have incompletely filled f subshells. Figure 8.3 distinguishes the groups of elements discussed here. For the representative elements, the The chemical reactivity of the elements is largely determined by their valence valence electrons are simply those electrons at the highest principal energy electrons, which are the outermost electrons. For the representative elements, the level n. valence electrons are those in the highest occupied n shell. All nonvalence electrons in an atom are referred to as core electrons. Looking at the electron configurations of the representative elements once again, a clear pattern emerges: all the elements in a given group have the same number and type of valence electrons. The similarity of the valence electron configurations is what makes the elements in the same group resemble one another in chemical behavior. Thus, for instance, the alkali metals (the Group 1A elements) all have the valence electron configuration of ns1 (Table 8.1) and they all tend to lose one electron to form the unipositive cations. Similarly, the alkaline earth metals (the Group 2A elements) all have the valence electron configuration of ns2, and they all tend to lose two electrons to form the dipositive cations. We must Table 8.1 be careful, however, in predicting element properties based solely on “group member- Electron Configurations ship.” For example, the elements in Group 4A all have the same valence electron of Group 1A and Group configuration ns2np2, but there is a notable variation in chemical properties among the 2A Elements elements: carbon is a nonmetal, silicon and germanium are metalloids, and tin and lead are metals. Group 1A Group 2A As a group, the noble gases behave very similarly. Helium and neon are Li [He]2s1 Be [He]2s2 chemically inert, and there are few examples of compounds formed by the other Na [Ne]3s1 Mg [Ne]3s2 noble gases. This lack of chemical reactivity is due to the completely filled ns K [Ar]4s1 Ca [Ar]4s2 and np subshells, a condition that often correlates with great stability. Although Rb [Kr]5s1 Sr [Kr]5s2 the valence electron configuration of the transition metals is not always the same Cs [Xe]6s1 Ba [Xe]6s2 within a group and there is no regular pattern in the change of the electron con- figuration from one metal to the next in the same period, all transition metals Fr [Rn]7s1 Ra [Rn]7s2 share many characteristics that set them apart from other elements. The reason is  8.2 Periodic Classification of the Elements 331 Zinc 1 Representative 18 Cadmium 1A elements 8A Mercury 1 2 2 Noble gases Lanthanides 13 14 15 16 17 H 2A 3A 4A 5A 6A 7A He 3 4 Transition 5 6 7 8 9 10 Li Be Actinides B C N O F Ne metals 11 12 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Figure 8.3 Classification of the elements. Note that the Group 2B elements are often classified as transition metals even though they do not exhibit the characteristics of the transition metals. that these metals all have an incompletely filled d subshell. Likewise, the lantha- nide (and the actinide) elements resemble one another because they have incom- pletely filled f subshells. Example 8.1 An atom of a certain element has 15 electrons. Without consulting a periodic table, answer the following questions: (a) What is the ground-state electron configuration of the element? (b) How should the element be classified? (c) Is the element diamagnetic or paramagnetic? Strategy (a) We refer to the building-up principle discussed in Section 7.9 and start writing the electron configuration with principal quantum number n 5 1 and continuing upward until all the electrons are accounted for. (b) What are the electron configuration characteristics of representative elements? transition elements? noble gases? (c) Examine the pairing scheme of the electrons in the outermost shell. What determines whether an element is diamagnetic or paramagnetic? Solution (a) We know that for n 5 1 we have a 1s orbital (2 electrons); for n 5 2 we have a 2s orbital (2 electrons) and three 2p orbitals (6 electrons); for n 5 3 we have a 3s orbital (2 electrons). The number of electrons left is 15 2 12 5 3 and these three electrons are placed in the 3p orbitals. The electron configuration is 1s22s22p63s23p3. (Continued) 332 Chapter 8 ■ Periodic Relationships Among the Elements (b) Because the 3p subshell is not completely filled, this is a representative element. Based on the information given, we cannot say whether it is a metal, a nonmetal, or a metalloid. (c) According to Hund’s rule, the three electrons in the 3p orbitals have parallel spins (three unpaired electrons). Therefore, the element is paramagnetic. Check For (b), note that a transition metal possesses an incompletely filled d subshell and a noble gas has a completely filled outer shell. For (c), recall that if the atoms of an element contain an odd number of electrons, then the element must Similar problem: 8.20. be paramagnetic. Practice Exercise An atom of a certain element has 20 electrons. (a) Write the ground-state electron configuration of the element, (b) classify the element, (c) determine whether the element is diamagnetic or paramagnetic. Representing Free Elements in Chemical Equations Having classified the elements according to their ground-state electron configurations, we can now look at the way chemists represent metals, metalloids, and nonmetals as free elements in chemical equations. Because metals do not exist in discrete molecu- lar units, we always use their empirical formulas in chemical equations. The empiri- cal formulas are the same as the symbols that represent the elements. For example, the empirical formula for iron is Fe, the same as the symbol for the element. For nonmetals there is no single rule. Carbon, for example, exists as an extensive three-dimensional network of atoms, and so we use its empirical formula (C) to rep- resent elemental carbon in chemical equations. But hydrogen, nitrogen, oxygen, and the halogens exist as diatomic molecules, and so we use their molecular formulas (H2, N2, O2, F2, Cl2, Br2, I2) in equations. The stable form of phosphorus is molecular (P4), and so we use P4. For sulfur, chemists often use the empirical formula (S) in chemi- cal equations, rather than S8, which is the stable form. Thus, instead of writing the equation for the combustion of sulfur as Note that these two equations for the S8 (s) 1 8O2 (g) ¡ 8SO2 (g) combustion of sulfur have identical stoichiometry. This correspondence should not be surprising, because both we usually write equations describe the same chemical system. In both cases, a number of sulfur S(s) 1 O2 (g) ¡ SO2 (g) atoms react with twice as many oxygen atoms. All the noble gases are monatomic species; thus we use their symbols: He, Ne, Ar, Kr, Xe, and Rn. The metalloids, like the metals, all have complex three-dimensional networks, and we represent them, too, with their empirical formulas, that is, their symbols: B, Si, Ge, and so on. Electron Configurations of Cations and Anions Because many ionic compounds are made up of monatomic anions and cations, it is helpful to know how to write the electron configurations of these ionic species. Just as for neutral atoms, we use the Pauli exclusion principle and Hund’s rule in writing the ground-state electron configurations of cations and anions. We will group the ions in two categories for discussion. Ions Derived from Representative Elements Ions formed from atoms of most representative elements have the noble-gas outer- electron configuration of ns2np6. In the formation of a cation from the atom of a representative element, one or more electrons are removed from the highest occupied  8.3 Periodic Variation in Physical Properties 333 n shell. The electron configurations of some atoms and their corresponding cations are as follows: Na: [Ne]3s1 Na1: [Ne] Ca: [Ar]4s2 Ca21: [Ar] Al: [Ne]3s23p1 Al31: [Ne] Note that each ion has a stable noble gas configuration. In the formation of an anion, one or more electrons are added to the highest partially filled n shell. Consider the following examples: H: 1s1 H2: 1s2 or [He] F: 1s22s22p5 F2: 1s22s22p6 or [Ne] O: 1s22s22p4 O22: 1s22s22p6 or [Ne] N: 1s22s22p3 N32: 1s22s22p6 or [Ne] All of these anions also have stable noble gas configurations. Notice that F2, Na1, and Ne (and Al31, O22, and N32) have the same electron configuration. They are said to be isoelectronic because they have the same number of electrons, and hence the same ground-state electron configuration. Thus, H2 and He are also isoelectronic. Cations Derived from Transition Metals In Section 7.9 we saw that in the first-row transition metals (Sc to Cu), the 4s orbital is always filled before the 3d orbitals. Consider manganese, whose electron configu- ration is [Ar]4s23d5. When the Mn21 ion is formed, we might expect the two electrons to be removed from the 3d orbitals to yield [Ar]4s23d3. In fact, the electron con- figuration of Mn21 is [Ar]3d5! The reason is that the electron-electron and electron- Bear in mind that the order of electron filling does not determine or predict the nucleus interactions in a neutral atom can be quite different from those in its ion. order of electron removal for transition Thus, whereas the 4s orbital is always filled before the 3d orbital in Mn, electrons metals. For these metals, the ns electrons are removed from the 4s orbital in forming Mn21 because the 3d orbital is more are lost before the (n 2 1)d electrons. stable than the 4s orbital in transition metal ions. Therefore, when a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n 2 1)d orbitals. Keep in mind that most transition metals can form more than one cation and that frequently the cations are not isoelectronic with the preceding noble gases. Review of Concepts Identify the elements that fit the following descriptions: (a) An alkaline earth metal ion that is isoelectronic with Kr. (b) An anion with a 23 charge that is isoelectronic with K1. (c) An ion with a 12 charge that is isoelectronic with Co31. 8.3 Periodic Variation in Physical Properties As we have seen, the electron configurations of the elements show a periodic variation with increasing atomic number. Consequently, there are also periodic variations in physical and chemical behavior. In this section and the next two, we will examine some physical properties of elements that are in the same group or period and addi- tional properties that influence the chemical behavior of the elements. First, let’s look at the concept of effective nuclear charge, which has a direct bearing on many atomic properties. 334 Chapter 8 ■ Periodic Relationships Among the Elements Effective Nuclear Charge In Chapter 7 we discussed the shielding effect that electrons close to the nucleus have on outer-shell electrons in many-electron atoms. The presence of other elec- trons in an atom reduces the electrostatic attraction between a given electron and 1A 8A the positively charged protons in the nucleus. The effective nuclear charge (Zeff) is 2A 3A 4A 5A 6A 7A the nuclear charge felt by an electron when both the actual nuclear charge (Z) and the repulsive effects (shielding) of the other electrons are taken into account. In general, Zeff is given by The increase in effective nuclear Zeff 5 Z 2 σ (8.2) charge from left to right across a period and from top to bottom in a group for representative elements. where σ (sigma) is called the shielding constant (also called the screening constant). The shielding constant is greater than zero but smaller than Z. One way to illustrate how electrons in an atom shield one another is to con- sider the amounts of energy required to remove the two electrons from a helium atom. Experiments show that it takes 3.94 3 10218 J to remove the first electron and 8.72 3 10218 J to remove the second electron. There is no shielding once the first electron is removed, so the second electron feels the full effect of the 12 nuclear charge. ⫺1 ⫺1 ⫺1 It takes 8.72 ⫻ 10⫺18 J to remove the It takes second electron 3.94 ⫻ 10⫺18 J ⫹2 to remove the ⫹2 first electron Because the core electrons are, on average, closer to the nucleus than valence electrons, core electrons shield valence electrons much more than valence electrons shield one another. Consider the second-period elements from Li to Ne. Moving See Figure 7.27 for radial probability from left to right, we find the number of core electrons (1s2) remains constant while plots of 1s and 2s orbitals. the nuclear charge increases. However, because the added electron is a valence electron and valence electrons do not shield each other well, the net effect of mov- ing across the period is a greater effective nuclear charge felt by the valence elec- trons, as shown here. Li Be B C N O F Ne Z 3 4 5 6 7 8 9 10 Zeff 1.28 1.91 2.42 3.14 3.83 4.45 5.10 5.76 The attractive force between the The effective nuclear charge also increases as we go down a particular periodic nucleus and a particular electron is directly proportional to the effective group. However, because the valence electrons are now added to increasingly large nuclear charge and inversely proportional shells as n increases, the electrostatic attraction between the nucleus and the valence to the square of the distance of separation. electrons actually decreases. Li Na K Rb Cs Z 3 11 19 37 55 Zeff 1.28 2.51 3.50 4.98 6.36  8.3 Periodic Variation in Physical Properties 335 Atomic Radius A number of physical properties, including density, melting point, and boiling Animation Atomic and Ionic Radius point, are related to the sizes of atoms, but atomic size is difficult to define. As we saw in Chapter 7, the electron density in an atom extends far beyond the nucleus, but we normally think of atomic size as the volume containing about 90 percent of the total electron density around the nucleus. When we must be even more specific, we define the size of an atom in terms of its atomic radius, which is one-half the distance between the two nuclei in two adjacent metal atoms or in a diatomic molecule. For atoms linked together to form an extensive three-dimensional network, atomic radius is simply one-half the distance between the nuclei in two neighboring atoms [Figure 8.4(a)]. For elements that exist as simple diatomic molecules, the atomic radius is one-half the distance between the nuclei of the two atoms in a particular (a) molecule [Figure 8.4(b)]. Figure 8.5 shows the atomic radius of many elements according to their posi- tions in the periodic table, and Figure 8.6 plots the atomic radii of these elements against their atomic numbers. Periodic trends are clearly evident. Consider the second-period elements. Because the effective nuclear charge increases from left to right, the added valence electron at each step is more strongly attracted by the nucleus than the one before. Therefore, we expect and indeed find the atomic radius (b) Figure 8.4 (a) In metals such as polonium, the atomic radius is defined as one-half the distance Increasing atomic radius between the centers of two adjacent atoms. (b) For elements that exist as diatomic molecules, 1A 2A 3A 4A 5A 6A 7A 8A such as iodine, the radius of the atom is defined as one-half the H He distance between the centers of the atoms in the molecule. 37 31 B C N O F Ne Li Be 152 112 85 77 75 73 72 70 Na Mg Al Si P S Cl Ar Increasing atomic radius 186 160 143 118 110 103 99 98 K Ca Ga Ge As Se Br Kr 227 197 135 123 120 117 114 112 Rb Sr In Sn Sb Te I Xe 248 215 166 140 141 143 133 131 Cs Ba Tl Pb Bi Po At Rn 265 222 171 175 155 164 142 140 Figure 8.5 Atomic radii (in picometers) of representative elements according to their positions in the periodic table. Note that there is no general agreement on the size of atomic radii. We focus only on the trends in atomic radii, not on their precise values. 336 Chapter 8 ■ Periodic Relationships Among the Elements 300 Cs 250 Rb K 200 Na Atomic radius (pm) Li Po 150 I Br 100 Cl F 50 0 10 20 30 40 50 60 70 80 90 Atomic number Figure 8.6 Plot of atomic radii (in picometers) of elements against their atomic numbers. decreases from Li to Ne. Within a group we find that atomic radius increases with atomic number. For the alkali metals in Group 1A, the valence electron resides in the ns orbital. Because orbital size increases with the increasing principal quantum number n, the size of the atomic radius increases even though the effective nuclear charge also increases from Li to Cs. 1A 8A 2A 3A 4A 5A 6A 7A N Example 8.2 Si P Referring to a periodic table, arrange the following atoms in order of increasing atomic radius: P, Si, N. Strategy What are the trends in atomic radii in a periodic group and in a particular period? Which of the preceding elements are in the same group? in the same period? Solution From Figure 8.1 we see that N and P are in the same group (Group 5A). Therefore, the radius of N is smaller than that of P (atomic radius increases as we go down a group). Both Si and P are in the third period, and Si is to the left of P. Therefore, the radius of P is smaller than that of Si (atomic radius decreases as we move from left to right across a period). Thus, the order of increasing radius is Similar problems: 8.37, 8.38. N , P , Si . Practice Exercise Arrange the following atoms in order of decreasing radius: C, Li, Be. Review of Concepts Compare the size of each pair of atoms listed here: (a) Be, Ba; (b) Al, S; (c) 12C, 13C.  8.3 Periodic Variation in Physical Properties 337 Ionic Radius Ionic radius is the radius of a cation or an anion. It can be measured by X-ray dif- fraction (see Chapter 11). Ionic radius affects the physical and chemical properties of an ionic compound. For example, the three-dimensional structure of an ionic com- pound depends on the relative sizes of its cations and anions. When a neutral atom is converted to an ion, we expect a change in size. If the atom forms an anion, its size (or radius) increases, because the nuclear charge remains the same but the repulsion resulting from the additional electron(s) enlarges the domain of the electron cloud. On the other hand, removing one or more elec- trons from an atom reduces electron-electron repulsion but the nuclear charge remains the same, so the electron cloud shrinks, and the cation is smaller than the atom. Figure 8.7 shows the changes in size that result when alkali metals are con- verted to cations and halogens are converted to anions; Figure 8.8 shows the changes in size that occur when a lithium atom reacts with a fluorine atom to form a LiF unit. Figure 8.9 shows the radii of ions derived from the familiar elements, arranged according to the elements’ positions in the periodic table. We can see parallel trends between atomic radii and ionic radii. For example, from top to bottom both the atomic radius and the ionic radius increase within a group. For ions derived from elements in different groups, a size comparison is meaningful only if the ions are isoelectronic. For isoelectronic ions, the size of the ion is based on the size of the electron cloud, not If we examine isoelectronic ions, we find that cations are smaller than anions. For on the number of protons in the nucleus. example, Na1 is smaller than F2. Both ions have the same number of electrons, but Na (Z 5 11) has more protons than F (Z 5 9). The larger effective nuclear charge of Na1 results in a smaller radius. 300 300 Figure 8.7 Comparison of atomic radii with ionic radii. Cs (a) Alkali metals and alkali metal cations. (b) Halogens 250 Rb 250 and halide ions. K I– 200 200 Br – Na Cl – Li Radius (pm) Radius (pm) Cs+ 150 150 Rb+ F– K+ I Br 100 100 Cl Na+ F 50 Li+ 50 0 10 20 30 40 50 60 0 10 20 30 40 50 60 Atomic number Atomic number (a) (b) Figure 8.8 Changes in the sizes of Li and F when they react to form LiF. + Li F Li + F– 338 Chapter 8 ■ Periodic Relationships Among the Elements Li+ Be2+ N3– O2– F– 78 34 171 140 133 Na+ Mg2+ Al3+ Fe3+ Fe2+ Cu2+ Cu+ 57 S2– Cl– 98 78 Ti3+ 3+ Cr Ni 2+ Sc3+ V5+ Mn2+ Co 2+ Zn 2+ Ga3+ 184 181 K+ Ca2+ 133 106 83 68 59 64 91 67 82 78 72 83 62 82 96 Sb5+ Se2– Br– In3+ Sn4+ Rb+ Sr2+ Ag+ Cd2+ 198 195 148 127 113 103 92 74 62 Pb4+ Te2– I– Cs+ Ba2+ Au+ Hg2+ Tl3+ 165 143 137 112 105 84 211 220 Figure 8.9 The radii (in picometers) of ions of familiar elements arranged according to the elements’ positions in the periodic table. Focusing on isoelectronic cations, we see that the radii of tripositive ions (ions that bear three positive charges) are smaller than those of dipositive ions (ions that bear two positive charges), which in turn are smaller than unipositive ions (ions that bear one positive charge). This trend is nicely illustrated by the sizes of three isoelec- tronic ions in the third period: Al31, Mg21, and Na1 (see Figure 8.9). The Al31 ion has the same number of electrons as Mg21, but it has one more proton. Thus, the electron cloud in Al31 is pulled inward more than that in Mg21. The smaller radius of Mg21 compared with that of Na1 can be similarly explained. Turning to isoelec- tronic anions, we find that the radius increases as we go from ions with uninegative charge (2) to those with dinegative charge (22), and so on. Thus, the oxide ion is larger than the fluoride ion because oxygen has one fewer proton than fluorine; the electron cloud is spread out more in O22. Example 8.3 For each of the following pairs, indicate which one of the two species is larger: (a) N32 or F2; (b) Mg21 or Ca21; (c) Fe21 or Fe31. Strategy In comparing ionic radii, it is useful to classify the ions into three categories: (1) isoelectronic ions, (2) ions that carry the same charges and are generated from atoms of the same periodic group, and (3) ions that carry different charges but are generated from the same atom. In case (1), ions that carry a greater negative charge are always larger; in case (2), ions from atoms having a greater atomic number are always larger; in case (3), ions having a smaller positive charge are always larger. (Continued)  8.3 Periodic Variation in Physical Properties 339 Solution (a) N32 and F2 are isoelectronic anions, both containing 10 electrons. Because N32 has only seven protons and F2 has nine, the smaller attraction exerted by the nucleus on the electrons results in a larger N32 ion. (b) Both Mg and Ca belong to Group 2A (the alkaline earth metals). Thus, Ca21 ion is larger than Mg21 because Ca’s valence electrons are in a larger shell (n 5 4) than are Mg’s (n 5 3). (c) Both ions have the same nuclear charge, but Fe21 has one more electron (24 electrons compared to 23 electrons for Fe31) and hence greater electron-electron repulsion. The radius of Fe21 is larger. Similar problems: 8.43, 8.45. 1 1 Practice Exercise Select the smaller ion in each of the following pairs: (a) K , Li ; (b) Au1, Au31; (c) P32, N32. Review of Concepts Identify the spheres shown here with each of the following: S22, Mg21, F2, Na1. Variation of Physical Properties Across a Period and Within a Group From left to right across a period there is a transition from metals to metalloids to nonmetals. Consider the third-period elements from sodium to argon (Figure 8.10). Sodium, the first element in the third period, is a very reactive metal, whereas chlo- rine, the second-to-last element of that period, is a very reactive nonmetal. In between, the elements show a gradual transition from metallic properties to nonmetallic proper- ties. Sodium, magnesium, and aluminum all have extensive three-dimensional atomic networks, which are held together by forces characteristic of the metallic state. Silicon is a metalloid; it has a giant three-dimensional structure in which the Si atoms are held together very strongly. Starting with phosphorus, the elements exist in simple, discrete molecular units (P4, S8, Cl2, and Ar) that have low melting points and boiling points. Within a periodic group the physical properties vary more predictably, especially if the elements are in the same physical state. For example, the melting points of argon and xenon are 2189.2°C and 2111.9°C, respectively. We can estimate the melting point of the intermediate element krypton by taking the average of these two values as follows: [(2189.2°C) 1 (2111.9°C)] melting point of Kr 5 5 2150.6°C 2 This value is quite close to the actual melting point of 2156.6°C. The Chemistry in Action essay on p. 341 illustrates one interesting application of periodic group properties. 340 Chapter 8 ■ Periodic Relationships Among the Elements Figure 8.10 The third-period elements. The photograph of argon, which is a colorless, odorless gas, shows the color emitted by the gas from a discharge tube. Sodium (Na) Magnesium (Mg) Aluminum (Al) Silicon (Si) Phosphorus (P4) Sulfur (S8) Chlorine (Cl2) Argon (Ar) 8.4 Ionization Energy Not only is there a correlation between electron configuration and physical prop- erties, but a close correlation also exists between electron configuration (a micro- scopic property) and chemical behavior (a macroscopic property). As we will see throughout this book, the chemical properties of any atom are determined by the configuration of the atom’s valence electrons. The stability of these outermost electrons is reflected directly in the atom’s ionization energies. Ionization energy (IE) is the minimum energy (in kJ/mol) required to remove an electron from a gaseous atom in its ground state. In other words, ionization energy is the amount of energy in kilojoules needed to strip 1 mole of electrons from 1 mole of gaseous atoms. Gaseous atoms are specified in this definition because an atom in the gas phase is virtually uninfluenced by its neighbors and so there are no intermolecu- lar forces (that is, forces between molecules) to take into account when measuring ionization energy. CHEMISTRY in Action The Third Liquid Element? O f the 118 known elements, 11 are gases under atmospheric conditions. Six of these are the Group 8A elements (the noble gases He, Ne, Ar, Kr, Xe, and Rn), and the other five are drops 81.4°; from sodium to potassium, 34.6°; from potassium to rubidium, 24°; from rubidium to cesium, 11°. On the basis of this trend, we can predict that the change from cesium to fran- hydrogen (H2), nitrogen (N2), oxygen (O2), fluorine (F2), and cium would be about 5°. If so, the melting point of francium chlorine (Cl2). Curiously, only two elements are liquids at 25°C: would be about 23°C, which would make it a liquid under atmo- mercury (Hg) and bromine (Br2). spheric conditions. We do not know the properties of all the known elements because some of them have never been prepared in quantities 180 large enough for investigation. In these cases, we must rely on Li periodic trends to predict their properties. What are the chances, 150 then, of discovering a third liquid element? Let us look at francium (Fr), the last member of Group 1A, Melting point (°C) to see if it might be a liquid at 25°C. All of francium’s isotopes 120 are radioactive. The most stable isotope is francium-223, which has a half-life of 21 minutes. (Half-life is the time it takes for Na 90 one-half of the nuclei in any given amount of a radioactive substance to disintegrate.) This short half-life means that only K 60 very small traces of francium could possibly exist on Earth. And although it is feasible to prepare francium in the laboratory, no Rb 30 Cs weighable quantity of the element has been prepared or isolated. Fr Thus, we know very little about francium’s physical and chemi- cal properties. Yet we can use the group periodic trends to predict some of those properties. 0 20 40 60 80 100 Take francium’s melting point as an example. The plot Atomic number shows how the melting points of the alkali metals vary with A plot of the melting points of the alkali metals versus their atomic numbers. atomic number. From lithium to sodium, the melting point By extrapolation, the melting point of francium should be 23°C. The magnitude of ionization energy is a measure of how “tightly” the electron is Note that while valence electrons are relatively easy to remove from the atom, held in the atom. The higher the ionization energy, the more difficult it is to remove core electrons are much harder to remove. the electron. For a many-electron atom, the amount of energy required to remove the Thus, there is a large jump in ionization energy between the last valence electron first electron from the atom in its ground state, and the first core electron. energy 1 X(g) ¡ X1 (g) 1 e2 (8.3) is called the first ionization energy (IE1). In Equation (8.3), X represents an atom of any element and e2 is an electron. The second ionization energy (IE2) and the third ionization energy (IE3) are shown in the following equations: energy 1 X1 (g) ¡ X21 (g) 1 e2    second ionization energy 1 X21 (g) ¡ X31 (g) 1 e2    third ionization The pattern continues for the removal of subsequent electrons. When an electron is removed from an atom, the repulsion among the remaining electrons decreases. Because the nuclear charge remains constant, more energy is 341 342 Chapter 8 ■ Periodic Relationships Among the Elements Table 8.2 The Ionization Energies (kJ/mol) of the First 20 Elements Z Element First Second Third Fourth Fifth Sixth 1 H 1,312 2 He 2,373 5,251 3 Li 520 7,300 11,815 4 Be 899 1,757 14,850 21,005 5 B 801 2,430 3,660 25,000 32,820 6 C 1,086 2,350 4,620 6,220 38,000 47,261 7 N 1,400 2,860 4,580 7,500 9,400 53,000 8 O 1,314 3,390 5,300 7,470 11,000 13,000 9 F 1,680 3,370 6,050 8,400 11,000 15,200 10 Ne 2,080 3,950 6,120 9,370 12,200 15,000 11 Na 495.9 4,560 6,900 9,540 13,400 16,600 12 Mg 738.1 1,450 7,730 10,500 13,600 18,000 13 Al 577.9 1,820 2,750 11,600 14,800 18,400 14 Si 786.3 1,580 3,230 4,360 16,000 20,000 15 P 1,012 1,904 2,910 4,960 6,240 21,000 16 S 999.5 2,250 3,360 4,660 6,990 8,500 17 Cl 1,251 2,297 3,820 5,160 6,540 9,300 18 Ar 1,521 2,666 3,900 5,770 7,240 8,800 19 K 418.7 3,052 4,410 5,900 8,000 9,600 20 Ca 589.5 1,145 4,900 6,500 8,100 11,000 1A 8A needed to remove another electron from the positively charged ion. Thus, ionization 2A 3A 4A 5A 6A 7A energies always increase in the following order: IE1 , IE2 , IE3 , . . . The increase in first ionization Table 8.2 lists the ionization energies of the first 20 elements. Ionization is always an energy from left to right across a endothermic process. By convention, energy absorbed by atoms (or ions) in the ioniza- period and from bottom to top in a tion process has a positive value. Thus, ionization energies are all positive quantities. group for representative elements. Figure 8.11 shows the variation of the first ionization energy with atomic number. The plot clearly exhibits the periodicity in the stability of the most loosely held elec- tron. Note that, apart from small irregularities, the first ionization energies of elements in a period increase with increasing atomic number. This trend is due to the increase in effective nuclear charge from left to right (as in the case of atomic radii variation). A larger effective nuclear charge means a more tightly held valence electron, and hence a higher first ionization energy. A notable feature of Figure 8.11 is the peaks, which correspond to the noble gases. We tend to associate full valence-shell electron configurations with an inherent degree of chemical stability. The high ionization ener- gies of the noble gases, stemming from their large effective nuclear charge, comprise one of the reasons for this stability. In fact, helium (1s2) has the highest first ioniza- tion energy of all the elements. At the bottom of the graph in Figure 8.11 are the Group 1A elements (the alkali metals), which have the lowest first ionization energies. Each of these metals has one valence electron (the outermost electron configuration is ns1), which is effectively shielded by the completely filled inner shells. Consequently, it is ener- getically easy to remove an electron from the atom of an alkali metal to form a  8.4 Ionization Energy 343 2500 He Ne 2000 First ionization energy (kJ/mol) Ar 1500 Kr Xe H Rn 1000 500 Li Na K Rb Cs 0 10 20 30 40 50 60 70 80 90 Atomic number (Z) Figure 8.11 Variation of the first ionization energy with atomic number. Note that the noble gases have high ionization energies, whereas the alkali metals and alkaline earth metals have low ionization energies. unipositive ion (Li1, Na1, K1, . . .). Significantly, the electron configurations of these cations are isoelectronic with those noble gases just preceding them in the periodic table. The Group 2A elements (the alkaline earth metals) have higher first ionization energies than the alkali metals do. The alkaline earth metals have two valence electrons (the outermost electron configuration is ns2). Because these two s elec- trons do not shield each other well, the effective nuclear charge for an alkaline earth metal atom is larger than that for the preceding alkali metal. Most alkaline earth compounds contain dipositive ions (Mg21, Ca21, Sr21, Ba21). The Be21 ion is isoelectronic with Li1 and with He, Mg21 is isoelectronic with Na1 and with Ne, and so on. As Figure 8.11 shows, metals have relatively low ionization energies compared to nonmetals. The ionization energies of the metalloids generally fall between those of metals and nonmetals. The difference in ionization energies suggests why metals always form cations and nonmetals form anions in ionic compounds. (The only important nonmetallic cation is the ammonium ion, NH14 .) For a given group, ion- ization energy decreases with increasing atomic number (that is, as we move down the group). Elements in the same group have similar outer electron configurations. However, as the principal quantum number n increases, so does the average distance of a valence electron from the nucleus. A greater separation between the electron and the nucleus means a weaker attraction, so that it becomes easier to remove the first electron as we go from element to element down a group even though the effective nuclear charge also increases in the same direction. Thus, the metallic character of the elements within a group increases from top to bottom. This trend is particularly noticeable for elements in Groups 3A to 7A. For example, in Group 4A, carbon is a nonmetal, silicon and germanium are metalloids, and tin and lead are metals. Although the general trend in the periodic table is for first ionization energies to increase from left to right, some irregularities do exist. The first exception occurs between Group 2A and 3A elements in the same period (for example, between Be 344 Chapter 8 ■ Periodic Relationships Among the Elements and B and between Mg and Al). The Group 3A elements have lower first ionization energies than 2A elements because they all have a single electron in the outermost p subshell (ns2np1), which is well shielded by the inner electrons and the ns2 electrons. Therefore, less energy is needed to remove a single p electron than to remove an s electron from the same principal energy level. The second irregularity occurs between Groups 5A and 6A (for example, between N and O and between P and S). In the Group 5A elements (ns2np3), the p electrons are in three separate orbitals according to Hund’s rule. In Group 6A (ns2np4), the additional electron must be paired with one of the three p electrons. The proximity of two electrons in the same orbital results in greater electrostatic repulsion, which makes it easier to ionize an atom of the Group 6A element, even though the nuclear charge has increased by one unit. Thus, the ionization energies for Group 6A elements are lower than those for Group 5A ele- ments in the same period. Example 8.4 compares the ionization energies of some elements. 1A 8A 2A 3A 4A 5A 6A 7A Li Be O Example 8.4 S (a) Which atom should have a lower first ionization energy: oxygen or sulfur? (b) Which atom should have a higher second ionization energy: lithium or beryllium? Strategy (a) First ionization energy decreases as we go down a group because the outermost electron is farther away from the nucleus and feels less attraction. (b) Removal of the outermost electron requires less energy if it is shielded by a filled inner shell. Solution (a) Oxygen and sulfur are members of Group 6A. They have the same valence electron configuration (ns2np4), but the 3p electron in sulfur is farther from the nucleus and experiences less nuclear attraction than the 2p electron in oxygen. Thus, we predict that sulfur should have a smaller first ionization energy. (b) The electron configurations of Li and Be are 1s22s1 and 1s22s2, respectively. The second ionization energy is the minimum energy required to remove an electron from a gaseous unipositive ion in its ground state. For the second ionization process, we write Li1 (g) ¡ Li21 (g) 1 e2 1s2 1s1 Be (g) ¡ Be21 (g) 1 e2 1 1s22s1 1s2 Because 1s electrons shield 2s electrons much more effectively than they shield each other, we predict that it should be easier to remove a 2s electron from Be1 than to remove a 1s electron from Li1. Check Compare your result with the data shown in Table 8.2. In (a), is your prediction consistent with the fact that the metallic character of the elements increases as we move down a periodic group? In (b), does your prediction account for the fact that alkali metals form 11 ions while alkaline earth metals form Similar problem: 8.55. 12 ions? Practice Exercise (a) Which of the following atoms should have a larger first ionization energy: N or P? (b) Which of the following atoms should have a smaller second ionization energy: Na or Mg?  8.5 Electron Affinity 345 Review of Concepts Label the plots shown here for the first, second, and third ionization energies for Mg, Al, and K. Ionization energy 1 2 3 Number of electrons removed 8.5 Electron Affinity Another property that greatly influences the chemical behavior of atoms is their abil- ity to accept one or more electrons. This property is called electron affinity (EA), which is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion. X(g) 1 e2 ¡ X2 (g) (8.4) Consider the process in which a gaseous fluorine atom accepts an electron: F(g) 1 e2 ¡ F2 (g) ¢H 5 2328 kJ/mol The electron affinity of fluorine is therefore assigned a value of 1328 kJ/mol. The Electron affinity is positive if the reaction is exothermic and negative if the reaction more positive is the electron affinity of an element, the greater is the affinity of an is endothermic. This convention is used atom of the element to accept an electron. Another way of viewing electron affinity in inorganic and physical chemistry texts. is to think of it as the energy that must be supplied to remove an electron from the anion. For fluorine, we write F2 (g) ¡ F(g) 1 e2 ¢H 5 1328 kJ/mol Thus, a large positive electron affinity means that the negative ion is very stable (that is, the atom has a great tendency to accept an electron), just as a high ionization energy of an atom means that the electron in the atom is very stable. Experimentally, electron affinity is determined by removing the additional elec- tron from an anion. In contrast to ionization energies, however, electron affinities are difficult to measure because the anions of many elements are unstable. Table 8.3 shows the electron affinities of some representative elements and the noble gases, and Figure 8.12 plots the electron affinities of the first 56 elements versus atomic number. The overall trend is an increase in the tendency to accept electrons (electron affinity values become more positive) from left to right across a period. The electron affinities of metals are generally lower than those of nonmetals. The values vary little within a given group. The halogens (Group 7A) have the highest electron affinity values. 346 Chapter 8 ■ Periodic Relationships Among the Elements Electron Affinities (kJ/mol) of Some Representative Elements Table 8.3 and the Noble Gases* 1A 2A 3A 4A 5A 6A 7A 8A H He 73 , 0 Li Be B C N O F Ne 60 #0 27 122 0 141 328 , 0 Na Mg Al Si P S Cl Ar 53 #0 44 134 72 200 349 , 0 K Ca Ga Ge As Se Br Kr 48 2.4 29 118 77 195 325 , 0 Rb Sr In Sn Sb Te I Xe 47 4.7 29 121 101 190 295 , 0 Cs Ba Tl Pb Bi Po At Rn 45 14 30 110 110 ? ? , 0 *The electron affinities of the noble gases, Be, and Mg have not been determined experimentally, but are believed to be close to zero or negative. There is a general correlation between electron affinity and effective nuclear charge, which also increases from left to right in a given period (see p. 334). How- ever, as in the case of ionization energies, there are some irregularities. For example, the electron affinity of a Group 2A element is lower than that for the corresponding Group 1A element, and the electron affinity of a Group 5A element is lower than that for the corresponding Group 4A element. These exceptions are due to the valence electron configurations of the elements involved. An electron added to a Figure 8.12 A plot of electron 400 affinity against atomic number from hydrogen to barium. Cl F Br 300 I Electron affinity (kJ/mol) S 200 Se Te O Si Ge Sn C 100 Sb H Li As Na P K Rb Cs Al Ga In Sr Ba 0 10 20 30 40 50 60 Atomic number (Z)  8.6 Variation in Chemical Properties of the Representative Elements 347 Group 2A element must end up in a higher-energy np orbital, where it is effectively shielded by the ns2 electrons and therefore experiences a weaker attraction to the nucleus. Therefore, it has a lower electron affinity than the corresponding Group 1A element. Likewise, it is harder to add an electron to a Group 5A element (ns2np3) than to the corresponding Group 4A element (ns2np2) because the electron added to the Group 5A element must be placed in a np orbital that already contains an elec- tron and will therefore experience a greater electrostatic repulsion. Finally, in spite of the fact that noble gases have high effective nuclear charge, they have extremely low electron affinities (zero or negative values). The reason is that an electron added to an atom with an ns2np6 configuration has to enter an (n 1 1)s orbital, where it The variation in electron affinities from top to bottom within a group is is well shielded by the core electrons and will only be very weakly attracted by the much less regular (see Table 8.3). nucleus. This analysis also explains why species with complete valence shells tend to be chemically stable. Example 8.5 shows why the alkaline earth metals do not have a great tendency to accept electrons. 1A 8A 2A 3A 4A 5A 6A 7A Example 8.5 Be Mg Ca Why are the electron affinities of the alkaline earth metals, shown in Table 8.3, either Sr Ba negative or small positive values? Strategy What are the electron configurations of alkaline earth metals? Would the added electron to such an atom be held strongly by the nucleus? Solution The valence electron configuration of the alkaline earth metals is ns2, where n is the highest principal quantum number. For the process M(g) 1 e2 ¡ M2 (g) ns2 ns2np1 where M denotes a member of the Group 2A family, the extra electron must enter the np subshell, which is effectively shielded by the two ns electrons (the ns electrons are more penetrating than the np electrons) and the inner electrons. Consequently, alkaline earth metals have little tendency to pick up an extra electron. Similar problem: 8.63. 2 Practice Exercise Is it likely that Ar will form the anion Ar ? Review of Concepts Why is it possible to measure the successive ionization energies of an atom until all the electrons are removed, but it becomes increasingly difficult and often impossible to measure the electron affinity of an atom beyond the first stage? 8.6 Variation in Chemical Properties of the Representative Elements Ionization energy and electron affinity help chemists understand the types of reactions that elements undergo and the nature of the elements’ compounds. On a conceptual level, these two measures are related in a simple way: Ionization energy measures the attraction of an atom for its own electrons, whereas electron affinity expresses the 348 Chapter 8 ■ Periodic Relationships Among the Elements attraction of an atom for an additional electron from some other source. Together they give us insight into the general attraction of an atom for electrons. With these con- cepts we can survey the chemical behavior of the elements systematically, paying particular attention to the relationship between their chemical properties and their electron configurations. We have seen that the metallic character of the elements decreases from left to right across a period and increases from top to bottom within a group. On the basis of these trends and the knowledge that metals usually have low ionization energies while nonmetals usually have high electron affinities, we can frequently predict the outcome of a reaction involving some of these elements. General Trends in Chemical Properties Before we study the elements in individual groups, let us look at some overall trends. We have said that elements in the same group resemble one another in chemical behavior because they have similar valence electron configurations. This statement, although correct in the general sense, must be applied with caution. Chemists have long known that the first member of each group (the element in the second period from lithium to fluorine) differs from the rest of the members of the same group. Lithium, for example, exhibits many, but not all, of the properties characteristic of the alkali metals. Similarly, beryllium is a somewhat atypical member of Group 2A, and so on. The difference can be attributed to the unusually small size of the first element in each group (see Figure 8.5). Another trend in the chemical behavior of the representative elements is the diagonal relationship. Diagonal relationships are similarities between pairs of ele- ments in different groups and periods of the periodic table. Specifically, the first three members of the second period (Li, Be, and B) exhibit many similarities to those ele- ments located diagonally below them in the periodic table (Figure 8.13). The reason for this phenomenon is the closeness of the charge densities of their cations. (Charge density is the charge of an ion divided by its volume.) Cations with comparable charge densities react similarly with anions and therefore form the same type of compounds. Thus, the chemistry of lithium resembles that of magnesium in some ways; the same holds for beryllium and aluminum and for boron and silicon. Each of these pairs is said to exhibit a diagonal relationship. We will see a number of examples of this relationship later. Bear in mind that a comparison of the properties of elements in the same group is most valid if we are dealing with elements of the same type with respect to their metallic character. This guideline applies to the elements in Groups 1A and 2A, which are all metals, and to the elements in Groups 7A and 8A, which are all nonmet- als. In Groups 3A through 6A, where the elements change either from nonmetals to metals or from nonmetals to metalloids, it is natural to expect greater variation in chemical properties even though the members of the same group have similar outer electron configurations. Now let us take a closer look at the chemical properties of the representative elements and the noble gases. (We will consider the chemistry of the transition met- als in Chapter 23.) 1A 2A 3A 4A Hydrogen (1s1) Li Be B C There is no totally suitable position for hydrogen in the periodic table. Traditionally hydrogen is shown in Group 1A, but it really could be a class by itself. Like the Na Mg Al Si alkali metals, it has a single s valence electron and forms a unipositive ion (H1), which is hydrated in solution. On the other hand, hydrogen also forms the hydride Figure 8.13 Diagonal ion (H2) in ionic compounds such as NaH and CaH2. In this respect, hydrogen relationships in the periodic table. resembles the halogens, all of which form uninegative ions (F2, Cl2, Br2, and I2)  8.6 Variation in Chemical Properties of the Representative Elements 349 in ionic compounds. Ionic hydrides react with water to produce hydrogen gas and the corresponding metal hydroxides: 2NaH(s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2 (g) CaH2 (s) 1 2H2O(l) ¡ Ca(OH) 2 (s) 1 2H2 (g) Of course, the most important compound of hydrogen is water, which forms when hydrogen burns in air: 2H2 (g) 1 O2 (g) ¡ 2H2O(l) Group 1A Elements (ns1, n $ 2) 1A 2A 3A 4A 5A 6A 7A 8A Li Figure 8.14 shows the Group 1A elements, the alkali metals. All of these elements Na K have low ionization energies and therefore a great tendency to lose the single Rb Cs valence electron. In fact, in the vast majority of their compounds they are uni- positive ions. These metals are so reactive that they are never found in the pure state in nature. They react with water to produce hydrogen gas and the correspond- ing metal hydroxide: 2M(s) 1 2H2O(l) ¡ 2MOH(aq) 1 H2 (g) where M denotes an alkali metal. When exposed to air, they gradually lose their shiny appearance as they combine with oxygen gas to form oxides. Lithium forms lithium oxide (containing the O22 ion): 4Li(s) 1 O2 (g) ¡ 2Li2O(s) The other alkali metals all form oxides and peroxides (containing the O222 ion). For example, 2Na(s) 1 O2 (g) ¡ Na2O2 (s) Figure 8.14 The Group 1A elements: the alkali metals. Francium (not shown) is radioactive. Lithium (Li) Sodium (Na) Potassium (K) Rubidium (Rb) Cesium (Cs) 350 Chapter 8 ■ Periodic Relationships Among the Elements Potassium, rubidium, and cesium also form superoxides (containing the O22 ion): K(s) 1 O2 (g) ¡ KO2 (s) The reason that different types of oxides are formed when alkali metals react with oxygen has to do with the stability of the oxides in the solid state. Because these oxides are all ionic compounds, their stability depends on how strongly the cations and anions attract one another. Lithium tends to form predominantly lithium oxide because this compound is more stable than lithium peroxide. The formation of other alkali metal oxides can be explained similarly. 1A 2A 3A 4A 5A 6A 7A 8A Group 2A Elements (ns2, n $ 2) Be Mg Figure 8.15 shows the Group 2A elements. As a group, the alkaline earth metals are Ca Sr somewhat less reactive than the alkali metals. Both the first and the second ionization Ba energies decrease from beryllium to barium. Thus, the tendency is to form M21 ions (where M denotes an alkaline earth metal atom), and hence the metallic character increases from top to bottom. Most beryllium compounds (BeH2 and beryllium halides, such as BeCl2) and some magnesium compounds (MgH2, for example) are molecular rather than ionic in nature. The reactivities of alkaline earth metals with water vary quite markedly. Beryllium does not react with water; magnesium reacts slowly with steam; calcium, strontium, and barium are reactive enough to attack cold water: Ba(s) 1 2H2O(l) ¡ Ba(OH) 2 (aq) 1 H2 (g) The reactivities of the alkaline earth metals toward oxygen also increase from Be to Ba. Beryllium and magnesium form oxides (BeO and MgO) only at elevated tem- peratures, whereas CaO, SrO, and BaO form at room temperature. Magnesium reacts with acids in aqueous solution, liberating hydrogen gas: Mg(s) 1 2H1 (aq) ¡ Mg21 (aq) 1 H2 (g) Beryllium (Be) Magnesium (Mg) Calcium (Ca) Strontium (Sr) Barium (Ba) Radium (Ra) Figure 8.15 The Group 2A elements: the alkaline earth metals.  8.6 Variation in Chemical Properties of the Representative Elements 351 Calcium, strontium, and barium also react with aqueous acid solutions to produce hydrogen gas. However, because these metals also attack water, two different reactions will occur simultaneously. The chemical properties of calcium and strontium provide an interesting exam- ple of periodic group similarity. Strontium-90, a radioactive isotope, is a major product of an atomic bomb explosion. If an atomic bomb is exploded in the atmo- sphere, the strontium-90 formed will eventually settle on land and water, and it will reach our bodies via a relatively short food chain. For example, if cows eat con- taminated grass and drink contaminated water, they will pass along strontium-90 in their milk. Because calcium and strontium are chemically similar, Sr21 ions can replace Ca21 ions in our bones. Constant exposure of the body to the high-energy radiation emitted by the strontium-90 isotopes can lead to anemia, leukemia, and other chronic illnesses. Group 3A Elements (ns2np1, n $ 2) 1A 2A 3A 4A 5A 6A 7A 8A B The first member of Group 3A, boron, is a metalloid; the rest are metals (Figure 8.16). Al Ga Boron does not form binary ionic compounds and is unreactive toward oxygen gas In Tl and water. The next element, aluminum, readily forms aluminum oxide when exposed to air: 4Al(s) 1 3O2 (g) ¡ 2Al2O3 (s) Aluminum that has a protective coating of aluminum oxide is less reactive than ele- mental aluminum. Aluminum forms only tripositive ions. It reacts with hydrochloric acid as follows: 2Al(s) 1 6H1 (aq) ¡ 2Al31 (aq) 1 3H2 (g) The other Group 3A metallic elements form both unipositive and tripositive ions. Moving down the group, we find that the unipositive ion becomes more stable than the tripositive ion. Figure 8.16 The Group 3A elements. The low melting point of gallium (29.8°C) causes it to melt when held in hand. Boron (B) Aluminum (Al) Gallium (Ga) Indium (In) 352 Chapter 8 ■ Periodic Relationships Among the Elements Carbon (graphite) Carbon (diamond) Silicon (Si) Germanium (Ge) Tin (Sn) Lead (Pb) Figure 8.17 The Group 4A elements. The metallic elements in Group 3A also form many molecular compounds. For example, aluminum reacts with hydrogen to form AlH3, which resembles BeH2 in its properties. (Here is an example of the diagonal relationship.) Thus, from left to right across the periodic table, we are seeing a gradual shift from metallic to nonmetallic character in the representative elements. 1A 2A 3A 4A 5A 6A 7A 8A Group 4A Elements (ns2np2, n $ 2) C Si The first member of Group 4A, carbon, is a nonmetal, and the next two members, Ge Sn silicon and germanium, are metalloids (Figure 8.17). The metallic elements of this Pb group, tin and lead, do not react with water, but they do react with acids (hydrochlo- ric acid, for example) to liberate hydrogen gas: Sn(s) 1 2H1 (aq) ¡ Sn21 (aq) 1 H2 (g) Pb(s) 1 2H1 (aq) ¡ Pb21 (aq) 1 H2 (g) The Group 4A elements form compounds in both the 12 and 14 oxidation states. For carbon and silicon, the 14 oxidation state is the more stable one. For example, CO2 is more stable than CO, and SiO2 is a stable compound, but SiO does not exist under normal conditions. As we move down the group, however, the trend in stability is reversed. In tin compounds the 14 oxidation state is only slightly more stable than the 12 oxidation state. In lead compounds the 12 oxidation state is unquestionably the more stable one. The outer electron configuration of lead is 6s26p2, and lead tends to lose only the 6p electrons (to form Pb21) rather than both the 6p and 6s electrons (to form Pb41). 1A 8A Group 5A Elements (ns2np3, n $ 2) 2A 3A 4A 5A 6A 7A N P In Group 5A, nitrogen and phosphorus are nonmetals, arsenic and antimony are met- As alloids, and bismuth is a metal (Figure 8.18). Thus, we expect a greater variation in Sb Bi properties within the group.  8.6 Variation in Chemical Properties of the Representative Elements 353 Figure 8.18 The Group 5A elements. Molecular nitrogen is a colorless, odorless gas. Liquid nitrogen (N2) White and red phosphorus (P) Arsenic (As) Antimony (Sb) Bismuth (Bi) Elemental nitrogen is a diatomic gas (N2). It forms a number of oxides (NO, N2O, NO2, N2O4, and N2O5), of which only N2O5 is a solid; the others are gases. Nitrogen has a tendency to accept three electrons to form the nitride ion, N32 (thus achieving the electron configuration 1s22s22p6, which is isoelectronic with neon). Most metallic nitrides (Li3N and Mg3N2, for example) are ionic compounds. Phos- phorus exists as P4 molecules. It forms two solid oxides with the formulas P4O6 and P4O10. The important oxoacids HNO3 and H3PO4 are formed when the following oxides react with water: N2O5 (s) 1 H2O(l) ¡ 2HNO3 (aq) P4O10 (s) 1 6H2O(l) ¡ 4H3PO4 (aq) Arsenic, antimony, and bismuth have extensive three-dimensional structures. Bismuth is a far less reactive metal than those in the preceding groups. Group 6A Elements (ns2np4, n $ 2) 1A 8A 2A 3A 4A 5A 6A 7A The first three members of Group 6A (oxygen, sulfur, and selenium) are nonmetals, O S and the last two (tellurium and polonium) are metalloids (Figure 8.19). Oxygen is Se Te a diatomic gas; elemental sulfur and selenium have the molecular formulas S8 and Po Se8, respectively; tellurium and polonium have more extensive three-dimensional structures. (Polonium, the last member, is a radioactive element that is difficult to study in the laboratory.) Oxygen has a tendency to accept two electrons to form the oxide ion (O22) in many ionic compounds. Sulfur, selenium, and tellurium also 354 Chapter 8 ■ Periodic Relationships Among the Elements Sulfur (S8) Selenium (Se8) Tellurium (Te) Figure 8.19 The Group 6A elements sulfur, selenium, and tellurium. Molecular oxygen is a colorless, odorless gas. Polonium (not shown) is radioactive. form dinegative anions (S22, Se22, and Te22). The elements in this group (espe- cially oxygen) form a large number of molecular compounds with nonmetals. The important compounds of sulfur are SO2, SO3, and H2S. The most important com- mercial sulfur compound is sulfuric acid, which is formed when sulfur trioxide reacts with water: SO3 (g) 1 H2O(l) ¡ H2SO4 (aq) 1A 2A 3A 4A 5A 6A 7A 8A Group 7A Elements (ns2np5, n $ 2) F Cl All the halogens are nonmetals with the general formula X2, where X denotes a Br I halogen element (Figure 8.20). Because of their great reactivity, the halogens are never At found in the elemental form in nature. (The last member of Group 7A, astatine, is a radioactive element. Little is known about its properties.) Fluorine is so reactive that it attacks water to generate oxygen: 2F2 (g) 1 2H2O(l) ¡ 4HF(aq) 1 O2 (g) Actually the reaction between molecular fluorine and water is quite complex; the products formed depend on reaction conditions. The reaction shown above is one of several possible chemical changes. The halogens have high ionization energies and large positive electron ­affinities. Anions derived from the halogens (F2, Cl2, Br2, and I2) are called halides. They are isoelectronic with the noble gases immediately to their right in Figure 8.20 The Group 7A elements chlorine, bromine, and iodine. Fluorine is a greenish- yellow gas that attacks ordinary glassware. Astatine is radioactive.  8.6 Variation in Chemical Properties of the Representative Elements 355 Helium (He) Neon (Ne) Argon (Ar) Krypton (Kr) Xenon (Xe) Figure 8.21 All noble gases are colorless and odorless. These pictures show the colors emitted by the gases from a discharge tube. the periodic table. For example, F2 is isoelectronic with Ne, Cl2 with Ar, and so on. The vast majority of the alkali metal halides and alkaline earth metal halides are ionic compounds. The halogens also form many molecular compounds among themselves (such as ICl and BrF3) and with nonmetallic elements in other groups (such as NF3, PCl5, and SF6). The halogens react with hydrogen to form hydrogen halides: H2 (g) 1 X2 (g) ¡ 2HX(g) When this reaction involves fluorine, it is explosive, but it becomes less and less violent as we substitute chlorine, bromine, and iodine. The hydrogen halides dissolve in water to form hydrohalic acids. Hydrofluoric acid (HF) is a weak acid (that is, it is a weak electrolyte), but the other hydrohalic acids (HCl, HBr, and HI) are all strong acids (strong electrolytes). Group 8A Elements (ns2np6, n $ 2) 1A 2A 8A 3A 4A 5A 6A 7A He Ne All noble gases exist as monatomic species (Figure 8.21). Their atoms have com- Ar Kr pletely filled outer ns and np subshells, which give them great stability. (Helium is Xe 1s2.) The Group 8A ionization energies are among the highest of all elements, and Rn these gases have no tendency to accept extra electrons. For years these elements were called inert gases, and rightly so. Until 1963 no one had been able to prepare a compound containing any of these elements. The British chemist Neil Bartlett† shattered chemists’ long-held views of these elements when he exposed xenon to platinum hexafluoride, a strong oxidizing agent, and brought about the following reaction (Figure 8.22): Xe(g) 1 2PtF6 (g) ¡ XeF 1 Pt2F11 2 (s) In 2000, chemists prepared a compound Since then, a number of xenon compounds (XeF4, XeO3, XeO4, XeOF4) and a few containing argon (HArF) that is stable krypton compounds (KrF2, for example) have been prepared (Figure 8.23). Despite only at very low temperatures. † Neil Bartlett (1932–2008). English chemist. Bartlett’s work was mainly in the preparation and study of compounds with unusual oxidation states and in solid-state chemistry. 356 Chapter 8 ■ Periodic Relationships Among the Elements Figure 8.22 (a) Xenon gas (colorless) and PtF6 (red gas) separated from each other. (b) When the two gases are allowed to mix, a yellow-orange solid compound is formed. Note that the product was initially given the incorrect formula XePtF6. the immense interest in the chemistry of the noble gases, however, their compounds do not have any major commercial applications, and they are not involved in natural In 2013 astronomers reported find- biological processes. No compounds of helium and neon are known. ing the emission of HAr1 in the Crab Nebula, making it the first molecular noble gas species to be Comparison of Group 1A and Group 1B Elements detected in space. When we compare the Group 1A elements (alkali metals) and the Group 1B elements (copper, silver, and gold), we arrive at an interesting conclusion. Although the metals 1A 8A 2A 3A 4A 5A 6A 7A in these two groups have similar outer electron configurations, with one electron in Li Na 1B the outermost s orbital, their chemical properties are quite different. K Rb Cu Ag The first ionization energies of Cu, Ag, and Au are 745 kJ/mol, 731 kJ/mol, and Cs Au 890 kJ/mol, respectively. Because these values are considerably larger than those of Fr the alkali metals (see Table 8.2), the Group 1B elements are much less reactive. The higher ionization energies of the Group 1B elements result from incomplete shielding of the nucleus by the inner d electrons (compared with the more effective shielding of the completely filled noble gas cores). Consequently the outer s electrons of these elements are more strongly attracted by the nucleus. In fact, copper, silver, and gold are so unreactive that they are usually found in the uncombined state in nature. The inertness and rarity of these metals make them valuable in the manufacture of coins and in jewelry. For this reason, these metals are also called “coinage metals.” The difference in chemical properties between the Group 2A elements (the alkaline earth metals) and the Group 2B metals (zinc, cadmium, and mercury) can be explained in a similar way. Properties of Oxides Across a Period Figure 8.23 Crystals of xenon tetrafluoride (XeF4 ). One way to compare the properties of the representative elements across a period is to examine the properties of a series of similar compounds. Because oxygen 1A 8A 2A 3A 4A 5A 6A 7A combines with almost all elements, we will compare the properties of oxides of the Na Mg Al Si P S Cl third-period elements to see how metals differ from metalloids and nonmetals. Some elements in the third period (P, S, and Cl) form several types of oxides, but for simplicity we will consider only those oxides in which the elements have the high- est oxidation number. Table 8.4 lists a few general characteristics of these oxides. We observed earlier that oxygen has a tendency to form the oxide ion. This tendency is greatly favored when oxygen combines with metals that have low ionization ener- gies, namely, those in Groups 1A and 2A, plus aluminum. Thus, Na2O, MgO, and Al2O3 are ionic compounds, as indicated by their high melting points and boiling points. They have extensive three-dimensional structures in which each cation is surrounded by a specific number of anions, and vice versa. As the ionization energies  8.6 Variation in Chemical Properties of the Representative Elements 357 Table 8.4 Some Properties of Oxides of the Third-Period Elements Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7 Type of compound —¬ ¬¬ Ionic ¬¬¬¡ —¬¬¬¬ Molecular ¬¬¬¡ Structure — Extensive three-dimensional ¡ —¬¬ Discrete ¬¬¡ molecular units Melting point (°C) 1275 2800 2045 1610 580 16.8 291.5 Boiling point (°C) ? 3600 2980 2230 ? 44.8 82 Acid-base nature Basic Basic Amphoteric —¬¬¬¬ Acidic ¬¬¬¬¡ of the elements increase from left to right, so does the molecular nature of the oxides that are formed. Silicon is a metalloid; its oxide (SiO2) also has a huge three- dimensional network, although no ions are present. The oxides of phosphorus, sul- fur, and chlorine are molecular compounds composed of small discrete units. The weak attractions among these molecules result in relatively low melting points and boiling points. Most oxides can be classified as acidic or basic depending on whether they produce acids or bases when dissolved in water or react as acids or bases in cer- tain processes. Some oxides are amphoteric, which means that they display both acidic and basic properties. The first two oxides of the third period, Na2O and MgO, are basic oxides. For example, Na2O reacts with water to form the base sodium hydroxide: Na2O(s) 1 H2O(l) ¡ 2NaOH(aq) Magnesium oxide is quite insoluble; it does not react with water to any appre- ciable extent. However, it does react with acids in a manner that resembles an acid-base reaction: MgO(s) 1 2HCl(aq) ¡ MgCl2 (aq) 1 H2O(l) Note that the products of this reaction are a salt (MgCl2) and water, the usual products of an acid-base neutralization. Aluminum oxide is even less soluble than magnesium oxide; it too does not react with water. However, it shows basic properties by reacting with acids: Al2O3 (s) 1 6HCl(aq) ¡ 2AlCl3 (aq) 1 3H2O(l) It also exhibits acidic properties by reacting with bases: Al2O3 (s) 1 2NaOH(aq) 1 3H2O(l) ¡ 2NaAl(OH) 4 (aq) Note that this acid-base neutralization produces a salt but no water. Thus, Al2O3 is classified as an amphoteric oxide because it has properties of both acids and bases. Other amphoteric oxides are ZnO, BeO, and Bi2O3. Silicon dioxide is insoluble and does not react with water. It has acidic properties, however, because it reacts with very concentrated bases: SiO2 (s) 1 2NaOH(aq) ¡ Na2SiO3 (aq) 1 H2O(l) For this reason, concentrated aqueous, strong bases such as NaOH(aq) should not be stored in Pyrex glassware, which is made of SiO2. CHEMISTRY in Action Discovery of the Noble Gases I n the late 1800s John William Strutt, Third Baron of Rayleigh, who was a professor of physics at the Cavendish Laboratory in Cambridge, England, accurately determined the atomic elements should be placed to the right of the halogens. The ap- parent discrepancy with the position of argon was resolved by Moseley, as discussed in the chapter. masses of a number of elements, but he obtained a puzzling Finally, the last member of the noble gases, radon, was result with nitrogen. One of his methods of preparing nitrogen discovered by the German chemist Frederick Dorn in 1900. A was by the thermal decomposition of ammonia: radioactive element and the heaviest elemental gas known, radon’s discovery not only completed the Group 8A elements, 2NH3 (g) ¡ N2 (g) 1 3H2 (g) but also advanced our understanding about the nature of radio- Another method was to start with air and remove from it oxygen, active decay and transmutation of elements. carbon dioxide, and water vapor. Invariably, the nitrogen from Lord Rayleigh and Ramsay both won Nobel prizes in 1904 air was a little denser (by about 0.5 percent) than the nitrogen for the discovery of argon. Lord Rayleigh received the prize in from ammonia. physics and Ramsay’s award was in chemistry. Lord Rayleigh’s work caught the attention of Sir William Ramsay, a professor of chemistry at the University College, London. In 1898 Ramsay passed nitrogen, which he had ob- tained from air by Rayleigh’s procedure, over red-hot magne- sium to convert it to magnesium nitride: 3Mg(s) 1 N2 (g) ¡ Mg3N2 (s) After all of the nitrogen had reacted with magnesium, Ramsay was left with an unknown gas that would not combine with anything. With the help of Sir William Crookes, the inventor of the discharge tube, Ramsay and Lord Rayleigh found that the emis- sion spectrum of the gas did not match any of the known ele- ments. The gas was a new element! They determined its atomic mass to be 39.95 amu and called it argon, which means “the lazy one” in Greek. Once argon had been discovered, other noble gases were quickly identified. Also in 1898 Ramsay isolated helium from uranium ores (see the Chemical Mystery essay on p. 324). From the atomic masses of helium and argon, their lack of chemical reactivity, and what was then known about the periodic table, Ramsay was convinced that there were other unreactive gases and that they were all members of one periodic group. He and his student Morris Travers set out to find the unknown gases. They used a refrigeration machine to first produce liquid air. Applying a technique called fractional distillation, they then allowed the liquid air to warm up gradually and collected com- ponents that boiled off at different temperatures. In this manner, they analyzed and identified three new elements—neon, krypton, and xenon—in only three months. Three new elements in three months is a record that may never be broken! The discovery of the noble gases helped to complete the periodic table. Their atomic masses suggested that these Sir William Ramsay (1852–1916). 358  Summary of Facts & Concepts 359 The remaining third-period oxides are acidic. They react with water to form phosphoric acid (H3PO4), sulfuric acid (H2SO4), and perchloric acid (HClO4): P4O10 (s) 1 6H2O(l) ¡ 4H3PO4 (aq) SO3 (g) 1 H2O(l) ¡ H2SO4 (aq) Cl2O7 (l) 1 H2O(l) ¡ 2HClO4 (aq) Certain oxides such as CO and NO are neutral; that is, they do not react with water to produce an acidic or basic solution. In general, oxides containing nonmetallic ele- ments are not basic. This brief examination of oxides of the third-period elements shows that as the metal- lic character of the elements decreases from left to right across the period, their oxides change from basic to amphoteric to acidic. Metallic oxides are usually basic, and most oxides of nonmetals are acidic. The intermediate properties of the oxides (as shown by the amphoteric oxides) are exhibited by elements whose positions are intermediate within the period. Note also that because the metallic character of the elements increases from top to bottom within a group of representative elements, we would expect oxides of elements with higher atomic numbers to be more basic than the lighter elements. This is indeed the case. 1A 8A 2A 3A 4A 5A 6A 7A Example 8.6 Be As Classify the following oxides as acidic, basic, or amphoteric: (a) Rb2O, (b) BeO, (c) As2O5. Rb Strategy What type of elements form acidic oxides? basic oxides? amphoteric oxides? Solution (a) Because rubidium is an alkali metal, we would expect Rb2O to be a basic oxide. (b) Beryllium is an alkaline earth metal. However, because it is the first member of Group 2A, we expect that it may differ somewhat from the other members of the group. In the text we saw that Al2O3 is amphoteric. Because beryllium and aluminum exhibit a diagonal relationship, BeO may resemble Al2O3 in properties. It turns out that BeO is also an amphoteric oxide. (c) Because arsenic is a nonmetal, we expect As2O5 to be an acidic oxide. Similar problem: 8.72. Practice Exercise Classify the following oxides as acidic, basic, or amphoteric: (a) ZnO, (b) P4O10, (c) CaO. Review of Concepts An oxide of an element was determined to be basic. Which of the following could be that element? (a) Ba, (b) Al, and (c) Sb. Key Equation Zeff 5 Z 2 σ (8.2) Definition of effective nuclear charge. Summary of Facts & Concepts 1. Nineteenth-century chemists developed the periodic 2. Electron configuration determines the properties of an table by arranging elements in the increasing order element. The modern periodic table classifies the ele- of  their atomic masses. Discrepancies in early ments according to their atomic numbers, and thus also versions of the periodic table were resolved by by their electron configurations. The configuration of arranging the elements in order of their atomic the valence electrons directly affects the properties of numbers. the atoms of the representative elements.  360 Chapter 8 ■ Periodic Relationships Among the Elements 3. Periodic variations in the physical properties of the ele- more positive the electron affinity, the greater the ten- ments reflect differences in atomic structure. The metal- dency for the atom to gain an electron. Metals usually lic character of elements decreases across a period from have low ionization energies, and nonmetals usually metals through the metalloids to nonmetals and in- have high electron affinities. creases from top to bottom within a particular group of 6. Noble gases are very stable because their outer ns and representative elements. np subshells are completely filled. The metals among 4. Atomic radius varies periodically with the arrangement the representative elements (in Groups 1A, 2A, and 3A) of the elements in the periodic table. It decreases from tend to lose electrons until their cations become isoelec- left to right and increases from top to bottom. tronic with the noble gases that precede them in the 5. Ionization energy is a measure of the tendency of an periodic table. The nonmetals in Groups 5A, 6A, and 7A atom to resist the loss of an electron. The higher the tend to accept electrons until their anions become iso- ionization energy, the stronger the attraction between electronic with the noble gases that follow them in the the nucleus and an electron. Electron affinity is a mea- periodic table. sure of the tendency of an atom to gain an electron. The Key Words Amphoteric oxide, p. 357 Effective nuclear charge Ionic radius, p. 337 Representative elements, Atomic radius, p. 335 (Zeff), p. 334 Ionization energy p. 330 Core electrons, p. 330 Electron affinity (IE), p. 340 Valence electrons, Diagonal relationship, p. 348 (EA), p. 345 Isoelectronic, p. 333 p. 330 Questions & Problems • Problems available in Connect Plus • 8.9 Without referring to a periodic table, write the name Red numbered problems solved in Student Solutions Manual and give the symbol for an element in each of the following groups: 1A, 2A, 3A, 4A, 5A, 6A, 7A, 8A, Development of the Periodic Table transition metals. Review Questions • 8.10 Indicate whether the following elements exist as atomic species, molecular species, or extensive three- 8.1 Briefly describe the significance of Mendeleev’s dimensional structures in their most stable states at periodic table. 25°C and 1 atm and write the molecular or empirical 8.2 What is Moseley’s contribution to the modern peri- formula for each one: phosphorus, iodine, magne- odic table? sium, neon, carbon, sulfur, cesium, and oxygen. 8.3 Describe the general layout of a modern periodic 8.11 You are given a dark shiny solid and asked to deter- table. mine whether it is iodine or a metallic element. Sug- 8.4 What is the most important relationship among gest a nondestructive test that would enable you to elements in the same group in the periodic table? arrive at the correct answer. 8.12 What are valence electrons? For representative ele- Periodic Classification of the Elements ments, the number of valence electrons of an ele- ment is equal to its group number. Show that this is Review Questions true for the following elements: Al, Sr, K, Br, P, S, C. • 8.5 Which of the following elements are metals, • 8.13 Write the outer electron configurations for the nonmetals, or metalloids? As, Xe, Fe, Li, B, Cl, Ba, (a) alkali metals, (b) alkaline earth metals, (c) halogens, P, I, Si. (d) noble gases. 8.6 Compare the physical and chemical properties of 8.14 Use the first-row transition metals (Sc to Cu) as an metals and nonmetals. example to illustrate the characteristics of the elec- 8.7 Draw a rough sketch of a periodic table (no details tron configurations of transition metals. are required). Indicate regions where metals, non- 8.15 The electron configurations of ions derived from metals, and metalloids are located. representative elements follow a common pattern. 8.8 What is a representative element? Give names and What is the pattern, and how does it relate to the symbols of four representative elements. stability of these ions?  Questions & Problems 361 8.16 What do we mean when we say that two ions or an • 8.28 Write the ground-state electron configurations of the atom and an ion are isoelectronic? following ions, which play important roles in bio- 8.17 What is wrong with the statement “The atoms of ele- chemical processes in our bodies: (a) Na1, (b) Mg21, ment X are isoelectronic with the atoms of element Y”? (c) Cl2, (d) K1, (e) Ca21, (f) Fe21, (g) Cu21, 8.18 Give three examples of first-row transition metal (h) Zn21. (Sc to Cu) ions whose electron configurations are • 8.29 Write the ground-state electron configurations of the represented by the argon core. following transition metal ions: (a) Sc31, (b) Ti41, (c) V51, (d) Cr31, (e) Mn21, (f) Fe21, (g) Fe31, Problems (h) Co21, (i) Ni21, (j) Cu1, (k) Cu21, (l) Ag1, (m) Au1, (n) Au31, (o) Pt21. 8.19 In the periodic table, the element hydrogen is some- times grouped with the alkali metals (as in this 8.30 Name the ions with 13 charges that have the fol- book) and sometimes with the halogens. Explain lowing electron configurations: (a) [Ar]3d3, why hydrogen can resemble the Group 1A and the (b) [Ar], (c) [Kr]4d 6, (d) [Xe]4f 145d 6. Group 7A elements. • 8.31 Which of the following species are isoelectronic 8.20 A neutral atom of a certain element has 17 elec- with each other? C, Cl2, Mn21, B2, Ar, Zn, Fe31, trons. Without consulting a periodic table, Ge21. (a) write the ground-state electron configuration 8.32 Group the species that are isoelectronic: Be21, F2, of the element, (b) classify the element, (c) deter- Fe21, N32, He, S22, Co31, Ar. mine whether this element is diamagnetic or paramagnetic. Periodic Variation in Physical Properties • 8.21 Group the following electron configurations in pairs Review Questions that would represent similar chemical properties of their atoms: 8.33 Define atomic radius. Does the size of an atom have (a) 1s22s22p63s2 a precise meaning? (b) 1s22s22p3 8.34 How does atomic radius change (a) from left to right (c) 1s22s22p63s23p64s23d104p6 across a period and (b) from top to bottom in a group? (d) 1s22s2 8.35 Define ionic radius. How does the size of an atom (e) 1s22s22p6 change when it is converted to (a) an anion and (b) a (f) 1s22s22p63s23p3 cation? • 8.22 Group the following electron configurations in pairs 8.36 Explain why, for isoelectronic ions, the anions are that would represent similar chemical properties of larger than the cations. their atoms: (a) 1s22s22p5 (b) 1s22s1 Problems (c) 1s22s22p6 • 8.37 On the basis of their positions in the periodic table, (d) 1s22s22p63s23p5 select the atom with the larger atomic radius in each (e) 1s22s22p63s23p64s1 of the following pairs: (a) Na, Cs; (b) Be, Ba; (c) N, Sb; (d) F, Br; (e) Ne, Xe. (f) 1s22s22p63s23p64s23d104p6 • 8.23 Without referring to a periodic table, write the elec- • 8.38 Arrange the following atoms in order of decreasing atomic radius: Na, Al, P, Cl, Mg. tron configuration of elements with the following atomic numbers: (a) 9, (b) 20, (c) 26, (d) 33. Clas- • 8.39 Which is the largest atom in Group 4A? sify the elements. • 8.40 Which is the smallest atom in Group 7A? 8.24 Specify the group of the periodic table in which 8.41 Why is the radius of the lithium atom considerably each of the following elements is found: (a) [Ne]3s1, larger than the radius of the hydrogen atom? (b) [Ne]3s23p3, (c) [Ne]3s23p6, (d) [Ar]4s23d8. 8.42 Use the second period of the periodic table as an • 8.25 A M21 ion derived from a metal in the first transition example to show that the size of atoms decreases as metal series has four electrons in the 3d subshell. we move from left to right. Explain the trend. What element might M be? • 8.43 Indicate which one of the two species in each of the • 8.26 A metal ion with a net 13 charge has five electrons following pairs is smaller: (a) Cl or Cl2; (b) Na or in the 3d subshell. Identify the metal. Na1; (c) O22 or S22; (d) Mg21 or Al31; (e) Au1 or Au31. • 8.27 Write the ground-state electron configurations of the following ions: (a) Li1, (b) H2, (c) N32, (d) F2, • 8.44 List the following ions in order of increasing ionic (e) S22, (f) Al31, (g) Se22, (h) Br2, (i) Rb1, (j) Sr21, radius: N32, Na1, F2, Mg21, O22. (k) Sn21, (l) Te22, (m) Ba21, (n) Pb21, (o) In31, • 8.45 Explain which of the following cations is larger, and (p) Tl1, (q) Tl31. why: Cu1 or Cu21.  362 Chapter 8 ■ Periodic Relationships Among the Elements • 8.46 Explain which of the following anions is larger, and atom could be stripped of its 80 electrons and therefore why: Se22 or Te22. would exist as Hg801. Use the equation in Problem 8.47 Give the physical states (gas, liquid, or solid) of the 8.57 to calculate the energy required for the last ioniza- representative elements in the fourth period (K, Ca, tion step, that is, Ga, Ge, As, Se, Br) at 1 atm and 25°C. Hg791 (g) ¡ Hg801 (g) 1 e2 8.48 Both H2 and He contain two 1s electrons. Which species is larger? Explain your choice. Electron Affinity Review Questions Ionization Energy 8.59 (a) Define electron affinity. (b) Electron affinity Review Questions measurements are made with gaseous atoms. Why? 8.49 Define ionization energy. Ionization energy mea- (c) Ionization energy is always a positive quantity, surements are usually made when atoms are in the whereas electron affinity may be either positive or gaseous state. Why? Why is the second ionization negative. Explain. energy always greater than the first ionization 8.60 Explain the trends in electron affinity from alumi- energy for any element? num to chlorine (see Table 8.3). 8.50 Sketch the outline of the periodic table and show group and period trends in the first ionization energy Problems of the elements. What types of elements have the highest ionization energies and what types the lowest • 8.61 Arrange the elements in each of the following ionization energies? groups in increasing order of the most positive electron affinity: (a) Li, Na, K; (b) F, Cl, Br, I; (c) O, Si, P, Ca, Ba. Problems • 8.62 Specify which of the following elements you • 8.51 Arrange the following in order of increasing first would expect to have the greatest electron affin- ionization energy: Na, Cl, Al, S, and Cs. ity and which would have the least: He, K, Co, 8.52 Arrange the following in order of increasing first S, Cl. ionization energy: F, K, P, Ca, and Ne. • 8.63 Considering their electron affinities, do you think it • 8.53 Use the third period of the periodic table as an example is possible for the alkali metals to form an anion like to illustrate the change in first ionization energies M2, where M represents an alkali metal? of the elements as we move from left to right. 8.64 Explain why alkali metals have a greater affinity for Explain the trend. electrons than alkaline earth metals. 8.54 In general, ionization energy increases from left to right across a given period. Aluminum, how- Variation in Chemical Properties of the ever, has a lower ionization energy than magne- Representative Elements sium. Explain. Review Questions 8.55 The first and second ionization energies of K are 419 kJ/mol and 3052 kJ/mol, and those of Ca are 8.65 What is meant by the diagonal relationship? Name 590 kJ/mol and 1145 kJ/mol, respectively. Compare two pairs of elements that show this relationship. their values and comment on the differences. 8.66 Which elements are more likely to form acidic 8.56 Two atoms have the electron configurations 1s22s22p6 oxides? Basic oxides? Amphoteric oxides? and 1s22s22p63s1. The first ionization energy of one is 2080 kJ/mol, and that of the other is 496 kJ/mol. Problems Match each ionization energy with one of the given 8.67 Use the alkali metals and alkaline earth metals as electron configurations. Justify your choice. examples to show how we can predict the chemical • 8.57 A hydrogenlike ion is an ion containing only one properties of elements simply from their electron electron. The energies of the electron in a hydrogen- configurations. like ion are given by 8.68 Based on your knowledge of the chemistry of the 1 alkali metals, predict some of the chemical proper- En 5 2(2.18 3 10218 J) Z2 a 2 b ties of francium, the last member of the group. n 8.69 As a group, the noble gases are very stable chemi- where n is the principal quantum number and Z is cally (only Kr and Xe are known to form com- the atomic number of the element. Calculate the ion- pounds). Use the concepts of shielding and the ization energy (in kJ/mol) of the He1 ion. effective nuclear charge to explain why the noble • 8.58 Plasma is a state of matter consisting of positive gas- gases tend to neither give up electrons nor accept eous ions and electrons. In the plasma state, a mercury additional electrons.  Questions & Problems 363 8.70 Why are Group 1B elements more stable than Group consult a handbook of chemistry for the melting- 1A elements even though they seem to have the same point values.) outer electron configuration, ns1, where n is the prin- • 8.81 Match each of the elements on the right with its cipal quantum number of the outermost shell? description on the left: 8.71 How do the chemical properties of oxides change (a) A dark-red liquid Calcium (Ca) from left to right across a period? From top to bottom (b) A colorless gas that burns Gold (Au) within a particular group? in oxygen gas Hydrogen (H2) • 8.72 Write balanced equations for the reactions between (c) A reactive metal that attacks Argon (Ar) each of the following oxides and water: (a) Li2O, water (b) CaO, (c) SO3. Bromine (Br2) (d) A shiny metal that is used 8.73 Write formulas for and name the binary hydrogen in jewelry compounds of the second-period elements (Li to F). (e) An inert gas Describe how the physical and chemical properties of these compounds change from left to right across 8.82 Arrange the following species in isoelectronic pairs: the period. O1, Ar, S22, Ne, Zn, Cs1, N32, As31, N, Xe. 8.74 Which oxide is more basic, MgO or BaO? Why? • 8.83 In which of the following are the species written in decreasing order by size of radius? (a) Be, Mg, Ba, (b) N32, O22, F2, (c) Tl31, Tl21, Tl1. Additional Problems 8.84 Which of the following properties show a clear peri- 8.75 State whether each of the following properties of odic variation? (a) first ionization energy, (b) molar the representative elements generally increases or mass of the elements, (c) number of isotopes of an decreases (a) from left to right across a period and element, (d) atomic radius. (b) from top to bottom within a group: metallic • 8.85 When carbon dioxide is bubbled through a clear cal- character, atomic size, ionization energy, acidity cium hydroxide solution, the solution appears milky. of oxides. Write an equation for the reaction and explain how • 8.76 With reference to the periodic table, name (a) a halo- this reaction illustrates that CO2 is an acidic oxide. gen element in the fourth period, (b) an element simi- 8.86 You are given four substances: a fuming red liq- lar to phosphorus in chemical properties, (c) the most uid, a dark metallic-looking solid, a pale-yellow reactive metal in the fifth period, (d) an element that gas, and a yellow-green gas that attacks glass. You has an atomic number smaller than 20 and is similar to are told that these substances are the first four strontium. members of Group 7A, the halogens. Name • 8.77 Write equations representing the following processes: each one. (a) The electron affinity of S2. 8.87 Calculate the change in energy for the following (b) The third ionization energy of titanium. processes: (c) The electron affinity of Mg21. (a) Na(g) 1 Cl(g) ¡ Na1(g) 1 Cl2(g) (d) The ionization energy of O22. (b) Ca(g) 1 2Br(g) ¡ Ca21(g) 1 2Br2(g) 8.78 List all the common ions of representative ele- 8.88 Calculate the change in energy for the following ments and transition metals that are isoelectronic processes: with Ar. (a) Mg(g) 1 2F(g) ¡ Mg21(g) 1 2F2(g) • 8.79 Write the empirical (or molecular) formulas of com- (b) 2Al(g) 1 3O(g) ¡ 2Al31(g) 1 3O22(g) pounds that the elements in the third period (sodium The electron affinity of O2 is 2844 kJ/mol. to chlorine) should form with (a) molecular oxygen 8.89 For each pair of elements listed, give three properties and (b) molecular chlorine. In each case indicate that show their chemical similarity: (a) sodium and whether you would expect the compound to be ionic potassium and (b) chlorine and bromine. or molecular in character. 8.90 Name the element that forms compounds, under ap- 8.80 Element M is a shiny and highly reactive metal propriate conditions, with every other element in the (melting point 63°C), and element X is a highly periodic table except He, Ne, and Ar. reactive nonmetal (melting point 27.2°C). They react to form a compound with the empirical for- 8.91 Explain why the first electron affinity of sulfur is mula MX, a colorless, brittle white solid that melts 200 kJ/mol but the second electron affinity is at 734°C. When dissolved in water or when in the 2649 kJ/mol. molten state, the substance conducts electricity. 8.92 The H 2 ion and the He atom have two 1s elec- When chlorine gas is bubbled through an aqueous trons each. Which of the two species is larger? solution containing MX, a reddish-brown liquid Explain. appears and Cl2 ions are formed. From these ob- • 8.93 Predict the products of the following oxides with servations, identify M and X. (You may need to water: Na2O, BaO, CO2, N2O5, P4O10, SO3. Write an  364 Chapter 8 ■ Periodic Relationships Among the Elements equation for each of the reactions. Specify whether 162 nm. Calculate the ionization energy of potas- the oxides are acidic, basic, or amphoteric. sium. How can you be sure that this ionization en- 8.94 Write the formulas and names of the oxides of the ergy corresponds to the electron in the valence shell second-period elements (Li to N). Identify the oxides (that is, the most loosely held electron)? as acidic, basic, or amphoteric. 8.103 Referring to the Chemistry in Action essay on • 8.95 State whether each of the following elements is a gas, p. 358, answer the following questions. (a) Why did it a liquid, or a solid under atmospheric conditions. Also take so long to discover the first noble gas (argon) state whether it exists in the elemental form as atoms, on Earth? (b) Once argon had been discovered, why as molecules, or as a three-dimensional network: Mg, did it take relatively little time to discover the rest of Cl, Si, Kr, O, I, Hg, Br. the noble gases? (c) Why was helium not isolated by 8.96 What factors account for the unique nature of the fractional distillation of liquid air? hydrogen? • 8.104 The energy needed for the following process is 1.96 3 104 kJ/mol: • 8.97 The air in a manned spacecraft or submarine needs to be purified of exhaled carbon dioxide. Write Li(g) ¡ Li31 (g) 1 3e2 equations for the reactions between carbon dioxide and (a) lithium oxide (Li2O), (b) sodium peroxide If the first ionization energy of lithium is 520 kJ/mol, (Na2O2), and (c) potassium superoxide (KO2). calculate the second ionization energy of lithium, that is, the energy required for the process 8.98 The formula for calculating the energies of an elec- tron in a hydrogenlike ion is given in Problem 8.57. Li1 (g) ¡ Li21 (g) 1 e2 This equation cannot be applied to many-electron atoms. One way to modify it for the more complex (Hint: You need the equation in Problem 8.57.) atoms is to replace Z with (Z 2 σ), where Z is • 8.105 An element X reacts with hydrogen gas at 200°C to the atomic number and σ is a positive dimensionless form compound Y. When Y is heated to a higher quantity called the shielding constant. Consider the temperature, it decomposes to the element X and helium atom as an example. The physical signifi- hydrogen gas in the ratio of 559 mL of H2 (mea- cance of σ is that it represents the extent of shielding sured at STP) for 1.00 g of X reacted. X also com- that the two 1s electrons exert on each other. Thus, the bines with chlorine to form a compound Z, which quantity (Z 2 σ) is appropriately called the “effective contains 63.89 percent by mass of chlorine. Deduce nuclear charge.” Calculate the value of σ if the first the identity of X. ionization energy of helium is 3.94 3 10218 J per • 8.106 A student is given samples of three elements, X, atom. (Ignore the minus sign in the given equation in Y, and Z, which could be an alkali metal, a mem- your calculation.) ber of Group 4A, and a member of Group 5A. She 8.99 Why do noble gases have negative electron affinity makes the following observations: Element X has values? a metallic luster and conducts electricity. It reacts slowly with hydrochloric acid to produce hydro- • 8.100 The atomic radius of K is 227 pm and that of K1 is gen gas. Element Y is a light-yellow solid that 133 pm. Calculate the percent decrease in volume that occurs when K(g) is converted to K1(g). [The does not conduct electricity. Element Z has a me- volume of a sphere is 1 43 2πr3 , where r is the radius of tallic luster and conducts electricity. When ex- the sphere.] posed to air, it slowly forms a white powder. A solution of the white powder in water is basic. • 8.101 The atomic radius of F is 72 pm and that of F2 is What can you conclude about the elements from 133 pm. Calculate the percent increase in volume these observations? that occurs when F(g) is converted to F2(g). (See Problem 8.100 for the volume of a sphere.) 8.107 Identify the ions whose orbital diagrams for the valence electrons are shown below. The charges of • 8.102 A technique called photoelectron spectroscopy is the ions are: (a) 11, (b) 31, (c) 41, (d) 21. used to measure the ionization energy of atoms. A sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies (a) hg hg hg hg hg of the ejected electrons are measured. Because the 4s 3d energy of the UV photon and the kinetic energy of the ejected electron are known, we can write (b) h h h h h 4s 3d hn 5 IE 1 12 mu2 (c) h h h where n is the frequency of the UV light, and m and 4s 3d u are the mass and velocity of the electron, respec- tively. In one experiment the kinetic energy of the (d) hg hg h h h ejected electron from potassium is found to be 5s 4d 5.34 3 10219 J using a UV source of wavelength  Questions & Problems 365 8.108 What is the electron affinity of the Na1 ion? Then, in 1986, a chemist reported that by reacting 8.109 The ionization energies of sodium (in kJ/mol), potassium hexafluoromanganate(IV) (K2MnF6) with starting with the first and ending with the elev- antimony pentafluoride (SbF5) at 150°C, he had enth, are 495.9, 4560, 6900, 9540, 13,400, 16,600, generated elemental fluorine. Balance the following 20,120, 25,490, 28,930, 141,360, 170,000. Plot equation representing the reaction: the log of ionization energy (y axis) versus the K2MnF6 1 SbF5 ¡ KSbF6 1 MnF3 1 F2 number of ionization (x axis); for example, log 495.9 is plotted versus 1 (labeled IE1, the first 8.117 Write a balanced equation for the preparation of ionization energy), log 4560 is plotted versus 2 (a) molecular oxygen, (b) ammonia, (c) carbon dioxide, (labeled IE2, the second ionization energy), and (d) molecular hydrogen, (e) calcium oxide. Indicate so on. (a) Label IE1 through IE11 with the electrons the physical state of the reactants and products in in orbitals such as 1s, 2s, 2p, and 3s. (b) What can each equation. you deduce about electron shells from the breaks 8.118 Write chemical formulas for oxides of nitrogen with in the curve? the following oxidation numbers: 11, 12, 13, 14, • 8.110 Experimentally, the electron affinity of an element 15. (Hint: There are two oxides of nitrogen with 14 can be determined by using a laser light to ionize the oxidation number.) anion of the element in the gas phase: 8.119 Most transition metal ions are colored. For example, a solution of CuSO4 is blue. How would you show that X2 (g) 1 hn ¡ X(g) 1 e2 the blue color is due to the hydrated Cu21 ions and not the SO422 ions? Referring to Table 8.3, calculate the photon wavelength (in nanometers) corresponding to the 8.120 In general, atomic radius and ionization energy have electron affinity for chlorine. In what region of opposite periodic trends. Why? the electromagnetic spectrum does this wave- 8.121 Explain why the electron affinity of nitrogen is ap- length fall? proximately zero, while the elements on either side, 8.111 Explain, in terms of their electron configurations, carbon and oxygen, have substantial positive electron why Fe21 is more easily oxidized to Fe31 than Mn21 affinities. to Mn31. • 8.122 Consider the halogens chlorine, bromine, and iodine. The melting point and boiling point of chlorine are • 8.112 The standard enthalpy of atomization of an element 2101.0°C and 234.6°C while those of iodine are is the energy required to convert one mole of an ele- ment in its most stable form at 25°C to one mole of 113.5°C and 184.4°C, respectively. Thus, chlorine is monatomic gas. Given that the standard enthalpy of a gas and iodine is a solid under room conditions. atomization for sodium is 108.4 kJ/mol, calculate Estimate the melting point and boiling point of the energy in kilojoules required to convert one bromine. Compare your values with those from a mole of sodium metal at 25°C to one mole of gas- handbook of chemistry. eous Na1 ions. 8.123 Write a balanced equation that predicts the reaction of 8.113 Write the formulas and names of the hydrides of the rubidium (Rb) with (a) H2O(l), (b) Cl2(g), (c) H2(g). following second-period elements: Li, C, N, O, F. 8.124 The successive IE of the first four electrons of a Predict their reactions with water. representative element are 738.1 kJ/mol, 1450 kJ/mol, 8.114 Based on knowledge of the electronic configuration 7730 kJ/mol, and 10,500 kJ/mol. Characterize the of titanium, state which of the following compounds element according to the periodic group. of titanium is unlikely to exist: K3TiF6, K2Ti2O5, • 8.125 Little is known of the chemistry of astatine, the last TiCl3, K2TiO4, K2TiF6. member of Group 7A. Describe the physical charac- 8.115 Name an element in Group 1A or Group 2A that is an teristics that you would expect this halogen to have. important constituent of each of the following sub- Predict the products of the reaction between sodium stances: (a) remedy for acid indigestion, (b) coolant astatide (NaAt) and sulfuric acid. (Hint: Sulfuric in nuclear reactors, (c) Epsom salt, (d) baking powder, acid is an oxidizing agent.) (e) gunpowder, (f) a light alloy, (g) fertilizer that also 8.126 As discussed in the chapter, the atomic mass of argon neutralizes acid rain, (h) cement, and (i) grit for icy is greater than that of potassium. This observation cre- roads. You may need to ask your instructor about ated a problem in the early development of the peri- some of the items. odic table because it meant that argon should be placed after potassium. (a) How was this difficulty resolved? • 8.116 In halogen displacement reactions a halogen element (b) From the following data, calculate the average can be generated by oxidizing its anions with a halo- gen element that lies above it in the periodic table. atomic masses of argon and potassium: Ar-36 (35.9675 This means that there is no way to prepare elemental amu; 0.337 percent), Ar-38 (37.9627 amu; 0.063 per- fluorine, because it is the first member of Group 7A. cent), Ar-40 (39.9624 amu; 99.60 percent); K-39 Indeed, for years the only way to prepare elemental (38.9637 amu; 93.258 percent), K-40 (39.9640 amu; fluorine was to oxidize F2 ions by electrolytic means. 0.0117 percent), K-41 (40.9618 amu; 6.730 percent).  366 Chapter 8 ■ Periodic Relationships Among the Elements • 8.127 Calculate the maximum wavelength of light (in nano- 8.138 One allotropic form of an element X is a colorless meters) required to ionize a single sodium atom. crystalline solid. The reaction of X with an excess 8.128 Predict the atomic number and ground-state electron amount of oxygen produces a colorless gas. This configuration of the next member of the alkali metals gas dissolves in water to yield an acidic solution. after francium. Choose one of the following elements that matches 8.129 Why do elements that have high ionization energies X: (a) sulfur, (b) phosphorus, (c) carbon, (d) boron, also have more positive electron affinities? Which and (e) silicon. group of elements would be an exception to this 8.139 When magnesium metal is burned in air, it forms generalization? two products A and B. A reacts with water to form • 8.130 The first four ionization energies of an element a basic solution. B reacts with water to form a are approximately 579 kJ/mol, 1980 kJ/mol, similar solution as that of A plus a gas with a pun- 2963 kJ/mol, and 6180 kJ/mol. To which periodic gent odor. Identify A and B and write equations group does this element belong? for the reactions. (Hint: See Chemistry in Action on p. 358.) 8.131 Some chemists think that helium should properly be called “helon.” Why? What does the ending in helium 8.140 The ionization energy of a certain element is 412 (-ium) suggest? kJ/mol. When the atoms of this element are in the 8.132 (a) The formula of the simplest hydrocarbon is CH4 first excited state, however, the ionization energy (methane). Predict the formulas of the simplest is only 126 kJ/mol. Based on this information, compounds formed between hydrogen and the fol- calculate the wavelength of light emitted in a tran- lowing elements: silicon, germanium, tin, and lead. sition from the first excited state to the ground (b) Sodium hydride (NaH) is an ionic compound. state. Would you expect rubidium hydride (RbH) to be 8.141 Use your knowledge of thermochemistry to calculate more or less ionic than NaH? (c) Predict the reaction the DH for the following processes: (a) Cl2(g) S between radium (Ra) and water. (d) When exposed Cl1(g) 1 2e2 and (b) K1(g) 1 2e2 S K2(g). to air, aluminum forms a tenacious oxide (Al2O3) 8.142 Referring to Table 8.2, explain why the first ion- coating that protects the metal from corrosion. ization energy of helium is less than twice the ion- Which metal in Group 2A would you expect to ization energy of hydrogen, but the second exhibit similar properties? Why? ionization energy of helium is greater than twice 8.133 Give equations to show that molecular hydrogen can the ionization energy of hydrogen. [Hint: Accord- act both as a reducing agent and an oxidizing agent. ing to Coulomb’s law, the energy between two 8.134 Both Mg21 and Ca21 are important biological ions. charges Q1 and Q2 separated by distance r is pro- One of their functions is to bind to the phosphate portional to (Q1Q2/r).] groups of ATP molecules or amino acids of proteins. 8.143 As mentioned in Chapter 3 (p. 105), ammonium ni- For Group 2A metals in general, the tendency trate (NH4NO3) is the most important nitrogen- for binding to the anions increases in the order containing fertilizer in the world. Describe how you Ba21 , Sr21 , Ca21 , Mg21. Explain the trend. would prepare this compound, given only air and • 8.135 Match each of the elements on the right with its water as the starting materials. You may have any description on the left: device at your disposal for this task. (a) A pale yellow gas that Nitrogen (N2) 8.144 One way to estimate the effective charge (Zeff) of a reacts with water. many-electron atom is to use the equation IE1 5 Boron (B) (1312 kJ/mol)(Z2eff/n2), where IE1 is the first ioniza- (b) A soft metal that reacts with Aluminum (Al) tion energy and n is the principal quantum number of water to produce hydrogen. Fluorine (F2) the shell in which the electron resides. Use this equa- (c) A metalloid that is hard Sodium (Na) tion to calculate the effective charges of Li, Na, and and has a high melting point. K. Also calculate Zeff/n for each metal. Comment on (d) A colorless, odorless gas. your results. (e) A metal that is more reactive than iron, but 8.145 To prevent the formation of oxides, peroxides, does not corrode in air. and superoxides, alkali metals are sometimes 8.136 Write an account on the importance of the periodic stored in an inert atmosphere. Which of the fol- table. Pay particular attention to the significance of the lowing gases should not be used for lithium: Ne, position of an element in the table and how the posi- Ar, N2, Kr? Explain. (Hint: As mentioned in the tion relates to the chemical and physical properties of chapter, Li and Mg exhibit a diagonal relation- the element. ship. Compare the common compounds of these 8.137 On the same graph, plot the effective nuclear charge two elements.) (see p. 334) and atomic radius (see Figure 8.5) versus 8.146 Describe the biological role of the elements in the atomic number for the second period elements Li to human body shown in the following periodic table. Ne. Comment on the trends.  Answers to Practice Exercises 367 (You may need to do research at websites such as 8.147 Recent theoretical calculations suggest that astatine www.webelements.com.) may be a monoatomic metal rather than a diatomic molecule like the other halogens. (a) Rationalize H this prediction based on astatine’s position in the C N O periodic table. (b) The energy required to remove an Na Mg P S Cl electron from one At atom was determined by laser K Ca Cr Mn Fe Co Cu Zn Br ionization to be 9.3175 eV. Given that 1 eV 5 1.602 3 10219 J, calculate the first ionization energy of asta- I tine in kJ/mol. (c) Comment on whether or not the following first ionization energies support your answer to part (a): Pb, 715.6 kJ/mol; Bi, 702.9 kJ/ mol; Po, 811.8 kJ/mol; Rn, 1037 kJ/mol. Interpreting, Modeling & Estimating 8.148 Consider the first 18 elements from hydrogen to by analyzing a victim’s hair. Where else might one argon. Would you expect the atoms of half of them look for the accumulation of the element if arsenic to be diamagnetic and half of them to be paramag- poisoning is suspected? netic? Explain. 8.152 The boiling points of neon and krypton are 8.149 Compare the work function for cesium (206 kJ/mol) 2245.9°C and 2152.9°C, respectively. Using these with its first ionization energy (376 kJ/mol). Explain data, estimate the boiling point of argon. the difference. 8.153 Using the following boiling-point data, estimate the 8.150 The only confirmed compound of radon is radon boiling point of francium: difluoride, RnF2. One reason that it is difficult to study the chemistry of radon is that all isotopes of Metal Li Na K Rb Cs Fr radon are radioactive so it is dangerous to handle the B.pt.(°C) 1347 882.9 774 688 678.4 ? substance. Can you suggest another reason why there are so few known radon compounds? 8.154 The energy gap between the 6s and 5d levels in gold (Hint: Radioactive decays are exothermic processes.) is 4.32 3 10219 J. Based on this information, predict 8.151 Arsenic (As) is not an essential element for the the perceived color of gold vapor. (Hint: You need to human body. (a) Based on its position in the periodic be familiar with the notion of complementary color; table, suggest a reason for its toxicity. (b) When see Figure 23.18.) arsenic enters a person’s body, it quickly shows up 8.155 Calculate the volume of 1 mole of K atoms (see Fig- in the follicle of the growing hair. This action has ure 8.5) and compare the result by using the density enabled detectives to solve many murder mysteries of K (0.856 g/cm3). Account for the difference. Answers to Practice Exercises 8.1 (a) 1s22s22p63s23p64s2, (b) it is a representative element, (c) diamagnetic. 8.2 Li . Be . C. 8.3 (a) Li1, (b) Au31, (c) N32. 8.4 (a) N, (b) Mg. 8.5 No. 8.6 (a) amphoteric, (b) acidic, (c) basic.  CHAPTER 9 Chemical Bonding I Basic Concepts Lewis developed many of the models we still use today to understand chemical bonding. CHAPTER OUTLINE A LOOK AHEAD 9.1 Lewis Dot Symbols  Our study of chemical bonds begins with an introduction to Lewis dot symbols, which shows the valence electrons on an atom. (9.1) 9.2 The Ionic Bond  We then study the formation of ionic bonds and learn how to determine lattice 9.3 Lattice Energy of Ionic energy, which is a measure of the stability of ionic compounds. (9.2 and 9.3) Compounds  Next we turn our attention to the formation of covalent bonds. We learn to 9.4 The Covalent Bond write Lewis structures, which are governed by the octet rule. (9.4) 9.5 Electronegativity  We see that electronegativity is an important concept in understanding the properties of molecules. (9.5) 9.6 Writing Lewis Structures  We continue to practice writing Lewis structures for molecules and ions and 9.7 Formal Charge and Lewis use formal charges to study the distribution of electrons in these species. Structure (9.6 and 9.7) 9.8 The Concept of Resonance  We learn further aspects of writing Lewis structures in terms of resonance 9.9 Exceptions to the Octet Rule structures, which are alternate Lewis structures for a molecule. We also see that there are exceptions to the octet rule. (9.8 and 9.9) 9.10 Bond Enthalpy  The chapter ends with an examination of the strength of covalent bonds, which leads to the use of bond enthalpies to determine the enthalpy of a reaction. (9.10) 368  9.1 Lewis Dot Symbols 369 W hy do atoms of different elements react? What are the forces that hold atoms together in molecules and ionic compounds? What shapes do they assume? These are some of the questions addressed in this chapter and in Chapter 10. We begin by looking at the two types of bonds—ionic and covalent—and the forces that stabilize them. 9.1 Lewis Dot Symbols The development of the periodic table and concept of electron configuration gave chemists a rationale for molecule and compound formation. This explanation, formu- lated by Gilbert Lewis,† is that atoms combine in order to achieve a more stable electron configuration. Maximum stability results when an atom is isoelectronic with a noble gas. When atoms interact to form a chemical bond, only their outer regions are in contact. For this reason, when we study chemical bonding, we are concerned primar- ily with the valence electrons of the atoms. To keep track of valence electrons in a chemical reaction, and to make sure that the total number of electrons does not change, chemists use a system of dots devised by Lewis called Lewis dot symbols. A Lewis dot symbol consists of the symbol of an element and one dot for each valence electron in an atom of the element. Figure 9.1 shows the Lewis dot symbols for the representative elements and the noble gases. Note that, except for helium, the number of valence electrons each atom has is the same as the group number of the element. For example, Li is a Group 1A element and has one dot for one valence electron; Be, a Group 2A element, has two valence electrons (two dots); and so on. Elements in the same group have similar outer electron configurations and hence similar Lewis dot symbols. The transition metals, lanthanides, and actinides all have incompletely filled inner shells, and in general, we cannot write simple Lewis dot symbols for them. In this chapter, we will learn to use electron configurations and the periodic table to predict the type of bond atoms will form, as well as the number of bonds an atom of a particular element can form and the stability of the product. † Gilbert Newton Lewis (1875–1946). American chemist. Lewis made many important contributions in the areas of chemical bonding, thermodynamics, acids and bases, and spectroscopy. Despite the significance of Lewis’s work, he was never awarded a Nobel prize. 1 18 1A 8A H 2 13 14 15 16 17 He 2A 3A 4A 5A 6A 7A Li Be B C N O F Ne Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar 3B 4B 5B 6B 7B 8B 1B 2B K Ca Ga Ge As Se Br Kr Rb Sr In Sn Sb Te I Xe Cs Ba Tl Pb Bi Po At Rn Fr Ra Fl Lv Figure 9.1 Lewis dot symbols for the representative elements and the noble gases. The number of unpaired dots corresponds to the number of bonds an atom of the element can form in a molecular compound without expanding the octet (p. 378). 370 Chapter 9 ■ Chemical Bonding I: Basic Concepts Review of Concepts What is the maximum number of dots that can be drawn around the atom of a representative element? 9.2 The Ionic Bond In Chapter 8 we saw that atoms of elements with low ionization energies tend to form cations, while those with high electron affinities tend to form anions. As a rule, the ele- ments most likely to form cations in ionic compounds are the alkali metals and alkaline earth metals, and the elements most likely to form anions are the halogens and oxygen. Consequently, a wide variety of ionic compounds combine a Group 1A or Group 2A metal with a halogen or oxygen. An ionic bond is the electrostatic force that holds ions together in an ionic compound. Consider, for example, the reaction between lithium and fluorine to form lithium fluoride, a poisonous white powder used in lowering the melting point of solders and in manufacturing ceramics. The electron configuration of lithium is 1s22s1, and that of fluorine is 1s22s22p5. When lithium and fluorine atoms come in con- tact with each other, the outer 2s1 valence electron of lithium is transferred to the fluorine atom. Using Lewis dot symbols, we represent the reaction like this: T Li  SO FT Q 88n Li SO F S Q (or LiF) (9.1) 2 1 2 2 5 2 2 2 6 1s 2s 1s 2s 2p 1s 1s 2s 2p For convenience, imagine that this reaction occurs in separate steps—first the ioniza- tion of Li: Lithium fluoride. Industrially, LiF (like most other ionic compounds) # Li ¡ Li1 1 e2 is obtained by purifying minerals containing the compound. and then the acceptance of an electron by F: O  e 88n SFS SFT O  Q Q Next, imagine the two separate ions joining to form a LiF unit: Li  SO F S 88n LiSO Q F S Q Note that the sum of these three equations is We normally write the empirical formulas F T 88n LiSO T Li  SO Q F S Q of ionic compounds without showing the charges. The 1 and 2 are shown here to emphasize the transfer of electrons. which is the same as Equation (9.1). The ionic bond in LiF is the electrostatic attrac- tion between the positively charged lithium ion and the negatively charged fluoride ion. The compound itself is electrically neutral. Many other common reactions lead to the formation of ionic bonds. For instance, calcium burns in oxygen to form calcium oxide: Animation 2Ca(s) 1 O2 (g) ¡ 2CaO(s) Reactions of Magnesium and Oxygen Assuming that the diatomic O2 molecule first splits into separate oxygen atoms (we will look at the energetics of this step later), we can represent the reaction with Lewis symbols: TCaT  O TO OS2 88n Ca2 SO QT Q [Ar]4s2 1s22s22p4 [Ar] [Ne]  9.2 The Ionic Bond 371 There is a transfer of two electrons from the calcium atom to the oxygen atom. Note that the resulting calcium ion (Ca21) has the argon electron configuration, the oxide ion (O22) is isoelectronic with neon, and the compound (CaO) is electrically neutral. In many cases, the cation and the anion in a compound do not carry the same charges. For instance, when lithium burns in air to form lithium oxide (Li2O), the balanced equation is 4Li(s) 1 O2 (g) ¡ 2Li2O(s) Using Lewis dot symbols, we write 2 TLi  TO O  O QT 88n 2Li SO QS (or Li2O) 2 1s22s1 1s22s22p4 [He] [Ne] In this process, the oxygen atom receives two electrons (one from each of the two lithium atoms) to form the oxide ion. The Li1 ion is isoelectronic with helium. When magnesium reacts with nitrogen at elevated temperatures, a white solid compound, magnesium nitride (Mg3N2), forms: 3Mg(s) 1 N2 (g) ¡ Mg3N2 (s) or OT 88n 3Mg2 3 TMgT  2 TR 2 SO QS (or Mg3N2) 3 N N [Ne]3s2 1s22s22p3 [Ne] [Ne] The reaction involves the transfer of six electrons (two from each Mg atom) to two nitrogen atoms. The resulting magnesium ion (Mg21) and the nitride ion (N32) are both isoelectronic with neon. Because there are three 12 ions and two 23 ions, the charges balance and the compound is electrically neutral. In Example 9.1, we apply the Lewis dot symbols to study the formation of an ionic compound. Example 9.1 Aluminum oxide, obtained from the mineral corundum, is used primarily for the production of aluminum metal. Use Lewis dot symbols to show the formation of aluminum oxide (Al2O3). Strategy We use electroneutrality as our guide in writing formulas for ionic compounds; that is, the total positive charges on the cations must be equal to the total negative charges on the anions. Solution According to Figure 9.1, the Lewis dot symbols of Al and O are The mineral corundum (Al2O3 ). R TAlT OT TO Q Because aluminum tends to form the cation (Al31) and oxygen the anion (O22) in ionic compounds, the transfer of electrons is from Al to O. There are three valence electrons in each Al atom; each O atom needs two electrons to form the O22 ion, which is isoelectronic with neon. Thus, the simplest neutralizing ratio of Al31 to O22 is 2:3; two Al31 ions have a total charge of 16, and three O22 ions have a total charge of 26. So the empirical formula of aluminum oxide is Al2O3, and the reaction is R 2 TAlT  OT 3 TO 88n 2Al3 3 SO OS2 (or Al2O3) Q Q [Ne]3s23p1 1s22s22p4 [Ne] [Ne] (Continued) 372 Chapter 9 ■ Chemical Bonding I: Basic Concepts Check Make sure that the number of valence electrons (24) is the same on both sides of the equation. Are the subscripts in Al2O3 reduced to the smallest possible Similar problems: 9.17, 9.18. whole numbers? Practice Exercise Use Lewis dot symbols to represent the formation of barium hydride. Review of Concepts Use Lewis dot symbols to represent the formation of rubidium sulfide. 9.3 Lattice Energy of Ionic Compounds We can predict which elements are likely to form ionic compounds based on ioniza- tion energy and electron affinity, but how do we evaluate the stability of an ionic compound? Ionization energy and electron affinity are defined for processes occur- ring in the gas phase, but at 1 atm and 25°C all ionic compounds are solids. The solid state is a very different environment because each cation in a solid is sur- rounded by a specific number of anions, and vice versa. Thus, the overall stability of a solid ionic compound depends on the interactions of all these ions and not merely on the interaction of a single cation with a single anion. A quantitative Lattice energy is determined by the charge of the ions and the distance between measure of the stability of any ionic solid is its lattice energy, defined as the energy the ions. required to completely separate one mole of a solid ionic compound into gaseous ions (see Section 6.7). The Born-Haber Cycle for Determining Lattice Energies Lattice energy cannot be measured directly. However, if we know the structure and composition of an ionic compound, we can calculate the compound’s lattice energy by using Coulomb’s† law, which states that the potential energy (E) between two ions is directly proportional to the product of their charges and inversely proportional to the distance of separation between them. For a single Li1 ion and a single F2 ion separated by distance r, the potential energy of the system is given by Because energy 5 force 3 distance, QLi1QF2 Coulomb’s law can also be stated as Er QLi 1 QF 2 r F5k r2 QLi1QF2 where F is the force between the ions. 5k (9.2) r where QLi and QF are the charges on the Li1 and F2 ions and k is the proportional- 1 2 ity constant. Because QLi is positive and QF is negative, E is a negative quantity, 1 2 and the formation of an ionic bond from Li1 and F2 is an exothermic process. Con- sequently, energy must be supplied to reverse the process (in other words, the lattice energy of LiF is positive), and so a bonded pair of Li1 and F2 ions is more stable than separate Li1 and F2 ions. We can also determine lattice energy indirectly, by assuming that the formation of an ionic compound takes place in a series of steps. This procedure, known as the Born-Haber cycle, relates lattice energies of ionic compounds to ionization energies, † Charles Augustin de Coulomb (1736–1806). French physicist. Coulomb did research in electricity and magnetism and applied Newton’s inverse square law to electricity. He also invented a torsion balance.  9.3 Lattice Energy of Ionic Compounds 373 electron affinities, and other atomic and molecular properties. It is based on Hess’s law (see Section 6.6). Developed by Max Born† and Fritz Haber,‡ the Born-Haber cycle defines the various steps that precede the formation of an ionic solid. We will illustrate its use to find the lattice energy of lithium fluoride. Consider the reaction between lithium and fluorine: Li(s) 1 12F2 (g) ¡ LiF(s) The standard enthalpy change for this reaction is 2594.1 kJ/mol. (Because the reac- tants and product are in their standard states, that is, at 1 atm, the enthalpy change is also the standard enthalpy of formation for LiF.) Keeping in mind that the sum of enthalpy changes for the steps is equal to the enthalpy change for the overall reaction (2594.1 kJ/mol), we can trace the formation of LiF from its elements through five separate steps. The process may not occur exactly this way, but this pathway enables us to analyze the energy changes of ionic compound formation, with the application of Hess’s law. 1. Convert solid lithium to lithium vapor (the direct conversion of a solid to a gas is called sublimation): Li(s) ¡ Li(g) ¢H°1 5 155.2 kJ/mol The energy of sublimation for lithium is 155.2 kJ/mol. 2. Dissociate 12 mole of F2 gas into separate gaseous F atoms: 1 2 F2 (g) ¡ F(g) ¢H°2 5 75.3 kJ/mol The F atoms in a F2 molecule are held together by a covalent bond. The energy required to break this bond is called the The energy needed to break the bonds in 1 mole of F2 molecules is 150.6 kJ. bond enthalpy (Section 9.10). Here we are breaking the bonds in half a mole of F2, so the enthalpy change is 150.6/2, or 75.3, kJ. 3. Ionize 1 mole of gaseous Li atoms (see Table 8.2): Li(g) ¡ Li1 (g) 1 e2 ¢H°3 5 520 kJ/mol This process corresponds to the first ionization of lithium. 4. Add 1 mole of electrons to 1 mole of gaseous F atoms. As discussed on page 345, the energy change for this process is just the opposite of electron affinity (see Table 8.3): F(g) 1 e2 ¡ F2 (g) ¢H°4 5 2328 kJ/mol 5. Combine 1 mole of gaseous Li 1 and 1 mole of F 2 to form 1 mole of solid LiF: Li1 (g) 1 F2 (g) ¡ LiF(s) ¢H°5 5 ? The reverse of step 5, energy 1 LiF(s) ¡ Li1 (g) 1 F2 (g) † Max Born (1882–1970). German physicist. Born was one of the founders of modern physics. His work covered a wide range of topics. He received the Nobel Prize in Physics in 1954 for his interpretation of the wave function for particles. ‡ Fritz Haber (1868–1934). German chemist. Haber’s process for synthesizing ammonia from atmospheric nitrogen kept Germany supplied with nitrates for explosives during World War I. He also did work on gas warfare. In 1918 Haber received the Nobel Prize in Chemistry. 374 Chapter 9 ■ Chemical Bonding I: Basic Concepts defines the lattice energy of LiF. Thus, the lattice energy must have the same magni- tude as ¢H°5 but an opposite sign. Although we cannot determine ¢H°5 directly, we can calculate its value by the following procedure. 1. Li(s) ¡ Li(g) ¢H°1 5 155.2 kJ/mol 1 2. 2 F2 (g) ¡ F(g) ¢H°2 5 75.3 kJ/mol 3. Li(g) ¡ Li1 (g) 1 e2 ¢H°3 5 520 kJ/mol 4. F(g) 1 e2 ¡ F2 (g) ¢H°4 5 2328 kJ/mol 5. Li (g) 1 F2 (g) ¡ 1 LiF(s) ¢H°5 5 ? Li(s) 1 12F2 (g) ¡ LiF(s) ¢H°overall 5 2594.1 kJ/mol According to Hess’s law, we can write ¢H°overall 5 ¢H°1 1 ¢H°2 1 ¢H°3 1 ¢H°4 1 ¢H°5 or 2594.1 kJ/mol 5 155.2 kJ/mol 1 75.3 kJ/mol 1 520 kJ/mol 2 328 kJ/mol 1 ¢H°5 Hence, ¢H°5 5 21017 kJ/mol and the lattice energy of LiF is 11017 kJ/mol. Figure 9.2 summarizes the Born-Haber cycle for LiF. Steps 1, 2, and 3 all require the input of energy. On the other hand, steps 4 and 5 release energy. Because ¢H°5 is a large negative quantity, the lattice energy of LiF is a large positive quantity, which accounts for the stability of solid LiF. The greater the lattice energy, the more stable the ionic compound. Keep in mind that lattice energy is always a positive quantity because the separation of ions in a solid into ions in the gas phase is, by Coulomb’s law, an endothermic process. Table 9.1 lists the lattice energies and the melting points of several common ionic compounds. There is a rough correlation between lattice energy and melting point. The larger the lattice energy, the more stable the solid and the more tightly held the ions. It takes more energy to melt such a solid, and so the solid has a higher melting point than one with a smaller lattice energy. Note that MgCl2, Na2O, and MgO have Figure 9.2 The Born-Haber cycle for the formation of 1 mole Li+(g) + F –(g) of solid LiF. Ionization –(Electron affinity) Δ H°3 = 520 kJ Δ H°4 = –328 kJ –(Lattice energy) Δ H°5 = –1017 kJ Li(g) + F(g) Sublimation Dissociation Δ H°1 = 155.2 kJ Δ H°2 = 75.3 kJ Δ H°overall = –594.1 kJ Li(s) + 1 F (g) LiF(s) 2 2  9.3 Lattice Energy of Ionic Compounds 375 Lattice Energies and Melting Points of Some Alkali Metal Table 9.1 and Alkaline Earth Metal Halides and Oxides Compound Lattice Energy (kJ/mol) Melting Point (8C) LiF 1017 845 LiCl 828 610 LiBr 787 550 LiI 732 450 NaCl 788 801 NaBr 736 750 NaI 686 662 KCl 699 772 KBr 689 735 KI 632 680 MgCl2 2527 714 Na2O 2570 Sub* MgO 3890 2800 *Na2O sublimes at 1275°C. unusually high lattice energies. The first of these ionic compounds has a doubly charged cation (Mg21) and the second a doubly charged anion (O22); in the third compound there is an interaction between two doubly charged species (Mg21 and O22). The coulombic attractions between two doubly charged species, or between a doubly charged ion and a singly charged ion, are much stronger than those between singly charged anions and cations. Lattice Energy and the Formulas of Ionic Compounds Because lattice energy is a measure of the stability of ionic compounds, its value can help us explain the formulas of these compounds. Consider magnesium chloride as an example. We have seen that the ionization energy of an element increases rapidly as successive electrons are removed from its atom. For example, the first ionization energy of magnesium is 738 kJ/mol, and the second ionization energy is 1450 kJ/mol, almost twice the first. We might ask why, from the standpoint of energy, magnesium does not prefer to form unipositive ions in its compounds. Why doesn’t magnesium chloride have the formula MgCl (containing the Mg1 ion) rather than MgCl2 (contain- ing the Mg21 ion)? Admittedly, the Mg21 ion has the noble gas configuration [Ne], which represents stability because of its completely filled shells. But the stability gained through the filled shells does not, in fact, outweigh the energy input needed to remove an electron from the Mg1 ion. The reason the formula is MgCl2 lies in the extra stability gained by the formation of solid magnesium chloride. The lattice energy of MgCl2 is 2527 kJ/mol, which is more than enough to compensate for the energy needed to remove the first two electrons from a Mg atom (738 kJ/mol 1 1450 kJ/mol 5 2188 kJ/mol). What about sodium chloride? Why is the formula for sodium chloride NaCl and not NaCl2 (containing the Na21 ion)? Although Na21 does not have the noble gas electron configuration, we might expect the compound to be NaCl2 because Na21 has a higher charge and therefore the hypothetical NaCl2 should have a greater lattice energy. Again, the answer lies in the balance between energy input (that is, ionization CHEMISTRY in Action Sodium Chloride—A Common and Important Ionic Compound W e are all familiar with sodium chloride as table salt. It is a typical ionic compound, a brittle solid with a high melt- ing point (801°C) that conducts electricity in the molten state and in aqueous solution. The structure of solid NaCl is shown in Figure 2.13. One source of sodium chloride is rock salt, which is found in subterranean deposits often hundreds of meters thick. It is also obtained from seawater or brine (a concentrated NaCl solu- tion) by solar evaporation. Sodium chloride also occurs in na- ture as the mineral halite. Sodium chloride is used more often than any other material in the manufacture of inorganic chemicals. World consumption of this substance is about 200 million tons per year. The major use of sodium chloride is in the production of other essential inorganic chemicals such as chlorine gas, sodium hydroxide, sodium metal, hydrogen gas, and sodium carbonate. It is also Solar evaporation process for obtaining sodium chloride. used to melt ice and snow on highways and roads. However, because sodium chloride is harmful to plant life and promotes corrosion of cars, its use for this purpose is of considerable en- vironmental concern. Meat processing, food canning, water softening, paper pulp, textiles Chlor-alkali process and dyeing, rubber (Cl2, NaOH, Na, H2) and oil industry 50% Na2CO3 10% 12% 4% Melting ice 3% on roads 4% 17% Other Domestic table salt chemical manufacture Animal feed Underground rock salt mining. Uses of sodium chloride. energies) and the stability gained from the formation of the solid. The sum of the first two ionization energies of sodium is 496 kJ/mol 1 4560 kJ/mol 5 5056 kJ/mol The compound NaCl2 does not exist, but if we assume a value of 2527 kJ/mol as its lattice energy (same as that for MgCl2), we see that the energy yield would be far too small to compensate for the energy required to produce the Na21 ion. 376  9.4 The Covalent Bond 377 What has been said about the cations applies also to the anions. In Section 8.5 we observed that the electron affinity of oxygen is 141 kJ/mol, meaning that the fol- lowing process releases energy (and is therefore favorable): O(g) 1 e2 ¡ O2 (g) As we would expect, adding another electron to the O2 ion O2 (g) 1 e 2 ¡ O22 (g) would be unfavorable in the gas phase because of the increase in electrostatic repul- sion. Indeed, the electron affinity of O2 is negative (2844 kJ/mol). Yet compounds containing the oxide ion (O22) do exist and are very stable, whereas compounds containing the O2 ion are not known. Again, the high lattice energy resulting from the O22 ions in compounds such as Na2O and MgO far outweighs the energy needed to produce the O22 ion. Review of Concepts Which of the following compounds has a larger lattice energy, LiCl or CsBr? 9.4 The Covalent Bond Although the concept of molecules goes back to the seventeenth century, it was not Animation Formation of a Covalent Bond until early in the twentieth century that chemists began to understand how and why molecules form. The first major breakthrough was Gilbert Lewis’s suggestion that a chemical bond involves electron sharing by atoms. He depicted the formation of a chemical bond in H2 as H #1 #H ¡ H:H This type of electron pairing is an example of a covalent bond, a bond in which two electrons are shared by two atoms. Covalent compounds are compounds that contain only covalent bonds. For the sake of simplicity, the shared pair of elec- trons is often represented by a single line. Thus, the covalent bond in the hydro- gen molecule can be written as H¬H. In a covalent bond, each electron in a shared pair is attracted to the nuclei of both atoms. This attraction holds the two atoms in H2 together and is responsible for the formation of covalent bonds in other molecules. Covalent bonding between many-electron atoms involves only the valence elec- This discussion applies only to trons. Consider the fluorine molecule, F2. The electron configuration of F is 1s22s22p5. representative elements. Remember that for these elements, the number of The 1s electrons are low in energy and stay near the nucleus most of the time. For valence electrons is equal to the group number (Groups 1A–7A). this reason they do not participate in bond formation. Thus, each F atom has seven valence electrons (the 2s and 2p electrons). According to Figure 9.1, there is only one unpaired electron on F, so the formation of the F2 molecule can be represented as follows: SO F T  TO Q Q F SO F S 88n SO Q FS Q or O SQ OS FOF Q Note that only two valence electrons participate in the formation of F2. The other, nonbonding electrons, are called lone pairs—pairs of valence electrons that are 378 Chapter 9 ■ Chemical Bonding I: Basic Concepts not involved in covalent bond formation. Thus, each F in F2 has three lone pairs of electrons: lone pairs F OO SO Q FS Q lone pairs The structures we use to represent covalent compounds, such as H2 and F2, are called Lewis structures. A Lewis structure is a representation of covalent bonding in which shared electron pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms. Only valence electrons are shown in a Lewis structure. Let us consider the Lewis structure of the water molecule. Figure 9.1 shows the Lewis dot symbol for oxygen with two unpaired dots or two unpaired electrons, so we expect that O might form two covalent bonds. Because hydrogen has only one electron, it can form only one covalent bond. Thus, the Lewis structure for water is OS H HSO or O HOOOH Q Q In this case, the O atom has two lone pairs. The hydrogen atom has no lone pairs because its only electron is used to form a covalent bond. In the F2 and H2O molecules, the F and O atoms achieve a noble gas configura- tion by sharing electrons: O F SO SQ FS OS H H SO Q Q 8e 8e 2e 8e 2e The formation of these molecules illustrates the octet rule, formulated by Lewis: An atom other than hydrogen tends to form bonds until it is surrounded by eight valence electrons. In other words, a covalent bond forms when there are not enough electrons for each individual atom to have a complete octet. By sharing electrons in a covalent bond, the individual atoms can complete their octets. The requirement for hydrogen is that it attain the electron configuration of helium, or a total of two electrons. The octet rule works mainly for elements in the second period of the periodic table. These elements have only 2s and 2p subshells, which can hold a total of eight electrons. When an atom of one of these elements forms a covalent com- pound, it can attain the noble gas electron configuration [Ne] by sharing electrons with other atoms in the same compound. Later, we will discuss a number of impor- tant exceptions to the octet rule that give us further insight into the nature of chemical bonding. Atoms can form different types of covalent bonds. In a single bond, two atoms are held together by one electron pair. Many compounds are held together by mul- tiple bonds, that is, bonds formed when two atoms share two or more pairs of electrons. If two atoms share two pairs of electrons, the covalent bond is called a double bond. Double bonds are found in molecules of carbon dioxide (CO2) and ethylene (C2H4): H H H H G D Shortly you will be introduced to the O QSSC SSO O O C SSC S S rules for writing proper Lewis structures. O O Q or OPCPO Q Q or CPC D G S S Here we simply want to become familiar H H H H with the language associated with them. 8e 8e 8e 8e 8e  9.4 The Covalent Bond 379 A triple bond arises when two atoms share three pairs of electrons, as in the nitrogen 74 pm 161 pm molecule (N2): SN OO NS or SNqNS O 8e 8e  H2 HI Figure 9.3 Bond length (in pm) in The acetylene molecule (C2H2) also contains a triple bond, in this case between two H2 and HI. carbon atoms: H SC OO CS H or HOCqCOH O 8e 8e  Note that in ethylene and acetylene all the valence electrons are used in bonding; there are no lone pairs on the carbon atoms. In fact, with the exception of carbon monoxide, stable molecules containing carbon do not have lone pairs on the car- bon atoms. Multiple bonds are shorter than single covalent bonds. Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule (Figure 9.3). Table 9.2 shows some experimentally determined bond lengths. For a given pair of atoms, such as carbon and nitrogen, triple bonds are shorter than dou- ble bonds, which, in turn, are shorter than single bonds. The shorter multiple bonds are also more stable than single bonds, as we will see later. Comparison of the Properties of Covalent and Ionic Compounds Animation Ionic vs. Covalent Bonding Ionic and covalent compounds differ markedly in their general physical properties because of differences in the nature of their bonds. There are two types of attractive Animation Ionic and Covalent Bonding forces in covalent compounds. The first type is the force that holds the atoms together in a molecule. A quantitative measure of this attraction is given by bond enthalpy, to be discussed in Section 9.10. The second type of attractive force oper- ates between molecules and is called an intermolecular force. Because intermolecu- If intermolecular forces are weak, it is relatively easy to break up aggregates of lar forces are usually quite weak compared with the forces holding atoms together molecules to form liquids (from solids) within a molecule, molecules of a covalent compound are not held together tightly. and gases (from liquids). Consequently covalent compounds are usually gases, liquids, or low-melting solids. On the other hand, the electrostatic forces holding ions together in an ionic com- pound are usually very strong, so ionic compounds are solids at room temperature and have high melting points. Many ionic compounds are soluble in water, and the resulting aqueous solutions conduct electricity, because the compounds are strong electrolytes. Most covalent compounds are insoluble in water, or if they do dissolve, their aqueous solutions generally do not conduct electricity, because the compounds are nonelectrolytes. Molten ionic compounds conduct electricity because they con- tain mobile cations and anions; liquid or molten covalent compounds do not conduct electricity because no ions are present. Table 9.3 compares some of the general properties of a typical ionic compound, sodium chloride, with those of a covalent compound, carbon tetrachloride (CCl4). Review of Concepts Why is it not possible for hydrogen to form double or triple bonds in covalent compounds? 380 Chapter 9 ■ Chemical Bonding I: Basic Concepts Table 9.2 Comparison of Some General Properties of an Ionic Compound Table 9.3 Average Bond Lengths of and a Covalent Compound Some Common Single, Double, and Triple Bonds Property NaCl CCl4 Bond Appearance White solid Colorless liquid Length Melting point (°C) 801 223 Bond Type (pm) Molar heat of fusion* (kJ/mol) 30.2 2.5 C¬H 107 Boiling point (°C) 1413 76.5 C¬O 143 Molar heat of vaporization* (kJ/mol) 600 30 C“O 121 Density (g/cm3) 2.17 1.59 C¬C 154 Solubility in water High Very low C“C 133 Electrical conductivity C‚C 120 Solid Poor Poor C¬N 143 Liquid Good Poor C“N 138 *Molar heat of fusion and molar heat of vaporization are the amounts of heat needed to melt 1 mole of the solid and to C‚N 116 vaporize 1 mole of the liquid, respectively. N¬O 136 N“O 122 O¬H 96 9.5 Electronegativity A covalent bond, as we have said, is the sharing of an electron pair by two atoms. In a molecule like H2, in which the atoms are identical, we expect the electrons to be equally shared—that is, the electrons spend the same amount of time in the vicinity of each atom. However, in the covalently bonded HF molecule, the H Hydrogen fluoride is a clear, fuming and F atoms do not share the bonding electrons equally because H and F are liquid that boils at 19.88C. It is used to make refrigerants and to prepare different atoms: hydrofluoric acid. OS H—F Q The bond in HF is called a polar covalent bond, or simply a polar bond, because the electrons spend more time in the vicinity of one atom than the other. Experi- mental evidence indicates that in the HF molecule the electrons spend more time near the F atom. We can think of this unequal sharing of electrons as a partial electron transfer or a shift in electron density, as it is more commonly described, from H to F (Figure 9.4). This “unequal sharing” of the bonding electron pair results in a relatively greater electron density near the fluorine atom and a correspondingly lower electron density near hydrogen. The HF bond and other polar bonds can be thought of as being intermediate between a (nonpolar) covalent bond, in which the sharing of electrons is exactly equal, and an ionic bond, in which the transfer of the electron(s) is nearly complete. A property that helps us distinguish a nonpolar covalent bond from a polar cova- lent bond is electronegativity, the ability of an atom to attract toward itself the elec- trons in a chemical bond. Elements with high electronegativity have a greater tendency to attract electrons than do elements with low electronegativity. As we might expect, Figure 9.4 Electrostatic potential electronegativity is related to electron affinity and ionization energy. Thus, an atom map of the HF molecule. The such as fluorine, which has a high electron affinity (tends to pick up electrons easily) distribution varies according to the and a high ionization energy (does not lose electrons easily), has a high electronega- colors of the rainbow. The most electron-rich region is red; the tivity. On the other hand, sodium has a low electron affinity, a low ionization energy, most electron-poor region is blue. and a low electronegativity.  9.5 Electronegativity 381 Increasing electronegativity 1A 8A H 2.1 2A 3A 4A 5A 6A 7A Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Increasing electronegativity Na Mg Al Si P S Cl 0.9 1.2 3B 4B 5B 6B 7B 8B 1B 2B 1.5 1.8 2.1 2.5 3.0 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.9 1.9 1.9 1.6 1.6 1.8 2.0 2.4 2.8 3.0 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 2.6 Cs Ba La-Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At 0.7 0.9 1.0-1.2 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.9 1.9 2.0 2.2 Fr Ra 0.7 0.9 Figure 9.5 The electronegativities of common elements. Electronegativity is a relative concept, meaning that an element’s electro- Electronegativity values have no units. negativity can be measured only in relation to the electronegativity of other ele- ments. Linus Pauling† devised a method for calculating relative electronegativities of most elements. These values are shown in Figure 9.5. A careful examination of this chart reveals trends and relationships among electronegativity values of different elements. In general, electronegativity increases from left to right across a period in the periodic table, as the metallic character of the elements decreases. Within each group, electronegativity decreases with increasing atomic number, and increasing metallic character. Note that the transition metals do not follow these trends. The most electronegative elements—the halogens, oxygen, nitrogen, and sulfur—are found in the upper right-hand corner of the periodic table, and the least electronegative elements (the alkali and alkaline earth metals) are clus- tered near the lower left-hand corner. These trends are readily apparent on a graph, as shown in Figure 9.6. Atoms of elements with widely different electronegativities tend to form ionic bonds (such as those that exist in NaCl and CaO compounds) with each other because the atom of the less electronegative element gives up its electron(s) to the atom of the more electronegative element. An ionic bond generally joins an atom of a metallic element and an atom of a nonmetallic element. Atoms of elements with comparable electronegativities tend to form polar covalent bonds with each other because the shift in electron density is usually small. Most covalent bonds involve atoms of nonmetallic elements. Only atoms of the same element, which have the same electronegativity, can be joined by a pure covalent bond. These trends and characteristics are what we would expect, given our knowledge of ionization ener- gies and electron affinities. † Linus Carl Pauling (1901–1994). American chemist. Regarded by many as the most influential chemist of the twentieth century, Pauling did research in a remarkably broad range of subjects, from chemical physics to molecular biology. Pauling received the Nobel Prize in Chemistry in 1954 for his work on protein structure, and the Nobel Peace Prize in 1962. He is the only person to be the sole recipient of two Nobel prizes. 382 Chapter 9 ■ Chemical Bonding I: Basic Concepts F 4 Cl 3 Br Electronegativity I Ru H 2 Mn Zn 1 Li Na Rb K 0 10 20 30 40 50 Atomic number Figure 9.6 Variation of electronegativity with atomic number. The halogens have the highest electronegativities, and the alkali metals the lowest. 100 There is no sharp distinction between a polar bond and an ionic bond, but the KBr LiF following general rule is helpful in distinguishing between them. An ionic bond forms KCl CsI KF when the electronegativity difference between the two bonding atoms is 2.0 or more. Percent ionic character 75 This rule applies to most but not all ionic compounds. Sometimes chemists use the KI CsCl CsF quantity percent ionic character to describe the nature of a bond. A purely ionic bond LiBr NaCl 50 LiI LiCl would have 100 percent ionic character, although no such bond is known, whereas a nonpolar or purely covalent bond has 0 percent ionic character. As Figure 9.7 shows, HF there is a correlation between the percent ionic character of a bond and the electro- 25 negativity difference between the bonding atoms. ICl IBr HCl Electronegativity and electron affinity are related but different concepts. Both HI HBr Cl2 indicate the tendency of an atom to attract electrons. However, electron affinity 0 0 1 2 3 refers to an isolated atom’s attraction for an additional electron, whereas electro- Electronegativity difference negativity signifies the ability of an atom in a chemical bond (with another atom) to attract the shared electrons. Furthermore, electron affinity is an experimentally Figure 9.7 Relation between percent ionic character and measurable quantity, whereas electronegativity is an estimated number that cannot electronegativity difference. be measured. Example 9.2 shows how a knowledge of electronegativity can help us determine whether a chemical bond is covalent or ionic. 1A 8A 2A 3A 4A 5A 6A 7A Example 9.2 Classify the following bonds as ionic, polar covalent, or covalent: (a) the bond in HCl, (b) the bond in KF, and (c) the CC bond in H3CCH3. The most electronegative elements Strategy We follow the 2.0 rule of electronegativity difference and look up the values are the nonmetals (Groups 5A–7A) in Figure 9.5. and the least electronegative elements are the alkali and alkaline Solution earth metals (Groups 1A–2A) and aluminum. Beryllium, the first (a) The electronegativity difference between H and Cl is 0.9, which is appreciable but member of Group 2A, forms not large enough (by the 2.0 rule) to qualify HCl as an ionic compound. Therefore, mostly covalent compounds. the bond between H and Cl is polar covalent. (Continued)  9.5 Electronegativity 383 (b) The electronegativity difference between K and F is 3.2, which is well above the 2.0 mark; therefore, the bond between K and F is ionic. (c) The two C atoms are identical in every respect—they are bonded to each other and each is bonded to three other H atoms. Therefore, the bond between them is purely covalent. Similar problems: 9.39, 9.40. Practice Exercise Which of the following bonds is covalent, which is polar covalent, and which is ionic? (a) the bond in CsCl, (b) the bond in H2S, (c) the NN bond in H2NNH2. Electronegativity and Oxidation Number In Chapter 4 we introduced the rules for assigning oxidation numbers of elements in their compounds. The concept of electronegativity is the basis for these rules. In essence, oxidation number refers to the number of charges an atom would have if electrons were transferred completely to the more electronegative of the bonded atoms in a molecule. Consider the NH3 molecule, in which the N atom forms three single bonds with the H atoms. Because N is more electronegative than H, electron density will be shifted from H to N. If the transfer were complete, each H would donate an electron to N, which would have a total charge of 23 while each H would have a charge of 11. Thus, we assign an oxidation number of 23 to N and an oxidation number of 11 to H in NH3. Oxygen usually has an oxidation number of 22 in its compounds, except in hydrogen peroxide (H2O2), whose Lewis structure is OOO HOO O Q QOH A bond between identical atoms makes no contribution to the oxidation number of those atoms because the electron pair of that bond is equally shared. Because H has an oxidation number of 11, each O atom has an oxidation number of 21. Can you see now why fluorine always has an oxidation number of 21? It is the most electronegative element known, and it always forms a single bond in its compounds. Therefore, it would bear a 21 charge if electron transfer were complete. Review of Concepts Identify the electrostatic potential maps shown here with HCl and LiH. In both diagrams, the H atom is on the left. 384 Chapter 9 ■ Chemical Bonding I: Basic Concepts 9.6 Writing Lewis Structures Although the octet rule and Lewis structures do not present a complete picture of covalent bonding, they do help to explain the bonding scheme in many compounds and account for the properties and reactions of molecules. For this reason, you should practice writing Lewis structures of compounds. The basic steps are as follows: 1. Write the skeletal structure of the compound, using chemical symbols and placing bonded atoms next to one another. For simple compounds, this task is fairly easy. For more complex compounds, we must either be given the information or make an intelligent guess about it. In general, the least electronegative atom occupies the central position. Hydrogen and fluorine usually occupy the terminal (end) positions in the Lewis structure. 2. Count the total number of valence electrons present, referring, if necessary, to Figure 9.1. For polyatomic anions, add the number of negative charges to that total. (For example, for the CO322 ion we add two electrons because the 22 charge indicates that there are two more electrons than are provided by the atoms.) For polyatomic cations, we subtract the number of positive charges from this total. (Thus, for NH14 we subtract one electron because the 11 charge indicates a loss of one electron from the group of atoms.) 3. Draw a single covalent bond between the central atom and each of the surrounding atoms. Complete the octets of the atoms bonded to the central atom. (Remember Hydrogen follows a “duet rule” when that the valence shell of a hydrogen atom is complete with only two electrons.) drawing Lewis structures. Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding. The total number of electrons to be used is that determined in step 2. 4. After completing steps 1–3, if the central atom has fewer than eight electrons, try adding double or triple bonds between the surrounding atoms and the central atom, using lone pairs from the surrounding atoms to complete the octet of the central atom. Examples 9.3, 9.4, and 9.5 illustrate the four-step procedure for writing Lewis structures of compounds and an ion. Example 9.3 Write the Lewis structure for nitrogen trifluoride (NF3) in which all three F atoms are bonded to the N atom. NF3 is a colorless, odorless, Solution We follow the preceding procedure for writing Lewis structures. unreactive gas. Step 1: The N atom is less electronegative than F, so the skeletal structure of NF 3 is F N F F Step 2: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, respectively. Thus, there are 5 1 (3 3 7), or 26, valence electrons to account for in NF3. (Continued)  9.6 Writing Lewis Structures 385 Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N: SO Q N OO F OO FS Q A FS SQ Because this structure satisfies the octet rule for all the atoms, step 4 is not required. Check Count the valence electrons in NF3 (in bonds and in lone pairs). The result is 26, the same as the total number of valence electrons on three F atoms (3 3 7 5 21) and one N atom (5). Similar problem: 9.45. Practice Exercise Write the Lewis structure for carbon disulfide (CS2). Example 9.4 Write the Lewis structure for nitric acid (HNO3) in which the three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms. Solution We follow the procedure already outlined for writing Lewis structures. Step 1: The skeletal structure of HNO3 is O N O H HNO3 is a strong electrolyte. O Step 2: The outer-shell electron configurations of N, O, and H are 2s22p3, 2s22p4, and 1s1, respectively. Thus, there are 5 1 (3 3 6) 1 1, or 24, valence electrons to account for in HNO3. Step 3: We draw a single covalent bond between N and each of the three O atoms and between one O atom and the H atom. Then we fill in electrons to comply with the octet rule for the O atoms: O O SOONOOOH Q Q A SQ OS Step 4: We see that this structure satisfies the octet rule for all the O atoms but not for the N atom. The N atom has only six electrons. Therefore, we move a lone pair from one of the end O atoms to form another bond with N. Now the octet rule is also satisfied for the N atom: O O OPNOOOH Q Q A SQ OS Check Make sure that all the atoms (except H) satisfy the octet rule. Count the valence electrons in HNO3 (in bonds and in lone pairs). The result is 24, the same as the total number of valence electrons on three O atoms (3 3 6 5 18), one N atom (5), and one H atom (1). Similar problem: 9.45. Practice Exercise Write the Lewis structure for formic acid (HCOOH). 386 Chapter 9 ■ Chemical Bonding I: Basic Concepts Example 9.5 Write the Lewis structure for the carbonate ion (CO22 3 ). Solution We follow the preceding procedure for writing Lewis structures and note that this is an anion with two negative charges. Step 1: We can deduce the skeletal structure of the carbonate ion by recognizing that C CO22 3 is less electronegative than O. Therefore, it is most likely to occupy a central position as follows: O O C O Step 2: The outer-shell electron configurations of C and O are 2s22p2 and 2s22p4, respectively, and the ion itself has two negative charges. Thus, the total number of electrons is 4 1 (3 3 6) 1 2, or 24. Step 3: We draw a single covalent bond between C and each O and comply with the octet rule for the O atoms: SO OS A O SOOCOOSO Q Q This structure shows all 24 electrons. Step 4: Although the octet rule is satisfied for the O atoms, it is not for the C atom. Therefore, we move a lone pair from one of the O atoms to form another bond with C. Now the octet rule is also satisfied for the C atom: We use the square brackets to indicate SO S 2 that the 22 charge is on the whole ion. B O SOOCOOS O Q Q Check Make sure that all the atoms satisfy the octet rule. Count the valence electrons in CO22 3 (in chemical bonds and in lone pairs). The result is 24, the same as the total number of valence electrons on three O atoms (3 3 6 5 18), one C atom (4), and two Similar problem: 9.44. negative charges (2). Practice Exercise Write the Lewis structure for the nitrite ion (NO22 ) . Review of Concepts The molecular model shown here represents guanine, a component of a DNA molecule. Only the connections between the atoms are shown in this model. Draw a complete Lewis structure of the molecule, showing all the multiple bonds and lone pairs. (For color code, see inside back endpaper.)  9.7 Formal Charge and Lewis Structure 387 9.7 Formal Charge and Lewis Structure By comparing the number of electrons in an isolated atom with the number of elec- trons that are associated with the same atom in a Lewis structure, we can determine the distribution of electrons in the molecule and draw the most plausible Lewis struc- ture. The bookkeeping procedure is as follows: In an isolated atom, the number of electrons associated with the atom is simply the number of valence electrons. (As usual, we need not be concerned with the inner electrons.) In a molecule, electrons associated with the atom are the nonbonding electrons plus the electrons in the bond- ing pair(s) between the atom and other atom(s). However, because electrons are shared in a bond, we must divide the electrons in a bonding pair equally between the atoms forming the bond. An atom’s formal charge is the electrical charge difference between the valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. To assign the number of electrons on an atom in a Lewis structure, we proceed as follows: • All the atom’s nonbonding electrons are assigned to the atom. • We break the bond(s) between the atom and other atom(s) and assign half of the bonding electrons to the atom. Let us illustrate the concept of formal charge using the ozone molecule (O3). Proceeding by steps, as we did in Examples 9.3 and 9.4, we draw the skeletal structure of O3 and then add bonds and electrons to satisfy the octet rule for the two end atoms: SO O OS OOOOO Q Q You can see that although all available electrons are used, the octet rule is not satisfied for the central atom. To remedy this, we convert a lone pair on one of the end atoms to a second bond between that end atom and the central atom, as follows: O O OS OPOOO Q Q Liquid ozone below its boiling The formal charge on each atom in O3 can now be calculated according to the fol- point (2111.3°C). Ozone is a toxic, lowing scheme: light-blue gas with a pungent odor. O O OS OPOOO Q Q Valence e 6 6 6 e assigned to atom 6 5 7 Assign half of the bonding electrons to each atom. Difference 0 1 1 (formal charge) where the wavy red lines denote the breaking of the bonds. Note that the breaking of a single bond results in the transfer of an electron, the breaking of a double bond results in a transfer of two electrons to each of the bonding atoms, and so on. Thus, the formal charges of the atoms in O3 are  O O OS OPOOO Q Q For single positive and negative charges, we normally omit the numeral 1. When you write formal charges, these rules are helpful: 1. For molecules, the sum of the charges must add up to zero because molecules In determining formal charges, does the atom in the molecule (or ion) have more are electrically neutral species. (This rule applies, for example, to the O 3 electrons than its valence electrons molecule.) (negative formal charge), or does the atom have fewer electrons than its 2. For cations, the sum of formal charges must equal the positive charge. For anions, valence electrons (positive formal charge)? the sum of formal charges must equal the negative charge. 388 Chapter 9 ■ Chemical Bonding I: Basic Concepts Note that formal charges help us keep track of valence electrons and gain a qualitative picture of charge distribution in a molecule. We should not interpret formal charges as actual, complete transfer of electrons. In the O 3 molecule, for example, experimental studies do show that the central O atom bears a partial positive charge while the end O atoms bear a partial negative charge, but there is no evidence that there is a complete transfer of electrons from one atom to another. Example 9.6 Write formal charges for the carbonate ion. Strategy The Lewis structure for the carbonate ion was developed in Example 9.5: SO S 2 B O SOOCOOS O Q Q The formal charges on the atoms can be calculated using the given procedure. Solution We subtract the number of nonbonding electrons and half of the bonding electrons from the valence electrons of each atom. The C atom: The C atom has four valence electrons and there are no nonbonding electrons on the atom in the Lewis structure. The breaking of the double bond and two single bonds results in the transfer of four electrons to the C atom. Therefore, the formal charge is 4 2 4 5 0. The O atom in C“O: The O atom has six valence electrons and there are four nonbonding electrons on the atom. The breaking of the double bond results in the transfer of two electrons to the O atom. Here the formal charge is 6 2 4 2 2 5 0. The O atom in C¬O: This atom has six nonbonding electrons and the breaking of the single bond transfers another electron to it. Therefore, the formal charge is 6 2 6 2 1 5 21. Thus, the Lewis structure for CO22 3 with formal charges is SO S B  O O  SOOCOOS Q Q Check Note that the sum of the formal charges is 22, the same as the charge on the Similar problem: 9.46. carbonate ion. Practice Exercise Write formal charges for the nitrite ion (NO22 ). Sometimes there is more than one acceptable Lewis structure for a given species. In such cases, we can often select the most plausible Lewis structure by using formal charges and the following guidelines: • For molecules, a Lewis structure in which there are no formal charges is prefer- able to one in which formal charges are present. • Lewis structures with large formal charges (12, 13, and/or 22, 23, and so on) are less plausible than those with small formal charges. • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.  9.7 Formal Charge and Lewis Structure 389 Example 9.7 shows how formal charges facilitate the choice of the correct Lewis structure for a molecule. Example 9.7 Formaldehyde (CH2O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound. Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity CH2O guidelines. Solution The two possible skeletal structures are H H C O H C O H (a) (b) First we draw the Lewis structures for each of these possibilities   H G O O HOCPOOH O CPO D Q H (a) (b) To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 2 5 5 21. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 2 5 5 11. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 2 4 5 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 2 6 5 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible? Similar problem: 9.47. Practice Exercise Draw the most reasonable Lewis structure of a molecule that contains a N atom, a C atom, and a H atom. Review of Concepts Consider three possible atomic arrangements for cyanamide (CH2N2): (a) H2CNN, (b) H2NCN, (c) HNNCH. Using formal charges as a guide, determine which is the most plausible arrangement. 390 Chapter 9 ■ Chemical Bonding I: Basic Concepts 9.8 The Concept of Resonance Our drawing of the Lewis structure for ozone (O3) satisfied the octet rule for the central atom because we placed a double bond between it and one of the two end O atoms. In fact, we can put the double bond at either end of the molecule, as shown by these two equivalent Lewis structures:   O O OS OPOOO  O O O SOOOPO Q Q Q Q Electrostatic potential map of O3. However, neither one of these two Lewis structures accounts for the known bond The electron density is evenly lengths in O3. distributed between the two end O atoms. We would expect the O¬O bond in O3 to be longer than the O“O bond because double bonds are known to be shorter than single bonds. Yet experimental evidence shows that both oxygen-to-oxygen bonds are equal in length (128 pm). We resolve Animation this discrepancy by using both Lewis structures to represent the ozone molecule: Resonance   O O OS mn OPOOO SO  O O OOOPO Q Q Q Q Each of these structures is called a resonance structure. A resonance structure, then, is one of two or more Lewis structures for a single molecule that cannot be repre- sented accurately by only one Lewis structure. The double-headed arrow indicates that the structures shown are resonance structures. The term resonance itself means the use of two or more Lewis structures to represent a particular molecule. Like the medieval European traveler to Africa who described a rhinoceros as a cross between a griffin and a unicorn, two familiar but imaginary animals, we describe ozone, a real molecule, in terms of two familiar but nonexistent structures. A common misconception about resonance is the notion that a molecule such as ozone somehow shifts quickly back and forth from one resonance structure to the other. Keep in mind that neither resonance structure adequately represents the actual molecule, which has its own unique, stable structure. “Resonance” is a human inven- tion, designed to address the limitations in these simple bonding models. To extend the animal analogy, a rhinoceros is a distinct creature, not some oscillation between mythical griffin and unicorn! The carbonate ion provides another example of resonance: SOS SO OS SO OS B A A  O OS mn OPCOO O O   O O SOOCOO Q Q Q QS mn SQ OOCPO Q According to experimental evidence, all carbon-to-oxygen bonds in CO22 3 are equiv- alent. Therefore, the properties of the carbonate ion are best explained by considering its resonance structures together. The concept of resonance applies equally well to organic systems. A good exam- ple is the benzene molecule (C6H6): H H A A H C H H C H H K H E H E N E C C C C A B mn B A CN EC C H KC E C HH E C HH H H A A H H  9.8 The Concept of Resonance 391 If one of these resonance structures corresponded to the actual structure of benzene, there would be two different bond lengths between adjacent C atoms, one charac- teristic of the single bond and the other of the double bond. In fact, the distance between all adjacent C atoms in benzene is 140 pm, which is shorter than a C¬C bond (154 pm) and longer than a C“C bond (133 pm). A simpler way of drawing the structure of the benzene molecule and other compounds containing the “benzene ring” is to show only the skeleton and not the carbon and hydrogen atoms. By this convention the resonance structures are represented by mn Note that the C atoms at the corners of the hexagon and the H atoms are all omit- ted, although they are understood to exist. Only the bonds between the C atoms are shown. Remember this important rule for drawing resonance structures: The positions of electrons, but not those of atoms, can be rearranged in different resonance structures. In other words, the same atoms must be bonded to one another in all the resonance structures for a given species. So far, the resonance structures shown in the examples all contribute equally to the real structure of the molecules and ion. This is not always the case, as we will see in Example 9.8. Example 9.8 Draw three resonance structures for the molecule nitrous oxide, N2O (the atomic arrangement is NNO). Indicate formal charges. Rank the structures in their relative importance to the overall properties of the molecule. Strategy The skeletal structure for N2O is N N O We follow the procedure used for drawing Lewis structures and calculating formal charges in Examples 9.5 and 9.6. Solution The three resonance structures are ⫺ ⫹ ⫹ ⫺ 2⫺ ⫹ ⫹ O NPNPO O SNqNOOSO O SNONqOS O O O O (a) (b) (c) We see that all three structures show formal charges. Structure (b) is the most important one because the negative charge is on the more electronegative oxygen atom. Structure (c) is the least important one because it has a larger separation of formal charges. Also, Resonance structures with formal the positive charge is on the more electronegative oxygen atom. charges greater than 12 or 22 are usually considered highly implausible and can be discarded. Check Make sure there is no change in the positions of the atoms in the structures. Because N has five valence electrons and O has six valence electrons, the total number of valence electrons is 5 3 2 1 6 5 16. The sum of formal charges is zero in each structure. Similar problems: 9.51, 9.56. 2 Practice Exercise Draw three resonance structures for the thiocyanate ion, SCN . Rank the structures in decreasing order of importance. 392 Chapter 9 ■ Chemical Bonding I: Basic Concepts Review of Concepts The molecular model shown here represents acetamide, which is used as an organic solvent. Only the connections between the atoms are shown in this model. Draw two resonance structures for the molecule, showing the positions of multiple bonds and formal charges. (For color code, see inside back endpaper.) 9.9 Exceptions to the Octet Rule As mentioned earlier, the octet rule applies mainly to the second-period elements. Exceptions to the octet rule fall into three categories characterized by an incomplete octet, an odd number of electrons, or more than eight valence electrons around the central atom. The Incomplete Octet In some compounds, the number of electrons surrounding the central atom in a stable molecule is fewer than eight. Consider, for example, beryllium, which is a Group 2A (and a second-period) element. The electron configuration of beryllium is 1s22s2; it Beryllium, unlike the other Group has two valence electrons in the 2s orbital. In the gas phase, beryllium hydride (BeH2) 2A elements, forms mostly exists as discrete molecules. The Lewis structure of BeH2 is covalent compounds of which BeH2 is an example. H¬Be¬H As you can see, only four electrons surround the Be atom, and there is no way to satisfy the octet rule for beryllium in this molecule. Elements in Group 3A, particularly boron and aluminum, also tend to form com- pounds in which they are surrounded by fewer than eight electrons. Take boron as an example. Because its electron configuration is 1s22s22p1, it has a total of three valence electrons. Boron reacts with the halogens to form a class of compounds having the general formula BX3, where X is a halogen atom. Thus, in boron trifluoride there are only six electrons around the boron atom: SOFS A SO F O Q A B SQFS  9.9 Exceptions to the Octet Rule 393 The following resonance structures all contain a double bond between B and F and satisfy the octet rule for boron: SO FS S F S⫹ SOFS A B A ⫹O ⫺ O ⫺ O ⫺ FP Q AB mn S FO Q AB mn S F O Q B B FS SQ SQFS S F S⫹ The fact that the B¬F bond length in BF3 (130.9 pm) is shorter than a single bond (137.3 pm) lends support to the resonance structures even though in each case the negative formal charge is placed on the B atom and the positive formal charge on the more electronegative F atom. Although boron trifluoride is stable, it readily reacts with ammonia. This reaction is better represented by using the Lewis structure in which boron has only six valence electrons around it: SOFS H SO FS H A A A A ⴙ SO F O B ⫹ S N OH 88n SO Q FOB⫺ON⫹OH Q A A A A SQFS H FS H SQ It seems that the properties of BF3 are best explained by all four resonance structures. The B¬N bond in the above compound is different from the covalent bonds discussed so far in the sense that both electrons are contributed by the N atom. This type of bond is called a coordinate covalent bond (also referred to as a dative bond), 8n defined as a covalent bond in which one of the atoms donates both electrons. Although the properties of a coordinate covalent bond do not differ from those of a normal covalent bond (because all electrons are alike no matter what their source), the distinc- tion is useful for keeping track of valence electrons and assigning formal charges. Odd-Electron Molecules Some molecules contain an odd number of electrons. Among them are nitric oxide (NO) and nitrogen dioxide (NO2): O O NPO O P ⫹OOS OPN O ⫺ R Q Q Q Because we need an even number of electrons for complete pairing (to reach NH3 1 BF3 ¡ H3N¬BF3 eight), the octet rule clearly cannot be satisfied for all the atoms in any of these molecules. Odd-electron molecules are sometimes called radicals. Many radicals are highly reactive. The reason is that there is a tendency for the unpaired electron to form a covalent bond with an unpaired electron on another molecule. For example, when two nitrogen dioxide molecules collide, they form dinitrogen tetroxide in which the octet rule is satisfied for both the N and O atoms: 1A 8A 2A 3A 4A 5A 6A 7A M M M M M S M S O O O O M M D M M NT ⫹ TN 88n NON D D M M M S M S M O O O O M M M M Yellow: second-period elements The Expanded Octet cannot have an expanded octet. Blue: third-period elements and Atoms of the second-period elements cannot have more than eight valence elec- beyond can have an expanded trons around the central atom, but atoms of elements in and beyond the third octet. Green: the noble gases usually period of the periodic table form some compounds in which more than eight only have an expanded octet. 394 Chapter 9 ■ Chemical Bonding I: Basic Concepts electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbit- als enable an atom to form an expanded octet. One compound in which there is an expanded octet is sulfur hexafluoride, a very stable compound. The electron configuration of sulfur is [Ne]3s23p4. In SF6, each of sulfur’s six valence electrons forms a covalent bond with a fluorine atom, so there are 12 electrons around the central sulfur atom: SO FS F A O SO Q FS H EQ S E H F A O SO Q FS FS Q SQ In Chapter 10 we will see that these 12 electrons, or six bonding pairs, are accom- modated in six orbitals that originate from the one 3s, the three 3p, and two of the five 3d orbitals. Sulfur also forms many compounds in which it obeys the octet rule. In sulfur dichloride, for instance, S is surrounded by only eight electrons: Sulfur dichloride is a toxic, foul-smelling O O OS SClOSOCl cherry-red liquid (boiling point: 59°C). Q Q Q Examples 9.9–9.11 concern compounds that do not obey the octet rule. Example 9.9 At high temperatures aluminum iodide (Al2I6) dissociates into AlI3 molecules. Draw the Lewis structure for AlI3. Strategy We follow the procedures used in Examples 9.5 and 9.6 to draw the Lewis structure and calculate formal charges. AlI3 has a tendency to dimerize or Solution The outer-shell electron configurations of Al and I are 3s23p1 and 5s25p5, combine two units to form Al2I6. respectively. The total number of valence electrons is 3 1 3 3 7 or 24. Because Al is less electronegative than I, it occupies a central position and forms three bonds with the I atoms: O SIS A O A SIOAl Q A SIS Q Note that there are no formal charges on the Al and I atoms. Similar problem: 9.62. Check Although the octet rule is satisfied for the I atoms, there are only six valence electrons around the Al atom. Thus, AlI3 is an example of the incomplete octet. Practice Exercise Draw the Lewis structure for BeF2. Example 9.10 Draw the Lewis structure for phosphorus pentafluoride (PF5), in which all five F atoms are bonded to the central P atom. Strategy Note that P is a third-period element. We follow the procedures given in Examples 9.5 and 9.6 to draw the Lewis structure and calculate formal charges. PF5 is a reactive gaseous (Continued) compound.  9.9 Exceptions to the Octet Rule 395 Solution The outer-shell electron configurations for P and F are 3s23p3 and 2s22p5, respectively, and so the total number of valence electrons is 5 1 (5 3 7), or 40. Phosphorus, like sulfur, is a third-period element, and therefore it can have an expanded octet. The Lewis structure of PF5 is SO FS SO F A QH O OFS Q EP F A SO Q FS SQ Note that there are no formal charges on the P and F atoms. Check Although the octet rule is satisfied for the F atoms, there are 10 valence electrons around the P atom, giving it an expanded octet. Similar problem: 9.64. Practice Exercise Draw the Lewis structure for arsenic pentafluoride (AsF5). Example 9.11 Draw a Lewis structure for the sulfate ion (SO22 4 ) in which all four O atoms are bonded to the central S atom. Strategy Note that S is a third-period element. We follow the procedures given in Examples 9.5 and 9.6 to draw the Lewis structure and calculate formal charges. Solution The outer-shell electron configurations of S and O are 3s23p4 and 2s22p4, respectively. Step 1: The skeletal structure of (SO22 4 ) is SO22 4 O O S O O Step 2: Both O and S are Group 6A elements and so have six valence electrons each. Including the two negative charges, we must therefore account for a total of 6 1 (4 3 6) 1 2, or 32, valence electrons in SO22 4 . Step 3: We draw a single covalent bond between all the bonding atoms: O SOS A O S OOS SOO O Q Q A SOS Q Next we show formal charges on the S and O atoms: O ⫺ SOS A ⫺ O 2⫹ O ⫺ SOO Q S OOSQ A SOS Q ⫺ (Continued) 396 Chapter 9 ■ Chemical Bonding I: Basic Concepts Note that we can eliminate some of the formal charges for SO22 4 by expanding the S atom’s octet as follows: Note that this structure is only one of the SOS six equivalent structures for SO422. B ⫺ O S OOS SOO O ⫺ Q Q B SOS The question of which of these two structures is more important, that is, the one in which the S atom obeys the octet rule but bears more formal charges or the one in which the S atom expands its octet, has been the subject of some debate among chemists. In many cases, only elaborate quantum mechanical calculations can provide a clearer answer. At this stage of learning, you should realize that both representations are valid Lewis structures and you should be able to draw both types of structures. One helpful rule is that in trying to minimize formal charges by expanding the central atom’s octet, only add enough double bonds to make the formal charge on the central atom zero. Thus, the following structure would give formal charges on S(22) and O(0) that are inconsistent with the electronegativities of these elements and should therefore not be included to represent the SO22 4 ion. Similar problem: 9.85. SOS B O 22 O O PS PO Q Q B SOS Practice Exercise Draw reasonable Lewis structures of sulfuric acid (H2SO4). A final note about the expanded octet: In drawing Lewis structures of compounds containing a central atom from the third period and beyond, sometimes we find that the octet rule is satisfied for all the atoms but there are still valence electrons left to place. In such cases, the extra electrons should be placed as lone pairs on the central atom. Example 9.12 shows this approach. Example 9.12 Draw a Lewis structure of the noble gas compound xenon tetrafluoride (XeF4) in which all F atoms are bonded to the central Xe atom. Strategy Note that Xe is a fifth-period element. We follow the procedures in Examples 9.5 and 9.6 for drawing the Lewis structure and calculating formal charges. XeF4 Solution Step 1: The skeletal structure of XeF4 is F F Xe F F Step 2: The outer-shell electron configurations of Xe and F are 5s25p6 and 2s22p5, respectively, and so the total number of valence electrons is 8 1 (4 3 7) or 36. (Continued) CHEMISTRY in Action Just Say NO N itric oxide (NO), the simplest nitrogen oxide, is an odd- electron molecule, and therefore it is paramagnetic. A col- orless gas (boiling point: 2152°C), NO can be prepared in the muscles to relax and allows the arteries to dilate. In this respect, it is interesting to note that Alfred Nobel, the inven- tor of dynamite (a mixture of nitroglycerin and clay that laboratory by reacting sodium nitrite (NaNO2) with a reducing stabilizes the explosive before use), who established the agent such as Fe21 in an acidic medium. prizes bearing his name, had heart trouble. But he refused his doctor’s recommendation to ingest a small amount of NO2 21 1 2 (aq) 1 Fe (aq) 1 2H (aq) ¡ nitroglycerin to ease the pain. NO(g) 1 Fe31 (aq) 1 H2O(l) That NO evolved as a messenger molecule is entirely appropriate. Nitric oxide is small and so can diffuse quickly Environmental sources of nitric oxide include the burning from cell to cell. It is a stable molecule, but under certain cir- of fossil fuels containing nitrogen compounds and the reaction cumstances it is highly reactive, which accounts for its protec- between nitrogen and oxygen inside the automobile engine at tive function. The enzyme that brings about muscle relaxation high temperatures contains iron for which nitric oxide has a high affinity. It is the binding of NO to the iron that activates the enzyme. Neverthe- N2 (g) 1 O2 (g) ¡ 2NO(g) less, in the cell, where biological effectors are typically very Lightning also contributes to the atmospheric concentration of large molecules, the pervasive effects of one of the smallest NO. Exposed to air, nitric oxide quickly forms brown nitrogen known molecules are unprecedented. dioxide gas: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) Nitrogen dioxide is a major component of smog. About 40 years ago scientists studying muscle relax- ation discovered that our bodies produce nitric oxide for use as a neurotransmitter. (A neurotransmitter is a small mole- cule that serves to facilitate cell-to-cell communications.) Since then, it has been detected in at least a dozen cell types in various parts of the body. Cells in the brain, the liver, the pancreas, the gastrointestinal tract, and the blood vessels can synthesize nitric oxide. This molecule also functions as a cel- lular toxin to kill harmful bacteria. And that’s not all: In 1996 it was reported that NO binds to hemoglobin, the oxygen- carrying protein in the blood. No doubt it helps to regulate blood pressure. The discovery of the biological role of nitric oxide has shed light on how nitroglycerin (C 3H 5N 3O 9) works as a drug. For many years, nitroglycerin tablets have been pre- scribed for heart patients to relieve pain (angina pectoris) caused by a brief interference in the flow of blood to the Colorless nitric oxide gas is produced by the action of Fe21 on an acidic heart, although how it worked was not understood. We now sodium nitrite solution. The gas is bubbled through water and immediately know that nitroglycerin produces nitric oxide, which causes reacts with oxygen to form the brown NO2 gas when exposed to air. 397 398 Chapter 9 ■ Chemical Bonding I: Basic Concepts Step 3: We draw a single covalent bond between all the bonding atoms. The octet rule is satisfied for the F atoms, each of which has three lone pairs. The sum of the lone pair electrons on the four F atoms (4 3 6) and the four bonding pairs (4 3 2) is 32. Therefore, the remaining four electrons are shown as two lone pairs on the Xe atom: M M SF FS MG M G M Xee MD M D M SF FS M M We see that the Xe atom has an expanded octet. There are no formal charges on Similar problem: 9.63. the Xe and F atoms. Practice Exercise Write the Lewis structure of sulfur tetrafluoride (SF4). Review of Concepts Both boron and aluminum tend to form compounds in which they are surrounded with fewer than eight electrons. However, aluminum is able to form compounds and polyatomic ions where it is surrounded by more than eight electrons (e.g., AlF2 6 ). Why is it possible for aluminum, but not boron, to expand the octet? 9.10 Bond Enthalpy Remember that it takes energy to break a A measure of the stability of a molecule is its bond enthalpy, which is the enthalpy bond so that energy is released when a bond is formed. change required to break a particular bond in 1 mole of gaseous molecules. (Bond enthalpies in solids and liquids are affected by neighboring molecules.) The experimentally determined bond enthalpy of the diatomic hydrogen molecule, for example, is H2 (g) ¡ H(g) 1 H(g) ¢H° 5 436.4 kJ/mol This equation tells us that breaking the covalent bonds in 1 mole of gaseous H2 mol- ecules requires 436.4 kJ of energy. For the less stable chlorine molecule, Cl2 (g) ¡ Cl(g) 1 Cl(g)  ¢H° 5 242.7 kJ/mol Bond enthalpies can also be directly measured for diatomic molecules containing unlike elements, such as HCl, HCl(g) ¡ H(g) 1 Cl(g) ¢H° 5 431.9 kJ/mol as well as for molecules containing double and triple bonds: .. .. The Lewis structure of O2 is O“O .. .. and O2 (g) ¡ O(g) 1 O(g) ¢H° 5 498.7 kJ/mol that for N2 is : N‚N :. N2 (g) ¡ N(g) 1 N(g)  ¢H° 5 941.4 kJ/mol Measuring the strength of covalent bonds in polyatomic molecules is more com- plicated. For example, measurements show that the energy needed to break  9.10 Bond Enthalpy 399 Some Bond Enthalpies of Diatomic Molecules* and Average Bond Table 9.4 Enthalpies for Bonds in Polyatomic Molecules Bond Enthalpy Bond Enthalpy Bond (kJ/mol) Bond (kJ/mol) H¬H 436.4 C¬I 240 H¬N 393 C¬P 263 H¬O 460 C¬S 255 H¬S 368 C“S 477 H¬P 326 N¬N 193 H¬F 568.2 N“N 418 H¬Cl 431.9 N‚N 941.4 H¬Br 366.1 N¬O 176 H¬I 298.3 N“O 607 C¬H 414 O¬O 142 C¬C 347 O“O 498.7 C“C 620 O¬P 502 C‚C 812 O“S 469 C¬N 276 P¬P 197 C“N 615 P“P 489 C‚N 891 S¬S 268 C¬O 351 S“S 352 C“O † 745 F¬F 156.9 C‚O 1076.5 Cl¬Cl 242.7 C¬F 450 Br¬Br 192.5 C¬Cl 338 I¬I 151.0 C¬Br 276 *Bond enthalpies for diatomic molecules (in color) have more significant figures than bond enthalpies for bonds in polyatomic molecules because the bond enthalpies of diatomic molecules are directly measurable quantities and not averaged over many compounds. † The C“O bond enthalpy in CO2 is 799 kJ/mol. the first O¬H bond in H2O is different from that needed to break the second O¬H bond: H2O(g) ¡ H(g) 1 OH(g) ¢H° 5 502 kJ/mol OH(g) ¡ H(g) 1 O(g) ¢H° 5 427 kJ/mol In each case, an O¬H bond is broken, but the first step is more endothermic than the second. The difference between the two ¢H° values suggests that the second O¬H bond itself has undergone change, because of the changes in the chemical environment. Now we can understand why the bond enthalpy of the same O¬H bond in two different molecules such as methanol (CH3OH) and water (H2O) will not be the same: Their environments are different. Thus, for polyatomic molecules we speak of the average bond enthalpy of a particular bond. For example, we can measure the energy of the O¬H bond in 10 different polyatomic molecules and obtain the average O¬H bond enthalpy by dividing the sum of the bond enthalpies by 10. Table 9.4 lists the average bond enthalpies of a number of diatomic and polyatomic molecules. As stated earlier, triple bonds are stronger than double bonds, which, in turn, are stronger than single bonds. 400 Chapter 9 ■ Chemical Bonding I: Basic Concepts Use of Bond Enthalpies in Thermochemistry A comparison of the thermochemical changes that take place during a number of reactions (Chapter 6) reveals a strikingly wide variation in the enthalpies of differ- ent reactions. For example, the combustion of hydrogen gas in oxygen gas is fairly exothermic: H2 (g) 1 12 O2 (g) ¡ H2O(l) ¢H° 5 2285.8 kJ/mol On the other hand, the formation of glucose (C6H12O6) from water and carbon diox- ide, best achieved by photosynthesis, is highly endothermic: 6CO2 (g) 1 6H2O(l) ¡ C6H12O6 (s) 1 6O2 (g) ¢H° 5 2801 kJ/mol We can account for such variations by looking at the stability of individual reactant and product molecules. After all, most chemical reactions involve the making and breaking of bonds. Therefore, knowing the bond enthalpies and hence the stability of molecules tells us something about the thermochemical nature of reactions that mol- ecules undergo. In many cases, it is possible to predict the approximate enthalpy of reaction by using the average bond enthalpies. Because energy is always required to break chem- ical bonds and chemical bond formation is always accompanied by a release of energy, we can estimate the enthalpy of a reaction by counting the total number of bonds broken and formed in the reaction and recording all the corresponding energy changes. The enthalpy of reaction in the gas phase is given by ¢H° 5 ©BE(reactants) 2 ©BE(products) (9.3) 5 total energy input 2 total energy released where BE stands for average bond enthalpy and © is the summation sign. As written, Equation (9.3) takes care of the sign convention for ¢H°. Thus, if the total energy input is greater than the total energy released, ¢H° is positive and the reaction is endothermic. On the other hand, if more energy is released than absorbed, ¢H° is negative and the reaction is exothermic (Figure 9.8). If reactants and products are all Figure 9.8 Bond enthalpy changes in (a) an endothermic Atoms Atoms reaction and (b) an exothermic reaction. – ∑ BE (products) ∑ BE (reactants) Enthalpy Enthalpy Product Reactant molecules molecules ∑ BE (reactants) – ∑ BE (products) Reactant Product molecules molecules (a) (b)  9.10 Bond Enthalpy 401 diatomic molecules, then Equation (9.3) will yield accurate results because the bond enthalpies of diatomic molecules are accurately known. If some or all of the reactants and products are polyatomic molecules, Equation  (9.3) will yield only approximate results because the bond enthalpies used will be averages. For diatomic molecules, Equation (9.3) is equivalent to Equation (6.18), so the results obtained from these two equations should correspond, as Example 9.13 illustrates. Example 9.13 Use Equation (9.3) to calculate the enthalpy of reaction for the process H2 (g) 1 Cl2 (g) ¡ 2HCl(g) Compare your result with that obtained using Equation (6.18). Strategy Keep in mind that bond breaking is an energy absorbing (endothermic) process and bond making is an energy releasing (exothermic) process. Therefore, the 6 overall energy change is the difference between these two opposing processes, as g described by Equation (9.3). Solution We start by counting the number of bonds broken and the number of bonds formed and the corresponding energy changes. This is best done by creating a table: Type of Number of Bond enthalpy Energy change bonds broken bonds broken (kJ/mol) (kJ/mol) H¬H (H2) 1 436.4 436.4 Cl¬Cl (Cl2) 1 242.7 242.7 Type of Number of Bond enthalpy Energy change bonds formed bonds formed (kJ/mol) (kJ/mol) H¬Cl (HCl) 2 431.9 863.8 Next, we obtain the total energy input and total energy released: total energy input 5 436.4 kJ/mol 1 242.7 kJ/mol 5 679.1 kJ/mol Refer to Table 9.4 for bond enthalpies of total energy released 5 863.8 kJ/mol these diatomic molecules. Using Equation (9.3), we write ¢H° 5 679.1 kJ/mol 2 863.8 kJ/mol 5 2184.7 kJ/mol Alternatively, we can use Equation (6.18) and the data in Appendix 3 to calculate the enthalpy of reaction: ¢H° 5 2¢H°f (HCl) 2 [¢H°f (H2 ) 1 ¢H°f (Cl2 )] 5 (2) (292.3 kJ/mol) 2 0 2 0 5 2184.6 kJ/mol Check Because the reactants and products are all diatomic molecules, we expect the results of Equations (9.3) and (6.18) to be the same. The small discrepancy here is due to different ways of rounding off. Similar problem: 9.104. Practice Exercise Calculate the enthalpy of the reaction H2 (g) 1 F2 (g) ¡ 2HF(g) using (a) Equation (9.3) and (b) Equation (6.18). 402 Chapter 9 ■ Chemical Bonding I: Basic Concepts Example 9.14 uses Equation (9.3) to estimate the enthalpy of a reaction involving a polyatomic molecule. Example 9.14 Estimate the enthalpy change for the combustion of hydrogen gas: 2H2 (g) 1 O2 (g) ¡ 2H2O(g) Strategy We basically follow the same procedure as that in Example 9.13. Note, however, that H2O is a polyatomic molecule, and so we need to use the average bond enthalpy value for the O¬H bond. 6 Solution We construct the following table: h Type of Number of Bond enthalpy Energy change bonds broken bonds broken (kJ/mol) (kJ/mol) H¬H (H2) 2 436.4 872.8 O“O (O2) 1 498.7 498.7 Type of Number of Bond enthalpy Energy change bonds formed bonds formed (kJ/mol) (kJ/mol) O¬H (H2O) 4 460 1840 Next, we obtain the total energy input and total energy released: total energy input 5 872.8 kJ/mol 1 498.7 kJ/mol 5 1371.5 kJ/mol total energy released 5 1840 kJ/mol Using Equation (9.3), we write ¢H° 5 1371.5 kJ/mol 2 1840 kJ/mol 5 2469 kJ/mol This result is only an estimate because the bond enthalpy of O¬H is an average quantity. Alternatively, we can use Equation (6.18) and the data in Appendix 3 to calculate the enthalpy of reaction: ¢H° 5 2¢H°f (H2O) 2 [2¢H°f (H2 ) 1 ¢H°f (O2 )] 5 2(2241.8 kJ/mol) 2 0 2 0 5 2483.6 kJ/mol Check Note that the estimated value based on average bond enthalpies is quite close to the value calculated using ¢H°f data. In general, Equation (9.3) works best for reactions that are either quite endothermic or quite exothermic, that is, reactions for which Similar problem: 9.72. ¢H°rxn . 100 kJ/mol or for which ¢H°rxn , 2100 kJ/mol. Practice Exercise For the reaction H2 (g) 1 C2H4 (g) ¡ C2H6 (g) (a) Estimate the enthalpy of reaction, using the bond enthalpy values in Table 9.4. (b) Calculate the enthalpy of reaction, using standard enthalpies of formation. ( ¢H°f for H2, C2H4, and C2H6 are 0, 52.3 kJ/mol, and 284.7 kJ/mol, respectively.) Review of Concepts Why does ¢H°rxn calculated using bond enthalpies not always agree with that calculated using ¢H°f values?  Questions & Problems 403 Key Equation ¢H° 5 ©BE(reactants) 2 ©BE(products) (9.3) Calculating enthalpy change of a reaction from bond enthalpies. Summary of Facts & Concepts 1. A Lewis dot symbol shows the number of valence elec- The arrangement of bonding electrons and lone pairs in trons possessed by an atom of a given element. Lewis dot a molecule is represented by a Lewis structure. symbols are useful mainly for the representative elements. 6. Electronegativity is a measure of an atom’s ability to 2. The elements most likely to form ionic compounds attract electrons in a chemical bond. have low ionization energies (such as the alkali metals 7. The octet rule predicts that atoms form enough covalent and the alkaline earth metals, which form cations) or bonds to surround themselves with eight electrons each. high electron affinities (such as the halogens and When one atom in a covalently bonded pair donates two oxygen, which form anions). electrons to the bond, the Lewis structure can include the 3. An ionic bond is the product of the electrostatic forces formal charge on each atom as a means of keeping track of attraction between positive and negative ions. An of the valence electrons. There are exceptions to the octet ionic compound consists of a large network of ions in rule, particularly for covalent beryllium compounds, which positive and negative charges are balanced. The elements in Group 3A, odd-electron molecules, and ele- structure of a solid ionic compound maximizes the net ments in the third period and beyond in the periodic table. attractive forces among the ions. 8. For some molecules or polyatomic ions, two or more 4. Lattice energy is a measure of the stability of an ionic Lewis structures based on the same skeletal structure solid. It can be calculated by means of the Born-Haber satisfy the octet rule and appear chemically reasonable. cycle, which is based on Hess’s law. Taken together, such resonance structures represent the 5. In a covalent bond, two electrons (one pair) are shared molecule or ion more accurately than any single Lewis by two atoms. In multiple covalent bonds, two or three structure does. pairs of electrons are shared by two atoms. Some cova- 9. The strength of a covalent bond is measured in terms of lently bonded atoms also have lone pairs, that is, pairs its bond enthalpy. Bond enthalpies can be used to esti- of valence electrons that are not involved in bonding. mate the enthalpy of reactions. Key Words Bond enthalpy, p. 398 Covalent bond, p. 377 Lewis dot symbol, p. 369 Polar covalent bond, p. 380 Bond length, p. 379 Covalent compound, p. 377 Lewis structure, p. 378 Resonance, p. 390 Born-Haber cycle, p. 372 Double bond, p. 378 Lone pair, p. 377 Resonance structure, p. 390 Coordinate covalent Electronegativity, p. 380 Multiple bond, p. 378 Single bond, p. 378 bond, p. 393 Formal charge, p. 387 Octet rule, p. 378 Triple bond, p. 379 Coulomb’s law, p. 372 Ionic bond, p. 370 Questions & Problems • Problems available in Connect Plus • 9.3 Without referring to Figure 9.1, write Lewis dot Red numbered problems solved in Student Solutions Manual symbols for atoms of the following elements: (a) Be, (b) K, (c) Ca, (d) Ga, (e) O, (f ) Br, (g) N, (h) I, Lewis Dot Symbols (i) As, ( j) F. Review Questions • 9.4 Write Lewis dot symbols for the following ions: (a) Li1, (b) Cl2, (c) S22, (d) Sr21, (e) N32. 9.1 What is a Lewis dot symbol? To what elements does the symbol mainly apply? • 9.5 Write Lewis dot symbols for the following atoms and ions: (a) I, (b) I2, (c) S, (d) S22, (e) P, (f) P32, (g) Na, 9.2 Use the second member of each group from Group (h) Na1, (i) Mg, ( j) Mg21, (k) Al, (l) Al31, (m) Pb, 1A to Group 7A to show that the number of valence (n) Pb21. electrons on an atom of the element is the same as its group number.  404 Chapter 9 ■ Chemical Bonding I: Basic Concepts The Ionic Bond be ionic or covalent. Write the empirical formula and Review Questions name of the compound: (a) I and Cl, (b) Mg and F. 9.6 Explain what an ionic bond is. • 9.20 For each of the following pairs of elements, state whether the binary compound they form is likely to 9.7 Explain how ionization energy and electron affinity be ionic or covalent. Write the empirical formula and determine whether atoms of elements will combine name of the compound: (a) B and F, (b) K and Br. to form ionic compounds. 9.8 Name five metals and five nonmetals that are very likely to form ionic compounds. Write formulas Lattice Energy of Ionic Compounds for compounds that might result from the combi- Review Questions nation of these metals and nonmetals. Name these 9.21 What is lattice energy and what role does it play in compounds. the stability of ionic compounds? 9.9 Name one ionic compound that contains only non- 9.22 Explain how the lattice energy of an ionic com- metallic elements. pound such as KCl can be determined using the 9.10 Name one ionic compound that contains a polyatomic Born-Haber cycle. On what law is this procedure cation and a polyatomic anion (see Table 2.3). based? 9.11 Explain why ions with charges greater than 3 are • 9.23 Specify which compound in the following pairs of seldom found in ionic compounds. ionic compounds has the higher lattice energy: 9.12 The term “molar mass” was introduced in Chapter 3. (a) KCl or MgO, (b) LiF or LiBr, (c) Mg3N2 or What is the advantage of using the term “molar NaCl. Explain your choice. mass” when we discuss ionic compounds? • 9.24 Compare the stability (in the solid state) of the fol- • 9.13 In which of the following states would NaCl be lowing pairs of compounds: (a) LiF and LiF2 (con- electrically conducting? (a) solid, (b) molten (that taining the Li21 ion), (b) Cs2O and CsO (containing is, melted), (c) dissolved in water. Explain your the O2 ion), (c) CaBr2 and CaBr3 (containing the answers. Ca31 ion). • 9.14 Beryllium forms a compound with chlorine that has the empirical formula BeCl2. How would you deter- Problems mine whether it is an ionic compound? (The com- pound is not soluble in water.) • 9.25 Use the Born-Haber cycle outlined in Section 9.3 for LiF to calculate the lattice energy of NaCl. [The heat of sublimation of Na is 108 kJ/mol and Problems ¢H°f (NaCl) 5 2411 kJ/mol. Energy needed to • 9.15 An ionic bond is formed between a cation A1 and an dissociate 12 mole of Cl2 into Cl atoms 5 121.4 kJ.] anion B2. How would the energy of the ionic bond • 9.26 Calculate the lattice energy of calcium chloride given [see Equation (9.2)] be affected by the following that the heat of sublimation of Ca is 121 kJ/mol and changes? (a) doubling the radius of A1, (b) tripling ¢H°f (CaCl2 ) 5 2795 kJ/mol. (See Tables 8.2 and the charge on A1, (c) doubling the charges on A1 8.3 for other data.) and B2, (d) decreasing the radii of A1 and B2 to half their original values. • 9.16 Give the empirical formulas and names of the com- The Covalent Bond pounds formed from the following pairs of ions: Review Questions (a) Rb1 and I2, (b) Cs1 and SO224 , (c) Sr 21 and N32, 9.27 What is Lewis’s contribution to our understanding 31 22 (d) Al and S . of the covalent bond? 9.17 Use Lewis dot symbols to show the transfer of elec- 9.28 Use an example to illustrate each of the following trons between the following atoms to form cations terms: lone pairs, Lewis structure, the octet rule, and anions: (a) Na and F, (b) K and S, (c) Ba and O, bond length. (d) Al and N. 9.29 What is the difference between a Lewis dot symbol • 9.18 Write the Lewis dot symbols of the reactants and and a Lewis structure? products in the following reactions. (First balance the equations.) • 9.30 How many lone pairs are on the underlined atoms in these compounds? HBr, H2S, CH4 (a) Sr 1 Se ¡ SrSe 9.31 Compare single, double, and triple bonds in a mol- (b) Ca 1 H2 ¡ CaH2 ecule, and give an example of each. For the same (c) Li 1 N2 ¡ Li3N bonding atoms, how does the bond length change (d) Al 1 S ¡ Al2S3 from single bond to triple bond? • 9.19 For each of the following pairs of elements, state 9.32 Compare the properties of ionic compounds and whether the binary compound they form is likely to covalent compounds.  Questions & Problems 405 Electronegativity and Bond Type • 9.44 Write Lewis structures for the following molecules Review Questions and ions: (a) OF2, (b) N2F2, (c) Si2H6, (d) OH2, (e) CH2ClCOO2, (f) CH3NH1 3. 9.33 Define electronegativity, and explain the difference between electronegativity and electron affinity. • 9.45 Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three Describe in general how the electronegativities of other P atoms), (d) H2S, (e) N2H4, (f) HClO3, the elements change according to position in the (g) COBr2 (C is bonded to O and Br atoms). periodic table. 9.34 What is a polar covalent bond? Name two compounds • 9.46 Write Lewis structures for the following ions: (a) O22 22 1 1 2 , (b) C2 , (c) NO , (d) NH4 . Show formal that contain one or more polar covalent bonds. charges. • 9.47 The following Lewis structures for (a) HCN, Problems (b) C2H2, (c) SnO2, (d) BF3, (e) HOF, (f) HCOF, and (g) NF3 are incorrect. Explain what is wrong with • 9.35 List the following bonds in order of increasing each one and give a correct structure for the molecule. ionic character: the lithium-to-fluorine bond in (Relative positions of atoms are shown correctly.) LiF, the potassium-to-oxygen bond in K2O, the nitrogen-to-nitrogen bond in N2, the sulfur-to- O O (a) HOCPN (f) H Q G oxygen bond in SO2, the chlorine-to-fluorine O O bond in ClF3. (b) HPCPCPH D QS COF (c) O O O QO • 9.36 Arrange the following bonds in order of increasing OOSnOO Q Q ionic character: carbon to hydrogen, fluorine to hy- (d) SO F Q OFS (g) SO F OFS drogen, bromine to hydrogen, sodium to chlorine, G DQ Q G DQ potassium to fluorine, lithium to chlorine. O B N A A • 9.37 Four atoms are arbitrarily labeled D, E, F, and G. FS SQ FS SQ Their electronegativities are as follows: D 5 3.8, E 5 3.3, F 5 2.8, and G 5 1.3. If the atoms of these O OS (e) HOOPF Q elements form the molecules DE, DG, EG, and DF, how would you arrange these molecules in order of 9.48 The skeletal structure of acetic acid shown below is increasing covalent bond character? correct, but some of the bonds are wrong. (a) Identify the incorrect bonds and explain what is wrong with • 9.38 List the following bonds in order of increasing ionic them. (b) Write the correct Lewis structure for acetic character: cesium to fluorine, chlorine to chlorine, acid. bromine to chlorine, silicon to carbon. • 9.39 Classify the following bonds as ionic, polar co- H S OS A A valent, or covalent, and give your reasons: HPCOCOOOH Q Q (a) the CC bond in H 3CCH3, (b) the KI bond in A KI, (c) the NB bond in H 3NBCl3, (d) the CF bond H in CF4. • 9.40 Classify the following bonds as ionic, polar cova- The Concept of Resonance lent, or covalent, and give your reasons: (a) the Review Questions SiSi bond in Cl3SiSiCl3, (b) the SiCl bond in Cl3SiSiCl3, (c) the CaF bond in CaF2, (d) the NH 9.49 Define bond length, resonance, and resonance bond in NH3. structure. What are the rules for writing resonance structures? 9.50 Is it possible to “trap” a resonance structure of a Lewis Structure and the Octet Rule compound for study? Explain. Review Questions Problems 9.41 Summarize the essential features of the Lewis octet rule. The octet rule applies mainly to the second- • 9.51 Write Lewis structures for the following species, in- period elements. Explain. cluding all resonance forms, and show formal 9.42 Explain the concept of formal charge. Do formal charges: (a) HCO22 , (b) CH2NO22 . Relative positions charges represent actual separation of charges? of the atoms are as follows: O H O Problems H C C N O H O • 9.43 Write Lewis structures for the following molecules and ions: (a) NCl3, (b) OCS, (c) H2O2, (d) CH3COO2, • 9.52 Draw three resonance structures for the chlorate ion, (e) CN2, (f) CH3CH2NH1 3. ClO2 3 . Show formal charges.  406 Chapter 9 ■ Chemical Bonding I: Basic Concepts • 9.53 Write three resonance structures for hydrazoic acid, Bond Enthalpy HN3. The atomic arrangement is HNNN. Show for- Review Questions mal charges. 9.67 What is bond enthalpy? Bond enthalpies of poly- • 9.54 Draw two resonance structures for diazomethane, atomic molecules are average values, whereas those CH2N2. Show formal charges. The skeletal structure of the molecule is of diatomic molecules can be accurately determined. Why? H 9.68 Explain why the bond enthalpy of a molecule is C N N usually defined in terms of a gas-phase reaction. H Why are bond-breaking processes always endo- thermic and bond-forming processes always • 9.55 Draw three resonance structures for the molecule exothermic? N2O3 (atomic arrangement is ONNO2). Show formal charges. Problems • 9.56 Draw three reasonable resonance structures for the OCN2 ion. Show formal charges. • 9.69 From the following data, calculate the average bond enthalpy for the N¬H bond: Exceptions to the Octet Rule NH3 (g) ¡ NH2 (g) 1 H(g)   ¢H° 5 435 kJ/mol Review Questions NH2 (g) ¡ NH(g) 1 H(g)   ¢H° 5 381 kJ/mol NH(g) ¡ N(g) 1 H(g)   ¢H° 5 360 kJ/mol 9.57 Why does the octet rule not hold for many com- pounds containing elements in the third period of • 9.70 For the reaction the periodic table and beyond? O(g) 1 O2 (g) ¡ O3 (g)  ¢H° 5 2107.2 kJ/mol 9.58 Give three examples of compounds that do not satisfy the octet rule. Write a Lewis structure for Calculate the average bond enthalpy in O3. each. • 9.71 The bond enthalpy of F2(g) is 156.9 kJ/mol. Calculate 9.59 Because fluorine has seven valence electrons ¢H°f for F(g). (2s22p5), seven covalent bonds in principle could • 9.72 For the reaction form around the atom. Such a compound might be 2C2H6 (g) 1 7O2 (g) ¡ 4CO2 (g) 1 6H2O(g) FH7 or FCl7. These compounds have never been prepared. Why? (a) Predict the enthalpy of reaction from the 9.60 What is a coordinate covalent bond? Is it different average bond enthalpies in Table 9.4. from a normal covalent bond? (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Problems Appendix 3) of the reactant and product molecules, and compare the result with your 9.61 The AlI3 molecule has an incomplete octet around answer for part (a). Al. Draw three resonance structures of the molecule in which the octet rule is satisfied for both the Al and the I atoms. Show formal charges. Additional Problems 9.62 In the vapor phase, beryllium chloride consists of • 9.73 Classify the following substances as ionic com- discrete BeCl2 molecules. Is the octet rule satisfied pounds or covalent compounds containing discrete for Be in this compound? If not, can you form an molecules: CH4, KF, CO, SiCl4, BaCl2. octet around Be by drawing another resonance 9.74 Which of the following are ionic compounds? structure? How plausible is this structure? Which are covalent compounds? RbCl, PF5, BrF3, • 9.63 Of the noble gases, only Kr, Xe, and Rn are known to KO2, CI4 form a few compounds with O and/or F. Write Lewis • 9.75 Match each of the following energy changes with structures for the following molecules: (a) XeF2, one of the processes given: ionization energy, elec- (b) XeF4, (c) XeF6, (d) XeOF4, (e) XeO2F2. In each tron affinity, bond enthalpy, and standard enthalpy case Xe is the central atom. of formation. • 9.64 Write a Lewis structure for SbCl5. Does this molecule (a) F(g) 1 e2 ¡ F2 (g) obey the octet rule? (b) F2 (g) ¡ 2F(g) • 9.65 Write Lewis structures for SeF4 and SeF6. Is the (c) Na(g) ¡ Na1 (g) 1 e2 octet rule satisfied for Se? (d) Na(s) 1 12F2 (g) ¡ NaF(s) • 9.66 Write Lewis structures for the reaction • 9.76 The formulas for the fluorides of the third-period AlCl3 1 Cl2 ¡ AlCl2 4 elements are NaF, MgF2, AlF3, SiF4, PF5, SF6, and ClF3. Classify these compounds as covalent What kind of bond joins Al and Cl in the product? or ionic.  Questions & Problems 407 • 9.77 Use ionization energy (see Table 8.2) and electron 9.89 Based on energy considerations, which of the fol- affinity values (see Table 8.3) to calculate the energy lowing reactions will occur more readily? change (in kJ/mol) for the following reactions: (a) Cl(g) 1 CH4 (g) ¡ CH3Cl(g) 1 H(g) (a) Li(g) 1 I(g) ¡ Li1 (g) 1 I2 (g) (b) Cl(g) 1 CH4 (g) ¡ CH3 (g) 1 HCl(g) (b) Na(g) 1 F(g) ¡ Na1 (g) 1 F2 (g) (Hint: Refer to Table 9.4, and assume that the (c) K(g) 1 Cl(g) ¡ K1 (g) 1 Cl2 (g) average bond enthalpy of the C¬Cl bond is 9.78 Describe some characteristics of an ionic compound 338 kJ/mol.) such as KF that would distinguish it from a covalent • 9.90 Which of the following molecules has the shortest compound such as benzene (C6H6). nitrogen-to-nitrogen bond? Explain. N2H4, N2O, N2, • 9.79 Write Lewis structures for BrF3, ClF5, and IF7. N2O4 Identify those in which the octet rule is not • 9.91 Most organic acids can be represented as RCOOH, obeyed. where COOH is the carboxyl group and R is the rest 9.80 Write three reasonable resonance structures for the of the molecule. (For example, R is CH3 in acetic azide ion N23 in which the atoms are arranged as acid, CH3COOH.) (a) Draw a Lewis structure for NNN. Show formal charges. the carboxyl group. (b) Upon ionization, the car- boxyl group is converted to the carboxylate group, • 9.81 The amide group plays an important role in deter- COO2. Draw resonance structures for the carboxyl- mining the structure of proteins: ate group. S OS • 9.92 Which of the following species are isoelectronic? B O ONOCO NH1 4 , C6H6, CO, CH4, N2, B3N3H6 A • 9.93 The following species have been detected in inter- H stellar space: (a) CH, (b) OH, (c) C2, (d) HNC, (e) HCO. Draw Lewis structures for these species Draw another resonance structure for this group. and indicate whether they are diamagnetic or Show formal charges. paramagnetic. 9.82 Give an example of an ion or molecule containing 9.94 The amide ion, NH2 2 , is a Brønsted base. Repre- Al that (a) obeys the octet rule, (b) has an expanded sent the reaction between the amide ion and octet, and (c) has an incomplete octet. water. • 9.83 Draw four reasonable resonance structures for the • 9.95 Draw Lewis structures for the following organic mol- PO3F22 ion. The central P atom is bonded to the ecules: (a) tetrafluoroethylene (C2F4), (b) propane three O atoms and to the F atom. Show formal (C3H8), (c) butadiene (CH2CHCHCH2), (d) propyne charges. (CH3CCH), (e) benzoic acid (C6H5COOH). (To draw 9.84 Attempts to prepare the compounds listed here as C6H5COOH, replace a H atom in benzene with a stable species under atmospheric conditions have COOH group.) failed. Suggest possible reasons for the failure. CF2, 9.96 The triiodide ion (I23 ) in which the I atoms are LiO2, CsCl2, PI5 arranged in a straight line is stable, but the corre- • 9.85 Draw reasonable resonance structures for the follow- sponding F23 ion does not exist. Explain. ing ions: (a) HSO2 32 2 22 4 , (b) PO4 , (c) HSO3 , (d) SO3 . 9.97 Compare the bond enthalpy of F2 with the energy (Hint: See comment on p. 396.) change for the following process: • 9.86 Are the following statements true or false? (a) Formal charges represent actual separation of charges. F2 (g) ¡ F1 (g) 1 F2 (g) (b) ¢H°rxn can be estimated from the bond enthalpies of reactants and products. (c) All second-period ele- Which is the preferred dissociation for F2, energeti- ments obey the octet rule in their compounds. cally speaking? (d) The resonance structures of a molecule can be separated from one another. 9.98 Methyl isocyanate (CH3NCO) is used to make certain pesticides. In December 1984, water 9.87 A rule for drawing plausible Lewis structures is that leaked into a tank containing this substance at a the central atom is invariably less electronegative chemical plant, producing a toxic cloud that than the surrounding atoms. Explain why this is so. killed thousands of people in Bhopal, India. Why does this rule not apply to compounds like H2O Draw Lewis structures for CH 3NCO, showing and NH3? formal charges. • 9.88 Using the following information and the fact that • 9.99 The chlorine nitrate molecule (ClONO2) is believed the average C¬H bond enthalpy is 414 kJ/mol, to be involved in the destruction of ozone in the Ant- estimate the standard enthalpy of formation of arctic stratosphere. Draw a plausible Lewis structure methane (CH4). for this molecule. C(s) ¡ C(g) ¢H°rxn 5 716 kJ/mol 9.100 Several resonance structures for the molecule CO2 2H2 (g) ¡ 4H(g) ¢H°rxn 5 872.8 kJ/mol are shown next. Explain why some of them are  408 Chapter 9 ■ Chemical Bonding I: Basic Concepts likely to be of little importance in describing the • 9.111 The resonance concept is sometimes described by bonding in this molecule. analogy to a mule, which is a cross between a horse ⫺ and a donkey. Compare this analogy with the one ⫹ (a) O O OPCPO O O (c) SOqC OS used in this chapter, that is, the description of a Q Q Q rhinoceros as a cross between a griffin and a uni- ⫹ ⫺ ⫺ 2⫹ ⫺ OS corn. Which description is more appropriate? (b) SOqCOO Q (d) SO O OOCOOS Q Q Why? 9.112 What are the other two reasons for choosing (b) in 9.101 For each of the following organic molecules draw a Example 9.7? Lewis structure in which the carbon atoms are bonded to each other by single bonds: (a) C2H6, (b) C4H10, 9.113 In the Chemistry in Action essay on p. 397, nitric (c) C5H12. For (b) and (c), show only structures in oxide is said to be one of about 10 of the smallest which each C atom is bonded to no more than two stable molecules known. Based on what you have other C atoms. learned in the course so far, write all the diatomic molecules you know, give their names, and show 9.102 Draw Lewis structures for the following chlorofluo- their Lewis structures. rocarbons (CFCs), which are partly responsible for the depletion of ozone in the stratosphere: (a) CFCl3, 9.114 The N¬O bond distance in nitric oxide is 115 pm, (b) CF2Cl2, (c) CHF2Cl, (d) CF3CHF2. which is intermediate between a triple bond (106  pm) and a double bond (120 pm). (a) Draw • 9.103 Draw Lewis structures for the following organic two resonance structures for NO and comment on molecules. In each there is one C“C bond, and the their relative importance. (b) Is it possible to draw rest of the carbon atoms are joined by C¬C bonds. a resonance structure having a triple bond between C2H3F, C3H6, C4H8 the atoms? • 9.104 Calculate ¢H° for the reaction 9.115 Write the formulas of the binary hydride for the H2 (g) 1 I2 (g) ¡ 2HI(g) second-period elements LiH to HF. Comment on the change from ionic to covalent character of using (a) Equation (9.3) and (b) Equation (6.18), these compounds. Note that beryllium behaves given that ¢H°f for I2(g) is 61.0 kJ/mol. differently from the rest of the Group 2A metals • 9.105 Draw Lewis structures for the following organic (see p. 348). molecules: (a) methanol (CH3OH); (b) ethanol 9.116 Hydrazine borane, NH2NH2BH3, has been proposed (CH3CH2OH); (c) tetraethyllead [Pb(CH2CH3)4], as a hydrogen storage material. When reacted which was used in “leaded gasoline”; (d) methyl- with lithium hydride (LiH), hydrogen gas is amine (CH3NH2), which is used in tanning; (e) mus- released tard gas (ClCH2CH2SCH2CH2Cl), a poisonous NH2NH2BH3 1 LiH ¡ LiNH2NHBH3 1 H2 gas used in World War I; (f ) urea [(NH2)2CO], a fertilizer; and (g) glycine (NH2CH2COOH), an Write Lewis structures for NH2NH2BH3 and amino acid. NH2NHBH2 3 and assign all formal charges. 9.106 Write Lewis structures for the following four 9.117 Although nitrogen dioxide (NO2) is a stable com- isoelectronic species: (a) CO, (b) NO1, (c) CN2, pound, there is a tendency for two such molecules to (d) N2. Show formal charges. combine to form dinitrogen tetroxide (N2O4). Why? • 9.107 Oxygen forms three types of ionic compounds in Draw four resonance structures of N2O4, showing which the anions are oxide (O22), peroxide (O22 2 ), formal charges. and superoxide (O22 ). Draw Lewis structures of 9.118 Another possible skeletal structure for the CO22 3 these ions. (carbonate) ion besides the one presented in Exam- 9.108 Comment on the correctness of the statement, “All ple 9.5 is O C O O. Why would we not use this compounds containing a noble gas atom violate the structure to represent CO223 ? octet rule.” • 9.119 Draw a Lewis structure for nitrogen pentoxide • 9.109 Write three resonance structures for (a) the cyanate (N 2O 5) in which each N is bonded to three O ion (NCO2) and (b) the isocyanate ion (CNO2). In atoms. each case, rank the resonance structures in order of • 9.120 In the gas phase, aluminum chloride exists as a increasing importance. dimer (a unit of two) with the formula Al2Cl6. Its • 9.110 (a) From the following data calculate the bond en- skeletal structure is given by thalpy of the F2 2 ion. Cl Cl Cl F2(g) ¡ 2F(g) ¢H°rxn 5 156.9 kJ/mol G D G D Al Al F2(g) ¡ F(g) 1 e2 ¢H°rxn 5 333 kJ/mol D G D G F2 2 Cl Cl Cl 2 (g) ¡ F2(g) 1 e ¢H°rxn 5 290 kJ/mol (b) Explain the difference between the bond Complete the Lewis structure and indicate the coor- enthalpies of F2 and F2 2. dinate covalent bonds in the molecule.  Questions & Problems 409 • 9.121 The hydroxyl radical (OH) plays an important role to break a single chemical bond. If 2.0 3 1029 N was in atmospheric chemistry. It is highly reactive and needed to break a C¬Si bond, estimate the bond has a tendency to combine with a H atom from other enthalpy in kJ/mol. Assume that the bond had to be compounds, causing them to break up. Thus, OH is stretched by a distance of 2 Å (2 3 10210 m) before sometimes called a “detergent” radical because it it is broken. helps to clean up the atmosphere. (a) Write the • 9.128 The American chemist Robert S. Mulliken sug- Lewis structure for the radical. (b) Refer to Table 9.4 gested a different definition for the electronegativity and explain why the radical has a high affinity for H (EN) of an element, given by atoms. (c) Estimate the enthalpy change for the fol- lowing reaction: IE 1 EA EN 5 2 OH(g) 1 CH4 (g) ¡ CH3 (g) 1 H2O(g) where IE is the first ionization energy and EA the (d) The radical is generated when sunlight hits electron affinity of the element. Calculate the elec- water vapor. Calculate the maximum wavelength tronegativities of O, F, and Cl using the above equa- (in nanometers) required to break an O¬H bond tion. Compare the electronegativities of these in H2O. elements on the Mulliken and Pauling scale. (To • 9.122 Experiments show that it takes 1656 kJ/mol to break convert to the Pauling scale, divide each EN value all the bonds in methane (CH4) and 4006 kJ/mol to by 230 kJ/mol.) break all the bonds in propane (C3H8). Based on • 9.129 Among the common inhaled anesthetics are: these data, calculate the average bond enthalpy of halothane: CF3CHClBr the C¬C bond. enflurane: CHFClCF2OCHF2 9.123 Calculate ¢H°rxn at 25°C of the reaction between isoflurane: CF3CHClOCHF2 carbon monoxide and hydrogen shown here using methoxyflurane: CHCl2CF2OCH3 both bond enthalpy and ¢H°f values. Draw Lewis structures of these molecules. 9.130 A student in your class claims that magnesium oxide actually consists of Mg1 and O2 ions, not Mg21 and O22 ions. Suggest some experiments one could do to 1 8n show that your classmate is wrong. 9.131 Shown here is a skeletal structure of borazine (B3N3H6). Draw two resonance structures of the molecule, showing all the bonds and formal charges. 9.124 Calculate ¢H°rxn at 25°C of the reaction between Compare its properties with the isoelectronic molecule ethylene and chlorine shown here using both bond benzene. enthalpy and ¢H°f values. ( ¢H°f for C2H4Cl2 is 2132 kJ/mol.) H H B H N N B B 1 8n H N H H 9.132 Calculate the wavelength of light needed to carry out the reaction 9.125 Draw three resonance structures of sulfur dioxide (SO2). Indicate the most plausible structure(s). H2 ¡ H 1 1 H2 9.126 Vinyl chloride (C2H3Cl) differs from ethylene 9.133 Sulfuric acid (H2SO4), the most important indus- (C2H4) in that one of the H atoms is replaced with a trial chemical in the world, is prepared by oxidiz- Cl atom. Vinyl chloride is used to prepare poly(vinyl ing sulfur to sulfur dioxide and then to sulfur chloride), which is an important polymer used in trioxide. Although sulfur trioxide reacts with water pipes. (a) Draw the Lewis structure of vinyl chlo- to form sulfuric acid, it forms a mist of fine drop- ride. (b) The repeating unit in poly(vinyl chloride) is lets of H2SO4 with water vapor that is hard to con- ¬CH2 ¬CHCl¬. Draw a portion of the molecule dense. Instead, sulfur trioxide is first dissolved in showing three such repeating units. (c) Calculate the 98 percent sulfuric acid to form oleum (H2S2O7). enthalpy change when 1.0 3 103 kg of vinyl chlo- On treatment with water, concentrated sulfuric acid ride forms poly(vinyl chloride). can be generated. Write equations for all the steps • 9.127 In 1998 scientists using a special type of electron and draw Lewis structures of oleum based on the microscope were able to measure the force needed discussion in Example 9.11.  410 Chapter 9 ■ Chemical Bonding I: Basic Concepts • 9.134 From the lattice energy of KCl in Table 9.1 and the attached to N with one single bond and one double ionization energy of K and electron affinity of Cl in bond. (b) Calculate the combined volume of the Tables 8.2 and 8.3, calculate the ¢H° for the gases at STP. (c) Assuming an initial explosion reaction temperature of 3000 K, estimate the pressure exerted by the gases using the result from (b). (The K(g) 1 Cl(g) ¡ KCl(s) standard enthalpy of formation of nitroglycerin is • 9.135 The species H1 3 is the simplest polyatomic ion. The 2371.1 kJ/mol.) geometry of the ion is that of an equilateral triangle. 9.139 Give a brief description of the medical uses of (a) Draw three resonance structures to represent the the following ionic compounds: AgNO3, BaSO4, ion. (b) Given the following information CaSO4, KI, Li2CO3, Mg(OH)2, MgSO4, NaHCO3, Na2CO3, NaF, TiO2, ZnO. You would need to do a 2H 1 H1 ¡ H31 ¢H° 5 2849 kJ/mol Web search of some of these compounds. and H2 ¡ 2H ¢H° 5 436.4 kJ/mol 9.140 Use Table 9.4 to estimate the bond enthalpy of the calculate ¢H° for the reaction C¬C, N¬N, and O¬O bonds in C2H6, N2H4, and H1 1 H2 ¡ H31 H2O2, respectively. What effect do lone pairs on adjacent atoms have on the strength of the particular • 9.136 The bond enthalpy of the C¬N bond in the amide bonds? group of proteins (see Problem 9.81) can be treated 9.141 The isolated O22 ion is unstable so it is not possible as an average of C¬N and C“N bonds. Calculate to measure the electron affinity of the O2 ion di- the maximum wavelength of light needed to break rectly. Show how you can calculate its value by the bond. using the lattice energy of MgO and the Born- 9.137 In 1999 an unusual cation containing only nitrogen Haber cycle. [Useful information: Mg(s) S Mg(g) (N15 ) was prepared. Draw three resonance structures ¢H° 5 148 kJ/mol.] of the ion, showing formal charges. (Hint: The N 9.142 When irradiated with light of wavelength 471.7 nm, atoms are joined in a linear fashion.) the chlorine molecule dissociates into chlorine at- 9.138 Nitroglycerin, one of the most commonly used oms. One Cl atom is formed in its ground electronic explosives, has the following structure state while the other is in an excited state that is CH2ONO2 10.5 kJ/mol above the ground state. What is the ƒ bond enthalpy of the Cl2 molecule? CHONO2 9.143 Recall from Chapter 8 that the product of the reac- ƒ tion between Xe(g) and PtF6(g) was originally CH2ONO2 thought to be an ionic compound composed of Xe1 The decomposition reaction is cations and PtF62 anions (see Figure 8.22). This prediction was based on the theoretical enthalpy of 4C3H5N3O9 (l) ¡ formation of XePtF6 calculated using a Born-Haber 12CO2 (g) 1 10H2O(g) 1 6N2 (g) 1 O2 (g) cycle. (a) The lattice energy for XePtF6 was esti- The explosive action is the result of the heat mated to be 460 kJ/mol. Explain whether or not released and the large increase in gaseous volume. this value is consistent with the lattice energies in (a) Calculate the ¢H° for the decomposition of one Table 9.1. (b) Calculate ¢H°f for XePtF6 given IE1 mole of nitroglycerin using both standard enthalpy for Xe(g) is 1170 kJ/mol and EA1 for PtF6(g) is of formation values and bond enthalpies. Assume 770 kJ/mol. Comment on the expected stability of that the two O atoms in the NO2 groups are XePtF6 based on your calculation. Interpreting, Modeling & Estimating 9.144 The reaction between fluorine (F2) with ethane used in rocket fuel.” (a) Draw a Lewis structure for (C2H6) produces predominantly CF4 rather than O4 and write a balanced chemical equation for the C2F6 molecules. Explain. reaction between ethane, C2H6(g), and O4(g) to 9.145 A new allotrope of oxygen, O4, has been reported. give carbon dioxide and water vapor. (b) Estimate The exact structure of O4 is unknown, but the sim- ¢H° for the reaction. (c) Write a chemical equa- plest possible structure would be a four-member tion illustrating the standard enthalpy of formation ring consisting of oxygen-oxygen single bonds. of O4(g) and estimate ¢H°f . (d) Assuming the oxy- The report speculated that the O4 molecule might gen allotropes are in excess, which will release be useful as a fuel “because it packs a lot of oxy- more energy when reacted with ethane (or any gen in a small space, so it might be even more other fuel): O2(g) or O4(g)? Explain using your energy-dense than the liquefied ordinary oxygen answers to parts (a)–(c).  Answers to Practice Exercises 411 9.146 Because bond formation is exothermic, when two hydrogen atom that absorbs the energy released gas-phase atoms come together to form a diatomic from this process? molecule it is necessary for a third atom or mole- 9.147 Estimate ¢H°f for sodium astatide (NaAt) according cule to absorb the energy that is released. Other- to the equation wise the molecule will undergo dissociation. If Na(s) 1 12At2 (s) ¡ NaAt(s) two atoms of hydrogen combine to form H2(g), what would be the increase in velocity of a third The information in Problem 8.147 may be useful. Answers to Practice Exercises Q 2 9.1 # Ba # 1 2 # H ¡ Ba21 2H :2  (or BaH2 ) SOS SOS [Xe]6s2 1s1 [Xe] [He] B A 21 O O 9.11 HOOOSOOOH O and HOOOSOOOH O 9.2 (a) Ionic, (b) polar covalent, (c) covalent. Q B Q Q A Q SOS SOS SQ OS B ⫺ 2 9.3 O O 9.4 HOCOOOH SPCPS O OPNOO 9.5 O O OS M M Q Q Q Q Q SF FS OPNOO O OS ⫺ MGMGM 9.6 O Q Q 9.7 HOCqN S 9.12 S MD D M ⫺ ⫺ ⫹ 2⫺ SF FS 9.8 O O mn SO SPCPN S OC qNSmn SS qC OO NS M M O 9.13 (a) 2543.1 kJ/mol, (b) 2543.2 kJ/mol. structure isOthe most important; the lastO The first O 9.14 (a) 2119 kJ/mol, (b) 2137.0 kJ/mol. structure is the least important. SO FS A OQFS 9.9 SO FOBeOF Q Q FOAs E OS 9.10 SO Q H A O QFS SQ FS  CHAPTER 10 Chemical Bonding II Molecular Geometry and Hybridization of Atomic Orbitals The shape of molecules plays an important role in complex biochemical reactions such as those between protein and DNA molecules. CHAPTER OUTLINE A LOOK AHEAD 10.1 Molecular Geometry  We firstReactions Redox examine the androle Electrochemical of chemical bonds Cellsand Equations lone pairs representing on the geometry redox reactions of a molecule can in beterms balanced of a simple using the approach ion-electron called the method. VSEPR These model.reactions (10.1) 10.2 Dipole Moments involve  We thenthe transfer learn of electrons the factors from a reducing that determine agent whether to an oxidizing a molecule agent.a possesses 10.3 Valence Bond Theory Using dipole separate moment compartments, such a reaction and how its measurement can be can help used us in thetostudy generate elec- of molec- trons that flow externally ular geometry. (10.2) in an arrangement called a galvanic cell. 10.4 Hybridization of Atomic Orbitals   Thermodynamics of Galvanic Next, we learn a quantum Cells The mechanical voltage called approach, measured the in a galvanic valence bond cell  can be broken down into the electrode potentials of the (VB) theory, in the study of chemical bonds. The VB theory explains anode (where why 10.5 Hybridization in Molecules oxidation takes place) andform cathode (where reduction takesoverlaps. place). This volt- and how chemical bonds in terms of atomic orbital (10.3) Containing Double age can be related to the Gibbs free energy change and the equilibrium and Triple Bonds  We see that constant theredox of the VB approach, in terms process. The ofequation Nernst the concept of mixing relates the cell or hybrid- voltage to ization the cellofvoltage atomicunder orbitals, accounts for standard-state both chemical conditions and thebond formation and concentrations of 10.6 Molecular Orbital Theory molecular geometry. (10.4 and 10.5) reacting species. 10.7 Molecular Orbital   We then examine Batteries another quantum are electrochemical mechanical cells that can supply treatment of the chemical direct current at a con- Configurations bond, called theThere stant voltage. molecular orbital are many (MO) theory. different types The MO theory of batteries considers used the in automi- 10.8 Delocalized Molecular formation of molecular biles, flashlights, and orbitals as a result pacemakers. Fuelof cells the overlap are aofspecial atomic orbitals, type of and is able to explain electrochemical the generates cell that paramagnetism of thebyoxygen electricity molecule. the oxidation (10.6) of hydrogen Orbitals  or hydrocarbons. We see that writing molecular orbital configuration is analogous to writing  electron configuration Corrosion for atoms is a spontaneous redoxinreaction that both theresults that Pauli inexclusion principle the formation of and Hund’s rust from rule iron, apply. silver Using sulfide fromhomonuclear diatomic silver, and patina molecules (copper as exam- carbonate) from ples, weCorrosion copper. can learn about causesthe strengthdamage enormous of a bond to as well as general buildings, magnetic structures, ships, properties frommethods and cars. many the molecular orbital have been configurations. devised to prevent (10.7) or minimize the effect  of Thecorrosion. concept of molecular orbital formation is extended to delocalized  molecular Electrolysisorbitals, which cover is the process threeelectrical in which or more energy atoms. We see that is used theseadelo- to cause non- calized orbitals spontaneous impart redox extra to reaction stability occur. to Themolecules quantitativelike relationship benzene. (10.8) between the current supplied and the products formed is provided by Faraday. Electrolysis is the major method for producing active metals and nonmetals and many essential industrial chemicals. 412  10.1 Molecular Geometry 413 I n Chapter 9, we discussed bonding in terms of the Lewis theory. Here we will study the shape, or geometry, of molecules. Geometry has an important influence on the physical and chemical properties of molecules, such as density, melting point, boiling point, and reactivity. We will see that we can predict the shapes of molecules with considerable accuracy using a simple method based on Lewis structures. The Lewis theory of chemical bonding, although useful and easy to apply, does not explain how and why bonds form. A proper understanding of bonding comes from quantum mechanics. Therefore, in the second part of the chapter we will apply quantum mechanics to the study of the geometry and stability of molecules. 10.1 Molecular Geometry Molecular geometry is the three-dimensional arrangement of atoms in a molecule. A molecule’s geometry affects its physical and chemical properties, such as melting point, boiling point, density, and the types of reactions it undergoes. In general, bond lengths and bond angles must be determined by experiment. However, there is a simple procedure that enables us to predict with considerable success the overall geometry of a molecule or ion if we know the number of electrons surrounding a central atom in its Lewis structure. The basis of this approach is the assumption that The term “central atom” means an atom that is not a terminal atom in a polyatomic electron pairs in the valence shell of an atom repel one another. The valence shell is molecule. the outermost electron-occupied shell of an atom; it holds the electrons that are usu- ally involved in bonding. In a covalent bond, a pair of electrons, often called the bonding pair, is responsible for holding two atoms together. However, in a polyatomic molecule, where there are two or more bonds between the central atom and the sur- rounding atoms, the repulsion between electrons in different bonding pairs causes them to remain as far apart as possible. The geometry that the molecule ultimately assumes (as defined by the positions of all the atoms) minimizes the repulsion. This approach to the study of molecular geometry is called the valence-shell electron-pair repulsion (VSEPR) model, because it accounts for the geometric arrangements of VSEPR is pronounced “vesper.” electron pairs around a central atom in terms of the electrostatic repulsion between electron pairs. Two general rules govern the use of the VSEPR model: Animation VSEPR 1. As far as electron-pair repulsion is concerned, double bonds and triple bonds can be treated like single bonds. This approximation is good for qualitative purposes. Animation VSEPR Theory However, you should realize that in reality multiple bonds are “larger” than sin- gle bonds; that is, because there are two or three bonds between two atoms, the electron density occupies more space. 2. If a molecule has two or more resonance structures, we can apply the VSEPR model to any one of them. Formal charges are usually not shown. With this model in mind, we can predict the geometry of molecules (and ions) in a systematic way. For this purpose, it is convenient to divide molecules into two catego- ries, according to whether or not the central atom has lone pairs. Molecules in Which the Central Atom Has No Lone Pairs For simplicity we will consider molecules that contain atoms of only two ele- ments, A and B, of which A is the central atom. These molecules have the general formula ABx, where x is an integer 2, 3, . . . . (If x 5 1, we have the diatomic 414 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Arrangement of Electron Pairs About a Central Atom (A) in a Table 10.1 Molecule and Geometry of Some Simple Molecules and Ions in Which the Central Atom Has No Lone Pairs Number of Arrangement Electron of Electron Molecular Pairs Pairs* Geometry* Examples 180° 2 BeCl2, HgCl2 A B A B Linear Linear B 120° BF3 A A 3 B B Trigonal planar Trigonal planar B 109.5° A A B CH4, NH4 4 B B Tetrahedral Tetrahedral B 90° B A B 5 120° A PCl5 B B Trigonal bipyramidal Trigonal bipyramidal B 90° B B A A B B 6 SF6 90° B Octahedral Octahedral *Bonds coming out of the page are represented as solid wedges. Bonds going into the page are represented as dashed wedges. Bonds in the plane of the page are represented as solid lines. molecule AB, which is linear by definition.) In the vast majority of cases, x is between 2 and 6. Table 10.1 shows five possible arrangements of electron pairs around the central atom A. As a result of mutual repulsion, the electron pairs stay as far from one another as possible. Note that the table shows arrangements of the electron pairs but not the positions of the atoms that surround the central atom. Molecules in which the central atom has no lone pairs have one of these five arrangements of bonding pairs. Using Table 10.1 as a reference, let us take a close look at the geometry of molecules with the formulas AB2, AB3, AB4, AB5, and AB6.  10.1 Molecular Geometry 415 AB2: Beryllium Chloride (BeCl2) The Lewis structure of beryllium chloride in the gaseous state is O OS SClOBeOCl Q Q Because the bonding pairs repel each other, they must be at opposite ends of a straight line in order for them to be as far apart as possible. Thus, the ClBeCl angle is predicted to be 180°, and the molecule is linear (see Table 10.1). The “ball-and-stick” model of BeCl2 is The blue and yellow spheres are for atoms in general. AB3: Boron Trifluoride (BF3) Boron trifluoride contains three covalent bonds, or bonding pairs. SO FS A B D G O SF Q O QFS In the most stable arrangement, the three BF bonds point to the corners of an equi- lateral triangle with B in the center of the triangle. According to Table 10.1, the geometry of BF3 is trigonal planar because all four atoms lie in the same plane and the three end atoms form an equilateral triangle: Thus, each of the three FBF angles is 120°. AB4: Methane (CH4) The Lewis structure of methane is H A HOCOH A H Because there are four bonding pairs, the geometry of CH4 is tetrahedral (see Table 10.1). A tetrahedron has four sides (the prefix tetra means “four”), or faces, all of which are equilateral triangles. In a tetrahedral molecule, the central atom (C in this case) 416 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals is located at the center of the tetrahedron and the other four atoms are at the corners. The bond angles are all 109.5°. AB5: Phosphorus Pentachloride (PCl5) The Lewis structure of phosphorus pentachloride (in the gas phase) is OS O SCl SCl Q HA O EP OCl QS O SCl A Q SClS Q The only way to minimize the repulsive forces among the five bonding pairs is to arrange the PCl bonds in the form of a trigonal bipyramid (see Table 10.1). A trigo- nal bipyramid can be generated by joining two tetrahedrons along a common trian- gular base: The central atom (P in this case) is at the center of the common triangle with the surrounding atoms positioned at the five corners of the trigonal bipyramid. The atoms that are above and below the triangular plane are said to occupy axial positions, and those that are in the triangular plane are said to occupy equatorial positions. The angle between any two equatorial bonds is 120°; that between an axial bond and an equato- rial bond is 90°, and that between the two axial bonds is 180°. AB6: Sulfur Hexafluoride (SF6) The Lewis structure of sulfur hexafluoride is SO FS SO FHA EO Q FS Q S E H F A O SO Q FS FS Q SQ The most stable arrangement of the six SF bonding pairs is in the shape of an octahedron, as shown in Table 10.1. An octahedron has eight sides (the prefix octa means “eight”). It can be generated by joining two square pyramids on a common base. The central atom (S in this case) is at the center of the square base and the surrounding atoms are at the  10.1 Molecular Geometry 417 six corners. All bond angles are 90° except the one made by the bonds between the cen- tral atom and the pairs of atoms that are diametrically opposite each other. That angle is 180°. Because the six bonds are equivalent in an octahedral molecule, we cannot use the terms “axial” and “equatorial” as in a trigonal bipyramidal molecule. Molecules in Which the Central Atom Has One or More Lone Pairs Determining the geometry of a molecule is more complicated if the central atom has both lone pairs and bonding pairs. In such molecules there are three types of repulsive forces—those between bonding pairs, those between lone pairs, and those between a bonding pair and a lone pair. In general, according to the VSEPR model, the repulsive forces decrease in the following order: lone-pair vs. lone-pair . lone-pair vs. bonding- . bonding-pair vs. bonding- repulsion pair repulsion pair repulsion Electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms. These electrons have less “spatial distribution” than lone pairs; that is, they take up less space than lone-pair electrons, which are associated with only one particular atom. Because lone-pair electrons in a molecule occupy more space, they experience greater repulsion from neighboring lone pairs and bonding pairs. To keep track of the total number of bonding pairs and lone pairs, we designate molecules with lone pairs as ABxEy, where A is the central atom, B is a surrounding atom, and E is a For x 5 1 we have a diatomic molecule, which by definition has a linear geometry. lone pair on A. Both x and y are integers; x 5 2, 3, . . . , and y 5 1, 2, . . . . Thus, the values of x and y indicate the number of surrounding atoms and number of lone pairs on the central atom, respectively. The simplest such molecule would be a triatomic molecule with one lone pair on the central atom and the formula is AB2E. As the following examples show, in most cases the presence of lone pairs on the central atom makes it difficult to predict the bond angles accurately. AB2E: Sulfur Dioxide (SO2) The Lewis structure of sulfur dioxide is O Q O O OPSPO Q Because VSEPR treats double bonds as though they were single, the SO2 molecule can be viewed as consisting of three electron pairs on the central S atom. Of these, two are bonding pairs and one is a lone pair. In Table 10.1 we see that the overall arrangement of three electron pairs is trigonal planar. But because one of the electron pairs is a lone pair, the SO2 molecule has a “bent” shape. S O O SO2 418 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Because the lone-pair versus bonding-pair repulsion is greater than the bonding-pair versus bonding-pair repulsion, the two sulfur-to-oxygen bonds are pushed together slightly and the OSO angle is less than 120°. AB3E: Ammonia (NH3) The ammonia molecule contains three bonding pairs and one lone pair: O HONOH A H As Table 10.1 shows, the overall arrangement of four electron pairs is tetrahedral. But in NH3 one of the electron pairs is a lone pair, so the geometry of NH3 is trigonal pyramidal (so called because it looks like a pyramid, with the N atom at the apex). Because the lone pair repels the bonding pairs more strongly, the three NH bonding pairs are pushed closer together: N H H H Thus, the HNH angle in ammonia is smaller than the ideal tetrahedral angle of 109.5° (Figure 10.1). AB2E2: Water (H2O) A water molecule contains two bonding pairs and two lone pairs: O HOOOH Q The overall arrangement of the four electron pairs in water is tetrahedral, the same as in ammonia. However, unlike ammonia, water has two lone pairs on the central O atom. These lone pairs tend to be as far from each other as possible. Consequently, Figure 10.1 (a) The relative sizes H of bonding pairs and lone pairs in CH4, NH3, and H2O. (b) The bond angles in CH4, NH3, and H2O. Note that the solid wedges represent electron pairs (bonding C N O pairs or lone pairs) coming out above the plane of the paper, the H H H dashed wedges represent electron H H pairs going back behind the plane of the paper, and the solid lines H H H represent electron pairs in the (a) plane of the paper. H C N O H H H H 109.5° H 107.3° 104.5° H H H (b)  10.1 Molecular Geometry 419 the two O¬H bonding pairs are pushed toward each other, and we predict an even greater deviation from the tetrahedral angle than in NH3. As Figure 10.1 shows, the HOH angle is 104.5°. The geometry of H2O is bent: O H H AB4E: Sulfur Tetrafluoride (SF4) The Lewis structure of SF4 is SO F Q OFS G DQ O S D G O SQ F O QFS The central sulfur atom has five electron pairs whose arrangement, according to Table 10.1, is trigonal bipyramidal. In the SF4 molecule, however, one of the electron pairs is a lone pair, so the molecule must have one of the following geometries: F F F F S F S F F F (a) (b) SF4 In (a) the lone pair occupies an equatorial position, and in (b) it occupies an axial position. The axial position has three neighboring pairs at 90° and one at 180°, while the equatorial position has two neighboring pairs at 90° and two more at 120°. The repulsion is smaller for (a), and indeed (a) is the structure observed experimentally. This shape is sometimes described as a distorted tetrahedron (or seesaw if you turn the structure 90° to the right to view it). The angle between the axial F atoms and S is 173°, and that between the equatorial F atoms and S is 102°. Table 10.2 shows the geometries of simple molecules in which the central atom has one or more lone pairs, including some that we have not discussed. Geometry of Molecules with More Than One Central Atom So far we have discussed the geometry of molecules having only one central atom. The overall geometry of molecules with more than one central atom is difficult to define in most cases. Often we can only describe the shape around each of the central atoms. For example, consider methanol, CH3OH, whose Lewis structure is shown below: H A O HOCOOOH Q A H The two central (nonterminal) atoms in methanol are C and O. We can say that the three CH and the CO bonding pairs are tetrahedrally arranged about the C atom. The HCH and OCH bond angles are approximately 109°. The O atom here is like the one 420 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Geometry of Simple Molecules and Ions in Which the Central Atom Has One Table 10.2 or More Lone Pairs Class of Total Number of Number of Number of Arrangement of Geometry of Molecule Electron Pairs Bonding Pairs Lone Pairs Electron Pairs Molecule or Ion Examples AB2E 3 2 1 A Bent B B Trigonal planar SO2 AB3E 4 3 1 B A Trigonal B pyramidal B Tetrahedral NH3 AB2E2 4 2 2 A Bent B B Tetrahedral H 2O B B Distorted A AB4E 5 4 1 tetrahedron B (or seesaw) B Trigonal bipyramidal SF4 B AB3E2 5 3 2 B A T-shaped B Trigonal bipyramidal ClF3 B AB2E3 5 2 3 A Linear B Trigonal bipyramidal I3– B B B A Square AB5E 6 5 1 B B pyramidal Octahedral BrF5 B B AB4E2 6 4 2 A Square planar B B Octahedral XeF4  10.1 Molecular Geometry 421 in water in that it has two lone pairs and two bonding pairs. Therefore, the HOC portion of the molecule is bent, and the angle HOC is approximately equal to 105° (Figure 10.2). Guidelines for Applying the VSEPR Model Having studied the geometries of molecules in two categories (central atoms with and without lone pairs), let us consider some rules for applying the VSEPR model to all types of molecules: Figure 10.2 The geometry of CH3OH. 1. Write the Lewis structure of the molecule, considering only the electron pairs around the central atom (that is, the atom that is bonded to more than one other atom). 2. Count the number of electron pairs around the central atom (bonding pairs and lone pairs). Treat double and triple bonds as though they were single bonds. Refer to Table 10.1 to predict the overall arrangement of the electron pairs. 3. Use Tables 10.1 and 10.2 to predict the geometry of the molecule. 4. In predicting bond angles, note that a lone pair repels another lone pair or a bond- ing pair more strongly than a bonding pair repels another bonding pair. Remember that in general there is no easy way to predict bond angles accurately when the central atom possesses one or more lone pairs. The VSEPR model generates reliable predictions of the geometries of a variety of molecular structures. Chemists use the VSEPR approach because of its simplicity. Although there are some theoretical concerns about whether “electron-pair repulsion” actually determines molecular shapes, the assumption that it does leads to useful (and generally reliable) predictions. We need not ask more of any model at this stage in our study of chemistry. Example 10.1 illustrates the application of VSEPR. Example 10.1 Use the VSEPR model to predict the geometry of the following molecules and ions: (a) AsH3, (b) OF2, (c) AlCl24, (d) I23, (e) C2H4. Strategy The sequence of steps in determining molecular geometry is as follows: draw Lewis ¡ find arrangement of ¡ find arrangement ¡ determine geometry structure electron pairs of bonding pairs based on bonding pairs Solution (a) The Lewis structure of AsH3 is O HOAsOH A H There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral (see Table 10.1). Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the As and H atoms). Thus, AsH3 removing the lone pair leaves us with three bonding pairs and a trigonal pyramidal geometry, like NH3. We cannot predict the HAsH angle accurately, but we know that it is less than 109.5° because the repulsion of the bonding electron pairs in the As—H bonds by the lone pair on As is greater than the repulsion between the bonding pairs. (b) The Lewis structure of OF2 is O Q O Q O SFOOOFS Q (Continued) OF2 422 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral (see Table 10.1). Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the O and F atoms). Thus, removing the two lone pairs leaves us with two bonding pairs and a bent geometry, like H2O. We cannot predict the FOF angle accurately, but we know that it must be less than 109.5° because the repulsion of the bonding electron pairs in the O¬F bonds by the lone pairs on O is greater than the repulsion between the bonding pairs. (c) The Lewis structure of AlCl2 4 is Q  SClS A O O SClOAlOClS Q A Q SClS O There are four electron pairs around the central atom; therefore, the electron pair AlCl24 arrangement is tetrahedral. Because there are no lone pairs present, the arrangement of the bonding pairs is the same as the electron pair arrangement. Therefore, AlCl24 has a tetrahedral geometry and the ClAlCl angles are all 109.5°. (d) The Lewis structure of I23 is  SIO QOQI OO QI S S S I23 There are five electron pairs around the central I atom; therefore, the electron pair arrangement is trigonal bipyramidal. Of the five electron pairs, three are lone pairs and two are bonding pairs. Recall that the lone pairs preferentially occupy the equatorial positions in a trigonal bipyramid (see Table 10.2). Thus, removing the lone pairs leaves us with a linear geometry for I23; that is, all three I atoms lie in a straight line. (e) The Lewis structure of C2H4 is H H G D CPC D G H H The C“C bond is treated as though it were a single bond in the VSEPR model. Because there are three electron pairs around each C atom and there are no C2H4 lone pairs present, the arrangement around each C atom has a trigonal planar shape like BF3, discussed earlier. Thus, the predicted bond angles in C2H4 are all 120°. H 120 H G D CPC 120° D G H 120 H Comment (1) The I23 ion is one of the few structures for which the bond angle (180°) can be predicted accurately even though the central atom contains lone pairs. (2) In C2H4, all six atoms lie in the same plane. The overall planar geometry is not predicted by the VSEPR model, but we will see why the molecule prefers to be planar later. In reality, the angles are close, but not equal, to 120° because the bonds are not Similar problems: 10.7, 10.8, 10.9. all equivalent. Practice Exercise Use the VSEPR model to predict the geometry of (a) SiBr4, (b) CS2, and (c) NO23.  10.2 Dipole Moments 423 Review of Concepts Which of the following geometries has a greater stability for tin(IV) hydride (SnH4)? 10.2 Dipole Moments In Section 9.5 we learned that hydrogen fluoride is a covalent compound with a polar bond. There is a shift of electron density from H to F because the F atom is more electronegative than the H atom (see Figure 9.4). The shift of electron density is symbolized by placing a crossed arrow ( 888n ) above the Lewis structure to indicate the direction of the shift. For example, 888n OS HOF Q The consequent charge separation can be represented as ␦ ␦ Q HOFQS where δ (delta) denotes a partial charge. This separation of charges can be confirmed in an electric field (Figure 10.3). When the field is turned on, HF molecules orient their negative ends toward the positive plate and their positive ends toward the nega- tive plate. This alignment of molecules can be detected experimentally. A quantitative measure of the polarity of a bond is its dipole moment (μ), which is the product of the charge Q and the distance r between the charges: μ5Q3r (10.1) – Figure 10.3 Behavior of polar + molecules (a) in the absence of an + external electric field and (b) when – + – – – + the electric field is turned on. + – + + – Nonpolar molecules are not – + + affected by an electric field. – – + – – + + + + + – – – – + – + – + + – + – + – – + + – + + + – – – + + – – (a) (b) 424 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals In a diatomic molecule like HF, the charge To maintain electrical neutrality, the charges on both ends of an electrically neutral Q is equal to δ1 and δ2. diatomic molecule must be equal in magnitude and opposite in sign. However, in Equation (10.1), Q refers only to the magnitude of the charge and not to its sign, so μ is always positive. Dipole moments are usually expressed in debye units (D), named for Peter Debye.† The conversion factor is 1 D 5 3.336 3 10230 C m where C is coulomb and m is meter. Animation Diatomic molecules containing atoms of different elements (for example, HCl, Polarity of Molecules CO, and NO) have dipole moments and are called polar molecules. Diatomic mol- ecules containing atoms of the same element (for example, H2, O2, and F2) are Animation examples of nonpolar molecules because they do not have dipole moments. For a Influence of Shape on Polarity molecule made up of three or more atoms both the polarity of the bonds and the molecular geometry determine whether there is a dipole moment. Even if polar bonds are present, the molecule will not necessarily have a dipole moment. Carbon dioxide (CO2), for example, is a triatomic molecule, so its geometry is either linear or bent: 8 88 8m mKCN O O 888n m88 88n resultant OPCPO dipole moment linear molecule bent molecule (no dipole moment) (would have a dipole moment) Each carbon-to-oxygen bond is polar, The arrows show the shift of electron density from the less electronegative carbon with the electron density shifted toward the more electronegative oxygen atom. atom to the more electronegative oxygen atom. In each case, the dipole moment of However, the linear geometry of the the entire molecule is made up of two bond moments, that is, individual dipole molecule results in the cancellation of moments in the polar C“O bonds. The bond moment is a vector quantity, which the two bond moments. means that it has both magnitude and direction. The measured dipole moment is equal to the vector sum of the bond moments. The two bond moments in CO2 are equal in magnitude. Because they point in opposite directions in a linear CO2 molecule, the sum or resultant dipole moment would be zero. On the other hand, if the CO2 mol- ecule were bent, the two bond moments would partially reinforce each other, so that The VSEPR model predicts that CO2 is a the molecule would have a dipole moment. Experimentally it is found that carbon linear molecule. dioxide has no dipole moment. Therefore, we conclude that the carbon dioxide mol- ecule is linear. The linear nature of carbon dioxide has been confirmed through other experimental measurements. Next let us consider the NH3 and NF3 molecules shown in Figure 10.4. In both cases, the central N atom has a lone pair, whose charge density is away from the N atom. From Figure 9.5 we know that N is more electronegative than H, and F is more electronegative than N. For this reason, the shift of electron density in NH3 is toward N and so contributes a larger dipole moment, whereas the NF bond moments are directed away from the N atom and so together they offset the contribution of the lone pair to the dipole moment. Thus, the resultant dipole moment in NH3 is larger than that in NF3. Dipole moments can be used to distinguish between molecules that have the same formula but different structures. For example, the following molecules both exist; they † Peter Joseph William Debye (1884–1966). American chemist and physicist of Dutch origin. Debye made many significant contributions in the study of molecular structure, polymer chemistry, X-ray analysis, and electrolyte solution. He was awarded the Nobel Prize in Chemistry in 1936.  10.2 Dipole Moments 425 Resultant dipole Figure 10.4 Bond moments and moment = 1.46 D resultant dipole moments in NH3 and NF3. The electrostatic potential maps show the electron density distributions in these molecules. N N H F H F H F Resultant dipole moment = 0.24 D have the same molecular formula (C2H2Cl2), the same number and type of bonds, but different molecular structures: resultant 88 dipole moment n888 m m m m Cl Cl H Cl 88 G D G D 88 88 88 m m m m CPC CPC 88 D G D G 88 88 88 m H H Cl H cis-dichloroethylene trans-dichloroethylene ␮  1.89 D ␮0 Because cis-dichloroethylene is a polar molecule but trans-dichloroethylene is not, they can readily be distinguished by a dipole moment measurement. Additionally, as we will see in Chapter 11, the strength of intermolecular forces is partially determined by whether molecules possess a dipole moment. Table 10.3 lists the dipole moments of several polar molecules. Table 10.3 Dipole Moments of Some Polar Molecules In cis-dichloroethylene (top), the bond moments reinforce Molecule Geometry Dipole Moment (D) one another and the molecule is polar. The opposite holds for HF Linear 1.92 trans-dichloroethylene and the HCl Linear 1.08 molecule is nonpolar. HBr Linear 0.78 HI Linear 0.38 H2O Bent 1.87 H2S Bent 1.10 NH3 Trigonal pyramidal 1.46 SO2 Bent 1.60 CHEMISTRY in Action Microwave Ovens—Dipole Moments at Work I n the last 40 years the microwave oven has become a ubiqui- tous appliance. Microwave technology enables us to thaw and cook food much more rapidly than conventional appliances do. In Chapter 7 we saw that microwaves are a form of electro- magnetic radiation (see Figure 7.3). Microwaves are generated by a magnetron, which was invented during World War II How do microwaves heat food so quickly? when radar technology was being developed. The magnetron is + – of microwave Electric field Direction of wave – + (a) + + of microwave Electric field Direction of wave – – (b) Interaction between the electric field component of the microwave and a polar molecule. (a) The negative end of the dipole follows the propagation of the wave (the positive region) and rotates in a clockwise direction. (b) If, after the molecule has rotated to the new position the radiation has also moved along to its next cycle, the positive end of the dipole will move into the negative region of the wave while the negative end will be pushed up. Thus, the molecule will rotate faster. No such interaction can occur with nonpolar molecules. 426 a hollow cylinder encased in a horseshoe-shaped magnet. In the therefore reach different parts of food at the same time. center of the cylinder is a cathode rod. The walls of the cylinder (Depending on the amount of water present, microwaves can act as an anode. When heated, the cathode emits electrons that penetrate food to a depth of several inches.) In a conventional travel toward the anode. The magnetic field forces the electrons oven, heat can affect the center of foods only by conduction to move in a circular path. This motion of charged particles (that is, by transfer of heat from hot air molecules to cooler generates microwaves, which are adjusted to a frequency of molecules in food in a layer-by-layer fashion), which is a very 2.45 GHz (2.45 3 109 Hz) for cooking. A “waveguide” directs slow process. the microwaves into the cooking compartment. Rotating fan The following points are relevant to the operation of a mi- blades reflect the microwaves to all parts of the oven. crowave oven. Plastics and Pyrex glasswares do not contain po- The cooking action in a microwave oven results from the lar molecules and are therefore not affected by microwave interaction between the electric field component of the radiation radiation. (Styrofoam and certain plastics cannot be used in mi- with the polar molecules—mostly water—in food. All molecules crowaves because they melt from the heat of the food.) Metals, rotate at room temperature. If the frequency of the radiation and however, reflect microwaves, thereby shielding the food and that of the molecular rotation are equal, energy can be transferred possibly returning enough energy to the microwave emitter to from the microwave to the polar molecule. As a result, the mol- overload it. Because microwaves can induce a current in the ecule will rotate faster. This is what happens in a gas. In the con- metal, this action can lead to sparks jumping between the con- densed state (for example, in food), a molecule cannot execute the tainer and the bottom or walls of the oven. Finally, although free rotation. Nevertheless, it still experiences a torque (a force water molecules in ice are locked in position and therefore can- that causes rotation) that tends to align its dipole moment with the not rotate, we routinely thaw food in a microwave oven. The oscillating field of the microwave. Consequently, there is friction reason is that at room temperature a thin film of liquid water between the molecules, which appears as heat in the food. quickly forms on the surface of frozen food and the mobile The reason that a microwave oven can cook food so fast is molecules in that film can absorb the radiation to start the thaw- that the radiation is not absorbed by nonpolar molecules and can ing process. Rotating blades Magnetron Waveguide Anode Cathode Magnet A microwave oven. The microwaves generated by the magnetron are reflected to all parts of the oven by the rotating fan blades. 427 428 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Example 10.2 shows how we can predict whether a molecule possesses a dipole moment if we know its molecular geometry. Example 10.2 Predict whether each of the following molecules has a dipole moment: (a) BrCl, (b) BF3 (trigonal planar), (c) CH2Cl2 (tetrahedral). Strategy Keep in mind that the dipole moment of a molecule depends on both the difference in electronegativities of the elements present and its geometry. A molecule can have polar bonds (if the bonded atoms have different electronegativities), but it may not possess a dipole moment if it has a highly symmetrical geometry. Solution (a) Because bromine chloride is diatomic, it has a linear geometry. Chlorine is more electronegative than bromine (see Figure 9.5), so BrCl is polar with chlorine at the negative end Electrostatic potential map of BrCl 88n shows that the electron density is Br—Cl shifted toward the Cl atom. Thus, the molecule does have a dipole moment. In fact, all diatomic molecules containing different elements possess a dipole moment. (b) Because fluorine is more electronegative than boron, each B¬F bond in BF3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a trigonal planar shape means that the three bond moments exactly cancel one another: 88n F A B 88 88 D G m m F F An analogy is an object that is pulled in the directions shown by the three bond Electrostatic potential map moments. If the forces are equal, the object will not move. Consequently, BF3 has shows that the electron density is symmetrically distributed in no dipole moment; it is a nonpolar molecule. the BF3 molecule. (c) The Lewis structure of CH2Cl2 (methylene chloride) is Cl A HOCOH A Cl This molecule is similar to CH4 in that it has an overall tetrahedral shape. However, because not all the bonds are identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5°. Because chlorine is more electronegative than carbon, which is more electronegative than hydrogen, the bond moments do not cancel and the molecule possesses a dipole moment: resultant Cl dipole moment Electrostatic potential map of C CH2Cl2. The electron density is H Cl shifted toward the electronegative H Cl atoms. Similar problems: 10.21, 10.22, 10.23. Thus, CH2Cl2 is a polar molecule. Practice Exercise Does the AlCl3 molecule have a dipole moment?  10.3 Valence Bond Theory 429 Review of Concepts Carbon dioxide has a linear geometry and is nonpolar. Yet we know that the molecule executes bending and stretching motions that create a dipole moment. How would you reconcile these conflicting descriptions about CO2? 10.3 Valence Bond Theory The VSEPR model, based largely on Lewis structures, provides a relatively simple and straightforward method for predicting the geometry of molecules. But as we noted earlier, the Lewis theory of chemical bonding does not clearly explain why chemical bonds exist. Relating the formation of a covalent bond to the pairing of electrons was a step in the right direction, but it did not go far enough. For example, the Lewis theory describes the single bond between the H atoms in H2 and that between the F atoms in F2 in essentially the same way—as the pairing of two elec- trons. Yet these two molecules have quite different bond enthalpies and bond lengths (436.4 kJ/mol and 74 pm for H2 and 150.6 kJ/mol and 142 pm for F2). These and many other facts cannot be explained by the Lewis theory. For a more complete explanation of chemical bond formation we look to quantum mechanics. In fact, the quantum mechanical study of chemical bonding also provides a means for under- standing molecular geometry. At present, two quantum mechanical theories are used to describe covalent bond formation and the electronic structure of molecules. Valence bond (VB) theory assumes that the electrons in a molecule occupy atomic orbitals of the individual atoms. It enables us to retain a picture of individual atoms taking part in the bond formation. The second theory, called molecular orbital (MO) theory, assumes the formation of molecular orbitals from the atomic orbitals. Neither theory perfectly explains all aspects of bonding, but each has contributed something to our understanding of many observed molecular properties. Let us start our discussion of valence bond theory by considering the formation of a H2 molecule from two H atoms. The Lewis theory describes the H¬H bond in terms of the pairing of the two electrons on the H atoms. In the framework of valence bond theory, the covalent H¬H bond is formed by the overlap of the two 1s orbit- als in the H atoms. By overlap, we mean that the two orbitals share a common region in space. What happens to two H atoms as they move toward each other and form a bond? Initially, when the two atoms are far apart, there is no interaction. We say that the potential energy of this system (that is, the two H atoms) is zero. As the atoms Recall that an object has potential energy by virtue of its position. approach each other, each electron is attracted by the nucleus of the other atom; at the same time, the electrons repel each other, as do the nuclei. While the atoms are still separated, attraction is stronger than repulsion, so that the potential energy of the system decreases (that is, it becomes negative) as the atoms approach each other (Figure 10.5). This trend continues until the potential energy reaches a minimum value. At this point, when the system has the lowest potential energy, it is most stable. This condition corresponds to substantial overlap of the 1s orbitals and the formation of a stable H2 molecule. If the distance between nuclei were to decrease further, the potential energy would rise steeply and finally become positive as a result of the increased electron-electron and nuclear-nuclear repulsions. In accord with the law of conservation of energy, the decrease in potential energy as a result of H2 formation must be accompanied by a release of energy. Experiments show that as a H2 molecule is formed from two H atoms, heat is given off. The converse is also true. To break a H¬H bond, energy must be supplied to the molecule. Figure 10.6 is another way of viewing the formation of a H2 molecule. 430 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Figure 10.5 Change in potential energy of two H atoms with their distance of separation. At the point of minimum potential energy, the H2 molecule is in its most stable state and the bond length is 74 pm. The spheres represent + the 1s orbitals. Potential energy 0 Distance of separation ¬ – Figure 10.6 Top to bottom: As two H atoms approach each other, their 1s orbitals begin to interact and each electron begins to feel the attraction of the other proton. Gradually, the electron density builds up in the region between the two nuclei (red color). Eventually, a stable H2 molecule is formed with an internuclear distance of 74 pm.  10.4 Hybridization of Atomic Orbitals 431 Thus, valence bond theory gives a clearer picture of chemical bond formation than the Lewis theory does. Valence bond theory states that a stable molecule forms from reacting atoms when the potential energy of the system has decreased to a minimum; the Lewis theory ignores energy changes in chemical bond formation. The concept of overlapping atomic orbitals applies equally well to diatomic mol- ecules other than H2. Thus, a stable F2 molecule forms when the 2p orbitals (containing The orbital diagram of the F atom is shown on p. 306. the unpaired electrons) in the two F atoms overlap to form a covalent bond. Similarly, the formation of the HF molecule can be explained by the overlap of the 1s orbital in H with the 2p orbital in F. In each case, VB theory accounts for the changes in poten- tial energy as the distance between the reacting atoms changes. Because the orbitals involved are not the same kind in all cases, we can see why the bond enthalpies and bond lengths in H2, F2, and HF might be different. As we stated earlier, Lewis theory treats all covalent bonds the same way and offers no explanation for the differences among covalent bonds. Review of Concepts Compare the Lewis theory and the valence bond theory of chemical bonding. 10.4 Hybridization of Atomic Orbitals The concept of atomic orbital overlap should apply also to polyatomic molecules. However, a satisfactory bonding scheme must account for molecular geometry. We will discuss three examples of VB treatment of bonding in polyatomic molecules. sp3 Hybridization Consider the CH4 molecule. Focusing only on the valence electrons, we can represent the orbital diagram of C as hg h h 2s 2p Because the carbon atom has two unpaired electrons (one in each of the two 2p orbit- als), it can form only two bonds with hydrogen in its ground state. Although the species CH2 is known, it is very unstable. To account for the four C¬H bonds in methane, we can try to promote (that is, energetically excite) an electron from the 2s orbital to the 2p orbital: h h h h 2s 2p Now there are four unpaired electrons on C that could form four C¬H bonds. How- ever, the geometry is wrong, because three of the HCH bond angles would have to be 90° (remember that the three 2p orbitals on carbon are mutually perpendicular), and yet all HCH angles are 109.5°. To explain the bonding in methane, VB theory uses hypothetical hybrid Animation Hybridization orbitals, which are atomic orbitals obtained when two or more nonequivalent 432 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals z z z z y y y y x x x x 2s 2px 2py 2pz Hybridization sp3 sp3 sp3 sp3 Figure 10.7 Formation of four sp3 hybrid orbitals from one 2s and three 2p orbitals. The sp3 orbitals point to the corners of a tetrahedron. orbitals of the same atom combine in preparation for covalent bond formation. Animation Hybridization is the term applied to the mixing of atomic orbitals in an atom Molecular Shape and Orbital Hybridization (usually a central atom) to generate a set of hybrid orbitals. We can generate four equivalent hybrid orbitals for carbon by mixing the 2s orbital and the three 2p orbitals: h h h h 3 sp is pronounced “s-p three.” sp3 orbitals Because the new orbitals are formed from one s and three p orbitals, they are called sp3 hybrid orbitals. Figure 10.7 shows the shape and orientations of the sp3 orbitals. These four hybrid orbitals are directed toward the four corners of a regular tetrahe- dron. Figure 10.8 shows the formation of four covalent bonds between the carbon H sp3 hybrid orbitals and the hydrogen 1s orbitals in CH4. Thus CH4 has a tetrahedral shape, and all the HCH angles are 109.5°. Note that although energy is required to bring about hybridization, this input is more than compensated for by the energy released upon the formation of C¬H bonds. (Recall that bond formation is an exo- thermic process.) C H H The following analogy is useful for understanding hybridization. Suppose that we have a beaker of a red solution and three beakers of blue solutions and that the volume of each is 50 mL. The red solution corresponds to one 2s orbital, the blue solutions represent three 2p orbitals, and the four equal volumes symbolize four H separate orbitals. By mixing the solutions we obtain 200 mL of a purple solution, Figure 10.8 Formation of four which can be divided into four 50-mL portions (that is, the hybridization process bonds between the carbon sp3 generates four sp3 orbitals). Just as the purple color is made up of the red and blue hybrid orbitals and the hydrogen 1s orbitals in CH4. The smaller components of the original solutions, the sp3 hybrid orbitals possess both s and p lobes are not shown. orbital characteristics.  10.4 Hybridization of Atomic Orbitals 433 Another example of sp3 hybridization is ammonia (NH3). Table 10.1 shows that the arrangement of four electron pairs is tetrahedral, so that the bonding in NH3 can be explained by assuming that N, like C in CH4, is sp3-hybridized. The ground-state electron configuration of N is 1s22s22p3, so that the orbital diagram for the sp3 hybrid- N ized N atom is H H h h h hg H Figure 10.9 The sp3-hybridized sp3 orbitals N atom in NH3. Three sp3 hybrid orbitals form bonds with the H atoms. The fourth is occupied Three of the four hybrid orbitals form covalent N¬H bonds, and the fourth hybrid by nitrogen’s lone pair. orbital accommodates the lone pair on nitrogen (Figure 10.9). Repulsion between the lone-pair electrons and electrons in the bonding orbitals decreases the HNH bond angles from 109.5° to 107.3°. It is important to understand the relationship between hybridization and the VSEPR model. We use hybridization to describe the bonding scheme only when the arrangement of electron pairs has been predicted using VSEPR. If the VSEPR model predicts a tetrahedral arrangement of electron pairs, then we assume that one s and three p orbitals are hybridized to form four sp3 hybrid orbitals. The following are examples of other types of hybridization. sp Hybridization The beryllium chloride (BeCl2) molecule is predicted to be linear by VSEPR. The orbital diagram for the valence electrons in Be is hg 2s 2p We know that in its ground state, Be does not form covalent bonds with Cl because its electrons are paired in the 2s orbital. So we turn to hybridization for an explanation of Be’s bonding behavior. First, we promote a 2s electron to a 2p orbital, resulting in h h 2s 2p Now there are two Be orbitals available for bonding, the 2s and 2p. However, if two Cl atoms were to combine with Be in this excited state, one Cl atom would share a 2s electron and the other Cl would share a 2p electron, making two non- equivalent BeCl bonds. This scheme contradicts experimental evidence. In the actual BeCl2 molecule, the two BeCl bonds are identical in every respect. Thus, the 2s and 2p orbitals must be mixed, or hybridized, to form two equivalent sp hybrid orbitals: h h sp orbitals empty 2p orbitals 434 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals z z z z y y y y Hybridization x x x x 2s 2px sp sp Figure 10.10 Formation of sp hybrid orbitals. Figure 10.10 shows the shape and orientation of the sp orbitals. These two hybrid orbitals lie on the same line, the x-axis, so that the angle between them is 180°. Each of the BeCl bonds is then formed by the overlap of a Be sp hybrid orbital and a Cl  3p orbital, and the resulting BeCl2 molecule has a linear geometry (Figure 10.11). sp2 Hybridization Next we will look at the BF3 (boron trifluoride) molecule, known to have a planar geometry. Considering only the valence electrons, the orbital diagram of B is hg h 2s 2p First, we promote a 2s electron to an empty 2p orbital: h h h 2s 2p sp2 is pronounced “s-p two.” Mixing the 2s orbital with the two 2p orbitals generates three sp2 hybrid orbitals: h h h sp2 orbitals empty 2p orbital These three sp2 orbitals lie in the same plane, and the angle between any two of them is 120° (Figure 10.12). Each of the BF bonds is formed by the overlap of a boron sp2 hybrid orbital and a fluorine 2p orbital (Figure 10.13). The BF3 molecule is planar with all the FBF angles equal to 120°. This result conforms to experimental findings and also to VSEPR predictions. Cl Be Cl You may have noticed an interesting connection between hybridization and the octet rule. Regardless of the type of hybridization, an atom starting with one s and Figure 10.11 The linear geometry of BeCl2 can be three p orbitals would still possess four orbitals, enough to accommodate a total of explained by assuming that Be is eight electrons in a compound. For elements in the second period of the periodic table, sp-hybridized. The two sp hybrid eight is the maximum number of electrons that an atom of any of these elements can orbitals overlap with the two chlorine 3p orbitals to form two accommodate in the valence shell. This is the reason that the octet rule is usually covalent bonds. obeyed by the second-period elements.  10.4 Hybridization of Atomic Orbitals 435 z y x Hybridization z z sp2 sp2 2s y y x x sp2 2px 2py 2 Figure 10.12 Formation of sp hybrid orbitals. The situation is different for an atom of a third-period element. If we use only the 3s and 3p orbitals of the atom to form hybrid orbitals in a molecule, then the octet rule applies. However, in some molecules the same atom may use one or more 3d orbitals, in addition to the 3s and 3p orbitals, to form hybrid orbitals. In these cases, the octet rule does not hold. We will see specific examples of the participation of the 3d orbital in hybridization shortly. To summarize our discussion of hybridization, we note that • The concept of hybridization is not applied to isolated atoms. It is a theoretical model used only to explain covalent bonding. • Hybridization is the mixing of at least two nonequivalent atomic orbitals, for example, s and p orbitals. Therefore, a hybrid orbital is not a pure atomic orbital. Hybrid orbitals and pure atomic orbitals have very different shapes. • The number of hybrid orbitals generated is equal to the number of pure atomic orbitals that participate in the hybridization process. • Hybridization requires an input of energy; however, the system more than recov- ers this energy during bond formation. • Covalent bonds in polyatomic molecules and ions are formed by the overlap of hybrid orbitals, or of hybrid orbitals with unhybridized ones. Therefore, the F F hybridization bonding scheme is still within the framework of valence bond the- B ory; electrons in a molecule are assumed to occupy hybrid orbitals of the indi- vidual atoms. Table 10.4 summarizes sp, sp2, and sp3 hybridization (as well as other types that we F will discuss shortly). Procedure for Hybridizing Atomic Orbitals Figure 10.13 The sp2 hybrid Before going on to discuss the hybridization of d orbitals, let us specify what we need orbitals of boron overlap with the 2p orbitals of fluorine. The BF3 to know in order to apply hybridization to bonding in polyatomic molecules in gen- molecule is planar, and all the FBF eral. In essence, hybridization simply extends Lewis theory and the VSEPR model. angles are 120°. 436 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Table 10.4 Important Hybrid Orbitals and Their Shapes Pure Atomic Hybridiza- Orbitals of tion of the Number Shape the Central Central of Hybrid of Hybrid Atom Atom Orbitals Orbitals Examples 1808 s, p sp 2 BeCl2 Linear s, p, p sp2 3 BF3 1208 Trigonal planar 109.58 s, p, p, p sp3 4 CH4, NH 4 Tetrahedral 908 s, p, p, p, d sp3d 5 PCl5 1208 Trigonal bipyramidal 908 s, p, p, p, d, d sp3d2 6 SF6 908 Octahedral  10.4 Hybridization of Atomic Orbitals 437 To assign a suitable state of hybridization to the central atom in a molecule, we must have some idea about the geometry of the molecule. The steps are as follows: 1. Draw the Lewis structure of the molecule. 2. Predict the overall arrangement of the electron pairs (both bonding pairs and lone pairs) using the VSEPR model (see Table 10.1). 3. Deduce the hybridization of the central atom by matching the arrangement of the electron pairs with those of the hybrid orbitals shown in Table 10.4. Example 10.3 illustrates this procedure. Example 10.3 Determine the hybridization state of the central (underlined) atom in each of the following molecules: (a) BeH2, (b) AlI3, and (c) PF3. Describe the hybridization process and determine the molecular geometry in each case. Strategy The steps for determining the hybridization of the central atom in a molecule are: draw Lewis structure use VSEPR to determine the use Table 10.4 to of the molecule 88n electron pair arrangement 88n determine the surrounding the central hybridization state of atom (Table 10.1) the central atom Solution (a) The ground-state electron configuration of Be is 1s22s2 and the Be atom has two valence electrons. The Lewis structure of BeH2 is H¬Be¬H There are two bonding pairs around Be; therefore, the electron pair arrangement is BeH2 linear. We conclude that Be uses sp hybrid orbitals in bonding with H, because sp orbitals have a linear arrangement (see Table 10.4). The hybridization process can be imagined as follows. First, we draw the orbital diagram for the ground state of Be: hg 2s 2p By promoting a 2s electron to the 2p orbital, we get the excited state: h h 2s 2p The 2s and 2p orbitals then mix to form two hybrid orbitals: h h sp orbitals empty 2p orbitals The two Be¬H bonds are formed by the overlap of the Be sp orbitals with the 1s orbitals of the H atoms. Thus, BeH2 is a linear molecule. (Continued) 438 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals (b) The ground-state electron configuration of Al is [Ne]3s23p1. Therefore, the Al atom has three valence electrons. The Lewis structure of AlI3 is SO IS A SOIOAlOO IS O O There are three pairs of electrons around Al; therefore, the electron pair arrangement is trigonal planar. We conclude that Al uses sp2 hybrid orbitals in bonding with I because sp2 orbitals have a trigonal planar arrangement (see Table 10.4). The orbital diagram of the ground-state Al atom is AlI3 hg h 3s 3p By promoting a 3s electron into the 3p orbital we obtain the following excited state: h h h 3s 3p The 3s and two 3p orbitals then mix to form three sp2 hybrid orbitals: h h h sp2 orbitals empty 3p orbital The sp2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent Al¬I bonds. We predict that the AlI3 molecule is trigonal planar and all the IAlI angles are 120°. (c) The ground-state electron configuration of P is [Ne]3s23p3. Therefore, P atom has five valence electrons. The Lewis structure of PF3 is SO O OS FOPOF O A SFS O There are four pairs of electrons around P; therefore, the electron pair arrangement is tetrahedral. We conclude that P uses sp3 hybrid orbitals in bonding to F, because sp3 orbitals have a tetrahedral arrangement (see Table 10.4). The hybridization process can PF3 be imagined to take place as follows. The orbital diagram of the ground-state P atom is hg h h h 3s 3p By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid orbitals. h h h hg sp3 orbitals As in the case of NH3, one of the sp3 hybrid orbitals is used to accommodate the lone pair on P. The other three sp3 hybrid orbitals form covalent P¬F bonds with the 2p orbitals of F. We predict the geometry of the molecule to be trigonal Similar problems: 10.31, 10.33. pyramidal; the FPF angle should be somewhat less than 109.5°. Practice Exercise Determine the hybridization state of the underlined atoms in the following compounds: (a) SiBr4 and (b) BCl3.  10.4 Hybridization of Atomic Orbitals 439 Hybridization of s, p, and d Orbitals We have seen that hybridization neatly explains bonding that involves s and p orbitals. For elements in the third period and beyond, however, we cannot always account for molecular geometry by assuming that only s and p orbitals hybridize. To understand the formation of molecules with trigonal bipyramidal and octahedral geometries, for instance, we must include d orbitals in the hybridization concept. Consider the SF6 molecule as an example. In Section 10.1 we saw that this mol- ecule has octahedral geometry, which is also the arrangement of the six electron pairs. Table 10.4 shows that the S atom is sp3d2-hybridized in SF6. The ground-state electron configuration of S is [Ne]3s23p4. Focusing only on the valence electrons, we have the orbital diagram hg hg h h 3s 3p 3d Because the 3d level is quite close in energy to the 3s and 3p levels, we can promote 3s and 3p electrons to two of the 3d orbitals: SF6 h h h h h h 3s 3p 3d Mixing the 3s, three 3p, and two 3d orbitals generates six sp3d2 hybrid orbitals: sp3d 2 is pronounced “s-p three d two.” h h h h h h sp3d 2 orbitals empty 3d orbitals The six S¬F bonds are formed by the overlap of the hybrid orbitals of the S atom with the 2p orbitals of the F atoms. Because there are 12 electrons around the S atom, the octet rule is violated. The use of d orbitals in addition to s and p orbitals to form an expanded octet (see Section 9.9) is an example of valence-shell expansion. Second- period elements, unlike third-period elements, do not have 2d energy levels, so they can never expand their valence shells. (Recall that when n 5 2, l 5 0 and 1. Thus, we can only have 2s and 2p orbitals.) Hence atoms of second-period elements can never be surrounded by more than eight electrons in any of their compounds. Example 10.4 deals with valence-shell expansion in a third-period element. Example 10.4 Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr5). Strategy Follow the same procedure shown in Example 10.3. Solution The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atom has five valence electrons. The Lewis structure of PBr5 is Q O SBrS SBr Q HA O O EP OBrS Q Q A SBr SBrS Q There are five pairs of electrons around P; therefore, the electron pair arrangement is trigonal bipyramidal. We conclude that P uses sp3d hybrid orbitals in bonding to Br, (Continued) PBr5 440 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals because sp3d hybrid orbitals have a trigonal bipyramidal arrangement (see Table 10.4). The hybridization process can be imagined as follows. The orbital diagram of the ground-state P atom is hg h h h 3s 3p 3d Promoting a 3s electron into a 3d orbital results in the following excited state: h h h h h 3s 3p 3d Mixing the one 3s, three 3p, and one 3d orbitals generates five sp3d hybrid orbitals: h h h h h sp3d orbitals empty 3d orbitals These hybrid orbitals overlap the 4p orbitals of Br to form five covalent P¬Br bonds. Similar problem: 10.40. Because there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal. Practice Exercise Describe the hybridization state of Se in SeF6. Review of Concepts What is the hybridization of Xe in XeF4 (see Example 9.12 on p. 396)? 10.5 Hybridization in Molecules Containing Double Ground and Triple Bonds state 2s 2p The concept of hybridization is useful also for molecules with double and triple Promotion bonds. Consider the ethylene molecule, C2H4, as an example. In Example 10.1 we of electron saw that C2H4 contains a carbon-carbon double bond and has planar geometry. Both 2s 2p the geometry and the bonding can be understood if we assume that each carbon sp2- Hybridized atom is sp2-hybridized. Figure 10.14 shows orbital diagrams of this hybridization state process. We assume that only the 2px and 2py orbitals combine with the 2s orbital, ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 2pz sp 2 orbitals and that the 2pz orbital remains unchanged. Figure 10.15 shows that the 2pz orbital Figure 10.14 The sp2 hybrid- is perpendicular to the plane of the hybrid orbitals. Now how do we account for ization of a carbon atom. The 2s the bonding of the C atoms? As Figure 10.16(a) shows, each carbon atom uses the orbital is mixed with only two 2p orbitals to form three equivalent three sp2 hybrid orbitals to form two bonds with the two hydrogen 1s orbitals and sp2 hybrid orbitals. This process one bond with the sp2 hybrid orbital of the adjacent C atom. In addition, the two leaves an electron in the unhybridized 2pz orbitals of the C atoms form another bond by overlapping side- unhybridized orbital, the 2pz orbital. ways [Figure 10.16(b)]. A distinction is made between the two types of covalent bonds in C2H4. The three Animation bonds formed by each C atom in Figure 10.16(a) are all sigma bonds (σ bonds), Sigma and Pi Bonds covalent bonds formed by orbitals overlapping end-to-end, with the electron density concentrated between the nuclei of the bonding atoms. The second type is called a pi bond (π bond), which is defined as a covalent bond formed by sideways overlapping orbitals with electron density concentrated above and below the plane of the nuclei of the bonding atoms. The two C atoms form a pi bond as shown in Figure 10.16(b).  10.5 Hybridization in Molecules Containing Double and Triple Bonds 441 It is this pi bond formation that gives ethylene its planar geometry. Figure 10.16(c) shows the orientation of the sigma and pi bonds. Figure 10.17 is yet another way of 90° looking at the planar C2H4 molecule and the formation of the pi bond. Although we normally represent the carbon-carbon double bond as C“C (as in a Lewis structure), it is important to keep in mind that the two bonds are different types: One is a sigma bond and the other is a pi bond. In fact, the bond enthalpies of the carbon-carbon pi and sigma bonds are about 270 kJ/mol and 350 kJ/mol, respectively. The acetylene molecule (C2H2) contains a carbon-carbon triple bond. Because the 120° molecule is linear, we can explain its geometry and bonding by assuming that each C atom is sp-hybridized by mixing the 2s with the 2px orbital (Figure 10.18). As Figure 10.19 shows, the two sp hybrid orbitals of each C atom form one sigma bond with a hydrogen 1s orbital and another sigma bond with the other C atom. In addition, Figure 10.15 Each carbon atom two pi bonds are formed by the sideways overlap of the unhybridized 2py and 2pz in the C2 H4 molecule has three sp2 hybrid orbitals (green) and one orbitals. Thus, the C‚C bond is made up of one sigma bond and two pi bonds. unhybridized 2pz orbital (gray), The following rule helps us predict hybridization in molecules containing mul- which is perpendicular to the tiple bonds: If the central atom forms a double bond, it is sp2-hybridized; if it forms plane of the hybrid orbitals. two double bonds or a triple bond, it is sp-hybridized. Note that this rule applies only to atoms of the second-period elements. Atoms of third-period elements and beyond that form multiple bonds present a more complicated picture and will not be dealt with here. H 1s H 1s 2pz 2pz π H H σ σ C C C C C σ C σ σ H H π H 1s H 1s (a) (b) (c) Figure 10.16 Bonding in ethylene, C2H4. (a) Top view of the sigma bonds between carbon atoms and between carbon and hydrogen atoms. All the atoms lie in the same plane, making C2 H4 a planar molecule. (b) Side view showing how the two 2pz orbitals on the two carbon atoms overlap, leading to the formation of a pi bond. The solid, dashed, and wedged lines show the directions of the sigma bonds. (c) The interactions in (a) and (b) lead to the formation of the sigma bonds and the pi bond in ethylene. Note that the pi bond lies above and below the plane of the molecule. Ground state 2s 2p π Promotion of electron 2s 2p H H sp- C C Hybridized H H state ⎧ ⎪ ⎨ ⎪ ⎩ 2py 2pz sp orbitals π Figure 10.18 The sp hybridization of a carbon atom. The 2s orbital is mixed with only (a) (b) one 2p orbital to form two sp hybrid orbitals. This process Figure 10.17 (a) Another view of the pi bond in the C2 H4 molecule. Note that all six atoms are in leaves an electron in each of the the same plane. It is the overlap of the 2pz orbitals that causes the molecule to assume a planar two unhybridized 2p orbitals, structure. (b) Electrostatic potential map of C2H4. namely, the 2py and 2pz orbitals. 442 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals z x y H σ 2pz 2pz 2py C H 1s H 1s 2py π C C C C C π C π C π σ H (a) (b) (c) Figure 10.19 Bonding in acetylene, C2 H2. (a) Top view showing the overlap of the sp orbitals between the C atoms and the overlap of the sp orbital with the 1s orbital between the C and H atoms. All the atoms lie along a straight line; therefore, acetylene is a linear molecule. (b) Side view showing the overlap of the two 2py orbitals and of the two 2pz orbitals of the two carbon atoms, which leads to the formation of two pi bonds. (c) Formation of the sigma and pi bonds as a result of the interactions in (a) and (b). (d) Electrostatic potential map of C2H2. (d) Example 10.5 Describe the bonding in the formaldehyde molecule whose Lewis structure is H G O Q CPO D H Assume that the O atom is sp2-hybridized. Strategy Follow the procedure shown in Example 10.3. CH2O Solution There are three pairs of electrons around the C atom; therefore, the electron pair arrangement is trigonal planar. (Recall that a double bond is treated as a single bond in the VSEPR model.) We conclude that C uses sp2 hybrid orbitals in bonding, because sp2 hybrid orbitals have a trigonal planar arrangement (see Table 10.4). We can imagine the hybridization processes for C and O as follows: C hg h h n h h h h 2s 2p sp2 orbitals 2pz O hg hg h h n h hg hg h 2s 2p sp2 orbitals 2pz (Continued)  10.6 Molecular Orbital Theory 443 Figure 10.20 Bonding in the H 1s formaldehyde molecule. A sigma π bond is formed by the overlap of σ the sp2 hybrid orbital of carbon C σ O and the sp2 hybrid orbital of oxygen; a pi bond is formed by the overlap of the 2pz orbitals of the σ π carbon and oxygen atoms. The H 1s two lone pairs on oxygen are placed in the other two sp2 orbitals of oxygen. Carbon has one electron in each of the three sp2 orbitals, which are used to form sigma bonds with the H atoms and the O atom. There is also an electron in the 2pz orbital, which forms a pi bond with oxygen. Oxygen has two electrons in two of its sp2 hybrid orbitals. These are the lone pairs on oxygen. Its third sp2 hybrid orbital with one electron is used to form a sigma bond with carbon. The 2pz orbital (with one electron) overlaps with the 2pz orbital of C to form a pi bond (Figure 10.20). Similar problems: 10.36, 10.37, 10.39. Practice Exercise Describe the bonding in the hydrogen cyanide molecule, HCN. Assume that N is sp-hybridized. Review of Concepts Which of the following pairs of atomic orbitals on adjacent nuclei can overlap to form a sigma bond? a pi bond? Which cannot overlap (no bond)? Consider the x axis to be the internuclear axis. (a) 1s and 2s, (b) 1s and 2px, (c) 2py and 2py, (d) 3py and 3pz, (e) 2px and 3px. 10.6 Molecular Orbital Theory Valence bond theory is one of the two quantum mechanical approaches that explain bonding in molecules. It accounts, at least qualitatively, for the stability of the covalent bond in terms of overlapping atomic orbitals. Using the concept of hybrid- ization, valence bond theory can explain molecular geometries predicted by the VSEPR model. However, the assumption that electrons in a molecule occupy atomic orbitals of the individual atoms can only be an approximation, because each bond- ing electron in a molecule must be in an orbital that is characteristic of the molecule as a whole. In some cases, valence bond theory cannot satisfactorily account for observed properties of molecules. Consider the oxygen molecule, whose Lewis structure is O Q O OPO Q According to this description, all the electrons in O2 are paired and oxygen should therefore be diamagnetic. But experiments have shown that the oxygen molecule has two unpaired electrons (Figure 10.21). This finding suggests a fundamental deficiency in valence bond theory, one that justifies searching for an alternative bonding approach that accounts for the properties of O2 and other molecules that do not match the predictions of valence bond theory. Figure 10.21 Liquid oxygen Magnetic and other properties of molecules are sometimes better explained by caught between the poles of a magnet, because the O2 another quantum mechanical approach called molecular orbital (MO) theory. Molec- molecules are paramagnetic, ular orbital theory describes covalent bonds in terms of molecular orbitals, which having two parallel spins. 444 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals result from interaction of the atomic orbitals of the bonding atoms and are associated with the entire molecule. The difference between a molecular orbital and an atomic orbital is that an atomic orbital is associated with only one atom. Review of Concepts One way to account for the fact that an O2 molecule contains two unpaired electrons is to draw the following Lewis structure: O Q O TOOOT Q Suggest two reasons why this structure is unsatisfactory. Bonding and Antibonding Molecular Orbitals According to MO theory, the overlap of the 1s orbitals of two hydrogen atoms leads to the formation of two molecular orbitals: one bonding molecular orbital and one antibonding molecular orbital. A bonding molecular orbital has lower energy and greater stability than the atomic orbitals from which it was formed. An antibonding molecular orbital has higher energy and lower stability than the atomic orbitals from which it was formed. As the names “bonding” and “antibonding” suggest, placing electrons in a bonding molecular orbital yields a stable covalent bond, whereas plac- ing electrons in an antibonding molecular orbital results in an unstable bond. In the bonding molecular orbital, the electron density is greatest between the nuclei of the bonding atoms. In the antibonding molecular orbital, on the other hand, the electron density decreases to zero between the nuclei. We can understand this distinction if we recall that electrons in orbitals have wave characteristics. A property unique to waves enables waves of the same type to interact in such a way that the resultant wave has either an enhanced amplitude or a diminished amplitude. In the Wave 1 Wave 1 former case, we call the interaction constructive interference; in the latter case, it is destructive interference (Figure 10.22). The formation of bonding molecular orbitals corresponds to constructive interfer- ence (the increase in amplitude is analogous to the buildup of electron density between the two nuclei). The formation of antibonding molecular orbitals corresponds to destructive interference (the decrease in amplitude is analogous to the decrease in Wave 2 Wave 2 electron density between the two nuclei). The constructive and destructive interactions between the two 1s orbitals in the H2 molecule, then, lead to the formation of a sigma bonding molecular orbital σ1s and a sigma antibonding molecular orbital σw1s: a sigma bonding a sigma antibonding molecular orbital molecular orbital 88 88 n n ␴1s ␴夹1s n n 88 88 formed from formed from Sum of 1 and 2 Sum of 1 and 2 1s orbitals 1s orbitals where the star denotes an antibonding molecular orbital. (a) (b) In a sigma molecular orbital (bonding or antibonding) the electron density is Figure 10.22 Constructive concentrated symmetrically around a line between the two nuclei of the bonding atoms. interference (a) and destructive Two electrons in a sigma molecular orbital form a sigma bond (see Section 10.5). interference (b) of two waves of the same wavelength and Remember that a single covalent bond (such as H¬H or F¬F) is almost always a amplitude. sigma bond.  10.6 Molecular Orbital Theory 445 Molecule Destructive Antibonding sigma (σ 1s ★) interference molecular orbital σ 1s ★ Atom Atom Energy 1s 1s Constructive Bonding sigma (σ 1s ) interference molecular orbital σ 1s (a) (b) Figure 10.23 (a) Energy levels of bonding and antibonding molecular orbitals in the H2 molecule. Note that the two electrons in the σ1s orbital must have opposite spins in accord with the Pauli exclusion principle. Keep in mind that the higher the energy of the molecular orbital, the less stable the electrons in that molecular orbital. (b) Constructive and destructive interferences between the two hydrogen 1s orbitals lead to the formation of a bonding and an antibonding molecular orbital. In the bonding molecular orbital, there is a buildup between the nuclei of electron density, which acts as a negatively charged “glue” to hold the positively charged nuclei together. In the antibonding molecular orbital, there is a nodal plane between the nuclei, where the electron density is zero. Figure 10.23 shows the molecular orbital energy level diagram—that is, the The two electrons in the sigma molecular relative energy levels of the orbitals produced in the formation of the H2 molecule— orbital are paired. The Pauli exclusion principle applies to molecules as well as and the constructive and destructive interferences between the two 1s orbitals. Notice to atoms. that in the antibonding molecular orbital there is a nodal plane between the nuclei that signifies zero electron density. The nuclei are repelled by each other’s positive charges, rather than held together. Electrons in the antibonding molecular orbital have higher energy (and less stability) than they would have in the isolated atoms. On the other hand, electrons in the bonding molecular orbital have less energy (and hence greater stability) than they would have in the isolated atoms. Although we have used the hydrogen molecule to illustrate molecular orbital formation, the concept is equally applicable to other molecules. In the H2 molecule, we consider only the interaction between 1s orbitals; with more complex molecules we need to consider additional atomic orbitals as well. Nevertheless, for all s orbit- als, the process is the same as for 1s orbitals. Thus, the interaction between two 2s or 3s orbitals can be understood in terms of the molecular orbital energy level diagram and the formation of bonding and antibonding molecular orbitals shown in Figure 10.23. For p orbitals, the process is more complex because they can interact with each other in two different ways. For example, two 2p orbitals can approach each other end-to-end to produce a sigma bonding and a sigma antibonding molecular orbital, as shown in Figure 10.24(a). Alternatively, the two p orbitals can overlap sideways to generate a bonding and an antibonding pi molecular orbital [Figure 10.24(b)]. a pi bonding a pi antibonding molecular orbital molecular orbital 88 88 n n ␲2p ␲夹2p n n 88 88 formed from formed from 2p orbitals 2p orbitals In a pi molecular orbital (bonding or antibonding), the electron density is concen- trated above and below a line joining the two nuclei of the bonding atoms. Two 446 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Molecule Destructive interference Antibonding sigma (σ 2p ★) σ 2p ★ molecular orbital Atom Atom + 2p 2p Energy Constructive interference Bonding sigma (σ 2p ) σ 2p molecular orbital + (a) Molecule Destructive interference Antibonding pi (π 2p ★) molecular orbital π 2p ★ Atom Atom + 2p 2p Energy π 2p + Constructive interference Bonding pi (π 2p) molecular orbital (b) Figure 10.24 Two possible interactions between two equivalent p orbitals and the corresponding molecular orbitals. (a) When the p orbitals overlap end-to-end, a sigma bonding and a sigma antibonding molecular orbital form. (b) When the p orbitals overlap side- to-side, a pi bonding and a pi antibonding molecular orbital form. Normally, a sigma bonding molecular orbital is more stable than a pi bonding molecular orbital, because side-to-side interaction leads to a smaller overlap of the p orbitals than does end-to-end interaction. We assume that the 2px orbitals take part in the sigma molecular orbital formation. The 2py and 2pz orbitals can interact to form only π molecular orbitals. The behavior shown in (b) represents the interaction between the 2py orbitals or the 2pz orbitals. In both cases, the dash line represents a nodal plane between the nuclei, where the electron density is zero. electrons in a pi molecular orbital form a pi bond (see Section 10.5). A double bond is almost always composed of a sigma bond and a pi bond; a triple bond is always a sigma bond plus two pi bonds. 10.7 Molecular Orbital Configurations To understand properties of molecules, we must know how electrons are distributed among molecular orbitals. The procedure for determining the electron configuration of a molecule is analogous to the one we use to determine the electron configurations of atoms (see Section 7.8). Rules Governing Molecular Electron Configuration and Stability In order to write the electron configuration of a molecule, we must first arrange the molecular orbitals in order of increasing energy. Then we can use the following guide- lines to fill the molecular orbitals with electrons. The rules also help us understand the stabilities of the molecular orbitals.  10.7 Molecular Orbital Configurations 447 1. The number of molecular orbitals formed is always equal to the number of atomic orbitals combined. 2. The more stable the bonding molecular orbital, the less stable the corresponding antibonding molecular orbital. 3. The filling of molecular orbitals proceeds from low to high energies. In a stable molecule, the number of electrons in bonding molecular orbitals is always greater than that in antibonding molecular orbitals because we place electrons first in the lower-energy bonding molecular orbitals. 4. Like an atomic orbital, each molecular orbital can accommodate up to two elec- trons with opposite spins in accordance with the Pauli exclusion principle. 5. When electrons are added to molecular orbitals of the same energy, the most stable arrangement is predicted by Hund’s rule; that is, electrons enter these molecular orbitals with parallel spins. 6. The number of electrons in the molecular orbitals is equal to the sum of all the electrons on the bonding atoms. Hydrogen and Helium Molecules Later in this section we will study molecules formed by atoms of the second-period elements. Before we do, it will be instructive to predict the relative stabilities of the simple species H12 , H2, He12 , and He2, using the energy-level diagrams shown in Figure 10.25. The σ1s and σw1s orbitals can accommodate a maximum of four elec- trons. The total number of electrons increases from one for H12 to four for He2. The Pauli exclusion principle stipulates that each molecular orbital can accommodate a maximum of two electrons with opposite spins. We are concerned only with the ground-state electron configurations in these cases. To evaluate the stabilities of these species we determine their bond order, defined as 1 number of electrons number of electrons bond order 5 a 2 b (10.2) 2 in bonding MOs in antibonding MOs The bond order indicates the approximate strength of a bond. For example, if there The quantitative measure of the strength of a bond is bond enthalpy (see Section 9.10). are two electrons in the bonding molecular orbital and none in the antibonding molec- ular orbital, the bond order is one, which means that there is one covalent bond and that the molecule is stable. Note that the bond order can be a fraction, but a bond order of zero (or a negative value) means the bond has no stability and the molecule cannot exist. Bond order can be used only qualitatively for purposes of comparison. For example, a bonding sigma molecular orbital with two electrons and a bonding pi molecular orbital with two electrons would each have a bond order of one. Yet, these two bonds must differ in bond strength (and bond length) because of the differences in the extent of atomic orbital overlap. σ 1s ★ σ 1s ★ σ 1s ★ σ 1s ★ Figure 10.25 Energy levels of the bonding and antibonding molecular orbitals in H 12 , H2, He12 , and He2. In all these species, the Energy molecular orbitals are formed by the interaction of two 1s orbitals. σ 1s σ 1s σ 1s σ 1s H +2 H2 He +2 He 2 448 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals We are ready now to make predictions about the stability of H1 1 2 , H2, He 2 , and 1 He2 (see Figure 10.25). The H 2 molecular ion has only one electron in the σ1s orbital. Because a covalent bond consists of two electrons in a bonding molecular 1 orbital, H 12 has only half of one bond, or a bond order of 2 . Thus, we predict that the H2 molecule may be a stable species. The electron configuration of H1 1 2 is writ- The superscript in (σ1s)1 indicates that ten as (σ1s)1. there is one electron in the sigma bonding molecular orbital. The H2 molecule has two electrons, both of which are in the σ1s orbital. Accord- ing to our scheme, two electrons equal one full bond; therefore, the H2 molecule has a bond order of one, or one full covalent bond. The electron configuration of H2 is (σ1s)2. As for the He12 molecular ion, we place the first two electrons in the σ1s orbital and the third electron in the σw1s orbital. Because the antibonding molecular orbital is destabilizing, we expect He12 to be less stable than H2. Roughly speaking, the instabil- ity resulting from the electron in the σw1s orbital is balanced by one of the σ1s electrons. The bond order is 12 (2 2 1) 5 12 and the overall stability of He1 2 is similar to that of the H12 molecule. The electron configuration of He12 is (σ1s)2(σw1s)1. In He2 there would be two electrons in the σ1s orbital and two electrons in the σw1s orbital, so the molecule would have a bond order of zero and no net stability. The electron configuration of He2 would be (σ1s)2(σw1s)2. To summarize, we can arrange our examples in order of decreasing stability: H2 . H 1 1 2 , He 2 . He2 We know that the hydrogen molecule is a stable species. Our simple molecular orbital method predicts that H1 1 2 and He 2 also possess some stability, because both 1 have bond orders of 2 . Indeed, their existence has been confirmed by experiment. It turns out that H1 1 2 is somewhat more stable than He 2 , because there is only one electron in the hydrogen molecular ion and therefore it has no electron-electron repulsion. Furthermore, H1 1 2 also has less nuclear repulsion than He 2 . Our prediction about He2 is that it would have no stability, but in 1993 He2 gas was found to exist. The “molecule” is extremely unstable and has only a transient existence under spe- cially created conditions. Review of Concepts Estimate the bond enthalpy (kJ/mol) of the H1 2 ion. Homonuclear Diatomic Molecules of Second-Period Elements We are now ready to study the ground-state electron configuration of molecules containing second-period elements. We will consider only the simplest case, that of homonuclear diatomic molecules, or diatomic molecules containing atoms of the same elements. Figure 10.26 shows the molecular orbital energy level diagram for the first mem- ber of the second period, Li2. These molecular orbitals are formed by the overlap of 1s and 2s orbitals. We will use this diagram to build up all the diatomic molecules, as we will see shortly. The situation is more complex when the bonding also involves p orbitals. Two p orbitals can form either a sigma bond or a pi bond. Because there are three p orbitals for each atom of a second-period element, we know that one sigma and two pi molec- ular orbitals will result from the constructive interaction. The sigma molecular orbital is formed by the overlap of the 2px orbitals along the internuclear axis, that is, the  10.7 Molecular Orbital Configurations 449 Molecule Figure 10.26 Molecular orbital σ 2s ★ energy level diagram for the Li2 molecule. The six electrons in Li2 (Li’s electron configuration 1s22s1) are in the σ1s, σw1s, and σ2s orbitals. Because there are two electrons each in σ1s and σw1s ( just as in He2 ), Atom Atom there is no net bonding or antibonding effect. Therefore, the single covalent bond in Li2 is formed by the two electrons in the 2s 2s bonding molecular orbital σ2s. Note that although the antibonding orbital (σw1s ) has Energy higher energy and is thus less stable than the bonding orbital (σ1s ), this antibonding orbital has σ 2s less energy and greater stability than the σ2s bonding orbital. σ 1s ★ 1s 1s σ 1s x-axis. The 2py and 2pz orbitals are perpendicular to the x-axis, and they will overlap sideways to give two pi molecular orbitals. The molecular orbitals are called σ2px , π2py , and π2pz orbitals, where the subscripts indicate which atomic orbitals take part in forming the molecular orbitals. As shown in Figure 10.24, overlap of the two p orbit- als is normally greater in a σ molecular orbital than in a π molecular orbital, so we would expect the former to be lower in energy. However, the energies of molecular orbitals actually increase as follows: σ1s , σw w w w w 1s , σ2s , σ2s , π2py 5 π2pz , σ2px , π2py 5 π2pz , σ2px The inversion of the σ2px orbital and the π2py and π2pz orbitals is due to the interaction between the 2s orbital on one atom with the 2p orbital on the other. In MO terminol- ogy, we say there is mixing between these orbitals. The condition for mixing is that the 2s and 2p orbitals must be close in energy. This condition is met for the lighter molecules B2, C2, and N2 with the result that the σ2px orbital is raised in energy rela- tive to the π2py and π2pz orbitals as already shown. The mixing is less pronounced for O2 and F2 so the σ2px orbital lies lower in energy than the π2py and π2pz orbitals in these molecules. With these concepts and Figure 10.27, which shows the order of increasing ener- gies for 2p molecular orbitals, we can write the electron configurations and predict the magnetic properties and bond orders of second-period homonuclear diatomic mol- ecules. We will consider a few examples. The Lithium Molecule (Li2) The electron configuration of Li is 1s22s1, so Li2 has a total of six electrons. According to Figure 10.26, these electrons are placed (two each) in the σ1s, σw 1s, and σ2s molecular orbitals. The electrons of σ1s and σw1s make no net contribution to the bonding in Li2. 450 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Figure 10.27 General molecular Molecule orbital energy level diagram for σ ★2px the second-period homonuclear diatomic molecules Li2, Be2, B2, C2, and N2. For simplicity, the σ1s and σ2s orbitals have been π ★2py π ★2pz omitted. Note that in these molecules, the σ2px orbital is higher in energy than either the π2py or the π2pz orbitals. This means that electrons in the σ2px Atom Atom orbitals are less stable than those Energy in π2py and π2pz. This abberation stems from the different interactions between the electrons 2px 2py 2pz 2px 2py 2pz in the σ2px orbital, on one hand, σ 2px and π2py and π2pz orbitals, on the other hand, with the electrons in the lower-energy σs orbitals. For O2 and F2, the σ2px orbital is lower π 2py π 2pz in energy than π2py and π2pz. Thus, the electron configuration of the molecular orbitals in Li2 is (σ1s)2(σw1s)2(σ2s)2. Since there are two more electrons in the bonding molecular orbitals than in antibonding orbitals, the bond order is 1 [see Equation (10.2)]. We conclude that the Li2 molecule is stable, and because it has no unpaired electron spins, it should be diamagnetic. Indeed, diamagnetic Li2 molecules are known to exist in the vapor phase. The Carbon Molecule (C2) The carbon atom has the electron configuration 1s22s22p2; thus, there are 12 electrons in the C2 molecule. Referring to Figures 10.26 and 10.27, we place the last four electrons in the π2py and π2pz orbitals. Therefore, C2 has the electron configuration (σ1s ) 2 (σw 2 2 w 2 2 1s ) (σ2s ) (σ2s ) (π2py ) (π2pz ) 2 Its bond order is 2, and the molecule has no unpaired electrons. Again, diamagnetic C2 molecules have been detected in the vapor state. Note that the double bonds in C2 are both pi bonds because of the four electrons in the two pi molecular orbitals. In most other molecules, a double bond is made up of a sigma bond and a pi bond. The Oxygen Molecule (O2) The ground-state electron configuration of O is 1s22s22p4; thus, there are 16 electrons in O2. Using the order of increasing energies of the molecular orbitals discussed above, we write the ground-state electron configuration of O2 as (σ1s ) 2 (σw 2 2 w 2 2 2 2 w 1 w 1 1s ) (σ2s ) (σ2s ) (σ2px ) (π2py ) (π2pz ) (π2py ) (π2pz ) According to Hund’s rule, the last two electrons enter the πw w 2py and π2pz orbitals with parallel spins. Ignoring the σ1s and σ2s orbitals (because their net effects on bonding are zero), we calculate the bond order of O2 using Equation (10.2): bond order 5 12 (6 2 2) 5 2 Therefore, the O2 molecule has a bond order of 2 and oxygen is paramagnetic, a prediction that corresponds to experimental observations.  10.7 Molecular Orbital Configurations 451 Table 10.5 Properties of Homonuclear Diatomic Molecules of the Second-Period Elements* Li2 B2 C2 N2 O2 F2 ␴夹 2px 夹 ␴2p x ␲夹 夹 2py, ␲2pz h h hg hg ␲夹 夹 2py, ␲2pz ␴2px hg hg hg hg hg ␲2py, ␲2pz ␲2py, ␲2pz h h hg hg hg hg hg hg ␴2px ␴夹 2s hg hg hg hg hg ␴夹 2s ␴2s hg hg hg hg hg hg ␴2s Bond order 1 1 2 3 2 1 Bond length (pm) 267 159 131 110 121 142 Bond enthalpy 104.6 288.7 627.6 941.4 498.7 156.9 (kJ/mol) Magnetic properties Diamagnetic Paramagnetic Diamagnetic Diamagnetic Paramagnetic Diamagnetic *For simplicity the σ1s and σw 1s orbitals are omitted. These two orbitals hold a total of four electrons. Remember that for O2 and F2, σ2px is lower in energy than π2py and π2pz. Table 10.5 summarizes the general properties of the stable diatomic molecules of the second period. Example 10.6 shows how MO theory can help predict molecular properties of ions. Example 10.6 The N12 ion can be prepared by bombarding the N2 molecule with fast-moving electrons. Predict the following properties of N1 2 : (a) electron configuration, (b) bond order, (c) magnetic properties, and (d) bond length relative to the bond length of N2 (is it longer or shorter?). Strategy From Table 10.5 we can deduce the properties of ions generated from the homonuclear molecules. How does the stability of a molecule depend on the number of electrons in bonding and antibonding molecular orbitals? From what molecular orbital is an electron removed to form the N12 ion from N2? What properties determine whether a species is diamagnetic or paramagnetic? Solution From Table 10.5 we can deduce the properties of ions generated from the homonuclear diatomic molecules. (a) Because N1 2 has one fewer electron than N2, its electron configuration is (σ1s ) 2 (σw 2 2 w 2 2 2 w 1 1s ) (σ2s ) (σ2s ) (π2py ) (π2pz ) (σ2px ) (b) The bond order of N12 is found by using Equation (10.2): bond order 5 12 (9 2 4) 5 2.5 (c) N1 2 has one unpaired electron, so it is paramagnetic. (Continued) 452 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals (d) Because the electrons in the bonding molecular orbitals are responsible for holding the atoms together, N12 should have a weaker and, therefore, longer bond than N2. (In fact, the bond length of N1 2 is 112 pm, compared with 110 pm for N2.) Check Because an electron is removed from a bonding molecular orbital, we expect the bond order to decrease. The N1 2 ion has an odd number of electrons (13), so it Similar problems: 10.57, 10.58. should be paramagnetic. Practice Exercise Which of the following species has a longer bond length: F2 or F22 ? 10.8 Delocalized Molecular Orbitals So far we have discussed chemical bonding only in terms of electron pairs. How- ever, the properties of a molecule cannot always be explained accurately by a single structure. A case in point is the O3 molecule, discussed in Section 9.8. There we overcame the dilemma by introducing the concept of resonance. In this section, we will tackle the problem in another way—by applying the molecular orbital approach. As in Section 9.8, we will use the benzene molecule and the carbonate ion as examples. Note that in discussing the bonding of polyatomic molecules or ions, it is convenient to determine first the hybridization state of the atoms present (a valence bond approach), followed by the formation of appropri- ate molecular orbitals. H The Benzene Molecule C Benzene (C6H6) is a planar hexagonal molecule with carbon atoms situated at the six H H C C corners. All carbon-carbon bonds are equal in length and strength, as are all carbon- hydrogen bonds, and the CCC and HCC angles are all 120°. Therefore, each carbon atom is sp2-hybridized; it forms three sigma bonds with two adjacent carbon atoms C C and a hydrogen atom (Figure 10.28). This arrangement leaves an unhybridized 2pz H H orbital on each carbon atom, perpendicular to the plane of the benzene molecule, or C benzene ring, as it is often called. So far the description resembles the configuration of ethylene (C2H4), discussed in Section 10.5, except that in this case there are six H unhybridized 2pz orbitals in a cyclic arrangement. Because of their similar shape and orientation, each 2pz orbital overlaps two oth- Figure 10.28 The sigma bond framework in the benzene ers, one on each adjacent carbon atom. According to the rules listed on p. 447, the molecule. Each carbon atom is interaction of six 2pz orbitals leads to the formation of six pi molecular orbitals, of sp2-hybridized and forms sigma which three are bonding and three antibonding. A benzene molecule in the ground bonds with two adjacent carbon atoms and another sigma bond state therefore has six electrons in the three pi bonding molecular orbitals, two elec- with a hydrogen atom. trons with paired spins in each orbital (Figure 10.29). Figure 10.29 (a) The six 2pz orbitals on the carbon atoms in benzene. (b) The delocalized Top view Side view molecular orbital formed by the overlap of the 2pz orbitals. The delocalized molecular orbital possesses pi symmetry and lies above and below the plane of the benzene ring. Actually, these 2pz orbitals can combine in six different ways to yield three bonding molecular orbitals and three antibonding molecular orbitals. The one shown here is the most stable. (a) (b)  10.8 Delocalized Molecular Orbitals 453 Unlike the pi bonding molecular orbitals in ethylene, those in benzene form delocalized molecular orbitals, which are not confined between two adjacent bonding atoms, but actually extend over three or more atoms. Therefore, electrons residing in any of these orbitals are free to move around the benzene ring. For this reason, the structure of benzene is sometimes represented as in which the circle indicates that the pi bonds between carbon atoms are not confined to individual pairs of atoms; rather, the pi electron densities are evenly distributed throughout the benzene molecule. The carbon and hydrogen atoms are not shown in the simplified diagram. Electrostatic potential map of benzene shows the electron We can now state that each carbon-to-carbon linkage in benzene contains a sigma density (red color) above and bond and a “partial” pi bond. The bond order between any two adjacent carbon atoms below the plane of the molecule. is therefore between 1 and 2. Thus, molecular orbital theory offers an alternative to For simplicity, only the framework the resonance approach, which is based on valence bond theory. (The resonance struc- of the molecule is shown. tures of benzene are shown on p. 390.) The Carbonate Ion Cyclic compounds like benzene are not the only ones with delocalized molecular orbitals. Let’s look at bonding in the carbonate ion (CO22 3 ). VSEPR predicts a trigo- nal planar geometry for the carbonate ion, like that for BF3. The planar structure of the carbonate ion can be explained by assuming that the carbon atom is sp2-hybridized. The C atom forms sigma bonds with three O atoms. Thus, the unhybridized 2pz orbital of the C atom can simultaneously overlap the 2pz orbitals of all three O atoms (Figure 10.30). The result is a delocalized molecular orbital that extends over all four nuclei in such a way that the electron densities (and hence the bond orders) in the carbon-to-oxygen bonds are all the same. Molecular orbital theory therefore provides an acceptable alternative explanation of the properties of the carbonate ion as com- pared with the resonance structures of the ion shown on p. 390. We should note that molecules with delocalized molecular orbitals are generally more stable than those containing molecular orbitals extending over only two atoms. For example, the benzene molecule, which contains delocalized molecular orbitals, is chemically less reactive (and hence more stable) than molecules containing “local- ized” C“C bonds, such as ethylene. Review of Concepts Describe the bonding in the nitrate ion (NO2 3 ) in terms of resonance structures and delocalized molecular orbitals. Figure 10.30 Bonding in the carbonate ion. The carbon atom forms three sigma bonds with the O three oxygen atoms. In addition, O the 2pz orbitals of the carbon and O C O C oxygen atoms overlap to form delocalized molecular orbitals, so O that there is also a partial pi bond O between the carbon atom and each of the three oxygen atoms. CHEMISTRY in Action Buckyball, Anyone? I n 1985 chemists at Rice University in Texas used a high- powered laser to vaporize graphite in an effort to create un- usual molecules believed to exist in interstellar space. Mass spectrometry revealed that one of the products was an unknown species with the formula C60. Because of its size and the fact 335 pm that it is pure carbon, this molecule has an exotic shape, which the researchers worked out using paper, scissors, and tape. Subsequent spectroscopic and X-ray measurements confirmed that C60 is shaped like a hollow sphere with a carbon atom at each of the 60 vertices. Geometrically, buckyball (short for “buckminsterfullerene”) is the most symmetrical molecule known. In spite of its unique features, however, its bonding scheme is straightforward. Each carbon is sp2-hybridized, and there are extensive delocalized molecular orbitals over the Graphite is made up of layers of six-membered rings of carbon. entire structure. The discovery of buckyball generated tremendous interest within the scientific community. Here was a new allotrope of to attach transition metals to buckyball. These derivatives show carbon with an intriguing geometry and unknown properties to promise as catalysts. Because of its unique shape, buckyball can investigate. Since 1985 chemists have created a whole class of be used as a lubricant. fullerenes, with 70, 76, and even larger numbers of carbon at- One fascinating discovery, made in 1991 by Japanese sci- oms. Moreover, buckyball has been found to be a natural com- entists, was the identification of structural relatives of buckyball. ponent of soot. These molecules are hundreds of nanometers long with a tubu- Buckyball and its heavier members represent a whole new lar shape and an internal cavity about 15 nm in diameter. concept in molecular architecture with far-reaching implications. Dubbed “buckytubes” or “nanotubes” (because of their size), For example, buckyball has been prepared with a helium atom these molecules have two distinctly different structures. One is trapped in its cage. Buckyball also reacts with potassium to give a single sheet of graphite that is capped at both ends with a K3C60, which acts as a superconductor at 18 K. It is also possible kind of truncated buckyball. The other is a scroll-like tube The geometry of a buckyball C60 (left) resembles a soccer ball (right). Computer-generated model of the binding of a buckyball derivative to the Scientists arrived at this structure by fitting together paper cutouts of site of HIV-protease that normally attaches to a protein needed for the enough hexagons and pentagons to accommodate 60 carbon atoms reproduction of HIV. The buckyball structure (purple color) fits tightly into at the points where they intersect. the active site, thus preventing the enzyme from carrying out its function. 454  The structure of a buckytube that consists of a single layer of carbon atoms. Note that the truncated buckyball “cap,” which has been separated from the rest of the buckytube in this view, has a different structure than the graphitelike cylindrical portion of the tube. Chemists have devised ways to open the cap in order to place other molecules inside the tube. having anywhere from 2 to 30 graphitelike layers. Nanotubes almost totally transparent yet the carbon atoms are packed so are many times stronger than steel wires of similar dimen- dense that not even helium, the smallest gaseous atom, can sions. Numerous potential applications have been proposed pass through it. It seems like many interesting and useful dis- for them, including conducting and high-strength materials, coveries will come from the study of this unusual substance in hydrogen storage media, molecular sensors, semiconductor the coming years. devices, and molecular probes. The study of these materials has created a new field called nanotechnology, so called be- cause scientists can manipulate materials on a molecular scale to create useful devices. In the first biological application of buckyball, chemists at the University of California at San Francisco and Santa Barbara made a discovery in 1993 that could help in designing drugs to treat AIDS. The human immunodeficiency virus (HIV) that causes AIDS reproduces by synthesizing a long protein chain, which is cut into smaller segments by an enzyme called HIV- protease. One way to stop AIDS, then, might be to inactivate the enzyme. When the chemists reacted a water-soluble derivative of buckyball with HIV-protease, they found that it binds to the portion of the enzyme that would ordinarily cleave the repro- ductive protein, thereby preventing the HIV from reproducing. Consequently the virus could no longer infect the human cells they had grown in the laboratory. The buckyball compound it- self is not a suitable drug for use against AIDS because of po- tential side effects and delivery difficulties, but it does provide a model for the development of such drugs. In a recent development, scientists used a piece of adhe- sive tape (like Scotch tape) to peel off a flake of carbon from a piece of graphite (as is found in pencils) with the thickness of just one atom. This new-found material, called graphene, is a two-dimensional crystal with unusual electrical and optical A micrograph of graphene showing honeycomb properties. It is an excellent heat conductor. Graphene is structure. 455 456 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Key Equations μ 5 Q 3 r (10.1) Expressing dipole moment in terms of charge (Q) and distance of separation (r) between charges. 1 number of electrons number of electrons bond order 5 a 2 b (10.2) 2 in bonding MOs in antibonding MOs Summary of Facts & Concepts 1. The VSEPR model for predicting molecular geometry is 7. In an sp2-hybridized atom (for example, carbon), the based on the assumption that valence-shell electron pairs one unhybridized p orbital can form a pi bond with an- repel one another and tend to stay as far apart as possible. other p orbital. A carbon-carbon double bond consists 2. According to the VSEPR model, molecular geometry of a sigma bond and a pi bond. In an sp-hybridized car- can be predicted from the number of bonding electron bon atom, the two unhybridized p orbitals can form two pairs and lone pairs. Lone pairs repel other pairs more pi bonds with two p orbitals on another atom (or atoms). forcefully than bonding pairs do and thus distort bond A carbon-carbon triple bond consists of one sigma bond angles from the ideal geometry. and two pi bonds. 3. Dipole moment is a measure of the charge separation in 8. Molecular orbital theory describes bonding in terms of the molecules containing atoms of different electronega- combination and rearrangement of atomic orbitals to form tivities. The dipole moment of a molecule is the resul- orbitals that are associated with the molecule as a whole. tant of whatever bond moments are present. Information 9. Bonding molecular orbitals increase electron density about molecular geometry can be obtained from dipole between the nuclei and are lower in energy than indi- moment measurements. vidual atomic orbitals. Antibonding molecular orbitals 4. There are two quantum mechanical explanations for co- have a region of zero electron density between the nu- valent bond formation: valence bond theory and molecu- clei, and an energy level higher than that of the indi- lar orbital theory. In valence bond theory, hybridized vidual atomic orbitals. atomic orbitals are formed by the combination and rear- 10. We write electron configurations for molecular orbitals rangement of orbitals from the same atom. The hybrid- as we do for atomic orbitals, filling in electrons in the ized orbitals are all of equal energy and electron density, order of increasing energy levels. The number of mo- and the number of hybridized orbitals is equal to the lecular orbitals always equals the number of atomic or- number of pure atomic orbitals that combine. bitals that were combined. The Pauli exclusion principle 5. Valence-shell expansion can be explained by assuming and Hund’s rule govern the filling of molecular orbitals. hybridization of s, p, and d orbitals. 11. Molecules are stable if the number of electrons in bond- 6. In sp hybridization, the two hybrid orbitals lie in a straight ing molecular orbitals is greater than that in antibonding line; in sp2 hybridization, the three hybrid orbitals are di- molecular orbitals. rected toward the corners of an equilateral triangle; in sp3 12. Delocalized molecular orbitals, in which electrons are hybridization, the four hybrid orbitals are directed toward free to move around a whole molecule or group of at- the corners of a tetrahedron; in sp3d hybridization, the five oms, are formed by electrons in p orbitals of adjacent hybrid orbitals are directed toward the corners of a trigo- atoms. Delocalized molecular orbitals are an alternative nal bipyramid; in sp3d2 hybridization, the six hybrid orbit- to resonance structures in explaining observed molecu- als are directed toward the corners of an octahedron. lar properties. Key Words Antibonding molecular Dipole moment (μ), p. 423 Pi bond (π bond), p. 440 Valence-shell electron-pair orbital, p. 444 Homonuclear diatomic Pi molecular orbital, p. 445 repulsion (VSEPR) Bond order, p. 447 molecule, p. 448 Polar molecule, p. 424 model, p. 413 Bonding molecular Hybrid orbital, p. 431 Sigma bond (σ bond), p. 440 orbital, p. 444 Hybridization, p. 432 Sigma molecular Delocalized molecular Molecular orbital, p. 443 orbital, p. 444 orbital, p. 453 Nonpolar molecule, p. 424 Valence shell, p. 413  Questions & Problems 457 Questions & Problems • Problems available in Connect Plus Dipole Moments Red numbered problems solved in Student Solutions Manual Review Questions Molecular Geometry 10.15 Define dipole moment. What are the units and sym- bol for dipole moment? Review Questions 10.16 What is the relationship between the dipole 10.1 How is the geometry of a molecule defined and why moment and the bond moment? How is it possible is the study of molecular geometry important? for a molecule to have bond moments and yet be 10.2 Sketch the shape of a linear triatomic molecule, a trigo- nonpolar? nal planar molecule containing four atoms, a tetrahedral 10.17 Explain why an atom cannot have a permanent di- molecule, a trigonal bipyramidal molecule, and an octa- pole moment. hedral molecule. Give the bond angles in each case. 10.18 The bonds in beryllium hydride (BeH2) molecules • 10.3 How many atoms are directly bonded to the central are polar, and yet the dipole moment of the molecule atom in a tetrahedral molecule, a trigonal bipyrami- is zero. Explain. dal molecule, and an octahedral molecule? 10.4 Discuss the basic features of the VSEPR model. Explain why the magnitude of repulsion decreases Problems in the following order: lone pair-lone pair . lone • 10.19 Referring to Table 10.3, arrange the following mol- pair-bonding pair . bonding pair-bonding pair. ecules in order of increasing dipole moment: H2O, 10.5 In the trigonal bipyramidal arrangement, why does a H2S, H2Te, H2Se. lone pair occupy an equatorial position rather than 10.20 The dipole moments of the hydrogen halides decrease an axial position? from HF to HI (see Table 10.3). Explain this trend. 10.6 The geometry of CH4 could be square planar, with • 10.21 List the following molecules in order of increasing the four H atoms at the corners of a square and the C dipole moment: H2O, CBr4, H2S, HF, NH3, CO2. atom at the center of the square. Sketch this geome- 10.22 Does the molecule OCS have a higher or lower try and compare its stability with that of a tetrahe- dipole moment than CS2? dral CH4 molecule. • 10.23 Which of the following molecules has a higher dipole moment? Problems • 10.7 Predict the geometries of the following species using Br G D H Br G D Br the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, CPC CPC (d) TeCl4. D G D G H Br H H • 10.8 Predict the geometries of the following species: (a) (b) (a) AlCl3, (b) ZnCl2, (c) ZnCl422. • 10.9 Predict the geometry of the following molecules • 10.24 Arrange the following compounds in order of in- and ion using the VSEPR model: (a) CBr4, (b) BCl3, creasing dipole moment: (c) NF3, (d) H2Se, (e) NO2 2. • 10.10 Predict the geometry of the following molecules Cl Cl Cl and ion using the VSEPR model: (a) CH3I, (b) ClF3, A A A (c) H2S, (d) SO3, (e) SO422. ClH ECl ECl ClH ECl • 10.11 Predict the geometry of the following molecules using the VSEPR method: (a) HgBr2, (b) N2O (arrangement of atoms is NNO), (c) SCN2 (arrangement of atoms A A A is SCN). Cl Cl Cl • 10.12 Predict the geometries of the following ions: (a) NH14 , (a) (b) (c) (d) (b) NH22, (c) CO322, (d) ICl22, (e) ICl42, (f ) AlH42, (g) SnCl2 1 22 5 , (h) H3O , (i) BeF4 . Valence Bond Theory • 10.13 Describe the geometry around each of the three cen- Review Questions tral atoms in the CH3COOH molecule. • 10.14 Which of the following species are tetrahedral? 10.25 What is valence bond theory? How does it differ SiCl4, SeF4, XeF4, CI4, CdCl22 4 from the Lewis concept of chemical bonding?  458 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals 10.26 Use valence bond theory to explain the bonding in diagrams to show the formation of sigma bonds and Cl2 and HCl. Show how the atomic orbitals overlap pi bonds in allene. when a bond is formed. • 10.40 Describe the hybridization of phosphorus in PF5. 10.27 Draw a potential energy curve for the bond forma- • 10.41 How many sigma bonds and pi bonds are there in tion in F2. each of the following molecules? H H Cl H A G D A Hybridization ClOCOCl CPC H 3COCPCOCqCOH Review Questions A D G A H H H H 10.28 (a) What is the hybridization of atomic orbitals? Why (a) (b) (c) is it impossible for an isolated atom to exist in the hybridized state? (b) How does a hybrid orbital differ • 10.42 How many pi bonds and sigma bonds are there in the from a pure atomic orbital? Can two 2p orbitals of an tetracyanoethylene molecule? atom hybridize to give two hybridized orbitals? NqC CqN • 10.29 What is the angle between the following two hybrid G D orbitals on the same atom? (a) sp and sp hybrid CPC D G orbitals, (b) sp2 and sp2 hybrid orbitals, (c) sp3 and sp3 NqC CqN hybrid orbitals 10.30 How would you distinguish between a sigma bond • 10.43 Give the formula of a cation comprised of iodine and fluorine in which the iodine atom is sp3d- and a pi bond? hybridized. 10.44 Give the formula of an anion comprised of iodine and fluorine in which the iodine atom is sp3d2- Problems hybridized. • 10.31 Describe the bonding scheme of the AsH3 molecule in terms of hybridization. • 10.32 What is the hybridization state of Si in SiH4 and in Molecular Orbital Theory H3Si¬SiH3 ? Review Questions • 10.33 Describe the change in hybridization (if any) of the 10.45 What is molecular orbital theory? How does it differ Al atom in the following reaction: from valence bond theory? AlCl3 1 Cl2 ¡ AlCl2 4 10.46 Sketch the shapes of the following molecular orbit- als: σ1s, σw w 1s, π2p, and π2p . How do their energies • 10.34 Consider the reaction compare? BF3 1 NH3 ¡ F3B¬NH3 10.47 Compare the Lewis theory, valence bond theory, and Describe the changes in hybridization (if any) of the molecular orbital theory of chemical bonding. B and N atoms as a result of this reaction. 10.48 Explain the significance of bond order. Can bond • 10.35 What hybrid orbitals are used by nitrogen atoms order be used for quantitative comparisons of the in the following species? (a) NH3, (b) H2N¬NH2, strengths of chemical bonds? (c) NO32 • 10.36 What are the hybrid orbitals of the carbon atoms in Problems the following molecules? (a) H3C¬CH3 10.49 Explain in molecular orbital terms the changes in (b) H3C¬CH“CH2 H¬H internuclear distance that occur as the molec- (c) CH3 ¬C‚C¬CH2OH ular H2 is ionized first to H1 21 2 and then to H2 . (d) CH3CH“O 10.50 The formation of H2 from two H atoms is an ener- getically favorable process. Yet statistically there is (e) CH3COOH less than a 100 percent chance that any two H atoms • 10.37 Specify which hybrid orbitals are used by carbon will undergo the reaction. Apart from energy consid- atoms in the following species: (a) CO, (b) CO2, erations, how would you account for this observation (c) CN2. based on the electron spins in the two H atoms? • 10.38 What is the hybridization state of the central N atom • 10.51 Draw a molecular orbital energy level diagram for in the azide ion, N23 ? (Arrangement of atoms: NNN.) each of the following species: He2, HHe, He1 2. • 10.39 The allene molecule H2C“C“CH2 is linear (the Compare their relative stabilities in terms of bond three C atoms lie on a straight line). What are orders. (Treat HHe as a diatomic molecule with the hybridization states of the carbon atoms? Draw three electrons.)  Questions & Problems 459 • 10.52 Arrange the following species in order of increasing of orbitals) for forming a delocalized molecular stability: Li2, Li1 2 2 , Li 2 . Justify your choice with a orbital? molecular orbital energy level diagram. 10.64 In Chapter 9 we saw that the resonance concept is 10.53 Use molecular orbital theory to explain why the Be2 useful for dealing with species such as the benzene molecule does not exist. molecule and the carbonate ion. How does molecu- • 10.54 Which of these species has a longer bond, B2 or B1 2? lar orbital theory deal with these species? Explain in terms of molecular orbital theory. • 10.55 Acetylene (C2H2) has a tendency to lose two protons Problems (H1) and form the carbide ion (C22 2 ), which is pres- 10.65 Both ethylene (C2H4) and benzene (C6H6) contain ent in a number of ionic compounds, such as CaC2 the C“C bond. The reactivity of ethylene is greater and MgC2. Describe the bonding scheme in the C22 2 than that of benzene. For example, ethylene readily ion in terms of molecular orbital theory. Compare reacts with molecular bromine, whereas benzene is the bond order in C22 2 with that in C2. normally quite inert toward molecular bromine and 10.56 Compare the Lewis and molecular orbital treatments many other compounds. Explain this difference in of the oxygen molecule. reactivity. • 10.57 Explain why the bond order of N2 is greater than 10.66 Explain why the symbol on the left is a better that of N12 , but the bond order of O2 is less than that representation of benzene molecules than that on of O12 . the right. • 10.58 Compare the relative stability of the following spe- cies and indicate their magnetic properties (that is, diamagnetic or paramagnetic): O2, O12 , O22 (super- oxide ion), O222 (peroxide ion). 10.59 Use molecular orbital theory to compare the relative 10.67 Determine which of these molecules has a more de- stabilities of F2 and F1 2. localized orbital and justify your choice. • 10.60 A single bond is almost always a sigma bond, and a double bond is almost always made up of a sigma bond and a pi bond. There are very few exceptions to this rule. Show that the B2 and C2 molecules are ex- amples of the exceptions. (Hint: Both molecules contain two benzene rings. In 10.61 In 2009 the ion N232 was isolated. Use a molecu- naphthalene, the two rings are fused together. In lar orbital diagram to compare its properties biphenyl, the two rings are joined by a single bond, (bond order and magnetism) with the isoelec- around which the two rings can rotate.) tronic ion O2 2. 10.62 The following potential energy curve represents the • 10.68 Nitryl fluoride (FNO2) is very reactive chemically. The fluorine and oxygen atoms are bonded to the formation of F2 from two F atoms. Describe the state nitrogen atom. (a) Write a Lewis structure for FNO2. of bonding at the marked regions. (b) Indicate the hybridization of the nitrogen atom. (c) Describe the bonding in terms of molecular or- bital theory. Where would you expect delocalized + 5 molecular orbitals to form? 10.69 Describe the bonding in the nitrate ion NO23 in terms Potential energy of delocalized molecular orbitals. 1 10.70 What is the state of hybridization of the central 0 r O atom in O3? Describe the bonding in O3 in terms 4 2 of delocalized molecular orbitals. Additional Problems – 3 • 10.71 Which of the following species is not likely to have a tetrahedral shape? (a) SiBr4, (b) NF1 4 , (c) SF4, (d) BeCl22 2 2 4 , (e) BF 4 , (f) AlCl 4 Delocalized Molecular Orbitals Review Questions • 10.72 Draw the Lewis structure of mercury(II) bromide. Is this molecule linear or bent? How would you estab- 10.63 How does a delocalized molecular orbital differ lish its geometry? from a molecular orbital such as that found in H2 10.73 Sketch the bond moments and resultant dipole mo- or C2H4? What do you think are the minimum con- ments for the following molecules: H2O, PCl3, XeF4, ditions (for example, number of atoms and types PCl5, SF6.  460 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals 10.74 Although both carbon and silicon are in Group 4A, 10.81 Briefly compare the VSEPR and hybridization ap- very few Si“Si bonds are known. Account for the proaches to the study of molecular geometry. instability of silicon-to-silicon double bonds in gen- • 10.82 Describe the hybridization state of arsenic in arsenic eral. (Hint: Compare the atomic radii of C and Si in pentafluoride (AsF5). Figure 8.5. What effect would the larger size have on pi bond formation?) • 10.83 Draw Lewis structures and give the other informa- tion requested for the following: (a) SO3. Polar or 10.75 Acetaminophen is the active ingredient in Tylenol. nonpolar molecule? (b) PF3. Polar or nonpolar mol- (a) Write the molecular formula of the compound. ecule? (c) F3SiH. Show the direction of the resultant (b) What is the hybridization state of each C, N, and dipole moment. (d) SiH32. Planar or pyramidal O atom? (c) Describe the geometry about each C, N, shape? (e) Br2CH2. Polar or nonpolar molecule? and O atom. • 10.84 Which of the following molecules and ions are lin- ear? ICl2 1 2 , IF 2 , OF2, SnI2, CdBr2 • 10.85 Draw the Lewis structure for the BeCl224 ion. Predict its geometry and describe the hybridization state of the Be atom. • 10.86 The N2F2 molecule can exist in either of the follow- ing two forms: F F F D G D NPN NPN D 10.76 Caffeine is a stimulant drug present in coffee. F (a) Write the molecular formula of the compound. (a) What is the hybridization of N in the molecule? (b) What is the hybridization state of each C, N, and O atom? (c) Describe the geometry about each C, N, (b) Which structure has a dipole moment? and O atom. 10.87 Cyclopropane (C3H6) has the shape of a triangle in which a C atom is bonded to two H atoms and two other C atoms at each corner. Cubane (C8H8) has the shape of a cube in which a C atom is bonded to one H atom and three other C atoms at each cor- ner. (a) Draw Lewis structures of these molecules. (b) Compare the CCC angles in these molecules with those predicted for an sp3-hybridized C atom. (c) Would you expect these molecules to be easy to make? 10.88 The compound 1,2-dichloroethane (C2H4Cl2) is nonpolar, while cis-dichloroethylene (C2H2Cl2) has a dipole moment: Cl Cl A A Cl Cl • 10.77 Predict the geometry of sulfur dichloride (SCl2) and HOCOCOH G D A A CPC the hybridization of the sulfur atom. D G H H H H • 10.78 Antimony pentafluoride, SbF5, reacts with XeF4 and 1,2-dichloroethane cis-dichloroethylene XeF6 to form ionic compounds, XeF31SbF62 and XeF1 2 5 SbF6 . Describe the geometries of the cations The reason for the difference is that groups con- and anion in these two compounds. nected by a single bond can rotate with respect to • 10.79 Draw Lewis structures and give the other informa- each other, but no rotation occurs when a double tion requested for the following molecules: (a) BF3. bond connects the groups. On the basis of bonding Shape: planar or nonplanar? (b) ClO32. Shape: pla- considerations, explain why rotation occurs in nar or nonplanar? (c) H2O. Show the direction of 1,2-dichloroethane but not in cis-dichloroethylene. the resultant dipole moment. (d) OF2. Polar or • 10.89 Does the following molecule have a dipole moment? nonpolar molecule? (e) NO2. Estimate the ONO bond angle. Cl H G D • 10.80 Predict the bond angles for the following molecules: D CPCPC G (a) BeCl2, (b) BCl3, (c) CCl4, (d) CH3Cl, (e) Hg2Cl2 H Cl (arrangement of atoms: ClHgHgCl), (f ) SnCl2, ( g) H2O2, (h) SnH4. (Hint: See the answer to Problem 10.39.)  Questions & Problems 461 • 10.90 So-called greenhouse gases, which contribute to • 10.95 Write the ground-state electron configuration for B2. global warming, have a dipole moment or can be Is the molecule diamagnetic or paramagnetic? bent or distorted into shapes that have a dipole • 10.96 What are the hybridization states of the C and N moment. Which of the following gases are green- atoms in this molecule? house gases? N2, O2, O3, CO, CO2, NO2, N2O, NH2 CH4, CFCl3 A 10.91 The bond angle of SO2 is very close to 1208, even C H K H E though there is a lone pair on S. Explain. N C A B 10.92 39-azido-39-deoxythymidine, shown here, commonly KCHNECH known as AZT, is one of the drugs used to treat ac- O H quired immune deficiency syndrome (AIDS). What A are the hybridization states of the C and N atoms in H this molecule? 10.97 Use molecular orbital theory to explain the differ- ence between the bond enthalpies of F2 and F2 2 (see O Problem 9.110). B 10.98 Referring to the Chemistry in Action essay on p. 426, HH ECH answer the following questions: (a) If you wanted to N C OCH3 A B cook a roast (beef or lamb), would you use a micro- C K HNE HC wave oven or a conventional oven? (b) Radar is a O H means of locating an object by measuring the time A HOOCH2 O A for the echo of a microwave from the object to return A A to the source and the direction from which it returns. C H H C Would radar work if oxygen, nitrogen, and carbon A A A A dioxide were polar molecules? (c) In early tests of H C C H radar at the English Channel during World War II, A A N H the results were inconclusive even though there B was no equipment malfunction. Why? (Hint: The N weather is often foggy in the region.) B 10.99 Which of the following molecules are polar? N • 10.93 The following molecules (AX4Y2) all have octahe- dral geometry. Group the molecules that are equiva- lent to each other. Y Y X X X Y A A (a) (b) (c) X X X X Y X 10.100 Which of the following molecules are polar? (a) (b) X X X Y X Y A A Y X X Y X X (c) (d) (a) (b) (c) 10.94 The compounds carbon tetrachloride (CCl4) and silicon tetrachloride (SiCl4) are similar in geometry 10.101 The stable allotropic form of phosphorus is P4, in and hybridization. However, CCl4 does not react which each P atom is bonded to three other P atoms. with water but SiCl4 does. Explain the difference in Draw a Lewis structure of this molecule and describe their chemical reactivities. (Hint: The first step of its geometry. At high temperatures, P4 dissociates to the reaction is believed to be the addition of a water form P2 molecules containing a P“P bond. Explain molecule to the Si atom in SiCl4.) why P4 is more stable than P2.  462 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals 10.102 Referring to Table 9.4, explain why the bond en- by heating or irradiation. (a) Starting with cis- thalpy for Cl2 is greater than that for F2. (Hint: The dichloroethylene, show that rotating the C“C bond lengths of F2 and Cl2 are 142 pm and 199 pm, bond by 180° will break only the pi bond but will respectively.) leave the sigma bond intact. Explain the formation 10.103 Use molecular orbital theory to explain the bonding of trans-dichloroethylene from this process. (Treat in the azide ion (N32). (Arrangement of atoms is the rotation as two stepwise 90° rotations.) NNN.) (b) Account for the difference in the bond enthal- pies for the pi bond (about 270 kJ/mol) and the • 10.104 The ionic character of the bond in a diatomic mol- sigma bond (about 350 kJ/mol). (c) Calculate the ecule can be estimated by the formula longest wavelength of light needed to bring about μ this conversion. 3 100% ed 10.109 Progesterone is a hormone responsible for female where μ is the experimentally measured dipole mo- sex characteristics. In the usual shorthand struc- ment (in C m), e the electronic charge, and d the ture, each point where lines meet represent a C bond length in meters. (The quantity ed is the hy- atom, and most H atoms are not shown. Draw the pothetical dipole moment for the case in which complete structure of the molecule, showing all C the transfer of an electron from the less electro- and H atoms. Indicate which C atoms are sp2- and negative to the more electronegative atom is com- sp3-hybridized. plete.) Given that the dipole moment and bond length of HF are 1.92 D and 91.7 pm, respec- CH3 tively, calculate the percent ionic character of the A C PO molecule. CH3A A • 10.105 Draw three Lewis structures for compounds with the CH3 formula C2H2F2. Indicate which of the compound(s) A are polar. 10.106 Greenhouse gases absorb (and trap) outgoing infra- K red radiation (heat) from Earth and contribute to O global warming. The molecule of a greenhouse gas either possesses a permanent dipole moment or has • 10.110 For each pair listed here, state which one has a higher a changing dipole moment during its vibrational first ionization energy and explain your choice: (a) H motions. Consider three of the vibrational modes of or H2, (b) N or N2, (c) O or O2, (d) F or F2. carbon dioxide • 10.111 The molecule benzyne (C6H4) is a very reactive species. It resembles benzene in that it has a six- m n n m n h h membered ring of carbon atoms. Draw a Lewis OPCPO OPCPO OPCPO g structure of the molecule and account for the mol- ecule’s high reactivity. where the arrows indicate the movement of the • 10.112 Assume that the third-period element phosphorus atoms. (During a complete cycle of vibration, the forms a diatomic molecule, P2, in an analogous way atoms move toward one extreme position and then as nitrogen does to form N2. (a) Write the electronic reverse their direction to the other extreme position.) configuration for P2. Use [Ne2] to represent the Which of the preceding vibrations are responsible electron configuration for the first two periods. for CO2 to behave as a greenhouse gas? Which of (b) Calculate its bond order. (c) What are its mag- the following molecules can act as a greenhouse gas: netic properties (diamagnetic or paramagnetic)? N2, O2, CO, NO2, and N2O? 10.113 Consider a N2 molecule in its first excited electronic • 10.107 Aluminum trichloride (AlCl3) is an electron-deficient state; that is, when an electron in the highest occu- molecule. It has a tendency to form a dimer (a mol- pied molecular orbital is promoted to the lowest ecule made of two AlCl3 units): empty molecular orbital. (a) Identify the molecular orbitals involved and sketch a diagram to show the AlCl3 1 AlCl3 S Al2Cl6 transition. (b) Compare the bond order and bond length of N2* with N2, where the asterisk denotes (a) Draw a Lewis structure for the dimer. (b) Describe the excited molecule. (c) Is N2* diamagnetic or the hybridization state of Al in AlCl3 and Al2Cl6. paramagnetic? (d) When N2* loses its excess energy (c) Sketch the geometry of the dimer. (d) Do these and converts to the ground state N2, it emits a photon molecules possess a dipole moment? of wavelength 470 nm, which makes up part of the 10.108 The molecules cis-dichloroethylene and trans- auroras lights. Calculate the energy difference be- dichloroethylene shown on p. 425 can be interconverted tween these levels.  Interpreting, Modeling & Estimating 463 10.114 As mentioned in the chapter, the Lewis structure for • 10.119 Write the electron configuration of the cyanide ion O2 is (CN2). Name a stable molecule that is isoelectronic with the ion. O O PO Q O Q 10.120 Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to Fe21 in the Use the molecular orbital theory to show that the hemoglobin molecule. The molecular orbitals of structure actually corresponds to an excited state of CO have the same energy order as those of the N2 the oxygen molecule. molecule. (a) Draw a Lewis structure of CO and 10.115 Referring to Problem 9.137, describe the hybridiza- assign formal charges. Explain why CO has a tion state of the N atoms and the overall shape of rather small dipole moment of 0.12 D. (b) Com- the ion. pare the bond order of CO with that from molecu- 10.116 Describe the geometry and hybridization for the re- lar orbital theory. (c) Which of the atoms (C or O) actants and product in the following reaction is more likely to form bonds with the Fe21 ion in ClF3 1 AsF5 ¡ [ClF12 ][AsF26 ] hemoglobin? 10.121 The geometries discussed in this chapter all lend • 10.117 Draw the Lewis structure of ketene (C2H2O) and de- themselves to fairly straightforward elucidation of scribe the hybridization states of the C atoms. The bond angles. The exception is the tetrahedron, be- molecule does not contain O¬H bonds. On separate cause its bond angles are hard to visualize. Consider diagrams, sketch the formation of sigma and pi the CCl4 molecule, which has a tetrahedral geometry bonds. and is nonpolar. By equating the bond moment • 10.118 TCDD, or 2,3,7,8-tetrachlorodibenzo-p-dioxin, is a of a particular C¬Cl bond to the resultant bond highly toxic compound moments of the other three C¬Cl bonds in opposite directions, show that the bond angles are all ClE EO E ECl equal to 109.5°. E E E • 10.122 Carbon suboxide (C3O2) is a colorless pungent- E smelling gas. Does it possess a dipole moment? Cl O Cl • 10.123 Which of the following ions possess a dipole mo- It gained considerable notoriety in 2004 when it was ment? (a) ClF12 , (b) ClF2 1 2 2 , (c) IF 4 , (d) IF 4 . implicated in the murder plot of a Ukrainian politi- 10.124 Given that the order of molecular orbitals for NO cian. (a) Describe its geometry and state whether the is similar to that for O2, arrange the following spe- molecule has a dipole moment. (b) How many pi cies in increasing bond orders: NO22, NO2, NO, bonds and sigma bonds are there in the molecule? NO1, NO21. Interpreting, Modeling & Estimating 10.125 Shown here are molecular models of SX4 for X 5 F, 10.126 Based on what you have learned from this chapter Cl, and Br. Comment on the trends in the bond angle and Chapter 9, name a diatomic molecule that has between the axial S¬X bonds in these molecules. the strongest known chemical bond and one with the weakest known chemical bond. 10.127 The stability of benzene is due to the fact that we can draw reasonable resonance structures for the molecule, which is equivalent to saying that there is electron delocalization. Resonance energy is a measure of how much more stable benzene is compared to the hypothetical molecule, which can be represented by just a single resonance structure. Shown on p. 464 are the enthalpies of hydrogenation (the addition of hydrogen) of cy- clohexene (C6H10) to cyclohexane (C6H12) and benzene to cyclohexane. SF4 SCl4 SBr4 464 Chapter 10 ■ Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals attached to a sp2-hybridized C atom and there are 1 H2 DH8 5 2120 kJ/mol two H atoms attached to a sp3-hybridized C atom.) Estimate the resonance energy of benzene from these data. 10.128 How many carbon atoms are contained in one 1 3H2 DH8 5 2208 kJ/mol square centimeter of graphene (see the Chemistry in Action essay on p. 454 for a description of gra- phene)? What would be the mass of a 1-cm2 section (In these simplified structures, each point where of graphene? lines meet represents a C atom. There is a H atom Answers to Practice Exercises 10.1 (a) Tetrahedral, (b) linear, (c) trigonal planar. unhybridized p orbitals on the C atom are used to form two 10.2 No. 10.3 (a) sp3, (b) sp2. 10.4 sp3d2. 10.5 The C atom pi bonds with the N atom. The lone pair on the N atom is is sp-hybridized. It forms a sigma bond with the H atom placed in the sp orbital. 10.6 F2 2. and another sigma bond with the N atom. The two  CHAPTER 11 Intermolecular A person throwing boiling water into the air at -51°C. Forces and Liquids and Solids CHAPTER OUTLINE A LOOK AHEAD 11.1 The Kinetic Molecular  We begin by applying the kinetic molecular theory to liquids and solids and Theory of Liquids compare their properties with those of gases. (11.1) and Solids  Next, we examine the different types of intermolecular forces between 11.2 Intermolecular Forces molecules and between ions and molecules. We also study a special type of intermolecular interaction called hydrogen bonding that involves hydrogen 11.3 Properties of Liquids and electronegative elements nitrogen, oxygen, and fluorine. (11.2) 11.4 Crystal Structure  We see that two important properties of liquids—surface tension and viscosity—can be understood in terms of intermolecular forces. (11.3) 11.5 X-Ray Diffraction by Crystals  We then move on to the world of solids and learn about the nature of crystals and ways of packing spheres to form different unit cells. (11.4) 11.6 Types of Crystals  We see that the best way to determine the dimensions of a crystal structure 11.7 Amorphous Solids is by X-ray diffraction, which is based on the scattering of X rays by the 11.8 Phase Changes atoms or molecules in a crystal. (11.5) 11.9 Phase Diagrams  The major types of crystals are ionic, covalent, molecular, and metallic. Intermolecular forces help us understand their structure and physical proper- ties such as density, melting point, and electrical conductivity. (11.6)  We learn that solids can also exist in the amorphous form, which lacks orderly three-dimensional arrangement. A well-known example of an amorphous solid is glass. (11.7)  We next study phase changes, or transitions among gas, liquids, and solids. We see that the dynamic equilibrium between liquid and vapor gives rise to equilibrium vapor pressure. The energy required for vaporization depends on the strength of intermolecular forces. We also learn that every substance has a critical temperature above which its vapor form cannot be liquefied. We then examine liquid-solid and solid-vapor transitions. (11.8)  The various types of phase transitions are summarized in a phase diagram, which helps us understand conditions under which a phase is stable and changes in pressure and/or temperature needed to bring about a phase transition. (11.9) 465 466 Chapter 11 ■ Intermolecular Forces and Liquids and Solids A lthough we live immersed in a mixture of gases that make up Earth’s atmosphere, we are more familiar with the behavior of liquids and solids because they are more visible. Every day we use water and other liquids for drinking, bathing, cleaning, and cooking, and we handle, sit upon, and wear solids. Molecular motion is more restricted in liquids than in gases; and in solids the atoms and molecules are packed even more tightly together. In fact, in a solid they are held in well-defined positions and are capable of little free motion relative to one another. In this chapter we will examine the structure of liquids and solids and discuss some of the fundamental properties of these two states of matter. We will also study the nature of transitions among gases, liquids, and solids. 11.1 The Kinetic Molecular Theory of Liquids and Solids In Chapter 5 we used the kinetic molecular theory to explain the behavior of gases in terms of the constant, random motion of gas molecules. In gases, the distances between molecules are so great (compared with their diameters) that at ordinary tem- peratures and pressures (say, 25°C and 1 atm), there is no appreciable interaction between the molecules. Because there is a great deal of empty space in a gas—that is, space that is not occupied by molecules—gases can be readily compressed. The lack of strong forces between molecules also allows a gas to expand to fill the volume of its container. Furthermore, the large amount of empty space explains why gases have very low densities under normal conditions. Liquids and solids are quite a different story. The principal difference between the condensed states (liquids and solids) and the gaseous state is the distance between molecules. In a liquid, the molecules are so close together that there is very little empty space. Thus, liquids are much more difficult to compress than gases, and they are also much denser under normal conditions. Molecules in a liquid are held together by one or more types of attractive forces, which will be discussed in Section 11.2. A liquid also has a definite volume, because molecules in a liquid do not break away from the attractive forces. The molecules can, however, move past one another freely, and so a liquid can flow, can be poured, and assumes the shape of its container. In a solid, molecules are held rigidly in position with virtually no freedom of motion. Many solids are characterized by long-range order; that is, the molecules are arranged in regular configurations in three dimensions. There is even less empty space in a solid than in a liquid. Thus, solids are almost incompressible and possess definite shape and volume. With very few exceptions (water being the most important), the density of the solid form is higher than that of the liquid form for a given substance. It is not uncommon for two states of a substance to coexist. An ice cube (solid) float- ing in a glass of water (liquid) is a familiar example. Chemists refer to the different states of a substance that are present in a system as phases. A phase is a homogeneous Table 11.1 Characteristic Properties of Gases, Liquids, and Solids State of Matter Volume/Shape Density Compressibility Motion of Molecules Gas Assumes the volume and Low Very compressible Very free motion shape of its container Liquid Has a definite volume High Only slightly compressible Slide past one another freely but assumes the shape of its container Solid Has a definite volume High Virtually incompressible Vibrate about fixed positions and shape  11.2 Intermolecular Forces 467 part of the system in contact with other parts of the system but separated from them by a well-defined boundary. Thus, our glass of ice water contains both the solid phase and the liquid phase of water. In this chapter we will use the term “phase” when talking about changes of state involving one substance, as well as systems containing more than one phase of a substance. Table 11.1 summarizes some of the characteris- tic properties of the three phases of matter. 11.2 Intermolecular Forces Intermolecular forces are attractive forces between molecules. Intermolecular forces are responsible for the nonideal behavior of gases described in Chapter 5. They exert even more influence in the condensed phases of matter—liquids and solids. As the temperature of a gas drops, the average kinetic energy of its molecules decreases. Eventually, at a sufficiently low temperature, the molecules no longer have enough energy to break away from the attraction of neighboring molecules. At this point, the molecules aggregate to form small drops of liquid. This transition from the gaseous to the liquid phase is known as condensation. In contrast to intermolecular forces, intramolecular forces hold atoms together in a molecule. (Chemical bonding, discussed in Chapters 9 and 10, involves intramolecular forces.) Intramolecular forces stabilize individual molecules, whereas intermolecular forces are primarily responsible for the bulk properties of matter (for example, melting point and boiling point). Generally, intermolecular forces are much weaker than intramolecular forces. It usually requires much less energy to evaporate a liquid than to break the bonds in the molecules of the liquid. For example, it takes about 41 kJ of energy to vaporize 1 mole of water at its boiling point; but about 930 kJ of energy are necessary to break the two O¬H bonds in 1 mole of water molecules. The boiling points of substances often reflect the strength of the intermolecular forces operating among the molecules. At the boiling point, enough energy must be supplied to overcome the attractive forces among molecules before they can enter the vapor phase. If it takes more energy to separate molecules of substance A than of substance B because A molecules are held together by stronger intermolecular forces, then the boiling point of A is higher than that of B. The same principle applies also to the melting points of the substances. In general, the melting points of substances increase with the strength of the intermolecular forces. To discuss the properties of condensed matter, we must understand the different types of intermolecular forces. Dipole-dipole, dipole-induced dipole, and dispersion forces make up what chemists commonly refer to as van der Waals forces, after the Dutch physicist Johannes van der Waals (see Section 5.8). Ions and dipoles are attracted to one another by electrostatic forces called ion-dipole forces, which are not van der Waals forces. Hydrogen bonding is a particularly strong type of dipole- dipole interaction. Because only a few elements can participate in hydrogen bond formation, it is treated as a separate category. Depending on the phase of a sub- stance, the nature of chemical bonds, and the types of elements present, more than one type of interaction may contribute to the total attraction between molecules, as we will see below. + – + – + – Dipole-Dipole Forces – + – + – + Dipole-dipole forces are attractive forces between polar molecules, that is, between + – + – + – molecules that possess dipole moments (see Section 10.2). Their origin is electrostatic, and they can be understood in terms of Coulomb’s law. The larger the dipole moment, Figure 11.1 Molecules that have the greater the force. Figure 11.1 shows the orientation of polar molecules in a solid. a permanent dipole moment tend to align with opposite polarities in In liquids, polar molecules are not held as rigidly as in a solid, but they tend to align the solid phase for maximum in a way that, on average, maximizes the attractive interaction. attractive interaction. 468 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Na+ Ion-Dipole Forces – + Coulomb’s law also explains ion-dipole forces, which attract an ion (either a I– cation or an anion) and a polar molecule to each other (Figure 11.2). The strength + – of this interaction depends on the charge and size of the ion and on the magnitude of the dipole moment and size of the molecule. The charges on cations are gener- Figure 11.2 Two types of ally more concentrated, because cations are usually smaller than anions. Therefore, ion-dipole interaction. a cation interacts more strongly with dipoles than does an anion having a charge of the same magnitude. Hydration, discussed in Section 4.1, is one example of ion-dipole interaction. Heat of hydration (see p. 259) is the result of the favorable interaction between the cations and anions of an ionic compound with water. Figure 11.3 shows the ion-dipole interaction between the Na1 and Mg21 ions with a water molecule, which has a large dipole moment (1.87 D). Because the Mg21 ion has a higher charge and a smaller ionic radius (78 pm) than that of the Na1 ion (98 pm), it interacts more strongly with water molecules. (In reality, each ion is surrounded by a number of water molecules in solution.) Consequently, the heats of hydration for the Na1 and Mg21 ions are 2405 kJ/mol and 21926 kJ/mol, respectively.† Similar differences exist for anions of different charges and sizes. Dispersion Forces What attractive interaction occurs in nonpolar substances? To learn the answer to this question, consider the arrangement shown in Figure 11.4. If we place an ion or a polar molecule near an atom (or a nonpolar molecule), the electron distribution of the atom (or molecule) is distorted by the force exerted by the ion or the polar molecule, resulting in a kind of dipole. The dipole in the atom (or nonpolar molecule) is said to be an induced dipole because the separation of positive and negative charges in the atom (or nonpolar molecule) is due to the proximity of an ion or a polar molecule. The attractive interaction between an ion and the induced dipole is called ion-induced dipole interaction, and the attractive interaction between a polar molecule and the induced dipole is called dipole- induced dipole interaction. † Heats of hydration of individual ions cannot be measured directly, but they can be reliably estimated. Weak interaction Na+ Strong interaction Mg2+ (a) (b) 1 21 Figure 11.3 (a) Interaction of a water molecule with a Na ion and a Mg ion. (b) In aqueous solutions, metal ions are usually surrounded by six water molecules in an octahedral arrangement.  11.2 Intermolecular Forces 469 The likelihood of a dipole moment being induced depends not only on the charge on the ion or the strength of the dipole but also on the polarizability of the atom or molecule—that is, the ease with which the electron distribution in the atom (or mol- ecule) can be distorted. Generally, the larger the number of electrons and the more (a) diffuse the electron cloud in the atom or molecule, the greater its polarizability. By diffuse cloud we mean an electron cloud that is spread over an appreciable volume, Induced dipole so that the electrons are not held tightly by the nucleus. Cation Polarizability enables gases containing atoms or nonpolar molecules (for example, + – + He and N2) to condense. In a helium atom the electrons are moving at some distance from the nucleus. At any instant it is likely that the atom has a dipole moment created (b) by the specific positions of the electrons. This dipole moment is called an instantaneous dipole because it lasts for just a tiny fraction of a second. In the next instant the Induced dipole electrons are in different locations and the atom has a new instantaneous dipole, Dipole – + – + and so on. Averaged over time (that is, the time it takes to make a dipole moment measurement), however, the atom has no dipole moment because the instantaneous (c) dipoles all cancel one another. In a collection of He atoms, an instantaneous dipole of one He atom can induce a dipole in each of its nearest neighbors (Figure 11.5). Figure 11.4 (a) Spherical charge distribution in a helium atom. At the next moment, a different instantaneous dipole can create temporary dipoles (b) Distortion caused by the in the surrounding He atoms. The important point is that this kind of interaction approach of a cation. (c) Distortion caused by the approach of a produces dispersion forces, attractive forces that arise as a result of temporary dipole. dipoles induced in atoms or molecules. At very low temperatures (and reduced atomic speeds), dispersion forces are strong enough to hold He atoms together, causing the gas to condense. The attraction between nonpolar molecules can be explained similarly. A quantum mechanical interpretation of temporary dipoles was provided by Fritz London† in 1930. London showed that the magnitude of this attractive interac- tion is directly proportional to the polarizability of the atom or molecule. As we For simplicity we use the term “intermolecular forces” for both might expect, dispersion forces may be quite weak. This is certainly true for helium, atoms and molecules. which has a boiling point of only 4.2 K, or 2269°C. (Note that helium has only two electrons, which are tightly held in the 1s orbital. Therefore, the helium atom has a very low polarizability.) Dispersion forces, which are also called London forces, usually increase with molar mass because molecules with larger molar mass tend to have more electrons, Table 11.2 and dispersion forces increase in strength with the number of electrons. Furthermore, Melting Points of Similar larger molar mass often means a bigger atom whose electron distribution is more eas- Nonpolar Compounds ily disturbed because the outer electrons are less tightly held by the nuclei. Table 11.2 Melting compares the melting points of similar substances that consist of nonpolar molecules. Compound Point (°C) As expected, the melting point increases as the number of electrons in the molecule increases. Because these are all nonpolar molecules, the only attractive intermolecular CH4 2182.5 forces present are the dispersion forces. CF4 2150.0 CCl4 223.0 CBr4 90.0 † Fritz London (1900–1954). German physicist. London was a theoretical physicist whose major work was CI4 171.0 on superconductivity in liquid helium. – + + + – – + – – + + – – + + – + + – – + – + + + + – – + – + – – – – – + + + – + + + + – + – + – – + – – – – – – + + + Figure 11.5 Induced dipoles interacting with each other. Such patterns exist only momentarily; new arrangements are formed in the next instant. This type of interaction is responsible for the condensation of nonpolar gases. 470 Chapter 11 ■ Intermolecular Forces and Liquids and Solids In many cases, dispersion forces are comparable to or even greater than the dipole-dipole forces between polar molecules. For a dramatic illustration, let us compare the boiling points of CH3F (278.4°C) and CCl4 (76.5°C). Although CH3F has a dipole moment of 1.8 D, it boils at a much lower temperature than CCl4, a nonpolar molecule. CCl4 boils at a higher temperature simply because it contains more electrons. As a result, the dispersion forces between CCl4 molecules are stronger than the dispersion forces plus the dipole-dipole forces between CH3F molecules. (Keep in mind that dispersion forces exist among species of all types, whether they are neutral or bear a net charge and whether they are polar or nonpolar.) Example 11.1 shows that if we know the kind of species present, we can readily determine the types of intermolecular forces that exist between the species. Example 11.1 What type(s) of intermolecular forces exist between the following pairs: (a) HBr and H2S, (b) Cl2 and CBr4, (c) I2 and NO23 , (d) NH3 and C6H6? Strategy Classify the species into three categories: ionic, polar (possessing a dipole moment), and nonpolar. Keep in mind that dispersion forces exist between all species. Solution (a) Both HBr and H2S are polar molecules Therefore, the intermolecular forces present are dipole-dipole forces, as well as dispersion forces. (b) Both Cl2 and CBr4 are nonpolar, so there are only dispersion forces between these molecules. (c) I2 is a homonuclear diatomic molecule and therefore nonpolar, so the forces between it and the ion NO23 are ion-induced dipole forces and dispersion forces. (d) NH3 is polar, and C6H6 is nonpolar. The forces are dipole-induced dipole forces and Similar problem: 11.10. dispersion forces. Practice Exercise Name the type(s) of intermolecular forces that exists between molecules (or basic units) in each of the following species: (a) LiF, (b) CH4, (c) SO2.  11.2 Intermolecular Forces 471 Figure 11.6 Boiling points of the hydrogen compounds of Groups 100 H2O 4A, 5A, 6A, and 7A elements. Although normally we expect the boiling point to increase as we Group 6A move down a group, we see that three compounds ( NH3, H2O, and HF ) behave differently. The HF anomaly can be explained in terms of intermolecular hydrogen 0 Boiling point (°C) H2Te bonding. Group 7A SbH3 NH3 H2Se HI H2S Group 5A AsH3 SnH4 HCl HBr –100 PH3 GeH4 SiH4 Group 4A CH 4 –200 2 3 4 5 Period The Hydrogen Bond Normally, the boiling points of a series of similar compounds containing elements in the same periodic group increase with increasing molar mass. This increase in boiling point is due to the increase in dispersion forces for molecules with more electrons. Hydrogen compounds of Group 4A follow this trend, as Figure 11.6 shows. The light- est compound, CH4, has the lowest boiling point, and the heaviest compound, SnH4, has the highest boiling point. However, hydrogen compounds of the elements in Groups 5A, 6A, and 7A do not follow this trend. In each of these series, the lightest compound (NH3, H2O, and HF) has the highest boiling point, contrary to our expec- tations based on molar mass. This observation must mean that there are stronger intermolecular attractions in NH3, H2O, and HF, compared to other molecules in the same groups. In fact, this particularly strong type of intermolecular attraction is called the hydrogen bond, which is a special type of dipole-dipole interaction between the 1A 8A hydrogen atom in a polar bond, such as N¬H, O¬H, or F¬H, and an electro- 2A 3A 4A 5A 6A 7A negative O, N, or F atom. The interaction is written N O F A¬H ? ? ? :B or A¬H ? ? ? :A A and B represent O, N, or F; A¬H is one molecule or part of a molecule and B is The three most electronegative a part of another molecule; and the dotted line represents the hydrogen bond. The elements that take part in three atoms usually lie in a straight line, but the angle AHB (or AHA) can deviate as hydrogen bonding. much as 30° from linearity. Note that the O, N, and F atoms all possess at least one lone pair that can interact with the hydrogen atom in hydrogen bonding. The average strength of a hydrogen bond is quite large for a dipole-dipole inter- action (up to 40 kJ/mol). Thus, hydrogen bonds have a powerful effect on the struc- tures and properties of many compounds. Figure 11.7 shows several examples of hydrogen bonding. The strength of a hydrogen bond is determined by the coulombic interaction between the lone-pair electrons of the electronegative atom and the hydrogen nucleus. For example, fluorine is more electronegative than oxygen, and so we would expect 472 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.7 Hydrogen bonding in water, ammonia, and hydrogen fluoride. Solid lines represent H H H covalent bonds, and dotted lines A A A OSZ HOO HOO OS HONSZ HONS OSZ HONS HOO represent hydrogen bonds. A A A A A A H H H H H H H H H A A A HON SZ HOO OS OS Z HON S HOF Q HON SZ HOF OS Q A A A A H H H H a stronger hydrogen bond to exist in liquid HF than in H2O. In the liquid phase, the HF molecules form zigzag chains: The boiling point of HF is lower than that of water because each H2O takes part in four intermolecular hydrogen bonds. Therefore, the forces holding the molecules together are stronger in H2O than in HF. We will return to this very important property of water in Section 11.3. Example 11.2 shows the type of species that can form hydrogen bonds with water. Example 11.2 Which of the following can form hydrogen bonds with water? CH3OCH3, CH4, F2, HCOOH, Na1. Strategy A species can form hydrogen bonds with water if it contains one of the three electronegative elements (F, O, or N) or it has a H atom bonded to one of these three elements. Solution There are no electronegative elements (F, O, or N) in either CH4 or Na1. Therefore, only CH3OCH3, F2, and HCOOH can form hydrogen bonds with water. S O S HCOOH forms hydrogen bonds D G with two H2O molecules. H H S OS J HOC H G D  OOH SO O SO OS FS HOO Q H3COOOS HOO OS G A A A H H H3C H Check Note that HCOOH (formic acid) can form hydrogen bonds with water in two Similar problem: 11.12. different ways. Practice Exercise Which of the following species are capable of hydrogen bonding among themselves? (a) H2S, (b) C6H6, (c) CH3OH.  11.3 Properties of Liquids 473 Review of Concepts Which of the following compounds is most likely to exist as a liquid at room temperature: ethane (C2H6), hydrazine (N2H4), fluoromethane (CH3F)? The intermolecular forces discussed so far are all attractive in nature. Keep in mind, though, that molecules also exert repulsive forces on one another. Thus, when two molecules approach each other, the repulsion between the electrons and between the nuclei in the molecules comes into play. The magnitude of the repulsive force rises very steeply as the distance separating the molecules in a condensed phase decreases. This is the reason that liquids and solids are so hard to compress. In these phases, the molecules are already in close contact with one another, and so they Figure 11.8 Intermolecular greatly resist being compressed further. forces acting on a molecule in the surface layer of a liquid and in the interior region of the liquid. 11.3 Properties of Liquids Intermolecular forces give rise to a number of structural features and properties of liquids. In this section we will look at two such phenomena associated with liquids in general: surface tension and viscosity. Then we will discuss the structure and properties of water. Surface Tension Molecules within a liquid are pulled in all directions by intermolecular forces; there is no tendency for them to be pulled in any one way. However, molecules at the surface are pulled downward and sideways by other molecules, but not upward away from the surface (Figure 11.8). These intermolecular attractions thus tend to pull the molecules into the liquid and cause the surface to tighten like an elastic film. Because there is little or no attraction between polar water molecules and, say, the nonpolar wax molecules on a freshly waxed car, a drop of water assumes the shape of a small round bead, because a sphere minimizes the surface area of a liquid. The waxy surface of a wet apple also produces this effect (Figure 11.9). A measure of the elastic force in the surface of a liquid is surface tension. The surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area (for example, by 1 cm2). Liquids that have strong intermolecular Surface tension enables the water forces also have high surface tensions. Thus, because of hydrogen bonding, water has a strider to “walk” on water. considerably greater surface tension than most other liquids. Another example of surface tension is capillary action. Figure 11.10(a) shows water rising spontaneously in a capillary tube. A thin film of water adheres to the wall of the glass tube. The surface tension of water causes this film to contract, and as it does, it pulls the water up the tube. Two types of forces bring about capillary action. One is cohesion, which is the intermolecular attraction between like mole- cules (in this case, the water molecules). The second force, called adhesion, is an attraction between unlike molecules, such as those in water and in the sides of a glass tube. If adhesion is stronger than cohesion, as it is in Figure 11.10(a), the contents of the tube will be pulled upward. This process continues until the adhe- sive force is balanced by the weight of the water in the tube. This action is by no means universal among liquids, as Figure 11.10(b) shows. In mercury, cohesion is greater than the adhesion between mercury and glass, so that when a capillary tube is dipped in mercury, the result is a depression or lowering, at the mercury level—that is, the height of the liquid in the capillary tube is below the surface Figure 11.9 Water beads on an of the mercury. apple, which has a waxy surface. 474 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.10 (a) When adhesion is greater than cohesion, the liquid (for example, water) rises in the capillary tube. (b) When cohesion is greater than adhesion, as it is for mercury, a depression of the liquid in the capillary tube results. Note that the meniscus in the tube of water is concave, or rounded downward, whereas that in the tube of mercury is convex, or rounded upward. (a) (b) Viscosity The expression “slow as molasses in January” owes its truth to another physical property of liquids called viscosity. Viscosity is a measure of a fluid’s resistance to flow. The greater the viscosity, the more slowly the liquid flows. The viscosity of a liquid usually decreases as temperature increases; thus, hot molasses flows much faster than cold molasses. Liquids that have strong intermolecular forces have higher viscosities than those that have weak intermolecular forces (Table 11.3). Water has a higher viscosity than many other liquids because of its ability to form hydrogen bonds. Interestingly, the viscosity of glycerol is significantly higher than that of all the other liquids listed in Table 11.3. Glycerol has the structure CH 2 OOH A CHOOH A CH 2 OOH Like water, glycerol can form hydrogen bonds. Each glycerol molecule has three ¬OH groups that can participate in hydrogen bonding with other glycerol molecules. Furthermore, because of their shape, the molecules have a great tendency to become Glycerol is a clear, odorless, syrupy liquid used to make entangled rather than to slip past one another as the molecules of less viscous liquids explosives, ink, and lubricants. do. These interactions contribute to its high viscosity. Table 11.3 Viscosity of Some Common Liquids at 20°C Viscosity Liquid (N s/m2)* Acetone (C3H6O) 3.16 3 1024 Benzene (C6H6) 6.25 3 1024 Blood 4 3 1023 Carbon tetrachloride (CCl4) 9.69 3 1024 Diethyl ether (C2H5OC2H5) 2.33 3 1024 Ethanol (C2H5OH) 1.20 3 1023 Glycerol (C3H8O3) 1.49 Mercury (Hg) 1.55 3 1023 Water (H2O) 1.01 3 1023 *The SI units of viscosity are newton-second per meter squared. CHEMISTRY in Action A Very Slow Pitch I n 1927, Thomas Parnell started what may be the longest- running laboratory experiment in the history of laboratory experiments. Professor Parnell wanted to show his physics stu- dents at the University of Queensland an interesting property of pitch, a derivative of tar. Pitch is viscoelastic, which means that it will break into pieces if struck with enough force, but like a viscous liquid, it also flows slowly. Very slowly! Parnell heated a sample of pitch to a temperature that al- lowed it to be poured into a funnel, and the funnel and a receiv- ing beaker were covered with a bell jar and placed on display outside of the lecture hall. And then they waited. The first drop fell 8–9 years after the pitch settled in the funnel and the stem was cut, but no one saw it fall. Drops fell at a rate of roughly one drop per decade, still with no witnesses. A few years after the third drop fell, Professor John Mainstone took over as curator and guardian of the experiment, but he did not get to witness any of the five drops that fell during his 52 years as curator, including several near misses and a webcam failure when the eighth drop fell on November 28, 2000. Based on the rate at which the drops have fallen over the past eight decades, the viscosity of the pitch used for this experiment is estimated to be 2.3 3 108 N∙s/m2, which makes it about 60 million times “slower than molasses.” John Mainstone watching the Pitch Drop Experiment. Sadly, Professor Mainstone died before he could see the ninth drop fall, but a website providing a live video feed to the experiment has been dedicated in his honor. Review of Concepts Why are motorists advised to use more viscous oils for their engines in the summer and less viscous oils in the winter? The Structure and Properties of Water Water is so common a substance on Earth that we often overlook its unique nature. If water did not have the ability to form hydrogen bonds, it would be a gas at All life processes involve water. Water is an excellent solvent for many ionic com- room temperature. pounds, as well as for other substances capable of forming hydrogen bonds with water. As Table 6.2 shows, water has a high specific heat. The reason is that to raise the temperature of water (that is, to increase the average kinetic energy of water molecules), we must first break the many intermolecular hydrogen bonds. Thus, water can absorb a substantial amount of heat while its temperature rises only slightly. The converse is also true: Water can give off much heat with only a slight decrease in its temperature. For this reason, the huge quantities of water that are present in our lakes and oceans can effectively moderate the climate of adjacent land areas by absorbing heat in the summer and giving off heat in the winter, with only small changes in the temperature of the body of water. The most striking property of water is that its solid form is less dense than its liquid form: ice floats at the surface of liquid water. The density of almost all other substances is greater in the solid state than in the liquid state (Figure 11.11). 475 476 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.11 Left: Ice cubes float on water. Right: Solid benzene sinks to the bottom of liquid benzene. To understand why water is different, we have to examine the electronic structure of the H2O molecule. As we saw in Chapter 9, there are two pairs of nonbonding electrons, or two lone pairs, on the oxygen atom: S S O D G H H Although many compounds can form intermolecular hydrogen bonds, the difference between H2O and other polar molecules, such as NH3 and HF, is that each oxygen atom Electrostatic potential map of water. can form two hydrogen bonds, the same as the number of lone electron pairs on the oxygen atom. Thus, water molecules are joined together in an extensive three-dimensional network in which each oxygen atom is approximately tetrahedrally bonded to four hydrogen atoms, two by covalent bonds and two by hydrogen bonds. This equality in the number of hydrogen atoms and lone pairs is not characteristic of NH3 or HF or, for that matter, of any other molecule capable of forming hydrogen bonds. Consequently, these other molecules can form rings or chains, but not three-dimensional structures. The highly ordered three-dimensional structure of ice (Figure 11.12) prevents the molecules from getting too close to one another. But consider what happens when ice Figure 11.12 The three- dimensional structure of ice. Each O atom is bonded to four H atoms. The covalent bonds are shown by short solid lines and the weaker hydrogen bonds by long dotted lines between O and H. The empty space in the structure accounts for the low density of ice. =O =H  11.4 Crystal Structure 477 melts. At the melting point, a number of water molecules have enough kinetic energy 1.00 to break free of the intermolecular hydrogen bonds. These molecules become trapped Density (g/mL) in the cavities of the three-dimensional structure, which is broken down into smaller 0.99 clusters. As a result, there are more molecules per unit volume in liquid water than in ice. Thus, because density 5 mass/volume, the density of water is greater than that of ice. With further heating, more water molecules are released from intermolecular 0.98 hydrogen bonding, so that the density of water tends to increase with rising tempera- ture just above the melting point. Of course, at the same time, water expands as it is 0.97 –20 0 20 40 60 80 being heated so that its density is decreased. These two processes—the trapping of Temperature (°C) free water molecules in cavities and thermal expansion—act in opposite directions. Figure 11.13 Plot of density From 0°C to 4°C, the trapping prevails and water becomes progressively denser. versus temperature for liquid Beyond 4°C, however, thermal expansion predominates and the density of water water. The maximum density of decreases with increasing temperature (Figure 11.13). water is reached at 4°C. The density of ice at 0°C is about 0.92 g/cm3. 11.4 Crystal Structure Solids can be divided into two categories: crystalline and amorphous. Ice is a crystal- line solid, which possesses rigid and long-range order; its atoms, molecules, or ions occupy specific positions. The arrangement of such particles in a crystalline solid is such that the net attractive intermolecular forces are at their maximum. The forces responsible for the stability of a crystal can be ionic forces, covalent bonds, van der Waals forces, hydrogen bonds, or a combination of these forces. Amorphous solids such as glass lack a well-defined arrangement and long-range molecular order. We will discuss them in Section 11.7. In this section, we will concentrate on the structure of crystalline solids. A unit cell is the basic repeating structural unit of a crystalline solid. Figure 11.14 Animation Cubic Unit Cells and Their Origins shows a unit cell and its extension in three dimensions. Each sphere represents an atom, ion, or molecule and is called a lattice point. In many crystals, the lattice point does not actually contain such a particle. Rather, there may be several atoms, ions, or molecules identically arranged about each lattice point. For simplicity, however, we can assume that each lattice point is occupied by an atom. This is certainly the case with most metals. Every crystalline solid can be described in terms of one of the seven types of unit cells shown in Figure 11.15. The geometry of the cubic unit cell is particularly simple because all sides and all angles are equal. Any of the unit cells, when repeated in space in all three dimensions, forms the lattice structure character- istic of a crystalline solid. Packing Spheres We can understand the general geometric requirements for crystal formation by con- Animation Packing Spheres sidering the different ways of packing a number of identical spheres (Ping-Pong balls, for example) to form an ordered three-dimensional structure. The way the spheres are arranged in layers determines what type of unit cell we have. Figure 11.14 (a) A unit cell and (b) its extension in three dimensions. The black spheres represent either atoms or molecules. (a) (b)  CHEMISTRY in Action Why Do Lakes Freeze from the Top Down? T he fact that ice is less dense than water has a profound ecological significance. Consider, for example, the tem- perature changes in the fresh water of a lake in a cold climate. As the temperature of the water near the surface drops, the density of this water increases. The colder water then sinks toward the bottom, while warmer water, which is less dense, rises to the top. This normal convection motion continues until the temperature throughout the water reaches 4°C. Below this temperature, the density of water begins to decrease with decreasing temperature (see Figure 11.13), so that it no longer sinks. On further cooling, the water begins to freeze at the surface. The ice layer formed does not sink because it is less dense than the liquid; it even acts as a thermal insulator for the water below it. Were ice heavier, it would sink to the bottom of the lake and eventually the water would freeze upward. Most living organisms in the body of water could not survive being frozen in ice. Fortunately, lake water does not freeze upward from the bottom. This unusual property of water makes the sport of ice fishing possible. Ice fishing. The ice layer that forms on the surface of a lake insulates the water beneath and maintains a high enough temperature to sustain aquatic life. Figure 11.15 The seven types of unit cells. Angle α is defined by edges b and c, angle β by edges b a and c, and angle γ by edges a a and b. c α β γ Simple cubic Tetragonal Orthorhombic Rhombohedral a=b=c a=b=c a=b=c a=b=c α = β = γ = 90° α = β = γ = 90° α = β = γ = 90° α = β = γ = 90° Monoclinic Triclinic Hexagonal a=b=c a=b=c a=b=c γ = α = β = 90° α = β = γ = 90° α = β = 90°, γ = 120° 478  11.4 Crystal Structure 479 Figure 11.16 Arrangement of identical spheres in a simple cubic cell. (a) Top view of one layer of spheres. (b) Definition of x a simple cubic cell. (c) Because each sphere is shared by eight unit cells and there are eight corners in a cube, there is the equivalent of one complete sphere inside a simple cubic (a) (b) (c) unit cell. In the simplest case, a layer of spheres can be arranged as shown in Figure 11.16(a). The three-dimensional structure can be generated by placing a layer above and below this layer in such a way that spheres in one layer are directly over the spheres in the layer below it. This procedure can be extended to generate many, many layers, as in the case of a crystal. Focusing on the sphere labeled with an “x,” we see that it is in contact with four spheres in its own layer, one sphere in the layer above, and one sphere in the layer below. Each sphere in this arrangement is said to have a coordination number of 6 because it has six immediate neighbors. The coordination number is defined as the number of atoms (or ions) surrounding an atom (or ion) in a crystal lattice. Its value gives us a measure of how tightly the spheres are packed together—the larger the coor- dination number, the closer the spheres are to each other. The basic, repeating unit in the array of spheres is called a simple cubic cell (scc) [Figure 11.16(b)]. The other types of cubic cells are the body-centered cubic cell (bcc) and the face-centered cubic cell (fcc) (Figure 11.17). A body-centered cubic arrangement differs from a simple cube in that the second layer of spheres fits into the depres- sions of the first layer and the third layer into the depressions of the second layer (Figure 11.18). The coordination number of each sphere in this structure is 8 (each sphere is in contact with four spheres in the layer above and four spheres in the layer below). In the face-centered cubic cell, there are spheres at the center of each of the six faces of the cube, in addition to the eight corner spheres. Figure 11.17 Three types of cubic cells. In reality, the spheres representing atoms, molecules, or ions are in contact with one another in these cubic cells. Simple cubic Body-centered cubic Face-centered cubic Figure 11.18 Arrangement of identical spheres in a body- centered cube. (a) Top view. (b) Definition of a body-centered cubic unit cell. (c) There is the equivalent of two complete spheres inside a body-centered cubic unit cell. (a) (b) (c) 480 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.19 (a) A corner atom in any cell is shared by eight unit cells. (b) An edge atom is shared by four unit cells. (c) A face-centered atom in a cubic cell is shared by two unit cells. (a) (b) (c) Because every unit cell in a crystalline solid is adjacent to other unit cells, most of a cell’s atoms are shared by neighboring cells. For example, in all types of cubic cells, each corner atom belongs to eight unit cells [Figure 11.19(a)]; an edge atom is shared by four unit cells [Figure 11.19(b)], and a face-centered atom is shared by two unit cells [Figure 11.19(c)]. Because each corner sphere is shared by eight unit cells and there are eight corners in a cube, there will be the equivalent of only one complete sphere inside a simple cubic unit cell (see Figure 11.17). A body-centered cubic cell contains the equivalent of two complete spheres, one in the center and eight shared corner spheres. A face-centered cubic cell contains four complete spheres—three from the six face-centered atoms and one from the eight shared corner spheres. Closest Packing Clearly there is more empty space in the simple cubic and body-centered cubic cells than in the face-centered cubic cell. Closest packing, the most efficient arrangement of spheres, starts with the structure shown in Figure 11.20(a), which we call layer A. Figure 11.20 (a) In a close- packed layer, each sphere is in contact with six others. (b) Spheres in the second layer fit into the depressions between the first-layer spheres. (c) In the hexagonal close-packed structure, each third-layer sphere is directly over a first-layer sphere. (d) In the cubic close-packed (a) structure, each third-layer sphere fits into a depression that is directly over a depression in the first layer. (b) (c) (d)  11.4 Crystal Structure 481 Focusing on the only enclosed sphere, we see that it has six immediate neighbors in that layer. In the second layer (which we call layer B), spheres are packed into the depressions between the spheres in the first layer so that all the spheres are as close together as possible [Figure 11.20(b)]. There are two ways that a third-layer sphere may cover the second layer to achieve closest packing. The spheres may fit into the depressions so that each third-layer sphere is directly over a first-layer sphere [Figure 11.20(c)]. Because there is no difference between the arrangement of the first and third layers, we also call the third layer layer A. Alternatively, the third-layer spheres may fit into the depressions that lie directly over the depressions in the first layer [Figure 11.20(d)]. In this case, we call the third layer layer C. Figure 11.21 shows the “exploded views” and the structures resulting from these two arrangements. The ABA arrangement is known as the hexagonal close- packed (hcp) structure, and the ABC arrangement is the cubic close-packed (ccp) struc- ture, which corresponds to the face-centered cube already described. Note that in the hcp structure, the spheres in every other layer occupy the same vertical position (ABA- BAB. . .), while in the ccp structure, the spheres in every fourth layer occupy the same vertical position (ABCABCA. . .). In both structures, each sphere has a coordination number of 12 (each sphere is in contact with six spheres in its own layer, three spheres in the layer above, and three spheres in the layer below). Both the hcp and ccp struc- tures represent the most efficient way of packing identical spheres in a unit cell, and there is no way to increase the coordination number to beyond 12. Many metals and noble gases, which are monatomic, form crystals with hcp or ccp structures. For example, magnesium, titanium, and zinc crystallize with their atoms in a hcp array, while aluminum, nickel, and silver crystallize in the ccp arrangement. All solid noble gases have the ccp structure except helium, which Figure 11.21 Exploded views of (a) a hexagonal close-packed structure and (b) a cubic close- packed structure. The arrow is tilted to show the face-centered cubic unit cell more clearly. Note that this arrangement is the same as the face-centered unit cell. Exploded view Hexagonal close-packed structure (a) Exploded view Cubic close-packed structure (b) 482 Chapter 11 ■ Intermolecular Forces and Liquids and Solids c b b a a a r r r scc bcc fcc a = 2r b2 = a2 + a2 b = 4r c2 = a2 + b2 b2 = a2 + a2 = 3a 2 16r 2 = 2a 2 c = √3a = 4r a = √8r a = 4r √3 Figure 11.22 The relationship between the edge length (a) and radius (r) of atoms in the simple cubic cell, body-centered cubic cell, and face-centered cubic cell. crystallizes in the hcp structure. It is natural to ask why a series of related sub- stances, such as the transition metals or the noble gases, would form different crys- tal structures. The answer lies in the relative stability of a particular crystal structure, which is governed by intermolecular forces. Thus, magnesium metal has the hcp structure because this arrangement of Mg atoms results in the greatest stability of the solid. Figure 11.22 summarizes the relationship between the atomic radius r and the edge length a of a simple cubic cell, a body-centered cubic cell, and a face-centered cubic cell. This relationship can be used to determine the atomic radius of a sphere if the density of the crystal is known, as Example 11.3 shows. Example 11.3 Gold (Au) crystallizes in a cubic close-packed structure (the face-centered cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of gold in picometers. Strategy We want to calculate the radius of a gold atom. For a face-centered cubic unit cell, the relationship between radius (r) and edge length (a), according to Figure 11.22, is a 5 18r. Therefore, to determine r of a Au atom, we need to find a. The volume of a 3 cube is V 5 a3 or a 5 1 V . Thus, if we can determine the volume of the unit cell, we can calculate a. We are given the density in the problem. need to find o mass density   given p volume r want to calculate The sequence of steps is summarized as follows: density of volume of edge length radius of — — — unit cell unit cell of unit cell Au atom (Continued)  11.5 X-Ray Diffraction by Crystals 483 Solution Step 1: We know the density, so in order to determine the volume, we find the mass of the unit cell. Each unit cell has eight corners and six faces. The total number of atoms within such a cell, according to Figure 11.19, is 1 1 a8 3 b 1 a6 3 b 5 4 8 2 The mass of a unit cell in grams is 4 atoms 1 mol 197.0 g Au m5 3 23 3 1 unit cell 6.022 3 10 atoms 1 mol Au 5 1.31 3 10221 g/unit cell From the definition of density (d 5 m/V), we calculate the volume of the unit Remember that density is an intensive cell as follows: property, so it is the same for one unit cell and 1 cm3 of the substance. m 1.31 3 10221 g V5 5 5 6.79 3 10223 cm3 d 19.3 g/cm3 Step 2: Because volume is length cubed, we take the cubic root of the volume of the unit cell to obtain the edge length (a) of the cell 3 a5 2 V 3 52 6.79 3 10223 cm3 5 4.08 3 1028 cm Step 3: From Figure 11.22 we see that the radius of an Au sphere (r) is related to the edge length by a 5 28 r Therefore, a 4.08 3 1028 cm r5 5 28 28 5 1.44 3 1028 cm 1 3 10 22 m 1 pm 5 1.44 3 10 28 cm 3 3 1 cm 1 3 10 212 m 5 144 pm Similar problem: 11.39. Practice Exercise When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 408.7 pm. Calculate the density of silver. Review of Concepts Tungsten crystallizes in a body-centered cubic lattice (the W atoms occupy only the lattice points). How many W atoms are present in a unit cell? 11.5 X-Ray Diffraction by Crystals Virtually all we know about crystal structure has been learned from X-ray diffraction studies. X-ray diffraction refers to the scattering of X rays by the units of a crystalline solid. The scattering, or diffraction, patterns produced are used to deduce the arrangement of particles in the solid lattice. 484 Chapter 11 ■ Intermolecular Forces and Liquids and Solids In Section 10.6 we discussed the interference phenomenon associated with waves (see Figure 10.22). Because X rays are one form of electromagnetic radia- tion, and therefore waves, we would expect them to exhibit such behavior under suitable conditions. In 1912 the German physicist Max von Laue† correctly sug- gested that, because the wavelength of X rays is comparable in magnitude to the distances between lattice points in a crystal, the lattice should be able to diffract X rays. An X-ray diffraction pattern is the result of interference in the waves asso- ciated with X rays. Figure 11.23 shows a typical X-ray diffraction setup. A beam of X rays is directed at a mounted crystal. Atoms in the crystal absorb some of the incoming radiation and then reemit it; the process is called the scattering of X rays. To understand how a diffraction pattern may be generated, consider the scat- tering of X rays by atoms in two parallel planes (Figure 11.24). Initially, the two incident rays are in phase with each other (their maxima and minima occur at the same positions). The upper wave is scattered, or reflected, by an atom in the first layer, while the lower wave is scattered by an atom in the second layer. In order for these two scattered waves to be in phase again, the extra distance traveled by the lower wave must be an integral multiple of the wavelength (λ) of the X ray; that is, BC 1 CD 5 2d sin θ 5 nλ n 5 1, 2, 3, . . . or 2d sin θ 5 nλ (11.1) where θ is the angle between the X rays and the plane of the crystal and d is the distance between adjacent planes. Equation (11.1) is known as the Bragg equation † Max Theodor Felix von Laue (1879–1960). German physicist von Laue received the Nobel Prize in Physics in 1914 for his discovery of X-ray diffraction. Shield Crystal X-ray beam Photographic plate X-ray tube (a) (b) Figure 11.23 (a) An arrangement for obtaining the X-ray diffraction pattern of a crystal. The shield prevents the strong undiffracted X rays from damaging the photographic plate. (b) X-ray diffraction pattern of crystalline lysozyme, a protein. The white "L" is a shadow of the sample holder and shield.  11.5 X-Ray Diffraction by Crystals 485 Incident rays Reflected rays Figure 11.24 Reflection of X rays from two layers of atoms. The lower wave travels a distance 2d sin θ longer than the upper wave does. For the two waves to be in phase again after reflection, it must be true that 2d sin θ 5 nλ, where λ is the wavelength of the A X ray and n 5 1, 2, 3. . . . The sharply θ θ defined spots in Figure 11.23 are observed only if the crystal is large θ θ enough to consist of hundreds of parallel layers. d B D d sin θ C d sin θ after William H. Bragg† and Sir William L. Bragg.‡ The reinforced waves produce a Reinforced waves are waves that have interacted constructively (see Figure 10.22). dark spot on a photographic film for each value of θ that satisfies the Bragg equation. Example 11.4 illustrates the use of Equation (11.1). Example 11.4 X rays of wavelength 0.154 nm strike an aluminum crystal; the rays are reflected at an angle of 19.3°. Assuming that n 5 1, calculate the spacing between the planes of aluminum atoms (in pm) that is responsible for this angle of reflection. The conversion factor is obtained from 1 nm 5 1000 pm. Strategy This is an application of Equation (11.1). Solution Converting the wavelength to picometers and using the angle of reflection (19.3°), we write nλ λ d5 5 2 sin θ 2 sin θ 1000 pm 0.154 nm 3 1 nm 5 2 sin 19.3° 5 233 pm Similar problems: 11.47, 11.48. Practice Exercise X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of 14.17°. Assuming that n 5 1, calculate the distance (in pm) between layers in the crystal. The X-ray diffraction technique offers the most accurate method for determining bond lengths and bond angles in molecules in the solid state. Because X rays are scattered by electrons, chemists can construct an electron-density contour map from the diffraction patterns by using a complex mathematical procedure. Basically, an electron-density con- tour map tells us the relative electron densities at various locations in a molecule. The densities reach a maximum near the center of each atom. In this manner, we can determine the positions of the nuclei and hence the geometric parameters of the molecule. † William Henry Bragg (1862–1942). English physicist. Bragg’s work was mainly in X-ray crystallography. He shared the Nobel Prize in Physics with his son Sir William Bragg in 1915. ‡ Sir William Lawrence Bragg (1890–1972). English physicist. Bragg formulated the fundamental equation for X-ray diffraction and shared the Nobel Prize in Physics with his father in 1915. 486 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Review of Concepts Why can the X-ray diffraction technique not be used to study molecular structure in a liquid? 11.6 Types of Crystals The structures and properties of crystals, such as melting point, density, and hardness, are determined by the kinds of forces that hold the particles together. We can classify any crystal as one of four types: ionic, covalent, molecular, or metallic. 564 pm Figure 11.25 Relation between Ionic Crystals the radii of Na1 and Cl2 ions and Ionic crystals have two important characteristics: (1) They are composed of charged the unit cell dimensions. Here the cell edge length is equal to twice species and (2) anions and cations are generally quite different in size. Knowing the the sum of the two ionic radii. radii of the ions is helpful in understanding the structure and stability of these com- pounds. There is no way to measure the radius of an individual ion, but sometimes it is possible to come up with a reasonable estimate. For example, if we know the radius of I2 in KI is about 216 pm, we can determine the radius of K1 ion in KI, and from that, the radius of Cl2 in KCl, and so on. The ionic radii in Figure 8.9 are average values derived from many different compounds. Let us consider the NaCl crystal, which has a face-centered cubic lattice (see Figure 2.13). Figure 11.25 shows that the edge length of the unit cell of NaCl is twice the sum of the ionic radii of Na1 and Cl2. Using the values given in Figure 8.9, we calculate the edge length to be 2(95 1 181) pm, or 552 pm. But the edge length shown in Figure 11.25 was determined by X-ray diffraction to be 564 pm. The discrepancy between these two values tells us that the radius of an ion actually varies slightly from one compound to another. Figure 11.26 shows the crystal structures of three ionic compounds: CsCl, ZnS, These giant potassium dihydrogen and CaF2. Because Cs1 is considerably larger than Na1, CsCl has the simple cubic phosphate crystals were grown in the laboratory. The largest one lattice. ZnS has the zincblende structure, which is based on the face-centered cubic weighs 701 lb! lattice. If the S22 ions occupy the lattice points, the Zn21 ions are located one-fourth of the distance along each body diagonal. Other ionic compounds that have the zincblende structure include CuCl, BeS, CdS, and HgS. CaF2 has the fluorite struc- ture. The Ca21 ions occupy the lattice points, and each F2 ion is tetrahedrally sur- rounded by four Ca21 ions. The compounds SrF2, BaF2, BaCl2, and PbF2 also have the fluorite structure. (a) (b) (c) Figure 11.26 Crystal structures of (a) CsCl, (b) ZnS, and (c) CaF2. In each case, the cation is the smaller sphere.  11.6 Types of Crystals 487 Examples 11.5 and 11.6 show how to calculate the number of ions in and the density of a unit cell. Example 11.5 How many Na1 and Cl2 ions are in each NaCl unit cell? Solution NaCl has a structure based on a face-centered cubic lattice. As Figure 2.13 shows, one whole Na1 ion is at the center of the unit cell, and there are twelve Na1 ions at the edges. Because each edge Na1 ion is shared by four unit cells [see Figure 11.19(b)], the total number of Na1 ions is 1 1 (12 3 14 ) 5 4. Similarly, there are six Cl2 ions at Cl– Na+ the face centers and eight Cl2 ions at the corners. Each face-centered ion is shared by two unit cells, and each corner ion is shared by eight unit cells [see Figures 11.19(a) and (c)], Figure 11.27 Portions of Na1 so the total number of Cl2 ions is (6 3 12 ) 1 (8 3 18 ) 5 4. Thus, there are four Na1 ions and Cl2 ions within a face-centered cubic unit cell. and four Cl2 ions in each NaCl unit cell. Figure 11.27 shows the portions of the Na1 and Cl2 ions within a unit cell. Check This result agrees with sodium chloride’s empirical formula. Similar problem: 11.41. Practice Exercise How many atoms are in a body-centered cube, assuming that all atoms occupy lattice points? Example 11.6 The edge length of the NaCl unit cell is 564 pm. What is the density of NaCl in g/cm3? Strategy To calculate the density, we need to know the mass of the unit cell. The volume can be calculated from the given edge length because V 5 a3. How many Na1 and Cl2 ions are in a unit cell? What is the total mass in amu? What are the conversion factors between amu and g and between pm and cm? Solution From Example 11.5 we see that there are four Na1 ions and four Cl2 ions in each unit cell. So the total mass (in amu) of a unit cell is mass 5 4(22.99 amu 1 35.45 amu) 5 233.8 amu Converting amu to grams, we write 1g 233.8 amu 3 5 3.882 3 10222 g 6.022 3 1023 amu The volume of the unit cell is V 5 a3 5 (564 pm)3. Converting pm3 to cm3, the volume is given by 1 3 10212 m 3 1 cm 3 V 5 (564 pm) 3 3 a b 3a b 5 1.794 3 10222 cm3 1 pm 1 3 1022 m Finally, from the definition of density mass 3.882 3 10222 g density 5 5 volume 1.794 3 10222 cm3 3 5 2.16 g/cm Similar problem: 11.42. Practice Exercise Copper crystallizes in a face-centered cubic lattice (the Cu atoms are at the lattice points only). If the density of the metal is 8.96 g/cm3, what is the unit cell edge length in pm? CHEMISTRY in Action High-Temperature Superconductors M etals such as copper and aluminum are good conduc- tors of electricity, but they do possess some electrical resistance. In fact, up to about 20 percent of electrical energy substances, called superconductors, for transmission of electric power because the cost of maintaining electrical cables at such low temperatures is prohibitive and would far exceed the sav- may be lost in the form of heat when cables made of these ings from more efficient electricity transmission. metals are used to transmit electricity. Wouldn’t it be marvel- In 1986 two physicists in Switzerland discovered a new ous if we could produce cables that possessed no electrical class of materials that are superconducting at around 30 K. resistance? Although 30 K is still a very low temperature, the improvement Actually it has been known for over a century that certain over the 4 K range was so dramatic that their work generated metals and alloys, when cooled to very low temperatures immense interest and triggered a flurry of research activity. (around the boiling point of liquid helium, or 4 K), lose their Within months, scientists synthesized compounds that are resistance totally. However, it is not practical to use these superconducting around 95 K, which is well above the boiling Cu O Y Ba Crystal structure of YBa2Cu3Ox (x 5 6 or 7). Because some of the O atom The levitation of a magnet above a high-temperature superconductor sites are vacant, the formula is not constant. immersed in liquid nitrogen. Most ionic crystals have high melting points, an indication of the strong cohesive forces holding the ions together. A measure of the stability of ionic crystals is the lattice energy (see Section 9.3); the higher the lattice energy, the more stable the compound. These solids do not conduct electricity because the ions are fixed in position. However, in the molten state (that is, when melted) or dissolved in water, the ions are free to move and the resulting liquid is electrically conducting. 488 point of liquid nitrogen (77 K). The figure on p. 488 shows the has several advantages as a high-temperature superconductor. crystal structure of one of these compounds, a mixed oxide of First, it is an inexpensive compound (about $2 per gram) so yttrium, barium, and copper with the formula YBa2Cu3Ox large quantities are available for testing. Second, the mecha- (where x 5 6 or 7). The accompanying figure shows a magnet nism of superconductivity in MgB2 is similar to the well- being levitated above such a superconductor, which is immersed understood metal alloy superconductors at 4 K. Third, it is in liquid nitrogen. much easier to fabricate this compound; that is, to make it into Despite the initial excitement, this class of high-temperature wires or thin films. With further research effort, it is hoped that superconductors has not fully lived up to its promise. After more someday soon different types of high-temperature supercon- than 30 years of intense research and development, scientists still ductors will be used to build supercomputers, whose speeds puzzle over how and why these compounds superconduct. It has are limited by how fast electric current flows, more powerful also proved difficult to make wires of these compounds, and other particle accelerators, efficient devices for nuclear fusion, and technical problems have limited their large-scale commercial more accurate magnetic resonance imaging (MRI) machines applications thus far. for medical use. The progress in high-temperature supercon- In another encouraging development, in 2001 scientists in ductors is just warming up! Japan discovered that magnesium diboride (MgB2) becomes superconducting at about 40 K. Although liquid neon (b.p. 27 K) must be used as coolant instead of liquid nitrogen, it is still much cheaper than using liquid helium. Magnesium diboride Crystal structure of MgB2. The Mg atoms (blue) form a hexagonal layer, An experimental levitation train that operates on superconducting material while the B atoms (gold) form a graphitelike honeycomb layer. at the temperature of liquid helium. Review of Concepts O2⫺ Shown here is a zinc oxide unit cell. How many Zn21 and O22 ions are in the unit cell? What is the formula of the compound? Zn2⫹ 489 490 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.28 (a) The structure of diamond. Each carbon is tetrahedrally bonded to four other carbon atoms. (b) The structure of graphite. The distance between successive layers is 335 pm. 335 pm (a) (b) Covalent Crystals In covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds. Well-known examples are the two allotropes of carbon: diamond and graphite (see Figure 8.17). In diamond, each carbon atom The central electrode in flashlight batter- is sp3-hybridized; it is bonded to four other atoms (Figure 11.28). The strong ies is made of graphite. covalent bonds in three dimensions contribute to diamond’s unusual hardness (it is the hardest material known) and very high melting point (3550°C). In graphite, carbon atoms are arranged in six-membered rings. The atoms are all sp2-hybridized; each atom is covalently bonded to three other atoms. The remaining unhybridized 2p orbital is used in pi bonding. In fact, each layer of graphite has the kind of delocalized molecular orbital that is present in benzene (see Section 10.8). Because electrons are free to move around in this extensively delocalized molec- ular orbital, graphite is a good conductor of electricity in directions along the planes of carbon atoms. The layers are held together by weak van der Waals forces. The covalent bonds in graphite account for its hardness; however, because the layers can slide over one another, graphite is slippery to the touch and is effective as a lubricant. It is also used in pencils and in ribbons made for com- puter printers and typewriters. Another covalent crystal is quartz (SiO2). The arrangement of silicon atoms in quartz is similar to that of carbon in diamond, but in quartz there is an oxygen atom between each pair of Si atoms. Because Si and O have different electronegativities, the Si¬O bond is polar. Nevertheless, SiO2 is similar to diamond in many respects, Quartz. such as hardness and high melting point (1610°C). Molecular Crystals In a molecular crystal, the lattice points are occupied by molecules, and the attractive forces between them are van der Waals forces and/or hydrogen bonding. An example of a molecular crystal is solid sulfur dioxide (SO2), in which the predominant attractive force is a dipole-dipole interaction. Intermolecular hydrogen bonding is mainly responsible for maintaining the three-dimensional lattice of ice (see Figure 11.12). Other examples of molecular crystals are I2, P4, and S8. In general, except in ice, molecules in molecular crystals are packed together as closely as their size and shape allow. Because van der Waals forces and hydrogen bonding are generally quite weak compared with covalent and ionic bonds, molecular crystals are more easily broken apart than ionic and covalent crystals. Indeed, most Sulfur. molecular crystals melt at temperatures below 100°C.  11.6 Types of Crystals 491 1 18 1A 8A 2 Hexagonal Body-centered 13 14 15 16 17 2A close-packed cubic 3A 4A 5A 6A 7A Li Be Face-centered Other structures cubic (see caption) Na Mg 3 4 5 6 7 8 9 10 11 12 Al 3B 4B 5B 6B 7B 8B 1B 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Figure 11.29 Crystal structures of metals. The metals are shown in their positions in the periodic table. Mn has a cubic structure, Ga an orthorhombic structure, In and Sn a tetragonal structure, and Hg a rhombohedral structure (see Figure 11.15). Metallic Crystals In a sense, the structure of metallic crystals is the simplest because every lattice point in a crystal is occupied by an atom of the same metal. Metallic crystals are generally body-centered cubic, face-centered cubic, or hexagonal close-packed (Figure 11.29). Consequently, metallic elements are usually very dense. The bonding in metals is quite different from that in other types of crystals. In a metal, the bonding electrons are delocalized over the entire crystal. In fact, metal atoms in a crystal can be imagined as an array of positive ions immersed in a sea of delocalized valence electrons (Figure 11.30). The great cohesive force resulting from Figure 11.30 A cross section of delocalization is responsible for a metal’s strength. The mobility of the delocalized a metallic crystal. Each circled positive charge represents the electrons makes metals good conductors of heat and electricity. nucleus and inner electrons of Table 11.4 summarizes the properties of the four different types of crystals a metal atom. The gray area discussed. surrounding the positive metal ions indicates the mobile sea of valence electrons. Table 11.4 Types of Crystals and General Properties Type Force(s) Holding of Crystal the Units Together General Properties Examples Ionic Electrostatic attraction Hard, brittle, high melting point, NaCl, LiF, MgO, CaCO3 poor conductor of heat and electricity Covalent Covalent bond Hard, high melting point, poor C (diamond),† SiO2 (quartz) conductor of heat and electricity Molecular* Dispersion forces, dipole-dipole Soft, low melting point, poor Ar, CO2, I2, H2O, C12H22O11 forces, hydrogen bonds conductor of heat and electricity (sucrose) Metallic Metallic bond Soft to hard, low to high melting point, All metallic elements; for good conductor of heat and electricity example, Na, Mg, Fe, Cu *Included in this category are crystals made up of individual atoms. † Diamond is a good thermal conductor. CHEMISTRY in Action And All for the Want of a Button I n June 1812, Napoleon’s mighty army, some 600,000 strong, marched into Russia. By early December, however, his forces were reduced to fewer than 10,000 men. An intriguing theory true, could be paraphrased in the Old English Nursery Rhyme: “And all for the want of a button.” for Napoleon’s defeat has to do with the tin buttons on his sol- diers’ coats! Tin has two allotropic forms called α (gray tin) and β (white tin). White tin, which has a cubic structure and a shiny metallic appearance, is stable at room temperature and above. Below 138C, it slowly changes into gray tin. The ran- dom growth of the microcrystals of gray tin, which has a tetragonal structure, weakens the metal and makes it crumble. Thus, in the severe Russian winter, the soldiers were probably more busy holding their coats together with their hands than carrying weapons. Actually, the so-called “tin disease” has been known for centuries. In the unheated cathedrals of medieval Europe, organ pipes made of tin were found to crumble as a result of the allo- tropic transition from white tin to gray tin. It is puzzling, there- fore, that Napoleon, a great believer in keeping his troops fit for Is Napoleon trying to instruct his soldiers how to keep their battle, would permit the use of tin for buttons. The tin story, if coats tight? 11.7 Amorphous Solids Solids are most stable in crystalline form. However, if a solid is formed rapidly (for example, when a liquid is cooled quickly), its atoms or molecules do not have time to align themselves and may become locked in positions other than those of a regular crystal. The resulting solid is said to be amorphous. Amorphous solids, such as glass, lack a regular three-dimensional arrangement of atoms. In this section, we will dis- cuss briefly the properties of glass. Glass is one of civilization’s most valuable and versatile materials. It is also one of the oldest—glass articles date back as far as 1000 b.c. Glass commonly refers to an optically transparent fusion product of inorganic materials that has cooled to a rigid state without crystallizing. By fusion product we mean that the glass is formed by mixing molten silicon dioxide (SiO2), its chief component, with com- pounds such as sodium oxide (Na2O), boron oxide (B2O3), and certain transition metal oxides for color and other properties. In some respects glass behaves more like a liquid than a solid. X-ray diffraction studies show that glass lacks long-range periodic order. There are about 800 different types of glass in common use today. Figure 11.31 shows two-dimensional schematic representations of crystalline quartz and amorphous quartz glass. Table 11.5 shows the composition and properties of quartz, Pyrex, and soda-lime glass. The color of glass is due largely to the presence of metal ions (as oxides). For example, green glass contains iron(III) oxide, Fe2O3, or copper(II) oxide, CuO; 492  11.8 Phase Changes 493 Figure 11.31 Two-dimensional representation of (a) crystalline quartz and (b) noncrystalline quartz glass. The small spheres represent silicon. In reality, the structure of quartz is three- dimensional. Each Si atom is tetrahedrally bonded to four O atoms. (a) (b) Table 11.5 Composition and Properties of Three Types of Glass Name Composition Properties and Uses Pure quartz glass 100% SiO2 Low thermal expansion, transparent to wide range of wavelengths. Used in optical research. Pyrex glass SiO2, 60–80% Low thermal expansion; transparent to visible and infrared, B2O3, 10–25% but not to UV radiation. Used mainly in laboratory and Al2O3, small amount household cooking glassware. Soda-lime glass SiO2, 75% Easily attacked by chemicals and sensitive to thermal shocks. Na2O, 15% Transmits visible light, but absorbs UV radiation. CaO, 10% Used mainly in windows and bottles. yellow glass contains uranium(IV) oxide, UO2; blue glass contains cobalt(II) and copper(II) oxides, CoO and CuO; and red glass contains small particles of gold and copper. Note that most of the ions mentioned here are derived from the transi- tion metals. 11.8 Phase Changes The discussions in Chapter 5 and in this chapter have given us an overview of the properties of the three phases of matter: gas, liquid, and solid. Phase changes, trans- formations from one phase to another, occur when energy (usually in the form of heat) is added or removed from a substance. Phase changes are physical changes character- ized by changes in molecular order; molecules in the solid phase have the greatest order, and those in the gas phase have the greatest randomness. Keeping in mind the relationship between energy change and the increase or decrease in molecular order will help us understand the nature of these physical changes. 494 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Figure 11.32 Kinetic energy distribution curves for molecules in a liquid (a) at a temperature T1 and Number of molecules Number of molecules (b) at a higher temperature T2. Note that at the higher temperature the curve flattens out. The shaded areas represent the number of molecules possessing T1 kinetic energy equal to or greater than a certain kinetic energy E1. T2 The higher the temperature, the greater the number of molecules with high kinetic energy. E1 E1 Kinetic energy E Kinetic energy E (a) (b) Liquid-Vapor Equilibrium Molecules in a liquid are not fixed in a rigid lattice. Although they lack the total freedom of gaseous molecules, these molecules are in constant motion. Because liq- uids are denser than gases, the collision rate among molecules is much higher in the liquid phase than in the gas phase. When the molecules in a liquid have sufficient energy to escape from the surface a phase change occurs. Evaporation, or vaporiza- tion, is the process in which a liquid is transformed into a gas. How does evaporation depend on temperature? Figure 11.32 shows the kinetic energy distribution of molecules in a liquid at two different temperatures. As we can see, the higher the temperature, the greater the kinetic energy, and hence more mol- ecules leave the liquid. Vapor Pressure The difference between a gas and a When a liquid evaporates, its gaseous molecules exert a vapor pressure. Consider vapor is explained on p. 174. the apparatus shown in Figure 11.33. Before the evaporation process starts, the mercury levels in the U-shaped manometer tube are equal. As soon as some mol- ecules leave the liquid, a vapor phase is established. The vapor pressure is measur- able only when a fair amount of vapor is present. The process of evaporation does not continue indefinitely, however. Eventually, the mercury levels stabilize and no further changes are seen. Figure 11.33 Apparatus for measuring the vapor pressure of a liquid. (a) Initially the liquid is frozen so there are no molecules Vacuum in the vapor phase. (b) On heating, a liquid phase is formed and evaporization begins. At Empty h equilibrium, the number of space molecules leaving the liquid is equal to the number of molecules returning to the liquid. The difference in the mercury levels (h) gives the equilibrium vapor pressure of the liquid at the Frozen liquid Liquid specified temperature. (a) (b)  11.8 Phase Changes 495 What happens at the molecular level during evaporation? In the beginning, the traffic is only one way: Molecules are moving from the liquid to the empty space. Soon the molecules in the space above the liquid establish a vapor phase. As the concentration of molecules in the vapor phase increases, some molecules condense, that is, they return to the liquid phase. Condensation, the change from the gas phase to the liquid phase, occurs because a molecule strikes the liquid surface and becomes trapped by intermolecular forces in the liquid. The rate of evaporation is constant at any given temperature, and the rate of condensation increases with the increasing concentration of molecules in the vapor phase. A state of dynamic equilibrium, in which the rate of a forward process is exactly balanced by the rate of the reverse process, is reached when the rates of condensation and evaporation become equal (Figure 11.34). The Equilibrium vapor pressure is indepen- dent of the amount of liquid as long as equilibrium vapor pressure is the vapor pressure measured when a dynamic there is some liquid present. equilibrium exists between condensation and evaporation. We often use the sim- pler term “vapor pressure” when we talk about the equilibrium vapor pressure Animation Equilibrium Vapor Pressure of a liquid. This practice is acceptable as long as we know the meaning of the abbreviated term. It is important to note that the equilibrium vapor pressure is the maximum vapor pressure of a liquid at a given temperature and that it is constant at a constant tem- perature. (It is independent of the amount of liquid as long as there is some liquid present.) From the foregoing discussion we expect the vapor pressure of a liquid to increase with temperature. Plots of vapor pressure versus temperature for three dif- ferent liquids in Figure 11.35 confirm this expectation. Rate of Dynamic evaporation equilibrium Molar Heat of Vaporization and Boiling Point established Rate A measure of the strength of intermolecular forces in a liquid is the molar heat of vaporization (≤Hvap), defined as the energy (usually in kilojoules) required to vapor- ize 1 mole of a liquid. The molar heat of vaporization is directly related to the Rate of condensation strength of intermolecular forces that exist in the liquid. If the intermolecular attrac- tion is strong, it takes a lot of energy to free the molecules from the liquid phase and the molar heat of vaporization will be high. Such liquids will also have a low Time vapor pressure. Figure 11.34 Comparison of The previous discussion predicts that the equilibrium vapor pressure (P) of a the rates of evaporation and condensation as the system liquid should increase with increasing temperature, as shown in Figure 11.35. approaches equilibrium at Analysis of this behavior reveals that the quantitative relationship between the vapor constant temperature. Figure 11.35 The increase in vapor pressure with temperature for three 2 Diethyl ether Water Mercury liquids. The normal boiling points of the liquids (at 1 atm) are shown on the horizontal axis. The strong Vapor pressure (atm) metallic bonding in mercury results in a much lower vapor pressure of the liquid at room temperature. 1 –100 0 34.6 100 200 357 400 Temperature (°C) 496 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Table 11.6 Molar Heats of Vaporization for Selected Liquids Substance Boiling Point* (8C) DHvap (kJ/mol) Argon (Ar) 2186 6.3 Benzene (C6H6) 80.1 31.0 Diethyl ether (C2H5OC2H5) 34.6 26.0 Ethanol (C2H5OH) 78.3 39.3 Mercury (Hg) 357 59.0 Methane (CH4) 2164 9.2 Water (H2O) 100 40.79 *Measured at 1 atm. pressure P of a liquid and the absolute temperature T is given by the Clausius†- Clapeyron‡ equation ¢Hvap ln P 5 2 1C (11.2) RT where ln is the natural logarithm, R is the gas constant (8.314 J/K ? mol), and C is a constant. The Clausius-Clapeyron equation has the form of the linear equation y 5 mx 1 b: ¢Hvap 1 ln P 5 a2 ba b 1 C R T 4 4 4 4 y5 m x 1 b By measuring the vapor pressure of a liquid at different temperatures (see Figure 11.35) and plotting ln P versus 1/T, we determine the slope, which is equal to 2≤Hvap /R. (≤Hvap is assumed to be independent of temperature.) This method is used to determine heats of vaporization (Table 11.6). Figure 11.36 shows plots of ln P versus 1/T for water and diethyl ether. Note that the straight line for water has a steeper slope because water has a larger ≤Hvap. If we know the values of ≤Hvap and P of a liquid at one temperature, we can use the Clausius-Clapeyron equation to calculate the vapor pressure of the liquid at a dif- C2H5OC2H5 ferent temperature. At temperatures T1 and T2, the vapor pressures are P1 and P2. From Equation (11.2) we can write ¢ Hvap ln P ln P1 5 2 1C (11.3) RT1 H2O ¢ Hvap ln P2 5 2 1C (11.4) RT2 1/T † Rudolf Julius Emanuel Clausius (1822–1888). German physicist. Clausius’s work was mainly in electricity, Figure 11.36 Plots of ln P versus kinetic theory of gases, and thermodynamics. 1/T for water and diethyl ether. The ‡ slope in each case is equal to Benoit Paul Emile Clapeyron (1799–1864). French engineer. Clapeyron made contributions to the thermo- 2≤Hvap/R. dynamic aspects of steam engines.  11.8 Phase Changes 497 Subtracting Equation (11.4) from Equation (11.3) we obtain ¢Hvap ¢Hvap ln P1 2 ln P2 5 2 2 a2 b RT1 RT2 ¢Hvap 1 1 5 a 2 b R T2 T1 Hence, P1 ¢Hvap 1 1 ln 5 a 2 b P2 R T2 T1 or P1 ¢Hvap T1 2 T2 ln 5 a b (11.5) P2 R T1 T2 Example 11.7 illustrates the use of Equation (11.5). Example 11.7 Diethyl ether is a volatile, highly flammable organic liquid that is used mainly as a solvent. The vapor pressure of diethyl ether is 401 mmHg at 18°C. Calculate its vapor pressure at 32°C. Strategy We are given the vapor pressure of diethyl ether at one temperature and asked to find the pressure at another temperature. Therefore, we need Equation (11.5). C2H5OC2H5 Solution Table 11.6 tells us that ≤Hvap 5 26.0 kJ/mol. The data are P1 5 401 mmHg       P2 5 ? T1 5 18°C 5 291 K    T2 5 32°C 5 305 K From Equation (11.5) we have 401 26,000 J/mol 291 K 2 305 K ln 5 c d P2 8.314 J/K ? mol (291 K) (305 K) 5 20.493 Taking the antilog of both sides (see Appendix 4), we obtain 401 5 e20.493 5 0.611 P2 Hence P2 5 656 mmHg Check We expect the vapor pressure to be greater at the higher temperature. Therefore, the answer is reasonable. Similar problem: 11.84. Practice Exercise The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 63.5°C? (≤Hvap for ethanol is 39.3 kJ/mol.) A practical way to demonstrate the molar heat of vaporization is by rubbing an alcohol such as ethanol (C2H5OH) or isopropanol (C3H7OH), or rubbing alcohol, on your hands. These alcohols have a lower ≤Hvap than water, so that the heat from your 498 Chapter 11 ■ Intermolecular Forces and Liquids and Solids hands is enough to increase the kinetic energy of the alcohol molecules and evaporate them. As a result of the loss of heat, your hands feel cool. This process is similar to perspiration, which is one of the means by which the human body maintains a constant temperature. Because of the strong intermolecular hydrogen bonding that exists in water, a considerable amount of energy is needed to vaporize the water in perspiration from the body’s surface. This energy is supplied by the heat generated in various metabolic processes. You have already seen that the vapor pressure of a liquid increases with tem- perature. Every liquid has a temperature at which it begins to boil. The boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure. The normal boiling point of a liquid is the temperature at which it boils when the external pressure is 1 atm. At the boiling point, bubbles form within the liquid. When a bubble forms, the liquid originally occupying that space is pushed aside, and the level of the liquid in the container is forced to rise. The pressure exerted on the bubble is largely atmo- spheric pressure, plus some hydrostatic pressure (that is, pressure due to the presence of liquid). The pressure inside the bubble is due solely to the vapor pressure of the liquid. When the vapor pressure becomes equal to the external pressure, the bubble rises to the surface of the liquid and bursts. If the vapor pressure in the bubble were lower than the external pressure, the bubble would collapse before it could rise. We can thus conclude that the boiling point of a liquid depends on the external pressure. (We usually ignore the small contribution due to the hydrostatic pressure.) For exam- ple, at 1 atm, water boils at 100°C, but if the pressure is reduced to 0.5 atm, water boils at only 82°C. Because the boiling point is defined in terms of the vapor pressure of the liquid, we expect the boiling point to be related to the molar heat of vaporization: The higher ≤Hvap, the higher the boiling point. The data in Table 11.6 roughly confirm our prediction. Ultimately, both the boiling point and ≤Hvap are determined by the strength of intermolecular forces. For example, argon (Ar) and methane (CH4), which have weak dispersion forces, have low boiling points and small molar heats of vaporization. Diethyl ether (C2H5OC2H5) has a dipole moment, and the dipole-dipole forces account for its moderately high boiling point and ≤Hvap. Both ethanol (C2H5OH) and water have strong hydrogen bonding, which accounts for their high boiling points and large ≤Hvap values. Strong metallic bonding causes mercury to have the highest boiling point and ≤Hvap of this group of liquids. Interestingly, the boiling point of benzene, which is nonpolar, is comparable to that of ethanol. Benzene has a high polarizability due to the distribution of its electrons in the delocalized pi molecular orbitals, and the dispersion forces among benzene molecules can be as strong as or even stronger than dipole-dipole forces and/or hydrogen bonds. Review of Concepts A student studies the ln P versus 1/T plots for two organic liquids: methanol (CH3OH) and dimethyl ether (CH3OCH3), such as those shown in Figure 11.36. The slopes are 22.32 3 103 K and 24.50 3 103 K, respectively. How should she assign the ≤Hvap values to these two compounds? Critical Temperature and Pressure The opposite of evaporation is condensation. In principle, a gas can be made to liquefy by either one of two techniques. By cooling a sample of gas we decrease the kinetic energy of its molecules, so that eventually molecules aggregate to form small drops  11.8 Phase Changes 499 (a) (b) (c) (d) Figure 11.37 The critical phenomenon of sulfur hexafluoride. (a) Below the critical temperature the clear liquid phase is visible. (b) Above the critical temperature the liquid phase has disappeared. (c) The substance is cooled just below its critical temperature. The fog represents the condensation of vapor. (d) Finally, the liquid phase reappears. of liquid. Alternatively, we can apply pressure to the gas. Compression reduces the average distance between molecules so that they are held together by mutual attrac- tion. Industrial liquefaction processes combine these two methods. Every substance has a critical temperature (Tc), above which its gas phase cannot be made to liquefy, no matter how great the applied pressure. This is also the highest temperature at which a substance can exist as a liquid. Putting it another way, above the critical temperature there is no fundamental distinction between a liquid and a gas—we simply have a fluid. Critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature. The existence of the critical temperature can be qualitatively explained as follows. Intermolecular attraction Intermolecular forces are independent of temperature; the kinetic energy of the is a finite quantity for any given substance and it is independent of temperature. Below molecules increases with temperature. Tc, this force is sufficiently strong to hold the molecules together (under some appro- priate pressure) in a liquid. Above Tc, molecular motion becomes so energetic that the molecules can break away from this attraction. Figure 11.37 shows what happens when sulfur hexafluoride is heated above its critical temperature (45.5°C) and then cooled down to below 45.5°C. Table 11.7 lists the critical temperatures and critical pressures of a number of common substances. The critical temperature of a substance reflects the strength of its intermolecular forces. Benzene, ethanol, mercury, and water, which have strong intermolecular forces, also have high critical temperatures compared with the other substances listed in the table. Liquid-Solid Equilibrium The transformation of liquid to solid is called freezing, and the reverse process is “Fusion” refers to the process of melting. Thus, a “fuse” breaks an electrical circuit called melting, or fusion. The melting point of a solid or the freezing point of a when a metallic strip melts due to the liquid is the temperature at which solid and liquid phases coexist in equilibrium. The heat generated by excessively high electrical current. normal melting (or freezing) point of a substance is the temperature at which a 500 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Critical Temperatures and Critical Pressures Table 11.7 of Selected Substances Substance Tc (8C) Pc (atm) Ammonia (NH3) 132.4 111.5 Argon (Ar) 2186 6.3 Benzene (C6H6) 288.9 47.9 Carbon dioxide (CO2) 31.0 73.0 Diethyl ether (C2H5OC2H5) 192.6 35.6 Ethanol (C2H5OH) 243 63.0 Mercury (Hg) 1462 1036 Methane (CH4) 283.0 45.6 Molecular hydrogen (H2) 2239.9 12.8 Molecular nitrogen (N2) 2147.1 33.5 Molecular oxygen (O2) 2118.8 49.7 Sulfur hexafluoride (SF6) 45.5 37.6 Water (H2O) 374.4 219.5 substance melts (or freezes) at 1 atm pressure. We generally omit the word “normal” when the pressure is at 1 atm. The most familiar liquid-solid equilibrium is that of water and ice. At 0°C and 1 atm, the dynamic equilibrium is represented by ice Δ water A practical illustration of this dynamic equilibrium is provided by a glass of ice water. As the ice cubes melt to form water, some of the water between ice cubes may freeze, thus joining the cubes together. This is not a true dynamic equilibrium, however, because the glass is not kept at 0°C; thus, all the ice cubes will eventually melt away. Figure 11.38 shows how the temperature of a substance changes as it absorbs heat from its surroundings. We see that as a solid is heated, its temperature increases until it reaches its melting point. At this temperature, the average kinetic energy of the molecules has become sufficiently large to begin overcoming the intermo- lecular forces that hold the molecules together in the solid state. A transition from the solid to liquid phase begins in which the absorption of heat is used to break apart more and more of the molecules in the solid. It is important to note that during this transition (A ¡ B) the average kinetic energy of the molecules does not change, so the temperature stays constant. Once the substance has completely melted, further absorption of heat increases its temperature until the boiling point is reached (B ¡ C). Here, the transition from the liquid to the gaseous phase occurs (C ¡ D) in which the absorbed heat is used to break the intermolecular forces holding the molecules in the liquid phase so the temperature again remains constant. Once this transition has been completed, the temperature of the gas increases on further heating. Molar heat of fusion (DHfus) is the energy (usually in kilojoules) required to melt 1 mole of a solid. Table 11.8 shows the molar heats of fusion for the sub- stances listed in Table 11.6. A comparison of the data in the two tables shows that for each substance ≤Hfus is smaller than ≤Hvap. This is consistent with the fact that molecules in a liquid are still fairly closely packed together, so that some  11.8 Phase Changes 501 Figure 11.38 A typical heating curve, from the solid phase through the liquid phase to the gas phase of a substance. Because ≤Hfus is smaller than ≤Hvap, a substance melts in less time than it takes to boil. This Vapor explains why AB is shorter than Boiling point CD. The steepness of the solid, Temperature C D liquid, and vapor heating lines is determined by the specific heat Liquid and vapor of the substance in each state. in equilibrium Melting point Solid and liquid Liquid in equilibrium A B Solid Time Table 11.8 Molar Heats of Fusion for Selected Substances Substance Melting Point* (8C) DHfus (kJ/mol) Argon (Ar) 2190 1.3 Benzene (C6H6) 5.5 10.9 Diethyl ether (C2H5OC2H5) 2116.2 6.90 Ethanol (C2H5OH) 2117.3 7.61 Mercury (Hg) 239 23.4 Methane (CH4) 2183 0.84 Water (H2O) 0 6.01 *Measured at 1 atm. energy is needed to bring about the rearrangement from solid to liquid. On the other hand, when a liquid evaporates, its molecules become completely separated from one another and considerably more energy is required to overcome the attrac- tive force. As we would expect, cooling a substance has the opposite effect of heating it. If we remove heat from a gas sample at a steady rate, its temperature decreases. As the liquid is being formed, heat is given off by the system, because its potential energy is decreasing. For this reason, the temperature of the system remains constant over the condensation period (D ¡ C). After all the vapor has condensed, the tempera- ture of the liquid begins to drop. Continued cooling of the liquid finally leads to freezing (B ¡ A). The phenomenon known as supercooling refers to the situation in which a liquid can be temporarily cooled to below its freezing point. Supercooling occurs when heat is removed from a liquid so rapidly that the molecules literally have no time to assume the ordered structure of a solid. A supercooled liquid is unstable; gentle stirring or the addition to it of a small “seed” crystal of the same substance will cause it to solidify quickly. 502 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Solid-Vapor Equilibrium Solids, too, undergo evaporation and, therefore, possess a vapor pressure. Consider the following dynamic equilibrium: solid Δ vapor Sublimation is the process in which molecules go directly from the solid into the vapor phase. Deposition is the reverse process, that is, molecules make the transition from vapor to solid directly. Naphthalene, which is the substance used to make mothballs, has a fairly high (equilibrium) vapor pressure for a solid (1 mmHg at 53°C); thus, its pungent vapor quickly permeates an enclosed space. Iodine also sublimes. Above room temperature, the violet color of iodine vapor is easily visible in a closed container. Because molecules are more tightly held in a solid, the vapor pressure of a solid is generally much less than that of the corresponding liquid. Molar heat of sublimation (DHsub) of a substance is the energy (usually in kilojoules) required to sublime 1 mole of a solid. It is equal to the sum of the molar heats of fusion and vaporization: Solid iodine in equilibrium with its vapor. ¢Hsub 5 ¢Hfus 1 ¢Hvap (11.6) Equation (11.6) is an illustration of Hess’s law (see Section 6.6). The enthalpy, or heat change, for the overall process is the same whether the substance changes directly from the solid to the vapor form or from the solid to the liquid and then to the vapor. Note that Equation (11.6) holds only if all the phase changes occur at the same tem- perature. If not, the equation can be used only as an approximation. Figure 11.39 summarizes the types of phase changes discussed in this section. When a substance is heated, its temperature will rise and eventually it will undergo a phase transition. To calculate the total energy change for such a process we must include all of the steps, shown in Example 11.8. Example 11.8 Gas Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water Condensation Vaporization from 0°C to 182°C. Assume that the specific heat of water is 4.184 J/g ? °C over the entire liquid range and that the specific heat of steam is 1.99 J/g ? °C. Strategy The heat change (q) at each stage is given by q 5 ms≤t (see p. 247), where m is the mass of water, s is the specific heat, and ≤t is the temperature change. If there Temperature Sublimation Deposition is a phase change, such as vaporization, then q is given by n≤Hvap, where n is the Liquid number of moles of water. Solution The calculation can be broken down in three steps. Freezing Melting Step 1: Heating water from 0°C to 100°C Using Equation (6.12) we write q1 5 ms¢t Solid 5 (346 g) (4.184 J/g ? °C) (100°C 2 0°C) 5 1.45 3 105 J 5 145 kJ Figure 11.39 The various phase changes that a substance can (Continued) undergo.  11.9 Phase Diagrams 503 Step 2: Evaporating 346 g of water at 100°C (a phase change) In Table 11.6 we see ≤Hvap 5 40.79 kJ/mol for water, so 1 mol H2O 40.79 kJ q2 5 346 g H2O 3 3 18.02 g H2O 1 mol H2O 5 783 kJ Step 3: Heating steam from 100°C to 182°C q3 5 ms¢t 5 (346 g) (1.99 J/g ? °C) (182°C 2 100°C) 5 5.65 3 104 J 5 56.5 kJ The overall energy required is given by qoverall 5 q1 1 q2 1 q3 5 145 kJ 1 783 kJ 1 56.5 kJ 5 985 kJ Check All the qs have a positive sign, which is consistent with the fact that heat is absorbed to raise the temperature from 0°C to 182°C. Also, as expected, much more heat is absorbed during the phase transition. Similar problem: 11.76. Practice Exercise Calculate the heat released when 68.0 g of steam at 124°C is converted to water at 45°C. 11.9 Phase Diagrams The overall relationships among the solid, liquid, and vapor phases are best repre- Animation Phase Diagrams and the States of Matter sented in a single graph known as a phase diagram. A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas. In this section we will briefly discuss the phase diagrams of water and carbon dioxide. Water Figure 11.40(a) shows the phase diagram of water. The graph is divided into three regions, each of which represents a pure phase. The line separating any two regions indicates conditions under which these two phases can exist in equilib- rium. For example, the curve between the liquid and vapor phases shows the variation of vapor pressure with temperature. (Compare this curve with Figure 11.35.) The other two curves similarly indicate conditions for equilibrium between ice and liquid water and between ice and water vapor. (Note that the solid-liquid boundary line has a negative slope.) The point at which all three curves meet is The negative slope for the solid-liquid boundary line is due to the fact that the called the triple point, which is the only condition under which all three phases molar volume of ice is greater than that can be in equilibrium with one another. For water, this point is at 0.01°C and of liquid water; hence water is denser than ice. An increase in pressure favors 0.006 atm. the liquid phase. Phase diagrams enable us to predict changes in the melting point and boiling point of a substance as a result of changes in the external pressure; we can also anticipate directions of phase transitions brought about by changes in temperature and pressure. The normal melting point and boiling point of water at 1 atm are 0°C and 504 Chapter 11 ■ Intermolecular Forces and Liquids and Solids 1 atm 1 atm Pressure Pressure Liquid Liquid Solid Solid 0.006 atm Vapor Vapor 0.01°C 0°C 100°C Temperature Temperature Decreased melting point Increased boiling point (a) (b) Figure 11.40 (a) The phase diagram of water. Each solid line between two phases specifies the conditions of pressure and temperature under which the two phases can exist in equilibrium. The point at which all three phases can exist in equilibrium (0.006 atm and 0.01°C) is called the triple point. (b) This phase diagram tells us that increasing the pressure on ice lowers its melting point and that increasing the pressure of liquid water raises its boiling point. 100°C, respectively. What would happen if melting and boiling were carried out at Liquid some other pressure? Figure 11.40(b) shows that increasing the pressure above 1 atm Solid will raise the boiling point and lower the melting point. A decrease in pressure will Pressure 5.2 atm lower the boiling point and raise the melting point. 1 atm Vapor Carbon Dioxide –78°C –57°C Temperature The phase diagram of carbon dioxide (Figure 11.41) is generally similar to that of water, with one important exception—the slope of the curve between solid and liquid Figure 11.41 The phase diagram of carbon dioxide. Note that the is positive. In fact, this holds true for almost all other substances. Water behaves dif- solid-liquid boundary line has a ferently because ice is less dense than liquid water. The triple point of carbon dioxide positive slope. The liquid phase is is at 5.2 atm and 257°C. not stable below 5.2 atm, so that only the solid and vapor phases An interesting observation can be made about the phase diagram in Figure 11.41. can exist under atmospheric As you can see, the entire liquid phase lies well above atmospheric pressure; therefore, conditions. it is impossible for solid carbon dioxide to melt at 1 atm. Instead, when solid CO2 is heated to 278°C at 1 atm, it sublimes. In fact, solid carbon dioxide is called dry ice because it looks like ice and does not melt (Figure 11.42). Because of this property, dry ice is useful as a refrigerant. Review of Concepts Which phase diagram corresponds to a substance that will sublime rather than melt as it is heated at 1 atm? 1 atm 1 atm 1 atm Figure 11.42 Under atmospheric conditions, solid carbon dioxide does not melt; it can only sublime. T T T The cold carbon dioxide gas causes nearby water vapor to (a) (b) (c) condense and form a fog. CHEMISTRY in Action Hard-Boiling an Egg on a Mountaintop, Pressure Cookers, and Ice Skating P hase equilibria are affected by external pressure. Depending on atmospheric conditions, the boiling point and freezing point of water may deviate appreciably from 1008C and 08C, ice decreases with increasing external pressure, as shown in Figure 11.40(b). This phenomenon helps to make ice skating possible. Because skates have very thin runners, a 130-lb respectively, as we see below. person can exert a pressure equivalent to 500 atm on the ice. (Remember that pressure is defined as force per unit area.) Consequently, at a temperature lower than 08C, the ice under Hard-Boiling an Egg on a Mountaintop the skates melts and the film of water formed under the run- Suppose you have just scaled Pike’s Peak in Colorado. To ner facilitates the movement of the skater over ice. help regain your strength following the strenuous work, you Calculations show that the melting point of ice decreases by decide to hard-boil an egg and eat it. To your surprise, water 7.4 3 1023 8C when the pressure increases by 1 atm. Thus, seems to boil more quickly than usual, but after 10 min in when the pressure exerted on the ice by the skater is 500 atm, boiling water, the egg is still not cooked. A little knowledge the melting point falls to 2(500 3 7.4 3 1023), or 23.78C. of phase equilibria could have saved you the disappointment Actually, it turns out that friction between the blades and the of cracking open an uncooked egg (especially if it is the only ice is the major cause for melting the ice. This explains why egg you brought with you). The summit of Pike’s Peak is it is possible to skate outdoors even when the temperature 14,000 ft above sea level. At this altitude, the atmospheric drops below 2208C. pressure is only about 0.6 atm. From Figure 11.40(b), we see that the boiling point of water decreases with decreasing pres- sure, so at the lower pressure water will boil at about 868C. However, it is not the boiling action but the amount of heat delivered to the egg that does the actual cooking, and the amount of heat delivered is proportional to the temperature of the water. For this reason, it would take considerably longer, perhaps 30 min, to hard-boil your egg. Pressure Cookers The effect of pressure on boiling point also explains why pres- sure cookers save time in the kitchen. A pressure cooker is a sealed container that allows steam to escape only when it exceeds a certain pressure. The pressure above the water in the cooker is the sum of the atmospheric pressure and the pressure of the steam. Consequently, the water in the pressure cooker will boil at a higher temperature than 1008C and the food in it will be hotter and cook faster. Ice Skating The pressure exerted by the skater on ice lowers its melting point, and the Let us now turn to the ice-water equilibrium. The negative film of water formed under the blades acts as a lubricant between the slope of the solid-liquid curve means that the melting point of skate and the ice. 505  CHEMISTRY in Action Liquid Crystals O rdinarily, there is a sharp distinction between the highly ordered state of a crystalline solid and the more ran- dom molecular arrangement of liquids. Crystalline ice and that the substance has the mechanical properties of a two- dimensional solid. Nematic liquid crystals are less ordered. Although the molecules in nematic liquid crystals are aligned liquid water, for example, differ from each other in this re- with their long axes parallel to one another, they are not sepa- spect. One class of substances, however, tends so greatly rated into layers. toward an ordered arrangement that a melting crystal first Thermotropic liquid crystals have many applications in sci- forms a milky liquid, called the paracrystalline state, with ence, technology, and medicine. The familiar black-and-white characteristically crystalline properties. At higher tempera- displays in timepieces and calculators are based on the properties tures, this milky fluid changes sharply into a clear liquid that of these substances. Transparent aligning agents made of tin behaves like an ordinary liquid. Such substances are known oxide (SnO2) applied to the lower and upper inside surfaces of the as liquid crystals. liquid crystal cell preferentially orient the molecules in the Molecules that exhibit liquid crystallinity are usually nematic phase by 908 relative to each other. In this way, the mol- long and rodlike. An important class of liquid crystals is called ecules become “twisted” through the liquid crystal phase. When thermotropic liquid crystals, which form when the solid is properly adjusted, this twist rotates the plane of polarization by heated. The two common structures of thermotropic liquid 908 and allows the light to pass through the two polarizers crystals are nematic and smectic. In smectic liquid crystals, (arranged at 908 to each other). When an electric field is applied, the long axes of the molecules are perpendicular to the plane the nematic molecules experience a torque (a torsion or rotation) of the layers. The layers are free to slide over one another so that forces them to align along the direction of the field. Now the Polarizer ⫹ Polarizer Electrode ⫺ Liquid crystal Glass plate Polarizer Mirror (a) (b) (c) A liquid crystal display (LCD) using nematic liquid crystals. Molecules in contact with the bottom and top cell surfaces are aligned at right angles to one another. (a) The extent of twist in the molecular orientation between the surfaces is adjusted so as to rotate the plane of polarized light by 908, allowing it to pass through the top polarizer. Consequently, the cell appears clear. (b) When the electric field is on, molecules orient along the direction of the field so the plane of polarized light can no longer pass through the top polarizer, and the cell appears black. (c) A cross section of a LCD such as that used in watches and calculators. 506 incident polarized light cannot pass through the top polarizer. In changes with temperature and therefore they are suitable for watches and calculators, a mirror is placed under the bottom use as sensitive thermometers. In metallurgy, for example, they polarizer. In the absence of an electric field, the reflected light are used to detect metal stress, heat sources, and conduction goes through both polarizers and the cell looks clear from the top. paths. Medically, the temperature of the body at specific sites When the electric field is turned on, the incident light from the top can be determined with the aid of liquid crystals. This tech- cannot pass through the bottom polarizer to reach the reflector nique has become an important diagnostic tool in treating in- and the cell becomes dark. Typically a few volts are applied fection and tumor growth (for example, breast tumors). Because across a nematic layer about 10 μm thick (1μm 5 1026 m). The localized infections and tumors increase metabolic rate and response time for molecules to align and relax when the electric hence temperature in the affected tissues, a thin film of liquid field is turned on and off is in the ms range (1 ms 5 1023 s). crystal can help a physician see whether an infection or tumor Another type of thermotropic liquid crystals is called cho- is present by responding to a temperature difference with a lesteric liquid crystals. The color of cholesteric liquid crystals change of color. Nematic Smectic The alignment of molecules in two types of liquid crystals. Nematic liquid crystals behave like a one-dimensional solid and smectic liquid crystals behave like a two-dimensional solid. A liquid crystal thermogram. The red color represents the highest tempera- ture and the blue color the lowest temperature. 507 508 Chapter 11 ■ Intermolecular Forces and Liquids and Solids Key Equations 2d sin θ 5 nλ (11.1) Bragg equation for calculating the distance between planes of atoms in a crystal lattice. ¢Hvap ln P 5 2 1 C (11.2) Clausius-Clapeyron equation for determining RT DHvap of a liquid. P1 ¢Hvap T1 2 T2 ln 5 a b (11.5) For calculating DHvap, vapor pressure, or P2 R T 1T 2 boiling point of a liquid. ¢Hsub 5 ¢Hfus 1 ¢Hvap (11.6) Application of Hess’s law. Summary of Facts & Concepts 1. All substances exist in one of three states: gas, liquid, absorbing substantial amounts of heat with only small or solid. The major difference between the condensed changes in the water temperature. state and the gaseous state is the distance separating 10. All solids are either crystalline (with a regular struc- molecules. ture of atoms, ions, or molecules) or amorphous (with- 2. Intermolecular forces act between molecules or between out a regular structure). Glass is an example of an molecules and ions. Generally, these attractive forces are amorphous solid. much weaker than bonding forces. 11. The basic structural unit of a crystalline solid is the unit 3. Dipole-dipole forces and ion-dipole forces attract cell, which is repeated to form a three-dimensional molecules with dipole moments to other polar mol- crystal lattice. X-ray diffraction has provided much of ecules or ions. our knowledge about crystal structure. 4. Dispersion forces are the result of temporary dipole 12. The four types of crystals and the forces that hold their moments induced in ordinarily nonpolar molecules. particles together are ionic crystals, held together by The extent to which a dipole moment can be induced ionic bonding; covalent crystals, covalent bonding; mo- in a molecule is called its polarizability. The term lecular crystals, van der Waals forces and/or hydrogen “van der Waals forces” refers to dipole-dipole, dipole- bonding; and metallic crystals, metallic bonding. induced dipole, and dispersion forces. 13. A liquid in a closed vessel eventually establishes a 5. Hydrogen bonding is a relatively strong dipole-dipole dynamic equilibrium between evaporation and con- interaction between a polar bond containing a hydrogen densation. The vapor pressure over the liquid under atom and an electronegative O, N, or F atom. Hydrogen these conditions is the equilibrium vapor pressure, bonds between water molecules are particularly strong. which is often referred to simply as “vapor pressure.” 6. Liquids tend to assume a geometry that minimizes surface 14. At the boiling point, the vapor pressure of a liquid area. Surface tension is the energy needed to expand a equals the external pressure. The molar heat of vapor- liquid surface area; strong intermolecular forces lead to ization of a liquid is the energy required to vaporize one greater surface tension. mole of the liquid. It can be determined by measuring 7. Viscosity is a measure of the resistance of a liquid to the vapor pressure of the liquid as a function of tem- flow; it decreases with increasing temperature. perature and using the Clausius-Clapeyron equation 8. Water molecules in the solid state form a three-dimensional [Equation (11.2)]. The molar heat of fusion of a solid is network in which each oxygen atom is covalently the energy required to melt one mole of the solid. bonded to two hydrogen atoms and is hydrogen-bonded 15. For every substance there is a temperature, called the to two hydrogen atoms. This unique structure accounts critical temperature, above which its gas phase cannot for the fact that ice is less dense than liquid water, a be made to liquefy. property that enables life to survive under the ice in 16. The relationships among the phases of a single sub- ponds and lakes in cold climates. stance are illustrated by a phase diagram, in which each 9. Water is also ideally suited for its ecological role by its region represents a pure phase and the boundaries high specific heat, another property imparted by its between the regions show the temperatures and pres- strong hydrogen bonding. Large bodies of water are sures at which the two phases are in equilibrium. At the able to moderate Earth’s climate by giving off and triple point, all three phases are in equilibrium.  Questions & Problems 509 Key Words Adhesion, p. 473 Deposition, p. 502 Intermolecular forces, p. 467 Phase changes, p. 493 Amorphous solid, p. 492 Dipole-dipole forces, p. 467 Intramolecular forces, p. 467 Phase diagram, p. 503 Boiling point, p. 498 Dispersion forces, p. 469 Ion-dipole forces, p. 468 Sublimation, p. 502 Closest packing, p. 480 Dynamic equilibrium, p. 495 Melting point, p. 499 Supercooling, p. 501 Cohesion, p. 473 Equilibrium vapor Molar heat of fusion Surface tension, p. 473 Condensation, p. 495 pressure, p. 495 (≤Hfus), p. 500 Triple point, p. 503 Coordination number, p. 479 Evaporation, p. 494 Molar heat of sublimation Unit cell, p. 477 Critical pressure (Pc ), p. 499 Freezing point, p. 499 (≤Hsub), p. 502 van der Waals forces, p. 467 Critical temperature Glass, p. 492 Molar heat of vaporization Vaporization, p. 494 (Tc ), p. 499 Hydrogen bond, p. 471 (≤Hvap), p. 495 Viscosity, p. 474 Crystalline solid, p. 477 Induced dipole, p. 468 Phase, p. 466 X-ray diffraction, p. 483 Questions & Problems • Problems available in Connect Plus 252°C. Explain the increase in boiling points from Red numbered problems solved in Student Solutions Manual CH4 to SnH4. 11.10 List the types of intermolecular forces that exist Intermolecular Forces between molecules (or basic units) in each of the Review Questions following species: (a) benzene (C6H6), (b) CH3Cl, (c) PF3, (d) NaCl, (e) CS2. 11.1 Give an example for each type of intermolecular 11.11 Ammonia is both a donor and an acceptor of hydrogen force. (a) dipole-dipole interaction, (b) dipole- in hydrogen-bond formation. Draw a diagram show- induced dipole interaction, (c) ion-dipole interaction, ing the hydrogen bonding of an ammonia molecule (d) dispersion forces, (e) van der Waals forces with two other ammonia molecules. 11.2 Explain the term “polarizability.” What kind of molecules tend to have high polarizabilities? What • 11.12 Which of the following species are capable of hydrogen-bonding among themselves? (a) C2H6, is the relationship between polarizability and inter- (b) HI, (c) KF, (d) BeH2, (e) CH3COOH molecular forces? 11.3 Explain the difference between a temporary dipole • 11.13 Arrange the following in order of increasing boiling point: RbF, CO2, CH3OH, CH3Br. Explain your rea- moment and the permanent dipole moment. soning. 11.4 Give some evidence that all atoms and molecules 11.14 Diethyl ether has a boiling point of 34.5°C, and exert attractive forces on one another. 1-butanol has a boiling point of 117°C: 11.5 What physical properties should you consider in comparing the strength of intermolecular forces in H H H H H H H H solids and in liquids? A A A A A A A A 11.6 Which elements can take part in hydrogen bond- HOCOCOOOCOCOH HOCOCOCOCOOH A A A A A A A A ing? Why is hydrogen unique in this kind of H H H H H H H H interaction? diethyl ether 1-butanol Problems Both of these compounds have the same numbers 11.7 The compounds Br2 and ICl have the same number and types of atoms. Explain the difference in their of electrons, yet Br2 melts at 27.2°C and ICl melts boiling points. at 27.2°C. Explain. • 11.15 Which member of each of the following pairs of 11.8 If you lived in Alaska, which of the following natu- substances would you expect to have a higher boil- ral gases would you keep in an outdoor storage tank ing point? (a) O2 and Cl2, (b) SO2 and CO2, in winter? Explain why. methane (CH4), propane (c) HF and HI (C3H8), or butane (C4H10) 11.16 Which substance in each of the following pairs 11.9 The binary hydrogen compounds of the Group 4A would you expect to have the higher boiling point? elements and their boiling points are: CH4, Explain why. (a) Ne or Xe, (b) CO2 or CS2, (c) CH4 2162°C; SiH4, 2112°C; GeH4, 288°C; and SnH4, or Cl2, (d) F2 or LiF, (e) NH3 or PH3  510 Chapter 11 ■ Intermolecular Forces and Liquids and Solids • 11.17 Explain in terms of intermolecular forces why (a) NH3 11.30 Outdoor water pipes have to be drained or insulated has a higher boiling point than CH4 and (b) KCl has a in winter in a cold climate. Why? higher melting point than I2. 11.18 What kind of attractive forces must be overcome in Problems order to (a) melt ice, (b) boil molecular bromine, (c) melt solid iodine, and (d) dissociate F2 into • 11.31 Predict which of the following liquids has greater surface tension: ethanol (C2H5OH) or dimethyl ether F atoms? (CH3OCH3). • 11.19 The following compounds have the same molecular 11.32 Predict the viscosity of ethylene glycol relative to formulas (C4H10). Which one would you expect to that of ethanol and glycerol (see Table 11.3). have a higher boiling point? CH2OOH A CH2OOH ethylene glycol Crystal Structure Review Questions 11.33 Define the following terms: crystalline solid, lattice 11.20 Explain the difference in the melting points of the point, unit cell, coordination number, closest packing. following compounds: 11.34 Describe the geometries of the following cubic cells: simple cubic, body-centered cubic, face-centered NO2 NO2 cubic. Which of these structures would give the A OH A highest density for the same type of atoms? Which E the lowest? 11.35 Classify the solid states in terms of crystal types of the A elements in the third period of the periodic table. Pre- OH dict the trends in their melting points and boiling m.p. 45⬚C m.p. 115⬚C points. • 11.36 The melting points of the oxides of the third-period (Hint: Only one of the two can form intramolecular elements are given in parentheses: Na2O (1275°C), hydrogen bonds.) MgO (2800°C), Al2O3 (2045°C), SiO2 (1610°C), P4O10 (580°C), SO3 (16.8°C), Cl2O7 (291.5°C). Classify these solids in terms of crystal types. Properties of Liquids Review Questions Problems 11.21 Explain why liquids, unlike gases, are virtually • 11.37 What is the coordination number of each sphere in incompressible. (a) a simple cubic cell, (b) a body-centered cubic 11.22 What is surface tension? What is the relationship cell, and (c) a face-centered cubic cell? Assume the between intermolecular forces and surface tension? spheres are all the same. How does surface tension change with temperature? 11.38 Calculate the number of spheres that would be found 11.23 Despite the fact that stainless steel is much denser within a simple cubic, a body-centered cubic, and a than water, a stainless-steel razor blade can be made face-centered cubic cell. Assume that the spheres to float on water. Why? are the same. 11.24 Use water and mercury as examples to explain • 11.39 Metallic iron crystallizes in a cubic lattice. The unit adhesion and cohesion. cell edge length is 287 pm. The density of iron is 7.87 g/cm3. How many iron atoms are within a unit 11.25 A glass can be filled slightly above the rim with cell? water. Explain why the water does not overflow. 11.40 Barium metal crystallizes in a body-centered cubic 11.26 Draw diagrams showing the capillary action of lattice (the Ba atoms are at the lattice points only). (a) water and (b) mercury in three tubes of differ- The unit cell edge length is 502 pm, and the density ent radii. of the metal is 3.50 g/cm3. Using this information, 11.27 What is viscosity? What is the relationship between calculate Avogadro’s number. [Hint: First calculate intermolecular forces and viscosity? the volume (in cm3) occupied by 1 mole of Ba atoms 11.28 Why does the viscosity of a liquid decrease with in the unit cells. Next calculate the volume (in cm3) increasing temperature? occupied by one Ba atom in the unit cell. Assume 11.29 Why is ice less dense than water? that 68% of the unit cell is occupied by Ba atoms.]  Questions & Problems 511 • 11.41 Vanadium crystallizes in a body-centered cubic lattice • 11.53 A solid is very hard and has a high melting point. (the V atoms occupy only the lattice points). How Neither the solid nor its melt conducts electricity. many V atoms are present in a unit cell? Classify the solid. • 11.42 Europium crystallizes in a body-centered cubic lattice 11.54 Which of the following are molecular solids and (the Eu atoms occupy only the lattice points). The den- which are covalent solids? Se8, HBr, Si, CO2, C, sity of Eu is 5.26 g/cm3. Calculate the unit cell edge P4O6, SiH4 length in pm. • 11.55 Classify the solid state of the following substances as • 11.43 Crystalline silicon has a cubic structure. The unit ionic crystals, covalent crystals, molecular crystals, cell edge length is 543 pm. The density of the solid or metallic crystals: (a) CO2, (b) B12, (c) S8, (d) KBr, is 2.33 g/cm3. Calculate the number of Si atoms in (e) Mg, (f) SiO2, (g) LiCl, (h) Cr. one unit cell. 11.56 Explain why diamond is harder than graphite. Why 11.44 A face-centered cubic cell contains 8 X atoms at the is graphite an electrical conductor but diamond corners of the cell and 6 Y atoms at the faces. What is not? is the empirical formula of the solid? Amorphous Solids X-Ray Diffraction of Crystals Review Questions Review Questions 11.57 What is an amorphous solid? How does it differ 11.45 Define X-ray diffraction. What are the typical wave- from crystalline solid? lengths (in nanometers) of X rays (see Figure 7.4)? 11.58 Define glass. What is the chief component of glass? 11.46 Write the Bragg equation. Define every term and Name three types of glass. describe how this equation can be used to measure interatomic distances. Phase Changes Problems Review Questions • 11.47 When X rays of wavelength 0.090 nm are dif- 11.59 What is a phase change? Name all possible changes fracted by a metallic crystal, the angle of first-order that can occur among the vapor, liquid, and solid diffraction (n 5 1) is measured to be 15.2°. What is phases of a substance. the distance (in pm) between the layers of atoms 11.60 What is the equilibrium vapor pressure of a liquid? responsible for the diffraction? How is it measured and how does it change with 11.48 The distance between layers in a NaCl crystal is 282 temperature? pm. X rays are diffracted from these layers at an 11.61 Use any one of the phase changes to explain what is angle of 23.0°. Assuming that n 5 1, calculate the meant by dynamic equilibrium. wavelength of the X rays in nm. 11.62 Define the following terms: (a) molar heat of vapor- ization, (b) molar heat of fusion, (c) molar heat of Types of Crystals sublimation. What are their units? Review Questions 11.63 How is the molar heat of sublimation related to the molar heats of vaporization and fusion? On what 11.49 Describe and give examples of the following types law are these relationships based? of crystals: (a) ionic crystals, (b) covalent crystals, 11.64 What can we learn about the intermolecular forces (c) molecular crystals, (d) metallic crystals. in a liquid from the molar heat of vaporization? 11.50 Why are metals good conductors of heat and 11.65 The greater the molar heat of vaporization of a liquid, electricity? Why does the ability of a metal to the greater its vapor pressure. True or false? conduct electricity decrease with increasing temperature? 11.66 Define boiling point. How does the boiling point of a liquid depend on external pressure? Referring to Table 5.3, what is the boiling point of water when Problems the external pressure is 187.5 mmHg? • 11.51 A solid is hard, brittle, and electrically nonconducting. 11.67 As a liquid is heated at constant pressure, its tem- Its melt (the liquid form of the substance) and an perature rises. This trend continues until the boiling aqueous solution containing the substance conduct point of the liquid is reached. No further rise in tem- electricity. Classify the solid. perature of the liquid can be induced by heating. 11.52 A solid is soft and has a low melting point (below Explain. 100°C). The solid, its melt, and an aqueous solution 11.68 What is critical temperature? What is the signifi- containing the substance are all nonconductors of cance of critical temperature in liquefaction of electricity. Classify the solid. gases?  512 Chapter 11 ■ Intermolecular Forces and Liquids and Solids 11.69 What is the relationship between intermolecular 11.84 The vapor pressure of benzene, C6H6, is 40.1 mmHg forces in a liquid and the liquid’s boiling point and at 7.6°C. What is its vapor pressure at 60.6°C? The critical temperature? Why is the critical temperature molar heat of vaporization of benzene is 31.0 kJ/mol. of water greater than that of most other substances? 11.85 The vapor pressure of liquid X is lower than that of 11.70 How do the boiling points and melting points of liquid Y at 20°C, but higher at 60°C. What can you water and carbon tetrachloride vary with pressure? deduce about the relative magnitude of the molar Explain any difference in behavior of these two heats of vaporization of X and Y? substances. 11.86 Explain why splashing a small amount of liquid 11.71 Why is solid carbon dioxide called dry ice? nitrogen (b.p. 77 K) is not as harmful as splashing 11.72 Wet clothes dry more quickly on a hot, dry day than boiling water on your skin. on a hot, humid day. Explain. 11.73 Which of the following phase transitions gives off Phase Diagrams more heat? (a) 1 mole of steam to 1 mole of water Review Questions at 100°C, or (b) 1 mole of water to 1 mole of ice at 0°C. 11.87 What is a phase diagram? What useful informa- 11.74 A beaker of water is heated to boiling by a Bunsen tion can be obtained from the study of a phase burner. Would adding another burner raise the boiling diagram? point of water? Explain. 11.88 Explain how water’s phase diagram differs from those of most substances. What property of water causes the difference? Problems • 11.75 Calculate the amount of heat (in kJ) required to Problems convert 74.6 g of water to steam at 100°C. • 11.76 How much heat (in kJ) is needed to convert 866 g • 11.89 The phase diagram of sulfur is shown here. (a) How many triple points are there? (b) Monoclinic and of ice at 210°C to steam at 126°C? (The specific heats rhombic are two allotropes of sulfur. Which is more of ice and steam are 2.03 J/g ? °C and 1.99 J/g ? °C, stable under atmospheric conditions? (c) Describe respectively.) what happens when sulfur at 1 atm is heated from • 11.77 How is the rate of evaporation of a liquid affected by 80°C to 200°C. (a) temperature, (b) the surface area of a liquid exposed to air, (c) intermolecular forces? 11.78 The molar heats of fusion and sublimation of mo- 154⬚C 1288 atm lecular iodine are 15.27 kJ/mol and 62.30 kJ/mol, respectively. Estimate the molar heat of vaporization Rhombic Liquid P (atm) of liquid iodine. Monoclinic • 11.79 The following compounds, listed with their boil- ing points, are liquid at 210°C: butane, 20.5°C; 1.0 ethanol, 78.3°C; toluene, 110.6°C. At 210°C, 10⫺4 atm which of these liquids would you expect to have 10⫺5 atm Vapor the highest vapor pressure? Which the lowest? Explain. 95.4⬚C 119⬚C t (⬚C) 11.80 Freeze-dried coffee is prepared by freezing brewed coffee and then removing the ice component with a 11.90 A length of wire is placed on top of a block of ice. vacuum pump. Describe the phase changes taking The ends of the wire extend over the edges of the ice, place during these processes. and a heavy weight is attached to each end. It is found 11.81 A student hangs wet clothes outdoors on a winter that the ice under the wire gradually melts, so that the day when the temperature is 215°C. After a few wire slowly moves through the ice block. At the same hours, the clothes are found to be fairly dry. Describe time, the water above the wire refreezes. Explain the the phase changes in this drying process. phase changes that accompany this phenomenon. 11.82 Steam at 100°C causes more serious burns than water 11.91 The boiling point and freezing point of sulfur dioxide at 100°C. Why? are 210°C and 272.7°C (at 1 atm), respectively. The 11.83 Vapor pressure measurements at several different triple point is 275.5°C and 1.65 3 1023 atm, and its temperatures are shown below for mercury. Deter- critical point is at 157°C and 78 atm. On the basis of mine graphically the molar heat of vaporization for this information, draw a rough sketch of the phase mercury. diagram of SO2. t (°C) 200 250 300 320 340 11.92 A phase diagram of water is shown at the end of this P (mmHg) 17.3 74.4 246.8 376.3 557.9 problem. Label the regions. Predict what would  Questions & Problems 513 happen as a result of the following changes: • 11.100 What is the vapor pressure of mercury at its normal (a) Starting at A, we raise the temperature at con- boiling point (357°C)? stant pressure. (b) Starting at C, we lower the 11.101 A flask of water is connected to a powerful vacuum temperature at constant pressure. (c) Starting at B, pump. When the pump is turned on, the water begins we lower the pressure at constant temperature. to boil. After a few minutes, the same water begins to freeze. Eventually, the ice disappears. Explain what happens at each step. B 11.102 The liquid-vapor boundary line in the phase diagram A of any substance always stops abruptly at a certain P point. Why? 11.103 The interionic distance of several alkali halide C crystals are: T NaCl NaBr NaI KCl KBr KI 282 pm 299 pm 324 pm 315 pm 330 pm 353 pm Plot lattice energy versus the reciprocal interionic Additional Problems distance. How would you explain the plot in terms • 11.93 Name the kinds of attractive forces that must be of the dependence of lattice energy on distance of overcome in order to (a) boil liquid ammonia, separation between ions? What law governs this in- (b) melt solid phosphorus (P4), (c) dissolve CsI in teraction? (For lattice energies, see Table 9.1.) liquid HF, (d) melt potassium metal. 11.104 Which has a greater density, crystalline SiO2 or • 11.94 Which of the following properties indicates very amorphous SiO2? Why? strong intermolecular forces in a liquid? (a) very low 11.105 In 2009, thousands of babies in China became ill surface tension, (b) very low critical temperature, from drinking contaminated milk. To falsely (c) very low boiling point, (d) very low vapor pressure boost the milk’s protein content, melamine 11.95 At 235°C, liquid HI has a higher vapor pressure (C3H6N6) was added to diluted milk because of than liquid HF. Explain. its high nitrogen composition. Unfortunately, 11.96 Based on the following properties of elemental bo- melamine forms a precipitate by hydrogen bond- ron, classify it as one of the crystalline solids dis- ing with cyanuric acid (C3H3N3O3), another con- cussed in Section 11.6: high melting point (2300°C), taminant present. The resulting stonelike particles poor conductor of heat and electricity, insoluble in caused severe kidney damage in many babies. water, very hard substance. Draw the hydrogen-bonded complex formed from • 11.97 Referring to Figure 11.41, determine the stable these two molecules. phase of CO2 at (a) 4 atm and 260°C and (b) 0.5 atm and 220°C. NH2 H 11.98 Classify the unit cell of molecular iodine. C O N O N N C C C C N N H2N N NH2 H C H O Melamine Cyanuric acid 11.106 The vapor pressure of a liquid in a closed container depends on which of the following? (a) The volume above the liquid, (b) the amount of liquid present, (c) temperature, (d) intermolecular forces between the molecules in the liquid. • 11.107 A student is given four solid samples labeled W, X, Y, and Z. All except Z have a metallic luster. 11.99 A CO2 fire extinguisher is located on the outside of She is told that the solids could be gold, lead a building in Massachusetts. During the winter sulfide, quartz (SiO2), and iodine. The results of months, one can hear a sloshing sound when the her investigations are: (a) W is a good electrical extinguisher is gently shaken. In the summertime conductor; X, Y, and Z are poor electrical conduc- there is often no sound when it is shaken. Explain. tors. (b) When the solids are hit with a hammer, W Assume that the extinguisher has no leaks and that flattens out, X shatters into many pieces, Y is it has not been used. smashed into a powder, and Z is cracked. (c) When 514 Chapter 11 ■ Intermolecular Forces and Liquids and Solids the solids are heated with a Bunsen burner, Y melts • 11.115 The fluorides of the second-period elements and with some sublimation, but X, W, and Z do not their melting points are: LiF, 845°C; BeF2, 800°C; melt. (d) In treatment with 6 M HNO3, X dis- BF3, 2126.7°C; CF4, 2184°C; NF3, 2206.6°C; solves; there is no effect on W, Y, or Z. On the OF 2, 2223.8°C; F 2, 2219.6°C. Classify the basis of these test results, identify the solids. type(s) of intermolecular forces present in each 11.108 Which of the following statements are false? compound. (a) Dipole-dipole interactions between molecules 11.116 The standard enthalpy of formation of gaseous are greatest if the molecules possess only temporary molecular iodine is 62.4 kJ/mol. Use this informa- dipole moments. (b) All compounds containing tion to calculate the molar heat of sublimation of hydrogen atoms can participate in hydrogen-bond molecular iodine at 25°C. formation. (c) Dispersion forces exist between all 11.117 The following graph shows approximate plots of atoms, molecules, and ions. (d) The extent of ion- ln P versus 1/T for three compounds: methanol induced dipole interaction depends only on the (CH3OH) methyl chloride (CH3Cl), and propane charge on the ion. (C3H8), where P is the vapor pressure. Match the 11.109 The diagram below shows a kettle of boiling water lines with these compounds. on a stove. Identify the phases in regions A and B. C B A B ln P A 1/T 11.110 The south pole of Mars is covered with dry ice, which partly sublimes during the summer. The CO2 11.118 Determine the final state and its temperature when vapor recondenses in the winter when the temper- 150.0 kJ of heat are added to 50.0 g of water at ature drops to 150 K. Given that the heat of subli- 20°C. The specific heat of steam is 1.99 J/g ? C. mation of CO2 is 25.9 kJ/mol, calculate the 11.119 The distance between Li1 and Cl2 is 257 pm in solid atmospheric pressure on the surface of Mars. LiCl and 203 pm in a LiCl unit in the gas phase. [Hint: Use Figure 11.41 to determine the normal Explain the difference in the bond lengths. sublimation temperature of dry ice and Equation 11.120 Heat of hydration, that is, the heat change that oc- (11.5), which also applies to sublimations.] curs when ions become hydrated in solution, is 11.111 The properties of gases, liquids, and solids differ in a largely due to ion-dipole interactions. The heats number of respects. How would you use the kinetic of hydration for the alkali metal ions are Li1, molecular theory (see Section 5.7) to explain the 2520 kJ/mol; Na1, 2405 kJ/mol; K1, 2321 kJ/mol. following observations? (a) Ease of compressibil- Account for the trend in these values. ity decreases from gas to liquid to solid. (b) Solids 11.121 If water were a linear molecule, (a) would it still retain a definite shape, but gases and liquids do not. be polar, and (b) would the water molecules still (c) For most substances, the volume of a given be able to form hydrogen bonds with one amount of material increases as it changes from another? solid to liquid to gas. 11.122 Calculate the ≤H ° for the following processes at 11.112 Select the substance in each pair that should have 25°C: (a) Br2(l) ¡ Br2(g) and (b) Br2(g) ¡ the higher boiling point. In each case identify the 2Br( g). Comment on the relative magnitudes of principal intermolecular forces involved and ac- these ≤H° values in terms of the forces involved in count briefly for your choice. (a) K2S or (CH3)3N, each case. {Hint: See Table 9.4, and given that (b) Br2 or CH3CH2CH2CH3 ≤H°f [Br2( g)] 5 30.7 kJ/mol.} 11.113 A small drop of oil in water assumes a spherical 11.123 Gaseous or highly volatile liquid anesthetics are of- shape. Explain. (Hint: Oil is made up of nonpolar ten preferred in surgical procedures because once molecules, which tend to avoid contact with inhaled, these vapors can quickly enter the blood- water.) stream through the alveoli and then enter the brain. 11.114 Under the same conditions of temperature and den- Shown here are several common gaseous anesthetics sity, which of the following gases would you expect with their boiling points. Based on intermolecular to behave less ideally: CH4, SO2? Explain. force considerations, explain the advantages of  Questions & Problems 515 using these anesthetics. (Hint: The brain barrier is of carbon made by the destructive distillation of made of membranes that have a nonpolar interior coal) at about 2000°C: region.) SiO2 (s) 1 2C(s) ¡ Si(l) 1 2CO(g) Br F F Cl F F F F Next, solid silicon is separated from other solid im- H C C F F C C O C F F C O C C Cl purities by treatment with hydrogen chloride at 350°C to form gaseous trichlorosilane (SiCl3H): Cl F F H H H F H Halothane Isoflurane Enflurane Si(s) 1 3HCl(g) ¡ SiCl3H(g) 1 H2 (g) 508C 48.58C 56.58C Finally, ultrapure Si can be obtained by reversing the above reaction at 1000°C: • 11.124 A beaker of water is placed in a closed container. Predict the effect on the vapor pressure of the water SiCl3H1g2 1 H2 1g2 ¡ Si1s2 1 3HCl1g2 when (a) its temperature is lowered, (b) the volume of the container is doubled, (c) more water is added (a) Trichlorosilane has a vapor pressure of 0.258 atm to the beaker. at 22°C. What is its normal boiling point? Is tri- 11.125 The phase diagram of helium is shown here. chlorosilane’s boiling point consistent with the type Helium is the only known substance that has two of intermolecular forces that exist among its mol- different liquid phases called helium-I and helium- ecules? (The molar heat of vaporization of trichlo- II. (a) What is the maximum temperature at which rosilane is 28.8 kJ/mol.) (b) What types of crystals helium-II can exist? (b) What is the minimum do Si and SiO2 form? (c) Silicon has a diamond pressure at which solid helium can exist? (c) What crystal structure (see Figure 11.28). Each cubic unit is the normal boiling point of helium-I? (d) Can cell (edge length a 5 543 pm) contains eight solid helium sublime? (e) How many triple points Si atoms. If there are 1.0 3 1013 boron atoms per are there? cubic centimeter in a sample of pure silicon, how many Si atoms are there for every B atom in the 100 sample? Does this sample satisfy the 1029 purity Solid requirement for the electronic grade silicon? 10 11.130 Carbon and silicon belong to Group 4A of the peri- Liquid odic table and have the same valence electron con- (helium-I) figuration (ns2np2). Why does silicon dioxide (SiO2) P (atm) 1 Liquid have a much higher melting point than carbon diox- (helium-II) ide (CO2)? 0.1 11.131 A pressure cooker is a sealed container that allows Vapor steam to escape when it exceeds a predetermined 0.01 pressure. How does this device reduce the time needed for cooking? 1 2 3 4 5 6 T (K) • 11.132 A 1.20-g sample of water is injected into an evac- uated 5.00-L flask at 65°C. What percentage of 11.126 Referring to Figure 11.26, determine the number of the water will be vapor when the system reaches each type of ion within the unit cells. equilibrium? Assume ideal behavior of water 11.127 Ozone (O3) is a strong oxidizing agent that can vapor and that the volume of liquid water is negli- oxidize all the common metals except gold and gible. The vapor pressure of water at 65°C is platinum. A convenient test for ozone is based on 187.5 mmHg. its action on mercury. When exposed to ozone, 11.133 What are the advantages of cooking the vegetable mercury becomes dull looking and sticks to glass broccoli with steam instead of boiling it in water? tubing (instead of flowing freely through it). • 11.134 A quantitative measure of how efficiently spheres Write a balanced equation for the reaction. What pack into unit cells is called packing efficiency, property of mercury is altered by its interaction which is the percentage of the cell space occupied with ozone? by the spheres. Calculate the packing efficiencies 11.128 A sample of limestone (CaCO3) is heated in a closed of a simple cubic cell, a body-centered cubic cell, vessel until it is partially decomposed. Write an and a face-centered cubic cell. (Hint: Refer to Fig- equation for the reaction and state how many phases ure 11.22 and use the relationship that the volume are present. of a sphere is 43 πr3, where r is the radius of the • 11.129 Silicon used in computer chips must have an impu- sphere.) rity level below 1029 (that is, fewer than one impu- 11.135 Provide an explanation for each of the following rity atom for every 109 Si atoms). Silicon is prepared phenomena: (a) Solid argon (m.p. 2189.2°C; b.p. by the reduction of quartz (SiO2) with coke (a form 2185.7°C) can be prepared by immersing a flask  516 Chapter 11 ■ Intermolecular Forces and Liquids and Solids containing argon gas in liquid nitrogen (b.p. 11.139 Swimming coaches sometimes suggest that a drop 2195.8°C) until it liquefies and then connecting of alcohol (ethanol) placed in an ear plugged with the flask to a vacuum pump. (b) The melting water “draws out the water.” Explain this action point of cyclohexane (C 6H 12) increases with from a molecular point of view. increasing pressure exerted on the solid cyclo- 11.140 Use the concept of intermolecular forces to explain hexane. (c) Certain high-altitude clouds contain why the far end of a walking cane rises when one water droplets at 210°C. (d) When a piece of dry raises the handle. ice is added to a beaker of water, fog forms above 11.141 Why do citrus growers spray their trees with water the water. to protect them from freezing? 11.142 What is the origin of dark spots on the inner glass walls of an old tungsten lightbulb? What is the purpose of filling these lightbulbs with argon gas? • 11.136Argon crystallizes in the face-centered cubic 11.143 The compound dichlorodifluoromethane (CCl2F2) arrangement at 40 K. Given that the atomic radius has a normal boiling point of 230°C, a critical tem- of argon is 191 pm, calculate the density of solid perature of 112°C, and a corresponding critical pres- argon. sure of 40 atm. If the gas is compressed to 18 atm at 11.137 A chemistry instructor performed the following 20°C, will the gas condense? Your answer should be mystery demonstration. Just before the students ar- based on a graphical interpretation. rived in class, she heated some water to boiling in an 11.144 A student heated a beaker of cold water (on a Erlenmeyer flask. She then removed the flask from tripod) with a Bunsen burner. When the gas is the flame and closed the flask with a rubber stopper. ignited, she noticed that there was water con- After the class commenced, she held the flask in densed on the outside of the beaker. Explain what front of the students and announced that she could happened. make the water boil simply by rubbing an ice cube 11.145 Sketch the cooling curves of water from about on the outside walls of the flask. To the amazement 110°C to about 210°C. How would you also show of everyone, it worked. Give an explanation for this the formation of supercooled liquid below 0°C phenomenon. which then freezes to ice? The pressure is at 1 atm 11.138 Given the phase diagram of carbon shown, answer throughout the process. The curves need not be the following questions: (a) How many triple drawn quantitatively. points are there and what are the phases that can 11.146 Iron crystallizes in a body-centered cubic lattice. coexist at each triple point? (b) Which has a The cell length as determined by X-ray diffraction is higher density, graphite or diamond? (c) Synthetic 286.7 pm. Given that the density of iron is 7.874 g/cm3, diamond can be made from graphite. Using the calculate Avogadro’s number. phase diagram, how would you go about making diamond? • 11.147 The boiling point of methanol is 65.0°C and the standard enthalpy of formation of methanol va- por is 2201.2 kJ/mol. Calculate the vapor pres- sure of methanol (in mmHg) at 25°C. (Hint: See Diamond Appendix 3 for other thermodynamic data of Liquid methanol.) P (atm) 2 × 104 11.148 An alkali metal in the form of a cube of edge length Graphite 0.171 cm is vaporized in a 0.843-L container at Vapor 1235 K. The vapor pressure is 19.2 mmHg. Iden- tify the metal by calculating the atomic radius in picometers and the density. (Hint: You need to con- 3300 sult Figures 8.5, 11.22, 11.29, and a chemistry t (°C) handbook.)  Answers to Practice Exercises 517 11.149 A closed vessel of volume 9.6 L contains 2.0 g of water. Calculate the temperature (in °C) at which only half of the water remains in the liquid phase. t (°C) (See Table 5.3 for vapor pressures of water at differ- ent temperatures.) • 11.150 A sample of water shows the following behavior as heat added heat added heat added heat added it is heated at a constant rate: (a) (b) (c) (d) (Used with permission from the Journal of Chemical Educa- tion, Vol. 79, No. 7, 2002, pp. 889–895; © 2002, Division of Chemical Education, Inc.) t (°C) 11.151 The electrical conductance of copper metal de- creases with temperature, but that of a CuSO4 solu- tion increases with temperature. Explain. heat added 11.152 Assuming ideal behavior, calculate the density of gas- If twice the mass of water has the same amount of eous HF at its normal boiling point (19.5°C). The ex- heat transferred to it, which of the following graphs perimentally measured density under the same best describes the temperature variation? Note that conditions is 3.10 g/L. Account for the discrepancy be- the scales for all the graphs are the same. tween your calculated value and the experimental result. Interpreting, Modeling & Estimating 11.153 Both calcium and strontium crystallize in face-centered 11.156 On a summer day the temperature and (relative) hu- cubic unit cells. Which metal has a greater density? midity were 95°F and 65 percent, respectively, in 11.154 Is the vapor pressure of a liquid more sensitive to Florida. What would be the volume of water in a changes in temperature if ≤H vap is small or large? typical student dormitory room if all of the water 11.155 Estimate the molar heat of vaporization of a liquid vapor were condensed to liquid? whose vapor pressure doubles when the temperature 11.157 Without the aid of instruments, give two examples is raised from 85°C to 95°C. of evidence that solids exhibit vapor pressure. Answers to Practice Exercises 11.1 (a) Ionic and dispersion forces, (b) dispersion forces, (c) dipole-dipole and dispersion forces. 11.2 Only (c). 11.3 10.50 g/cm3. 11.4 315 pm. 11.5 Two. 11.6 361 pm. 11.7 369 mmHg. 11.8 173 kJ.  CHAPTER 12 Physical Properties of Solutions A sugar cube dissolving in water. The properties of a solution are markedly different from those of its solvent. CHAPTER OUTLINE A LOOK AHEAD 12.1 Types of Solutions  We begin by examining different types of solutions that can be formed from the three states of matter: solid, liquid, and gas. We also characterize 12.2 A Molecular View of a solution by the amount of solute present as unsaturated, saturated, and the Solution Process supersaturated. (12.1) 12.3 Concentration Units  Next we study the formation of solutions at the molecular level and see 12.4 The Effect of Temperature how intermolecular forces affect the energetics of the solution process and on Solubility solubility. (12.2)  We study the four major types of concentration units—percent by mass, 12.5 The Effect of Pressure on mole fraction, molarity, and molality—and their interconversions. (12.3) the Solubility of Gases  Temperature in general has a marked effect on the solubility of gases as 12.6 Colligative Properties of well as liquids and solids. (12.4) Nonelectrolyte Solutions  We see that pressure has no influence on the solubility of liquids and solids, 12.7 Colligative Properties of but greatly affects the solubility of gases. The quantitative relationship Electrolyte Solutions between gas solubility and pressure is given by Henry’s law. (12.5) 12.8 Colloids  We learn that physical properties such as the vapor pressure, melting point, boiling point, and osmotic pressure of a solution depend only on the concentration and not the identity of the solute present. We first study these colligative properties and their applications for nonelectrolyte solutions. (12.6)  We then extend our study of colligative properties to electrolyte solutions and learn about the influence of ion pair formation on these properties. (12.7)  The chapter ends with a brief examination of colloids, which are particles larger than individual molecules that are dispersed in another medium. (12.8) 518  12.1 Types of Solutions 519 M ost chemical reactions take place, not between pure solids, liquids, or gases, but among ions and molecules dissolved in water or other solvents. In Chapters 5 and 11 we looked at the properties of gases, liquids, and solids. In this chapter we examine the properties of solutions, concentrating mainly on the role of intermolecular forces in solubility and other physical properties of solution. 12.1 Types of Solutions In Section 4.1 we noted that a solution is a homogeneous mixture of two or more substances. Because this definition places no restriction on the nature of the sub- stances involved, we can distinguish six types of solutions, depending on the original states (solid, liquid, or gas) of the solution components. Table 12.1 gives examples of each type. Our focus in this chapter will be on solutions involving at least one liquid component—that is, gas-liquid, liquid-liquid, and solid-liquid solutions. And, per- haps not too surprisingly, the liquid solvent in most of the solutions we will study is water. Chemists also characterize solutions by their capacity to dissolve a solute. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than it has the capacity to dissolve. A third type, a supersaturated solution, contains more solute than is present in a saturated solution. Supersaturated solu- tions are not very stable. In time, some of the solute will come out of a supersatu- rated solution as crystals. Crystallization is the process in which dissolved solute comes out of solution and forms crystals (Figure 12.1). Note that both precipitation Table 12.1 Types of Solutions State of Resulting Component 1 Component 2 Solution Examples Gas Gas Gas Air Gas Liquid Liquid Soda water (CO2 in water) Gas Solid Solid H2 gas in palladium Liquid Liquid Liquid Ethanol in water Solid Liquid Liquid NaCl in water Solid Solid Solid Brass (Cu/Zn), solder (Sn/Pb) Figure 12.1 In a supersaturated sodium acetate solution (left), sodium acetate crystals rapidly form when a small seed crystal is added. 520 Chapter 12 ■ Physical Properties of Solutions and crystallization describe the separation of excess solid substance from a super- saturated solution. However, solids formed by the two processes differ in appear- ance. We normally think of precipitates as being made up of small particles, whereas crystals may be large and well formed. 12.2 A Molecular View of the Solution Process In Section 6.6 we discussed the solution The intermolecular attractions that hold molecules together in liquids and solids also process from a macroscopic point of view. play a central role in the formation of solutions. When one substance (the solute) dissolves in another (the solvent), particles of the solute disperse throughout the solvent. The solute particles occupy positions that are normally taken by solvent molecules. The ease with which a solute particle replaces a solvent molecule depends on the relative strengths of three types of interactions: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction For simplicity, we can imagine the solution process taking place in three distinct steps (Figure 12.2). Step 1 is the separation of solvent molecules, and step 2 entails the separation of solute molecules. These steps require energy input to break attrac- tive intermolecular forces; therefore, they are endothermic. In step 3 the solvent and solute molecules mix. This process can be exothermic or endothermic. The heat of solution ≤Hsoln is given by This equation is an application of ¢Hsoln 5 ¢H1 1 ¢H2 1 ¢H3 Hess’s law. If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (≤Hsoln , 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (≤Hsoln . 0). You may wonder why a solute dissolves in a solvent at all if the attraction for its own molecules is stronger than the solute-solvent attraction. The solution pro- cess, like all physical and chemical processes, is governed by two factors. One is energy, which determines whether a solution process is exothermic or endothermic. The second factor is an inherent tendency toward disorder in all natural events. In much the same way that a deck of new playing cards becomes mixed up after it has been shuffled a few times, when solute and solvent molecules mix to form a Figure 12.2 A molecular view of the solution process portrayed as taking place in three steps: Step 1 Step 2 First the solvent and solute molecules are separated (steps 1 Δ H1 Δ H2 and 2). Then the solvent and Solvent Solute solute molecules mix (step 3). Step 3 Δ H3 Solution  12.2 A Molecular View of the Solution Process 521 solution, there is an increase in randomness, or disorder. In the pure state, the solvent and solute possess a fair degree of order, characterized by the more or less regular arrangement of atoms, molecules, or ions in three-dimensional space. Much of this order is destroyed when the solute dissolves in the solvent (see Figure 12.2). Therefore, the solution process is accompanied by an increase in disorder. It is the increase in disorder of the system that favors the solubility of any substance, even if the solution process is endothermic. Solubility is a measure of how much solute will dissolve in a solvent at a spe- Animation Dissolution of an Ionic and a Covalent cific temperature. The saying “like dissolves like” is helpful in predicting the solu- Compound bility of a substance in a given solvent. What this expression means is that two substances with intermolecular forces of similar type and magnitude are likely to be soluble in each other. For example, both carbon tetrachloride (CCl4) and benzene (C6H6) are nonpolar liquids. The only intermolecular forces present in these sub- stances are dispersion forces (see Section 11.2). When these two liquids are mixed, they readily dissolve in each other, because the attraction between CCl4 and C6H6 molecules is comparable in magnitude to the forces between CCl4 molecules and between C6H6 molecules. Two liquids are said to be miscible if they are completely soluble in each other in all proportions. Alcohols such as methanol, ethanol, and 1,2-ethylene glycol are miscible with water because they can form hydrogen bonds with water molecules: H H H H H CH3OH A A A A A HOCOOOH HOCOCOOOH HOOOCOCOOOH A A A A A H H H H H methanol ethanol 1,2-ethylene glycol When sodium chloride dissolves in water, the ions are stabilized in solution by hydration, which involves ion-dipole interaction. In general, we predict that ionic compounds should be much more soluble in polar solvents, such as water, C2H5OH liquid ammonia, and liquid hydrogen fluoride, than in nonpolar solvents, such as benzene and carbon tetrachloride. Because the molecules of nonpolar solvents lack a dipole moment, they cannot effectively solvate the Na1 and Cl2 ions. (Solvation is the process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner. The process is called hydration when the solvent is water.) The predominant intermolecular interaction between ions and nonpolar compounds is ion-induced dipole interaction, which is much weaker than ion- dipole interaction. Consequently, ionic compounds usually have extremely low solubility in nonpolar solvents. Example 12.1 illustrates how to predict solubility based on a knowledge of the intermolecular forces in the solute and the solvent. CH2(OH)CH2(OH) Example 12.1 Predict the relative solubilities in the following cases: (a) Bromine (Br2) in benzene (C6H6, μ 5 0 D) and in water (μ 5 1.87 D), (b) KCl in carbon tetrachloride (CCl4, μ 5 0 D) and in liquid ammonia (NH3, μ 5 1.46 D), (c) formaldehyde (CH2O) in carbon disulfide (CS2, μ 5 0 D) and in water. Strategy In predicting solubility, remember the saying: Like dissolves like. A nonpolar solute will dissolve in a nonpolar solvent; ionic compounds will generally dissolve in polar solvents due to favorable ion-dipole interaction; solutes that can form hydrogen bonds with the solvent will have high solubility in the solvent. (Continued) 522 Chapter 12 ■ Physical Properties of Solutions Solution (a) Br2 is a nonpolar molecule and therefore should be more soluble in C6H6, which is also nonpolar, than in water. The only intermolecular forces between Br2 and C6H6 are dispersion forces. (b) KCl is an ionic compound. For it to dissolve, the individual K1 and Cl2 ions must be stabilized by ion-dipole interaction. Because CCl4 has no dipole moment, KCl should be more soluble in liquid NH3, a polar molecule with a large dipole moment. (c) Because CH2O is a polar molecule and CS2 (a linear molecule) is nonpolar, CH2O the forces between molecules of CH2O and CS2 are dipole-induced dipole and dispersion. On the other hand, CH2O can form hydrogen bonds with water, so it Similar problem: 12.11. should be more soluble in that solvent. Practice Exercise Is iodine (I2) more soluble in water or in carbon disulfide (CS2)? Review of Concepts Which of the following would you expect to be more soluble in benzene than in water: C4H10, HBr, KNO3, P4? 12.3 Concentration Units Quantitative study of a solution requires knowing its concentration, that is, the amount of solute present in a given amount of solution. Chemists use several different con- centration units, each of which has advantages as well as limitations. Let us examine the four most common units of concentration: percent by mass, mole fraction, molar- ity, and molality. Types of Concentration Units Percent by Mass The percent by mass (also called percent by weight or weight percent) is the ratio of the mass of a solute to the mass of the solution, multiplied by 100 percent: mass of solute percent by mass 5 3 100% mass of solute 1 mass of solvent mass of solute or percent by mass 5 3 100% (12.1) mass of soln The percent by mass is a unitless number because it is a ratio of two similar quantities.  12.3 Concentration Units 523 Example 12.2 A sample of 0.892 g of potassium chloride (KCl) is dissolved in 54.6 g of water. What is the percent by mass of KCl in the solution? Strategy We are given the mass of a solute dissolved in a certain amount of solvent. Therefore, we can calculate the mass percent of KCl using Equation (12.1). Solution We write mass of solute percent by mass of KCl 5 3 100% mass of soln 0.892 g 5 3 100% 0.892 g 1 54.6 g 5 1.61% Similar problem: 12.15. Practice Exercise A sample of 6.44 g of naphthalene (C10H8) is dissolved in 80.1 g of benzene (C6H6). Calculate the percent by mass of naphthalene in this solution. Mole Fraction (X) The mole fraction was introduced in Section 5.6. The mole fraction of a component of a solution, say, component A, is written XA and is defined as moles of A mole fraction of component A 5 XA 5 sum of moles of all components The mole fraction is also unitless, because it too is a ratio of two similar quantities. Molarity (M) In Section 4.5 molarity was defined as the number of moles of solute in 1 L of For calculations involving molarity, see solution; that is, Examples 4.7 and 4.8 on pp. 146 and 147. moles of solute molarity 5 liters of soln Thus, the units of molarity are mol/L. Molality (m) Molality is the number of moles of solute dissolved in 1 kg (1000 g) of solvent— that is, moles of solute molality 5 (12.2) mass of solvent (kg) For example, to prepare a 1 molal, or 1 m, sodium sulfate (Na2SO4) aqueous solution, we need to dissolve 1 mole (142.0 g) of the substance in 1000 g (1 kg) of water. Depending on the nature of the solute-solvent interaction, the final volume of the solution will be either greater or less than 1000 mL. It is also possible, though very unlikely, that the final volume could be equal to 1000 mL. 524 Chapter 12 ■ Physical Properties of Solutions Example 12.3 shows how to calculate the molality of a solution. Example 12.3 Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is 98.09 g. Strategy To calculate the molality of a solution, we need to know the number of moles of solute and the mass of the solvent in kilograms. Solution The definition of molality (m) is H2SO4 moles of solute m5 mass of solvent (kg) First, we find the number of moles of sulfuric acid in 24.4 g of the acid, using its molar mass as the conversion factor. 1 mol H2SO4 moles of H2SO4 5 24.4 g H2SO4 3 98.09 g H2SO4 5 0.249 mol H2SO4 The mass of water is 198 g, or 0.198 kg. Therefore, 0.249 mol H2SO4 m5 0.198 kg H2O Similar problem: 12.17. 5 1.26 m Practice Exercise What is the molality of a solution containing 7.78 g of urea [(NH2)2CO] in 203 g of water? Comparison of Concentration Units The choice of a concentration unit is based on the purpose of the experiment. For instance, the mole fraction is not used to express the concentrations of solutions for titrations and gravimetric analyses, but it is appropriate for calculating partial pres- sures of gases (see Section 5.6) and for dealing with vapor pressures of solutions (to be discussed later in this chapter). The advantage of molarity is that it is generally easier to measure the volume of a solution, using precisely calibrated volumetric flasks, than to weigh the solvent, as we saw in Section 4.5. For this reason, molarity is often preferred over molality. On the other hand, molality is independent of temperature, because the concentra- tion is expressed in number of moles of solute and mass of solvent. The volume of a solution typically increases with increasing temperature, so that a solution that is 1.0 M at 25°C may become 0.97 M at 45°C because of the increase in volume on warming. This concentration dependence on temperature can significantly affect the accuracy of an experiment. Therefore, it is sometimes preferable to use molality instead of molarity. Percent by mass is similar to molality in that it is independent of temperature. Furthermore, because it is defined in terms of ratio of mass of solute to mass of solu- tion, we do not need to know the molar mass of the solute in order to calculate the percent by mass. Sometimes it is desirable to convert one concentration unit of a solution to another; for example, the same solution may be employed for different experiments that require different concentration units for calculations. Suppose we want to express  12.3 Concentration Units 525 the concentration of a 0.396 m glucose (C6H12O6) solution in molarity. We know there is 0.396 mole of glucose in 1000 g of the solvent and we need to determine the vol- ume of this solution to calculate molarity. First, we calculate the mass of the solution from the molar mass of glucose: 180.2 g a0.396 mol C6H12O6 3 b 1 1000 g H2O 5 1071 g 1 mol C6H12O6 The next step is to experimentally determine the density of the solution, which is found to be 1.16 g/mL. We can now calculate the volume of the solution in liters by writing mass volume 5 density 1071 g 1L 5 3 1.16 g/mL 1000 mL 5 0.923 L Finally, the molarity of the solution is given by moles of solute molarity 5 liters of soln 0.396 mol 5 0.923 L 5 0.429 mol/L 5 0.429 M As you can see, the density of the solution serves as a conversion factor between molality and molarity. Examples 12.4 and 12.5 show concentration unit conversions. Example 12.4 The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the solution? The molar mass of methanol is 32.04 g. Strategy To calculate the molality, we need to know the number of moles of methanol and the mass of solvent in kilograms. We assume 1 L of solution, so the number of moles of methanol is 2.45 mol. CH3OH given o moles of solute m p mass of solvent (kg) want to calculate r need to find Solution Our first step is to calculate the mass of water in 1 L of the solution, using density as a conversion factor. The total mass of 1 L of a 2.45 M solution of methanol is 1000 mL soln 0.976 g 1 L soln 3 3 5 976 g 1 L soln 1 mL soln (Continued) 526 Chapter 12 ■ Physical Properties of Solutions Because this solution contains 2.45 moles of methanol, the amount of water (solvent) in the solution is mass of H2O 5 mass of soln 2 mass of solute 32.04 g CH3OH 5 976 g 2 a2.45 mol CH3OH 3 b 1 mol CH3OH 5 898 g The molality of the solution can be calculated by converting 898 g to 0.898 kg: 2.45 mol CH3OH molality 5 0.898 kg H2O Similar problems: 12.18(a), 12.19. 5 2.73 m Practice Exercise Calculate the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL. Example 12.5 Calculate the molality of a 35.4 percent (by mass) aqueous solution of phosphoric acid (H3PO4). The molar mass of phosphoric acid is 97.99 g. Strategy In solving this type of problem, it is convenient to assume that we start with 100.0 g of the solution. If the mass of phosphoric acid is 35.4 percent, or 35.4 g, the percent by mass and mass of water must be 100.0% 2 35.4% 5 64.6% H3PO4 and 64.6 g. Solution From the known molar mass of phosphoric acid, we can calculate the molality in two steps, as shown in Example 12.3. First we calculate the number of moles of phosphoric acid in 35.4 g of the acid 1 mol H3PO4 moles of H3PO4 5 35.4 g H3PO4 3 97.99 g H3PO4 5 0.361 mol H3PO4 The mass of water is 64.6 g, or 0.0646 kg. Therefore, the molality is given by 0.361 mol H3PO4 molality 5 0.0646 kg H2O Similar problem: 12.18(b). 5 5.59 m Practice Exercise Calculate the molality of a 44.6 percent (by mass) aqueous solution of sodium chloride. Review of Concepts A solution is prepared at 20°C and its concentration is expressed in three different units: percent by mass, molality, and molarity. The solution is then heated to 88°C. Which of the concentration units will change (increase or decrease)?  12.4 The Effect of Temperature on Solubility 527 12.4 The Effect of Temperature on Solubility Recall that solubility is defined as the maximum amount of a solute that will dissolve in a given quantity of solvent at a specific temperature. Temperature affects the solu- bility of most substances. In this section we will consider the effects of temperature on the solubility of solids and gases. Solid Solubility and Temperature Figure 12.3 shows the temperature dependence of the solubility of some ionic com- pounds in water. In most but certainly not all cases, the solubility of a solid substance increases with temperature. However, there is no clear correlation between the sign of ≤Hsoln and the variation of solubility with temperature. For example, the solution process of CaCl2 is exothermic, and that of NH4NO3 is endothermic. But the solubility of both compounds increases with increasing temperature. In general, the effect of temperature on solubility is best determined experimentally. Fractional Crystallization The dependence of the solubility of a solid on temperature varies considerably, as Figure 12.3 shows. The solubility of NaNO3, for example, increases sharply with temperature, while that of NaCl changes very little. This wide variation provides a means of obtaining pure substances from mixtures. Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their dif- fering solubilities. Suppose we have a sample of 90 g of KNO3 that is contaminated with 10 g of NaCl. To purify the KNO3 sample, we dissolve the mixture in 100 mL of water at 60°C and then gradually cool the solution to 0°C. At this temperature, the solubilities of KNO3 and NaCl are 12.1 g/100 g H2O and 34.2 g/100 g H2O, respectively. Thus, (90 2 12) g, or 78 g, of KNO3 will crystallize out of the solution, but all of the NaCl will remain dissolved (Figure 12.4). In this manner, we can obtain about 90 percent of the original amount of KNO3 in pure form. The KNO3 crystals can be separated from the solution by filtration. 250 KNO3 Figure 12.3 Temperature dependence of the solubility of some ionic compounds in water. 200 Solubility (g solute/100 g H2O) NaNO3 150 NaBr 100 KBr KCl 50 NaCl Na 2SO4 Ce2(SO4)3 0 20 40 60 80 100 Temperature (°C) 528 Chapter 12 ■ Physical Properties of Solutions Figure 12.4 The solubilities of 150 KNO3 and NaCl at 0°C and 60°C. KNO3 The difference in temperature dependence enables us to isolate one of these compounds from a solution containing both of them, 112 g/100 g H2O through fractional crystallization. Solubility (g solute/100 g H2O) 100 50 38 g/100 g H2O NaCl 34.2 g/100 g H2O 12.1 g/100 g H2O 0 20 40 60 80 100 Temperature (°C) Many of the solid inorganic and organic compounds that are used in the labora- tory were purified by fractional crystallization. Generally, the method works best if the compound to be purified has a steep solubility curve, that is, if it is considerably more soluble at high temperatures than at low temperatures. Otherwise, much of it will remain dissolved as the solution is cooled. Fractional crystallization also works well if the amount of impurity in the solution is relatively small. Review of Concepts Using Figure 12.3, rank the potassium salts in increasing order of solubility at 40°C. Gas Solubility and Temperature The solubility of gases in water usually decreases with increasing temperature (Figure 12.5). When water is heated in a beaker, you can see bubbles of air form- Solubility (mol/ L) 0.002 ing on the side of the glass before the water boils. As the temperature rises, the dissolved air molecules begin to “boil out” of the solution long before the water 0.001 itself boils. The reduced solubility of molecular oxygen in hot water has a direct bearing on thermal pollution—that is, the heating of the environment (usually waterways) to temperatures that are harmful to its living inhabitants. It is estimated that every 0 20 40 60 80 100 Temperature (°C) year in the United States some 100,000 billion gallons of water are used for industrial cooling, mostly in electric power and nuclear power production. This Figure 12.5 Dependence on temperature of the solubility of O2 process heats the water, which is then returned to the rivers and lakes from which gas in water. Note that the it was taken. Ecologists have become increasingly concerned about the effect of solubility decreases as thermal pollution on aquatic life. Fish, like all other cold-blooded animals, have temperature increases. The pressure of the gas over much more difficulty coping with rapid temperature fluctuation in the environ- the solution is 1 atm. ment than humans do. An increase in water temperature accelerates their rate of  12.5 The Effect of Pressure on the Solubility of Gases 529 metabolism, which generally doubles with each 10°C rise. The speedup of metab- olism increases the fish’s need for oxygen at the same time that the supply of oxygen decreases because of its lower solubility in heated water. Effective ways to cool power plants while doing only minimal damage to the biological environ- ment are being sought. On the lighter side, a knowledge of the variation of gas solubility with tempera- ture can improve one’s performance in a popular recreational sport—fishing. On a hot summer day, an experienced fisherman usually picks a deep spot in the river or lake to cast the bait. Because the oxygen content is greater in the deeper, cooler region, most fish will be found there. 12.5 The Effect of Pressure on the Solubility of Gases For all practical purposes, external pressure has no influence on the solubilities of liquids and solids, but it does greatly affect the solubility of gases. The quantitative relationship between gas solubility and pressure is given by Henry’s† law, which states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution: crP c 5 kP  (12.3) Here c is the molar concentration (mol/L) of the dissolved gas; P is the pressure (in atm) of the gas over the solution at equilibrium; and, for a given gas, k is a constant Each gas has a different k value at a that depends only on temperature. The constant k has the units mol/L ? atm. You can given temperature. see that when the pressure of the gas is 1 atm, c is numerically equal to k. If several gases are present, P is the partial pressure. Henry’s law can be understood qualitatively in terms of the kinetic molecular theory. The amount of gas that will dissolve in a solvent depends on how frequently the gas molecules collide with the liquid surface and become trapped by the condensed phase. Suppose we have a gas in dynamic equilibrium with a solution [Figure 12.6(a)]. At every instant, the number of gas molecules entering the solution is equal to the number of dissolved molecules moving into the gas phase. If the partial pressure of the gas is increased [Figure 12.6(b)], more molecules dissolve in the liquid because † William Henry (1775–1836). English chemist. Henry’s major contribution to science was his discovery of the law describing the solubility of gases, which now bears his name. Figure 12.6 A molecular interpretation of Henry’s law. When the partial pressure of the gas over the solution increases from (a) to (b), the concentration of the dissolved gas also increases according to Equation (12.3). (a) (b) 530 Chapter 12 ■ Physical Properties of Solutions more molecules are striking the surface of the liquid. This process continues until the concentration of the solution is again such that the number of molecules leaving the solution per second equals the number entering the solution. Because of the higher concentration of molecules in both the gas and solution phases, this number is greater in (b) than in (a), where the partial pressure is lower. A practical demonstration of Henry’s law is the effervescence of a soft drink when the cap of the bottle is removed. Before the beverage bottle is sealed, it is pressurized with a mixture of air and CO2 saturated with water vapor. Because of the high partial pressure of CO2 in the pressurizing gas mixture, the amount dis- solved in the soft drink is many times the amount that would dissolve under normal atmospheric conditions. When the cap is removed, the pressurized gases escape, eventually the pressure in the bottle falls to atmospheric pressure, and the amount of CO2 remaining in the beverage is determined only by the normal atmospheric partial pressure of CO2, 0.0003 atm. The excess dissolved CO2 comes out of solu- tion, causing the effervescence. The effervescence of a soft drink. Example 12.6 applies Henry’s law to nitrogen gas. The bottle was shaken before being opened to dramatize the escape of CO2. Example 12.6 The solubility of nitrogen gas at 25°C and 1 atm is 6.8 3 1024 mol/L. What is the concentration (in molarity) of nitrogen dissolved in water under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm. Strategy The given solubility enables us to calculate Henry’s law constant (k), which can then be used to determine the concentration of the solution. Solution The first step is to calculate the quantity k in Equation (12.3): c 5 kP 6.8 3 1024 mol/L 5 k (1 atm) k 5 6.8 3 1024 mol/L ? atm Therefore, the solubility of nitrogen gas in water is c 5 (6.8 3 1024 mol/L ? atm) (0.78 atm) 5 5.3 3 1024 mol/L 5 5.3 3 1024 M The decrease in solubility is the result of lowering the pressure from 1 atm to 0.78 atm. Check The ratio of the concentrations [(5.3 3 1024 M/6.8 3 1024 M) 5 0.78] should Similar problem: 12.37. be equal to the ratio of the pressures (0.78 atm/1.0 atm 5 0.78). Practice Exercise Calculate the molar concentration of oxygen in water at 25°C for a partial pressure of 0.22 atm. The Henry’s law constant for oxygen is 1.3 3 1023 mol/L ? atm. Most gases obey Henry’s law, but there are some important exceptions. For example, if the dissolved gas reacts with water, higher solubilities can result. The solubility of ammonia is much higher than expected because of the reaction NH3 1 H2O Δ NH1 4 1 OH 2 Carbon dioxide also reacts with water, as follows: CO2 1 H2O Δ H2CO3 CHEMISTRY in Action The Killer Lake D isaster struck swiftly and without warning. On August 21, 1986, Lake Nyos in Cameroon, a small nation on the west coast of Africa, suddenly belched a dense cloud of carbon diox- ide. Speeding down a river valley, the cloud asphyxiated over 1700 people and many livestock. How did this tragedy happen? Lake Nyos is stratified into layers that do not mix. A boundary separates the freshwater at the surface from the deeper, denser solution containing dis- solved minerals and gases, including CO2. The CO2 gas comes from springs of carbonated groundwater that percolate upward into the bottom of the volcanically formed lake. Given the high water pressure at the bottom of the lake, the concentration of CO2 gradually accumulated to a dangerously high level, in ac- cordance with Henry’s law. What triggered the release of CO2 is not known for certain. It is believed that an earthquake, land- slide, or even strong winds may have upset the delicate balance within the lake, creating waves that overturned the water layers. When the deep water rose, dissolved CO2 came out of solution, just as a soft drink fizzes when the bottle is uncapped. Being heavier than air, the CO2 traveled close to the ground and liter- ally smothered an entire village 15 miles away. Now, more than 25 years after the incident, scientists are concerned that the CO2 concentration at the bottom of Lake Nyos is again reaching saturation level. To prevent a recurrence of the earlier tragedy, an attempt has been made to pump up the deep water, thus releasing the dissolved CO2. In addition to be- ing costly, this approach is controversial because it might disturb the waters near the bottom of the lake, leading to an uncontrollable release of CO2 to the surface. In the meantime, a Deep waters in Lake Nyos are pumped to the surface to remove dissolved natural time bomb is ticking away. CO2 gas. Another interesting example is the dissolution of molecular oxygen in blood. Normally, oxygen gas is only sparingly soluble in water (see Practice Exercise in Example 12.6). However, its solubility in blood is dramatically greater because of the high content of hemoglobin (Hb) molecules. Each hemoglobin molecule can bind up to four oxygen molecules, which are eventually delivered to the tissues for use in metabolism: Hb 1 4O2 Δ Hb(O2 ) 4 It is this process that accounts for the high solubility of molecular oxygen in blood. The above Chemistry in Action essay explains a natural disaster with Henry’s law. Review of Concepts Which of the following gases has the greatest Henry’s law constant in water at 25°C: CH4, Ne, HCl, H2? 531 532 Chapter 12 ■ Physical Properties of Solutions 12.6 Colligative Properties of Nonelectrolyte Solutions Colligative properties (or collective properties) are properties that depend only on the number of solute particles in solution and not on the nature of the solute par- ticles. These properties are bound together by a common origin—they all depend on the number of solute particles present, regardless of whether they are atoms, ions, or molecules. The colligative properties are vapor-pressure lowering, boiling- point elevation, freezing-point depression, and osmotic pressure. For our discussion of colligative properties of nonelectrolyte solutions it is important to keep in mind that we are talking about relatively dilute solutions, that is, solutions whose con- centrations are # 0.2 M. Vapor-Pressure Lowering To review the concept of equilibrium vapor If a solute is nonvolatile (that is, it does not have a measurable vapor pressure), the vapor pressure as it applies to pure liquids, see pressure of its solution is always less than that of the pure solvent. Thus, the relation- Section 11.8. ship between solution vapor pressure and solvent vapor pressure depends on the con- centration of the solute in the solution. This relationship is expressed by Raoult’s† law, which states that the vapor pressure of a solvent over a solution, P1, is given by the vapor pressure of the pure solvent, P°1, times the mole fraction of the solvent in the solution, X1: P1 5 X1P°1 (12.4) In a solution containing only one solute, X1 5 1 2 X2, where X2 is the mole fraction of the solute. Equation (12.4) can therefore be rewritten as P1 5 (1 2 X2 )P°1 or P1 5 P°1 2 X2P°1 so that P°1 2 P1 5 ¢P 5 X2P°1 (12.5) We see that the decrease in vapor pressure, ≤P, is directly proportional to the solute concentration (measured in mole fraction). Example 12.7 illustrates the use of Raoult’s law [Equation (12.5)]. Example 12.7 Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass 5 180.2 g/mol) in 460 mL of water at 30°C. What is the vapor-pressure lowering? The vapor pressure of pure water at 30°C is given in Table 5.3 (p. 199). Assume the density of the solvent is 1.00 g/mL. Strategy We need Raoult’s law [Equation (12.4)] to determine the vapor pressure of a solution. Note that glucose is a nonvolatile solute. C6H12O6 (Continued) † François Marie Raoult (1830–1901). French chemist. Raoult’s work was mainly in solution properties and electrochemistry.  12.6 Colligative Properties of Nonelectrolyte Solutions 533 Solution The vapor pressure of a solution (P1) is need to find o P1 X1P°1 p r want to calculate given First we calculate the number of moles of glucose and water in the solution: 1.00 g 1 mol n1 (water) 5 460 mL 3 3 5 25.5 mol 1 mL 18.02 g 1 mol n2 (glucose) 5 218 g 3 5 1.21 mol 180.2 g The mole fraction of water, X1, is given by n1 X1 5 n1 1 n2 25.5 mol 5 5 0.955 25.5 mol 1 1.21 mol From Table 5.3, we find the vapor pressure of water at 30°C to be 31.82 mmHg. Therefore, the vapor pressure of the glucose solution is P1 5 0.955 3 31.82 mmHg 5 30.4 mmHg Finally, the vapor-pressure lowering (≤P) is (31.82 2 30.4) mmHg, or 1.4 mmHg. Check We can also calculate the vapor pressure lowering by using Equation (12.5). Because the mole fraction of glucose is (1 2 0.955), or 0.045, the vapor pressure lowering is given by (0.045)(31.82 mmHg) or 1.4 mmHg. Similar problems: 12.49, 12.50. Practice Exercise Calculate the vapor pressure of a solution made by dissolving 82.4 g of urea (molar mass 5 60.06 g/mol) in 212 mL of water at 35°C. What is the vapor-pressure lowering? Why is the vapor pressure of a solution less than that of the pure solvent? As was mentioned in Section 12.2, one driving force in physical and chemical processes is an increase in disorder—the greater the disorder, the more favorable the process. Vaporization increases the disorder of a system because molecules in a vapor have less order than those in a liquid. Because a solution is more disordered than a pure solvent, the difference in disorder between a solution and a vapor is less than that between a pure solvent and a vapor. Thus, solvent molecules have less of a tendency to leave a solution than to leave the pure solvent to become vapor, and the vapor pressure of a solution is less than that of the solvent. If both components of a solution are volatile (that is, have measurable vapor pressure), the vapor pressure of the solution is the sum of the individual partial pres- sures. Raoult’s law holds equally well in this case: PA 5 XAP°A PB 5 XBP°B where PA and PB are the partial pressures over the solution for components A and B; PA° and PB° are the vapor pressures of the pure substances; and XA and XB are their 534 Chapter 12 ■ Physical Properties of Solutions 800 mole fractions. The total pressure is given by Dalton’s law of partial pressure (see PT = Pbenzene + Ptoluene Section 5.6): 600 PT 5 PA 1 PB Pressure (mmHg) 400 or Pbenzene PT 5 XAP°A 1 XBP°B 200 Ptoluene For example, benzene and toluene are volatile components that have similar structures and therefore similar intermolecular forces: 0.0 0.2 0.4 0.6 0.8 1.0 Xbenzene CH3 Figure 12.7 The dependence of A the partial pressures of benzene and toluene on their mole fractions in a benzene-toluene solution (Xtoluene 5 1 2 Xbenzene ) at 80°C. This solution is said to be ideal benzene toluene because the vapor pressures obey Raoult’s law. In a solution of benzene and toluene, the vapor pressure of each component obeys Raoult’s law. Figure 12.7 shows the dependence of the total vapor pressure (PT) in a benzene-toluene solution on the composition of the solution. Note that we need only express the composition of the solution in terms of the mole fraction of one component. For every value of Xbenzene, the mole fraction of toluene, Xtoluene, is given by (1 2 Xbenzene). The benzene-toluene solution is one of the few examples of an ideal solution, which is any solution that obeys Raoult’s law. One characteristic of an ideal solution is that the heat of solution, ≤Hsoln, is zero. Most solutions do not behave ideally in this respect. Designating two volatile substances as A and B, we can consider the following two cases: Case 1: If the intermolecular forces between A and B molecules are weaker than those between A molecules and between B molecules, then there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution. Con- sequently, the vapor pressure of the solution is greater than the sum of the vapor pressures as predicted by Raoult’s law for the same concentration. This behavior gives rise to the positive deviation [Figure 12.8(a)]. In this case, the heat of solution is positive (that is, mixing is an endothermic process). Case 2: If A molecules attract B molecules more strongly than they do their own kind, the vapor pressure of the solution is less than the sum of the vapor Figure 12.8 Nonideal solutions. (a) Positive deviation occurs when PT is greater than that predicted PT by Raoult’s law (the solid black line). (b) Negative deviation. Here, PT PT is less than that predicted by Raoult’s law (the solid black line). Pressure Pressure PA PA PB PB 0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 XA XA (a) (b)  12.6 Colligative Properties of Nonelectrolyte Solutions 535 pressures as predicted by Raoult’s law. Here we have a negative deviation [Figure 12.8(b)]. In this case, the heat of solution is negative (that is, mixing is an exothermic process). Review of Concepts A solution contains equal molar amounts of liquids A and B. The vapor pressures of pure A and B are 120 mmHg and 180 mmHg, respectively, at a certain temperature. If the vapor pressure of the solution is 164 mmHg, what can you deduce about the intermolecular forces between A and B molecules compared to the intermolecular forces between A molecules and between B molecules? Fractional Distillation Solution vapor pressure has a direct bearing on fractional distillation, a procedure for separating liquid components of a solution based on their different boiling points. Fractional distillation is somewhat analogous to fractional crystallization. Suppose we want to separate a binary system (a system with two components), say, benzene- toluene. Both benzene and toluene are relatively volatile, yet their boiling points are appreciably different (80.1°C and 110.6°C, respectively). When we boil a solution containing these two substances, the vapor formed is somewhat richer in the more volatile component, benzene. If the vapor is condensed in a separate container and that liquid is boiled again, a still higher concentration of benzene will be obtained in the vapor phase. By repeating this process many times, it is possible to separate ben- zene completely from toluene. In practice, chemists use an apparatus like that shown in Figure 12.9 to separate volatile liquids. The round-bottomed flask containing the benzene-toluene solution is Figure 12.9 An apparatus for Thermometer small-scale fractional distillation. The fractionating column is packed with tiny glass beads. The longer the fractionating column, the more complete the separation of the volatile liquids. Condenser Adapter Water Water Fractionating column Receiving flask Distilling flask Heating mantle 536 Chapter 12 ■ Physical Properties of Solutions fitted with a long column packed with small glass beads. When the solution boils, the vapor condenses on the beads in the lower portion of the column, and the liquid falls back into the distilling flask. As time goes on, the beads gradually heat up, allowing the vapor to move upward slowly. In essence, the packing material causes the benzene- toluene mixture to be subjected continuously to numerous vaporization-condensation steps. At each step the composition of the vapor in the column will be richer in the more volatile, or lower boiling-point, component (in this case, benzene). The vapor that rises to the top of the column is essentially pure benzene, which is then condensed and collected in a receiving flask. Fractional distillation is as important in industry as it is in the laboratory. The petroleum industry employs fractional distillation on a large scale to separate the components of crude oil. More will be said of this process in Chapter 24. Boiling-Point Elevation The boiling point of a solution is the temperature at which its vapor pressure equals the external atmospheric pressure (see Section 11.8). Because the presence of a non- volatile solute lowers the vapor pressure of a solution, it must also affect the boiling point of the solution. Figure 12.10 shows the phase diagram of water and the changes that occur in an aqueous solution. Because at any temperature the vapor pressure of the solution is lower than that of the pure solvent regardless of temperature, the liquid- vapor curve for the solution lies below that for the pure solvent. Consequently, the dashed solution curve intersects the horizontal line that marks P 5 1 atm at a higher temperature than the normal boiling point of the pure solvent. This graphical analysis shows that the boiling point of the solution is higher than that of water. The boiling- point elevation (DTb) is defined as the boiling point of the solution (Tb) minus the boiling point of the pure solvent (T °b ): ¢Tb 5 Tb 2 T °b Because Tb . T °b, ¢Tb is a positive quantity. In calculating the new boiling point, add The value of ≤Tb is proportional to the vapor-pressure lowering, and so it is also ≤Tb to the normal boiling point of the proportional to the concentration (molality) of the solution. That is, solvent. ¢Tb r m ¢Tb 5 Kbm (12.6) Figure 12.10 Phase diagram illustrating the boiling-point elevation and freezing-point depression of aqueous solutions. The dashed curves pertain to the 1 atm solution, and the solid curves to the pure solvent. As you can see, Liquid Pressure the boiling point of the solution is higher than that of water, and the freezing point of the solution is lower than that of water. Solid Vapor ΔTf ΔTb Temperature Freezing Freezing Boiling Boiling point of point of point of point of solution water water solution  12.6 Colligative Properties of Nonelectrolyte Solutions 537 Table 12.2 Molal Boiling-Point Elevation and Freezing-Point Depression Constants of Several Common Liquids Normal Freezing Kf Normal Boiling Kb Solvent Point (8C)* (8C/m) Point (8C)* (8C/m) Water 0 1.86 100 0.52 Benzene 5.5 5.12 80.1 2.53 Ethanol −117.3 1.99 78.4 1.22 Acetic acid 16.6 3.90 117.9 2.93 Cyclohexane 6.6 20.0 80.7 2.79 *Measured at 1 atm. where m is the molality of the solution and Kb is the molal boiling-point elevation constant. The units of Kb are °C/m. It is important to understand the choice of con- centration unit here. We are dealing with a system (the solution) whose temperature is not constant, so we cannot express the concentration units in molarity because molarity changes with temperature. Table 12.2 lists values of Kb for several common solvents. Using the boiling-point elevation constant for water and Equation (12.6), you can see that if the molality of an aqueous solution is 1.00 m, the boiling point will be 100.52°C. Freezing-Point Depression A nonscientist may remain forever unaware of the boiling-point elevation phenom- enon, but a careful observer living in a cold climate is familiar with freezing-point depression. Ice on frozen roads and sidewalks melts when sprinkled with salts such as NaCl or CaCl2. This method of thawing succeeds because it depresses the freez- ing point of water. Figure 12.10 shows that lowering the vapor pressure of the solution shifts the solid-liquid curve to the left. Consequently, this line intersects the horizontal line at De-icing of airplanes is based on a temperature lower than the freezing point of water. The freezing-point depression freezing-point depression. (DTf ) is defined as the freezing point of the pure solvent (T °) f minus the freezing point of the solution (Tf): ¢Tf 5 T °f 2 Tf Because T °f . Tf, ¢Tf is a positive quantity. Again, ≤Tf is proportional to the con- In calculating the new freezing point, centration of the solution: subtract ≤Tf from the normal freezing point of the solvent. ¢Tf r m ¢Tf 5 Kf m (12.7) where m is the concentration of the solute in molality units, and Kf is the molal freezing-point depression constant (see Table 12.2). Like Kb, Kf has the units °C/m. A qualitative explanation of the freezing-point depression phenomenon is as fol- lows. Freezing involves a transition from the disordered state to the ordered state. For this to happen, energy must be removed from the system. Because a solution has greater disorder than the solvent, more energy needs to be removed from it to create order than in the case of a pure solvent. Therefore, the solution has a lower freezing point than its solvent. Note that when a solution freezes, the solid that separates is the pure solvent component. 538 Chapter 12 ■ Physical Properties of Solutions In order for boiling-point elevation to occur, the solute must be nonvolatile, but no such restriction applies to freezing-point depression. For example, methanol (CH3OH), a fairly volatile liquid that boils at only 65°C, has sometimes been used as an antifreeze in automobile radiators. A practical application of the freezing-point depression is described in Example 12.8. Example 12.8 Ethylene glycol (EG), CH2(OH)CH2(OH), is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C). Calculate the freezing point of a solution containing 651 g of this substance in 2505 g of water. Would you keep this substance in your car radiator during the summer? The molar mass of ethylene glycol is 62.01 g. Strategy This question asks for the depression in freezing point of the solution. constant o Tf Kf m p r want to calculate need to find The information given enables us to calculate the molality of the solution and we refer In cold climate regions, antifreeze to Table 12.2 for the Kf of water. must be used in car radiators in winter. Solution To solve for the molality of the solution, we need to know the number of moles of EG and the mass of the solvent in kilograms. We find the molar mass of EG, and convert the mass of the solvent to 2.505 kg, and calculate the molality as follows: 1 mol EG 651 g EG 3 5 10.5 mol EG 62.07 g EG moles of solute m5 mass of solvent (kg) 10.5 mol EG 5 5 4.19 mol EG/kg H2O 2.505 kg H2O 5 4.19 m From Equation (12.7) and Table 12.2 we write ¢Tf 5 Kf m 5 (1.86°C/m) (4.19 m) 5 7.79°C Because pure water freezes at 0°C, the solution will freeze at (0 2 7.79)°C or 27.79°C. We can calculate boiling-point elevation in the same way as follows: ¢Tb 5 Kbm 5 (0.52°C/m) (4.19 m) 5 2.2°C Because the solution will boil at (100 1 2.2)°C, or 102.2°C, it would be preferable to leave the antifreeze in your car radiator in summer to prevent the solution from Similar problems: 12.56, 12.59. boiling. Practice Exercise Calculate the boiling point and freezing point of a solution containing 478 g of ethylene glycol in 3202 g of water.  12.6 Colligative Properties of Nonelectrolyte Solutions 539 Osmotic pressure Semipermeable membrane Solute molecule Solvent molecule (a) (b) Figure 12.11 Osmotic pressure. (a) The levels of the pure solvent (left) and of the solution (right) are equal at the start. (b) During osmosis, the level on the solution side rises as a result of the net flow of solvent from left to right. The osmotic pressure is equal to the hydrostatic pressure exerted by the column of fluid in the right tube at equilibrium. Basically the same effect occurs when the pure solvent is replaced by a more dilute solution than that on the right. Review of Concepts Sketch a phase diagram like that shown in Figure 12.10 for nonaqueous solutions such as naphthalene dissolved in benzene. Would the freezing-point depression and boiling-point elevation still apply in this case? Osmotic Pressure Many chemical and biological processes depend on osmosis, the selective passage Animation Osmosis of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. Figure 12.11 illustrates this phenomenon. The left compartment of the apparatus contains pure solvent; the right compartment contains a solution. The two compartments are separated by a semipermeable membrane, which allows the passage of solvent molecules but blocks the passage of solute molecules. At the start, the water levels in the two tubes are equal [see Figure 12.11(a)]. After some time, the level in the right tube begins to rise and continues to go up until equi- librium is reached, that is, until no further change can be observed. The osmotic pressure (π) of a solution is the pressure required to stop osmosis. As shown in Figure 12.11(b), this pressure can be measured directly from the difference in the final fluid levels. What causes water to move spontaneously from left to right in this case? The situation depicted in Figure 12.12 helps us understand the driving force behind Figure 12.12 (a) Unequal vapor pressures inside the container lead to a net transfer of water Net transfer of solvent from the left beaker (which contains pure water) to the right one (which contains a solution). (b) At equilibrium, all the water in the left beaker has been transferred to the right beaker. This driving force for solvent transfer is analogous to the osmotic phenomenon that is shown in Figure 12.11. (a) (b) 540 Chapter 12 ■ Physical Properties of Solutions osmosis. Because the vapor pressure of pure water is higher than the vapor pressure of the solution, there is a net transfer of water from the left beaker to the right one. Given enough time, the transfer will continue until no more water remains in the left beaker. A similar driving force causes water to move from the pure solvent into the solution during osmosis. The osmotic pressure of a solution is given by π 5 MRT (12.8) where M is the molarity of solution, R is the gas constant (0.0821 L ? atm/K ? mol), and T is the absolute temperature. The osmotic pressure, π, is expressed in atm. Because osmotic pressure measurements are carried out at constant temperature, we express the concentration in terms of the more convenient units of molarity rather than molality. Like boiling-point elevation and freezing-point depression, osmotic pressure is directly proportional to the concentration of solution. This is what we would expect, because all colligative properties depend only on the number of solute particles in solution. If two solutions are of equal concentration and, hence, have the same osmotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pres- sures, the more concentrated solution is said to be hypertonic and the more dilute solution is described as hypotonic (Figure 12.13). Although osmosis is a common and well-studied phenomenon, relatively little is known about how the semipermeable membrane stops some molecules yet allows others to pass. In some cases, it is simply a matter of size. A semipermeable membrane Water molecules Solute molecules (a) (b) (c) (d) Figure 12.13 A cell in (a) an isotonic solution, (b) a hypotonic solution, and (c) a hypertonic solution. The cell remains unchanged in (a), swells in (b), and shrinks in (c). (d) From left to right: a red blood cell in an isotonic solution, in a hypotonic solution, and in a hypertonic solution.  12.6 Colligative Properties of Nonelectrolyte Solutions 541 may have pores small enough to let only the solvent molecules through. In other cases, a different mechanism may be responsible for the membrane’s selectivity—for exam- ple, the solvent’s greater “solubility” in the membrane. The osmotic pressure phenomenon manifests itself in many interesting applica- tions. To study the contents of red blood cells, which are protected from the external environment by a semipermeable membrane, biochemists use a technique called hemo- lysis. The red blood cells are placed in a hypotonic solution. Because the hypotonic solution is less concentrated than the interior of the cell, water moves into the cells, as shown in the middle photo of Figure 12.13(d). The cells swell and eventually burst, releasing hemoglobin and other molecules. Home preserving of jam and jelly provides another example of the use of osmotic pressure. A large quantity of sugar is actually essential to the preservation process because the sugar helps to kill bacteria that may cause botulism. As Fig- ure  12.13(c)  shows, when a bacterial cell is in a hypertonic (high-concentration) sugar solution, the intracellular water tends to move out of the bacterial cell to the more concentrated solution by osmosis. This process, known as crenation, causes the cell to shrink and, eventually, to cease functioning. The natural acidity of fruits also inhibits bacteria growth. Osmotic pressure also is the major mechanism for transporting water upward in plants. Because leaves constantly lose water to the air, in a process called transpira- tion, the solute concentrations in leaf fluids increase. Water is pulled up through the trunk, branches, and stems of trees by osmotic pressure. Up to 10 to 15 atm pressure is necessary to transport water to the leaves at the tops of California’s redwoods, which reach about 120 m in height. (The capillary action discussed in Section 11.3 is responsible for the rise of water only up to a few centimeters.) Example 12.9 shows that an osmotic pressure measurement can be used to find the concentration of a solution. California redwoods. Example 12.9 The average osmotic pressure of seawater, measured in the kind of apparatus shown in Figure 12.11, is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous solution of sucrose (C12H22O11) that is isotonic with seawater. Strategy When we say the sucrose solution is isotonic with seawater, what can we conclude about the osmotic pressures of these two solutions? Solution A solution of sucrose that is isotonic with seawater must have the same osmotic pressure, 30.0 atm. Using Equation (12.8). π 5 MRT π 30.0 atm M5 5 RT (0.0821 L ? atm/K ? mol) (298 K) 5 1.23 mol/L 5 1.23 M Similar problem: 12.63. Practice Exercise What is the osmotic pressure (in atm) of a 0.884 M urea solution at 16°C? Review of Concepts What does it mean when we say that the osmotic pressure of a sample of seawater is 25 atm at a certain temperature? 542 Chapter 12 ■ Physical Properties of Solutions Using Colligative Properties to Determine Molar Mass The colligative properties of nonelectrolyte solutions provide a means of determin- ing the molar mass of a solute. Theoretically, any of the four colligative properties is suitable for this purpose. In practice, however, only freezing-point depression and osmotic pressure are used because they show the most pronounced changes. The procedure is as follows. From the experimentally determined freezing-point depression or osmotic pressure, we can calculate the molality or molarity of the solution. Knowing the mass of the solute, we can readily determine its molar mass, as Examples 12.10 and 12.11 demonstrate. Example 12.10 A 7.85-g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are the molar mass and molecular formula of this compound? Strategy Solving this problem requires three steps. First, we calculate the molality of the solution from the depression in freezing point. Next, from the molality we determine the number of moles in 7.85 g of the compound and hence its molar mass. Finally, comparing the experimental molar mass with the empirical molar mass enables us to write the molecular formula. Solution The sequence of conversions for calculating the molar mass of the compound is freezing-point ¡ molality ¡ number of ¡ molar mass depression moles Our first step is to calculate the molality of the solution. From Equation (12.7) and Table 12.2 we write ¢Tf 1.05°C molality 5 5 5 0.205 m Kf 5.12°C/m Because there is 0.205 mole of the solute in 1 kg of solvent, the number of moles of solute in 301 g, or 0.301 kg, of solvent is 0.205 mol 0.301 kg 3 5 0.0617 mol 1 kg Thus, the molar mass of the solute is grams of compound molar mass 5 moles of compound 7.85 g 5 5 127 g/mol 0.0617 mol Now we can determine the ratio C10H8 molar mass 127 g/mol 5 <2 empirical molar mass 64 g/mol Similar problem: 12.57. Therefore, the molecular formula is (C5H4)2 or C10H8 (naphthalene). Practice Exercise A solution of 0.85 g of an organic compound in 100.0 g of benzene has a freezing point of 5.16°C. What are the molality of the solution and the molar mass of the solute?  12.6 Colligative Properties of Nonelectrolyte Solutions 543 Example 12.11 A solution is prepared by dissolving 35.0 g of hemoglobin (Hb) in enough water to make up 1 L in volume. If the osmotic pressure of the solution is found to be 10.0 mmHg at 25°C, calculate the molar mass of hemoglobin. Strategy We are asked to calculate the molar mass of Hb. The steps are similar to those outlined in Example 12.10. From the osmotic pressure of the solution, we calculate the molarity of the solution. Then, from the molarity, we determine the number of moles in 35.0 g of Hb and hence its molar mass. What units should we use for π and temperature? Solution The sequence of conversions is as follows: osmotic pressure ¡ molarity ¡ number of moles ¡ molar mass First we calculate the molarity using Equation (12.8) π 5 MRT π M5 RT 1 atm 10.0 mmHg 3 760 mmHg 5 (0.0821 L ? atm/K ? mol) (298 K) 5 5.38 3 1024 M The volume of the solution is 1 L, so it must contain 5.38 3 1024 mol of Hb. We use this quantity to calculate the molar mass: mass of Hb moles of Hb 5 molar mass of Hb mass of Hb molar mass of Hb 5 moles of Hb 35.0 g 5 5.38 3 1024 mol 5 6.51 3 104 g/mol Similar problems: 12.64, 12.66. Practice Exercise A 202-mL benzene solution containing 2.47 g of an organic polymer has an osmotic pressure of 8.63 mmHg at 21°C. Calculate the molar mass of the polymer. A pressure of 10.0 mmHg, as in Example 12.11, can be measured easily and The density of mercury is 13.6 g/mL. accurately. For this reason, osmotic pressure measurements are very useful for deter- Therefore, 10 mmHg corresponds to a column of water 13.6 cm in height. mining the molar masses of large molecules, such as proteins. To see how much more practical the osmotic pressure technique is than freezing-point depression would be, let us estimate the change in freezing point of the same hemoglobin solution. If an aqueous solution is quite dilute, we can assume that molarity is roughly equal to molality. (Molarity would be equal to molality if the density of the aqueous solution were 1 g/mL.) Hence, from Equation (12.7) we write ¢Tf 5 (1.86°C/m)(5.38 3 1024 m) 5 1.00 3 1023°C The freezing-point depression of one-thousandth of a degree is too small a temperature change to measure accurately. For this reason, the freezing-point depression technique 544 Chapter 12 ■ Physical Properties of Solutions is more suitable for determining the molar mass of smaller and more soluble mol- ecules, those having molar masses of 500 g or less, because the freezing-point depres- sions of their solutions are much greater. 12.7 Colligative Properties of Electrolyte Solutions The study of colligative properties of electrolytes requires a slightly different approach than the one used for the colligative properties of nonelectrolytes. The reason is that electrolytes dissociate into ions in solution, and so one unit of an electrolyte compound separates into two or more particles when it dissolves. (Remember, it is the total number of solute particles that determines the colligative properties of a solution.) For example, each unit of NaCl dissociates into two ions— Na1 and Cl2. Thus, the colligative properties of a 0.1 m NaCl solution should be twice as great as those of a 0.1 m solution containing a nonelectrolyte, such as sucrose. Similarly, we would expect a 0.1 m CaCl2 solution to depress the freezing point by three times as much as a 0.1 m sucrose solution because each CaCl2 pro- duces three ions. To account for this effect we define a quantity called the van’t Hoff † factor, given by actual number of particles in soln after dissociation i5 (12.9) number of formula units initially dissolved in soln Every unit of NaCl or KNO3 that dissociates Thus, i should be 1 for all nonelectrolytes. For strong electrolytes such as NaCl yields two ions (i 5 2); every unit of and KNO3, i should be 2, and for strong electrolytes such as Na2SO4 and CaCl2, Na2SO4 or MgCl2 that dissociates produces three ions (i 5 3). i should be 3. Consequently, the equations for colligative properties must be modified as ¢Tb 5 iKbm (12.10) ¢Tf 5 iKf m (12.11) + π 5 iMRT (12.12) – – + In reality, the colligative properties of electrolyte solutions are usually smaller + + than anticipated because at higher concentrations, electrostatic forces come into – play and bring about the formation of ion pairs. An ion pair is made up of one or – more cations and one or more anions held together by electrostatic forces. The presence of an ion pair reduces the number of particles in solution, causing a (a) reduction in the colligative properties (Figure 12.14). Electrolytes containing mul- ticharged ions such as Mg21, Al31, SO422, and PO432 have a greater tendency to – + form ion pairs than electrolytes such as NaCl and KNO3, which are made up of + singly charged ions. – – Table 12.3 shows the experimentally measured values of i and those calculated + + – assuming complete dissociation. As you can see, the agreement is close but not per- fect, indicating that the extent of ion-pair formation in these solutions at that concen- (b) tration is appreciable. Figure 12.14 (a) Free ions and (b) ion pairs in solution. Such an † ion pair bears no net charge and Jacobus Hendricus van’t Hoff (1852–1911). Dutch chemist. One of the most prominent chemists of his therefore cannot conduct time, van’t Hoff did significant work in thermodynamics, molecular structure and optical activity, and electricity in solution. solution chemistry. In 1901 he received the first Nobel Prize in Chemistry.  12.7 Colligative Properties of Electrolyte Solutions 545 Table 12.3 The van’t Hoff Factor of 0.0500 M Electrolyte Solutions at 258C Electrolyte i (Measured) i (Calculated) Sucrose* 1.0 1.0 HCl 1.9 2.0 NaCl 1.9 2.0 MgSO4 1.3 2.0 MgCl2 2.7 3.0 FeCl3 3.4 4.0 *Sucrose is a nonelectrolyte. It is listed here for comparison only. Review of Concepts Indicate which compound in each of the following groups has a greater tendency to form ion pairs in water: (a) NaCl or Na2SO4, (b) MgCl2 or MgSO4, (c) LiBr or KBr. Example 12.12 The osmotic pressure of a 0.010 M potassium iodide (KI) solution at 25°C is 0.465 atm. Calculate the van’t Hoff factor for KI at this concentration. Strategy Note that KI is a strong electrolyte, so we expect it to dissociate completely in solution. If so, its osmotic pressure would be 2(0.010 M) (0.0821 L ? atm/K ? mol) (298 K) 5 0.489 atm However, the measured osmotic pressure is only 0.465 atm. The smaller than predicted osmotic pressure means that there is ion-pair formation, which reduces the number of solute particles (K1 and I2 ions) in solution. Solution From Equation (12.12) we have π i5 MRT 0.465 atm 5 (0.010 M) (0.0821 L ? atm/K ? mol) (298 K) 5 1.90 Similar problem: 12.77. Practice Exercise The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Calculate the van’t Hoff factor of MgSO4 at this concentration. Review of Concepts The osmotic pressure of blood is about 7.4 atm. What is the approximate concentration of a saline solution (NaCl solution) a physician should use for intravenous injection? Use 37°C for physiological temperature. The Chemistry in Action essay on p. 546 describes dialysis, a medical procedure through which a person’s blood is cleansed of toxins. CHEMISTRY in Action Dialysis T he function of the kidneys is to filter metabolic waste prod- ucts such as urea, toxins, excess mineral salts, and water from the blood. In humans, the two kidneys have a combined mass of only about 11 oz. Despite their small size, a large vol- ume of blood (about 1 L/min) flows through the kidneys. Each kidney contains several million filtering units called nephrons, which carry blood from the renal artery to the glomeruli, minute networks of capillaries. In the glomeruli, the normal blood pres- sure forces water and dissolved substances through the pores of the capillaries, but not proteins and red blood cells, because they are too large. The filtered fluid contains many substances the blood cannot lose in large quantities: salts, sugars, amino acids, and water. Most of the substances are reabsorbed into the blood by a process called active transport (movement of a substance from a low-concentration region to a higher concentration one). A patient undergoing dialysis. Much of the water also returns to the blood by osmosis. Eventually, the excess water, mineral salts, and waste substances are dis- charged from the body as urine. from a fluid (the dialysate) that contains many of the blood’s Kidney malfunction can be life threatening. When the kid- components at similar concentrations so that there is no net neys fail, the blood must be purified by a dialysis procedure in passage of them between the blood and the dialysate through which waste products are removed from the body by artificial the membrane. The size of the membrane pores is such that means. In hemodialysis, which is one type of treatment, blood is only small waste products can pass through them into the di- removed from the body and pumped into a machine (called the alysate to be washed away. Again, proteins, cells, and other dialyzer) that filters the toxic substances out of the blood and important blood components are prevented by their size from then returns the purified blood to the patient’s bloodstream. By passing through the membrane and remain in the purified a simple surgical procedure, blood is pumped out of the patient’s (dialyzed) blood. Typically a hemodialysis procedure lasts artery, through a dialyzer, and returned to the vein. Inside the 4 to 6 h. In many cases, one weekly treatment is sufficient to dialyzer, an artificial porous membrane separates the blood purify the blood. 12.8 Colloids The solutions discussed so far are true homogeneous mixtures. Now consider what happens if we add fine sand to a beaker of water and stir. The sand particles are suspended at first but then gradually settle to the bottom. This is an example of a heterogeneous mixture. Between these two extremes is an intermediate state called a colloidal suspension, or simply, a colloid. A colloid is a dispersion of particles of one substance (the dispersed phase) throughout a dispersing medium made of another substance. Colloidal particles are much larger than the normal solute mol- ecules; they range from 1 3 103 pm to 1 3 106 pm. Also, a colloidal suspension lacks the homogeneity of an ordinary solution. The dispersed phase and the dispers- ing medium can be gases, liquids, solids, or a combination of different phases, as shown in Table 12.4. A number of colloids are familiar to us. An aerosol consists of liquid droplets or solid particles dispersed in a gas. Examples are fog and smoke. Mayonnaise, which 546  12.8 Colloids 547 Table 12.4 Types of Colloids Dispersing Dispersed Medium Phase Name Example Gas Liquid Aerosol Fog, mist Gas Solid Aerosol Smoke Liquid Gas Foam Whipped cream Liquid Liquid Emulsion Mayonnaise Liquid Solid Sol Milk of magnesia Solid Gas Foam Plastic foams Figure 12.15 Three beams of Solid Liquid Gel Jelly, butter white light, passing through a Solid Solid Solid sol Certain alloys colloid of sulfur particles in water, change to orange, pink, and (steel), opal bluish-green. The colors produced depend on the size of the particles and also on the position of the viewer. The smaller the dispersed is made by breaking oil into small droplets in water, is an example of emulsion, which particles, the shorter (and bluer) the wavelengths. consists of liquid droplets dispersed in another liquid. Milk of magnesia is an exam- ple of sol, a suspension of solid particles in a liquid. One way to distinguish a solution from a colloid is by the Tyndall† effect. When a beam of light passes through a colloid, it is scattered by the dispersed phase (Figure  12.15). No such scattering is observed with ordinary solutions because the solute molecules are too small to interact with visible light. Another demonstration of the Tyndall effect is the scattering of sunlight by dust or smoke in the air (Figure 12.16). Figure 12.16 Sunlight scattered by dust particles in the air. Hydrophilic and Hydrophobic Colloids Among the most important colloids are those in which the dispersing medium is water. Such colloids are divided into two categories called hydrophilic, or water-loving, and hydrophobic, or water-fearing. Hydrophilic colloids are usually solutions containing extremely large molecules such as proteins. In the aqueous phase, a protein like hemo- globin folds in such a way that the hydrophilic parts of the molecule, the parts that can interact favorably with water molecules by ion-dipole forces or hydrogen-bond formation, are on the outside surface (Figure 12.17). † John Tyndall (1820–1893). Irish physicist. Tyndall did important work in magnetism, and explained glacier motion. –O O Figure 12.17 Hydrophilic groups C on the surface of a large molecule +NH such as protein stabilizes the 3 molecule in water. Note that all these groups can form hydrogen bonds with water. Protein NH2 HO C HO O 548 Chapter 12 ■ Physical Properties of Solutions + + + – + – – Colloidal – Repulsion – – Colloidal + particle particle – + – + + – – + + Figure 12.18 Diagram showing the stabilization of hydrophobic colloids. Negative ions are adsorbed onto the surface and the repulsion between like charges prevents the clumping of the particles. A hydrophobic colloid normally would not be stable in water, and the particles would clump together, like droplets of oil in water merging to form a film of oil at water’s surface. They can be stabilized, however, by adsorption of ions on their sur- face (Figure 12.18). (Adsorption refers to adherence onto a surface. It differs from absorption in that the latter means passage to the interior of the medium.) These adsorbed ions can interact with water, thus stabilizing the colloid. At the same time, the electrostatic repulsion between the particles prevents them from clumping together. Soil particles in rivers and streams are hydrophobic particles stabilized in this way. When the freshwater enters the sea, the charges on the particles are neutralized by the high-salt medium, and the particles clump together to form the silt that is seen at the mouth of the river. Another way hydrophobic colloids can be stabilized is by the presence of other hydrophilic groups on their surfaces. Consider sodium stearate, a soap molecule that has a polar head and a long hydrocarbon tail that is nonpolar (Figure 12.19). The cleansing action of soap is the result of the dual nature of the hydrophobic tail and the hydrophilic end group. The hydrocarbon tail is readily soluble in oily substances, O CH2 CH2 C H2 C H2 C H2 C H2 C H2 C H2 C CH3 CH2 CH2 C H2 C H2 C H2 C H2 C H2 C H2 O – Na+ Sodium stearate (C17H35 COO – Na+) (a) Hydrophilic head Hydrophobic tail (b) Figure 12.19 (a) A sodium stearate molecule. (b) The simplified representation of the molecule that shows a hydrophilic head and a hydrophobic tail.  Summary of Facts & Concepts 549 Grease (a) (b) (c) Figure 12.20 The cleansing action of soap. (a) Grease (oily substance) is not soluble in water. (b) When soap is added to water, the nonpolar tails of soap molecules dissolve in grease. (c) Finally, the grease is removed in the form of an emulsion. Note that each oily droplet now has an ionic exterior that is hydrophilic. which are also nonpolar, while the ionic ¬COO 2 group remains outside the oily surface. When enough soap molecules have surrounded an oil droplet, as shown in Figure 12.20, the entire system becomes solubilized in water because the exterior portion is now largely hydrophilic. This is how greasy substances are removed by the action of soap. Key Equations mass of solute percent by mass 5 3 100% (12.1) Calculating the percent by mass of a solution. mass of soln moles of solute molality (m) 5 (12.2) Calculating the molality of a solution. mass of solvent (kg) c 5 kP (12.3) Henry’s law for calculating solubility of gases. P1 5 X1P°1 (12.4) Raoult’s law relating the vapor pressure of a liquid to its vapor pressure in a solution. ¢P 5 X2P°1 (12.5) Vapor pressure lowering in terms of the concentration of solution. ¢Tb 5 Kb m (12.6) Boiling-point elevation. ¢Tf 5 Kf m (12.7) Freezing-point depression. π 5 MRT (12.8) Osmotic pressure of a solution. actual number of particles in soln after dissociation i5 (12.9) Calculating the van’t Hoff factor for an number of formula units initially dissolved in soln electrolyte solution. Summary of Facts & Concepts 1. Solutions are homogeneous mixtures of two or more 4. Increasing temperature usually increases the solubility substances, which may be solids, liquids, or gases. of solid and liquid substances and usually decreases the 2. The ease of dissolving a solute in a solvent is governed solubility of gases in water. by intermolecular forces. Energy and the disorder that 5. According to Henry’s law, the solubility of a gas in a results when molecules of the solute and solvent mix to liquid is directly proportional to the partial pressure of form a solution are the forces driving the solution process. the gas over the solution. 3. The concentration of a solution can be expressed as per- 6. Raoult’s law states that the partial pressure of a sub- cent by mass, mole fraction, molarity, and molality. The stance A over a solution is equal to the mole fraction choice of units depends on the circumstances. (XA) of A times the vapor pressure (P°A) of pure A. An  550 Chapter 12 ■ Physical Properties of Solutions ideal solution obeys Raoult’s law over the entire range provides a measure of the extent of dissociation of elec- of concentration. In practice, very few solutions exhibit trolytes in solution. ideal behavior. 9. A colloid is a dispersion of particles (about 1 3 103 7. Vapor-pressure lowering, boiling-point elevation, freezing- pm to 1 3 106 pm) of one substance in another sub- point depression, and osmotic pressure are colligative stance. A colloid is distinguished from a regular solu- properties of solutions; that is, they depend only on the tion by the Tyndall effect, which is the scattering of number of solute particles that are present and not on visible light by colloidal particles. Colloids in water their nature. are classified as hydrophilic colloids and hydrophobic 8. In electrolyte solutions, the interaction between ions colloids. leads to the formation of ion pairs. The van’t Hoff factor Key Words Boiling-point elevation Freezing-point depression Molality, p. 523 Semipermeable (≤Tb), p. 536 (≤Tf), p. 537 Nonvolatile, p. 532 membrane, p. 539 Colligative properties, p. 532 Henry’s law, p. 529 Osmosis, p. 539 Solvation, p. 521 Colloid, p. 546 Hydrophilic, p. 547 Osmotic pressure Supersaturated Crystallization, p. 519 Hydrophobic, p. 547 (π), p. 539 solution, p. 519 Fractional Ideal solution, p. 534 Percent by mass, p. 522 Unsaturated solution, p. 519 crystallization, p. 527 Ion pair, p. 544 Raoult’s law, p. 532 van’t Hoff factor (i), p. 544 Fractional distillation, p. 535 Miscible, p. 521 Saturated solution, p. 519 Volatile, p. 533 Questions & Problems • Problems available in Connect Plus 12.7 Explain why the solution process usually leads to an Red numbered problems solved in Student Solutions Manual increase in disorder. • 12.8 Describe the factors that affect the solubility of a Types of Solutions solid in a liquid. What does it mean to say that two Review Questions liquids are miscible? 12.1 Distinguish between an unsaturated solution, a satu- Problems rated solution, and a supersaturated solution. 12.9 Why is naphthalene (C10H8) more soluble than CsF 12.2 From which type of solution listed in Question 12.1 in benzene? does crystallization or precipitation occur? How does a crystal differ from a precipitate? • 12.10 Explain why ethanol (C2H5OH) is not soluble in cyclohexane (C6H12). A Molecular View of the Solution Process • 12.11 Arrange the following compounds in order of in- Review Questions creasing solubility in water: O2, LiCl, Br2, methanol (CH3OH). • 12.3 Briefly describe the solution process at the molecu- 12.12 Explain the variations in solubility in water of the lar level. Use the dissolution of a solid in a liquid as alcohols listed here: an example. • 12.4 Basing your answer on intermolecular force consid- Compound Solubility in Water (g/100 g) at 208C erations, explain what “like dissolves like” means. 12.5 What is solvation? What factors influence the CH3OH q extent to which solvation occurs? Give two exam- CH3CH2OH q ples of solvation; include one that involves ion- CH3CH2CH2OH q dipole interaction and one in which dispersion forces come into play. CH3CH2CH2CH2OH 9 12.6 As you know, some solution processes are endother- CH3CH2CH2CH2CH2OH 2.7 mic and others are exothermic. Provide a molecular (Note: q means that the alcohol and water are completely miscible in interpretation for the difference. all proportions.)  Questions & Problems 551 Concentration Units • 12.24 The density of an aqueous solution containing Review Questions 10.0 percent of ethanol (C2H5OH) by mass is 0.984 g/mL. (a) Calculate the molality of this 12.13 Define the following concentration terms and solution. (b) Calculate its molarity. (c) What vol- give their units: percent by mass, mole fraction, ume of the solution would contain 0.125 mole of molarity, molality. Compare their advantages and ethanol? disadvantages. 12.14 Outline the steps required for conversion between molarity, molality, and percent by mass. The Effect of Temperature on Solubility Review Questions Problems 12.25 How do the solubilities of most ionic compounds in water change with temperature? With pressure? • 12.15 Calculate the percent by mass of the solute in 12.26 Describe the fractional crystallization process and each of the following aqueous solutions: (a) 5.50 g of NaBr in 78.2 g of solution, (b) 31.0 g of KCl its application. in 152 g of water, (c) 4.5 g of toluene in 29 g of benzene. Problems • 12.16 Calculate the amount of water (in grams) that must • 12.27 A 3.20-g sample of a salt dissolves in 9.10 g of be added to (a) 5.00 g of urea (NH2)2CO in the prep- water to give a saturated solution at 25°C. What aration of a 16.2 percent by mass solution, and is the solubility (in g salt/100 g of H 2O) of the (b) 26.2 g of MgCl2 in the preparation of a 1.5 per- salt? cent by mass solution. • 12.17 Calculate the molality of each of the following solu- • 12.28 The solubility of KNO3 is 155 g per 100 g of water at 75°C and 38.0 g at 25°C. What mass (in grams) tions: (a) 14.3 g of sucrose (C12H22O11) in 676 g of of KNO3 will crystallize out of solution if exactly water, (b) 7.20 moles of ethylene glycol (C2H6O2) in 100 g of its saturated solution at 75°C is cooled 3546 g of water. to 25°C? • 12.18 Calculate the molality of each of the following aque- • 12.29 A 50-g sample of impure KClO3 (solubility 5 7.1 g per ous solutions: (a) 2.50 M NaCl solution (density of 100 g H2O at 20°C) is contaminated with 10 percent solution 5 1.08 g/mL), (b) 48.2 percent by mass of KCl (solubility 5 25.5 g per 100 g of H2O at KBr solution. 20°C). Calculate the minimum quantity of 20°C • 12.19 Calculate the molalities of the following aqueous water needed to dissolve all the KCl from the solutions: (a) 1.22 M sugar (C12H22O11) solution sample. How much KClO3 will be left after this (density of solution 5 1.12 g/mL), (b) 0.87 M treatment? (Assume that the solubilities are unaf- NaOH solution (density of solution 5 1.04 g/mL), fected by the presence of the other compound.) (c) 5.24 M NaHCO3 solution (density of solution 5 1.19 g/mL). 12.20 For dilute aqueous solutions in which the density Gas Solubility of the solution is roughly equal to that of the Review Questions pure solvent, the molarity of the solution is equal to its molality. Show that this statement is 12.30 Discuss the factors that influence the solubility of a correct for a 0.010 M aqueous urea (NH 2) 2CO gas in a liquid. solution. 12.31 What is thermal pollution? Why is it harmful to • 12.21 The alcohol content of hard liquor is normally aquatic life? given in terms of the “proof,” which is defined as 12.32 What is Henry’s law? Define each term in the equa- twice the percentage by volume of ethanol tion, and give its units. How would you account for (C2H5OH) present. Calculate the number of grams the law in terms of the kinetic molecular theory of of alcohol present in 1.00 L of 75-proof gin. The gases? Give two exceptions to Henry’s law. density of ethanol is 0.798 g/mL. 12.33 A student is observing two beakers of water. One • 12.22 The concentrated sulfuric acid we use in the labora- beaker is heated to 30°C, and the other is heated to tory is 98.0 percent H2SO4 by mass. Calculate the 100°C. In each case, bubbles form in the water. Are molality and molarity of the acid solution. The den- these bubbles of the same origin? Explain. sity of the solution is 1.83 g/mL. 12.34 A man bought a goldfish in a pet shop. Upon return- • 12.23 Calculate the molarity and the molality of ing home, he put the goldfish in a bowl of recently an NH3 solution made up of 30.0 g of NH 3 in boiled water that had been cooled quickly. A few 70.0 g of water. The density of the solution is minutes later the fish was found dead. Explain what 0.982 g/mL. happened to the fish.  552 Chapter 12 ■ Physical Properties of Solutions Problems Problems 12.35 A beaker of water is initially saturated with dissolved • 12.49 A solution is prepared by dissolving 396 g of su- air. Explain what happens when He gas at 1 atm is crose (C12H22O11) in 624 g of water. What is the bubbled through the solution for a long time. vapor pressure of this solution at 30°C? (The vapor 12.36 A miner working 260 m below sea level opened a pressure of water is 31.8 mmHg at 30°C.) carbonated soft drink during a lunch break. To his • 12.50 How many grams of sucrose (C12H22O11) must be surprise, the soft drink tasted rather “flat.” Shortly added to 552 g of water to give a solution with a afterward, the miner took an elevator to the surface. vapor pressure 2.0 mmHg less than that of pure During the trip up, he could not stop belching. Why? water at 20°C? (The vapor pressure of water at 20°C • 12.37 The solubility of CO2 in water at 25°C and 1 atm is is 17.5 mmHg.) 0.034 mol/L. What is its solubility under atmospheric • 12.51 The vapor pressure of benzene is 100.0 mmHg at conditions? (The partial pressure of CO2 in air is 26.1°C. Calculate the vapor pressure of a solution 0.0003 atm.) Assume that CO2 obeys Henry’s law. containing 24.6 g of camphor (C10H16O) dissolved • 12.38 The solubility of N2 in blood at 37°C and at a partial in 98.5 g of benzene. (Camphor is a low-volatility pressure of 0.80 atm is 5.6 3 1024 mol/L. A deep- solid.) sea diver breathes compressed air with the partial • 12.52 The vapor pressures of ethanol (C2H5OH) and pressure of N2 equal to 4.0 atm. Assume that the to- 1-propanol (C3H7OH) at 35°C are 100 mmHg and tal volume of blood in the body is 5.0 L. Calculate 37.6 mmHg, respectively. Assume ideal behavior the amount of N2 gas released (in liters at 37°C and and calculate the partial pressures of ethanol and 1 atm) when the diver returns to the surface of the 1-propanol at 35°C over a solution of ethanol in water, where the partial pressure of N2 is 0.80 atm. 1-propanol, in which the mole fraction of ethanol is 0.300. Colligative Properties of • 12.53 The vapor pressure of ethanol (C2H5OH) at 20°C Nonelectrolyte Solutions is 44 mmHg, and the vapor pressure of methanol Review Questions (CH3OH) at the same temperature is 94 mmHg. A mixture of 30.0 g of methanol and 45.0 g of etha- 12.39 What are colligative properties? What is the mean- nol is prepared (and can be assumed to behave as ing of the word “colligative” in this context? an ideal solution). (a) Calculate the vapor pressure 12.40 Write the equation representing Raoult’s law, and of methanol and ethanol above this solution at express it in words. 20°C. (b) Calculate the mole fraction of methanol 12.41 Use a solution of benzene in toluene to explain what and ethanol in the vapor above this solution at is meant by an ideal solution. 20°C. (c) Suggest a method for separating the two 12.42 Write the equations relating boiling-point elevation components of the solution. and freezing-point depression to the concentration • 12.54 How many grams of urea [(NH2)2CO] must be of the solution. Define all the terms, and give their added to 450 g of water to give a solution with a units. vapor pressure 2.50 mmHg less than that of pure 12.43 How is vapor-pressure lowering related to a rise in water at 30°C? (The vapor pressure of water at 30°C the boiling point of a solution? is 31.8 mmHg.) 12.44 Use a phase diagram to show the difference in freez- • 12.55 What are the boiling point and freezing point of a ing points and boiling points between an aqueous 2.47 m solution of naphthalene in benzene? (The urea solution and pure water. boiling point and freezing point of benzene are 12.45 What is osmosis? What is a semipermeable mem- 80.1°C and 5.5°C, respectively.) brane? • 12.56 An aqueous solution contains the amino acid gly- 12.46 Write the equation relating osmotic pressure to the cine (NH2CH2COOH). Assuming that the acid does concentration of a solution. Define all the terms and not ionize in water, calculate the molality of the so- specify their units. lution if it freezes at 21.1°C. 12.47 Explain why molality is used for boiling-point el- • 12.57 Pheromones are compounds secreted by the evation and freezing-point depression calcula- females of many insect species to attract males. tions and molarity is used in osmotic pressure One of these compounds contains 80.78 percent calculations. C, 13.56 percent H, and 5.66 percent O. A solu- 12.48 Describe how you would use freezing-point depres- tion of 1.00 g of this pheromone in 8.50 g of sion and osmotic pressure measurements to deter- benzene freezes at 3.37°C. What are the molecu- mine the molar mass of a compound. Why are lar formula and molar mass of the compound? boiling-point elevation and vapor-pressure lowering (The normal freezing point of pure benzene is normally not used for this purpose? 5.50°C.)  Questions & Problems 553 • 12.58 The elemental analysis of an organic solid ex- 12.68 What is the van’t Hoff factor? What information tracted from gum arabic (a gummy substance used does it provide? in adhesives, inks, and pharmaceuticals) showed that it contained 40.0 percent C, 6.7 percent H, and 53.3 percent O. A solution of 0.650 g of the solid in Problems 27.8 g of the solvent diphenyl gave a freezing-point • 12.69 Which of the following aqueous solutions has (a) the depression of 1.56°C. Calculate the molar mass higher boiling point, (b) the higher freezing point, and molecular formula of the solid. (Kf for diphe- and (c) the lower vapor pressure: 0.35 m CaCl2 or nyl is 8.00°C/m.) 0.90 m urea? Explain. Assume CaCl2 to undergo • 12.59 How many liters of the antifreeze ethylene glycol complete dissociation. [CH2(OH)CH2(OH)] would you add to a car radiator 12.70 Consider two aqueous solutions, one of sucrose containing 6.50 L of water if the coldest winter tem- (C12H22O11) and the other of nitric acid (HNO3). perature in your area is 220°C? Calculate the boil- Both solutions freeze at 21.5°C. What other proper- ing point of this water-ethylene glycol mixture. (The ties do these solutions have in common? density of ethylene glycol is 1.11 g/mL.) • 12.71 Arrange the following solutions in order of de- • 12.60 A solution is prepared by condensing 4.00 L of a creasing freezing point: 0.10 m Na3PO4, 0.35 m gas, measured at 27°C and 748 mmHg pressure, into NaCl, 0.20 m MgCl2, 0.15 m C6H12O6, 0.15 m 58.0 g of benzene. Calculate the freezing point of CH3COOH. this solution. • 12.72 Arrange the following aqueous solutions in order 12.61 The molar mass of benzoic acid (C6H5COOH) of decreasing freezing point, and explain your determined by measuring the freezing-point depres- reasoning: 0.50 m HCl, 0.50 m glucose, 0.50 m sion in benzene is twice what we would expect for acetic acid. the molecular formula, C7H6O2. Explain this appar- ent anomaly. • 12.73 What are the normal freezing points and boiling points of the following solutions? (a) 21.2 g NaCl in 12.62 A solution of 2.50 g of a compound having the 135 mL of water and (b) 15.4 g of urea in 66.7 mL empirical formula C6H5P in 25.0 g of benzene is of water observed to freeze at 4.3°C. Calculate the molar mass of the solute and its molecular formula. • 12.74 At 25°C the vapor pressure of pure water is 23.76 mmHg and that of seawater is 22.98 mmHg. • 12.63 What is the osmotic pressure (in atm) of a 1.36 M Assuming that seawater contains only NaCl, esti- aqueous solution of urea [(NH2)2CO] at 22.0°C? mate its molal concentration. 12.64 A solution containing 0.8330 g of a polymer of 12.75 Both NaCl and CaCl2 are used to melt ice on roads unknown structure in 170.0 mL of an organic and sidewalks in winter. What advantages do these solvent was found to have an osmotic pressure of substances have over sucrose or urea in lowering the 5.20 mmHg at 25°C. Determine the molar mass of freezing point of water? the polymer. 12.76 A 0.86 percent by mass solution of NaCl is called • 12.65 A quantity of 7.480 g of an organic compound is “physiological saline” because its osmotic pressure dissolved in water to make 300.0 mL of solution. is equal to that of the solution in blood cells. Calcu- The solution has an osmotic pressure of 1.43 atm at late the osmotic pressure of this solution at normal 27°C. The analysis of this compound shows that it body temperature (37°C). Note that the density of contains 41.8 percent C, 4.7 percent H, 37.3 percent the saline solution is 1.005 g/mL. O, and 16.3 percent N. Calculate the molecular for- mula of the compound. • 12.77 The osmotic pressure of 0.010 M solutions of CaCl2 and urea at 25°C are 0.605 atm and 0.245 atm, • 12.66 A solution of 6.85 g of a carbohydrate in 100.0 g of respectively. Calculate the van’t Hoff factor for the water has a density of 1.024 g/mL and an osmotic CaCl2 solution. pressure of 4.61 atm at 20.0°C. Calculate the molar mass of the carbohydrate. • 12.78 Calculate the osmotic pressure of a 0.0500 M MgSO4 solution at 25°C. (Hint: See Table 12.3.) Colligative Properties of Electrolyte Solutions Review Questions Colloids Review Questions 12.67 What are ion pairs? What effect does ion-pair for- mation have on the colligative properties of a solu- 12.79 What are colloids? Referring to Table 12.4, why is tion? How does the ease of ion-pair formation there no colloid in which both the dispersed phase depend on (a) charges on the ions, (b) size of the and the dispersing medium are gases? ions, (c) nature of the solvent (polar versus nonpo- 12.80 Describe how hydrophilic and hydrophobic colloids lar), (d) concentration? are stabilized in water.  554 Chapter 12 ■ Physical Properties of Solutions Additional Problems 12.90 Calculate the mass of naphthalene (C10H8) that must be added to 250 g of benzene (C6H6) to give a solu- 12.81 Aqueous solutions A and B both contain urea at dif- tion with a freezing point 2.00°C below that of pure ferent concentrations. On standing exposed to air, benzene. the vapor pressure of A remains constant while that of B gradually decreases. (a) Which solution has a • 12.91 Consider the three mercury manometers shown higher boiling point? (b) Eventually the two solu- here. One of them has 1 mL of water on top of the tions have the same vapor pressure. Explain. mercury, another has 1 mL of a 1 m urea solution on top of the mercury, and the third one has 1 mL of a 12.82 Water and methanol are miscible with each other 1  m NaCl solution placed on top of the mercury. but they are immiscible with octane (C8H18). Which of these solutions is in the tube labeled X, Which of the following shows the correct picture which is in Y, and which is in Z? when equal volumes of these three liquids are mixed in a test tube at 20°C? Assume volumes to X Y Z be additive. (The densities of the liquids are meth- anol: 0.792 g/mL; octane: 0.703 g/mL; water: 0.998 g/mL.) (a) (b) (c) (d) 12.92 A forensic chemist is given a white powder for anal- • 12.83 Lysozyme is an enzyme that cleaves bacterial cell ysis. She dissolves 0.50 g of the substance in 8.0 g of walls. A sample of lysozyme extracted from egg benzene. The solution freezes at 3.9°C. Can the white has a molar mass of 13,930 g. A quantity of chemist conclude that the compound is cocaine 0.100 g of this enzyme is dissolved in 150 g of water (C17H21NO4)? What assumptions are made in the at 25°C. Calculate the vapor-pressure lowering, the analysis? depression in freezing point, the elevation in boiling 12.93 “Time-release” drugs have the advantage of releas- point, and the osmotic pressure of this solution. (The ing the drug to the body at a constant rate so that the vapor pressure of water at 25°C is 23.76 mmHg.) drug concentration at any time is not too high as to • 12.84 Solutions A and B have osmotic pressures of 2.4 atm have harmful side effects or too low as to be ineffec- and 4.6 atm, respectively, at a certain temperature. tive. A schematic diagram of a pill that works on this What is the osmotic pressure of a solution prepared basis is shown below. Explain how it works. by mixing equal volumes of A and B at the same temperature? 12.85 A cucumber placed in concentrated brine (salt water) Elastic shrivels into a pickle. Explain. impermeable • 12.86 Two liquids A and B have vapor pressures of membrane 76 mmHg and 132 mmHg, respectively, at 25°C. Semipermeable What is the total vapor pressure of the ideal solution membrane Saturated made up of (a) 1.00 mole of A and 1.00 mole of B Drug NaCl solution and (b) 2.00 moles of A and 5.00 moles of B? • 12.87 Calculate the van’t Hoff factor of Na3PO4 in a 0.40 m solution whose freezing point is 22.6°C. Rigid wall containing • 12.88 A 262-mL sample of a sugar solution containing tiny holes 1.22 g of the sugar has an osmotic pressure of 30.3 mmHg at 35°C. What is the molar mass of the sugar? • 12.94 A solution of 1.00 g of anhydrous aluminum chlo- 12.89 An aqueous solution of a 0.10 M monoprotic acid ride, AlCl3, in 50.0 g of water freezes at 21.11°C. HA has an osmotic pressure of 3.22 atm at 25°C. Does the molar mass determined from this freezing What is the percent ionization of the acid at this point agree with that calculated from the formula? concentration? Why?  Questions & Problems 555 12.95 Desalination is a process of removing dissolved of the volume of O2 collected to the initial volume of salts from seawater. (a) Briefly describe how you the H2O2 solution. would apply distillation and freezing for this pur- • 12.100 State which of the alcohols listed in Problem 12.12 pose. (b) Desalination can also be accomplished by you would expect to be the best solvent for each of reverse osmosis, which uses high pressure to force the following substances, and explain why: (a) I2, water from a more concentrated solution to a less (b) KBr, (c) CH3CH2CH2CH2CH3. concentrated one. Assuming a sample of seawater is 12.101 Before a carbonated beverage bottle is sealed, it is 0.50 M in NaCl, calculate the minimum pressure pressurized with a mixture of air and carbon diox- that needs to be applied for reverse osmosis at 25°C. ide. (a) Explain the effervescence that occurs when What is the main advantage of reverse osmosis over the cap of the bottle is removed. (b) What causes the distillation and freezing? fog to form near the mouth of the bottle right after the cap is removed? Pressure 12.102 Iodine (I2) is only sparingly soluble in water (left photo). Yet upon the addition of iodide ions (for ex- ample, from KI), iodine is converted to the triiodide Semipermeable ion, which readily dissolves (right photo): membrane I2 (s) 1 I2 (aq) Δ I2 3 (aq) Describe the change in solubility of I2 in terms of the Freshwater Seawater change in intermolecular forces. 12.96 Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxy- gen and nitrogen in air to be 0.20 atm and 0.80 atm, respectively, calculate the mole fractions of oxygen and nitrogen in water at 298 K. Comment on your results. See Example 12.6 for Henry’s law constants. • 12.97 A protein has been isolated as a salt with the formula Na20P (this notation means that there are 20 Na1 • 12.103 Two beakers, one containing a 50-mL aqueous ions associated with a negatively charged protein 1.0 M glucose solution and the other a 50-mL P202). The osmotic pressure of a 10.0-mL solution aqueous 2.0 M glucose solution, are placed un- containing 0.225 g of the protein is 0.257 atm at der a tightly sealed bell jar at room temperature. 25.0°C. (a) Calculate the molar mass of the protein What are the volumes in these two beakers at from these data. (b) Calculate the actual molar mass equilibrium? of the protein. • 12.104 In the apparatus shown here, what will happen if the • 12.98 A nonvolatile organic compound Z was used to membrane is (a) permeable to both water and the make up two solutions. Solution A contains 5.00 g Na1 and Cl2 ions, (b) permeable to water and Na1 of Z dissolved in 100 g of water, and solution B ions but not to Cl2 ions, (c) permeable to water but contains 2.31 g of Z dissolved in 100 g of ben- not to Na1 and Cl2 ions? zene. Solution A has a vapor pressure of 754.5 mmHg at the normal boiling point of water, and solution B has the same vapor pressure at the nor- mal boiling point of benzene. Calculate the molar Membrane mass of Z in solutions A and B and account for the difference. • 12.99 Hydrogen peroxide with a concentration of 3.0 per- cent (3.0 g of H2O2 in 100 mL of solution) is sold in 0.01 M 0.1 M drugstores for use as an antiseptic. For a 10.0-mL NaCl NaCl 3.0 percent H2O2 solution, calculate (a) the oxygen gas produced (in liters) at STP when the compound undergoes complete decomposition and (b) the ratio  556 Chapter 12 ■ Physical Properties of Solutions 12.105 Explain why it is essential that fluids used in intra- • 12.113 A solution contains two volatile liquids A and B. venous injections have approximately the same Complete the following table, in which the symbol osmotic pressure as blood. · indicates attractive intermolecular forces. 12.106 Concentrated hydrochloric acid is usually available at a concentration of 37.7 percent by mass. What is Deviation from its molar concentration? (The density of the solution Attractive Forces Raoult’s Law DHsoln is 1.19 g/mL.) A · A, B · B . 12.107 Explain each of the following statements: (a) The A·B boiling point of seawater is higher than that of pure water. (b) Carbon dioxide escapes from the Negative solution when the cap is removed from a carbon- Zero ated soft-drink bottle. (c) Molal and molar con- centrations of dilute aqueous solutions are 12.114 The concentration of commercially available con- approximately equal. (d) In discussing the colli- centrated sulfuric acid is 98.0 percent by mass, or gative properties of a solution (other than osmotic 18 M. Calculate the density and the molality of the pressure), it is preferable to express the concen- solution. tration in units of molality rather than in molarity. (e) Methanol (b.p. 65°C) is useful as an antifreeze, • 12.115 The concentration of commercially available con- centrated nitric acid is 70.0 percent by mass, or but it should be removed from the car radiator 15.9 M. Calculate the density and the molality of during the summer season. the solution. • 12.108 A mixture of NaCl and sucrose (C12H22O11) of com- • 12.116 A mixture of ethanol and 1-propanol behaves ideally bined mass 10.2 g is dissolved in enough water to at 36°C and is in equilibrium with its vapor. If the make up a 250 mL solution. The osmotic pressure of mole fraction of ethanol in the solution is 0.62, calcu- the solution is 7.32 atm at 23°C. Calculate the mass late its mole fraction in the vapor phase at this tem- percent of NaCl in the mixture. perature. (The vapor pressures of pure ethanol and 12.109 A 0.050 M hydrofluoric acid (HF) solution is 11 per- 1-propanol at 36°C are 108 mmHg and 40.0 mmHg, cent ionized at 25°C. Calculate the osmotic pressure respectively.) of the solution. 12.117 For ideal solutions, the volumes are additive. This 12.110 Shown here is a plot of vapor pressures of two liq- means that if 5 mL of A and 5 mL of B form an uids A and B at different concentrations at a certain ideal solution, the volume of the solution is 10 mL. temperature. Which of the following statements are Provide a molecular interpretation for this observa- false? (a) The solutions exhibit negative deviation tion. When 500 mL of ethanol (C2H5OH) are mixed from Raoult’s law. (b) A and B molecules attract with 500 mL of water, the final volume is less than each other more weakly than they do their own kind. 1000 mL. Why? (c) ≤Hsoln is positive. (d) At XA 5 0.20, the solution has a higher boiling point than liquid B and a lower 12.118 Ammonia (NH3) is very soluble in water, but nitro- boiling point than liquid A. gen trichloride (NCl3) is not. Explain. 12.119 Aluminum sulfate [Al2(SO4)3] is sometimes used in municipal water treatment plants to remove undesirable particles. Explain how this process o PA works. • 12.120 Acetic acid is a weak acid that ionizes in solution as Pressure follows: P Bo CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq) If the freezing point of a 0.106 m CH3COOH solu- tion is 20.203°C, calculate the percent of the acid 0 0.2 0.4 0.6 0.8 1.0 that has undergone ionization. XA 12.121 Making mayonnaise involves beating oil into • 12.111 A 1.32-g sample of a mixture of cyclohexane (C6H12) small droplets in water, in the presence of egg and naphthalene (C10H8) is dissolved in 18.9 g of yolk. What is the purpose of the egg yolk? (Hint: benzene (C6H6). The freezing point of the solution is Egg yolk contains lecithins, which are molecules 2.2°C. Calculate the mass percent of the mixture. with a polar head and a long nonpolar hydrocar- (See Table 12.2 for constants.) bon tail.) • 12.112 How does each of the following affect the solubility • 12.122 Acetic acid is a polar molecule and can form hydro- of an ionic compound? (a) Lattice energy, (b) solvent gen bonds with water molecules. Therefore, it has a (polar versus nonpolar), (c) enthalpies of hydration high solubility in water. Yet acetic acid is also solu- of cation and anion. ble in benzene (C6H6), a nonpolar solvent that lacks  Questions & Problems 557 the ability to form hydrogen bonds. A solution of A has a vapor pressure of 95 mmHg and B has a 3.8 g of CH3COOH in 80 g C6H6 has a freezing vapor pressure of 42 mmHg. A solution is pre- point of 3.5°C. Calculate the molar mass of the sol- pared by mixing equal masses of A and B. (a) Cal- ute and suggest what its structure might be. (Hint: culate the mole fraction of each component in the Acetic acid molecules can form hydrogen bonds solution. (b) Calculate the partial pressures of A between themselves.) and B over the solution at 55°C. (c) Suppose that • 12.123 A 2.6-L sample of water contains 192 μg of lead. some of the vapor described in (b) is condensed to Does this concentration of lead exceed the safety a liquid in a separate container. Calculate the mole limit of 0.050 ppm of lead per liter of drinking wa- fraction of each component in this liquid and the ter? [Hint: 1 μg 5 1 3 1026 g. Parts per million vapor pressure of each component above this (ppm) is defined as (mass of component/mass of liquid at 55°C. solution) 3 106.] • 12.131 A very long pipe is capped at one end with a 12.124 Certain fishes in the Antarctic Ocean swim in water semipermeable membrane. How deep (in meters) at about 22°C. (a) To prevent their blood from must the pipe be immersed into the sea for freshwa- freezing, what must be the concentration (in molal- ter to begin to pass through the membrane? Assume ity) of the blood? Is this a reasonable physiological the water to be at 20°C and treat it as a 0.70 M NaCl concentration? (b) In recent years scientists have solution. The density of seawater is 1.03 g/cm3 and discovered a special type of protein in these fishes’ the acceleration due to gravity is 9.81 m/s2. blood which, although present in quite low concen- 12.132 Two beakers, 1 and 2, containing 50 mL of 0.10 M trations (# 0.001 m), has the ability to prevent the urea and 50 mL of 0.20 M urea, respectively, are blood from freezing. Suggest a mechanism for its placed under a tightly sealed container (see Figure action. 12.12) at 298 K. Calculate the mole fraction of 12.125 As we know, if a soft drink can is shaken and then urea in the solutions at equilibrium. Assume ideal opened, the drink escapes violently. However, if behavior. after shaking the can we tap it several times with a • 12.133 A mixture of liquids A and B exhibits ideal behavior. metal spoon, no such “explosion” of the drink At 84°C, the total vapor pressure of a solution occurs. Why? containing 1.2 moles of A and 2.3 moles of B is 12.126 Why are ice cubes (for example, those you see in 331 mmHg. Upon the addition of another mole of the trays in the freezer of a refrigerator) cloudy B to the solution, the vapor pressure increases to inside? 347 mmHg. Calculate the vapor pressures of pure A and B at 84°C. 12.127 Two beakers are placed in a closed container. Bea- ker A initially contains 0.15 mole of naphthalene 12.134 Use Henry’s law and the ideal gas equation to prove (C10H8) in 100 g of benzene (C6H6) and beaker B the statement that the volume of a gas that dissolves initially contains 31 g of an unknown compound in a given amount of solvent is independent of the dissolved in 100 g of benzene. At equilibrium, bea- pressure of the gas. (Hint: Henry’s law can be modi- ker A is found to have lost 7.0 g of benzene. fied as n 5 kP, where n is the number of moles of Assuming ideal behavior, calculate the molar mass the gas dissolved in the solvent.) of the unknown compound. State any assumptions 12.135 (a) Derive the equation relating the molality (m) of a made. solution to its molarity (M) • 12.128 At 27°C, the vapor pressure of pure water is 23.76 M mmHg and that of an urea solution is 22.98 mmHg. m5 Mm Calculate the molality of solution. d2 1000 12.129 An example of the positive deviation shown in Figure 12.8(a) is a solution made of acetone where d is the density of the solution (g/mL) and m (CH3COCH3) and carbon disulfide (CS2). (a) Draw is the molar mass of the solute (g/mol). (Hint: Lewis structures of these molecules. Explain the Start by expressing the solvent in kilograms in deviation from ideal behavior in terms of intermo- terms of the difference between the mass of the lecular forces. (b) A solution composed of 0.60 solution and the mass of the solute.) (b) Show mole of acetone and 0.40 mole of carbon disulfide that, for dilute aqueous solutions, m is approxi- has a vapor pressure of 615 mmHg at 35.2°C. What mately equal to M. would be the vapor pressure if the solution behaved • 12.136 At 298 K, the osmotic pressure of a glucose solution ideally? The vapor pressure of the pure solvents at is 10.50 atm. Calculate the freezing point of the the same temperature are: acetone: 349 mmHg; solution. The density of the solution is 1.16 g/mL. carbon disulfide: 501 mmHg. (c) Predict the sign 12.137 A student carried out the following procedure to of ≤Hsoln. measure the pressure of carbon dioxide in a soft • 12.130 Liquids A (molar mass 100 g/mol) and B (molar drink bottle. First, she weighed the bottle (853.5 g). mass 110 g/mol) form an ideal solution. At 55°C, Next, she carefully removed the cap to let the CO2 558 Chapter 12 ■ Physical Properties of Solutions gas escape. She then reweighed the bottle with the 12.139 Often the determination of the molar mass of a cap (851.3 g). Finally, she measured the volume of compound by osmotic pressure measurement is the soft drink (452.4 mL). Given that Henry’s law carried out at several different concentrations to constant for CO2 in water at 25°C is 3.4 3 1022 get a more reliable average value. From the follow- mol/L ? atm, calculate the pressure of CO2 in the ing data for the osmotic pressure of poly(methyl original bottle. Why is this pressure only an estimate methacrylate) in toluene at 25°C, determine graph- of the true value? ically the molar mass of the polymer. [Hint: Rearrange 12.138 Valinomycin is an antibiotic. It functions by binding Equation (12.8) so that π is expressed in terms of K1 ions and transporting them across the mem- c, which is the number of grams of the solute per brane into cells to offset the ionic balance. The liter of solution.] molecule is represented here by its skeletal struc- ture in which the end of each straight line corre- π (atm) 8.40 3 1.72 3 2.52 3 3.23 3 7.75 3 sponds to a carbon atom (unless a N or an O atom 1024 1023 1023 1023 1023 is shown at the end of the line). There are as many c (g/L) 8.10 12.31 15.00 18.17 28.05 H atoms attached to each C atom as necessary to give each C atom a total of four bonds. Use the 12.140 Here is an after-dinner trick. With guests still sit- “like dissolves like” guideline to explain its func- ting at the table, the host provided each of them tion. (Hint: The ¬CH3 groups at the two ends of with a glass of water containing an ice cube float- the Y shape are nonpolar.) ing on top and a piece of string about 2–3 in. in O length. He then asked them to find a way to lift the O ice cube without touching it by hand or using any N O NH O H other objects such as a spoon or fork. Explain how O O O this task can be accomplished. (Hint: The table had O not been cleared so the salt and pepper shakers HN O were still there.) O NH O O O O O HN O H N O O Interpreting, Modeling & Estimating 12.141 The molecule drawn here has shown promise as an of oxygen in water under 1 atmosphere of air. Com- agent for cleaning up oil spills in water. Instead of ment on the prospect for our survival without hemo- dispersing the oil into water as soap molecules globin molecules. (Recall from previous problems would do (see Figures 12.19 and 12.20), these mol- that the total volume of blood in an adult human is ecules bind with the oil to form a gel, which can be about 5 L.) easily separated from the body of water. Suggest an 12.143 The diagram shows the vapor pressure curves for explanation for the ability of this compound to pure benzene and a solution of a nonvolatile solute remove oil from water. in benzene. Estimate the molality of the benzene solution. OH OH O H15C7 O O C7H15 O O O 1.0 P (atm) 12.142 The Henry’s law constant of oxygen in water at 75 80 85 25°C is 1.3 3 1023 mol/L ? atm. Calculate the molarity t (8C)  Answers to Practice Exercises 559 12.144 A common misconception is that adding salt to the liquids. Given that A is more volatile than B, match water used to cook spaghetti will decrease the cook- the curves with the pure liquids and the solution. ing time, presumably because it increases the boil- ing point of the water. Calculate the boiling point of a typical salted water solution used to cook spa- ghetti. Do you think this increase in temperature will make much difference in the cooking time for spaghetti? P 12.145 Estimate the volume of the oil droplet that would be formed by the compound sodium stearate shown in Figure 12.19. 12.146 The diagram here shows the vapor pressure curves of two liquids A and B and a solution of the two T Answers to Practice Exercises 12.1 Carbon disulfide. 12.2 7.44 percent. 12.3 0.638 m. 12.4 8.92 m. 12.5 13.8 m. 12.6 2.9 3 1024 M. 12.7 37.8 mmHg; 4.4 mmHg. 12.8 Tb: 101.3°C; Tf: 24.48°C. 12.9 21.0 atm. 12.10 0.066 m and 1.3 3 102 g/mol. 12.11 2.60 3 104 g. 12.12 1.21. CHEMICAL M YS TERY The Wrong Knife† D r. Thomas Noguchi, the renowned Los Angeles coroner, was performing an autopsy on a young man in his twenties who had been stabbed to death. A Los Angeles Police Department homicide detective entered the room, carrying a brown bag that held the fatal weapon. “Do you want to take a look at it?” he asked. “No,” Dr. Noguchi said. “I’ll tell you exactly what it looks like.” Dr. Noguchi wasn’t showing off. He wanted to demonstrate an important forensic technique to the pathology residents who were observing the autopsy. The traditional method of measuring a knife was to pour barium sulfate (BaSO4) solution into the wound and X ray it. Dr. Noguchi thought he had found a better way. He lit a little Bunsen burner and melted some Wood’s metal over it while the detec- tive and the residents watched. (Wood’s metal is an alloy of bismuth, lead, tin, and cadmium that has a low melting point of 718C.) Then he selected a wound in the victim’s chest above the location of the liver and poured the liquid metal into it. The metal slid down through the wound into the punctured liver. When it was cool he removed an exact mold of the tip of the murder weapon. He added the length of this tip to the distance between the liver and the skin surface of the chest. Then he said to the homicide detec- tive, “It’s a knife five and a half inches long, one inch wide, and one sixteenth of an inch thick.” The detective smiled and reached into his bag. “Sorry, Dr. Noguchi.” He pulled out a much smaller pocketknife, only about three inches in length. “That’s the wrong knife,” Dr. Noguchi said at once. “Oh, now, come on,” the detective said. “We found the knife that killed him right at the scene.” “You don’t have the murder weapon,” Dr. Noguchi insisted. The detective didn’t believe him. But two days later police found a blood-stained knife in a trash can two blocks from the crime scene. That weapon was exactly five and a half inches long, one inch wide, and one sixteenth of an inch thick. And the blood on its blade matched the victim’s. It turned out to be the murder weapon. The pocketknife the police discovered at the scene had been used by the victim in self-defense. And two knives indicated a knife fight. Was it part of a gang war? The police investigated and found out that the victim was a member of a gang that was at war with another gang. By interrogating members of the rival gang, they eventually identified the murderer. † Adapted from Simon & Schuster from “Coroner,” by Thomas T. Noguchi, M.D., Copyright 1984 by Thomas Noguchi and Joseph DiMona. 560  Composition of Wood’s Metal* Component Melting Point (8C) Bismuth (50%) 271 Cadmium (12.5%) 321 Lead (25%) 328 Tin (12.5%) 232 *The components are shown in percent by mass, and the melting point is that of the pure metal. Chemical Clues 1. What is the function of the BaSO4 solution as a traditional method for measuring a knife wound in a homicide victim’s body? Describe a medical application of BaSO4. 2. As the table shows, the melting points of the pure metals are much higher than that of Wood’s metal. What phenomenon accounts for its low melting point? 3. Wood’s metal is used in automatic sprinklers in the ceilings of hotels and stores. Explain how such a sprinkling system works. 4. The melting point of an alloy can be altered by changing the composition. Certain organic materials have also been developed for the same purpose. Shown here is a simplified diagram of the pop-up thermometer used in cooking turkeys. Describe how this thermometer works. Thermometer indicator Alloy Compressed spring A pop-up thermometer used for cooking turkeys. 561  CHAPTER 13 Chemical Kinetics The rates of chemical reactions vary greatly. The conversion of graphite to diamond in Earth’s crust may take millions of years to complete. Explosive reactions such as those of dynamite and TNT, on the other hand, are over in a fraction of a second. CHAPTER OUTLINE A LOOK AHEAD 13.1 The Rate of a Reaction  We begin by studying the rate of a reaction expressed in terms of the con- centrations of reactants and products and how the rate is related to the stoi- 13.2 The Rate Law chiometry of a reaction. (13.1) 13.3 The Relation Between  We then see how the rate law of a reaction is defined in terms of the rate Reactant Concentration constant and reaction order. (13.2) and Time  Next, we examine the relationship between reactant concentration and time 13.4 Activation Energy and for three types of reactions: zero order, first order, and second order. The Temperature Dependence half-life, which is the time required for the concentration of a reactant to of Rate Constants decrease to half of its initial value, is useful for distinguishing between reac- tions of different orders. (13.3) 13.5 Reaction Mechanisms  We see that the rate of a reaction usually increases with temperature. 13.6 Catalysis Activation energy, which is the minimum amount of energy required to initi- ate a chemical reaction, also influences the rate. (13.4)  We examine the mechanism of a reaction in terms of the elementary steps and see that we can determine the rate law from the slowest or rate- determining step. We learn how chemists verify mechanisms by experi- ments. (13.5)  Finally, we study the effect of catalyst on the rate of a reaction. We learn the characteristics of heterogeneous catalysis, homogeneous catalysis, and enzyme catalysis. (13.6) 562  13.1 The Rate of a Reaction 563 I n previous chapters, we studied basic definitions in chemistry, and we examined the properties of gases, liquids, solids, and solutions. We have discussed molecular proper- ties and looked at several types of reactions in some detail. In this chapter and in subse- quent chapters, we will look more closely at the relationships and the laws that govern chemical reactions. How can we predict whether or not a reaction will take place? Once started, how fast does the reaction proceed? How far will the reaction go before it stops? The laws of thermodynam- ics (to be discussed in Chapter 17) help us answer the first question. Chemical kinetics, the subject of this chapter, provides answers to the question about the speed of a reaction. The last question is one of many answered by the study of chemical equilibrium, which we will consider in Chapters 14, 15, and 16. 13.1 The Rate of a Reaction Chemical kinetics is the area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs. The word “kinetic” suggests movement or change; in Chapter 5 we defined kinetic energy as the energy available because of the motion of an object. Here kinetics refers to the rate of a reaction, or the reaction rate, which is the change in the concentration of a reactant or a product with time (M/s). There are many reasons for studying the rate of a reaction. To begin with, there is intrinsic curiosity about why reactions have such vastly different rates. Some processes, such as the initial steps in vision and photosynthesis and nuclear chain reactions, take place on a time scale as short as 10212 s to 1026 s. Others, like the curing of cement and the conversion of graphite to diamond, take years or millions of years to complete. On a practical level, a knowledge of reaction rates is useful in drug design, in pollution control, and in food processing. Industrial chemists often place more emphasis on speeding up the rate of a reaction rather than on maximiz- ing its yield. We know that any reaction can be represented by the general equation reactants ¡ products This equation tells us that during the course of a reaction, reactants are consumed while products are formed. As a result, we can follow the progress of a reaction by monitoring either the decrease in concentration of the reactants or the increase in concentration of the products. Figure 13.1 shows the progress of a simple reaction in which A molecules are converted to B molecules: A ¡ B The decrease in the number of A molecules and the increase in the number of B molecules with time are shown in Figure 13.2. In general, it is more convenient to Figure 13.1 The progress of reaction A ¡ B at 10-s intervals over a period of 60 s. Initially, only A molecules (gray spheres) are present. As time progresses, B molecules (red spheres) are formed. 564 Chapter 13 ■ Chemical Kinetics Figure 13.2 The rate of reaction 40 A ¡ B, represented as the A molecules Number of molecules decrease of A molecules with time and as the increase of B molecules 30 with time. B molecules 20 10 0 10 20 30 40 50 60 t (s) express the reaction rate in terms of the change in concentration with time. Thus, for the reaction A ¡ B we can express the rate as ¢[A] ¢[B] rate 5 2     or    rate 5 ¢t ¢t Recall that D denotes the difference where D[A] and D[B] are the changes in concentration (molarity) over a time period between the final and initial states. Dt. Because the concentration of A decreases during the time interval, D[A] is a negative quantity. The rate of a reaction is a positive quantity, so a minus sign is needed in the rate expression to make the rate positive. On the other hand, the rate of product formation does not require a minus sign because D[B] is a positive quan- tity (the concentration of B increases with time). These rates are average rates because they are averaged over a certain time period Dt. Our next step is to see how the rate of a reaction is obtained experimentally. By definition, we know that to determine the rate of a reaction we have to monitor the concentration of the reactant (or product) as a function of time. For reactions in solu- tion, the concentration of a species can often be measured by spectroscopic means. If ions are involved, the change in concentration can also be detected by an electrical conductance measurement. Reactions involving gases are most conveniently followed by pressure measurements. We will consider two specific reactions for which different methods are used to measure the reaction rates. Reaction of Molecular Bromine and Formic Acid In aqueous solutions, molecular bromine reacts with formic acid (HCOOH) as follows: Br2(aq) 1 HCOOH(aq) ¡ 2Br2 (aq) 1 2H1 (aq) 1 CO2(g) Molecular bromine is reddish-brown in color. All the other species in the reaction are colorless. As the reaction progresses, the concentration of Br2 steadily decreases and its color fades (Figure 13.3). This loss of color and hence concentration can be mon- itored easily with a spectrometer, which registers the amount of visible light absorbed by bromine (Figure 13.4). Measuring the change (decrease) in bromine concentration at some initial time and then at some final time enables us to determine the average rate of the reaction during that interval: ¢[Br2] average rate 5 2 ¢t [Br2]final 2 [Br2]initial 52 tfinal 2 tinitial  13.1 The Rate of a Reaction 565 t1 Absorption t2 t3 300 400 500 600 Wavelength (nm) Figure 13.4 Plot of absorption of bromine versus wavelength. The maximum absorption of visible light by bromine occurs at 393 nm. As the reaction progresses (t1 to t3), the absorption, which is proportional to [Br2], decreases. Figure 13.3 From left to right: The decrease in bromine concentration as time elapses shows up as a loss of color (from left to right). Using the data provided in Table 13.1 we can calculate the average rate over the first 50-s time interval as follows: (0.0101 2 0.0120) M average rate 5 2 5 3.80 3 1025 M/s 50.0 s If we had chosen the first 100 s as our time interval, the average rate would then be given by: (0.00846 2 0.0120) M average rate 5 2 5 3.54 3 1025 M/s 100.0 s These calculations demonstrate that the average rate of the reaction depends on the time interval we choose. By calculating the average reaction rate over shorter and shorter intervals, we can obtain the rate for a specific instant in time, which gives us the instantaneous rate of the reaction at that time. Figure 13.5 shows the plot of [Br2] versus time, based on the data shown in Table 13.1. Graphically, the instantaneous rate at 100 s after the Table 13.1 Rates of the Reaction Between Molecular Bromine and Formic Acid at 25°C rate 21 Time (s) [Br2] (M) Rate (M/s) k5 (s ) [Br2 ] 0.0 0.0120 4.20 3 1025 3.50 3 1023 50.0 0.0101 3.52 3 1025 3.49 3 1023 100.0 0.00846 2.96 3 1025 3.50 3 1023 150.0 0.00710 2.49 3 1025 3.51 3 1023 200.0 0.00596 2.09 3 1025 3.51 3 1023 250.0 0.00500 1.75 3 1025 3.50 3 1023 300.0 0.00420 1.48 3 1025 3.52 3 1023 350.0 0.00353 1.23 3 1025 3.48 3 1023 400.0 0.00296 1.04 3 1025 3.51 3 1023 566 Chapter 13 ■ Chemical Kinetics 0.0120 0.0100 Rate at 100 s: 2.96 × 10 –5 M/s 0.00800 Rate at 200 s: [Br2] (M ) 2.09 × 10 –5 M/s 0.00600 Rate at 300 s: 1.48 × 10 –5 M/s 0.00400 0.00200 0 100 200 300 400 t (s) Figure 13.5 The instantaneous rates of the reaction between molecular bromine and formic acid at t 5 100 s, 200 s, and 300 s are given by the slopes of the tangents at these times. start of the reaction, say, is given by the slope of the tangent to the curve at that instant. The instantaneous rate at any other time can be determined in a similar man- ner. Note that the instantaneous rate determined in this way will always have the same value for the same concentrations of reactants, as long as the temperature is kept constant. We do not need to be concerned with what time interval to use. Unless otherwise stated, we will refer to the instantaneous rate at a specific time merely as “the rate” at that time. The following travel analogy helps to distinguish between average rate and instantaneous rate. The distance by car from San Francisco to Los Angeles is 512 mi along a certain route. If it takes a person 11.4 h to go from one city to the other, the average speed is 512 mi/11.4 h or 44.9 mph. But if the car is traveling at 55.3 mph 3 h and 26 min after departure, then the instantaneous speed of the car is 55.3 mph at that time. In other words, instantaneous speed is the speed that you would read from the speedometer. Note that the speed of the car in our example can increase or decrease during the trip, but the instantaneous rate of a reaction always decreases with time. The rate of the bromine-formic acid reaction also depends on the concentra- tion of formic acid. However, by adding a large excess of formic acid to the reaction mixture we can ensure that the concentration of formic acid remains virtually constant throughout the course of the reaction. Under this condition the change in the amount of formic acid present in solution has no effect on the measured rate. Let’s consider the effect that the bromine concentration has on the rate of reac- tion. Look at the data in Table 13.1. Compare the concentration of Br2 and the reac- tion rate at t 5 50 s and t 5 250 s. At t 5 50 s, the bromine concentration is 0.0101 M and the rate of reaction is 3.52 3 1025 M/s. At t 5 250 s, the bromine concentration is 0.00500 M and the rate of reaction is 1.75 3 1025 M/s. The concentration at t 5 50 s is double the concentration at t 5 250 s (0.0101 M versus 0.00500 M), and the rate of reaction at t 5 50 s is double the rate at t 5 250 s (3.52 3 1025 M/s versus 1.75 3 1025 M/s). We see that as the concentration of bromine is doubled, the  13.1 The Rate of a Reaction 567 5.00 × 10 –5 Figure 13.6 Plot of rate versus molecular bromine concentration for the reaction between 4.00 × 10 –5 molecular bromine and formic acid. The straight-line relationship shows that the rate of reaction is Rate (M/s) 3.00 × 10 –5 directly proportional to the molecular bromine concentration. 2.00 × 10 –5 1.00 × 10 –5 0 0.00200 0.00600 0.0100 0.0140 [Br2] (M ) rate of reaction also doubles. Thus, the rate is directly proportional to the Br2 con- centration, that is rate r [Br2] 5 k[Br2] where the term k is known as the rate constant, a constant of proportionality between the reaction rate and the concentration of reactant. This direct proportionality between Br2 concentration and rate is also supported by plotting the data. Figure 13.6 is a plot of the rate versus Br2 concentration. The fact that this graph is a straight line shows that the rate is directly proportional to the concentration; the higher the concentration, the higher the rate. Rearranging the last equation gives rate k5 [Br2] Because reaction rate has the units M/s, and [Br2] is in M, the unit of k is 1/s, or As we will see, for a given reaction, k is s21 in this case. It is important to understand that k is not affected by the concentration affected only by a change in temperature. of Br2. To be sure, the rate is greater at a higher concentration and smaller at a lower concentration of Br2, but the ratio of rate/[Br2] remains the same provided the tem- perature does not change. From Table 13.1 we can calculate the rate constant for the reaction. Taking the data for t 5 50 s, we write rate k5 [Br2] 3.52 3 1025 M/s 5 5 3.49 3 1023 s21 8n 0.0101 M We can use the data for any t to calculate k. The slight variations in the values of k listed in Table 13.1 are due to experimental deviations in rate measurements. Decomposition of Hydrogen Peroxide If one of the products or reactants is a gas, we can use a manometer to find the reaction rate. Consider the decomposition of hydrogen peroxide at 208C: 2H2O2 (aq) ¡ 2H2O(l) 1 O2 (g) 2H2O2 ¡ 2H2O 1 O2 568 Chapter 13 ■ Chemical Kinetics In this case, the rate of decomposition can be determined by monitoring the rate of oxygen evolution with a manometer (Figure 13.7). The oxygen pressure can be readily converted to concentration by using the ideal gas equation: PV 5 nRT or n P5 RT 5 [O2]RT V where n/V gives the molarity of oxygen gas. Rearranging the equation, we get 1 [O2] 5 P RT The reaction rate, which is given by the rate of oxygen production, can now be written as ¢[O2] 1 ¢P rate 5 5 ¢t RT ¢t Figure 13.8 shows the increase in oxygen pressure with time and the determination of an instantaneous rate at 400 min. To express the rate in the normal units of M/s, After conversion, the rate is 1.1 3 1027 M/s. we convert mmHg/min to atm/s, then multiply the slope of the tangent (DP/Dt) by 1/RT, as shown in the previous equation. Reaction Rates and Stoichiometry We have seen that for stoichiometrically simple reactions of the type A ¡ B, the rate can be either expressed in terms of the decrease in reactant concentration with 120 100 80 P (mmHg) Slope = 0.12 mmHg/min 60 40 20 Figure 13.7 The rate of hydrogen peroxide decomposition can be measured with a manometer, 0 200 400 600 800 1000 1200 which shows the increase in the t (min) oxygen gas pressure with time. The arrows show the mercury Figure 13.8 The instantaneous rate for the decomposition of hydrogen peroxide at 400 min is levels in the U tube. given by the slope of the tangent multiplied by 1/RT.  13.1 The Rate of a Reaction 569 time, 2D[A]/Dt, or the increase in product concentration with time, D[B]/Dt. For more complex reactions, we must be careful in writing the rate expressions. Consider, for example, the reaction 2A ¡ B Two moles of A disappear for each mole of B that forms; that is, the rate at which B forms is one-half the rate at which A disappears. Thus, the rate can be expressed as 1 ¢[A] ¢[B] rate 5 2     or    rate 5 2 ¢t ¢t In general, for the reaction aA 1 bB ¡ cC 1 dD the rate is given by 1 ¢[A] 1 ¢[B] 1 ¢[C] 1 ¢[D] rate 5 2 52 5 5 a ¢t b ¢t c ¢t d ¢t Examples 13.1 and 13.2 show writing the reaction rate expressions and calculating rates of product formation and reactant disappearance. Example 13.1 Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products: (a) I 2 (aq) 1 OCl 2 (aq) ¡ Cl 2 (aq) 1 OI 2 (aq) (b) 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) (c) CH4 (g) 1 12Br2 (g) ¡ CH3Br(g) 1 HBr(g) Strategy To express the rate of the reaction in terms of the change in concentration of a reactant or product with time, we need to use the proper sign (minus or plus) and the reciprocal of the stoichiometric coefficient. Solution (a) Because each of the stoichiometric coefficients equals 1, ¢[I 2 ] ¢[OCl 2 ] ¢[Cl 2 ] ¢[OI 2 ] rate 5 2 52 5 5 ¢t ¢t ¢t ¢t (b) Here the coefficients are 4, 5, 4, and 6, so 1 ¢[NH3] 1 ¢[O2] 1 ¢[NO] 1 ¢[H2O] rate 5 2 52 5 5 Similar problems: 13.5, 13.6. 4 ¢t 5 ¢t 4 ¢t 6 ¢t (c) The coefficient for Br2 is 12 , so to express the rate in terms of the disappearance of dibromine, we multiply by the reciprocal of the coefficient [( 12 ) 21 5 2], giving ¢[CH4] ¢[Br2] ¢[CH3Br] ¢[HBr] rate 5 2 5 22 5 5 ¢t ¢t ¢t ¢t Practice Exercise Write the rate expressions for the following reaction: CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(g) 570 Chapter 13 ■ Chemical Kinetics Example 13.2 Consider the reaction 4NO2 (g) 1 O2 (g) ¡ 2N2O5 (g) Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s. (a) At what rate is N2O5 being formed? (b) At what rate is NO2 reacting? Strategy To calculate the rate of formation of N2O5 and disappearance of NO2, we need to express the rate of the reaction in terms of the stoichiometric coefficients as in Example 13.1: 1 ¢[NO2] ¢[O2] 1 ¢[N2O5] rate 5 2 52 5 4 ¢t ¢t 2 ¢t We are given ¢[O2] 5 20.024 M/s ¢t where the minus sign shows that the concentration of O2 is decreasing with time. Solution (a) From the preceding rate expression we have ¢[O2] 1 ¢[N2O5] 2 5 ¢t 2 ¢t Therefore, ¢[N2O5] 5 22(20.024 M/s) 5 0.048 M/s ¢t (b) Here we have 1 ¢[NO2] ¢[O2] 2 52 4 ¢t ¢t so ¢[NO2] Similar problems: 13.7, 13.8. 5 4(20.024 M/s) 5 20.096 M/s ¢t Practice Exercise Consider the reaction 4PH3 (g) ¡ P4 (g) 1 6H2 (g) Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of 0.078 M/s. (a) At what rate is P4 being formed? (b) At what rate is PH3 reacting? Review of Concepts Write a balanced equation for a gas-phase reaction whose rate is given by 1 ¢[NOCl] 1 ¢[NO] ¢[Cl2] rate 5 2 5 5 2 ¢t 2 ¢t ¢t  13.2 The Rate Law 571 13.2 The Rate Law So far we have learned that the rate of a reaction is proportional to the concentra- tion of reactants and that the proportionality constant k is called the rate constant. The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. For the general reaction aA 1 bB ¡ cC 1 dD the rate law takes the form rate 5 k[A]x[B]y (13.1) where x and y are numbers that must be determined experimentally. Note that, in general, x and y are not equal to the stoichiometric coefficients a and b. When we know the values of x, y, and k, we can use Equation (13.1) to calculate the rate of the reaction, given the concentrations of A and B. The exponents x and y specify the relationships between the concentrations of reactants A and B and the reaction rate. Added together, they give us the overall reaction order, defined as the sum of the powers to which all reactant concentrations appearing in the rate law are raised. For Equation (13.1) the overall reaction order is x 1 y. Alternatively, we can say that the reaction is xth order in A, yth order in B, and (x 1 y)th order overall. To see how to determine the rate law of a reaction, let us consider the reaction between fluorine and chlorine dioxide: F2 (g) 1 2ClO2 (g) ¡ 2FClO2 (g) One way to study the effect of reactant concentration on reaction rate is to determine how the initial rate depends on the starting concentrations. It is preferable to measure 8n the initial rates because as the reaction proceeds, the concentrations of the reactants decrease and it may become difficult to measure the changes accurately. Also, there may be a reverse reaction of the type products ¡ reactants which would introduce error into the rate measurement. Both of these complications are virtually absent during the early stages of the reaction. Table 13.2 shows three rate measurements for the formation of FClO2. Looking at entries 1 and 3, we see that as we double [F2] while holding [ClO2] constant, the reac- F2 1 2ClO2 ¡ 2FClO2 tion rate doubles. Thus, the rate is directly proportional to [F2]. Similarly, the data in entries 1 and 2 show that as we quadruple [ClO2] at constant [F2], the rate increases by Table 13.2 Rate Data for the Reaction Between F2 and ClO2 [F2] (M) [ClO2] (M) Initial Rate (M/s) 1. 0.10 0.010 1.2 3 1023 2. 0.10 0.040 4.8 3 1023 3. 0.20 0.010 2.4 3 1023 572 Chapter 13 ■ Chemical Kinetics four times, so that the rate is also directly proportional to [ClO2]. We can summarize our observations by writing the rate law as rate 5 k[F2][ClO2] Because both [F2] and [ClO2] are raised to the first power, the reaction is first order in F2, first order in ClO2, and (1 1 1) or second order overall. Note that [ClO2] is raised to the power of 1 whereas its stoichiometric coefficient in the overall equation is 2. The equality of reaction order (first) and stoichiometric coefficient (1) for F2 is coincidental in this case. From the reactant concentrations and the initial rate, we can also calculate the rate constant. Using the first entry of data in Table 13.2, we can write rate k5 [F2][ClO2] 1.2 3 1023 M/s 5 (0.10 M)(0.010 M) 5 1.2/M ? s Reaction order enables us to understand how the reaction depends on reactant con- centrations. Suppose, for example, that for the general reaction aA 1 bB ¡ cC 1 dD we have x 5 1 and y 5 2. The rate law for the reaction is [see Equation (13.1)] rate 5 k[A][B]2 This reaction is first order in A, second order in B, and third order overall (1 1 2 5 3). Let us assume that initially [A] 5 1.0 M and [B] 5 1.0 M. The rate law tells us that if we double the concentration of A from 1.0 M to 2.0 M at constant [B], we also double the reaction rate: for [A] 5 1.0 M rate1 5 k(1.0 M)(1.0 M) 2 5 k(1.0 M3 ) for [A] 5 2.0 M rate2 5 k(2.0 M)(1.0 M) 2 5 k(2.0 M3 ) Hence, rate2 5 2(rate1) On the other hand, if we double the concentration of B from 1.0 M to 2.0 M at constant [A] 5 1 M, the rate will increase by a factor of 4 because of the power 2 in the exponent: for [B] 5 1.0 M rate1 5 k(1.0 M)(1.0 M) 2 5 k(1.0 M3 ) for [B] 5 2.0 M rate2 5 k(1.0 M)(2.0 M) 2 5 k(4.0 M3 ) Hence, rate2 5 4(rate1 ) If, for a certain reaction, x 5 0 and y 5 1, then the rate law is rate 5 k[A]0[B] 5 k[B]  13.2 The Rate Law 573 This reaction is zero order in A, first order in B, and first order overall. The exponent Zero order does not mean that the rate is zero. It just means that the rate is zero tells us that the rate of this reaction is independent of the concentration of A. independent of the concentration of Note that reaction order can also be a fraction. A present. The following points summarize our discussion of the rate law: 1. Rate laws are always determined experimentally. From the concentrations of reac- tants and the initial reaction rates we can determine the reaction order and then the rate constant of the reaction. 2. Reaction order is always defined in terms of reactant (not product) concentrations. 3. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the overall balanced equation. Example 13.3 illustrates the procedure for determining the rate law of a reaction. Example 13.3 The reaction of nitric oxide with hydrogen at 12808C is 2NO(g) 1 2H2 (g) ¡ N2 (g) 1 2H2O(g) From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant, and (c) the rate of the reaction when [NO] 5 12.0 3 1023 M and [H2] 5 6.0 3 1023 M. Experiment [NO] (M) [H2] (M) Initial Rate (M/s) 1 5.0 10 3 2.0 10 3 1.3 10 5 8n 2 10.0 10 3 2.0 10 3 5.0 10 5 3 10.0 10 3 4.0 10 3 10.0 10 5 Strategy We are given a set of concentration and reaction rate data and asked to determine the rate law and the rate constant. We assume that the rate law takes the form rate 5 k[NO]x[H2]y How do we use the data to determine x and y? Once the orders of the reactants are known, we can calculate k from any set of rate and concentrations. Finally, the rate law enables us to calculate the rate at any concentrations of NO and H2. Solution 2NO 1 2H2 ¡ N2 1 2H2O (a) Experiments 1 and 2 show that when we double the concentration of NO at constant concentration of H2, the rate quadruples. Taking the ratio of the rates from these two experiments rate2 5.0 3 1025 M/s k(10.0 3 1023 M) x (2.0 3 1023 M) y 5 < 4 5 rate1 1.3 3 1025 M/s k(5.0 3 1023M) x (2.0 3 1023M) y Therefore, (10.0 3 1023M) x 5 2x 5 4 (5.0 3 1023M) x or x 5 2; that is, the reaction is second order in NO. Experiments 2 and 3 indicate that doubling [H2] at constant [NO] doubles the rate. Here we write the ratio as rate3 10.0 3 1025M/s k(10.0 3 1023 M) x (4.0 3 1023M) y 5 525 rate2 25 5.0 3 10 M/s k(10.0 3 1023M) x (2.0 3 1023 M) y (Continued) 574 Chapter 13 ■ Chemical Kinetics Therefore, (4.0 3 1023M) y 5 2y 5 2 (2.0 3 1023M) y or y 5 1; that is, the reaction is first order in H2. Hence the rate law is given by rate 5 k[NO]2[H2] which shows that it is a (2 1 1) or third-order reaction overall. (b) The rate constant k can be calculated using the data from any one of the experiments. Rearranging the rate law, we get rate k5 [NO]2[H2] The data from experiment 2 give us 5.0 3 1025 M/s k5 (10.0 3 1023 M) 2 (2.0 3 1023 M) 5 2.5 3 102/M2 ? s (c) Using the known rate constant and concentrations of NO and H2, we write rate 5 (2.5 3 102/M2 ? s) (12.0 3 1023 M) 2 (6.0 3 1023 M) 5 2.2 3 1024 M/s Comment Note that the reaction is first order in H2, whereas the stoichiometric coefficient for H2 in the balanced equation is 2. The order of a reactant is not related to Similar problem: 13.15. the stoichiometric coefficient of the reactant in the overall balanced equation. Practice Exercise The reaction of peroxydisulfate ion (S2O282) with iodide ion (I2) is S2O22 2 22 2 8 (aq) 1 3I (aq) ¡ 2SO4 (aq) 1 I3 (aq) From the following data collected at a certain temperature, determine the rate law and calculate the rate constant. Experiment [S2O82 ] (M) [l ] (M ) Initial Rate (M/s) 1 0.080 0.034 2.2 10 4 2 0.080 0.017 1.1 10 4 3 0.16 0.017 2.2 10 4 Review of Concepts The relative rates of the reaction 2A 1 B ¡ products shown in the diagrams (a)–(c) are 1:2:4. The red spheres represent A molecules and the green spheres represent B molecules. Write a rate law for this reaction. (a) (b) (c)  13.3 The Relation Between Reactant Concentration and Time 575 13.3 The Relation Between Reactant Concentration and Time Rate law expressions enable us to calculate the rate of a reaction from the rate constant and reactant concentrations. The rate laws can also be used to determine the concentrations of reactants at any time during the course of a reaction. We will illustrate this application by first considering two of the most common rate laws—those applying to reactions that are first order overall and those applying to reactions that are second order overall. First-Order Reactions A first-order reaction is a reaction whose rate depends on the reactant concentration raised to the first power. In a first-order reaction of the type A ¡ product the rate is ¢[A] rate 5 2 ¢t From the rate law we also know that rate 5 k[A] To obtain the units of k for this rate law, we write rate M/s k5 5 5 1/s or s21 [A] M Combining the first two equations for the rate we get ¢[A] 2 5 k[A] (13.2) ¢t Using calculus, we can show from Equation (13.2) that In differential form, Equation (13.2) becomes d[A] k[A] [A]t dt ln 5 2kt (13.3) Rearranging, we get [A]0 d[A] kdt [A] where ln is the natural logarithm, and [A]0 and [A]t are the concentrations of A at times Integrating between t 0 and t t gives t 5 0 and t 5 t, respectively. It should be understood that t 5 0 need not correspond to [A] t d[A] t k dt the beginning of the experiment; it can be any time when we choose to start monitoring [A]0 [A] 0 ln [A]t ln [A]0 kt the change in the concentration of A. Equation (13.3) can be rearranged as follows: or ln [A]t kt [A]0 ln [A]t 5 2kt 1 ln [A]0 (13.4) Equation (13.4) has the form of the linear equation y 5 mx 1 b, in which m is the slope of the line that is the graph of the equation: ln [A]t 5 (2k) 1t) 1 ln [A]0 4 4 4 4 y 5 m x 1 b 576 Chapter 13 ■ Chemical Kinetics Figure 13.9 First-order reaction characteristics: (a) the exponential ln [A]0 decrease of reactant concentration with time; (b) a plot of ln [A]t versus t. The slope of the line is slope  k ln [A]t equal to 2k. [A]t t t (a) (b) Consider Figure 13.9. As we would expect during the course of a reaction, the con- centration of the reactant A decreases with time [Figure 13.9(a)]. For a first-order reaction, if we plot ln [A]t versus time (y versus x), we obtain a straight line with a slope equal to 2k and a y intercept equal to ln [A]0 [Figure 13.9(b)]. Thus, we can calculate the rate constant from the slope of this plot. There are many first-order reactions. An example is the decomposition of ethane (C2H6) into highly reactive fragments called methyl radicals (CH3): C2H6 ¡ 2CH3 The decomposition of N2O5 is also a first-order reaction 2N2O5 (g) ¡ 4NO2 (g) 1 O2 (g) In Example 13.4 we apply Equation (13.3) to an organic reaction. Example 13.4 The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 3 1024 s21 at 5008C. CH2 D G CH2OCH2 88n CH3OCHPCH2 cyclopropane propene 88n (a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? (b) How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (c) How long (in minutes) will it take to convert 74 percent of the starting material? Strategy The relationship between the concentrations of a reactant at different times in a first-order reaction is given by Equation (13.3) or (13.4). In (a) we are given [A]0 5 0.25 M and asked for [A]t after 8.8 min. In (b) we are asked to calculate the time it takes for cyclopropane to decrease in concentration from 0.25 M to 0.15 M. No concentration values are given for (c). However, if initially we have 100 percent of the compound and 74 percent has reacted, then what is left must be (100% 2 74%), or 26%. Thus, the ratio of the percentages will be equal to the ratio of the actual concentrations; that is, [A]t /[A]0 5 26%/100%, or 0.26/1.00. (Continued)  13.3 The Relation Between Reactant Concentration and Time 577 Solution (a) In applying Equation (13.4), we note that because k is given in units of s21, we must first convert 8.8 min to seconds: 60 s 8.8 min 3 5 528 s 1 min We write ln [A]t 5 2kt 1 ln [A]0 5 2(6.7 3 1024 s21 ) (528 s) 1 ln (0.25) 5 21.74 Hence, [A]t 5 e21.74 5 0.18 M Note that in the ln [A]0 term, [A]0 is expressed as a dimensionless quantity (0.25) because we cannot take the logarithm of units. (b) Using Equation (13.3), 0.15 M ln 5 2(6.7 3 1024 s21 )t 0.25 M 1 min t 5 7.6 3 102 s 3 60 s 5 13 min (c) From Equation (13.3), 0.26 ln 5 2(6.7 3 1024 s21 )t 1.00 1 min t 5 2.0 3 103 s 3 5 33 min Similar problem: 13.94. 60 s Practice Exercise The reaction 2A ¡ B is first order in A with a rate constant of 2.8 3 1022 s21 at 808C. How long (in seconds) will it take for A to decrease from 0.88 M to 0.14 M? N2O5 Now let us determine graphically the order and rate constant of the decomposition of dinitrogen pentoxide in carbon tetrachloride (CCl4) solvent at 458C: 2N2O5 (CCl4 ) ¡ 4NO2 (g) 1 O2 (g) The following table shows the variation of N2O5 concentration with time, and the corresponding ln [N2O5] values. t (s) [N2O5] (M) ln [N2O5] 0 0.91 20.094 300 0.75 20.29 600 0.64 20.45 1200 0.44 20.82 3000 0.16 21.83 Applying Equation (13.4) we plot ln [N2O5] versus t, as shown in Figure 13.10. The N2O5 decomposes to give NO2 fact that the points lie on a straight line shows that the rate law is first order. Next, (brown color). 578 Chapter 13 ■ Chemical Kinetics Figure 13.10 Plot of ln [N2O5]t 0 versus time. The rate constant can be determined from the slope of the straight line. (400 s, 0.34) –0.50 ln [N2O5 ]t Δy –1.00 Δx (2430 s, 1.50) –1.50 –2.00 0 500 1000 1500 2000 2500 3000 3500 t (s) we determine the rate constant from the slope. We select two points far apart on the line and subtract their y and x values as follows: ¢y slope (m) 5 ¢x 21.50 2 (20.34) 5 (2430 2 400) s 5 25.7 3 1024 s21 Because m 5 2k, we get k 5 5.7 3 1024 s21. For gas-phase reactions we can replace the concentration terms in Equation (13.3) with the pressures of the gaseous reactant. Consider the first-order reaction A(g) ¡ product Using the ideal gas equation we write PV 5 nART or nA P 5 [A] 5 V RT Substituting [A] 5 P/RT in Equation (13.3), we get [A]t Pt /RT Pt ln 5 ln 5 ln 5 2kt [A]0 P0/RT P0 The equation corresponding to Equation (13.4) now becomes ln Pt 5 2kt 1 ln P0 (13.5)  13.3 The Relation Between Reactant Concentration and Time 579 Example 13.5 shows the use of pressure measurements to study the kinetics of a first-order reaction. Example 13.5 The rate of decomposition of azomethane (C2H6N2) is studied by monitoring the partial pressure of the reactant as a function of time: CH3 ¬N“N¬CH3 (g) ¡ N2 (g) 1 C2H6 (g) The data obtained at 3008C are shown in the following table: 88n Partial Pressure of Time (s) Azomethane (mmHg) 0 284 100 220 150 193 200 170 250 150 300 132 Are these values consistent with first-order kinetics? If so, determine the rate constant. Strategy To test for first-order kinetics, we consider the integrated first-order rate law that has a linear form, which is Equation (13.4) C2H6N2 ¡ N2 1 C2H6 ln [A]t 5 2kt 1 ln [A]0 If the reaction is first order, then a plot of ln [A]t versus t ( y versus x) will produce a straight line with a slope equal to 2k. Note that the partial pressure of azomethane at any time is directly proportional to its concentration in moles per liter (PV 5 nRT, so P ~ n /V ). Therefore, we substitute partial pressure for concentration [Equation (13.5)]: ln Pt 5 2kt 1 ln P0 where P0 and Pt are the partial pressures of azomethane at t 5 0 and t 5 t, respectively. Solution First we construct the following table of t versus ln Pt. t (s) ln Pt 0 5.649 100 5.394 150 5.263 200 5.136 250 5.011 300 4.883 Figure 13.11, which is based on the data given in the table, shows that a plot of ln Pt versus t yields a straight line, so the reaction is indeed first order. The slope of the line is given by 5.05 2 5.56 slope 5 5 22.55 3 1023 s21 (233 2 33) s According to Equation (13.4), the slope is equal to 2k, so k 5 2.55 3 1023 s21 . Similar problems: 13.19, 13.20. (Continued) 580 Chapter 13 ■ Chemical Kinetics Figure 13.11 Plot of ln Pt versus 5.80 time for the decomposition of azomethane. 5.60 (33 s, 5.56) 5.40 1n Pt Δy 5.20 (233 s, 5.05) 5.00 Δx 4.80 0 100 200 300 t (s) Practice Exercise Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) ¡ C2H4 (g) 1 HI(g) From the following data determine the order of the reaction and the rate constant. Time (min) [C2H5I] (M) 0 0.36 15 0.30 30 0.25 48 0.19 75 0.13 Reaction Half-life As a reaction proceeds, the concentration of the reactant(s) decreases. Another mea- sure of the rate of a reaction, relating concentration to time, is the half-life, t12 , which is the time required for the concentration of a reactant to decrease to half of its initial concentration. We can obtain an expression for t12 for a first-order reaction as follows. Equation (13.3) can be rearranged to give 1 [A]0 t5 ln k [A]t By the definition of half-life, when t 5 t 12 , [A]t 5 [A]0 /2, so 1 [A]0 t 12 5 ln k [A]0/2   or 1 0.693 t 12 5 ln 2 5 (13.6) k k Equation (13.6) tells us that the half-life of a first-order reaction is independent of the initial concentration of the reactant. Thus, it takes the same time for the concentration of the reactant to decrease from 1.0 M to 0.50 M, say, as it does for a  13.3 The Relation Between Reactant Concentration and Time 581 Figure 13.12 A plot of [A]t versus time for the first-order reaction A ¡ products. The half-life of the reaction is 1 min. After the elapse of each half-life, the concentration of A is halved. [A]0 [A]t t12 [A]0/2 t12 [A]0/4 [A]0/8 t12 0 0 1 2 3 4 Time (min) decrease in concentration from 0.10 M to 0.050 M (Figure 13.12). Measuring the half- life of a reaction is one way to determine the rate constant of a first-order reaction. The following analogy may be helpful for understanding Equation (13.6). If a college student takes 4 yr to graduate, the half-life of his or her stay at the college is 2 yr. Thus, half-life is not affected by how many other students are present. Similarly, the half-life of a first-order reaction is concentration independent. The usefulness of t12 is that it gives us a measure of the magnitude of the rate constant—the shorter the half-life, the larger the k. Consider, for example, two radio- active isotopes used in nuclear medicine: 24Na (t 12 5 14.7 h) and 60Co (t 12 5 5.3 yr). It is obvious that the 24Na isotope decays faster because it has a shorter half-life. If we started with 1 mole each of the isotopes, most of the 24Na would be gone in a week while the 60Co sample would be mostly intact. In Example 13.6 we calculate the half-life of a first-order reaction. 8n Example 13.6 The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 3 1024 s21 at 7008C: C2H6 (g) ¡ 2CH3 (g) Calculate the half-life of the reaction in minutes. Strategy To calculate the half-life of a first-order reaction, we use Equation (13.6). A conversion is needed to express the half-life in minutes. (Continued) C2H6 ¡ 2CH3 582 Chapter 13 ■ Chemical Kinetics Solution For a first-order reaction, we only need the rate constant to calculate the half-life of the reaction. From Equation (13.6) 0.693 t 12 5 k 0.693 5 5.36 3 1024 s21 1 min 5 1.29 3 103 s 3 60 s Similar problem: 13.26. 5 21.5 min Practice Exercise Calculate the half-life of the decomposition of N2O5, discussed on p. 577. Review of Concepts Consider the first-order reaction A ¡ B in which A molecules (blue spheres) are converted to B molecules (orange spheres). (a) What are the half-life and rate constant for the reaction? (b) How many molecules of A and B are present at t 5 20 s and t 5 30 s? t0s t  10 s Second-Order Reactions A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the concentrations of two different reactants, each raised to the first power. The simpler type involves only one kind of reactant molecule: A ¡ product where ¢[A] rate 5 2 ¢t From the rate law, rate 5 k[A]2 As before, we can determine the units of k by writing rate M/s k5 5 2 5 1/M ? s [A]2 M Another type of second-order reaction is A 1 B ¡ product  13.3 The Relation Between Reactant Concentration and Time 583 and the rate law is given by rate 5 k[A][B] The reaction is first order in A and first order in B, so it has an overall reaction order of 2. Using calculus, we can obtain the following expressions for “A ¡ product” Equation (13.7) is the result of [A] t d[A] t second-order reactions: k dt 2 [A] 0 [A] 0 1 1 5 kt 1 (13.7) [A]t [A]0 Equation (13.7) has the form of a linear equation. As Figure 13.13 shows, a plot of 1/[A]t versus t gives a straight line with slope 5 k and y intercept 5 1/[A]0. Pseudo-First-Order Reactions The other type of second-order reaction A 1 B ¡ product and the corresponding rate law rate 5 k[A][B] is actually more common than the k[A]2 second-order kinetics already shown. How- ever, it is considerably more difficult to treat mathematically. While it is possible to solve the integrated form of the rate law, a common approach is to measure the second-order reaction rates under pseudo-first-order kinetics conditions. If the above reaction is carried out under the conditions where one of the reactants is in large excess over the other, then the concentration of the excess reactant will not change appreciably over the course of the reaction. For example, if [B] @ [A], then [B] will be essentially constant and we have rate 5 k[A][B] 5 kobs[A] Note that the rate law now has the appearance of a first-order reaction. The rate con- stant kobs, called the pseudo-first-order rate constant, is given by kobs 5 k[B], where the subscript “obs” denotes observed and k is the second-order rate constant. If we measure kobs for many different initial concentrations of B, then a plot of kobs versus [B] will yield a straight line with a slope equal to k. Previously, we saw that the reaction between bromine and formic acid can be treated as a first-order reaction because formic acid is present in excess (see p. 564). Another well-studied example is the hydrolysis (meaning reaction with water) of ethyl acetate to yield acetic acid and ethanol: CH3COOC2H5 1 H2O ¡ CH3COOH 1 C2H5OH Because the concentration of water, the solvent, is about 55.5 M† compared to 1 M slope  k OO [A]t 1 or less for ethyl acetate, [H2O] can be treated as a constant so the rate is given by rate 5 k[CH3COOC2H5][H2O] 5 kobs[CH3COOC2H5] 1 OO [A]0 where kobs 5 k[H2O]. t Figure 13.13 A plot of 1/[A]t versus t for the second-order † In 1 L of a relatively dilute solution, the mass of water is approximately 1000 g so there are reaction A ¡ products. The 1000 g/(18.02 g/mol) or 55.5 mole of water. Thus, the concentration of water is 55.5 M. slope of the line is equal to k. 584 Chapter 13 ■ Chemical Kinetics Reaction Half-life We can obtain an equation for the half-life of a second-order reaction of the type A ¡ product by setting [A]t 5 [A]0 /2 in Equation (13.7) 1 1 5 kt 12 1 [A]0 /2 [A]0 Solving for t 12 we obtain 1 t 12 5 (13.8) k[A]0 Note that the half-life of a second-order reaction is inversely proportional to the initial reactant concentration. This result makes sense because the half-life should be shorter in the early stage of the reaction when more reactant molecules are present to collide with each other. Measuring the half-lives at different ini- tial concentrations is one way to distinguish between a first-order and a second- order reaction. The kinetic analysis of a second-order reaction is shown in Example 13.7. Example 13.7 Iodine atoms combine to form molecular iodine in the gas phase I(g) 1 I(g) ¡ I2 (g) This reaction follows second-order kinetics and has the high rate constant 7.0 3 109/M ? s at 238C. (a) If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. Strategy (a) The relationship between the concentrations of a reactant at different 88n times is given by the integrated rate law. Because this is a second-order reaction, we use Equation (13.7). (b) We are asked to calculate the half-life. The half-life for a second-order reaction is given by Equation (13.8). Solution (a) To calculate the concentration of a species at a later time of a second-order reaction, we need the initial concentration and the rate constant. Applying Equation (13.7) 1 1 5 kt 1 [A]t [A]0 I 1 I ¡ I2 1 60 s 1 5 (7.0 3 109/M ? s)a2.0 min 3 b1 [A]t 1 min 0.086 M where [A]t is the concentration at t 5 2.0 min. Solving the equation, we get [A]t 5 1.2 3 10212 M This is such a low concentration that it is virtually undetectable. The very large rate constant for the reaction means that practically all the I atoms combine after only 2.0 min of reaction time. (Continued)  13.3 The Relation Between Reactant Concentration and Time 585 (b) We need Equation (13.8) for this part. [A]0 For [I]0 5 0.60 M 1 t 12 5 [A]t k[A]0 1 slope  k 5 9 (7.0 3 10 /M ? s) (0.60 M) 5 2.4 3 10210 s For [I]0 5 0.42 M t 1 Figure 13.14 A plot of [A]t versus t12 5 t for a zero-order reaction. The (7.0 3 109/M ? s) (0.42 M) slope of the line is equal 5 3.4 3 10210 s to 2k. Check These results confirm that the half-life of a second-order reaction, unlike that of a first-order reaction, is not a constant but depends on the initial concentration of the reactant(s). Does it make sense that a larger initial concentration should have a shorter half-life? Similar problems: 13.27, 13.28. Practice Exercise The reaction 2A ¡ B is second order with a rate constant of 51/M ? min at 248C. (a) Starting with [A]0 5 0.0092 M, how long will it take for [A]t 5 3.7 3 1023 M? (b) Calculate the half-life of the reaction. Review of Concepts Consider the reaction A ¡ products. The half-life of the reaction depends on the initial concentration of A. Which of the following statements is inconsistent with the given information? (a) The half-life of the reaction decreases as the initial concentration increases. (b) A plot of ln [A]t versus t yields a straight line. (c) Doubling the concentration of A quadruples the rate. Zero-Order Reactions First- and second-order reactions are the most common reaction types. Reactions whose order is zero are rare. For a zero-order reaction A ¡ product the rate law is given by rate 5 k[A]0 Recall that any number raised to the power zero is equal to one. 5k Thus, the rate of a zero-order reaction is a constant, independent of reactant concen- tration. Using calculus, we can show that [A]t 5 2kt 1 [A]0 (13.9) Equation (13.9) is the result of [A] t t d[A] k dt Equation (13.9) has the form of a linear equation. As Figure 13.14 shows, a plot of [A]0 0 [A]t versus t gives a straight line with slope 5 2k and y intercept 5 [A]0. To calculate the half-life of a zero-order reaction, we set [A]t 5 [A]0 /2 in Equation (13.9) and obtain [A]0 t 12 5 (13.10) 2k  CHEMISTRY in Action Radiocarbon Dating H ow do scientists determine the ages of artifacts from archaeo- logical excavations? If someone tried to sell you a manuscript supposedly dating from 1000 b.c., how could you be certain of its authenticity? Is a mummy found in an Egyptian pyramid really 3000 years old? Is the so-called Shroud of Turin truly the burial cloth of Jesus Christ? The answers to these and other similar ques- tions can usually be found by applying chemical kinetics and the radiocarbon dating technique. Earth’s atmosphere is constantly being bombarded by cos- mic rays of extremely high penetrating power. These rays, which originate in outer space, consist of electrons, neutrons, and atomic nuclei. One of the important reactions between the atmosphere and cosmic rays is the capture of neutrons by atmospheric nitrogen (nitrogen-14 isotope) to produce the radioactive carbon-14 isotope and hydrogen. The unstable carbon atoms eventually form 14CO2, which mixes with the ordinary carbon dioxide (12CO2) in the air. As the carbon-14 isotope decays, it emits β particles (electrons). The rate of decay (as measured by the number of electrons emitted per second) obeys first-order kinetics. It is customary in the study of radioactive decay to write the rate law as rate 5 kN where k is the first-order rate constant and N the number of 14C nuclei present. The half-life of the decay, t 12 , is 5.73 3 103 yr, so that from Equation (13.6) we write 0.693 The Shroud of Turin. For generations there has been controversy about k5 5 1.21 3 1024 yr21 whether the Shroud, a piece of linen bearing the image of a man, was the 5.73 3 103 yr burial cloth of Jesus Christ. Many of the known zero-order reactions take place on a metal surface. An exam- ple is the decomposition of nitrous oxide (N2O) to nitrogen and oxygen in the presence of platinum (Pt): Keep in mind that [A]0 and [A]t in 2N2O(g) ¡ 2N2 (g) 1 O2 (g) Equation (13.9) refer to the concentration of N2O in the gas phase. When all the binding sites on Pt are occupied, the rate becomes constant regardless of the amount of N2O present in the gas phase. As we will see in Section 13.6, another well-studied zero-order reaction occurs in enzyme catalysis. Third-order and higher order reactions are quite complex; they are not presented in this book. Table 13.3 summarizes the kinetics of zero-order, first-order, and second- order reactions. The above Chemistry in Action essay describes the application of chemical kinetics to estimating the ages of objects. 586  The carbon-14 isotopes enter the biosphere when carbon 1 N0 dioxide is taken up in plant photosynthesis. Plants are eaten by t5 ln k Nt animals, which exhale carbon-14 in CO2. Eventually, carbon-14 1 decay rate at t 5 0 participates in many aspects of the carbon cycle. The 14C lost by 5 24 21 ln 1.21 3 10 yr decay rate at t 5 t radioactive decay is constantly replenished by the production of 1 decay rate of fresh sample new isotopes in the atmosphere. In this decay-replenishment 5 ln 24 21 process, a dynamic equilibrium is established whereby the ratio 1.21 3 10 yr decay rate of old sample of 14C to 12C remains constant in living matter. But when an individual plant or an animal dies, the carbon-14 isotope in it is Knowing k and the decay rates for the fresh sample and the old no longer replenished, so the ratio decreases as 14C decays. sample, we can calculate t, which is the age of the old sample. This same change occurs when carbon atoms are trapped in This ingenious technique is based on a remarkably simple idea. coal, petroleum, or wood preserved underground, and, of Its success depends on how accurately we can measure the rate course, in Egyptian mummies. After a number of years, there of decay. In fresh samples, the ratio 14Cy12C is about 1y1012, so are proportionately fewer 14C nuclei in, say, a mummy than in the equipment used to monitor the radioactive decay must be a living person. very sensitive. Precision is more difficult with older samples In 1955, Willard F. Libby† suggested that this fact could be because they contain even fewer 14C nuclei. Nevertheless, radio- used to estimate the length of time the carbon-14 isotope in a carbon dating has become an extremely valuable tool for esti- particular specimen has been decaying without replenishment. mating the age of archaeological artifacts, paintings, and other Rearranging Equation (13.3), we can write objects dating back 1000 to 50,000 years. A well-publicized application of radiocarbon dating was N0 the determination of the age of the Shroud of Turin. In 1988 ln 5 kt three laboratories in Europe and the United States, working on Nt samples of less than 50 mg of the Shroud, showed by carbon-14 where N0 and Nt are the number of 14C nuclei present at t 5 0 dating that the Shroud dates from between a.d. 1260 and a.d. and t 5 t, respectively. Because the rate of decay is directly 1390. These findings seem to indicate that the Shroud could not proportional to the number of 14C nuclei present, the preceding have been the burial cloth of Christ. However, recent research equation can be rewritten as reported new evidence suggesting the finding was invalid be- cause the dating analysis was based on contaminants introduced by repairs to the Shroud in later years. It seems the controversy † Willard Frank Libby (1908–1980). American chemist. Libby received the Nobel will continue for some time and further testing on the Shroud is Prize in Chemistry in 1960 for his work on radiocarbon dating. warranted. Summary of the Kinetics of Zero-Order, First-Order, Table 13.3 and Second-Order Reactions Concentration- Order Rate Law Time Equation Half-Life [A]0 0 Rate 5 k [A]t 5 2kt 1 [A]0 2k [A]t 0.693 1 Rate 5 k[A] ln 5 2kt [A]0 k 1 1 1 2† Rate 5 k[A]2 5 kt 1 [A]t [A]0 k[A]0 † A ¡ product. 587  588 Chapter 13 ■ Chemical Kinetics 13.4 Activation Energy and Temperature Dependence of Rate Constants Rate constant With very few exceptions, reaction rates increase with increasing temperature. For example, the time required to hard-boil an egg in water is much shorter if the “reaction” is carried out at 1008C (about 10 min) than at 808C (about 30 min). Conversely,  an effective way to preserve foods is to store them at subzero tem- peratures, thereby slowing the rate of bacterial decay. Figure 13.15 shows a typi- cal example of the relationship between the rate constant of a reaction and Temperature temperature. In order to explain this behavior, we must ask how reactions get started in the first place. Figure 13.15 Dependence of rate constant on temperature. The rate constants of most reactions The Collision Theory of Chemical Kinetics increase with increasing temperature. The kinetic molecular theory of gases (p. 202) postulates that gas molecules frequently collide with one another. Therefore, it seems logical to assume—and it is generally true—that chemical reactions occur as a result of collisions between reacting molecules. In terms of the collision theory of chemical kinetics, then, we expect the rate of a reac- tion to be directly proportional to the number of molecular collisions per second, or to the frequency of molecular collisions: number of collisions rate r s This simple relationship explains the dependence of reaction rate on concentration. Consider the reaction of A molecules with B molecules to form some product. Suppose that each product molecule is formed by the direct combination of an A molecule and a B molecule. If we doubled the concentration of A, then the number (a) of A-B collisions would also double, because there would be twice as many A mol- ecules that could collide with B molecules in any given volume (Figure 13.16). Con- sequently, the rate would increase by a factor of 2. Similarly, doubling the concentration of B molecules would increase the rate twofold. Thus, we can express the rate law as rate 5 k[A][B] (b) The reaction is first order in both A and B and obeys second-order kinetics. The collision theory is intuitively appealing, but the relationship between rate and molecular collisions is more complicated than you might expect. The implication of the collision theory is that a reaction always occurs when an A and a B molecule col- lide. However, not all collisions lead to reactions. Calculations based on the kinetic molecular theory show that, at ordinary pressures (say, 1 atm) and temperatures (say, (c) 298 K), there are about 1 3 1027 binary collisions (collisions between two molecules) in 1 mL of volume every second in the gas phase. Even more collisions per second Figure 13.16 Dependence of occur in liquids. If every binary collision led to a product, then most reactions would number of collisions on concentration. We consider here be complete almost instantaneously. In practice, we find that the rates of reactions only A-B collisions, which can lead differ greatly. This means that, in many cases, collisions alone do not guarantee that a to formation of products. (a) There reaction will take place. are four possible collisions among two A and two B molecules. Any molecule in motion possesses kinetic energy; the faster it moves, the greater (b) Doubling the number of either the kinetic energy. But a fast-moving molecule will not break up into fragments on type of molecule (but not both) its own. To react, it must collide with another molecule. When molecules collide, part increases the number of collisions to eight. (c) Doubling both the A of their kinetic energy is converted to vibrational energy. If the initial kinetic energies and B molecules increases the are large, then the colliding molecules will vibrate so strongly as to break some of number of collisions to sixteen. In the chemical bonds. This bond fracture is the first step toward product formation. If each case, the collision between a red sphere and a gray sphere the initial kinetic energies are small, the molecules will merely bounce off each other can only be counted once. intact. Energetically speaking, there is some minimum collision energy below which  13.4 Activation Energy and Temperature Dependence of Rate Constants 589 Figure 13.17 Potential energy profiles for (a) exothermic and AB‡ AB‡ (b) endothermic reactions. These plots show the change in potential energy as reactants A and B are converted to products C and D. Potential energy Potential energy Ea The activated complex (AB‡) is a highly unstable species with a Ea high potential energy. The A+B C+D activation energy is defined for the forward reaction in both (a) and (b). Note that the products C and D are more stable than the reactants in (a) and less stable C+D A+B than those in (b). Reaction progress Reaction progress (a) (b) no reaction occurs. Lacking this energy, the molecules remain intact, and no change results from the collision. We postulate that in order to react, the colliding molecules must have a total Animation Activation Energy kinetic energy equal to or greater than the activation energy (Ea), which is the mini- mum amount of energy required to initiate a chemical reaction. When molecules collide they form an activated complex (also called the transition state), a temporary species formed by the reactant molecules as a result of the collision before they form the product. Figure 13.17 shows two different potential energy profiles for the reaction A 1 B ¡ AB‡ ¡ C 1 D where AB‡ denotes an activated complex formed by the collision between A and B. If the products are more stable than the reactants, then the reaction will be accompanied by a release of heat; that is, the reaction is exothermic [Figure 13.17(a)]. On the other hand, if the products are less stable than the reactants, then heat will be absorbed by the reacting mixture from the surroundings and we have an endothermic reaction [Fig- ure 13.17(b)]. In both cases we plot the potential energy of the reacting system versus the progress of the reaction. Qualitatively, these plots show the potential energy changes as reactants are converted to products. We can think of activation energy as a barrier that prevents less energetic mol- ecules from reacting. Because the number of reactant molecules in an ordinary reaction is very large, the speeds, and hence also the kinetic energies of the mole- cules, vary greatly. Normally, only a small fraction of the colliding molecules—the fastest-moving ones—have enough kinetic energy to exceed the activation energy. These molecules can therefore take part in the reaction. The increase in the rate (or the rate constant) with temperature can now be explained: The speeds of the mol- ecules obey the Maxwell distributions shown in Figure 5.17. Compare the speed distributions at two different temperatures. Because more high-energy molecules are present at the higher temperature, the rate of product formation is also greater at the higher temperature. The Arrhenius Equation The dependence of the rate constant of a reaction on temperature can be expressed by the following equation, known as the Arrhenius equation: k 5 Ae2Ea/RT (13.11) 590 Chapter 13 ■ Chemical Kinetics where Ea is the activation energy of the reaction (in kJ/mol), R the gas constant (8.314 J/K ? mol), T the absolute temperature, and e the base of the natural logarithm scale (see Appendix 4). The quantity A represents the collision frequency and is called the frequency factor. It can be treated as a constant for a given reacting system over a fairly wide temperature range. Equation (13.11) shows that the rate constant is directly proportional to A and, therefore, to the collision frequency. In addition, because of the minus sign associated with the exponent Ea/RT, the rate constant decreases with increasing activation energy and increases with increasing tempera- ture. This equation can be expressed in a more useful form by taking the natural logarithm of both sides: ln k 5 ln Ae2Ea/RT Ea or ln k 5 ln A 2 (13.12) RT Equation (13.12) can be rearranged to a linear equation: Ea 1 ln k 5 a2 b a b 1 ln A (13.13) 4 R T 4 4 4 y 5 m x 1 b Thus, a plot of ln k versus 1/T gives a straight line whose slope m is equal to 2Ea /R and whose intercept b with the y axis is ln A. Example 13.8 demonstrates a graphical method for determining the activation energy of a reaction. Example 13.8 The rate constants for the decomposition of acetaldehyde CH3CHO(g) ¡ CH4 (g) 1 CO(g) were measured at five different temperatures. The data are shown in the table. Plot ln k versus 1/T, and determine the activation energy (in kJ/mol) for the 1 reaction. Note that the reaction is “ 32 ” order in CH3CHO, so k has the units of 1/M 2 ? s. k (1/M2 # s) 1 T (K) 88n 0.011 700 0.035 730 0.105 760 0.343 790 0.789 810 Strategy Consider the Arrhenius equation written as a linear equation Ea 1 ln k 5 a2 b a b 1 ln A R T A plot of ln k versus 1/T (y versus x) will produce a straight line with a slope equal to 2Ea /R. Thus, the activation energy can be determined from the slope of the plot. CH3CHO ¡ CH4 1 CO (Continued)  13.4 Activation Energy and Temperature Dependence of Rate Constants 591 0.00 Figure 13.18 Plot of ln k versus (1.24 × 10 –3 K–1, 0.45) 1/T. The slope of the line is equal to 2Ea /R. –1.00 –2.00 Δy 1n k –3.00 Δx – 4.00 (1.41 × 10 –3 K–1, 4.00) –5.00 1.20 × 10 –3 1.30 × 10 –3 1.40 × 10 –3 1/T (K–1) Solution First we convert the data to the following table ln k 1/T (K 1) 4.51 1.43  103 3.35 1.37  103 2.254 1.32  103 1.070 1.27  103 0.237 1.23  103 A plot of these data yields the graph in Figure 13.18. The slope of the line is calculated from two pairs of coordinates: 24.00 2 (20.45) slope 5 5 22.09 3 104 K (1.41 2 1.24) 3 1023 K21 From the linear form of Equation (13.13) Ea slope 5 2 5 22.09 3 104 K R Ea 5 (8.314 J/K ? mol) (2.09 3 104 K) 5 1.74 3 105 J/mol 5 1.74 3 102 kJ/mol Check 1 It is important to note that although the rate constant itself has the units 1/M 2 ? s, the quantity ln k has no units (we cannot take the logarithm of a unit). Similar problem: 13.40. Practice Exercise The second-order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures: k (1/M s) t (°C) 1.87  103 600 0.0113 650 0.0569 700 0.244 750 Determine graphically the activation energy for the reaction. An equation relating the rate constants k1 and k2 at temperatures T1 and T2 can be used to calculate the activation energy or to find the rate constant at another 592 Chapter 13 ■ Chemical Kinetics temperature if the activation energy is known. To derive such an equation we start with Equation (13.12): Ea ln k1 5 ln A 2 RT1 Ea ln k2 5 ln A 2 RT2 Subtracting ln k2 from ln k1 gives Ea 1 1 ln k1 2 ln k2 5 a 2 b R T2 T1 k1 Ea 1 1 ln 5 a 2 b k2 R T2 T1 k1 Ea T1 2 T2 ln 5 a b (13.14) k2 R T1 T2 Example 13.9 illustrates the use of the equation we have just derived. Example 13.9 The rate constant of a first-order reaction is 3.46 3 1022 s21 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? Strategy A modified form of the Arrhenius equation relates two rate constants at two different temperatures [see Equation (13.14)]. Make sure the units of R and Ea are consistent. Solution The data are k1 5 3.46 3 1022 s21   k2 5 ? T1 5 298 K    T2 5 350 K Substituting in Equation (13.14), 3.46 3 1022 s21 50.2 3 103 J/mol 298 K 2 350 K ln 5 c d k2 8.314 J/K ? mol (298 K) (350 K) We convert Ea to units of J/mol to match the units of R. Solving the equation gives 3.46 3 1022 s21 ln 5 23.01 k2 22 21 3.46 3 10 s 5 e23.01 5 0.0493 k2 k2 5 0.702 s21 Check The rate constant is expected to be greater at a higher temperature. Therefore, Similar problem: 13.42. the answer is reasonable. Practice Exercise The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 3 10210 s21 at 258C. Calculate the rate constant at 408C if the activation energy is 116 kJ/mol.  13.4 Activation Energy and Temperature Dependence of Rate Constants 593 8n  CO NO2 CO2 NO (a) 8n  CO NO2 CO NO2 (b) Figure 13.19 The orientations of the molecules shown in (a) are effective and will likely lead to formation of products. The orientations shown in (b) are ineffective and no products will be formed. For simple reactions (for example, reactions between atoms), we can equate the frequency factor (A) in the Arrhenius equation with the frequency of colli- sion between the reacting species. For more complex reactions, we must also consider the “orientation factor,” that is, how reacting molecules are oriented Animation Orientation of Collision relative to each other. The reaction between carbon monoxide (CO) and nitrogen dioxide (NO2) to form carbon dioxide (CO2) and nitric oxide (NO) illustrates this point: CO(g) 1 NO2 (g) ¡ CO2 (g) 1 NO(g) This reaction is most favorable when the reacting molecules approach each other according to that shown in Figure 13.19(a). Otherwise, few or no products are formed [Figure 13.19(b)]. The quantitative treatment of orientation factor is to modify Equa- tion (13.11) as follows: k 5 pAe2Ea /RT (13.15) where p is the orientation factor. The orientation factor is a unitless quantity; its value ranges from 1 for reactions involving atoms such as I 1 I ¡ I2 to 1026 or smaller for reactions involving molecules. Review of Concepts (a) What can you deduce about the magnitude of the activation energy of a reaction if its rate constant changes appreciably with a small change in temperature? (b) If a reaction occurs every time two reacting molecules collide, what can you say about the orientation factor and the activation energy of the reaction? 594 Chapter 13 ■ Chemical Kinetics 13.5 Reaction Mechanisms As we mentioned earlier, an overall balanced chemical equation does not tell us much about how a reaction actually takes place. In many cases, it merely repre- sents the sum of several elementary steps, or elementary reactions, a series of simple reactions that represent the progress of the overall reaction at the molec- ular level. The term for the sequence of elementary steps that leads to product formation is reaction mechanism. The reaction mechanism is comparable to the route of travel followed during a trip; the overall chemical equation specifies only the origin and destination. As an example of a reaction mechanism, let us consider the reaction between nitric oxide and oxygen: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) We know that the products are not formed directly from the collision of two NO molecules with an O2 molecule because N2O2 is detected during the course of the reaction. Let us assume that the reaction actually takes place via two elementary steps as follows: 2NO(g) ¡ N2O2 (g) ⴙ 8n N2O2 (g) 1 O2 (g) ¡ 2NO2 (g) ⴙ 8n ⴙ In the first elementary step, two NO molecules collide to form a N2O2 molecule. This event is followed by the reaction between N2O2 and O2 to give two molecules of NO2. The net chemical equation, which represents the overall change, is given by the sum of the elementary steps: Step 1: NO NO 88n N2O2 Step 2: N2O2 O2 88n 2NO2 The sum of the elementary steps must give the overall balanced equation. Overall reaction: 2NO N2O2 O2 ¡ N2O2 2NO2 Species such as N2O2 are called intermediates because they appear in the mechanism of the reaction (that is, the elementary steps) but not in the overall balanced equation. Keep in mind that an intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elemen- tary step. These molecules may be of the same or different types. Each of the elemen- tary steps discussed above is called a bimolecular reaction, an elementary step that involves two molecules. An example of a unimolecular reaction, an elementary step in which only one reacting molecule participates, is the conversion of cyclopropane to propene discussed in Example 13.4. Very few termolecular reactions, reactions that involve the participation of three molecules in one elementary step, are known, because the simultaneous encounter of three molecules is a far less likely event than a bimolecular collision.  13.5 Reaction Mechanisms 595 Rate Laws and Elementary Steps Knowing the elementary steps of a reaction enables us to deduce the rate law. Suppose we have the following elementary reaction: A ¡ products Because there is only one molecule present, this is a unimolecular reaction. It follows that the larger the number of A molecules present, the faster the rate of product for- mation. Thus, the rate of a unimolecular reaction is directly proportional to the con- centration of A, or is first order in A: rate 5 k[A] For a bimolecular elementary reaction involving A and B molecules A 1 B ¡ product the rate of product formation depends on how frequently A and B collide, which in turn depends on the concentrations of A and B. Thus, we can express the rate as rate 5 k[A][B] Similarly, for a bimolecular elementary reaction of the type A 1 A ¡ products or 2A ¡ products the rate becomes rate 5 k[A]2 The preceding examples show that the reaction order for each reactant in an elemen- Note that the rate law can be written directly from the coefficients of an tary reaction is equal to its stoichiometric coefficient in the chemical equation for that elementary step. step. In general, we cannot tell by merely looking at the overall balanced equation whether the reaction occurs as shown or in a series of steps. This determination is made in the laboratory. When we study a reaction that has more than one elementary step, the rate law for the overall process is given by the rate-determining step, which is the slowest step in the sequence of steps leading to product formation. An analogy for the rate-determining step is the flow of traffic along a narrow road. Assuming the cars cannot pass one another on the road, the rate at which the cars travel is governed by the slowest-moving car. Experimental studies of reaction mechanisms begin with the collection of data (rate measurements). Next, we analyze the data to determine the rate constant and order of the reaction, and we write the rate law. Finally, we suggest a plausible mechanism for the reaction in terms of elementary steps (Figure 13.20). The elemen- tary steps must satisfy two requirements: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law as is determined experimentally. Remember that for a proposed reaction scheme, we must be able to detect the pres- ence of any intermediate(s) formed in one or more elementary steps.  596 Chapter 13 ■ Chemical Kinetics Postulating Measuring Formulating a reasonable the rate of the rate law reaction a reaction mechanism Figure 13.20 Sequence of steps in the study of a reaction mechanism. The decomposition of hydrogen peroxide and the formation of hydrogen iodide from molecular hydrogen and molecular iodine illustrate the elucidation of reaction mechanisms by experimental studies. Hydrogen Peroxide Decomposition The decomposition of hydrogen peroxide is facilitated by iodide ions (Figure 13.21). The overall reaction is 2H2O2 (aq) ¡ 2H2O(l) 1 O2 (g) By experiment, the rate law is found to be rate 5 k[H2O2][I2] Thus, the reaction is first order with respect to both H2O2 and I2. You can see that H2O2 decomposition does not occur in a single elementary step corresponding to the overall balanced equation. If it did, the reaction would be second order in H2O2 (as a result of the collision of two H2O2 molecules). What’s more, the I2 ion, which is not even part of the overall equation, appears in the rate law expres- sion. How can we reconcile these facts? First, we can account for the observed rate law by assuming that the reaction takes place in two separate elementary steps, each Figure 13.21 The decomposition of which is bimolecular: of hydrogen peroxide is catalyzed by the iodide ion. A few drops of k liquid soap have been added to Step 1: H 2O 2 I ¡1 H 2O IO the solution to dramatize the k Step 2: H 2O 2 IO ¡ 2 H 2O O2 I evolution of oxygen gas. (Some of the iodide ions are oxidized to molecular iodine, which then If we further assume that step 1 is the rate-determining step, then the rate of the reac- reacts with iodide ions to form the tion can be determined from the first step alone: brown triiodide I 2 3 ion.) rate 5 k1[H2O2][I 2 ] where k1 5 k. Note that the IO2 ion is an intermediate because it does not appear in the overall balanced equation. Although the I2 ion also does not appear in the overall equa- Intermediate tion, I2 differs from IO2 in that the former is present at the start of the reaction and at its completion. The function of I2 is to speed up the reaction—that is, it is a catalyst. Potential energy Ea We will discuss catalysis in Section 13.6. Figure 13.22 shows the potential energy profile (Step 1) for a reaction like the decomposition of H2O2. We see that the first step, which is rate R Ea determining, has a larger activation energy than the second step. The intermediate, (Step 2) although stable enough to be observed, reacts quickly to form the products. P The Hydrogen Iodide Reaction Reaction progress A common reaction mechanism is one that involves at least two elementary steps, the Figure 13.22 Potential energy first of which is very rapid in both the forward and reverse directions compared with profile for a two-step reaction the second step. An example is the reaction between molecular hydrogen and molec- in which the first step is rate- determining. R and P represent ular iodine to produce hydrogen iodide: reactants and products, respectively. H2 (g) 1 I2 (g) ¡ 2HI(g)  13.5 Reaction Mechanisms 597 Experimentally, the rate law is found to be rate 5 k[H2][I2] For many years it was thought that the reaction occurred just as written; that is, it is a bimolecular reaction involving a hydrogen molecule and an iodine molecule, as shown on p. 596. However, in the 1960s chemists found that the actual mechanism is more complicated. A two-step mechanism was proposed: k1 Step 1: I2 Δ k 2I 88n 1 k2 Step 2: H2 2I ¡ 2HI where k1, k21, and k2 are the rate constants for the reactions. The I atoms are the intermediate in this reaction. When the reaction begins, there are very few I atoms present. But as I2 dis- sociates, the concentration of I2 decreases while that of I increases. Therefore, the forward rate of step 1 decreases and the reverse rate increases. Soon the two rates become equal, and a chemical equilibrium is established. Because the elementary reactions in step 1 are much faster than the one in step 2, equilibrium is reached H2 1 I2 ¡ 2HI before any significant reaction with hydrogen occurs, and it persists throughout the reaction. In the equilibrium condition of step 1 the forward rate is equal to the reverse rate; Chemical equilibrium will be discussed in Chapter 14. that is, k1[I2] 5 k21[I]2 k1 or [I]2 5 [I2] k21 The rate of the reaction is given by the slow, rate-determining step, which is step 2: rate 5 k2[H2][I]2 Substituting the expression for [I]2 into this rate law, we obtain k1k2 rate 5 [H2][I2] k21 5 k[H2][I2] where k 5 k1k2/k21. As you can see, this two-step mechanism also gives the correct rate law for the reaction. This agreement along with the observation of intermediate I atoms provides strong evidence that the mechanism is correct. Finally, we note that not all reactions have a single rate-determining step. A reac- tion may have two or more comparably slow steps. The kinetic analysis of such reac- tions is generally more involved. Example 13.10 concerns the mechanistic study of a relatively simple reaction. Example 13.10 The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps: k Step 1: N2O ¡1 N2 O k Step 2: N2O O¡2 N2 O2 (Continued) 598 Chapter 13 ■ Chemical Kinetics Experimentally the rate law is found to be rate 5 k[N2O]. (a) Write the equation for the overall reaction. (b) Identify the intermediate. (c) What can you say about the relative rates of steps 1 and 2? Strategy (a) Because the overall reaction can be broken down into elementary steps, knowing the elementary steps would enable us to write the overall reaction. (b) What are the characteristics of an intermediate? Does it appear in the overall reaction? (c) What determines which elementary step is rate determining? How does a knowledge of the rate-determining step help us write the rate law of a reaction? 88n Solution (a) Adding the equations for steps 1 and 2 gives the overall reaction 2N2O ¡ 2N2 1 O2 (b) Because the O atom is produced in the first elementary step and it does not appear in the overall balanced equation, it is an intermediate. (c) If we assume that step 1 is the rate-determining step, then the rate of the overall reaction is given by rate 5 k1[N2O] and k 5 k1. 2N2O ¡ 2N2 1 O2 Check There are two criteria that must be met for a proposed reaction mechanism to be plausible. (1) The individual steps (elementary steps) must sum to the corrected overall reaction. (2) The rate-determining step (the slow step) must have the same rate Similar problem: 13.55. law as the experimentally determined rate law. Practice Exercise The reaction between NO2 and CO to produce NO and CO2 is believed to occur via two steps: Step 1: NO2 NO2 88n NO NO3 Step 2: NO3 CO 88n NO2 CO2 The experimental rate law is rate 5 k[NO2]2. (a) Write the equation for the overall reaction. (b) Identify the intermediate. (c) What can you say about the relative rates of steps 1 and 2? Review of Concepts The rate law for the reaction H2 1 2IBr ¡ I2 1 2HBr is rate 5 k[H2][IBr]. Given that HI is an intermediate, write a two-step mechanism for the reaction. Experimental Support for Reaction Mechanisms How can we find out whether the proposed mechanism for a particular reaction is correct? In the case of hydrogen peroxide decomposition we might try to detect the presence of the IO2 ions by spectroscopic means. Evidence of their presence would support the reaction scheme. Similarly, for the hydrogen iodide reaction, detection of iodine atoms would lend support to the two-step mechanism. For example, I2 dis- sociates into atoms when it is irradiated with visible light. Thus, we might predict that the formation of HI from H2 and I2 would speed up as the intensity of light is increased because that should increase the concentration of I atoms. Indeed, this is just what is observed.  13.6 Catalysis 599 In another case, chemists wanted to know which C¬O bond is broken in the reaction between methyl acetate and water in order to better understand the reaction mechanism O O B B CH 3OCOOOCH 3  H 2O 8888n CH 3OCOOH  CH 3OH methyl acetate acetic acid methanol The two possibilities are O O B B † CH 3OCOOOCH 3 † CH 3OCOOOCH 3 (a) (b) To distinguish between schemes (a) and (b), chemists used water containing the oxygen-18 isotope instead of ordinary water (which contains the oxygen-16 isotope). When the oxygen-18 water was used, only the acetic acid formed contained oxygen-18: O B CH 3OCO 18 OOH Thus, the reaction must have occurred via bond-breaking scheme (a), because the product formed via scheme (b) would retain both of its original oxygen atoms. Another example is photosynthesis, the process by which green plants produce glucose from carbon dioxide and water 6CO2 1 6H2O ¡ C6H12O6 1 6O2 A question that arose early in studies of photosynthesis was whether the molecular oxygen was derived from water, from carbon dioxide, or from both. By using water containing the oxygen-18 isotope, it was demonstrated that the evolved oxygen came from water, and none came from carbon dioxide, because the O2 contained only the 18O isotopes. This result supported the mechanism in which water mole- cules are “split” by light: 2H2O 1 hv ¡ O2 1 4H 1 1 4e 2 where hv represents the energy of a photon. The protons and electrons are used to drive energetically unfavorable reactions that are necessary for plant growth and function. These examples give some idea of how inventive chemists must be in studying reaction mechanisms. For complex reactions, however, it is virtually impossible to prove the uniqueness of any particular mechanism. 13.6 Catalysis For the decomposition of hydrogen peroxide we saw that the reaction rate depends Animation on the concentration of iodide ions even though I2 does not appear in the overall Catalysis equation. We noted that I2 acts as a catalyst for that reaction. A catalyst is a substance that increases the rate of a reaction by lowering the activation energy. It does so by A rise in temperature also increases the rate of a reaction. However, at high providing an alternative reaction pathway. The catalyst may react to form an inter- temperatures, the products formed may mediate with the reactant, but it is regenerated in a subsequent step so it is not con- undergo other reactions, thereby reduc- ing the yield. sumed in the reaction. 600 Chapter 13 ■ Chemical Kinetics In the laboratory preparation of molecular oxygen, a sample of potassium chlorate is heated, as shown in Figure 4.13(b). The reaction is 2KClO3 (s) ¡ 2KCl(s) 1 3O2 (g) However, this thermal decomposition process is very slow in the absence of a catalyst. The rate of decomposition can be increased dramatically by adding a small amount of the catalyst manganese(IV) dioxide (MnO2), a black powdery substance. All of the MnO2 can be recovered at the end of the reaction, just as all the I2 ions remain fol- lowing H2O2 decomposition. To extend the traffic analogy, adding a A catalyst speeds up a reaction by providing a set of elementary steps with more catalyst can be compared with building a tunnel through a mountain to connect favorable kinetics than those that exist in its absence. From Equation (13.11) we know two towns that were previously linked by that the rate constant k (and hence the rate) of a reaction depends on the frequency factor a winding road over the mountain. A and the activation energy Ea—the larger A or the smaller Ea, the greater the rate. In many cases, a catalyst increases the rate by lowering the activation energy for the reaction. Let us assume that the following reaction has a certain rate constant k and an activation energy Ea. k A1B ¡ C1D In the presence of a catalyst, however, the rate constant is kc (called the catalytic rate constant): kc A1B ¡ C1D By the definition of a catalyst, ratecatalyzed . rateuncatalyzed A catalyst lowers the activation energy for Figure 13.23 shows the potential energy profiles for both reactions. Note that the both the forward and reverse reactions. total energies of the reactants (A and B) and those of the products (C and D) are unaffected by the catalyst; the only difference between the two is a lowering of the activation energy from Ea to E9a. Because the activation energy for the reverse reaction is also lowered, a catalyst enhances the rates of the reverse and forward reactions equally. There are three general types of catalysis, depending on the nature of the rate- increasing substance: heterogeneous catalysis, homogeneous catalysis, and enzyme catalysis. Figure 13.23 Comparison of the activation energy barriers of an uncatalyzed reaction and the same reaction with a catalyst. The catalyst lowers the energy Ea barrier but does not affect the Potential energy Potential energy actual energies of the reactants E'a or products. Although the reactants and products are the same in both cases, the reaction A+B A+B mechanisms and rate laws are different in (a) and (b). C+D C+D Reaction progress Reaction progress (a) (b)  13.6 Catalysis 601 1A Heterogeneous Catalysis 3A In heterogeneous catalysis, the reactants and the catalyst are in different phases. Usu- 4B 5B 6B 7B 8B 1B 2B Al K Ti V Cr Mn Fe Co Ni Cu Zn ally the catalyst is a solid and the reactants are either gases or liquids. Heterogeneous Zr Mo Ru Rh Pd W Re Os Ir Pt Au catalysis is by far the most important type of catalysis in industrial chemistry, espe- cially in the synthesis of many key chemicals. Here we describe three specific exam- Metals and compounds of metals ples of heterogeneous catalysis that account for millions of tons of chemicals produced that are most frequently used in annually on an industrial scale. heterogeneous catalysis. The Haber Synthesis of Ammonia Ammonia is an extremely valuable inorganic substance used in the fertilizer industry, the manufacture of explosives, and many other applications. Around the turn of the twentieth century, many chemists strove to synthesize ammonia from nitrogen and hydrogen. The supply of atmospheric nitrogen is virtually inexhaustible, and hydrogen gas can be produced readily by passing steam over heated coal: H2O(g) 1 C(s) ¡ CO(g) 1 H2 (g) Hydrogen is also a by-product of petroleum refining. The formation of NH3 from N2 and H2 is exothermic: N2 (g) 1 3H2 (g) ¡ 2NH3 (g) ¢H° 5 292.6 kJ/mol But the reaction rate is extremely slow at room temperature. To be practical on a large scale, a reaction must occur at an appreciable rate and it must have a high yield of the desired product. Raising the temperature does accelerate the above reaction, but at the same time it promotes the decomposition of NH3 molecules into N2 and H2, thus lowering the yield of NH3. In 1905, after testing literally hundreds of compounds at various temperatures and pressures, Fritz Haber discovered that iron plus a few percent of oxides of potas- sium and aluminum catalyze the reaction of hydrogen with nitrogen to yield ammonia at about 5008C. This procedure is known as the Haber process. In heterogeneous catalysis, the surface of the solid catalyst is usually the site of the reaction. The initial step in the Haber process involves the dissociation of N2 and H2 on the metal surface (Figure 13.24). Although the dissociated species are not truly free atoms because they are bonded to the metal surface, they are highly reactive. The two reactant molecules behave very differently on the catalyst surface. Studies show that H2 dissociates into atomic hydrogen at temperatures as low as 21968C (the boil- ing point of liquid nitrogen). Nitrogen molecules, on the other hand, dissociate at 8n 8n Figure 13.24 The catalytic action in the synthesis of ammonia. First the H2 and N2 molecules bind to the surface of the catalyst. This interaction weakens the covalent bonds within the molecules and eventually causes the molecules to dissociate. The highly reactive H and N atoms combine to form NH3 molecules, which then leave the surface. 602 Chapter 13 ■ Chemical Kinetics Figure 13.25 Platinum-rhodium catalyst used in the Ostwald process. about 5008C. The highly reactive N and H atoms combine rapidly at high temperatures to produce the desired NH3 molecules: N 1 3H ¡ NH3 The Manufacture of Nitric Acid Nitric acid is one of the most important inorganic acids. It is used in the production of fertilizers, dyes, drugs, and explosives. The major industrial method of producing nitric acid is the Ostwald† process. The starting materials, ammonia and molecular oxygen, are heated in the presence of a platinum-rhodium catalyst (Figure 13.25) to about 8008C: 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) The nitric oxide readily oxidizes (without catalysis) to nitrogen dioxide: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) When dissolved in water, NO2 forms both nitrous acid and nitric acid: 2NO2 (g) 1 H2O(l) ¡ HNO2 (aq) 1 HNO3 (aq) On heating, nitrous acid is converted to nitric acid as follows: 3HNO2 (aq) ¡ HNO3 (aq) 1 H2O(l) 1 2NO(g) The NO generated can be recycled to produce NO2 in the second step. Catalytic Converters At high temperatures inside a running car’s engine, nitrogen and oxygen gases react to form nitric oxide: N2 (g) 1 O2 (g) ¡ 2NO(g) † Wilhelm Ostwald (1853–1932). German chemist. Ostwald made important contributions to chemical kinetics, thermodynamics, and electrochemistry. He developed the industrial process for preparing nitric acid that now bears his name. He received the Nobel Prize in Chemistry in 1909.  13.6 Catalysis 603 Figure 13.26 A two-stage catalytic converter for an Exhaust manifold automobile. Exhaust pipe Tail pipe Air compressor: source of secondary air Catalytic converters When released into the atmosphere, NO rapidly combines with O2 to form NO2. Nitrogen dioxide and other gases emitted by an automobile, such as carbon monoxide (CO) and various unburned hydrocarbons, make automobile exhaust a major source of air pollution. Most new cars are equipped with catalytic converters (Figure 13.26). An efficient catalytic converter serves two purposes: It oxidizes CO and unburned hydrocarbons to CO2 and H2O, and it reduces NO and NO2 to N2 and O2. Hot exhaust gases into which air has been injected are passed through the first chamber of one converter to accelerate the complete burning of hydrocarbons and to decrease CO emission. (A cross section of the catalytic converter is shown in Figure 13.27.) However, because high temperatures increase NO production, a second chamber containing a different catalyst (a transition metal or a transition metal oxide such as CuO or Cr2O3) and operating at a lower temperature are required to dissociate NO into N2 and O2 before the exhaust is discharged through the tailpipe. Homogeneous Catalysis In homogeneous catalysis the reactants and catalyst are dispersed in a single phase, usually liquid. Acid and base catalyses are the most important types of homogeneous catalysis in liquid solution. For example, the reaction of ethyl acetate with water to form acetic acid and ethanol normally occurs too slowly to be measured. O O B B CH 3OCOOOC 2H 5  H 2O 88n CH 3OCOOH  C 2H 5OH ethyl acetate acetic acid ethanol In the absence of the catalyst, the rate law is given by This is a pseudo-first-order reaction rate 5 k[CH3COOC2H5] discussed earlier. Figure 13.27 A cross-sectional view of a catalytic converter. The beads contain platinum, palladium, and rhodium, which catalyze the conversion of CO and hydrocarbons to carbon dioxide and water. 604 Chapter 13 ■ Chemical Kinetics However, the reaction can be catalyzed by an acid. In the presence of hydrochloric acid, the rate is faster and the rate law is given by rate 5 kc[CH3COOC2H5][H 1 ] Note that because kc . k, the rate is determined solely by the catalyzed portion of the reaction. Homogeneous catalysis can also take place in the gas phase. A well-known example of catalyzed gas-phase reactions is the lead chamber process, which for many years was the primary method of manufacturing sulfuric acid. Starting with sulfur, we would expect the production of sulfuric acid to occur in the following steps: S(s) 1 O2 (g) ¡ SO2 (g) 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g) H2O(l) 1 SO3 (g) ¡ H2SO4 (aq) In reality, however, sulfur dioxide is not converted directly to sulfur trioxide; rather, the oxidation is more efficiently carried out in the presence of the catalyst nitrogen dioxide: 2SO2 (g) 1 2NO2 (g) ¡ 2SO3 (g) 1 2NO(g) 2NO(g) 1 O2 (g) ¡ 2NO2 (g) Overall reaction: 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g) Note that there is no net loss of NO2 in the overall reaction, so that NO2 meets the criteria for a catalyst. In recent years, chemists have devoted much effort to developing a class of transition metal compounds to serve as homogeneous catalysts. These compounds are soluble in various organic solvents and therefore can catalyze reactions in the same phase as the dissolved reactants. Many of the processes they catalyze are organic. For example, a red-violet compound of rhodium, [(C6H5)3P]3RhCl, cata- lyzes the conversion of a carbon-carbon double bond to a carbon-carbon single bond as follows: This reaction is important in the food A A A A industry. It converts “unsaturated fats” CPC  H2 88n OCOCO (compounds containing many C“C bonds) to “saturated fats” (compounds A A A A containing few or no C“C bonds). H H Homogeneous catalysis has several advantages over heterogeneous catalysis. For one thing, the reactions can often be carried out under atmospheric conditions, thus reducing production costs and minimizing the decomposition of products at high temperatures. In addition, homogeneous catalysts can be designed to func- tion selectively for a particular type of reaction, and homogeneous catalysts cost less than the precious metals (for example, platinum and gold) used in heteroge- neous catalysis. Enzyme Catalysis Of all the intricate processes that have evolved in living systems, none is more strik- ing or more essential than enzyme catalysis. Enzymes are biological catalysts. The amazing fact about enzymes is that not only can they increase the rate of biochem- ical reactions by factors ranging from 106 to 1018, but they are also highly specific. An enzyme acts only on certain molecules, called substrates (that is, reactants), while leaving the rest of the system unaffected. It has been estimated that an average living  13.6 Catalysis 605 Figure 13.28 The lock-and-key model of an enzyme’s specificity for substrate molecules. Substrate Products + + Enzyme Enzyme-substrate Enzyme complex cell may contain some 3000 different enzymes, each of them catalyzing a specific reaction in which a substrate is converted into the appropriate products. Enzyme catalysis is usually homogeneous because the substrate and enzyme are present in aqueous solution. An enzyme is typically a large protein molecule that contains one or more active sites where interactions with substrates take place. These sites are structurally compat- ible with specific substrate molecules, in much the same way as a key fits a particu- lar lock. In fact, the notion of a rigid enzyme structure that binds only to molecules whose shape exactly matches that of the active site was the basis of an early theory of enzyme catalysis, the so-called lock-and-key theory developed by the German chemist Emil Fischer† in 1894 (Figure 13.28). Fischer’s hypothesis accounts for the specificity of enzymes, but it contradicts research evidence that a single enzyme binds to substrates of different sizes and shapes. Chemists now know that an enzyme mol- ecule (or at least its active site) has a fair amount of structural flexibility and can modify its shape to accommodate more than one type of substrate. Figure 13.29 shows a molecular model of an enzyme in action. † Emil Fischer (1852–1919). German chemist. Regarded by many as the greatest organic chemist of the nineteenth century, Fischer made many significant contributions in the synthesis of sugars and other impor- tant molecules. He was awarded the Nobel Prize in Chemistry in 1902. 8n Figure 13.29 Left to right: The binding of a glucose molecule (red) to hexokinase (an enzyme in the metabolic pathway). Note how the region at the active site closes around glucose after binding. Frequently, the geometries of both the substrate and the active site are altered to fit each other. CHEMISTRY in Action Pharmacokinetics C hemical kinetics is very important in understanding the 0.10 Ethanol concentration in blood (g/100 mL) absorption, distribution, metabolism, and excretion of drugs in the body. In this sense, pharmacokinetics is the study of what the body does to a drug (as opposed to pharmacodynamics, 0.08 which is the study of what a drug does to the body). Knowledge of the rates of drug absorption and distribution in the body is 0.06 essential to achieving and maintaining proper dosages as well as understanding the mechanisms of action. Drug concentrations are typically measured in blood 0.04 plasma or urine at various times to give a drug concentration versus time plot. As the drug is absorbed into the bloodstream, 0.02 it is distributed to the various tissues and organs in the body and simultaneously eliminated by a combination of excretion and metabolism (biotransformation), all of which occur at dif- 0.00 ferent rates depending on the drug. The sum of all these pro- 0 1 2 3 4 5 6 7 8 cesses is the mechanism of drug delivery and distribution. t (hours) Because the drug must be distributed between different organs Concentration of ethanol in blood versus time after oral administration of and cross between aqueous (blood and urine) and lipid (fat) various doses: red (14 g), yellow (28 g), green (42 g), blue (56 g). tissue, and because many biological processes involve enzymes, kinetic behavior that is zero order in the drug is much more common in pharmacokinetics than it is in homogeneous solu- tion reaction kinetics. For example, the decomposition of etha- the more gradual decline. The minimum effective concentra- nol by the enzyme alcohol dehydrogenase is zero order in tion (MEC) is the minimum concentration required for the ethanol. drug to provide the desired therapeutic effect. The minimum Usually the absorption of a drug is more rapid than the toxic concentration (MTC) is the drug concentration at which elimination, giving a steeper rise in concentration compared to the drug becomes toxic or other undesired side effects outweigh The mathematical treatment of enzyme kinetics is quite complex, even when we know the basic steps involved in the reaction. A simplified scheme is given by the following elementary steps: k1 E⫹SΔ k ES 1 k2 ES ¡ E ⫹ P where E, S, and P represent enzyme, substrate, and product, and ES is the enzyme- substrate intermediate. It is often assumed that the formation of ES and its decompo- sition back to enzyme and substrate molecules occur rapidly and that the rate- determining step is the formation of product. (This is similar to the formation of HI discussed on p. 596.) In general, the rate of such a reaction is given by the equation ¢[P] rate 5 ¢t 5 k2[ES] 606 the benefit of the drug. Taken together, the MEC and MTC define a therapeutic index, and one of the goals of pharmaco- Drug concentration in blood kinetics is to determine a dosing regimen that keeps the drug MTC Therapeutic index concentration within the therapeutic index; that is, above the MEC but below the MTC. For example, most antibiotics such as amoxicillin have a fairly wide therapeutic index, but anti- coagulant (blood thinner) medications such as Coumadin® have a narrow therapeutic index. Determination of the correct dosage is based on the kinetics of the drug’s delivery as well MEC as the rate of disappearance due to decomposition, biotrans- formation, and excretion. Often this dosage will depend on the body weight of the person, because blood volume will be roughly proportional to body weight and the drug concentra- Initial Onset Next Next dose time dose dose tion will depend on the volume of distribution (blood in this Time case) as well as the amount of drug administered. Doctors and nurses refer to dosage charts in references such as the Drug concentration in blood as a function of time. The concentration of the drug increases after the drug is administered, and then decreases Physician’s Desk Reference (PDR), which are based on phar- as the drug is metabolized and excreted. This process is repeated when macokinetic studies. the next dose is administered, giving the plot its characteristic The onset time is the time required after the drug is “sawtooth” shape. administered for the concentration to reach the MEC and enter the therapeutic range. Sometimes a doctor will pre- scribe a higher first dose of the drug (loading dose) to re- long as the drug is needed; however, physiological adapta- duce the onset time. After that, the drug must be administered tions to the drug may require an adjustment in the regimen. at intervals to keep the concentration within the therapeutic For some drugs (for example, certain steroids), the drug dos- index, giving a characteristic “sawtooth” profile of drug age is tapered off rather than abruptly stopped in order to concentration versus time. The dosage is continued for as avoid shock to the system. The concentration of the ES intermediate is itself proportional to the amount of the All active sites substrate present, and a plot of the rate versus the concentration of substrate typically Rate of product formation are occupied yields a curve like that shown in Figure 13.30. Initially the rate rises rapidly with at and beyond increasing substrate concentration. However, above a certain concentration all the this substrate concentration active sites are occupied, and the reaction becomes zero order in the substrate. In other words, the rate remains the same even though the substrate concentration increases. At and beyond this point, the rate of formation of product depends only on how fast the ES intermediate breaks down, not on the number of substrate mole- cules present. [S] Figure 13.30 Plot of the rate of product formation versus substrate concentration in an Review of Concepts enzyme-catalyzed reaction. Which of the following is false regarding catalysis? (a) Ea is lower for a catalyzed reaction. (b) ¢H°rxn is lower for a catalyzed reaction. (c) A catalyzed reaction has a different mechanism. 607 608 Chapter 13 ■ Chemical Kinetics Key Equations rate 5 k[A]x[B]y (13.1) Rate law expressions. The sum (x 1 y) gives the overall order of the reaction. [A]t Relationship between concentration and time ln 5 2kt (13.3) for a first-order reaction. [A]0 ln [A]t 5 2kt 1 ln [A]0 (13.4) Equation for the graphical determination of k for a first-order reaction. 0.693 t12  (13.6) Half-life for a first-order reaction. k 1 1 Relationship between concentration and time 5 kt 1 (13.7) for a second-order reaction. [A]t [A]0 k 5 Ae2Ea/RT (13.11) The Arrhenius equation expressing the dependence of the rate constant on activation energy and temperature. Ea 1 Equation for the graphical determination of ln k 5 a2 b a b 1 ln A (13.13) activation energy. R T k1 Ea T1 2 T 2 Relationships of rate constants at two ln 5 a b (13.14) different temperatures. k2 R T 1T 2 Summary of Facts & Concepts 1. The rate of a chemical reaction is the change in the con- 5. In terms of collision theory, a reaction occurs when centration of reactants or products over time. The rate is molecules collide with sufficient energy, called the acti- not constant, but varies continuously as concentrations vation energy, to break the bonds and initiate the reac- change. tion. The rate constant and the activation energy are 2. The rate law expresses the relationship of the rate of a related by the Arrhenius equation. reaction to the rate constant and the concentrations of 6. The overall balanced equation for a reaction may be the the reactants raised to appropriate powers. The rate sum of a series of simple reactions, called elementary constant k for a given reaction changes only with steps. The complete series of elementary steps for a temperature. reaction is the reaction mechanism. 3. Reaction order is the power to which the concentration 7. If one step in a reaction mechanism is much slower than of a given reactant is raised in the rate law. Overall all other steps, it is the rate-determining step. reaction order is the sum of the powers to which reac- 8. A catalyst speeds up a reaction usually by lowering the tant concentrations are raised in the rate law. The rate value of Ea. A catalyst can be recovered unchanged at law and the reaction order cannot be determined from the end of a reaction. the stoichiometry of the overall equation for a reac- 9. In heterogeneous catalysis, which is of great industrial tion; they must be determined by experiment. For a importance, the catalyst is a solid and the reactants are zero-order reaction, the reaction rate is equal to the gases or liquids. In homogeneous catalysis, the catalyst rate constant. and the reactants are in the same phase. Enzymes are 4. The half-life of a reaction (the time it takes for the catalysts in living systems. concentration of a reactant to decrease by one-half) can be used to determine the rate constant of a first- order reaction.  Questions & Problems 609 Key Words Activated complex, p. 589 Enzyme, p. 604 Rate constant (k), p. 567 Second-order Activation energy (Ea), p. 589 First-order reaction, p. 575 Rate-determining step, p. 595 reaction, p. 582 Bimolecular reaction, p. 594 Half-life (t12), p. 580 Rate law, p. 571 Termolecular reaction, p. 594 Catalyst, p. 599 Intermediate, p. 594 Reaction mechanism, p. 594 Transition state, p. 589 Chemical kinetics, p. 563 Molecularity of a Reaction order, p. 571 Unimolecular Elementary step, p. 594 reaction, p. 594 Reaction rate, p. 563 reaction, p. 594 Questions & Problems • Problems available in Connect Plus 0.074 M/s. (a) At what rate is ammonia being formed? Red numbered problems solved in Student Solutions Manual (b) At what rate is molecular nitrogen reacting? The Rate of a Reaction The Rate Law Review Questions Review Questions 13.1 What is meant by the rate of a chemical reaction? 13.9 Explain what is meant by the rate law of a reaction. What are the units of the rate of a reaction? 13.10 What are the units for the rate constants of zero-order, 13.2 Distinguish between average rate and instantaneous first-order, and second-order reactions? rate. Which of the two rates gives us an unambigu- 13.11 Consider the zero-order reaction: A ¡ product. ous measurement of reaction rate? Why? (a) Write the rate law for the reaction. (b) What are 13.3 What are the advantages of measuring the initial rate the units for the rate constant? (c) Plot the rate of the of a reaction? reaction versus [A]. 13.4 Can you suggest two reactions that are very slow (take 13.12 On which of the following properties does the rate days or longer to complete) and two reactions that are constant of a reaction depend? (a) reactant concen- very fast (reactions that are over in minutes or seconds)? trations, (b) nature of reactants, (c) temperature. Problems Problems • 13.5 Write the reaction rate expressions for the following • 13.13 The rate law for the reaction reactions in terms of the disappearance of the reac- tants and the appearance of products: NH41 (aq) 1 NO2 2 (aq) ¡ N2 (g) 1 2H2O(l) (a) H2 (g) 1 I2 (g) ¡ 2HI(g) is given by rate 5 k[NH1 2 4 ][NO2 ]. At 258C, the rate (b) 5Br2 (aq) 1 BrO2 1 3 (aq) 1 6H (aq) ¡ 24 constant is 3.0 3 10 /M ? s. Calculate the rate of the 3Br2 (aq) 1 3H2O(l) reaction at this temperature if [NH14 ] 5 0.26 M and • 13.6 Write the reaction rate expressions for the following [NO2 2 ] 5 0.080 M. reactions in terms of the disappearance of the reac- 13.14 Use the data in Table 13.2 to calculate the rate of the tants and the appearance of products: reaction at the time when [F2] 5 0.010 M and [ClO2] (a) 2H2 (g) 1 O2 (g) ¡ 2H2O(g) 5 0.020 M. (b) 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g) • 13.15 Consider the reaction • 13.7 Consider the reaction A 1 B ¡ products 2NO(g) 1 O2 (g) ¡ 2NO2 (g) From the following data obtained at a certain tem- Suppose that at a particular moment during the perature, determine the order of the reaction and reaction nitric oxide (NO) is reacting at the rate of calculate the rate constant: 0.066 M/s. (a) At what rate is NO2 being formed? (b) At what rate is molecular oxygen reacting? [A] (M) [B] (M ) Rate (M/s) • 13.8 Consider the reaction 1.50 1.50 3.20  101 N2 (g) 1 3H2 (g) ¡ 2NH3 (g) 1.50 2.50 3.20  101 Suppose that at a particular moment during the reac- 3.00 1.50 6.40  101 tion molecular hydrogen is reacting at the rate of  610 Chapter 13 ■ Chemical Kinetics • 13.16 Consider the reaction Time (s) P (mmHg) X1Y ¡ Z 0 15.76 From the following data, obtained at 360 K, 181 18.88 (a) determine the order of the reaction, and 513 22.79 (b) determine the initial rate of disappearance of X 1164 27.08 when the concentration of X is 0.30 M and that of Y is 0.40 M. where P is the total pressure. Initial Rate of Disappearance of X (M/s) [X] (M) [Y] (M) The Relation Between Reactant 0.053 0.10 0.50 Concentration and Time 0.127 0.20 0.30 Review Questions 1.02 0.40 0.60 13.21 Write an equation relating the concentration of a re- 0.254 0.20 0.60 actant A at t 5 0 to that at t 5 t for a first-order re- 0.509 0.40 0.30 action. Define all the terms and give their units. Do the same for a second-order reaction. • 13.17 Determine the overall orders of the reactions to which 13.22 Define half-life. Write the equation relating the half- the following rate laws apply: (a) rate 5 k[NO2]2, life of a first-order reaction to the rate constant. 1 (b) rate 5 k, (c) rate 5 k[H2][Br 2]2, (d) rate 5 2 13.23 Write the equations relating the half-life of a second- k[NO] [O2]. order reaction to the rate constant. How does it differ • 13.18 Consider the reaction from the equation for a first-order reaction? A ¡ B 13.24 For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its The rate of the reaction is 1.6 3 1022 M/s when original value? Express your answer in terms of the the concentration of A is 0.35 M. Calculate the rate half-life (t12) and in terms of the rate constant k. constant if the reaction is (a) first order in A and (b) second order in A. Problems • 13.19 Cyclobutane decomposes to ethylene according to the equation • 13.25 What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in C4H8 (g) ¡ 2C2H4 (g) 60 min? Assume first-order kinetics. Determine the order of the reaction and the rate con- • 13.26 The thermal decomposition of phosphine (PH3) into stant based on the following pressures, which were phosphorus and molecular hydrogen is a first-order recorded when the reaction was carried out at 4308C reaction: in a constant-volume vessel. 4PH3 (g) ¡ P4 (g) 1 6H2 (g) The half-life of the reaction is 35.0 s at 6808C. Cal- Time (s) PC4H8 (mmHg) culate (a) the first-order rate constant for the reaction 0 400 and (b) the time required for 95 percent of the phos- phine to decompose. 2,000 316 4,000 248 • 13.27 The rate constant for the second-order reaction 6,000 196 2NOBr(g) ¡ 2NO(g) 1 Br2 (g) 8,000 155 is 0.80/M ? s at 108C. (a) Starting with a concentra- 10,000 122 tion of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when 13.20 The following gas-phase reaction was studied at [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M. 2908C by observing the change in pressure as a func- 13.28 The rate constant for the second-order reaction tion of time in a constant-volume vessel: 2NO2 (g) ¡ 2NO(g) 1 O2 (g) ClCO2CCl3 (g) ¡ 2COCl2 (g) is 0.54/M ? s at 3008C. How long (in seconds) would Determine the order of the reaction and the rate con- it take for the concentration of NO2 to decrease from stant based on the following data: 0.62 M to 0.28 M?  Questions & Problems 611 • 13.29 Consider the first-order reaction A ¡ B shown constant and T is the absolute temperature. Which re- here. (a) What is the rate constant of the reaction? action has a greater activation energy? (2) The diagram (b) How many A (yellow) and B (blue) molecules in (b) shows the plots for a first-order reaction at two are present at t 5 20 s and 30 s? different temperatures. Which plot corresponds to a higher temperature? ln [A]t ln k t0s t  10 s 13.30 The reaction X ¡ Y shown here follows first-order kinetics. Initially different amounts of X molecules are placed in three equal-volume containers at the 1/T t same temperature. (a) What are the relative rates of (a) (b) the reaction in these three containers? (b) How would the relative rates be affected if the volume of each container were doubled? (c) What are the relative • 13.38 Given the same reactant concentrations, the reaction half-lives of the reactions in (i) to (iii)? CO(g) 1 Cl2 (g) ¡ COCl2 (g) at 2508C is 1.50 3 103 times as fast as the same reaction at 1508C. Calculate the activation energy for this reac- tion. Assume that the frequency factor is constant. 13.39 Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants. An example is (i) (ii) (iii) k1 A ¡ B k2 Activation Energy and A ¡ C Review Questions The activation energies are 45.3 kJ/mol for k1 and 13.31 Define activation energy. What role does activation 69.8 kJ/mol for k2. If the rate constants are equal at energy play in chemical kinetics? 320 K, at what temperature will k1/k2 5 2.00? 13.32 Write the Arrhenius equation and define all terms. • 13.40 Variation of the rate constant with temperature for the first-order reaction 13.33 Use the Arrhenius equation to show why the rate constant of a reaction (a) decreases with increasing 2N2O5 (g) ¡ 2N2O4 (g) 1 O2 (g) activation energy and (b) increases with increasing is given in the following table. Determine graphi- temperature. cally the activation energy for the reaction. 13.34 The burning of methane in oxygen is a highly exo- thermic reaction. Yet a mixture of methane and oxy- gen gas can be kept indefinitely without any apparent T (K) k (s 1) change. Explain. 5 13.35 Sketch a potential energy versus reaction progress 298 1.74  10 5 plot for the following reactions: 308 6.61  10 4 (a) S(s) 1 O2 (g) ¡ SO2 (g) ¢H° 5 318 2.51  10 4 2296 kJ/mol 328 7.59  10 3 338 2.40  10 (b) Cl2 (g) ¡ Cl(g) 1 Cl(g) ¢H° 5 243 kJ/mol 13.36 The reaction H 1 H2 ¡ H2 1 H has been studied for many years. Sketch a potential energy versus • 13.41 For the reaction reaction progress diagram for this reaction. NO(g) 1 O3 (g) ¡ NO2 (g) 1 O2 (g) Problems the frequency factor A is 8.7 3 1012 s21 and the acti- 13.37 (1) The diagram in (a) shows the plots of ln k versus vation energy is 63 kJ/mol. What is the rate constant 1/T for two first-order reactions, where k is the rate for the reaction at 758C?  612 Chapter 13 ■ Chemical Kinetics • 13.42 The rate constant of a first-order reaction is • 13.49 Classify each of the following elementary steps as 4.60 3 1024 s21 at 3508C. If the activation energy unimolecular, bimolecular, or termolecular. is 104 kJ/mol, calculate the temperature at which its rate constant is 8.80 3 1024 s21.   8n  • 13.43 The rate constants of some reactions double with (a) every 10-degree rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to dou- 8n  ble as described? (b) 13.44 Consider the first-order reaction CH3NC(g) ¡ CH3CN(g)  8n  (c) Given that the frequency factor and activation energy for the reaction are 3.98 3 1013 s21 and 161 kJ/mol, 13.50 Reactions can be classified as unimolecular, bimo- respectively, calculate the rate constant at 6008C. lecular, and so on. Why are there no zero-molecular 13.45 Consider the second-order reaction reactions? Explain why termolecular reactions are NO(g) 1 Cl2(g) ¡ NOCl(g) 1 Cl(g) rare. Given that the frequency factor and activation • 13.51 Determine the molecularity and write the rate law for each of the following elementary steps: energy for the reaction are 4.0 3 109/M ? s and (a) X ¡ products 85 kJ/mol, respectively, calculate the rate constant at 5008C. (b) X 1 Y ¡ products (c) X 1 Y 1 Z ¡ products • 13.46 The rate at which tree crickets chirp is 2.0 3 102 per minute at 278C but only 39.6 per minute at 58C. (d) X 1 X ¡ products From these data, calculate the “activation energy” (e) X 1 2Y ¡ products for the chirping process. (Hint: The ratio of rates is 13.52 What is the rate-determining step of a reaction? equal to the ratio of rate constants.) Give an everyday analogy to illustrate the meaning • 13.47 The diagram here describes the initial state of the of “rate determining.” reaction A2 1 B2 ¡ 2AB. 13.53 The equation for the combustion of ethane (C2H6) is 2C2H6 (g) 1 7O2 (g) ¡ 4CO2 (g) 1 6H2O(l) A2 Explain why it is unlikely that this equation also B2 represents the elementary step for the reaction. AB 13.54 Specify which of the following species cannot be isolated in a reaction: activated complex, product, intermediate. Suppose the reaction is carried out at two temperatures as shown below. Which picture represents the result at Problems the higher temperature? (The reaction proceeds for the • 13.55 The rate law for the reaction same amount of time at both temperatures.) 2NO(g) 1 Cl2 (g) ¡ 2NOCl(g) is given by rate 5 k[NO][Cl2]. (a) What is the order of the reaction? (b) A mechanism involving the fol- lowing steps has been proposed for the reaction: NO(g) 1 Cl2 (g) ¡ NOCl2 (g) NOCl2 (g) 1 NO(g) ¡ 2NOCl(g) (a) (b) If this mechanism is correct, what does it imply about the relative rates of these two steps? Reaction Mechanisms 13.56 For the reaction X2 1 Y 1 Z ¡ XY 1 XZ it is found that doubling the concentration of X2 Review Questions doubles the reaction rate, tripling the concentra- 13.48 What do we mean by the mechanism of a reaction? tion of Y triples the rate, and doubling the concen- What is an elementary step? What is the molecular- tration of Z has no effect. (a) What is the rate law ity of a reaction? for this reaction? (b) Why is it that the change in  Questions & Problems 613 the concentration of Z has no effect on the rate? Problems (c) Suggest a mechanism for the reaction that is consistent with the rate law. 13.65 The diagram shown here represents a two-step mechanism. (a) Write the equation for each step • 13.57 The rate law for the decomposition of ozone to and the overall reaction. (b) Identify the intermedi- molecular oxygen ate and catalyst. The color codes are A 5 green and 2O3 (g) ¡ 3O2 (g) B 5 red. is [O3]2 8n 8n rate 5 k [O2] The mechanism proposed for this process is k1 • 13.66 Consider the following mechanism for the enzyme- O3 Δ k 1 O  O2 catalyzed reaction: k k1 O  O3 ¡2 2O2 E1SΔ ES (fast equilibrium) k 1 Derive the rate law from these elementary steps. k2 ES ¡ E 1 P (slow) Clearly state the assumptions you use in the deriva- tion. Explain why the rate decreases with increasing Derive an expression for the rate law of the reaction O2 concentration. in terms of the concentrations of E and S. (Hint: To 13.58 The rate law for the reaction solve for [ES], make use of the fact that, at equilib- rium, the rate of forward reaction is equal to the rate 2H2 (g) 1 2NO(g) ¡ N2 (g) 1 2H2O(g) of the reverse reaction.) is rate 5 k[H2][NO]2. Which of the following mech- anisms can be ruled out on the basis of the observed Additional Problems rate expression? • 13.67 The following diagrams represent the progress of the Mechanism I reaction A ¡ B, where the red spheres represent A H2 1 NO 88n H2O 1 N (slow) molecules and the green spheres represent B mole- N 1 NO 88n N2 1 O (fast) cules. Calculate the rate constant of the reaction. O 1 H2 88n H2O (fast) Mechanism II H2 1 2NO 88n N2O 1 H2O (slow) N2O 1 H2 88n N2 1 H2O (fast) Mechanism III 2NO 34 N2O2 (fast equilibrium) t0s t  20 s t  40 s N2O2 1 H2 88n N2O 1 H2O (slow) N2O 1 H2 88n N2 1 H2O (fast) • 13.68 The following diagrams show the progress of the reaction 2A ¡ A2. Determine whether the reac- Catalysis tion is first order or second order and calculate the Review Questions rate constant. 13.59 How does a catalyst increase the rate of a reaction? 13.60 What are the characteristics of a catalyst? 13.61 A certain reaction is known to proceed slowly at room temperature. Is it possible to make the reac- tion proceed at a faster rate without changing the temperature? t  0 min t  15 min t  30 min 13.62 Distinguish between homogeneous catalysis and het- erogeneous catalysis. Describe three important indus- • 13.69 Suggest experimental means by which the rates of trial processes that utilize heterogeneous catalysis. the following reactions could be followed: 13.63 Are enzyme-catalyzed reactions examples of homo- (a) CaCO3 (s) ¡ CaO(s) 1 CO2 (g) geneous or heterogeneous catalysis? Explain. (b) Cl2 (g) 1 2Br2 (aq) ¡ Br2 (aq) 1 2Cl2 (aq) 13.64 The concentrations of enzymes in cells are usually (c) C2H6 (g) ¡ C2H4 (g) 1 H2 (g) quite small. What is the biological significance of (d) C2H5I(g) 1 H2O(l) ¡ this fact? C2H5OH(aq) 1 H1 (aq) 1 I2 (aq)  614 Chapter 13 ■ Chemical Kinetics 13.70 List four factors that influence the rate of a reaction. 13.71 “The rate constant for the reaction Concentration NO2 (g) 1 CO(g) ¡ NO(g) 1 CO2 (g) is 1.64 3 1026/M ? s.” What is incomplete about this statement? • 13.72 In a certain industrial process involving a hetero- geneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm3. Calculate the sur- face area of the catalyst. If the sphere is broken t (s) down into eight spheres, each having a volume of 1.25 cm3, what is the total surface area of the • 13.78 The reaction 2A 1 3B ¡ C is first order with spheres? Which of the two geometric configura- respect to A and B. When the initial concentrations tions of the catalyst is more effective? (The surface are [A] 5 1.6 3 1022 M and [B] 5 2.4 3 1023 M, area of a sphere is 4πr2, where r is the radius the rate is 4.1 3 1024 M/s. Calculate the rate con- of the sphere.) Based on your analysis here, explain stant of the reaction. why it is sometimes dangerous to work in grain • 13.79 The bromination of acetone is acid-catalyzed: elevators. H CH3COCH3 Br2 8 8S catalyst CH3COCH2Br H Br 13.73 Use the data in Example 13.5 to determine graphi- cally the half-life of the reaction. The rate of disappearance of bromine was measured • 13.74 The following data were collected for the reaction for several different concentrations of acetone, bro- between hydrogen and nitric oxide at 7008C: mine, and H1 ions at a certain temperature: 2H2 (g) 1 2NO(g) ¡ 2H2O(g) 1 N2 (g) Rate of Disappearance Experiment [H2] [NO] Initial Rate (M/s) [CH3COCH3] [Br2] [H ] of Br2 (M/s) 5 1 0.010 0.025 2.4  10 6 (1) 0.30 0.050 0.050 5.7  10 5 2 0.0050 0.025 1.2  10 6 (2) 0.30 0.10 0.050 5.7  10 4 3 0.010 0.0125 0.60  10 6 (3) 0.30 0.050 0.10 1.2  10 4 (4) 0.40 0.050 0.20 3.1  10 5 (5) 0.40 0.050 0.050 7.6  10 (a) Determine the order of the reaction. (b) Calcu- late the rate constant. (c) Suggest a plausible mech- (a) What is the rate law for the reaction? (b) Deter- anism that is consistent with the rate law. (Hint: mine the rate constant. (c) The following mechanism Assume that the oxygen atom is the intermediate.) has been proposed for the reaction: 13.75 When methyl phosphate is heated in acid solution, it reacts with water:  O OH B B CH3OPO3H2 1 H2O ¡ CH3OH 1 H3PO4 CH3OCOCH3  H3O 34 CH3OCOCH3  H2O If the reaction is carried out in water enriched (fast equilibrium) with 18O, the oxygen-18 isotope is found in the OH  OH phosphoric  acid product but not in the methanol. B A What does this tell us about the mechanism of the CH3OCOCH3  H2O 88n CH3OCPCH2  H3O (slow) reaction? OH O 13.76 The rate of the reaction A B CH3OCPCH2  Br2 88n CH3OCOCH2Br  HBr (fast) CH3COOC2H5 (aq) 1 H2O(l2 ¡ CH3COOH(aq) 1 C2H5OH(aq) Show that the rate law deduced from the mechanism shows first-order characteristics—that is, rate 5 is consistent with that shown in (a). k[CH3COOC2H5]—even though this is a second-order • 13.80 The decomposition of N2O to N2 and O2 is a first- reaction (first order in CH3COOC2H5 and first order order reaction. At 7308C the half-life of the reaction in H2O). Explain. is 3.58 3 103 min. If the initial pressure of N2O is 13.77 Which of the following equations best describes the 2.10 atm at 7308C, calculate the total gas pressure diagram shown above: (a) A ¡ B, (b) A ¡ 3B, after one half-life. Assume that the volume remains (c) 3A ¡ B? constant.  Questions & Problems 615 13.81 The reaction S2O22 8 1 2I 2 ¡ 2SO22 4 1 I2 pro- ceeds slowly in aqueous solution, but it can be cata- lyzed by the Fe31 ion. Given that Fe31 can oxidize III I2 and Fe21 can reduce S2O 822, write a plausible two-step mechanism for this reaction. Explain why the uncatalyzed reaction is slow. 13.82 What are the units of the rate constant for a third- t50s t 5 50 s order reaction? 13.83 The integrated rate law for the zero-order reaction 13.87 Referring to Example 13.5, explain how you would A ¡ B is [A]t 5 [A]0 2 kt. (a) Sketch the following measure the partial pressure of azomethane experi- plots: (i) rate versus [A]t and (ii) [A]t versus t. mentally as a function of time. (b) Derive an expression for the half-life of the reaction. 13.88 The rate law for the reaction 2NO2 (g) ¡ N2O4(g) (c) Calculate the time in half-lives when the integrated is rate 5 k[NO2]2. Which of the following changes rate law is no longer valid, that is, when [A]t 5 0. will change the value of k? (a) The pressure of NO2 • 13.84 A flask contains a mixture of compounds A and B. is doubled. (b) The reaction is run in an organic Both compounds decompose by first-order kinetics. solvent. (c) The volume of the container is doubled. The half-lives are 50.0 min for A and 18.0 min for B. (d) The temperature is decreased. (e) A catalyst is If the concentrations of A and B are equal initially, added to the container. how long will it take for the concentration of A to be 13.89 The reaction of G2 with E2 to form 2EG is exother- four times that of B? mic, and the reaction of G2 with X2 to form 2XG is 13.85 Shown here are plots of concentration of reactant endothermic. The activation energy of the exother- versus time for two first-order reactions at the same mic reaction is greater than that of the endothermic temperature. In each case, determine which reaction reaction. Sketch the potential energy profile dia- has a greater rate constant. grams for these two reactions on the same graph. 13.90 In the nuclear industry, workers use a rule of thumb that the radioactivity from any sample will be rela- tively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after ln [A]t [A]t this time period. (Hint: Radioactive decays obey first-order kinetics.) 13.91 Briefly comment on the effect of a catalyst on each of the following: (a) activation energy, (b) reaction t t mechanism, (c) enthalpy of reaction, (d) rate of for- (a) (b) ward step, (e) rate of reverse step. 13.86 The diagrams here represent the reaction • 13.92 When 6 g of granulated Zn is added to a solution of A 1 B ¡ C carried out under different initial 2 M HCl in a beaker at room temperature, hydrogen concentrations of A and B. Determine the rate law gas is generated. For each of the following changes of the reaction. (The color codes are A 5 red, (at constant volume of the acid) state whether the B 5 green, C 5 blue.) rate of hydrogen gas evolution will be increased, de- creased, or unchanged: (a) 6 g of powdered Zn is used; (b) 4 g of granulated Zn is used; (c) 2 M acetic acid is used instead of 2 M HCl; (d) temperature is raised to 408C. I 13.93 Strictly speaking, the rate law derived for the reaction in Problem 13.74 applies only to certain concentrations of H2. The general rate law for the reaction takes the form t50s t 5 50 s k1[NO]2[H2] rate 5 1 1 k2[H2] where k1 and k2 are constants. Derive rate law ex- II pressions under the conditions of very high and very low hydrogen concentrations. Does the result from Problem 13.74 agree with one of the rate expres- t50s t 5 50 s sions here?  616 Chapter 13 ■ Chemical Kinetics • 13.94 A certain first-order reaction is 35.5 percent com- The chlorine radical then reacts with ozone as follows: plete in 4.90 min at 258C. What is its rate constant? Cl 1 O3 ¡ ClO 1 O2 13.95 The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent (CCl4) ClO 1 O ¡ Cl 1 O2 at a certain temperature: The O atom is from the photochemical decomposi- 2N2O5 ¡ 4NO2 1 O2 tion of O2 molecules. (a) Write the overall reaction for the last two steps. (b) What are the roles of Cl and ClO? (c) Why is the [N2O5] Initial Rate (M/s) fluorine radical not important in this mechanism? 0.92 0.95 ⫻ 10 5 (d) One suggestion to reduce the concentration of 1.23 1.20 ⫻ 10 5 chlorine radicals is to add hydrocarbons such as ethane (C2H6) to the stratosphere. How will this 1.79 1.93 ⫻ 10 5 work? (e) Draw potential energy versus reaction 2.00 2.10 ⫻ 10 5 progress diagrams for the uncatalyzed and cata- 2.21 2.26 ⫻ 10 5 lyzed (by Cl) destruction of ozone: O3 1 O ¡ 2O2. Use the thermodynamic data in Appendix 3 to de- Determine graphically the rate law for the reaction termine whether the reaction is exothermic or and calculate the rate constant. endothermic. • 13.96 The thermal decomposition of N2O5 obeys first-order • 13.102 Chlorine oxide (ClO), which plays an important role kinetics. At 458C, a plot of ln [N2O5] versus t gives a in the depletion of ozone (see Problem 13.101), slope of 26.18 3 1024 min21. What is the half-life decays rapidly at room temperature according to the of the reaction? equation 13.97 When a mixture of methane and bromine is exposed 2ClO(g) ¡ Cl2 (g) 1 O2 (g) to visible light, the following reaction occurs slowly: From the following data, determine the reaction CH4 (g) 1 Br2 (g) ¡ CH3Br(g) 1 HBr(g) order and calculate the rate constant of the reaction Suggest a reasonable mechanism for this reac- tion. (Hint: Bromine vapor is deep red; methane is Time (s) [ClO] (M) colorless.) 0.12 ⫻ 10 3 8.49 ⫻ 10 6 13.98 The rate of the reaction between H2 and I2 to form 0.96 ⫻ 10 3 7.10 ⫻ 10 6 HI (discussed on p. 596) increases with the inten- sity of visible light. (a) Explain why this fact sup- 2.24 ⫻ 10 3 5.79 ⫻ 10 6 ports the two-step mechanism given. (The color 3.20 ⫻ 10 3 5.20 ⫻ 10 6 of I2 vapor is shown on p. 502.) (b) Explain why 4.00 ⫻ 10 3 4.77 ⫻ 10 6 the visible light has no effect on the formation of H atoms. • 13.103 A compound X undergoes two simultaneous first- order reactions as follows: X ¡ Y with rate con- • 13.99 The carbon-14 decay rate of a sample obtained stant k1 and X ¡ Z with rate constant k2. The ratio from a young tree is 0.260 disintegration per second per gram of the sample. Another wood sample pre- of k1/k2 at 408C is 8.0. What is the ratio at 3008C? pared from an object recovered at an archaeological Assume that the frequency factors of the two reac- excavation gives a decay rate of 0.186 disintegration tions are the same. per second per gram of the sample. What is the age 13.104 Consider a car fitted with a catalytic converter. The of the object? (Hint: See the Chemistry in Action first 5 minutes or so after it is started are the most essay on p. 586.) polluting. Why? • 13.100 Consider the following elementary step: 13.105 The following scheme in which A is converted to B, which is then converted to C is known as a consecu- X 1 2Y ¡ XY2 tive reaction. (a) Write a rate law for this reaction. (b) If the initial A ¡ B ¡ C rate of formation of XY2 is 3.8 3 1023 M/s and the Assuming that both steps are first order, sketch on initial concentrations of X and Y are 0.26 M and the same graph the variations of [A], [B], and [C] 0.88 M, what is the rate constant of the reaction? with time. 13.101 In recent years ozone in the stratosphere has been 13.106 Hydrogen and iodine monochloride react as follows: depleted at an alarmingly fast rate by chlorofluoro- carbons (CFCs). A CFC molecule such as CFCl3 is H2 (g) 1 2ICl(g) ¡ 2HCl(g) 1 I2 (g) first decomposed by UV radiation: The rate law for the reaction is rate 5 k[H2][ICl]. CFCl3 ¡ CFCl2 1 Cl Suggest a possible mechanism for the reaction.  Questions & Problems 617 13.107 The rate law for the following reaction 13.112 The first-order rate constant for the decomposition CO(g) 1 NO2 (g) ¡ CO2 (g) 1 NO(g) of dimethyl ether is rate 5 k[NO2]2. Suggest a plausible mechanism (CH3 ) 2O(g) ¡ CH4 (g) 1 H2 (g) 1 CO(g) for the reaction, given that the unstable species NO3 is 3.2 3 1024 s21 at 4508C. The reaction is carried is an intermediate. out in a constant-volume flask. Initially only dimethyl • 13.108 Radioactive plutonium-239 1t12 5 2.44 3 105 yr2 is ether is present and the pressure is 0.350 atm. What is used in nuclear reactors and atomic bombs. If there the pressure of the system after 8.0 min? Assume are 5.0 3 102 g of the isotope in a small atomic ideal behavior. bomb, how long will it take for the substance to 13.113 At 258C, the rate constant for the ozone-depleting decay to 1.0 3 102  g, too small an amount for an reaction O(g) 1 O3(g) ¡ 2O2(g) is 7.9 3 10215 effective bomb? cm3/molecule ? s. Express the rate constant in units 13.109 Many reactions involving heterogeneous catalysts are of 1/M ? s. zero order; that is, rate 5 k. An example is the decom- 13.114 Consider the following elementary steps for a con- position of phosphine (PH3) over tungsten (W): secutive reaction: 4PH3 (g) ¡ P4 (g) 1 6H2 (g) k1 k2 A ¡ B ¡ C It is found that the reaction is independent of [PH3] (a) Write an expression for the rate of change of B. as long as phosphine’s pressure is sufficiently high (b) Derive an expression for the concentration of B un- ($ 1 atm). Explain. der steady-state conditions; that is, when B is decom- • 13.110 Thallium(I) is oxidized by cerium(IV) as follows: posing to C at the same rate as it is formed from A. Tl1 1 2Ce41 ¡ Tl31 1 2Ce31 13.115 Ethanol is a toxic substance that, when consumed in excess, can impair respiratory and cardiac functions The elementary steps, in the presence of Mn(II), are by interference with the neurotransmitters of the as follows: nervous system. In the human body, ethanol is me- Ce41 1 Mn21 ¡ Ce31 1 Mn31 tabolized by the enzyme alcohol dehydrogenase to Ce41 1 Mn31 ¡ Ce31 1 Mn41 acetaldehyde, which causes “hangovers.” (a) Based Tl1 1 Mn41 ¡ Tl31 1 Mn21 on your knowledge of enzyme kinetics, explain why binge drinking (that is, consuming too much (a) Identify the catalyst, intermediates, and the alcohol too fast) can prove fatal. (b) Methanol is rate-determining step if the rate law is rate 5 even more toxic than ethanol. It is also metabolized k[Ce41][Mn21]. (b) Explain why the reaction is slow by alcohol dehydrogenase, and the product, formal- without the catalyst. (c) Classify the type of catalysis dehyde, can cause blindness or death. An antidote (homogeneous or heterogeneous). to methanol poisoning is ethanol. Explain how this • 13.111 Sucrose (C12H22O11), commonly called table sugar, procedure works. undergoes hydrolysis (reaction with water) to pro- duce fructose (C6H12O6) and glucose (C6H12O6): • 13.116 Strontium-90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half- C12H22O11 1 H2O ¡ C6H12O6 1 C6H12O6 life of 28.1 yr. (a) Calculate the first-order rate con- fructose glucose stant for the nuclear decay. (b) Calculate the fraction This reaction is of considerable importance in the of 90Sr that remains after 10 half-lives. (c) Calculate candy industry. First, fructose is sweeter than su- the number of years required for 99.0 percent of 90 crose. Second, a mixture of fructose and glucose, Sr to disappear. called invert sugar, does not crystallize, so the • 13.117 Consider the potential energy profiles for the fol- candy containing this sugar would be chewy rather lowing three reactions (from left to right). (1) Rank than brittle as candy containing sucrose crystals the rates (slowest to fastest) of the reactions. would be. (a) From the following data determine (2) Calculate DH for each reaction and determine the order of the reaction. (b) How long does it take which reaction(s) are exothermic and which to hydrolyze 95 percent of sucrose? (c) Explain reaction(s) are endothermic. Assume the reactions why the rate law does not include [H2O] even have roughly the same frequency factors. though water is a reactant. 20 kJ/mol Potential energy 30 kJ/mol Time (min) [C12H22O11] 0 0.500 40 kJ/mol 60.0 0.400 50 kJ/mol 40 kJ/mol 20 kJ/mol 96.4 0.350 Reaction progress Reaction progress Reaction progress 157.5 0.280 (a) (b) (c)  618 Chapter 13 ■ Chemical Kinetics • 13.118 Consider the following potential energy profile for 13.123 At a certain elevated temperature, ammonia decom- the A ¡ D reaction. (a) How many elementary poses on the surface of tungsten metal as follows: steps are there? (b) How many intermediates are 2NH3 ¡ N2 1 3H2 formed? (c) Which step is rate determining? (d) Is the overall reaction exothermic or endothermic? From the following plot of the rate of the reaction versus the pressure of NH3, describe the mechanism of the reaction. Potential energy A Rate B C D Reaction progress PNH3 13.119 A factory that specializes in the refinement of transi- 13.124 The following expression shows the dependence of tion metals such as titanium was on fire. The firefight- the half-life of a reaction ( t12 ) on the initial reactant ers were advised not to douse the fire with water. Why? concentration [A]0: • 13.120 The activation energy for the decomposition of hy- 1 drogen peroxide t12 r [A]0n21 2H2O2 (aq) ¡ 2H2O2 (l) 1 O2 (g) where n is the order of the reaction. Verify this depen- is 42 kJ/mol, whereas when the reaction is catalyzed dence for zero-, first-, and second-order reactions. by the enzyme catalase, it is 7.0 kJ/mol. Calculate 13.125 Polyethylene is used in many items, including the temperature that would cause the uncatalyzed water pipes, bottles, electrical insulation, toys, reaction to proceed as rapidly as the enzyme- and mailer envelopes. It is a polymer, a molecule catalyzed decomposition at 208C. Assume the with a very high molar mass made by joining frequency factor A to be the same in both cases. many ethylene molecules together. (Ethylene is • 13.121 The activity of a radioactive sample is the number of the basic unit, or monomer for polyethylene.) nuclear disintegrations per second, which is equal to The initiation step is the first-order rate constant times the number of k1 R2 ¡ 2R ?      initiation radioactive nuclei present. The fundamental unit of radioactivity is the curie (Ci), where 1 Ci corre- The R ? species (called a radical) reacts with an eth- sponds to exactly 3.70 3 1010 disintegrations per ylene molecule (M) to generate another radical second. This decay rate is equivalent to that of 1 g of R ? 1 M ¡ M1 ? radium-226. Calculate the rate constant and half-life for the radium decay. Starting with 1.0 g of the ra- Reaction of M1 ? with another monomer leads to the dium sample, what is the activity after 500 yr? The growth or propagation of the polymer chain: molar mass of Ra-226 is 226.03 g/mol. kp M1 ? 1 M ¡ M2 ?       propagation 13.122 To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin (HbO2) This step can be repeated with hundreds of monomer according to the simplified equation units. The propagation terminates when two radicals k combine Hb(aq) 1 O2 (aq) ¡ HbO2 (aq) kt M¿ ? 1 M– ? ¡ M¿¬M–      termination where the second-order rate constant is 2.1 3 106/M ? s at 378C. (The reaction is first order in Hb The initiator frequently used in the polymerization and O2.) For an average adult, the concentrations of ethylene is benzoyl peroxide [(C6H5COO)2]: of Hb and O2 in the blood at the lungs are 8.0 3 [(C6H5COO) 2] ¡ 2C6H5COO ? 1026 M and 1.5 3 1026 M, respectively. (a) Calcu- late the rate of formation of HbO2. (b) Calculate This is a first-order reaction. The half-life of benzoyl the rate of consumption of O2. (c) The rate of for- peroxide at 1008C is 19.8 min. (a) Calculate the rate mation of HbO2 increases to 1.4 3 1024 M/s dur- constant (in min21) of the reaction. (b) If the half- ing exercise to meet the demand of increased life of benzoyl peroxide is 7.30 h, or 438 min, at metabolism rate. Assuming the Hb concentration 708C, what is the activation energy (in kJ/mol) for to remain the same, what must be the oxygen con- the decomposition of benzoyl peroxide? (c) Write centration to sustain this rate of HbO2 formation? the rate laws for the elementary steps in the above  Questions & Problems 619 polymerization process, and identify the reactant, Which of the following mechanisms is consistent product, and intermediates. (d) What condition with the experimental data? would favor the growth of long, high-molar-mass (a) A ¡ B, A ¡ C polyethylenes? (b) A ¡ B 1 C • 13.126 The rate constant for the gaseous reaction (c) A ¡ B, B ¡ C 1 D H2 (g) 1 I2 (g) ¡ 2HI(g) (d) A ¡ B, B ¡ C is 2.42 3 1022/M ? s at 4008C. Initially an equimolar sample of H2 and I2 is placed in a vessel at 4008C and the total pressure is 1658 mmHg. (a) What is the ␭1 Light absorption initial rate (M/min) of formation of HI? (b) What are the rate of formation of HI and the concentration of HI (in molarity) after 10.0 min? ␭2 13.127 A protein molecule, P, of molar mass m dimerizes when it is allowed to stand in solution at room ␭3 temperature. A plausible mechanism is that the protein molecule is first denatured (that is, loses Time its activity due to a change in overall structure) before it dimerizes: • 13.132 A gas mixture containing CH3 fragments, C2H6 mol- k ecules, and an inert gas (He) was prepared at 600 K P ¡ P*(denatured)      slow with a total pressure of 5.42 atm. The elementary 2P* ¡ P2 fast reaction where the asterisk denotes a denatured protein CH3 1 C2H6 ¡ CH4 1 C2H5 molecule. Derive an expression for the average molar mass (of P and P2), m, in terms of the initial has a second-order rate constant of 3.0 3 104/M ? s. protein concentration [P]0 and the concentration at Given that the mole fractions of CH3 and C2H6 are time t, [P]t, and m. Describe how you would deter- 0.00093 and 0.00077, respectively, calculate the ini- mine k from molar mass measurements. tial rate of the reaction at this temperature. • 13.128 When the concentration of A in the reaction 13.133 To prevent brain damage, a drastic medical proce- A ¡ B was changed from 1.20 M to 0.60 M, the dure is to lower the body temperature of someone half-life increased from 2.0 min to 4.0 min at 258C. who has suffered cardiac arrest. What is the physio- Calculate the order of the reaction and the rate con- chemical basis for this treatment? stant. (Hint: Use the equation in Problem 13.124.) • 13.134 The activation energy (Ea) for the reaction 13.129 At a certain elevated temperature, ammonia decom- poses on the surface of tungsten metal as follows: 2N2O(g) ¡ 2N2 (g) 1 O2 (g) ¢H° 5 2164 kJ/mol NH3 ¡ 12 N2 1 32 H2 is 240 kJ/mol. What is Ea for the reverse reaction? 13.135 The rate constants for the first-order decomposition The kinetic data are expressed as the variation of the of an organic compound in solution are measured at half-life with the initial pressure of NH3: several temperatures: P (mmHg) 264 130 59 16 k (s21) 0.00492 0.0216 0.0950 0.326 1.15 t12 (s) 456 228 102 60 T (K) 278 288 298 308 318 (a) Determine the order of the reaction. (b) How Determine graphically the activation energy and does the order depend on the initial pressure? frequency factor for the reaction. (c) How does the mechanism of the reaction vary 13.136 Assume that the formation of nitrogen dioxide: with pressure? (Hint: You need to use the equation in Problem 13.124 and plot log t12 versus log P.) 2NO(g) 1 O2 (g) ¡ 2NO2 (g) 13.130 The activation energy for the reaction is an elementary reaction. (a) Write the rate law for N2O(g) ¡ N2 (g) 1 O(g) this reaction. (b) A sample of air at a certain tempera- ture is contaminated with 2.0 ppm of NO by volume. is 2.4 3 102 kJ/mol at 600 K. Calculate the percent- Under these conditions, can the rate law be simpli- age of the increase in rate from 600 K to 606 K. fied? If so, write the simplified rate law. (c) Under the Comment on your results. condition described in (b), the half-life of the reac- • 13.131 The rate of a reaction was followed by the absorp- tion has been estimated to be 6.4 3 103 min. What tion of light by the reactants and products as a func- would be the half-life if the initial concentration of tion of wavelengths (l1, l2, l3) as time progresses. NO were 10 ppm? 620 Chapter 13 ■ Chemical Kinetics Interpreting, Modeling & Estimating 13.137 An instructor performed a lecture demonstration of 13.141 What are the shortest and longest times (in years) the thermite reaction (see p. 258). He mixed alumi- that can be estimated by carbon-14 dating? num with iron(III) oxide in a metal bucket placed on 13.142 In addition to chemical and biological systems, a block of ice. After the extremely exothermic reac- kinetic treatments can sometimes be applied to be- tion started, there was an enormous bang, which was havioral and social processes such as the evolution not characteristic of thermite reactions. Give a plau- of technology. For example, in 1965, Gordon Moore, sible chemical explanation for the unexpected sound a co-founder of Intel, described a trend that the num- effect. The bucket was open to air. ber of transistors on an integrated circuit (N) roughly 13.138 Account for the variation of the rate of an enzyme- doubles every 1.5 yr. Now referred to as Moore’s catalyzed reaction versus temperature shown here. law, this trend has persisted for the past several de- What is the approximate temperature that corre- cades. A plot of ln N versus year is shown here. (a) sponds to the maximum rate in the human body? Determine the rate constant for the growth in the number of transistors on an integrated circuit. (b) Based on the rate constant, how long does it take for N to double? (c) If Moore’s law continues until the end of the century, what will be the number of Rate transistors on an integrated circuit in the year 2100? Comment on your result. 25 20 T ln Nt 15 13.139 Is the rate constant (k) of a reaction more sensitive to 10 changes in temperature if Ea is small or large? 5 13.140 Shown here is a plot of [A]t versus t for the reaction 0 A ¡ product. (a) Determine the order and the 1950 1960 1970 1980 1990 2000 2010 rate constant of the reaction. (b) Estimate the initial t (yr) rate and the rate at 30 s. 1.0 0.8 [A]t (M) 0.6 0.4 0.2 0.0 0 10 20 30 40 50 60 t (s) Answers to Practice Exercises 13.1 13.5 First order. 1.4 3 1022 min21. 13.6 1.2 3 103 s. ¢[CH4] 1 ¢[O2] ¢[CO2] 1 ¢[H2O] 13.7 (a) 3.2 min. (b) 2.1 min. 13.8 240 kJ/mol. rate 5 2 52 5 5 ¢t 2 ¢t ¢t 2 ¢t 13.9 3.13 3 1029 s21. 13.10 (a) NO2 1 CO ¡ NO 1 13.2 (a) 0.013 M/s. (b) 20.052 M/s. 13.3 rate 5 k[S2O822][I2]; CO2. (b) NO3. (c) The first step is rate-determining. k 5 8.1 3 1022/M ? s. 13.4 66 s.  CHAPTER 14 Chemical Equilibrium Chemical equilibrium is an example of dynamic equilibrium, much like what the juggler is trying to establish here. CHAPTER OUTLINE A LOOK AHEAD 14.1 The Concept of Equilibrium  We begin by discussing the nature of equilibrium and the difference and the Equilibrium Constant between chemical and physical equilibrium. We define the equilibrium constant in terms of the law of mass action. (14.1) 14.2 Writing Equilibrium Constant Expressions  We then learn to write the equilibrium constant expression for homoge- neous and heterogeneous equilibria. We see how to express equilibrium 14.3 The Relationship Between constants for multiple equilibria. (14.2) Chemical Kinetics and  Next, we examine the relationship between the rate constants and equilib- Chemical Equilibrium rium constant of a reaction. This exercise shows why the equilibrium con- 14.4 What Does the Equilibrium stant is a constant and why it varies with temperature. (14.3) Constant Tell Us?  We see that knowing the equilibrium constant enables us to predict the 14.5 Factors That Affect direction of a net reaction toward equilibrium and to calculate equilibrium concentrations. (14.4) Chemical Equilibrium  The chapter concludes with a discussion of the four factors that can pos- sibly affect the position of an equilibrium: concentration, volume or pres- sure, temperature, and catalyst. We learn to use Le Châtelier’s principle to predict the changes. (14.5) 621 622 Chapter 14 ■ Chemical Equilibrium E quilibrium is a state in which there are no observable changes as time goes by. When a chemical reaction has reached the equilibrium state, the concentrations of reactants and products remain constant over time, and there are no visible changes in the system. However, there is much activity at the molecular level because reactant molecules continue to form prod- uct molecules while product molecules react to yield reactant molecules. This dynamic equi- librium situation is the subject of this chapter. Here we will discuss different types of equilibrium reactions, the meaning of the equilibrium constant and its relationship to the rate constant, and factors that can disrupt a system at equilibrium. 14.1 The Concept of Equilibrium and the Equilibrium Constant Few chemical reactions proceed in only one direction. Most are reversible, at least to some extent. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from prod- Animation uct molecules. Chemical equilibrium is achieved when the rates of the forward and Chemical Equilibrium reverse reactions are equal and the concentrations of the reactants and products remain constant. Chemical equilibrium is a dynamic process. As such, it can be likened to the movement of skiers at a busy ski resort, where the number of skiers carried up the mountain on the chair lift is equal to the number coming down the slopes. Although there is a constant transfer of skiers, the number of people at the top and the number at the bottom of the slope do not change. Note that chemical equilibrium involves different substances as reactants and products. Equilibrium between two phases of the same substance is called physical equilibrium because the changes that occur are physical processes. The vaporization of water in a closed container at a given temperature is an example of physical equi- h6 6g librium. In this instance, the number of H2O molecules leaving and the number return- ing to the liquid phase are equal: H2O(l) Δ H2O(g) (Recall from Chapter 4 that the double arrow means that the reaction is reversible.) Liquid water in equilibrium with its The study of physical equilibrium yields useful information, such as the equi- vapor in a closed system at room librium vapor pressure (see Section 11.8). However, chemists are particularly inter- temperature. ested in chemical equilibrium processes, such as the reversible reaction involving nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4) (Figure 14.1). The progress of the reaction N2O4(g) Δ 2NO2(g) can be monitored easily because N2O4 is a colorless gas, whereas NO2 has a dark- brown color that makes it sometimes visible in polluted air. Suppose that N2O4 is injected into an evacuated flask. Some brown color appears immediately, indicating NO2 and N2O4 gases at equilibrium. the formation of NO2 molecules. The color intensifies as the dissociation of N2O4 continues until eventually equilibrium is reached. Beyond that point, no further change in color is evident because the concentrations of both N2O4 and NO2 remain constant. We can also bring about an equilibrium state by starting with pure NO2. As some of the NO2 molecules combine to form N2O4, the color fades. Yet another way to create an equilibrium state is to start with a mixture of NO2 and N2O4 and monitor the system until the color stops changing. These studies demonstrate that the preceding reaction is indeed reversible, because a pure component (N2O4 or  14.1 The Concept of Equilibrium and the Equilibrium Constant 623 Figure 14.1 A reversible reaction between N2O4 and NO2 molecules. NO2) reacts to give the other gas. The important thing to keep in mind is that at equilibrium, the conversions of N2O4 to NO2 and of NO2 to N2O4 are still going on. We do not see a color change because the two rates are equal—the removal of NO2 molecules takes place as fast as the production of NO2 molecules, and N2O4 molecules are formed as quickly as they dissociate. Figure 14.2 summarizes these three situations. The Equilibrium Constant Table 14.1 shows some experimental data for the reaction just described at 25°C. The gas concentrations are expressed in molarity, which can be calculated from the num- ber of moles of gases present initially and at equilibrium and the volume of the flask in liters. Note that the equilibrium concentrations of NO2 and N2O4 vary, depending on the starting concentrations. We can look for relationships between [NO2] and [N2O4] present at equilibrium by comparing the ratio of their concentrations. The simplest ratio, that is, [NO2]/[N2O4], gives scattered values. But if we examine other possible mathematical relationships, we find that the ratio [NO2]2/[N2O4] at equilibrium gives a nearly constant value that averages 4.63 3 1023, regardless of the initial concentrations present: [NO2]2 K5 5 4.63 3 1023 (14.1) [N2O4] N2O4 N2O4 Concentration Concentration Concentration N2O4 NO2 NO2 NO2 Time Time Time (a) (b) (c) Figure 14.2 Change in the concentrations of NO2 and N2O4 with time, in three situations. (a) Initially only NO2 is present. (b) Initially only N2O4 is present. (c) Initially a mixture of NO2 and N2O4 is present. In each case, equilibrium is established to the right of the vertical line. 624 Chapter 14 ■ Chemical Equilibrium Table 14.1 The NO2–N2O4 System at 258C Initial Equilibrium Ratio of Concentrations Concentrations Concentrations (M) (M) at Equilibrium [NO2 ] [NO2 ] 2 [NO2] [N2O4] [NO2] [N2O4] [N2O4 ] [N2O4 ] 0.000 0.670 0.0547 0.643 0.0851 4.65 3 1023 0.0500 0.446 0.0457 0.448 0.102 4.66 3 1023 0.0300 0.500 0.0475 0.491 0.0967 4.60 3 1023 0.0400 0.600 0.0523 0.594 0.0880 4.60 3 1023 0.200 0.000 0.0204 0.0898 0.227 4.63 3 1023 where K is a constant. Note that the exponent 2 for [NO2] in this expression is the same as the stoichiometric coefficient for NO2 in the reversible reaction. We can generalize this phenomenon with the following reaction at equilibrium: aA 1 bB Δ cC 1 dD where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. For the reaction at a particular temperature [C]c[D]d Equilibrium concentrations must be used K5 (14.2) in this equation. [A]a[B]b where K is the equilibrium constant. Equation (14.2) was formulated by two Norwegian chemists, Cato Guldberg† and Peter Waage,‡ in 1864. It is the mathematical expression of their law of mass action, which holds that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K (the equilibrium constant). Note that although the concentrations may vary, as long as a given reaction is at equilibrium and the temperature does not change, according to the law of mass action, the value of K remains constant. The validity of Equation (14.2) and the law of mass action has been established by studying many reversible reactions. The equilibrium constant, then, is defined by a quotient, the numerator of which is obtained by multiplying together the equilibrium concentrations of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation. Applying the same procedure to the equilibrium concentrations of reactants gives the denominator. The magnitude of the equilibrium constant tells us whether an equilibrium The signs @ and ! mean “much greater reaction favors the products or reactants. If K is much greater than 1 (that is, K @ 1), than” and “much smaller than,” respectively. the equilibrium will lie to the right and favors the products. Conversely, if the equi- librium constant is much smaller than 1 (that is, K ! 1), the equilibrium will lie to the left and favor the reactants (Figure 14.3). In this context, any number greater than 10 is considered to be much greater than 1, and any number less than 0.1 is much less than 1. † Cato Maximilian Guldberg (1836–1902). Norwegian chemist and mathematician. Guldberg’s research was mainly in thermodynamics. ‡ Peter Waage (1833–1900). Norwegian chemist. Like that of his coworker, Guldberg, Waage’s research was primarily in thermodynamics.  14.2 Writing Equilibrium Constant Expressions 625 Although the use of the words “reactants” and “products” may seem confusing because any substance serving as a reactant in the forward reaction also is a prod- uct of the reverse reaction, it is in keeping with the convention of referring to Products substances on the left of the equilibrium arrows as “reactants” and those on the right K >> 1 as “products.” Reactants 34 (a) Review of Concepts Consider the equilibrium X Δ Y, where the forward reaction rate constant is greater than the reverse reaction rate constant. Which of the following is true Reactants about the equilibrium constant? (a) Kc . 1, (b) Kc , 1, (c) Kc 5 1. K << 1 34 Products (b) 14.2 Writing Equilibrium Constant Expressions Figure 14.3 (a) At equilibrium, there are more products than reactants, and the equilibrium is The concept of equilibrium constants is extremely important in chemistry. As you will said to lie to the right. (b) In the soon see, equilibrium constants are the key to solving a wide variety of stoichiometry opposite situation, when there are problems involving equilibrium systems. For example, an industrial chemist who more reactants than products, the equilibrium is said to lie to the left. wants to maximize the yield of sulfuric acid, say, must have a clear understanding of the equilibrium constants for all the steps in the process, starting from the oxidation of sulfur and ending with the formation of the final product. A physician specializing in clinical cases of acid-base imbalance needs to know the equilibrium constants of weak acids and bases. And a knowledge of equilibrium constants of pertinent gas- phase reactions will help an atmospheric chemist better understand the process of ozone destruction in the stratosphere. To use equilibrium constants, we must express them in terms of the reactant and product concentrations. Our only guide is the law of mass action [Equation (14.2)], which is the general formula for finding equilibrium concentrations. However, because the concentrations of the reactants and products can be expressed in differ- ent units and because the reacting species are not always in the same phase, there may be more than one way to express the equilibrium constant for the same reaction. To begin with, we will consider reactions in which the reactants and products are in the same phase. Homogeneous Equilibria The term homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. An example of homogeneous gas-phase equilib- rium is the dissociation of N2O4. The equilibrium constant, as given in Equa- tion  (14.1), is [NO2]2 Kc 5 [N2O4] Note that the subscript in Kc indicates that the concentrations of the reacting species are expressed in molarity or moles per liter. The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. From Equation (5.8) we see that at constant temperature the pressure P of a gas is directly related to the concentration in mol/L of the gas; that is, P 5 (n/V)RT. Thus, for the equilibrium process N2O4 (g) Δ 2NO2 (g) 626 Chapter 14 ■ Chemical Equilibrium we can write P2NO2 KP 5 (14.3) PN2O4 where PNO2 and PN2O4 are the equilibrium partial pressures (in atm) of NO2 and N2O4, respectively. The subscript in KP tells us that equilibrium concentrations are expressed in terms of pressure. In general, Kc is not equal to KP, because the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter. A simple relationship between KP and Kc can be derived as follows. Let us consider the follow- ing equilibrium in the gas phase: aA(g) Δ bB(g) where a and b are stoichiometric coefficients. The equilibrium constant Kc is given by [B]b Kc 5 [A]a PBb and the expression for KP is KP 5 PAa where PA and PB are the partial pressures of A and B. Assuming ideal gas behavior, PAV 5 nART nART PA 5 V where V is the volume of the container in liters. Also PBV 5 nBRT nBRT PB 5 V Substituting these relations into the expression for KP, we obtain nBRT b nB b a b a b V V KP 5 5 (RT) b2a nART a nA a a b a b V V Now both nA/V and nB/V have units of mol/L and can be replaced by [A] and [B], so that [B]b KP 5 (RT) ¢n [A]a 5 Kc (RT) ¢n (14.4) where ¢n 5 b 2 a 5 moles of gaseous products 2 moles of gaseous reactants  14.2 Writing Equilibrium Constant Expressions 627 Because pressures are usually expressed in atm, the gas constant R is given by 0.0821 L ? atm/K ? mol, and we can write the relationship between KP and Kc as KP 5 Kc (0.0821T) ¢n (14.5) To use this equation, the pressures in KP must be in atm. In general, KP fi Kc except in the special case in which ≤n 5 0 as in the equilibrium mixture of molecular hydrogen, molecular bromine, and hydrogen bromide: H2 (g) 1 Br2 (g) Δ 2HBr(g) In this case, Equation (14.5) can be written as KP 5 Kc (0.0821T ) 0 Any number raised to the zero power is equal to 1. 5 Kc As another example of homogeneous equilibrium, let us consider the ionization of acetic acid (CH3COOH) in water: CH3COOH(aq) 1 H2O(l) Δ CH3COO 2 (aq) 1 H3O 1 (aq) The equilibrium constant is [CH3COO 2 ][H3O 1 ] K¿c 5 [CH3COOH][H2O] (We use the prime for Kc here to distinguish it from the final form of equilibrium con- stant to be derived below.) In 1 L, or 1000 g, of water, there are 1000 g/(18.02 g/mol), or 55.5 moles, of water. Therefore, the “concentration” of water, or [H2O], is 55.5 mol/L, or 55.5 M (see p. 583). This is a large quantity compared to the concentrations of other species in solution (usually 1 M or smaller), and we can assume that it does not change appreciably during the course of a reaction. Thus, we can treat [H2O] as a constant and rewrite the equilibrium constant as [CH3COO 2 ][H3O 1 ] Kc 5 [CH3COOH] where Kc 5 K¿c [H2O] Equilibrium Constant and Units Note that it is general practice not to include units for the equilibrium constant. In thermodynamics, the equilibrium constant is defined in terms of activities For nonideal systems, the activities are not exactly numerically equal to rather than concentrations. For an ideal system, the activity of a substance is concentrations. In some cases, the the ratio of its concentration (or partial pressure) to a standard value, which is differences can be appreciable. Unless otherwise noted, we will treat 1 M (or 1 atm). This procedure eliminates all units but does not alter the numer- all systems as ideal. ical parts of the concentration or pressure. Consequently, K has no units. We will extend this practice to acid-base equilibria and solubility equilibria in Chapters 15 and 16. Examples 14.1 through 14.3 illustrate the procedure for writing equilibrium constant expressions and calculating equilibrium constants and equilibrium concentrations. 628 Chapter 14 ■ Chemical Equilibrium Example 14.1 Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium: (a) HF(aq) 1 H2O(l) Δ H3O 1 (aq) 1 F 2 (aq) (b) 2NO(g) 1 O2 (g) Δ 2NO2 (g) (c) CH3COOH(aq) 1 C2H5OH(aq) Δ CH3COOC2H5 (aq) 1 H2O(l) Strategy Keep in mind the following facts: (1) the KP expression applies only to gaseous reactions and (2) the concentration of solvent (usually water) does not appear in the equilibrium constant expression. Solution (a) Because there are no gases present, KP does not apply and we have only Kc. [H3O 1 ][F 2 ] K¿c 5 [HF][H2O] HF is a weak acid, so that the amount of water consumed in acid ionizations is negligible compared with the total amount of water present as solvent. Thus, we can rewrite the equilibrium constant as [H3O 1 ][F 2 ] Kc 5 [HF] [NO2]2 P2NO2 (b) Kc 5     KP 5 [NO]2[O2] P2NOPO2 (c) The equilibrium constant K9c is given by [CH3COOC2H5][H2O] K¿c 5 [CH3COOH][C2H5OH] Because the water produced in the reaction is negligible compared with the water solvent, the concentration of water does not change. Thus, we can write the new equilibrium constant as [CH3COOC2H5] Similar problems: 14.7, 14.8. Kc 5 [CH3COOH][C2H5OH] Practice Exercise Write Kc and KP for the decomposition of dinitrogen pentoxide: 2N2O5 (g) Δ 4NO2 (g) 1 O2 (g) Example 14.2 The following equilibrium process has been studied at 230°C: 2NO(g) 1 O2 (g) Δ 2NO2 (g) In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO] 5 0.0542 M, [O2] 5 0.127 M, and [NO2] 5 15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature. (Continued)  14.2 Writing Equilibrium Constant Expressions 629 Strategy The concentrations given are equilibrium concentrations. They have units of mol/L, so we can calculate the equilibrium constant (Kc) using the law of mass action [Equation (14.2)]. Solution The equilibrium constant is given by [NO2]2 Kc 5 [NO]2[O2] 43 Substituting the concentrations, we find that (15.5) 2 Kc 5 5 6.44 3 105 (0.0542) 2 (0.127) Check Note that Kc is given without units. Also, the large magnitude of Kc is consistent with the high product (NO2) concentration relative to the concentrations of the reactants (NO and O2). Practice Exercise Carbonyl chloride (COCl2), also called phosgene, was used in 2NO 1 O2 Δ 2NO2 World War I as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride Similar problem: 14.16. CO(g) 1 Cl2 (g) Δ COCl2 (g) at 74°C are [CO] 5 1.2 3 1022 M, [Cl2] 5 0.054 M, and [COCl2] 5 0.14 M. Calculate the equilibrium constant (Kc). Example 14.3 The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C? Strategy The concentrations of the reacting gases are given in atm, so we can express 43 the equilibrium constant in KP. From the known KP value and the equilibrium pressures of PCl3 and PCl5, we can solve for PCl2 . Solution First, we write KP in terms of the partial pressures of the reacting species PPCl3PCl2 KP 5 PPCl5 Knowing the partial pressures, we write (0.463) (PCl2 ) 1.05 5 (0.875) (1.05) (0.875) or PCl2 5 5 1.98 atm PCl5 Δ PCl3 1 Cl2 (0.463) (Continued) 630 Chapter 14 ■ Chemical Equilibrium Similar problem: 14.19. Check Note that we have added atm as the unit for PCl . 2 Practice Exercise The equilibrium constant KP for the reaction 2NO2 (g) Δ 2NO(g) 1 O2 (g) is 158 at 1000 K. Calculate PO2 if PNO2 5 0.400 atm and PNO 5 0.270 atm. Example 14.4 Methanol (CH3OH) is manufactured industrially by the reaction CO(g) 1 2H2 (g) Δ CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of KP at this temperature? Strategy The relationship between Kc and KP is given by Equation (14.5). What is the 43 change in the number of moles of gases from reactants to product? Recall that ¢n 5 moles of gaseous products 2 moles of gaseous reactants What unit of temperature should we use? Solution The relationship between Kc and KP is KP 5 Kc (0.0821T ) ¢n CO 1 2H2 Δ CH3OH Because T 5 273 1 220 5 493 K and ≤n 5 1 2 3 5 22, we have KP 5 (10.5) (0.0821 3 493) 22 5 6.41 3 1023 Check Note that KP, like Kc, is a dimensionless quantity. This example shows that we can get a quite different value for the equilibrium constant for the same reaction, depending on whether we express the concentrations in moles per liter or in Similar problems: 14.17, 14.18. atmospheres. Practice Exercise For the reaction N2 (g) 1 3H2 (g) Δ 2NH3 (g) KP is 4.3 3 1024 at 375°C. Calculate Kc for the reaction. Heterogeneous Equilibria As you might expect, a heterogeneous equilibrium results from a reversible reac- tion involving reactants and products that are in different phases. For example, when calcium carbonate is heated in a closed vessel, the following equilibrium The mineral calcite is made of is attained: calcium carbonate, as are chalk and marble. CaCO3 (s) Δ CaO(s) 1 CO2 (g)  14.2 Writing Equilibrium Constant Expressions 631 The two solids and one gas constitute three separate phases. At equilibrium, we might write the equilibrium constant as [CaO][CO2] K¿c 5 (14.6) [CaCO3] (Again, the prime for Kc here is to distinguish it from the final form of equilibrium constant to be derived shortly.) However, the “concentration” of a solid, like its den- sity, is an intensive property and does not depend on how much of the substance is present. For example, the “molar concentration” of copper (density: 8.96 g/cm3) at 20°C is the same, whether we have 1 gram or 1 ton of the metal: 8.96 g 1 mol [Cu] 5 3 3 5 0.141 mol/cm3 5 141 mol/L 1 cm 63.55 g For this reason, the terms [CaCO3] and [CaO] are themselves constants and can be combined with the equilibrium constant. We can simplify Equation (14.6) by writing [CaCO3] K¿c 5 Kc 5 [CO2] (14.7) [CaO] where Kc, the “new” equilibrium constant, is conveniently expressed in terms of a single concentration, that of CO2. Note that the value of Kc does not depend on how much CaCO3 and CaO are present, as long as some of each is present at equilibrium (Figure 14.4). The situation becomes simpler if we replace concentrations with activities. In ther- modynamics, the activity of a pure solid is 1. Thus, the concentration terms for CaCO3 and CaO are both unity, and from the preceding equilibrium equation, we can immedi- ately write Kc 5 [CO2]. Similarly, the activity of a pure liquid is also 1. Thus, if a reactant or a product is a liquid, we can omit it in the equilibrium constant expression. Alternatively, we can express the equilibrium constant as KP 5 PCO2 (14.8) The equilibrium constant in this case is numerically equal to the pressure of CO2 gas, an easily measurable quantity. Figure 14.4 In (a) and (b) the equilibrium pressure of CO2 is the same at the same temperature, despite the presence of different amounts of CaCO3 and CaO. CaO CaCO3 CaCO3 CaO (a) (b) 632 Chapter 14 ■ Chemical Equilibrium Example 14.5 Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous systems: (a) (NH4 ) 2Se(s) Δ 2NH3 (g) 1 H2Se(g) (b) AgCl(s) Δ Ag 1 (aq) 1 Cl 2 (aq) (c) P4 (s) 1 6Cl2 (g) Δ 4PCl3 (l) Strategy We omit any pure solids or pure liquids in the equilibrium constant expression because their activities are unity. Solution (a) Because (NH4)2Se is a solid, the equilibrium constant Kc is given by Kc 5 [NH3]2[H2Se] Alternatively, we can express the equilibrium constant KP in terms of the partial pressures of NH3 and H2Se: KP 5 P2NH3PH2Se (b) Here AgCl is a solid so the equilibrium constant is given by Kc 5 [Ag 1 ][Cl 2 ] Because no gases are present, there is no KP expression. (c) We note that P4 is a solid and PCl3 is a liquid, so they do not appear in the equilibrium constant expression. Thus, Kc is given by 1 Kc 5 [Cl2]6 Alternatively, we can express the equilibrium constant in terms of the pressure of Cl2: 1 Similar problem: 14.8. KP 5 P6Cl2   Practice Exercise Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) 1 4CO(g) Δ Ni(CO) 4 (g) Example 14.6 Consider the following heterogeneous equilibrium: CaCO3 (s) Δ CaO(s) 1 CO2 (g) At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature. Strategy Remember that pure solids do not appear in the equilibrium constant expression. The relationship between KP and Kc is given by Equation (14.5). (Continued)  14.2 Writing Equilibrium Constant Expressions 633 Solution (a) Using Equation (14.8) we write KP 5 PCO2 5 0.236 (b) From Equation (14.5), we know KP 5 Kc (0.0821 T) ¢n In this case, T 5 800 1 273 5 1073 K and ≤n 5 1, so we substitute these values in the equation and obtain 0.236 5 Kc (0.0821 3 1073) Kc 5 2.68 3 1023 Similar problem: 14.22. Practice Exercise Consider the following equilibrium at 395 K: NH4HS(s) Δ NH3 (g) 1 H2S(g) The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction. Review of Concepts For which of the following reactions is Kc equal to KP? (a) 4NH3 (g) 1 5O2 (g) Δ 4NO(g) 1 6H2O(g) (b) 2H2O2 (aq) Δ 2H2O(l) 1 O2 (g) (c) PCl3 (g) 1 3NH3 (g) Δ 3HCl(g) 1 P(NH2 ) 3 (g) Multiple Equilibria The reactions we have considered so far are all relatively simple. A more complicated situation is one in which the product molecules in one equilibrium system are involved in a second equilibrium process: A1B Δ C1D C1D Δ E1F The products formed in the first reaction, C and D, react further to form products E and F. At equilibrium we can write two separate equilibrium constants: [C][D] K¿c 5 [A][B] [E][F] and K–c 5 [C][D] The overall reaction is given by the sum of the two reactions A1B Δ C1 D K¿c C 1D Δ E 1 F K–c Overall reaction: A1B Δ E1F Kc 634 Chapter 14 ■ Chemical Equilibrium and the equilibrium constant Kc for the overall reaction is [E][F] Kc 5 [A][B] We obtain the same expression if we take the product of the expressions for K¿c and K–c : [C][D] [E][F] [E][F] K¿c K–c 5 3 5 [A][B] [C][D] [A][B] Therefore, Kc 5 K¿c K–c (14.9) We can now make an important statement about multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Among the many known examples of multiple equilibria is the ionization of diprotic acids in aqueous solution. The following equilibrium constants have been determined for carbonic acid (H2CO3) at 25°C: [H1][HCO2 3] H2CO3 (aq) Δ H1 (aq) 1 HCO2 3 (aq)      K¿c 5 5 4.2 3 1027 [H2CO3] [H 1 ][CO22 3 ] HCO2 1 22 3 (aq) Δ H (aq) 1 CO3 (aq)     K– c 5 2 5 4.8 3 10211 [HCO3 ] The overall reaction is the sum of these two reactions H2CO3 (aq) Δ 2H 1 (aq) 1 CO22 3 (aq) and the corresponding equilibrium constant is given by [H1]2[CO22 3 ] Kc 5 [H2CO3] Using Equation (14.9) we arrive at Kc 5 K¿c K–c 5 (4.2 3 1027 )(4.8 3 10211 ) 5 2.0 3 10217 Review of Concepts You are given the equilibrium constant for the reaction N2 (g) 1 O2 (g) Δ 2NO(g) Suppose you want to calculate the equilibrium constant for the reaction N2 (g) 1 2O2 (g) Δ 2NO2 (g) What additional equilibrium constant value (for another reaction) would you need for this calculation? Assume all the equilibria are studied at the same temperature. The Form of K and the Equilibrium Equation Before concluding this section, let us look at two important rules for writing equilib- rium constants:  14.2 Writing Equilibrium Constant Expressions 635 1. When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium con- The reciprocal of x is 1/x. stant. Thus, if we write the NO2–N2O4 equilibrium as N2O4 (g) Δ 2NO2 (g) then at 25°C, [NO2]2 Kc 5 5 4.63 3 1023 [N2O4] However, we can represent the equilibrium equally well as 2NO2 (g) Δ N2O4 (g) and the equilibrium constant is now given by [N2O4] 1 1 K¿c 5 2 5 5 5 216 [NO2] Kc 4.63 3 1023 You can see that Kc 5 1/K¿c or KcK¿c 5 1.00. Either Kc or K¿c is a valid equilib- rium constant, but it is meaningless to say that the equilibrium constant for the NO2–N2O4 system is 4.63 3 1023, or 216, unless we also specify how the equi- librium equation is written. 2. The value of K also depends on how the equilibrium equation is balanced. Consider the following ways of describing the same equilibrium: 1 [NO2] 2 N2O4 (g) Δ NO2 (g)   K¿c 5 1 [N2O4] 2 [NO2]2 N2O4 (g) Δ 2NO2 (g)    Kc 5 [N2O4] Looking at the exponents we see that K¿c 5 2Kc . In Table 14.1 we find Kc 5 4.63 3 1023; therefore K9c 5 0.0680. According to the law of mass action, each concentration term in the equilibrium constant expression is raised to a power equal to its stoichiometric coefficient. Thus, if you double a chemical equation throughout, the corresponding equilibrium constant will be the square of the original value; if you triple the equation, the equi- librium constant will be the cube of the original value, and so on. The NO2–N2O4 example illustrates once again the need to write the chemical equation when quot- ing the numerical value of an equilibrium constant. Example 14.7 deals with the relationship between the equilibrium constants for differently balanced equations describing the same reaction. Example 14.7 The reaction for the production of ammonia can be written in a number of ways: (a) N2 (g) 1 3H2 (g) Δ 2NH3 (g) (b) 12N2 (g) 1 32H2 (g) Δ NH3 (g) (c) 13N2 (g) 1 H2 (g) Δ 23NH3 (g) (Continued) 636 Chapter 14 ■ Chemical Equilibrium Write the equilibrium constant expression for each formulation. (Express the concentrations of the reacting species in mol/L.) (d) How are the equilibrium constants related to one another? Strategy We are given three different expressions for the same reacting system. Remember that the equilibrium constant expression depends on how the equation is balanced, that is, on the stoichiometric coefficients used in the equation. Solution [NH3]2 (a) Ka [N2][H2]3 [NH3] (b) Kb 1 3 [N2]2[H2]2 2 [NH3]3 (c) Kc 1 [N2]3[H2] (d) Ka K 2b Ka K3c 3 Similar problem: 14.20. K2b K3c or Kb Kc2 Practice Exercise Write the equilibrium expression (Kc) for each of the following reactions and show how they are related to each other: (a) 3O2 (g) Δ 2O3 (g) , (b) O2 (g) Δ 23O3 (g) . Review of Concepts From the following equilibrium constant expression, write a balanced chemical equation for the gas-phase reaction. [NH3]2[H2O]4 Kc 5 [NO2]2[H2]7 Summary of Guidelines for Writing Equilibrium Constant Expressions 1. The concentrations of the reacting species in the condensed phase are expressed in mol/L; in the gaseous phase, the concentrations can be expressed in mol/L or in atm. Kc is related to KP by a simple equation [Equation (14.5)]. 2. The concentrations of pure solids, pure liquids (in heterogeneous equilibria), and solvents (in homogeneous equilibria) do not appear in the equilibrium constant expressions. 3. The equilibrium constant (Kc or KP) is a dimensionless quantity. 4. In quoting a value for the equilibrium constant, we must specify the balanced equation and the temperature. 5. If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium con- stants of the individual reactions.  14.3 The Relationship Between Chemical Kinetics and Chemical Equilibrium 637 14.3 The Relationship Between Chemical Kinetics and Chemical Equilibrium We have seen that K, defined in Equation (14.2), is constant at a given temperature regardless of variations in individual equilibrium concentrations (review Table 14.1). We can find out why this is so and at the same time gain insight into the equilibrium process by considering the kinetics of chemical reactions. Let us suppose that the following reversible reaction occurs via a mechanism To review reaction mechanisms, see Section 13.5. consisting of a single elementary step in both the forward and reverse directions: kf A 2B Δ k AB2 r The forward rate is given by ratef 5 kf[A][B]2 and the reverse rate is given by rater 5 kr[AB2] where kf and kr are the rate constants for the forward and reverse directions. At equi- librium, when no net changes occur, the two rates must be equal: ratef 5 rater or kf[A][B]2 5 kr[AB2] kf [AB2] 5 kr [A][B]2 Because both kf and kr are constants at a given temperature, their ratio is also a con- stant, which is equal to the equilibrium constant Kc. kf [AB2] 5 Kc 5 kr [A][B]2 So Kc is always a constant regardless of the equilibrium concentrations of the reacting species because it is always equal to kf/kr, the quotient of two quantities that are themselves constant at a given temperature. Because rate constants are temperature- dependent [see Equation (13.11)], it follows that the equilibrium constant must also change with temperature. Now suppose the same reaction has a mechanism with more than one elementary step. Suppose it occurs via a two-step mechanism as follows: k¿f Step 1: 2B Δ k¿ B2 r k–f Step 2: A 1 B2 Δ k– AB2 r Overall reaction: A 1 2B Δ AB2 This is an example of multiple equilibria, discussed in Section 14.2. We write the expressions for the equilibrium constants: k¿f [B2] K¿ 5 5 (14.10) k¿r [B]2 k–f [AB2] K– 5 5 (14.11) k–r [A][B2] 638 Chapter 14 ■ Chemical Equilibrium Multiplying Equation (14.10) by Equation (14.11), we get [B2][AB2] [AB2] K¿K– 5 5 [B]2[A][B2] [A][B]2 For the overall reaction, we can write [AB2] Kc 5 5 K¿K– [A][B]2 Because both K¿ and K– are constants, Kc is also a constant. This result lets us generalize our treatment of the reaction aA 1 bB Δ cC 1 dD Regardless of whether this reaction occurs via a single-step or a multistep mechanism, we can write the equilibrium constant expression according to the law of mass action shown in Equation (14.2): [C]c[D]d K5 [A]a[B]b In summary, we see that in terms of chemical kinetics, the equilibrium constant of a reaction can be expressed as a ratio of the rate constants of the forward and reverse reactions. This analysis explains why the equilibrium constant is a constant and why its value changes with temperature. Review of Concepts The equilibrium constant (Kc) for reaction A Δ B 1 C is 4.8 3 1022 at 80°C. If the forward rate constant is 3.2 3 102 s21, calculate the reverse rate constant. 14.4 What Does the Equilibrium Constant Tell Us? We have seen that the equilibrium constant for a given reaction can be calculated from known equilibrium concentrations. Once we know the value of the equilibrium con- stant, we can use Equation (14.2) to calculate unknown equilibrium concentrations— remembering, of course, that the equilibrium constant has a constant value only if the temperature does not change. In general, the equilibrium constant helps us to predict the direction in which a reaction mixture will proceed to achieve equilib- rium and to calculate the concentrations of reactants and products once equilib- rium has been reached. These uses of the equilibrium constant will be explored in this section. Predicting the Direction of a Reaction The equilibrium constant Kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase 43 H2 (g) 1 I2 (g) Δ 2HI(g) is 54.3 at 430°C. Suppose that in a certain experiment we place 0.243 mole of H2, H2 1 I2 Δ 2HI 0.146 mole of I2, and 1.98 moles of HI all in a 1.00-L container at 430°C. Will there  14.4 What Does the Equilibrium Constant Tell Us? 639 Qc Figure 14.5 The direction of a reversible reaction to reach equilibrium depends on the relative Kc Qc Kc Kc magnitudes of Qc and Kc. Note that Kc is a constant at a given temperature, but Qc varies according to the relative amounts Qc of reactants and products present. Reactants n Products Equilibrium : no net change Reactants m Products be a net reaction to form more H2 and I2 or more HI? Inserting the starting concentra- tions in the equilibrium constant expression, we write [HI]20 (1.98) 2 5 5 111 [H2]0[I2]0 (0.243)(0.146) where the subscript 0 indicates initial concentrations (before equilibrium is reached). Because the quotient [HI]02 /[H2]0[I2]0 is greater than Kc, this system is not at equilibrium. For reactions that have not reached equilibrium, such as the formation of HI Keep in mind that the method for calculating Q is the same as that for K, except that considered above, we obtain the reaction quotient (Qc), instead of the equilibrium nonequilibrium concentrations are used. constant by substituting the initial concentrations into the equilibrium constant expression. To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of Qc and Kc. The three possible cases are as follows: • Qc , Kc The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. • Qc 5 Kc The initial concentrations are equilibrium concentrations. The system is at equilibrium. • Qc . Kc The ratio of initial concentrations of products to reactants is too large. To reach equilibrium, products must be converted to reactants. The system proceeds from right to left (consuming products, forming reactants) to reach equilibrium. Figure 14.5 shows a comparison of Kc with Qc. Example 14.8 shows how the value of Qc can help us determine the direction of net reaction toward equilibrium. Example 14.8 At the start of a reaction, there are 0.249 mol N2, 3.21 3 1022 mol H2, and 6.42 3 1024 mol NH3 in a 3.50-L reaction vessel at 375°C. If the equilibrium constant (Kc) for 43 the reaction N2 (g) 1 3H2 (g) Δ 2NH3 (g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed. N2 1 3H2 Δ 2NH3 (Continued) 640 Chapter 14 ■ Chemical Equilibrium Strategy We are given the initial amounts of the gases (in moles) in a vessel of known volume (in liters), so we can calculate their molar concentrations and hence the reaction quotient (Qc). How does a comparison of Qc with Kc enable us to determine if the system is at equilibrium or, if not, in which direction will the net reaction proceed to reach equilibrium? Solution The initial concentrations of the reacting species are 0.249 mol [N2]0 5 5 0.0711 M 3.50 L 3.21 3 1022 mol [H2]0 5 5 9.17 3 1023 M 3.50 L 6.42 3 1024 mol [NH3]0 5 5 1.83 3 1024 M 3.50 L Next we write [NH3]20 (1.83 3 1024 ) 2 Qc 5 5 5 0.611 [N2]0[H2]30 (0.0711) (9.17 3 1023 ) 3 Because Qc is smaller than Kc (1.2), the system is not at equilibrium. The net result will be an increase in the concentration of NH3 and a decrease in the concentrations of N2 and H2. That is, the net reaction will proceed from left to right until equilibrium Similar problems: 14.39, 14.40. is reached. Practice Exercise The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) 1 Cl2 (g) Δ 2NOCl(g) is 6.5 3 104 at 35°C. In a certain experiment, 2.0 3 1022 mole of NO, 8.3 3 1023 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium? Review of Concepts The equilibrium constant (Kc) for the A2 1 B2 Δ 2AB reaction is 3 at a certain temperature. Which of the diagrams shown here corresponds to the reaction at equilibrium? For those mixtures that are not at equilibrium, will the net reaction move in the forward or reverse direction to reach equilibrium? (a) (b) (c) Calculating Equilibrium Concentrations If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations. Commonly, only the initial reactant concentrations are given. Let us consider the following system  14.4 What Does the Equilibrium Constant Tell Us? 641 involving two organic compounds, cis-stilbene and trans-stilbene, in a nonpolar hydro- carbon solvent (Figure 14.6): cis-stilbene Δ trans-stilbene The equilibrium constant (Kc) for this system is 24.0 at 200°C. Suppose that initially only cis-stilbene is present at a concentration of 0.850 mol/L. How do we calculate the concentrations of cis- and trans-stilbene at equilibrium? From the stoichiometry of the reaction we see that for every mole of cis-stilbene converted, 1 mole of trans- stilbene is formed. Let x be the equilibrium concentration of trans-stilbene in mol/L; therefore, the equilibrium concentration of cis-stilbene must be (0.850 2 x) mol/L. It is useful to summarize the changes in concentration as follows: cis-stilbene Δ trans-stilbene Initial (M): 0.850 0 This procedure of solving equilibrium Change (M): x x concentrations is sometimes referred to as the ICE method, where the acronym Equilibrium (M): (0.850 x) x stands for Initial, Change, and Equilibrium. A positive (1) change represents an increase and a negative (2) change a decrease in concentration at equilibrium. Next, we set up the equilibrium constant expression [trans-stilbene] Kc 5 [cis-stilbene] x 24.0 5 0.850 2 x x 5 0.816 M Having solved for x, we calculate the equilibrium concentrations of cis-stilbene and trans-stilbene as follows: [cis-stilbene] 5 (0.850 2 0.816) M 5 0.034 M [trans-stilbene] 5 0.816 M To check the results we could use the equilibrium concentrations to calculate Kc. We summarize our approach to solving equilibrium constant problems as follows: 1. Express the equilibrium concentrations of all species in terms of the initial con- centrations and a single unknown x, which represents the change in concentration. 2. Write the equilibrium constant expression in terms of the equilibrium concentra- tions. Knowing the value of the equilibrium constant, solve for x. 3. Having solved for x, calculate the equilibrium concentrations of all species. Examples 14.9 and 14.10 illustrate the application of this three-step procedure. Example 14.9 Figure 14.6 The equilibrium between cis-stilbene and trans- A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask stilbene. Note that both molecules at 430°C. The equilibrium constant Kc for the reaction H2 (g) 1 I2 (g) Δ 2HI(g) is have the same molecular formula 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at equilibrium. (C14H12) and also the same type of bonds. However, in cis-stilbene, Strategy We are given the initial amounts of the gases (in moles) in a vessel of known the benzene rings are on one side of the C“C bond and the H volume (in liters), so we can calculate their molar concentrations. Because initially no atoms are on the other side HI was present, the system could not be at equilibrium. Therefore, some H2 would react whereas in trans-stilbene the with the same amount of I2 (why?) to form HI until equilibrium was established. benzene rings (and the H atoms) are across from the C“C bond. (Continued) These compounds have different melting points and dipole moments. 642 Chapter 14 ■ Chemical Equilibrium Solution We follow the preceding procedure to calculate the equilibrium concentrations. Step 1: The stoichiometry of the reaction is 1 mol H2 reacting with 1 mol I2 to yield 2 mol HI. Let x be the depletion in concentration (mol/L) of H2 and I2 at equilibrium. It follows that the equilibrium concentration of HI must be 2x. We summarize the changes in concentrations as follows: H2 I2 34 2HI Initial (M): 0.500 0.500 0.000 Change (M): x x 2x Equilibrium (M): (0.500 x) (0.500 x) 2x Step 2: The equilibrium constant is given by [HI]2 Kc 5 [H2][I2] Substituting, we get (2x) 2 54.3 5 (0.500 2 x) (0.500 2 x) Taking the square root of both sides, we get 2x 7.37 5 0.500 2 x x 5 0.393 M Step 3: At equilibrium, the concentrations are [H2] 5 (0.500 2 0.393) M 5 0.107 M [I2] 5 (0.500 2 0.393) M 5 0.107 M [HI] 5 2 3 0.393 M 5 0.786 M Check You can check your answers by calculating Kc using the equilibrium concentrations. Remember that Kc is a constant for a particular reaction at a Similar problem: 14.48. given temperature. Practice Exercise Consider the reaction in Example 14.9. Starting with a concentration of 0.040 M for HI, calculate the concentrations of HI, H2, and I2 at equilibrium. Example 14.10 For the same reaction and temperature as in Example 14.9, suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium. Strategy From the initial concentrations we can calculate the reaction quotient (Qc) to see if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium. A comparison of Qc with Kc also enables us to determine if there will be a depletion in H2 and I2 or HI as equilibrium is established. (Continued)  14.4 What Does the Equilibrium Constant Tell Us? 643 Solution First we calculate Qc as follows: [HI]20 (0.0224) 2 Qc 5 5 5 19.5 [H2]0[I2]0 (0.00623) (0.00414) Because Qc (19.5) is smaller than Kc (54.3), we conclude that the net reaction will proceed from left to right until equilibrium is reached (see Figure 14.4); that is, there will be a depletion of H2 and I2 and a gain in HI. Step 1: Let x be the depletion in concentration (mol/L) of H2 and I2 at equilibrium. From the stoichiometry of the reaction it follows that the increase in concentration for HI must be 2x. Next we write H2 1 I2 34 2HI Initial (M): 0.00623 0.00414 0.0224 Change (M): 2x 2x 12x Equilibrium (M): (0.00623 2 x) (0.00414 2 x) (0.0224 1 2x) Step 2: The equilibrium constant is [HI]2 Kc 5 [H2][I2] Substituting, we get (0.0224 1 2x) 2 54.3 5 (0.00623 2 x) (0.00414 2 x) It is not possible to solve this equation by the square root shortcut, as the starting concentrations [H2] and [I2] are unequal. Instead, we must first carry out the multiplications 54.3(2.58 3 1025 2 0.0104x 1 x2 ) 5 5.02 3 1024 1 0.0896x 1 4x2 Collecting terms, we get 50.3x2 2 0.654x 1 8.98 3 1024 5 0 This is a quadratic equation of the form ax2 1 bx 1 c 5 0. The solution for a quadratic equation (see Appendix 4) is 2b 6 2b2 2 4ac x5 2a Here we have a 5 50.3, b 5 20.654, and c 5 8.98 3 1024, so that 0.654 6 2(20.654) 2 2 4(50.3) (8.98 3 1024 ) x5 2 3 50.3 x 5 0.0114 M    or    x 5 0.00156 M The first solution is physically impossible because the amounts of H2 and I2 reacted would be more than those originally present. The second solution gives the correct answer. Note that in solving quadratic equations of this type, one answer is always physically impossible, so choosing a value for x is easy. (Continued) 644 Chapter 14 ■ Chemical Equilibrium Step 3: At equilibrium, the concentrations are [H2] 5 (0.00623 2 0.00156) M 5 0.00467 M [I2] 5 (0.00414 2 0.00156) M 5 0.00258 M [HI] 5 (0.0224 1 2 3 0.00156) M 5 0.0255 M Check You can check the answers by calculating Kc using the equilibrium concentrations. Remember that Kc is a constant for a particular reaction at a given Similar problem: 14.90. temperature. Practice Exercise At 1280°C the equilibrium constant (Kc) for the reaction Br2 (g) Δ 2Br(g) is 1.1 3 1023. If the initial concentrations are [Br2] 5 6.3 3 1022 M and [Br] 5 1.2 3 1022 M, calculate the concentrations of these species at equilibrium. Examples 14.9 and 14.10 show that we can calculate the concentrations of all the reacting species at equilibrium if we know the equilibrium constant and the initial concentrations. This information is valuable if we need to estimate the yield of a reaction. For example, if the reaction between H2 and I2 to form HI were to go to completion, the number of moles of HI formed in Example 14.9 would be 2 3 0.500 mol, or 1.00 mol. However, because of the equilibrium process, the actual amount of HI formed can be no more than 2 3 0.393 mol, or 0.786 mol, a 78.6 percent yield. 14.5 Factors That Affect Chemical Equilibrium Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in experimental conditions may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. When we say that an equilibrium position shifts to the right, for example, we mean that the net reaction is now from left to right. Variables that can be controlled experimentally are concentration, pressure, vol- ume, and temperature. Here we will examine how each of these variables affects a reacting system at equilibrium. In addition, we will examine the effect of a catalyst on equilibrium. Le Châtelier’s Principle There is a general rule that helps us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature Animation occurs. The rule, known as Le Châtelier’s† principle, states that if an external stress Le Châtelier’s Principle is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. The word “stress” here means a change in concentration, pressure, volume, or temperature that removes the system from the equilibrium state. We will use Le Châtelier’s principle to assess the effects of such changes. † Henri Louis Le Châtelier (1850–1936). French chemist. Le Châtelier did work on metallurgy, cements, glasses, fuels, and explosives. He was also noted for his skills in industrial management.  14.5 Factors That Affect Chemical Equilibrium 645 (a) (b) (c) (d) Figure 14.7 Effect of concentration change on the position of equilibrium. (a) An aqueous Fe(SCN)3 solution. The color of the solution is due to both the red FeSCN 21 and the yellow Fe31 ions. (b) After the addition of some NaSCN to the solution in (a), the equilibrium shifts to the left. (c) After the addition of some Fe(NO3 )3 to the solution in (a), the equilibrium shifts to the left. (d) After the addition of some H2C2O4 to the solution in (a), the equilibrium shifts to the right. The yellow color is due to the Fe(C2O4 )332 ions. Changes in Concentration Iron(III) thiocyanate [Fe(SCN)3] dissolves readily in water to give a red solution. The red color is due to the presence of hydrated FeSCN21 ion. The equilibrium between undissociated FeSCN21 and the Fe31 and SCN2 ions is given by FeSCN2 (aq) 34 Fe3 (aq) SCN (aq) red pale yellow colorless What happens if we add some sodium thiocyanate (NaSCN) to this solution? In this case, the stress applied to the equilibrium system is an increase in the concentration of SCN2 (from the dissociation of NaSCN). To offset this stress, some Fe31 ions react with the added SCN2 ions, and the equilibrium shifts from right to left: FeSCN21 (aq) — Fe31 (aq) 1 SCN 2 (aq) Consequently, the red color of the solution deepens (Figure 14.7). Similarly, if we Both Na1 and NO23 are colorless spectator ions. added iron(III) nitrate [Fe(NO3)3] to the original solution, the red color would also deepen because the additional Fe31 ions [from Fe(NO3)3] would shift the equilibrium from right to left. Now suppose we add some oxalic acid (H2C2O4) to the original solution. Oxalic acid ionizes in water to form the oxalate ion, C2O422, which binds strongly to the Fe31 ions. The formation of the stable yellow ion Fe(C2O4)323 removes free Fe 31 ions 21 in solution. Consequently, more FeSCN units dissociate and the equilibrium shifts from left to right: FeSCN21 (aq) ¡ Fe31 (aq) 1 SCN 2 (aq) Oxalic acid is sometimes used to The red solution will turn yellow due to the formation of Fe(C2O4 ) 32 remove bathtub rings that consist 3 ions. of rust, or Fe2O3. This experiment demonstrates that all reactants and products are present in the reacting system at equilibrium. Second, increasing the concentrations of the products (Fe31 or SCN2) shifts the equilibrium to the left, and decreasing the concentration of the product Fe31 shifts the equilibrium to the right. These results are just as predicted Le Châtelier’s principle simply summarizes by Le Châtelier’s principle. the observed behavior of equilibrium systems; therefore, it is incorrect to say The effect of a change in concentration on the equilibrium position is shown in that a given equilibrium shift occurs Example 14.11. “because of” Le Châtelier’s principle.  646 Chapter 14 ■ Chemical Equilibrium Initial Final equilibrium Change equilibrium Example 14.11 At 720°C, the equilibrium constant Kc for the reaction N2 (g) 1 3H2 (g) Δ 2NH3 (g) H2 is 2.37 3 1023. In a certain experiment, the equilibrium concentrations are [N2] 5 0.683 M, [H2] 5 8.80 M, and [NH3] 5 1.05 M. Suppose some NH3 is added to Concentration the mixture so that its concentration is increased to 3.65 M. (a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. (b) Confirm your prediction by calculating the reaction quotient Qc and comparing NH3 its value with Kc. Strategy (a) What is the stress applied to the system? How does the system adjust to offset the stress? (b) At the instant when some NH3 is added, the system is no N2 longer at equilibrium. How do we calculate the Qc for the reaction at this point? How does a comparison of Qc with Kc tell us the direction of the net reaction to reach equilibrium? Time Solution Figure 14.8 Changes in concentration of H2, N2, and NH3 (a) The stress applied to the system is the addition of NH3. To offset this stress, some after the addition of NH3 to the NH3 reacts to produce N2 and H2 until a new equilibrium is established. The net equilibrium mixture. When the new reaction therefore shifts from right to left; that is, equilibrium is established, all the concentrations are changed but N2 (g) 1 3H2 (g) — 2NH3 (g) Kc remains the same because temperature remains constant. (b) At the instant when some of the NH3 is added, the system is no longer at equilibrium. The reaction quotient is given by [NH3]20 Qc 5 [N2]0[H2]30 (3.65) 2 5 (0.683) (8.80) 3 5 2.86 3 1022 Because this value is greater than 2.37 3 1023, the net reaction shifts from right to left until Qc equals Kc. Similar problem: 14.46. Figure 14.8 shows qualitatively the changes in concentrations of the reacting species. Practice Exercise At 430°C, the equilibrium constant (KP) for the reaction 2NO(g) 1 O2 (g) Δ 2NO2 (g) is 1.5 3 105. In one experiment, the initial pressures of NO, O2, and NO2 are 2.1 3 1023 atm, 1.1 3 1022 atm, and 0.14 atm, respectively. Calculate QP and predict the direction that the net reaction will shift to reach equilibrium. Changes in Volume and Pressure Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases (say, in an aqueous solution) because liquids and solids are virtually incompressible. On the other hand, concentrations of gases are greatly affected by changes in pressure. Let us look again at Equation (5.8): PV 5 nRT n P 5 a b RT V  14.5 Factors That Affect Chemical Equilibrium 647 Note that P and V are related to each other inversely: The greater the pressure, the smaller the volume, and vice versa. Note, too, that the term (n/V) is the concentration of the gas in mol/L, and it varies directly with pressure. Suppose that the equilibrium system N2O4 (g) Δ 2NO2 (g) is in a cylinder fitted with a movable piston. What happens if we increase the pressure on the gases by pushing down on the piston at constant temperature? Because the volume decreases, the concentration (n/V) of both NO2 and N2O4 increases. Note that the concentration of NO2 is squared in the equilibrium constant expression, so the increase in pressure increases the numerator more than the denominator. The system is no longer at equilibrium and we write [NO2]20 Qc 5 [N2O4]0 Thus, Qc . Kc and the net reaction will shift to the left until Qc 5 Kc (Figure 14.9). Conversely, a decrease in pressure (increase in volume) would result in Qc , Kc, and the net reaction would shift to the right until Qc 5 Kc. (This conclusion is also pre- dicted by Le Châtelier’s principle.) In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases (the reverse reaction, in this case), and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases (here, the forward reaction). For reactions in which there is no change in the number of moles of gases, a pressure (or volume) change Figure 14.9 The effect of an has no effect on the position of equilibrium. increase in pressure on the It is possible to change the pressure of a system without changing its volume. N2O4(g) Δ 2NO2(g) equilibrium. Suppose the NO2–N2O4 system is contained in a stainless-steel vessel whose vol- ume is constant. We can increase the total pressure in the vessel by adding an inert gas (helium, for example) to the equilibrium system. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of both NO2 and N2O4; but the partial pressure of each gas, given by the product of its mole fraction and total pressure (see Section 5.6), does not change. Thus, the presence of an inert gas in such a case does not affect the equilibrium. Example 14.12 illustrates the effect of a pressure change on the equilibrium position. Example 14.12 Consider the following equilibrium systems: (a) 2PbS(s) 1 3O2 (g) Δ 2PbO(s) 1 2SO2 (g) (b) PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) (c) H2 (g) 1 CO2 (g) Δ H2O(g) 1 CO(g) Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature. Strategy A change in pressure can affect only the volume of a gas, but not that of a solid because solids (and liquids) are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier’s principle, the system will adjust to (Continued) 648 Chapter 14 ■ Chemical Equilibrium partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: PV 5 nRT so P r n. Solution (a) Consider only the gaseous molecules. In the balanced equation, there are 3 moles of gaseous reactants and 2 moles of gaseous products. Therefore, the net reaction will shift toward the products (to the right) when the pressure is increased. (b) The number of moles of products is 2 and that of reactants is 1; therefore, the net reaction will shift to the left, toward the reactant. (c) The number of moles of products is equal to the number of moles of reactants, so a change in pressure has no effect on the equilibrium. Similar problem: 14.56. Check In each case, the prediction is consistent with Le Châtelier’s principle. Practice Exercise Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide, and molecular chlorine 2NOCl(g) Δ 2NO(g) 1 Cl2 (g) Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at constant temperature. Review of Concepts The diagram here shows the gaseous reaction 2A Δ A2 at equilibrium. If the pressure is decreased by increasing the volume at constant temperature, how would the concentrations of A and A2 change when a new equilibrium is established? Changes in Temperature A change in concentration, pressure, or volume may alter the equilibrium position, that is, the relative amounts of reactants and products, but it does not change the value of the equilibrium constant. Only a change in temperature can alter the equilibrium constant. To see why, let us consider the reaction N2O4 (g) Δ 2NO2 (g) The forward reaction is endothermic (absorbs heat, ≤H° . 0): heat 1 N2O4 (g) ¡ 2NO2 (g) ¢H° 5 58.0 kJ/mol so the reverse reaction is exothermic (releases heat, ≤H° , 0): 2NO2 (g) ¡ N2O4 (g) 1 heat   ¢H° 5 258.0 kJ/mol At equilibrium at a certain temperature, the heat effect is zero because there is no net reaction. If we treat heat as though it were a chemical reagent, then a rise in temperature  14.5 Factors That Affect Chemical Equilibrium 649 Figure 14.10 (a) Two bulbs containing a mixture of NO2 and N2O4 gases at equilibrium. (b) When one bulb is immersed in ice water (left), its color becomes lighter, indicating the formation of colorless N2O4 gas. When the other bulb is immersed in hot water, its color darkens, indicating an increase in NO2. (a) (b) “adds” heat to the system and a drop in temperature “removes” heat from the system. As with a change in any other parameter (concentration, pressure, or volume), the system shifts to reduce the effect of the change. Therefore, a temperature increase favors the endothermic direction of the reaction (from left to right of the equilibrium equation), which decreases [N2O4] and increases [NO2]. A temperature decrease favors the exothermic direction of the reaction (from right to left of the equilibrium equation), which decreases [NO2] and increases [N2O4]. Consequently, the equilibrium constant, given by [NO2]2 Kc 5 [N2O4] increases when the system is heated and decreases when the system is cooled (Figure 14.10). As another example, consider the equilibrium between the following ions: CoCl24 6H2O Δ Co(H2O) 26 4Cl blue pink The formation of CoCl22 4 is endothermic. On heating, the equilibrium shifts to the left and the solution turns blue. Cooling favors the exothermic reaction [the formation of Co(H2O) 216 ] and the solution turns pink (Figure 14.11). In summary, a temperature increase favors an endothermic reaction, and a tem- perature decrease favors an exothermic reaction. Figure 14.11 (Left) Heating favors the formation of the blue CoCl422 ion. (Right) Cooling favors the formation of the pink Co(H2O)621 ion. 650 Chapter 14 ■ Chemical Equilibrium Review of Concepts The diagrams shown here represent the reaction X2 1 Y2 Δ 2XY at equilibrium at two temperatures (T2 . T1). Is the reaction endothermic or exothermic? T1 T2 The Effect of a Catalyst We know that a catalyst enhances the rate of a reaction by lowering the reaction’s activation energy (Section 13.6). However, as Figure 13.23 shows, a catalyst lowers the activation energy of the forward reaction and the reverse reaction to the same extent. We can therefore conclude that the presence of a catalyst does not alter the equilibrium constant, nor does it shift the position of an equilibrium system. Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait much longer for it to happen. Summary of Factors That May Affect the Equilibrium Position We have considered four ways to affect a reacting system at equilibrium. It is impor- tant to remember that, of the four, only a change in temperature changes the value of the equilibrium constant. Changes in concentration, pressure, and volume can alter the equilibrium concentrations of the reacting mixture, but they cannot change the equilibrium constant as long as the temperature does not change. A catalyst will speed up the process, but it has no effect on the equilibrium constant or on the equilibrium concentrations of the reacting species. Two processes that illustrate the effects of changed conditions on equilibrium processes are discussed in Chemistry in Action essays on pp. 651 and 652. The effects of temperature, concentration, and pressure change, as well as the addition of an inert gas, on an equilibrium system are treated in Example 14.13. Example 14.13 Consider the following equilibrium process between dinitrogen tetrafluoride (N2F4) and nitrogen difluoride (NF2): 43 N2F4 (g) Δ 2NF2 (g) ¢H° 5 38.5 kJ/mol Predict the changes in the equilibrium if (a) the reacting mixture is heated at constant volume; (b) some N2F4 gas is removed from the reacting mixture at constant temperature and volume; (c) the pressure on the reacting mixture is decreased at constant temperature; and (d) a catalyst is added to the reacting mixture. N2F4 Δ 2NF2 (Continued) CHEMISTRY in Action Life at High Altitudes and Hemoglobin Production I n the human body, countless chemical equilibria must be maintained to ensure physiological well-being. If environ- mental conditions change, the body must adapt to keep func- equilibrium will then gradually shift back toward the forma- tion of oxyhemoglobin. It takes two to three weeks for the in- crease in hemoglobin production to meet the body’s basic tioning. The consequences of a sudden change in altitude needs adequately. A return to full capacity may require several dramatize this fact. Flying from San Francisco, which is at years to occur. Studies show that long-time residents of high- sea level, to Mexico City, where the elevation is 2.3 km (1.4 mi), altitude areas have high hemoglobin levels in their blood— or scaling a 3-km mountain in two days can cause head- sometimes as much as 50 percent more than individuals living ache, nausea, extreme fatigue, and other discomforts. These at sea level! conditions are all symptoms of hypoxia, a deficiency in the amount of oxygen reaching body tissues. In serious cases, the victim may slip into a coma and die if not treated quickly. And yet a person living at a high altitude for weeks or months gradually recovers from altitude sickness and adjusts to the low oxygen content in the atmosphere, so that he or she can function normally. The combination of oxygen with the hemoglobin (Hb) molecule, which carries oxygen through the blood, is a complex reaction, but for our purposes it can be represented by a simpli- fied equation: Hb(aq) 1 O2 (aq) Δ HbO2 (aq) where HbO2 is oxyhemoglobin, the hemoglobin-oxygen com- plex that actually transports oxygen to tissues. The equilibrium constant is [HbO2] Kc 5 [Hb][O2] At an altitude of 3 km the partial pressure of oxygen is only about 0.14 atm, compared with 0.2 atm at sea level. According to Le Châtelier’s principle, a decrease in oxygen concentration will shift the equilibrium shown in the equation above from right to left. This change depletes the supply of oxyhemoglo- bin, causing hypoxia. Given enough time, the body copes with Mountaineers need weeks or even months to become acclimatized before this problem by producing more hemoglobin molecules. The scaling summits such as Mount Everest. Strategy (a) What does the sign of ≤H° indicate about the heat change (endothermic or exothermic) for the forward reaction? (b) Would the removal of some N2F4 increase or decrease the Qc of the reaction? (c) How would the decrease in pressure change the volume of the system? (d) What is the function of a catalyst? How does it affect a reacting system not at equilibrium? at equilibrium? (Continued) 651 CHEMISTRY in Action The Haber Process K nowing the factors that affect chemical equilibrium has great practical value for industrial applications, such as the synthesis of ammonia. The Haber process for synthesizing am- 100 80 Mole percent of NH3 monia from molecular hydrogen and nitrogen uses a heteroge- neous catalyst to speed up the reaction (see p. 601). Let us look at the equilibrium reaction for ammonia synthesis to determine 60 whether there are factors that could be manipulated to enhance the yield. 40 Suppose, as a prominent industrial chemist at the turn of the twentieth century, you are asked to design an efficient pro- 20 cedure for synthesizing ammonia from hydrogen and nitrogen. Your main objective is to obtain a high yield of the product 0 while keeping the production costs down. Your first step is to 1000 2000 3000 4000 take a careful look at the balanced equation for the production Pressure (atm) of ammonia: Mole percent of NH3 as a function of the total pressure of the gases at 425°C. N2 (g) 1 3H2 (g) Δ 2NH3 (g)  ¢H° 5 292.6 kJ/mol Two ideas strike you: First, because 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3, a higher yield of NH3 can be obtained at equilibrium if the reaction is carried yield of ammonia increases with decreasing temperature. A out under high pressures. This is indeed the case, as shown by low-temperature operation (say, 220 K or 253°C) is desirable the plot of mole percent of NH3 versus the total pressure of in other respects too. The boiling point of NH3 is 233.5°C, so the reacting system. Second, the exothermic nature of the for- as it formed it would quickly condense to a liquid, which ward reaction tells you that the equilibrium constant for the could be conveniently removed from the reacting system. reaction will decrease with increasing temperature. Thus, for (Both H2 and N2 are still gases at this temperature.) maximum yield of NH3, the reaction should be run at the low- Consequently, the net reaction would shift from left to right, est possible temperature. The graph on p. 653 shows that the just as desired. Solution (a) The stress applied is the heat added to the system. Note that the N2F4 ¡ 2NF2 reaction is an endothermic process (≤H° . 0), which absorbs heat from the surroundings. Therefore, we can think of heat as a reactant heat 1 N2F4 (g) Δ 2NF2 (g) The system will adjust to remove some of the added heat by undergoing a decomposition reaction (from left to right). The equilibrium constant [NF2]2 Kc 5 [N2F4] will therefore increase with increasing temperature because the concentration of NF2 has increased and that of N2F4 has decreased. Recall that the equilibrium constant is (Continued) 652  H2 + N2 Reaction Compressor chamber 100 (catalysts) NH3 N2 + H2 80 NH3 + H2 + N2 Mole percent 60 Unreacted Ammonia H2 + N2 condenser 40 Liquid NH3 20 Storage 0 tanks 200 300 400 500 Temperature (°C) The composition (mole percent) of H2 1 N2 and NH3 at equilibrium (for a Schematic diagram of the Haber process for ammonia synthesis. The heat certain starting mixture) as a function of temperature. generated from the reaction is used to heat the incoming gases. On paper, then, these are your conclusions. Let us carried out at about 500°C. This high-temperature opera- compare your recommendations with the actual conditions tion is costly and the yield of NH 3 is low. The justification in an industrial plant. Typically, the operating pressures are for this choice is that the rate of NH3 production increases between 500 atm and 1000 atm, so you are right to advocate with increasing temperature. Commercially, faster produc- high pressure. Furthermore, in the industrial process the tion of NH 3 is preferable even if it means a lower yield and NH3 never reaches its equilibrium value but is constantly higher operating cost. For this reason, the combination of removed from the reaction mixture in a continuous process high-pressure, high-temperature conditions, and the proper operation. This design makes sense, too, as you had antici- catalyst is the most efficient way to produce ammonia on a pated. The only discrepancy is that the operation is usually large scale. a constant only at a particular temperature. If the temperature is changed, then the equilibrium constant will also change. (b) The stress here is the removal of N2F4 gas. The system will shift to replace some of the N2F4 removed. Therefore, the system shifts from right to left until equilibrium is reestablished. As a result, some NF2 combines to form N2F4. Comment The equilibrium constant remains unchanged in this case because temperature is held constant. It might seem that Kc should change because NF2 combines to produce N2F4. Remember, however, that initially some N2F4 was removed. The system adjusts to replace only some of the N2F4 that was removed, so that overall the amount of N2F4 has decreased. In fact, by the time the equilibrium is reestablished, the amounts of both NF2 and N2F4 have decreased. Looking at the equilibrium constant expression, we see that dividing a smaller numerator by a smaller denominator gives the same value of Kc. (Continued) 653 654 Chapter 14 ■ Chemical Equilibrium (c) The stress applied is a decrease in pressure (which is accompanied by an increase in gas volume). The system will adjust to remove the stress by increasing the pressure. Recall that pressure is directly proportional to the number of moles of a gas. In the balanced equation we see that the formation of NF2 from N2F4 will increase the total number of moles of gases and hence the pressure. Therefore, the system will shift from left to right to reestablish equilibrium. The equilibrium constant will remain unchanged because temperature is held constant. (d) The function of a catalyst is to increase the rate of a reaction. If a catalyst is added to a reacting system not at equilibrium, the system will reach equilibrium faster than if left undisturbed. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of NF2 and N2F4 or the Similar problems: 14.57, 14.58. equilibrium constant. Practice Exercise Consider the equilibrium between molecular oxygen and ozone 3O2 (g) Δ 2O3 (g) ¢H° 5 284 kJ/mol What would be the effect of (a) increasing the pressure on the system by decreasing the volume, (b) adding O2 to the system at constant volume, (c) decreasing the temperature, and (d) adding a catalyst? Key Equations [C]c[D]d K5 (14.2) Law of mass action. General expression of [A]a[B]b equilibrium constant. KP 5 Kc (0.0821T) ¢n (14.5) Relationship between KP and Kc. Kc 5 K¿cK–c (14.9) The equilibrium constant for the overall reaction is given by the product of the equilibrium constants for the individual reactions. Summary of Facts & Concepts 1. Dynamic equilibria between phases are called physical do not appear in the equilibrium constant expression of equilibria. Chemical equilibrium is a reversible process a reaction. in which the rates of the forward and reverse reactions 5. If a reaction can be expressed as the sum of two or more are equal and the concentrations of reactants and prod- reactions, the equilibrium constant for the overall reac- ucts do not change with time. tion is given by the product of the equilibrium constants 2. For the general chemical reaction of the individual reactions. 6. The value of K depends on how the chemical equation aA 1 bB Δ cC 1 d D is balanced, and the equilibrium constant for the reverse the concentrations of reactants and products at equilib- of a particular reaction is the reciprocal of the equilib- rium (in moles per liter) are related by the equilibrium rium constant of that reaction. constant expression [Equation (14.2)]. 7. The equilibrium constant is the ratio of the rate con- 3. The equilibrium constant for gases, KP, expresses the stant for the forward reaction to that for the reverse relationship of the equilibrium partial pressures (in atm) reaction. of reactants and products. 8. The reaction quotient Q has the same form as the equi- 4. A chemical equilibrium process in which all reactants librium constant, but it applies to a reaction that may and products are in the same phase is homogeneous. If not be at equilibrium. If Q . K, the reaction will pro- the reactants and products are not all in the same phase, ceed from right to left to achieve equilibrium. If Q , K, the equilibrium is heterogeneous. The concentrations of the reaction will proceed from left to right to achieve pure solids, pure liquids, and solvents are constant and equilibrium.  Questions & Problems 655 9. Le Châtelier’s principle states that if an external stress in concentration, pressure, or volume may change the is applied to a system at chemical equilibrium, the sys- equilibrium concentrations of reactants and products. tem will adjust to partially offset the stress. The addition of a catalyst hastens the attainment of 10. Only a change in temperature changes the value of the equilibrium but does not affect the equilibrium concen- equilibrium constant for a particular reaction. Changes trations of reactants and products. Key Words Chemical equilibrium, p. 622 Heterogeneous Law of mass action, p. 624 Physical equilibrium, p. 622 Equilibrium constant equilibrium, p. 630 Le Châtelier’s Reaction quotient (K), p. 624 Homogeneous principle, p. 644 (Qc), p. 639 equilibrium, p. 625 Questions & Problems • Problems available in Connect Plus • 14.9 Write the equilibrium constant expressions for Kc Red numbered problems solved in Student Solutions Manual and KP, if applicable, for the following reactions: (a) 2NO2 (g) 1 7H2 (g) Δ 2NH3 (g) 1 4H2O(l) The Concept of Equilibrium and (b) 2ZnS(s) 1 3O2 (g) Δ 2ZnO(s) 1 2SO2 (g) the Equilibrium Constant (c) C(s) 1 CO2 (g) Δ 2CO(g) Review Questions (d) C6H5COOH(aq) Δ 14.1 Define equilibrium. Give two examples of a dynamic C6H5COO2(aq) 1 H1(aq) equilibrium. 14.10 Write the equation relating Kc to KP, and define all 14.2 Explain the difference between physical equilibrium the terms. and chemical equilibrium. Give two examples of each. 14.11 What is the rule for writing the equilibrium con- 14.3 What is the law of mass action? stant for the overall reaction involving two or more 14.4 Briefly describe the importance of equilibrium in reactions? the study of chemical reactions. 14.12 Give an example of a multiple equilibria reaction. Equilibrium Constant Expressions Problems Review Questions • 14.13 The equilibrium constant for the reaction A Δ B 14.5 Define homogeneous equilibrium and heterogeneous is Kc 5 10 at a certain temperature. (1) Starting equilibrium. Give two examples of each. with only reactant A, which of the diagrams shown 14.6 What do the symbols Kc and KP represent? here best represents the system at equilibrium? • 14.7 Write the expressions for the equilibrium constants (2) Which of the diagrams best represents the sys- KP of the following thermal decomposition reactions: tem at equilibrium if Kc 5 0.10? Explain why you (a) 2NaHCO3 (s) Δ can calculate Kc in each case without knowing the Na2CO3 (s) 1 CO2 (g) 1 H2O(g) volume of the container. The gray spheres represent the A molecules and the green spheres represent (b) 2CaSO4 (s) Δ the B molecules. 2CaO(s) 1 2SO2 (g) 1 O2 (g) • 14.8 Write equilibrium constant expressions for Kc, and for KP, if applicable, for the following processes: (a) 2CO2 (g) Δ 2CO(g) 1 O2 (g) (b) 3O2 (g) Δ 2O3 (g) (c) CO(g) 1 Cl2 (g) Δ COCl2 (g) (d) H2O(g) 1 C(s) Δ CO(g) 1 H2 (g) (e) HCOOH(aq) Δ H 1 (aq) 1 HCOO 2 (aq) (a) (b) (c) (d) (f ) 2HgO(s) Δ 2Hg(l) 1 O2 (g)  656 Chapter 14 ■ Chemical Equilibrium • 14.14 The following diagrams represent the equilibrium • 14.21 The equilibrium constant Kc for the reaction state for three different reactions of the type I2 (g) Δ 2I(g) A 1 X Δ AX (X 5 B, C, or D): 25 is 3.8 3 10 at 727°C. Calculate Kc and KP for the equilibrium 2I(g) Δ I2 (g) at the same temperature. 14.22 At equilibrium, the pressure of the reacting mixture CaCO3 (s) Δ CaO(s) 1 CO2 (g) is 0.105 atm at 350°C. Calculate KP and Kc for this x x x reaction. A ⫹ B w AB A ⫹ C w AC A ⫹ D w AD • 14.23 The equilibrium constant KP for the reaction PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) (a) Which reaction has the largest equilibrium constant? (b) Which reaction has the smallest is 1.05 at 250°C. The reaction starts with a mixture equilibrium constant? of PCl5, PCl3, and Cl2 at pressures 0.177 atm, • 14.15 The equilibrium constant (Kc) for the reaction 0.223 atm, and 0.111 atm, respectively, at 250°C. When the mixture comes to equilibrium at that 2HCl(g) Δ H2 (g) 1 Cl2 (g) temperature, which pressures will have decreased is 4.17 3 10234 at 25°C. What is the equilibrium and which will have increased? Explain why. constant for the reaction • 14.24 Ammonium carbamate, NH4CO2NH2, decomposes as follows: H2 (g) 1 Cl2 (g) Δ 2HCl(g) NH4CO2NH2 (s) Δ 2NH3 (g) 1 CO2 (g) at the same temperature? 14.16 Consider the following equilibrium process at Starting with only the solid, it is found that at 40°C 700°C: the total gas pressure (NH3 and CO2) is 0.363 atm. Calculate the equilibrium constant KP. 2H2 (g) 1 S2 (g) Δ 2H2S(g) • 14.25 Consider the following reaction at 1600°C. Analysis shows that there are 2.50 moles of H2, 1.35 3 Br2 (g) Δ 2Br(g) 1025 mole of S2, and 8.70 moles of H2S present in a 12.0-L flask. Calculate the equilibrium constant Kc for When 1.05 moles of Br 2 are put in a 0.980-L the reaction. flask, 1.20 percent of the Br 2 undergoes dissoci- ation. Calculate the equilibrium constant Kc for • 14.17 What is KP at 1273°C for the reaction the reaction. 2CO(g) 1 O2 (g) Δ 2CO2 (g) • 14.26 Pure phosgene gas (COCl2), 3.00 3 1022 mol, was if Kc is 2.24 3 1022 at the same temperature? placed in a 1.50-L container. It was heated to 800 K, and at equilibrium the pressure of CO was found to 14.18 The equilibrium constant KP for the reaction be 0.497 atm. Calculate the equilibrium constant KP 2SO3 (g) Δ 2SO2 (g) 1 O2 (g) for the reaction is 1.8 3 1025 at 350°C. What is Kc for this reaction? CO(g) 1 Cl2 (g) Δ COCl2 (g) • 14.19 Consider the following reaction: • 14.27 Consider the equilibrium N2 (g) 1 O2 (g) Δ 2NO(g) 2NOBr(g) Δ 2NO(g) 1 Br2 (g) If the equilibrium partial pressures of N2, O2, and If nitrosyl bromide, NOBr, is 34 percent dissoci- NO are 0.15 atm, 0.33 atm, and 0.050 atm, respec- ated at 25°C and the total pressure is 0.25 atm, cal- tively, at 2200°C, what is KP? culate KP and Kc for the dissociation at this • 14.20 A reaction vessel contains NH3, N2, and H2 at equi- temperature. librium at a certain temperature. The equilibrium • 14.28 A 2.50-mole quantity of NOCl was initially in a concentrations are [NH3] 5 0.25 M, [N2] 5 0.11 M, 1.50-L reaction chamber at 400°C. After equilib- and [H2] 5 1.91 M. Calculate the equilibrium con- rium was established, it was found that 28.0 percent stant Kc for the synthesis of ammonia if the reaction of the NOCl had dissociated: is represented as (a) N2 (g) 1 3H2 (g) Δ 2NH3 (g) 2NOCl(g) Δ 2NO(g) 1 Cl2 (g) (b) 12N2 (g) 1 32H2 (g) Δ NH3 (g) Calculate the equilibrium constant Kc for the reaction.  Questions & Problems 657 14.29 The following equilibrium constants have been (a) If k1 5 2.4 3 1025 s21 and k21 5 1.3 3 1011/M ? s, determined for hydrosulfuric acid at 25°C: calculate the equilibrium constant K where K 5 H2S(aq) Δ H 1 (aq) 1 HS 2 (aq) [H1][OH2]/[H2O]. (b) Calculate the product K¿c 5 9.5 3 1028 [H1][OH2] and [H1] and [OH2]. HS (aq) Δ H (aq) 1 S22 (aq) 2 1 • 14.36 Consider the following reaction, which takes place K–c 5 1.0 3 10219 in a single elementary step: Calculate the equilibrium constant for the following k1 2A 1 B Δ k 1 A2B reaction at the same temperature: H2S(aq) Δ 2H 1 (aq) 1 S22 (aq) If the equilibrium constant Kc is 12.6 at a certain temperature and if kr 5 5.1 3 1022 s21, calculate the 14.30 The following equilibrium constants have been de- value of kf. termined for oxalic acid at 25°C: H2C2O4 (aq) Δ H 1 (aq) 1 HC2O42 (aq) What Does the Equilibrium Constant Tell Us? K¿c 5 6.5 3 1022 2 HC2O4 (aq) Δ H (aq) 1 C2O22 1 Review Questions 4 (aq) K–c 5 6.1 3 1025 14.37 Define reaction quotient. How does it differ from Calculate the equilibrium constant for the following equilibrium constant? reaction at the same temperature: 14.38 Outline the steps for calculating the concentrations 1 of reacting species in an equilibrium reaction. H2C2O4 (aq) Δ 2H (aq) 1 C2O22 4 (aq) • 14.31 The following equilibrium constants were deter- Problems mined at 1123 K: • 14.39 The equilibrium constant KP for the reaction C(s) 1 CO2 (g) Δ 2CO(g) K¿P 5 1.3 3 1014 CO(g) 1 Cl2 (g) Δ COCl2 (g) K–P 5 6.0 3 1023 2SO2 (g) 1 O2 (g) Δ 2SO3 (g) Write the equilibrium constant expression KP, and is 5.60 3 104 at 350°C. The initial pressures of SO2 calculate the equilibrium constant at 1123 K for and O2 in a mixture are 0.350 atm and 0.762 atm, respectively, at 350°C. When the mixture equili- C(s) 1 CO2 (g) 1 2Cl2 (g) Δ 2COCl2 (g) brates, is the total pressure less than or greater than • 14.32 At a certain temperature the following reactions the sum of the initial pressures (1.112 atm)? have the constants shown: • 14.40 For the synthesis of ammonia S(s) 1 O2 (g) Δ SO2 (g) K¿c 5 4.2 3 1052 N2 (g) 1 3H2 (g) Δ 2NH3 (g) 2S(s) 1 3O2 (g) Δ 2SO3 (g) K–c 5 9.8 3 10128 the equilibrium constant Kc at 375°C is 1.2. Starting Calculate the equilibrium constant Kc for the follow- with [H2]0 5 0.76 M, [N2]0 5 0.60 M, and [NH3]0 5 ing reaction at that temperature: 0.48 M, which gases will have increased in concen- 2SO2 (g) 1 O2 (g) Δ 2SO3 (g) tration and which will have decreased in concentra- tion when the mixture comes to equilibrium? • 14.41 For the reaction The Relationship Between Chemical Kinetics H2 (g) 1 CO2 (g) Δ H2O(g) 1 CO(g) and Chemical Equilibrium at 700°C, Kc 5 0.534. Calculate the number of Review Questions moles of H2 that are present at equilibrium if a mix- 14.33 Based on rate constant considerations, explain why ture of 0.300 mole of CO and 0.300 mole of H2O is the equilibrium constant depends on temperature. heated to 700°C in a 10.0-L container. 14.34 Explain why reactions with large equilibrium con- • 14.42 At 1000 K, a sample of pure NO2 gas decomposes: stants, such as the formation of rust (Fe2O3), may have very slow rates. 2NO2 (g) Δ 2NO(g) 1 O2 (g) The equilibrium constant KP is 158. Analysis shows Problems that the partial pressure of O2 is 0.25 atm at equilib- rium. Calculate the pressure of NO and NO2 in the • 14.35 Water is a very weak electrolyte that undergoes the mixture. following ionization (called autoionization): • 14.43 The equilibrium constant Kc for the reaction k1 H2O(l) Δ k 1 H1(aq) 1 OH2(aq) H2 (g) 1 Br2 (g) Δ 2HBr(g)  658 Chapter 14 ■ Chemical Equilibrium is 2.18 3 106 at 730°C. Starting with 3.20 moles of Problems HBr in a 12.0-L reaction vessel, calculate the con- centrations of H2, Br2, and HBr at equilibrium. • 14.53 Consider the following equilibrium system involv- ing SO2, Cl2, and SO2Cl2 (sulfuryl dichloride): • 14.44 The dissociation of molecular iodine into iodine atoms is represented as SO2 (g) 1 Cl2 (g) Δ SO2Cl2 (g) I2 (g) Δ 2I(g) Predict how the equilibrium position would change if (a) Cl2 gas were added to the system; (b) SO2Cl2 At 1000 K, the equilibrium constant Kc for the were removed from the system; (c) SO2 were reaction is 3.80 3 1025. Suppose you start with removed from the system. The temperature remains 0.0456 mole of I2 in a 2.30-L flask at 1000 K. What constant. are the concentrations of the gases at equilibrium? • 14.45 The equilibrium constant Kc for the decomposition • 14.54 Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: of phosgene, COCl2, is 4.63 3 1023 at 527°C: 2NaHCO3 (s) Δ Na2CO3 (s) 1 H2O(g) 1 CO2 (g) COCl2 (g) Δ CO(g) 1 Cl2 (g) What would happen to the equilibrium position if Calculate the equilibrium partial pressure of all the (a) some of the CO2 were removed from the sys- components, starting with pure phosgene at 0.760 atm. tem; (b) some solid Na2CO3 were added to the • 14.46 Consider the following equilibrium process at system; (c) some of the solid NaHCO3 were re- 686°C: moved from the system? The temperature remains constant. CO2 (g) 1 H2 (g) Δ CO(g) 1 H2O(g) • 14.55 Consider the following equilibrium systems: The equilibrium concentrations of the reacting spe- (a) A Δ 2B ¢H° 5 20.0 kJ/mol cies are [CO] 5 0.050 M, [H2] 5 0.045 M, [CO2] 5 (b) A 1 B Δ C ¢H° 5 25.4 kJ/mol 0.086 M, and [H2O] 5 0.040 M. (a) Calculate Kc for (c) A Δ B ¢H° 5 0.0 kJ/mol the reaction at 686°C. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the con- Predict the change in the equilibrium constant Kc centrations of all the gases be when equilibrium is that would occur in each case if the temperature of reestablished? the reacting system were raised. • 14.47 Consider the heterogeneous equilibrium process: • 14.56 What effect does an increase in pressure have on each of the following systems at equilibrium? The C(s) 1 CO2 (g) Δ 2CO(g) temperature is kept constant and, in each case, the reactants are in a cylinder fitted with a movable At 700°C, the total pressure of the system is found to be piston. 4.50 atm. If the equilibrium constant KP is 1.52, calcu- (a) A(s) Δ 2B(s) late the equilibrium partial pressures of CO2 and CO. (b) 2A(l) Δ B(l) • 14.48 The equilibrium constant Kc for the reaction (c) A(s) Δ B(g) H2 (g) 1 CO2 (g) Δ H2O(g) 1 CO(g) (d) A(g) Δ B(g) is 4.2 at 1650°C. Initially 0.80 mol H2 and 0.80 mol (e) A(g) Δ 2B(g) CO2 are injected into a 5.0-L flask. Calculate the • 14.57 Consider the equilibrium concentration of each species at equilibrium. 2I(g) Δ I2 (g) Factors That Affect Chemical Equilibrium What would be the effect on the position of equilib- Review Questions rium of (a) increasing the total pressure on the sys- tem by decreasing its volume; (b) adding gaseous I2 14.49 Explain Le Châtelier’s principle. How can this prin- to the reaction mixture; and (c) decreasing the tem- ciple help us maximize the yields of reactions? perature at constant volume? 14.50 Use Le Châtelier’s principle to explain why the • 14.58 Consider the following equilibrium process: equilibrium vapor pressure of a liquid increases with increasing temperature. PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) ¢H° 5 92.5 kJ/mol 14.51 List four factors that can shift the position of an Predict the direction of the shift in equilibrium when equilibrium. Only one of these factors can alter (a) the temperature is raised; (b) more chlorine gas is the value of the equilibrium constant. Which one added to the reaction mixture; (c) some PCl3 is is it? removed from the mixture; (d) the pressure on the 14.52 Does the addition of a catalyst have any effects on gases is increased; (e) a catalyst is added to the reac- the position of an equilibrium? tion mixture.  Questions & Problems 659 • 14.59 Consider the reaction 14.66 Diagram (a) shows the reaction A2 (g) 1 B2 (g) Δ 2AB(g) at equilibrium at a certain 2SO2 (g) 1 O2 (g) Δ 2SO3 (g) temperature, where the blue spheres represent A and ¢H° 5 2198.2 kJ/mol the yellow spheres represent B. If each sphere repre- Comment on the changes in the concentrations sents 0.020 mole and the volume of the container is of SO2, O2, and SO3 at equilibrium if we were to 1.0 L, calculate the concentration of each species (a) increase the temperature; (b) increase the pres- when the reaction in (b) reaches equilibrium. sure; (c) increase SO2; (d) add a catalyst; (e) add helium at constant volume. • 14.60 In the uncatalyzed reaction N2O4 (g) Δ 2NO2 (g) the pressure of the gases at equilibrium are PN2O4 5 0.377 atm and PNO2 5 1.56 atm at 100°C. What (a) (b) would happen to these pressures if a catalyst were added to the mixture? • 14.67 The equilibrium constant (KP) for the formation of • 14.61 Consider the gas-phase reaction the air pollutant nitric oxide (NO) in an automobile engine at 530°C is 2.9 3 10211: 2CO(g) 1 O2 (g) Δ 2CO2 (g) N2 (g) 1 O2 (g) Δ 2NO(g) Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture (a) at (a) Calculate the partial pressure of NO under these constant pressure and (b) at constant volume. conditions if the partial pressures of nitrogen and • 14.62 Consider the following equilibrium reaction in a oxygen are 3.0 atm and 0.012 atm, respectively. closed container: (b) Repeat the calculation for atmospheric condi- tions where the partial pressures of nitrogen and CaCO3 (s) Δ CaO(s) 1 CO2 (g) oxygen are 0.78 atm and 0.21 atm and the tempera- What will happen if (a) the volume is increased; ture is 25°C. (The KP for the reaction is 4.0 3 10231 (b) some CaO is added to the mixture; (c) some CaCO3 at this temperature.) (c) Is the formation of NO is removed; (d) some CO2 is added to the mixture; endothermic or exothermic? (d) What natural phe- (e) a few drops of a NaOH solution are added to the nomenon promotes the formation of NO? Why? mixture; (f) a few drops of a HCl solution are added 14.68 Baking soda (sodium bicarbonate) undergoes ther- to the mixture (ignore the reaction between CO2 and mal decomposition as follows: water); (g) temperature is increased? 2NaHCO3 (s) Δ Na2CO3 (s) 1 CO2 (g) 1 H2O(g) Additional Problems Would we obtain more CO2 and H2O by adding extra 14.63 Consider the statement: “The equilibrium con- baking soda to the reaction mixture in (a) a closed stant of a reacting mixture of solid NH4Cl and vessel or (b) an open vessel? gaseous NH3 and HCl is 0.316.” List three impor- • 14.69 Consider the following reaction at equilibrium: tant pieces of information that are missing from A(g) Δ 2B(g) this statement. • 14.64 Pure nitrosyl chloride (NOCl) gas was heated to From the data shown here, calculate the equilibrium 240°C in a 1.00-L container. At equilibrium the total constant (both KP and Kc) at each temperature. Is the pressure was 1.00 atm and the NOCl pressure was reaction endothermic or exothermic? 0.64 atm. Temperature (°C) [A] (M) [B] (M) 2NOCl(g) Δ 2NO(g) 1 Cl2 (g) 200 0.0125 0.843 (a) Calculate the partial pressures of NO and Cl2 300 0.171 0.764 in the system. (b) Calculate the equilibrium 400 0.250 0.724 constant KP. 14.65 Determine the initial and equilibrium concentrations 14.70 The equilibrium constant KP for the reaction of HI if the initial concentrations of H2 and I2 are 2H2O(g) Δ 2H2 (g) 1 O2 (g) both 0.16 M and their equilibrium concentrations 242 are both 0.072 M at 430°C. The equilibrium constant is 2 3 10 at 25°C. (a) What is Kc for the reaction at (Kc) for the reaction H2 (g) 1 I2 (g) Δ 2HI(g) is the same temperature? (b) The very small value of KP 54.2 at 430°C. (and Kc) indicates that the reaction overwhelmingly  660 Chapter 14 ■ Chemical Equilibrium favors the formation of water molecules. Explain In one experiment, a chemist finds that when why, despite this fact, a mixture of hydrogen and 0.054 mole of I2 was placed in a flask of volume oxygen gases can be kept at room temperature with- 0.48 L at 587 K, the degree of dissociation (that is, out any change. the fraction of I2 dissociated) was 0.0252. Calculate • 14.71 Consider the following reacting system: Kc and KP for the reaction at this temperature. 2NO(g) 1 Cl2 (g) Δ 2NOCl(g) • 14.78 One mole of N2 and three moles of H2 are placed in a flask at 375°C. Calculate the total pressure of the What combination of temperature and pressure (high system at equilibrium if the mole fraction of NH3 is or low) would maximize the yield of nitrosyl chlo- 0.21. The KP for the reaction is 4.31 3 1024. ride (NOCl)? [Hint: ¢H°f (NOCl) 5 51.7 kJ/mol. • 14.79 At 1130°C the equilibrium constant (Kc) for the You will also need to consult Appendix 3.] reaction • 14.72 At a certain temperature and a total pressure of 2H2S(g) Δ 2H2 (g) 1 S2 (g) 1.2 atm, the partial pressures of an equilibrium mixture is 2.25 3 1024. If [H2S] 5 4.84 3 1023 M and [H2] 5 1.50 3 1023 M, calculate [S2]. 2A(g) Δ B(g) • 14.80 A quantity of 6.75 g of SO2Cl2 was placed in a are PA 5 0.60 atm and PB 5 0.60 atm. (a) Calculate 2.00-L flask. At 648 K, there is 0.0345 mole of SO2 the KP for the reaction at this temperature. (b) If present. Calculate Kc for the reaction the total pressure were increased to 1.5 atm, what SO2Cl2 (g) Δ SO2 (g) 1 Cl2 (g) would be the partial pressures of A and B at equilibrium? • 14.81 The formation of SO3 from SO2 and O2 is an inter- • 14.73 The decomposition of ammonium hydrogen sulfide mediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. NH4HS(s) Δ NH3 (g) 1 H2S(g) The equilibrium constant KP for the reaction is an endothermic process. A 6.1589-g sample of the 2SO2 (g) 1 O2 (g) Δ 2SO3 (g) solid is placed in an evacuated 4.000-L vessel at ex- actly 24°C. After equilibrium has been established, is 0.13 at 830°C. In one experiment 2.00 mol SO2 the total pressure inside is 0.709 atm. Some solid and 2.00 mol O2 were initially present in a flask. NH4HS remains in the vessel. (a) What is the KP for What must the total pressure at equilibrium be in the reaction? (b) What percentage of the solid has order to have an 80.0 percent yield of SO3? decomposed? (c) If the volume of the vessel were 14.82 Consider the dissociation of iodine: doubled at constant temperature, what would hap- pen to the amount of solid in the vessel? I2 (g) Δ 2I(g) • 14.74 Consider the reaction A 1.00-g sample of I2 is heated to 1200°C in a 500-mL flask. At equilibrium the total pressure is 2NO(g) 1 O2 (g) Δ 2NO2 (g) 1.51 atm. Calculate KP for the reaction. [Hint: At 430°C, an equilibrium mixture consists of Use the result in 14.117(a). The degree of disso- 0.020 mole of O2, 0.040 mole of NO, and 0.96 mole ciation α can be obtained by first calculating the of NO2. Calculate KP for the reaction, given that the ratio of observed pressure over calculated pres- total pressure is 0.20 atm. sure, assuming no dissociation.] • 14.75 When heated, ammonium carbamate decomposes as 14.83 Eggshells are composed mostly of calcium carbon- follows: ate (CaCO3) formed by the reaction NH4CO2NH2 (s) Δ 2NH3 (g) 1 CO2 (g) Ca21 (aq) 1 CO22 3 (aq) Δ CaCO3 (s) At a certain temperature the equilibrium pressure of The carbonate ions are supplied by carbon dioxide the system is 0.318 atm. Calculate KP for the reaction. produced as a result of metabolism. Explain why eggshells are thinner in the summer when the rate of 14.76 A mixture of 0.47 mole of H2 and 3.59 moles of HCl panting by chickens is greater. Suggest a remedy for is heated to 2800°C. Calculate the equilibrium par- this situation. tial pressures of H2, Cl2, and HCl if the total pressure is 2.00 atm. For the reaction • 14.84 The equilibrium constant KP for the following reac- tion is 4.31 3 1024 at 375°C: H2 (g) 1 Cl2 (g) Δ 2HCl(g) N2 (g) 1 3H2 (g) Δ 2NH3 (g) KP is 193 at 2800°C. In a certain experiment a student starts with 0.862 atm • 14.77 When heated at high temperatures, iodine vapor dis- of N2 and 0.373 atm of H2 in a constant-volume ves- sociates as follows: sel at 375°C. Calculate the partial pressures of all I2 (g) Δ 2I(g) species when equilibrium is reached.  Questions & Problems 661 • 14.85 A quantity of 0.20 mole of carbon dioxide was where P is the total pressure. Calculate the equilib- heated to a certain temperature with an excess of rium constant KP of this reaction. graphite in a closed container until the following 14.92 When a gas was heated under atmospheric condi- equilibrium was reached: tions, its color deepened. Heating above 150°C C(s) 1 CO2 (g) Δ 2CO(g) caused the color to fade, and at 550°C the color was barely detectable. However, at 550°C, the Under these conditions, the average molar mass of color was partially restored by increasing the pres- the gases was 35 g/mol. (a) Calculate the mole frac- sure of the system. Which of the following best fits tions of CO and CO2. (b) What is KP if the total the above description? Justify your choice. (a) A pressure is 11 atm? (Hint: The average molar mass mixture of hydrogen and bromine, (b) pure bro- is the sum of the products of the mole fraction of mine, (c) a mixture of nitrogen dioxide and dinitro- each gas and its molar mass.) gen tetroxide. (Hint: Bromine has a reddish color • 14.86 When dissolved in water, glucose (corn sugar) and and nitrogen dioxide is a brown gas. The other fructose (fruit sugar) exist in equilibrium as follows: gases are colorless.) fructose Δ glucose 14.93 In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it A chemist prepared a 0.244 M fructose solution at speeds up both the forward and reverse rates to the 25°C. At equilibrium, it was found that its concen- same extent. To test this statement, consider a situa- tration had decreased to 0.113 M. (a) Calculate the tion in which an equilibrium of the type equilibrium constant for the reaction. (b) At equilib- rium, what percentage of fructose was converted 2A(g) Δ B(g) to glucose? is established inside a cylinder fitted with a weight- 14.87 At room temperature, solid iodine is in equilibrium less piston. The piston is attached by a string to the with its vapor through sublimation and deposition cover of a box containing a catalyst. When the piston (see p. 502). Describe how you would use radioac- moves upward (expanding against atmospheric pres- tive iodine, in either solid or vapor form, to show sure), the cover is lifted and the catalyst is exposed to that there is a dynamic equilibrium between these the gases. When the piston moves downward, the two phases. box is closed. Assume that the catalyst speeds up the • 14.88 At 1024°C, the pressure of oxygen gas from the de- forward reaction (2A ¡ B) but does not affect composition of copper(II) oxide (CuO) is 0.49 atm: the reverse process (B ¡ 2A). Suppose the cata- lyst is suddenly exposed to the equilibrium system as 4CuO(s) Δ 2Cu2O(s) 1 O2 (g) shown here. Describe what would happen subse- (a) What is KP for the reaction? (b) Calculate the quently. How does this “thought” experiment con- fraction of CuO that will decompose if 0.16 mole of vince you that no such catalyst can exist? it is placed in a 2.0-L flask at 1024°C. (c) What would the fraction be if a 1.0 mole sample of CuO were used? (d) What is the smallest amount of CuO (in moles) that would establish the equilibrium? • 14.89 A mixture containing 3.9 moles of NO and 0.88 mole of CO2 was allowed to react in a flask at a certain temperature according to the equation String NO(g) 1 CO2 (g) Δ NO2 (g) 1 CO(g) 2A B At equilibrium, 0.11 mole of CO2 was present. Calcu- late the equilibrium constant Kc of this reaction. Catalyst • 14.90 The equilibrium constant Kc for the reaction H2 (g) 1 I2 (g) Δ 2HI(g) is 54.3 at 430°C. At the start of the reaction there are 0.714 mole of H2, 0.984 mole of I2, and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium. 14.94 The equilibrium constant Kc for the following reac- • 14.91 When heated, a gaseous compound A dissociates as tion is 1.2 at 375°C. follows: N2 (g) 1 3H2 (g) Δ 2NH3 (g) A(g) Δ B(g) 1 C(g) (a) What is the value of KP for this reaction? In an experiment, A was heated at a certain tempera- (b) What is the value of the equilibrium constant Kc ture until its equilibrium pressure reached 0.14P, for 2NH3 (g) Δ N2 (g) 1 3H2 (g)?  662 Chapter 14 ■ Chemical Equilibrium (c) What is the value of Kc for 12N2 (g) 1 32H2 (g) 14.100 The equilibrium constant for the reaction Δ NH3 (g)? 4X 1 Y Δ 3Z is 33.3 at a certain temperature. Which diagram shown here corresponds to the sys- (d) What are the values of KP for the reactions tem at equilibrium? If the system is not at equilib- described in (b) and (c)? rium, predict the direction of the net reaction to reach • 14.95 A sealed glass bulb contains a mixture of NO2 and equilibrium. Each molecule represents 0.20 mole N2O4 gases. Describe what happens to the follow- and the volume of the container is 1.0 L. The color ing properties of the gases when the bulb is heated codes are X 5 blue, Y 5 green, and Z 5 red. from 20°C to 40°C: (a) color, (b) pressure, (c) aver- age molar mass, (d) degree of dissociation (from N2O4 to NO2), (e) density. Assume that volume re- mains constant. (Hint: NO2 is a brown gas; N2O4 is colorless.) • 14.96 At 20°C, the vapor pressure of water is 0.0231 atm. Calculate KP and Kc for the process (a) (b) (c) H2O(l) Δ H2O(g) 14.97 Industrially, sodium metal is obtained by electro- • 14.101 About 75 percent of hydrogen for industrial use is produced by the steam-reforming process. This pro- lyzing molten sodium chloride. The reaction at the cess is carried out in two stages called primary and cathode is Na 1 1 e 2 ¡ Na. We might expect secondary reforming. In the primary stage, a mix- that potassium metal would also be prepared by ture of steam and methane at about 30 atm is heated electrolyzing molten potassium chloride. However, over a nickel catalyst at 800°C to give hydrogen and potassium metal is soluble in molten potassium carbon monoxide: chloride and therefore is hard to recover. Further- more, potassium vaporizes readily at the operating CH4 (g) 1 H2O(g) Δ CO(g) 1 3H2 (g) temperature, creating hazardous conditions. Instead, ¢H° 5 260 kJ/mol potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor The secondary stage is carried out at about 1000°C, at 892°C: in the presence of air, to convert the remaining meth- ane to hydrogen: Na(g) 1 KCl(l) Δ NaCl(l) 1 K(g) CH4 (g) 1 12O2 (g) Δ CO(g) 1 2H2 (g) In view of the fact that potassium is a stronger re- ¢H° 5 35.7 kJ/mol ducing agent than sodium, explain why this ap- proach works. (The boiling points of sodium and (a) What conditions of temperature and pressure potassium are 892°C and 770°C, respectively.) would favor the formation of products in both the primary and secondary stage? (b) The equilibrium • 14.98 In the gas phase, nitrogen dioxide is actually a mix- constant Kc for the primary stage is 18 at 800°C. ture of nitrogen dioxide (NO2) and dinitrogen te- troxide (N2O4). If the density of such a mixture is (i) Calculate KP for the reaction. (ii) If the partial 2.3 g/L at 74°C and 1.3 atm, calculate the partial pressures of methane and steam were both 15 atm at pressures of the gases and KP for the dissociation the start, what are the pressures of all the gases at of N2O4. equilibrium? 14.99 The equilibrium constant for the reaction • 14.102 Photosynthesis can be represented by A 1 2B Δ 3C is 0.25 at a certain tempera- 6CO2 (g) 1 6H2O(l) Δ C6H12O6 (s) 1 6O2 (g) ture. Which diagram shown here corresponds to ¢H° 5 2801 kJ/mol the system at equilibrium? If the system is not at equilibrium, predict the direction of the net reac- Explain how the equilibrium would be affected tion to reach equilibrium. Each molecule repre- by the following changes: (a) partial pressure of CO2 sents 0.40 mole and the volume of the container is increased, (b) O2 is removed from the mixture, is 2.0 L. The color codes are A 5 green, B 5 red, (c) C6H12O6 (glucose) is removed from the mixture, C 5 blue. (d) more water is added, (e) a catalyst is added, (f) temperature is decreased. • 14.103 Consider the decomposition of ammonium chloride at a certain temperature: NH4Cl(s) Δ NH3 (g) 1 HCl(g) Calculate the equilibrium constant KP if the total (a) (b) (c) pressure is 2.2 atm at that temperature.  Questions & Problems 663 • 14.104At 25°C, the equilibrium partial pressures of NO2 14.111 Consider the potential energy diagrams for two and N2O4 are 0.15 atm and 0.20 atm, respectively. If types of reactions A Δ B. In each case, an- the volume is doubled at constant temperature, cal- swer the following questions for the system at culate the partial pressures of the gases when a new equilibrium. (a) How would a catalyst affect the equilibrium is established. forward and reverse rates of the reaction? (b) 14.105 In 1899 the German chemist Ludwig Mond devel- How would a catalyst affect the energies of the oped a process for purifying nickel by converting it reactant and product? (c) How would an increase to the volatile nickel tetracarbonyl [Ni(CO)4] in temperature affect the equilibrium constant? (b.p. 5 42.2°C): (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse Ni(s) 1 4CO(g) Δ Ni(CO) 4 (g) reactions, show that the equilibrium constant re- mains unchanged if a catalyst is added to the re- (a) Describe how you can separate nickel and its acting mixture. solid impurities. (b) How would you recover nickel? [ ¢H°f for Ni(CO)4 is 2602.9 kJ/mol.] 14.106 Consider the equilibrium reaction described in Potential energy Potential energy Problem 14.23. A quantity of 2.50 g of PCl5 is A B A placed in an evacuated 0.500-L flask and heated to 250°C. (a) Calculate the pressure of PCl5, assuming B it does not dissociate. (b) Calculate the partial pres- sure of PCl5 at equilibrium. (c) What is the total Reaction progress Reaction progress pressure at equilibrium? (d) What is the degree of dissociation of PCl5? (The degree of dissociation is given by the fraction of PCl5 that has undergone 14.112 The equilibrium constant K c for the reaction dissociation.) 2NH3 (g) Δ N2 (g) 1 3H2 (g) is 0.83 at 375°C. A 14.6-g sample of ammonia is placed in a 4.00-L 14.107 Consider the equilibrium system 3A Δ B. flask and heated to 375°C. Calculate the concen- Sketch the changes in the concentrations of A and trations of all the gases when equilibrium is B over time for the following situations: (a) ini- reached. tially only A is present; (b) initially only B is pres- 14.113 A quantity of 1.0 mole of N2O4 was introduced into ent; (c) initially both A and B are present (with A in an evacuated vessel and allowed to attain equilib- higher concentration). In each case, assume that rium at a certain temperature the concentration of B is higher than that of A at equilibrium. N2O4 (g) Δ 2NO2 (g) 14.108 The vapor pressure of mercury is 0.0020 mmHg at 26°C. (a) Calculate Kc and KP for the process The average molar mass of the reacting mixture Hg(l) Δ Hg(g). (b) A chemist breaks a ther- was 70.6 g/mol. (a) Calculate the mole fractions mometer and spills mercury onto the floor of a labo- of the gases. (b) Calculate KP for the reaction if ratory measuring 6.1 m long, 5.3 m wide, and 3.1 m the total pressure was 1.2 atm. (c) What would be high. Calculate the mass of mercury (in grams) va- the mole fractions if the pressure were increased porized at equilibrium and the concentration of mer- to 4.0 atm by reducing the volume at the same cury vapor in mg/m3. Does this concentration exceed temperature? the safety limit of 0.05 mg/m3? (Ignore the volume 14.114 The equilibrium constant (KP) for the reaction of furniture and other objects in the laboratory.) C(s) 1 CO2 (g) Δ 2CO(g) 14.109 At 25°C, a mixture of NO2 and N2O4 gases are in equilibrium in a cylinder fitted with a movable pis- is 1.9 at 727°C. What total pressure must be applied ton. The concentrations are [NO2] 5 0.0475 M and to the reacting system to obtain 0.012 mole of CO2 [N2O4] 5 0.487 M. The volume of the gas mixture is and 0.025 mole of CO? halved by pushing down on the piston at constant 14.115 The forward and reverse rate constants for the reac- temperature. Calculate the concentrations of the tion A(g) 1 B(g) Δ C(g) are 3.6 3 1023/M ? s gases when equilibrium is reestablished. Will the and 8.7 3 1024 s21, respectively, at 323 K. Calculate color become darker or lighter after the change? the equilibrium pressures of all the species starting [Hint: Kc for the dissociation of N2O4 to NO2 is at PA 5 1.6 atm and PB 5 0.44 atm. 4.63 3 1023. N2O4(g) is colorless and NO2(g) has a 14.116 The equilibrium constant (KP) for the reaction brown color.] PCl3 (g) 1 Cl2 (g) Δ PCl5 (g) is 2.93 at 127°C. 14.110 A student placed a few ice cubes in a drinking glass Initially there were 2.00 moles of PCl3 and 1.00 mole with water. A few minutes later she noticed that of Cl2 present. Calculate the partial pressures of some of the ice cubes were fused together. Explain the gases at equilibrium if the total pressure is what happened. 2.00 atm.  664 Chapter 14 ■ Chemical Equilibrium 14.117 Consider the reaction between NO2 and N2O4 in a In this representation, the H atoms are omitted and a closed container: C atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each N2O4 (g) Δ 2NO2 (g) direction. The activation energy for the chair S boat Initially, 1 mole of N2O4 is present. At equilibrium, conversion is 41 kJ/mol. If the frequency factor is α mole of N2O4 has dissociated to form NO2. 1.0 3 1012 s21, what is k1 at 298 K? The equilibrium (a) Derive an expression for KP in terms of α and P, constant Kc for the reaction is 9.83 3 103 at 298 K. the total pressure. (b) How does the expression in (a) 14.122 Consider the following reaction at a certain help you predict the shift in equilibrium due to an temperature increase in P? Does your prediction agree with Le Châtelier’s principle? A2 1 B2 Δ 2AB 14.118 The dependence of the equilibrium constant of a re- The mixing of 1 mole of A2 with 3 moles of B2 gives action on temperature is given by the van’t Hoff rise to x mole of AB at equilibrium. The addition of equation: 2 more moles of A2 produces another x mole of AB. What is the equilibrium constant for the reaction? ¢H° ln K 5 2 1C 14.123 Iodine is sparingly soluble in water but much more RT so in carbon tetrachloride (CCl4). The equilibrium where C is a constant. The following table gives the constant, also called the partition coefficient, for the equilibrium constant (KP) for the reaction at various distribution of I2 between these two phases temperatures I2 (aq) Δ I2 (CCl4 ) 2NO(g) 1 O2 (g) Δ 2NO2 (g) is 83 at 20°C. (a) A student adds 0.030 L of CCl4 to KP 138 5.12 0.436 0.0626 0.0130 0.200 L of an aqueous solution containing 0.032 g I2. T(K) 600 700 800 900 1000 The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of I2 re- Determine graphically the ≤H° for the reaction. maining in the aqueous phase. (b) The student now 14.119 (a) Use the van’t Hoff equation in Problem 14.118 to repeats the extraction of I2 with another 0.030 L of derive the following expression, which relates the CCl4. Calculate the fraction of the I2 from the origi- equilibrium constants at two different temperatures nal solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction K1 ¢H° 1 1 ln 5 a 2 b using 0.060 L of CCl4. Comment on the difference. K2 R T2 T1 14.124 Consider the following equilibrium system: How does this equation support the prediction based N2O4 (g) Δ 2NO2 (g) ¢H° 5 58.0 kJ/mol on Le Châtelier’s principle about the shift in equilib- rium with temperature? (b) The vapor pressures of (a) If the volume of the reacting system is changed water are 31.82 mmHg at 30°C and 92.51 mmHg at constant temperature, describe what a plot of P at 50°C. Calculate the molar heat of vaporization versus 1/V would look like for the system. (Hint: of water. See Figure 5.7.) (b) If the temperatures of the react- 14.120 The KP for the reaction ing system is changed at constant pressure, describe what a plot of V versus T would look like for the SO2Cl2 (g) Δ SO2 (g) 1 Cl2 (g) system. (Hint: See Figure 5.9.) is 2.05 at 648 K. A sample of SO2Cl2 is placed in a 14.125 At 1200°C, the equilibrium constant (Kc) for the re- container and heated to 648 K while the total pres- action I2(g) Δ 2I(g) is 2.59 3 1023. Calculate the sure is kept constant at 9.00 atm. Calculate the par- concentrations of I2 and I after the stopcock is tial pressures of the gases at equilibrium. opened and the system reestablishes equilibrium at • 14.121 The “boat” form and “chair” form of cyclohexane the same temperature. (C6H12) interconverts as shown here: k1 1L 2L kⴚ1 0.100 mol I2 Boat Chair 0.0161 mol I  Answers to Practice Exercises 665 Interpreting, Modeling & Estimating 14.126 Estimate the vapor pressure of water at 60°C (see in decomposition of the entire 0.01 mole of XY2(s) Problem 14.119). according to the above reaction. Estimate Kc and KP 14.127 A compound XY2(s) decomposes to form X(g) and for the reaction at 500°C. Y(g) according to the following chemical equation: 14.128 Using the simplified chemical equilibrium given in the Chemistry in Action essay on p. 651, by how XY2 (s) ¡ X(g) 1 2Y(g) much would the concentration of hemoglobin, Hb, A 0.01-mol sample of XY2(s) was placed in a 1-L in a person’s blood need to increase if she moved to vessel, which was sealed and heated to 500°C. The an altitude of 2 km above sea level, in order to give reaction was allowed to reach equilibrium, at which the same concentration of HbO2 as when she was point some XY2(s) remained in the vessel. The ex- living at sea level? periment was repeated, this time using a 2-L vessel, 14.129 The equilibrium constant (KP) for the reaction and again some XY2(s) remained in the vessel after I2 (g) ¡ 2I(g) equilibrium was established. This process was re- –4 peated, each time doubling the volume of the vessel, is 1.8 3 10 at 872 K and 0.048 at 1173 K. From until finally a 16-L vessel was used, at which point these data, estimate the bond enthalpy of I2. (Hint: heating the vessel and its contents to 500°C resulted See van’t Hoff’s equation in Problem 14.119.) Answers to Practice Exercises [NO2]4[O2] P4NO2PO2 14.8 From right to left. 14.9 [HI] 5 0.031 M, [H2] 5 4.3 3 14.1 Kc 5 2 ; KP 5 14.2 2.2 3 102 1023 M, [I2] 5 4.3 3 1023 M 14.10 [Br2] 5 0.065 M, [N2O5] P2N2O5 [Br] 5 8.4 3 1023 M 14.11 QP 5 4.0 3 105; the net 14.3 347 atm 14.4 1.2 reaction will shift from right to left. 14.12 Left to right. [Ni(CO) 4] PNi(CO)4 14.13 The equilibrium will shift from (a) left to right, 14.5 Kc 5 ; KP 5 [CO] 4 P4CO (b) left to right, and (c) right to left. (d) A catalyst has no 14.6 KP 5 0.0702; Kc 5 6.68 3 1025 effect on the equilibrium. 2 [O3]2 [O3]3 14.7 (a) Ka 5 , (b) Kb 5 ; Ka 5 K3b [O2]3 [O2]  CHAPTER 15 Acids and Bases Many organic acids occur in the vegetable kingdom. Lemons, oranges, and tomatoes contain ascorbic acid, also known as vitamin C (C6H8O6), and citric acid (C6H8O7), and rhubarb and spinach contain oxalic acid (H2C2O4). CHAPTER OUTLINE A LOOK AHEAD 15.1 Brønsted Acids and Bases  We start by reviewing and extending Brønsted’s definitions of acids and bases (in Chapter 4) in terms of acid-base conjugate pairs. (15.1) 15.2 The Acid-Base Properties of Water  Next, we examine the acid-base properties of water and define the ion-product constant for the autoionization of water to give H1 and OH2 ions. (15.2) 15.3 pH—A Measure of Acidity  We define pH as a measure of acidity and also introduce the pOH scale. 15.4 Strength of Acids and Bases We see that the acidity of a solution depends on the relative concentrations 15.5 Weak Acids and Acid of H1 and OH2 ions. (15.3) Ionization Constants  Acids and bases can be classified as strong or weak, depending on the extent of their ionization in solution. (15.4) 15.6 Weak Bases and Base Ionization Constants  We learn to calculate the pH of a weak acid solution from its concentration and ionization constant and to perform similar calculations for weak bases. 15.7 The Relationship Between (15.5 and 15.6) the Ionization Constants  We derive an important relationship between the acid and base ionization of Acids and Their constants of a conjugate pair. (15.7) Conjugate Bases  We then study diprotic and polyprotic acids. (15.8) 15.8 Diprotic and Polyprotic Acids  We continue by exploring the relationship between acid strength and molecular structure. (15.9) 15.9 Molecular Structure and  The reactions between salts and water can be studied in terms of acid and base the Strength of Acids ionizations of the individual cations and anions making up the salt. (15.10) 15.10 Acid-Base Properties of Salts  Oxides and hydroxides can be classified as acidic, basic, and ampho- 15.11 Acid-Base Properties of teric. (15.11) Oxides and Hydroxides  The chapter concludes with a discussion of Lewis acids and Lewis bases. A Lewis acid is an electron acceptor and a Lewis base is an electron donor. 15.12 Lewis Acids and Bases (15.12) 666  15.1 Brønsted Acids and Bases 667 S ome of the most important processes in chemical and biological systems are acid-base reactions in aqueous solutions. In this first of two chapters on the properties of acids and bases, we will study the definitions of acids and bases, the pH scale, the ionization of weak acids and weak bases, and the relationship between acid strength and molecular structure. We will also look at oxides that can act as acids and bases. 15.1 Brønsted Acids and Bases In Chapter 4 we defined a Brønsted acid as a substance capable of donating a proton, and a Brønsted base as a substance that can accept a proton. These defini- tions are generally suitable for a discussion of the properties and reactions of acids and bases. An extension of the Brønsted definition of acids and bases is the concept of the conjugate acid-base pair, which can be defined as an acid and its con- Conjugate means “joined together.” jugate base or a base and its conjugate acid. The conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid. Conversely, a conjugate acid results from the addition of a proton to a Brønsted base. Every Brønsted acid has a conjugate base, and every Brønsted base has a conjugate acid. For example, the chloride ion (Cl2) is the conjugate base formed from the acid HCl, and H3O1 (hydronium ion) is the conjugate acid of the base H2O. HCl 1 H2O ¡ H3O1 1 Cl2 Similarly, the ionization of acetic acid can be represented as The proton is always associated with water molecules in aqueous solution. The H3O1 ion is the H SOS H SOS simplest formula of a hydrated A B A B  proton. O HOCOCOOOH  O 34 HOCOCOOS HOOS O   HOOOH O A Q A A Q A H H H H CH3COOH(aq)  H2O(l) 34 CH3COO(aq)  H3O(aq) acid1 base2 base1 acid2 The subscripts 1 and 2 designate the two conjugate acid-base pairs. Thus, the ace- tate ion (CH3COO2) is the conjugate base of CH3COOH. Both the ionization of HCl (see Section 4.3) and the ionization of CH3COOH are examples of Brønsted acid-base reactions. The Brønsted definition also enables us to classify ammonia as a base because of its ability to accept a proton:  H A O HONOH O 34 HONOH  HOOS O   HOOS A A A Q H H H NH3(aq)  H2O(l) 34 NH 4(aq)  OH(aq) base1 acid2 acid1 base2 In this case, NH14 is the conjugate acid of the base NH3, and the hydroxide ion OH2 is the conjugate base of the acid H2O. Note that the atom in the Brønsted base that accepts a H1 ion must have a lone pair. In Example 15.1, we identify the conjugate pairs in an acid-base reaction. 668 Chapter 15 ■ Acids and Bases Example 15.1 Identify the conjugate acid-base pairs in the reaction between ammonia and hydrofluoric acid in aqueous solution NH3 (aq) 1 HF(aq) Δ NH14 (aq) 1 F2 (aq) Strategy Remember that a conjugate base always has one fewer H atom and one more negative charge (or one fewer positive charge) than the formula of the corresponding acid. Solution NH3 has one fewer H atom and one fewer positive charge than NH14. F2 has one fewer H atom and one more negative charge than HF. Therefore, the conjugate acid- Similar problem: 15.5. base pairs are (1) NH14 and NH3 and (2) HF and F2. Practice Exercise Identify the conjugate acid-base pairs for the reaction CN2 1 H2O Δ HCN 1 OH2 Review of Concepts Which of the following does not constitute a conjugate acid-base pair? (a) HNO2–NO22. (b) H2CO3–CO22 1 3 . (c) CH3NH 3–CH3NH2. It is acceptable to represent the proton in aqueous solution either as H1 or as H3O . The formula H1 is less cumbersome in calculations involving hydrogen ion 1 concentrations and in calculations involving equilibrium constants, whereas H3O1 is more useful in a discussion of Brønsted acid-base properties. 15.2 The Acid-Base Properties of Water Water, as we know, is a unique solvent. One of its special properties is its ability to act either as an acid or as a base. Water functions as a base in reactions with acids such as HCl and CH3COOH, and it functions as an acid in reactions with bases such Tap water and water from underground as NH3. Water is a very weak electrolyte and therefore a poor conductor of electricity, sources do conduct electricity because but it does undergo ionization to a small extent: they contain many dissolved ions. H2O(l) Δ H1 (aq) 1 OH2 (aq) This reaction is sometimes called the autoionization of water. To describe the acid- base properties of water in the Brønsted framework, we express its autoionization as follows (also shown in Figure 15.1):  O  HOOS HOOS O 34 HOOOH O O   HOOS A A A Q H H H or H2O  H2O 34 H3O  OH (15.1) acid1 base2 acid2 base1 The acid-base conjugate pairs are (1) H2O (acid) and OH2 (base) and (2) H3O1 (acid) and H2O (base). The Ion Product of Water Recall that in pure water, [H2O] 5 55.5 M In the study of acid-base reactions, the hydrogen ion concentration is key; its value (see p. 583). indicates the acidity or basicity of the solution. Because only a very small fraction of  15.2 The Acid-Base Properties of Water 669 Figure 15.1 Reaction between two water molecules to form  34  hydronium and hydroxide ions. water molecules are ionized, the concentration of water, [H2O], remains virtually unchanged. Therefore, the equilibrium constant for the autoionization of water, accord- ing to Equation (15.1), is Kc 5 [H3O1][OH2] Because we use H1(aq) and H3O1(aq) interchangeably to represent the hydrated pro- ton, the equilibrium constant can also be expressed as Kc 5 [H1][OH2] To indicate that the equilibrium constant refers to the autoionization of water, we replace Kc with Kw Kw 5 [H3O1][OH2] 5 [H1][OH2] (15.2) where Kw is called the ion-product constant, which is the product of the molar con- centrations of H1 and OH2 ions at a particular temperature. In pure water at 25°C, the concentrations of H1 and OH2 ions are equal and If you could randomly remove and examine 10 particles (H2O, H1, or OH2) per second found to be [H1] 5 1.0 3 1027 M and [OH2] 5 1.0 3 1027 M. Thus, from Equation from a liter of water, on average it would (15.2), at 25°C take you two years, working nonstop, to find one H1 ion! Kw 5 (1.0 3 1027 )(1.0 3 1027 ) 5 1.0 3 10214 Whether we have pure water or an aqueous solution of dissolved species, the follow- ing relation always holds at 25°C: Kw 5 [H1][OH2] 5 1.0 3 10214 (15.3) Whenever [H1] 5 [OH2], the aqueous solution is said to be neutral. In an acidic solution there is an excess of H1 ions and [H1] . [OH2]. In a basic solution there is an excess of hydroxide ions, so [H1] , [OH2]. In practice we can change the concentration of either H1 or OH2 ions in solution, but we cannot vary both of them independently. If we adjust the solution so that [H1] 5 1.0 3 1026 M, the OH2 concentration must change to Kw 1.0 3 10214 [OH2] 5 5 5 1.0 3 1028 M [H1] 1.0 3 1026 An application of Equation (15.3) is given in Example 15.2. Example 15.2 The concentration of OH2 ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H1 ions. Strategy We are given the concentration of the OH2 ions and asked to calculate [H1]. The relationship between [H1] and [OH2] in water or an aqueous solution is given by the ion-product of water, Kw [Equation (15.3)]. (Continued) 670 Chapter 15 ■ Acids and Bases Solution Rearranging Equation (15.3), we write Kw 1.0 3 10214 [H1] 5 2 5 5 4.0 3 10212 M [OH ] 0.0025 Check Because [H1] , [OH2], the solution is basic, as we would expect from the Similar problems: 15.15, 15.16. earlier discussion of the reaction of ammonia with water. Practice Exercise Calculate the concentration of OH2 ions in a HCl solution whose hydrogen ion concentration is 1.3 M. Review of Concepts If the H1 ion concentration in an aqueous solution is 0.0010 M, why is it not possible for the OH2 ion concentration to be 1.0 3 10210 M? 15.3 pH—A Measure of Acidity Because the concentrations of H1 and OH2 ions in aqueous solutions are frequently very small numbers and therefore inconvenient to work with, Soren Sorensen† in 1909 proposed a more practical measure called pH. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L): pH 5 2log [H3O1]  or  pH 5 2log [H1] (15.4) Keep in mind that Equation (15.4) is simply a definition designed to give us convenient The pH of concentrated acid solutions can numbers to work with. The negative logarithm gives us a positive number for pH, be negative. For example, the pH of a 2.0 M HCl solution is 20.30. which otherwise would be negative due to the small value of [H1]. Furthermore, the term [H1] in Equation (15.4) pertains only to the numerical part of the expression for hydrogen ion concentration, for we cannot take the logarithm of units. Thus, like the equilibrium constant, the pH of a solution is a dimensionless quantity. Because pH is simply a way to express hydrogen ion concentration, acidic and basic solutions at 25°C can be distinguished by their pH values, as follows: Acidic solutions: [H1] . 1.0 3 1027 M, pH , 7.00 Basic solutions: [H1] , 1.0 3 1027 M, pH . 7.00 Neutral solutions: [H1] 5 1.0 3 1027 M, pH 5 7.00 Notice that pH increases as [H1] decreases. Sometimes we may be given the pH value of a solution and asked to calculate the H1 ion concentration. In that case, we need to take the antilog of Equation (15.4) as follows: [H3O1] 5 102pH  or  [H1] 5 102pH (15.5) Be aware that the definition of pH just shown, and indeed all the calculations involving solution concentrations (expressed either as molarity or molality) discussed in previous chapters, are subject to error because we have implicitly assumed ideal behavior. In reality, ion-pair formation and other types of intermolecular interactions † Soren Peer Lauritz Sorensen (1868–1939). Danish biochemist. Sorensen originally wrote the symbol as pH and called p the “hydrogen ion exponent” (Wasserstoffionexponent); it is the initial letter of Potenz (German), puissance (French), and power (English). It is now customary to write the symbol as pH.  15.3 pH—A Measure of Acidity 671 Figure 15.2 A pH meter is commonly used in the laboratory to determine the pH of a solution. Although many pH meters have scales marked with values from 1 to 14, pH values can, in fact, be less than 1 and greater than 14. may affect the actual concentrations of species in solution. The situation is analogous to the relationships between ideal gas behavior and the behavior of real gases dis- cussed in Chapter 5. Depending on temperature, volume, and amount and type of gas present, the measured gas pressure may differ from that calculated using the ideal gas equation. Similarly, the actual or “effective” concentration of a solute may not be what we think it is, knowing the amount of substance originally dissolved in solution. Just as we have the van der Waals and other equations to reconcile discrepancies between the ideal gas and nonideal gas behavior, we can account for nonideal behavior in solution. One way is to replace the concentration term with activity, which is the effective concentration. Strictly speaking, then, the pH of solution should be defined as pH 5 2log aH1 (15.6) where aH1 is the activity of the H1 ion. As mentioned in Chapter 14 (see p. 627), Table 15.1 for an ideal solution activity is numerically equal to concentration. For real solu- tions, activity usually differs from concentration, sometimes appreciably. Knowing The pHs of Some the solute concentration, there are reliable ways based on thermodynamics for esti- Common Fluids mating its activity, but the details are beyond the scope of this text. Keep in mind, Sample pH Value therefore, that, except for dilute solutions, the measured pH is usually not the same Gastric juice in 1.0–2.0 as that calculated from Equation (15.4) because the concentration of the H1 ion in the stomach molarity is not numerically equal to its activity value. Although we will continue Lemon juice 2.4 to use concentration in our discussion, it is important to know that this approach Vinegar 3.0 will give us only an approximation of the chemical processes that actually take place in the solution phase. Grapefruit juice 3.2 In the laboratory, the pH of a solution is measured with a pH meter (Figure 15.2). Orange juice 3.5 Table 15.1 lists the pHs of a number of common fluids. As you can see, the pH of Urine 4.8–7.5 body fluids varies greatly, depending on location and function. The low pH (high Water exposed 5.5 acidity) of gastric juices facilitates digestion whereas a higher pH of blood is neces- to air* sary for the transport of oxygen. These pH-dependent actions will be illustrated in Saliva 6.4–6.9 Chemistry in Action essays in this chapter and Chapter 16. Milk 6.5 A pOH scale analogous to the pH scale can be devised using the negative loga- Pure water 7.0 rithm of the hydroxide ion concentration of a solution. Thus, we define pOH as Blood 7.35–7.45 Tears 7.4 pOH 5 2log [OH2] (15.7) Milk of 10.6 magnesia If we are given the pOH value of a solution and asked to calculate the OH2 ion Household 11.5 concentration, we can take the antilog of Equation (15.7) as follows ammonia *Water exposed to air for a long period [OH2] 5 102pOH (15.8) of time absorbs atmospheric CO2 to form carbonic acid, H2CO3. 672 Chapter 15 ■ Acids and Bases Now consider again the ion-product constant for water at 25°C: [H1][OH2] 5 Kw 5 1.0 3 10214 Taking the negative logarithm of both sides, we obtain 2(log [H1] 1 log [OH2]) 5 2log (1.0 3 10214 ) 2log [H1] 2 log [OH2] 5 14.00 From the definitions of pH and pOH we obtain pH 1 pOH 5 14.00 (15.9) Equation (15.9) provides us with another way to express the relationship between the H1 ion concentration and the OH2 ion concentration. Examples 15.3, 15.4, and 15.5 illustrate calculations involving pH. Example 15.3 The concentration of H1 ions in a bottle of table wine was 3.2 3 1024 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 3 1023 M. Calculate the pH of the wine on these two occasions. Strategy We are given the H1 ion concentration and asked to calculate the pH of the solution. What is the definition of pH? Solution According to Equation (15.4), pH 5 2log [H1]. When the bottle was first opened, [H1] 5 3.2 3 1024 M, which we substitute in Equation (15.4) pH 5 2log [H1] 5 2log (3.2 3 1024 ) 5 3.49 In each case the pH has only two significant On the second occasion, [H1] 5 1.0 3 1023 M, so that figures. The two digits to the right of the decimal in 3.49 tell us that there are two pH 5 2log (1.0 3 1023 ) 5 3.00 significant figures in the original number (see Appendix 4). Comment The increase in hydrogen ion concentration (or decrease in pH) is largely the result of the conversion of some of the alcohol (ethanol) to acetic acid, a reaction Similar problems: 15.17, 15.18. that takes place in the presence of molecular oxygen. Practice Exercise Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.76 M. Example 15.4 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. Calculate the H1 ion concentration of the rainwater. Strategy Here we are given the pH of a solution and asked to calculate [H1]. Because pH is defined as pH 5 2log [H1], we can solve for [H1] by taking the antilog of the pH; that is, [H1] 5 102pH, as shown in Equation (15.5). Solution From Equation (15.4) pH 5 2log [H1] 5 4.82 (Continued)  15.4 Strength of Acids and Bases 673 Therefore, log [H1] 5 24.82 To calculate [H1], we need to take the antilog of 24.82 Scientific calculators have an antilog function that is sometimes labeled 1 24.82 25 [H ] 5 10 5 1.5 3 10 M INV log or 10x. Check Because the pH is between 4 and 5, we can expect [H1] to be between 1 3 1024 M and 1 3 1025 M. Therefore, the answer is reasonable. Similar problem: 15.19. Practice Exercise The pH of a certain orange juice is 3.33. Calculate the H1 ion concentration. Example 15.5 In a NaOH solution [OH2] is 2.9 3 1024 M. Calculate the pH of the solution. Strategy Solving this problem takes two steps. First, we need to calculate pOH using Equation (15.7). Next, we use Equation (15.9) to calculate the pH of the solution. Solution We use Equation (15.7): pOH 5 2log [OH2] 5 2log (2.9 3 1024 ) 5 3.54 Now we use Equation (15.9): pH 1 pOH 5 14.00 pH 5 14.00 2 pOH 5 14.00 2 3.54 5 10.46 Alternatively, we can use the ion-product constant of water, Kw 5 [H1][OH2] to calculate [H1], and then we can calculate the pH from the [H1]. Try it. Check The answer shows that the solution is basic (pH . 7), which is consistent with a NaOH solution. Similar problem: 15.18. 2 27 Practice Exercise The OH ion concentration of a blood sample is 2.5 3 10 M. What is the pH of the blood? Review of Concepts Which is more acidic: a solution where [H1] 5 2.5 3 1023 M or a solution with a pOH 5 11.6? 15.4 Strength of Acids and Bases Strong acids are strong electrolytes that, for practical purposes, are assumed to In reality, no acids are known to ionize completely in water. ionize completely in water (Figure 15.3). Most of the strong acids are inorganic acids: hydrochloric acid (HCl), nitric acid (HNO3), perchloric acid (HClO4), and Animation The Dissociation of Strong and Weak Acids sulfuric acid (H2SO4): HCl(aq) 1 H2O(l) ¡ H3O1 (aq) 1 Cl2 (aq) HNO3 (aq) 1 H2O(l) ¡ H3O1 (aq) 1 NO23 (aq) HClO4 (aq) 1 H2O(l) ¡ H3O1 (aq) 1 ClO24 (aq) H2SO4 (aq) 1 H2O(l) ¡ H3O1 (aq) 1 HSO24 (aq) 674 Chapter 15 ■ Acids and Bases Before At Before At Ionization Equilibrium Ionization Equilibrium HCl H+ Cl– HF HF H+ F – Figure 15.3 The extent of ionization of a strong acid such as Cl– H2O HCl (left) and a weak acid such as HF (right). Initially, there were HF H3O+ 6 HCl and 6 HF molecules present. The strong acid is assumed to F– be completely ionized in solution. The proton exists in solution as the hydronium ion (H3O1). Note that H2SO4 is a diprotic acid; we show only the first stage of ionization here. At equilibrium, solutions of strong acids will not contain any nonionized acid molecules. Most acids are weak acids, which ionize only to a limited extent in water. At equilibrium, aqueous solutions of weak acids contain a mixture of nonionized acid molecules, H3O1 ions, and the conjugate base. Examples of weak acids are hydro- fluoric acid (HF), acetic acid (CH3COOH), and the ammonium ion (NH1 4 ). The limited ionization of weak acids is related to the equilibrium constant for ionization, which we will study in the next section. Like strong acids, strong bases are strong electrolytes that ionize completely in water. Hydroxides of alkali metals and certain alkaline earth metals are strong bases. [All alkali metal hydroxides are soluble. Of the alkaline earth hydroxides, Be(OH)2 and Mg(OH)2 are insoluble; Ca(OH)2 and Sr(OH)2 are slightly soluble; and Ba(OH)2 is soluble.] Some examples of strong bases are HO NaOH(s) ¡ 2 Na1 (aq) 1 OH2 (aq) H2O KOH(s) ¡ K1 (aq) 1 OH2 (aq) H2O Ba(OH) 2 (s) ¡ Ba21 (aq) 1 2OH2 (aq) Zn reacts more vigorously with a Strictly speaking, these metal hydroxides are not Brønsted bases because they cannot strong acid like HCl (left) than with accept a proton. However, the hydroxide ion (OH2) formed when they ionize is a a weak acid like CH3COOH (right) of the same concentration Brønsted base because it can accept a proton: because there are more H1 ions in the former solution. H3O1 (aq) 1 OH2 (aq) ¡ 2H2O(l)  15.4 Strength of Acids and Bases 675 Table 15.2 Relative Strengths of Conjugate Acid-Base Pairs Acid Conjugate Base ⎧ HClO4 (perchloric acid) ClO24 (perchlorate ion)  ↑     HI (hydroiodic acid) I2 (iodide ion)  Strong acids     HBr (hydrobromic acid) Br2 (bromide ion)   ⎨    HCl (hydrochloric acid) Cl2 (chloride ion)       H2SO4 (sulfuric acid) HSO24 (hydrogen sulfate ion)  Base strength increases Acid strength increases  ⎩ HNO3 (nitric acid) NO23 (nitrate ion)     H3O1 (hydronium ion) H2O (water)     ⎧ HSO24 (hydrogen sulfate ion) SO22 4 (sulfate ion)  HF (hydrofluoric acid)   2 F (fluoride ion)       HNO2 (nitrous acid) NO22 (nitrite ion)    Weak acids HCOO2 (formate ion)    HCOOH (formic acid)   ⎨ CH3COOH (acetic acid) CH3COO2 (acetate ion)      1   NH4 (ammonium ion) NH3 (ammonia)   HCN (hydrocyanic acid) CN2 (cyanide ion)        H2O (water) OH2 (hydroxide ion)  ↑  ⎩ NH3 (ammonia) NH2 2 (amide ion) Thus, when we call NaOH or any other metal hydroxide a base, we are actually refer- ring to the OH2 species derived from the hydroxide. Weak bases, like weak acids, are weak electrolytes. Ammonia is a weak base. It ionizes to a very limited extent in water: NH3 (aq) 1 H2O(l) Δ NH14 (aq) 1 OH2 (aq) Note that, unlike acids, NH3 does not donate a proton to water. Rather, NH3 behaves as a base by accepting a proton from water to form NH1 2 4 and OH ions. Table 15.2 lists some important conjugate acid-base pairs, in order of their relative strengths. Conjugate acid-base pairs have the following properties: 1. If an acid is strong, its conjugate base has no measurable strength. Thus, the Cl2 ion, which is the conjugate base of the strong acid HCl, is an extremely weak base. 2. H3O1 is the strongest acid that can exist in aqueous solution. Acids stronger than H3O1 react with water to produce H3O1 and their conjugate bases. Thus, HCl, which is a stronger acid than H3O1, reacts with water completely to form H3O1 and Cl2: HCl(aq) 1 H2O(l) ¡ H3O1 (aq) 1 Cl2 (aq) Acids weaker than H3O1 react with water to a much smaller extent, producing H3O1 and their conjugate bases. For example, the following equilibrium lies primarily to the left: HF(aq) 1 H2O(l) Δ H3O1 (aq) 1 F2 (aq) 3. The OH2 ion is the strongest base that can exist in aqueous solution. Bases stronger than OH2 react with water to produce OH2 and their conjugate acids. 676 Chapter 15 ■ Acids and Bases For example, the oxide ion (O22) is a stronger base than OH2, so it reacts with water completely as follows: O22 (aq) 1 H2O(l) ¡ 2OH2 (aq) For this reason the oxide ion does not exist in aqueous solutions. Example 15.6 shows calculations of pH for a solution containing a strong acid and a solution of a strong base. Example 15.6 Calculate the pH of (a) a 1.0 3 1023 M HCl solution and (b) a 0.020 M Ba(OH)2 solution. Strategy Keep in mind that HCl is a strong acid and Ba(OH)2 is a strong base. Thus, these species are completely ionized and no HCl or Ba(OH)2 will be left in solution. Solution (a) The ionization of HCl is 1 1 Recall that H (aq) is the same as H3O (aq). HCl(aq) ¡ H1 (aq) 1 Cl2 (aq) The concentrations of all the species (HCl, H1, and Cl2) before and after ionization can be represented as follows: We use the ICE method for solving HCl(aq) ¡ H⫹(aq) ⫹ Cl⫺(aq) equilibrium concentrations as shown Initial (M): 1.0 ⫻ 10⫺3 0.0 0.0 in Section 14.4 (p. 641). Change (M): ⫺1.0 ⫻ 10⫺3 ⫹1.0 ⫻ 10⫺3 ⫹1.0 ⫻ 10⫺3 Final (M): 0.0 1.0 ⫻ 10⫺3 1.0 ⫻ 10⫺3 A positive (1) change represents an increase and a negative (2) change indicates a decrease in concentration. Thus, [H1] 5 1.0 3 1023 M pH 5 2log (1.0 3 1023 ) 5 3.00 (b) Ba(OH)2 is a strong base; each Ba(OH)2 unit produces two OH2 ions: Ba(OH) 2 (aq) Δ Ba21 (aq) 1 2OH2 (aq) The changes in the concentrations of all the species can be represented as follows: Ba(OH)2(aq) ¡ Ba2⫹(aq) ⫹ 2OH⫺(aq) Initial (M): 0.020 0.00 0.00 Change (M): ⫺0.020 ⫹0.020 ⫹2(0.020) Final (M): 0.00 0.020 0.040 Thus, [OH2] 5 0.040 M pOH 5 2log 0.040 5 1.40 Therefore, from Equation (15.8), pH 5 14.00 2 pOH 5 14.00 2 1.40 5 12.60 (Continued)  15.5 Weak Acids and Acid Ionization Constants 677 Check Note that in both (a) and (b) we have neglected the contribution of the autoionization of water to [H1] and [OH2] because 1.0 3 1027 M is so small compared with 1.0 3 1023 M and 0.040 M. Similar problem: 15.18. Practice Exercise Calculate the pH of a 1.8 3 1022 M Ba(OH)2 solution. If we know the relative strengths of two acids, we can predict the position of equilibrium between one of the acids and the conjugate base of the other, as illustrated in Example 15.7. Example 15.7 Predict the direction of the following reaction in aqueous solution: HNO2 (aq) 1 CN2 (aq) Δ HCN(aq) 1 NO22 (aq) Strategy The problem is to determine whether, at equilibrium, the reaction will be shifted to the right, favoring HCN and NO22 , or to the left, favoring HNO2 and CN2. Which of the two is a stronger acid and hence a stronger proton donor: HNO2 or HCN? Which of the two is a stronger base and hence a stronger proton acceptor: CN2 or NO22 ? Remember that the stronger the acid, the weaker its conjugate base. Solution In Table 15.2 we see that HNO2 is a stronger acid than HCN. Thus, CN2 is a stronger base than NO2 2 . The net reaction will proceed from left to right as written because HNO2 is a better proton donor than HCN (and CN2 is a better proton acceptor than NO22 ). Similar problem: 15.37. Practice Exercise Predict whether the equilibrium constant for the following reaction is greater than or smaller than 1: CH3COOH(aq) 1 HCOO2 (aq) Δ CH3COO2 (aq) 1 HCOOH(aq) Review of Concepts (a) List in order of decreasing concentration of all the ionic and molecular species in the following acid solutions: (i) HNO3 and (ii) HF. (b) List in order of decreasing concentration of all the ionic and molecular species in the following base solutions: (i) NH3 and (ii) KOH. 15.5 Weak Acids and Acid Ionization Constants As we have seen, there are relatively few strong acids. The vast majority of acids are weak acids. Consider a weak monoprotic acid, HA. Its ionization in water is repre- sented by HA(aq) 1 H2O(l) Δ H3O1 (aq) 1 A2 (aq) or simply HA(aq) Δ H1 (aq) 1 A2 (aq) 678 Chapter 15 ■ Acids and Bases The equilibrium expression for this ionization is [H3O1][A2] [H1][A2] All concentrations in this equation are Ka 5   or  Ka 5 (15.10) equilibrium concentrations. [HA] [HA] Animation where Ka, the acid ionization constant, is the equilibrium constant for the ionization Acid Ionization of an acid. At a given temperature, the strength of the acid HA is measured quanti- tatively by the magnitude of Ka. The larger Ka, the stronger the acid—that is, the greater the concentration of H1 ions at equilibrium due to its ionization. Keep in mind, however, that only weak acids have Ka values associated with them. The back endpaper gives an index to all Table 15.3 lists a number of weak acids and their Ka values at 25°C in order of the useful tables and figures in this text. decreasing acid strength. Although all these acids are weak, within the group there is great variation in their strengths. For example, Ka for HF (7.1 3 1024) is about 1.5 million times that for HCN (4.9 3 10210). Generally, we can calculate the hydrogen ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its K a value. Alternatively, if we know the pH of a weak acid solution and its initial concentration, we can determine its Ka. The basic approach for solving these Table 15.3 Ionization Constants of Some Weak Acids and Their Conjugate Bases at 258C Name of Acid Formula Structure Ka Conjugate Base K b† 24 2 Hydrofluoric acid HF H¬F 7.1 3 10 F 1.4 3 10211 Nitrous acid HNO2 O“N¬O¬H 4.5 3 1024 NO22 2.2 3 10211 Acetylsalicylic acid C9H8O4 O 3.0 3 1024 C9H7O2 4 3.3 3 10211 (aspirin) B OCOOOH OOOCOCH3 B O Formic acid HCOOH O 1.7 3 1024 HCOO2 5.9 3 10211 B HOCOOOH Ascorbic acid* C6H8O6 HOOH EOH 8.0 3 1025 C6H7O2 6 1.3 3 10210 CP P PC H G C CPO D CHOH O A CH2OH Benzoic acid C6H5COOH O 6.5 3 1025 C6H5COO2 1.5 3 10210 B OCOOOH Acetic acid CH3COOH O 1.8 3 1025 CH3COO2 5.6 3 10210 B CH3OCOOOH Hydrocyanic acid HCN H¬C‚N 4.9 3 10210 CN2 2.0 3 1025 Phenol C6H5OH 1.3 3 10210 C6H5O2 7.7 3 1025 OOOH *For ascorbic acid it is the upper left hydroxyl group that is associated with this ionization constant. † The base ionization constant Kb is discussed in Section 15.6.  15.5 Weak Acids and Acid Ionization Constants 679 problems, which deal with equilibrium concentrations, is the same one outlined in Chapter 14. However, because acid ionization represents a major category of chemical equilibrium in aqueous solution, we will develop a systematic procedure for solving this type of problem that will also help us to understand the chem- istry involved. Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25°C. The ionization of HF is given by HF(aq) Δ H1 (aq) 1 F2 (aq) From Table 15.3 we write [H1][F2] Ka 5 5 7.1 3 1024 [HF] The first step is to identify all the species present in solution that may affect its pH. Because weak acids ionize to a small extent, at equilibrium the major spe- cies present are nonionized HF and some H1and F2 ions. Another major species is H2O, but its very small Kw (1.0 3 10214) means that water is not a significant contributor to the H1 ion concentration. Therefore, unless otherwise stated, we will always ignore the H1 ions produced by the autoionization of water. Note that we need not be concerned with the OH2 ions that are also present in solution. The OH2 concentration can be determined from Equation (15.3) after we have calcu- lated [H1]. We can summarize the changes in the concentrations of HF, H1, and F2 accord- ing to the steps shown on p. 641 as follows: HF(aq) Δ H1(aq) 1 F2(aq) Initial (M): 0.50 0.00 0.00 Change (M): 2x 1x 1x Equilibrium (M): 0.50 2 x x x The equilibrium concentrations of HF, H1, and F2, expressed in terms of the unknown x, are substituted into the ionization constant expression to give (x)(x) Ka 5 5 7.1 3 1024 0.50 2 x Rearranging this expression, we write x2 1 7.1 3 1024x 2 3.6 3 1024 5 0 This is a quadratic equation which can be solved using the quadratic formula (see Appendix 4). Or we can try using a shortcut to solve for x. Because HF is a weak acid and weak acids ionize only to a slight extent, we reason that x must be small compared to 0.50. Therefore, we can make the approximation 0.50 2 x < 0.50 The sign ¯ means “approximately equal to.” An analogy of the approximation is a truck loaded with coal. Losing a few Now the ionization constant expression becomes lumps of coal on a delivery trip will not appreciably change the overall mass of the load. x2 x2 < 5 7.1 3 1024 0.50 2 x 0.50 680 Chapter 15 ■ Acids and Bases Rearranging, we get x2 5 (0.50)(7.1 3 1024 ) 5 3.55 3 1024 x 5 23.55 3 1024 5 0.019 M Thus, we have solved for x without having to use the quadratic equation. At equilibrium, we have [HF] 5 (0.50 2 0.019) M 5 0.48 M [H1] 5 0.019 M [F2] 5 0.019 M and the pH of the solution is pH 5 2log (0.019) 5 1.72 How good is this approximation? Because Ka values for weak acids are generally known to an accuracy of only 65%, it is reasonable to require x to be less than 5% of 0.50, the number from which it is subtracted. In other words, the approximation is valid if the following expression is equal to or less than 5%: 0.019 M 3 100% 5 3.8% 0.50 M Thus, the approximation we made is acceptable. Now consider a different situation. If the initial concentration of HF is 0.050 M, and we use the above procedure to solve for x, we would get 6.0 3 1023 M. However, the following test shows that this answer is not a valid approximation because it is greater than 5 percent of 0.050 M: 6.0 3 1023 M 3 100% 5 12% 0.050 M In this case, we can get an accurate value for x by solving the quadratic equation. The Quadratic Equation We start by writing the ionization expression in terms of the unknown x: x2 5 7.1 3 1024 0.050 2 x x2 1 7.1 3 1024x 2 3.6 3 1025 5 0 This expression fits the quadratic equation ax2 1 bx 1 c 5 0. Using the quadratic formula, we write 2b 6 2b2 2 4ac x5 2a 27.1 3 10 24 6 2(7.1 3 10 24 ) 2 2 4(1)(23.6 3 10 25 ) 5 2112 27.1 3 1024 6 0.012 5 2 5 5.6 3 10 23 M  or  26.4 3 10 23 M  15.5 Weak Acids and Acid Ionization Constants 681 The second solution (x 5 26.4 3 1023 M) is physically impossible because the concentration of ions produced as a result of ionization cannot be negative. Choosing x 5 5.6 3 1023 M, we can solve for [HF], [H1], and [F2] as follows: [HF] 5 (0.050 2 5.6 3 1023 ) M 5 0.044 M [H1] 5 5.6 3 1023 M [F2] 5 5.6 3 1023 M The pH of the solution, then, is pH 5 2log (5.6 3 1023 ) 5 2.25 In summary, the main steps for solving weak acid ionization problems are: 1. Identify the major species that can affect the pH of the solution. In most cases we can ignore the ionization of water. We omit the hydroxide ion because its concentration is determined by that of the H1 ion. 2. Express the equilibrium concentrations of these species in terms of the initial concentration of the acid and a single unknown x, which represents the change in concentration. 3. Write the weak acid ionization and express the ionization constant Ka in terms of the equilibrium concentrations of H1, the conjugate base, and the unionized acid. First solve for x by the approximate method. If the approximate method is not valid, use the quadratic equation to solve for x. 4. Having solved for x, calculate the equilibrium concentrations of all species and/or the pH of the solution. Example 15.8 provides another illustration of this procedure. Example 15.8 Calculate the pH of a 0.036 M nitrous acid (HNO2) solution: HNO2 (aq) Δ H1 (aq) 1 NO2 2 (aq) HNO2 Strategy Recall that a weak acid only partially ionizes in water. We are given the initial concentration of a weak acid and asked to calculate the pH of the solution at equilibrium. It is helpful to make a sketch to keep track of the pertinent species. As in Example 15.6, we ignore the ionization of H2O so the major source of H1 ions is the acid. The concentration of OH2 ions is very small as we would expect from an acidic solution so it is present as a minor species. (Continued) 682 Chapter 15 ■ Acids and Bases Solution We follow the procedure already outlined. Step 1: The species that can affect the pH of the solution are HNO2, H1, and the conjugate base NO22 . We ignore water’s contribution to [H1]. Step 2: Letting x be the equilibrium concentration of H1 and NO22 ions in mol/L, we summarize: HNO2(aq) 34 H (aq) NO2 (aq) Initial (M): 0.036 0.00 0.00 Change (M): x x x Equilibrium (M): 0.036 x x x Step 3: From Table 15.3 we write [H1][NO22] Ka 5 [HNO2] x2 4.5 3 1024 5 0.036 2 x Applying the approximation 0.036 2 x < 0.036, we obtain x2 x2 4.5 3 1024 5 < 0.036 2 x 0.036 x2 5 1.62 3 1025 x 5 4.0 3 1023 M To test the approximation, 4.0 3 1023 M 3 100% 5 11% 0.036 M Because this is greater than 5 percent, our approximation is not valid and we must solve the quadratic equation, as follows: x2 1 4.5 3 1024x 2 1.62 3 1025 5 0 24.5 3 1024 6 2(4.5 3 1024 ) 2 2 4(1) (21.62 3 1025 ) x5 2(1) 5 3.8 3 1023 M  or  24.3 3 1023 M The second solution is physically impossible, because the concentration of ions produced as a result of ionization cannot be negative. Therefore, the solution is given by the positive root, x 5 3.8 3 1023 M. Step 4: At equilibrium [H1] 5 3.8 3 1023 M pH 5 2log (3.8 3 1023 ) 5 2.42 Check Note that the calculated pH indicates that the solution is acidic, which is what we would expect for a weak acid solution. Compare the calculated pH with that of a 0.036 M strong acid solution such as HCl to convince yourself of the difference Similar problem: 15.43. between a strong acid and a weak acid. Practice Exercise What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 3 1024?  15.5 Weak Acids and Acid Ionization Constants 683 One way to determine Ka of an acid is to measure the pH of the acid solution of known concentration at equilibrium. Example 15.9 shows this approach. Example 15.9 The pH of a 0.10 M solution of formic acid (HCOOH) is 2.39. What is the Ka of the acid? Strategy Formic acid is a weak acid. It only partially ionizes in water. Note that the concentration of formic acid refers to the initial concentration, before ionization has started. The pH of the solution, on the other hand, refers to the equilibrium state. To HCOOH calculate Ka, then, we need to know the concentrations of all three species: [H1], [HCOO2], and [HCOOH] at equilibrium. As usual, we ignore the ionization of water. The following sketch summarizes the situation. Solution We proceed as follows. Step 1: The major species in solution are HCOOH, H1, and the conjugate base HCOO2. Step 2: First we need to calculate the hydrogen ion concentration from the pH value pH 5 2log [H1] 2.39 5 2log [H1] Taking the antilog of both sides, we get [H1] 5 1022.39 5 4.1 3 1023 M Next we summarize the changes: HCOOH(aq) Δ H1(aq) 1 HCOO2(aq) Initial (M ): 0.10 0.00 0.00 Change (M ): 24.1 3 1023 14.1 3 1023 14.1 3 1023 Equilibrium (M ): (0.10 2 4.1 3 1023) 4.1 3 1023 4.1 3 1023 Note that because the pH and hence the H1 ion concentration is known, it follows that we also know the concentrations of HCOOH and HCOO2 at equilibrium. Step 3: The ionization constant of formic acid is given by [H1][HCOO2] Ka 5 [HCOOH] (4.1 3 1023 ) (4.1 3 1023 ) 5 (0.10 2 4.1 3 1023 ) 5 1.8 3 1024 (Continued)  684 Chapter 15 ■ Acids and Bases Check The Ka value differs slightly from the one listed in Table 15.3 because of the Similar problem: 15.45. rounding-off procedure we used in the calculation. Practice Exercise The pH of a 0.060 M weak monoprotic acid is 3.44. Calculate the Ka of the acid. Percent Ionization We have seen that the magnitude of Ka indicates the strength of an acid. Another measure of the strength of an acid is its percent ionization, which is defined as ionized acid concentration at equilibrium We can compare the strengths of acids percent ionization 5 3 100% (15.11) in terms of percent ionization only if initial concentration of acid concentrations of the acids are the same. The stronger the acid, the greater the percent ionization. For a monoprotic acid HA, the concentration of the acid that undergoes ionization is equal to the concentration of the H1 ions or the concentration of the A2 ions at equilibrium. Therefore, we can write the percent ionization as [H1] percent ionization 5 3 100% [HA]0 where [H1] is the concentration at equilibrium and [HA]0 is the initial concentration. Referring to Example 15.8, we see that the percent ionization of a 0.036 M HNO2 solution is 3.8 3 1023 M percent ionization 5 3 100% 5 11% 0.036 M Thus, only about one out of every nine HNO2 molecules has ionized. This is consis- tent with the fact that HNO2 is a weak acid. The extent to which a weak acid ionizes depends on the initial concentration of the acid. The more dilute the solution, the greater the percentage ionization (Figure 15.4). In qualitative terms, when an acid is diluted, the concentration of the “particles” in the solution is reduced. According to Le Châtelier’s principle (see Section 14.5), this reduction in particle concentration (the stress) is counteracted by shifting the reaction to the side with more particles; that is, the equilibrium shifts from the nonionized acid side (one particle) to the side containing H1 ions and the conjugate base (two particles): HA Δ H1 1 A2 . Consequently, the concentra- tion of “particles” increases in the solution. The dependence of percent ionization on initial concentration can be illustrated Strong acid by the HF case discussed on page 680: 100 0.50 M HF % Ionization 0.019 M percent ionization 5 3 100% 5 3.8% Weak acid 0.50 M 0 0.050 M HF Initial concentration of acid Figure 15.4 Dependence of 5.6 3 1023 M percent ionization on initial percent ionization 5 3 100% 5 11% 0.050 M concentration of acid. Note that at very low concentrations, all acids (weak and strong) are We see that, as expected, a more dilute HF solution has a greater percent ionization almost completely ionized. of the acid.  15.6 Weak Bases and Base Ionization Constants 685 Review of Concepts The “concentration” of water is 55.5 M. Calculate the percent ionization of water. 15.6 Weak Bases and Base Ionization Constants The ionization of weak bases is treated in the same way as the ionization of weak acids. When ammonia dissolves in water, it undergoes the reaction The lone pair (red color) on the NH3 (aq) 1 H2O(l) Δ NH14 (aq) 1 OH2 (aq) N atom accounts for ammonia’s basicity. The equilibrium constant is given by [NH14 ][OH2] K5 [NH3][H2O] Compared with the total concentration of water, very few water molecules are consumed by this reaction, so we can treat [H2O] as a constant. Thus, we can write the base ion- Animation Base Ionization ization constant (Kb), which is the equilibrium constant for the ionization reaction, as [NH14 ][OH2] Kb 5 K[H2O] 5 [NH3] 5 1.8 3 1025 Table 15.4 lists a number of common weak bases and their ionization constants. Note that the basicity of all these compounds is attributable to the lone pair of electrons on the nitrogen atom. The ability of the lone pair to accept a H1 ion makes these substances Brønsted bases. In solving problems involving weak bases, we follow the same procedure we used for weak acids. The main difference is that we calculate [OH2] first, rather than [H1]. Example 15.10 shows this approach. Example 15.10 What is the pH of a 0.40 M ammonia solution? Strategy The procedure here is similar to the one used for a weak acid (see Example 15.8). From the ionization of ammonia, we see that the major species in solution at equilibrium are NH3, NH14, and OH2. The hydrogen ion concentration is very small as we would expect from a basic solution, so it is present as a minor species. As before, we ignore the ionization of water. We make a sketch to keep track of the pertinent species as follows: (Continued) 686 Chapter 15 ■ Acids and Bases Table 15.4 Ionization Constants of Some Weak Bases and Their Conjugate Acids at 25°C Conjugate Name of Base Formula Structure Kb* Acid Ka 1 Ethylamine C2H5NH2 O CH3OCH2ONOH 5.6 3 10 24 C2H5NH3 1.8 3 10211 A H 1 Methylamine CH3NH2 O CH3ONOH 4.4 3 1024 CH3NH3 2.3 3 10211 A H Ammonia NH3 O HONOH 1.8 3 1025 NH14 5.6 3 10210 A H 1 Pyridine C5H5N 1.7 3 1029 C5H5NH 5.9 3 1026 NS 1 Aniline C6H5NH2 3.8 3 10210 C6H5NH3 2.6 3 1025 O ONOH A H 1 Caffeine C8H10N4O2 O 5.3 3 10214 C8H11N4O2 0.19 B H3C ECH3 H ECH N N C A B COH CH EC K N N Q O A CH3 1 Urea (NH2)2CO O 1.5 3 10214 H2NCONH3 0.67 B O O HONOCONOH A A H H *The nitrogen atom with the lone pair accounts for each compound’s basicity. In the case of urea, Kb can be associated with either nitrogen atom. Solution We proceed according to the following steps. Step 1: The major species in an ammonia solution are NH3, NH14, and OH2. We ignore the very small contribution to OH2 concentration by water. Step 2: Letting x be the equilibrium concentration of NH14 and OH2 ions in mol/L, we summarize: NH3(aq) H2O(l) 34 NH 4(aq) OH (aq) Initial (M): 0.40 0.00 0.00 Change (M): x x x Equilibrium (M): 0.40 x x x Step 3: Table 15.4 gives us Kb: [NH1 2 4 ][OH ] Kb 5 [NH3] 25 x2 1.8 3 10 5 0.40 2 x (Continued)  15.7 The Relationship Between the Ionization Constants of Acids and Their Conjugate Bases 687 Applying the approximation 0.40 2 x < 0.40, we obtain x2 x2 1.8 3 1025 5 < 0.40 2 x 0.40 x2 5 7.2 3 1026 x 5 2.7 3 1023 M To test the approximation, we write The 5 percent rule (p. 680) also applies to bases. 23 2.7 3 10 M 3 100% 5 0.68% 0.40 M Therefore, the approximation is valid. Step 4: At equilibrium, [OH2] 5 2.7 3 1023 M. Thus, pOH 5 2log (2.7 3 1023 ) 5 2.57 pH 5 14.00 2 2.57 5 11.43 Check Note that the pH calculated is basic, which is what we would expect from a weak base solution. Compare the calculated pH with that of a 0.40 M strong base solution, such as KOH, to convince yourself of the difference between a strong base and a weak base. Similar problem: 15.55. Practice Exercise Calculate the pH of a 0.26 M methylamine solution (see Table 15.4). Review of Concepts Consider the following three solutions of equal concentration. Using data in Table 15.4, rank the three solutions from most basic to least basic: (a) aniline, (b) methylamine, (c) caffeine. 15.7 The Relationship Between the Ionization Constants of Acids and Their Conjugate Bases An important relationship between the acid ionization constant and the ionization con- stant of its conjugate base can be derived as follows, using acetic acid as an example: CH3COOH(aq) Δ H1 (aq) 1 CH3COO2 (aq) [H1][CH3COO2] Ka 5 [CH3COOH] The conjugate base, CH3COO2, supplied by a sodium acetate (CH3COONa) solution, reacts with water according to the equation CH3COO2 (aq) 1 H2O(l) Δ CH3COOH(aq) 1 OH2 (aq) and we can write the base ionization constant as [CH3COOH][OH2] Kb 5 [CH3COO2] The product of these two ionization constants is given by [H1][CH3COO2] [CH3COOH][OH2] KaKb 5 3 [CH3COOH] [CH3COO2] 1 2 5 [H ][OH ] 5 Kw 688 Chapter 15 ■ Acids and Bases This result may seem strange at first, but if we add the two equations we see that the sum is simply the autoionization of water. (1) CH3COOH(aq) Δ H1 (aq) 1 CH3COO2 (aq) Ka (2) CH3COO (aq) 1 H2O(l) Δ CH3COOH(aq) 1 OH2 (aq) 2 Kb (3) H2O(l) Δ H1 (aq) 1 OH2 (aq) Kw This example illustrates one of the rules for chemical equilibria: When two reac- tions are added to give a third reaction, the equilibrium constant for the third reaction is the product of the equilibrium constants for the two added reactions (see Section 14.2). Thus, for any conjugate acid-base pair it is always true that KaKb 5 Kw (15.12) Expressing Equation (15.12) as Kw Kw Ka 5     Kb 5 Kb Ka enables us to draw an important conclusion: The stronger the acid (the larger Ka), the weaker its conjugate base (the smaller Kb), and vice versa (see Tables 15.3 and 15.4). We can use Equation (15.12) to calculate the Kb of the conjugate base (CH3COO2) of CH3COOH as follows. We find the Ka value of CH3COOH in Table 15.3 and write Kw Kb 5 Ka 1.0 3 10214 5 1.8 3 1025 5 5.6 3 10210 Review of Concepts Consider the following two acids and their ionization constants: HCOOH  Ka 5 1.7 3 1024 HCN  Ka 5 4.9 3 10210 Which conjugate base (HCOO2 or CN2) is stronger? 15.8 Diprotic and Polyprotic Acids The treatment of diprotic and polyprotic acids is more involved than that of mono- protic acids because these substances may yield more than one hydrogen ion per molecule. These acids ionize in a stepwise manner; that is, they lose one proton at a time. An ionization constant expression can be written for each ionization stage. Consequently, two or more equilibrium constant expressions must often be used to calculate the concentrations of species in the acid solution. For example, for carbonic acid, H2CO3, we write [H1][HCO23 ] H2CO3 (aq) Δ H1 (aq) 1 HCO23 (aq)     Ka1 5 [H2CO3] [H1][CO22 3 ] Top to bottom: H2CO3, HCO23, HCO2 1 22 3 (aq) Δ H (aq) 1 CO3 (aq)   Ka2 5 2 and CO223 . [HCO3 ]  15.8 Diprotic and Polyprotic Acids 689 Note that the conjugate base in the first ionization stage becomes the acid in the second ionization stage. Table 15.5 on p. 690 shows the ionization constants of several diprotic acids and one polyprotic acid. For a given acid, the first ionization constant is much larger than the second ionization constant, and so on. This trend is reasonable because it is easier to remove a H1 ion from a neutral molecule than to remove another H1 ion from a negatively charged ion derived from the molecule. In Example 15.11 we calculate the equilibrium concentrations of all the species of a diprotic acid in aqueous solution. Example 15.11 Oxalic acid (H2C2O4) is a poisonous substance used chiefly as a bleaching and cleansing agent (for example, to remove bathtub rings). Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution. Strategy Determining the equilibrium concentrations of the species of a diprotic acid in aqueous solution is more involved than for a monoprotic acid. We follow the same H2C2O4 procedure as that used for a monoprotic acid for each stage, as in Example 15.8. Note that the conjugate base from the first stage of ionization becomes the acid for the second stage ionization. Solution We proceed according to the following steps. Step 1: The major species in solution at this stage are the nonionized acid, H1 ions, and the conjugate base, HC2O24 . Step 2: Letting x be the equilibrium concentration of H1 and HC2O24 ions in mol/L, we summarize: H2C2O4(aq) Δ H1(aq) 1 HC2O24(aq) Initial (M): 0.10 0.00 0.00 Change (M): 2x 1x 1x Equilibrium (M): 0.10 2 x x x Step 3: Table 15.5 gives us [H1][HC2O2 4] Ka 5 [H2C2O4] x2 6.5 3 1022 5 0.10 2 x Applying the approximation 0.10 2 x < 0.10, we obtain x2 x2 6.5 3 1022 5 < 0.10 2 x 0.10 x2 5 6.5 3 1023 x 5 8.1 3 1022 M To test the approximation, 8.1 3 1022 M 3 100% 5 81% 0.10 M (Continued) 690 Chapter 15 ■ Acids and Bases Table 15.5 Ionization Constants of Some Diprotic Acids and a Polyprotic Acid and Their Conjugate Bases at 258C Conjugate Name of Acid Formula Structure Ka Base Kb O B Sulfuric acid H2SO4 HOOOSOOOH very large HSO24 very small B O O B Hydrogen sulfate ion HSO24 HOOOSOO 1.3 3 1022 SO22 4 7.7 3 10213 B O O O B B Oxalic acid H2C2O4 HOOOCOCOOOH 6.5 3 1022 HC2O24 1.5 3 10213 O O B B Hydrogen oxalate ion HC2O24 HOOOCOCOO 6.1 3 1025 C2O22 4 1.6 3 10210 O B Sulfurous acid* H2SO3 HOOOSOOOH 1.3 3 1022 HSO23 7.7 3 10213 O B Hydrogen sulfite ion HSO23 HOOOSOO 6.3 3 1028 SO22 3 1.6 3 1027 O B Carbonic acid H2CO3 HOOOCOOOH 4.2 3 1027 HCO23 2.4 3 1028 O B Hydrogen carbonate ion HCO23 HOOOCOO 4.8 3 10211 CO22 3 2.1 3 1024 Hydrosulfuric acid H2S H¬S¬H 9.5 3 1028 HS2 1.1 3 1027 Hydrogen sulfide ion† HS2 H¬S2 1 3 10219 S22 1 3 105 O B Phosphoric acid H3PO4 HOOOPOOOH 7.5 3 1023 H2PO24 1.3 3 10212 A O A H O B Dihydrogen phosphate ion H2PO24 HOOOPOO 6.2 3 1028 HPO22 4 1.6 3 1027 A O A H O B Hydrogen phosphate ion HPO22 4 HOOOPOO 4.8 3 10213 PO32 4 2.1 3 1022 A O *H2SO3 has never been isolated and exists in only minute concentration in aqueous solution of SO2. The Ka value here refers to the process SO2 (g) 1 H2O(l) Δ H 1 (aq) 1 HSO23 (aq). † The ionization constant of HS2 is very low and difficult to measure. The value listed here is only an estimate.  15.8 Diprotic and Polyprotic Acids 691 Clearly the approximation is not valid. Therefore, we must solve the quadratic equation x2 1 6.5 3 1022x 2 6.5 3 1023 5 0 The result is x 5 0.054 M. Step 4: When the equilibrium for the first stage of ionization is reached, the concen- trations are [H1] 5 0.054 M [HC2O24 ] 5 0.054 M [H2C2O4] 5 (0.10 2 0.054) M 5 0.046 M Next we consider the second stage of ionization. Step 1: At this stage, the major species are HC2O24, which acts as the acid in the second stage of ionization, H1, and the conjugate base C2O22 4 . Step 2: Letting y be the equilibrium concentration of H1 and C2O22 4 ions in mol/L, we summarize: HC2O24 (aq) Δ H1(aq) 1 C2O22 4 (aq) Initial (M): 0.054 0.054 0.00 Change (M): 2y 1y 1y Equilibrium (M): 0.054 2 y 0.054 1 y y Step 3: Table 15.5 gives us [H1][C2O224 ] Ka 5 [HC2O2 4] (0.054 1 y) (y) 6.1 3 1025 5 (0.054 2 y) Applying the approximation 0.054 1 y < 0.054 and 0.054 2 y < 0.054, we obtain (0.054) (y) 5 y 5 6.1 3 1025 M (0.054) and we test the approximation, 6.1 3 1025 M 3 100% 5 0.11% 0.054 M The approximation is valid. Step 4: At equilibrium, [H2C2O4] 5 0.046 M [HC2O24 ] 5 (0.054 2 6.1 3 1025 ) M 5 0.054 M [H1] 5 (0.054 1 6.1 3 1025 ) M 5 0.054 M [C2O22 4 ] 5 6.1 3 1025 M [OH2] 5 1.0 3 10214y0.054 5 1.9 3 10213 M Similar problem: 15.66. Practice Exercise Calculate the concentrations of H2C2O4, HC2O24, C2O22 4 , and H 1 ions in a 0.20 M oxalic acid solution. 692 Chapter 15 ■ Acids and Bases Review of Concepts Which of the diagrams shown here represents a solution of sulfuric acid? Water molecules have been omitted for clarity.   H2SO4  HSO4  SO 24  H3O (a) (b) (c) Example 15.11 shows that for diprotic acids, if Ka1 @ Ka2 , then we can assume that the concentration of H1 ions is the product of only the first stage of ionization. Furthermore, the concentration of the conjugate base for the second-stage ionization is numerically equal to Ka2 . Phosphoric acid (H3PO4) is a polyprotic acid with three ionizable hydrogen atoms: [H1][H2PO24 ] H3PO4 (aq) Δ H1 (aq) 1 H2PO24 (aq)   Ka1 5 5 7.5 3 1023 [H3PO4] [H1][HPO224 ] H2PO24 (aq) Δ H1 (aq) 1 HPO22 4 (aq)   Ka2 5 5 6.2 3 1028 [H2PO24 ] [H1][PO32 4 ] HPO22 1 32 4 (aq) Δ H (aq) 1 PO4 (aq)   Ka3 5 22 5 4.8 3 10213 [HPO4 ] H3PO4 We see that phosphoric acid is a weak polyprotic acid and that its ionization con- stants decrease markedly for the second and third stages. Thus, we can predict that, in a solution containing phosphoric acid, the concentration of the nonionized acid is the highest, and the only other species present in significant concentrations are H1 and H2PO24 ions. 15.9 Molecular Structure and the Strength of Acids The strength of an acid depends on a number of factors, such as the properties of the solvent, the temperature, and, of course, the molecular structure of the acid. When we compare the strengths of two acids, we can eliminate some variables by considering their properties in the same solvent and at the same temperature and concentration. Then we can focus on the structure of the acids. Let us consider a certain acid HX. The strength of the acid is measured by its tendency to ionize: HX ¡ H1 1 X2  15.9 Molecular Structure and the Strength of Acids 693 Table 15.6 Bond Enthalpies for Hydrogen Halides and Acid Strengths for Hydrohalic Acids Bond Bond Enthalpy (kJ/mol) Acid Strength H¬F 568.2 weak H¬Cl 431.9 strong H¬Br 366.1 strong H¬I 298.3 strong Two factors influence the extent to which the acid undergoes ionization. One is the strength of the H¬X bond—the stronger the bond, the more difficult it is for the HX molecule to break up and hence the weaker the acid. The other factor is the polarity of the H¬X bond. The difference in the electronegativities between H and X results in a polar bond like δ1 δ2 H¬X If the bond is highly polarized, that is, if there is a large accumulation of positive and negative charges on the H and X atoms, HX will tend to break up into H1 and X2 ions. So a high degree of polarity characterizes a stronger acid. Below we will con- sider some examples in which either bond strength or bond polarity plays a prominent role in determining acid strength. Hydrohalic Acids The halogens form a series of binary acids called the hydrohalic acids (HF, HCl, HBr, and HI). Of this series, which factor (bond strength or bond polarity) is the pre- dominant factor in determining the strength of the binary acids? Consider first the strength of the H¬X bond in each of these acids. Table 15.6 shows that HF has the highest bond enthalpy of the four hydrogen halides, and HI has the lowest bond enthalpy. It takes 568.2 kJ/mol to break the H¬F bond and only 298.3 kJ/mol to break the H¬I bond. Based on bond enthalpy, HI should be the strongest acid because it is easiest to break the bond and form the H1 and I2 ions. Second, consider the polarity of the H¬X bond. In this series of acids, the polarity of the bond decreases from HF to HI because F is the most electronegative of the halogens (see Figure 9.5). Based on bond polarity, then, HF should be the strongest acid because of the largest accumulation of positive and negative charges on the H and F atoms. Thus, we have 1A 8A two competing factors to consider in determining the strength of binary acids. The 2A 3A 4A 5A 6A 7A fact that HI is a strong acid and that HF is a weak acid indicates that bond enthalpy F Cl is the predominant factor in determining the acid strength of binary acids. In this Br I series of binary acids, the weaker the bond, the stronger the acid so that the strength of the acids increases as follows: Strength of hydrohalic acids HF ! HCl , HBr , HI increases from HF to HI. Oxoacids Now let us consider the oxoacids. Oxoacids, as we learned in Chapter 2, contain To review the nomenclature of inorganic acids, see Section 2.8 (p. 62). hydrogen, oxygen, and one other element Z, which occupies a central position. Figure 15.5 shows the Lewis structures of several common oxoacids. As you can 694 Chapter 15 ■ Acids and Bases Figure 15.5 Lewis structures of some common oxoacids. For SO S SO S simplicity, the formal charges have B B O O OCO O HO Q OOH HO O OOO NP O O O O ONO O HO Q OS been omitted. Q Q Q Q Carbonic acid Nitrous acid Nitric acid SO S SO S SO S B B B O O PO O O OO PO OO OO SOO HO O O HOO Q A QOH HO O Q A QOH Q B QOH H SO S SO S A H Phosphorous acid Phosphoric acid Sulfuric acid see, these acids are characterized by the presence of one or more O¬H bonds. The central atom Z might also have other groups attached to it: G OZOOOH D As the oxidation number of an atom If Z is an electronegative element, or is in a high oxidation state, it will attract becomes larger, its ability to draw electrons in a bond toward itself electrons, thus making the Z¬O bond more covalent and the O¬H bond more increases. polar. Consequently, the tendency for the hydrogen to be donated as a H1 ion increases: G ␦ ␦ G OZOOOH 88n OZOO  H D D To compare their strengths, it is convenient to divide the oxoacids into two groups. 1. Oxoacids Having Different Central Atoms That Are from the Same Group of the Periodic Table and That Have the Same Oxidation Number. Within this group, acid strength increases with increasing electronegativity of the central atom, as HClO3 and HBrO3 illustrate: OS SO SO OS 1A 2A 3A 4A 5A 6A 7A 8A A A O O O O Cl Q Q QS HOOOClOO Q Q QS HOOOBrOO Br I Cl and Br have the same oxidation number, 15. However, because Cl is more electronegative than Br, it attracts the electron pair it shares with oxygen (in the Strength of halogen-containing Cl¬O¬H group) to a greater extent than Br does. Consequently, the O¬H oxoacids having the same number bond is more polar in chloric acid than in bromic acid and ionizes more readily. of O atoms increases from bottom Thus, the relative acid strengths are to top. HClO3 . HBrO3 2. Oxoacids Having the Same Central Atom but Different Numbers of Attached Groups. Within this group, acid strength increases as the oxidation number of the central atom increases. Consider the oxoacids of chlorine shown in Figure 15.6. In this series the ability of chlorine to draw electrons away from the OH group (thus making the O¬H bond more polar) increases with the number of electro- negative O atoms attached to Cl. Thus, HClO4 is the strongest acid because it  15.9 Molecular Structure and the Strength of Acids 695 Figure 15.6 Lewis structures O OO Cl O O ClO O OS of the oxoacids of chlorine. The Q QS HO O HO O Q Q O Q oxidation number of the Cl atom Hypochlorous acid (ⴙ1) Chlorous acid (ⴙ3) is shown in parentheses. For simplicity, the formal charges have been omitted. Note that although hypochlorous acid is written as HClO, the H atom OS SO SO OS is bonded to the O atom. A A O HO Q O O ClO OS O HO O O OS Q O Q Q O ClO A Q SO S O Chloric acid (ⴙ5) Perchloric acid (ⴙ7) has the largest number of O atoms attached to Cl, and the acid strength decreases as follows: HClO4 . HClO3 . HClO2 . HClO Example 15.12 compares the strengths of acids based on their molecular structures. Example 15.12 Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HBrO, and HIO; (b) HNO3 and HNO2. Strategy Examine the molecular structure. In (a) the two acids have similar structure but differ only in the central atom (Cl, Br, and I). Which central atom is the most electronegative? In (b) the acids have the same central atom (N) but differ in the number of O atoms. What is the oxidation number of N in each of these two acids? Solution (a) These acids all have the same structure, and the halogens all have the same oxidation number (11). Because the electronegativity decreases from Cl to I, the Cl atom attracts the electron pair it shares with the O atom to the greatest extent. Consequently, the O¬H bond is the most polar in HClO and least polar in HIO. Thus, the acid strength decreases as follows: HClO . HBrO . HIO (b) The structures of HNO3 and HNO2 are shown in Figure 15.5. Because the oxidation number of N is 15 in HNO3 and 13 in HNO2, HNO3 is a stronger acid than HNO2. Similar problem: 15.70. Practice Exercise Which of the following acids is weaker: HClO2 or HClO3? Carboxylic Acids So far the discussion has focused on inorganic acids. A group of organic acids that also deserves attention is the carboxylic acids, whose Lewis structures can be represented by SOS B O ROCOOOH Q 696 Chapter 15 ■ Acids and Bases where R is part of the acid molecule and the shaded portion represents the carboxyl group, ¬COOH. The strength of carboxylic acids depends on the nature of the R group. Consider, for example, acetic acid and chloroacetic acid: H SOS Cl SOS A B A B O HOCOCOOOH Q HOCOCOOOHO Q A A H H acetic acid (K a  1.8  10ⴚ5) chloroacetic acid (K a  1.4  10ⴚ3) The presence of the electronegative Cl atom in chloroacetic acid shifts electron density toward the R group, thereby making the O¬H bond more polar. Consequently, there is a greater tendency for the acid to ionize: CH2ClCOOH(aq) Δ CH2ClCOO2 (aq) 1 H1 (aq) The conjugate base of the carboxylic acid, called the carboxylate anion (RCOO2), can exhibit resonance: SOS OS⫺ SO B A ROCOOO QS⫺ m8n ROCPOO Q In the language of molecular orbital theory, we attribute the stability of the anion to its ability to spread or delocalize the electron density over several atoms. The greater the extent of electron delocalization, the more stable the anion and the greater the tendency for the acid to undergo ionization. Thus, benzoic acid (C6H5COOH, Electrostatic potential map of the Ka 5 6.5 3 1025) is a stronger acid than acetic acid because the benzene ring (see acetate ion. The electron density is evenly distributed between the p. 453) facilitates electron delocalization, so that the benzoate anion (C6H5COO2) is two O atoms. more stable than the acetate anion (CH3COO2). 15.10 Acid-Base Properties of Salts As defined in Section 4.3, a salt is an ionic compound formed by the reaction between an acid and a base. Salts are strong electrolytes that completely dissociate into ions The word “hydrolysis” is derived from the in water. The term salt hydrolysis describes the reaction of an anion or a cation of Greek words hydro, meaning “water,” and lysis, meaning “to split apart.” a salt, or both, with water. Salt hydrolysis usually affects the pH of a solution. Salts That Produce Neutral Solutions In reality, all positive ions give acid It is generally true that salts containing an alkali metal ion or alkaline earth metal ion solutions in water. (except Be21) and the conjugate base of a strong acid (for example, Cl2, Br2, and NO23 ) do not undergo hydrolysis to an appreciable extent, and their solutions are assumed to be neutral. For instance, when NaNO3, a salt formed by the reaction of NaOH with HNO3, dissolves in water, it dissociates completely as follows: H2O NaNO3 (s) ¡   Na1 (aq) 1 NO23 (aq) The hydrated Na1 ion neither donates nor accepts H1 ions. The NO23 ion is the con- jugate base of the strong acid HNO3, and it has no affinity for H1 ions. Consequently, a solution containing Na1 and NO23 ions is neutral, with a pH of about 7. Salts That Produce Basic Solutions The solution of a salt derived from a strong base and a weak acid is basic. For example, the dissociation of sodium acetate (CH3COONa) in water is given by H2O CH3COONa(s) ¡   Na1 (aq) 1 CH3COO2 (aq)  15.10 Acid-Base Properties of Salts 697 The hydrated Na1 ion has no appreciable acidic or basic properties. The acetate ion The mechanism by which metal ions CH3COO2, however, is the conjugate base of the weak acid CH3COOH and therefore produce acid solutions is discussed on p. 699. has an affinity for H1 ions. The hydrolysis reaction is given by CH3COO2 (aq) 1 H2O(l) Δ CH3COOH(aq) 1 OH2 (aq) Because this reaction produces OH2 ions, the sodium acetate solution will be basic. The equilibrium constant for this hydrolysis reaction is the same as the base ionization constant expression for CH3COO2, so we write (see p. 687) [CH3COOH][OH2] Kb 5 5 5.6 3 10210 [CH3COO2] Because each CH3COO2 ion that hydrolyzes produces one OH2 ion, the concentration of OH2 at equilibrium is the same as the concentration of CH3COO2 that hydrolyzed. We can define the percent hydrolysis as [CH3COO2]hydrolyzed % hydrolysis 5 3 100% [CH3COO2]initial 2 [OH ]equilibrium 5 3 100% [CH3COO2]initial A calculation based on the hydrolysis of CH3COONa is illustrated in Example 15.13. In solving salt hydrolysis problems, we follow the same procedure we used for weak acids and weak bases. Example 15.13 Calculate the pH of a 0.15 M solution of sodium acetate (CH3COONa). What is the percent hydrolysis? Strategy What is a salt? In solution, CH3COONa dissociates completely into Na1 and CH3COO2 ions. The Na1 ion, as we saw earlier, does not react with water and has no effect on the pH of the solution. The CH3COO2 ion is the conjugate base of the weak acid CH3COOH. Therefore, we expect that it will react to a certain extent with water to produce CH3COOH and OH2, and the solution will be basic. Solution Step 1: Because we started with a 0.15 M sodium acetate solution, the concentrations of the ions are also equal to 0.15 M after dissociation: CH3COONa(aq) 88n Na (aq) CH3COO (aq) Initial (M): 0.15 0 0 Change (M): 0.15 0.15 0.15 Final (M): 0 0.15 0.15 Of these ions, only the acetate ion will react with water CH3COO2 (aq) 1 H2O(l) Δ CH3COOH(aq) 1 OH2 (aq) At equilibrium, the major species in solution are CH3COOH, CH3COO2, and OH2. The concentration of the H1 ion is very small as we would expect for a basic solution, so it is treated as a minor species. We ignore the ionization of water. (Continued) 698 Chapter 15 ■ Acids and Bases Step 2: Let x be the equilibrium concentration of CH3COOH and OH2 ions in mol/L, we summarize: CH3COO (aq) H2O(l) 34 CH3COOH(aq) OH (aq) Initial (M ): 0.15 0.00 0.00 Change (M): x x x Equilibrium (M): 0.15 x x x Step 3: From the preceding discussion and Table 15.3 we write the equilibrium constant of hydrolysis, or the base ionization constant, as [CH3COOH][OH2] Kb 5 [CH3COO2] x2 5.6 3 10210 5 0.15 2 x Because Kb is very small and the initial concentration of the base is large, we can apply the approximation 0.15 2 x < 0.15: x2 x2 5.6 3 10210 5 < 0.15 2 x 0.15 x 5 9.2 3 1026 M Step 4: At equilibrium: [OH2] 5 9.2 3 1026 M pOH 5 2log (9.2 3 1026 ) 5 5.04 pH 5 14.00 2 5.04 5 8.96 Thus the solution is basic, as we would expect. The percent hydrolysis is given by 9.2 3 1026 M % hydrolysis 5 3 100% 0.15 M 5 0.0061% Check The result shows that only a very small amount of the anion undergoes hydrolysis. Note that the calculation of percent hydrolysis takes the same form as the Similar problem: 15.81. test for the approximation, which is valid in this case. Practice Exercise Calculate the pH of a 0.24 M sodium formate solution (HCOONa). Salts That Produce Acidic Solutions When a salt derived from a strong acid such as HCl and a weak base such as NH3 dissolves in water, the solution becomes acidic. For example, consider the process H2O NH4Cl(s) ¡   NH14 (aq) 1 Cl2 (aq) The Cl2 ion, being the conjugate base of a strong acid, has no affinity for H1 and no tendency to hydrolyze. The ammonium ion NH14 is the weak conjugate acid of the weak base NH3 and ionizes as follows: NH1 1 4 (aq) 1 H2O(l) Δ NH3 (aq) 1 H3O (aq)  15.10 Acid-Base Properties of Salts 699 – + Al(H2O)63+ + H2O Al(OH)(H2O)52+ + H3O+ Figure 15.7 The six H2O molecules surround the Al 31 ion octahedrally. The attraction of the small Al 31 ion for the lone pairs on the oxygen atoms is so great that the O ¬ H bonds in a H2O molecule attached to the metal cation are weakened, allowing the loss of a proton (H1) to an incoming H2O molecule. This hydrolysis of the metal cation makes the solution acidic. or simply NH14 (aq) Δ NH3 (aq) 1 H1 (aq) Note that this reaction also represents the hydrolysis of the NH14 ion. Because H1 ions are produced, the pH of the solution decreases. The equilibrium constant (or ionization constant) for this process is given by [NH3][H1] Kw 1.0 3 10214 Ka 5 5 5 5 5.6 3 10210 By coincidence, Ka of NH14 has the same [NH14] Kb 1.8 3 1025 numerical value as Kb of CH3COO2. and we can calculate the pH of an ammonium chloride solution following the same procedure used in Example 15.13. In principle, all metal ions react with water to produce an acidic solution. However, because the extent of hydrolysis is most pronounced for the small and highly charged metal cations such as Al31, Cr31, Fe31, Bi31, and Be21, we gener- ally neglect the relatively small interaction of alkali metal ions and most alkaline earth metal ions with water. When aluminum chloride (AlCl3) dissolves in water, the Al31 ions take the hydrated form Al(H2O)631 (Figure 15.7). Let us consider one bond between the metal ion and an oxygen atom from one of the six water mol- ecules in Al(H2O)631: H 88 m88 m Al O m 88 H The positively charged Al31 ion draws electron density toward itself, increasing the polarity of the O¬H bonds. Consequently, the H atoms have a greater tendency to ionize than those in water molecules not involved in hydration. The resulting ionization process can be written as Al(H2O) 31 21 1 6 (aq) 1 H2O(l) Δ Al(OH)(H2O) 5 (aq) 1 H3O (aq) The hydrated Al31 qualifies as a proton donor and thus a Brønsted acid in this reaction. or simply Al(H2O) 31 21 1 6 (aq) Δ Al(OH)(H2O) 5 (aq) 1 H (aq) 700 Chapter 15 ■ Acids and Bases The equilibrium constant for the metal cation hydrolysis is given by [Al(OH)(H2O) 21 1 5 ][H ] Note that Al(OH)361 is roughly as strong Ka 5 5 1.3 3 1025 an acid as CH3COOH. [Al(H2O) 31 6 ] Note that Al(OH) (H2O) 21 5 can undergo further ionization Al(OH)(H2O) 21 1 1 5 (aq) Δ Al(OH) 2 (H2O) 4 (aq) 1 H (aq) and so on. However, it is generally sufficient to take into account only the first stage of hydrolysis. The extent of hydrolysis is greatest for the smallest and most highly charged ions because a “compact” highly charged ion is more effective in polarizing the O¬H bond and facilitating ionization. This is why relatively large ions of low charge such as Na1 and K1 do not undergo appreciable hydrolysis. Salts in Which Both the Cation and the Anion Hydrolyze So far we have considered salts in which only one ion undergoes hydrolysis. For salts derived from a weak acid and a weak base, both the cation and the anion hydrolyze. However, whether a solution containing such a salt is acidic, basic, or neutral depends on the relative strengths of the weak acid and the weak base. Because the mathematics associated with this type of system is rather involved, we will focus on making qualitative predictions about these solutions based on the following guidelines: • Kb > Ka. If Kb for the anion is greater than Ka for the cation, then the solution must be basic because the anion will hydrolyze to a greater extent than the cation. At equilibrium, there will be more OH2 ions than H1 ions. • Kb < Ka. Conversely, if Kb for the anion is smaller than Ka for the cation, the solution will be acidic because cation hydrolysis will be more extensive than anion hydrolysis. • K b ≈ K a. If K a is approximately equal to K b, the solution will be nearly neutral. Table 15.7 summarizes the behavior in aqueous solution of the salts discussed in this section. Example 15.14 illustrates how to predict the acid-base properties of salt solutions. Table 15.7 Acid-Base Properties of Salts lons That Undergo pH of Type of Salt Examples Hydrolysis Solution Cation from strong base; anion from strong acid NaCl, KI, KNO3, RbBr, BaCl2 None < 7 Cation from strong base; anion from weak acid CH3COONa, KNO2 Anion . 7 Cation from weak base; anion from strong acid NH4Cl, NH4NO3 Cation , 7 Cation from weak base; anion from weak acid NH4NO2, CH3COONH4, NH4CN Anion and cation , 7 if Kb , Ka < 7 if Kb < Ka . 7 if Kb . Ka Small, highly charged cation; anion from strong acid AlCl3, Fe(NO3)3 Hydrated cation , 7  15.10 Acid-Base Properties of Salts 701 Example 15.14 Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) NH4I, (b) NaNO2, (c) FeCl3, (d) NH4F. Strategy In deciding whether a salt will undergo hydrolysis, ask yourself the following questions: Is the cation a highly charged metal ion or an ammonium ion? Is the anion the conjugate base of a weak acid? If yes to either question, then hydrolysis will occur. In cases where both the cation and the anion react with water, the pH of the solution will depend on the relative magnitudes of Ka for the cation and Kb for the anion (see Table 15.7). Solution We first break up the salt into its cation and anion components and then examine the possible reaction of each ion with water. (a) The cation is NH14 , which will hydrolyze to produce NH3 and H1. The I2 anion is the conjugate base of the strong acid HI. Therefore, I2 will not hydrolyze and the solution is acidic. (b) The Na1 cation does not hydrolyze. The NO22 is the conjugate base of the weak acid HNO2 and will hydrolyze to give HNO2 and OH2. The solution will be basic. (c) Fe31 is a small metal ion with a high charge and hydrolyzes to produce H1 ions. The Cl2 does not hydrolyze. Consequently, the solution will be acidic. (d) Both the NH14 and F2 ions will hydrolyze. From Tables 15.3 and 15.4 we see that the Ka of NH14 (5.6 3 10210) is greater than the Kb for F2 (1.4 3 10211). Therefore, the solution will be acidic. Similar problems: 15.77, 15.78. Practice Exercise Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) LiClO4, (b) Na3PO4, (c) Bi(NO3)3, (d) NH4CN. Review of Concepts The diagrams shown here represent solutions of three salts NaX (X 5 A, B, or C). (a) Which X2 has the weakest conjugate acid? (b) Arrange the three X2 anions in order of increasing base strength. The Na1 ion and water molecules have been omitted for clarity.  HA, HB, or HC  A, B, or C  OH NaA NaB NaC Finally we note that some anions can act either as an acid or as a base. For example, the bicarbonate ion (HCO23 ) can ionize or undergo hydrolysis as follows (see Table 15.5): HCO23 (aq) 1 H2O(l) Δ H3O 1 (aq) 1 CO23 2 (aq)     Ka 5 4.8 3 10211 HCO23 (aq) 1 H2O(l) Δ H2CO3 (aq) 1 OH2 (aq)   Kb 5 2.4 3 1028 Because Kb . Ka, we predict that the hydrolysis reaction will outweigh the ionization process. Thus, a solution of sodium bicarbonate (NaHCO3) will be basic. 702 Chapter 15 ■ Acids and Bases 15.11 Acid-Base Properties of Oxides and Hydroxides As we saw in Chapter 8, oxides can be classified as acidic, basic, or amphoteric. Our discussion of acid-base reactions would be incomplete if we did not examine the properties of these compounds. Figure 15.8 shows the formulas of a number of oxides of the representative ele- ments in their highest oxidation states. Note that all alkali metal oxides and all alkaline earth metal oxides except BeO are basic. Beryllium oxide and several metallic oxides in Groups 3A and 4A are amphoteric. Nonmetallic oxides in which the oxidation number of the representative element is high are acidic (for example, N2O5, SO3, and Cl2O7), but those in which the oxidation number of the representative element is low (for example, CO and NO) show no measurable acidic properties. No nonmetallic oxides are known to have basic properties. The basic metallic oxides react with water to form metal hydroxides: H2O Na2O(s) ¡ 2NaOH(aq) H2O BaO(s) ¡ Ba(OH) 2 (aq) The reactions between acidic oxides and water are as follows: CO2 (g) 1 H2O(l) Δ H2CO3 (aq) SO3 (g) 1 H2O(l) Δ H2SO4 (aq) N2O5 (g) 1 H2O(l) Δ 2HNO3 (aq) P4O10 (s) 1 6H2O(l) Δ 4H3PO4 (aq) Cl2O7 (l) 1 H2O(l) Δ 2HClO4 (aq) The reaction between CO2 and H2O explains why when pure water is exposed to air We will look at the causes and effects of (which contains CO2) it gradually reaches a pH of about 5.5 (Figure 15.9). The reaction acid rain in Chapter 20. between SO3 and H2O is largely responsible for acid rain (Figure 15.10). 1 Basic oxide 18 1A 8A 2 Acidic oxide 13 14 15 16 17 2A 3A 4A 5A 6A 7A Li2O BeO Amphoteric oxide B2O3 CO2 N2O5 OF2 Na2O MgO 3 4 5 6 7 8 9 10 11 12 Al2O3 SiO2 P4O10 SO3 Cl2O7 3B 4B 5B 6B 7B 8B 1B 2B K2O CaO Ga2O3 GeO2 As2O5 SeO3 Br2O7 Rb2O SrO In2O3 SnO2 Sb2O5 TeO3 I2O7 Cs2O BaO Tl2O3 PbO2 Bi2O5 PoO3 At2O7 Figure 15.8 Oxides of the representative elements in their highest oxidation states.  15.11 Acid-Base Properties of Oxides and Hydroxides 703 Figure 15.9 (Left) A beaker of water to which a few drops of bromothymol blue indicator have been added. (Right) As dry ice is added to the water, the CO2 reacts to form carbonic acid, which turns the solution acidic and changes the color from blue to yellow. Figure 15.10 A forest damaged by acid rain. Reactions between acidic oxides and bases and those between basic oxides and acids resemble normal acid-base reactions in that the products are a salt and water: CO2(g) 2NaOH(aq) 88n Na2CO3(aq) H2O(l) acidic oxide base salt water BaO(s) 2HNO3(aq) 88n Ba(NO3)2(aq) H2O(l) basic oxide acid salt water As Figure 15.8 shows, aluminum oxide (Al2O3) is amphoteric. Depending on the reaction conditions, it can behave either as an acidic oxide or as a basic oxide. For example, Al2O3 acts as a base with hydrochloric acid to produce a salt (AlCl3) and water: Al2O3 (s) 1 6HCl(aq) ¡ 2AlCl3 (aq) 1 3H2O(l) and acts as an acid with sodium hydroxide: Al2O3 (s) 1 2NaOH(aq) ¡ 2NaAlO2 (aq) 1 H2O(l) Some transition metal oxides in which the metal has a high oxidation number act The higher the oxidation number of the metal, the more covalent the compound; as acidic oxides. Two familiar examples are manganese(VII) oxide (Mn2O7) and the lower the oxidation number, the more chromium(VI) oxide (CrO3), both of which react with water to produce acids: ionic the compound. Mn2O7 (l) H2O(l) 88n 2HMnO4(aq) permanganic acid CrO3(s) H2O(l) 88n H2CrO4(aq) chromic acid Basic and Amphoteric Hydroxides We have seen that the alkali and alkaline earth metal hydroxides [except Be(OH)2] are basic in properties. The following hydroxides are amphoteric: Be(OH)2, Al(OH)3, 704 Chapter 15 ■ Acids and Bases Sn(OH)2, Pb(OH)2, Cr(OH)3, Cu(OH)2, Zn(OH)2, and Cd(OH)2. For example, alumi- num hydroxide reacts with both acids and bases: Al(OH) 3 (s) 1 3H1 (aq) ¡ Al31 (aq) 1 3H2O(l) Al(OH) 3 (s) 1 OH2 (aq) Δ Al(OH) 24 (aq) All amphoteric hydroxides are insoluble. It is interesting that beryllium hydroxide, like aluminum hydroxide, exhibits amphoterism: Be(OH) 2 (s) 1 2H1 (aq) ¡ Be21 (aq) 1 2H2O(l) Be(OH) 2 (s) 1 2OH2 (aq) Δ Be(OH) 22 4 (aq) This is another example of the diagonal relationship between beryllium and aluminum (see p. 348). Review of Concepts Arrange the following oxides in order of increasing basicity: K2O, Al2O3, BaO. 15.12 Lewis Acids and Bases So far we have discussed acid-base properties in terms of the Brønsted theory. To behave as a Brønsted base, for example, a substance must be able to accept protons. By this definition both the hydroxide ion and ammonia are bases: O H  SOOH O 88n HOOOH Q Q  H H A A H  SNOH 88n HONOH A A H H In each case, the atom to which the proton becomes attached possesses at least one unshared pair of electrons. This characteristic property of OH2, NH3, and other Brønsted bases suggests a more general definition of acids and bases. Lewis acids are either deficient in electrons In 1932 the American chemist G. N. Lewis formulated such a definition. He defined (cations) or the central atom has a vacant valence orbital. what we now call a Lewis base as a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons. For example, in the pro- tonation of ammonia, NH3 acts as a Lewis base because it donates a pair of electrons to the proton H1, which acts as a Lewis acid by accepting the pair of electrons. A Lewis acid-base reaction, therefore, is one that involves the donation of a pair of electrons from one species to another. Such a reaction does not produce a salt and water. The significance of the Lewis concept is that it is more general than other defini- tions. Lewis acid-base reactions include many reactions that do not involve Brønsted acids. Consider, for example, the reaction between boron trifluoride (BF3) and ammo- nia to form an adduct compound (Figure 15.11): F H F H Figure 15.11 A Lewis acid-base A A A A reaction involving BF3 and NH3. FOB  SNOH 88n FOBONOH A A A A F H F H acid base  15.12 Lewis Acids and Bases 705 In Section 10.4 we saw that the B atom in BF3 is sp2-hybridized. The vacant, unhy- A coordinate covalent bond (see p. 393) is always formed in a Lewis acid-base bridized 2pz orbital accepts the pair of electrons from NH3. So BF3 functions as an reaction. acid according to the Lewis definition, even though it does not contain an ionizable proton. Note that a coordinate covalent bond is formed between the B and N atoms, as is the case in all Lewis acid-base reactions. Another Lewis acid containing boron is boric acid. Boric acid (a weak acid used in eyewash) is an oxoacid with the following structure: H A SOS A O O HOOOBOOOH Q Q H3BO3 Boric acid does not ionize in water to produce a H1 ion. Its reaction with water is B(OH) 3 (aq) 1 H2O(l) Δ B(OH)24 (aq) 1 H1 (aq) In this Lewis acid-base reaction, boric acid accepts a pair of electrons from the hydroxide ion that is derived from the H2O molecule. The hydration of carbon dioxide to produce carbonic acid CO2 (g) 1 H2O(l) Δ H2CO3 (aq) can be understood in the Lewis framework as follows: The first step involves donation of a lone pair on the oxygen atom in H2O to the carbon atom in CO2. An orbital is vacated on the C atom to accommodate the lone pair by removal of the electron pair in the C¬O pi bond. These shifts of electrons are indicated by the curved arrows. O OS SO OS H B A G O C 888n HOOOC S O D B A B S H H SOS SOS Therefore, H2O is a Lewis base and CO2 is a Lewis acid. Next, a proton is transferred onto the O atom bearing a negative charge to form H2CO3. SO OS HOOOS A A O HOOOC 888n SO OOC A B A B H SOS H SOS Other examples of Lewis acid-base reactions are Ag (aq) 2NH3(aq) Δ Ag(NH3)2 (aq) acid base Cd2 (aq) 4I (aq) Δ CdI24 (aq) acid base Ni(s) 4CO(g) Δ Ni(CO) 4 (g) acid base It is important to note that the hydration of metal ions in solution is in itself a Lewis acid-base reaction (see Figure 15.7). Thus, when copper(II) sulfate (CuSO4) dissolves in water, each Cu21 ion is associated with six water molecules as Cu(H2O)216 . In this case, the Cu21 ion acts as the acid and the H2O molecules act as the base. CHEMISTRY in Action Antacids and the pH Balance in Your Stomach A n average adult produces between 2 and 3 L of gastric juice daily. Gastric juice is a thin, acidic digestive fluid secreted by glands in the mucous membrane that lines the Food Mucous stomach. It contains, among other substances, hydrochloric membrane acid. The pH of gastric juice is about 1.5, which corresponds to a hydrochloric acid concentration of 0.03 M—a concentration strong enough to dissolve zinc metal! What is the purpose of Blood this highly acidic medium? Where do the H1 ions come from? plasma What happens when there is an excess of H1 ions present in the stomach? A simplified diagram of the stomach is shown here. The HCl(aq) inside lining is made up of parietal cells, which are fused to- gether to form tight junctions. The interiors of the cells are pro- tected from the surroundings by cell membranes. These membranes allow water and neutral molecules to pass in and out Blood plasma of the stomach, but they usually block the movement of ions such as H1, Na1, K1, and Cl2. The H1 ions come from carbonic To intestines Cl– H+ (active transport) acid (H2CO3) formed as a result of the hydration of CO2, an end product of metabolism: A simplified diagram of the human stomach. CO2 (g) 1 H2O(l) Δ H2CO3 (aq) H2CO2 (aq) Δ H1 (aq) 1 HCO2 3 (aq) membrane back to the blood plasma can cause muscle contrac- tion, pain, swelling, inflammation, and bleeding. These reactions take place in the blood plasma bathing the cells One way to temporarily reduce the H1 ion concentration in in the mucosa. By a process known as active transport, H1 ions the stomach is to take an antacid. The major function of antacids move across the membrane into the stomach interior. (Active is to neutralize excess HCl in gastric juice. The table on p. 707 transport processes are aided by enzymes.) To maintain electri- lists the active ingredients of some popular antacids. The reac- cal balance, an equal number of Cl2 ions also move from the tions by which these antacids neutralize stomach acid are as blood plasma into the stomach. Once in the stomach, most of follows: these ions are prevented from diffusing back into the blood plasma by cell membranes. NaHCO3 (aq) 1 HCl(aq) ¡ The purpose of the highly acidic medium within the NaCl(aq) 1 H2O(l) 1 CO2 (g) stomach is to digest food and to activate certain digestive en- CaCO3 (s) 1 2HCl(aq) ¡ CaCl2 (aq) 1 H2O(l) 1 CO2 (g) zymes. Eating stimulates H1 ion secretion. A small fraction of these ions normally are reabsorbed by the mucosa, causing MgCO3 (s) 1 2HCl(aq) ¡ many tiny hemorrhages. About half a million cells are shed by MgCl2 (aq) 1 H2O(l) 1 CO2 (g) the lining every minute, and a healthy stomach is completely Mg(OH) 2 (s) 1 2HCl(aq) ¡ MgCl2 (aq) 1 2H2O(l) relined every three days or so. However, if the acid content is Al(OH) 2NaCO3 (s) 1 4HCl(aq) ¡ excessively high, the constant influx of H1 ions through the AlCl3 (aq) 1 NaCl(aq) 1 3H2O(l) 1 CO2 (g) 706  The CO2 released by most of these reactions increases gas pres- sure in the stomach, causing the person to belch. The fizzing that takes place when an Alka-Seltzer tablet dissolves in water is caused by carbon dioxide, which is released by the reaction between citric acid and sodium bicarbonate: C4H7O5(COOH)(aq) NaHCO3(aq) 88n citric acid C4H7O5COONa(aq) H2O(l) CO2(g) sodium citrate This action helps to disperse the ingredients and even enhances the taste of the solution. The mucosa of the stomach is also damaged by the action of aspirin, the chemical name of which is acetylsalicylic acid. Aspirin is itself a moderately weak acid: O O B B When an Alka-Seltzer tablet dissolves in water, the bicarbonate ions in it OOOCOCH3 OOOCOCH3 react with the acid component in the tablet to produce carbon dioxide gas. 3:4 OCOOH OCOO  H B B O O acetylsalicylic acid acetylsalicylate ion In the presence of the high concentration of H1 ions in the Some Common Commercial Antacid Preparations stomach, this acid remains largely nonionized. A relatively nonpolar molecule, acetylsalicylic acid has the ability to Commercial Name Active Ingredients penetrate membrane barriers that are also made up of nonpo- Alka-2 Calcium carbonate lar molecules. However, inside the membrane are many Alka-Seltzer Aspirin, sodium bicarbonate, small water pockets, and when an acetylsalicylic acid mol- citric acid ecule enters such a pocket, it ionizes into H1 and acetylsalicy- late ions. These ionic species become trapped in the interior Bufferin Aspirin, magnesium carbonate, regions of the membrane. The continued buildup of ions in aluminum glycinate this fashion weakens the structure of the membrane and Buffered aspirin Aspirin, magnesium carbonate, eventually causes bleeding. Approximately 2 mL of blood aluminum hydroxide-glycine are usually lost for every aspirin tablet taken, an amount not Milk of magnesia Magnesium hydroxide generally considered harmful. However, the action of aspirin Rolaids Dihydroxy aluminum sodium can result in severe bleeding in some individuals. It is inter- carbonate esting to note that the presence of alcohol makes acetylsali- Tums Calcium carbonate cylic acid even more soluble in the membrane, and so further promotes the bleeding. 707 708 Chapter 15 ■ Acids and Bases Although the Lewis definition of acids and bases has greater significance because of its generality, we normally speak of “an acid” and “a base” in terms of the Brøn- sted definition. The term “Lewis acid” usually is reserved for substances that can accept a pair of electrons but do not contain ionizable hydrogen atoms. Example 15.15 classifies Lewis acids and Lewis bases. Example 15.15 Identify the Lewis acid and Lewis base in each of the following reactions: (a) C2H5OC2H5 1 AlCl3 Δ (C2H5 ) 2OAlCl3 (b) Hg21 (aq) 1 4CN2 (aq) Δ Hg(CN) 22 4 (aq) Strategy In Lewis acid-base reactions, the acid is usually a cation or an electron-deficient molecule, whereas the base is an anion or a molecule containing an atom with lone pairs. (a) Draw the molecular structure for C2H5OC2H5. What is the hybridization state of Al in AlCl3? (b) Which ion is likely to be an electron acceptor? An electron donor? Solution (a) The Al is sp2-hybridized in AlCl3 with an empty 2pz orbital. It is electron-deficient, sharing only six electrons. Therefore, the Al atom has a tendency to gain two electrons to complete its octet. This property makes AlCl3 a Lewis acid. On the other hand, the lone pairs on the oxygen atom in C2H5OC2H5 make the compound a Lewis base: What are the formal charges on Al and O in the product? (b) Here the Hg21 ion accepts four pairs of electrons from the CN2 ions. Therefore, Similar problem: 15.94. Hg21 is the Lewis acid and CN2 is the Lewis base. Practice Exercise Identify the Lewis acid and Lewis base in the reaction Co31 (aq) 1 6NH3 (aq) Δ Co(NH3 ) 31 6 (aq) Review of Concepts Which of the following cannot behave as a Lewis base? (a) NH3, (b) OF2, (c) CH4, (d) OH2, (e) Fe31. Key Equations Kw 5 [H1][OH2] (15.3) Ion-product constant of water. pH 5 2log [H1] (15.4) Definition of pH of a solution. 1 2pH [H ] 5 10 (15.5) Calculating H1 ion concentration from pH. pOH 5 2log [OH2] (15.7) Definition of pOH of a solution. 2 2pOH [OH ] 5 10 (15.8) Calculating OH2 ion concentration from pOH. pH 1 pOH 5 14.00 (15.9) Another form of Equation (15.3). ionized acid concentration at equilibrium percent ionization 5 3 100% (15.11) initial concentration of acid KaKb 5 Kw (15.12) Relationship between the acid and base ionization constants of a conjugate acid-base pair.  Questions & Problems 709 Summary of Facts & Concepts 1. Brønsted acids donate protons, and Brønsted bases ac- 7. The product of the ionization constant of an acid and cept protons. These are the definitions that normally the ionization constant of its conjugate base is equal to underlie the use of the terms “acid” and “base.” the ion-product constant of water. 2. The acidity of an aqueous solution is expressed as its 8. The relative strengths of acids can be explained qualita- pH, which is defined as the negative logarithm of the tively in terms of their molecular structures. hydrogen ion concentration (in mol/L). 9. Most salts are strong electrolytes that dissociate com- 3. At 25°C, an acidic solution has pH , 7, a basic solution pletely into ions in solution. The reaction of these ions has pH . 7, and a neutral solution has pH 5 7. with water, called salt hydrolysis, can produce acidic or 4. In aqueous solution, the following are classified as basic solutions. In salt hydrolysis, the conjugate bases strong acids: HClO4, HI, HBr, HCl, H2SO4 (first stage of weak acids yield basic solutions, and the conjugate of ionization), and HNO3. Strong bases in aqueous solu- acids of weak bases yield acidic solutions. tion include hydroxides of alkali metals and of alkaline 10. Small, highly charged metal ions, such as Al31 and earth metals (except beryllium). Fe31, hydrolyze to yield acidic solutions. 5. The acid ionization constant Ka increases with acid 11. Most oxides can be classified as acidic, basic, or ampho- strength. Kb similarly expresses the strengths of teric. Metal hydroxides are either basic or amphoteric. bases. 12. Lewis acids accept pairs of electrons and Lewis bases 6. Percent ionization is another measure of the strength of donate pairs of electrons. The term “Lewis acid” is gen- acids. The more dilute a solution of a weak acid, the erally reserved for substances that can accept electron greater the percent ionization of the acid. pairs but do not contain ionizable hydrogen atoms. Key Words Acid ionization constant Conjugate acid-base Lewis base, p. 704 Strong acid, p. 673 (Ka), p. 678 pair, p. 667 Percent ionization, p. 684 Strong base, p. 674 Base ionization constant Ion-product constant, p. 669 pH, p. 670 Weak acid, p. 674 (Kb), p. 685 Lewis acid, p. 704 Salt hydrolysis, p. 696 Weak base, p. 675 Questions & Problems† • Problems available in Connect Plus • 15.4 Write the formulas of the conjugate bases of the Red numbered problems solved in Student Solutions Manual following acids: (a) HNO2, (b) H2SO4, (c) H2S, (d) HCN, (e) HCOOH (formic acid). Brønsted Acids and Bases • 15.5 Identify the acid-base conjugate pairs in each of the Review Questions following reactions: (a) CH3COO2 1 HCN Δ CH3COOH 1 CN2 15.1 Define Brønsted acids and bases. Give an example of a conjugate pair in an acid-base reaction. (b) HCO2 2 3 1 HCO3 Δ H2CO3 1 CO3 22 2 22 1 15.2 In order for a species to act as a Brønsted base, an (c) H2PO4 1 NH3 Δ HPO4 1 NH4 atom in the species must possess a lone pair of elec- (d) HClO 1 CH3NH2 Δ CH3NH1 3 1 ClO 2 22 2 2 trons. Explain why this is so. (e) CO3 1 H2O Δ HCO3 1 OH • 15.6 Write the formula for the conjugate acid of each of Problems the following bases: (a) HS2, (b) HCO2 22 3 , (c) CO3 , (d) H2PO4 , (e) HPO4 , (f) PO4 , (g) HSO4 , (h) SO22 2 22 32 2 4 , • 15.3 Classify each of the following species as a Brønsted (i) SO22 acid or base, or both: (a) H2O, (b) OH2, (c) H3O1, 3 . (d) NH3, (e) NH1 2 2 22 4 , (f) NH2 , (g) NO3 , (h) CO3 , (i) HBr, ( j) HCN. † Unless otherwise stated, the temperature is assumed to be 25°C.  710 Chapter 15 ■ Acids and Bases • 15.7 Oxalic acid (H2C2O4) has the following structure: 15.20 Calculate the hydrogen ion concentration in mol/L for each of the following solutions: (a) a solution whose OPCOOH pH is 5.20, (b) a solution whose pH is 16.00, (c) a solu- A OPCOOH tion whose hydroxide concentration is 3.7 3 1029 M. • 15.21 Complete the following table for a solution: An oxalic acid solution contains the following spe- cies in varying concentrations: H2C2O4, HC2O42, pH [H1] Solution is C2O422, and H1. (a) Draw Lewis structures of ,7 HC2O42, and C2O422. (b) Which of the above four ,1.0 3 1027 M species can act only as acids, which can act only Neutral as bases, and which can act as both acids and bases? • 15.22 Fill in the word acidic, basic, or neutral for the fol- • 15.8 Write the formula for the conjugate base of each of lowing solutions: the following acids: (a) CH2ClCOOH, (b) HIO4, (a) pOH . 7; solution is (c) H3PO4, (d) H2PO2 22 4 , (e) HPO4 , (f ) H2SO4 , (b) pOH 5 7; solution is (g) HSO4 , (h) HIO3, (i) HSO3 , ( j) NH41 , (k) H2S, 2 2 (l) HS2, (m) HClO. (c) pOH , 7; solution is • 15.23 The pOH of a strong base solution is 1.88 at 25°C. Calculate the concentration of the base (a) if the The Acid-Base Properties of Water base is KOH and (b) if the base is Ba(OH)2. Review Questions • 15.24 Calculate the number of moles of KOH in 5.50 mL 15.9 What is the ion-product constant for water? of a 0.360 M KOH solution. What is the pOH of the 15.10 Write an equation relating [H1] and [OH2] in solu- solution? tion at 25°C. • 15.25 How much NaOH (in grams) is needed to prepare 15.11 The ion-product constant for water is 1.0 3 10214 at 546 mL of solution with a pH of 10.00? 25°C and 3.8 3 10214 at 40°C. Is the forward process 15.26 A solution is made by dissolving 18.4 g of HCl in 662 mL of water. Calculate the pH of the solution. H2O(l) Δ H1 (aq) 1 OH2 (aq) (Assume that the volume remains constant.) endothermic or exothermic? Strength of Acids and Bases pH—A Measure of Acidity Review Questions Review Questions 15.27 Explain what is meant by the strength of an acid. 15.12 Define pH. Why do chemists normally choose to 15.28 Without referring to the text, write the formulas of discuss the acidity of a solution in terms of pH rather four strong acids and four weak acids. than hydrogen ion concentration, [H1]? • 15.29 What are the strongest acid and strongest base that 15.13 The pH of a solution is 6.7. From this statement can exist in water? alone, can you conclude that the solution is acidic? 15.30 H2SO4 is a strong acid, but HSO24 is a weak acid. If not, what additional information would you need? Account for the difference in strength of these two Can the pH of a solution be zero or negative? If so, related species. give examples to illustrate these values. Problems 15.14 Define pOH. Write the equation relating pH and pOH. • 15.31 Which of the following diagrams best represents a Problems strong acid, such as HCl, dissolved in water? Which represents a weak acid? Which represents a • 15.15 Calculate the concentration of OH2 ions in a very weak acid? (The hydrated proton is shown as 1.4 3 1023 M HCl solution. a hydronium ion. Water molecules are omitted • 15.16 Calculate the concentration of H1 ions in a 0.62 M for clarity.) NaOH solution. • 15.17 Calculate the pH of each of the following solutions: (a) 0.0010 M HCl, (b) 0.76 M KOH. • 15.18 Calculate the pH of each of the following solutions: (a) 2.8 3 1024 M Ba(OH)2, (b) 5.2 3 1024 M HNO3. • 15.19 Calculate the hydrogen ion concentration in mol/L for solutions with the following pH values: (a) 2.42, (b) 11.21, (c) 6.96, (d) 15.00. (a) (b) (c) (d)  Questions & Problems 711 • 15.32 (1) Which of the following diagrams represents a Problems solution of a weak diprotic acid? (2) Which dia- grams represent chemically implausible situations? • 15.43 The Ka for benzoic acid is 6.5 3 1025. Calculate the (The hydrated proton is shown as a hydronium ion. pH of a 0.10 M benzoic acid solution. Water molecules are omitted for clarity.) 15.44 A 0.0560-g quantity of acetic acid is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H1, CH3COO2, and CH3COOH at equilibrium. (Ka for acetic acid 5 1.8 3 1025.) • 15.45 The pH of an acid solution is 6.20. Calculate the Ka for the acid. The initial acid concentration is 0.010 M. 15.46 What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium? (a) (b) (c) (d) • 15.47 Calculate the percent ionization of benzoic acid having the following concentrations: (a) 0.20 M, • 15.33 Classify each of the following species as a weak (b) 0.00020 M. or strong acid: (a) HNO3, (b) HF, (c) H2SO4, (d) HSO24 , (e) H2CO3, (f ) HCO32, (g) HCl, (h) HCN, 15.48 Calculate the percent ionization of hydrofluoric (i) HNO2. acid at the following concentrations: (a) 0.60 M, 15.34 Classify each of the following species as a weak or (b) 0.0046 M, (c) 0.00028 M. Comment on the trends. strong base: (a) LiOH, (b) CN2, (c) H2O, (d) ClO2 4, • 15.49 A 0.040 M solution of a monoprotic acid is 14 percent (e) NH22. ionized. Calculate the ionization constant of the acid. • 15.35 Which of the following statements is/are true for a • 15.50 (a) Calculate the percent ionization of a 0.20 M solu- 0.10 M solution of a weak acid HA? tion of the monoprotic acetylsalicylic acid (aspirin) (a) The pH is 1.00. for which Ka 5 3.0 3 1024. (b) The pH of gastric juice in the stomach of a certain individual is 1.00. (b) [H1] @ [A2] After a few aspirin tablets have been swallowed, the (c) [H1] 5 [A2] concentration of acetylsalicylic acid in the stomach is (d) The pH is less than 1. 0.20 M. Calculate the percent ionization of the acid • 15.36 Which of the following statements is/are true under these conditions. What effect does the nonion- regarding a 1.0 M solution of a strong acid HA? ized acid have on the membranes lining the stomach? (a) [A2] . [H1] (Hint: See the Chemistry in Action essay on p. 706.) (b) The pH is 0.00. (c) [H1] 5 1.0 M Weak Bases and Base Ionization Constants (d) [HA] 5 1.0 M Review Questions • 15.37 Predict the direction that predominates in this reaction: 15.51 Use NH3 to illustrate what we mean by the strength of a base. F2 (aq) 1 H2O(l) Δ HF(aq) 1 OH2 (aq) 15.52 Which of the following has a higher pH? (a) 0.20 M 15.38 Predict whether the following reaction will proceed NH3, (b) 0.20 M NaOH from left to right to any measurable extent: CH3COOH(aq) 1 Cl2 (aq) ¡ Problems 15.53 Calculate the pH of a 0.24 M solution of a weak base with a Kb of 3.5 3 1026. Weak Acids and Acid Ionization Constants 15.54 The diagrams here represent three different weak Review Questions base solutions of equal concentration. List the bases 15.39 What does the ionization constant tell us about the in order of increasing Kb value. (Water molecules strength of an acid? are omitted for clarity.) 15.40 List the factors on which the Ka of a weak acid B HB1 OH2 depends. 15.41 Why do we normally not quote Ka values for strong acids such as HCl and HNO3? Why is it necessary to specify temperature when giving Ka values? • 15.42 Which of the following solutions has the highest pH? (a) 0.40 M HCOOH, (b) 0.40 M HClO4, (c) 0.40 M CH3COOH. (a) (b) (c)  712 Chapter 15 ■ Acids and Bases • 15.55 Calculate the pH for each of the following solutions: • 15.71 Which of the following is the stronger acid: (a) 0.10 M NH3, (b) 0.050 M C5H5N (pyridine). CH2ClCOOH or CHCl2COOH? Explain your choice. • 15.56 The pH of a 0.30 M solution of a weak base is 10.66. 15.72 Consider the following compounds: What is the Kb of the base? • 15.57 What is the original molarity of a solution of ammo- OOOH CH3OOOH nia whose pH is 11.22? phenol methanol • 15.58 In a 0.080 M NH3 solution, what percent of the NH3 Experimentally, phenol is found to be a stronger acid is present as NH14? than methanol. Explain this difference in terms of the structures of the conjugate bases. (Hint: A more The Relationship Between the Ionization stable conjugate base favors ionization. Only one of Constants of Acids and Their Conjugate Bases the conjugate bases can be stabilized by resonance.) Review Questions 15.59 Write the equation relating Ka for a weak acid and Kb Acid-Base Properties of Salts for its conjugate base. Use NH3 and its conjugate acid Review Questions NH1 4 to derive the relationship between Ka and Kb. 15.73 Define salt hydrolysis. Categorize salts according to 15.60 From the relationship KaKb 5 Kw, what can you de- how they affect the pH of a solution. duce about the relative strengths of a weak acid and 15.74 Explain why small, highly charged metal ions are its conjugate base? able to undergo hydrolysis. 15.75 Al31 is not a Brønsted acid but Al(H2O) 31 6 is. Ex- Diprotic and Polyprotic Acids plain. Review Questions • 15.76 Specify which of the following salts will undergo 15.61 Carbonic acid is a diprotic acid. Explain what that hydrolysis: KF, NaNO3, NH4NO2, MgSO4, KCN, means. C6H5COONa, RbI, Na2CO3, CaCl2, HCOOK. 15.62 Write all the species (except water) that are present in a phosphoric acid solution. Indicate which species can Problems act as a Brønsted acid, which as a Brønsted base, and • 15.77 Predict the pH (. 7, , 7, or < 7) of aqueous which as both a Brønsted acid and a Brønsted base. solutions containing the following salts: (a) KBr, (b) Al(NO3)3, (c) BaCl2, (d) Bi(NO3)3. Problems • 15.78 Predict whether the following solutions are acidic, basic, or nearly neutral: (a) NaBr, (b) K2SO3, • 15.63 The first and second ionization constants of a di- (c) NH4NO2, (d) Cr(NO3)3. protic acid H2A are Ka1 and Ka2 at a certain tempera- ture. Under what conditions will [A22 ] 5 Ka2 ? 15.79 A certain salt, MX (containing the M1 and X2 ions), 15.64 Compare the pH of a 0.040 M HCl solution with that is dissolved in water, and the pH of the resulting of a 0.040 M H2SO4 solution. (Hint: H2SO4 is a solution is 7.0. Can you say anything about the strong acid; Ka for HSO2 22 strengths of the acid and the base from which the 4 5 1.3 3 10 .) salt is derived? • 15.65 What are the concentrations of HSO2 1 22 4 , SO4 , and • 15.80 In a certain experiment a student finds that the pHs H in a 0.20 M KHSO4 solution? of 0.10 M solutions of three potassium salts KX, • 15.66 Calculate the concentrations of H1, HCO32, and KY, and KZ are 7.0, 9.0, and 11.0, respectively. Ar- CO223 in a 0.025 M H2CO3 solution. range the acids HX, HY, and HZ in the order of in- creasing acid strength. Molecular Structure and the Strength of Acids • 15.81 Calculate the pH of a 0.36 M CH3COONa solution. Review Questions 15.82 Calculate the pH of a 0.42 M NH4Cl solution. 15.67 List four factors that affect the strength of an acid. • 15.83 Predict the pH (. 7, , 7, < 7) of a NaHCO3 15.68 How does the strength of an oxoacid depend on the solution. electronegativity and oxidation number of the cen- • 15.84 Predict whether a solution containing the salt tral atom? K2HPO4 will be acidic, neutral, or basic. Problems Acidic and Basic Oxides and Hydroxides • 15.69 Predict the acid strengths of the following com- Review Questions pounds: H2O, H2S, and H2Se. • 15.85 Classify the following oxides as acidic, basic, ampho- • 15.70 Compare the strengths of the following pairs of acids: teric, or neutral: (a) CO2, (b) K2O, (c) CaO, (d) N2O5, (a) H2SO4 and H2SeO4, (b) H3PO4 and H3AsO4. (e) CO, (f) NO, (g) SnO2, (h) SO3, (i) Al2O3, (j) BaO.  Questions & Problems 713 • 15.86 Write equations for the reactions between (a) CO2 bases in increasing order of Kb. (c) Calculate the per- and NaOH(aq), (b) Na2O and HNO3(aq). cent ionization of each acid. (d) Which of the 0.1 M sodium salt solutions (NaX, NaY, or NaZ) has the Problems lowest pH? (The hydrated proton is shown as a hydro- nium ion. Water molecules are omitted for clarity.) 15.87 Explain why metal oxides tend to be basic if the oxidation number of the metal is low and acidic if the oxidation number of the metal is high. (Hint: Metallic compounds in which the oxidation num- bers of the metals are low are more ionic than those in which the oxidation numbers of the met- als are high.) • 15.88 Arrange the oxides in each of the following groups in order of increasing basicity: (a) K2O, Al2O3, BaO, (b) CrO3, CrO, Cr2O3. HX HY HZ • 15.89 Zn(OH)2 is an amphoteric hydroxide. Write balanced ionic equations to show its reaction with (a) HCl, • 15.100 A typical reaction between an antacid and the hy- (b) NaOH [the product is Zn(OH)22 4 ]. drochloric acid in gastric juice is 15.90 Al(OH)3 is an insoluble compound. It dissolves in NaHCO3 (s) 1 HCl(aq) Δ excess NaOH in solution. Write a balanced ionic NaCl(aq) 1 H2O(l) 1 CO2 (g) equation for this reaction. What type of reaction is this? Calculate the volume (in L) of CO2 generated from 0.350 g of NaHCO3 and excess gastric juice at Lewis Acids and Bases 1.00 atm and 37.0°C. Review Questions • 15.101 To which of the following would the addition of an equal volume of 0.60 M NaOH lead to a solution 15.91 What are the Lewis definitions of an acid and a having a lower pH? (a) water, (b) 0.30 M HCl, base? In what way are they more general than the (c) 0.70 M KOH, (d) 0.40 M NaNO3. Brønsted definitions? 15.102 The pH of a 0.0642 M solution of a monoprotic acid 15.92 In terms of orbitals and electron arrangements, what is 3.86. Is this a strong acid? must be present for a molecule or an ion to act as a 15.103 Like water, liquid ammonia undergoes autoionization: Lewis acid (use H1 and BF3 as examples)? What must be present for a molecule or ion to act as a NH3 1 NH3 Δ NH1 2 4 1 NH2 Lewis base (use OH2 and NH3 as examples)? (a) Identify the Brønsted acids and Brønsted bases in this reaction. (b) What species correspond to Problems H1 and OH2 and what is the condition for a • 15.93 Classify each of the following species as a Lewis neutral solution? acid or a Lewis base: (a) CO2, (b) H2O, (c) I2, (d) SO2, 15.104 HA and HB are both weak acids although HB is the (e) NH3, (f) OH2, (g) H1, (h) BCl3. stronger of the two. Will it take a larger volume of a 15.94 Describe the following reaction in terms of the 0.10 M NaOH solution to neutralize 50.0 mL of 0.10 M Lewis theory of acids and bases: HB than would be needed to neutralize 50.0 mL of 0.10 M HA? AlCl3 (s) 1 CI2 (aq) ¡ AlCl2 4 (aq) 15.105 A solution contains a weak monoprotic acid HA and • 15.95 Which would be considered a stronger Lewis acid: its sodium salt NaA both at 0.1 M concentration. (a) BF3 or BCl3, (b) Fe21 or Fe31? Explain. Show that [OH2] 5 Kw/Ka. 15.96 All Brønsted acids are Lewis acids, but the reverse is 15.106 The three common chromium oxides are CrO, Cr2O3, not true. Give two examples of Lewis acids that are and CrO3. If Cr2O3 is amphoteric, what can you say not Brønsted acids. about the acid-base properties of CrO and CrO3? • 15.107 Use the data in Table 15.3 to calculate the equilib- Additional Problems rium constant for the following reaction: 15.97 Determine the concentration of a NaNO2 solution HCOOH(aq) 1 OH2 (aq) Δ that has a pH of 8.22. HCOO2 (aq) 1 H2O(l2 15.98 Determine the concentration of a NH4Cl solution that has a pH of 5.64. • 15.108 Use the data in Table 15.3 to calculate the equilib- rium constant for the following reaction: • 15.99 The diagrams here show three weak acids HA (A 5 X, Y, or Z) in solution. (a) Arrange the acids CH3COOH(aq) 1 NO2 2 (aq) Δ in order of increasing Ka. (b) Arrange the conjugate CH3COO2 (aq) 1 HNO2 (aq)  714 Chapter 15 ■ Acids and Bases 15.109 Most of the hydrides of Group 1A and Group 2A • 15.121 Calculate the pH of a 2.00 M NH4CN solution. metals are ionic (the exceptions are BeH2 and • 15.122 Calculate the concentrations of all species in a MgH2, which are covalent compounds). (a) De- 0.100 M H3PO4 solution. scribe the reaction between the hydride ion (H2) 15.123 Identify the Lewis acid and Lewis base that lead to and water in terms of a Brønsted acid-base reac- the formation of the following species: (a) AlCl24 , tion. (b) The same reaction can also be classified (b) Cd(CN) 22 2 4 , (c) HCO3 , (d) H2SO4. as a redox reaction. Identify the oxidizing and reducing agents. 15.124 Very concentrated NaOH solutions should not be stored in Pyrex glassware. Why? (Hint: See • 15.110 Calculate the pH of a 0.20 M ammonium acetate Section 11.7.) (CH3COONH4) solution. 15.111 Novocaine, used as a local anesthetic by dentists, is • 15.125 In the vapor phase, acetic acid molecules associate to a certain extent to form dimers: a weak base (Kb 5 8.91 3 1026). What is the ratio of the concentration of the base to that of its acid in the 2CH3COOH(g) Δ (CH3COOH) 2 (g) blood plasma (pH 5 7.40) of a patient? At 51°C the pressure of a certain acetic acid vapor 15.112 Which of the following is the stronger base: NF3 or system is 0.0342 atm in a 360-mL flask. The vapor NH3? (Hint: F is more electronegative than H.) is condensed and neutralized with 13.8 mL of 15.113 Which of the following is a stronger base: NH3 or 0.0568 M NaOH. (a) Calculate the degree of disso- PH3? (Hint: The N¬H bond is stronger than the ciation (α) of the dimer under these conditions: P¬H bond.) • 15.114 The ion product of D2O is 1.35 3 10215 at 25°C. (CH3COOH) 2 Δ 2CH3COOH (a) Calculate pD where pD 5 2log [D1]. (b) For (Hint: See Problem 14.117 for general procedure.) what values of pD will a solution be acidic in D2O? (b) Calculate the equilibrium constant KP for the (c) Derive a relation between pD and pOD. reaction in (a). 15.115 Give an example of (a) a weak acid that contains oxygen atoms, (b) a weak acid that does not contain • 15.126 Calculate the concentrations of all the species in a 0.100 M Na2CO3 solution. oxygen atoms, (c) a neutral molecule that acts as a Lewis acid, (d) a neutral molecule that acts as a • 15.127 Henry’s law constant for CO2 at 38°C is 2.28 3 1023 Lewis base, (e) a weak acid that contains two ioniz- mol/L ? atm. Calculate the pH of a solution of CO2 able H atoms, (f) a conjugate acid-base pair, both of at 38°C in equilibrium with the gas at a partial pres- which react with HCl to give carbon dioxide gas. sure of 3.20 atm. 15.128 Hydrocyanic acid (HCN) is a weak acid and a deadly • 15.116 What is the pH of 250.0 mL of an aqueous solution poisonous compound—in the gaseous form (hydro- containing 0.616 g of the strong acid trifluorometh- ane sulfonic acid (CF3SO3H)? gen cyanide) it is used in gas chambers. Why is it dangerous to treat sodium cyanide with acids (such 15.117 (a) Use VSEPR to predict the geometry of the hy- as HCl) without proper ventilation? dronium ion, H3O1. (b) The O atom in H2O has two lone pairs and in principle can accept two H1 ions. • 15.129 How many grams of NaCN would you need to dis- Explain why the species H4O21 does not exist. What solve in enough water to make exactly 250 mL of would be its geometry if it did exist? solution with a pH of 10.00? 15.118 HF is a weak acid, but its strength increases with • 15.130 A solution of formic acid (HCOOH) has a pH of concentration. Explain. (Hint: F2 reacts with HF to 2.53. How many grams of formic acid are there in form HF2 2 . The equilibrium constant for this reac- 100.0 mL of the solution? tion is 5.2 at 25°C.) • 15.131 Calculate the pH of a 1-L solution containing 0.150 15.119 When chlorine reacts with water, the resulting mole of CH3COOH and 0.100 mole of HCl. solution is weakly acidic and reacts with AgNO3 • 15.132 A 1.87-g sample of Mg reacts with 80.0 mL of a to give a white precipitate. Write balanced equa- HCl solution whose pH is 20.544. What is the pH tions to represent these reactions. Explain why of the solution after all the Mg has reacted? Assume manufacturers of household bleaches add bases constant volume. such as NaOH to their products to increase their 15.133 You are given two beakers, one containing an aque- effectiveness. ous solution of strong acid (HA) and the other an • 15.120 When the concentration of a strong acid is not sub- aqueous solution of weak acid (HB) of the same stantially higher than 1.0 3 1027 M, the ionization concentration. Describe how you would compare of water must be taken into account in the calcula- the strengths of these two acids by (a) measuring tion of the solution’s pH. (a) Derive an expression the pH, (b) measuring electrical conductance, for the pH of a strong acid solution, including the (c) studying the rate of hydrogen gas evolution contribution to [H1] from H2O. (b) Calculate the pH when these solutions are reacted with an active of a 1.0 3 1027 M HCl solution. metal such as Mg or Zn.  Questions & Problems 715 15.134 Use Le Châtelier’s principle to predict the effect of 15.143 About half of the hydrochloric acid produced annu- the following changes on the extent of hydrolysis of ally in the United States (3.0 billion pounds) is used sodium nitrite (NaNO2) solution: (a) HCl is added, in metal pickling. This process involves the removal (b) NaOH is added, (c) NaCl is added, (d) the solu- of metal oxide layers from metal surfaces to prepare tion is diluted. them for coating. (a) Write the overall and net ionic 15.135 Describe the hydration of SO2 as a Lewis acid-base equations for the reaction between iron(III) oxide, reaction. (Hint: Refer to the discussion of the hydra- which represents the rust layer over iron, and HCl. tion of CO2 on p. 705.) Identify the Brønsted acid and base. (b) Hydrochlo- ric acid is also used to remove scale (which is mostly 15.136 The disagreeable odor of fish is mainly due to or- CaCO3) from water pipes (see p. 126). Hydrochloric ganic compounds (RNH2) containing an amino acid reacts with calcium carbonate in two stages; the group, ¬NH2, where R is the rest of the molecule. first stage forms the bicarbonate ion, which then Amines are bases just like ammonia. Explain why reacts further to form carbon dioxide. Write equa- putting some lemon juice on fish can greatly reduce tions for these two stages and for the overall reac- the odor. tion. (c) Hydrochloric acid is used to recover oil • 15.137 A solution of methylamine (CH3NH2) has a pH of from the ground. It dissolves rocks (often CaCO3) so 10.64. How many grams of methylamine are there in that the oil can flow more easily. In one process, a 100.0 mL of the solution? 15 percent (by mass) HCl solution is injected into • 15.138 A 0.400 M formic acid (HCOOH) solution freezes at an oil well to dissolve the rocks. If the density of 20.758°C. Calculate the Ka of the acid at that tem- the acid solution is 1.073 g/mL, what is the pH of perature. (Hint: Assume that molarity is equal to the solution? molality. Carry your calculations to three significant • 15.144 Which of the following does not represent a Lewis figures and round off to two for Ka.) acid-base reaction? 15.139 Both the amide ion (NH22) and the nitride ion (N32) (a) H2O 1 H1 ¡ H3O1 are stronger bases than the hydroxide ion and (b) NH3 1 BF3 ¡ H3NBF3 hence do not exist in aqueous solutions. (a) Write (c) PF3 1 F2 ¡ PF5 equations showing the reactions of these ions with water, and identify the Brønsted acid and (d) Al(OH) 3 1 OH2 ¡ Al(OH) 2 4 base in each case. (b) Which of the two is the 15.145 True or false? If false, explain why the statement stronger base? is wrong. (a) All Lewis acids are Brønsted acids, 15.140 The atmospheric sulfur dioxide (SO2) concentration (b) the conjugate base of an acid always carries a over a certain region is 0.12 ppm by volume. Calcu- negative charge, (c) the percent ionization of a base late the pH of the rainwater due to this pollutant. increases with its concentration in solution, (d) a solu- Assume that the dissolution of SO2 does not affect tion of barium fluoride is acidic. its pressure. 15.146 How many milliliters of a strong monoprotic acid 15.141 Calcium hypochlorite [Ca(OCl)2] is used as a disin- solution at pH 5 4.12 must be added to 528 mL fectant for swimming pools. When dissolved in wa- of the same acid solution at pH 5 5.76 to change ter it produces hypochlorous acid its pH to 5.34? Assume that the volumes are additive. Ca(OCl) 2 (s) 1 2H2O(l) Δ 2HClO(aq) 1 Ca(OH) 2 (s) • 15.147 Calculate the pH and percent ionization of a 0.80 M HNO2 solution. which ionizes as follows: 15.148 Consider the two weak acids HX (molar mass 5 180 g/mol) and HY (molar mass 5 78.0 g/mol). If a HClO(aq) Δ H1 (aq) 1 ClO2 (aq) solution of 16.9 g/L of HX has the same pH as one Ka 5 3.0 3 1028 containing 9.05 g/L of HY, which is the stronger As strong oxidizing agents, both HClO and ClO2 acid at these concentrations? can kill bacteria by destroying their cellular compo- 15.149 Hemoglobin (Hb) is a blood protein that is respon- nents. However, too high a HClO concentration is sible for transporting oxygen. It can exist in the pro- irritating to the eyes of swimmers and too high a tonated form as HbH1. The binding of oxygen can concentration of ClO2 will cause the ions to decom- be represented by the simplified equation pose in sunlight. The recommended pH for pool water is 7.8. Calculate the percent of these species HbH1 1 O2 Δ HbO2 1 H1 present at this pH. (a) What form of hemoglobin is favored in the lungs 15.142 Explain the action of smelling salt, which is ammo- where oxygen concentration is highest? (b) In body nium carbonate [(NH4)2CO3]. (Hint: The thin film tissues, where the cells release carbon dioxide pro- of aqueous solution that lines the nasal passage is duced by metabolism, the blood is more acidic due slightly basic.) to the formation of carbonic acid. What form of  716 Chapter 15 ■ Acids and Bases hemoglobin is favored under this condition? 15.156 A 10.0-g sample of white phosphorus was burned in (c) When a person hyperventilates, the concentra- an excess of oxygen. The product was dissolved in tion of CO2 in his or her blood decreases. How does enough water to make 500 mL of solution. Calculate this action affect the above equilibrium? Frequently the pH of the solution at 25°C. a person who is hyperventilating is advised to 15.157 Calculate the pH of a 0.20 M NaHCO3 solution. breathe into a paper bag. Why does this action help (Hint: As an approximation, calculate hydrolysis the individual? and ionization separately first, followed by partial 15.150 A 1.294-g sample of a metal carbonate (MCO3) is neutralization.) reacted with 500 mL of a 0.100 M HCl solution. The 15.158 (a) Shown here is a solution containing hydroxide excess HCl acid is then neutralized by 32.80 mL of ions and hydronium ions. What is the pH of the solu- 0.588 M NaOH. Identify M. tion? (b) How many H3O1 ions would you need to 15.151 Prove the statement that when the concentration of a draw for every OH2 ion if the pH of the solution is weak acid HA decreases by a factor of 10, its per- 5.0? The color codes are H3O1 (red) and OH2 cent ionization increases by a factor of 110. State (green). Water molecules and counter ions are omit- any assumptions. ted for clarity. 15.152 Calculate the pH of a solution that is 1.00 M HCN and 1.00 M HF. Compare the concentration (in mo- larity) of the CN2 ion in this solution with that in a 1.00 M HCN solution. Comment on the difference. 15.153 Teeth enamel is hydroxyapatite [Ca3(PO4)3OH]. When it dissolves in water (a process called demin- eralization), it dissociates as follows: Ca5 (PO4 ) 3OH ¡ 5Ca21 1 3PO32 4 1 OH 2 The reverse process, called remineralization, is the body’s natural defense against tooth decay. Acids 15.159 In this chapter, HCl, HBr, and HI are all listed as produced from food remove the OH2 ions and strong acids because they are assumed to be ion- thereby weaken the enamel layer. Most toothpastes ized completely in water. If, however, we choose a contain a fluoride compound such as NaF or SnF2. solvent such as acetic acid that is a weaker Brønsted What is the function of these compounds in prevent- base than water, it is possible to rank the acids in ing tooth decay? increasing strength as HCl , HBr , HI. (a) Write 15.154 Use the van’t Hoff equation (see Problem 14.119) equations showing proton transfer between the and the data in Appendix 3 to calculate the pH of acids and CH3COOH. Describe how you would water at its normal boiling point. compare the strength of the acids in this solvent experimentally. (b) Draw a Lewis structure of the • 15.155 At 28°C and 0.982 atm, gaseous compound HA has conjugate acid CH3COOH21 . a density of 1.16 g/L. A quantity of 2.03 g of this compound is dissolved in water and diluted to 15.160 Use the data in Appendix 3 to calculate the ¢H°rxn for exactly 1 L. If the pH of the solution is 5.22 (due to the following reactions: (a) NaOH(aq) 1 HCl(aq) S the ionization of HA) at 25°C, calculate the Ka of NaCl(aq) 1 H2O(l) and (b) KOH(aq) 1 HNO3(aq) S the acid. KNO3(aq) 1 H2O(l). Comment on your results.  Answers to Practice Exercises 717 Interpreting, Modeling & Estimating 15.161 Malonic acid [CH2(COOH)2] is a diprotic acid. Explain why H3PO4(aq) is a triprotic acid, but Compare its two Ka values with that of acetic acid H3PO3(aq) is only a diprotic acid. (CH3COOH) (Ka ), and account for the differences 15.164 Chicken egg shells are composed primarily of cal- in the three Ka values. cium carbonate, CaCO3. In a classic demonstration 15.162 Look up the contents of a Tums tablet. How many carried out in chemistry and biology classes, vine- tablets are needed to increase the pH of the gastric gar is used to remove the shell from an uncooked juice in a person’s stomach from 1.2 to 1.5? egg, revealing the semipermeable membrane that 15.163 Phosphorous acid, H3PO3(aq), is a diprotic acid with surrounds the egg keeping it intact. Refer to the Ka1 5 3 3 1022. (a) After examining the Ka values in Chemical Mystery on p. 774 to see a schematic dia- Table 15.5, estimate Ka2 for H3PO3(aq) and calculate gram of a chicken egg. Estimate the minimum the pH of a 0.10 M solution of Na2HPO3(aq). amount of vinegar required to remove the entire (b) The structure of H3PO3 is given in Figure 15.5. shell from the egg. Answers to Practice Exercises 15.1 (1) H2O (acid) and OH2 (base); (2) HCN (acid) and CN2 (base). 15.2 7.7 3 10215 M. 15.3 0.12. 15.4 4.7 3 1024 M. 15.5 7.40. 15.6 12.56. 15.7 Smaller than 1. 15.8 2.09. 15.9 2.2 3 1026. 15.10 12.03. 15.11 [H2C2O4] 5 0.11 M, [HC2O2 4 ] 5 0.086 M, [C2O22 4 ] 5 6.1 3 10 25 M, [H1] 5 0.086 M. 15.12 HClO2. 15.13 8.58. 15.14 (a) pH < 7, (b) pH . 7, (c) pH , 7, (d) pH . 7. 15.15 Lewis acid: Co31; Lewis base: NH3. CHEMICAL M YS TERY Decaying Papers L ibrarians are worried about their books. Many of the old books in their collections are crumbling. The situation is so bad, in fact, that about one-third of the books in the U.S. Library of Congress cannot be circulated because the pages are too brittle. Why are the books deteriorating? Until the latter part of the eighteenth century, practically all paper produced in the West- ern Hemisphere was made from rags of linen or cotton, which is mostly cellulose. Cellulose is a polymer comprised of glucose (C6H12O6) units joined together in a specific fashion: 冢 冣 H H H H EA E EA HO EA AE GOH OE CH2OH O CH2OH O HO G O G A H A H H A H A H A GA H HO E G HO E G E OE OH AE OH O A A OH A A CH2OH H H H H H Glucose A portion of cellulose As the demand for paper grew, wood pulp was substituted for rags as a source of cellulose. Wood pulp also contains lignin, an organic polymer that imparts rigidity to the paper, but lignin oxidizes easily, causing the paper to discolor. Paper made from wood pulp that has not been treated to remove the lignin is used for books and newspapers for which a long life is not an important consideration. Another problem with paper made from wood pulp is that it is porous. Tiny holes in the surface of the paper soak up ink from a printing press, spreading it over a larger area than is intended. To prevent ink creep, a coating of aluminum sulfate [Al2(SO4)3] and rosin is applied to some paper to seal the holes. This process, called sizing, results in a smooth surface. You can readily tell the difference between papers with and without sizing by feeling the surface of a newspaper and this page. (Or try to write on them with a felt-tip pen.) Aluminum sulfate was chosen for the treatment because it is colorless and cheap. Because paper without sizing does not crumble, aluminum sulfate must be responsible for the slow decay. But how? Chemical Clues 1. When books containing “sized” paper are stored in a high-humidity environment, Al2(SO4)3 absorbs moisture, which eventually leads to the production of H1 ions. The H1 ions catalyze the hydrolysis of cellulose by attaching to the shaded O atoms in cellulose. The long chain unit of glucose units breaks apart, resulting in the crumbling of the paper. Write equations for the production of H1 ions from Al2(SO4)3. 2. To prevent papers from decaying, the obvious solution is to treat them with a base. However, both NaOH (a strong base) and NH3 (a weak base) are unsatisfactory choices. Suggest how you could use these substances to neutralize the acid in the paper and describe their drawbacks. 718  Acid-damaged paper. 3. After much testing, chemists developed a compound that stabilizes paper: diethylzinc [Zn(C2H5)2]. Diethylzinc is volatile so it can be sprayed onto books. It reacts with water to form zinc oxide (ZnO) and gaseous ethane (C2H6). (a) Write an equation for this reaction. (b) ZnO is an amphoteric oxide. What is its reaction with H1 ions? 4. One disadvantage of diethylzinc is that it is extremely flammable in air. Therefore, oxygen must not be present when this compound is applied. How would you remove oxygen from a room before spraying diethylzinc onto stacks of books in a library? 5. Nowadays papers are sized with titanium dioxide (TiO2), which, like ZnO, is a non- toxic white compound that will prevent the hydrolysis of cellulose. What advantage does TiO2 have over ZnO? 719  CHAPTER 16 Acid-Base Equilibria and Solubility Equilibria Downward-growing icicle-like stalactites and upward- growing, columnar stalagmites. It may take thousands of years for these structures, which are mostly calcium carbonate, to form. CHAPTER OUTLINE A LOOK AHEAD 16.1 Homogeneous versus  We continue our study of acid-base properties and reactions from Chapter 15 Heterogeneous Solution by considering the effect of common ions on the extent of acid ionization and Equilibria hence the pH of the solution. (16.2) 16.2 The Common Ion Effect  We then extend our discussion to buffer solutions, whose pH remains largely unchanged upon the addition of small amounts of acids and bases. (16.3) 16.3 Buffer Solutions  We conclude our study of acid-base chemistry by examining acid-base 16.4 Acid-Base Titrations titration in more detail. We learn to calculate the pH during any stage of 16.5 Acid-Base Indicators titration involving strong and/or weak acids and bases. In addition, we see how acid-base indicators are used to determine the end point of a titration. 16.6 Solubility Equilibria (16.4 and 16.5) 16.7 Separation of Ions by  Next, we explore a type of heterogeneous equilibrium, which deals with Fractional Precipitation the solubility of sparingly soluble substances. We learn to express the solubility of these substances in terms of solubility product. We see that 16.8 The Common Ion Effect different types of metal ions can be effectively separated depending on and Solubility their differing solubility products. (16.6 and 16.7) 16.9 pH and Solubility  We then see how Le Châtelier’s principle helps us explain the effects of 16.10 Complex Ion Equilibria common ions and pH on solubility. (16.8 and 16.9) and Solubility  We learn how complex ion formation, which is a type of Lewis acid-base reaction, can enhance the solubility of an insoluble compound. (16.10) 16.11 Application of the Solubility Product Principle to  Finally, we apply the solubility product principle to qualitative analysis, Qualitative Analysis which is the identification of ions in solution. (16.12) 720  16.2 The Common Ion Effect 721 I n this chapter we will continue the study of acid-base reactions with a discussion of buffer action and titrations. We will also look at another type of aqueous equilibrium—that between slightly soluble compounds and their ions in solution. 16.1 Homogeneous versus Heterogeneous Solution Equilibria In Chapter 15, we saw that weak acids and weak bases do not ionize completely in water. Thus, at equilibrium a weak acid solution, for example, contains nonion- ized acid as well as H1 ions and the conjugate base. Nevertheless, all of these species are dissolved so the system is an example of homogeneous equilibrium (see Chapter 14). Another type of equilibrium, which we will consider in the second half of this chapter, involves the dissolution and precipitation of slightly soluble substances. These processes are examples of heterogeneous equilibria—that is, they pertain to reactions in which the components are in more than one phase. 16.2 The Common Ion Effect Our discussion of acid-base ionization and salt hydrolysis in Chapter 15 was limited to solutions containing a single solute. In this section, we will consider the acid-base properties of a solution with two dissolved solutes that contain the same ion (cation or anion), called the common ion. The presence of a common ion suppresses the ionization of a weak acid or a The common ion effect is simply an weak base. If sodium acetate and acetic acid are dissolved in the same solution, for application of Le Châtelier’s principle. example, they both dissociate and ionize to produce CH3COO2 ions: H2O CH3COONa(s) ¡ CH3COO2 (aq) 1 Na1 (aq) CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq) CH3COONa is a strong electrolyte, so it dissociates completely in solution, but CH3COOH, a weak acid, ionizes only slightly. According to Le Châtelier’s principle, the addition of CH3COO2 ions from CH3COONa to a solution of CH3COOH will suppress the ionization of CH3COOH (that is, shift the equilibrium from right to left), thereby decreasing the hydrogen ion concentration. Thus, a solution containing both CH3COOH and CH3COONa will be less acidic than a solution containing only CH3COOH at the same concentration. The shift in equilibrium of the acetic acid ionization is caused by the acetate ions from the salt. CH3COO2 is the common ion because it is supplied by both CH3COOH and CH3COONa. The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The common ion effect plays an important role in determining the pH of a solution and the solubility of a slightly soluble salt (to be discussed later in this chapter). Here we will study the common ion effect as it relates to the pH of a solution. Keep in mind that despite its distinctive name, the common ion effect is simply a special case of Le Châtelier’s principle. Let us consider the pH of a solution containing a weak acid, HA, and a soluble salt of the weak acid, such as NaA. We start by writing HA(aq) 1 H2O(l) Δ H3O1 (aq) 1 A2 (aq) or simply HA(aq) Δ H1 (aq) 1 A2 (aq) 722 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria The ionization constant Ka is given by [H1][A2] Ka 5 (16.1) [HA] Rearranging Equation (16.1) gives Ka[HA] [H1] 5 [A2] Taking the negative logarithm of both sides, we obtain [HA] 2log [H1] 5 2log Ka 2 log [A2] [A2] or 2log [H1] 5 2log Ka 1 log [HA] [A2] So pH 5 pKa 1 log (16.2) [HA] pKa is related to Ka as pH is related to where pKa 5 2log Ka (16.3) [H1]. Remember that the stronger the acid (that is, the larger the Ka), the smaller the pKa. Equation (16.2) is called the Henderson-Hasselbalch equation. A more general form of this expression is [conjugate base] Keep in mind that pKa is a constant, pH 5 pKa 1 log (16.4) but the ratio of the two concentration [acid] terms in Equation (16.4) depends on a particular solution. In our example, HA is the acid and A2 is the conjugate base. Thus, if we know Ka and the concentrations of the acid and the salt of the acid, we can calculate the pH of the solution. It is important to remember that the Henderson-Hasselbalch equation is derived from the equilibrium constant expression. It is valid regardless of the source of the conjugate base (that is, whether it comes from the acid alone or is supplied by both the acid and its salt). In problems that involve the common ion effect, we are usually given the starting concentrations of a weak acid HA and its salt, such as NaA. As long as the concentrations of these species are reasonably high ($ 0.1 M), we can neglect the ionization of the acid and the hydrolysis of the salt. This is a valid approximation because HA is a weak acid and the extent of the hydrolysis of the A2 ion is generally very small. Moreover, the presence of A2 (from NaA) further suppresses the ionization of HA and the presence of HA further suppresses the hydrolysis of A2. Thus, we can use the starting concentrations as the equilibrium concentrations in Equation (16.1) or Equation (16.4). In Example 16.1 we calculate the pH of a solution containing a common ion. Example 16.1 (a) Calculate the pH of a 0.20 M CH3COOH solution. (b) What is the pH of a solution containing both 0.20 M CH3COOH and 0.30 M CH3COONa? The Ka of CH3COOH is 1.8 3 1025. (Continued)  16.2 The Common Ion Effect 723 Strategy (a) We calculate [H1] and hence the pH of the solution by following the procedure in Example 15.8 (p. 681). (b) CH3COOH is a weak acid 1CH3COOH Δ CH3COO2 1 H1 2 , and CH3COONa is a soluble salt that is completely dissociated in solution (CH3COONa ¡ Na1 1 CH3COO2 ) . The common ion here is the acetate ion, CH3COO2. At equilibrium, the major species in solution are CH3COOH, CH3COO2, Na1, H1, and H2O. The Na1 ion has no acid or base properties and we ignore the ionization of water. Because Ka is an equilibrium constant, its value is the same whether we have just the acid or a mixture of the acid and its salt in solution. Therefore, we can calculate [H1] at equilibrium and hence pH if we know both [CH3COOH] and [CH3COO2] at equilibrium. Solution (a) In this case, the changes are CH3COOH(aq) Δ H1(aq) 1 CH3COO2(aq) Initial (M): 0.20 0 0 Change (M): 2x 1x 1x Equilibrium (M): 0.20 2 x x x [H1][CH3COO2] Ka 5 [CH3COOH] x2 1.8 3 1025 5    0.20 2 x Assuming 0.20 2 x < 0.20, we obtain x2 x2 1.8 3 1025 5 <       0.20 2 x 0.20 or x 5 [H1] 5 1.9 3 1023 M Thus, pH 5 2log (1.9 3 1023 ) 5 2.72 (b) Sodium acetate is a strong electrolyte, so it dissociates completely in solution: CH3COONa(aq) ¡ Na1 (aq) 1 CH3COO2 (aq) 0.30 M 0.30 M The initial concentrations, changes, and final concentrations of the species involved in the equilibrium are CH3COOH(aq) Δ H1(aq) 1 CH3COO2(aq) Initial (M): 0.20 0 0.30 Change (M): 2x 1x 1x Equilibrium (M): 0.20 2 x x 0.30 1 x From Equation (16.1), [H1][CH3COO2] Ka 5 [CH3COOH] 25 (x) (0.30 1 x) 1.8 3 10 5 0.20 2 x (Continued) 724 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Assuming that 0.30 1 x < 0.30 and 0.20 2 x < 0.20, we obtain (x) (0.30 1 x) (x) (0.30) 1.8 3 1025 5 < 0.20 2 x 0.20 or x 5 [H1] 5 1.2 3 1025 M Thus, pH 5 2log [H1] 5 2log (1.2 3 1025 ) 5 4.92 Check Comparing the results in (a) and (b), we see that when the common ion (CH3COO2) is present, according to Le Châtelier’s principle, the equilibrium shifts from right to left. This action decreases the extent of ionization of the weak acid. Consequently, fewer H1 ions are produced in (b) and the pH of the solution is higher than that in (a). As always, you should check the validity of the Similar problem: 16.5. assumptions. Practice Exercise What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Compare your result with the pH of a 0.30 M HCOOH solution. The common ion effect also operates in a solution containing a weak base, such as NH3, and a salt of the base, say NH4Cl. At equilibrium, NH14 (aq) Δ NH3 (aq) 1 H1 (aq) [NH3][H1] Ka 5 [NH14 ] We can derive the Henderson-Hasselbalch equation for this system as follows. Rear- ranging the above equation we obtain Ka[NH1 4] [H1] 5 [NH3] Taking the negative logarithm of both sides gives [NH14 ] 2log [H1] 5 2log Ka 2 log [NH3] [NH3] 2log [H1] 5 2log Ka 1 log [NH14 ] or [NH3] pH 5 pKa 1 log [NH14 ] A solution containing both NH3 and its salt NH4Cl is less basic than a solution con- taining only NH3 at the same concentration. The common ion NH1 4 suppresses the ionization of NH3 in the solution containing both the base and the salt. 16.3 Buffer Solutions A buffer solution is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers are very important to chemical and biological systems. The pH in the human body varies greatly from Fluids for intravenous injection must include buffer systems to one fluid to another; for example, the pH of blood is about 7.4, whereas the gastric maintain the proper blood pH. juice in our stomachs has a pH of about 1.5. These pH values, which are crucial for  16.3 Buffer Solutions 725 proper enzyme function and the balance of osmotic pressure, are maintained by buf- Animation Buffer Solutions fers in most cases. A buffer solution must contain a relatively large concentration of acid to react with Animation any OH2 ions that are added to it, and it must contain a similar concentration of base Properties of Buffers to react with any added H1 ions. Furthermore, the acid and the base components of the buffer must not consume each other in a neutralization reaction. These requirements are satisfied by an acid-base conjugate pair, for example, a weak acid and its conjugate base (supplied by a salt) or a weak base and its conjugate acid (supplied by a salt). A simple buffer solution can be prepared by adding comparable molar amounts of acetic acid (CH3COOH) and its salt sodium acetate (CH3COONa) to water. The equilibrium concentrations of both the acid and the conjugate base (from CH3COONa) are assumed to be the same as the starting concentrations (see p. 722). A solution containing these two substances has the ability to neutralize either added acid or added base. Sodium acetate, a strong electrolyte, dissociates completely in water: H2O CH3COONa(s) ¡ CH3COO2 (aq) 1 Na1 (aq) If an acid is added, the H1 ions will be consumed by the conjugate base in the buffer, CH3COO2, according to the equation CH3COO2 (aq) 1 H1 (aq) ¡ CH3COOH(aq) If a base is added to the buffer system, the OH2 ions will be neutralized by the acid in the buffer: CH3COOH(aq) 1 OH2 (aq) ¡ CH3COO2 (aq) 1 H2O(l) As you can see, the two reactions that characterize this buffer system are identical to those for the common ion effect described in Example 16.1. The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. The larger the amount, the greater the buffering capacity. In general, a buffer system can be represented as salt-acid or conjugate base-acid. Thus, the sodium acetate–acetic acid buffer system discussed above can be written as CH3COONa/CH3COOH or simply CH3COO2/CH3COOH. Figure 16.1 shows this buf- fer system in action. Example 16.2 distinguishes buffer systems from acid-salt combinations that do not function as buffers. (a) (b) (c) (d) Figure 16.1 The acid-base indicator bromophenol blue (added to all solutions shown) is used to illustrate buffer action. The indicator’s color is blue-purple above pH 4.6 and yellow below pH 3.0. (a) A buffer solution made up of 50 mL of 0.1 M CH3COOH and 50 mL of 0.1 M CH3COONa. The solution has a pH of 4.7 and turns the indicator blue-purple. (b) After the addition of 40 mL of 0.1 M HCl solution to the solution in (a), the color remains blue-purple. (c) A 100-mL CH3COOH solution whose pH is 4.7. (d) After the addition of 6 drops (about 0.3 mL) of 0.1 M HCl solution, the color turns yellow. Without buffer action, the pH of the solution decreases rapidly to less than 3.0 upon the addition of 0.1 M HCl. 726 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Example 16.2 Which of the following solutions can be classified as buffer systems? (a) KH2PO4/H3PO4, (b) NaClO4/HClO4, (c) C5H5N/C5H5NHCl (C5H5N is pyridine; its Kb is given in Table 15.4). Explain your answer. Strategy What constitutes a buffer system? Which of the preceding solutions contains a weak acid and its salt (containing the weak conjugate base)? Which of the preceding solutions contains a weak base and its salt (containing the weak conjugate acid)? Why is the conjugate base of a strong acid not able to neutralize an added acid? Solution The criteria for a buffer system is that we must have a weak acid and its salt (containing the weak conjugate base) or a weak base and its salt (containing the weak conjugate acid). (a) H3PO4 is a weak acid, and its conjugate base, H2PO24 , is a weak base (see Table 15.5). Therefore, this is a buffer system. (b) Because HClO4 is a strong acid, its conjugate base, ClO2 4 , is an extremely weak base. This means that the ClO2 4 ion will not combine with a H1 ion in solution to form HClO4. Thus, the system cannot act as a buffer system. 1 (c) As Table 15.4 shows, C5H5N is a weak base and its conjugate acid, C5H5N H (the cation of the salt C5H5NHCl), is a weak acid. Therefore, this is a buffer Similar problems: 16.9, 16.10. system. Practice Exercise Which of the following couples are buffer systems? (a) KF/HF, (b) KBr/HBr, (c) Na2CO3/NaHCO3. The effect of a buffer solution on pH is illustrated by Example 16.3. Example 16.3 (a) Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume that the volume of the solution does not change when the HCl is added. Strategy (a) The pH of the buffer system before the addition of HCl can be calculated with the procedure described in Example 16.1, because it is similar to the common-ion problem. The Ka of CH3COOH is 1.8 3 1025 (see Table 15.3). (b) It is helpful to make a sketch of the changes that occur in this case. Solution (a) We summarize the concentrations of the species at equilibrium as follows: CH3COOH(aq) Δ H1(aq) 1 CH3COO2(aq) Initial (M): 1.0 0 1.0 Change (M): 2x 1x 1x Equilibrium (M): 1.0 2 x x 1.0 1 x (Continued)  16.3 Buffer Solutions 727 [H1][CH3COO2] Ka 5 [CH3COOH] 25 (x) (1.0 1 x) 1.8 3 10 5 (1.0 2 x) Assuming 1.0 1 x < 1.0 and 1.0 2 x < 1.0, we obtain (x) (1.0 1 x) x(1.0) 1.8 3 1025 5 < (1.0 2 x) 1.0 or x 5 [H1] 5 1.8 3 1025 M When the concentrations of the acid Thus, pH 5 2log (1.8 3 1025 ) 5 4.74 and the conjugate base are the same, the pH of the buffer is equal to the pKa of the acid. (b) When HCl is added to the solution, the initial changes are HCl(aq) ¡ H1(aq) 1 Cl2(aq) Initial (mol): 0.10 0 0 Change (mol): 20.10 10.10 10.10 Final (mol): 0 0.10 0.10 The Cl2 ion is a spectator ion in solution because it is the conjugate base of a strong acid. The H1 ions provided by the strong acid HCl react completely with the conjugate base of the buffer, which is CH3COO2. At this point it is more convenient to work with moles rather than molarity. The reason is that in some cases the volume of the solution may change when a substance is added. A change in volume will change the molarity, but not the number of moles. The neutralization reaction is summarized next: CH3COO2(aq) 1 H1(aq) ¡ CH3COOH(aq) Initial (mol): 1.0 0.10 1.0 Change (mol): 20.10 20.10 10.10 Final (mol): 0.90 0 1.1 Finally, to calculate the pH of the buffer after neutralization of the acid, we convert back to molarity by dividing moles by 1.0 L of solution. CH3COOH(aq) Δ H1(aq) 1 CH3COO2(aq) Initial (M): 1.1 0 0.90 Change (M): 2x 1x 1x Equilibrium (M): 1.1 2 x x 0.90 1 x [H1][CH3COO2] Ka 5 [CH3COOH] 25 (x) (0.90 1 x) 1.8 3 10 5 1.1 2 x Assuming 0.90 1 x < 0.90 and 1.1 2 x < 1.1, we obtain (x) (0.90 1 x) x(0.90) 1.8 3 1025 5 < 1.1 2 x 1.1 or x 5 [H1] 5 2.2 3 1025 M Thus, pH 5 2log (2.2 3 1025 ) 5 4.66 (Continued) 728 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Check The pH decreases by only a small amount upon the addition of HCl. This is Similar problem: 16.17. consistent with the action of a buffer solution. Practice Exercise Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? In the buffer solution examined in Example 16.3, there is a decrease in pH (the solution becomes more acidic) as a result of added HCl. We can also compare the changes in H1 ion concentration as follows: Before addition of HCl: [H1] 5 1.8 3 1025 M After addition of HCl: [H1] 5 2.2 3 1025 M Thus, the H1 ion concentration increases by a factor of 2.2 3 1025 M 5 1.2 1.8 3 1025 M To appreciate the effectiveness of the CH3COONa/CH3COOH buffer, let us find out what would happen if 0.10 mol HCl were added to 1 L of water, and compare the increase in H1 ion concentration. 7 Before addition of HCl: [H1] 5 1.0 3 1027 M 6 Buffer solution After addition of HCl: [H1] 5 0.10 M 5 pH 4 3 Water As a result of the addition of HCl, the H 1 ion concentration increases by a 2 factor of 1 0 0 0.02 0.04 0.06 0.08 0.1 0.10 M 5 1.0 3 106 Mole of HCl added 1.0 3 1027 M Figure 16.2 A comparison of the change in pH when 0.10 mol HCl amounting to a millionfold increase! This comparison shows that a properly cho- is added to pure water and to an acetate buffer solution as sen buffer solution can maintain a fairly constant H1 ion concentration, or pH described in Example 16.3. (Figure 16.2). Review of Concepts The following diagrams represent solutions containing a weak acid HA and/or its sodium salt NaA. Which solutions can act as a buffer? Which solution has the greatest buffer capacity? The Na1 ions and water molecules are omitted for clarity. HA A⫺ (a) (b) (c) (d)  16.3 Buffer Solutions 729 Preparing a Buffer Solution with a Specific pH Now suppose we want to prepare a buffer solution with a specific pH. How do we go about it? Equation (16.4) indicates that if the molar concentrations of the acid and its conjugate base are approximately equal; that is, if [acid] < [conjugate base], then [conjugate base] log <0 [acid] or pH < pKa Thus, to prepare a buffer solution, we work backwards. First we choose a weak acid whose pKa is close to the desired pH. Next, we substitute the pH and pKa values in Equation (16.4) to obtain the ratio [conjugate base]/[acid]. This ratio can then be converted to molar quantities for the preparation of the buffer solution. Example 16.4 shows this approach. Example 16.4 Describe how you would prepare a “phosphate buffer” with a pH of about 7.40. Strategy For a buffer to function effectively, the concentrations of the acid component must be roughly equal to the conjugate base component. According to Equation (16.4), when the desired pH is close to the pKa of the acid, that is, when pH < pKa, [conjugate base] log <0 [acid] [conjugate base] or <1 [acid] Solution Because phosphoric acid is a triprotic acid, we write the three stages of ionization as follows. The Ka values are obtained from Table 15.5 and the pKa values are found by applying Equation (16.3). H3PO4 (aq) Δ H1 (aq) 1 H2PO2 23 4 (aq)  Ka1 5 7.5 3 10 ; pKa1 5 2.12 H2PO24 (aq) Δ H 1 (aq) 1 HPO 22 4 (aq)  K a2 5 6.2 3 1028 ; pKa2 5 7.21 HPO4 (aq) Δ H (aq) 1 PO4 (aq)   Ka3 5 4.8 3 10213; pKa3 5 12.32 22 1 32 The most suitable of the three buffer systems is HPO422/H2PO24 , because the pKa of the acid H2PO2 4 is closest to the desired pH. From the Henderson-Hasselbalch equation we write [conjugate base] pH 5 pKa 1 log [acid] [HPO22 4 ] 7.40 5 7.21 1 log [H2PO24 ] [HPO22 4 ] log 5 0.19 [H2PO24] (Continued) 730 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Taking the antilog, we obtain [HPO22 4 ] 5 100.19 5 1.5 [H2PO24 ] Thus, one way to prepare a phosphate buffer with a pH of 7.40 is to dissolve disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) in a mole ratio of 1.5:1.0 in water. For example, we could dissolve 1.5 moles of Na2HPO4 and Similar problems: 16.19, 16.20. 1.0 mole of NaH2PO4 in enough water to make up a 1-L solution. Practice Exercise How would you prepare a liter of “carbonate buffer” at a pH of 10.10? You are provided with carbonic acid (H2CO3), sodium hydrogen carbonate (NaHCO3), and sodium carbonate (Na2CO3). See Table 15.5 for Ka values. The Chemistry in Action essay on p. 732 illustrates the importance of buffer systems in the human body. 16.4 Acid-Base Titrations Animation Having discussed buffer solutions, we can now look in more detail at the quantita- Acid-Base Titrations tive aspects of acid-base titrations (see Section 4.6). We will consider three types of reactions: (1) titrations involving a strong acid and a strong base, (2) titrations involving a weak acid and a strong base, and (3) titrations involving a strong acid and a weak base. Titrations involving a weak acid and a weak base are complicated by the hydrolysis of both the cation and the anion of the salt formed. It is difficult to determine the equivalence point in these cases. Therefore, these titrations will not be dealt with here. Figure 16.3 shows the arrangement for monitoring the pH during the course of a titration. Strong Acid–Strong Base Titrations The reaction between a strong acid (say, HCl) and a strong base (say, NaOH) can be represented by NaOH(aq) 1 HCl(aq) ¡ NaCl(aq) 1 H2O(l) Figure 16.3 A pH meter is used to monitor an acid-base titration.  16.4 Acid-Base Titrations 731 14 Volume NaOH 13 added (mL) pH 12 0.0 1.00 11 5.0 1.18 10 10.0 1.37 15.0 1.60 9 20.0 1.95 8 22.0 2.20 pH 7 Equivalence 24.0 2.69 6 point 25.0 7.00 5 26.0 11.29 28.0 11.75 4 30.0 11.96 3 35.0 12.22 2 40.0 12.36 1 45.0 12.46 50.0 12.52 0 10 20 30 40 50 Volume of NaOH added (mL) Figure 16.4 pH profile of a strong acid–strong base titration. A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M HCl solution in an Erlenmeyer flask (see Figure 4.21). This curve is sometimes referred to as a titration curve. or in terms of the net ionic equation H1 (aq) 1 OH2 (aq) ¡ H2O(l) Consider the addition of a 0.100 M NaOH solution (from a buret) to an Erlenmeyer flask containing 25.0 mL of 0.100 M HCl. For convenience, we will use only three significant figures for volume and concentration and two significant figures for pH. Figure 16.4 shows the pH profile of the titration (also known as the titration curve). Before the addition of NaOH, the pH of the acid is given by 2log (0.100), or 1.00. When NaOH is added, the pH of the solution increases slowly at first. Near the equivalence point the pH begins to rise steeply, and at the equivalence point (that is, the point at which equimolar amounts of acid and base have reacted) the curve rises almost vertically. In a strong acid–strong base titration both the hydrogen ion and hydroxide ion concentrations are very small at the equivalence point (approximately 1 3 1027 M); consequently, the addition of a single drop of the base can cause a large increase in [OH2] and in the pH of the solution. Beyond the equivalence point, the pH again increases slowly with the addition of NaOH. It is possible to calculate the pH of the solution at every stage of titration. Here are three sample calculations. 1. After the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. The total volume of the solution is 35.0 mL. The number of moles of NaOH in 10.0 mL is 0.100 mol NaOH 1L 10.0 mL 3 3 5 1.00 3 1023 mol A faster way to calculate the number 1 L NaOH 1000 mL of moles of NaOH is to write 0.100 mol 10.0 mL 3 5 1.0 3 10 23 mol 1000 mL The number of moles of HCl originally present in 25.0 mL of solution is 0.100 mol HCl 1L 25.0 mL 3 3 5 2.50 3 1023 mol 1 L HCl 1000 mL CHEMISTRY in Action Maintaining the pH of Blood A ll higher animals need a circulatory system to carry fuel and oxygen for their life processes and to remove wastes. In the human body this vital exchange takes place in the versatile fluid known as blood, of which there are about 5 L (10.6 pints) in an average adult. Blood circulating deep in the tissues carries oxygen and nutrients to keep cells alive, and removes carbon dioxide and other waste materials. Using several buffer systems, nature has provided an extremely efficient method for the delivery of oxygen and the removal of carbon dioxide. Blood is an enormously complex system, but for our pur- poses we need look at only two essential components: blood plasma and red blood cells, or erythrocytes. Blood plasma contains many compounds, including proteins, metal ions, and inorganic phosphates. The erythrocytes contain hemoglobin molecules, as well as the enzyme carbonic anhydrase, which catalyzes both the formation of carbonic acid (H2CO3) and its decomposition: Electron micrograph of red blood cells in a small branch of an artery. CO2 (aq) 1 H2O(l) Δ H2CO3 (aq) As the figure on p. 733 shows, carbon dioxide produced by The substances inside the erythrocyte are protected from extra- metabolic processes diffuses into the erythrocyte, where it is cellular fluid (blood plasma) by a cell membrane that allows rapidly converted to H2CO3 by carbonic anhydrase: only certain molecules to diffuse through it. The pH of blood plasma is maintained at about 7.40 by CO2 (aq) 1 H2O(l) Δ H2CO3 (aq) several buffer systems, the most important of which is the HCO2 The ionization of the carbonic acid 3 / H2CO3 system. In the erythrocyte, where the pH is 7.25, the principal buffer systems are HCO2 3 /H2CO3 and hemoglobin. H2CO3 (aq) Δ H1 (aq) 1 HCO2 3 (aq) The hemoglobin molecule is a complex protein molecule (molar mass 65,000 g) that contains a number of ionizable protons. As has two important consequences. First, the bicarbonate ion dif- a very rough approximation, we can treat it as a monoprotic acid fuses out of the erythrocyte and is carried by the blood plasma of the form HHb: to the lungs. This is the major mechanism for removing carbon dioxide. Second, the H1 ions shift the equilibrium in favor of HHb(aq) Δ H1 (aq) 1 Hb2 (aq) the nonionized oxyhemoglobin molecule: where HHb represents the hemoglobin molecule and Hb2 the H1 (aq) 1 HbO2 2 (aq) Δ HHbO2 (aq) conjugate base of HHb. Oxyhemoglobin (HHbO2), formed by Because HHbO2 releases oxygen more readily than does its the combination of oxygen with hemoglobin, is a stronger acid conjugate base (HbO2 2 ), the formation of the acid promotes the than HHb: following reaction from left to right: HHbO2 (aq) Δ H1 (aq) 1 HbO2 2 (aq) HHbO2 (aq) Δ HHb(aq) 1 O2 (aq) 732  Capillary Capillary ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ Tissues Lungs Erythrocyte Erythrocyte O2 O2 + HHb O2 O2 + HHb HHbO2 HbO 2– + H+ HHbO2 HbO 2– + H+ CA CA CO2 CO2 + H2O H2CO3 H+ + HCO –3 CO2 CO2 + H2O H2CO3 H+ + HCO –3 HCO –3 HCO –3 Plasma Plasma (a) (b) Oxygen–carbon dioxide transport and release by blood. (a) The partial pressure of CO2 is higher in the metabolizing tissues than in the plasma. Thus, it diffuses into the blood capillaries and then into erythrocytes. There it is converted to carbonic acid by the enzyme carbonic anhydrase (CA). The protons provided by the carbonic acid then combine with the HbO 22 anions to form HHbO2, which eventually dissociates into HHb and O2. Because the partial pressure of O2 is higher in the erythrocytes than in the tissues, oxygen molecules diffuse out of the erythrocytes and then into the tissues. The bicarbonate ions also diffuse out of the erythrocytes and are carried by the plasma to the lungs. (b) In the lungs, the processes are exactly reversed. Oxygen molecules diffuse from the lungs, where they have a higher partial pressure, into the erythrocytes. There they combine with HHb to form HHbO2. The protons provided by HHbO2 combine with the bicarbonate ions diffused into the erythrocytes from the plasma to form carbonic acid. In the presence of carbonic anhydrase, carbonic acid is converted to H2O and CO2. The CO2 then diffuses out of the erythrocytes and into the lungs, where it is exhaled. The O2 molecules diffuse out of the erythrocyte and are taken up The carbon dioxide diffuses to the lungs and is eventually by other cells to carry out metabolism. exhaled. The formation of the Hb2 ions (due to the reaction When the venous blood returns to the lungs, the above between HHb and HCO2 3 shown in the left column) also favors processes are reversed. The bicarbonate ions now diffuse into the uptake of oxygen at the lungs the erythrocyte, where they react with hemoglobin to form car- bonic acid: Hb2 (aq) 1 O2 (aq) Δ HbO2 2 (aq) HHb(aq) 1 HCO2 2 3 (aq) Δ Hb (aq) 1 H2CO3 (aq) because Hb2 has a greater affinity for oxygen than does HHb. When the arterial blood flows back to the body tissues, the Most of the acid is then converted to CO2 by carbonic anhydrase: entire cycle is repeated. H2CO3 (aq) Δ H2O(l) 1 CO2 (aq) 733 734 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Keep in mind that 1 mol NaOH ∞ 1 mol Thus, the amount of HCl left after partial neutralization is (2.50 3 1023) 2 HCl. (1.00 3 1023), or 1.50 3 1023 mol. Next, the concentration of H1 ions in 35.0 mL of solution is found as follows: 1.50 3 1023 mol HCl 1000 mL 3 5 0.0429 mol HCl/L 35.0 mL 1L 5 0.0429 M HCl Thus, [H1] 5 0.0429 M, and the pH of the solution is pH 5 2log 0.0429 5 1.37 2. After the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. This is a simple calculation, because it involves a complete neutralization reac- Neither Na1 nor Cl2 ions undergo tion and the salt (NaCl) does not undergo hydrolysis. At the equivalence point, hydrolysis. [H1] 5 [OH2] 5 1.00 3 1027 M and the pH of the solution is 7.00. 3. After the addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. The total volume of the solution is now 60.0 mL. The number of moles of NaOH added is 0.100 mol NaOH 1L 35.0 mL 3 3 5 3.50 3 1023 mol 1 L NaOH 1000 mL The number of moles of HCl in 25.0 mL solution is 2.50 3 1023 mol. After com- plete neutralization of HCl, the number of moles of NaOH left is (3.50 3 1023) 2 (2.50 3 1023), or 1.00 3 1023 mol. The concentration of NaOH in 60.0 mL of solution is 1.00 3 1023 mol NaOH 1000 mL 3 5 0.0167 mol NaOH/L 60.0 mL 1L 5 0.0167 M NaOH Thus, [OH2] 5 0.0167 M and pOH 5 2log 0.0167 5 1.78. Hence, the pH of the solution is pH 5 14.00 2 pOH 5 14.00 2 1.78 5 12.22 Weak Acid–Strong Base Titrations Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base): CH3COOH(aq) 1 NaOH(aq) ¡ CH3COONa(aq) 1 H2O(l) This equation can be simplified to CH3COOH(aq) 1 OH2 (aq) ¡ CH3COO2 (aq) 1 H2O(l) The acetate ion undergoes hydrolysis as follows: CH3COO2 (aq) 1 H2O(l) Δ CH3COOH(aq) 1 OH2 (aq)  16.4 Acid-Base Titrations 735 14 Volume NaOH 13 added (mL) pH 12 0.0 2.87 11 5.0 4.14 10 10.0 4.57 15.0 4.92 9 Equivalence 20.0 5.35 8 point 22.0 5.61 pH 7 24.0 6.12 6 25.0 8.72 5 26.0 10.29 28.0 11.75 4 30.0 11.96 3 35.0 12.22 2 40.0 12.36 1 45.0 12.46 50.0 12.52 0 10 20 30 40 50 Volume of NaOH added (mL) Figure 16.5 pH profile of a weak acid–strong base titration. A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M CH3COOH solution in an Erlenmeyer flask. Due to the hydrolysis of the salt formed, the pH at the equivalence point is greater than 7. Therefore, at the equivalence point, when we only have sodium acetate present, the pH will be greater than 7 as a result of the excess OH2 ions formed (Figure 16.5). Note that this situation is analogous to the hydrolysis of sodium acetate (CH3COONa) (see p. 697). Example 16.5 deals with the titration of a weak acid with a strong base. Example 16.5 Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution of (a) 10.0 mL of 0.100 M NaOH, (b) 25.0 mL of 0.100 M NaOH, (c) 35.0 mL of 0.100 M NaOH. Strategy The reaction between CH3COOH and NaOH is CH3COOH(aq) 1 NaOH(aq) ¡ CH3COONa(aq) 1 H2O(l) We see that 1 mol CH3COOH ∞ 1 mol NaOH. Therefore, at every stage of the titration we can calculate the number of moles of base reacting with the acid, and the pH of the solution is determined by the excess acid or base left over. At the equivalence point, however, the neutralization is complete and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is CH3COONa. Solution (a) The number of moles of NaOH in 10.0 mL is 0.100 mol NaOH 1L 10.0 mL 3 3 5 1.00 3 1023 mol 1 L NaOH soln 1000 mL The number of moles of CH3COOH originally present in 25.0 mL of solution is 0.100 mol CH3COOH 1L 25.0 mL 3 3 5 2.50 3 1023 mol 1 L CH3COOH soln 1000 mL (Continued) 736 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria We work with moles at this point because when two solutions are mixed, the solution volume increases. As the volume increases, molarity will change but the number of moles will remain the same. The changes in number of moles are summarized next: CH3COOH(aq) 1 NaOH(aq) ¡ CH3COONa(aq) 1 H2O(l) Initial (mol): 2.50 3 1023 1.00 3 1023 0 Change (mol): 21.00 3 1023 21.00 3 1023 11.00 3 1023 Final (mol): 1.50 3 1023 0 1.00 3 1023 At this stage we have a buffer system made up of CH3COOH and CH3COO2 (from the salt, CH3COONa). To calculate the pH of the solution, we write [H1][CH3COO2] Ka 5 [CH3COOH] 1 [CH3COOH]Ka Because the volume of the solution is [H ] 5 the same for CH3COOH and [CH3COO2] CH3COO2 (35 mL), the ratio of the (1.50 3 1023 ) (1.8 3 1025 ) 5 5 2.7 3 1025 M number of moles present is equal to 1.00 3 1023 the ratio of their molar concentrations. Therefore, pH 5 2log (2.7 3 1025 ) 5 4.57 (b) These quantities (that is, 25.0 mL of 0.100 M NaOH reacting with 25.0 mL of 0.100 M CH3COOH) correspond to the equivalence point. The number of moles of NaOH in 25.0 mL of the solution is 0.100 mol NaOH 1L 25.0 mL 3 3 5 2.50 3 1023 mol 1 L NaOH soln 1000 mL The changes in number of moles are summarized next: CH3COOH(aq) 1 NaOH(aq) ¡ CH3COONa(aq) 1 H2O(l ) Initial (mol): 2.50 3 1023 2.50 3 1023 0 Change (mol): 22.50 3 1023 22.50 3 1023 12.50 3 1023 Final (mol): 0 0 2.50 3 1023 At the equivalence point, the concentrations of both the acid and the base are zero. The total volume is (25.0 1 25.0) mL or 50.0 mL, so the concentration of the salt is 2.50 3 1023 mol 1000 mL [CH3COONa] 5 3 50.0 mL 1L 5 0.0500 mol/L 5 0.0500 M The next step is to calculate the pH of the solution that results from the hydrolysis of the CH3COO2 ions. Following the procedure described in Example  15.13 and looking up the base ionization constant (Kb) for CH3COO2 in Table 15.3, we write [CH3COOH][OH2] x2 Kb 5 5.6 3 10210 5    2 5 [CH3COO ] 0.0500 2 x x 5 [OH2] 5 5.3 3 1026 M, pH 5 8.72 (c) After the addition of 35.0 mL of NaOH, the solution is well past the equivalence point. The number of moles of NaOH originally present is 0.100 mol NaOH 1L 35.0 mL 3 3 5 3.50 3 1023 mol 1 L NaOH soln 1000 mL (Continued)  16.4 Acid-Base Titrations 737 The changes in number of moles are summarized next: CH3COOH(aq) 1 NaOH(aq) ¡ CH3COONa(aq) 1 H2O(l) Initial (mol): 2.50 3 1023 3.50 3 1023 0 23 Change (mol): 22.50 3 10 22.50 3 1023 12.50 3 1023 Final (mol): 0 1.00 3 1023 2.50 3 1023 At this stage we have two species in solution that are responsible for making the solution basic: OH2 and CH3COO2 (from CH3COONa). However, because OH2 is a much stronger base than CH3COO2, we can safely neglect the hydrolysis of the CH3COO2 ions and calculate the pH of the solution using only the concentration of the OH2 ions. The total volume of the combined solutions is (25.0 1 35.0) mL or 60.0 mL, so we calculate OH2 concentration as follows: 1.00 3 1023 mol 1000 mL [OH2] 5 3 60.0 mL 1L 5 0.0167 mol/L 5 0.0167 M pOH 5 2log [OH2] 5 2log 0.0167 5 1.78 pH 5 14.00 2 1.78 5 12.22 Similar problem: 16.35. Practice Exercise Exactly 100 mL of 0.10 M nitrous acid (HNO2) are titrated with a 0.10 M NaOH solution. Calculate the pH for (a) the initial solution, (b) the point at which 80 mL of the base has been added, (c) the equivalence point, (d) the point at which 105 mL of the base has been added. Strong Acid–Weak Base Titrations Consider the titration of HCl, a strong acid, with NH3, a weak base: HCl(aq) 1 NH3 (aq) ¡ NH4Cl(aq) or simply H1 (aq) 1 NH3 (aq) ¡ NH14 (aq) The pH at the equivalence point is less than 7 due to the hydrolysis of the NH14 ion: NH14 (aq) 1 H2O(l) Δ NH3 (aq) 1 H3O1 (aq) or simply NH14 (aq) Δ NH3 (aq) 1 H1 (aq) Because of the volatility of an aqueous ammonia solution, it is more convenient to add hydrochloric acid from a buret to the ammonia solution. Figure 16.6 shows the titration curve for this experiment. Example 16.6 Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution. Strategy The reaction between NH3 and HCl is NH3 (aq) 1 HCl(aq) ¡ NH4Cl(aq) (Continued) 738 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria 12 Volume HCl 11 added (mL) pH 10 0.0 11.13 9 5.0 9.86 10.0 9.44 8 15.0 9.08 7 20.0 8.66 22.0 8.39 pH 6 24.0 7.88 5 Equivalence 25.0 5.28 point 26.0 2.70 4 28.0 2.22 3 30.0 2.00 35.0 1.70 2 40.0 1.52 1 45.0 1.40 50.0 1.30 0 10 20 30 40 50 Volume of HCl added (mL) Figure 16.6 pH profile of a strong acid–weak base titration. A 0.100 M HCl solution is added from a buret to 25.0 mL of a 0.100 M NH3 solution in an Erlenmeyer flask. As a result of salt hydrolysis, the pH at the equivalence point is lower than 7. We see that 1 mol NH3 ∞ 1 mol HCl. At the equivalence point, the major species in solution are the salt NH4Cl (dissociated into NH14 and Cl2 ions) and H2O. First, we determine the concentration of NH4Cl formed. Then we calculate the pH as a result of the NH14 ion hydrolysis. The Cl2 ion, being the conjugate base of a strong acid HCl, does not react with water. As usual, we ignore the ionization of water. Solution The number of moles of NH3 in 25.0 mL of 0.100 M solution is 0.100 mol NH3 1L 25.0 mL 3 3 5 2.50 3 1023 mol 1 L NH3 1000 mL At the equivalence point the number of moles of HCl added equals the number of moles of NH3. The changes in number of moles are summarized below: NH3(aq) 1 HCl(aq) ¡ NH4Cl(aq) Initial (mol): 2.50 3 1023 2.50 3 1023 0 Change (mol): 22.50 3 1023 22.50 3 1023 12.50 3 1023 Final (mol): 0 0 2.50 3 1023 At the equivalence point, the concentrations of both the acid and the base are zero. The total volume is (25.0 1 25.0) mL, or 50.0 mL, so the concentration of the salt is 2.50 3 1023 mol 1000 mL [NH4Cl] 5 3 50.0 mL 1L 5 0.0500 mol/L 5 0.0500 M The pH of the solution at the equivalence point is determined by the hydrolysis of NH14 ions. We follow the procedure on p. 697. (Continued)  16.5 Acid-Base Indicators 739 Step 1: We represent the hydrolysis of the cation NH14 , and let x be the equilibrium concentration of NH3 and H1 ions in mol/L: NH14 (aq) Δ NH3(aq) 1 H1(aq) Initial (M): 0.0500 0.000 0.000 Change (M): 2x 1x 1x Equilibrium (M): (0.0500 2 x) x x Step 2: From Table 15.4 we obtain the Ka for NH14 : [NH3][H1] Ka 5 [NH14 ] x2 5.6 3 10210 5    0.0500 2 x Applying the approximation 0.0500 2 x < 0.0500, we get Always check the validity of the approximation. x2 x2 5.6 3 10210 5 <       0.0500 2 x 0.0500 x 5 5.3 3 1026 M Thus, the pH is given by pH 5 2log (5.3 3 1026 ) 5 5.28 Check Note that the pH of the solution is acidic. This is what we would expect from the hydrolysis of the ammonium ion. Similar problem: 16.33. Practice Exercise Calculate the pH at the equivalence point in the titration of 50 mL of 0.10 M methylamine (see Table 15.4) with a 0.20 M HCl solution. Review of Concepts For which of the following titrations will the pH at the equivalence point not be neutral? (a) HNO2 by NaOH, (b) KOH by HClO4, (c) HCOOH by KOH, (d) CH3NH2 by HNO3. 16.5 Acid-Base Indicators The equivalence point, as we have seen, is the point at which the number of moles of OH2 ions added to a solution is equal to the number of moles of H1 ions originally present. To determine the equivalence point in a titration, then, we must know exactly how much volume of a base to add from a buret to an acid in a flask. One way to achieve this goal is to add a few drops of an acid-base indicator to the acid solution at the start of the titration. You will recall from Chapter 4 that an indicator is usually a weak organic acid or base that has distinctly different colors in its nonionized and ionized forms. These two forms are related to the pH of the solution in which the indicator is dissolved. The end point of a titration occurs when the indicator changes The end point is where the color of the color. However, not all indicators change color at the same pH, so the choice of indicator changes. The equivalence point is where neutralization is complete. indicator for a particular titration depends on the nature of the acid and base used in Experimentally we use the end point to the titration (that is, whether they are strong or weak). By choosing the proper indica- estimate the equivalence point. tor for a titration, we can use the end point to determine the equivalence point, as shown on p. 740. 740 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Let us consider a weak monoprotic acid that we will call HIn. To be an effective indicator, HIn and its conjugate base, In2, must have distinctly different colors. In solution, the acid ionizes to a small extent: HIn(aq) Δ H1 (aq) 1 In2 (aq) If the indicator is in a sufficiently acidic medium, the equilibrium, according to Le Châtelier’s principle, shifts to the left and the predominant color of the indicator is that of the nonionized form (HIn). On the other hand, in a basic medium the equilib- rium shifts to the right and the color of the solution will be due mainly to that of the conjugate base (In2). Roughly speaking, we can use the following concentration ratios to predict the perceived color of the indicator: [HIn] $ 10   color of acid (HIn) predominates [In2] [HIn] # 0.1  color of conjugate base (In2 ) predominates [In2] If [HIn] < [In2], then the indicator color is a combination of the colors of HIn and In2. Typical indicators change color over the The end point of an indicator does not occur at a specific pH; rather, there is a pH range given by pH 5 pKa 6 1, where range of pH within which the end point will occur. In practice, we choose an indica- Ka is the acid ionization constant of the indicator. tor whose end point lies on the steep part of the titration curve. Because the equiva- lence point also lies on the steep part of the curve, this choice ensures that the pH at the equivalence point will fall within the range over which the indicator changes color. In Section 4.6 we mentioned that phenolphthalein is a suitable indicator for the titra- tion of NaOH and HCl. Phenolphthalein is colorless in acidic and neutral solutions, but reddish pink in basic solutions. Measurements show that at pH , 8.3 the indica- tor is colorless but that it begins to turn reddish pink when the pH exceeds 8.3. As shown in Figure 16.4, the steepness of the pH curve near the equivalence point means that the addition of a very small quantity of NaOH (say, 0.05 mL, which is about the volume of a drop from the buret) brings about a large rise in the pH of the solution. What is important, however, is the fact that the steep portion of the pH profile includes the range over which phenolphthalein changes from colorless to reddish pink. When- ever such a correspondence occurs, the indicator can be used to locate the equivalence point of the titration (Figure 16.7). Figure 16.7 The titration curve of 14 a strong acid with a strong base. Because the regions over which 13 the indicators methyl red and 12 phenolphthalein change color occur along the steep portion of 11 the curve, they can be used to 10 monitor the equivalence point of Phenolphthalein the titration. Thymol blue, on the 9 other hand, cannot be used for 8 the same purpose because the pH color change does not match the 7 steep portion of the titration curve 6 (see Table 16.1). Methyl red 5 4 3 Thymol blue 2 1 0 10 20 30 40 50 Volume of NaOH added (mL)  16.5 Acid-Base Indicators 741 Figure 16.8 Solutions containing extracts of red cabbage (obtained by boiling the cabbage in water) produce different colors when treated with an acid and a base. The pH of the solutions increases from left to right. Many acid-base indicators are plant pigments. For example, by boiling chopped red cabbage in water we can extract pigments that exhibit many different colors at various pHs (Figure 16.8). Table 16.1 lists a number of indicators commonly used in acid-base titrations. The choice of a particular indicator depends on the strength of the acid and base to be titrated. Example 16.7 illustrates this point. Example 16.7 Which indicator or indicators listed in Table 16.1 would you use for the acid-base titrations shown in (a) Figure 16.4, (b) Figure 16.5, and (c) Figure 16.6? Strategy The choice of an indicator for a particular titration is based on the fact that its pH range for color change must overlap the steep portion of the titration curve. Otherwise we cannot use the color change to locate the equivalence point. (Continued) Table 16.1 Some Common Acid-Base Indicators Color Indicator In Acid In Base pH Range* Thymol blue Red Yellow 1.2–2.8 Bromophenol blue Yellow Bluish purple 3.0–4.6 Methyl orange Orange Yellow 3.1–4.4 Methyl red Red Yellow 4.2–6.3 Chlorophenol blue Yellow Red 4.8–6.4 Bromothymol blue Yellow Blue 6.0–7.6 Cresol red Yellow Red 7.2–8.8 Phenolphthalein Colorless Reddish pink 8.3–10.0 *The pH range is defined as the range over which the indicator changes from the acid color to the base color. 742 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Solution (a) Near the equivalence point, the pH of the solution changes abruptly from 4 to 10. Therefore, all the indicators except thymol blue, bromophenol blue, and methyl orange are suitable for use in the titration. (b) Here the steep portion covers the pH range between 7 and 10; therefore, the suitable indicators are cresol red and phenolphthalein. (c) Here the steep portion of the pH curve covers the pH range between 3 and 7; therefore, the suitable indicators are bromophenol blue, methyl orange, methyl red, Similar problem: 16.43. and chlorophenol blue. Practice Exercise Referring to Table 16.1, specify which indicator or indicators you would use for the following titrations: (a) HBr versus CH3NH2, (b) HNO3 versus NaOH, (c) HNO2 versus KOH. Review of Concepts Under what conditions will the end point of an acid-base titration accurately represent the equivalence point? 16.6 Solubility Equilibria Precipitation reactions are important in industry, medicine, and everyday life. For example, the preparation of many essential industrial chemicals such as sodium car- bonate (Na2CO3) is based on precipitation reactions. The dissolving of tooth enamel, which is mainly made of hydroxyapatite [Ca5(PO4)3OH], in an acidic medium leads to tooth decay. Barium sulfate (BaSO4), an insoluble compound that is opaque to X rays, is used to diagnose ailments of the digestive tract. Stalactites and stalagmites, which consist of calcium carbonate (CaCO3), are produced by a precipitation reaction, and so are many foods, such as fudge. The general rules for predicting the solubility of ionic compounds in water were introduced in Section 4.2. Although useful, these solubility rules do not enable us to make quantitative predictions about how much of a given ionic compound will dis- solve in water. To develop a quantitative approach, we start with what we already know about chemical equilibrium. Unless otherwise stated, in the following discussion BaSO4 imaging of human large the solvent is water and the temperature is 25°C. intestine. Solubility Product Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s) Δ Ag1(aq) 1 Cl2(aq) Silver chloride is an insoluble salt (see Table 4.2). The small amount of solid AgCl that dissolves in water is assumed to dissociate completely into Ag1 and Cl2 ions. Recall that the activity of the solid is We know from Chapter 14 that for heterogeneous reactions the concentration of the one (p. 631). solid is a constant. Thus, we can write the equilibrium constant for the dissolution of AgCl (see Example 14.5) as Ksp 5 [Ag1][Cl2]  16.6 Solubility Equilibria 743 where Ksp is called the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentra- tions of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. Because each AgCl unit contains only one Ag1 ion and one Cl2 ion, its solubil- ity product expression is particularly simple to write. The following cases are more complex: • MgF2 MgF2 (s) Δ Mg21(aq) 1 2F2(aq)   Ksp 5 [Mg21][F2]2 • Ag2CO3 Ag2CO3 (s) Δ 2Ag1(aq) 1 CO22 3 (aq)   Ksp 5 [Ag1]2[CO22 3 ] • Ca3(PO4)2 Ca3 (PO4 ) 2 (s) Δ 3Ca21(aq) 1 2PO32 21 3 32 2 4 (aq)   Ksp 5 [Ca ] [PO4 ] Table 16.2 lists the solubility products for a number of salts of low solubility. Soluble salts such as NaCl and KNO3, which have very large Ksp values, are not listed in the table for essentially the same reason that we did not include Ka values for strong acids in Table 15.3. The value of Ksp indicates the solubility of an ionic compound— the smaller the value, the less soluble the compound in water. However, in using Ksp values to compare solubilities, you should choose compounds that have similar for- mulas, such as AgCl and ZnS, or CaF2 and Fe(OH)2. Table 16.2 Solubility Products of Some Slightly Soluble Ionic Compounds at 25°C Compound Ksp Compound Ksp Aluminum hydroxide [Al(OH)3] 1.8 3 10233 Lead(II) chromate (PbCrO4) 2.0 3 10214 Barium carbonate (BaCO3) 8.1 3 1029 Lead(II) fluoride (PbF2) 4.1 3 1028 Barium fluoride (BaF2) 1.7 3 1026 Lead(II) iodide (PbI2) 1.4 3 1028 Barium sulfate (BaSO4) 1.1 3 10210 Lead(II) sulfide (PbS) 3.4 3 10228 Bismuth sulfide (Bi2S3) 1.6 3 10272 Magnesium carbonate (MgCO3) 4.0 3 1025 Cadmium sulfide (CdS) 8.0 3 10228 Magnesium hydroxide [Mg(OH)2] 1.2 3 10211 Calcium carbonate (CaCO3) 8.7 3 1029 Manganese(II) sulfide (MnS) 3.0 3 10214 Calcium fluoride (CaF2) 4.0 3 10211 Mercury(I) chloride (Hg2Cl2) 3.5 3 10218 Calcium hydroxide [Ca(OH)2] 8.0 3 1026 Mercury(II) sulfide (HgS) 4.0 3 10254 Calcium phosphate [Ca3(PO4)2] 1.2 3 10226 Nickel(II) sulfide (NiS) 1.4 3 10224 Chromium(III) hydroxide [Cr(OH)3] 3.0 3 10229 Silver bromide (AgBr) 7.7 3 10213 Cobalt(II) sulfide (CoS) 4.0 3 10221 Silver carbonate (Ag2CO3) 8.1 3 10212 Copper(I) bromide (CuBr) 4.2 3 1028 Silver chloride (AgCl) 1.6 3 10210 Copper(I) iodide (CuI) 5.1 3 10212 Silver iodide (AgI) 8.3 3 10217 Copper(II) hydroxide [Cu(OH)2] 2.2 3 10220 Silver sulfate (Ag2SO4) 1.4 3 1025 Copper(II) sulfide (CuS) 6.0 3 10237 Silver sulfide (Ag2S) 6.0 3 10251 Iron(II) hydroxide [Fe(OH)2] 1.6 3 10214 Strontium carbonate (SrCO3) 1.6 3 1029 Iron(III) hydroxide [Fe(OH)3] 1.1 3 10236 Strontium sulfate (SrSO4) 3.8 3 1027 Iron(II) sulfide (FeS) 6.0 3 10219 Tin(II) sulfide (SnS) 1.0 3 10226 Lead(II) carbonate (PbCO3) 3.3 3 10214 Zinc hydroxide [Zn(OH)2] 1.8 3 10214 Lead(II) chloride (PbCl2) 2.4 3 1024 Zinc sulfide (ZnS) 3.0 3 10223 744 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria A cautionary note: In Chapter 15 (p. 670) we assumed that dissolved substances exhibit ideal behavior for our calculations involving solution concentrations, but this assumption is not always valid. For example, a solution of barium fluoride (BaF2) may contain both neutral and charged ion pairs, such as BaF2 and BaF1, in addition to free Ba21 and F2 ions. Furthermore, many anions in the ionic compounds listed in Table 16.2 are conjugate bases of weak acids. Consider copper sulfide (CuS). The S22 ion can hydrolyze as follows S22(aq) 1 H2O(l) Δ HS2(aq) 1 OH2(aq) HS2(aq) 1 H2O(l) Δ H2S(aq) 1 OH2(aq) And highly charged small metal ions such as Al31 and Bi31 will undergo hydrolysis as discussed in Section 15.10. Both ion-pair formation and salt hydrolysis decrease the concentrations of the ions that appear in the Ksp expression, but we need not be concerned with the deviations from ideal behavior here. For the dissolution of an ionic solid in aqueous solution, any one of the following conditions may exist: (1) the solution is unsaturated, (2) the solution is saturated, or (3) the solution is supersaturated. For concentrations of ions that do not correspond to equilibrium conditions we use the reaction quotient (see Section 14.4), which in this case is called the ion product (Q), to predict whether a precipitate will form. Note that Q has the same form as Ksp except that the concentrations of ions are not equi- librium concentrations. For example, if we mix a solution containing Ag1 ions with one containing Cl2 ions, then the ion product is given by Q 5 [Ag1]0[Cl2]0 The subscript 0 reminds us that these are initial concentrations and do not neces- sarily correspond to those at equilibrium. The possible relationships between Q and Ksp are Depending on how a solution is made Q , Ksp Unsaturated solution (no precipitation) up, [Ag1] may or may not be equal to [Ag1]0[Cl2]0 , 1.6 3 10210 [Cl2]. Q 5 Ksp Saturated solution (no precipitation) [Ag1][Cl2] 5 1.6 3 10210 Q . Ksp Supersaturated solution; AgCl will [Ag1]0[Cl2]0 . 1.6 3 10210 precipitate out until the product of the ionic concentrations is equal to 1.6 3 10210 Review of Concepts The following diagrams represent solutions of AgCl, which may also contain ions such as Na1 and NO2 3 (not shown) that do not affect the solubility of AgCl. If (a) represents a saturated solution of AgCl, classify the other solutions as unsaturated, saturated, or supersaturated. ⫽ Ag⫹ ⫽ Cl⫺ (a) (b) (c) (d)  16.6 Solubility Equilibria 745 Figure 16.9 Sequence of Molar Concentrations steps (a) for calculating Ksp Solubility of Ksp of solubility of of cations from solubility data and (b) for compound compound compound and anions calculating solubility from Ksp data. (a) Concentrations Molar Ksp of Solubility of of cations solubility of compound compound and anions compound (b) Molar Solubility and Solubility There are two other ways to express a substance’s solubility: molar solubility, which is the number of moles of solute in 1 L of a saturated solution (mol/L), and solubility, which is the number of grams of solute in 1 L of a saturated solution (g/L). Note that both these expressions refer to the concentration of saturated solutions at some given temperature (usually 25°C). Both molar solubility and solubility are convenient to use in the laboratory. We can use them to determine Ksp by following the steps outlined in Figure 16.9(a). Example 16.8 illustrates this procedure. Example 16.8 The solubility of calcium sulfate (CaSO4) is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. Strategy We are given the solubility of CaSO4 and asked to calculate its Ksp. The sequence of conversion steps, according to Figure 16.9(a), is solubility of molar solubility [Ca21] and K of ¡ ¡ ¡ sp CaSO4 in g/L of CaSO4 [SO22 4 ] CaSO4 Solution Consider the dissociation of CaSO4 in water. Let s be the molar solubility Calcium sulfate is used as a dry- (in mol/L) of CaSO4. ing agent and in the manufacture of paints, ceramics, and paper. CaSO4 (s) Δ Ca21 (aq) 1 SO22 4 (aq) A hydrated form of calcium sul- Initial (M): 0 0 fate, called plaster of Paris, is used Change (M): 2s 1s 1s to make casts for broken bones. Equilibrium (M): s s The solubility product for CaSO4 is Ksp 5 [Ca21][SO22 4 ] 5 s 2 First, we calculate the number of moles of CaSO4 dissolved in 1 L of solution: 0.67 g CaSO4 1 mol CaSO4 3 5 4.9 3 1023 mol/L 5 s 1 L soln 136.2 g CaSO4 From the solubility equilibrium we see that for every mole of CaSO4 that dissolves, 1 mole of Ca21 and 1 mole of SO224 are produced. Thus, at equilibrium, [Ca21] 5 4.9 3 1023 M  and  [SO22 4 ] 5 4.9 3 10 23 M (Continued) 746 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Now we can calculate Ksp: Ksp 5 [Ca21][SO22 4 ] 5 (4.9 3 1023 ) (4.9 3 1023 ) Similar problem: 16.56. 5 2.4 3 1025 Practice Exercise The solubility of lead chromate (PbCrO4) is 4.5 3 1025 g/L. Calculate the solubility product of this compound. Sometimes we are given the value of Ksp for a compound and asked to calculate the compound’s molar solubility. For example, the Ksp of silver bromide (AgBr) is 7.7 3 10213. We can calculate its molar solubility by the same procedure as that for acid ionization constants. First we identify the species present at equilibrium. Here we have Ag1 and Br2 ions. Let s be the molar solubility (in mol/L) of AgBr. Because one unit of AgBr yields one Ag1 and one Br2 ion, at equilibrium both [Ag1] and [Br2] are equal to s. We summarize the changes in concentrations as follows: AgBr(s) Δ Ag1(aq) 1 Br2(aq) Initial (M): 0.00 0.00 Change (M): 2s 1s 1s Silver bromide is used in photo- graphic emulsions. Equilibrium (M): s s From Table 16.2 we write Ksp 5 [Ag1][Br2] 7.7 3 10213 5 (s)(s) s 5 27.7 3 10213 5 8.8 3 1027 M Therefore, at equilibrium [Ag1] 5 8.8 3 1027 M [Br2] 5 8.8 3 1027 M Thus, the molar solubility of AgBr also is 8.8 3 1027 M. Example 16.9 makes use of this approach. Example 16.9 Using the data in Table 16.2, calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L. Strategy We are given the Ksp of Cu(OH)2 and asked to calculate its solubility in g/L. The sequence of conversion steps, according to Figure 16.9(b), is Ksp of [Cu21] and molar solubility solubility of ¡ ¡ of Cu(OH) ¡ Cu(OH)2 [OH2] 2 Cu(OH)2 in g/L Copper(II) hydroxide is used as a Solution Consider the dissociation of Cu(OH)2 in water: pesticide and to treat seeds. Cu(OH) 2 (s) Δ Cu21 (aq) 1 2OH2 (aq) Initial (M): 0 0 Change (M): 2s 1s 12s Equilibrium (M): s 2s (Continued)  16.6 Solubility Equilibria 747 Note that the molar concentration of OH2 is twice that of Cu21. The solubility product of Cu(OH)2 is Ksp 5 [Cu21][OH2]2 5 (s) (2s) 2 5 4s3 From the Ksp value in Table 16.2, we solve for the molar solubility of Cu(OH)2 as follows: 2.2 3 10220 5 4s3 2.2 3 10220 s3 5 5 5.5 3 10221 4 Hence s 5 1.8 3 1027 M Finally, from the molar mass of Cu(OH)2 and its molar solubility, we calculate the solubility in g/L: 1.8 3 1027 mol Cu(OH) 2 97.57 g Cu(OH) 2 solubility of Cu(OH) 2 5 3 1 L soln 1 mol Cu(OH) 2 5 1.8 3 1025 g/L Similar problem: 16.58. Practice Exercise Calculate the solubility of silver chloride (AgCl) in g/L. As Examples 16.8 and 16.9 show, solubility and solubility product are related. If we know one, we can calculate the other, but each quantity provides different infor- mation. Table 16.3 shows the relationship between molar solubility and solubility product for a number of ionic compounds. When carrying out solubility and/or solubility product calculations, keep in mind the following important points: 1. Solubility is the quantity of a substance that dissolves in a certain quantity of water to produce a saturated solution. In solubility equilibria calculations, it is usually expressed as grams of solute per liter of solution. Molar solubility is the number of moles of solute per liter of solution. 2. Solubility product is an equilibrium constant. 3. Molar solubility, solubility, and solubility product all refer to a saturated solution. Table 16.3 Relationship Between Ksp and Molar Solubility (s) Compound Ksp Expression Cation Anion Relation Between Ksp and s 1 AgCl [Ag1][Cl2] s s Ksp 5 s2 ; s 5 (Ksp ) 2 1 BaSO4 [Ba21][SO22 4 ] s s Ksp 5 s2 ; s 5 (Ksp ) 2 1 Ksp 3 Ag2CO3 [Ag1]2[CO22 3 ] 2s s Ksp 5 4s3 ; s 5 a b 4 1 K sp 3 PbF2 [Pb21][F2]2 s 2s Ksp 5 4s3 ; s 5 a b 4 1 Ksp 4 Al(OH)3 [Al31][OH2]3 s 3s Ksp 5 27s4 ; s 5 a b 27 1 Ksp 5 Ca3(PO4)2 [Ca21]3[PO32 4 ] 2 3s 2s Ksp 5 108s5 ; s 5 a b 108 748 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Predicting Precipitation Reactions From a knowledge of the solubility rules (see Section 4.2) and the solubility products listed in Table 16.2, we can predict whether a precipitate will form when we mix two solutions or add a soluble compound to a solution. This ability often has practical value. In industrial and laboratory preparations, we can adjust the concentrations of ions until the ion product exceeds Ksp in order to obtain a given compound (in the form of a precipitate). The ability to predict precipitation reactions is also useful in medicine. For A kidney stone. example, kidney stones, which can be extremely painful, consist largely of calcium oxalate, CaC2O4 (Ksp 5 2.3 3 1029). The normal physiological concentration of calcium ions in blood plasma is about 5 mM (1 mM 5 1 3 1023 M). Oxalate ions (C2O22 4 ), derived from oxalic acid present in many vegetables such as rhubarb and spinach, react with the calcium ions to form insoluble calcium oxalate, which can gradually build up in the kidneys. Proper adjustment of a patient’s diet can help to reduce precipitate for- mation. Example 16.10 illustrates the steps involved in predicting precipitation reactions. Example 16.10 Exactly 200 mL of 0.0040 M BaCl2 are mixed with exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? Strategy Under what condition will an ionic compound precipitate from solution? The ions in solution are Ba21, Cl2, K1, and SO422. According to the solubility rules listed in Table 4.2 (p. 122), the only precipitate that can form is BaSO4. From the information given, we can calculate [Ba21] and [SO422] because we know the number of moles of the ions in the original solutions and the volume of the combined solution. Next, we calculate the ion product Q (Q 5 [Ba21]0[SO422]0) and compare the value of Q with Ksp of BaSO4 to see if a precipitate will form, that is, if the solution is supersaturated. It is helpful to make a sketch of the situation. Solution The number of moles of Ba21 present in the original 200 mL of solution is 0.0040 mol Ba21 1L 200 mL 3 3 5 8.0 3 1024 mol Ba21 1 L soln 1000 mL We assume that the volumes are The total volume after combining the two solutions is 800 mL. The concentration of additive. Ba21 in the 800-mL volume is 8.0 3 1024 mol 1000 mL [Ba21] 5 3 800 mL 1 L soln 5 1.0 3 1023 M (Continued)  16.7 Separation of Ions by Fractional Precipitation 749 The number of moles of SO422 in the original 600-mL solution is 0.0080 mol SO22 4 1L 600 mL 3 3 5 4.8 3 1023 mol SO22 4 1 L soln 1000 mL The concentration of SO422 in the 800 mL of the combined solution is 4.8 3 1023 mol 1000 mL [SO22 4 ] 5 3 800 mL 1 L soln 5 6.0 3 1023 M Now we must compare Q and Ksp. From Table 16.2, BaSO4 (s) Δ Ba21 (aq) 1 SO22 4 (aq)    Ksp 5 1.1 3 10 210 As for Q, Q 5 [Ba21]0[SO22 23 23 4 ]0 5 (1.0 3 10 ) (6.0 3 10 ) 5 6.0 3 1026 Therefore, Q . Ksp The solution is supersaturated because the value of Q indicates that the concentrations of the ions are too large. Thus, some of the BaSO4 will precipitate out of solution until [Ba21][SO22 4 ] 5 1.1 3 10 210 Similar problem: 16.61. Practice Exercise If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will precipitation occur? 16.7 Separation of Ions by Fractional Precipitation In chemical analysis, it is sometimes desirable to remove one type of ion from solu- tion by precipitation while leaving other ions in solution. For instance, the addition of sulfate ions to a solution containing both potassium and barium ions causes BaSO4 to precipitate out, thereby removing most of the Ba21 ions from the solution. The other “product,” K2SO4, is soluble and will remain in solution. The BaSO4 precipitate can be separated from the solution by filtration. Even when both products are insoluble, we can still achieve some degree of Compound Ksp separation by choosing the proper reagent to bring about precipitation. Consider a AgCl 1.6 3 10210 AgBr 7.7 3 10213 solution that contains Cl2, Br2, and I2 ions. One way to separate these ions is to Agl 8.3 3 10217 convert them to insoluble silver halides. As the Ksp values in the margin show, the solubility of the halides decreases from AgCl to AgI. Thus, when a soluble compound such as silver nitrate is slowly added to this solution, AgI begins to precipitate first, followed by AgBr and then AgCl. Example 16.11 describes the separation of only two ions (Cl2 and Br2), but the procedure can be applied to a solution containing more than two different types of ions if precipitates of differing solubility can be formed. 750 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Example 16.11 A solution contains 0.020 M Cl2 ions and 0.020 M Br2 ions. To separate the Cl2 ions from the Br2 ions, solid AgNO3 is slowly added to the solution without changing the volume. What concentration of Ag1 ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl? Strategy In solution, AgNO3 dissociates into Ag1 and NO23 ions. The Ag1 ions then combine with the Cl2 and Br2 ions to form AgCl and AgBr precipitates. Because AgBr is less soluble (it has a smaller Ksp than that of AgCl), it will precipitate first. Therefore, this is a fractional precipitation problem. Knowing the concentrations of Cl2 and Br2 ions, we can calculate [Ag1] from the Ksp values. Keep in mind that Ksp refers to a saturated solution. To initiate precipitation, [Ag1] must exceed the concentration in the saturated solution in each case. Solution The solubility equilibrium for AgBr is Suspension of AgCl precipitate (left) and AgBr precipitate (right). AgBr(s) Δ Ag1 (aq) 1 Br2 (aq)    Ksp 5 [Ag1][Br2] Because [Br2] 5 0.020 M, the concentration of Ag1 that must be exceeded to initiate the precipitation of AgBr is Ksp 7.7 3 10213 [Ag1] 5 5 [Br2] 0.020 5 3.9 3 10211 M Thus, [Ag1] . 3.9 3 10211 M is required to start the precipitation of AgBr. The solubility equilibrium for AgCl is AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq)    Ksp 5 [Ag1][Cl2] so that Ksp 1.6 3 10210 [Ag1] 5 2 5 [Cl ] 0.020 5 8.0 3 1029 M Therefore, [Ag1] . 8.0 3 1029 M is needed to initiate the precipitation of AgCl. To precipitate the Br2 ions as AgBr without precipitating the Cl2 ions as AgCl, Similar problems: 16.63, 16.64. then, [Ag1] must be greater than 3.9 3 10211 M and lower than 8.0 3 1029 M. Practice Exercise The solubility products of AgCl and Ag3PO4 are 1.6 3 10210 and 1.8 3 10218, respectively. If Ag1 is added (without changing the volume) to 1.00 L of a solution containing 0.10 mol Cl2 and 0.10 mol PO32 4 , calculate the concentration of Ag1 ions (in mol/L) required to initiate (a) the precipitation of AgCl and (b) the precipitation of Ag3PO4. Example 16.11 raises the question, What is the concentration of Br2 ions remain- ing in solution just before AgCl begins to precipitate? To answer this question we let [Ag1] 5 8.0 3 1029 M. Then Ksp [Br2] 5 [Ag1] 7.7 3 10213 5 8.0 3 1029 5 9.6 3 1025 M  16.8 The Common Ion Effect and Solubility 751 The percent of Br2 remaining in solution (the unprecipitated Br2) at the critical con- centration of Ag1 is [Br2]unppt’d % Br2 5 3 100% [Br2]original 9.6 3 1025 M 5 3 100% 0.020 M 5 0.48% unprecipitated Thus, (100 2 0.48) percent, or 99.52 percent, of Br2 will have precipitated as AgBr just before AgCl begins to precipitate. By this procedure, the Br2 ions can be quan- titatively separated from the Cl2 ions. Review of Concepts AgNO3 is slowly added to a solution that contains 0.1 M each of Br2, CO22 3 , and SO224 ions. Which compound will precipitate first and which compound will precipitate last? (Use the Ksp of each compound to calculate [Ag1] needed to produce a saturated solution.) 16.8 The Common Ion Effect and Solubility In Section 16.2 we discussed the effect of a common ion on acid and base ioniza- tions. Here we will examine the relationship between the common ion effect and solubility. As we have noted, the solubility product is an equilibrium constant; precipitation of an ionic compound from solution occurs whenever the ion product exceeds Ksp for that substance. In a saturated solution of AgCl, for example, the ion product [Ag1][Cl2] is, of course, equal to Ksp. Furthermore, simple stoichiometry tells us that [Ag1] 5 [Cl2]. But this equality does not hold in all situations. Suppose we study a solution containing two dissolved substances that share a common ion, say, AgCl and AgNO3. In addition to the dissociation of AgCl, the fol- lowing process also contributes to the total concentration of the common silver ions in solution: H2O AgNO3 (s) ¡ Ag1 (aq) 1 NO2 3 (aq) The solubility equilibrium of AgCl is AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) If AgNO3 is added to a saturated AgCl solution, the increase in [Ag1] will make the ion product greater than the solubility product: Q 5 [Ag1]0[Cl2]0 . Ksp To reestablish equilibrium, some AgCl will precipitate out of the solution, as Le At a given temperature, only the Châtelier’s principle would predict, until the ion product is once again equal to Ksp. solubility of a compound is altered (decreased) by the common ion effect. The effect of adding a common ion, then, is a decrease in the solubility of the salt Its solubility product, which is an (AgCl) in solution. Note that in this case [Ag1] is no longer equal to [Cl2] at equi- equilibrium constant, remains the librium; rather, [Ag1] . [Cl2]. same whether or not other substances Example 16.12 shows the common ion effect on solubility. are present in the solution. 752 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Example 16.12 Calculate the solubility of silver chloride (in g/L) in a 6.5 3 1023 M silver nitrate solution. Strategy This is a common-ion problem. The common ion here is Ag1, which is supplied by both AgCl and AgNO3. Remember that the presence of the common ion will affect only the solubility of AgCl (in g/L), but not the Ksp value because it is an equilibrium constant. Solution Step 1: The relevant species in solution are Ag1 ions (from both AgCl and AgNO3) and Cl2 ions. The NO2 3 ions are spectator ions. Step 2: Because AgNO3 is a soluble strong electrolyte, it dissociates completely: H2O AgNO3 (s) ¡    Ag1 (aq)    1   NO23 (aq) 6.5 3 1023 M 6.5 3 1023 M Let s be the molar solubility of AgCl in AgNO3 solution. We summarize the changes in concentrations as follows: AgCl(s) Δ Ag1(aq) 1 Cl2(aq) Initial (M): 6.5 3 1023 0.00 Change (M): 2s 1s 1s Equilibrium (M): (6.5 3 1023 1 s) s Step 3: Ksp 5 [Ag1][Cl2] 1.6 3 10210 5 (6.5 3 1023 1 s) (s) Because AgCl is quite insoluble and the presence of Ag1 ions from AgNO3 further lowers the solubility of AgCl, s must be very small compared with 6.5 3 1023. Therefore, applying the approximation 6.5 3 1023 1 s < 6.5 3 1023, we obtain 1.6 3 10210 5 (6.5 3 1023 )s s 5 2.5 3 1028 M Step 4: At equilibrium [Ag1] 5 (6.5 3 1023 1 2.5 3 1028 ) M < 6.5 3 1023 M [Cl2] 5 2.5 3 1028 M and so our approximation was justified in step 3. Because all the Cl2 ions must come from AgCl, the amount of AgCl dissolved in AgNO3 solution also is 2.5 3 1028 M. Then, knowing the molar mass of AgCl (143.4 g), we can calculate the solubility of AgCl as follows: 2.5 3 1028 mol AgCl 143.4 g AgCl solubility of AgCl in AgNO3 solution 5 3 1 L soln 1 mol AgCl 26 5 3.6 3 10 g/L Check The solubility of AgCl in pure water is 1.9 3 1023 g/L (see the Practice Exercise in Example 16.9). Therefore, the lower solubility (3.6 3 1026 g/L) in the presence of AgNO3 is reasonable. You should also be able to predict the lower solubility using Le Châtelier’s principle. Adding Ag1 ions shifts the equilibrium to Similar problem: 16.68. the left, thus decreasing the solubility of AgCl. Practice Exercise Calculate the solubility in g/L of AgBr in (a) pure water and (b) 0.0010 M NaBr.  16.9 pH and Solubility 753 Review of Concepts For each pair of solutions, determine the one in which PbI2(s) will be more soluble: (a) NaClO3(aq) or NaI(aq), (b) Pb(NO3)2(aq) or Ba(NO3)2(aq). 16.9 pH and Solubility The solubilities of many substances also depend on the pH of the solution. Consider the solubility equilibrium of magnesium hydroxide: Mg(OH) 2 (s) Δ Mg21 (aq) 1 2OH2 (aq) Adding OH2 ions (increasing the pH) shifts the equilibrium from right to left, thereby decreasing the solubility of Mg(OH)2. (This is another example of the common ion effect.) On the other hand, adding H1 ions (decreasing the pH) shifts the equilibrium from left to right, and the solubility of Mg(OH)2 increases. Thus, insoluble bases tend to dissolve in acidic solutions. Similarly, insoluble acids dissolve in basic solutions. To explore the quantitative effect of pH on the solubility of Mg(OH)2, let us first calculate the pH of a saturated Mg(OH)2 solution. We write Ksp 5 [Mg21][OH2]2 5 1.2 3 10211 Let s be the molar solubility of Mg(OH)2. Proceeding as in Example 16.9, Ksp 5 (s)(2s) 2 5 4s3 4s3 5 1.2 3 10211 s3 5 3.0 3 10212 s 5 1.4 3 1024 M At equilibrium, therefore, [OH2] 5 2 3 1.4 3 1024 M 5 2.8 3 1024 M pOH 5 2log (2.8 3 1024 ) 5 3.55 pH 5 14.00 2 3.55 5 10.45 In a medium with a pH of less than 10.45, the solubility of Mg(OH)2 would increase. This follows from the fact that a lower pH indicates a higher [H1] and thus a lower [OH2], as we would expect from Kw 5 [H1][OH2]. Consequently, [Mg21] rises to maintain the equilibrium condition, and more Mg(OH)2 dissolves. The dissolution process and the effect of extra H1 ions can be summarized as follows: Mg(OH) 2 (s) Δ Mg21 (aq) 1 2OH2 (aq) 2H (aq) 1 2OH2 (aq) Δ 2H2O(l) 1 Overall:   Mg(OH) 2 (s) 1 2H1 (aq) Δ Mg21 (aq) 1 2H2O(l) If the pH of the medium were higher than 10.45, [OH2] would be higher and the solubility of Mg(OH)2 would decrease because of the common ion (OH2) effect. Milk of magnesia, which contains The pH also influences the solubility of salts that contain a basic anion. For Mg(OH)2, is used to treat acid example, the solubility equilibrium for BaF2 is indigestion. BaF2 (s) Δ Ba21 (aq) 1 2F2 (aq) and Ksp 5 [Ba21][F2]2 754 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria In an acidic medium, the high [H1] will shift the following equilibrium from left to right: Because HF is a weak acid, its conjugate H1 (aq) 1 F2 (aq) Δ HF(aq) base, F2, has an affinity for H1. As [F2] decreases, [Ba21] must increase to maintain the equilibrium condition. Thus, more BaF2 dissolves. The dissolution process and the effect of pH on the solubility of BaF2 can be summarized as follows: BaF2 (s) Δ Ba21 (aq) 1 2F2 (aq) 2H (aq) 1 2F2 (aq) Δ 2HF(aq) 1 Overall:   BaF2 (s) 1 2H1 (aq) Δ Ba21 (aq) 1 2HF(aq) The solubilities of salts containing anions that do not hydrolyze are unaffected by pH. Examples of such anions are Cl2, Br2, and I2. Examples 16.13 and 16.14 deal with the effect of pH on solubility. Example 16.13 Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO4? Strategy In each case, write the dissociation reaction of the salt into its cation and anion. The cation will not interact with the H1 ion because they both bear positive charges. The anion will act as a proton acceptor only if it is the conjugate base of a weak acid. How would the removal of the anion affect the solubility of the salt? Solution (a) The solubility equilibrium for CuS is CuS(s) Δ Cu21 (aq) 1 S22 (aq) The sulfide ion is the conjugate base of the weak acid HS2. Therefore, the S22 ion reacts with the H1 ion as follows: S22 (aq) 1 H1 (aq) ¡ HS2 (aq) This reaction removes the S22 ions from solution. According to Le Châtelier’s principle, the equilibrium will shift to the right to replace some of the S22 ions that were removed, thereby increasing the solubility of CuS. (b) The solubility equilibrium is AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) Because Cl2 is the conjugate base of a strong acid (HCl), the solubility of AgCl is not affected by an acid solution. (c) The solubility equilibrium for PbSO4 is PbSO4 (s) Δ Pb21 (aq) 1 SO22 4 (aq) The sulfate ion is a weak base because it is the conjugate base of the weak acid HSO2 22 1 4 . Therefore, the SO 4 ion reacts with the H ion as follows: SO22 1 2 4 (aq) 1 H (aq) ¡ HSO4 (aq) This reaction removes the SO 422 ions from solution. According to Le Châtelier’s principle, the equilibrium will shift to the right to replace some of the SO 422 ions Similar problem: 16.72. that were removed, thereby increasing the solubility of PbSO4. Practice Exercise Is the solubility of the following compounds increased in an acidic solution? (a) Ca(OH)2, (b) Mg3(PO4)2, (c) PbBr2.  16.9 pH and Solubility 755 Example 16.14 Calculate the concentration of aqueous ammonia necessary to initiate the precipitation of iron(II) hydroxide from a 0.0030 M solution of FeCl2. Strategy For iron(II) hydroxide to precipitate from solution, the product [Fe21][OH2]2 must be greater than its Ksp. First, we calculate [OH2] from the known [Fe21] and the Ksp value listed in Table 16.2. This is the concentration of OH2 in a saturated solution of Fe(OH)2. Next, we calculate the concentration of NH3 that will supply this concentration of OH2 ions. Finally, any NH3 concentration greater than the calculated value will initiate the precipitation of Fe(OH)2 because the solution will become supersaturated. Solution Ammonia reacts with water to produce OH2 ions, which then react with Fe21 to form Fe(OH)2. The equilibria of interest are NH3 (aq) 1 H2O(l) Δ NH1 2 4 (aq) 1 OH (aq) Fe21 (aq) 1 2OH2 (aq) Δ Fe(OH) 2 (s) First we find the OH2 concentration above which Fe(OH)2 begins to precipitate. We write Ksp 5 [Fe21][OH2]2 5 1.6 3 10214 Because FeCl2 is a strong electrolyte, [Fe21] 5 0.0030 M and 1.6 3 10214 [OH2]2 5 5 5.3 3 10212 0.0030 [OH2] 5 2.3 3 1026 M Next, we calculate the concentration of NH3 that will supply 2.3 3 1026 M OH2 ions. Let x be the initial concentration of NH3 in mol/L. We summarize the changes in concentrations resulting from the ionization of NH3 as follows. NH3(aq) 1 H2O(l) Δ NH14(aq) 1 OH2(aq) Initial (M): x 0.00 0.00 Change (M): 22.3 3 1026 12.3 3 1026 12.3 3 1026 Equilibrium (M): (x 2 2.3 3 1026) 2.3 3 1026 2.3 3 1026 Substituting the equilibrium concentrations in the expression for the ionization constant (see Table 15.4), [NH1 2 4 ][OH ] Kb 5 [NH3] (2.3 3 1026 ) (2.3 3 1026 ) 1.8 3 1025 5 (x 2 2.3 3 1026 ) Solving for x, we obtain x 5 2.6 3 1026 M Therefore, the concentration of NH3 must be slightly greater than 2.6 3 1026 M to initiate the precipitation of Fe(OH)2. Similar problem: 16.76. Practice Exercise Calculate whether or not a precipitate will form if 2.0 mL of 0.60 M NH3 are added to 1.0 L of 1.0 3 1023 M ZnSO4. 756 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria 16.10 Complex Ion Equilibria and Solubility Lewis acids and bases are discussed Lewis acid-base reactions in which a metal cation combines with a Lewis base result in Section 15.12. in the formation of complex ions. Thus, we can define a complex ion as an ion con- taining a central metal cation bonded to one or more molecules or ions. Complex ions are crucial to many chemical and biological processes. Here we will consider the effect of complex ion formation on solubility. In Chapter 23 we will discuss the chemistry of complex ions in more detail. Transition metals have a particular tendency to form complex ions because they have incompletely filled d subshells. This property enables them to act effectively as Lewis acids in reactions with many molecules or ions that serve as electron donors, or as Lewis bases. For example, a solution of cobalt(II) chloride is pink because of According to our definition, Co(H2O)621 the presence of the Co(H2O) 621 ions (Figure 16.10). When HCl is added, the solution itself is a complex ion. When we write turns blue as a result of the formation of the complex ion CoCl 422: Co(H2O)621, we mean the hydrated Co21 ion. Co21 (aq) 1 4Cl2 (aq) Δ CoCl22 4 (aq) Copper(II) sulfate (CuSO4) dissolves in water to produce a blue solution. The hydrated copper(II) ions are responsible for this color; many other sulfates (Na2SO4, for example) are colorless. Adding a few drops of concentrated ammonia solution to a CuSO4 solution causes the formation of a light-blue precipitate, copper(II) hydroxide: Cu21 (aq) 1 2OH2 (aq) ¡ Cu(OH) 2 (s) The OH2 ions are supplied by the ammonia solution. If more NH3 is added, the blue precipitate redissolves to produce a beautiful dark-blue solution, this time due to the formation of the complex ion Cu(NH3)21 4 (Figure 16.11): Cu(OH) 2 (s) 1 4NH3 (aq) Δ Cu(NH3 ) 21 2 4 (aq) 1 2OH (aq) Thus, the formation of the complex ion Cu(NH3) 421 increases the solubility of Cu(OH)2. A measure of the tendency of a metal ion to form a particular complex ion is given by the formation constant Kf (also called the stability constant), which is the equilibrium constant for the complex ion formation. The larger Kf is, the more stable the complex ion is. Table 16.4 lists the formation constants of a number of complex ions. The formation of the Cu(NH3)214 ion can be expressed as Cu21 (aq) 1 4NH3 (aq) Δ Cu(NH3 ) 21 4 (aq) Figure 16.10 (Left) An aqueous cobalt(II) chloride solution. The pink color is due to the presence of Co(H2O)621 ions. (Right) After the addition of HCl solution, the solution turns blue because of the formation of the complex CoCl422 ions.  16.10 Complex Ion Equilibria and Solubility 757 Figure 16.11 Left: An aqueous solution of copper(II) sulfate. Center: After the addition of a few drops of a concentrated aqueous ammonia solution, a light-blue precipitate of Cu(OH)2 is formed. Right: When more concentrated aqueous ammonia solution is added, the Cu(OH)2 precipitate dissolves to form the dark-blue complex ion Cu(NH3)421. for which the formation constant is [Cu(NH3 ) 21 4 ] Kf 5 [Cu21][NH3]4 5 5.0 3 1013 The very large value of Kf in this case indicates that the complex ion is quite stable in solution and accounts for the very low concentration of copper(II) ions at equilibrium. Example 16.15 A 0.20-mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is the concentration of Cu21 ions at equilibrium? Strategy The addition of CuSO4 to the NH3 solution results in complex ion formation Cu21 (aq) 1 4NH3 (aq) Δ Cu(NH3 ) 21 4 (aq) (Continued) Table 16.4 Formation Constants of Selected Complex Ions in Water at 258C Formation Complex Ion Equilibrium Expression Constant (Kf) Ag(NH3)1 2 Ag1 1 2NH3 Δ Ag(NH3 )1 2 1.5 3 107 Ag(CN)22 Ag1 1 2CN2 Δ Ag(CN)22 1.0 3 1021 Cu(CN)22 4 Cu21 1 4CN2 Δ Cu(CN) 22 4 1.0 3 1025 Cu(NH3)214 Cu21 1 4NH3 Δ Cu(NH3 ) 21 4 5.0 3 1013 Cd(CN)22 4 Cd21 1 4CN2 Δ Cd(CN) 22 4 7.1 3 1016 CdI22 4 Cd21 1 4I2 Δ CdI 22 4 2.0 3 106 HgCl22 4 Hg21 1 4Cl2 Δ HgCl 22 4 1.7 3 1016 HgI22 4 Hg21 1 4I2 Δ HgI 22 4 2.0 3 1030 Hg(CN)22 4 Hg21 1 4CN2 Δ Hg(CN) 22 4 2.5 3 1041 Co(NH3)316 Co31 1 6NH3 Δ Co(NH3 ) 31 6 5.0 3 1031 Zn(NH3)21 4 Zn21 1 4NH3 Δ Zn(NH3 ) 21 4 2.9 3 109 758 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria From Table 16.4 we see that the formation constant (Kf) for this reaction is very large; therefore, the reaction lies mostly to the right. At equilibrium, the concentration of Cu21 will be very small. As a good approximation, we can assume that essentially all the dissolved Cu21 ions end up as Cu(NH3)21 4 ions. How many moles of NH3 will react with 0.20 mole of Cu21? How many moles of Cu(NH3)21 4 will be produced? A very small amount of Cu21 will be present at equilibrium. Set up the Kf expression for the preceding equilibrium to solve for [Cu21]. Solution The amount of NH3 consumed in forming the complex ion is 4 3 0.20 mol, or 0.80 mol. (Note that 0.20 mol Cu21 is initially present in solution and four NH3 molecules are needed to form a complex ion with one Cu21 ion.) The concentration of NH3 at equilibrium is therefore (1.20 2 0.80) mol/L soln or 0.40 M, and that of Cu(NH3)214 is 0.20 mol/L soln or 0.20 M, the same as the initial concentration of Cu21. [There is a 1:1 mole ratio between Cu21 and Cu(NH3)21 21 4 .] Because Cu(NH3)4 does dissociate to a slight extent, we call the concentration of Cu21 at equilibrium x and write [Cu(NH3 ) 21 4 ] Kf 5 [Cu21][NH3]4 0.20 5.0 3 1013 5 x(0.40) 4 Solving for x and keeping in mind that the volume of the solution is 1 L, we obtain x 5 [Cu21] 5 1.6 3 10213 M Check The small value of [Cu21] at equilibrium, compared with 0.20 M, certainly Similar problem: 16.79. justifies our approximation. Practice Exercise If 2.50 g of CuSO4 are dissolved in 9.0 3 102 mL of 0.30 M NH3, what are the concentrations of Cu21, Cu(NH3)21 4 , and NH3 at equilibrium? The effect of complex ion formation generally is to increase the solubility of a substance, as Example 16.16 shows. Example 16.16 Calculate the molar solubility of AgCl in a 1.0 M NH3 solution. Strategy AgCl is only slightly soluble in water AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) The Ag1 ions form a complex ion with NH3 (see Table 16.4) Ag1 (aq) 1 2NH3 (aq) Δ Ag(NH3 )12 Combining these two equilibria will give the overall equilibrium for the process. Solution Step 1: Initially, the species in solution are Ag1 and Cl2 ions and NH3. The reaction between Ag1 and NH3 produces the complex ion Ag(NH3)12 . (Continued)  16.10 Complex Ion Equilibria and Solubility 759 Step 2: The equilibrium reactions are AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) Ksp 5 [Ag1][Cl2] 5 1.6 3 10210 Ag1 (aq) 1 2NH3 (aq) Δ Ag(NH3 )12 (aq) [Ag(NH3 )12 ] Kf 5 5 1.5 3 107 [Ag1][NH3]2 Overall:   AgCl(s) 1 2NH3 (aq) Δ Ag(NH3 )12 (aq) 1 Cl2 (aq) The equilibrium constant K for the overall reaction is the product of the equilibrium constants of the individual reactions (see Section 14.2): [Ag(NH3 )12 ][Cl2] K 5 KspKf 5 [NH3]2 5 (1.6 3 10210 ) (1.5 3 107 ) 5 2.4 3 1023 Let s be the molar solubility of AgCl (mol/L). We summarize the changes in concentrations that result from formation of the complex ion as follows: AgCl(s) 1 2NH3 (aq) Δ Ag(NH3 )12 (aq) 1 Cl2 (aq) Initial (M): 1.0 0.0 0.0 Change (M): 2s 22s 1s 1s Equilibrium (M): (1.0 2 2s) s s The formation constant for Ag(NH3)12 is quite large, so most of the silver ions exist in the complexed form. In the absence of ammonia we have, at equilibrium, [Ag1] 5 [Cl2]. As a result of complex ion formation, however, we can write [Ag(NH3)12 ] 5 [Cl2]. (s) (s) Step 3: K5 (1.0 2 2s) 2 s2 2.4 3 1023 5 (1.0 2 2s) 2 Taking the square root of both sides, we obtain s 0.049 5 1.0 2 2s s 5 0.045 M Step 4: At equilibrium, 0.045 mole of AgCl dissolves in 1 L of 1.0 M NH3 solution. Check The molar solubility of AgCl in pure water is 1.3 3 1025 M. Thus, the formation of the complex ion Ag(NH3)12 enhances the solubility of AgCl (Figure 16.12). Similar problem: 16.82. Practice Exercise Calculate the molar solubility of AgBr in a 1.0 M NH3 solution. Review of Concepts Which compound, when added to water, will increase the solubility of CdS? (a) LiNO3, (b) Na2SO4, (c) KCN, (d) NaClO3.  CHEMISTRY in Action How an Eggshell Is Formed T he formation of the shell of a hen’s egg is a fascinating example of a natural precipitation process. An average eggshell weighs about 5 g and is 40 percent calcium. Most of the calcium in an eggshell is laid down within a 16-h period. This means that it is deposited at a rate of about 125 mg per hour. No hen can consume calcium fast enough to meet this demand. Instead, it is supplied by special bony masses in the hen’s long bones, which accumulate large reserves of calcium for eggshell formation. [The inorganic calcium compo- nent of the bone is calcium phosphate, Ca3(PO4)2, an insoluble compound.] If a hen is fed a low-calcium diet, her eggshells become progressively thinner; she might have to mobilize 10 percent of the total amount of calcium in her bones just to lay one egg! When the food supply is consistently low in calcium, Chicken eggs. egg production eventually stops. X-ray micrograph of an eggshell, The eggshell is largely composed of calcite, a crystalline showing columns of calcite. form of calcium carbonate (CaCO3). Normally, the raw materi- als, Ca21 and CO223 , are carried by the blood to the shell gland. The calcification process is a precipitation reaction: Carbonic acid ionizes stepwise to produce carbonate ions: Ca21 (aq) 1 CO22 3 (aq) Δ CaCO3 (s) H2CO3 (aq) Δ H1 (aq) 1 HCO2 3 (aq) HCO2 1 22 3 (aq) Δ H (aq) 1 CO3 (aq) In the blood, free Ca21 ions are in equilibrium with calcium ions bound to proteins. As the free ions are taken up by the shell Chickens do not perspire and so must pant to cool them- gland, more are provided by the dissociation of the protein- selves. Panting expels more CO2 from the chicken’s body than bound calcium. normal respiration does. According to Le Châtelier’s principle, The carbonate ions necessary for eggshell formation are a panting will shift the CO2–H2CO3 equilibrium shown from right metabolic byproduct. Carbon dioxide produced during metabo- to left, thereby lowering the concentration of the CO22 3 ions in lism is converted to carbonic acid (H2CO3) by the enzyme car- solution and resulting in thin eggshells. One remedy for this prob- bonic anhydrase (CA): lem is to give chickens carbonated water to drink in hot weather. The CO2 dissolved in the water adds CO2 to the chicken’s body CA CO2 (g) 1 H2O(l2 Δ H2CO3(aq) fluids and shifts the CO2–H2CO3 equilibrium to the right. Figure 16.12 From left to right: Formation of AgCl precipitate when AgNO3 solution is added to NaCl solution. With the addition of NH3 solution, the AgCl precipitate dissolves as the soluble Ag(NH3 )1 2 forms. 760  16.11 Application of the Solubility Product Principle to Qualitative Analysis 761 Finally, we note that there is a class of hydroxides, called amphoteric hydroxides, All amphoteric hydroxides are which can react with both acids and bases. Examples are Al(OH)3, Pb(OH)2, Cr(OH)3, insoluble compounds. Zn(OH)2, and Cd(OH)2. Thus, Al(OH)3 reacts with acids and bases as follows: Al(OH) 3 (s) 1 3H1 (aq) ¡ Al31 (aq) 1 3H2O(l) Al(OH) 3 (s) 1 OH2 (aq) Δ Al(OH)24 (aq) The increase in solubility of Al(OH)3 in a basic medium is the result of the formation of the complex ion Al(OH)24 in which Al(OH)3 acts as the Lewis acid and OH2 acts as the Lewis base. Other amphoteric hydroxides behave in a similar manner. 16.11 Application of the Solubility Product Principle to Qualitative Analysis In Section 4.6, we discussed the principle of gravimetric analysis, by which we measure the amount of an ion in an unknown sample. Here we will briefly discuss qualitative analysis, the determination of the types of ions present in a solution. We will focus on the cations. There are some 20 common cations that can be analyzed readily in aqueous solu- Do not confuse the groups in Table 16.5, tion. These cations can be divided into five groups according to the solubility products which are based on solubility products, with those in the periodic table, which of their insoluble salts (Table 16.5). Because an unknown solution may contain from are based on the electron configurations one to all 20 ions, any analysis must be carried out systematically from group 1 of the elements. Table 16.5 Separation of Cations into Groups According to Their Precipitation Reactions with Various Reagents Group Cation Precipitating Reagents Insoluble Compound Ksp 1 Ag1 HCl AgCl 1.6 3 10210 Hg21 2 Hg2Cl2 3.5 3 10218 Pb21 PbCl2 2.4 3 1024 2 Bi31 H2S Bi2S3 1.6 3 10272 Cd21 in acidic CdS 8.0 3 10228 Cu21 solutions CuS 6.0 3 10237 Hg21 HgS 4.0 3 10254 Sn21 SnS 1.0 3 10226 3 Al31 H2S Al(OH)3 1.8 3 10233 Co21 in basic CoS 4.0 3 10221 Cr31 solutions Cr(OH)3 3.0 3 10229 Fe21 FeS 6.0 3 10219 Mn21 MnS 3.0 3 10214 Ni21 NiS 1.4 3 10224 Zn21 ZnS 3.0 3 10223 4 Ba21 Na2CO3 BaCO3 8.1 3 1029 Ca21 CaCO3 8.7 3 1029 Sr21 SrCO3 1.6 3 1029 5 K1 No precipitating None Na1 reagent None NH1 4 None 762 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria through group 5. Let us consider the general procedure for separating these 20 ions by adding precipitating reagents to an unknown solution. • Group 1 Cations. When dilute HCl is added to the unknown solution, only the Ag1, Hg221, and Pb21 ions precipitate as insoluble chlorides. The other ions, whose chlorides are soluble, remain in solution. • Group 2 Cations. After the chloride precipitates have been removed by filtration, hydrogen sulfide is reacted with the unknown acidic solution. Under this condi- tion, the concentration of the S22 ion in solution is negligible. Therefore, the precipitation of metal sulfides is best represented as M21 (aq) 1 H2S(aq) Δ MS(s) 1 2H1 (aq) Adding acid to the solution shifts this equilibrium to the left so that only the least soluble metal sulfides, that is, those with the smallest Ksp values, will precipitate out of solution. These are Bi2S3, CdS, CuS, HgS, and SnS (see Table 16.5). • Group 3 Cations. At this stage, sodium hydroxide is added to the solution to make it basic. In a basic solution, the above equilibrium shifts to the right. Therefore, the more soluble sulfides (CoS, FeS, MnS, NiS, ZnS) now precipitate out of solution. Note that the Al31 and Cr31 ions actually precipitate as the hydroxides Al(OH)3 and Cr(OH)3, rather than as the sulfides, because the hydrox- ides are less soluble. The solution is then filtered to remove the insoluble sulfides and hydroxides. • Group 4 Cations. After all the group 1, 2, and 3 cations have been removed from solution, sodium carbonate is added to the basic solution to precipitate Ba21, Ca21, and Sr21 ions as BaCO3, CaCO3, and SrCO3. These precipitates too are removed from solution by filtration. • Group 5 Cations. At this stage, the only cations possibly remaining in solution are Na1, K1, and NH14. The presence of NH14 can be determined by adding sodium hydroxide: NaOH(aq) 1 NH41 (aq) ¡ Na1 (aq) 1 H2O(l) 1 NH3 (g) The ammonia gas is detected either by noting its characteristic odor or by observ- ing a piece of wet red litmus paper turning blue when placed above (not in Because NaOH is added in group 3 and contact with) the solution. To confirm the presence of Na1 and K1 ions, we Na2CO3 is added in group 4, the flame usually use a flame test, as follows: A piece of platinum wire (chosen because test for Na1 ions is carried out using the original solution. platinum is inert) is moistened with the solution and is then held over a Bunsen burner flame. Each type of metal ion gives a characteristic color when heated in this manner. For example, the color emitted by Na1 ions is yellow, that of K1 ions is violet, and that of Cu21 ions is green (Figure 16.13). Figure 16.14 summarizes this scheme for separating metal ions. Two points regarding qualitative analysis must be mentioned. First, the separa- tion of the cations into groups is made as selective as possible; that is, the anions that are added as reagents must be such that they will precipitate the fewest types of cations. For example, all the cations in group 1 also form insoluble sulfides. Thus, if H2S were reacted with the solution at the start, as many as seven different sulfides might precipitate out of solution (group 1 and group 2 sulfides), an undesir- able outcome. Second, the removal of cations at each step must be carried out as completely as possible. For example, if we do not add enough HCl to the unknown solution to remove all the group 1 cations, they will precipitate with the group 2 cations as insoluble sulfides, interfering with further chemical analysis and leading to erroneous conclusions.  Key Equations 763 Figure 16.13 Left to right: Flame colors of lithium, sodium, potassium, and copper. Figure 16.14 A flow chart for the Solution containing ions separation of cations in qualitative of all cation groups analysis. +HCl Group 1 precipitates Filtration AgCl, Hg2Cl2, PbCl2 Solution containing ions of remaining groups +H2S Group 2 precipitates Filtration CuS, CdS, HgS, SnS, Bi2S3 Solution containing ions of remaining groups +NaOH Group 3 precipitates CoS, FeS, MnS, NiS Filtration ZnS, Al(OH)3, Cr(OH)3 Solution containing ions of remaining groups +Na2CO3 Group 4 precipitates Filtration BaCO3, CaCO3, SrCO3 Solution contains Na+, K+, NH +4 ions Key Equations pKa 5 2log Ka (16.3) Definition of pKa. [conjugate base] pH 5 pKa 1 log (16.4) Henderson-Hasselbalch equation. [acid]  764 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria Summary of Facts & Concepts 1. The common ion effect tends to suppress the ionization 5. The solubility product Ksp expresses the equilibrium be- of a weak acid or a weak base. This action can be ex- tween a solid and its ions in solution. Solubility can be plained by Le Châtelier’s principle. found from Ksp and vice versa. 2. A buffer solution is a combination of either a weak acid 6. The presence of a common ion decreases the solubility and its weak conjugate base (supplied by a salt) or a of a slightly soluble salt. weak base and its weak conjugate acid (supplied by a 7. The solubility of slightly soluble salts containing basic salt); the solution reacts with small amounts of added anions increases as the hydrogen ion concentration in- acid or base in such a way that the pH of the solution creases. The solubility of salts with anions derived from remains nearly constant. Buffer systems play a vital role strong acids is unaffected by pH. in maintaining the pH of body fluids. 8. Complex ions are formed in solution by the combina- 3. The pH at the equivalence point of an acid-base titration tion of a metal cation with a Lewis base. The formation depends on hydrolysis of the salt formed in the neutral- constant Kf measures the tendency toward the formation ization reaction. For strong acid–strong base titrations, of a specific complex ion. Complex ion formation can the pH at the equivalence point is 7; for weak acid– increase the solubility of an insoluble substance. strong base titrations, the pH at the equivalence point is 9. Qualitative analysis is the identification of cations and greater than 7; for strong acid–weak base titrations, the anions in solution. pH at the equivalence point is less than 7. 4. Acid-base indicators are weak organic acids or bases that change color near the equivalence point in an acid- base neutralization reaction. Key Words Buffer solution, p. 724 End point, p. 739 Molar solubility, p. 745 Solubility product Common ion effect, p. 721 Formation constant Qualitative analysis, p. 761 (Ksp), p. 743 Complex ion, p. 756 (Kf), p. 756 Solubility, p. 745 Questions & Problems† • Problems available in Connect Plus 16.4 The pKas of two monoprotic acids HA and HB are Red numbered problems solved in Student Solutions Manual 5.9 and 8.1, respectively. Which of the two is the stronger acid? The Common Ion Effect Review Questions Problems 16.1 Use Le Châtelier’s principle to explain how the • 16.5 Determine the pH of (a) a 0.40 M CH3COOH solu- common ion effect affects the pH of a solution. tion, (b) a solution that is 0.40 M CH3COOH and 0.20 M CH3COONa. • 16.2 Describe the effect on pH (increase, decrease, or no change) that results from each of the following • 16.6 Determine the pH of (a) a 0.20 M NH3 solution, additions: (a) potassium acetate to an acetic acid (b) a solution that is 0.20 M in NH3 and 0.30 M solution; (b) ammonium nitrate to an ammonia so- NH4Cl. lution; (c) sodium formate (HCOONa) to a formic acid (HCOOH) solution; (d) potassium chloride to Buffer Solutions a hydrochloric acid solution; (e) barium iodide to a Review Questions hydroiodic acid solution. 16.3 Define pKa for a weak acid. What is the relationship 16.7 What is a buffer solution? What constitutes a buffer between the value of the pKa and the strength of the solution? acid? Do the same for a weak base. † The temperature is assumed to be 25°C for all the problems.  Questions & Problems 765 • 16.8 Which of the following has the greatest buffer ca- ⫽ H2A ⫽ HA⫺ ⫽ A2 ⫺ pacity? (a) 0.40 M CH3COONa/0.20 M CH3COOH, (b) 0.40 M CH3COONa/0.60 M CH3COOH, and (c) 0.30 M CH3COONa/0.60 M CH3COOH. Problems 16.9 Which of the following solutions can act as a buffer? (a) KCl/HCl, (b) KHSO4/H2SO4, (c) Na2HPO4/ NaH2PO4, (d) KNO2/HNO2. (a) (b) (c) (d) • 16.10 Which of the following solutions can act as a buf- fer? (a) KCN/HCN, (b) Na2SO4/NaHSO4, (c) NH3/ 16.22 The diagrams shown here represent solutions con- NH4NO3, (d) NaI/HI. taining a weak acid HA (pKa 5 5.00) and its so- dium salt NaA. (1) Calculate the pH of the • 16.11 Calculate the pH of the buffer system made up of solutions. (2) What is the pH after the addition of 0.15 M NH3/0.35 M NH4Cl. 0.1 mol H1 ions to solution (a)? (3) What is the pH • 16.12 Calculate the pH of the following two buffer solu- after the addition of 0.1 mol OH2 ions to solution tions: (a) 2.0 M CH3COONa/2.0 M CH3COOH, (d)? Treat each sphere as 0.1 mol. (b) 0.20 M CH3COONa/0.20 M CH3COOH. Which is the more effective buffer? Why? ⫽ HA ⫽ A⫺ • 16.13 The pH of a bicarbonate-carbonic acid buffer is 8.00. Calculate the ratio of the concentration of carbonic acid (H2CO3) to that of the bicarbonate ion (HCO32). • 16.14 What is the pH of the buffer 0.10 M Na2HPO4/0.15 M KH2PO4? • 16.15 The pH of a sodium acetate–acetic acid buffer is 4.50. Calculate the ratio [CH3COO2]/[CH3COOH]. 16.16 The pH of blood plasma is 7.40. Assuming the prin- (a) (b) (c) (d) cipal buffer system is HCO32/H2CO3, calculate the ratio [HCO32]/[H2CO3]. Is this buffer more effective 16.23 How much NaOH (in moles) must be added to 1 L of against an added acid or an added base? a buffer solution that is 1.8 M in acetic acid and 1.2 M • 16.17 Calculate the pH of the 0.20 M NH3/0.20 M NH4Cl in sodium acetate to result in a buffer solution of pH buffer. What is the pH of the buffer after the addi- 5.22? Assume volume to remain constant. tion of 10.0 mL of 0.10 M HCl to 65.0 mL of the 16.24 How much HCl (in moles) must be added to 1 L of a buffer? buffer solution that is 0.84 M in ammonia and 0.96 M • 16.18 Calculate the pH of 1.00 L of the buffer 1.00 M in ammonium chloride to result in a buffer solution CH3COONa/1.00 M CH3COOH before and after the of pH 8.56? Assume volume to remain constant. addition of (a) 0.080 mol NaOH, (b) 0.12 mol HCl. (Assume that there is no change in volume.) Acid-Base Titrations 16.19 A diprotic acid, H2A, has the following ionization Review Questions constants: Ka1 5 1.1 3 1023 and Ka2 5 2.5 3 1026. 16.25 Briefly describe what happens in an acid-base In order to make up a buffer solution of pH 5.80, titration. which combination would you choose? NaHA/H2A or Na2A/NaHA. 16.26 Sketch titration curves for the following acid-base titrations: (a) HCl versus NaOH, (b) HCl versus 16.20 A student is asked to prepare a buffer solution at CH3NH2, (c) CH3COOH versus NaOH. In each pH 5 8.60, using one of the following weak acids: case, the base is added to the acid in an Erlenmeyer HA (Ka 5 2.7 3 1023), HB (Ka 5 4.4 3 1026), flask. Your graphs should show pH on the y-axis and HC (Ka 5 2.6 3 1029). Which acid should she volume of base added on the x-axis. choose? Why? • 16.21 The diagrams shown above contain one or more of Problems the compounds: H2A, NaHA, and Na2A, where H2A is a weak diprotic acid. (1) Which of the • 16.27 A 0.2688-g sample of a monoprotic acid neutralizes solutions can act as buffer solutions? (2) Which 16.4 mL of 0.08133 M KOH solution. Calculate the solution is the most effective buffer solution? molar mass of the acid. Water molecules and Na1 ions have been omitted • 16.28 A 5.00-g quantity of a diprotic acid was dissolved for clarity. in water and made up to exactly 250 mL. Calculate  766 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria the molar mass of the acid if 25.0 mL of this solu- 16.38 The diagrams shown here represent solutions at tion required 11.1 mL of 1.00 M KOH for neutral- various stages in the titration of a weak base B ization. Assume that both protons of the acid were (such as NH3) with HCl. Identify the solution that titrated. corresponds to (1) the initial stage before the • 16.29 In a titration experiment, 12.5 mL of 0.500 M H2SO4 addition of HCl, (2) halfway to the equivalence neutralize 50.0 mL of NaOH. What is the concentra- point, (3) the equivalence point, (4) beyond the tion of the NaOH solution? equivalence point. Is the pH greater than, less than, or equal to 7 at the equivalence point? Water • 16.30 In a titration experiment, 20.4 mL of 0.883 M molecules and Cl2 ions have been omitted for HCOOH neutralize 19.3 mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution? clarity. • 16.31 A 0.1276-g sample of an unknown monoprotic acid ⫽B ⫽ BH⫹ ⫽ H3O⫹ was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base re- quired to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added dur- ing the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid? • 16.32 A solution is made by mixing 5.00 3 102 mL of 0.167 M NaOH with 5.00 3 102 mL of 0.100 M CH3COOH. Calculate the equilibrium concentra- (a) (b) (c) (d) tions of H1, CH3COOH, CH3COO2, OH2, and Na1. 16.39 A 0.054 M HNO2 solution is titrated with a KOH • 16.33 Calculate the pH at the equivalence point for the fol- solution. What is [H1] at half way to the equivalence lowing titration: 0.20 M HCl versus 0.20 M methyl- point? amine (CH3NH2). (See Table 15.4.) 16.40 A student titrates an unknown monoprotic acid with 16.34 Calculate the pH at the equivalence point for the a NaOH solution from a buret. After the addition of following titration: 0.10 M HCOOH versus 0.10 M 12.35 mL of NaOH, the pH of the solution read NaOH. 5.22. The equivalence point is reached at 24.70 mL • 16.35 A 25.0-mL solution of 0.100 M CH3COOH is titrated of NaOH. What is the Ka of the acid? with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, Acid-Base Indicators (e) 15.0 mL. Review Questions • 16.36 A 10.0-mL solution of 0.300 M NH3 is titrated with 16.41 Explain how an acid-base indicator works in a titra- a 0.100 M HCl solution. Calculate the pH after the tion. What are the criteria for choosing an indicator following additions of the HCl solution: (a) 0.0 mL, for a particular acid-base titration? (b) 10.0 mL, (c) 20.0 mL, (d) 30.0 mL, (e) 40.0 mL. 16.42 The amount of indicator used in an acid-base titra- • 16.37 The diagrams shown here represent solutions at dif- tion must be small. Why? ferent stages in the titration of a weak acid HA with NaOH. Identify the solution that corresponds to Problems (1) the initial stage before the addition of NaOH, (2) halfway to the equivalence point, (3) the equiva- 16.43 Referring to Table 16.1, specify which indicator or lence point, (4) beyond the equivalence point. Is the indicators you would use for the following titra- pH greater than, less than, or equal to 7 at the equiv- tions: (a) HCOOH versus NaOH, (b) HCl versus alence point? Water molecules and Na1 ions have KOH, (c) HNO3 versus CH3NH2. been omitted for clarity. 16.44 A student carried out an acid-base titration by add- ing NaOH solution from a buret to an Erlenmeyer ⫽ HA ⫽ A⫺ ⫽ OH⫺ flask containing HCl solution and using phenol- phthalein as indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened? 16.45 The ionization constant Ka of an indicator HIn is 1.0 3 1026. The color of the nonionized form is red and that of the ionized form is yellow. What is the color of this indicator in a solution whose pH (a) (b) (c) (d) is 4.00?  Questions & Problems 767 • 16.46 The Ka of a certain indicator is 2.0 3 1026. The Fractional Precipitation color of HIn is green and that of In2 is red. A few Problems drops of the indicator are added to a HCl solution, which is then titrated against a NaOH solution. At • 16.63 Solid NaI is slowly added to a solution that is what pH will the indicator change color? 0.010 M in Cu1 and 0.010 M in Ag1. (a) Which compound will begin to precipitate first? (b) Cal- culate [Ag1] when CuI just begins to precipitate. Solubility Equilibria (c) What percent of Ag1 remains in solution at Review Questions this point? 16.47 Use BaSO4 to distinguish between solubility, molar 16.64 Find the approximate pH range suitable for the sepa- solubility, and solubility product. ration of Fe31 and Zn21 ions by precipitation of Fe(OH)3 from a solution that is initially 0.010 M in 16.48 Why do we usually not quote the Ksp values for sol- both Fe31 and Zn21. Assume a 99 percent precipita- uble ionic compounds? tion of Fe(OH)3. • 16.49 Write balanced equations and solubility product ex- pressions for the solubility equilibria of the follow- ing compounds: (a) CuBr, (b) ZnC2O4, (c) Ag2CrO4, The Common Ion Effect and Solubility (d) Hg2Cl2, (e) AuCl3, (f) Mn3(PO4)2. Review Questions • 16.50 Write the solubility product expression for the ionic 16.65 How does the common ion effect influence solubil- compound AxBy. ity equilibria? Use Le Châtelier’s principle to ex- 16.51 How can we predict whether a precipitate will form plain the decrease in solubility of CaCO3 in a when two solutions are mixed? Na2CO3 solution. 16.52 Silver chloride has a larger Ksp than silver car- bonate (see Table 16.2). Does this mean that • 16.66 The molar solubility of AgCl in 6.5 3 1023 M AgNO3 is 2.5 3 1028 M. In deriving Ksp from these data, AgCl also has a larger molar solubility than which of the following assumptions are reasonable? Ag2CO3? (a) Ksp is the same as solubility. (b) Ksp of AgCl is the same in 6.5 3 1023 M Problems AgNO3 as in pure water. • 16.53 Calculate the concentration of ions in the following (c) Solubility of AgCl is independent of the saturated solutions: (a) [I2] in AgI solution with concentration of AgNO3. [Ag1] 5 9.1 3 1029 M, (b) [Al31] in Al(OH)3 solu- (d) [Ag1] in solution does not change significantly tion with [OH2] 5 2.9 3 1029 M. upon the addition of AgCl to 6.5 3 1023 M • 16.54 From the solubility data given, calculate the solubil- AgNO3. ity products for the following compounds: (a) SrF2, (e) [Ag1] in solution after the addition of AgCl to 7.3 3 1022 g/L, (b) Ag3PO4, 6.7 3 1023 g/L. 6.5 3 1023 M AgNO3 is the same as it would • 16.55 The molar solubility of MnCO3 is 4.2 3 1026 M. be in pure water. What is Ksp for this compound? • 16.56 The solubility of an ionic compound MX (molar Problems mass 5 346 g) is 4.63 3 1023 g/L. What is Ksp for the compound? • 16.67 How many grams of CaCO3 will dissolve in • 16.57 The solubility of an ionic compound M2X3 (molar 3.0 3 102 mL of 0.050 M Ca(NO3)2? mass 5 288 g) is 3.6 3 10217 g/L. What is Ksp for 16.68 The solubility product of PbBr2 is 8.9 3 1026. De- the compound? termine the molar solubility (a) in pure water, 16.58 Using data from Table 16.2, calculate the molar sol- (b) in 0.20 M KBr solution, (c) in 0.20 M Pb(NO3)2 ubility of CaF2. solution. • 16.59 What is the pH of a saturated zinc hydroxide • 16.69 Calculate the molar solubility of AgCl in a 1.00-L solution? solution containing 10.0 g of dissolved CaCl2. • 16.60 The pH of a saturated solution of a metal hydrox- • 16.70 Calculate the molar solubility of BaSO4 (a) in water, ide MOH is 9.68. Calculate the K sp for the (b) in a solution containing 1.0 M SO22 4 ions. compound. • 16.61 If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL pH and Solubility of 0.10 M Na2CO3, will BaCO3 precipitate? Problems • 16.62 A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of 0.15 M Sr(NO3)2. Calculate the concentra- • 16.71 Which of the following ionic compounds will be tions in the final solution of NO2 1 21 3 , Na , Sr , and more soluble in acid solution than in water? 2 210 F . (Ksp for SrF2 5 2.0 3 10 .) (a) BaSO4, (b) PbCl2, (c) Fe(OH)3, (d) CaCO3.  768 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria 16.72 Which of the following will be more soluble in acid reagent that would enable her to separate AgCl(s) solution than in pure water? (a) CuI, (b) Ag2SO4, from PbCl2(s). (c) Zn(OH)2, (d) BaC2O4, (e) Ca3(PO4)2. 16.88 In a group 1 analysis, a student adds HCl acid to the • 16.73 Compare the molar solubility of Mg(OH)2 in water unknown solution to make [Cl2] 5 0.15 M. Some and in a solution buffered at a pH of 9.0. PbCl2 precipitates. Calculate the concentration of • 16.74 Calculate the molar solubility of Fe(OH)2 in a solu- Pb21 remaining in solution. tion buffered at (a) pH 8.00, (b) pH 10.00. 16.89 Both KCl and NH4Cl are white solids. Suggest one • 16.75 The solubility product of Mg(OH)2 is 1.2 3 10211. reagent that would enable you to distinguish between What minimum OH2 concentration must be attained these two compounds. (for example, by adding NaOH) to decrease the 16.90 Describe a simple test that would enable you to Mg21 concentration in a solution of Mg(NO3)2 to distinguish between AgNO3(s) and Cu(NO3)2(s). less than 1.0 3 10210 M? 16.76 Calculate whether or not a precipitate will form if Additional Problems 2.00 mL of 0.60 M NH3 are added to 1.0 L of 16.91 To act as an effective buffer, the concentrations of 1.0 3 1023 M FeSO4. the acid and the conjugate base should not differ by more than a factor of 10, that is, Complex Ion Equilibria and Solubility [conjugate base] Review Questions 10 $ $ 0.1 [acid] 16.77 Explain the formation of complexes in Table 16.4 in (a) Show that the buffer range, that is, the range of terms of Lewis acid-base theory. the concentration ratio over which the buffer is 16.78 Give an example to illustrate the general effect of effective, is given by pH 5 pKa 6 1. (b) Calculate complex ion formation on solubility. the pH range for the following buffer systems: (a) acetate, (b) nitrite, (c) bicarbonate, (d) phosphate. 16.92 The pKa of the indicator methyl orange is 3.46. Over Problems what pH range does this indicator change from • 16.79 If 2.50 g of CuSO4 are dissolved in 9.0 3 102 mL of 90 percent HIn to 90 percent In2? 0.30 M NH3, what are the concentrations of Cu21, 16.93 The iodide impurity in a 4.50-g sample of a metal Cu(NH3)21 4 , and NH3 at equilibrium? nitrate is precipitated as silver iodide. If 5.54 mL of 16.80 Calculate the concentrations of Cd21, Cd(CN)422, 0.186 M AgNO3 solution is needed for the precipi- and CN2 at equilibrium when 0.50 g of Cd(NO3)2 tation, calculate the mass percent of iodide in the dissolves in 5.0 3 102 mL of 0.50 M NaCN. sample. • 16.81 If NaOH is added to 0.010 M Al31, which will be the 16.94 A sodium acetate-acetic acid buffer solution was pre- predominant species at equilibrium: Al(OH)3 or pared by adding a 0.020 M HCl solution to 500 mL of Al(OH)2 4 ? The pH of the solution is 14.00. [Kf for 0.020 M CH3COONa and then diluting the mixed Al(OH)2 33 4 5 2.0 3 10 .] solution to 1.0 L. Calculate the original volume of • 16.82 Calculate the molar solubility of AgI in a 1.0 M NH3 the HCl solution needed to prepare a buffer solution solution. of pH 5.00. 16.83 Both Ag1 and Zn21 form complex ions with NH3. 16.95 Sketch the titration curve of a weak acid versus a Write balanced equations for the reactions. How- strong base like the one shown in Figure 16.5. On ever, Zn(OH)2 is soluble in 6 M NaOH, and AgOH is your graph indicate the volume of base used at not. Explain. the equivalence point and also at the half- equivalence point, that is, the point at which half • 16.84 Explain, with balanced ionic equations, why (a) CuI2 of the acid has been neutralized. Show how you dissolves in ammonia solution, (b) AgBr dissolves in NaCN solution, (c) HgCl2 dissolves in KCl solution. can measure the pH of the solution at the half- equivalence point. Using Equation (16.4), ex- plain how you can determine the pKa of the acid Qualitative Analysis by this procedure. Review Questions 16.96 A 200-mL volume of NaOH solution was added to 400 mL of a 2.00 M HNO2 solution. The pH of the 16.85 Outline the general procedure of qualitative analysis. mixed solution was 1.50 units greater than that of 16.86 Give two examples of metal ions in each group the original acid solution. Calculate the molarity of (1 through 5) in the qualitative analysis scheme. the NaOH solution. Problems • 16.97 The pKa of butyric acid (HBut) is 4.7. Calculate Kb for the butyrate ion (But2). • 16.87 In a group 1 analysis, a student obtained a precipi- 16.98 A solution is made by mixing 5.00 3 102 mL of tate containing both AgCl and PbCl2. Suggest one 0.167 M NaOH with 5.00 3 102 mL 0.100 M  Questions & Problems 769 HCOOH. Calculate the equilibrium concentrations 16.109 Barium is a toxic substance that can seriously of H1, HCOOH, HCOO2, OH2, and Na1. impair heart function. For an X ray of the gastro- 16.99 Cd(OH)2 is an insoluble compound. It dissolves in intestinal tract, a patient drinks an aqueous sus- excess NaOH in solution. Write a balanced ionic pension of 20 g BaSO4. If this substance were to equation for this reaction. What type of reaction is equilibrate with the 5.0 L of the blood in the pa- this? tient’s body, what would be [Ba21]? For a good estimate, we may assume that the temperature is • 16.100 A student mixes 50.0 mL of 1.00 M Ba(OH) 2 at 25°C. Why is Ba(NO3)2 not chosen for this with 86.4 mL of 0.494 M H 2SO4. Calculate the mass of BaSO 4 formed and the pH of the mixed procedure? solution. 16.110 The pKa of phenolphthalein is 9.10. Over what pH • 16.101 For which of the following reactions is the equilib- range does this indicator change from 95 percent rium constant called a solubility product? HIn to 95 percent In2? (a) Zn(OH) 2 (s) 1 2OH2 (aq) Δ • 16.111 Solid NaBr is slowly added to a solution that is Zn(OH) 22 4 (aq) 0.010 M in Cu1 and 0.010 M in Ag1. (a) Which 21 32 compound will begin to precipitate first? (b) Calcu- (b) 3Ca (aq) 1 2PO4 (aq) Δ Ca3 (PO4 ) 2 (s) late [Ag1] when CuBr just begins to precipitate. (c) CaCO3 (s) 1 2H1 (aq) Δ (c) What percent of Ag1 remains in solution at this Ca21 (aq) 1 H2O(l) 1 CO2 (g) point? (d) PbI2 (s) Δ Pb21 (aq) 1 2I2 (aq) • 16.112 Cacodylic acid is (CH3)2AsO2H. Its ionization con- 16.102 A 2.0-L kettle contains 116 g of boiler scale stant is 6.4 3 1027. (a) Calculate the pH of 50.0 mL (CaCO3). How many times would the kettle have to of a 0.10 M solution of the acid. (b) Calculate the pH be completely filled with distilled water to remove of 25.0 mL of 0.15 M (CH3)2AsO2Na. (c) Mix the all of the deposit? solutions in part (a) and part (b). Calculate the pH of • 16.103 Equal volumes of 0.12 M AgNO3 and 0.14 M ZnCl2 the resulting solution. solution are mixed. Calculate the equilibrium con- • 16.113 Radiochemical techniques are useful in estimat- centrations of Ag1, Cl2, Zn21, and NO2 3. ing the solubility product of many compounds. In • 16.104 Calculate the solubility (in g/L) of Ag2CO3. one experiment, 50.0 mL of a 0.010 M AgNO3 16.105 Find the approximate pH range suitable for sepa- solution containing a silver isotope with a radio- rating Mg21 and Zn21 by the precipitation of activity of 74,025 counts per min per mL were Zn(OH)2 from a solution that is initially 0.010 M in mixed with 100 mL of a 0.030 M NaIO3 solution. Mg21 and Zn21. The mixed solution was diluted to 500 mL and • 16.106 A volume of 25.0 mL of 0.100 M HCl is titrated filtered to remove all of the AgIO3 precipitate. against a 0.100 M CH3NH2 solution added to it from The remaining solution was found to have a radio- a buret. Calculate the pH values of the solution activity of 44.4 counts per min per mL. What is (a) after 10.0 mL of CH3NH2 solution have been the Ksp of AgIO3? added, (b) after 25.0 mL of CH3NH2 solution have 16.114 The molar mass of a certain metal carbonate, been added, (c) after 35.0 mL of CH3NH2 solution MCO3, can be determined by adding an excess of have been added. HCl acid to react with all the carbonate and then • 16.107 The molar solubility of Pb(IO3)2 in a 0.10 M NaIO3 “back titrating” the remaining acid with a NaOH solution is 2.4 3 10211 mol/L. What is Ksp for solution. (a) Write equations for these reactions. Pb(IO3)2? (b) In a certain experiment, 18.68 mL of 5.653 M 16.108 When a KI solution was added to a solution of HCl were added to a 3.542-g sample of MCO3. The mercury(II) chloride, a precipitate [mercury(II) excess HCl required 12.06 mL of 1.789 M NaOH for iodide] formed. A student plotted the mass of the neutralization. Calculate the molar mass of the car- precipitate versus the volume of the KI solution bonate and identify M. added and obtained the following graph. Explain the 16.115 Acid-base reactions usually go to completion. Con- appearance of the graph. firm this statement by calculating the equilibrium constant for each of the following cases: (a) A strong acid reacting with a strong base. (b) A strong acid reacting with a weak base (NH3). (c) A weak acid Mass of HgI2 formed (CH3COOH) reacting with a strong base. (d) A weak acid (CH3COOH) reacting with a weak base (NH3). (Hint: Strong acids exist as H1 ions and strong bases exist as OH2 ions in solution. You need to look up Ka, Kb, and Kw.) 16.116 Calculate x, which is the number of molecules of Volume of KI added water in oxalic acid hydrate, H2C2O4 ? xH2O, from  770 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria the following data: 5.00 g of the compound is made 16.127 One of the most common antibiotics is penicillin G up to exactly 250 mL solution, and 25.0 mL of this (benzylpenicillinic acid), which has the structure solution requires 15.9 mL of 0.500 M NaOH solution shown next: for neutralization. 16.117 Describe how you would prepare a 1-L 0.20 M O CH3COONa/0.20 M CH3COOH buffer system by B H COOH (a) mixing a solution of CH3COOH with a solu- G D J O tion of CH3COONa, (b) reacting a solution of H3C C G D NOC H CH3COOH with a solution of NaOH, and (c) re- C A A A D G COC ONOCOCH2O acting a solution of CH3COONa with a solution H3C S A A B of HCl. H H O 16.118 Phenolphthalein is the common indicator for the titration of a strong acid with a strong base. (a) If the It is a weak monoprotic acid: pKa of phenolphthalein is 9.10, what is the ratio of HP Δ H1 1 P2    Ka 5 1.64 3 1023 the nonionized form of the indicator (colorless) to the ionized form (reddish pink) at pH 8.00? (b) If where HP denotes the parent acid and P2 the conju- 2 drops of 0.060 M phenolphthalein are used in a gate base. Penicillin G is produced by growing titration involving a 50.0-mL volume, what is the molds in fermentation tanks at 25°C and a pH range concentration of the ionized form at pH 8.00? of 4.5 to 5.0. The crude form of this antibiotic is (Assume that 1 drop 5 0.050 mL.) obtained by extracting the fermentation broth with 16.119 Oil paintings containing lead(II) compounds as an organic solvent in which the acid is soluble. constituents of their pigments darken over the (a) Identify the acidic hydrogen atom. (b) In one years. Suggest a chemical reason for the color stage of purification, the organic extract of the crude change. penicillin G is treated with a buffer solution at 16.120 What reagents would you employ to separate the pH 5 6.50. What is the ratio of the conjugate base of following pairs of ions in solution? (a) Na1 and penicillin G to the acid at this pH? Would you ex- Ba21, (b) K1 and Pb21, (c) Zn21 and Hg21. pect the conjugate base to be more soluble in water than the acid? (c) Penicillin G is not suitable for oral 16.121 Look up the Ksp values for BaSO4 and SrSO4 in administration, but the sodium salt (NaP) is because Table 16.2. Calculate the concentrations of Ba21, it is soluble. Calculate the pH of a 0.12 M NaP solu- Sr21, and SO22 4 in a solution that is saturated with tion formed when a tablet containing the salt is dis- both compounds. solved in a glass of water. 16.122 In principle, amphoteric oxides, such as Al2O3 16.128 Which of the following solutions has the highest and BeO, can be used to prepare buffer solutions [H1]? (a) 0.10 M HF, (b) 0.10 M HF in 0.10 M NaF, because they possess both acidic and basic prop- (c) 0.10 M HF in 0.10 M SbF5. (Hint: SbF5 reacts erties (see Section 15.11). Explain why these with F2 to form the complex ion SbF2 6 .) compounds are of little practical use as buffer components. 16.129 Distribution curves show how the fractions of nonionized acid and its conjugate base vary as a 16.123 CaSO4 (Ksp 5 2.4 3 1025) has a larger Ksp value function of pH of the medium. Plot distribution than that of Ag2SO4 (Ksp 5 1.4 3 1025). Does it curves for CH3COOH and its conjugate base follow that CaSO4 also has greater solubility CH3COO2 in solution. Your graph should show (g/L)? fraction as the y axis and pH as the x axis. What 16.124 When lemon juice is squirted into tea, the color be- are the fractions and pH at the point where these comes lighter. In part, the color change is due to di- two curves intersect? lution, but the main reason for the change is an 16.130 Water containing Ca21 and Mg21 ions is called acid-base reaction. What is the reaction? (Hint: Tea hard water and is unsuitable for some household contains “polyphenols” which are weak acids and and industrial use because these ions react with lemon juice contains citric acid.) soap to form insoluble salts, or curds. One way to • 16.125 How many milliliters of 1.0 M NaOH must be added remove the Ca21 ions from hard water is by adding to a 200 mL of 0.10 M NaH2PO4 to make a buffer washing soda (Na2CO3 ? 10H2O). (a) The molar solution with a pH of 7.50? solubility of CaCO3 is 9.3 3 1025 M. What is its • 16.126 The maximum allowable concentration of Pb21 molar solubility in a 0.050 M Na2CO3 solution? ions in drinking water is 0.05 ppm (that is, 0.05 g (b) Why are Mg21 ions not removed by this proce- of Pb21 in 1 million g of water). Is this guideline dure? (c) The Mg21 ions are removed as Mg(OH)2 exceeded if an underground water supply is at by adding slaked lime [Ca(OH)2] to the water to equilibrium with the mineral anglesite, PbSO4 produce a saturated solution. Calculate the pH of a (Ksp 5 1.6 3 1028)? saturated Ca(OH)2 solution. (d) What is the  Questions & Problems 771 concentration of Mg21 ions at this pH? (e) In gen- 16.137 Histidine is one of the 20 amino acids found in pro- eral, which ion (Ca21 or Mg21) would you remove teins. Shown here is a fully protonated histidine first? Why? molecule where the numbers denote the pKa values 16.131 Consider the ionization of the following acid-base of the acidic groups. indicator: 9.17 O HIn(aq) Δ H1 (aq) 1 In2 (aq) ⫹ B 1.82 H3NOCHOCOOH A The indicator changes color according to the ratios CH2 of the concentrations of the acid to its conjugate 6.00 A ⫹ base as described on p. 740. Show that the pH range HN over which the indicator changes from the acid color NH to the base color is pH 5 pKa 6 1, where Ka is the ionization constant of the acid. (a) Show stepwise ionization of histidine in solution. 16.132 Amino acids are building blocks of proteins. These (Hint: The H 1 ion will first come off from the compounds contain at least one amino group (¬NH2) strongest acid group followed by the next stron- and one carboxyl group (¬COOH). Consider gest acid group and so on.) (b) A dipolar ion is glycine (NH2CH2COOH). Depending on the pH of one in which the species has an equal number the solution, glycine can exist in one of three possi- of  positive and negative charges. Identify the ble forms: dipolar ion in (a). (c) The pH at which the dipo- ⫹ lar ion predominates is called the isoelectric Fully protonated: NH3—CH2—COOH point, denoted by pI. The isoelectric point is the ⫹ Dipolar ion: NH —CH —COO– 3 2 average of the pK a values leading to and follow- ing the formation of the dipolar ion. Calculate Fully ionized: NH2—CH2—COO– the pI of histidine. (d) The histidine group plays an important role in buffering blood (see Chem- Predict the predominant form of glycine at pH 1.0, istry in Action on p. 732). Which conjugate acid- 7.0, and 12.0. The pKa of the carboxyl group is 2.3 base pair shown in (a) is responsible for this and that of the ammonium group (¬NH1 3 ) is 9.6. action? 16.133 (a) Referring to Figure 16.6, describe how you 16.138 A sample of 0.96 L of HCl at 372 mmHg and 22°C would determine the pKb of the base. (b) Derive an is bubbled into 0.034 L of 0.57 M NH3. What is the analogous Henderson-Hasselbalch equation relat- pH of the resulting solution? Assume the volume of ing pOH to pKb of a weak base B and its conjugate solution remains constant and that the HCl is totally acid HB1. Sketch a titration curve showing the vari- dissolved in the solution. ation of the pOH of the base solution versus the 16.139 (a) Assuming complete dissociation and no ion- volume of a strong acid added from a buret. De- pair formation, calculate the freezing point of a scribe how you would determine the pKb from this 0.50 m NaI solution. (b) What is the freezing curve. (Hint: pKb 5 2log Kb.) point after the addition of sufficient HgI2, an in- 16.134 A 25.0-mL of 0.20 M HF solution is titrated with soluble compound, to the solution to react with all a 0.20 M NaOH solution. Calculate the volume of the free I2 ions in solution? Assume volume to NaOH solution added when the pH of the solu- remain constant. tion is (a) 2.85, (b) 3.15, (c) 11.89. Ignore salt 16.140 Calculate the maximum mass (in grams) of each of the hydrolysis. following soluble salts that can be added to 200 mL of 16.135 Draw distribution curves for an aqueous carbonic 0.100 M MgCl2 without causing a precipitate to acid solution. Your graph should show fraction of form: (a) Na2CO3, (b) AgNO3, (c) KOH. Assume species present as the y axis and pH as the x axis. volume to remain constant. Note that at any pH, only two of the three species 16.141 A 1.0-L saturated silver carbonate solution at 5°C (H2CO3, HCO32, and CO322) are present in appre- is treated with enough hydrochloric acid to decom- ciable concentrations. Use the pK a values in pose the compound. The carbon dioxide generated Table 15.5. is collected in a 19-mL vial and exerts a pressure 16.136 One way to distinguish a buffer solution with an of 114 mmHg at 25°C. What is the Ksp of Ag2CO3 acid solution is by dilution. (a) Consider a buffer at 5°C? solution made of 0.500 M CH3COOH and 0.500 M 16.142 The two curves shown on p. 772 represent the titra- CH3COONa. Calculate its pH and the pH after it has tion of two weak acids of the same concentration been diluted 10-fold. (b) Compare the result in (a) with a strong base such as NaOH. Use three obser- with the pHs of a 0.500 M CH3COOH solution vations to determine which of the two acids is before and after it has been diluted 10-fold. stronger. 772 Chapter 16 ■ Acid-Base Equilibria and Solubility Equilibria 14 14 14 14 12 12 12 12 10 10 10 10 8 8 8 8 pH pH pH pH 6 6 6 6 4 4 4 4 2 2 2 2 0 0 0 0 0 10 20 30 40 0 10 20 30 40 0 10 20 30 40 0 10 20 30 40 Volume of NaOH added (mL) Volume of NaOH added (mL) Volume of HCl added (mL) Volume of HCl added (mL) (a) (b) (a) (b) 16.143 The two curves shown here represent the titration of 16.144 A 100-mL 0.100 M CuSO4 solution is mixed with two weak bases of the same concentration with a a 100-mL 0.100 M Ba(OH)2 solution. Calculate strong acid such as HCl. Use three observations to the concentrations of the ions in the combined determine which of the two bases is stronger. solution. Interpreting, Modeling & Estimating 16.145 The titration curve shown here represents the titration 16.147 Use appropriate equations to account for the solubil- of a weak diprotic acid (H2A) versus NaOH. Identify ity of the amphoteric aluminum hydroxide [Al(OH)3)] the major species present at the marked points and at low and high pHs. estimate the pKa1 and pKa2 values of the acid. 14 Molar solubility 12 10 8 pH 6 4 3 4 5 6 7 8 9 10 11 12 13 2 pH 0 0 10 20 30 40 50 60 70 16.148 From Table 16.2 we see that silver bromide (AgBr) Volume of NaOH added (mL) has a larger solubility product than iron(II) hydrox- ide [Fe(OH)2]. Does this mean that AgBr is more 16.146 The titration curve shown here represents the titration soluble than Fe(OH)2? of a weak dibasic base (for example, a compound that 16.149 Aspirin is a weak acid with pKa 5 3.5. What is the contains two ¬NH2 groups) versus HCl. Identify the ratio of neutral (protonated) aspirin to deproton- major species present at the marked points and esti- ated aspirin in the following body fluids: (a) saliva, mate the pKb1 and pKb2 values of the base. (b) gastric juices in the stomach, and (c) blood? 14 12 10 8 pH 6 4 2 0 0 10 20 30 40 50 60 70 Volume of HCl added (mL)  Answers to Practice Exercises 773 Answers to Practice Exercises 16.1 4.01; 2.15. 16.2 (a) and (c). 16.3 9.17; 9.20. phenolphthalein. 16.8 2.0 3 10214. 16.9 1.9 3 1023 g/L. 16.4 Weigh out Na2CO3 and NaHCO3 in mole ratio of 16.10 No. 16.11 (a) . 1.6 3 1029 M, (b) . 2.6 3 1026 M. 0.60 to 1.0. Dissolve in enough water to make up a 1-L 16.12 (a) 1.7 3 1024 g/L, (b) 1.4 3 1027 g/L. 16.13 (a) More solution. 16.5 (a) 2.19, (b) 3.95, (c) 8.02, (d) 11.39. soluble in acid solution, (b) more soluble in acid solution, 16.6 5.92. 16.7 (a) Bromophenol blue, methyl orange, (c) about the same. 16.14 Zn(OH)2 precipitate will form. methyl red, and chlorophenol blue; (b) all except thymol blue, 16.15 [Cu21] 5 1.2 3 10213 M, [Cu(NH3)421] 5 0.017 M, bromophenol blue, and methyl orange; (c) cresol red and [NH3] 5 0.23 M. 16.16 3.5 3 1023 mol/L. CHEMICAL M YS TERY A Hard-Boiled Snack M ost of us have eaten hard-boiled eggs. They are easy to cook and nutritious. But when was the last time you thought about the process of boiling an egg or looked carefully at a hard-boiled egg? A lot of interesting chemical and physical changes occur while an egg cooks. A hen’s egg is a complicated biochemical system, but here we will focus on the three major parts that we see when we crack open an egg: the shell, the egg white or albumen, and the yolk. The shell protects the inner components from the outside environment, but it has many microscopic pores through which air can pass. The albumen is about 88 per- cent water and 12 percent protein. The yolk contains 50 percent water, 34 percent fat, 16 percent protein, and a small amount of iron in the form of Fe21 ions. Proteins are polymers made up of amino acids. In solution, each long chain of a protein molecule folds in such a way that the hydrophobic parts of the molecule are bur- ied inside and the hydrophilic parts are on the exterior, in contact with the solution. This is the stable or native state of a protein which allows it to perform normal physiological functions. Heat causes protein molecules to unfold, or denature. Chemicals such as acids and salt (NaCl) can also denature proteins. To avoid contact with water, the hydrophobic parts of denatured proteins will clump together, or coagulate to form a semirigid opaque white solid. Heating also decomposes some proteins so that the sulfur in them combines with hydrogen to form hydrogen sulfide (H2S), an unpleasant smelling gas that can some- times be detected when the shell of a boiled egg is cracked. The accompanying photo of hard-boiled eggs shows an egg that has been boiled for about 12 minutes and one that has been overcooked. Note that the outside of the over- cooked yolk is green. What is the chemical basis for the changes brought about by boiling an egg? Chemical Clues 1. One frequently encountered problem with hard-boiled eggs is that their shells crack in water. The recommended procedure for hard boiling eggs is to place the eggs in cold water and then bring the water to a boil. What causes the shells to crack in this case? How does pin holing, that is, piercing the shell with a needle, prevent the shells Schematic diagram of an egg. The chalazae are the cords that anchor the yolk to the shell and keep it centered. Shell Membrane Albumen Yolk membrane Yolk Chalaza Air space 774  from cracking? A less satisfactory way of hard boiling eggs is to place room- temperature eggs or cold eggs from the refrigerator in boiling water. What additional mechanism might cause the shells to crack? 2. When an eggshell cracks during cooking, some of the egg white leaks into the hot water to form unsightly “streamers.” An experienced cook adds salt or vinegar to the water prior to heating eggs to minimize the formation of streamers. Explain the chemical basis for this action. 3. Identify the green substance on the outer layer of the yolk of an overcooked egg and write an equation representing its formation. The unsightly “green yolk” can be elim- inated or minimized if the overcooked egg is rinsed with cold water immediately after it has been removed from the boiling water. How does this action remove the green substance? 4. The way to distinguish a raw egg from a hard-boiled egg, without cracking the shells, is to spin the eggs. How does this method work? A 12-minute egg (left) and an overcooked hard-boiled egg (right). Iron(II) sulfide. 775  CHAPTER 17 Entropy, Free Energy, and Equilibrium The laws of thermodynamics set an upper limit on how much heat can be converted to work, as in the case of a steam locomotive. CHAPTER OUTLINE A LOOK AHEAD 17.1 The Three Laws of  This chapter begins with a discussion of the three laws of thermodynamics Thermodynamics and the nature of spontaneous processes. (17.1 and 17.2) 17.2 Spontaneous Processes  We then see that entropy is the thermodynamic function for predicting the spontaneity of a reaction. On a molecular level, the entropy of a system can in 17.3 Entropy principle be calculated from the number of microstates associated with the 17.4 The Second Law of system. We learn that in practice entropy is determined by calorimetric methods Thermodynamics and standard entropy values are known for many substances. (17.3)  The second law of thermodynamics states that the entropy of the universe 17.5 Gibbs Free Energy increases in a spontaneous process and remains unchanged in an equilibrium 17.6 Free Energy and Chemical process. We learn ways to calculate the entropy change of a system and of the Equilibrium surroundings, which together make up for the change in the entropy of the universe. We also discuss the third law of thermodynamics, which enables us 17.7 Thermodynamics to determine the absolute value of entropy of a substance. (17.4) in Living Systems  We see that a new thermodynamic function called Gibbs free energy is needed to focus on the system. The change in Gibbs free energy can be used to predict spontaneity and equilibrium. For changes carried out under standard- state conditions, the change in Gibbs free energy is related to the equilib- rium constant of a reaction. (17.5 and 17.6)  The chapter concludes with a discussion of how thermodynamics is applied to living systems. We see that the principle of coupled reactions plays a crucial role in many biological processes. (17.7) 776  17.2 Spontaneous Processes 777 T hermodynamics is an extensive and far-reaching scientific discipline that deals with the interconversion of heat and other forms of energy. Thermodynamics enables us to use information gained from experiments on a system to draw conclusions about other aspects of the same system without further experimentation. For example, we saw in Chapter 6 that it is possible to calculate the enthalpy of reaction from the standard enthalpies of formation of the reactant and product molecules. This chapter introduces the second law of thermodynamics and the Gibbs free-energy function. It also discusses the relationship between Gibbs free energy and chemical equilibrium. 17.1 The Three Laws of Thermodynamics In Chapter 6 we encountered the first of three laws of thermodynamics, which says that energy can be converted from one form to another, but it cannot be created or destroyed. One measure of these changes is the amount of heat given off or absorbed by a system during a constant-pressure process, which chemists define as a change in enthalpy (DH). The second law of thermodynamics explains why chemical processes tend to favor one direction. The third law is an extension of the second law and will be examined briefly in Section 17.4. 17.2 Spontaneous Processes One of the main objectives in studying thermodynamics, as far as chemists are con- cerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (for example, at a certain temperature, pressure, and concentration). This knowledge is important whether one is synthesizing compounds in a research laboratory, manufacturing chemicals on an industrial scale, or trying to understand the intricate biological processes in a cell. A reaction that does occur under the given set of conditions is called a spontaneous A spontaneous reaction does not neces- sarily mean an instantaneous reaction. reaction. If a reaction does not occur under specified conditions, it is said to be non- spontaneous. We observe spontaneous physical and chemical processes every day, including many of the following examples: • A waterfall runs downhill, but never up, spontaneously. • A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form. • Water freezes spontaneously below 0°C, and ice melts spontaneously above 0°C (at 1 atm). • Heat flows from a hotter object to a colder one, but the reverse never happens spontaneously. • The expansion of a gas into an evacuated bulb is a spontaneous process [Figure 17.1(a)]. The reverse process, that is, the gathering of all the molecules into one bulb, is not spontaneous [Figure 17.1(b)]. • A piece of sodium metal reacts violently with water to form sodium hydroxide and hydrogen gas. However, hydrogen gas does not react with sodium hydroxide to form water and sodium. • Iron exposed to water and oxygen forms rust, but rust does not spontaneously change back to iron. These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction. If we assume that spontaneous processes occur so as to decrease the energy of a system, we can explain why a ball rolls downhill and why springs in a clock unwind. 778 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Figure 17.1 (a) A spontaneous process. After the valve is opened, the molecules distribute evenly between the two bulbs. (b) A nonspontaneous process. After the valve is opened, the molecules preferentially gather in one bulb. (a) (b) Similarly, a large number of exothermic reactions are spontaneous. An example is the Because of the activation energy barrier, combustion of methane: an input of energy is needed to get this reaction going at an observable rate. CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l)  ¢H° 5 2890.4 kJ/mol Another example is the acid-base neutralization reaction: H1 (aq) 1 OH2 (aq) ¡ H2O(l)   ¢H° 5 256.2 kJ/mol But consider a solid-to-liquid phase transition such as H2O(s) ¡ H2O(l)            ¢H° 5 6.01 kJ/mol   In this case, the assumption that spontaneous processes always decrease a system’s energy fails. Experience tells us that ice melts spontaneously above 0°C even though the process is endothermic. Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: H2O NH4NO3 (s) ¡ NH 1 2 4 (aq) 1 NO 3 (aq) ¢H° 5 25 kJ/mol This process is spontaneous, and yet it is also endothermic. The decomposition of mercury(II) oxide is an endothermic reaction that is nonspontaneous at room temperature, but it becomes spontaneous when the temperature is raised: 2HgO(s) ¡ 2Hg(l) 1 O2 (g)      ¢H° 5 90.7 kJ/mol From a study of the examples mentioned and many more cases, we come to the fol- lowing conclusion: Exothermicity favors the spontaneity of a reaction but does not guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is possible for an exothermic reaction to be nonspontaneous. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis When heated, HgO decomposes of energy changes in the system. To make this kind of prediction we need another to give Hg and O2. thermodynamic quantity, which turns out to be entropy. 17.3 Entropy In order to predict the spontaneity of a process, we need to introduce a new thermo- dynamic quantity called entropy. Entropy (S) is often described as a measure of how spread out or dispersed the energy of a system is among the different possible ways  17.3 Entropy 779 that system can contain energy. The greater the dispersal, the greater is the entropy. Most processes are accompanied by a change in entropy. A cup of hot water has a certain amount of entropy due to the dispersal of energy among the various energy states of the water molecules (for example, energy states associated with the translational, rotational, and vibrational motions of the water molecules). If left standing on a table, the water loses heat to the cooler surroundings. Consequently, there is an overall increase in entropy because of the dispersal of energy over a great many energy states of the air molecules. As another example, consider the situation depicted in Figure 17.1. Before the valve is opened, the system possesses a certain amount of entropy. Upon opening the valve, the gas molecules now have access to the combined volume of both bulbs. Quantum mechanical analysis shows that the spacing between translational energy A larger volume for movement results in a narrowing of the gap between transla- levels is inversely proportional to the tional energy levels of the molecules. Consequently, the entropy of the system volume of the container and the mass of the molecules. increases because closely spaced energy levels leads to a greater dispersal among the energy levels. Microstates and Entropy Before we introduce the second law of thermodynamics, which relates entropy change (increase) to spontaneous processes, it is useful to first provide a proper definition of entropy. To do so let us consider a simple system of four molecules distributed between two equal compartments, as shown in Figure 17.2. There is only one way to arrange all the molecules in the left compartment, four ways to have three molecules in the left compartment and one in the right compartment, and six ways to have two molecules in each of the two compartments. The eleven possible ways of distributing the molecules are called microscopic states or microstates and each set of similar microstates is called a distribution.† As you can see, distribution III † Actually there are still other possible ways to distribute the four molecules between the two compartments. We can have all four molecules in the right compartment (one way) and three molecules in the right compartment and one molecule in the left compartment (four ways). However, the distributions shown in Figure 17.2 are sufficient for our discussion. Distribution Microstates Figure 17.2 Some possible ways of distributing four molecules between two equal compartments. I Distribution I can be achieved in only one way (all four molecules in the left compartment) and has one microstate. Distribution II can be achieved in four ways and has four microstates. Distribution III can be achieved in six ways and II has six microstates. III 780 Chapter 17 ■ Entropy, Free Energy, and Equilibrium is the most probable because there are six microstates or six ways to achieve it and distribution I is the least probable because it has one microstate and therefore there is only one way to achieve it. Based on this analysis, we conclude that the probabil- ity of occurrence of a particular distribution (state) depends on the number of ways (microstates) in which the distribution can be achieved. As the number of molecules approaches macroscopic scale, it is not difficult to see that they will be evenly dis- tributed between the two compartments because this distribution has many, many more microstates than all other distributions. In 1868 Boltzmann showed that the entropy of a system is related to the natural log of the number of microstates (W ): S 5 k ln W (17.1) where k is called the Boltzmann constant (1.38 3 10223 J/K). Thus, the larger the W, the greater is the entropy of the system. Like enthalpy, entropy is a state function (see Section 6.3). Consider a certain process in a system. The entropy change for the process, DS, is ¢S 5 Sf 2 Si (17.2) where Si and Sf are the entropies of the system in the initial and final states, respec- tively. From Equation (17.1) we can write ¢S 5 k ln Wf 2 k ln Wi Engraved on Ludwig Boltzmann’s Wf tombstone in Vienna is his famous 5 k ln (17.3) equation. The “log” stands for Wi “loge ,” which is the natural logarithm or ln. where Wi and Wf are the corresponding numbers of microstates in the initial and final states. Thus, if Wf . Wi, DS . 0 and the entropy of the system increases. Review of Concepts Referring to the footnote on p. 779, draw the missing distributions in Figure 17.2. Changes in Entropy Earlier we described the increase in entropy of a system as a result of the increase in the dispersal of energy. There is a connection between the qualitative description of entropy in terms of dispersal of energy and the quantitative definition of entropy in terms of microstates given by Equation (17.1). We conclude that • A system with fewer microstates (smaller W) among which to spread its energy (small dispersal) has a lower entropy. • A system with more microstates (larger W) among which to spread its energy (large dispersal) has a higher entropy. Next, we will study several processes that lead to a change in entropy of a system in terms of the change in the number of microstates of the system. Consider the situations shown in Figure 17.3. In a solid, the atoms or molecules are confined to fixed positions and the number of microstates is small. Upon melting, these atoms or molecules can occupy many more positions as they move away from the lattice points. Consequently, the number of microstates increases because there are now many more ways to arrange the particles. Therefore, we predict this “order ¡ disorder” phase transition to result in an increase in entropy because the number of microstates has increased. Similarly, we predict the vaporization process will also lead to an  17.3 Entropy 781 Figure 17.3 Processes that lead to an increase in entropy of the system: (a) melting: Sliquid . Ssolid ; (b) vaporization: Svapor . Sliquid ; Solid Liquid (c) dissolving: in general, Ssoln . (a) Ssolute 1 Ssolvent (d) heating: ST2 . ST1. Liquid Vapor (b) Solvent Solute Solution (c) System at T1 System at T2 (T2 > T1 ) (d) increase in the entropy of the system. The increase will be considerably greater than that for melting, however, because molecules in the gas phase occupy much more space, and therefore there are far more microstates than in the liquid phase. The solution pro- cess usually leads to an increase in entropy. When a sugar crystal dissolves in water, the highly ordered structure of the solid and part of the ordered structure of water break down. Thus, the solution has a greater number of microstates than the pure solute and pure solvent combined. When an ionic solid such as NaCl dissolves in water, there are two contributions to entropy increase: the solution process (mixing of solute with sol- vent) and the dissociation of the compound into ions: H2O NaCl(s) ¡ Na1 (aq) 1 Cl2 (aq) More particles lead to a greater number of microstates. However, we must also consider hydration, which causes water molecules to become more ordered around the ions. This process decreases entropy because it reduces the number of microstates of the solvent molecules. For small, highly charged ions such as Al31 and Fe31, the decrease in entropy due to hydration can outweigh the increase in entropy due to mixing and dissociation so that the entropy change for the overall process can actually be negative. Heating also increases the entropy of a system. In addition to translational motion, molecules can also execute rotational motions and vibrational motions (Figure 17.4). As the tempera- ture is increased, the energies associated with all types of molecular motion increase. This increase in energy is distributed or dispersed among the quantized energy levels. Consequently, more microstates become available at a higher temperature; therefore, the entropy of a system always increases with increasing temperature. Standard Entropy Equation (17.1) provides a useful molecular interpretation of entropy, but is normally not used to calculate the entropy of a system because it is difficult to determine the number of microstates for a macroscopic system containing many 782 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Figure 17.4 (a) A vibrational motion in a water molecule. The atoms are displaced as shown by h the arrows and then reverse their directions to complete a cycle of vibration. (b) A rotational motion of a water molecule about an axis through the oxygen atom. The molecule can also vibrate and h h rotate in other ways. (a) (b) molecules. Instead, entropy is obtained by calorimetric methods. In fact, as we will Table 17.1 see shortly, it is possible to determine the absolute value of entropy of a substance, Standard Entropy Values called absolute entropy, something we cannot do for energy or enthalpy. Standard (S°) for Some Substances entropy is the absolute entropy of a substance at 1 atm and usually quoted with its at 25°C value at 25°C. (Recall that the standard state refers only to 1 atm. The reason for S° specifying 25°C is that many processes are carried out at room temperature.) Table Substance (J/K ? mol) 17.1 lists standard entropies of a few elements and compounds; Appendix 3 pro- H2O(l) 69.9 vides a more extensive listing.† The units of entropy are J/K or J/K ? mol for H2O(g) 188.7 1 mole of the substance. We use joules rather than kilojoules because entropy values are typically quite small. Entropies of elements and compounds are all positive Br2(l) 152.3 (that is, S° . 0). By contrast, the standard enthalpy of formation (DH°f ) for elements Br2(g) 245.3 in their stable form is arbitrarily set equal to zero, and for compounds, it may be I2(s) 116.7 positive or negative. I2(g) 260.6 Referring to Table 17.1, we see that the standard entropy of water vapor is C (diamond) 2.4 greater than that of liquid water. Similarly, bromine vapor has a higher standard C (graphite) 5.69 entropy than liquid bromine, and iodine vapor has a greater standard entropy than CH4 (methane) 186.2 solid iodine. For different substances in the same phase, molecular complexity deter- C2H6 (ethane) 229.5 mines which ones have higher entropies. Both diamond and graphite are solids, but He(g) 126.1 diamond has a more ordered structure and hence a smaller number of microstates Ne(g) 146.2 (see Figure 11.28). Therefore, diamond has a smaller standard entropy than graphite. Consider the natural gases methane and ethane. Ethane has a more complex structure and hence more ways to execute molecular motions, which also increase its micro- states. Therefore, ethane has a greater standard entropy than methane. Both helium The spinning motion of an atom about its and neon are monatomic gases, which cannot execute rotational or vibrational own axis does not constitute a rotational motion because it does not displace the motions, but neon has a greater standard entropy than helium because its molar mass position of the nucleus. is greater. Heavier atoms have more closely spaced energy levels so there is a greater distribution of the atoms’ energy among the levels. Consequently, there are more microstates associated with these atoms. Example 17.1 Predict whether the entropy change is greater or less than zero for each of the following processes: (a) freezing ethanol, (b) evaporating a beaker of liquid bromine at room temperature, (c) dissolving glucose in water, (d) cooling nitrogen gas from 80°C to 20°C. (Continued) † Because the entropy of an individual ion cannot be studied experimentally, chemists arbitrarily assign a zero value of entropy for the hydrogen ion in solution. Based on this scale, one can then determine the entropy of the chloride ion (from measurements on HCl), which in turn enables one to determine the entropy of the sodium ion (from measurements on NaCl), and so on. From Appendix 3 you will note that some ions have positive entropy values, while others have negative values. The signs are determined by the extent of hydration relative to the hydrogen ion. If an ion has a greater extent of hydration than the hydrogen ion, then the entropy of the ion has a negative value. The opposite holds for ions with positive entropies.  17.4 The Second Law of Thermodynamics 783 Strategy To determine the entropy change in each case, we examine whether the number of microstates of the system increases or decreases. The sign of DS will be positive if there is an increase in the number of microstates and negative if the number of microstates decreases. Solution (a) Upon freezing, the ethanol molecules are held rigid in position. This phase transition reduces the number of microstates and therefore the entropy decreases; that is, DS , 0. (b) Evaporating bromine increases the number of microstates because the Br2 molecules can occupy many more positions in nearly empty space. Therefore, DS . 0. (c) Glucose is a nonelectrolyte. The solution process leads to a greater dispersal of matter due to the mixing of glucose and water molecules so we expect DS . 0. Bromine is a fuming liquid at room temperature. (d) The cooling process decreases various molecular motions. This leads to a decrease in microstates and so DS , 0. Similar problem: 17.5. Practice Exercise How does the entropy of a system change for each of the following processes? (a) condensing water vapor, (b) forming sucrose crystals from a supersaturated solution, (c) heating hydrogen gas from 60°C to 80°C, and (d) subliming dry ice. Review of Concepts For which of the following physical changes is DS positive? (a) condensing ether vapor, (b) melting iron, (c) subliming solid iodine, (d) freezing benzene. 17.4 The Second Law of Thermodynamics The connection between entropy and the spontaneity of a reaction is expressed by the second law of thermodynamics: The entropy of the universe increases in a Just talking about entropy increases its value in the universe. spontaneous process and remains unchanged in an equilibrium process. Because the universe is made up of the system and the surroundings, the entropy change in the universe (DSuniv) for any process is the sum of the entropy changes in the system (DSsys) and in the surroundings (DSsurr). Mathematically, we can express the second law of thermodynamics as follows: For a spontaneous process: ¢Suniv 5 ¢Ssys 1 ¢Ssurr . 0 (17.4) For an equilibrium process: ¢Suniv 5 ¢Ssys 1 ¢Ssurr 5 0 (17.5) For a spontaneous process, the second law says that DSuniv must be greater than zero, but it does not place a restriction on either DSsys or DSsurr. Thus, it is possible for either DSsys or DSsurr to be negative, as long as the sum of these two quantities is greater than zero. For an equilibrium process, DSuniv is zero. In this case, DSsys and DSsurr must be equal in magnitude, but opposite in sign. What if for some hypothetical process we find that DSuniv is negative? What this means is that the process is not spontaneous in the direction described. Rather, it is spontaneous in the opposite direction. Entropy Changes in the System To calculate DSuniv, we need to know both DSsys and DSsurr. Let us focus first on DSsys. Suppose that the system is represented by the following reaction: aA 1 bB ¡ cC 1 d D 784 Chapter 17 ■ Entropy, Free Energy, and Equilibrium As is the case for the enthalpy of a reaction [see Equation (6.18)], the standard entropy of reaction DS8rxn is given by the difference in standard entropies between products and reactants: ¢S°rxn 5 [cS°(C) 1 dS°(D)] 2 [aS°(A) 1 bS°(B)] (17.6) or, in general, using © to represent summation and m and n for the stoichiometric coefficients in the reaction ¢S°rxn 5 ©nS°(products) 2 ©mS°(reactants) (17.7) The standard entropy values of a large number of compounds have been measured in J/K ? mol. To calculate DS°rxn (which is DSsys), we look up their values in Appendix 3 and proceed according to Example 17.2. Example 17.2 From the standard entropy values in Appendix 3, calculate the standard entropy changes for the following reactions at 25°C. (a) CaCO3 (s) ¡ CaO(s) 1 CO2 (g) (b) N2 (g) 1 3H2 (g) ¡ 2NH3 (g) (c) H2 (g) 1 Cl2 (g) ¡ 2HCl(g) Strategy To calculate the standard entropy of a reaction, we look up the standard entropies of reactants and products in Appendix 3 and apply Equation (17.7). As in the calculation of enthalpy of reaction [see Equation (6.18)], the stoichiometric coefficients have no units, so DS°rxn is expressed in units of J/K ? mol. Solution (a) ¢S°rxn 5 [S°(CaO) 1 S°(CO2 )] 2 [S°(CaCO3 )] 5 [(39.8 J/K ? mol) 1 (213.6 J/K ? mol)] 2 (92.9 J/K ? mol) 5 160.5 J/K ? mol Thus, when 1 mole of CaCO3 decomposes to form 1 mole of CaO and 1 mole of gaseous CO2, there is an increase in entropy equal to 160.5 J/K ? mol. (b) ¢S°rxn 5 [2S°(NH3 )] 2 [S°(N2 ) 1 3S°(H2 )] 5 (2) (193 J/K ? mol) 2 [(192 J/K ? mol) 1 (3) (131 J/K ? mol)] 5 2199 J/K ? mol This result shows that when 1 mole of gaseous nitrogen reacts with 3 moles of gaseous hydrogen to form 2 moles of gaseous ammonia, there is a decrease in entropy equal to 2199 J/K ? mol. (c) ¢S°rxn 5 [2S°(HCl)] 2 [S°(H2 ) 1 S°(Cl2 )] 5 (2) (187 J/K ? mol) 2 [(131 J/K ? mol) 1 (223 J/K ? mol)] 5 20 J/K ? mol Thus, the formation of 2 moles of gaseous HCl from 1 mole of gaseous H2 and 1 mole of gaseous Cl2 results in a small increase in entropy equal to 20 J/K ? mol. Similar problems: 17.11 and 17.12. Comment The DS°rxn values all apply to the system. (Continued)  17.4 The Second Law of Thermodynamics 785 Practice Exercise Calculate the standard entropy change for the following reactions at 25°C: (a) 2CO(g) 1 O2 (g) ¡ 2CO2 (g) (b) 3O2 (g) ¡ 2O3 (g) (c) 2NaHCO3(s) ¡ Na2CO3(s) 1 H2O(l) 1 CO2(g) The results of Example 17.2 are consistent with those observed for many other reactions. Taken together, they support the following general rules: • If a reaction produces more gas molecules than it consumes [Example 17.2(a)], We omit the subscript rxn for simplicity. DS° is positive. • If the total number of gas molecules diminishes [Example 17.2(b)], DS° is negative. • If there is no net change in the total number of gas molecules [Example 17.2(c)], then DS° may be positive or negative, but will be relatively small numerically. These conclusions make sense, given that gases invariably have greater entropy than liquids and solids. For reactions involving only liquids and solids, predicting the sign of DS° is more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase in entropy. Example 17.3 shows how knowing the nature of reactants and products makes it possible to predict entropy changes. Example 17.3 Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H2 (g) 1 O2 (g) ¡ 2H2O(l) (b) NH4Cl(s) ¡ NH3 (g) 1 HCl(g) (c) H2 (g) 1 Br2 (g) ¡ 2HBr(g) Strategy We are asked to predict, not calculate, the sign of entropy change in the reactions. The factors that lead to an increase in entropy are: (1) a transition from a condensed phase to the vapor phase and (2) a reaction that produces more product molecules than reactant molecules in the same phase. It is also important to compare the relative complexity of the product and reactant molecules. In general, the more complex the molecular structure, the greater the entropy of the compound. Solution (a) Two reactant molecules combine to form one product molecule. Even though H2O is a more complex molecule than either H2 and O2, the fact that there is a net decrease of one molecule and gases are converted to liquid ensures that the number of microstates will be diminished and hence DS° is negative. (b) A solid is converted to two gaseous products. Therefore, DS° is positive. (c) The same number of molecules is involved in the reactants as in the product. Furthermore, all molecules are diatomic and therefore of similar complexity. As a result, we cannot predict the sign of DS°, but we know that the change must be quite small in magnitude. Similar problems: 17.13 and 17.14. Practice Exercise Discuss qualitatively the sign of the entropy change expected for each of the following processes: (a) I2 (s) ¡ 2I(g) (b) 2Zn(s) 1 O2 (g) ¡ 2ZnO(s) (c) N2 (g) 1 O2 (g) ¡ 2NO(g) 786 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Review of Concepts Consider the gas-phase reaction of A2 (blue) and B2 (orange) to form AB3. (a) Write a balanced equation for the reaction. (b) What is the sign of DS for the reaction? Entropy Changes in the Surroundings Next we see how DSsurr is calculated. When an exothermic process takes place in the system, the heat transferred to the surroundings enhances motion of the molecules in the surroundings. Consequently, there is an increase in the number of microstates and the entropy of the surroundings increases. Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the sur- roundings because molecular motion decreases (Figure 17.5). For constant-pressure processes the heat change is equal to the enthalpy change of the system, DHsys. There- fore, the change in entropy of the surroundings, DSsurr, is proportional to DHsys: ¢Ssurr r 2¢Hsys The minus sign is used because if the process is exothermic, DHsys is negative and DSsurr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, DHsys is positive and the negative sign ensures that the entropy of the surroundings decreases. The change in entropy for a given amount of heat absorbed also depends on the temperature. If the temperature of the surroundings is high, the molecules are S Surroundings urr Sur Surroundings Entropy decreases Entropy increases Heat Heat System System (a) (b) Figure 17.5 (a) An exothermic process transfers heat from the system to the surroundings and results in an increase in the entropy of the surroundings. (b) An endothermic process absorbs heat from the surroundings and thereby decreases the entropy of the surroundings.  17.4 The Second Law of Thermodynamics 787 already quite energetic. Therefore, the absorption of heat from an exothermic pro- cess in the system will have relatively little impact on molecular motion and the resulting increase in entropy of the surroundings will be small. However, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a more drastic increase in molecular motion and hence a larger increase in entropy. By analogy, someone coughing in a crowded restaurant will not disturb too many people, but someone coughing in a library definitely will. From the inverse relationship between DSsurr and temperature T (in kelvins)—that is, the higher the temperature, the smaller the DSsurr and vice versa—we can rewrite the above relationship as This equation, which can be derived from 2¢Hsys the laws of thermodynamics, assumes ¢Ssurr 5 (17.8) that both the system and the surround- T ings are at temperature T. Let us now apply the procedure for calculating DSsys and DSsurr to the synthesis of ammonia and ask whether the reaction is spontaneous at 25°C: N2 (g) 1 3H2 (g) ¡ 2NH3 (g) ¢H°rxn 5 292.6 kJ/mol 43 From Example 17.2(b) we have DSsys 5 2199 J/K ? mol, and substituting DHsys (292.6 kJ/mol) in Equation (17.8), we obtain 2(292.6 3 1000) J/mol ¢Ssurr 5 5 311 J/K ? mol 298 K The synthesis of NH3 from N2 The change in entropy of the universe is and H2. ¢Suniv 5 ¢Ssys 1 ¢Ssurr 5 2199 J/K ? mol 1 311 J/K ? mol 5 112 J/K ? mol Because DSuniv is positive, we predict that the reaction is spontaneous at 25°C. It is important to keep in mind that just because a reaction is spontaneous does not mean that it will occur at an observable rate. The synthesis of ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur. Reaction rates are the subject of chemical kinetics (see Chapter 13). The Third Law of Thermodynamics and Absolute Entropy Finally, it is appropriate to consider the third law of thermodynamics briefly in con- nection with the determination of entropy values. So far we have related entropy to microstates—the greater the number of microstates a system possesses, the larger is the entropy of the system. Consider a perfect crystalline substance at absolute zero (0 K). Under these conditions, molecular motions are kept at a minimum and the number of microstates (W ) is one (there is only one way to arrange the atoms or molecules to form a perfect crystal). From Equation (17.1) we write S 5 k ln W 5 k ln 1 5 0 According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at the absolute zero of temperature. As the temperature increases, the freedom of motion increases and hence also the number of microstates. Thus, the entropy of any substance at a temperature above 0 K is greater than zero. Note also 788 Chapter 17 ■ Entropy, Free Energy, and Equilibrium that if the crystal is impure or if it has defects, then its entropy is greater than zero even at 0 K, because it would not be perfectly ordered and the number of microstates would be greater than one. The important point about the third law of thermodynamics is that it enables us to determine the absolute entropies of substances. Starting with the knowledge that the entropy of a pure crystalline substance is zero at absolute zero, we can measure the increase in entropy of the substance when it is heated from 0 K to, say, 298 K. The change in entropy, DS, is given by ¢S 5 Sf 2 Si 5 Sf The entropy increase can be calculated because Si is zero. The entropy of the substance at 298 K, then, is given by DS or from the temperature change and heat capacity of the substance, plus any Sf, which is called the absolute entropy because this is the true value and not a value phase changes. derived using some arbitrary reference as in the case of standard enthalpy of forma- tion. Thus, the entropy values quoted so far and those listed in Appendix 3 are all absolute entropies. Because measurements are carried out at 1 atm, we usually refer to absolute entropies as standard entropies. In contrast, we cannot have the absolute energy or enthalpy of a substance because the zero of energy or enthalpy is undefined. Figure 17.6 shows the change (increase) in entropy of a substance with temperature. At absolute zero, it has a zero entropy value (assuming that it is a perfect crystalline substance). As it is heated, its entropy increases gradually because of greater molecu- lar motion. At the melting point, there is a sizable increase in entropy as the liquid state is formed. Further heating increases the entropy of the liquid again due to enhanced molecular motion. At the boiling point there is a large increase in entropy as a result of the liquid-to-vapor transition. Beyond that temperature, the entropy of the gas continues to rise with increasing temperature. Figure 17.6 Entropy increase of a substance as the temperature rises from absolute zero. Gas Boiling S° (J/K• mol) (DSvap) Liquid Solid Melting (DSfus) Temperature (K)  17.5 Gibbs Free Energy 789 17.5 Gibbs Free Energy The second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, DSuniv . 0. In order to determine the sign of DSuniv for a reaction, however, we would need to calculate both DSsys and DSsurr. In general, we are usually concerned only with what happens in a particular system. Therefore, we need another thermodynamic function to help us determine whether a reaction will occur spontaneously if we consider only the system itself. From Equation (17.4), we know that for a spontaneous process, we have ¢Suniv 5 ¢Ssys 1 ¢Ssurr . 0 Substituting 2DHsys/T for DSsurr, we write ¢Hsys ¢Suniv 5 ¢Ssys 2 . 0 T Multiplying both sides of the equation by T gives T¢Suniv 5 2¢Hsys 1 T¢Ssys . 0 Now we have a criterion for a spontaneous reaction that is expressed only in terms of the properties of the system (DHsys and DSsys). For convenience, we can change the preceding equation by multiplying it throughout by 21 and replacing the . sign with ,: 2T¢Suniv 5 ¢Hsys 2 T¢Ssys , 0 The change in unequal sign when we multiply the equation by 21 follows from the fact that 1 . 0 and 21 , 0. This equation says that for a process carried out at constant pressure and temperature T, if the changes in enthalpy and entropy of the system are such that DHsys 2 TDSsys is less than zero, the process must be spontaneous. In order to express the spontaneity of a reaction more directly, we introduce another thermodynamic function called Gibbs† free energy (G), or simply free energy: G 5 H 2 TS (17.9) All quantities in Equation (17.9) pertain to the system, and T is the temperature of the system. You can see that G has units of energy (both H and TS are in energy units). Like H and S, G is a state function. The change in free energy (DG) of a system for a constant-temperature process is ¢G 5 ¢H 2 T¢S (17.10) In this context, free energy is the energy available to do work. Thus, if a particular The word “free” in the term “free energy” does not mean without cost. reaction is accompanied by a release of usable energy (that is, if DG is negative), this fact alone guarantees that it is spontaneous, and there is no need to worry about what happens to the rest of the universe. Note that we have merely organized the expression for the entropy change of the universe and equating the free-energy change of the system (DG) with 2TDSuniv, so † Josiah Willard Gibbs (1839–1903). American physicist. One of the founders of thermodynamics, Gibbs was a modest and private individual who spent almost all of his professional life at Yale University. Because he published most of his works in obscure journals, Gibbs never gained the eminence that his contemporary and admirer James Maxwell did. Even today, very few people outside of chemistry and physics have ever heard of Gibbs.  CHEMISTRY in Action The Efficiency of Heat Engines A n engine is a machine that converts energy to work; a heat engine is a machine that converts thermal energy to work. Heat engines play an essential role in our technological society; The figure on p. 791 shows the heat transfer processes in a heat engine. Initially, a certain amount of heat flows from the heat reservoir (at temperature Th) into the engine. As the engine they range from automobile engines to the giant steam turbines does work, some of the heat is given off to the surroundings, or that run generators to produce electricity. Regardless of the type of heat engine, its efficiency level is of great importance; that is, for a given amount of heat input, how much useful work can we get out of the engine? The second law of thermodynamics helps T1 T2 T1 us answer this question. The figure shows a simple form of a heat engine. A cyl- inder fitted with a weightless piston is initially at temperature T1. Next, the cylinder is heated to a higher temperature T2. The gas in the cylinder expands and pushes up the piston. Finally the cylinder is cooled down to T1 and the apparatus returns to its original state. By repeating this cycle, the up- and-down movement of the piston can be made to do me- chanical work. A unique feature of heat engines is that some heat must be given off to the surroundings when they do work. With the piston in the up position, no further work can be done if we do not cool the cylinder back to T1. The cooling process re- (a) (b) (c) moves some of the thermal energy that could otherwise be A simple heat engine. (a) The engine is initially at T1. (b) When heated to T2, converted to work and thereby places a limit on the efficiency the piston is pushed up due to gas expansion. (c) When cooled to T1, the of heat engines. piston returns to the original position. that we can focus on changes in the system. We can now summarize the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of DG as follows: ¢G , 0 The reaction is spontaneous in the forward direction. ¢G . 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction. Table 17.2 ¢G 5 0 The system is at equilibrium. There is no net change. Conventions for Standard States Standard Free-Energy Changes State of Standard Matter State The standard free-energy of reaction (DG8rxn ) is the free-energy change for a reaction when it occurs under standard-state conditions, when reactants in their standard Gas 1 atm pressure states are converted to products in their standard states. Table 17.2 summarizes the Liquid Pure liquid conventions used by chemists to define the standard states of pure substances as well Solid Pure solid as solutions. To calculate DG°rxn we start with the equation Elements* DG°f 5 0 Solution 1 molar aA 1 bB ¡ cC 1 dD concentration The standard free-energy change for this reaction is given by *The most stable allotropic form at 25°C and 1 atm. ¢G°rxn 5 [c¢G°f (C) 1 d¢G°f (D)] 2 [a¢G°f (A) 1 b¢G°f (B)] (17.11) 790  Tc Heat reservoir Th efficiency 5 a1 2 b 3 100% Th Th 2 Tc 5 3 100% Th Heat engine Work Thus, the efficiency of a heat engine is given by the difference in temperatures between the heat reservoir and the heat sink (both in kelvins), divided by the temperature of the heat reservoir. In practice, we can make (Th 2 Tc) as large as possible, but because Tc cannot be zero and Th cannot be infinite, the efficiency of a Heat sink Tc heat engine can therefore never be 100 percent. At a power plant, superheated steam at about 560°C (833 K) Heat transfers during the operation of a heat engine. is used to drive a turbine for electricity generation. The temperature of the heat sink is about 38°C (or 311 K). The efficiency is given by 833 K 2 311 K efficiency 5 3 100% the heat sink (at temperature Tc). By definition, the efficiency of 833 K a heat engine is 5 63% useful work output In practice, because of friction, heat loss, and other complica- efficiency 5 3 100% energy input tions, the maximum efficiency of a steam turbine is only about 40 percent. Therefore, for every ton of coal used at a power Analysis based on the second law shows that efficiency can also plant, 0.40 ton generates electricity while the rest of it ends up be expressed as warming the surroundings! or, in general, ¢G°rxn 5 ©n¢G°f (products) 2 ©m¢G°f (reactants) (17.12) where m and n are stoichiometric coefficients. The term DG8f is the standard free- energy of formation of a compound, that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states. For the combustion of graphite: C(graphite) 1 O2 (g) ¡ CO2 (g) the standard free-energy change [from Equation (17.12)] is ¢G°rxn 5 ¢G°f (CO2 ) 2 [¢G°f (C, graphite) 1 ¢G°f (O2 )] As in the case of the standard enthalpy of formation (p. 253), we define the standard free-energy of formation of any element in its stable allotropic form at 1 atm and 25°C as zero. Thus, ¢G°f (C, graphite) 5 0    and    ¢G°f (O2 ) 5 0 791 792 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Therefore, the standard free-energy change for the reaction in this case is equal to the standard free energy of formation of CO2: ¢G°rxn 5 ¢G°f (CO2 ) Appendix 3 lists the values of DG°f for a number of compounds. Calculations of standard free-energy changes are handled as shown in Example 17.4. Example 17.4 Calculate the standard free-energy changes for the following reactions at 25°C. (a) CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l) (b) 2MgO(s) ¡ 2Mg(s) 1 O2 (g) Strategy To calculate the standard free-energy change of a reaction, we look up the standard free energies of formation of reactants and products in Appendix 3 and apply Equation (17.12). Note that all the stoichiometric coefficients have no units so DG°rxn is expressed in units of kJ/mol, and DG°f for O2 is zero because it is the stable allotropic element at 1 atm and 25°C. Solution (a) According to Equation (17.12), we write ¢G°rxn 5 [¢G°f (CO2 ) 1 2¢G°f (H2O)] 2 [¢G°f (CH4 ) 1 2¢G°f (O2 )] We insert the appropriate values from Appendix 3: ¢G°rxn 5 [(2394.4 kJ/mol) 1 (2)(2237.2 kJ/mol)] 2 [(250.8 kJ/mol) 1 (2) (0 kJ/mol)] 5 2818.0 kJ/mol (b) The equation is ¢G°rxn 5 [2¢G°f (Mg) 1 ¢G°f (O2 )] 2 [2¢G°f (MgO)] From data in Appendix 3 we write ¢G°rxn 5 [(2) (0 kJ/mol) 1 (0 kJ/mol)] 2 [(2) (2569.6 kJ/mol)] Similar problems: 17.17 and 17.18. 5 1139 kJ/mol Practice Exercise Calculate the standard free-energy changes for the following reactions at 25°C: (a) H2 (g) 1 Br2 (l) ¡ 2HBr(g) (b) 2C2H6 (g) 1 7O2 (g) ¡ 4CO2 (g) 1 6H2O(l) Applications of Equation (17.10) In order to predict the sign of DG, according to Equation (17.10), we need to know both DH and DS. A negative DH (an exothermic reaction) and a positive DS (a reac- tion that results in an increase in the microstates of the system) tend to make DG negative, although temperature may also influence the direction of a spontaneous reaction. The four possible outcomes of this relationship are • If both DH and DS are positive, then DG will be negative only when the TDS term is greater in magnitude than DH. This condition is met when T is large. • If DH is positive and DS is negative, DG will always be positive, regardless of temperature.  17.5 Gibbs Free Energy 793 Table 17.3 Factors Affecting the Sign of DG in the Relationship DG 5 DH 2 TDS DH DS DG Example 1 1 Reaction proceeds spontaneously at high temperatures. At low 2HgO(s) ¡ 2Hg(l) 1 O2 (g) temperatures, reaction is spontaneous in the reverse direction. 1 2 DG is always positive. Reaction is spontaneous in the reverse 3O2 (g) ¡ 2O3 (g) direction at all temperatures. 2 1 DG is always negative. Reaction proceeds spontaneously at all 2H2O2 (aq) ¡ 2H2O(l) 1 O2 (g) temperatures. 2 2 Reaction proceeds spontaneously at low temperatures. At high NH3 (g) 1 HCl(g) ¡ NH4Cl(s) temperatures, the reverse reaction becomes spontaneous. • If DH is negative and DS is positive, then DG will always be negative regardless of temperature. • If DH is negative and DS is negative, then DG will be negative only when TDS is smaller in magnitude than DH. This condition is met when T is small. The temperatures that will cause DG to be negative for the first and last cases depend on the actual values of DH and DS of the system. Table 17.3 summarizes the effects of the possibilities just described. Review of Concepts (a) Under what circumstances will an endothermic reaction proceed spontaneously? (b) Explain why, in many reactions in which both the reactant and product species are in the solution phase, DH often gives a good hint about the spontaneity of a reaction at 298 K. Before we apply the change in free energy to predict reaction spontaneity, it is useful to distinguish between DG and DG°. Suppose we carry out a reaction in solu- tion with all the reactants in their standard states (that is, all at 1 M concentration). As soon as the reaction starts, the standard-state condition no longer exists for the reactants or the products because their concentrations are different from 1 M. Under In Section 17.6 we will see an equation relating DG8 to the equilibrium constant K. nonstandard state conditions, we must use the sign of DG rather than that of DG° to predict the direction of the reaction. The sign of DG°, on the other hand, tells us whether the products or the reactants are favored when the reacting system reaches equilibrium. Thus, a negative value of DG° indicates that the reaction favors product formation whereas a positive value of DG° indicates that there will be more reactants than products at equilibrium. We will now consider two specific applications of Equation (17.10). Temperature and Chemical Reactions Calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic sub- stance used in steelmaking, production of calcium metal, the paper industry, water treatment, and pollution control. It is prepared by decomposing limestone (CaCO3) in a kiln at a high temperature: CaCO3 (s) Δ CaO(s) 1 CO2 (g) The production of quicklime The reaction is reversible, and CaO readily combines with CO2 to form CaCO3. The (CaO) from limestone (CaCO3) pressure of CO2 in equilibrium with CaCO3 and CaO increases with temperature. In in a rotary kiln. 794 Chapter 17 ■ Entropy, Free Energy, and Equilibrium the industrial preparation of quicklime, the system is never maintained at equilibrium; rather, CO2 is constantly removed from the kiln to shift the equilibrium from left to right, promoting the formation of calcium oxide. The important information for the practical chemist is the temperature at which the decomposition of CaCO3 becomes appreciable (that is, the temperature at which the reaction begins to favor products). We can make a reliable estimate of that tem- perature as follows. First we calculate DH° and DS° for the reaction at 25°C, using the data in Appendix 3. To determine DH° we apply Equation (6.18): ¢H° 5 [¢H°f (CaO) 1 ¢H°f (CO2 )] 2 [¢H°f (CaCO3 )] 5 [(2635.6 kJ/mol) 1 (2393.5 kJ/mol)] 2 (21206.9 kJ/mol) 5 177.8 kJ/mol Next we apply Equation (17.6) to find DS°: ¢S° 5 [S°(CaO) 1 S°(CO2 )] 2 S°(CaCO3 ) 5 [(39.8 J/K ? mol) 1 (213.6 J/K ? mol)] 2 (92.9 J/K ? mol) 5 160.5 J/K ? mol From Equation (17.10), ¢G° 5 ¢H° 2 T¢S° we obtain 1 kJ ¢G° 5 177.8 kJ/mol 2 (298 K)(160.5 J/K ? mol)a b 1000 J 5 130.0 kJ/mol Because DG° is a large positive quantity, we conclude that the reaction is not favored for product formation at 25°C (or 298 K). Indeed, the pressure of CO2 is so low at room temperature that it cannot be measured. In order to make DG° negative, we first have to find the temperature at which DG° is zero; that is, 0 5 ¢H° 2 T¢S° ¢H° or T5 ¢S° (177.8 kJ/mol)(1000 J/1 kJ) 5 160.5 J/K ? mol 5 1108 K or 835°C At a temperature higher than 835°C, DG° becomes negative, indicating that the reac- tion now favors the formation of CaO and CO2. For example, at 840°C, or 1113 K, ¢G° 5 ¢H° 2 T¢S° 1 kJ 5 177.8 kJ/mol 2 (1113 K)(160.5 J/K ? mol)a b 1000 J 5 20.8 kJ/mol Two points are worth making about such a calculation. First, we used the DH° and DS° values at 25°C to calculate changes that occur at a much higher temperature. Because both DH° and DS° change with temperature, this approach will not give us an accurate value of DG°, but it is good enough for “ballpark” estimates. Second, we should not be misled into thinking that nothing happens below 835°C and that at 835°C CaCO3  17.5 Gibbs Free Energy 795 suddenly begins to decompose. Far from it. The fact that DG° is a positive value at some temperature below 835°C does not mean that no CO2 is produced, but rather that the pressure of the CO2 gas formed at that temperature will be below 1 atm (its standard-state value; see Table 17.2). As Figure 17.7 shows, the pressure of CO2 at first increases very The equilibrium constant of this reaction slowly with temperature; it becomes easily measurable above 700°C. The significance of is Kp 5 PCO2. 835°C is that this is the temperature at which the equilibrium pressure of CO2 reaches 1 atm. Above 835°C, the equilibrium pressure of CO2 exceeds 1 atm. 3 Phase Transitions PCO2 (atm) 2 At the temperature at which a phase transition occurs (that is, at the melting point or boiling point) the system is at equilibrium (DG 5 0), so Equation (17.10) becomes 1 835°C ¢G 5 ¢H 2 T¢S 0 5 ¢H 2 T¢S 0 200 400 600 800 t (°C) ¢H or ¢S 5 Figure 17.7 Equilibrium pressure T of CO2 from the decomposition of CaCO3, as a function of Let us first consider the ice-water equilibrium. For the ice S water transition, DH is temperature. This curve is the molar heat of fusion (see Table 11.8), and T is the melting point. The entropy calculated by assuming that DH° and DS° of the reaction do not change is therefore change with temperature. 6010 J/mol ¢SiceSwater 5 The melting of ice is an endothermic 273 K process (DH is positive), and the freezing 5 22.0 J/K ? mol of water is exothermic (DH is negative). Thus, when 1 mole of ice melts at 0°C, there is an increase in entropy of 22.0 J/K ? mol. The increase in entropy is consistent with the increase in microstates from solid to liquid. Conversely, for the water S ice transition, the decrease in entropy is given by 26010 J/mol ¢SwaterSice 5 273 K 5 222.0 J/K ? mol In the laboratory, we normally carry out unidirectional changes, that is, either ice to water or water to ice transition. We can calculate entropy change in each case using the equa- tion DS 5 DH/T as long as the temperature remains at 0°C. The same procedure can be applied to the water S steam transition. In this case DH is the heat of vaporization and T is the boiling point of water. Example 17.5 examines the phase transitions in benzene. Example 17.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid S liquid and liquid S vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. Strategy At the melting point, liquid and solid benzene are at equilibrium, so DG 5 0. From Equation (17.10) we have DG 5 0 5 DH 2 TDS or DS 5 DH/T. To calculate the entropy change for the solid benzene S liquid benzene transition, we write DSfus 5 DHfus /Tf. Here DHfus is positive for an endothermic process, so DSfus is also positive, as expected for a solid to liquid transition. The same procedure applies to the liquid Liquid and solid benzene in benzene S vapor benzene transition. What temperature unit should be used? equilibrium at 5.5°C. (Continued) 796 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Solution The entropy change for melting 1 mole of benzene at 5.5°C is ¢Hfus ¢Sfus 5 Tf (10.9 kJ/mol) (1000 J/1 kJ) 5 (5.5 1 273) K 5 39.1 J/K ? mol Similarly, the entropy change for boiling 1 mole of benzene at 80.1°C is ¢Hvap ¢Svap 5 Tb (31.0 kJ/mol) (1000 J/1 kJ) 5 (80.1 1 273) K 5 87.8 J/K ? mol Check Because vaporization creates more microstates than the melting process, Similar problem: 17.64. DSvap . DSfus. Practice Exercise The molar heats of fusion and vaporization of argon are 1.3 kJ/mol and 6.3 kJ/mol, and argon’s melting point and boiling point are 2190°C and 2186°C, respectively. Calculate the entropy changes for fusion and vaporization. Review of Concepts Consider the sublimation of iodine (I2) at 45°C in a closed flask shown here. If the enthalpy of sublimation is 62.4 kJ/mol, what is the DS for sublimation? 17.6 Free Energy and Chemical Equilibrium As mentioned earlier, during the course of a chemical reaction not all the reactants and products will be at their standard states. Under this condition, the relationship between DG and DG°, which can be derived from thermodynamics, is Note that the units of DG and DG8 are ¢G 5 ¢G° 1 RT ln Q (17.13) kJ/mol, where the “per mole” unit can- cels that in R. where R is the gas constant (8.314 J/K ? mol), T is the absolute temperature of the reaction, and Q is the reaction quotient (see p. 639). We see that DG depends on two quantities: DG° and RT ln Q. For a given reaction at temperature T the value of DG°  17.6 Free Energy and Chemical Equilibrium 797 is fixed but that of RT ln Q is not, because Q varies according to the composition of the reaction mixture. Let us consider two special cases: Case 1: A large negative value of DG° will tend to make DG also negative. Thus, the net reaction will proceed from left to right until a significant amount of product has been formed. At that point, the RT ln Q term will become positive enough to match the negative DG° term in magnitude. Case 2: A large positive DG° term will tend to make DG also positive. Thus, the net reaction will proceed from right to left until a significant amount of reactant has been formed. At that point the RT ln Q term will become negative enough to match the positive DG° term in magnitude. At equilibrium, by definition, DG 5 0 and Q 5 K, where K is the equilibrium Sooner or later a reversible reaction will reach equilibrium. constant. Thus, 0 5 ¢G° 1 RT ln K or ¢G° 5 2RT ln K (17.14) In this equation, KP is used for gases and Kc for reactions in solution. Note that the larger the K is, the more negative DG° is. For chemists, Equation (17.14) is one of the most important equations in thermodynamics because it enables us to find the equilibrium constant of a reaction if we know the change in standard free energy and vice versa. It is significant that Equation (17.14) relates the equilibrium constant to the standard free-energy change DG° rather than to the actual free-energy change DG. The actual free-energy change of the system varies as the reaction progresses and becomes zero at equilibrium. On the other hand, DG° is a constant for a particular reaction at a given temperature. Figure 17.8 shows plots of the free energy of a reacting system versus the extent of the reaction for two types of reactions. As you can see, if DG° , 0, the products are favored over reactants at equilibrium. Conversely, G°(reactants) G°(products) Free energy (G) of the reacting system Free energy (G) of the reacting system Δ G°  G°(products) – Δ G°  G°(products) – G°(reactants) < 0 G°(reactants) > 0 Q<K Q>K G°(products) G°(reactants) ΔG < 0 ΔG > 0 Q>K Q<K QK QK ΔG > 0 ΔG < 0 ΔG  0 ΔG  0 Equilibrium Equilibrium position position Pure Pure Pure Pure reactants products reactants products Extent of reaction Extent of reaction (a) (b) Figure 17.8 (a) DG° , 0. At equilibrium, there is a significant conversion of reactants to products. (b) DG° . 0. At equilibrium, reactants are favored over products. In both cases, the net reaction toward equilibrium is from left to right (reactants to products) if Q , K and right to left (products to reactants) if Q . K. At equilibrium, Q 5 K. 798 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Relation Between DG8 and K as Predicted by the Equation Table 17.4 DG8 5 2RT ln K K ln K DG8 Comments .1 Positive Negative Products are favored over reactants at equilibrium. 51 0 0 Products and reactants are equally favored at equilibrium. ,1 Negative Positive Reactants are favored over products at equilibrium. if DG° . 0, there will be more reactants than products at equilibrium. Table 17.4 summarizes the three possible relations between DG° and K, as predicted by Equation (17.14). Remember this important distinction: It is the sign of DG and not that of DG° that determines the direction of reaction spontaneity. The sign of DG° only tells us the relative amounts of products and reactants when equilibrium is reached, not the direction of the net reaction. For reactions having very large or very small equilibrium constants, it is generally very difficult, if not impossible, to measure the K values by monitoring the concentrations of all the reacting species. Consider, for example, the formation of nitric oxide from molecular nitrogen and molecular oxygen: N2 (g) 1 O2 (g) Δ 2NO(g) At 25°C, the equilibrium constant KP is P2NO KP 5 5 4.0 3 10231 PN2PO2 The very small value of KP means that the concentration of NO at equilibrium will be exceedingly low. In such a case the equilibrium constant is more conveniently obtained from DG°. (As we have seen, DG° can be calculated from DH° and DS°.) On the other hand, the equilibrium constant for the formation of hydrogen iodide from molecular hydrogen and molecular iodine is near unity at room temperature: H2 (g) 1 I2 (g) Δ 2HI(g) For this reaction, it is easier to measure KP and then calculate DG° using Equation (17.14) than to measure DH° and DS° and use Equation (17.10). Examples 17.6–17.8 illustrate the use of Equations (17.13) and (17.14). Example 17.6 Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C: 2H2O(l) Δ 2H2 (g) 1 O2 (g) Strategy According to Equation (17.14), the equilibrium constant for the reaction is related to the standard free-energy change; that is, DG° 5 2RT ln K. Therefore, we first need to calculate DG° by following the procedure in Example 17.4. Then we can calculate KP. What temperature unit should be used? (Continued)  17.6 Free Energy and Chemical Equilibrium 799 Solution According to Equation (17.12), ¢G°rxn 5 [2¢G°f (H2 ) 1 ¢G°f (O2 )] 2 [2¢G°f (H2O)] 5 [(2) (0 kJ/mol) 1 (0 kJ/mol)] 2 [(2) (2237.2 kJ/mol)] 5 474.4 kJ/mol Using Equation (17.14) ¢G°rxn 5 2RT ln KP 1000 J 474.4 kJ/mol 3 5 2(8.314 J/K ? mol) (298 K) ln KP 1 kJ ln KP 5 2191.5 To calculate KP, enter 2191.5 on your KP 5 e2191.5 5 7 3 10284 calculator and then press the key labeled “e” or “inv(erse) ln x.” Comment This extremely small equilibrium constant is consistent with the fact that water does not spontaneously decompose into hydrogen and oxygen gases at 25°C. Thus, a large positive DG° favors reactants over products at equilibrium. Similar problems: 17.23 and 17.26. Practice Exercise Calculate the equilibrium constant (KP) for the reaction 2O3 (g) ¡ 3O2 (g) at 25°C. Example 17.7 In Chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility product of silver chloride at 25°C (1.6 3 10210), calculate DG° for the process AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) Strategy According to Equation (17.14), the equilibrium constant for the reaction is related to standard free-energy change; that is, DG° 5 2RT ln K. Because this is a heterogeneous equilibrium, the solubility product (Ksp) is the equilibrium constant. We calculate the standard free-energy change from the Ksp value of AgCl. What temperature unit should be used? Solution The solubility equilibrium for AgCl is AgCl(s) Δ Ag1 (aq) 1 Cl2 (aq) Ksp 5 [Ag1][Cl2] 5 1.6 3 10210 Using Equation (17.14) we obtain ¢G° 5 2(8.314 J/K ? mol) (298 K) ln (1.6 3 10210 ) 5 5.6 3 104 J/mol 5 56 kJ/mol Check The large, positive DG° indicates that AgCl is slightly soluble and that the Similar problem: 17.25. equilibrium lies mostly to the left. Practice Exercise Calculate DG° for the following process at 25°C: BaF2 (s) Δ Ba21 (aq) 1 2F2 (aq) The Ksp of BaF2 is 1.7 3 1026. 800 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Example 17.8 The equilibrium constant (KP) for the reaction N2O4 (g) Δ 2NO2 (g) is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 5 0.122 atm and PN2O4 5 0.453 atm. Calculate DG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium. Strategy From the information given we see that neither the reactant nor the product is at its standard state of 1 atm. To determine the direction of the net reaction, we need to calculate the free-energy change under nonstandard-state conditions (DG) using Equation (17.13) and the given DG° value. Note that the partial pressures are expressed as dimensionless quantities in the reaction quotient QP because they are divided by the standard-state value of 1 atm (see p. 627 and Table 17.2). Solution Equation (17.13) can be written as ¢G 5 ¢G° 1 RT ln QP P2NO2 5 ¢G° 1 RT ln PN2O4 (0.122) 2 5 5.40 3 103 J/mol 1 (8.314 J/K ? mol) (298 K) 3 ln 0.453 5 5.40 3 103 J/mol 2 8.46 3 103 J/mol 5 23.06 3 103 J/mol 5 23.06 kJ/mol Because DG , 0, the net reaction proceeds from left to right to reach equilibrium. Check Note that although DG° . 0, the reaction can be made to favor product formation initially by having a small concentration (pressure) of the product compared Similar problems: 17.27 and 17.28. to that of the reactant. Confirm the prediction by showing that QP , KP. Practice Exercise The KP for the reaction N2(g) 1 3H2(g) Δ 2NH3(g) is 4.3 3 1024 at 375°C. In one experiment, the initial pressures are: PH2 5 0.40 atm, PN2 5 0.86 atm, and PNH3 5 4.4 3 1023 atm. Calculate DG for the reaction and predict the direction of the net reaction. Review of Concepts A reaction has a positive DH° and a negative DS°. Is the equilibrium constant (K ) for this reaction greater than 1, equal to 1, or less than 1? 17.7 Thermodynamics in Living Systems Many biochemical reactions have a positive DG° value, yet they are essential to the maintenance of life. In living systems, these reactions are coupled to an energetically favorable process, one that has a negative DG° value. The principle of coupled reactions is based on a simple concept: We can use a thermodynamically favorable reaction to drive an unfavorable one. Consider an industrial process. Suppose we wish to extract zinc from the ore sphalerite (ZnS). The following reaction will not work because it has a large positive DG° value: ZnS(s) ¡ Zn(s) 1 S(s)       ¢G° 5 198.3 kJ/mol CHEMISTRY in Action The Thermodynamics of a Rubber Band W e all know how useful a rubber band can be. But not every- one is aware that a rubber band has some very interesting thermodynamic properties based on its structure. You can easily perform the following experiments with a rubber band that is at least 0.5 cm wide. Quickly stretch the rubber band and then press it against your lips. You will feel a slight warming effect. Next, reverse the process. Stretch a rubber band (a) (b) and hold it in position for a few seconds. Then quickly release the tension and press the rubber band against your lips. This time you (a) Rubber molecules in their normal state. Note the high degree of entan- glement (large number of microstates and a high entropy). (b) Under ten- will feel a slight cooling effect. A thermodynamic analysis of these sion, the molecules line up and the arrangement becomes much more two experiments can tell us something about the molecular struc- ordered (a small number of microstates and a low entropy). ture of rubber. Rearranging Equation (17.10) (DG 5 DH 2 TDS) gives natural state is more entangled (has more microstates) than when T¢S 5 ¢H 2 ¢G it is under tension. When the tension is removed, the stretched rubber band The warming effect (an exothermic process) due to stretch- spontaneously snaps back to its original shape; that is, DG is ing means that DH , 0, and since stretching is nonspontaneous negative and 2DG is positive. The cooling effect means that it (that is, DG . 0 and 2DG , 0) TDS must be negative. Because is an endothermic process (DH . 0), so that TDS is positive. T, the absolute temperature, is always positive, we conclude that Thus, the entropy of the rubber band increases when it goes DS due to stretching must be negative, and therefore rubber in its from the stretched state to the natural state. On the other hand, the combustion of sulfur to form sulfur dioxide is favored because of its large negative DG° value: S(s) 1 O2 (g) ¡ SO2 (g) ¢G° 5 2300.1 kJ/mol By coupling the two processes we can bring about the separation of zinc from zinc sulfide. In practice, this means heating ZnS in air so that the tendency of S to form SO2 will promote the decomposition of ZnS: ZnS(s) ¡ Zn(s) 1 S(s)     ¢G° 5 198.3 kJ/mol      S(s) 1 O2 (g) ¡ SO2 (g) ¢G° 5 2300.1 kJ/mol ZnS(s) 1 O2 (g) ¡ Zn(s) 1 SO2 (g) ¢G° 5 2101.8 kJ/mol Coupled reactions play a crucial role in our survival. In biological systems, enzymes facilitate a wide variety of nonspontaneous reactions. For example, in the human body, food molecules, represented by glucose (C6H12O6), are converted to carbon dioxide and water during metabolism with a substantial release of free energy: C6H12O6 (s) 1 6O2 (g) ¡ 6CO2 (g) 1 6H2O(l) ¢G° 5 22880 kJ/mol A mechanical analog for coupled reactions. We can make the smaller weight move upward In a living cell, this reaction does not take place in a single step (as burning glucose (a nonspontaneous process) by in a flame would); rather, the glucose molecule is broken down with the aid of coupling it with the falling of a enzymes in a series of steps. Much of the free energy released along the way is used larger weight. 801 802 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Figure 17.9 Structure of ATP and NH2 NH2 ADP in ionized forms. The adenine group is in blue, the ribose group C N C N in black, and the phosphate group N C N C CH CH in red. Note that ADP has one HC C HC C fewer phosphate group than ATP. N N N N O O O A O O A B B B A B B A  OOPOOOPOOOPOOCH2 O A  OOPOOOPOOCH2 O A A A A A A A A A A A H A A H A      O O O H O O H H A A H H A A H A A A A HO OH HO OH Adenosine triphosphate Adenosine diphosphate (ATP) (ADP) to synthesize adenosine triphosphate (ATP) from adenosine diphosphate (ADP) and phosphoric acid (Figure 17.9): ADP 1 H3PO4 ¡ ATP 1 H2O     ¢G° 5 131 kJ/mol The function of ATP is to store free energy until it is needed by cells. Under appropri- ate conditions, ATP undergoes hydrolysis to give ADP and phosphoric acid, with a release of 31 kJ/mol of free energy, which can be used to drive energetically unfavor- able reactions, such as protein synthesis. Proteins are polymers made of amino acids. The stepwise synthesis of a protein molecule involves the joining of individual amino acids. Consider the formation of the dipeptide (a two-amino-acid unit) alanylglycine from alanine and glycine. This reaction represents the first step in the synthesis of a protein molecule: Alanine 1 Glycine ¡ Alanylglycine  ¢G° 5 129 kJ/mol As you can see, this reaction does not favor the formation of product, and so only a little of the dipeptide would be formed at equilibrium. However, with the aid of an enzyme, the reaction is coupled to the hydrolysis of ATP as follows: ATP 1 H2O 1 Alanine 1 Glycine ¡ ADP 1 H3PO4 1 Alanylglycine The overall free-energy change is given by DG° 5 231 kJ/mol 1 29 kJ/mol 5 22 kJ/mol, which means that the coupled reaction now favors the formation of prod- uct, and an appreciable amount of alanylglycine will be formed under this condition. Figure 17.10 shows the ATP-ADP interconversions that act as energy storage (from metabolism) and free-energy release (from ATP hydrolysis) to drive essential reactions. Figure 17.10 Schematic representation of ATP synthesis Glucose ATP Proteins and coupled reactions in living systems. The conversion of glucose to carbon dioxide and water during metabolism releases Free Energy free energy. The released free energy is used to convert ADP into ATP. The ATP molecules are then used as an energy source to drive unfavorable reactions, such as protein synthesis from amino acids. ADP CO2  H2O Amino acids  Key Words 803 Key Equations S 5 k ln W (17.1) Relating entropy to number of microstates. ¢Suniv 5 ¢Ssys 1 ¢Ssurr . 0 (17.4) The second law of thermodynamics (spontaneous process). ¢Suniv 5 ¢Ssys 1 ¢Ssurr 5 0 (17.5) The second law of thermodynamics (equilibrium process). ¢S°rxn 5 ©nS°(products) 2 ©mS°(reactants) (17.7) Standard entropy change of a reaction. G 5 H 2 TS (17.9) Definition of Gibbs free energy. ¢G 5 ¢H 2 T¢S (17.10) Free-energy change at constant temperature. ¢G°rxn 5 ©n¢G°f (products) (17.12) Standard free-energy change 2 ©m¢G°f (reactants) of a reaction. ¢G 5 ¢G° 1 RT ln Q (17.13) Relationship between free-energy change and standard free-energy change and reaction quotient. ¢G° 5 2RT ln K (17.14) Relationship between standard free-energy change and the equilibrium constant. Summary of Facts & Concepts 1. Entropy is described as a measure of the different ways 5. For a chemical or physical process at constant a system can disperse its energy. Any spontaneous temperature and pressure, DG 5 DH 2 TDS. This process must lead to a net increase in entropy in the equation can be used to predict the spontaneity of a universe (second law of thermodynamics). process. 2. The standard entropy of a chemical reaction can be 6. The standard free-energy change for a reaction, DG°, calculated from the absolute entropies of reactants and can be calculated from the standard free energies of products. formation of reactants and products. 3. The third law of thermodynamics states that the en- 7. The equilibrium constant of a reaction and the standard tropy of a perfect crystalline substance is zero at 0 K. free-energy change of the reaction are related by the This law enables us to measure the absolute entropies equation DG° 5 2RT ln K. of substances. 8. Many biological reactions are nonspontaneous. They 4. Under conditions of constant temperature and pressure, are driven by the hydrolysis of ATP, for which DG° is the free-energy change DG is less than zero for a sponta- negative. neous process and greater than zero for a nonspontaneous process. For an equilibrium process, DG 5 0. Key Words Entropy (S), p. 778 Second law of Standard free-energy of Third law of Free energy (G), p. 789 thermodynamics, p. 783 formation (DG°f ), p. 791 thermodynamics, p. 787 Gibbs free energy Standard entropy of reaction Standard free-energy of (G), p. 789 (DS°rxn), p. 784 reaction (DG°rxn), p. 790  804 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Questions & Problems • Problems available in Connect Plus same molar amount is used in the comparison. Ex- Red numbered problems solved in Student Solutions Manual plain the basis for your choice. (a) Li(s) or Li(l); (b) C2H5OH(l) or CH3OCH3(l) (Hint: Which molecule Spontaneous Processes and Entropy can hydrogen-bond?); (c) Ar(g) or Xe(g); (d) CO(g) or Review Questions CO2(g); (e) O2(g) or O3(g); (f) NO2(g) or N2O4(g) 17.10 Arrange the following substances (1 mole each) 17.1 Explain what is meant by a spontaneous process. in order of increasing entropy at 25°C: (a) Ne(g), Give two examples each of spontaneous and non- (b) SO2(g), (c) Na(s), (d) NaCl(s), (e) H2(g). Give spontaneous processes. the reasons for your arrangement. 17.2 Which of the following processes are spontaneous • 17.11 Using the data in Appendix 3, calculate the stan- and which are nonspontaneous? (a) dissolving table dard entropy changes for the following reactions salt (NaCl) in hot soup; (b) climbing Mt. Everest; at 25°C: (c) spreading fragrance in a room by removing the (a) S(s) 1 O2 (g) ¡ SO2 (g) cap from a perfume bottle; (d) separating helium and neon from a mixture of the gases (b) MgCO3 (s) ¡ MgO(s) 1 CO2 (g) 17.3 Which of the following processes are spontaneous and • 17.12 Using the data in Appendix 3, calculate the stan- which are nonspontaneous at a given temperature? dard entropy changes for the following reactions at H2O 25°C: (a) NaNO3 (s) ¡ NaNO3 (aq) saturated soln H2O (a) H2 (g) 1 CuO(s) ¡ Cu(s) 1 H2O(g) (b) NaNO3 (s) ¡ NaNO3 (aq) unsaturated soln H2O (b) 2Al(s) 1 3ZnO(s) ¡ Al2O3(s) 1 3Zn(s) (c) NaNO3 (s) ¡ NaNO3 (aq) supersaturated soln (c) CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l) 17.4 Define entropy. What are the units of entropy? • 17.13 Without consulting Appendix 3, predict whether the entropy change is positive or negative for each Problems of the following reactions. Give reasons for your predictions. 17.5 How does the entropy of a system change for each of (a) 2KClO4 (s) ¡ 2KClO3 (s) 1 O2 (g) the following processes? (b) H2O(g) ¡ H2O(l) (a) A solid melts. (c) 2Na(s) 1 2H2O(l) ¡ (b) A liquid freezes. 2NaOH(aq) 1 H2 (g) (c) A liquid boils. (d) N2 (g) ¡ 2N(g) (d) A vapor is converted to a solid. 17.14 State whether the sign of the entropy change ex- (e) A vapor condenses to a liquid. pected for each of the following processes will be (f ) A solid sublimes. positive or negative, and explain your predictions. (g) Urea dissolves in water. (a) PCl3 (l) 1 Cl2 (g) ¡ PCl5 (s) 17.6 Consider the arrangement in Figure 17.1. Because the (b) 2HgO(s) ¡ 2Hg(l) 1 O2 (g) volume of the two bulbs is the same, the probability (c) H2 (g) ¡ 2H(g) of finding a molecule in either bulb is 12 . Calculate the (d) U(s) 1 3F2 (g) ¡ UF6 (s) probability of all the molecules ending up in the same bulb if the number is (a) 2, (b) 100, and (c) 6 3 1023. Gibbs Free Energy Based on your results, explain why the situation shown in Figure 17.1(b) will not be observed for a Review Questions macroscopic system. 17.15 Define free energy. What are its units? The Second Law of Thermodynamics • 17.16 Why is it more convenient to predict the direction of a reaction in terms of DGsys instead of DSuniv? Under Review Questions what conditions can DGsys be used to predict the 17.7 State the second law of thermodynamics in words spontaneity of a reaction? and express it mathematically. Problems 17.8 State the third law of thermodynamics and explain its usefulness in calculating entropy values. • 17.17 Calculate DG° for the following reactions at 25°C: (a) N2 (g) 1 O2 (g) ¡ 2NO(g) Problems (b) H2O(l) ¡ H2O(g) • 17.9 For each pair of substances listed here, choose the one (c) 2C2H2 (g) 1 5O2 (g) ¡ having the larger standard entropy value at 25°C. The 4CO2 (g) 1 2H2O(l)  Questions & Problems 805 (Hint: Look up the standard free energies of • 17.28 The equilibrium constant (KP) for the reaction formation of the reactants and products in Appendix 3.) H2 (g) 1 CO2 (g) Δ H2O(g) 1 CO(g) • 17.18 Calculate DG° for the following reactions at 25°C: is 4.40 at 2000 K. (a) Calculate DG° for the reaction. (a) 2Mg(s) 1 O2 (g) ¡ 2MgO(s) (b) Calculate DG for the reaction when the partial (b) 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g) pressures are PH2 5 0.25 atm, PCO2 5 0.78 atm, PH2O (c) 2C2H6 (g) 1 7O2 (g) ¡ 5 0.66 atm, and PCO 5 1.20 atm. 4CO2 (g) 1 6H2O(l) • 17.29 Consider the decomposition of calcium carbonate: See Appendix 3 for thermodynamic data. CaCO3 (s) Δ CaO(s) 1 CO2 (g) • 17.19 From the values of DH and DS, predict which of the Calculate the pressure in atm of CO2 in an equilib- following reactions would be spontaneous at 25°C: Reaction A: DH 5 10.5 kJ/mol, DS 5 30 J/K ? mol; rium process (a) at 25°C and (b) at 800°C. Assume reaction B: DH 5 1.8 kJ/mol, DS 5 2113 J/K ? mol. that DH° 5 177.8 kJ/mol and DS° 5 160.5 J/K ? mol If either of the reactions is nonspontaneous at for the temperature range. 25°C, at what temperature might it become • 17.30 The equilibrium constant KP for the reaction spontaneous? CO(g) 1 Cl2 (g) Δ COCl2 (g) 17.20 Find the temperatures at which reactions with the following DH and DS values would become is 5.62 3 1035 at 25°C. Calculate DG°f for COCl2 spontaneous: (a) DH 5 2126 kJ/mol, DS 5 84 at 25°C. J/K ? mol; (b) DH 5 211.7 kJ/mol, DS 5 2105 17.31 At 25°C, DG° for the process J/K ? mol. H2O(l) Δ H2O(g) is 8.6 kJ/mol. Calculate the vapor pressure of water Free Energy and Chemical Equilibrium at this temperature. Review Questions 17.32 Calculate DG° for the process 17.21 Explain the difference between DG and DG°. C(diamond) ¡ C(graphite) 17.22 Explain why Equation (17.14) is of great importance in chemistry. Is the formation of graphite from diamond favored at 25°C? If so, why is it that diamonds do not become graphite on standing? Problems • 17.23 Calculate KP for the following reaction at 25°C: H2 (g) 1 I2 (g) Δ 2HI(g)  ¢G° 5 2.60 kJ/mol Thermodynamics in Living Systems Review Questions • 17.24 For the autoionization of water at 25°C, 17.33 What is a coupled reaction? What is its importance H2O(l) Δ H1 (aq) 1 OH2 (aq) in biological reactions? Kw is 1.0 3 10214. What is DG° for the process? 17.34 What is the role of ATP in biological reactions? • 17.25 Consider the following reaction at 25°C: Fe(OH) 2 (s) Δ Fe21 (aq) 1 2OH2 (aq) Problems Calculate DG° for the reaction. Ksp for Fe(OH)2 is 17.35 Referring to the metabolic process involving glucose 1.6 3 10214. on p. 801, calculate the maximum number of moles • 17.26 Calculate DG° and KP for the following equilibrium of ATP that can be synthesized from ADP from the reaction at 25°C. breakdown of one mole of glucose. 17.36 In the metabolism of glucose, the first step is the 2H2O(g) Δ 2H2 (g) 1 O2 (g) conversion of glucose to glucose 6-phosphate: • 17.27 (a) Calculate DG° and KP for the following equilib- glucose 1 H3PO4 ¡ glucose 6-phosphate 1 H2O rium reaction at 25°C. The DG°f values are 0 for ¢G° 5 13.4 kJ/mol Cl2(g), 2286 kJ/mol for PCl3(g), and 2325 kJ/mol for PCl5(g). Because DG° is positive, this reaction does not favor PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) the formation of products. Show how this reaction can be made to proceed by coupling it with the hy- (b) Calculate DG for the reaction if the partial pres- drolysis of ATP. Write an equation for the coupled sures of the initial mixture are PPCl5 5 0.0029 atm, reaction and estimate the equilibrium constant for PPCl3 5 0.27 atm, and PCl2 5 0.40 atm. the coupled process.  806 Chapter 17 ■ Entropy, Free Energy, and Equilibrium Additional Problems • 17.47 Calculate the equilibrium pressure of CO2 due to 17.37 Explain the following nursery rhyme in terms of the the decomposition of barium carbonate (BaCO3) second law of thermodynamics. at 25°C. Humpty Dumpty sat on a wall; 17.48 (a) Trouton’s rule states that the ratio of the molar Humpty Dumpty had a great fall. heat of vaporization of a liquid (DHvap) to its boiling point in kelvins is approximately 90 J/K ? mol. Use All the King’s horses and all the King’s men the following data to show that this is the case and Couldn’t put Humpty together again. explain why Trouton’s rule holds true: • 17.38 Calculate DG for the reaction H2O(l) Δ H1 (aq) 1 OH2 (aq) tbp(8C) DHvap(kJ/mol) at 25°C for the following conditions: Benzene 80.1 31.0 (a) [H 1] 5 1.0 3 1027 M, [OH2] 5 1.0 3 1027 M Hexane 68.7 30.8 (b) [H 1] 5 1.0 3 1023 M, [OH2] 5 1.0 3 1024 M Mercury 357 59.0 (c) [H 1] 5 1.0 3 10212 M, [OH2] 5 2.0 3 1028 M Toluene 110.6 35.2 (d) [H 1] 5 3.5 M, [OH2] 5 4.8 3 1024 M 17.39 Calculate the DS°soln for the following processes: (b) Use the values in Table 11.6 to calculate the same (a) NH4NO3(s) ¡ NH14 (aq) 1 NO23 (aq) and ratio for ethanol and water. Explain why Trouton’s (b) FeCl3(s) ¡ Fe31(aq) 1 3Cl2(aq). Give a rule does not apply to these two substances as well as qualitative explanation for the signs. it does to other liquids. 17.40 The following reaction is spontaneous at a certain 17.49 Referring to Problem 17.48, explain why the ratio is temperature T. Predict the sign of DSsurr. considerably smaller than 90 J/K ? mol for liquid HF. • 17.50 Carbon monoxide (CO) and nitric oxide (NO) are polluting gases contained in automobile exhaust. Under suitable conditions, these gases can be made 8n to react to form nitrogen (N2) and the less harmful carbon dioxide (CO2). (a) Write an equation for this reaction. (b) Identify the oxidizing and reduc- ing agents. (c) Calculate the KP for the reaction at 17.41 Which of the following thermodynamic functions are 25°C. (d) Under normal atmospheric conditions, associated only with the first law of thermodynamics: the partial pressures are PN2 5 0.80 atm, PCO2 5 S, U, G, and H? 3.0 3 1024 atm, PCO 5 5.0 3 1025 atm, and PNO 5 17.42 A student placed 1 g of each of three compounds 5.0 3 1027 atm. Calculate QP and predict the direc- A, B, and C in a container and found that after 1 tion toward which the reaction will proceed. (e) week no change had occurred. Offer some possi- Will raising the temperature favor the formation of ble explanations for the fact that no reactions took N2 and CO2? place. Assume that A, B, and C are totally misci- • 17.51 For reactions carried out under standard-state ble liquids. conditions, Equation (17.10) takes the form DG° 5 17.43 Use the data in Appendix 3 to calculate the equilibrium DH° 2 TDS°. (a) Assuming DH° and DS° are inde- constant for the reaction AgI(s) Δ Ag1 (aq) 1 pendent of temperature, derive the equation I2 (aq) at 25°C. Compare your result with the Ksp K2 ¢H° T2 2 T1 value in Table 16.2. ln 5 a b K1 R T1T2 17.44 Predict the signs of DH, DS, and DG of the system for the following processes at 1 atm: (a) ammonia where K1 and K2 are the equilibrium constants at T1 melts at 260°C, (b) ammonia melts at 277.7°C, and T2, respectively. (b) Given that at 25°C Kc is (c) ammonia melts at 2100°C. (The normal melting 4.63 3 1023 for the reaction point of ammonia is 277.7°C.) N2O4 (g) Δ 2NO2 (g)  ¢H° 5 58.0 kJ/mol 17.45 Consider the following facts: Water freezes sponta- neously at 25°C and 1 atm, and ice has a more or- calculate the equilibrium constant at 65°C. dered structure than liquid water. Explain how a 17.52 Use the thermodynamic data in Appendix 3 to calcu- spontaneous process can lead to a decrease in late the Ksp of AgCl. entropy. 17.53 Consider the reaction A ¡ B 1 C at 298 K. Given 17.46 Ammonium nitrate (NH4NO3) dissolves spontane- that the forward rate constant (k f) is 0.46 s21 and the ously and endothermically in water. What can you reverse rate constant (kr) is 1.5 3 1022/M ? s, calcu- deduce about the sign of DS for the solution process? late DG° of the reaction.  Questions & Problems 807 17.54 The Ksp of AgCl is given in Table 16.2. What is its • 17.64 The molar heat of vaporization of ethanol is 39.3 value at 60°C? [Hint: You need the result of Problem kJ/mol and the boiling point of ethanol is 78.3°C. 17.51(a) and the data in Appendix 3 to calculate Calculate DS for the vaporization of 0.50 mol DH°.] ethanol. 17.55 Under what conditions does a substance have a 17.65 A certain reaction is known to have a DG° value of standard entropy of zero? Can a substance ever 2122 kJ/mol. Will the reaction necessarily occur if have a negative standard entropy? the reactants are mixed together? • 17.56 Water gas, a mixture of H2 and CO, is a fuel made by • 17.66 In the Mond process for the purification of nickel, reacting steam with red-hot coke (a by-product of carbon monoxide is reacted with heated nickel to coal distillation): produce Ni(CO)4, which is a gas and can therefore be separated from solid impurities: H2O(g) 1 C(s) Δ CO(g) 1 H2 (g) Ni(s) 1 4CO(g) Δ Ni(CO) 4 (g) From the data in Appendix 3, estimate the temperature at which the reaction begins to favor the formation of Given that the standard free energies of formation of products. CO(g) and Ni(CO)4(g) are 2137.3 kJ/mol and • 17.57 Consider the following Brønstead acid-base reaction 2587.4 kJ/mol, respectively, calculate the equilib- at 25°C: rium constant of the reaction at 80°C. Assume that DG°f is temperature independent. HF(aq) 1 Cl2 (aq) Δ HCl(aq) 1 F2 (aq) • 17.67 Calculate DG° and KP for the following processes at (a) Predict whether K will be greater or smaller 25°C: than unity. (b) Does DS° or DH° make a greater (a) H2 (g) 1 Br2 (l) Δ 2HBr(g) contribution to DG°? (c) Is DH° likely to be posi- (b) 12H2 (g) 1 12Br2 (l) Δ HBr(g) tive or negative? Account for the differences in DG° and KP obtained • 17.58 Crystallization of sodium acetate from a supersatu- for (a) and (b). rated solution occurs spontaneously (see p. 519). • 17.68 Calculate the pressure of O2 (in atm) over a sample What can you deduce about the signs of DS and DH? of NiO at 25°C if DG° 5 212 kJ/mol for the reaction • 17.59 Consider the thermal decomposition of CaCO3: NiO(s) Δ Ni(s) 1 12 O2 (g) CaCO3 (s) Δ CaO(s) 1 CO2 (g) 17.69 Comment on the statement: “Just talking about The equilibrium vapor pressures of CO 2 are entropy increases its value in the universe.” 22.6 mmHg at 700°C and 1829 mmHg at 950°C. 17.70 For a reaction with a negative DG° value, which of Calculate the standard enthalpy of the reaction. the following statements is false? (a) The equilib- [Hint: See Problem 17.51(a).] rium constant K is greater than one, (b) the reaction • 17.60 A certain reaction is spontaneous at 72°C. If the is spontaneous when all the reactants and products enthalpy change for the reaction is 19 kJ/mol, what are in their standard states, and (c) the reaction is is the minimum value of DS (in J/K ? mol) for the always exothermic. reaction? 17.71 Consider the reaction • 17.61 Predict whether the entropy change is positive or N2 (g) 1 O2 (g) Δ 2NO(g) negative for each of these reactions: (a) Zn(s) 1 2HCl(aq) ¡ ZnCl2 (aq) 1 H2 (g) Given that DG° for the reaction at 25°C is 173.4 kJ/ (b) O(g) 1 O(g) ¡ O2 (g) mol, (a) calculate the standard free energy of forma- tion of NO, and (b) calculate KP of the reaction. (c) NH4NO3 (s) ¡ N2O(g) 1 2H2O(g) (c) One of the starting substances in smog formation (d) 2H2O2 (l) ¡ 2H2O(l) 1 O2 (g) is NO. Assuming that the temperature in a running 17.62 The reaction NH3 (g) 1 HCl(g) ¡ NH4Cl(s) automobile engine is 1100°C, estimate KP for the proceeds spontaneously at 25°C even though there is above reaction. (d) As farmers know, lightning helps a decrease in the number of microstates of the system to produce a better crop. Why? (gases are converted to a solid). Explain. • 17.72 Heating copper(II) oxide at 400°C does not produce • 17.63 Use the following data to determine the normal boil- any appreciable amount of Cu: ing point, in kelvins, of mercury. What assumptions CuO(s) Δ Cu(s) 1 12 O2(g)  ¢G° 5 127.2 kJ/mol must you make in order to do the calculation? However, if this reaction is coupled to the conver- Hg(l): ¢H°f 5 0 (by definition) sion of graphite to carbon monoxide, it becomes S° 5 77.4 J/K ? mol spontaneous. Write an equation for the coupled Hg(g): ¢H°f 5 60.78 kJ/mol process and calculate the equilibrium constant for S° 5 174.7 J/K ? mol the coupled reaction.  808 Chapter 17 ■ Entropy, Free Energy, and Equilibrium 17.73 The internal engine of a 1200-kg car is designed to 25°C and comment on whether this method is feasi- run on octane (C8H18), whose enthalpy of combus- ble for removing SO2. (c) Would this procedure be- tion is 5510 kJ/mol. If the car is moving up a slope, come more or less effective at a higher temperature? calculate the maximum height (in meters) to which 17.79 Describe two ways that you could measure DG° of a the car can be driven on 1.0 gallon of the fuel. As- reaction. sume that the engine cylinder temperature is 2200°C 17.80 The following reaction represents the removal of and the exit temperature is 760°C, and neglect all ozone in the stratosphere: forms of friction. The mass of 1 gallon of fuel is 3.1 kg. [Hint: See the Chemistry in Action essay on 2O3 (g) Δ 3O2 (g) p. 790. The work done in moving the car over a ver- Calculate the equilibrium constant (KP) for the re- tical distance is mgh, where m is the mass of the car action. In view of the magnitude of the equilibrium in kg, g the acceleration due to gravity (9.81 m/s2), constant, explain why this reaction is not consid- and h the height in meters.] ered a major cause of ozone depletion in the ab- 17.74 Consider the decomposition of magnesium carbonate: sence of man-made pollutants such as the nitrogen MgCO3 (s) Δ MgO(s) 1 CO2 (g) oxides and CFCs. Assume the temperature of the stratosphere to be 230°C and DG°f to be tempera- Calculate the temperature at which the decomposition ture independent. begins to favor products. Assume that both DH° and DS° are independent of temperature. • 17.81 A 74.6-g ice cube floats in the Arctic Sea. The tem- perature and pressure of the system and surround- 17.75 (a) Over the years there have been numerous ings are at 1 atm and 0°C. Calculate DSsys, DSsurr, claims about “perpetual motion machines,” ma- and DSuniv for the melting of the ice cube. What can chines that will produce useful work with no input you conclude about the nature of the process from of energy. Explain why the first law of thermody- the value of DSuniv? (The molar heat of fusion of namics prohibits the possibility of such a machine water is 6.01 kJ/mol.) existing. (b) Another kind of machine, sometimes called a “perpetual motion of the second kind,” • 17.82 Comment on the feasibility of extracting copper from its ore chalcocite (Cu2S) by heating: operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then ex- Cu2S(s2 ¡ 2Cu(s) 1 S(s) tracting heat from the water, converting the heat to electric power to run the ship, and dumping the Calculate the DG° for the overall reaction if the water back into the ocean. This process does not above process is coupled to the conversion of sul- violate the first law of thermodynamics, for no en- fur to sulfur dioxide. Given that DG°f (Cu2S) 5 ergy is created—energy from the ocean is just con- 286.1 kJ/mol. verted to electrical energy. Show that the second 17.83 Active transport is the process in which a substance law of thermodynamics prohibits the existence of is transferred from a region of lower concentration such a machine. to one of higher concentration. This is a nonspon- 17.76 The activity series in Section 4.4 shows that reaction taneous process and must be coupled to a sponta- (a) is spontaneous while reaction (b) is nonsponta- neous process, such as the hydrolysis of ATP. The neous at 25°C: concentrations of K1 ions in the blood plasma and in nerve cells are 15 mM and 400 mM, respectively (a) Fe(s) 1 2H1 ¡ Fe21 (aq) 1 H2 (g) (1 mM 5 1 3 1023 M). Use Equation (17.13) to (b) Cu(s) 1 2H1 ¡ Cu21 (aq) 1 H2 (g) calculate DG for the process at the physiological Use the data in Appendix 3 to calculate the equilib- temperature of 37°C: rium constant for these reactions and hence confirm that the activity series is correct. K1 (15 mM) ¡ K1 (400 mM) • 17.77 The rate constant for the elementary reaction In this calculation, the DG° term can be set to zero. What is the justification for this step? 2NO(g) 1 O2 (g) ¡ 2NO2 (g) • 17.84 Large quantities of hydrogen are needed for the is 7.1 3 109/M2 ? s at 25°C. What is the rate constant synthesis of ammonia. One preparation of hydrogen for the reverse reaction at the same temperature? involves the reaction between carbon monoxide and 17.78 The following reaction is the cause of sulfur deposits steam at 300°C in the presence of a copper-zinc formed at volcanic sites (see p. 911): catalyst: 2H2S(g) 1 SO2 (g) Δ 3S(s) 1 2H2O(g) CO(g) 1 H2O(g) Δ CO2 (g) 1 H2 (g) It may also be used to remove SO2 from powerplant Calculate the equilibrium constant (KP) for the reac- stack gases. (a) Identify the type of redox reaction it tion and the temperature at which the reaction favors is. (b) Calculate the equilibrium constant (KP) at the formation of CO and H2O. Will a larger KP be  Questions & Problems 809 attained at the same temperature if a more efficient 17.89 Use the thermodynamic data in Appendix 3 to deter- catalyst is used? mine the normal boiling point of liquid bromine. 17.85 Shown here are the thermodynamic data for ethanol: Assume the values are independent of temperature. 17.90 In each of the following reactions, there is one species DHf8(kJ/mol) S8(J/K ? mol) for which the standard entropy value is not listed in Appendix 3. Determine the S° for that species. (a) The liquid 2276.98 161.0 DS°rxn for the reaction Na(s) ¡ Na(l) is 48.64 J/K ? vapor 2235.1 282.7 mol. (b) The DS°rxn for the reaction 2S(monoclinic) 1 Cl2(g) ¡ S2Cl2(g) is 43.4 J/K ? mol. (c) The DS°rxn for the reaction FeCl2(s) ¡ Fe21(aq) 1 2Cl2(aq) is Calculate the vapor pressure of ethanol at 25°C. 2118.3 J/K ? mol. Assume the thermodynamic values are independent of temperature. 17.91 A rubber band is stretched vertically by attach- ing a weight to one end and holding the other end 17.86 The reaction shown here is spontaneous at a certain by hand. On heating the rubber band with a hot- temperature T. What is the sign of DSsurr? air blower, it is observed to shrink slightly in length. Give a thermodynamic analysis for this behavior. (Hint: See the Chemistry in Action es- say on p. 801.) 8n 17.92 One of the steps in the extraction of iron from its ore (FeO) is the reduction of iron(II) oxide by carbon monoxide at 900°C: FeO(s) 1 CO(g) Δ Fe(s) 1 CO2 (g) 17.87 Consider two carboxylic acids (acids that contain If CO is allowed to react with an excess of FeO, the ¬COOH group): CH3COOH (acetic acid, calculate the mole fractions of CO and CO2 at equi- Ka 5 1.8 3 1025) and CH2ClCOOH (chloroacetic librium. State any assumptions. acid, Ka 5 1.4 3 1023). (a) Calculate DG° for the 17.93 Derive the following equation ionization of these acids at 25°C. (b) From the equa- tion DG° 5 DH° 2 TDS°, we see that the contribu- ¢G 5 RT ln (Q/K) tions to the DG° term are an enthalpy term (DH°) where Q is the reaction quotient and describe how you and a temperature times entropy term (TDS°). These would use it to predict the spontaneity of a reaction. contributions are listed below for the two acids: 17.94 The sublimation of carbon dioxide at 278°C is CO2 (s) ¡ CO2 (g) ¢Hsub 5 62.4 kJ/mol DH8(kJ/mol) TDS8(kJ/mol) Calculate DSsub when 84.8 g of CO2 sublimes at this CH3COOH 20.57 227.6 temperature. CH2ClCOOH 24.7 221.1 17.95 Entropy has sometimes been described as “time’s arrow” because it is the property that determines the Which is the dominant term in determining the value forward direction of time. Explain. of DG° (and hence Ka of the acid)? (c) What pro- 17.96 Referring to Figure 17.1, we see that the probabil- cesses contribute to DH°? (Consider the ionization ity of finding all 100 molecules in the same bulb is of the acids as a Brønsted acid-base reaction.) (d) 8 3 10231. Assuming that the age of the universe is Explain why the TDS° term is more negative for 13 billion years, calculate the time in seconds dur- CH3COOH. ing which this event can be observed. 17.88 Many hydrocarbons exist as structural isomers, 17.97 A student looked up the ¢G°f , ¢H°f , and S° values which are compounds that have the same molecular for CO2 in Appendix 3. Plugging these values into formula but different structures. For example, both Equation (17.10), he found that ¢G°f ? ¢H°f 2 TS° butane and isobutane have the same molecular for- at 298 K. What is wrong with his approach? mula of C4H10 (see Problem 11.19). Calculate the 17.98 Consider the following reaction at 298 K: mole percent of these molecules in an equilibrium mixture at 25°C, given that the standard free energy 2H2 (g) 1 O2 (g) ¡ 2H2O(l)  ¢H° 5 2571.6 kJ/mol of formation of butane is 215.9 kJ/mol and that of Calculate DSsys, DSsurr, and DSuniv for the reaction. isobutane is 218.0 kJ/mol. Does your result sup- 17.99 As an approximation, we can assume that proteins port the notion that straight-chain hydrocarbons exist either in the native (or physiologically func- (that is, hydrocarbons in which the C atoms are tioning) state and the denatured state joined along a line) are less stable than branch- chain hydrocarbons? native Δ denatured  810 Chapter 17 ■ Entropy, Free Energy, and Equilibrium The standard molar enthalpy and entropy of the M because the physiological pH is about 7. denaturation of a certain protein are 512 kJ/mol Consequently, the change in the standard Gibbs and 1.60 kJ/K ? mol, respectively. Comment on the free energy according to these two conventions signs and magnitudes of these quantities, and cal- will be different involving uptake or release of H1 culate the temperature at which the process favors ions, depending on which convention is used. We the denatured state. will therefore replace DG° with DG°9, where the 17.100 Which of the following are not state functions: S, H, prime denotes that it is the standard Gibbs free- q, w, T ? energy change for a biological process. (a) Con- 17.101 Which of the following is not accompanied by an sider the reaction increase in the entropy of the system? (a) mixing of A 1 B ¡ C 1 xH1 two gases at the same temperature and pressure, (b) mixing of ethanol and water, (c) discharging a where x is a stoichiometric coefficient. Use Equa- battery, (d) expansion of a gas followed by com- tion (17.13) to derive a relation between DG° and pression to its original temperature, pressure, and DG°9, keeping in mind that DG is the same for a volume. process regardless of which convention is used. Re- 17.102 Hydrogenation reactions (for example, the process peat the derivation for the reverse process: of converting C“C bonds to C¬C bonds in food C 1 xH1 ¡ A 1 B industry) are facilitated by the use of a transition metal catalyst, such as Ni or Pt. The initial step is the (b) NAD1 and NADH are the oxidized and reduced adsorption, or binding, of hydrogen gas onto the forms of nicotinamide adenine dinucleotide, two metal surface. Predict the signs of DH, DS, and DG key compounds in the metabolic pathways. For the when hydrogen gas is adsorbed onto the surface of oxidation of NADH: Ni metal. NADH 1 H1 ¡ NAD1 1 H2 17.103 Give a detailed example of each of the following, with an explanation: (a) a thermodynamically DG° is 221.8 kJ/mol at 298 K. Calculate DG°9. spontaneous process; (b) a process that would vio- Also calculate DG using both the chemical and bi- late the first law of thermodynamics; (c) a process ological conventions when [NADH] 5 1.5 3 1022 that would violate the second law of thermody- M, [H1] 5 3.0 3 1025 M, [NAD] 5 4.6 3 1023 M, namics; (d) an irreversible process; (e) an equilib- and PH2 5 0.010 atm. rium process. 17.108 The following diagram shows the variation of the 17.104 At 0 K, the entropy of carbon monoxide crystal is equilibrium constant with temperature for the reaction not zero but has a value of 4.2 J/K ? mol, called the I2 (g) Δ 2I(g) residual entropy. According to the third law of ther- modynamics, this means that the crystal does not Calculate DG°, DH°, and DS° for the reaction at 872 have a perfect arrangement of the CO molecules. K. (Hint: See Problem 17.51.) (a) What would be the residual entropy if the ar- rangement were totally random? (b) Comment on the difference between the result in (a) and 4.2 J/K ? mol. [Hint: Assume that each CO molecule has two choices for orientation and use Equation (17.1) to K2  0.0480 calculate the residual entropy.] 17.105 Comment on the correctness of the analogy some- K1  1.80  104 ln K times used to relate a student’s dormitory room be- coming untidy to an increase in entropy. • 17.106 The standard enthalpy of formation and the standard T2  1173 K entropy of gaseous benzene are 82.93 kJ/mol and T1  872 K 269.2 J/K ? mol, respectively. Calculate DH°, DS°, and DG° for the process at 25°C. 1D T C6H6 (l) ¡ C6H6 (g) Comment on your answers. • 17.109 Consider the gas-phase reaction between A2 (green) 17.107 In chemistry, the standard state for a solution is 1 and B2 (red) to form AB at 298 K: M (see Table 17.2). This means that each solute A2 (g) 1 B2 (g) Δ 2AB(g)  ¢G° 5 23.4 kJ/mol concentration expressed in molarity is divided by 1 M. In biological systems, however, we define (1) Which of the following reaction mixtures is at the standard state for the H1 ions to be 1 3 1027 equilibrium?  Answers to Practice Exercises 811 (2) Which of the following reaction mixtures has a 17.110 The KP for the reaction negative DG value? N2 1 3H2 Δ 2NH3 (3) Which of the following reaction mixtures has a 23 positive DG value? is 2.4 3 10 at 720°C. What is the minimum partial The partial pressures of the gases in each frame are pressure of N2 required for the reaction to be sponta- equal to the number of A2, B2, and AB molecules neous in the forward direction if the partial pres- times 0.10 atm. Round your answers to two signifi- sures of H2 and NH3 are 1.52 atm and 2.1 3 1022 cant figures. atm, respectively? 17.111 The table shown here lists the ion-product constant (Kw) of water at several temperatures. Determine graphically the DH° for the ionization of water. Kw 0.113 3 0.292 3 1.008 3 2.917 3 5.4743 10214 10214 10214 10214 10214 t(°C) 0 10 25 40 50 (Hint: See Problem 14.118.) (a) (b) (c) Interpreting, Modeling & Estimating 17.112 The reaction NH3(g) 1 HCl(g) ¡ NH4Cl(s) is 17.115 Estimate DS for the process depicted in Figure spontaneous at room temperature (see Figure 5.20). 17.1(a) if the apparatus contained 20 molecules in Estimate the temperature at which the reaction is the flask on the left in the initial distribution, and no longer spontaneous under standard conditions. each flask contained 10 molecules in the final distri- 17.113 The boiling point of diethyl ether is 34.6°C. Esti- bution. Useful information: The number of ways to mate (a) its molar heat of vaporization and (b) its distribute n objects between two bins such that r par- vapor pressure at 20°C. (Hint: See Problems 17.48 ticles are in one bin is called the number of combi- and 17.51.) nations (C) and is given by the equation 17.114 Nicotine is the compound in tobacco responsible for n! C(n, r) 5 addiction to smoking. While most of the nicotine in r!(n 2 r)! tobacco exists in the neutral form, roughly 90 percent of the nicotine in the bloodstream is protonated, as where n! (“n factorial”) 5 1 3 2 3 3 3 . . . 3 n, and represented in the following chemical equation. Esti- 0! is defined to be 1. mate DG° for the reaction. 17.116 At what point in the series H¬On ¬H(g) (n 5 1, 2, 3, . . .) does formation of the compound from the CH3 1 CH3 elements H2(g) and O2(g) become nonspontaneous? N N H 1 H1 34 N N Answers to Practice Exercises 17.1 (a) Entropy decreases, (b) entropy decreases, (c) entropy 17.4 (a) 2106.4 kJ/mol, (b) 22935.0 kJ/mol. increases, (d) entropy increases. 17.2 (a) 2173.6 J/K ? mol, 17.5 DSfus 5 16 J/K ? mol; DSvap 5 72 J/K ? mol. (b) 2139.8 J/K ? mol, (c) 215.3 J/K ? mol. 17.6 2 3 1057. 17.7 33 kJ/mol. 17.8 DG 5 21.0 kJ/mol; 17.3 (a) ¢S . 0, (b) ¢S , 0, (c) ¢S < 0. direction is from left to right.  CHAPTER 18 Electrochemistry Michael Faraday at work in his laboratory. Faraday is regarded by many as the greatest experimental scientist of the nineteenth century. CHAPTER OUTLINE A LOOK AHEAD 18.1 Redox Reactions  We begin with a review of redox reactions and learn how to balance equations describing these processes. (18.1) 18.2 Galvanic Cells  Next, we examine the essentials of galvanic cells. (18.2) 18.3 Standard Reduction Potentials  We learn to determine the standard reduction potentials based on the stan- 18.4 Thermodynamics of Redox dard hydrogen electrode reference and use them to calculate the emf of a Reactions cell and hence the spontaneity of a cell reaction. A relationship exists 18.5 The Effect of Concentration between a cell’s emf, the change in the standard Gibbs free energy, and the equilibrium constant for the cell reaction. (18.3 and 18.4) on Cell Emf  We see that the emf of a cell under nonstandard state conditions can be cal- 18.6 Batteries culated using the Nernst equation. (18.5) 18.7 Corrosion  We examine several common types of batteries and the operation of fuel 18.8 Electrolysis cells. (18.6)  We then study a spontaneous electrochemical process—corrosion—and learn ways to prevent it. (18.7)  Finally, we explore a nonspontaneous electrochemical process— electrolysis—and learn the quantitative aspects of electrolytic processes. (18.8) 812  18.1 Redox Reactions 813 O ne form of energy that has tremendous practical significance is electrical energy. A day without electricity from either the power company or batteries is unimaginable in our technological society. The area of chemistry that deals with the interconversion of electrical energy and chemical energy is electrochemistry. Electrochemical processes are redox reactions in which the energy released by a spontane- ous reaction is converted to electricity or in which electricity is used to drive a nonspontaneous chemical reaction. The latter type is called electrolysis. This chapter explains the fundamental principles and applications of galvanic cells, the thermodynamics of electrochemical reactions, and the cause and prevention of corrosion by electrochemical means. Some simple electrolytic processes and the quantitative aspects of elec- trolysis are also discussed. 18.1 Redox Reactions Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy. Electrochemical processes are redox (oxidation- reduction) reactions in which the energy released by a spontaneous reaction is con- verted to electricity or in which electrical energy is used to cause a nonspontaneous reaction to occur. Although redox reactions were discussed in Chapter 4, it is helpful to review some of the basic concepts that will come up again in this chapter. In redox reactions, electrons are transferred from one substance to another. The reaction between magnesium metal and hydrochloric acid is an example of a redox reaction: 0 11 12 0 Mg(s) 1 2HCl(aq) ¡ MgCl2 (aq) 1 H2 (g) Rules for assigning oxidation numbers are presented in Section 4.4. Recall that the numbers above the elements are the oxidation numbers of the elements. The loss of electrons by an element during oxidation is marked by an increase in the element’s oxidation number. In reduction, there is a decrease in oxidation number resulting from a gain of electrons by an element. In the preceding reaction, Mg metal is oxidized and H1 ions are reduced; the Cl2 ions are spectator ions. Balancing Redox Equations Equations for redox reactions like the preceding one are relatively easy to balance. However, in the laboratory we often encounter more complex redox reactions involv- ing oxoanions such as chromate (CrO22 22 4 ), dichromate (Cr2O7 ), permanganate (MnO4 ), 2 2 22 nitrate (NO3 ), and sulfate (SO4 ). In principle, we can balance any redox equation using the procedure outlined in Section 3.7, but there are some special techniques for handling redox reactions, techniques that also give us insight into electron transfer processes. Here we will discuss one such procedure, called the ion-electron method. In this approach, the overall reaction is divided into two half-reactions, one for oxida- tion and one for reduction. The equations for the two half-reactions are balanced separately and then added together to give the overall balanced equation. Suppose we are asked to balance the equation showing the oxidation of Fe21 ions to Fe31 ions by dichromate ions (Cr2O22 7 ) in an acidic medium. As a result, the Cr2O7 22 31 ions are reduced to Cr ions. The following steps will help us balance the equation. Step 1: Write the unbalanced equation for the reaction in ionic form. Fe21 1 Cr2O22 7 ¡ Fe31 1 Cr31 Step 2: Separate the equation into two half-reactions. 12 13 Oxidation: Fe21 ¡ Fe31 16 13 Reduction: Cr2O22 7 ¡ Cr31 814 Chapter 18 ■ Electrochemistry Step 3: Balance each half-reaction for number and type of atoms and charges. For reactions in an acidic medium, add H2O to balance the O atoms and H1 to balance the H atoms. In an oxidation half-reaction, electrons Oxidation half-reaction: The atoms are already balanced. To balance the charge, we appear as a product; in a reduction half- reaction, electrons appear as a reactant. add an electron to the right-hand side of the arrow: Fe21 ¡ Fe31 1 e2 Reduction half-reaction: Because the reaction takes place in an acidic medium, we add seven H2O molecules to the right-hand side of the arrow to balance the O atoms: Cr2O22 7 ¡ 2Cr31 1 7H2O To balance the H atoms, we add 14 H1 ions on the left-hand side: 14H1 1 Cr2O22 7 ¡ 2Cr31 1 7H2O There are now 12 positive charges on the left-hand side and only six positive charges on the right-hand side. Therefore, we add six electrons on the left: 14H1 1 Cr2O22 7 1 6e 2 ¡ 2Cr31 1 7H2O Step 4: Add the two half-equations together and balance the final equation by inspec- tion. The electrons on both sides must cancel. If the oxidation and reduction half-reactions contain different numbers of electrons, we need to multiply one or both half-reactions to equalize the number of electrons. Here we have only one electron for the oxidation half-reaction and six electrons for the reduction half-reaction, so we need to multiply the oxidation half-reaction by 6 and write 6(Fe21 ¡ Fe31 1 e2 ) 1 14H 1 Cr2O22 7 1 6e2 ¡ 2Cr31 1 7H2O 6Fe21 1 14H1 1 Cr2O22 7 1 6e 2 ¡ 6Fe31 1 2Cr31 1 7H2O 1 6e2 The electrons on both sides cancel, and we are left with the balanced net ionic equation: 6Fe21 1 14H1 1 Cr2O22 7 ¡ 6Fe31 1 2Cr31 1 7H2O Step 5: Verify that the equation contains the same type and numbers of atoms and the same charges on both sides of the equation. A final check shows that the resulting equation is “atomically” and “electrically” balanced. For reactions in a basic medium, we proceed through step 4 as if the reaction were carried out in an acidic medium. Then, for every H1 ion we add an equal num- ber of OH2 ions to both sides of the equation. Where H1 and OH2 ions appear on the same side of the equation, we combine the ions to give H2O. Example 18.1 illus- trates this procedure. Example 18.1 Write a balanced ionic equation to represent the oxidation of iodide ion (I2) by permanganate ion (MnO2 4 ) in basic solution to yield molecular iodine (I2) and manganese(IV) oxide (MnO2). (Continued)  18.1 Redox Reactions 815 Strategy We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium. Solution Step 1: The unbalanced equation is MnO2 4 1 I 2 ¡ MnO2 1 I2 Step 2: The two half-reactions are 21 0 2 Oxidation: I ¡ I2 17 14 Reduction: MnO2 4 ¡ MnO2 Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We first balance the I atoms: 2I2 ¡ I2 To balance charges, we add two electrons to the right-hand side of the equation: 2I2 ¡ I2 1 2e2 Reduction half-reaction: To balance the O atoms, we add two H2O molecules on the right: MnO2 4 ¡ MnO2 1 2H2O To balance the H atoms, we add four H1 ions on the left: MnO2 4 1 4H 1 ¡ MnO2 1 2H2O There are three net positive charges on the left, so we add three electrons to the same side to balance the charges: MnO2 1 4 1 4H 1 3e 2 ¡ MnO2 1 2H2O Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows: 3(2I2 ¡ I2 1 2e2 ) 2(MnO2 4 1 4H 1 3e2 ¡ MnO2 1 2H2O) 1 6I2 1 2MnO2 1 4 1 8H 1 6e 2 ¡ 3I2 1 2MnO2 1 4H2O 1 6e2 The electrons on both sides cancel, and we are left with the balanced net ionic equation: 6I2 1 2MnO2 4 1 8H 1 ¡ 3I2 1 2MnO2 1 4H2O This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H1 ion we need to add equal number of OH2 ions to both sides of the equation: 6I2 1 2MnO2 1 4 1 8H 1 8OH 2 ¡ 3I2 1 2MnO2 1 4H2O 1 8OH2 Finally, combining the H1 and OH2 ions to form water, we obtain 6I2 1 2MnO2 4 1 4H2O ¡ 3I2 1 2MnO2 1 8OH 2 (Continued) 816 Chapter 18 ■ Electrochemistry Step 5: A final check shows that the equation is balanced in terms of both atoms and Similar problems: 18.1, 18.2. charges. Practice Exercise Balance the following equation for the reaction in an acidic medium by the ion-electron method: Fe21 1 MnO2 4 ¡ Fe 31 1 Mn21 Review of Concepts For the following reaction in acidic solution, what is the coefficient for NO2 when the equation is balanced? Sn 1 NO23 ¡ SnO2 1 NO2 18.2 Galvanic Cells In Section 4.4 we saw that when a piece of zinc metal is placed in a CuSO4 solution, Zn is oxidized to Zn21 ions while Cu21 ions are reduced to metallic copper (see Figure 4.10): Zn(s) 1 Cu21 (aq) ¡ Zn21 (aq) 1 Cu(s) The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu21) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire). As the reaction progresses, it sets up a constant flow of elec- trons and hence generates electricity (that is, it produces electrical work such as driv- ing an electric motor). The experimental apparatus for generating electricity through the use of a spontane- Animation ous reaction is called a galvanic cell or voltaic cell, after the Italian scientists Luigi Galvanic Cells Galvani and Alessandro Volta, who constructed early versions of the device. Figure 18.1 Animation shows the essential components of a galvanic cell. A zinc bar is immersed in a ZnSO4 Current Generation from a Voltaic Cell solution, and a copper bar is immersed in a CuSO4 solution. The cell operates on the Animation principle that the oxidation of Zn to Zn21 and the reduction of Cu21 to Cu can be made The Cu/Zn Voltaic Cell to take place simultaneously in separate locations with the transfer of electrons between them occurring through an external wire. The zinc and copper bars are called electrodes. Alphabetically anode precedes cathode This particular arrangement of electrodes (Zn and Cu) and solutions (ZnSO4 and CuSO4) and oxidation precedes reduction. Therefore, anode is where oxidation is called the Daniell cell. By definition, the anode in a galvanic cell is the electrode at occurs and cathode is where reduction which oxidation occurs and the cathode is the electrode at which reduction occurs. takes place. For the Daniell cell, the half-cell reactions, that is, the oxidation and reduction reactions at the electrodes, are Half-cell reactions are similar to the Zn electrode (anode): Zn(s) ¡ Zn21 (aq) 1 2e2 half-reactions discussed earlier. Cu electrode (cathode): Cu21 (aq) 1 2e2 ¡ Cu(s) Note that unless the two solutions are separated from each other, the Cu21 ions will react directly with the zinc bar: Cu21 (aq) 1 Zn(s) ¡ Cu(s) 1 Zn21 (aq) and no useful electrical work will be obtained. To complete the electrical circuit, the solutions must be connected by a conducting Animation medium through which the cations and anions can move from one electrode compartment Operation of Voltaic Cell to the other. This requirement is satisfied by a salt bridge, which, in its simplest form, is an inverted U tube containing an inert electrolyte solution, such as KCl or NH4NO3, whose ions will not react with other ions in solution or with the electrodes (see Figure 18.1).  18.2 Galvanic Cells 817 e– e– Zinc Copper anode Cl– K+ cathode Salt bridge Cotton Zn2+ plugs Cu2⫹ SO42– SO42– ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn 2+ Zn Cu Zn is oxidized Cu2+ is reduced to Zn2+ at anode. Net reaction to Cu at cathode. Zn(s) Zn2+(aq) + 2e– Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 2e– + Cu2+(aq) Cu(s) Figure 18.1 A galvanic cell. The salt bridge (an inverted U tube) containing a KCl solution provides an electrically conducting medium between two solutions. The openings of the U tube are loosely plugged with cotton balls to prevent the KCl solution from flowing into the containers while allowing the anions and cations to move across. The lightbulb is lit as electrons flow externally from the Zn electrode (anode) to the Cu electrode (cathode). During the course of the overall redox reaction, electrons flow externally from the anode (Zn electrode) through the wire to the cathode (Cu electrode). In the solution, the cations (Zn21, Cu21, and K1) move toward the cathode, while the anions (SO22 2 4 and Cl ) move toward the anode. Without the salt bridge connecting the two solutions, the buildup of positive charge in the anode compartment (due to the formation of Zn21 ions) and neg- ative charge in the cathode compartment (created when some of the Cu21 ions are reduced to Cu) would quickly prevent the cell from operating. An electric current flows from the anode to the cathode because there is a differ- ence in electrical potential energy between the electrodes. This flow of electric current is analogous to that of water down a waterfall, which occurs because there is a differ- ence in gravitational potential energy, or the flow of gas from a high-pressure region to a low-pressure region. Experimentally, the difference in electrical potential between the anode and the cathode is measured by a voltmeter (Figure 18.2). The voltage across the electrodes of a galvanic cell is called the cell voltage, or cell potential. Another common term for cell voltage is the electromotive force or emf (E), which, despite its name, is a measure of voltage, not force. We will see that the voltage of a cell depends not only on the nature of the electrodes and the ions, but also on the concentrations of the ions and the temperature at which the cell is operated. The conventional notation for representing galvanic cells is the cell diagram. For the Daniell cell shown in Figure 18.1, if we assume that the concentrations of Zn21 and Cu21 ions are 1 M, the cell diagram is Zn(s) 0 Zn21 (1 M) 0 0 Cu21 (1 M) 0 Cu(s) 818 Chapter 18 ■ Electrochemistry Figure 18.2 Practical setup of the galvanic cell described in Figure 18.1. Note the U tube (salt bridge) connecting the two beakers. When the concentrations of ZnSO4 and CuSO4 are 1 molar (1 M) at 25°C, the cell voltage is Salt bridge 1.10 V. No current flows between the electrodes during a voltage measurement. The single vertical line represents a phase boundary. For example, the zinc electrode is a solid and the Zn21 ions (from ZnSO4) are in solution. Thus, we draw a line between Zn and Zn21 to show the phase boundary. The double vertical lines denote the salt bridge. By convention, the anode is written first, to the left of the double lines and the other components appear in the order in which we would encounter them in moving from the anode to the cathode. Review of Concepts Write the cell diagram for the following redox reaction, where the concentrations of the Fe21 and Al31 ions are both 1 M. 3Fe21 (aq) 1 2Al(s) ¡ 3Fe(s) 1 2Al31 (aq) 18.3 Standard Reduction Potentials When the concentrations of the Cu21 and Zn21 ions are both 1.0 M, we find that the The choice of an arbitrary reference for voltage or emf of the Daniell cell is 1.10 V at 25°C (see Figure 18.2). This voltage measuring electrode potential is analogous to choosing the surface of the ocean as must be related directly to the redox reactions, but how? Just as the overall cell reac- the reference for altitude, calling it zero tion can be thought of as the sum of two half-cell reactions, the measured emf of the meters, and then referring to any terrestrial altitude as being a certain number of cell can be treated as the sum of the electrical potentials at the Zn and Cu electrodes. meters above or below sea level. Knowing one of these electrode potentials, we could obtain the other by subtraction (from 1.10 V). It is impossible to measure the potential of just a single electrode, but if we arbitrarily set the potential value of a particular electrode at zero, we can use it to determine the relative potentials of other electrodes. The hydrogen electrode, shown H2 gas at in Figure 18.3, serves as the reference for this purpose. Hydrogen gas is bubbled into 1 atm a hydrochloric acid solution at 25°C. The platinum electrode has two functions. First, it provides a surface on which the dissociation of hydrogen molecules can take place: H2 ¡ 2H1 1 2e2 Second, it serves as an electrical conductor to the external circuit. Pt electrode Under standard-state conditions (when the pressure of H2 is 1 atm and the con- centration of the HCl solution is 1 M; see Table 17.2), the potential for the reduction 1 M HCl of H1 at 25°C is taken to be exactly zero: Figure 18.3 A hydrogen electrode 2H1 (1 M) 1 2e2 ¡ H2 (1 atm) E° 5 0 V operating under standard-state conditions. Hydrogen gas at The superscript “°” denotes standard-state conditions, and E° is the standard reduction 1 atm is bubbled through a 1 M HCl solution. The platinum electrode is potential, or the voltage associated with a reduction reaction at an electrode when part of the hydrogen electrode. all solutes are 1 M and all gases are at 1 atm. Thus, the standard reduction potential  18.3 Standard Reduction Potentials 819 Voltmeter Voltmeter 0.76 V 0.34 V Zn H2 gas at 1 atm H2 gas at 1 atm Cu Salt bridge Salt bridge Pt electrode Pt electrode 1 M ZnSO4 1 M HCl 1 M HCl 1 M CuSO4 Zinc electrode Hydrogen electrode Hydrogen electrode Copper electrode (a) (b) Figure 18.4 (a) A cell consisting of a zinc electrode and a hydrogen electrode. (b) A cell consisting of a copper electrode and a hydrogen electrode. Both cells are operating under standard-state conditions. Note that in (a) the SHE acts as the cathode, but in (b) it acts as the anode. As mentioned in Figure 18.2, no current flows between the electrodes during a voltage measurement. of the hydrogen electrode is defined as zero. The hydrogen electrode is called the standard hydrogen electrode (SHE). We can use the SHE to measure the potentials of other kinds of electrodes. For example, Figure 18.4(a) shows a galvanic cell with a zinc electrode and a SHE. In this case, the zinc electrode is the anode and the SHE is the cathode. We deduce this fact from the decrease in mass of the zinc electrode during the operation of the cell, which is consistent with the loss of zinc to the solution caused by the oxidation reaction: Zn(s) ¡ Zn21 (aq) 1 2e2 The cell diagram is Zn(s) 0 Zn21(1 M) 0 0 H1(1 M) 0 H2 (1 atm) 0 Pt(s) As mentioned earlier, the Pt electrode provides the surface on which the reduction takes place. When all the reactants are in their standard states (that is, H2 at 1 atm, H1 and Zn21 ions at 1 M), the emf of the cell is 0.76 V at 25°C. We can write the half-cell reactions as follows: Anode (oxidation): Zn(s) ¡ Zn21(1 M) 1 2e2 Cathode (reduction): 2H1(1 M) 1 2e2 ¡ H2 (1 atm) Overall: Zn(s) 1 2H1(1 M) ¡ Zn21(1 M) 1 H2 (1 atm) By convention, the standard emf of the cell, E8cell , which is composed of a contribu- tion from the anode and a contribution from the cathode, is given by E°cell 5 E°cathode 2 E°anode (18.1) where both E°cathode and E°anode are the standard reduction potentials of the electrodes. For the Zn-SHE cell, we write E°cell 5 E°H1/H2 2 E°Zn21/Zn 0.76 V 5 0 2 E°Zn21/Zn where the subscript H1/H2 means 2H1 1 2e2 S H2 and the subscript Zn21/Zn means Zn21 1 2e2 S Zn. Thus, the standard reduction potential of zinc, E°Zn21/Zn , is 20.76 V. The standard electrode potential of copper can be obtained in a similar fashion, by using a cell with a copper electrode and a SHE [Figure 18.4(b)]. In this case, the 820 Chapter 18 ■ Electrochemistry copper electrode is the cathode because its mass increases during the operation of the cell, as is consistent with the reduction reaction: Cu21 (aq) 1 2e2 ¡ Cu(s) The cell diagram is Pt(s) 0 H2 (1 atm) 0 H1 (1 M) 0 0 Cu21 (1 M) 0 Cu(s) and the half-cell reactions are Anode (oxidation): H2 (1 atm) ¡ 2H1(1 M) 1 2e2 Cathode (reduction): Cu21(1 M) 1 2e2 ¡ Cu(s) Overall: H2 (1 atm) 1 Cu21(1 M) ¡ 2H1(1 M) 1 Cu(s) Under standard-state conditions and at 25°C, the emf of the cell is 0.34 V, so we write E°cell 5 E°cathode 2 E°anode 0.34 V 5 E°Cu21/Cu 2 E°H1/H2 5 E°Cu21/Cu 2 0 In this case, the standard reduction potential of copper, E°Cu21/Cu , is 0.34 V, where the subscript means Cu21 1 2e2 S Cu. For the Daniell cell shown in Figure 18.1, we can now write Anode (oxidation): Zn(s) ¡ Zn21 (1 M) 1 2e2 Cathode (reduction): Cu21 (1 M) 1 2e2 ¡ Cu(s) Overall: Zn(s) 1 Cu21 (1 M) ¡ Zn21 (1 M) 1 Cu(s) The emf of the cell is E°cell 5 E°cathode 2 E°anode 5 E°Cu21/Cu 2 E°Zn21/Zn 5 0.34 V 2 (20.76 V) 5 1.10 V As in the case of DG° (p. 793), we can use the sign of E° to predict the extent of a redox reaction. A positive E° means the redox reaction will favor the formation of products at equilibrium. Conversely, a negative E° means that more reactants than products will be formed at equilibrium. We will examine the relationships among E°cell, DG°, and K later in this chapter. The activity series in Figure 4.16 is based Table 18.1 lists standard reduction potentials for a number of half-cell reactions. on data given in Table 18.1. By definition, the SHE has an E° value of 0.00 V. Below the SHE the negative standard reduction potentials increase, and above it the positive standard reduction potentials increase. It is important to know the following points about the table in calculations: 1. The E° values apply to the half-cell reactions as read in the forward (left to right) direction. 2. The more positive E° is, the greater the tendency for the substance to be reduced. For example, the half-cell reaction F2 (1 atm) 1 2e2 ¡ 2F2 (1 M) E° 5 2.87 V has the highest positive E° value among all the half-cell reactions. Thus, F2 is the strongest oxidizing agent because it has the greatest tendency to be reduced. At the other extreme is the reaction Li1 (1 M) 1 e2 ¡ Li(s) E° 5 23.05 V  18.3 Standard Reduction Potentials 821 Table 18.1 Standard Reduction Potentials at 25˚C* Half-Reaction E°(V) F2(g) 1 2e2 ¡ 2F2(aq) 12.87 O3(g) 1 2H1(aq) 1 2e2 ¡ O2(g) 1 H2O 12.07 Co31(aq) 1 e2 ¡ Co21(aq) 11.82 H2O2(aq) 1 2H1(aq) 1 2e2 ¡ 2H2O 11.77 PbO2(s) 1 4H1(aq) 1 SO22 2 4 (aq) 1 2e ¡ PbSO4(s) 1 2H2O 11.70 41 2 31 Ce (aq) 1 e ¡ Ce (aq) 11.61 MnO2 1 2 21 4 (aq) 1 8H (aq) 1 5e ¡ Mn (aq) 1 4H2O 11.51 Au31(aq) 1 3e2 ¡ Au(s) 11.50 Cl2(g) 1 2e2 ¡ 2Cl2(aq) 11.36 Cr2O722(aq) 1 14H1(aq) 1 6e2 ¡ 2Cr31(aq) 1 7H2O 11.33 MnO2(s) 1 4H1(aq) 1 2e2 ¡ Mn21(aq) 1 2H2O 11.23 O2(g) 1 4H1(aq) 1 4e2 ¡ 2H2O 11.23 Br2(l) 1 2e2 ¡ 2Br2(aq) 11.07 NO2 1 2 3 (aq) 1 4H (aq) 1 3e ¡ NO(g) 1 2H2O 10.96 21 2 2Hg (aq) 1 2e ¡ Hg21 2 (aq) 10.92 21 2 Hg2 (aq) 1 2e ¡ 2Hg(l) 10.85 Ag1(aq) 1 e2 ¡ Ag(s) 10.80 Fe31(aq) 1 e2 ¡ Fe21(aq) 10.77 O2(g) 1 2H1(aq) 1 2e2 ¡ H2O2(aq) 10.68 Increasing strength as oxidizing agent Increasing strength as reducing agent MnO2 2 2 4 (aq) 1 2H2O 1 3e ¡ MnO2(s) 1 4OH (aq) 10.59 2 2 I2(s) 1 2e ¡ 2I (aq) 10.53 O2(g) 1 2H2O 1 4e2 ¡ 4OH2(aq) 10.40 Cu21(aq) 1 2e2 ¡ Cu(s) 10.34 AgCl(s) 1 e2 ¡ Ag(s) 1 Cl2(aq) 10.22 SO22 1 2 4 (aq) 1 4H (aq) 1 2e ¡ SO2(g) 1 2H2O 10.20 21 2 1 Cu (aq) 1 e ¡ Cu (aq) 10.15 Sn41(aq) 1 2e2 ¡ Sn21(aq) 10.13 2H1(aq) 1 2e2 ¡ H2(g) 0.00 Pb21(aq) 1 2e2 ¡ Pb(s) 20.13 Sn21(aq) 1 2e2 ¡ Sn(s) 20.14 Ni21(aq) 1 2e2 ¡ Ni(s) 20.25 Co21(aq) 1 2e2 ¡ Co(s) 20.28 PbSO4(s) 1 2e2 ¡ Pb(s) 1 SO422(aq) 20.31 Cd21(aq) 1 2e2 ¡ Cd(s) 20.40 Fe21(aq) 1 2e2 ¡ Fe(s) 20.44 Cr31(aq) 1 3e2 ¡ Cr(s) 20.74 Zn21(aq) 1 2e2 ¡ Zn(s) 20.76 2H2O 1 2e2 ¡ H2(g) 1 2OH2(aq) 20.83 Mn21(aq) 1 2e2 ¡ Mn(s) 21.18 Al31(aq) 1 3e2 ¡ Al(s) 21.66 Be21(aq) 1 2e2 ¡ Be(s) 21.85 Mg21(aq) 1 2e2 ¡ Mg(s) 22.37 Na1(aq) 1 e2 ¡ Na(s) 22.71 Ca21(aq) 1 2e2 ¡ Ca(s) 22.87 Sr21(aq) 1 2e2 ¡ Sr(s) 22.89 Ba21(aq) 1 2e2 ¡ Ba(s) 22.90 K1(aq) 1 e2 ¡ K(s) 22.93 Li1(aq) 1 e2 ¡ Li(s) 23.05 *For all half-reactions the concentration is 1 M for dissolved species and the pressure is 1 atm for gases. These are the standard-state values. 822 Chapter 18 ■ Electrochemistry which has the most negative E° value. Thus, Li1 is the weakest oxidizing agent because it is the most difficult species to reduce. Conversely, we say that F2 is the weakest reducing agent and Li metal is the strongest reducing agent. Under standard-state conditions, the oxidizing agents (the species on the left-hand side of the half-reactions in Table 18.1) increase in strength from bottom to top and the reducing agents (the species on the right-hand side of the half-reactions) increase in strength from top to bottom. 3. The half-cell reactions are reversible. Depending on the conditions, any elec- trode can act either as an anode or as a cathode. Earlier we saw that the SHE is the cathode (H1 is reduced to H2) when coupled with zinc in a cell and that it becomes the anode (H2 is oxidized to H1) when used in a cell with copper. 4. Under standard-state conditions, any species on the left of a given half-cell reac- tion will react spontaneously with a species that appears on the right of any half-cell reaction located below it in Table 18.1. This principle is sometimes called the diagonal rule. In the case of the Daniell cell The diagonal red line shows that Cu21 is Cu21 (1 M) 1 2e2 ¡ Cu(s) E° 5 0.34 V the oxidizing agent and Zn is the reduc- ing agent. Zn21 (1 M) 1 2e2 ¡ Zn(s) E° 5 20.76 V We see that the substance on the left of the first half-cell reaction is Cu21 and the substance on the right in the second half-cell reaction is Zn. Therefore, as we saw earlier, Zn spontaneously reduces Cu21 to form Zn21 and Cu. 5. Changing the stoichiometric coefficients of a half-cell reaction does not affect the value of E° because electrode potentials are intensive properties. This means that the value of E° is unaffected by the size of the electrodes or the amount of solu- tions present. For example, I2 (s) 1 2e2 ¡ 2I2 (1 M) E° 5 0.53 V but E° does not change if we multiply the half-reaction by 2: 2I2 (s) 1 4e2 ¡ 4I2 (1 M) E° 5 0.53 V 6. Like DH, DG, and DS, the sign of E° changes but its magnitude remains the same when we reverse a reaction. As Examples 18.2 and 18.3 show, Table 18.1 enables us to predict the outcome of redox reactions under standard-state conditions, whether they take place in a gal- vanic cell, where the reducing agent and oxidizing agent are physically separated from each other, or in a beaker, where the reactants are all mixed together. Example 18.2 Predict what will happen if molecular bromine (Br2) is added to a solution containing NaCl and NaI at 25°C. Assume all species are in their standard states. Strategy To predict what redox reaction(s) will take place, we need to compare the standard reduction potentials of Cl2, Br2, and I2 and apply the diagonal rule. (Continued)  18.3 Standard Reduction Potentials 823 Solution From Table 18.1, we write the standard reduction potentials as follows: Cl2 (1 atm) 1 2e2 ¡ 2Cl2 (1 M) E° 5 1.36 V Br2 (l) 1 2e2 ¡ 2Br2 (1 M) E° 5 1.07 V I2 (s) 1 2e2 ¡ 2I2 (1 M) E° 5 0.53 V Applying the diagonal rule we see that Br2 will oxidize I2 but will not oxidize Cl2. Therefore, the only redox reaction that will occur appreciably under standard-state conditions is Oxidation: 2I2 (1 M) ¡ I2 (s) 1 2e2 Reduction: Br2 (l) 1 2e2 ¡ 2Br2 (1 M ) Overall: 2I2 (1 M) 1 Br2 (l) ¡ I2 (s) 1 2Br2 (1 M) Check We can confirm our conclusion by calculating E°cell. Try it. Note that the Na1 ions are inert and do not enter into the redox reaction. Similar problems: 18.14, 18.17. 21 Practice Exercise Can Sn reduce Zn (aq) under standard-state conditions? Example 18.3 A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag electrode in a 1.0 M AgNO3 solution. Calculate the standard emf of this cell at 25°C. Strategy At first it may not be clear how to assign the electrodes in the galvanic cell. From Table 18.1 we write the standard reduction potentials of Ag and Mg and apply the diagonal rule to determine which is the anode and which is the cathode. Solution The standard reduction potentials are Ag1 (1.0 M) 1 e2 ¡ Ag(s) E° 5 0.80 V Mg21 (1.0 M) 1 2e2 ¡ Mg(s) E° 5 22.37 V Applying the diagonal rule, we see that Ag1 will oxidize Mg: Anode (oxidation): Mg(s) ¡ Mg21 (1.0 M) 1 2e2 Cathode (reduction): 2Ag1 (1.0 M) 1 2e2 ¡ 2Ag(s) Overall: Mg(s) 1 2Ag1 (1.0 M) ¡ Mg21 (1.0 M) 1 2Ag(s) Note that in order to balance the overall equation we multiplied the reduction of Ag1 by 2. We can do so because, as an intensive property, E° is not affected by this procedure. We find the emf of the cell by using Equation (18.1) and Table 18.1: E°cell 5 E°cathode 2 E°anode 5 E°Ag1/Ag 2 E°Mg21/Mg 5 0.80 V 2 (22.37 V) 5 3.17 V Check The positive value of E° shows that the forward reaction is favored. Similar problems: 18.11, 18.12. Practice Exercise What is the standard emf of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution at 25°C? 824 Chapter 18 ■ Electrochemistry Review of Concepts Which of the following metals will react with (that is, be oxidized by) HNO3, but not with HCl: Cu, Zn, Ag? 18.4 Thermodynamics of Redox Reactions Our next step is to see how E°cell is related to thermodynamic quantities such as DG° and K. In a galvanic cell, chemical energy is converted to electrical energy to do electrical work such as running an electric motor. Electrical energy, in this case, is the product of the emf of the cell and the total electrical charge (in coulombs) that passes through the cell: electrical energy 5 coulombs 3 volts 5 joules The equality means that 1J51C31V The total charge is determined by the number of electrons that pass through the cell, so we have total charge 5 number of e2 3 charge of one e2 In general, it is more convenient to express the total charge in molar quantities. The charge of one mole of electrons is called the Faraday constant (F), after the English chemist and physicist Michael Faraday,† where In most calculations, we round the 1 F 5 6.022 3 1023 e2/mol e2 3 1.602 3 10219 C/e2 Faraday constant to 96,500 C/mol e2. 5 9.647 3 104 C/mol e2 Therefore, the total charge can now be expressed as nF, where n is the number of moles of electrons exchanged between the oxidizing agent and reducing agent in the overall redox equation for the electrochemical process. The measured emf (Ecell) is the maximum voltage the cell can achieve. Therefore, the electrical work done, wele, which is the maximum work that can be done (wmax), is given by the product of the total charge and the emf of the cell: wmax 5 wele 5 2nFEcell The sign convention for electrical work is The negative sign indicates that the electrical work is done by the system (galvanic the same as that for P-V work, discussed in Section 6.3. cell) on the surroundings. In Chapter 17 we defined free energy as the energy avail- able to do work. Specifically, the change in free energy (DG) represents the maximum amount of useful work that can be obtained in a reaction: ¢G 5 wmax 5 wele † Michael Faraday (1791–1867). English chemist and physicist. Faraday is regarded by many as the greatest experimental scientist of the nineteenth century. He started as an apprentice to a bookbinder at the age of 13, but became interested in science after reading a book on chemistry. Faraday invented the electric motor and was the first person to demonstrate the principle governing electrical generators. Besides making notable contributions to the fields of electricity and magnetism, Faraday also worked on optical activity, and discovered and named benzene.  18.4 Thermodynamics of Redox Reactions 825 Therefore, we can write ¢G 5 2nFEcell (18.2) For a spontaneous reaction, DG is negative. Because both n and F are positive quan- tities, it follows that Ecell must also be positive. For reactions in which reactants and products are in their standard states (1 M or 1 atm), Equation (18.2) becomes ¢G° 5 2nFE°cell (18.3) Now we can relate E°cell to the equilibrium constant (K ) of a redox reaction. In Section 17.5 we saw that the standard free-energy change DG° for a reaction is related to its equilibrium constant as follows [see Equation (17.14)]: ¢G° 5 2RT ln K If we combine Equations (17.14) and (18.3), we obtain 2nFE°cell 5 2RT ln K Solving for E°cell RT E°cell 5 ln K (18.4) nF When T 5 298 K, Equation (18.4) can be simplified by substituting for R and F: (8.314 J/K ? mol)(298 K) E°cell 5 ln K In calculations involving F, we sometimes n(96,500 J/V ? mol) omit the symbol e2. 0.0257 V or E°cell 5 ln K (18.5) n Alternatively, Equation (18.5) can be written using the base-10 logarithm of K: 0.0592 V E°cell 5 log K (18.6) n Thus, if any one of the three quantities DG°, K, or E°cell is known, the other two can be calculated using Equation (17.14), Equation (18.3), or Equation (18.4) (Figure 18.5). We summarize the relationships among DG°, K, and E°cell and characterize the spon- taneity of a redox reaction in Table 18.2. For simplicity, we sometimes omit the subscript “cell” in E° and E. Examples 18.4 and 18.5 apply Equations (18.3) and (18.5). E°cell Example 18.4 E °ce l c°el FE ll = nF –n _R_T Calculate the equilibrium constant for the following reaction at 25°C: °= _ ln K ΔG Sn(s) 1 2Cu21 (aq) Δ Sn21 (aq) 1 2Cu1 (aq) Strategy The relationship between the equilibrium constant K and the standard emf ΔG° = –RT lnK ΔG° K is given by Equation (18.5): E°cell 5 (0.0257 V/n)ln K. Thus, if we can determine the (Continued) Figure 18.5 Relationships among E °, K, and DG°. 826 Chapter 18 ■ Electrochemistry Table 18.2 Relationships Among DG°, K, and E°cell Reaction Under DG° K E°cell Standard-State Conditions Negative .1 Positive Favors formation of products. 0 51 0 Reactants and products are equally favored. Positive ,1 Negative Favors formation of reactants. standard emf, we can calculate the equilibrium constant. We can determine the E°cell of a hypothetical galvanic cell made up of two couples (Sn21/Sn and Cu21/Cu1) from the standard reduction potentials in Table 18.1. Solution The half-cell reactions are Anode (oxidation): Sn(s) ¡ Sn21 (aq) 1 2e2 Cathode (reduction): 2Cu (aq) 1 2e2 ¡ 2Cu1 (aq) 21 E°cell 5 E°cathode 2 E°anode 5 E°Cu21/Cu1 2 E°Sn21/Sn 5 0.15 V 2 (20.14 V) 5 0.29 V Equation (18.5) can be written nE° ln K 5 0.0257 V In the overall reaction we find n 5 2. Therefore, (2) (0.29 V) ln K 5 5 22.6 0.0257 V Similar problems: 18.23, 18.24. K 5 e22.6 5 7 3 109 Practice Exercise Calculate the equilibrium constant for the following reaction at 25°C: Fe21 (aq) 1 2Ag(s) Δ Fe(s) 1 2Ag1 (aq) Example 18.5 Calculate the standard free-energy change for the following reaction at 25°C: 2Au(s) 1 3Ca21 (1.0 M) ¡ 2Au31 (1.0 M) 1 3Ca(s) Strategy The relationship between the standard free-energy change and the standard emf of the cell is given by Equation (18.3): DG° 5 2nFE°cell. Thus, if we can determine E°cell, we can calculate DG°. We can determine the E°cell of a hypothetical galvanic cell made up of two couples (Au31/Au and Ca21/Ca) from the standard reduction potentials in Table 18.1. (Continued)  18.5 The Effect of Concentration on Cell Emf 827 Solution The half-cell reactions are Anode (oxidation): 2Au(s) ¡ 2Au31 (1.0 M) 1 6e2 Cathode (reduction): 3Ca (1.0 M) 1 6e2 ¡ 3Ca(s) 21 E°cell 5 E°cathode 2 E°anode 5 E°Ca21/Ca 2 E°Au31/Au 5 22.87 V 2 1.50 V 5 24.37 V Now we use Equation (18.3): ¢G° 5 2nFE° The overall reaction shows that n 5 6, so ¢G° 5 2(6) (96,500 J/V ? mol) (24.37 V) 5 2.53 3 106 J/mol 5 2.53 3 103 kJ/mol Check The large positive value of DG° tells us that the reaction favors the reactants at equilibrium. The result is consistent with the fact that E° for the galvanic cell is negative. Similar problem: 18.26. Practice Exercise Calculate DG° for the following reaction at 25°C: 2Al31 (aq) 1 3Mg(s) Δ 2Al(s) 1 3Mg21 (aq) Review of Concepts Compare the ease of measuring the equilibrium constant of a reaction electrochemically with that by chemical means in general [see Equation (17.14)]. 18.5 The Effect of Concentration on Cell Emf So far we have focused on redox reactions in which reactants and products are in their standard states, but standard-state conditions are often difficult, and sometimes impossible, to maintain. However, there is a mathematical relationship between the emf of a galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions. This equation is derived next. The Nernst Equation Consider a redox reaction of the type aA 1 bB ¡ c C 1 d D From Equation (17.13), ¢G 5 ¢G° 1 RT ln Q Because DG 5 2nFE and DG° 5 2nFE°, the equation can be expressed as 2nFE 5 2nFE° 1 RT ln Q 828 Chapter 18 ■ Electrochemistry Dividing the equation through by 2nF, we get RT E 5 E° 2 ln Q (18.7) nF Note that the Nernst equation is used to where Q is the reaction quotient (see Section 14.4). Equation (18.7) is known as the calculate the cell voltage under non- standard-state conditions. Nernst† equation. At 298 K, Equation (18.7) can be rewritten as 0.0257 V E 5 E° 2 ln Q (18.8) n or, expressing Equation (18.8) using the base-10 logarithm of Q: 0.0592 V E 5 E° 2 log Q (18.9) n During the operation of a galvanic cell, electrons flow from the anode to the cathode, resulting in product formation and a decrease in reactant concentration. Thus, Q increases, which means that E decreases. Eventually, the cell reaches equilibrium. At equilibrium, there is no net transfer of electrons, so E 5 0 and Q 5 K, where K is the equilibrium constant. The Nernst equation enables us to calculate E as a function of reactant and prod- uct concentrations in a redox reaction. For example, for the Daniell cell in Figure 18.1 Zn(s) 1 Cu21 (aq) ¡ Zn21 (aq) 1 Cu(s) The Nernst equation for this cell at 25°C can be written as 0.0257 V [Zn21] Remember that concentrations of pure E 5 1.10 V 2 ln solids (and pure liquids) do not appear in 2 [Cu21] the expression for Q. If the ratio [Zn21]/[Cu21] is less than 1, ln ([Zn21]/[Cu21]) is a negative number, so that the second term on the right-hand side of the preceding equation is positive. Under this condition E is greater than the standard emf E°. If the ratio is greater than 1, E is smaller than E°. Example 18.6 illustrates the use of the Nernst equation. Example 18.6 Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) 1 Fe21 (aq) ¡ Co21 (aq) 1 Fe(s) given that [Co21] 5 0.15 M and [Fe21] 5 0.68 M. Strategy Because the reaction is not run under standard-state conditions (concentrations are not 1 M), we need Nernst’s equation [Equation (18.8)] to calculate the emf (E ) of a (Continued) † Walther Hermann Nernst (1864–1941). German chemist and physicist. Nernst’s work was mainly on elec- trolyte solution and thermodynamics. He also invented an electric piano. Nernst was awarded the Nobel Prize in Chemistry in 1920 for his contribution to thermodynamics.  18.5 The Effect of Concentration on Cell Emf 829 hypothetical galvanic cell and determine the spontaneity of the reaction. The standard emf (E°) can be calculated using the standard reduction potentials in Table 18.1. Remember that solids do not appear in the reaction quotient (Q) term in the Nernst equation. Note that 2 moles of electrons are transferred per mole of reaction, that is, n 5 2. Solution The half-cell reactions are Anode (oxidation): Co(s) ¡ Co21 (aq) 1 2e2 Cathode (reduction): Fe (aq) 1 2e2 ¡ Fe(s) 21 E°cell 5 E°cathode 2 E°anode 5 E°Fe21/ Fe 2 E°Co21/Co 5 20.44 V 2 (20.28 V) 5 20.16 V From Equation (18.8) we write 0.0257 V E 5 E° 2 ln Q n 0.0257 V [Co21] 5 E° 2 ln n [Fe21] 0.0257 V 0.15 5 20.16 V 2 ln 2 0.68 5 20.16 V 1 0.019 V 5 20.14 V Because E is negative, the reaction is not spontaneous in the direction written. Similar problems: 18.31, 18.32. Practice Exercise Will the following reaction occur spontaneously at 25°C, given that [Fe21] 5 0.60 M and [Cd21] 5 0.010 M? Cd(s) 1 Fe21 (aq) ¡ Cd21 (aq) 1 Fe(s) Now suppose we want to determine at what ratio of [Co21] to [Fe21] the reaction in Example 18.6 would become spontaneous. We can use Equation (18.8) as follows: 0.0257 V E 5 E° 2 ln Q n We first set E equal to zero, which corresponds to the equilibrium situation. When E 5 0, Q 5 K. 0.0257 V [Co21] 0 5 20.16 V 2 ln 2 [Fe21] [Co21] ln 5 212.5 [Fe21] [Co21] 5 e212.5 5 K [Fe21] or K 5 4 3 1026 Thus, for the reaction to be spontaneous, the ratio [Co21]/[Fe21] must be smaller than 4 3 1026 so that E would become positive. As Example 18.7 shows, if gases are involved in the cell reaction, their concen- trations should be expressed in atm. 830 Chapter 18 ■ Electrochemistry Example 18.7 Consider the galvanic cell shown in Figure 18.4(a). In a certain experiment, the emf (E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn21] 5 1.0 M and PH2 5 1.0 atm. Calculate the molar concentration of H1. Strategy The equation that relates standard emf and nonstandard emf is the Nernst equation. The overall cell reaction is Zn(s) 1 2H1 (? M) ¡ Zn21 (1.0 M) 1 H2 (1.0 atm) Given the emf of the cell (E), we apply the Nernst equation to solve for [H1]. Note that 2 moles of electrons are transferred per mole of reaction; that is, n 5 2. Solution As we saw earlier (p. 819), the standard emf (E°) for the cell is 0.76 V. From Equation (18.8) we write 0.0257 V E 5 E° 2 ln Q n 21 0.0257 V [Zn ]PH2 The concentrations in Q are divided by 5 E° 2 ln their standard-state value of 1 M and n [H1]2 pressure is divided by 1 atm. 0.0257 V (1.0) (1.0) 0.54 V 5 0.76 V 2 ln 2 [H1]2 0.0257 V 1 20.22 V 5 2 ln 1 2 2 [H ] 1 17.1 5 ln 1 2 [H ] 17.1 1 e 5 12 [H ] 1 1 [H ] 5 5 2 3 1024 M A 3 3 107 Check The fact that the nonstandard-state emf (E) is given in the problem means that not all the reacting species are in their standard-state concentrations. Thus, because both Similar problem: 18.34. Zn21 ions and H2 gas are in their standard states, [H1] is not 1 M. Practice Exercise What is the emf of a galvanic cell consisting of a Cd21/Cd half-cell and a Pt/H1/H2 half-cell if [Cd21] 5 0.20 M, [H1] 5 0.16 M, and PH2 5 0.80 atm? Review of Concepts Consider the following cell diagram: Mg(s) 0 MgSO4 (0.40 M) 0 0 NiSO4 (0.60 M) 0 Ni(s) Calculate the cell voltage at 25°C. How does the cell voltage change when (a) [Mg21] is decreased by a factor of 4 and (b) [Ni21] is decreased by a factor of 3? Ag—AgCl electrode Example 18.7 shows that a galvanic cell whose cell reaction involves H1 ions Thin-walled glass membrane can be used to measure [H1] or pH. The pH meter described in Section 15.3 is HCl solution based on this principle. However, the hydrogen electrode (see Figure 18.3) is nor- mally not employed in laboratory work because it is awkward to use. Instead, it is Figure 18.6 A glass electrode that is used in conjunction with a replaced by a glass electrode, shown in Figure 18.6. The electrode consists of a reference electrode in a pH meter. very thin glass membrane that is permeable to H1 ions. A silver wire coated with  18.5 The Effect of Concentration on Cell Emf 831 silver chloride is immersed in a dilute hydrochloric acid solution. When the elec- trode is placed in a solution whose pH is different from that of the inner solution, the potential difference that develops between the two sides of the membrane can be monitored using a reference electrode. The emf of the cell made up of the glass electrode and the reference electrode is measured with a voltmeter that is calibrated in pH units. Concentration Cells Because electrode potential depends on ion concentrations, it is possible to construct a galvanic cell from two half-cells composed of the same material but differing in ion concentrations. Such a cell is called a concentration cell. Consider a situation in which zinc electrodes are put into two aqueous solutions of zinc sulfate at 0.10 M and 1.0 M concentrations. The two solutions are connected by a salt bridge, and the electrodes are joined by a piece of wire in an arrangement like that shown in Figure 18.1. According to Le Châtelier’s principle, the tendency for the reduction Zn21 (aq) 1 2e2 ¡ Zn(s) increases with increasing concentration of Zn21 ions. Therefore, reduction should occur in the more concentrated compartment and oxidation should take place on the more dilute side. The cell diagram is Zn(s) 0 Zn21 (0.10 M) 0 0 Zn21 (1.0 M) 0 Zn(s) and the half-reactions are Oxidation: Zn(s) ¡ Zn21(0.10 M) 1 2e2 Reduction: Zn21(1.0 M) 1 2e2 ¡ Zn(s) Overall: Zn21(1.0 M) ¡ Zn21(0.10 M) The emf of the cell is 0.0257 V [Zn21]dil E 5 E° 2 ln 2 [Zn21]conc where the subscripts “dil” and “conc” refer to the 0.10 M and 1.0 M concentrations, respectively. The E° for this cell is zero (the same electrode and the same type of ions are involved), so 0.0257 V 0.10 E502 ln 2 1.0 5 0.0296 V The emf of concentration cells is usually small and decreases continually during the operation of the cell as the concentrations in the two compartments approach each other. When the concentrations of the ions in the two compartments are the same, E becomes zero, and no further change occurs. A biological cell can be compared to a concentration cell for the purpose of calculating its membrane potential. Membrane potential is the electrical potential that exists across the membrane of various kinds of cells, including muscle cells and nerve cells. It is responsible for the propagation of nerve impulses and heartbeat. A mem- brane potential is established whenever there are unequal concentrations of the same type of ion in the interior and exterior of a cell. For example, the concentrations of K1 ions in the interior and exterior of a nerve cell are 400 mM and 15 mM, respectively. 1 mM 5 1 3 1023 M. 832 Chapter 18 ■ Electrochemistry Treating the situation as a concentration cell and applying the Nernst equation for just one kind of ion, we can write 0.0257 V [K1]ex E 5 E° 2 ln 1 1 [K ]in 15 5 2(0.0257 V) ln 400 5 0.084 V or 84 mV where “ex” and “in” denote exterior and interior. Note that we have set E° 5 0 because the same type of ion is involved. Thus, an electrical potential of 84 mV exists across the membrane due to the unequal concentrations of K1 ions. 18.6 Batteries A battery is a galvanic cell, or a series of combined galvanic cells, that can be used as a source of direct electric current at a constant voltage. Although the operation of a battery is similar in principle to that of the galvanic cells described in Section 18.2, a battery has the advantage of being completely self-contained and requiring no aux- iliary components such as salt bridges. Here we will discuss several types of batteries that are in widespread use. The Dry Cell Battery The most common dry cell, that is, a cell without a fluid component, is the Leclanché + cell used in flashlights and transistor radios. The anode of the cell consists of a zinc can or container that is in contact with manganese dioxide (MnO2) and an electrolyte. The electrolyte consists of ammonium chloride and zinc chloride in water, to which Paper spacer starch is added to thicken the solution to a pastelike consistency so that it is less likely Moist paste of to leak (Figure 18.7). A carbon rod serves as the cathode, which is immersed in the ZnCl2 and NH4Cl electrolyte in the center of the cell. The cell reactions are Layer of MnO2 Graphite cathode Anode: Zn(s) ¡ Zn21(aq) 1 2e2 Cathode: 2NH1 4 (aq)1 2MnO2 (s) 1 2e2 ¡ Mn2O3 (s) 1 2NH3 (aq) Zinc anode 1H2O(l) Figure 18.7 Interior section of Overall: 1 21 Zn(s) 1 2NH 4 (aq) 1 2MnO2 (s) ¡ Zn (aq) 1 2NH3 (aq) a dry cell of the kind used in flashlights and transistor radios. 1 H2O(l) 1 Mn2O3 (s) Actually, the cell is not completely dry, as it contains a moist Actually, this equation is an oversimplification of a complex process. The voltage electrolyte paste. produced by a dry cell is about 1.5 V. The Mercury Battery The mercury battery is used extensively in medicine and electronic industries and is Cathode (steel) more expensive than the common dry cell. Contained in a stainless steel cylinder, the Insulation Anode (Zn can) mercury battery consists of a zinc anode (amalgamated with mercury) in contact with a strongly alkaline electrolyte containing zinc oxide and mercury(II) oxide (Figure 18.8). The cell reactions are Anode: Zn(Hg) 1 2OH2(aq) ¡ ZnO(s) 1 H2O(l) 1 2e2 Cathode: HgO(s) 1 H2O(l) 1 2e2 ¡ Hg(l) 1 2OH2(aq) Electrolyte solution containing KOH Overall: Zn(Hg) 1 HgO(s) ¡ ZnO(s) 1 Hg(l) and paste of Zn(OH)2 and HgO Figure 18.8 Interior section of Because there is no change in electrolyte composition during operation—the overall a mercury battery. cell reaction involves only solid substances—the mercury battery provides a more  18.6 Batteries 833 constant voltage (1.35 V) than the Leclanché cell. It also has a considerably higher capacity and longer life. These qualities make the mercury battery ideal for use in pacemakers, hearing aids, electric watches, and light meters. The Lead Storage Battery The lead storage battery commonly used in automobiles consists of six identical cells joined together in series. Each cell has a lead anode and a cathode made of lead dioxide (PbO2) packed on a metal plate (Figure 18.9). Both the cathode and the anode are immersed in an aqueous solution of sulfuric acid, which acts as the electrolyte. The cell reactions are Anode: Pb(s) 1 SO22 4 (aq) ¡ PbSO4(s) 1 2e 2 1 22 2 Cathode: PbO2(s) 1 4H (aq) 1 SO4 (aq) 1 2e ¡ PbSO4(s) 1 2H2O(l) Overall: Pb(s) 1 PbO2(s) 1 4H1(aq) 1 2SO22 4 (aq) ¡ 2PbSO4(s) 1 2H2O(l) Under normal operating conditions, each cell produces 2 V; a total of 12 V from the six cells is used to power the ignition circuit of the automobile and its other electrical systems. The lead storage battery can deliver large amounts of current for a short time, such as the time it takes to start up the engine. Unlike the Leclanché cell and the mercury battery, the lead storage battery is rechargeable. Recharging the battery means reversing the normal electrochemical reac- tion by applying an external voltage at the cathode and the anode. (This kind of process is called electrolysis, see p. 841.) The reactions that replenish the original materials are PbSO4(s) 1 2e2 ¡ Pb(s) 1 SO22 4 (aq) PbSO4(s) 1 2H2O(l) ¡ PbO2(s) 1 4H1(aq) 1 SO22 4 (aq) 1 2e 2 Overall: 2PbSO4(s) 1 2H2O(l) ¡ Pb(s) 1 PbO2(s) 1 4H1(aq) 1 2SO22 4 (aq) The overall reaction is exactly the opposite of the normal cell reaction. Two aspects of the operation of a lead storage battery are worth noting. First, because the electrochemical reaction consumes sulfuric acid, the degree to which the battery has been discharged can be checked by measuring the density of the electro- lyte with a hydrometer, as is usually done at gas stations. The density of the fluid in a “healthy,” fully charged battery should be equal to or greater than 1.2 g/mL. Second, people living in cold climates sometimes have trouble starting their cars because the battery has “gone dead.” Thermodynamic calculations show that the emf of many galvanic cells decreases with decreasing temperature. However, for a lead storage Removable cap Figure 18.9 Interior section of Anode Cathode a lead storage battery. Under normal operating conditions, the – concentration of the sulfuric acid solution is about 38 percent by mass. + H2SO4 electrolyte Negative plates (lead grills filled with spongy lead) Positive plates (lead grills filled with PbO2) 834 Chapter 18 ■ Electrochemistry battery, the temperature coefficient is about 1.5 3 1024 V/°C; that is, there is a decrease in voltage of 1.5 3 1024 V for every degree drop in temperature. Thus, even allowing for a 40°C change in temperature, the decrease in voltage amounts to only 6 3 1023 V, which is about 6 3 1023 V 3 100% 5 0.05% 12 V of the operating voltage, an insignificant change. The real cause of a battery’s appar- ent breakdown is an increase in the viscosity of the electrolyte as the temperature decreases. For the battery to function properly, the electrolyte must be fully conduct- ing. However, the ions move much more slowly in a viscous medium, so the resistance of the fluid increases, leading to a decrease in the power output of the battery. If an apparently “dead battery” is warmed to near room temperature on a frigid day, it recov- ers its ability to deliver normal power. The Lithium-Ion Battery Figure 18.10 shows a schematic diagram of a lithium-ion battery. The anode is made of a conducting carbonaceous material, usually graphite, which has tiny spaces in its structure that can hold both Li atoms and Li1 ions. The cathode is made of a transi- tion metal oxide such as CoO2, which can also hold Li1 ions. Because of the high reactivity of the metal, nonaqueous electrolyte (organic solvent plus dissolved salt) must be used. During the discharge of the battery, the half-cell reactions are Anode (oxidation): Li(s) ¡ Li1 1 e2 Cathode (reduction): Li 1 CoO2 1 e2 ¡ LiCoO2 (s) 1 Overall: Li (s) 1 CoO2 ¡ LiCoO2 (s) Ecell 5 3.4 V The advantage of the battery is that lithium has the most negative standard reduc- tion potential (see Table 18.1) and hence the greatest reducing strength. Furthermore, lithium is the lightest metal so that only 6.941 g of Li (its molar mass) are needed to Figure 18.10 A lithium-ion battery. Lithium atoms (represented by green spheres) are embedded between sheets of graphene (a single layer of graphite), which serve as the anode. A metal oxide material (gray and red spheres) is the e− Anode (−) Cathode (+) e− cathode. During operation, Li1 ions migrate through an electrolyte solution from the anode to the cathode while electrons flow externally through the wire from e− e− the anode to the cathode to complete the circuit. e− e− e− e− e− e− e− Li Li+ + e− Li+ + MO2 + e− LiMO2  18.6 Batteries 835 produce 1 mole of electrons. A lithium-ion battery can be recharged literally hundreds of times without deterioration. These desirable characteristics make it suitable for use in cellular telephones, digital cameras, and laptop computers. Recent progress in the manufacture of electric and hybrid automobiles, and increasing demand for these vehicles, has in turn created an intense demand for lithium-ion batteries. The batteries used in most highway-rated electric vehicles and some power tools are lithium iron phosphate (LFP) batteries. The design of LFP batteries is functionally the same as that shown in Figure 18.10, except that the cathode is FePO4, and LiFePO4 is formed at the cathode as the battery is discharged. These batteries share many of the advantages of other lithium-ion batteries (low weight, greater tendency for the metal to become oxidized at the anode), but they also have the additional advantage of extremely high chemical and thermal stability. As such, LFP batteries can be recharged many times and withstand very high tem- peratures without significant decomposition, and they avoid the problems with fires caused by the arrays of conventional lithium-ion batteries used in early prototypes of electric vehicles. Some other advantages of LFP batteries include reduced envi- ronmental concerns and a greater ability to retain a charge compared to other batter- ies. LFP batteries have a somewhat lower energy density than traditional lithium-ion batteries, but that trade-off is considered acceptable in applications that require a more robust battery. Early LFP batteries suffered from poor conductivity, but that problem has been addressed by “doping” the batteries with compounds that improve the conductivity. The increasing demand for lithium caused by the rapidly growing battery market raises questions about the world supply of this important alkali metal. It is projected that over the next few years the demand for lithium will quickly outpace the supply, which predominantly comes from Chile, Argentina, and China. The discovery in 2010 of a massive lithium deposit in Afghanistan may help to address this growing demand. Review of Concepts How many Leclanché cells are contained in a 9-volt battery? Fuel Cells Fossil fuels are a major source of energy, but conversion of fossil fuel into electrical energy is a highly inefficient process. Consider the combustion of methane: CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l) 1 energy To generate electricity, heat produced by the reaction is first used to convert water to steam, which then drives a turbine that drives a generator. An appreciable fraction of the energy released in the form of heat is lost to the surroundings at each step; even the most efficient power plant converts only about 40 percent of the original chemical energy into electricity. Because combustion reactions are redox reactions, it is more desirable to carry them out directly by electrochemical means, thereby greatly increas- ing the efficiency of power production. This objective can be accomplished by a device known as a fuel cell, a galvanic cell that requires a continuous supply of reactants to keep functioning. In its simplest form, a hydrogen-oxygen fuel cell consists of an electrolyte solu- tion, such as potassium hydroxide solution, and two inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode compartments (Figure 18.11), where the following reactions take place: Anode: 2H2 (g) 1 4OH2 (aq) ¡ 4H2O(l) 1 4e2 Cathode: O2 (g) 1 2H2O(l) 1 4e2 ¡ 4OH2 (aq) A car powered by hydrogen fuel cells manufactured by General Overall: 2H2 (g) 1 O2 (g) ¡ 2H2O(l) Motors. 836 Chapter 18 ■ Electrochemistry Figure 18.11 A hydrogen-oxygen fuel cell. The Ni and NiO embedded in the porous carbon electrodes are electrocatalysts. e– e– Anode Cathode H2 O2 Porous carbon electrode Porous carbon electrode containing Ni containing Ni and NiO Hot KOH solution Oxidation Reduction 2H2 (g) + 4OH– (aq) 4H2 O(l ) + 4e – O2 (g) + 2H2O(l) + 4e – 4OH– (aq) The standard emf of the cell can be calculated as follows, with data from Table 18.1: E°cell 5 E°cathode 2 E°anode 5 0.40 V 2 (20.83 V) 5 1.23 V Thus, the cell reaction is spontaneous under standard-state conditions. Note that the reaction is the same as the hydrogen combustion reaction, but the oxidation and reduction are carried out separately at the anode and the cathode. Like platinum in the standard hydrogen electrode, the electrodes have a twofold function. They serve as electrical conductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, prior to electron trans- fer. They are electrocatalysts. Metals such as platinum, nickel, and rhodium are good electrocatalysts. In addition to the H2-O2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The half-cell reactions are Anode: C3H8 (g) 1 6H2O(l) ¡ 3CO2 (g) 1 20H1 (aq) 1 20e2 Cathode: 5O2 (g) 1 20H1 (aq) 1 20e2 ¡ 10H2O(l) Overall: C3H8 (g) 1 5O2 (g) ¡ 3CO2 (g) 1 4H2O (l) The overall reaction is identical to the burning of propane in oxygen. Unlike batteries, fuel cells do not store chemical energy. Reactants must be constantly resupplied, and products must be constantly removed from a fuel cell. In this respect, a fuel cell resembles an engine more than it does a battery. However, the fuel cell does not operate like a heat engine and therefore is not subject to the same kind of thermodynamic limitations in energy conversion (see the Chemistry in Action essay on p. 791). Properly designed fuel cells may be as much as 70 percent efficient, about twice as efficient as an internal combustion engine. In addition, fuel-cell generators are free of the noise, vibration, heat transfer, thermal pollution, and other problems normally Figure 18.12 A hydrogen- associated with conventional power plants. Nevertheless, fuel cells are not yet in oxygen fuel cell once used in the widespread use. A major problem lies in the lack of cheap electrocatalysts able to space program. The pure water produced by the cell was function efficiently for long periods of time without contamination. The most success- consumed by the astronauts. ful application of fuel cells to date has been in space vehicles (Figure 18.12). CHEMISTRY in Action Bacteria Power U sable electricity generated from bacteria? Yes, it is possible. Scientists at the University of Massachusetts at Amherst have discovered an organism known as the Geobacter species and then flow externally to the graphite cathode. Here, the elec- tron acceptor is oxygen. So far, the current generated by such a fuel cell is small. that do exactly that. The ubiquitous Geobacter bacteria normally With proper development, however, it can someday be used to grow at the bottom of rivers or lakes. They get their energy by generate electricity for cooking, lighting, and powering electri- oxidizing the decaying organic matter to produce carbon dioxide. cal appliances and computers in homes, and in remote sensing The bacteria possess tentacles 10 times the length of their own devices. This is also a desirable way to clean the environment. size to reach the electron acceptors [mostly iron(III) oxide] in the Although the end product in the redox process is carbon diox- overall anaerobic redox process. ide, a greenhouse gas, the same product would be formed from The Massachusetts scientists constructed a bacterial fuel the normal decay of the organic wastes. cell using graphite electrodes. The Geobacter grow naturally on The oxidizing action of Geobacter has another beneficial the surface of the electrode, forming a stable “biofilm.” The effect. Tests show that uranium salts can replace iron(III) oxide overall reaction is as the electron acceptor. Thus, by adding acetate ions and the bacteria to the groundwater contaminated with uranium, it is CH3COO2 1 2O2 1 H1 ¡ 2CO2 1 2H2O possible to reduce the soluble uranium(VI) salts to the insoluble where the acetate ion represents organic matter. The electrons uranium(IV) salts, which can be readily removed before the are transferred directly from Geobacter to the graphite anode water ends up in households and farmlands. e– e– Fritted disc CH3COO– + 2H2O O2 + 4H+ + 4e – 2CO2 + 7H+ + 8e – 2H2O A bacterial fuel cell. The blowup shows the scanning electron micrograph of the bacteria growing on a graphite anode. The fritted disc allows the ions to pass between the compartments. 837 838 Chapter 18 ■ Electrochemistry 18.7 Corrosion Corrosion is the term usually applied to the deterioration of metals by an electrochemi- cal process. We see many examples of corrosion around us. Rust on iron, tarnish on silver, and the green patina formed on copper and brass are a few of them (Figure 18.13). Corrosion causes enormous damage to buildings, bridges, ships, and cars. The cost of metallic corrosion to the U.S. economy has been estimated to be well over 200 billion Figure 18.13 Examples of corrosion: (a) a rusted bridge, (b) a half-tarnished silver dish, and (c) the Statue of Liberty coated with patina before its restoration in 1986. (a) (b) (c)  18.7 Corrosion 839 dollars a year! This section discusses some of the fundamental processes that occur in corrosion and methods used to protect metals against it. By far the most familiar example of corrosion is the formation of rust on iron. Oxygen gas and water must be present for iron to rust. Although the reactions involved are quite complex and not completely understood, the main steps are believed to be as follows. A region of the metal’s surface serves as the anode, where oxidation occurs: Fe(s) ¡ Fe21 (aq) 1 2e2 The electrons given up by iron reduce atmospheric oxygen to water at the cathode, which is another region of the same metal’s surface: O2 (g) 1 4H1 (aq) 1 4e2 ¡ 2H2O(l) The overall redox reaction is 2Fe(s) 1 O2 (g) 1 4H1 (aq) ¡ 2Fe21 (aq) 1 2H2O(l) With data from Table 18.1, we find the standard emf for this process: E°cell 5 E°cathode 2 E°anode The positive standard emf means that the process will favor rust formation. 5 1.23 V 2 (20.44 V) 5 1.67 V Note that this reaction occurs in an acidic medium; the H1 ions are supplied in part by the reaction of atmospheric carbon dioxide with water to form H2CO3. The Fe21 ions formed at the anode are further oxidized by oxygen: 4Fe21 (aq) 1 O2 (g) 1 (4 1 2x)H2O(l) ¡ 2Fe2O3 ? xH2O(s) 1 8H1 (aq) This hydrated form of iron(III) oxide is known as rust. The amount of water associ- ated with the iron oxide varies, so we represent the formula as Fe2O3 ? xH2O. Figure 18.14 shows the mechanism of rust formation. The electrical circuit is completed by the migration of electrons and ions; this is why rusting occurs so rapidly in salt water. In cold climates, salts (NaCl or CaCl2) spread on roadways to melt ice and snow are a major cause of rust formation on automobiles. Metallic corrosion is not limited to iron. Consider aluminum, a metal used to make many useful things, including airplanes and beverage cans. Aluminum has a much greater tendency to oxidize than iron does; in Table 18.1 we see that Al has a more negative standard reduction potential than Fe. Based on this fact alone, we might expect to see airplanes slowly corrode away in rainstorms, and soda cans transformed into piles of corroded aluminum. These processes do not occur because the layer of insoluble aluminum oxide (Al2O3) that forms on its surface when the Air O2 Figure 18.14 The electro- chemical process involved in Water Rust rust formation. The H1 ions are supplied by H2CO3, which forms when CO2 dissolves in water. Fe2+ Fe3+ Iron e– Anode Cathode Fe(s) Fe2+(aq) + 2e – O2(g) + 4H +(aq) + 4e – 2H2O(l) Fe2+(aq) Fe3+(aq) + e – 840 Chapter 18 ■ Electrochemistry metal is exposed to air serves to protect the aluminum underneath from further cor- rosion. The rust that forms on the surface of iron, however, is too porous to protect the underlying metal. Coinage metals such as copper and silver also corrode, but much more slowly. Cu(s) ¡ Cu21 (aq) 1 2e2 Ag(s) ¡ Ag1 (aq) 1 e2 In normal atmospheric exposure, copper forms a layer of copper carbonate (CuCO3), a green substance also called patina, that protects the metal underneath from further corrosion. Likewise, silverware that comes into contact with foodstuffs develops a layer of silver sulfide (Ag2S). A number of methods have been devised to protect metals from corrosion. Most of these methods are aimed at preventing rust formation. The most obvious approach is to coat the metal surface with paint. However, if the paint is scratched, pitted, or dented to expose even the smallest area of bare metal, rust will form under the paint layer. The surface of iron metal can be made inactive by a process called passivation. A thin oxide layer is formed when the metal is treated with a strong oxidizing agent such as concentrated nitric acid. A solution of sodium chromate is often added to cooling systems and radiators to prevent rust formation. The tendency for iron to oxidize is greatly reduced when it is alloyed with certain other metals. For example, in stainless steel, an alloy of iron and chromium, a layer of chromium oxide forms that protects the iron from corrosion. An iron container can be covered with a layer of another metal such as tin or zinc. A “tin” can is made by applying a thin layer of tin over iron. Rust formation is prevented as long as the tin layer remains intact. However, once the surface has been scratched, rusting occurs rapidly. If we look up the standard reduction potentials, according to the diagonal rule, we find that iron acts as the anode and tin as the cathode in the corrosion process: Sn21 (aq) 1 2e2 ¡ Sn(s) E° 5 20.14 V Fe21 (aq) 1 2e2 ¡ Fe(s) E° 5 20.44 V The protective process is different for zinc-plated, or galvanized, iron. Zinc is more easily oxidized than iron (see Table 18.1): Zn21 (aq) 1 2e2 ¡ Zn(s) E° 5 20.76 V So even if a scratch exposes the iron, the zinc is still attacked. In this case, the zinc metal serves as the anode and the iron is the cathode. Cathodic protection is a process in which the metal that is to be protected from corrosion is made the cathode in what amounts to a galvanic cell. Figure 18.15 shows Figure 18.15 An iron nail that is cathodically protected by a piece of zinc strip does not rust in water, while an iron nail without such protection rusts readily.  18.8 Electrolysis 841 Figure 18.16 Cathodic protection of an iron storage tank (cathode) by magnesium, a more electropositive metal (anode). Because only the magnesium is Mg e– depleted in the electrochemical process, it is sometimes called the Iron storage tank sacrificial anode. Oxidation: Mg(s) Mg 2+(aq) + 2e– Reduction: O2(g) + 4H +(aq) + 4e– 2H2 O(l ) how an iron nail can be protected from rusting by connecting the nail to a piece of zinc. Without such protection, an iron nail quickly rusts in water. Rusting of underground iron pipes and iron storage tanks can be prevented or greatly reduced by connecting them to metals such as zinc and magnesium, which oxidize more readily than iron (Figure 18.16). Review of Concepts Which of the following metals can act as a sacrificial anode to protect iron? Sr, Ni, Pb, Co. The Chemistry in Action essay on p. 846 shows that dental filling discomfort can result from an electrochemical phenomenon. 18.8 Electrolysis In contrast to spontaneous redox reactions, which result in the conversion of chemical energy into electrical energy, electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. An electrolytic cell is an apparatus for carrying out electrolysis. The same principles underlie electrolysis and the processes that take place in galvanic cells. Here we will discuss three examples of electrolysis based on those principles. Then we will look at the quantitative aspects of electrolysis. Electrolysis of Molten Sodium Chloride In its molten state, sodium chloride, an ionic compound, can be electrolyzed to form sodium metal and chlorine. Figure 18.17(a) is a diagram of a Downs cell, which is used for large-scale electrolysis of NaCl. In molten NaCl, the cations and anions are the Na1 and Cl2 ions, respectively. Figure 18.17(b) is a simplified diagram showing the reactions that occur at the electrodes. The electrolytic cell contains a pair of elec- trodes connected to the battery. The battery serves as an “electron pump,” driving electrons to the cathode, where reduction occurs, and withdrawing electrons from the anode, where oxidation occurs. The reactions at the electrodes are Anode (oxidation): 2Cl2 (l) ¡ Cl2 (g) 1 2e2 Cathode (reduction): 2Na1 (l) 1 2e2 ¡ 2Na(l) Overall: 2Na1 (l) 1 2Cl2 (l) ¡ 2Na(l) 1 Cl2 (g) 842 Chapter 18 ■ Electrochemistry Figure 18.17 (a) A practical Cl2 gas Battery arrangement called a Downs cell NaCl + – for the electrolysis of molten NaCl (m.p. 5 801°C). The sodium e– e– metal formed at the cathodes is Molten NaCl Anode Cathode in the liquid state. Since liquid sodium metal is lighter than Liquid Na Liquid Na molten NaCl, the sodium floats to the surface, as shown, and is collected. Chlorine gas forms at the anode and is collected at the top. (b) A simplified diagram Molten NaCl showing the electrode reactions during the electrolysis of molten NaCl. The battery is needed to drive the nonspontaneous Iron cathode Iron cathode Oxidation Reduction reactions. 2Cl– Cl2(g) + 2e– 2Na+ + 2e– 2Na(l) Carbon anode (a) (b) This process is a major source of pure sodium metal and chlorine gas. Theoretical estimates show that the E° value for the overall process is about 24 V, which means that this is a nonspontaneous process. Therefore, a minimum of 4 V must be supplied by the battery to carry out the reaction. In practice, a higher voltage is necessary because of inefficiencies in the electrolytic process and because of overvolt- age, to be discussed shortly. Electrolysis of Water Water in a beaker under atmospheric conditions (1 atm and 25°C) will not spontane- ously decompose to form hydrogen and oxygen gas because the standard free-energy change for the reaction is a large positive quantity: 2H2O(l) ¡ 2H2 (g) 1 O2 (g) ¢G° 5 474.4 kJ/mol However, this reaction can be induced in a cell like the one shown in Figure 18.18. This electrolytic cell consists of a pair of electrodes made of a nonreactive metal, such as platinum, immersed in water. When the electrodes are connected to the battery, nothing happens because there are not enough ions in pure water to carry much of an electric current. (Remember that at 25°C, pure water has only 1 3 1027 M H1 ions and H2 O2 1 3 1027 M OH2 ions.) On the other hand, the reaction occurs readily in a 0.1 M H2SO4 solution because there are a sufficient number of ions to conduct electricity. Immedi- ately, gas bubbles begin to appear at both electrodes. Figure 18.19 shows the electrode reactions. The process at the anode is 2H2O(l) ¡ O2 (g) 1 4H1 (aq) 1 4e2 while at the cathode we have 2H1 (aq) 1 2e2 ¡ H2 (g) The overall reaction is given by Anode (oxidation): 2H2O(l) ¡ O2 (g) 1 4H1 (aq) 1 4e2 Figure 18.18 Apparatus for small-scale electrolysis of water. Cathode (reduction): 2[2H (aq) 1 2e2 ¡ H2 (g)] 1 The volume of hydrogen gas generated at the cathode is twice Overall: 2H2O(l) ¡ 2H2 (g) 1 O2 (g) that of oxygen gas generated at the anode. Note that no net H2SO4 is consumed.  18.8 Electrolysis 843 Battery Figure 18.19 A diagram showing + – the electrode reactions during the electrolysis of water. Note that the e– e– signs of the electrodes are Anode Cathode opposite to those of a galvanic cell. In a galvanic cell, the anode is negative because it supplies electrons to the external circuit. In an electrolytic cell, the anode is positive because electrons are withdrawn from it by the battery. Dilute H2SO4 solution Oxidation Reduction 2H2O(l) O2(g) + 4H+(aq) + 4e– 4H+(aq) + 4e– 2H2(g) Review of Concepts What is the minimum voltage needed for the electrolytic process shown above? Electrolysis of an Aqueous Sodium Chloride Solution This is the most complicated of the three examples of electrolysis considered here because aqueous sodium chloride solution contains several species that could be oxi- dized and reduced. The oxidation reactions that might occur at the anode are (1) 2Cl2 (aq) ¡ Cl2 (g) 1 2e2 (2) 2H2O(l) ¡ O2 (g) 1 4H1 (aq) 1 4e2 Referring to Table 18.1, we find Cl2 (g) 1 2e2 ¡ 2Cl2 (aq) E° 5 1.36 V O2 (g) 1 4H1 (aq) 1 4e2 ¡ 2H2O(l) E° 5 1.23 V The standard reduction potentials of (1) and (2) are not very different, but the Because Cl2 is more easily reduced than O2, it follows that it would be more difficult values do suggest that H2O should be preferentially oxidized at the anode. How- to oxidize Cl2 than H2O at the anode. ever, by experiment we find that the gas liberated at the anode is Cl2, not O2! In studying electrolytic processes, we sometimes find that the voltage required for a reaction is considerably higher than the electrode potential indicates. The overvolt- age is the difference between the electrode potential and the actual voltage required to cause electrolysis. The overvoltage for O2 formation is quite high. Therefore, under normal operating conditions Cl 2 gas is actually formed at the anode instead of O2. The reductions that might occur at the cathode are (3) 2H1 (aq) 1 2e2 ¡ H2 (g) E° 5     0.00 V (4) 2H2O(l) 1 2e2 ¡ H2 (g) 1 2OH2 (aq) E° 5 20.83 V (5) Na1 (aq) 1 e2 ¡ Na(s) E° 5 22.71 V Reaction (5) is ruled out because it has a very negative standard reduction potential. Reaction (3) is preferred over (4) under standard-state conditions. At a pH of 7 (as is the case for a NaCl solution), however, they are equally probable. We generally use 844 Chapter 18 ■ Electrochemistry (4) to describe the cathode reaction because the concentration of H1 ions is too low (about 1 3 1027 M) to make (3) a reasonable choice. Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are Anode (oxidation): 2Cl2 (aq) ¡ Cl2 (g) 1 2e2 Cathode (reduction): 2H2O(l) 1 2e2 ¡ H2 (g) 1 2OH2 (aq) Overall: 2H2O(l) 1 2Cl2 (aq) ¡ H2 (g) 1 Cl2 (g) 1 2OH2 (aq) As the overall reaction shows, the concentration of the Cl2 ions decreases during electrolysis and that of the OH2 ions increases. Therefore, in addition to H2 and Cl2, the useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of the electrolysis. Keep in mind the following from our analysis of electrolysis: cations are likely to be reduced at the cathode and anions are likely to be oxidized at the anode, and in aqueous solutions water itself may be oxidized and/or reduced. The outcome depends on the nature of other species present. Example 18.8 deals with the electrolysis of an aqueous solution of sodium sul- fate (Na2SO4). Example 18.8 An aqueous Na2SO4 solution is electrolyzed, using the apparatus shown in Figure 18.18. If the products formed at the anode and cathode are oxygen gas and hydrogen gas, respectively, describe the electrolysis in terms of the reactions at the electrodes. The SO422 ion is the conjugate base of Strategy Before we look at the electrode reactions, we should consider the the weak acid HSO2 22 4 (Ka 5 1.3 3 10 ). following facts: (1) Because Na2SO4 does not hydrolyze, the pH of the solution is However, the extent to which SO422 hydrolyzes is negligible. Also, the SO422 close to 7. (2) The Na1 ions are not reduced at the cathode and the SO422 ions are ion is not oxidized at the anode. not oxidized at the anode. These conclusions are drawn from the electrolysis of water in the presence of sulfuric acid and in aqueous sodium chloride solution, as discussed earlier. Therefore, both the oxidation and reduction reactions involve only water molecules. Solution The electrode reactions are Anode: 2H2O(l) ¡ O2 (g) 1 4H 1 (aq) 1 4e2 Cathode: 2H2O(l) 1 2e2 ¡ H2 (g) 1 2OH2 (aq) The overall reaction, obtained by doubling the cathode reaction coefficients and adding the result to the anode reaction, is 6H2O(l) ¡ 2H2 (g) 1 O2 (g) 1 4H1 (aq) 1 4OH2 (aq) If the H1 and OH2 ions are allowed to mix, then 4H1 (aq) 1 4OH2 (aq) ¡ 4H2O(l) and the overall reaction becomes Similar problem: 18.46. 2H2O(l) ¡ 2H2 (g) 1 O2 (g) Practice Exercise An aqueous solution of Mg(NO3)2 is electrolyzed. What are the gaseous products at the anode and cathode?  18.8 Electrolysis 845 Review of Concepts Complete the following electrolytic cell by labeling the electrodes and showing the half-cell reactions. Explain why the signs of the anode and cathode are opposite to those in a galvanic cell. Battery + – Molten MgCl2 Electrolysis has many important applications in industry, mainly in the extrac- tion and purification of metals. We will discuss some of these applications in Chapter 21. Current (amperes) and time (seconds) Quantitative Aspects of Electrolysis Product of The quantitative treatment of electrolysis was developed primarily by Faraday. He current and time observed that the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar Charge mass of the substance in question. For example, in the electrolysis of molten NaCl, in the cathode reaction tells us that one Na atom is produced when one Na1 ion accepts coulombs an electron from the electrode. To reduce 1 mole of Na1 ions, we must supply Avo- Divide by gadro’s number (6.02 3 1023) of electrons to the cathode. On the other hand, the the Faraday stoichiometry of the anode reaction shows that oxidation of two Cl2 ions yields one constant chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of Number 2 moles of electrons from the Cl2 ions to the anode. Similarly, it takes 2 moles of of electrons to reduce 1 mole of Mg21 ions and 3 moles of electrons to reduce 1 mole moles of electrons of Al31 ions: Use mole ratio in half-cell reaction Mg21 1 2e2 ¡ Mg Al31 1 3e2 ¡ Al Moles of substance reduced In an electrolysis experiment, we generally measure the current (in amperes, A) or oxidized that passes through an electrolytic cell in a given period of time. The relationship between charge (in coulombs, C) and current is Use molar mass or ideal gas equation 1C51A31s Grams or liters of that is, a coulomb is the quantity of electrical charge passing any point in the circuit product in 1 second when the current is 1 ampere. Figure 18.20 shows the steps involved in calculating the quantities of sub- Figure 18.20 Steps involved in stances produced in electrolysis. Let us illustrate the approach by considering calculating amounts of substances molten CaCl2 in an electrolytic cell. Suppose a current of 0.452 A is passed reduced or oxidized in electrolysis. CHEMISTRY in Action Dental Filling Discomfort I n modern dentistry one of the most commonly used mate- rials to fill decaying teeth is known as dental amalgam. (An amalgam is a substance made by combining mercury with another metal or metals.) Dental amalgam actually consists of three solid phases having stoichiometries ap- proximately corresponding to Ag2Hg3, Ag3Sn, and Sn8Hg. The standard reduction potentials for these solid phases are: Gold inlay Hg 221/Ag 2Hg 3, 0.85 V; Sn 21/Ag 3Sn, 20.05 V; and Sn 21/ Sn8Hg, 20.13 V. O2(g) + 4H+(aq) + 4e– 2H2O(l ) Anyone who bites a piece of aluminum foil (such as that used for wrapping candies) in such a way that the foil presses e– against a dental filling will probably experience a momentary Sn8Hg Sn2+ sharp pain. In effect, an electrochemical cell has been created in the mouth, with aluminum (E° 5 21.66 V) as the anode, the filling as the cathode, and saliva as the electrolyte. Contact be- tween the aluminum foil and the filling short-circuits the cell, Dental filling causing a weak current to flow between the electrodes. This current stimulates the sensitive nerve of the tooth, causing an unpleasant sensation. Another type of discomfort results when a less electroposi- Corrosion of a dental filling brought about by contact with a gold inlay. tive metal touches a dental filling. For example, if a filling makes contact with a gold inlay in a nearby tooth, corrosion of the filling will occur. In this case, the dental filling acts as the anode and the gold inlay as the cathode. Referring to the E° the mouth produces an unpleasant metallic taste. Prolonged cor- values for the three phases, we see that the Sn8Hg phase is most rosion will eventually result in another visit to the dentist for a likely to corrode. When that happens, release of Sn(II) ions in replacement filling. through the cell for 1.50 h. How much product will be formed at the anode and at the cathode? In solving electrolysis problems of this type, the first step is to determine which species will be oxidized at the anode and which species will be reduced at the cathode. Here the choice is straightforward because we only have Ca21 and Cl2 ions in molten CaCl2. Thus, we write the half- and overall cell reactions as Anode (oxidation): 2Cl2 (l) ¡ Cl2 (g) 1 2e2 Cathode (reduction): Ca (l) 1 2e2 ¡ Ca(l) 21 Overall: Ca21 (l) 1 2Cl2 (l) ¡ Ca(l) 1 Cl2 (g) The quantities of calcium metal and chlorine gas formed depend on the number of electrons that pass through the electrolytic cell, which in turn depends on current 3 time, or charge: 3600 s 1C ? C 5 0.452 A 3 1.50 h 3 3 5 2.44 3 103 C 1h 1A?s 846  18.8 Electrolysis 847 Because 1 mole e2 5 96,500 C and 2 mol e2 are required to reduce 1 mole of Ca21 ions, the mass of Ca metal formed at the cathode is calculated as follows: 1 mol e2 1 mol Ca 40.08 g Ca ? g Ca 5 2.44 3 103 C 3 3 3 5 0.507 g Ca 96,500 C 2 mol e2 1 mol Ca The anode reaction indicates that 1 mole of chlorine is produced per 2 mol e2 of electricity. Hence the mass of chlorine gas formed is 1 mol e2 1 mol Cl2 70.90 g Cl2 ? g Cl2 5 2.44 3 103 C 3 3 3 5 0.896 g Cl2 96,500 C 2 mol e2 1 mol Cl2 Example 18.9 applies this approach to the electrolysis in an aqueous solution. Example 18.9 A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases generated at STP. Strategy Earlier (see p. 842) we saw that the half-cell reactions for the process are Anode (oxidation): 2H2O(l) ¡ O2 (g) 1 4H1 (aq) 1 4e2 Cathode (reduction): 2[2H (aq) 1 2e2 ¡ H2 (g)] 1 Overall: 2H2O(l) ¡ 2H2 (g) 1 O2 (g) According to Figure 18.20, we carry out the following conversion steps to calculate the quantity of O2 in moles: current 3 time S coulombs S moles of e2 S moles of O2 Then, using the ideal gas equation we can calculate the volume of O2 in liters at STP. A similar procedure can be used for H2. Solution First we calculate the number of coulombs of electricity that pass through the cell: 3600 s 1C ? C 5 1.26 A 3 7.44 h 3 3 5 3.37 3 104 C 1h 1A?s Next, we convert number of coulombs to number of moles of electrons 1 mol e2 3.37 3 104 C 3 5 0.349 mol e2 96,500 C From the oxidation half-reaction we see that 1 mol O2 ∞ 4 mol e2. Therefore, the number of moles of O2 generated is 1 mol O2 0.349 mol e2 3 5 0.0873 mol O2 4 mol e2 The volume of 0.0873 mol O2 at STP is given by nRT V5 P (0.0873 mol) (0.0821 L ? atm/K ? mol) (273 K) 5 5 1.96 L 1 atm (Continued) 848 Chapter 18 ■ Electrochemistry The procedure for hydrogen is similar. To simplify, we combine the first two steps to calculate the number of moles of H2 generated: 1 mol e2 1 mol H2 3.37 3 104 C 3 3 5 0.175 mol H2 96,500 C 2 mol e2 The volume of 0.175 mol H2 at STP is given by nRT V5 P (0.175 mol) (0.0821 L ? atm/K ? mol) (273 K) 5 1 atm 5 3.92 L Check Note that the volume of H2 is twice that of O2 (see Figure 18.18), which is what we would expect based on Avogadro’s law (at the same temperature and pressure, Similar problem: 18.51. volume is directly proportional to the number of moles of gases). Practice Exercise A constant current is passed through an electrolytic cell containing molten MgCl2 for 18 h. If 4.8 3 105 g of Cl2 are obtained, what is the current in amperes? Review of Concepts In the electrolysis of molten CaCl2, a current of 1.24 A is passed through the cell for 2.0 h. What is the mass of Ca produced at the cathode? Key Equations E°cell 5 E°cathode 2 E°anode (18.1) Calculating the standard emf of a galvanic cell. ¢G 5 2nFEcell (18.2) Relating free-energy change to the emf of the cell. ¢G° 5 2nFE°cell (18.3) Relating the standard free-energy change to the standard emf of the cell. 0.0257 V E°cell 5 ln K (18.5) Relating the standard emf of the cell to the n equilibrium constant. 0.0592 V E°cell 5 log K (18.6) Relating the standard emf of the cell to the n equilibrium constant. 0.0257 V E 5 E° 2 ln Q (18.8) Relating the emf of the cell to the concentrations n under nonstandard state conditions. 0.0592 V E 5 E° 2 log Q (18.9) Relating the emf of the cell to the concentrations n under nonstandard state conditions. Summary of Facts & Concepts 1. Redox reactions involve the transfer of electrons. Equa- take place separately at the anode and cathode, re- tions representing redox processes can be balanced spectively, and the electrons flow through an exter- using the ion-electron method. nal circuit. 2. All electrochemical reactions involve the transfer of 4. The two parts of a galvanic cell are the half-cells, and electrons and therefore are redox reactions. the reactions at the electrodes are the half-cell reac- 3. In a galvanic cell, electricity is produced by a sponta- tions. A salt bridge allows ions to flow between the neous chemical reaction. Oxidation and reduction half-cells.  Questions & Problems 849 5. The electromotive force (emf) of a cell is the voltage 10. The Nernst equation gives the relationship between the difference between the two electrodes. In the exter- cell emf and the concentrations of the reactants and nal circuit, electrons flow from the anode to the products under nonstandard-state conditions. cathode in a galvanic cell. In solution, the anions 11. Batteries, which consist of one or more galvanic cells, move toward the anode and the cations move toward are used widely as self-contained power sources. Some the cathode. of the better-known batteries are the dry cell, such as the 6. The quantity of electricity carried by 1 mole of elec- Leclanché cell, the mercury battery, and the lead stor- trons is equal to 96,500 C. age battery used in automobiles. Fuel cells produce 7. Standard reduction potentials show the relative likeli- electrical energy from a continuous supply of reactants. hood of half-cell reduction reactions and can be used to 12. The corrosion of metals, such as the rusting of iron, is predict the products, direction, and spontaneity of redox an electrochemical phenomenon. reactions between various substances. 13. Electric current from an external source is used to drive 8. The decrease in free energy of the system in a spon- a nonspontaneous chemical reaction in an electrolytic taneous redox reaction is equal to the electrical cell. The amount of product formed or reactant con- work done by the system on the surroundings, or sumed depends on the quantity of electricity transferred DG 5 2nFE. at the electrodes. 9. The equilibrium constant for a redox reaction can be found from the standard electromotive force of a cell. Key Words Anode, p. 816 Electrochemistry, p. 813 Faraday constant (F), p. 824 Overvoltage, p. 843 Battery, p. 832 Electrolysis, p. 841 Fuel cell, p. 835 Standard emf (E°cell), p. 819 Cathode, p. 816 Electrolytic cell, p. 841 Galvanic cell, p. 816 Standard reduction Cell voltage, p. 817 Electromotive force (emf) Half-cell reaction, p. 816 potential, p. 818 Corrosion, p. 838 (E), p. 817 Nernst equation, p. 828 Questions & Problems • Problems available in Connect Plus (b) Bi1OH2 3 1 SnO22 2 ¡ SnO223 1 Bi (in Red numbered problems solved in Student Solutions Manual basic solution) (c) Cr2O227 1 C2O4 22 ¡ Cr31 1 CO2 (in Balancing Redox Equations acidic solution) Problems (d) ClO23 1 Cl 2 ¡ Cl2 1 ClO2 (in acidic solution) • 18.1 Balance the following redox equations by the ion- electron method: Galvanic Cells and Standard Emfs (a) H2O2 1 Fe21 ¡ Fe31 1 H2O (in acidic Review Questions solution) 18.3 Define the following terms: anode, cathode, cell (b) Cu 1 HNO3 ¡ Cu21 1 NO 1 H2O (in voltage, electromotive force, standard reduction acidic solution) potential. (c) CN2 1 MnO24 ¡ CNO2 1 MnO2 (in 18.4 Describe the basic features of a galvanic cell. Why basic solution) are the two components of the cell separated from (d) Br2 ¡ BrO2 2 3 1 Br (in basic solution) each other? (e) S2O3 1 I2 ¡ I 1 S4O22 22 2 6 (in acidic 18.5 What is the function of a salt bridge? What kind of solution) electrolyte should be used in a salt bridge? 18.2 Balance the following redox equations by the ion- 18.6 What is a cell diagram? Write the cell diagram for a electron method: galvanic cell consisting of an Al electrode placed in (a) Mn21 1 H2O2 ¡ MnO2 1 H2O (in basic a 1 M Al(NO3)3 solution and a Ag electrode placed solution) in a 1 M AgNO3 solution.  850 Chapter 18 ■ Electrochemistry 18.7 What is the difference between the half-reactions 18.20 The E°cell for the following cell is 1.54 V at 25°C discussed in redox processes in Chapter 4 and the U(s) 0 U31 (aq) 0 0 Ni21 (aq) 0 Ni(s) half-cell reactions discussed in Section 18.2? 18.8 After operating a Daniell cell (see Figure 18.1) for a Calculate the standard reduction potential for the few minutes, a student notices that the cell emf be- U31/U half-cell. gins to drop. Why? 18.9 Use the information in Table 2.1, and calculate the Spontaneity of Redox Reactions Faraday constant. Review Questions 18.10 Discuss the spontaneity of an electrochemical reac- 18.21 Write the equations relating DG° and K to the stan- tion in terms of its standard emf (E°cell ). dard emf of a cell. Define all the terms. 18.22 The E° value of one cell reaction is positive and that Problems of another cell reaction is negative. Which cell reac- • 18.11 Calculate the standard emf of a cell that uses the tion will proceed toward the formation of more Mg/Mg21and Cu/Cu21 half-cell reactions at 25°C. products at equilibrium? Write the equation for the cell reaction that occurs under standard-state conditions. Problems 18.12 Calculate the standard emf of a cell that uses Ag/ • 18.23 What is the equilibrium constant for the following Ag1 and Al/Al31 half-cell reactions. Write the reaction at 25°C? cell reaction that occurs under standard-state conditions. Mg(s) 1 Zn21 (aq) Δ Mg21 (aq) 1 Zn(s) 18.13 Predict whether Fe31 can oxidize I2 to I2 under • 18.24 The equilibrium constant for the reaction standard-state conditions. Sr(s) 1 Mg21 (aq) Δ Sr21 (aq) 1 Mg(s) 18.14 Which of the following reagents can oxidize H2O to O2(g) under standard-state conditions? H1(aq), is 2.69 3 1012 at 25°C. Calculate E° for a cell made Cl2(aq), Cl2(g), Cu21(aq), Pb21(aq), MnO24 (aq) (in up of Sr/Sr21 and Mg/Mg21 half-cells. acid). • 18.25 Use the standard reduction potentials to find the • 18.15 Consider the following half-reactions: equilibrium constant for each of the following reac- tions at 25°C: MnO2 1 4 (aq) 1 8H (aq) 1 5e 2 ¡ Mn21 (aq) 1 4H2O(l) (a) Br2 (l) 1 2I2 (aq) Δ 2Br2 (aq) 1 I2 (s) (b) 2Ce41 (aq) 1 2Cl2 (aq) Δ NO2 1 3 (aq) 1 4H (aq) 1 3e 2 ¡ Cl2 (g2 1 2Ce31 (aq) NO(g) 1 2H2O(l) (c) 5Fe21 (aq) 1 MnO2 1 4 (aq) 1 8H (aq) Δ Mn21 (aq) 1 4H2O(l) 1 5Fe31 (aq) Predict whether NO23 ions will oxidize Mn21 to MnO24 under standard-state conditions. • 18.26 Calculate DG° and Kc for the following reactions at 25°C: 18.16 Predict whether the following reactions would occur spontaneously in aqueous solution at 25°C. Assume (a) Mg(s) 1 Pb21 (aq) Δ Mg21 (aq) 1 Pb(s) that the initial concentrations of dissolved species (b) Br2 (l) 1 2I2 (aq) Δ 2Br2 (aq) 1 I2 (s) are all 1.0 M. (c) O2 (g2 1 4H1 (aq) 1 4Fe21 (aq) Δ (a) Ca(s) 1 Cd21 (aq) ¡ Ca21 (aq) 1 Cd(s) 2H2O(l) 1 4Fe31 (aq) (b) 2Br2 (aq) 1 Sn21 (aq) ¡ Br2 (l) 1 Sn(s) (d) 2Al(s) 1 3I2 (s) Δ 2Al31 (aq) 1 6I2 (aq) (c) 2Ag(s) 1 Ni21 (aq) ¡ 2Ag1 (aq) 1 Ni(s) • 18.27 Under standard-state conditions, what spontaneous (d) Cu1 (aq) 1 Fe31 (aq) ¡ reaction will occur in aqueous solution among the Cu21 (aq) 1 Fe21 (aq) ions Ce41, Ce31, Fe31, and Fe21? Calculate DG° and Kc for the reaction. • 18.17 Which species in each pair is a better oxidizing agent under standard-state conditions? (a) Br2 or Au31, • 18.28 Given that E° 5 0.52 V for the reduction (b) H2 or Ag1, (c) Cd21 or Cr31, (d) O2 in acidic Cu1 (aq) 1 e2 S Cu(s), calculate E°, DG°, and K media or O2 in basic media. for the following reaction at 25°C: 18.18 Which species in each pair is a better reducing agent 2Cu1 (aq) ¡ Cu21 (aq) 1 Cu(s) under standard-state conditions? (a) Na or Li, (b) H2 or I2, (c) Fe21 or Ag, (d) Br2 or Co21. The Effect of Concentration on Cell Emf 18.19 Consider the electrochemical reaction Sn21 1 X S Review Questions Sn 1 X21. Given that E°cell 5 0.14 V, what is the E° for the X21/X half-reaction? 18.29 Write the Nernst equation and explain all the terms.  Questions & Problems 851 18.30 Write the Nernst equation for the following pro- 18.40 Calculate the standard emf of the propane fuel cell cesses at some temperature T: discussed on p. 836 at 25°C, given that DG°f for (a) Mg(s) 1 Sn21(aq) ¡ Mg21(aq) 1 Sn(s) propane is 223.5 kJ/mol. (b) 2Cr(s) 1 3Pb21(aq) ¡ 2Cr31(aq) 1 3Pb(s) Corrosion Problems Review Questions • 18.31 What is the potential of a cell made up of Zn/Zn 21 18.41 Steel hardware, including nuts and bolts, is often and Cu/Cu21 half-cells at 25°C if [Zn21] 5 0.25 M coated with a thin plating of cadmium. Explain the and [Cu21] 5 0.15 M? function of the cadmium layer. • 18.32 Calculate E°, E, and DG for the following cell 18.42 “Galvanized iron” is steel sheet that has been coated reactions. with zinc; “tin” cans are made of steel sheet coated (a) Mg(s) 1 Sn21(aq) ¡ Mg21(aq) 1 Sn(s) with tin. Discuss the functions of these coatings and [Mg21] 5 0.045 M, [Sn21] 5 0.035 M the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface (b) 3Zn(s) 1 2Cr31(aq) ¡ 3Zn21(aq) 1 2Cr(s) of a galvanized iron sheet or a tin can. [Cr31] 5 0.010 M, [Zn21] 5 0.0085 M 18.43 Tarnished silver contains Ag2S. The tarnish can be • 18.33 Calculate the standard potential of the cell consisting removed by placing silverware in an aluminum pan of the Zn/Zn21 half-cell and the SHE. What will the containing an inert electrolyte solution, such as emf of the cell be if [Zn21] 5 0.45 M, PH2 5 2.0 atm, NaCl. Explain the electrochemical principle for this and [H1] 5 1.8 M? procedure. [The standard reduction potential for the • 18.34 What is the emf of a cell consisting of a Pb21/Pb half-cell reaction Ag2S(s) 1 2e2 S 2Ag(s) 1 half-cell and a Pt/H1/H2 half-cell if [Pb21] 5 0.10 M, S22(aq) is 20.71 V.] [H1] 5 0.050 M, and PH2 5 1.0 atm? 18.44 How does the tendency of iron to rust depend on the • 18.35 Referring to the arrangement in Figure 18.1, calcu- pH of solution? late the [Cu21]/[Zn21] ratio at which the following reaction is spontaneous at 25°C: Electrolysis Cu(s) 1 Zn21 (aq) ¡ Cu21 (aq) 1 Zn(s) Review Questions • 18.36 Calculate the emf of the following concentration 18.45 What is the difference between a galvanic cell (such cell: as a Daniell cell) and an electrolytic cell? Mg(s) 0 Mg21 (0.24 M) 0 0 Mg21 (0.53 M) 0 Mg(s) 18.46 Describe the electrolysis of an aqueous solution of KNO3. Batteries and Fuel Cells Problems Review Questions • 18.47 The half-reaction at an electrode is 18.37 Explain the differences between a primary galvanic Mg21 (molten) 1 2e2 ¡ Mg(s) cell—one that is not rechargeable—and a storage cell (for example, the lead storage battery), which is Calculate the number of grams of magnesium that can rechargeable. be produced by supplying 1.00 F to the electrode. 18.38 Discuss the advantages and disadvantages of fuel • 18.48 Consider the electrolysis of molten barium chloride, cells over conventional power plants in producing BaCl2. (a) Write the half-reactions. (b) How many electricity. grams of barium metal can be produced by supplying 0.50 A for 30 min? Problems 18.49 Considering only the cost of electricity, would it be cheaper to produce a ton of sodium or a ton of alumi- • 18.39 The hydrogen-oxygen fuel cell is described in Sec- num by electrolysis? tion 18.6. (a) What volume of H2(g), stored at 25°C at a pressure of 155 atm, would be needed to run an • 18.50 If the cost of electricity to produce magnesium by the electrolysis of molten magnesium chloride is electric motor drawing a current of 8.5 A for 3.0 h? $155 per ton of metal, what is the cost (in dollars) of (b) What volume (liters) of air at 25°C and 1.00 atm the electricity necessary to produce (a) 10.0 tons will have to pass into the cell per minute to run the of aluminum, (b) 30.0 tons of sodium, (c) 50.0 tons motor? Assume that air is 20 percent O2 by volume of calcium? and that all the O2 is consumed in the cell. The other components of air do not affect the fuel-cell reac- • 18.51 One of the half-reactions for the electrolysis of water is tions. Assume ideal gas behavior. 2H2O (l) ¡ O2 (g) 1 4H1 (aq) 1 4e2  852 Chapter 18 ■ Electrochemistry If 0.076 L of O2 is collected at 25°C and 755 mmHg, metal X was deposited in another cell (containing an how many moles of electrons had to pass through the aqueous XCl3 solution) in series with the AgNO3 solution? cell. Calculate the molar mass of X. • 18.52 How many moles of electrons are required to pro- • 18.62 One of the half-reactions for the electrolysis of duce (a) 0.84 L of O2 at exactly 1 atm and 25°C water is from aqueous H2SO4 solution; (b) 1.50 L of Cl2 at 750 mmHg and 20°C from molten NaCl; (c) 6.0 g of 2H1 (aq) 1 2e2 ¡ H2 (g) Sn from molten SnCl2? If 0.845 L of H2 is collected at 25°C and 782 mmHg, • 18.53 Calculate the amounts of Cu and Br2 produced in how many moles of electrons had to pass through 1.0 h at inert electrodes in a solution of CuBr2 by a the solution? current of 4.50 A. 18.63 A steady current of 10.0 A is passed through three • 18.54 In the electrolysis of an aqueous AgNO3 solution, electrolytic cells for 10.0 min. Calculate the mass of 0.67 g of Ag is deposited after a certain period of the metals formed if the solutions are 0.10 M time. (a) Write the half-reaction for the reduction AgNO3, 0.10 M Cu(NO3)2, and 0.10 M Au(NO3)3. of Ag1. (b) What is the probable oxidation half- 18.64 Industrially, copper metal can be purified electro- reaction? (c) Calculate the quantity of electricity lytically according to the following arrangement. used, in coulombs. The anode is made of the impure Cu electrode and • 18.55 A steady current was passed through molten the cathode is the pure Cu electrode. The electrodes CoSO4 until 2.35 g of metallic cobalt was pro- are immersed in a CuSO4 solution. (a) Write the duced. Calculate the number of coulombs of elec- half-cell reactions at the electrodes. (b) Calculate tricity used. the mass (in g) of Cu purified after passing a cur- • 18.56 A constant electric current flows for 3.75 h through rent of 20 A for 10 h. (c) Explain why impurities two electrolytic cells connected in series. One con- such as Zn, Fe, Au, and Ag are not deposited at the tains a solution of AgNO3 and the second a solution electrodes. of CuCl2. During this time 2.00 g of silver are depos- ited in the first cell. (a) How many grams of copper Battery are deposited in the second cell? (b) What is the cur- rent flowing, in amperes? Impure Pure • 18.57 What is the hourly production rate of chlorine gas copper copper (in kg) from an electrolytic cell using aqueous NaCl anode cathode electrolyte and carrying a current of 1.500 3 103A? The anode efficiency for the oxidation of Cl2 is 93.0 percent. 18.58 Chromium plating is applied by electrolysis to objects suspended in a dichromate solution, Cu2+ according to the following (unbalanced) half- SO42– reaction: Cr2O22 2 1 7 (aq) 1 e 1 H (aq) ¡ Cr(s) 1 H2O(l) How long (in hours) would it take to apply a Additional Problems chromium plating 1.0 3 1022 mm thick to a car 18.65 A Daniell cell consists of a zinc electrode in 1.00 L bumper with a surface area of 0.25 m2 in an electro- of 1.00 M ZnSO4 and a Cu electrode in 1.00 L of lytic cell carrying a current of 25.0 A? (The density 1.00 M CuSO4 at 25°C. A steady current of 10.0 A is of chromium is 7.19 g/cm3.) drawn from the cell. Calculate the Ecell after 1.00 h. • 18.59 The passage of a current of 0.750 A for 25.0 min Assume volumes to remain constant. deposited 0.369 g of copper from a CuSO4 solution. 18.66 A concentration cell is constructed having Cu elec- From this information, calculate the molar mass of trodes in two CuSO4 solutions A and B. At 25°C, the copper. osmotic pressures of the two solutions are 48.9 atm 18.60 A quantity of 0.300 g of copper was deposited from and 4.89 atm, respectively. Calculate the Ecell, a CuSO4 solution by passing a current of 3.00 A assuming no ion-pair formation. through the solution for 304 s. Calculate the value of • 18.67 For each of the following redox reactions, (i) write the Faraday constant. the half-reactions, (ii) write a balanced equation for • 18.61 In a certain electrolysis experiment, 1.44 g of Ag the whole reaction, (iii) determine in which direc- were deposited in one cell (containing an aqueous tion the reaction will proceed spontaneously under AgNO3 solution), while 0.120 g of an unknown standard-state conditions:  Questions & Problems 853 (a) H2(g) 1 Ni21(aq) ¡ H1(aq) 1 Ni(s) 18.74 Calcium oxalate (CaC2O4) is insoluble in water. (b) MnO24 (aq) 1 Cl2(aq) ¡ This property has been used to determine the amount Mn21(aq) 1 Cl2(g) (in acid solution) of Ca21 ions in blood. The calcium oxalate isolated (c) Cr(s) 1 Zn (aq) ¡ Cr31(aq) 1 Zn(s) 21 from blood is dissolved in acid and titrated against a standardized KMnO4 solution as described in Prob- • 18.68 The oxidation of 25.0 mL of a solution contain- lem 18.72. In one test, it is found that the calcium ing Fe21 requires 26.0 mL of 0.0250 M K2Cr2O7 oxalate isolated from a 10.0-mL sample of blood in acidic solution. Balance the following equa- requires 24.2 mL of 9.56 3 1024 M KMnO4 for tion and calculate the molar concentration titration. Calculate the number of milligrams of cal- of Fe21: cium per milliliter of blood. Cr2O22 7 1 Fe 21 1 H1 ¡ Cr31 1 Fe31 18.75 From the following information, calculate the solu- • 18.69 The SO2 present in air is mainly responsible for the bility product of AgBr: phenomenon of acid rain. The concentration of SO2 Ag1 (aq) 1 e2 ¡ Ag(s) E° 5 0.80 V can be determined by titrating against a standard AgBr(s) 1 e2 ¡ Ag(s) 1 Br2 (aq) E° 5 0.07 V permanganate solution as follows: 5SO2 1 2MnO2 4 1 2H2O ¡ • 18.76 Consider a galvanic cell composed of the SHE and a 5SO22 4 1 2Mn 21 1 4H1 half-cell using the reaction Ag1(aq) 1 e2 S Ag(s). (a) Calculate the standard cell potential. (b) What is Calculate the number of grams of SO2 in a sample of the spontaneous cell reaction under standard-state air if 7.37 mL of 0.00800 M KMnO4 solution are conditions? (c) Calculate the cell potential when required for the titration. [H1] in the hydrogen electrode is changed to • 18.70 A sample of iron ore weighing 0.2792 g was dis- (i) 1.0 3 1022 M and (ii) 1.0 3 1025 M, all other solved in an excess of a dilute acid solution. All the reagents being held at standard-state conditions. iron was first converted to Fe(II) ions. The solution (d) Based on this cell arrangement, suggest a design then required 23.30 mL of 0.0194 M KMnO4 for for a pH meter. oxidation to Fe(III) ions. Calculate the percent by • 18.77 A galvanic cell consists of a silver electrode in con- mass of iron in the ore. tact with 346 mL of 0.100 M AgNO3 solution and a • 18.71 The concentration of a hydrogen peroxide solution magnesium electrode in contact with 288 mL of can be conveniently determined by titration against 0.100 M Mg(NO3)2 solution. (a) Calculate E for the a standardized potassium permanganate solution in cell at 25°C. (b) A current is drawn from the cell an acidic medium according to the following unbal- until 1.20 g of silver have been deposited at the sil- anced equation: ver electrode. Calculate E for the cell at this stage of MnO2 21 operation. 4 1 H2O2 ¡ O2 1 Mn 18.78 Explain why chlorine gas can be prepared by elec- (a) Balance the above equation. (b) If 36.44 mL of a trolyzing an aqueous solution of NaCl but fluorine 0.01652 M KMnO4 solution are required to com- gas cannot be prepared by electrolyzing an aqueous pletely oxidize 25.00 mL of a H2O2 solution, calcu- solution of NaF. late the molarity of the H2O2 solution. • 18.72 Oxalic acid (H2C2O4) is present in many plants and • 18.79 Calculate the emf of the following concentration cell at 25°C: vegetables. (a) Balance the following equation in acid solution: Cu(s) 0 Cu21 (0.080 M) 0 0 Cu21 (1.2 M) 0 Cu(s) MnO24 1 C2O22 ¡ Mn 21 1 CO2 4 • 18.80 The cathode reaction in the Leclanché cell is (b) If a 1.00-g sample of H2C2O4 requires 24.0 mL given by of 0.0100 M KMnO4 solution to reach the equiva- 2MnO2 (s) 1 Zn21 (aq) 1 2e2 ¡ ZnMn2O4 (s) lence point, what is the percent by mass of H2C2O4 in the sample? If a Leclanché cell produces a current of 0.0050 18.73 Complete the following table. State whether the A, calculate how many hours this current supply cell reaction is spontaneous, nonspontaneous, or at will last if there are initially 4.0 g of MnO2 pres- equilibrium. ent in the cell. Assume that there is an excess of Zn21 ions. E DG Cell Reaction 18.81 Suppose you are asked to verify experimentally the electrode reactions shown in Example 18.8. In addi- .0 tion to the apparatus and the solution, you are also .0 given two pieces of litmus paper, one blue and the other red. Describe what steps you would take in this 50 experiment.  854 Chapter 18 ■ Electrochemistry 18.82 For a number of years it was not clear whether • 18.89 When an aqueous solution containing gold(III) salt mercury(I) ions existed in solution as Hg1 or as is electrolyzed, metallic gold is deposited at the Hg212 . To distinguish between these two possibili- cathode and oxygen gas is generated at the anode. ties, we could set up the following system: (a) If 9.26 g of Au is deposited at the cathode, calcu- late the volume (in liters) of O2 generated at 23°C Hg(l) 0 soln A 0 0 soln B 0 Hg(l) and 747 mmHg. (b) What is the current used if the where soln A contained 0.263 g mercury(I) nitrate electrolytic process took 2.00 h? per liter and soln B contained 2.63 g mercury(I) 18.90 In an electrolysis experiment, a student passes the nitrate per liter. If the measured emf of such a cell is same quantity of electricity through two electrolytic 0.0289 V at 18°C, what can you deduce about the cells, one containing a silver salt and the other a gold nature of the mercury(I) ions? salt. Over a certain period of time, she finds that 18.83 An aqueous KI solution to which a few drops of phe- 2.64 g of Ag and 1.61 g of Au are deposited at the nolphthalein have been added is electrolyzed using cathodes. What is the oxidation state of gold in the an apparatus like the one shown here: gold salt? 18.91 People living in cold-climate countries where there is plenty of snow are advised not to heat their garages in the winter. What is the electrochemical basis for this recommendation? • 18.92 Given that 2Hg21 (aq) 1 2e2 ¡ Hg21 2 (aq) E° 5 0.92 V Hg21 2 (aq) 1 2e 2 ¡ 2Hg(l) E° 5 0.85 V calculate DG° and K for the following process at 25°C: Hg21 21 2 (aq) ¡ Hg (aq) 1 Hg(l) Describe what you would observe at the anode and the cathode. (Hint: Molecular iodine is only (The preceding reaction is an example of a dispro- slightly soluble in water, but in the presence of I2 portionation reaction in which an element in one ions, it forms the brown color of I32 ions. See Prob- oxidation state is both oxidized and reduced.) lem 12.102.) 18.93 A galvanic cell with E°cell 5 0.30 V can be con- 18.84 A piece of magnesium metal weighing 1.56 g is structed using an Fe electrode in a 1.0 M Fe(NO3)2 placed in 100.0 mL of 0.100 M AgNO3 at 25°C. Cal- solution, and either a Sn electrode in a 1.0 M culate [Mg21] and [Ag1] in solution at equilibrium. Sn(NO3)2 solution, or a Cr electrode in a 1.0 M What is the mass of the magnesium left? The vol- Cr(NO 3) 3 solution, even though Sn 21/Sn and ume remains constant. Cr31/Cr have different standard reduction poten- 18.85 Describe an experiment that would enable you to tials. Explain. determine which is the cathode and which is the 18.94 Shown here is a galvanic cell connected to an elec- anode in a galvanic cell using copper and zinc trolytic cell. Label the electrodes (anodes and cath- electrodes. odes) and show the movement of electrons along 18.86 An acidified solution was electrolyzed using copper the wires and cations and anions in solution. For electrodes. A constant current of 1.18 A caused the simplicity, the salt bridge is not shown for the gal- anode to lose 0.584 g after 1.52 3 103 s. (a) What is vanic cell. the gas produced at the cathode and what is its vol- ume at STP? (b) Given that the charge of an electron is 1.6022 3 10219 C, calculate Avogadro’s number. Assume that copper is oxidized to Cu21 ions. • 18.87 In a certain electrolysis experiment involving Al31 ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many minutes did the elec- trolysis last? 18.88 Consider the oxidation of ammonia: 4NH3 (g) 1 3O2 (g) ¡ 2N2 (g) 1 6H2O(l) (a) Calculate the DG° for the reaction. (b) If this reaction were used in a fuel cell, what would the standard cell potential be? Galvanic cell Electrolytic cell  Questions & Problems 855 • 18.95 Fluorine (F2) is obtained by the electrolysis of liquid (a) Which is the strongest oxidizing agent and hydrogen fluoride (HF) containing potassium fluo- which is the strongest reducing agent? (b) Which ride (KF). (a) Write the half-cell reactions and the substances can be oxidized by B2? (c) Which sub- overall reaction for the process. (b) What is the pur- stances can be reduced by B2? (d) Write the overall pose of KF? (c) Calculate the volume of F2 (in liters) equation for a cell that delivers a voltage of 2.59 V collected at 24.0°C and 1.2 atm after electrolyzing under standard-state conditions. the solution for 15 h at a current of 502 A. 18.104 Consider a concentration cell made of the following • 18.96 A 300-mL solution of NaCl was electrolyzed for two compartments: Cl2(0.20 atm) 0 Cl2(1.0 M ) and 6.00 min. If the pH of the final solution was 12.24, Cl2(2.0 atm) 0 Cl2(1.0 M). Platinum is used as the calculate the average current used. inert electrodes. Draw a cell diagram for the cell and 18.97 Industrially, copper is purified by electrolysis. The calculate the emf of the cell at 25°C. impure copper acts as the anode, and the cathode is • 18.105 A silver rod and a SHE are dipped into a saturated made of pure copper. The electrodes are immersed aqueous solution of silver oxalate, Ag2C2O4, at 25°C. in a CuSO4 solution. During electrolysis, copper at The measured potential difference between the rod the anode enters the solution as Cu21 while Cu21 and the SHE is 0.589 V, the rod being positive. ions are reduced at the cathode. (a) Write half-cell Calculate the solubility product constant for silver reactions and the overall reaction for the electrolytic oxalate. process. (b) Suppose the anode was contaminated • 18.106 Zinc is an amphoteric metal; that is, it reacts with with Zn and Ag. Explain what happens to these im- both acids and bases. The standard reduction poten- purities during electrolysis. (c) How many hours tial is 21.36 V for the reaction will it take to obtain 1.00 kg of Cu at a current of 18.9 A? Zn(OH) 22 4 (aq) 1 2e 2 ¡ Zn(s) 1 4OH2 (aq) 18.98 An aqueous solution of a platinum salt is elec- trolyzed at a current of 2.50 A for 2.00 h. As a result, Calculate the formation constant (Kf) for the 9.09 g of metallic Pt are formed at the cathode. Cal- reaction culate the charge on the Pt ions in this solution. 18.99 Consider a galvanic cell consisting of a magne- Zn21 (aq) 1 4OH2 (aq) Δ Zn(OH2 22 4 (aq) sium electrode in contact with 1.0 M Mg(NO3)2 and a cadmium electrode in contact with 1.0 M 18.107 Use the data in Table 18.1 to determine whether or Cd(NO3)2. Calculate E° for the cell, and draw a not hydrogen peroxide will undergo disproportion- diagram showing the cathode, anode, and direc- ation in an acid medium: 2H2O2 S 2H2O 1 O2 . tion of electron flow. 18.108 The magnitudes (but not the signs) of the standard 18.100 A current of 6.00 A passes through an electrolytic reduction potentials of two metals X and Y are cell containing dilute sulfuric acid for 3.40 h. If the Y21 1 2e2 ¡ Y 0 E°0 5 0.34 V volume of O2 gas generated at the anode is 4.26 L X21 1 2e2 ¡ X 0 E° 0 5 0.25 V (at STP), calculate the charge (in coulombs) on an electron. where the 0 0 notation denotes that only the magni- 18.101 Gold will not dissolve in either concentrated nitric tude (but not the sign) of the E° value is shown. acid or concentrated hydrochloric acid. However, When the half-cells of X and Y are connected, elec- the metal does dissolve in a mixture of the acids trons flow from X to Y. When X is connected to a (one part HNO3 and three parts HCl by volume), SHE, electrons flow from X to SHE. (a) Are the E° called aqua regia. (a) Write a balanced equation for values of the half-reactions positive or negative? this reaction. (Hint: Among the products are HAuCl4 (b) What is the standard emf of a cell made up of and NO2.) (b) What is the function of HCl? X and Y? 18.102 Explain why most useful galvanic cells give voltages • 18.109 A galvanic cell is constructed as follows. One half- of no more than 1.5 to 2.5 V. What are the prospects cell consists of a platinum wire immersed in a solu- for developing practical galvanic cells with voltages tion containing 1.0 M Sn21 and 1.0 M Sn41; the of 5 V or more? other half-cell has a thallium rod immersed in a 18.103 The table here shows the standard reduction poten- solution of 1.0 M Tl1. (a) Write the half-cell reac- tials of several half-reactions: tions and the overall reaction. (b) What is the equi- librium constant at 25°C? (c) What is the cell Half-Reactions E8 (V) voltage if the T11 concentration is increased tenfold? A21 1 2e2 ¡ A 21.46 (E°Tl1/Tl 5 20.34 V.) B2 1 2e2 ¡ 2B2 0.33 18.110 Given the standard reduction potential for Au31 in C31 1 3e2 ¡ C 1.13 Table 18.1 and D1 1 e2 ¡ D 20.87 Au1 (aq) 1 e2 ¡ Au(s) E° 5 1.69 V  856 Chapter 18 ■ Electrochemistry answer the following questions. (a) Why does gold 18.118 Consider a Daniell cell operating under nonstandard- not tarnish in air? (b) Will the following dispropor- state conditions. Suppose that the cell’s reaction is tionation occur spontaneously? multiplied by 2. What effect does this have on each of the following quantities in the Nernst equation? 3Au1 (aq) ¡ Au31 (aq) 1 2Au(s) (a) E, (b) E°, (c) Q, (d) ln Q, and (e) n? (c) Predict the reaction between gold and fluorine gas. 18.119 An electrolysis cell was constructed similar to 18.111 The ingestion of a very small quantity of mercury is the one shown in Figure 18.18, except 0.1 M not considered too harmful. Would this statement MgCl2(aq) was used as the electrolyte solution. still hold if the gastric juice in your stomach were Under these conditions, a clear gas was formed at mostly nitric acid instead of hydrochloric acid? one electrode and a very pale green gas was 18.112 When 25.0 mL of a solution containing both Fe21 formed at the other electrode in roughly equal and Fe31 ions is titrated with 23.0 mL of 0.0200 M volumes. (a) What gases are formed at these elec- KMnO4 (in dilute sulfuric acid), all of the Fe21 ions trodes? (b) Write balanced half-reactions for each are oxidized to Fe31 ions. Next, the solution is treated electrode. Account for any deviation from the with Zn metal to convert all of the Fe31 ions to Fe21 normally expected results. ions. Finally, 40.0 mL of the same KMnO4 solution 18.120 Comment on whether F2 will become a stronger are added to the solution in order to oxidize the Fe21 oxidizing agent with increasing H1 concentration. ions to Fe31. Calculate the molar concentrations of 18.121 In recent years there has been much interest in elec- Fe21 and Fe31 in the original solution. tric cars. List some advantages and disadvantages of 18.113 Consider the Daniell cell in Figure 18.1. When electric cars compared to automobiles with internal viewed externally, the anode appears negative and combustion engines. the cathode positive (electrons are flowing from the 18.122 Calculate the pressure of H2 (in atm) required to anode to the cathode). Yet in solution anions are maintain equilibrium with respect to the following moving toward the anode, which means that it must reaction at 25°C: appear positive to the anions. Because the anode Pb(s) 1 2H1 (aq) Δ Pb21 (aq) 1 H2 (g) cannot simultaneously be negative and positive, give an explanation for this apparently contradic- Given that [Pb21] 5 0.035 M and the solution is tory situation. buffered at pH 1.60. 18.114 Use the data in Table 18.1 to show that the decom- 18.123 A piece of magnesium ribbon and a copper wire are position of H2O2 (a disproportionation reaction) is partially immersed in a 0.1 M HCl solution in a spontaneous at 25°C: beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at 2H2O2 (aq) ¡ 2H2O(l) 1 O2 (g) both the Mg and Cu surfaces. (a) Write equations 18.115 Consider two electrolytic cells A and B. Cell A con- representing the reactions occurring at the metals. tains a 0.20 M CoSO4 solution and platinum elec- (b) What visual evidence would you seek to show trodes. Cell B differs from cell A only in that cobalt that Cu is not oxidized to Cu21? (c) At some stage, metals are used as electrodes. In each case, a current of NaOH solution is added to the beaker to neutralize 0.20 A is passed through the cell for 1.0 h. (a) Write the HCl acid. Upon further addition of NaOH, a equations for the half-cell and overall cell reactions white precipitate forms. What is it? for these cells. (b) Calculate the products formed 18.124 The zinc-air battery shows much promise for electric (in grams) at the anode and cathode in each case. cars because it is lightweight and rechargeable: 18.116 A galvanic cell consists of a Mg electrode in a 1 M Mg(NO3)2 solution and another metal electrode X in a 1 M X(NO3)2 solution. Listed here are the E°cell values of four such galvanic cells. In each case, identify X from Table 18.1. (a) E°cell 5 2.12 V, (b) E°cell 5 2.24 V, (c) E°cell 5 1.61 V, (d) E°cell 5 1.93 V. • 18.117 The concentration of sulfuric acid in the lead-storage battery of an automobile over a period of time has decreased from 38.0 percent by mass (density 5 1.29 g/mL) to 26.0 percent by mass (1.19 g/mL). Assume the volume of the acid remains constant at 724 mL. (a) Calculate the total charge in coulombs supplied by the battery. (b) How long (in hours) The net transformation is Zn(s) 1 12O2 (g) S ZnO(s). will it take to recharge the battery back to the orig- (a) Write the half-reactions at the zinc-air electrodes inal sulfuric acid concentration using a current of and calculate the standard emf of the battery at 22.4 amperes? 25°C. (b) Calculate the emf under actual operating  Questions & Problems 857 conditions when the partial pressure of oxygen is energy from the battery. (Hint: Assume all of the 0.21 atm. (c) What is the energy density (measured lead will be used up in the electrochemical reaction as the energy in kilojoules that can be obtained from and refer to the electrode reactions on p. 833.) 1 kg of the metal) of the zinc electrode? (d) If a cur- (c) Calculate E°cell and DG° for the battery. rent of 2.1 3 105 A is to be drawn from a zinc-air 18.131 Use Equations (17.10) and (18.3) to calculate the battery system, what volume of air (in liters) would emf values of the Daniell cell at 25°C and 80°C. need to be supplied to the battery every second? As- Comment on your results. What assumptions are sume that the temperature is 25°C and the partial used in the derivation? (Hint: You need the thermody- pressure of oxygen is 0.21 atm. namic data in Appendix 3.) 18.125 Calculate E° for the reactions of mercury with (a) 1 M • 18.132 A construction company is installing an iron cul- HCl and (b) 1 M HNO3. Which acid will oxidize Hg vert (a long cylindrical tube) that is 40.0 m long to Hg 221 under standard-state conditions? Can you with a radius of 0.900 m. To prevent corrosion, the identify which test tube shown contains HNO3 and culvert must be galvanized. This process is carried Hg and which contains HCl and Hg? out by first passing an iron sheet of appropriate dimensions through an electrolytic cell containing Zn21 ions, using graphite as the anode and the iron sheet as the cathode. If the voltage is 3.26 V, what is the cost of electricity for depositing a layer 0.200 mm thick if the efficiency of the process is 95 percent? The electricity rate is $0.12 per kilo- watt hour (kWh), where 1 W 5 1 J/s and the den- sity of Zn is 7.14 g/cm3. • 18.133 A 9.00 3 102-mL 0.200 M MgI2 was electrolyzed. 18.126 Because all alkali metals react with water, it is not As a result, hydrogen gas was generated at the possible to measure the standard reduction potentials cathode and iodine was formed at the anode. The of these metals directly as in the case of, say, zinc. volume of hydrogen collected at 26°C and 779 An indirect method is to consider the following hy- mmHg was 1.22 3 103 mL. (a) Calculate the pothetical reaction charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a cur- Li1 (aq) 1 12H2 (g) ¡ Li(s) 1 H1 (aq) rent of 7.55 A was used? (c) A white precipitate Using the appropriate equation presented in this was formed in the process. What was it and what chapter and the thermodynamic data in Appendix 3, was its mass in grams? Assume the volume of the calculate E° for Li1 (aq) 1 e2 S Li(s) at 298 K. solution was constant. Compare your result with that listed in Table 18.1. • 18.134 Based on the following standard reduction poten- (See back endpaper for the Faraday constant.) tials: • 18.127 A galvanic cell using Mg/Mg21 and Cu/Cu21 half- Fe21 (aq) 1 2e2 ¡ Fe(s) E°1 5 20.44 V cells operates under standard-state conditions at Fe31 (aq) 1 e2 ¡ Fe21 (aq) E°2 5 0.77 V 25°C and each compartment has a volume of 218 mL. The cell delivers 0.22 A for 31.6 h. (a) How many calculate the standard reduction potential for the grams of Cu are deposited? (b) What is the [Cu21] half-reaction remaining? Fe31 (aq) 1 3e2 ¡ Fe(s) E°3 5 ? 18.128 Given the following standard reduction potentials, calculate the ion-product, Kw, for water at 25°C: 18.135 A galvanic cell is constructed by immersing a piece of copper wire in 25.0 mL of a 0.20 M CuSO4 solu- 2H1 (aq) 1 2e2 ¡ H2 (g) E° 5 0.00 V tion and a zinc strip in 25.0 mL of a 0.20 M ZnSO4 2H2O(l) 1 2e2 ¡ H2 (g) 1 2OH2 (aq) solution. (a) Calculate the emf of the cell at 25°C E° 5 20.83 V and predict what would happen if a small amount of concentrated NH3 solution were added to (i) the 18.129 Compare the pros and cons of a fuel cell, such as the CuSO4 solution and (ii) the ZnSO4 solution. Assume hydrogen-oxygen fuel cell, and a coal-fired power that the volume in each compartment remains con- station for generating electricity. stant at 25.0 mL. (b) In a separate experiment, 18.130 Lead storage batteries are rated by ampere hours, 25.0 mL of 3.00 M NH3 are added to the CuSO4 that is, the number of amperes they can deliver in an solution. If the emf of the cell is 0.68 V, calculate the hour. (a) Show that 1 A ? h 5 3600 C. (b) The lead formation constant (Kf) of Cu(NH3)21 4 . anodes of a certain lead-storage battery have a total 18.136 Calculate the equilibrium constant for the following mass of 406 g. Calculate the maximum theoretical reaction at 298 K: capacity of the battery in ampere hours. Explain why in practice we can never extract this much Zn(s) 1 Cu21 (aq) ¡ Zn21 (aq) 1 Cu(s) 858 Chapter 18 ■ Electrochemistry 18.137 To remove the tarnish (Ag2S) on a silver spoon, a reactions. Also label the signs (1 or 2) on the bat- student carried out the following steps. First, she tery terminals. (b) What is the minimum voltage to placed the spoon in a large pan filled with water so drive the reaction? (c) After the passage of 10.0 A the spoon was totally immersed. Next, she added a for 2.00 h the battery is replaced with a voltmeter few tablespoonful of baking soda (sodium bicar- and the electrolytic cell now becomes a galvanic bonate), which readily dissolved. Finally, she cell. Calculate Ecell. Assume volumes to remain con- placed some aluminum foil at the bottom of the pan stant at 1.00 L in each compartment. in contact with the spoon and then heated the solu- tion to about 80°C. After a few minutes, the spoon Battery was removed and rinsed with cold water. The tar- nish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the Salt bridge electrochemical basis for the procedure. (b) Adding NaCl instead of NaHCO3 would also work because both compounds are strong electrolytes. What is the added advantage of using NaHCO3? (Hint: Con- sider the pH of the solution.) (c) What is the pur- pose of heating the solution? (d) Some commercial tarnish removers contain a fluid (or paste) that is a dilute HCl solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disad- vantages of using this procedure compared to the 18.140 Fluorine is a highly reactive gas that attacks water to one described above. form HF and other products. Follow the procedure 18.138 The nitrite ion (NO2 2 ) in soil is oxidized to nitrate in Problem 18.126 to show how you can determine ion (NO2 3 ) by the bacteria Nitrobacter agilis in the indirectly the standard reduction for fluorine as presence of oxygen. The half-reduction reactions shown in Table 18.1. are 18.141 Show a sketch of a galvanic concentration cell. Each compartment consists of a Co electrode in a NO23 1 2H1 1 2e2 ¡ NO22 1 H2O E° 5 0.42 V Co(NO3)2 solution. The concentrations in the com- O2 1 4H1 1 4e2 ¡ 2H2O E° 5 1.23 V partments are 2.0 M and 0.10 M, respectively. Label Calculate the yield of ATP synthesis per mole of ni- the anode and cathode compartments. Show the trite oxidized. (Hint: See Section 17.7.) direction of electron flow. (a) Calculate the Ecell at 18.139 The diagram here shows an electrolytic cell consist- 25°C. (b) What are the concentrations in the com- ing of a Co electrode in a 2.0 M Co(NO3)2 solution partments when the Ecell drops to 0.020 V? Assume and a Mg electrode in a 2.0 M Mg(NO3)2 solution. volumes to remain constant at 1.00 L in each (a) Label the anode and cathode and show the half-cell compartment. Interpreting, Modeling & Estimating 18.142 The emf of galvanic cells varies with temperature 18.144 It has been suggested that a car can be powered from (either increases or decreases). Starting with Equa- the hydrogen generated by reacting aluminum soda tion (18.3), derive an equation that expresses E°cell in cans with a solution of lye (sodium hydroxide) terms of DH° and DS°. Predict whether E°cell will according to the following reaction: increase or decrease if the temperature of a Daniell cell increases. Assume both DH° and DS° to be tem- 2Al(s) 1 2OH2 (aq) 1 6H2O(l) ¡ perature independent. 2 Al(OH) 2 4 (aq) 1 3H2 (g) 18.143 A concentration cell ceases to operate when the con- centrations of the two cell compartments are equal. How many aluminum soda cans would be re- At this stage, is it possible to generate an emf from quired to generate the same amount of chemical the cell by adjusting another parameter without energy as contained in one tank of gasoline? Read changing the concentrations? Explain. the Chemistry in Action on aluminum recycling in  Answers to Practice Exercises 859 Chapter 21 (p. 950), and comment on the cost and was measured at several concentrations of Mn1(aq), environmental impact of powering a car with alu- giving the following plot. What is the value of n in minum cans. the half-reaction? 18.145 Estimate how long it would take to electroplate a 0.655 teaspoon with silver from a solution of AgNO3, assuming a constant current of 2 A. 0.650 Ecell (V) 18.146 The potential for a cell based on the standard hydro- 0.645 gen electrode and the half-reaction 0.640 Mn1 (aq) 1 ne2 ¡ M(s) 0.635 0 0.2 0.4 0.6 0.8 log (1/[Mn1]) Answers to Practice Exercises 18.1 5Fe21 1 MnO24 1 8H1 S 5Fe31 1 Mn21 1 4H2O. 18.2 No. 18.3 0.34 V. 18.4 1 3 10242. 18.5 DG° 5 24.1 3 102 kJ/mol. 18.6 Yes, E 5 10.01 V. 18.7 0.38 V. 18.8 Anode, O2; cathode, H2. 18.9 2.0 3 104 A. CHEMICAL M YS TERY Tainted Water† T he salesman was persuasive and persistent. “Do you realize what’s in your drinking water?” he asked Tom. Before Tom could answer, he continued: “Let me demonstrate.” First he filled a glass of water from the kitchen faucet, then he produced an electrical device that had a pair of probes and a lightbulb. It resembled a standard conductivity tester. He inserted the probes into the water and the bulb immediately beamed brightly. Next the salesman poured some water from a jar labeled “distilled water” into another glass. This time when he inserted the probes into the water, the bulb did not light. “Okay, can you explain the difference?” the salesman looked at Tom with a trium- phant smile. “Sure,” Tom began to recall an experiment he did in high school long ago, “The tap water contains minerals that caused . . . .” † Adapted from “Tainted Water,” by Joseph J. Hesse, CHEM MATTERS, February, 1988, p. 13. Copyright 1988 American Chemical Society. The precipitator with its electrodes immersed in tap water. Left: Before electrolysis has started. Right: 15 minutes after electrolysis commenced. 860  “Right on!” the salesman interrupted. “But I’m not sure if you realize how harmful the nation’s drinking water has become.” He handed Tom a booklet entitled The Miracle of Distilled Water. “Read the section called ‘Heart Conditions Can Result from Mineral Depos- its,’” he told Tom. “The tap water may look clear, although we know it contains dissolved minerals. What most people don’t realize is that it also contains other invisible substances that are harmful to our health. Let me show you.” The salesman proceeded to do another demonstration. This time he produced a device that he called a “precipitator,” which had two large electrodes attached to a black box. “Just look what’s in our tap water,” he said, while filling another large glassful from the faucet. The tap water appeared clean and pure. The salesman plugged the precipitator into the ac (alternating current) outlet. Within seconds, bubbles rose from both electrodes. The tap water took on a yellow hue. In a few minutes a brownish scum covered the surface of the water. After 15 minutes the glass of water was filled with a black- brown precipitate. When he repeated the experiment with distilled water, nothing happened. Tom was incredulous. “You mean all this gunk came from the water I drink?” “Where else?” beamed the salesman. “What the precipitator did was to bring out all the heavy metals and other undesirable substances. Now don’t worry. There is a remedy for this problem. My company makes a distiller that will convert tap water to distilled water, which is the only safe water to drink. For a price of $600 you will be able to produce distilled water for pennies with the distiller instead of paying 80 cents for a gallon of water from the supermarket.” Tom was tempted but decided to wait. After all, $600 is a lot to pay for a gadget that he only saw briefly. He decided to consult his friend Sarah, the chemistry teacher at the local high school, before making the investment. The salesman promised to return in a few days and left the precipitator with Tom so that he could do further testing. Chemical Clues 1. After Sarah examined the precipitator, she concluded that it was an electrolytic device consisting of what seemed like an aluminum electrode and an iron electrode. Because electrolysis cannot take place with ac (why not?), the precipitator must contain a rectifier, a device that converts ac to dc (direct current). Why does the water heat up so quickly during electrolysis? 2. From the brown color of the electrolysis product(s), deduce which metal acts as the cathode and which electrode acts as the anode. 3. Write all possible reactions at the anode and the cathode. Explain why there might be more than one type of reaction occurring at an electrode. 4. In analyzing the solution, Sarah detected aluminum. Suggest a plausible structure for the aluminum-containing ion. What property of aluminum causes it to dissolve in the solution? 5. Suggest two tests that would confirm Sarah’s conclusion that the precipitate originated from the electrodes and not from the tap water. 861  CHAPTER 19 Nuclear Chemistry The Large Hadron Collider (LHC) is the largest particle accelerator in the world. By colliding protons moving at nearly the speed of light, scientists hope to create conditions that existed right after the Big Bang. CHAPTER OUTLINE A LOOK AHEAD 19.1 The Nature of Nuclear  We begin by comparing nuclear reactions with ordinary chemical reactions. Reactions We learn to balance nuclear equations in terms of elementary particles like electrons, protons, neutrons, and alpha particles. (19.1) 19.2 Nuclear Stability  Next, we examine the stability of a nucleus in terms of the neutron-to- 19.3 Natural Radioactivity proton ratio. We use the Einstein mass-energy equation to calculate 19.4 Nuclear Transmutation nuclear binding energy. (19.2) 19.5 Nuclear Fission  We then study the decay of 238U as an example of natural radioactivity. We also see how radioactive decays, which are all first-order rate pro- 19.6 Nuclear Fusion cesses, are used to date objects. (19.3) 19.7 Uses of Isotopes  Nuclear transmutations are nuclear reactions induced by the bombardment 19.8 Biological Effects of a nucleus by particles such as neutrons, alpha particles, or other small of Radiation nuclei. Transuranium elements are all created in this way in a particle accelerator. (19.4)  In nuclear fission, a heavy nucleus splits into two smaller nuclei when bombarded with a neutron. The process releases large amounts of energy and additional neutrons, which can lead to a chain reaction if critical mass is present. Nuclear fission reactions are employed in atomic bombs and nuclear reactors. (19.5)  In nuclear fusion, two small nuclei fuse to yield a larger nucleus with the release of large amounts of energy. Nuclear fusion reactions are used in hydrogen or thermonuclear bombs, but nuclear fusion reactors for energy generation are still not commercially available. (19.6)  Isotopes, especially radioactive isotopes, find many applications in struc- tural determination and mechanistic studies as well as in medicine. (19.7)  The chapter concludes with a discussion of the biological effects of radiation. (19.8) 862  19.1 The Nature of Nuclear Reactions 863 N uclear chemistry is the study of reactions involving changes in atomic nuclei. This branch of chemistry began with the discovery of natural radioactivity by Antoine Becquerel and grew as a result of subsequent investigations by Pierre and Marie Curie and many others. Nuclear chemistry is very much in the news today. In addition to applications in the manufac- ture of atomic bombs, hydrogen bombs, and neutron bombs, even the peaceful use of nuclear energy has become controversial, in part because of safety concerns about nuclear power plants and also because of problems with radioactive waste disposal. In this chapter, we will study nuclear reactions, the stability of the atomic nucleus, radioactivity, and the effects of radiation on biological systems. 19.1 The Nature of Nuclear Reactions With the exception of hydrogen (11H), all nuclei contain two kinds of fundamental particles, called protons and neutrons. Some nuclei are unstable; they emit particles and/or electromagnetic radiation spontaneously (see Section 2.2). The name for this phenomenon is radioactivity. All elements having an atomic number greater than 83 are radioactive. For example, the isotope of polonium, polonium-210 (21084Po), decays spontaneously to 20682Pb by emitting an α particle. Another type of radioactivity, known as nuclear transmutation, results from the bombardment of nuclei by neutrons, protons, or other nuclei. An example of a nuclear transmutation is the conversion of atmospheric 147N to 146C and 11H, which results when the nitrogen isotope captures a neutron (from the sun). In some cases, heavier elements are synthesized from lighter elements. This type of transmutation occurs naturally in outer space, but it can also be achieved artificially, as we will see in Section 19.4. Radioactive decay and nuclear transmutation are nuclear reactions, which differ significantly from ordinary chemical reactions. Table 19.1 summarizes the differences. Balancing Nuclear Equations To discuss nuclear reactions in any depth, we need to understand how to write and balance the equations. Writing a nuclear equation differs somewhat from writing equa- tions for chemical reactions. In addition to writing the symbols for various chemical elements, we must also explicitly indicate protons, neutrons, and electrons. In fact, we must show the numbers of protons and neutrons present in every species in such an equation. The symbols for elementary particles are as follows: 1 1p or 11H 1 0n 0 21e or 210β 0 11eor 110β 4 2He or 42α proton neutron electron positron α particle Table 19.1 Comparison of Chemical Reactions and Nuclear Reactions Chemical Reactions Nuclear Reactions 1. Atoms are rearranged by the breaking and 1. Elements (or isotopes of the same elements) are converted forming of chemical bonds. from one to another. 2. Only electrons in atomic or molecular orbitals are 2. Protons, neutrons, electrons, and other elementary particles involved in the breaking and forming of bonds. may be involved. 3. Reactions are accompanied by absorption or release 3. Reactions are accompanied by absorption or release of of relatively small amounts of energy. tremendous amounts of energy. 4. Rates of reaction are influenced by temperature, 4. Rates of reaction normally are not affected by temperature, pressure, concentration, and catalysts. pressure, and catalysts. 864 Chapter 19 ■ Nuclear Chemistry In accordance with the notation used in Section 2.3, the superscript in each case denotes the mass number (the total number of neutrons and protons present) and the subscript is the atomic number (the number of protons). Thus, the “atomic number” of a proton is 1, because there is one proton present, and the “mass number” is also 1, because there is one proton but no neutrons present. On the other hand, the “mass number” of a neutron is 1, but its “atomic number” is zero, because there are no protons present. For the electron, the “mass number” is zero (there are neither protons nor neutrons present), but the “atomic number” is 21, because the electron possesses a unit negative charge. The symbol 210e represents an electron in or from an atomic orbital. The symbol 0 21β represents an electron that, although physically identical to any other electron, comes from a nucleus (in a decay process in which a neutron is converted to a proton and an electron) and not from an atomic orbital. The positron has the same mass as the electron, but bears a 11 charge. The α particle has two protons and two neutrons, A positron is the antiparticle of the electron. In 2007 physicists so its atomic number is 2 and its mass number is 4. prepared dipositronium (Ps2), In balancing any nuclear equation, we observe the following rules: which contains only electrons and • The total number of protons plus neutrons in the products and in the reactants positrons. The diagram here shows the central nuclear positions must be the same (conservation of mass number). containing positrons (red) • The total number of nuclear charges in the products and in the reactants must be surrounded by electrons (green). The Ps2 species exists for less the same (conservation of atomic number). than a nanosecond before the If we know the atomic numbers and mass numbers of all the species but one in a electron and positron annihilate each other with the emission nuclear equation, we can identify the unknown species by applying these rules, as of γ rays. shown in Example 19.1, which illustrates how to balance nuclear decay equations. Example 19.1 Balance the following nuclear equations (that is, identify the product X): 212 208 (a) 84Po ¡ 82Pb 1 X 137 137 (b) 55Cs ¡ 56Ba 1 X Keep in mind that nuclear equations are Strategy In balancing nuclear equations, note that the sum of atomic numbers and that often not balanced electrically. of mass numbers must match on both sides of the equation. Solution (a) The mass number and atomic number are 212 and 84, respectively, on the left-hand side and 208 and 82, respectively, on the right-hand side. Thus, X must have a mass number of 4 and an atomic number of 2, which means that it is an α particle. The balanced equation is 212 208 84 Po ¡ 82 Pb 1 42α (b) In this case, the mass number is the same on both sides of the equation, but the atomic number of the product is 1 more than that of the reactant. Thus, X must have a mass number of 0 and an atomic number of 21, which means that it is a β particle. The only way this change can come about is to have a neutron in the Cs nucleus transformed into a proton and an electron; that is, 10n ¡ 11p 1 210β (note that this process does not alter the mass number). Thus, the balanced equation is We use the 210β notation here because 137 137 the electron came from the nucleus. 55 Cs ¡ 56 Ba 1 210 β (Continued)  19.2 Nuclear Stability 865 Check Note that the equation in (a) and (b) are balanced for nuclear particles but not for electrical charges. To balance the charges, we would need to add two electrons on the right-hand side of (a) and express barium as a cation (Ba1) in (b). Similar problems: 19.7, 19.8. Practice Exercise Identify X in the following nuclear equation: 78 0 33As ¡ 21β 1X Review of Concepts Write a nuclear equation depicting the formation of a positron from a proton. 19.2 Nuclear Stability The nucleus occupies a very small portion of the total volume of an atom, but it contains most of the atom’s mass because both the protons and the neutrons reside there. In studying the stability of the atomic nucleus, it is helpful to know something about its density, because it tells us how tightly the particles are packed together. As a sample calculation, let us assume that a nucleus has a radius of 5 3 1023 pm and a mass of 1 3 10222 g. These figures correspond roughly to a nucleus containing 30 protons and 30 neutrons. Density is mass/volume, and we can calculate the volume from the known radius (the volume of a sphere is 43πr3 , where r is the radius of the sphere). First, we convert the pm units to cm. Then we calculate the density in g/cm3: 1 3 10212 m 100 cm r 5 5 3 1023 pm 3 3 5 5 3 10213 cm 1 pm 1m mass 1 3 10222 g 1 3 10222 g density 5 5 4 3 54 213 volume 3 πr 3 π(5 3 10 cm) 3 5 2 3 1014 g/cm3 This is an exceedingly high density. The highest density known for an element is To dramatize the almost incomprehensibly 22.6 g/cm3, for osmium (Os). Thus, the average atomic nucleus is roughly 9 3 1012 high density, it has been suggested that it is equivalent to packing the mass of all (or 9 trillion) times more dense than the densest element known! the world’s automobiles into one thimble. The enormously high density of the nucleus prompts us to wonder what holds the particles together so tightly. From Coulomb’s law we know that like charges repel and unlike charges attract one another. We would thus expect the protons to repel one another strongly, particularly when we consider how close they must be to each other. This indeed is so. However, in addition to the repulsion, there are also short-range attractions between proton and proton, proton and neutron, and neutron and neutron. The stability of any nucleus is determined by the difference between coulombic repul- sion and the short-range attraction. If repulsion outweighs attraction, the nucleus dis- integrates, emitting particles and/or radiation. If attractive forces prevail, the nucleus is stable. The principal factor that determines whether a nucleus is stable is the neutron- to-proton ratio (n/p). For stable atoms of elements having low atomic number, the n/p value is close to 1. As the atomic number increases, the neutron-to-proton ratios of the stable nuclei become greater than 1. This deviation at higher atomic numbers arises because a larger number of neutrons is needed to counteract the strong repulsion among the protons and stabilize the nucleus. The following rules are useful in predict- ing nuclear stability: 1. Nuclei that contain 2, 8, 20, 50, 82, or 126 protons or neutrons are generally more stable than nuclei that do not possess these numbers. For example, there 866 Chapter 19 ■ Nuclear Chemistry Number of Stable Isotopes with Even and Odd Numbers Table 19.2 of Protons and Neutrons Protons Neutrons Number of Stable Isotopes Odd Odd 4 Odd Even 50 Even Odd 53 Even Even 164 are 10 stable isotopes of tin (Sn) with the atomic number 50 and only 2 stable isotopes of antimony (Sb) with the atomic number 51. The numbers 2, 8, 20, 50, 82, and 126 are called magic numbers. The significance of these numbers for nuclear stability is similar to the numbers of electrons associated with the very stable noble gases (that is, 2, 10, 18, 36, 54, and 86 electrons). 2. Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of these particles (Table 19.2). 3. All isotopes of the elements with atomic numbers higher than 83 are radioactive. All isotopes of technetium (Tc, Z 5 43) and promethium (Pm, Z 5 61) are radioactive. Figure 19.1 shows a plot of the number of neutrons versus the number of protons in various isotopes. The stable nuclei are located in an area of the graph known as the belt of stability. Most radioactive nuclei lie outside this belt. Above the stability belt, the nuclei have higher neutron-to-proton ratios than those within the belt (for the same number of protons). To lower this ratio (and hence move Figure 19.1 Plot of neutrons versus protons for various stable isotopes, represented by dots. 120 The straight line represents the points at which the neutron- to-proton ratio equals 1. The shaded area represents the belt of stability. 100 80 Number of neutrons Belt of stability 60 Neutrons/Protons = 1 40 20 0 20 40 60 80 Number of protons  19.2 Nuclear Stability 867 down toward the belt of stability), these nuclei undergo the following process, called β-particle emission: 1 0n ¡ 11p 1 210β Beta-particle emission leads to an increase in the number of protons in the nucleus and a simultaneous decrease in the number of neutrons. Some examples are 14 14 0 6C ¡ 7 N 1 21β 40 40 0 19K ¡ 20Ca 1 21β 97 97 0 40Zr ¡ 41Nb 1 21β Below the stability belt the nuclei have lower neutron-to-proton ratios than those in the belt (for the same number of protons). To increase this ratio (and hence move up toward the belt of stability), these nuclei either emit a positron 1 1p ¡ 10n 1 0 11β or undergo electron capture. An example of positron emission is 38 38 0 19K ¡ 18Ar 1 11β Electron capture is the capture of an electron—usually a 1s electron—by the nucleus. The captured electron combines with a proton to form a neutron so that the atomic number decreases by one while the mass number remains the same. This process has the same net effect as positron emission: 37 18Ar 1 210e ¡ 37 17Cl We use 210e rather than 210β here because the electron came from an atomic orbital 55 26Fe 1 210e ¡ 55 25Mn and not from the nucleus. Review of Concepts The following isotopes are unstable. Use Figure 19.1 to predict whether they will undergo beta decay or positron emission. (a) 13B. (b) 188Au. Write a nuclear equation for each case. Nuclear Binding Energy A quantitative measure of nuclear stability is the nuclear binding energy, which is the energy required to break up a nucleus into its component protons and neutrons. This quantity represents the conversion of mass to energy that occurs during an exo- thermic nuclear reaction. The concept of nuclear binding energy evolved from studies of nuclear properties showing that the masses of nuclei are always less than the sum of the masses of the nucleons, which is a general term for the protons and neutrons in a nucleus. For example, the 199F isotope has an atomic mass of 18.9984 amu. The nucleus has 9 protons and 10 neutrons and therefore a total of 19 nucleons. Using the known masses of the 11H atom (1.007825 amu) and the neutron (1.008665 amu), we can carry out the following analysis. The mass of 9 11H atoms (that is, the mass of 9 protons and 9 electrons) is 9 3 1.007825 amu 5 9.070425 amu and the mass of 10 neutrons is 10 3 1.008665 amu 5 10.08665 amu 868 Chapter 19 ■ Nuclear Chemistry Therefore, the atomic mass of a 199F atom calculated from the known numbers of electrons, protons, and neutrons is 9.070425 amu 1 10.08665 amu 5 19.15708 amu There is no change in the electron’s mass which is larger than 18.9984 amu (the measured mass of 199F) by 0.1587 amu. because it is not a nucleon. The difference between the mass of an atom and the sum of the masses of its pro- tons, neutrons, and electrons is called the mass defect. Relativity theory tells us that the loss in mass shows up as energy (heat) given off to the surroundings. Thus, the forma- tion of 199 F is exothermic. Einstein’s mass-energy equivalence relationship states that This is the only equation listed in E 5 mc2 (19.1) Bartlett’s Familiar Quotations. where E is energy, m is mass, and c is the speed of light. We can calculate the amount of energy released by writing ¢E 5 ( ¢m)c2 (19.2) where ¢E and ¢m are defined as follows: ¢E 5 energy of product 2 energy of reactants ¢m 5 mass of product 2 mass of reactants Thus, the change in mass is given by ¢m 5 18.9984 amu 2 19.15708 amu 5 20.1587 amu Because 199F has a mass that is less than the mass calculated from the number of electrons and nucleons present, ¢m is a negative quantity. Consequently, ¢E is also a negative quantity; that is, energy is released to the surroundings as a result of the formation of the fluorine-19 nucleus. So we calculate ¢E as follows: ¢E 5 (20.1587 amu)(3.00 3 108 m/s) 2 5 21.43 3 1016 amu m2/s2 With the conversion factors 1 kg 5 6.022 3 1026 amu 1 J 5 1 kg m2/s2 we obtain amu ? m2 1.00 kg 1J When you apply Equation (19.2), remem- ¢E 5 a21.43 3 1016 2 b3a 26 b3a b ber to express the mass defect in kilo- s 6.022 3 10 amu 1 kg ? m2/s2 grams because 1 J 5 1 kg ? m2/s2. 5 22.37 3 10211 J The nuclear binding energy is a positive This is the amount of energy released when one fluorine-19 nucleus is formed from quantity. 9 protons and 10 neutrons. The nuclear binding energy of the nucleus is 2.37 3 10211 J, which is the amount of energy needed to decompose the nucleus into separate protons and neutrons. In the formation of 1 mole of fluorine nuclei, for instance, the energy released is ¢E 5 (22.37 3 10211 J)(6.022 3 1023/mol) 5 21.43 3 1013 J/mol 5 21.43 3 1010 kJ/mol The nuclear binding energy, therefore, is 1.43 3 1010 kJ for 1 mole of fluorine-19 nuclei, which is a tremendously large quantity when we consider that the enthalpies  19.2 Nuclear Stability 869 Figure 19.2 Plot of nuclear 56Fe binding energy per nucleon Nuclear binding energy per nucleon (J) 4He versus mass number. 1.5 × 10–12 238U 1.2 × 10–12 9 × 10–13 6 × 10–13 3 × 10–13 2H 0 20 40 60 80 100 120 140 160 180 200 220 240 260 Mass number of ordinary chemical reactions are of the order of only 200 kJ. The procedure we have followed can be used to calculate the nuclear binding energy of any nucleus. As we have noted, nuclear binding energy is an indication of the stability of a nucleus. However, in comparing the stability of any two nuclei we must account for the fact that they have different numbers of nucleons. For this reason it is more mean- ingful to use the nuclear binding energy per nucleon, defined as nuclear binding energy nuclear binding energy per nucleon 5 number of nucleons For the fluorine-19 nucleus, 2.37 3 10211 J nuclear binding energy per nucleon 5 19 nucleons 5 1.25 3 10212 J/nucleon The nuclear binding energy per nucleon enables us to compare the stability of all nuclei on a common basis. Figure 19.2 shows the variation of nuclear binding energy per nucleon plotted against mass number. As you can see, the curve rises rather steeply. The highest binding energies per nucleon belong to elements with intermedi- ate mass numbers—between 40 and 100—and are greatest for elements in the iron, cobalt, and nickel region (the Group 8B elements) of the periodic table. This means that the net attractive forces among the particles (protons and neutrons) are greatest for the nuclei of these elements. Nuclear binding energy and nuclear binding energy per nucleon are calculated for an iodine nucleus in Example 19.2. Example 19.2 The atomic mass of 127 53I is 126.9004 amu. Calculate the nuclear binding energy of this nucleus and the corresponding nuclear binding energy per nucleon. Strategy To calculate the nuclear binding energy, we first determine the difference between the mass of the nucleus and the mass of all the protons and neutrons, which gives us the mass defect. Next, we apply Equation (19.2) [¢E 5 (¢m)c2]. (Continued) 870 Chapter 19 ■ Nuclear Chemistry Solution There are 53 protons and 74 neutrons in the iodine nucleus. The mass of 53 11H atom is 53 3 1.007825 amu 5 53.41473 amu and the mass of 74 neutrons is 74 3 1.008665 amu 5 74.64121 amu 127 Therefore, the predicted mass for 53I is 53.41473 1 74.64121 5 128.05594 amu, and the mass defect is ¢m 5 126.9004 amu 2 128.05594 amu 5 21.1555 amu The energy released is ¢E 5 ( ¢m)c2 5 (21.1555 amu) (3.00 3 108 m/s) 2 5 21.04 3 1017 amu ? m2/s2 Let’s convert to a more familiar energy unit of joules. Recall that 1 J 5 1 kg ? m2/s2. Therefore, we need to convert amu to kg: amu ? m2 1.00 g 1 kg ¢E 5 21.04 3 1017 2 3 23 3 s 6.022 3 10 amu 1000 g 210 kg ? m2 210 5 21.73 3 10 5 21.73 3 10 J s2 The neutron-to-proton ratio is 1.4, which Thus, the nuclear binding energy is 1.73 3 10210 J . The nuclear binding energy per places iodine-127 in the belt of stability. nucleon is obtained as follows: 1.73 3 10210 J Similar problems: 19.21, 19.22. 5 1.36 3 10212 J/nucleon 127 nucleons Practice Exercise Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of 209 83Bi (208.9804 amu). Review of Concepts What is the change in mass (in kg) for the following reaction? CH4 (g) 1 2O2 (g) ¡ CO2 (g) 1 2H2O(l)    ¢H° 5 2890.4 kJ/mol 19.3 Natural Radioactivity Nuclei outside the belt of stability, as well as nuclei with more than 83 protons, tend to be unstable. The spontaneous emission by unstable nuclei of particles or electro- magnetic radiation, or both, is known as radioactivity. The main types of radiation are: α particles (or doubly charged helium nuclei, He21); β particles (or electrons); γ rays, which are very-short-wavelength (0.1 nm to 1024 nm) electromagnetic waves; positron emission; and electron capture. Animation The disintegration of a radioactive nucleus is often the beginning of a radioactive Radioactive Decay decay series, which is a sequence of nuclear reactions that ultimately result in the formation of a stable isotope. Table 19.3 shows the decay series of naturally occurring  19.3 Natural Radioactivity 871 Table 19.3 The Uranium Decay Series* 238 92 U 8888n ␣ 66 4.51  109 yr 66 g 90 Th 8888n ␤ 234 66 24.1 days 66 g 91 Pa 8888n ␤ 234 66 1.17 min 66 g 92 U 8888n ␣ 234 66 2.47  105 yr 66 g 90 Th 8888n ␣ 230 66 7.5  104 yr 66 g 88 Ra 8888n ␣ 226 66 1.60  103 yr 66 g 86 Rn 8888n ␣ 222 66 3.82 days 66 g 84Po 8888n ␣ ␤ m8888 218 888 3.05 min 0.04% 888 8n 8n 888n ␣ m8888 218 85At 8 214 82Pb 8888n ␤ 888 8 2s n 26.8 min 83Bi 8888n ␣ ␤ m8888 214 99.96%888 19.7 min 888 8n 8n 888n ␣ m8888 214 84Po88 210 81Tl 8888n ␤ 88n 8 1.6  104 s 1.32 min 8888n ␤ 210 82Pb 66 20.4 yr 66 g 83Bi 8888n ␣ ␤ m8888 210 ⬃100% 888 5.01 days 888 8n 8n 888n ␣ m8888 210 84 Po8 206 81 Tl 8888n ␤ 888 8 138 days n 4.20 min 206 82 Pb *The times denote the half-lives. uranium-238, which involves 14 steps. This decay scheme, known as the uranium decay series, also shows the half-lives of all the intermediate products. It is important to be able to balance the nuclear reaction for each of the steps in a radioactive decay series. For example, the first step in the uranium decay series is the decay of uranium-238 to thorium-234, with the emission of an α particle. Hence, the reaction is 238 234 92U ¡ 90Th 1 42α The next step is represented by 234 234 90Th ¡ 91Pa 1 210β and so on. In a discussion of radioactive decay steps, the beginning radioactive isotope is called the parent and the product, the daughter. 872 Chapter 19 ■ Nuclear Chemistry Kinetics of Radioactive Decay All radioactive decays obey first-order kinetics. Therefore, the rate of radioactive decay at any time t is given by rate of decay at time t 5 λN (19.3) where λ is the first-order rate constant and N is the number of radioactive nuclei pres- ent at time t. (We use λ instead of k for rate constant in accord with the notation used by nuclear scientists.) According to Equation (13.3), the number of radioactive nuclei at time zero (N0) and time t (Nt) is Nt ln 5 2λt N0 and the corresponding half-life of the reaction is given by Equation (13.5): 0.693 t12 5 λ The half-lives (hence the rate constants) of radioactive isotopes vary greatly from nucleus to nucleus. For example, looking at Table 19.3, we find two extreme cases: 238 234 We do not have to wait 4.51 3 109 yr 92U ¡ 90Th 1 42α    t 12 5 4.51 3 109 yr to make a half-life measurement of 214 210 uranium-238. Its value can be calculated 84Po ¡ 82Pb 1 42α    t 12 5 1.6 3 1024 s from the rate constant using Equation (13.5). The ratio of these two rate constants after conversion to the same time unit is about 1 3 1021, an enormously large number. Furthermore, the rate constants are unaf- fected by changes in environmental conditions such as temperature and pressure. These highly unusual features are not seen in ordinary chemical reactions (see Table 19.1). Review of Concepts Iron-59 (yellow spheres) decays to cobalt (blue spheres) via beta decay with a half-life of 45.1 d. (a) Write a balanced nuclear equation for the process. (b) From the following diagram, determine how many half-lives have elapsed. Dating Based on Radioactive Decay The half-lives of radioactive isotopes have been used as “atomic clocks” to determine the ages of certain objects. Some examples of dating by radioactive decay measure- ments will be described here. Radiocarbon Dating The carbon-14 isotope is produced when atmospheric nitrogen is bombarded by cosmic rays: 14 7N 1 10n ¡ 14 6C 1 11H  19.3 Natural Radioactivity 873 The radioactive carbon-14 isotope decays according to the equation 14 14 6C ¡ 7N 1 210β The preceding decay series is the basis of the radiocarbon dating technique described on p. 586. In 2009, scientists in Sweden applied the technique to settle a controversy regarding heart muscle regeneration. The long-held view was that the heart cannot produce new muscle cells so people die with the same heart they were born with. After a heart attack, part of the cardiac muscle is lost and the heart heals principally through the formation of scar tissues. Carbon-14 concentration in the atmosphere remained relatively stable until 1955, when aboveground nuclear bomb tests caused a sharp increase until a test ban treaty was signed in 1963, after which it gradually decreased. The newly generated carbon-14 isotopes from nuclear blasts were incorporated into plants and animals. Through diets these isotopes finally appeared in the DNA of new cells in our bodies and stay unchanged for the life of the cell. It was found that for people born before 1955 the concentration of carbon-14 in their heart muscle cells exceeded the atmospheric concentration of carbon-14 at the time of their birth. This finding suggests that they continued to A human heart. build new heart muscle cells after 1955 when atmospheric carbon-14 concentration increased. Those born during or after the atmospheric testing had a lower concentra- tion due to the steady decline in atmospheric carbon-14 after the tests were halted. Because the level of carbon-14 in the atmosphere falls each year, the amount of carbon-14 in the DNA can serve to indicate the cell’s birth date. The results show that about 1 percent of heart muscle cells are replaced every year at age 25, and the rate gradually falls to less than half a percent per year by age 75. Knowing that heart muscle cells do regenerate opens up possibilities of regulating that process. It is hoped that someday drugs will be developed to increase heart muscle cells in heart attack patients. Dating Using Uranium-238 Isotopes Because some of the intermediate products in the uranium decay series have very long half-lives (see Table 19.3), this series is particularly suitable for estimating the age of rocks in the earth and of extraterrestrial objects. The half-life for the first step We can think of the first step as the rate- (238 234 9 92U to 90Th) is 4.51 3 10 yr. This is about 20,000 times the second largest value determining step in the overall process. (that is, 2.47 3 10 yr), which is the half-life for 234 5 230 92U to 90Th. Therefore, as a good approximation we can assume that the half-life for the overall process (that is, from 238 206 92U to 82Pb) is governed solely by the first step: 238 206 92 U ¡ 82Pb 1 842α 1 6210β   t 12 5 4.51 3 109 yr In naturally occurring uranium minerals we should and do find some lead-206 isotopes formed by radioactive decay. Assuming that no lead was present when the mineral was formed and that the mineral has not undergone chemical changes that would allow the lead-206 isotope to be separated from the parent uranium-238, it is possible to estimate the age of the rocks from the mass ratio of 206 238 82Pb to 92U. The 238U previous equation tells us that for every mole, or 238 g, of uranium that undergoes complete decay, 1 mole, or 206 g, of lead is formed. If only half a mole of uranium-238 has undergone decay, the mass ratio 206Pb/238U becomes 238U 206Pb t 1_ 206 g/2 2 5 0.866 238 g/2 4.51 × 109 yr 9 and the process would have taken a half-life of 4.51 3 10 yr to complete (Figure 19.3). Ratios lower than 0.866 mean that the rocks are less than 4.51 3 109 yr old, and higher ratios suggest a greater age. Interestingly, studies based on the uranium series as Figure 19.3 After one half-life, well as other decay series put the age of the oldest rocks and, therefore, probably the half of the original uranium-238 is age of Earth itself at 4.5 3 109, or 4.5 billion, years. converted to lead-206. 874 Chapter 19 ■ Nuclear Chemistry Dating Using Potassium-40 Isotopes This is one of the most important techniques in geochemistry. The radioactive potassium-40 isotope decays by several different modes, but the relevant one as far as dating is concerned is that of electron capture: 40 19K 1 210 e ¡ 40 18Ar t 12 5 1.2 3 109 yr The accumulation of gaseous argon-40 is used to gauge the age of a specimen. When a potassium-40 atom in a mineral decays, argon-40 is trapped in the lattice of the min- eral and can escape only if the material is melted. Melting, therefore, is the procedure for analyzing a mineral sample in the laboratory. The amount of argon-40 present can be conveniently measured with a mass spectrometer (see p. 84). Knowing the ratio of argon-40 to potassium-40 in the mineral and the half-life of decay makes it possible to establish the ages of rocks ranging from millions to billions of years old. 19.4 Nuclear Transmutation The scope of nuclear chemistry would be rather narrow if study were limited to natu- ral radioactive elements. An experiment performed by Rutherford in 1919, however, suggested the possibility of producing radioactivity artificially. When he bombarded a sample of nitrogen with α particles, the following reaction took place: 14 7N 1 42α ¡ 17 8O 1 11p Note that the 17O isotope is not radioactive. An oxygen-17 isotope was produced with the emission of a proton. This reaction demonstrated for the first time the feasibility of converting one element into another, by the process of nuclear transmutation. Nuclear transmutation differs from radioac- tive decay in that the former is brought about by the collision of two particles. The preceding reaction can be abbreviated as 147N(α,p) 178O. Note that in the parentheses the bombarding particle is written first, followed by the ejected particle. Example 19.3 illustrates the use of this notation to represent nuclear transmutations. Example 19.3 56 54 Write the balanced equation for the nuclear reaction 26Fe(d,α)25Mn, where d represents the deuterium nucleus (that is, 21H). Strategy To write the balanced nuclear equation, remember that the first isotope 56 26Fe is the reactant and the second isotope 54 25Mn is the product. The first symbol in parentheses (d) is the bombarding particle and the second symbol in parentheses (α) is the particle emitted as a result of nuclear transmutation. Solution The abbreviation tells us that when iron-56 is bombarded with a deuterium nucleus, it produces the manganese-54 nucleus plus an α particle. Thus, the equation for this reaction is 56 26Fe 1 21H ¡ 42α 1 54 25Mn Check Make sure that the sum of mass numbers and the sum of atomic numbers are Similar problems: 19.37, 19.38. the same on both sides of the equation. 106 109 Practice Exercise Write a balanced equation for 46Pd(α,p) 47Ag. Although light elements are generally not radioactive, they can be made so by bombarding their nuclei with appropriate particles. As we saw earlier, the radioactive  19.4 Nuclear Transmutation 875 carbon-14 isotope can be prepared by bombarding nitrogen-14 with neutrons. Tritium, 3 1H, is prepared according to the following bombardment: 6 3Li 1 10n ¡ 31H 1 42α Tritium decays with the emission of β particles: 3 1H ¡ 32He 1 210β t12 5 12.5 yr Particle Accelerator Many synthetic isotopes are prepared by using neutrons as projectiles. This approach is particularly convenient because neutrons carry no charges and there- fore are not repelled by the targets—the nuclei. In contrast, when the projectiles are positively charged particles (for example, protons or α particles), they must have considerable kinetic energy to overcome the electrostatic repulsion between themselves and the target nuclei. The synthesis of phosphorus from aluminum is one example: 27 13Al 1 42α ¡ 30 15P 1 10n A particle accelerator uses electric and magnetic fields to increase the kinetic ener- gy of charged species so that a reaction will occur (Figure 19.4). Alternating the polarity (that is, 1 and 2) on specially constructed plates causes the particles to accelerate along a spiral path. When they have the energy necessary to initiate the desired nuclear reaction, they are guided out of the accelerator into a collision with a target substance. Various designs have been developed for particle accelerators. Located 300 ft below ground along the border between France and Switzerland, the world’s largest accelera- tor, the Large Hadron Collider (LHC) shown on p. 862, is housed inside a circular tun- nel 17 mi around. It is now possible to accelerate particles to a speed close to that of light. (According to Einstein’s theory of relativity, it is impossible for a particle to move at the speed of light. The only exception is the photon, which has a zero rest mass.) The extremely energetic particles produced in accelerators are employed by physicists Alternating-voltage Figure 19.4 Schematic diagram source of a cyclotron particle accelerator. The particle (an ion) to be accelerated starts at the center and is forced to move in a spiral 1 2 path through the influence of Vacuum Dees chamber electric and magnetic fields until it emerges at a high velocity. The magnetic fields are perpendicular to the plane of the dees (so-called because of their shape), which are Path of hollow and serve as electrodes. particle Magnet (top magnet To vacuum not shown) pump Particle Target source 876 Chapter 19 ■ Nuclear Chemistry to smash atomic nuclei to fragments. Studying the debris from such disintegrations provides valuable information about nuclear structure and binding forces. The Transuranium Elements Particle accelerators made it possible to synthesize the so-called transuranium elements, elements with atomic numbers greater than 92. Neptunium (Z 5 93) was first prepared in 1940. Since then, 25 other transuranium elements have been synthesized. All isotopes of these elements are radioactive. Table 19.4 lists the transuranium elements up to copernicium (Z 5 112) and the reactions through which they were formed. Island of Stability The latest synthesis of element 117 in 2010 filled the periodic table up to already- created element 118. Nuclear scientists believe that some heavier elements may occupy an “island of stability” in which atoms have longer half-lives (Figure 19.5). Analogous to the electronic structure of atoms, atomic nuclei can be thought of as concentric shells of protons and neutrons. The most stable nuclei occur when the outermost shells are filled. Some theories predict this will happen with 184 neutrons and 114, 120, or 126 protons, the presumed center of the island of stability. Despite considerable tech- nical difficulties, scientists are hopeful that elements 119 and 120 and beyond will someday be synthesized. Isotopes of these elements may have half-lives of seconds, days, or even years. (In contrast, super-heavy elements made so far have half-lives of fractions of a second.) They are likely to find applications in industry and medicine. The intriguing question for scientists is this: Does the periodic table come to an end, and if so, where does it end? Table 19.4 The Transuranium Elements Atomic Number Name Symbol Preparation 238 1 239 0 93 Neptunium Np 92U 1 0n ¡ 93Np 1 21β 239 239 0 94 Plutonium Pu 93Np ¡ 94Pu 1 21β 239 1 240 0 95 Americium Am 94Pu 1 0n ¡ 95Am 1 21β 239 4 242 1 96 Curium Cm 94Pu 1 2α ¡ 96Cm 1 0n 241 4 243 1 97 Berkelium Bk 95 Am 1 2α ¡ 97Bk 1 20n 242 4 245 1 98 Californium Cf 96Cm 1 2α ¡ 98Cf 1 0n 238 1 253 0 99 Einsteinium Es 92U 1 150n ¡ 99Es 1 721β 238 1 255 0 100 Fermium Fm 92U 1 170n ¡ 100Fm 1 821β 253 4 256 1 101 Mendelevium Md 99Es 1 2α ¡ 101Md 1 0n 246 12 254 1 102 Nobelium No 96Cm 1 6C ¡ 102No 1 40n 252 10 257 1 103 Lawrencium Lr 98Cf 1 5B ¡ 103Lr 1 50n 249 12 257 1 104 Rutherfordium Rf 98Cf 1 6C ¡ 104Rf 1 40n 249 15 260 1 105 Dubnium Db 98Cf 1 7N ¡ 105Db 1 40n 249 18 263 1 106 Seaborgium Sg 98Cf 1 8O ¡ 106Sg 1 40n 209 54 262 1 107 Bohrium Bh 83Bi 1 24Cr ¡ 107Bh 1 0n 208 58 265 1 108 Hassium Hs 82Pb 1 26Fe ¡ 108Hs 1 0n 209 58 266 1 109 Meitnerium Mt 83Bi 1 26Fe ¡ 109Mt 1 0n 208 62 269 1 110 Darmstadtium Ds 82Pb 1 28Ni ¡ 110Ds 1 0n 209 64 272 1 111 Roentgenium Rg 83Bi 1 28Ni ¡ 111Rg 1 0n 208 70 277 1 112 Copernicium Cn 82Pb 1 30Zn ¡ 112Cn 1 0n 244 48 289 1 114 Flerovium Fl 94 Pu 1 20Ca ¡ 114Fl 1 30n 248 48 293 1 116 Livermorium Lv 96 Cm 1 20Ca ¡ 116Lv 1 30n  19.5 Nuclear Fission 877 Figure 19.5 Island of stability. Review of Concepts Element 118, known currently by its IUPAC systematic name ununoctium (symbol: Uuo), was first created in 2006 in Dubna, Russia. The nuclear reaction used to produce this element was 249 48 294 98Cf(20Ca,X)118Uuo. Determine the product X and write the balanced equation for this nuclear reaction. 19.5 Nuclear Fission Nuclear fission is the process in which a heavy nucleus (mass number . 200) divides to form smaller nuclei of intermediate mass and one or more neutrons. Because the Animation Nuclear Fission heavy nucleus is less stable than its products (see Figure 19.2), this process releases a large amount of energy. The first nuclear fission reaction to be studied was that of uranium-235 bombarded with slow neutrons, whose speed is comparable to that of air molecules at room tempera- ture. Under these conditions, uranium-235 undergoes fission, as shown in Figure 19.6. Actually, this reaction is very complex: More than 30 different elements have been found among the fission products (Figure 19.7). A representative reaction is 235 92 U 1 10n ¡ 90 38Sr 1 143 1 54 Xe 1 30n Relative amounts of fission product 8n 90 38 Sr 8n 8n 8n 8n 235 92 U 8 n 80 100 120 140 160 236 92 U 143 Mass number 54 Xe Figure 19.7 Relative yields of the Figure 19.6 Nuclear fission of 235U. When a 235U nucleus captures a neutron (green sphere), it products resulting from the fission undergoes fission to yield two smaller nuclei. On the average, 2.4 neutrons are emitted for every of 235U as a function of mass 235 U nucleus that divides. number. 878 Chapter 19 ■ Nuclear Chemistry Table 19.5 Although many heavy nuclei can be made to undergo fission, only the fission of naturally occurring uranium-235 and of the artificial isotope plutonium-239 has Nuclear Binding Energies any practical importance. Table 19.5 shows the nuclear binding energies of of 235U and Its Fission Products uranium-235 and its fission products. As the table shows, the binding energy per nucleon for uranium-235 is less than the sum of the binding energies for Nuclear Binding strontium-90 and xenon-143. Therefore, when a uranium-235 nucleus is split into Energy two smaller nuclei, a certain amount of energy is released. Let us estimate the 235 U 2.82 3 10210 J magnitude of this energy. The difference between the binding energies of the 90 Sr 1.23 3 10210 J reactants and products is (1.23 3 10210 1 1.92 3 10210) J 2 (2.82 3 10210) J, 143 Xe 1.92 3 10210 J or 3.3 3 10211 J per uranium-235 nucleus. For 1 mole of uranium-235, the energy released would be (3.3 3 10211) 3 (6.02 3 1023), or 2.0 3 1013 J. This is an extremely exothermic reaction, considering that the heat of combustion of 1 ton of coal is only about 5 3 107 J. The significant feature of uranium-235 fission is not just the enormous amount of energy released, but the fact that more neutrons are produced than are origi- nally captured in the process. This property makes possible a nuclear chain reac- tion, which is a self-sustaining sequence of nuclear fission reactions. The neutrons generated during the initial stages of fission can induce fission in other uranium-235 nuclei, which in turn produce more neutrons, and so on. In less than a second, the reaction can become uncontrollable, liberating a tremendous amount of heat to the surroundings. For a chain reaction to occur, enough uranium-235 must be present in the sample to capture the neutrons. Otherwise, many of the neutrons will escape from the sample and the chain reaction will not occur. In this situation the mass of the sample is said to be subcritical. Figure 19.8 shows what happens when the amount of the fissionable material is equal to or greater than the critical mass, the minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction. In this case, most of the neutrons will be captured by uranium-235 nuclei, and a chain reaction will occur. Figure 19.8 If a critical mass 132 51 Sb 235 92 U is present, many of the neutrons emitted during the fission process will be captured by other 235U 143 235 54 Xe 92 U nuclei and a chain reaction will occur. 235 101 92 U 41 Nb 235 92 U 141 56 Ba 235 92 U 90 38 Sr 235 92 U 92 36 Kr 235 92 U  19.5 Nuclear Fission 879 The Atomic Bomb The first application of nuclear fission was in the development of the atomic bomb. How is such a bomb made and detonated? The crucial factor in the bomb’s design is the determination of the critical mass for the bomb. A small atomic bomb is Subcritical equivalent to 20,000 tons of TNT (trinitrotoluene). Because 1 ton of TNT releases U-235 mass about 4 3 109 J of energy, 20,000 tons would produce 8 3 1013 J. Earlier we saw that 1 mole, or 235 g, of uranium-235 liberates 2.0 3 1013 J of energy when it undergoes fission. Thus, the mass of the isotope present in a small bomb must be at least 8 3 1013 J 235 g 3 < 1 kg 2.0 3 1013 J Subcritical U-235 mass For obvious reasons, an atomic bomb is never assembled with the critical mass already present. Instead, the critical mass is formed by using a conventional explosive, such TNT explosive as TNT, to force the fissionable sections together, as shown in Figure 19.9. Neutrons from a source at the center of the device trigger the nuclear chain reaction. Figure 19.9 Schematic diagram Uranium-235 was the fissionable material in the bomb dropped on Hiroshima, Japan, of an atomic bomb. The TNT on August 6, 1945. Plutonium-239 was used in the bomb exploded over Nagasaki 3 explosives are set off first. The explosion forces the sections of days later. The fission reactions generated were similar in these two cases, as was the fissionable material together to extent of the destruction. form an amount considerably larger than the critical mass. Nuclear Reactors A peaceful but controversial application of nuclear fission is the generation of elec- tricity using heat from a controlled chain reaction in a nuclear reactor. Currently, nuclear reactors provide about 20 percent of the electrical energy in the United States. This is a small but by no means negligible contribution to the nation’s energy production. Several different types of nuclear reactors are in operation; we will briefly discuss the main features of three of them, along with their advantages and disadvantages. Light Water Reactors Most of the nuclear reactors in the United States are light water reactors. Figure 19.10 is a schematic diagram of such a reactor, and Figure 19.11 shows the refueling process in the core of a nuclear reactor. An important aspect of the fission process is the speed of the neutrons. Slow neutrons split uranium-235 nuclei more efficiently than do fast ones. Because fis- sion reactions are highly exothermic, the neutrons produced usually move at high velocities. For greater efficiency they must be slowed down before they can be used to induce nuclear disintegration. To accomplish this goal, scientists use mod- erators, which are substances that can reduce the kinetic energy of neutrons. A good moderator must satisfy several requirements: It should be nontoxic and inex- pensive (as very large quantities of it are necessary); and it should resist conversion into a radioactive substance by neutron bombardment. Furthermore, it is advanta- geous for the moderator to be a fluid so that it can also be used as a coolant. No substance fulfills all these requirements, although water comes closer than many others that have been considered. Nuclear reactors that use light water (H2O) as a moderator are called light water reactors because 11H is the lightest isotope of the element hydrogen. The nuclear fuel consists of uranium, usually in the form of its oxide, U3O8 (Figure 19.12). Naturally occurring uranium contains about 0.7 percent of the uranium- 235 isotope, which is too low a concentration to sustain a small-scale chain reaction. For effective operation of a light water reactor, uranium-235 must be enriched to 880 Chapter 19 ■ Nuclear Chemistry Figure 19.10 Schematic Shield diagram of a nuclear fission reactor. The fission process is controlled by cadmium or boron To steam turbine rods. The heat generated by the process is used to produce steam for the generation of electricity via a heat exchange system. Steam Shield Control rod Uranium fuel Water Pump a concentration of 3 or 4 percent. In principle, the main difference between an atomic bomb and a nuclear reactor is that the chain reaction that takes place in a nuclear reactor is kept under control at all times. The factor limiting the rate of the reaction is the number of neutrons present. This can be controlled by lowering cad- mium or boron control rods between the fuel elements. These rods capture neutrons according to the equations 113 48 Cd 1 10n ¡ 114 48 Cd 1 γ 10 5B 1 10n ¡ 73Li 1 42α where γ denotes gamma rays. Without the control rods the reactor core would melt from the heat generated and release radioactive materials into the environment. Nuclear reactors have rather elaborate cooling systems that absorb the heat given off by the nuclear reaction and transfer it outside the reactor core, where it is used to Figure 19.11 Refueling the core produce enough steam to drive an electric generator. In this respect, a nuclear power of a nuclear reactor. plant is similar to a conventional power plant that burns fossil fuel. In both cases, large quantities of cooling water are needed to condense steam for reuse. Thus, most nuclear power plants are built near a river or a lake. Unfortunately this method of cooling causes thermal pollution (see Section 12.4). Heavy Water Reactors Another type of nuclear reactor uses D2O, or heavy water, as the moderator, rather than H2O. Deuterium absorbs neutrons much less efficiently than does ordinary hydrogen. Because fewer neutrons are absorbed, the reactor is more efficient and does not require enriched uranium. The fact that deuterium is a less efficient moderator has a negative impact on the operation of the reactor, because more neutrons leak out of the reactor. However, this is not a serious disadvantage. The main advantage of a heavy water reactor is that it eliminates the need for Figure 19.12 Uranium oxide, building expensive uranium enrichment facilities. However, D2O must be prepared U3O8. by either fractional distillation or electrolysis of ordinary water, which can be very  19.5 Nuclear Fission 881 expensive considering the amount of water used in a nuclear reactor. In countries where hydroelectric power is abundant, the cost of producing D2O by electrolysis can be reasonably low. Canada is currently one of a few nations successfully using heavy water nuclear reactors. The fact that no enriched uranium is required in a heavy water reactor enables a country to enjoy the benefits of nuclear power without under- taking work that is closely associated with weapons technology. Breeder Reactors A breeder reactor uses uranium fuel, but unlike a conventional nuclear reactor, it produces more fissionable materials than it uses. We know that when uranium-238 is bombarded with fast neutrons, the following reactions take place: Figure 19.13 The red glow of the radioactive plutonium oxide, 238 PuO2. 92 U 1 10n ¡ 239 92 U Plutonium-239 forms plutonium oxide, 239 239 0 92 U ¡ 93 Np 1 21β t 12 5 23.4 min which can be readily separated from 239 239 0 uranium. 93 Np ¡ 94 Pu 1 21β t 12 5 2.35 days In this manner the nonfissionable uranium-238 is transmuted into the fissionable iso- tope plutonium-239 (Figure 19.13). In a typical breeder reactor, nuclear fuel containing uranium-235 or plutonium-239 is mixed with uranium-238 so that breeding takes place within the core. For every uranium-235 (or plutonium-239) nucleus undergoing fission, more than one neutron is captured by uranium-238 to generate plutonium-239. Thus, the stockpile of fission- able material can be steadily increased as the starting nuclear fuels are consumed. It takes about 7 to 10 yr to regenerate the sizable amount of material needed to refuel the original reactor and to fuel another reactor of comparable size. This interval is called the doubling time. Another fertile isotope is 232 90Th. Upon capturing slow neutrons, thorium is trans- muted to uranium-233, which, like uranium-235, is a fissionable isotope: 232 90 Th 1 10n ¡ 233 90 Th 233 233 0 90 Th ¡ 91 Pa 1 21β t 12 5 22 min 233 233 0 91 Pa ¡ 92 U 1 21β t 12 5 27.4 days Uranium-233 (t12 5 1.6 3 105 yr) is stable enough for long-term storage. Although the amounts of uranium-238 and thorium-232 in Earth’s crust are rela- tively plentiful (4 ppm and 12 ppm by mass, respectively), the development of breed- er reactors has been very slow. To date, the United States does not have a single operating breeder reactor, and only a few have been built in other countries, such as France and Russia. One problem is economics; breeder reactors are more expensive to build than conventional reactors. There are also more technical difficulties associ- ated with the construction of such reactors. As a result, the future of breeder reactors, in the United States at least, is rather uncertain. Hazards of Nuclear Energy Many people, including environmentalists, regard nuclear fission as a highly undesirable method of energy production. Many fission products such as strontium-90 are dangerous radioactive isotopes with long half-lives. Plutonium-239, used as a nuclear fuel and Plutonium is chemically toxic in addition produced in breeder reactors, is one of the most toxic substances known. It is an alpha to being radioactive. emitter with a half-life of 24,400 yr. Accidents, too, present many dangers. An accident at the Three Mile Island reac- tor in Pennsylvania in 1979 first brought the potential hazards of nuclear plants to CHEMISTRY in Action Nature’s Own Fission Reactor I t all started with a routine analysis in May 1972 at the nu- clear fuel processing plant in Pierrelatte, France. A staff member was checking the isotope ratio of U-235 to U-238 in a uranium ore and obtained a puzzling result. It had long been known that the relative natural occurrence of U-235 and U-238 is 0.7202 percent and 99.2798 percent, respectively. In this case, however, the amount of U-235 present was only 0.7171 percent. This may seem like a very small deviation, but the measurements were so precise that this difference was considered highly significant. The ore had come from the Oklo mine in the Gabon Republic, a small country on the west coast of Africa. Subsequent analyses of other samples showed that some contained even less U-235, in some cases as little as 0.44 percent. The logical explanation for the low percentages of U-235 was that a nuclear fission reaction at the mine must have con- sumed some of the U-235 isotopes. But how did this happen? There are several conditions under which such a nuclear fission reaction could take place. In the presence of heavy water, for example, a chain reaction is possible with unenriched uranium. Without heavy water, such a fission reaction could still occur if the uranium ore and the moderator were arranged according to some specific geometric constraints at the site of the reaction. Both of the possibilities seem rather farfetched. The most plau- sible explanation is that the uranium ore originally present in the mine was enriched with U-235 and that a nuclear fission reaction took place with light water, as in a conventional nuclear reactor. As mentioned earlier, the natural abundance of U-235 is Photo showing the open-pit mining of the Oklo uranium deposit in Gabon 0.7202 percent, but it has not always been that low. The half- revealed more than a dozen zones where nuclear fission had once taken lives of U-235 and U-238 are 700 million and 4.51 billion place. years, respectively. This means that U-235 must have been more abundant in the past, because it has a shorter half-life. In fact, at the time Earth was formed, the natural abundance of moderator present. It appears that as a result of a geological U-235 was as high as 17 percent! Because the lowest concen- transformation, uranium ore was continually being washed into tration of U-235 required for the operation of a fission reactor the Oklo region to yield concentrated deposits. The moderator is 1 percent, a nuclear chain reaction could have taken place as needed for the fission process was largely water, present as recently as 400 million years ago. By analyzing the amounts of water of crystallization in the sedimentary ore. radioactive fission products left in the ore, scientists concluded Thus, in a series of extraordinary events, a natural nuclear that the Gabon “reactor” operated about 2 billion years ago. fission reactor operated at the time when the first life forms ap- Having an enriched uranium sample is only one of the re- peared on Earth. As is often the case in scientific endeavors, quirements for starting a controlled chain reaction. There must humans are not necessarily the innovators but merely the imita- also have been a sufficient amount of the ore and an appropriate tors of nature. 882  19.6 Nuclear Fusion 883 public attention. In this instance, very little radiation escaped the reactor, but the plant remained closed for more than a decade while repairs were made and safety issues addressed. Only a few years later, on April 26, 1986, a reactor at the Chernobyl nuclear plant in Ukraine surged out of control. The fire and explosion that followed released much radioactive material into the environment. People working near the plant died within weeks as a result of the exposure to the intense radiation. Twenty- five years later, it is still not known how many people died from radiation-induced cancer cases. Estimates vary between a few thousand and hundreds of thousands. The latest large-scale nuclear plant accident occurred in Fukushima, Japan, on March 11, 2011. A powerful earthquake, followed by a destructive tsunami, severely damaged the nuclear reactors at the plant. The long-term harmful effects of the radiation leak- age to the environment are not yet fully assessed, but it is believed to be comparable to that at Chernobyl. In addition to the risk of accidents, the problem of radioactive waste disposal has not been satisfactorily resolved even for safely operated nuclear plants. Many sugges- tions have been made as to where to store or dispose of nuclear waste, including burial underground, burial beneath the ocean floor, and storage in deep geologic for- mations. But none of these sites has proved absolutely safe in the long run. Leakage of radioactive wastes into underground water, for example, can endanger nearby com- munities. The ideal disposal site would seem to be the sun, where a bit more radiation would make little difference, but this kind of operation requires 100 percent reliabil- ity in space technology. Because of the hazards, the future of nuclear reactors is clouded. What was once hailed as the ultimate solution to our energy needs in the twenty-first century is now being debated and questioned by both the scientific community and laypeople. It seems likely that the controversy will continue for some time. Molten glass is poured over nuclear waste before burial. Review of Concepts Why are boron compounds often added to badly damaged nuclear reactors such as those in Fukushima, Japan? 19.6 Nuclear Fusion In contrast to the nuclear fission process, nuclear fusion, the combining of small nuclei into larger ones, is largely exempt from the waste disposal problem. Figure 19.2 showed that for the lightest elements, nuclear stability increases with increasing mass number. This behavior suggests that if two light nuclei combine or fuse together to form a larger, more stable nucleus, an appreciable amount of energy will be released in the process. This is the basis for ongoing research into the harness- ing of nuclear fusion for the production of energy. Nuclear fusion occurs constantly in the sun. The sun is made up mostly of hydro- gen and helium. In its interior, where temperatures reach about 15 million degrees Celsius, the following fusion reactions are believed to take place: 1 2 1H 1 1H ¡ 32He 3 3 2He 1 2He ¡ 42He 1 211H 1 1 1H 1 1H ¡ 21H 1 110β Nuclear fusion keeps the Because fusion reactions take place only at very high temperatures, they are often temperature in the interior of called thermonuclear reactions. the sun at about 15 million °C. 884 Chapter 19 ■ Nuclear Chemistry Fusion Reactors A major concern in choosing the proper nuclear fusion process for energy production is the temperature necessary to carry out the process. Some promising reactions are Reaction Energy Released 2 1H 1 21H ¡ 31H 1 11H 4.9 3 10213 J 2 1H 1 31H ¡ 42He 1 10n 2.8 3 10212 J 6 3Li 1 21H ¡ 242He 3.6 3 10212 J These reactions take place at extremely high temperatures, on the order of 100 million degrees Celsius, to overcome the repulsive forces between the nuclei. The first reaction is particularly attractive because the world’s supply of deuterium is virtually inexhaustible. The total volume of water on Earth is about 1.5 3 1021 L. Because the natural abundance of deuterium is 1.5 3 1022 percent, the total amount of deuterium present is roughly 4.5 3 1021 g, or 5.0 3 1015 tons. The cost of preparing deuterium is minimal compared with the value of the energy released by the reaction. In contrast to the fission process, nuclear fusion looks like a very promising energy source, at least “on paper.” Although thermal pollution would be a problem, fusion has the following advantages: (1) The fuels are cheap and almost inexhaust- ible and (2) the process produces little radioactive waste. If a fusion machine were turned off, it would shut down completely and instantly, without any danger of a meltdown. If nuclear fusion is so great, why isn’t there even one fusion reactor producing energy? Although we command the scientific knowledge to design such a reactor, the technical difficulties have not yet been solved. The basic problem is finding a way to hold the nuclei together long enough, and at the appropriate temperature, for fusion to occur. At temperatures of about 100 million degrees Celsius, molecules cannot exist, and most or all of the atoms are stripped of their electrons. This state of matter, a gaseous mixture of positive ions and electrons, is called plasma. The problem of containing this plasma is a formidable one. What solid container can exist at such temperatures? None, unless the amount of plasma is small; but then the solid surface would immediately cool the sample and quench the fusion reaction. One approach to solving this problem is to use magnetic confinement. Because a plasma consists of charged particles moving at high speeds, a magnetic field will exert force on it. As Figure 19.14 shows, the plasma moves through a doughnut-shaped tunnel, confined Figure 19.14 A magnetic plasma confinement design called tokamak. Plasma Magnet  19.6 Nuclear Fusion 885 Figure 19.15 The reaction chamber at the National Ignition Facility, where 192 lasers are used to initiate fusion in a small pellet of deuterium and tritium. by a complex magnetic field. Thus, the plasma never comes in contact with the walls of the container. Another design employs lasers to ignite the nuclear reaction. One approach focus- es 192 high-power laser beams on a small fuel pellet containing deuterium and tritium (Figure 19.15). The energy from the lasers heats the pellet to an extremely high temperature, causing it to implode, that is, to collapse inward from all sides into a tiny volume. Under this condition, the fusion process is initiated. The U.S. National Ignition Facility reported in February 2014 that for the first time the energy produced by fusion in the laboratory exceeded the amount used to induce the process, including one experiment that doubled the amount of energy required for ignition. Although many technical difficulties still need to be solved before nuclear fusion can be put to practical use on a large scale, scientists are excited by the first positive result after many years of hard work. The Hydrogen Bomb The technical problems inherent in the design of a nuclear fusion reactor do not affect the production of a hydrogen bomb, also called a thermonuclear bomb. In this case, the objective is all power and no control. Hydrogen bombs do not contain gaseous hydrogen or gaseous deuterium; they contain solid lithium deuteride (LiD), which can be packed very tightly. The detonation of a hydrogen bomb occurs in two stages—first a fission reaction and then a fusion reaction. The required tem- perature for fusion is achieved with an atomic bomb. Immediately after the atomic bomb explodes, the following fusion reactions occur, releasing vast amounts of energy (Figure 19.16): 6 3Li 1 21H ¡ 242α 2 1H 1 21H ¡ 31H 1 11H There is no critical mass in a fusion bomb, and the force of the explosion is limited only by the quantity of reactants present. Thermonuclear bombs are described as being “cleaner” than atomic bombs because the only radioactive isotopes they produce are tritium, which is a weak β-particle emitter 1t 12 5 12.5 yr2 , and the products of the fission starter. Their damaging effects on the environment can be aggravated, Figure 19.16 Explosion of a however, by incorporating in the construction some nonfissionable material such as thermonuclear bomb. 886 Chapter 19 ■ Nuclear Chemistry cobalt. Upon bombardment by neutrons, cobalt-59 is converted to cobalt-60, which is a very strong γ-ray emitter with a half-life of 5.2 yr. The presence of radioactive cobalt isotopes in the debris or fallout from a thermonuclear explosion would be fatal to those who survived the initial blast. 19.7 Uses of Isotopes Radioactive and stable isotopes alike have many applications in science and medicine. We have previously described the use of isotopes in the study of reaction mechanisms (see Section 13.5) and in dating artifacts (p. 586 and Section 19.3). In this section we will discuss a few more examples. Structural Determination The formula of the thiosulfate ion is S2O22 3 . For some years chemists were uncertain as to whether the two sulfur atoms occupied equivalent positions in the ion. The thiosulfate ion is prepared by treatment of the sulfite ion with elemental sulfur: SO22 22 3 (aq) 1 S(s) ¡ S2O3 (aq) When thiosulfate is treated with dilute acid, the reaction is reversed. The sulfite ion is reformed and elemental sulfur precipitates: H1 S2O22 22 3 (aq) ¡ SO3 (aq) 1 S(s) (19.4) If this sequence is started with elemental sulfur enriched with the radioactive sulfur-35 isotope, the isotope acts as a “label” for S atoms. All the labels are found in the sulfur precipitate in Equation (19.4); none of them appears in the final sulfite ions. Clearly, then, the two atoms of sulfur in S2O322 are not structurally equivalent, as would be the case if the structure were 2 O Q O Q O Q O Q O SOOSOOOSOOS Q Otherwise, the radioactive isotope would be present in both the elemental sulfur pre- cipitate and the sulfite ion. Based on spectroscopic studies, we now know that the struc- ture of the thiosulfate ion is 2 SSS B O SOOSOOS Q O Q B SOS S2O22 3 Study of Photosynthesis The study of photosynthesis is also rich with isotope applications. The overall photo- synthesis reaction can be represented as 6CO2 1 6H2O ¡ C6H12O6 1 6O2 In Section 13.5 we learned that the 18O isotope was used to determine the source of O2. The radioactive 14C isotope helped to determine the path of carbon in photosynthesis. Starting with 14CO2, it was possible to isolate the intermediate products during photosyn- thesis and measure the amount of radioactivity of each carbon-containing compound. In this manner, the path from CO2 through various intermediate compounds to carbohydrate could be clearly charted. Isotopes, especially radioactive isotopes that are used to trace the path of the atoms of an element in a chemical or biological process, are called tracers.  19.7 Uses of Isotopes 887 Isotopes in Medicine Tracers are used also for diagnosis in medicine. Sodium-24 (a β emitter with a half- life of 14.8 h) injected into the bloodstream as a salt solution can be monitored to trace the flow of blood and detect possible constrictions or obstructions in the circu- latory system. Iodine-131 (a β emitter with a half-life of 8 days) has been used to test the activity of the thyroid gland. A malfunctioning thyroid can be detected by giving the patient a drink of a solution containing a known amount of Na131I and measuring the radioactivity just above the thyroid to see if the iodine is absorbed at the normal rate. Of course, the amounts of radioisotope used in the human body must always be kept small; otherwise, the patient might suffer permanent damage from the high-energy radiation. A radioactive isotope of fluorine, fluorine-18, emits positrons, which are annihilated by electrons, forming γ-rays used to image the brain (Figure 19.17). Table 19.6 shows some of the radioactive isotopes used in medicine. Technetium, the first artificially prepared element, is one of the most useful ele- ments in nuclear medicine. Although technetium is a transition metal, all its isotopes are radioactive. In the laboratory it is prepared by the nuclear reactions 98 42Mo 1 10n ¡ 99 42Mo Image of a person’s skeleton 99 42Mo ¡ 99m 43Tc 1 210β obtained using 99m 43Tc. where the superscript m denotes that the technetium-99 isotope is produced in its excited nuclear state. This isotope has a half-life of about 6 h, decaying by γ radiation Figure 19.17 This patient was injected with a small dose of a radioactive tracer that binds to glucose molecules in the blood. The glucose concentrates in the more active regions of the brain that can then be identified by gamma emission (shown as red in the figure), allowing one to determine which parts of the brain are associated with different activities. Table 19.6 Some Radioactive Isotopes Used in Medicine Isotope Half-Life Uses 18 F 1.8 h Brain imaging, bone scan 24 Na 15 h Monitoring blood circulation 32 P 14.3 d Location of ocular, brain, and skin tumors 43 K 22.4 h Myocardial scan 47 Ca 4.5 d Study of calcium metabolism 51 Cr 27.8 d Determination of red blood cell volume, spleen imaging, placenta localization 60 Co 5.3 yr Sterilization of medical equipment, cancer treatment 99m Tc 6h Imaging of various organs, bones, placenta location 125 I 60 d Study of pancreatic function, thyroid imaging, liver function 131 I 8d Brain imaging, liver function, thyroid activity 888 Chapter 19 ■ Nuclear Chemistry Figure 19.18 Schematic diagram Cathode Anode of a Geiger counter. Radiation (α Insulator or β particles, or γ rays) entering through the window ionized the argon gas to generate a small current flow between the Window electrodes. This current is amplified and is used to flash a light or operate a counter with a clicking sound. Argon gas Amplifier and counter High voltage to technetium-99 in its nuclear ground state. Thus, it is a valuable diagnostic tool. The patient either drinks or is injected with a solution containing 99mTc. By detecting the γ rays emitted by 99mTc, doctors can obtain images of organs such as the heart, liver, and lungs. A major advantage of using radioactive isotopes as tracers is that they are easy to detect. Their presence even in very small amounts can be detected by photograph- ic techniques or by devices known as counters. Figure 19.18 is a diagram of a Geiger counter, an instrument widely used in scientific work and medical laboratories to detect radiation. 19.8 Biological Effects of Radiation In this section, we will examine briefly the effects of radiation on biological systems. But first let us define quantitative measures of radiation. The fundamental unit of radioactivity is the curie (Ci); 1 Ci corresponds to exactly 3.70 3 1010 nuclear disin- tegrations per second. This decay rate is equivalent to that of 1 g of radium. A millicurie (mCi) is one-thousandth of a curie. Thus, 10 mCi of a carbon-14 sample is the quantity that undergoes (10 3 1023 )(3.70 3 1010 ) 5 3.70 3 108 disintegrations per second. The intensity of radiation depends on the number of disintegrations as well as on the energy and type of radiation emitted. One common unit for the absorbed dose of radiation is the rad (radiation absorbed dose), which is the amount of radiation that results in the absorption of 1 3 1022 J per kilogram of irradiated material. The bio- logical effect of radiation also depends on the part of the body irradiated and the type of radiation. For this reason, the rad is often multiplied by a factor called RBE (relative biological effectiveness). The RBE is approximately 1 for beta and gamma radiation and about 10 for alpha radiation. To measure the biological damage, which depends on dose rate, total dose, and the type of tissue affected, we introduce anoth- er term called a rem (roentgen equivalent for man), given by number of rems 5 (number of rads)(RBE) (19.5) Of the three types of nuclear radiation, alpha particles usually have the least penetrating power. Beta particles are more penetrating than alpha particles, but less so than gamma rays. Gamma rays have very short wavelengths and high energies. Furthermore, because they carry no charge, they cannot be stopped by shielding mate- rials as easily as alpha and beta particles. However, if alpha or beta emitters are  19.8 Biological Effects of Radiation 889 Table 19.7 Average Yearly Radiation Doses for Americans Source Dose (mrem/yr)* Cosmic rays 20–50 Ground and surroundings 25 Human body† 26 Medical and dental X rays 50–75 Air travel 5 Fallout from weapons tests 5 Nuclear waste 2 Total 133–188 *1 mrem 5 1 millirem 5 1 3 1023 rem. † The radioactivity in the body comes from food and air. ingested, their damaging effects are greatly aggravated because the organs will be constantly subject to damaging radiation at close range. For example, strontium-90, a beta emitter, can replace calcium in bones, where it does the greatest damage. Table 19.7 lists the average amounts of radiation an American receives every year. It should be pointed out that for short-term exposures to radiation, a dosage of 50–200 rem will cause a decrease in white blood cell counts and other complications, while a dosage of 500 rem or greater may result in death within weeks. Current safety standards permit nuclear workers to be exposed to no more than 5 rem per year and specify a maximum of 0.5 rem of human-made radiation per year for the general public. The chemical basis of radiation damage is that of ionizing radiation. Radiation of either particles or gamma rays can remove electrons from atoms and molecules in its path, leading to the formation of ions and radicals. Radicals (also called free radicals) are molecular fragments having one or more unpaired electrons; they are usually short-lived and highly reactive. For example, when water is irradiated with gamma rays, the following reactions take place: radiation H2O     H2O1 1 e2 H2O 1 H2O         H3O1 1 1 ? OH hydroxyl radical The electron (in the hydrated form) can subsequently react with water or with a hydrogen ion to form atomic hydrogen, and with oxygen to produce the superoxide ion, O2 2 (a radical): e2 1 O2 ¡ ? O2 2 In the tissues the superoxide ions and other free radicals attack cell membranes and a host of organic compounds, such as enzymes and DNA molecules. Organic com- pounds can themselves be directly ionized and destroyed by high-energy radiation. It has long been known that exposure to high-energy radiation can induce can- cer in humans and other animals. Cancer is characterized by uncontrolled cellular growth. On the other hand, it is also well established that cancer cells can be destroyed by proper radiation treatment. In radiation therapy, a compromise is sought. The radiation to which the patient is exposed must be sufficient to destroy cancer cells without killing too many normal cells and, it is hoped, without inducing another form of cancer. Radiation damage to living systems is generally classified as somatic or genetic. Somatic injuries are those that affect the organism during its own lifetime. Sunburn, skin rash, cancer, and cataracts are examples of somatic damage. Genetic damage Chromosomes are parts of the cell structure that contain the genetic material (DNA). means inheritable changes or gene mutations. For example, a person whose chromo- somes have been damaged or altered by radiation may have deformed offspring.  CHEMISTRY in Action Food Irradiation I f you eat processed food, you have probably eaten ingredients exposed to radioactive rays. In the United States, up to 10 per- cent of herbs and spices are irradiated to control mold, zapped with X rays at a dose equal to 60 million chest X rays. Although food irradiation has been used in one way or another for more than 50 yr, it faces an uncertain future in this country. Back in 1953, the U.S. Army started an experimental pro- gram of food irradiation so that deployed troops could have fresh food without refrigeration. The procedure is a simple one. Food is exposed to high levels of radiation to kill insects and harmful bacteria. It is then packaged in airtight containers, in which it can be stored for months without deterioration. The radiation sources for most food preservation are cobalt-60 and cesium-137, both of which are γ emitters, although X rays and electron beams can also be used to irradiate food. The benefits of food irradiation are obvious—it reduces energy demand by eliminating the need for refrigeration, and it prolongs the shelf life of various foods, which is of vital impor- Strawberries irradiated at 200 kilorads (right) are still fresh after 15 days tance for poor countries. Yet there is considerable opposition to storage at 4°C; those not irradiated are moldy. this procedure. First, there is a fear that irradiated food may it- self become radioactive. No such evidence has been found. A more serious objection is that irradiation can destroy the nutri- radical, which then react with the organic molecules to produce ents such as vitamins and amino acids. Furthermore, the ioniz- potentially harmful substances. Interestingly, the same effects ing radiation produces reactive species, such as the hydroxyl are produced when food is cooked by heat. Food Irradiation Dosages and Their Effects* Dosage Effect Low dose (Up to 100 kilorad) Inhibits sprouting of potatoes, onions, garlics. Inactivates trichinae in pork. Kills or prevents insects from reproducing in grains, fruits, and vegetables after harvest. Medium dose (100 to 1000 kilorads) Delays spoilage of meat, poultry, and fish by killing spoilage microorganism. Reduces salmonella and other food-borne pathogens in meat, fish, and poultry. Extends shelf life by delaying mold growth on strawberries and some other fruits. High dose (1000 to 10,000 kilorads) Sterilizes meat, poultry, fish, and some other foods. Kills microorganisms and insects in spices and seasoning. *Source: Chemical & Engineering News, May 5 (1986). Key Equations E 5 mc2 (19.1) Einstein’s mass-energy equivalence relationship 2 ¢E 5 (¢m)c (19.2) Relation between mass defect and energy released 890 CHEMISTRY in Action Boron Neutron Capture Therapy E ach year more than half a million people in the world con- tract brain tumors and about 2000 die from the disease. Treatment of a brain tumor is one of the most challenging of biological effectiveness. However, a rapidly expanding tumor frequently depletes its blood supply and hence also the oxygen content. BNCT does not require oxygen and therefore does not cancer cases because of the site of the malignant growth, which suffer from this limitation. BNCT is currently an active research makes surgical excision difficult and often impossible. Like- area involving collaborations of chemists, nuclear physicists, wise, conventional radiation therapy using X rays or γ rays from and physicians. outside the skull is seldom effective. An ingenious approach to this problem is called boron neu- tron capture therapy (BNCT). This technique brings together two components, each of which separately has minimal harmful effects on the cells. The first component uses a compound containing a stable boron isotope (10B) that can be concentrated in tumor cells. The second component is a beam of low-energy neutrons. Upon capturing a neutron, the following nuclear reaction takes place: 10 5B 1 10n ¡ 73Li 1 42α The recoiling α particle and the lithium nucleus together carry about 3.8 3 10213 J of energy. Because these high-energy par- ticles are confined to just a few μm (about the diameter of a cell), they can preferentially destroy tumor cells without damag- ing the surrounding tissues. 10B has a large neutron absorption cross section and is therefore particularly suited for this applica- Setup for a lateral BNCT brain irradiation using the fission converter-based tion. Ionizing radiation like X rays requires oxygen to produce epithermal neutron beam at the Massachusetts Institute of Technology with reactive hydroxyl and superoxide radicals to enhance their a 12-cm-diameter aperture. Summary of Facts & Concepts 1. For stable nuclei of low atomic number, the neutron-to- 6. Nuclear fission is the splitting of a large nucleus into proton ratio is close to 1. For heavier stable nuclei, the two smaller nuclei and one or more neutrons. When the ratio becomes greater than 1. All nuclei with 84 or more free neutrons are captured efficiently by other nuclei, a protons are unstable and radioactive. Nuclei with even chain reaction can occur. atomic numbers tend to have a greater number of stable 7. Nuclear reactors use the heat from a controlled nuclear isotopes than those with odd atomic numbers. fission reaction to produce power. The three important 2. Nuclear binding energy is a quantitative measure of nu- types of reactors are light water reactors, heavy water clear stability. Nuclear binding energy can be calculated reactors, and breeder reactors. from a knowledge of the mass defect of the nucleus. 8. Nuclear fusion, the type of reaction that occurs in the 3. Radioactive nuclei emit α particles, β particles, posi- sun, is the combination of two light nuclei to form one trons, or γ rays. The equation for a nuclear reaction in- heavy nucleus. Fusion takes place only at very high cludes the particles emitted, and both the mass numbers temperatures, so high that controlled large-scale nuclear and the atomic numbers must balance. fusion has so far not been achieved. 4. Uranium-238 is the parent of a natural radioactive decay 9. Radioactive isotopes are easy to detect and thus make series that can be used to determine the ages of rocks. excellent tracers in chemical reactions and in medical 5. Artificial radioactive elements are created by bombard- practice. ing other elements with accelerated neutrons, protons, 10. High-energy radiation damages living systems by caus- or α particles. ing ionization and the formation of free radicals. 891  892 Chapter 19 ■ Nuclear Chemistry Key Words Breeder reactor, p. 881 Nuclear chain reaction, p. 878 Nucleon, p. 867 Radioactive decay series, p. 870 Critical mass, p. 878 Nuclear fission, p. 877 Plasma, p. 884 Thermonuclear reaction, p. 883 Mass defect, p. 868 Nuclear fusion, p. 883 Positron, p. 864 Tracer, p. 886 Moderators, p. 879 Nuclear transmutation, p. 863 Radical, p. 889 Transuranium elements, p. 876 Nuclear binding energy, p. 867 Questions & Problems • Problems available in Connect Plus Nuclear Stability Red numbered problems solved in Student Solutions Manual Review Questions Nuclear Reactions 19.9 State the general rules for predicting nuclear stability. Review Questions 19.10 What is the belt of stability? 19.1 How do nuclear reactions differ from ordinary chem- 19.11 Why is it impossible for the isotope 22He to exist? ical reactions? 19.12 Define nuclear binding energy, mass defect, and 19.2 What are the steps in balancing nuclear equations? nucleon. 19.3 What is the difference between 210e and 210β? 19.13 How does Einstein’s equation, E 5 mc2, enable us to 19.4 What is the difference between an electron and a calculate nuclear binding energy? positron? 19.14 Why is it preferable to use nuclear binding energy 19.5 Which of the following nuclear decays produces per nucleon for a comparison of the stabilities of a daughter nucleus with a higher atomic number? different nuclei? γ, 110 β, 210β, α. 19.6 The table here is a summary of different modes of Problems nuclear decay. Fill in the changes in atomic number (Z), number of neutrons (N), and mass number (A) • 19.15 The radius of a uranium-235 nucleus is about in each case. Use “1” sign for increase, “2” sign 7.0 3 1023 pm. Calculate the density of the nucleus for decrease, and “0” for no change. in g/cm3. (Assume the atomic mass is 235 amu.) • 19.16 For each pair of isotopes listed, predict which one Decay Mode Change in is less stable: (a) 63Li or 93Li, (b) 23 25 11Na or 11Na, (c) 48 20 Ca or 48 21 Sc. Z N A • 19.17 For each pair of elements listed, predict which one α decay has more stable isotopes: (a) Co or Ni, (b) F or Se, 0 21β decay (c) Ag or Cd. 0 11β decay • 19.18 In each pair of isotopes shown, indicate which one 2 e capture you would expect to be radioactive: (a) 20 10Ne and 17 40 45 95 92 195 10Ne, (b) 20Ca and 20Ca, (c) 42Mo and 43Tc, (d) 80Hg 196 209 242 Problems and 80Hg, (e) 83Bi and 96Cm. • 19.19 Given that • 19.7 Complete the following nuclear equations and iden- tify X in each case: H(g) 1 H(g) ¡ H2 (g)  ¢H° 5 2436.4 kJ/mol (a) 26 1 4 12Mg 1 1p ¡ 2α 1 X calculate the change in mass (in kg) per mole of H2 (b) 27Co 1 1H ¡ 60 59 2 27Co 1 X formed. (c) 235 92 U 1 1 94 139 0 ¡ 36Kr 1 56 Ba 1 3X n • 19.20 Estimates show that the total energy output of the (d) 53 4 1 sun is 5 3 1026 J/s. What is the corresponding mass 24Cr 1 2α ¡ 0n 1 X 20 20 loss in kg/s of the sun? (e) 8O ¡ 9F 1 X • 19.8 Complete the following nuclear equations and iden- • 19.21 Calculate the nuclear binding energy (in J) and the binding energy per nucleon of the following isotopes: tify X in each case: (a) 37Li (7.01600 amu) and (b) 17 35 Cl (34.95952 amu). (a) 135 135 53 I ¡ 54 Xe 1 X (b) 19K ¡ 210β 1 X 40 • 19.22 Calculate the nuclear binding energy (in J) and the binding energy per nucleon of the following (c) 59 1 56 27Co 1 0n ¡ 25Mn 1 X isotopes: (a) 42He (4.0026 amu) and (b) 184 74W (d) 92U 1 0n ¡ 40Zr 1 135 235 1 99 52Te 1 2X (183.9510 amu).  Questions & Problems 893 Natural Radioactivity times as many X as there are Y. If the half-life of X Review Questions is 2.0 d, calculate the half-life of Y in days. 19.34 Determine the symbol AZX for the parent nucleus 19.23 Discuss factors that lead to nuclear decay. whose α decay produces the same daughter as the 19.24 Outline the principle for dating materials using 0 220 21β decay of 85At. radioactive isotopes. Problems Nuclear Transmutation Review Questions • 19.25 Fill in the blanks in the following radioactive decay 19.35 What is the difference between radioactive decay series: α β β and nuclear transmutation? (a) 232Th ¡ ¡ ¡ 228Th α β α 19.36 How is nuclear transmutation achieved in practice? (b) 235U ¡ ¡ ¡ 227Ac α 233 β α (c) ¡ Pa ¡ ¡ Problems • 19.26 A radioactive substance undergoes decay as follows: • 19.37 Write balanced nuclear equations for the following reactions and identify X: Time (days) Mass (g) (a) X(p,α)126C, (b) 27 55 13Al(d,α)X, (c) 25Mn(n,γ)X 0 500 • 19.38 Write balanced nuclear equations for the following 1 389 reactions and identify X: 2 303 (a) 80 9 10 34Se(d,p)X, (b) X(d,2p) 3Li, (c) 5B(n,α)X 3 236 19.39 Describe how you would prepare astatine-211, start- 4 184 ing with bismuth-209. 5 143 • 19.40 A long-cherished dream of alchemists was to produce 6 112 gold from cheaper and more abundant elements. This dream was finally realized when 198 80Hg was converted Calculate the first-order decay constant and the half- into gold by neutron bombardment. Write a balanced life of the reaction. equation for this reaction. • 19.27 The radioactive decay of Tl-206 to Pb-206 has a half-life of 4.20 min. Starting with 5.00 3 1022 at- Nuclear Fission oms of Tl-206, calculate the number of such atoms Review Questions left after 42.0 min. • 19.28 A freshly isolated sample of 90Y was found to have • 19.41 Define nuclear fission, nuclear chain reaction, and critical mass. an activity of 9.8 3 105 disintegrations per minute at 1:00 p.m. on December 3, 2003. At 2:15 p.m. on 19.42 Which isotopes can undergo nuclear fission? December 17, 2003, its activity was redetermined 19.43 Explain how an atomic bomb works. and found to be 2.6 3 104 disintegrations per min- 19.44 Explain the functions of a moderator and a control ute. Calculate the half-life of 90Y. rod in a nuclear reactor. 19.29 Why do radioactive decay series obey first-order 19.45 Discuss the differences between a light water and a kinetics? heavy water nuclear fission reactor. What are the • 19.30 In the thorium decay series, thorium-232 loses a to- advantages of a breeder reactor over a conventional tal of 6 α particles and 4 β particles in a 10-stage nuclear fission reactor? process. What is the final isotope produced? 19.46 No form of energy production is without risk. • 19.31 Strontium-90 is one of the products of the fission of Make a list of the risks to society involved in fuel- uranium-235. This strontium isotope is radioactive, ing and operating a conventional coal-fired electric with a half-life of 28.1 yr. Calculate how long (in yr) power plant, and compare them with the risks of it will take for 1.00 g of the isotope to be reduced to fueling and operating a nuclear fission-powered 0.200 g by decay. electric plant. • 19.32 Consider the decay series A ¡ B ¡ C ¡ D Nuclear Fusion Review Questions where A, B, and C are radioactive isotopes with half- lives of 4.50 s, 15.0 days, and 1.00 s, respectively, 19.47 Define nuclear fusion, thermonuclear reaction, and and D is nonradioactive. Starting with 1.00 mole of plasma. A, and none of B, C, or D, calculate the number of 19.48 Why do heavy elements such as uranium undergo moles of A, B, C, and D left after 30 days. fission while light elements such as hydrogen and 19.33 Two radioactive isotopes X and Y have the same lithium undergo fusion? molar amount at t 5 0. A week later, there are four 19.49 How does a hydrogen bomb work? 894 Chapter 19 ■ Nuclear Chemistry 19.50 What are the advantages of a fusion reactor over a an odd number of protons and/or an odd number of fission reactor? What are the practical difficulties in neutrons. What is the significance of the even num- operating a large-scale fusion reactor? bers of protons and neutrons in this case? • 19.59 Tritium, 3H, is radioactive and decays by electron Uses of Isotopes emission. Its half-life is 12.5 yr. In ordinary water the ratio of 1H to 3H atoms is 1.0 3 1017 to 1. Problems (a) Write a balanced nuclear equation for tritium 19.51 Describe how you would use a radioactive iodine decay. (b) How many disintegrations will be isotope to demonstrate that the following process is observed per minute in a 1.00-kg sample of water? in dynamic equilibrium: • 19.60 (a) What is the activity, in millicuries, of a 0.500-g 21 2 sample of 237 93Np? (This isotope decays by α-particle PbI2 (s) Δ Pb (aq) 1 2I (aq) emission and has a half-life of 2.20 3 106 yr.) 19.52 Consider the following redox reaction: (b) Write a balanced nuclear equation for the decay of 237 93Np. IO2 2 4 (aq) 1 2I (aq) 1 H2O(l2 ¡ • 19.61 The following equations are for nuclear reactions I2 (s) 1 IO2 2 3 (aq) 1 2OH (aq) that are known to occur in the explosion of an atomic When KIO4 is added to a solution containing iodide bomb. Identify X. ions labeled with radioactive iodine-128, all the ra- (a) 235 1 140 92U 1 0n ¡ 56Ba 1 30n 1 X 1 dioactivity appears in I2 and none in the IO2 3 ion. (b) 235 1 144 90 92U 1 0n ¡ 55Cs 1 37Rb 1 2X What can you deduce about the mechanism for the (c) 92U 1 0n ¡ 35Br 1 310n 1 X 235 1 87 redox process? (d) 235 1 160 72 92U 1 0n ¡ 62Sm 1 30Zn 1 4X 19.53 Explain how you might use a radioactive tracer to • 19.62 Calculate the nuclear binding energies, in J/nucleon, show that ions are not completely motionless in for the following species: (a) 10B (10.0129 amu), crystals. (b) 11B (11.00931 amu), (c) 14N (14.00307 amu), 19.54 Each molecule of hemoglobin, the oxygen carrier in (d) 56Fe (55.9349 amu). blood, contains four Fe atoms. Explain how you • 19.63 Write complete nuclear equations for the following would use the radioactive 59 26Fe (t12 5 46 days) to processes: (a) tritium, 3H, undergoes β decay; show that the iron in a certain food is converted into (b) 242Pu undergoes α-particle emission; (c) 131I un- hemoglobin. dergoes β decay; (d) 251Cf emits an α particle. 19.64 The nucleus of nitrogen-18 lies above the stability Additional Problems belt. Write an equation for a nuclear reaction by 19.55 In the chapter, we saw that the unit curie corre- which nitrogen-18 can achieve stability. sponds to exactly 3.70 3 1010 nuclear disintegration 19.65 Why is strontium-90 a particularly dangerous iso- per second for 1 g of radium. Derive this unit given tope for humans? that the half-life of 226 3 88Ra is 1.6 3 10 yr. 19.66 How are scientists able to tell the age of a fossil? 19.56 Manganese-50 (red spheres) decays via 110 β particle 19.67 After the Chernobyl accident, people living close to emission with a half-life of 0.282 s. (a) Write a bal- the nuclear reactor site were urged to take large anced nuclear equation for the process. (b) From the amounts of potassium iodide as a safety precaution. diagram shown here, determine how many half-lives What is the chemical basis for this action? have elapsed. (The green spheres represent the de- 19.68 Astatine, the last member of Group 7A, can be pre- cay product.) pared by bombarding bismuth-209 with α particles. (a) Write an equation for the reaction. (b) Represent the equation in the abbreviated form, as discussed in Section 19.4. 19.69 To detect bombs that may be smuggled onto air- planes, the Federal Aviation Administration (FAA) will soon require all major airports in the United States to install thermal neutron analyzers. The ther- mal neutron analyzer will bombard baggage with low-energy neutrons, converting some of the nitrogen-14 nuclei to nitrogen-15, with simultaneous emission of γ rays. Because nitrogen content is usu- ally high in explosives, detection of a high dosage of γ rays will suggest that a bomb may be present. 19.57 How does a Geiger counter work? (a) Write an equation for the nuclear process. 19.58 Nuclei with an even number of protons and an even (b) Compare this technique with the conventional number of neutrons are more stable than those with X-ray detection method.  Questions & Problems 895 19.70 Explain why achievement of nuclear fusion in the • 19.78 As a result of being exposed to the radiation re- laboratory requires a temperature of about 100 mil- leased during the Chernobyl nuclear accident, the lion degrees Celsius, which is much higher than dose of iodine-131 in a person’s body is 7.4 mCi that in the interior of the sun (15 million degrees (1 mCi 5 1 3 1023 Ci). Use the relationship rate 5 lN Celsius). to calculate the number of atoms of iodine-131 to 19.71 Tritium contains one proton and two neutrons. There which this radioactivity corresponds. (The half-life is no proton-proton repulsion present in the nucleus. of 131I is 8.1 d.) Why, then, is tritium radioactive? 19.79 Referring to the Chemistry in Action essay on p. 890, • 19.72 The carbon-14 decay rate of a sample obtained from why is it highly unlikely that irradiated food would a young tree is 0.260 disintegration per second per become radioactive? gram of the sample. Another wood sample prepared • 19.80 From the definition of curie, calculate Avogadro’s from an object recovered at an archaeological exca- number, given that the molar mass of 226Ra is vation gives a decay rate of 0.186 disintegration per 226.03 g/mol and that it decays with a half-life of second per gram of the sample. What is the age of 1.6 3 103 yr. the object? • 19.81 As of 2011, elements 113 through 118 have all been • 19.73 The usefulness of radiocarbon dating is limited to synthesized. Element 113 (Uut) was formed by the objects no older than 50,000 yr. What percent of the alpha decay of element 115 (Uup); element 114 (Uuq) carbon-14, originally present in the sample, remains was created by bombarding 244Pu with 48Ca; element after this period of time? 115 (Uup) was created by bombarding 243Am with 48 • 19.74 The radioactive potassium-40 isotope decays to Ca; element 116 (Uuh) was created by bombard- argon-40 with a half-life of 1.2 3 109 yr. (a) Write a ing 248Cm with 48Ca; element 117 (Uus) was created balanced equation for the reaction. (b) A sample of by bombarding 249Bk with 48Ca; element 118 (Uuo) moon rock is found to contain 18 percent potassium-40 was created by bombarding 249Cf with 48Ca. Write and 82 percent argon by mass. Calculate the age of an equation for each synthesis. Predict the chemi- the rock in years. cal properties of these elements. (Before transura- 19.75 Both barium (Ba) and radium (Ra) are members nium elements are given proper names, they are of Group 2A and are expected to exhibit similar temporarily assigned three-letter symbols all starting chemical properties. However, Ra is not found in with U.) barium ores. Instead, it is found in uranium ores. 19.82 Sources of energy on Earth include fossil fuels, Explain. geothermal, gravitational, hydroelectric, nuclear fission, nuclear fusion, solar, wind. Which of • 19.76 Nuclear waste disposal is one of the major concerns these have a “nuclear origin,” either directly or of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration indirectly? must be given to the heat released during nuclear 19.83 A person received an anonymous gift of a decora- decay. As an example, consider the β decay of 90Sr tive box, which he placed on his desk. A few (89.907738 amu): months later he became ill and died shortly after- ward. After investigation, the cause of his death 90 38Sr ¡ 90 39Y 1 210β t 12 5 28.1 yr was linked to the box. The box was air-tight and had no toxic chemicals on it. What might have The 90Y (89.907152 amu) further decays as follows: killed the man? 19.84 Identify two of the most abundant radioactive ele- 90 90 39Y ¡ 40 Zr 1 210 β t 12 5 64 h ments that exist on Earth. Explain why they are still present. (You may need to consult a handbook of Zirconium-90 (89.904703 amu) is a stable isotope. chemistry.) (a) Use the mass defect to calculate the energy released (in joules) in each of the above two decays. • 19.85 (a) Calculate the energy released when an U-238 isotope decays to Th-234. The atomic masses are (The mass of the electron is 5.4857 3 1024 amu.) given by: U-238: 238.0508 amu; Th-234: (b) Starting with one mole of 90Sr, calculate the 234.0436 amu; He-4: 4.0026 amu. (b) The energy number of moles of 90Sr that will decay in a year. released in (a) is transformed into the kinetic (c) Calculate the amount of heat released (in kilo- energy of the recoiling Th-234 nucleus and the α joules) corresponding to the number of moles of 90Sr particle. Which of the two will move away faster? decayed to 90Zr in (b). Explain. 19.77 Calculate the energy released (in joules) from the following fusion reaction: • 19.86 Cobalt-60 is an isotope used in diagnostic medicine and cancer treatment. It decays with γ ray emission. 2 1H 1 31H ¡ 42He 1 10n Calculate the wavelength of the radiation in nano- meters if the energy of the γ ray is 2.4 3 10213 The atomic masses are 12H 5 2.0140 amu, 13H 5 3.01603 J/photon. amu, 42He 5 4.00260 amu, 10n 5 1.008665 amu.  896 Chapter 19 ■ Nuclear Chemistry 19.87 Americium-241 is used in Current 19.94 The diagram here shows part of the thorium decay smoke detectors because it has series. Write a nuclear equation for each step of de- a long half-life (458 yr) and its cay. Use the ZAX symbol for each isotope. emitted α particles are ener- getic enough to ionize air mol- ecules. Given the schematic 144 diagram of a smoke detector, 241Am 142 explain how it works. Battery 19.88 The constituents of wine contain, among others, car- N 140 bon, hydrogen, and oxygen atoms. A bottle of wine 138 was sealed about 6 yr ago. To confirm its age, which of the isotopes would you choose in a radioactive 136 dating study? The half-lives of the isotopes are: 13 C: 5730 yr; 15O: 124 s; 3H: 12.5 yr. Assume that 86 88 90 92 the activities of the isotopes were known at the time Z the bottle was sealed. 19.89 Name two advantages of a nuclear-powered subma- • 19.95 The half-life of 27Mg is 9.50 min. (a) Initially there were 4.20 3 1012 27Mg nuclei present. How many rine over a conventional submarine. 27 Mg nuclei are left 30.0 min later? (b) Calculate the 19.90 In 1997, a scientist at a nuclear research center in 27 Mg activities (in Ci) at t 5 0 and t 5 30.0 min. Russia placed a thin shell of copper on a sphere of (c) What is the probability that any one 27Mg nu- highly enriched uranium-235. Suddenly, there cleus decays during a 1-s interval? What assumption was a huge burst of radiation, which turned the air is made in this calculation? blue. Three days later, the scientist died of radia- 19.96 The radioactive isotope 238Pu, used in pacemakers, tion damage. Explain what caused the accident. decays by emitting an alpha particle with a half-life (Hint: Copper is an effective metal for reflecting of 86 yr. (a) Write an equation for the decay process. neutrons.) (b) The energy of the emitted alpha particle is 9.0 3 • 19.91 A radioactive isotope of copper decays as follows: 10213 J, which is the energy per decay. Assuming 64 64 that all the alpha particle energy is used to run the Cu ¡ Zn 1 210β t 12 5 12.8 h pacemaker, calculate the power output at t 5 0 and Starting with 84.0 g of 64Cu, calculate the quantity t 5 10 yr. Initially 1.0 mg of 238Pu was present in of 64Zn produced after 18.4 h. the pacemaker. (Hint: After 10 yr, the activity of • 19.92 A 0.0100-g sample of a radioactive isotope with a half- the isotope decreases by 8.0 percent. Power is mea- life of 1.3 3 109 yr decays at the rate of 2.9 3 104 dpm. sured in watts or J/s.) Calculate the molar mass of the isotope. • 19.97 (a) Assuming nuclei are spherical in shape, show that 19.93 In each of the diagrams shown here, identify the its radius (r) is proportional to the cube root of mass isotopes involved and the type of decay process. Use number (A). (b) In1 general, the radius of a nucleus is the ZAX symbol for each isotope. given by r 5 r0 A3, where r0, the proportionality con- stant, is given by 1.2 3 10215 m. Calculate the volume of the 238U nucleus. 40 • 19.98 The quantity of a radioactive material is often mea- N sured by its activity (measured in curies or millicu- 39 ries) rather than by its mass. In a brain scan procedure, a 70-kg patient is injected with 20.0 mCi of 99mTc, which decays by emitting γ-ray photons 32 33 with a half-life of 6.0 h. Given that the RBE of these Z photons is 0.98 and only two-thirds of the photons (a) are absorbed by the body, calculate the rem dose received by the patient. Assume all of the 99mTc nu- clei decay while in the body. The energy of a gamma 66 136 photon is 2.29 3 10214 J. N N 135 • 19.99 Describe, with appropriate equations, nuclear pro- 65 cesses that lead to the formation of the noble gases 134 He, Ne, Ar, Kr, Xe, and Rn. (Hint: Helium is formed from radioactive decay, neon is formed from the 47 48 84 85 86 positron emission of 22Na, the formation of Ar, Xe, Z Z and Rn are discussed in the chapter, and Kr is pro- (b) (c) duced from the fission of 235U.)  Answers to Practice Exercises 897 19.100 Modern designs of atomic bombs contain, in addition 19.104 An electron and a positron are accelerated to nearly to uranium or plutonium, small amounts of tritium the speed of light before colliding in a particle ac- and deuterium to boost the power of explosion. What celerator. The ensuing collision produces an exotic is the role of tritium and deuterium in these bombs? particle having a mass many times that of a proton. 19.101 What is the source of heat for volcanic activities on Does the result violate the law of conservation of Earth? mass? • 19.102 Alpha particles produced from radioactive decays 19.105 The volume of an atom’s nucleus is 1.33 3 10242 m3. eventually pick up electrons from the surroundings The nucleus contains 110 neutrons. Identify the to form helium atoms. Calculate the volume (mL) of atom and write the symbol of the atom as ZAX. He collected at STP when 1.00 g of pure 226Ra is (Hint: See Problem 19.97.) stored in a closed container for 100 yr. (Hint: Focus- 19.106 In the chapter, we learned to calculate the nuclear ing only on half-lives that are short compared to 100 binding energy, which pertains to the stability of a years and ignoring minor decay schemes in Table particular nucleus. It is also possible to estimate the 19.3, first show that there are 5 α particles generated binding energy of a single nucleon (neutron or pro- per 226Ra decay to 206Pb.) ton) to the remainder of the nucleus. (a) From the • 19.103 In 2006, an ex-KGB agent was murdered in London. following nuclear equation and nuclear masses, cal- Subsequent investigation showed that the cause of culate the binding energy of a single neutron: death was poisoning with the radioactive isotope 14 13 210 7N ¡ 7N 1 10n Po, which was added to his drinks/food. (a) 210Po is prepared by bombarding 209Bi with neutrons. Write an (Useful information: 147N: 14.003074 amu; 137N: equation for the reaction. (b) Who discovered the ele- 13.005738 amu; 10n: 1.00866 amu.) (b) By a similar ment polonium? (Hint: See Appendix 1.) (c) The half- procedure, we can calculate the binding energy of a life of 210Po is 138 d. It decays with the emission of an single proton according to the equation α particle. Write an equation for the decay process. (d) Calculate the energy of an emitted α particle. 14 7N ¡ 13 6C 1 11p Assume both the parent and daughter nuclei to have zero kinetic energy. The atomic masses are: (Useful information: 136C: 13.003355 amu; 1 1p: 210 Po (209.98285 amu), 206Pb (205.97444 amu), 42α 1.00794 amu.) Comment on your results. (4.00150 amu). (e) Ingestion of 1 μg of 210Po could prove fatal. What is the total energy released by this quantity of 210Po? Interpreting, Modeling & Estimating 19.107 Which of the following poses a greater health haz- a certain period of time, the car consumed 0.2 g of ard: a radioactive isotope with a short half-life or a deuterium fuel. How many gallons of gasoline would radioactive isotope with a long half-life? Assume have to be burned to equal the energy generated by equal molar amounts and the same type of radiation the deuterium fuel? (For useful information about and comparable energies per particle emitted. gasoline energy content, see Problem 17.73.) 19.108 To start a deuterium-deuterium fusion reaction, it 19.110 The leakage of radioactive materials to the environ- has been estimated that each nucleus needs an initial ment when a nuclear reactor core malfunctions is kinetic energy of about 4 3 10214 J. What would be often made worse by explosions at the nuclear plant the corresponding temperature for the process? Why caused by hydrogen gas, as was the case in Fuku- is this temperature value an overestimate? shima, Japan, in 2011. Explain what caused the hy- 19.109 In a science fiction novel a nuclear engineer designed drogen explosion. (Useful information: The nuclear a car powered by deuterium-deuterium fusion. Over fuel rods are held in zirconium alloy tubes.) Answers to Practice Exercises 78 19.1 34Se. 19.2 2.63 3 10210 J; 1.26 3 10212 J/nucleon. 19.3 46Pd 1 42α ¡ 109 106 1 47Ag 1 1p. CHEMICAL M YS TERY The Art Forgery of the Twentieth Century H an van Meegeren must be one of the few forgers ever to welcome technical analy- sis of his work. In 1945 he was captured by the Dutch police and accused of sell- ing a painting by the Dutch artist Jan Vermeer (1632–1675) to Nazi Germany. This was a crime punishable by death. Van Meegeren claimed that not only was the painting in question, entitled The Woman Taken in Adultery, a forgery, but he had also produced other “Vermeers.” To prove his innocence, van Meegeren created another Vermeer to demonstrate his skill at imitating the Dutch master. He was acquitted of charges of collaboration with the enemy, but was convicted of forgery. He died of a heart attack before he could serve the 1-yr sentence. For 20 yr after van Meegeren’s death, art scholars debated whether at least one of his alleged works, Christ and His Disciples at Emmaus, was a fake or a real Vermeer. The mystery was solved in 1968 using a radiochemical technique. White lead—lead hydroxy carbonate [Pb3(OH)2(CO3)2]—is a pigment used by artists for centuries. The metal in the compound is extracted from its ore, galena (PbS), which contains uranium and its daughter products in radioactive equilibrium with it. By radioac- tive equilibrium we mean that a particular isotope along the decay series is formed from its precursor as fast as it breaks down by decay, and so its concentration (and its radio- activity) remains constant with time. This radioactive equilibrium is disturbed in the chemical extraction of lead from its ore. Two isotopes in the uranium decay series are of particular importance in this process: 226Ra (t 12 5 1600 yr) and 210Pb (t 12 5 21 yr) . (See Table 19.3.) Most 226Ra is removed during the extraction of lead from its ore, but 210Pb eventually ends up in the white lead, along with the stable isotope of lead (206Pb). No longer supported by its relatively long-lived ancestor, 226Ra, 210Pb begins to decay without replenishment. This process continues until the 210Pb activity is once more in equilibrium with the much smaller quantity of 226Ra that survived the separation process. Assuming the concentration ratio of 210Pb to 226Ra is 100:1 in the sample after extraction, it would take 270 yr to reestablish radioactive equilibrium for 210Pb. If Vermeer did paint Emmaus around the mid-seventeenth century, the radioactive equilibrium would have been restored in the white lead pigment by 1960. But this was not the case. Radiochemical analysis showed that the paint used was less than 100 yr old. Therefore, the painting could not have been the work of Vermeer. Chemical Clues 226 210 1. Write equations for the decay of Ra and Pb. 2. Consider the following consecutive decay series: A ¡ B ¡ C 898  “Christ and His Disciples at Emmaus,” a painting attributed to Han van Meegeren. where A and B are radioactive isotopes and C is a stable isotope. Given that the half- life of A is 100 times that of B, plot the concentrations of all three species versus time on the same graph. If only A was present initially, which species would reach radioactive equilibrium? 3. The radioactive decay rates for 210Pb and 226Ra in white lead paint taken from Emmaus in 1968 were 8.5 and 0.8 disintegrations per minute per gram of lead (dpm/g), respec- tively. (a) How many half-lives of 210Pb had elapsed between 1660 and 1968? (b) If Vermeer had painted Emmaus, what would have been the decay rate of 210Pb in 1660? Comment on the reasonableness of this rate value. 4. To make his forgeries look authentic, van Meegeren re-used canvases of old paintings. He rolled one of his paintings to create cracks in the paint to resemble old works. X-ray examination of this painting showed not only the underlying painting, but also the cracks in it. How did this discovery reveal to the scientists that the painting on top was of a more recent origin? 899  CHAPTER 20 Chemistry in the Atmosphere Lightning causes atmospheric nitrogen and oxygen to form nitric oxide, which is eventually converted to nitrates. CHAPTER OUTLINE A LOOK AHEAD 20.1 Earth’s Atmosphere  We begin by examining the regions and composition of Earth’s atmos- phere. (20.1) 20.2 Phenomena in the Outer Layers of the Atmosphere  We then study a natural phenomenon—aurora borealis—and a human-made phenomenon—the glow of space shuttles—in the outer layers of the atmos- 20.3 Depletion of Ozone phere. (20.2) in the Stratosphere  Next, we study the depletion of ozone in the stratosphere and its detrimental 20.4 Volcanoes effects and ways to slow the progress. (20.3) 20.5 The Greenhouse Effect  Focusing on events in the troposphere, we first examine volcanic erup- tions. (20.4) 20.6 Acid Rain  We study the cause and effect of greenhouse gases and ways to curtail the 20.7 Photochemical Smog emission of carbon dioxide and other harmful gases. (20.5) 20.8 Indoor Pollution  We see that acid rain is largely caused by human activities such as the burning of fossil fuels and roasting of metal sulfides. We discuss ways to minimize sulfur dioxide and nitrogen oxides productions. (20.6)  Another human-made pollution is smog formation, which is the result of the heavy use of automobiles. We examine mechanisms of smog formation and ways to reduce the pollution. (20.7)  Finally, we consider some examples of indoor pollutants such as radon, carbon dioxide and carbon monoxide, and formaldehyde. (20.8) 900  20.1 Earth’s Atmosphere 901 W e have studied basic definitions in chemistry, and we have examined the properties of gases, liquids, solids, and solutions. We have discussed chemical bonding and intermolecular forces and seen how chemical kinetics and chemical equilibrium concepts help us understand the nature of chemical reactions. It is appropriate at this stage to apply our knowledge to the study of one extremely important system: the atmosphere. Although Earth’s atmosphere is fairly simple in composition, its chemistry is very complex and not fully understood. The chemical processes that take place in our atmosphere are induced by solar radiation, but they are intimately connected to natural events and human activities on Earth’s surface. In this chapter, we will discuss the structure and composition of the atmosphere, together with some of the chemical processes that occur there. In addition, we will take a look at the major sources of air pollution and prospects for controlling them. 20.1 Earth’s Atmosphere Earth is unique among the planets of our solar system in having an atmosphere that is chemically active and rich in oxygen. Mars, for example, has a much thinner atmosphere that is about 90 percent carbon dioxide. Jupiter, on the other hand, has no solid surface; it is made up of 90 percent hydrogen, 9 percent helium, and 1 percent other substances. It is generally believed that 3 to 4 billion years ago, Earth’s atmosphere con- sisted mainly of ammonia, methane, and water. There was little, if any, free oxygen present. Ultraviolet (UV) radiation from the sun probably penetrated the atmosphere, rendering the surface of Earth sterile. However, the same UV radiation may have triggered the chemical reactions (perhaps beneath the surface) that eventually led to life on Earth. Primitive organisms used energy from the sun to break down carbon dioxide (produced by volcanic activity) to obtain carbon, which they incorporated in their own cells. The major by-product of this process, called photosynthesis, is oxygen. Another important source of oxygen is the photodecomposition of water vapor by UV light. Over time, the more reactive gases such as ammonia and meth- ane have largely disappeared, and today our atmosphere consists mainly of oxygen and nitrogen gases. Biological processes determine to a great extent the atmospheric concentrations of these gases, one of which is reactive (oxygen) and the other unre- active (nitrogen). Table 20.1 shows the composition of dry air at sea level. The total mass of the atmosphere is about 5.3 3 1018 kg. Water is excluded from this table because its concentration in air can vary drastically from location to location. Figure 20.1 shows the major processes involved in the cycle of nitrogen in nature. Molecular nitrogen, with its triple bond, is a very stable molecule. How- Table 20.1 ever, through biological and industrial nitrogen fixation, the conversion of molec- Composition of Dry Air ular nitrogen into nitrogen compounds, atmospheric nitrogen gas is converted at Sea Level into nitrates and other compounds suitable for assimilation by algae and plants. Another important mechanism for producing nitrates from nitrogen gas is light- Composition ning. The steps are Gas (% by Volume) N2 78.09 electrical N2 (g) 1 O2 (g) ¬¬¡ energy 2NO(g) O2 20.95 2NO(g) 1 O2 (g) ¬¬¡ 2NO2 (g) Ar 0.93 2NO2 (g) 1 H2O(l) ¬¬¡ HNO2 (aq) 1 HNO3 (aq) CO2 0.040 Ne 0.0018 About 30 million tons of HNO3 are produced this way annually. Nitric acid is He 0.000524 converted to nitrate salts in the soil. These nutrients are taken up by plants, which Kr 0.00011 in turn are ingested by animals. Animals use the nutrients from plants to make Xe 0.000006 proteins and other essential biomolecules. Denitrification reverses nitrogen fixation 902 Chapter 20 ■ Chemistry in the Atmosphere Atmospheric nitrogen Atmospheric fixation Fixed juvenile nitrogen Industrial fixation Biological fixation Protein Denitrification Igneous rocks Plant and animal wastes, dead organisms Nitrate reduction Nitrous oxide Ammonium Nitrite Nitrate To groundwater Figure 20.1 The nitrogen cycle. Although the supply of nitrogen in the atmosphere is virtually inexhaustible, it must be combined with hydrogen or oxygen before it can be assimilated by higher plants, which in turn are consumed by animals. Juvenile nitrogen is nitrogen that has not previously participated in the nitrogen cycle. to complete the cycle. For example, certain anaerobic organisms decompose ani- mal wastes as well as dead plants and animals to produce free molecular nitrogen from nitrates. The main processes of the global oxygen cycle are shown in Figure 20.2. This cycle is complicated by the fact that oxygen takes so many different chemical forms. Atmospheric oxygen is removed through respiration and various industrial processes (mostly combustion), which produce carbon dioxide. Photosynthesis is the major mechanism by which molecular oxygen is regenerated from carbon diox- ide and water. Scientists divide the atmosphere into several different layers according to tem- perature variation and composition (Figure 20.3). As far as visible events are concerned, the most active region is the troposphere, the layer of the atmosphere that contains about 80 percent of the total mass of air and practically all of the  atmosphere’s water vapor. The troposphere is the thinnest layer of the atmo- sphere (10 km), but it is where all the dramatic events of weather—rain, lightning, hurricanes—occur. Temperature decreases almost linearly with increasing altitude in this region.  20.1 Earth’s Atmosphere 903 O High-energy ultraviolet radiation H H2O H2O O2 O2 O2 ⫹ 2CO 2CO2 OH O3 Ozone screen O O2 CO O2 O2 Volcanism CO2 CO2 CO2 Oxidative weathering Phytoplankton 4FeO ⫹ O2 2Fe2O3 Photic zone H2O ⫹ CO2 H2CO3 HCO3ⴚ ⫹ H⫹ 2HCOⴚ 3 CO2ⴚ 3 CaCO3 Ca2⫹ Sediments Sediments H2O Figure 20.2 The oxygen cycle. The cycle is complicated because oxygen appears in so many chemical forms and combinations, primarily as molecular oxygen, in water, and in organic and inorganic compounds. Above the troposphere is the stratosphere, which consists of nitrogen, oxygen, and ozone. In the stratosphere, the air temperature rises with altitude. This warming effect is the result of exothermic reactions triggered by UV radiation from the sun (to be discussed in Section 20.3). One of the products of this reaction sequence is ozone (O3), which, as we will see shortly, serves to prevent harmful UV rays from reaching Earth’s surface. In the mesosphere, which is above the stratosphere, the concentration of ozone and other gases is low, and the temperature decreases with increasing altitude. The thermosphere, or ionosphere, is the uppermost layer of the atmosphere. The rise in temperature in this region is the result of the bombardment of molecular oxygen and nitrogen and atomic species by energetic particles, such as electrons and protons, from the sun. Typical reactions are N2 ¡ 2N ¢H° 5 941.4 kJ/mol N ¡ N1 1 e2 ¢H° 5 1400 kJ/mol O2 ¡ O 1 2 1 e 2 ¢H° 5 1176 kJ/mol In reverse, these processes liberate the equivalent amount of energy, mostly as heat. Ionized particles are responsible for the reflection of radio waves back toward Earth. 904 Chapter 20 ■ Chemistry in the Atmosphere 500 km 1000˚C 400 Space station 300 950˚C Thermosphere 200 900˚C Aurora 700˚C borealis 100 165˚C Shooting star –80˚C 80 Mesosphere 50 0˚C Stratosphere –50˚C Ozone layer Concorde Troposphere 10 1˚C Mt. Pinatubo 0 Figure 20.3 Regions of Earth’s atmosphere. Notice the variation of temperature with altitude. Most of the phenomena shown here are discussed in the chapter.  20.2 Phenomena in the Outer Layers of the Atmosphere 905 20.2 Phenomena in the Outer Layers of the Atmosphere In this section, we will discuss two dazzling phenomena that occur in the outer regions of the atmosphere. One is a natural event. The other is a curious by-product of human space travel. Aurora Borealis and Aurora Australis Violent eruptions on the surface of the sun, called solar flares, result in the ejection of myriad electrons and protons into space, where they disrupt radio transmission and provide us with spectacular celestial light shows known as auroras (Figure 20.4). These electrons and protons collide with the molecules and atoms in Earth’s upper atmosphere, causing them to become ionized and electronically excited. Eventually, the excited molecules and ions return to the ground state with the emission of light. For example, an excited oxygen atom emits photons at wavelengths of 558 nm (green) and between 630 nm and 636 nm (red): O* ¡ O 1 hv where the asterisk denotes an electronically excited species and hv the emitted photon (see Section 7.2). Similarly, the blue and violet colors often observed in auroras result from the transition in the ionized nitrogen molecule: N1 1 2 * ¡ N2 1 hv The wavelengths for this transition fall between 391 and 470 nm. The incoming streams of solar protons and electrons are oriented by Earth’s magnetic field so that most auroral displays occur in doughnut-shaped zones about 2000 km in diameter centered on the North and South Poles. Aurora bore- alis is the name given to this phenomenon in the Northern Hemisphere. In the Southern Hemisphere, it is called aurora australis. Sometimes, the number of solar particles is so immense that auroras are also visible from other locations on Earth. Figure 20.4 Aurora borealis, commonly referred to as the northern lights. 906 Chapter 20 ■ Chemistry in the Atmosphere Example 20.1 The bond enthalpy of O2 is 498.7 kJ/mol. Calculate the maximum wavelength (nm) of a photon that can cause the dissociation of an O2 molecule. Strategy We want to calculate the wavelength of a photon that will break an O“O bond. Therefore, we need the amount of energy in one bond. The bond enthalpy of O2 is given in units of kJ/mol. The units needed for the energy of one bond are J/molecule. Once we know the energy in one bond, we can calculate the minimum frequency and maximum wavelength needed to dissociate one O2 molecule. The conversion steps are kJ/mol ¡ J/molecule ¡ frequency of photon ¡ wavelength of photon Solution First we calculate the energy required to break one O“O bond: 498.7 3 103 J 1 mol J 3 23 5 8.281 3 10219 1 mol 6.022 3 10 molecules molecule The energy of the photon is given by E 5 hv [Equation (7.2)]. Therefore, E 8.281 3 10219 J v5 5 h 6.63 3 10234 J ? s 5 1.25 3 1015 s21 Finally, we calculate the wavelength of the photon, given by l 5 c/v [see Equation (7.1)], as follows: 3.00 3 108 mys λ5 1.25 3 1015 s21 5 2.40 3 1027 m 5 240 nm Comment In principle, any photon with a wavelength of 240 nm or shorter can Similar problem: 20.11. dissociate an O2 molecule. Practice Exercise Calculate the wavelength (in nm) of a photon needed to dissociate an O3 molecule: O3 ¡ O 1 O2    ¢H° 5 107.2 kJ/mol The Mystery Glow of Space Shuttles A human-made light show that baffled scientists for several years is produced by space shuttles orbiting Earth. In 1983, astronauts first noticed an eerie orange glow on the outside surface of their spacecraft at an altitude about 300 km above Earth (Figure 20.5). The light, which usually extends about 10 cm away from the protec- tive silica heat tiles and other surface materials, is most pronounced on the parts of the shuttle facing its direction of travel. This fact led scientists to postulate that collision between oxygen atoms in the atmosphere and the fast-moving shuttle somehow produced the orange light. Spectroscopic measurements of the glow, as well as laboratory tests, strongly suggested that nitric oxide (NO) and nitrogen Figure 20.5 The glowing tail dioxide (NO2) also played a part. It is believed that oxygen atoms interact with section of the space shuttle viewed from inside the vehicle. nitric oxide adsorbed on (that is, bound to) the shuttle’s surface to form electroni- cally excited nitrogen dioxide: O 1 NO ¡ NO2* As the NO2* leaves the shell of the spacecraft, it emits photons at a wavelength of 680 nm (orange). NO2* ¡ NO2 1 hv  20.3 Depletion of Ozone in the Stratosphere 907 Support for this explanation came inadvertently in 1991, when astronauts aboard Discovery released various gases, including carbon dioxide, neon, xenon, and nitric oxide, from the cargo bay in the course of an unrelated experiment. Expelled one at a time, these gases scattered onto the surface of the shuttle’s tail. The nitric oxide caused the normal shuttle glow to intensify markedly, but the other gases had no effect on it. What is the source of the nitric oxide on the outside of the spacecraft? Scientists believe that some of it may come from the exhaust gases emitted by the shuttle’s rockets and that some of it is present in the surrounding atmosphere. The shuttle glow does not harm the vehicle, but it does interfere with spectroscopic measurements on distant objects made from the spacecraft. 20.3 Depletion of Ozone in the Stratosphere As mentioned earlier, ozone in the stratosphere prevents UV radiation emitted by the sun from reaching Earth’s surface. The formation of ozone in this region begins with the photodissociation of oxygen molecules by solar radiation at wavelengths below 240 nm: Photodissociation is the breaking of chemical bonds by radiant energy. UV O2 ¬¬¡ , 240 nm O1O (20.1) The highly reactive O atoms combine with oxygen molecules to form ozone as follows: O 1 O 2 1 M ¡ O3 1 M (20.2) where M is some inert substance such as N2. The role of M in this exothermic reaction is to absorb some of the excess energy released and prevent the spontaneous decom- position of the O3 molecule. The energy that is not absorbed by M is given off as heat. (As the M molecules themselves become de-excited, they release more heat to the surroundings.) In addition, ozone itself absorbs UV light between 200 and 300 nm: UV O3 ¡ O 1 O 2 (20.3) The process continues when O and O2 recombine to form O3 as shown in Equation (20.2), further warming the stratosphere. If all the stratospheric ozone were compressed into a single layer at STP on Earth, that layer would be only about 3 mm thick! Although the concentration of ozone in the stratosphere is very low, it is sufficient to filter out (that is, absorb) solar radiation Recycling feasible in the 200- to 300-nm range [see Equation (20.3)]. In the stratosphere, it acts as our Recycling not feasible protective shield against UV radiation, which can induce skin cancer, cause genetic mutations, and destroy crops and other forms of vegetation. The formation and destruction of ozone by natural processes is a dynamic equi- Auto air librium that maintains a constant concentration of ozone in the stratosphere. Since the conditioning Foam 21% mid-1970s scientists have been concerned about the harmful effects of certain chlo- insulation 20% rofluorocarbons (CFCs) on the ozone layer. The CFCs, which are generally known Commercial by the trade name Freons, were first synthesized in the 1930s. Some of the common refrigeration Other foam 17% ones are CFCl3 (Freon 11), CF2Cl2 (Freon 12), C2F3Cl3 (Freon 113), and C2F4Cl2 uses 13% Solvent (Freon 114). Because these compounds are readily liquefied, relatively inert, nontoxic, cleaning noncombustible, and volatile, they have been used as coolants in refrigerators and air 4% 11% 14% conditioners, in place of highly toxic liquid sulfur dioxide (SO2) and ammonia (NH3). Aerosols Large quantities of CFCs are also used in the manufacture of disposable foam products such as cups and plates, as aerosol propellants in spray cans, and as solvents to clean Others (sterilization, household refrigeration) newly soldered electronic circuit boards (Figure 20.6). In 1977, the peak year of production, nearly 1.5 3 106 tons of CFCs were produced in the United States. Most Figure 20.6 Uses of CFCs. Since 1978, the use of aerosol of the CFCs produced for commercial and industrial use are eventually discharged propellants has been banned in into the atmosphere. the United States. 908 Chapter 20 ■ Chemistry in the Atmosphere Because of their relative inertness, the CFCs slowly diffuse unchanged up to the stratosphere, where UV radiation of wavelengths between 175 nm and 220 nm causes them to decompose: CFCl3 ¡ CFCl2 1 Cl CF2Cl2 ¡ CF2Cl 1 Cl It can take years for CFCs to reach the The reactive chlorine atoms then undergo the following reactions: stratosphere. Cl is a homogeneous catalyst. Cl 1 O3 ¡ ClO 1 O2 (20.4) ClO 1 O ¡ Cl 1 O2 (20.5) The overall result [sum of Equations (20.4) and (20.5)] is the net removal of an O3 molecule from the stratosphere: O3 1 O ¡ 2O2 (20.6) The oxygen atoms in Equation (20.5) are supplied by the photochemical decomposition of molecular oxygen and ozone described earlier. Note that the Cl atom plays the role of a catalyst in the reaction mechanism scheme represented by Equations (20.4) and (20.5) because it is not used up and therefore can take part in many such reactions. One Cl atom can destroy up to 100,000 O3 molecules before it is removed by some other reaction. The ClO (chlorine monoxide) species is an intermediate because it is produced in the first elementary step [Equation (20.4)] and consumed in the second step [Equation (20.5)]. The preceding mechanism for the destruction of ozone has been supported by the detection of ClO in the stratosphere in recent years (Figure 20.7). As can be seen, the concentration of O3 decreases in regions that have high amounts of ClO. Another group of compounds that can destroy stratospheric ozone are the nitrogen oxides, generally denoted as NOx. (Examples of NOx are NO and NO2.) These com- pounds come from the exhausts of high-altitude supersonic aircraft and from human and natural activities on Earth. Solar radiation decomposes a substantial amount of the other nitrogen oxides to nitric oxide (NO), which participates in the destruction of ozone as follows: O3 ¡ O2 1 O NO 1 O3 ¡ NO2 1 O2 NO2 1 O ¡ NO 1 O2 Overall: 2O3 ¡ 3O2 Figure 20.7 The variations in the concentrations of ClO and O3 2.5 with latitude. O3 Chlorine monoxide (ppb by volume) 1.0 2.0 Ozone (ppm by volume) 1.5 0.5 1.0 ClO 0.5 0 0 63°S 72°S Latitude  20.3 Depletion of Ozone in the Stratosphere 909 Figure 20.8 In recent years, scientists have found that the ozone layer in the stratosphere over the South Pole has become thinner. This map, based on data collected over a number of years, shows the depletion of ozone in red. (Source: NASA/Goddard Space Flight Center) In this case, NO is the catalyst and NO2 is the intermediate. Nitrogen dioxide also reacts with chlorine monoxide to form chlorine nitrate: ClO 1 NO2 ¡ ClONO2 Chlorine nitrate is relatively stable and serves as a “chlorine reservoir,” which plays a role in the depletion of the stratospheric ozone over the North and South Poles. Polar Ozone Holes In the mid-1980s, evidence began to accumulate that an “Antarctic ozone hole” devel- oped in late winter, depleting the stratospheric ozone over Antarctica by as much as 50 percent (Figure 20.8). In the stratosphere, a stream of air known as the “polar vortex” circles Antarctica in winter. Air trapped within this vortex becomes extremely cold during the polar night. This condition leads to the formation of ice particles known as polar stratospheric clouds (PSCs) (Figure 20.9). Acting as a heterogeneous Figure 20.9 Polar stratospheric clouds containing ice particles can catalyze the formation of Cl atoms and lead to the destruction of ozone. 910 Chapter 20 ■ Chemistry in the Atmosphere catalyst, these PSCs provide a surface for reactions converting HCl (emitted from Earth) and chlorine nitrate to more reactive chlorine molecules: HCl 1 ClONO2 ¡ Cl2 1 HNO3 By early spring, the sunlight splits molecular chlorine into chlorine atoms Cl2 1 hn ¡ 2Cl which then attack ozone as shown earlier. The situation is not as severe in the warmer Arctic region, where the vortex does not persist quite as long. Studies have shown that ozone levels in this region have declined between 4 and 8 percent in the past decade. Volcanic eruptions, such as that of Mount Pinatubo in the Philippines in 1991, inject large quantities of dust-sized particles and sulfuric acid aerosols (see p. 547) into the atmosphere. These particles can perform the same catalytic function as the ice crystals at the South Pole. As a result, the Arctic hole is expected to grow larger during the next few years. Recognizing the serious implications of the loss of ozone in the stratosphere, nations throughout the world have acknowledged the need to drastically curtail or totally stop the production of CFCs. In 1978 the United States was one of the few countries to ban the use of CFCs in hair sprays and other aerosols. An international treaty—the Montreal protocol—was signed by most industrialized nations in 1987, setting targets for cutbacks in CFC production and the complete elimination of these substances by the year 2000. While some progress has been made in this respect, many nations have not been able to abide by the treaty because of the importance of CFCs to their economies. Recycling could play a significant supplementary role in preventing CFCs already in appliances from escaping into the atmosphere. As Figure 20.6 shows, more than half of the CFCs in use are recoverable. An intense effort is under way to find CFC substitutes that are effective refriger- ants but not harmful to the ozone layer. One of the promising candidates is hydro- chlorofluorocarbon 134a, or HCFC-134a (CH2FCF3). The presence of the hydrogen atoms makes the compound more susceptible to oxidation in the lower atmosphere, The OH radical is formed by a series of so that it never reaches the stratosphere. Specifically, it is attacked by the hydroxyl complex reactions in the troposphere that are driven by sunlight. radical in the troposphere: CH2FCF3 1 OH ¡ CHFCF3 1 H2O The CHFCF3 fragments react with oxygen, eventually decomposing to CO2, water, and hydrogen fluoride that are removed by rainwater. Although it is not clear whether the CFCs already released to the atmosphere will eventually result in catastrophic damage to life on Earth, it is conceivable that the depletion of ozone can be slowed by reducing the availability of Cl atoms. Indeed, some chemists have suggested sending a fleet of planes to spray 50,000 tons of ethane (C2H6) or propane (C3H8) high over the South Pole in an attempt to heal the hole in the ozone layer. Being a reactive species, the chlorine atom would react with the hydrocarbons as follows: Cl 1 C2H6 ¡ HCl 1 C2H5 Cl 1 C3H8 ¡ HCl 1 C3H7 The products of these reactions would not affect the ozone concentration. A less realistic plan is to rejuvenate the ozone layer by producing large quantities of ozone and releas- ing it into the stratosphere from airplanes. Technically this solution is feasible, but it would be enormously costly and it would require the collaboration of many nations.  20.4 Volcanoes 911 Having discussed the chemistry in the outer regions of Earth’s atmosphere, we will focus in Sections 20.4 through 20.8 on events closer to us, that is, in the ­troposphere. 20.4 Volcanoes Volcanic eruptions, Earth’s most spectacular natural displays of energy, are instru- mental in forming large parts of Earth’s crust. The upper mantle, immediately under the crust, is nearly molten. A slight increase in heat, such as that generated by the movement of one crustal plate under another, melts the rock. The molten rock, called magma, rises to the surface and generates some types of volcanic eruptions (Figure 20.10). An active volcano emits gases, liquids, and solids. The gases spewed into the atmosphere include primarily N2, CO2, HCl, HF, H2S, and water vapor. It is estimated that volcanoes are the source of about two-thirds of the sulfur in the air. On the slopes of Mount St. Helens, which last erupted in 1980, deposits of elemental sulfur are visible near the eruption site. At high temperatures, the hydrogen sulfide gas given off by a volcano is oxidized by air: 2H2S(g) 1 3O2 (g) ¡ 2SO2 (g) 1 2H2O(g) Some of the SO2 is reduced by more H2S from the volcano to elemental sulfur and water: Sulfur deposits at a volcanic site. 2H2S(g) 1 SO2 (g) ¡ 3S(s) 1 2H2O(g) The rest of the SO2 is released into the atmosphere, where it reacts with water to form acid rain (see Section 20.6). The tremendous force of a volcanic eruption carries a sizable amount of gas into the stratosphere. There SO2 is oxidized to SO3, which is eventually converted to sul­ furic acid aerosols in a series of complex mechanisms. In addition to destroying ozone in the stratosphere (see p. 910), these aerosols can also affect climate. Because the stratosphere is above the atmospheric weather patterns, the aerosol clouds often ­persist for more than a year. They absorb solar radiation and thereby cause a drop in tem- perature at Earth’s surface. However, this cooling effect is local rather than global, because it depends on the site and frequency of volcanic eruptions. Figure 20.10 A volcanic eruption on the island of Hawaii. 912 Chapter 20 ■ Chemistry in the Atmosphere 20.5 The Greenhouse Effect A dramatic illustration of the greenhouse Although carbon dioxide is only a trace gas in Earth’s atmosphere, with a concentra- effect is found on Venus where the atmosphere is 97 percent CO2 and the tion of about 0.033 percent by volume (see Table 20.1), it plays a critical role in atmospheric pressure is 9 3 106 Pa controlling our climate. The so-called greenhouse effect describes the trapping of heat (equivalent to 89 atm). The surface temperature of Venus is about 730 K! near Earth’s surface by gases in the atmosphere, particularly carbon dioxide. The glass roof of a greenhouse transmits visible sunlight and absorbs some of the outgo- ing infrared (IR) radiation, thereby trapping the heat. Carbon dioxide acts somewhat like a glass roof, except that the temperature rise in the greenhouse is due mainly to the restricted air circulation inside. Calculations show that if the atmosphere did not contain carbon dioxide, Earth would be 30°C cooler! Figure 20.11 shows the carbon cycle in our global ecosystem. The transfer of carbon dioxide to and from the atmosphere is an essential part of the carbon cycle. Carbon dioxide is produced when any form of carbon or a carbon-containing com- pound is burned in an excess of oxygen. Many carbonates give off CO2 when heated, and all give off CO2 when treated with acid: CaCO3 (s) ¡ CaO(s) 1 CO2 (g) CaCO3 (s) 1 2HCl(aq) ¡ CaCl2 (aq) 1 H2O(l) 1 CO2 (g) Carbon dioxide in atmosphere Plant respiration Assimilation by plants Soil Animal respiration respiration Litter Dead organisms Root respiration Decomposition Figure 20.11 The carbon cycle.  20.5 The Greenhouse Effect 913 Figure 20.12 The incoming radiation from the sun and the outgoing radiation from Earth’s Incoming solar radiation surface. Energy Outgoing terrestrial radiation 5000 15,000 25,000 Wavelength (nm) Carbon dioxide is also a by-product of the fermentation of sugar: yeast C6H12O6 (aq) ¡ 2C2H5OH(aq) 1 2CO2 (g) glucose ethanol Carbohydrates and other complex carbon-containing molecules are consumed by ani- mals, which respire and release CO2 as an end product of metabolism: C6H12O6 (aq) 1 6O2 (g) ¡ 6CO2 (g) 1 6H2O(l) As mentioned earlier, another major source of CO2 is volcanic activity. Carbon dioxide is removed from the atmosphere by photosynthetic plants and certain microorganisms: 6CO2 (g) 1 6H2O(l) ¡ C6H12O6 (aq) 1 6O2 (g) This reaction requires radiant energy (visible light). After plants and animals die, the carbon in their tissues is oxidized to CO2 and returns to the atmosphere. In addition, there is a dynamic equilibrium between atmospheric CO2 and carbonates in the oceans and lakes. The solar radiant energy received by Earth is distributed over a band of wave- lengths between 100 and 5000 nm, but much of it is concentrated in the 400- to Stable form 700-nm range, which is the visible region of the spectrum (Figure 20.12). By contrast, the thermal radiation emitted by Earth’s surface is characterized by wavelengths longer than 4000 nm (IR region) because of the much lower average surface temperature Stretched compared to that of the sun. The outgoing IR radiation can be absorbed by water and carbon dioxide, but not by nitrogen and oxygen. All molecules vibrate, even at the lowest temperatures. The energy associated with molecular vibration is quantized, much like the electronic energies of atoms and Compressed molecules. To vibrate more energetically, a molecule must absorb a photon of a specific wavelength in the IR region. First, however, its dipole moment must change Figure 20.13 Vibrational motion of a diatomic molecule. Chemical during the course of a vibration. [Recall that the dipole moment of a molecule is the bonds can be stretched and product of the charge and the distance between charges (see p. 423).] Figure 20.13 compressed like a spring. 914 Chapter 20 ■ Chemistry in the Atmosphere (a) (b) Figure 20.15 Two of the four ways a carbon dioxide molecule can vibrate. The vibration in (a) does not result in a change in dipole moment, but the vibration in (b) renders the molecule IR active. shows how a diatomic molecule can vibrate. If the molecule is homonuclear like N2 and O2, there can be no change in the dipole moment; the molecule has a zero dipole moment no matter how far apart or close together the two atoms are. We call such molecules IR-inactive because they cannot absorb IR radiation. On the other hand, all heteronuclear diatomic molecules are IR-active; that is, they all can absorb IR radiation because their dipole moments constantly change as the bond lengths change. A polyatomic molecule can vibrate in more than one way. Water, for example, can vibrate in three different ways, as shown in Figure 20.14. Because water is a polar molecule, it is easy to see that any of these vibrations results in a change in dipole moment because there is a change in bond length. Therefore, a H2O molecule is IR-active. Carbon dioxide has a linear geometry and is nonpolar. Figure 20.15 shows two of the four ways a CO2 molecule can vibrate. One of Figure 20.14 The three different them [Figure 20.15(a)] symmetrically displaces atoms from the center of gravity modes of vibration of a water and will not create a dipole moment, but the other vibration [Figure 20.15(b)] is molecule. Each mode of vibration IR-active because the dipole moment changes from zero to a maximum value in can be imagined by moving the atoms along the arrows and then one direction and then reaches the same maximum value when it changes to the reversing their directions. other extreme position. Upon receiving a photon in the IR region, a molecule of H2O or CO2 is promoted to a higher vibrational energy level: H2O 1 hn ¡ H2O* CO2 1 hn ¡ CO2* (the asterisk denotes a vibrationally excited molecule). These energetically excited molecules soon lose their excess energy either by collision with other molecules or by spontaneous emission of radiation. Part of this radiation is emitted to outer space Electricity and part returns to Earth’s surface. production 35% Although the total amount of water vapor in our atmosphere has not altered Cars and noticeably over the years, the concentration of CO2 has been rising steadily since trucks the turn of the twentieth century as a result of the burning of fossil fuels (petro- 30% leum, natural gas, and coal). Figure 20.16 shows the percentages of CO2 emitted Industry 24% due to human activities in the United States in 1998, and Figure 20.17 shows the variation of carbon dioxide concentration over a period of years, as measured in 11% Hawaii. In the Northern Hemisphere, the seasonal oscillations are caused by removal of carbon dioxide by photosynthesis during the growing season and its Residential heating buildup during the fall and winter months. Clearly, the trend is toward an increase in CO2. The current rate of increase is about 1 ppm (1 part CO2 per million parts Figure 20.16 Sources of carbon dioxide emission in the air) by volume per year, which is equivalent to 9 3 109 tons of CO2! Scientists United States. Note that not all have estimated that by the year 2014 the CO2 concentration will exceed preindus- of the emitted CO2 enters the trial levels by about 40 percent. atmosphere. Some of it is taken up by carbon dioxide “sinks,” In addition to CO2 and H2O, other greenhouse gases, such as the CFCs, CH4, such as the ocean. NOx, and N2O also contribute appreciably to the warming of the atmosphere.  20.5 The Greenhouse Effect 915 400 CO2 concentration (ppm by volume) 380 360 340 320 1960 1970 1980 1990 2000 2010 Figure 20.17 Yearly variation of carbon dioxide concentration at Mauna Loa, Hawaii. The general trend clearly points to an increase of carbon dioxide in the atmosphere. In May 2013, the mean concentration recorded was 400 ppm. Figure 20.18 shows the gradual increase in temperature over the years and Figure 20.19 shows the relative contributions of the greenhouse gases to global warming. It is predicted by some meteorologists that should the buildup of greenhouse gases The difference in global temperatures between today and the last ice age is continue at its current rate, Earth’s average temperature will increase by about 1° to only 4–5°C. 3°C in this century. Although a temperature increase of a few degrees may seem insignificant, it is actually large enough to disrupt the delicate thermal balance on Earth and could cause glaciers and icecaps to melt. Consequently, the sea level would rise and coastal areas would be flooded. To combat the greenhouse effect, we must lower carbon dioxide emission. This As more nations industrialize, the production of CO2 will increase can be done by improving energy efficiency in automobiles and in household heating appreciably. and lighting, and by developing nonfossil fuel energy sources, such as photovoltaic 0.6 Figure 20.18 The change in Annual average global temperature from 1850 to Five year average 2008. (Source: NASA Goddard Institute for 0.4 Space Studies) Temperature deviation (⬚C) 0.2 0 ⫺0.2 ⫺0.4 ⫺0.6 1860 1880 1900 1920 1940 1960 1980 2000 916 Chapter 20 ■ Chemistry in the Atmosphere cells. Nuclear energy is a viable alternative, but its use is highly controversial due to the difficulty of disposing of radioactive waste and the fact that nuclear power stations CO2 are more prone to accidents than conventional power stations (see Chapter 19). The 55% proposed phasing out of CFCs, the most potent greenhouse gases, will help to slow down the warming trend. The recovery of methane gas generated at landfills and the N2O reduction of natural gas leakages are other steps we could take to control CO2 emis- 6% sion. Finally, the preservation of the Amazon jungle, tropical forests in Southeast Asia, CH4 CFCs 15% and other large forests is vital to maintaining the steady-state concentration of CO2 24% in the atmosphere. Converting forests to farmland for crops and grassland for cattle may do irreparable damage to the delicate ecosystem and permanently alter the cli- mate pattern on Earth. Figure 20.19 Contribution to global warming by various greenhouse gases. The concentrations of CFCs and Example 20.2 methane are much lower than that of carbon dioxide. However, because they can absorb IR Which of the following gases qualify as a greenhouse gas: CO, NO, NO2, Cl2, H2, Ne? radiation much more effectively than CO2, they make an Strategy To behave as a greenhouse gas, either the molecule must possess a dipole appreciable contribution to moment or some of its vibrational motions must generate a temporary dipole moment. the overall warming effect. These conditions immediately rule out homonuclear diatomic molecules and atomic species. Solution Only CO, NO, and NO2, which are all polar molecules, qualify as greenhouse gases. Both Cl2 and H2 are homonuclear diatomic molecules, and Ne is Similar problem: 20.36. atomic. These three species are all IR-inactive. Practice Exercise Which of the following is a more effective greenhouse gas: CO or H2O? 20.6 Acid Rain Scientists have known about acid rain Every year acid rain causes hundreds of millions of dollars’ worth of damage to stone since the late nineteenth century, but it has been a public issue for only about buildings and statues throughout the world. The term “stone leprosy” is used by some 30 years. environmental chemists to describe the corrosion of stone by acid rain (Figure 20.20). Acid rain is also toxic to vegetation and aquatic life. Many well-documented cases show dramatically how acid rain has destroyed agricultural and forest lands and killed aquatic organisms (see Figure 15.10). Figure 20.20 The effect of acid rain on the marble statue of George Washington in Washington Square, New York City. The photos were taken 50 years apart (1944–1994).  20.6 Acid Rain 917 Figure 20.21 Mean precipitation pH in the United States in 2009. Most SO2 comes from the midwestern states. Prevailing winds carry the acid droplets formed over the Northeast. Nitrogen oxides also contribute to the acid rain formation. Precipitation in the northeastern United States has an average pH of about 4.3 (Figure 20.21). Because atmospheric CO2 in equilibrium with rainwater would not be expected to result in a pH less than 5.5, sulfur dioxide (SO2) and, to a lesser extent, nitrogen oxides from auto emissions are believed to be responsible for the high acidity of rainwater. Acidic oxides, such as SO2, react with water to give the corresponding acids. There are several sources of atmospheric SO2. Nature itself contributes much SO2 in the form of volcanic eruptions. Also, many metals exist combined with sulfur in nature. Extracting the metals often entails smelting, or roast- ing, the ores—that is, heating the metal sulfide in air to form the metal oxide and SO2. For example, 2ZnS(s) 1 3O2 (g) ¡ 2ZnO(s) 1 2SO2 (g) The metal oxide can be reduced more easily than the sulfide (by a more reactive metal or in some cases by carbon) to the free metal. Although smelting is a major source of SO2, the burning of fossil fuels in industry, in power plants, and in homes accounts for most of the SO2 emitted to the atmosphere. The sulfur content of coal ranges from 0.5 to 5 percent by mass, depending on the source of the coal. The sulfur content of other fossil fuels is similarly variable. Oil from the Middle East, for instance, is low in sulfur, while that from Venezuela has a high sulfur content. To a lesser extent, the nitrogen- containing compounds in oil and coal are converted to nitrogen oxides, which can also acidify rainwater. All in all, some 50 million to 60 million tons of SO2 are released into the atmo- sphere each year! In the troposphere, SO2 is almost all oxidized to H2SO4 in the form of aerosol, which ends up in wet precipitation or acid rain. The mechanism for the conversion of SO2 to H2SO4 is quite complex and not fully understood. The reaction is believed to be initiated by the hydroxyl radical (OH): OH 1 SO2 ¡ HOSO2 The HOSO2 radical is further oxidized to SO3: HOSO2 1 O2 ¡ HO2 1 SO3 918 Chapter 20 ■ Chemistry in the Atmosphere Figure 20.22 Common procedure for removing SO2 from Mostly CO2 and air burning fossil fuel. Powdered limestone decomposes into CaO, which reacts with SO2 to form CaSO3. The remaining SO2 is Smokestack reacted with an aqueous suspension of CaO to form CaSO3. S  O2 SO2 Purification chamber CaCO3 CaO  CO2 CaO  SO2 CaSO3 Aqueous suspension of CaO CaCO3 SO2, CO2 Furnace Air Air Coal CaSO3 The sulfur trioxide formed would then rapidly react with water to form sulfuric acid: SO3 1 H2O ¡ H2SO4 SO2 can also be oxidized to SO3 and then converted to H2SO4 on particles by hetero- geneous catalysis. Eventually, the acid rain can corrode limestone and marble (CaCO3). A typical reaction is CaCO3 (s) 1 H2SO4 (aq) ¡ CaSO4 (s) 1 H2O(l) 1 CO2 (g) Sulfur dioxide can also attack calcium carbonate directly: 2CaCO3 (s) 1 2SO2 (g) 1 O2 (g) ¡ 2CaSO4 (s) 1 2CO2 (g) There are two ways to minimize the effects of SO2 pollution. The most direct approach is to remove sulfur from fossil fuels before combustion, but this is techno- logically difficult to accomplish. A cheaper but less efficient way is to remove SO2 as it is formed. For example, in one process powdered limestone is injected into the power plant boiler or furnace along with coal (Figure 20.22). At high temperatures the following decomposition occurs: CaCO3 (s) ¡ CaO(s) 1 CO2 (g) limestone quicklime The quicklime reacts with SO2 to form calcium sulfite and some calcium sulfate: CaO(s) 1 SO2 (g) ¡ CaSO3 (s) 2CaO(s) 1 2SO2 (g) 1 O2 (g) ¡ 2CaSO4 (s) To remove any remaining SO2, an aqueous suspension of quicklime is injected into a purification chamber prior to the gases’ escape through the smokestack. Quicklime is also added to lakes and soils in a process called liming to reduce their acidity (Figure 20.23). Installing a sulfuric acid plant near a metal ore refining site is also an effective way to cut SO2 emission because the SO2 produced by roasting metal sulfides can be captured for use in the synthesis of sulfuric acid. This is a very sensible way to turn what is a pollutant in one process into a starting material for another process!  20.7 Photochemical Smog 919 Figure 20.23 Spreading calcium oxide (CaO) over acidified soil. This process is called liming. 20.7 Photochemical Smog The word “smog” was coined to describe the combination of smoke and fog that shrouded London during the 1950s. The primary cause of this noxious cloud was sulfur dioxide. Today, however, we are more familiar with photochemical smog, which is formed by the reactions of automobile exhaust in the presence of sunlight. Automobile exhaust consists mainly of NO, CO, and various unburned hydrocar- bons. These gases are called primary pollutants because they set in motion a series of photochemical reactions that produce secondary pollutants. It is the secondary pollutants—chiefly NO2 and O3—that are responsible for the buildup of smog. The heavy use of automobiles is Nitric oxide is the product of the reaction between atmospheric nitrogen and the cause of photochemical smog oxygen at high temperatures inside an automobile engine: formation. N2 (g) 1 O2 (g) ¡ 2NO(g) Once released into the atmosphere, nitric oxide is oxidized to nitrogen dioxide: 2NO(g) 1 O2 (g) ¡ 2NO2 (g) Sunlight causes the photochemical decomposition of NO2 (at a wavelength shorter than 400 nm) into NO and O: NO2 (g) 1 hn ¡ NO(g) 1 O(g) Atomic oxygen is a highly reactive species that can initiate a number of important reactions, one of which is the formation of ozone: O(g) 1 O2 (g) 1 M ¡ O3 (g) 1 M where M is some inert substance such as N2. Ozone attacks the C“C linkage in Ozone plays a dual role in the atmosphere: It is Dr. Jekyll in the stratosphere and rubber: Mr. Hyde in the troposphere. R R R R R R G D G DO G D H2O G D CPC  O3 88n C C 88n CPO  OPC  H2O2 D G DG G D G R R R OOOD R R R 920 Chapter 20 ■ Chemistry in the Atmosphere Figure 20.24 Typical variations with time in concentration of air pollutants on a smoggy day. Relative concentrations Hydrocarbons NO2 O3 NO 4 6 8 10 12 2 4 6 A.M. Noon P.M. where R represents groups of C and H atoms. In smog-ridden areas, this reaction can cause automobile tires to crack. Similar reactions are also damaging to lung tissues and other biological substances. Ozone can be formed also by a series of very complex reactions involving unburned hydrocarbons, nitrogen oxides, and oxygen. One of the products of these reactions is peroxyacetyl nitrate, PAN: CH3OCOOOOONO2 B O PAN is a powerful lachrymator, or tear producer, and causes breathing difficulties. Figure 20.24 shows typical variations with time of primary and secondary pol- lutants. Initially, the concentration of NO2 is quite low. As soon as solar radiation penetrates the atmosphere, more NO2 is formed from NO and O2. Note that the PAN concentration of ozone remains fairly constant at a low level in the early morning hours. As the concentration of unburned hydrocarbons and aldehydes increases in the air, the concentrations of NO2 and O3 also rise rapidly. The actual amounts, of course, depend on the location, traffic, and weather conditions, but their presence is always accompanied by haze (Figure 20.25). The oxidation of hydrocarbons produces various organic intermediates, such as alcohols and carboxylic acids, which are all less volatile than the hydrocarbons themselves. These substances eventually condense into small droplets of liquid. The dispersion of these droplets in air, called aerosol, scatters sunlight and reduces visibility. This interaction also makes the air look hazy. As the mechanism of photochemical smog formation has become better under- stood, major efforts have been made to reduce the buildup of primary pollutants. Most automobiles now are equipped with catalytic converters designed to oxidize CO and unburned hydrocarbons to CO2 and H2O and to reduce NO and NO2 to N2 and O2 (see Section 13.6). More efficient automobile engines and better public transportation systems would also help to decrease air pollution in urban areas. A recent techno- logical innovation to combat photochemical smog is to coat automobile radiators and air conditioner compressors with a platinum catalyst. So equipped, a running car can purify the air that flows under the hood by converting ozone and carbon monoxide The haze over the Smoky Mountains to oxygen and carbon dioxide: is caused by aerosols produced by the oxidation of hydrocarbons Pt emitted by pine trees. O3 (g) 1 CO(g) ¡ O2 (g) 1 CO2 (g)  20.8 Indoor Pollution 921 Figure 20.25 A smoggy day in Beijing. In a city like Los Angeles, where the number of miles driven in one day equals nearly 300 million, this approach would significantly improve the air quality and reduce the “high-ozone level” warnings frequently issued to its residents. In fact, a drive on the freeway would help to clean up the air! 20.8 Indoor Pollution Difficult as it is to avoid air pollution outdoors, it is no easier to avoid indoor pollution. The air quality in homes and in the workplace is affected by human activities, by con- struction materials, and by other factors in our immediate environment. The common indoor pollutants are radon, carbon monoxide and carbon dioxide, and formaldehyde. The Risk from Radon In a highly publicized case in the mid-1980s, an employee reporting for work at the Limerick Nuclear Power Plant in Pennsylvania set off the plant’s radiation monitor. Astonishingly, the source of his contamination turned out not to be the plant, but radon in his home! A lot has been said and written about the potential dangers of radon as an air pol- lutant. Just what is radon? Where does it come from? And how does it affect our health? Radon is a member of Group 8A (the noble gases). It is an intermediate product of the radioactive decay of uranium-238. All isotopes of radon are radioactive, but The uranium decay series is discussed in Chapter 19. radon-222 is the most hazardous because it has the longest half-life—3.8 days. Radon, 922 Chapter 20 ■ Chemistry in the Atmosphere ● > 8 pCi/L ● 4-8 pCi/L ● 2-4 pCi/L ● < 2 pCi/L Figure 20.26 Map of radon emission in the United States measured in picocuries per liter of air. which accounts for slightly over half the background radioactivity on Earth, is gener- ated mostly from the phosphate minerals of uranium (Figure 20.26). Since the 1970s, high levels of radon have been detected in homes built on reclaimed land above uranium mill tailing deposits. The colorless, odorless, and tasteless radon After cigarette smoking, radon is the gas enters a building through tiny cracks in the basement floor (Figure 20.27). It is leading cause of lung cancer in the United States. It is responsible for slightly soluble in water, so it can be spread in different media. Radon-222 is an perhaps 20,000 deaths per year. α-emitter. When it decays, it produces radioactive polonium-214 and polonium-218, which can build up to high levels in an enclosed space. These solid radioactive particles can adhere to airborne dust and smoke, which are inhaled into the lungs and deposited in the respiratory tract. Over a long period of time, the α particles emitted by polonium and its decay products, which are also radioactive, can cause lung cancer. What can be done to combat radon pollution indoors? The first step is to measure the radon level in the basement with a reliable test kit. Short-term and long-term kits are available (Figure 20.28). The short-term tests use activated charcoal (that is, heat- treated charcoal) to collect the decay products of radon over a period of several days. The container is sent to a laboratory where a technician measures the radioactivity (γ rays) from radon-decay products lead-214 and bismuth-214. Knowing the length of exposure, the lab technician back-calculates to determine radon concentration. The long- term test kits use a piece of special polymer film on which an α particle will leave a “track.” After several months’ exposure, the film is etched with a sodium hydroxide solution and the number of tracks counted. Knowing the length of exposure enables the technician to calculate the radon concentration. If the radon level is unacceptably high, Basement Radon h Uranium n Radium Figure 20.27 Radon usually enters houses through the foundation or basement walls. Figure 20.28 Home radon detectors: Long-term track etch (left) and short-term charcoal canister (right).  20.8 Indoor Pollution 923 then the house must be regularly ventilated. This precaution is particularly important in recently built houses, which are well insulated. A more effective way to prevent radon pollution is to reroute the gas before it gets into the house, for example, by installing a ventilation duct to draw air from beneath the basement floor to the outside. Currently there is considerable controversy regarding the health effects of radon. The first detailed studies of the effects of radon on human health were carried out in the 1950s when it was recognized that uranium miners suffered from an abnormally high incidence of lung cancer. Some scientists have challenged the validity of these studies because the miners were also smokers. It seems quite likely that there is a syn- ergistic effect between radon and smoking on the development of lung cancer. Radon- decay products will adhere not only to tobacco tar deposits in the lungs, but also to the solid particles in cigarette smoke, which can be inhaled by smokers and nonsmokers. More systematic studies are needed to evaluate the environmental impact of radon. In the meantime, the Environmental Protection Agency (EPA) has recommended remedial action where the radioactivity level due to radon exceeds 4 pico-curies (pCi) per liter of air. (A curie corresponds to 3.70 3 1010 disintegrations of radioactive nuclei per second; a picocurie is a trillionth of a curie, or 3.70 3 1022 disintegrations per second.) Example 20.3 The half-life of Rn-222 is 3.8 days. Starting with 1.0 g of Rn-222, how much will be left after 10 half-lives? Recall that radioactive decays obey first-order kinetics. Strategy All radioactive decays obey first-order kinetics. Therefore, its half-life is independent of the initial concentration. Solution After one half-life, the amount of Rn left is 0.5 3 1.0 g, or 0.5 g. After two half-lives, only 0.25 g of Rn remains. Generalizing the fraction of the isotope left after n half lives as (1/2)n, where n 5 10, we write 1 10 quantity of Rn-222 left 5 1.0 g 3 a b 2 5 9.8 3 1024 g An alternative solution is to calculate the first-order rate constant from the half-life. Next, use Equation (13.3) to calculate the concentration of radon after 10 half-lives. Try it. Similar problem: 20.73. Practice Exercise The concentration of Rn-222 in the basement of a house is 1.8 3 1026 mol/L. Assume the air remains static and calculate the concentration of the radon after 2.4 days. Carbon Dioxide and Carbon Monoxide Both carbon dioxide (CO2) and carbon monoxide (CO) are products of combustion. In the presence of an abundant supply of oxygen, CO2 is formed; in a limited supply of oxygen, both CO and CO2 are formed. The indoor sources of these gases are gas cooking ranges, woodstoves, space heaters, tobacco smoke, human respiration, and exhaust fumes from cars (in garages). Carbon dioxide is not a toxic gas, but it does have an asphyxiating effect (see Chemistry in Action on p. 531). In airtight buildings, the concentration of CO2 can reach as high as 2000 ppm by volume (compared with 300 ppm outdoors). Workers exposed to high concentrations of CO2 in skyscrapers and other sealed environments become fatigued more easily and have difficulty con- centrating. Adequate ventilation is the solution to CO2 pollution. Like CO2, CO is a colorless and odorless gas, but it differs from CO2 in that it is highly poisonous. The toxicity of CO lies in its unusual ability to bind very strongly to hemoglobin, the oxygen carrier in blood. Both O2 and CO bind to the Fe(II) ion 924 Chapter 20 ■ Chemistry in the Atmosphere in hemoglobin, but the affinity of hemoglobin for CO is about 200 times greater than that for O2 (see Chapter 25). Hemoglobin molecules with tightly bound CO (called carboxyhemoglobin) cannot carry the oxygen needed for metabolic processes. A small amount of CO intake can cause drowsiness and headache; death may result when about half the hemoglobin molecules are complexed with CO. The best first-aid response to CO poisoning is to remove the victim immediately to an area with a plentiful oxygen supply or to give mouth-to-mouth resuscitation. Formaldehyde Formaldehyde (CH2O) is a rather disagreeable-smelling liquid used as a preservative for laboratory specimens. Industrially, formaldehyde resins are used as bonding agents in building and furniture construction materials such as plywood and particle board. In addition, urea-formaldehyde insulation foams are used to fill wall cavities. The resins and foams slowly break down to release free formaldehyde, especially under acid and humid conditions. Low concentrations of formaldehyde in the air can cause drowsiness, nausea, headaches, and other respiratory ailments. Laboratory tests show that breathing high concentrations of formaldehyde can induce cancers in animals, CH2O and it is now also classified as a human carcinogen. The safe standard of formaldehyde in indoor air has been set at 0.1 ppm by volume. Because formaldehyde is a reducing agent, devices have been constructed to remove it by means of a redox reaction. Indoor air is circulated through an air purifier contain- ing an oxidant such as Al2O3/KMnO4, which converts formaldehyde to the less harmful and less volatile formic acid (HCOOH). Proper ventilation is the best way to remove formaldehyde. However, care should be taken not to remove the air from a room too quickly without replenishment, because a reduced pressure would cause the formalde- hyde resins to decompose faster, resulting in the release of more formaldehyde. Summary of Facts & Concepts 1. Earth’s atmosphere is made up mainly of nitrogen and 5. Carbon dioxide’s ability to absorb infrared radiation en- oxygen, plus a number of other trace gases. The chem- ables it to trap some of the outgoing heat from Earth, ical processes that go on in the atmosphere are influ- warming its surface. Other gases such as the CFCs and enced by solar radiation, volcanic eruption, and human methane also contribute to global warming. activities. 6. Sulfur dioxide, and to a lesser extent nitrogen oxides, 2. In the outer regions of the atmosphere the bombard- generated mainly from the burning of fossil fuels and ment of molecules and atoms by solar particles gives from the roasting of metal sulfides, causes acid rain. rise to auroras. The glow on space shuttles is caused 7. Photochemical smog is formed by the photochemical by excitation of molecules adsorbed on the shuttles’ reaction of automobile exhaust in the presence of sun- surface. light. It is a complex reaction involving nitrogen oxides, 3. Ozone in the stratosphere absorbs harmful UV radiation ozone, and hydrocarbons. in the 200- to 300-nm range and protects life under- 8. Indoor air pollution is caused by radon, a radioactive neath. For many years, chlorofluorocarbons have been gas formed during uranium decay; carbon monoxide destroying the ozone layer. and carbon dioxide, products of combustion; and form- 4. Volcanic eruptions can lead to air pollution, deplete aldehyde, a volatile organic substance released from ozone in the stratosphere, and affect climate. resins used in construction materials. Key Words Greenhouse effect, p. 912 Mesosphere, p. 903 Photochemical smog, p. 919 Thermosphere, p. 903 Ionosphere, p. 903 Nitrogen fixation, p. 901 Stratosphere, p. 903 Troposphere, p. 902  Questions & Problems 925 Questions & Problems • Problems available in Connect Plus excited oxygen atom at 558 nm. Calculate the en- Red numbered problems solved in Student Solutions Manual ergy difference between the two levels involved in the emission process. Earth’s Atmosphere Review Questions Depletion of Ozone in the Stratosphere • 20.1 Describe the regions of Earth’s atmosphere. Review Questions 20.2 Briefly outline the main processes of the nitrogen 20.13 Briefly describe the absorption of solar radiation in and oxygen cycles. the stratosphere by O2 and O3 molecules. 20.3 Explain why, for maximum performance, super- 20.14 Explain the processes that have a warming effect on sonic airplanes need to fly at a high altitude (in the the stratosphere. stratosphere). 20.15 List the properties of CFCs, and name four major 20.4 Jupiter’s atmosphere consists mainly of hydrogen uses of these compounds. (90 percent) and helium (9 percent). How does this mixture of gases contrast with the composition 20.16 How do CFCs and nitrogen oxides destroy ozone in of Earth’s atmosphere? Why does the composition the stratosphere? differ? 20.17 What causes the polar ozone holes? 20.18 How do volcanic eruptions contribute to ozone Problems destruction? • 20.5 Referring to Table 20.1, calculate the mole fraction 20.19 Describe ways to curb the destruction of ozone in of CO2 and its concentration in parts per million by the stratosphere. volume. 20.20 Discuss the effectiveness of some of the CFC • 20.6 Calculate the partial pressure of CO2 (in atm) in dry substitutes. air when the atmospheric pressure is 754 mmHg. 20.7 Describe the processes that result in the warming of Problems the stratosphere. • 20.21 Given that the quantity of ozone in the strato- • 20.8 Calculate the total mass (in kilograms) of nitrogen, sphere is equivalent to a 3.0-mm-thick layer of oxygen, and carbon dioxide gases in the atmosphere. ozone on Earth at STP, calculate the number of (Hint: See Problem 5.106 and Table 20.1. Use ozone molecules in the stratosphere and their 29.0 g/mol for the molar mass of air.) mass in kilograms. (Hint: The radius of Earth is 6371 km and the surface area of a sphere is 4πr2, Phenomena in the Outer Layers where r is the radius.) of the Atmosphere • 20.22 Referring to the answer in Problem 20.21, and as- Review Questions suming that the level of ozone in the stratosphere has already fallen 6.0 percent, calculate the number 20.9 What process gives rise to aurora borealis and aurora of kilograms of ozone that would have to be manu- australis? factured on a daily basis so that we could restore the 20.10 Why can astronauts not release oxygen atoms to test ozone to the original level in 100 yr. If ozone is the mechanism of shuttle glow? made according to the process 3O2(g) ¡ 2O3(g), how many kilojoules of energy would be required? Problems 20.23 Both Freon-11 and Freon-12 are made by the reac- • 20.11 The highly reactive OH radical (a species with an tion of carbon tetrachloride (CCl4) with hydrogen unpaired electron) is believed to be involved in fluoride. Write equations for these reactions. some atmospheric processes. Table 9.4 lists the 20.24 Why are CFCs not decomposed by UV radiation in bond enthalpy for the oxygen-to-hydrogen bond the troposphere? in OH as 460 kJ/mol. What is the longest wave- 20.25 The average bond enthalpies of the C¬Cl and length (in nm) of radiation that can bring about C¬F bonds are 340 kJ/mol and 485 kJ/mol, respec- the reaction tively. Based on this information, explain why the C¬Cl bond in a CFC molecule is preferentially OH(g) ¡ O(g) 1 H(g) broken by solar radiation at 250 nm. • 20.12 The green color observed in aurora borealis is pro- • 20.26 Like CFCs, certain bromine-containing compounds duced by the emission of a photon by an electronically such as CF3Br can also participate in the destruction  926 Chapter 20 ■ Chemistry in the Atmosphere of ozone by a similar mechanism starting with the • 20.41 The molar heat capacity of a diatomic molecule is Br atom: 29.1 J/K ? mol. Assuming the atmosphere contains only nitrogen gas and there is no heat loss, calculate CF3Br ¡ CF3 1 Br the total heat intake (in kilojoules) if the atmosphere Given that the average C¬Br bond enthalpy is warms up by 3°C during the next 50 yr. Given that 276 kJ/mol, estimate the longest wavelength required there are 1.8 3 1020 moles of diatomic molecules to break this bond. Will this compound be decom- present, how many kilograms of ice (at the North and posed in the troposphere only or in both the tropo- South Poles) will this quantity of heat melt at 0°C? sphere and stratosphere? (The molar heat of fusion of ice is 6.01 kJ/mol.) • 20.27 Draw Lewis structures for chlorine nitrate (ClONO2) 20.42 As mentioned in the chapter, spraying the stratosphere and chlorine monoxide (ClO). with hydrocarbons such as ethane and propane • 20.28 Draw Lewis structures for HCFC-123 (CF3CHCl2) should eliminate Cl atoms. What is the drawback of and CF3CFH2. this procedure if used on a large scale for an extended period of time? Volcanoes Review Questions Acid Rain Review Questions 20.29 What are the effects of volcanic eruptions on climate? 20.30 Classify the reaction between H2S and SO2 that 20.43 Name the gas that is largely responsible for the acid leads to the formation of sulfur at the site of a volcanic rain phenomenon. eruption. 20.44 List three detrimental effects of acid rain. 20.45 Briefly discuss two industrial processes that lead to The Greenhouse Effect acid rain. Review Questions 20.46 Discuss ways to curb acid rain. 20.31 What is the greenhouse effect? What is the criterion 20.47 Water and sulfur dioxide are both polar molecules for classifying a gas as a greenhouse gas? and their geometry is similar. Why is SO2 not con- 20.32 Why is more emphasis placed on the role of carbon sidered a major greenhouse gas? dioxide in the greenhouse effect than on that of water? 20.48 Describe the removal of SO2 by CaO (to form CaSO3) 20.33 Describe three human activities that generate carbon in terms of a Lewis acid-base reaction. dioxide. List two major mechanisms for the uptake of carbon dioxide. Problems 20.34 Deforestation contributes to the greenhouse effect in • 20.49 An electric power station annually burns 3.1 3 107 kg two ways. What are they? of coal containing 2.4 percent sulfur by mass. Calcu- 20.35 How does an increase in world population enhance late the volume of SO2 emitted at STP. the greenhouse effect? • 20.50 The concentration of SO2 in the troposphere over a 20.36 Is ozone a greenhouse gas? If so, sketch three ways certain region is 0.16 ppm by volume. The gas dis- an ozone molecule can vibrate. solves in rainwater as follows: 20.37 What effects do CFCs and their substitutes have on SO2 (g) 1 H2O(l ) Δ H1 (aq) 1 HSO23 (aq) Earth’s temperature? 20.38 Why are CFCs more effective greenhouse gases than Given that the equilibrium constant for the preceding methane and carbon dioxide? reaction is 1.3 3 1022, calculate the pH of the rain- water. Assume that the reaction does not affect the Problems partial pressure of SO2. • 20.39 The annual production of zinc sulfide (ZnS) is 4.0 3 104 tons. Estimate the number of tons of SO2 Photochemical Smog produced by roasting it to extract zinc metal. Review Questions • 20.40 Calcium oxide or quicklime (CaO) is used in steel- 20.51 What is photochemical smog? List the factors that making, cement manufacture, and pollution control. favor the formation of photochemical smog. It is prepared by the thermal decomposition of cal- cium carbonate: 20.52 What are primary and secondary pollutants? 20.53 Identify the gas that is responsible for the brown CaCO3 (s) ¡ CaO(s) 1 CO2 (g) color of photochemical smog. Calculate the yearly release of CO2 (in kilograms) to 20.54 The safety limits of ozone and carbon monoxide are the atmosphere if the annual production of CaO in 120 ppb by volume and 9 ppm by volume, respec- the United States is 1.7 3 1010 kg. tively. Why does ozone have a lower limit?  Questions & Problems 927 20.55 Suggest ways to minimize the formation of photo- • 20.66 A volume of 5.0 L of polluted air at 18.0°C and chemical smog. 747 mmHg is passed through lime water [an aqueous • 20.56 In which region of the atmosphere is ozone benefi- suspension of Ca(OH)2], so that all the carbon diox- cial? In which region is it detrimental? ide present is precipitated as CaCO3. If the mass of the CaCO3 precipitate is 0.026 g, calculate the percentage by volume of CO2 in the air sample. Problems • 20.57 Assume that the formation of nitrogen dioxide: Additional Problems 2NO(g) 1 O2 (g) ¡ 2NO2 (g) 20.67 Briefly describe the harmful effects of the following is an elementary reaction. (a) Write the rate law for substances: O3, SO2, NO2, CO, CH3COOONO2 this reaction. (b) A sample of air at a certain tem- (PAN), Rn. perature is contaminated with 2.0 ppm of NO by • 20.68 The equilibrium constant (KP) for the reaction volume. Under these conditions, can the rate law N2 (g) 1 O2 (g) Δ 2NO(g) be simplified? If so, write the simplified rate law. (c) Under the conditions described in (b), the half- is 4.0 3 10 at 25°C and 2.6 3 1026 at 1100°C, the 231 life of the reaction has been estimated to be temperature of a running car’s engine. Is this an en- 6.4 3 103 min. What would the half-life be if the dothermic or exothermic reaction? initial concentration of NO were 10 ppm? • 20.69 As stated in the chapter, carbon monoxide has a • 20.58 The gas-phase decomposition of peroxyacetyl nitrate much higher affinity for hemoglobin than oxygen (PAN) obeys first-order kinetics: does. (a) Write the equilibrium constant expression (Kc) for the following process: CH3COOONO2 ¡ CH3COOO 1 NO2 CO(g) 1 HbO2 (aq) Δ O2 (g) 1 HbCO(aq) with a rate constant of 4.9 3 1024 s21. Calculate the rate of decomposition in M/s if the concentration of where HbO2 and HbCO are oxygenated hemoglo- PAN is 0.55 ppm by volume. Assume STP conditions. bin and carboxyhemoglobin, respectively. (b) The composition of a breath of air inhaled by a person • 20.59 On a smoggy day in a certain city the ozone con- smoking a cigarette is 1.9 3 1026 mol/L CO and centration was 0.42 ppm by volume. Calculate the partial pressure of ozone (in atm) and the number 8.6 3 1023 mol/L O2. Calculate the ratio of [HbCO] of ozone molecules per liter of air if the tempera- to [HbO2], given that Kc is 212 at 37°C. ture and pressure were 20.0°C and 748 mmHg, 20.70 Instead of monitoring carbon dioxide, suggest an- respectively. other gas that scientists could study to substantiate 20.60 Which of the following settings is the most suitable the fact that CO2 concentration is steadily increasing for photochemical smog formation? (a) Gobi desert in the atmosphere. at noon in June, (b) New York City at 1 p.m. in July, • 20.71 In 1991 it was discovered that nitrous oxide (N2O) is (c) Boston at noon in January. Explain your choice. produced in the synthesis of nylon. This compound, which is released into the atmosphere, contributes both to the depletion of ozone in the stratosphere and Indoor Pollution to the greenhouse effect. (a) Write equations repre- Review Questions senting the reactions between N2O and oxygen atoms in the stratosphere to produce nitric oxide (NO), which 20.61 List the major indoor pollutants and their sources. is then oxidized by ozone to form nitrogen dioxide. (b) 20.62 What is the best way to deal with indoor pollution? Is N2O a more effective greenhouse gas than carbon 20.63 Why is it dangerous to idle a car’s engine in a poorly dioxide? Explain. (c) One of the intermediates in ny- ventilated place, such as the garage? lon manufacture is adipic acid [HOOC(CH2)4COOH]. 20.64 Describe the properties that make radon an indoor About 2.2 3 109 kg of adipic acid are consumed every pollutant. Would radon be more hazardous if 222Rn year. It is estimated that for every mole of adipic acid had a longer half-life? produced, 1 mole of N2O is generated. What is the maximum number of moles of O3 that can be de- stroyed as a result of this process per year? Problems • 20.72 A glass of water initially at pH 7.0 is exposed to • 20.65 A concentration of 8.00 3 102 ppm by volume of dry air at sea level at 20°C. Calculate the pH of the CO is considered lethal to humans. Calculate the mini- water when equilibrium is reached between atmo- mum mass of CO in grams that would become a lethal spheric CO2 and CO2 dissolved in the water, given concentration in a closed room 17.6 m long, 8.80 m that Henry’s law constant for CO2 at 20°C is wide, and 2.64 m high. The temperature and pressure 0.032 mol/L ? atm. (Hint: Assume no loss of water are 20.0°C and 756 mmHg, respectively. due to evaporation and use Table 20.1 to calculate  928 Chapter 20 ■ Chemistry in the Atmosphere the partial pressure of CO2. Your answer should 20.79 What is funny about the following cartoon? correspond roughly to the pH of rainwater.) • 20.73 A 14-m by 10-m by 3.0-m basement had a high ra- don content. On the day the basement was sealed off from its surroundings so that no exchange of air could take place, the partial pressure of 222Rn was 1.2 3 1026 mmHg. Calculate the number of 222Rn isotopes (t12 5 3.8 d) at the beginning and end of 31 days. Assume STP conditions. • 20.74 Ozone in the troposphere is formed by the follow- ing steps: NO2 ¡ NO 1 O (1) O 1 O2 ¡ O3 (2) The first step is initiated by the absorption of visible light (NO2 is a brown gas). Calculate the longest • 20.80 Calculate the standard enthalpy of formation (DH°f ) of ClO from the following bond enthalpies: Cl2: wavelength required for step (1) at 25°C. [Hint: You 242.7 kJ/mol; O2: 498.7 kJ/mol; ClO: 206 kJ/mol. need to first calculate DH and hence DU for (1). Next, determine the wavelength for decomposing NO2 20.81 Methyl bromide (CH3Br, b.pt. 3.6°C) is used as a from DU.] soil fumigant to control insects and weeds. It is also a marine by-product. Photodissociation of the 20.75 Although the hydroxyl radical (OH) is present only C¬Br bond produces Br atoms that can react with in a trace amount in the troposphere, it plays a cen- ozone similar to Cl, except more effectively. Do tral role in its chemistry because it is a strong oxidiz- you expect CH3Br to be photolyzed in the tropo- ing agent and can react with many pollutants as well sphere? The bond enthalpy of the C¬Br bond is as some CFC substitutes (see p. 910). The hydroxyl about 293 kJ/mol. radical is formed by the following reactions: • 20.82 The effective incoming solar radiation per unit area λ , 320 nm on Earth is 342 W/m2. Of this radiation, 6.7 W/m2 is O3 ¬¬¡ O* 1 O2 absorbed by CO2 at 14,993 nm in the atmosphere. O 1 H2O ¬¬¡ 2OH How many photons at this wavelength are absorbed where O* denotes an electronically excited atom. per second in 1 m2 by CO2? (1 W 5 1 J/s) (a) Explain why the concentration of OH is so small • 20.83 As stated in the chapter, about 50 million tons of even though the concentrations of O3 and H2O are sulfur dioxide are released into the atmosphere every quite large in the troposphere. (b) What property year. (a) If 20 percent of the SO2 is eventually con- makes OH a strong oxidizing agent? (c) The reaction verted to H2SO4, calculate the number of 1000-lb between OH and NO2 contributes to acid rain. Write marble statues the resulting acid rain can damage. As an equation for this process. (d) The hydroxyl radical an estimate, assume that the acid rain only destroys can oxidize SO2 to H2SO4. The first step is the for- the surface layer of each statue, which is made up of mation of a neutral HSO3 species, followed by its 5 percent of its total mass. (b) What is the other unde- reaction with O2 and H2O to form H2SO4 and the sirable result of the acid rain damage? hydroperoxyl radical (HO2). Write equations for these processes. • 20.84 Peroxyacetyl nitrate (PAN) undergoes thermal de- composition as follows: • 20.76 The equilibrium constant (KP) for the reaction CH3 (CO)OONO2 ¡ CH3 (CO)OO 1 NO2 2CO(g) 1 O2 (g) Δ 2CO2 (g) 90 The rate constant is 3.0 3 1024 s21 at 25°C. At the is 1.4 3 10 at 25°C. Given this enormous value, boundary between the troposphere and stratosphere, why doesn’t CO convert totally to CO2 in the where the temperature is about 240°C, the rate troposphere? constant is reduced to 2.6 3 1027 s21. (a) Calculate 20.77 A person was found dead of carbon monoxide poi- the activation energy for the decomposition of PAN. soning in a well-insulated cabin. Investigation (b) What is the half-life of the reaction (in minutes) showed that he had used a blackened bucket to heat at 25°C? water on a butane burner. The burner was found to 20.85 How are past temperatures determined from ice function properly with no leakage. Explain, with an cores obtained from the Artic or Antarctica? (Hint: appropriate equation, the cause of his death. Look up the stable isotopes of hydrogen and oxy- 20.78 The carbon dioxide level in the atmosphere today is gen. How does energy required for vaporization de- often compared with that in preindustrial days. Ex- pend on the masses of H2O molecules containing plain how scientists use tree rings and air trapped in different isotopes? How would you determine the polar ice to arrive at the comparison. age of an ice core?)  Answers to Practice Exercises 929 • 20.86 The balance between SO2 and SO3 is important in 20.88 The HO3 radical was once thought of as a temporary understanding acid rain formation in the troposphere. reservoir of atmospheric OH radicals. Draw a Lewis From the following information at 25°C structure of the species. S(s) 1 O2 (g) Δ SO2 (g) K1 5 4.2 3 1052 20.89 What is the difference between weather and cli- 128 mate? In the winter months of 2010, record snows 2S(s) 1 3O2 (g) Δ 2SO3 (g) K2 5 9.8 3 10 fell on the East Coast from Washington, D.C., to the calculate the equilibrium constant for the reaction Deep South. Can this occurrence be taken as evi- dence against global warming? Can you suggest one 2SO2 (g) 1 O2 (g) Δ 2SO3 (g) way in which global warming might actually be 20.87 Draw Lewis structures of the species in each step in responsible for what happened? the conversion of SO2 to H2SO4 discussed on p. 917. Interpreting, Modeling & Estimating 20.90 Estimate the annual production of carbon dioxide First, without doing any calculations, predict (in kilograms) by an average passenger car in the whether each of the reactions is endothermic or exo- United States. thermic. Then estimate DH° for each reaction, and 20.91 The following reactions are common in the strato- comment on your predictions and estimates. sphere: (a) NO2(g) ¡ NO(g) 1 O(g) (b) N2O(g) ¡ N2(g) 1 O(g) (c) H2(g) 1 O(g) ¡ OH(g) 1 H(g) (d) CH4(g) 1 O(g) ¡ OH(g) 1 CH3(g) Answers to Practice Exercises 20.1 1.12 3 103 nm. 20.2 H2O. 20.3 1.2 3 1026 mol/L.  CHAPTER 21 Metallurgy and the Chemistry of Metals Crystals of salt composed of sodium anion and a complex sodium cation with an organic compound called crown ether. CHAPTER OUTLINE A LOOK AHEAD 21.1 Occurrence of Metals  We first survey the occurrence of ores containing various metals. (21.1) 21.2 Metallurgical Processes  We then study the sequence of steps from the preparation of the ores to the production of metals. We focus mainly on the metallurgy of iron 21.3 Band Theory of Electrical and the making of steel. We also examine several methods of metal Conductivity purification. (21.2) 21.4 Periodic Trends in Metallic  Next, we study the properties of solids and see how the band theory Properties explains the difference between conductors (metals) and insulators. We learn the special properties of semiconductors. (21.3) 21.5 The Alkali Metals  We briefly examine the periodic trends in metallic properties. (21.4) 21.6 The Alkaline Earth Metals  For alkali metals we discuss sodium and potassium and focus on their 21.7 Aluminum preparations, properties, compounds, and uses. (21.5)  For alkaline earth metals we discuss magnesium and calcium and focus on their preparations, properties, compounds, and uses. (21.6)  Finally, we study the preparation, properties, compounds, and uses of a Group 3A metal—aluminum. (21.7) 930  21.1 Occurrence of Metals 931 U p to this point we have concentrated mainly on fundamental principles: theories of chemical bonding, intermolecular forces, rates and mechanisms of chemical reactions, equilibrium, the laws of thermodynamics, and electrochemistry. An understanding of these topics is necessary for the study of the properties of representative metallic elements and their compounds. The use and refinement of metals date back to early human history. For example, arche- ologists have found evidence that in the first millennium a.d. inhabitants of Sri Lanka used monsoon winds to run iron-smelting furnaces to produce high-carbon steel. Through the years, these furnaces could have been sources of steel for the legendary Damascus swords, known for their sharpness and durability. In this chapter, we will study the methods for extracting, refining, and purifying metals and examine the properties of metals that belong to the representative elements. We will emphasize (1) the occurrence and preparation of metals, (2) the physical and chemical proper- ties of some of their compounds, and (3) their uses in modern society and their roles in biological systems. 21.1 Occurrence of Metals Most metals come from minerals. A mineral is a naturally occurring substance with a range of chemical composition. A mineral deposit concentrated enough to allow economical recovery of a desired metal is known as ore. Table 21.1 lists the prin- cipal types of minerals, and Figure 21.1 shows a classification of metals according to their minerals. The most abundant metals, which exist as minerals in Earth’s crust, are aluminum, iron, calcium, magnesium, sodium, potassium, titanium, and manganese (see p. 49). Seawater is a rich source of some metal ions, including Na1, Mg21, and Ca21. Fur- thermore, vast areas of the ocean floor are covered with manganese nodules, which are made up mostly of manganese, along with iron, nickel, copper, and cobalt in a chemically combined state (Figure 21.2). Table 21.1 Principal Types of Minerals Type Minerals Uncombined metals Ag, Au, Bi, Cu, Pd, Pt Carbonates BaCO3 (witherite), CaCO3 (calcite, limestone), MgCO3 (magnesite), CaCO3 ? MgCO3 (dolomite), PbCO3 (cerussite), ZnCO3 (smithsonite) Halides CaF2 (fluorite), NaCl (halite), KCl (sylvite), Na3AlF6 (cryolite) Oxides Al2O3 ? 2H2O (bauxite), Al2O3 (corundum), Fe2O3 (hematite), Fe3O4 (magnetite), Cu2O (cuprite), MnO2 (pyrolusite), SnO2 (cassiterite), TiO2 (rutile), ZnO (zincite) Phosphates Ca3(PO4)2 (phosphate rock), Ca5(PO4)3OH (hydroxyapatite) Silicates Be3Al2Si6O18 (beryl), ZrSiO4 (zircon), NaAlSi3O8 (albite), Mg3(Si4O10)(OH)2 (talc) Sulfides Ag2S (argentite), CdS (greenockite), Cu2S (chalcocite), FeS2 (pyrite), HgS (cinnabar), PbS (galena), ZnS (sphalerite) Sulfates BaSO4 (barite), CaSO4 (anhydrite), PbSO4 (anglesite), SrSO4 (celestite), MgSO4 ? 7H2O (epsomite) 932 Chapter 21 ■ Metallurgy and the Chemistry of Metals 1 Sulfides Uncombined 18 1A 8A 2 Other compounds; 13 14 15 16 17 Chlorides see caption 2A 3A 4A 5A 6A 7A Li Be Oxides Na Mg 3 4 5 6 7 8 9 10 11 12 Al 3B 4B 5B 6B 7B 8B 1B 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Figure 21.1 Metals and their best-known minerals. Lithium is found in spodumene ( LiAlSi2O6 ), and beryllium in beryl (see Table 21.1). The rest of the alkaline earth metals are found in minerals that are carbonates and sulfates. The minerals for Sc, Y, and La are the phosphates. Some metals have more than one type of important mineral. For example, in addition to the sulfide, iron is found as the oxides hematite ( Fe2O3 ) and magnetite ( Fe3O4 ); and aluminum, in addition to the oxide, is found in beryl ( Be3 Al2Si6O18 ). Technetium ( Tc) is a synthetic element. 21.2 Metallurgical Processes Metallurgy is the science and technology of separating metals from their ores and of compounding alloys. An alloy is a solid solution either of two or more metals, or of a metal or metals with one or more nonmetals. The three principal steps in the recovery of a metal from its ore are (1) prepara- tion of the ore, (2) production of the metal, and (3) purification of the metal. Figure 21.2 Manganese nodules on the ocean floor. Preparation of the Ore In the preliminary treatment of an ore, the desired mineral is separated from waste materials—usually clay and silicate minerals—which are collectively called the gangue. One very useful method for carrying out such a separation is called flotation. In this process, the ore is finely ground and added to water containing oil and detergent. The liquid mixture is then beaten or blown to form a froth. The oil preferentially wets the mineral particles, which are then carried to the top in the froth, while the gangue settles to the bottom. The froth is skimmed off, allowed to collapse, and dried to recover the mineral particles. Another physical separation process makes use of the magnetic properties of certain minerals. Ferromagnetic metals are strongly attracted to magnets. The mineral magnetite (Fe3O4), in particular, can be separated from the gangue by using a strong electromagnet. Cobalt is another ferromagnetic metal. Mercury forms amalgams with a number of metals. An amalgam is an alloy of Unreactive metals such as gold and silver mercury with another metal or metals. Mercury can therefore be used to extract metal can be leached from the ores using the cyanide ions (see Section 22.3). from ore. Mercury dissolves the silver and gold in an ore to form a liquid amalgam, which is easily separated from the remaining ore. The gold or silver is recovered by distilling off mercury.  21.2 Metallurgical Processes 933 Table 21.2 Reduction Processes for Some Common Metals Metal Reduction Process activity of metals Lithium, sodium, magnesium, calcium Electrolytic reduction of the molten chloride Decreasing Aluminum Electrolytic reduction of anhydrous oxide (in molten cryolite) Chromium, manganese, titanium, Reduction of the metal oxide with a more electropositive vanadium, iron, zinc metal, or reduction with coke and carbon monoxide Mercury, silver, platinum, copper, gold These metals occur in the free (uncombined) state or can be obtained by roasting their sulfides Production of Metals Because metals in their combined forms always have positive oxidation numbers, the production of a free metal is a reduction process. Preliminary operations may be necessary to convert the ore to a chemical state more suitable for reduction. For example, an ore may be roasted to drive off volatile impurities and at the same time to convert the carbonates and sulfides to the corresponding oxides, which can be reduced more conveniently to yield the pure metals: CaCO3(s) ¡ CaO(s) 1 CO2(g) 2PbS(s) 1 3O2(g) ¡ 2PbO(s) 1 2SO2(g) This last equation points out the fact that the conversion of sulfides to oxides is a major source of sulfur dioxide, a notorious air pollutant (p. 917). How a pure metal is obtained by reduction from its combined form depends on the standard reduction potential of the metal (see Table 18.1). Table 21.2 outlines the reduction processes for several metals. Most major metallurgical processes now in use involve pyrometallurgy, procedures carried out at high temperatures. The reduction in these procedures may be accomplished either chemically or electrolytically. Chemical Reduction We can use a more electropositive metal as a reducing agent to separate a less elec- A more electropositive metal has a more negative standard reduction potential tropositive metal from its compound at high temperatures: (see Table 18.1). V2O5 (s) 1 5Ca(l) ¡ 2V(l) 1 5CaO(s) TiCl4 (g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2 (l) Cr2O3 (s) 1 2Al(s) ¡ 2Cr(l) 1 Al2O3 (s) 3Mn3O4 (s) 1 8Al(s) ¡ 9Mn(l) 1 4Al2O3 (s) In some cases, even molecular hydrogen can be used as a reducing agent, as in the preparation of tungsten (used as filaments in lightbulbs) from tungsten(VI) oxide: WO3 (s) 1 3H2 (g) ¡ W(s) 1 3H2O(g) Electrolytic Reduction Electrolytic reduction is suitable for very electropositive metals, such as sodium, mag- nesium, and aluminum. The process is usually carried out on the anhydrous molten oxide or halide of the metal: 2MO(l) ¡ 2M (at cathode) 1 O2 (at anode) 2MCl(l) ¡ 2M (at cathode) 1 Cl2 (at anode) We will describe the specific procedures later in this chapter. 934 Chapter 21 ■ Metallurgy and the Chemistry of Metals Figure 21.3 A blast furnace. CO, CO2 Charge (ore, limestone, coke) Iron ore, limestone, and coke are introduced at the top of the furnace. Iron is obtained from the ore by reduction with carbon. 200°C 3Fe2O3 + CO 2Fe3O4 + CO2 CaCO3 CaO + CO2 Solid charge descends Fe3O4 + CO 3FeO + CO2 Hot gases rise 700°C C + CO2 2CO FeO + CO Fe + CO2 1200°C Iron melts Molten slag forms 1500°C 2C + O2 2CO 2000°C Hot air blast Slag Molten iron The Metallurgy of Iron The extraction of iron from FeS2 leads Iron exists in Earth’s crust in many different minerals, such as iron pyrite (FeS2), sider- to SO2 production and acid rain (see Section 20.6). ite (FeCO3), hematite (Fe2O3), and magnetite (Fe3O4, often represented as FeO ? Fe2O3). Of these, hematite and magnetite are particularly suitable for the extraction of iron. The metallurgical processing of iron involves the chemical reduction of the minerals by carbon (in the form of coke) in a blast furnace (Figure 21.3). The concentrated iron ore, limestone (CaCO3), and coke are introduced into the furnace from the top. A blast of hot air is forced up the furnace from the bottom—hence the name blast furnace. The oxygen gas reacts with the carbon in the coke to form mostly carbon monoxide and some carbon dioxide. These reactions are highly exothermic, and as the hot CO and CO2 gases rise, they react with the iron oxides in different temperature zones, as shown in Figure 21.3. The key steps in the extraction of iron are 3Fe2O3 (s) 1 CO(g) ¡ 2Fe3O4 (s) 1 CO2 (g) Fe3O4 (s) 1 CO(g) ¡ 3FeO(s) 1 CO2 (g) FeO(s) 1 CO(g) ¡ Fe(l) 1 CO2 (g) CaCO3 and other compounds that are The limestone decomposes in the furnace as follows: used to form a molten mixture with the impurities in the ore for easy removal are called flux. CaCO3 (s) ¡ CaO(s) 1 CO2 (g) The calcium oxide then reacts with the impurities in the iron, which are mostly sand (SiO2) and aluminum oxide (Al2O3): CaO(s) 1 SiO2 (s) ¡ CaSiO3 (l) CaO(s) 1 Al2O3 (s) ¡ Ca(AlO2 ) 2 (l)  21.2 Metallurgical Processes 935 The mixture of calcium silicate and calcium aluminate that remains molten at the furnace temperature is known as slag. By the time the ore works its way down to the bottom of the furnace, most of it has already been reduced to iron. The temperature of the lower part of the furnace is above the melting point of impure iron, and so the molten iron at the lower level can be run off to a receiver. The slag, because it is less dense, forms the top layer above the molten iron and can be run off at that level, as shown in Figure 21.3. Iron extracted in this way contains many impurities and is called pig iron; it may contain up to 5 percent carbon and some silicon, phosphorus, manganese, and sulfur. Some of the impurities stem from the silicate and phosphate minerals, while carbon and sulfur come from coke. Pig iron is granular and brittle. It has a relatively low melting point (about 1180°C), so it can be cast in various forms; for this reason it is also called cast iron. Steelmaking Steel manufacturing is one of the most important metal industries. In the United States, the annual consumption of steel is well above 100 million tons. Steel is an iron alloy that contains from 0.03 to 1.4 percent carbon plus various amounts of other elements. The wide range of useful mechanical properties associated with steel is primarily a function of chemical composition and heat treatment of a particular type of steel. Whereas the production of iron is basically a reduction process (converting iron oxides to metallic iron), the conversion of iron to steel is essentially an oxidation process in which the unwanted impurities are removed from the iron by reaction with oxygen gas. One of several methods used in steelmaking is the basic oxygen process. Because of its ease of operation and the relatively short time (about 20 minutes) required for each large-scale (hundreds of tons) conversion, the basic oxygen process is by far the most common means of producing steel today. Figure 21.4 shows the basic oxygen process. Molten iron from the blast furnace is poured into an upright cylindrical vessel. Pressurized oxygen gas is introduced via a water-cooled tube above the molten metal. Under these conditions, manganese, O2 Figure 21.4 The basic oxygen process of steelmaking. The capacity of a typical vessel is CO2, SO2 CaO or SiO2 100 tons of cast iron. Slag Molten Horizontal position steel Molten steel + slag Vertical position 936 Chapter 21 ■ Metallurgy and the Chemistry of Metals Figure 21.5 Steelmaking. phosphorus, and silicon, as well as excess carbon, react with oxygen to form oxides. These oxides are then reacted with the appropriate fluxes (for example, CaO or SiO2) to form slag. The type of flux chosen depends on the composition of the iron. If the main impurities are silicon and phosphorus, a basic flux such as CaO is added to the iron: SiO2 (s) 1 CaO(s) ¡ CaSiO3 (l) P4O10 (l) 1 6CaO(s) ¡ 2Ca3 (PO4 ) 2 (l) On the other hand, if manganese is the main impurity, then an acidic flux such as SiO2 is needed to form the slag: MnO(s) 1 SiO2 (s) ¡ MnSiO3 (l) The molten steel is sampled at intervals. When the desired blend of carbon and other impurities has been reached, the vessel is rotated to a horizontal position so that the molten steel can be tapped off (Figure 21.5). The properties of steel depend not only on its chemical composition but also on the heat treatment. At high temperatures, iron and carbon in steel combine to form iron carbide, Fe3C, called cementite: 3Fe(s) 1 C(s) Δ Fe3C(s) The forward reaction is endothermic, so that the formation of cementite is favored at high temperatures. When steel containing cementite is cooled slowly, the preced- ing equilibrium shifts to the left, and the carbon separates as small particles of graphite, which give the steel a gray color. (Very slow decomposition of cementite also takes place at room temperature.) If the steel is cooled rapidly, equilibrium is not attained and the carbon remains largely in the form of cementite, Fe3C. Steel containing cementite is light in color, and it is harder and more brittle than that containing graphite. Heating the steel to some appropriate temperature for a short time and then cooling it rapidly in order to give it the desired mechanical properties is known as  21.2 Metallurgical Processes 937 Table 21.3 Types of Steel Composition (Percent by Mass)* Type C Mn P S Si Ni Cr Others Uses Plain 1.35 1.65 0.04 0.05 0.06 — — Cu (0.2–0.6) Sheet products, tools High-strength 0.25 1.65 0.04 0.05 0.15–0.9 0.4–1.0 0.3–1.3 Cu (0.01–0.08) Construction, steam turbines Stainless 0.03–1.2 1.0–10 0.04–0.06 0.03 1–3 1–22 4.0–27 — Kitchen utensils, razor blades *A single number indicates the maximum amount of the substance present. “tempering.” In this way, the ratio of carbon present as graphite and as cementite can be varied within rather wide limits. Table 21.3 shows the composition, properties, and uses of various types of steel. Purification of Metals Metals prepared by reduction usually need further treatment to remove impurities. The extent of purification, of course, depends on how the metal will be used. Three com- mon purification procedures are distillation, electrolysis, and zone refining. Distillation Metals that have low boiling points, such as mercury, magnesium, and zinc, can be separated from other metals by fractional distillation. One well-known method of fractional distillation is the Mond† process for the purification of nickel. Carbon mon- oxide gas is passed over the impure nickel metal at about 70°C to form the volatile tetracarbonylnickel (b.p. 43°C), a highly toxic substance, which is separated from the less volatile impurities by distillation: Ni(s) 1 4CO(g) ¡ Ni(CO) 4 (g) Pure metallic nickel is recovered from Ni(CO)4 by heating the gas at 200°C: Ni(CO) 4 (g) ¡ Ni(s) 1 4CO(g) The carbon monoxide that is released is recycled back into the process. Battery Electrolysis + – Electrolysis is another important purification technique. The copper metal obtained Impure Pure copper copper by roasting copper sulfide usually contains impurities such as zinc, iron, silver, and anode cathode gold. The more electropositive metals are removed by an electrolysis process in which the impure copper acts as the anode and pure copper acts as the cathode in a sulfuric acid solution containing Cu21 ions (Figure 21.6). The reactions are Anode (oxidation): Cu(s) ¡ Cu21 (aq) 1 2e2 Cathode (reduction): Cu (aq) 1 2e2 ¡ Cu(s) 21 Cu2+ SO42– † Ludwig Mond (1839–1909). British chemist of German origin. Mond made many important contributions to industrial chemistry. His method for purifying nickel by converting it to the volatile Ni(CO)4 compound Figure 21.6 Electrolytic has been described as having given “wings” to the metal. purification of copper. 938 Chapter 21 ■ Metallurgy and the Chemistry of Metals Figure 21.7 Copper cathodes used in the electrorefining process. Reactive metals in the copper anode, such as iron and zinc, are also oxidized at the anode and enter the solution as Fe21 and Zn21 ions. They are not reduced at the The metal impurities separated from the cathode, however. The less electropositive metals, such as gold and silver, are not copper anode are valuable by-products, the sale of which often pays for the oxidized at the anode. Eventually, as the copper anode dissolves, these metals fall to electricity needed to drive the electrolysis. the bottom of the cell. Thus, the net result of this electrolysis process is the transfer of copper from the anode to the cathode. Copper prepared this way has a purity greater than 99.5 percent (Figure 21.7). Zone Refining Another often-used method of obtaining extremely pure metals is zone refining. In this process, a metal rod containing a few impurities is drawn through an electrical heating coil that melts the metal (Figure 21.8). Most impurities dissolve in the molten metal. As the metal rod emerges from the heating coil, it cools and the pure metal crystallizes, leaving the impurities in the molten metal portion that is still in the Figure 21.8 Zone-refining Heating coil Metal rod technique for purifying metals. Top to bottom: An impure metal rod is moved slowly through a heating coil. As the metal rod moves forward, the impurities dissolve in the molten portion of the metal while pure metal crystallizes out in front of the molten zone.  21.3 Band Theory of Electrical Conductivity 939 heating coil. (This is analogous to the freezing of seawater, in which the solid that separates is mostly pure solvent—water. In zone refining, the liquid metal acts as the solvent and the impurities as the solutes.) When the molten zone carrying the impuri- ties, now at increased concentration, reaches the end of the rod, it is allowed to cool and is then cut off. Repeating this procedure a number of times results in metal with a purity greater than 99.99 percent. 21.3 Band Theory of Electrical Conductivity In Section 11.6 we saw that the ability of metals to conduct heat and electricity can be explained with molecular orbital theory. To gain a better understanding of the conductivity properties of metals we must also apply our knowledge of quantum mechanics. The model we will use to study metallic bonding is band theory, so called because it states that delocalized electrons move freely through “bands” formed by overlapping molecular orbitals. We will also apply band theory to certain elements that are semiconductors. Conductors Metals are characterized by high electrical conductivity. Consider magnesium, for example. The electron configuration of Mg is [Ne]3s2, so each atom has two valence electrons in the 3s orbital. In a metallic crystal, the atoms are packed closely together, so the energy levels of each magnesium atom are affected by the immediate neighbors of the atom as a result of orbital overlaps. In Chapter 10 we saw that, in terms of molecular orbital theory, the interaction between two atomic orbitals leads to the formation of a bonding and an antibonding molecular orbital. Because the number of atoms in even a small piece of magnesium is enormously large (on the order of 1020 atoms), the number of molecular orbitals they form is also very large. These molecular orbitals are so closely spaced on the energy scale that they are more appro- priately described as a “band” (Figure 21.9). The closely spaced filled energy levels make up the valence band. The upper half of the energy levels corresponds to the empty, delocalized molecular orbitals formed by the overlap of the 3p orbitals. This set of closely spaced empty levels is called the conduction band. We can imagine a metallic crystal as an array of positive ions immersed in a sea of delocalized valence electrons (see Figure 11.30). The great cohesive force resulting from the delocalization is partly responsible for the strength noted in most metals. Because the valence band and the conduction band are adjacent to each other, the amount of energy needed to promote a valence electron to the conduction ⎧ Figure 21.9 Formation of 3p ⎨ Conduction band ⎩ conduction bands in magnesium. ⎧ The electrons in the 1s, 2s, and 3s ⎨ Valence band 2p orbitals are localized on each ⎩ Mg atom. However, the 3s and Energy 3p orbitals overlap to form delocalized molecular orbitals. 2p Electrons in these orbitals can travel throughout the metal, and 2s this accounts for the electrical conductivity of the metal. 1s 12 + 12 + 12 + 12 + 12 + Mg Mg Mg Mg Mg 940 Chapter 21 ■ Metallurgy and the Chemistry of Metals Figure 21.10 Comparison of the energy gaps between the valence Conduction band band and the conduction band in Conduction band Conduction band a metal, a semiconductor, and an Energy Energy Energy insulator. In a metal, the energy gap is virtually nonexistent; in a Energy gap Energy gap semiconductor the energy gap is small; and in an insulator the Valence band energy gap is very large, thus Valence band making the promotion of an Valence band electron from the valence band to the conduction band difficult. Metal Semiconductor Insulator band is negligible. There, the electron can travel freely through the metal, because the conduction band is void of electrons. This freedom of movement accounts for the fact that metals are good conductors, that is, they are capable of conducting electric current. Why don’t substances like wood and glass conduct electricity as metals do? Figure 21.10 provides an answer to this question. Basically, the electrical conduc- tivity of a solid depends on the spacing and the state of occupancy of the energy bands. In magnesium and other metals, the valence bands are adjacent to the con- duction bands, and, therefore, these metals readily act as conductors. In wood and glass, on the other hand, the gap between the valence band and the conduction band is considerably greater than that in a metal. Consequently, much more energy is needed to excite an electron into the conduction band. Lacking this energy, electrons cannot move freely. Therefore, glass and wood are insulators, ineffective conductors of electricity. Semiconductors A number of elements are semiconductors, that is, they normally are not conductors, but will conduct electricity at elevated temperatures or when combined with a small amount of certain other elements. The Group 4A elements silicon and germanium are especially suited for this purpose. The use of semiconductors in transistors and solar cells, to name two applications, has revolutionized the electronic industry in recent decades, leading to increased miniaturization of electronic equipment. The energy gap between the filled and empty bands of these solids is much smaller than that for insulators (see Figure 21.10). If the energy needed to excite electrons from the valence band into the conduction band is provided, the solid becomes a conductor. Note that this behavior is opposite that of the metals. A metal’s ability to conduct electricity decreases with increasing temperature, because the enhanced vibration of atoms at higher temperatures tends to disrupt the flow of electrons. The ability of a semiconductor to conduct electricity can also be enhanced by adding small amounts of certain impurities to the element, a process called doping. Let us consider what happens when a trace amount of boron or phosphorus is added to solid silicon. (Only about five out of every million Si atoms are replaced by B or P atoms.) The structure of solid silicon is similar to that of diamond; each Si atom is covalently bonded to four other Si atoms. Phosphorus ([Ne]3s23p3) has one more valence electron than silicon ([Ne]3s23p2), so there is a valence electron left over after four of them are used to form covalent bonds with silicon (Figure 21.11). This extra electron can be removed from the phosphorus atom by applying a voltage across the solid. The free electron can move through the structure and function as a conduction electron. Impurities of this type are known as donor impurities, because they provide conduction electrons. Solids containing donor impurities are called n-type semicon- ductors, where n stands for negative (the charge of the “extra” electron).  21.4 Periodic Trends in Metallic Properties 941 Figure 21.11 (a) Silicon crystal doped with phosphorus. (b) Silicon e– + crystal doped with boron. Note the formation of a negative center in (a) and of a positive center in (b). (a) (b) The opposite effect occurs if boron is added to silicon. A boron atom has three valence electrons (1s22s22p1). Thus, for every boron atom in the silicon crystal there is a single vacancy in a bonding orbital. It is possible to excite a valence electron from a nearby Si into this vacant orbital. A vacancy created at that Si atom can then be filled by an electron from a neighboring Si atom, and so on. In this manner, elec- trons can move through the crystal in one direction while the vacancies, or “positive holes,” move in the opposite direction, and the solid becomes an electrical conductor. Impurities that are electron deficient are called acceptor impurities. Semiconductors that contain acceptor impurities are called p-type semiconductors, where p stands for positive. In both the p-type and n-type semiconductors, the energy gap between the valence band and the conduction band is effectively reduced, so that only a small amount of energy is needed to excite the electrons. Typically, the conductivity of a semiconductor is increased by a factor of 100,000 or so by the presence of impurity atoms. The growth of the semiconductor industry since the early 1960s has been truly remarkable. Today semiconductors are essential components of nearly all electronic equipment, ranging from radios and television sets to pocket calculators and computers. One of the main advantages of solid-state devices over vacuum-tube electronics is that the former can be made on a single “chip” of silicon no larger than the cross section of a pencil eraser. Consequently, much more equipment can be packed into a small volume—a point of particular importance in space travel, as well as in handheld calculators and microprocessors (computers-on-a-chip). 21.4 Periodic Trends in Metallic Properties Metals are lustrous in appearance, solid at room temperature (with the exception of mercury), good conductors of heat and electricity, malleable (can be hammered flat), and ductile (can be drawn into wire). Figure 21.12 shows the positions of the representative metals and the Group 2B metals in the periodic table. (The transition metals are discussed in Chapter 23.) As we saw in Figure 9.5, the electronegativity of elements increases from left to right across a period and from bottom to top in a group. The metallic character of metals increases in just the opposite directions, that is, from right to left across a period and from top to bottom in a group. Because metals generally have low electronegativities, they tend to form cations and almost always have positive oxidation numbers in their compounds. However, beryllium and magnesium in Group 2A and metals in Group 3A and beyond also form cova- lent compounds. In Sections 21.5, 21.6, and 21.7 we will study the chemistry of selected metals from Group 1A (the alkali metals), Group 2A (the alkaline earth metals), and Group 3A (aluminum). 942 Chapter 21 ■ Metallurgy and the Chemistry of Metals 1 18 1A 8A H 2 13 14 15 16 17 He 2A 3A 4A 5A 6A 7A Li Be B C N O F Ne Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar 3B 4B 5B 6B 7B 8B 1B 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Figure 21.12 Representative metals and Group 2B metals according to their positions in the periodic table. 21.5 The Alkali Metals As a group, the alkali metals (the Group 1A elements) are the most electropositive (or the least electronegative) elements known. They exhibit many similar properties, some of which are listed in Table 21.4. From their electron configurations we expect the oxidation number of these elements in their compounds to be 11 because the cations would be isoelectronic with the noble gases. This is indeed the case. The alkali metals have low melting points and are soft enough to be sliced with a knife (see Figure 8.14). These metals all possess a body-centered crystal structure (see Figure 11.29) with low packing efficiency. This accounts for their low densities among metals. In fact, lithium is the lightest metal known. Because of their great chemical reactivity, the alkali metals never occur naturally in elemental form; they are found combined with halide, sulfate, carbonate, and silicate ions. In this section we will describe the chemistry of two members of Group 1A—sodium and potassium. Table 21.4 Properties of Alkali Metals Li Na K Rb Cs 1 1 1 1 Valence electron configuration 2s 3s 4s 5s 6s1 Density (g/cm3) 0.534 0.97 0.86 1.53 1.87 Melting point (°C) 179 97.6 63 39 28 Boiling point (°C) 1317 892 770 688 678 Atomic radius (pm) 152 186 227 248 265 Ionic radius (pm)* 78 98 133 148 165 Ionization energy (kJ/mol) 520 496 419 403 375 Electronegativity 1.0 0.9 0.8 0.8 0.7 Standard reduction potential (V)† 23.05 22.71 22.93 22.93 22.92 *Refers to the cation M1, where M denotes an alkali metal atom. † The half-reaction M1 (aq) 1 e2 ¡ M(s).  21.5 The Alkali Metals 943 The chemistry of lithium, rubidium, and cesium is less important; all isotopes of francium, the last member of the group, are radioactive. Sodium and potassium are about equally abundant in nature. They occur in sili- cate minerals such as albite (NaAlSi3O8) and orthoclase (KAlSi3O8). Over long peri- ods of time (on a geologic scale), silicate minerals are slowly decomposed by wind and rain, and their sodium and potassium ions are converted to more soluble com- pounds. Eventually rain leaches these compounds out of the soil and carries them to the sea. Yet when we look at the composition of seawater, we find that the concentra- Figure 21.13 Halite (NaCl). tion ratio of sodium to potassium is about 28 to 1. The reason for this uneven distri- bution is that potassium is essential to plant growth, while sodium is not. Plants take up many of the potassium ions along the way, while sodium ions are free to move on to the sea. Other minerals that contain sodium or potassium are halite (NaCl), shown in Figure 21.13, Chile saltpeter (NaNO3), and sylvite (KCl). Sodium chloride is also obtained from rock salt (see p. 376). Metallic sodium is most conveniently obtained from molten sodium chloride by electrolysis in the Downs cell (see Section 18.8). The melting point of sodium chloride is rather high (801°C), and much energy is needed to keep large amounts of the sub- Remember that Ca21 is harder to reduce than Na1. stance molten. Adding a suitable substance, such as CaCl2, lowers the melting point to about 600°C—a more convenient temperature for the electrolysis process. Metallic potassium cannot be easily prepared by the electrolysis of molten KCl because it is too soluble in the molten KCl to float to the top of the cell for col- lection. Moreover, it vaporizes readily at the operating temperatures, creating haz- ardous conditions. Instead, it is usually obtained by the distillation of molten KCl in the presence of sodium vapor at 892°C. The reaction that takes place at this temperature is Na(g) 1 KCl(l) Δ NaCl(l) 1 K(g) Note that this is a chemical rather than electrolytic reduction. This reaction may seem strange given that potassium is a stronger reducing agent than sodium (see Table 21.4). However, potassium has a lower boiling point (770°C) than sodium (892°C), so it is more volatile at 892°C and distills off more easily. Accord- ing to Le Châtelier’s principle, constant removal of potassium vapor shifts the above equilibrium from left to right, assuring recovery of metallic potassium. Sodium and potassium are both extremely reactive, but potassium is the more reactive of the two. Both react with water to form the corresponding hydroxides. In a limited supply of oxygen, sodium burns to form sodium oxide (Na2O). However, in the presence of excess oxygen, sodium forms the pale-yellow peroxide: 2Na(s) 1 O2 (g) ¡ Na2O2 (s) Sodium peroxide reacts with water to give an alkaline solution and hydrogen peroxide: Na2O2 (s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2O2 (aq) Like sodium, potassium forms the peroxide. In addition, potassium also forms the superoxide when it burns in air: K(s) 1 O2 (g) ¡ KO2 (s) When potassium superoxide reacts with water, oxygen gas is evolved: 2KO2 (s) 1 2H2O(l) ¡ 2KOH(aq) 1 O2 (g) 1 H2O2 (aq) This reaction is utilized in breathing equipment (Figure 21.14). Exhaled air contains both moisture and carbon dioxide. The moisture reacts with KO2 in the apparatus to 944 Chapter 21 ■ Metallurgy and the Chemistry of Metals Figure 21.14 Self-contained breathing apparatus. generate oxygen gas as shown in the preceding equation. Furthermore, KO2 also reacts with exhaled CO2, which produces more oxygen gas: 4KO2 (s) 1 2CO2 (g) ¡ 2K2CO3 (s) 1 3O2 (g) Thus, a person using the apparatus can continue to breathe oxygen without being exposed to toxic fumes outside. Sodium and potassium metals dissolve in liquid ammonia to produce a beautiful blue solution: NH3 1 2 Na ¡ Na 1 e NH3 1 2 K ¡ K 1e Both the cation and the electron exist in the solvated form; the solvated electrons are responsible for the characteristic blue color of such solutions. Metal-ammonia solutions are powerful reducing agents (because they contain free electrons); they are useful in synthesizing both organic and inorganic compounds. It was discovered that the hitherto unknown alkali metal anions, M2, are also formed in such solutions. This means that an ammonia solution of an alkali metal contains ion pairs such as Na1Na2 and K1K2! (Keep in mind that in each case the metal cation exists as a complex ion with crown ether, an organic compound with a high affinity for cations.) In fact, these “salts” are so stable that they can be isolated in crystalline form (see p. 930). This finding is of considerable theoretical interest, for it shows clearly that the alkali metals can have an oxidation number of 21, although 21 is not found in ordinary compounds. Sodium and potassium are essential elements of living matter. Sodium ions and potassium ions are present in intracellular and extracellular fluids, and they are essen- tial for osmotic balance and enzyme functions. We now describe the preparations and uses of several of the important compounds of sodium and potassium. Sodium Chloride The source, properties, and uses of sodium chloride were discussed in Chapter 9 (see p. 376).  21.5 The Alkali Metals 945 Sodium Carbonate Sodium carbonate (called soda ash) is used in all kinds of industrial processes, includ- ing water treatment and the manufacture of soaps, detergents, medicines, and food additives. Today about half of all Na2CO3 produced is used in the glass industry (in soda-lime glass; see Section 11.7). Sodium carbonate ranks eleventh among the chem- icals produced in the United States (11 million tons in 2010). For many years Na2CO3 was produced by the Solvay† process, in which ammonia is first dissolved in a satu- rated solution of sodium chloride. Bubbling carbon dioxide into the solution results in the precipitation of sodium bicarbonate as follows: NH3 (aq) 1 NaCl(aq) 1 H2CO3 (aq) ¡ NaHCO3 (s) 1 NH4Cl(aq) Sodium bicarbonate is then separated from the solution and heated to give sodium carbonate: 2NaHCO3 (s) ¡ Na2CO3 (s) 1 CO2 (g) 1 H2O(g) However, the rising cost of ammonia and the pollution problem resulting from by- The last plant using the Solvay process in the United States closed in 1986. products have prompted chemists to look for other sources of sodium carbonate. One is the mineral trona [Na5(CO3)2(HCO3) ? 2H2O], large deposits of which have been found in Wyoming. When trona is crushed and heated, it decomposes as follows: 2Na5 (CO3 ) 2 (HCO3 ) ? 2H2O(s) ¡ 5Na2CO3 (s) 1 CO2 (g) 1 3H2O(g) The sodium carbonate obtained this way is dissolved in water, the solution is filtered to remove the insoluble impurities, and the sodium carbonate is crystallized as Na2CO3 ? 10H2O. Finally, the hydrate is heated to give pure, anhydrous sodium carbonate. Sodium Hydroxide and Potassium Hydroxide The properties of sodium hydroxide and potassium hydroxide are very similar. These hydroxides are prepared by the electrolysis of aqueous NaCl and KCl solutions (see Section 18.8); both hydroxides are strong bases and very soluble in water. Sodium hydroxide is used in the manufacture of soap and many organic and inorganic com- pounds. Potassium hydroxide is used as an electrolyte in some storage batteries, and aqueous potassium hydroxide is used to remove carbon dioxide and sulfur dioxide from air. Sodium Nitrate and Potassium Nitrate Large deposits of sodium nitrate (chile saltpeter) are found in Chile. It decomposes with the evolution of oxygen at about 500°C: 2NaNO3 (s) ¡ 2NaNO2 (s) 1 O2 (g) Potassium nitrate (saltpeter) is prepared beginning with the “reaction” KCl(aq) 1 NaNO3 (aq) ¡ KNO3 (aq) 1 NaCl(aq) † Ernest Solvay (1838–1922). Belgian chemist. Solvay’s main contribution to industrial chemistry was the development of the process for the production of sodium carbonate that now bears his name. 946 Chapter 21 ■ Metallurgy and the Chemistry of Metals This process is carried out just below 100°C. Because KNO3 is the least soluble salt at room temperature, it is separated from the solution by fractional crystallization. Like NaNO3, KNO3 decomposes when heated. Gunpowder consists of potassium nitrate, wood charcoal, and sulfur in the approx- imate proportions of 6:1:1 by mass. When gunpowder is heated, the reaction is 2KNO3 (s) 1 S(l) 1 3C(s) ¡ K2S(s) 1 N2 (g) 1 3CO2 (g) The sudden formation of hot expanding gases causes an explosion. 21.6 The Alkaline Earth Metals The alkaline earth metals are somewhat less electropositive and less reactive than the alkali metals. Except for the first member of the family, beryllium, which resem- bles aluminum (a Group 3A metal) in some respects, the alkaline earth metals have similar chemical properties. Because their M21 ions attain the stable electron con- figuration of the preceding noble gas, the oxidation number of alkaline earth metals in the combined form is almost always 12. Table 21.5 lists some common proper- ties of these metals. Radium is not included in the table because all radium isotopes are radioactive and it is difficult and expensive to study the chemistry of this Group 2A element. Magnesium Magnesium (see Figure 8.15) is the sixth most plentiful element in Earth’s crust (about 2.5 percent by mass). Among the principal magnesium ores are brucite, Mg(OH)2; dolomite, CaCO3 ? MgCO3 (Figure 21.15); and epsomite, MgSO4 ? 7H2O. Seawater is a good source of magnesium; there are about 1.3 g of magnesium in each kilogram of seawater. As is the case with most alkali and alkaline earth metals, metallic magnesium is obtained by electrolysis, in this case from its molten chloride, MgCl2 (obtained from seawater, see p. 156). Table 21.5 Properties of Alkaline Earth Metals Be Mg Ca Sr Ba Valence electron configuration 2s2 3s2 4s2 5s2 6s2 Density (g/cm3) 1.86 1.74 1.55 2.6 3.5 Melting point (8C) 1280 650 838 770 714 Boiling point (8C) 2770 1107 1484 1380 1640 Atomic radius (pm) 112 160 197 215 222 Ionic radius (pm)* 34 78 106 127 143 First and second ionization 899 738 590 548 502 energies (kJ/mol) 1757 1450 1145 1058 958 Electronegativity 1.5 1.2 1.0 1.0 0.9 Standard reduction potential (V)† 21.85 22.37 22.87 22.89 22.90 Figure 21.15 Dolomite *Refers to the cation M21, where M denotes an alkali earth metal atom. (CaCO3 ? MgCO3 ). † The half-reaction is M21 (aq) 1 2e2 ¡ M(s).  21.6 The Alkaline Earth Metals 947 The chemistry of magnesium is intermediate between that of beryllium and the heavier Group 2A elements. Magnesium does not react with cold water but does react slowly with steam: Mg(s) 1 H2O(g) ¡ MgO(s) 1 H2(g) It burns brilliantly in air to produce magnesium oxide and magnesium nitride (see Figure 4.9): 2Mg(s) 1 O2(g) ¡ 2MgO(s) 3Mg(s) 1 N2(g) ¡ Mg3N2(s) This property makes magnesium (in the form of thin ribbons or fibers) useful in flash photography and flares. Magnesium oxide reacts very slowly with water to form magnesium hydroxide, a white solid suspension called milk of magnesia (see p. 707), which is used to treat acid indigestion: MgO(s) 1 H2O(l) ¡ Mg(OH)2(s) Magnesium is a typical alkaline earth metal in that its hydroxide is a strong base. [The only alkaline earth hydroxide that is not a strong base is Be(OH)2, which is amphoteric.] The major uses of magnesium are in lightweight structural alloys, for cathodic protection (see Section 18.7), in organic synthesis, and in batteries. Magnesium is essential to plant and animal life, and Mg21 ions are not toxic. It is estimated that the average adult ingests about 0.3 g of magnesium ions daily. Magnesium plays several important biological roles. It is present in intracellular and extracellular fluids. Magne- sium ions are essential for the proper functioning of a number of enzymes. Magnesium is also present in the green plant pigment chlorophyll, which plays an important part in photosynthesis. Calcium Earth’s crust contains about 3.4 percent calcium (see Figure 8.15) by mass. Calcium occurs in limestone, calcite, chalk, and marble as CaCO3; in dolomite as CaCO3 ? MgCO3 (see Figure 21.15); in gypsum as CaSO4 ? 2H2O; and in fluorite as CaF2 (Figure 21.16). Metallic calcium is best prepared by the electrolysis of molten calcium chloride (CaCl2). As we read down Group 2A from beryllium to barium, we note an increase in metallic properties. Unlike beryllium and magnesium, calcium (like strontium and barium) reacts with cold water to yield the corresponding hydroxide, although the rate of reaction is much slower than those involving the alkali metals (see Figure 4.14): Ca(s) 1 2H2O(l) ¡ Ca(OH) 2 (aq) 1 H2 (g) Calcium hydroxide [Ca(OH)2] is commonly known as slaked lime or hydrated lime. Lime (CaO), which is also referred to as quicklime, is one of the oldest materials known to mankind. Quicklime is produced by the thermal decomposition of calcium carbonate (see Section 17.5): CaCO3 (s) ¡ CaO(s) 1 CO2 (g) Figure 21.16 Fluorite (CaF2 ). 948 Chapter 21 ■ Metallurgy and the Chemistry of Metals while slaked lime is produced by the reaction between quicklime and water: CaO(s) 1 H2O(l) ¡ Ca(OH) 2 (aq) Quicklime is used in metallurgy (see Section 21.2) and the removal of SO2 when fossil fuel is burned (see p. 918). Slaked lime is used in water treatment. For many years, farmers have used lime to lower the acidity of soil for their crops (a process called liming). Nowadays lime is also applied to lakes affected by acid rain (see Section 20.6). Metallic calcium has rather limited uses. It serves mainly as an alloying agent for metals like aluminum and copper and in the preparation of beryllium metal from its compounds. It is also used as a dehydrating agent for organic solvents. Calcium is an essential element in living matter. It is the major component of bones and teeth; the calcium ion is present in a complex phosphate salt, hydroxyapa- tite, Ca5(PO4)3OH. A characteristic function of Ca21 ions in living systems is the activation of a variety of metabolic processes. Calcium plays a vital role in heart action, blood clotting, muscle contraction, and nerve impulse transmission. 21.7 Aluminum Aluminum (see Figure 8.16) is the most abundant metal and the third most plentiful element in Earth’s crust (7.5 percent by mass). The elemental form does not occur in nature; its principal ore is bauxite (Al2O3 ? 2H2O). Other minerals containing alumi- num are orthoclase (KAlSi3O8), beryl (Be3Al2Si6O18), cryolite (Na3AlF6), and corun- dum (Al2O3) (Figure 21.17). Aluminum is usually prepared from bauxite, which is frequently contaminated with silica (SiO2), iron oxides, and titanium(IV) oxide. The ore is first heated in sodium hydroxide solution to convert the silica into soluble silicates: SiO2 (s) 1 2OH2 (aq) ¡ SiO22 3 (aq) 1 H2O(l) At the same time, aluminum oxide is converted to the aluminate ion (AlO2 2 ): Al2O3 (s) 1 2OH2 (aq) ¡ 2AlO2 2 (aq) 1 H2O(l) Iron oxide and titanium oxide are unaffected by this treatment and are filtered off. Figure 21.17 Corundum (Al2O3 ). Next, the solution is treated with acid to precipitate the insoluble aluminum hydroxide: AlO2 1 2 (aq) 1 H3O (aq) ¡ Al(OH) 3 (s) Carbon anodes After filtration, the aluminum hydroxide is heated to obtain aluminum oxide: 2Al(OH) 3 (s) ¡ Al2O3 (s) 1 3H2O(g) Carbon cathode Anhydrous aluminum oxide, or corundum, is reduced to aluminum by the Hall† process. Figure 21.18 shows a Hall electrolytic cell, which contains a series of car- bon anodes. The cathode is also made of carbon and constitutes the lining inside the Molten Al2O3 in † aluminum Charles Martin Hall (1863–1914). American inventor. While Hall was an undergraduate at Oberlin College, molten cryolite he became interested in finding an inexpensive way to extract aluminum. Shortly after graduation, when Figure 21.18 Electrolytic he was only 22 years old, Hall succeeded in obtaining aluminum from aluminum oxide in a backyard production of aluminum based woodshed. Amazingly, the same discovery was made at almost the same moment in France, by Paul Héroult, on the Hall process. another 22-year-old inventor working in a similar makeshift laboratory.  21.7 Aluminum 949 cell. The key to the Hall process is the use of cryolite, or Na3AlF6 (m.p. 1000°C), Molten cryolite provides a good conducting medium for electrolysis. as the solvent for aluminum oxide (m.p. 2045°C). The mixture is electrolyzed to produce aluminum and oxygen gas: Anode (oxidation): 3[2O22 ¡ O2 (g) 1 4e2] 31 Cathode (reduction): 4[Al 1 3e2 ¡ Al(l)] Overall: 2Al2O3 ¡ 4Al(l) 1 3O2 (g) Oxygen gas reacts with the carbon anodes (at elevated temperatures) to form carbon monoxide, which escapes as a gas. The liquid aluminum metal (m.p. 660.2°C) sinks to the bottom of the vessel, from which it can be drained from time to time during the procedure. Aluminum is one of the most versatile metals known. It has a low density (2.7  g/cm3) and high tensile strength (that is, it can be stretched or drawn out). Movie Aluminum Production Aluminum is ­malleable, it can be rolled into thin foils, and it is an excellent elec- trical conductor. Its conductivity is about 65 percent that of copper. However, because aluminum is cheaper and lighter than copper, it is widely used in high- voltage transmission lines. Although aluminum’s chief use is in aircraft construc- tion, the pure metal itself is too soft and weak to withstand much strain. Its mechanical properties are greatly improved by alloying it with small amounts of metals such as copper, magnesium, and manganese, as well as silicon. ­Aluminum is not used by living systems and is generally considered to be nontoxic. As we read across the periodic table from left to right in a given period, we note a gradual decrease in metallic properties. Thus, although aluminum is considered an active metal, it does not react with water as do sodium and calcium. Aluminum reacts with hydrochloric acid and with strong bases as follows: 2Al(s) 1 6HCl(aq) ¡ 2AlCl3 (aq) 1 3H2 (g) 2Al(s) 1 2NaOH(aq) 1 2H2O(l) ¡ 2NaAlO2 (aq) 1 3H2 (g) Aluminum readily forms the oxide Al2O3 when exposed to air: 4Al(s) 1 3O2 (g) ¡ 2Al2O3 (s) A tenacious film of this oxide protects metallic aluminum from further corrosion and accounts for some of the unexpected inertness of aluminum. Aluminum oxide has a very large exothermic enthalpy of formation (DH8f 5 21670 kJ/mol). This property makes aluminum suitable for use in solid propellants for rockets such as those used for some space shuttles. When a mixture of aluminum and ammonium perchlorate (NH4ClO4) is ignited, aluminum is oxidized to Al2O3, and the heat liberated in the reaction causes the gases that are formed to expand with great force. This action lifts the rocket. The great affinity of aluminum for oxygen is illustrated nicely by the reaction of aluminum powder with a variety of metal oxides, particularly the transition metal oxides, to produce the corresponding metals. A typical reaction is 2Al(s) 1 Fe2O3 (s) ¡ Al2O3 (s) 1 2Fe(l)  ¢H° 5 2822.8 kJ/mol which can result in temperatures approaching 3000°C. This reaction, which is used in the welding of steel and iron, is called the thermite reaction (Figure 21.19). Aluminum chloride exists as a dimer: Cl Cl Cl G D q D Figure 21.19 The temperature of Al Al D r D G a thermite reaction can reach Cl Cl Cl 30008C.  CHEMISTRY in Action Recycling Aluminum A luminum beverage cans were virtually unknown in 1960; yet by the early 1970s over 1.3 billion pounds of aluminum had been used for these containers. The reasons for aluminum’s popu- of the metal cans and foils are discarded in the United States annually. They litter the countryside and clog landfills. The best solution to this environmental problem, and the way to prevent larity in the beverage industry are that it is nontoxic, odorless, the rapid depletion of a finite resource, is recycling. tasteless, and lightweight. Furthermore, it is thermally conducting, What are the economic benefits of recycling aluminum? so the fluid inside the container can be chilled rapidly. Let us compare the energy consumed in the production of alu- The tremendous increase in the demand for aluminum does minum from bauxite with that consumed when aluminum is have a definite drawback, however. More than 3 billion pounds recycled. The overall reaction for the Hall process can be Left: Collecting aluminum cans for recycling. Right: Melting and purifying recycled aluminum. Ground Each of the bridging chlorine atoms forms a normal covalent bond and a coordinate state covalent bond (indicated by S) with two aluminum atoms. Each aluminum atom is 3s 3p assumed to be sp3-hybridized, so the vacant sp3 hybrid orbital can accept a lone pair Promotion from the chlorine atom (Figure 21.20). Aluminum chloride undergoes hydrolysis as of electron 3s 3p follows: sp3- Hybridized AlCl3 (s) 1 3H2O(l) ¡ Al(OH) 3 (s) 1 3HCl(aq) state ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ sp 3 orbitals Aluminum hydroxide, like Be(OH)2, is amphoteric: Figure 21.20 The sp3 hybridization of an Al atom in Al2Cl6. Each Al atom has one vacant sp3 hybrid Al(OH) 3 (s) 1 3H1 (aq) ¡ Al31 (aq) 1 3H2O(l) orbital that can accept a lone pair Al(OH) 3 (s) 1 OH2 (aq) ¡ Al(OH) 24 (aq) from the bridging Cl atom. In contrast to the boron hydrides, which are a well-defined series of compounds, In 2002, chemists prepared the first aluminum hydride is a polymer in which each aluminum atom is surrounded octahe- member of aluminum hydride (Al2H6), which possesses bridging H atoms like drally by bridging hydrogen atoms (Figure 21.21). diborane, B2H6. When an aqueous mixture of aluminum sulfate and potassium sulfate is evapo- rated slowly, crystals of KAl(SO4)2 ? 12H2O are formed. Similar crystals can be 950 represented as where m is the molar mass, s is the specific heat of Al, and Dt is the temperature change. Thus, the total energy needed to recycle Al2O3 (in molten cryolite) 1 3C(s) ¡ 2Al(l) 1 3CO(g) 1 mole of Al is given by for which DH° 5 1340 kJ/mol and DS° 5 586 J/K ? mol. At total energy 5 15.4 kJ 1 10.7 kJ 1000°C, which is the temperature of the process, the standard 5 26.1 kJ free-energy change for the reaction is given by To compare the energy requirements of the two methods we write ¢G° 5 ¢H° 2 T¢S° 586 J 1 kJ energy needed to recycle 1 mol Al 5 1340 kJ/mol 2 (1273 K)a ba b K ? mol 1000 J energy needed to produce 1 mol Al by electrolysis 5 594 kJ/mol 26.1 kJ 5 3 100% 297 kJ Equation (18.3) states that DG° 5 2nFE°; therefore, the amount 5 8.8% of electrical energy needed to produce 1 mole of Al from bauxite is 594 kJ/2, or 297 kJ. Thus, by recycling aluminum cans we can save about 91 percent Recycling aluminum requires only enough energy to heat of the energy required to extract the metal from bauxite. the metal to its melting point (660°C) plus the heat of fusion Recycling most of the aluminum cans thrown away each year (10.7 kJ/mol). The heat change where 1 mole of aluminum is saves 20 billion kilowatt-hours of electricity—about 1 percent heated from 25°C to 660°C is of the electric power used in the United States annually. (Watt is the unit for power, 1 watt 5 1 joule per second.) heat input 5 ms¢t 5 (27.0 g) (0.900 J/g ? °C) (660 2 25)°C 5 15.4 kJ formed by substituting Na1 or NH1 1 4 for K , and Cr 31 or Fe31 for Al31. These com- pounds are called alums, and they have the general formula M1M31 (SO4 ) 2 ? 12H2O M1: K1, Na1, NH14 M31 : Al31 , Cr31 , Fe31 Alums are examples of double salts, that is, salts that contain two different cations. Figure 21.21 Structure of aluminum hydride. Note that this compound is a polymer. Each Al atom is surrounded octahedrally by six bridging H atoms. 951  952 Chapter 21 ■ Metallurgy and the Chemistry of Metals Summary of Facts & Concepts 1. Depending on their reactivities, metals exist in nature in temperatures, and therefore conductivity increases with either the free or combined state. increasing temperature as more electrons are able to 2. Recovering a metal from its ore is a three-stage process. reach the conduction band. First, the ore must be prepared. The metal is then separated, 6. n-Type semiconductors contain donor impurities and usually by a reduction process, and finally, it is purified. extra electrons. p-Type semiconductors contain accep- 3. The methods commonly used for purifying metals are tor impurities and “positive holes.” distillation, electrolysis, and zone refining. 7. The alkali metals are the most reactive of all the metal- 4. Metallic bonds can be thought of as the force between lic elements. They have an oxidation number of 11 in positive ions immersed in a sea of electrons. In terms of their compounds. Under special conditions, some of band theory, the atomic orbitals merge to form energy them also form uninegative ions. bands. A substance is a conductor when electrons can 8. The alkaline earth metals are somewhat less reactive be readily promoted to the conduction band, where they than the alkali metals. They almost always have an oxi- are free to move through the substance. dation number of 12 in their compounds. The properties 5. In insulators, the energy gap between the valence band of the alkaline earth elements become increasingly me- and the conduction band is so large that electrons can- tallic from top to bottom in their periodic group. not be promoted into the conduction band. In semicon- 9. Aluminum does not react with water due to the forma- ductors, electrons can cross the energy gap at higher tion of a protective oxide; its hydroxide is amphoteric. Key Words Acceptor impurity, p. 941 Conductor, p. 940 Metallurgy, p. 932 p-Type semiconductor, p. 941 Alloy, p. 932 Donor impurity, p. 940 Mineral, p. 931 Pyrometallurgy, p. 933 Amalgam, p. 932 Ferromagnetic, p. 932 n-Type semiconductor, p. 940 Semiconductors, p. 940 Band theory, p. 939 Insulator, p. 940 Ore, p. 931 Questions & Problems • Problems available in Connect Plus 21.7 Describe with examples the chemical and electro- Red numbered problems solved in Student Solutions Manual lytic reduction processes used in the production of metals. Occurrence of Metals 21.8 Describe the main steps used to purify metals. Review Questions 21.9 Describe the extraction of iron in a blast furnace. 21.1 Define mineral, ore, and metallurgy. 21.10 Briefly discuss the steelmaking process. 21.2 List three metals that are usually found in an uncom- bined state in nature and three metals that are always Problems found in a combined state in nature. • 21.11 In the Mond process for the purification of • 21.3 Write chemical formulas for the following minerals: nickel, CO is passed over metallic nickel to give (a) calcite, (b) dolomite, (c) fluorite, (d) halite, Ni(CO)4: (e) corundum, (f) magnetite, (g) beryl, (h) galena, (i) epsomite, ( j) anhydrite. Ni(s) 1 4CO(g) Δ Ni(CO) 4 (g) • 21.4 Name the following minerals: (a) MgCO3, (b) Na3AlF6, Given that the standard free energies of formation of (c) Al2O3, (d) Ag2S, (e) HgS, (f) ZnS, (g) SrSO4, (h) PbCO3, (i) MnO2, ( j) TiO2. CO(g) and Ni(CO)4(g) are 2137.3 kJ/mol and 2587.4 kJ/mol, respectively, calculate the equilib- Metallurgical Processes rium constant of the reaction at 80°C. (Assume DG°f to be independent of temperature.) Review Questions • 21.12 Copper is purified by electrolysis (see Figure 21.6). 21.5 Describe the main steps involved in the preparation A 5.00-kg anode is used in a cell where the current of an ore. is 37.8 A. How long (in hours) must the current run 21.6 What does roasting mean in metallurgy? Why is to dissolve this anode and electroplate it onto the roasting a major source of air pollution and acid rain? cathode?  Questions & Problems 953 • 21.13 Consider the electrolytic procedure for purifying (c) Na(s) 1 O2 (g) ¡ copper described in Figure 21.6. Suppose that a (d) K(s) 1 O2 (g) ¡ sample of copper contains the following impurities: 21.28 Write a balanced equation for each of the following Fe, Ag, Zn, Au, Co, Pt, and Pb. Which of the metals reactions: (a) sodium reacts with water; (b) an aque- will be oxidized and dissolved in solution and which ous solution of NaOH reacts with CO2; (c) solid will be unaffected and simply form the sludge that Na2CO3 reacts with a HCl solution; (d) solid NaHCO3 accumulates at the bottom of the cell? reacts with a HCl solution; (e) solid NaHCO3 is 21.14 How would you obtain zinc from sphalerite (ZnS)? heated; (f ) solid Na2CO3 is heated. 21.15 Starting with rutile (TiO2), explain how you would 21.29 Sodium hydride (NaH) can be used as a drying agent obtain pure titanium metal. (Hint: First convert TiO2 for many organic solvents. Explain how it works. to TiCl4. Next, reduce TiCl4 with Mg. Look up physi- cal properties of TiCl4, Mg, and MgCl2 in a chemistry • 21.30 Calculate the volume of CO2 at 10.0°C and 746 mmHg pressure obtained by treating 25.0 g of Na2CO3 with an handbook.) excess of hydrochloric acid. • 21.16 A certain mine produces 2.0 3 108 kg of copper from chalcopyrite (CuFeS2) each year. The ore con- Alkaline Earth Metals tains only 0.80 percent Cu by mass. (a) If the density of the ore is 2.8 g/cm3, calculate the volume (in cm3) Review Questions of ore removed each year. (b) Calculate the mass (in 21.31 List the common ores of magnesium and calcium. kg) of SO2 produced by roasting (assume chalcopy- 21.32 How are the metals magnesium and calcium obtained rite to be the only source of sulfur). commercially? 21.17 Which of the following compounds would require electrolysis to yield the free metals? Ag2S, CaCl2, Problems NaCl, Fe2O3, Al2O3, TiCl4. 21.18 Although iron is only about two-thirds as abun- • 21.33 From the thermodynamic data in Appendix 3, calculate the DH° values for the following decompositions: dant as aluminum in Earth’s crust, mass for mass it costs only about one-quarter as much to pro- (a) MgCO3 (s) ¡ MgO(s) 1 CO2 (g) duce. Why? (b) CaCO3 (s) ¡ CaO(s) 1 CO2 (g) Which of the two compounds is more easily decom- Band Theory of Electrical Conductivity posed by heat? Review Questions • 21.34 Starting with magnesium and concentrated nitric acid, describe how you would prepare magnesium 21.19 Define the following terms: conductor, insulator, oxide. [Hint: First convert Mg to Mg(NO3)2. Next, semiconducting elements, donor impurities, ac- MgO can be obtained by heating Mg(NO3)2.] ceptor impurities, n-type semiconductors, p-type 21.35 Describe two ways of preparing magnesium semiconductors. chloride. 21.20 Briefly discuss the nature of bonding in metals, in- 21.36 The second ionization energy of magnesium is sulators, and semiconducting elements. only about twice as great as the first, but the third 21.21 Describe the general characteristics of n-type and ionization energy is 10 times as great. Why does it p-type semiconductors. take so much more energy to remove the third • 21.22 State whether silicon would form n-type or p-type electron? semiconductors with the following elements: Ga, 21.37 List the sulfates of the Group 2A metals in order Sb, Al, As. of increasing solubility in water. Explain the trend. (Hint: You need to consult a chemistry Alkali Metals handbook.) Review Questions 21.38 Helium contains the same number of electrons in its outer shell as do the alkaline earth metals. Explain 21.23 How is sodium prepared commercially? why helium is inert whereas the Group 2A metals 21.24 Why is potassium usually not prepared electrolyti- are not. cally from one of its salts? • 21.39 When exposed to air, calcium first forms calcium 21.25 Describe the uses of the following compounds: oxide, which is then converted to calcium hydroxide, NaCl, Na2CO3, NaOH, KOH, KO2. and finally to calcium carbonate. Write a balanced 21.26 Under what conditions do sodium and potassium equation for each step. form Na2 and K2 ions? 21.40 Write chemical formulas for (a) quicklime, (b) slaked lime, (c) limewater. Problems Aluminum • 21.27 Complete and balance the following equations: (a) K(s) 1 H2O(l) ¡ Review Questions (b) NaH(s) 1 H2O(l) ¡ 21.41 Describe the Hall process for preparing aluminum.  954 Chapter 21 ■ Metallurgy and the Chemistry of Metals 21.42 What action renders aluminum inert? 21.55 An early view of metallic bonding assumed that bonding in metals consisted of localized, shared Problems electron-pair bonds between metal atoms. What evidence would help you to argue against this 21.43 Before Hall invented his electrolytic process, alumi- viewpoint? num was produced by the reduction of its chloride with an active metal. Which metals would you use 21.56 Referring to Figure 21.6, would you expect H2O and for the production of aluminum in that way? H1 to be reduced at the cathode and H2O oxidized at the anode? • 21.44 With the Hall process, how many hours will it take to deposit 664 g of Al at a current of 32.6 A? • 21.57 A 0.450-g sample of steel contains manganese as an impurity. The sample is dissolved in acidic so- 21.45 Aluminum forms the complex ions AlCl2 32 4 and AlF6 . 32 lution and the manganese is oxidized to the per- Describe the shapes of these ions. AlCl6 does not manganate ion MnO42. The MnO42 ion is reduced form. Why? (Hint: Consider the relative sizes of Al31, to Mn21 by reacting with 50.0 mL of 0.0800 M F2, and Cl2 ions.) FeSO4 solution. The excess Fe21 ions are then • 21.46 The overall reaction for the electrolytic production oxidized to Fe31 by 22.4 mL of 0.0100 M K2Cr2O7. of aluminum by means of the Hall process may be Calculate the percent by mass of manganese in the represented as sample. Al2O3 (s) 1 3C (s) ¡ 2Al(l) 1 3CO(g) • 21.58 Given that DG°f (Fe2O3) 5 2741.0 kJ/mol and that DG°f (Al2O3) 5 21576.4 kJ/mol, calculate DG° for At 1000°C, the standard free-energy change for the following reactions at 25°C: this process is 594 kJ/mol. (a) Calculate the mini- (a) 2Fe2O3 (s) ¡ 4Fe(s) 1 3O2 (g) mum voltage required to produce 1 mole of alumi- num at this temperature. (b) If the actual voltage (b) 2Al2O3 (s) ¡ 4Al(s) 1 3O2 (g) applied is exactly three times the ideal value, cal- 21.59 Use compounds of aluminum as an example to culate the energy required to produce 1.00 kg of explain what is meant by amphoterism. the metal. 21.60 When an inert atmosphere is needed for a metallur- • 21.47 In basic solution, aluminum metal is a strong re- gical process, nitrogen is frequently used. However, ducing agent and is oxidized to AlO22. Give bal- in the reduction of TiCl4 by magnesium, helium is anced equations for the reaction of Al in basic used. Explain why nitrogen is not suitable for this solution with the following: (a) NaNO3, to give process. ammonia; (b) water, to give hydrogen; (c) Na2SnO3, 21.61 It has been shown that Na2 species form in the vapor to give metallic tin. phase. Describe the formation of the “disodium • 21.48 Write a balanced equation for the thermal decompo- molecule” in terms of a molecular orbital energy- sition of aluminum nitrate to form aluminum oxide, level diagram. Would you expect the alkaline earth nitrogen dioxide, and oxygen gas. metals to exhibit a similar property? 21.49 Describe some of the properties of aluminum that 21.62 Explain each of the following statements: (a) An make it one of the most versatile metals known. aqueous solution of AlCl3 is acidic. (b) Al(OH)3 is 21.50 The pressure of gaseous Al2Cl6 increases more rap- soluble in NaOH solutions but not in NH3 solution. idly with temperature than predicted by the ideal gas • 21.63 Write balanced equations for the following reac- equation even though Al2Cl6 behaves like an ideal tions: (a) the heating of aluminum carbonate; (b) the gas. Explain. reaction between AlCl3 and K; (c) the reaction 21.51 Starting with aluminum, describe with balanced equa- between solutions of Na2CO3 and Ca(OH)2. tions how you would prepare (a) Al2Cl6, (b) Al2O3, 21.64 Write a balanced equation for the reaction between (c) Al2(SO4)3, (d) NH4Al(SO4)2 ? 12H2O. calcium oxide and dilute HCl solution. 21.52 Explain the change in bonding when Al2Cl6 dissoci- • 21.65 What is wrong with the following procedure for ates to form AlCl3 in the gas phase. obtaining magnesium? Additional Problems MgCO3 ¡ MgO(s) 1 CO2 (g) 21.53 In steelmaking, nonmetallic impurities such as P, S, MgO(s) 1 CO(g) ¡ Mg(s) 1 CO2 (g) and Si are removed as the corresponding oxides. The inside of the furnace is usually lined with • 21.66 Explain why most metals have a flickering CaCO3 and MgCO3, which decompose at high appearance. temperatures to yield CaO and MgO. How do CaO 21.67 Predict the chemical properties of francium, the last and MgO help in the removal of the nonmetallic member of Group 1A. oxides? 21.68 Describe a medicinal or health-related application • 21.54 When 1.164 g of a certain metal sulfide was roasted for each of the following compounds: NaF, Li2CO3, in air, 0.972 g of the metal oxide was formed. If the Mg(OH)2, CaCO3, BaSO4, Al(OH)2NaCO3. (You oxidation number of the metal is 12, calculate the would need to do a Web search for some of these molar mass of the metal. compounds.)  Questions & Problems 955 21.69 The following are two reaction schemes involving 21.73 After heating, a metal surface (such as that of a magnesium. Scheme I: When magnesium burns in cooking pan or skillet) develops a color pattern like oxygen, a white solid (A) is formed. A dissolves in an oil slick on water. Explain. 1 M HCl to give a colorless solution (B). Upon ad- • 21.74 Chemical tests of four metals A, B, C, and D show dition of Na2CO3 to B, a white precipitate is the following results. formed (C). On heating, C decomposes to D and a (a) Only B and C react with 0.5 M HCl to give H2 colorless gas is generated (E). When E is passed gas. through limewater [an aqueous suspension of (b) When B is added to a solution containing the Ca(OH)2], a white precipitate appears (F). Scheme II: ions of the other metals, metallic A, C, and D Magnesium reacts with 1 M H2SO4 to produce a are formed. colorless solution (G). Treating G with an excess of NaOH produces a white precipitate (H). H dis- (c) A reacts with 6 M HNO3 but D does not. solves in 1 M HNO3 to form a colorless solution. Arrange the metals in the increasing order as re- When the solution is slowly evaporated, a white ducing  agents. Suggest four metals that fit these solid (I) appears. On heating I, a brown gas is descriptions. given off. Identify A–I and write equations repre- 21.75 The electrical conductance of copper metal de- senting the reactions involved. creases with temperature, but that of a CuSO4 solu- 21.70 Lithium and magnesium exhibit a diagonal relation- tion increases with temperature. Explain. ship in some chemical properties. How does lithium • 21.76 As stated in the chapter, potassium superoxide resemble magnesium in its reaction with oxygen and (KO2) is a useful source of oxygen employed in nitrogen? Consult a handbook of chemistry and breathing equipment. Calculate the pressure at compare the solubilities of carbonates, fluorides, which oxygen gas stored at 20°C would have the and phosphates of these metals. same density as the oxygen gas provided by KO2. • 21.71 To prevent the formation of oxides, peroxides, and The density of KO2 at 20°C is 2.15 g/cm3. superoxides, alkali metals are sometimes stored in an • 21.77 A sample of 10.00 g of sodium reacts with oxygen inert atmosphere. Which of the following gases to form 13.83 g of sodium oxide (Na2O) and sodium should not be used for lithium? Why? Ne, Ar, N2, Kr. peroxide (Na2O2). Calculate the percent composi- 21.72 Which of the following metals is not found in the tion of the mixture. free state in nature: Ag, Cu, Zn, Au, Pt?  CHAPTER 22 Nonmetallic Elements and Their Compounds The nose cone of the space shuttle is made of graphite and silicon carbide and can withstand the tremendous heat generated when the vehicle enters Earth’s atmosphere. CHAPTER OUTLINE A LOOK AHEAD 22.1 General Properties  This chapter starts by examining the general properties of the nonmetals. (22.1) of Nonmetals  We see that hydrogen does not have a unique position in the periodic table. 22.2 Hydrogen We learn the preparation of hydrogen and study several different types of compounds containing hydrogen. We also discuss the hydrogenation reaction 22.3 Carbon and the role hydrogen plays in energy production. (22.2) 22.4 Nitrogen and Phosphorus  Next, we consider the inorganic world of carbon in terms of carbides, cyanides, 22.5 Oxygen and Sulfur and carbon monoxide and carbon dioxide. (22.3) 22.6 The Halogens  Nitrogen is the most abundant element in the atmosphere. Its major com- pounds are ammonia, hydrazine, and several oxides. Nitric acid, a strong oxidizing agent, is a major industrial chemical. Phosphorus is the other important element in Group 5A. It is a major component of teeth and bones and in genetic materials like deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Phosphorus compounds include hydride and oxides. Phosphoric acid has many commercial applications. (22.4)  Oxygen is the most abundant element in Earth’s crust. It forms compounds with most other elements as oxides, peroxides, and superoxides. Its allotropic form, ozone, is a strong oxidizing agent. Sulfur, the second member of Group 6A, also forms many compounds with metals and nonmetals. Sulfuric acid is the most important industrial chemical in the world. (22.5)  The halogens are the most electronegative and most reactive of the nonmetals. We study their preparations, properties, reactions, and applications of their compounds. (22.6) 956  22.1 General Properties of Nonmetals 957 O f the 118 elements known, only 25 are nonmetallic elements. Unlike the metals, the chemistry of these elements is diverse. Despite their relatively small number, most of the essential elements in biological systems are nonmetals (H, C, N, P, O, S, Cl, and I). This group of nonmetallic elements also includes the most unreactive of the elements—the noble gases. The unique properties of hydrogen set it aside from the rest of the elements in the periodic table. A whole branch of chemistry—organic chemistry—is based on carbon compounds. In this chapter, we continue our survey of the elements by concentrating on the nonmetals. The emphasis will again be on important chemical properties and on the roles of nonmetals and their compounds in industrial, chemical, and biological processes. 22.1 General Properties of Nonmetals Properties of nonmetals are more varied than those of metals. A number of nonmet- als are gases in the elemental state: hydrogen, oxygen, nitrogen, fluorine, chlorine, and the noble gases. Only one, bromine, is a liquid. All the remaining nonmetals are solids at room temperature. Unlike metals, nonmetallic elements are poor con- ductors of heat and electricity; they exhibit both positive and negative oxidation numbers. A small group of elements, called metalloids, have properties characteristic of both metals and nonmetals. The metalloids boron, silicon, germanium, and arsenic are semiconducting elements (see Section 21.3). Nonmetals are more electronegative than metals. The electronegativity of ele- ments increases from left to right across any period and from bottom to top in any group in the periodic table (see Figure 9.5). With the exception of hydrogen, the Recall that there is no totally suitable position for hydrogen in the periodic nonmetals are concentrated in the upper right-hand corner of the periodic table table. (Figure 22.1). Compounds formed by a combination of metals and nonmetals tend to be ionic, having a metallic cation and a nonmetallic anion. In this chapter, we will discuss the chemistry of a number of common and impor- tant nonmetallic elements: hydrogen; carbon (Group 4A); nitrogen and phosphorus (Group 5A); oxygen and sulfur (Group 6A); and fluorine, chlorine, bromine, and iodine (Group 7A). 1 18 1A 8A 1 2 2 13 14 15 16 17 H 2A 3A 4A 5A 6A 7A He 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv Figure 22.1 Representative nonmetallic elements (in blue) and metalloids (gray). 958 Chapter 22 ■ Nonmetallic Elements and Their Compounds 22.2 Hydrogen Hydrogen is the simplest element known—its most common atomic form contains only one proton and one electron. The atomic form of hydrogen exists only at very high temperatures, however. Normally, elemental hydrogen is a diatomic molecule, the product of an exothermic reaction between H atoms: H(g) 1 H(g) ¡ H2 (g) ¢H° 5 2436.4 kJ/mol Molecular hydrogen is a colorless, odorless, and nonpoisonous gas. At 1 atm, liquid hydrogen has a boiling point of 2252.9°C (20.3 K). Hydrogen is the most abundant element in the universe, accounting for about 70 percent of the universe’s total mass. It is the tenth most abundant element in Earth’s crust, where it is found in combination with other elements. Unlike Jupiter and Saturn, Earth does not have a strong enough gravitational pull to retain the lightweight H2 molecules, so hydrogen is not found in our atmosphere. The ground-state electron configuration of H is 1s1. It resembles the alkali metals Hydrogen typically has 11 oxidation in that it can be oxidized to the H1 ion, which exists in aqueous solutions in the number in its compounds, but in ionic hydrated form. On the other hand, hydrogen resembles the halogens in that it forms hydrides it has a 21 oxidation number. the uninegative hydride ion (H2), which is isoelectronic with helium (1s2). Hydrogen is found in a large number of covalent compounds. It also has the unique capacity for hydrogen-bond formation (see Section 11.2). Hydrogen gas plays an important role in industrial processes. About 95 percent of the hydrogen produced is used captively; that is, it is produced at or near the plant where it is used for industrial processes, such as the synthesis of ammonia. The large- scale industrial preparation is the reaction between propane (from natural gas and also as a product of oil refineries) and steam in the presence of a catalyst at 900°C: C3H8 (g) 1 3H2O(g) ¡ 3CO(g) 1 7H2 (g) In another process, steam is passed over a bed of red-hot coke: C(s) 1 H2O(g) ¡ CO(g) 1 H2 (g) The mixture of carbon monoxide and hydrogen gas produced in this reaction is com- monly known as water gas. Because both CO and H2 burn in air, water gas was used as a fuel for many years. But because CO is poisonous, water gas has been replaced by natural gases, such as methane and propane. Small quantities of hydrogen gas can be prepared conveniently in the laboratory by reacting zinc with dilute hydrochloric acid (Figure 22.2): Zn(s) 1 2HCl(aq) ¡ ZnCl2 (aq) 1 H2 (g) Hydrogen gas can also be produced by the reaction between an alkali metal or an alkaline earth metal (Ca or Ba) and water (see Section 4.4), but these reactions are too violent to be suitable for the laboratory preparation of hydrogen gas. Very pure hydrogen gas can be obtained by the electrolysis of water, but this method consumes too much energy to be practical on a large scale. Binary Hydrides Binary hydrides are compounds containing hydrogen and another element, either a metal or a nonmetal. Depending on structure and properties, these hydrides are broadly divided into three types: (1) ionic hydrides, (2) covalent hydrides, and (3)  interstitial hydrides.  22.2 Hydrogen 959 Figure 22.2 Apparatus for the laboratory preparation of hydrogen HCl gas. The gas is collected over water, as is also the case of oxygen gas (see Figure 5.12). H2 gas Water Zn Ionic Hydrides Ionic hydrides are formed when molecular hydrogen combines directly with any alkali metal or with the alkaline earth metals Ca, Sr, or Ba: 2Li(s) 1 H2 (g) ¡ 2LiH(s) Ca(s) 1 H2 (g) ¡ CaH2 (s) All ionic hydrides are solids that have the high melting points characteristic of ionic compounds. The anion in these compounds is the hydride ion, H2, which is a very strong Brønsted base. It readily accepts a proton from a proton donor such as water: H2 (aq) 1 H2O(l) ¡ OH2 (aq) 1 H2 (g) Due to their high reactivity with water, ionic hydrides are frequently used to remove traces of water from organic solvents. Covalent Hydrides In covalent hydrides, the hydrogen atom is covalently bonded to the atom of another element. There are two types of covalent hydrides—those containing discrete molec- ular units, such as CH4 and NH3, and those having complex polymeric structures, such as (BeH2)x and (AlH3)x, where x is a very large number. This is an example of the diagonal relationship between Be and Al Figure 22.3 shows the binary ionic and covalent hydrides of the representative (see p. 348). elements. The physical and chemical properties of these compounds change from ionic to covalent across a given period. Consider, for example, the hydrides of the second- period elements: LiH, BeH2, B2H6, CH4, NH3, H2O, and HF. LiH is an ionic com- pound with a high melting point (680°C). The structure of BeH2 (in the solid state) is polymeric; it is a covalent compound. The molecules B2H6 and CH4 are nonpolar. In contrast, NH3, H2O, and HF are all polar molecules in which the hydrogen atom is the positive end of the polar bond. Of this group of hydrides (NH3, H2O, and HF), only HF is acidic in water. As we move down any group in Figure 22.3, the compounds change from cova- lent to ionic. In Group 2A, for example, BeH2 and MgH2 are covalent, but CaH2, SrH2, and BaH2 are ionic. 960 Chapter 22 ■ Nonmetallic Elements and Their Compounds 1 Discrete molecular units 18 1A 8A 2 Polymeric structure; covalent compound 13 14 15 16 17 2A 3A 4A 5A 6A 7A LiH BeH2 Ionic compound B2H6 CH4 NH3 H2O HF NaH MgH2 3 4 5 6 7 8 9 10 11 12 AlH3 SiH4 PH3 H2S HCl 3B 4B 5B 6B 7B 8B 1B 2B KH CaH2 GaH3 GeH4 AsH3 H2Se HBr RbH SrH2 InH3 SnH4 SbH3 H2Te HI CsH BaH2 TlH3 PbH4 BiH3 Figure 22.3 Binary hydrides of the representative elements. In cases in which hydrogen forms more than one compound with the same element, only the formula of the simplest hydride is shown. The properties of many of the transition metal hydrides are not well characterized. Interstitial Hydrides Molecular hydrogen forms a number of hydrides with transition metals. In some of these compounds, the ratio of hydrogen atoms to metal atoms is not a constant. Such Interstitial compounds are sometimes compounds are called interstitial hydrides. For example, depending on conditions, the called nonstoichiometric compounds. formula for titanium hydride can vary between TiH1.8 and TiH2. Note that they do not obey the law of definite proportions (see Section 2.1). Many of the interstitial hydrides have metallic properties such as electrical con- ductivity. Yet it is known that hydrogen is definitely bonded to the metal in these compounds, although the exact nature of the bonding is often not clear. Molecular hydrogen interacts in a unique way with palladium (Pd). Hydrogen gas is readily adsorbed onto the surface of the palladium metal, where it dissociates into atomic hydrogen. The H atoms then “dissolve” into the metal. On heating and under the pressure of H2 gas on one side of the metal, these atoms diffuse through the metal and recombine to form molecular hydrogen, which emerges as the gas from the other side. Because no other gas behaves in this way with palladium, this process has been used to separate hydrogen gas from other gases on a small scale. Isotopes of Hydrogen The 11Hisotope is also called protium. Hydrogen has three isotopes: 11H (hydrogen), 21H (deuterium, symbol D), and 31H (tritium, Hydrogen is the only element whose isotopes are given different names. symbol T). The natural abundances of the stable hydrogen isotopes are hydrogen, 99.985 percent; and deuterium, 0.015 percent. Tritium is a radioactive isotope with a half-life of about 12.5 years. Table 22.1 compares some of the common properties of H2O with those of D2O. Deuterium oxide, or heavy water as it is commonly called, is used in some nuclear reactors as a coolant and a moderator of nuclear reactions (see Chapter 19). D2O can be separated from H2O by fractional distillation because H2O boils at a lower temperature, as Table 22.1 shows. Another technique for its separation is electrolysis of water. Because H2 gas is formed about eight times as fast as D2 during electrolysis, the water remaining in the electrolytic cell becomes progressively enriched with D2O. Interestingly, the Dead Sea, which for thousands of years has entrapped water that has  22.2 Hydrogen 961 Table 22.1 Properties of H2O and D2O Property H2O D2O Molar mass (g/mol) 18.02 20.03 Melting point (°C) 0 3.8 Boiling point (°C) 100 101.4 Density at 4°C (g/cm3 ) 1.000 1.108 no outlet other than through evaporation, has a higher [D2O]/[H2O] ratio than water found elsewhere. Although D2O chemically resembles H2O in most respects, it is a toxic substance. The reason is that deuterium is heavier than hydrogen; thus, its compounds often react more slowly than those of the lighter isotope. Regular drinking of D2O instead of H2O could prove fatal because of the slower rate of transfer of D1 compared with that of H1 in the acid-base reactions involved in enzyme catalysis. This kinetic isotope effect is also manifest in acid ionization constants. For example, the ionization constant of acetic acid CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq) Ka 5 1.8 3 1025 is about three times as large as that of deuterated acetic acid: CH3COOD(aq) Δ CH3COO2 (aq) 1 D1 (aq) Ka 5 6 3 1026 Hydrogenation Hydrogenation is the addition of hydrogen to compounds containing multiple bonds, especially C“C and C‚C bonds. A simple hydrogenation reaction is the conversion of ethylene to ethane: H H H H A A G D H2 ⫹ CPC 88n HOCOCOH D G A A H H H H ethylene ethane This reaction is quite slow under normal conditions, but the rate can be greatly increased by the presence of a catalyst such as nickel or platinum. As in the Haber Platinum catalyst on alumina synthesis of ammonia (see Section 13.6), the main function of the catalyst is to (Al2O3) used in hydrogenation. weaken the H—H bond and facilitate the reaction. Hydrogenation is an important process in the food industry. Vegetable oils have considerable nutritional value, but some oils must be hydrogenated before we can use them because of their unsavory flavor and their inappropriate molecular structures (that is, there are too many C“C bonds present). Upon exposure to air, these polyunsatu- rated molecules (that is, molecules with many C“C bonds) undergo oxidation to yield unpleasant-tasting products (oil that has oxidized is said to be rancid). In the hydroge- nation process, a small amount of nickel (about 0.1 percent by mass) is added to the oil and the mixture is exposed to hydrogen gas at high temperature and pressure. Afterward, the nickel is removed by filtration. Hydrogenation reduces the number of double bonds in the molecule but does not completely eliminate them. If all the double bonds are eliminated, the oil becomes hard and brittle. Under controlled conditions, suitable cooking oils and margarine may be prepared by hydrogenation from vegetable oils extracted from cottonseed, corn, and soybeans.  CHEMISTRY in Action Metallic Hydrogen S cientists have long been interested in how nonmetallic sub- stances, including hydrogen, behave under exceedingly high pressure. It was predicted that when atoms or molecules Insulating molecular hydrogen Metallic molecular hydrogen are compressed, their bonding electrons might be delocalized, producing a metallic state. In 1996, physicists at the Lawrence Metallic atomic hydrogen Livermore Laboratory used a 60-foot-long gun to generate a Rock core shock compression onto a thin (0.5-mm) layer of liquid hydro- gen. For an instant, at pressures between 0.9 and 1.4 million atm, they were able to measure the electrical conductivity of the hydrogen sample and found that it was comparable to that of cesium metal at 2000 K. (The temperature of the hydrogen sam- ple rose as a result of compression, although it remained in the molecular form.) As the pressure fell rapidly, the metallic state of hydrogen disappeared. Interior composition of Jupiter. The Livermore experiment suggested that metallic hydro- gen, if it can be kept in a stable state, may act as a room- due to the heat-driven motion of liquid iron within its core.) temperature superconductor. The fact that hydrogen becomes Jupiter is composed of an outer layer of nonmetallic molecular metallic at pressures lower than previously thought possible hydrogen that continuously transforms hydrogen within the also has provided new insight into planetary science. For many core to metallic fluid hydrogen. It is now believed that this me- years scientists were puzzled by Jupiter’s strong magnetic field, tallic layer is much closer to the surface (because the pressure which is 20 times greater than that of Earth. A planet’s magnetic needed to convert molecular hydrogen to metallic hydrogen is field results from the convection motion of electrically conduc- not as high as previously thought), which would account for tive fluid in its interior. (For example, Earth’s magnetic field is Jupiter’s unusually strong magnetic field. The Hydrogen Economy The world’s fossil fuel reserves are being depleted at an alarmingly fast rate. Faced with this dilemma, scientists have made intensive efforts in recent years to develop a method of obtaining hydrogen gas as an alternative energy source. Hydrogen gas could replace gasoline to power automobiles (after considerable modification of the engine, of course) or be used with oxygen gas in fuel cells to generate electricity (see p. 835). One major advantage of using hydrogen gas in these ways is that the reactions are essentially free of pollutants; the end product formed in a hydrogen- powered engine or in a fuel cell would be water, just as in the burning of hydrogen gas in air: 2H2 (g) 1 O2 (g) ¡ 2H2O(l) Of course, success of a hydrogen economy would depend on how cheaply we could produce hydrogen gas and how easily we could store it. Although electrolysis of water consumes too much energy for large-scale appli- cation, if scientists can devise a more practical method of “splitting” water mol- The total volume of ocean water is about 1 3 1021 L. Thus, the ocean contains an ecules, we could obtain vast amounts of hydrogen from seawater. One approach that almost inexhaustible supply of hydrogen. is currently in the early stages of development would use solar energy. In this 962  22.3 Carbon 963 scheme, a catalyst (a complex molecule containing one or more transition metal atoms, such as ruthenium) absorbs a photon from solar radiation and becomes ener- getically excited. In its excited state, the catalyst is capable of reducing water to molecular hydrogen. Some of the interstitial hydrides we have discussed would make suitable storage compounds for hydrogen. The reactions that form these hydrides are usually revers- ible, so hydrogen gas can be obtained simply by reducing the pressure of the hydro- gen gas above the metal. The advantages of using interstitial hydrides are as follows: (1) many metals have a high capacity to take up hydrogen gas—sometimes up to three times as many hydrogen atoms as there are metal atoms; and (2) because these hydrides are solids, they can be stored and transported more easily than gases or liquids. The Chemistry in Action essay on p. 962 describes what happens to hydrogen under pressure. 22.3 Carbon Although it constitutes only about 0.09 percent by mass of Earth’s crust, carbon is an The carbon cycle is discussed on p. 912. essential element of living matter. It is found free in the form of diamond and graph- ite (see Figure 8.17), and it is also a component of natural gas, petroleum, and coal. (Coal is a natural dark-brown to black solid used as a fuel; it is formed from fossilized plants and consists of amorphous carbon with various organic and some inorganic compounds.) Carbon combines with oxygen to form carbon dioxide in the atmosphere and occurs as carbonate in limestone and chalk. Diamond and graphite are allotropes of carbon. Figure 22.4 shows the phase The structures of diamond and graphite are shown in Figure 11.28. diagram of carbon. Although graphite is the stable form of carbon at 1 atm and 25°C, owners of diamond jewelry need not be alarmed, for the rate of the spontaneous process C(diamond) ¡ C(graphite) ¢G° 5 22.87 kJ/mol is extremely slow. Millions of years may pass before a diamond turns to graphite. Synthetic diamond can be prepared from graphite by applying very high pres- sures and temperatures. Figure 22.5 shows a synthetic diamond and its starting material, graphite. Synthetic diamonds generally lack the optical properties of natu- ral diamonds. They are useful, however, as abrasives and in cutting concrete and many other hard substances, including metals and alloys. The uses of graphite are described on p. 490. Carbon has the unique ability to form long chains (consisting of more than 50 C atoms) and stable rings with five or six members. This phenomenon is called catena- tion, the linking of like atoms. Carbon’s versatility is responsible for the millions of Figure 22.4 Phase diagram Diamond of carbon. Note that under atmospheric conditions, graphite Liquid is the stable form of carbon. P (atm) 2 × 104 Graphite Vapor 3300 t (°C) 964 Chapter 22 ■ Nonmetallic Elements and Their Compounds Figure 22.5 Synthetic diamonds and the starting material—graphite. organic compounds (made up of carbon and hydrogen and other elements such as oxygen, nitrogen, and halogens) found on Earth. The chemistry of organic compounds is discussed in Chapter 24. Carbides and Cyanides Carbon combines with metals to form ionic compounds called carbides, such as CaC2 and Be2C, in which carbon is in the form of C22 2 or C 42 ions. These ions are strong Brønsted bases and react with water as follows: C22 2 2 (aq) 1 2H2O(l) ¡ 2OH (aq) 1 C2H2(g) C42(aq) 1 4H2O(l) ¡ 4OH2(aq) 1 CH4(g) Carbon also forms a covalent compound with silicon. Silicon carbide, SiC, is called carborundum and is prepared as follows: SiO2 (s) 1 3C(s) ¡ SiC(s) 1 2CO(g) Carborundum is also formed by heating silicon with carbon at 1500°C. Carborundum is almost as hard as diamond and it has the diamond structure; each carbon atom is tetrahedrally bonded to four Si atoms, and vice versa. It is used mainly for cutting, grinding, and polishing metals and glasses. Another important class of carbon compounds, the cyanides, contain the anion group :C‚N:2. Cyanide ions are extremely toxic because they bind almost irre- versibly to the Fe(III) ion in cytochrome oxidase, a key enzyme in metabolic HCN is the gas used in gas execution processes. Hydrogen cyanide, which has the aroma of bitter almonds, is even more chambers. dangerous because of its volatility (b.p. 26°C). A few tenths of 1 percent by volume of HCN in air can cause death within minutes. Hydrogen cyanide can be prepared by treating sodium cyanide or potassium cyanide with acid: NaCN(s) 1 HCl(aq) ¡ NaCl(aq) 1 HCN(aq) Because HCN (in solution, called hydrocyanic acid) is a very weak acid (Ka 5 4.9 3 10210), most of the HCN produced in this reaction is in the nonionized form and leaves the solution as hydrogen cyanide gas. For this reason, acids should never be mixed with metal cyanides in the laboratory without proper ventilation.  22.3 Carbon 965 Figure 22.6 Cyanide ponds for extracting gold from metal ore. Cyanide ions are used to extract gold and silver. Although these metals are usu- ally found in the uncombined state in nature, in other metal ores they may be present in relatively small concentrations and are more difficult to extract. In a typical process, the crushed ore is treated with an aqueous cyanide solution in the presence of air to dissolve the gold by forming the soluble complex ion [Au(CN) 2]2 : 4Au(s) 1 8CN2 (aq) 1 O2 (g) 1 2H2O(l) ¡ 4[Au(CN) 2]2 (aq) 1 4OH2 (aq) The complex ion [Au(CN)2]2 (along with some cation, such as Na1) is separated from other insoluble materials by filtration and treated with an electropositive metal such as zinc to recover the gold: Zn(s) 1 2[Au(CN) 2]2 (aq) ¡ [Zn(CN) 4]22 (aq) 1 2Au(s) Figure 22.6 shows an aerial view of cyanide ponds used for the extraction of gold. Oxides of Carbon Of the several oxides of carbon, the most important are carbon monoxide, CO, and carbon dioxide, CO2. Carbon monoxide is a colorless, odorless gas formed by the incomplete combustion of carbon or carbon-containing compounds: 2C(s) 1 O2 (g) ¡ 2CO(g) Carbon monoxide is used in metallurgical process for extracting nickel (see p. 937), The role of CO as an indoor air pollutant in organic synthesis, and in the production of hydrocarbon fuels with hydrogen. Indus- is discussed on p. 923. trially, it is prepared by passing steam over heated coke. Carbon monoxide burns readily in oxygen to form carbon dioxide: 2CO(g) 1 O2 (g) ¡ 2CO2 (g) ¢H° 5 2566 kJ/mol Carbon monoxide is not an acidic oxide (it differs from carbon dioxide in that regard), and it is only slightly soluble in water. Carbon dioxide is a colorless and odorless gas. Unlike carbon monoxide, CO2 Carbon dioxide is the primary greenhouse is nontoxic. It is an acidic oxide (see p. 703). Carbon dioxide is used in beverages, gas (see p. 912). in fire extinguishers, and in the manufacture of baking soda, NaHCO3, and soda ash, Na2CO3. Solid carbon dioxide, called dry ice, is used as a refrigerant (see Figure 11.42). CHEMISTRY in Action Synthetic Gas from Coal T he very existence of our technological society depends on an abundant supply of energy. Although the United States has only 5 percent of the world’s population, we consume about formation of sulfur dioxide (SO2) from the sulfur-containing compounds. This process leads to the formation of “acid rain,” discussed on p. 916. 20 percent of the world’s energy! At present, the two major One of the most promising methods for making coal a sources of energy are nuclear fission and fossil fuels (discussed more efficient and cleaner fuel involves the conversion of coal in Chapters 19 and 24, respectively). Coal, oil (which is also to a gaseous form, called syngas for “synthetic gas.” This pro- known as petroleum), and natural gas (mostly methane) are col- cess is called coal gasification. In the presence of very hot lectively called fossil fuels because they are the end result of the steam and air, coal decomposes and reacts according to the fol- decomposition of plants and animals over tens or hundreds of lowing simplified scheme: millions of years. Oil and natural gas are cleaner-burning and more efficient fuels than coal, so they are preferred for most C(s) 1 H2O(g) ¡ CO(g) 1 H2 (g) purposes. However, supplies of oil and natural gas are being C(s) 1 2H2 (g) ¡ CH4 (g) depleted at an alarming rate, and research is under way to make The main component of syngas is methane. In addition, the first coal a more versatile source of energy. reaction yields hydrogen and carbon monoxide gases and other Coal consists of many high-molar-mass carbon compounds useful by-products. Under suitable conditions, CO and H2 com- that also contain oxygen, hydrogen, and small amounts of nitro- bine to form methanol: gen and sulfur. Coal constitutes about 90 percent of the world’s fossil fuel reserves. For centuries coal has been used as a fuel CO(g) 1 2H2 (g) ¡ CH3OH(l) both in homes and in industry. However, underground coal mining is expensive and dangerous, and strip mining (that is, Methanol has many uses, for example, as a solvent and a start- mining in an open pit after removal of the overlaying earth and ing material for plastics. Syngas is easier than coal to store and rock) is tremendously harmful to the environment. Another transport. What’s more, it is not a major source of air pollution problem, this one associated with the burning of coal, is the because sulfur is removed in the gasification process. Underground coal mining. 966  22.4 Nitrogen and Phosphorus 967 22.4 Nitrogen and Phosphorus Nitrogen About 78 percent of air by volume is nitrogen. The most important mineral sources The nitrogen cycle is discussed on p. 901. of nitrogen are saltpeter (KNO3) and Chile saltpeter (NaNO3). Nitrogen is an essential element of life; it is a component of proteins and nucleic acids. Molecular nitrogen is obtained by fractional distillation of air (the boiling points of liquid nitrogen and liquid oxygen are 2196°C and 2183°C, respectively). In the Molecular nitrogen will boil off before molecular oxygen does during the laboratory, very pure nitrogen gas can be prepared by the thermal decomposition of fractional distillation of liquid air. ammonium nitrite: NH4NO2 (s) ¡ 2H2O(g) 1 N2 (g) The N2 molecule contains a triple bond and is very stable with respect to disso- ciation into atomic species. However, nitrogen forms a large number of com- pounds with hydrogen and oxygen in which the oxidation number of nitrogen varies from 23 to 15 (Table 22.2). Most nitrogen compounds are covalent; however, when heated with certain metals, nitrogen forms ionic nitrides containing the N32 ion: 6Li(s) 1 N2 (g) ¡ 2Li3N(s) Table 22.2 Common Compounds of Nitrogen Oxidation Number Compound Formula Structure 23 Ammonia NH3 O HONOH A H 22 Hydrazine N2H4 O O HONONOH A A H H 21 Hydroxylamine NH2OH O O HONOOOH Q A H 0 Nitrogen* (dinitrogen) N2 SNqNS 11 Nitrous oxide (dinitrogen monoxide) N2O O SNqNOOS Q N O 12 Nitric oxide (nitrogen monoxide) NO SNPO Q 13 Nitrous acid HNO2 Q O O O OPNOOOH Q N O 14 Nitrogen dioxide NO2 O SOONPO Q Q 15 Nitric acid HNO3 O O OPNOOOH Q Q A SO QS *We list the element here as a reference. 968 Chapter 22 ■ Nonmetallic Elements and Their Compounds The nitride ion is a strong Brønsted base and reacts with water to produce ammonia and hydroxide ions: N32 (aq) 1 3H2O(l) ¡ NH3 (g) 1 3OH2 (aq) Ammonia Ammonia is one of the best-known nitrogen compounds. It is prepared industrially from nitrogen and hydrogen by the Haber process (see Section 13.6 and p. 601). It can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide: NH4Cl(aq) 1 NaOH(aq) ¡ NaCl(aq) 1 H2O(l) 1 NH3 (g) Ammonia is a colorless gas (b.p. 233.4°C) with an irritating odor. About three- quarters of the ammonia produced annually in the United States (about 18 million tons in 2010) is used in fertilizers. Liquid ammonia, like water, undergoes autoionization: 2NH3 (l) Δ NH14 1 NH2 2 or simply NH3 (l) Δ H1 1 NH2 2 The amide ion is a strong Brønsted base where NH22 is called the amide ion. Note that both H1 and NH22 are solvated with and does not exist in water. the NH3 molecules. (Here is an example of ion-dipole interaction.) At 250°C, the ion product [H1][NH2 2 ] is about 1 3 10 233 , considerably smaller than 1 3 10214 for water at 25°C. Nevertheless, liquid ammonia is a suitable solvent for many electrolytes, especially when a more basic medium is required or if the solutes react with water. The ability of liquid ammonia to dissolve alkali metals was discussed in Section 21.5. Hydrazine Another important hydride of nitrogen is hydrazine: H H G D O O NON D G H H N2H4 Each N atom is sp3-hybridized. Hydrazine is a colorless liquid that smells like ammo- nia. It melts at 2°C and boils at 114°C. Hydrazine is a base that can be protonated to give the N2H1 21 5 and N2H6 ions. As 31 21 2 21 2 a reducing agent, it can reduce Fe to Fe , MnO4 to Mn , and I2 to I . Its reaction with oxygen is highly exothermic: N2H4(l) 1 O2(g) ¡ N2(g) 1 2H2O(l) ¢H° 5 2666.6 kJ/mol Hydrazine and its derivative methylhydrazine, N2H3(CH3), together with the oxidizer dinitrogen tetroxide (N2O4), are used as rocket fuels. Hydrazine also plays a role in polymer synthesis and in the manufacture of pesticides. Oxides and Oxoacids of Nitrogen There are many nitrogen oxides, but the three particularly important ones are: nitrous oxide, nitric oxide, and nitrogen dioxide.  22.4 Nitrogen and Phosphorus 969 Nitrous oxide, N2O, is a colorless gas with a pleasing odor and sweet taste. It is prepared by heating ammonium nitrate to about 270°C: NH4NO3 (s) ¡ N2O(g) 1 2H2O(g) Nitrous oxide resembles molecular oxygen in that it supports combustion. It does so because it decomposes when heated to form molecular nitrogen and molecular oxygen: 2N2O(g) ¡ 2N2 (g) 1 O2 (g) It is chiefly used as an anesthetic in dental procedures and other minor surgery. Nitrous oxide is also called “laughing gas” because a person inhaling the gas becomes some- what giddy. No satisfactory explanation has yet been proposed for this unusual physi- ological response. Nitrous oxide is also used as the propellant in cans of whipped cream Figure 22.7 The production of NO2 gas when copper reacts with due to its high solubility in the whipped cream mixture. concentrated nitric acid. Nitric oxide, NO, is a colorless gas. The reaction of N2 and O2 in the atmosphere N2 (g) 1 O2 (g) Δ 2NO(g) ¢G° 5 173.4 kJ/mol is a form of nitrogen fixation (see p. 901). The equilibrium constant for the above According to Le Châtelier’s principle, the reaction is very small at room temperature: KP is only 4.0 3 10231 at 25°C, so forward endothermic reaction is favored by heating. very little NO will form at that temperature. However, the equilibrium constant increases rapidly with temperature, for example, in a running auto engine. An appreciable amount of nitric oxide is formed in the atmosphere by the action of lightning. In the laboratory, the gas can be prepared by the reduction of dilute nitric acid with copper: 3Cu(s) 1 8HNO3 (aq) ¡ 3Cu1NO3 2 2 (aq) 1 4H2O(l) 1 2NO(g) The nitric oxide molecule is paramagnetic, containing one unpaired electron. It can be represented by the following resonance structures: P Q O Q mn NPO O P NPO Q Q  As we noted in Chapter 9, this molecule does not obey the octet rule. The properties of nitric oxide are discussed on p. 397. Unlike nitrous oxide and nitric oxide, nitrogen dioxide is a highly toxic yellow- The role of NO2 in smog formation is brown gas with a choking odor. In the laboratory nitrogen dioxide is prepared by the discussed on p. 920. action of concentrated nitric acid on copper (Figure 22.7): Cu(s) 1 4HNO3 (aq) ¡ Cu(NO3 ) 2 (aq) 1 2H2O(l) 1 2NO2 (g) Nitrogen dioxide is paramagnetic. It has a strong tendency to dimerize to dinitrogen tetroxide, which is a diamagnetic molecule: 2NO2 Δ N2O4 This reaction occurs in both the gas phase and the liquid phase. Nitrogen dioxide is an acidic oxide; it reacts rapidly with cold water to form both nitrous acid, HNO2, and nitric acid: 2NO2 (g) 1 H2O(l) ¡ HNO2 (aq) 1 HNO3 (aq) Neither N2O nor NO reacts with water. This is a disproportionation reaction (see p. 142) in which the oxidation number of nitrogen changes from 14 (in NO2) to 13 (in HNO2) and 15 (in HNO3). Note that this reaction is quite different from that between CO2 and H2O, in which only one acid (carbonic acid) is formed. 970 Chapter 22 ■ Nonmetallic Elements and Their Compounds Nitric acid is one of the most important inorganic acids. It is a liquid (b.p. 82.6°C), but it does not exist as a pure liquid because it decomposes spontaneously to some extent as follows: On standing, a concentrated nitric acid 4HNO3 (l) ¡ 4NO2 (g) 1 2H2O(l) 1 O2 (g) solution turns slightly yellow as a result of NO2 formation. The major industrial method of producing nitric acid is the Ostwald process, discussed in Section 13.6. The concentrated nitric acid used in the laboratory is 68 percent HNO3 by mass (density 1.42 g/cm3), which corresponds to 15.7 M. Nitric acid is a powerful oxidizing agent. The oxidation number of N in HNO3 is 15. The most common reduction products of nitric acid are NO2 (oxidation num- ber of N 5 14), NO (oxidation number of N 5 12), and NH1 4 (oxidation number of N 5 23). Nitric acid can oxidize metals both below and above hydrogen in the activity series (see Figure 4.16). For example, copper is oxidized by concentrated nitric acid, as discussed earlier. In the presence of a strong reducing agent, such as zinc metal, nitric acid can be reduced all the way to the ammonium ion: 4Zn(s) 1 10H1 (aq) 1 NO2 21 1 3 (aq) ¡ 4Zn (aq) 1 NH4 (aq) 1 3H2O(l) Concentrated nitric acid does not oxidize gold. However, when the acid is added to concentrated hydrochloric acid in a 1:3 ratio by volume (one part HNO3 to three parts HCl), the resulting solution, called aqua regia, can oxidize gold, as follows: Au(s) 1 3HNO3 (aq) 1 4HCl(aq) ¡ HAuCl4 (aq) 1 3H2O(l) 1 3NO2 (g) The oxidation of Au is promoted by the complexing ability of the Cl2 ion (to form the AuCl2 4 ion). Concentrated nitric acid also oxidizes a number of nonmetals to their correspond- ing oxoacids: P4 (s) 1 20HNO3 (aq) ¡ 4H3PO4 (aq) 1 20NO2 (g) 1 4H2O(l) S(s) 1 6HNO3 (aq) ¡ H2SO4 (aq) 1 6NO2 (g) 1 2H2O(l) Nitric acid is used in the manufacture of fertilizers, dyes, drugs, and explosives. The Chemistry in Action essay on p. 974 describes a nitrogen-containing fertilizer that can be highly explosive. Phosphorus Like nitrogen, phosphorus is a member of the Group 5A family; in some respects the chemistry of phosphorus resembles that of nitrogen. Phosphorus occurs most com- monly in nature as phosphate rocks, which are mostly calcium phosphate, Ca3(PO4)2, and fluoroapatite, Ca5(PO4)3F (Figure 22.8). Elemental phosphorus can be obtained by heating calcium phosphate with coke and silica sand: 2Ca3(PO4)2(s) 1 10C(s) 1 6SiO2(s) ¡ 6CaSiO3(s) 1 10CO(g) 1 P4(s) There are several allotropic forms of phosphorus, but only white phosphorus and red phosphorus (see Figure 8.18) are of importance. White phosphorus consists of discrete tetrahedral P4 molecules (Figure 22.9). A solid (m.p. 44.2°C), white phospho- rus is insoluble in water but quite soluble in carbon disulfide (CS2) and in organic solvents such as chloroform (CHCl3). White phosphorus is a highly toxic substance. It bursts into flames spontaneously when exposed to air; hence it is used in incendiary bombs and grenades: P4 (s) 1 5O2 (g) ¡ P4O10 (s)  22.4 Nitrogen and Phosphorus 971 Figure 22.8 Phosphate mining. The high reactivity of white phosphorus is attributed to structural strain: The P—P bonds are compressed in the tetrahedral P4 molecule. White phosphorus was once used in matches, but because of its toxicity it has been replaced by tetraphosphorus trisulfide, P4S3. When heated in the absence of air, white phosphorus is slowly converted to red phosphorus at about 300°C: nP4 (white phosphorus) ¡ (P4 2 n (red phosphorus) Red phosphorus has a polymeric structure (see Figure 22.9) and is more stable and less volatile than white phosphorus. Hydride of Phosphorus The most important hydride of phosphorus is phosphine, PH3, a colorless, very poi- sonous gas formed by heating white phosphorus in concentrated sodium hydroxide: P4 (s) 1 3NaOH(aq) 1 3H2O(l) ¡ 3NaH2PO2 (aq) 1 PH3 (g) Figure 22.9 The structures of white and red phosphorus. Red phosphorus is believed to have a chain structure, as shown. White phosphorus Red phosphorus 972 Chapter 22 ■ Nonmetallic Elements and Their Compounds Phosphine is moderately soluble in water and more soluble in carbon disulfide and organic solvents. Its aqueous solution is neutral, unlike that of ammonia. In liquid ammonia, phosphine dissolves to give NH41PH2 2 . Phosphine is a strong reducing agent; it reduces many metal salts to the corresponding metal. The gas burns in air: PH3 (g) 1 2O2 (g) ¡ H3PO4 (s) Halides of Phosphorus Phosphorus forms binary compounds with halogens: the trihalides, PX3, and the pen- tahalides, PX5, where X denotes a halogen atom. In contrast, nitrogen can form only trihalides (NX3). Unlike nitrogen, phosphorus has a 3d subshell, which can be used for valence-shell expansion. We can explain the bonding in PCl5 by assuming that phos- phorus undergoes sp3d hybridization of its 3s, 3p, and 3d orbitals (see Example 10.4). The five sp3d hybrid orbitals also account for the trigonal bipyramidal geometry of the PCl5 molecule (see Table 10.4). Phosphorus trichloride is prepared by heating white phosphorus in chlorine: P4 (l) 1 6Cl2 (g) ¡ 4PCl3 (g) A colorless liquid (b.p. 76°C), PCl3 is hydrolyzed according to the equation: PCl3 (l) 1 3H2O(l) ¡ H3PO3 (aq) 1 3HCl(g) In the presence of an excess of chlorine gas, PCl3 is converted to phosphorus penta- chloride, which is a light-yellow solid: PCl3 (l) 1 Cl2 (g) ¡ PCl5 (s) X-ray studies have shown that solid phosphorus pentachloride exists as [PCl14 ][PCl 26 ], in which the PCl1 2 4 ion has a tetrahedral geometry and the PCl6 ion has an octahedral geometry. In the gas phase, PCl5 (which has trigonal bipyramidal geometry) is in equilibrium with PCl3 and Cl2: PCl5 (g) Δ PCl3 (g) 1 Cl2 (g) Phosphorus pentachloride reacts with water as follows: PCl5 (s) 1 4H2O(l) ¡ H3PO4 (aq) 1 5HCl(aq) Oxides and Oxoacids of Phosphorus The two important oxides of phosphorus are tetraphosphorus hexaoxide, P4O6, and tetraphosphorus decaoxide, P4O10 (Figure 22.10). The oxides are obtained by burning white phosphorus in limited and excess amounts of oxygen gas, respectively: P4 (s) 1 3O2 (s) ¡ P4O6 (s) P4 (s) 1 5O2 (g) ¡ P4O10 (s) Both oxides are acidic; that is, they are converted to acids in water. The compound P4O10 is a white flocculent powder (m.p. 420°C) that has a great affinity for water: P4O10 (s) 1 6H2O(l) ¡ 4H3PO4 (aq) For this reason, it is often used for drying gases and for removing water from solvents.  22.4 Nitrogen and Phosphorus 973 Phosphorus Figure 22.10 The structures of P4O6 and P4O10. Note the Oxygen tetrahedral arrangement of the P atoms in P4O10. P4O6 P4O10 There are many oxoacids containing phosphorus. Some examples are phospho- Phosphoric acid is the most important phosphorus-containing oxoacid. rous acid, H3PO3; phosphoric acid, H3PO4; hypophosphorous acid, H3PO2; and tri- phosphoric acid, H5P3O10 (Figure 22.11). Phosphoric acid, also called orthophosphoric acid, is a weak triprotic acid (see p. 692). It is prepared industrially by the reaction of calcium phosphate with sulfuric acid: Ca3 (PO4 ) 2 (s) 1 3H2SO4 (aq) ¡ 2H3PO4 (aq) 1 3CaSO4 (s) In the pure form phosphoric acid is a colorless solid (m.p. 42.2°C). The phos- phoric acid we use in the laboratory is usually an 82 percent H3PO4 solution (by mass). Phosphoric acid and phosphates have many commercial applications in detergents, fertilizers, flame retardants, and toothpastes, and as buffers in car- bonated beverages. Like nitrogen, phosphorus is an element that is essential to life. It constitutes only about 1 percent by mass of the human body, but it is a very important 1 per- cent. About 23 percent of the human skeleton is mineral matter. The phosphorus content of this mineral matter, calcium phosphate, Ca3(PO4)2, is 20 percent. Our teeth are basically Ca3(PO4)2 and Ca5(PO4)3OH. Phosphates are also important components of the genetic materials deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Figure 22.11 Structures of some common phosphorus-containing S OS SOS oxoacids. B B O HO O O P O O O OH O HOO O P OH M A M M A H H Phosphorous acid (H 3PO 3 ) Hypophosphorous acid (H 3 PO 2 ) SOS SOS SOS SOS B B B B HO O O P O O O O OH HO O O P O O O OO POO O O POO O OH M A M M A M A M A M SOS S OS SOS SOS A A A A H H H H Phosphoric acid (H 3 PO 4 ) Triphosphoric acid (H 5 P 3 O 10 ) CHEMISTRY in Action Ammonium Nitrate—The Explosive Fertilizer A mmonium nitrate is the most important fertilizer in the world (see p. 105). It ranked fifteenth among the industrial chemicals produced in the United States in 2009 (8 million tons). threefold. Ammonium nitrate can also be mixed with charcoal, flour, sugar, sulfur, rosin, and paraffin to form an explosive. Intense heat from the explosion causes the gases to expand Unfortunately, it is also a powerful explosive. In 1947 an explo- rapidly, generating shock waves that destroy most objects in sion occurred aboard a ship being loaded with the fertilizer in their path. Texas. The fertilizer was in paper bags and apparently blew up Federal law regulates the sale of explosive-grade ammo- after sailors tried to stop a fire in the ship’s hold by closing a nium nitrate, which is used for 95 percent of all commercial hatch, thereby creating the compression and heat necessary for an blasting in road construction and mining. However, the wide explosion. More than 600 people died as a result of the accident. availability of large quantities of ammonium nitrate and other More recent disasters involving ammonium nitrate took place at substances that enhance its explosive power make it possible for the World Trade Center in New York City in 1993 and at the anyone who is so inclined to construct a bomb. The bomb that Alfred P. Murrah Federal Building in Oklahoma City in 1995. destroyed the federal building in Oklahoma City is estimated to A strong oxidizer, ammonium nitrate is stable at room tem- have contained 4000 pounds of ammonium nitrate and fuel oil, perature. At 250°C, it begins to decompose as follows: which was set off by another small explosive device. How can the use of ammonium nitrate by terrorists be pre- NH4NO3 (g) ¡ N2O(g) 1 2H2O(g) vented? The most logical approach is to desensitize or neutral- ize the compound’s ability to act as an explosive, but to date no At 300°C, different gaseous products and more heat are satisfactory way has been found to do so without diminishing its produced: value as a fertilizer. A more passive method is to add to the fer- 2NH4NO3 (g) ¡ 2N2 (g) 1 4H2O(g) 1 O2 (g) tilizer an agent known as a taggant, which would allow law en- forcement to trace the source of an ammonium nitrate explosive. About 1.46 kJ of heat are generated per gram of the compound A number of European countries now forbid the sale of ammo- decomposed. When it is combined with a combustible mate- nium nitrate without taggants, although the U.S. Congress has rial, such as fuel oil, the energy released increases almost yet to pass such a law. A bag of ammonium nitrate fertilizer, which is labeled as an explosive. The Alfred P. Murrah building after a deadly explosion caused by an ammonium nitrate bomb. 974  22.5 Oxygen and Sulfur 975 22.5 Oxygen and Sulfur Oxygen Oxygen is by far the most abundant element in Earth’s crust, constituting about 46 per- The oxygen cycle is discussed on p. 902. cent of its mass. In addition, the atmosphere contains about 21 percent molecular oxygen by volume (23 percent by mass). Like nitrogen, oxygen in the free state is a diatomic molecule (O2). In the laboratory, oxygen gas can be obtained by heating potassium chlorate (see Figure 5.15): 2KClO3 (s) ¡ 2KCl(s) 1 3O2 (g) The reaction is usually catalyzed by manganese(IV) dioxide, MnO2. Pure oxygen gas can be prepared by electrolyzing water (p. 842). Industrially, oxygen gas is prepared by the fractional distillation of liquefied air (p. 535). Oxygen gas is color- less and odorless. Oxygen is a building block of practically all biomolecules, accounting for about a fourth of the atoms in living matter. Molecular oxygen is the essential oxidant in the metabolic breakdown of food molecules. Without it, a human being cannot survive for more than a few minutes. Properties of Diatomic Oxygen Although oxygen has two allotropes, O2 and O3, when we speak of molecular oxygen, we normally mean O2. Ozone, O3, is less stable than O2. The O2 molecule is para- magnetic because it contains two unpaired electrons (see Section 10.7). A strong oxidizing agent, molecular oxygen is one of the most widely used industrial chemicals. Its main uses are in the steel industry (see Section 21.2) and in sewage treatment. Oxygen is also used as a bleaching agent for pulp and paper, in medicine to ease breathing difficulties, in oxyacetylene torches, and as an oxidizing agent in many inorganic and organic reactions. Oxides, Peroxides, and Superoxides Oxygen forms three types of oxides: the normal oxide (or simply the oxide), which contains the O22 ion; the peroxide, which contains the O22 2 ion; and the superoxide, which contains the O22 ion: T SO OS2 Q SO OSO Q OS2 Q SO OS OSQ Q oxide peroxide superoxide The ions are all strong Brønsted bases and react with water as follows: Oxide: O22(aq) 1 H2O(l) ¡ 2OH2(aq) Peroxide: 2O22 2 2 (aq) 1 2H2O(l) ¡ O2(g) 1 4OH (aq) Superoxide: 4O2 2 2 (aq) 1 2H2O(l) ¡ 3O2(g) 1 4OH (aq) Note that the reaction of O22 with water is a hydrolysis reaction, but those involving O22 2 2 and O2 are redox processes. The nature of bonding in oxides changes across any period in the periodic table (see Figure 15.8). Oxides of elements on the left side of the periodic table, such as those of the alkali metals and alkaline earth metals, are generally ionic solids with high melting points. Oxides of the metalloids and of the metallic elements toward the middle of the periodic table are also solids, but they have much less ionic char- acter. Oxides of nonmetals are covalent compounds that generally exist as liquids or gases at room temperature. The acidic character of the oxides increases from left to right. 976 Chapter 22 ■ Nonmetallic Elements and Their Compounds Consider the oxides of the third-period elements (see Table 8.4): Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7 basic amphoteric acidic The basicity of the oxides increases as we move down a particular group. MgO does 97° 86° not react with water but reacts with acid as follows: MgO(s) 1 2H1 (aq) ¡ Mg21 (aq) 1 H2O(l) Figure 22.12 The structure of On the other hand, BaO, which is more basic, undergoes hydrolysis to yield the cor- H2 O2. responding hydroxide: BaO(s) 1 H2O(l) ¡ Ba(OH) 2 (aq) The best-known peroxide is hydrogen peroxide (H2O2). It is a colorless, syrupy liquid (m.p. 20.9°C), prepared in the laboratory by the action of cold dilute sulfuric acid on barium peroxide octahydrate: BaO2 ? 8H2O(s) 1 H2SO4 (aq) ¡ BaSO4 (s) 1 H2O2 (aq) 1 8H2O(l) The structure of hydrogen peroxide is shown in Figure 22.12. Using the VSEPR method we see that the H¬O and O¬O bonds are bent around each oxygen atom in a configuration similar to the structure of water. The lone-pair–bonding-pair repulsion is greater in H2O2 than in H2O, so that the HOO angle is only 97° (com- pared with 104.5° for HOH in H2O). Hydrogen peroxide is a polar molecule ( μ 5 2.16 D). Hydrogen peroxide readily decomposes when heated or exposed to sunlight or even in the presence of dust particles or certain metals, including iron and copper: 2H2O2 (l) ¡ 2H2O(l) 1 O2 (g) ¢H° 5 2196.4 kJ/mol Note that this is a disproportionation reaction. The oxidation number of oxygen changes from 21 to 22 and 0. Hydrogen peroxide is miscible with water in all proportions due to its ability to hydrogen-bond with water. Dilute hydrogen peroxide solutions (3 percent by mass), available in drugstores, are used as mild antiseptics; more concentrated H2O2 solutions are employed as bleaching agents for textiles, fur, and hair. The high heat of decom- position of hydrogen peroxide also makes it a suitable component in rocket fuel. Hydrogen peroxide is a strong oxidizing agent; it can oxidize Fe21 ions to Fe31 ions in an acidic solution: H2O2 (aq) 1 2Fe21 (aq) 1 2H1 (aq) ¡ 2Fe31 (aq) 1 2H2O(l) It also oxidizes SO22 22 3 ions to SO4 ions: H2O2 (aq) 1 SO22 22 3 (aq) ¡ SO4 (aq) 1 H2O(l) In addition, hydrogen peroxide can act as a reducing agent toward substances that are stronger oxidizing agents than itself. For example, hydrogen peroxide reduces silver oxide to metallic silver: H2O2 (aq) 1 Ag2O(s) ¡ 2Ag(s) 1 H2O(l) 1 O2 (g) and permanganate, MnO2 4 , to manganese(II) in an acidic solution: 5H2O2 (aq) 1 2MnO2 1 21 4 (aq) 1 6H (aq) ¡ 2Mn (aq) 1 5O2 (g) 1 8H2O(l)  22.5 Oxygen and Sulfur 977 Outer tube Metal foil on outer tube O3 plus some unreacted O2 Figure 22.13 The preparation of O3 from O2 by electrical discharge. The outside of the outer tube and the inside of the inner tube are coated with metal O2 foils that are connected to a high-voltage source. (The metal foil on the inside of the inner tube is not shown.) During the electrical discharge, O2 gas is passed through the tube. The O3 Inner tube gas formed exits from the upper right-hand tube, along with some unreacted O2 gas. High-voltage source If we want to determine hydrogen peroxide concentration, this reaction can be carried out as a redox titration, using a standard permanganate solution. There are relatively few known superoxides, that is, compounds containing the O22 ion. In general, only the most reactive alkali metals (K, Rb, and Cs) form superoxides. We should take note of the fact that both the peroxide ion and the superoxide ion are by-products of metabolism. Because these ions are highly reactive, they can inflict great damage on living cells. Fortunately, our bodies are equipped with the enzymes catalase, peroxidase, and superoxide dismutase which convert these toxic substances to water and molecular oxygen. Ozone Ozone is a rather toxic, light-blue gas (b.p. 2111.3°C). Its pungent odor is noticeable around sources of significant electrical discharges (such as a subway train). Ozone can be prepared from molecular oxygen, either photochemically or by subjecting O2 to an electrical discharge (Figure 22.13): 3O2 (g) ¡ 2O3 (g) ¢G° 5 326.8 kJ/mol Liquid ozone. Because the standard free energy of formation of ozone is a large positive quantity [¢G°f 5 (326.8/2) kJ/mol or 163.4 kJ/mol], ozone is less stable than molecular oxygen. The ozone molecule has a bent structure in which the bond angle is 116.5°: O O O O D M mn J G S S S S O O O O S S S S S S Ozone is mainly used to purify drinking water, to deodorize air and sewage gases, and to bleach waxes, oils, and textiles. Ozone is a very powerful oxidizing agent—its oxidizing power is exceeded only by that of molecular fluorine (see Table 18.1). For example, ozone can oxidize sulfides of many metals to the corresponding sulfates: 4O3 (g) 1 PbS(s) ¡ PbSO4 (s) 1 4O2 (g) Ozone oxidizes all the common metals except gold and platinum. In fact, a convenient test for ozone is based on its action on mercury. When exposed to ozone, mercury 978 Chapter 22 ■ Nonmetallic Elements and Their Compounds loses its metallic luster and sticks to glass tubing (instead of flowing freely through it). This behavior is attributed to the change in surface tension caused by the forma- tion of mercury(II) oxide: O3 (g) 1 3Hg(l) ¡ 3HgO(s) The beneficial effect of ozone in the stratosphere and its undesirable action in smog formation were discussed in Chapter 20. Sulfur Although sulfur is not a very abundant element (it constitutes only about 0.06 percent of Earth’s crust by mass), it is readily available because it occurs commonly in nature in the elemental form. The largest known reserves of sulfur are found in sedimentary deposits. In addition, sulfur occurs widely in gypsum (CaSO4 ? 2H2O) and various Figure 22.14 Pyrite (FeS2 ), sulfide minerals such as pyrite (FeS2) (Figure 22.14). Sulfur is also present in natural commonly called “fool’s gold” gas as H2S, SO2, and other sulfur-containing compounds. because of its gold luster. Sulfur is extracted from underground deposits by the Frasch† process, shown in Figure 22.15. In this process, superheated water (liquid water heated to about 160°C under high pressure to prevent it from boiling) is pumped down the outermost pipe to melt the sulfur. Next, compressed air is forced down the innermost pipe. Liquid sulfur mixed with air forms an emulsion that is less dense than water and therefore rises to the surface as it is forced up the middle pipe. Sulfur produced in this manner, which amounts to about 10 million tons per year, has a purity of about 99.5 percent. † Herman Frasch (1851–1914). German chemical engineer. Besides inventing the process for obtaining pure sulfur, Frasch developed methods for refining petroleum. Figure 22.15 The Frasch process. Three concentric pipes Compressed air are inserted into a hole drilled down to the sulfur deposit. Superheated water is forced Sulfur down the outer pipe into the sulfur, causing it to melt. Molten sulfur is then forced up the Superheated water middle pipe by compressed air. Molten sulfur  22.5 Oxygen and Sulfur 979 There are several allotropic forms of sulfur, the most important being the rhom- bic and monoclinic forms. Rhombic sulfur is thermodynamically the most stable form; it has a puckered S8 ring structure: S S S S S S SH ESH ES S S S S S S S S ESH S S S8 S S S S It is a yellow, tasteless, and odorless solid (m.p. 112°C) (see Figure 8.19) that is insoluble in water but soluble in carbon disulfide. When heated, it is slowly converted to monoclinic sulfur (m.p. 119°C), which also consists of the S8 units. When liquid sulfur is heated above 150°C, the rings begin to break up, and the entangling of the sulfur chains results in a sharp increase in the liquid’s viscosity. Further heating tends to rupture the chains, and the viscosity decreases. Like nitrogen, sulfur shows a wide variety of oxidation numbers in its compounds (Table 22.3). The best-known hydrogen compound of sulfur is hydrogen sulfide, which is prepared by the action of an acid on a sulfide; for example, FeS(s) 1 H2SO4 (aq) ¡ FeSO4 (aq) 1 H2S(g) Table 22.3 Common Compounds of Sulfur Oxidation Number Compound Formula Structure 22 Hydrogen sulfide H2S S S S D G H H SSH ESS S S S S 0 Sulfur* S8 ESH S S S S S S S S ESH S S S S S S O 11 Disulfur dichloride S2Cl2 EClS Q O S O Q Q O S O E SCl Q S S 12 Sulfur dichloride SCl2 S D G S S Cl Cl S S S S 14 Sulfur dioxide SO2 O S E E E E S S O O S S 16 Sulfur trioxide SO3 SOS B S D G S S O O S S S S *We list the element here as a reference. 980 Chapter 22 ■ Nonmetallic Elements and Their Compounds Today, hydrogen sulfide used in qualitative analysis (see Section 16.11) is prepared by the hydrolysis of thioacetamide: S O J J CH3OC  2H2O  H 888n CH3OC  H2S  NH4 G G NH2 OOH thioacetamide acetic acid Hydrogen sulfide is a colorless gas (b.p. 260.2°C) that smells like rotten eggs. (The odor of rotten eggs actually does come from hydrogen sulfide, which is formed by the bacterial decomposition of sulfur-containing proteins.) Hydrogen sulfide is a highly toxic substance that, like hydrogen cyanide, attacks respiratory enzymes. It is a very weak diprotic acid (see Table 15.5). In basic solution, H2S is a reducing agent. For example, it is oxidized by permanganate to elemental sulfur: 3H2S (aq) 1 2MnO2 2 4 (aq) ¡ 3S(s) 1 2MnO2 (s) 1 2H2O(l) 1 2OH (aq) Oxides of Sulfur Sulfur has two important oxides: sulfur dioxide, SO2; and sulfur trioxide, SO3. Sulfur dioxide is formed when sulfur burns in air: S(s) 1 O2 (g) ¡ SO2 (g) In the laboratory, it can be prepared by the action of an acid on a sulfite; for example, 2HCl(aq) 1 Na2SO3 (aq) ¡ 2NaCl(aq) 1 H2O(l) 1 SO2 (g) or by the action of concentrated sulfuric acid on copper: Cu(s) 1 2H2SO4 (aq) ¡ CuSO4 (aq) 1 2H2O(l) 1 SO2 (g) Sulfur dioxide (b.p. 210°C) is a pungent, colorless gas that is quite toxic. As an acidic oxide, it reacts with water as follows: There is no evidence for the formation of SO2 (g) 1 H2O(l) Δ H1 (aq) 1 HSO23 (aq) sulfurous acid, H2SO3, in water. Sulfur dioxide is slowly oxidized to sulfur trioxide, but the reaction rate can be greatly enhanced by a platinum or vanadium oxide catalyst (see Section 13.6): 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g) Sulfur trioxide dissolves in water to form sulfuric acid: SO3 (g) 1 H2O(l) ¡ H2SO4 (aq) The contributing role of sulfur dioxide to acid rain is discussed on p. 917. Sulfuric Acid Approximately 50 million tons of sulfuric Sulfuric acid is the world’s most important industrial chemical. It is prepared industri- acid are produced annually in the United States. ally by first burning sulfur in air: S(s) 1 O2 (g) ¡ SO2 (g) Next is the key step of converting sulfur dioxide to sulfur trioxide: 2SO2 (g) 1 O2 (g) ¡ 2SO3 (g)  22.5 Oxygen and Sulfur 981 Vanadium(V) oxide (V2O5) is the catalyst used for the second step. Because the sul- fur dioxide and oxygen molecules react in contact with the surface of solid V2O5, the process is referred to as the contact process. Although sulfur trioxide reacts with water to produce sulfuric acid, it forms a mist of fine droplets of H2SO4 with water vapor that is hard to condense. Instead, sulfur trioxide is first dissolved in 98 percent sulfuric acid to form oleum (H2S2O7): SO3 (g) 1 H2SO4 (aq) ¡ H2S2O7 (aq) On treatment with water, concentrated sulfuric acid can be generated: H2S2O7 (aq) 1 H2O(l) ¡ 2H2SO4 (aq) Vanadium oxide on alumina Sulfuric acid is a diprotic acid (see Table 15.5). It is a colorless, viscous liquid (Al2O3 ). (m.p. 10.4°C). The concentrated sulfuric acid we use in the laboratory is 98 percent H2SO4 by mass (density: 1.84 g/cm3), which corresponds to a concentration of 18 M. The oxidizing strength of sulfuric acid depends on its temperature and concentration. A cold, dilute sulfuric acid solution reacts with metals above hydrogen in the activity series (see Figure 4.15), thereby liberating molecular hydrogen in a displacement reaction: Mg(s) 1 H2SO4 (aq) ¡ MgSO4 (aq) 1 H2 (g) This is a typical reaction of an active metal with an acid. The strength of sulfuric acid as an oxidizing agent is greatly enhanced when it is both hot and concentrated. In such a solution, the oxidizing agent is actually the sulfate ion rather than the hydrated proton, H1(aq). Thus, copper reacts with concentrated sulfuric acid as follows: Cu(s) 1 2H2SO4 (aq) ¡ CuSO4 (aq) 1 SO2 (g) 1 2H2O(l) Depending on the nature of the reducing agents, the sulfate ion may be further reduced to elemental sulfur or the sulfide ion. For example, reduction of H2SO4 by HI yields H2S and I2: 8HI(aq) 1 H2SO4 (aq) ¡ H2S(aq) 1 4I2 (s) 1 4H2O(l) Concentrated sulfuric acid oxidizes nonmetals. For example, it oxidizes carbon to carbon dioxide and sulfur to sulfur dioxide: C(s) 1 2H2SO4 (aq) ¡ CO2 (g) 1 2SO2 (g) 1 2H2O(l) S(s) 1 2H2SO4 (aq) ¡ 3SO2 (g) 1 2H2O(l) Other Compounds of Sulfur Carbon disulfide, a colorless, flammable liquid (b.p. 46°C), is formed by heating carbon and sulfur to a high temperature: C(s) 1 2S(l) ¡ CS2 (l) It is only slightly soluble in water. Carbon disulfide is a good solvent for sulfur, phosphorus, iodine, and nonpolar substances such as waxes and rubber. Another interesting compound of sulfur is sulfur hexafluoride, SF6, which is pre- pared by heating sulfur in an atmosphere of fluorine: S(l) 1 3F2 (g) ¡ SF6 (g) Sulfur hexafluoride is a nontoxic, colorless gas (b.p. 263.8°C). It is the most inert of all sulfur compounds; it resists attack even by molten KOH. The structure and bonding 982 Chapter 22 ■ Nonmetallic Elements and Their Compounds of SF6 were discussed in Chapters 9 and 10 and its critical phenomenon illustrated in Chapter 11 (see Figure 11.37). 22.6 The Halogens The halogens—fluorine, chlorine, bromine, and iodine—are reactive nonmetals (see Figure 8.20). Table 22.4 lists some of the properties of these elements. Although all halogens are highly reactive and toxic, the magnitude of reactivity and toxicity gen- Recall that the first member of a group erally decreases from fluorine to iodine. The chemistry of fluorine differs from that usually differs in properties from the rest of the members of the group (see p. 348). of the rest of the halogens in the following ways: 1. Fluorine is the most reactive of all the halogens. The difference in reactivity between fluorine and chlorine is greater than that between chlorine and bromine. Table 22.4 shows that the F¬F bond is considerably weaker than the Cl¬Cl bond. The weak bond in F2 can be explained in terms of the lone pairs on the F atoms: SO OS F—F Q Q The small size of the F atoms (see Table 22.4) allows a close approach of the three lone pairs on each of the F atoms, resulting in a greater repulsion than that found in Cl2, which consists of larger atoms. 2. Hydrogen fluoride, HF, has a high boiling point (19.5°C) as a result of strong intermolecular hydrogen bonding, whereas all other hydrogen halides have much lower boiling points (see Figure 11.6). 3. Hydrofluoric acid is a weak acid, whereas all other hydrohalic acids (HCl, HBr, and HI) are strong acids. 4. Fluorine reacts with cold sodium hydroxide solution to produce oxygen difluoride as follows: 2F2 (g) 1 2NaOH(aq) ¡ 2NaF(aq) 1 H2O(l) 1 OF2 (g) Table 22.4 Properties of the Halogens Property F Cl Br I 2 5 2 5 2 5 Valence electron 2s 2p 3s 3p 4s 4p 5s25p5 configuration Melting point (°C) 2223 2102 27 114 Boiling point (°C) 2187 235 59 183 Appearance* Pale- Yellow- Red- Dark-violet vapor yellow green brown Dark metallic- gas gas liquid looking solid Atomic radius (pm) 72 99 114 133 Ionic radius (pm)† 133 181 195 220 Ionization energy (kJ/mol) 1680 1251 1139 1003 Electronegativity 4.0 3.0 2.8 2.5 Standard reduction 2.87 1.36 1.07 0.53 potential (V)* Bond enthalpy (kJ/mol)* 150.6 242.7 192.5 151.0 *These values and descriptions apply to the diatomic species X2, where X represents a halogen atom. The half-reaction is X2(g) 1 2e2 ¡ 2X2(aq). † Refers to the anion X2.  22.6 The Halogens 983 The same reaction with chlorine or bromine, on the other hand, produces a halide and a hypohalite: X2 (g) 1 2NaOH(aq) ¡ NaX(aq) 1 NaXO(aq) 1 H2O(l) where X stands for Cl or Br. Iodine does not react under the same conditions. 5. Silver fluoride, AgF, is soluble. All other silver halides (AgCl, AgBr, and AgI) are insoluble (see Table 4.2). The element astatine also belongs to the Group 7A family. However, all isotopes of astatine are radioactive; its longest-lived isotope is astatine-210, which has a half-life of 8.3 h. Therefore, it is both difficult and expensive to study astatine in the laboratory. The halogens form a very large number of compounds. In the elemental state they form diatomic molecules, X2. In nature, however, because of their high reactivity, halogens are always found combined with other elements. Chlorine, bromine, and iodine occur as halides in seawater, and fluorine occurs in the minerals fluorite (CaF2) (see Figure 21.16) and cryolite (Na3AlF6). Preparation and General Properties of the Halogens Because fluorine and chlorine are strong oxidizing agents, they must be prepared by electrolysis rather than by chemical oxidation of the fluoride and chloride ions. Elec- trolysis does not work for aqueous solutions of fluorides, however, because fluorine is a stronger oxidizing agent than oxygen. From Table 18.1 we find that F2 (g) 1 2e2 ¡ 2F2 (aq) E° 5 2.87 V O2 (g) 1 4H (aq) 1 4e2 ¡ 2H2O(l) 1 E° 5 1.23 V If F2 were formed by the electrolysis of an aqueous fluoride solution, it would imme- diately oxidize water to oxygen. For this reason, fluorine is prepared by electrolyzing liquid hydrogen fluoride containing potassium fluoride to increase its conductivity, at about 70°C (Figure 22.16): Anode (oxidation): 2F2 ¡ F2(g) 1 2e2 Cathode (reduction): 2H 1 2e2 ¡ H2(g) 1 Overall reaction: 2HF(l) ¡ H2(g) 1 F2(g) F2 gas Figure 22.16 Electrolytic cell for the preparation of fluorine gas. Carbon anode Note that because H2 and F2 form an explosive mixture, these H2 gas H2 gas gases must be separated from each other. Diaphragm to prevent mixing of H2 and F2 gases Steel cathode Liquid HF 984 Chapter 22 ■ Nonmetallic Elements and Their Compounds Figure 22.17 Mercury cell used Cl2 Graphite anode in the chlor-alkali process. The cathode contains mercury. The sodium-mercury amalgam is treated with water outside the cell to produce sodium hydroxide and Brine Brine hydrogen gas. Hg cathode Hg plus Na/Hg Chlorine gas, Cl2, is prepared industrially by the electrolysis of molten NaCl (see Section 18.8) or by the chlor-alkali process, the electrolysis of a concentrated aque- ous NaCl solution (called brine). (Chlor denotes chlorine and alkali denotes an alkali metal, such as sodium.) Two of the common cells used in the chlor-alkali process are the mercury cell and the diaphragm cell. In both cells the overall reaction is electrolysis 2NaCl(aq) 1 2H2O(l) ¬¬¬¡ 2NaOH(aq) 1 H2(g) 1 Cl2(g) As you can see, this reaction yields two useful by-products, NaOH and H2. The cells are designed to separate the molecular chlorine from the sodium hydroxide solution and the molecular hydrogen to prevent side reactions such as 2NaOH(aq) 1 Cl2 (g) ¡ NaOCl(aq) 1 NaCl(aq) 1 H2O(l) H2 (g) 1 Cl2 (g) ¡ 2HCl(g) These reactions consume the desired products and can be dangerous because a mixture of H2 and Cl2 is explosive. Figure 22.17 shows the mercury cell used in the chlor-alkali process. The cathode is a liquid mercury pool at the bottom of the cell, and the anode is made of either graphite or titanium coated with platinum. Brine is continuously passed through the cell as shown in the diagram. The electrode reactions are Anode (oxidation): 2Cl2(aq) ¡ Cl2(g) 1 2e2 Hg(l) Cathode (reduction): 2Na1 (aq) 1 2e2 ¡ 2Na/Hg Overall reaction: 2NaCl(aq) ¡ 2Na/Hg 1 Cl2(g) where Na/Hg denotes the formation of sodium amalgam. The chlorine gas generated this way is very pure. The sodium amalgam does not react with the brine solution but decomposes as follows when treated with pure water outside the cell: 2Na/Hg 1 2H2O(l) ¡ 2NaOH(aq) 1 H2 (g) 1 2Hg(l) The by-products are sodium hydroxide and hydrogen gas. Although the mercury is cycled back into the cell for reuse, some of it is always discharged with waste solu- tions into the environment, resulting in mercury pollution. This is a major drawback of the mercury cell. Figure 22.18 shows the industrial manufacture of chlorine gas. The half-cell reactions in a diaphragm cell are shown in Figure 22.19. The asbes- tos diaphragm is permeable to the ions but not to the hydrogen and chlorine gases and so prevents the gases from mixing. During electrolysis a positive pressure is applied on the anode side of the compartment to prevent the migration of the OH2 ions from the cathode compartment. Periodically, fresh brine solution is added to the cell and the sodium hydroxide solution is run off as shown. The diaphragm cell presents no  22.6 The Halogens 985 Figure 22.18 The industrial manufacture of chlorine gas. pollution problems. Its main disadvantage is that the sodium hydroxide solution is contaminated with unreacted sodium chloride. The preparation of molecular bromine and iodine from seawater by oxidation with From Table 18.1 we see that the oxidizing strength decreases from Cl2 to Br2 to I2. chlorine was discussed in Section 4.4. In the laboratory, chlorine, bromine, and iodine can be prepared by heating the alkali halides (NaCl, KBr, or KI) in concentrated sulfuric acid in the presence of manganese(IV) oxide. A representative reaction is MnO2 (s) 1 2H2SO4 (aq) 1 2NaCl(aq) ¡ MnSO4 (aq) 1 Na2SO4 (aq) 1 2H2O(l) 1 Cl2 (g) Compounds of the Halogens Most of the halides can be classified into two categories. The fluorides and chlorides of many metallic elements, especially those belonging to the alkali metal and alkaline earth metal (except beryllium) families, are ionic compounds. Most of the halides of nonmetals such as sulfur and phosphorus are covalent compounds. As Figure 4.10 shows, the oxidation numbers of the halogens can vary from 21 to 17. The only exception is fluorine. Because it is the most electronegative element, fluorine can have only two oxidation numbers, 0 (as in F2) and 21, in its compounds. The Hydrogen Halides The hydrogen halides, an important class of halogen compounds, can be formed by the direct combination of the elements: H2 (g) 1 X2 (g) Δ 2HX(g) Battery Figure 22.19 Diaphragm cell used in the chlor-alkali process. e– e– Asbestos Anode diaphragm Cathode Brine NaOH solution Oxidation Reduction 2Cl–(aq) Cl2( g) + 2e – 2H2O(l) + 2e – H2(g) + 2OH–(aq) 986 Chapter 22 ■ Nonmetallic Elements and Their Compounds where X denotes a halogen atom. These reactions (especially the ones involving F2 and Cl2) can occur with explosive violence. Industrially, hydrogen chloride is pro- duced as a by-product in the manufacture of chlorinated hydrocarbons: C2H6 (g) 1 Cl2 (g) ¡ C2H5Cl(g) 1 HCl(g) In the laboratory, hydrogen fluoride and hydrogen chloride can be prepared by react- ing the metal halides with concentrated sulfuric acid: CaF2 (s) 1 H2SO4 (aq) ¡ 2HF(g) 1 CaSO4 (s) 2NaCl(s) 1 H2SO4 (aq) ¡ 2HCl(g) 1 Na2SO4 (aq) Hydrogen bromide and hydrogen iodide cannot be prepared this way because they are oxidized to elemental bromine and iodine. For example, the reaction between NaBr and H2SO4 is 2NaBr(s) 1 2H2SO4 (aq) ¡ Br2 (l) 1 SO2 (g) 1 Na2SO4 (aq) 1 2H2O(l) Instead, hydrogen bromide is prepared by first reacting bromine with phosphorus to form phosphorus tribromide: P4 (s) 1 6Br2 (l) ¡ 4PBr3 (l) Next, PBr3 is treated with water to yield HBr: PBr3 (l) 1 3H2O(l) ¡ 3HBr(g) 1 H3PO3 (aq) Hydrogen iodide can be prepared in a similar manner. The high reactivity of HF is demonstrated by the fact that it attacks silica and silicates: 6HF(aq) 1 SiO2 (s) ¡ H2SiF6 (aq) 1 2H2O(l) This property makes HF suitable for etching glass and is the reason that hydro- gen fluoride must be kept in plastic or inert metal (for example, Pt) containers. Hydrogen fluoride is used in the manufacture of Freons (see Chapter 20); for example, CCl4 (l) 1 HF(g) ¡ CFCl3 (g) 1 HCl(g) CFCl3 (g) 1 HF(g) ¡ CF2Cl2 (g) 1 HCl(g) It is also important in the production of aluminum (see Section 21.7). Hydrogen chloride is used in the preparation of hydrochloric acid, inorganic chlorides, and in various metallurgical processes. Hydrogen bromide and hydrogen iodide do not have any major industrial uses. Aqueous solutions of hydrogen halides are acidic. The strength of the acids increases as follows: HF ! HCl , HBr , HI Oxoacids of the Halogens The halogens also form a series of oxoacids with the following general formulas: HXO HXO2 HXO3 HXO4 hypohalous halous halic perhalic acid acid acid acid  22.6 The Halogens 987 Chlorous acid, HClO2, is the only known halous acid. All the halogens except fluorine form halic and perhalic acids. The Lewis structures of the chlorine oxo- acids are SO OS Q O O Q O O Q O O Q O Q OS O O HSOSClS Q Q SQS HSOSClSO HSOSClSO HSOSClSO Q QS SO OS SQO OS Q hypochlorous chlorous chloric perchloric acid acid acid acid For a given halogen, the acid strength decreases from perhalic acid to hypohalous acid; the explanation of this trend is discussed in Section 15.9. Table 22.5 lists some of the halogen compounds. Periodic acid, HIO4, does not appear because this compound cannot be isolated in the pure form. Instead the formula H5IO6 is often used to represent periodic acid. Uses of the Halogens Fluorine Applications of the halogens and their compounds are widespread in industry, health care, and other areas. One is fluoridation, the practice of adding small quan- tities of fluorides (about 1 ppm by mass) such as NaF to drinking water to reduce dental caries. One of the most important inorganic fluorides is uranium hexafluoride, UF6, which is essential to the gaseous diffusion process for separating isotopes of uranium (U-235 and U-238). Industrially, fluorine is used to produce polytetrafluoroethylene, a polymer better known as Teflon: ¬(CF2 ¬CF2 )¬n where n is a large number. Teflon is used in electrical insulators, high-temperature plastics, cooking utensils, and so on. Chlorine Chlorine plays an important biological role in the human body, where the chloride ion is the principal anion in intracellular and extracellular fluids. Chlorine is widely used as an industrial bleaching agent for paper and textiles. Ordinary household laundry bleach contains the active ingredient sodium hypochlorite (about 5 percent Table 22.5 Common Compounds of Halogens* Compound F Cl Br I Hydrogen halide HF (21) HCl (21) HBr (21) HI (21) Oxides OF2 (21) Cl2O (11) Br2O (11) I2O5 (15) ClO2 (14) BrO2 (14) Cl2O7 (17) Oxoacids HFO (21) HClO (11) HBrO (11) HIO (11) HClO2 (13) HClO3 (15) HBrO3 (15) HIO3 (15) HClO4 (17) HBrO4 (17) H5IO6 (17) *The number in parentheses indicates the oxidation number of the halogen. 988 Chapter 22 ■ Nonmetallic Elements and Their Compounds by mass), which is prepared by reacting chlorine gas with a cold solution of sodium hydroxide: Cl2 (g) 1 2NaOH(aq) ¡ NaCl(aq) 1 NaClO(aq) 1 H2O(l) Chlorine is also used to purify water and disinfect swimming pools. When chlorine dissolves in water, it undergoes the following reaction: Cl2 (g) 1 H2O(l) ¡ HCl(aq) 1 HClO(aq) It is thought that the ClO2 ions destroy bacteria by oxidizing life-sustaining com- pounds within them. Chlorinated methanes, such as carbon tetrachloride and chloroform, are useful organic solvents. Large quantities of chlorine are used to produce insecticides, such as DDT. However, in view of the damage they inflict on the environment, the use of many of these compounds is either totally banned or greatly restricted in the United States. Chlorine is also used to produce polymers such as poly(vinyl chloride). Bromine So far as we know, bromine compounds occur naturally only in some marine organ- isms. Seawater is about 1 3 1023 M Br2; therefore, it is the main source of bromine. Bromine is used to prepare ethylene dibromide (BrCH2CH2Br), which is used as an insecticide and as a scavenger for lead (that is, to combine with lead) in gasoline to keep lead deposits from clogging engines. Studies have shown that ethylene dibromide is a very potent carcinogen. Bromine combines directly with silver to form silver bromide (AgBr), which is used in photographic films. Iodine Iodine is not used as widely as the other halogens. A 50 percent (by mass) alcohol solution of iodine, known as tincture of iodine, is used medicinally as an antiseptic. Iodine is an essential constituent of the thyroid hormone thyroxine: I I G G H O A J HOO OOO OCH2OCOC A G D D NH2 OH I I Iodine deficiency in the diet may result in enlargement of the thyroid gland (known as goiter). Iodized table salt sold in the United States usually contains 0.01 percent KI or NaI, which is more than sufficient to satisfy the 1 mg of iodine per week required for the formation of thyroxine in the human body. A compound of iodine that deserves mention is silver iodide, AgI. It is a pale- yellow solid that darkens when exposed to light. In this respect it is similar to silver bromide. Silver iodide is sometimes used in cloud seeding, a process for inducing rainfall on a small scale (Figure 22.20). The advantage of using silver iodide is that enormous numbers of nuclei (that is, small particles of silver iodide on which ice crystals can form) become available. About 1015 nuclei are produced from 1 g of AgI by vaporizing an acetone solution of silver iodide in a hot flame. The nuclei are then dispersed into the clouds from an airplane.  Key Words 989 Figure 22.20 Cloud seeding using AgI particles. Summary of Facts & Concepts 1. Hydrogen atoms contain one proton and one electron. The phosphates are the most important phosphorus They are the simplest atoms. Hydrogen combines with compounds. many metals and nonmetals to form hydrides; some hy- 6. Elemental oxygen, O2, is paramagnetic and contains drides are ionic and some are covalent. two unpaired electrons. Oxygen forms ozone (O3), 2. There are three isotopes of hydrogen: 11H, 21H (deuterium), oxides (O22), peroxides (O222), and superoxides (O2 2 ). and 31H (tritium). Heavy water contains deuterium. The most abundant element in Earth’s crust, oxygen is 3. The important inorganic compounds of carbon are the essential for life on Earth. carbides; the cyanides, most of which are extremely 7. Sulfur is taken from Earth’s crust by the Frasch process toxic; carbon monoxide, also toxic and a major air as a molten liquid. Sulfur exists in a number of allo- pollutant; the carbonates and bicarbonates; and carbon tropic forms and has a variety of oxidation numbers in dioxide, an end product of metabolism and a compo- its compounds. nent of the global carbon cycle. 8. Sulfuric acid is the cornerstone of the chemical indus- 4. Elemental nitrogen, N2, contains a triple bond and is try. It is produced from sulfur via sulfur dioxide and very stable. Compounds in which nitrogen has oxida- sulfur trioxide by means of the contact process. tion numbers from 23 to 15 are formed between nitro- 9. The halogens are toxic and reactive elements that are gen and hydrogen and/or oxygen atoms. Ammonia, found only in compounds with other elements. Fluorine NH3, is widely used in fertilizers. and chlorine are strong oxidizing agents and are pre- 5. White phosphorus, P4, is highly toxic, very reactive, and pared by electrolysis. flammable; the polymeric red phosphorus, (P4)n, is 10. The reactivity, toxicity, and oxidizing ability of the hal- more stable. Phosphorus forms oxides and halides with ogens decrease from fluorine to iodine. The halogens all oxidation numbers of 13 and 15 and several oxoacids. form binary acids (HX) and a series of oxoacids. Key Words Carbide, p. 964 Chlor-alkali Cyanide, p. 964 Catenation, p. 963 process, p. 984 Hydrogenation, p. 961  990 Chapter 22 ■ Nonmetallic Elements and Their Compounds Questions & Problems • Problems available in Connect Plus Assume that deuterium abundance is 0.015 percent Red numbered problems solved in Student Solutions Manual and that recovery is 80 percent. 22.17 Predict the outcome of the following reactions: General Properties of Nonmetals (a) CuO(s) 1 H2 (g) ¡ Review Questions (b) Na2O(s) 1 H2 (g) ¡ • 22.1 Without referring to Figure 22.1, state whether each 22.18 Starting with H2, describe how you would prepare of the following elements are metals, metalloids, or (a) HCl, (b) NH3, (c) LiOH. nonmetals: (a) Cs, (b) Ge, (c) I, (d) Kr, (e) W, (f) Ga, (g) Te, (h) Bi. Carbon 22.2 List two chemical and two physical properties that Review Questions distinguish a metal from a nonmetal. 22.3 Make a list of physical and chemical properties of 22.19 Give an example of a carbide and a cyanide. chlorine (Cl2) and magnesium. Comment on their 22.20 How are cyanide ions used in metallurgy? differences with reference to the fact that one is a 22.21 Briefly discuss the preparation and properties of car- metal and the other is a nonmetal. bon monoxide and carbon dioxide. 22.4 Carbon is usually classified as a nonmetal. However, 22.22 What is coal? the graphite used in “lead” pencils conducts electric- 22.23 Explain what is meant by coal gasification. ity. Look at a pencil and list two nonmetallic proper- 22.24 Describe two chemical differences between CO ties of graphite. and CO2. Hydrogen Review Questions Problems 22.5 Explain why hydrogen has a unique position in the 22.25 Describe the reaction between CO2 and OH2 in periodic table. terms of a Lewis acid-base reaction such as that shown on p. 705. 22.6 Describe two laboratory and two industrial prepara- tions for hydrogen. • 22.26 Draw a Lewis structure for the C222 ion. 22.7 Hydrogen exhibits three types of bonding in its • 22.27 Balance the following equations: compounds. Describe each type of bonding with an (a) Be2C(s) 1 H2O(l) ¡ example. (b) CaC2 (s) 1 H2O(l) ¡ • 22.8 What are interstitial hydrides? • 22.28 Unlike CaCO3, Na2CO3 does not readily yield CO2 22.9 Give the name of (a) an ionic hydride and (b) a cova- when heated. On the other hand, NaHCO3 under- lent hydride. In each case describe the preparation and goes thermal decomposition to produce CO2 and give the structure of the compound. Na2CO3. (a) Write a balanced equation for the reac- 22.10 Describe what is meant by the “hydrogen economy.” tion. (b) How would you test for the CO2 evolved? [Hint: Treat the gas with limewater, an aqueous so- Problems lution of Ca(OH)2.] 22.29 Two solutions are labeled A and B. Solution A con- • 22.11 Elements number 17 and 20 form compounds with tains Na2CO3 and solution B contains NaHCO3. De- hydrogen. Write the formulas for these two com- scribe how you would distinguish between the two pounds and compare their chemical behavior in water. solutions if you were provided with a MgCl2 solu- 22.12 Give an example of hydrogen as (a) an oxidizing tion. (Hint: You need to know the solubilities of agent and (b) a reducing agent. MgCO3 and MgHCO3.) 22.13 Compare the physical and chemical properties of the 22.30 Magnesium chloride is dissolved in a solution con- hydrides of each of the following elements: Na, Ca, taining sodium bicarbonate. On heating, a white pre- C, N, O, Cl. cipitate is formed. Explain what causes the 22.14 Suggest a physical method that would enable you to precipitation. separate hydrogen gas from neon gas. • 22.31 A few drops of concentrated ammonia solution • 22.15 Write a balanced equation to show the reaction added to a calcium bicarbonate solution cause a between CaH2 and H2O. How many grams of CaH2 white precipitate to form. Write a balanced equation are needed to produce 26.4 L of H2 gas at 20°C and for the reaction. 746 mmHg? 22.32 Sodium hydroxide is hygroscopic—that is, it ab- • 22.16 How many kilograms of water must be processed to sorbs moisture when exposed to the atmosphere. obtain 2.0 L of D2 at 25°C and 0.90 atm pressure? A student placed a pellet of NaOH on a watch glass.  Questions & Problems 991 A few days later, she noticed that the pellet was cov- 22.48 Explain why, under normal conditions, the reac- ered with a white solid. What is the identity of this tion of zinc with nitric acid does not produce solid? (Hint: Air contains CO2.) hydrogen. 22.33 A piece of red-hot magnesium ribbon will continue • 22.49 Potassium nitrite can be produced by heating a to burn in an atmosphere of CO2 even though CO2 mixture of potassium nitrate and carbon. Write a does not support combustion. Explain. balanced equation for this reaction. Calculate the 22.34 Is carbon monoxide isoelectronic with nitrogen theoretical yield of KNO2 produced by heating (N2)? 57.0 g of KNO3 with an excess of carbon. • 22.50 Predict the geometry of nitrous oxide, N2O, by the VSEPR method and draw resonance structures for Nitrogen and Phosphorus the molecule. (Hint: The atoms are arranged as Review Questions NNO.) 22.35 Describe a laboratory and an industrial preparation • 22.51 Consider the reaction of nitrogen gas. N2 (g) 1 O2 (g) Δ 2NO(g) 22.36 What is meant by nitrogen fixation? Describe a process for fixation of nitrogen on an industrial Given that the ¢G° for the reaction at 298 K is scale. 173.4 kJ/mol, calculate (a) the standard free energy 22.37 Describe an industrial preparation of phosphorus. of formation of NO, (b) KP for the reaction, and (c) Kc for the reaction. 22.38 Why is the P4 molecule unstable? • 22.52 From the data in Appendix 3, calculate ¢H° for the synthesis of NO (which is the first step in the manu- Problems facture of nitric acid) at 25°C: • 22.39 Nitrogen can be obtained by (a) passing ammonia 4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(l) over red-hot copper(II) oxide and (b) heating am- monium dichromate [one of the products is 22.53 Explain why two N atoms can form a double bond Cr(III) oxide]. Write a balanced equation for or a triple bond, whereas two P atoms normally can each preparation. form only a single bond. 22.40 Write balanced equations for the preparation of 22.54 When 1.645 g of white phosphorus are dissolved in sodium nitrite by (a) heating sodium nitrate and 75.5 g of CS2, the solution boils at 46.709°C, whereas (b) heating sodium nitrate with carbon. pure CS2 boils at 46.300°C. The molal boiling-point • 22.41 Sodium amide (NaNH2) reacts with water to produce elevation constant for CS2 is 2.34°C/m. Calculate the sodium hydroxide and ammonia. Describe this reac- molar mass of white phosphorus and give the mo- tion as a Brønsted acid-base reaction. lecular formula. • 22.42 Write a balanced equation for the formation of urea, 22.55 Starting with elemental phosphorus, P4, show how (NH2)2CO, from carbon dioxide and ammonia. you would prepare phosphoric acid. Should the reaction be run at a high or low pressure • 22.56 Dinitrogen pentoxide is a product of the reaction to maximize the yield? between P4O10 and HNO3. Write a balanced equa- 22.43 Some farmers feel that lightning helps produce a tion for this reaction. Calculate the theoretical better crop. What is the scientific basis for this yield of N2O5 if 79.4 g of P4O10 are reacted with belief? an excess of HNO3. (Hint: One of the products is HPO3.) 22.44 At 620 K the vapor density of ammonium chloride relative to hydrogen (H2) under the same conditions 22.57 Explain why (a) NH3 is more basic than PH3, (b) NH3 of temperature and pressure is 14.5, although, has a higher boiling point than PH3, (c) PCl5 exists but according to its formula mass, it should have a vapor NCl5 does not, (d) N2 is more inert than P4. density of 26.8. How would you account for this 22.58 What is the hybridization of phosphorus in the phos- discrepancy? phonium ion, PH1 4? 22.45 Explain, giving one example in each case, why ni- trous acid can act both as a reducing agent and as an Oxygen and Sulfur oxidizing agent. Review Questions 22.46 Explain why nitric acid can be reduced but not oxidized. 22.59 Describe one industrial and one laboratory prepara- 22.47 Write a balanced equation for each of the follow- tion of O2. ing processes: (a) On heating, ammonium nitrate 22.60 Give an account of the various kinds of oxides that produces nitrous oxide. (b) On heating, potassium exist and illustrate each type by two examples. nitrate produces potassium nitrite and oxygen gas. 22.61 Hydrogen peroxide can be prepared by treating bar- (c) On heating, lead nitrate produces lead(II) oxide, ium peroxide with sulfuric acid. Write a balanced nitrogen dioxide (NO2), and oxygen gas. equation for this reaction.  992 Chapter 22 ■ Nonmetallic Elements and Their Compounds 22.62 Describe the Frasch process for obtaining sulfur. 22.77 Describe two reactions in which sulfuric acid acts as 22.63 Describe the contact process for the production of an oxidizing agent. sulfuric acid. 22.78 Concentrated sulfuric acid reacts with sodium 22.64 How is hydrogen sulfide generated in the laboratory? iodide to produce molecular iodine, hydrogen sulfide, and sodium hydrogen sulfate. Write a balanced Problems equation for the reaction. 22.65 Draw molecular orbital energy level diagrams for O2, O2 22 2 , and O2 . The Halogens 22.66 One of the steps involved in the depletion of Review Questions ozone in the stratosphere by nitric oxide may be 22.79 Describe an industrial method for preparing each of represented as the halogens. NO(g) 1 O3 (g) ¡ NO2 (g) 1 O2 (g) 22.80 Name the major uses of the halogens. From the data in Appendix 3 calculate ¢G°, KP, and Kc for the reaction at 25°C. Problems • 22.67 Hydrogen peroxide is unstable and decomposes 22.81 Metal chlorides can be prepared in a number of readily: ways: (a) direct combination of metal and molecular 2H2O2 (aq) ¡ 2H2O(l) 1 O2 (g) chlorine, (b) reaction between metal and hydro- chloric acid, (c) acid-base neutralization, (d) metal This reaction is accelerated by light, heat, or a cata- carbonate treated with hydrochloric acid, (e) precipi- lyst. (a) Explain why hydrogen peroxide sold in tation reaction. Give an example for each type of drugstores comes in dark bottles. (b) The concentra- preparation. tions of aqueous hydrogen peroxide solutions are 22.82 Sulfuric acid is a weaker acid than hydrochloric normally expressed as percent by mass. In the de- acid. Yet hydrogen chloride is evolved when con- composition of hydrogen peroxide, how many liters centrated sulfuric acid is added to sodium chloride. of oxygen gas can be produced at STP from 15.0 g Explain. of a 7.50 percent hydrogen peroxide solution? 22.83 Show that chlorine, bromine, and iodine are very • 22.68 What are the oxidation numbers of O and F in much alike by giving an account of their behavior HFO? (a) with hydrogen, (b) in producing silver salts, 22.69 Oxygen forms double bonds in O2, but sulfur forms (c) as oxidizing agents, and (d) with sodium hydrox- single bonds in S8. Explain. ide. (e) In what respects is fluorine not a typical • 22.70 In 2008, about 48 million tons of sulfuric acid were halogen element? produced in the United States. Calculate the amount • 22.84 A 375-gallon tank is filled with water containing of sulfur (in grams and moles) used to produce that 167 g of bromine in the form of Br2 ions. How amount of sulfuric acid. many liters of Cl2 gas at 1.00 atm and 20°C will be 22.71 Sulfuric acid is a dehydrating agent. Write balanced required to oxidize all the bromide to molecular equations for the reactions between sulfuric acid and bromine? the following substances: (a) HCOOH, (b) H3PO4, • 22.85 Draw structures for (a) (HF)2 and (b) HF2 2. (c) HNO3, (d) HClO3. (Hint: Sulfuric acid is not de- 22.86 Hydrogen fluoride can be prepared by the action of composed by the dehydrating action.) sulfuric acid on sodium fluoride. Explain why hydro- • 22.72 Calculate the amount of CaCO3 (in grams) that gen bromide cannot be prepared by the action of the would be required to react with 50.6 g of SO2 emit- same acid on sodium bromide. ted by a power plant. 22.87 Aqueous copper(II) sulfate solution is blue. When 22.73 SF6 exists but OF6 does not. Explain. aqueous potassium fluoride is added to the CuSO4 22.74 Explain why SCl6, SBr6, and SI6 cannot be prepared. solution, a green precipitate is formed. If aqueous 22.75 Compare the physical and chemical properties of potassium chloride is added instead, a bright-green H2O and H2S. solution is formed. Explain what happens in each case. • 22.76 The bad smell of water containing hydrogen sulfide can be removed by the action of chlorine. The reac- • 22.88 What volume of bromine (Br2) vapor measured at tion is 100°C and 700 mmHg pressure would be obtained if 2.00 L of dry chlorine (Cl2), measured at 15°C and H2S(aq) 1 Cl2 (aq) ¡ 2HCl(aq) 1 S(s) 760 mmHg, were absorbed by a potassium bromide If the hydrogen sulfide content of contaminated solution? water is 22 ppm by mass, calculate the amount of • 22.89 Use the VSEPR method to predict the geometries Cl2 (in grams) required to remove all the H2S from of the following species: (a) I23 , (b) SiCl4, (c) PF5, 2.0 3 102 gallons of water. (1 gallon 5 3.785 L.) (d) SF4.  Questions & Problems 993 • 22.90 Iodine pentoxide, I2O5, is sometimes used to remove 22.100 Lubricants used in watches usually consist of long- carbon monoxide from the air by forming carbon di- chain hydrocarbons. Oxidation by air forms solid oxide and iodine. Write a balanced equation for this polymers that eventually destroy the effectiveness of reaction and identify species that are oxidized and the lubricants. It is believed that one of the initial reduced. steps in the oxidation is removal of a hydrogen atom (hydrogen abstraction). By replacing the hydrogen Additional Problems atoms at reactive sites with deuterium atoms, it is 22.91 Write a balanced equation for each of the following possible to substantially slow down the overall oxi- reactions: (a) Heating phosphorous acid yields dation rate. Why? (Hint: Consider the kinetic isotope phosphoric acid and phosphine (PH3). (b) Lithium effect.) carbide reacts with hydrochloric acid to give lithium 22.101 How are lightbulbs frosted? (Hint: Consider the ac- chloride and methane. (c) Bubbling HI gas through tion of hydrofluoric acid on glass, which is made of an aqueous solution of HNO2 yields molecular silicon dioxide.) iodine and nitric oxide. (d) Hydrogen sulfide is 22.102 Life evolves to adapt to its environment. In this oxidized by chlorine to give HCl and SCl2. respect, explain why life most frequently needs • 22.92 (a) Which of the following compounds has the oxygen for survival, rather than the more abundant greatest ionic character? PCl5, SiCl4, CCl4, BCl3 nitrogen. (b) Which of the following ions has the smallest 22.103 As mentioned in Chapter 3, ammonium nitrate is the ionic radius? F2, C42, N32, O22 (c) Which of the most important nitrogen-containing fertilizer in the following atoms has the highest ionization energy? world. Given only air and water as starting materials F, Cl, Br, I (d) Which of the following oxides is most and any equipment and catalyst at your disposal, acidic? H2O, SiO2, CO2 describe how you would prepare ammonium nitrate. 22.93 Both N2O and O2 support combustion. Suggest one State conditions under which you can increase the physical and one chemical test to distinguish be- yield in each step. tween the two gases. 22.104 As we saw in Section 21.2, the reduction of iron • 22.94 What is the change in oxidation number for the fol- oxides is accomplished by using carbon monoxide lowing reaction? as a reducing agent. Starting with coke in a blast furnace, the following equilibrium plays a key role 3O2 ¡ 2O3 in the extraction of iron: 22.95 Describe the bonding in the C222 ion in terms of the C(s) 1 CO2 (g) Δ 2CO(g) molecular orbital theory. 22.96 Starting with deuterium oxide (D2O), describe how Use the data in Appendix 3 to calculate the equilib- you would prepare (a) NaOD, (b) DCl, (c) ND3, rium constant at 25°C and 1000°C. Assume ¢H° (d) C2D2, (e) CD4, (f) D2SO4. and ¢S° to be independent of temperature. 22.97 Solid PCl5 exists as [PCl 14 ][PCl2 6 ]. Draw Lewis 22.105 Assuming ideal behavior, calculate the density of structures for these ions. Describe the hybridization gaseous HF at its normal boiling point (19.5°C). The state of the P atoms. experimentally measured density under the same 22.98 Consider the Frasch process. (a) How is it possible to conditions is 3.10 g/L. Account for the discrepancy heat water well above 100°C without turning it into between your calculated value and the experimental steam? (b) Why is water sent down the outermost result. pipe? (c) Why would excavating a mine and digging • 22.106 A 10.0-g sample of white phosphorus was burned in for sulfur be a dangerous procedure for obtaining the an excess of oxygen. The product was dissolved in element? enough water to make 500 mL of solution. Calculate 22.99 Predict the physical and chemical properties of asta- the pH of the solution at 25°C. tine, a radioactive element and the last member of Group 7A.  CHAPTER 23 Transition Metals Chemistry and Copper ions implanted in beta-alumina emit visible Coordination radiation when excited by UV light. The color of light can be changed by adding other elements in small amounts. Compounds CHAPTER OUTLINE A LOOK AHEAD 23.1 Properties of the Transition  We first survey the general properties of transition metals, focusing on their Metals electron configurations and oxidation states. (23.1) 23.2 Chemistry of Iron  Next, we study the chemistry of two representative transition metals—iron and and Copper copper. (23.2)  We then consider the general characteristics of coordination compounds in 23.3 Coordination Compounds terms of the nature of ligands and also cover the nomenclature of these com- 23.4 Structure of Coordination pounds. (23.3) Compounds  We see that the structure of coordination compounds can give rise to geometric 23.5 Bonding in Coordination and/or optical isomers. We become acquainted with the use of a polarimeter Compounds: Crystal Field in studying optical isomers. (23.4) Theory  Crystal field theory can satisfactorily explain the origin of color in and magnetic properties of octahedral, tetrahedral, and square-planar complexes. (23.5) 23.6 Reactions of Coordination Compounds  We examine the reactivity of coordination compounds and learn that they can be classified as labile or inert with regard to ligand exchange reactions. (23.6) 23.7 Applications of Coordination Compounds  This chapter concludes with a discussion of several applications of coordina- tion compounds. (23.7) 994  23.1 Properties of the Transition Metals 995 T he series of elements in the periodic table in which the d and f subshells are gradually filled are called the transition elements. There are about 50 transition elements, and they have widely varying and fascinating properties. To present even one interesting feature of each transition element is beyond the scope of this book. We will therefore limit our discussion to the transition elements that have incompletely filled d subshells and to their most commonly encountered property—the tendency to form complex ions. 23.1 Properties of the Transition Metals Transition metals typically have incompletely filled d subshells or readily give rise to ions with incompletely filled d subshells (Figure 23.1). (The Group 2B metals—Zn, Cd, and Hg—do not have this characteristic electron configuration and so, although they are sometimes called transition metals, they really do not belong in this category.) This attribute is responsible for several notable properties, including distinctive color- ing, formation of paramagnetic compounds, catalytic activity, and especially a great tendency to form complex ions. In this chapter we focus on the first-row elements from scandium to copper, the most common transition metals. Table 23.1 lists some of their properties. As we read across any period from left to right, atomic numbers increase, elec- trons are added to the outer shell, and the nuclear charge increases by the addition of protons. In the third-period elements—sodium to argon—the outer electrons weakly shield one another from the extra nuclear charge. Consequently, atomic radii decrease rapidly from sodium to argon, and the electronegativities and ionization energies increase steadily (see Figures 8.5, 8.11, and 9.5). For the transition metals, the trends are different. Looking at Table 23.1 we see that the nuclear charge, of course, increases from scandium to copper, but electrons are being added to the inner 3d subshell. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than outer-shell electrons can shield one another, so the atomic radii decrease less rapidly. For the same reason, electronegativities and ionization energies increase only slightly from scandium across to copper compared with the increases from sodium to argon. 1 18 1A 8A 1 2 2 13 14 15 16 17 H 2A 3A 4A 5A 6A 7A He 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 Na Mg 3B 4B 5B 6B 7B 8B 1B 2B Al Si P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Fl Lv Figure 23.1 The transition metals (blue squares). Note that although the Group 2B elements ( Zn, Cd, Hg) are described as transition metals by some chemists, neither the metals nor their ions possess incompletely filled d subshells. 996 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Table 23.1 Electron Configurations and Other Properties of the First-Row Transition Metals Sc Ti V Cr Mn Fe Co Ni Cu Electron configuration M 4s23d1 4s23d2 4s23d3 4s13d5 4s23d5 4s23d 6 4s23d7 4s23d 8 4s13d10 M21 — 3d2 3d3 3d 4 3d5 3d 6 3d7 3d 8 3d 9 M31 [Ar] 3d1 3d2 3d3 3d 4 3d5 3d 6 3d7 3d 8 Electronegativity 1.3 1.5 1.6 1.6 1.5 1.8 1.9 1.9 1.9 Ionization energy (kJ/mol) First 631 658 650 652 717 759 760 736 745 Second 1235 1309 1413 1591 1509 1561 1645 1751 1958 Third 2389 2650 2828 2986 3250 2956 3231 3393 3578 Radius (pm) M 162 147 134 130 135 126 125 124 128 M21 — 90 88 85 91 82 82 78 72 M31 83 68 74 64 66 67 64 — — Standard reduction potential (V)* 22.08 21.63 21.2 20.74 21.18 20.44 20.28 20.25 0.34 *The half-reaction is M21(aq) 1 2e2 ¡ M(s) (except for Sc and Cr, where the ions are Sc31 and Cr31, respectively). Although the transition metals are less electropositive (or more electronegative) than the alkali and alkaline earth metals, the standard reduction potentials of the first-row transition metals suggest that all of them except copper should react with strong acids such as hydrochloric acid to produce hydrogen gas. However, most transition metals are inert toward acids or react slowly with them because of a protective layer of oxide. A case in point is chromium: Despite a rather negative standard reduction potential, it is quite inert chemically because of the formation on its surfaces of chromium(III) oxide, Cr2O3. Consequently, chromium is com- monly used as a protective and noncorrosive plating on other metals. On the bum- pers and trim of vintage automobiles, chromium plating serves a decorative as well as a functional purpose. General Physical Properties Most of the transition metals have a close-packed structure (see Figure 11.29) in which each atom has a coordination number of 12. Furthermore, these elements have rela- tively small atomic radii. The combined effect of closest packing and small atomic size results in strong metallic bonds. Therefore, transition metals have higher densities, higher melting points and boiling points, and higher heats of fusion and vaporization than the Group 1A, 2A, and 2B metals (Table 23.2). Electron Configurations The electron configurations of the first-row transition metals were discussed in Section 7.9. Calcium has the electron configuration [Ar]4s2. From scandium across to copper, electrons are added to the 3d orbitals. Thus, the outer electron configura- tion of scandium is 4s23d1, that of titanium is 4s23d 2, and so on. The two exceptions are chromium and copper, whose outer electron configurations are 4s13d 5 and 4s13d10, respectively. These irregularities are the result of the extra stability associated with half-filled and completely filled 3d subshells.  23.1 Properties of the Transition Metals 997 Table 23.2 Physical Properties of Elements K to Zn 1A 2A Transition Metals 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic radius (pm) 227 197 162 147 134 130 135 126 125 124 128 138 Melting point (°C) 63.7 838 1539 1668 1900 1875 1245 1536 1495 1453 1083 419.5 Boiling point (°C) 760 1440 2730 3260 3450 2665 2150 3000 2900 2730 2595 906 Density (g/cm3) 0.86 1.54 3.0 4.51 6.1 7.19 7.43 7.86 8.9 8.9 8.96 7.14 When the first-row transition metals form cations, electrons are removed first from the 4s orbitals and then from the 3d orbitals. (This is the opposite of the order in which orbitals are filled in atoms.) For example, the outer electron configuration of Fe21 is 3d 6, not 4s23d 4. Review of Concepts Locate the transition metal atoms and ions in the periodic table shown here. Atoms: (1) [Kr]5s24d 5. (2) [Xe]6s24f 145d 4. Ions: (3) [Ar]3d 3 (a 14 ion). (4) [Xe]4f 145d 8 (a 13 ion). (See Table 7.3.) Oxidation States Transition metals exhibit variable oxidation states in their compounds. Figure 23.2 shows the oxidation states from scandium to copper. Note that the common oxidation states for each element include 12, 13, or both. The 13 oxidation states are more stable at the beginning of the series, whereas toward the end the 12 oxidation states are more stable. The reason for this trend can be understood by examining the ioniza- tion energy plots in Figure 23.3. In general, the ionization energies increase gradually from left to right. However, the third ionization energy (when an electron is removed from the 3d orbital) increases more rapidly than the first and second ionization ener- gies. Because it takes more energy to remove the third electron from the metals near the end of the row than from those near the beginning, the metals near the end tend to form M21 ions rather than M31 ions. The highest oxidation state for a transition metal is 17, for manganese (4s23d5). Recall that oxides in which the metal has a high oxidation number are covalent For elements to the right of Mn (Fe to Cu), oxidation numbers are lower. Transition and acidic, whereas those in which the metals usually exhibit their highest oxidation states in compounds with very electro- metal has a low oxidation number are ionic and basic (see Section 15.11). negative elements such as oxygen and fluorine—for example, V2O5, CrO3, and Mn2O7. 998 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Figure 23.2 Oxidation states of the first-row transition metals. The Sc Ti V Cr Mn Fe Co Ni Cu most stable oxidation numbers are shown in color. The zero oxidation state is encountered in some +7 compounds, such as Ni(CO)4 and Fe(CO)5. +6 +6 +6 +5 +5 +5 +5 +4 +4 +4 +4 +4 +4 +3 +3 +3 +3 +3 +3 +3 +3 +3 +2 +2 +2 +2 +2 +2 +2 +2 +1 Figure 23.3 Variation of the 4000 first, second, and third ionization energies for the first-row transition Third metals. Ionization energy (kJ/mol) 3000 2000 Second 1000 First 0 Sc Ti V Cr Mn Fe Co Ni Cu Element 23.2 Chemistry of Iron and Copper Figure 23.4 shows the first-row transition metals. In this section, we will briefly survey the chemistry of two of these elements—iron and copper—paying particular attention to their occurrence, preparation, uses, and important compounds. Iron After aluminum, iron is the most abundant metal in Earth’s crust (6.2 percent by mass). It is found in many ores; some of the important ones are hematite, Fe2O3; siderite, FeCO3; and magnetite, Fe3O4 (Figure 23.5). The preparation of iron in a blast furnace and steelmaking were discussed in Section 21.2. Pure iron is a gray metal and is not particularly hard. It is an essential element in living systems. Iron reacts with hydrochloric acid to give hydrogen gas: Fe(s) 1 2H1 (aq) ¡ Fe21 (aq) 1 H2 (g)  23.2 Chemistry of Iron and Copper 999 Scandium (Sc) Titanium (Ti) Vanadium (V) Chromium (Cr) Manganese (Mn) Iron (Fe) Cobalt (Co) Nickel (Ni) Copper (Cu) Figure 23.4 The first-row transition metals. Concentrated sulfuric acid oxidizes the metal to Fe31, but concentrated nitric acid renders the metal “passive” by forming a thin layer of Fe3O4 over the surface. One of the best-known reactions of iron is rust formation (see Section 18.7). The two oxidation states of iron are 12 and 13. Iron(II) compounds include FeO (black), FeSO4 ? 7H2O (green), FeCl2 (yellow), and FeS (black). In the presence of oxygen, Fe21 ions in solution are readily oxidized to Fe31 ions. Iron(III) oxide is reddish brown, and iron(III) chloride is brownish black. Figure 23.5 The iron ore magnetite, Fe3O4. Copper Copper, a rare element (6.8 3 1023 percent of Earth’s crust by mass), is found in nature in the uncombined state as well as in ores such as chalcopyrite, CuFeS2 (Figure 23.6). The reddish-brown metal is obtained by roasting the ore to give Cu2S and then metallic copper: 2CuFeS2 (s) 1 4O2 (g) ¡ Cu2S(s) 1 2FeO(s) 1 3SO2 (g) Cu2S(s) 1 O2 (g) ¡ 2Cu(l) 1 SO2 (g) Impure copper can be purified by electrolysis (see Section 21.2). After silver, which Figure 23.6 Chalcopyrite, is too expensive for large-scale use, copper has the highest electrical conductivity. It CuFeS2. 1000 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds is also a good thermal conductor. Copper is used in alloys, electrical cables, plumbing (pipes), and coins. Copper reacts only with hot concentrated sulfuric acid and nitric acid (see Figure 22.7). Its two important oxidation states are 11 and 12. The 11 state is less stable and disproportionates in solution: 2Cu1 (aq) ¡ Cu(s) 1 Cu21 (aq) All compounds of Cu(I) are diamagnetic and colorless except for Cu2O, which is red. The Cu(II) compounds are all paramagnetic and colored. The hydrated Cu21 ion is blue. Some important Cu(II) compounds are CuO (black), CuSO4 ? 5H2O (blue), and CuS (black). 23.3 Coordination Compounds Transition metals have a distinct tendency to form complex ions (see p. 756). A coor- Recall that a complex ion contains a dination compound typically consists of a complex ion and counter ion. [Note that central metal ion bonded to one or more ions or molecules (see Section 16.10). some coordination compounds such as Fe(CO)5 do not contain complex ions.] Our understanding of the nature of coordination compounds stems from the classic work of Alfred Werner,† who prepared and characterized many coordination compounds. In 1893, at the age of 26, Werner proposed what is now commonly referred to as Werner’s coordination theory. Nineteenth-century chemists were puzzled by a certain class of reactions that seemed to violate valence theory. For example, the valences of the elements in cobalt(III) chloride and in ammonia seem to be completely satisfied, and yet these two substances react to form a stable compound having the formula CoCl3 ? 6NH3. To explain this behavior, Werner postulated that most elements exhibit two types of valence: primary valence and secondary valence. In modern terminology, primary valence corresponds to the oxidation number and secondary valence to the coordina- tion number of the element. In CoCl3 ? 6NH3, according to Werner, cobalt has a primary valence of 3 and a secondary valence of 6. Today we use the formula [Co(NH3)6]Cl3 to indicate that the ammonia molecules and the cobalt atom form a complex ion; the chloride ions are not part of the complex but are held to it by ionic forces. Most, but not all, of the metals in coordination compounds are transition metals. The molecules or ions that surround the metal in a complex ion are called ligands (Table 23.3). The interactions between a metal atom and the ligands can be thought Ligands act as Lewis bases by donating of as Lewis acid-base reactions. As we saw in Section 15.12, a Lewis base is a sub- pairs of electrons to metals, which act as Lewis acids. stance capable of donating one or more electron pairs. Every ligand has at least one unshared pair of valence electrons, as these examples show: O N O H H SClS Q ⫺ SCqOS H H H Therefore, ligands play the role of Lewis bases. On the other hand, a transition metal atom (in either its neutral or positively charged state) acts as a Lewis acid, accepting (and sharing) pairs of electrons from the Lewis bases. Thus, the metal-ligand bonds are usually coordinate covalent bonds (see Section 9.9). † Alfred Werner (1866–1919). Swiss chemist. Werner started as an organic chemist but became interested in coordination chemistry. For his theory of coordination compounds, Werner was awarded the Nobel Prize in Chemistry in 1913.  23.3 Coordination Compounds 1001 Table 23.3 Some Common Ligands Name Structure Monodentate ligands Ammonia O HONOH A H Carbon monoxide SCqOS Chloride ion O SClS Q ⫺ Cyanide ion [SCqNS]⫺ Thiocyanate ion O [SSOCqNS Q ]⫺ Water O HOOOH Q Bidentate ligands Ethylenediamine H2O O 2 NOCH2OCH2ONH 2⫺ S S O O M J S S COC D G S S Oxalate ion O O S S S S Polydentate ligand SOS SOS 4⫺ B B C C D G D G S S O CH2 CH O G D 2 S S S S Ethylenediaminetetraacetate O O NOCH 2OCH 2ON D G S S ion (EDTA) O CH 2 CH2 O M D G J S S C C A A SOS Q SOS Q The atom in a ligand that is bound directly to the metal atom is known as the donor atom. For example, nitrogen is the donor atom in the [Cu(NH3)4]21 complex ion. The coordination number in coordination compounds is defined as In a crystal lattice, the coordination number of an atom (or ion) is defined as the number of donor atoms surrounding the central metal atom in a complex ion. the number of atoms (or ions) surround- For example, the coordination number of Ag1 in [Ag(NH3)2]1 is 2, that of Cu21 ing the atom (or ion). in [Cu(NH3)4]21 is 4, and that of Fe31 in [Fe(CN)6]32 is 6. The most common coordination numbers are 4 and 6, but coordination numbers such as 2 and 5 are also known. Depending on the number of donor atoms present, ligands are classified as mono- dentate, bidentate, or polydentate (see Table 23.3). H2O and NH3 are monodentate ligands with only one donor atom each. One bidentate ligand is ethylenediamine (sometimes abbreviated “en”): H2O O 2 NOCH2OCH2ONH The two nitrogen atoms can coordinate with a metal atom, as shown in Figure 23.7. 1002 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Figure 23.7 (a) Structure of a CH2 metal-ethylenediamine complex H2N CH2 cation, such as [Co(en)3 ] 21. H2 C NH2 Each ethylenediamine molecule NH2 H2C provides two N donor atoms and M M is therefore a bidentate ligand. H2N NH2 (b) Simplified structure of the same complex cation. H2N CH2 CH2 (a) (b) Bidentate and polydentate ligands are also called chelating agents because of their ability to hold the metal atom like a claw (from the Greek chele, meaning “claw”). One example is ethylenediaminetetraacetate ion (EDTA), a polydentate ligand used to treat metal poisoning (Figure 23.8). Six donor atoms enable EDTA to form a very stable complex ion with lead. In this form, it is removed from the blood and tissues and excreted from the body. EDTA is also used to clean up spills of radioactive metals. Review of Concepts What is the difference between these two compounds: CrCl3 ? 6H2O and [Cr(H2O)6]Cl3? Oxidation Numbers of Metals in Coordination Compounds The term “oxidation state” is commonly Another important property of coordination compounds is the oxidation number of used when referring to the oxidation number of the transition metal in the central metal atom. The net charge of a complex ion is the sum of the charges on coordination compounds. the central metal atom and its surrounding ligands. In the [PtCl6]22 ion, for example, each chloride ion has an oxidation number of 21, so the oxidation number of Pt must be 14. If the ligands do not bear net charges, the oxidation number of the metal is Figure 23.8 (a) EDTA complex of lead. The complex bears a net charge of 22 because each of the six O donor atoms has a charge of 12 and the lead ion carries a charge of 21. Only the lone pairs that participate in bonding are shown. Note the octahedral geometry around the Pb21 ion. (b) Molecular model of O the Pb21–EDTA complex. The green sphere is the Pb21 ion. C O O CH2 C CH2 O N CH2 Pb N CH2 O O C CH2 CH 2 O C O (a) (b)  23.3 Coordination Compounds 1003 equal to the charge of the complex ion. Thus, in [Cu(NH3)4]21 each NH3 is neutral, so the oxidation number of Cu is 12. Example 23.1 deals with oxidation numbers of metals in coordination compounds. Example 23.1 Specify the oxidation number of the central metal atom in each of the following compounds: (a) [Ru(NH3)5(H2O)]Cl2, (b) [Cr(NH3)6](NO3)3, (c) [Fe(CO)5], and (d) K4[Fe(CN)6]. Strategy The oxidation number of the metal atom is equal to its charge. First we examine the anion or the cation that electrically balances the complex ion. This step gives us the net charge of the complex ion. Next, from the nature of the ligands (charged or neutral species) we can deduce the net charge of the metal and hence its oxidation number. Solution (a) Both NH3 and H2O are neutral species. Because each chloride ion carries a 21 charge, and there are two Cl2 ions, the oxidation number of Ru must be 12. (b) Each nitrate ion has a charge of 21; therefore, the cation must be [Cr(NH3)6]31. NH3 is neutral, so the oxidation number of Cr is 13. (c) Because the CO species are neutral, the oxidation number of Fe is zero. (d) Each potassium ion has a charge of 11; therefore, the anion is [Fe(CN)6]42. Next, we know that each cyanide group bears a charge of 21, so Fe must have an oxidation number of 12. Similar problems: 23.13, 23.14. Practice Exercise Write the oxidation numbers of the metals in the compound K[Au(OH)4]. Naming Coordination Compounds Now that we have discussed the various types of ligands and the oxidation numbers of metals, our next step is to learn what to call these coordination compounds. The rules for naming coordination compounds are as follows: 1. The cation is named before the anion, as in other ionic compounds. The rule holds regardless of whether the complex ion bears a net positive or a negative charge. For example, in K3[Fe(CN)6] and [Co(NH3)4Cl2]Cl compound, we name the K1 and [Co(NH3)4Cl2]1 cations first, respectively. 2. Within a complex ion the ligands are named first, in alphabetical order, and the metal ion is named last. 3. The names of anionic ligands end with the letter o, whereas a neutral ligand is usually called by the name of the molecule. The exceptions are H2O (aqua), CO (carbonyl), and NH3 (ammine). Table 23.4 lists some common ligands. 4. When several ligands of a particular kind are present, we use the Greek prefixes di-, tri-, tetra-, penta-, and hexa- to name them. Thus, the ligands in the cation [Co(NH3)4Cl2]1 are “tetraamminedichloro.” (Note that prefixes are ignored when alphabetizing ligands.) If the ligand itself contains a Greek prefix, we use the prefixes bis (2), tris (3), and tetrakis (4) to indicate the number of ligands present. For example, the ligand ethylenediamine already contains di; therefore, if two such ligands are present the name is bis(ethylenediamine). 5. The oxidation number of the metal is written in Roman numerals following the name of the metal. For example, the Roman numeral III is used to indicate the 1004 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Table 23.4 Names of Common Ligands in Coordination Compounds Name of Ligand in Ligand Coordination Compound Bromide, Br2 Bromo Chloride, Cl2 Chloro Cyanide, CN2 Cyano Hydroxide, OH2 Hydroxo Oxide, O22 Oxo Carbonate, CO322 Carbonato Nitrite, NO2 2 Nitro Oxalate, C2O22 4 Oxalato Ammonia, NH3 Ammine Carbon monoxide, CO Carbonyl Water, H2O Aqua Ethylenediamine Ethylenediamine Ethylenediaminetetraacetate Ethylenediaminetetraacetato 13 oxidation state of chromium in [Cr(NH3)4Cl2]1, which is called tetraammine- dichlorochromium(III) ion. 6. If the complex is an anion, its name ends in -ate. For example, in K4[Fe(CN)6] the anion [Fe(CN)6]42 is called hexacyanoferrate(II) ion. Note that the Roman numeral II indicates the oxidation state of iron. Table 23.5 gives the names of anions con- taining metal atoms. Examples 23.2 and 23.3 deal with the nomenclature of coordination compounds. Table 23.5 Names of Anions Containing Metal Atoms Example 23.2 Name of Write the systematic names of the following coordination compounds: (a) Ni(CO)4, Metal in (b) NaAuF4, (c) K3[Fe(CN)6], (d) [Cr(en)3]Cl3. Anionic Metal Complex Strategy We follow the preceding procedure for naming coordination compounds and refer to Tables 23.4 and 23.5 for names of ligands and anions containing metal atoms. Aluminum Aluminate Chromium Chromate Solution Cobalt Cobaltate (a) The CO ligands are neutral species and therefore the Ni atom bears no net Copper Cuprate charge. The compound is called tetracarbonylnickel(0) , or more commonly, Gold Aurate nickel tetracarbonyl . Iron Ferrate (b) The sodium cation has a positive charge; therefore, the complex anion has a negative charge (AuF24 ). Each fluoride ion has a negative charge so the oxidation Lead Plumbate number of gold must be 13 (to give a net negative charge). The compound is called Manganese Manganate sodium tetrafluoroaurate(III) . Molybdenum Molybdate (c) The complex ion is the anion and it bears three negative charges because each Nickel Nickelate potassium ion bears a 11 charge. Looking at [Fe(CN)6]32, we see that the Silver Argentate oxidation number of Fe must be 13 because each cyanide ion bears a 21 charge Tin Stannate (26 total). The compound is potassium hexacyanoferrate(III) . This compound is commonly called potassium ferricyanide . Tungsten Tungstate Zinc Zincate (Continued)  23.4 Structure of Coordination Compounds 1005 (d) As we noted earlier, en is the abbreviation for the ligand ethylenediamine. Because there are three chloride ions each with a 21 charge, the cation is [Cr(en)3]31. The en ligands are neutral so the oxidation number of Cr must be 13. Because there are three en groups present and the name of the ligand already contains di (rule 4), the compound is called tris(ethylenediamine)chromium(III) chloride . Similar problems: 23.15, 23.16. Practice Exercise What is the systematic name of [Cr(H2O)4Cl2]Cl? Example 23.3 Write the formulas for the following compounds: (a) pentaamminechlorocobalt(III) chloride, (b) dichlorobis(ethylenediamine)platinum(IV) nitrate, (c) sodium hexanitrocobaltate(III). Strategy We follow the preceding procedure and refer to Tables 23.4 and 23.5 for names of ligands and anions containing metal atoms. Solution (a) The complex cation contains five NH3 groups, a Cl2 ion, and a Co ion having a 13 oxidation number. The net charge of the cation must be 12, [Co(NH3)5Cl]21. Two chloride anions are needed to balance the positive charges. Therefore, the formula of the compound is [Co(NH3)5Cl]Cl2 . (b) There are two chloride ions (21 each), two en groups (neutral), and a Pt ion with an oxidation number of 14. The net charge on the cation must be 12, [Pt(en)2Cl2]21. Two nitrate ions are needed to balance the 12 charge of the complex cation. Therefore, the formula of the compound is [Pt(en)2Cl2](NO3)2 . (c) The complex anion contains six nitro groups (21 each) and a cobalt ion with an oxidation number of 13. The net charge on the complex anion must be 23, [Co(NO2)6]32. Three sodium cations are needed to balance the 23 charge of the complex anion. Therefore, the formula of the compound is Na3[Co(NO2)6] . Similar problems: 23.17, 23.18. Practice Exercise Write the formula for the following compound: tris(ethylene- diamine)cobalt(III) sulfate. Review of Concepts A student writes the name of the compound [Cr(H2O)4Cl2]Cl as dichlorotetraaquachromium chloride. Is this correct? If not, provide a proper systematic name. 23.4 Structure of Coordination Compounds In studying the geometry of coordination compounds, we often find that there is more than one way to arrange ligands around the central atom. Compounds rearranged in this fashion have distinctly different physical and chemical properties. Figure 23.9 shows four different geometric arrangements for metal atoms with monodentate ligands. In these diagrams, we see that structure and coordination number of the metal atom relate to each other as follows: Coordination number Structure 2 Linear 4 Tetrahedral or square planar 6 Octahedral 1006 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Figure 23.9 Common L L geometries of complex ions. In L L L L each case, M is a metal and L is L M L M M M L a monodentate ligand. L L L L L L L Linear Tetrahedral Square planar Octahedral Stereoisomers are compounds that are made up of the same types and numbers of atoms bonded together in the same sequence but with different spatial arrangements. There are two types of stereoisomers: geometric isomers and optical isomers. Coor- dination compounds may exhibit one or both types of isomerism. Note, however, that many coordination compounds do not have stereoisomers. Geometric Isomers Geometric isomers are stereoisomers that cannot be interconverted without breaking a chemical bond. Geometric isomers usually come in pairs. We use the terms “cis” and “trans” to distinguish one geometric isomer of a compound from the other. Cis means that two particular atoms (or groups of atoms) are adjacent to each other, and trans means that the atoms (or groups of atoms) are on opposite sides in the structural for- mula. The cis and trans isomers of coordination compounds generally have quite dif- ferent colors, melting points, dipole moments, and chemical reactivities. Figure 23.10 shows the cis and trans isomers of diamminedichloroplatinum(II). Note that although cis-tetraamminedichlorocobalt( III ) the types of bonds are the same in both isomers (two Pt¬N and two Pt¬Cl bonds), chloride (left) and trans- the spatial arrangements are different. Another example is tetraamminedichloro- tetraamminedichlorocobalt( III ) chloride (right). cobalt(III) ion, shown in Figure 23.11. Optical Isomers Optical isomers are nonsuperimposable mirror images. (“Superimposable” means that if one structure is laid over the other, the positions of all the atoms will match.) Like geometric isomers, optical isomers come in pairs. However, the optical isomers of a H3N Cl Cl NH3 Pt Pt H3N Cl H3N Cl (a) (b) Figure 23.10 The (a) cis and (b) trans isomers of diamminedichloroplatinum(II). Note that the two Cl atoms are adjacent to each other in the cis isomer and diagonally across from each other in the trans isomer. NH3 NH3 Cl Cl Cl NH3 Cl NH3 Cl NH3 H3N NH3 Co Co Co Co Cl NH3 H3N Cl H3N NH3 H3N NH3 NH3 NH3 NH3 Cl (a) (b) (c) (d) Figure 23.11 The (a) cis and (b) trans isomers of the tetraamminedichlorocobalt(III) ion, [ Co( NH3 )4 Cl2 ] 1. The structure shown in (c) can be generated by rotating that in (a), and the structure shown in (d) can be generated by rotating that in (b). The ion has only two geometric isomers, (a) [or (c)] and (b) [or (d)].  23.4 Structure of Coordination Compounds 1007 compound have identical physical and chemical properties, such as melting point, Mirror boiling point, dipole moment, and chemical reactivity toward molecules that are not optical isomers themselves. Optical isomers differ from each other in their interactions with plane-polarized light, as we will see. The structural relationship between two optical isomers is analogous to the rela- tionship between your left and right hands. If you place your left hand in front of a mirror, the image you see will look like your right hand (Figure 23.12). We say that your left hand and right hand are mirror images of each other. However, they are nonsuperimposable, because when you place your left hand over your right hand (with both palms facing down), they do not match. Figure 23.13 shows the cis and trans isomers of dichlorobis(ethylenediamine)- cobalt(III) ion and their images. Careful examination reveals that the trans isomer and its mirror image are superimposable, but the cis isomer and its mirror Figure 23.12 A left hand and image are not. Therefore, the cis isomer and its mirror image are optical isomers. its mirror image, which looks the Optical isomers are described as chiral (from the Greek word for “hand”) because, same as the right hand. like your left and right hands, chiral molecules are nonsuperimposable. Isomers that are superimposable with their mirror images are said to be achiral. Chiral molecules play a vital role in enzyme reactions in biological systems. Many drug molecules are chiral. It is interesting to note that frequently only one of a pair of chiral isomers is biologically effective. Chiral molecules are said to be optically active because of their ability to rotate Animation Chirality the plane of polarization of polarized light as it passes through them. Unlike ordinary light, which vibrates in all directions, plane-polarized light vibrates only in a single plane. We use a polarimeter to measure the rotation of polarized light by optical Polaroid sheets are used to make Polaroid glasses. isomers (Figure 23.14). A beam of unpolarized light first passes through a Polaroid sheet, called the polarizer, and then through a sample tube containing a solution of an optically active, chiral compound. As the polarized light passes through the sample tube, its plane of polarization is rotated either to the right or to the left. This rotation can be measured directly by turning the analyzer in the appropriate direction until minimal light transmission is achieved (Figure 23.15). If the plane of polarization is rotated to the right, the isomer is said to be dextrorotatory (d); it is levorotatory (l) if the rotation is to the left. The d and l isomers of a chiral substance, called enantiomers, always rotate the light by the same amount, but in opposite directions. Thus, in an equimolar mixture of two enantiomers, called a racemic mixture, the net rotation is zero. Review of Concepts How many geometric isomers of the [CoBr2(en)(NH3)2]1 ion are possible? Mirror Mirror Cl Cl Cl Cl M M M M Cl Cl Cl Cl (a) (b) Figure 23.13 The (a) cis and (b) trans isomers of the dichlorobis(ethylenediamine)cobalt(III) ion and their mirror images. If you could rotate the mirror image in (b) 90° clockwise about the vertical position and place the ion over the trans isomer, you would find that the two are superimposable. No matter how you rotated the cis isomer and its mirror image in (a), however, you could not superimpose one on the other. 1008 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds + Analyzer 0° – Degree scale +90° –90° 180° Polarimeter tube Plane of polarization Fixed polarizer Optically active substance in solution Light source Figure 23.14 Operation of a polarimeter. Initially, the tube is filled with an achiral compound. The analyzer is rotated so that its plane of polarization is perpendicular to that of the polarizer. Under this condition, no light reaches the observer. Next, a chiral compound is placed in the tube as shown. The plane of polarization of the polarized light is rotated as it travels through the tube so that some light reaches the observer. Rotating the analyzer (either to the left or to the right) until no light reaches the observer again enables the angle of optical rotation to be measured. Figure 23.15 With one Polaroid sheet over a picture, light passes through. With a second sheet of Polaroid placed over the first so that the axes of polarization of the sheets are perpendicular, little or no light passes through. If the axes of polarization of the two sheets were parallel, light would pass through.  23.5 Bonding in Coordination Compounds: Crystal Field Theory 1009 23.5 Bonding in Coordination Compounds: Crystal Field Theory A satisfactory theory of bonding in coordination compounds must account for proper- ties such as color and magnetism, as well as stereochemistry and bond strength. No single theory as yet does all this for us. Rather, several different approaches have been applied to transition metal complexes. We will consider only one of them here—crystal field theory—because it accounts for both the color and magnetic properties of many coordination compounds. We will begin our discussion of crystal field theory with the most straightforward case, namely, complex ions with octahedral geometry. Then we will see how it is applied to tetrahedral and square-planar complexes. Crystal Field Splitting in Octahedral Complexes Crystal field theory explains the bonding in complex ions purely in terms of elec- The name “crystal field” is associated with the theory used to explain the trostatic forces. In a complex ion, two types of electrostatic interaction come into properties of solid, crystalline materials. play. One is the attraction between the positive metal ion and the negatively The same theory is used to study coordination compounds. charged ligand or the negatively charged end of a polar ligand. This is the force that binds the ligands to the metal. The second type of interaction is electrostatic repulsion between the lone pairs on the ligands and the electrons in the d orbitals of the metals. As we saw in Chapter 7, d orbitals have different orientations, but in the absence of external disturbance they all have the same energy. In an octahedral complex, a central metal atom is surrounded by six lone pairs of electrons (on the six ligands), so all five d orbitals experience electrostatic repulsion. The magnitude of this repul- sion depends on the orientation of the d orbital that is involved. Take the dx2 2y2 orbital as an example. In Figure 23.16, we see that the lobes of this orbital point toward corners of the octahedron along the x and y axes, where the lone-pair electrons are z Figure 23.16 The five d orbitals in an octahedral environment. The metal atom (or ion) is at the x center of the octahedron, and the six lone pairs on the donor atoms of the ligands are at the corners. y dz2 dx2 – y2 dxy dyz dxz 1010 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Figure 23.17 Crystal field splitting between d orbitals in an octahedral complex. dx 2 – y 2 dz2 Energy Crystal field splitting dxy dyz dxz positioned. Thus, an electron residing in this orbital would experience a greater repul- sion from the ligands than an electron would in, say, the dxy orbital. For this reason, the energy of the dx2 2y2 orbital is increased relative to the dxy , dyz, and dxz orbitals. The dz2 orbital’s energy is also greater, because its lobes are pointed at the ligands along the z-axis. As a result of these metal-ligand interactions, the five d orbitals in an octahedral complex are split between two sets of energy levels: a higher level with two orbitals (dx2 2y2 and dz2 ) having the same energy and a lower level with three equal-energy orbitals (dxy, dyz, and dxz ), as shown in Figure 23.17. The crystal field splitting (D) is the energy difference between two sets of d orbitals in a metal atom when ligands are present. The magnitude of D depends on the metal and the nature of the ligands; it has a direct effect on the color and magnetic properties of complex ions. Color Animation In Chapter 7 we learned that white light, such as sunlight, is a combination of all Absorption of Color colors. A substance appears black if it absorbs all the visible light that strikes it. If it absorbs no visible light, it is white or colorless. An object appears green if it absorbs all light but reflects the green component. An object also looks green if it reflects all colors except red, the complementary color of green (Figure 23.18). What has been said of reflected light also applies to transmitted light (that is, the light that passes through the medium, for example, a solution). Consider the hydrated cupric ion, [Cu(H2O)6]21, which absorbs light in the orange region of the spectrum so that a solution of CuSO4 appears blue to us. Recall from Chapter 7 that when the energy of a photon is equal to the difference between the ground state and an excited state, absorption occurs as the photon strikes the atom (or ion or compound), and an electron is promoted to a higher level. This knowledge enables us to calculate the energy change involved in the electron transition. The energy of a photon, given by Equation (7.2), is E 5 hv 650 nm 580 nm where h represents Planck’s constant (6.63 3 10234 J ? s) and v is the frequency of the 700 nm radiation, which is 5.00 3 1014/s for a wavelength of 600 nm. Here E 5 D, so we have 560 nm 400 nm ¢ 5 hv 490 nm 5 (6.63 3 10234 J ? s)(5.00 3 1014 ys) 430 nm 5 3.32 3 10219 J Figure 23.18 A color wheel with appropriate wavelengths. (Note that this is the energy absorbed by one ion.) If the wavelength of the photon A compound that absorbs in the green region will appear red, the absorbed by an ion lies outside the visible region, then the transmitted light looks the complementary color of green. same (to us) as the incident light—white—and the ion appears colorless.  23.5 Bonding in Coordination Compounds: Crystal Field Theory 1011 Figure 23.19 (a) The process of photon absorption and (b) the Photon of energy h␯ dx 2 – y 2 dz2 dx 2 – y 2 dz2 absorption spectrum of [Ti(H2O)6]31. The energy of the incoming photon is equal to the crystal field splitting. The maximum absorption peak in the dxy dyz dxz dxy dyz dxz visible region occurs at 498 nm. (a) Absorption 400 500 600 700 Wavelength (nm) (b) The best way to measure crystal field splitting is to use spectroscopy to determine A d-to-d transition must occur for a the wavelength at which light is absorbed. The [Ti(H2O)6]31 ion provides a straightfor- transition metal complex to show color. Therefore, ions with d0 or d10 electron ward example, because Ti31 has only one 3d electron [Figure 23.19(a)]. The [Ti(H2O)6]31 configurations are usually colorless. ion absorbs light in the visible region of the spectrum (Figure 23.20). The wavelength corresponding to maximum absorption is 498 nm [Figure 23.19(b)]. This information enables us to calculate the crystal field splitting as follows. We start by writing ¢ 5 hv (23.1) Also c v5 λ where c is the speed of light and λ is the wavelength. Therefore, hc (6.63 3 10234 J ? s)(3.00 3 108 mys) ¢5 5 Equation (7.3) shows that E 5 hc/λ. λ (498 nm) (1 3 1029 my1 nm) 5 3.99 3 10219 J Figure 23.20 Colors of some of the first-row transition metal ions in solution. From left to right: Ti 31, Cr 31, Mn 21, Fe 31, Co 21, Ni 21, Cu 21. The Sc 31 and V 51 ions are colorless. 1012 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds This is the energy required to excite one [Ti(H2O)6]31 ion. To express this energy difference in the more convenient units of kilojoules per mole, we write ¢ 5 (3.99 3 10219 Jyion)(6.02 3 1023 ionsymol) 5 240,000 Jymol 5 240 kJymol Aided by spectroscopic data for a number of complexes, all having the same metal ion but different ligands, chemists calculated the crystal splitting for each ligand and established a spectrochemical series, which is a list of ligands arranged in increasing order of their abilities to split the d orbital energy levels: The order in the spectrochemical series I2 , Br2 , Cl2 , OH2 , F2 , H2O , NH3 , en , CN2 , CO is the same no matter which metal atom (or ion) is present. These ligands are arranged in the order of increasing value of D. CO and CN2 are called strong-field ligands, because they cause a large splitting of the d orbital energy levels. The halide ions and hydroxide ion are weak-field ligands, because they split the d orbitals to a lesser extent. Review of Concepts The Cr31 ion forms octahedral complexes with two neutral ligands X and Y. The color of CrX631 is blue while that of CrY631 is yellow. Which is a stronger field ligand? Magnetic Properties The magnitude of the crystal field splitting also determines the magnetic properties of a complex ion. The [Ti(H2O)6]31 ion, having only one d electron, is always paramag- netic. However, for an ion with several d electrons, the situation is less clearcut. Con- sider, for example, the octahedral complexes [FeF6]32 and [Fe(CN)6]32 (Figure 23.21). The electron configuration of Fe31 is [Ar]3d5, and there are two possible ways to dis- tribute the five d electrons among the d orbitals. According to Hund’s rule (see Section 7.8), maximum stability is reached when the electrons are placed in five separate orbit- als with parallel spins. But this arrangement can be achieved only at a cost; two of the five electrons must be promoted to the higher-energy dx2 2y2 and dz2 orbitals. No such energy investment is needed if all five electrons enter the dxy, dyz, and dxz orbitals. According to Pauli’s exclusion principle (p. 305), there will be only one unpaired elec- tron present in this case. Figure 23.22 shows the distribution of electrons among d orbitals that results in low- and high-spin complexes. The actual arrangement of the electrons is determined by the amount of stability gained by having maximum parallel spins versus the invest- ment in energy required to promote electrons to higher d orbitals. Because F2 is a weak-field ligand, the five d electrons enter five separate d orbitals with parallel spins to create a high-spin complex (see Figure 23.21). On the other hand, the cyanide ion is a strong-field ligand, so it is energetically preferable for all five electrons to be in the lower orbitals and therefore a low-spin complex is formed. High-spin complexes are more paramagnetic than low-spin complexes. The magnetic properties of a complex ion depend on the number of unpaired The actual number of unpaired electrons (or spins) in a complex ion can be found electrons present. by magnetic measurements, and in general, experimental findings support predictions  23.5 Bonding in Coordination Compounds: Crystal Field Theory 1013 dx 2 – y 2 dz2 dx 2 – y 2 dz2 Energy Fe3+ ion dxy dyz dxz [FeF6 ]3– dxy dyz dxz (high spin) [Fe(CN)6 ] 3– (low spin) Figure 23.21 Energy-level diagrams for the Fe31 ion and for the [FeF6]32 and [Fe(CN)6]32 complex ions. High spin Low spin dx 2 – y 2 dz2 d4 dx 2 – y 2 dz2 dxy dyz dxz dxy dyz dxz d5 d6 d7 Figure 23.22 Orbital diagrams for the high-spin and low-spin octahedral complexes corresponding to the electron configurations d4, d5, d6, and d7. No such distinctions can be made for d1, d2, d3, d8, d9, and d10. 1014 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds based on crystal field splitting. However, a distinction between low- and high-spin complexes can be made only if the metal ion contains more than three and fewer than eight d electrons, as shown in Figure 23.22. Example 23.4 Predict the number of unpaired spins in the [Cr(en)3]21 ion. Strategy The magnetic properties of a complex ion depend on the strength of the ligands. Strong-field ligands, which cause a high degree of splitting among the d orbital energy levels, result in low-spin complexes. Weak-field ligands, which cause a small degree of splitting among the d orbital energy levels, result in high-spin complexes. Solution The electron configuration of Cr21 is [Ar]3d 4. Because en is a strong-field ligand, we expect [Cr(en)3]21 to be a low-spin complex. According to Figure 23.22, all four electrons will be placed in the lower-energy d orbitals (dxy, dyz, and dxz) and there Similar problem: 23.35. will be a total of two unpaired spins. Practice Exercise How many unpaired spins are in [Mn(H2O)6]21? (H2O is a weak- field ligand.) Tetrahedral and Square-Planar Complexes So far we have concentrated on octahedral complexes. The splitting of the d orbital energy levels in two other types of complexes—tetrahedral and square-planar—can also be accounted for satisfactorily by the crystal field theory. In fact, the splitting pattern for a tetrahedral ion is just the reverse of that for octahedral complexes. In this case, the dxy, dyz, and dxz orbitals are more closely directed at the ligands and therefore have more energy than the dx2 2y2 and dz2 orbitals (Figure 23.23). Most tetrahedral complexes are high-spin complexes. Presumably, the tetrahedral arrange- ment reduces the magnitude of metal-ligand interactions, resulting in a smaller D value compared to the octahedral case. This is a reasonable assumption because the number of ligands is smaller in a tetrahedral complex. As Figure 23.24 shows, the splitting pattern for square-planar complexes is the most complicated of the three cases. Clearly, the dx2 2y2 orbital possesses the highest energy (as in the octahedral case), and the dxy orbital the next highest. However, the relative placement of the dz2 and the dxz and dyz orbitals cannot be determined simply by inspection and must be calculated. Figure 23.23 Crystal field splitting between d orbitals in a tetrahedral complex. dxy dyz dxz Energy Crystal field splitting dx 2 – y 2 dz2  23.6 Reactions of Coordination Compounds 1015 Figure 23.24 Energy-level diagram for a square-planar complex. Because there are dx 2 – y 2 more than two energy levels, we cannot define crystal field splitting as we can for octahedral and tetrahedral complexes. dxy Energy dz2 dxz dyz 23.6 Reactions of Coordination Compounds Complex ions undergo ligand exchange (or substitution) reactions in solution. The rates of these reactions vary widely, depending on the nature of the metal ion and the ligands. In studying ligand exchange reactions, it is often useful to distinguish between the stability of a complex ion and its tendency to react, which we call kinetic lability. Stability in this context is a thermodynamic property, which is measured in terms of the species’ formation constant Kf (see p. 756). For example, we say that the complex ion tetracyanonickelate(II) is stable because it has a large formation constant (Kf < 1 3 1030 ) Ni21 1 4CN2 Δ [Ni(CN) 4]22 By using cyanide ions labeled with the radioactive isotope carbon-14, chemists have shown that [Ni(CN)4]22 undergoes ligand exchange very rapidly in solution. The following equilibrium is established almost as soon as the species are mixed: [Ni(CN) 4]22 1 4*CN2 Δ [Ni(*CN) 4]22 1 4CN2 At equilibrium, there is a distribution of *CN2 ions in the complex ion. where the asterisk denotes a 14C atom. Complexes like the tetracyanonickelate(II) ion are termed labile complexes because they undergo rapid ligand exchange reactions. Thus, a thermodynamically stable species (that is, one that has a large formation constant) is not necessarily unreactive. (In Section 13.4 we saw that the smaller the activation energy, the larger the rate constant, and hence the greater the rate.) A complex that is thermodynamically unstable in acidic solution is [Co(NH3)6]31. The equilibrium constant for the following reaction is about 1 3 1020: [Co(NH3 ) 6]31 1 6H1 1 6H2O Δ [Co(H2O) 6]31 1 6NH1 4 When equilibrium is reached, the concentration of the [Co(NH3)6]31 ion is very low. However, this reaction requires several days to complete because of the inertness of the [Co(NH3)6]31 ion. This is an example of an inert complex, a complex ion that undergoes very slow exchange reactions (on the order of hours or even days). It shows that a thermodynamically unstable species is not necessarily chemically reac- tive. The rate of reaction is determined by the energy of activation, which is high in this case. CHEMISTRY in Action Coordination Compounds in Living Systems C oordination compounds play many important roles in ani- mals and plants. They are essential in the storage and trans- port of oxygen, as electron transfer agents, as catalysts, and in The heme group in hemoglobin. The Fe21 ion is coordinated with the nitrogen atoms of the heme group. The ligand below the porphyrin is photosynthesis. Here we focus on coordination compounds con- H H the histidine group, which is taining iron and magnesium. O attached to the protein. The sixth Because of its central function as an oxygen carrier for N N ligand is a water molecule. metabolic processes, hemoglobin is probably the most studied Fe of all the proteins. The molecule contains four folded long N N chains called subunits. Hemoglobin carries oxygen in the blood N from the lungs to the tissues, where it delivers the oxygen mol- HN ecules to myoglobin. Myoglobin, which is made up of only one subunit, stores oxygen for metabolic processes in muscle. The porphine molecule forms an important part of the Protein hemoglobin structure. Upon coordination to a metal, the H1 ions that are bonded to two of the four nitrogen atoms in por- phine are displaced. Complexes derived from porphine are O O O O O O N N N N N N Fe Fe Fe N N N N N N N N N N N N N H HN HN HN Fe H N N N N (a) (b) (c) Three possible ways for molecular oxygen to bind to the heme group in hemoglobin. The structure shown in (a) would have a coordination number Porphine Fe2ⴙ-porphyrin of 7, which is considered unlikely for Fe(II) complexes. Although the end-on arrangement in (b) seems the most reasonable, evidence points Simplified structures of the porphine molecule and the Fe21-porphyrin to the structure in (c) as the correct one. The structure shown in (c) is the complex. most plausible. Most complex ions containing Co31, Cr31, and Pt21 are kinetically inert. Because they exchange ligands very slowly, they are easy to study in solution. As a result, our knowledge of the bonding, structure, and isomerism of coordination compounds has come largely from studies of these compounds. 23.7 Applications of Coordination Compounds Coordination compounds are found in living systems and have many uses in the home, in industry, and in medicine. We describe a few examples here and in the above Chemistry in Action essay. 1016 called porphyrins, and the iron-porphyrin combination is called The heme group in cytochrome c. the heme group. The iron in the heme group has an oxidation The ligands above and below the CH2 porphyrin are the methionine group number of 12; it is coordinated to the four nitrogen atoms in the and histidine group of the protein, porphine group and also to a nitrogen donor atom in a ligand H3C CH2 respectively. that is attached to the protein. The sixth ligand is a water mol- S ecule, which binds to the Fe21 ion on the other side of the ring N N Fe to complete the octahedral complex. This hemoglobin molecule is called deoxyhemoglobin and imparts a bluish tinge to venous N N blood. The water ligand can be replaced readily by molecular N oxygen to form red oxyhemoglobin found in arterial blood. HN Each subunit contains a heme group, so each hemoglobin mol- ecule can bind up to four O2 molecules. There are three possible structures for oxyhemoglobin. For Protein a number of years, the exact arrangement of the oxygen mol- ecule relative to the porphyrin group was not clear. Most ex- perimental evidence suggests that the bond between O and Fe is bent relative to the heme group. The porphyrin group is a very effective chelating agent, and not surprisingly, we find it in a number of biological systems. The iron-heme complex is present in another class of proteins, C C C called the cytochromes. The iron forms an octahedral complex in C C C C these proteins, but because both the histidine and the methionine groups are firmly bound to the metal ion, they cannot be dis- C N N C placed by oxygen or other ligands. Instead, the cytochromes act as electron carriers, which are essential to metabolic processes. HC Mg CH In cytochromes, iron undergoes rapid reversible redox reactions: C N N C 31 2 21 Fe 1 e Δ Fe C C C which are coupled to the oxidation of organic molecules such as C C the carbohydrates. The chlorophyll molecule, which is necessary for plant The porphyrin structure in chlorophyll. The dotted lines indicate the coordi- photosynthesis, also contains the porphyrin ring, but in this case nate covalent bonds. The electron-delocalized portion of the molecule is the metal ion is Mg21 rather than Fe21. shown in color. Metallurgy The extraction of silver and gold by the formation of cyanide complexes (p. 965) and the purification of nickel (p. 937) by converting the metal to the gaseous compound Ni(CO)4 are typical examples of the use of coordination compounds in metallurgical processes. Therapeutic Chelating Agents Earlier we mentioned that the chelating agent EDTA is used in the treatment of lead poisoning. Certain platinum-containing compounds can effectively inhibit the growth of cancerous cells. A specific case is discussed on p. 1018. 1017  CHEMISTRY in Action Cisplatin—The Anticancer Drug L uck often plays a role in major scientific breakthroughs, but it takes an alert and well-trained person to recognize the sig- nificance of an accidental discovery and to take full advantage dividing. It did not take long for the group to determine that a platinum-containing substance extracted from the bacterial cul- ture inhibited cell division. of it. Such was the case when, in 1964, the biophysicist Barnett Rosenberg reasoned that the platinum compound might be Rosenberg and his research group at Michigan State University useful as an anticancer agent, because cancer involves uncon- were studying the effect of an electric field on the growth of bac- trolled division of affected cells, so he set out to identify the teria. They suspended a bacterial culture between two platinum substance. Given the presence of ammonia and chloride ions in electrodes and passed an electric current through it. To their sur- solution during electrolysis, Rosenberg synthesized a number of prise, they found that after an hour or so the bacteria cells ceased platinum compounds containing ammonia and chlorine. The one that proved most effective at inhibiting cell division was cis-diamminedichloroplatinum(II) [Pt(NH3)2Cl2], also called cisplatin. The mechanism for the action of cisplatin is the chelation of DNA (deoxyribonucleic acid), the molecule that contains the genetic code. During cell division, the double-stranded DNA splits into two single strands, which must be accurately copied in order for the new cells to be identical to their parent cell. X-ray studies show that cisplatin binds to DNA by forming cross-links in which the two chlorides on cisplatin are replaced by nitrogen atoms in the adjacent guanine bases on the same strand of the DNA. (Guanine is one of the four bases in DNA. See Figure 25.17.) Consequently, the double-stranded structure assumes a bent configuration at the binding site. Scientists be- lieve that this structural distortion is a key factor in inhibiting Cisplatin, a bright yellow compound, is administered intravenously to cancer replication. The damaged cell is then destroyed by the body’s patients. immune system. Because the binding of cisplatin to DNA Chemical Analysis Although EDTA has a great affinity for a large number of metal ions (especially 21 and 31 ions), other chelates are more selective in binding. For example, dimethylglyoxime, H3C G CPNOOH A CPNOOH D H3C forms an insoluble brick-red solid with Ni21 and an insoluble bright-yellow solid with Pd21. These characteristic colors are used in qualitative analysis to identify An aqueous suspension of nickel and palladium. Further, the quantities of ions present can be determined by bis(dimethylglyoximato)nickel(II). gravimetric analysis (see Section 4.6) as follows: To a solution containing Ni21 ions, 1018 requires both Cl atoms to be on the same side of the complex, the trans isomer of the compound is totally ineffective as an anticancer drug. Unfortunately, cisplatin can cause serious side effects, including severe kidney damage. Therefore, ongoing research efforts are directed toward finding related complexes that destroy cancer cells with less harm to healthy tissues. Cisplatin 33° Pt N H3 N H3 Cisplatin destroys the cancer cells’ ability to reproduce by changing the configuration of their DNA. It binds to two sites on a strand of DNA, causing it to bend about 33° away from the rest of the strand. The structure of this DNA adduct was elucidated by Professor Stephen Lippard’s group at MIT. cis-Pt(NH3 )2Cl2 say, we add an excess of dimethylglyoxime reagent, and a brick-red precipitate forms. The precipitate is then filtered, dried, and weighed. Knowing the formula of the complex (Figure 23.25), we can readily calculate the amount of nickel present in the original solution. Figure 23.25 Structure of nickel O HZ Z O dimethylglyoxime. Note that the overall structure is stabilized by hydrogen bonds. H3C CH 3 N N C C Ni C C N N H 3C CH 3 OZ Z H O 1019 1020 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds Detergents SOS SOS SOS 5ⴚ The cleansing action of soap in hard water is hampered by the reaction of the Ca21 B B B O O O SOOPOOOPOOOPOOS Q Q Q O Q ions in the water with the soap molecules to form insoluble salts or curds. In the late A A A SOS Q SOS Q SOS Q 1940s the detergent industry introduced a “builder,” usually sodium tripolyphosphate, to circumvent this problem. The tripolyphosphate ion is an effective chelating agent Tripolyphosphate ion. that forms stable, soluble complexes with Ca21 ions. Sodium tripolyphosphate revo- lutionized the detergent industry. However, because phosphates are plant nutrients, waste waters containing phosphates discharged into rivers and lakes cause algae to grow, resulting in oxygen depletion. Under these conditions most or all aquatic life eventually succumbs. This process is called eutrophication. Consequently, many states have banned phosphate detergents since the 1970s, and manufacturers have reformu- lated their products to eliminate phosphates. Key Equation D 5 hν (23.1) Crystal-field splitting Summary of Facts & Concepts 1. Transition metals usually have incompletely filled d field theory, the d orbitals are split into two higher- subshells and a pronounced tendency to form com- energy and three lower-energy orbitals in an octahedral plexes. Compounds that contain complex ions are called complex. The energy difference between these two sets coordination compounds. of d orbitals is the crystal field splitting. 2. The first-row transition metals (scandium to copper) are 6. Strong-field ligands cause a large crystal field splitting, the most common of all the transition metals; their and weak-field ligands cause a small splitting. Electron chemistry is characteristic, in many ways, of the entire spins tend to be parallel with weak-field ligands and group. paired with strong-field ligands, where a greater invest- 3. Complex ions consist of a metal ion surrounded by li- ment of energy is required to promote electrons into the gands. The donor atoms in the ligands each contribute high-lying d orbitals. an electron pair to the central metal ion in a complex. 7. Complex ions undergo ligand exchange reactions in 4. Coordination compounds may display geometric and/or solution. optical isomerism. 8. Coordination compounds find application in many dif- 5. Crystal field theory explains bonding in complexes in ferent areas, for example, as antidotes for metal poison- terms of electrostatic interactions. According to crystal ing and in chemical analysis. Key Words Chelating agent, p. 1002 Crystal field splitting Inert complex, p. 1015 Racemic mixture, p. 1007 Chiral, p. 1007 (D), p. 1010 Labile complex, p. 1015 Spectrochemical Coordination Donor atom, p. 1001 Ligand, p. 1000 series, p. 1012 compound, p. 1000 Enantiomers, p. 1007 Optical isomers, p. 1006 Stereoisomers, p. 1006 Coordination number, p. 1001 Geometric isomers, p. 1006 Polarimeter, p. 1007  Questions & Problems 1021 Questions & Problems • Problems available in Connect Plus • 23.16 What are the systematic names for the following ion Red numbered problems solved in Student Solutions Manual and compounds? (a) [cis-Co(en)2Cl2]1 Properties of the Transition Metals (b) [Pt(NH3)5Cl]Cl3 Review Questions (c) [Co(NH3)5Cl]Cl2 23.1 What distinguishes a transition metal from a repre- • 23.17 Write the formulas for each of the following ions sentative metal? and compounds: (a) tetrahydroxozincate(II), (b) penta- 23.2 Why is zinc not considered a transition metal? aquachlorochromium(III) chloride, (c) tetrabromo- cuprate(II), (d) ethylenediaminetetraacetatoferrate(II). 23.3 Explain why atomic radii decrease very gradually from scandium to copper. • 23.18 Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)- 23.4 Without referring to the text, write the ground-state dichlorochromium(III), (b) pentacarbonyliron(0), electron configurations of the first-row transition (c) potassium tetracyanocuprate(II), (d) tetraammine- metals. Explain any irregularities. aquachlorocobalt(III) chloride. 23.5 Write the electron configurations of the following ions: V51, Cr31, Mn21, Fe31, Cu21, Sc31, Ti41. Structure of Coordination Compounds 23.6 Why do transition metals have more oxidation states Review Questions than other elements? 23.7 Give the highest oxidation states for scandium to 23.19 Define the following terms: stereoisomers, geo- copper. metric isomers, optical isomers, plane-polarized 23.8 Why does chromium seem to be less reactive than light. its standard reduction potential suggests? 23.20 Specify which of the following structures can exhibit geometric isomerism: (a) linear, (b) square-planar, Coordination Compounds (c) tetrahedral, (d) octahedral. Review Questions 23.21 What determines whether a molecule is chiral? How does a polarimeter measure the chirality of a molecule? 23.9 Define the following terms: coordination com- 23.22 Explain the following terms: (a) enantiomers, pound, ligand, donor atom, coordination number, (b) racemic mixtures. chelating agent. 23.10 Describe the interaction between a donor atom and a Problems metal atom in terms of a Lewis acid-base reaction. • 23.23 The complex ion [Ni(CN)2Br2]22 has a square-planar Problems geometry. Draw the structures of the geometric isomers of this complex. • 23.11 Complete the following statements for the complex • 23.24 How many geometric isomers are in the following ion [Co(en)2(H2O)CN]21. (a) en is the abbreviation species? (a) [Co(NH3)2Cl4]2, (b) [Co(NH3)3Cl3] for . (b) The oxidation number of Co is . 23.25 Draw structures of all the geometric and optical (c) The coordination number of Co is . isomers of each of the following cobalt (d) is a bidentate ligand. complexes: 23.12 Complete the following statements for the complex (a) [Co(NH3)6]31 ion [Cr(C2O4)2(H2O)2]2. (a) The oxidation number of Cr is . (b) The coordination number of Cr is (b) [Co(NH3)5Cl]21 . (c) is a bidentate ligand. (c) [Co(C2O4)3]32 • 23.13 Give the oxidation numbers of the metals in the fol- 23.26 Draw structures of all the geometric and optical lowing species: (a) K3[Fe(CN)6], (b) K3[Cr(C2O4)3], isomers of each of the following cobalt complexes: (c) [Ni(CN)4]22. (a) [Co(NH3)4Cl2]1, (b) [Co(en)3]31. • 23.14 Give the oxidation numbers of the metals in the Bonding in Coordination Compounds following species: (a) Na2MoO4, (b) MgWO4, (c) Fe(CO)5. Review Questions • 23.15 What are the systematic names for the following 23.27 Briefly describe crystal field theory. ions and compounds? 23.28 Define the following terms: crystal field splitting, (a) [Co(NH3)4Cl2]1 (c) [Co(en)2Br2]1 high-spin complex, low-spin complex, spectrochemi- (b) Cr(NH3)3Cl3 (d) [Co(NH3)6]Cl3 cal series.  1022 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds 23.29 What is the origin of color in a coordination 23.43 Aqueous copper(II) sulfate solution is blue in color. compound? When aqueous potassium fluoride is added, a green 23.30 Compounds containing the Sc31 ion are colorless, precipitate is formed. When aqueous potassium chlo- whereas those containing the Ti31 ion are colored. ride is added instead, a bright-green solution is Explain. formed. Explain what is happening in these two 23.31 What factors determine whether a given complex cases. will be diamagnetic or paramagnetic? 23.44 When aqueous potassium cyanide is added to a so- 23.32 For the same type of ligands, explain why the crystal lution of copper(II) sulfate, a white precipitate, field splitting for an octahedral complex is always soluble in an excess of potassium cyanide, is greater than that for a tetrahedral complex. formed. No precipitate is formed when hydrogen sulfide is bubbled through the solution at this point. Problems Explain. 23.45 A concentrated aqueous copper(II) chloride solution 23.33 The [Ni(CN)4]22 ion, which has a square-planar is bright green in color. When diluted with water, the geometry, is diamagnetic, whereas the [NiCl4]22 solution becomes light blue. Explain. ion, which has a tetrahedral geometry, is paramag- netic. Show the crystal field splitting diagrams for • 23.46 In a dilute nitric acid solution, Fe31 reacts with thio- cyanate ion (SCN2) to form a dark-red complex: those two complexes. 23.34 Transition metal complexes containing CN2 ligands [Fe(H2O) 6]31 1 SCN2 Δ are often yellow in color, whereas those containing H2O 1 [Fe(H2O) 5NCS]21 H2O ligands are often green or blue. Explain. The equilibrium concentration of [Fe(H2O)5NCS]21 • 23.35 Predict the number of unpaired electrons in may be determined by how darkly colored the solution the following complex ions: (a) [Cr(CN)6]42, is (measured by a spectrometer). In one such experi- (b) [Cr(H2O)6]21. ment, 1.0 mL of 0.20 M Fe(NO3)3 was mixed with • 23.36 The absorption maximum for the complex ion 1.0 mL of 1.0 3 1023 M KSCN and 8.0 mL of dilute [Co(NH3)6]31 occurs at 470 nm. (a) Predict the color HNO3. The color of the solution quantitatively indi- of the complex and (b) calculate the crystal field cated that the [Fe(H2O)5NCS]21 concentration was splitting in kJ/mol. 7.3 3 1025 M. Calculate the formation constant for • 23.37 From each of the following pairs, choose the com- [Fe(H2O)5NCS]21. plex that absorbs light at a longer wavelength: (a) [Co(NH3)6]21, [Co(H2O)6]21; (b) [FeF6]32, [Fe(CN)6]32; (c) [Cu(NH3)4]21, [CuCl4]22. Additional Problems 23.47 As we read across the first-row transition metals • 23.38 A solution made by dissolving 0.875 g of from left to right, the 12 oxidation state becomes Co(NH3)4Cl3 in 25.0 g of water freezes at 20.568C. more stable in comparison with the 13 state. Why is Calculate the number of moles of ions produced this so? when 1 mole of Co(NH3)4Cl3 is dissolved in water and suggest a structure for the complex ion present in 23.48 Which is a stronger oxidizing agent in aqueous solu- this compound. tion, Mn31 or Cr31? Explain your choice. 23.49 Carbon monoxide binds to Fe in hemoglobin some Reactions of Coordination Compounds 200 times more strongly than oxygen. This is the Review Questions reason why CO is a toxic substance. The metal-to- ligand sigma bond is formed by donating a lone 23.39 Define the terms (a) labile complex, (b) inert pair from the donor atom to an empty sp3d2 orbital complex. on Fe. (a) On the basis of electronegativities, would 23.40 Explain why a thermodynamically stable species you expect the C or O atom to form the bond to Fe? may be chemically reactive and a thermodynami- (b) Draw a diagram illustrating the overlap of the cally unstable species may be unreactive. orbitals involved in the bonding. 23.50 What are the oxidation states of Fe and Ti in the ore Problems ilmenite, FeTiO3? (Hint: Look up the ionization en- 23.41 Oxalic acid, H2C2O4, is sometimes used to clean rust ergies of Fe and Ti in Table 23.1; the fourth ioniza- stains from sinks and bathtubs. Explain the chemis- tion energy of Ti is 4180 kJ/mol.) try underlying this cleaning action. 23.51 A student has prepared a cobalt complex that has 23.42 The [Fe(CN)6]32 complex is more labile than the one of the following three structures: [Co(NH3)6]Cl3, [Fe(CN)6]42 complex. Suggest an experiment that [Co(NH3)5Cl]Cl2, or [Co(NH3)4Cl2]Cl. Explain would prove that [Fe(CN)6]32 is a labile complex. how the student would distinguish between these  Questions & Problems 1023 possibilities by an electrical conductance experi- 23.58 The Co21-porphyrin complex is more stable than the ment. At the student’s disposal are three strong Fe21-porphyrin complex. Why, then, is iron the electrolytes—NaCl, MgCl2, and FeCl3—which may metal ion in hemoglobin (and other heme-containing be used for comparison purposes. proteins)? • 23.52 Chemical analysis shows that hemoglobin contains 23.59 What are the differences between geometric isomers 0.34 percent of Fe by mass. What is the minimum and optical isomers? possible molar mass of hemoglobin? The actual mo- 23.60 Oxyhemoglobin is bright red, whereas deoxyhemo- lar mass of hemoglobin is about 65,000 g. How do globin is purple. Show that the difference in color you account for the discrepancy between your mini- can be accounted for qualitatively on the basis of mum value and the actual value? high-spin and low-spin complexes. (Hint: O2 is a 23.53 Explain the following facts: (a) Copper and iron strong-field ligand; see the Chemistry in Action essay have several oxidation states, whereas zinc has only on p. 1016.) one. (b) Copper and iron form colored ions, whereas 23.61 Hydrated Mn21 ions are practically colorless (see zinc does not. Figure 23.20) even though they possess five 3d elec- • 23.54 A student in 1895 prepared three coordination com- trons. Explain. (Hint: Electronic transitions in which pounds containing chromium, with the following there is a change in the number of unpaired elec- properties: trons do not occur readily.) • 23.62 Which of the following hydrated cations are color- less: Fe21(aq), Zn21(aq), Cu1(aq), Cu21(aq), Cl2 Ions in V51(aq), Ca21(aq), Co21(aq), Sc31(aq), Pb21(aq)? Solution per Explain your choice. Formula Color Formula Unit 23.63 Aqueous solutions of CoCl2 are generally either (a) CrCl3 ? 6H2O Violet 3 light pink or blue. Low concentrations and low tem- (b) CrCl3 ? 6H2O Light green 2 peratures favor the pink form while high concentra- (c) CrCl3 ? 6H2O Dark green 1 tions and high temperatures favor the blue form. Adding hydrochloric acid to a pink solution of CoCl2 causes the solution to turn blue; the pink color is restored by the addition of HgCl2. Account Write modern formulas for these compounds and for these observations. suggest a method for confirming the number of Cl2 ions present in solution in each case. (Hint: 23.64 Suggest a method that would allow you to distinguish Some of the compounds may exist as hydrates and between cis-Pt(NH3)2Cl2 and trans-Pt(NH3)2Cl2. Cr has a coordination number of 6 in all the • 23.65 You are given two solutions containing FeCl2 and compounds.) FeCl3 at the same concentration. One solution is • 23.55 The formation constant for the reaction Ag1 1 light yellow and the other one is brown. Identify 2NH3 Δ [Ag(NH3 ) 2]1 is 1.5 3 107 and that these solutions based only on color. for the reaction Ag1 1 2CN2 Δ [Ag(CN) 2]2 23.66 The label of a certain brand of mayonnaise lists is 1.0 3 1021 at 258C (see Table 16.3). Calculate EDTA as a food preservative. How does EDTA pre- the equilibrium constant and DG8 at 258C for the vent the spoilage of mayonnaise? reaction 23.67 The compound 1,1,1-trifluoroacetylacetone (tfa) is a [Ag(NH3 ) 2]1 1 2CN2 Δ [Ag(CN) 2]2 1 2NH3 bidentate ligand: • 23.56 From the standard reduction potentials listed in O O Table 18.1 for Zn/Zn21 and Cu1/Cu21, calculate B B DG8 and the equilibrium constant for the reaction CF3CCH2CCH3 Zn(s) 1 2Cu21 (aq) ¡ Zn21 (aq) 1 2Cu1 (aq) It forms a tetrahedral complex with Be21 and a square planar complex with Cu21. Draw structures 23.57 Using the standard reduction potentials listed in of these complex ions and identify the type of isom- Table 18.1 and the Handbook of Chemistry and erism exhibited by these ions. Physics, show that the following reaction is favor- 23.68 How many geometric isomers can the following able under standard-state conditions: square planar complex have? 2Ag(s) 1 Pt21 (aq) ¡ 2Ag1 (aq) 1 Pt(s) a b G D What is the equilibrium constant of this reaction at Pt D G 258C? d c  1024 Chapter 23 ■ Transition Metals Chemistry and Coordination Compounds 23.69 [Pt(NH3)2Cl2] is found to exist in two geometric iso- • 23.74 (a) The free Cu(I) ion is unstable in solution and has mers designated I and II, which react with oxalic a tendency to disproportionate: acid as follows: 2Cu1 (aq2 Δ Cu21 (aq) 1 Cu(s) I 1 H2C2O4 ¡ [Pt(NH3 ) 2C2O4] Use the information in Table 18.1 (p. 821) to calculate II 1 H2C2O4 ¡ [Pt(NH3 ) 2 (HC2O4 ) 2] the equilibrium constant for the reaction. (b) Based Comment on the structures of I and II. on your result in (a), explain why most Cu(I) com- 23.70 The Kf for the complex ion formation between Pb21 pounds are insoluble. and EDTA42 • 23.75 Consider the following two ligand exchange 21 42 22 reactions: Pb 1 EDTA Δ Pb(EDTA) [Co(H2O) 6]31 1 6NH3 Δ [Co(NH3 ) 6]31 1 6H2O is 1.0 3 10 at 258C. Calculate [Pb21] at equilib- 18 [Co(H2O) 6]31 1 3en Δ [Co(en) 3]31 1 6H2O rium in a solution containing 1.0 3 1023 M Pb21 and 2.0 3 1023 M EDTA42. (a) Which of the reactions should have a larger DS8? • 23.71 Manganese forms three low-spin complex ions with (b) Given that the Co¬N bond strength is approxi- the cyanide ion with the formulas [Mn(CN)6]52, mately the same in both complexes, which reaction [Mn(CN)6]42, and [Mn(CN)6]32. For each complex will have a larger equilibrium constant? Explain ion, determine the oxidation number of Mn and the your choices. number of unpaired d electrons present. 23.76 Copper is also known to exist in the 13 oxidation • 23.72 Commercial silver-plating operations frequently use state, which is believed to be involved in some a solution containing the complex Ag(CN)2 2 ion. Be- biological electron-transfer reactions. (a) Would you cause the formation constant (Kf) is quite large, this expect this oxidation state of copper to be stable? procedure ensures that the free Ag1 concentration in Explain. (b) Name the compound K3CuF6 and pre- solution is low for uniform electrodeposition. In one dict the geometry of the complex ion and its mag- process, a chemist added 9.0 L of 5.0 M NaCN to netic properties. (c) Most of the known Cu(III) 90.0 L of 0.20 M AgNO3. Calculate the concentra- compounds have square planar geometry. Are these tion of free Ag1 ions at equilibrium. See Table 16.4 compounds diamagnetic or paramagnetic? for Kf value. 23.73 Draw qualitative diagrams for the crystal field split- tings in (a) a linear complex ion ML2, (b) a trigonal- planar complex ion ML3, and (c) a trigonal-bipyramidal complex ion ML5. Answers to Practice Exercises 23.1 K: 11; Au: 13. 23.2 Tetraaquadichlorochromium(III) chloride. 23.3 [Co(en)3]2(SO4)3. 23.4 5.  CHAPTER 24 Organic Chemistry A chemical plant. Many small organic compounds such as acetic acid, benzene, ethylene, formaldehyde, and methanol form the basis of multi-billion-dollar pharmaceutical and polymer industries. CHAPTER OUTLINE A LOOK AHEAD 24.1 Classes of Organic  We begin by defining the scope and nature of organic chemistry. (24.1) Compounds  Next, we examine aliphatic hydrocarbons. First we study the nomenclature 24.2 Aliphatic Hydrocarbons and reactions of alkanes. We examine the optical isomerism of substituted alkanes and also the properties of cycloalkanes. We then study unsaturated 24.3 Aromatic Hydrocarbons hydrocarbons, molecules that contain carbon-to-carbon double bonds and 24.4 Chemistry of the Functional triple bonds. We focus on their nomenclature, properties, and geometric Groups isomers. (24.2)  Aromatic compounds all contain one or more benzene rings. They are in general more stable than aliphatic hydrocarbons. (24.3)  Finally, we see that the reactivity of organic compounds can be largely accounted for by the presence of functional groups. We classify the oxygen- and nitrogen-containing functional groups in alcohols, ethers, aldehydes and ketones, carboxylic acids, esters, and amines. (24.4) 1025 1026 Chapter 24 ■ Organic Chemistry O rganic chemistry is the study of carbon compounds. The word “organic” was originally used by eighteenth-century chemists to describe substances obtained from living sources— plants and animals. These chemists believed that nature possessed a certain vital force and that only living things could produce organic compounds. This romantic notion was disproved in 1828 by Friedrich Wohler, a German chemist who prepared urea, an organic compound, from the reaction between inorganic compounds lead cyanate and aqueous ammonia: Pb(OCN)2 1 2NH3 1 2H2O ¡ 2(NH2)2CO 1 Pb(OH)2 urea Today, well over 20 million synthetic and natural organic compounds are known. This number is significantly greater than the 100,000 or so known inorganic compounds. 24.1 Classes of Organic Compounds Recall that the linking of like atoms is Carbon can form more compounds than any other element because carbon atoms are called catenation. The ability of carbon to catenate is discussed in Section 22.3. able not only to form single, double, and triple carbon-carbon bonds, but also to link up with each other in chains and ring structures. The branch of chemistry that deals 1A 8A with carbon compounds is organic chemistry. H 2A 3A 4A 5A 6A 7A B C N O F Classes of organic compounds can be distinguished according to functional Si P S Cl Br groups they contain. A functional group is a group of atoms that is largely respon- I sible for the chemical behavior of the parent molecule. Different molecules containing the same kind of functional group or groups undergo similar reactions. Thus, by Common elements in organic learning the characteristic properties of a few functional groups, we can study and compounds. understand the properties of many organic compounds. In the second half of this chapter, we will discuss the functional groups known as alcohols, ethers, aldehydes and ketones, carboxylic acids, and amines. Most organic compounds are derived from a group of compounds known as Note that the octet rule is satisfied for all hydrocarbons because they are made up of only hydrogen and carbon. On the basis hydrocarbons. of structure, hydrocarbons are divided into two main classes—aliphatic and aromatic. Aliphatic hydrocarbons do not contain the benzene group, or the benzene ring, whereas aromatic hydrocarbons contain one or more benzene rings. 24.2 Aliphatic Hydrocarbons Aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes, discussed next (Figure 24.1). Figure 24.1 Classification of hydrocarbons. Hydrocarbons Aromatic Aliphatic Alkanes Cycloalkanes Alkenes Alkynes  24.2 Aliphatic Hydrocarbons 1027 Alkanes Alkanes have the general formula CnH2n 1 2, where n 5 1, 2, . . . . The essential characteristic of alkane hydrocarbon molecules is that only single covalent bonds are present. The alkanes are known as saturated hydrocarbons because they contain the maximum number of hydrogen atoms that can bond with the number of carbon atoms present. The simplest alkane (that is, with n 5 1) is methane CH4, which is a natural product of the anaerobic bacterial decomposition of vegetable matter under water. Because it was first collected in marshes, methane became known as “marsh gas.” A rather improbable but proven source of methane is termites. When these vora- cious insects consume wood, the microorganisms that inhabit their digestive sys- tem break down cellulose (the major component of wood) into methane, carbon dioxide, and other compounds. An estimated 170 million tons of methane are produced annually by termites! It is also produced in some sewage treatment processes. Commercially, methane is obtained from natural gas. The Chemistry in Action essay on p. 1038 describes an interesting compound formed by methane and water molecules. Figure 24.2 shows the structures of the first four alkanes (n 5 1 to n 5 4). Natural gas is a mixture of methane, ethane, and a small amount of propane. We discussed the bonding scheme of methane in Chapter 10. Indeed, the carbon atoms in all the alkanes can be assumed to be sp3-hybridized. The structures of ethane and propane are straightforward, for there is only one way to join the carbon atoms in these molecules. Butane, however, has two possible bonding schemes resulting in the structural isomers n-butane (n stands for normal) and isobutane, molecules that have Termites are a natural source of methane. the same molecular formula, but different structures. Alkanes such as the structural isomers of butane are described as having the straight chain or branched chain struc- tures. n-Butane is a straight-chain alkane because the carbon atoms are joined along one line. In a branched-chain alkane like isobutane, one or more carbon atoms are bonded to at least three other carbon atoms. In the alkane series, as the number of carbon atoms increases, the number of structural isomers increases rapidly. For example, butane, C4H10, has two isomers; decane, C10H22, has 75 isomers; and the alkane C30H62 has over 400 million, or 4 3 108, possible isomers! Obviously, most of these isomers do not exist in nature nor have they been synthesized. Nevertheless, the numbers help to explain why carbon is found in so many more compounds than any other element. Example 24.1 deals with the number of structural isomers of an alkane. Figure 24.2 Structures of the H H H H H H first four alkanes. Note that butane A A A A A A can exist in two structurally H O CO H HOC OC OH HOC OC OC OH different forms, called structural A A A A A A isomers. H H H H H H Methane Ethane Propane H A HOC OH A H H H H H A H A A A A A A A HOC OCO COC OH HO C OC OCOH A A A A A A A H H H H H H H n-Butane Isobutane 1028 Chapter 24 ■ Organic Chemistry Example 24.1 How many structural isomers can be identified for pentane, C5H12? Strategy For small hydrocarbon molecules (eight or fewer C atoms), it is relatively easy to determine the number of structural isomers by trial and error. Solution The first step is to write the straight-chain structure: H H H H H A A A A A n-pentane HOCOCOCOCOCOH A A A A A H H H H H n-pentane (b.p. 36.1°C) The second structure, by necessity, must be a branched chain: H CH3 H H A A A A C OOCOCOH HOCOCO A A A A H H H H 2-methylbutane 2-methylbutane (b.p. 27.9°C) Yet another branched-chain structure is possible: H CH3 H A A A HOCOCOOOCOH A A A H CH3 H 2,2-dimethylpropane (b.p. 9.5°C) We can draw no other structure for an alkane having the molecular formula C5H12. 2,2-dimethylpropane Thus, pentane has three structural isomers, in which the numbers of carbon and hydrogen atoms remain unchanged despite the differences in structure. Similar problem: 24.11. Practice Exercise How many structural isomers are there in the alkane C6H14? Table 24.1 shows the melting and boiling points of the straight-chain isomers of the first 10 alkanes. The first four are gases at room temperature; and pentane through decane are liquids. As molecular size increases, so does the boiling point, because of the increasing dispersion forces (see Section 11.2). Alkane Nomenclature The nomenclature of alkanes and all other organic compounds is based on the recom- mendations of the International Union of Pure and Applied Chemistry (IUPAC). The first four alkanes (methane, ethane, propane, and butane) have nonsystematic names. As Table 24.1 shows, the number of carbon atoms is reflected in the Greek prefixes for the alkanes containing five to ten carbons. We now apply the IUPAC rules to the following examples: 1. The parent name of the hydrocarbon is that given to the longest continuous chain of carbon atoms in the molecule. Thus, the parent name of the following com- pound is heptane because there are seven carbon atoms in the longest chain CH 3 1 2 3 4A 5 6 7 CH 3OCH 2OCH 2OCHOCH 2OCH 2OCH 3  24.2 Aliphatic Hydrocarbons 1029 Table 24.1 The First 10 Straight-Chain Alkanes Name of Molecular Number of Melting Boiling Hydrocarbon Formula Carbon Atoms Point (8C) Point (8C) Methane CH4 1 2182.5 2161.6 Ethane CH3¬CH3 2 2183.3 288.6 Propane CH3¬CH2¬CH3 3 2189.7 242.1 Butane CH3¬(CH2)2¬CH3 4 2138.3 20.5 Pentane CH3¬(CH2)3¬CH3 5 2129.8 36.1 Hexane CH3¬(CH2)4¬CH3 6 295.3 68.7 Heptane CH3¬(CH2)5¬CH3 7 290.6 98.4 Octane CH3¬(CH2)6¬CH3 8 256.8 125.7 Nonane CH3¬(CH2)7¬CH3 9 253.5 150.8 Decane CH3¬(CH2)8¬CH3 10 229.7 174.0 2. An alkane less one hydrogen atom is an alkyl group. For example, when a hydrogen atom is removed from methane, we are left with the CH3 fragment, which is called a methyl group. Similarly, removing a hydrogen atom from the ethane molecule gives an ethyl group, or C2H5. Table 24.2 lists the names of several common alkyl groups. Any chain branching off the longest chain is named as an alkyl group. 3. When one or more hydrogen atoms are replaced by other groups, the name of the compound must indicate the locations of carbon atoms where replacements are made. The procedure is to number each carbon atom on the longest chain in the direction that gives the smaller numbers for the locations of all branches. Consider the two different systems for the same compound shown here: CH 3 CH 3 1 2A 3 4 5 1 2 3 4A 5 CH 3OCHOCH 2OCH 2OCH 3 CH 3OCH 2OCH 2OCHOCH 3 2-methylpentane 4-methylpentane The compound on the left is numbered correctly because the methyl group is located at carbon 2 of the pentane chain; in the compound on the right, the methyl Table 24.2 Common Alkyl Groups Name Formula Methyl ¬CH3 Ethyl ¬CH2¬CH3 n-Propyl ¬CH2¬CH2¬CH3 n-Butyl ¬CH2¬CH2¬CH2¬CH3 CH 3 A Isopropyl OCOH A CH 3 CH 3 A t-Butyl* OCOCH 3 A CH 3 *The letter t stands for tertiary. 1030 Chapter 24 ■ Organic Chemistry Table 24.3 group is located at carbon 4. Thus, the name of the compound is 2-methylpentane, and not 4-methylpentane. Note that the branch name and the parent name are Names of Common written as a single word, and a hyphen follows the number. Substituent Groups 4. When there is more than one alkyl branch of the same kind present, we use a Functional prefix such as di-, tri-, or tetra- with the name of the alkyl group. Consider the Group Name following examples: ¬NH2 Amino CH 3 CH 3 CH 3 2A 3A 3A ¬F Fluoro 1 4 5 6 1 2 4 5 6 ¬Cl Chloro CH 3OCHOCHOCH 2OCH 2OCH 3 CH 3OCH 2OCOCH 2OCH 2OCH 3 ¬Br Bromo A CH 3 ¬I Iodo 2,3-dimethylhexane 3,3-dimethylhexane ¬NO2 Nitro ¬CH“CH2 Vinyl When there are two or more different alkyl groups, the names of the groups are listed alphabetically. For example, CH 3 C 2 H 5 1 2 4A 3A 5 6 7 CH 3OCH 2OCHOCHOCH 2OCH 2OCH 3 4-ethyl-3-methylheptane 5. Of course, alkanes can have many different types of substituents. Table 24.3 lists the names of some substituents, including nitro and bromo. Thus, the compound NO 2 Br 1 2A 3A 4 5 6 CH 3OCHOCHOCH 2OCH 2OCH 3 is called 3-bromo-2-nitrohexane. Note that the substituent groups are listed alpha- betically in the name, and the chain is numbered in the direction that gives the lowest number to the first substituted carbon atom. Example 24.2 Give the IUPAC name of the following compound: CH3 CH3 A A CH 3OCOCH 2OCHOCH 2OCH 3 A CH3 Strategy We follow the IUPAC rules and use the information in Table 24.2 to name the compound. How many C atoms are there in the longest chain? Solution The longest chain has six C atoms so the parent compound is called hexane. Note that there are two methyl groups attached to carbon number 2 and one methyl group attached to carbon number 4. CH3 CH3 1 2A 3 4A 5 6 CH 3OCOCH 2OCHOCH 2OCH 3 A CH3 Similar problem: 24.26. Therefore, we call the compound 2,2,4-trimethylhexane. Practice Exercise Give the IUPAC name of the following compound: CH3 C2H5 C2H5 A A A CH 3OCHOCH 2OCHOCH 2OCHOCH 2OCH 3  24.2 Aliphatic Hydrocarbons 1031 Example 24.3 shows that prefixes such as di-, tri-, and tetra- are used as needed, but are ignored when alphabetizing. Example 24.3 Write the structural formula of 3-ethyl-2,2-dimethylpentane. Strategy We follow the preceding procedure and the information in Table 24.2 to write the structural formula of the compound. How many C atoms are there in the longest chain? Solution The parent compound is pentane, so the longest chain has five C atoms. There are two methyl groups attached to carbon number 2 and one ethyl group attached to carbon number 3. Therefore, the structure of the compound is CH3 C2H5 1 2A 3A 4 5 CH 3OCOOCHOCH 2OCH 3 A Similar problem: 24.27. CH3 Practice Exercise Write the structural formula of 5-ethyl-2,4,6-trimethyloctane. Reactions of Alkanes Alkanes are generally not considered to be very reactive substances. However, under suitable conditions they do react. For example, natural gas, gasoline, and fuel oil are alkanes that undergo highly exothermic combustion reactions: CH4(g) 1 2O2(g) ¡ CO2(g) 1 2H2O(l) DH° 5 2890.4 kJ/mol 2C2H6(g) 1 7O2(g) ¡ 4CO2(g) 1 6H2O(l) DH° 5 23119 kJ/mol These, and similar combustion reactions, have long been utilized in industrial pro- cesses and in domestic heating and cooking. Halogenation of alkanes—that is, the replacement of one or more hydrogen atoms by halogen atoms—is another type of reaction that alkanes undergo. When a mixture of methane and chlorine is heated above 100°C or irradiated with light of a suitable wavelength, methyl chloride is produced: The systematic names of methyl chloride, CH4(g) 1 Cl2(g) ¡ CH3Cl(g) 1 HCl(g) methylene chloride, and chloroform are methyl chloride monochloromethane, dichloromethane, and trichloromethane, respectively. If an excess of chlorine gas is present, the reaction can proceed further: CH3Cl(g) 1 Cl2(g) ¡ CH2Cl2(l) 1 HCl(g) methylene chloride CH2Cl2(l) 1 Cl2(g) ¡ CHCl3(l) 1 HCl(g) chloroform CHCl3(l) 1 Cl2(g) ¡ CCl4(l) 1 HCl(g) carbon tetrachloride A great deal of experimental evidence suggests that the initial step of the first halo- genation reaction occurs as follows: Cl2 1 energy ¡ Cl ? 1 Cl ? Thus, the covalent bond in Cl2 breaks and two chlorine atoms form. We know it is the Cl¬Cl bond that breaks when the mixture is heated or irradiated because the bond 1032 Chapter 24 ■ Organic Chemistry enthalpy of Cl2 is 242.7 kJ/mol, whereas about 414 kJ/mol are needed to break C¬H bonds in CH4. A chlorine atom is a radical, which contains an unpaired electron (shown by a single dot). Chlorine atoms are highly reactive and attack methane molecules accord- ing to the equation CH4 1 Cl ? ¡ ? CH3 1 HCl This reaction produces hydrogen chloride and the methyl radical ? CH3. The methyl radical is another reactive species; it combines with molecular chlorine to give methyl chloride and a chlorine atom: ? CH3 1 Cl2 ¡ CH3Cl 1 Cl ? The production of methylene chloride from methyl chloride and any further reactions can be explained in the same way. The actual mechanism is more complex than the scheme we have shown because “side reactions” that do not lead to the desired prod- ucts often take place, such as Cl ? 1 Cl ? ¡ Cl2 ? CH3 1 ? CH3 ¡ C2H6 Alkanes in which one or more hydrogen atoms have been replaced by a halogen atom are called alkyl halides. Among the large number of alkyl halides, the best known are chloroform (CHCl3), carbon tetrachloride (CCl4), methylene chloride (CH2Cl2), and the chlorofluorohydrocarbons. Chloroform is a volatile, sweet-tasting liquid that was used for many years as an anesthetic. However, because of its toxicity (it can severely damage the liver, kidneys, and heart) it has been replaced by other compounds. Carbon tetrachloride, also a toxic substance, serves as a cleaning liquid, for it removes grease stains from clothing. Methylene chloride was used as a solvent to decaffeinate coffee and as a paint remover. The preparation of chlorofluorocarbons and the effect of these compounds on ozone in the stratosphere were discussed in Chapter 20. Optical Isomerism of Substituted Alkanes Optical isomerism was first discussed in Optical isomers are compounds that are nonsuperimposable mirror images of each Section 23.4. other. Figure 24.3 shows perspective drawings of the substituted methanes CH2ClBr and CHFClBr and their mirror images. The mirror images of CH2ClBr are superim- Animation posable but those of CHFClBr are not, no matter how we rotate the molecules. Thus, Chirality the CHFClBr molecule is chiral. Most simple chiral molecules contain at least one asymmetric carbon atom—that is, a carbon atom bonded to four different atoms or groups of atoms. Example 24.4 Is the following molecule chiral? Cl A HOCOCH 2OCH 3 A CH3 (Continued)  24.2 Aliphatic Hydrocarbons 1033 Mirror Mirror Figure 24.3 (a) The CH2ClBr molecule and its mirror image. Br Br Because the molecule and its Br Br mirror image are superimposable, the molecule is said to be achiral. (b) The CHFClBr molecule and its mirror image. Because the H H H H F H H F molecule and its mirror image are Cl Cl Cl Cl not superimposable, no matter how we rotate one with respect to the other, the molecule is said to be chiral. Br Br H H F H Cl Cl Br Br H H H F Cl Cl (a) (b) Strategy Recall the condition for chirality. Is the central C atom asymmetric; that is, does it have four different atoms or different groups attached to it? Solution We note that the central carbon atom is bonded to a hydrogen atom, a chlorine atom, a ¬CH3 group, and a ¬CH2¬CH3 group. Therefore, the central carbon atom is asymmetric and the molecule is chiral. Similar problem: 24.25. Practice Exercise Is the following molecule chiral? Br A IOCOCH 2OCH 3 A Br Cycloalkanes Alkanes whose carbon atoms are joined in rings are known as cycloalkanes. They have the general formula CnH2n, where n 5 3, 4, . . . . The simplest cycloalkane is cyclopropane, C3H6 (Figure 24.4). Many biologically significant substances such as cholesterol, testosterone, and progesterone contain one or more such ring systems. Theoretical analysis shows that cyclohexane can assume two different geometries that are relatively free of strain (Figure 24.5). By “strain” we mean that bonds are com- pressed, stretched, or twisted out of normal geometric shapes as predicted by sp3 hybridization. The most stable geometry is the chair form. Alkenes The alkenes (also called olefins) contain at least one carbon-carbon double bond. Alkenes have the general formula CnH2n, where n 5 2, 3, . . . . The simplest alkene 1034 Chapter 24 ■ Organic Chemistry Figure 24.4 Structures of the first four cycloalkanes and their H H H H HH H H H H simplified forms. H H C C C C C C H C C H C H H H H H H H H C C H C C H H C C H C C C H H HH H H H H H H Cyclopropane Cyclobutane Cyclopentane Cyclohexane Figure 24.5 The cyclohexane Axial molecule can exist in various shapes. The most stable shape is Equatorial the chair form and a less stable one is the boat form. Two types of H atoms are labeled axial and equatorial, respectively. Chair form Boat form is C2H4, ethylene, in which both carbon atoms are sp2-hybridized and the double bond is made up of a sigma bond and a pi bond (see Section 10.5). Alkene Nomenclature In naming alkenes, we indicate the positions of the carbon-carbon double bonds. The names of compounds containing C“C bonds end with -ene. As with the alkanes, the name of the parent compound is determined by the number of carbon atoms in the longest chain (see Table 24.1), as shown here: CH 2PCHOCH 2OCH 3 H 3COCHPCHOCH 3 1-butene 2-butene The numbers in the names of alkenes refer to the lowest numbered carbon atom in the chain that is part of the C“C bond of the alkene. The name “butene” means that there are four carbon atoms in the longest chain. Alkene nomenclature must specify whether a given molecule is cis or trans if it is a geometric isomer, such as 1 In the cis isomer, the two H atoms are on CH 3 H 3C H the same side of the C“C bond; in the 1 4A 5 6 G2 3D trans isomer, the two H atoms are across H 3C CHOCH 2OCH 3 CPC from each other. Geometric isomerism G 2 3 D D G4 5 6 was introduced in Section 23.4. CPC H CHOCH 2OCH 3 D G A H H CH 3 4-methyl-cis-2-hexene 4-methyl-trans-2-hexene Properties and Reactions of Alkenes Ethylene is an extremely important substance because it is used in large quantities for the manufacture of organic polymers (to be discussed in Chapter 25) and in the  24.2 Aliphatic Hydrocarbons 1035 preparation of many other organic chemicals. Ethylene is prepared industrially by the cracking process, that is, the thermal decomposition of a large hydrocarbon into smaller molecules. When ethane is heated to about 800°C, it undergoes the follow- ing reaction: Pt catalyst C2H6 (g) ¬ ¬¬¡ CH2 “CH2 (g) 1 H2 (g) Other alkenes can be prepared by cracking the higher members of the alkane family. Alkenes are classified as unsaturated hydrocarbons, compounds with double or triple carbon-carbon bonds that enable them to add hydrogen atoms. Unsaturated hydrocarbons commonly undergo addition reactions, in which one molecule adds to another to form a single product. Hydrogenation (see p. 961) is an example of addi- tion reaction. Other addition reactions to the C“C bond include C2H4(g) 1 HX(g) ¡ CH3¬CH2X(g) The addition reaction between C2H4(g) 1 X2(g) ¡ CH2X¬CH2X(g) HCl and ethylene. The initial inter- action is between the positive end of HCl (blue) and the electron-rich where X represents a halogen (Cl, Br, or I). region of ethylene (red), which is The addition of a hydrogen halide to an unsymmetrical alkene such as propene associated with the pi electrons of is more complicated because two products are possible: the C“C bond. H 3C H H H H H G D A A A A CPC  HBr 888n H 3 COCOCOH and/or H 3 COCOCOH D G A A A A H H H Br Br H propene 1-bromopropane 2-bromopropane In reality, however, only 2-bromopropane is formed. This phenomenon was observed in all reactions between unsymmetrical reagents and alkenes. In 1871, Vladimir Markovnikov† postulated a generalization that enables us to predict the outcome of such an addition reaction. This generalization, now known as Markovnikov’s rule, states that in the addition of unsymmetrical (that is, polar) reagents to alkenes, the positive portion of the reagent (usually hydrogen) adds to the carbon atom that already The electron density is higher on has the most hydrogen atoms. the carbon atom in the CH2 group in propene. Geometric Isomers of Alkenes In a compound such as ethane, C2H6, the rotation of the two methyl groups about the carbon-carbon single bond (which is a sigma bond) is quite free. The situation is different for molecules that contain carbon-carbon double bonds, such as ethyl- ene, C2H4. In addition to the sigma bond, there is a pi bond between the two carbon atoms. Rotation about the carbon-carbon linkage does not affect the sigma bond, but it does move the two 2pz orbitals out of alignment for overlap and, hence, partially or totally destroys the pi bond (see Figure 10.16). This process requires an input of energy on the order of 270 kJ/mol. For this reason, the rotation of a carbon-carbon double bond is considerably restricted, but not impossible. Conse- quently, molecules containing carbon-carbon double bonds (that is, the alkenes) may have geometric isomers, which cannot be interconverted without breaking a chemical bond. † Vladimir W. Markovnikov (1838–1904). Russian chemist. Markovnikov’s observations of the addition reactions to alkenes were published a year after his death. 1036 Chapter 24 ■ Organic Chemistry The molecule dichloroethylene, ClHC“CHCl, can exist as one of the two geo- metric isomers called cis-dichloroethylene and trans-dichloroethylene: resultant 88dipole moment m n888 m m m Cl Cl H Cl 88 G D G D 88 88 88 m m m m CPC CPC 88 D G D G 88 88 88 m H H Cl H cis-dichloroethylene trans-dichloroethylene m  1.89 D m0 b.p. 60.3C b.p. 47.5C where the term cis means that two particular atoms (or groups of atoms) are adjacent to each other, and trans means that the two atoms (or groups of atoms) are across from each other. Generally, cis and trans isomers have distinctly dif- ferent physical and chemical properties. Heat or irradiation with light is commonly used to bring about the conversion of one geometric isomer to another, a process called cis-trans isomerization, or geometric isomerization. As the above data show, In cis-dichloroethylene (top), dipole moment measurements can be used to distinguish between geometric the bond moments reinforce one isomers. In general, cis isomers possess a dipole moment, whereas trans isomers another and the molecule is polar. The opposite holds for do not. trans-dichloroethylene and the molecule is nonpolar. Cis-Trans Isomerization in the Vision Process. The molecules in the retina that respond to light are rhodopsin, which has two components called 11-cis retinal and opsin (Figure 24.6). Retinal is the light-sensitive component and opsin is a protein molecule. Upon receiving a photon in the visible region, the 11-cis retinal isomerizes to the all-trans retinal by breaking a carbon-carbon pi bond. With the pi bond broken, the remaining carbon-carbon sigma bond is free to rotate and transforms into the all-trans retinal. At this point an electrical impulse is generated and transmitted to the brain, which forms a visual image. The all-trans retinal does not fit into the binding site on opsin and eventually separates from the protein. In time, the trans isomer is converted back to 11-cis retinal by an enzyme (in the absence of light) and rhodopsin is regenerated by the binding of the cis isomer to opsin and the visual cycle can begin again. all-trans isomer An electron micrograph of 11-cis isomer rod-shaped cells (containing rhodopsins) in the retina. 11 11 12 12 light Opsin Opsin Figure 24.6 The primary event in the vision process is the conversion of 11-cis retinal to the all-trans isomer on rhodopsin. The double bond at which the isomerization occurs is between carbon-11 and carbon-12. For simplicity, most of the H atoms are omitted. In the absence of light, this transformation takes place about once in a thousand years!  24.2  Aliphatic Hydrocarbons 1037 Alkynes Alkynes contain at least one carbon-carbon triple bond. They have the general for- mula CnH2n 2 2, where n 5 2, 3, . . . . Alkyne Nomenclature Names of compounds containing C ‚ C bonds end with -yne. Again, the name of the parent compound is determined by the number of carbon atoms in the longest chain (see Table 24.1 for names of alkane counterparts). As in the case of alkenes, the names of alkynes indicate the position of the carbon-carbon triple bond, as, for example, in HC‚C¬CH2 ¬CH3   H3C¬C‚C¬CH3 1-butyne 2-butyne Properties and Reactions of Alkynes The simplest alkyne is ethyne, better known as acetylene (C2H2). The structure and bond- ing of C2H2 were discussed in Section 10.5. Acetylene is a colorless gas (b.p. 284°C) prepared by the reaction between calcium carbide and water: CaC2 (s) 1 2H2O(l) ¡ C2H2 (g) 1 Ca(OH) 2 (aq) Acetylene has many important uses in industry. Because of its high heat of combustion 2C2H2 (g) 1 5O2 (g) ¡ 4CO2 (g) 1 2H2O(l) ¢H° 5 22599.2 kJ/mol acetylene burned in an “oxyacetylene torch” gives an extremely hot flame (about 3000°C). Thus, oxyacetylene torches are used to weld metals (see p. 257). The standard free energy of formation of acetylene is positive (DG°f 5 209.2 kJ/mol), unlike that of the alkanes. This means that the molecule is unstable (relative to its ele- ments) and has a tendency to decompose: C2H2 (g) ¡ 2C(s) 1 H2 (g) The reaction of calcium carbide with water produces acetylene, a In the presence of a suitable catalyst or when the gas is kept under pressure, this flammable gas. reaction can occur with explosive violence. To be transported safely, the gas must be dissolved in an organic solvent such as acetone at moderate pressure. In the liquid state, acetylene is very sensitive to shock and is highly explosive. Acetylene, an unsaturated hydrocarbon, can be hydrogenated to yield ethylene: C2H2 (g) 1 H2 (g) ¡ C2H4 (g) It undergoes the following addition reactions with hydrogen halides and halogens: C2H2 (g) 1 HX(g) ¡ CH2 “CHX(g) C2H2 (g) 1 X2 (g) ¡ CHX“CHX(g) C2H2 (g) 1 2X2 (g) ¡ CHX2 ¬CHX2 (g) Methylacetylene (propyne), CH3¬C ‚ C¬H, is the next member in the alkyne ­family. It undergoes reactions similar to those of acetylene. The addition reactions of propyne also obey Markovnikov’s rule: H 3C H G D CH 3 OC O O COH O  HBr 888n CPC D G Br H Propyne. Can you account for propyne 2-bromopropene Markovnikov’s rule in this molecule? CHEMISTRY in Action Ice That Burns I ce that burns? Yes, there is such a thing. It is called methane hydrate, and there is enough of it to meet America’s energy needs for years. But scientists have yet to figure out how to mine gas on land. However, harvesting the energy stored in meth- ane hydrate presents a tremendous engineering challenge. It is believed that methane hydrate acts as a kind of cement to it without causing an environmental disaster. keep the ocean floor sediments together. Tampering with the Bacteria in the sediments on the ocean floor consume hydrate deposits could cause underwater landslides, leading organic material and generate methane gas. Under high- to the discharge of methane into the atmosphere. This event pressure and low-temperature conditions, methane forms could have serious consequences for the environment, be- methane hydrate, which consists of single molecules of the cause methane is a potent greenhouse gas (see Section 20.5). natural gas trapped within crystalline cages formed by In fact, scientists have speculated that the abrupt release of frozen water molecules. A lump of methane hydrate looks methane hydrates may have hastened the end of the last like a gray ice cube, but if one puts a lighted match to it, it ice age about 10,000 years ago. As the great blanket of will burn. continental ice melted, global sea levels swelled by more Oil companies have known about methane hydrate since than 90 m, submerging Arctic regions rich in hydrate deposits. the 1930s, when they began using high-pressure pipelines to The relatively warm ocean water would have melted the transport natural gas in cold climates. Unless water is carefully hydrates, unleashing tremendous amounts of methane, which removed before the gas enters the pipeline, chunks of methane led to global warming. hydrate will impede the flow of gas. The total reserve of the methane hydrate in the world’s oceans is estimated to be 1013 tons of carbon content, about twice the amount of carbon in all the coal, oil, and natural Methane hydrate. The methane molecule is trapped in a cage of frozen water molecules (blue spheres) held together by hydrogen bonds. Methane hydrate burning in air. 1038  24.3 Aromatic Hydrocarbons 1039 24.3 Aromatic Hydrocarbons Benzene, the parent compound of this large family of organic substances, was dis- covered by Michael Faraday in 1826. Over the next 40 years, chemists were preoc- cupied with determining its molecular structure. Despite the small number of atoms in the molecule, there are quite a few ways to represent the structure of benzene without violating the tetravalency of carbon. However, most proposed structures were rejected because they did not explain the known properties of benzene. Finally, in 1865, August Kekulé† deduced that the benzene molecule could be best represented by a ring structure—a cyclic compound consisting of six carbon atoms: H H A A H C H HH C H H K H E E N E C C C C A B or B A CN EC CH KC H E C HH E C H H H A A H H As we saw in Section 9.8, the properties of benzene are best represented by both of the above resonance structures. Alternatively, the properties of benzene can be explained in terms of delocalized molecular orbitals (see p. 452): Nomenclature of Aromatic Compounds The naming of monosubstituted benzenes, that is, benzenes in which one H atom has been replaced by another atom or a group of atoms, is quite straightforward, as shown here: CH2CH3 Cl NH2 NO2 A A A A ethylbenzene chlorobenzene aminobenzene nitrobenzene (aniline) If more than one substituent is present, we must indicate the location of the second group relative to the first. The systematic way to accomplish this is to number the carbon atoms as follows: 1 6 2 5 3 4 † August Kekulé (1829–1896). German chemist. Kekulé was a student of architecture before he became interested in chemistry. He supposedly solved the riddle of the structure of the benzene molecule after having a dream in which dancing snakes bit their own tails. Kekulé’s work is regarded by many as the crowning achievement of theoretical organic chemistry of the nineteenth century. 1040 Chapter 24 ■ Organic Chemistry Three different dibromobenzenes are possible: Br Br Br A Br A A E H Br A Br 1,2-dibromobenzene 1,3-dibromobenzene 1,4-dibromobenzene (o-dibromobenzene) (m-dibromobenzene) (p-dibromobenzene) The prefixes o- (ortho-), m- (meta-), and p- (para-) are also used to denote the rela- tive positions of the two substituted groups, as shown above for the dibromobenzenes. Compounds in which the two substituted groups are different are named accordingly. Thus, NO2 A H Br is named 3-bromonitrobenzene, or m-bromonitrobenzene. Finally, we note that the group containing benzene minus a hydrogen atom (C6H5) is called the phenyl group. Thus, the following molecule is called 2-phenylpropane: This compound is also called isopropyl benzene (see Table 24.2). A CH3OCHOCH3 Properties and Reactions of Aromatic Compounds Benzene is a colorless, flammable liquid obtained chiefly from petroleum and coal tar. Perhaps the most remarkable chemical property of benzene is its relative inert- ness. Although it has the same empirical formula as acetylene (CH) and a high degree of unsaturation, it is much less reactive than either ethylene or acetylene. The stability of benzene is the result of electron delocalization. In fact, benzene can be hydrogenated, but only with difficulty. The following reaction is carried out at significantly higher temperatures and pressures than are similar reactions for the alkenes: H H H A H GD H H H G DH H E Pt HO O  3H2 8888n E HH catalyst HO OH G H D A H D G H H H H cyclohexane We saw earlier that alkenes react readily with halogens to form addition prod- ucts, because the pi bond in C“C can be broken easily. The most common reac- tion of halogens with benzene is the substitution reaction, in which an atom or  24.3 Aromatic Hydrocarbons 1041 group of atoms replaces an atom or group of atoms in another molecule. For example, H Br A A H EH H H FeBr3 H EH  Br2 8888n catalyst  HBr E HH E HH H H A A H H bromobenzene Note that if the reaction were an addition reaction, electron delocalization would be destroyed in the product H A Br HH D OH OBr HE A G H H and the molecule would not have the aromatic characteristic of chemical unreactivity. Alkyl groups can be introduced into the ring system by allowing benzene to react with an alkyl halide using AlCl3 as the catalyst: CH2CH3 A AlCl3  CH3CH2Cl 8888n catalyst  HCl ethyl chloride ethylbenzene An enormously large number of compounds can be generated from substances in which benzene rings are fused together. Some of these polycyclic aromatic hydrocar- bons are shown in Figure 24.7. The best known of these compounds is naphthalene, which is used in mothballs. These and many other similar compounds are present in coal tar. Some of the compounds with several rings are powerful carcinogens—they can cause cancer in humans and other animals. Figure 24.7 Some polycyclic aromatic hydrocarbons. Compounds denoted by * are potent carcinogens. An enormous number of such compounds exist in nature. Naphthalene Anthracene Phenanthrene Naphthacene Benz(a)anthracene* Dibenz(a,h)anthracene* Benzo(a)pyrene 1042 Chapter 24 ■ Organic Chemistry 24.4 Chemistry of the Functional Groups We now examine in greater depth some organic functional groups, groups that are responsible for most of the reactions of the parent compounds. In particular, we focus on oxygen-containing and nitrogen-containing compounds. Alcohols All alcohols contain the hydroxyl functional group, ¬OH. Some common alcohols are shown in Figure 24.8. Ethyl alcohol, or ethanol, is by far the best known. It is produced biologically by the fermentation of sugar or starch. In the absence of oxygen, the enzymes present in bacterial cultures or yeast catalyze the reaction enzymes C6H12O6 (aq) ¬¬¡ 2CH3CH2OH(aq) 1 2CO2 (g) ethanol C2H5OH This process gives off energy, which microorganisms, in turn, use for growth and other functions. Commercially, ethanol is prepared by an addition reaction in which water is com- bined with ethylene at about 280°C and 300 atm: 2 4 H SO CH2 “ CH2 (g) 1 H2O(g) ¬¡ CH3CH2OH(g) Ethanol has countless applications as a solvent for organic chemicals and as a start- ing compound for the manufacture of dyes, synthetic drugs, cosmetics, and explo- sives. It is also a constituent of alcoholic beverages. Ethanol is the only nontoxic (more properly, the least toxic) of the straight-chain alcohols; our bodies produce an enzyme, called alcohol dehydrogenase, which helps metabolize ethanol by oxidizing it to acetaldehyde: alcohol dehydrogenase CH3CH2OH ¬¬¬¬¬¬¬¡ CH3CHO 1 H2 acetaldehyde This equation is a simplified version of what actually takes place; the H atoms are taken up by other molecules, so that no H2 gas is evolved. See Chemistry in Action on p. 144. Ethanol can also be oxidized by inorganic oxidizing agents, such as acidified dichromate, to acetaldehyde and acetic acid: Cr2O22 7 Cr2O22 7 CH3CH2OH ¬¡ H1 CH3CHO ¬¡ H1 CH3COOH Figure 24.8 Common alcohols. Note that all the compounds H H H H H H A A A A A A contain the OH group. The HOCOOH HOCOC OOH HOC OCOC OH properties of phenol are quite A A A A A A different from those of the H H H H OH H aliphatic alcohols. Methanol Ethanol 2-Propanol (methyl alcohol) (ethyl alcohol) (isopropyl alcohol) OH H H A A H O CO CO H A A OH OH Phenol Ethylene glycol  24.4 Chemistry of the Functional Groups 1043 Ethanol is called an aliphatic alcohol because it is derived from an alkane (ethane). The simplest aliphatic alcohol is methanol, CH3OH. Called wood alcohol, it was prepared at one time by the dry distillation of wood. It is now synthesized industrially by the reaction of carbon monoxide and molecular hydrogen at high temperatures and pressures: Fe2O3 CO(g) 1 2H2 (g) ¬¡ catalyst CH3OH(l) methanol Methanol is highly toxic. Ingestion of only a few milliliters can cause nausea and blindness. Ethanol intended for industrial use is often mixed with methanol to prevent people from drinking it. Ethanol containing methanol or other toxic substances is called denatured alcohol. The alcohols are very weakly acidic; they do not react with strong bases, such as NaOH. The alkali metals react with alcohols to produce hydrogen: 2CH3OH 1 2Na ¡ 2CH3ONa 1 H2 sodium methoxide However, the reaction is much less violent than that between Na and water: 2H2O 1 2Na ¡ 2NaOH 1 H2 Two other familiar aliphatic alcohols are 2-propanol (or isopropanol), commonly known as rubbing alcohol, and ethylene glycol, which is used as an antifreeze. Note that ethylene glycol has two ¬OH groups and so can form hydrogen bonds with Alcohols react more slowly with water molecules more effectively than compounds that have only one ¬OH group sodium metal than water does. (see Figure 24.8). Most alcohols—especially those with low molar masses—are highly flammable. Ethers Ethers contain the R¬O¬R9 linkage, where R and R9 are a hydrocarbon (aliphatic or aromatic) group. They are formed by the reaction between an alkoxide (containing the RO2 ion) and an alkyl halide: NaOCH3 1 CH3Br ¡ CH3OCH3 1 NaBr sodium methoxide methyl bromide dimethyl ether Diethyl ether is prepared on an industrial scale by heating ethanol with sulfuric acid at 140°C CH3OCH3 C2H5OH 1 C2H5OH ¡ C2H5OC2H5 1 H2O This reaction is an example of a condensation reaction, which is characterized by the joining of two molecules and the elimination of a small molecule, usually water. Like alcohols, ethers are extremely flammable. When left standing in air, they have a tendency to slowly form explosive peroxides: CH3 A C2H5OC2H5  O2 88n C2H5OOCOOOOOH A H diethyl ether 1-ethyoxyethyl hydroperoxide 1044 Chapter 24 ■ Organic Chemistry Peroxides contain the ¬O¬O¬ linkage; the simplest peroxide is hydrogen perox- ide, H2O2. Diethyl ether, commonly known as “ether,” was used as an anesthetic for many years. It produces unconsciousness by depressing the activity of the central nervous system. The major disadvantages of diethyl ether are its irritating effects on the respiratory system and the occurrence of postanesthetic nausea and vomiting. “Neothyl,” or methyl propyl ether, CH3OCH2CH2CH3, is currently favored as an anes- thetic because it is relatively free of side effects. Aldehydes and Ketones Under mild oxidation conditions, it is possible to convert alcohols to aldehydes and ketones: CH3OH  12 O2 88n H2CPO  H2O formaldehyde H3C G C2H5OH  12 O2 88n CPO  H2O D H acetaldehyde CH3CHO H H3C A G CH3OCOCH3  2 O2 88n 1 CPO  H2O A D OH H3C acetone The functional group in these compounds is the carbonyl group, ≈ √ C“O. In an aldehyde at least one hydrogen atom is bonded to the carbon in the carbonyl group. In a ketone, the carbon atom in the carbonyl group is bonded to two hydrocarbon groups. The simplest aldehyde, formaldehyde (H2C“O) has a tendency to polymerize; that is, the individual molecules join together to form a compound of high molar mass. This action gives off much heat and is often explosive, so formaldehyde is usually prepared and stored in aqueous solution (to reduce the concentration). This rather disagreeable-smelling liquid is used as a starting material in the polymer industry (see Chapter 25) and in the laboratory as a preservative for animal specimens. Interestingly, the higher molar mass aldehydes, such as cinnamic aldehyde Cinnamic aldehyde gives cinnamon its H characteristic aroma. D OCHPCHOC M O have a pleasant odor and are used in the manufacture of perfumes. Ketones generally are less reactive than aldehydes. The simplest ketone is ace- tone, a pleasant-smelling liquid that is used mainly as a solvent for organic compounds and nail polish remover. Carboxylic Acids Under appropriate conditions both alcohols and aldehydes can be oxidized to carbox- ylic acids, acids that contain the carboxyl group, ¬COOH: CH3CH2OH 1 O2 ¡ CH3COOH 1 H2O CH3COOH CH3CHO 1 12O2 ¡ CH3COOH  24.4 Chemistry of the Functional Groups 1045 Figure 24.9 Some common carboxylic acids. Note that they O H O H H H O O B A B A A A B B all contain the COOH group. HOCOOH HOC OCOOH HO COC OC OC OOH COOH (Glycine is one of the amino A A A A acids found in proteins.) H H H H Formic acid Acetic acid Butyric acid Benzoic acid O O H OH H O H H O B B A A A B A A B C OOH HOO COC OC OC OCOOH NO COC OOH A A A A A A C OOH H C H H H B J G O O OH Glycine Oxalic acid Citric acid These reactions occur so readily, in fact, that wine must be protected from atmospheric The oxidization of ethanol to acetic acid in wine is catalyzed by enzymes. oxygen while in storage. Otherwise, it would soon turn to vinegar due to the formation of acetic acid. Figure 24.9 shows the structure of some of the common carboxylic acids. Carboxylic acids are widely distributed in nature; they are found in both the plant and animal kingdoms. All protein molecules are made of amino acids, a special kind of carboxylic acid containing an amino group (¬NH2) and a carboxyl group (¬COOH). Unlike the inorganic acids HCl, HNO3, and H2SO4, carboxylic acids are usually weak. They react with alcohols to form pleasant-smelling esters: O B CH3COOH  HOCH2CH3 88n CH3OCOOOCH2CH3  H2O This is a condensation reaction. acetic acid ethanol ethyl acetate Other common reactions of carboxylic acids are neutralization CH3COOH 1 NaOH ¡ CH3COONa 1 H2O and formation of acid halides, such as acetyl chloride CH3COOH 1 PCl5 ¡ CH3COCl 1 HCl 1 POCl3 acetyl phosphoryl chloride chloride Acid halides are reactive compounds used as intermediates in the preparation of many other organic compounds. They hydrolyze in much the same way as many nonmetal- lic halides, such as SiCl4: CH3COCl(l) 1 H2O(l) ¡ CH3COOH(aq) 1 HCl(g) SiCl4(l) 1 3H2O(l) ¡ H2SiO3(s) 1 4HCl(g) silicic acid Esters Esters have the general formula R9COOR, where R9 can be H or a hydrocarbon group and R is a hydrocarbon group. Esters are used in the manufacture of perfumes and as flavoring agents in the confectionery and soft-drink industries. Many fruits owe their characteristic smell and flavor to the presence of small quantities of esters. For example, bananas contain 3-methylbutyl acetate [CH3COOCH2CH2CH(CH3)2], oranges contain octyl acetate (CH3COOCHCH3C6H13), and apples contain methyl The odor of fruits is mainly due to butyrate (CH3CH2CH2COOCH3). the ester compounds they contain. 1046 Chapter 24 ■ Organic Chemistry The functional group in esters is the ¬COOR group. In the presence of an acid catalyst, such as HCl, esters undergo hydrolysis to yield a carboxylic acid and an alcohol. For example, in acid solution, ethyl acetate hydrolyzes as follows: CH3COOC2H5 1 H2O Δ CH3COOH 1 C2H5OH ethyl acetate acetic acid ethanol However, this reaction does not go to completion because the reverse reaction, that is, the formation of an ester from an alcohol and an acid, also occurs to an appre- ciable extent. On the other hand, when NaOH solution is used in hydrolysis the sodium acetate does not react with ethanol, so this reaction does go to completion from left to right: CH3COOC2H5 1 NaOH ¡ CH3COO2 Na1 1 C2H5OH ethyl acetate sodium acetate ethanol For this reason, ester hydrolysis is usually carried out in basic solutions. Note that NaOH does not act as a catalyst; rather, it is consumed by the reaction. The term The action of soap is discussed on p. 548. saponification (meaning soapmaking) was originally used to describe the alkaline hydrolysis of fatty acid esters to yield soap molecules (sodium stearate): C17H35COOC2H5 1 NaOH ¡ C17H35COO2 Na1 1 C2H5OH ethyl stearate sodium stearate Saponification has now become a general term for alkaline hydrolysis of any type of ester. Amines Amines are organic bases having the general formula R3N, where R may be H or a hydrocarbon group. As with ammonia, the reaction of amines with water is RNH2 1 H2O ¡ RNH13 1 OH2 where R represents a hydrocarbon group. Like all bases, the amines form salts when allowed to react with acids: CH3CH2NH2 1 HCl ¡ CH3CH2NH1 3 Cl 2 ethylamine ethylammonium CH3NH2 chloride These salts are usually colorless, odorless solids. Aromatic amines are used mainly in the manufacture of dyes. Aniline, the sim- plest aromatic amine, itself is a toxic compound; a number of other aromatic amines such as 2-naphthylamine and benzidine are potent carcinogens: NH2 A ENH2 H2NO OO ONH2 aniline 2-naphthylamine benzidine Summary of Functional Groups Table 24.4 summarizes the common functional groups, including the C“C and C‚C groups. Organic compounds commonly contain more than one functional group. Gen- erally, the reactivity of a compound is determined by the number and types of func- tional groups in its makeup. Example 24.5 shows how we can use the functional groups to predict reactions.  24.4 Chemistry of the Functional Groups 1047 Table 24.4 Important Functional Groups and Their Reactions Functional Group Name Typical Reactions G D Carbon-carbon Addition reactions with halogens, hydrogen CPC double bond halides, and water; hydrogenation to yield alkanes D G OCqCO Carbon-carbon Addition reactions with halogens, hydrogen triple bond halides; hydrogenation to yield alkenes and alkanes OXOS Halogen Exchange reactions: Q CH3CH2Br 1 KI ¡ CH3CH2I 1 KBr (X  F, Cl, Br, I) O OOOH Hydroxyl Esterification (formation of an ester) with carboxylic acids; Q oxidation to aldehydes, ketones, and carboxylic acids G O CPO Carbonyl Reduction to yield alcohols; oxidation of aldehydes to D Q yield carboxylic acids SOS B O OCOOOH Carboxyl Esterification with alcohols; reaction with phosphorus Q pentachloride to yield acid chlorides SOS B OCOOORO Ester Hydrolysis to yield acids and alcohols Q (R  hydrocarbon) R D ONO Amine Formation of ammonium salts with acids G R (R  H or hydrocarbon) Example 24.5 Cholesterol is a major component of gallstones, and it is believed that the cholesterol level in the blood is a contributing factor in certain types of heart disease. From the following structure of the compound, predict its reaction with (a) Br2, (b) H2 (in the presence of a Pt catalyst), (c) CH3COOH. C8H17 An artery becoming blocked by CH3 cholesterol. A CH3 A E HO Strategy To predict the type of reactions a molecule may undergo, we must first identify the functional groups present (see Table 24.4). Solution There are two functional groups in cholesterol: the hydroxyl group and the carbon-carbon double bond. (a) The reaction with bromine results in the addition of bromine to the double-bonded carbons, which become single-bonded. (Continued) CHEMISTRY in Action The Petroleum Industry I n 2010 an estimated 40 percent of the energy needs of the United States were supplied by oil or petroleum. The rest was provided by natural gas (approximately 25 percent), coal Gas (23 percent), hydroelectric power (4 percent), nuclear power (8 percent), and other sources (0.5 percent). In addition to en- ergy, petroleum is the source of numerous organic chemicals used to manufacture drugs, clothing, and many other products. Gasoline Unrefined petroleum, a viscous, dark-brown liquid, is often 30°C–180°C called crude oil. A complex mixture of alkanes, alkenes, cycloal- kanes, and aromatic compounds, petroleum was formed in Naphtha Earth’s crust over the course of millions of years by the anaerobic 110°C–195°C decomposition of animal and vegetable matter by bacteria. Petroleum deposits are widely distributed throughout the world, but they are found mainly in North America, Mexico, Kerosene Russia, China, Venezuela, and, of course, the Middle East. The 170°C–290°C actual composition of petroleum varies with location. In the United States, for example, Pennsylvania crude oils are mostly aliphatic hydrocarbons, whereas the major components of west- Heating oil ern crude oils are aromatic in nature. 260°C–350°C Although petroleum contains literally thousands of hydro- carbon compounds, we can classify its components according to the range of their boiling points. These hydrocarbons can be Lubricating oil 300°C–370°C Heated crude oil at 370°C Residue A fractional distillation column for separating the components of petroleum crude oil. As the hot vapor moves upward, it condenses and the various components of the crude oil are separated according to their boiling points and are drawn off as shown. Crude oil. Major Fractions of Petroleum Fraction Carbon Atoms* Boiling Point Range (8C) Uses Natural gas C1–C4 2161 to 20 Fuel and cooking gas Petroleum ether C5–C6 30–60 Solvent for organic compounds Ligroin C7 20–135 Solvent for organic compounds Gasoline C6–C12 30–180 Automobile fuels Kerosene C11–C16 170–290 Rocket and jet engine fuels, domestic heating Heating fuel oil C14–C18 260–350 Domestic heating and fuel for electricity production Lubricating oil C15–C24 300–370 Lubricants for automobiles and machines *The entries in this column indicate the numbers of carbon atoms in the compounds involved. For example, C1–C4 tells us that in natural gas the compounds contain 1 to 4 carbon atoms, and so on. 1048  Intake valve open Spark plug fires Exhaust valve open (a) ( b) (c) (d) The four stages of operation of an internal combustion engine. This is the type of engine used in practically all automobiles and is described technically as a four-stroke Otto cycle engine. (a) The intake valve opens to let in a gasoline-air mixture. (b) During the compression stage the two valves are closed. (c) The spark plug fires and the piston is pushed outward. (d) Finally, as the piston is pushed downward, the exhaust valve opens to let out the exhaust gas. separated on the basis of molar mass by fractional distillation. rather than a smooth, strong push. This action produces a Heating crude oil to about 400°C converts the viscous oil into “knocking” or “pinging” sound, as well as a decrease in effi- hot vapor and fluid. In this form it enters the fractionating ciency in the conversion of combustion energy to mechanical tower. The vapor rises and condenses on various collecting energy. It turns out that straight-chain hydrocarbons have the trays according to the temperatures at which the various com- greatest tendency to produce knocking, whereas the branched- ponents of the vapor liquefy. Some gases are drawn off at the chain and aromatic hydrocarbons give the desired smooth push. top of the column, and the unvaporized residual oil is collected Gasolines are usually rated according to the octane number, at the bottom. a measure of their tendency to cause knocking. On this scale, a Gasoline is probably the best-known petroleum product. A branched C8 compound (2,2,4-trimethylpentane, or isooctane) mixture of volatile hydrocarbons, gasoline contains mostly has been arbitrarily assigned an octane number of 100, and that alkanes, cycloalkanes, and a few aromatic hydrocarbons. Some of n-heptane, a straight-chain compound, is zero. The higher the of these compounds are far more suitable for fueling an auto- octane number of the hydrocarbon, the better its performance in mobile engine than others, and herein lies the problem of the the internal combustion engine. Aromatic hydrocarbons such as further treatment and refinement of gasoline. benzene and toluene have high octane numbers (106 and 120, Most automobiles employ the four-stroke operation of respectively), as do aliphatic hydrocarbons with branched chains. the Otto cycle engine. A major engineering concern is to con- The octane rating of hydrocarbons can be improved by the trol the burning of the gasoline-air mixture inside each cylin- addition of small quantities of compounds called antiknocking der to obtain a smooth expansion of the gas mixture. If the mixture burns too rapidly, the piston receives a hard jerk (Continued) 1049  CHEMISTRY in Action (Continued) agents. Among the most widely used antiknocking agents are discharge of automobile exhaust into the atmosphere has the following: become a serious environmental problem. Federal regulations require that all automobiles made after 1974 use “unleaded” CH3 gasolines. The catalytic converters with which late-model A automobiles are equipped can be “poisoned” by lead, another CH3 CH2 reason for its exclusion from gasoline. To minimize knocking, A A CH3OPbOCH3 CH3OCH2OPbOCH2OCH3 unleaded gasolines contain methyl tert-butyl ether (MTBE), A A which minimizes knocking and increases the oxygen content CH3 CH2 of gasoline, making the fuel burn cleaner. Unfortunately, in A CH3 the late 1990s MTBE was found in drinking water supplies, tetramethyllead tetraethyllead primarily because of leaking gasoline storage tanks. The sub- stance makes water smell and taste foul and is a possible hu- The addition of 2 to 4 g of either of these compounds to man carcinogen. At this writing, some states have begun to a gallon of gasoline increases the octane rating by 10 or phase out the use of MTBE in gasoline, although no suitable more. However, lead is a highly toxic metal, and the constant substitute has been found. (b) This is a hydrogenation reaction. Again, the carbon-carbon double bond is converted to a carbon-carbon single bond. (c) The acid reacts with the hydroxyl group to form an ester and water. Figure 24.10 shows the products of these reactions. Figure 24.10 The products CH3 C8H17 CH3 C8H17 CH3 C8H17 formed by the reaction of cholesterol with (a) molecular bromine, (b) molecular hydrogen, CH3 CH3 CH3 and (c) acetic acid. O B HO Br HO H3CO COO Br Similar problem: 24.41. (a) (b) (c) Practice Exercise Predict the products of the following reaction: CH3OH 1 CH3CH2COOH ¡ ? The Chemistry in Action essay on p. 1048 shows the key organic compounds present in petroleum. Summary of Facts & Concepts 1. Because carbon atoms can link up with other carbon 3. Methane, CH4, is the simplest of the alkanes, a family atoms in straight and branched chains, carbon can form of hydrocarbons with the general formula CnH2n12. more compounds than any other element. Cyclopropane, C3H6, is the simplest of the cycloal- 2. Organic compounds are derived from two types of kanes, a family of alkanes whose carbon atoms are hydrocarbons: aliphatic hydrocarbons and aromatic joined in a ring. Alkanes and cycloalkanes are saturated hydrocarbons. hydrocarbons. 1050  Questions & Problems 1051 4. Ethylene, CH2 “CH2, is the simplest of the olefins, or 6. Compounds that contain one or more benzene rings alkenes, a class of hydrocarbons containing carbon- are called aromatic hydrocarbons. These compounds carbon double bonds and having the general formula undergo substitution by halogens and alkyl groups. CnH2n. 7. Functional groups impart specific types of chemical 5. Acetylene, CH‚CH, is the simplest of the alkynes, reactivity to molecules. Classes of compounds characterized which are compounds that have the general formula by their functional groups include alcohols, ethers, alde- CnH2n22 and contain carbon-carbon triple bonds. hydes and ketones, carboxylic acids and esters, and amines. Key Words Addition reactions, p. 1035 Alkyne, p. 1037 Ester, p. 1045 Saponification, p. 1046 Alcohol, p. 1042 Amine, p. 1046 Ether, p. 1043 Saturated hydrocarbon, p. 1027 Aldehyde, p. 1044 Aromatic hydrocarbon, p. 1026 Functional group, p. 1026 Structural isomer, p. 1027 Aliphatic hydrocarbon, p. 1026 Carboxylic acid, p. 1044 Hydrocarbon, p. 1026 Substitution reaction, p. 1040 Alkane, p. 1027 Condensation reaction, p. 1043 Ketone, p. 1044 Unsaturated Alkene, p. 1033 Cycloalkane, p. 1033 Organic chemistry, p. 1026 hydrocarbon, p. 1035 Questions & Problems • Problems available in Connect Plus 24.12 How many distinct chloropentanes, C5H11Cl, Red numbered problems solved in Student Solutions Manual could be produced in the direct chlorination of n-pentane, CH3(CH2)3CH3? Draw the structure of Classes of Organic Compounds each molecule. Review Questions • 24.13 Draw all possible isomers for the molecule C4H8. 24.1 Explain why carbon is able to form so many more • 24.14 Draw all possible isomers for the molecule C3H5Br. compounds than any other element. 24.15 The structural isomers of pentane, C5H12, have quite 24.2 What is the difference between aliphatic and aromatic different boiling points (see Example 24.1). Explain hydrocarbons? the observed variation in boiling point, in terms of Aliphatic Hydrocarbons structure. Review Questions 24.16 Discuss how you can determine which of the fol- lowing compounds might be alkanes, cycloal- 24.3 What do “saturated” and “unsaturated” mean when kanes, alkenes, or alkynes, without drawing their applied to hydrocarbons? Give examples of a satu- formulas: (a) C6H12, (b) C4H6, (c) C5H12, (d) C7H14, rated hydrocarbon and an unsaturated hydrocarbon. (e) C3H4. 24.4 Give three sources of methane. • 24.17 Draw the structures of cis-2-butene and trans-2-butene. 24.5 Alkenes exhibit geometric isomerism because rota- Which of the two compounds would have the higher tion about the C“C bond is restricted. Explain. heat of hydrogenation? Explain. 24.6 Why is it that alkanes and alkynes, unlike alkenes, 24.18 Would you expect cyclobutadiene to be a stable have no geometric isomers? molecule? Explain. 24.7 What is Markovnikov’s rule? H 24.8 Describe reactions that are characteristic of alkanes, H EH alkenes, and alkynes. COC B B 24.9 What factor determines whether a carbon atom in a compound is chiral? ECOCH H H 24.10 Give examples of a chiral substituted alkane and an achiral substituted alkane. 24.19 How many different isomers can be derived from ethylene if two hydrogen atoms are replaced by Problems a fluorine atom and a chlorine atom? Draw their • 24.11 Draw all possible structural isomers for the following structures and name them. Indicate which are struc- alkane: C7H16. tural isomers and which are geometric isomers.  1052 Chapter 24 ■ Organic Chemistry 24.20 Suggest two chemical tests that would help you dis- • 24.27 Write structural formulas for the following organic tinguish between these two compounds: compounds: (a) 3-methylhexane, (b) 1,3,5-trichloro- (a) CH3CH2CH2CH2CH3 cyclohexane, (c) 2,3-dimethylpentane, (d) 2-bromo- 4-phenylpentane, (e) 3,4,5-trimethyloctane. (b) CH3CH2CH2CH“CH2 24.28 Write structural formulas for the following com- • 24.21 Sulfuric acid (H2SO4) adds to the double bond of pounds: (a) trans-2-pentene, (b) 2-ethyl-1-butene, alkenes as H1 and 2OSO3H. Predict the products (c) 4-ethyl-trans-2-heptene, (d) 3-phenyl-butyne. when sulfuric acid reacts with (a) ethylene and (b) propene. 24.22 Acetylene is an unstable compound. It has a ten- Aromatic Hydrocarbons dency to form benzene as follows: Review Questions 3C2H2 (g) ¡ C6H6 (l) 24.29 Comment on the extra stability of benzene compared to ethylene. Why does ethylene undergo addition re- Calculate the standard enthalpy change in kilojoules actions while benzene usually undergoes substitution per mole for this reaction at 25°C. reactions? • 24.23 Predict products when HBr is added to (a) 1-butene 24.30 Benzene and cyclohexane molecules both contain and (b) 2-butene. six-membered rings. Benzene is a planar molecule, 24.24 Geometric isomers are not restricted to com- and cyclohexane is nonplanar. Explain. pounds containing the C“C bond. For example, certain disubstituted cycloalkanes can exist in the Problems cis and the trans forms. Label the following mol- ecules as the cis and trans isomer, of the same • 24.31 Write structures for the following compounds: compound: (a) 1-bromo-3-methylbenzene, (b) 1-chloro-2-propyl- benzene, (c) 1,2,4,5-tetramethylbenzene. H H • 24.32 Name the following compounds: A A A A Cl NO2 (a) H (b) H A A Cl Cl Cl H A A A A (a) (b) A A A A H H H Cl H H A Cl A CH2 CH3 CH3 NO2 • 24.25 Which of the following amino acids are chiral: CH3 (a) CH3CH(NH2)COOH, (b) CH2(NH2)COOH, A CH (c) CH2(OH)CH(NH2)COOH? E 3 (c) • 24.26 Name the following compounds: E H3 C A CH 3 CH3 A (a) CH 3 OCHOCH 2 OCH 2 OCH 3 C 2 H 5 CH 3 CH 3 Chemistry of the Functional Groups A A A (b) CH 3 OCHOOCHOCHOCH 3 Review Questions (c) CH 3 OCH 2 OCHOCH 2 OCH 3 24.33 What are functional groups? Why is it logical and A useful to classify organic compounds according to CH 2 OCH 2 OCH 3 their functional groups? CH 3 • 24.34 Draw the Lewis structure for each of the following A functional groups: alcohol, ether, aldehyde, ketone, (d) CH 2 PCHOCHOCHPCH 2 carboxylic acid, ester, amine. (e) CH 3 OC O O COCH 2 OCH 3 O Problems • 24.35 Draw structures for molecules with the follow- A ing formulas: (a) CH4O, (b) C2H6O, (c) C3H6O2, (f ) CH 3 OCH 2 OCHOCHPCH 2 (d) C3H8O.  Questions & Problems 1053 • 24.36 Classify each of the following molecules as alco- Additional Problems hol, aldehyde, ketone, carboxylic acid, amine, or • 24.43 Draw all the possible structural isomers for the mol- ether: ecule having the formula C7H7Cl. The molecule (a) CH3¬O¬CH2¬CH3 contains one benzene ring. (b) CH3¬CH2¬NH2 • 24.44 Given these data O J C2H4 (g) 1 3O2 (g) ¡ 2CO2 (g) 1 2H2O(l) (c) CH3 OCH2 OC ¢H° 5 21411 kJ/mol G H 2C2H2 (g) 1 5O2 (g) ¡ 4CO2 (g) 1 2H2O(l) (d) CH3 OCOCH2 OCH3 ¢H° 5 22599 kJ/mol B O H2 (g) 1 12O2 (g) ¡ H2O(l) O ¢H° 5 2285.8 kJ/mol B (e) HOCOOH calculate the heat of hydrogenation for acetylene: (f) CH3¬CH2¬CH2¬OH C2H2 (g) 1 H2 (g) ¡ C2H4 (g) NH2 O A B • 24.45 State which member of each of the following pairs (g) OCH2OCOOCOOH of compounds is the more reactive and explain why: A (a) propane and cyclopropane, (b) ethylene and H methane, (c) acetaldehyde and acetone. 24.37 Generally aldehydes are more susceptible to oxi- 24.46 State which of the following types of compounds can dation in air than are ketones. Use acetaldehyde form hydrogen bonds with water molecules: (a) car- and acetone as examples and show why ketones boxylic acids, (b) alkenes, (c) ethers, (d) aldehydes, such as acetone are more stable than aldehydes in (e) alkanes, (f) amines. this respect. • 24.47 An organic compound is found to contain 37.5 per- • 24.38 Complete the following equation and identify the cent carbon, 3.2 percent hydrogen, and 59.3 percent products: fluorine by mass. The following pressure and vol- ume data were obtained for 1.00 g of this substance HCOOH 1 CH3OH ¡ at 90°C: • 24.39 A compound has the empirical formula C5H12O. Upon controlled oxidation, it is converted into a P (atm) V (L) compound of empirical formula C5H10O, which be- haves as a ketone. Draw possible structures for the 2.00 0.332 original compound and the final compound. 1.50 0.409 • 24.40 A compound having the molecular formula C4H10O 1.00 0.564 does not react with sodium metal. In the presence of 0.50 1.028 light, the compound reacts with Cl2 to form three compounds having the formula C4H9OCl. Draw a structure for the original compound that is consistent The molecule is known to have no dipole moment. with this information. (a) What is the empirical formula of this sub- • 24.41 Predict the product or products of each of the fol- stance? (b) Does this substance behave as an ideal lowing reactions: gas? (c) What is its molecular formula? (d) Draw the Lewis structure of this molecule and describe (a) CH3CH2OH 1 HCOOH ¡ its geometry. (e) What is the systematic name of (b) HOCO OCOCH3  H2 88n this compound? (c) C 2 H 5 H 24.48 State at least one commercial use for each of the fol- G D CPC  HBr 888n lowing compounds: (a) 2-propanol (isopropanol), D G (b) acetic acid, (c) naphthalene, (d) methanol, (e) etha- H H nol, (f) ethylene glycol, (g) methane, (h) ethylene. 24.42 Identify the functional groups in each of the following molecules: • 24.49 How many liters of air (78 percent N2, 22 percent O2 by volume) at 20°C and 1.00 atm are needed for (a) CH3CH2COCH2CH2CH3 the complete combustion of 1.0 L of octane, C8H18, (b) CH3COOC2H5 a typical gasoline component that has a density of (c) CH3CH2OCH2CH2CH2CH3 0.70 g/mL?  1054 Chapter 24 ■ Organic Chemistry 24.50 How many carbon-carbon sigma bonds are pres- • 24.59 Draw structures for the following compounds: ent in each of the following molecules? (a) cyclopentane, (b) cis-2-butene, (c) 2-hexanol, (a) 2-butyne, (b) anthracene (see Figure 24.7), (d) 1,4-dibromobenzene, (e) 2-butyne. (c) 2,3-dimethylpentane 24.60 Name the classes to which the following compounds • 24.51 How many carbon-carbon sigma bonds are present belong: in each of the following molecules? (a) benzene, (a) C4H9OH (b) cyclobutane, (c) 3-ethyl-2-methylpentane (b) CH3OC2H5 • 24.52 The combustion of 20.63 mg of compound Y, which (c) C2H5CHO contains only C, H, and O, with excess oxygen gave (d) C6H5COOH 57.94 mg of CO2 and 11.85 mg of H2O. (a) Calcu- late how many milligrams of C, H, and O were pres- (e) CH3NH2 ent in the original sample of Y. (b) Derive the 24.61 Ethanol, C2H5OH, and dimethyl ether, CH3OCH3, empirical formula of Y. (c) Suggest a plausible are structural isomers. Compare their melting points, structure for Y if the empirical formula is the same boiling points, and solubilities in water. as the molecular formula. 24.62 Amines are Brønsted bases. The unpleasant smell • 24.53 Draw all the structural isomers of compounds with of fish is due to the presence of certain amines. the formula C4H8Cl2. Indicate which isomers are Explain why cooks often add lemon juice to sup- chiral and give them systematic names. press the odor of fish (in addition to enhancing 24.54 The combustion of 3.795 mg of liquid B, which con- the flavor). tains only C, H, and O, with excess oxygen gave 24.63 You are given two bottles, each containing a colorless 9.708 mg of CO2 and 3.969 mg of H2O. In a molar liquid. You are told that one liquid is cyclohexane and mass determination, 0.205 g of B vaporized at the other is benzene. Suggest one chemical test that 1.00 atm and 200.0°C and occupied a volume of would allow you to distinguish between these two 89.8 mL. Derive the empirical formula, molar mass, liquids. and molecular formula of B and draw three plausible 24.64 Give the chemical names of the following organic structures. compounds and write their formulas: marsh gas, grain 24.55 Beginning with 3-methyl-1-butyne, show how you alcohol, wood alcohol, rubbing alcohol, antifreeze, would prepare the following compounds: mothballs, chief ingredient of vinegar. 24.65 The compound CH3¬C‚C¬CH3 is hydrogenated Br CH 3 A A to an alkene using platinum as the catalyst. Predict (a) CH 2 PCOCHOCH 3 whether the product is the pure trans isomer, the pure cis isomer, or a mixture of cis and trans iso- CH 3 mers. Based on your prediction, comment on the A mechanism of the heterogeneous catalysis. (b) CH 2 BrOCBr 2 OCHOCH 3 • 24.66 How many asymmetric carbon atoms are present in each of the following compounds? Br CH 3 A A H H H (c) CH 3 OCHOCHOCH 3 A A A (a) HOCOCOCOCl • 24.56 Indicate the asymmetric carbon atoms in the follow- A A A ing compounds: H Cl H CH 3 O OH CH 3 A B A A (a) CH 3 OCH 2 OCHOCHOCONH 2 (b) H 3COCOOCOCH 2OH A A A NH 2 H H H CH 2OH A A A H C O OH (b) Br A A A H H H A A (c) C OH H C A A A A A A H Br HO C C H • 24.57 Suppose benzene contained three distinct single A A bonds and three distinct double bonds. How many H OH different isomers would there be for dichloro- • 24.67 Isopropanol is prepared by reacting propylene benzene (C 6H 4Cl 2)? Draw all your proposed (CH3CHCH2) with sulfuric acid, followed by treat- structures. ment with water. (a) Show the sequence of steps 24.58 Write the structural formula of an aldehyde that is a leading to the product. What is the role of sulfuric structural isomer of acetone. acid? (b) Draw the structure of an alcohol that is an  Answers to Practice Exercises 1055 isomer of isopropanol. (c) Is isopropanol a chiral where R, R9, and R– represent long hydrocarbon molecule? (d) What property of isopropanol makes chains. (a) Suggest a reaction that leads to the for- it useful as a rubbing alcohol? mation of a triglyceride molecule, starting with 24.68 When a mixture of methane and bromine vapor is ex- glycerol and carboxylic acids (see p. 474 for struc- posed to light, the following reaction occurs slowly: ture of glycerol). (b) In the old days, soaps were made by hydrolyzing animal fat with lye (a sodium CH4 (g) 1 Br2 (g) ¡ CH3Br(g) 1 HBr(g) hydroxide solution). Write an equation for this re- Suggest a mechanism for this reaction. (Hint: Bro- action. (c) The difference between fats and oils mine vapor is deep red; methane is colorless.) is that at room temperature, the former are solids 24.69 Under conditions of acid catalysis, alkenes react with and the latter are liquids. Fats are usually produced water to form alcohols. As in the case with hydrogen by animals, whereas oils are commonly found in halides, the addition reaction in the formation of plants. The melting points of these substances are alcohols is also governed by Markovnikov’s rule. An determined by the number of C“C bonds (or the alkene of approximate molar mass of 42 g reacts extent of unsaturation) present—the larger the with water and sulfuric acid to produce a compound number of C“C bonds, the lower the melting point that reacts with acidic potassium dichromate solution and the more likely that the substance is a liquid. to produce a ketone. Identify all the compounds in Explain. (d) One way to convert liquid oil to solid fat the preceding steps. is to hydrogenate the oil, a process by which some or all of the C“C bonds are converted to C—C • 24.70 2-Butanone can be reduced to 2-butanol by reagents bonds. This procedure prolongs shelf life of the oil such as lithium aluminum hydride (LiAlH4). (a) Write by removing the more reactive C“C group and fa- the formula of the product. Is it chiral? (b) In reality, cilitates packaging. How would you carry out such a the product does not exhibit optical activity. Explain. process (that is, what reagents and catalyst would • 24.71 Write the structures of three alkenes that yield you employ)? (e) The degree of unsaturation of oil 2-methylbutane on hydrogenation. can be determined by reacting the oil with iodine, • 24.72 An alcohol was converted to a carboxylic acid with which reacts with the C“C bond as follows: acidic potassium dichromate. A 4.46-g sample of the acid was added to 50.0 mL of 2.27 M NaOH and the I I A A A A A A A A excess NaOH required 28.7 mL of 1.86 M HCl for O C O C P C O C O + I2 88n O CO CO CO CO neutralization. What is the molecular formula of the A A A A A A alcohol? 24.73 Write the structural formulas of the alcohols with The procedure is to add a known amount of iodine the formula C6H13O and indicate those that are chi- to the oil and allow the reaction to go to comple- ral. Show only the C atoms and the ¬OH groups. tion. The amount of excess (unreacted) iodine is determined by titrating the remaining iodine with 24.74 Fat and oil are names for the same class of compounds, a standard sodium thiosulfate (Na2S2O3) solution: called triglycerides, which contain three ester groups I2 1 2Na2S2O3 ¡ Na2S4O6 1 2NaI O B The number of grams of iodine that react with CH2OOOCOR 100 grams of oil is called the iodine number. In one A A O case, 43.8 g of I2 were treated with 35.3 g of corn oil. A B The excess iodine required 20.6 mL of a 0.142 M CHOOOCOR Na2S2O3 for neutralization. Calculate the iodine A A O number of the corn oil. A B CH2OOOCOR A fat or oil Answers to Practice Exercises 24.1 5. 24.4 No. 24.2 4,6-diethyl-2-methyloctane. 24.5 CH3CH2COOCH3 and H2O. CH 3 CH 3 C 2 H 5 CH 3 A A A A 24.3 CH 3 OCHOCH 2 OCHOCHOCHOCH 2 OCH 3 CHEMICAL M YS TERY The Disappearing Fingerprints† I n 1993, a young girl was abducted from her home and taken away in a car. Later she managed to escape from her attacker and was rescued by a local resident and safely returned home unharmed. A few days later the police arrested a suspect and recovered the car. In building the case against the man, the law officers found that they lacked some crucial evidence. The girl’s detailed description indicated that she must have been in the car, yet none of her fingerprints could be found. Fortunately, the police were able to link the girl to the car and its owner by matching fibers found in the car with those from the girl’s nightgown. What are fingerprints? Our fingertips are studded with sweat pores. When a finger touches something, the sweat from these pores is deposited on the surface, providing a mir- ror image of the ridge pattern, called a fingerprint. No two individuals have the same fingerprints. This fact makes fin- gerprint matching one of the most powerful methods for iden- tifying crime suspects. Why were the police not able to find the girl’s finger- prints in the car? The residue that is deposited by fingerprints is about 99 percent water. The other 1 percent contains oils and fatty acids, esters, amino acids, and salts. Fingerprint samples from adults contain heavy oils and long carbon chains linked together by ester groups, but children’s samples contain mostly unesterified and shorter fatty chains that are light and more volatile. (The hydrogen atoms are omitted for clarity.) O J COCOCOCOCOCOCOCOCOCOCOCOC G OH from a child’s fingerprint OOCOCOCOCOCOCOCOCOCOCOCOCOCOCOCOC D COCOCOCOCOCOCOCOCOCOCOCOCOCOCOC M O from an adult’s fingerprint In general, adult fingerprints last at least several days, but children’s fingerprints often vanish within 24 h. For this reason, in cases involving children, crime scene investigation must be done very quickly. † Adapted from “The Disappearing Fingerprints” by Deborah Noble, CHEM MATTERS, February, 1997, p. 9. Copyright 1997 American Chemical Society. 1056 Chemical Clues When a finger touches a surface, it leaves an invisible pattern of oil called a latent fingerprint. Forensic investigators must develop a latent fingerprint into a visible print that can be pho- tographed, then scanned and stored for matching purposes. The following are some of the common methods for developing latent fingerprints. 1. The dusting powder method: This is the traditional method in which fine powder (usually carbon black, which is an amorphous form of carbon obtained by the thermal decom- position of hydrocarbons) is brushed onto nonporous sur- faces. The powder sticks to the sweat, making the ridge pattern visible. An improvement on this method is the use of fluorescent powders. What are the advantages of this modification? 2. The iodine method: When heated, iodine sublimes and its vapor reacts with the carbon-carbon double bonds in fats and oils, turning the ridge pattern to a yellow-brown color. This method is particularly well suited for fingerprints on porous objects like papers and cardboard. Write an equation showing the reaction of I2 with fats and oils. 3. The ninhydrin method: This is one of the most popular methods for developing latent fingerprints on porous, absor- bent surfaces like paper and wood. This method is based on a complex reaction between ninhydrin and amino acids (see Table 25.2) in the presence of a base to produce a com- pound, which turns purple when heated. The unbalanced equation is O O O B B A OH   H3NCHCOO  OH 88n PNO OH A B R B B O O O ninhydrin amino acid Ruhemann’s purple where R is a substituent. Because the amino acids in sweat do not interact with the cel- lulose content of paper or wood, this technique enables prints that may be years old to be developed. Draw resonance structures of Ruhemann’s purple, showing the movement of electrons with curved arrows. 1057  CHAPTER 25 Synthetic and Natural Organic Polymers University of Michigan researchers have developed a faster, more efficient way to produce nanoparticle drug delivery systems, using DNA molecules to bind the particles together. CHAPTER OUTLINE A LOOK AHEAD 25.1 Properties of Polymers  We begin with a discussion of the general properties of organic polymers. (25.1) 25.2 Synthetic Organic Polymers  We then study the synthesis of organic polymers by addition reactions and 25.3 Proteins condensation reactions. We examine both natural and synthetic rubber and 25.4 Nucleic Acids other synthetic polymers. (25.2)  Next, we learn that proteins are polymers of amino acids. We examine the structure of a protein molecule in terms of its primary, secondary, tertiary, and quaternary structures. We also study the stability of a protein molecule, the cooperativity effect, and protein denaturation. (25.3)  The chapter ends with a brief discussion of the structure and composition of the genetic materials deoxyribonucleic acids (DNA) and ribonucleic acids (RNA). (25.4) 1058  25.2 Synthetic Organic Polymers 1059 P olymers are very large molecules containing hundreds or thousands of atoms. People have been using polymers since prehistoric time, and chemists have been synthesizing them for the past century. Natural polymers are the basis of all life processes, and our technological society is largely dependent on synthetic polymers. This chapter discusses some of the preparation and properties of important synthetic organic polymers in addition to two naturally occurring polymers that are vital to living systems—proteins and nucleic acids. 25.1 Properties of Polymers A polymer is a molecular compound distinguished by a high molar mass, ranging into thousands and millions of grams, and made up of many repeating units. The physical properties of these so-called macromolecules differ greatly from those of small, ordinary molecules, and special techniques are required to study them. Naturally occurring polymers include proteins, nucleic acids, cellulose (polysac- charides), and rubber (polyisoprene). Most synthetic polymers are organic compounds. Familiar examples are nylon, poly(hexamethylene adipamide); Dacron, poly(ethylene terephthalate); and Lucite or Plexiglas, poly(methyl methacrylate). The development of polymer chemistry began in the 1920s with the investiga- tion into a puzzling behavior of certain materials, including wood, gelatin, cotton, and rubber. For example, when rubber, with the known empirical formula of C5H8, was dissolved in an organic solvent, the solution displayed several unusual properties—high viscosity, low osmotic pressure, and negligible freezing-point depression. These observations strongly suggested the presence of solutes of very high molar mass, but chemists were not ready at that time to accept the idea that such giant molecules could exist. Instead, they postulated that materials such as rubber consist of aggre- gates of small molecular units, like C5H8 or C10H16, held together by intermolec- ular forces. This misconception persisted for a number of years, until Hermann Staudinger† clearly showed that these so-called aggregates are, in fact, enormously large molecules, each of which contains many thousands of atoms held together by covalent bonds. Once the structures of these macromolecules were understood, the way was open for manufacturing polymers, which now pervade almost every aspect of our daily lives. About 90 percent of today’s chemists, including biochemists, work with polymers. 25.2 Synthetic Organic Polymers Because of their size, we might expect molecules containing thousands of carbon and hydrogen atoms to form an enormous number of structural and geometric isomers (if C“C bonds are present). However, these molecules are made up of monomers, simple repeating units, and this type of composition severely restricts the number of possible isomers. Synthetic polymers are created by joining monomers together, one at a time, by means of addition reactions and condensation reactions. Addition Reactions Addition reactions involve unsaturated compounds containing double or triple bonds, Addition reactions were described on p. 1035. particularly C“C and C‚C. Hydrogenation and reactions of hydrogen halides and halogens with alkenes and alkynes are examples of addition reactions. † Hermann Staudinger (1881–1963). German chemist. One of the pioneers in polymer chemistry. Staudinger was awarded the Nobel Prize in Chemistry in 1953. 1060 Chapter 25 ■ Synthetic and Natural Organic Polymers Figure 25.1 Structure of polyethylene. Each carbon atom is sp3-hybridized. Polyethylene, a very stable polymer used in packaging wraps, is made by joining ethylene monomers via an addition-reaction mechanism. First an initiator molecule (R2) is heated to produce two radicals: R2 ¡ 2R ? The reactive radical attacks an ethylene molecule to generate a new radical: R ? 1 CH2 “CH2 ¡ R¬CH2 ¬CH2 ? which further reacts with another ethylene molecule, and so on: R¬CH2 ¬CH2 ? 1 CH2 “CH2 ¡ R¬CH2 ¬CH2 ¬CH2 ¬CH2 ? Very quickly a long chain of CH2 groups is built. Eventually, this process is ter- minated by the combination of two long-chain radicals to give the polymer called polyethylene: R ¬CH ( ) n CH2CH2 ? 1 R ¬CH 2 ¬CH2 ¬ ( 2 ¬CH2 ¬) n CH2CH2 ? ¡ R ¬CH ( 2 ¬CH ) ¬ 2 n CH 2 CH 2 ¬CH 2CH2 ¬CH ( ) nR 2 ¬CH2 ¬ where ¬ ( CH2¬CH2 ¬ ) n is a convenient shorthand convention for representing the repeating unit in the polymer. The value of n is understood to be very large, on the order of hundreds. The individual chains of polyethylene pack together well and so account for the substance’s crystalline properties (Figure 25.1). Polyethylene is mainly used in films, in frozen food packaging and other product wrappings. A specially treated type of Common mailing envelopes made polyethylene called Tyvek is used for home insulation. of Tyvek. Polyethylene is an example of a homopolymer, which is a polymer made up of only one type of monomer. Other homopolymers that are synthesized by the radical mechanism are Teflon, polytetrafluoroethylene (Figure 25.2) and poly(vinyl chloride) (PVC): OCF ( 2 OCF 2 O )n OCH ( 2 OCHO)n A Cl Teflon PVC The chemistry of polymers is more complex if the starting units are asymmetric: 冢 冣 H3C H CH3 H G D A A CPC OCOOOCO C D G A A H H H Hn propene polypropene Several geometric isomers can result from an addition reaction of propenes (Figure 25.3). If the additions occur randomly, we obtain atactic polypropenes, which do not pack Figure 25.2 A cooking utensil coated with Silverstone, which together well. These polymers are rubbery, amorphous, and relatively weak. Two other contains polytetrafluoroethylene. possibilities are an isotactic structure, in which the R groups are all on the same side  25.2 Synthetic Organic Polymers 1061 (a) (b) (c) Figure 25.3 Stereoisomers of polymers. When the R group (green sphere) is CH3, the polymer is polypropene. (a) When the R groups are all on one side of the chain, the polymer is said to be isotactic. (b) When the R groups alternate from side to side, the polymer is said to be syndiotactic. (c) When the R groups are disposed at random, the polymer is atactic. of the asymmetric carbon atoms, and a syndiotactic form, in which the R groups alternate to the left and right of the asymmetric carbons. Of these, the isotactic isomer has the highest melting point and greatest crystallinity and is endowed with superior mechanical properties. A major problem that the polymer industry faced in the beginning was how to synthesize either the isotactic or syndiotactic polymer selectively without having it contaminated by other products. The solution came from Giulio Natta† and Karl Ziegler,‡ who demonstrated that certain catalysts, including triethylaluminum [Al(C2H5)3] and titanium trichloride (TiCl3), promote the formation only of specific isomers. Using Natta-Ziegler catalysts, chemists can design polymers to suit any purpose. Rubber is probably the best known organic polymer and the only true hydrocarbon polymer found in nature. It is formed by the radical addition of the monomer isoprene. Actually, polymerization can result in either poly-cis-isoprene or poly-trans-isoprene— or a mixture of both, depending on reaction conditions: 冢 冣 冢 冣 CH3 CH3 H OCH2 H A G D G D nCH2PCOCHPCH2 88n CPC and/or CPC D G D G OCH2 CH2On CH3 CH2On isoprene poly-cis-isoprene poly-trans-isoprene Note that in the cis isomer the two CH2 groups are on the same side of the C“C bond, whereas the same groups are across from each other in the trans isomer. Nat- ural rubber is poly-cis-isoprene, which is extracted from the tree Hevea brasiliensis (Figure 25.4). † Giulio Natta (1903–1979). Italian chemist. Natta received the Nobel Prize in Chemistry in 1963 for discovering stereospecific catalysts for polymer synthesis. ‡ Karl Ziegler (1898–1976). German chemist. Ziegler shared the Nobel Prize in Chemistry in 1963 with Natta for his work in polymer synthesis. 1062 Chapter 25 ■ Synthetic and Natural Organic Polymers Figure 25.4 Latex (aqueous suspension of rubber particles) being collected from a rubber tree. An unusual and very useful property of rubber is its elasticity. Rubber will stretch up to 10 times its length and, if released, will return to its original size. In contrast, a piece of copper wire can be stretched only a small percentage of its length and still return to its original size. Unstretched rubber has no regular X-ray diffraction pattern and is therefore amorphous. Stretched rubber, however, possesses a fair amount of crystallinity and order. (a) The elastic property of rubber is due to the flexibility of its long-chain molecules. In the bulk state, however, rubber is a tangle of polymeric chains, and if the external force is strong enough, individual chains slip past one another, thereby causing the rubber to lose most of its elasticity. In 1839, Charles Goodyear† discovered that nat- ural rubber could be cross-linked with sulfur (using zinc oxide as the catalyst) to prevent chain slippage (Figure 25.5). His process, known as vulcanization, paved the way for many practical and commercial uses of rubber, such as in automobile tires and dentures. (b) During World War II a shortage of natural rubber in the United States prompted an intensive program to produce synthetic rubber. Most synthetic rubbers (called elas- tomers) are made from petroleum products such as ethylene, propene, and butadiene. For example, chloroprene molecules polymerize readily to form polychloroprene, commonly known as neoprene, which has properties that are comparable or even superior to those of natural rubber: 冢 冣 OCH2 H G D H2CPCClOCHPCH2 CPC D G (c) Cl CH2On Figure 25.5 Rubber molecules chloroprene polychloroprene ordinarily are bent and convoluted. Parts (a) and (b) represent the long chains before and after Another important synthetic rubber is formed by the addition of butadiene to vulcanization, respectively; styrene in a 3:1 ratio to give styrene-butadiene rubber (SBR). Because styrene and (c) shows the alignment of molecules when stretched. Without vulcanization these molecules † would slip past one another, Charles Goodyear (1800–1860). American chemist. Goodyear was the first person to realize the potential and rubber’s elastic properties of natural rubber. His vulcanization process made rubber usable in countless ways and opened the way for would be gone. the development of the automobile industry.  25.2 Synthetic Organic Polymers 1063 H 2 NO (CH 2 )6 ONH 2 ⴙ HOOC O (CH 2 ) 4 O COOH Hexamethylenediamine Adipic acid Condensation O B H 2 NO (CH 2 ) 6 ONOC O (CH 2 ) 4 OCOOH ⴙ H2O A H Further condensation reactions O O O B B B O (CH 2 ) 4 OCONO (CH 2 )6 ONOC O (CH 2 ) 4 OC ONO (CH 2 ) 6 O A A A H H H Figure 25.7 The nylon rope trick. Adding a solution of adipoyl chloride (an adipic acid derivative Figure 25.6 The formation of nylon by the condensation reaction between hexamethylenediamine in which the OH groups have and adipic acid. been replaced by Cl groups) in cyclohexane to an aqueous solution of hexamethylenediamine causes nylon to form at the interface of the two solutions, butadiene are different monomers, SBR is called a copolymer, which is a poly- which do not mix. It can then be mer containing two or more different monomers. Table 25.1 shows a number of drawn off. common and familiar homopolymers and one copolymer produced by addition reactions. Condensation Reactions One of the best-known polymer condensation processes is the reaction between hexa- Condensation reaction was defined on p. 1043. methylenediamine and adipic acid, shown in Figure 25.6. The final product, called nylon 66 (because there are six carbon atoms each in hexamethylenediamine and adipic acid), was first made by Wallace Carothers† at Du Pont in 1931. The versatil- ity of nylons is so great that the annual production of nylons and related substances now amounts to several billion pounds. Figure 25.7 shows how nylon 66 is prepared in the laboratory. Condensation reactions are also used in the manufacture of Dacron (polyester) 冢 冣 O O O O B B B B nHOOCO OCOOH ⫹ nHOO(CH2)2OOH 88n OCO OCOOOCH2CH2OOO ⫹ nH2O n terephthalic acid 1,2-ethylene glycol Dacron Polyesters are used in fibers, films, and plastic bottles. † Wallace H. Carothers (1896–1937). American chemist. Besides its enormous commercial success, Carothers’ work on nylon is ranked with that of Staudinger in clearly elucidating macromolecular structure and properties. Depressed by the death of his sister and convinced that his life’s work was a failure, Carothers committed suicide at the age of 41. 1064 Chapter 25 ■ Synthetic and Natural Organic Polymers Table 25.1 Some Monomers and Their Common Synthetic Polymers Monomer Polymer Formula Name Name and Formula Uses H2CPCH2 Ethylene Polyethylene Plastic piping, O( CH2OCH2O )n bottles, electrical H insulation, toys A H2CPC Propene Polypropene Packaging film, A 冢 冣 CH3 OCHOCH2OCHOCH2O carpets, crates A A for soft-drink CH3 CH3 n bottles, lab H wares, toys A H2CPC Vinyl chloride Poly(vinyl chloride) (PVC) Piping, siding, A O( CH2OCHO )n gutters, floor Cl A tile, clothing, H Cl toys A H2CPC Acrylonitrile Polyacrylonitrile (PAN) Carpets, knitwear A 冢 冣 CN OCH2OCHO A CN n F2CPCF2 Tetrafluoro- Polytetrafluoroethylene Coating on ethylene (Teflon) cooking O( CF2OCF2O )n utensils, electrical insulation, COOCH3 bearings A H2CPC Methyl Poly(methyl methacrylate) Optical A methacrylate (Plexiglas) equipment, CH3 COOCH3 home A furnishings O( CH2OCO )n A H CH3 A H2CPC Styrene Polystyrene Containers, A O( CH2OCHO )n thermal A insulation (ice buckets, water coolers), toys H H Butadiene Polybutadiene Tire tread, coating A A O( CH2CHPCHCH2O )n resin H2CPCOCPCH2 See above Butadiene and Styrene-butadiene rubber Synthetic rubber structures styrene (SBR) O( CHOCH2OCH2OCHP CHO CH2O )n A Bubble gums contain synthetic styrene-butadiene rubber.  25.3 Proteins 1065 25.3 Proteins Proteins are polymers of amino acids; they play a key role in nearly all biological processes. Enzymes, the catalysts of biochemical reactions, are mostly proteins. Pro- teins also facilitate a wide range of other functions, such as transport and storage of vital substances, coordinated motion, mechanical support, and protection against dis- eases. The human body contains an estimated 100,000 different kinds of proteins, each of which has a specific physiological function. As we will see in this section, the chemical composition and structure of these complex natural polymers are the basis of their specificity. Amino Acids Proteins have high molar masses, ranging from about 5000 g to 1 3 107 g, and yet the 1A 8A H 2A 3A 4A 5A 6A 7A percent composition by mass of the elements in proteins is remarkably constant: carbon, C N O S 50 to 55 percent; hydrogen, 7 percent; oxygen, 23 percent; nitrogen, 16 percent; and sulfur, 1 percent. The basic structural units of proteins are amino acids. An amino acid is a com- pound that contains at least one amino group (¬NH2 ) and at least one carboxyl Elements in proteins. group (¬COOH): H O D J ON OC G G H O OH amino group carboxyl group Twenty different amino acids are the building blocks of all the proteins in the human body. Table 25.2 shows the structures of these vital compounds, along with their three-letter abbreviations. Amino acids in solution at neutral pH exist as dipolar ions, meaning that the proton on the carboxyl group has migrated to the amino group. Consider glycine, the simplest amino acid. The un-ionized form and the dipolar ion of glycine are shown below: NH2 NH⫹ 3 A A HOCOCOOH HOCOCOO⫺ A A H H un-ionized form dipolar ion The first step in the synthesis of a protein molecule is a condensation reaction between an amino group on one amino acid and a carboxyl group on another amino acid. The molecule formed from the two amino acids is called a dipeptide, and the bond joining them together is a peptide bond: H O H O H O H O A B A B A B A B  H3NOCOCOO   H 3NOCOCOO  34  H3NOCOOCONOCOCOO   H2O A A A A A n It is interesting to compare this reaction 8888 R1 R2 R1 H R2 with the one shown in Figure 25.6. peptide bond where R1 and R2 represent a H atom or some other group; ¬CO¬NH¬ (the shaded area in the above reaction) is also called the amide group. Because the equilibrium of the reaction joining two amino acids lies to the left, the process is coupled to the hydrolysis of ATP (see p. 802). 1066 Chapter 25 ■ Synthetic and Natural Organic Polymers Table 25.2 The 20 Amino Acids Essential to Living Organisms* Name Abbreviation Structure H A Alanine Ala H3COCOCOO⫺ A NH⫹ 3 H H A A Arginine Arg H2NOCONOCH2OCH2OCH2OCOCOO⫺ B A NH NH⫹ 3 O H B A Asparagine Asn H2NOCOCH2OCOCOO⫺ A NH⫹ 3 H A Aspartic acid Asp HOOCOCH2OCOCOO⫺ A NH⫹ 3 H A Cysteine Cys HSOCH2OCOCOO⫺ A NH⫹ 3 H A Glutamic acid Glu HOOCOCH2OCH2OCOCOO⫺ A NH⫹ 3 O H B A Glutamine Gln H2NOCOCH2OCH2OCOCOO⫺ A NH⫹ 3 H A Glycine Gly HOCOCOO⫺ A NH⫹ 3 H A Histidine His HCPCOCH 2 OCOCOO ⫺ A A A ⫹ N NH NH M D 3 C H CH3 H A A Isoleucine Ile H 3 COCH 2 OCOOCOCOO ⫺ A A H NH⫹ 3 (Continued) *The shaded portion is the R group of the amino acid.  25.3 Proteins 1067 Table 25.2 The 20 Amino Acids Essential to Living Organisms—Cont. Name Abbreviation Structure H3C H G A Leucine Leu CHOCH 2 OCOCOO ⫺ D A H 3C NH⫹ 3 H A Lysine Lys H 2 NOCH 2 OCH 2 OCH 2 OCH 2 OCOCOO ⫺ A NH⫹ 3 H A Methionine Met H 3 COSOCH 2 OCH 2 OCOCOO ⫺ A NH⫹ 3 H A Phenylalanine Phe OCH 2 OCOCOO ⫺ A NH⫹ 3 H ⫹ A Proline Pro H 2 NOOOCOCOO ⫺ A A H2C CH2 G D CH2 H A Serine Ser HOOCH 2 OCOCOO ⫺ A NH⫹ 3 OH H A A Threonine Thr H 3 COCOOCOCOO ⫺ A A H NH⫹ 3 H A Tryptophan Trp OOOCOCH 2 OCOCOO ⫺ B A OO NH⫹ D CH 3 N H H A Tyrosine Tyr HOO OCH 2 OCOCOO ⫺ A NH⫹ 3 H3C H G A Valine Val CHOCOCOO ⫺ D A H 3C NH⫹ 3 1068 Chapter 25 ■ Synthetic and Natural Organic Polymers H O H O A B A B ⫹ H 3 NOC OC OO⫺ ⫹ H 3 N O C O C O O⫺ A A CH 3 H Alanine Glycine H O H O H O H O A B A B A B A B ⫹ H 3 NO COCONOC OC OO⫺ ⫹ H 3 NO C OCONOC OCO O⫺ A A A A A A CH 3 H H H H CH 3 Alanylglycine Glycylalanine Figure 25.8 The formation of two dipeptides from two different amino acids. Alanylglycine is different from glycylalanine in that in alanylglycine the amino and methyl groups are bonded to the same carbon atom. Either end of a dipeptide can engage in a condensation reaction with another amino acid to form a tripeptide, a tetrapeptide, and so on. The final product, the protein molecule, is a polypeptide; it can also be thought of as a polymer of amino acids. An amino acid unit in a polypeptide chain is called a residue. Typically, a poly- peptide chain contains 100 or more amino acid residues. The sequence of amino acids in a polypeptide chain is written conventionally from left to right, starting with the amino-terminal residue and ending with the carboxyl-terminal residue. Let us consider a dipeptide formed from glycine and alanine. Figure 25.8 shows that alanylglycine and glycylalanine are different molecules. With 20 different amino acids to choose from, 202, or 400, different dipeptides can be generated. Even for a very small protein such as insulin, which contains only 50 amino acid residues, the number of chemically different structures that is possible is of the order of 2050 or 1065! This is an incred- ibly large number when you consider that the total number of atoms in our galaxy is about 1068. With so many possibilities for protein synthesis, it is remarkable that generation after generation of cells can produce identical proteins for specific physi- ological functions. Protein Structure The type and number of amino acids in a given protein along with the sequence or order in which these amino acids are joined together determine the protein’s structure. In the 1930s, Linus Pauling and his coworkers conducted a systematic investigation of protein structure. First, they studied the geometry of the basic repeating group, that is, the amide group, which is represented by the following resonance structures: Q Q SO SOS⫺ B Q A ⫹ OCONO mn OCPNO Figure 25.9 The planar amide A A group in protein. Rotation about H H the peptide bond in the amide group is hindered by its double- Because it is more difficult (that is, it would take more energy) to twist a double bond character. The black atoms bond than a single bond, the four atoms in the amide group become locked in the represent carbon; blue, nitrogen; red, oxygen; green, R group; and same plane (Figure 25.9). Figure 25.10 depicts the repeating amide group in a poly- gray, hydrogen. peptide chain.  25.3 Proteins 1069 H O H O H O H O A B A B A B A B O C OCONOC OCONO C OCONOCO CONO A A A A A A A A R H R H R H H H Figure 25.10 A polypeptide chain. Note the repeating units of the amide group. The symbol R represents part of the structure characteristic of the individual amino acids. For glycine, R is simply a H atom. On the basis of models and X-ray diffraction data, Pauling deduced that there are two common structures for protein molecules, called the α helix and the β-pleated sheet. The α-helical structure of a polypeptide chain is shown in Figure 25.11. The helix is stabilized by intramolecular hydrogen bonds between the NH and CO groups of the main chain, giving rise to an overall rodlike shape. The CO group of each amino acid is hydrogen-bonded to the NH group of the amino acid that is four residues away in the sequence. In this manner all the main-chain CO and NH groups take part in hydrogen bonding. X-ray studies have shown that the structure of a number of proteins, including myoglobin and hemoglobin, is to a great extent α-helical in nature. The β-pleated structure is markedly different from the α helix in that it is like a sheet rather than a rod. The polypeptide chain is almost fully extended, and each chain forms many intermolecular hydrogen bonds with adjacent chains. Figure 25.12 shows the two different types of β-pleated structures, called parallel and antiparallel. Silk molecules possess the β structure. Because its polypeptide chains are already in extended form, silk lacks elasticity and extensibility, but it is quite strong due to the Figure 25.11 The α-helical structure of a polypeptide chain. many intermolecular hydrogen bonds. The gray spheres are hydrogen It is customary to divide protein structure into four levels of organization. The atoms. The structure is held in position by intramolecular primary structure refers to the unique amino acid sequence of the polypeptide chain. hydrogen bonds, shown as The secondary structure includes those parts of the polypeptide chain that are sta- dotted lines. For color key, see bilized by a regular pattern of hydrogen bonds between the CO and NH groups of Fig. 25.9. the backbone, for example, the α helix. The term tertiary structure applies to the three-dimensional structure stabilized by dispersion forces, hydrogen bonding, and other intermolecular forces. It differs from secondary structure in that the amino acids taking part in these interactions may be far apart in the polypeptide chain. A protein molecule may be made up of more than one polypeptide chain. Thus, in addition to the various interactions within a chain that give rise to the secondary and tertiary structures, we must also consider the interaction between chains. The overall arrangement of the polypeptide chains is called the quaternary structure. For example, the hemoglobin molecule consists of four separate polypeptide chains, or subunits. These subunits are held together by van der Waals forces and ionic forces (Figure 25.13). Pauling’s work was a great triumph in protein chemistry. It showed for the Intermolecular forces play an important role in the secondary, tertiary, and first time how to predict a protein structure purely from a knowledge of the geom- quaternary structure of proteins. etry of its fundamental building blocks—amino acids. However, there are many proteins whose structures do not correspond to the α-helical or β structure. Chemists now know that the three-dimensional structures of these biopolymers are maintained by several types of intermolecular forces in addition to hydrogen bonding (Figure 25.14). The delicate balance of the various interactions can be appreciated by considering an example: When glutamic acid, one of the amino acid residues in two of the four polypeptide chains in hemoglobin, is replaced by 1070 Chapter 25 ■ Synthetic and Natural Organic Polymers Parallel Antiparallel (a) (b) Figure 25.12 Hydrogen bonds (a) in a parallel β-pleated sheet structure, in which all the polypeptide chains are oriented in the same direction, and (b) in an antiparallel β-pleated sheet, in which adjacent polypeptide chains run in opposite directions. For color key, see Fig. 25.9. valine, another amino acid, the protein molecules aggregate to form insoluble polymers, causing the disease known as sickle cell anemia (see the Chemistry in Action essay on p. 1072). In spite of all the forces that give proteins their structural stability, most proteins have a certain amount of flexibility. Enzymes, for example, are flexible enough to change their geometry to fit substrates of various sizes and shapes. Another interesting example of protein flexibility is found in the binding of hemoglobin to oxygen. Each of the four polypeptide chains in hemoglobin contains a heme group that can bind to an oxygen molecule (see Section 23.7). In deoxyhemoglobin, the affinity of each of the heme groups for oxygen is about the same. However, as soon as one of the heme groups becomes oxygenated, the affinity of the other three hemes for oxygen is greatly enhanced. This phenomenon, called cooperativity, makes hemoglobin a particularly suitable substance for the uptake of oxygen in the lungs. By the same token, once a fully oxygenated hemoglobin molecule releases an oxygen molecule (to myoglobin in the tissues), the other three oxygen molecules will depart with increasing ease. The cooperative nature of the binding is such that information about the presence (or absence) of oxygen molecules is transmitted from one subunit to another along the polypeptide chains, a process made possible by the flexibility of the three-dimensional  25.3 Proteins 1071 Ala His Val Pro Tertiary structure Quaternary structure Primary structure Secondary structure Figure 25.13 The primary, secondary, tertiary, and quaternary structure of the hemoglobin molecule. structure (Figure 25.15). It is believed that the Fe21 ion has too large a radius to fit into the porphyrin ring of deoxyhemoglobin. When O2 binds to Fe21, however, the ion shrinks somewhat so that it can fit into the plane of the ring. As the ion slips into the ring, it pulls the histidine residue toward the ring and thereby sets off a sequence of structural changes from one subunit to another. Although the details of the changes are not clear, biochemists believe that this is how the binding of an oxygen molecule Figure 25.14 Intermolecular NH3 forces in a protein molecule: (a) ionic forces, (b) hydrogen + bonding, (c) dispersion forces, (a) and (d) dipole-dipole forces. (c) (b) – O O H O O C O C C NH2 CH + CH2 CH3 CH3 NH2 CH3 CH3 CH3 (c) (c) C O C O (d) (a) CH3 – CH3 CH2 CH3 O CH C O CHEMISTRY in Action Sickle Cell Anemia—A Molecular Disease S ickle cell anemia is a hereditary disease in which abnormally shaped red blood cells restrict the flow of blood to vital organs in the human body, causing swelling, severe pain, and in many soluble in water. Therefore, the aggregated HbS molecules eventually precipitate out of solution. The precipitate causes normal disk-shaped red blood cells to assume a warped crescent cases a shortened life span. There is currently no cure for this con- or sickle shape (see figure on p. 292). These deformed cells clog dition, but its painful symptoms are known to be caused by a defect the narrow capillaries, thereby restricting blood flow to organs in hemoglobin, the oxygen-carrying protein in red blood cells. of the body. It is the reduced blood flow that gives rise to the The hemoglobin molecule is a large protein with a molar mass symptoms of sickle cell anemia. Sickle cell anemia has been of about 65,000 g. Normal human hemoglobin (HbA) consists of termed a molecular disease by Linus Pauling, who did some of two α chains, each containing 141 amino acids, and two β chains the early important chemical research on the nature of the afflic- made up of 146 amino acids each. These four polypeptide chains, tion, because the destructive action occurs at the molecular level or subunits, are held together by ionic and van der Waals forces. and the disease is, in effect, due to a molecular defect. There are many mutant hemoglobin molecules—molecules Some substances, such as urea and the cyanate ion, with an amino acid sequence that differs somewhat from the sequence in HbA. Most mutant hemoglobins are harmless, but H2NOCONH2 OPCPN⫺ B sickle cell hemoglobin (HbS) and others are known to cause O serious diseases. HbS differs from HbA in only one very small urea cyanate ion detail. A valine molecule replaces a glutamic acid molecule on each of the two β chains: can break up the hydrophobic interaction between HbS mole- cules and have been applied with some success to reverse the H “sickling” of red blood cells. This approach may alleviate the A HOOCOCH2OCH2OCOCOO⫺ pain and suffering of sickle cell patients, but it does not prevent A the body from making more HbS. To cure sickle cell anemia, ⫹ NH3 researchers must find a way to alter the genetic machinery that glutamic acid directs the production of HbS. H3C H G A CHOCOCOO⫺ D A H 3C ⫹ NH 3 valine Yet this small change (two amino acids out of 292) has a profound effect on the stability of HbS in solution. The valine groups are located at the bottom outside of the molecule to form a protruding “key” on each of the β chains. The nonpolar portion of valine H3C G CHO D H3C can attract another nonpolar group in the α chain of an adjacent HbS molecule through dispersion forces. Biochemists often refer to this kind of attraction between nonpolar groups as hydrophobic (see Chapter 12) interaction. Gradually, enough HbS molecules will aggregate to form a “superpolymer.” A general rule about the solubility of a substance is that the The overall structure of hemoglobin. Each hemoglobin molecule contains two α chains and two β chains. Each of the four chains is similar to a myo- larger its molecules, the lower its solubility because the solva- globin molecule in structure, and each also contains a heme group for tion process becomes unfavorable with increasing molecular binding oxygen. In sickle cell hemoglobin, the defective regions (the valine surface area. For this reason, proteins generally are not very groups) are located near the ends of the β chains, as indicated by the dots. 1072  25.4 Nucleic Acids 1073 Histidine Rate Porphyrin ring Fe 2+ Fe 2+ Optimum temperature Oxygen molecule Temperature (a) (b) Figure 25.16 Dependence of the rate of an enzyme-catalyzed Figure 25.15 The structural changes that occur when the heme group in hemoglobin binds to an reaction on temperature. Above oxygen molecule. (a) The heme group in deoxyhemoglobin. (b) Oxyhemoglobin. the optimum temperature at which an enzyme is most effective, its activity drops off as a consequence of denaturation. to one heme group affects another heme group. The structural changes drastically affect the affinity of the remaining heme groups for oxygen molecules. When proteins are heated above body temperature or when they are subjected to Hard-boiling an egg denatures the proteins in the egg white. unusual acid or base conditions or treated with special reagents called denaturants, they lose some or all of their tertiary and secondary structure. Called denatured pro- teins, proteins in this state no longer exhibit normal biological activities. Figure 25.16 shows the variation of rate with temperature for a typical enzyme-catalyzed reaction. Initially, the rate increases with increasing temperature, as we would expect. Beyond the optimum temperature, however, the enzyme begins to denature and the rate falls rapidly. If a protein is denatured under mild conditions, its original structure can often be regenerated by removing the denaturant or by restoring the temperature to normal conditions. This process is called reversible denaturation. 25.4 Nucleic Acids Nucleic acids are high molar mass polymers that play an essential role in protein synthesis. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acid. DNA molecules are among the largest molecules known; they have molar masses of up to tens of billions of grams. On the other hand, RNA molecules vary greatly in size, some having a molar mass of about 25,000 g. Compared with proteins, which are made of up to 20 different amino acids, nucleic acids are fairly simple in composition. A DNA or RNA molecule contains only four types of building blocks: purines, pyrimidines, furanose sugars, and phosphate groups (Figure 25.17). Each purine or pyrimidine is called a base. In the 1940s, Erwin Chargaff† studied DNA molecules obtained from various sources and observed certain regularities. Chargaff’s rules, as his findings are now known, describe these patterns: 1. The amount of adenine (a purine) is equal to that of thymine (a pyrimidine); that is, A 5 T, or A/T 5 1. 2. The amount of cytosine (a pyrimidine) is equal to that of guanine (a purine); that is, C 5 G, or C/G 5 1. 3. The total number of purine bases is equal to the total number of pyrimidine bases; that is, A 1 G 5 C 1 T. † Erwin Chargaff (1905–2002). American biochemist of Austrian origin. Chargaff was the first to show that different biological species contain different DNA molecules. 1074 Chapter 25 ■ Synthetic and Natural Organic Polymers Figure 25.17 The components of the nucleic acids DNA and RNA. Based on chemical analyses and information obtained from X-ray diffraction measurements, James Watson† and Francis Crick‡ formulated the double-helical structure for the DNA molecule in 1953. Watson and Crick determined that the DNA molecule has two helical strands. Each strand is made up of nucleotides, which consist of a base, a deoxyribose, and a phosphate group linked together (Figure 25.18). The key to the double-helical structure of DNA is the formation of hydrogen bonds between bases in the two strands of a molecule. Although hydrogen bonds can form between any two bases, called base pairs, Watson and Crick found that the most favorable couplings are between adenine and thymine and between cytosine and An electron micrograph of a DNA molecule. The double-helical structure is evident. If the DNA † molecules from all the cells in a James Dewey Watson (1928– ). American biologist. Watson shared the 1962 Nobel Prize in Physiology human were stretched and joined or Medicine with Crick and Maurice Wilkins for their work on the DNA structure, which is considered by end to end, the length would be many to be the most significant development in biology in the twentieth century. about 100 times the distance from ‡ Francis Harry Compton Crick (1916–2004). British biologist. Crick started as a physicist but became Earth to the sun! interested in biology after reading the book What Is Life? by Erwin Schrödinger (see Chapter 7). In addi- tion to elucidating the structure of DNA, for which he was a corecipient of the Nobel Prize in Physiology or Medicine in 1962, Crick made many significant contributions to molecular biology.  25.4 Nucleic Acids 1075 NH 2 N N N N Adenine unit Oⴚ A OP P O O CH 2 O A Oⴚ H H H H Phosphate unit OH H Deoxyribose unit Figure 25.18 Structure of a nucleotide, one of the repeating units in DNA. guanine (Figure 25.19). Note that this scheme is consistent with Chargaff’s rules, because every purine base is hydrogen-bonded to a pyrimidine base, and vice versa (A 1 G 5 C 1 T). Other attractive forces such as dipole-dipole interactions and van der Waals forces between the base pairs also help to stabilize the double helix. The structure of RNA differs from that of DNA in several respects. First, as shown in Figure 25.17, the four bases found in RNA molecules are adenine, cytosine, gua- nine, and uracil. Second, RNA contains the sugar ribose rather than the 2-deoxyribose of DNA. Third, chemical analysis shows that the composition of RNA does not obey Chargaff’s rules. In other words, the purine-to-pyrimidine ratio is not equal to 1 as in the case of DNA. This and other evidence rule out a double-helical structure. In fact, the RNA molecule exists as a single-strand polynucleotide. There are actually three AT CG GC TA Adenine H Thymine OH AT O N H O CH3 H2C CG O N O P GC P H O TA O H OH CH2 N O N H N H O CG H H H N N H GC H O TA O H2C OH AT O O H O P H H H P O O CG O H N H O OH CH2 Cytosine Guanine GC O H TA H N N H AT H H H O N H N O N H2C OH O H O P H P O N H O O N OH CH2 O H (a) (b) Figure 25.19 (a) Base-pair formation by adenine and thymine and by cytosine and guanine. (b) The double-helical strand of a DNA molecule held together by hydrogen bonds (and other intermolecular forces) between base pairs A-T and C-G.  CHEMISTRY in Action DNA Fingerprinting T he human genetic makeup, or genome, consists of about 3 billion nucleotides. These 3 billion units compose the 23 pairs of chromosomes, which are continuous strands of DNA DNA. The DNA is extracted from cell nuclei and cut into frag- ments by the addition of so-called restriction enzymes. These fragments, which are negatively charged, are separated by an ranging in length from 50 million to 500 million nucleotides. electric field in gel. The smaller fragments move faster than Encoded in this DNA and stored in units called genes are the larger ones, so they eventually separate into bands. The bands of instructions for protein synthesis. Each of about 100,000 genes DNA fragments are transferred from the gel to a plastic mem- is responsible for the synthesis of a particular protein. In addi- brane, and their position is thereby fixed. Then a DNA probe—a tion to instructions for protein synthesis, each gene contains a DNA fragment that has been tagged with a radioactive label—is sequence of bases, repeated several times, that has no known added. The probe binds to the fragments that have a comple- function. What is interesting about these sequences, called mentary DNA sequence. An X-ray film is laid directly over the minisatellites, is that they appear many times in different loca- plastic sheet, and bands appear on the exposed film in the posi- tions, not just in a particular gene. Furthermore, each person has tions corresponding to the fragments recognized by the probe. a unique number of repeats. Only identical twins have the same About four different probes are needed to obtain a profile that is number of minisatellite sequences. unique to just one individual. It is estimated that the probability In 1985 a British chemist named Alec Jeffreys suggested that of finding identical patterns in the DNA of two randomly se- minisatellite sequences provide a means of identification, much lected individuals is on the order of 1 in 10 billion. like fingerprints. DNA fingerprinting has since gained prominence The first U.S. case in which a person was convicted of a with law enforcement officials as a way to identify crime suspects. crime with the help of DNA fingerprints was tried in 1987. To make a DNA fingerprint, a chemist needs a sample of Today, DNA fingerprinting has become an indispensable tool of any tissue, such as blood or semen; even hair and saliva contain law enforcement. – Bloodstain + DNA is extracted A restriction en- Fragments are sepa- The DNA band pattern in from blood cells zyme cuts DNA rated into bands by the gel is transferred to a into fragments gel electrophoresis nylon membrane Replicate Pattern patterns, from Radioactive DNA probe binds X-ray film detects same another to specific DNA sequences radioactive pattern person person Procedure for obtaining a DNA fingerprint. The developed film shows the DNA fingerprint, which is compared with patterns from known subjects. 1076  Questions & Problems 1077 types of RNA molecules—messenger RNA (mRNA), ribosomal RNA (rRNA), and In the 1980s chemists discovered that certain RNAs can function as enzymes. transfer RNA (tRNA). These RNAs have similar nucleotides but differ from one another in molar mass, overall structure, and biological functions. DNA and RNA molecules direct the synthesis of proteins in the cell, a subject that is beyond the scope of this book. Introductory texts in biochemistry and molecular biology explain this process. The Chemistry in Action essay on p. 1076 describes a technique in crime investiga- tion that is based on our knowledge of DNA sequence. Summary of Facts & Concepts 1. Polymers are large molecules made up of small, repeating 7. The primary structure of a protein is its amino acid units called monomers. sequence. Secondary structure is the shape defined by 2. Proteins, nucleic acids, cellulose, and rubber are natural hydrogen bonds joining the CO and NH groups of the polymers. Nylon, Dacron, and Lucite are examples of amino acid backbone. Tertiary and quaternary structures synthetic polymers. are the three-dimensional folded arrangements of pro- 3. Organic polymers can be synthesized via addition re- teins that are stabilized by hydrogen bonds and other actions or condensation reactions. intermolecular forces. 4. Stereoisomers of a polymer made up of asymmetric 8. Nucleic acids—DNA and RNA—are high-molar-mass monomers have different properties, depending on how polymers that carry genetic instructions for protein the starting units are joined together. synthesis in cells. Nucleotides are the building blocks of DNA and RNA. DNA nucleotides each contain a 5. Synthetic rubbers include polychloroprene and styrene- purine or pyrimidine base, a deoxyribose molecule, butadiene rubber, which is a copolymer of styrene and and a phosphate group. RNA nucleotides are similar butadiene. but contain different bases and ribose instead of 6. Structure determines the function and properties of deoxyribose. proteins. To a great extent, hydrogen bonding and other intermolecular forces determine the structure of proteins. Key Words Amino acid, p. 1065 Deoxyribonucleic acid Nucleic acid, p. 1073 Ribonucleic acid Copolymer, p. 1063 (DNA), p. 1073 Nucleotide, p. 1074 (RNA), p. 1073 Denatured Homopolymer, p. 1060 Polymer, p. 1059 protein, p. 1073 Monomer, p. 1059 Protein, p. 1065 Questions & Problems • Problems available in Connect Plus • 25.3 Calculate the molar mass of a particular polyethyl- Red numbered problems solved in Student Solutions Manual ene sample, ¬( CH2¬CH2 ¬ )n, where n 5 4600. 25.4 Describe the two major mechanisms of organic Synthetic Organic Polymers polymer synthesis. Review Questions 25.5 What are Natta-Ziegler catalysts? What is their role in polymer synthesis? • 25.1 Define the following terms: monomer, polymer, • 25.6 In Chapter 12 you learned about the colligative homopolymer, copolymer. properties of solutions. Which of the colligative 25.2 Name 10 objects that contain synthetic organic properties is suitable for determining the molar polymers. mass of a polymer? Why?  1078 Chapter 25 ■ Synthetic and Natural Organic Polymers Problems 25.20 Draw the structures of the dipeptides that can be formed from the reaction between the amino acids • 25.7 Teflon is formed by a radical addition reaction glycine and lysine. involving the monomer tetrafluoroethylene. Show the mechanism for this reaction. • 25.21 The amino acid glycine can be condensed to form a polymer called polyglycine. Draw the repeating 25.8 Vinyl chloride, H2C“CHCl, undergoes copolymer- monomer unit. ization with 1,1-dichloroethylene, H2C“CCl2, to form a polymer commercially known as Saran. Draw 25.22 The following are data obtained on the rate of product the structure of the polymer, showing the repeating formation of an enzyme-catalyzed reaction: monomer units. 25.9 Kevlar is a copolymer used in bullet-proof vests. It Rate of Product is formed in a condensation reaction between the Temperature (8C) Formation (M/s) following two monomers: 10 0.0025 20 0.0048 O O B B 30 0.0090 H2NO ONH2 HOOCO OCOOH 35 0.0086 45 0.0012 Sketch a portion of the polymer chain showing sev- eral monomer units. Write the overall equation for Comment on the dependence of rate on temperature. the condensation reaction. (No calculations are required.) 25.10 Describe the formation of polystyrene. • 25.11 Deduce plausible monomers for polymers with the Nucleic Acids following repeating units: Review Questions (a) ¬( CH2¬CF2 ¬ )n • 25.23 Describe the structure of a nucleotide. 冢 (b) OCOO OCONHO ONHO 冣n 25.24 What is the difference between ribose and deoxy- ribose? • 25.12 Deduce plausible monomers for polymers with the • 25.25 What are Chargaff’s rules? following repeating units: 25.26 Describe the role of hydrogen bonding in maintaining the double-helical structure of DNA. (a) ¬( CH2¬CH“CH¬CH2 ¬ )n (b) ¬( CO¬ ( CH2 ¬)6 NH¬)n Additional Problems 25.27 Discuss the importance of hydrogen bonding in bio- Proteins logical systems. Use proteins and nucleic acids as Review Questions examples. 25.13 Discuss the characteristics of an amide group and its • 25.28 Proteins vary widely in structure, whereas nucleic importance in protein structure. acids have rather uniform structures. How do you account for this major difference? 25.14 What is the α-helical structure in proteins? 25.29 If untreated, fevers of 104°F or higher may lead to 25.15 Describe the β-pleated structure present in some brain damage. Why? proteins. 25.16 Discuss the main functions of proteins in living • 25.30 The “melting point” of a DNA molecule is the tem- perature at which the double-helical strand breaks systems. apart. Suppose you are given two DNA samples. One 25.17 Briefly explain the phenomenon of cooperativity sample contains 45 percent C-G base pairs while the exhibited by the hemoglobin molecule in binding other contains 64 percent C-G base pairs. The total oxygen. number of bases is the same in each sample. Which of 25.18 Why is sickle cell anemia called a molecular the two samples has a higher melting point? Why? disease? 25.31 When fruits such as apples and pears are cut, the exposed parts begin to turn brown. This is the result of an oxidation reaction catalyzed by enzymes pres- Problems ent in the fruit. Often the browning action can be • 25.19 Draw the structures of the dipeptides that can be prevented or slowed by adding a few drops of lemon formed from the reaction between the amino acids juice to the exposed areas. What is the chemical basis glycine and alanine. for this treatment?  Questions & Problems 1079 • 25.32 “Dark meat” and “white meat” are one’s choices • 25.43 Nylon was designed to be a synthetic silk. (a) The when eating a turkey. Explain what causes the meat average molar mass of a batch of nylon 66 is to assume different colors. (Hint: The more active 12,000 g/mol. How many monomer units are there muscles in a turkey have a higher rate of metabolism in this sample? (b) Which part of nylon’s structure and need more oxygen.) is similar to a polypeptide’s structure? (c) How 25.33 Nylon can be destroyed easily by strong acids. many different tripeptides (made up of three amino Explain the chemical basis for the destruction. acids) can be formed from the amino acids alanine (Hint: The products are the starting materials of the (Ala), glycine (Gly), and serine (Ser), which ac- polymerization reaction.) count for most of the amino acids in silk? 25.34 Despite what you may have read in science fiction • 25.44 The enthalpy change in the denaturation of a novels or seen in horror movies, it is extremely un- certain protein is 125 kJ/mol. If the entropy likely that insects can ever grow to human size. Why? change is 397 J/K ? mol, calculate the minimum (Hint: Insects do not have hemoglobin molecules in temperature at which the protein would denature their blood.) spontaneously. • 25.35 How many different tripeptides can be formed by 25.45 When deoxyhemoglobin crystals are exposed to lysine and alanine? oxygen, they shatter. On the other hand, deoxymyo- 25.36 Chemical analysis shows that hemoglobin contains globin crystals are unaffected by oxygen. Explain. 0.34 percent Fe by mass. What is the minimum (Myoglobin is made up of only one of the four possible molar mass of hemoglobin? The actual subunits, or polypeptide chains, in hemoglobin.) molar mass of hemoglobin is four times this mini- 25.46 The α-helix and β-sheet structures are prevalent in mum value. What conclusion can you draw from proteins. What is the common feature that they these data? have which makes them suitable for this role? 25.37 The folding of a polypeptide chain depends not only 25.47 In protein synthesis, the selection of a particular amino on its amino acid sequence but also on the nature of acid is determined by the so-called genetic code, or a the solvent. Discuss the types of interactions that sequence of three bases in DNA. Will a sequence of might occur between water molecules and the amino only two bases unambiguously determine the selec- acid residues of the polypeptide chain. Which groups tion of 20 amino acids found in proteins? Explain. would be exposed on the exterior of the protein in • 25.48 Consider the fully protonated amino acid valine: contact with water and which groups would be buried in the interior of the protein? ⫹ 9.62 CH3 NH3 • 25.38 What kind of intermolecular forces are responsible 2.32 for the aggregation of hemoglobin molecules that H C C COOH leads to sickle cell anemia? (Hint: See the Chemis- try in Action essay on p. 1072.) CH3 H • 25.39 Draw structures of the nucleotides containing the fol- lowing components: (a) deoxyribose and cytosine, (b) where the numbers denote the pKa values. (a) Which 1 ribose and uracil. of the two groups (¬NH3 or ¬COOH) is more 25.40 When a nonapeptide (containing nine amino acid acidic? (b) Calculate the predominant form of valine residues) isolated from rat brains was hydrolyzed, it at pH 1.0, 7.0, and 12.0. (c) Calculate the isoelectric gave the following smaller peptides as identifiable point of valine. (Hint: See Problem 16.137.) products: Gly-Ala-Phe, Ala-Leu-Val, Gly-Ala-Leu, • 25.49 Consider the formation of a dimeric protein Phe-Glu-His, and His-Gly-Ala. Reconstruct the 2P ¡ P2 amino acid sequence in the nonapeptide, giving your reasons. (Remember the convention for writ- At 25°C, we have DH° 5 17 kJ/mol and DS° 5 ing peptides.) 65 J/K ? mol. Is the dimerization favored at this tem- • 25.41 At neutral pH amino acids exist as dipolar ions. perature? Comment on the effect of lowering the Using glycine as an example, and given that the pKa temperature. Does your result explain why some of the carboxyl group is 2.3 and that of the ammo- enzymes lose their activities under cold conditions? nium group is 9.6, predict the predominant form of 25.50 Molar mass measurements play an important role the molecule at pH 1, 7, and 12. Justify your answers in characterizing polymer solutions. Number- using Equation (16.4). average molar mass (Mn ) is defined as the total molar 25.42 In Lewis Carroll’s tale “Through the Looking Glass,” mass (given by g NiMi) divided by the total number of Alice wonders whether “looking-glass milk” on the molecules: other side of the mirror would be fit to drink. Based on your knowledge of chirality and enzyme action, a NiMi Mn 5 comment on the validity of Alice’s concern. a Ni 1080 Chapter 25 ■ Synthetic and Natural Organic Polymers where Ni is the number of molecules with molar protein structure is maintained in part by the disul- mass Mi. Another important definition is the weight- fide bonds (¬S¬S¬) between the amino acid average molar mass (Mw ) where residues (each color sphere represents an S atom). Using certain denaturants, the compact structure is 2 a Ni M i destroyed and the disulfide bonds are converted to Mw 5 sulfhydryl groups (¬SH) shown on the right of the a Ni Mi arrow. (a) Describe the bonding scheme in the disul- The difference between these two definitions is that fide bond in terms of hybridization. (b) Which Mw is based on experimental measurements that are amino acid in Table 25.2 contains the ¬SH group? affected by the size of molecules. (a) Consider a (c) Predict the signs of DH and DS for the denatur- solution containing five molecules of molar masses ation process. If denaturation is induced by a change 1.0, 3.0, 4.0, 4.0, and 6.0 kg/mol. Calculate both Mn and in temperature, show why a rise in temperature Mw . (b) Mw is always greater than Mn because of the would favor denaturation. (d) The sulfhydryl groups square term in the definition. However, if all the mole- can be oxidized (that is, removing the H atoms) to cules have identical molar mass, then we have form the disulfide bonds. If the formation of the di- Mn 5 Mw . Show that this is the case if we have four sulfide bonds is totally random between any two molecules having the same molar mass of 5 kg/mol. ¬SH groups, what is the fraction of the regenerated (c) Explain how a comparison of these two average protein structures that corresponds to the native molar masses gives us information about the distribu- form? (e) An effective remedy to deodorize a dog tion of the size of synthetic polymers like polyethylene that has been sprayed by a skunk is to rub the af- and poly(vinyl chloride). (d) Proteins like myoglobin fected areas with a solution of an oxidizing agent and cytochrome c have the same Mn and Mw , while such as hydrogen peroxide. What is the chemical this is not the case for hemoglobin. Explain. basis for this action? (Hint: An odiferous compo- nent of a skunk’s secretion is 2-butene-1-thiol, 25.51 The diagram (left) shows the structure of the enzyme CH3CH“CHCH2SH.2 ribonuclease in its native form. The three-dimensional H H H H H H H H Native form Denatured form  Questions & Problems 1081 Interpreting, Modeling & Estimating 25.52 Assume the energy of hydrogen bonds per base pair 25.53 The average distance between base pairs measured to be 10 kJ/mol. Given two complementary strands of parallel to the axis of a DNA molecule is 3.4 Å DNA containing 10 base pairs each, estimate the ratio (see Figure 25.19). The average molar mass of a of two separate strands to hydrogen-bonded double pair of nucleotides is 650 g/mol. Estimate the helix in solution at 300 K. The ratio for a single base length in centimeters of a DNA molecule of molar pair is given by the formula exp(2DE/RT), where DE mass 5.0 3 109 g/mol. is the energy of hydrogen bond per base pair, R is the gas constant, and T is the temperature in kelvins. CHEMICAL M YS TERY A Story That Will Curl Your Hair S ince ancient times people have experimented with ways to change their hair. Today, getting a permanent wave is a routine procedure that can be done either in a hair- dresser shop or at home. Changing straight hair to curly hair is a practical application of protein denaturation and renaturation. Hair contains a special class of proteins called keratins, which are also present in wool, nails, hoofs, and horns. X-ray studies show that keratins are made of α-helices coiled to form a superhelix. The disulfide bonds (¬S¬S¬) linking the α-helices together are largely responsible for the shape of the hair. The figure on p. 1083 shows the basic steps involved in a permanent wave process. Starting with straight hair, the disulfide bonds are first reduced to the sulfhydryl groups (¬SH) 2HS¬CH2COO2 1 d¬S¬S¬d ¡ 2 OOCCH2¬S¬S¬CH2COO2 1 2 d¬SH where the red spheres represent different protein molecules joined by the disulfide bonds and thioglycolate (HS¬CH2COO2) is the common reducing agent. The reduced hair is then wrapped around curlers and set in the desired pattern. Next, the hair is treated with an oxi- dizing agent to reform the disulfide bonds. Because the S¬S linkages are now formed between different positions on the polypeptide chains, the result is a new hairdo of wavy hairs. This process involves the denaturation and renaturation of keratins. Although disulfide bonds are formed at different positions in the renatured proteins, there is no biological consequence because keratins in hair do not have any specific functions. The word “per- manent” applies only to the portion of hair treated with the reducing and oxidizing agents, and the wave lasts until new and untreated keratins replace it. Chemical Clues 1. Describe the bonding in the ¬S¬S¬ linkage. 2. What are the oxidation numbers of S in the disulfide bond and in the sulfhydryl group? 1082  –S—S– –S –S — H –S — S– HS – S– –S—S– 8n –S — S– 8n –S H HS – 8n –S — S– –S—S– –S — –S –S — H HS – S– S– Straight hair Wet hair Reduced hair Oxidized hair on curler on curler off curler (permanent wave) 3. In addition to the disulfide bonds, the α helices are joined together by hydrogen bonds. Based on this information, explain why hair swells a bit when it is wet. 4. Hair grows at the approximate rate of 6 in per year. Given that the vertical distance for a complete turn of an α helix is 5.4 Å (1 Å 5 1028 cm), how many turns are spun off every second? 5. In the 1980s an English heiress died from a long illness. Autopsy showed that the cause of death was arsenic poisoning. The police suspected that her husband had administered the poison. The year prior to her death the heiress had taken three 1-month trips to America to visit friends on her own. Discuss how forensic analysis eventually helped the law enforcement build their case against her husband. [Hint: Arsenic poisoning was discussed in another chemical mystery in Chapter 4 (see p. 170). Studies show that within hours of ingesting as little as 3 mg of arsenic trioxide (As2O3), arsenic enters in the blood and becomes trapped and carried up the follicle in the growing hair. At the time of her death, the heiress had shoulder-length hair.] 1083 Appendix 1 Derivation of the Names of Elements* Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Actinium Ac 89 (227) 1899 A. Debierne (Fr.) Gr. aktis, beam or ray Aluminum Al 13 26.98 1827 F. Woehler (Ge.) Alum, the aluminum compound in which it was discovered; derived from L. alumen, astringent taste Americium Am 95 (243) 1944 A. Ghiorso (USA) The Americas R. A. James (USA) G. T. Seaborg (USA) S. G. Thompson (USA) Antimony Sb 51 121.8 Ancient L. antimonium (anti, opposite of; monium, isolated condition), so named because it is a tangible (metallic) substance which combines readily; symbol L. stibium, mark Argon Ar 18 39.95 1894 Lord Raleigh (GB) Gr. argos, inactive Sir William Ramsay (GB) Arsenic As 33 74.92 1250 Albertus Magnus (Ge.) Gr. aksenikon, yellow pigment; L. arsenicum, orpiment; the Greeks once used arsenic trisulfide as a pigment Astatine At 85 (210) 1940 D. R. Corson (USA) Gr. astatos, unstable K. R. MacKenzie (USA) E. Segre (USA) Barium Ba 56 137.3 1808 Sir Humphry Davy (GB) barite, a heavy spar, derived from Gr. barys, heavy Berkelium Bk 97 (247) 1950 G. T. Seaborg (USA) Berkeley, Calif. S. G. Thompson (USA) A. Ghiorso (USA) Beryllium Be 4 9.012 1828 F. Woehler (Ge.) Fr. L. beryl, sweet A. A. B. Bussy (Fr.) Source: Reprinted with permission from “The Elements and Derivation of Their Names and Symbols,” G. P. Dinga, Chemistry 41 (2), 20–22 (1968). Copyright by the American Chemical Society. *At the time this table was drawn up, only 103 elements were known to exist. † The atomic masses given here correspond to the 1961 values of the Commission on Atomic Weights. Masses in parentheses are those of the most stable or most common isotopes. ‡ The abbreviations are (Ar.) Arabic; (Au.) Austrian; (Du.) Dutch; (Fr.) French; (Ge.) German; (GB) British; (Gr.) Greek; (H.) Hungarian; (I.) Italian; (L.) Latin; (P.) Polish; (R.) Russian; (Sp.) Spanish; (Swe.) Swedish; (USA) American. (Continued) A-1  Appendix 1 A-2 Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Bismuth Bi 83 209.0 1753 Claude Geoffroy (Fr.) Ge. bismuth, probably a distortion of weisse masse (white mass) in which it was found Boron B 5 10.81 1808 Sir Humphry Davy (GB) The compound borax, J. L. Gay-Lussac (Fr.) derived from Ar. L. J. Thenard (Fr.) buraq, white Bromine Br 35 79.90 1826 A. J. Balard (Fr.) Gr. bromos, stench Cadmium Cd 48 112.4 1817 Fr. Stromeyer (Ge.) Gr. kadmia, earth; L. cadmia, calamine (because it is found along with calamine) Calcium Ca 20 40.08 1808 Sir Humphry Davy (GB) L. calx, lime Californium Cf 98 (249) 1950 G. T. Seaborg (USA) California S. G. Thompson (USA) A. Ghiorso (USA) K. Street, Jr. (USA) Carbon C 6 12.01 Ancient L. carbo, charcoal Cerium Ce 58 140.1 1803 J. J. Berzelius (Swe.) Asteroid Ceres William Hisinger (Swe.) M. H. Klaproth (Ge.) Cesium Cs 55 132.9 1860 R. Bunsen (Ge.) L. caesium, blue G. R. Kirchhoff (Ge.) (cesium was discovered by its spectral lines, which are blue) Chlorine Cl 17 35.45 1774 K. W. Scheele (Swe.) Gr. chloros, light green Chromium Cr 24 52.00 1797 L. N. Vauquelin (Fr.) Gr. chroma, color (because it is used in pigments) Cobalt Co 27 58.93 1735 G. Brandt (Ge.) Ge. kobold, goblin (because the ore yielded cobalt instead of the expected metal, copper, it was attributed to goblins) Copper Cu 29 63.55 Ancient L. cuprum, copper, derived from cyprium, Island of Cyprus, the main source of ancient copper Curium Cm 96 (247) 1944 G. T. Seaborg (USA) Pierre and Marie Curie R. A. James (USA) A. Ghiorso (USA) Dysprosium Dy 66 162.5 1886 Lecoq de Boisbaudran Gr. dysprositos, hard (Fr.) to get at Einsteinium Es 99 (254) 1952 A. Ghiorso (USA) Albert Einstein Erbium Er 68 167.3 1843 C. G. Mosander (Swe.) Ytterby, Sweden, where many rare earths were discovered Europium Eu 63 152.0 1896 E. Demarcay (Fr.) Europe (Continued) A-3 Appendix 1 Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Fermium Fm 100 (253) 1953 A. Ghiorso (USA) Enrico Fermi Fluorine F 9 19.00 1886 H. Moissan (Fr.) Mineral fluorspar, from L. fluere, flow (because fluorspar was used as a flux) Francium Fr 87 (223) 1939 Marguerite Perey (Fr.) France Gadolinium Gd 64 157.3 1880 J. C. Marignac (Fr.) Johan Gadolin, Finnish rare earth chemist Gallium Ga 31 69.72 1875 Lecoq de Boisbaudran L. Gallia, France (Fr.) Germanium Ge 32 72.59 1886 Clemens Winkler (Ge.) L. Germania, Germany Gold Au 79 197.0 Ancient L. aurum, shining dawn Hafnium Hf 72 178.5 1923 D. Coster (Du.) L. Hafnia, G. von Hevesey (H.) Copenhagen Helium He 2 4.003 1868 P. Janssen (spectr) (Fr.) Gr. helios, sun Sir William Ramsay (because it was first (isolated) (GB) discovered in the sun’s spectrum) Holmium Ho 67 164.9 1879 P. T. Cleve (Swe.) L. Holmia, Stockholm Hydrogen H 1 1.008 1766 Sir Henry Cavendish Gr. hydro, water; genes, (GB) forming (because it produces water when burned with oxygen) Indium In 49 114.8 1863 F. Reich (Ge.) Indigo, because of its T. Richter (Ge.) indigo blue lines in the spectrum Iodine I 53 126.9 1811 B. Courtois (Fr.) Gr. iodes, violet Iridium Ir 77 192.2 1803 S. Tennant (GB) L. iris, rainbow Iron Fe 26 55.85 Ancient L. ferrum, iron Krypton Kr 36 83.80 1898 Sir William Ramsay (GB) Gr. kryptos, hidden M. W. Travers (GB) Lanthanum La 57 138.9 1839 C. G. Mosander (Swe.) Gr. lanthanein, concealed Lawrencium Lr 103 (257) 1961 A. Ghiorso (USA) E. O. Lawrence (USA), T. Sikkeland (USA) inventor of the A. E. Larsh (USA) cyclotron R. M. Latimer (USA) Lead Pb 82 207.2 Ancient Symbol, L. plumbum, lead, meaning heavy Lithium Li 3 6.941 1817 A. Arfvedson (Swe.) Gr. lithos, rock (because it occurs in rocks) Lutetium Lu 71 175.0 1907 G. Urbain (Fr.) Luteria, ancient name C. A. von Welsbach (Au.) for Paris Magnesium Mg 12 24.31 1808 Sir Humphry Davy (GB) Magnesia, a district in Thessaly; possibly derived from L. magnesia Manganese Mn 25 54.94 1774 J. G. Gahn (Swe.) L. magnes, magnet (Continued)  Appendix 1 A-4 Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Mendelevium Md 101 (256) 1955 A. Ghiorso (USA) Mendeleev, Russian G. R. Choppin (USA) chemist who prepared G. T. Seaborg (USA) the periodic chart and B. G. Harvey (USA) predicted properties of S. G. Thompson (USA) undiscovered elements Mercury Hg 80 200.6 Ancient Symbol, L. hydrargyrum, liquid silver Molybdenum Mo 42 95.94 1778 G. W. Scheele (Swe.) Gr. molybdos, lead Neodymium Nd 60 144.2 1885 C. A. von Welsbach Gr. neos, new; (Au.) didymos, twin Neon Ne 10 20.18 1898 Sir William Ramsay (GB) Gr. neos, new M. W. Travers (GB) Neptunium Np 93 (237) 1940 E. M. McMillan (USA) Planet Neptune P. H. Abelson (USA) Nickel Ni 28 58.69 1751 A. F. Cronstedt (Swe.) Swe. kopparnickel, false copper; also Ge. nickel, referring to the devil that prevented copper from being extracted from nickel ores Niobium Nb 41 92.91 1801 Charles Hatchett (GB) Gr. Niobe, daughter of Tantalus (niobium was considered identical to tantalum, named after Tantalus, until 1884) Nitrogen N 7 14.01 1772 Daniel Rutherford (GB) Fr. nitrogene, derived from L. nitrum, native soda, or Gr. nitron, native soda, and Gr. genes, forming Nobelium No 102 (253) 1958 A. Ghiorso (USA) Alfred Nobel T. Sikkeland (USA) J. R. Walton (USA) G. T. Seaborg (USA) Osmium Os 76 190.2 1803 S. Tennant (GB) Gr. osme, odor Oxygen O 8 16.00 1774 Joseph Priestley (GB) Fr. oxygene, generator C. W. Scheele (Swe.) of acid, derived from Gr. oxys, acid, and L. genes, forming (because it was once thought to be a part of all acids) Palladium Pd 46 106.4 1803 W. H. Wollaston (GB) Asteroid Pallas Phosphorus P 15 30.97 1669 H. Brandt (Ge.) Gr. phosphoros, light bearing Platinum Pt 78 195.1 1735 A. de Ulloa (Sp.) Sp. platina, silver 1741 Charles Wood (GB) (Continued) A-5 Appendix 1 Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Plutonium Pu 94 (242) 1940 G. T. Seaborg (USA) Planet Pluto E. M. McMillan (USA) J. W. Kennedy (USA) A. C. Wahl (USA) Polonium Po 84 (210) 1898 Marie Curie (P.) Poland Potassium K 19 39.10 1807 Sir Humphry Davy (GB) Symbol, L. kalium, potash Praseodymium Pr 59 140.9 1885 C. A. von Welsbach Gr. prasios, green; (Au.) didymos, twin Promethium Pm 61 (147) 1945 J. A. Marinsky (USA) Gr. mythology, L. E. Glendenin (USA) Prometheus, the Greek C. D. Coryell (USA) Titan who stole fire from heaven Protactinium Pa 91 (231) 1917 O. Hahn (Ge.) Gr. protos, first; L. Meitner (Au.) actinium (because it disintegrates into actin- ium) Radium Ra 88 (226) 1898 Pierre and Marie Curie L. radius, ray (Fr., P.) Radon Rn 86 (222) 1900 F. E. Dorn (Ge.) Derived from radium Rhenium Re 75 186.2 1925 W. Noddack (Ge.) L. Rhenus, Rhine I. Tacke (Ge.) Otto Berg (Ge.) Rhodium Rh 45 102.9 1804 W. H. Wollaston (GB) Gr. rhodon, rose (because some of its salts are rose-colored) Rubidium Rb 37 85.47 1861 R. W. Bunsen (Ge.) L. rubidus, dark red G. Kirchhoff (Ge.) (discovered with the spectroscope, its spectrum shows red lines) Ruthenium Ru 44 101.1 1844 K. K. Klaus (R.) L. Ruthenia, Russia Samarium Sm 62 150.4 1879 Lecoq de Boisbaurdran Samarskite, after (Fr.) Samarski, a Russian engineer Scandium Sc 21 44.96 1879 L. F. Nilson (Swe.) Scandinavia Selenium Se 34 78.96 1817 J. J. Berzelius (Swe.) Gr. selene, moon (because it resembles tellurium, named for the earth) Silicon Si 14 28.09 1824 J. J. Berzelius (Swe.) L. silex, silicis, flint Silver Ag 47 107.9 Ancient Symbol, L. argentum, silver Sodium Na 11 22.99 1807 Sir Humphry Davy (GB) L. sodanum, headache remedy; symbol, L. natrium, soda Strontium Sr 38 87.62 1808 Sir Humphry Davy (GB) Strontian, Scotland, derived from mineral strontionite (Continued)  Appendix 1 A-6 Atomic Atomic Date of Discoverer and Element Symbol No. Mass† Discovery Nationality‡ Derivation Sulfur S 16 32.07 Ancient L. sulphurium (Sanskrit, sulvere) Tantalum Ta 73 180.9 1802 A. G. Ekeberg (Swe.) Gr. mythology, Tantalus, because of difficulty in isolating it Technetium Tc 43 (99) 1937 C. Perrier (I.) Gr. technetos, artificial (because it was the first artificial element) Tellurium Te 52 127.6 1782 F. J. Müller (Au.) L. tellus, earth Terbium Tb 65 158.9 1843 C. G. Mosander (Swe.) Ytterby, Sweden Thallium Tl 81 204.4 1861 Sir William Crookes Gr. thallos, a budding (GB) twig (because its spectrum shows a bright green line) Thorium Th 90 232.0 1828 J. J. Berzelius (Swe.) Mineral thorite, derived from Thor, Norse god of war Thulium Tm 69 168.9 1879 P. T. Cleve (Swe.) Thule, early name for Scandinavia Tin Sn 50 118.7 Ancient Symbol, L. stannum, tin Titanium Ti 22 47.88 1791 W. Gregor (GB) Gr. giants, the Titans, and L. titans, giant deities Tungsten W 74 183.9 1783 J. J. and F. de Swe. tung sten, heavy Elhuyar (Sp.) stone; symbol, wolframite, a mineral Uranium U 92 238.0 1789 M. H. Klaproth (Ge.) Planet Uranus 1841 E. M. Peligot (Fr.) Vanadium V 23 50.94 1801 A. M. del Rio (Sp.) Vanadis, Norse 1830 N. G. Sefstrom (Swe.) goddess of love and beauty Xenon Xe 54 131.3 1898 Sir William Ramsay Gr. xenos, stranger (GB) M. W. Travers (GB) Ytterbium Yb 70 173.0 1907 G. Urbain (Fr.) Ytterby, Sweden Yttrium Y 39 88.91 1843 C. G. Mosander (Swe.) Ytterby, Sweden Zinc Zn 30 65.39 1746 A. S. Marggraf (Ge.) Ge. zink, of obscure origin Zirconium Zr 40 91.22 1789 M. H. Klaproth (Ge.) Zircon, in which it was found, derived from Ar. zargum, gold color Appendix 2 Units for the Gas Constant In this appendix, we will see how the gas constant R can be expressed in units J/K ? mol. Our first step is to derive a relationship between atm and pascal. We start with force pressure 5 area mass 3 acceleration 5 area volume 3 density 3 acceleration 5 area 5 length 3 density 3 acceleration By definition, the standard atmosphere is the pressure exerted by a column of mercury exactly 76 cm high of density 13.5951 g/cm3, in a place where acceleration due to gravity is 980.665 cm/s2. However, to express pressure in N/m2 it is necessary to write density of mercury 5 1.35951 3 104 kgym3 acceleration due to gravity 5 9.80665 mys2 The standard atmosphere is given by 1 atm 5 (0.76 m Hg)(1.35951 3 104 kgym3)(9.80665 mys2) 5 101,325 kg mym2 ? s2 5 101,325 Nym2 5 101,325 Pa From Section 5.4 we see that the gas constant R is given by 0.082057 L ? atm/K ? mol. Using the conversion factors 1 L 5 1 3 1023 m3 1 atm 5 101,325 Nym2 we write 2 L atm 1 3 1023 m3 101,325 Nym R 5 a0.082057 ba ba b K mol 1L 1 atm Nm 5 8.314 K mol J 5 8.314 K mol and 1 L ? atm 5 (1 3 1023 m3)(101,325 Nym2) 5 101.3 N m 5 101.3 J A-7  Appendix 3 Thermodynamic Data at 1 atm and 25°C* Inorganic Substances Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) Ag(s) 0 0 42.7 Ag1(aq) 105.9 77.1 73.9 AgCl(s) 2127.0 2109.7 96.1 AgBr(s) 299.5 295.9 107.1 AgI(s) 262.4 266.3 114.2 AgNO3(s) 2123.1 232.2 140.9 Al(s) 0 0 28.3 Al31(aq) 2524.7 2481.2 2313.38 AlCl3(s) 2705.6 2630.0 109.3 Al2O3(s) 21669.8 21576.4 50.99 As(s) 0 0 35.15 AsO342(aq) 2870.3 2635.97 2144.77 AsH3(g) 171.5 H3AsO4(s) 2900.4 Au(s) 0 0 47.7 Au2O3(s) 80.8 163.2 125.5 AuCl(s) 235.2 AuCl3(s) 2118.4 B(s) 0 0 6.5 B2O3(s) 21263.6 21184.1 54.0 H3BO3(s) 21087.9 2963.16 89.58 H3BO3(aq) 21067.8 2963.3 159.8 Ba(s) 0 0 66.9 Ba21(aq) 2538.4 2560.66 12.55 BaO(s) 2558.2 2528.4 70.3 BaCl2(s) 2860.1 2810.66 125.5 BaSO4(s) 21464.4 21353.1 132.2 BaCO3(s) 21218.8 21138.9 112.1 Be(s) 0 0 9.5 BeO(s) 2610.9 2581.58 14.1 Br2(l) 0 0 152.3 Br2(g) 30.91 3.11 245.3 Br2(aq) 2120.9 2102.8 80.7 HBr(g) 236.2 253.2 198.48 C(graphite) 0 0 5.69 C(diamond) 1.90 2.87 2.4 CO(g) 2110.5 2137.3 197.9 CO2(g) 2393.5 2394.4 213.6 CO2(aq) 2412.9 2386.2 121.3 *The thermodynamic quantities of ions are based on the reference states that DH°f [H1(aq)] 5 0, DG°f [H1(aq)] 5 0, and S°[H1(aq)] 5 0 (see p. 782). (Continued) A-8 A-9 Appendix 3 Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) CO232(aq) 2676.3 2528.1 253.1 HCO2 3 (aq) 2691.1 2587.1 94.98 H2CO3(aq) 2699.7 2623.2 187.4 CS2(g) 115.3 65.1 237.8 CS2(l) 87.3 63.6 151.0 HCN(aq) 105.4 112.1 128.9 CN2(aq) 151.0 165.69 117.99 (NH2)2CO(s) 2333.19 2197.15 104.6 (NH2)2CO(aq) 2319.2 2203.84 173.85 Ca(s) 0 0 41.6 Ca21(aq) 2542.96 2553.0 255.2 CaO(s) 2635.6 2604.2 39.8 Ca(OH)2(s) 2986.6 2896.8 83.4 CaF2(s) 21214.6 21161.9 68.87 CaCl2(s) 2794.96 2750.19 113.8 CaSO4(s) 21432.69 21320.3 106.69 CaCO3(s) 21206.9 21128.8 92.9 Cd(s) 0 0 51.46 Cd21(aq) 272.38 277.7 261.09 CdO(s) 2254.6 2225.06 54.8 CdCl2(s) 2389.1 2342.59 118.4 CdSO4(s) 2926.17 2820.2 137.2 Cl2(g) 0 0 223.0 Cl2(aq) 2167.2 2131.2 56.5 HCl(g) 292.3 295.27 187.0 Co(s) 0 0 28.45 Co21(aq) 267.36 251.46 155.2 CoO(s) 2239.3 2213.38 43.9 Cr(s) 0 0 23.77 Cr21(aq) 2138.9 Cr2O3(s) 21128.4 21046.8 81.17 CrO422(aq) 2863.16 2706.26 38.49 Cr2O722(aq) 21460.6 21257.29 213.8 Cs(s) 0 0 82.8 Cs1(aq) 2247.69 2282.0 133.05 Cu(s) 0 0 33.3 Cu1(aq) 51.88 50.2 226.4 Cu21(aq) 64.39 64.98 299.6 CuO(s) 2155.2 2127.2 43.5 Cu2O(s) 2166.69 2146.36 100.8 CuCl(s) 2134.7 2118.8 91.6 CuCl2(s) 2205.85 ? ? CuS(s) 248.5 249.0 66.5 CuSO4(s) 2769.86 2661.9 113.39 F2(g) 0 0 203.34 F2(aq) 2329.1 2276.48 29.6 HF(g) 2271.6 2270.7 173.5 Fe(s) 0 0 27.2 (Continued)  Appendix 3 A-10 Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) Fe21(aq) 287.86 284.9 2113.39 Fe31(aq) 247.7 210.5 2293.3 FeCl3(s) 2400 2334 142.3 FeO(s) 2272.0 2255.2 60.8 Fe2O3(s) 2822.2 2741.0 90.0 Fe(OH)2(s) 2568.19 2483.55 79.5 Fe(OH)3(s) 2824.25 ? ? H(g) 218.2 203.2 114.6 H2(g) 0 0 131.0 H1(aq) 0 0 0 OH2(aq) 2229.94 2157.30 210.5 H2O(l) 2285.8 2237.2 69.9 H2O(g) 2241.8 2228.6 188.7 H2O2(l) 2187.6 2118.1 ? Hg(l) 0 0 77.4 Hg21(aq) 2164.38 HgO(s) 290.7 258.5 72.0 HgCl2(s) 2230.1 Hg2Cl2(s) 2264.9 2210.66 196.2 HgS(s) 258.16 248.8 77.8 HgSO4(s) 2704.17 Hg2SO4(s) 2741.99 2623.92 200.75 I2(s) 0 0 116.7 I2(g) 62.25 19.37 260.6 I(g) 106.6 70.16 180.7 I2(aq) 255.9 251.67 109.37 HI(g) 25.9 1.30 206.3 K(s) 0 0 63.6 K1(aq) 2251.2 2282.28 102.5 KOH(s) 2425.85 KCl(s) 2435.87 2408.3 82.68 KClO3(s) 2391.20 2289.9 142.97 KClO4(s) 2433.46 2304.18 151.0 KBr(s) 2392.17 2379.2 96.4 KI(s) 2327.65 2322.29 104.35 KNO3(s) 2492.7 2393.1 132.9 Li(s) 0 0 28.0 Li1(aq) 2278.46 2293.8 14.2 Li2O(s) 2595.8 ? ? LiOH(s) 2487.2 2443.9 50.2 Mg(s) 0 0 32.5 Mg21(aq) 2461.96 2456.0 2117.99 MgO(s) 2601.8 2569.6 26.78 Mg(OH)2(s) 2924.66 2833.75 63.1 MgCl2(s) 2641.8 2592.3 89.5 MgSO4(s) 21278.2 21173.6 91.6 MgCO3(s) 21112.9 21029.3 65.69 Mn(s) 0 0 31.76 (Continued) A-11 Appendix 3 Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) Mn21(aq) 2218.8 2223.4 283.68 MnO2(s) 2520.9 2466.1 53.1 N2(g) 0 0 191.5 N23 (aq) 245.18 ? ? NH3(g) 246.3 216.6 193.0 NH1 4 (aq) 2132.80 279.5 112.8 NH4Cl(s) 2315.39 2203.89 94.56 NH4NO3(s) 2365.6 2184.0 151 NH3(aq) 280.3 226.5 111.3 N2H4(l) 50.4 NO(g) 90.4 86.7 210.6 NO2(g) 33.85 51.8 240.46 N2O4(g) 9.66 98.29 304.3 N2O(g) 81.56 103.6 219.99 HNO2(aq) 2118.8 253.6 HNO3(l) 2173.2 279.9 155.6 NO2 3 (aq) 2206.57 2110.5 146.4 Na(s) 0 0 51.05 Na1(aq) 2239.66 2261.87 60.25 Na2O(s) 2415.9 2376.56 72.8 NaCl(s) 2411.0 2384.0 72.38 NaI(s) 2288.0 Na2SO4(s) 21384.49 21266.8 149.49 NaNO3(s) 2466.68 2365.89 116.3 Na2CO3(s) 21130.9 21047.67 135.98 NaHCO3(s) 2947.68 2851.86 102.09 Ni(s) 0 0 30.1 Ni21(aq) 264.0 246.4 2159.4 NiO(s) 2244.35 2216.3 38.58 Ni(OH)2(s) 2538.06 2453.1 79.5 O(g) 249.4 230.1 160.95 O2(g) 0 0 205.0 O3(aq) 212.09 16.3 110.88 O3(g) 142.2 163.4 237.6 P(white) 0 0 44.0 P(red ) 218.4 13.8 29.3 PO342(aq) 21284.07 21025.59 2217.57 P4O10(s) 23012.48 PH3(g) 9.25 18.2 210.0 HPO422(aq) 21298.7 21094.1 235.98 H2PO2 4 (aq) 21302.48 21135.1 89.1 Pb(s) 0 0 64.89 Pb21(aq) 1.6 224.3 21.3 PbO(s) 2217.86 2188.49 69.45 PbO2(s) 2276.65 2218.99 76.57 PbCl2(s) 2359.2 2313.97 136.4 PbS(s) 294.3 292.68 91.2 PbSO4(s) 2918.4 2811.2 147.28 (Continued)  Appendix 3 A-12 Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) Pt(s) 0 0 41.84 PtCl242(aq) 2516.3 2384.5 175.7 Rb(s) 0 0 69.45 Rb1(aq) 2246.4 2282.2 124.27 S(rhombic) 0 0 31.88 S(monoclinic) 0.30 0.10 32.55 SO2(g) 2296.4 2300.4 248.5 SO3(g) 2395.2 2370.4 256.2 SO232(aq) 2624.25 2497.06 43.5 SO242(aq) 2907.5 2741.99 17.15 H2S(g) 220.15 233.0 205.64 HSO2 3 (aq) 2627.98 2527.3 132.38 HSO2 4 (aq) 2885.75 2752.87 126.86 H2SO4(l) 2811.3 ? ? SF6(g) 21096.2 ? ? Si(s) 0 0 18.70 SiO2(s) 2859.3 2805.0 41.84 Sr(s) 0 0 54.39 Sr21(aq) 2545.5 2557.3 239.33 SrCl2(s) 2828.4 2781.15 117.15 SrSO4(s) 21444.74 21334.28 121.75 SrCO3(s) 21218.38 21137.6 97.07 Zn(s) 0 0 41.6 Zn21(aq) 2152.4 2147.2 2106.48 ZnO(s) 2348.0 2318.2 43.9 ZnCl2(s) 2415.89 2369.26 108.37 ZnS(s) 2202.9 2198.3 57.7 ZnSO4(s) 2978.6 2871.6 124.7 Organic Substances Substance Formula DH°f (kJ/mol) DG°f (kJ/mol) S° (J/K ? mol) Acetic acid(l) CH3COOH 2484.2 2389.45 159.8 Acetaldehyde(g) CH3CHO 2166.35 2139.08 264.2 Acetone(l) CH3COCH3 2246.8 2153.55 198.7 Acetylene(g) C2H2 226.6 209.2 200.8 Benzene(l) C6H6 49.04 124.5 172.8 Butane(g) C4H10 2124.7 215.7 310.0 Ethanol(l) C2H5OH 2276.98 2174.18 161.0 Ethanol(g) C2H5OH 2235.1 2168.5 282.7 Ethane(g) C2H6 284.7 232.89 229.5 Ethylene(g) C2H4 52.3 68.1 219.5 Formic acid(l) HCOOH 2409.2 2346.0 129.0 Glucose(s) C6H12O6 21274.5 2910.56 212.1 Methane(g) CH4 274.85 250.8 186.2 Methanol(l) CH3OH 2238.7 2166.3 126.8 Propane(g) C3H8 2103.9 223.5 269.9 Sucrose(s) C12H22O11 22221.7 21544.3 360.2 Appendix 4 Mathematical Operations Logarithms Common Logarithms The concept of the logarithms is an extension of the concept of exponents, which is discussed in Chapter 1. The common, or base-10, logarithm of any number is the power to which 10 must be raised to equal the number. The following examples illustrate this relationship: Logarithm Exponent log 150 100 5 1 log 10 5 1 101 5 10 log 100 5 2 102 5 100 log 1021 5 21 1021 5 0.1 log 1022 5 22 1022 5 0.01 In each case, the logarithm of the number can be obtained by inspection. Because the logarithms of numbers are exponents, they have the same properties as expo- nents. Thus, we have Logarithm Exponent log AB 5 log A 1 log B 10A 3 10B 5 10A1B A 10A log 5 log A 2 log B 5 10A2B B 10B Furthermore, log An 5 n log A. Now suppose we want to find the common logarithm of 6.7 3 1024. On most electronic calculators, the number is entered first and then the log key is pressed. This operation gives us log 6.7 3 1024 5 23.17 Note that there are as many digits after the decimal point as there are significant figures in the original number. The original number has two significant figures and the “17” in 23.17 tells us that the log has two significant figures. The “3” in 23.17 serves only to locate the decimal point in the number 6.7 3 1024. Other examples are Number Common Logarithm 62 1.79 0.872 20.0595 1.0 3 1027 27.00 Sometimes (as in the case of pH calculations) it is necessary to obtain the number whose logarithm is known. This procedure is known as taking the antilogarithm; it is simply the reverse of taking the logarithm of a number. Suppose in a certain calculation we have pH 5 1.46 and are asked to calculate [H1]. From the definition of pH (pH 5 2log [H1]) we can write [H1] 5 1021.46 Many calculators have a key labeled log21 or INV log to obtain antilogs. Other calculators have a 10x or y x key (where x corresponds to 21.46 in our example and y is 10 for base-10 loga- rithm). Therefore, we find that [H1] 5 0.035 M. A-13  Appendix 4 A-14 Natural Logarithms Logarithms taken to the base e instead of 10 are known as natural logarithms (denoted by ln or loge); e is equal to 2.7183. The relationship between common logarithms and natural loga- rithms is as follows: log 10 5 1 101 5 10 ln 10 5 2.303 e2.303 5 10 Thus, ln x 5 2.303 log x To find the natural logarithm of 2.27, say, we first enter the number on the electronic calculator and then press the ln key to get ln 2.27 5 0.820 If no ln key is provided, we can proceed as follows: 2.303 log 2.27 5 2.303 3 0.356 5 0.820 Sometimes we may be given the natural logarithm and asked to find the number it represents. For example, ln x 5 59.7 On many calculators, we simply enter the number and press the e key: x 5 e59.7 5 8 3 1025 The Quadratic Equation A quadratic equation takes the form ax2 1 bx 1 c 5 0 If coefficients a, b, and c are known, then x is given by 2b 6 2b2 2 4ac x5 2a Suppose we have the following quadratic equation: 2x2 1 5x 2 12 5 0 Solving for x, we write 25 6 2(5) 2 2 4(2) (212) x5 2(2) 25 6 225 1 96 5 4 Therefore, 25 1 11 3 x5 5 4 2 and 25 2 11 x5 5 24 4 Glossary The number in parentheses is the number of alkynes. Hydrocarbons that contain one or Aufbau principle. As protons are added one the section in which the term first appears. more carbon-carbon triple bonds. They by one to the nucleus to build up the have the general formula CnH2n22, where elements, electrons similarly are added to A n 5 2,3, . . . . (24.2) allotropes. Two or more forms of the same the atomic orbitals. (7.9) Avogadro’s law. At constant pressure and absolute temperature scale. A temperature element that differ significantly in temperature, the volume of a gas is scale that uses the absolute zero of chemical and physical properties. (2.6) directly proportional to the number of temperature as the lowest temperature. (5.3) alloy. A solid solution composed of two or moles of the gas present. (5.3) absolute zero. Theoretically the lowest more metals, or of a metal or metals with Avogadro’s number (NA). 6.022 3 1023; the attainable temperature. (5.3) one or more nonmetals. (21.2) number of particles in a mole. (3.2) acceptor impurities. Impurities that alpha particles. See alpha rays. can accept electrons from semiconductors. (21.3) alpha (a) rays. Helium ions with a positive charge of 12. (2.2) B accuracy. The closeness of a measurement to amalgam. An alloy of mercury with another band theory. Delocalized electrons move the true value of the quantity that is metal or metals. (21.2) freely through “bands” formed by measured. (1.8) amines. Organic bases that have the functional overlapping molecular orbitals. (21.3) acid. A substance that yields hydrogen ions group —NR2, where R may be H, an alkyl barometer. An instrument that measures (H1) when dissolved in water. (2.7) group, or an aromatic group. (24.4) atmospheric pressure. (5.2) acid ionization constant (Ka). The amino acids. A compound that contains at base. A substance that yields hydroxide ions equilibrium constant for the acid least one amino group and at least one (OH2) when dissolved in water. (2.7) ionization. (15.5) carboxyl group. (25.3) base ionization constant (Kb). The actinide series. Elements that have amorphous solid. A solid that lacks a regular equilibrium constant for the base incompletely filled 5f subshells or readily three-dimensional arrangement of atoms ionization. (15.6) give rise to cations that have incompletely or molecules. (11.7) battery. A galvanic cell, or a series of filled 5f subshells. (7.9) amphoteric oxide. An oxide that exhibits combined galvanic cells, that can be used activated complex. The species temporarily both acidic and basic properties. (8.6) as a source of direct electric current at a formed by the reactant molecules as a amplitude. The vertical distance from the constant voltage. (18.6) result of the collision before they form the middle of a wave to the peak or trough. (7.1) beta particles. See beta rays. product. (13.4) anion. An ion with a net negative charge. (2.5) beta (b) rays. Electrons. (2.2) activation energy (Ea). The minimum anode. The electrode at which oxidation bimolecular reaction. An elementary step amount of energy required to initiate a occurs. (18.2) that involves two molecules. (13.5) chemical reaction. (13.4) antibonding molecular orbital. A molecular binary compounds. Compounds formed activity series. A summary of the results of orbital that is of higher energy and lower from just two elements. (2.7) many possible displacement reactions. (4.4) stability than the atomic orbitals from boiling point. The temperature at which the actual yield. The amount of product actually which it was formed. (10.6) vapor pressure of a liquid is equal to the obtained in a reaction. (3.10) aqueous solution. A solution in which the external atmospheric pressure. (11.8) addition reaction. A reaction in which one solvent is water. (4.1) boiling-point elevation (DTb). The boiling molecule adds to another. (24.2) aromatic hydrocarbon. A hydrocarbon that point of the solution (Tb) minus the boiling adhesion. Attraction between unlike contains one or more benzene rings. (24.1) point of the pure solvent (T°b). (12.6) molecules. (11.3) atmospheric pressure. The pressure exerted bond enthalpy. The enthalpy change alcohol. An organic compound containing the by Earth’s atmosphere. (5.2) required to break a bond in a mole of hydroxyl group —OH. (24.4) atom. The basic unit of an element that can gaseous molecules. (9.10) aldehydes. Compounds with a carbonyl enter into chemical combination. (2.2) bond length. The distance between the nuclei functional group and the general formula atomic mass. The mass of an atom in atomic of two bonded atoms in a molecule. (9.4) RCHO, where R is an H atom, an alkyl, or mass units. (3.1) bond order. The difference between the num- an aromatic group. (24.4) atomic mass unit (amu). A mass exactly bers of electrons in bonding molecular or- aliphatic hydrocarbons. Hydrocarbons that equal to 121 th the mass of one carbon-12 bitals and antibonding molecular do not contain the benzene group or the atom. (3.1) orbitals, divided by two. (10.7) benzene ring. (24.1) atomic number (Z). The number of protons bonding molecular orbital. A molecular alkali metals. The Group 1A elements (Li, in the nucleus of an atom. (2.3) orbital that is of lower energy and greater Na, K, Rb, Cs, and Fr). (2.4) atomic orbital. The wave function (Ψ) of an stability than the atomic orbitals from alkaline earth metals. The Group 2A electron in an atom. (7.5) which it was formed. (10.6) elements (Be, Mg, Ca, Sr, Ba, and Ra). (2.4) atomic radius. One-half the distance be- Born-Haber cycle. The cycle that relates alkanes. Hydrocarbons having the general for- tween the two nuclei in two adjacent at- lattice energies of ionic compounds to mula CnH2n12, where n 5 1,2, . . . . (24.2) oms of the same element in a metal. For ionization energies, electron affinities, alkenes. Hydrocarbons that contain one or elements that exist as diatomic units, the heats of sublimation and formation, and more carbon-carbon double bonds. They atomic radius is one-half the distance be- bond enthalpies. (9.3) have the general formula CnH2n, where tween the nuclei of the two atoms in a boundary surface diagram. Diagram of the n 5 2,3, . . . . (24.2) particular molecule. (8.3) region containing a substantial amount of G-1  Glossary G-2 the electron density (about 90 percent) in converting the substance into some other coordination compound. A neutral species an orbital. (7.7) substance. (1.6) containing one or more complex ions. Boyle’s law. The volume of a fixed amount of chemical reaction. A process in which a (23.3) gas maintained at constant temperature is substance (or substances) is changed into coordination number. In a crystal lattice it is inversely proportional to the gas one or more new substances. (3.7) defined as the number of atoms (or ions) pressure. (5.3) chemistry. The study of matter and the surrounding an atom (or ion) (11.4). In breeder reactor. A nuclear reactor that changes it undergoes. (1.1) coordination compounds it is defined as produces more fissionable materials than chiral. Compounds or ions that are not the number of donor atoms surrounding it uses. (19.5) superimposable with their mirror the central metal atom in a complex. (23.3) Brønsted acid. A substance capable of images. (23.4) copolymer. A polymer containing two or donating a proton. (4.3) chlor-alkali process. The production of more different monomers. (25.2) Brønsted base. A substance capable of chlorine gas by the electrolysis of aqueous core electrons. All nonvalence electrons in an accepting a proton. (4.3) NaCl solution. (22.6) atom. (8.2) buffer solution. A solution of (a) a weak acid closed system. A system that enables the corrosion. The deterioration of metals by an or base and (b) its salt; both components exchange of energy (usually in the form electrochemical process. (18.7) must be present. The solution has the ability of heat) but not mass with its Coulomb’s law. The potential energy to resist changes in pH upon the addition of surroundings. (6.2) between two ions is directly proportional small amounts of either acid or base. (16.3) closest packing. The most efficient to the product of their charges and arrangements for packing atoms, inversely proportional to the distance C molecules, or ions in a crystal. (11.4) cohesion. The intermolecular attraction between them. (9.3) covalent bond. A bond in which two calorimetry. The measurement of heat between like molecules. (11.3) electrons are shared by two atoms. (9.4) changes. (6.5) colligative properties. Properties of solutions covalent compounds. Compounds containing carbides. Ionic compounds containing the that depend on the number of solute only covalent bonds. (9.4) C222 or C42 ion. (22.3) particles in solution and not on the nature critical mass. The minimum mass of carboxylic acids. Acids that contain the of the solute particles. (12.6) fissionable material required to generate carboxyl group —COOH. (24.4) colloid. A dispersion of particles of one a self-sustaining nuclear chain catalyst. A substance that increases the rate substance (the dispersed phase) throughout reaction. (19.5) of a chemical reaction without itself being a dispersing medium made of another critical pressure (Pc). The minimum pressure consumed. (13.6) substance. (12.8) necessary to bring about liquefaction at catenation. The ability of the atoms of combination reaction. A reaction in which the critical temperature. (11.8) an element to form bonds with one two or more substances combine to form a critical temperature (Tc). The temperature another. (22.3) single product. (4.4) above which a gas will not liquefy. (11.8) cathode. The electrode at which reduction combustion reaction. A reaction in which a crystal field splitting (D). The energy occurs. (18.2) substance reacts with oxygen, usually with difference between two sets of d orbitals cation. An ion with a net positive charge. (2.5) the release of heat and light, to produce a in a metal atom when ligands are cell voltage. Difference in electrical potential flame. (4.4) present. (23.5) between the anode and the cathode of a common ion effect. The shift in equilibrium crystalline solid. A solid that possesses rigid galvanic cell. (18.2) caused by the addition of a compound and long-range order; its atoms, molecules, Charles’ and Gay-Lussac’s law. See having an ion in common with the or ions occupy specific positions. (11.4) Charles’ law. dissolved substances. (16.2) crystallization. The process in which Charles’ law. The volume of a fixed amount complex ion. An ion containing a central dissolved solute comes out of solution of gas maintained at constant pressure is metal cation bonded to one or more and forms crystals. (12.1) directly proportional to the absolute molecules or ions. (16.10) cyanides. Compounds containing the temperature of the gas. (5.3) compound. A substance composed of atoms CN2 ion. (22.3) chelating agent. A substance that forms of two or more elements chemically united cycloalkanes. Alkanes whose carbon atoms complex ions with metal ions in in fixed proportions. (1.4) are joined in rings. (24.2) solution. (23.3) concentration of a solution. The amount of chemical energy. Energy stored within the structural units of chemical substances. (6.1) solute present in a given quantity of solvent or solution. (4.5) D chemical equation. An equation that uses condensation. The phenomenon of going from Dalton’s law of partial pressures. The total chemical symbols to show what happens the gaseous state to the liquid state. (11.8) pressure of a mixture of gases is just the during a chemical reaction. (3.7) condensation reaction. A reaction in which sum of the pressures that each gas would chemical equilibrium. A state in which the two smaller molecules combine to form a exert if it were present alone. (5.6) rates of the forward and reverse reactions larger molecule. Water is invariably one of decomposition reaction. The breakdown are equal. (14.1) the products of such a reaction. (24.4) of a compound into two or more chemical formula. An expression showing conductor. Substance capable of conducting components. (4.4) the chemical composition of a compound electric current. (21.3) delocalized molecular orbitals. Molecular in terms of the symbols for the atoms of conjugate acid-base pair. An acid and its orbitals that are not confined between two the elements involved. (2.6) conjugate base or a base and its conjugate adjacent bonding atoms but actually chemical kinetics. The area of chemistry acid. (15.1) extend over three or more atoms. (10.8) concerned with the speeds, or rates, at coordinate covalent bond. A bond in which denatured protein. Protein that does not which chemical reactions occur. (13.1) the pair of electrons is supplied by one of exhibit normal biological activities. (25.3) chemical property. Any property of a the two bonded atoms; also called a dative density. The mass of a substance divided by substance that cannot be studied without bond. (9.9) its volume. (1.6) G-3 Glossary deoxyribonucleic acids (DNA). A type of electrolyte. A substance that, when dissolved enzyme. A biological catalyst. (13.6) nucleic acid. (25.4) in water, results in a solution that can equilibrium constant (K). A number equal to deposition. The process in which the conduct electricity. (4.1) the ratio of the equilibrium concentrations molecules go directly from the vapor into electrolytic cell. An apparatus for carrying of products to the equilibrium concentrations the solid phase. (11.8) out electrolysis. (18.8) of reactants, each raised to the power of its diagonal relationship. Similarities between electromagnetic radiation. The emission and stoichiometric coefficient. (14.1) pairs of elements in different groups and transmission of energy in the form of elec- equilibrium vapor pressure. The vapor pres- periods of the periodic table. (8.6) tromagnetic waves. (7.1) sure measured under dynamic diamagnetic. Repelled by a magnet; a electromagnetic wave. A wave that has equilibrium of condensation and diamagnetic substance contains only an electric field component and a evaporation at some temperature. (11.8) paired electrons. (7.8) mutually perpendicular magnetic field equivalence point. The point at which the diatomic molecule. A molecule that consists component. (7.1) acid has completely reacted with or been of two atoms. (2.5) electromotive force (emf) (E). The voltage neutralized by the base. (4.7) diffusion. The gradual mixing of molecules difference between electrodes. (18.2) esters. Compounds that have the general of one gas with the molecules of another electron. A subatomic particle that has a very formula R9COOR, where R9 can be H or by virtue of their kinetic properties. (5.7) low mass and carries a single negative an alkyl group or an aromatic group and dilution. A procedure for preparing a less electric charge. (2.2) R is an alkyl group or an aromatic concentrated solution from a more electron affinity (EA). The negative of the en- group. (24.4) concentrated solution. (4.5) thalpy change when an electron is ether. An organic compound containing the dipole moment (μ). The product of charge accepted by an atom in the gaseous state R¬O¬R9 linkage, where R and R9 are and the distance between the charges in a to form an anion. (8.5) alkyl and/or aromatic groups. (24.4) molecule. (10.2) electron configuration. The distribution of evaporation. The process in which a liquid is dipole-dipole forces. Forces that act between electrons among the various orbitals in an transformed into a gas; also called polar molecules. (11.2) atom or molecule. (7.8) vaporization. (11.8) diprotic acid. Each unit of the acid yields electron density. The probability that an elec- excess reagents. One or more reactants two hydrogen ions upon ionization. (4.3) tron will be found at a particular present in quantities greater than necessary dispersion forces. The attractive forces that region in an atomic orbital. (7.5) to react with the quantity of the limiting arise as a result of temporary dipoles electronegativity. The ability of an atom to reagent. (3.9) induced in the atoms or molecules; also attract electrons toward itself in a chemical excited state (or level). A state that has called London forces. (11.2) bond. (9.5) higher energy than the ground displacement reaction. An atom or an ion in element. A substance that cannot be state. (7.3) a compound is replaced by an atom of separated into simpler substances by exothermic processes. Processes that give off another element. (4.4) chemical means. (1.4) heat to the surroundings. (6.2) disproportionation reaction. A reaction in elementary steps. A series of simple reactions extensive property. A property that which an element in one oxidation state is that represent the progress of the overall depends on how much matter is being both oxidized and reduced. (4.4) reaction at the molecular level. (13.5) considered. (1.6) donor atom. The atom in a ligand that is emission spectra. Continuous or line spectra bonded directly to the metal atom. (23.3) donor impurities. Impurities that emitted by substances. (7.3) empirical formula. An expression showing F provide conduction electrons to the types of elements present and the family. The elements in a vertical column of semiconductors. (21.3) simplest ratios of the different kinds of the periodic table. (2.4) double bond. Two atoms are held together by atoms. (2.6) Faraday constant. Charge contained in two pairs of electrons. (9.4) enantiomers. Optical isomers, that is, 1 mole of electrons, equivalent to dynamic equilibrium. The condition in compounds and their nonsuperimposable 96,485.3 coulombs. (18.4) which the rate of a forward process is mirror images. (23.4) ferromagnetic. Attracted by a magnet. The exactly balanced by the rate of a reverse endothermic processes. Processes that unpaired spins in a ferromagnetic process. (11.8) absorb heat from the surroundings. (6.2) substance are aligned in a common end point. The pH at which the indicator direction. (21.2) E changes color. (16.5) energy. The capacity to do work or to first law of thermodynamics. Energy can be converted from one form to another, but effective nuclear charge (Zeff). The nuclear produce change. (6.1) cannot be created or destroyed. (6.3) charge felt by an electron when both the enthalpy (H). A thermodynamic quantity first-order reaction. A reaction whose rate actual charge (Z ) and the repulsive effect used to describe heat changes taking place depends on reactant concentration raised (shielding) of the other electrons are taken at constant pressure. (6.4) to the first power. (13.3) into account. (8.3) enthalpy of reaction (DHrxn). The difference formal charge. The difference between the effusion. A process by which a gas under between the enthalpies of the products and valence electrons in an isolated atom and pressure escapes from one compartment of the enthalpies of the reactants. (6.4) the number of electrons assigned to that a container to another by passing through a enthalpy of solution (DHsoln). The heat atom in a Lewis structure. (9.7) small opening. (5.7) generated or absorbed when a certain formation constant (Kf). The equilibrium electrochemistry. The branch of chemistry that amount of solute is dissolved in a certain constant for the complex ion deals with the interconversion of electrical amount of solvent. (6.7) formation. (16.10) energy and chemical energy. (18.1) entropy (S). A measure of how dispersed fractional crystallization. The separation electrolysis. A process in which electrical the energy of a system is among the dif- of a mixture of substances into pure energy is used to cause a nonspontaneous ferent ways that system can contain components on the basis of their different chemical reaction to occur. (18.8) energy. (17.3) solubilities. (12.4)  Glossary G-4 fractional distillation. A procedure for half-life (t 12 ). The time required for the hydrocarbons. Compounds made up only of separating liquid components of a solution concentration of a reactant to decrease to carbon and hydrogen. (24.1) that is based on their different boiling half of its initial concentration. (13.3) hydrogen bond. A special type of dipole- points. (12.6) half-reaction. A reaction that explicitly dipole interaction between the hydrogen free energy (G). The energy available to do shows electrons involved in either atom bonded to an atom of a very useful work. (17.5) oxidation or reduction. (4.4) electronegative element (F, N, O) and freezing point. The temperature at which the halogens. The nonmetallic elements in Group another atom of one of the three solid and liquid phases of a substance 7A (F, Cl, Br, I, and At). (2.4) electronegative elements. (11.2) coexist at equilibrium. (11.8) heat. Transfer of energy between two bodies hydrogenation. The addition of hydrogen, freezing-point depression (DTf ). The that are at different temperatures. (6.2) especially to compounds with double and freezing point of the pure solvent (T°f ) heat capacity (C ). The amount of heat triple carbon-carbon bonds. (22.2) minus the freezing point of the solution required to raise the temperature of a given hydronium ion. The hydrated proton, (Tf). (12.6) quantity of the substance by one degree H3O1. (4.3) frequency (ν). The number of waves that Celsius. (6.5) hydrophilic. Water-liking. (12.8) pass through a particular point per unit heat of dilution. The heat change associated hydrophobic. Water-fearing. (12.8) time. (7.1) with the dilution process. (6.7) hypothesis. A tentative explanation for a set fuel cell. A galvanic cell that requires a heat of hydration (DHhydr). The heat of observations. (1.3) continuous supply of reactants to keep change associated with the hydration functioning. (18.6) functional group. That part of a molecule process. (6.7) heat of solution. See enthalpy of solution. I characterized by a special arrangement Heisenberg uncertainty principle. It is ideal gas. A hypothetical gas whose pressure- of atoms that is largely responsible for impossible to know simultaneously both volume-temperature behavior can be the chemical behavior of the parent the momentum and the position of a completely accounted for by the ideal molecule. (24.1) particle with certainty. (7.5) gas equation. (5.4) Henry’s law. The solubility of a gas in a ideal gas equation. An equation expressing G liquid is proportional to the pressure of the gas over the solution. (12.5) the relationships among pressure, volume, temperature, and amount of galvanic cell. The experimental apparatus for Hess’s law. When reactants are converted to gas (PV 5 nRT, where R is the gas generating electricity through the use of a products, the change in enthalpy is the constant). (5.4) spontaneous redox reaction. (18.2) same whether the reaction takes place in ideal solution. Any solution that obeys gamma (g) rays. High-energy radiation. (2.2) one step or in a series of steps. (6.6) Raoult’s law. (12.6) gas constant (R). The constant that appears heterogeneous equilibrium. An equilibrium indicators. Substances that have distinctly in the ideal gas equation. It is usually state in which the reacting species are not different colors in acidic and basic expressed as 0.08206 L ? atm/K ? mol, or all in the same phase. (14.2) media. (4.7) 8.314 J/K ? mol. (5.4) heterogeneous mixture. The individual induced dipole. The separation of positive geometric isomers. Compounds with the components of a mixture remain and negative charges in a neutral atom same type and number of atoms and the physically separated and can be seen as (or a nonpolar molecule) caused by same chemical bonds but different spatial separate components. (1.4) the proximity of an ion or a polar arrangements; such isomers cannot be homogeneous equilibrium. An equilibrium molecule. (11.2) interconverted without breaking a state in which all reacting species are in inert complex. A complex ion that undergoes chemical bond. (23.4) the same phase. (14.2) very slow ligand exchange reactions. Gibbs free energy. See free energy. homogeneous mixture. The composition of a (23.6) glass. The optically transparent fusion mixture, after sufficient stirring, is the inorganic compounds. Compounds other product of inorganic materials that has same throughout the solution. (1.4) than organic compounds. (2.7) cooled to a rigid state without homonuclear diatomic molecule. A insulator. A substance incapable of crystallizing. (11.7) diatomic molecule containing atoms of the conducting electricity. (21.3) Graham’s law of diffusion. Under the same same element. (10.7) intensive property. A property that does not conditions of temperature and pressure, homopolymer. A polymer that is made from depend on how much matter is being rates of diffusion for gases are inversely only one type of monomer. (25.2) considered. (1.6) proportional to the square roots of their Hund’s rule. The most stable arrangement intermediate. A species that appears in the molar masses. (5.7) of electrons in subshells is the one with mechanism of the reaction (that is, the gravimetric analysis. An experimental the greatest number of parallel spins. elementary steps) but not in the overall procedure that involves the measurement (7.8) balanced equation. (13.5) of mass. (4.6) hybrid orbitals. Atomic orbitals obtained intermolecular forces. Attractive forces that greenhouse effect. Carbon dioxide and when two or more nonequivalent orbitals exist among molecules. (11.2) other gases’ influence on Earth’s of the same atom combine. (10.4) International System of Units (SI). A temperature. (20.5) hybridization. The process of mixing the system of units based on metric units. (1.7) ground state (or level). The lowest energy atomic orbitals in an atom (usually the intramolecular forces. Forces that hold state of a system. (7.3) central atom) to generate a set of new atoms together in a molecule. (11.2) group. The elements in a vertical column of atomic orbitals. (10.4) ion. An atom or a group of atoms that has a the periodic table. (2.4) hydrates. Compounds that have a specific net positive or negative charge. (2.5) number of water molecules attached to ion pair. One or more cations and one or H them. (2.7) hydration. A process in which an ion or a more anions held together by electrostatic forces. (12.7) half-cell reactions. Oxidation and reduction molecule is surrounded by water molecules ionic bond. The electrostatic force that holds reactions at the electrodes. (18.2) arranged in a specific manner. (4.1) ions together in an ionic compound. (9.2) G-5 Glossary ionic compound. Any neutral compound phenomena that is always the same under masses of its protons, neutrons, and containing cations and anions. (2.5) the same conditions. (1.3) electrons. (19.2) ionic equation. An equation that shows law of conservation of energy. The total mass number (A). The total number of dissolved species as free ions. (4.2) quantity of energy in the universe is neutrons and protons present in the ionic radius. The radius of a cation or constant. (6.1) nucleus of an atom. (2.3) an anion as measured in an ionic law of conservation of mass. Matter can be matter. Anything that occupies space and compound. (8.3) neither created nor destroyed. (2.1) possesses mass. (1.4) ionization energy (IE). The minimum energy law of definite proportions. Different melting point. The temperature at which required to remove an electron from an samples of the same compound always solid and liquid phases coexist in isolated atom (or an ion) in its ground contain its constituent elements in the equilibrium. (11.8) state. (8.4) same proportions by mass. (2.1) mesosphere. A region between the ion-dipole forces. Forces that operate law of mass action. For a reversible reaction stratosphere and the ionosphere. (20.1) between an ion and a dipole. (11.2) at equilibrium and a constant temperature, metalloid. An element with properties ion-product constant. Product of hydrogen a certain ratio of reactant and product intermediate between those of metals and ion concentration and hydroxide ion concentrations has a constant value, K (the nonmetals. (2.4) concentration (both in molarity) at a equilibrium constant). (14.1) metallurgy. The science and technology of particular temperature. (15.2) law of multiple proportions. If two elements separating metals from their ores and of ionosphere. The uppermost layer of the can combine to form more than one type compounding alloys. (21.2) atmosphere. (20.1) of compound, the masses of one element metals. Elements that are good conductors isoelectronic. Ions, or atoms and ions, that that combine with a fixed mass of the of heat and electricity and have the possess the same number of electrons, other element are in ratios of small whole tendency to form positive ions in ionic and hence the same ground-state numbers. (2.1) compounds. (2.4) electron configuration, are said to be Le Châtelier’s principle. If an external metathesis reaction. A reaction that involves isoelectronic. (8.2) stress is applied to a system at equilib- the exchange of parts between two isolated system. A system that does not allow rium, the system will adjust itself in such compounds. (4.2) the transfer of either mass or energy to or a way as to partially offset the stress as microscopic properties. Properties that from its surroundings. (6.2) the system reaches a new equilibrium po- cannot be measured directly without the isotopes. Atoms having the same sition. (14.5) aid of a microscope or other special atomic number but different mass Lewis acid. A substance that can accept a instrument. (1.7) numbers. (2.3) pair of electrons. (15.12) mineral. A naturally occurring substance with Lewis base. A substance that can donate a a range of chemical composition. (21.1) pair of electrons. (15.12) miscible. Two liquids that are completely J Lewis dot symbol. The symbol of an element soluble in each other in all proportions are with one or more dots that represent the said to be miscible. (12.2) Joule (J). Unit of energy given by newtons 3 number of valence electrons in an atom of mixture. A combination of two or more meters. (5.7) the element. (9.1) substances in which the substances retain Lewis structure. A representation of covalent their identity. (1.4) bonding using Lewis symbols. Shared moderator. A substance that can reduce the K electron pairs are shown either as lines or kinetic energy of neutrons. (19.5) kelvin. The SI base unit of temperature. (1.7) as pairs of dots between two atoms, and molality. The number of moles of solute Kelvin temperature scale. See absolute lone pairs are shown as pairs of dots on dissolved in one kilogram of solvent. temperature scale. individual atoms. (9.4) (12.3) ketones. Compounds with a carbonyl ligand. A molecule or an ion that is bonded to molar concentration. See molarity. functional group and the general formula the metal ion in a complex ion. (23.3) molar heat of fusion (DHfus). The energy (in RR9CO, where R and R9 are alkyl and/or limiting reagent. The reactant used up first in kilojoules) required to melt one mole of a aromatic groups. (24.4) a reaction. (3.9) solid. (11.8) kinetic energy (KE). Energy available line spectra. Spectra produced when radiation molar heat of sublimation (DHsub). The because of the motion of an object. (5.7) is absorbed or emitted by substances only energy (in kilojoules) required to sublime kinetic molecular theory of gases. Treatment at some wavelengths. (7.3) one mole of a solid. (11.8) of gas behavior in terms of the random liter. The volume occupied by one cubic deci- molar heat of vaporization (DHvap). The motion of molecules. (5.7) meter. (1.7) energy (in kilojoules) required to vaporize lone pairs. Valence electrons that are not one mole of a liquid. (11.8) involved in covalent bond formation. (9.4) molar mass (m). The mass (in grams or L kilograms) of one mole of atoms, labile complex. Complexes that undergo M molecules, or other particles. (3.2) molar solubility. The number of moles of rapid ligand exchange reactions. (23.6) macroscopic properties. Properties that can solute in one liter of a saturated solution lanthanide (rare earth) series. Elements that be measured directly. (1.7) (mol/L). (16.6) have incompletely filled 4f subshells or manometer. A device used to measure the molarity (M ). The number of moles of solute readily give rise to cations that have pressure of gases. (5.2) in one liter of solution. (4.5) incompletely filled 4f subshells. (7.9) many-electron atoms. Atoms that contain mole (mol). The amount of substance that lattice energy. The energy required to com- two or more electrons. (7.5) contains as many elementary entities pletely separate one mole of a solid ionic mass. A measure of the quantity of matter (atoms, molecules, or other particles) as compound into gaseous ions. (6.7) contained in an object. (1.6) there are atoms in exactly 12 grams law. A concise verbal or mathematical mass defect. The difference between the (or 0.012 kilograms) of the carbon-12 statement of a relationship between mass of an atom and the sum of the isotope. (3.2)  Glossary G-6 mole fraction. Ratio of the number of moles nonpolar molecule. A molecule that does not direction indicated by the difference in of one component of a mixture to the total possess a dipole moment. (10.2) electronegativity. (4.4) number of moles of all components in the nonvolatile. Does not have a measurable oxidation reaction. The half-reaction that mixture. (5.6) vapor pressure. (12.6) involves the loss of electrons. (4.4) mole method. An approach for determining n-type semiconductors. Semiconductors that oxidation-reduction reaction. A reaction the amount of product formed in a contain donor impurities. (21.3) that involves the transfer of electron(s) or reaction. (3.8) nuclear binding energy. The energy required the change in the oxidation state of molecular equations. Equations in which the to break up a nucleus into its protons and reactants. (4.4) formulas of the compounds are written as neutrons. (19.2) oxidation state. See oxidation number. though all species existed as molecules or nuclear chain reaction. A self-sustaining oxidizing agent. A substance that can accept whole units. (4.2) sequence of nuclear fission reactions. (19.5) electrons from another substance or molecular formula. An expression showing nuclear fission. A heavy nucleus (mass increase the oxidation numbers in another the exact numbers of atoms of each number . 200) divides to form smaller substance. (4.4) element in a molecule. (2.6) nuclei of intermediate mass and one or oxoacid. An acid containing hydrogen, molecular mass. The sum of the atomic masses more neutrons. (19.5) oxygen, and another element (the central (in amu) present in the molecule. (3.3) nuclear fusion. The combining of small element). (2.7) molecular orbital. An orbital that results nuclei into larger ones. (19.6) oxoanion. An anion derived from an from the interaction of the atomic orbitals nuclear transmutation. The change oxoacid. (2.7) of the bonding atoms. (10.6) undergone by a nucleus as a result of molecularity of a reaction. The number of molecules reacting in an elementary bombardment by neutrons or other particles. (19.1) P step. (13.5) nucleic acids. High molar mass polymers paramagnetic. Attracted by a magnet. A molecule. An aggregate of at least two atoms that play an essential role in protein paramagnetic substance contains one or in a definite arrangement held together by synthesis. (25.4) more unpaired electrons. (7.8) special forces. (2.5) nucleon. A general term for the protons and partial pressure. Pressure of one component monatomic ion. An ion that contains only neutrons in a nucleus. (19.2) in a mixture of gases. (5.6) one atom. (2.5) nucleotide. The repeating unit in each strand pascal (Pa). A pressure of one newton per monomer. The single repeating unit of a of a DNA molecule which consists of a square meter (1 N/m2). (5.2) polymer. (25.2) base-deoxyribose-phosphate linkage. Pauli exclusion principle. No two electrons monoprotic acid. Each unit of the acid yields (25.4) in an atom can have the same four one hydrogen ion upon ionization. (4.3) nucleus. The central core of an atom. (2.2) quantum numbers. (7.8) multiple bonds. Bonds formed when two percent by mass. The ratio of the mass of a atoms share two or more pairs of electrons. (9.4) O solute to the mass of the solution, multiplied by 100 percent. (12.3) octet rule. An atom other than hydrogen percent composition by mass. The percent by N tends to form bonds until it is surrounded by eight valence electrons. (9.4) mass of each element in a compound. (3.5) percent ionization. Ratio of ionized acid Nernst equation. The relation between the open system. A system that can exchange concentration at equilibrium to the initial emf of a galvanic cell and the standard emf mass and energy (usually in the form of concentration of acid. (15.5) and the concentrations of the oxidizing and heat) with its surroundings. (6.2) percent yield. The ratio of actual yield to reducing agents. (18.5) optical isomers. Compounds that are theoretical yield, multiplied by 100 net ionic equation. An equation that nonsuperimposable mirror images. (23.4) percent. (3.10) indicates only the ionic species that ore. The material of a mineral deposit in a period. A horizontal row of the periodic actually take part in the reaction. (4.2) sufficiently concentrated form to allow table. (2.4) neutralization reaction. A reaction between economical recovery of a desired periodic table. A tabular arrangement of the an acid and a base. (4.3) metal. (21.1) elements. (2.4) neutron. A subatomic particle that bears no organic chemistry. The branch of chemistry pH. The negative logarithm of the hydrogen net electric charge. Its mass is slightly that deals with carbon compounds. (24.1) ion concentration. (15.3) greater than a proton’s. (2.2) organic compounds. Compounds that phase. A homogeneous part of a system in newton (N). The SI unit for force. (5.2) contain carbon, usually in combination contact with other parts of the system but nitrogen fixation. The conversion of molecu- with elements such as hydrogen, oxygen, separated from them by a well-defined lar nitrogen into nitrogen compounds. nitrogen, and sulfur. (2.7) boundary. (11.1) (20.1) osmosis. The net movement of solvent phase change. Transformation from one noble gas core. The electron configuration of molecules through a semipermeable phase to another. (11.8) the noble gas element that most nearly membrane from a pure solvent or from a phase diagram. A diagram showing the precedes the element being considered. (7.9) dilute solution to a more concentrated conditions at which a substance exists as a noble gases. Nonmetallic elements in Group solution. (12.6) solid, liquid, or vapor. (11.9) 8A (He, Ne, Ar, Kr, Xe, and Rn). (2.4) osmotic pressure (π). The pressure required photochemical smog. Formation of smog by node. The point at which the amplitude of the to stop osmosis. (12.6) the reactions of automobile exhaust in the wave is zero. (7.4) overvoltage. The difference between the elec- presence of sunlight. (20.7) nonelectrolyte. A substance that, when dis- trode potential and the actual voltage re- photoelectric effect. A phenomenon in which solved in water, gives a solution that is not quired to cause electrolysis. (18.8) electrons are ejected from the surface of electrically conducting. (4.1) oxidation number. The number of charges an certain metals exposed to light of at least a nonmetals. Elements that are usually poor atom would have in a molecule if electrons certain minimum frequency. (7.2) conductors of heat and electricity. (2.4) were transferred completely in the photon. A particle of light. (7.2) G-7 Glossary physical equilibrium. An equilibrium in which quantitative. Comprising numbers reducing agent. A substance that can donate only physical properties change. (14.1) obtained by various measurements of electrons to another substance or decrease physical property. Any property of a the system. (1.3) the oxidation numbers in another substance that can be observed without quantitative analysis. The determination of substance. (4.4) transforming the substance into some other the amount of substances present in a reduction reaction. The half-reaction that substance. (1.6) sample. (4.5) involves the gain of electrons. (4.4) pi bond (p). A covalent bond formed by side- quantum. The smallest quantity of energy representative elements. Elements in Groups ways overlapping orbitals; its electron that can be emitted (or absorbed) in the 1A through 7A, all of which have density is concentrated above and below form of electromagnetic radiation. (7.1) incompletely filled s or p subshell of the plane of the nuclei of the bonding quantum numbers. Numbers that describe highest principal quantum number. (8.2) atoms. (10.5) the distribution of electrons in hydrogen resonance. The use of two or more Lewis pi molecular orbital. A molecular orbital in and other atoms. (7.6) structures to represent a particular which the electron density is concentrated molecule. (9.8) above and below the plane of the two nuclei of the bonding atoms. (10.6) R resonance structure. One of two or more alternative Lewis structures for a molecule plasma. A gaseous mixture of positive ions racemic mixture. An equimolar mixture of that cannot be described fully with a single and electrons. (19.6) the two enantiomers. (23.4) Lewis structure. (9.8) polar covalent bond. In such a bond, the radiant energy. Energy transmitted in the reversible reaction. A reaction that can occur electrons spend more time in the vicinity form of waves. (6.1) in both directions. (4.1) of one atom than the other. (9.5) radiation. The emission and transmission of ribonucleic acid (RNA). A form of nucleic polar molecule. A molecule that possesses a energy through space in the form of acid. (25.4) dipole moment. (10.2) particles and/or waves. (2.2) root-mean-square (rms) speed (urms). A polarimeter. The instrument for measuring radical. Any neutral fragment of a molecule measure of the average molecular speed at the rotation of polarized light by optical containing an unpaired electron. (19.8) a given temperature. (5.7) isomers. (23.4) radioactive decay series. A sequence of polyatomic ion. An ion that contains more than one atom. (2.5) nuclear reactions that ultimately result in the formation of a stable isotope. (19.3) S polyatomic molecule. A molecule that radioactivity. The spontaneous breakdown of salt. An ionic compound made up of a cation consists of more than two atoms. (2.5) an atom by emission of particles and/or other than H1 and an anion other than OH2 polymer. A compound distinguished by a radiation. (2.2) or O22. (4.3) high molar mass, ranging into thousands Raoult’s law. The vapor pressure of the salt hydrolysis. The reaction of the anion or and millions of grams, and made up of solvent over a solution is given by the cation, or both, of a salt with water. (15.10) many repeating units. (25.1) product of the vapor pressure of the pure saponification. Soapmaking. (24.4) positron. A particle that has the same mass solvent and the mole fraction of the saturated hydrocarbons. Hydrocarbons that as the electron, but bears a 11 charge. solvent in the solution. (12.6) contain the maximum number of hydrogen (19.1) rare earth series. See lanthanide series. atoms that can bond with the number of potential energy. Energy available by virtue rate constant (k). Constant of proportionality carbon atoms present. (24.2) of an object’s position. (6.1) between the reaction rate and the saturated solution. At a given temperature, precipitate. An insoluble solid that separates concentrations of reactants. (13.1) the solution that results when the from the solution. (4.2) rate law. An expression relating the rate of a maximum amount of a substance has precipitation reaction. A reaction that results reaction to the rate constant and the dissolved in a solvent. (12.1) in the formation of a precipitate. (4.2) concentrations of the reactants. (13.2) scientific method. A systematic approach to precision. The closeness of agreement of two rate-determining step. The slowest step in research. (1.3) or more measurements of the same the sequence of steps leading to the second law of thermodynamics. The entropy quantity. (1.8) formation of products. (13.5) of the universe increases in a spontaneous pressure. Force applied per unit area. (5.2) reactants. The starting substances in a process and remains unchanged in an product. The substance formed as a result of chemical reaction. (3.7) equilibrium process. (17.4) a chemical reaction. (3.7) reaction mechanism. The sequence of second-order reaction. A reaction whose protein. Polymers of amino acids. (25.3) elementary steps that leads to product rate depends on reactant concentration proton. A subatomic particle having a single formation. (13.5) raised to the second power or on the positive electric charge. The mass of a pro- reaction order. The sum of the powers to concentrations of two different reactants, ton is about 1840 times that of an electron. which all reactant concentrations each raised to the first power. (13.3) (2.2) appearing in the rate law are raised. (13.2) semiconductors. Elements that normally p-type semiconductors. Semiconductors that reaction quotient (Qc). A number equal to the cannot conduct electricity, but can have contain acceptor impurities. (21.3) ratio of product concentrations to reactant their conductivity greatly enhanced either pyrometallurgy. Metallurgical processes that concentrations, each raised to the power of by raising the temperature or by adding are carried out at high temperatures. (21.2) its stoichiometric coefficient at some point certain impurities. (21.3) other than equilibrium. (14.4) semipermeable membrane. A membrane Q reaction rate. The change in the concentration of reactant or product with that enables solvent molecules to pass through, but blocks the movement of qualitative. Consisting of general time. (13.1) solute molecules. (12.6) observations about the system. (1.3) redox reaction. A reaction in which there is sigma bond (s). A covalent bond formed by qualitative analysis. The determination either a transfer of electrons or a change in orbitals overlapping end-to-end; its of the types of ions present in a the oxidation numbers of the substances electron density is concentrated between solution. (16.11) taking part in the reaction. (4.4) the nuclei of the bonding atoms. (10.5)  Glossary G-8 sigma molecular orbital. A molecular orbital in which the electron density is standard reduction potential. The voltage measured as a reduction reaction occurs at T concentrated around a line between the the electrode when all solutes are 1 M and termolecular reaction. An elementary step two nuclei of the bonding atoms. (10.6) all gases are at 1 atm. (18.3) that involves three molecules. (13.5) significant figures. The number of standard solution. A solution of accurately ternary compounds. Compounds consisting meaningful digits in a measured or known concentration. (4.7) of three elements. (2.7) calculated quantity. (1.8) standard state. The condition of 1 atm of theoretical yield. The amount of product single bond. Two atoms are held together by pressure. (6.6) predicted by the balanced equation when all one electron pair. (9.4) standard temperature and pressure (STP). of the limiting reagent has reacted. (3.10) solubility. The maximum amount of solute 08C and 1 atm. (5.4) theory. A unifying principle that explains a that can be dissolved in a given quantity of state function. A property that is determined body of facts and the laws that are based solvent at a specific temperature. (4.2, 16.6) by the state of the system. (6.3) on them. (1.3) solubility product (Ksp). The product of the state of a system. The values of all pertinent thermal energy. Energy associated molar concentrations of the constituent macroscopic variables (for example, with the random motion of atoms and ions, each raised to the power of its composition, volume, pressure, and molecules. (6.1) stoichiometric coefficient in the temperature) of a system. (6.3) thermochemical equation. An equation that equilibrium equation. (16.6) stereoisomers. Compounds that are made up shows both the mass and enthalpy solute. The substance present in smaller of the same types and numbers of atoms relations. (6.4) amount in a solution. (4.1) bonded together in the same sequence but thermochemistry. The study of heat changes solution. A homogeneous mixture of two or with different spatial arrangements. (23.4) in chemical reactions. (6.2) more substances. (4.1) stoichiometric amounts. The exact molar thermodynamics. The scientific study of the solvation. The process in which an ion or a amounts of reactants and products that interconversion of heat and other forms of molecule is surrounded by solvent appear in the balanced chemical energy. (6.3) molecules arranged in a specific equation. (3.9) thermonuclear reactions. Nuclear fusion manner. (12.2) stoichiometry. The quantitative study of reactions that occur at very high solvent. The substance present in larger reactants and products in a chemical temperatures. (19.6) amount in a solution. (4.1) reaction. (3.8) thermosphere. The region of the atmosphere specific heat (s). The amount of heat energy stratosphere. The region of the atmosphere in which the temperature increases required to raise the temperature of one extending upward from the troposphere to continuously with altitude. (20.1) gram of a substance by one degree about 50 km from Earth. (20.1) third law of thermodynamics. The entropy Celsius. (6.5) strong acids. Strong electrolytes which are of a perfect crystalline substance is zero at spectator ions. Ions that are not involved in assumed to ionize completely in the absolute zero of temperature. (17.4) the overall reaction. (4.2) water. (15.4) titration. The gradual addition of a solution spectrochemical series. A list of ligands strong bases. Strong electrolytes which are of accurately known concentration to arranged in increasing order of their abilities assumed to ionize completely in another solution of unknown concentration to split the d-orbital energy levels. (23.5) water. (15.4) until the chemical reaction between the standard atmospheric pressure (1 atm). structural formula. A chemical formula that two solutions is complete. (4.7) The pressure that supports a column of shows how atoms are bonded to one tracers. Isotopes, especially radioactive mercury exactly 76 cm high at 08C at sea another in a molecule. (2.6) isotopes, that are used to trace the path of level. (5.2) structural isomers. Molecules that have the the atoms of an element in a chemical or standard emf (E8). The difference of the same molecular formula but different biological process. (19.7) standard reduction potential of the substance structures. (24.2) transition metals. Elements that have that undergoes reduction and the standard sublimation. The process in which molecules incompletely filled d subshells or readily reduction potential of the substance that go directly from the solid into the vapor give rise to cations that have incompletely undergoes oxidation. (18.3) phase. (11.8) filled d subshells. (7.9) standard enthalpy of formation (DH8f ). The substance. A form of matter that has a transition state. See activated complex. heat change that results when one mole of definite or constant composition (the transuranium elements. Elements with a compound is formed from its elements in number and type of basic units present) atomic numbers greater than 92. (19.4) their standard states. (6.6) and distinct properties. (1.4) triple bond. Two atoms are held together by standard enthalpy of reaction (DH8rxn). The substitution reaction. A reaction in which an three pairs of electrons. (9.4) enthalpy change when the reaction is atom or group of atoms replaces an atom or triple point. The point at which the vapor, carried out under standard-state groups of atoms in another molecule. (24.3) liquid, and solid states of a substance are conditions. (6.6) supercooling. Cooling of a liquid below its in equilibrium. (11.9) standard entropy of reaction (DS8rxn). The freezing point without forming the triprotic acid. Each unit of the acid yields entropy change when the reaction is solid. (11.8) three protons upon ionization. (4.3) carried out under standard-state supersaturated solution. A solution that troposphere. The layer of the atmosphere conditions. (17.4) contains more of the solute than is present which contains about 80 percent of the standard free-energy of formation (DG8f ). in a saturated solution. (12.1) total mass of air and practically all of the The free-energy change when 1 mole of a surface tension. The amount of energy atmosphere’s water vapor. (20.1) compound is synthesized from its elements required to stretch or increase the surface in their standard states. (17.5) standard free-energy of reaction (DG8rxn). of a liquid by a unit area. (11.3) surroundings. The rest of the universe U The free-energy change when the reaction outside a system. (6.2) unimolecular reaction. An elementary step is carried out under standard-state system. Any specific part of the universe that in which only one reacting molecule conditions. (17.5) is of interest to us. (6.2) participates. (13.5) G-9 Glossary unit cell. The basic repeating unit of the arrangement of atoms, molecules, or ions and unshared electron pairs around a central atom in terms of the repulsions W in a crystalline solid. (11.4) between electron pairs. (10.1) wave. A vibrating disturbance by which unsaturated hydrocarbons. Hydrocarbons van der Waals equation. An equation that energy is transmitted. (7.1) that contain carbon-carbon double bonds describes the P, V, and T of a nonideal wavelength (λ). The distance between or carbon-carbon triple bonds. (24.2) gas. (5.8) identical points on successive waves. (7.1) unsaturated solution. A solution that van der Waals forces. The dipole-dipole, weak acids. Weak electrolytes that ionize contains less solute than it has the capacity dipole-induced dipole, and dispersion only to a limited extent in water. (15.4) to dissolve. (12.1) forces. (11.2) weak bases. Weak electrolytes that ionize van’t Hoff factor (i). The ratio of actual only to a limited extent in water. (15.4) V number of particles in solution after dissociation to the number of formula weight. The force that gravity exerts on an object. (1.7) valence electrons. The outer electrons of an units initially dissolved in solution. (12.7) work. Directed energy change resulting from atom, which are those involved in vaporization. The escape of molecules from a process. (6.1) chemical bonding. (8.2) the surface of a liquid; also called valence shell. The outermost electron- occupied shell of an atom, which holds evaporation. (11.8) viscosity. A measure of a fluid’s resistance to X the electrons that are usually involved in flow. (11.3) X-ray diffraction. The scattering of X rays bonding. (10.1) volatile. Has a measurable vapor by the units of a regular crystalline valence-shell electron-pair repulsion pressure. (12.6) solid. (11.5) (VSEPR) model. A model that accounts volume. It is the length cubed. (1.6) for the geometrical arrangements of shared  Answers to Even-Numbered Problems Chapter 1 (d) HI(aq). (e) Na2(NH4)PO4. (f) PbCO3. (g) SnF2. (h) P4S10. (i) HgO. (j) Hg2I2. (k) SeF6. 2.62 (a) Dinitrogen pentoxide (N2O5). 1.4 (a) Hypothesis. (b) Law. (c) Theory. 1.12 (a) Physical change. (b) Boron trifluoride (BF3). (c) Dialuminum hexabromide (Al2Br6). (b) Chemical change. (c) Physical change. (d) Chemical change. 2.64 (a) 52 22 107 127 239 25Mn. (b) 10Ne. (c) 47 Ag. (d) 53 I. (e) 94 Pu. (e) Physical change. 1.14 (a) Cs. (b) Ge. (c) Ga. (d) Sr. (e) U. 2.66 (c) Changing the electrical charge of an atom usually has a (f) Se. (g) Ne. (h) Cd. 1.16 (a) Homogeneous mixture. (b) Element. major effect on its chemical properties. 2.68 I2. 2.70 NaCl is an (c) Compound. (d) Homogeneous mixture. (e) Heterogeneous ionic compound. It does not form molecules. 2.72 Element: mixture. (f) Heterogeneous mixture. (g) Element. 1.22 71.2 g. (b), (c), (e), (f), (g), (j), (k). Molecules but not compounds: (b), (f), 1.24 (a) 418C. (b) 11.38F. (c) 1.1 3 1048F. (d) 2338C. (g), (k). Compounds but not molecules: (i), (l). Compounds and 1.26 (a) 21968C. (b) 22698C. (c) 3288C. 1.30 (a) 0.0152. molecules: (a), (d), (h). 2.74 (a) Ne: 10 p, 10 n. (b) Cu: 29 p, 34 n. (b) 0.0000000778. 1.32 (a) 1.8 3 1022. (b) 1.14 3 1010. (c) 25 3 (c) Ag: 47 p, 60 n. (d) W: 74 p, 108 n. (e) Po: 84 p, 119 n. (f) Pu: 104. (d) 1.3 3 103. 1.34 (a) One. (b) Three. (c) Three. (d) Four. 94 p, 140 n. 2.76 (a) Cu. (b) P. (c) Kr. (d) Cs. (e) Al. (f) Sb. (g) Cl. (e) Two or three. (f) One. (g) One or two. 1.36 (a) 1.28. (b) 3.18 3 (h) Sr. 2.78 (a) The magnitude of a particle scattering depends on 1023 mg. (c) 8.14 3 107 dm. (d) 3.8 m/s. 1.38 Tailor X’s the number of protons present. (b) Density of nucleus: 3.25 3 measurements are the most precise. Tailor Y’s measurements are 1014 g/cm3; density of space occupied by electrons: 3.72 3 the least accurate and least precise. Tailor Z’s measurements are 1024 g/cm3. The result supports Rutherford’s model. 2.80 The the most accurate. 1.40 (a) 1.10 3 108 mg. (b) 6.83 3 1025 m3. empirical and molecular formulas of acetaminophen are both (c) 7.2 3 103 L. (d) 6.24 3 1028 lb. 1.42 3.1557 3 107 s. C8H9NO2. 2.82 (a) Tin(IV) chloride. (b) Copper(I) oxide. (c) 1.44 (a) 118 in/s. (b) 1.80 3 102 m/min. (c) 10.8 km/h. Cobalt(II) nitrate. (d) Sodium dichromate. 2.84 (a) Ionic 1.46 178 mph. 1.48 3.7 3 1023 g Pb. 1.50 (a) 1.5 3 102 lb. compounds formed between metallic and nonmetallic elements. (b) 4.4 3 1017 s. (c) 2.3 m. (d) 8.86 3 104 L. 1.52 6.25 3 (b) Transition metals, lanthanides, and actinides. 2.86 23Na. 1024 g/cm3. 1.54 (a) Chemical. (b) Chemical. (c) Physical. 2.88 Hg and Br2. 2.90 H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn. (d) Physical. (e) Chemical. 1.56 2.6 g/cm3. 1.58 9.20 cm. 2.92 Unreactive. He, Ne, and Ar are chemically inert. 2.94 Ra is a 1.60 767 mph. 1.62 Liquid must be less dense than ice; radioactive decay product of U-238. 2.96 77Se22. 2.98 (a) NaH, temperature below 08C. 1.64 2.3 3 103 cm3. 1.66 6.4¢. 1.68 738S. sodium hydride. (b) B2O3, diboron trioxide. (c) Na2S, sodium 1.70 (a) 8.6 3 103 L air/day. (b) 0.018 L CO/day. 1.72 26,700,000 sulfide. (d) AlF3, aluminum fluoride. (e) OF2, oxygen difluoride. basketballs. 1.74 7.0 3 1020 L. 1.76 88 lb; 40 kg. 1.78 O: 4.0 3 104 g; (f) SrCl2, strontium chloride. 2.100 NF3 (nitrogen trifluoride), PBr5 C: 1.1 3 104 g; H: 6.2 3 103 g; N: 2 3 103 g; Ca: 9.9 3 102 g; (phosphorus pentabromide), SCl2 (sulfur dichloride). 2.102 1st row: P: 7.4 3 102 g. 1.80 4.6 3 1028C; 8.6 3 1028F. 1.82 $2.4 3 1012. Mg21, HCO2 21 2 3 , Mg(HCO3)2. 2nd row: Sr , Cl , strontium chloride. 1.84 5.4 3 1022 Fe atoms. 1.86 29 times. 1.88 1.450 3 1022 mm. 3rd row: Fe(NO2)3, iron(III) nitrite. 4th row: Mn21, ClO2 3, 1.90 1.3 3 103 mL. 1.92 (a) 11.063 mL. (b) 0.78900 g/mL. Mn(ClO3)2. 5th row: Sn41, Br2, tin(IV) bromide. 6th row: (c) 7.140 g/mL. 1.94 0.88 s. 1.96 (a) 327 L CO. (b) 5.0 3 1028 g/L. Co3(PO4)2, cobalt(II) phosphate. 7th row: Hg2I2, mercury(I) iodide. (c) 1.20 3 103 μg/mL. 1.98 0.853 cm. 1.100 4.97 3 104 g. 8th row: Cu1, CO22 1 32 3 , copper(I) carbonate. 9th row: Li , N , Li3N. 1.102 2.413 g/mL. 1.104 The glass bottle would crack. 10th row: Al2S3, aluminum sulfide. 2.104 1.91 3 1028 g. Mass is too small to be detected. 2.106 (a) Volume of a sphere is given by Chapter 2 V 5 (4/3)πr3. Volume is also proportional to the number of 2.8 0.12 mi. 2.14 145. 2.16 N(7,8,7); S(16,17,16); Cu(29,34,29); neutrons and protons present, or the mass number A. Therefore, Sr(38,46,38); Ba(56,74,56); W(74,112,74); Hg(80,122,80). r3 ~ A or r ~ A1/3. (b) 5.1 3 10244 m3. (c) The nucleus occupies only 2.18 (a) 186 201 74 W. (b) 80 Hg. 2.24 (a) Metallic character increases 3.5 3 10213% of the atom’s volume. The result supports down a group. (b) Metallic character decreases from left to right. Rutherford’s model. 2.108 (a) Yes. (b) Ethane: CH3 and C2H6. 2.26 F and Cl; Na and K; P and N. 2.32 (a) Diatomic molecule and Acetylene: CH and C2H2. 2.110 Manganese (Mn). 2.112 From left compound. (b) Polyatomic molecule and compound. (c) Polyatomic to right: chloric acid, nitrous acid, hydrocyanic acid, and sulfuric molecule and element. 2.34 (a) H2 and F2. (b) HCl and CO. (c) S8 acid. 2.114 XY2. X is likely in Group 4B or Group 4A and Y is and P4. (d) H2O and C12H22O11 (sucrose). 2.36 (protons, electrons): likely in Group 6A. Examples: titanium(IV) oxide (TiO2), tin(IV) K1(19,18); Mg21(12,10); Fe31(26,23); Br2(35,36); Mn21(25,23); oxide (SnO2), and lead(IV) oxide (PbO2). C42(6,10); Cu21(29,27). 2.44 (a) CuBr. (b) Mn2O3. (c) Hg2I2. (d) Mg3(PO4)2. 2.46 (a) AlBr3. (b) NaSO2. (c) N2O5. (d) K2Cr2O7. 2.48 C2H6O. 2.50 Ionic: NaBr, BaF2, CsCl. Molecular: CH4, CCl4, Chapter 3 ICl, NF3. 2.58 (a) Potassium hypochlorite. (b) Silver carbonate. 3.6 7.5% and 92.5%. 3.8 5.1 3 1024 amu. 3.12 5.8 3 103 light-yr. (c) Iron(II) chloride. (d) Potassium permanganate. (e) Cesium 3.14 9.96 3 10215 mol Co. 3.16 3.01 3 103 g Au. 3.18 (a) 1.244 3 chlorate. (f) Hypoiodous acid. (g) Iron(II) oxide. (h) Iron(III) 10222 g/As atom. (b) 9.746 3 10223 g/Ni atom. 3.20 6.0 3 1020 Cu oxide. (i) Titanium(IV) chloride. (j) Sodium hydride. (k) Lithium atoms. 3.22 Pb. 3.24 (a) 73.89 g. (b) 76.15 g. (c) 119.37 g. nitride. (l) Sodium oxide. (m) Sodium peroxide. (n) Iron(III) (d) 176.12 g. (e) 101.11 g. (f) 100.95 g. 3.26 6.69 3 1021 C2H6 chloride hexahydrate. 2.60 (a) CuCN. (b) Sr(ClO2)2. (c) HBrO4. molecules. 3.28 C: 1.10 3 1026 atoms; S: 5.50 3 1025 atoms; AP-1 AP-2 Answers to Even-Numbered Problems H: 3.30 3 1026 atoms; O: 5.50 3 1025 atoms. 3.30 8.56 3 1022 Oxidizing agent: Cl2; reducing agent: Br2. (c) Si ¡ Si41 1 4e2; molecules. 3.34 7. 3.40 C: 10.06%; H: 0.8442%; Cl: 89.07%. F2 1 2e2 ¡ 2F2. Oxidizing agent: F2; reducing agent: Si. 3.42 NH3. 3.44 C2H3NO5. 3.46 39.3 g S. 3.48 5.97 g F. 3.50 (a) (d) H2 ¡ 2H1 1 2e2; Cl2 1 2e2 ¡ 2Cl2. Oxidizing agent: CH2O. (b) KCN. 3.52 C6H6. 3.54 C5H8O4NNa. 3.60 (a) 2N2O5 ¡ Cl2; reducing agent: H2. 4.46 (a) 15. (b) 11. (c) 13. (d) 15. 2N2O4 1 O2. (b) 2KNO3 ¡ 2KNO2 1 O2. (c) NH4NO3 ¡ (e) 15. (f) 15. 4.48 All are zero. 4.50 (a) 23. (b) 21/2. (c) 21. N2O 1 2H2O. (d) NH4NO2 ¡ N2 1 2H2O. (e) 2NaHCO3 ¡ (d) 14. (e) 13. (f) 22. (g) 13. (h) 16. 4.52 Li and Ca. 4.54 (a) No Na2CO3 1 H2O 1 CO2. (f) P4O10 1 6H2O ¡ 4H3PO4. reaction. (b) No reaction. (c) Mg 1 CuSO4 ¡ MgSO4 1 Cu. (g) 2HCl 1 CaCO3 ¡ CaCl2 1 H2O 1 CO2. (h) 2Al 1 (d) Cl2 1 2KBr ¡ Br2 1 2KCl. 4.56 (a) Combination. 3H2SO4 ¡ Al2(SO4)3 1 3H2. (i) CO2 1 2KOH ¡ K2CO3 1 (b) Decomposition. (c) Displacement. (d) Disproportionation. H2O. (j) CH4 1 2O2 ¡ CO2 1 2H2O. (k) Be2C 1 4H2O ¡ 4.58 O12 . 4.62 Dissolve 15.0 g NaNO3 in enough water to make up 2Be(OH)2 1 CH4. (l) 3Cu 1 8HNO3 ¡ 3Cu(NO3)2 1 2NO 1 250 mL. 4.64 10.8 g. 4.66 (a) 1.37 M. (b) 0.426 M. (c) 0.716 M. 4H2O. (m) S 1 6HNO3 ¡ H2SO4 1 6NO2 1 2H2O. (n) 2NH3 1 4.68 (a) 6.50 g. (b) 2.45 g. (c) 2.65 g. (d) 7.36 g. (e) 3.95 g. 3CuO ¡ 3Cu 1 N2 1 3H2O. 3.64 (d). 3.66 1.01 mol. 3.68 20 mol. 4.70 11.83 g. 4.74 0.0433 M. 4.76 126 mL. 4.78 1.09 M. 3.70 (a) 2NaHCO3 ¡ Na2CO3 1 CO2 1 H2O. (b) 78.3 g. 4.82 35.73%. 4.84 0.00975 M. 4.90 0.217 M. 4.92 (a) 6.00 mL. 3.72 255.9 g; 0.324 L. 3.74 0.294 mol. 3.76 (a) NH4NO3 ¡ (b) 8.00 mL. 4.96 9.44 3 1023 g. 4.98 0.06020 M. 4.100 6.15 mL. N2O 1 2H2O. (b) 20 g. 3.78 18.0 g. 3.82 1 mole H2 left and 4.102 0.232 mg. 4.104 (i) Only oxygen supports combustion. 6 moles NH3 produced. 3.84 (a) 2NH3 1 H2SO4 ¡ (NH4)2SO4. (ii) Only CO2 reacts with Ca(OH)2(aq) to form CaCO3 (white (b) 5.23 g NH3; 21.0 g H2SO4. 3.86 HCl; 23.4 g. 3.90 (a) 7.05 g. precipitate). 4.106 1.26 M. 4.108 (a) 15.6 g Al(OH)3. (b) [Al31] 5 (b) 92.9%. 3.92 3.48 3 103 g. 3.94 8.55 g; 76.6%. 3.96 85Rb: 0.250 M, [NO2 1 3 ] 5 2.25 M, [K ] 1.50 M. 4.110 0.171 M. 72.1%; 87Rb: 27.9%. 3.98 (b). 3.100 (a) C5H12 1 8O2 ¡ 5CO2 1 4.112 0.115 M. 4.114 Ag: 1.25 g; Zn: 2.12 g. 4.116 0.0721 M 6H2O. (b) NaHCO3 1 HCl ¡ CO2 1 NaCl 1 H2O. (c) 6Li 1 NaOH. 4.118 24.0 g/mol; Mg. 4.120 2. 4.122 1.72 M. 4.124 Only N2 ¡ 2Li3N. (d) PCl3 1 3H2O ¡ H3PO3 1 3HCl. (e) 3CuO 1 Fe(II) is oxidized by KMnO4 solution and can therefore change the 2NH3 ¡ 3Cu 1 N2 1 3H2O. 3.102 Cl2O7. 3.104 18.7 g. purple color to colorless. 4.126 Ions are removed as the BaSO4 3.106 (a) 0.212 mol. (b) 0.424 mol. 3.108 18. 3.110 2.4 3 precipitate. 4.128 FeCl2 ? 4H2O. 4.130 (i) Conductivity test. 1023 atoms. 3.112 65.4 amu; Zn. 3.114 89.5%. 3.116 CH2O; (ii) Only NaCl reacts with AgNO3 to form AgCl precipitate. C6H12O6. 3.118 51.9 g/mol; Cr. 3.120 1.6 3 104 g/mol. 3.122 NaBr: 4.132 The Cl2 ion cannot accept any electrons. 4.134 Reaction is 24.03%; Na2SO4: 75.97%. 3.124 C3H8 1 5O2 ¡ 3CO2 1 4H2O. too violent. 4.136 Use sodium bicarbonate: HCO2 3 1H ¡ 1 3.126 Ca: 38.76%; P: 19.97%; O: 41.27%. 3.128 Yes. 3.130 2.01 3 H2O 1 CO2. NaOH is a caustic substance and unsafe to use in this 1021 molecules. 3.132 16.00 amu. 3.134 (e). 3.136 PtCl2; PtCl4. manner. 4.138 (a) Conductivity. Reaction with AgNO3 to form 3.138 (a) 12 g; 28 mL. (b) 15 g. 3.140 (a) X: MnO2; Y: Mn3O4. AgCl. (b) Soluble in water. Nonelectrolyte. (c) Possesses (b) 3MnO2 ¡ Mn3O4 1 O2. 3.142 6.1 3 105 tons. properties of acids. (d) Soluble. Reacts with acids to give CO2. 3.144 C3H2ClF5O. C: 19.53%; H: 1.093%; Cl: 19.21%; F: 51.49%; (e) Soluble, strong electrolyte. Reacts with acids to give CO2. O: 8.672%. 3.146 Mg3N2 (magnesium nitride). 3.148 PbC8H20. (f) Weak electrolyte and weak acid. (g) Soluble in water. Reacts 3.150 (a) 4.3 3 1022 atoms. (b) 1.6 3 102 pm. 3.152 28.97 g/mol. with NaOH to produce Mg(OH)2 precipitate. (h) Strong electrolyte 3.154 (a) Fe2O3 1 6HCl ¡ 2FeCl3 1 3H2O. (b) 396 g FeCl3. and strong base. (i) Characteristic odor. Weak electrolyte and weak 3.156 (a) C3H8 1 3H2O ¡ 3CO 1 7H2. (b) 9.09 3 102 kg. base. (j) Insoluble. Reacts with acids. (k) Insoluble. Reacts with 3.158 (a) There is only one reactant so the use of “limiting reagent” acids to produce CO2. 4.140 NaCl: 44.11%; KCl: 55.89%. 4.142 is unnecessary. (b) The term “limiting reagent” usually applies (a) AgOH(s) 1 HNO3(aq) ¡ AgNO3(aq) 1 H2O(l). only to one reactant. 3.160 (a) $0.47/kg. (b) 0.631 kg K2O. 4.144 1.33 g. 4.146 56.18%. 4.148 (a) 1.40 M. (b) 4.96 g. 3.162 BaBr2. 3.164 NaCl: 32.17%; Na2SO4: 20.09%; NaNO3: 47.75%. 4.150 (a) NH1 2 4 1 OH ¡ NH3 1 H2O. (b) 97.99%. 4.152 Zero. 4.154 0.224%. Yes. 4.156 (a) Zn 1 H2SO4 ¡ Chapter 4 ZnSO4 1 H2. (b) 2KClO3 ¡ 2KCl 1 3O2. (c) Na2CO3 1 4.8 (c). 4.10 (a) Strong electrolyte. (b) Nonelectrolyte. (c) Weak 2HCl ¡ 2NaCl 1 CO2 1 H2O. (d) NH4NO2 ¡ N2 1 2H2O. electrolyte. (d) Strong electrolyte. 4.12 (b) and (c). 4.14 HCl does 4.158 Yes. 4.160 (a) 8.316 3 1027 M. (b) 3.286 3 1025 g. not ionize in benzene. 4.18 (b). 4.20 (a) Insoluble. (b) Soluble. 4.162 [Fe21] 5 0.0920 M, [Fe31] 5 0.0680 M. (c) Soluble. (d) Insoluble. (e) Soluble. 4.22 (a) Ionic: 2Na1 1 S22 1 4.164 (a) Precipitation: Mg21 1 2OH2 ¡ Mg(OH)2; acid-base: Zn21 1 2Cl2 ¡ ZnS 1 2Na1 1 2Cl2. Net ionic: Zn21 1 Mg(OH)2 1 2HCl ¡ MgCl2 1 2H2O; redox: MgCl2 ¡ Mg 1 S22 ¡ ZnS. (b) Ionic: 6K1 1 2PO432 1 3Sr21 1 6NO2 3 ¡ Cl2. (b) NaOH is more expensive than CaO. (c) Dolomite provides Sr3(PO4)2 1 6KNO3. Net ionic: 3Sr21 1 2PO432¡ Sr3(PO4)2. additional Mg. 4.166 D , A , C , B. D 5 Au, A 5 Cu, C 5 Zn, (c) Ionic: Mg21 1 2NO2 1 2 3 1 2Na 1 2OH ¡ Mg(OH)2 1 B 5 Mg. 4.168 (a) Cu21 1 SO422 1 Ba21 1 2OH2 ¡ 2Na1 1 2NO2 3 . Net ionic: Mg 21 1 2OH 2 ¡ Mg(OH)2. Cu(OH)2 1 BaSO4. (b) 14.6 g Cu(OH)2, 35.0 g BaSO4. 4.24 (a) Add chloride ions. (b) Add hydroxide ions. (c) Add [Cu21] 5 [SO22 4 ] 5 0.0417 M. carbonate ions. (d) Add sulfate ions. 4.32 (a) Brønsted base. (b) Brønsted base. (c) Brønsted acid. (d) Brønsted base and Chapter 5 Brønsted acid. 4.34 (a) Ionic: CH3COOH 1 K1 1 OH2 ¡ K1 1 5.14 0.797 atm; 80.8 kPa. 5.18 (1) b. (2) a. (3) c. (4) a. 5.20 53 atm. CH3COO2 1 H2O; Net ionic: CH3COOH 1 OH2 ¡ 5.22 (a) 0.69 L. (b) 61 atm. 5.24 1.3 3 102 K. 5.26 ClF3. 5.32 6.2 CH3COO2 1 H2O. (b) Ionic: H2CO3 1 2Na1 1 2OH2 ¡ atm. 5.34 745 K. 5.36 1.9 atm. 5.38 0.82 L. 5.40 45.1 L. 5.42 6.1 3 2Na1 1 CO322 1 2H2O; Net ionic: H2CO3 1 2OH2 ¡ CO322 1 1023 atm. 5.44 35.1 g/mol. 5.46 N2: 2.1 3 1022; O2: 5.7 3 1021; 2H2O. (c) Ionic: 2H1 1 2NO2 3 1 Ba 21 1 2OH2 ¡ Ba21 1 Ar: 3 3 1020. 5.48 2.98 g/L. 5.50 SF4. 5.52 F2: 59.7%; Cl2: 40.3%. 2 1 2 2NO3 1 2H2O; Net ionic: H 1 OH ¡ H2O. 4.44 (a) Fe ¡ 5.54 370 L. 5.56 88.9%. 5.58 M 1 3HCl ¡ (3/2)H2 1 MCl3; Fe31 1 3e2; O2 1 4e2 ¡ 2O22. Oxidizing agent: O2; reducing M2O3, M2(SO4)3. 5.60 2.84 3 1022 mol CO2; 94.7%. agent: Fe. (b) 2Br2 ¡ Br2 1 2e2; Cl2 1 2e2 ¡ 2Cl2. The impurities must not react with HCl to produce CO2.  Answers to Even-Numbered Problems AP-3 5.62 1.71 3 103 L. 5.64 86.0%. 5.68 (a) 0.89 atm. (b) 1.4 L. 6.142 (a) Heating water at room temperature to its boiling point. 5.70 349 mmHg. 5.72 19.8 g. 5.74 H2: 650 mmHg; N2: 217 (b) Heating water at its boiling point. (c) A chemical reaction mmHg. 5.76 (a) Box on right. (b) Box on left. 5.82 N2: 472 m/s; taking place in a bomb calorimeter (an isolated system) where O2: 441 m/s; O3: 360 m/s. 5.84 2.8 m/s; 2.7 m/s. Squaring favors there is no heat exchange with the surroundings. 6.144 2101.3 J. the larger values. 5.86 1.0043. 5.88 4. 5.94 No. 5.96 Ne. Yes, because in a cyclic process, the change in a state function 5.98 C6H6. 5.100 445 mL. 5.102 (a) 9.53 atm. (b) Ni(CO)4 must be zero. 6.146 (a) Exothermic. (b) No clear conclusion. It is a decomposes to give CO, which increases the pressure. balance between the energy needed to break the ionic bond and the 5.104 1.30 3 1022 molecules; CO2, O2, N2, H2O. 5.106 5.25 3 energy released during hydration. (c) No clear conclusion. It is a 1018 kg. 5.108 0.0701 M. 5.110 He: 0.16 atm; Ne: 2.0 atm. balance between the energy needed to break the A—B bond and the 5.112 HCl dissolves in the water, creating a partial vacuum. energy released when the A¬C bond is formed. (d) Endothermic. 5.114 7. 5.116 (a) 61.2 m/s. (b) 4.58 3 1024 s. (c) 328 m/s; 366 m/s. The velocity 328 m/s is that of a particular atom and urms is Chapter 7 an average value. 5.118 2.09 3 104 g; 1.58 3 104 L. 5.120 Higher 7.8 (a) 6.58 3 1014/s. (b) 1.22 3 108 nm. 7.10 2.5 min. partial pressure of C2H4 inside the paper bag. 5.122 To equalize 7.12 4.95 3 1014/s. 7.16 (a) 4.0 3 102 nm. (b) 5.0 3 10219 J. the pressure as the amount of ink decreases. 5.124 (a) NH4NO3 ¡ 7.18 1.2 3 102 nm (UV). 7.20 (a) 3.70 3 102 nm. (b) UV. N2O 1 2H2O. (b) 0.0821 L ? atm/K ? mol. 5.126 C6H6. 5.128 The (c) 5.38 3 10219 J. 7.22 8.16 3 10219 J. 7.26 Use a prism. low atmospheric pressure caused the harmful gases (CO, CO2, 7.28 Compare the emission spectra with those on Earth of known CH4) to flow out of the mine, and the man suffocated. 5.130 Br2 elements. 7.30 3.027 3 10219 J. 7.32 6.17 3 1014/s. 486 nm. (159.8 g/mol; red); SO3 (80.07 g/mol; yellow); N2 (28.02 g/mol; 7.34 5. 7.40 1.37 3 1026 nm. 7.42 1.7 3 10223 nm. 7.56 / 5 2: green); CH4 (16.04 g/mol; blue). 5.132 (a) 5 3 10222 atm. (b) 5 3 m/ 5 22, 21, 0, 1, 2. / 5 1: m/ 5 21, 0, 1. / 5 0: m/ 5 0. 1020 L/g H. 5.134 91%. 5.136 1.7 3 1012 molecules. 5.138 4.66 L. 7.58 (a) n 5 3, / 5 0, m/ 5 0. (b) n 5 4, / 5 1, m/ 5 21, 0, 1. 5.140 3.8 3 1023 m/s; 1.0 3 10230 J. 5.142 2.3 3 103 L. (c) n 5 3, / 5 2, m/ 5 22, 21, 0, 1, 2. In all cases, ms 5 11/2 or 5.144 1.8 3 102 mL. 5.146 (a) 1.09 3 1044 molecules. (b) 1.18 3 21/2. 7.60 Differ in orientation only. 7.62 6s, 6p, 6d, 6f, 6g, and 1022 molecules/breath. (c) 2.60 3 1030 molecules. (d) 2.39 3 10214; 6h. 7.64 2n2. 7.66 (a) 3. (b) 6. (c) 0. 7.68 There is no shielding in 3 3 108 molecules. (e) Complete mixing of air; no molecules an H atom. 7.70 (a) 2s , 2p. (b) 3p , 3d. (c) 3s , 4s. (d) 4d , 5f. escaped to the outer atmosphere; no molecules used up during 7.76 Al: 1s22s22p63s23p1. B: 1s22s22p1. F: 1s22s22p5. 7.78 B(1), metabolism, nitrogen fixation, etc. 5.148 3.7 nm; 0.31 nm. Ne(0), P(3), Sc(1), Mn(5), Se(2), Kr(0), Fe(4), Cd(0), I(1), Pb(2). 5.150 0.54 atm. 5.152 H2: 0.5857; D2: 0.4143. 5.154 53.4%. 7.88 [Kr]5s24d5. 7.90 Ge: [Ar]4s23d104p2. Fe: [Ar]4s23d 6. Zn: [Ar] 5.156 CO: 54.4%; CO2: 45.6%. 5.158 CH4: 0.789; C2H6: 0.211. 4s23d10. Ni: [Ar]4s23d 8. W: [Xe]6s24f 145d 4. Tl: [Xe]6s24f 145d106p1. 5.160 (a) 8(4πr3/3). (b) (16/3)NAπr3. The excluded volume is 4 7.92 S1. 7.94 6.68 3 1016 photons. 7.96 (a) Incorrect. (b) Correct. times the volumes of the atoms. 5.162 CH4. 5.164 NO. 5.166 (b). (c) Incorrect. 7.98 (a) 4e: An e in a 2s and an e in each 2p orbital. 5.168 (i) (b) 8.0 atm. (c) 5.3 atm. (ii) PT 5 5.3 atm. PA 5 2.65 atm. (b) 6e: 2e each in a 4p, a 4d, and a 4f orbital. (c) 10e: 2e in each of PB 5 2.65 atm. 5.170 CH4: 2.3 atm. C2H6: 0.84 atm. C3H8: 1.4 atm. the five 3d orbitals. (d) 1e: An e in a 2s orbital. (e) 2e: 2e in a 4f orbital. 7.100 Wave properties. 7.102 (a) 1.05 3 10225 nm. Chapter 6 (b) 8.86 nm. 7.104 (a) n 5 2. The possible / values are from 0 to 6.16 (a) 0. (b) 29.5 J. (c) 218 J. 6.18 48 J. 6.20 23.1 3 103 J. (n 2 1) integer values. (b) Possible / values are 0, 1, 2, or 3. 6.26 1.57 3 104 kJ. 6.28 2553.8 kJ/mol. 6.32 0.237 J/g ? 8C. Possible m/ values range from 2/ to 1/ integer values. 6.34 3.31 kJ. 6.36 98.6 g. 6.38 22.398C. 6.46 O2. 7.106 (a) 1.20 3 1018 photons. (b) 3.76 3 108 W. 7.108 419 nm. 6.48 (a) DHf8[Br2(l)] 5 0; DHf8[Br2(g)] . 0. (b) DHf8[I2(s)] 5 0; Yes. 7.110 Ne. 7.112 He1: 164 nm, 121 nm, 109 nm, 103 nm DHf8[I2(g)] . 0. 6.50 Measure DH8 for the formation of (all in the UV region). H: 657 nm, 487 nm, 434 nm, 411 nm Ag2O from Ag and O2 and of CaCl2 from Ca and Cl2. (all in the visible region). 7.114 1.2 3 102 photons. 7.116 2.5 3 1017 6.52 (a) 2167.2 kJ/mol. (b) 256.2 kJ/mol. 6.54 (a) 21411 kJ/mol. photons. 7.118 Yellow light will generate more electrons; blue (b) 21124 kJ/mol. 6.56 218.2 kJ/mol. 6.58 71.58 kJ/g. 6.60 2.70 3 light will generate electrons with greater kinetic energy. 102 kJ. 6.62 284.6 kJ/mol. 6.64 2847.6 kJ/mol. 6.72 11 kJ. 7.120 (a) He. (b) N. (c) Na. (d) As. (e) Cl. See Table 7.3 for 6.74 22.90 3 102 kJ/mol. 6.76 (a) 2336.5 kJ/mol. (b) NH3. ground-state electron configurations. 7.122 They might have 6.78 26.5 kJ/mol. 6.80 43.6 kJ. 6.82 0. 6.84 2350.7 kJ/mol. discovered the wave properties of electrons. 7.124 7.39 3 1022 6.86 2558.2 kJ/mol. 6.88 0.492 J/g ? 8C. 6.90 The first (exothermic) nm. 7.126 (a) False. (b) False. (c) True. (d) False. (e) True. reaction can be used to promote the second (endothermic) 7.128 2.0 3 1025 m/s. 7.130 (a) and (f) violate Pauli exclusion reaction. 6.92 1.09 3 104 L. 6.94 4.10 L. 6.96 5.60 kJ/mol. principle; (b), (d), and (e) violate Hund’s rule. 7.132 2.8 3 106 K. 6.98 (a). 6.100 (a) 0. (b) 29.1 J. (c) 2.4 L; 248 J. 6.102 (a) A more 7.134 2.76 3 10211 m. 7.136 17.4 pm. 7.138 0.929 pm; 3.23 3 fully packed freezer has a greater mass and hence a larger heat 1020/s. 7.140 ni 5 5 to nf 5 3. 7.142 (a) B: 4 ¡ 2; C: 5 ¡ 2. capacity. (b) Tea or coffee has a greater amount of water, which (b) A: 41.1 nm; B: 30.4 nm. (c) 2.18 3 10218 J. (d) At high values has a higher specific heat than noodles. 6.104 1.84 3 103 kJ. of n, the energy levels are very closely spaced, leading to a 6.106 3.0 3 109. 6.108 5.35 kJ/8C. 6.110 25.2 3 106 kJ. continuum of lines. 7.144 n 5 1: 1.96 3 10217 J; n 5 5: 7.85 3 6.112 (a) 3.4 3 105 g. (b) 22.0 3 108 J. 6.114 286.7 kJ/mol. 10219 J. 10.6 nm. 7.146 9.5 3 103 m/s. 7.148 3.87 3 105 m/s. 6.116 (a) 1.4 3 102 kJ. (b) 3.9 3 102 kJ. 6.118 (a) 265.2 kJ/mol. 7.150 Photosynthesis and vision. 7.152 1.06 nm. 7.154 (a) 1.12 pm. (b) 29.0 kJ/mol. 6.120 2110.5 kJ/mol. It will form both CO and (b) Smaller than the molecule. CO2. 6.122 (a) 0.50 J. (b) 32 m/s. (c) 0.128C. 6.124 2277.0 kJ/mol. 6.126 104 g. 6.128 296 kJ. 6.130 9.9 3 108 J; 3048C. Chapter 8 6.132 (a) CaC2 1 2H2O ¡ C2H2 1 Ca(OH)2. (b) 1.51 3 103 kJ. 8.20 (a) 1s22s22p63s23p5. (b) Representative. (c) Paramagnetic. 6.134 DU 5 25153 kJ/mol; DH 5 25158 kJ/mol. 8.22 (a) and (d); (b) and (e); (c) and (f). 8.24 (a) Group 1A. 6.136 2564.2 kJ/mol. 6.138 96.21%. 6.140 (a) CH. (b) 49 kJ/mol. (b) Group 5A. (c) Group 8A. (d) Group 8B. 8.26 Fe. 8.28 (a) [Ne]. AP-4 Answers to Even-Numbered Problems (b) [Ne]. (c) [Ar]. (d) [Ar]. (e) [Ar]. (f) [Ar]3d6. (g) [Ar]3d9. 9.20 (a) BF3, covalent. Boron triflouride. (b) KBr, ionic. Potassium (h) [Ar]3d10. 8.30 (a) Cr31. (b) Sc31. (c) Rh31. (d) Ir31. 8.32 Be21 bromide. 9.26 2195 kJ/mol. 9.36 C¬H , Br¬H , F¬H , and He; F2 and N32; Fe21 and Co31; S22 and Ar. 8.38 Na . Li¬Cl , Na¬Cl , K¬F. 9.38 Cl¬Cl , Br¬Cl , Si¬C , Mg . Al . P . Cl. 8.40 F. 8.42 The effective nuclear charge that Cs¬F. 9.40 (a) Covalent. (b) Polar covalent. (c) Ionic. (d) Polar the outermost electrons feel increases across the period. covalent. 8.44 Mg21 , Na1 , F2 , O22 , N32. 8.46 Te22. 8.48 H2 is O O Q O Q O 9.44 (a) SFOOOFS Q Q O O O (b) SFONPNOFS Q larger. 8.52 K , Ca , P , F , Ne. 8.54 The single 3p electron in Al is well shielded by the 1s, 2s, and 3s electrons. 8.56 1s22s22p6: H H 2080 kJ/mol. 8.58 8.40 3 106 kJ/mol. 8.62 Greatest: Cl; least: He. A A 8.64 The ns1 configuration enables them to accept another O (c) HOSiOSiOH (d) SOOH Q A A electron. 8.68 Fr should be the most reactive toward water and H H oxygen, forming FrOH and Fr2O, Fr2O2, and FrO2. 8.70 The Group 1B elements have higher ionization energies due to the H SOS H H A B A A incomplete shielding of the inner d electrons. 8.72 (a) Li2O 1 (e) HOCOCOOS O  (f ) HOCON OH Q H2O ¡ 2LiOH. (b) CaO 1 H2O ¡ Ca(OH)2. (c) SO3 1 A A A H2O ¡ H2SO4. 8.74 BaO. 8.76 (a) Bromine. (b) Nitrogen. SClS H H (c) Rubidium. (d) Magnesium. 8.78 P32, S22, Cl2, K1, Ca21, Sc31, O Ti41, V51, Cr61, Mn71. 8.80 M is K; X is Br. 8.82 N and O1; Ne  O  Q O    9.46 (a) SOOOSQ (b) SCqCS (c)SNqOS and N32; Ar and S22; Zn and As31; Cs1 and Xe. 8.84 (a) and (d). 8.86 Yellow-green gas: F2; yellow gas: Cl2; red liquid: Br2; dark 9.48 (a) The double bond between C and H; the single bond solid: I2. 8.88 (a) DH 5 1532 kJ/mol. (b) DH 5 12,405 kJ/mol. between C and the end O; the lone pair on C atom. 8.90 Fluorine. 8.92 H2. 8.94 Li2O (basic); BeO (amphoteric); (b) H SOS B2O3 (acidic); CO2 (acidic); N2O5 (acidic). 8.96 It forms both the A B H1 and H2 ions; H1 is a single proton. 8.98 0.65. 8.100 79.9%. HOCOCOOOH O Q A 8.102 418 kJ/mol. Use maximum wavelength. 8.104 7.28 3 103 H kJ/mol. 8.106 X: Sn or Pb; Y: P; Z: alkali metal. 8.108 495.9 kJ/mol. 8.110 343 nm (UV). 8.112 604.3 kJ/mol. 8.114 K2TiO4. O  O  SOS SOS SOS 8.116 2K2MnF6 1 4SbF5 ¡ 4KSbF6 1 2MnF3 1 F2. A B A 8.118 N2O (11), NO (12), N2O3 (13), NO2 and N2O4 (14), 9.52 O OPClOOS Q O  mn SOOClOOS O Q M O   O Q M O Q mn SOOClPO N2O5 (15). 8.120 The larger the effective nuclear charge, the more M Q   Q  tightly held are the electrons. The atomic radius will be small and H H the ionization energy will be large. 8.122 m.p.: 6.38C; b.p.: 74.98C. A  A  O  8.124 An alkaline earth metal. 8.126 (a) It was discovered that the Q mn HOCONqNS 9.54 HOCPNPN M periodic table was based on atomic number, not atomic mass.  (b) Ar: 39.95 amu; K: 39.10 amu. 8.128 Z 5 119;  9.56 O Q O O  O Q mn SOOCqNSmnSOqCONS OPCPN Q Q 2 [Rn]7s25f146d107p68s1. 8.130 Group 3A. 8.132 (a) SiH4, GeH4, SnH4, PbH4. (b) RbH more ionic. (c) Ra 1 2H2O ¡ Ra(OH)2 1 2 H2. (d) Be. 8.134 Mg21 is the smallest cation and has the largest O Q O 9.62 ClPBePCl Q Not plausible. charge density and is closest to the negative ion. Ba21 is just the opposite. Thus, Mg21 binds the tightest and Ba21 the least. Cl A 8.136 See chapter. 8.138 Carbon (diamond). 8.140 419 nm. 9.64 ClOSbOCl The octet rule is not obeyed. 8.142 The first ionization energy of He is less than twice the ionization D G of H because the radius of He is greater than that of H and the Cl Cl shielding in He makes Zeff less than 2. In He1, there is no shielding Cl and the greater nuclear attraction makes the second ionization of He A  greater than twice the ionization energy of H. 8.144 Zeff: Li (1.26); 9.66 ClOAlOCl Coordinate covalent bond. A Na (1.84); K (2.26). Zeff/n: Li (0.630); Na (0.613); K (0.565). Zeff Cl increases as n increases. Thus, Zeff/n remains fairly constant. 8.146 Go to the recommended website. Click on “Biology” tab 9.70 303.0 kJ/mol. 9.72 (a) 22759 kJ/mol. (b) 22855 kJ/mol. above the periodic table and then click on each of the listed elements. 9.74 Ionic: RbCl, KO2; covalent: PF5, BrF3, CI4. 9.76 Ionic: NaF, A brief summary of the biological role of each element is provided. MgF2, AlF3; covalent: SiF4, PF5, SF6, ClF3. 9.78 KF: ionic, high melting point, soluble in water, its melt and solution conduct Chapter 9 electricity. C6H6: covalent and discrete molecule, low melting 9.16 (a) RbI, rubidium iodide. (b) Cs2SO4, cesium sulfate. point, insoluble in water, does not conduct electricity. (c) Sr3N2, strontium nitride. (d) Al2S3, aluminum sulfide.     2 2  9.80 O O NPNPN Q O Q mn SNqNONS Q mn O SNONqNS Q O Q 88n Sr SSeS 9.18 (a) TSrTTSeT O Q 2 2 9.82 (a) AlCl2 32 4 . (b) AlF6 . (c) AlCl3. 9.84 CF2: violates the octet  (b) TCaT 2HT 88n Ca 2 2HS rule; LiO2: lattice energy too low; CsCl2: second ionization energy too high to produce Cs21; PI5: I atom too bulky to fit around P. O  O 3 R 88n 3Li SNS (c) 3LiTTNT Q 9.86 (a) False. (b) True. (c) False. (d) False. 9.88 267 kJ/mol. T 9.90 N2. 9.92 NH1 O (d) 2TAlT 3TST 3 O Q 88n 2Al 3SSS Q 2 4 and CH4; CO and N2; B3N3H6 and C6H6.  Answers to Even-Numbered Problems AP-5 Q Q Q 9.94 HONS  HOOS 88n HONOH  SOOH O 10.34 B: sp2 to sp3; N: remains at sp3. 10.36 From left to right. Q (a) sp3. (b) sp3, sp2, sp2. (c) sp3, sp, sp, sp3. (d) sp3, sp2. (e) sp3, sp2. A A A H H H 10.38 sp. 10.40 sp3d. 10.42 9 pi bonds and 9 sigma bonds. 9.96 F32 violates the octet rule. 10.44 IF42. 10.50 Electron spins must be paired in H2.   10.52 Li22 5 Li1 1 2 , Li2. 10.54 B2 . 10.56 MO theory predicts O2 is O O 9.98 CH3ONPCPO O Q mn CH3ONqCOOS Q paramagnetic. 10.58 O222 , O2 2 , O2 , O1 2 . 10.60 B2 contains a pi bond; C2 contains 2 pi bonds. 10.62 (1) Atoms far apart. No 9.100 (c) No bond between C and O. (d) Large formal charges. interaction. (2) The 2p orbitals begin to overlap. Attractive forces Cl Cl F F H operating. (3) The system at its most stable state. The potential A A A A A energy reaches a minimum. (4) As the distance decreases further, 9.102 (a) FOCOCl (b) FOCOF (c) HOCOF (d) FOCOCOF nuclear-nuclear and electron-electron repulsion increase. A A A A A Cl Cl Cl F F (5) Further decrease in distance leads to instability of F2 molecule. 10.66 The circle shows electron delocalization. 9.104 (a) 29.2 kJ/mol. (b) 29.2 kJ/mol. 9.106 (a) 2:C‚O:1 (b) :N‚O:1 (c) 2:C‚N: (d) :N‚N: 9.108 True. O SFS O SFS 9.110 (a) 114 kJ/mol. (b) Extra electron increases repulsion A A 10.68 (a) O OPNOOS Q O mn SOONPO Q   O Q O (b) sp2. (c) N forms Q between F atoms. 9.112 Lone pair on C and negative formal   charge on C. sigma bonds with F and O atoms. There is a pi molecular orbital  a O mn O 9.114 (a) SNPO a delocalized over N and O atoms. 10.70 sp2. 10.72 Linear. Dipole Q Q NPO Q (b) No. moment measurement. 10.74 The large size of Si results in poor 9.116 H H H H H sideways overlap of p orbitals to form pi bonds. 10.76 (a) C8H10N4O2. A A A A A (b) C atoms in the ring and O are sp2. C atom in CH3 group is sp3. SNONOBOH SNONOBOH O A A A A A A Double-bonded N is sp2; single-bonded N is sp3. (c) Geometry H H H H H H about the sp2 C and N atoms is trigonal planar. Geometry about sp3 C and N atom is tetrahedral. 10.78 XeF1 3 : T-shaped; 9.118 The OCOO structure leaves a lone pair and a negative XeF51: square pyramidal; SbF62: octahedral. 10.80 (a) 1808. charge on C. (b) 1208. (c) 109.58. (d) About 109.58. (e) 1808. (f) About 1208. Cl Cl Cl (g) About 109.58. (h) 109.58. 10.82 sp3d. 10.84 ICl2 2 and CdBr2. G D q D 9.120 Al Al The arrows indicate coordinate covalent 10.86 (a) sp2. (b) Molecule on the right. 10.88 The pi bond in D r D G bonds. cis-dichloroethylene prevents rotation. 10.90 O3, CO, CO2, NO2, Cl Cl Cl N2O, CH4, CFCl3. 10.92 C: all single-bonded C atoms are sp3, the 9.122 347 kJ/mol. double-bonded C atoms are sp2; N: single-bonded N atoms are sp3, 9.124 From bond enthalpies: 2140 kJ/mol; from standard N atoms that form one double bond are sp2, N atom that forms two enthalpies of formation: 2184 kJ/mol. double bonds is sp. 10.94 Si has 3d orbitals so water can add to H H H H H H Si (valence shell expansion). 10.96 C: sp2; N: N atom that forms a A A A A A A double bond is sp2, the others are sp3. 10.98 (a) Use a conventional 9.126 (a) CPC (b) OCOCOCOCO (c) 21.2 3 106 kJ. oven. (b) No. Polar molecules would absorb microwaves. (c) Water A A A A A A H Cl H Cl H Cl molecules absorb part of microwaves. 10.100 (a) and (b) are polar. 10.102 The small size of F results in a shorter bond and greater 9.128 O: 3.16; F: 4.37; Cl: 3.48. 9.130 (1) The MgO solid lone pair repulsion. 10.104 43.6%. 10.106 Second and third containing Mg1 and O2 ions would be paramagnetic. (2) The vibrations. CO, NO2, N2O. 10.108 (a) The two 908 rotations will lattice energy would be like NaCl (too low). 9.132 71.5 nm. break and make the pi bond and convert cis-dichloroethylene to 9.134 2629 kJ/mol. 9.136 268 nm. 9.138 (a) From bond trans-dichloroethylene. (b) The pi bond is weaker because of the enthalpies: 21937 kJ/mol; from standard enthalpies of formation: lesser extent of sideways orbital overlap. (c) 444 nm. 10.110 (a) H2. 21413.9 kJ/mol. (b) 162 L. (c) 11.0 atm. 9.140 The repulsion The electron is removed from the more stable bonding molecular between lone pairs on adjacent atoms weakens the bond. There are orbital. (b) N2. Same as (a). (c) O. The atomic orbital in O is more two lone pairs on each O atom in H2O2. The repulsion is the stable than the antibonding molecular orbital in O2. (d) The atomic greatest; it has the smallest bond enthalpy (about 142 kJ/mol). orbital in F is more stable than the antibonding molecular orbital There is one lone pair on each N atom in N2H4; it has the in F2. 10.112 (a) [Ne2](σ3s)2(σ w 2 2 2 2 3 s) (π3py ) (π3pz ) (σ3px ) . (b) 3. intermediate bond enthalpy (about 193 kJ/mol). There are no lone (c) Diamagnetic. 10.114 For all the electrons to be paired in O2 pairs on the C atoms in C2H6; it has the greatest bond enthalpy (see Table 10.5), energy is needed to flip the spin of one of the (about 347 kJ/mol). 9.142 244 kJ/mol. electrons in the antibonding molecular orbitals. This arrangement is less stable according to Hund’s rule. 10.116 ClF3: T-shaped; Chapter 10 sp3d. AsF5: Trigonal bipyramidal; sp3d; ClF1 3 2 2 : bent; sp ; AsF6 : 3 2 10.8 (a) Trigonal planar. (b) Linear. (c) Tetrahedral. Octahedral; sp d . 10.118 (a) Planar and no dipole moment. 10.10 (a) Tetrahedral. (b) T-shaped. (c) Bent. (d) Trigonal planar. (b) 20 sigma bonds and 6 pi bonds. 10.120 (a) The negative formal (e) Tetrahedral. 10.12 (a) Tetrahedral. (b) Bent. (c) Trigonal planar. charge is placed on the less electronegative carbon, so there is less (d) Linear. (e) Square planar. (f) Tetrahedral. (g) Trigonal charge separation and a smaller dipole moment. (b) Both the bipyramidal. (h) Trigonal pyramidal. (i) Tetrahedral. 10.14 SiCl4, Lewis structure and the molecular orbital treatment predicts CI4, CdCl422. 10.20 Electronegativity decreases from F to I. a triple bond. (c) C. 10.122 O“C“C“C“O. The molecule is 10.22 Larger. 10.24 (b) 5 (d) , (c) , (a). 10.32 sp3 for both. linear and nonpolar. 10.124 NO22 , NO2 , NO 5 NO21 , NO1. AP-6 Answers to Even-Numbered Problems Chapter 11 acetic acid . 0.50 m HCl. 12.74 0.9420 m. 12.76 7.6 atm. 12.78 1.6 atm. 12.82 (c). 12.84 3.5 atm. 12.86 (a) 104 mmHg. 11.8 Methane. 11.10 (a) Dispersion forces. (b) Dispersion and (b) 116 mmHg. 12.88 2.95 3 103 g/mol. 12.90 12.5 g. 12.92 No. dipole-dipole forces. (c) Same as (b). (d) Dispersion and ion-ion 12.94 No. AlCl3 dissociates into Al31 and 3 Cl2 ions. 12.96 O2: forces. (e) Same as (a). 11.12 (e). 11.14 Only 1-butanol can form 4.7 3 1026; N2: 9.7 3 106. The mole fraction of O2 compared to hydrogen bonds. 11.16 (a) Xe. (b) CS2. (c) Cl2. (d) LiF. (e) NH3. the mole fraction of N2 in water is greater compared to that in 11.18 (a) Hydrogen bond and dispersion forces. (b) Dispersion air. 12.98 The molar mass in B (248 g/mol) is twice as large as that forces. (c) Dispersion forces. (d) Covalent bond. 11.20 The in A (124 g/mol). A dimerization process. 12.100 (a) Last alcohol. compound on the left can form an intramolecular hydrogen bond, (b) Methanol. (c) Last alcohol. 12.102 I2-water: weak reducing intermolecular hydrogen bonding. 11.32 Between ethanol dipole2induced dipole; I32-water: favorable ion-dipole interaction. and glycerol. 11.38 scc: 1; bcc: 2; fcc: 4. 11.40 6.20 3 1023 12.104 (a) Same NaCl solution on both sides. (b) Only water Ba atoms/mol. 11.42 458 pm. 11.44 XY3. 11.48 0.220 nm. would move from left to right. (c) Normal osmosis. 12.106 12.3 M. 11.52 Molecular solid. 11.54 Molecular solids: Se8, HBr, CO2, 12.108 14.2%. 12.110 (a) and (d). 12.112 (a) Decreases with P4O6, SiH4. Covalent solids: Si, C. 11.56 Each C atom in diamond increasing lattice energy. (b) Increases with increasing polarity of is covalently bonded to four other C atoms. Graphite has solvent. (c) Increases with increasing enthalpy of hydration. delocalized electrons in two dimensions. 11.76 2.67 3 103 kJ. 12.114 1.80 g/mL. 5.0 3 102 m. 12.116 0.815. 12.118 NH3 can 11.78 47.03 kJ/mol. 11.80 Freezing, sublimation. 11.82 When form hydrogen bonds with water. 12.120 3%. 12.122 1.2 3 102 steam condenses at 1008C, it releases heat equal to heat of g/mol. It forms a dimer in benzene. 12.124 (a) 1.1 m. (b) The protein vaporization. 11.84 331 mmHg. 11.86 The small amount of liquid prevents the formation of ice crystals. 12.126 It is due to the nitrogen will evaporate quickly, extracting little heat from the skin. precipitated minerals that refract light and create an opaque Boiling water will release much more heat to the skin as it cools. appearance. 12.128 1.9 m. 12.130 (a) XA 5 0.524, XB 5 0.476. Water has a high specific heat. 11.90 Initially ice melts because of (b) A: 50 mmHg; B: 20 mmHg. (c) XA 5 0.71, XB 5 0.29. PA 5 67 the increase in pressure. As the wire sinks into the ice, the water mmHg. PB 5 12 mmHg. 12.132 2.7 3 1023. 12.134 From n 5 kP above it refreezes. In this way, the wire moves through the ice and PV 5 nRT, show that V 5 kRT. 12.136 20.7378C. 12.138 The without cutting it in half. 11.92 (a) Ice melts. (b) Water vapor polar groups 1C“O2 can bind the K1 ions. The exterior is condenses to ice. (c) Water boils. 11.94 (d). 11.96 Covalent crystal. nonpolar (due to the ¬CH3 groups), which enables the molecule 11.98 Orthorhombic. 11.100 760 mmHg. 11.102 It is the critical to pass through the cell membranes containing nonpolar lipids. point. 11.104 Crystalline SiO2. 11.106 (c) and (d). 12.140 The string is wetted and laid on top of the ice cube. Salt is 11.108 (a), (b), (d). 11.110 8.3 3 1023 atm. 11.112 (a) K2S. Ionic. shaken onto the top of the ice cube and the moistened string. The (b) Br2. Dispersion. 11.114 SO2. It is a polar molecule. presence of salt lowers the freezing point of the ice, resulting in 11.116 62.4 kJ/mol. 11.118 3048C. 11.120 Small ions have the melting of the ice on the surface. Melting is an endothermic more concentrated charges and are more effective in ion-dipole process. The water in the moist string freezes, and the string interaction, resulting in a greater extent of hydration. The distance becomes attached to the ice cube. The ice cube can now be lifted of separation between cation and anion is also shorter. out of the glass. 11.122 (a) 30.7 kJ/mol. (b) 192.5 kJ/mol. 11.124 (a) Decreases. (b) No change. (c) No change. 11.126 (a) 1 Cs1 ion and 1 Cl2 ion. (b) 4 Zn21 ions and 4 S22 ions. (c) 4 Ca21 ions and 8 F2 ions. Chapter 13 11.128 CaCO3(s) ¡ CaO(s) 1 CO2(g). Three phases. 11.130 SiO2 13.6 (a) Rate 5 2(1/2)D[H2]/Dt 5 2D[O2]/Dt 5 (1/2)D[H2O]/Dt. is a covalent crystal. CO2 exists as discrete molecules. 11.132 66.8%. (b) Rate 5 2(1/4)D[NH3]/Dt 5 2(1/5)D[O2]/Dt 5 (1/4)D[NO]/Dt 5 11.134 scc: 52.4%; bcc: 68.0%; fcc: 74.0%. 11.136 1.69 g/cm3. (1/6)D[H2O]/Dt. 13.8 (a) 0.049 M/s. (b) 0.025 M/s. 13.14 2.4 3 11.138 (a) Two (diamond/graphite/liquid and graphite/liquid/ 1024 M/s. 13.16 (a) Third order. (b) 0.38 M/s. 13.18 (a) 0.046 s21. vapor). (b) Diamond. (c) Apply high pressure at high temperature. (b) 0.13/M ? s. 13.20 First order. 1.08 3 1023 s21. 13.26 (a) 0.0198 11.140 Molecules in the cane are held together by intermolecular s21. (b) 151 s. 13.28 3.6 s. 13.30 (a) The relative rates for (i), (ii), forces. 11.142 When the tungsten filament is heated to a high and (iii) are 4:3:6. (b) The relative rates would be unaffected, but temperature (ca. 30008C), it sublimes and condenses on the inside each of the absolute rates would decrease by 50%. (c) The relative walls. The inert pressurized Ar gas retards sublimation. 11.144 When half-lives are 1:1:1. 13.38 135 kJ/mol. 13.40 103 kJ/mol. methane burns in air, it forms CO2 and water vapor. The latter 13.42 644 K. 13.44 9.25 3 103 s21. 13.46 51.0 kJ/mol. condenses on the outside of the cold beaker. 11.146 6.019 3 1023 13.56 (a) Rate 5 k[X2][Y]. (b) Reaction is zero order in Z. Fe atoms/mol. 11.148 Na (186 pm and 0.965 g/cm3). (c) X2 1 Y ¡ XY 1 X (slow). X 1 Z ¡ XZ (fast). 11.150 (d). 11.152 0.833 g/L. Hydrogen bonding in the gas phase. 13.58 Mechanism I. 13.66 Rate 5 (k1k2/k21)[E][S]. 13.68 This is a first-order reaction. The rate constant is 0.046 min21. 13.70 Temperature, energy of activation, concentration of Chapter 12 reactants, catalyst. 13.72 22.6 cm2; 44.9 cm2. The large surface 12.10 Cyclohexane cannot form hydrogen bonds. 12.12 The longer area of grain dust can result in a violent explosion. 13.74 (a) Third chains become more nonpolar. 12.16 (a) 25.9 g. (b) 1.72 3 103 g. order. (b) 0.38/M2 ? s. (c) H2 1 2NO ¡ N2 1 H2O 1 O (slow); 12.18 (a) 2.68 m. (b) 7.82 m. 12.20 0.010 m. 12.22 5.0 3 102 m; O 1 H2 ¡ H2O (fast). 13.76 Water is present in excess so its 18.3 M. 12.24 (a) 2.41 m. (b) 2.13 M. (c) 0.0587 L. concentration does not change appreciably. 13.78 10.7/M ? s. 13.80 12.28 45.9 g. 12.36 CO2 pressure is greater at the bottom of the 2.63 atm. 13.82 M22 s21. 13.84 56.4 min. 13.86 rate 5 mine. 12.38 0.28 L. 12.50 1.3 3 103 g. 12.52 Ethanol: 30.0 mmHg; k[A][B]2. 13.88 (b), (d), (e). 13.90 9.8 3 1024. 13.92 (a) Increase. 1-propanol: 26.3 mmHg. 12.54 128 g. 12.56 0.59 m. 12.58 120 g/mol. (b) Decrease. (c) Decrease. (d) Increase. 13.94 0.0896 min21. C4H8O4. 12.60 28.68C. 12.62 4.3 3 102 g/mol. C24H20P4. 13.96 1.12 3 103 min. 13.98 (a) I2 absorbs visible light to form I 12.64 1.75 3 104 g/mol. 12.66 343 g/mol. 12.70 Boiling point, atoms. (b) UV light is needed to dissociate H2. 13.100 (a) Rate 5 vapor pressure, osmotic pressure. 12.72 0.50 m glucose . 0.50 m k[X][Y]2. (b) 1.9 3 1022/M2 ? s. 13.102 Second order.  Answers to Even-Numbered Problems AP-7 2.4 3 107/M ? s. 13.104 Because the engine is relatively cold so Chapter 15 the exhaust gases will not fully react with the catalytic converter. 13.106 H2(g) 1 ICl(g) ¡ HCl(g) 1 HI(g) (slow). HI(g) 1 15.4 (a) NO2 2 2 2 2 . (b) HSO4 . (c) HS . (d) CN . (e) HCOO . 2 ICl(g) ¡ HCl(g) 1 I2(g) (fast). 13.108 5.7 3 105 yr. 15.6 (a) H2S. (b) H2CO3. (c) HCO2 3 . (d) H 3PO 4 . (e) H2 PO 2 4. 13.110 (a) Mn21; Mn31; first step. (b) Without the catalyst, (f) HPO4 . (g) H2SO4. (h) HSO4 . (i) HSO3 . 15.8 (a) CH2ClCOO2. 22 2 2 reaction would be termolecular. (c) Homogeneous. (b) IO2 2 22 32 2 4 . (c) H2PO4 . (d) HPO4 . (e) PO4 . (f) HSO4 . (g) SO4 . 22 2 22 2 22 2 13.112 0.45 atm. 13.114 (a) k1[A] 2 k2[B]. (b) [B] 5 (k1/k2)[A]. (h) IO3 . (i) SO3 . (j) NH3. (k) HS . (l) S . (m) OCl . 13.116 (a) 2.47 3 1022 yr21. (b) 9.8 3 1024. (c) 186 yr. 15.16 1.6 3 10214 M. 15.18 (a) 10.74. (b) 3.28. 15.20 (a) 6.3 3 13.118 (a) 3. (b) 2. (c) C ¡ D. (d) Exothermic. 13.120 1.8 3 1026 M. (b) 1.0 3 10216 M. (c) 2.7 3 1026 M. 15.22 (a) Acidic. 103 K. 13.122 (a) 2.5 3 1025 M/s. (b) Same as (a). (c) 8.3 3 1026 (b) Neutral. (c) Basic. 15.24 1.98 3 1023 mol. 0.444. 15.26 0.118. M. 13.126 (a) 1.13 3 1023 M/min. (b) 6.83 3 1024 M/min; 8.8 3 15.32 (1) c. (2) b and d. 15.34 (a) Strong. (b) Weak. (c) Weak. 1023 M. 13.128 Second order. 0.42/M ? min. 13.130 60% increase. (d) Weak. (e) Strong. 15.36 (b) and (c). 15.38 No. 15.44 [H1] 5 The result shows the profound effect of an exponential [CH3COO2] 5 5.8 3 1024 M, [CH3COOH] 5 0.0181 M. dependence. 13.132 2.6 3 1024 M/s. 13.134 404 kJ/mol. 15.46 2.3 3 1023 M. 15.48 (a) 3.5%. (b) 33%. (c) 79%. Percent 13.136 (a) Rate 5 k[NO]2[O2]. (b) Rate 5 kobs[NO]2. ionization increases with dilution. 15.50 (a) 3.9%. (b) 0.30%. (c) 1.3 3 103 min. 15.54 (c) , (a) , (b). 15.56 7.1 3 1027. 15.58 1.5%. 15.64 HCl: 1.40; H2SO4: 1.31. 15.66 [H1] 5 [HCO2 3 ] 5 1.0 3 10 24 M, [CO322] 5 4.8 3 10211 M. 15.70 (a) H2SO4 . H2SeO4. (b) H3PO4 . Chapter 14 H3AsO4. 15.72 The conjugate base of phenol can be stabilized by 14.14 (a) A 1 C Δ AC. (b) A 1 D Δ AD. 14.16 1.08 3 resonance. 15.78 (a) Neutral. (b) Basic. (c) Acidic. (d) Acidic. 107. 14.18 3.5 3 1027. 14.20 (a) 0.082. (b) 0.29. 14.22 0.105; 15.80 HZ , HY , HX. 15.82 4.82. 15.84 Basic. 15.88 (a) Al2O3 , 2.05 3 1023. 14.24 7.09 3 1023. 14.26 3.3. 14.28 0.0353. BaO , K2O. (b) CrO3 , Cr2O3 , CrO. 15.90 Al(OH)3 1 OH2 ¡ 14.30 4.0 3 1026. 14.32 5.6 3 1023. 14.36 0.64/M2 ? s. Al(OH)2 4 . Lewis acid-base reaction. 15.94 AlCl3 is the Lewis acid, 14.40 [NH3] will increase and [N2] and [H2] will decrease. Cl2 is the Lewis base. 15.96 CO2 and BF3. 15.98 0.0094 M. 14.42 NO: 0.50 atm; NO2: 0.020 atm. 14.44 [I] 5 8.58 3 1024 M; 15.100 0.106 L. 15.102 No. 15.104 No, volume is the same. [I2] 5 0.0194 M. 14.46 (a) 0.52. (b) [CO2] 5 0.48 M, [H2] 5 15.106 CrO is basic and CrO3 is acidic. 15.108 4.0 3 1022. 0.020 M, [CO] 5 0.075 M, [H2O] 5 0.065 M. 14.48 [H2] 5 15.110 7.00. 15.112 NH3. 15.114 (a) 7.43. (b) pD , 7.43. [CO2] 5 0.05 M, [H2O] 5 [CO] 5 0.11 M. 14.54 (a) Shift position (c) pD 1 pOD 5 14.87. 15.116 1.79. 15.118 F2 reacts with HF to of equilibrium to the right. (b) No effect. (c) No effect. form HF2 2 , thereby shifting the ionization of HF to the right. 14.56 (a) No effect. (b) No effect. (c) Shift the position of 15.120 (b) 6.80. 15.122 [H1] 5 [H2PO2 4 ] 5 0.0239 M, [H3PO4] 5 equilibrium to the left. (d) No effect. (e) To the left. 14.58 (a) To 0.076 M, [HPO422] 5 6.2 3 1028 M, [PO432] 5 1.2 3 10218 M. the right. (b) To the left. (c) To the right. (d) To the left. (e) No 15.124 Pyrex glass contains 10–25% B2O3, an acidic oxide. effect. 14.60 No change. 14.62 (a) More CO2 will form. (b) No 15.126 [Na1] 5 0.200 M, [HCO2 2 3 ] 5 [OH ] 5 4.6 3 10 23 M, change. (c) No change. (d) Some CO2 will combine with CaO to [H2CO3] 5 2.4 3 1028 M, [H1] 5 2.2 3 10212 M. 15.128 The H1 form CaCO3. (e) Some CO2 will react with NaOH so equilibrium ions convert CN2 to HCN, which escapes as a gas. 15.130 0.25 g. will shift to the right. (f) HCl reacts with CaCO3 to produce CO2. 15.132 20.20. 15.134 (a) Equilibrium will shift to the right. (b) To Equilibrium will shift to the left. (g) Equilibrium will shift to the the left. (c) No effect. (d) To the right. 15.136 The amines are right. 14.64 (a) NO: 0.24 atm; Cl2: 0.12 atm. (b) 0.017. converted to their salts RNH1 24 3 . 15.138 1.4 3 10 . 15.140 4.40. 14.66 [A2] 5 [B2] 5 0.040 M. [AB] 5 0.020 M. 14.68 (a) No 15.142 In a basic medium, the ammonium salt is converted to the effect. (b) More CO2 and H2O will form. 14.70 (a) 8 3 10244. pungent-smelling ammonia. 15.144 (c). 15.146 21 mL. 15.148 HX (b) The reaction has a very large activation energy. 14.72 (a) 1.7. is the stronger acid. 15.150 Mg. 15.152 1.57. [CN2] 5 1.8 3 1028 (b) A: 0.69 atm, B: 0.81 atm. 14.74 1.5 3 105. 14.76 H2: 0.28 atm, M in 1.00 M HF and 2.2 3 1025 M in 1.00 M HCN. HF is a stronger Cl2: 0.049 atm, HCl: 1.67 atm. 14.78 5.0 3 101 atm. 14.80 3.84 3 acid than HCN. 15.154 6.02. 15.156 1.18. 15.158 (a) pH 5 7.24. 1022. 14.82 3.13. 14.84 N2: 0.860 atm; H2: 0.366 atm; NH3: (b) 10,000 H3O1 ions for every OH2 ion. 15.160 Both are 4.40 3 1023 atm. 14.86 (a) 1.16. (b) 53.7%. 14.88 (a) 0.49 atm. 255.9 kJ/mol because they have the same net ionic equation. (b) 0.23. (c) 0.037. (d) Greater than 0.037 mol. 14.90 [H2] 5 0.070 M, [I2] 5 0.182 M, [HI] 5 0.825 M. 14.92 (c). 14.94 (a) 4.2 3 1024. (b) 0.83. (c) 1.1. (d) In (b): 2.3 3 103; in Chapter 16 (c): 0.021. 14.96 0.0231; 9.60 3 1024. 14.98 NO2: 1.2 atm; N2O4: 16.6 (a) 11.28. (b) 9.08. 16.10 (a), (b), and (c). 16.12 4.74 for both. 0.12 atm. KP 5 12. 14.100 (a) Kc 5 33.3. (b) Qc 5 2.8. Shift to the (a) is more effective because it has a higher concentration. right. (c) Qc 5 169. Shift to the left. 14.102 (a) The equilibrium 16.14 7.03. 16.16 10. More effective against the acid. will shift to the right. (b) To the right. (c) No change. (d) No 16.18 (a) 4.82. (b) 4.64. 16.20 HC. 16.22 (l) (a): 5.10. (b): 4.82. change. (e) No change. (f) To the left. 14.104 NO2: 0.100 atm; (c): 5.22. (d): 5.00. (2) 4.90. (3) 5.22. 16.24 0.53 mole. N2O4: 0.09 atm. 14.106 (a) 1.03 atm. (b) 0.39 atm. (c) 1.67 atm. 16.28 90.1 g/mol. 16.30 0.467 M. 16.32 [H1] 5 3.0 3 10213 M, (d) 0.620. 14.108 (a) KP 5 2.6 3 1026; Kc 5 1.1 3 1027. [OH2] 5 0.0335 M, [Na1] 5 0.0835 M, [CH3COO2] 5 0.0500 M, (b) 22 mg/m3. Yes. 14.110 Temporary dynamic equilibrium [CH3COOH] 5 8.4 3 10210 M. 16.34 8.23. 16.36 (a) 11.36. between the melting ice cubes and the freezing of the water (b) 9.55. (c) 8.95. (d) 5.19. (e) 1.70. 16.38 (1) (c). (2) (a). (3) between the ice cubes. 14.112 [NH3] 5 0.042 M, [N2] 5 0.086 M, (d). (4) (b). pH , 7 at the equivalence point. 16.40 6.0 3 1026. [H2] 5 0.26 M. 14.114 1.3 atm. 14.116 PCl5: 0.683 atm; PCl3: 16.44 CO2 dissolves in water to form H2CO3, which neutralizes 1.11 atm; Cl2: 0.211 atm. 14.118 2115 kJ/mol. 14.120 SO2: 2.71 NaOH. 16.46 5.70. 16.54 (a) 7.8 3 10210. (b) 1.8 3 10218. atm; Cl2: 2.71 atm; SO2Cl2: 3.58 atm. 14.122 4.0. 14.124 (a) The 16.56 1.80 3 10210. 16.58 2.2 3 1024 M. 16.60 2.3 3 1029. plot curves toward higher pressure at low values of 1/V. (b) The 16.62 [Na1] 5 0.045 M, [NO2 21 3 ] 5 0.076 M, [Sr ] 5 0.016 M, plot curves toward higher volume as T increases. [F2] 5 1.1 3 1024 M. 16.64 pH greater than 3.34 and less than AP-8 Answers to Even-Numbered Problems 8.11. 16.68 (a) 0.013 M. (b) 2.2 3 1024 M. (c) 3.3 3 1023 M. (c) False. 17.72 C 1 CuO Δ CO 1 Cu. 6.1. 17.74 673.2 K. 16.70 (a) 1.0 3 1025 M. (b) 1.1 3 10210 M. 16.72 (b), (c), (d), and 17.76 (a) 7.6 3 1014. (b) 4.1 3 10212. 17.78 (a) A reverse (e). 16.74 (a) 0.016 M. (b) 1.6 3 1026 M. 16.76 Yes. 16.80 [Cd21] 5 disproportionation reaction. (b) 8.2 3 1015. Yes, a large K makes 1.1 3 10218 M, [Cd(CN)422] 5 4.2 3 1023 M, [CN2] 5 0.48 M. this an efficient process. (c) Less effective. 17.80 1.8 3 1070. 16.82 3.5 3 1025 M. 16.84 (a) Cu21 1 4NH3 Δ Cu(NH3)21 4 . Reaction has a large activation energy. 17.82 Heating the ore alone (b) Ag1 1 2CN2 Δ Ag(CN)2 2 . (c) Hg 21 1 4Cl2 Δ HgCl22 4 . is not a feasible process. 2214.3 kJ/mol. 17.84 KP 5 36. 981 K. 16.88 0.011 M. 16.90 Use Cl2 ions or flame test. 16.92 From 2.51 No. 17.86 Negative. 17.88 Mole percents: butane 5 30%; to 4.41. 16.94 1.8 3 102 mL. 16.96 1.28 M. 16.98 [H1] 5 3.0 3 isobutane 5 70%. Yes. 17.90 (a) Na(l): 99.69 J/K ? mol. 10213 M, [OH2] 5 0.0335 M, [HCOO2] 5 0.0500 M, [HCOOH] 5 (b) S2Cl2(g): 331.5 J/K ? mol. (c) FeCl2(s): 117.9 J/K ? mol. 8.8 3 10211 M, [Na1] 5 0.0835 M. 16.100 9.97 g. pH 5 13.04. 17.92 Mole fractions are: CO 5 0.45, CO2 5 0.55. Use DGf8 16.102 6.0 3 103. 16.104 0.036 g/L. 16.106 (a) 1.37. (b) 5.97. values at 258C for 9008C. 17.94 617 J/K. 17.96 3 3 10213 s. (c) 10.24. 16.108 Original precipitate was HgI2. In the presence 17.98 DSsys 5 2327 J/K ? mol, DSsurr 5 1918 J/K ? mol, DSuniv 5 of excess KI, it redissolves as HgI422. 16.110 7.82 2 10.38. 1591 J/K ? mol. 17.100 q, w. 17.102 DH , 0, DS , 0, DG , 0. 16.112 (a) 3.60. (b) 9.69. (c) 6.07. 16.114 (a) MCO3 1 2HCl ¡ 17.104 (a) 5.76 J/K ? mol. (b) The orientation is not totally MCl2 1 H2O 1 CO2. HCl 1 NaOH ¡ NaCl 1 H2O. random. 17.106 DH8 5 33.89 kJ/mol; DS8 5 96.4 J/K ? mol; (b) 24.3 g/mol. Mg. 16.116 2. 16.118 (a) 12.6. (b) 8.8 3 1026 M. DG8 5 5.2 kJ/mol. This is an endothermic liquid to vapor process 16.120 (a) Sulfate. (b) Sulfide. (c) Iodide. 16.122 They are so both DH8 and DS8 are positive. DG8 is positive because the insoluble. 16.124 The ionized polyphenols have a dark color. The temperature is below the boiling point of benzene (80.18C). H1 ions from lemon juice shift the equilibrium to the light color 17.108 DG8 5 62.5 kJ/mol; DH8 5 157.8 kJ/mol; DS8 5 109 acid. 16.126 Yes. 16.128 (c). 16.130 (a) 1.7 3 1027 M. (b) MgCO3 J/K ? mol. 17.110 Slightly larger than 0.052 atm. is more soluble than CaCO3. (c) 12.40. (d) 1.9 3 1028 M. (e) Ca21 because it is present in larger amount. 16.132 pH 5 1.0, fully Chapter 18 protonated; pH 5 7.0, dipolar ion; pH 5 12.0, fully ionized. 18.2 (a) Mn21 1 H2O2 1 2OH2 ¡ MnO2 1 2H2O. 16.134 (a) 8.4 mL. (b) 12.5 mL. (c) 27.0 mL. 16.136 (a) 4.74 (b) 2Bi(OH)3 1 3SnO22 2 ¡ 2Bi 1 3H2O 1 3SnO3 . 22 22 1 22 31 before and after dilution. (b) 2.52 before and 3.02 after dilution. (c) Cr2O7 1 14H 1 3C2O4 ¡ 2Cr 1 6CO2 1 7H2O. 16.138 4.75. 16.140 (a) 0.0085 g. (b) 2.7 3 1028 g. (c) 1.2 3 1024 g. (d) 2Cl2 1 2ClO23 1 4H1 ¡ Cl2 1 2ClO2 1 2H2O. 18.12 2.46 V. 16.142 (1) The initial pH of acid (a) is lower. (2) The pH at half- Al 1 3Ag1 ¡ 3Ag 1 Al31. 18.14 Cl2(g) and MnO2 4 (aq). way to the equivalence point is lower for (a). (3) The pH at the 18.16 Only (a) and (d) are spontaneous. 18.18 (a) Li. (b) H2. equivalence point is lower for acid (a), indicating that (a) forms a (c) Fe21. (d) Br2. 18.20 21.79 V. 18.24 0.368 V. weaker conjugate base than (b). Thus, (a) is the stronger acid. 18.26 (a) 2432 kJ/mol, 5 3 1075. (b) 2104 kJ/mol, 2 3 1018. 16.144 [Cu21] 5 1.8 3 1027 M. [OH2] 5 3.6 3 1027 M. [Ba21] 5 (c) 2178 kJ/mol, 1 3 1031. (d) 21.27 3 103 kJ/mol, 8 3 10211. [SO224 ] 5 1.0 3 10 25 M. 18.28 0.37 V, 236 kJ/mol, 2 3 106. 18.32 (a) 2.23 V, 2.23 V, 2430 kJ/mol. (b) 0.02 V, 0.04 V, 223 kJ/mol. 18.34 0.083 V. 18.36 0.010 Chapter 17 V. 18.40 1.09 V. 18.48 (b) 0.64 g. 18.50 (a) $2.10 3 103. (b) $2.46 17.6 (a) 0.25. (b) 8 3 10231. (c) < 0. For a macroscopic system, 3 103. (c) $4.70 3 103. 18.52 (a) 0.14 mol. (b) 0.121 mol. (c) 0.10 the probability is practically zero that all the molecules will be mol. 18.54 (a) Ag1 1 e2 ¡ Ag. (b) 2H2O ¡ O2 1 4H1 1 found only in one bulb. 17.10 (c) , (d) , (e) , (a) , (b). Solids 4e2. (c) 6.0 3 102 C. 18.56 (a) 0.589 Cu. (b) 0.133 A. 18.58 2.3 h. have smaller entropies than gases. More complex structures have 18.60 9.66 3 104 C. 18.62 0.0710 mol. 18.64 (a) Anode: Cu(s) higher entropies. 17.12 (a) 47.5 J/K ? mol. (b) 212.5 J/K ? mol. ¡ Cu21(aq) 1 2e2. Cathode: Cu21(aq) 1 2e2 ¡ Cu(s). (c) 2242.8 J/K ? mol. 17.14 (a) DS , 0. (b) DS . 0. (c) DS . 0. (b) 2.4 3 102 g. (c) Copper is more easily oxidized than Ag and (d) DS , 0. 17.18 (a) 21139 kJ/mol. (b) 2140.0 kJ/mol. Au. Copper ions (Cu21) are more easily reduced than Fe21 and (c) 22935.0 kJ/mol. 17.20 (a) At all temperatures. (b) Below Zn21. 18.66 0.0296 V. 18.68 0.156 M. Cr2O22 7 1 6Fe 21 1 1 31 31 111 K. 17.24 8.0 3 101 kJ/mol. 17.26 4.572 3 102 kJ/mol. 7.2 3 14H ¡ 2Cr 1 6Fe 1 7H2O. 18.70 45.1%. 18.72 (a) 10281. 17.28 (a) 224.6 kJ/mol. (b) 21.33 kJ/mol. 17.30 2341 2MnO2 1 22 4 1 16H 1 5C2O4 ¡ 2Mn 21 1 10CO2 1 8H2O. 21 kJ/mol. 17.32 22.87 kJ/mol. The process has a high activation (b) 5.40%. 18.74 0.231 mg Ca /mL blood. 18.76 (a) 0.80 V. energy. 17.36 1 3 103. glucose 1 ATP ¡ glucose 6-phosphate 1 (b) 2Ag1 1 H2 ¡ 2Ag 1 2H1. (c) (i) 0.92 V. (ii) 1.10 V. ADP. 1 3 103. 17.38 (a) 0. (b) 4.0 3 104 J/mol. (c) 23.2 3 104 (d) The cell operates as a pH meter. 18.78 Fluorine gas reacts with J/mol. (d) 6.4 3 104 J/mol. 17.40 Positive. 17.42 (a) No reaction is water. 18.80 2.5 3 102 h. 18.82 Hg21 21 2 . 18.84 [Mg ] 5 0.0500 M, 1 255 possible because DG . 0. (b) The reaction has a very large [Ag ] 5 7 3 10 M, 1.44 g. 18.86 (a) 0.206 L H2. (b) 6.09 activation energy. (c) Reactants and products already at their 3 1023/mol e2. 18.88 (a) 21356.8 kJ/mol. (b) 1.17 V. 18.90 13. equilibrium concentrations. 17.44 In all cases DH . 0 and DS . 18.92 6.8 kJ/mol, 0.064. 18.94 In both cells, the anode is on 0. DG , 0 for (a), 5 0 for (b), and . 0 for (c). 17.46 DS . 0. the left and the cathode is on the right. In the galvanic cell, the 17.48 (a) Most liquids have similar structure so the changes in anode is negatively charged and the cathode is positively charged. entropy from liquid to vapor are similar. (b) DSvap are larger for The opposite holds for the electrolytic cell. Electrons flow from ethanol and water because of hydrogen bonding (there are fewer the anode in the galvanic cell to the cathode in the electrolytic microstates in these liquids). 17.50 (a) 2CO 1 2NO ¡ 2CO2 1 cell and electrons flow from the anode in the electrolytic cell N2. (b) Oxidizing agent: NO; reducing agent: CO. (c) 3 3 10120. to the cathode in the galvanic cell. 18.96 1.4 A. 18.98 14. 18.100 (d) 1.2 3 1018. From left to right. (e) No. 17.52 2 3 10210. 1.60 3 10219 C/e2. 18.102 A cell made of Li1/Li and F2/F2 17.54 2.6 3 1029. 17.56 976 K. 17.58 DS , 0; DH , 0. gives the maximum voltage of 5.92 V. Reactive oxidizing and 17.60 55 J/K ? mol. 17.62 Increase in entropy of the surroundings reducing agents are hard to handle. 18.104 0.030 V. 18.106 offsets the decrease in entropy of the system. 17.64 56 J/K. 2 3 1020. 18.108 (a) E8 for X is negative; E8 for Y is positive. 17.66 4.5 3 105. 17.68 4.8 3 10275 atm. 17.70 (a) True. (b) True. (b) 0.59 V. 18.110 (a) The reduction potential of O2 is insufficient  Answers to Even-Numbered Problems AP-9 to oxidize gold. (b) Yes. (c) 2Au 1 3F2 ¡ 2AuF3. 18.112 20.42 Ethane and propane are greenhouse gases. 20.50 4.34. [Fe21] 5 0.0920 M, [Fe31] 5 0.0680 M. 18.114 E8 5 1.09 V. 20.58 1.2 3 10211 M/s. 20.60 (b). 20.66 0.12%. 20.68 Endothermic. Spontaneous. 18.116 (a) Ni. (b) Pb. (c) Zn. (d) Fe. 18.118 (a) 20.70 O2. 20.72 5.72. 20.74 394 nm. 20.76 It has a high activation Unchanged. (b) Unchanged. (c) Squared. (d) Doubled. (e) energy. 20.78 Size of tree rings are related to CO2 content. Age Doubled. 18.120 Stronger. 18.122 4.4 3 102 atm. 18.124 (a) Zn of CO2 in ice can be determined by radiocarbon dating. ¡ Zn21 1 2e2; (1/2)O2 1 2e2 ¡ O22. 1.65 V. (b) 1.63 V. 20.80 165 kJ/mol. 20.82 5.1 3 1020 photons. 20.84 (a) 62.6 (c) 4.87 3 103 kJ/kg. (d) 62 L. 18.126 23.05 V. 18.128 1 3 O Q O kJ/mol. (b) 38 min. 20.86 5.6 3 1023. 20.88 HOOOOOOT Q O Q 10214. 18.130 (b) 104 A ? h. The concentration of H2SO4 keeps decreasing. (c) 2.01 V; 23.88 3 102 kJ/mol. 18.132 Chapter 21 $217. 18.134 20.037 V. 18.136 2 3 1037. 18.138 5 mol ATP. 21.12 111 h. 21.14 Roast the sulfide followed by reduction of the 18.140 2.87 V. oxide with coke or carbon monoxide. 21.16 (a) 8.9 3 1012 cm3. (b) 4.0 3 108 kg. 21.18 Iron does not need to be produced Chapter 19 electrolytically. 21.28 (a) 2Na 1 2H2O ¡ 2NaOH 1 H2. 19.6 (Z,N,A) 42α decay: (22, 22, 24). 210β decay: (11, 21, 0). (b) 2NaOH 1 CO2 ¡ Na2CO3 1 H2O. (c) Na2CO3 1 0 2 0 2HCl ¡ 2NaCl 1 CO2 1 H2O. (d) NaHCO3 1 HCl ¡ 11β decay: (21, 11, 0). e capture: (21, 11, 0). 19.8 (a) 21β. (b) 40 Ca. (c) 4 α. (d) 1 n. 19.16 (a) 9 Li. (b) 25 Na. (c) 48 Sc. NaCl 1 CO2 1 H2O. (e) 2NaHCO3 ¡ Na2CO3 1 CO2 1 H2O. 20 2 0 3 11 21 19.18 (a) 17 45 92 195 242 (f) No reaction. 21.30 5.59 L. 21.34 First react Mg with HNO3 to 10Ne. (b) 20Ca. (c) 43Tc. (d) 80 Hg. (e) 96 Cm. 19.20 6 3 9 212 10 kg/s. 19.22 (a) 4.55 3 10 J; 1.14 3 10 J/nucleon. 212 form Mg(NO3)2. On heating, 2Mg(NO3)2 ¡ 2MgO 1 4NO2 1 (b) 2.36 3 10210 J; 1.28 3 10212 J/nucleon. 19.26 0.251 d21. 2.77 d. O2. 21.36 The third electron is removed from the neon core. 19.28 2.7 d. 19.30 208 21.38 Helium has a closed-shell noble gas configuration. 82 Pb. 19.32 A: 0; B: 0.25 mole; C: 0; D: 0.75 mole. 19.34 224 80 2 1 81 21.40 (a) CaO. (b) Ca(OH)2. (c) An aqueous suspension of 88 Ra. 19.38 (a) 34Se 1 1H ¡ 1p 1 34Se. (b) 94Be 1 21H ¡ 211p 1 93Li. (c) 105B 1 10n ¡ 42α 1 73Li. Ca(OH)2. 21.44 60.7 h. 21.46 (a) 1.03 V. (b) 3.32 3 104 kJ/mol. 19.40 198 1 198 1 2 21.48 4Al(NO3)3 ¡ 2Al2O3 1 12NO2 1 3O2. 21.50 Because 80 Hg 1 0n ¡ 79 Au 1 1p. 19.52 IO3 is only formed from IO2 . 19.54 Incorporate Fe-59 into a person’s body. After a Al2Cl6 dissociates to form AlCl3. 21.52 From sp3 to sp2. 4 few days isolate red blood cells and monitor radioactivity from the 21.54 65.4 g/mol. 21.56 No. 21.58 (a) 1482 kJ/mol. hemoglobin molecules. 19.56 (a) 50 50 0 (b) 3152.8 kJ/mol. 21.60 Magnesium reacts with nitrogen to form 25Mn ¡ 24Cr 1 11β. (b) Three half-lives. 19.58 An analogous Pauli exclusion principle magnesium nitride. 21.62 (a) Al31 hydrolyzes in water to produce for nucleons. 19.60 (a) 0.343 mCi. (b) 237 4 233 H1 ions. (b) Al(OH)3 dissolves in a strong base to form Al(OH)2 4. 93 Np ¡ 2α 1 91 Pa. 212 19.62 (a) 1.040 3 10 J/nucleon. (b) 1.111 3 10 J/nucleon. 212 21.64 CaO 1 2HCl ¡ CaCl2 1 H2O. 21.66 Electronic (c) 1.199 3 10212 J/nucleon. (d) 1.410 3 10212 J/nucleon. transitions (in the visible region) between closely spaced energy 19.64 187N ¡ 188O 1 210β. 19.66 Radioactive dating. levels. 21.68 NaF: toothpaste additive; Li2CO3: to treat mental 19.68 (a) 209 4 211 1 209 211 illness; Mg(OH)2: antacid; CaCO3: antacid; BaSO4: for X-ray 83 Bi 1 2α ¡ 85 At 1 20n. (b) 83 Bi(α,2n) 85 At. 19.70 The sun exerts a much greater gravity on the particles. diagnostic of digestive system; Al(OH)2NaCO3: antacid. 19.72 2.77 3 103 yr. 19.74 (a) 40 40 0 21.70 (i) Both Li and Mg form oxides. (ii) Like Mg, Li forms 19K ¡ 18Ar 1 11β. (b) 3.0 3 10 yr. 19.76 (a) Sr: 5.59 3 10 J; Y: 2.84 3 10213 J. 9 90 215 90 nitride. (iii) The carbonates, fluorides, and phosphates of Li and (b) 0.024 mole. (c) 4.26 3 106 kJ. 19.78 2.7 3 1014 I-131 atoms. Mg have low solubilities. 21.72 Zn. 21.74 D , A , C , B. 19.80 5.9 3 1023/mol. 19.82 All except gravitational. 19.84 U-238 21.76 727 atm. and Th-232. Long half-lives. 19.86 8.3 3 1024 nm. 19.88 31H. 19.90 The reflected neutrons induced a nuclear chain reaction. Chapter 22 19.92 2.1 3 102 g/mol. 19.94 First step: 234 234 90 Th ¡ 91 Pa 1 21β. 0 22.12 (a) Hydrogen reacts with alkali metals to form hydrides. 234 234 Second step: 91 Pa ¡ 92 U 1 21β. Third step: 0 (b) Hydrogen reacts with oxygen to form water. 22.14 Use 234 230 4 230 92 U ¡ 90 Th 1 2α. Fourth step: 90 Th ¡ 88 Ra 1 2α. 226 4 palladium metal to separate hydrogen from other gases. Fifth step: 88 Ra ¡ 86 Rn 1 2α. 19.96 (a) 94 Pu ¡ 2α 1 234 226 222 4 238 4 92 U. 22.16 11 kg. 22.18 (a) H2 1 Cl2 ¡ 2HCl. (b) N2 1 3H2 ¡ (b) t 5 0: 0.58 mW; t 5 10 yr: 0.53 mW. 19.98 0.49 rem. 2NH3. (c) 2Li 1 H2 ¡ 2LiH, LiH 1 H2O ¡ LiOH 1 H2. 19.100 The high temperature attained during the chain reaction 22.26 :C‚C:22 . 22.28 (a) 2NaHCO3 ¡ Na2CO3 1 H2O 1 causes a small-scale nuclear fusion: 21H 1 31H ¡ 42He 1 10n. The CO2. (b) CO2 reacts with Ca(OH)2 solution to form a white additional neutrons will result in a more powerful fission bomb. precipitate (CaCO3). 22.30 On heating, the bicarbonate ion 19.102 21.5 mL. 19.104 No. According to Equation (19.1), energy decomposes: 2HCO2 22 3 ¡ CO3 1 H2O 1 CO2. Mg 21 ions and mass are interconvertible. 19.106 (a) 1.69 3 10212 J. (b) 1.23 3 combine with CO22 3 ions to form MgCO3. 22.32 First, 2NaOH 1 10212 J. Because a proton feels the repulsion from other protons, CO2 ¡ Na2CO3 1 H2O. Then, Na2CO3 1 CO2 1 H2O ¡ it has a smaller binding energy than a neutron. 2NaHCO3. 22.34 Yes. 22.40 (a) 2NaNO3 ¡ 2NaNO2 1 O2. (b) NaNO3 1 C ¡ NaNO2 1 CO. 22.42 2NH3 1 CO2 ¡ Chapter 20 (NH2)2CO 1 H2O. At high pressures. 22.44 NH4Cl decomposes to form NH3 and HCl. 22.46 N is in its highest oxidation state (15) 20.6 3.3 3 1024 atm. 20.8 N2: 3.96 3 1018 kg; O2: 1.22 3 1018 kg; in HNO3. 22.48 Favored reaction: 4Zn 1 NO2 3 1 10H ¡ 1 CO2: 2.63 3 1015 kg. 20.12 3.57 3 10219 J. 20.22 5.2 3 106 21 1 4Zn 1 NH4 1 3H2O. 22.50 Linear. 22.52 21168 kJ/mol. kg/day. 5.6 3 1014 kJ. 20.24 The wavelength is not short enough. 22.54 P4. 125 g/mol. 22.56 P4O10 1 4HNO3 ¡ 2N2O5 1 20.26 434 nm. Both. 4HPO3. 60.4 g. 22.58 sp3. 22.66 2198.3 kJ/mol, 6 3 1034, F H F H 6 3 1034. 22.68 0; 21. 22.70 4.4 3 1011 mol; 1.4 3 1013 g. A A A A 20.28 FOCOCOCl FOCOCOH 20.40 1.3 3 1010 kg. 22.72 79.1 g. 22.74 Cl, Br, and I atoms are too bulky around the A A A A S atom. 22.76 35 g. 22.78 9H2SO4 1 8NaI ¡ H2S 1 4I2 1 F Cl F F 4H2O 1 8NaHSO4. 22.82 H2SO4 1 NaCl ¡ HCl 1 NaHSO4. AP-10 Answers to Even-Numbered Problems The HCl gas escapes, driving the equilibrium to the right. measurements. 23.66 EDTA sequesters essential metal ions (Ca21, 22.84 25.3 L. 22.86 Sulfuric acid oxidizes sodium bromide to Mg21). 23.68 3. 23.70 1.0 3 10218 M. 23.72 2.2 3 10220 M. molecular bromine. 22.88 2.81 L. 22.90 I2O5 1 5CO ¡ I2 1 23.74 (a) 2.7 3 106. (b) Cu1 ions are unstable in solution. 5CO2. C is oxidized; I is reduced. 22.92 (a) SiCl4. (b) F2. (c) F. 23.76 (a) Cu31 is unstable in solution because it can be easily (d) CO2. 22.94 No change. 22.96 (a) 2Na 1 D2O ¡ 2NaOD 1 reduced. (b) Potassium hexafluorocuprate(III). Octahedral. D2. (b) 2D2O ¡ 2D2 1 O2 (electrolysis). D2 1 Cl2 ¡ Paramagnetic. (c) Diamagnetic. 2DCl. (c) Mg3N2 1 6D2O ¡ 3Mg(OD)2 1 2ND3. (d) CaC2 1 2D2O ¡ C2D2 1 Ca(OD)2. (e) Be2C 1 4D2O ¡ Chapter 24 2Be(OD)2 1 CD4. (f) SO3 1 D2O ¡ D2SO4. 22.98 (a) At 24.12 CH3CH2CH2CH2CH2Cl. CH3CH2CH2CHClCH3. elevated pressure, water boils above 1008C. (b) So the water is CH3CH2CHClCH2CH3. able to melt a larger area of sulfur deposit. (c) Sulfur deposits are structurally weak. Conventional mining would be dangerous. H CH3 Br CH3 G D G D 22.100 The C¬D bond breaks at a slower rate. 22.102 Molecular 24.14 CPC CPC D G D G oxygen is a powerful oxidizing agent, reacting with substances Br H H H such as glucose to release energy for growth and function. Molecular nitrogen (containing the nitrogen-to-nitrogen triple H H G D bond) is too unreactive at room temperature to be of any practical H CH2Br H C H G D G D G D use. 22.104 258C: 9.61 3 10222; 10008C: 138. High temperature CPC COOC favors the formation of CO. 22.106 1.18. D G D G H H H Br Chapter 23 24.16 (a) Alkene or cycloalkane. (b) Alkyne. (c) Alkane. (d) Like 23.12 (a) 13. (b) 6. (c) oxalate. 23.14 (a) Na: 11, Mo: 16. (a). (e) Alkyne. 24.18 No, too much strain. 24.20 (a) is alkane and (b) Mg: 12, W: 16. (c) Fe: 0. 23.16 (a) cis-dichlorobis(ethylene- (b) is alkene. Only an alkene reacts with a hydrogen halide and diamine)cobalt(III). (b) pentaamminechloroplatinum(IV) chloride. hydrogen. 24.22 2630.8 kJ/mol. 24.24 (a) cis-1,2- (c) pentaamminechlorocobalt(III) chloride. 23.18 (a) [Cr(en)2Cl2]1. dichlorocylopropane. (b) trans-1,2-dichlorocylopropane. (b) Fe(CO)5. (c) K2[Cu(CN)4]. (d) [Co(NH3)4(H2O)Cl]Cl2. 24.26 (a) 2-methylpentane. (b) 2,3,4-trimethylhexane. 23.24 (a) 2. (b) 2. 23.26 (a) Two geometric isomers: (c) 3-ethylhexane. (d) 3-methyl-1,4-pentadiene. (e) 2-pentyne. (f) 3-phenyl-1-pentene. Cl Cl H3N≈ A ∞NH3 H3N≈ A ∞Cl CH3 H H C2H5 G D G D )Co- )Co- 24.28 (a) CPC (b) CPC H3N A NH3 H3N A NH3 D G D G Cl NH3 H C2H5 H C2H5 trans cis CH3 H G D HOCqCOCHOCH3 (b) Two optical isomers: CPC A D G (c) H CHOC3H7 (d) A C2H5 24.32 (a) 1,3-dichloro-4-methylbenzene. (b) 2-ethyl- Co Co 1,4-dinitrobenzene. (c) 1,2,4,5-tetramethylbenzene. 24.36 (a) Ether. (b) Amine. (c) Aldehyde. (d) Ketone. (e) Carboxylic acid. (f) Alcohol. (g) Amino acid. 24.38 HCOOH 1 CH3OH ¡ HCOOCH3 1 H2O. Methyl formate. 24.40 (CH3)2CH¬O¬CH3. 24.42 (a) Ketone. (b) Ester. (c) Ether. 24.44 2174 kJ/mol. 24.46 (a), (c), (d), (f). 23.34 CN2 is a strong-field ligand. Absorbs near UV (blue) 24.48 (a) Rubbing alcohol. (b) Vinegar. (c) Moth balls. so appears yellow. 23.36 (a) Orange. (b) 255 kJ/mol. (d) Organic synthesis. (e) Organic synthesis. (f) Antifreeze. 23.38 [Co(NH3)4Cl2]Cl. 2 moles. 23.42 Use 14CN2 label (g) Natural gas. (h) Synthetic polymer. 24.50 (a) 3. (b) 16. (c) 6. (in NaCN). 23.44 First Cu(CN)2 (white) is formed. It redissolves 24.52 (a) C: 15.81 mg, H: 1.33 mg, O: 3.49 mg. (b) C6H6O. as Cu(CN)422. 23.46 1.4 3 102. 23.48 Mn31. The 3d3 electron configuration of Cr31 is stable. 23.50 Ti: 13; Fe: 13. 23.52 Four (c) Phenol. OH A Fe atoms per hemoglobin molecule. 1.6 3 104 g/mol. 23.54 (a) [Cr(H2O)6]Cl3. (b) [Cr(H2O)5Cl]Cl2 ? H2O. (c) [Cr(H2O)4Cl2]Cl ? 2H2O. Compare electrical conductance with solutions of NaCl, 24.54 Empirical and molecular formula: C5H10O. 88.7 g/mol. MgCl2, and FeCl3 of the same molar concentration. 23.56 21.8 3 102 kJ/mol; 6 3 1030. 23.58 Iron is more abundant. CH H2COOOCH2 E H2 A A H2C CH C 2 23.60 Oxyhemoglobin is low spin and therefore absorbs higher A A H2C CH(CH3) energy light. 23.62 All except Fe21, Cu21, and Co21. The colorless H2CH ECH2 G D O ions have electron configurations d0 and d10. 23.64 Dipole moment O  Answers to Even-Numbered Problems AP-11 CH2 “CH¬CH2 ¬O¬CH2 ¬CH3 . 24.56 (a) The C atoms bonded 25.12 (a) CH2 “CH¬CH“CH2 . (b) HO2C(CH2)6NH2. to the methyl group and the amino group and the H atom. (b) The 25.22 At 358C the enzyme begins to denature. 25.28 Proteins are C atoms bonded to Br. 24.58 CH3CH2CHO. 24.60 (a) Alcohol. made of 20 amino acids. Nucleic acids are made of four building (b) Ether. (c) Aldehyde. (d) Carboxylic acid. (e) Amine. 24.62 The blocks (purines, pyrimidines, sugar, phosphate group) only. acids in lemon juice convert the amines to the ammonium salts, 25.30 C-G base pairs have three hydrogen bonds and higher which have very low vapor pressures. 24.64 Methane (CH4), boiling point; A-T base pairs have two hydrogen bonds. 25.32 Leg ethanol (C2H5OH), methanol (CH3OH), isopropanol (C3H7OH), muscles are active, have a high metabolic rate and hence a high ethylene glycol (CH2OHCH2OH), naphthalene (C10H8), acetic acid concentration of myoglobin. The iron content in Mb makes the (CH3COOH). 24.66 (a) 1. (b) 2. (c) 5. 24.68 Br2 dissociates into Br meat look dark. 25.34 Insects have blood that contains no atoms, which react with CH4 to form CH3Br and HBr. hemoglobin. It is unlikely that a human-sized insect could obtain sufficient oxygen for metabolism by diffusion. 25.36 There are OH A four Fe atoms per hemoglobin molecule. 1.6 3 104 g/mol. 24.70 (a) CH3OCOCH C 2OCH3 . The compound is chiral. 25.38 Mostly dispersion forces. 25.40 Gly-Ala-Phe-Glu-His-Gly- A Ala-Leu-Val. 25.42 No. Enzymes only act on one of the two H optical isomers of a compound. 25.44 315 K. (b) The product is a racemic mixture. 25.46 Hydrogen bonding. 25.48 (a) The ¬COOH group. (b) pH 5 1.0: The valine is in the fully protonated form. OH pH 5 7.0: Only the ¬COOH group is ionized. pH 5 12.0: A 24.72 CH3CH2CH2OH or CH3OCHOCH3 . 24.74 (a) Reaction Both groups are ionized. (c) 5.97. 25.50 (a) Mn 5 3.6 kg/mol; between glycerol and carboxylic acid (formation of an ester). Mw 5 4.3 kg/mol. (b) Mn 5 5 kg/mol; Mw 5 5 kg/mol. (c) If Mn (b) Fat or oil (shown in problem) 1 NaOH ¡ Glycerol 1 and Mw are close in value, that indicates a small spread in the 3RCOO2Na1 (soap). (c) Molecules having more C“C bonds are distribution of polymer sizes. (d) The four subunits in hemoglobin harder to pack tightly. Consequently, they have a lower melting molecule dissociate in solution, giving a distribution of molar point. (d) H2 gas with a homogeneous or heterogeneous catalyst. masses. There are no subunits in myoglobin or cytochrome c, so (e) 123. no distribution of molar masses. Chapter 25 25.8 ¬( CH2¬CHCl¬CH2¬CCl2 ¬. ) 25.10 By an addition reaction involving styrene monomers.  Photo Credits © McGraw-Hill Higher Education Inc./Ken Front Matter Chapter 2 Karp, Photographer; p. 155 (bottom): © p. iii (top): Courtesy Margaret A. Chang; Opener: © Mary Evans/Science Source; 2.4 McGraw-Hill Higher Education Inc./Ken Karp, p. iii (bottom): Courtesy Nancy Goldsby; p. v: (all): © The McGraw-Hill Companies, Inc./ Photographer; p. 156: Historical photo courtesy © Robert R. Johnson, Institute for Computational Charles D. 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Barrie, Jr.; tal Photographs; 20.23: © Owen Franken; 23.5: © Dirk Wiersma/Science Source; 23.6:  Index A Adipic acid, 1063 Aerosols, 547, 910, 921 ionization constant of, 685 ion product, 968 Absolute entropy, 782, 788 AIDS, 455 as Lewis base, 704 Absolute temperature scale, 15, 182 Air, composition of, 901 molecular geometry, 418, 433 Absolute zero, 182 Air pollution preparation of, 601, 652 Absorption spectrum, 564, 1011 carbon monoxide and, 924 solubility of, 530 Acceptor impurity, 941 radon and, 921 as solvent, 944 Accuracy, 22 smog and, 919 Ammonium chloride (NH4Cl), 698 Acetaldehyde (CH3CHO), 1044 sulfur dioxide and, 917 Ammonium ion, 56, 129 Acetic acid (CH3COOH), 121, 667, 734, 1043 Alcohol(s), 1042 Ammonium nitrate (NH4NO3), 105, 974 ionization constant of, 678 condensation reactions of, 1043 Amorphous solids, 492 titrations of, 734 denatured, 1043 Ampere (A), 845 Acetic acid-sodium acetate system, 721, 725 oxidation of, 1042 Amphoteric hydroxide, 703, 761 Acetone (CH3COCH3), 1044 Alcohol dehydrogenase, 606, 1042 Amphoteric oxides, 357, 702 Acetyl chloride (CH3COCl), 1045 Aldehydes, 1044 Amplitude of wave, 275 Acetylene (C2H2), 1037 Aliphatic alcohols, 1042 Anaerobic organism, 1027 bonding in, 379, 441 Aliphatic hydrocarbons. See Alkanes Analytical chemistry. See Chemical properties and reactions of, 1037 Alkali metal(s), 50, 349, 942 analysis Acetylsalicylic acid (aspirin), 678, 707 coinage metals compared with, 356 Angstrom (Å), 44 Achiral molecules, 1007 electronegativity, 942 Angular momentum quantum number (l), 296 Acid(s), 62, 126, 667 group trends of, 349 Anhydrous compounds, 64 Arrhenius, 127 ionization energy, 942 Aniline (aminobenzene), 1046 Brønsted, 127, 667 properties of, 349, 942 Anions, 51 diprotic, 128, 688 reactions of, with oxygen, 349, 943 containing metal atoms, 1004 general properties of, 127 Alkali metal hydroxides, 674, 945 electron configuration of, 333 ionization constants of. See Ionization Alkaline earth metal(s), 50, 350, 946 hydrolysis, 697 constants properties of, 350, 946 names of, 58, 1004 Lewis, 704 Alkaline earth metal hydroxides, 674 radius of, 337 monoprotic, 128, 677 amphoterism of, 703 Anode, 41, 816 polyprotic, 128, 688 Alkanes (aliphatic hydrocarbons), 65, 1027 sacrificial, 841 strength of, 673, 692 nomenclature of, 1028 Antacids, 706 strong and weak, defined, 673 optical isomerism of substituted, 1032 Antibonding molecular orbitals, 444 triprotic, 128, 692 reactions of, 1031 Antifreeze, 538 Acid ionization constants (Ka), 678 Alkenes (olefins), 1033 Antiknocking agent, 1049 of diprotic and polyprotic acids, 690 geometric isomers of, 1035 Antitumor agents, 1018 of monoprotic acids, 678 nomenclature of, 1034 Aqua regia, 970 relation between base ionization properties and reactions of, 1034 Aqueous solution, 119 constants and, 687 Alkyl group, 1029 Aqua ligand, 1004 Acid paper, 718 Alkyl halides, 1032 Argon, 355, 358 Acid rain, 702, 916 Alkynes, 1037 Aristotle, 39 Acid strength, 673, 692 Allotropes, 52, 253 Aromatic hydrocarbons, 1039 Acid-base indicators, 152, 739, 741 (table) carbon, 52, 253, 454, 963 nomenclature of, 1039 Acid-base properties, 127 oxygen, 52, 975 properties and reactions of, 1040 of hydroxides, 703 phosphorus, 970 Arrhenius, Svante, 127 of oxides, 702 sulfur, 979 Arrhenius acid-base theory, 127 of salt solutions, 696 tin, 492 Arrhenius equation, 589 of water, 668 Alloys, 932 Arsenic, 170, 1083 Acid-base reactions, 130, 151, 730 Alpha helix, 1069 Art forgery, 898 Acid-base theory Alpha (α) particles, 43 Artificial radioactivity, 874 Arrhenius, 127 Alpha (α) rays. See Alpha particles Artificial snow, 240 Brønsted, 127 Alum, 951 Ascorbic acid. See Vitamin C Lewis, 704 Aluminum, 351, 948 Aspirin (acetylsalicylic acid), 678, 707 Acid-base titrations, 151, 730 metallurgy of, 948 Astatine, 354 Acidic oxides, 357, 702 recovery of, 950 Aston, Francis, 84 Actinide series, 311 Aluminum chloride (AlCl3), 699, 949 Atactic polymers, 1060 Activated complex, 589 Aluminum hydride (AlH3), 950, 959 Atmospheric composition, 901 Activation energy (Ea), 589, 1015 Aluminum hydroxide [Al(OH)3], 703, 950 Atmospheric pollution. See Air pollution Active site, 605 Aluminum oxide (Al2O3), 351, 948 Atmospheric pressure, 175 Active transport, 706 Aluminum sulfate [Al2(SO4)3], 718 boiling point and, 504 Activity, 627, 671 Amalgams, 846, 932 freezing point and, 504 Activity series, 140 Amide group, 1068 standard, 176 Actual yield, 103 Amide ion, 675, 968 Atom, 40 Addition reactions, 604, 1035, 1059 Amines, 1046 Dalton’s theory of, 39 Adenine, 802, 1073 Amino acids, 1065, 1066 (table) emission spectrum of, 282 Adenosine diphosphate, 802 Aminobenzene (aniline), 1046 Greek theories of, 39 Adenosine triphosphate, 802 Ammonia (NH3), 968 Rutherford’s model of, 44 Adhesion, 473 as base, 129 structure of, 45 Adiabatic process, 240 in fertilizers, 105 Thomson’s model of, 43 I-1 I-2 Index Atomic bomb, 47, 879 Beta (β) rays. See Beta particles Bromine, 142, 354, 982, 988 Atomic mass, 76 Bidentate ligands, 1001 Bromine-formic acid reaction, 564 Atomic mass unit (amu), 76 Bimolecular reaction, 594 Brønsted, Johannes N., 127 Atomic nucleus, 44 Binary compounds, 56 Brønsted acid, 127, 667 Atomic number (Z), 46, 328 Binary hydrides, 958 Brønsted base, 127, 667 Atomic orbitals, 295, 297 Binding energy. See Nuclear binding energy Brønsted acid-base theory, 127, 667 electron assignment to, 306 Biological effects of radiation, 888 Buckminsterfullerine. See Buckyball energies of, 300 Biological nitrogen fixation, 901 Buckyball, 454 hybrid. See Hybrid orbitals Biosphere II, 228 Buffer solutions, 724 relationship between quantum numbers Blast furnace, 934 Buret, 12, 152 and, 295 Blood Butadiene, 1063 Atomic radii, 335 oxygen in, 531, 651, 732 Atomic theory. See Atom Atomic weight. See Atomic mass pH of, 732 Body-centered cubic cell (bcc), 479 C Aufbau principle, 308 Bohr, Niels, 282 Calcite. See Calcium carbonate Aurora borealis, 904 Bohr model, 282 Calcium, 350, 947 Autoionization of water, 668 Boiler scale, 126 Calcium carbide (CaC2), 964, 1037 Automotive emissions, 602, 919 Boiling point, 498 Calcium carbonate (CaCO3), 760, 918, 947 Average atomic mass, 76 and intermolecular forces, 498 decomposition of, 630, 793 Average bond enthalpies, 399 pressure and, 498, 504 production of iron, 934 Avogadro, Amedeo, 78, 183 vapor pressure and, 498 sulfur dioxide removal with, 918 Avogadro’s law, 183 Boiling-point elevation, 536 Calcium hydroxide [Ca(OH)2; slaked Avogadro’s number, 78 Boltzmann, Ludwig, 202 lime], 947 Axial position, 416 Boltzmann constant, 780 Calcium oxide (CaO; quicklime), 370, Boltzmann equation, 780 918, 947 B Bomb calorimeter, 248 Bombardier beetle, 256 Calcium phosphate, 760, 973 Calorie, 250 Bacteria fuel cell, 837 Bond(s) Calorimeter Balancing equations, 92, 813 coordinate covalent, 393 constant-pressure, 249 equilibrium constant and, 635 of coordination compounds, 1009 constant-volume bomb, 247 nuclear reactions, 863 covalent. See Covalent bonds Calorimetry, 246 redox reactions, 813 dative, 393 Cancer, 891, 1018 Ball-and-stick model, 52 double. See Double bonds See also Carcinogenicity Balmer series, 285 electronegativity and, 380 Capillary action, 473 Band theory, 939 enthalpy, 398 Carbides, 964, 1037 Barium, 350, 947 hydrogen. See Hydrogen bond Carbon, 352, 963 Barium hydroxide [Ba(OH)2], 129, 674 ionic, 370, 372 allotropes of, 52, 253, 454, 963. Barium sulfate (BaSO4), 742 length, 379 See also Diamond; Graphite Barometer, 176 in metals, 491, 939 atomic mass of, 76 Bartlett, Neil, 355 multiple, 378 in inorganic compounds, 56 Base(s), 64, 127, 667 pi, 440 phase diagram of, 963 Arrhenius, 127 polar covalent, 380 in steelmaking, 936 Brønsted, 127, 667 sigma, 440 Carbon cycle, 912 general properties of, 127 single. See Single bonds Carbon dioxide (CO2), 965 ionization constant of, 685 in solids, 486 acidic properties, 703 Lewis, 704 triple. See Triple bonds bond moments of, 424 strength of, 674 Bond angles, 414, 418 climate and, 912 Base ionization constants (Kb), 685 Bond enthalpy, 398, 399 (table) enthalpy of formation of, 254 relationship between acid ionization Bond length, 379 indoor pollutant, 924 constants and, 687 Bond moments, dipole, 423 phase diagram of, 504 Base pairs, 1074 Bond order, 447 photosynthesis and, 599, 913 Base strength, 674 Bond polarity, 380 solid (dry ice), 504 Basic oxides, 357, 702 Bond strength, acid strength and, 693 solubility of, 531 Basic oxygen process, 935 Bonding molecular orbitals, 444 toxicity of, 531 Batteries, 832 Bonding pairs, 413, 417 Carbon disulfide (CS2), 981 dry cell, 832 Boric acid, 705 Carbon-12, 76 fuel cell, 835 Born, Max, 373 Carbon-14, 586, 873 lead storage, 833 Born-Haber cycle, 372 Carbon-14 dating, 586, 873 mercury, 832 Boron, 351 Carbon monoxide (CO), 965 lithium ion, 834 Boron neutron capture therapy, 891 enthalpy of formation, 255 Bauxite, 948 Boron trifluoride (BF3), 392, 434, 704 from automotive emissions, 603, 919 Becquerel, Antoine, 43 Bose, Satyendra, 206 hemoglobin affinity for, 924 Belt of stability, 866 Bose-Einstein condensate, 206 indoor pollutant, 924 Benzene (C6H6), 1039 Boson(s), 6 metal purification with, 937 bonding in, 390, 452 Higgs. See Higgs boson toxicity of, 924 electron micrograph of, 1039 Boundary surface diagrams, 298 Carbon tetrachloride (CCl4), 379, 1032 structure of, 390, 452, 1039 Boyle, Robert, 178 Carbonate ion, 386, 390, 453 Benzoic acid, 696, 1045 Boyle’s law, 178 Carbonic acid (H2CO3), 634, 688, 705 Beryl, 931 Bragg, Sir William L., 485 formation, 705, 732 Beryllium, 350, 946 Bragg, William H., 485 ionization constants, 690 Beryllium chloride (BeCl2), 415, 433 Bragg equation, 484 Carbonic anhydrase, 732, 760 Beryllium hydride (BeH2), 392, 959 Brass, 519 Carbonyl group, 1044 Beta (β) particles, 43 Breathalyzer, 144 Carborundum, 964 Beta pleated sheet, 1069 Breeder reactors, 881 Carboxyhemoglobin, 924  Index I-3 Carboxyl group, 1044 Chemical reactions, 90 Color Carboxylic acids, 1044 acid-base, 130, 151, 730 of glass, 493 acid strength, 695 addition, 604, 1035, 1059 of indicators, 741 Carcinogenicity of alkanes, 1031 of transition metal ions, 1010 of amines, 1046 of alkenes, 1034 wavelength and, 278, 1010 of ethylene dibromide, 988 of alkynes, 1037 Color wheel, 1010 of polycyclic aromatic of aromatic compounds, 1040 Combination reaction, 137 hydrocarbons, 1041 bimolecular, 594 Combustion, 139 of radiation, 889 combination, 137 of acetylene, 257, 1037 Carothers, Wallace, 1063 combustion, 139 of alkanes, 1031 Cast (pig) iron, 935 condensation, 1043, 1063, 1065 of hydrogen, 11, 232 Catalysis, 599 of coordination compounds, 1015 of methane, 243, 1031 air pollution reduction by, 602 Dalton’s definition of, 39 of sulfur, 137, 235 enzyme, 604 decomposition, 139 Common ion effect heterogeneous, 601 displacement, 139 acid-base equilibria and, 721 homogeneous, 603 disproportionation, 142 solubility and, 751 Catalysts, 599 first-order, 575 Complex ion(s), 756, 1000 in catalytic converters, 602 gases in, 193 magnetic properties of, 1012 effects of, on equilibrium, 650 half, 133 solubility equilibria and, 756 enzymes as, 604 half-cell, 816 Complex ion formation, 756 heterogeneous, 601 metathesis, 121 Compounds, 8 homogeneous, 603 neutralization, 130, 151, 730 anhydrous, 64 Natta-Ziegler, 1061 nuclear reactions compared with, 863 aromatic. See Aromatic hydrocarbons Catalytic converters, 602 oxidation-reduction. See Oxidation- coordination. See Coordination Catalytic rate constant (kc), 600 reduction reactions compounds Catenation, 963, 1026 precipitation, 121, 748 in Dalton’s theory, 39 Cathode, 41, 816 rate of. See Rate of reaction inorganic, 56 Cathode ray(s), 41 second-order, 582 ionic, 51, 54, 372 Cathode ray tube, 41 spontaneous, 777, 783, 789 molecular, 59 Cathodic protection, 840 substitution, 1040 nonstoichiometric, 960 Cations, 51 termolecular, 594 organic, 56, 65, 1026 electron configuration of, 332 thermite, 949 Concentration, 145, 522 hydrolysis of, 698 unimolecular, 594 chemical equilibria and identification of, 761 zero-order, 585, 607 changes in, 645 ionic radius of, 337 Chemistry, 2 effects on emf, 827 nomenclature of, 58 Chernobyl, 883 Concentration cells, 831 Caustic soda. See Sodium hydroxide Chile saltpeter (NaNO3), 945 Concentration of solution, 145, 522 Cell diagram, 817 Chiral molecules, 1007, 1032 Concentration units, 145, 522 Cell potential, 817 Chlor-alkali process, 984 compared, 524 Cell voltage, 817 Chlorine, 354, 982 molality, 523 See also Electromotive force preparation of, 984 molarity, 145, 523 Cellulose, 718 uses of, 987 mole fraction, 197, 523 Celsius temperature scale, 15 Chlorine monoxide (ClO), 907 percent by mass, 522 Cementite, 936 Chlorofluorohydrocarbons (CFCs), 907 Condensation, 495 CERN, 6 Chloroform (CHCl3), 1032 Condensation reactions, 1043, Cesium, 349 Chlorophyll, 1017 1063, 1065 Chadwick, James, 45 Chlorous acid (HClO2), 63, 987 Conduction band, 939 Chain reaction, nuclear, 878 Cholesterol, 1047 Conductivity Chalcopyrite (CuFeS2), 999 Chromium, 310, 996 of metals, 491, 939 Chalk, 947 Chromosomes, 889 of nonmetallic elements, 940 Chargaff, E., 1073 Cinnamic aldehyde, 1044 Conductor, 940 Chargaff’s rule, 1073 Cisplatin, 1018 Conjugate acid, 667 Charge cloud (electron charge cloud), 295 Cis-trans isomers Conjugate acid-base pair, 667, 687 Charge-to-mass ratio (e/m), 41 of alkenes, 1034 Conjugate base, 667 Charles, Jacques, 181 of coordination compounds, 1006 Constant-pressure calorimeter, 249 Charles’ law (Charles’ and Gay-Lussac’s law), Clapeyron, Benoit, 496 Constant-volume bomb calorimeter, 247 182 Clausius, Rudolf, 496 Constructive interference, 444, 485 Chelating agents, 1002 Clausius-Clapeyron equation, 496 Contact process, 981 Chemical analysis Climate Control rods, 880 See also Qualitative analysis; carbon dioxide and, 912 Cooling curve, 501 Quantitative analysis effects of water on, 475 Cooperativity, 1070 with coordination compounds, 1018 Closed system, 232 Coordinate covalent bonds, 393, 704 Chemical energy, 231 Closest packing, 480 Coordination compounds, 1000 Chemical equations, 90 Cloud seeding, 988 applications of, 1016 balanced. See Balancing equations Coal, 963 bonding in, 1009 free elements in, 332 Coal gasification, 966 in living systems, 1016 interpretation of, 91 Cohesion, 473 magnetic properties, 1012 Chemical equilibrium, 121, 622 Coinage metals, 356 naming, 1003 Chemical formulas, 52 Coke, 934 oxidation number, 1002 empirical, 53, 88 Colligative properties reactions of, 1015 molecular, 52 of electrolyte solutions, 544 stereochemistry of, 1006 structural, 53 of nonelectrolyte solutions, 532 Coordination number, 479, 1001 Chemical kinetics, 563 Collision theory, 588 Coordination theory of Werner, 1000 Chemical properties, 11 Colloids, 546 Copolymer, 1063 I-4 Index Copper, 999 Davisson, Clinton, 291 fingerprinting by, 1076 corrosion of, 840 de Broglie, Louis, 287 structure of, 1075 electron configuration of, 310 de Broglie’s hypothesis, 287 Dolomite, 946 ionization energy of, 356 Debye (D), 424 Donor atom, 1001 metallurgy of, 999 Debye, Peter J., 424 Donor impurity, 940 purification of, 937 Decay series. See Radioactive Doping, 940 Copper carbonate (CuCO3; patina), 840 decay series Double bonds, 379, 440 Copper sulfate (CuSO4), 64 Decomposition reactions, 139 Doubling time, 881 Core Definite proportions, law of, 40 Downs cell, 841 atomic. See Nucleus Delocalized molecular orbitals, 452 Dry cell batteries, 832 noble gas, 308 of benzene, 452 Dry ice, 65, 504 nuclear reactor, 880 of carbonate ion, 453 Dynamic equilibrium, 495 Core electrons, 330 of metals, 491, 939 Corona, 324 Corrosion, 838 Democritus, 39 Denaturant, 1073 E Corundum (Al2O3), 948 Denatured alcohol, 1043 Earth Coulomb (C), 824, 845 Denatured proteins, 774, 1073 age of, 874 Coulomb, Charles, 372 Denitrification, 902 composition of, 49 (table) Coulomb’s law, 372, 865 Density, 11 EDTA (ethylenediaminetetraacetate), 1001 Coupled reactions, 800 gas, 190 treatment of metal poisoning with, 1002 Covalent bonds, 377 of nucleus, 865 Effective nuclear charge, 334 coordinate, 393 water, 476 Efficiency, 791 polar, 380 Dental amalgam, 846 Effusion, gaseous, 209 Covalent compounds, 377 Deoxyhemoglobin, 1017, 1070 Egg Covalent crystals, 490 Deoxyribonucleic acid (DNA). See DNA formation, 760 Covalent hydrides, 959 Deposition, 502 hard boiling, 505, 774 Cracking process, 1035 Derived SI units, 14 Einstein, Albert, 40, 208, 279, 868 Crenation, 541 Destructive interference, 444 Einstein’s mass-energy equation, 868 Crick, Francis, 1074 Detergents, 1020 Einstein’s relativity theory, 868, 875 Critical mass, 878 Deuterium, 46, 960 Elastomers (synthetic rubber), 1062 Critical pressure (Pc), 499 Deuterium oxide (D2O; heavy water), 880, 960 Electrical work, 824 Critical temperature (Tc), 499, 500 (table) Deviation from ideal gas behavior, 210 Electrocatalysts, 836 Crown ether, 930, 944 Dextrorotatory isomers, 1007 Electrochemical series, 140 Crude oil, 1048 Diagonal relationship, 348, 704, 959 Electrochemistry, 813 Cryolite (Na3AlF6), 948 Diagonal rule, 822 Electrode(s), 816 Crystal(s), 486, 491 (table) Diamagnetism, 304 anode, 816 covalent, 490 Diamond cathode, 816 ionic, 486 as allotrope of carbon, 52, 253, 963 Electrode potential. See Standard metallic, 491 entropy of, 782 reduction potential molecular, 490 structure of, 490 Electrolysis, 841 X-ray diffraction by, 484 synthetic, 963 of aqueous sodium chloride, 843 Crystal field splitting, 1010 Diaphragm cell, 984 metal purification by, 937 Crystal field theory, 1009 Dialysis, 546 of molten sodium chloride, 841 Crystal structure, 476 Diatomic molecules, 50 quantitative aspects of, 845 Crystalline solids, 476 heteronuclear, 914 of water, 842 Crystallization, 519 homonuclear, 448, 914 Electrolyte(s), 119 fractional, 527 Dichloroethylene, 425, 1036 strong, 120 Cubic close-packed (ccp) structure, 481 Dichromate ion, 155, 813 weak, 120 Cubic unit cell, 478 Diethyl ether, 1043 Electrolyte solutions, colligative properties Curie (Ci), 888 Diffusion, gaseous, 207 of, 544 Curie, Marie, 43 Dilution of solutions, 147 Electrolytic cell, 841 Curie, Pierre, 43 Dimensional analysis, 23 Electromagnetic radiation, 277 Cyanide, 964 Dimethylglyoxine, 1018 Electromagnetic wave, 276 Cycloalkanes, 1033 Dinitrogen pentoxide (N2O5), 577 Electromotive force (emf), 817 Cyclohexane, 1033 Dinitrogen tetroxide (N2O4), 622, 648 effects of concentration on, 827 Cyclotron, 875 Dinosaurs, 36 standard, 819 Cytochrome c, 1017 Dipeptide, 1065 Electron(s), 41 Cytochrome oxidase, 964 Dipolar ion, 1065 charge-to-mass ratio of, 41 Cytosine, 1073 Dipole moments (μ), 423, 425 (table) nonbonding. See Lone pairs Dipole-dipole forces, 467 probability distribution of, 298 D Dipole-induced dipole, 468 Dipositive ions, 338 valence, 330 Electron affinity, 345, 346 (table) d Orbitals, 299, 1009 Diprotic acids, 128, 688 Electron capture, 867 and crystal field theory, 1009 ionization constant of, 690 Electron charge, 41 hybridization of, 439 Dispersion forces, 469 Electron charge cloud, 295 Dacron, 1063 Displacement reactions, 139 Electron configuration, 302 Dalton (atomic mass unit), 76 Disproportionation reactions, 142 anions, 333 Dalton, John, 39 Distillation Aufbau principle and, 308 Dalton’s atomic theory, 39 fractional, 535, 1048 cations, 332 Dalton’s law of partial pressures, 196 metal purification by, 937 diamagnetism and paramagnetism in, 303 Daniel cell, 816 Distribution, 779 electron assignment to orbitals in, 302 Data, 4 DNA (deoxyribonucleic acid), 1073 ground state, 302, 309 Dating, radionuclear, 586, 873 cisplatin binding to, 1018 Hund’s rule and, 305 Dative bonds, 393 electron micrograph, 1074 and molecular orbitals, 446  Index I-5 Pauli exclusion principle and, 303 Enzyme(s), 604 Excited level (excited state), 284 and shielding effect, 304 alcohol dehydrogenase, 606, 1042 Exothermic processes, 233 Electron density, 295 carbonic anhydrase, 732, 760 Expanded octet, 393 Electron microscope, 292 catalysis of, 604 Expanded valence shell, 393, 439 Electron probability, 294, 298 cytochrome oxidase, 964 Explosives, 879, 946, 974 Electron spin, 296, 302 hexokinase, 605 Exponential notation. See Scientific notation in coordination compounds, 1012 HIV-protease, 455 Extensive properties, 11 Hund’s rule and, 305 lock-and-key model of, 605 Pauli exclusion principle and, 303 Electron spin quantum number (ms), 296 Enzyme-substrate intermediate (ES), 606 Equation F Electron subshell, 296 Arrhenius, 589 f Orbitals, 299, 310 Electron-dot symbols, 369 Boltzmann, 780 Face-centered cubic unit cell (fcc), 479, 481 Electronegativity, 380 chemical, 90 Factor label method, 23 Elementary particles, 863 Clausius-Clapeyron, 496 Fahrenheit temperature scale. See Elementary steps, 594 Einstein, 868 Temperature scale Elements, 7 Henderson-Hasselbach, 722 Family of elements, 48 abundance, 49 ideal gas, 184 Faraday, Michael, 824, 1039 atomic radii of, 335 ionic, 124 Faraday constant (F), 824 classification of, 48, 329 molecular, 123 Fat cells, 250 derivation of names and symbols, A-1 Nernst, 828 Fermentation, 913, 1042 electron affinity of, 345 net ionic, 124 Fermion(s), 6 electronegativity of, 380 nuclear, 863 Ferromagnetic substances, 932 essential, 49 redox, 813 Fertilizers, 105 ground state electron configurations of, Schrödinger, 294 Fingerprints, 1056 309 (table), 329 thermochemical, 243 First law of thermodynamics, 234 ionization energies of, 342 (table) van der Waals, 212 First-order reactions, 575 periodic and group properties of, 347 Equatorial position, 416 Fischer, Emil, 605 representative, 330 Equilibrium, 121, 622 Fission reactions, 877 symbols of, 8 (table) catalyst effect on, 650 Fission reactors, 879 transuranium. See Transuranium elements and chemical kinetics, 637 Flame test, 762 Emf. See Electromotive force and concentration changes, 645 Flotation method, 932 Emission spectra, 282 dynamic, 495 Fluorapatite, 105 Empirical formula, 53, 88 free energy and, 796 Fluoridation, 987 Emulsion, 547 heterogeneous, 630 Fluorine, 354, 982 Enantiomers, 1007 homogeneous, 625 fluoridation with, 987 End point, 739 liquid-solid, 499 mass defect of, 868 Endothermic process, 233 liquid-vapor, 494 oxidation number of, 136, 383 Energy, 231 multiple, 633 preparation of, 983 chemical, 231 solid-vapor, 502 uses, 987 crystal field splitting, 1010 and temperature changes, 648 Fluorite (CaF2), 487, 947 of hydrogen atom, 284 volume and pressure changes and, 646 Flux, 934 ionization, 340 Equilibrium constant (K), 624, 797 Food irradiation, 890 kinetic. See Kinetic energy balanced equation and, 635 Force, 175 lattice. See Lattice energy and equilibrium concentration adhesive, 473 law of conservation of, 231 calculations, 641 dispersion, 469 mass-energy conversion, 868 in heterogeneous equilibrium, 631 intermolecular. See Intermolecular forces molecular orbital energy level in homogeneous equilibrium, 625 intramolecular, 467 diagram, 445 and law of mass action, 624 unit of, 175 nuclear binding. See Nuclear binding energy in multiple equilibria, 633 van der Waals, 467 potential. See Potential energy units, 627 Formal charge, 387 solar. See Solar radiation Equilibrium vapor pressure, 495 Formaldehyde (CH2O), 442, 924, 1044 thermal. See Heat Equivalence point Formation constant (Kf), 756, 757 (table) unit of, 202 in acid-base titrations, 152, 730 Formic acid (HCOOH), 564, 678, 1045 Energy changes in redox titrations, 155 Formula mass, 83 in chemical reactions, 242 Erythrocytes. See Red blood cells Formulas. See Chemical formulas and first law of thermodynamics, 234 Escape velocity, 207 Fossil fuels, 962, 966, 1048 Englert, François, 6 Essential elements, 49 Fractional crystallization, 527 Enthalpy (H), 241 Esters, 1045 Fractional distillation, 535, 1048 and Born-Haber cycle, 372 Ethane (C2H6), 1027 Fractional precipitation, 749 standard, 253 Ethanol (C2H5OH), 88, 1042 Fractionating column, 535, 1048 Enthalpy of reaction, 242 Ethers, 1043 Francium, 341 Enthalpy of solution, 259 Ethyl acetate (CH3COOC2H5), 603, 1046 Frasch, Herman, 978 Entropy (S), 778 Ethyl group (C2H5), 1029 Frasch process, 978 absolute, 782, 788 Ethylene (C2H4), 1033 Fraunhofer, Josef, 324 changes, 780 bonding in, 379, 440 Free energy (G), 789 and microstate, 779 in polymerization, 1060 chemical equilibria and, 796 phase transition, 780, 795 Ethylene dibromide, 988 and electrical work, 824 standard, 782 Ethylene glycol [CH2(OH)CH2(OH)], in phase transition, 795 Environmental pollution 538, 1043 spontaneity and, 790 acid rain, 702, 916 Ethylenediamine, 1001 standard free energy of reaction, 790 Freon, 907 Ethylenediaminetetraacetate. See EDTA temperature and, 793 nuclear wastes, 883 Eutrophication, 1020 Free radicals, 889, 1032 sulfur dioxide, 917 Evaporation. See Vaporization Freezing point, 499 thermal, 528, 880 Excess reagent, 99 Freezing-point depression, 537 I-6 Index Freons, 907, 986 Group (periodic), 48 intermolecular forces in, 469 Frequency (v), 275 Guanine, 1073 ionization energy of, 342 Frequency factor (A), 590 Guldberg, Cato, 624 Hematite (Fe2O3), 998 Fuel, fossil. See Fossil fuels Gunpowder, 946 Heme group, 1016, 1070 Fuel cell, 835 Gypsum (CaSO4 · 2H2O), 947, 978 Hemodialysis, 546 Fuel value, 250 Hemoglobin (Hb) Functional groups, 65, 1026, 1047 (table) Fusion H binding of oxygen, 531, 651, 1016, 1070 as buffer, 732 entropy and, 795 H2. See also Hydrogen; Hydrogen atom carbon monoxide affinity for, 924 molar heat of, 501 (table) Lewis structure of, 377 production of, 651 nuclear, 883 molecular orbitals of, 445 structure of, 1016, 1071 potential energy of, 429 Hemolysis, 540 G Haber, Fritz, 373, 601 Haber process, 601, 652 Henderson-Hasselbach equation, 722 Henry, William, 529 Gallium, 327 Hair, 1082 Henry’s law, 529, 531 Galvanic cells, 816 Half-cell potential. See Standard reduction Hertz (Hz), 276 Galvanized iron, 840 potential Hess, Germain H., 255 Gamma (γ) rays, 43 Half-cell reactions, 816 Hess’s law, 255, 259, 374 Gamow, George, 6 Half-life, 341, 580 Heterogeneous catalysis, 601 Gangue, 932 of carbon-14, 586 Heterogeneous equilibria, 630 Gas(es), 9, 173 of cobalt-60, 581, 885 Heterogeneous mixture, 7 Avogadro’s law, 183 of first-order reactions, 580 Heteronuclear diatomic molecules, 914 Boyle’s law, 178 of francium-223, 341 Hexagonal close-packed (hcp) Charles’ law, 182 of iodine-131, 887 structure, 481 in chemical reactions, 193 of plutonium-239, 881 Hexamethylenediamine, 1063 Dalton’s law of partial pressure of, 196 of potassium-40, 874 Hexokinase, 605 density of, 190 of radon-222, 922 High-spin complexes, 1012 diffusion of. See Diffusion of isotopes used in medicine, 887 (table) High-temperature superconductor, 407 effusion of. See Effusion of second-order reactions, 584 Higgs boson, 6 emission spectrum of, 282 of sodium-24, 581, 886 Higgs, Peter, 6 kinetic molecular theory of, 202 of technetium-99, 887 Hindenburg, 233 monatomic, 173 of tritium, 875, 885, 960 Hiroshima, 879 noble. See Noble gases of uranium-238, 873 HIV, 455 pressure of, 174 of zero-order reactions, 585 Homogeneous catalysis, 603 solubility of, 528, 529, 531 Half-reaction, 133 Homogeneous equilibria, 625 Gas constant (R), 184 Halic acids, 693, 986 Homogeneous mixture, 7 units of, 185, A-7 Halides, 355, 985 Homonuclear diatomic molecules, van der Waals, 212 (table) alkali metal, lattice energy and, 375 448, 914 Gasoline, 1049 alkyl, 1032 Homopolymers, 1060 antiknocking agents in, 1049 hydrogen. See Hydrogen halides Human immunodeficiency virus. See HIV Gastric juice, 706 phosphorus, 972 Hund, Fredrick, 305 Gauge pressure, 272 solubility of, 122 Hund’s rule, 305, 447, 450, 1012 Gay-Lussac, Joseph, 181 Hall, Charles, 948 Hybrid orbitals, 431, 436 (table) Geiger, Hans, 44 Hall process, 948 of molecules with double and triple Geiger counter, 888 Halogen(s), 50, 354, 982 bonds, 440 Genetic effects of radiation, 889 displacement of, 141 sp, 433, 441 2 Geobacter, 837 electronegativity, 982 sp , 434, 440 Geometric isomer(s), 1006, 1035 industrial and biological roles of, 987 sp3, 431 Geometric shapes of orbitals, 297, 436 ionization energy, 982 sp3d, 439 Gerlach, Walther, 297 oxoacids, 62, 986 sp3d2, 439 Germer, Lester, 291 preparation of, 983 Hybridization, 432 Gibbs, Josiah, 789 properties of, 983 Hydrate, 64, 1038 Gibbs free energy. See Free energy Halogenation of alkanes, 1031 Hydration, 120, 259 Glass, 492, 493 (table) Hard water, 126 heat of, 259 Glass electrode, 830 Heat, 232, 239 of ions, 120, 259, 781 Glucose (C6H12O6), 718, 887, 1040 of dilution, 260 of protons, 128, 667 Glutamic acid, 1066, 1072 of fusion, 500 Hydrazine (N2H4), 968 Glycerol, 474 of hydration, 259 Hydrides Glycine, 1066, 1068 of solution, 259 binary, 958 Gold of vaporization, 495 covalent, 959 extraction of, 965 Heat capacity (C), 246 interstitial, 959 ionization energy of, 356 Heat content. See Enthalpy ionic, 959 oxidation of, 970 Heat engine, 790 phosphorus, 971 Goodyear, Charles, 1062 Heating curve, 501 Hydrocarbons, 65, 1026 Gram (g), 13 Heavy water. See Deuterium oxide aliphatic. See Alkanes Graham, Thomas, 209 Heavy water reactor, 880 alkynes. See Alkynes Graham’s law of diffusion, 209 Heisenberg, Werner, 293 aromatic. See Aromatic hydrocarbons Graphene, 455, 834 Heisenberg uncertainty cycloalkanes, 1033 Graphite, 52, 253, 490, 963 principle, 293 saturated, 1027 as covalent crystal, 490 Helium, 355 unsaturated, 1033, 1037, 1039 entropy of, 782 boiling point of, 469 Hydrochloric acid (HCl), 128, 693 Gravimetric analysis, 149 discovery of, 324 in acid-base titrations, 730, 737 Greenhouse effect, 912 escape velocity of, 207 as monoprotic acid, 128 Ground state (ground level), 284 formation of, 871 preparation of, 986  Index I-7 Hydrocyanic acid (HCN), 678, 964 Hydrofluoric acid (HF) I Ionosphere, 904 Iridium, 37 ionization constant of, 678 Ice, 475 Iron, 998 as weak acid, 679 ICE method, 641 corrosion of, 839 Hydrogen, 348, 958 Ice skating, 505 ferromagnetic properties of, 932 atomic orbitals of, 297 Ice-water equilibrium, 500 galvanized, 840 combustion of, 11, 232 Ideal gas, 185 metallurgy of, 934 displacement of, 139 Ideal gas equation, 184 Iron sulfide (FeS), 775 isotopes of, 46, 960 Ideal solution, 534 Isoelectronic ions, 333 metallic, 962 Impurities Isolated system, 232 oxidation number of, 136 acceptor, 941 Isomer(s) preparation of, 958 donor, 940 geometric, 1006, 1035 properties of, 348, 958 Incomplete octet, 392 optical, 1007, 1032 Hydrogen atom Indicators. See Acid-base indicators of polymers, 1060 Bohr’s theory of, 282 Induced dipole, 468 structural, 1027 emission spectrum of, 282 Inert complexes, 1015 Isomerism. See Isomer(s) energy of, 284 Infrared active, 914 Isoprene, 1061 Schrödinger equation and, 294 Initial rate, 571 Isopropanol, 1043 Hydrogen bomb, 885 Inorganic compounds, 56 Isotactic polymers, 1060 Hydrogen bond, 471, 1069, 1075 Instantaneous rate, 565 Isotonic solution, 540 Hydrogen bromide (HBr), 986 Insulators, 940 Isotopes, 46, 886, 960 Hydrogen chloride (HCl), 986 Intensive properties, 11 applications of, 599, 886, 1015 Hydrogen cyanide (HCN), 174, 964 Interference of waves, 444, 485 IUPAC, 48, 330, 1028 Hydrogen economy, 962 Intermediates, 594 Hydrogen fluoride (HF), 380, 423, 986 Hydrogen halides, 985 Intermolecular forces, 174, 467 J dipole-dipole forces, 467 acid strength of, 693 dispersion forces, 468 Jeffreys, Alec, 1076 dipole moments of, 425 ion-dipole forces, 468 Joule (J), 202 Hydrogen iodide (HI), 986 ion-induced dipole, 468 Joule, James Prescott, 202 kinetics of formation, 596 van der Waals forces, 467 Jupiter, 207, 962 Hydrogen ion Internal energy, 235 hydrated, 128, 667 pH and concentration, 671 International System of Units (SI units), 12 K International Union of Pure and Applied Hydrogen molecule (H2) Chemistry. See IUPAC Kekule, August, 390, 1039 combustion, 11, 232 Interstitial hydrides, 960 Kelvin, Lord (William Thomson), 182 Lewis structure, 377 Intramolecular forces, 467 Kelvin temperature scale, 15, 182 molecular orbital, 445 Iodine, 354, 982 Keratin, 1082 Hydrogen peroxide (H2O2), 976 nuclear stability of, 869 Ketones, 1044 decomposition of, 143, 568, 596 preparation of, 142, 985 Kilogram (kg), 13 disproportionation, 143 sublimation of, 502 Kinetic energy, 202, 231 as oxidizing agent, 976 uses of, 988 Kinetic isotope effect, 961 percent composition by mass of, 85 Iodine-131, 887 Kinetic molecular theory as reducing agent, 976 Ion(s), 50 of gases, 202 Hydrogen sulfide (H2S), 979 dipositive, 338 liquids and solids in, 466 as diprotic acid, 690 electron configuration of, 333 Kinetics. See Chemical kinetics preparation of, 979 hydrated, 120, 259 Krypton, 355 in qualitative analysis, 762 hydrolysis, 696 Hydrogenation, 961 Hydrogen-oxygen fuel cell, 835, 962 monatomic, 51 L polyatomic, 51 Hydrohalic acids, 693, 986 separation of, by fractional precipitation, 749 Labile complexes, 1015 Hydrolysis spectator, 124 Lachrymator, 920 alkaline (saponification; base transition metal, 57, 996, 1011 Lake Nyos, 531 hydrolysis), 1046 tripositive, 338 Lanthanide series. See Rare earth elements of anions, 697, 734 unipositive, 338 Large Hadron Collider (LHC), 6, 862, 875 of esters, 599, 1046 Ion pairs, 544 Laser, 290, 885 metal ion, 699 Ion product constant, 669 Lattice energy (U), 259, 372, 375 (table) salt, 696 Ion-dipole forces, 468 of alkali metal halides, 372 Hydrometer, 834 Ion-electron method, 813 and Born-Haber cycle, 372 Hydronium ion (H3O+), 128, 667 Ion-induced dipole, 468 and chemical formulas, 372 Hydrophilic interaction, 547 Ionic bond, 370, 372, 382 Lattice point, 477 Hydrophobic interaction, Ionic compounds, 51, 54 Laue, Max von, 484 547, 1072 nomenclature, 56 Laughing gas (nitrous oxide), 65, 969 Hydroxides Ionic crystals, 486 Law, 4 alkali metal, 674, 703, 945 Ionic equation, 124 Law(s) alkaline earth metal, 674, 703 Ionic hydrides, 959 Avogadro’s, 183 amphoteric, 703 Ionic radii, 337 Boyle’s, 178 Hydroxyapatite, 742, 948 Ionic solids (ionic crystals), 486 Charles’, 181 Hydroxyl groups (OH groups), 1042 Ionization constants of conservation of energy, 231 Hydroxyl radical, 889, of bases, 686 of conservation of mass, 40 910, 917 of diprotic and polyprotic acids, 690 Coulomb’s, 372, 824 Hypertonic solution, 540 of monoprotic acid, 678 Dalton’s, of partial pressures, 195 Hypochlorous acid, 63, 987 Ionization energy, 340, 342 (table) of definite proportions, 40 Hypothesis, 4 Ionizing radiation, 889 first law of thermodynamics, 235 Hypotonic solution, 540 Ion-product constant of water (Kw), 669 Graham’s, of diffusion, 209 I-8 Index Law(s)—(cont.) London forces. See Dispersion forces Melting point, 499 Henry’s, 529, 530 London, Fritz, 469 of alkali metal halides, 372 Hess’s, 255, 256, 257, 372 Lone pairs, 378 of alkali metals, 341 of mass action, 624 Low-spin complexes, 1012 of diamond, 490 of multiple proportions, 40 Lucite (Plexiglas; polymethyl of francium, 341 of octaves, 327 methacrylate), 1059 pressure and, 504 Raoult’s, 532 of quartz, 490 rate, 571 second law of thermodynamics, 783 M Membrane potential, 831 Mendeleev, Dmitri, 327 third law of thermodynamics, 787 Macromolecules. See Polymers Mercury Le Chatelier, Henry L., 644 Macroscopic properties, 12 in amalgam, 846, 932 Le Chatelier’s principle, 644 Magic number, 865 in barometers, 176 acid ionization and, 684, 740 Magnesium, 156, 351, 947 mineral extraction with, 932 chemical equilibrium and, 644 band theory of, 939 Mercury batteries, 832 common ion effect and, 722, 751 cathodic protection with, 840 Mercury oxide (HgO), 139, 233, 778 and eggshell formation, 760 combustion, 133 Mesosphere, 904 solubility equilibria and, 751 preparation, 156 Metabolism, 801 Lead, 352 Magnesium hydroxide [Mg(OH)2], 156, Metal(s), 48, 491, 930 tetraethyl, 1050 707, 947 alkali. See Alkali metal(s) tetramethyl, 1050 Magnesium nitride (Mg3N2), 947 alkaline earth. See Alkaline earth metal(s) treatment of, 1002 Magnesium oxide (MgO), 133, 947 bonding in, 491, 939 Lead-206, 873, 898 Magnetic confinement, 884 coinage, 356 Lead chamber process, 604 Magnetic field displacement reactions, 140 Lead storage batteries, 833 of electromagnetic waves, 277 corrosion. See Corrosion Leclanche cell, 832 electron spin and, 296, 303 in ionic compounds, 58 Length, SI base unit of, 13 Magnetic quantum number (ml), 296 occurrence of, 932 Levorotatory isomers, 1007 Magnetism, 303 preparation of, 933 Lewis acid, 704 of complex ions, 1012 properties of, 48, 941 Lewis base, 704 diamagnetism, 304, 1012 purification of, 937 Lewis acid-base theory, 704 ferromagnetism, 932 Metal hydrides, 959 Lewis dot symbols, 369 paramagnetism, 304, 444, 1012 Metal ion Lewis, Gilbert N., 369 of transition metals, 1012 electron configurations, 332 Lewis structures, 378 Magnetite (Fe3O4), 999 hydrolysis of, 699 formal charge and, 387 Main group elements, 330 radii, 338 octet rule and, 378 Mainstone, John, 475 Metallic bonds, 491, 939 and resonance concept, 390 Manganese dioxide (MnO2), 198, 600 Metallic crystals, 491 Libby, Willard F., 586 Manganese nodules, 932 Metallic elements, 48, 491, 930. See also Ligands, 1000, 1001 (table) Manometer, 177 Metal(s) strong-field, 1012 Many-electron atoms, 295 Metalloids, 48 weak-field, 1012 Marble, 947 Metallurgy, 932 Light Markovnikov, Vladimir, 1035 coordination compounds in, 937, 965 absorption of, and crystal field Markovnikov’s rule, 1035 pyrometallurgy, 933 theory, 1010 Marsden, Ernest, 44 Metathesis reaction, 121 electromagnetic theory of, 276 Marsh, James, 170 Meter, 13 particle-wave duality of, 279, 287 Marsh gas. See Methane Methane (CH4), 1027 plane-polarized, 1007 Marsh test, 170 combustion of, 243, 1031 speed of, 277 Martian Climate Orbiter, 17 hydrate, 1038 Light water reactors, 879 Mass, 11 molecular geometry of, 416, 432 Lime, 793, 919 atomic. See Atomic mass Methane hydrate, 1038 Limestone. See Calcium carbonate critical, 878 Methanol (CH3OH), 421, 1042 Liming, 919 defect, 868 Methyl acetate, 599 Limiting reagents, 99 electron, 46 Methyl chloride, 1031 Line spectra, 282 formula, 83 Methyl group, 1029 Linear molecule, 415, 433 molar, 78, 191, 542 Methyl radical, 1032 Liquid(s), 10, 473 molecular, 81 Methyl propyl ether (neothyl), 1044 properties of, 466 (table) number (A), 46 Methylene chloride, 1031 solutions of liquids in, 519 percent composition by. See Percent Methyl-tert-butyl ether (MTBE), 1050 solutions of solids in, 519 composition Metric unit, 12 surface tension in, 473 SI base unit of, 13 Meyer, Lothar, 327 viscosity of, 474 of subatomic particles, 46 Microscopic properties, 12 Liquid crystals, 506 subcritical, 878 Microstate, 779 Liquid-solid equilibrium, 499 Mass action, law of, 624 Microwave oven, 426 Liquid-vapor equilibrium, 494 Mass defect, 868 Microwaves, 278, 426 Liter (L), 14 Mass number (A), 46 Milk of magnesia, 707, 753, 947 Lithium, 349 Mass spectrometer, 84 Millikan, Robert A., 42 Lithium deuteride (LiD), 885 Mass-energy conversion, 40, 868 Mineral, 931 (table) Lithium fluoride (LiF), 372 Matter, 6 Miscible liquids, 521 Lithium oxide (Li2O), 349 classification of, 6 Mixture, 7 Litmus, 127 conservation of, 40 gas, law of partial pressures and, 195 Living systems Maxwell, James, 203 heterogeneous, 7 coordination compounds in, 1016 Maxwell speed distribution, 204 homogeneous, 7 thermodynamics of, 800 Mean square speed, 206 racemic, 1007 Lock-and-key theory, 605 Mechanical work, 237 Moderator, 879 Logarithm, A-13 Melting, entropy and, 795 Molal boiling-point elevation constant, 537  Index I-9 Molal freezing-point depression constant, 537 Neon, 84, 355 Nuclear chemistry, 862 Molality (m), 523 Neoprene (polychloroprene), 1062 Nuclear decay series, 870 Molar concentration, 145 Neothyl, 1044 Nuclear energy Molar heat Neptunium, 881 from fission reactors, 879 of fusion, 500 (table) Nernst, Walther, 828 from fusion reactors, 883 sublimation, 502 Nernst equation, 828 hazards of, 881 of vaporization, 495, 496 (table) Net ionic equation, 124 Nuclear equation, 863 Molar mass, 78, 191, 542 Neutralization reactions, 130, 151, 730 Nuclear fission, 877 Molar solubility, 745 Neutron, 45, 863 reactions, 877 Molarity (M), 145, 523 Neutron activation analysis, 171 reactors, 879 Mole (mol), 77 Newlands, John, 327 Nuclear fusion, 883 Mole fraction (X), 197, 523 Newton (N), 4, 175 Nuclear reactions, 863 Mole method, 95 Newton, Sir Isaac, 175 balancing, 863 Molecular compounds, 59 Newton’s second law of motion, 4, 17, 175 and decay series, 870 Molecular crystals, 491 Nickel, 996 fission, 877 Molecular equation, 123 chemical analysis of, 1018 fusion, 883 Molecular formula, 52, 89 extraction of, 937 moderator of, 879 Molecular geometry, 413 Nitric acid (HNO3), 675, 970 nature of, 863 of coordinating compounds, 1005 Oswald process in production of, 602 by transmutation, 874, 876 of cycloalkanes, 1033 as oxidizing agent, 970 Nuclear reactors, 879, 884 Molecular mass, 81 as strong acid, 675 breeder, 881 Molecular models, 52 Nitric oxide (NO), 397, 969 fission, 879 Molecular orbital theory, 443 Nitride ion, 967 fusion, 884 Molecular orbitals, 443 Nitrogen, 353, 967 heavy water, 880 bonding and antibonding, 444 bonding in, 379, 451 light water, 879 configurations of, 446 common compounds of, 967 (table) natural, 882 delocalized, 452 preparation of, 967 thermal pollution and, 880 energy level diagram of, 445, 447, 449, Nitrogen cycle, 902 Nuclear stability, 865 450, 451 Nitrogen dioxide (NO2), 623, 649, 969 Nuclear transmutation, 863, 876 Molecular rotation, 782 in smog formation, 920 Nuclear wastes, 883 Molecular shapes. See Molecular geometry Nitrogen fixation, 901 Nucleic acids, 1073 Molecular speed, 206 Nitrogen narcosis, 201 Nucleons, 46, 867 distribution of, 203 Nitrogen pentoxide (N2O5), 577 Nucleotide, 1074 root-mean-square, 206 Nitroglycerin, 397 Nucleus, 44 Molecular vibration, 782, 914 Nitrous oxide N2O (laughing gas), 65, 969 density of, 865 Molecular weight. See Molecular mass Noble gas core, 308 radius of, 45, 865 Molecularity, 594 Noble gases, 50, 355, 358 Nylon (polyhexamethylene Molecules, 50 Node, 287, 444 adipamide), 1063 chemical formulas and, 52 Noguchi, Thomas, 560 Nylon rope trick, 1063 chiral, 1007, 1032 Nomenclature diatomic, 50 linear, 415, 433 of acids, 56 of acids and their conjugate bases, 675 (table) O nonpolar, 424 of alkanes, 66, 1034 O2. See also Oxygen odd-electron, 393 of alkenes, 1034 preparation of, 975 planar, 47, 434, 440 of alkynes, 1037 properties of, 975 polar, 424 of anions, 57 (table), 1004 (table) solubility, 528, 531 polyatomic, 50 of aromatic compounds, 1039 O3. See Ozone Monatomic gases, 173 of bases, 64 Octahedron, 417 Monatomic ions, 51 of cations, 58 (table) Octane number, 1049 Mond, Ludwig, 937 of common compounds, 65 (table) Octaves, law of, 327 Mond process, 937 of coordination compounds, 1003 Octet rule, 378 Monodentate ligands, 1001 of hydrates, 64 exceptions to, 392 Monomers, 1059, 1064 (table) of inorganic compounds, 56, 61 Odd-electron molecules, 393 Monoprotic acids, 128, 677 of molecular compounds, 59, 61 Oil Moseley, Henry, 328 of oxoacids, 63 (table) as fossil fuel, 962, 1048 Most probable speed, 204 of oxoanions, 63 (table) in ore preparation, 932 Multiple bonds, 378, 440 of simple acids, 62 (table) Oil, hydrogenated, 604, 962 Multiple equilibria, 633 Nonbonding electrons, 378 Olefins. See Alkenes Multiple proportions, law of, 40 Nonelectrolyte(s), 119 Oleum, 981 Myoglobin, 1016 Nonelectrolyte solutions, colligative properties Open system, 232 of, 532 Optical isomers, 1007, 1032 N Nonideal gas behavior, 210 Nonmetal, 48, 956 Orbitals. See Atomic orbitals; Hybrid orbitals; Molecular orbitals N2. See Nitrogen Nonmetallic elements, 48, 956 Ores, 931 n-type semiconductors, 940 Nonmetallic oxides, 356, 975 preparation of, 932 Nagasaki, 879 Nonpolar molecule, 424 roasting of, 917, 933 Naming compounds. See Nomenclature Nonspontaneous reactions, 777 Organic chemistry, 1025 Naphthalene (C10H8), 1041 Nonstoichiometric compounds, 960 Organic compounds, 56, 65, 1025 Napoleon, 170, 492 Nonvolatile solutes, 532 Organic polymers. See Polymers National Ignition Facility, 885 Nuclear binding energy, 867 Orientation factor, 594 Natta, Giulio, 1061 nuclear stability and, 867 Orthoclase. See Phosphoric acid Natural gas, 1027 per nucleon, 869 Osmosis, 539 Natural polymers, 1061, 1065 of uranium, 878 Osmotic pressure (π), 539 Negative deviation, 535 Nuclear chain reaction, 878 Ostwald, Wilhelm, 602 I-10 Index Oswald process, 602 Pentane (C5H12), 1028, 1029 pKa, 722 Otto cycle, 1049 Peptide bond, 1066 Planck, Max, 275, 278 Overlap Percent composition by mass, 85, 522 Planck constant (h), 279 in hybridization of atomic orbitals, 432 Percent hydrolysis, 697 Plane-polarized light, 1007 in molecular orbitals, 444 Percent ionic character, 382 Plants in valence bond theory, 429 Percent ionization, 684 in carbon cycle, 912 Overvoltage, 843 Percent yield, 103 osmotic pressure in, 541 Oxalic acid, 645 Perchloric acid (HClO4), 63, 675, 695, 986 Plasma, 884 Oxidation numbers, 135 Perhalic acids, 986 Platinum assignment of, 136, 383 Period, 48 as catalyst, 602, 921 and electronegativity, 383 Periodic group, 48 as electrocatalyst, 818 of halogens, 38 Periodic table, 48, 327 therapeutic uses of complexes of, 1017 of metals in coordination compounds, 1002 atomic radii trends in, 335 Plato, 39 of nonmetallic elements, 138 electron affinity trends in, 346 Plutonium-239, 879, 881 of transition elements, 138, 998 electronegativity trends in, 381 pOH, 672 Oxidation reactions, 133 families in, 48 Polar bonds, 380 Oxidation states. See Oxidation numbers groups in, 48 Polar covalent bonds, 380 Oxidation-reduction reactions (redox historical development of, 327 Polar molecules, 424 reactions), 132 ionization energy trends in, 343 Polar ozone hole, 908 balancing equations of, 813 periods of, 48 Polar solvent, 120 quantitative aspects of, 155 Permanent wave, 1082 Polarimeter, 1007 spontaneous, 824 Permanganate ion, as oxidizing agent, 155 Polarizability, 469 Oxides Peroxide, 349, 975, 1043 Polaroid film, 1007 acidic, 357, 702, 975 Peroxyacetyl nitrate (PAN), 920 Pollution. See Environmental pollution amphoteric, 357, 702, 975 Petroleum, 1048 Polyatomic ions, 51 basic, 357, 702, 975 pH, 671 Polyatomic molecules, 50 Oxidizing agent, 134 of acid-base titrations, 731, 735, 738 Polychloroprene (neoprene), 1062 Oxoacid, 62, 695, 986 of acid rain, 917 Poly-cis-isoprene, 1061 Oxoanion, 63 blood, 732 Polycyclic aromatic hydrocarbons, 1041 Oxyacetylene torch, 257, 1037 of buffer solutions, 728 Polydentate ligands, 1002 Oxygen, 354, 975 common ion effect on, 721 Polyester, 1063 alkali metal reactions with, 349, 943 solubility equilibria and, 753 Polyethylene, 1060 allotropes of, 52, 253, 975 pH meter, 671, 730, 830 Polyisopropene. See Rubber in blood, 531, 651, 732 Pharmacokinetics, 606 Polymer(s), 1059, 1064 (table) hemoglobin and, 531, 651, 733, Phase, 466 Polymerization 1016, 1073 Phase changes, 493 by addition, 1060 molecular orbital theory of, 444, 450 effects of pressure on, 504 by condensation, 1063, 1065 oxidation number of, 138, 383 and entropy, 795 Polypeptide, 1068 paramagnetism, 444 liquid-solid, 499 Polypropenes, 1060 and photosynthesis, 599, 901 liquid-vapor, 494 Polyprotic acids, 128, 688 preparation of, 975 solid-vapor, 502 Polytetrafluoroethylene (Teflon), 987, 1060 Oxygen cycle, 902 Phase diagrams, 503, 504, 963 Poly(vinyl chloride), 1060 Oxygen-hydrogen fuel cell, 835 Phenolphthalein, 152, 741 Porphine, 1016 Oxygen-propane fuel cell, 836 Phenyl group, 1040 Porphyrins, 1016 Oxyhemoglobin, 531, 651, 733, 1016, 1073 Phosphate buffer, 729 Positive deviation, 535 Ozone, 52, 977 Phosphate rocks, 105, 970 Positron, 864 depletion of, 906 Phosphine, 971 Potassium, 349, 943 preparation of, 977 Phosphoric acid (H3PO4), 128, 692, 973 Potassium-40, 874 properties of, 977 ionization constants of, 692 Potassium chlorate (KClO3), 198, 600 resonance structure of, 390 Phosphorus, 353, 970 Potassium dichromate (K2Cr2O7), 144, 155, 813 in smog formation, 920 allotropes of, 971 Potassium hydrogen phthalate, 152 in fertilizers, 105 Potassium hydroxide (KOH), 945 P Phosphorus acid (H3PO3), 973 Phosphorus(V) oxide (P4O10), 972 Potassium nitrate (KNO3), 945 Potassium permanganate (KMnO4), 155, 814 p Orbitals, 298 Phosphorus(III) oxide (P4O6), 972 Potassium superoxide (KO2), 349, 943 P4, structure of, 971 Phosphorus pentachloride (PCl5), 972 Potential. See Standard reduction potential p-type semiconductors, 941 Phosphorus trichloride (PCl3), 972 Potential energy, 231 Packing efficiency, 480 Photochemical smog, 919 Precipitate, 121 Palladium, 960, 1018 Photodissociation, 906 Precipitation reaction, 121, 748 Paramagnetism, 304, 444, 1012 Photoelectric effect, 279 ion separation by fractional, 749 Parnell, Thomas, 475 Photons, 279 Precision, 22 Partial pressure, 195 Photosynthesis, 599, 901, 913 Prefixes Dalton’s law of, 196 carbon dioxide and, 599, 913 nomenclature, 60 (table) Particle accelerators, 875 chlorophyll in, 1017 SI unit, 13 (table) Particle theory of light, 281 isotope applications to, 599, 913 Pressure, 175 Particle-wave duality, 281, 289 oxygen and, 599, 901 atmospheric. See Atmospheric pressure Pascal (Pa), 175 Physical equilibrium, 622 chemical equilibrium and changes in, 646 Pascal, Blaise, 175 Physical properties, 10 critical, 499 Passivation, 840 Pi (π) bond, 440 gas, 174 Patina, 840 Pi (π) molecular orbitals, 445 osmotic, 539 Pauli, Wolfgang, 303 Pig (cast) iron, 935 partial, 195 Pauli exclusion principle, 303, 445, 447, 1012 Pipet, 12 phase changes and, 504 Pauling, Linus, 381, 1068, 1072 Pitch, 475 SI unit, 175 Penetrating power, 304 Pitch Drop Experiment, 475 vapor. See Vapor pressure  Index I-11 Pressure cookers, 505 Radioactivity, 43 Rotation Pressure-volume relationship of gas, 178 artificial, 875 about bonds, 1036 Primary pollutant, 919 biological effects of, 888 molecular, 782 Primary structure, 1069 natural, 870 of plane-polarized light, 1007 Primary valence, 1000 nuclear stability and, 865 Rotational motion, 782 Principal quantum number (n), 284, 295 Radiocarbon dating, 586, 873 Rubber (poly-cis-isopropene), 1061 Probability, in electron distribution, Radiotracers, 886 natural, 1061 294, 298 Radium, 888, 898 structure, 801, 1062 Problem solving, 25, 27, 79, 680 Radius synthetic, 1062 Product, 91 atomic, 335 thermodynamics of, 801 Propane, 1027 ionic, 338 vulcanization, 1062 Propane-oxygen fuel cell, 836 nuclear, 865 Rubbing (isopropyl) alcohol, 1043 Propene, 1035 Radon, 358, 921 Ruby laser, 288 Properties Ramsay, Sir William, 358 Rust, 4, 839 chemical, 11 Raoult, Francois M., 532 Rutherford, Ernest, 44, 328, 874 extensive, 11 Raoult’s law, 532 Rydberg, Johannes, 284 intensive, 11 Rare earth series, 310 Rydberg constant (RH), 284 macroscopic, 162 Rate constant, 567 microscopic, 12 physical, 10 Rate-determining step, 595 Rate law, 571 S Propyne (methylacetylene), 1037 Rate of reaction, 563 s Orbitals, 297 Protein, 1068 and stoichiometry, 569 S8, structure of, 979 denatured, 774, 1073 Rays Sacrificial anode, 841 structure of, 1068 alpha, 43 Salt(s), 130 Protium, 960 beta, 43 hydrolysis of, 696 Proton, 44, 863 gamma, 43 Salt bridge, 816 Proust, Joseph L., 40 RBE (relative biological effectiveness), 888 Salt hydrolysis, 696 Pseudo first-order reaction, 583 Reactants, 91 Saltpeter (KNO3), 945 Pyrex glass, 493 Reaction. See Chemical reactions; Nuclear Saponification, 1046 Pyrite, 978 reactions; Thermochemical reactions Saturated hydrocarbons, 1027 Pyrometallurgy, 933 Reaction mechanisms, 594 See also Alkanes elementary steps, 594 Saturated solutions, 519 Q experimental study, 598 and molecularity of reaction, 594 SBR (styrene-butadiene rubber), 1063 Scanning electron microscope, 292 Quadratic equation, 680, A-14 Reaction order, 571 Scattering experiment, 44 Qualitative analysis, 761 determination of, 571 Schrödinger, Erwin, 294 Qualitative data, 4 first-order, 575 Schrödinger equation, 294 Quantitative analysis, 149. See also Acid-base second-order, 582 Scientific method, 4, 6 titrations zero-order, 585, 607 Scientific notation, 18 gravimetric, 149 Reaction quotient (Qc), 639, 744, 796, 828 Scuba diving, 200 of redox reactions, 155 Reaction rate, 563 Seawater, 200 Quantitative data, 4 Reaction yield, 103 Second law of motion, 4, 17, 175 Quantum, 278 Reactors. See Nuclear reactors Second law of thermodynamics, 783 Quantum dot, 312 Red blood cells (erythrocytes), 732, 1072 Secondary pollutant, 919 Quantum mechanics, 294 Red cabbage, 741 Secondary structure, 1069 Quantum numbers, 295 Red phosphorus, 971 Secondary valence, 1000 angular momentum, 296 Redox reactions. See Oxidation-reduction Second-order reactions, 582 electron spin, 296 reactions Seed crystals, 519 magnetic, 296 Redox titration, 155 Semiconductors, 940 principal, 284, 295 Reducing agent, 134 Semipermeable membrane, 539 Quantum theory, 275 Reduction potential. See Standard reduction SHE (standard hydrogen Quartz potential electrode), 818 crystalline, 490 Reduction reaction, 133 Shell, 296 melting point of, 490 electrolytic, 933 Shielding constant, 334 structure of, 493 of minerals, 933 Shielding effect, 328, 334 Quaternary structure, 1069 Refining of metals, 937 Shroud of Turin, 587 Quicklime. See Calcium oxide Relative biological effectiveness SI units (International System of (RBE), 888 Units), 12 R Relativity, theory of, 868, 875 Rem, 888 Sickle cell anemia, 292, 1072 Sigma (σ) bonds, 440 Racemic mixture, 1007 Representative (main group) elements, 330 Sigma (σ) molecular orbital, 444 Rad, 888 Residue, 1068 Significant figures, 19, 672, A-13 Radiant energy, 231 Resonance, 390 Silica glass. See Quartz Radiation, 41 Resonance structure, 390 Silicon, 352 biological effect of, 888 Reversible reaction, 121 doping of, 940 climate and, 913 Reversible renaturation, 1073 purification of, 938 electromagnetic, 277 Ribonucleic acid. See RNA Silicon carbide (SiC; carborundum), 964 ionizing, 889 RNA, 1073 Silicon dioxide (SiO2), 490, 493 solar. See also Solar radiation Roasting of ores, 917, 933 Silk, 1069 Radiation dose, 889 Rocks Silver Radicals, 393, 889, 1032 age determination of, 873 corrosion of, 840 Radioactive decay series, 870 phosphate, 105, 971 extraction of, 965 Radioactive isotopes, 886 Röntgen, Wilhelm, 42 ionization energy of, 356 Radioactive waste disposal, 883 Root-mean-square speed, 206 Silver bromide (AgBr), 746, 988 I-12 Index Silver chloride (AgCl) isotonic, hypertonic, and hypotonic, 540 State functions, 234 fractional precipitation of, 749 nonelectrolyte, colligative properties State of a system, 234 gravimetric analysis of, 149 of, 532 Staudinger, Hermann, 1059 solubility and, 742 saturated, 519 Steel, 935 Silver iodide (AgI), 988 standard, 151 Stereoisomers, 1006, 1032, 1061 Simple cubic cell (scc), 479 stock, 147 Stern, Otto, 297 Simplest formula, 53, 88 supersaturated, 519 Stock, Alfred, 57 Single bond, 378 types of, 519 Stock solution, 147 Slag, 935 unsaturated, 519 Stock system, 57 Slaked lime [calcium hydroxide, Solution process, 520 Stoichiometric amounts, 99 Ca(OH)2], 948 Solution stoichiometry, 149, 151, 155, 730 Stoichiometry, 95 Smelting of ores, 917, 933 Solvation, 521 actual, theoretical, and percentage yields Smog, 919 Solvay, Ernest, 945 in, 103 Snowmaking, 240 Solvay process, 945 and gas reactions, 193 Soap, 548 Solvent, 119 rate of reaction and, 569 Soda ash (sodium carbonate, Na2CO3), 945 Somatic effects of radiation, 889 Stone leprosy, 916 Soda lime glass, 493 Sorensen, Soren P., 671 STP (standard temperature and pressure), 185 Sodium, 349, 943 sp Hybridization, 433, 441 Straight-chain alkanes, 66, 1027 production of, 943 sp2 Hybridization, 434, 440 Stratosphere, 904 reaction with water, 139 sp3 Hybridization, 431 Strength Sodium acetate (CH3COONa), 519, 721, 725 sp3d Hybridization, 439 of acids and bases, 131, 673 Sodium acetate-acetic acid system, 721, 725 sp3d2 Hybridization, 439 molecular structure and acid, 692 Sodium carbonate (Na2CO3; soda ash), 945 Space shuttle glow, 906 Strong acids, 673 Sodium chloride (NaCl), 54, 376 Space-filling model, 53 Strong bases, 674 electrolysis of aqueous, 843 Specific heat (s), 246 Strong-field ligands, 1012 electrolysis of molten, 841 Spectator ions, 124 Strontium, 351 melting ice with, 537 Spectrochemical series, 1012 Strontium-90, 351 structure of, 54 Spectrum Structural formula, 53 Sodium fluoride, 987 absorption, 565, 1011 Structural isomers, 1027 Sodium hydroxide (NaOH; caustic soda), 945 emission, 282 Structure, acid strength and, 691 in saponification, 1046 visible. See Visible spectrum Strutt, John William (Lord Rayleigh), 358 in titrations, 152, 731 Speed Styrene-butadiene rubber (SBR), 1063 Sodium nitrate (NaNO3), 945 of electromagnetic waves, 277 Subatomic particles, 46, 863 Sodium peroxide, 350 of light, 277 Subcritical mass, 878 Sodium stearate, 548 Maxwell speed distribution, 204 Sublimation, 502 Sodium tripolyphosphate, 1020 Most probable, 204 Subshell, 296 Soft water, 126 Root-mean-square, 206 Substance, 7 Solar energy, 231 Spin. See Electron spin Substituted alkanes, optical isomerism Solar radiation Spontaneous processes, 777, 790 of, 1032 as energy source, 231 Square planar complex, 1006 Substitution reactions, 1040 in hydrogen preparation, 962 Stability Substrates, 605 oxygen balance and, 906 belt of, 988 Subunits, 1016, 1069 ozone protecting from, 907 nuclear, 988 Sulfur, 354, 491, 978 Solder, 519 Stability constant. See Formation constant allotropes of, 979 Solids Stable nucleus, 988 combustion of, 138, 235 characteristic properties of, 10, 466 Stainless steel, 937 common compounds of, 981 solutions of, in liquids, 519 Stalactites, 720 deposits at volcanic sites, 911 temperature and solubility of, 527 Stalagmites, 720 extraction by Frasch process, 978 See also Crystal(s) Standard atmospheric pressure, 176 in vulcanization process, 1062 Solid-vapor equilibrium, 502 Standard cell potential, 818 Sulfur dioxide (SO2), 980 Solubility, 122, 745 Standard electrode potential, 818 in acid rain, 917 common ion effect and, 751 See also Standard reduction potential Lewis structure of, 417 gas, 528, 529, 531 Standard emf, 818 Sulfur hexafluoride (SF6), 394, 417, 499, 981 molar, 745 Standard enthalpy of formation (ΔHfo ), Sulfur tetrafluoride (SF4), 419 rules of, 122 253, A-8 Sulfur trioxide (SO3), 917, 980 and temperature, 527 Standard enthalpy of reaction, 253 Sulfuric acid (H2SO4), 128, 980 Solubility equilibria, 742 Standard entropies (So), 782, A-8 in batteries, 833 common ion effect and, 751 Standard entropy of reaction, 784 as diprotic acid, 128, 690, 980 complex ions and, 756 Standard free energy of formation (ΔGfo), heat of dilution, 260 in fractional precipitation, 750 791, A-8 as oxidizing agent, 981 pH and, 753 Standard free energy of reaction, 790 production of, 980 Solubility product, 742, 743 (table) Standard hydrogen electrode (SHE), 818 as strong acid, 673 molar solubility and, 747 (table) Standard reduction potential, 818, 821 (table) Sun. See also Solar radiation qualitative analysis of, 761 of transition elements, 997 emission spectrum of, 324, 913 Solubility rules, 122 Standard solution, 151 nuclear fusion in, 883 Solutes, 119 Standard state, 253, 790 Superconductors, 488 nonvolatile, 533 Standard temperature and pressure (STP), 185 Supercooling, 501 volatile, 533 Standing waves, 287 Superoxide ion, 349, 975 Solution(s), 119 State Supersaturated solution, 519 concentration units, 145, 522 excited, 284 Surface tension, 473 dilution of, 147 ground, 284 Surroundings, 232, 786 electrolyte, colligative properties of, 544 oxidation. See Oxidation numbers Syndiotactic polymers, 1061 heat of, 259 standard, 253, 790 Syngas, 966 ideal, 534 thermodynamic, 234, 790 Synthetic rubbers (elastomers), 1062  Index I-13 System closed, 232 Torricelli, Evangelista, 176 Toxicity V defined, 232 of arsenic, 170, 1083 Valence, 1000 isolated, 232 of carbon dioxide, 531, 924 Valence band, 939 open, 232 of carbon monoxide, 924 Valence bond theory, 429 state of, 234 of carbon tetrachloride, 1032 Valence electrons, 330 of chloroform, 1032 Valence shell, 413 T of cyanide, 964 of deuterium oxide, 961 Valence shell expansion, 439 Valence-shell electron-pair repulsion (VSEPR) Technetium-99, 887 of formaldehyde, 924 model, 413 Teflon (polytetrafluoroethylene), 987, 1060 of gases, 174 and molecules in which central atom has no Temperature of hydrogen sulfide, 980 lone pairs, 413 chemical equilibria and changes, 648 of methanol, 1043 and molecules in which central atom has critical, 499 of ozone, 920, 978 one or more lone pairs, 417 and rate of reaction, 588 of plutonium-239, 881 Valine, 1074 solubility and, 527 of radon-222, 922 Van der Waals, Johannes D., 211 and water vapor pressure, 199 (table) of smog, 919 Van der Waals constants, 212 (table) Temperature scales of strontium-90, 351 Van der Waals equation, 212 Celsius, 15, 182 of sulfur dioxide, 980 Van der Waals forces, 467 Fahrenheit, 15 of tetracarbonylnickel, 937 Van Meegeren, Han, 898 Kelvin, 15, 182 of white phosphorus, 971 Vanadium oxide (V2O5), 981 Temporary dipole, 469 Tracers, 886 van’t Hoff, Jacobus, 534 Termites, 1027 Trans isomers. See Cis-trans isomers van’t Hoff factor (i), 544 Termolecular reactions, 594 Transition metal(s), 57, 310, Vapor, 174 Ternary compound, 57 329, 995 Vapor pressure, 199, 494 Tertiary structure, 1069 electron configuration of, Vaporization (evaporation), 494 Tetracarbonylnickel [Ni(CO4)], 937 333, 996 entropy and, 781 Tetraethyllead [(C2H5)4Pb], 1050 oxidation numbers of, 138, 998 molar heat of, 495, 496 (table) Tetrahedral complex, 1006 properties of, 996 Vapor-pressure lowering, 532 Tetrahedron, 416 Transition metal oxides, 703 Vector, 424 Theoretical yield, 103 Transition state, 589 Vermeer, Jan, 898 Theory, 5 Translational motion, 782 Vibrational motion, 782, 913 Therapeutic chelating agents, 1018 Transmutation, nuclear, 874, 876 Viscoelastic, 475 Thermal energy, 231 Transpiration, 541 Viscosity, 474 Thermal motion, 205 Transuranium elements, 876 (table) Visible spectrum, 278, 1011 Thermal (slow) neutrons, 877 Triethylaluminum [Al(C2H5)3], 1061 Vision, 1036 Thermal pollution, 528, 880 Trigonal bipyramid, 416 Vitamin C, 678 Thermite reaction, 258, 949 Trinitrotoluene (TNT), 879 Volatile solutes, 533 Thermochemical equation, 243 Triple bonds, 379, 441 Volcanoes, 911 Thermochemistry, 232 Triple point, 503 Volt, 824 Thermodynamic efficiency, 791, 836 Tripolyphosphate, 1020 Voltage, 817. See also Electromotive force Thermodynamics, 234, 776 Tripositive ion, 338 Voltaic (galvanic) cell, 816 first law of, 234 Triprotic acid, 128, 692 Voltmeter, 817 in living systems, 800 Tritium, 46, 875, 960 Volume, 11 second law of, 783 Trona, 945 chemical equilibria and changes in, 646 third law of, 787 Troposphere, 904 constant, 247 Thermonuclear bomb, 885 Tyndall, John, 547 SI unit of, 14 Thermonuclear reactions, 883 Tyndall effect, 547 Volumetric flask, 12, 146 Thermosphere, 904 Tyvek, 1060 VSEPR. See Valence-shell electron-pair Thioacetamide, 980 repulsion model Thiosulfate ions, 886 Third law of thermodynamics, 787 U Vulcanization, 1062 Thomson, George P., 291 Uncertainty principle, 293 Thomson, Joseph J., 42, 43 Thorium-232, 881 Unimolecular reaction, 594 Unipositive ion, 338 W Three Mile Island nuclear reactor, 883 Unit, SI, 12 Waage, Peter, 624 Threshold frequency, 280 Unit cells, 477 Waste disposal, radioactive waste, 883 Thymine, 1074 Unsaturated hydrocarbons, 1034 Water Thyroid gland, 887 Unsaturated solutions, 519 acid-base properties of, 668 Thyroxine, 988 Unshared electron pairs, 378 autoionization of, 669 Time Uranium density of, 476 SI unit of, 13 fission product of, 877 dipole moment of, 425 Tin, 352, 492 isotopes of, 47 electrolysis of, 842 Tincture of iodine, 988 Uranium decay series, 871 fluoridation of, 987 Titanium dioxide, 719 Uranium oxide (U3O8), 880 hard, 126 Titanium(III) chloride (TiCl3), 1061 Uranium-235, 47, 877, 878, 879 hydrogen bonds in, 475 Titration Uranium-238, 47, 881 ion product constant (Kw) of, 669 acid-base, 151, 730 abundance of, 881 as moderator, 879 redox, 155 dating with, 873 phase diagram of, 504 Titration curve, 731, 735, 738 decay of, 871 soft, 126 Tokamak, 884 Urea specific heat of, 247, 475 Toluene, 534 in fertilizer, 105 structure of, 475 Tooth decay, 846 preparation of, 1026 surface tension of, 473 Torr, 176 treatment of sickle-cell anemia, 1072 vapor pressure of, 199 (table) I-14 Index Water—(cont.) vibrational motions, 782, 913 ionization constants of, 685 strong acid reactions with, 737 Y viscosity of, 474 Weak-field ligands, 1012 Yields Water gas, 958 Weight, 13 actual, 103 Water vapor, pressure of, 199 (table) atomic. See Atomic mass percent, 103 Watson, James, 1074 molecular. See Molecular mass theoretical, 103 Watt, 951 percentage, composition by. See Wave function, 294 Wave mechanics, 294 Percentage composition Werner, Alfred, 1000 Z Wavelength, 275 White lead [Pb3(OH)2(CO3)2], 898 Zero electron density (node), 287, 444 color and, 278, 1011 White phosphorus, 971 Zero-order reactions, 585, 607 radiation and, 277 Wohler, F., 1026 Ziegler, Karl, 1061 Wave-particle duality, 279, 287 Wood alcohol. See Methanol Zincblende, 486 Waves, 275 Wood’s metal, 561 Zinc amplitude, 275 Work, 231, 236 in batteries, 832 electromagnetic, 277 electrical, 824 cathodic protection with, 840 frequency, 275 free energy and, 824 Zinc sulfide (ZnS), 42 interference, 444 and gas expansion, 237 Zone refining, 938 length, 275 Work function, 280 properties of, 275 standing, 288 Weak acids defined, 674 X ionization constants of, 678, 690 X rays, 42 strong base reactions with, 734 diffraction of, 484 Weak bases periodic table and, 328 defined, 675 Xenon, 355