Chemistry: The Science in Context

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W. W. Norton & Company, Inc. • www.NortonEbooks.com

SECOND

EDITION

.------. --C-b-e Thomas R. Gilbert NORTHEASTERN

UNIVERSITY

Rein V. Kirss NORTHEASTERN

UNIVERSITY

Natalie Foster LEHIGH

UNIVERSITY

Geoffrey Davies NORTHEASTERN

W.

w.

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NORTON & COMPANY

NEW YORK·

LONDON

•

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Cover images: Galaxy: Science Faction/Getty Images; Earth from Space: PLIIGetty Images; Bees: Getty Images; DNA: 3DClinic/Getty Images Library of Congress Cataloging-in-Publication Data Chemistry: the science in context. -

2nd ed. / Thomas R. Gilbert ... [et al.l.

p. cm. Includes index. ISBN 978-0-393-92649-1 (hardcover) l. Chemistry. I. Gilbert, Thomas R. Chemistry.

QP33.2.G55 2009 540-dc22 2007050388 W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, NY 10110 www.wwnorton.com W. W. Norton & Company Ltd., Castle House, 75/76 Wells Street, London WIT 3QT 1234567890

Brief Contents 1 Matter, Energy, and the Origins of the Universe

1

2

Atoms, Ions, and Compounds

41

3

Chemical Reactions and Earth's Composition

80

4

Solution Chemistry and the Hydrosphere

129

5 Thermochemistry

183

6

Properties of Gases: The Air We Breathe

237

7

Electrons in Atoms and Periodic Properties

297

8

Chemical Bonding and Atmospheric Molecules

355

9

Molecular Geometry and Bonding Theories

405

Forces between Ions and Molecules and Colligative Properties

461

10

11 The Chemistry of Solids

523

12

568

Organic Chemistry: Fuels and Materials

13 Thermodynamics:

Spontaneous Entropy, and Free Energy

Processes,

633

14

Chemical Kinetics

673

15

Chemical Equilibrium

735

16

Equilibrium in the Aqueous Phase

789

17 The Colorful Chemistry of Transition Metals

847

18

Electrochemistry

and Electric Vehicles

890

19

Biochemistry: The Compounds of Life

930

20

Nuclear Chemistry

980

21

Life and the Periodic Table

1023

iii

Blank Page

Contents

Chemical Bonding and Atmospheric Molecules A Look Ahead: Take a Deep Breath

8.1 8.2

Ionic, Covalent, and Metallic Bonds An Introduction to Lewis Theory Lewis Structures for Molecules with Single Bonds Lewis Structures for Molecules with Multiple Bonds Lewis Structures for Ionic Compounds

8.3

Unequal Sharing, Electronegativity, and Other Periodic Properties Electronegativity Bond Polarity

8.4 8.5 8.6

Resonance Formal Charges: Choosing among Lewis Structures Exceptions to the Octet Rule Molecules with Less than an Octet Molecules with More than an Octet Odd-Electron Molecules

8.7

The Length and Strength of Bonds Bond Length Bond Energies

392 394 394 395 397

Summary

Questions and Problems

Molecular Geometry and Bonding Theories

405

A Look Ahead: Aromas, Pharmaceuticals,

406 406 408 408

and the Greenhouse Effect

Molecular Shape Valence-Shell Electron-Pair Repulsion Theory (VSEPR) Central Atoms with No Lone Pairs Central Atoms with Lone Pairs and Bonding Pairs of Electrons

9.3 9.4

Polar Bonds and Polar Molecules Valence Bond Theory Orbital Overlap and Hybridization Tetrahedral Geometry: Sp3 Hybrid Orbitals Trigonal Geometry: Sp2 Hybrid Orbitals Linear Geometry: sp Hybrid Orbitals Octahedral and Trigonal Bipyramidal Geometry: Sp3 d2 and Sp3 d Hybrid Orbitals

9.5 9.6

Shapes and Bonding in Larger Molecules Molecular Orbital (MO) Theory Molecular Molecular Molecular Molecular

9.7

367 367 371 374 377 381 381 382 385 388 388 389

Fluorine and Oxygen: Location, Location, Location

Visual Problems

9.2

356 357 358 358 363 366

Summary Problem-Solving

9.1

355

Orbitals Orbitals Orbitals Orbitals

for for for for

H2 Homonuclear Diatomic Molecules NO and Heteronuclear Diatomic Species N/ and Spectra of Auroras

A Bonding Retrospective From Alcohol to Asparagus: The Nose Knows Summary Problem-Solving

Summary

Visual Problems Questions and Problems

413

420 424 425 426 428 430 432 435 438 439 441 446 448 449 450 452 453 453 454

I

ix

Contents

List of Descriptive

Chemistry

Boxes

List of ChemTours About the Authors Preface

Matter,

Energy, and the Origins of the Universe

A Look Ahead: Blinded by the Sun

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Chemistry and the Classes of Matter Matter: An Atomic View Mixtures and How to Separate Them Properties of Matter A Method for Answering Qyestions and Solving Problems States of Matter The Scientific Method: Starting Off with a Bang Making Measurements and Expressing the Results SI Units Significant Figures Significant Figures in Calculations Precision and Accuracy 1.9 Unit Conversions and Dimensional Analysis 1.10 Testing a Hypothesis: The Big Bang Revisited Temperature Scales An Echo of the Big Bang

11 13


41.

A Look Ahead: When Artillery Shells Bounced off Tissue Paper

42 42 44 46 46 48 49 50 52 54 54 55

Summary

Visual Problems

2.4 2.5 2.6

2 3 6 8 10


Problem-Solving

2.1 2.2 2.3

1.

16 18 18 20 21 26 27 30 30 33 35 35 36 37

Summary

II

xvii xviii xix xx

Elements of the Solar System Compounds and Earth's Early Atmosphere The Rutherford Model of Atomic Structure Electrons Radioactivity and the Nuclear Atom Protons and Neutrons Isotopes Average Atomic Masses The Periodic Table of the Elements Mendeleev's Periodic Table Navigating the Modern Periodic Table

I

v

vi

I

Contents

2.7 2.8

Trends in Compound Formation Naming Compounds and Writing Their Formulas Binary Molecular Compounds Binary Ionic Compounds Binary Compounds of Transition Metals Polyatomic Ions Acids

2.9

Nucleosynthesis Early Nucleosynthesis Stellar Nucleosynthesis Summary Problem-Solving

Summary


Chemical

Reactions

and Earth's Composition

A Look Ahead: Purple Salt and Bottled Water 3.1 3.2

Composition of Earth The Mole Molar Mass Molecular Masses and Formula Masses Moles and Chemical Equations

3.3 3.4

3.5 3.6 3.7

Writing Balanced Chemical Equations Combustion Reactions Stoichiometric Calculations and the Carbon Cycle Determining Empirical Formulas from Percent Composition Empirical and Molecular Formulas Compared Mass Spectrometry and Molecular Mass Using Percent Composition and Mass Spectra to Determine Molecular Formulas

3.8 3.9

Combustion Analysis Limiting Reactants and Percent Yield

Solution

Summary

Chemistry

and the Hydrosphere

A Look Ahead: The Constant Ocean

4.5 4.6

69 72 72 73 74

80 81 81 82 83 85 87

88 91 94 97 103 105 107 109 112

116 119 120 121 121


4.4

67 67

Hydrogen and Helium: The Bulk of the Universe Problem-Solving

4.3

64 66

113 115

Visual Problems

4.2

61 63

Calculations Involving Limiting Reactants Actual Yields versus Theoretical Yields Summary

4.1

58

60 60

Earth: The Water Planet Solution Concentration and Molarity Dilutions Electrolytes and Nonelectrolytes Acid-Base Reactions Precipitation Reactions Making Insoluble Salts Using Precipitation in Analysis Saturated Solutions and Supersaturation

129 130 130 134

140 143 145 148 149 151 153

Contents

4.7 4.8

4.9

Ion Exchange Oxidation-Reduction Reactions Oxidation Numbers Examples of Redox Reactions Balancing Redox Reactions Using Half-Reactions Redox in Nature Titrations Calcium: In the Limelight Summary Problem-Solving

Summary


II.

183

Thermochemistry A Look Ahead: The Sunlight Unwinding

5.1

5.2

5.3 5.4

5.5

5.6 5.7

5.8

Energy: Basic Concepts and Definitions Heat Transfer and Temperature Work, Potential Energy, and Kinetic Energy Kinetic Energy and Potential Energy at the Molecular Level Systems, Surroundings, and the Flow of Energy Isolated, Closed, and Open Systems Heat Flow Units of Energy Enthalpy and Enthalpy Changes Heating Curves and Heat Capacity Hot Soup on a Cold Day Cold Drinks on a Hot Day Calorimetry: Measuring Heat Capacity and Calorimeter Constants Determining Heat Capacity and Specific Heat Calorimeter Constants Enthalpies of Formation and Enthalpies of Reaction Fuel Values and Food Values Fuel Value Food Value Hess's Law

184 184 185 185 188 191 191 192 197 200 201 201 206


209 209 211 213 218 218 219 221 226 228 228 229 231

Properties

237

Carbon: Diamonds, Graphite, and the Molecules

of Life


Summary

Visual Problems

of Gases: The Air We Breathe

A Look Ahead: An Invisible Necessity

6.1 6.2 6.3

154 157 157 159 161 164 167 170 172 172 173 174

The Gas Phase Atmospheric Pressure The Gas Laws Boyle's Law: Relating Pressure and Volume Charles's Law: Relating Volume and Temperature Avogadro's Law: Relating Volume and Moles of Gas Amontons's Law: Relating Pressure and Temperature

238 238 240 246 246 249 252 252

I

I

vii

viii

I

Contents

6.4 6.5 6.6 6.7

6.8


255 260 265 270 271 272 273 276 278 278 280 282 284 284 285 288

Electrons in Atoms and Periodic Properties

297

A Look Ahead: Can Nature Be as Absurd as It Seems?

298 299 299 301 302 303 304 306 306 307 310 310 311 313 315 315 318 320 325 325 327

The Ideal Gas Law Gas Density Dalton's Law and Mixtures of Gases The Kinetic Molecular Theory of Gases and Graham's Law Explaining Boyle's, Dalton's, and Avogadro's Laws Explaining Charles's and Amontons's Laws Molecular Speeds and Kinetic Energy Graham's Law: Effusion and Diffusion Real Gases Deviations from Ideality The van der Waals Equation for Real Gases Nitrogen: Feeding Plants and Inflating Air Bags Summary Problem-Solving

Summary

Visual Problems

7.1

Waves of Light Properties of Waves The Behavior of Waves Colors Missing from Sunlight The Expanding Universe 7.2 Atomic Spectra 7.3 Particles of Light and Qyantum Theory Qyantum Theory The Photoelectric Effect 7.4 The Hydrogen Spectrum and the Bohr Model Lines in the Hydrogen Spectrum The Bohr Model of Hydrogen Energies of Photons and Electron Transitions 7.5 Electrons as Waves De Broglie Wavelengths The Heisenberg Uncertainty Principle 7.6 Quantum Numbers and Electron Spin 7.7 The Sizes and Shapes of Atomic Orbitals s Orbitals p and d Orbitals 7.8 The Periodic Table and Filling the Orbitals of Multi electron Atoms 7.9 Electron Configurations oflons Ions of Main Group Elements Transition Metal Ions 7.10 The Sizes of Atoms and Ions Orbital Penetration and Effective Nuclear Charge Trends in Atom and Ion Sizes 7.11 Ionization Energies A Noble Family: Special Status for Special Behavior Summary Problem-Solving

Summary


328 333 333 335 337 338 338 340 344 347 348 349 349

x

I

Contents

II

Forces between Ions and Molecules and Colligative Properties A Look Ahead: A Breath of Fresh Air

Sea Spray and Salts Ion-Ion Interactions and Lattice Energy Calculating Lattice Energies Using the Born-Haber Cycle 10.2 Interactions Involving Polar Molecules Ion-Dipole Interactions Dipole-Dipole Interactions 10.3 Interactions Involving Nonpolar Molecules 10.4 Real Gases Revisited 10.5 Polarity and Solubility Calculating the Solubility of Gases in Water: Henry's Law Combinations of Intermolecular Forces 10.6 Vapor Pressure Seawater versus Pure Water Vapor Pressures of Solutions: Raoult's Law 10.7 Phase Diagrams: Intermolecular Forces at Work Physical States and Phase Transformations Reading a Phase Diagram 10.8 The Remarkable Behavior of Water and Properties of Liquids Surface Tension and Viscosity Water and Aquatic Life 10.9 Colligative Properties of Solutions Boiling Point Elevation and Freezing Point Depression The van't Hoff Factor Osmosis and Osmotic Pressure Reverse Osmosis 10.10 Measuring the Molar Mass of a Solute Using Colligative Properties 10.1

The Halogens: The Salt of the Earth Summary Problem-Solving

Summary


III

The Chemistry

of Solids

A Look Ahead: Stronger, Tougher, Harder 11.1 11.2 11.3

11.4

11.5 11.6 11.7

Metallic Bonds and Conduction Bands Semimetals and Semiconductors Metallic Crystals Stacking Patterns Packing Structures and Unit Cells Unit-Cell Dimensions Alloys Substitutional Alloys Interstitial Alloys Aluminum and Its Alloys: Light Weight and High Performance The Structures of Some Nonmetals Salt Crystals: Ionic Solids

461.

462 463 464 468 472 472 473 476 480 481 482 484 485 486 488 490 490 491 492 493 495 496 496 500 504 507 509 512 514 514 516 517 523

524 525 527 528 528 529 531 535 535 536 540 542 544

Contents

11.8

11.9

Ceramics: Insulators to Superconductors Polymorphs of Silica Ionic Silicates From Clay to Ceramic Structure and Bonding: Superconductors X-ray Diffraction: How We Know Crystal Structures Silicon, Silica, Silicates,

Silicone: What's in a Name?


Summary


Organic Chemistry:

Fuels and Materials

A Look Ahead: The Stuff of Daily Life

12.1

12.2

12.3

12.4

12.5

12.6

12.7

12.8

Carbon: The Scope of Organic Chemistry Families Based on Functional Groups Small Molecules, Oligomers, and Polymers Small Molecules versus Polymers: Physical Properties Hydrocarbons Alkanes: Physical Properties and Structure Drawing Organic Molecules Structural Isomers Naming Alkanes Cycloalkanes Sources of Alkanes Volatility and the Clausius-Clapeyron Equation Volatility at Work: Fractionating Crude Oil Cuts of Crude Oil Alkenes and Alkynes Chemical Reactivities of Alkenes and Alkynes Isomers Naming Alkenes and Alkynes Polymers of Alkenes Aromatic Compounds Structural Isomers of Aromatic Compounds Polymers Containing Aromatic Rings Alcohols, Ethers, and Reformulated Gasoline Alcohols: Methanol and Ethanol Ethers: Diethyl Ether Polymers of Alcohols and Ethers More Oxygen-Containing Functional Groups: Aldehydes, Ketones, Carboxylic Acids, Esters, Amides Aldehydes and Ketones Carboxylic Acids, Esters, and Amides Polyesters and Polyamides Hydrocarbon Fuels from Biomass Methane from Plants Amines A Final Note Summary Problem-Solving

Summary


547 547 548 550 551 553 556 558 558 559 561 568 569 570 570 570 572 573 575 576 578 581 583 584 584 586 592 592 593 594 596 597 600 601 602 603 603 607 608 610 610 611 612 618 618 618 619 620 620 621 623

I

I

xi

xii

I

Contents

II

Thermodynamics: and Free Energy

Spontaneous

Processes,

Entropy,

633

A Look Ahead: The Game of Energy 13.1

13.2

13.3 13.4 13.5

13.6

Spontaneous Processes and Entropy Entropy and Microstates: Flat Tires Statistical Entropy: A Mathematical View of Microstates Thermodynamic Entropy Isothermal and Nonisothermal Processes A Closer Look at Reversible Processes Entropy Changes for Other Common Processes Absolute Entropy, the Third Law of Thermodynamics, and Structure Calculating Entropy Changes in Chemical Reactions Free Energy and Free-Energy Change Determining the Entropy Change in the Universe The Meaning of Free Energy Calculating Free-Energy Changes Temperature, Spontaneity, and Free Energy Driving the Human Engine: Coupled Reactions


646 650 654 654 655 656 658 660 666 666 667 668

Chemical

673


Summary

Visual Problems

III

Kinetics

A Look Ahead: How Quickly Can We Clean the Air? 14.1 14.2

14.3

14.4 14.5

14.6

634 634 635 638 639 639 642 643

Cars, Trucks, and Air Quality Reaction Rates Rate Units and Relative Reaction Rates Experimentally Determined Rates: Actual Values Average Reaction Rates and the Formation of NO Instantaneous Reaction Rates and the Formation ofNOz Effect of Concentration on Reaction Rate Reaction Order and Rate Constants Deriving a Rate Law from a Single Experiment Reaction Half-Lives Second-Order Reactions Pseudo-First-Order Reactions Zero-Order Reactions Reaction Rates, Temperature, and the Arrhenius Equation Reaction Mechanisms Elementary Steps Rate Laws and Reaction Mechanisms Mechanisms and Zero-Order Reactions Catalysis Catalysts and the Ozone Layer Catalysts and Catalytic Converters The Platinum Group: Catalysts, Summary Problem-Solving

Summary


Jewelry, and Investment

674 675 677 677 679 681 682 683 684 690 693 695 699 702 703 708 708 710 714 714 714 719 721 722 722 723 725

Contents

II

I Chemical Equilibrium


736 737 739 741 745 746 746 748 750 752 754 759 761 761 762 764 766 766 772 772 778 778 779 780 781


789

A Look Ahead: A Balancing Act

790 791 792 792 794 795 796 797 799 800 801 802 807 809 813 815 820 823 824 826 828

A Look Ahead: The Eyes Have It

15.1 15.2

15.3

15.4 15.5 15.6 15.7

15.8 15.9

Understanding Equilibrium: A Qyalitative View The Equilibrium Constant Expression and K Writing Equilibrium Constant Expressions Relationship between K c and K p Manipulating Equilibrium Constant Expressions K for Reverse Reactions K for Equations Multiplied by a Number Combining KValues Equilibrium Constants and Reaction Quotients Equilibrium and Thermodynamics Heterogeneous Equilibria Le Chatelier's Principle Effects of Adding or Removing Reactants or Products Effects of Changes in Pressure and Volume Effect of Temperature Changes Catalysts and Equilibrium Calculations Based on K Changing K with Changing Temperature Temperature, K, and tlGo Using K and Tto Determine tlHo and tlSo Summary Problem-Solving

Summary

Visual Problems

II 16.1 16.2

16.3

16.4

16.5 16.6 16.7 16.8 16.9

735

Acids and Bases The Bronsted-Lowry Model Strong and Weak Acids Conjugate Acid-Base Pairs Strong and Weak Bases Relative Strengths of Acids and Bases pH and the Autoionization of Water The pH Scale pOH Calculations with pH and K Weak Acids Weak Bases Polyprotic Acids Acid Strength and Molecular Structure pH of Salt Solutions The Common-Ion Effect pH Buffers An Environmental Buffer A Physiological Buffer Buffer Range and Capacity

I

xiii

xiv

I

Contents

16.10

Acid-Base Titrations and Indicators Acid-Base Titrations Alkalinity Titrations

828

The Chemistry of Two Strong Acids: Sulfuric and Nitric Acids


836 837 838 839 840

The Colorful Chemistry of Transition Metals

847

Summary Problem-Solving Summary Visual Problems

A Look Ahead: The Company They Keep

Lewis Acids and Bases 17.2 Complex Ions 17.3 Complex- Ion Equilibria 17.4 Hydrated Metal Ions as Acids 17.5 Solubilities oflonic Compounds and K sp The Solubility Product K sp Complexation, pH, and Solubility 17.6 Polydentate Ligands 17.7 Ligand Strength and the Chelation Effect 17.8 Crystal Field Theory 17.9 Magnetism and Spin States 17.10 Naming Complex Ions and Coordination Compounds Complexes with a Positive Charge Complexes with a Negative Charge Coordination Compounds 17.11 Isomerism in Coordination Compounds Geometric Isomers Enantiomers 17.12 Metal Complexes in Biomolecules 17.1

Summary Problem-Solving Summary Visual Problems Questions and Problems

Electrochemistry

18.1 18.2 18.3 18.4 18.5 18.6

18.7

18.8 18.9


829 833

848 848 851 854 856 858 858 860 863 866 868 872

875 875 876 876 877

878 879

880 883 884 884 885 890

A Look Ahead: 100 mpg: Charging Up Hybrid Cars

891

Redox Chemistry Revisited Electrochemical Cells Chemical Energy and Electrical Work Standard Potentials (EO) A Reference Point: The Standard Hydrogen Electrode The Effect of Concentration on E cell The Nernst Equation Equilibrium and Dead Batteries Relating Battery Capacity to Quantities of Reactants Nickel-Metal Hydride Batteries Lithium-Ion Batteries Electrolytic Cells and Rechargeable Batteries Fuel Cells

891 894 897 900 903 904 904 906 909 909 910 912

916

About the Authors

Thomas R. Gilbert has a BS in chemistry from Clarkson and a PhD in analytical chemistry from MIT. After 10 years with the Research Department of the New England Aquarium in Boston, he joined the faculty of Northeastern University, where he is currently an associate professor of chemistry and chemical biology and Associate Dean of the Graduate School of Education. His research interests are in chemical and science education. He teaches general chemistry and science education courses and conducts professional development workshops for K-12 teachers. He has won Northeastern's Excellence in Teaching Award and Outstanding Teacher of First-Year Engineering Students Award. Rein V. Kirss received both a BS in chemistry and a BA in history as well as an MA in chemistry from SUNY Buffalo. He received his PhD in inorganic chemistry from the University of Wisconsin, Madison, where the seeds for this textbook were undoubtedly planted. After two years of postdoctoral study at the University of Rochester, he spent a year at Advanced Technology Materials, Inc., before returning to academics at Northeastern University in 1989. He is an associate professor of chemistry with an active research interest in organometallic chemistry. Natalie Foster is an associate professor of chemistry at Lehigh University in Bethlehem, PA. She received a BS in chemistry from Muhlenberg College and MS, DA, and PhD degrees from Lehigh University. Her research interests include studying poly(vinyl alcohol) gels by NMR as part of a larger interest in porphyrins and phthalocyanines as candidate contrast enhancement agents for MRI. She teaches the introductory chemistry class every fall to engineering, biology, and other non-chemistry majors. Natalie also regularly teaches a spectral analysis course at the graduate level. Geoffrey Davies holds BSc, PhD, and DSc degrees in chemistry from Birmingham University, England. He joined the faculty at Northeastern University in 1971 after postdoctoral research on the kinetics of very rapid reactions at Brandeis University, Brookhaven National Laboratory, and the University of Kent at Canterbury. He is now a Matthews Distinguished University Professor at Northeastern University. His research group has explored experimental and theoretical redox chemistry, alternative fuels, transmetalation reactions, tunable metal-zeolite catalysts and, most recently, the chemistry of humic substances, the essential brown animal and plant metabolites in sediments, soils, and water. He edits a column on experiential and study-abroad education in the Journal of Chemical Education and a book series on humic substances. He is a Fellow of the Royal Society of Chemistry and was awarded Northeastern's Excellence in Teaching Award in 1981, 1993, and 1999, and its first Lifetime Achievement in Teaching Award in 2004. xix

Contents

Power, Pigments, and Phosphors: The Group 12 Elements Summary Problem-Solving Summary Visual Problems Questions and Problems

The Compounds of Life

930

A Look Ahead: The Long and the Short of It

931 932 932 934 935 935 939 942 942 945 947 948 949 950 951 953 954 955 958 960 960 964 965 965 967

Biochemistry:

19.1

19.2

19.3

19.4

19.5

19.6

19.7

Biomolecules: Building Blocks of Life Small-Molecule Subunits Formation of Biomolecules and Organized Assemblies Chirality Optical Isomerism Chirality in Nature Protein Composition Amino Acids Peptides Protein Structure and Function Primary Structure Secondary Structure Tertiary and O1taternary Structure Proteins as Catalysts: Enzymes Carbohydrates Molecular Structures of Glucose and Fructose Disaccharides and Polysaccharides Photosynthesis and Biomass Lipids Function and Metabolism of Lipids Other Types of Lipids Nucleotides and Nucleic Acids Components and Structure From DNA to Protein An Additional Role for Ribonucleotide Monomers: Energy and Metabolism

Visual Problems

968 969 970 970


972

Summary Problem-Solving Summary

II

Nuclear Chemistry A Look Ahead: Alchemy in the 21st Century

20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9

918 921 921 922 923

Nuclear Chemistry Radioactive Decay Hydrogen Fusion Nuclear Binding Energies Synthesis of Heavy Elements The Belt of Stability Artificial Nuclides Nuclear Fission Measuring Radioactivity

980

981 981 981 984 985 987 990 994 997 999

I

xv

xvi

I

Contents

I

20.10 Biological Effects of Radioactivity Radiation Dosage Assessing the Risks of Radiation 20.11 Medical Applications of Radionuc1ides Radiation Therapy Medical Imaging 20.12 Radiochemical Dating Radon, Radium, and Uranium: Too Hot to Handle Summary Problem-Solving

Summary


y

II


1.023

A Look Ahead: What Elements Do We Find in Our Bodies?

21.1 21.2

21.3

21.4

21.5

The Periodic Table of Life Essential Major Elements Alkali Metals Alkaline Earth Metals Chlorine Essential Trace and Ultratrace Elements Main Group Elements Transition Metals Diagnosis and Therapy Diagnosis of Disease Therapeutic Applications Medical Devices and Materials Toxic Elements Beryllium Lead Mercury Cadmium Summary Problem-Solving

Summary


Glossary Answers to Concept Tests and Practice Exercises Photo Credits Index

1024 1024 1026 1027 1028 1030 1032 1032 1035 1039 1040 1047 1052 1055 1055 1056 1056 1058 1059 1059 1060 1062

A-I

Appendices

Answers to Selected End-of-Chapter

1002 1003 1005 1007 1007 1007 1009 1012 1013 1014 1014 1016


A-32 A-46 A-56 A-93 A-96

Descriptive Chemistry Boxes Hydrogen and Helium: The Bulk of the Universe

116

Calcium: In the Limelight

170

Carbon: Diamonds, Graphite, and the Molecules of Life

226

Nitrogen: Feeding Plants and Inflating Air Bags

282

A Noble Family: Special Status for Special Behavior

344

Fluorine and Oxygen: Location, Location, Location

392

From Alcohol to Asparagus: The Nose Knows

450

The Halogens: The Salt of the Earth

512

Silicon, Silica, Silicates, Silicone: What's in a Name?

556

The Platinum Group: Catalysts, Jewelry, and Investment

721

The Chemistry of Two Strong Acids: Sulfuric and Nitric Acids

836

Power, Pigments, and Phosphors: The Group 12 Elements

918

Radon, Radium, and Uranium: Too Hot to Handle

1012

xvii

ChemTours Big Bang Significant Figures Scientific Notation Dimensional Analysis Temperature Conversion Rutherford Experiment Synthesis of Elements Avogadro's Number Balancing Equations Carbon Cycle Percent Composition Limiting Reactant Molarity Saturated Solutions State Functions and Path Functions Internal Energy Pressure-Volume Work Heating Curves Calorimetry Hess's Law The Ideal Gas Law Dalton's Law Molecular Speed Electromagnetic Radiation Light Diffraction Doppler Effect Light Emission and Absorption Bohr Model of the Atom De Broglie Wavelength Quantum Numbers The Periodic Table Bonding Lewis Dot Structures Resonance Estimating Enthalpy Changes VSEPR Model Partial Charges and Bond Dipoles Greenhouse Effect Vibrational Modes Expanded Valence Shells Hybridization Molecular Orbitals Lattice Energy Intermolecular Forces Molecular Motion Raoult's Law Phase Diagrams Hydrogen Bonding in Water Capillary Action xviii

Boiling and Freezing Points

17 21 21 28 30 49 69 83 89 95 97 113 136 153 187 196 198 202 211 221 255 265 273 299 301 303 304 311 316 321 341 357 359 375 390 411 420 424 424 432 434 444 466 473 487 488 491 492 494 500

Osmotic Pressure Crystal Packing Unit Cell Allotropes of Carbon Superconductors X-ray Diffraction Structure of Cyclohexane Cyclohexane in 3-D Fractional Distillation Structure of Benzene Polymers Entropy Dissolution of Ammonium Nitrate Gibbs Free Energy Reaction Rate Reaction Order Arrhenius Equation Collision Theory Reaction Mechanisms Equilibrium Equilibrium in the Gas Phase Equilibrium and Thermodynamics Le Chatelier's Principle Solving Equilibrium Problems Acid Rain Acid-Base

Ionization

Self-Ionization of Water pH Scale Acid Strength and Molecular Structure Buffers Strong Acid and Strong Base Titration Titrations of Weak Acids Crystal Field Splitting Zinc-Copper Cell Free Energy Cell Potential Alkaline Battery Fuel Cell Chirality Chiral Centers Condensation of Biological Polymers Fiber Strength and Elasticity Formation of Sucrose Half-Life Fusion of Hydrogen Radioactive Decay Modes Balancing Nuclear Reactions

505 529 530 542 552 553 583 583 586 601 612 635 643 655 677 684 703 704 709 737 741 754 761 766 791 792 798 799 814 824 831 831 868 894 897 900 901 916 936 938 945 949 955 982 984 993 993

Preface The challenge facing instructors of general chemistry is to make connections between chemical principles and students' lives. We continue to learn how the world works at the molecular level, and the chemist's point of view-the molecular point of view-is valuable as we seek to understand and change our world. Whatever disciplines our students ultimately pursue in their professional lives, the substance of that discipline will be expressed in the molecular language of chemistry. Health and disease are being understood at the molecular level, as are corrosion, adhesion, conductivity, memory, and thought. We explain how drugs work in terms of their molecular structure; we make better contact lenses because we understand how the molecules in the natural lenses in our eyes function; we produce tougher, stronger, and lighter materials to make cars more fuel-efficient, to protect athletes from injury, and to enable the construction of environmentally friendly and beautiful buildings because we understand how molecular structure determines physical properties. This text introduces chemical principles in the context of other disciplines, but its heartland is the molecular language of chemistry. Chemists see the world differently from the way others do, and that different point of view-the molecular point of view-is the keystone of this text. We gauge students' understanding of the molecular point of view largely by their ability to solve problems. They must recall the chemical principles they have learned and apply them to situations which may not seem familiar. Sometimes the hardest part of solving problems is recognizing the necessary information provided, identifying what is needed, and understanding where to begin to work out the answer. We ask our students to do this in homework assignments, on exams, and in the laboratory. Chemistry: 1he Science in Context, Second Edition, introduces chemical principles within the context of issues and topics from fields as diverse as biology, environmental science, materials science, astronomy, medicine, and geology. Our experience is that the context makes the chemistry more interesting, more relevant, and more memorable to students. By placing the concepts and facts of chemistry in a meaningful framework, we make them easier for students to recall and draw upon when solving problems. To complement this framework, we provide an easy-to-remember and consistent problem-solving approach for students to use as they develop their problem-solving skills. In designing Chemistry: 1he Science in Context, Second Edition to be an effective learning tool for all students, we began by recognizing that students use textbooks in different ways. Some students begin with page one of a chapter, and for them we have started each chapter with A Look Ahead. This feature is written to capture the students' attention by providing a glimpse of how the upcoming material connects to the world around them. We have used topics familiar to most students, but we place them in chemical contexts that may surprise them. As students read, they find key terms in boldface and defined in the text as well as in the margin. We have deliberately duplicated this material so that students xx

Preface

I

can continue reading without interruption, while those reviewing a chapter can quickly find the important terms and definitions. All key terms are also defined in the Glossary in the back of the book. We have worked with artists to create molecular art that is both appealing and informative. In most chapters we have chosen specific concepts and figures to illustrate "What a Chemist Sees" (see, for example, Figure 5.13 on page 196). Chemists look at the world at the molecular level, and students need to begin to incorporate this perspective as they learn. @@ CONNECTION In class, we often tie a concept we're teaching to a concept taught earlier in the course. We make these same important connections between topics in the book with Connection icons. ~II CHEMTOUR Students looking for additional help may take advantage of the approximately 100 ChemTours available on StudySpace (see p. xxiii) for the second edition. Denoted by the ChemTour icon, ChemTours demonstrate dynamic processes and help students visualize events at the molecular level. In many cases, students can manipulate variables and observe the resulting changes to a system. O1testions at the end of tutorials guide students through step-by-step problem-solving exercises and offer useful feedback. Concept Tests are short, conceptual questions that are integrated throughout each chapter. They serve as a self-check by asking students to Which contains more atoms-l oz of gold (Au) or 1 oz of silver (Ag)? stop and answer a question about something they have just read. Concept (Answm to Conupt Tl!stsare in fhl!back ofthl! book.) Tests are designed so students can see for themselves whether they have grasped a key concept. The answers to all Concept Tests can be found in the back of the book. The parts of the chapter used most often by students are the Sample Exercises. For students reading the entire chapter, the Sample Exercise takes the concept being discussed and shows them how to apply it to a problem. For Calculating Molarity from Mass and Volume students who begin a chapter by solving the end-of-chapter Problems, the PVC is widely used in constructing homes and office buildings, but because vinyl Sample Exercises are the models they use when they get stuck. Because probchloride may leach from PVC pipe, it is used for drain pipes, not for drinking water supplies. The maximum concentration of vinyl chloride (C H CI) allowed lem solving is so important, Sample Exercises were a focus of this revision. in drinking water in the United States is 0.002 mg/L. What is this concentration in terms of molarity? As beginning problem-solvers, students need a consistent approach COLLECT AND ORGANIZE We are asked to convert a concentration from milthey can use. In Section 1.5 we introduce the acronym COAST (Collect ligrams of solute per liter of solution to moles of solute per liter of solution. This conversion involves changing mass in milligrams into mass in grams and then and Organize, Analyze, Solve, and Think about it) to represent a fourinto an equivalent number of moles. step method for answering questions and solving problems. We then use ANALYZE The conversion factors we need are the molar mass ofC2H3Cl and the conversion factor that 1000 mg 1g: these steps in every Sample Exercise throughout the book: 1

3

=

COLLECT AND ORGANIZE helps students understand where to begin. In this step we restate the problem, identify the information given, and articulate what the student is asked to find. ANALYZE

shows students

how to map out a strategy

for solving the

problem.

+ l~X

35.45 gel l~

~62.49g/mol

Setting up the unit conversion steps so that the initial set of units cancels out, we have:

SOLVE

0.002 ~ X _l_g_ X 1 mol L 1000 fflg 62.49 g

=

3.2 X 10-8 mol L

=

3 X 10-8 M or 0.03 µ.M

ABOUT IT This number may at first seem very small, but remember that the mass of PVC (0.002 mg, or 2 X 1O-~ g) in 1 L of water is also very small. Sometimes it is helpful to do an "order of magnitude" estimation in a problem like this: estimate the concentration as 1 X 10- giL, the molar mass as 100 g/mol (1 X 1()2 g/mol), and the approximate concentration is then

takes the information and the strategy and shows students each step of working through the problem. We show and explain each step so students can follow the logic used and each mathematical step required by that logic.

THINK

reminds students to check their answer and think about the result. Does it seem realistic? Are the units correct? Is the number of significant figures appropriate?

which is very dose to the number we calculated. Our number seems reasonable based on an estimation you can do quite quickly in your head. The extremely low limit of only 0.002 mg/L or 3 X 10-8 M reflects recognition of the danger this substance poses to human health.

SOLVE

THINK

ABOUT

IT

We believe that providing an easy-to-remember and consistent model will help students develop good habits as they use the book and that COAST will become the process they use as they solve problems on their own.

6

6

1 X 10- giL 1 X 10] g/mol

=

1 X 10-8 mollL

Practice Exercise When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier than ordinary seawater) is evaporated, 179 grams ofMgCI] are recovered. What is the molar concentration ofMgCl2 in the original sample? (Answas

to Practice Exacim

are in tlx hack

ifthe

hook.)

xxi

I

xxii

Preface

Problem·Solving Summary Comp.';nl;onconc,"""rio", ;n.qo.cc.URr>.it;. ~N 10 ~ 0\lNQ;, \\.l.\;~ \ t;(l'iHNT \WliI' NONE of '(oUR R~9\\'l.&f" (~~ gl;; ~~R9£9 ~~ "'Cc.uR~Ib.

1.51. A brand new silver dollar weighs 0.934 ounces. Express this mass in grams and kilograms. 1 oz = 28.35 g. 1.52. A dime weighs 2.5 g. What is the dollar value of 1 kg of dimes? 1.53. What volume of gold would be equal in mass to a piece of copper with a volume of 125 mL? *1.54. A small hot-air balloon is filled with 1.00 X 106 L of air (d = 1.20 giL). As the air in the balloon is heated, it expands to 1.09 X 106 L. What is the density of the heated air in the balloon? 1.55. What is the volume of1.00 Ib (pound) of mercury? 1.56. A student wonders whether a piece of jewelry is made of pure silver. She determines that its mass is 3.17 g. Then she drops it into a 10 mL graduated cylinder partially filled with

8na1: S ... that have natural abundances (expressed as decimals) a1, a2, a3, .... We use Equation 2.1 to calculate average atomic masses in the following Sample Exercise.

Calculating an Average Atomic Mass from Natural Abundances The precious metal platinum has six isotopes with the following natural abundances:

Symbol

Mass (amu)

Natural Abundance

190Pt

189.96

0.00014

192Pt

191.96

0.00783

194Pt

193.96

0.32967

195Pt

194.96

0.33832

196Pt

195.96

0.25242

198Pt

197.97

0.07163

Use these data to calculate the average atomic mass of platinum. AND ORGANIZE The average atomic mass of an element is the weighted average of the masses of its isotopes. To calculate a weighted average, we need to find the product of the mass of each isotope times its natural abundance and then sum the products (Equation 2.1).

COLLECT

Notice that this table is the same as the previous data table except that we have added a column to contain the product of the mass of each isotope times its natural abundance. These are the aimi terms to be added together in Equation 2.1. At the same time let's add a row at the bottom of the table for the sums of the natural abundance values as a check that our data set is complete. ANALYZE

Symbol

Mass (amu)

Natural Abundance

190Pt

189.96

0.00014

?

192Pt

191.96

0.00783

?

194Pt

193.96

0.32967

?

195Pt

194.96

0.33832

?

196Pt

195.96

0.25242

?

198Pt

197.97

0.07163

?

?

?

Sum

Mass x Natural Abundance

(amu)

I

Average Atomic Masses

I

53

54

I

Chapter Two

I


N ow let's complete the table by calculating the product of the mass of each isotope times its natural abundance and the sum of these products (in blue):

SOL V E

Symbol

Mass (amu)

190Pt

189.96

0.00014

0.027

192Pt

191.96

0.00783

1.50

194Pt

193.96

0.32967

63.943

!95Pt

194.96

0.33832

65.959

196Pt

195.96

0.25242

49.464

!98Pt

197.97

0.07163

14.18

Sum

Natural Abundance

0.9997 ~ 1.000

Mass x Natural Abundance

(amu)

195.07

The value in the lower right-hand corner of the table, 195.07 amu, is the one we seek: the average atomic mass of all platinum atoms. ABOUT IT Notice that the natural abundances add up to a value very close to 1, as they should if we have included all the platinum isotopes in our table (we have). The slight difference in the total arises from measurement errors and rounding. The result of the calculation-195.07-is the weighted average atomic mass of platinum. The value given in the inside front cover-195.084 amu-is the most precise value currently available. Remember, however, that no individual Pt atom has that mass; it is the weighted average of all Pt atoms.

THINK

Practice Exercise Silver (Ag) has only two stable isotopes: !07Ag, with a mass of106.90 amu, and !09Ag, with a mass of 108.90 amu. If the atomic mass of silver is 107.87 amu, what is the natural abundance of each of the isotopes? Hint: Let x be the natural abundance of one of the isotopes. Then 1 - x is the natural abundance of the other. (Answers

..

to Practice Exercises are in the back of the book.)

The Periodic Table of the Elements

The Rutherford model of the internal structure of the atom was one of several major advances in the early 20th century that enabled scientists to understand better the nature of matter on an atomic scale and to explain why the elements exhibit their particular physical and chemical properties. Actually, chemists had for many years looked for patterns in the properties of elements, even before they were aware of the existence of subatomic particles and the concept of atomic numbers. Among the pioneers in this effort was Russian chemist Dmitrii Ivanovich Mendeleev (1834-1907).

Mendeleev's Periodic Table Mendeleev published his version of the periodic table of the elements (Figure 2.10) in 1869. It was a remarkable accomplishment because many elements were still unknown in the mid-19th century. Being a circumspect scientist, Mendeleev knew that there was still much he did not know, so he left open spaces in his periodic table to account for elements not yet discovered. He was able to predict the chemical properties of these missing elements based on the locations of the blank spaces with respect to known elements. These insights greatly facilitated the subsequent discov-

2.6

Group Number I Row 1 2 3 4

II

7

IV

V

VI VII VIII

7

9.4 11

12

14

16

19

Li

Be

C

N

0

F

23

B

24 27.3 28

31

32 35.5

Na Mg

AI

Si

P

S

Cl

39

40

44

48

51

52

55

K

Ca

?

Ti

V

Cr Mn

68

72 ?

75

78

80

As

Se

Br

85

87

Rb

Sr

?

56

59

Fe

Co

----

59

1

Ni

:;;:;;:00-

104 104 106

88 90 94 96 100 Y Zr Nb Mo ?

Ru

Rh

Pd 1

108 112 113 118 122 125 127 Ag

Cd

In

Sn

Sb

Te

I

Ba

Di

Ce

9

178 180 182 184 Er

La

------

W

Ta

11 Au Hg Tl

Pb

-

195 197 198

199 200 204 207 208 ~

Bi

231

240

Th

U

Os

Ir

Pt

1

FIGURE 2.10 Mendeleevorganized elements in his periodic table on the basis of their chemical and physical properties and atomic masses. He assigned three elements with similar properties to group VIII in rows 4, 6, and 10.Because he did this, the elements in the rows that followed lined up in appropriate groups. In this way, rows 4 and 5 combined, for example, in Mendeleev's table contains spaces for 18 elements, corresponding to the 18 groups in the modern periodic table.

ery of the missing elements by other scientists. Mendeleev arranged the elements in his periodic table in order of increasing atomic mass, which could be estimated in that era, and not atomic number, which was not identified until over 50 years later.

CONCEPT TEST

Find two pairs of elements in the modern periodic table that would have been out of sequence in Mendeleev's version, which was based on atomic mass. (Answers

55

1

8 Cs

12

I

)

III

133 137 138 140

10I

The Periodic Table of the Elements

H

63 65 5 Cu Zn 6

I

to Concept Tests are in the back of the book.)

Navigating the Modern Periodic Table The modern periodic table (Figure 2.11) is also based on a classification of elements in terms of their physical and chemical properties. As we have already indicated, elements appear in the table in order of increasing atomic number, or number of protons in the nucleus of the atom. Further, the table is organized as a series of horizontal rows (called periods) and columns (called groups or families). Elements in the same family have similar chemical properties. The periods are numbered at the far left of each row. The first period consists of only two elements-hydrogen and helium. The next two periods have eight elements each: lithium through neon, and sodium through argon. The fourth and fifth periods have 18 elements each; elements 21 through 30 and 39 through 48 are the first two rows of the block known collectively as the transition metals. The sixth period is actually 32 elements long. In part to accommodate the periodic

o Period 3 o Group 14 The horizontal rows in the periodic table are called periods. Columns elements Elements chemical

in the periodic table define in the same family, or group. in the same family have similar properties.

Transition metals are the elements in groups 3 through 12 in the periodic table.

56

I

I

Chapter Two


18

1

r--

1 2 3

5 6 7

17W,

1

H

2

3

4

Li

Be

11

12

f- Atomic

5

f-

B

Na Mg

number Symbol for element

~

4 19 20 21 22

7 5 6 23 24 25

K

V

Ca

3

Sc

Ti

37 38 39 40 Rb

Cr Mn

44 45

55 56 57

72 Hf

73 74 75

76 77

Ta

Os

Re

1 H

Rh Ir

46 47 48

49 50 51

52 53 54

Pd

In

Te

Ag

Cd

Sn

78 79

80 81 82

Pt

Hg

Au

TI

Pb

Sb

I

Po

Rn

At

Fr

Rf Db

Bh

Hs

Mt

Ds

lG ~ 17 ~ Cl

83 84 85 86 Bi

104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Sg

3 Li

Xe

87 88 89 Ra Ac

-

9

Ru

W

10 Ne

}.'

41 42 43

Ba La

9

F

13 14 15 16 17 18 10 11 12 AI Si P S CI Ar 28 29 30 31 32 33 34 35 36 Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Nb Mo

Tc

7 N

9 26 27

Zr

Sr

16 8 0

15

8

y

Cs

14 6 C

13

Rg

1

~t

35

S~

llt

,,~

S3

I 6 Lanthanides 7

Actinides

58

59 60 61

Ce

Pr

62 63 64

Nd Pm Sm Eu

65 66 67

Gd Tb

Dy Ho

68 69 Er

70 71

Tm Yb

93 94 95

98 99 100 101 102 103

1h

Np

Cf

Pa

U

(a)

Es

Fm Md No

I~

10

Na

18

19

J\r

K

36

37 Rb

38

I+

0.00794

Cl-

19.353

545.88

559.40

SO,>-

2.712

28.23

28.93

0.126

2.06

2.11

Br-

0.0673

0.844

0.865

B(OH)3

0.0257

0.416

0.426

0.00130

0.068

35.169

Total

1119.87

0.070 1147.59

W

FIGURE 4.4 (a) Very high concentrations (called "blooms") of a phytoplankton called coccolithophores in the Bering Sea were captured by a NASA satellite in April 1998. During blooms,

coccolithophores

chlorophyll

turn the color of the ocean a milky aquamarine.

in the phytoplankton,

many intricately These coccoliths

the milkiness

from sunlight

scattering

intense

The color comes from off the organisms.

(b) The

plates (or coccoliths) made of CaC03 surround each cell of the organism. are so tiny that 500 of them placed end-to-end would fit in a length of 1 mm.

designed

of two or

The solutes of a solution are the components present in smaller amounts than the solvent. A solution may contain one or more solutes.

'Molar solutions are discussed in Section 4.2.

w

mixture

I 133

The solvent of a solution is the component present in the largest amount. In an aqueous solution, water is the solvent.

0.0928

HC0 3

F-

Earth: The Water Planet

A solution is a uniform more substances.

mmoljkg

0.0906

I

134

I

I Chapter Four


FIGURE 4.5 Particles

in suspension,

such as phytoplankton

in seawater,

separated

Scientists

by filtration.

the material organic

trapped

on the filter with an

solvent such as acetone that dissolves

the phytoplankton's pigments.

can be may extract

chlorophyll

The intensity

acetone extracts concentration

and other

Phytoplankton in filter

Phytoplankton in filter

of the color of these

provides

a measure

of phytoplankton

of the

! _____

Water

in the original

Acetone containing dissolved chlorophyll

Filtration

sample.

Chlorophyll

.. ~

Extraction

Acetone

by filtering samples of seawater to collect the suspended matter. During filtration, particulate matter including phytoplankton is trapped on the filter, while salts and molecular solutes pass through it (Figure 4.5).

CONCEPT A ball-and-stick

model of acetone.

TEST

Which, if any, of the following forms of matter are solutions-that is, homogeneous mixtures? (a) muddy river water; (b) helium gas; (c) clear cough syrup; (d) filtered air. (Answers to Concept Tests are in the back of the book.)

~

Solution Concentration and Molarity

Table 4.1 lists the major ions in seawater and their concentrations expressed using three sets of units. The left column of data is expressed in grams of solute ion per kilogram (g/kg) of seawater. Many scientists, including those involved in environmental studies, use mass-to-mass ratios such as milligrams per kilogram (mg/kg), or mass-to-volume ratios such as milligrams per liter (mg/L), to express concentrations. When environmental regulatory agencies establish limits! on the concentrations of contaminants that are permitted in drinking water, those limits are usually expressed using the mass-to-volume ratio of milligrams of contaminant (solute) per liter of drinking water (solution). These agencies often express the maximum recommended human dosage of these contaminants in units of milligrams of contaminant per kilogram of body mass. Another convenient set of units for expressing very small concentrations is based on parts per million (ppm) and parts per billion (ppb). A solution with a concentration of 1 ppm contains 1 part solute for every million parts of solution, for example, 1 gram of solute for every million (106) grams of solution. Put an'The

u.s. Environmental

Protection Agency (EPA) sets contaminant

concentration

limits called maxi-

mum contaminant levels (MCLs) for air and water, including drinking water. The MCLs of all contaminants in drinking water are given in mg/L. Similar limits have been established by Environment the World Health Organization, and other environmental agencies.

Canada,

4.2

other way, each gram of a 1 ppm solution contains one-millionth 10-6 g = 1 µg) of solute, as shown by the following conversions:

of a gram (or

19 10-6 10-6 g 1 µg 1 ppm = 106 g X 10-6 = -g= g A concentration of 1 ppm (1 µg/g) is also the same as 1 mg/kg: t*>g 1 mg 1-X--X--=-g 103t*>g

103g

1 mg

kg

kg

A concentration of 1 ppb is 1/1000 as concentrated as a 1 ppm solution. Since a 1 ppm solution is the same as 1 mg/kg, a 1 ppb solution is equivalent to 1 µg/kg: 1 ppm 10-6 g 1 µg 1 ppb = 1000 = 1000 g = kg Let's express the smallest concentration value in Table 4.1 (0.00130 g F- Ikg seawater) in ppm. To do so, we need to convert grams of F- ion into milligrams ofF- ion because 1 mg/kg is the same as 1 ppm: g

0.00130 -

kg

F

_

103 mg mg _ _ X -= 1.30 - F = 1.30 ppm F g

kg

As you can see, parts per million is a particularly convenient unit for expressing the concentration of F- ion in seawater because it avoids the use of exponents or lots of zeroes to set the decimal place. Comparing Ion Concentrations in Aqueous Solutions The average concentration of chloride ion in seawater is 19.353 g/kg. The maximum allowable concentration of chloride ion in drinking water is 250 ppm. How many times as much chloride ion is there in seawater than in the maximum amount allowed in safe drinking water? COLLECT AND ORGANIZE The concentrations ofCI- ion in the two samples are given in different units. To compare them, we must express them in the same units.

A N A L Y Z E A link between ppm and mg/kg is the fact that mg/kg has the same

denominator as the units in which the seawater concentration is expressed. We can convert the seawater concentration units of g/kg to mg/kg using the conversion factor 1000 mg/g. SOLVE

1. We convert the concentration of CI- ion in seawater to ppm: g CI-

19.353--

kg

103 mg mg ClX -",= 19,353 = 19,353 ppm CI0

kg

2. Next we take the ratio of the two concentrations: (19,353ppm Cl-)l(250 ppm CI-) = 77.4 times as much chloride ion in seawater as in acceptable drinking water. This question could also be answered by converting the suggested standard for chloride ion in drinking water from parts per million to grams per kilogram and comparing that number to 19.353 g/kg. It doesn't matter what

THINK

ABOUT

IT

I

Solution Concentration

and Molarity

I 135

136

I Chapter Four

I


Molarity (M) is a concentration unit; it is the amount of solute (in moles) divided by the volume of solution (in liters): M = nlV. A 1.0 M solution contains 1.0 mole of solute per liter of solution.

units you choose in order to make the comparison, as long as the units of both numbers are the same.

Practice Exercise The drinking water standard of the World Health Organization (WHO) for arsenic is 10.0 µg/L. Some water from tube wells in Bangladesh was found to contain as much as 1.2 mg/L of arsenic per liter. How many times above the WHO standard is this level? (Answers to Practice Exercises are in the back of the book.)

~II

CHEMTOUR

Molarity

Most scientists interested in studying chemical processes in natural waters or in solutions in the laboratory prefer to work with concentrations based on quantities of solute expressed in moles, not mass. When quantities are expressed in moles, comparisons of effects and behavior are based on the numbers of particles of substances. For these scientists, the preferred concentration unit is moles of solute per liter of solution volume. This ratio is called molarity (M). A 1.0 molar (1.0 M) solution contains 1.0 mol of solute for every liter of solution. We can express this definition in equation form:

.

M oIanty

moles of solute = _.,----"".:..::.:.:....::.....:....::.....:....::....c.:.:.:..::.......,

__

volume of solution in liters

or

M=!f;

(4.1)

For a 1.0 M solution,

If we know the volume (V) of a solution in liters and its molar concentration (M), we can readily calculate the number of moles (n) of the solute in the solution. First we rearrange the terms in Equation 4.1 as follows: n= VXM

(4.2)

Next, we convert the number of moles of solute into an equivalent mass in grams by multiplying n by the molar mass (.At): Mass of solute (g) Combining

= n X .At

(4.3)

Equations 4.2 and 4.3 yields this equation: Mass of solute (g)

= .At

X

VX M

(4.4)

Equation 4.4 is useful when we need to calculate the mass of a solute needed to prepare a solution of a desired volume and concentration, or when we have a solution of known concentration and want to know the mass of solute present in a given volume of the solution. Before we get to several Sample Exercises involving calculations of molarity, we should note that in many environmental and biological systems, a 1.0 mollL solution represents a very high concentration of solute. Consider, for example, the major salts in seawater. The concentrations of these salts make seawater unfit for us to drink, yet the right-most column in Table 4.1 expresses these concentrations not in moles per liter but in millimoles per liter (mmollL), which for simplicity are denoted mM (1 mM = 10-3 M). The concentrations of the minor and trace elements in surface seawater are often expressed in µM ("micro molar," which is

4.2

I


and Molarity

I 137

equivalent to 10-6 mollL), nM ("nanomolar," or 10-9 mollL), or even pM ("picomolar," or 10-12 mollL). The concentration ranges of many biologically active substances in blood, urine, and other biological liquids are also so small that they are often expressed in units such as these. The middle column in Table 4.1 contains values based on the number of millimoles of each ion per kilogram of seawater. Oceanographers prefer concentration units based on mass of seawater rather than volume because the volume of a given mass of water decreases with increasing depth on account of increasing pressure.

Calculating Molarity from Mass and Volume

SAMPLE EXERCISE 4.2

PVC is widely used in constructing homes and office buildings, but because vinyl chloride may leach from PVC pipe, it is used for drain pipes, not for drinking water supplies. The maximum concentration of vinyl chloride (C2H3Cl) allowed in drinking water in the United States is 0.002 mg/L. What is this concentration in terms of molarity? We are asked to convert a concentration from milligrams of solute per liter of solution to moles of solute per liter of solution. This conversion involves changing mass in milligrams into mass in grams and then into an equivalent number of moles. COLLECT

AND

ORGANIZE

The conversion factors we need are the molar mass ofC2H3CI conversion factor that 1000 mg = 1 g:

ANALYZE

.Mc

H Cl 23

= 2 ffl6l-G

12.01 g X

1

C

ffl6l-G

1.008 g

n

H

H

H

Cl

\ / C=C / \

Vinyl cWoride

..(i-i)··.. H

CI n

Polyvinyl chloride

and the

H

+ 3 fflel...M X lfflel...M +

1 tttel-Gl X

35.45 g CI __

=

62.49 g/mol

Setting up the unit conversion steps so that the initial set of units cancels out, we have:

SOL V E

0.002 tttg X

L

1g X 1 mol 1000 tttg 62.49 g

= 3.2

X 10-8 mol

L

=3

X 10-8

M or 0.03 µM

ABOUT IT This number may at first seem very small, but remember that the mass of PVC (0.002 mg, or 2 X 10-6 g) in 1 L of water is also very small. Sometimes it is helpful to do an "order of magnitude" estimation in a problem like this: estimate the concentration as 1 X 10-6 giL, the molar mass as 100 g/mol (1 X 102 g/mol), and the approximate concentration is then

THINK

1 X 10-6 giL

----

1 X 102 g/mol

=

1 X 10-8 mollL

which is very close to the number we calculated. Our number seems reasonable based on an estimation you can do quite quickly in your head. The extremely low limit of only 0.002 mg/L or 3 X 10-8 M reflects recognition of the danger this substance poses to human health.

Vinyl chloride, a suspected carcinogen, enters drinking water by leaching from pipes made of polyvinyl chloride (PVC). Vinyl chloride is the common name of a small molecule, called a monomer (Greek for "one unit"), from which the very large molecule called a polymer ("many units") is formed. PVC usually consists of hundreds of molecules of vinyl chloride bonded together.

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Practice Exercise When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier than ordinary seawater) is evaporated, 179 grams ofMgCl2 are recovered. What is the molar concentration ofMgCl2 in the original sample? (Answers


Calculating Molarity from Density A water sample from the Great Salt Lake in Utah contains 83.6 milligrams of Na+ per 1.000 gram oflake water. What is the molarity ofNa+ if the water has a density of 1.160 g/mL? COLLECT AND ORGAN IZE We are given a concentration in terms of a mass of solute per gram of solution. To convert this concentration into molarity (moles of solute per liter of solution), we need to convert the mass of solute into an equivalent number of moles, and the mass of solution into an equivalent volume of solution in liters.

Starting with the solute, we need to convert milligrams of Na+ to grams ofNa+ and then into moles ofNa+. For this calculation, we need the molar mass of sodium ion: 22.99 g/mol. As for the solution, we need to convert its mass in grams into a volume in milliliters by dividing by the density given in the problem, and then convert milliliters into liters. ANALYZE

SOLVE

1. We find the number of moles of solute (sodium ion): 83.6

Na+ X tttg

1g X 1 mol ~t8:+ 1000 tttg 22.99 ~

=

3.64 X 10-3 mol Na+

2. Then we calculate the volume of solution in liters, given that the sample of solution has a mass of 1.000 g and a density of 1.160 giL: 1.000

g

X

1~ X 1L 1.160 g 1000 ~

= 8.621

X 10-4

L

The molarity of sodium ion in the water from the Great Salt Lake is moles

L

=

3.64 X 10-3 mol Na+ 8.621 X 10-4 L

=

4.22 M

ABOUT IT The data in Table 4.1 indicate that there are about 10 g ofNa+ for every 1 kg of seawater; this is the equivalent to 10 mg of Na+ for every 1 g of water, which is about 10 mg/mL. The concentration given for sodium in the Great Salt Lake is over 8 times that number, so the concentration should be more than 8 times the concentration given in the table. Our answer is reasonable in light of that comparison. Only the Dead Sea (on the boundary between Israel and Jordan) is a saltier body of water, with Na+ ion concentrations almost 11 times the average concentration of seawater. THINK

Practice Exercise The density of ocean water depends on depth because pressure compresses the water. If the density of ocean water at a depth of 10,000 m is 1.071 g/mL and if 25.0 g of water at that depth contains 190 mg of potassium chloride, what is the molar concentration (Answers

of potassium


chloride in the sample?

4.2

I


and Molarity

I 139

Calculating the Quantity of Solute Needed to Prepare a Solution

SAMPLE EXERCISE 4.4

In setting up a saltwater aquarium, you need a stock solution of calcium hydroxide called Kalkwasser. How many grams of Ca(OH\ do you need in order to make 500.0 mL of a 0.0225 M solution of calcium hydroxide? AND ORGANIZE We know the volume (V) and molar concentration (M) of the solution. We also know the identity of the solute, which enables us to calculate its molar mass (.At). We can then use Equation 4.4

COLLECT

Mass (g) of solute

= .At X V X M

to calculate the mass of solute Ca(OH\ needed. We also need to convert the volume in milliliters to liters to obtain the mass in grams of solute required. ANALYZE

1 mol Ga

We calculate the molar mass ofCa(OH)2 X

40.08 g 1G mo a

+ 2 mel-G

X

16.00 g mel-G

as follows:

+ 2 mel-H

X'

1 008

g = 74.10 g/mol mel-H

and 500.0 mL converts to 0.5000 L. We combine the molar mass and volume values with the desired concentrations using Equation 4.4 to calculate the mass of Ca(OH)2 needed:

SOLVE

Mass (g) of Ca(OH)2

= .At

X

= 74.10

VX M

!t

X 0.0225

7

X 0.5000 ±,

= 0.834 g

THINK ABOUT IT To solve a problem like this, all you really need to remember is the definition of molarity: moles of solute per liter of solution. The solution you want has a concentration of 0.0225 M, or 0.0225 mole/L. You do not need an entire liter of solution; you only need 0.5000 L (~ L). This means that you do not need 0.0225 mole of Ca(OH)2; you only need ~ that amount (0.0113 mole). Finally, just calculate how many grams of calcium hydroxide are in 0.0113 mole of

(a)

(b)

(c)

The steps for preparing 500.0 mL of 0.0225 M Ca(OH\ include (a) weighing out and transferring the desired mass of Ca(OH)2 to a 500 mL volumetric flask; (b) adding 200 to 300 mL of water to the flask; (c) swirling the flask to make sure all the solute is dissolved; and then (d) diluting the solution to exactly 500 mL. Stopper the flask and invert it several times to thoroughly mix the solution.

(d)

140

I Chapter Four

I


calcium hydroxide (0.834 g). Calcium hydroxide not only supplies calcium ions, which are essential to the health of creatures in the aquarium, but it also helps to maintain the correct levels of alkalinity (total quantity of basic substances) in the water.

Practice Exercise A solution known as Ringer's lactate is administered intravenously to trauma victims suffering from blood loss or severe burns. The solution contains the chloride salts of sodium, potassium, and calcium and is also 4.0 mM in sodium lactate (NaC3HsO). How many grams of sodium lactate are needed to prepare 10.0 liters of Ringer's lactate? (Answers


CONCEPT

TEST

According to the data in Table 4.1, the concentration of CI- ion in seawater is 545.88 mmollkg and 559.40 mmollL. Does a 1.000 liter sample of seawater weigh more or less than 1.000 kilogram? (Answers


~

~II

CHEMTOUR

Dilution

Dilutions

In laboratories that test the quality of drinking water, standard stock solutions of regulated substances, such as pesticides and toxic metals, are available commercially in concentrations that are too high (typically 1000 mg/L) to be used directly in analytical procedures. Instead, these stock solutions are diluted to prepare working standards, which have concentrations of the substances of interest close to those found in actual samples. Dilutions lower concentration by adding more solvent so that the solute in a sample of the stock solution is distributed throughout a larger volume of solvent than before. We can use Equation 4.2 and the concentration of a stock solution to determine how to make a working standard with a desired concentration by dilution. Suppose, for example, we need to prepare 250.0 mL of a working standard with a concentration of 3.18 X 10-3 Min Cu2+ by starting with a standard solution of Cu2+ that has a concentration of 0.1000 M. What volume of the standard solution do we need? Using Equation 4.2 we can calculate the number of moles of Cu2+ needed in the final solution: nfinal=

v;,nalX Mfinal= (0.2500 L) X (3.18 X 10-3 mollL)

= 7.95 X 10-4 mol

The subscript "final" in the above equation refers to the final solution. Using the same equation, we can calculate the initial quantities, that is, what volume of the initial stock solution we need to deliver that number of moles: ninitial

A stock solution is a concentrated solution of a substance used to prepare solutions oflower concentration. Dilution is the process oflowering the concentration of a solution by adding more solvent.

=

~nitial

X

Minitial

4

- 7.95 X 10- mol ~ni'ial- 0.1000 mollL

or

ninitial

~nitial= Mini'ial

= 7.95 X 10-3 L

=

7.95 mL

To make the desired working standard, add enough solvent (water) to 7.95 mL of the stock solution to make a solution with a final volume of250.0 mL. The images in Figure 4.6 illustrate how this dilution is carried out.

4.3

I

Dilutions

\;

(a)

Multiply by molarity of concentrated solution (Minitial) Moles of solute in concentrated solution

Moles of solute in dilute solution

(b)

Divide by total volume of dilute solution (Vfinal) Molarity of (Mfinal)

dilute solution (c)

FIGURE 4.6 To prepare 250.0 mL of a solution 3.18 X 10-3 Min Cu2+, (a) a pipet is used to withdraw 7.95 mL of a 0.1000 M standard solution. (b) The volume is transferred into a 250.0 mL volumetric flask. (c) Water is then added to bring the total volume of the new solution to 250.0 mL.

I 141

142

I Chapter Four

I


Because the number of moles of solute is the same in both calculations (ninitial= nfinal),we can combine these two steps into one calculation in the form of an equation that applies to all situations in which we know any three of the .. I); an initial solute concentration four following variables: an initial volume (Vmltla (M mltla .. I = C100tia .. I); a final volume Wfi I); and a final solute concentration (Mfi na I = na Cfinal).These variables are related by the following equation: V:nitial

X

Minitial

=

=

ninitial

=

nfinal

Vfinal

X

Mfina1

or for the most general case, V:nitialX Cinitial= (amount of solute\nitial Simplifying

= (amount of solute)final = ~nal X Cfinal

this produces (4.5)

V:nitialX Cinitial= ~nal X C final

Equation 4.5 works because each side of the equation represents a specific quantity of solute, and that quantity does not change because of dilution. Furthermore, this equation can be used for any units of volume and concentration, as long as the units used to express the initial and final volumes are the same, and the units for the initial and final concentrations are the same. The usefulness ofEquation 4.5 becomes clear when we apply it to the following calculation for a system on a much larger scale: determining the concentration ofNa+ ions in a coastal bay when the salinity is reduced by fresh water flowing in from rivers. Where large rivers flow into the sea, the salinity of seawater is reduced because it is diluted by the volume of fresh river water that enters it. Suppose a coastal bay contains 3.8 X 1012 L of ocean water mixed with 1.2 X 1012 L of river water for a total volume of 5.0 X 1012 L. What is the concentration ofNa+ ions in the bay water, assuming the open ocean water is 0.48 M Na+? Assume that the fresh river water has an insignificant amount of sodium ion in it and as such does not contribute to the Na+ ion content of the bay water. Solving Equation 4.5 for Cfinaland inserting the values just given, we have: C

= V:nitialX Cinitial= (3.8 final

12

L) X 0.48 M 5.0 X 1012 L

X 10

Vfinal

= 0.36 M

The concentration of sodium ion in the bay is 0.36 M. It is lower than the concentration of sodium ion in ocean water (0.48 M) because bay water is diluted by the addition of fresh water from the rivers.

Calculating

Dilutions Using

Vinitial X Cinitial

=

Vllnal X Cllnal

The solution used in hospitals for intravenous infusion is called physiological saline ("saline solution"). It is 0.155 Min NaCl. It may be purchased as a more concentrated stock solution: typically 5.28 L of stock are diluted with water to make 60.0 L of 0.155 M NaCl. What is the initial concentration of NaCI in the stock solution? AND ORGANIZE We know the volume (Vfinal= 60.0 L) and concen(Cfinal= 0.155 M) of the diluted solution of NaCI, and the volume of the solution (V:nitial= 5.28 L) before dilution. We are asked to calculate the concentration of the stock solution before dilution. These four variables are by Equation 4.5:

COLLECT

tration initial initial related

V:nitialX Cinitial

VfinalX Cfinal

4.4

I

Electrolytes and Nonelectrolytes

I

143

A N A L Y Z E The two volumes are expressed in the same units so no unit conver-

sions are needed. We do need to rearrange the terms in Equation 4.5 to solve for Cinitial"

SOLVE Cfina1

Solving Equation 4.5 for gives us the following: C.

initial

=

v;,nal X

C

Vmitial

final

Cinitial

and inserting the values of

60.0 b

X

0.155

Mfina1

=

1.76

V;nitiaI' v;,naI'

and

Minitial

5.28b

TH IN K ABOUT IT The answer makes sense because the volume of the stock solu-

tion is about one-tenth the volume of the diluted solution, so its concentration should be about 10 times greater than 0.155 M.

Practice Exercise The concentration of Pb2+ in a commercially available standard solution is 1.000 mg/mL. What volume of this solution should be diluted to 500.0 mL to produce a solution in which the concentration

of Pb2+ is

0.0575 mg/L? (Answers to Practice Exercises are in the back

CONCEPT

of the book.)

TEST

Bags of physiological saline (see Sample Exercise 4.5) are usually labeled "0.90%

w/V sodium chloride," where w/V means that the percentage defines the mass of NaCI per unit volume of solution (g/100 mL). Show that this concentration is approximately 0.155 M. (Answers to Concept Tests are in the back of the book.)

~

Electrolytes and Nonelectrolytes

The high concentrations of NaCI and other salts in seawater (Table 4.1) make it a good conductor of electricity, as demonstrated by the following experiment. Suppose we immerse two electrodes (metal wires) in a sample of distilled water (Figure 4.7a) and connect these wires to a battery and a lightbulb. For electricity to flow and the bulb to illuminate, the circuit must be completed by mobile charge carriers (ions) in the solution. The lightbulb does not light up when the electrodes are placed in distilled water, because there are very few ions in distilled water. If the electrodes are immersed in 0.50 M NaCl (used in this demonstration to approximate seawater), the bulb lights up as shown in Figure 4.7(b) because the abundant ions in the solution carry the electric current between the two electrodes. The saline solution conducts electricity because Na+ ions are attracted to and migrate toward the electrode connected to the negative terminal of the battery, while CI- ions migrate toward the positive electrode. These migrating ions take their electric charges with them, and the migrating charges carry electric current through the water. The electric charge passes through the connecting wires and lightbulb in the form of a flow of electrons. Any solute that imparts electrical conductivity to an aqueous solution is called an electrolyte. Sodium chloride is considered a strong electrolyte because it completely dissociates in water into its component ions when it dissolves, creating an abundance of charge carriers. However, not all solutes dissociate completely

An electrolyte is a substance that dissociates into ions when it dissolves, enhancing the conductivity of the solvent. A strong electrolyte is a substance that dissociates completely into ions when it dissolves in water.

144

I

Chapter Four

I


(b)

Cl-

Acetate ion (CH3COO-)

a Hydrogen FIGURE

4.7

ion (H+)

What a Chemist Sees

The solutions in this demonstration the same molar concentration

all contain

of solute. (a)

Pure water (no solute) conducts very poorly because water is a molecular

material and

contains very few ions. (b) A 0.50 M solution of sodium chloride (NaCl), approximating seawater, conducts electricity very well. (c) A

0.50 M solution of ethanol (CH3CH20H) does not improve the conductivity

of water. (d)

A 0.50 M solution of acetic acid (CH3COOH) conducts electricity

better than pure water or

into ions upon dissolution. Solutions of substances like sugar or an alcohol such as ethanol (CH3CH20H) in water (Figure 4.7c) conduct electricity no better than pure water. These solutes do not form ions in aqueous solutions; they are nonelectrolytes in water. If we pour vinegar (an aqueous solution of acetic acid) in the apparatus in Figure 4.7(d) to prepare a concentration of acetic acid in the solution equivalent to that of the other solutes shown in the figure, the lightbulb lights up but not very brightly. The acetic acid molecules ionize, forming acetate ions and hydrogen ions, but the process does not go to completion:

the ethanol solution but to a much smaller

(4.6)

extent than the NaCI solution. A small fraction of the acetic acid molecules dissociate into acetate and H + ions; this partial ionization

is characteristic

of weak electrolytes

like acetic acid.

~II CHEMTOUR Migration in Solution

oflons

A nonelectrolyte is a substance that does not form ions and does not enhance the conductivity of water when it dissolves. A weak electrolyte only partly ionizes when it dissolves in water.

The small proportion of acetic acid molecules that ionize makes acetic acid a weak electrolyte. The arrow with the top half pointing right and the bottom half pointing left in Equation 4.6 indicates that all species are present in solution: the hydrogen ion, the acetate ion, and un-ionized, molecular acetic acid. This type of arrow is always used to symbolize the incomplete ionization of a weak electrolyte.

CONCEPT

TEST

Why does the demonstration in Figure 4.7 specify equivalent molar concentrations of solutes in the solutions whose conductivity is being compared? (Answers to Concept Tests are in the back of the book.)

4.5

I

Acid-Base

Reactions

I 145

11III Acid-Base Reactions Volcanic activity introduces a variety of gases into the atmosphere, including volatile oxides of sulfur, nitrogen, and other nonmetals. When these oxides dissolve in rainwater, they produce hydrogen ions in solution. We define acids as materials that increase the concentration of hydrogen ions in aqueous solution. To understand the chemistry of acids more deeply, we focus in this section on the processes that occur in solution when hydrogen ions are produced. Let's look more closely at the behavior of a binary acid like hydrochloric acid (HCI). Pure hydrogen chloride is a molecular gas. However, when it dissolves in water, it behaves like a strong electrolyte and ionizes completely to give a solution containing H+(aq) and Cl-(aq). Generally we use H+(aq) to describe the cation produced in an aqueous medium by an acid. However, the hydrogen ion (H+, also called the proton) does not have an independent existence in water because it associates very strongly with a water molecule to form H3 0+. The hydronium ion (H30+) also associates with other water molecules, so it is written H30+(aq). We write the hydrogen ion in water this way when we want to emphasize the role of water, and we describe the ionization of HCI by saying that it donates a proton to a water molecule. Acids are proton donors. Because the water molecule accepts the proton, water in this case is a proton acceptor and so is a base. These definitions were originally proposed by two chemists, Johannes Bf0nsted and Thomas Lowry, and the acids and bases so identified are called BrrJnsted-Lowry acids and bases. We will develop this concept much more completely later, but for now it provides a convenient way to define acids and bases in aqueous solutions of ions. For the reaction between HCI and water, HCI(g) proton donor (acid)

+

(4.7)

H20(e) proton acceptor (base)

The simplest definition of a base is a substance that produces hydroxide ions in an aqueous solution. With the new definition of base as a proton acceptor, the hydroxide ion is still classified as a base, but other substances can now be categorized as bases as well. By focusing on the behavior of the hydrogen ion, we have expanded our definition of acids and bases. If we look at the reaction that occurs when an aqueous solution of HCI reacts with an aqueous solution of NaOH [which is completely dissociated to give Na+(aq) and OH-(aq) ions] HCI(aq)

+ NaOH(aq)

~ NaCI(aq)

@Xi) CONNECTION We defined hydrolysis reactions in Chapter 3. The nonmetal oxides not only dissolve in water; they also react with water (hydrolyze) to produce ions. Water is both a solvent and a co-reactant.

+ H20(e)

(4.8)

we see that the HCI functions as an acid by donating a proton to the hydroxide ion, which functions as a base and accepts the proton; the two ions combine chemically to form a molecule of water. The remaining two ions-the anion characteristic of the acid (CI-) and the cation characteristic of the base (Na+)-form a salt, sodium chloride, which remains dissolved in the aqueous solution. This reaction is called a neutralization reaction because both the acid and the base have been eliminated as a result of the reaction. The products of the neutralization of a strong acid like HCI with a strong base like NaOH are always water and a salt. This provides us with a new definition of a salt as a substance formed along with water as a product of a neutralization reaction.

The hydronium ion (H30+) is the form in which the hydrogen ion is found in an aqueous solution. Hydronium ion, hydrogen ion, and proton are synonymous; all are used to describe the hydrogen ion produced by acids in aqueous solution. A Bmnsted-Lowry

acid is a proton donor.

A Bmnsted-Lowry acceptor.

base is a proton

A neutralization reaction takes place when an acid reacts with a base and produces a solution of a salt in water. A salt is the product of a neutralization reaction; a salt is made up of the cation characteristic of the base and the anion characteristic of the acid in that reaction.

146

I Chapter Four

I


A molecular equation describes a reaction in solution in which the reactants are written as undissociated molecules. An overall ionic equation shows all the species, both ionic and molecular, present in a reaction occurring in aqueous solution. A net ionic equation describes the actual reaction taking place in aqueous solution, and is found by eliminating the spectator ions from the overall ionic equation. Spectator ions are present in, but remain unchanged by, the reaction taking place.

Equation 4.8 shows a neutralization reaction written in the form of a molecular equation. Each of the participants in the reaction is written as a neutral molecule. The subscript aq after HCI, NaOH, and NaCI indicates that these substances are dissolved in water. Any undissociated molecules in aqueous solution are also typically written with an aq subscript to indicate they are hydrated (surrounded by water molecules). Molecular equations are sometimes the easiest equations to balance, so reactions involving substances that dissociate to form ions in solution are frequently written first as molecular equations to indicate clearly the formulas of the compounds involved and to simplify balancing the equation. Another way to write Equation 4.8 indicates the species actually present in solution. Called an overall ionic equation, Equation 4.9 shows the ions formed by substances in the reaction that dissociate in water:

+

HCI(aq)

NaOH(aq)

NaCI(aq)

Overall ionic equations clearly distinguish ionic substances from molecular substances in a chemical reaction taking place in solution. Now think of Equation 4.9 as you would an algebraic equation. Notice that chloride ions and sodium ions appear on both sides as hydrated ions. If this were an algebraic equation, we would cross out the terms that are the same on both sides of the equation. When we do the same thing in chemistry, the result is called a net ionic equation. The ions that are removed (in this case, sodium ion and chloride ion) are spectator ions, and the resulting equation (4.10) focuses only on the actual chemical change that occurs: Overall ionic equation:

H+(aq)

+ CI-~fl+, is incorporated kaolinite

structure

then the difference kaolinite

ions.

ion,

charge. This negative

by the inclusion

such as Na+ in the spaces between of hydroxide

AI3+

in their charges gives the

a net negative

charge is balanced

into the

in place of an

of a cation the layers

Thus the overall result of the ion-exchange process is the replacement of Na+(aq) and Cl-(aq) in the original water supply by H20(t'). Similar reactions can be written for other ionic species. The liquid product that flows from the ion-exchange cartridge is essentially pure, deionized water. The ion-exchange reaction proceeds as long as both H+[RC]- and [RA]+OH- reaction sites are available. The cartridges are replaced when most of the sites have been consumed by the ion-exchange process (often indicated by a change in the color of the resin). Some naturally occurring minerals known as aluminosilicates function as ion exchangers, too. Zeolites are porous materials in this family that consist of threedimensional networks of channels made of aluminum, silicon, and oxygen. The channels also contain sodium ions or other 1+ cations. The 1+ cations are not part of the rigid structure and are exchangeable for other ions. When hard water passes through a zeolite bed, one 2+ ion can exchange for two sodium ions and thereby soften the water. Synthetic zeolites are being added to some detergents now for exactly this purpose. The structure of many other aluminosilicates found in Earth's crust feature layers of two-dimensional sheets made up of silicon and oxygen, or aluminum and oxygen, bonded together (Figure 4.13). The multilayered structure of these minerals gives them distinctive physical and chemical properties. Other properties may be related to imperfections in their molecular structures due to the presence of cations oflower charge in sites normally occupied by AP+ ions. In many aluminosilicates, Fe2+ or Mg2+ ions substitute for AP+ in the AI-O layer (Figure 4.14). These substitutions mean that the structure is deficient in positive charge and thus has a negative charge overall. In nature this charge is balanced by additional positively charged ions, such as Na+ or K+, that reside in the gaps between aluminosilicate layers, similar to the way 1+ ions reside in the three-dimensional channels of zeolites. Because these ions are not held tightly in place, they can be replaced by other positive ions. For example, H+ ions in acidic groundwater can replace Na+ ions in an ion-exchange reaction that removes hydrogen ions from solution, which also makes it a neutralization reaction.

4.8

~

Oxidation-Reduction

Reactions

One of the earliest descriptions of the properties of oxygen came from Leonardo da Vinci in the 15th century: "Where flame cannot live, no animal that draws breath can live." Although oxygen was not isolated and recognized as an element until over 250 years later, da Vinci captured an important property of the substance in his observation. Oxygen makes up about 50% by mass of Earth's crust, it is almost 89% by mass of the water that covers 70% of Earth's surface, and it is about 20% of the volume of Earth's atmosphere. It is essential to the metabolism of all creatures that "draw breath" just as it is essential for combustion reactions. Its name is applied to perhaps the most important type of chemical processes implied in da Vinci's statement: oxidation-reduction reactions. On early Earth the reactions of oxygen with sulfur and nitrogen to produce acid anhydrides in the atmosphere and with metallic elements to produce the minerals of the crust are illustrations of this type of reaction. Oxidation was first defined as a reaction that increased the oxygen content of a substance. Hence two reactions involving oxygen-with methane [CH4(g)] in the atmosphere of early Earth to produce CO2(g) and H20(e), and with elemental iron as Earth's crust both oxidations; the products contain more oxyformed to produce Fe203(s)-are gen than CH4 or Fe:

CH/g) 4Fe(s)

I

Oxidation-Reduction

Oxidation is a chemical change in which a species loses electrons. The oxidation number of the species increases. Reduction is a chemical change in which a species gains electrons. The oxidation number of the species decreases.

3 O2 (g) ~ 2 Fe203(s)

Reduction reactions were originally defined in a similar fashion-as reactions in which the oxygen content of a substance is reduced. A classic example of a reduction reaction is the processing of iron ore to produce pure metallic iron:

The oxygen content of the ore [Fe203(s)] is reduced as a result of its reaction with CO(g). Note that in this process the carbon atom in CO(g) by this definition is oxidized-the oxygen content of the carbon species has been increased. Ultimately, these first definitions of oxidation as the addition of oxygen to another substance and reduction as the removal of oxygen proved too restrictive. Oxidation now refers to any chemical reaction in which a substance loses electrons. Similarly, reduction is a gain of electrons.2 Oxidation and reduction always occur together: If one substance loses electrons, then another substance must gain them; if one species is oxidized, then another must be reduced. The crucial event in oxidation-reduction is this movement of electrons from one atom or ion to another. The recognition that oxidation and reduction must occur together gives rise to the common way of referring to them as redox reactions.

Oxidation Numbers Some redox reactions are easy to identify: all combustion reactions, all reactions that generate electricity in batteries or that take place when an electric current is passed through a solution, and all reactions of elements that combine to make compounds are redox reactions. Beyond that, an unambiguous way to determine whether a reaction involves redox is by assigning an oxidation number (O.N.) to each element in the All fires are redox reactions. 2 The mnemonic OIL Reduction Is Gain."

RIG

may be helpful for remembering

these expanded definitions: "Qxidation Is Loss;

I 157

Zeolites are natural crystalline minerals or synthetic materials consisting of threedimensional networks of channels that contain sodium or other 1 + cations. They may function as ion exchangers.

+ 202(g) ~ CO2(g) + 2H20(t') +

Reactions

158

I Chapter Four

I


The oxidation number (O.N.) of an element in a molecule or ion, also called its oxidation state, is a positive or negative number based on the number of electrons that each atom of the element gains or loses when it forms an ion, or that it shares when it forms a covalent bond with another element. Pure elements have an oxidation number of zero.

reactants and products. If the oxidation number of an element changes as a result of a chemical reaction, then the reaction is definitely redox. The oxidation number (O.N.), or oxidation state, of an element in a molecule or ion is a positive or negative number whose value depends on the number of electrons that each atom of the element gains or loses when it forms an ion, or the number of electrons that it shares when it forms a covalent bond with another element. The O.N. value represents the charge an atom has, or appears to have, as determined by the following set of guidelines:

1. The oxidation numbers of the elements in a neutral molecule sum to zero; those on the elements in an ion sum to the charge on the ion.

2. Each atom in a pure element has an oxidation number of zero. For example:

@Xi)

CONNECTION In Chapter 2 we discussed the naming of compounds containing ions of transition metals that could have more than one charge. The roman numerals in symbols such as Fe(II) and Fe(III) identify the different oxidation numbers of the iron ion.

F2: O.N. ofF

=0

Fe: O.N. of Fe

=0

O2: O.N. of 0

=0

Na: O.N. ofNa

=0

3. For monatomic

(one-element) charge on the ion. Examples:

ions, the oxidation number is equal to the

F-: O.N.

= -1

Fe3+: O.N.

= +3

02-: O.N.

= -2

Na+: O.N.

= +1

Note that the number precedes the sign of the charge in, for example, Fe3+, but the sign comes first in writing an oxidation number, as in "the O.N. of Fe in Fe203 is +3." 4. In compounds with other elements, fluorine is always -1. Examples: KF: Fis -1; Kis +1 OF2: each F atom is -1; 0 is +2 CF4: each F atom is -1; Cis +4 5. The oxidation number of hydrogen is +1 and oxygen is -2 in most compounds. Common exceptions are the hydrogen in metal hydrides (for example: LiH) where it is -1; the oxygen in the peroxide ion (022- as in hydrogen peroxide: H202) where it is -1. 6. Unless they are combined with oxygen or fluorine, the halogens chlorine, fluorine, and bromine are -1. Examples: CaCI2: O.N. ofCI

= -1; O.N. ofCa = +2

CI04-:

= -2; O.N. ofCI = +7

O.N. of 0

IFs: O.N. ofF

= -1; O.N. of I = +5

Determining What is the oxidation (b) Na2S; (c) CaSO 4

Oxidation Numbers

number of sulfur in each of these compounds?

(a) S02;

AND ORGANIZE We are asked to assign O.N. values to sulfur in three of its compounds. To do so we apply the guidelines on this page for determining the oxidation states of the elements in these compounds.

COLLECT

One compound, S02' is molecular because its components are both nonmetals. Na2S is a simple binary ionic compound, and the oxidation number of an element in an ionic compound equals the charge of its ion. Sulfur in CaSO 4 is

ANALYZE

4.8

I

Oxidation-Reduction

Reactions

I 159

part of the sulfate ion, so the O.N. values of the elements in the ion must add up to the charge of the ion. SOLVE

a. For the oxygen in S02' we automatically assign an O.N. value of -2. For sulfur, let O.N. = x.The sum of the O.N. values of the sulfur atom and the two oxygen atoms must add up to zero: x

+ 2(-2) =

0

or

x=

+4

b. Sodium is present in all its compounds as Na+. This means that the O.N. ofNa is +1. To balance the O.N. values in Na2S, we can use the following equality, and solve for y, the O.N. of sulfur: 2(+1)

+y =

or

0

y =-2

c. The charge on the calcium ion is always 2+. This means that the charge on the sulfate ion must be 2-. Assigning O.N. values of -2 for oxygen and z for sulfur, we have the following equality for the sulfate ion: z

+ 4(-2) =

-2

z = +6

or

Sulfur has oxidation numbers that range from -2 to +6 in various molecules and ions. Unlike H, Na, and Ca, there is no oxidation number rule for S (or for most other elements). We must derive their O.N. values by applying the guidelines and considering the oxidation numbers of the other elements in the molecule or ion. THINK

ABOUT

IT

Practice Exercise Determine the oxidation number of nitrogen in the following compounds: (a) N02; (b) N20; (c) HN03 (Answers to Practice Exercises are in the back

of the

book.)

Examples of Redox Reactions The combustion of hydrocarbon fuels is a classic redox reaction because it involves oxygen as a reactant. In the complete combustion of a fuel like methane (CH4), the carbon is completely converted into CO/g) and the hydrogen to H20(t'). Combustion reactions of hydrocarbons can often be balanced by first balancing the moles of carbon on both sides of the reaction arrow, then the moles of hydrogen, and finally the moles of oxygen. However, like all redox reactions, hydrocarbon combustion reactions can also be balanced by balancing the changes in oxidation numbers of the elements in the reactants and products. Let's use this second approach with the combustion of methane. First we assign O.N. values to the atoms in the reactants and products:

-4

I

oxidation

+4

CH.(g)+ 2 0E:(g)

o

• +

(4.15)

reduction

These values tell us that carbon is oxidized because each carbon atom loses 8 electrons in going from a -4 oxidation state in methane to a +4 oxidation state

@@> CONNECTION In Chapter 3 we defined complete combustion of a hydrocarbon as a reaction with oxygen that converts a compound of carbon and hydrogen into carbon dioxide and water.

160

I Chapter Four

I


The oxidizing agent in a redox reaction accepts electrons from another species, thereby oxidizing that species (increasing its oxidation number). The oxidizing agent itself is reduced (decreasing its oxidation number). The reducing agent in a redox reaction gives up electrons to another species, thereby reducing that species (reducing its oxidation number). The reducing agent itself is oxidized (increasing its oxidation number).

in carbon dioxide. Remember that electrons have a charge of 1-; so, going from -4 to +4 involves a loss of 8 electrons. Oxygen, on the other hand, is reduced: The O.N. value for oxygen changes from 0 to -2 since each of the 4 oxygen atoms gains 2 electrons. To balance the gains and losses of electrons, we need 4 atoms of 0 for every atom ofC. The coefficient of2 in front of O2 in Equation 4.15 gives us the 4:1 ratio we need. We also need a coefficient of2 in front of H20 to balance the moles of oxygen and hydrogen. Note that hydrogen is neither oxidized nor reduced in this reaction. Because oxygen in Equation 4.15 is the substance that takes the electrons from the carbon in methane, oxygen is called the oxidizing agent in the combustion reaction. We say that "oxygen oxidizes methane." In the process of functioning as an oxidizing agent, the oxygen is reduced. This is always the case: the agent that takes electrons away from some material is reduced in the process. Any species that is reduced experiences a reduction in oxidation number; oxygen goes from O.N. = 0 to O.N. = -2. By the same token, the carbon in methane is the species being oxidized; as such, it gives electrons to another substance and reduces it. Methane in this reaction is called the reducing agent because it gives electrons to another species (oxygen in this case), thus causing that species (oxygen) to be reduced. The reducing agent is always oxidized. The material that is oxidized experiences an increase in oxidation number; in this reaction, carbon goes from O.N. = -4 to O.N. = +4. Every redox reaction has an oxidizing agent and a reducing agent. Another classic redox reaction that can be balanced by inspection is the reduction of iron ore [as Fe20/s)] to metallic iron. Iron ore is heated with carbon monoxide, and the products of the reaction are elemental iron and carbon dioxide: +3

I

reduction

Fe20/s)

+ 3 CO(g) I +2

0

+

~ 2 Fe(s) + 3 CO/g)

t oxidation

+4

The iron in the iron ore is reduced from +3 to 0; the carbon in carbon monoxide is oxidized from +2 to +4. The iron ore is the oxidizing agent; the carbon monoxide is the reducing agent. Two iron atoms with O.N. = +3 become 2 iron atoms with O.N. = 0, for a total gain of 6 electrons; 3 carbon atoms at +2 become 3 carbon atoms at +4, for a total loss of 6 electrons. The same number of electrons is gained as lost. Note that the oxygen atoms are not involved in this redox reaction; carbon is oxidized, iron is reduced.

Identifying Oxidizing Agents and Reducing Agents One of the reactions of oxygen in the atmosphere of early Earth was with S02(g) to produce SO/g). Identify the species oxidized and the species reduced in this reaction. Also identify the oxidizing agent and the reducing agent. COL LEe TAN

D 0 R GA N I Z E

product is sulfur trioxide.

The reactants

are oxygen and sulfur dioxide; the

4.8

A N A L Y Z E The equation

I

Oxidation-Reduction

Reactions

I 161

of the reaction is: SO/g)

+

O/g) ~ SO/g)

2 S02(g)

+

02(g) ~ 2 S03(g)

Unbalanced: Balanced:

By inspection we can see that oxygen is the oxidizing agent because S02 is oxidized. That means that oxygen is reduced and S02 is the reducing agent. We verify this by assigning oxidation numbers: SOLVE

+4 oxidation +6 I • 2 SO/g) + 02(g) ~ 2 S03(g)

I

o reduction

+

(4.16)

- 2

ABO UTI T Note that not all of the atoms in a reaction participate in the transfer of electrons. We tend to identify the specific atoms that are oxidized or reduced, while whole molecules or ions are identified as the oxidizing agents or reducing agents. In Equation 4.16, sulfur is oxidized but sulfur dioxide is the reducing agent.

T H INK

Practice Exercise The following reaction is used to remove dissolved oxygen gas from aqueous solutions:

Identify the species oxidized and the species reduced in this reaction. Also identify the oxidizing agent and the reducing agent. (Answers


Balancing Redox Reactions Using Half-Reactions Consider the following reaction in aqueous medium. When a piece of copper wire [elemental copper: Cu(s)] is placed in a colorless solution of silver nitrate [Ag+(aq) + N03 -(aq)], the solution gradually turns blue, the color of a solution ofCu2+(aq), and beautiful branchlike structures of pure silver [Ag(s)] form in the medium (Figure 4.15). The process can be summarized as follows:

+1

I

reduction

Ag+(aq)

in a solution of AgN03, Cu metal starts to oxidize to Cu2+ ions as Ag+

•

+ Cu(s) ~ Ag(s) + Cu2+(aq)

o

(4.17)

+

oxidation

FIGURE 4.15 (a) When a helix ofCu wire is immersed

0

I

(b)

(a)

+2

At first glance, this reaction may seem balanced-and in a material sense it is: the number of silver and copper atoms or ions is the same on each side. However, the charges are not because one electron is involved in the reduction reaction while two are involved in the oxidation reaction. Equal numbers of electrons must be lost and gained, so we must develop a method to balance redox reactions like this one that accounts for electron transfer. The method we apply is called the half-reaction method. Think of the reaction in Equation 4.17 as consisting of two reactions: one oxidation and one reduction. Each of these reactions represents half of the overall reaction; hence, each is called a half-reaction.

ions are reduced to Ag metal. (b) A day later the solution has the blue color of a solution ofCu(NO)2

and the wire is coated with Ag

metal.

Oxidation-reduction reactions consist of two half·reactions, one for the oxidation component and one for the reduction component.

162

I Chapter Four

I


1. Write the separated reactants and products: Cu(s) ~ Cu2+(aq)

Oxidation:

Ag+(aq) ~ Ag(s)

Reduction:

2. Balance the number ofparticles on both sides ofboth equations: This example requires no changes at this point, because both equations are balanced in terms of mass. Each species has an understood coefficient of 1:

Cu(s) ~ Cu2+(aq)

Oxidation:

Ag+(aq) ~ Ag(s)

Reduction:

3. Balance each half-reaction in terms ofcharge by adding electrons to the appropriate side to make the charges on both sides equal. Always add electrons; never subtract them. Remember that electrons have a negative charge:

Cu(s) ~ Cu2+(aq) + 2e-

Oxidation:

total charge

total charge

= (2+) + (2-) = 0

1 e- + Ag+(aq) ~ Ag(s)

Reduction: total charge

=0

= (1-) + (1+) = 0

total charge

=0

Note that electrons are added on the reactant side of the reduction halfreaction and on the product side of the oxidation reaction. This will always be the case in this step. 4. The number of electrons lost in oxidation must equal the number gained in reduction. Multiply each half-reaction by the appropriate whole number to make the number of electrons in the oxidation half-reaction equal the number in the reduction half-reaction: 1 X [Cu(s) ~

Oxidation: Reduction:

2

X

[1 e-

Cu2+(aq)

+ 2e-]

+ Ag+(aq) ~ Ag(s)]

As a result we have the following: Oxidation: Reduction:

2e- + 2 Ag+(aq) ~ 2 Ag(s)

5. Add the two half-reactions together to generate the overall reaction. The electrons must cancel:

Cu(s) ~ Cu2+(aq)

Oxidation: Reduction: Overall reaction:

~

+~

+ 2 Ag+(aq) ~ 2 Ag(s)

2 Ag+(aq)

+ Cu(s) ~ 2 Ag(s) + Cu2+(aq)

The reaction is now balanced in terms of mass and electron transfer. The final equation is a net ionic equation. Only those species that are involved in the redox reaction-those losing or gaining electrons-are included. The nitrate ion mentioned at the beginning of this subsection is a spectator ion in this process because it does not participate in the chemical reaction; hence it does not show up in Equation 4.17 or any of the subsequent half-reactions we used to balance Equation 4.17.

4.8

Balancing Redox Reactions by the Half-Reaction Method Iodine is slightly soluble in water and makes a pale yellow-brown solution of I2(aq). When Sn2+(aq) is dissolved in an iodine-containing solution, the solution turns colorless as I-(aq) forms and Sn2+(aq) is converted into the Sn4+(aq) ion. (a) Is this a redox reaction? (b) Balance the equation that describes the process taking place. COLLECT

AND

The reactants in this process are I2(aq) and Sn2+(aq); and Sn4+(aq).

ORGANIZE

the products are I-(aq) ANALYZE

(O.N.

=

The oxidation numbers change from Sn2+(aq) (O.N. = +2) to Sn4+(aq) +4), and from I2(aq) (O.N. = 0) to I-(aq) (O.N. = -1).

SOLVE

a. Because the oxidation numbers change, this is definitely a redox reaction. b. The unbalanced equation showing reactants and products is Sn2+(aq)

+ I2(aq)

~ I-(aq)

+

Sn4+(aq)

We will balance the equation following the steps for the half-reaction method: (1)

(2)

Oxidation:

Sn2+(aq)

Reduction:

I2(aq)

I2(aq)

~ 2 I-(aq)

Sn2+(aq) ~ Sn4+(aq) + 2 e-


~ I-(aq)

Sn2+(aq) ~ Sn4+(aq)


(3)

~ Sn4+(aq)

2e- + I2(aq)

~ 2 I-(aq)

(4)

(No multiplication needed: numbers of electrons are the same in both reactions.)

(5)


Sn2+(aq) ~ Sn4+(aq) + ~ ~

+ I2(aq) ~ 2 I-(aq)

Overall reaction: Sn2+(aq) + I2(aq) ~ 2 I-(aq)

+ Sn4+(aq)

Notice in Step 4 that the total charge in the oxidation halfreaction does not have to be zero, but it must be the same on both sides. Always check the final overall reaction for mass balance (here, 1 Sn + 2 I = 2 I + 1 Sn) and the charge balance [here the reactant side = 2+, and the product side = 2(-1) + (+4) = +2].

THINK

ABOUT

IT

Practice Exercise If a nail made of elemental iron [Fe(s)] is placed in an aqueous solution of a soluble palladium(II) salt [Pd2+(aq)], the nail will gradually disappear as the iron enters the solution as Fe3+(aq) and palladium metal [Pd(s)] forms. (a) Is this a redox reaction? (b) Balance the equation that describes this reaction. (Answers

to Practice Exercises are in the back

of the

book.)

I

Oxidation-Reduction

Reactions

I 163

164

I

Chapter Four I Solution Chemistry and the Hydrosphere

Redox in Nature

FIGURE 4.16 The orange-red strata in Bryce Canyon consist of oxides of the iron(III) ion that result from weathering of the rock.

(a)

(b)

FIGURE 4.17 Wetland soils: (a) the soil in this wetland has a blue-gray color because of the presence ofiron(II) compounds. (b) Orange mottling indicates the presence ofFe(III) oxides formed as a result of02 permeation through channels made by plant roots.

Just as acid-base reactions and precipitation reactions are involved in the chemical weathering of rocks, redox processes also playa major role in determining the character of rocks and soils. For example, oxygen reacts with iron to form iron oxide minerals such as hematite. The red-colored rocks like those seen in Bryce Canyon National Park in Utah are examples of iron(III) oxidecontaining minerals (Figure 4.16). Iron(III) oxide is also the form of iron known as rust, the crumbly red-brown solid that forms on objects made of iron that are exposed to water and oxygen. Rocks containing hematite are relatively weak and easily broken, a fact which gives rise to the fascinating shapes of such deposits. The color of the soil in an area is also influenced by its mineral content. Wetlands, areas that are saturated or flooded with water during much of the year, are protected environments in several areas of the United States. Defining a given area as a wetland relies in part on the characteristics of the soils in the area, and the color of the soil is an important diagnostic feature in identifying an area as a true wetland (Figure 4.17). The color of soil is strongly influenced by the frequency and duration of periods of water saturation, primarily because soil saturated with water is a reducing environment in which free oxygen is prevented from contacting the minerals. Iron-containing components of the soil are reduced in such an environment: the soil is gray with bluish-green tints (Figure 4.17a), in contrast to the red color associated with Fe(III) compounds. When the roots of plants penetrate this soil, they create pores in the soil. This allows iron in its reduced form to come into contact with oxygen diffusing through the pores; as a result the iron becomes oxidized as the familiar orange- or red-brown iron(III) compounds (Figure 4.17b). The redox reaction of iron (II) and iron(III) compounds in the soils of wetlands can be modeled in the laboratory. Figure 4.18(a) shows a solution of ferrous ammonium sulfate [(NH4\Fe(SO 4\(aq)]. The greenish color of the solution is characteristic of the iron(II) ion in its compounds and in an aqueous solution. When a solution of sodium hydroxide is added, Fe(OH)2(s) precipitates out of solution (Figure 4.18b) as a grayish olive-green solid, very similar in color to the wetland soil shown in Figure 4.17(a). The solid remains that color as long as it is submerged in water. When the solution is poured onto a paper towel (Figure 4.18c), however, the iron(II) compound is exposed to free oxygen in the air. Within about 20 minutes (Figure 4.18d) the green precipitate turns the orangebrown color of iron (III) hydroxide. The iron(II) has been oxidized-exactly the way the iron(II) compounds in a wetland soil are oxidized when exposed to free oxygen. The color of iron-containing soil is a reliable indicator of the extent to which soil in an area is exposed to free oxygen. Because oxygen gas is not very soluble in water, the soil color in an area is therefore a good index of the saturation level of water in the ground and is useful data for assisting in the classification of true wetlands. Balancing the reaction for the conversion ofiron(II) to iron(III) in an aqueous base requires us to revisit the half-reaction method and add some further steps. Many redox reactions take place in acids and bases, and the acid [as H+(aq)], the base [as OH-(aq)], or even the water in the medium may playa role in the reaction. Let's look at a reaction used in analyzing solutions containing iron before we tackle the one demonstrated in Figure 4.18. A method for determining the concentration of Fe2+(aq) in an acidic solution involves its oxidation by adding the intensely purple-colored permanganate ion

4.8

I

Oxidation-Reduction

Reactions

I 165

[MnO 4 -(aq)] to Fe3+(aq). In the course of the reaction, the permanganate ion is reduced to Mn2+(aq), which is very pale pink. Here is the unbalanced reaction: Fe2+(aq)

~ Fe3+(aq)

+ Mn04-(aq)

+ Mn2+(aq)

To balance the equation for this reaction, we follow the steps outlined in the half-reaction method. The first step is the familiar step 1 from before: (1)

Fe2+(aq) ~ Fe3+(aq)

Oxidation:

Mn04-(aq)

Reduction:

(b) Fe(OHMs)

~ Mn2+(aq)

Step 2 needs to be broken down into substeps to account for any role played by the aqueous acid: First we balance all the elements except hydrogen and oxygen: Fe2+(aq) ~ Fe3+(aq)

(2a) Oxidation:

Mn04-(aq)

Reduction:

In this case the iron and the manganese are already balanced, so we move to the second substep: balancing any oxygen present by adding water to the reaction. This sub step is appropriate because it accounts for any role played by water in the reaction:

(No oxygen, so this half-reaction

Mn04-(aq)

Reduction:

stays the same.)

~ Mn2+(aq)

+ 4 H20(e)

~ Mn2+(aq)

(. .. so we need to add water.) In the final sub step we balance any hydrogen by adding H+(aq); this is appropriate because the reaction is run in acidic medium: Fe2+(aq) ~ Fe3+(aq)

(2c) Oxidation: (No hydrogen,

so this half-reaction

Mn04-(aq)

Reduction:

(Hydrogen 8H+(aq) (...

+ Mn04-(aq)

stays the same.)

+ 4 H20(e)

~ Mn2+(aq)

is present ... )

+ 4 H20(e)

~ Mn2+(aq)

so we need to add hydrogen ions.)

Steps 3, 4, and 5 are the same as before: (3)

Reduction: (4)

5 e-

+

8 H+(aq)

+ MnO 5


(5)

Fe2+(aq) ~ Fe3+(aq)

Oxidation:

1 X [5 e-

+

8 H+(aq)

4 -(aq)

~ Fe3+(aq)

+ 1e-]

~ Mn2+(aq)

5 Fe2+(aq) ~ 5 Fe3+(aq)

&-e=-

Overall reaction: 8H+(aq) + Mn04-(aq)

+

8H+(aq)

+ Mn04-(aq)

+ 5 Fe2+(aq)

~5Fe3+(aq)

~ Mn2+(aq)

1e-

+ 4 H20(e)

-(aq) ~ Mn2+(aq)

X [Fe2+(aq)

+ MnO


4

+

+ 4 H20(e)] + &-e=+ 4 H20(e)

+ Mn2+(aq) + 4H20(e)

(a) A solution offerrous

sulfate. (b) The same solution

after the addition resulting

of NaOH(aq).

suspension

immediately

(c) The after being

poured onto a paper towel. (d) The same to air.

(Oxygen is present ... )

Mn04-(aq)

FIGURE 4.18 ammonium

precipitate

Fe2+(aq) ~ Fe3+(aq)

(2b) Oxidation:

(d) Fe(OH)is)

(c) Fe(OHMs)

~ Mn2+(aq)

after about 20 minutes of exposure

166

I Chapter Four

Acidic mine drainage

I


flowing into this

stream in western Pennsylvania high concentrations

contains

of iron that precipitate

as orange Fe(OH)3 when the drainage

mixes

Whenever you balance an equation, always check the mass balance and a charge balance as a final check. In this equation, the reactant side contains 8 H, 1 Mn, 40, and 5 Fe atoms and the product side shows 8 H, 1 Mn, 4 0, and 5 Fe, so the mass balance is correct. Now let's consider charge: in a net ionic equation the charges on both sides must be equal. On the reactant side the total charge is (8+) + (1-) + (10+) = 17+, and on the product side (15+) + (2+) = 17+, so the reaction is correctly balanced. This method works for balancing any redox reaction taking place in an aqueous acid. The demonstration of the oxidation ofFe(OH)2(s) to Fe(OH)/s) by oxygen in the air (Figure 4.18), however, was carried out in a basic medium. How do we account for that? Balancing a redox reaction in a basic medium is simple: we carry out the steps outlined above as though the reaction were run in an acidic medium. Then we add a final step in which sufficient hydroxide ion is added to both sides of the step converts the medium to an equation to convert any H+(aq) ions to H20(e).This aqueous basic solution, as specified in the reaction. Let's see how this works. As always, we start with an unbalanced equation to clearly define the reactants and products:

with less-acidic stream water.

Fe(OH)/s)

+

O/g) ~ Fe(OH)/s)

This reaction takes place in the solid phase in the presence of hydroxide ions in water. The oxidation half-reaction is clear: iron(II) hydroxide [Fe(OH)2(s)] is oxidized to iron(III) hydroxide [Fe(OH)/s)]. What is the reduction half-reaction? The oxygen must be converted into the additional OH- ion in the iron(III) hydroxide: (1)

Fe(OH)/s)

~ Fe(OH)/s)

02(g) ~ OH-(aq) H20(f)

(2a)

+ Fe(OH)/s) O/g)

H20(f)

(2b)

+ Fe(OH)/s)

3 H+(aq) (3)

H20(f)

+

4 X [H20(f)

+ Fe(OH)/s)

+ Fe(OH)/s)

1 X [4e- + 3 H+(aq) + O/g) 4 H20(f)

(5)

+ H20(f)

~ OH-(aq)

~ Fe(OH)3(s)

+ 4 Fe(OH)/s)

~ Fe(OH)/s) ~ OH-(aq)

+ H+(aq) + 1 e+ H20(f)

~ Fe(OH)/s) ~ OH-(aq)

+ H+(aq)

+ H20(f)

02(g) ~ OH-(aq)

4e- + 3 H+(aq) + O/g) (4)

~ Fe(OH)/s)

+ H+(aq) + 1 e-] + H20(f)]

~ 4 Fe(OH)/s)

4e= + 3 H+(aq) + 02(g) ~ OH-(aq)

+ 4 H+(aq) + 4e=

+ H20(f)

Overall reaction: 3

4H20(f)

+ 4 Fe(OH)/s)

+ 3 H+(aqj + 02(g) ~ 4 Fe(OH)/s)

1

+ 4H+(aq) + OH-(aq)

+ H,O(€)

We can simplify the equation the same way we would an algebraic one by cancelling out the same species on both sides: 2

JH20(f)

+ 4 Fe(OH\(s)

+ 02(g) ~ 4 Fe(OH)3(s) + ~ ,

+ OH-(affl

4.9

I

Titrations

I 167

This gives us the final balanced equation:

Checking this equation for mass, we find it is correct:

12 H

+ 12 0 + 4 Fe = 12 H + 12 0 + 4 Fe

as is the charge balance: 0 = O. The reaction in wetland soil is essentially the same as the one we solved here. The absence of oxygen in soil heavily saturated or flooded with water establishes a reducing environment that favors iron(II) compounds. When free oxygen becomes available, either because of dry conditions or because of porosity caused by the action of plant roots, the iron(II) species oxidizes quickly to red- or orangebrown iron(III) compounds.

~

Titrations

In Section 4.6 we discussed a method for determining the dissolved concentration of chloride ions in a sample of drinking water by precipitating Cl-(aq) ions as insoluble AgCI(s) and then weighing the precipitate. Another analytical approach, called a titration, is based on the same precipitation reaction, but titrations are based on measured volumes instead of masses. In a precipitation titration to analyze the concentration of chloride ion, we precisely measure the volume of a standard solution of AgN03(aq) of known concentration, called the titrant, needed to completely react with the chloride in a known volume of sample. This information is then used to calculate the concentration of chloride in the sample. The calculations involved are described in Sample Exercise 4.12.

Calculating a Concentration from the Results of a Titration In climates with subfreezing winters, elevated concentrations ofNaCI in groundwater are often the result of contamination from runoff containing salt used on roads. To determine the concentration of CI-(aq) in a sample of drinking water from a well, a sample is titrated with a standard solution of Ag+(aq). Determining the chloride concentration of a sample in this way is a widely used application of a precipitation titration. The titration reaction is given in Equation 4.12: NaCI(aq)

+ AgNO/aq)

~ AgCI(s)

+ NaNO/aq)

What is the concentration of CI-(aq) in a 100.0 mL sample of the well water if 24.1 mL of 0.100 M AgNO/aq) are needed to react with all the Cl-(aq) in the sample? Express your answer in molarity (as a chemist would) and in mg Cl-(aq)1 L (as environmental scientists tend to do). We know the volume (24.1 mL) and concentration (0.100 M) of the AgN03 solution and the volume of the sample (100.0 mL). We need to calculate the concentration ofCI-(aq) ion in the sample. The link between the quantity of AgN03(aq) used and the quantity of CI-(aq) ions in the sample is the stoichiometry of the titration reaction, in which 1 mol Ag+(aq) reacts with 1 mol Cl-(aq) as they precipitate as AgCI(s).

COLLECT

AND

ORGANIZE

Let's first calculate the quantity of Ag+(aq) consumed from the volume and concentration of the AgN03 solution, then convert that to an equivalent

ANALYZE

A titration is an analytical method for precisely determining the concentration of a solute in a sample by reacting it with a standard solution of known concentration. A standard solution is a solution of known concentration. The titrant is the standard solution added to the sample in a titration.

168

I Chapter Four

I


quantity of Cl-(aq), and finally divide that quantity of CI-(aq) by the volume of the sample to obtain the values of CI-(aq) concentration in molarity and then in milligrams per liter. SOLVE

The quantity of Ag+(aq) consumed is

0.100 mol Ag+(aq)

24.1 m±:' X The quantity ofCI-(aq)

= 2.41 mmol Ag+(aq)

in the sample must have been 1 mol CI-(aq)

2.41 mfl'lel Al!+(a~ X ---1 ffiel Ag+ and the concentration ofCI-(aq)

2.41 ffimol Cl-(aq)

---.---:::---

100 ffiL

241 .

X

1Gl10-2 ffie L

X

in the sample was

= 2.41

Expressing this concentration

= 2.41 mmol Cl-

X

mol CI-(aq)

10-2----

L

= 2.41

X

10-2 M

of CI- (aq) in milligrams per liter, we get

35.45 g Cl-(aq) ffiel 0-

X

1000 mg

g

= 854 m g CI-(a q)/L

ABOUT IT To estimate the answer, consider that the balanced equation says that 1 mole ofCI-(aq) requires 1 mole of Ag+(aq) to form AgCI(s). The volume of Ag+(aq) solution (about 25 mL) used to titrate the sample was about one-quarter the size of the sample (100 mL), which means that the concentration ofCI-(aq) in the sample was about one-quarter the concentration of the Ag+(aq) solution used in the titration. Our answer (2.41 X 10-2 M = 0.0241 M) is about one-quarter the value of the Ag+(aq) concentration, so it is a very reasonable answer; it is also well above the maximum concentration allowed by drinking-water standards. THINK

Practice Exercise The concentration of SO /-(aq)

in a sample of river water can be determined using a precipitation titration in which a salt ofBa2+(aq) is the standard solution and BaSO 4(S) is the precipitate formed. What is the concentration of SO /- (aq) in a 50.0 mL sample if 6.55 mL of a 0.00100 M Ba2+(aq) solution is needed to precipitate all the SO /-(aq) in the sample? (Answers to Practice Exercises are in the back of the book.)

One of the most common types of titration is acid-base neutralization (see Section 4.5). A typical analysis to which an acid-base titration is often applied consists of quantifying the environmental problem of sulfuric acid contamination of the waters draining from abandoned coal mines. Suppose we have a 100.0 mL sample of drainage water containing an unknown concentration of sulfuric acid. We decide to use an acid-base titration in which the sulfuric acid in the sample is neutralized with a standard solution of 0.00100 M NaOH. The following neutralization reaction takes place: H2S04(aq)

+ 2 NaOH(aq)

~ Na2S04(aq)

+ 2 H20(e)

Suppose that exactly 22.40 mL of the NaOH solution is required to completely react with the H2S04 in the 100.0 mL sample. Because we know the volume (V) and molar concentration (M) of the NaOH solution, we can calculate the number of moles (n) ofNaOH consumed in the neutralization reaction using Equation 4.2:

4.9

n= VXM

=

0.02240 b

x

0.00100 mol ~aOH

=

0.00002240 mol NaOH

We know from the stoichiometry of the neutralization reaction that 2 moles of NaOH are required to neutralize 1 mole of H2SO 4' so the number of moles of H2SO 4 in the sample must be as follows: 1 mol H2S04 0.00002240 mol NaOH X ---2 mol ~l"aOH

= 0.00001120 mol H2S04

If the 100.0 mL sample of mine effiuent contains 0.00001120 mol H2S04, the concentration of H2SO 4 is 0.00001120 mol H2S04 ---0-.-10-0--0-L---

= 1.120 X 10-4

mol H2S04 L

I

Titrations

I

169

The equivalence point in a titration is that point where the number of moles of titrant added is stoichiometrically equal to the number of moles of the substance being analyzed. In a titration, the end point is reached when just enough standard solution has been added to cause the indicator to change color.

then

= 1.12 X 10-4 M H2S04

Volume ofNaOH

How did we know that exactly 22.40 mL of the NaOH solution were needed to react with the sulfuric acid in the sample? A setup for doing an acid-base titration is illustrated in Figure 4.19. The standard solution ofNaOH is poured into a buret, a narrow glass cylinder with volume markings. The solution is gradually added to the sample until a visual indicator that changes color in response to the concentration ofH+(aq) ion in the solution signals that the reaction is complete. The point in the titration when just enough standard solution has been added to completely react with all the solute of interest is called the equivalence point. If the correct indicator has been chosen for the reaction, the equivalence point is very close to the end point, the point at which the indicator changes color. To detect the end point in our sulfuric acid-sodium hydroxide titration, we might use a color indicator called phenolphthalein, which is colorless in acidic solutions but changes to pink when a slight excess of base is present. The part of the titration that requires the most skill is adding just enough NaOH solution to reach the end point, which is indicated by a persistent pink color in the solution being titrated. To catch this end point, the standard solution must be added no faster than one drop at a time with thorough mixing in between drops.

used (L)

Multiply by MNaOH

Moles ofNaOH 1 mol H2S04) (2 mol NaOH

Divide by volume (L) of H2S04 solution

Concentration of H2S04 (MH,so)

FIGURE 4.19 The concentration

ofa

sulfuric acid solution can be determined through

titration with a standard

sodium hydroxide. the H2S04

solution of

(a) A known volume of

solution is placed into the flask.

The buret is filled with an aqueous solution of sodium hydroxide of known concentration. A few drops of phenolphthalein

indicator

solution are added to the flask. (b) Sodium hydroxide is carefully added until the indicator changes from colorless to pink. The concentration

of H2SO

4

is calculated by using

the steps in the flowchart above.

(a)

(b)

Calcium: In the Limelight Calcium is the fifth most abundant element in Earth's crust. Vast deposits of CaC03 occur over large areas of the planet as limestone, marble, and chalk. These minerals are the fossilized remains of earlier life in the oceans. Calcium carbonate is called a biomineral because it is an inorganic substance synthesized as an integral part of a living organism. Coccolithophores, which produce the coccoliths pictured in Figure 4.4, are among the largest producers of calcium carbonate on Earth today, and probably have been since the Jurassic period (150 million years ago). Corals and sea shells

(a)

(b)

FIGURE 4.20 Prized examples of calcium minerals are (a) pearls (heterogeneous mixtures of calcium carbonate and water) and (b) alabaster (fine-grained calcium sulfate).

are mainly calcium carbonate, and islands like the Florida Keys and the Bahamas are made of large beds of CaC03. Calcium sulfate, normally called gypsum, is another important calcium-containing mineral. If you are reading this in a room with plaster walls, you are surrounded by gypsum, as it is the primary component in sheet rock and plaster. Although mostly known for their use as bulk chemicals in heavy industry and construction, special forms of these two chemicals are valued for their beauty. Pearls (Figure 4.20a) are gemstones consisting mainly of calcium carbonate, in particular a heterogeneous mixture of CaCO/s) and a small amount of water suspended within the solid. Light travels differently through the solid material and the dispersed water in a fine pearl. The same phenomenon occurs when a thin layer of oil or gas floats on the surface of water: we see an array of colors. Diffraction of the light gives a high-quality pearl its special iridescence. The substance known as alabaster, prized for its translucence and subtle patterns, is a finegrained form of calcium sulfate (Figure 4.20b). Compounds of calcium have myriad other small uses. The sulfate and carbonate of calcium are two common materials used in desiccants, those small packets of granules included in boxes of electronic devices, shoes, pharmaceuticals, and foods. Desiccants (from the Latin siccus, "dry") remove water from the air, thereby keeping dry the items with which they are packed. About 0.5% by mass of calcium silicate (CaSi03) is added to solid table salt to keep it freely flowing in damp weather. It is estimated that calcium silicate can absorb almost 600 times its mass in water and still remain a free-flowing powder. Calcium hypochlorite [Ca(OCI\J is one of the sources of chlorine used to disinfect the water in swimming pools. In terms of bulk consumption, calcium carbonate is an industrial chemical of great utility. High-quality paper contains CaCO/s), which improves brightness as well as the paper's ability to bind ink. The pharmaceutical industry sells it as an antacid and in dietary supplements, but CaC03 has an even greater industrial use as the precursor of two other important materials: quicklime (CaO) and slaked lime [Ca(OH)J To produce lime, calcium carbonate (derived usually from mining it or from oyster shells) is roasted. CaCO/s)

heat)

CaO(s)

+

CO/g)

Slaked lime is produced from the hydrolysis of calcium oxide:

171

To be "in the limelight," is nowoften used to describe anyone who is the center of attention. The expression derives from the use of lighting systems Furnace Scrubber (coal burned here) based on calcium oxide in 19th-century theaters. A British engineer discovered that calcium oxide emits a brilliant but soft white light when heated in a flame gases of hydrogen mixed with oxygen. By the mid-1860s, limelights fitted with reflectors to direct and spread illumination were in wide use in theaters. In modern times, lime has moved out of Gypsum the entertainment business and into centrifuge - m2, are poised at the starting gate of a downhill course. Is the potential energy (PE) of skier 1 different from or the same as the PE of skier 2? If you think they are different, which skier has more potential energy? (Answers


CONCEPT

TEST

Two skiers with masses m1 and m2, where m1 > m2, go past the same point on parallel racecourses at the same time. (a) At that moment, is the potential energy of skier m2 more than, less than, or equal to the PE of skier m1? (b) If the two are moving at the same speed, which has the greater kinetic energy (KE), or do they have the same KE? (Answers


Kinetic Energy and Potential Energy at the Molecular Level

Thermal energy is the kinetic energy of atoms and molecules.

The relation just described between kinetic and potential energy holds for atoms and molecules as well. The analogy with the kinetic and potential energies of the skier breaks down, however, because the gravitational forces that control the skier play virtually no role in the interactions of very small objects. Temperature and charge dominate the relation between kinetic and potential energy at the microscopic level. Temperature governs motion on the atomic-molecular scale. Chemical bonds and differential electric charges cause interactions between particles that give rise to the potential energy stored in the arrangements of atoms and ions in matter. The kinetic energy of a particle at the microscopic level still depends on its mass and its velocity, but because its velocity depends on temperature, its kinetic energy does, too. As the temperature of a population of molecules increases, the average kinetic energy of the molecules also increases. Consequently, if we have two portions of water vapor (.Att = 18.02 g/mol) at room temperature, the two populations of molecules have the same average kinetic energy. The average velocities of the molecules are the same because their masses are identical. If the temperature of one portion of water vapor is increased, that population of molecules will have a higher kinetic energy because the average velocity of the molecules will be higher. An equivalent population of vapor-phase ethanol molecules (.Att = 46.07 g/mol) at room temperature has the same average kinetic energy as the water molecules at room temperature, but the average velocity of the ethanol molecules is lower because their mass is higher (Figure 5.5). This kinetic energy associated with the random motion of molecules is called thermal energy, and the thermal energy of a given material is proportional to temperature. However, thermal energy in a sample also depends on the number of particles in that sample. The water in a swimming pool and in a cup of water taken from the pool have the same temperature, so their molecules have the same

5.1

I

Energy: Basic Concepts and Definitions

FIGURE .M

=

18.02 g/mol

.M Twater

=

5.5

Two populations

have the same temperature

46.07 g/mol

average kinetic

= TethanoI

molecules

I

189

of molecules

and the same

energy. Because ethanol

have a greater

mass than water

molecules (moth,no] > mw,,)' the average velocity of the water molecules in the vapor

KEwater = KEethanol

phase above the liquid is greater than the average velocity of the gaseous ethanol molecules

(uwater

> Uethanol).

average kinetic energy. The water in the pool has much more thermal energy than the water in the cup, simply because there is a larger number of molecules in the pool. A large number of particles at a given temperature has a higher total thermal energy than a small number of particles at the same temperature.

CONCEPT

TEST

If we heat a cup of water from the swimming pool to a temperature close to the boiling point (100°C), do you think its thermal energy will be more than, less than, or the same as the water in the pool? (Answers


One of the most important forms of potential energy at the atomic-molecular level arises from electrostatic interactions due to charges on particles. The magnitude of this electrostatic potential energy (Eel)' also known as coulombic interaction, is directly proportional to (ex:) the product of the charges on the particles (QI and Q) and inversely proportional to the distance (d) between them:

E ex: cI

Q1

X d

Q2

(5.4)

Ifboth charges in Equation 5.4 are the same (either both positive or both negative), the product of the charges is positive, the particles repel each other, and Eel

Electrostatic potential energy (E) is the energy of a particle because of its position with respect to another particle. It is directly proportional to the product of the charges of the particles and inversely proportional to the distance between them.

190

I Chapter Five

I

Thermochemistry

t ~

I

FIGURE 5.6 Electrostatic potential energy (E): (a) A positive ion and a negative ion are

.+:

so far apart (d is large) that they do not interact at all. (b) As the particles

move closer together

(d decreases), the electrostatic

potential

energy between the two ions becomes more negative. (c) At some point the attraction between oppositely charged particles produces an arrangement

that is the most favorable

energetically

because the particles in that

combination

have the lowest potential

The minimum

energy.

in the curve corresponds

most favorable arrangement. forced even closer together,

to that

(d) If the ions are they repel each

other as their nuclei begin to interact.

Combustion

of hydrogen

launch rockets.

can be used to

is positive. If one particle is positive and the other negative, however, the product of the charges is negative, particles attract each other, and Eel in this case is negative (Figure S.6a-c). Lower energy (a more-negative value of E) corresponds to greater stability, so particles that attract each other because of their charge form an arrangement characterized by a lower electrostatic potential energy than particles that repel each other. There is a limit, however, to how close two particles can be to each other. Remember that ions, whatever their charge, have positively charged nuclei that begin to repel each other if they are pushed together too closely (Figure S.6d). At the microscopic level, potential energy is stored in the position of particles with respect to each other, just as the potential energy of skiers is determined by their positions on a slope. Neutral molecules such as water attract each other as well because of distortions in the distribution of electrons about the nuclei of their atoms, and the same analysis applies in terms of their relative potential energy. Coulombic interactions, not gravitational forces, determine the potential energy of matter at the atomic level. The total energy of matter at the microscopic level is the sum of the kinetic energy due to the random motion of particles and the potential energy in their arrangement. The energy given off or absorbed during a chemical reaction is due to the difference in potential energy stored in the arrangement of atoms in molecules of reactants and products. For example, when hydrogen molecules are burned in oxygen, the products of the combustion reaction are water and a considerable amount of released energy. The energy given off by this reaction has been used frequently to power rockets to launch satellites and is now being applied to run vehicles such as buses or automobiles. Because significant energy is given off in this process, the molecules of the product must be at a lower potential energy than the molecules of the reactants, as shown graphically in Figure 5.7. The difference between the energy of the two arrangements of atoms is the energy released. In the case of hydrogen combustion this energy is sufficient to run a variety of vehicles now powered by the combustion of fossil fuels.

5.2

I

Systems, Surroundings,

+

+

~Energy

~

I 191

and the Flow of Energy

energy

released

2 HP(g)

Progress of the reaction (b)

(a) FIGU RE 5.7 (a) Fueling a hydrogen-powered

vehicle. (b) Hydrogen

reacts with oxygen to

produce water. The molecules in the product are at a lower energy than those in the reactants; hence the reaction releases energy.

~

Systems, Surroundings, and the Flow of Energy

The study of thermodynamics requires us to define precisely what part of the universe we are evaluating. In thermochemistry and thermodynamics, the specific and limited part of the universe we are studying is called the system. Everything else that is not part of the system is called the surroundings. Typically, we limit our concern about surroundings to that part of the universe that can exchange energy (or energy and matter) with the system we have defined. Even for a system as small as a red blood cell, the surroundings are the entire universe, but we conventionally limit our consideration to the immediate region around the cell. In evaluating the energy gained or lost in chemical reactions, the system may include the reaction vessel and its contents, or it may be just the atoms, molecules, or ions that are involved in the reaction.

Isolated, Closed, and Open Systems Although systems can be as large as a galaxy or as small as a living cell, most of the systems we examine quantitatively in this chapter are small and are of a size that fits on a laboratory bench. We limit our concern about surroundings to that part of the universe that can exchange energy with the system we have defined. In discussions involving the transfer of energy as heat, work, or both, three types of systems are common: isolated systems, closed systems, and open systems (Figure 5.8). These designations are important because they define the system we are dealing with and the part of the universe with which the system interacts. From the point of view of systems and surroundings, hot soup in a thermos bottle is an example of an isolated system, which exchanges neither energy nor matter with its surroundings. The insulated bottle does not allow heat or matter to be transferred away from the soup (the system). From the soup's point of view, the soup itself is the entire universe; in an isolated system, the system has no surroundings. Of course, even the best thermos cannot maintain the soup over

A system is the part of the universe that is the focus of a thermodynamic study. The surroundings include everything is not part of the system.

that

An isolated system exchanges neither energy nor matter with the surroundings.

192

I

Chapter Five

I

Thermochemistry Water vapor

FIGURE 5.8 The transfer of energy and matter in isolated, closed, and open systems: (a) Hot soup in a thermos bottle approximates an isolated system: no vapor escapes, no escapes to the surroundings

Heat~

Heat~.

matter is added or removed, and no heat

..

(as long as the

thermos behaves as a perfect insulator). An

(a) Isolated System

(b) Closed System

ideal thermos bottle completely isolates the soup from the rest of the universe. (b) Hot soup in a cup with a lid is a closed system; the soup transfers heat to the surroundings

Pepper

mill

as it

cools; however, no matter escapes and none is added. (c) Hot soup in a cup is an open system; it transfers both matter (water vapor) and energy (heat) to the surroundings cools. Matter

as it

in the form of pepper, grated

cheese, crackers, or other matter from the surroundings

may also be added to the soup

(the system).

A thermos bottle with the lid screwed on tight

A cup of soup with a lid

Grated cheese

Crackers

An open cup of soup

time as a perfectly isolated system, because heat will leak out and the contents will cool, but for short time periods the soup in a good thermos approximates an isolated system. Soup in a cup with a lid is an example of a closed system. A closed system (the soup) exchanges energy but not matter with the surroundings (the cup, the air, the tabletop). Because the cup has a lid, no vapor escapes from the soup and nothing else can be added. Only heat is exchanged with the surroundings, not matter. The heat from the soup is transferred to the walls of the cup, then into the air and the tabletop, and the soup gradually cools. Soup in an open cup is an example of an open system. An open system exchanges energy and matter with the surroundings. The heat from the soup is transferred to the walls of the cup, to the air, and to the tabletop, and matter from the soup in the form of water vapor leaves the system and enters the air. We may also add matter from the surroundings to the system by sprinkling on a little grated cheese, some ground pepper, or adding a few crackers. Most of the systems we deal with in introductory thermochemistry are either isolated systems or closed systems, because they are the easiest to study quantitatively. However, many interesting, important systems are open-including cells, living organisms, engines, and Earth itself (Figure 5.9)

Heat Flow A closed system exchanges energy but not matter with the surroundings. An open system exchanges both energy and matter with the surroundings. In an exothermic process, heat flows from a system into its surroundings. In an endothermic process, heat flows from the surroundings into the system.

A chemical reaction or a physical change of state that produces a net flow of heat energy from a system to its surroundings is exothermic from the system's point of view. This flow can be detected by an increase in the temperature of the surroundings. Combustion reactions that release heat and light are examples of exothermic reactions. In contrast, chemical reactions and physical changes that absorb heat from their surroundings are endothermic from the system's point of view. For example, ice cubes (the system) in a glass of warm water (the surroundings) absorb heat from the water, which causes them to melt. From the point of view of the system (ice cubes), the process is endothermic.

I

5.2


Tissues

Lungs


I 193

Heat

~xygenout

Oxygen in/

(a) Red Blood Cells Physical environment

Decomposers break down the bodies of dead organisms, thereby returning nutrients to the physical environment.

NUTRIENT CYCLING

Producers (plants) Energy lost as metabolic heat

Consumers (herbivores, predators, parasites, decomposers) Energy lost as metabolic heat

(b) Flow of Energy through an Ecosystem

(c) A Massive Exchange of Energy and Matter

FIGURE 5.9 Living thermodynamic

systems and beyond: (a) Red blood cells are open systems. Matter

blood cells. Red blood cells also exchange heat with their surroundings. CO2 from the atmosphere is transferred

(d) Black Holes: One-way Streets in the Cosmos

(b) Energy flows through

are absorbed by green plants, which use these to produce glucose and other plant materials. The energy stored in glucose

to living systems, which lose heat as they oxidize plant materials back to CO,. When living organisms

chemical components

of their bodies to the physical environment.

Decomposition

Earth,

collides with interstellar

Sixty-five million years ago, an asteroid colliding with Earth is believed to have caused massive destruction, (d) Black holes have gravitational

die, decomposition

returns the

in the absence of oxygen produces the highly reduced forms of

carbon in fossil fuels. (c) Massive exchanges of energy and matter occur when our biosphere, dinosaurs.

(as O2 and water) enters and leaves red

an ecosystem as heat and light from the sun and

matter such as asteroids.

climate change, and the extinction

of the

fields so strong that matter and light are drawn in and not released. They are one-way systems, where the

flow of matter and energy is inexorably in, never out. Black holes stretch our convenient

definitions

of earthly thermodynamic

systems.

194

I Chapter Five

I

Thermochemistry

In another phase change, water vapor (the system) from the air on a humid day condensing into droplets of liquid water on the outside surface of a glass containing an ice-cold drink (the surroundings) requires that heat flow from the system (the water vapor) to the surroundings (the glass). From the point of view of the water vapor, this process is exothermic. If we suddenly put hot water in the glass, the water droplets on the outside surface would absorb heat and vaporize; from the point of view of the droplets, that process is endothermic. This illustrates an important concept: an exothermic process in one direction (vapor ~ liquid: condensation) is endothermic in the reverse direction (liquid ~ vapor: vaporization) (Figure 5.10).

Liquid

Vapor

Liquid

System

Vapor

System

FIGURE 5.10 Reversing a process changes the sign but not the magnitude

Heat out to surroundings

of the heat

Heat in from surroundings

Exothermic

change.

Endothermic

The flow of heat during physical changes is summarized in Figure 5.11. We use the symbol q to represent heat produced or consumed by a chemical reaction or a physical change. If the reaction or process is exothermic, then q is negative, meaning that the system loses heat to its surroundings. If the reaction or process is endothermic, then q is positive, indicating that heat is gained as it flows from its surroundings into the system. In Figure 5.11, endothermic changes of state are represented by arrows pointing upward: solid ~ liquid, liquid ~ gas, and solid ~ gas. The opposite changes represented by arrows pointing downward are exothermic. To summarize: Exothermic:

q0

Gas

Heat Heat

o

,g Liquid

FIGURE 5.11 Matter can be transformed

from

one of the three physical states into another by adding or removing heat. Upward-pointing endothermic

arrows represent

processes (heat enters the system; q

Downward-pointing

arrows represent exothermic

processes (heat leaves the system and enters the surroundings;

q < 0).

> 0).

-' Heat

o ~. o. o ;:l

I

5.2

(a)

I

I

~ Iii

I

Boiling flask

I 195


(c)

(b)

II


Water out

FIGURE 5.12 A laboratory setup for the distillation

of water: (a) Water is heated to

the boiling point in the distillation

flask.

(b) Vapors rise and enter the condenser,

where

they are liquefied. (c) The liquid is collected in the receiving flask.

Identifying Exothermic and Endothermic Processes Describe the flow of heat during the purification of water by distillation 5.12). Consider the water to be the system.

(Figure

COLLECT AND ORGANIZE Since the water is the system, we must evaluate how the water gains or loses heat during distillation.

A N A L Y ZEIn

distillation, heat flows in two distinct steps: (a) Liquid water is heated to the boiling point. (b) The resulting vapors are cooled as they pass into the condenser and are then collected.

SOL V E (a) Heat flows into the water to make it boil. The liquid water is converted into water vapor. Heat flows from the surroundings (the heat source) into the system (the water); the process is endothermic. (b) The vapors from the boiling water enter the condenser. Heat flows from the system (the water vapor) into the surroundings (the walls of the condenser). The process is exothermic. ABOUT IT Being able to define the system and the surroundings important in dealing with heat flow.

THINK

is very

Practice Exercise What is the sign of q as (a) a match burns, (b) drops of molten candle wax solidify, and (c) perspiration (water) evaporates? In each case, define the system clearly and indicate which of these processes are endothermic and which are exothermic. (Answers


of the

book.)

196

I

Chapter Five

I

Thermochemistry

(a) Molecules close together; same nearest neighbor over time.

FIGURE

Sees

5.13

Solid

What a Chemist

Phase changes: (a) Solid ice absorbs

heat and is converted

to liquid water. The

molecules in the solid are held together rigid three-dimensional

arrangement;

have the same nearest neighbors

in a they

over time.

(b) Molecules close together but moving; exchanging nearest neighbors.

Liquid

(b) When the solid melts, the molecules in the liquid phase exchange nearest neighbors and occupy many more positions with respect to each other than were possible in the solid. (c) The liquid absorbs heat and is converted to a vapor. The molecules in a gas are widely separated

and move very rapidly. The reverse

(c) Molecules widely separated; moving rapidly.

Gas

of these processes occurs when water in the vapor phase loses heat and condenses to a liquid. The liquid also loses heat when it is converted

~II

to a solid.

CHEMTOUR Internal Energy

Some of the types of molecular contribute

to the overall internal

a system: (a) translational

motion that energy of

motion, motion

from place to place along a path; (b) rotational motion, motion about a fixed axis; and (c) vibrational

motion, movement back and

Consider the flow of energy when a tray of ice cubes is inadvertently left on a kitchen counter (Figure 5.13). Let's focus on the ice cubes (the system). As the ice cubes warm and start to melt, the attractive forces that hold molecules of H20 in place in the solid ice are broken. These molecules now have more positions they can occupy, so they have more potential energy. After all the ice at QOChas melted into liquid water at QOC,the temperature of the water (the system) slowly rises to room temperature. Higher temperature is also the result of increasing the kinetic energy of the molecules of water in the tray. This kinetic energy is part of the overall internal energy (E) of the system. The absolute internal energy of the system is the sum of the kinetic and potential energies of all the components of the system. The values of these energies are difficult to determine, but changes in internal energy (t:,E) are fairly easy to measure, because a change in a system's physical state or temperature provides a measure of the change in its internal energy. (The capital "delta", t:" is the standard way scientists symbolize change in a quantity.) The change in internal energy (t:,E) is simply the difference between the final internal energy (Efina1) of the system and the initial internal energy (Einitial) of the system:

t:,E =

Efina1 -

(b)

(5.5)

Internal energy is another example of a state function, because the value of the change in internal energy depends only on the initial state and the final state. How the change occurs in the system does not matter.

forth from some central position.

(a)

Einitial

(c)

5.2

I


Top closed

~ Vat compression

I

-

,

i~, --lli

~hlr- ~1

Fuel explodes

"""-

I

I

----\0,.

I 197

The internal energy (E) of a system is the sum of all the kinetic and potential energies of all of the components of the system.

~

{


V at

1

"""'"';'0

The first law of thermodynamics states that the energy gained or lost by a system must equal the energy lost or gained by the surroundings. The calorie (cal) is the amount of heat necessary to raise the temperature of 1 g of water I°C. The joule 0) is the SI unit of energy; 4.184 J = 1 cal.

(b)

(a)

FIGU RE 5.14 Work performed by changing the volume of a gas. (a) Highest position of the piston after it compresses the gas in the cylinder, causing the fuel to ignite in a diesel engine. (b) The exploding

fuel gives offheat,

causing the gas in the cylinder to expand and

push the piston down.

The law of conservation of energy applies to processes involving the transfer of energy in materials. The total energy change experienced by a given system must be balanced by the total energy change experienced by its surroundings. This is the first law of thermodynamics, and it is another statement of the law of conservation of energy. The energy changes of the system and the surroundings are equal in magnitude but opposite in sign, so their sum is zero. Energy is neither created nor destroyed; it is conserved.

Units of Energy We need to express quantities of energy transferred in terms of some unit. Energy changes that accompany chemical reactions and changes in physical state have been traditionally measured in calories. A calorie (cal) is the quantity of heat required to raise the temperature ofl g of water from 14.5 to 15SC. In the Systeme International the unit of energy is the joule 0); 1 cal = 4.184 J. What is called a "Calorie" (Cal; note capital C) in nutrition is actually the kilocalorie (kcal): 1 Cal = 1 kcal = 1000 cals. The energy value in food is still frequently expressed in Calories in the United States. We routinely use joules throughout this text. Doing work on a system is a way to add to its internal energy. For example (Figure 5.14), compressing a quantity of gas into a smaller volume, like the action of a piston to compress a mixture of fuel vapors and air in a diesel engine, is work done on the gases that causes their temperature to rise. The total increase in the energy of a system is the sum of the work done on it (w) and the heat (q) that flows into it:

t:,E

=

q

+

w

(5.6)

When diesel fuel in the cylinder ignites and produces hot gases, the gases (the system) expand and do work by pushing on the piston. As another example of a system doing work on its surroundings, consider the enormous balloons used by adventurers to fly around the world. These balloons typically have a chamber of hot air heated by burners fueled by ethane and propane (Figure 5.15). The burners allow a balloonist to adjust the temperature of the air to control the buoyancy of the balloon and its altitude. Burning propane or ethane causes the air in the balloon above the burner to expand. As the temperature of the gas filling the balloon goes up, the gas expands, lowering its density. The increase in volume of

FIGU RE 5.15 In a deflated balloon, V = o. the balloon does work (PI'1V) as the

Inflating

incoming gas causes the balloon to push against the surrounding

atmosphere.

198

I

Chapter Five

I

Thermochemistry

(P ~V) work is Pressure-volume work associated with the expansion compression of gases.

or

Surroundings

System Heat in ••

+q

Workin.+w

,t out

-q

,rkout

-w

the balloon against the pressure of the atmosphere outside the balloon is a type of work known as pressure-volume (P AV) work. The pressure of the atmosphere on the balloon and indeed on all objects is a result of the Earth's gravity pulling the atmosphere toward the surface of Earth, where it exerts a force on all things because ofits mass. The pressure at sea level on a nice dry, sunny day is approximately 1 atmosphere (1 atm) of pressure, which equals a force of 14.7lb/in2 pressing on each square inch of surface area. Atmospheric pressure varies slightly with the weather and decreases with elevation above sea level. Atmospheric pressure (P) remains constant throughout the process of inflating the balloon. If we define the system as the air in the balloon, then the work is done to expand the air trapped in the balloon above the burner. This is P AV work done by the system on its surroundings; the surroundings in this case are the atmosphere. The energy of the system decreases as it performs this P AV work. We can relate the change in internal energy (A E) of the system to the heat lost by the system (q) and the work done by the system (PAV) with this equation:

AE = q

M=q+w

FIGURE 5.16 Heat absorbed by a system from the surroundings

and work done on a

system by the surroundings

(arrows entering

the system) are both positive quantities because both increase the internal energy of a system. Heat released by a system to the surroundings

and work done by a system on

the surroundings

(arrows leaving the system)

are both negative quantities

because both

+ (- P AV) = q -

P AV

(5.7)

The negative sign in front of P AV is appropriate because when the system expands (positive change in volume AV), it loses energy (negative change in internal energy AE) as it does work on its surroundings. When the surroundings do work on the system (for example, when a gas is compressed), then PAVhas a positive sign. By the same token, the sign of the heat (q) may be either positive or negative. If heat flows from the surroundings into the system, heat is positive (+q). Heat has a positive sign in the case of the balloon because the heat flows from the surroundings (the burner) into the air in the balloon (the system). When heat flows from the system into the surroundings, its sign is negative (-q) (Figure 5.16). Considering Equation 5.7, the change in internal energy of a system is positive when more energy enters the system than leaves, and negative when more energy leaves the system than enters.

decrease the internal energy of a system.

Calculating

Changes in Internal Energy

Think of the figure below as a simplified version of a piston and cylinder in the engine of an automobile. When the fuel in the cylinder burns, it produces heat. Suppose the gases in the cylinder (the system) absorb 155 J of heat from the combustion reaction of the fuel and expand, thereby pushing the piston down. In doing so, the gases do 93 J of work. What is the change in internal energy (AE) of the gases?

~II

CHEMTOUR

Pressure-Volume

Work

Change in internal energy is related to the work done by or on a system and the heat gained or lost by the system (Equation 5.6). First we have to decide whether heat and work are positive or negative from the point of view of the system (the gases in the cylinder). COL LEe TAN

D 0 RGA N IZ E

In this example, the system (the gases in the cylinder) absorbs heat, so (the piston), so w < 0 (negative).

ANALYZE

q > 0 (positive). The system also does work on the surroundings

5.2

I


The change in internal energy is the sum of the heat and the work, so

SOL V E

l1E

=q

=

+ w

(+155])

=

+ (-93])

+62J

ABOUT IT In this example, more energy enters the system than leaves it, so the change in internal energy (l1E) is positive.

THINK

Practice Exercise In a different experiment the piston in Sample Exercise 5.2 compresses the gases in the cylinder and does 64 J of work on the system. As a result, the system gives off 132 J of heat to the surroundings. What is the change in internal energy of the system? (Answers to Practice Exercises are in the back

of the

book.)

Calculating P llV Work

SAMPLE EXERCISE 1.3

A tank of compressed helium is used to inflate balloons for sale at a carnival on a nice day when the atmospheric pressure is 1.01 atm. If 100 balloons are each inflated from an initial volume of 0.0 L to a final volume of 4.8 L, how much P l1V work is done to inflate the balloons? Assume that the pressure remains constant during the filling process. COLLECT AND ORGANIZE We know that one balloon goes from being completely flat (V = 0.0 L) to a volume of 4. 8 L. We are given a value for the pressure, which remains constant, so work is P l1V.

We are asked about the Pl1Vwork done by 100 such balloons, so we have to multiply the volume change of one balloon by 100 to get the total change in volume (l1V). ANALYZE

SOLVE

The total volume change (l1V) as all the balloons are inflated is 100 balloons X 4.8 Llballoon

=

480 L

The work (w) to inflate this volume at a pressure of 1.01 atm is w

= Pl1V= 1.01 atm X 480 L = 480 L· atm

Because the system did work on the surroundings by pushing sphere as the balloons inflated, the work has a negative sign: w

back the atmo-480 L . atm.

=

"L . atm" (liters times atmospheres) may seem strange units for expressing work. The SI unit for energy (the ability to do work) is the joule (J). If we had a conversion factor that related joules to liters times atmospheres, we could convert the answer into the units we are used to seeing for work. From calculations involving gases, this conversion factor has been determined as 101.32 JI(L . atm). Using this factor for the work done in filling the balloons, we get THINK

ABOUT

w

=

IT

-480

L . atm [101.32 JI(L . atm)]

Practice Exercise The Spirit

of Freedom,

=

-4.9

X 104 J

=

-49 kJ

the balloon flown around the world by American Steve Fossett in June and July of 2002, contained 550,000 cubic feet of helium. How much P l1Vwork was done to inflate the balloon, assuming 1.00 atm of pressure? Express your answer in liters times atmospheres (L . atm) and then convert it to joules, given that: 1 m3 = 1000 L = 35.3 ft3, and 1.00 L . atm = 101.32 J. (Answers to Practice Exercises are in the back

of the

book.)

Spirit of Freedom


I

199

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I Chapter Five

I

Thermochemistry

Enthalpy (H) is the sum of the internal energy and the pressure-volume product of a system. The change in enthalpy (!!H) for a reaction carried out at constant pressure is the heat gained or lost by the system.

~

Enthalpy and Enthalpy Changes

Many of the physical and chemical changes discussed so far, as well as many others, take place at constant atmospheric pressure (P). An important thermodynamic variable that relates heat flow into or out of a system during chemical reactions or physical changes at constant pressure is called enthalpy. Enthalpy (H) is the sum of a system's internal energy and the product of its pressure and its volume: H=E+

(5.8)

PV

Suppose a chemical change takes place at constant pressure. companied by a change in internal energy (t..E) and by work changes volume (t..V) at constant pressure (P). These changes in enthalpy (t..H), described mathematically by inserting t.. in priate variables in Equation 5.8: t..H=

t..E

This change is acdone if the system produce a change front of the appro-

+ Pt..V

(5.9)

Let's use Equation 5.9 to define enthalpy in terms of work and heat by substituting the mathematical definition of change in internal energy from Equation 5.7 (q - P t..V) for t..E: t..H Simplifying

= (q -

Pt..V)

+ Pt..V

this expression, we get

(5.10) where qp stands for heat flow at constant pressure (the usual condition in a laboratory). Equation 5.10 tells us that the enthalpy change of a reaction at constant pressure is equivalent to the heat gained or lost by the system during the reaction. When heat flows from the system to the surroundings, the change is exothermic and the enthalpy change is negative: t..H < o. When heat flows from the surroundings into the system, the change is endothermic and t..H > O. For example, at a constant atmospheric pressure, heat flows into a melting ice cube from its surroundings; this process is endothermic (t..H > 0). However, to make ice cubes in a freezer, heat must be removed from the water in the ice-cube tray (t..H < 0). The changes in enthalpy for the two processes-melting and freezing-have different signs but, for a given amount of water, they have the same absolute value. For example, t..Hf us = +6.01 kJ if a mole of ice melts, but t..H 501'.d = -6.01 kJ if a mole of water freezes. We add the subscriptfus or solid to t..H to identify which process is occurring: melting (fusion) or freezing (solidification). The part of the universe of interest (system or surroundings) may also be indicated as t..H sys or t..H surr .

Determining

the Value and Sign of I1H

In between periods of a hockey game a Zamboni water across the surface of a hockey rink.

machine

spreads 855 liters of

a. If the system is the water, what is the sign of t..H sys as it freezes? b. To completely freeze this volume of water at O°C, what quantity of heat is lost? Assume that 6.01 kJ of heat must be removed to freeze 1 mol of water, and the density of water is 1.00 g/mL. The water from the refinishing machine is identified as the system, so we determine the sign of t..H with respect to heat transfer into or out of the water. Because t..H values are reported in units of joules per mole (J/mol or kJlmol) or joules per gram (Jig), we are also given a specific volume of water and its density, so we can calculate the mass of water involved in the phase change. COLLECT

AND

ORGANIZE

5.4

A N A L Y Z E (a) We may assume that the freezing process takes place at constant

pressure, so the sign of the enthalpy change is the same as the sign of the quantity of heat transferred. (b) To calculate the amount of heat lost, we must convert 855 L into an equivalent number of moles, because the conversion factor between the quantity of heat removed and the quantity of water that freezes is 6.01 kJlmol. SOLVE

a. The system (the water) changes from a liquid to a solid-it freezes. For that to happen, heat must be removed from the system. Heat lost is -q; therefore the sign of f}.H sys is also negative. b. We convert the volume of water into number of moles:

855 L X Calculating

1.00 g X 1 mol 1000 HH". 1L X HH". 18.02 g

X 104 mol

q

6.01 kJ 4.74 X 104 ffiel X -1ffiel f}.H sys

= 4.74

=

2.85 X 105 kJ of heat lost by the water

= -q

-q because pressure is constant.

Therefore, f}.H sys

= -2.85 X 105 kJ

T HI N K ABO UTI T The value given in the problem for the molar heat of fusion

of water applies specifically to conditions of 1 atm of pressure. The value of f}.H changes with pressure, but the magnitude of the change is insignificant for pressures around normal atmospheric pressure.

Practice Exercise The flame in a torch used to cut metal is produced by acetylene burning in an atmosphere of pure oxygen. Assuming the combustion of a mole of acetylene (C2H) releases 1251 kJ of heat, what mass of acetylene is needed to cut through a piece of steel if the process requires 5.42 X 104 kJ of heat? (Answers


of the

book.)

.:!:. Heating Curves and Heat Capacity Winter hikers and high-altitude mountain climbers use portable stoves fueled by propane or butane to prepare hot meals where the only source of water may be ice or snow. In this section, we use this scenario to examine the flow of heat into water that begins as snow and ends up as vapor.

Hot Soup on a Cold Day Let's consider the changes in temperature and changes of state that water undergoes as some hikers set out to prepare hot soup using dry soup mix and melted snow. Suppose they start with a saucepan filled with snow at O°F (-18°C). They place it above the flame of a portable stove, and heat begins to flow into the snow. The temperature of the snow immediately begins to rise. If the flame of the stove is steady (so that the flow of heat is constant), then the temperature of the snow should follow the series of connected lines segments shown in Figure 5.17. First, heat from the stove warms the snow to its melting point, O°c. This temperature rise and the accompanying heat flow into the snow are represented by line segment AB in Figure 5.17. When the temperature reaches O°C, it remains steady

I

Heating Curves and Heat Capacity

I 201

202

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Chapter Five

I

Thermochemistry

u o

1:f 100

E ~

r

Boiling point

Melting

ID

Liquid and vapor in equilibrium

•

E

point

E

~

Solid and liquid in equilibrium

Heat added

FIGURE 5.17 The heat required to melt snow and boil the resultant water is illustrated by the four line segments on the heating curve of water: heating snow to its melting point (AB); melting the snow to form liquid water (BC); heating the water to its boiling point (CD); and boiling the water to convert it to vapor (DE).

~II

CHEMTOUR Heating Curves

while the snow continues to absorb heat and the snow melts. This process is a phase change that yields liquid water from snow. Melting, also called fusion, is represented by the constant temperature (flat) line segment BC. When all the snow has melted, the temperature of the water again rises (line segment CD in the figure) until it reaches 100°C. At lOOT another phase change takes place: liquid water absorbs a quantity of energy (denoted by the length of line segment DE) and is vaporized. If all of the liquid water in the pot were converted into vapor, the temperature of whatever vapor remained in the pot would begin to rise again as more heat was added. The differences in the x-axis coordinates of the points along the curve in Figure 5.17 indicate how much heat is required in each step in this heating process. In the first step (along line segment AB), the heat required to raise the temperature of snow from -18°C to O°C can be calculated if we know how many moles of snow we have and what quantity of heat is required to change the temperature of a mole of snow. This latter quantity, called molar heat capacity, has been determined for many materials; Table 5.1 lists a few of these. Molar heat capacity (c) is the quantity of heat required to raise the temperature of 1 mole of a substance by 1°C. Water in the solid phase has a molar heat capacity of37.1 JI(mol . 0C). From the units of molar heat capacity, JI(mol . °C), we see that if we know the number of moles of snow and the temperature change it experiences, we can calculate q for the process:

(5.11) Molar heat capacity (c) is the heat required to raise the temperature of 1 mole of a substance by I°C. When the system is at constant pressure, the symbol used is c p. Specific heat (c) is the heat required to raise the temperature of 1 gram of a substance by I°C at constant pressure. Heat capacity (C p) is the quantity of heat needed to raise the temperature of some specific object by I°C at constant pressure.

where q represents heat, n is the number of moles, t:,T is the temperature change in degrees Celsius, and cp is the molar heat capacity of snow or ice at constant pressure in JI(mol . 0C). In addition to the molar heat capacity, there are other measures of the quantity of heat required to increase the temperature of a substance, each referring to a specific object or a specific amount of the substance. For example, some tables of thermodynamic data list a parameter related to the molar heat capacity, called speCific heat (c), which is the heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius and has units of JI(g . 0C). Later in this chapter, we make use of the parameter heat capacity (Cp), which is the quantity of energy required to increase the heat of a particular object by 1 degree Celsius.

5.4

Substance

cp [J/(mol


I

I

Molar Heat Capacities of Selected Substances

TABLE 5.1

I

. °C)]

Water(s)

37.1

Water(C)

75.3 I

113.1

Ethanol(C): CH1CHPH Carbon(s): graphite

I I

8.54

I

Al(s)

24.4

Cu(s)

24.5

Fe(s)

25.1

Let's assume that the hikers decide to cook their meal with 270.0 g of snow. Dividing the mass of snow by the molar mass of water (18.02 g/mol), we find that they are using 14.98 mol of snow. To calculate how much heat is needed to raise the temperature of14.98 mol ofH20(s) from -18°C to OT, we use Equation 5.11 and insert the quantity of snow (14.98 mol), the temperature change (+ 18T), and the molar heat capacity of snow, [+37.1 JI(mol· T)]:

q

=

nCp!JT

=

14.98 fHel X

37.1 J fHel·

or-.

X (+18::G)

=

1.0 X 104J

=

+10 kJ

Notice that this value is positive, which means that the system (the snow) gains energy as it warms up. During the next phase of the heating process, the snow melts, or fuses, into liquid water. The heat absorbed during this step, represented by line segment Be in Figure 5.17, can be calculated using the amount of heat required to melt 1 mole of snow, called the molar heat of fusion (D.Hfus) of water (6.01 kJlmol), in the following equation:

q = n D.Hfus Substituting

(5.12)

the given information into the equation, we get

q = 14.98 fHel X

6.01 kJ fHel

= +90.2 kJ

This value is positive, because the snow requires energy to break the attractive forces between the water molecules in the solid phase and become a freely flowing liquid. No factor for temperature appears in this calculation; phase changes of pure substances take place ideally at constant temperature. Snow at O°C becomes liquid water at OT. When all the snow has melted, the temperature of the water rises from its melting point of O°C to its boiling point of 100°C as more heat is added. During this stage, the relation between temperature and heat absorbed is defined by Equation 5.11 for liquid water, which has a molar heat capacity of75.3 JI(mol . T):

q = nCpD.T

=

14.98 fHel X

75.3J fHel·

or-.

The molar heat of fusion (f'..Hf)

X 100::G

=

+1.13 X 105 J

=

+1.13 X 102 kJ

This value is positive because the system takes in energy as its temperature

rises.

of a substance is the heat required to convert 1 mole of a solid substance at its melting point to 1 mole of liquid.

203

204

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Chapter Five

I

Thermochemistry

The molar heat of vaporization (!::'Hvop) is the heat required to convert 1 mole of a substance at its boiling point to 1 mole of vapor.

At this point in our story, let's assume that our hiker-chefs accidentally leave the boiling water unattended and it is completely vaporized. The conversion of liquid water into vapor is depicted by line segment DE in Figure 5.17. The temperature of the water remains at 100°C until enough energy (represented by the length of horizontal line segment DE) has been absorbed to vaporize all of it. The enthalpy change associated with changing 1 mole of a liquid to a gas is the molar heat of vaporization (6.Hva/ This quantity of energy absorbed in this part of the process is the product of the number of moles of water times the value of 6.Hvap for water, which is 40.67 kJlmol:

q = n6.H yap

(5.13)

= (14.98 mol) (40.67 kJlmol) = 610 kJ

OT

20T

(a)

(b)

FIGURE 5.18 Macroscopic

and molecular-level

(a) ice, (b) water, and (c) water vapor.

This value is positive, because the system takes up heat in converting a liquid into a vapor. Again no factor for temperature appears because the phase change takes place ideally at constant temperature. Only after all the water has completely vaporized does its temperature increase above 100°C along the line above point E. The molar heat capacity of steam [43.1 JI(mol . OC)] is used to calculate the energy required by the system to heat the vapor to a temperature above 100°C. Note that the cp values of solid and liquid water and water vapor are all different. This trend is common: nearly all substances have different molar heat capacities in their different physical states. In Figure 5.17 notice that the horizontal line segment DE representing the boiling of water is much longer than the line segment representing the melting of ice or snow. The relative lengths of these lines indicate that the molar heat of vaporization of water (40.67 kJlmol) is much larger than the molar heat of fusion of ice (6.01 kJlmol). Why does it take more energy to boil a mole of water than to melt a mole of ice? The answer is related to the extent to which attractive forces between molecules must be overcome in each process and how the potential energy (PE) of the system changes as heat is added (Figure 5.18). Attractive forces determine the organization of molecules and their relation to their nearest neighbors. In snow or ice, the values of KE (kinetic energy) and PE are low and attractive forces are strong enough to hold molecules in place relative to each other. Melting ice-changing the solid into a liquid-requires that some of these attractive forces between water molecules be >100T overcome. The heat energy provided to the (c) system at the melting point of the solid is sufficient to overcome the attractive forces and change the arrangement (and hence PE) of the molecules. The attractive forces still exist between molecules of H20 in liquid water, but the energy added at the melting point goes into increasing the potential energy of the molviews of ecules and enables them to occupy positions

5.4

other than those within the rigid framework of the solid with respect to their nearest neighbors. The molecules are still as close together in the liquid as in the solid, but their positions relative to each other change; they no longer have the same nearest neighbor over time. The molecules have higher potential energy in the liquid than in the solid. Once the solid has completely melted, additional heat causes the temperature to rise, increasing values of both KE and PE. When water vaporizes at the boiling point, essentially all the attractive forces between water molecules must be overcome to separate the molecules by increasing the distance between them as they enter the gas phase. Separating the molecules widely in space is a much larger change in position-in potential energy-for the molecules, and hence more energy is required.

Calculating the Heat Needed to Raise the Temperature Calculate the amount of heat required to raise the temperature (about 1 cup) from lS.0°C to 100.0°C.

of 237 g of water

The amount of heat required can be determined using the molar heat capacity of water (given in Table 5.1) and the number of moles of water (which can be calculated from the mass of water specified). We have the initial and final temperatures of the water, so we can calculate the temperature change (!1T). COLLECT

AND

ORGANIZE

A N A LY Z E: We can calculate the number of moles in 237 grams of water using the

molar mass of water (.Att = 18.02 g/mol). The molar heat capacity ofliquid water is 75.3 JI(mol . 0c). The water increases in temperature from lS.0°C to 100.0°C. SOLVE

Calculate the number of moles. n

=

237

1 mol H20

.=...bl....Cl.

~

X ----

18.02

g---Up

=

13.2 mol H 2

°

Using the number of moles of H20, its molar heat capacity, and the temperature change, we solve for the heat required using Equation 5.11:

q

=

nCp!1T

= (13.2 mel) ~.~~

(100.0 - lS.0)~

= 8.45 X 104 J = 84.5 kJ ABOUT IT Using the definition of molar heat capacity, we can solve this exercise even without Equation 5.11. Molar heat capacity (cp) defines the amount of heat needed to raise the temperature of 1 mole of a substance by 1°C. Multiplying that value by the number of moles of substance at issue (in this case, 13.2 moles) gives the amount of heat needed to raise the temperature of 13.2 moles of water by 1°C. We want to raise the temperature by 8ST, so multiplying by 85°C gives us the amount of heat needed to raise 13.2 moles of water from 1ST to lOOT.

THINK

Practice Exercise Calculate the heat released when 125 g of water vapor at 100.0°C condenses to liquid water and then cools to 2S.0°C. (Answers


of the

book.)

I


I 205

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I Chapter Five

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Thermochemistry

Water is an extraordinary substance for many reasons, but its high molar heat capacity is one of the more important. The ability of water to absorb large quantities of heat is one reason it is used as a heat sink, both in the radiators of cars and in our own bodies. Weather and climate are driven in large part by cycles involving the retention of heat by large bodies of water.

Cold Drinks on a Hot Day Let's consider another case in heat transfer. Suppose we throw a party in the summer and plan to chill three cases (72 aluminum cans, each containing 355 mL) of our favorite beverages by placing them in the bottom of an insulated cooler and covering them with ice cubes. If the temperature of the ice (sold in 10-pound bags) is -8.0°C and the temperature of the beverages is initially 25.0°C, how many bags of ice do we need to chill the cans and their contents to O.O°C (as in "ice cold")? Perhaps you have confronted this situation in the past and already have an idea that more than 1 bag, but probably fewer than 10, will be needed. With a little mathematical analysis, we can predict more exactly how much ice is required. In doing so, we assume that whatever heat is absorbed by the ice is lost by the beverage cans. As ice absorbs heat from the cans, its temperature will increase from -8.0°C to O.O°C, and the resulting liquid water should remain at O.OOCuntil all the ice has melted (see the heating curve in Figure 5.17). We need enough ice so that the last portion melts just as the temperature of the beverage in the cans reaches that of the melting ice, 0.0°c. The cooling process is a little simpler because there is no phase change, only the cooling of liquid beverage and the solid aluminum cans. Let's consider the heat lost in the cooling process first. Two materials are to be chilled: 72 aluminum cans and 72 X 355 mL = 25,560 mL of beverage. The beverages are mostly water. The other ingredients are present in such small concentrations that they will not significantly affect our calculation, so we assume that we need to reduce the temperature of what is essentially 25,560 mL of water by 25.0OC. We can calculate the amount of heat lost with Equation 5.11 if we first calculate the number of moles of water in 25,560 mL of water: 1 mol H20 X ---18.02 g-l=Ip

25,560 ffil" H20 X 1.00 ~

= 1418 mol H20

We can use the molar heat capacity of water and Equation 5.11 to calculate the heat lost by 1418 moles of water as its temperature decreases from 25.0°C to O.O°C:

q

=

nCpD.T

= 1418 mel X ~~

=

-2.66

X 106

k

X (-25.0~)

J

We must also consider the energy needed to lower the temperature of 72 aluminum cans by 25.0°C. Suppose the mass of each can is 12.5 g. The molar heat capacity of solid aluminum (Table 5.1) is 24.4 (J/mol . OC) and its molar mass is 26.98 glmol. Using these values in Equation 5.11:

5.4

= 72 €QH& X

=

-2.03

12.5

g-Al x 1 mel

X

26.98 g

eaR

24.4 J

mel . ~

X (-25.0~)

X 104 J

The total quantity of energy that must be removed from the cans of beverage is the sum of the heat removed from the beverage and the heat removed from the cans: q total

lost

= q beverage + q cans = [(-2.66 X 106) + (-2.03 X 104)] J = (-2.66 -0.0203)

=

X 106 J

= -2.68

X 106 J

X 103 kJ

-2.68

This quantity of heat must be absorbed by the ice as it warms to its melting point and then melts. The calculation of the amount of ice that is needed is an algebra problem. Let x be the number of moles of ice needed. Then the heat absorbed is the sum of the heat needed (1) to raise the temperature of x moles of ice from -8.0 to O.OT and (2) to melt x moles of ice. The quantities of heat required can be calculated with Equation 5.11 for step 1 and Equation 5.12 for step 2. Note that cpis in units of joules, while 6.Hfus is in kilojoules. We need to make sure both terms are expressed in the same units: 0

qtotalgained

= qj + q2 nCp6.T

+

n

6.Hfus

37.1 J ) T (8.0T) mol·

+x

= x( =x

(6.01 kJlmol)

(6.31 kJlmol)

The heat lost by the cans of beverage balances the heat gained by the ice:

-q

= +qtotal gained

total lost

2.68 X 103 kJ x

= x (6.31 kJlmol) = 4.25 X 102 mol ice

Converting 4.25 X 102 moles of ice into pounds gives us (4.25 X 102 mel) X

18.02.Q:

mel

0

X

lIb 453.6 g

= 171b ofice

Thus we will need at least two 10-pound bags of ice to chill three cases of our favorite beverages. CONCEPT

TEST

The heat lost by the beverage inside the 72 cans in the preceding illustration was more than 100 times the heat lost by the cans themselves. What factors contributed to this large difference between the heat lost by the cans and the heat lost by their contents? (Answers to Concept Tests are in the back of the book.)

I


I 207

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Thermochemistry

Calculating Temperature at Thermal Equilibrium If you add 250.0 g ofice that is initially at -18.0T to 237 g (1 cup) offreshly brewed tea initially at 100.0T, and the ice melts, what would be the final temperature of the tea? Assume that the mixture is an isolated system (in an insulated container) and that 237 g of tea has the same molar specific heat as 237 g of water. COLLECT AND ORGANIZE Before solving this problem, we need to think about the changes that take place during the process in which the hot tea and the ice cubes come into contact. Assuming the ice cubes melt completely, three heat-

flows

(q" q2' and q3) consume the heat lost by the hot tea:

q,: raising the temperature of the ice cubes to their melting point (OT); q2: melting the ice cubes; 3. q3: bringing the mixture to thermal equilibrium, in which the temperature

1. 2.

of the melted ice rises and the temperature of the tea (TIOltia .. I = 100.0T) falls to the final temperature of the mixture (Tfina1 = ?). Let's begin by noting that the heat gained by the ice cubes equals the heat lost by the tea:

ANALYZE

qicc cubes

Based on our organization

=

-qtca

of the information, we know that

qice = q, + q2 + q3 SOLVE

The heat lost by the tea as it cools from 100.0°C to its final temperature

(Tfinal ) is

= 237 g X = (990

1ffiGl 75.3J 80 X 00 X 1.2g ffiGl.

JrC)

(Tfina1 -

0

(Tfi

Ina

100.0 C)

100.0°C)

The transfer of this heat is responsible for the changes in the ice. We can treat the process of heat transfer from the hot tea to the ice cubes in terms of the processes we identified. In step 1, the ice is warmed from -18.0°C (Tlfi1t1a .. I) to its melting point (OT; Tfinal in this step):

q, = nCpt..T =2500

. g

X 1ffiGl X 37.1J 18.02g ffiGl·~

X[0~-(-18.0~)]

= 9.26 X 103 J In step 2, the ice melts, requiring the absorption of the following quantity of heat based on t..Hfus:

= 250.0g

X 1 ffiGl X 6.01 kJ _ 18.02 g ffiGl - 83.4 kJ

In step 3, the melted ice water warms to the final temperature (where t..T = Tfi na I - TIfllt1a .. I = Tfi na I - O°C):

5.5

q3

=

nH20

CH20(C)

I

Calorimetry: Measuring Heat Capacity and Calorimeter Constants

I

llTH20 1 mel

75.3 J

°

= 250.0 g X 18.02 g X mel. 0C X (Tfina1- 0 C)

=

1045 JloC X Tfina1

As we indicated at the outset, qicc' the sum of the amount of heat absorbed by the ice during steps 1 through 3, must balance the heat lost by the hot tea:

9260 J Expressing 9.26 kJ

+

83.4 kJ

+

(1045 Jl°C)(Tfinal)

= - [(990 Jl0C) (Tfina1- lOOT)]

all values in the same units (in this case, in kilojoules), we get:

+

Rearranging

83.4 kJ

+

(1.045 kJlT)(~nal)

= -

[(0.990 kJlT)

(Tfina1- lOOT)]

the terms to solve for ~naI' we have:

(2.04 kJlT)Tfina1

= -9.26 kJ - 83.4 kJ + 99.0 kJ = 6.34 kJ

= 3°C Tfina1 ABOUT IT In the statement of the problem, the setup was described as an isolated system. If you were to carry out this experiment in a real ceramic mug instead of an ideal insulated container, do you think the final temperature of the solution at thermal equilibrium would be 3°C, as your calculation predicted? THINK

Practice Exercise Calculate the final temperature ice cubes at -18°C (Answers

~

of a mixture of 350.0 g of

and 237 g of water at 100.0°C.


of the

book.)


Up to this point, we have discussed heats of reaction and heats associated with phase changes, but we have not broached the issue of how we know these values. The experimental approach for accurately measuring the quantities of heat associated with chemical reactions and physical changes is called calorimetry. The device used to measure the heat released or absorbed during a process is a calorimeter.

Determining Heat Capacity and Specific Heat When we determined the amount of ice needed to cool 72 cans of a beverage, we used the molar heat capacity of aluminum to determine the quantity of heat taken up by the cans. How are heat capacities determined? We can apply the first law of thermodynamics and design an experiment to determine the specific heat of aluminum, from which we can calculate the molar heat capacity. The units of specific heat tell us exactly what we have to do. "Determine the specific heat of aluminum" means determine the amount of heat required to change the temperature of 1 g of aluminum by 1°C. If we determine how much heat is required to change the temperature of any known mass of aluminum by any measured number of

Calorimetry is the measurement of the change in heat that occurs during a physical change or chemical process. A calorimeter is the device used to measure the absorption or release of heat by a physical or chemical process.

209

210

I Chapter Five

I

Thermochemistry

100.0T-

26.0T-

23.0T-

FIGURE 5.19 Experimental setup to determine heat capacity of a metal: (a) a pile of pure aluminum beads with a mass of23.5 g is heated to 100.OT in boiling water; (b) 130.0 g of water at a temperature of 23.0°C is in an insulated beaker. (c) The hot Al beads are dropped into the water, and the temperature when thermal equilibrium is reached is 26.0T.

(a)

Insulation

Insulation

(b)

(c)

degrees, we can calculate how much heat is required to change the temperature of 1 g of aluminum by 1°C. Suppose we have beads of pure aluminum with a total mass of23.5 g and want to transfer a known amount of heat to or from the aluminum. We can rely on one fact and two definitions to help design the experiment: (1) the specific heat of water [4.184 JI(g . 0C)], (2) the definition of thermal equilibrium, and (3) the first law of thermodynamics. Refer to Figure 5.19 for our discussion. First, we bring a quantity of water to its boiling point (100.0T) in a beaker containing a test tube holding the aluminum beads (the system). We wait a few minutes while the beads and the boiling water come to the same temperature: 100.0°C (Figure 5.19a). While the metal is heating, we place a measured amount of water (in this case, 130.0 g) in a Styrofoam block with a cavity and with a Styrofoam lid (Figure 5.19b). The Styrofoam insulation effectively isolates our experiment from the rest of the universe and enables us to define the water in the device as the surroundings of the metal (the system). We place a thermometer in the water, read the temperature (room temperature = 23.0°C), and leave the thermometer in it. When we have waited long enough for the Al beads (the system) to heat up to 100.0°C, we remove the test tube from the boiling water, drop the aluminum beads into the water (the surroundings) in the insulated container, and close the lid (Figure 5.19c). The temperature of the water rises because it is now in contact with the hot aluminum. Because of the insulation, we assume that all the heat coming from the aluminum beads goes into the water. Suppose the temperature of the aluminum-water mixture rises to 26.0°C, at which point it does not change anymore, so the system is at thermal equilibrium. Heat is supplied from the hot beads of aluminum to the water. From the first law of thermodynamics, we can state that -qaluminum

qwater

Now we have everything we need to calculate the specific heat of aluminum. Note that the water in Figure 5.19(b) is insulated from the surroundings. The boiling water in Figure 5.19(a) was just the vehicle we used to bring the aluminum to some known temperature. First, we determine the amount of heat gained by the water:

5.5

where

f:,Twatcr

=

Tfina1 -

=

(130.0 g H20)

I


The enthalpy of reaction (!!'H ) is the heat absorbed or given off by a'~'hemical reaction. It is also called the heat of reaction.

= (26.0 - 23.0rC = + 3.0°C

I;nitial

= +1600 J

J) (3.0°C)

4.184 ( g' C -0-

I 211

A bomb calorimeter is a constant-volume device used to measure the heat of a combustion reaction.

The heat gained by the water has a positive sign. The water (the surroundings) absorbed heat from the system. The heat that changed the temperature of the water came from the beads. Therefore, nCs!1~luminum

-qa!uminum

where f:,Ta1. umlOum -qaluminum

= Tfi na 1 - T10itia .. 1 = (26.0 -100.0rC

=

= -74.0°C

(23.5 g)(c)( -74.0°C)

where cs' the specific heat of aluminum, is the unknown. The heat lost by the system (the aluminum) has a negative sign. In terms of the heat-transfer part of the experiment, the aluminum starts out at 100.0°C and drops in temperature when it transfers heat to the water. Recalling the first law of thermodynamics, we can write (23.5 g)(c)(-74.0°C)

~II

CHEMTOUR

Calorimetry

= -(+1600 J)

and solve for cs :

0.920 J c =--S g' OC The molar heat capacity of aluminum is the specific heat multiplied by the mass of 1 mole of aluminum: 0.910 24.8 c X .Aft = ( (26.98 g/mol) = _ g' C mol. --0

-

J)

J

S

CONCEPT TEST

In experiments to calculate very precise values of specific heats, thermometers capable of measuring to the nearest O.OOl°C are used. What impact would this change in equipment have in the numerical values reported for specific heats? (Answers to Concept Tests are in the back of the book.)

Thermometer Wire for ignition

Calorimeter Constants The heat change accompanying a chemical reaction, an important characteristic, also called is defined by a quantity known as the enthalpy of reaction (f:,Hrx)' the heat of reaction. The subscript "rxn" stands for "reaction." The subscript may be changed to reflect the specific type of reaction studied; for example, f:,Hcomb stands for the enthalpy change of a combustion reaction. Heats of reaction are measured with a device called a bomb calorimeter (Figure 5.20). The sample is placed in a sealed vessel (called a bomb) capable of withstanding high pressures. Oxygen is introduced into the vessel, and the mixture is ignited with an electric spark. As combustion occurs, heat generated by the reaction flows into the walls of the bomb and the water surrounding the bomb, all of which is inside a heavily insulated container. A good bomb calorimeter contains the system (the chemical reaction) within the bomb and ensures that all heat generated by the reaction is taken up by the calorimeter. From the point of view of the reactants (the system),

Stirrer

Steel "bomb"

FIGURE 5.20 A bomb calorimeter.

212

I

Chapter Five

I

Thermochemistry

The heat capacity of a calorimeter calorimeter

constant (C"10,;m,,,).

is its

the surroundings (the rest of the universe that needs to be considered) consists of the bomb, the water, the vessel in which they are contained, and anything else in that vessel. Insulation ensures that no heat escapes from the calorimeter; thus from the system's point of view, the rest of the universe (the surroundings) consists of only the calorimeter and its contents. The calorimeter is defined as everything outside the system. The heat gained by the calorimeter is determined by measuring the temperature of the water before and after the reaction takes place. The water is in thermal equilibrium with the parts of the calorimeter it contacts-the walls of the bomb, the thermometer, and the stirrer-so the temperature change of the water takes the entire calorimeter into account. Measuring the change in temperature, however, is not the whole story. We also need to know the heat capacity of the surroundings (the calorimeter). Because this is a value unique to every calorimeter, it is frequently referred to as the calorimeter constant (C caI.onmetcr ). If we know the value of Cca 1.onmetcr and if we can measure the change in temperature (t:,T), we can calculate the quantity of heat (q) that flows from the reaction into the calorimeter:

q=

Ccalorimeter

t:,T

(5.14)

Rearranging the terms in Equation 5.14, we have the following expression for the heat capacity of the surroundings: q Ccalorimeter

= t:,T

(5.15)

Equation 5.15 indicates that heat capacity is expressed in units of heat/temperature, or kilojoules per degree Celsius (kJrC). Equation 5.15 can be used to determine the heat capacity of a bomb calorimeter. To determine Ccalorimeter' we must completely burn a quantity of material in the calorimeter that produces a known quantity of heat when it burns. Benzoic acid (C7H602) is often used for this purpose, because it can be obtained in a form that is both very pure and very dry. The combustion of 1 gram of benzoic acid is complete and produces 26.38 kJ of heat. Once the heat capacity of a calorimeter has been determined, the calorimeter can be used to determine t:,Hcomb for other substances, and the observed increases in temperature as a result of combustion reactions can be used to calculate the quantities of heat that they produce on a per-gram or per-mole basis. Because there is no change in the volume of the reaction mixture, this technique is referred to as constant-volume calorimetry. No P t:,V work is done, so according to Equation 5.15 the heat lost by the system equals the internal energy lost by the system during combustion: qsys

= t:,E comb

(5.16)

Other types of calorimeters allow the volume to change while the pressure remains constant, so qsys in those cases equals the enthalpy of reaction (t:,H rxn ). The pressure inside a bomb calorimeter may change as a result of a combustion same as t:,H com b' However, reaction, and q sys (t:,E com b) in that case is not exactl11the J the pressure effects are usually so small that t:,Hcomb is nearly the same as t:,Ecomb' and we do not need to be concerned about the very small differences. Hence we shall discuss enthalpies of reactions (t:,HrxJ throughout and apply the approximate relation

= !:1Hcomb

q calorimeter or even more generally

qrxn

-

qcalorimeter

(5.17)

5.6

I

Enthalpies of Formation and Enthalpies of Reaction

I 213

Determining a Calorimeter Constant Before we can determine the heat of reaction, we must determine the heat capacity of the calorimeter being used. What is the calorimeter constant, Ccalorimeter' of a bomb calorimeter assuming that combustion of1.000 g of benzoic acid causes the temperature of the calorimeter to rise by 7.248°C? AND ORGANIZE We are asked to find the heat capacity of a calorimeter. That means we need to determine the amount of heat required to raise the temperature of the calorimeter by 1°C. We are given data describing how much the temperature of the calorimeter rises when benzoic acid is burned in it. To proceed, we need the heat of reaction of benzoic acid; it was given earlier in this section as 26.38 kJlg. COLLECT

A N A L Y Z E The heat capacity of a calorimeter

can be calculated using Equation 5.15 and the knowledge that the combustion of 1.000 g of benzoic acid produces 26.38 kJ of heat.

q

SOLVE

Ccalorimeter

= f:..T 26.38

kJ

7.248T

=

3.640 kJlT

Now this calorimeter can be used to determine the heat of combustion of any material. T HI N K ABO UTI T The calorimeter constant is determined for a specific calorimeter.

If anything changes-if

the thermometer breaks and has to be replaced, or if the calorimeter loses any of the water it contains-a new constant must be determined.

Practice Exercise Assume that when 0.500 g of a mixture of volatile hydrocarbons is burned in the bomb calorimeter described in Sample Exercise 5.7, the temperature rises by 6.76°C. How much energy (in kilojoules) is released during combustion? How much energy would be released by the combustion of 1.000 g of the same mixture? (Answers

~



As noted in Section 5.2, it is impossible to measure the absolute value of the internal energy of a substance. The same is true for the enthalpy of a substance. However, we can establish relative enthalpy values that are referenced to a convenient standard. This approach is similar to using the freezing point of water as the zero point on the Celsius temperature scale or sea level as the zero point for expressing altitude. The zero point of enthalpy values is the standard enthalpy of formation (f:..H;), defined as the enthalpy change that takes place when 1 mole of a substance in its standard state is formed from its constituent elements in their standard states. A reaction that fits this description is known as a formation reaction. The standard state of an element is its most stable physical form under standard conditions of pressure (1 bar) and some specified temperature. This value of pressure is very close to 1 atmosphere; for the level of precision

The standard enthalpy of formation (AH;), also called the standard heat offormation (or simply the enthalpy of formation or heat offormation), is the enthalpy change of a formation reaction. A formation reaction is the process of forming 1 mole of a substance in its standard state from its component elements in their standard states. The standard state of a substance is its most stable form under 1 bar pressure and some specified temperature (assumed to be 25.0°C unless otherwise stated).

214

I Chapter Five

I

Thermochemistry

required in the data used in this book, a standard pressure of 105 pascals will be considered equivalent to 1 atm. Enthalpy change is equivalent to the flow of heat at constant pressure, so 6.H; is frequently called the standard heat offormation or simply the heat offormation. By definition, 6.H; of a pure element in its most stable form under standard conditions is zero. In their standard states, oxygen is a gas, water is a liquid, hydrogen is a gas, carbon dioxide is a gas, and carbon is the solid graphite. Because the definition of a formation reaction specifies 1 mole of product, writing balanced equations for formation reactions may require the use of something we avoided in Chapter 3: fractional coefficients. For example, the reaction for the production of ammonia from nitrogen and hydrogen is usually written O

6.H

rxn

=

-92.2

kJ

Although all reactants and products in this equation are in their standard states, it is not a formation reaction because 2 moles of product are formed, which is why we denote the heat of this reaction by 6.H~xn rather than 6.H;. To write the formation reaction of ammonia, all coefficients in the equation are divided by 2 (to make the coefficient ofNH/g) equal to 1). Because heat is stoichiometric, the heat of reaction is divided by 2 as well to determine 6.H;: 6.H;

Recognizing

= -46.1 kJ

Formation

Reactions

Which of the following reactions are formation reactions at 25°C? For those that are not formation reactions, explain why they are not. a. H2(g)

+ ~ 02(g)

~ H20(g)

b. CgraphitJs) +32 H2(g) + ~ o/g) ~ CH30H(e) c. CH4(g) + 2" 02(g) ~ CO2(g) + 2 H20(e) d. P/s) + 2 (g) + 6 CI2(g) ~ 4 pOCI/e) (POCI3 is a liquid at room temperature.)

°2

AND ORGANIZE For a reaction to be a formation reaction, it must be a reaction that produces 1 mole of a substance in its standard state from its component elements in their standard states. The standard state is the most stable form of a substance at 1 atm pressure and some specified temperature.

COLLECT

A N A LY Z E Each reaction must be evaluated for the quantity of product and for the

state of the substances involved. (a) The reaction shows 1 mole of water being formed from its constituent elements, but water appears as a gas. The standard state of water at 2S.0°C and 1 atm pressure is liquid, so this is not a formation reaction; it is not at standard conditions, and the heat of this reaction would be symbolized 6.Hrxn. (b) Methanol (CH30H) is a liquid under standard conditions; 1 mole of it is formed from its constituent elements in their standard states. This reaction therefore is a formation reaction, and its heat change would be symbolized 6.H;. (c) This is not a formation reaction because the reactants are not elements in their standard states, and more than one product is being formed. The change in heat of this reaction would be symbolized 6.Hcomb'

SOL V E

5.6

I


(d) This is not a formation reaction because the product is 4 moles of POCl3, whereas in a formation reaction, by definition, only 1 mole of product is formed. Note, however, that all of the constituent elements and the product are in their standard states and that only one compound is formed. We could therefore call o this reaction, as written, a standard heat of reaction (!1H rxn ). We could easily convert this into a formation reaction by dividing all the coefficients in the balanced equation by 4; the standard heat of reaction (!1H:x)' also divided by 4, would then be equal to the standard heat off ormation (!1H;).

The standard

enthalpy

I

215

of reaction

(!1.Ho ) is the heat associated with a react'(~n that takes place under standard conditions. Also called standard heat of reaction.

Just because we can write formation reactions for substances like methanol does not mean that anyone would ever use that reaction to make methanol. Remember that formation reactions are defined to provide a standard against which other reactions can be compared when their thermochemistry is evaluated. T HI N K ABO UTI T

Thermodynamic Properties at 25°C

TABLE 5.2

Practice Exercise Write formation reactions for the following substances: (a) CaCO/s); (Answers

(b) CH3COOH(t')

(acetic acid); (c) KMnO/s).


of the

book.)

Regarding the symbol !1H;, the subscript f stands for "formation," but the meaning of the superscript is not as obvious. Throughout thermochemistry this superscript refers to the value of a parameter, such as !1Hrxn' under standard conditions. The symbol of the enthalpy change associated with a reaction that takes place under standard conditions is thus !1H;xn, and the value is called a standard enthalpy of reaction. Implied in our notion of standard states and standard conditions is the assumption that parameters such as !1H change with temperature and pressure. That assumption is correct, although the changes are so small that we ignore them in many of the calculations in this chapter and those that follow. Table 5.2 lists thermochemical values for some compounds, primarily hydrocarbons. A more complete list of these values for a wide variety of compounds can be found in Appendix 4. Note that the standard enthalpies of formation of the acetylene and ethylene in Table 5.2 are actually positive, which means that the reactions in which these compounds are formed from their component elements are endothermic. The other hydrocarbon fuels in Table 5.2 have negative enthalpies of formation, but the values for all the fuels are less negative than those for the formation of water and carbon dioxide. Recall from Chapter 3 that the products of the complete combustion of hydrocarbons are carbon dioxide and water (C02 and H20). The more negative the heat of formation, the more stable the substance. The implication of this fact is something stated earlier: the reaction of fuels with oxygen to produce CO2 and H20 is exothermic. The reaction produces heat, which is why hydrocarbons are useful as fuels. Note also that the values for the C2 to C 4 hydrocarbons become more negative with increasing molar mass. Knowing the standard enthalpies of formation of substances is useful for predicting heats of reaction. We can calculate enthalpy changes by determining the difference in !1H; between products and reactants in a balanced chemical equation. To see how this approach works, let's calculate !1H:omb for the combustion of methane

~c ..:::"u::u

lUI

Compounds

I

A. STANDARD ENTHALPIES OF FORMATION Substance

0

6H~(kJjmol)

Water(g)

-241.8

Water(C)

-285.8

CH4(g):

--

methane

-74.8

C2H/g):

acetylene

C2H.(g):

ethylene

C2H/g):

ethane

226.7

--

52.26 -84.68

C1Hg(g): propane

-103.8

C 4H,0(g): butane

-125.6

CO2(g)

-393.5

CO(g)

-110.5

--

NH3(g):

ammonia

N2H.(g):

hydrazine

N2H.(C):

hydrazine

CHPH(C):

-46.1 95.4 50.63 -238.7

methanol

CH1CHPH(C):

-277.7

ethanol

CH1COOH(C): acetic acid

-484.5

B. STANDARD ENTHALPIES OF COMBUSTION Substance

6H~omb (kJjmol)

CO(g) C5H12(C):

-283.0 pentane

-3535

C9H20(C) (avg. gasoline molecule)

I

C14H30(C) (avg. diesel molecule)

-6160

-

-7940

-

-

216

I

Chapter Five

I

Thermochemistry

using the values from Table 5.2. In this calculation, we make use of the following equation:

t..H

I

O

CH4 Methane

"]

C: HI:

4

(CH4)

is a renewable source of

t..H;xn

swamps, hence commonly Experiments methane

called swamp gas.

=

[(-393.5 kJ) + (-483.6 kJ)] - [(-74.8 kJ) + (-0.0 kJ)]

CO2)(-393.5 kJl~) CH4)(-74.8 kJl~)

+ (2 ~ H20)(-241.8 + (2 ~ 02)(0.0 kJl~)]

kJl~)]

=

-

-803.3 kJ

of

from natural sources, like the one

shown here of a plastic tarp covering a lagoon of animal waste, are being carried out worldwide

(5.18)

[(1 ~ [(1 ~

is produced in

in the bulk production

m t..Hof,reactants

=

of organic matter by methano-

genic bacteria. Methane

I

n t..Hof,products -

Equation 5.18 states that t..H;xn equals the sum (I) of the t..H; values for the number of moles of each product minus the sum of the t..H; values for the number of moles of each reactant in the balanced chemical equation. Thus we multiply the values of t..H; for 02(g) by 2 before summing with t..H; for CH/g) because the coefficient for 02 is 2 in the balanced equation. We do the same with t..H; for water in the product. Inserting values for the products (C02 and H20) and the reactants (CH4 and 0) and their coefficients from the balanced chemical equation into Equation 5.18, yields the following:

energy because it can be produced by degradation

rxn

Calculations of t..Hr:n from t..H; values can be carried out for all kinds of chemical reactions. Consider the following overall reaction between methane and steam, which yields a mixture of hydrogen and carbon monoxide known as water gas:

to augment the fuel supply.

This reaction is used to synthesize the hydrogen used in the steel industry to remove impurities from molten iron and in the chemical industry to make literally hundreds of compounds including, for example, ammonia, NH}' for use in fertilizers and industrial refrigerants. It is also important in the manufacture of hydrogen fuel for fuel cells, which are used to generate electricity. Inserting the appropriate values for the compounds from Table 5.2 into Equation 5.18 along with the coefficient 3 for H2, we have the following:

t..H;xn

=

[(1 ~ [(1 ~

kJl~) kJl~)

CO)(-1l0.5 CH4)(-74.8

H)(O.O kJl~) H20)(-241.8 kJl~)]

+ (3 ~ + (1 ~

= +206.1 kJ The positive enthalpy change tells us that this reaction is endothermic. Therefore, heat must be added to make the reaction take place. Remember that values for changes in enthalpy are independent of pathway, as Figure 5.21 illustrates: It

CO(g)

CO (g) +3 Hh)

t

r

tlHrxn

CH4(g)

+

H20(g)

reactants

(a) Progress of reaction

products =

+206.1 kJ

t

+

3 H2(g)

reactants

1

Mfrxn = -206.1 kJ

CH4(g)

+

H20 (g)

products

~

(b) Progress of reaction

~

FIGURE 5.21 (a) The reaction of methane with water to produce carbon monoxide and hydrogen

is endothermic.

produce methane

(b) The reverse of the reaction,

and water, is exothermic.

has the same magnitude

hydrogen

plus carbon monoxide

The value of the enthalpy

but is opposite in sign.

to

change of the two reactions

5.6

I


does not matter what path we take to get from the reactants to the products; the enthalpy difference between them will always be the same. And as we saw in Section 5.2 with phase changes, once we know the value of the heat associated with running a reaction in one direction, we also know the value of the heat associated with running the reaction in the reverse direction because it has the same magnitude but the opposite sign:

The steam-reforming (or steam-methane reforming) process, as it is called in industry, is typically conducted at temperatures near 1000°C. Although hydrogen is attractive as a fuel because it burns vigorously and produces only water as a product, its production by steam reforming does not reduce the use of fossil fuels but merely shifts them to an early step in the production process. Fossil fuels other than methane (natural gas) are also used as the starting materials in steam reforming; the reaction is still endothermic and requires the input of energy, which means more fossil fuels are consumed; and CO2 is released into the environment as a result of the production of hydrogen in this way.

Calculating Standard Heats of Reaction Calculate I1H;xn for the combustion of propane (C3Hg) in air using the appropriate values from Table 5.2. COLLECT AND ORGANIZE A typical combustion reaction, by definition, involves oxygen as a reactant. Propane is a hydrocarbon, so the products are water vapor (steam) and carbon dioxide. The values we need from Table 5.2 are the standard enthalpies of formation of the reactants and products in the equation for the combustion of propane. We also need Equation 5.18, which gives the relation between enthalpies of formation of reactants and products and the standard enthalpy of reaction.

A N A L Y Z E The balanced equation for combustion of propane is

Inserting values for the products [C02(g) and H20(g)] and reactants [C3H/g) and a/g)] from Table 5.2 and the coefficients in the balanced chemical equation into Equation 5.18:

SOLVE

kJlffiel) + (4 ffiel H20)(-241.8 kJlffiel)] [(1 ffiel C3Hg)( -103.8 kJlffiel) + (5 ffiel 0)(0.0 kJlffiel)]

I1H;xn = [(3 ffiel CO2)(-393.5

+

= -2043.9 kJ T HI N K ABO UTI T Hydrocarbons such as methane, ethane, and propane are components of natural gas and are excellent fuels. They are, however, nonrenewable fuels, and their combustion produces carbon dioxide.

Practice Exercise Calculate I1H;xn for the water-gas shift reaction, in which the carbon monoxide formed by the steam-reforming process is reacted with more steam, producing CO2 and H2:

(Answers


of the

book.)

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CONCEPT

TEST

What is the standard heat of reaction for the production ofC3H8 and oxygen from CO/g) and H20(f)? (See information in Sample Exercise 5.9.) (Answers to Concept Tests are in the back of the book.)

.aI Fuel Values and Food Values The results of calculations in Section 5.6 show that the enthalpy change is much greater and much more heat is released in the combustion of a mole of propane (-2043.9 kJ) than in the combustion of a mole of methane (-803.3 kJ). Does this make propane an inherently better (that is, higher-energy) fuel? Not necessarily. Expressing 6.H values on a per-mole basis is the only way to insure that we are talking about the same number of molecules. That approach can be very important in certain analyses. However, we do not purchase fuels, or anything for that matter, in units of moles. Depending on the fuel, we buy it in units of mass or volume.

Fuel Value To calculate the enthalpy change that takes place when 1 gram of methane or 1 gram of propane burns in air, we divide the absolute value of 6.H;xn (in units of kilojoules per mole) for each of the combustion reactions by the number of grams in 1 mole of the compound to determine the number of kilo joules of heat released per gram of substance:

803.3 kJ ffiel

X

1 ffiel = 50.08 kJlg 16.04 g

kJ

X

1 ffiel = 46.35 kJlg 44.10 g

2043.9 ffiel

These quantities of energy per gram of fuel are called fuel values. If we carry out similar calculations for the other hydrocarbons in natural gas, we can determine the fuel values for the compounds listed in Table 5.2. Fuel values of hydrocarbons decrease with increasing molar mass. Why should this be the case? The answer lies in the hydrogen-to-carbon ratio in these compounds. As the number of carbon atoms per molecule increases, the hydrogento-carbon ratio decreases. Consider that there are 12 times as many H atoms in a gram of hydrogen as there are C atoms in a gram of carbon; this means that, gram for gram, 1 g of hydrogen can form six times as many moles of H20 as 1 g of carbon can form moles of CO2, More energy is released in a combustion reaction by the formation of a mole of CO2 (393.5 kJ) than by a mole ofH20(g) (241.8 kJ), but even if we take this difference into account, gram for gram hydrogen has many times the fuel value of carbon.

CONCEPT

TEST

Without doing any calculations, predict which of the following quantities of fuel releases more heat during combustion in air: (a) 1 mole of CH4, or 1 mole of H2; (b) 1 gram ofCH4, or 1 gram ofH2. Fuel value is the energy released during complete combustion of 1 g of a substance.

(Answers to Concept Tests are in the back of the book.)

5.7

I

Fuel Values and Food Values

I 219

Comparing Fuel Values The great majority of automobiles run on either gasoline or diesel fuel. Although both fuels are mixtures, the energy content in regular gasoline can be approximated by considering it to be composed of hydrocarbons with the formula C9H20 (d = 0.718 g/mL), while diesel fuel may be considered to be C14H30 (d = 0.763 g/mL). Using these two formulas as approximations of the two fuels, compare the fuel values (a) per gram of, and the fuel density (b) per liter of, gasoline and diesel fuels. Fuel values can be used to calculate the amount of heat energy available from a given quantity of fuel. The heats of combustion in Table 5.2 are given in units of kilojoules per mole, so to answer this question in terms of grams and liters, some conversions are necessary. For (a), we need molar masses to convert moles to grams and for (b) we need density (d = mIV), typically in grams per milliliter, to convert grams to liters. COLLECT

AND

ORGANIZE

Taking C9H20 and C14H30 as the average molecules in regular gasoline and diesel fuels, respectively, we can determine their molar masses. Then the fuel value per gram can be calculated from the fuel values given on a per-mole basis. Using the densities given in the problem, we can calculate fuel values on a per-liter basis.

ANALYZE

SOLVE

kJ ffiel

a. Gasoline as C9H20: 6160 -

.

DIesel as C14H30:

.

b. Gasolme as C9H20:

.

DIesel as C14H30:

X

1ffiel 128.25 g

= 48.0

kJlg

kJ 1ffiel 7940 X 98 8 = 40.0 kJlg ffiel 1 .3 g 48.0 kJ g 1000 ~ -g- X 0.718 ~ X • = 34,500 kJlL T

40.0

kJ

gX

g

0.763 ~

X

1000~ •

T

= 30,500 kJlL

ABOUT IT In the United States, fuel for cars and trucks is bought by the gallon and its consumption is rated in terms of miles per gallon, whereas in most countries it is bought by the liter (1 gal = 3.7854 L). Because of the way we buy fuels, it makes sense to compare fuel values in terms of volume, which is also called fuel density.

THINK

Practice Exercise Another hydrocarbon fuel derived from crude oil is kerosene, used as jet fuel to power high-performance aircraft and also in space heaters to heat homes. Kerosene is intermediate in composition between gasoline and diesel fuel, and may be approximated as C12H26 (d = 0.750 g/mL; Micomb = -7050 kJlmol). Estimate the fuel value of kerosene on both a per-gram and a per-liter basis. (Answers to Practice Exercises are in the back

of the

book.)

Food Value Food serves in part the same purpose in living systems as fuel does in mechanical systems. The chemical reactions that convert food into heat and energy resemble combustion but consist of many more steps that are much more highly controlled. Carbon dioxide and water are the end products, so in a fundamental way, metabolism of food by a living system and combustion of fuel in an engine are the same process. The food value of the material we eat-the amount of heat produced when food is burned completely-can be determined using the same equipment

Food value is the amount of heat produced when a material consumed by an organism for sustenance is burned completely. It is typically reported in Calories (kilocalories).

220

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Thermochemistry

and applying the same concepts of thermochemistry that we have developed to discuss fuels for vehicles. We can analyze the relative food value of the items we consume the same way we analyzed fuel value: by burning material in a bomb calorimeter and quantifying the heat released. As an illustration, let's consider the food value of a jelly doughnut. To determine its caloric content, we simply have to burn a jelly doughnut in a calorimeter. Fresh jelly doughnuts may be hard to incinerate, however, because of their high water content, so we prepare the doughnut first by drying it. To get an accurate value of the heat of combustion, we need to make sure that the great majority of the mass we attribute to the sample is due to its carbon-containing components that will burn completely and produce CO2 and H20. After drying completely, suppose our jelly doughnut weighs 55 g. We then ut P it in a large calorimeter, whose calorimeter constant is known (Cca 1.onmctcr = 41.8 kJrC) and burn it completely in excess oxygen. We observe that the temperature of the calorimeter (/:,T) rises by 25.0°C in this experiment. What is the food value of the jelly doughnut? Using Equations 5.14 and 5.17 yields the answer in SI-derived units. First we insert the given values into Equation 5.14 to get the heat value of the calorimeter: qcalorimeter

=

/:,T

Ccalorimeter

=

(41.8 kJl°C)(25.0°C)

=

1040 kJ

Now we insert this result into Equation 5.17 to find the food value of the jelly doughnut:

-q comb,

doughnut

= qcalorimeter = 1040 kJ

Notice that the heat flow from the jelly doughnut has the same value-but opposite sign-as the heat absorbed by the calorimeter. Most of us still think in terms of nutritional Calories (kilocalories), and we can use the definition of a Calorie (abbreviated Cal) from Section 5.1 to convert the heat in kilojoules into that more familiar unit used by nutritionists: 1040

kJ

----------X

jelly doughnut

1 Cal 4.184

kJ

249 Cal = ----------

jelly doughnut

Between 8 and 10 jelly doughnuts would provide all the calories needed by an adult human being with normal activity in one day. In contrast, a bicyclist riding the Tour de France would need the caloric equivalent of about 32 jelly doughnuts a day.

Calculating

Food Value

Glucose (C6H1206) is a simple sugar formed in plants during photosynthesis. The complete combustion of 0.5763 g of glucose in a calorimeter (Cca 1.onmeter = 6.20 kJrC) raises the temperature of the calorimeter by 1.45°C. What is the food value of glucose in Calories per gram? COLLECT AND ORGANIZE We are asked to determine the food value of glucose, which means the heat given off when 1 g is burned completely. We have the mass of glucose burned and the calorimeter constant. Equation 5.17 relates the heat given off by the system (the reaction) to heat gained by the surroundings (the calorimeter). We also have Equation 5.14, which relates the heat given off by the reaction to the temperature change and the calorimeter constant.

5.8

I

Hess's Law

We use the data from the calorimetry experiment to determine how much heat in kilojoules is given off when the stated amount of glucose is burned. We can convert that quantity into Calories by using the equality 1 Cal = 4.184 J. ANALYZE

SOLVE qcalorimctcr

=

Ccalorimctcr

f:,T

=

(6.20 kJlT)(1.45°C)

=

8.99

kJ

qg 1ucose = -q ca 1·onmeter = -8.99 kJ (for combustion of 0.5763 g of glucose) To convert this energy to a fuel value, we divide by the sample mass: -8.99 kJ 0.5763 g

=

-15.6 kJlg

and finally into calories: (15.6 kJ/g)

1 Cal ) ( 4.184 kJ

=

3.73 Cal/g

T The value of the heat calculated with the data from the calorimetry experiment has a negative sign because combustion of a food material, just like combustion of a fuel, is an exothermic process. T HI N K ABO UTI

Practice Exercise Sucrose (table sugar) has the formula C12H22011 (.Att = 342.30 g) and a food value of 6.49 kJlg. Determine the calorimeter constant of the calorimeter in which the combustion of 1.337 g of sucrose raises the temperature by 1.96°C. (Answers

~


of the

book.)

Hess's Law

In Section 5.6 we described a process for synthesizing hydrogen gas from methane and steam at high temperatures. The overall reaction is carried out on the industrial scale in two steps. The first step is an endothermic reaction between methane and a limited supply of high-temperature steam, producing carbon monoxide and hydrogen gas:

In the second step the carbon monoxide from the first reaction is allowed to react with more steam, producing carbon dioxide and more hydrogen gas: f:,H;

=

-41

kJ

We can write an overall reaction equation that combines steps 1 and 2: CH/g)

+ H20(g)

+ CO(g) + H20(g)

~ CO2(g) + H2(g) + CO(g) + 3 H/g)

which simplifies to the following overall reaction:

Just as we obtained the overall chemical equation by adding the equations for steps 1 and 2, we obtain the enthalpy change for the overall reaction by adding the values of f:,Ho for steps 1 and 2. The thermochemical equation for the overall reaction is f:,H;

= +165

kJ

~II

CHEMTOUR

Hess's Law

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Chapter Five

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Thermochemistry

Hess's law states that the enthalpy change of a reaction that is the sum of two or more reactions is equal to the sum of the enthalpy changes of the constituent reactions.

In summary: (1)

+ H20(g)

CH4(g)

+

(2)

row + H 0(g) 2

~

row + 3 H (g)

t..H~= +206kJ

2

~ CO2(g)

t..H; = -41 kJ

+ H2(g)

t..H; = +165 kJ

Overall reaction:

This calculation of t..Ho for a given reaction by summing the enthalpy changes of the steps within the overall process is an application of Hess's law. Also known by the longer name Hess's law ofconstant heat ofsummation, this law states that the enthalpy change of a reaction that is the sum of two or more other reactions is equal to the sum of the enthalpy changes of the constituent reactions. This relation is illustrated in Figure 5.22.

CO(g) + 3 H2(g)~41

FIGU RE 5.22

and CO2(g) from CH4(g) sum of the enthalpies formation

4 H (g) + COh) 2

+H20~

Hess's law predicts that the

enthalpy change for the production

kJ

ofH2(g)

and HP(g)

is the

of two reactions: the

Mil

=

+206 kJ

of carbon monoxide and hydrogen

from methane

and water (f'..H,), and the

reaction of carbon monoxide with water to produce carbon dioxide and hydrogen

(f'..H).

Let's consider another case. Hess's law is perhaps most useful for calculating energy changes that are difficult to measure directly. For example, CO2 is the principal product of the combustion of carbon (as in charcoal):

C(s)

+ 02(g)

~ CO2(g)

However, CO, a highly toxic gas, is also produced when the oxygen supply is limited:

C(s)

+ ~ a/g)

~ CO(g)

It is difficult to directly measure the enthalpy of combustion of carbon to carbon monoxide, because as long as any oxygen is present, some of the CO(g) formed reacts with it to form CO2(g). A mixture of CO and CO2 results as the product. Because we can run the reaction of carbon with excess oxygen and thereby force it to go to completion, we can measure the enthalpy of combustion of carbon to form carbon dioxide: Reaction A:

t..H~omb = -393.5 kJ

Because pure samples of CO(g) are available, it is possible to measure the enthalpy of combustion of CO(g) in oxygen to form carbon dioxide:

t..H~omb = -283.0 kJ Hess's law gives us a way to calculate the heat of reaction for the process we cannot measure: Reaction C:

t..H~omb = ?

5.8

To calculate the unknown heat of reaction, we must find a way to combine these three equations so that any two of them sum to equal the third. Once we have that combination, because we know two of the enthalpies, we can calculate the one we do not know. One way to approach this task is to focus on the reactants and products in the reaction whose enthalpy is unknown. Reaction (C) has carbon and oxygen as reactants and carbon monoxide as a product. Reaction B has carbon monoxide as a reactant. If we add reaction B to reaction C, the carbon monoxide cancels out, and we end up with reaction A as the sum of Band C:

+

Reaction C:

C(s)

+ ~ a/g)

~ CO(g)

t...H~omb = ?

Reaction B:

CO(g)

+ ~ a/g)

~ CO2(g)

t...H~omb = -283.0 t...H~omb

Reaction A: As written,

kJ

= -393.5 kJ

Therefore,

t...HO(C)

+ (-283.0 kJ) = -393.5 kJ t...HO(C)

= -110.5 kJ

Always remember that t...H is a state function. This means we can manipulate equations in two important ways, should the need arise when applying Hess's law. (1) We may multiply the coefficients in a balanced equation and the heat of reaction by a factor to change the quantity of material we are dealing with. (2) We may reverse an equation (make the reactants the products and the products the reactants) if we also change the sign on t...Hrxn' The use of both techniques is illustrated in Sample Exercise 5.12.

Calculating Enthalpies Using Hess's Law

of Reaction

When hydrocarbons burn in a limited supply of air, they may not burn completely, and CO(g) may be generated. One of the reasons why furnaces and hotwater heaters fueled by natural gas need to be properly vented is that incomplete combustion can produce toxic carbon monoxide:

Using the following two thermochemical Reaction B: Reaction C:

CH4(g)

+ 2 02(g)

2 CO(g)

+

reactions

~ CO/g)

02(g) ~ 2 CO/g)

+ 2 H20(g)

t...HO = -802 t...Ho = -566

kJ kJ

calculate the enthalpy of the reaction for the incomplete combustion of methane to produce carbon monoxide and water. AND ORGANIZE We are given two equations (B and C) with thermochemical data and a third (A) for which we are asked to find t...H. We may manipulate the two thermochemical equations so that they sum to give the equation for which t...H is unknown. Then we can calculate t...Hby applying Hess's law.

COLLECT

I

Hess's Law

I 223

224

I Chapter Five

I

Thermochemistry

The reaction of interest (A) has methane on the reactant side. Reaction B shows methane as a reactant. We can use B as written. Reaction A has CO as a product. Reaction C involves CO, but as a reactant, so we have to reverse C in order to get CO on the product side and change the sign of t...Ho when we do. If the coefficients as given will not enable us to sum the two reactions to yield reaction A, we may have to multiply one or both reactions by other factors.

ANALYZE

Start with reaction B as written, and add the reverse of reaction C, remembering to change the sign on t...H;xn(C):

SOLVE

B:

CH/g)

+ 2 O/g) ~ CO/g) + 2 H20(g) 2 CO/g) ~ 2 CO(g)

C:

t...H;xn(B) = -802 kJ

+ O/g)

t...H;xn(C) = +566 kJ

Look at

A: and note that methane has a coefficient of 2. That means we should multiply all the terms in reaction B, including t...H;xn (B), by 2:

2[t...H;xn (B) = -802 kJ] The carbon monoxide in reaction A has a coefficient of 2 as written, so we do not need to multiply reaction C by any factor. Here is the result of adding 2 X B to C and canceling out common terms:

2B: 2 CH4(g) C:

3

+ 4 02(g) ~ 2 CO2(g) + 4 H20(g) 2 COJ¥) ~ 2 CO(g)

t...H;xn(2B)

+~

The value of t...H;xn(A) for the incomplete

t...H;xn(C)

combustion

of methane

= -1604 kJ = +566 kJ to CO is thus

-1038 kJ. ABOUT IT It does not matter how you assemble the equations. In this Sample Exercise, we could have assembled reactions A and C so that the sum would give reaction B. The calculation would still have resulted in the same value for the unknown t...H;xn(A). THINK

Practice Exercise At high temperatures, such as those in the combustion chambers of automobile engines, molecular nitrogen and oxygen combine in an endothermic reaction to form nitrogen monoxide: t...Ho When NO is released into the environment, N02 in the following exothermic reaction:

= +180 kJ

it may undergo further oxidation to

t...Ho = -112 kJ Is the overall reaction between N2 and O2 forming N02

t...Ho =? exothermic

or endothermic?

What is t...Ho for this reaction?

(Answers to Practice Exercises are in the back of the book.)

5.8

When we used standard enthalpies of formation (t...H;) to determine standard heats of reaction (t...H;xn) in Section 5.6, we were actually applying Hess's law. To see that this is the case, consider the reaction of ammonia [NH3(g)] with oxygen to make NO(g) and water in the first step of the industrial process for the synthesis of nitric acid:

First we write the formation reactions for each reactant and product that is not an element in its standard state (remember t...H; = 0 for elements in their standard states):

t...H; = -46.1 kJlmol t...H; = +90.3 kJlmol t...H; = -285.8 kJlmol When we manipulate the equations

4 [NH/g)

1

+ H2(g)]

~ ~ N2(g)

=

4(t...H;

+46.1 kJlmol)

4 [~N2(g)

+ ~ 02(g) ~ NO(g)]

4(t...H; = +90.3 kJlmol)

6 [H2(g)

+ ~ 02(g) ~ H20(e)]

6(t...H; = -285.8 kJlmol)

their sum yields the reaction of interest:

4 NH3(g) 2 NJe-)

~ 2 R(p')

+ 2 02(g) ~

+

6 H,(p')

4 NO(g)

+ 6 HJe-) + 3 02(g) ~ 6 H20(e) This is exactly the same mathematical tion 5.18:

t...Hrxn o

kJ t...H;xn = +361.2 kJ t...H;xn = -1714.8 kJ t...H;xn = +184.4

operation that results when we use Equa-

'" 0 '" 0 L n t...H[,products -L m t...H[,rcactants

=

[4 mol(t...H;)NO(g)

+ 6 mol (t...H;)H20(e)]

[4 mol(t...H;)NH/g)

+5 mol (t...H;)O/g)]

= [4 mol(+90.3 kJlmol) + 6 mol(-285.8 kJlmol)] [4 mol(-46.1

kJlmol) +5 mol (0 kJlmol)]

= -1169.2 kJlmol

Hess's law is simply a consequence of the fact that enthalpy is a state function. This statement means that for a particular set of reactants and products, the enthalpy change of the reaction is the same whether the reaction takes place in one step or a series of steps. The concept of enthalpy and its expression in Hess's law are very useful because the heats associated with a large number of reactions can be calculated from a few that are actually measured.

I

Hess's Law

I 225

Carbon: Diamonds, Graphite, and the Molecules of Life Carbon is an element central to our existence both because we are carbon-based life-forms and because carbon is the central element in hydrocarbons-currently our most significant sources of energy and environmental changes. We also value carbon in its crystalline forms. One of these is diamond; we prize diamonds for their beauty and take advantage of its hardness by using it in drill bits and as abrasives. Another crystalline form, graphite, is black and opaque, and prized for its strength and flexibility (fibers are used in golf clubs and tennis rackets), and for its softness and smoothness (flakes are used as lubricants, and mixed and baked with clay to make the "lead" in pencils). A large variety of highly specialized noncrystalline materials known as activated carbons are made industrially for wastewater treatment, gas purification, and sugar refining (where activated carbon decolorizes solutions of raw sugar by selectively binding colored impurities to produce white crystals of sugar). In 1996 the Nobel Prize in chemistry was awarded to Robert Curl, Harold Kroto, and Richard Smalley for their

1985 discovery of a third crystalline form of carbon, buckminsterfullerene. Each molecule consists of a cluster of carbon atoms with the formula C60; they are also known as "buckyballs" because they look like soccer balls and are reminiscent of the architectural designs of Buckminster Fuller. Buckminsterfullerene is the most abundant form of a class of carbon-clustering molecules called fullerenes. Their discovery has spawned research into applications ranging from drugs to treat cancer and AIDS to rocket fuels. Carbon is found in minor amounts in Earth's crust as the free element. Diamonds are found in ancient volcanic pipes (openings in the crust through which lava flows); the exact way they were formed is still an item of debate. Graphite deposits occur worldwide. Over half of the carbon on Earth is held in the form of carbonate minerals (limestone, dolomite, and chalk), in which carbon is in its highest oxidation state as the COt ion (O.N. +4), and in the fossil fuels (petroleum, coal, natural gas), in which carbon is in its fully reduced state (O.N. -4). These two reservoirs of carbon are interconnected by a dynamic system we introduced you to in Chapter 3, namely the carbon cycle, which links free CO/g) in the atmosphere and the carbon-based life-forms on Earth's surface with the carbonate minerals and fossil fuels. Atmospheric CO2 is the source of all carbon-containing compounds produced in the leaves of green plants by photosynthesis: Sunlight

Diamond

+ 6 CO2 (g) + 6H20(e) ~ C6HI206(aq) + 60/g)

The most fundamental carbon-containing product of photosynthesis is the sugar glucose (C6HI206), which is the main food for biological systems. Glucose is also the primary structural subunit of cellulose, which is the chief component of plant fibers such as wood and cotton. A single molecule of cellulose may contain over 1500 glucose units bonded together; plants produce over 100 billion tons of cellulose

Graphite

Buckyball C60 Three crystalline

forms of carbon: diamond,

buckminsterfullerene

("buckyball").

graphite and Cotton

is a major source of cellulose.

227

each year. The oxidation number of carbon in glucose is 0, so green plants may be thought of as carbon-reducing machines. Other life-forms eat plants and derive large amounts of energy from metabolism through a series of oxidation reactions that are the reverse of photosynthesis, which takes carbon at O.N. 0 (glucose) back to O.N. +4 (carbon dioxide). Life-forms are carbon-oxidizing machines. When plants and animals die on the surface of Earth, their carbon-containing molecules are oxidized back to CO2 and water if decomposition is complete. However, plant material deprived of oxygen may be further reduced by specialized bacteria that produce the hydrocarbon fossil fuels petroleum and coal. The hydrocarbons that make up these fuels consist of carbon in its most reduced form. When a hydrocarbon fuel reacts with oxygen in a combustion reaction, energy is released because carbon at O.N. -4 becomes carbon at O.N. +4 when the hydrocarbons are converted into CO2, Some CO2 is returned to the atmosphere as a result of respiration by animals and plants, but our burning of fos-

hensible variety of structures that in turn can link together to form gigantic molecules containing hundreds and even thousands of carbon atoms. Over 16 million compounds of carbon are known, and over 90% of the thousands of new materials synthesized each year contain carbon. The number of compounds that contain carbon is far greater than the number of compounds that do not. An entire branch of chemistry, called organic chemistry, is devoted to the study of their structures and properties. The chemistry of carbon is the chemistry of pharmaceuticals, soft contact lenses, paints, synthetic fibers, plastics, and gasoline. It is the chemistry of the food we eat and the fuel that powers our cars and airplanes. Fossil fuels such as petroleum are the primary source of molecules that industry modifies to make the items we regard as part of modern life, so petroleum can be regarded as fuel for the manufacturing industry as well as for the transportation industry. Combined in precise ways with a small number of other elements-such as sulfur, phosphorous, oxygen, nitrogen, and hydrogen-carbon serves as the foundation oflife.

Atmospheric

Sunflowers

CO2 affects coral reefs.

sil fuels inserts carbon that was formerly sequestered underground into the atmosphere as CO2, which adds to the carbon dioxide produced by respiration. Atmospheric CO2 dissolves in seawater to form carbonic acid, which increases the acidity of the ocean. Changing the acidity affects the solubility of CaC03 in seawater, which in turn has potentially disastrous consequences for marine life, including the many oceanic-life forms that build shells, reefs, and other structures from CaC03 and other carbonates. In addition to its critical role in the biosphere, carbon forms a vast number of compounds. Carbon atoms attach themselves to other carbon atoms to an extent not possible for atoms of any other element. The result is the formation of chains and rings that combine to create an almost incompre-

are the source of many useful compounds.

Corn provides alternative

fuels.

228

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Chapter Five

I Thermochemistry

Summary 5.1 Energy is the capacity to transfer heat or to do work (w). A thermochemical equation includes the heat energy absorbed or released in a chemical reaction or physical change. Thermodynamics is the study of energy and its transformations. Differences of temperature result in heat transfer from one object to another. Thermal equilibrium is established when objects have the same temperature and net heat transfer ceases. Potential energy (PE) is the energy of position and is a state function. Kinetic energy (KE) is the energy of motion. The law of conservation of energy states that energy cannot be created or destroyed. Heating a sample of molecules increases their average kinetic and thermal energies. Electrostatic potential energy (E) results from interaction of charged particles. Energy is stored in compounds, and heat is absorbed or released when they are transformed into different compounds. SECTION

5.2 A system is the part of the universe under study. Everything not part of the system is called the surroundings. Isolated systems exchange neither heat nor matter with their surroundings, closed systems exchange only energy, and open systems exchange energy and matter. In an exothermic process the system loses heat (-q) to its surroundings; in an endothermic process the system absorbs heat (+q) from its surroundings. The sum of the kinetic and potential energies of a system is called its internal energy (E). The first law of thermodynamics states that the energy gained or lost by a system equals the energy lost or gained by the surroundings. Common energy units are calories (cal) and joules (J). The internal energy of a system is increased (I::!.E = Efina1 - E;n;t;al is positive) when it is heated (+q) and work is done on it (+w). SECTION

5.3 The enthalpy (H) of a system is given by H = E + The enthalpy change (I::!.H) of a system is equal to the heat (qp) added to or removed from the system at constant pressure. From the system's point of view, I::!.H > 0 for endothermic reactions and I::!.H < 0 for exothermic reactions.

SECTION

pv.

Several concepts are useful in discussing the amounts of heat involved in heating and cooling objects at constant pressure. The heat capacity (CI') of an object is the amount of heat required to SECTION

5.4

increase the temperature of the object by 1.00°C. The molar heat capacity (c) of a substance is the amount of heat required to increase the temperature of 1 mole by 1.00T. The specific heat (c,) of an object is the amount of heat required to increase the temperature of 1 gram of the object by 1.00T. The molar heat offusion (I::!.HfJ of a substance is the amount of heat required to convert 1 mole of the solid substance to a liquid at the melting point of the substance. The molar heat of vaporization (I::!.Hvap) of a substance is the amount of heat required to convert 1 mole of the liquid substance to a vapor at the boiling point of the substance. 5.5 A calorimeter is a device used to measure the amounts of heat involved in physical and chemical processes. A bomb calorimeter is a device used to measure the heat of a combustion reaction (I::!.Hcomb)' The quantity of heat associated with a process is called the enthalpy change or enthalpy of reaction (I::!.HrxJ

SECTION

5.6 Standard conditions are 1 atm of pressure and some specified temperature. The standard enthalpy of formation of a substance, I::!.H;, is the amount of heat involved in a formation reaction, in which 1 mole of the substance is made from its constituent elements in their standard states (under standard conditions). The values of the standard enthalpies of formation of all elements are defined as zero. Enthalpy changes in processes such as combustion can be calculated from the enthalpies of formation of the reactants and products.

SECTION

SECTION 5.7 The fuel value is the amount of heat released on complete combustion of 1 gram of the fuel. Methane, CH4 (fuel value 50 kJ/g), has the highest chemical fuel value of the hydrocarbons. Food value is the amount of heat released when a material consumed by an organism for sustenance is burned completely; nutritionists often express food values in Calories (kilocalories) rather than the SI-derived unit kilojoules. 5.8 Hess's law states that the enthalpy change of a reaction that is the sum of two or more reactions is equal to the sum of the enthalpy changes of the constituent reactions. It can be used to calculate the heat changes of reactions that are hard or impossible to measure.

SECTION

Problem-Solving Summary TYPE OF PROBLEM

CONCEPTS

Identifying exothermic calculating (I::!.E) and P

For the system: I::!.E = q + w (where w

endothermic and processes, and internal energy change I::!.V work

Determining the value and sign of I::!.H associated with changing the temperature of a substance or phase change

AND EQUATIONS

SAMPLE

5.1-5.3 =

-PI::!. V).

Freezing a liquid at its freezing point: I::!.H = -n I::!.Hf (where I::!.Hf = heat of fusion, and n = number of moles). Heating a liquid: "' q = nc pI::!.T (where cp = heat capacity of the liquid, and I::!.T = temperature change).

5.4-5.6

EXERCISES

Visual Problems TYPE OF PROBLEM

CONCEPTS

Measuring the heat capacity (calorimeter constant) of a calorimeter

C .I. = ql I::!. T (where C .I. = heat capacity of calorimeter, q ~O~;~unt of heat release~n;y':'standard combustion reaction, and I::!. T = temperature change of calorimeter).

5.7

Recognizing formation reactions

In a formation reaction, the reactants are elements in their standard states and the product is one mole of a single compound in its standard state.

5.8

Calculating standard heats of reaction from heats of formation

I1H;xn

5.9

Calculating the fuel values of fuels in mass and volume units; calculating food value of foods

The total fuel value or food value of a sample is its mass or volume times the appropriate term defining heat content, such as k]/unit mass or Callunit volume.

5.10-5.11

Calculating enthalpies of reaction using Hess's law

Reorganize the information so that the reactions add together as desired. Reversing a reaction changes the sign of the heat change. Multiplying the coefficients in a reaction by a factor has the same effect on the standard heat change involved.

5.12

=

L

n

AND EQUATIONS

I1H;,producrs

-

L

m

SAMPLE

I1H;,reacrants

I

229

EXERCISES

Visual Problems 5.1.

A brick lies perilously half off the edge of the flat roof of a building (Figure P5.1). The roof edge is 50 feet above street level, and the brick has 500 J of potential energy with respect to street level. Someone (we don't know who) edges the brick off the roof, and it begins to fall. What is the brick's kinetic energy when it is 35 feet above street level? What is its kinetic energy when it hits the street surface?

5.2.

(a)

Figure P5.2 diagrams pairs of cations and anions in contact. The energy of each interaction is proportional to QjQ/d, where Qj and Q2 are the charges on the cation and anion, and d is the distance between their nuclei. Which pair has the greatest interaction energy and which has the smallest?

(b)

(c)

FIGURE PS.2 5.3.

Figure P5.3 shows a sectional view of an assembly consisting of a gas trapped inside a stainless steel cylinder with a stainless steel piston and a block of iron on top of the piston to keep it in place. (a) Sketch the situation after the assembly has been heated for a few seconds with a blowtorch. (b) Is the piston higher or lower in the cylinder? (c) Has heat (q) been added to the system? (d) Has the system done work (w) on its surroundings, or have the surroundings done work on the system? Iron

FIGURE PS.l --

FIGURE PS.3

Cylinder

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I Chapter Five

5.4.

I Thermochemistry 300

Figure P5.4 represents the energy change in a chemical reaction. The reaction results in a very small decrease in volume. (a) Has the internal energy of the system decreased or increased as a result of the reaction? (b) What sign is given to I::!.Efor the reaction system? (c) Is the reaction endothermic or exothermic?

C2Hh)

200I 100) I::!.H;, k]lmol

Reactants

a)

C(s), Hig), 02(g)

-1000

CO(g)

-2000

•

H2O(g)

-3000

Products

CO2(g)

-400J

FIGURE PS.4 5.5.

a

(

The closed rigid metal box lying on the wooden kitchen table in Figure P5.5 is about to be involved in an accident. (a) Are the contents of the box an isolated system, a closed system, or an open system? (b) What will happen to the internal energy of the system if the table catches fire and burns? (c) Will the system do any work on the surroundings while the fire is burning or after the table has burned away and collapsed?

FIGURE PS.7 5.8.

The process illustrated in Figure P5.8 takes place at constant pressure. (The figure is not to scale; the molecules are actually much, much smaller than shown here.) (a) Write a balanced equation for the process. (b) Is w positive, negative, or zero for this reaction? (c) Using data from Appendix 4, calculatel::!.H;xnfor the formation of 1 mole of the product.

FIGURE PS.S Reaction 5.6.

~

The diagram in Figure P5.6 shows how a chemical reaction ina cyli nder with a piston affects the volume of the system. (a) In this reaction, does the system do work on the surroundings? (b) If the reaction is endothermic, does the internal energy of the system increase or decrease when the reaction is proceeding?

S 0 FIGURE PS.8

;., " .V •

1

Piston 5.9.

Reaction

-system-!

..

Explain how the information in Figure P5.9 can be used to calculate the heat change labeled I::!.Hc Write an equation that corresponds to your answer.

,

FIGURE PS.6 5.7.

The enthalpy diagram in Figure P5.7 indicates the enthalpies of formation of four compounds made from the elements listed on the "zero" line of the vertical axis. (a) Why are the elements all put on the same horizontal line? (b) Why is C2H2(g) sometimes called an "endothermic" compound? (c) How could the data be used to calculate the heat of the reaction that converts a stoichiometric mixture of C2H2(g) and 02(g) to CO(g) and HP(g)?

FIGURE PS.9


I

231

Questions and Problems Energy: Basic Concepts and Definitions CONCEPT

REVIEW

5.10. How are energy and work related? 5.11. Explain the difference between potential energy and kinetic energy. 5.12. Explain what is meant by a state function. 5.13. Are kinetic and potential energies both state functions? 5.14. Explain the nature of the potential energy in the following: a. the new battery for your portable CD player. b. a gallon of gasoline. c. the crest of a wave before it crashes onto shore. 5.15. Explain the kinetic energy in a stationary ice cube.

Systems, Surroundings, and the flow of Energy CONCEPT

5.29. Calculatel::!.E for the combustion of a gas that releases 210.0 kJ of heat to its surroundings and does 65.5 kJ of work on its surroundings. 5.30. Calculatel::!.E for a chemical reaction that produces 90.7 kJ of heat but does no work on its surroundings. *5.31. The following reactions take place in a cylinder equipped with a movable piston at atmospheric pressure (Figure P5.31). Which reactions will result in work being done on the surroundings? Assume the system returns to an initial temperature of 110°C. Hint: The volume of a gas is proportional to number of moles (n) at constant temperature and pressure. a. CH/g) b. C3Hs(g) c. N2(g)

+ 2 O/g) ~ + 5 02(g) ~

+ 2 a/g)

~

CO/g)

+ 2 HP(g) + 4 HP(g)

3 CO2(g) 2 NO/g)

REVIEW

5.16. What is meant by the terms system and surroundings? 5.17. What is the difference between an exothermic process and an endothermic one? 5.18. Give two ways ofincreasing the internal energy of a gas sample. 5.19. In each of the following processes describe the system and give the sign of qsysrcm: (a) combustion of methane; (b) freezing water to make ice; (c) touching a hot stove. 5.20. In each of the following processes describe the system and give the sign of qsysrcm: (a) driving an automobile; (b) applying ice to a sprained ankle; (c) cooking a hot dog. PROBLEMS

5.21. Which of the following processes are exothermic, and which are endothermic? (a) a match burns; (b) a molten metal solidifies; (c) rubbing alcohol feels cold on the skin 5.22. Which of the following processes are exothermic, and which are endothermic? (a) ice cubes solidify in the freezer; (b) water evaporates from a glass left on a windowsill; (c) dew forms on grass overnight 5.23. What happens to the internal energy of a liquid at its boiling point when it vaporizes? 5.24. What happens to the internal energy of a gas when it expands (with no heat flow)? 5.25. How much P I::!.V work does a gas system do on its surroundings at a constant pressure of 1.00 atm if the volume of gas triples from 250.0 mL to 750.0 mL? Express your answer in L . atm and joules (J). 5.26. An expanding gas does 150.0 J of work on its surroundings at a constant pressure of 1.01 atm. If the gas initially occupied 68 mL, what is the final volume of the gas? 5.27. Calculatel::!.E for the following situations: a. q = 100.0J; w = -50.0J b. q = 6.2 kJ; w = 0.70J c. q = -615J; w = -325J 5.28. Calculatel::!.E for a system that absorbs 726 kJ of heat from its surroundings and does 526 kJ of work on its surroundings.

FIGURE PS.3i *5.32. In which direction will the piston described in Figure 5.31 move when the following reactions are carried out at atmospheric pressure inside the cylinder and after the system has returned to its initial temperature of 110°C? Hint: The volume of a gas is proportional to number of moles (n) at constant temperature and pressure.

+ 3 H/g) C(s) + 02(g) ~ C2HPH(g) +

a. N2(g) b. c.

~ 2 NH3(g) CO2(g) 3 02(g) ~

2 CO2(g)

+

3 HP(g)

Enthalpy and Enthalpy Changes CONCEPT

REVIEW

5.33. What is meant by an enthalpy change? 5.34. Describe the difference between an internal energy change (I::!.E) and an enthalpy change (I::!.H). 5.35. Why is the sign of I::!.H negative for an exothermic process? 5.36. What happens to the magnitude and sign of the enthalpy change when a process is reversed? PROBLEMS

5.37. A Clogged Sink Adding Drano to a clogged sink causes the drainpipe to get warm. What is the sign of I::!.H for this process? 5.38. A Cold Pack Breaking a small pouch of water inside a larger bag containing ammonium nitrate activates chemical cold packs. What is the sign of I::!.H for the process taking place in the cold pack?

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I Thermochemistry

5.39. Break a Bond The stable form of oxygen at room temperature and pressure is the diatomic molecule 02" What is the sign of I::!.H for the following process?

point of water is 93°C. Is this amount of energy sufficient to heat the water to its boiling point? 5.51. Use the following data to sketch a heating curve for 1 mole of methanol. Start the curve at -100°C and end it at 100°C.

O/g) -72 O(g) 5.40. Plaster of Paris Gypsum is the common name of calcium sulfate dihydrate (CaS04 • 2H20). When gypsum is heated to 150°C, it loses most of the water in its formula and forms plaster of Paris (CaS04 • 0.5 H20): 2 CaSO

4 •

2 HP(s) -7 2 CaS04

•

0.5 HP(s)

Boiling point

81.1 J/(mol . T)

(g)

43.9 J/(mol . T)

(s)

48.7 J/(mol . T)

Boiling point

125TC

Melting point

-56.8T

REVIEW

5.43. What is the difference between specific heat and heat capacity? 5.44. What happens to the heat capacity of a material ifits mass is doubled? Is the same true for the specific heat? 5.45. Are the heats of fusion and vaporization of a given substance usually the same? 5.46. An equal amount of heat is added to pieces of metal A and metal B having the same mass. Does the metal with the larger heat capacity reach the higher temperature? *5.47. Cooling an Automobile Engine Most automobile engines are cooled by water circulating through them and a radiator. However, the original Volkswagen Beetle had an air-cooled engine. Why might car designers choose water cooling over air cooling? *5.48. Nuclear Reactor Coolants The reactor-core cooling systems in some nuclear power plants use liquid sodium as the coolant. Sodium has a thermal conductivity of 1.42 J/(cm . s . K), which is quite high compared with that of water [6.1 X 10-3 JI(cm . s . K)]. The respective molar heat capacities are 28.28 JI(mol . K) and 75.31 JI(mol . K). What is the advantage of using liquid sodium over water in this application? PROBLEMS

5.49. How much heat is needed to raise the temperature of 100.0 g of water from 30.0°C to 100.0T? 5.50. 100.0 grams of water at 30°C absorbs 290.0 kJ of heat from a mountain climber's stove at an elevation where the boiling

20.7 kJlmol

I::!.Hfu,

Molar heat capacity (C) (g)

+ CO/g) + HP(C)


41.5 kJlmol

I::!.H,,"

-7

If the contents of the bottle are considered the system, is work being done on the surroundings or on the system?

CONCEPT

Molar heat capacity (C)

5.52. Use the following data to sketch a heating curve for 1 mole of octane. Start the curve at -5TC and end it at 150°C.

5.42. Kitchen Chemistry A simple "kitchen chemistry" experiment requires placing some vinegar in a soda bottle. A deflated balloon containing baking soda is stretched over the mouth of the bottle. Adding the baking soda to the vinegar starts the following reaction and inflates the balloon:

CH3C02Na(aq)

3.18 kJlmol

I::!.Hfu,

H2(g) -7 H/s)

+ CH3C02H(aq)

37 kJ/mol

I::!.H"m

What is the sign of I::!.H for making plaster of Paris from gypsum?

NaHC03(aq)

-94T

Melting point

+ 3 HP(g)

5.41. Metallic Hydrogen A solid with metallic properties is formed when hydrogen gas is compressed under extremely high pressures. Predict the sign of the enthalpy change for the following reaction:

65T

254.6 JI(mol . T) 316.9 kJl(mol . °C)

5.53. Keeping an Athlete Cool During a strenuous workout, an athlete generates 2000.0 kJ of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat? 5.54. The same quantity of energy is added to 10.00 g pieces of gold, magnesium, and platinum, all initially at 25°C. The molar heat capacities of these three metals are 25.41 JI(mol . T), 24.79 JI(mol· °C), and 25.95 JI(mol . 0c), respectively. Which piece of metal has the highest final temperature? *5.55. Exactly 10.0 mL of water at 25.0°C are added to a hot iron skillet. All of the water is converted into steam at 100.0T. The mass of the pan is 1.20 kg and the molar heat capacity of iron is 25.19 JI(mol . °C). What is the temperature change of the skillet? *5.56. A 20.0 g piece ofiron and a 20.0 g piece of gold at 100.0°C were dropped into 1.00 L of water at 20.0°C. The molar heat capacities ofiron and gold are 25.19 J/(mol . T) and 25.41 J/(mol' 0c), respectively. What is the final temperature of the water and pieces of metal?

Calorimetry: Measuring Heat Capacity and Calorimeter Constants CONCEPT

REVIEW

5.57. Why is it necessary to know the heat capacity of a calorimeter? 5.58. Could an endothermic reaction be used to measure the heat capacity of a calorimeter? 5.59. If we replace the water in a bomb calorimeter with another liquid, do we need to redetermine the heat capacity of the calorimeter? 5.60. When measuring the heat of combustion of a very small amount of material, would you prefer to use a calorimeter having a heat capacity that is small or large?

Questions and Problems PROBLEMS

5.61. Calculate the heat capacity of a calorimeter if the combustion of 5.000 g of benzoic acid led to a temperature increase of 16.39TC. 5.62. Calculate the heat capacity of a calorimeter if the combustion of 4.663 g of benzoic acid led to an increase in temperature of7.149°C. 5.63. The complete combustion of 1.200 g of cinnamaldeyde (C9HsO, one of the compounds in cinnamon) in a bomb calorimeter (C"lodm"" = 3.640 kJI"C) produced an increase in temperature of 12. 79°C. Calculate the molar enthalpy of combustion of cinnamaldehyde (I::!.Hcomb) in kilojoules per mole of cinnamaldehyde. 5.64. Aromatic Spice The aromatic hydrocarbon cymene (C1oH14)is found in nearly 100 spices and fragrances including coriander, anise, and thyme. The complete combustion of 1.608 g of cymene in a bomb calorimeter (Cca1o,;mctc, = 3.640 kJloC) produced an increase in temperature of 19.35°C. Calculate the molar enthalpy of combustion of cymene (I::!.Hcomb) in kilojoules per mole of cymene. 5.65. Phthalate Plasticizers Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of I::!.H:omb for dimethylphthalate (CloH1004) is 4685 kJlmol. Assume that 1.00 g of dimethylphthalate is combusted in a calorimeter whose heat capacity (Cca1o,;mcr) is 7.854 kJloC at 20.215°C. What is the final temperature of the calori meter? 5.66. Flavorings The flavor of anise is due to anethole, a compound with the molecular formula C1OH120. The I::!.H:omb value for anethole is 5541 kJ/mol. Assume 0.950 g of anethole is combusted in a calorimeter whose heat capacity (Cca1o,;mcr) is 7.854 kVC at 20.611°C. What is the final temperature of the calorimeter?

REVIEW

5.67. Oxygen and ozone are both forms of elemental oxygen. Are the standard enthalpies of formation of oxygen and ozone the same? 5.68. Why are the standard enthalpies of formation of elements in their standard states assigned a value of zero? PROBLEMS

5.69. For which of the following reactions does I::!.H;xn represent an enthalpy of formation? a. C(s) + O/g) ~ CO/g) b. CO/g) + C(s) ~ 2 CO(g) c. CO/g) + H/g) ~ HP(g) + CO(g) d. 2 H/g) + C(s) ~ CH/g) 5.70. For which of the following reactions does I::!.H,o,nalso represent an enthalpy of formation? a. 2 N2(g) + 3 a/g) ~ 2 N02(g) + 2 NO(g) b. N/g) + O/g) ~ 2 NO(g) c. 2 NO/g) ~ NP/g) d. N/g) + 2 02(g) ~ 2 NO/g)

233

5.71. Methanogenesis Use enthalpies of formation to calculate the standard enthalpy of reaction for the following methane-generating reaction of methanogenic bacteria: 4 H2(g)

+

CO2(g) ~ CH/g)

+

2 HP(g)

5.72. Methanogenesis Use enthalpies offormation to calculate the standard enthalpy of reaction for the following methane-generating reaction of methanogenic bacteria, given I::!.HrofCH3NH2(g) = -22.97J/mol: 4 CH3NH/g)

+

2 HP(C)

~

3 CH/g)

+

CO/g)

+

4 NH3(g)

5.73. Ammonium nitrate decomposes to N20 and water vapor at temperatures between 250 and 300°C. Write a balanced chemical reaction describing the decomposition of ammonium nitrate, and calculate the standard heat of reaction using the appropriate enthalpies of formation. 5.74. Above 300°C, ammonium nitrate decomposes to N2, 02' and H20. Write a balanced chemical reaction describing the decomposition of ammonium nitrate, and determine the standard heat of reaction by using the appropriate enthalpies of formation. 5.75. Explosives Mixtures of fertilizer (ammonium nitrate) and fuel oil (a mixture oflong-chain hydrocarbons similar to decane, C1oH22) are the basis for a powerful explosion. Determine the standard enthalpy change of the following explosive reaction by using the appropriate enthalpies of formation (I::!.H;c H = 249.7 kJlmo\): , 10 22 3 NH4N03(s) + C1oH2/C) + 14 a/g) ~ 3 N2(g) + 17 HP(g) + 10 CO/g) *5.76. A Little TNT Trinitrotoluene (TNT) is a highly explosive compound. The thermal decomposition of TNT is described by the following chemical equation: 2 C7HsN306(s)

~ 12 CO(g)

+

5 H2(g)

+

3 N2(g)

+

2 C(s)

If I::!.H"n for this reaction is -10,153 kJlmol, how much TNT is needed to equal the explosive power of 1 mol of ammonium nitrate in Problem 5.75?

Enthalpies of Formation and Enthalpies of Reaction CONCEPT

I

Fuel Values and Food Values CONCEPT

5.77. 5.78. 5.79. 5.80.

REVIEW

What is meant by fuel value? What are the units of fuel values? How are fuel values calculated from molar heats of combustion? Is the fuel value ofliquid propane the same as that of propane gas?

PROBLEMS

5.81. If all the energy obtained from burning 1.00 pound of propane is used to heat water, how many kilograms of water can be heated from 20.0°C to 45.0°C? 5.82. A 1995 article in Discover magazine on world-class sprinters contained the following statement: "In one race, a field of eight runners releases enough energy to boil a gallon jug of ice at O.O°C in ten seconds!" How much "energy" do the runners release in 10 seconds? Assume that the ice weighs 128 ounces.

234

I Chapter Five

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5.83. The Joys of Camping Lightweight camping stoves typically use white gas, a mixture of Cs and C6 hydrocarbons. a. Calculate the fuel value ofCsH12' given that I::!.H:omb = -3535 kJlmol. b. How much heat is released during the combustion of 1.00 kg ofCsH1/ c. How many grams ofCsH12 must be burned to heat 1.00 kg of water from 20.0°C to 90.0°C? Assume that all the heat released during combustion is used to heat the water. 5.84. The heavier hydrocarbons in white gas are hexanes (C6H14). a. Calculate the fuel value of C 6Hw given that I::!.H;omb = -4163 kJlmol. b. How much heat is released during the combustion of 1.00 kg ofC6H1/ c. How many grams ofC6H14 are needed to heat 1.00 kg of water from 25.0°C to 85.0°C? Assume that all of the heat released during combustion is used to heat the water. d. Assume white gas is 25% Cs hydrocarbons and 75% C6 hydrocarbons; how many grams of white gas are needed to heat 1.00 kg of water from 25.0T to 85.0°C?

5.93. The mineral spodumene (LiAISi206) exists in two crystalline forms called a and {3. Use Hess's law and the following information to calculatel::!.H;xn for the conversion of a-spodumene into (3-spodumene: LiP(s)

1

+ 2 AI(s) + 4 Si02(s) + 02(g) ~ 2 a-LiAISiP6(s) I::!.Ho = -1870.6kJ

1

LiP(s)

+ 2 AI(s) + 4 Si02(s) + 02(g) ~ 2 (3-LiAISiP6(s) I::!.Ho = -1814.6kJ 5.94. Use the following data to determine whether the conversion of diamond into graphite is exothermic or endothermic: o I::!.H = -395.4 kJ Cd;,mond(S) + 02(g) ~ CO2(g) 2 CO/g)

~ 2 CO(g)

+

O/g)

2 CO(g) ~ Cgmph;)s) + CO/g)

o I::!.H = 566.0 kJ I::!.Ho = -172.5

kJ

I::!.H =? O

5.95. You are given the following data: ~ N/g)

Hess's Law CONCEPT

+ ~ O/g) ~ NO(g)

NO(g) + ~ CI/g) ~ NOCI(g)

REVIEW

5.85. How is Hess's law consistent with the law of conservation of energy? 5.86. Why is the enthalpy offormation ofCO(g) difficult to measure experimentally? 5.87. Explain how the use of I::!.H~ to calculatel::!.H;xn is an example of Hess's law. 5.88. Why is it important for Hess's law that enthalpy is a state function?

2 NOCI(g) ~ N/g) a. Which of formation? b. Determine 5.96. The enthalpy the following N02CI from

+

O/g)

5.90. How can the enthalpy offormation ofCO(g) be calculated from the enthalpy of formation of CO/g) and the heat of combustion ofCO(g)? 5.91. Calculate the enthalpy offormation of S02(g) from the standard enthalpy changes of the following reactions:

+

02(g) ~ 2 SO/g)

I::!.H;,n = -196 kJ

~ S8(S)+ 3 02(g) ~ 2 SO/g)

I::!.H;xn = -790 kJ

2 SO/g)

~ S8(S)+ 02(g) ~ SO/g)

I::!.H; =?

5.92. Ozone Layer The destruction of the ozone layer by chlorofluorocarbons (CFCs) can be described by the following reactions: CIO(g)

+

03(g) ~ CI(g)

+

I::!.H;xn = -29.90 kJ

20/g)

I::!.H;,n = 24.18 kJ

2 03(g) ~ 3 O/g) Determine CI(g)

the value of heat of reaction for the following:

+

O/g) ~ CIO(g)

+

O/g)

I::!.H =? O

~ N/g)

+

O/g)

I::!.H;xn = -38.6

kJ

+ CI2(g) I::!.H;xn = ?

I::!.Hcxn for the decomposition of NOCI. of decomposition ofN02CI is -114 kJ. Use data to calculate the heat of formation of N2, O2, and C12: N02CI(g) ~ NO/g)

~ N/g)

kJ

the I::!.Hcxn values represent enthalpies of

PROBLEMS

5.89. How can the first two of the following reactions be combined to obtain the third reaction? a. CO(g) + ~ O/g) ~ CO/g) b. C(s) + O/g) ~ CO/g) c. C(s) + ~ O/g) ~ CO(g)

I::!.H;xn = -90.3

+ ~ CI/g)

+ O/g) ~ N02(g)

+ ~ CI/g)

~ N02CI(g)

I::!.H;xn = -114 kJ I::!.H; = +33.2 kJ I::!.H; = ?

Additional Problems *5.97. Laundry forgotten and left outside to dry on a clothesline in the winter slowly dries by "ice vaporization" (sublimation). The increase in internal energy of water vapor produced by sublimation is less that the amount of heat absorbed. Explain. 5.98. Chlorofluorocarbons (CFCs) such as CF2Cl2 are refrigerants whose use has been phased out because of their destructive effect on Earth's ozone layer. The standard enthalpy of evaporation ofCF2CI2 is 17.4 kJ/mol, compared with I::!.H:,p = 41 kJ/mol for liquid water. How many grams ofliquid CF2Cl2 are needed to cool 200.0 g of water from 50.0 to 40.0°C? The specific heat of water is 4.184 JI(g . T). 5.99. A 100.0 mL sample of 1.0 M NaOH is mixed with 50.0 mL of 1.0 M H2S04 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 22.3°C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is


5.lO0.

*5.10l.

5.102.

5.103.

*5.104.

5.105.

*5.106.

31.4°C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g . °C), and that no heat is lost to the surroundings. (a) Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. (b) Is any NaOH or H2SO 4 left in the Styrofoam cup when the reaction is over? (c) Calculate the enthalpy change per mole of H2S04 in the reaction. Varying the scenario in Problem 5.99 assumes this time that 65.0 mL of 1.0 M H2S04 are mixed with 100.0 mL of 1.0 M NaOH and that both solutions are initially at 25.0°C. Assume that the mixed solutions in the Styrofoam cup have the same density and specific heat as in Problem 5.99 and that no heat is lost to the surroundings. What is the maximum measured temperature in the Styrofoam cup? An insulated container is used to hold 50.0 g of water at 25.0°C. A sample of copper weighing 7.25 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.1°C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 JI(g . °C), calculate the maximum temperature of the water in the insulated container after the copper metal is added. The mineral magnetite (Fe304) is magnetic, whereas iron(II) oxide is not. (a) Write and balance the chemical equation for the formation of magnetite from iron (II) oxide and oxygen. (b) Given that 318 kJ of heat are released for each mole of Fe304 formed, what is the enthalpy change of the balanced reaction of formation ofFe30 4 from iron(II) oxide and oxygen? Which of the following substances has a standard heat of formation I::!.H; of zero? (a) Pb at 1000°C; (b) C3H8 (g) at 25.0°C and 1 atm pressure; (c) solid glucose at room temperature; (d) N/g) at 25.0°C and 1 atm pressure The standard molar heat of formation of water is -285.8 kJlmol. (a) What is the significance of the negative sign associated with this value? (b) Why is the magnitude of this value so much larger than the heat of vaporization of water (I::!.H:,p = 41 kJlmo\)? (c) Calculate the amount of heat produced in making 50.0 mL of water from its elements under standard conditions. Endothermic compounds are unusual: they have positive heats of formation. An example is acetylene, C2H2 (I::!.H~ = 226.7 kJlmo\). The combustion of acetylene is used to melt and weld steel. Use Appendix 4 to answer the following questions. (a) Give the chemical formulas and names of three other endothermic compounds. (b) Calculate the standard molar heat of combustion of acetylene. Balance the following chemical equation, name the reactants and products, and calculate the standard enthalpy change using the data in Appendix 4. FeO(s)

+

a/g)

-7

Fep3(s)

*5.107. Add reactions 1 and 2, and label the resulting reaction 3. Consult Appendix 4 to find the standard enthalpy change for balanced reaction 3. (1)

Zn(s)

(2) ZnS(s)

i S/s)

-7

ZnS(s)

+ 2 02(g)

-7

ZnSO/s)

+

+

S02(g)

I 235

5.108. Conversion of 0.90 g ofliquid water to steam at lOO.O°C requires 2.0 kJ of heat. Calculate the molar enthalpy of evaporation of water at 100.0°C. 5.109. The specific heat of solid copper is 0.385 JI(g . °C). What thermal energy change occurs when the temperature of a 35.3 g sample of copper is cooled from 35.0T to 15.0°C? Be sure to give your answer the proper sign. This amount of heat is used to melt solid ice at O.O°C.The molar heat of fusion of ice is 6.01 kJlmol. How many moles of ice are melted? *5.110. Methanol decomposes according to the following reaction: 90.2 kJ

5.11l.

+

CHPH(g)

-7

+

CO(g)

2 H/g)

a. Is this reaction endothermic or exothermic? b. What is the value of I::!.H"n for this reaction? c. How much heat would be absorbed or released if 60.0 g of methanol were decomposed in this reaction? d. Would you expect I::!.H"n for the decomposition of CHPH(C) to give CO(g) and H2(g) to be larger or smaller than 128 kJ? Using the information given below, write an equation for the enthalpy change in reaction 3: (1)

B(g)

+

(2)

C(g)

-7

(3)

C(s)

-7

A(s)

C(g)

-7

C(s)

I::!.H1 I::!.H2

+

A(s)

B(g)

I::!.H3

5.112. From the information given below, write an equation for the enthalpy change in reaction 3: (1)

F(g)

(2)

F(s)

-7

F(g)

(3)

D(s)

+

E(g)

-7

+

D(s)

E(g) I::!.H1 I::!.H2 F(s) I::!.H3

-7

5.113. The reaction ofCHPH(g) with N/g) to give HCN(g) and NH3(g) requires 164 kJ/mole of heat. a. Write a balanced chemical equation for this reaction. b. Should the thermal energy involved be written as a reactant or as a product? c. How much heat is involved in the reaction of 60.0 g ofCHPH(g) with excess N/g) to give HCN(g) and NH3(g) in this reaction? 5.114. Calculatel::!.H;xnfor the reaction 2 Ni(s)

+ ~ S8(S) +

O/g)

3

-72

NiS03(s)

from the following information: (1)

NiS03(s)

(2)

i S8(S) + O/g)

(3)

Ni(s)

-7

+

NiO(s)

I::!.H;xn = 156 kJ

S02(g)

-7

SO/g)

I::!.H;xn = -297 kJ

-7

NiO(s)

I::!.H;xn = -241 kJ

+ ~ 02(g)

5.115. Use the following information to calculate the amount of heat involved in the complete reaction of 3.0 g of carbon to form PbC03 (s) in reaction 4. Be sure to give the proper sign (positive or negative) with your answer.

+ ~ O/g)

(1)

Pb(s)

(2)

C(s)

(3)

PbC03(s)

(4)

Pb(s)

+ +

02(g) -7

C(s)

-7

-7

I::!.H;xn = -219 kJ

PbO(s)

I::!.H;xn = -394 kJ

CO/g)

PbO(s)

+

CO/g)

+ 1.5 02(g)

-7

PbCO/s)

I::!.H;xn = 86 kJ

236

I

Chapter Five

I

Thermochemistry

*5.116. Ethanol as Automobile Fuel Brazilians are quite familiar with fueling their automobiles with ethanol, a fermentation product from sugarcane. Calculate the standard molar enthalpy for the complete combustion of liquid ethanol using the standard enthalpies of formation of the reactants and products as given in Appendix 4. 5.117. Formation of CO2 Baking soda decomposes on heating as follows, creating the holes in baked bread: 2 NaHC03(s)

-7

Na2C03(s)

+ CO/g)

This reaction is much too slow for the enthalpy change to be measured directly using a temperature change. Instead, the enthalpy change for the reaction may be calculated from the following data: Compound

+ HP(t')

Calculate the standard molar enthalpy of formation of NaHC03(s) from the following information: I::!.H;xn = -129.3 kJ I::!.H; [CO/g)]

= -394 kJ/mol

I::!.H~ [Na2C03(s)]

= -1131 kJ/mol

I::!.H; [HP(t')]

= -286 kJlmol

*5.118. Specific Heats of Metals In 1819, Pierre Dulong and Alexis Petit reported that the product of the atomic mass of a metal times its specific heat is approximately constant, an observation called the law of Dulong and Petit. Use the following data to answer the following questions. Element

M (g/mol)

Bismuth

c, [J/(g . °C)]

M x c,

207.2

0.123

Gold

197.0

0.125

Platinum

195.1

Tin

118.7

Zinc

65.4

0.388

Copper

63.5

0.397 0.433

Iron

55.8

Sulfur

32.1

0.460

Average value: a. Complete each row in the table by multiplying each given molar mass and specific heat pair (one result has been entered in the table). What are the units of the resulting values in column 4? b. Next, calculate the average of the values in column 4. c. Use the mean value from part b to calculate the missing atomic masses in the table. Do you feel confident in identifying the element from the calculated atomic mass? d. Use the average value from part (b) to predict the missing specific heat values in the table. *5.119. Unpleasant Smells Urine odor gets worse with time because urine contains the metabolic product urea [CO(NH)2]' a compound that is slowly converted to ammonia and carbon dioxide:

CO/aq)

-412.9

HP(t')

-285.8 -80.3

Calculate the standard molar enthalpy for the reaction. Experiment with a Metal Explain how the specific heat of a metal sample could be measured in the lab. Hint: you'll need a test tube, a boiling water bath, a Styrofoam cup calorimeter containing a weighed amount of water, a calibrated thermometer, and a weighed sample of the metal. *5.121. Rocket Fuels The payload of a rocket includes a fuel and oxygen for combustion of the fuel. Reactions 1 and 2 describe the combustion of dimethylhydrazine and hydrogen, respectively. Pound for pound, which is the better rocket fuel, dimethylhydrazine or hydrogen? (1)

0.233

-319.2

5.120.

25.5

0.215

Urea(aq)

NH3(aq)

0.120

Lead

dH;(kJ/mol)

(CH)2NNH/t') N/g)

(2)

H2(g)

+

+

+ 4 O/g)

4 HP(g)

i 02(g)

-7

-7

+

2 CO2(g) HP(g)

I::!.H;xn = -1694 kJ I::!.H;xn = -286 kJ

Carbon: Diamonds, Graphite, and the Molecules of Life *5.122. Industrial Use of Cellulose Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings, and plastics. Cellulose consists oflong chains of glucose molecules (C6H1206), so for the purposes of modeling the reaction we can consider the conversion of glucose to formaldehyde. a. Is the reaction to convert glucose into formaldehyde an oxidation or a reduction? b. Calculate the heat of reaction for the conversion of 1 mole of glucose into formaldehyde, given the following thermochemical data: I::!.H:omb of formaldehyde gas I::!.H~ of solid glucose

Glucose 5.123.

-572.9 kJlmol -1274.4 kJlmol

Formaldehyde

Converting Diamond to Graphite The standard state of carbon is graphite. I::!.H~,dbmondis +1.896 kJlmol. Diamonds are normally weighed in units of carats, where 1 carat = 0.20 g. Determine the standard enthalpy of the reaction for the conversion of a 4-carat diamond into an equivalent mass of graphite. Is this reaction endothermic or exothermic?

Properties of Gases: The Air We Breathe

A LOOK AHEAD: An Invisible Necessity 6.1

The Gas Phase

6.2

Atmospheric

6.3

The Gas Laws

6.4

The Ideal Gas Law

6.5

Gas Density

6.6

Dalton's Law and Mixtures of Gases

6.7

The Kinetic Molecular Theory of Gases and Graham's Law

6.8

Real Gases

Pressure

SCUBA DIVER AND SEA GOLDIES 1he gas exhaled by the diver forms bubbles that increase in size as they rise to the surface.

238

I Chapter Six

I

Properties


A LOOK AHEAD: An Invisible Necessity Gases are easy to overlook. We do not think much about breathing, but the exchange of gases between our lungs and the surrounding atmosphere is crucial to our lives. One reason we may not notice gases like oxygen-a colorless, odorless, tasteless, invisible gas-is that we do not have to pay for the oxygen needed to stay alive. We pay for the fuel to drive cars, but the air required to burn the fuel is free, so we rarely think about it. For every gallon of gasoline an engine burns, it consumes the oxygen in an average-sized room. The same quantity of air would supply the oxygen needed for an entire day by three people breathing normally. In a hospital operating room, managing the delicate balance of gases entering and leaving a patient can mean the difference between transitioning to a normal conscious life or irreversible coma. Having the right mixture of gases in our bodies is so important that during surgery, anesthesiologists constantly and carefully monitor blood levels of oxygen and carbon dioxide to ensure that safe levels of these gases are maintained. Life as we know it in our biopshere requires oxygen. Fish, insects, birds, mammals, and even plants must bring 02(g) into their systems and expel CO2(g). All the food we eat depends ultimately on plants' use of sunlight; during photosynthesis they combine carbon dioxide and water vapor to form sugars that contain the chemical energy required to support most of the life on our planet. The thin shell of gases that surrounds Earth and forms its atmosphere is a unique blend not found on other planets in our solar system. Mercury, the planet closest to the sun, has no significant atmosphere. The atmosphere of Venus consists mostly of carbon dioxide (96%) and nitrogen (3%) and features thick clouds of sulfuric acid. The atmospheric pressure on Venus is 90 times that on Earth. The Earth's early atmosphere was modified to its current composition (21% oxygen and 78% nitrogen) when plant life appeared and began to give off oxygen as a product of photosynthesis. Mars' atmosphere, with a pressure only about 1/100 of Earth's, is composed of carbon dioxide (95%), nitrogen (3%), and less than 0.2% oxygen. Beyond Mars, the gas giants-Jupiter, Saturn, Uranus, and Neptune-are essentially huge balls of gas with frozen cores. The atmosphere of each gas giant is mostly hydrogen, helium, and small amounts of methane and other hydrocarbons, with ammonia and hydrogen sulfide particles suspended by perpetual winds. Only Earth has significant free oxygen in its atmosphere. Our dependence on our atmosphere is so great that if we venture very far away from Earth's surface, we have to take the right blend of gases with us to sustain life. Hence, a commercial airplane has a pressurized cabin, an undersea explorer carries a mixture of compressed gases in a scuba tank, and a mountain climber takes canisters of oxygen to breathe when ascending far above sea level. Because of their physical properties as well as their chemical reactivity, gases are used in many ways that touch our lives. Our ability to understand how gases fit into our lives-whether in living rooms, scuba tanks, tires, airplanes, or balloons-is based on our knowledge of how gases behave in response to changing volume, temperature, and pressure-the subject of this chapter.

II11II The Gas Phase Earth is surrounded by a layer of gases 50 km thick. By volume it is composed primarily of nitrogen (78%), oxygen (21%), and lesser amounts of other gases (Table 6.1). To put the thickness of the atmosphere in perspective, if Earth were

6.1

TABLE 6.1

% (by volume)

The Gas Phase

I

Composition of Earth's Atmosphere

Compound

I

Mole Fraction"

Nitrogen

78.08

0.7808

Oxygen

20.95

0.2095 -----i

Argon Carbon

dioxide

0.934

0.00934

0.033

0.00033

I I -----i

Methane

2

X

10-4

2

X

10-6

Hydrogen

5

X

10-5

5

X

10-7

'Mole fraction is defined in Section 6.6.

-

the size of an apple, the atmosphere would be about as thick as the apple's skin. This layer of gases, which is held around our planet by gravity, exerts pressure on Earth's surface and on all sides of objects on the surface. This pressure decreases with elevation above sea level. As discussed in Chapter 1, gases have neither a definite volume nor a definite shape and expand to occupy the entire volume and shape of their container. This property is only one of the differences between gases and other states of matter. Let's review some of the other properties of gases that distinguish them from liquids and solids under conditions we might encounter in our everyday lives: 1. Unlike liquids and solids, the volume occupied by a gas changes significantly with pressure. If we were to carry a balloon from sea level (0 m) to the top of a 1600 m (5249 ft) mountain, the volume of the balloon would increase by about 20%. The volume of a liquid or solid would be unchanged under these conditions. 2. The volume of a gas also changes with temperature. For example, a balloon filled with room-temperature air decreases in size when it is taken outside on a cold winter's day. A temperature decrease of 20°C leads to a volume decrease of about 7%, whereas the volume of a liquid or solid remains practically unchanged by such modest temperature changes. 3. Gases are miscible with each other, which means that they can be mixed together in any proportion (unless they chemically react with each other). A hospital patient experiencing respiratory difficulties may be given a mixture of nitrogen and oxygen in which the relative amount of oxygen is much higher than its proportion in the air we ordinarily breathe. A scuba diver may leave the ocean's surface with a tank of air containing 17% oxygen, 34% nitrogen, and 49% helium, which form a homogeneous mixture. In contrast, many pairs ofliquids are immiscible, such as the oil and water in salad dressing. 4. Gases are typically much less dense than liquids or solids. The density of dry air at 20°C at typical atmospheric pressures is 1.20 giL. Notice that the density of gases is expressed in units of grams per liter. The density of liquids, such as water, is expressed in units of grams per milliliter. Thus, the density ofliquid water, 0.9986 glmL (998.6 giL) at 20°C, is over 800 times greater than the density of dry air. These four observations about gases are consistent with the idea that the molecules or atoms are further apart in gases than in solids and liquids. The larger spaces between molecules in air, for example, account for the ready compressibility of air

~ CONNECTION The physical properties of gases were introduced in Chapter 1.

I

239

240 I Chapter Six I Properties of Gases: The Air We Breathe

into cylinders used by scuba divers and into tank trucks that transport gases for industrial use. Greater distances between molecules account for the lower densities and observed miscibility of gases. We will explore each of these general properties of gases in more detail, including mathematical relations that allow us to determine exactly how pressure and temperature affect volume and density.

Pressure (P) is defined as the ratio of force to surface area. Atmospheric pressure (P ) is the force exerted by the gases surro~t~ding Earth on Earth's surface and on all surfaces of all objects. A barometer is an instrument that measures atmospheric pressure.

~

The standard atmosphere (1 atm) is the pressure capable of supporting a column of mercury 760 mm high in a barometer. It is also called simply 1 atmosphere.

Atmospheric Pressure

The gases in Earth's atmosphere are pulled toward gravity. The resulting weight of these gases exerts the surface (Figure 6.1). The ratio of the force (F) pressure (P), and the force exerted by the gases atmospheric pressure (P atm ):

Pressure is measured in millimeters of mercury (mmHg), also known as torr, where 1 atm = 760 mmHg = 760 torr.

the surface of the planet by a force that is spread across to surface area (A) is called on Earth's surface is called

P= F A

(6.1)

Atmospheric pressure is routinely measured with an instrument called a barometer. A simple but effective barometer design is shown in Figure 6.2. It consists of a tube nearly a meter long and closed at one end. The tube is filled with mercury, and its open end is inverted in a pool of mercury open to the atmosphere. Gravity pulls the mercury in the tube downward, creating a vacuum at the top of the tube, but atmospheric pressure pushes the mercury in the pool up into the tube. The net effect of these opposing forces is indicated by the height of the mercury in the tube, which provides a measure of atmospheric pressure. () Atmospheric pressure varies from place to place and with changing weather conditions. For reference purposes, the standard atmosphere (1 atm) of pressure is the pressure capable of supporting a column of mercury 760 mm high in a barometer. The pressure unit millimeters of mercury (mmHg) is also called the torr in honor of Evangelista Torricelli (1608-1647), the Italian mathematician and physicist who invented the barometer. Thus we have the following relation:

1 atm

The SI unit of pressure is the pascal (Pa), named in honor of French mathematician and physicist Blaise Pascal (1623-1662). He was the first to propose that atmospheric pressure decreases with increasing altitude. We can explain this phenomenon by noting that pressure is related to the total mass of molecules of gas above a given altitude (Figure 6.3). As altitude increases, the mass of the gases remaining above it must decrease, and less mass means less pressure. The pascal is an example of a derived SI unit; it is defined using the following SI base units (see Table 1.1):

(a) FIG resu atm (b) I upw YouJ text the to al

= 760 mmHg = 760 torr

(b)

IPa=

1kg -m'

S2

6.2

To understand the logic of this combination of units, consider the law in physics that states that pushing an object of mass (m) with a force (F) causes the object to accelerate at a rate a:

F= ma

(6.2)

Combining Equations 6.1 and 6.2 we have F __

ma

--p-A A

(6.3)

I

Atmospheric

Pressure

I 241

:~~~;\ 1 Vacuum

Weight of

Atmospheric pressur~atm)

Height of mercury column

If the units of the terms in the right-hand side of Equation 6.3 are kilograms (kg) for mass (m), square meters (m2) for area (A), and meters per square second (m/s2) for acceleration (a), then pressure (P) is given in units of ~'m

-2-2 m ·s

which simplifies to

~

--2 m's

or Pa

The relation between atmospheres and pascals is

1 atm

= 101,325 Pa

Clearly, 1 pascal is a tiny quantity of pressure. In many applications it is more convenient to express pressure in kilopascals (1 kPa = 1000 Pa); the above relation now becomes

1 atm = 101.325 kPa

10,000

8,000

6. 5,000 c::

0

.~

>

~

4,000

2,000

00.2 Atmospheric pressure, atm FIGU RE 6.3 Atmospheric pressure decreases with increasing altitude as the mass of the column of air above a given area decreases with increasing altitude. One atmosphere of pressure at sea level corresponds to 0.83 atm in Denver and 0.35 atm at the summit ofMt. Everest.

Mercury pool FIGURE 6.2 The height of the mercury column in this simple barometer designed by Evangelista Torricelli is proportional to atmospheric pressure.

242

I Chapter Six

I

Properties


For many years, meteorologists have expressed atmospheric pressure in millibars (mbar). The weather maps prepared by the U.S. National Weather Service show changes in atmospheric pressure by using contour lines of constant pressure, called isobars, that are 4 mbar apart (Figure 6.4). There are exactly 10 mbar in 1 kPa; thus at standard atmospheric pressure,

1 atm

= (101.325 kI!a)(10 mbarlkl!a) = 1013.25

mbar

The relations between different units for pressure are summarized

Calculating Atmospheric Pressure

SAMPLE EXERCISE 1.2.

The mass of Earth's atmosphere is estimated to be 5.3 X 1018 kg. The surface area of Earth is 5.1 X 1014 m2. The acceleration due to gravity is 9.8 m/s2. From these values calculate an average atmospheric pressure in kilopascals (kPa).

FIGU RE 6.4 Small differences in atmospheric

in Table 6.2.

pressure are associated with

major changes in weather. Adjacent

isobars

on this map of the eye of Hurricane

Katrina

differ by 4 mbar.

COLLECT AND ORGANIZE We are given the total mass of the atmosphere in kilograms, the surface area of Earth in square meters, and the acceleration due to gravity in meters per squared seconds (m/s2). These units can be combined to calculate atmospheric pressure in pascals. ANALYZE

(5.3

The force of gravity is exerted

on the mass of the atmosphere

1018 kg). The area is the surface area of Earth, 5.1 X 1014 m2. We can use Equation 6.3 and the known acceleration due to gravity to calculate atmospheric X

pressure. SOLVE

Substituting

the values given in Equation 6.3 gives us the following: F=

ma

A (5.3

X

1018 kg) (9.8 mN)

5.1 X 1014 m';'

= 1.0 -

5 X 10 kg

m's

2

= 1.0

X 105 Pa

= 1.0 X 102 kPa T HI N K ABO UTI T The results of this calculation are consistent with the average

value of atmospheric pressure at sea level: 101.325 kPa.

Practice Exercise Calculate the pressure exerted on a tabletop by a cube of iron that is 1.00 cm on each side and weighs 7.87 g. (Answers to Practice Exercises are in the back of the book.)

Other units of pressure are derived from masses and areas in the English system. The pressure of 1 atm is equivalent to 14.7 pounds per square inch (lb/in2; also abbreviated psi). Tire pressure in the United States is most often expressed in psi units. Weather reports in the United States also tend to give readings of atmospheric pressure in inches of Hg, rather than mmHg. A pressure of 1 atm is equivalent to 29.92 inches ofHg. The relation between different units for pressure is illustrated in Sample Exercise 6.2 and in Table 6.2. The subsequent Sample Exercise illustrates how a barometer measures atmospheric pressure.

6.2

TABLE 6.2

I

Units for Expressing Pressure Unit

Value

Atmosphere (atm)

1 atm

Pascal (Pa)

1 atm

=

1.01325 X 105 Pa

Kilopascal (kPa)

1 atm

101.325 kPa

Millimeter of mercury (mmHg)

1 atm

= =

Torr

1 atm

Bar

1 atm

Millibar (mbar)

1 atm

Pounds per square inch (psi)

1 atm

I

I I

760 mmHg

I

Expressing

= =

760 mmHg

=

= =

1013.25 mbar

760 torr

1.01325 bar I

14.7 psi

I

Pressure Using Different

Units

The lowest atmospheric pressure ever recorded at sea level was 25.69 inches ofHg. This pressure was measured on October 12, 1979, in the eye of typhoon Tip, northwest of Guam in the Pacific Ocean. What was the pressure in typhoon Tip in units of (a) mmHg, (b) atm, (c) kPa, and (d) torr? There are 2.54 cm in 1 inch. AND ORGANIZE We are given the atmospheric pressure in inches of mercury and asked to express the pressure in a variety of other units. We need selected conversion factors from Table 6.2.

COLLECT

A N A L Y Z E This is an exercise in converting units of pressure. (a) We can convert the

pressure given in inches of mercury to millimeters of mercury with the conversion factor given in the statement of the problem. (b) From there we can calculate pressure in atmospheres. (c) Pressure in atmospheres is easily converted to pressure in torr, (d) from which we can change the pressure units to pascals and kilopascals (kPa). SOL V E

The following conversions provide the desired new units of pressure:

a. Converting from inches ofHg to mmHg requires us to convert from inches to millimeters. We can do this in two steps, using the conversion factor relating inches and centimeters (1 in = 2.54 cm), followed by the conversion factor for centimeters to millimeters (10 mm/cm):

(25.69 ffiHg) (2.~~)

(1~ :m)

= 652.5

mmHg

b. Using the conversion factor 1 atm = 760 mmHg, we can convert the units of a pressure measurement from millimeters of mercury to atmospheres:

(652.5 mmHj?;)(7 1 atm H ) = 0.8586 atm 60mm g c. Earlier, 1 atm was defined as 101.325 kPa, so multiplying the pressure in atmospheres by 101.325 kPa/atm yields the pressure in kilopascals:

(0.8586

atfR)( 101.~

kPa)

= 87.00 kPa

d. Finally, we recognize that pressure units of mmHg are the same as those in torr, so the conversion factor between these units is 1 mmHg = 1 torr:

(652.5 mmHg)(

1 torr ) ImmHg

= 652.5

torr

I

Atmospheric

Pressure

I

243

244

I Chapter Six

I

Properties


A manometer is an instrument for measuring the pressure exerted by a gas.

Let's think about the unit conversions we just completed. Typhoons and hurricanes are associated with extremely low pressures, so we would expect each answer to be less than the value for the atmospheric pressure we calculated in the text just prior to this Sample Exercise: 1 atm = 760 mmHg = 760 torr = 101.325 kPa. Looking at the results of our calculations, we find that the pressure in each case is indeed lower-652.5 mmHg < 760 mmHg, 0.8586 atm < 1 atm, 87.00 kPa < 101.325 kPa, and 652.5 torr < 760 torr-so our answers all make sense. THINK

ABOUT

IT

Practice Exercise Atmospheric pressure in the eye of Hurricane Katrina, which devastated New Orleans in August 2005, was 90.2 kPa. What is this pressure in (a) atm, (b) mmHg, and (c) torr? Scientists conducting experiments with gases must often measure the pressures exerted by gases. A manometer is an instrument used to measure gas pressures in closed systems. Two types of manometers are illustrated in Figure 6.5. In each case, a U-shaped tube filled with mercury (or another dense liquid) is attached to a flask containing a gas. The difference between the two manometers is whether the other end of the tube is sealed or open to the atmosphere. In a closed-end manometer, the difference in the height of the columns on either side of the U-tube is a direct measure of the gas pressure. In a manometer in which the U-tube is open, the difference in height represents the difference between the pressure in the flask and atmospheric pressure. When flask pressure is greater than atmospheric pressure, the level in the arm attached to the flask is lower than

A satellite infrared image of Hurricane Katrina.

Closed end

ah

=

0

FIGURE 6.5 A closed-end manometer (a,b): (a) The air has been removed from the flask and the mercury levels in the arms of the V-shaped

tube are equal, because both ends of

Mercury Gas

the V-tube are exposed to a vacuum. (b) Gas is allowed to enter the flask; it exerts pressure

(a) I::!.h = 0

on the mercury, causing the mercury level to drop in the left-hand right-hand

arm and rise in the

arm. The difference in the height

of the mercury columns (ah) is the pressure exerted by the gas. An open-end (c-e): (c) The right-hand atmosphere,

manometer

arm is open to the

while the left is connected

to the

flask. The gas in the flask is at atmospheric pressure,

so the mercury level is the same

in both arms. (d) The pressure of gas in the flask is less than atmospheric mercury level in the left-hand

pressure,

so the

arm is higher

than the right: Pg .., = P"m - I::!.h. (e) Here the pressure of gas in the flask is greater than atmospheric

pressure, so the mercury level is

lower in the left-hand

Pgas

=

Patm

+ ah.

arm than in the right:

Open end

(b)Pgas = I::!.h

I

6.2

Atmospheric

Pressure

I 245

the level in the arm exposed to atmospheric pressure; when the pressure in the flask is lower, the level in the arm attached to the flask is higher.

Measuring Gas Pressure with a Manometer Oyster shells are composed of calcium carbonate (CaCO). When they are roasted (heated) on an industrial scale to produce calcium oxide [CaO(s), quicklime], CO2(g) is also a product of the reaction. A chemist roasts an oyster shell in a flask (from which most of the air has been removed) that is attached to a closed-end manometer (see part a of the accompanying figure). When the roasting is complete and the system has cooled to room temperature, the difference in levels of mercury (~h) in the arms of the manometer is 143.7 mm (part b of the figure). Calculate the pressure ofC02(g) in (a) torr, (b) atmospheres, and (c) kilopascals. COL L E C TAN D 0 R G A N I Z E The pressure of the gas in the flask equals the differ-

ence in the height of the mercury in the two columns in a closed manometer. We are given that in the problem.

.--

I::!.h .-~

~ \: (b)

(a) Roasting

CaC03

I::!.h = 143.7 mm

in a closed manometer:

(a) An evacuated flask containing

oyster

shells. (b) The same setup after roasting the shells, which releases CO2(g) in the flask.

A N A L Y Z E The difference is measured in millimeters

of mercury, so we must convert that to the requested units of torr and atmospheres. The appropriate conversion factors are 1 mmHg = 1 torr and 760 torr = 1 atm. SOLVE

a. First we convert millimeters of Hg to torr: 143.7 ffiffiHe= X

1torr H = 143.7 torr ffiffi e=

b. Then we convert torr to atmospheres: 143.7

teH

X

1 atm 760 teH

= 0.1891 atm

c. Finally we convert atmospheres to kilopascals: 0.1891

iHffi X

101/~kPa

=

19.16 kPa

THINK ABOUT IT Because millimeters ofHg and torr are equivalent, the numerical value of a measurement expressed in one unit is the same as in the other. The other numbers make sense: a pressure less than 760 torr should also be less than 1 atm; by the same token, a pressure less than 1 atm should also be less than 101.325 kPa. The answers fit these criteria.

Practice Exercise If the same quantity of CO2(g) is produced in an open-end manometer (see accompanying figure) when the atmospheric pressure is 760 torr, indicate where the mercury levels should be at the conclusion of the experiment. What is the numerical value of ~h? Before roasting

CONCEPT

TEST

If you are running an experiment using an open-ended manometer, how do you determine the value of atmospheric pressure? (Answers to Concept Tests are in the back of the book.)

~

246

I Chapter Six

I

Properties


~

Boyle's law states that the volume of a given amount of gas at constant temperature is inversely proportional to its pressure.

The Gas Laws

In Section 6.1, we summarized

some of the properties of gases and described the effect of pressure (P) and temperature (T) on volume (V) mostly in qualitative terms. Our knowledge of the quantitative relationships between P, V, and T goes back more than three centuries, to a time before the field of chemistry as we now recognize it even existed. Many years of experiments led to an understanding of the behavior of gases in terms of their response to changes in temperature, pressure, and amount of gas present. Some of these experiments were driven by human interest in flight, specifically using hot-air balloons to rise above Earth's surface. Conclusions from these experiments have stood the test of time and scientific scrutiny. Centuries later, we now understand the relationships between P, V, and T on a molecular level. Balloons, which were so much a part of early experiments in responses of gases to changing conditions, are still used extensively in the study of weather and atmospheric phenomena (Figure 6.6), and their successful and safe use requires applying an understanding of gas properties that was first gained in the 17th and 18th centuries.

Boyle's Law: Relating Pressure and Volume Air, like any gas or mixture of gases, is compressible. The compressibility of air allows us to pump up bicycle or automobile tires and store large amounts of air in relatively small metal cylinders in compressed form. The ability to compress air is the cornerstone of the compressed-gas industry because it makes the packaging and transport oflarge quantities of gas in a small space economical (Figure 6.7). The relation between the pressure (P) and volume (V) of a fixed quantity of gas (constant value of n, where n is the number of moles) at constant temperature (T) was investigated by British chemist Robert Boyle (1627-1691) and is known as Boyle's law. Boyle observed that the volume of a given amount of a gas is inversely proportional to its pressure when kept at a constant temperature, as shown in Figure 6.8 and symbolized below:

FIGU RE 6.6 Weather balloons are used to carry instruments

aloft in the atmosphere.

P rx 1 V

FIGU RE 6.7 Because gases are important commercial

items, large quantities

are compressed

of gas

in trucks and railcars.

p= 1 atm

= 760 mmHg

The change in the size of the balloon demonstrates 1 atm (760 mmHg),

comparable

pressure to 266 mmHg,

to atmospheric

comparable

(6.4)

In other words, as the pressure on a constant amount of gas increases, the volume of the gas decreases, and conversely, as the pressure on the gas decreases, the volume of a given amount of gas increases.

into small volumes for more

efficient transport

(T and n fixed)

p= 266 mmHg

the relation between pressure and volume. (a) The balloon inside the bell jar is at a pressure of

pressure on a nice day at sea level. (b) A slight vacuum applied to the bell jar drops the interior

to the pressure at the top ofMt.

Everest (over 29,000 ft above sea level), and the volume of the balloon increases.

6.3

'"

I

247

'"

oS

~ p:;'"

The Gas Laws

§

§ '"'

I

'J

1.0

0.5

~ p:;'"

,,, ,, ,

nnlm

FIGU RE 6.8 The graphs show the relation between pressure and volume. (a) At a constant temperature,

0.5

the pressure of a given quantity

gas is inversely proportional

of

to the volume that

it occupies. This relation means that (b) a plot ,

•

22.4 44.8

of Pversus l/Vis

pc

0.0223

Volume,L

0.0446

a straight line.

__ 1_ L-1 Volume'

(a)

(b)

Boyle's law describes why a balloon expands (increases its volume) when Atmospheric Atmospheric carried up a mountain. How can we use Boyle's law to predict exactly how pressure pressure much a balloon will expand when the pressure decreases? In his experiments Hg added Boyle used a U-shaped tube like the one shown in the manometer in Figure (pressure ~ 6.5. One end of the tube was sealed, and mercury was poured into the open increases) Closed end arm of the tube. The pressure exerted on the gas trapped in the tube depends Volume of on the mass of the mercury added. The amount of gas in his experiment VOlumeof\ gas after remained constant, and the temperature of the apparatus was also held congas before adding Hg ........ stant (Figure 6.9). The pressure on the gas in the tube was changed by varyadding Hg ........ ing the amount of mercury poured into the tube. Remember that the more mercury used, the greater the force exerted by the mercury (F = ma), and the greater the force per unit area (P = FIA). As Boyle added mercury to the FIGURE 6.9 Boyle used aU-shaped U-tube, the volume occupied by the gas decreased in a way described by Equation tube similar to the one shown here for his 6.4. Mathematically we can replace the proportionality symbol with an equals experiments on the relation between P and sign and a constant,

u

V. The volume of gas in the closed end of

PV

= constant

(6.5)

the U-tube changes in proportion

to the

difference in height of the mercury columns as

where the value of the constant depends on the mass of gas in the sample and the temperature. Because the value of the product PV in Equation 6.5 does not change for a given mass of gas at constant temperature, any two combinations of pressure and volume are related as follows: Pl~

= P2~

(6.6)

This relationship is illustrated by the dotted lines in Figure 6.8. When 44.8 L (~) of gas is held at a pressure of 0.500 atm (PI)' then Pl~

=

(0.500 atm)(44.8 L)

=

22.4 L· atm

Equation 6.6 enables us to calculate the pressure required to compress this quantity of gas to a volume, for example, of22.4 L: Pl~

= P2~ (0.500 atm)(44.8I:.)

Pl~ --=P ~

2

22.4b

= 1.00 atm The products of P and Vare the same for each point on the graph. Equation 6.6 can also be used to calculate the change in volume that takes place at constant

mercury is added to the open arm of the tube.

248

I Chapter Six

I

Properties


temperature when the pressure of a quantity of gas changes. The result in the following Sample Exercise confirms the statement in Section 6.1 that taking a balloon from sea level to 1600 m leads to an increase in volume of some 20%.

Using Boyle's Law A balloon is partly inflated with 5.00 liters of helium at sea level, where the atmospheric pressure is 1.00 atm. The balloon ascends to an altitude of 1600 meters, where the pressure is 0.83 atm. (a) Describe what happens to the balloon as it rises. (b) What is the volume of the helium in the balloon at the higher altitude? (c) What is the percent increase in the volume of the balloon? Assume that the temperature of the gas in the balloon does not change in the ascent.

W;

AND ORGANIZE We are given the volume = 5.00 L) of a gas at a given pressure (PI = 1.00 atm), and we are asked to find the volume that same quantity of gas would occupy (~ = unknown) when the pressure changes (P2 = 0.83 atm). The problem further states that the temperature of the gas does not change and that the amount of gas in the balloon remains constant. COLLECT

The balloon contains a fixed amount of gas and its temperature is constant, so we have a pressure-volume relation, or an application of Boyle's law. Boyle's law tells us that pressure and volume are inversely proportional, and that PV for a quantity of gas is constant as long as nand T are constant.

ANALYZE

SOLVE

a. Since the pressure decreases as the balloon rises, its volume should increase. b. Equation 6.6 (Pl~ = P2~) relates the pressure and volume of a given quantity of gas under two sets of conditions. We let the pressure at sea level be PI and the volume inside the balloon at sea level ~, so the pressure at 1600 m is P2 and the volume of the balloon at that altitude ~. We know the values for PI' V;, and P2, and we need to calculate ~. Rearranging Equation 6.6 to solve for ~ and inserting the given values for PI' P2, and ~ gives us the following: Pl~ P

V=2

V

2

2

=

(1.00 afffi)(5.00 L) __

=

6.0 L

c. To calculate the percent increase in the volume we must determine the difference between VI and ~ and compare the result with ~: ~ - VI X 100 VI THINK

ABOUT

IT

=

6.0 L - 5.0 L X 100 5.0L

=

20%

The prediction we made in part a is confirmed by the calcula-

tion in part b.

Practice Exercise A scuba diver exhales 3.5 L of air while swimming at a depth of 20 m where the sum of atmospheric pressure and water pressure is 3.0 atm. By the time the bubbles of air rise to the surface, where the pressure is 1.0 atm, what is their total volume?

6.3

At the time Boyle discovered the relationship between P and V in the 17th century, scientists lacked any understanding of atoms or molecules. Given what we now know about matter, how can Boyle's law be explained on a molecular basis? Consider the effects of compressing a collection of molecules of gas into a smaller space. The molecules in the gas phase are constantly in random motion, which means that they collide with one another and with the interior surface of their container. The collisions with the walls of its container are responsible for the pressure exerted by a volume of gas. If molecules are squeezed into a smaller space-if the volume of the container is decreased-more collisions occur per unit time (Figure 6.10). The more frequent the collisions, the greater the force exerted by the molecules against the walls of their container, and thus the greater the pressure. Dividing the force (F) of the collisions by the area (A) over which they occur yields the pressure (P) (Equation 6.1). The area (A) of the walls of the container decreases when the size of the container is decreased, which in turn causes FIA to increase. Therefore the value of P increases as the size of the container decreases; as volume goes down, pressure goes up, because the frequency of collisions between gas molecules and container walls increases. The converse is also true: as the volume of a container is increased, A increases, the frequency of collisions between the molecules and the walls decreases, and pressure drops; as volume goes up, pressure goes down.

I

The Gas Laws

I

249

FIGURE 6.10 Gas molecules are in constant random motion, exerting pressure through

collisions with their surroundings.

When a quantity of gas is squeezed into half its original volume, the frequency of collisions per unit of surface area on the inside of the container

increases by a factor of 2, and so

does the pressure.

Charles's Law: Relating Volume and Temperature Nearly a century after Boyle's discovery of the inverse relation between pressure and volume, French scientist Jacques Charles (1746-1823) first documented the linear relation between the volume and temperature of a fixed quantity of gas at constant pressure. Now known as Charles's law, the relation states that the volume (V) of a fixed quantity of gas is directly proportional to the absolute temperature (T, in kelvins) of the gas at a constant pressure:

Vet:: T

(P and n fixed)

(6.7)

273 K

293 K

373 K

3.2 3.0 ....1 ~ 2.8

..3 ~

2.6

The effects of Charles's law can be seen in Figure 6.11, where a bal2.4 loon has been attached to a flask, trapping a fixed amount of gas in the 2.2 apparatus. Heating the flask causes the gas to expand, inflating the 260 balloon. The higher the temperature, the greater the volume occupied by the gas, and the bigger the balloon. As in the case of Boyle's law, we can replace the proportionality symbol in Equation 6.7 by an equals sign if we include a proportionality constant: I

,.

I

280

300

320

340

Temperature,

FIGURE 6.11

Sees

360

380

What a Chemist

A deflated balloon attached to a flask

inflates as the temperature

V = constant X T

V

=

constant

of the gas inside

the flask increases from 273 K to 373 K. This

This time, let's solve for the constant (by dividing each side of the equation by T):

T

400

K

behavior is described by Charles's law.

(6.8)

The value of the constant in Equation 6.8 depends on the pressure (P) of the gas and the number of moles of gas (n) present in the sample. The ratio VIT does not change for a given quantity of gas at constant pressure, and any two combinations of volume and temperature are related as follows:

(6.9)

Charles's law states that the volume of a fixed quantity of gas at constant pressure is directly proportional to its absolute temperature.

250

I Chapter Six

I

Properties


Applying Charles's law, as described by Equation 6.9, allows us to make the statement in Section 6.1 that a decrease in temperature of 20°C leads to a decrease in volume of about 7% for a balloon. Let's see how we arrive at these numbers. Consider what happens when a balloon containing 2.00 L (V) of air at 20°C (~) is taken outside on a winter's day when the temperature is O°C (T2). The amount of gas in the balloon is fixed, and the atmospheric pressure is constant. Our own experience tells us that the volume of the balloon will decrease; Charles's law enables us to quantify that change. Substituting the values we are given into Equation 6.9, we can calculate the final volume (~), provided that we express the temperatures ~ and I; in kelvins, not degrees Celsius. We must use absolute temperatures in gas equations. Because the lowest temperature on the absolute scale (Kelvin scale) is absolute zero (0 K), temperatures expressed in kelvins are always positive numbers:

~=~~= ~

(20

+ 273)

~

K

0

+ 273

K

Solving for ~:

~ =

2.00L X 273K 293 K

= 1.86L

As predicted, the volume of the balloon decreased when the temperature decreased. The percent change in the volume of the balloon is calculated as: % decrease

= ~ - ~ ~

X 100

= 2.00b - 1.86b 2.00b

X 100

=

7%

Sample Exercise 6.5 provides another example of calculations using Charles's law.

Using Charles's

Law

Charles and his compatriot Joseph Gay-Lussac (1778-1850) were drawn to the study of gases because of the interest in hot-air balloons in the late 18th century. We can imagine an early experiment these two scientists might have performed: what temperature is required to increase the volume of a sealed balloon from 2.00 L to 3.00 L if the initial temperature is 15T and the atmospheric pressure is constant? COLLECT AND ORGANIZE We are given the volume of a gas at an initial temperature and asked to calculate the temperature needed to make the gas reach a different volume. The container (a balloon) is sealed, so n is constant, as is atmospheric pressure.

We are given a volume (~ = 2.00 L) at a specific temperature (~ = 15°C), and are asked to calculate the temperature needed (T2 = unknown) to increase the volume of the gas to a new value (~ = 3.00 L). The relation between volume and temperature is summarized by Charles's law. One form of Charles's law is given by Equation 6.9 when temperatures are expressed in kelvins: ANALYZE

6.3

SOLVE

First rearrange Equation 6.9 to solve for

I

The Gas Laws

I

251

7;:

~~ VI

T=2

Substitute for VI' ~, and ~, and remember to convert temperature to kelvins:

T = ~~ = (3.00 b)(288 K) = 432 K

V;

2

2.00 b

To convert the final temperature the temperature in kelvins:

7; =

to degrees Celsius, we must subtract 273 from

432 K - 273

=

159T

IT Notice that to increase the volume by 1.5 times, the temperature in kelvins must also increase 1.5 times [432 K = (1.5)(288K)], as expected from the statement that volume is directly proportional to temperature. The volume of the gas is not directly proportional to temperature in degrees Celsius. THINK ABOUT

1---1'~,

Practice Exercise Hot expanding gases can be used to move a piston and to perform useful work in a cylinder with a movable piston. If the temperature of the gas is raised from 245°C to 560°C, what is the ratio of the initial volume to the final volume if pressure remains constant? Charles's law helps explain why a balloon inflated with hot air rises. A given mass of hot air occupies more volume than does the same mass of cooler ambient air. Therefore a balloon filled with hot air displaces a mass of ambient air larger than its own mass, giving the balloon buoyancy. Put another way, the density of a given quantity of gas decreases with increasing temperature. In the following way, Charles's law also allows us to determine that the value of absolute zero (0 K) is equal to -273.15T. Consider what happens if we plot the volume of a fixed quantity of gas at constant pressure as a function of temperature. Some typical results for 28 cm3 (28 mL) of gas initially at 293 K (20T) are shown in Figure 6.12. The decrease in volume with decreasing temperature is linear. In the laboratory we are limited by what low temperatures we can reach before the gas liquefies; at this point the 50 gas laws no longer describe its behavior because it is no longer a gas. However, we can extrapolate from our data to the point where the volume of a gas would reach zero if liquefaction 40 did not occur. As we saw in Chapter 1, that temperature is defined as absolute zero.

FI G U R E 6.12

The volume of a gas (n and P constant) is

plotted against temperature scales. As predicted temperature

in both the Celsius and Kelvin

decreases; the relationship

The dashed line shows the extrapolation corresponding

to a zero volume.

as absolute zero: -273°C Kelvin scale. Numerically, temperature

;' ,,

by Charles's law, volume decreases as the is linear on both scales.

10

This temperature

,,

is known

o

on the Celsius scale and 0 K on the volume is directly proportional

only on the Kelvin scale.

,,;,,;,

from the data to a point

to

Temperature,oC: Temperature,

K:

I

1.#

!

-273 -200 0

73

-100

o

100

200

300

173

273

373

473

573

•

252

I Chapter Six

I

Properties


Avogadro's law states that the volume of a gas at a given temperature and pressure is proportional to the quantity of the gas. Amontons's law states that, as the absolute temperature of a fixed amount of gas increases, the pressure increases as long as the volume and quantity of gas remain constant.

1 mole = 1 V

4· ;,..

:>

c:

c:

~

~

'.p

'.p

..:s

o:i'" 0:::

0:::'" Speed

Speed

(a)

(b)

FIGU RE 6.23 speed

(uJ,

The speeds of gas atoms or molecules at a constant temperature

associated with the highest point in the plot, characterizes

cover a range of values. (a) The most probable

the largest fraction of molecules in the population;

more molecules have the most probable speed than have any other speed. The average speed (u,vg)' a little higher than the most probable speed, is simply the arithmetic directly proportional

average of all the speeds. The root-mean-square

to the square root of the temperature

of its molar mass. (b) The curves describing temperature

show that the lower the molar mass, the higher the root-mean-square

of speeds. (Approximate

speed (u,m)' a little higher still, is

of the gas (in kelvins) and inversely proportional

the relative numbers of molecules in a population

u,m, values are indicated by dashed lines.)

Speed

CONNECTION The equation for kinetic energy was introduced in Chapter 7, where m was the mass of an object. Here m is always the molar mass of the gas being described.

'" '0 E

..!:l

273

speed (u ) is the The root-mean-square square root of the average of the ';quared speeds of all the molecules in a population of gas molecules. A molecule possessing the average kinetic energy moves at this speed.

(6.24)

-3'"u

I

to the square root

plotted against speed at constant

speed (u,m) and the broader the distribution

274 I Chapter Six

I

Properties


The value of urms is typically is expressed in kelvins, but Equation 6.24. What value (see Table 6.3) that includes R ./2000 K

Speed

FIGURE 6.24 The most probable speed (um' dashed lines) increases with increasing temperature.

As the temperature

notice that the distributions smaller portion

increases,

broaden, a

of atoms or molecules moves

at any given speed, and more speeds are represented population.

by a significant

portion of the

expressed in meters per second and the temperature care must be taken in choosing the units for R in of the gas constant R should we use? One value of R meters among its units is

= 8.314 kg' m2/(s2. mol· K)

Using this value for R, however, requires that we express the molar mass of the gas in units of kilograms per mole rather than in the more common units of grams per mole. Equation 6.24 indicates that, because different gases have different molar masses, heavier molecules, such as N2 (28.01 g/mol = 2.801 X 10-2 kg/mol) move more slowly at a given temperature than lighter molecules, such as helium (4.003 g/mol = 4.003 X 10-3 kg/mol). Figure 6.23(b) shows the different distributions of speeds for several gases at a constant temperature. Figure 6.24 shows how the most probable speed and the kinetic energy of gases increase with temperature. How fast do gas molecules move? Gas molecules move very rapidly at or slightly above room temperature-up to 1000 mi/hr and even faster (depending on temperature). Calculating

Root-Mean-Square

Speeds

Calculate the root-mean-square speed of nitrogen at 300.0 K. Express the answer in (a) meters per second and (b) miles per hour. COL LEe TAN 0 0 R G A N I Z E Given only the temperature and identity of a gas (N 2 in this case), we are asked to calculate the root-mean-square speed of nitrogen. According to Equation 6.24, we need the molar mass of the gas in question in addition to the temperature.

In this case, the gas is nitrogen (N), which has a molar mass of 28.01 g/mol. We also need the correct value of R. The second value listed in Table 6.3, R = 8.314 kg· m2/(s2. mol K), is the correct one for this calculation because it will give us a speed in meters per second. Having chosen this value of R, we then need to express the molar mass in kilograms; for N2 this value is 0.02801 kg/mol.

ANALYZE

SOLVE

a. Inserting the appropriate values into Equation 6.24 yields the root-meansquare speed in meters per second: 3[8'314*g.m2] U ,m.,N,

= ~ 3~T =

" f(me!

0 ~)

0.02801~

~

=

(300,

,~)

1 Ht6l

516.9 m/s

b. To convert speed from meters per second into miles per hour (mi/hr) requires the use of several conversion factors: 1 mi

=

1.6093 km

=

1.6093 X 103 m

and

1 hr

= 3600 s

Using these factors gives us the requested result in miles per hour: U N2

ffi)( 1.60931 mi )(3600~) X 10 ffi hr

=

(516.9

=

1156 mi/hr

3

6

=

1.156 X 103 mi/hr

6.7

THINK


It's quite amazing to find that the average N2 molecule at a close to room temperature moves at more than 1000 mi/hr.

ABOUT

temperature

I

IT

Practice Exercise Calculate the root-mean-square speed of helium at 300 K. Express the answer in meters per second, and compare your result with the rootmean-square speed of nitrogen.

CONCEPT

TEST

Put the following gas molecules in order of increasing root-mean-square a temperature of 20°C: H2, CO2, Ar, SF6, UF6, Kr.

speed at

It is not immediately obvious why the pressure exerted by two different gases, such as N2 and He, is the same at a constant temperature when their root-meansquare speeds are quite different. The answer lies in the assumption in the kinetic molecular theory that all populations of gas molecules have the same average kinetic energy at a given temperature, independent of their molar mass. Can we show that two gases, such as nitrogen and helium, have the same average kinetic energy at 300 K? Let's calculate the average kinetic energy of 1 mole of N2 and 1 mole of He at 300 K. The solution to Sample Exercise 6.15 reveals that the urms of N2 molecules at 300 K is 517 m/s. The corresponding root-mean-square speed for helium atoms is 1370 m/s (the result you should have gotten for the last Practice Exercise). For 1 mole of each gas we can substitute into Equation 6.23 as follows: KEN,

= ~(2.801 X 10-2 kg/mol)(517 m/s)2

= ~mN,(urms,Nl

= 3.75 X 103 kg' m2/(s2. mol) KEHc

=

~mHC(urms,HY= ~(4.003

X 10-3 kg/mol)(1370

m/s)2

= 3.75 X 103 kg' m2/(s2. mol) This calculation demonstrates that the same quantities (same values of n) of the two diff'erent gases have the identical average kinetic energy. Therefore at the same temperature (T) they exert the same pressure (P). The slower nitrogen molecules collide with the container walls less often but exert a greater force than the faster helium atoms because nitrogen molecules are heavier. Equation 6.24 enables us to compare the relative root-mean-square speeds of two gases at the same temperature. Consider an equimolar mixture (one with the same number of moles) of nitrogen and helium. We have just demonstrated that at a constant temperature (300 K) their average kinetic energies are the same: KE N, = 1 )2 = 1 )2 2 m N,(u rms,N, 2 m He(u rms,He

= KE He

or m (U N2

rms,N2

)2

m

He

(U rms,Hc )2

Rearranging the terms to express the ratio of the root-mean-square speeds of the two gases in terms of the ratio of their molar masses, and then taking the square root of each side, gives (u

rms,He

)2

Urms,N2

N2

(urms,N )2

mH

= ~mN2 =

28 01 .1.~4.003 .

2

Urms,He

m

mHc

e

g g

=

2.65

I 275

276

I

Chapter Six

I

Properties


The root-mean-square speed of helium atoms at 300 K is 2.65 times higher than that of nitrogen molecules. We can test this using the values for urms from the calculation of average kinetic energies:

= 2 65

1370 m/s 517 m/s

urms,Hc urms,N,

We can write a generic form of the preceding gases (x and y) to their masses:

(m

U rms,x

. equation that relates to any pair of

)112

Y

(6.25)

mx

Urms,y

Equation 6.25 holds only when we have the same number of particles of each gas, say 1 mol of gas x and 1 mol of gas y. Equation 6.25 leads to the conclusion that the root-mean-square speed of more massive particles is lower than urms of lighter particles.

Graham's Law: Effusion and Diffusion Let's observe what happens to two balloons, one filled with nitrogen and another with helium (Figure 6.25). The volume, temperature, and pressure of the gas in each balloon are identical. The pressure inside each balloon, however, is greater than the external atmospheric pressure. Over time we observe that the volume of the helium balloon decreases significantly while the nitrogen-filled balloon has nearly the same volume as it had at the start of our experiment. The skin of any balloon is slightly permeable; that is, it has microscopic holes that allow gas to escape, reducing the pressure inside the balloon. Why does the helium balloon leak faster than the balloon filled with nitrogen? We now know the relation between the root-mean-square speeds and molar masses of two different gases, and we have calculated that helium atoms move about 2.65 times faster than nitrogen atoms. It appears that the gas with the greater root-mean-square speed (He) leaks out of the balloon at a higher rate. This process of moving out of a space at higher pressure through a small opening into a region of lower pressure is called effusion. Helium effuses through the imperfections in a rubber balloon into a region oflower pressure more rapidly than nitrogen. In the 19th century, Scottish chemist Thomas Graham (1805-1869) recognized that the rate of effusion of a gas is related to its molar mass. Today this relation is known as Graham's law of effusion, which states that the rate of effusion of gases is inversely proportional to the square root of their molar masses. We can derive a mathematical representation of Graham's law by substituting rate (r) for root-mean-square speed in Equation 6.25:

ir

=

U urms,x

y

From Equation

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

By substituting

~

mY x

6.24 we know that: urms

Effusion is the process by which a gas escapes from its container through a tiny hole into a region oflower pressure.

=

rms,y

r

for

urms

= ~3~T

in the above expression we obtain

(6.26)

6.7

I


I 277

which simplifies to the usual statement of Graham's law:

r~ =

-.2

_Y

ry

M

(6.27)

x

Using Equation 6.27 in the example of the helium (MHe = 4.003 g/mol) and nitrogen (MN2 = 28.01 g/mol) balloons, we can show that the rate of effusion of helium is 2.64 times that of nitrogen:

rHe

-r N2

_

~MN2

-

_

MHe

-

~28.01 g/mol _ -2.64 4.003 g/mol

Notice that this expression has the same form-and hence the ratio has the same value-as the ratio of the root-mean-square speeds of the two gas molecules. How does the kinetic molecular theory explain Graham's law of effusion? The escape of a gas molecule from a balloon (see Figure 6.25) requires that the molecule collide with the pinhole. The faster a gas molecule moves, the more likely it is to find the hole. The effusion of gas through a tiny hole in a container is related to the process of diffusion, the spread of one substance through another. A couple of examples are the diffusion of perfume molecules from a person throughout a room, and the smell of bread baking from the kitchen throughout a house. The rates of diffusion depend on the average speeds of the molecules of gas and on their molar masses. Graham's law applies equally well to the diffusion of gases. FIGU RE 6.25 Two balloons at the same temperature

SAMPLE EXERCISE 8.18

Applying Graham's Law to the Diffusion of Gases

An odorous gas emitted by a hot spring was found to diffuse 2.92 times faster than helium. What is the molar mass of this gas? We are asked to determine the molar mass of an unknown gas based on its rate of diffusion relative to helium. AND

ORGANIZE

(N2) and helium (He)

both decrease in volume over time as gas effuses through

small pores in the membranes

of the balloons. The helium atoms are lighter than the nitrogen

molecules and therefore

have higher root-mean-square consequence,

COLLECT

and pressure and filled with equal

volumes of nitrogen

speeds. As a

the helium atoms effuse faster,

causing the helium-containing

balloon

to shrink more quickly than the nitrogen balloon.

Graham's law (Equation 6.27) relates the relative rate of diffusion of two gases to their molar masses. We know that helium (M _ = 4.003 g/mol) dif' 'e fuses 2.92 times faster than the unidentified gas (y).

ANALYZE

We insert the atomic mass of He and the ratio of their diffusion rates into Equation 6.27 and solve for molar mass of the unknown gas:

SOL V E

2.92

My

= =

My 4.003 g/mol (4.003 g/mol)(2.92)2

=

Diffusion is the spread of one substance (usually a gas or liquid) through another.

34.1 g/mol

278

I Chapter Six

I

Properties


Two things about Equation 6.27 for Graham's law are worth noting: (1) like the root-me an-square speed in Equation 6.25, Graham's law involves a square root. (2) With labels x and y for two gases in a mixture, the left side of Equation 6.27 has the x term in the numerator and the y term in the denominator, while the right side of the equation has the opposite: y term in the numerator and x term in the denominator. Keeping the labels straight is the key to solving problems involving Graham's law. The molar mass of the unidentified gas is 34.1 g/mol. Coupled with other data, such as the distinctive odor of the compound, its density, and its source, a scientist might suspect it to be hydrogen sulfide (H2S).

THINK

ABOUT

IT

Practice Exercise Helium effuses 3.16 times as fast as which other noble gas?

CONCEPT

TEST

Put the following gas molecules in order of increasing rate of effusion from a container at a temperature of 20°C: H2' CO2, Ar, SF6, UF6, Kr.

~

Real Gases

Up to now we have treated all gases as ideal gases. Under typical atmospheric pressures and temperatures, most gases do behave as ideal gases. We have also assumed, according to the kinetic molecular theory, that the volume occupied by individual gas molecules is negligible compared with the total volume occupied by the gas. In addition, we have assumed that no interactions occur between gas molecules other than random elastic collisions. This situation changes, however, when we begin to compress gases into increasingly smaller volumes or cool them to lower temperatures. Gas properties then begin to deviate from ideal behavior.

Deviations from Ideality Let's start by considering the behavior of 1.0 mole of a gas as we substantially increase the pressure on the gas. From the ideal gas law, we know that PV/RT = n, so for 1 mole of gas, PV/RT should remain equal to 1.0 regardless of how we change the pressure. This relation between PV/ RT and P for an ideal gas is shown by the purple line in Figure 6.26. However, plots of PV/RTversus P for CH4, H2, and CO2 at very high pressure do not have flat, straight lines. Not

2.0

h 1.5

r

~

7/

"'" 1.°

FIGURE 6.26 The effect of high pressure on

7'

.c

Ideal gas

0.5

the behavior of real and ideal gases: The curves diverge from ideal behavior in a manner unique to each gas. When gases deviate from ideal behavior, their identity matters.

o

200

400

600

800

P, atm

1000 1200

6.8

I

Real Gases

I

279

FIGURE 6.27 At high pressures, more of the volume of a gas is occupied by the atoms or molecules of the gas. The greater density of particles between

also results in more interactions

them (black arrows in expanded

view). These interactions

reduce the frequency

and force of collisions with the walls of the container pressure.

only do the curves diverge from the ideal value of PV/RT but also the curve is different for each gas, indicating that in these circumstances the identity of the gas does matter. Why don't real gases behave like ideal gases at high pressure? One reason is that the ideal gas law considers gas molecules to have so little volume compared with the volume of their container that they are assumed to have no volume at all. However, at high pressures, more molecules are squeezed into a given volume of space (Figure 6.27). Under these conditions the volume of the individual molecules can become significant. What is the impact of this occurrence on the relation PV/RT = 1? In the expression PV/RT, the volume (V) actually refers to the free volume WffCC vo Iume ), the empty space not occupied by gas molecules. Because VffCC vo Iume is difficult to measure, we instead measure V;o,al' the total volume of the container holding the gas: V:otal

~ree

volume

+

Vmolecules

In an ideal gas the molecules have "no" volume (Vmo Ieeu Ies = 0), so Vtota 1= Vfree vo Iume . In a real gas, when the external pressure is low and the volume occupied by the gas is large, the volume contributed by all the molecules collectively is so minuscule that Vmolcculcs is essentially O. Under these conditions, then, the ideal gas equation is a reasonable predictor of the behavior of a real gas. Now consider what happens to a real gas in a closed but flexible container (like a balloon) as the conditions are changed. Increasing the external pressure on the container causes the container-and hence the volumes occupied by the gas-to shrink. As the external pressure increases, the assumption that V;uolecules = 0 becomes less and less valid because the proportion of the container's volume taken up by the molecules themselves becomes increasingly significant (Vrno Iceu Ics > 0). The relation V;o,al

=

~rec volume

by V;otal' Since an ideal gas:

+

V;otal

Vmolcculcs

> ~ree

still applies, but ~rce volumc can no longer be approximated the ratio PV/RT is also larger for a real gas than for

volume'

PV ~ RT

> PVfrcc

volume

RT

As a consequence, the curve for a real gas in Figure 6.26 diverges upward from the ideal gas line. A second situation also causes the PV/RT ratio to diverge from 1. Kinetic molecular theory assumes that molecules do not interact, but molecules do attract

(red arrow), thereby reducing the

280

I

Chapter Six

I

Properties


The van der Waals equation attempts to account for the behavior of real gases by including experimentally determined factors that quantify the contributions of molecular volume and intermolecular interactions to the properties of gases.

each other. The attractive forces are much weaker than the bonds that hold atoms together in molecules, but they arise in all matter because of imbalances in the distribution of electrons. These attractive forces function over short distances, so the assumption that molecules behave independently is a good one as long as the molecules are far apart. As pressure increases on a population of gas molecules, pushing them closer and closer together, intermolecular attractive forces can become significant. This causes the molecules to associate with each other, which decreases the force of their collisions with the walls of their container, thereby decreasing the pressure exerted by the gas. If the value of Pin PV/RT is smaller than the ideal gas value, then the value of the ratio decreases and the curve for a real gas diverges from the ideal gas line, moving below it on the graph.

The van der Waals Equation for Real Gases Because the ideal gas equation does not hold at high pressures, it is useful to find a relation that can be used under nonideal conditions. In determining a new equation for real gases, we must consider that 1. the true volume will be less than the volume of an ideal gas since the molecules themselves occupy significant space; and 2. the observed pressure will be less than the pressure of an ideal gas because of molecular attractions. The van der Waals equation (P

n2a) (V + V2

nb)

= nRT

(6.28)

includes terms to correct for the pressure (n2a/V2) and for the volume (nb). The symbols a and b are called van der Waals constants; their values have been determined experimentally for each gas. Table 6.5 lists the van der Waals constants for common atmospheric gases. Both a and b increase with increasing molar mass and the number of atoms in the molecule.

TABLE 6.5

Van der Waals Constants of Selected Gases

Substance f--

f--

f--

He

'--

b (Ljmol)

0.0341

0.02370

I

Ar

1.34

0.0322

H,

0.244

0.0266

N2

1.39

0.0391

°2

1.36

0.0318

I

I

I

CH

2.25

0.0428

CO2

3.59

0.0427

CO

1.45

0.0395

Hp NO

5.46

0.0305

1.34

0.02789

N02 HCl

5.28

0.04424

I

3.67

0.04081

I

S02

6.71

0.05636

4

f--

a (L2 . atmjmol2)

I I

I I ---- //

'Rb

• Sr'

'!n- 'Sn' ·~b· -t;·

'Cs

'Ba'

'TI-

'~b'

.. ..

:.I: :~f:

.. .. :~f· :IW: '~i' .~~.

FIGURE

electron

'Ra'

8.3

group elements elements

• Fr

I

A single bond results when two atoms share one pair of electrons.

0

--.~.

An Introduction to Lewis Theory

A Lewis symbol is the chemical symbol for an atom surrounded by one or more dots representing the valence electrons.

.. -g-.. :f.-.. :~1: .. .~..~.

//

'K

16

I

Lewis symbols for the main of the periodic

table. Because

in a family have the same outer-shell configuration,

they have the same

Lewis symbol. The dots represent

the valence

electrons.

electron pairs, which are represented by pairs of dots around an atomic symbol. A Lewis structure helps us consider where the electrons are in a molecule by showing how the valence electrons are distributed among its atoms. Lewis structures are sometimes sufficient for understanding the structure of a molecule; frequently, they are just the starting point for more extensive analyses of structure. The starting point for drawing the Lewis structure for a molecule is to draw Lewis symbols for the atoms involved, in which electrons are portrayed using dots. The Lewis symbol for an atom is the atomic symbol surrounded by the number of dots equal to the number of valence electrons in the neutral atom. The dots are placed around the symbol one at a time on the four sides of the symbol (top, bottom, right, and left). The order in which they are placed does not matter as long as one dot is placed on each side before any dots are paired. The number of unpaired dots indicates the capacity of the atom to form bonds. Figure 8.3 shows the Lewis symbols for the main group elements. Because the valenceelectron configuration of each element in a group is the same, all atoms in a group have the same Lewis symbol. When Lewis structures are drawn for molecules, an electron pair is symbolized like this:

so the Lewis structure looks like this:

for a hydrogen molecule containing

Both hydrogen atoms in the

two hydrogen atoms

H1 molecule have achieved a duet by sharing:

@® The circles enclose the electrons associated with each atom. They overlap to indicate that electrons in bonds are shared by two atoms. The second convention in standard notation is to use a line (-) for a single bond involving the sharing of two electrons, so the hydrogen molecule is also symbolized in a line-bond formula as

H-H

Hydrogen

~II

CHEMTOUR

Lewis Dot Structures

359

360

I

Chapter Eight

I

Chemical Bonding and Atmospheric

A lone pair, or unshared pair, of electrons is a pair of electrons that is not shared. A bonding pair of electrons is a pair of electrons shared between two atoms.

Molecules

Starting with these two conventions for constructing Lewis structures, let's develop a stepwise approach to drawing Lewis structures by looking at three covalent molecules: HCI(g), aqueous solutions of which are called hydrochloric acid; CH/g), a major hydrocarbon fuel; and NH/g), widely used as a fertilizer. Essentially, we consider the number of valence electrons available and distribute them according to a set of guidelines. Notice our use of the word guideline. The statements are suggestions, not rules. We will modify and extend them as we deal with larger molecules with more diverse options for arrangements. Here are the first two guidelines arising directly from Lewis's model: 1. Hydrogen tends to form one single bond containing 2 electrons. 2. Other elements tend to form bonds to achieve an octet of electrons. Using a system of electron bookkeeping, we determine the number of electrons each element brings to the molecule and compare that to the number of electrons each atom needs to complete its outermost shell. The difference between what an atom has and what it needs is related to the number of covalent bonds it must form to achieve a complete outer shell. EXAMPLE

A: HCI (HYDROGEN

CHLORIDE)

a. Initial electron bookkeeping: Look at the number of valence electrons each atom has and compare that with the number each atom needs to have either a duet (hydrogen) or octet (all others) of electrons. For HCI, we use this step to define the number of covalent bonds in the molecule: Need:

(lH X 2e-)

+

Have:

(1 H X 1 e-)

+ (1 CI

(lCI X 8e-) X 7 e-)

=

10e-

= 8 e-

The difference between what the atoms have and what Lewis theory suggests they need is achieved by sharing 2 electrons, which corresponds to one covalent bond: 2e2e-/bon

-

=

1bond

Putting those two electrons between the hydrogen and the chlorine makes 1 covalent bond:

H:Cl or H-Cl b. We have 6 electrons left over that are not involved in bonding. These leftover electrons are distributed around the atoms in pairs to satisfy the octet (or duet) rule. Hydrogen already has its duet through sharing. The leftover electrons are given to chlorine to complete its octet:

H:¢):

or

H-¢):

c. Final electron bookkeeping: We have used all 8 electrons. The chlorine atom has achieved an octet and the hydrogen atom a duet by sharing two electrons in a covalent bond:

Hydrogen

chloride

The electron pairs on chlorine that are not involved in bonding are called lone pairs (or unshared pairs) to distinguish them from bonding pairs that are shared by two atoms.

8.2

EXAMPLE

I


B: CH. (METHANE)

Methane has more atoms than HCI, but we carry out exactly the same steps. In a polyatomic molecule or ion, we need to identify the "central" atom-the atom to which the others are bound. Another guideline is: 3. The first atom in a molecular formula is often the central atom. a. Initial electron bookkeeping: To develop the Lewis structure for methane, we calculate the number of covalent bonds required. To accomplish this, we first total the number of valence electrons in all the atoms in the molecule:

Next we apply guidelines 1 and 2: The carbon atom needs an octet: Each hydrogen atom needs a duet:

1 C = 1 X 8e- = 4H = 4 X 2e- =

8e8e-

Total electrons wanted: The atoms in the structure need a total of 16 e-, but only 8 e- are available. The difference is: 16e- - 8e- = 8e-. That means the Lewis structure needs 4 covalent bonds: 4 bonds b. We use a pair of electrons to form a bond between each pair of bound atoms. How can we tell which atoms are bound? C comes first in the formula, so we assign it the position of central atom. Also remember the first two guidelines: H tends to form 1 bond to achieve a duet; other elements tend to share as many electrons as they need to achieve an octet. The carbon atom has 4 electrons; it tends to pick up 4 more electrons by sharing, so it tends to make 4 bonds:

H

I I

H-C-H

H Hydrogens achieve duets; carbon achieves octet.

Lewis structures Methane

c. Final electron bookkeeping: The atoms contributed a total of 8 electrons; the structure contains 8 electrons (2 in each single bond). The carbon has achieved an octet; each hydrogen has a duet; therefore all the electrons have been used. EXAMPLE

C: NH3 (AMMONIA)

a. Initial electron bookkeeping: Need: (IN X 8e-)

+

(3H X 2e-)

=

14e-

Have: (IN X 5e-)

+

(3H X le-)

=

8e-

Difference:

6e-

=

3 bonds

I

361

362

I Chapter Eight

I


Molecules

b. Three covalent bonds require 6e-. We have 2e-Ieft over. Each hydrogen atom has a duet, but the nitrogen atom only has 6 e- at this point:

Therefore the remaining 2e- must be a lone pair on the nitrogen: or

or

H-N-H

I

H Ammonia

c. Final electron bookkeeping: We started with 8 electrons and the final structure has 8 electrons. Nitrogen has an octet (three bonding pairs of electrons plus one lone pair); each hydrogen atom has a duet.

Drawing Lewis Symbols for Atoms Draw the Lewis symbol for argon, the third most abundant gas in the atmosphere. COLLECT AN D ORGANIZE We are asked to draw the Lewis symbol for argon, an element in group 18 of the periodic table.

We need to determine the number of valence electrons in an atom of argon and then arrange them as Lewis dots around the argon symbol (Ar).

ANALYZE

The condensed electron configuration of argon is [Ne]3s23p6, telling us that argon has 8 valence electrons. Arranging these 8 electrons around the symbol for an argon atom gives the result:

SOLVE

ABOUT IT Argon is in group 18, the noble gases. It has a full outer shell and a complete octet of electrons.

THINK

Practice Exercise The interaction of chlorine-containing compounds with light in the atmosphere can lead to the formation of chlorine atoms. Draw the Lewis symbol for a chlorine atom. (Answers to Practice Exercises are in the back of the book.)

Drawing Lewis Structures

for Molecules

Chloroform (CHCI) is a low-boiling liquid that was used as an early anesthetic in surgery. It does not interact with hemoglobin, but breathing its vapors induces unconsciousness. What is the Lewis structure for chloroform? We need to locate each element in the chloroform molecule on the periodic table and determine the number of valence electrons it has. We can then use the step-by-step process outlined earlier in this section to determine the Lewis structure. COLLECT

AND

ORGANIZE

Carbon is written first in the formula, so we assume it is the central atom. This makes sense because there are four other atoms in chloroform (1 H atom and 3 CI atoms) and carbon tends to make 4 bonds.

ANALYZE

We apply the conventions for Lewis structures and take the steps developed in Examples A through C to get the following:

SOLVE

8.2

I


a. Initial electron bookkeeping: Need: (lC X 8e-)

+

(lH X 2e-)

+

(3Cl X 8e-)

Have: (lC X 4e-)

+

(lH X le-)

+

(3CI X 7e-) = 26e-

=

34e-

8e- = 4 bonds

Difference:

b. If carbon is the central atom, then four pairs of electrons (4 covalent bonds) can be located about it. Each bond extends from the carbon to one of the other atoms: CI CI I or H-C-Cl H:C:CI I CI CI We have now used 8 electrons, leaving us 26e- - 8e- = 18e- to accommodate. The hydrogen atom already has its duet of electrons, and the carbon has its octet, so we put the remaining electrons in pairs around the chlorine atoms to achieve octets for the three chlorines: :CI:

:CI: H:C:CI:

I I

H-C-Cl:

:¢):"

.. ..

:QI:

c. Final electron bookkeeping: The atoms in CHCl3 contributed a total of 26 electrons to the structure for the molecule. The Lewis structure as drawn contains 26 electrons. The hydrogen atom has a duet of electrons; each carbon and chlorine atom has achieved an octet of electrons. THINK

ABOUT

IT

The carbon atom has 4 valence electrons, so it tends to make

4 covalent bonds to achieve an octet. It makes sense to draw the carbon atom in the center of the chloroform molecule that contains four other atoms in addition to the carbon. The chlorine atoms each contain 7 electrons, so they only need to make 1 covalent bond to achieve an octet. Noticing how atoms tend to combine in simple molecules like this will help us draw more challenging molecules.

Practice Exercise Hydrogen peroxide, H102, is used to disinfect cuts and to bleach hair. The H201 molecule contains an 0-0 bond. Draw the Lewis structure for H201•

Lewis Structures for Molecules with Multiple Bonds Since Chapter 2 we have shown structures for molecules such as acetic acid (CH3COOH) and formaldehyde (CH10) that have a pair of lines connecting some of the atoms:

I

II

'0:

H

I

~-H .'

H

H H-C-C

H

Acetic acid (CH3COOH)

.

\ C=O' / .'

Formaldehyde

(CH10)

Lewis structures also describe molecules such as these with multiple bonds between pairs of atoms. When two atoms share two pairs of electrons (4 electrons, 2 covalent

Chloroform

I

363

364

I

Chapter Eight

I

Chemical Bonding and Atmospheric Molecules

A double bond results when two atoms share two pairs of electrons. A triple bond results when two atoms share three pairs of electrons. Bond length is the distance between the nuclear centers of two atoms joined together in a bond.

bonds), the bond is called a double bond (=); when two atoms share three pairs of electrons, the bond is called a triple bond (=). In Section 8.7 we discuss the characteristics of these bonds in detail, but for now we just need to understand the concept of bond length. Bond length is the distance between the nuclear centers of two atoms that are joined together in a bond. If we compare single, double, and triple bonds between the same two atoms, the bond lengths decrease in the following order: Single bond> EXAMPLE

double bond>

triple bond

D: O2 (OXYGEN)

a. Initial electron bookkeeping: Need: 20

X 8e-

= 16e-

Have: 20

X 6e-

= 12e4e-

Difference:

=

2 bonds

b. Two covalent bonds require that 4 electrons be shared between the two oxygen atoms in a double bond: 0::0

or

0=0

That leaves 12e- - 4e- = 8 electrons (four electron pairs) to be arranged about the oxygen atoms to achieve an octet:

'0::0' .. .. Oxygen

~ ~

'0=0' .. ..

c. Final electron bookkeeping: All 12 electrons have been used, and each oxygen atom has achieved an octet using two lone pairs of electrons and two bonding pairs. Notice that we have drawn the lone pairs on the oxygen atoms at an angle of about 120 from each other and the bonding pairs. Electrons are negatively charged and repel each other, so we draw them as far apart from each other as possible. This logic will be an important part of drawing three-dimensional structures of molecules in Chapter 9. 0

EXAMPLE

E: C2H. (ETHYLENE)

a. Initial electron bookkeeping: (2C X 8e-)

+

(4H X 2e-)

=

24e-

Have: (2C X 4e-)

+

(4H X 1e-)

=

12e-

Need:

12e-

Difference:

2.5

§

2.0

.~ ..,

~

I

1.5 1.0 0.5

o

main spheres [easing 1a

family

lUmber heights) Nith lumn Ig atomic trends :tive 1

\.1

370

I Chapter Eight

I


Molecules

4

4

3

I-

D Ionization energy D Electronegativity

3

"0

E

'-

:;c M

0

::0 2

to c": c": 0

.p oj

1

LiIBe

BeN

0

F

0

AI Si P 5 CI

Li

Be

B

C

N

0

F

N

·c

.£

0

Element 4

I

1

3 C.> .p

4

D Ionization energy D Electronegativity

3

FIGURE 8.7

The trend in ionization

:;c M

0

::0

follows the trend in their electronegativities. The columns represent

on the left for each element

the first ionization

The columns

energies.

bfJ

g"u c:

E

'-

oj

energies among the main group elements

'i

2

2

Q"

to c": c": 0

.p oj

1

1

N

·c

.£

on the right represent

electronegativity

values. Elements

in the

second and third row are shown.

o

Na

Mg

AI

Si

p

S

Cl

o

Element

N

CONNECTION We defined ionization energy in Chapter 7 as the amount of energy required to remove 1 mole of electrons from 1 mole of atoms or molecules in the gas phase.

ments on the right-hand side than in atoms on the left-hand side of the periodic table. Greater attraction to the nucleus draws the valence electrons of an atom inward, making the size of the atoms decrease from left to right across a row. Greater attractive force is also experienced by bonding electrons, so electronegativity increases from left to right across a row.

CONCEPT

TEST

Why does Figure 8.5 not include electronegativity values of the noble gases?

N

CONNECTION The periodic variation in electronegativity is another illustration of trends in properties as a result of shielding and Z"p as discussed in Chapter 7.

Within a group of elements, differences in electronegativity are related to increasing atomic size with increasing atomic number. From the top down within a group, the size of the outermost orbitals increases as the value of the principal quantum number (n) increases. The valence electrons of the elements toward the bottom of a group in the periodic table are shielded from their nuclei by greater distances and by more layers of inner-shell electrons than are the elements toward the top of the group. More shielding and greater atomic size lead to less pull on bonding pairs of electrons; consequently electronegativity decreases with increasing atomic number. The correlation between electronegativity and ionization energy is easy to rationalize. The same factors that make it more difficult to remove an outer-shell

8.3

I

Unequal Sharing, Electronegativity, and Other Periodic Properties

electron from an atom also make that atom more effective at attracting shared valence electrons. Electronegativity values follow the same pattern as ionization energies.

Bond Polarity Electronegativity allows us to determine which end of a covalent bond is electron-rich and which is electron-poor because we can quantify the difference in electronegativity between the two atoms involved in a bond. It also allows us to estimate the covalent character of a bond by determining a relative bond polarity as a measure of the degree of unequal sharing of electrons in a covalent bond. The greater the difference in electro negativity, the more uneven the distribution of electrons between the two atoms, and the more polar the bond (Figure 8.8). There is no simple rule for all compounds, but in general we assume that ionic compounds are formed between two elements when the absolute value of the electronegativity difference between the two atoms is 2.0 or greater. Such differences frequently exist in compounds formed between metallic and nonmetallic elements. For example, calcium oxide is considered an ionic compound because the difference in electro negativity between Ca (1.0) and 0 (3.5) is 2.5. On the other hand, the bonds formed between different nonmetals are covalent bonds. These bonds vary in polarity as a function of the difference in the electronegativities of the elements. For example, a H-C bond is not as polar as a H-O bond because the difference in electronegativity between C and H (2.5 - 2.1 = 0.4) is less than the difference in electronegativity between 0 and H (3.5 - 2.1 = 1.4). Electronegativity differences make covalent bonds polar. Figure 8.8 illustrates the impact of decreasing electronegativity on polarity in the series of hydrogen halides. From top to bottom, electronegativity decreases from F to CI to Br to I, and the polarity of the covalent bond also decreases from H-F to H-Cl to H-Br to H-I. Figure 8.9 lists values for the polarity of bonds and introduces the use of some new symbolism: The arrow with the plus sign embedded in its tail (~) is placed over a bond to indicate the direction ofpolarity: it points toward the more electronegative, electron-rich atom in the bond, and the position of the plus sign indicates the more positive, electron-poor atom. The polarity of the bond may also be symbolized by placing 8- (lowercase delta minus) next to the electron-rich atom and 8+ (lowercase delta plus) next to the electron-poor atom in the structure. The presence of a 8 means that the charges are less than a full 1+ or 1-, indicating some polarity in the bond (uneven distribution) but not complete transfer of an electron.

o

Hel

HF FIGURE

8.8

The main group elements

1-

HBr whose atoms have the highest

tend to be at the top of a group. Electronegativity most electronegative

1)-

decreases

HI electro negativity

values

down the group. In this case, F is the

element in group 17, with I the least electronegative.

I

371

Bond polarity is a measure of the extent to which bonding electrons are shared between two atoms in a covalent bond. The less equally they are shared, the more uneven the distribution, the more polar the bond. Differences in electro negativity determine bond polarity.

372

I Chapter Eight

I


Molecules

(b) Methane

(a) Hydrogen

(c) Methanol

H

H

If--+ H-C-H I

H-H

I H-C-O-H I

H

H

I

f--+

H-C-O-H

H

C: 2.5 H: 2.1 II = 2.5 - 2.1 = 0.4

0: 3.5 H: 2.1 II = 3.5 -2.1 = 1.4

Pure covalent

Polar covalent

Polar covalent

8.9

0+

I

H

H: 2.1 H: 2.1 II = 2.1 - 2.1 = 0.0 FIGURE

bp-bp. Repulsions caused by a lone pair are also greater than the repulsions associated with a double bond, so the interactions of a lone pair with the rest of the electrons dominate the predictions of deviations from ideal structures. Sulfur dioxide has two resonance forms, but both resonance forms have the same steric number and hence the same molecular geometry. Predicting

Relative Sizes of Bond Angles

The central atoms in S03 (sulfur trioxide) and 03 (ozone) both have SN = 3. Define the geometry of each molecule and compare the size of the O-S-O bond angle to that of the 0-0-0 bond angle. AN DORGAN IZE We are given the steric number of both central atoms, so we can draw structures with the correct electron-pair geometry and, from those, determine the molecular geometry. COLLECT

In S03' three oxygen atoms are bonded to the central sulfur atom, so the sulfur has no lone pairs of electrons. Because only two oxygen atoms are bound to the central oxygen in 03, there must be one lone pair on the central oxygen.

ANALYZE

Sulfur trioxide has three resonance structures, and ozone has two. However, for both substances the resonance structures are equivalent in terms of steric number of the central atom, so we need only deal with one resonance structure, as shown in the accompanying figure:

SOLVE

:0:

I

• -::7 S" '.

.9:

'Q

SN

=

3

SN

=

3

Three atoms bonded to the central atom

Two atoms and one lone pair bonded to the central atom

Electron-pair geometry trigonal planar

Electron-pair planar

Molecular planar

geometry

=

= trigonal

Molecular

geometry

geometry

= trigonal

= bent

The theoretical bond angle in trigonal planar arrangements is 120°. Since sulfur trioxide has no lone pairs, we predict the O-S-O bond angle in sulfur trioxide to be 120°. We expect the bond angle in 03 to be slightly smaller than the ideal angle because a lone pair occupies more space than a bonding pair and forces the bonding pairs closer together; therefore, we predict that the 0-0-0 bond angle in ozone is slightly smaller than the ideal value of 120°. In many cases the presence oflone pairs on a central atom pushes atoms bonded to the central atom closer together and decreases the bond angles.

T H INK ABO UTI T

(tXt)

CONNECTION In Chapter 8 we discussed the value of the ozone layer that protects life on Earth from potentially damaging UV radiation.

Practice Exercise The central sulfur atom in sulfur dioxide (S02) has SN = 3. Predict how the O-S-O bond angle in S02 compares in size to the bond angles in S03 and 03,

CONCEPT

TEST

Why is CO2 a linear molecule, but S02 is bent? (Answers

to Concept Tests are in the back

of the

book.)

9.2

I

Valence-Shell Electron-Pair Repulsion Theory (VSEPR)

For SN = 4, there are three possible combinations of atoms and lone pairs about the central atom: four atoms and no lone pairs (CH4), three atoms and one lone pair, and two atoms and two lone pairs. SN

=

4, ELECTRON·PAIR GEOMETRY = TETRAHEDRAL

I--

No. of Bonded Atoms

No. of Lone Pairs

Molecular

Geometry

4

0

Tetrahedral

3

1

Trigonal

2

2

Bent

pyramidal

To illustrate the second situation, consider ammonia (NH). The Lewis structure for ammonia shows three atoms and one lone pair of electrons about the central nitrogen atom, so SN = 4. That means the electron-pair geometry of ammonia is tetrahedral.

H-N-H I H Lewis structure SN =4

Electron-pair geometry tetrahedral

Molecular geometry trigonal pyramidal

=

Predicted

bond angle

-. band continuous band of available energies. Because the band comes from overlapping ~ +-----(partially valence-shell orbitals, we call it a valence band. 4s filled) Band theory explains the conductivity of copper and many other metals by assuming that there is essentially no gap between the energy of the occupied Cu lower portion of the valence band and the empty upper portion. Therefore, valence electrons can move easily from the lower portion to the upper portion, FIGURE 11.2 What a Chemist where they are free to move from one empty orbital to the next and flow throughSees As the half-filled 4s orbitals of an increasing number of Cu atoms overlap, their out the solid. energies are split into a half-filled valence This partially filled valence-shell model explains the conductivity of many band. Electrons can move from the filled half metals, but not all. Consider one of copper's next-door neighbors in the peri(purple) of the valence band to the slightly odic table, zinc. Its electron configuration ([Ar]3d!04s2) tells us that all of its higher-energy upper half (red), where they valence-shell electrons reside in filled orbitals. This means that the valence band are free to migrate from one empty orbital to in solid zinc is completely filled, too. With no empty orbitals to move into, the another across the entire solid. valence-shell electrons in Zn might appear to be immobile. However, they actually do migrate; zinc is a good conductor of heat and electricity. Band theory explains why. It assumes all atomic orbitals of comparable shape and energy, including zinc's empty 4p orbitals, can mix ..... together to form energy bands. The energy band produced by splitting empty 4p orbitals is broad enough to overlap Conduction band the valence band produced by splitting the filled 4s orbit(empty) 4p als, as shown in Figure 11.3. This means that it is easy for electrons from the filled valence band to move up to the empty orbitals of what is called a conduction band, where P they are free to migrate and conduct electricity. Overlapping occupied (valence) and unoccupied (con-' -Valence band 4s duction) bands make metals good conductors of electricity (full) and also heat. For example, when atoms on the outside of a copper pot are heated, they vibrate faster. Electrons absorb this thermal energy and are raised from the filled portion of the valence band to the unfilled portion, where they are FIGURE 11.3 What a Chemist Sees As the filled 4s orbitals free to migrate. The higher the temperature, the more elecof an increasing number of Zn atoms overlap, they form a filled valence trons move into the unfilled portion. Thermal energy in band (shown in purple). An empty conduction band (gray) is produced by the form of a large population of mobile electrons moves the overlap of empty 4p orbitals on a large number (N) of Zn atoms. Once rapidly through a metal pot from the source of heat to the N is large enough so that these bands overlap each other, electrons can move easily from the filled valence band to the empty conduction band. contents of the pot. Band theory is an extension orbital theory that describes solids.

~I

of molecular bonding in

* **

-----

** * * * * * ----

PCi)b

11.2

CONCEPT

TEST

Is the electrical conductivity of magnesium metal best explained in terms of overlapping conduction and valence bands, or in terms of a partially filled valence band? Explain your answer. (Answers to Concept Tests are in the back of the book.)

.:ID Semimetals and Semiconductors To the right of the metals in the periodic table there is a "staircase" of elements that tend to have the physical properties of metals and chemical properties of nonmetals. These semimetals (or metalloids) are not as good at conducting electricity as metals, but they are much better at it than the nonmetals. We can explain this "in between" behavior using band theory. Let's begin with the most abundant of the semimetals, silicon. When we apply band theory to the bonding in pure silicon we find that there is an energy gap of about 100 kJlmol between the occupied valence band and the empty conduction band (Figure 11.4a). The difference is called, logically, a band gap (Eg). Generally, only a few of the electrons in the valence band of solid Si have sufficient energy to move to the conduction band, which limits silicon's ability to conduct heat and electricity. However, as we are about to see, the conductivity of silicon and the other semi metals can be enhanced by many orders of magnitude with relatively small changes in chemical composition. This varying ability to conduct electricity, in between that of metals and nonmetals, makes the semimetals semiconductors. We can change the conductivity of solid Si (or any elemental semiconductor) by replacing some of the Si atoms in its solid structure with atoms of another element of similar atomic radius but with a different number of valence electrons. The replacement process is called doping, and the added element is called a dopant. Suppose the dopant is a group 15 element such as phosphorus: each phosphorus atom adds one extra valence electron to the structure. These electrons have a different energy from the silicon electrons and populate a band within the silicon band gap (Figure 11.4b). This arrangement effectively diminishes the size of the band gap, making it easier for valence electrons to reach the conduction band. This n-type semiconductor is so called because the dopant contributes extra negative charges (electrons) to the structure of the host element. The conductivity of solid silicon can also be enhanced by substituting Si atoms with atoms of a group 13 element, such as gallium. These substitutions mean that there are fewer valence electrons in the solid. The result is an empty band in the band gap (Figure 11Ac). Less energy is needed to promote an electron from the filled valence band of silicon to the new empty (conduction) band created by the group 13 element, increasing electrical conductivity. This type of semiconductor, in which there is a reduction in the number of valence electrons (or negative charges) is called a p-type semiconductor because a reduction in the number of negatively charged electrons is equivalent to the presence of positively charged holes. The transistors used in solid-state electronics are combinations of n- and p-type semiconductors. Semiconductors can also be prepared from mixtures of group 13 and group 15 elements. For example, gallium arsenide (GaAs) is a semiconductor that emits red (A = 874 nm) light when connected to an electrical circuit. This emission is

I

I

Semimetals and Semiconductors

527

The energy gap between the valence and conduction bands is called the band gap (E/ A semiconductor is a substance whose conductivity can be made to vary over several orders of magnitude by altering its chemical composition. contains excess An n·type semiconductor electrons contributed by electron-rich dopant atoms . A p·type semiconductor contains electronpoor dopant atoms that cause a reduction in the number of electrons, which is equivalent to the presence of positively charged holes.

@@ CONNECTION We introduced the semi metals (metalloids) in Section 2.6 when we described the structure of the periodic table.

I

Conduction band

Conduction band

Conduction band

Eg Eg Valence band

Valence band

Valence band

(a) Pure

(b) n-type

(c) p-type

FIGURE 11.4 Semiconductors

such as

silicon have variable conductivities

that can be

greatly enhanced

by adding other elements.

(a) Pure Si has a band gap (Eg) of about 100

kJ/mol. (b) Adding valence electrons

an element

that has more

per atom than Si creates

a filled band in the band gap of an n-type semiconductor.

(c) Adding

fewer valence electrons

an element with

than the host material

creates an empty conduction gap of a p-type

semiconductor.

band in the band

528

I

Chapter Eleven

I

The Chemistry of Solids

used in devices such as laser pointers, bar-code readers in stores, and CD players. The energy of each photon of red light corresponds to the energy gap between the valence and conduction bands in GaAs. When electrical energy is applied to the material, electrons are raised to the conduction band. When they fall back to the valence band, they emit light. If aluminum is substituted for gallium in GaAs, the band gap increases, and predictably the wavelength of emitted light decreases. For example, a material with the empirical formula AIGaAs2 emits yellow-orange light (A = 620 nm). Many of the multicolored indicator lights in electronic devices such as sound system components rely on light-emitting diodes (LEDs) based on aluminum gallium arsenide semiconductors.

Distinguishing Between n-Type Semiconductors What kind of semiconductor-n-type arsenic create? COL LEe TAN

or p-type-does

p- and

doping germanium

with

We are asked to determine the type of semiconductor (Ge, group 14) is doped with arsenic (As, group 15).

D 0 R GA N I Z E

formed when germanium

The fact that As is in group 15 means that its atoms have one more valence electron than the atoms of Ge. ANALYZE

When the dopant has more valence electrons than the host semimetal, the two form an n-type semiconductor.

SOL V E

Arsenic is a good candidate for making an n-type semiconductor because its atoms are nearly the same size as those of germanium and so fit easily into the structure of solid Ge. T HI N K ABO UTI T

Practice Exercise Gallium arsenide (GaAs) cal scanners in retail stores. GaAs can be made ductor by replacing some of the As with another Sn-would form an n-type semiconductor with p-type semiconductor? (Answers


of the

is a semiconductor used in optian n-type or a p-type semiconelement. Which element-Se or GaAs, and which would form a

book.)

mD Metallic Crystals In Section 11.1 we observed that each Cu atom in a piece of copper metal touches 12 other copper atoms. Let's explore how the atoms in different metals adopt characteristic stacking patterns.

Stacking Patterns

A crystalline solid is made of an ordered array of atoms, ions, or molecules.

When copper is heated above its melting point and then slowly allowed to cool, its atoms arrange themselves in ordered three-dimensional arrays, forming a crystalline solid. Think of these arrays as stacked layers (two-dimensional planes) of Cu atoms packed together as tightly as possible. Each atom in a layer touches 6 others in that layer. The atoms in the layer directly above nestle into the spaces created by the atoms below, as shown in Figure 11.5. Similarly, the atoms in a third layer nestle among those in the second. However, two different alignments are possible

I

11.3

I

Metallic Crystals

529

Layera------.:..

~'t!fiiI~

Layer b Third layer --'--same as layer a (a)

(a) Displays

(b) of (a) oranges in a grocery store or (b) cannonballs

closest-packed

arrays of spherical

ababab ...

=!'1l~~

Layer a at an 18th-century

fort illustrate

Layer b

objects.

for the atoms in the third layer. They could be located directly above the atoms in the first layer (Figure 11.5a). However, they could fit just as snugly above the atoms of the second layer in such a way that the third layer of atoms is not directly above the first layer (Figure 11.5b). When a fourth layer is nestled into the spaces of the third layer in Figure l1.5(b), the atoms of the fourth layer do lie directly above those in the first. The three layers with their distinctive alignments are labeled a, b, and c. Together they form a repeating layering pattern abcabcabc ... throughout the crystal. On the other hand, the layers in Figure 11.5(a), where the third layer is aligned with the first (layer a), have a simpler ababab ... repeating pattern.

1 Layer a Layer b Layer c

rt,..',,,

===::::t ~

------u __

Fourth layer same

----l-

as layer a

layers of solid spheres; the difference

Let's take a closer look at the ababab ... stacking pattern, and focus on a cluster of atoms in this pattern (Figure 11.6). The cluster forms a hexagonal (six-sided) prism of closely packed spheres. In fact, they are as tightly packed as they can be, so the structure is called hexagonal closest-packed (hcp). The atoms in 16 metallic elements (see Figure 11.7) have hcp structures. Within these structures the 15-atom cluster in Figure l1.6(c) serves as an atomic-scale building block-a pattern of atoms repeated over and over again in all three dimensions inside the solid. We

them stems from the arrangement

hexagonal

unit cells represent

metals, including

closest-packed

closest-packed

structure

(hcp) structures

(c) Hexagonal

and (c) their corresponding

a highly efficient way to pack atoms in crystalline

all those in groups 3 and 4, have hcp structures.

unit cell

solids. Sixteen

between

of the

third layer: (a) The spheres in the third layer are directly above those in the first, giving an ababab ...

stacking

pattern.

(b) The alignment

of spheres in the third layer is different

from

either of the two layers just below. A fourth layer does align with the first, generating abcabc ...

~II

FIGURE 11.6 (a,b) Hexagonal

abcabc ...

FIGURE 11.5 Two efficient ways to stack

Packing Structures and Unit Cells

(b) Hexagonal

.....

......

(b)

(a) ababab ... layering

." -

stacking

CHEMTOUR

an

pattern.

Crystal

Packing

Hexagonal closest-packed (hep) describes a crystal structure in which the layers of atoms or ions in hexagonal unit cells have an ababab ... stacking pattern.

530

I

Chapter

Eleven

I


The unit cell is the basic repeati ng un it of the arrangement of atoms, ions, or molecules in a crystalline solid. A crystal lattice is the three-dimensional array of particles (atoms, ions, or molecules) in a crystalline solid. Cubic closest-packed (ccp) describes a crystal structure in which the layers of atoms, ions, or molecules in face-centered cubic unit cells have an abcabcabc. stacking pattern. A face-centered cubic (fcc) unit cell has the same kind of particle (atom, ion, or molecule) at the 8 corners of a cube and at the center of each face.

1

•

Hexagonal rlosest-oacked

•

n2

18

Body-centered cubic

,----

13

Diamond

Li

Be

Na

Mg

K

Ca

Sc

Ti

V

Cr

Mn

Rh

Sr

Y

Zr

Nb

Mo

Cs

Ba

La

Hf

Ta

W

I FIGURE

10

11

12

Fe

Co

Ni

Cu

Zn

Ga

Ge

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sh

Re

Os

Ir

Pt

Au

Hg

TI

Ph

Bi

I

I

I

I

I

I

I

I

FIGURE

I

in the color key) have structures

more complicated

"unit

is the minimum

set

of squares that defines the pattern repeated over the entire board. (b) This wrapping has a more complex pattern. highlighted.

paper

One unit cell is

Can you locate others?

than can easily be described

(Sb), and bismuth (Bi).

call these crystalline building blocks unit cells. A unit cell represents the minimum repeating pattern that describes the three-dimensional array of atoms in a crystalline structure, or crystal lattice, of a solid. Think of unit cells as microscopic analogues of the repeating patterns you may have seen in fabrics, wrapping paper, or even a checkerboard, as illustrated in Figure 11.8. Look carefully at the image in Figure 11.8a to confirm that the outlined portion represents the minimum repeating pattern. A unit cell serves a similar function: it is the threedimensional pattern of atoms, ions, or molecules that is repeated throughout a crystalline solid. The six-sided cell in Figure 11.6 is called a hexagonal unit cell. Atoms in solid copper and in a dozen other metals adopt the abcabc ... stacking pattern when they solidify. To understand the sort of crystals they form, consider what happens when we take a cluster of 14 spheres stacked in an abcabc ... pattern and turn the cluster about 45° as shown in Figure 11.9. The atoms form a cube: 8 of them at the 8 corners of the cube and 6 at the centers of each of the 6 faces. Because the atoms are stacked together as closely as possible, this structure is called cubic closest-packed (ccp), and the corresponding unit cell is called a face-centered cubic (fcc) unit cell. In the fcc unit cell, the atoms at the corners do not touch each other, but they do touch along the face diagonal, a line that connects the centers of the atoms at opposite corners of any face of the cube. The edges of the fcc unit cell are all of equal length, and the angle between any two edges is 90°.

Unit Cell

11.8 (a) The highlighted

Po

11.7 Summary of the crystal structures of metals in the periodic table. Five metals

Unit cell

cell" of a checkerboard

16 17

AI

9

3

in this book: mercury (Hg), gallium (Ga), indium (In), antimony

CHEMTOUR

15

Other structure

("other structure"

~II

14

(a)

(b)

11.3

FIGURE

11.9

I

Metallic Crystals

I

53:1..

Solids with cubic c1osest-

packed (ccp) structures

have face-centered

cubic (fcc) unit cells, which are more easily seen if the abcabc ... 45°. The packing

layers are tipped about

efficiency in a ccp structure

is the same as in a hexagonal (b) Cubic closest-packed

(a) abcabc ... layering

structure

(c) Face-centered

cubic unit cell

(hcp) structure

(see Figure

closest-packed

11.6).

So far we have introduced two closest-packed lattice structures-hexagonal and ccp-along with their associated unit cells (hexagonal and fcc). There are also structures in which the atoms or ions are arranged close together but not as efficientlyas hcp and ccp lattices. Figure 11.10 shows two other types of cubic unit cells. The body-centered cubic (bcc) unit cell has an atom at each of the 8 corners and 1 in the middle of the cell (Figure 11.lOb). All the group 1 metals and many transition metals have bcc unit cells. The simple cubic unit cell consists of just 8 corner atoms (Figure 11.10c). It is the least efficiently packed of the three structures in the cubic crystal system and quite rare among metals: only radioactive polonium (Po) forms a simple cubic unit cell. You may be wondering why some metals adopt one kind of unit cell and other metals another. We will address that question later in the chapter. For now, keep in mind that crystalline solids with cubic unit cells form crystals that are cubic in appearance on a macroscopic scale, whereas crystalline solids with hexagonal unit cells tend to form hexagonal crystals.

Unit-Cell Dimensions The cutaway views in Figure 11.10 show the geometric and mathematical relationships between the edge length (f), face-diagonal, and body-diagonal dimensions in fcc and bcc unit cells. These views also provide us with a way to determine how many atoms are actually inside each type of cubic unit cell. Let's start with the simplest unit cell, the cutaway view of the simple cubic unit cell (Figure 11.10c). Note how only a fraction of each corner atom is actually inside the unit-cell boundaries. Each corner atom is shared by the 8 unit cells meeting at that corner, and so only one-eighth of each corner atom is inside each unit cell. There are 8 corners in a cube, so there is a total of

i

X 8

=

1 corner atom per simple cubic unit cell

i

(Actually, the calculation X 8 = 1 applies to the corner atoms in any cubic unit cell.) Note that the atoms in Figure 11.10(c) touch along each edge. Therefore the edge length (f) in the simple cubic unit cell of an element is equal to twice the atomic radius of the element.

A body-centered cubic (bee) unit cell has atoms at the 8 corners of a cube and at the center of the cell. A simple cubic unit cell has atoms only at the 8 corners of a cube.

532

I

Chapter Eleven

I


(a) Face-centered

cubic

(b) Body-centered

cubic

FIGURE 11.10 Cubic crystals may form from a variety of cubic unit cells. Shown here are whole-atom of (a) face-centered

and cutaway views

cubic (fcc), (b) body-centered

cubic (bcc), and

(c) simple cubic unit cells. Note how the atoms in an fcc unit cell touch along the face diagonal,

which forms a right triangle

with

C. This relationship means that the length of the face diagonal is -VC2 + C2 = c{2. In a bcc cell the two adjoining

edges oflength

atoms touch along a body diagonal. from an edge, a face diagonal, body diagonal

..J3 times

Another

right triangle

and a body diagonal,

the unit-cell

edge length.

making

forms the (c) Simple cubic

In an fcc unit cell (Figure l1.lOa) there are 8 corner atoms and 1 atom in the center of each of the 6 faces. Each face atom is shared by the 2 unit cells that abut each other at that face. Therefore each cell gets credit for half of each face atom, making a total of ~ X 6

=

3 face atoms

and a grand total of

(i X 8 = 1) corner

atom

+ 3 face

=

atoms

4 atoms per fcc unit cell

To relate the size of these atoms to the dimensions of the cell, note that the atoms touch along the face diagonal in the cutaway view of Figure 11.10(a). Look carefully and see how the diagonal spans a radius of 2 corner atoms and the diameter (2 radii) of the atom in the center of the face. Therefore the face diagonal is equivalent to 1 + 2 + 1 = 4 atomic radii (r). The face diagonal connects the ends of two edges and forms a right triangle. Therefore, according to the Pythagorean theorem, the face diagonal is the square root of the sum of the squares of the two edge-length (e) distances:

=

Face diagonal so

4r

=

r =

-ve + e efI 2

4

2

=

efI

= 0.3536e

(11.1)

Now let's focus on the cutaway view of the bcc unit cell in Figure 11.10(b). In addition to the corner atoms there is also 1 atom in the center of the cell that is entirely within the cell. This means that there are a total of:

(i X 8 = 1) corner

atom

+

1 center atom

=

2 atoms per bcc unit cell

11.3

Relating the unit-cell edge length to the atomic radius of the atoms in a bcc cell is complicated by the fact that the atoms do not touch along the edges, nor do they touch along face diagonals. However, each corner atom does touch the atom in the center of the cell. This means that the body diagonal spans (1) the radius of a corner atom, (2) the diameter (2 radii) of the atom in the center, and (3) the radius of the atom in the opposite corner. This makes the body diagonal equivalent to 1 + 2 + 1 = 4 atomic radii (r). We can again use the Pythagorean theorem to define the edge length (£) with respect to the body diagonal. A right triangle is formed by an edge, a face diagonal, and a body diagonal serving as the hypotenuse of the triangle. In such a triangle the body diagonal relates to the edge length as follows: Body diagonal

=

so

4r

=

~(edge length)2

r

=

£YJ 4

+

(face diagona1)2

=

~£2

+ (£-12)2

= 0.4330£

=

£YJ

(11.2)

In the following Sample Exercises we use Equations 11.1 and 11.2 to calculate the radii of the atoms in a unit cell and to calculate the densities of elements that crystallize in the unit cells depicted in Figure 11.10. For both calculations we need the edge lengths of the unit cell, which are obtained using an analytical technique called X-ray diffraction (described in detail later in Section 11.9). As you do these calculations, you may wish to consult the summary in Table 11.1 of how atoms in different locations in fcc, bcc, and simple cubic unit cells contribute to the contents of those cells.

Calculating Atomic Radius from Crystal Structure At room temperature the most stable crystal structure of iron has a body-centered cubic unit cell with an edge length of 287 pm (1 pm = 10-12 m). What is the radius in picometers of the iron atoms? N D 0 R G A N I Z E The crystal structure is based on a bcc unit cell (Figure 11.10b). This means that the iron atoms do not touch along cell edges or face diagonals but do touch along the body diagonals. We know the edge length of the cell: 287 pm. CO LL E eTA

Atoms touch along the body diagonals of a bcc unit cell, so the radius (r) of the atoms is related to the cell's edge length (£) by Equation 11.2:

A N A LY Z E

r SOLVE

We insert the edge-length r

=

= OA330£ value into this equation to solve for r:

OA330 X 287 pm

=

124 pm

T The average atomic radius of iron atoms is 126 pm, so the result of this calculation is reasonable. It should be because the values for atomic radii are derived from analyses of crystal structures in the first place.

T HI N K ABO UTI

Practice Exercise At 1070°C the most stable structure of iron has an fcc unit cell with an edge length of 361 pm. What is the atomic radius of iron at this temperature?

I

Metallic Crystals

I

533

534

I

Chapter Eleven

I


Contributions of Atoms to Cubic Unit Cells

TABLE 11.1

Position of Atom in Unit Cell

Unit-Cell Type

Contribution to Unit Cell

Center

f-Face

1

bcc

1

fcc

1

fcc, bcc, simple cubic

-2:

Corner

"8

L

Using Crystal Structure to Calculate Density Calculate the density of iron in grams per cubic centimeter (glcm3) at 25°C, given that its crystal structure is based on a bcc unit cell, and that the edge length of the cell is 287 pm. COL LEe TAN D 0 R G A N I Z E We are asked to calculate a density of iron knowing its unit-cell geometry and edge length. We assume that the density of the unit cell is the same as the density of solid Fe. The fact that Fe has a bcc unit cell means that there are 2 atoms of Fe per cell. Recall that density is the ratio of mass to volume. The molar mass of Fe is 55.84 glmol. The volume of a cubic cell is the cube of its edge length: V= £3.

The density of an Fe bcc unit cell is the mass of2 Fe atoms divided by the volume of the cell (287 pm)3. We need to calculate the mass of 2 Fe atoms starting with the molar mass (55.84 g/mol).The conversion includes dividing by Avogadro's number to calculate the mass of each Fe atom in grams. To obtain a cell volume in cubic centimeters (cm3), we make use of the definitions for pico- and centi-: ANALYZE

1 pm

=

10-12 m

=

1 cm

10-2 m

These may be combined to produce the conversion factor 1 pm

=

10-10 cm

SOLVE

We calculate the mass (m) of2 Fe atoms first-

m

=

55.84 g~

X

~

***

1 6.022 X 1023~

X 2~

= 1.855

X 10-22 g

and then the volume of the cell in cubic centimeters:

V = £3 = (287 F)3

X

(10-10 cm)3 3 fffi"

=

2.364 X 10-23 cm3

Taking the ratio of mass to volume, we have: d

=

m

V

=

22 1.855 X 10- g 2.364 X 10-23 cm3

3

= 7.85 g/cm

ABOUT IT According to the data in Appendix 3, the density of iron is 7.874 glmL, so the result of the calculation is reasonable.

THINK

11.4

I

Alloys

I

535

Practice Exercise Silver and gold both crystallize in a face-centered cubic unit cell with very similar edge lengths: 407.7 and 407.0 pm. Calculate the density of each metal and compare your answers with the densities listed in Appendix 3.

mD Alloys People have been using metals for tools, weapons, currency, and decorations (particularly jewelry) for tens of thousands of years. For most of that time the available metals were only three group 11 elements-Cu, Ag and Au. These were the only ones found as free metals in Earth's crust. Even so, most silver and copper occurs not as Ag and Cu but in ores such as argentite (Ag2S), chalcopyrite (CuFeS2), and chalcocite (Cu2S). Ores are naturally occurring compounds or mixtures of compounds from which elements can be extracted. Around 6000 years ago metal technology took a giant leap forward when artisans in the Tigris-Euphrates river valley (present-day Iraq) and the Indus River valley (now Pakistan) discovered how to convert copper ore, principally CuFeS2, to copper metal. The process involved pulverizing the ore and then baking it in an oven. Baking initiated a chemical reaction with O2 (from air), which converted the Cu in CuFeS2 to CuO. The second step in the process allows CuO to react with carbon monoxide, which was produced by burning wood or charcoal (mostly carbon) in a furnace with an insufficient supply of air: CuO(s)

+

CO(g)

---?

Cu(s)

+

CO2(g)

(11.3)

One disadvantage of making tools and weapons from copper is that the metal is very malleable, which means that copper tools and weapons are easily bent and damaged. We can explain the malleability of Cu (and other metals) in terms of the diffuse metallic bonds between its atoms (Section 11.1) and their cubic closest-packed (ccp) structure (Section 11.3). A solid piece of copper has an enormous number oflayers of tightly packed, two-dimensional planes of copper atoms that are only weakly bonded to each of their nearest neighbors. This arrangement gives the atoms in one plane the ability, under stress, to slip past atoms in adjacent planes (Figure 11.11). When the stress is relieved and the atoms stop slipping, many have different atoms as nearest neighbors but the overall crystal structure is still cubic closest-packed. The ease with which copper atoms slip past each other made it easy for prehistoric metalworkers to hammer copper metal into spear points and shields, but it also meant that those objects could easily be damaged in the heat of battle.

External

force

Metal is deformed

FIGURE 11.11 Metals such as copper are malleable

because their atoms are stacked

in two-dimensional

planes that can slip past

each other under stress. Slippage is possible because of the diffuse nature of metallic bonds and the relatively weak interactions pairs of atoms in adjoining

between

planes.

Substitutional Alloys About 5500 years ago, people living around the Aegean Sea discovered that mixing molten tin and copper together produced bronze, a new material that was much stronger than either tin or copper alone. Its discovery ushered in the Bronze Age. Bronze is an alloy, a metallic material made when a host metal is blended with one or more other elements, which mayor may not be metals, changing the properties of the host metal. Alloys can be classified into categories according to their composition. Bronze is an example of a homogeneous alloy, a solid solution (homogeneous mixture) in which the atoms of the added e1ement(s) (in this case tin) are randomly but uniformly distributed among the atoms of the host (cop-

Ores are naturally occurring compounds or mixtures of compounds from which elements can be extracted. An alloy is a blend of a host metal and one or more other elements (which mayor may not be metals) that are added to change the properties of the host metal.

536

I

Chapter Eleven

I


per, for example). Some alloys are heterogeneous alloys because a sample has small "islands" within the larger matrix where different metals are in contact. Still others are intermetallic compounds because they have a reproducible stoichiometry and constant composition. At the concentrations of tin used in Bronze Age tools and weapons (up to 30% by mass), the alloy formed by tin and copper is also known as a substitutional alloy because the atoms of one metal replace atoms in the crystal lattice of the other. Substitutional alloys may form between metals that have the same crystal structure and atomic radii that are within about 15% of each other.

(a)

CONCEPT

TEST

Is it accurate to call bronze a solution of tin dissolved in copper? Explain why or why not.

(b)

FIGURE 11.12 Two atomic-scale of one type of bronze,

a substitutional

(a) A plane (layer) of close-packed with a random

substitution

(b) An fcc unit cell ofCu atoms have replaced

views alloy:

Cu atoms

ofSn atoms. atoms in which Sn

two of the Cu atoms.

Figure 11.12 illustrates the structure of bronze in which tin (Sn) atoms have replaced some of the copper atoms in what would have been a face-centered cubic unit cell of Cu. The radii of copper and tin atoms are similar-128 pm and 140 pm, respectively. Inserting the slightly larger Sn atoms in the cubic closestpacked structure of copper atoms disturbs the Cu lattice structure a little, making the planes of copper atoms "bumpy" instead of uniform (Figure 11.13). This atomic-scale roughness makes it more difficult for the copper atoms to slip past one another. Less slippage makes bronze less malleable than copper, but being less malleable also means that bronze is harder and stronger.

CONCEPT

FIGURE 11.13 The larger Sn atoms in bronze perturb atoms, producing

the crystal structure atomic-scale

slip plane (wavy line) between

of Cu

layers ofCu

for Cu atoms to slide by each other when an external

force is applied,

Does the face-centered cube ofCu and Sn atoms in Figure 11.12(b) represent the unit cell of the alloy?

bumps in the

atoms. These bumps make it more difficult

bronze tools and weapons

TEST

and that makes harder and stronger

than those made of pure copper.

There are many other substitutional alloys besides bronze. Some are also copper based, including brass (zinc alloyed with copper) and pewter (tin alloyed with copper and antimony). However, most substitutional alloys in the modern world are ferrous alloys, which means that the host metal is iron. An important class of these alloys are rust-resistant stainless steels, which contain about 10% nickel and up to 20% chromium. When atoms ofCr on the surface of a piece of stainless steel combine with oxygen, they form a layer of Cr 203 that tightly bonds to the surface and protects the metallic material beneath from further oxidation. This resistance to surface discoloration due to corrosion means that the surfaces of these alloys "stain less" than those of pure iron.

Interstitial Alloys The Bronze Age began to fade about 3000 years ago with the discovery that iron oxides could be reduced to iron metal in wood or charcoal fires by limiting the air supplied to the fire. As in copper smelting, the reducing agent for iron oxide is carbon monoxide: (11A) Iron quickly replaced bronze as the metallic material of choice for fabricating tools and weapons both because iron ore is much more abundant in Earth's crust

11.4

than the ores of copper and tin, and because tools and weapons made of iron and iron-containing alloys are much stronger than those made of bronze. Today, the reduction of iron ore is done in blast furnaces, enormous reaction vessels (some more than 50 m tall) that operate continuously at about 1600°C (Figure 11.14). Iron ore, hot carbon (coke), and limestone are added to the top of the vessel and molten iron and solid by-products (called slag) are harvested from the bottom (Figure 11.14a). Blast furnaces get their name from blasts of air preheated to about 1000°C that are injected through nozzles near the bottom and that suspend the reactants until the iron reduction process is complete. It may take as long as eight hours for the reactants to fall to the bottom of a blast furnace. On their way down, O2 in the hot air partially oxidizes coke (carbon) to carbon monoxide, and the CO reduces the iron in iron ore as described in Equation 11A. Limestone (CaCO) decomposes to calcium oxide (CaO) or lime: (11.5) CaO in turn reacts with silica impurities in the ore to form calcium silicate:

CaO(s)

+ Si02(s)

---?

(11.6)

CaSiO/t')

Calcium silicate becomes part of the slag that floats on the denser molten iron at the bottom of the furnace. The largest blast furnaces, operating continuously, produce 10,000 to 15,000 metric tons of iron a week in response to increasing Exhaust gases Iron are, coke, and limestone

,,1t

~

Hotair~ Slag~ ~Molten

iron

(a) Blast furnace Oxygen

1

FIGURE 11.14 (a) Blast furnaces operate continuously

at temperatures

near 1600°C

to convert iron are into iron (see Equation 11.4) at rates as high as 2000 tons of iron

Steel shell

per day. Injected Refractory

lining

O2 converts

C to CO2, and

lime (CaO) is added to react with Si and P impurities.

The products

of these reactions

become part of the slag layer. (b) In the basic oxygen process the carbon content

Slag Molten

iron

is reduced by injecting

vessel pivots on its side arms, allowing purified

(b) The basic oxygen process

of steel

oxygen. The reaction

layer.

iron to be tapped offbelow

the

the slag

I

Alloys

I

537

538

I

I

Chapter Eleven


DWorldwide

1200

•

1000

:=

B .~

800

)j

'"E

"'0

600

~t::

'":g

0

P

400

200

o 2000

2001

2002

2003 Year

FIGURE 11.15 Worldwide

production

of steel has risen steadily in recent years due mainly to the rapidly growing

steel industry

in China.

C FIGURE 11.16 interstitial

Carbon

form of iron (austentite) temperatures

steel is an

alloy of carbon in iron. The fcc that forms at high

can accommodate

in its octahedral

carbon atoms

holes.

FIGURE 11.17 Close-packed

China

world demand for products made of steel. Figure 11.15 shows annual worldwide steel production since the beginning of the 21st century. Molten iron produced in a blast furnace may contain up to 5% carbon. To reduce the carbon content, the molten iron is transferred to a second furnace into which hot O2 and additional CaO are injected. In this basic oxygen process (Figure 11.14b), some of the carbon is oxidized to CO2 and any remaining silicon impurities form a slag of CaSiOr Sulfur and phosphorus impurities in the fuels used in the blast furnace also end up in the slag. Worldwide about 60% of the iron used to make steel is refined using the basic oxygen process. When molten iron cools to its melting point of 1538°C, it crystallizes in a face-centered cubic structure called austentite. With further cooling the structure changes into a body-centered 2004 2005 cubic structure called ferrite. The spaces, or holes, between iron atoms in austentite can accommodate carbon atoms forming an interstitial alloy because the carbon atoms occupy spaces, or interstices, between the iron atoms (Figure 11.16). There are two kinds of holes between the atoms in closest-packed crystal structures (Figure 11.17). The smaller of the two are located between clusters of 4 host atoms and are called tetrahedral holes. The larger holes are surrounded by clusters of 6 host atoms in the shape of an octahedron and are called octahedral holes. The data in Table 11.2 provide guidelines on which holes are more likely to be occupied based on the relative sizes of the smaller atoms in the holes and larger host atoms. According to the data in Appendix 3, the atomic radii of C and Fe are 77 and 124 pm, respectively. The ratio of the two is 77/124 = 0.62. According to the guidelines in Table 11.2, this ratio means that C atoms should fit in the octahedral holes of austentite, as shown in Figure 11.16, but not in the tetrahedral holes. As austentite iron cools to room temperature, it turns into body-centered cubic ferrite iron. The octahedral holes in ferrite are smaller than those in austentite: too small to accommodate carbon atoms. As a result, carbon precipitates as clusters of carbon atoms, or it may react with iron forming iron carbide, Fe3 C. The clusters of carbon and Fe3C disrupt ferrite's crystal structure and inhibit the host iron atoms from slipping past each other when a stress is applied. This resistance to slippage, which is much like that experienced by the copper atoms in bronze (Figure 11.12), makes iron-carbon alloys, known as carbon steel, much harder and stronger than pure iron. In general, the higher the carbon concentration, the stronger the steel, as indicated by the information in Table 11.3. Note from the values in the table that increased strength and hardness comes at the cost of increased brittleness.

atoms in

adjacent layers of a crystal lattice produce tetrahedral

and octahedral

holes between

clusters of 4 and 6 adjacent spheres, respectively. tetrahedral

Octahedral

holes are larger than

holes and so can accommodate

larger atoms in interstitial

alloys.

Tetrahedral

holes

11.4

TABLE 11.2

Atomic Size Ratios and the Location of Atoms in Unit Cells Type of Hole for Smaller Atom

Radius Ratio'

hcp or ccp

Tetrahedral

0.22-0.41

hcp or ccp

Octahedral

0.41-0.73

Cubic

0.73-1.00

Packing

t=

Simple cubic

---

"Using radius ratios as predictors of solid-state strucrures is limited because atoms are not truly solid spheres with a constant radius; the radius of an atom may differ in different compounds, so these ranges are approximate.

L-

__

TABLE 11.3

Effect of Carbon Content on the Properties of Steel

Carbon Content (%)

t=

Designation

Properties

0.05-0.19

Low carbon

Malleable, ductile

Nails, cables

0.20-0.49

Medium carbon

High strength

Construction girders

High carbon

Hard but brittle

Cutting tools

0.5-3.0

-

Predicting the Structure a Two-Element Alloy

Used to Make

I

of

As noted at the beginning of this chapter, pure gold is too soft to use in jewelry. So too is pure silver. However, sterling silver, which is 93% Ag and 7% Cu by mass, is widely used in jewelry. The presence of Cu inhibits tarnishing and strengthens the alloy. Is this copper-silver alloy a substitutional or an interstitial alloy? Silver has a cubic closest-packed (ccp) structure with face-centered cubic (fcc) unit cells. COLLECT AN DORGAN IZE We are asked to decide whether Cu alloys withAg by forming a substitutional alloy or an interstitial alloy. Elements form substitutional alloys when the atoms of the elements have similar sizes. According to Table 11.2, interstitial alloys may form only when atoms of an alloying element are less than about 73% the size of the atoms in a closest-packed lattice of the host element. We are given that the atoms in solid Ag form fcc unit cells and are closest-packed. The atomic radii of Ag and Cu are 144 and 128 pm, respectively. ANALYZE

The ratio of the atomic radii ofCu

to Ag is 128 pm/144 pm

=

0.89,

or 89%. SOL V E Atoms of Cu can substitute for atoms of Ag in a ccp structure of the latter with some room to spare. Both elements form fcc unit cells, so little disruption to the Ag lattice should result from incorporating Cu atoms, which are nearly the same size. Copper atoms are too big, however, to fit into the octahedral holes in the lattice of silver atoms. Thus, the Cu-Ag alloy, sterling silver, is a substitutional alloy. ABOUT IT The guideline for two metals forming a substitutional alloy is that their atomic radii are within 15% of each other. The radii of Ag and Cu differ by only 11%, so we would expect copper and silver to form a substitutional alloy. Moreover, copper and silver are in the same group in the periodic table and have the same crystal structures (Figure 11. 7). THINK

Practice Exercise Would you expect gold (atomic radius = 144 pm) to form substitutional alloys with silver (atomic radius = 144 pm) or copper (atomic radius = 128 pm)?

I

Alloys

I

539

540

I Chapter Eleven

I


mD

Aluminum and Its Alloys: Light Weight and High Performance

Over the last century, aluminum and its alloys have replaced steel in applications where high strength-to-weight ratios and corrosion resistance are paramount, as in airplanes, motor vehicles, and the facades of buildings. The most important aluminum ore is called bauxite. It is a mixture of aluminum minerals containing aluminum(III) oxide. To convert bauxite to aluminum metal, the ore is crushed and extracted with a concentrated solution ofNaOH in a reaction called the Bayer process. The product of this reaction is a soluble complex ion that has an AP+ ion at its center surrounded by 4 OH- ions: (11. 7) Impurities in bauxite, such as CaC03, Fe203, and Ti02, are insoluble in concentrated NaOH and can be removed by filtration. Acid is then added to neutralize the NaOH in the filtered solution, resulting in the precipitation of high-purity AI(OH\(s): (11.8) The precipitate is then converted to pure Al203 by heating, which drives off steam: (11.9) In 1886, two 23-year-old chemists, Charles Hall and Paul-Louis-Toussaint Heroult (both 1863-1914), working independently in the United States (Hall) and France (Heroult), developed the same industrial process for converting Al203 (called alumina) to aluminum. The process is based on passing an electric current between carbon electrodes immersed in a solution of aluminum oxide dissolved in molten cryolite (Na3AIF6), as shown in Figure 11.18. As electricity passes through the solution, aluminum ions are reduced to aluminum metal at the negative electrode while the positively charged carbon electrode is oxidized to carbon dioxide. The process is described by the following reaction: 2 AIP3

(in molten Na3AIF6)

+ 3 C(s) ~

4 AI(C)

+ 3 CO/g)

(11.10)

The high cost of the electricity needed to produce aluminum in this way makes recycling aluminum metal a good idea both economically and environmentally. Let's compare the energy requirements of producing Al from Al203 with recycling products made of Al metal. The principal energy cost of the Hall-Heroult process is the electricity needed to reduce A1203. The major cost in recycling is the energy required to melt aluminum metal. We can estimate the energy required for the Hall-Heroult process by calculating the standard enthalpy change for the reaction in Equation 11.10. Starting with standard enthalpies of formation, we have: I:l.H;xn

independently

electrolytic

developed

the same

process for producing

aluminum

metal from alumina. chemistry

Hall's

major at Oberlin

her brother

a financial

eventually of America,

assisted

in the lab, and her business

helped make their aluminum company

sister, Julia, also a College,

skills

production

+ 4(I:l.H;.Al)]

- [2(I:l.H;.AI20)

+ 3(I:l.H;,c)]

2

_[(

mol CO2

Ll o)t-1675.7kJ) 2 mol,..1 2 3 mol Al203

mel-Al J

)] + (OU_10)(0.Ok 3~ _1 0 oU

~

= 2171 kJ

success. The company

became the Aluminum shortened

[3(I:l.H;.co)

= [(3 mol CO )( -393.5 kJ) + (4 mel-Al)( 0.0 kJ )]

Charles M. Hall and Paul-Louis-Toussaint Heroult

=

Company

to Alcoa, Inc.

Dividing this value by the 4 moles of Al produced by the reaction, we get 542.7 kJ per mole of AI.

11.5

I

Aluminum and Its Alloys: Light Weight and High Performance

I

54:1..

Now let's calculate the energy required to recycle 1 mole of Al by heating it from 25°C to its melting point (660°C) until all the Al melts. The energy needed to heat 1mole of aluminum from 25°C to 660°C can be calculated from its molar heat capacity [24.31 JI(mol- 0C)] and Equation 5.11:

q

= nCpD..T =

1.00 mel X

24.20

mel-

J o~

X (660 - 25)~

= 15.37 kJ Once the Al is at its melting point, the heat required to melt 1 mole of Al is its heat of fusion (D..Hfus = 10.79 kJlmol): 10.79 kJ

mel

X 1.00

mel

=

10.79 kJ

Thus the total energy needed to heat and melt 1.00 mole of Al is 15.37 kJ

+

10.79 kJ

=

26.16 kJ

This value represents 16 26. 14. X 100% 542.714

= 4.82%

of the energy needed to electrochemically reduce 1 mole of AP+ to Alo. There are other energy costs associated with aluminum production and recycling, but overall, recycling saves aluminum manufacturers about 95% of the energy required to produce the metal from aluminum ore. This energy savings has inspired the rapid growth of a global aluminum recycling industry. (Aluminum recycling in the United States is a $1 billion per year business.) Currently, junkyards in the United States recycle 85% of the aluminum in cars (5 X lOB kg/yr). Over 50% of the aluminum in food and beverage containers is recycled. The value of scrap aluminum is about 10 times that of scrap steel because of these energy considerations and because it takes less energy to melt aluminum than it does to melt the same mass of scrap steel. Aluminum forms alloys with many elements. Alloys with Mg, Si, Cu, and Zn are the most common and are widely used in aircraft construction. The tabs on beverage cans are manufactured from an aluminum alloy containing Mg and Mn.

FIGURE 11.18 This sketch of Hall's first aluminum

production

facility shows

large trays made of cast iron that held the molten reaction carbon electrodes

mixture.

Positively

were attached

rods to copper plates suspended trays. Molten charged

aluminum

carbon electrodes

charged

by copper over the

formed at negatively at the bottom

of

the trays and was poured into the ingot molds shown on the floor.

542

I

Chapter Eleven

I


Aluminum alloys containing Li are attractive for applications where light weight is essential, because Li has the smallest molar mass (6.941 g/mol) and lowest density (0.534 g/mL) of all the metallic elements. Aluminum and aluminum alloys are widely used as building materials and in fuselages of airplanes because they are corrosion resistant. This resistance is noteworthy because aluminum is actually a very reactive metal. However, when the surface of a piece of aluminum starts to oxidize in air, a thin layer of Al203 is produced that adheres strongly to the aluminum metal below it and acts as a protective shield against further oxidation. This corrosion protection is similar to the way a thin coating of Cr203 protects stainless steel from rusting.

A covalent network solid consists of atoms held together by extended arrays of covalent bonds.

mD The Structures of Some Nonmetals

@@

CONNECTION Allotropes are structurally different forms of the same physical state of an element, as explained Chapter 8.

~II

CHEMTOUR

FIGURE 11.19 (a) diamond, network

Allotropes

Carbon

of Carbon

crystallizes

a three-dimensional

as

covalent

solid made of carbon atoms,

each connected

by

(J'

bonds to 4 adjacent

carbon atoms; and (b) graphite, carbon atoms are connected delocalized allotrope

in

1T

by

in which (J'

bonds. The structure

bonds and of a third

of carbon is shown in Figure

11.20.

Figure 11.7 describes the crystal structures of metallic elements. Notice (from the color key) that it includes a structure type labeled "diamond." Diamonds are not metals. They are a rare and thus expensive crystalline form of the nonmetal carbon, although they owe their distinctive colors and some of their value to the presence of metal impurities in their crystal structures. Diamond is one of three allotropes of carbon (Figures 11.19 and 11.20), the other two being graphite and fullerenes. Diamond is a crystalline covalent network solid, because it consists of atoms held together in an extended three-dimensional network of covalent bonds (Figure 11.19a). Each carbon atom in diamond is bonded by overlapping Sp3 hybrid orbitals with 4 neighboring carbon atoms, creating a network of carbon tetrahedra. The carbon atoms in these tetrahedra are connected by localized (J' bonds, making diamonds poor electrical conductors. The three-dimensional network of these (J' bonds is extremely rigid, making diamond the hardest natural material known. The atoms of other group 14 elements, particularly silicon, germanium, and tin, also form covalent network solids based on the diamond structure. Natural diamond forms from graphite under intense heat (> 1700 K) and pressure (> 50,000 atm) deep in Earth. Industrial diamonds are also synthesized at high temperatures and pressures from graphite or any source rich in carbon. Synthetic diamonds are used as abrasives and for coating the tips and edges of cutting tools. Diamond has the highest thermal conductivity of any natural substance (five times higher than copper and silver, the most thermally conductive metals), so tools made from diamond do not become overheated. Most industrial diamonds lack the size and optical clarity of gemstones. By far the most abundant allotrope of carbon is graphite, another covalent network solid, frequently the principal ingredient in soot and smoke and used to make pencils, lubricants, and gunpowder. Graphite contains sheets of carbon atoms, in which each atom is connected by overlapping Sp2 hybrid orbitals to 3 neighboring carbon atoms in a two-dimensional covalent network of 6membered rings (Figure l1.19b). Each carbon-carbon bond is 142 pm, shorter than the C-C bond in diamond. Overlapping unhybridized p orbitals on the carbon atoms form a network of '7T bonds that are delocalized

f 1----=~~4.a~ct-w

335 pm

(a)

(b)

11.6

across the plane defined by the rings. The mobility of these delocalized electrons makes graphite a conductor of electricity. As shown in Figure 11.19(b), the two-dimensional sheets in graphite are 335 pm apart. This distance is much too long to be a covalent bonding distance. Instead, the sheets are held together only by London (dispersion) forces. These intermolecular interactions are much weaker than covalent bonds. This weakness allows adjacent sheets to slide past each other, making graphite soft, flexible, and a good lubricant.

CONCEPT

I

The Structures

I

of Some Nonmetals

543

Crystals of molecular solids are formed by neutral covalently bonded molecules held together by intermolecular attractive forces.

@@ London Chapter

CONNECTION forces between

We introduced molecules in

10.

TEST

a

b

This phase diagram of carbon shows the temperatures and pressures at which carbon vapor, liquid carbon, and two solid forms (graphite and diamond) are stable. Which region (a, b, c, or d) represents diamond?

d c

A third allotrope of carbon was discovered in the 1980s. Networks of 5- and 6-atom carbon rings form molecules of 60, 70, or more carbon atoms that look like miniature soccer balls (Figure 11.20a). They are called fullerenes because their shape resembles the geodesic domes designed by American architect R. Buckminster Fuller (1895-1983). Many chemists call them "bucky balls" for the same reason. In terms of their size and their properties, bucky balls are too small to be covalent network solids but too large to be molecular solids (discussed in the following paragraphs). They fall in an ambiguous zone between small molecules and large networks and are classified as clusters. When fullerenes were first discovered they were believed to be a form of carbon rarely found in nature. In recent years, analyses of soot and emission spectra from giant stars have disclosed that fullerenes are present in trace amounts throughout the universe. Other nonmetals have cagelike crystal structures too, and may be considered clusters, including a crystalline form of boron that is made of closest-packed arrays of 12-sided molecules called icosahedra (singular icosahedron; Figure 11.20b). Other nonmetals form crystalline molecular solids. Molecular solids consist of neutral, covalently bonded molecules held together by intermolecular forces. Ice, CO2, glucose, and all organic molecules that crystallize are molecular solids. The structures of the two most common allotropes of phosphorus, white and red phosphorus (Figure 11.21), are both based on tetrahedral P4 molecules that crystallize in ordered arrays. White phosphorus is a molecular solid, consisting of discrete P4 tetrahedra arranged in a cubic array. It is a waxy material that oxidizes rapidly under standard conditions and gives off a yellow-green light called phosphorescence. Red phosphorus is a covalent network solid made of chains of P4 tetrahedra connected by covalent phosphorus-phosphorus bonds. Both red and white phosphorus melt to give the same liquid consisting of symmetrical P4 tetrahedral molecules.

3300 T,OC Phase diagram

of carbon.

(b)B12

(a)C60 FIGURE 11.20

Some solids are described (C60) is a

as clusters: (a) Buckminsterfullerene molecular

solid containing

60 sp2-hybridized

carbon atoms. Both 5- and 6-membered rings are required spherical

to construct

molecule.

the nearly

(b) One crystalline

of boron consists of a close-packed

form

array ofB'2

molecules.

FIGURE 11.21 allotropes

The two most common

of phosphorus

based on P4 tetrahedra: consists of discrete

are molecular (a) White

P4 molecules;

solids

phosphorus it is waxy

and soft enough to be cut with a knife. (b) Red phosphorus

is a polymer formed from

white phosphorus P4 tetrahedron (a)

(b)

when one bond in each

is broken and a covalent P-P

bond is formed between

adjacent

tetrahedra.

544

I Chapter Eleven

I


Ionic solids consist of monatomic or polyatomic ions held together by ionic bonds.

FIGURE 11.22 molecular

Oneformofsulfurisa

solid based on puckered

S8 rings. Different different

stacking

Sulfur has more allotropic forms than any other element. Most are molecular solids. This variety of forms arises because sulfur atoms form cyclic (ring) compounds of different sizes, which in turn gives rise to the possibility of different crystalline arrangements of the molecules. The most common allotropes of sulfur consist of puckered rings (Figure 11.22) containing 8 sulfur atoms covalently bonded together. Dispersion forces hold one ring to another in solid sulfur. The weakness of these interactions is the reason why elemental sulfur is soft and melts at only 119.6°C. The packing ofSB rings together in the crystalline molecular solid results in a very complex structure that cannot be described in terms of any of the lattice arrangements we have discussed.

allotropes patterns

8-membered

of sulfur have

mD Salt Crystals: Ionic Solids

of these rings.

Most of Earth's crust is composed of ionic solids consisting of monatomic or polyatomic ions held together by ionic bonds. Most of these solids are crystalline. The simplest structures are those of binary salts, such as NaCI (Figure 11.23). The cubic shape of salt crystals (Figure 11.23a) is reflected in the cubic shape of the NaCI unit cell, which is based on a face-centered cubic array of the larger CI- ions (Figure 11.23b). Smaller Na+ ions occupy the 12 octahedral holes along the edges of the unit cell and the single octahedral hole in the middle of the cell. Let's take an inventory of the ions in a unit cell of NaCI. Like the metal atoms in the fcc structure in Figure 11.10(a), there are the equivalent of 4 Cl- ions in the unit cell in Figure 11.23(c). As for Na+ ions, there is 1 in the central octahedral hole, and 1 in each of the 12 octahedral holes along the edges of the cell. Each N a+ ion along an edge is shared by 4 unit cells so only one-fourth of it is in each cell. Only the Na+ in the center belongs completely to 1 unit cell. Therefore the total number ofNa+ ions in the unit cell is

(a)

(12 X ~) + 1 = 4 Na+ ions

(b)

(c)

FIGURE 11.23

(a) Sodium

cubic crystals that sometimes

chloride

forms

reach impressive

size. (b) The NaCI crystal structure

is based

on an fcc unit cell that the CI- ions adopt, with Na+ ions occupying

the octahedral

holes.

(c) Ions at the corners and faces and along the edges are shared with other unit cells. As shown in this cutaway view, only fractional portions

of ions at these positions

to a single unit cell.

are assigned

The ratio ofNa+ to Cl- ions in the unit cell is 4:4 (or 1:1), which is consistent with the chemical formula ofNaCI. Because the 4 Na+ ions occupy all the octahedral holes in the unit cell, the result of this calculation also means that each fcc unit cell contains the equivalent of 4 octahedral holes. Note in Figure 11.23(c) that the Cl- ions do not touch along the face diagonal the way the metal atoms do in Figure 11.10(a). This is because the CI- ions have to spread out a little to accommodate the Na+ ions in the octahedral holes. Sodium and chloride ions touch along each edge of the unit cell in NaCI. This means that each Na+ ion in NaCI touches 6 CI- ions and each CI- ion touches 6 Na+ ions. This arrangement of positive and negative ions is common enough among binary ionic compounds to be assigned its own name: the rock salt structure. In other binary ionic solids, the smaller ion is small enough to fit into the tetrahedral holes formed by the larger ions. For example, in the unit cell of the mineral sphalerite (zinc sulfide) the larger S2- anions (ionic radius, 184 pm) are arranged in an fcc structure (Figure 11.24), and half of the eight tetrahedral holes inside the cell are occupied by smaller Zn2+ cations (74 pm). Therefore the unit cell contains four Zn2+ ions which balance the charges on the four S2- ions in the unit cell. This pattern of half-filled tetrahedral holes in an fcc structure is sometimes called the sphalerite structure. Many other compounds, particularly those

11.7

I

Salt Crystals: Ionic Solids

FIGURE 11.24

Sphalerite:

crystals of the mineral

I

545

(a) Many

sphalerite,

like

the largest one in this photograph,

have a

tetrahedral

shape. (b) The crystal structure

sphalerite

is based on an fcc unit cell of S2-

of

ions with Zn2+ ions in 4 of the 8 tetrahedral holes.

(a)

formed between transition metal cations with a 2+ charge and anions of the group 6 elements with a 2- charge, have sphalerite structures. The structure of the mineral fluorite (CaF2) is based on an fcc unit cell of larger Ca2+ ions in which all 8 tetrahedral holes are filled by smaller F- ions. Because there is a total of 4 Ca2+ ions in each unit cell, this arrangement satisfies the 1:2 mole ratio of Ca2+ ions to F- ions. This structure is so common that it too has its own name: the fluorite structure (Figure 11.25). Other compounds having this structure include SrF2' BaCI2, CaF2' and PbF2. Another group of compounds with a cation-to-anion mole ratio of 2:1 has an antijluorite structure. In the crystal lattices of these compounds, which include Li20 and K2S, the smaller cations occupy the tetrahedral holes formed by cubic closest packing of the larger anions.

SAMPLE EXERCISE I1.S

Calculating an Ionic Radius from a Unit-Cell Dimension (a)

Lithium chloride (LiCI) crystallizes with the same structure as NaCI except the Li+ ions are small enough to allow the Cl- ions to touch along the face diagonals in LiCI. If the edge length (f) of the fcc cell is 513 pm, what is the radius of the Cl- ion in LiCl? AND ORGANIZE We know that the CI- ions are in a face-centered cubic unit cell such as that shown in Figure 11.10(a). We also know the edge length of the unit cell: 513 pm. COLLECT

In a unit cell where the particles (CI- ions in this case) touch along the face diagonal, the relationship between edge length (513 pm) and the radius (r) of the ions is given by Equation 11.1:

ANALYZE

r

=

0.3536 f FIGURE 11.25

(a) The cubic structure

of the ions in the mineral

SOLVE

produce beautiful,

r = 0.3536 X 513 pm = 181 pm

crystals.

fluorite (CaF2)

nearly transparent

(b) The crystal structure

can

cubic

of CaF2

is based on an fcc array of Ca2+ ions, with

ABOUT IT The average ionic radii values in Figure 10.2 were determined from crystal structures. The value in Figure 10.2 for CI- ions is 181 pm, so our calculation is correct.

F- ions occupying all 8 tetrahedral holes. However, F- ions are bigger than Ca2+ ions,

Practice Exercise Assuming the radius of the CI- ion in NaCI is also 181 pm, what is the radius of the Na+ ion in NaCl? Assume that the edge length of the unit cell of NaCI shown in Figure 11.23(c) is 564 pm.

Ca2+ ions. Instead,

THINK

so they do not fit in the normal tetrahedral holes of a cubic closest-packed to accommodate maintaining

array of

the Ca2+ ions spread out

the larger F- ions while

an fcc structure.

546

I Chapter Eleven

I


Calculating the Density of a Salt from Its Unit-Cell Dimensions What is the density of LiCI at 513 pm?

25°C if the edge length of its fcc unit cell is

AND ORGANIZE We are given the edge length of an fcc unit cell of a crystalline salt and are asked to calculate its density. As in Sample Exercise 11.3, we have to assume that the density of the unit cell is the same as the density of the overall solid. The fact that LiCI has an fcc unit cell means that there are 4 CIions and 4 Li+ ions in the cell. The molar masses of these ions are 35A5 and 6.941 glmol, respectively. Density is the ratio of mass to volume, and the volume of a cubic cell is the cube of its edge length: V = t:3• COLLECT

The density of an fcc unit cell ofLiCI is the sum of the masses of 4 Clions and 4 Li+ ions divided by the volume of the cell (513 pm)3. As in Sample Exercise 11.3, we need to convert molar masses into the masses of individual ions by dividing by Avogadro's number. To obtain a cell volume in cubic centimeters, we use the conversion factor: ANALYZE

1pm SOLVE

=

10-10 cm

Calculating the masses of 4 Cl- ions, we get:

m

=

35A5 g ~ a'lOl Cl-

1 fHel X 6.022 X 1023

***' 4 ***' X

=

2.355 X

10-22

g

and the masses of 4 Li+ ions: - 6.941 g I,,+!:. X 1 fHel. X 4 a'lOl Li+ 6.022 X 1023 ***'

m -

***' = OA61

X 10-22

g

Next we combine the masses of the two kinds of ions in the unit cell: 2.355 X 10-22 g

+ 0.461

X 10-22

g

2.816 X 10-22 g The volume of the cell in cubic centimeters is

V = t:3

=

(513

f*A)3

X

(10-10 cm)3 =

1.350 X

10-22

cm3

~ Taking the ratio of mass to volume, we have d

= m = 2.816 V

X 10-

22

g

1.350 X 10-22 cm3

=

3 2.08 glcm

The result is reasonable in that most minerals are more dense than water but less dense than common metals.

THINK

ABOUT

IT

Practice Exercise What is the density of NaCI at its fcc unit cell is 564 pm?

25°C if the edge length of

11.8

~

Ceramics: Insulators to Superconductors

A ceramic is a solid inorganic compound or mixture of compounds that has been heated to transform it to a harder and more heat-resistant material. The use of ceramic materials preceded metal technology by many thousands of years. The first ceramics were probably made of clay: the very fine-grained products of the physical and chemical weathering of igneous rocks (rocks of volcanic origin). Moist clay is easily molded into the desired shape and then hardened over fires or in wood-burning kilns. This ancient process is still in use today. In this section we examine some of the physical properties of both primitive earthenware (ceramics fired at low temperatures) and modern ceramic materials (typically fired at high temperatures), and relate those properties to the chemical structures of these materials and to the chemical changes that occur when they are heated.

Polymorphs of Silica

I

547

A ceramic is a solid inorganic compound or mixture that has been transformed into a harder, more heat-resistant material by heating.

(a)

One of the most abundant families of minerals found in igneous rocks has the chemical composition Si02 and is called silica or, to be chemically correct, silicon dioxide. Silica is a covalent network solid in which each silicon atom is bonded to 4 oxygen atoms, forming oxygen tetrahedra with silicon atoms at their centers and O-Si-O bond angles of 109° (Figure 11.26). Each oxygen atom is bonded to 2 silicon atoms, thereby linking the tetrahedra into extended three-dimensional networks. Because each corner oxygen atom is bonded to 2 silicon atoms, each silicon atom gets only half "ownership" of the 4 oxygen atoms to which it is bonded-hence the formula Si02• At least eight different minerals have the chemical composition Si02• The members of a family of substances with the same composition but different structures and properties are called polymorphs. The most abundant silica polymorph is quartz, a type of Si02 that can form impressively large, nearly transparent crystals (Figure 11.26a). Note how the hexagonal ordering of the Si02 tetrahedra (Figure 11.26b) translates into hexagonal crystals. Most, but not all, silica is crystalline. When lava containing molten Si02 flows from a volcano into the sea or a lake, it cools quickly. During rapid cooling, Si and o atoms may not have enough time to achieve an ordered structure as the viscous (slow-moving) lava solidifies. A product of rapid solidification is an amorphous (disordered, noncrystalline) form of silica known as volcanic glass, or obsidian (Figure 11.27). Glass is a term scientists and engineers use to describe any solid that has no crystalline structure or only very tiny crystals surrounded by disor-

(b)

FIGURE 11.26

(b)

Qyartz:

(a) The hexagonal

shape of quartz crystals where (b) the hexagonal

arrays of silicon-oxygen

tetrahedra

share oxygen atoms in their crystalline structures.

(Here and in Figures

11.27-11.30,

the Si [silver] and 0 [red] atoms are drawn undersized

to make the arrangement

tetrahedra

easier to see.) These "hexagons

tetrahedra"

are covalently

bonded

of

dimensional

network

of

to others

above and below them in an extended

three-

(not shown).

FIGURE 11.27

(a) Obsidian

glass) is an unusual

form of silica in that

it is not crystalline.

(b) It contains

amorphous (a)

I


silica with random

of silicon and oxygen atoms.

(volcanic mostly

arrangements

548 I Chapter Eleven

I


(a) FIGURE 11.28 of a-cristobalite

Cristobalite:

(a) Deposits

(white) formed here on pieces

of volcanic glass. (b) In the crystal structure of a-cristobalite, within

four silica tetrahedra

an expanded

reside

fcc unit cell of silicon

atoms.

dered arrays of atoms. This definition applies to laboratory glassware and the drinking glasses we use at home, as well as to solids of the type discussed here that have no crystalline structure. Other members of the silica group have slightly different crystal structures. For example, in the unit cell of the mineral a-cristobalite (Figure 11.28), a common ingredient of volcanic rocks, four silicon-oxygen tetrahedra reside within a face-centered cubic lattice of silicon atoms. There are also f3-cristobalite and f3-quartz structures that differ from the a-cristobalite and quartz structures described here. (b) As you may have noticed, there are many structural forms of silica in nature. As solid silica minerals form from cooling molten Si02, they take on different structures depending on the pressures and temperatures at which they crystallize. Denser forms, such as f3-cristobalite and f3-quartz, are more likely to form at higher pressures; less dense forms occur at lower pressures. The presence of these different structures provides geologists with clues about the conditions under which igneous rocks in a region formed.

Ionic Silicates In addition to covalent silica, igneous rocks also contain ionic minerals made of silicon and oxygen. These minerals have some of the tetrahedral structure of silica, but not all the oxygen corner atoms are bonded to 2 Si atoms. Instead some of these 0 atoms have an extra electron. The result is a silicate anion. One of the most common kinds of silicates consists of sheets of linked silicon-oxygen tetrahedra that form hexagonal clusters of six tetrahedra each (Figure 11.29). Each tetrahedron has 3 0 atoms that it shares with other tetrahedra and 1 0 atom-the one with the extra electron-that it does not share. Thus the basic tetrahedral repeating unit consists of 1 Si atom and [1 + 3(~) 1 = 2.5 0 atoms as well as a negative charge. This gives the sheet the empirical formula Si02.s-. We generally use whole-number subscripts in chemical formulas when possible. In this case, multiplying Si02.s - by 2 gives us the empirical formula Si20/- for these silicate sheets. The subscript n in the formula in Figure 11.29 indicates that there are many empirical formula units in a single crystal.

FIGURE 11.29 the two principal

(a) Chrysotile minerals

is one of

in asbestos. The

ease with which thin fibers of chrysotile can flake off is related to (b) its layered crystalline

structure

intermolecular Chrysotile

and the relatively weak

interactions

between

and apparently

noncarcinogenic.

the second and less commonly asbestos,

layers.

fibers are curled, slightly soluble,

makes straighter,

Crodidolite, used form of

insoluble

fibers

(not shown); these do pose a health risk when breathed

in regularly,

and lung cancer.

leading to asbestosis

(a)

11.8

Of course, silicate minerals are neutral materials, so they must contain cations to neutralize the negative charges on the Si20s2-layers. When this cation is AP+, the minerals are called aluminosilicates. One of the most common examples of an alumino silicate is a clay mineral called kaolinite (Figure 11.30). At least a little kaolinite is found in practically every soil, but rich deposits of nearly pure, brilliantly white kaolinite are found in highly weathered soils. For centuries these deposits have been mined to make fine china and white porcelain. Today the greatest demand for kaolinite is in the production of glossy white paper, which is used in most magazines and books (including this one). Common metal ions found in igneous rocks-including N a+, K+, Ca 2+,Mg2+, and Fe3+ -are largely absent in kaolinite. Their absence indicates that kaolinite deposits form under acidic weathering conditions. Under these conditions H+ ions displace other cations from ion-exchange sites. For example, -O-Na+ sites exchange H+ for Na+, leaving behind -OH groups such as those shown in Figure 11.30. Kaolinite has the formula AI/Si20s)(OH)4. The foundation of its crystal structure is a silicate layer that fits the Si20/formula and structure shown in Figure 11.29(b). Aluminum(III) ions form a second layer in which each ion resides at the center of an octahedron of 6 oxygen atoms: 3 from the Si-O tetrahedra in the layer below, and 3 from hydroxide ions (OH-) that form a third layer above. The OH- ions are in turn hydrogen-bonded to 0 atoms of a second layer of silicate tetrahedra. This entire layering pattern repeats itself, forming an extended three-dimensional structure. The strengths of the ionic interactions and hydrogen bonds between layers are so strong that the layers in kaolinite are not easily separated; unlike with most clays, water molecules cannot penetrate between them. This property has made kaolinite pots handy vessels for carrying water for thousands of years. Because water cannot penetrate between its layers, kaolinite does not expand when water is added. For the same reason, it does not shrink as much as most clays when dehydrated at high temperatures. This is another property that has made kaolinite a desirable starting material for ceramics. High-purity kaolinite is also bright white, making it the material of choice for the manufacture of

I

I


riXi>

CONNECTION Ion-exchange reactions were discussed in Chapter 4.

FIGURE 11.30 This edge-on view of the layered structure

of kaolini te shows a bottom

layer of silicate (Si20/-) represented and red spheres, the top-down

by silver

much like the bottom

view in Figure

layer in

11.29(b). Next

comes a layer of AP+ ions (gray spheres) that interact

with lone pairs of valence electrons

on the 0 atoms of the silicate layer below and with a layer of hydroxide

ions in the top

layer. Each AP+ ion interacts

with 3 0 atoms

from each layer, and so resides in a kind of octahedral

hole surrounded

The em pi rical formula AI2(SiP5)(OH)4·

This enormous trillion ceramics

kaolinite

mine in Bulgaria

(1012) metric tons of the mineral to glossy white paper.

contributes

to a worldwide

per year. The mineral

production

of about 44

is used in many products,

from

549

by

60

atoms.

of this structure

is

550

I

Chapter Eleven

I


the porcelain used in bathroom fixtures, fine china, and other ceramics. Finally, moist kaolinite is plastic, meaning that it can be molded into a shape, and it keeps that shape during heating and cooling.

From Clay to Ceramic Creating ceramic objects from kaolinite and other clays takes several steps. First, moist clay is formed into pots, bricks, and other objects on a potter's wheel or in molds or presses. Drying at just above 100°C removes much of the water that made the clay plastic. Further heating to about 450°C removes water that was adsorbed onto the surfaces of the clay particles or between the layers of nonkaolinite clays. Between 450°C and 650°C the hydroxide ions of the octahedra surrounding the AP+ ions react with each other to form water and a solid that is richer in Al and Si: (11.11) The next structural change occurs just below 1000°C when Al2Si207 verted into another alumino silicate, AI4Si3012: 2 AI2Si20/s)

-9S0"C)

AI4SiPl/s)

+

Si02(s)

is con-

(11.12)

In this transformation, Si02 is excluded from the alumino silicate structure, forming a second solid phase. At even higher temperatures Al4Si3012 continues to lose Si02: AI4Si3012(s) 3 AI2SiOs(s)

~ 13S0'C )

2 AI2SiOs(s) AI Si 0 /s) 6

2

1

+ SiO/s) + Si02(s)

(11.13) (11.14)

The last product, A16Si2013, is called mullite. The formula of mullite is sometimes written as 3A1203·2Si02 indicating that it is a blend of Al203 (alumina) and Si02 (silica) structures. Mullite is a widely used ceramic material in the manufacture of products that must tolerate temperatures as high as 1700°C: furnaces, boilers, ladles, and kilns. These products are used as containers of molten metals and in the glass, chemical, and cement industries. Mullite is also very hard and is widely used as an abrasive.

FIGURE 11.31 About 25,000 heatresistant tiles cover much of the surface of each U.S. space shuttle, protecting re-entry temperatures

it from

of over 1l00°C. Most

tiles are made of high-purity,

amorphous

Si02 fibers. Ninety percent of each tile is open space between the fibers, which means that the tiles are both lightweight resistant to high temperatures.

and highly

11.8

Other ceramics are used in high-temperature applications ranging from cookware to fireplace bricks to the tiles on the U.S. space shuttles (Figure 11.31). Ceramics are well suited to these uses because of their high melting points and because they are good thermal and electrical insulators. For example, the thermal conductivity of aluminum metal at 100°C is over eight times that of alumina

(AI20). Structure and Bonding: Superconductors Band theory can be used to describe the properties of, and bonding in, all solids, including the compounds in ceramic materials. Earlier in the chapter we noted that metals are good conductors because of the ease with which their valence electrons can move to conduction bands, whereas semimetals are semiconductors because they have significant energy gaps between their valence and conduction bands (though these gaps can be reduced by the process of doping). Similarly, the insulating properties of ceramics can be explained by the large energy gaps between their valence and conduction bands (see Figure 11.4). As a result, their valence electrons have very limited mobility. In conventional metallic conductors the vibration of metal atoms in their crystalline lattices can interfere with the flow of free electrons: the higher the temperature, the greater the atomic vibration and the greater the resistance to electron flow. However, at temperatures approaching absolute zero these vibrations become very small. In the early 20th century, scientists discovered that the lattice vibrations in mercury and some metal alloys are so small at temperatures below about 20 K, called critical temperatures (-z:), that electrons can pass freely through these materials, and they become superconductors. Today superconducting alloys such as Nb3Sn are widely used in devices that require very high electrical currents, such as the electromagnets of magnetic resonance imaging (MRI) instruments in hospitals. Unfortunately it is difficult and expensive to chill materials to 20 K. In 1986 it was discovered that YBa2Cu307 and several other ceramic materials become superconductors when cooled to temperatures just above the boiling point of liquid nitrogen (77 K). This represented a tremendous increase in the critical temperature for the onset of superconductivity. Since then, scientists from around the world have produced a variety of superconducting ceramic materials with critical temperatures as high as 133 K (-140°C). It is much easier and cheaper to cool a material with liquid nitrogen than with liquid helium (boiling point = 4 K), which is used to achieve superconductivity in Hg and metal alloys. Let's take a closer look at the structure of one of the so-called "High-~" superconductors, YBa2Cu30T We will start with the unit cell shown in Figure 11.32(a). It is called aperovskite unit cell, and is adopted by many compounds that have the generic formula ABO]" An ion of metal "A," which is represented by the Ba2+ ion in Figure l1.32(a), is in the center of the cell; eight ions of a metal "B" (Cu2+ in this case) occupy the corners, and 12 oxide ions are located along the edges. Figure l1.32(b) contains a stack of three such perovskite structures, two containing a Ba2+ ion and one containing a Y4+ion at its center. Applying our usual practice of assigning fractions of atoms and ions to unit cells, we find that the array of 3 unit cells in Figure 11.32(b) has 9 oxide ions (~ of the 20 that occupy edges of the 3-cell array + ~ of the 8 ions that occupy faces-9 in all) and 3 copper ions (~ of the 8 ions on the top and bottom surfaces + ~ of the 8 ions around the middle-3 in all). Thus, the material with the structure shown

I


I

The critical temperature (T) is the temperature below which a ;;'aterial becomes a superconductor. A superconductor is a material that has zero resistance to the flow of electric current.

551

552

I

Chapter Eleven

I


FIGURE 11.32 The crystal structure ofYBa2Cu307,

a high-temperature

superconductor:

(a) A perovskite

(b) The process of building copper superconductor

unit cell.

begins by stacking

unit cells with a central barium and bottom

modified

structure

oxide ions in edge positions, 2 the number produces

~ II

assignable

°

3

ion in

(c) This has 8 fewer which reduces by

to the structure,

and (a)

the formula YBa2Cu307•

CHEMTOUR

Cu

corresponding

is YBa2Cus09•

perovskite

Y

ion in the top

cells, and a central yttrium

the middle cell. The formula to this structure

Ba

a yttrium-barium-

(b)

(c)

in Figure 11.32(b) has the chemical formula YBa2Cu309, but it is not superconducting. However, it becomes superconducting when it loses 8 oxygen atoms from edge sites to form YBa2Cu307 (Figure 11.32c). Yttrium-barium-copper oxides and related materials behave as superconductors because of the formation of electron pairs called Cooper pairs in their crystals (Figure 11.33). When an electron is introduced into a superconductor, electrostatic forces produce a slight displacement in the positively charged cations in the crystal. The attraction between electrons and these positively charged cations can bring two electrons into close proximity, resulting in formation of a Cooper pair. The movement of Cooper pairs rather than individual electrons through the material accounts for its superconductivity. To understand the superconductivity of a Cooper pair, consider a pair of horses harnessed together to pull a wagon

Superconductors

Overall density of

charge = 16 x2

T 1 x

FIGURE 11.33 (a) An electron moving through

a lattice of positively

causes (b) a distortion cations are attracted electrical

charged

to the electron's

charge. (c) This localized

positive charge brings another the first, creating

cations

in the lattice as the negative increase in

electron

toward

/(c)

~electron /

(a)

a Cooper pair.

Overall density of

(b)

Add second electron

charge = 16

l

11.9

I

X-ray Diffraction: How We Know Crystal Structures

(or a three-legged race at a child's birthday party). The team of horses will travel forward, avoiding obstacles in their path but never separating from each other. Single animals, however, will take divergent paths, scattering over a wide area. Similarly, single electrons are easily deflected by the atoms or ions in a solid, but electrons scatter much less when harnessed in a Cooper pair. Less scattering lowers electrical resistance. Superconductors are of technological interest because, in principle, high currents can flow through them with no resistance. This property is attractive for the design of extremely fast computers. Superconductors also have the ability to exclude magnetic fields-a property known as the Meissner effect (Figure 11.34). A small cylinder of superconductive material cooled below its critical temperature floats in a magnetic field because magnetic lines of force are excluded from it. Because of this exclusion, the magnet below the superconductor repels it, suspending it in air. This magnetic levitation could in principle be used to both float a train above electromagnetic tracks and propel it along the tracks at relatively low cost and at much higher speeds than a conventional train. Levitating trains, however, remain a goal for the future.

(background)

cannot

a result, the magnet

effect:

penetrate

the small

material

that has

XY _ YZ = sine d

e + dsin e = 2dsin e

As

repels the superconductor,

causing it to float above the magnet.

Earlier in the chapter we pointed out that the dimensions of unit cells are determined using X-ray diffraction (XRD). In an XRD instrument a narrow beam of X-rays is directed at the surface of a crystalline sample, as shown in Figure 11.35. The angle between the surface and the beam is called the angle of incidence and given the symbol e. Some of the X-rays hit atoms on the surface of the crystal and bounce off them. These reflected X-rays leave the surface with an angle of reflection that is also e. The total change in direction of the reflected beam is the sum of the angles of incidence and reflections, or 2e. This value is called the angle of diffraction. Now suppose some of the incident X-rays pass through the surface and bounce off atoms (or ions) in a second layer a distance (d) deeper into the crystal. They will also undergo an angle of diffraction 2e. These X-rays and those reflected from the first row of atoms are much like the rays of light passing through two narrow slits (see Figure 7.5). This means they can undergo constructive or destructive interference and so produce an interference pattern. To interfere constructively, the two beams of reflected X-rays must be in phase, as shown in Figure 11.35(b). To be in phase, the greater distance traveled by the beam reflecting off the second row must be some multiple of the wavelength. The greater distance is the sum of the lengths of line segments XY and YZ. The two right triangles incorporating these line segments share a hypotenuse equal to d. Geometry tells us that the angles opposite XY and YZ are not only the same but also equal to e. According to trigonometry, the ratio of either XY or YZ to d is the sine of e:

Rearranging these terms, we find that the difference in distance traveled (XY YZ) is

553

by the magnet

been chilled below its critical temperature.

How We Know Crystal Structures

dsin

The Meissner

field produced

cylinder of superconductive

II!D X-ray Diffraction:

d-

FIGURE 11.34 The magnetic

I

+

~II

CHEMTOUR

X-ray DifFraction

X-ray diffraction (XRD) is a technique for determining the arrangement of atoms or ions in a crystal by analyzing the pattern that results when X-rays are scattered after bombarding the crystal.

554

I

Chapter Eleven

I


(a)

-

- ~

..

t '" '"

1l >-
-
-

SECTION 11.9 X-ray diffraction is an analytical method that records constructive and destructive interference of X-rays reflecting off different layers of atoms or ions in crystals. The distances between layers can be calculated using the Bragg equation. X-ray diffraction makes it possible to determine the solid-state structure of crystalline solids.

Summary

TYPE OF PROBLEM

CONCEPTS AND EQUATIONS

SAMPLE EXERCISES

Distinguishing between p- and n-type semiconductors

Determine whether the dopant has more (n-type) or fewer (p-type) valence electrons than the host semiconductor

11.1

Calculating atomic radii from crystal structures

Determine the length of a unit-cell edge, face diagonal, or body diagonal along which the atoms touch. Calculate the number of radii that correspond to this distance.

11.2

d= m/V

11.3

Using crystal structure to calculate the density of a solid

Determine the mass of the atoms in the unit cell from their molar mass and the volume of the unit cell from the unit-cell edge length.

Predicting the type of alloy two elements form

Compare the radii of the alloying elements: similarly sized radii (within 15%) predict a substitutional alloy. When the radius of the smaller atom is < 73% of the radius of the larger atom, an interstitial alloy forms.

11.4

Calculating ionic radii from unit-cell dimensions

Determine the dimension (edge or diagonal) along which the ions touch. Use the geometric relationship between the edge or diagonal and the ionic radius (r) to calculate the value of r.

11.5

Visual Problems

TYPE OF PROBLEM

CONCEPTS

AND EQUATIONS

Calculating the density of a salt from its unit-cell dimensions

Determine the mass of the ions in the unit cell and the volume of the unit cell from the unit-cell edge length. Then apply d = mlV.

Determining interlayer distances by X-ray diffraction

Apply the Bragg equation: nA

=

SAMPLE

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559

EXERCISES

11.6

2d sin e.

11.7


In Figure P11.1, which drawing is analogous to a crystalline solid, and which is analogous to an amorphous solid?

11.4.

How many total cations (A and B) and anions (X) are there in the unit cell in Figure P11.4?

=A

=B (a)

(b)

(c)

=X

(d)

FIGURE P11.1 11.2.

The unit cells in Figure P11.2 continue infinitely in two dimensions. Draw a box around the unit cell in each pattern. How many light squares and how many dark squares are in each unit cell?

FIGURE P11.4 11.5.

How many equivalent atoms of elements A and B are there in the portion of a unit cell in Figure P11.5?

• • • • • • • • •

=A

=B

(a)

(b)

FIGURE P11.2 11.3.

FIGURE P11.5

The pattern in Figure P11.3 continues indefinitely in three dimensions. Draw a box around the unit cell. If the red circles represent element A and the blue circles element B, what is the chemical formula of the compound?

a:::

Y-

It::

lI:::

~

What is the chemical formula of the compound a portion of whose unit cell is shown in Figure P11.6?

n. A

'1,

tk::

11.6.

..

A

."

-

A

~,

=A

=B =X

~?'

~

FIGURE P11.3

FIGURE P11.6

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11. 7. What is the chemical formula of the ionic compound a portion of whose unit cell is shown in Figure P11.7? (A and B are cations, X is an anion.)

11.11. What is the formula of the compound that crystallizes with aluminum ions occupying half of the octahedral holes and magnesium ions occupying one-eighth of the tetrahedral holes in a cubic closest-packed arrangement of oxide ions? See Figure P11.11.

=A =B

=X FIGURE P11.7

11.8.

When amorphous red phosphorus is heated at high pressure, it is transformed into the allotrope black phosphorus, which can exist in one of several forms. One form consists of 6-membered rings of phosphorus atoms (Figure P11.8a). Why are the 6-atom rings in black phosphorus puckered, whereas the 6-atom rings in graphite are planar (Figure P11.8b)?

FIGURE P11.11 11.12. What is the formula of the compound that crystallizes with zinc atoms occupying half of the tetrahedral holes in a cubic closest-packed arrangement of sulfur atoms? See Figure P11.12.

(b) Graphite

FIGURE P11.8

11.9.

The distance between atoms in a cubic form of phosphorus is 238 pm (Figure P11.9). Calculate the density of this form of phosphorus.

f238 pm

----±

11.13. What is the formula of the compound that crystallizes with lithium ions occupying all of the tetrahedral holes in a cubic closest-packed arrangement of sulfide ions? See Figure P11.13.

FIGURE P11.9

11.10. How many of the large spheres and how many of the small ones are assignable to the unit cell in Figure P11.10?

FIGURE P11.13

FIGURE P11.10

11.14. Figure P11.14 shows the unit cell ofCsCl. From the information given and the radius of the chloride (corner ions) of 181 pm, calculate the radius of Cs+ ions.


I

56:1

*11.17. Superconducting Materials In 2000, magnesium boride was observed to behave as a superconductor. Its unit cell is shown in Figure P11.17. What is the formula of magnesium boride? A boron atom is in the center of the unit cell (on the left), which is part of the hexagonal structure (right).

FIGURE P11.14 11.15. A number of metal chlorides adopt the rock salt structure in which the metal ions occupy all the octahedral holes in a face-centered cubic array of chloride ions. For at least one (maybe more) of the metallic elements highlighted in Figure P11.1S, this crystal structure is not possible. Which one(s) are they? 1234

S 6 7 8 9101112131415161718

FIGURE P11.17 *11.18. Superconducting Materials The 1987 Nobel Prize in physics was awarded to G. Bednarz and K. A. Miiller for their discovery of superconducting ceramic materials such as YBa2Cu307" Figure P11.18 shows the unit cell of another yttrium-barium-copper oxide. a. What is the chemical formula of this compound? b. Eight oxygen atoms must be removed from the unit cell shown here to produce the unit cell ofYBa2Cu307" Does it make a difference which oxygen atoms are removed?

FIGURE P11.15 11.16. A number of metal fluorides adopt the fluorite structure in which the fluoride ions occupy all the tetrahedral holes in a face-centered cubic array of metal ions. For at least one (maybe more) of the highlighted metallic elements in Figure P11.16, this crystal structure is not possible. Which one(s)?

=

Ba

=Y =

Cu

1 2 3 4 5 6 7 8 9 10 1112 13 1415 16 17 18

FIGURE P11.18 FIGURE P11.16

Questions and Problems Metallic Bonds and Conduction Bands CONCEPT

REVIEW

11.19. How does the sea-of-electrons model (Chapter 8) explain the high electrical conductivity of gold? *11.20. How does band theory explain the high electrical conductivity of mercury?

11.21. The melting and boiling points of sodium metal are much lower than those of sodium chloride. What does this difference reveal about the relative strengths of metallic bonds and ionic bonds? *11.22. Which metal do you expect to have the higher melting point-Alar Na? Explain your answer.

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11.23. Some scientists believe that the solid hydrogen that forms at very low temperatures and high pressures may conduct electricity. Is this hypothesis supported by band theory? 11.24. Would you expect solid helium to conduct electricity?

Semimetals and Semiconductors CONCEPT

REVIEW

11.25. Which groups in the periodic table contain metals with filled valence bands? 11.26. Insulators are materials that do not conduct electricity; conductors are substances that allow electricity to flow through them easily. Rank the following in order of increasing band gap: semiconductor, insulator, conductor. 11.27. Why is it important to keep phosphorus out of silicon chips during their manufacture? 11.28. How might doping of silicon with germanium affect the conductivity of silicon? *11.29. Antimony (Sb) combines with sulfur to form the semiconductor compound Sb2Sy In which group of the periodic table might you find elements for doping Sb2S3 to form a p-type semiconductor? *11.30. In which group of the periodic table might you find elements for doping Sb2S3 to form an n-type semiconductor? PROBLEMS

11.31. Thin films of doped diamond hold promise as semiconductor materials. Trace amounts of nitrogen impart a yellow color to otherwise colorless pure diamonds. a. Are nitrogen-doped diamonds examples of semiconductors that are p-type or n-type? b. Draw a picture of the band structure of diamond to indicate the difference between pure diamond and N-doped (nitrogen-doped) diamond. *c. N-doped diamonds absorb violet light at about 425 nm. What is the magnitude of Eg that corresponds to this wavelength? 11.32. Trace amounts of boron give diamonds (including the Smithsonian's Hope Diamond) a blue color (Figure P11.32). a. Are boron-doped diamonds examples of semiconductors that are p-type or n-type? b. Draw a picture of the band structure of diamond to indicate the difference between pure diamond and Bdoped diamond. *c. What is the band gap in energy ifblue diamonds absorb orange light with a wavelength of 675 nm?

*11.33. The nitride ceramics AlN, GaN, and InN are all semiconductors used in the microelectronics industry. Their band gaps are 580.6,322.1, and 192.9 kJlmol, respectively. Which if any of these energies correspond to radiation in the visible region of the spectrum? *11.34. Calculate the wavelengths oflight emitted by the semiconducting phosphides AlP, GaP, and InP, which have band gaps of241.1, 216.0, and 122.5 kJlmol used in LEDs and are like those in Figure P11.34.

FIGURE P11.34

Metallic Crystals CONCEPT

REVIEW

11.35. Explain the difference between cubic closest-packed and hexagonal closest-packed arrangements of identical spheres. *11.36. Is it possible to have a closest-packed structure with four different repeating layers, abcdabcd ... ? 11.37. Which unit cell has the greater packing efficiency, simple cubic or body-centered cubic? 11.38. Consult Figure 11.10 to predict which unit cell has the greater packing efficiency: body-centered cubic, or facecentered cubic. 11.39. Iron metal adopts either an fcc or a bcc structure, depending on its temperature (see Sample Exercise 11.2). Do these two crystal structures represent allotropes of iron? Explain your answer. *11.40. Calcium metal crystallizes in two different unit cells, an fcc and a bcc. Why might calcium adopt a closest-packed arrangement at low temperature (fcc unit ceIl) but the nonclosest-packed structure at higher temperature (bcc)? PROBLEMS

11.41. Derive the edge length in bcc and fcc unit cells in terms of the radius (r) of the atoms in the cells in Figure 11.10. *11.42. Derive the length of the body diagonal in simple cubic and fcc unit cells in terms of the radius (r) of the atoms in the unit cells in Figure 11.10. 11.43. Europium is one of the lanthanide elements used in television screens. Europium crystallizes in a bcc structure with a unit-cell edge of240.6 pm. Calculate the radius of a europium atom. 11.44. Nickel has an fcc unit cell with an edge length of350.7 pm. Calculate the radius of a nickel atom.

FIGURE P11.32

11.45. What is the length of an edge of the unit cell when barium (atomic radius = 222 pm) crystallizes in a bcc structure? 11.46. What is the length of an edge of the unit cell when aluminum (atomic radius = 143 pm) crystallizes in an fcc structure?

Questions and Problems 11.47. A crystalline form of copper has a density of8.95 glcm3. If the radius of copper atoms is 127.8 pm, what is the copper unit ceIl? (a) simple cubic; (b) body-centered cubic; or (c) face-centered cubic 11.48. A crystalline form of molybdenum has a density 10.28 glcm3 at a temperature at which the radius of a molybdenum atom is 139 pm. Which unit cell is consistent with these data? (a) simple cubic; (b) body-centered cubic; or (c) face-centered cubic

Alloys CONCEPT

REVIEW

11.49. Is there a difference between a solid solution and a homogeneous alloy? 11.50. White gold was originally developed to give the appearance of platinum. One formulation of white gold contains 25% nickel and 75% gold. Which is more malleable, white gold or pure gold? *11.51. Explain why an alloy that is 28% Cu and 72% Ag melts at a lower temperature than the melting points of either Cu orAg. 11.52. Is it possible for an alloy to be both substitutional and interstitial? 11.53. The interstitial alloy tungsten carbide (WC) is one of the hardest materials known. It is used on the tips of cutting tools. Without consulting a table of radii, decide which element you think is the host and which occupies the holes. 11.54. Why are the alloys that second-row nonmetals-such as B, C, and N-form with transition metals more likely to be interstitial than substitutional? PROBLEMS

11.55. The unit cell of a substitutional alloy consists of a facecentered cube that has an atom of element X at each corner and an atom of element Y at the center of each face. a. What is the formula of the alloy? b. What would the formula of the alloy be if the positions of the two elements were reversed in the unit cell? 11.56. The bcc unit cell of a substitutional alloy has atoms of element A at the corners of the unit cell and an atom of element B at the center of the unit cell. a. What is the formula of the alloy? b. What would the formula of the alloy be if the positions of the two elements were reversed in the unit cell? 11.57. Vanadium reacts with carbon to form vanadium carbide, an interstitial alloy. Given the atomic radii of V (135 pm) and C (77 pm), which holes in a cubic closest-packed array of vanadium atoms do you think the carbon atoms are more likely to occupy-octahedral or tetrahedral? 11.58. What is the minimum atomic radius required for a cubic closest-packed metal to accommodate boron atoms (radius 88 pm) in its octahedral holes? 11.59. Dental Fillings Dental fillings are mixtures of several alloys including one with the formula Ag3Sn. Silver (radius 144 pm) and tin (140 pm) both crystallize in an fcc unit cell. Is this alloy likely to be a substitutional alloy or an interstitial alloy? 11.60. An alloy used in dental fillings has the formula SnsHg. The radii of tin and mercury atoms are 140 pm and 151 pm,

I

563

respectively. Which alloy has a smaller mismatch (percent difference in atomic radii), Sn3Hg or bronze (Cu/Sn alloys)? *11.61. Hardening Metal Surfaces Plasma nitriding is a process for embedding nitrogen atoms in the surfaces of metals that hardens the surfaces and makes them more corrosion resistant. Do the nitrogen atoms in the nitrided surface of a sample of cubic closest-packed iron fit in the octahedral holes of the crystal lattice? (Assume that the atomic radii of N and Fe are 75 and 126 pm, respectively.) *11.62. Hydrogen Storage A number of transition metals (including titanium, zirconium, and hafnium) can store hydrogen as metal hydrides for use as fuel in a hydrogenpowered vehicle. Which metal or metals are most likely to accommodate H atoms (radius 37 pm) with the least distortion, given that their atomic radii are 147, 160, and 159 pm, respectively? 11.63. An interstitial alloy is prepared from metals A and B where B has the smaller atomic radius. Metal A crystallizes in an fcc structure. What is the formula of the alloy ifB occupies the following? (a) all of the octahedral holes; (b) half of the octahedral holes; (c) half of the tetrahedral holes 11.64. An interstitial alloy was prepared from two metals. Metal A with the larger atomic radius has a hexagonal closest-packed structure. What is the formula of the alloy if atoms of metal B occupy the following? (a) all of the tetrahedral holes; (b) half of the tetrahedral holes; (c) half of the octahedral holes 11.65. An interstitial alloy contains 1 atom ofB for every 5 atoms of host element A, which has an fcc structure. What fraction of the octahedral holes is occupied in this alloy? 11.66. If the B atoms in the alloy described in Problem 11.65 occupy tetrahedral holes in A, what percentage of the holes would they occupy?

The Structures of Some Nonmetals CONCEPT

REVIEW

11.67. Ss is not a flat octagon-why? *11.68. Selenium exists as Ses rings or in a crystalline structure with helical chains of Se atoms. Are these two structures of selenium allotropes? Explain your answer. *11.69. If the carbon atoms in graphite are replaced by alternating Band N atoms, would the resulting structure contain puckered rings (see Figure P11.8) like black phosphorus or flat ones like graphite? *11.70. Cyclic allotropes of sulfur containing up to 20 sulfur atoms have been isolated and characterized. Propose a reason why the bond angles in S, (where n = 10,12,18, and 20) are all close to 106'? PROBLEMS

11.71. Ice is a network solid. However, theory predicts that, under high pressure, ice (solid H20) becomes an ionic compound composed ofH+ and 02- ions. The proposed structure of ice under these conditions is a body-centered cubic arrangement of oxygen ions with hydrogen ions in holes. a. How many H+ and 02- ions are in each unit ceIl? b. Draw a Lewis structure for "ionic" ice.

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11.72. Ice Under Pressure Kurt Vonnegut's novel Cat's Cradle describes an imaginary, high-pressure form of ice called "ice nine." With the assumption that ice nine has a cubic closest-packed arrangement of oxygen atoms with hydrogen atoms in the appropriate holes, what type of hole will accommodate the H atoms? 11.73. A chemical reaction between H2S4 and S2Cl2 produces cyclic S6' What are the bond angles in S6? 11.74. Reaction between S8 and six equivalents of AsFs yields [S/+][AsF6-]2 by the reaction S8 + 6 AsFs ~ 2 [S/+][AsF6

-L

+ AsF3

(The brackets identify complex ions in the structure, in this case indicating that the S4 unit has a 2+ charge and the AsF6 unit has a 1- charge.) The S/+ ion has a cyclic structure. Are all 4 sulfur atoms in one plane?

Salt Crystals: Ionic Solids CONCEPT

REVIEW

11.75. Crystals of both LiCI and KCI have the rock salt crystal structure. In the unit cell ofLiCI the Cl-ions touch each other. In KCI they don't. Why? 11.76. Can CaCl2 have the rock salt crystal structure? *11.77. In some books the unit cell ofCsCI is described as being body-centered cubic (Figure P11.77); in others, as simple cubic (see Figure 11.10c). Explain how CsCI might be assigned to both structures.

11.84. As the cation-anion radius ratio increases for an ionic compound with the rock salt structure, is the length of the unit-cell edge calculated from ionic radii likely to be greater than, or less than, the observed unit-cell edge length? PROBLEMS

11.85. What is the formula of the oxide that crystallizes with Fe3+ ions in one-fourth of the octahedral holes, Fe3+ ions in oneeighth of the tetrahedral holes, and Mg2+ in one-fourth of the octahedral holes of a cubic closest-packed arrangement of oxide ions (02-)? 11.86. What is the formula of the compound that crystallizes with Ba2+ ions occupying half of the cubic holes in a simple cubic arrangement of fluoride ions? 11.87. The Vinland Map At Yale University there is a map, believed to date from the 1400s, of a landmass labeled "Vinland" (Figure Pl1.87). The map is thought to be evidence of early Viking exploration of North America. Debate over the map's authenticity centers on yellow stains on the map paralleling the black ink lines. One analysis suggests the yellow color is from the mineral anatase, a form ofTi02 that was not used in 15th-century inks. a. The structure of anatase is approximated by a ccp arrangement of oxide ions with titanium(IV) ions in holes. Which type of hole are Ti4+ ions likely to occupy? b. What fraction of these holes are likely to be occupied? (The radius ofTi4+ is 60.5 pm.)

-,

41[ FIGURE P11.77 11.78. In the crystals of ionic compounds, how do the relative sizes of the ions influence the location of the smaller ions? *11.79. Instead of describing the unit cell ofNaCI as an fcc array ofCI- ions with Na+ ions in octahedral holes, might we describe it as an fcc array ofNa+ ions with CI- ions in octahedral holes? Explain why or why not. 11.80. Why isn't crystalline sodium chloride considered a network solid? 11.81. If the unit cell of a substitutional alloy of copper and tin has the same unit-cell edge as the unit cell of copper, will the alloy have a greater density than copper? 11.82. If the unit cell of an interstitial alloy of vanadium and carbon has the same unit-cell edge as the unit cell of vanadium, will the alloy have a greater density than vanadium? *11.83. As the cation-anion radius ratio increases for an ionic compound with the rock salt structure, is the calculated density more likely to be greater than, or less than, the measured value?

FIGURE P11.87 *11.88. The structure of olivine-M2SiO 4 (M = Mg, Fe)-can be viewed as a ccp arrangement of oxide ions atoms with Si(IV) in tetrahedral holes and the metal ions in octahedral holes. a. What fraction of each type of hole is occupied? b. The unit-cell volumes ofMg2Si04 and Fe2Si04 are 2.91 X 10-26 cm3 and 3.08 X 10-26 cm3• Why is the unitcell volume ofFe2SiO 4 larger? *11.89. In nature, cadmium(II) sulfide (CdS) exists as two minerals. One of them, called hawleyite, has a sphalerite crystal structure, and its density at 25'C is 4.83 g/cm3• A hypothetical form of CdS with the rock salt structure would have a density of 5.72 glcm3. Why should the rock salt structure of CdS be denser? The ionic radii ofCd2+ and S2are 95 pm and 184 pm, respectively.


11.90. There are two crystalline forms of manganese(II) sulfide (MnS): the a form has a NaCl structure; the f3 form has a sphalerite structure. a. Describe the differences between the two structures of MnS. b. The ionic radii ofMn2+ and S2- are 67 and 184 pm. Which type of cubic hole in a ccp lattice of sulfide ions could theoretically accommodate an Mn2+ ion? consists of a cube with rhenium atoms at the corners and an oxygen atom on each of the 12 edges. The atoms touch along the edge of the atoms in Re03 are 137 and unit cell. The radii ofRe and 73 pm. Calculate the density ofReOr 11.92. With reference to Figure P11.77, calculate the density of simple cubic CsCl.

11.93. Magnesium oxide crystallizes in the rock salt structure. Its density is 3.60 g/cm3• What is the edge length of the fcc unit cell ofMgO? 11.94. Crystalline potassium bromide (KBr) has a rock salt structure and a density of2.75 glcm3• Calculate its unit-cell edge length.

Ceramics: Insulators to Superconductors CONCEPT

Anorthite

Ca2Al2Si07 Gehlenite

+

CaSi03

~

Ca3AI2(Si04\

Wollastonite

Grossular

a. Balance this chemical equation. b. Express the composition of gehlenite the way mineralogists often do: as the percentage of the metal and semimetal oxides in it, that is, %CaO, %AI203, and %Si02· The ceramic material barium titanate (BaTi03) is used in devices that measure pressure. The radii ofBa2+, Ti4+, and 02- are 135, 60.5, and 140 pm, respectively. If the 02- ions are in a closest-packed structure, which hole(s) can accommodate the metal cations? 11.104. The mixed metal oxide LiMnTi04 has a structure with cubic closest-packed oxide ions and metal ions in both octahedral and tetrahedral holes. Which metal ion is most likely to be found in the tetrahedral holes? The ionic radii ofLi+, Mn2+, Ti4+, and 02- are 76, 67, 60.5, and 140 pm, respectively.

11.103.

CONCEPT

REVIEW

Why does an amorphous solid not produce an XRD scan with sharp peaks? 11.106. X-ray diffraction cannot be used to determine the structures of compounds in solution-why? 11.107. Why are X-rays rather than microwaves chosen for diffraction studies of crystalline solids? 11.108. The radiation sources used in X-ray diffraction can be changed. Figure P11.108 shows a diffraction pattern made by a short-wavelength source. How would changing to a longer-wavelength source affect the pattern?

11.105.

800

PROBLEMS

Kaolinite [A12(Sips)(OH)4] is formed by weathering of the mineral KAISi30s in the presence of carbon dioxide and water, as described by the following unbalanced reaction: KAISips(s)

+

HP(t') + CO/aq) ~ + SiO/s) AI/Sips)(OH)/s)

+

K2C03(aq)

Balance the reaction and determine whether or not this is a redox reaction. 11.100. Albite, a feldspar mineral with an ideal composition of KAlSi30s' can be converted to jadeite (NaAISi206) and quartz. Write a balanced chemical equation describing this transformation. 11.101.

+

Ca3Al2Si20s

X-ray Diffraction: How We Know Crystal Structures

REVIEW

11.95. Which of the following properties are associated with ceramics and which are associated with metals? Ductile; thermal insulator; electrically conductive; malleable 11.96. Many ceramics such as Ti02 are electrical insulators. What differences are there in the band structure ofTi02 compared with Ti metal that account for the different electrical properties? 11.97. Replacement of AP+ ions in kaolinite [A12(Sips)(OH)4] with Mg2+ ions yields the mineral antigorite. What is its formula? 11.98. What is the formula of the silicate mineral talc, obtained by the replacement of AP+ ions in pyrophyllite [AI4SiP2o (OH)4] with Mg2+ ions?

11.99.

565

*11.102. The calcium silicate mineral grossular is also formed under pressure in a reaction between anorthite (Ca3AI2Si20s)' gehlenite (Ca2AI2Si07), and wollastonite (CaSiOJ

11.91. The unit cell of rhenium trioxide (Re03)

°

I

Under the high pressures in Earth's crust, the mineral anorthite (CaAI2Si20s) is converted to a mixture of three minerals: grossular [Ca3Al/SiO )3]' kyanite (AI2SiOs)' and quartz (SiO). (a) Write a balanced chemical equation describing this transformation. (b) Determine the charges and formulas of the silicate anions in anorthite, grossular, and kyanite.

700 600 500

t t::

1l

,.s

400 300 200 100

a -100

a

10

20

30

40 28

FIGURE Pll.l08

50

60

70

80

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*11.109. Why might a crystallographer (a scientist who studies crystal structures) use different X-ray wavelengths to determine a crystal structure? (Hint: consider what mechanical limits are inherent in the design of the instrument depicted in Figure 11.35, and how those limits impact the 28 scanning range.) *11.110. Where in earlier chapters have we seen diffraction used to acquire structural information? PROBLEMS

11.111. The spacing between the layers of ions in sylvite (the mineral form of KCI) is larger than in halite (NaC\). Which crystal will diffract X-rays of a given wavelength through larger 28 values? 11.112. Silver halides are used in black-and-white photography. In which compound would you expect to see a larger distance between ion layers, AgCI or AgBr? Which compound would you expect to diffract X-rays through larger values of 2 8 if the same wavelength of X-ray were used? 11.113.

Galena, Illinois, is named for the rich deposits oflead(II) sulfide (PbS) found nearby. When PbS is exposed to X-rays with A = 71.2 pm, strong reflections from a single crystal of PbS are observed at 13.98° and 21.25°. Determine the values of n to which these reflections correspond, and calculate the spacing between the crystal layers. 11.114. Pigments in Ceramics Cobalt(II) oxide is used as a pigment in ceramics. It has the same type of crystal structure as NaCl. When cobalt(II) oxide is exposed to X-rays with A = 154 pm, reflections are observed at 42.38°,65.68°, and 92.60°. Determine the values of n to which these reflections correspond, and calculate the spacing between the crystal layers.

11.115.

Pyrophyllite [AI2Si40lO(OH)2J is a silicate mineral with a layered structure. The distances between the layers is 1855 pm. What is the smallest angle of diffraction of X-rays with A = 154 pm from this solid? 11.116. Minnesotaite [Fe3Si4010(OH)2J is a silicate mineral with a layered structure similar to that of kaolinite. The distance between the layers in minnesotaite is 1940 ± 10 pm. What is the smallest angle of diffraction of X-rays with A = 154 pm from this solid?

Additional Problems 11.117. A unit cell consists of a cube that has an ion of element X at each corner, an ion of element Y at the center of the cube, and an ion of element Z at the center of each face. What is the formula of the compound? 11.118. The unit cell of an oxide of uranium consists of cubic closest-packed uranium ions with oxide ions in all the tetrahedral holes. What is the formula of the oxide? *11.119. The density of pure silicon is 2.33 g/mL. What is the packing efficiency (percentage of the space inside the unit cell occupied by atoms) of the Si atoms in pure Si if the radius of one Si atom is 117 pm? Packing efficiency (%) = volume occupied by Si atoms volume of unit cell

X 100

*11.120. The Composition of Light-Emitting Diodes The colored lights on many electronic devices are light-emitting diodes (LEDs). One of the compounds used to make them is aluminum phosphide (AlP), which crystallizes in a sphalerite structure. a. If AlP were an ionic compound, would the ionic radii of AI3+ and P3- be consistent with the size requirements of the ions in a sphalerite structure? b. If AlP were a covalent compound, would the atomic radii of Al and P be consistent with the size requirements of atoms in a sphalerite structure? 11.121. Under the appropriate reaction conditions, small cubes of molybdenum, 4.8 nm on a side, can be deposited on carbon surfaces. These "nanocubes" are made ofbcc arrays ofMo atoms. a. If the edge of each nanocube corresponds to 15 unit-cell lengths, what is the effective radius of a molybdenum atom in these structures? b. What is the density of each molybdenum nanocube? c. How many Mo atoms are in each nanocube? 11.122. In one form ofthe fullerene C60, molecules of C60 form a cubic closest-packed array of spheres with a unit-cell edge length of 1410 pm. a. What is the density of crystalline C6/ b. What is the radius of the C60 molecule? c. C60 reacts with alkali metals to form M3C60 (where M = Na or K). The structure ofM3C60 contains cubic closest-packed spheres of C60 with metal ions in holes. If the radius of a K+ ion is 138 pm, which type of hole is a K+ ion likely to occupy? What fraction of the holes will be occupied? d. Under certain conditions, a different substance, M6C60, can be formed in which the C60 molecules have a bcc structure. Calculate the density of a crystal of M6 C 60. 11.123. The center of Earth is composed of a solid iron core within a molten iron outer core. When molten iron cools, it crystallizes in different ways depending on pressure. It crystallizes in a bcc unit cell at low pressure and in an hcp array at high pressures like those at Earth's center. a. Calculate the density ofbcc iron given that the radius of iron is 126 pm. b. Calculate the density ofhcp iron given a unit-cell volume of 5.414 X 10-23 cm3. *c. Seismic studies suggest that the density of Earth's solid core is only about 90% of that of hcp Fe. Laboratory studies have shown that up to 4% by mass of Si can be substituted for Fe without changing the hcp crystal structure. Calculate the density of such a crystal. 11.124. An alloy with a 1:1 ratio of magnesium and strontium crystallizes in the bcc CsCI structure (see Problem 11.77). The unit-cell edge ofMgSr is 390 pm. What is the density ofMgSr? 11.125. Gold and silver can be separately alloyed with zinc to form AuZn (unit-cell edge = 319 pm) and AgZn (unit-cell edge = 316 pm). The two alloys have the same crystalline structure. Which alloy is more dense?


*11.126. Removing two electrons from cyclo-Ss yields the dication cyclo-Ss2+. Will all of the sulfur atoms be in one plane in the Ss2+ cation? 11.127. The packing efficiency for a unit cell can be calculated by the following equation: Packing efficiency (%) =

volume occupied by spheres If' II va ume 0 Ullit ce

X

100

What is the packing efficiency in a simple cubic unit cell? 11.128. Using the equation in Problem 11.127, calculate the packing efficiency in body-centered cubic unit cells. 11.129. Manganese steels are a mixture of iron, manganese, and carbon. Is the manganese likely to occupy holes in the austentite structure or are manganese steels substitutional alloys? 11.130. Aluminum forms alloys with lithium (LiAI), gold (AuAI), and titanium (AI3Ti). Based on their crystal structures, each of these alloys is considered to be a substitutional alloy. a. Do these alloys fit the general size requirements for substitutional alloys? The atomic radii for Li, AI, Au, and Ti are 152, 143, 144, and 147 pm, respectively. b. If LiAl adopts the bcc CsCI structure (see Problem 11.77), what is the density ofLiAl? *11.131. The aluminum alloy Cu3Al crystallizes in a bcc unit cell. Propose a way that the Cu and AI atoms could be allocated between bcc unit cells that is consistent with the formula of the alloy.

I

567

Silicon, Silica, Silicates, Silicone: What's in a Name? 11.132. Write a balanced chemical equation for each of the following reactions: (a) silicon dioxide reacts with carbon; (b) silicon reacts with chlorine; (c) germanium reacts with bromine. 11.133. Write balanced chemical equations for each of the following reactions: (a) tin reacts with chlorine to make a tetrahalide; (b) lead reacts with chlorine to make a dihalide; (c) silicon(IV) chloride is heated above 700°C. 11.134. Calcium silicide (CaSi) has a structure consisting of graphitelike layers of Si atoms with Ca atoms between the layers. In an X-ray diffraction analysis of a sample of CaSi2, X-rays with a wavelength of 154 nm produced signals at 28 = 29.86°,45.46°, and 62.00° for n = 2, 3, and 4. a. What is the distance between the Si layers? b. If the Ca2+ ion lies exactly halfway between the layers, what is the Ca-Si distance? c. CaSi2 is sometimes used as a "deoxidizer" in converting iron are to steel. Explain how CaSi2 might fill this role. (Hint: What is the most common oxidation state of Si?) 11.135. The ionization energies and electron affinities of the group 14 elements have values close to the corresponding average values for all the elements in their rows in the periodic table. How does this "average" behavior explain the tendency for the group 14 elements to form covalent bonds rather than ionic bonds?

Organic Chemistry: Fuels and Materials

A LOOK AHEAD: The Stuff of Daily Life

12.1 Carbon: The Scope of Organic Chemistry

12.2 Hydrocarbons 12.3 Sources of Alkanes 12.4 Alkenes and Alkynes 12.5 Aromatic Compounds 12.6 Alcohols, Ethers, and Reformulated Gasoline

12.7 More Oxygen-Containing Functional Groups: Aldehydes, Ketones, Carboxylic Acids, Esters, Amides

12.8 Hydrocarbon Fuels from Biomass

OLYMPIC POLYMERS The skis and body armor used by ski racers are made of synthetic organic polymers.

A Look Ahead: The Stuff of Daily Life

A LOOK AHEAD: The Stuff of Daily Life Think for a moment about the stuff of everyday life. We drive cars powered by gasoline or diesel fuel. We cook chicken on a grill that burns propane. We live in homes heated directly by the burning of natural gas or fuel oil, or in electrically heated homes, for which most electricity is produced by the combustion of a fossil fuel. All of these fuels are organic compounds, and they are used as fuels because their combustion reactions with oxygen to produce CO2 and H20 are highly exothermic. The great majority of medicines used to treat everything from headaches to strokes, to control diabetes, and to combat cancer are organic compounds. Whether they are natural products derived from living plants or animals, or synthetic materials prepared in the laboratory, the products of the pharmaceutical industry are compounds of carbon. Your favorite soft T-shirt is made of cotton; the soles of your running shoes are made of a synthetic rubber. Cotton is an organic natural fiber produced by a plant. Because of the composition, shape, and orientation of the large molecules that form cotton fibers, the material is soft and absorbent. Synthetic rubber is a polymer, manufactured from compounds of carbon and hydrogen derived from crude oil. Because of the size, shape, and orientation of the polymer molecules in synthetic rubber, the material is springy, nonabsorbent, and resistant to abrasive contact with surfaces. Fans at a hockey game sit on molded plastic seats and watch goalies wearing helmets of impact-resistant Kevlar. We pour sodas out of plastic bottles and drink them from Styrofoam cups while eating sandwiches wrapped in plastic wrap. Plastics, Kevlar, Styrofoam, and cling wrap are made from compounds derived from crude oil. The plastic seat is strong, light, and capable of bearing considerable weight without breaking. The goalie's helmet must be light-weight and able to withstand the impact of a hockey puck moving at speeds close to 100 miles per hour. Soda bottles must resist punctures and prevent CO2 from escaping, the Styrofoam cup must be a poor conductor of heat so that beverages in it stay hot or cold, and the sandwich wrap must be flexible and prevent oxygen from reaching the sandwich. All these properties can be designed into the materials at the molecular level, because the physical properties of these materials are directly linked to their structure and composition. Apart from our bones and teeth, and the water and electrolytes that form the basis of bodily fluids, we humans are all composed of carbon compounds. The great bulk of the food we eat is made up of carbon-containing molecules, too. Compounds of carbon are everywhere, and they are so varied in size, shape, and properties that an entire field within chemistry is devoted to their study. This chapter is a brief introduction to that field, which is known as organic chemistry. A knowledge of organic chemistry is fundamental and essential for a scientific understanding of fuels, foods, pharmaceutical agents, plastics, fibers, living creatures, plants, indeed almost everything we are, almost everything we need to survive, and almost everything we have produced that makes our lives easier in the modern world. This chapter introduces a systematic study of the incredible range of organic compounds, which are classified in accordance with their structure, reactivity, and physical properties. We also discuss the relationships between the structures of organic compounds and their function.

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Carbon: The Scope of Organic Chemistry

The designation "organic" for carbon-containing compounds was once limited to substances produced only by living organisms, but that definition has been broadened for two fundamental reasons. First, since Friedrich Wohler (1800-1882) discovered how to prepare urea in the laboratory "without the intervention of a kidney" (1828), scientists have learned to synthesize many materials in vitro that were previously thought to be the products only ofliving systems. Second, chemists have also learned how to synthesize many materials based on carbon that have never been produced in vivo. The study of organic chemistry today encompasses the chemistry of all carbon compounds, regardless of their origin. Urea

Families Based on Functional Groups

N

CONNECTION In Chapter saw that compounds with the same formula but different arrangements differed in physical properties such point and boiling point.

10, we chemical of atoms as melting

The term in vivo is Latin for "in life." When applied to the origin of a molecule, it means the molecule was produced in a living system. The term in vitro literally means "in glass" and refers to materials produced in a test tube, that is, in the laboratory. Organic chemistry is the study of the compounds of carbon, in which carbon atoms are bound to other carbon atoms, to hydrogen atoms, as well as to atoms of other elements. A functional group is an atom or group of atoms within a molecule that imparts characteristic properties to the organic compounds containing that group. The symbol R in a general formula stands for an organic group that has one available bond. It is used to indicate the variable part of a structure so that the focus is placed on the functional group. Polymers or macromolecules are very large molecules with high molar masses. Polymer literally means "many units."

Carbon is an integral part of a vast number of compounds because of its ability to bond to itself as well as to other elements in several different ways. Much of the variety in the chemistry of carbon arises from a carbon atom binding to another carbon atom, which in turn binds to another, to make compounds containing a few to many thousands of carbon atoms. Additional variety is introduced because alternative arrangements are possible for the carbon atoms in these molecules; they do not simply form long, straight chains. Just as we saw in Chapter 9, threedimensional structure is as important as composition in determining physical and chemical properties. The combination of wide-ranging structural variety for a given atomic composition with almost unending variability of composition creates a scenario in which the number of existing and possible carbon compounds vastly surpasses the possibilities for any other element. Managing the wealth of information about millions of organic compounds requires some organizing concepts, and scientists have developed two major principles to aid them in this task. For the first, recall that the periodic table groups elements in families on the basis of similar chemical and physical properties. Using the table, we can predict the behavior of elements by using our knowledge of periodic trends. Chemists handle the overwhelming number of organic compounds in a similar fashion by grouping them in families or classes of structure of compounds defined on the basis of functional groups-subunits that confer upon molecules particular chemical and physical properties. We have already discussed some classes of compounds and functional groups: alkanes, composed of carbon and hydrogen atoms connected by single bonds; the -OH group in alcohols; the -NH2 group in amines; and the -COOH group in carboxylic acids. In this chapter, we deal with these and the other groups summarized in Table 12.1. When functional groups are discussed, it is conventional to use the symbol R to indicate the remaining hydrocarbon part of the molecule so that we can focus on the identity of the functional group. Unless otherwise specified R stands for any other organic group(s). It literally means the rest of the molecule.

Small Molecules, Oligomers, and Polymers A second organizing principle for organic compounds is based on a physical property as opposed to a chemical one. Scientists divide the universe of carbon compounds into one of two categories based simply on size. Small molecules, with molar masses typically less than 500-1000 glmol, are one category; polymers or macromolecules are the second. These size boundaries are somewhat arbi-

12.1

TABLE 12.1

Name

Alkane

Alkene

Alkyne

Aromatic

Alcohol Ether

I

Carbon: The Scope of Organic Chemistry

I 571

Functional Groups of Organic Compounds Structural Formula of Group

Example and Name

R-H

CH3CH2CH3 H

\ / C=C / \

H

0

0

C

C

"---H

HC/ 3

0

0

C

C

"---R

H C/ 3

"---CH

0

C

C

3

II

"---OH

H C/ 3

Acetic acid "---OH

0

0

C

C

C

"---H

Acetone

0

II

"---OR

HC/ 3

Methyl acetate "---OCH

3

0

II

R/

Acetaldehyde

II

0 Amide

Diethyl ether

II

II

R/

Ethanol

CH3CH2OCH2CH3

II

Ester

Benzene

CH3CHPH

II

R/

(ethyne)

0

II

Carboxylic acid

Acetylene ----i

R-O-R

R/

(ethene)

H

---

R-OH

Ketone

Ethylene

H-C-C-H

,g,O R/

H

\ / C=C / \

-C-C-

Aldehyde

Propane

II

"---NH

2

-

H C/ 3

C

Acetamide "---NH

2

R-NH2 Amine

R-NHR

H3C-NH2

Methylamine

R-NR2

trary, and in between small molecules and polymers is the realm of oligomers, which are materials that are not quite polymers but do contain multiple structural units. We have already seen several examples of polymers in previous chapters: poly(vinylchloride) used to make drain pipes and other building materials; Teflon used in nonstick cookware and lubricants; cellulose from plants used to make paper. Polymers are large molecules with high molar masses ranging from several thousand to over 1,000,000 glmol. Polymers are composed of small structural units called monomers.

Oligomers are molecules that contain a

few mers; "oligos" is Greek for "few." The boundary between small molecules and polymers is not distinct, and oligomers occupy the middle ground. Monomers are small molecules that bond

together to form polymers. The root word meros is Greek for "part" or "unit," so a monomer is one unit.

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Synthetic polymers are macromolecules

made in the laboratory and often produced industrially for commercial use.

Fuels and Materials

The properties of polymers depend on the functional groups in the monomers from which they are made, but the relationship of their size and shape to their behavior distinguishes them as a class from small molecules. Polymers may be formed from more than one type of monomer unit, they may have more than one functional group, and they may have shapes other than long, straight chains. However, the single feature that distinguishes them as a class is their large size.

Small Molecules versus Polymers: Physical Properties Polymers are extraordinarily useful materials. Much of what we say about their behavior in this chapter reflects the influence of composition and structure on their properties. With that concept in mind, let's first discuss the difference between small organic molecules and organic polymers. For now, we restrict our discussion to synthetic polymers made in the laboratory. We have already seen examples of many important compounds made of small organic molecules. All of them have well-defined properties: for example, they have constant composition, and their phase transitions take place at well-defined temperatures (their melting points and boiling points). These features are typical characteristics of pure samples of substances made of small molecules. ~ CONNECTION We have already On the other hand, many commercially produced polymers are not materials introduced many organic molecules in with constant composition and well-defined properties. Most synthetic polymers Chapters 4 through 10: acetic acid; methane, are really mixtures of very similar, large molecules that differ only in the number ethane, and other hydrocarbons as fuels; of monomers in their chains. Their physical properties depend heavily on the range acetylene used in welding; ethanol and other alcohols. and distribution of molecular weights of the molecules in the bulk sample. Figure 12.1 shows two examples of common items made from synthetic polymers. In Chapter 10 we discussed the influence of mass and intermolecular attractive forces on melting and boiling points. Polymers are made of molecules so large that a molecule cannot take on sufficient kinetic energy to enable it to enter the gas phase. Polymers typically do not have well-defined melting points and boiling points. As temperature increases, some polymers gradually soften and become more malleable. Some may eventually melt and become very viscous liquids, while others remain solids up to temperatures at which their covalent bonds break, causing the material to decompose. For example, plastic grocery bags are made of polyethylene, the monomer unit for which is ethylene (CH2 = CH2). The melting point of ethylene is -169°C; its boiling point is -104°C. Polyethylene is a tough and flexible solid at room temperature. It softens gradually over a range of temperatures from 85°C to 110°C, and at higher temperatures tends to decompose into molecules oflow molar-mass organic gases. For the purposes of our initial discussion, it is sufficient for you to understand that matter composed of polymers has a physical behavior very different from matter composed of small molecules, and that this difference is mostly because the molecules are so much larger. No element rivals carbon for the sheer number and variety of compounds it forms. From small molecules to macromolecules, carbon is the element beyond all others that builds the fabric of our lives. In this chapter, we use two means of clasFIG U R E 12.1 Items made of synthetic polymers are ubiquitous in modern society. sification to describe the huge array of carbon compounds. We

12.2

start with small molecules and define the physical and chemical properties of most of the important families of compounds based on functional groups. We then describe the polymers that arise when monomers containing these functional groups combine to make materials ranging from plastic wrap and coffee cups to artifical grass and drain pipes.

I

Alkanes are hydrocarbons bonds are single bonds.

Hydrocarbons

The first three functional groups discussed here define the families of compounds called alkanes, alkenes, and alkynes (Table 12.1). We saw examples of these families in Chapter 5 among the compounds in hydrocarbon fuels. Alkanes consist of carbon in its most reduced state. Methane, propane, octane, and other alkanes are the principal ingredients in several common fuels. When they are converted from alkanes to carbon dioxide and water during complete combustion, they give off large amounts of energy in the process. All alkanes, alkenes, and alkynes have only carbon and hydrogen in their structures. The three classes are distinguished by the types of carbon-carbon bonds in their molecules: alkanes have only carbon-carbon single bonds; alkenes have one or more carbon-carbon double bonds; alkynes have one or more carbon-carbon triple bonds. The double and triple bonds in alkenes and alkynes (between Sp2 and sp hybridized carbons, respectively) are the functional groups that differentiate them from alkanes, which have only Sp3 hybridized carbon atoms. Alkanes are also known as saturated hydrocarbons because they have the maximum ratio of hydrogen atoms to carbon atoms in their structure. The generic molecular formula of alkanes is C"H2n+2, where n is the number of carbon atoms per molecule. The hydrogen-to-carbon ratio in alkanes [(2n + 2)/n] is as large as it can be, given the limit of four bonds per carbon atom. The influence of the double and triple bonds on the hydrogen-to-carbon ratio can be seen by looking at similar molecules in the three families. The two-carbon alkane is ethane, CH3 -CH3' in which the H-to-C ratio = 3:1; the simplest alkene is ethene (more commonly in which the H-to-C ratio = 2:1; the simplest known as ethylene), CH2=CH2, alkyne is ethyne (more commonly known as acetylene), CH-CH, in which the H-to-C ratio = 1:1. Alkenes and alkynes are unsaturated hydrocarbons, because their double or triple bonds can react with hydrogen to incorporate more hydrogen into their molecular structures. The reaction in which hydrogen is added to a molecule is called hydrogenation. No general formula can be written to describe the hydrogen-to-carbon ratios in alkenes and alkynes, because that ratio will vary depending on how many unsaturations are in each molecule. However, all unsaturated hydrocarbons react with hydrogen and can be converted into saturated hydrocarbons (Figure 12.2). Hydrogenation reactions result in the addition of one or more molecules of hydrogen to each molecule of unsaturated hydrocarbon. A molecule with one double bond is described as having one unsaturation and combines with one molecule of H2; a molecule with one triple bond has two unsaturations and combines with two molecules of H2. If a molecule had one double bond and one triple bond, it would have three unsaturations. This feature is referred to as the degree of unsaturation, and determining the degree of unsaturation of an unknown compound using a hydrogenation reaction can help identify its structure.

573

in which all the

Alkenes are hydrocarbons containing one or more carbon-carbon double bonds. Alkynes are hydrocarbons containing or more carbon-carbon triple bonds.

~

I

Hydrocarbons

one

A saturated hydrocarbon is one in which each carbon is bonded to four other atoms. Alkenes and alkynes are unsaturated hydrocarbons. They contain one or more carbon-carbon double or carbon-carbon triple bonds and therefore contain less than the maximum amount of hydrogen possible per carbon atom. The reaction of an unsaturated hydrocarbon with hydrogen is called hydrogenation.

N

CONNECTION In Chapter 5, we defined hydrocarbons as compounds composed of only carbon and hydrogen atoms.

~

+H2

~

~

Propene

Propane

+

2H2~

Propyne

FIGURE 12.2 Carbon-skeleton

~ Propane structures

of propene, propyne, and propane each contain 3 carbon atoms. One mole of the alkene reacts completely with 1 mole of hydrogen

to produce 1 mole of propane;

propyne reacts completely with 2 moles of hydrogen

to produce 1 mole of propane.

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Distinguishing between Alkanes, Alkenes, and Alkynes We are given 1 mole quantities of three unidentified hydrocarbons, each having 4 carbon atoms in their structures. We are told one is an alkane, one an alkene (with one double bond), one an alkyne (with one triple bond). Design an experiment based on their reactivity with hydrogen to determine which is the alkane, which is the alkene, and which is the alkyne. COLLECT AND ORGANIZE We are asked to classify three hydrocarbons based on their reactivity with hydrogen. We have 1 mole of each compound, and we are directed to use their reactivities with hydrogen gas to classify them.

The alkane is a saturated hydrocarbon; it already contains the maximum number of hydrogen atoms possible in its structure. It does not react with hydrogen. The alkene and alkyne are unsaturated, so they can both react with hydrogen. One mole of the alkene, with one double bond, will react with 1 mole of hydrogen. One mole of the alkyne, with one triple bond, will react with 2 moles of hydrogen.

ANALYZE

We can classify the compounds by measuring the amount of hydrogen gas they consume in a reaction. If we allow 1 mole of each substance to react with 2 moles of hydrogen and measure how much hydrogen is consumed in the process, we can identify the compounds. We pick 2 moles because that is the maximum amount that any of the unknowns require for complete reaction. SOLVE

1 mol alkane

+2

mol H2

no reaction

) 2 mol H2 left

1 mol alkene

+2

1 mol alkyne

+ 2 mol H2 ~ 1 mol alkane + 0 mol H2 left

mol H2 ~ 1 mol alkane

+

1 mol H2 left

We can classify the compounds because the alkane is the unknown that does not react at all with hydrogen, while 1 mole of the alkene consumes 1 mole of hydrogen, and 1 mole of the alkyne consumes 2 moles of hydrogen. T HI N K ABO UTI T Measuring the amount of hydrogen that reacts with a hydrocar-

bon is a way to distinguish saturated from unsaturated hydrocarbons. Unsaturated hydrocarbons may be distinguished if their degree of saturation differs.

Practice Exercise The labels have fallen off two containers in the lab. One label has the structure of compound A printed on it; the other, the structure of compound B. Can you distinguish between the two compounds in the unlabeled bottles using hydrogenation reactions? What are the structures of the products of the hydrogenation reactions in both cases? CH -CH 3

-CH=CH-CH

2

CH-CH

3

Compound A

CH3-CH=CH-CH=CH-CH CompoundB (Answers


CH2

12.2

CONCEPT

I

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TEST

Could we use information from hydrogenation reactions to distinguish a hydrocarbon with one triple bond from a hydrocarbon with two double bonds? (Answers to Concept Tests are in the back of the book.)

Alkanes: Physical Properties and Structure Alkanes are also known as paraffins, a name derived from the Latin meaning "little affinity." This is a perfect description of the alkanes, which tend to be rather unreactive compared to the other hydrocarbon families. Despite their lack of reactivity, alkanes are compounds of great importance because of their use as fuels, oils, and lubricants. "Unreactive" may not immediately seem like the correct term to apply to compounds that are fuels, but alkanes do not react readily, even with oxygen. Most fuels need some source of energy, such as a spark from a spark plug in an engine, to initiate combustion. In Chapter 10 we discussed how to predict relative boiling points of compounds based on molar mass and the strengths of intermolecular forces. In doing so, we looked at a series of alkanes as a function of their molecular weight and their shape. Table 12.2 contains a data set for a homologous series. A homologous series of alkanes consists of compounds in which one member differs from the next by the addition of a -CH2 - unit, called a methylene group. The terminal -CH3 group is called a methyl group. You can think of a methyl group as simply

TABLE 12.2

Melting Points and Boiling Points for Selected Alkanes

Condensed Structural Formula"

j

CH3CH2CH3

-

-190

Normal Boiling Point (OC)

-42

-138

-0.5

CH3(CH2)3CH3

-130

36

CH3(CH)4CH3

-95

69

CH3(CH2)sCH3

-91

98

-57

126

-54

151

-10

216

28

316

37

343

72-75

na

72-76

na

Gaseous fuels

CH3(CH)2CH3

I

CH3(CH)SH3 ,

Melting Point (OC)

Use

I

Gasoline

-

CH3(CH2),CH3 CH3(CH2),oCH3 through CH3(CH)16CH3

Diesel fuel and heating oil

CH3(CH)ISCH3 through CH3(CH2)32CH3

Paraffin candle wax

CH3(CH2)34CH3 and higher homologs

Asphalt

:

~

~--

"See Figure 12.3. na = not available; compound decomposes before boiling at 1 atm pressure.

-

I

I

A homologous series is a set of related organic compounds that differ from one another by the number of subgroups, such as -CH2-, in their molecular structures. A methylene group (-CH2-) is a structural unit that can make two bonds. A methyl group (-CH) is a structural unit that can make only one bond.

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A straight-chain alkane is a hydrocarbon in which the carbon atoms are bonded together in one continuous line. Linear hydrocarbon chains have a methyl group at each end with methylene groups connecting them.

N

CONNECTION InChapterlOwe identified dispersion forces as the main attractive force between molecules such as the hydrocarbons, and we related these to the differences in boiling points as a function of linear or branched structures.

Fuels and Materials

methane (CHwith ) one hydrogen atom missing. All of the alkanes in Table 12.2 4 are straight-chain alkanes with two terminal methyl groups and a continuous zig-zag chain with an ever-increasing number of methylene groups. Straightchain alkanes are identified by "n-" (for normal alkanes) preceding the chemical name. As we will see in the next section on drawing organic molecules, linear chains of Sp3 hybridized carbon atoms are kinked because of the carbon-carbon bond angles. The term straight-chain alkane simply means a continuous, linear sequence of carbon atoms. The data sets given in Table 12.2 show similar trends in melting and boiling points: as alkanes increase in molar mass, their melting and boiling points increase. This trend in physical properties is typical of all homologous series.

Drawing Organic Molecules It soon gets tedious to write Lewis structures for alkanes or any organic molecule. Figure 12.3 shows examples of several systems chemists use to convey structures in organic chemistry. For example, a condensed structural formula does not show individual C-H or single C-C bonds the way Lewis structures do, as illustrated below for n-octane: H H H H H H H H Lewis structure:

I I

I I

I I

I I

I I

I I

I I

I I

H-C-C-C-C-C-C-C-C-H H H H H H H H H

Alkene

Alkane

(a)

H

H

H

H

H

H

H

H

H

H

I I I H-C-C-C-C-C-H I I I

I I

Alkyne

I I

H

H

H

H

H

H

I H-C==C-C-C-C-H I

I I

(b)

(c)

H

\

(d)

H

/

C=C / H

\/ CH

CH '-....2 CH

2

3

(e) FI G U R E 12.3

These structures

(a) Lewis structures;

of alkanes, alkenes, and alkynes illustrate common ways to represent molecular structures:

(b), (c), and (d) are forms of condensed structural

formulas; and (e) carbon-skeleton

structures.

I I

12.2

Condensed structures can be shortened even further by indicating the number of times a particular subgroup is repeated in a structure, such as the -CH2 - groups in octane:. CH3(CH2)6CH3 The numerical subscript after the parenthetical methylene group means that 6 of these groups separate the 2 methyl groups. Eight carbon atoms form the backbone of this molecule. Another common short-hand system does not even use the symbols of the elements. Short, connected lines symbolize the carbon-carbon bonds of a molecule in a carbon-skeleton structure, as shown here for butane: Condensed strucutural formula for butane: Carbon-skeleton

CH3CH2CH2CH3

structure for butane:

or CH3(CH2)2CH3

~

The group at the end of a single line in a skeletal structure is a -CH3 group and the group at the intersection of two single lines is a -CH2group. Using these conventions, octane is:

Carbon-skeleton structures are favored for drawing the simplest representation of a molecule and especially to indicate the bond angles between the carbon atoms. VSEPR theory and valence bond theory both predict a tetrahedral bond orientation around each carbon atom. The complete three-dimensional structure of butane showing all carbon and hydrogen atoms is

The three-dimensional structure is much more complicated to draw but the same information about the carbon framework is conveyed by the carbon-skeleton structure. Hydrogen atoms are not shown in the skeletal structure because all carbon atoms are known to make four bonds, and any bonds not shown are understood to be C - H bonds. In the carbon-skeleton structures of the alkanes in Figure 12.3, all the carbon atoms are drawn in the plane of the paper, but some hydrogen atoms are understood to be either in front of (solid wedge) or in back of (dashed wedge) that plane, as shown above in the more complete threedimensional structure of butane.

Drawing Structures Write a condensed structural formula and draw the carbon-skeleton structure for the 3-carbon straight-chain hydrocarbon propane and the 12-carbon straightchain hydrocarbon n-dodecane. COL L E C TAN D 0 R G A N I Z E Condensed structural formulas do not show C - H or

single C-C bonds, and the carbon-skeleton structures use short lines to represent the carbon-carbon bonds in molecules.

I

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The compounds are straight-chain atoms will be in one continuous line.

ANALYZE

hydrocarbons,

so all the carbon

Propane only has 3 carbon atoms, and there is only one way to connect them. Two short straight lines suffice for the skeleton structure; both ends are methyl groups and there is one methylene group in the middle.

SOLVE

Condensed structural formula: CH3CH2CH3

Carbon-skeleton

structure:

~

n-Dodecane is a continuous chain of 12 carbon atoms. There must be 2 methyl groups at the ends with 10 methylene groups between them. Condensed structural formula:

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 or CH/CH2\oCH3

Carbon -skeleton structure: ABOUT IT When drawing structures, we pick the style that best illustrates the point we want to make.

THINK

Practice

Exercise

CH3(CH)4CHV

Draw the carbon-skeleton structure of n-hexane, and write two condensed structural formulas of n-heptane.

n-Heptane

Structural Isomers The family of alkanes would be huge ifit consisted of only straight-chain molecules. However, another structural possibility arises that makes the family even larger. We first saw this possibility in Chapter 10 when we compared the boiling points of compounds having the same molecular formula but different shapes. Butane, for example, has the formula C 4HJO'but the straight-chain arrangement of atoms TABLE 12.3 Comparing Structural Isomers we discussed (Table 12.3, column 1) is n-Butane 2-Methylpropane not the only one possible. An alternative structure (Table 12.3, column 2) is not Condensed structural formula: a straight chain but rather is branched. or A branch is a side chain attached to the main carbon skeleton. Note that the formula of the side chain (a methyl group in this structure) is written in parentheses Carbon-skeleton structure: in the condensed structural formula. The formula of both structures is C 4HJO' but they represent different -138 -160 Melting point CC) compounds with different properties. In recognition of these differences, the two Normal boiling point CC) o -12 compounds have different formal names 0.5788 0.5934 Density of gas (giL) (n-butane and 2-methylpropane), but

y

I

12.2

the issue for you is not to be able to name them but to be able to recognize them as structural isomers: compounds with the same formula but with atoms that are connected in different ways. This difference makes them chemically distinct species. 2-Methylpropane is an example of a branched-chain hydrocarbon, which means that its molecular structure contains at least one side chain. To determine whether or not two condensed structural formulas or carbonskeleton structures represent hydrocarbons that are identical or structural isomers of each other, follow these steps: • Translate the structural information into molecular formulas and compare them. If they are different, then the compounds are different . • If their molecular formulas are the same, then compare their structures. You may have to reverse or rotate the structures to determine whether they really are structural isomers. For example, structures 1 and 2 below may seem to be different:

(1)

(2)

1800 rotation

After rotation

However, if we rotate structure (2) 180°, we see that it is the same as structure (1). The two represent the same compound: 2-methylbutane. To make it easier to compare hydrocarbon structures, the longest chain is drawn from left to right with any branches drawn above or below it. To determine whether two structures are identical, draw the structures with the longest chain horizontal, and then check to see whether the same side chains are attached at the same positions along the longest chain. For example, structures (1) and (2) both have a four-carbon chain with a methyl group on the second carbon when we number the carbon atoms starting with the end closer to the side chain.

Recognizing

Structural

Isomers

Do the two structures in each set describe the same compound, two isomers, or compounds with different molecular formulas?

(a)

(b)

(c)

II II

I

Hydrocarbons

I 579

Structural isomers have the same formula but different structures. They are different compounds and have different chemical and physical properties. A branched-chain hydrocarbon is an organic molecule in which the chain of carbon atoms is not linear.

580

I Chapter Twelve

I

Organic Chemistry:

Fuels and Materials

COL L E C TAN D 0 R GA N I Z E We are asked to determine if the structures represent

compounds that are identical (same formula, same connectivity of the atoms), are isomers (same formula but different connectivity), or are compounds with different formulas. In each pair, we check first to see if the formulas are the same. That means we count the number of carbon atoms and hydrogen atoms. If they are different, they represent different compounds. If they are the same, we may have one compound drawn two ways, or two isomers with different structures. If the carbon skeletons of the molecules are the same, they represent the same hydrocarbon.

ANALYZE

SOLVE

a. The formulas are the same: C7H16• Converting the condensed structural formula of the compound on the left to a carbon-skeleton structure:

we get a structure that is identical to the one on the right. Therefore the formula and structure represent the same molecule. b. Both have the same formula: CSH1S' Both have six carbon atoms in the longest continuous chains. Both structures have a methyl group attached to the second carbon from one end of the chain, and another methyl group is attached to the third carbon atom from the other end. Therefore, they must describe the same compound. c. The first structure contains nine carbon atoms; the second, eight. Therefore, they have different formulas and represent different compounds. When we draw structures of organic compounds, such as carbon-skeleton structures, we usually draw them with the longest carbon sequence horizontal to make it easy to compare structures.

THINK

ABOUT

IT

Practice Exercise Do the two structures in each set describe the same compound, two isomers, or compounds with completely different formulas?

(a)

(b)

(c)

12.2

CONCEPT

I

Hydrocarbons

I 581

TEST

When the heats of combustion of the isomers in Table 12.3 are estimated using bond energies, the two isomers have the same value. The experimentally determined values of heat of combustion differ. Why?

Naming Alkanes Now that we know how to draw the structures of alkanes, let's look at a few simple principles for naming them systematically. Since the system follows the same pattern for all families of organic compounds, we begin with rules for the alkanes and will add a few additional rules when we discuss alkenes and alkynes. These rules are called IUPAC rules after the International Union of Pure and Applied Chemistry, the organization that defined them. Appendix 7 contains a more extensive treatment of nomenclature for organic compounds that you may consult as required. The first four members of the alkane family are methane (1 carbon atom: a Cl alkane), ethane (2 carbon atoms: a C2 alkane), propane (a C3 alkane), and butane (a C 4 alkane). The names of alkanes with more than 4 carbon atoms are derived from the Latin or Greek prefix for the number of carbon atoms per molecule followed by -ane. Table 12.4 lists the prefixes for Cl through CIO alkanes. Molecules with more than 10 carbon atoms in their parent structure are named with the appropriate Latin prefix for the number: Cll is undecane, C12 dodecane, and so forth. Because every alkane larger than propane has structural isomers, we need an unambiguous way to name specific structures. We have already introduced the use of the prefix n- before a name to indicate a straight-chain alkane. To name branched-chain alkanes, follow these steps to build up a systematic name: 1. Select the longest continuous chain and use the prefixes in Table 12.4 to name this as the parent structure. For the structure that follows, the parent name is pentane because the longest chain is 5 carbon atoms long.

TABLE 12.4

Prefix

Prefixes for Naming n-Alkanes

Formula

Name --

Meth-

CH4

Eth-

CH3CH3

Ethane

Parent structure

Prop-

CH3CH2CH3

Propane

Branch

But-

CH3(CH')2CH3

Butane

Pent-

CH3(CH2)3CH3

Pentane

Hex-

CH3(CH')4CH3

Hexane

Hept-

CH3(CH2\CH3

Heptane

Oct-

CH3(CH)SH3

Octane

Non-

CH3(CH2),CH3

Nonane

Dec-

CH3(CH')sCH3

Decane

2. Identify the branches and name them with the prefix from Table 12.4 that defines the number of carbons in the branch; append the suffix -yl to the prefix. A -CH3 group is a methyl group, a -CH2 CH3 group is an ethyl group, and so forth. The structure in step 1 has a -CH3 (methyl) group as a branch; the name of the branch comes before the name of the parent structure, and the two are written together as one word: methylpentane.

--

Methane

I I I :

582

I Chapter Twelve

I

Organic Chemistry:

Fuels and Materials

3. To indicate the point where the branch is attached, we number the carbons in the parent chain so that the branch (or branches, if more than one is present) has the lowest possible number. In the case of the molecule in step 1, we start numbering from the left so that the methyl group is on carbon atom 2 in the parent chain.

1

3

5

~ We show the position of the branch by a 2 followed by a hyphen (2-), in front of the name generated in step 2. The unambiguous name for this compound is 2-methylpentane. 4. If the same group is attached more than once to the parent structure, we use the prefix di-, tri-, tetra- and so forth to indicate the number of groups present. The position of each group is indicated by the appropriate number before the group name. These numbers are separated by commas. The structure that follows is the compound 2,4-dimethylpentane.

CH3 CH( CH)CH2 CH( CH)CH3 1 2 3 4 5 CH3 /CH2 /CH3 "-::CH ~CH

I

I

CH3 1

CH3 3

5

5. If different groups are attached to a parent chain, they are named in alphabetical order. The following compound is 3-ethyl-4-methylheptane

or drawn as condensed structural formulas:

CH3 CH2 CH( CH2 CH)CH( CH)CH2 CH2 CH3 123

4

567

12.2

Note that we begin numbering the parent chain in this structure at the carbon on the far left, in accord with the instruction in step 2 so that the branch locations have the lowest possible numbers. In naming the compound, the side groups are written in alphabetical order. Prefixes are disregarded in determining alphabetical order: ethyl comes before methyl, and ethyl also comes before dimethyl.

CONCEPT

I

Hydrocarbons

Cycloalkanes are ring-containing with the general formula C"H2".

I

alkanes

TEST

A useful way of determining whether structures are the same is to name the compound according to the IUPAC rules. Why is this the case?

Cycloalkanes Alkanes can also form ring structures (Figure 12.4). Compounds with this structure are called cyc1oalkanes. They have the general formula C"H2n, which is different from the general formula for straight-chain alkanes because cycloalkanes have one more carbon-carbon bond and 2 fewer hydrogen atoms per molecule than the n-alkanes with the same number of carbon atoms. Although these compounds with rings are alkanes, and all the carbon atoms in cycloalkane rings are spj hybridized, the compounds are actually unsaturated. In principle, 1 mole of a cycloalkane can react with 1 mole of hydrogen to make 1 mole of a straight-chain compound. In practice, only the three-carbon and four-carbon rings actually react with hydrogen, but all the cycloalkanes are considered unsaturated. The most abundant cycloalkane in the fraction of crude oil used as gasoline is cyclohexane, C6H12. Figure 12.4 shows the condensed structural formula of cyclohexane drawn in two dimensions, in which the C -C -C bond angles appear to be 120°. Because the carbon atoms have spj hybrid orbitals, we expect the angles to be 109.5°, and indeed they are in this molecule. A more accurate representation of the structure of cyclohexane is shown in the adjacent structures, in which the ring is puckered instead of flat and the bond angles are 109.5°. There are two possible pucker patterns: one is called the chair form, and the other is called the boat form. In the boat form, 2 hydrogen atoms across the ring from each other are pointed toward each other. Repulsion between these 2 atoms causes this form to be less stable than the chair form, and so the chair form is the preferred, lower-energy form. Figure 12.4 also shows the carbon-skeleton forms of the three structures.

HC/

Condensed structural formulas

CH2 ~CH

21

1 2

H2C ........../CH2 CH2 Two-dimensional

Carbon-skeleton structures

o

Two-dimensional

FIGURE 12.4 The six-membered its three-dimensional

puckered rings.

Chair

Boat

Chair

Boat

ring of cyclohexane is drawn either flat or in styles that show

~II

CHEMTOUR Structure of

Cyclohexane

~II

583

CHEMTOUR Cyclohexane in 3-D

584

I Chapter Twelve

I

Organic Chemistry:

Fuels and Materials

Ring structures are possible for other alkanes, but smaller rings are less stable than the six-membered one. One of the important features that determines the relative stability of rings is the size of the C-C-C bond angle. For example, cyclopropane (a C3 ring) exists, but it is a very reactive species because the interior

600 Cyclopropane

Cyclobutane

Cyclopentane

angle in an equilateral triangle is only 60°, which is very far from the ideal bond angle of 109.5° for an Sp3 hybridized carbon atom. No puckering is possible in a three-membered ring to relieve the strain in this system, and cyclopropane tends to react in a fashion that opens up the ring and relieves the strain. Rings of four and five carbons (cyclobutane and cyclopentane) are progressively less strained. Rings of six Sp3 hybridized carbons and beyond are essentially the same as straightchain alkanes in terms of bond angle and have no ring strain. Six-membered rings are the most favored, because other thermodynamic factors destabilize rings with seven or more carbons.

~

Sources of Alkanes

As noted in Chapter 5, carbon compounds have been accumulating in Earth's crust since the onset of biological activity. When decaying aquatic plant and animal matter is buried in rapidly accumulating sediments, its carbon and hydrogen content is shielded from oxidation to CO2 and H20. The matter is slowly converted into natural gas and a complex mixture of water-insoluble organic compounds called kerogen. As kerogen is buried by more sediment, increasing pressure and temperature promotes additional chemical transformations. When these transformations take place at depths between about 1 and 4 km below Earth's surface and at temperatures between 75°C and 150°C, the products include crude oil, a mixture of hundreds of compounds composed mostly of carbon and hydrogen. Crude oil contains hydrocarbons with 5 or more carbon atoms in their molecular structures. The hydrocarbons with 1 to 4 carbon atoms are usually found in deposits of natural gas, although they are also dissolved in deposits of crude oil. Kerogen is a mixture of a large number of liquid organic compounds formed during the conversion of plant matter into crude oil in the absence of oxygen. Crude oil is a combustible liquid mixture of hundreds of hydrocarbons and other organic molecules formed under the Earth's surface by the processing of plant matter under conditions of high temperature and pressure.

The Clausius-Clapeyron

equation relates the vapor pressure of a substance at different temperatures to its heat of vaporization.

CONCEPT TEST

The nonpolar constituents of natural gas may dissolve in crude oil, which consists of mostly nonpolar liquid components. As crude oil is pumped to the surface from thousands of meters deep in Earth's crust, how is the solubility of natural gas affected?

Volatility and the Clausius-Clapeyron

Equation

After crude oil has been extracted from the ground, it is separated at refineries into fractions, or cuts, on the basis of the volatilities of the components in the solution. As we learned in Chapter 10, the vapor pressure of a liquid increases as tempera-

12.3

I

Sources of Alkanes

I 585

7.00

In Pvap (PinmmHg)

6.00

FIGU RE 12.5 Plotting the natural log of the vapor pressure (here in torr or mmHg) versus the reciprocal of temperature in K gives a straight line described by the ClausiusClapeyron equation. The graph shows the plot for n-pentane.

5.00

3.25

3.50 3.75 (lIT) X 103 (K-1)

ture increases. Before we can understand what happens when a solution of volatile substances is distilled, we need to examine the vapor pressure of a pure substance as a function of temperature. The relationship between the vapor pressure of a pure substance and absolute temperature is not linear. However, if we graph the natural logarithm of the vapor pressure [In (Pvap)] versus liT, we get a straight line (Figure 12.5). This line can be described by the equation

(1) + C

__

tJ.Hvap R T

In(Pvap) -

where tJ.Hvap is the enthalpy of vaporization, R is the gas constant, and C is a constant that depends on the identity of the liquid. We can solve this equation for C and write it in terms of two different temperatures:

InP(vap,T)

yap

+(tJ.H -Rr;

yap

) =C=lnP (vap,T,)

(tJ.H +-RT2

)

which can then be rearranged to In

I;a p

p

1

,T

)

=

tJ.H vap R ( ~ _ ~)

vap,T2

2

(12.1)

1

Equation 12.1 is called the Clausius-Clapeyron equation, and it can be used to calculate tJ.Hvap if the vapor pressures at two temperatures are known, or to calculate the vapor pressure (P) at any given temperature (T2) or the temperature (T) at any given pressure (P2) if tJ.Hvap and PI and are known. Because the units of tJ.Hvap are typically joules (J) or kilojoules (kJ), the value of the gas constant R used in the Clausius-Clapeyron equation is 8.314 Jlmol· K.

r;

Calculating

Vapor Pressure

At its normal boiling point of 126°C, octane [CH/CH)6CHJ has a vapor pressure of 760 torr. What is its vapor pressure at room temperature (25°C)? The enthalpy of vaporization (tJ.Hvap) of octane is 39.07 kJlmol. COLLECT AND ORGANIZE We are asked to determine the vapor pressure of octane at 25°C. We are given its enthalpy of vaporization and its vapor pressure at its normal boiling point.

N

CONNECTION In Chapter 10 we defined the normal boiling point as the temperature at which the vapor pressure of a liquid reaches 1 atm (760 torr).

586

I Chapter Twelve

I

Organic Chemistry:

Fuels and Materials

We can use the Clausius-Clapeyron equation (Equation 12.1) to answer this question. The temperatures must be in K, and the form of the gas constant R used must be 8.314 Jlmol· K. Because of the units of R, t:..Hvap must be expressed in joules.

ANALYZE

SOL V E The Clausius-Clapeyron

equation relates vapor pressure, temperature,

and heat of vaporization:

I

pp 1 ) In ;a ,T

vap,T,

=

t:..H vap R (

i - i) 2

1

Considering the values we are given, the only unknown in this expression is the vapor pressure at 25°C. Expressed in K, the two temperatures given are

~ = 126"C + 273 = 399 K and the value of t:..H

yap

and

in J' oules is ~ 1000J X -39.07 mol 1~

= 39,070-

J mol

39,070~ In(760 torr) Pyap, T ,

=_ 8314-.~

ffl6l (_1

f

298

*

1_) 399

*

Solving the equation for P2 yields 14.1 torr as the vapor pressure of octane at 25°C. T HI N K ABO UTI T We expect octane to have a low vapor pressure at a tempera-

ture much lower than its boiling point, so this number seems reasonable.

Practice Exercise Pentane [CH3(CH)3CH3] boils at 36"C. What is its molar heat of vaporization in kilojoules per mole if its vapor pressure at room temperature (25°C) is 505 torr?

CONCEPT

TEST

Diesel fuel is made of hydrocarbons with an average of 13 carbon atoms per molecule, and gasoline is made of hydrocarbons with an average of 7 carbon atoms per molecule. Which fuel do you predict has the higher vapor pressure at room temperature?

Volatility at Work: Fractionating Crude Oil

N

CONNECTION distillation in Chapter water. ~II

CHEMTOUR

We saw simple 1 as a way to purify sea

Fractional

Distillation

Fractional distillation is a method of separating a mixture of compounds on the basis of their different boiling points.

A process called fractional distillation (Figure 12.6) is used to separate the components of crude oil. Fractional distillation can separate a mixture of volatile substances because the vapors that arise from a solution of volatile substances are always enriched in the component with the lowest boiling point. If the vapors from a boiling solution are collected and then redistilled, the products of that distillation will be even more enriched in the component with the lowest boiling point. In a fractional distillation apparatus, repeated distillation steps allow components with only slightly different boiling points to be separated from one another. To see how fractional distillation works, let's examine a series of graphs describing the behavior of pure substances and of a solution of two volatile liquids

12.3

I

Sources of Alkanes

I 587

Gas Gasoline

II

Kerosene

Thermometer

Trays

Heating oil

Lubricating

Vapor condensed, collected, and measured

oil

Vapors rise, condense, vaporize, condense ...

,

Fractionating column

Solution being distilled

Crude oil

~-----~f-Asphalts and tars

~ (b)

(a)

FIGURE 12.6 Fractional and (b) on the laboratory they repeatedly

distillation

condense and vaporize.

higher in the column; components scale, the more volatile components components.

is carried out to separate mixtures

scale. In both cases, vapors rise through On the industrial

of the mixture

on (a) an industrial

a fractionating

scale

column where

scale, the more volatile components

rise

are collected at different levels. On the laboratory

distill first, followed by increasingly

The identity of the material distilling

is determined

Heat source

less volatile, higher-boiling

by the boiling point observed.

in simple distillation. The graph in Figure 12.7(a) is an idealized description of the behavior of a pure substance when boiled. The temperature remains constant during the phase change, and the composition of the vapor is constant because the substance is pure. If we carry out a simple distillation of a mixture of two volatile liquids, the graph in Figure 12.7(b) shows that the temperature is not constant during the course of the distillation. The two components co-distill over a range of temperatures. Figure 12.7(c) shows the composition of the vapor phase above a series of boiling solutions with a range of concentrations, and herein lies the key to how distillation can separate mixtures. The vapor (red line) in equilibrium with a solution of two volatile substances is enriched in the lower boiling (more volatile) component. The vapor has a different composition than the liquid (blue line). Now let's carry out a thought experiment to understand how fractional distillation works. Suppose we heat 100 mL of a solution that is 50% by volume n-heptane (bp 98°C) and 50% n-octane (bp 126"C) in a miniature version of the

To power supply

588

I Chapter Twelve

I

Organic Chemistry:

Fuels and Materials

175

t: 150 I

126°

1: 125

B

100

"

75

~

50

0.-

E

25

(1)

(2) Heat added

(a) Distillation of n-octane (1) Octane begins to distill (2) Octane finishes distilling

175

I

~ ~

150 125

:

B

100

1:L ~

uuuu_ u+uuuu_ I

75

:

50 25

I

: (1)

(2) Heat added

(b) Distillation of a mixture of n-heptane and n-octane (1) Solution begins to distill (2) Solution finishes distilling

distillation column in Figure 12.6. The graph in Figure 12.7(c) tells us that the boiling point of the solution will be initially nearly 110°C (point 1 on the graph). The graph also tells us that the vapor that comes from the solution the moment it begins to boil (point 2) is not a 50:50 mixture of n-heptane and n-octane, but rather it is enriched in the component with the lower boiling point: it is about 67% n-heptane and only 33% n-octane. Suppose this vapor rises up in the distillation column, cools, and condenses in the first tray, a process represented by the red arrow from points 2 to 3 in Figure 12.7(c).The liquid in the tray is also 67% n-heptane and 33% n-octane. Continued heating of the column warms the liquid in the tray and it vaporizes. The vapor that initially forms from the solution in the first tray would be even more enriched in the n-heptane: about 80%, according to the coordinates of point 4 in Figure 12.7(c). The vapor rises up to the second tray, where it cools and condenses, and the redistillation process is repeated. If we continue this process of redistilling mixtures with increasing concentrations of n-heptane and then cooling and condensing the vapors they produce in trays higher and higher in the distillation column, we will eventually obtain a condensate that is pure n-heptane in the uppermost tray. If we could monitor the temperature at which vapors condense at the very top of our distillation column, we would see a profile of temperature vs. volume of distillate produced like that shown in Figure 12.8. The first liquid to be produced would be pure n-heptane, which has a boiling point of 98°C. Over time all 50 mL of n-heptane in the original sample would be recovered. As we continued to add heat at the bottom of the column, the temperature at the top would rise to 126°C, signaling that the second component (pure n-octane, bp 126°C) is being collected. Figure 12.8 is highly idealized. Fractional distillation does not work as well as this graph implies, but the graph indicates the best separation one could hope for if the system behaved perfectly. This thought experiment using two alkanes describes the basis of the fractional distillation of crude oil. The cuts taken when crude oil is fractionated are not pure materials but solutions of several liquids that have desirable properties. The cuts are taken over a range of temperatures where mixtures of similar molecules distill together. The most volatile fraction from crude oil after any natural gas is removed is used as gasoline. The next most volatile cut is used as cleaning

FIGURE 12.7 Simple and fractional

140

distillation

130 ~

2"

distillation

120

of a pure substance

constant temperature.

110

of a mixture

oj

takes place at a

(b) Simple distillation

of two volatile substances

place over a range of temperatures.

~ 100

E ~

graphs (a) and (b) describe

the behavior of boiling liquids. (a) Simple

takes

Graph

(c) shows the results of many experiments 90

o 100

50 50

100% Octane 0% Heptane

(c) Boiling points of octanelheptane mixtures (blue curve) and the composition of the vapors produced at those boiling points (red curve).

CD Solution boils at this temperature CD Vapor has this composition

in which the composition

of vapor from a

boiling liquid was analyzed. shows the temperatures given composition

boil. The red line shows the

G) Vapor

condenses, then boils

8) Vapor

has this composition

solutions. The stair-step

0), the dissolution

process is spontaneous

because the result is

a positive change in entropy (I1S

> 0).

644

I Chapter Thirteen

I

Thermodynamics:

Spontaneous

Processes,

Entropy, and Free Energy

related to a flow of heat into, or out of, the solution. If there is no significant heat flow, then there is no significant change in the dispersion of energy in the surroundings. This means there is little change in the entropy of the surroundings. Therefore, the spontaneity of the process (~SuniY > 0) must be linked to the increase in entropy that happens when solute and solvent molecules mix together, dispersing into a larger overall volume (~S sys > 0). Spontaneous processes must result in an increase in the entropy of the universe. No violation of this concept has ever been observed, and it is so widely accepted as true that most people and virtually all scientists have stopped looking for violations. This is why so-called perpetual motion machines (also called freeenergy devices), which store or put out more energy than is put into them, are all hoaxes; everything ultimately runs down. The combinations of ~S sys and ~S surr that produce a positive ~S unlV. are shown on the left side of Figure 13.7 and summarized in mathematical terms as follows: 1. If

~SSYS

2. If ~S

sys

> 0 and

tude that

~SSys

> 0,

~Ssurr

< 0, then ~S

SUff

then

~SUniY

+ ~Ssurr > 0 (Figure

3. If ~Ssys > 0, then (Figure 13.7c).

~Ssurr

>0

(Figure 13.7a).

must be greater than zero, and of such a magni13.7b).

may be less than zero as long as ~Ssys

+ ~Ssurr > 0

Note that the right side of the figure shows various scenarios for nonspontaneous processes (Figure 13.7d-f).

Predicting

the Sign of Entropy Changes

Predict whether ~Ssys is positive or negative for each of these spontaneous processes taking place at constant temperature: a. H20(t') ~ H20(g)

+ HCI(g) ~

b. NH/g)

c. Pb2+(aq)

+ 2CI-(aq)

d. C12H2201l(S)

NH4CI(s)

~ PbCI/s)

Hp) C12H22011(aq)

We are asked to predict whether the entropy change for the system is positive or negative. COL LEe TAN

D 0 R GA N I Z E

ANALYZE

a. A mole of water molecules becomes a mole of vapor molecules, which increases their freedom of motion. b. Two moles of gas form one mole of solid, which results in a decrease in the freedom of motion. c. Three moles of ions in solution form one mole of a solid compound, which represents a decrease in freedom of motion. d. One mole of a solid compound dissolves, forming a mole of molecules dispersed in an aqueous solution. SOLVE

a. ~S

~

>0

b. ~S

~

.

g

J:>.

0

I

,,,

'

r

,

I:

g>.

J:>.

0 I

I

,

I

,

,

I

I:

~

~

(a)

6::

g

(d)

6::

0 I

I

e

I"

,

I:

~

g

0 I

I:

I

~

•

(e)

(b)

FIGURE 13.7 The relationships between I1S,y" I1S,u,,' and I1Sun;v expressed graphically,

~ ~ 6::

g

with spontaneous

~ 0 I

I:

~

I

•

10

"

ulln.

nonspontaneous

~ l:

~I: 0 I

processes on the left and processes on the right. A

reaction or process is spontaneous I

I

,

,

(a) if I1S,y,

and I1S,u" are both positive; (b) if I1S,u"

>

0,

I1S,y, < 0, and II1S,u" 1 > 1-I1S,y,l; or (c) if I1S,y, > 0, I1Sm" < 0, and II1S,y.l > II1S,uJ A reaction or process is nonspontaneous

if

(d) I1S,y, and I1S,"" are both negative; (e) if (c)

(f)

I1S,y, < 0, I1S

> 0, and II1S,y,1 > II1S,u"l; < 0, I1S,y, > 0, and II1S,u,,1 >

M,

or (f) if I1S,u"

II1S,yJ

646

I

Chapter Thirteen

I

Thermodynamics:

The third law of thermodynamics states that the entropy of a perfect crystal is zero at absolute zero. entropy is the entropy change of a substance taken from S = 0 (at T = 0 K) to some other temperature.

Absolute

molar entropy (SO) is the absolute entropy of 1 mole of a substance in its standard state.

Standard

~ CONNECTION In Chapter 5 we defined the standard state of a pure substance in its most stable form at 1 bar (100 kPa) of pressure and some specified temperature, often 25.0°C. However, most thermodynamic data in the U.S. (and those given in Appendix 4) are for the slightly higher pressure of 1 atm.

Spontaneous

Processes,


mD Absolute Entropy, the Third Law of Thermodynamics,

and Structure

The entropy of a substance depends on its temperature. We have seen this relation in terms of the examples given earlier in the chapter; this is also predicted by the kinetic molecular theory: higher temperatures mean higher kinetic energies of particles. Higher kinetic energy means more energy of motion: more vibrational, rotational, and translational motion; more accessible microstates, and therefore more entropy. Conversely, lower temperatures mean less of all these quantities. If we lower the temperature of a substance to absolute zero, in principle all motion ceases. If the particles of a crystalline solid are perfectly aligned and motionless at 0 K, then only one microstate exists, and from Equation 13.1 we know that entropy is zero (S = kIn W = kIn 1 = 0). This situation is addressed by the third law of thermodynamics, which states that the entropy of a perfect crystal is zero at absolute zero (0 K). By perftct we mean that all particles are exactly aligned with one another in a crystal structure (Figure 13.8) and there are no gaps or imperfections in the crystal. Only one arrangement that is perfect can exist. Also, by definition, at absolute zero molecular motion ceases, so no changes in the position of particles occur because there is no movement. By setting a zero point on the entropy scale, scientists can compute absolute entropy values of pure substances. Table 13.1 lists standard molar entropy (SO) values at 298 K for several solids, liquids, and gases. The values for liquid water [69.9 JI(mol· K)] and water vapor [188.8 JI(mol· K)] illustrate an important difference in the entropies of liquids and gases: The molecules in a gas under standard conditions are much more dispersed. On this basis alone we should expect water vapor to have more entropy at the same temperature than liquid water. All of the gases in Table 13.1 have higher standard molar entropies than their corresponding liquids. The entropies of the different phases of a given substance at a given temperature follow the order S so IOd 1

< Slo

TABLE

°d

IqUi

< S gas . Select Standard Molar Entropy Values (1 atm, 298 K)a

13.1

I Formula f--

--

Br/C) Cdiamonis)

~,,(s)

CO(g)

--

152.2

-

CH3CH3(g) CH3OH(g)

5.7

CH3OH(C)

197.7

CO/g)

213.8

H2(g)

130.6

N/g)

-

Methane Ethane

--

CH3CHPH(g)

126.8

-

n-Butane

O,(g)

205.0

CH3(CH2),CH3(f)

HP(g)

188.8

C6H6(g)

HP(f)

69.9

C6H6(f)

NH3(g)

192.3

-

C12H22011(S)

--t

160.7

CH3(CH2)2CH/g)

"Valuesfor additionalsubstancesaregivenin Appendix4.

-
0). b. This reaction describes ammonia gas dissolves of gas to the condensed, a loss in system entropy pack reaction in that it low-entropy solid.

the chemical changes that take place when in water. The key factor here is the loss of 1mole aqueous phase. Therefore this reaction results in (;:"S,y, < 0). This example differs from the coldentails the dissolution of a high-entropy gas, not a

THINK ABOUT IT To determine whether the entropy of a system (S,y) increases when that system is a chemical reaction, we need to evaluate changes in temperature, changes in volume, and changes in the number of independently moving particles that occur as reactants are converted into products.

Practice Exercise Which process or processes result in an increase in entropy of the system? (a) Amorphous sulfur crystallizes. (b) Solid carbon dioxide sublimes at room temperature. (c) Iron rusts.

mD

Calculating Entropy Changes in Chemical Reactions

We can calculate the change in entropy under standard conditions for any chemical reaction (;:..sorxn ) from the difference in the standard molar entropies of m moles of reactants and n moles of products: ;:"So

rxn

I nSOproducts - I mSOreactants

(13.4)

The symbol I represents the sum of all of the entropies of the reactants or products under standard conditions. Each standard entropy SO is multiplied by the appropriate coefficient for that substance in the balanced chemical equation. For example, 2 moles of a reactant or product have twice the entropy of 1 mole. In other words, just as we discovered about ;:"Ho in Chapter 5, entropy is an extensive thermodynamic property that depends on the amount of a substance present or the amount consumed or produced in a reaction. Standard molar entropies for selected substances are tabulated in Appendix 4.

Calculating Changes in Entropy Calculate ;:"S;xn for the dissolution standard entropy values:

NH4NO/s) SO[JI(mol· K)]

151.0

of ammonium

~

nitrate,

NH/(aq)

given the following

+

112.8

AND ORGANIZE We are given a balanced chemical equation for the reaction and asked to calculate the entropy change associated with the reaction. We are also given the standard molar entropies for the species involved in the reaction.

COLLECT

ANALYZE

We can calculate ;:"S;xnusing the given So values and Equation

13.4.

13.4

I


I

651

SOLVE

t:.Sorxn

I nS products O

_ - [ l+He!x

I mS

O

-

(112.8 -- J)

+He!.K

reactants

+ 1 +He!X

(146.4J)] --

+He!.K

-l+He!x--

(151.0J)

+He!.K

= 108.2J1K The value of t:.S:xn is positive, so the entropy of the system increases. T The reaction is the dissolution of a solid salt in water. Because the number of independently moving particles increases, we expect the entropy of the system to increase, so our answer is logical.

T HI N K ABO UTI

NH/(aq)

Practice Exercise Calculate the standard entropy change (t:.S:x) for the combustion of methane gas using the appropriate So values in Appendix 4. Before

+

carrying out the calculation, predict whether the entropy of the system increases or decreases. Assume that liquid water is one of the products. The entropy changes calculated using Equation 13.4 are all entropy changes of the system. As we have repeatedly seen, the surroundings matter as well, even though we tend to treat them as a large heat source or sink, depending on the direction of heat flow during a process. The entropy change experienced by the surroundings of a chemical reaction depends on whether the reaction is exothermic (heat flows into the surroundings) or endothermic (heat flows from the surroundings into the system). If we assume the reactions we are studying are isothermal processes occurring at constant pressure, then the heat flow is reversible (qrJ, allowing us to use Equation 13.2 to determine the change in entropy of the surroundings. Under these conditions, qrey is equal to the enthalpy change (t:.H) of the reaction: (13.5) We can calculate t:.H for the reaction in Sample Exercise 13.3 by using values from Appendix 4 in Equation 5.18 from Chapter 5, which we repeat here:

I n t:.HO[,products - I m t:.H

t:.HOrxn

O

(13.6)

f,reactants

For the dissolution of1 mole ofNH4N03(s)-recall that ammonium nitrate is the solid in a cold pack-Equation 13.6 gives a value of t:.Ho = +26.3 kJlmoi. The process is endothermic. To calculate the entropy change of the surroundings for the dissolution taking place under standard conditions at 25°C (298 K): 3

t:.S

26.3 X 10 J) (1md:) X ( ---md: SUff

T

T

298 K

= -88.2J1K Note that the sign of the entropy change for the surroundings is negative (-88.2 JlK) while that for the system (calculated in Sample Exercise 13.3) is positive (108.2 JlK). Both terms must be considered to determine how the entropy of the universe changes during this process:

t:.Suniy = t:.S

SYS

+ t:.Ssurr

= 20.0JlK

= (108.2 JlK) + (-88.2 JlK)

......... ~

N

....•.••.

CONNECTION In Chapter 5 we defined a change in enthalpy (I1H) as the heat gained or lost in a reaction carried out at constant pressure.

652

I

Chapter Thirteen

~ ~

6:

~§ .:: 0 I

'

'

..

I

Thermodynamics:

' ~.... ,

Spontaneous

Processes,


The entropy change of the universe has a positive sign, so entropy increases. The process is spontaneous even though it is endothermic. For this to be the case, the decrease in entropy of the surroundings must be offset by an increase in entropy of the system (Figure 13.11). Notice two other features of this discussion. First, because the reaction is endothermic, it takes heat energy from the surroundings. As discussed in Chapter 5, the amount of heat that the reaction absorbs equals the amount of heat taken from the surroundings: ~H sys

=

-~H

surr

As we learned earlier in the chapter, the entropy change in the system is not the opposite of the entropy change in the surroundings: FIGURE 13.11 If the surroundings experience

~s

a decrease in entropy, the system

must experience

an increase in entropy that

more than offsets the decrease if the process is spontaneous.

sys

=1=

~S surr

Second, enthalpy changes are typically expressed in kilojoules (kJ) whereas entropy changes in many chemical reactions are routinely expressed in joules per kelvin. For reactions with absolute ~S values of hundreds of JlK and that occur at temperatures of hundreds of kelvins, T~S values, like ~H values, are more conveniently expressed in kJ.

Predicting the Spontaneity of a Reaction

+

Consider the reaction of nitrogen gas and hydrogen gas at 298 K to make ammonia at the same temperature: N2(g)

1

+ 3 H2(g)

~ 2 NH/g)

a. Before doing any calculations, predict the sign of the entropy change for the system. b. Use the data in Table 13.1 and/or Appendix 4 to calculate ~So for the system. c. Use data in Appendix 4 to calculate ~S surr . d. Is the reaction, as written, spontaneous at 298 K and 1 atm pressure? We are asked to predict the sign and calculate the value of the entropy change for a reaction. We are also asked to calculate the entropy change for the surroundings and to determine whether the reaction is spontaneous under the stated conditions. COLLECT

AND

ORGANIZE

(a) We have a balanced equation, so we can consider the volume and number of particles to determine the sign of the entropy change for the system. (b) We need to look up ~So for the reactants and products and use Equation 13.4 to calculate ~So of the system. (c) To calculate ~Ssurr we need to know the heat the reaction gives off to the surroundings. We can calculate that from the heats of formation (~H;) of the reactants and products found in Appendix 4. (d) We can use Equation 13.3 to calculate the total entropy change for the universe and predict spontaneity based on the sign of that value.

ANALYZE

SOLVE

a. Four volumes of gas (4 moles of particles) on the reactant side of the chemical equation are converted into 2 volumes of gas (2 moles of particles) on the product side. We predict that this decrease in volume

13.4

I


and decrease in number of particles will result in a negative value of the entropy change for the reaction. b. Using data from Table 13.1 we calculate the entropy change for the system (the reaction) as ~S:xn using Equation 13.4:

I nS products

~SOrxn

O

_ [

- 2 ffiel

X

I mS reactants O

-

(192.3 J)] _ [ ffiel.

K

1 ffiel

(191.5 J )

X ffiel.

K + 3 ffiel

X

(130.6 J)] ffiel.

K

= -198.7 JlK The entropy change for the reaction is negative, as predicted in part a. c. If we consider the process to be carried out very slowly so that the exchange of heat with the surroundings is reversible, then we can calculate the entropy change of the surroundings from the enthalpy change of the reaction.

~HO sun

= - ~Ho rxn = - [InHO

= - {[2 mol = 92.2 JlK

X ( -46.1

f,products

:~J]-

-

[1 mol

I mHO[,reactants 1 X

(0.0

:~J

+ 3 mol

X

(0.0

:~J]}

heat absorbed by the surroundings

Since we were given the temperature (298 K), we now have all the values we need to calculate the entropy change of the surroundings. Combining Equations 13.2 and 13.6 from the perspective of the surroundings:

~S The surroundings

surr

~H T

= ~

='

92 200J = 309 J/K 298 K

experience an increase in entropy.

d. We can use Equation 13.3 to calculate the total entropy change for the universe and predict spontaneity based on the sign of that value: ~S sys

+

~S sun

~S unlV.

-198.7 JlK

+

309 JlK

1l0JIK

The entropy change of the universe is positive, so the reaction is spontaneous as written. As is always the case, just because a reaction is spontaneous does not mean that it actually proceeds rapidly. Even though the reaction is spontaneous, nitrogen and hydrogen at room temperature and 1 atm pressure do not react without a significant "push" in terms of heat and pressure.

THINK

ABOUT

IT

Practice Exercise Consider the reaction of hydrogen gas and oxygen gas at 298 K to produce water:

a. Predict the sign of the entropy change for this process. b. Use the data in Table 13.1 and Appendix 4 to calculate ~So for the system. c. Use the data in Appendix 4 to calculate ~Ssurr' d. Is the reaction as written spontaneous at 298 K and 1 atm pressure?

I

653

I

654

Chapter Thirteen

I

Thermodynamics:

Spontaneous

Processes,


mD Free Energy and Free-Energy Change So far in this text, we have identified two driving forces that make chemical reactions happen: ~

CONNECTION Determining the enthalpy change of reactions was discussed in detail in Chapter 5.

1. The formation oflow-energy products from high-energy exothermic reactions (!::..H < 0). 2. The formation (!::..S> 0).

g6::

0 I

r

,

"

'!

"

,

c

~

(a)

reactants in

of products that have more entropy than the reactants

A reaction that is both exothermic and has a positive entropy (!::..Ssys> 0) must be spontaneous. The heat flow produced by the exothermic reaction leads to a positive value for!::"S SUff . When!::..S sys and!::..S SUff are both positive, then!::"S umv. is also positive (see Equation 13.3 and Figure 13.12a); according to the second law of thermodynamics, a reaction under these conditions is spontaneous. On the other hand, a reaction that cools its surroundings (!::..Ssurr < 0) because it is endothermic (!::..H > 0) and that also results in a loss of entropy on the part of the system (!::..Ssys < 0) will never be spontaneous, because!::..S unlv. is negative in this case (Figure 13.12b). Figure 13.7 summarizes all the possible combinations of positive and negative entropies of systems and surroundings. The key point to remember is that spontaneous reactions increase the entropy of the universe (!::..Ssys +!::..S SUff > 0; Figure 13.7a-c), whereas nonspontaneous reactions decrease the entropy of the universe (!::..Ssys +!::..S surr < 0; Figure 13.7d-f).

Determining the Entropy Change in the Universe §

-

6::

~ 0 I

r:';

•

i

.

~

,

•

(b)

FIGURE

13.12

The magnitudes

of I1S,y, and I1S,"" determine magnitude

of I1Snn;,

and signs

the sign and

(a) When

I1S,y, and I1S,""

are both positive, I1Snn;v is always positive and the process is spontaneous.

(b) When I1S,y,

and I1S,"" are both negative, I1Sn"iV is always negative and the process is nonspontaneous.

Left unaddressed in this analysis is whether an endothermic reaction with a positive entropy (!::..S > 0) or an exothermic reaction with a negative entropy (!::..S < 0) ~ ~ is spontaneous. Either combination may be spontaneous or not, depending on the relative magnitudes of !::..Ssurr and !::..SSYS (see Figure 13.7). Because it is not practical to measure changes in the entropy of the universe, we need a parameter based on the system that enables us to predict when a process will produce a positive value for !::..Suniy and therefore will be spontaneous. The American mathematical physicist J. Willard Gibbs (1839-1903) is credited with developing the mathematical foundation of modern thermodynamics. Among his accomplishments, he developed a way to use!::..H sys and!::..S sys to predict the spontaneity of a reaction occurring at constant pressure and temperature. In doing so, he defined a new state function, for which in 1877 he coined the term and symbol free energy (G). The free in free energy does not mean "at no cost"; it means the energy that is available to do useful work-that portion of energy that can be dispersed into the universe. Free energy provides a way of predicting the spontaneity of a reaction by relating G to the entropy and enthalpy changes of the system. Let's see how by deriv= ing that relation. Based on our discussion of entropy, we know that !::..Suniv !::..S +!::..S . Furthermore, we learned in Chapter 5 that -q = +q and have sys SUff sys surr learned in this chapter that !::..S

surr

Free energy (G) is an indication of the energy available to do useful work. Free energy is a thermodynamic state function that provides a criterion for spontaneous change.

Therefore, it is mathematically

= -qrcv

-!::..H

T

T

sys

correct to write

!::..S _ (-!::..H ) uniy-!::..S sys+ sys

(13.7)

13.5

Equation 13.7 is important because it identifies the change in entropy of the universe exclusively in terms of the system. To reformat Equation 13.7 and eliminate the quotient, we multiply both sides by -T: -T!::.S.

-T!::.S

limv

=!::.H

sys

sys

+!::.H

I

Free Energy and Free-Energy Change

sys

- T!::.S sys

T!::.S

(13.8)

This is the expression for Gibbs free energy. The lack of subscripts means that all the terms refer to the system. If !::.G is negative, then !::.Suniv must be positive (remember the definition: !::.G = -T!::.S umv.). Therefore the following conditions hold: 1. If !::.G for a process is negative (!::.G spontaneous.

< 0), the process as written is

2. If !::.G for a process is positive (!::.G spontaneous.

> 0), the

3. If !::.G is zero (!::.G

=

process as written is not

0), no net change is occurring.

Equation 13.8 mathematically expresses our understanding of the two thermodynamic forces that drive chemical reactions: enthalpy and entropy. Note that !::.G is negative (and thus a reaction is spontaneous) if !::.H has a large negative value, if T!::.S has a large positive value, or both. Table 13.2 summarizes the effects of the signs of !::.H and !::.S on !::.G and on reaction spontaneity.

TABLE 13.2

Effects of

aH, as,

and Ton

aG and Spontaneity

SIGN OF ~H

+ + +

~G

I1S

+

Spontaneity

Always -

Always spontaneous

- at low temperatures

Spontaneous

at low temperatures

- at high temperatures

Spontaneous

at high temperatures

Always

+

Never spontaneous

The Meaning of Free Energy We have already defined free energy as the energy available to do useful work. Let's explore what this means using the combustion of gasoline in an automobile engine. The combustion reaction releases energy (!::.E) in the form of heat (q) and work (w) as defined in Chapter 5 by Equation 5.6: !::.E

=

q

+

w

Much of the heat q produced by combustion of fuel is wasted energy that flows from the engine to its surroundings as heat. Automobile engines have cooling systems to manage this heat, but the key issue is that this heat energy does nothing to move the car. The useful energy that propels the car is derived from the rapid expansion of the gaseous products of combustion. The product of the pressure (P) that is exerted by these gases on the engine's pistons times the volume change

655

Free-energy change (I1G) is the change in free energy of a process. For spontaneous processes at a constant temperature and pressure, I1G < O.

The free-energy change (dG) for a process occurring at constant pressure and temperature is defined as the term -T!::.Suniv' and !::.G= !::.H-

I

~II

CHEMTOUR Gibbs Free Energy

656

I Chapter Thirteen

I

Thermodynamics:

I1V=Al1h

FIGU RE 13.13

Thermal expansion

of the gases in a cylinder in a car engine pushes down on a piston with a pressure (P) represented

by the blue arrow. The product

of P times the change in volume of the gases

(11V) is the work done by the expanding gases that propels the car.

Spontaneous

Processes,


(~V) produced by expansion of the gases (as the pistons are pushed down) is useful (P ~ V) work (Figure 13.13). The concept of free energy relates to our attempts to obtain useful work from chemical reactions. The free energy released by a spontaneous chemical reaction at constant T and P is a measure of the maximum amount of energy that is free to do useful work. To see how this amount compares with the total energy released, let's rearrange the terms in Equation 13.8 to show this:

~H=

~G

+ T~S

(13.9)

Equation 13.9 tells us that the enthalpy change (~H) that comes from the making and breaking of bonds during a chemical process may be divided into two parts. One part, ~G, is the portion of the energy that devices such as internal combustion engines-or steam generators, batteries, and fuel cells-convert into motion, light, or some other manifestation of the interaction of energy with matter. This is what the statement means that ~G is energy available to do work. The other part of the enthalpy change, T ~S, is not usable: it is the temperaturedependent change in entropy, the portion of energy that "spreads out," thereby increasing the entropy of the universe. In the case of an automobile engine (or any engine for that matter), the conversion of chemical energy stored in bonds (~H) into useful mechanical energy to move a car (~G) is never 100% efficient. Some portion of the energy released by spontaneous processes such as burning fuels is always wasted. Remember the axiom given at the outset of the chapter: in the game of energy, we cannot win (energy cannot be created) nor can we even break even (every time we use energy, some is wasted). We can think of the efficiency of an engine as the ratio of work done to energy consumed. Efficiency is usually inversely related to the rate at which a spontaneous reaction takes place. The slower the reaction, the greater the amount of free energy likely to be harvested; the faster the reaction, the lesser the amount of free energy likely to be used to do work and the more that will be wasted in heating up the environment. Recall our earlier discussion of reversible processes: the more slowly a process is carried out, the more closely it approaches complete reversibility, at which point it would be at maximum efficiency. Reporting the efficiencies of energy-using devices in the media is becoming increasingly common. We can now understand what these numbers mean in terms of our discussion here. Mechanical efficiency is frequently expressed in relative percentages of work done per amount of energy consumed: water wheels are 90%; fuel cells, 80%; jet engines, 60%; car engines, 30%; fluorescent lights, 20%; and incandescent lightbulbs, 5%. To reiterate: not only is all of the T~S energy from a reaction wasted, but the amount of ~G that is actually used must always be less than 100% of the available free energy. Additional heat losses always flow into the surroundings without doing anything useful.

Calculating Free-Energy Changes The standard free energy of formation (I1C;) is the change in free energy associated with the formation of 1 mole of a compound in its standard state from its elements.

The free-energy change in a reaction can be calculated from Equation 13.8. It can also be calculated from the standard free energies of formation of the products and reactants. The standard free energy of formation (~G;) is the change in free energy associated with the formation of 1 mole of a compound in its standard state from its elements in their standard states. See Appendix 4 for a list of values.

~G;

13.5

We can use free energies of formation to calculate ~G:xn using the following equation: ~GO

In~Go

rxn

- Im~Go

f,products

f,reactants

(13.10)

Note the similarities between Equation 13.10 and other expressions for ~H:xn (Equation 5.18/13.6) and ~S:xn (Equation 13.4): ~HO

rxn

~SOrxn

=

In~Ho

= InS

- Im~Ho

f,products

o

f,reactants

o

products

-

ImS

reactants

According to Equation 13.10, the free-energy change for a chemical reaction forming n moles of products from m moles of reactants is the difference in the sums of the free energies of formation of the products and the reactants. By definition, the free energy of formation of an element in its standard state is zero.

Predicting aGo of Reactions The physical properties of structural isomers of C8 alkanes (see Chapter 12) are significantly different. So, too, are their free energies of formation, as shown in the following data:

c. Alkane

.1G~ (kJjmol)

n-Octane

16.3

2-Methylheptane

11.7

3,3-Dimethylhexane

12.6

Structure

Each compound burns according to the following chemical equation:

Predict whether the ~Gorxn values for the combustion reactions of these isomers are all the same or different. AND ORGANIZE We are asked to predict whether the values of ~Go for the combustion reaction of structural isomers are the same or different.

COLLECT

A N A L Y Z E The free energy

of combustion can be calculated according to Equation 13.10; however, there is actually no need to do the calculation. The Products of the reaction are the same for each isomer, so ~Gof,pro ducts will be the same in each combustion reaction.

I


I

657

658

I Chapter Thirteen

I

Thermodynamics:

Spontaneous

Processes,

~G; of 02 is zero

~G;

by definition, and is different for each isomer, ~G;,rcactants and ~G:xn for the combustion reaction are different for each isomer:

SOLVE

Because


~GOrxn =

I n~GO f,products - I m~GO f,reactants

~G:xn = [same products] - [~G;,isomcr

I1H

263

!)]

T HI N K ABO UTI T The free energy is different for each isomer, so changes in both

I1S I

273 Temperature (K)

(25 mol X 0.0

The notation "[same products]" means that each isomer produces the same products when it is burned; the balanced equations for the combustion of the isomers yield the same quantities of CO/g) and H20(g).

kJ

~

+

283

enthalpy and entropy must arise because the arrangement molecule is unique.

of atoms within each

Practice Exercise Using data from the table in Sample Exercise 13.5 and Appendix 4, calculate ~G:xn for the combustion of 1 mole of n-octane.

(a)

Temperature, Spontaneity, and Free Energy

TI1S

(kJ)

263

273 Temperature (K)

283

(b)

Spontaneous I1G < 0

I1H

263

273 Temperature (K)

283

(c)

FIGURE 13.14 Changes in entropy (I1S), enthalpy (I1H), the quantity TI1S, and free energy (I1G) for ice melting (fusion) at temperatures ranging from -lo°C (263 K) to +lOT (283 K).

(iXj)

CONNECTION In ChapterlO we discussed equilibrium between phases of a substance and used phase diagrams to illustrate which physical states are stable at various combinations of temperature and pressure.

Let's reconsider the isothermal process of ice melting, this time in light of the full complement of thermodynamic quantities discussed in this chapter. The influence of temperature on ~H, ~S, T ~S, and ~G as ice is warmed from -10°C to +lOT (263 to 283 K) is presented graphically in Figure 13.14. The first graph (Figure 13.14a) shows that the enthalpy of fusion (green line) changes little, while the entropy of the system (red line) gradually increases as the temperature rises. In the second graph (Figure 13.14b) T~S increases rapidly as the temperature increases. The third graph (Figure 13.14c) shows the combined influence of ~H and T~S on ~G as temperature changes (remember: ~G = ~H - T~S). This graph defines conditions under which melting is either nonspontaneous or spontaneous. ice melts spontaneously. In the language of therAt temperatures above modynamics, the larger T~S term in Equation 13.8 for the process of melting more than offsets the positive ~H term. As a result, ~G is negative, indicating that ice melts spontaneously at temperatures above O°e. At temperatures below ice does not melt spontaneously, but the opposite process-liquid water freezing-is spontaneous. Again, in thermodynamic terms: when the T~S term in Equation 13.8 is smaller than the positive ~H term, the free-energy change for melting is positive. Thus, below melting is not spontaneous while freezing is. The important point here is the same as with other thermodynamic quantities: the free-energy change of a process or reaction is equal in magnitude but opposite in sign to the free-energy change of the reverse process or reaction. Another significant aspect of Figure 13.14(c) is the point on the graph where ~G = O. The temperature value at that point defines the melting (or freezing) point, which by definition is the temperature at which the solid melts at the same rate as the liquid freezes. The two phases are at equilibrium. At no net change takes place and both phases coexist. A similar situation exists for water at 100°C and 1 atm pressure, the temperature and pressure at which the ~G values for the vaporization of water and the condensation of water vapor both equal zero. At 100°C, liquid water vaporizes and water vapor condenses at the same rate; the two phases exist in equilibrium with each other. The temperature at which the free-energy change for a process equals zero and the process achieves equilibrium can be calculated from Equation 13.8 if

ooe,

ooe,

ooe,

ooe,

13.5

we know the values of tl.H and tl.S for the process in question. For example, the changes in standard enthalpy and standard entropy for the fusion of water (melting of ice; forward reaction as written) are

(tl.Ho = 6.01 X 103 J; tl.So = 22.1 JlK) We can use this information and Equation 13.8 to calculate the temperature at which tl.G for ice melting is zero. In doing this calculation, we assume that the values of tl.Ho and tl.So do not change significantly with small changes in temperature. For this process and for most physical and chemical changes, such an assumption is acceptable. Therefore we can assume that tl.G = tl.H - Ttl.S tl.Ho - Ttl.So. Inserting the values of tl.Ho and So and using the fact that tl.G = 0 for a process at equilibrium, we get the following:

=

tl.G = (6.01 X 103 J) - T(22.1 JlK) = 0 Solving for T3

6.01 X 10

T= -we

22.1J1K

J=

273 K

= 00C

obtain the familiar value for the melting point of ice.

Calculating Free-Energy Changes Calculate tl.Go for the dissolution of 1 mole of ammonium nitrate (NH4NO) in water (total volume = 1 liter) at 298 K, given tl.Ho = 26.4 kJlmol and tl.So = 108.2 JI(mol· K). We are asked to find the free-energy change for NH4N03 dissolving in water. We are given the change in enthalpy and the change in entropy for the process under standard state conditions. COLLECT

AND

ORGANIZE

ANALYZE

We can use Equation 13.8 for this calculation.

Substituting the values of tl.Ho and tl.So for tl.H and tl.S in Equation 13.8 allows us to calculate tl.Go:

SOLVE

*

0.108 kJ) tl.Go = tl.Ho - Ttl.So = 26.4 kJ - (298 *)(

= -5.8 kJ THINK

ABOUT

IT

tl.Go is negative, and the reaction is spontaneous. The decrease

in standard free-energy is the result of an unfavorable increase in enthalpy that is more than offset by a favorable increase in entropy: the endothermic enthalpy and the positive entropy change.

Practice Exercise Predict the signs of tl.Ho, tl.So, and tl.Go for the combustion of methane:

Like physical processes, chemical reactions also can have zero change in free energy. As a spontaneous chemical reaction proceeds (tl.G < 0), reactants become products and tl.G becomes less negative (for reasons that will become clear in a later chapter). In fact, tl.G may reach zero before all the reactants are consumed. In this case, no more products form and no more reactants are consumed. Rather, reactants and products coexist in equilibrium with each other. Reactants still

I


I 659

660

I Chapter Thirteen

I

Thermodynamics:

Spontaneous

Processes,


react and products are still formed, but the reverse reaction, in which products reform reactants, proceeds at the same rate as the forward reaction. This state, called chemical equilibrium, is the subject of Chapters 15 and 16. No net change in the amounts of reactants and products occurs once equilibrium is reached.

Relating Reaction Spontaneity to and for a Chemical Reaction

as

aH

A chemical reaction is spontaneous at low temperatures but not spontaneous at high temperatures. Use Equation 13.8 (~G = ~H - T~S) to determine whether ~H and ~S of the reactions are greater or less than zero. COLLECT AND ORGANIZE We are asked to determine the signs of ~H and ~S based on the change in spontaneity of a reaction as a function of temperature. This means we need to think about how ~G varies with temperature depending on the signs of ~H and ~S.

Following the relations summarized in Table 13.2, a reaction is spontaneous at all temperatures when ~H < 0 and ~S > 0, and a reaction will never be spontaneous when ~H > 0 and ~S < O. There are only two other possibilities for this reaction: (1) both ~H and ~S are positive, or (2) both ~H and ~S are negative. The importance of ~S increases with increasing temperature, because the product of the two parameters (T~S) appears in Equation 13.8. If the reaction is not spontaneous at high temperatures where the magnitude of T ~S is more likely to overwhelm an unfavorable ~H value, then entropy works against the reaction being spontaneous. On the other hand, the reaction is spontaneous at low temperatures where the impact of an unfavorable (negative) ~S is more than offset by a favorable decrease in enthalpy (~H < 0). ANALYZE

SOLVE

The reaction must have negative ~S and negative ~Hvalues.

ABOUT IT The summary in Table 13.2 confirms our prediction that a process that is spontaneous only at high temperatures is one in which there is a decrease in both entropy and enthalpy.

THINK

Practice Exercise The synthesis of ammonia from N2 and H2 is described by the following chemical equation N/g)

+ 3 H/g)

~ 2 NH3(g)

Use the appropriate thermodynamic data in Appendix 4 to determine whether (a) the reaction is exothermic or endothermic; (b) the reaction is spontaneous at all temperatures.

CONCEPT

TEST

How can a reaction be spontaneous when ~S:xn is negative?

mm

Driving the Human Engine: Coupled Reactions

The laws of thermodynamics describe the chemical reactions that power the human engine, and the same rules apply in living systems as apply in the inanimate world (Figure 13.15). Organisms transform the energy contained in the

13.6

I


molecules that make up food and use that energy to do work. Just like mechanical engines, humans and other life forms are far from 100% efficient. Life requires the continuous input of energy in terms of the caloric content of the food we eat. Living systems are therefore thermodynamically open systems that continuously take in food (starting materials, or reactants) and give off products of metabolism (C02, water, and heat). Because of this constant input of reactants, elimination of products, and exchange of heat, metabolic reactions in living systems do not come to equilibrium. We as humans constantly absorb energy in the form of food and release heat and waste products to our surroundings. Two concepts are used to describe reactions depending on the sign of !lG: spontaneous reactions (!lG < 0) involving a decrease in free energy are called exergonic reactions, while nonspontaneous processes (!lG > 0) requiring the constant input of energy to proceed are called endergonic reactions. Exergonic reactions are typical of those dealing with breaking down foods, while endergonic reactions are typical of those involved in building substances needed by the body from the components supplied by food. Living systems use the energy provided from exergonic reactions to run endergonic reactions; we say that exergonic reactions are coupled (linked or paired) to endergonic reactions in living organisms. Part of the study of biochemistry involves deciphering the molecular mechanisms that enable the coupling of exergonic reactions-like the breakdown of glucose-to the endergonic reactions our bodies carry out every moment of our lives, from replicating DNA to building proteins in muscle to making memories. This topic will be dealt with in more detail in Chapter 19, but for now it is sufficient to know that elegant molecular processes have evolved to enable living systems to manage the energy of exergonic reactions and direct that energy to drive the endergonic reactions that create and maintain the structures and functions we associate with life. In Chapter 5 we looked at the energy content of a jelly doughnut, which contains a little protein and a lot of fats and carbohydrates. These three chemical groups-carbohydrates, fats, and proteins-are the primary constituents of the human diet. Young women have an average daily nutritional need of 2100 kilocalories (kcal, or Cal); for young men the figure is 2900 kcal, or a little more than 12,000 kJ each day. This level of caloric intake provides the energy we need in order to function at all levels, from getting out of bed in the morning to thinking, and the wasted energy our bodies give off as heat makes our normal temperature about 3TC. All of the food we consume to sustain our bodies has sunlight as its ultimate source. Energy flows into the living world in the form of sunlight, which green plants store in their tissues in molecules, such as glucose, that they produce from CO2 and H20 during photosynthesis. Animals then consume plant products, use the energy stored in the chemical bonds of glucose and other molecules, and release CO2 and H20 back into the environment. They also release heat, which increases the entropy of their surroundings. Because the heat is wasted energyenergy not available to do useful work-all living things participate in the ultimate spreading out of energy. With all its organization, life increases the entropy of the universe by converting chemical and physical energy into heat, and then releasing that heat into the surroundings. The breakdown of glucose-

o

-is highly exergonic; !lG for this reaction is -880 kJlmol. The production of 1 mole of glucose by a plant is endergonic (nonspontaneous), which is why green

Exergonic reactions. Endergonic reactions.

I 661

reactions are spontaneous

reactions are nonspontaneous

~ CONNECTION An open thermodynamic system is one that freely exchanges matter and energy with its surroundings, as we discussed in Chapter 5.

662

I

I Chapter Thirteen

Thermodynamics:

Spontaneous

Processes,


Heat and light are dissipated Heat

Electricity

Combustion

Incandescent light: 5% efficiency

of coal

(a)

+

+

(b)

(c) FIGURE 13.15 (a) A coal-fueled

(d)

The rules of thermodynamics

that apply in the physical world at all levels also apply to all living systems.

power plant that generates electricity

into the environment.

The electricity

increases entropy as it burns fuel, releasing CO2,

it generates contributes

appliances powered by electricity lose energy whenever they use energy. (b) Living organisms, energy from nutrients

released, increasing the entropy of the universe. (c) Microscopic nutrients

the universe. (d) Growth

organisms-such

and

such as honey bees, extract

heat, water, and carbon dioxide are

as the E. coli that live in our gastrointestinal

to live and generate heat in the process. Their expenditure of the human population

as fossil fuels are increasingly of Earth.

and waste heat

to support life. Bees store energy as honey, a mixture of sugars they produce; honey is then consumed

by bees, human beings, and other animals to generate energy. When honey is consumed, tracts-consume

H20,

further to the entropy of the universe, because all machines

of energy also increases the entropy of

in the last 150 years has been accompanied

burned to power vehicles and produce electricity,

by dramatic increases in entropy

as reflected in this composite night-time

photo

13.6

I


plants require the energy of sunlight to produce glucose. When we consume glucose and convert it into energy, CO2, and H20, the process is exergonic. This spontaneous process in living systems is highly controlled, so that the energy it produces can be directed into the endergonic molecular processes essential to organisms that don't photosynthesize. Millions of years of evolution have resulted in a complex series of chemical reactions that mediate the conversion of sugar (glucose) into carbon dioxide and water in our cells. Several sequences of chemical reactions are required to manage this process and to use the considerable energy it releases to drive endergonic reactions. We will look at a portion of one of those sequences to discuss coupled reactions. In the sequence known as glycolysis, each mole of glucose (C6H1206) is converted into 2 moles of pyruvic acid (Figure 13.16). Pyruvic acid-or more precisely, the pyruvate ion (CH3COCOO-)-is a reactant in the Krebs cycle, a series of reactions in which carbon in molecules or ions is converted into CO2, The hydrogen atoms in these molecules are eventually converted into H20. To understand how coupled reactions work, let's first consider an early stage in glycolysis: the conversion of glucose into glucose 6-phosphate. The sequences of chemical reactions that are part of metabolism are not presented here for you to memorize. Rather, the intent is to aid your understanding of how changes in free energy allow spontaneous reactions to drive nonspontaneous reactions, and to illustrate operationally what "coupled reactions" actually means. The conversion of glucose into glucose 6-phosphate is an example of a phosphorylation reaction (Figure 13.17). Glucose reacts with hydrogen phosphate ion (HPO/-), producing glucose 6-phosphate and water. This reaction is not spontao neous (!::..G = +13.8 kJ). The energy needed to make this reaction happen comes from a compound called adenosine triphosphate (ATP). ATP functions in our cells both as a storehouse of energy and as an energy-transfer agent: it hydrolyzes to adenosine diphosphate (ADP) in a reaction (Figure 13.18) that produces a hydrogen phosphate ion and energy: !::..G:xn = -30.5 kJ. (Note that ATP and ADP are both anions; the phosphate groups are ionized under typical conditions in the living system. ATP has a charge of 4-, ADP a charge of3-. The presence of the phosphate group in glucose 6-phosphate gives the compound a 2- charge under conditions in the living system.)

OH HI I CH2 H

..... C__ '\

1__

0

HO.....\ --C-\ HO_C";"'-C\,-c-H I H OHI H OH

Glucose(aq)

+

HPOi-

+

HPOi-(aq)

~

(glucose 6-phosphate)2-(aq)

+

H20(C)

o Glu(aq)

~

Glu-O-P-O-

FIGURE 13.17 The conversion of glucose into glucose 6-phosphate (nonspontaneous). I1G:,n = 17.8 kJ/mol.

This reaction is endergonic

II I 0-

+

H20(C)

is an early step in glycolysis.

I

663

Glycolysis is a series of reactions that converts glucose into pyruvate. It is a major anaerobic (no oxygen required) pathway for the metabolism of glucose in the cells of almost all living organisms. A phosphorylation reaction results in the addition of a phosphate group to an organic molecule.

H

OH

Glucose

~

Glycolysis

~

( ~ /?)

2 H3C-C-C'o_ Pyruvate ion

FIGURE 13.16 In glycolysis, one molecule of glucose is converted

into two pyruvate ions.

664

I

Chapter Thirteen

I

Thermodynamics:

Spontaneous

Processes,


NH2

o

P~N

0

II II -O-P-O-P-O-P-O~O I I 0-

0-

<JL)

0

II I

N

,..-

0-

II I 0-

+

H

II I 0-

II I 0-

-O-P-O-P-O-P-O-A

'-
O. The entropy change of the universe equals the sum of the entropy change of the system and the entropy change of the surroundings: I:!.Sun;, = I:!.Ssys + I:!.Ssuc," Reversible processes are ideal processes that take place in very small steps and very slowly. No change in entropy takes place in a reversible process. This concept enables us to calculate maximum efficiency, which is the most work available from a given quantity of energy. If I:!.Ssys < 0 in a spontaneous process, then I:!.Ssm, > 0 and I:!.Ssm, must be large enough to insure that I:!.Ssys + I:!.Ssuc, > O. If I:!.Ssys > 0, then I:!.Ssm, must have a value so that I:!.Ssys + I:!.Ssm, > O.

SECTION

13.3 According to the third law of thermodynamics, perfect crystals of a pure substance have zero entropy at absolute zero.

SECTION

All substances have positive entropies at temperatures above absolute zero. Standard molar entropies (SO) are entropy values under standard state conditions. The entropy of a system increases with increasing molecular complexity and temperature. 13.4 The entropy change in a reaction under standard conditions can be calculated from the standard entropies of the products and reactants and their coefficients in the balanced chemical equation. If we study isothermal processes carried out reversibly, we can calculate the entropy change of the surroundings by dividing the heat exchanged (q,,) by the temperature at which the process occurs. We can then also calculate the entropy change of the universe and determine whether the process is spontaneous by noting the sign of SECTION

I:!.Sun;v· 13.5 The free-energy change (.1G) in a process is a state function giving the maximum useful work the system can do on its surroundings. Spontaneous processes have negative I:!.G values. Reversing a process changes the sign of I:!.G. For a system at equilibo rium, I:!.G = O.The standard I:!.G free-energy change in a process can be calculated from the standard free energies of formation (.1G;) of the products and reactants or from the enthalpy and entropy changes. Values for I:!.G~ refer to standard state conditions, and for the formation of elements, I:!.G~ = O. The temperature range over which a process is spontaneous depends on the values of I:!.H and I:!.S. SECTION

13.6 Exergonic reactions are spontaneous and endergonic processes are not. Many important biochemical processes, including glycolysis and phosphorylation, are made possible by coupled exergonic and endergonic reactions. The free energy released in such processes (such as the hydrolysis of ATP to form ADP) is used in our bodies to drive endergonic processes.

SECT ION



SAMPLE

EXERCISES

Predicting the sign of the entropy change in a physical or chemical process

Look for the net removal of gas molecules or precipitation of a solute from solution (I1S < 0 for both). The reverse processes have positive entropy changes.

13.1 and 13.2

Visual Problems

TYPE OF PROBLEM

CONCEPTS

AND EQUATIONS

Calculating the entropy change of a chemical reaction

Use Equation 13.4

SAMPLE

EXERCISES

13.3 I1S rxn = InS 0

o products

-

o

ImS

reactants

where S;mduw and S:",,,n,, are the standard molar entropies of the products and reactants and nand m are the stoichiometric coefficients for the process. Predicting the spontaneity of a process

A process is spontaneous if

13.4

Calculating the free-energy change of a physical or chemical process

Use Equation 13.10

13.5 and 13.6

~G:xn = L n .6.G;,products

-

L m .6.G;,reactants

where I1G~,pmduwand I1G~,"""n" are the standard molar free energies of formation of the products and reactants, and nand m are the stoichiometric coefficients for the process. Use Equation 13.8,

where I1H and I1S are the standard enthalpy and entropy changes for the reaction, respectively. O

0

Relating reaction spontaneity to I1Hand I1S

Use Equation 13.8:

13.7

Calculating the free-energy change of coupled reactions

Like enthalpy and entropy changes, free-energy changes are additive. If adding two reactions gives the desired reaction, add the respective free-energy changes to obtain the free-energy change of the desired reaction.

I1G

=

I1H -

ns 13.8


More air is added to the partially filled party balloon in Figure P13.1. Does the entropy of the balloon and its contents increase or decrease?

but the sample in (a) is cooled so that it solidifies, which sample has the higher entropy?

FIGURE P13.1 13.2.

The two cubic containers on the right (Figure P13.2) contain two gas samples at the same temperature and pressure. Which has the higher entropy-the sample in (a) or the sample in (b)? If the sample in (b) is left unchanged

(b)

(a)

FIGURE P13.2

I

667

668

I Chapter Thirteen

13.3.

I

Thermodynamics:

Spontaneous

Processes,


Figure P13.3 shows a glass tube connecting two bulbs containing a mixture of ideal gases A (red spheres) and B (blue spheres). Is the probability high or low that gas A will collect in the left-hand bulb and gas B will collect in the right-hand bulb? Explain your answer in terms of the entropy changes involved. I1H

0.0

FIGURE P13.3 13.4.

The box in Figure P13.4(a) contains a gas. Complete the box in Figure P13.4(b) to illustrate deposition of the gas as a solid. Suggest whether I:!.S, I:!.H, and/or I:!.G are positive or negative for this spontaneous process. Does the entropy of the surroundings increase or decrease?

20.0

40.0 60.0 Temperature (OC)

80.0

100.0

FIGURE P13.5 13.6.

Figure P13.6 is based on the I1G; values of elements and compounds selected from Appendix 4. Which of the following conversions are spontaneous? (a) C6H6(C) to CO/g) and HP(C); (b) CO/g) to C2H2(g); (c) H/g) and 02(g) to HP(C). Explain your reasoning. 300 --C2Hh:)

200

~

"0

......E

~

100 0,

--C6H6(C) Hh),Oig)

o

B" Tl

theory, some fraction of a population

of

molecules [the shaded areas in (a)] has kinetic energies equal to or greater than the energy (E) of the reaction. As temperature

increases from

T, to T;, the

number of

molecules with energies exceeding E, increases, leading to an increase in reaction rate. (b) The rate constant

for any reaction increases with increasing

temperature.

point on a high plateau to a

on a lower plain.

CHEMTOUR

Arrhenius

Equation

(K)

Fraction of molecules in sample with sufficient energy to react at Tl

reactant

to (b)

the hill that a bicyclist must climb en route from a starting

= Ae-E,IRT

The energy profile (a) of

an exergonic reaction includes an activation

Activation energy (E ) is the minimum energy molecules need to react when they collide. The Arrhenius equation relates the rate constant of a reaction to absolute temperature (T), the activation energy of the reaction (E), and the frequency factor (A).

704

I

Chapter Fourteen

I

Chemical Kinetics

The term A is called the frequency factor. The frequency factor is the product of collision frequency and a term that corrects for the fact that not every collision results in a chemical reaction. Some collisions do not lead to products because the colliding molecules are not oriented with respect to each other in the right way. To examine the importance of molecular orientation during collisions, let's revisit a reaction (Equation 14.7) that sustains the level of N02 in polluted urban air after the evening rush hour:

Reactants

1

Effective collision

1 Products

(a)

Reactants

*

Ineffective collision No products

(b)

FIGURE 14.15 The importance

of

molecular

orientation

illustrated

by the reaction between NO and

03• (a) Collisions

on reaction rate is

in which the nitrogen

on

NO contacts an oxygen atom of 03 may lead to an activated complex that yields N02 Or (b) Collisions

+ NO(g)

O/g)

Activated complex

and

between oxygen atoms on

NO and 03 molecules produce no activated complex and no reaction.

~ 02(g)

+ N02(g)

Two of the ways in which ozone and nitric oxide molecules might approach each other are shown in Figure 14.15. Only one of these orientations, the one in which an oxygen atom of an 03 molecule approaches the nitrogen atom of NO, leads to a chemical reaction between the two molecules. A collision between 03 and NO molecules with the correct orientation and enough kinetic energy may result in the formation of an activated complex in which some of the bonds that hold these molecules together break and the bonds that hold O2 and N02 together form. Activated complexes are formed by reacting species that have acquired enough potential energy to react with each other. The energies of these complexes are much higher than those of reactants and products, and represent transition states between reactants and products. In fact, the energies of these transition states define the heights of the activation energy barriers of chemical reactions. Consider the energy profile for the reaction between nitric oxide and ozone in Figure 14.16. The x-axis of the right-hand profiles represents the progress of the reaction. Distances along the y-axis represent changes in chemical energy. The size of the activation energy barrier depends on the direction from which it is o approached. If a reaction is spontaneous (t1G < 0) in the forward direction, such as the one illustrated in Figure 14.16(a), then Ea is smaller in the forward direction than in the reverse direction (Figure 14.16b). A smaller activation energy barrier means that the spontaneous reaction proceeds at a higher rate than the reverse, nonspontaneous reaction. One of the many uses of the Arrhenius equation is to calculate the value of Ea for a chemical reaction. To see how this procedure works, we must first rewrite Equation 14.28 by taking the natural logarithm of both sides: In k

= - ; (~) + In A

(14.29)

y = mx + b ~II

CHEMTOUR Collision Theory

The frequency factor (A) is the product of the frequency of molecular collisions and a factor that expresses the probability that the orientation of the molecules is appropriate for a reaction to occur. An activated complex is a species formed in a chemical reaction when molecules have enough potential energy to react with each other. A transition state in a chemical reaction is a high-energy state between reactants and products.

Equation 14.29 fits the general equation of a straight line (y = mx + b) if we make (In k) the y variable and (1IT) the x variable. We can calculate Ea by determining the rate constants for a reaction at several temperatures. Plotting Ink versus IIT should give a straight line, the slope of which is -E/R. Table 14.6 and Figure 14.17 show the results of measurements of the rate of the reaction between NO and 03 at six different temperatures. The straight line that best fits the points in Figure 14.17 has a slope of -1260 K. There are no units for the y values (they are logarithms), and the units of the x values are 11K. Therefore, the units of the slope t1ylt1xwill be 1/(I/K), or simply K.The slope equals -E/R, so

Ea

=

-slope

= -(-1260

= 10.5

X

R 8.314 K) X (-mol·K

kJlmol

J) = 1.05 X 10 Jlmol 4

14.4 I Reaction Rates, Temperature, and the Arrhenius Equation I 705 Activated complex

50 Ea

=

~

oI

10.5 kJlmol

--=-;;.:::::::::::

t

Reactants

E

=

10.5 k]/mol

a

- 200 ~---------------------------------------~"'-------

o

Products

-250'~-------------------------------

10.5

Progress of reaction

~

Kinetic energy (a) Activated complex

50 Ea

=

o

210 k]/mol

-200 I

/.. Reactants

-250'~------------------------------Progress of reaction

~

Kinetic energy (b)

FIGURE 14.16 (a) The energy profile for the spontaneous reaction 03(g) + NO(g) -7 02(g) + N02(g) includes an activation energy barrier of 10.5 k]/mol. (b) The reverse reaction has an activation energy of 210 k]/mol. In the left-hand plots of kinetic energy versus fraction of molecules, the shaded areas indicate the molecules with sufficient energy (E) to react.

25.0

As shown by the different sizes of the shaded areas under the kinetic energy curves, fewer molecules have the kinetic energy needed to overcome the larger activation

energy of the

l

reverse reaction. The reverse reaction proceeds much more slowly than the forward reaction.

TABLE 14 ••

I

NO(g) T (K)

300

-

-

350

+

03(g) ~

k (M-1 S-1)

1.21 X 1010

-

1.67 x 1010

325 -

Temperature Dependence of the Rate of Reaction for the Reaction

--

2.20 x 1010

N02(g) In k

23.216

+ -

-

23.814

-

23.5 1/T (K-1)

3.33 x 10-3

2.79 x 1010

24.052

2.67 X 10-3

400

3.45 X 1010

24.264

2.50 x 10-3

-

4.15 x 1010

-

24.449

-

23.0 ,'-0.0022

2.86 x 10-3

375

425

In k 24.0

02(g)

3.08 X 10-3

23.539

24.5

2.35 X 10-3

-'-'0.0026 0.0030 1/T(K-1)

---" 0.0034

FIGURE 14.17 A graph of the natural logarithm of (In k) versus liT yields a straight line with a slope equal to -E/R and a y intercept equal to the natural logarithm of the frequency factor (In A). the rate constant

706

I

Chapter Fourteen

I

Chemical Kinetics

The y intercept (11T = 0) of the line that best fits the data is 27.4. This value represents In A. The value of A is

A = e27.4 = 7.9 X 1011 We can use the values of Ea and A to calculate k at any temperature. at T= 250 K k

For example,

= Ae-E,IRT =

4 -(1.05 x 10 t/nml)

7.9 X 1011 e [ (8.314j1nm1·*)(250*)

]

Calculating the Activation Energy from Rate Constants The following data were collected in a study of the effect of temperature rate of decomposition ofCIO into Cl2 and O2 in the reaction

Determine

on the

the activation energy for the reaction. k (M-1 S-l)

T(K)

1.9 x 109

238

3.1 x 109

258

4.9 x 109

278

7.2 x 10

298

9

CO II EeTA N D 0 R GA N I Z E We are asked to determine the activation energy (E) for a reaction. We are given values of the rate constant k as a function of absolute temperature.

The activation energy for any reaction can be calculated from rate constants for the reaction at a series of temperatures by using Equation 14.29:

ANALYZE

In k

1)

= - Ea R ( T + In A

The y variable is (In k), so we need to convert k values to (In k) values first. The x and y variables in the Arrhenius equation are (1IT) and (In k). We must convert temperature and rate constant values to these variables before plotting the data. SOL V E Expanding

the data table to include columns of (1IT) and (In k) yields k (M-l S-l)

In k

T (K)

1fT (K-l)

1.9 x 109

21.365

238

4.20 x 10-3

3.1 X 109

21.855

258

3.90 x 10-3

4.9 X 109

22.313

278

3.60 x 10-3

7.2 X 109

22.697

298

3.36 x 10-3

14.4

I

Reaction Rates, Temperature, and the Arrhenius Equation

A plot of (In k) versus (1IT) yields a straight line: 23

In k 22

21LI

-L

0.0032

~

~

0.0036 0.0040 1/T(K-1)

0.0044

the slope of which is -1564 K. According to the Arrhenius equation, the slope is equal to -E/R, where Ea is the activation energy and R is the ideal gas constant with units ofJl(mol· K). By using the values of the slope (-1564 K) and R [8.314 JI(mol· K)], we can calculate the value of E a :

Ea = -slope

X

R

= -(-1564*)

8.314

X (--

mol·*

J) = 1.30 X 10 J1mol 4

= 13.0 kJlmol T HI N K ABO UTI T The activation energy is the height of a barrier that has to be overcome before a reaction can proceed. Generally speaking, the higher the barrier, the lower the rate constant and the slower the reaction. The activation energy for CIO decomposition is relatively low, consistent with its very large rate constants. The values of the rate constant are very large, so the reaction is very rapid.

Practice Exercise Atomic bromine reacts with ozone to form BrO and O2:

The rate constant for the reaction was determined at four temperatures from 238 to 298 K. Calculate the activation energy for this reaction. T(K)

k [cm3/(molecule·

258

x 7.7 x

278

9.6

298

1.2

238

5.9

x x

ranging

s)]

10-13 10-13 10-13 10-12

Earlier in this section, we noted that activation energy barriers tend to be lower when approached from the reaction direction that has a negative t1G. This observation raises a question: Do spontaneous reactions always proceed more

I 707

I

708

I

Chapter Fourteen

Chemical Kinetics

Ea

=

\...2 NO

102 kJlmol

laG 41 =

12NOsY

(a)

1

Bimolecular collision

Progress of reaction ~

FIGURE 14.18 Spontaneous

+

Cl2

'" k]/mol

NO + CO2 "-Progress of reaction ~

(b)

reactions are not necessarily rapid reactions.

and C12, which has an activation energy of102 k]/mol, and (b) the spontaneous CO --? NO

+ CO

2,

Here are reaction-

(aG > 0) decomposition

energy profiles for two reactions: (a) the nonspontaneous

ofNOCI

to NO

reaction N02

+

which has an activation energy of 134 k]/mol. The spontaneous

actually be the slower of the two because its activation

reaction may

energy barrier is higher.

Activated complex

1 Product

Intermediate

1 Product

Activated complex

l

unimolecular decomposition

Product

Products

rapidly than nonspontaneous reactions? The answer is: not necessarily. Consider the reaction-energy profiles in Figure 14.18. The nonspontaneous (t1G > 0) decomposition of NOCI to NO and Cl2 will occur only if we add energy. On the other hand, the reaction ofN02 with CO to form NO and CO2 is spontaneous under standard conditions and 298 K. However, if both reactions are run at a temperature at which NOCI decomposes, the NOCI reaction may proceed faster than the reaction between N02 and CO because the latter reaction has a larger activation energy barrier: 134 kJlmol vs. 102 kJlmol. Thus, a reaction with a large decrease in free energy may be "spontaneous" from a thermodynamic point of view, but spontaneous does not necessarily mean the reaction occurs rapidly. In fact, the rates of some spontaneous processes, such as the transformation of diamond into graphite, are so slow that they do not proceed at all: they are favored thermodynamically, but not kinetically.

mI1I Reaction Mechanisms To think about how a reaction proceeds at the molecular level, let's revisit the chemical equation describing the thermal decomposition ofN02:

Overall reaction:

+

+

~

+

2NO

FIGURE 14.19 The decomposition ofN02

begins when two N02

collide, producing decomposes

molecules

NO and NOr

N03 rapidly

into NO and O2• The second

step is faster than the first. The first step is the rate-determining

step, and the overall

reaction is second order in N02.

We noted in Section 14.3 that this reaction is second order in N02 because it takes place as a result of the collisions of pairs of molecules of NO 2. How exactly do we picture the atoms inside two colliding molecules ofN02 rearranging themselves to form two molecules of NO and a single molecule of O} The answer to this question is contained in the mechanism of the reaction. A reaction mechanism describes the stepwise manner in which the bonds in molecules of reactants break and the bonds that hold together molecules of the product form.

Elementary Steps A mechanism proposed for the thermal decomposition of N02 is shown in Figure 14.19. In the first step of the mechanism, a collision between a pair of N02 molecules produces a very short-lived activated complex in which the two

14.5

molecules share an oxygen atom. Activated complexes such as this one represent midway points in chemical reactions. They have extremely brieflifetimes and rapidly fall apart, either forming products or re-forming reactants. To form products, the shared oxygen atom is transferred from one N02 molecule to the other, forming a molecule of NO and a molecule ofN03 that rapidly decomposes by forming another activated complex. The bonds in this second complex rearrange so that two oxygen atoms become bonded together, forming a molecule of O2 and leaving behind a molecule of NO. N03 is considered an intermediate in this mechanism because it is produced in one step and then consumed in the next. Intermediates are not considered reactants or products and do not appear in the overall equation describing a reaction. The reaction mechanism in Figure 14.19 is a combination of two elementary steps that provide detailed molecular views of the reaction. Elementary steps may involve one or more molecules. Those that involve a single molecule are called unimolecular. Elementary steps based on collisions between two molecules are called bimolecular. They are much more common than termolecular (threemolecule) reactions because the chances of three molecules colliding simultaneously are much less than those of bimolecular collisions. The terms uni-, bi-, and termolecular are used by chemists to describe the molecularity of an elementary step, which refers to the number of atomic-scale particles (free atoms, ions, or molecules) that collide with one another in that step. A valid mechanism must be consistent with the stoichiometry of the overall reaction. In other words, the sum of the processes shown in Figure 14.19 must be consistent with the observed proportions of reactants and products. In this case, the sum of the two elementary steps matches the overall stoichiometry: (1)

2 NO/g) ~ NO(g)

NO/g) ~ NO(g)

(2)

I

Reaction Mechanisms

I

A reaction mechanism proposes a set of steps that describe how the reaction occurs. The mechanism must be consistent with the rate law for the reaction. An intermediate is a species produced in one step of a reaction and consumed in a following step. An elementary step is a molecular-level view of a single process taking place in a chemical reaction. A unimolecular step in a mechanism reaction involves only one molecule.

or

A bimolecular step in a mechanism reaction involves a collision between molecules.

or two

A termolecular step in a mechanism or reaction involves a collision between three molecules. The molecularity of a step in a mechanism refers to the number of particles (ions, atoms, or molecules) that collide in that step.

+ NO/g) + 02(g)

Summing steps 1 and 2 yields

2 NO/g)

+ N03(g) ~ 2NO(g) + N03(g) + 02(g)

Simplifying, we get the overall reaction

Nitrogen trioxide, N03, is an intermediate in this mechanism. It appears in both elementary steps but not in the overall reaction. Before ending our discussion of the thermal decomposition ofN02, let's consider how the concept of activation energy applies to a two-step reaction such as this one. The two elementary steps produce an energy profile with two maxima, as shown in Figure 14.20. In elementary step 1, collisions between pairs ofN02 molecules result in the formation of an activated complex associated with the first transition state in Figure 14.20. As this activated complex transforms into NO and NO}, the energy of the system drops, reaching the bottom of the valley between the two maxima. The top of the second energy barrier represents the transition state associated with a second activated complex in which the bonds in double bond forms to give N03 rearrange. Two N-O bonds break and an 0-0 the final products, NO and O2. For the reaction in the forward direction, the activation energy for elementary step 1 is much greater than that for elementary step 2. This difference is consistent with the relative rates of the two elementary steps: the first elementary step is slower than the second one. When the reaction proceeds in the reverse direction (as NO and O2 react, forming N02) the first energy barrier is the smaller of

709

~II

CHEMTOUR

Reaction

Mechanisms

710

I Chapter Fourteen

I

Chemical Kinetics

the two, so we expect a faster first elementary step in the reverse process. Experimental evidence supports this expectation.

Rate Laws and Reaction Mechanisms The observed rate law for the decomposition second order in N02: Rate

=

of N02

is

k[N02F

For a mechanism to be accepted as providing a possible view of a chemical reaction, it must be consistent with the rate law derived from experimental data. Is the proposed mechanism in Figure 14.20 consistent with the rate law? For it to be consistent, the molecularity of one of the elementary steps in the mechanism must be the same as the order of the reaction expressed in the rate law. Remember that we cannot write a rate law from a Progress of reaction ~ balanced chemical equation. The individual elementary FIGU RE 14.20 The reaction-energy profile for the decomposition ofN02 steps that show how a reaction proceeds and account for to NO and O2 features activation energy barriers for both elementary steps the formation and reaction of intermediates, however, in the overall reaction mechanism. The activation energy of the first step is are different from the overall balanced equation. A rate larger than that of the second step, so the first step is the slower of the two law for an elementary step can be written directly from and thus is the rate-determining step. its balanced equation. For the decomposition of N02, therefore, the rate law for the first elementary step is Ea,step 1 (slow)

Ea, step 2 (fast)

and for the second step Rate2

The rate-determining step is the slowest one in a multi-step chemical reaction.

=

k2[N03]

where kj and k2 represent the rate constants for the respective elementary steps. Comparing the rate laws for the two elementary steps with the observed rate law for the overall reaction shows that the rate law for elementary step 1 is exactly the same as the experimentally determined rate law. This means that step 1 defines how rapidly the entire reaction proceeds. It is the rate-determining step in the reaction. The rate-determining step is the slowest elementary step in a chemical reaction. The rate of the slowest elementary step controls the rate of the overall reaction. In the case of the decomposition ofN02, elementary step 1 is the slower of the two steps. One way of visualizing the concept of a rate-determining step is to analyze the flow of people through a busy airport. Typically, the number of people in the line outside the security point greatly exceeds the number of available security gates. The time required to make it to your flight depends on the time needed to pass through security. Security screening is an example of the rate-determining step on your way to your flight. Which step is rate determining in the decomposition of NO/ The observed rate law is identical to the rate law written for step 1. Therefore, we conclude that the first step is the rate-determining step that governs how rapidly the overall reaction proceeds. The rate constant (kj) is smaller than the rate constant for step 2 (k). There is actual experimental evidence that kj « k2• Chemists have studied the kinetics of the formation and decomposition of N03 extensively and have

I

14.5

determined that the rate constant of step 1 is only about 1 X 10-10 M-1 S-I at 300 K. The rate constant of step 2 is enormous by comparison: 6.3 X 104 S-I. Therefore, as soon as any N03 is formed in step 1, it rapidly falls apart, forming NO and O2 in step 2. Similarly, at the airport, as soon as people get through security (the rate-determining step), they move rapidly to the gates for their flights. Now let's consider the reverse of the thermal decomposition ofN02-namely, the formation ofN02 by the reaction of NO and O2, a key step in the formation of photochemical smog:

Reaction Mechanisms

(1) Formation

of intermediate

Jf

Formation

kf

I 711

Decom:,osition

(2) Reaction of intermediate

The observed rate law for this reaction is second order in NO and first order in O2:

One proposed mechanism includes two elementary steps, as shown in Figure 14.21. (1)

NO(g)

+

(2)

N03(g)

+ NO(g)

02(g) ~ NO/g) ~ 2 NO/g)

Rate

=

Rate

= kJN03][NO]

kl[NO][02]

If the first step were the rate-determining

step, then the reaction would be first order in NO and O2, but that is not what the experimentally determined rate law indicates. If the second step were the rate-determining step, then the reaction would be first order in NO and NO}, but that is not what the rate law stipulates either. Some other option must be suggested that gives the same picture as the rate law. Consider what happens if the first step is fast and reversible. In that case, during the reaction, N03 is rapidly produced from NO and O2 but decomposes rapidly back into NO and O2. In other words, the rate of step 1 in the forward direction matches the rate of step 1 in reverse, and both are rapid. Expressed mathematically, we have

Overall reaction:

+

FIGURE 14.21 A mechanism formation

ofN02

steps: (1) in a fast, reversible bimolecular reaction, NO and O2 form N03,

Rate forward reaction

=

k[[NO][02]

=

a slower rate-limiting

fast

for the

from NO and O2 has two

bimolecular

and (2) in reaction,

N03 reacts with a second molecule of NO,

Rate of reverse reaction

= kJNOJ

= equally fast

forming two molecules of NOr

where kf and kr are the rate constants for the forward and reverse reactions, respectively. Combining the two rate-law expressions, we have

Solving for [N03],

we have [NOJ

=

kf

I[NO][02]

(14.30)

r

Remember that the rate law for the rate-determining step in the mechanism must be the same as the experimentally determined rate law. The rate law for the first step does not match the observed rate law. What about the rate law for step 2? The rate law for step 2 is (14.31) Equation 14.31 still doesn't match the observed rate law for the reaction:

712

I Chapter Fourteen

I

Chemical Kinetics

However, if we combine Equations 14.30 and 14.31 by replacing the [N03] in Equation 14.31 with the right-hand side of Equation 14.30, we get Rate

=

k2

term

kf

k [NOP[02] r

The three rate constants can be combined into one overall rate constant

and the rate-law expression for the reaction becomes Rate

= koverall[NOP[02]

This expression matches the observed order of the reaction, so the proposed mechanism, a fast reversible first step followed by a slow second step, may be valid. The following exercises provide additional practice in relating experimental rate laws to proposed reaction mechanisms.

Linking Reaction Mechanisms to Experimental Rate Laws At high temperature, vapor:

The experimentally

NO reacts with hydrogen, producing nitrogen and water

determined rate law for the reaction is

It has been proposed that this reaction takes place in two elementary steps: (1)

2NO(g)

(2)

N20(g)

+ H2(g) ~ N20(g) + H20(g)

+ H2(g)

~ N2(g)

+

H20(g)

Is the proposed reaction mechanism consistent with the stoichiometry of the reaction and the rate-law expression? If so, which step in the proposed mechanism is the rate-determining step? AND ORGANIZE We are first asked if the proposed mechanism is consistent with the stoichiometry of the reaction. We are given the two steps in the proposed mechanism, and we can use them to answer that question. We are then asked if the mechanism is consistent with the experimentally determined rate law. We can look at the molecularity of the steps and the rate law and answer that question. Finally, we are asked to identify the rate-determining step in the mechanism.

COLLECT

ANALYZE

To answer the questions posed in this exercise we need to do the

following: 1. Determine whether the chemical equations of the elementary steps add up to the overall reaction equation. 2. Compare the rate-law expressions of the elementary steps to the experimental rate law of the overall reaction to determine which step is rate determining.

14.5

SOLVE

Let's test whether the elementary steps add up to the overall reaction: (1)

2 NO(g)

(2)

N20(g)

+ H/g) ~ N20(g) + H20(g) + H/g) ~ N2(g) + H20(g)

Summing these reactions gives

2NO(g) + 2H2(g) + N20(g) ~ N20(g) + N2(g) + 2H20(g) Simplifying leads to

2NO(g) + 2H/g)

+ N20(g) ~ N20(g) + N2(g) + 2H20(g)

or

which is indeed the same equation as that of the overall reaction. The elementary steps are consistent with the stoichiometry of the overall reaction. Next we need to focus on the reaction mechanism. The first step involves two molecules of NO colliding with one molecule of H2 in a termolecular reaction. The second reaction is a bimolecular reaction between the N 20 produced in the first step and another molecule of H2. Now we make use of the fact that the rate laws for elementary steps can be written directly from their stoichiometries. Therefore: (1) Rate = kj[H2][NOP (2)

Rate

= k2[H2][N20]

The rate law of step 1 matches the observed rate law of the overall reaction. Therefore, the proposed two-step mechanism is consistent with the experimental rate law, and step 1, by definition, is the rate-determining step. ABOUT IT Remember that the mechanism in this Sample Exercise is only a proposed mechanism. We do not have direct proof that it is the correct mechanism. Nevertheless, it is consistent with the available data. An acceptable mechanism for a reaction must contain a balanced equation whose stoichiometry is consistent with the rate law for the overall process. This step is the ratedetermining step in the reaction. Notice that no intermediates can appear in the overall rate law but may appear in individual steps.

THINK

Practice Exercise Here is another proposed mechanism for the reduction of NObyH2:

(1)

H2(g)

+ NO(g) ~ N(g) + H20(g)

(2)

N(g)

+ NO(g) ~ N2(g) + O(g)

(3)

H/g)

+ O(g) ~ H20(g)

Is this a valid mechanism?

As a final point regarding reaction mechanisms, even though the proposed reaction mechanism is consistent with the reaction's overall stoichiometry and with the experimentally derived rate-law expression, these consistencies do not prove that the proposed mechanism is correct. We need more experimental evidence, such as detecting the presence of an intermediate (such as N20 in Sample Exercise 14.12) in the reaction mixture. On the other hand, not finding a reactive (and transient) intermediate would not necessarily disprove a reaction mechanism.

I

Reaction Mechanisms

I 713

714 I Chapter

Fourteen

I

Chemical

Kinetics

Mechanisms and Zero-Order Reactions Before we leave this section on reaction mechanisms, let's consider the mechanism of the reaction between N02 and CO producing NO and CO2 that we last saw at the end of Section 14.3: N02(g)

+

CO(g) ~ NO(g) Rate

=

+

CO/g)

k[N02F

The observed rate law for this reaction is zero order in CO, second order in N02, and second order overall. What do these experimentally determined facts tell us about the reaction? First, remember that the rate law describes the process taking place in the rate-determining step. The rate law does not include a term for CO; that means that CO is not involved in that step in the mechanism. CO is clearly involved in the reaction-it is converted into COr But whatever step involves CO must be fast compared to the rate-determining step. This leads us to the conclusion that the reaction must involve at least two elementary steps, one rate determining and one not. The elementary step that is rate determining must involve two molecules of NO 2 -the rate law tells us that because it is second order in N02. In light of these facts, it has been proposed that this reaction takes place in two elementary steps: (1) (2)

2 NO/g) N03(g)

+

~ N03(g)

CO(g) ~ N02(g)

+ NO(g)

Rate

= k1[N02F

+

Rate

= kJNOJ[CO]

CO2 (g)

The experimental rate law matches the rate law for the first step, which must be the slower, rate-determining step.

mm

Catalysis

We noted in Section 14.4 that spontaneous reactions may be slow if they have a high activation energy. Suppose we wanted to increase the rate of such a reaction. How could we do it? One way is to increase the temperature of the reaction mixture. However, in some chemical reactions, elevated temperatures can lead to undesired products or lower product yields. Another way is to add another substance, called a catalyst, to the reaction mixture. A catalyst is a substance that increases the rate of a reaction but is not consumed by it. In this section, we consider catalysts that exist in the atmosphere and have an impact on the levels of ozone in the stratosphere over Antarctica. We also examine the catalysts that were developed by chemists to reduce air pollution from vehicle exhaust, thereby addressing the problems with photochemical smog that we described in Section 14.1. Later in this book, we return to catalysis in the biological context when we discuss the roles enzymes play in all life forms.

Catalysts and the Ozone Layer

A catalyst

is a substance added to a reaction that increases the rate of the reaction but is not consumed in the process.

As we discussed in Chapter 8, the way we think about ozone depends on where the ozone is located. Ozone in the stratosphere between 10 and 40 km above Earth's surface is necessary to protect life from harmful UV radiation; ozone at ground level is hazardous to human health and life in general. In this section, we discuss the role of catalysis in the breakdown of ozone in the stratosphere that has led to the formation of the ozone hole.

I

14.6

Catalysis

I 715

In 1974, two American

scientists, Sherwood Rowland and Mario Molina, predicted significant depletion of stratospheric ozone because chlorofluorocarbons (CFCs) released at ground level ultimately enter the stratosphere. We will examine how CFCs can deplete stratospheric ozone later in this section. For now just note that Rowland and Molina's predictions were later supported by experimental evidence of a thinning of the ozone layer over Antarctica during springtime in the Southern Hemisphere. This region oflowered ozone concentrations is known as the ozone hole. By 2000, stratospheric ozone concentrations over Antarctica were less than half of what they were in the late 1970s (Figure 14.22), and the ozone hole covered nearly all of Antarctica and the tip of South America. Less severe thinning of stratospheric ozone was observed in the Northern Hemisphere. Antarctic ozone minimum (lat 60 -90 0

0

S)

200

? ·2

180

:>

c:

0

..D0 160

8 0)

c:

0 N

140

E :> E ·2

120

0

~ 100

80

1980

1985

1990

1995

2000

2005

Year Average area of Antarctic ozone hole

30

Area of North America

25

]

:;r c: .s

::;l

20

15

Area of Antarctica

S 0)

.~

Cf)

10 TQlijIOZQ""

:::=:=.a::::= 5

0

Area having low stratospheric ozone « 220 Dobson units) Average over 30 days Vertical line extends from minimum to maximum area 1980

1985

1990

1995

2000

2005

Year FIGU RE 14.22 Graphs describing the Antarctic ozone hole. (a) The change in the minimum 03 concentration over time from 1979 to early 2005 in Dobson units, a measure of ozone in the atmosphere. (b) The average area experiencing a decreased ozone concentration over the same period. Areas of North America and Antarctica shown by dashed Iines for reference.

100

CU

500

NASA photos showing the change in the ozone hole from September 1979 through October 2005. The colorized scale is in Dobson units (DU), where purple is low ozone (100 DU) and red is high ozone (500 DU). The ozone hole is considered to be wherever the ozone concentration is less than 220 Dobson units.

716

I

Chapter Fourteen

I

Chemical Kinetics

Ozone hole minimum (lat 40 -90 0

FIGURE 14.23 Graph showing annual variations in stratospheric ozone concentration during Antarctic winter and spring from 1990 through 2006.

280

---

260

~.~.,

•

0

S)

1990-2001 2003 2006 2007

'">'"

·2

:>

c:

0

..D0 200

8 0)

c:

180

0 N

0

0;1

i§

~ 1200 c:

B

u

'S

6

800

S

600

·5

400

§ c: o :>

100 80 Jun

200

U µ..

U

0198788 89 90 91 92 93 94 95 96 Year

(a)

.s

~

e

CFCI/g)

Nov

Dec

:~ 0.0~9929'3 ~

~ CFCI2(g)

+

(14.32)

CI(g)

Free chlorine atoms react with ozone to make chlorine monoxide ifboth reactants are in the gas phase:

0.2

c: o

(b)

Sep

In 1987, an international agreement known as the Montreal Protocol called for reduced production of ozone-depleting compounds, including CFCs. Revisions in 1992 called for an end to the production of most of these compounds by 1996. The Montreal Protocol has had a dramatic effect on CFC production and emission into the atmosphere. However, these compounds last for many years in the atmosphere, and recovery of the ozone layer may take most of the twenty-first century. Furthermore, despite all these precautionary measures, in September and October of 2006, a colder spring than usual resulted in the largest ozone hole on record (Figure 14.23). Three of the more widely used CFCs were CF2CI2, CFCI3, and CF3CI (Figure 14.24). The chemical inertness of CFCs means that they persist in the environment and so can be swept up to the stratosphere by wind and air currents in the troposphere. In the stratosphere, CFCs encounter UV rays with enough energy to break their C-CI bonds, releasing chlorine atoms:

B ]

Aug

Jul

Month

.."

~

140 120

1000

c: o

160

94

95 Year

96

(14.33)

97 ="98

@

FIGURE 14.24 (a) Worldwide production ofCFCs. (b) Release ofCFCs into the atmosphere. Both production and release of CFCs decreased since ratification of the Montreal Protocol in 1986. However, the chemical stability of these volatile compounds means that their concentrations in the atmosphere decrease much more slowly.

Chlorine monoxide then reacts with more ozone to yield oxygen and regenerate atomic chlorine: CIO(g)

+

O/g) ~ CI(g)

+ 2 O2 (g)

(14.34)

If we add Equations 14.33 and 14.34 together, we get CI(g)

+ 20/g)

+

CIO(g) ~ CIO(g)

+

CI(g)

+ 3 O2 (g)

(14.35)

Cancelling species that appear on both sides of the reaction arrow gives (14.36) The rates of the above reactions increase in the presence of fluid droplets such as those present in the clouds that cover much of Antarctica each spring. Actually,

14.6

I

Catalysis

I 717

these clouds form as crystals of ice and tiny droplets of nitric acid in the winter months, when there is no sunlight and temperatures in the lower stratosphere dip to -80°C. The clouds become collection sites for HCI, CIO, and other compounds containing chlorine. When the sun returns in August (at the end of the Antarctic winter), sunlight melts the ice and CIO is free to migrate to the surface of the droplets ofliquid water. These droplets catalyze reactions by adsorbing (binding to the surface) the reactants. The droplets are not reactants in themselves, but they provide a surface on which the reactants collect. The resulting proximity of the reactants increases the likelihood of interaction between them and decreases activation energies for the process. Evidence supporting the occurrence of these reactions in the stratosphere is presented in Figure 14.25. The concentration ofCIO decreases sharply in satellite images of the South Pole taken only one month apart, consistent with the reaction between CIO and 03 shown in Equation 14.34. Although it is difficult to track the concentration of CI atoms directly, some of them end up as the hydrogen chloride, HCI, present in the icy polar clouds that form during the winter and melt in August. The images in Figure 14.25 confirm that a decrease in CIO concentration between the end of Antarctic winter (late August) and the Antarctic spring (late September) is accompanied by an increase in HCI concentration, which reflects the rise in CI concentration.

(a) HCI: August 25, 2004

(c) HCI: September 24, 2004

(b) CIO: August 25, 2004

(d) CIO: September 24, 2004

FIGU RE 14.25 NASA satellite images of the South Pole showing the stratospheric concentrations of (a) hydrogen chloride (HC!) and (b) chlorine monoxide (CIO) in late August 2004. One month later, as spring arrives in the Southern Hemisphere, the concentrations of (c) HCI and (d) CIO have changed. The concentrations of the two compounds are inversely correlated. The concentration of CIO decreases as it reacts with ozone; the reaction yields chlorine atoms that in turn lead to an increased concentration ofHCl. The blue color indicates a low concentration and red indicates a high concentration.

718

I

I

Chapter Fourteen

Chemical Kinetics

A homogeneous catalyst phase as the reactants. A heterogeneous catalyst phase than the reactants.

is in the same

In terms of reactants and products, Equation 14.36 is the same as the one describing the naturally occurring photodecomposition of ozone in the stratosphere. The natural pathway for ozone destruction involves light:

is in a different

hv

203 ~302 The natural photochemical conversion of ozone to oxygen begins with the absorption of a photon ofUV radiation and the generation of atomic oxygen: (1)

03(g) ~ O/g)

+

O(g)

Oxygen atoms generated in this fashion can react with additional ozone molecules in a second step to form two more molecules of oxygen: (2) Ea, uncatalyzed

= 17.7 kJ/mol

~

"
(15.4)

The square brackets indicate molar concentrations of the species involved in the reaction, and each concentration is raised to a power equal to its coefficient in the balanced chemical equation. Products appear in the numerator and reactants appear in the denominator. If the substances involved in the reaction are all gases (as they are in the steam-reforming and the water-gas shift reactions), then the equilibrium constant can also be expressed in terms of partial pressures: (15.5) As we have seen in Chapter 6 and Chapter 10, it is common practice to describe amounts of gases in a mixture in terms of their partial pressures. Equation 15.5 is the generalized mass action expression written with respect to partial pressures.

Writing Equilibrium Constant

Expressions

At high pressures the dark brown gas N02 exists in equilibrium with the colorless gas N204, which is a dimer ofN02. Write expressions for Kc and Kp for the reaction ofN02 to produce N204. AND ORGANIZE We are given the reactant, N02, and the product, N 2 4' and asked to write equilibrium constant expressions for Kc and Kp for this system. These expressions relate the concentrations (K) or partial pressures (Kp) of products and reactants. COLLECT

°

To write expressions for K, we need a balanced chemical equation. For this reaction, the equation is

ANALYZE

SOLVE

The expression for Kc written in terms of molar concentrations

is

~II CHEMTOUR Gas Phase

Equilibrium

in the

742 I Chapter Fifteen

I


The expression for K p written in terms of partial pressures is K

P

=

(PN20)

-)2

(PN02

The mass action expression can be written for any reaction directly from the balanced chemical equation. Kc is used when concentrations are of interest. K p is used specifically for gas-phase reactions, when partial pressures of gases are of interest.

THINK

ABOUT

IT

Practice Exercise Write the equilibrium constant expressions, Kc and Kp' for the steam-reforming

reaction (Equation 15.1):


CONCEPT

TEST

With reference to the water-gas shift reaction on page 739, suppose an equal number of moles of steam, carbon dioxide, carbon monoxide, and hydrogen gas are injected into a rigid, sealed reaction vessel that is heated to 500 K. The mixture of gases is allowed to come to chemical equilibrium. Which of the following expressions about the equilibrium reaction mixture is true? a. [CO] b. [C02] C. [C02] d. [H20] e. [H20]

= [H2] = [C02] = [H20] = [H2] > [CO] = [H20]

= [H2] < [CO] = [H20] = [H2] > [C02] = [CO] = [H2] < [C02] = [CO]

(Answers to Concept Tests are in the back of the book.)

@Xi)

CONNECTION In Chapter 5, we deli ned the standard pressure as 1 bar, but stated that, because that value is so close to 1 atmosphere, any differences in values resulting from the substitution of 1 atm for 1 bar are almost always so small that they may be ignored.

@Xi) CONNECTION In Chapter 10 we discussed nonideal behavior of salts in solution because of attractive forces between ions, and we calculated the i factor to quantify this non ideal behavior in terms of colligative properties.

Remember that the value of K for a given system at a given temperature is constant. As is almost always the case with actual experimental data, you must be prepared for slight variation in calculated answers because of variations in the precision of the experiment and rounding. It is important to note at this point that equilibrium constants are reported without units because they involve measurements that account for nonideal behavior of substances participating in the reaction. Concentrations and partial pressures corrected for nonideal behavior are called activities, and when equilibrium constants are determined from thermodynamic data, they are defined in terms of activities. For concentration, activity is the ratio of the molar concentration of a substance to a standard concentration of 1 M. For pressure, activity is a ratio of the partial pressure of a gas in atmospheres to a standard pressure of 1 atm. Therefore each concentration or pressure in an equilibrium expression is actually a ratio in which the units cancel. We use concentrations and partial pressures, not activities, in this text, which means we assume that the behavior of substances is ideal. Because each term is understood to be a ratio of a concentration or partial pressure to a standard value, the units on each term cancel, which results in the value of K having no units.

15.2

I

Calculating Kc Table 15.2 contains data from four experiments at 100°C for the dimerization of NOz in a rigid, closed container at a high pressure. Calculate the value of Kc for each experiment.

INITIAL CONCENTRATION (M)

EQUILIBRIUM

CONCENTRATION (M)

Experiment

-

-

[N021

[N2041

[N021

[N2O.1

1

0.0200

0.0000

0.0172

0.00140

2

0.0300

3

0.0400

4

0.0000

0.0000

-

-

0.0244

0.0000

0.0310

0.0200

0.0310

--

0.00280 ----i

0.00452 --

0.00452

~

COLLECT AND ORGAN IZE We are given a data set that contains initial and equilibrium concentrations of a reactant and product. We are asked to determine the value of the equilibrium constant in each experiment.

We balanced the equation and wrote the equilibrium constant expression for this reaction in Sample Exercise 15.1:

ANALYZE

Because we are given data in terms of concentrations,

we use the expression for

K c here. SOLVE

Experiment

K = [Nz04] c [NOY 1: K c

Experiment 2: K c

Experiment

3: K c

Experiment

4: K c

= 0.00140

(0.0172)Z

=

= 4.73

0.00280 (0.0244)2

= 0.00452

(0.0310)Z

= 0.00452

(0.0310)2

=

4.70

= 4.70 = 4.70

Kc are nearly the same. We expect them to be the same because they are all based on experiments run at the same temperature.

THINK ABOUT IT The values of

Practice Exercise Mixtures

of gaseous CO and Hz called synthesis gas can be used commercially to prepare methanol (CH30H), a compound considered an alternative fuel to gasoline. Under equilibrium conditions at 700 K, [Hz] = 0.074 mollL, [CO] = 0.025 mollL, and [CH30H] = 0.040 mollL. What is the value of K c for this reaction at 700 K?

The Equilibrium Constant Expression and K

I

743


I


If all reactants and products are gases and we know their partial pressures in the system at equilibrium, we can use those values to calculate the equilibrium constant (K)p for the reaction.

Calculating Kp from Partial Pressures A sealed chamber contains an equilibrium mixture ofN02 and N204 at 300°C. The partial pressures of the two gases are P NO, = 0.101 atm and P N,O. = 0.074 atm. What is the value of Kp for the equilibrium between N02 and N204 under these conditions? COL LEe TAN D 0 R G A N I Z E We are asked to determine the value of the equilibrium constant for the reaction ofN02 to form N204• We are given partial pressures at equilibrium of both gases, so the constant is symbolized K p .

The equation for the reaction is

ANALYZE

SO

the mass action expression for the reaction based on partial pressures is

P

SOL V E

0)

(P

= __N 2_4

K

(P

N02

)2

Inserting the given partial pressures and solving for K p we have K p

=

0.074 (0.101)2

=

7.3

T The mass action expression in terms of K has the same form p as that for Kc; partial pressures for products are in the numerator and partial pressures for reactants are in the denominator. K p is calculated using partial pressures instead of concentrations.

T HI N K ABO UTI

Practice

Exercise A reaction vessel contains an equilibrium mixture of S02' 02' and S03' Given the following partial pressures of these gases: Pso, = 0.0018 atm, Po, = 0.0032 atm, and PS03 = 0.0166 atm, calculate the value of Kp for the reaction:

The size of K gives an indication of how far a reaction proceeds toward making products under a stated set of conditions. Values for equilibrium constants span a huge range. The ones we have seen thus far-Kc for the water-gas shift reaction at 500 K and both K p and K c for the dimerization ofN02 at lOOT-are considered intermediate in size. Their values are close to 1, and comparable concentrations of reactants and products are present at equilibrium. In contrast, the reaction between H2 and 02 to form water proceeds explosively, and when the reaction is over, very little starting material is left. As a result, K at 25°C is very large. Kc

= 3 X 1081

On the other side of the range, the decomposition of CO2 into CO and 02 at 25°C proceeds hardly at all. K for this equilibrium is very small and virtually no product is formed. Kc

= 3 X 10-92

15.2

I

Relationship between K c and K p The values of Kc and Kp for a given reaction may differ numerically, but they are related to each other via the ideal gas equation, PV = nRT If volume (V) is expressed in liters, we can rearrange this equation to express pressure (P) in terms of molar concentration (n/V = M): P=!!:...RT=MRT

V

For N02 and N204 in the equilibrium P

NO,

in Sample Exercise 15.3

n

=

= [NO 2 ]RT

N°'RT

V

and

Substituting

these values into the expression for K p we get _ (PN20)

or

[N204]RT

_

K---------p (P NO, )2 ([NO 2]RT)2

_

Kc RT

= Kc

K(RT) p

This equation defines the specific relationship between Kc and Kp for this reaction, but a more general expression can be derived from Equation 15.5

=

K

C

D

(PA)w(P B)x

P

In terms of concentrations,

(P )Y(P )z

the general expression is K

([C]RT)Y([D]RT)z ([A]RT)w([B]RT)x

P

[C)Y[D] [C]Y[O]z (RT)y+z :--_._-X --[A]W[B]x (RT)w+ x = K (RT)(Y+z)-(w+x)

(15.6)

c

To understand what this means, look back at the general equation. The term (y + z) is the sum of the coefficients of the gaseous products, and the term (w + x) is the corresponding value for the gaseous reactants. The difference between these two numbers is the power to which RT is raised in Equation 15.6. We can symbolize that term as !:J.n (change in the number of moles of gas) and write Equation 15.6 as K Looking

at the steam-reforming

CH/g) H20(g)

P

=

K(RT)6.>i

(15.7)

c

and water-gas

shift reactions

+ H20(g) ~ CO(g) + 3H/g) + CO(g) ~ H/g) + CO/g)

we see that steam-reforming of methane has 4 moles of products 2 moles of reactants (1 + 1), while the water-gas shift reaction has the ber of moles of products (1 + 1 = 2) as reactants (1 + 1 = 2). For steam !:J.n = 2, so K pcp= K (RT)2, and for the shift reaction!:J.n = 0 and K = (RT)O = 1.

(1 + 3) and same numreforming, K,c because

The Equilibrium Constant Expression and K

I

745


I


Calculating Kc from Kp In Sample Exercise 15.3, we calculated Kp for the dimerization ofN02 to N204 in a sealed chamber at 300°C. What is the value of K c for this reaction at 300°C (K p = 7.2)? AND ORGANIZE We are given K p and asked to calculate K c for the reaction under the same conditions. COLLECT

We need a balanced chemical equation to determine the change in the number of moles of gas in going from reactant to product.

ANALYZE

We can determine Cln and use Equation 15.7 to calculate Kc' SOL V E

The balanced equation shows 1 mole of product and 2 moles of reactant, so Cln

= (moles of product - moles of reactant) = 1 - 2 = -1

K

=

P

K(RT)D.U c

T= 300

+ 273 = 573K K

7.2 = K (0.08206 X 573t1 = ( c)( c 0.08206 573 Kc

= (7.2)(0.08206)(573) = 339

T The value of K c differs from the value of K p because the total number of moles of gaseous product is not the same as the total number of moles of gaseous reactant.

T HI N K ABO UTI

Practice Exercise For the reaction that produces ammonia from hydrogen and nitrogen at 30°C, the value of K c is 2.8 X 10-9• What is K p for this reaction under the same conditions?

~ @Xi)

CONNECTION We carried out many of these mathematical operations in Chapter 5 in calculating how much heat is generated or absorbed by chemical reactions.

Manipulating Equilibrium Constant Expressions

Equilibrium constant expressions may be written in different ways for a given reaction just as chemical equations may be written differently. We can write equilibrium constant expressions for reactions running in reverse, for chemical equations that have been multiplied or divided by a value that gives a key component a coefficient of one, and for overall reactions that are combinations of other reactions.

K for Reverse Reactions The K values for forward and reverse reactions in an equilibrium are related. For example, an equilibrium system involving A and B could be written

A~B or

15.3

I

Manipulating

For example, in Sample Exercise 15.1, we derived the mass action expression for the dimerization reaction

as

[N204]

K=-c [NOY

The equation for the reverse reaction in which N02 forms from N204

gives rise to the following mass action expression for the decomposition of the dimer:

reaction

Let's compare the two mass action expressions: the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Expressing this relation mathematically gives 1 K=-

K

f

(15.8)

r

where the subscripts "f" and "r" indicate the forward and reverse reaction directions. If a reaction has a large equilibrium constant, then the system at equilibrium contains mostly product and little reactant. In the reverse reaction, there should be little product and mostly reactant at equilibrium and a correspondingly small equilibrium constant. For example, the equilibrium constant (K) for the dimerization ofNO/o N204 is 4.7 at lOOT. Using Equation 15.8, the value of the equilibrium constant (K) for the reverse reaction, the decomposition ofN204 to N02, at 100°C is

= -1 = - 1 = 0.21

K r

Kf

4.7

The equilibrium constant for the reverse of a reaction is the reciprocal of the equilibrium constant for the reaction in the forward direction. We must always specify the reactants, products, and temperature when reporting Kc or Kp'

Calculating K for a Reverse Reaction When NO is formed during combustion in automobile engines, it reacts with available 02 to form N02. The reverse reaction is the decomposition of N02 to produce NO and 0r At 184°C, the value of Kc for the forward reaction is 1.48 X 104• Write the mass action expressions for both reactions and determine the value of K c for the reverse reaction. AND ORGANIZE We are asked to write mass action expressions for a forward and reverse reaction for a process at equilibrium and to calculate the equilibrium constant for the reverse reaction. We are given the reactants and products and the value of Kc for the forward reaction. COLLECT

A N A L Y Z E We need a balanced

chemical equation describing the formation of N02 from the reaction of NO and 02' The reverse reaction is the decomposition ofN02 to reform the reactants. Forward reaction: Reverse reaction:

2 NO(g)

+ 02(g) ~

2 N02(g)

2 N02(g) ~ 2 NO(g)

+ a/g)

Equilibrium Constant Expressions

I

747


I


SOL V E

The mass action expressions for the forward and reverse reactions are [N02P

K = .,,-----:-::~

Forward reaction:

f

Reverse reaction:

= -.l =

K r

T HI N K ABO UTI

K

1.48

Kr = 1

X

104

[NOP[02J

[NOP[02J ---::::[N-=-O-=-2-=-=P-

= 6.8

X 10-5

T The value of Kf is large, so it makes sense that its reciprocal,

Kr' is small. Practice Exercise At 303 K, the equilibrium constant for the gas phase reaction ofN2 and H2 to form NH3 is 2.8 X 10-9. What is the equilibrium constant for the decomposition and nitrogen?

of ammonia

at the same temperature

to produce hydrogen

K for Equations Multiplied by a Number As we saw in Chapter 5, sometimes reactions are written with fractional coefficients to enable calculation of properties that reflect the behavior of 1 mole of a substance. As an illustration of this, consider the forward reaction in Sample Exercise 15.5:

We could also write this reaction to reflect the preparation multiplying each coefficient by ~ to give

NO(g)

of 1 mole of N02 by

+ ~O/g) ~ NO/g)

(15.9)

Because the mass action expression must be written to match the balanced chemical equation, its form for Equation 15.9 is

If we compare this expression to the mass action expression written ward reaction in Sample Exercise 15.5:

we see that the equilibrium constant expression for Equation expression for Kf to the ~ power (the square root of Kf):

for the for-

15.9 is equal to the

In general terms, if the balanced equation for a reaction is multiplied by some factor n, then the value of K for that reaction is raised to the nth power. In the illustration for the reaction ofN02, the original equation is multiplied by a factor of~, so the equilibrium constant of the original equation is raised to the ~ power.

15.3

I

Manipulating

Calculating K for Different Coefficients An important reaction in the formation of atmospheric aerosols of sulfuric acid (and acid rain) is the oxidation ofS02 to S03' which combines with H20 to form H2S04, One way to write a chemical equation for the oxidation reaction is Equation

1:

The value of K c for this reaction at 298 K is 2.8 X 1012• What is the value of K c at 298 K for the following reaction? Equation

2:

COLLECT AND ORGANIZE We are given Kc for a reaction that contains fractional coefficients and describes the production of 1 mole of SO 3' The reaction is rewritten so that the coefficients are all whole numbers. We are asked to calculate K c for the rewritten reaction.

A N A L Y Z E The first equation

is multiplied by 2 to generate the second equation. This means that Kc for the first equation must be squared (raised to the second power) to generate Kc for the rewritten equation. SOLVE

K2=(Kl =

(2.8 X 1012)2

=

7.8 X 1024

IT The concentrations (or partial pressures) of reactants and products are raised to a power defined by the coefficients in a balanced chemical equation. Therefore, doubling the coefficients means squaring the value of K.

TH IN K ABOUT

Practice Exercise At 1000 K, the reaction

has a K c value of2.4 reaction:

X

10-3. What is the value of K c at 1000 K for the following

Writing these equations may cause you to wonder how the same reaction having the same reactants and products can have two or more equilibrium constant values. Surely the same ingredients should be present in the same proportions at equilibrium no matter how we choose to write a balanced equation for their reaction. They are. The difference in K values is not chemical, it is mathematical. It is related to how we use the equilibrium concentrations in calculating their ratios. It is, for example, about choosing to use [NO] and not [NOP in such a calculation. That choice does not affect the actual concentration of NO at equilibrium. Always remember that the description of an equilibrium must be defined with respect to a specific balanced equation, and that the coefficients in that equation are the powers to which concentrations or partial pressures are raised in the specific expression for K in the system described.


I

749

750

I Chapter Fifteen

I


Combining K Values In Chapter 5, we applied Hess's law to calculate the enthalpies of combined reactions. We carry out a similar process with equilibrium constants to determine the overall K for a reaction that is the sum of two or more other reactions. This technique is especially useful in situations such as the one we faced in the chapter on kinetics involving photochemical smog, where the product of one reaction was frequently the reactant in another reaction occurring at the same time. Consider two reactions that we discussed in Chapter 14 that are involved in the formation of photochemical smog. The NO produced from N2 and 02 by car and truck engines may be further oxidized with additional 02 to form NOr If we sum the two reactions to obtain the overall reaction for the formation ofN02, we get (1)

N/g)

(2) 2NO(;R) Air quality in Los Angeles, California

+

a/g)

+

02(g) ~ 2N02(g)

~ 2NO(y)

has

improved since this photo of "brown LA

The mass action expression for the overall reaction is

haze" was taken. The color was caused by high concentrations

ofN02•

We can derive this expression from the equilibrium tions 1 and 2: [NOF

K = ."---.,,.,,--~ j

[N2] [02]

constant expressions for reac-

and

if we multiply Kj by K2

This approach works for all series of reactions, and as a general rule

(15.10) The overall equilibrium constant for a combination of two or more reactions is the product of the equilibrium constants of the reactions. The value of Kc for the overall reaction for the formation ofN02 from N2 and 02 at 1000 K is the product of the equilibrium constants for reactions 1 and 2:

K = j

[NOP [N2] [02]

[N02]2 K2 = [NOP[02]

= 72 X 10-9 .

= 0.020

Using Equation 15.10, Koverall = Kj X K2 = 7.2 X 10-9 X 0.020 = 1.4 X 10-10. Remember that the equilibrium constant expression for the overall reaction must have the appropriate product and reactant terms for that reaction. Just as with Hess's law in thermodynamics, we may need to reverse an equation or multiply an equation by a factor when we combine it with another to create the equation of interest. If we reverse a reaction, we must take the reciprocal of its K; if we multiply a reaction by a constant, we must raise its K to that power.

15.3

Calculating

I

Manipulating

K by Combining Equilibria

At 1000 K, the equilibrium constant Kc of the reaction N204(g)

~ 2N02(g)

is 1.5 X 106. On the basis of this value and the value of the equilibrium constant for the reaction

+ 20/g)

N/g)

~ 2 NO/g)

at 1000 K (Kc = 1.4 X 10-10), calculate the value of the equilibrium constant at 1000 K for the reaction

+2

N2(g)

°2

(g) ~ N204(g)

COL LEe TAN D 0 R G A N I Z E We are given two reactions and the values of their equilibrium constants. We need to combine the two reactions in such a way that

°

N 2 and side of the overall equation and N 2 2 are on the reactant product side.

°

4

is on the

A N A l Y Z E The overall reaction is the sum of the reverse of the first reaction and

the second reaction as written: 2Jl>JO(p) ~

N20/g)

N2(g)

+2

?' ll'JO

Overall N2(g)

+2

°2 °2

(g) ~

(o)

(g) ~ N204(g)

SOLVE When a reaction is the sum of two other reactions, then Koverall = KJ X K2. We are given K2 in the problem for the reaction as used in the combination, but we reversed the first reaction. Therefore we must use the reciprocal of the value of Kl in our calculation:

1 Koverall =-XK K

1 1.5 X 106 X 1.4 X 10-10 = 9.3 X 10-17

2

J

T Remember that values of K always relate to a specific reaction and how it is written. If we alter a reaction by writing its reverse, or multiply it by some factor, we must always make corresponding changes in the value of the equilibrium constant.

T HI N K ABO UTI

Practice

Exercise

Calculate the value at K c for the reaction

Q(g)

+ X(g)

~ M(g)

from the following information: 2M(g) ~

Kc

Z(g)

Z(g) ~ 2Q(g)

+ 2X(g)

To summarize the key points in manipulating

= 6.2

X 10-4

K c = 5.6 X 10-2 equilibrium constants:

• K values are unitless. • The value of K for a reverse reaction in an equilibrium is the reciprocal of the K of the forward reaction. • If the original equation describing an equilibrium is multiplied by some factor n, then K for the new expression is calculated by raising the original K to the nth power.


I 751

752

I

Chapter Fifteen

I


The reaction quotient (Q) is the numerical value of the mass action expression for any values of the concentrations (or partial pressures) of reactants and products. At equilibrium, the value of the reaction quotient (Q) equals that of the equilibrium constant (K).

~

Equilibrium Constants and Reaction Quotients

If we were in the business of making hydrogen gas from the water-gas reaction,

shift

we could start with any combination of the products and reactants and let them come to equilibrium. In point of fact, if we were using this reaction to produce hydrogen, we would start with a reaction mixture that does not contain any Hr If the concentration of one of the reactants or products in a reversible reaction is zero, the reaction will proceed in the direction that produces some of that substance. If all the reactants and products are present, however, we must apply the law of mass action to determine the direction in which the reaction will proceed. Let's examine the water-gas shift [H20] and [CO] No change in [H2] and [C02] reaction (Figure 15.3a) with this idea in mind. If we react to form more ~ concentrations of +-react to form more sampled the reaction mixture at some time point after reactants and products [H20] and [CO] [H2] and [C02] CO and H20 were initially mixed but before equilibrium was achieved, the concentrations ofH2 and CO2 would be less than their equilibrium concentrations. If we were to solve the mass action expression using the concentrations of substances present at this point, the numerator would be smaller than at equilibrium ~ ~ and the denominator would be larger. Consequently, Equilibrium: the value of the expression would be smaller than the Reactants ~ Products Reactants +-Products no net change equilibrium constant. The value of the mass action expression written for nonequilibrium concentrations /[H20] or [CO] is called the reaction quotient (Q), and its size relative to K indicates the direction of a reaction on its way c: c: ~o to equilibrium. In the example of the water-gas shift Q=K .~ I Q>K X Q K, and example, when we start with only products in the products react to form more reactants. reaction vessel. Under those conditions, the numerator in the mass action equation is too large and the reaction runs in the reverse of the written direction TABLE 15.3 Comparison of Q and K Values (Figure 15.3c), reducing the concentration ofH2 and CO2 and increasing the concentrations of H20 and Value of Q What It Means CO until equilibrium is achieved. The system shifts Reaction proceeds in the forward direction (~) QK marized in Table 15.3.

iI\.

i oj

1\

15.4

I

Equilibrium Constants and Reaction Quotients

Sample Exercise 15.8 shows how we can tell whether a system is at equilibrium or not. If it is not at equilibrium, we can predict the direction in which the system will shift to attain equilibrium.

Using Q and K A reaction vessel at 2300 K contains the following molar concentrations of gases: [N21

[0,1

[NOI

0.50

0.25

0.0042

If K c = 1.5 X 10-3 for the reaction

at 2300 K, is the reaction mixture at equilibrium? If not, in what direction will the reaction proceed to reach equilibrium? We are asked to evaluate if a reaction is at equilibrium given molar concentrations of the substances involved and the value of Kc under the stated conditions. COLLECT

AND

ORGANIZE

A N A L Y Z E The mass action expression for this reaction based on concentrations

is

If we insert the given concentration values into the mass action expression, we can determine the value of Q. Comparing Q to K will enable us to determine if the reaction is at equilibrium or the direction in which the reaction will proceed if it is not. SOLVE

Q = c

(0.0042)2 (0.50)(0.25)

= 1.4 X 10-

4

The value of Qc is less than the value of Kc' so the reaction mixture is not at equilibrium. To achieve equilibrium more of the product must form and the concentration of the reactants must decrease, which happens if the reaction proceeds in the forward direction. ABOUT IT Determining the reaction quotient Qc and comparing it to Kc enables us to predict if a reaction proceeds and whether it proceeds in the forward or reverse direction under a given set of concentrations.

THINK

Practice Exercise The value of K c for the reaction 2N02(g) ~ N20/g) is 4.7 at 373 K. Is a mixture of the two gases in which [N02] = 0.025 M and [N204] = 0.0014 M in chemical equilibrium? If not, in which direction does the reaction proceed to achieve equilibrium?

I 753

754

I

Chapter Fifteen

I


.mw Equilibrium

~II CHEMTOUR Equilibrium and Thermodynamics

@Xi) CONNECTION In Chapter 13 we defined Gibbs free energy (G) as the energy available to do useful work. t::.G is a state function that provides a criterion for spontaneous change.

and Thermodynamics

We have seen that the value of the equilibrium constant for a reaction does not depend on the kinetics of the reaction. Knowing how rapidly a reaction proceeds tells us nothing about the position of equilibrium; correspondingly, knowing the value of the equilibrium constant tells us nothing about how rapidly equilibrium is achieved. However, the equilibrium constant is directly related to the thermodynamics of the reaction and, in particular, to the change in free energy, t::.G. In Chapter 13 we introduced the change in Gibbs free energy, t::.G, as a thermodynamic function that indicates the tendency of a system to change under conditions of constant temperature and pressure. For example, when two phases are at equilibrium, as are the solid and liquid phases of a pure material at its melting point, the solid melts as rapidly as the liquid freezes, no net change takes place, and the system is at equilibrium. In this situation, t::.G = O. The same considerations apply to chemical systems, where the rates of forward and reverse reactions at equilibrium are equal, no net change in concentrations of reactants and products takes place, and t::.G = O. Because the sign of t::.G indicates whether a process as written is spontaneous, the sign of t::.G for a reaction mixture tells us the direction in which the reaction proceeds to achieve equilibrium. If t::.G is negative, then a reaction is spontaneous as written and proceeds in the forward direction. If t::.G is positive, then the reaction as written is not spontaneous; the reverse of the reaction has a negative t::.G value and is spontaneous, and the reaction of products to form reactants proceeds. As a spontaneous reaction under constant temperature and pressure proceeds, the concentrations of reactants and products change, and the free energy of the system changes as well. At equilibrium, t::.G is zero, no free energy remains to do useful work, and the reaction has achieved equilibrium. The magnitude of t::.G-how far it is from zero in either a negative or positive direction-indicates how far a system is from its equilibrium position. Whether Q is larger or smaller than K also tells us how far a system is from equilibrium, so it is reasonable to think that Q and t::.G may be related. They are, and their mathematical relationship is one of the most important connections in chemistry because it enables us to relate experimentally measurable properties of substances to the equilibrium composition of a chemical system. The thermodynamic view of equilibrium and the relationship between t::.G and Q are described by the following equation:

t::.G = t::.G

o

+ RT

In

Q

(15.11)

Let's explore what this means by looking at a reaction we have examined several times in this chapter, the dissociation ofN204 to N02:

We can calculate the change in standard free energy for the reaction (t::.G:xJ the conversion ofN204 into N02 as we did in Chapter 13 using the formula

t::.G rxn = It::.Go O

f,prod

-

It::.Go

f,react

for

(15.12)

= 2 mel (51.3 kJlmel) - 1 mel (97.8 kJlmel) = +4.8 kJ o

The positive value of t::.G rxn indicates that the reaction, as written, is not spontaneous at 298 K; but understand that this value of t::.G:xn applies only under standard conditions when PNO and PN 0 equal 1 atm. That is not the case when N02 and 2 2 4 N204 partial pressures vary.

15.5

t

I

Equilibrium and Thermodynamics

I 755

MolesN02 Let's explore what happens in a reaction vessel 0 0.4 0.8 1.2 1.6 2.0 at 298 K that initially contains 1 mole ofN204 at a partial pressure of 1 atm. While 1 mole of N204 102.6 k] ~----------------------------------------------------------------J (J) will never completely turn into 2 moles of N02 under these conditions, some fraction (x) of 1 mole ofN204 will decompose into 2x moles of NOr In t::..GO= other words, the decomposition reaction proceeds G 4.8 kJ a little and then stops as chemical equilibrium is reached. Figure 15.4(a) shows how free energy (G) changes as the amounts of product and reactant 97.8 k] . CD change in the reaction mixture during the conversionofN204 to N02 at25°C and 1 atm pressure. The minimum in the plot of free energy for the reaction 1.0 0.8 0.2 o (point ®) corresponds to a free energy lower than (a) that of either pure reactant (point CD)or pure product (point@). In this system under the stated condi10 tions, this minimum occurs for a gaseous mixture of 0.828 moles ofN204 and 0.345 moles ofN02. Extent of reaction To achieve such a mixture, (1.00 - 0.828) or 0.172 moles ofN204 decompose into 0.345 moles ofNOy ~ -10 In other words, x = 0.172 mol. 2 c.:J Now let's continue the story by asking what 0).

N2(g)

+

3 H/g)

~ 2 NH3(g)

+ heat

and the reverse reaction is endothermic. If we think about heat as a product in this reaction, we can use Le Chatelier's principle to predict the result of a temperature change. Increasing the temperature favors the reverse reaction; decreasing the temperature favors the forward reaction. One major difference arises, however; unlike changes in concentration or pressure, temperature changes cause changes in the value of K. An increase in the temperature would cause the concentration of ammonia to drop and the concentrations of nitrogen and hydrogen to rise. A new equilibrium characterized by a smaller K would be established. In general, the value of K for exothermic reactions decreases with increasing temperature and increases if the temperature decreases. The equilibrium constant for endothermic reactions has the opposite behavior: K increases with increasing temperature and decreases as the temperature falls. At the end of this chapter, we will look in greater detail at the influence of temperature changes on K values, but for now this general analysis will enable us to explain stress on equilibria caused by changing temperature.

Predicting Changes in Equilibria with Temperature The color of an aqueous acidic solution of CoCl2 depends on the temperature. In aqueous HCI, the solution is pink when ice-cold, magenta at room temperature, and dark blue at elevated temperature. The equilibrium can be described as Pink ~ blue Is the reaction as written exothermic or endothermic? COLLECT AND ORGANIZE We are asked to determine if an equilibrium as written is exothermic or endothermic (is heat a reactant or product?). Temperature

= 5°C

Temperature

= 75°C

y

y

Pink

Royal blue

Two different forms of cobalt, one pink and one blue, are in equilibrium

in aqueous hydrochloric

The position of equilibrium changed,

acid solution.

shifts as the temperature

causing the color of the solution to change.

is

A N A L Y Z E If heat is a product,

the reaction is exothermic; if it is a reactant, the reaction is endothermic. If we examine how temperature changes alter the position of equilibrium, we can determine the role of heat in the reaction.

SOLVE When heat is added to the system, the pink color turns blue. That means heat is a reactant, because increasing it pushes the reaction to the right. The reaction as written must be endothermic:

Heat

+ pink

~

blue

THINK ABOUT IT To determine the effect of increasing or decreasing temperature on an equilibrium, we can treat heat as a reactant or product if we know whether the reaction as written is endothermic or exothermic.

Practice Exercise Predict how the value of the equilibrium following reaction changes with increasing temperature: O

tl.H

=

181 kJ

constant of the

15.7

Table 15.4 lists examples of stress on chemical equilibria and how a system at equilibrium responds for an exothermic reaction in which 2 moles of gas "A:' combine to make 1 mole of gas "B".

'ABLE 15.4

Responses of an Exothennic Reaction

=--l

[2 A(g) ~ B(g)l at Equilibrium to Different Kinds of Stress

I Kind of Stress

How Stress Is Relieved

Add A

Direction

of Shift

Remove A

To the right

Remove A

Add A

To the left

Remove B

AddB

To the right

I---

I---

AddB Increase temperature

by

Remove B

To the left

Remove some of the heat

To the left

Add heat

To the right

Remove A to relieve pressure

To the right

adding heat Decrease temperature removing heat I---

by

---

Increase pressure I---

Decrease pressure

--

increase Add A to maintain

equilibrium

To the left

pressure

CONCEPT TEST

Describe the effects of removing or adding a reactant to an equilibrium mixture. Indicate how the composition of the mixture changes as the system relieves the stresses caused by these changes.

Let's return to the industrial preparation of ammonia O

;:"H

rxn

= -92.2 kJ

in which manufacturers take advantage of changes in concentration, pressure, and temperature and apply Le Chatelier's principle to obtain more of the desired product. The process typically uses reaction vessels that operate at 200 atm of pressure. To understand why, note that a total of 4 moles of reactant gases combines to make only 2 moles of product. When the reaction is carried out in a rigid, sealed chamber, the pressure inside the chamber decreases as the reaction proceeds because the total number of moles of gas in the chamber decreases. Running the reaction at high pressure puts a stress on the system that is relieved by making more ammonia. The reaction is also typically run at high temperatures, which may seem counterintuitive considering the effect of temperature on the equilibrium. The reaction is exothermic, so we might initially assume that cooling the reaction would produce more product. Remember this analysis is based solely on equilibrium. However, kinetics also plays an important role. At room temperature the rate of the reaction is extremely slow. Because the yield of product (equilibrium) and the rate of production of product (kinetics) must both be considered, a compromise must be found between kinetics and equilibrium.

I

Le Chatelier's Principle

I

765

766

I Chapter Fifteen

I


To make a commercially viable amount of ammonia, a hot (400°C) reaction mixture is pumped through a system of chilled condensers, where ammonia, which has a much higher boiling point than either hydrogen or nitrogen, condenses as a liquid and is removed from the gaseous reaction mixture. The removal of ammonia stresses the equilibrium toward the production of more ammonia; the high pressure stresses the equilibrium in the same direction; the high temperature stresses the equilibrium in the reverse direction but means that more molecules collide with enough energy to overcome the activation energy barrier.

Catalysts and Equilibrium Uncatalyzed Catalyzed

Progress of reaction ~

FIGURE 15.8 The graph shows the effect of the presence of a catalyst on a reaction. A catalyst lowers the activation

energy barrier

by lowering the energy of the transition

state;

therefore the rate of the reaction increases. Both forward and reverse reactions occur more rapidly, but the position of equilibrium does not change. The system simply comes to equilibrium

more rapidly.

@Xi)

CONNECTION In Chapter 14 we discovered that a catalyst simultaneously increases the rate of a reaction in both the forward and the reverse directions.

The industrial production of ammonia described in the previous section was developed by the German chemists Fritz Haber (1868-1934) reaction and Carl Bosch (1874-1940) in the early 20th century and is still widely referred to as the Haber-Bosch process. What makes the process commercially feasible is the use of special catalysts that accelerate the reaction. As discussed in Chapter 14, a catalyst increases the rate Products of a chemical reaction by lowering its activation energy. The question is this: If a catalyst increases the rate of a reaction, does a catalyst affect the equilibrium constant of a reaction? To answer this question, consider the catalyzed and uncatalyzed energy profiles of the exothermic reaction in Figure 15.8. The catalyst increases the rate of the reaction by decreasing the height of the activation energy barrier. The height of the barrier is reduced by the same amount whether the reaction proceeds in the forward direction or in reverse. As a result, the increases in reaction rates produced by the catalyst are the same in both directions. Therefore a catalyst has no effect on the equilibrium constant of a reaction or the composition of an equilibrium reaction mixture, but it does decrease the time it takes for a reaction to reach equilibrium. reaction

Calculations Based on K

~

Reference books and the tables in Appendix 5 of this book contain lists of equilibrium constants for chemical reactions. These values are used in several kinds of calculations, including those in which 1. We know the equilibrium concentrations of the reactants and products and we wish to calculate the equilibrium constant, as we did for Kc in the formation ofN204 from N02 (Sample Exercise 15.2) 2. We want to determine whether a reaction mixture has reached equilibrium, as in Sample Exercise 15.8 3. We know the value of K and the starting concentrations or partial pressures of reactants and products, and we want to calculate the reactant and product concentrations or pressures at equilibrium. ~II CHEMTOUR Problems

Solving Equilibrium

In this section we'll focus on the third kind of calculation. In our first example, we will calculate how much nitrogen monoxide (NO) forms in a sample of air heated to a temperature at which the value of K p is 1.00 X 10-5 for NO formation from N2 and 02:

The initial partial pressures ofN2 and 02 are 0.79 and 0.21 atm, respectively, and we assume there is no NO present initially.

15.8

The equilibrium

constant expression for the formation of NO from N2 and

02 is

An ICE table is an excellent tool for calculating equilibrium concentrations. The acronym ICE stands for the initial partial pressures or concentrations of reactants and products, the changes in their pressures or concentrations as the reaction proceeds to equilibrium, and their pressures or concentrations after equilibrium has been achieved. I, C, and E are the rows of the table, and each reactant and product has a column. Our ICE table for the reaction to produce NO has three rows and three columns. We start by filling in the initial partial pressures of the reactants and product in the first row from the information given in the example: PN2 (atm)

P02 (atm)

PNO (atm)

0.79

0.21

0

Initial Change Equilibrium

We know the equilibrium will shift from left to right because there is no NO initially present, so Q = 0 and Q < < K. We need to do some basic algebra to fill in the remaining rows. We don't know how much N2 or 02 will be consumed or how much NO will be made. We can define the change in N2 as -x, because N2 is consumed during the reaction. The mole ratio ofN2 to 02 in the reaction is 1:1; so the change in 02 is also -x. Two moles of NO are produced from each mole of N2 and 02' so the change in P NO is +2x. Inserting these values in the second row of the ICE table, we have PN2 (atm)

P02 (atm)

0.79

0.21

0

-x

-x

+2x

Initial Change

PNO (atm)

Equilibrium

Combining the initial and change rows of the ICE table, we obtain the following partial pressures of all three gases at equilibrium: PN2 (atm)

P02 (atm)

PNO (atm)

0.79

0.21

0

-x

-x

+2x

0.79 - x

0.21 - x

2x


The last row of partial-pressure the reaction:

terms can be substituted into the K p equation for (P NO)2 Kp

= (P NYP 0) (2X)2 (0.79 - x)(0.21 - x)

= 1.00

X

10-5

I

Calculations

Based on K

I 767

768

I Chapter Fifteen

I


Multiplication of the binomial terms in the denominator fraction equal to the value of K p gives 4X2

5

Kp = 1.00 X 10- = -0-.16-5-9---1.00x Cross-multiplying,

the

+ X2

we get

1.659 X 10-6 Combining

and setting the resulting

X2

(1.00 X 1O-5)x

-

terms and rearranging,

+

3.99999x2

+

4X2

we have

(1.00 X 1O-5)x - 1.659 X 10-6

You may recognize this equation equation:

=

(1.00 X 1O-5)x2

=

0

as one that fits the general form of a quadratic

+ bx + c = 0

ax2

which can be solved for x with a scientific calculator or using the quadratic formula -b ± -vb2 2a

x=

-

4ac

Two values are possible for x, but only one is positive: 6.428 X 10-4 atm. A gas can't have a negative partial pressure, so we use only the positive value to calculate equilibrium partial pressures: Po,

= 0.21 - x = 0.21 - (6.428 X 10-4) = 0.21 atm (to two significant figures)

PN,=0.79-x

PNO

=

0.79 - (6.428 X 10-4)

=

2x

= 2(6.428

=

0.79 atm

= 1.2855

X 10-4)

X 10-3

= 0.0013 atm

The equilibrium partial pressures of the reactants hardly differ from their initial partial pressures. The tiny decreases make sense given the small value (1.0 X 10-5) of the equilibrium constant. Because the x terms in the denominator of the equilibrium constant equation are small compared with the initial partial pressures, a much simpler approach to calculating P NO is possible: we ignore the x terms in the denominator and use the initial values for P N, and Po, instead, to give K p

=

5

1.0 X 10-

=

(P NO? (P )(P ) N,

0,

4x2

_

(0.79 - x)(0.21 - x) 4X2

=

x

_

4X2

- 0.1659

= (0.79)(0.21)(1.0 X 10-5)

= X2

4X2

(0.79)(0.21)

1.659 X 10-6

= 4.148 X 10-7

=

6.440 X 10-4 atm

15.8

Note that the difference between this value of x and that obtained from the solution to the quadratic equation (6.428 X 10-4) is (6.440 X 10-4)

(6.428 X 10-4)

-

= 1.2 X 10-6 atm

Dividing this difference by the original value of x, we have a relative difference of only 6

1.2 X 10- iHflT X 100% 6.428 X 10-4 iHflT

=

0.19%

A small difference like this is not significant when we know the initial partial pressures to only two significant figures. Generally speaking, we can ignore the x component of the concentration or partial-pressure terms for reactants if the value of x is less than 5% of the smallest initial value. Based on the small value of K p (1.00 X 10-5) for this reaction, we can say that 0.79 - x = 0.79 and 0.21 - x = 0.21 to avoid more complicated math and get very close to the correct answer. Another situation where the math can be simplified is when the initial concentrations of the reactants are the same. For example, suppose we have a vessel containing a mixture of 0.100 M concentrations of N2 and 02 at a temperature where Kc for the same reaction is 0.100. What is the equilibrium concentration of NO?

+

N2(g)

02(g) ~ 2 NO(g)

Our ICE table looks like this: [N21 (M)

[021 (M)

[NOI (M)

0.100

0.100

0

-x

-x

+2x

0.100 - x

0.100 - x

2x


Inserting the values from the E line into the expression for K p gives

K =

(2x)2

c

(0.100 - x)(0.100 - x)

= 0.100

There are two possibilities for simplifying the math and getting to the value of x. One is to "drop the x," which means we would put 0.10 - x = 0.10 in the denominator of the Kc expression. This should not be done, considering that Kc is 1.00 X 10-1, which means that the equilibrium is further to the right than it is with the K c = 1.00 X 10-5 we had before. Another alternative is to note that the Kc expression has the form (X)2/(Y)2 = Z, where X = 2x, Y = 0.10 - x and Z = Kc' Taking the square root of each side gives

2x

=

0.316

Collecting the x terms leads to 2.316 x = 0.0316, which gives x = 0.0136 M. The equilibrium concentrations are thus [N2] = [02] = 0.086 M and [NO] = 2x = 0.0272 M. It's a good idea to check that these concentrations are consistent with the known value of Kc' Substituting into the Kc expression we get

=

K c

(0.0272)2 (0.086)2

=

0.100

I

Calculations

Based on K

I 769


I


Using an ICE Table to Calculate Equilibrium Concentrations Much of the H2 used in the Haber-Bosch process is produced by steam-reforming of methane followed by the water-gas shift reaction:

If a reaction vessel at 400°C is filled with an equimolar mixture of CO and steam such that Peo = P H,O = 2.00 atm, what will be the partial pressure of H2 at equilibrium? The equilibrium constant K p = 10 at 400T. COLLECT AND ORGANIZE We are asked to find the partial pressure of a gas at equilibrium given the chemical equation and the initial concentrations of reactants. One way to approach this problem involves (1) setting up an ICE table, (2) using the partial pressures from the E line of the ICE table in the equilibrium constant expression, and (3) solving for PH,'

Let's first look at the reactant and product quantities. This system initially contains no product. This means that the reaction quotient Q is equal to zero, which is less than K. Therefore, the reaction proceeds from left to right, decreasing the reactant concentrations and increasing the product concentrations. We can set up the ICE table for this reaction with the initial partial pressures specified.

ANALYZE

SOL V E Let x be the increase in partial pressure of H2 as a result of the reaction. The change in Peo, will also be x, and the changes in Peo and PH, will be -x. Inserting these numbers into the ICE table and using them to develop equilibrium terms for the partial pressures of the reactants and products, we have

Initial Change

Equilibrium

Pco (atm)

PH20 (atm)

PC02 (atm)

PH2 (atm)

2.00

2.00

0.00

0.00

-x

-x

+x

+x

2.00 - x

2.00 - x

x

x

Inserting these equilibrium terms into the equilibrium constant expression for the reaction gives

The equation has the form X2/Y2 square root of both sides:

=

Z, and so it can be simplified by taking the

x 2.00 - x Solving for x after rearrangement partial pressure ofH2 and COr

= -YfO = 3.16

gives x

= 1.52 atm, which is the equilibrium

ABOUT IT We should always check the result when the calculation is finished. Substituting P H,O = Peo = 2.00 - 1.52 = 0.48 atm and PH, = Peo, = 1.52 atm into the equilibrium constant expression gives K = (1.52)2/ (0.48)2 = P 10.03, which is close enough to show we made no errors in the calculation. It is

THINK

15.8

also a good idea to look for an X2/Y2 quadratic equation to find x.

=

Z situation to avoid having to use the

Practice Exercise The balanced chemical equation for the formation of hydrogen iodide from H2 and 12is H2(g)

+ 12(g) ~

2HI(g)

The value of K p for the reaction is 50 at 450°C. What is the partial pressure of HI in a sealed reaction vessel at 450°C if the initial partial pressures ofH2 and 12are both 0.100 atm and there is no HI present?

Calculating

Equilibrium Concentrations

Let's revisit the water-gas shift reaction:

Suppose a reaction vessel at 400°C contains a mixture of CO, steam, and H2 at the following initial partial pressures: Peo = 2.00 atm, PH,o = 2.00 atm, and PH, = 0.15 atm. What is the partial pressure of H2 at equilibrium, given K p = 10 at 400°C? COLLECT AND ORGANIZE We are asked to calculate the partial pressure of a product at equilibrium. We know the initial partial pressures of CO, steam, and H2 and the value of Kp' The difference between this and the preceding Sample Exercise is that some H2 (a product) is already present before the reaction starts. Comparing the reaction quotient Q with the value of K lets us know the direction (left or right) that the reaction moves to attain equilibrium.

Because there is no CO2 initially present, the reaction quotient Q is zero. Therefore, the reaction proceeds in the forward direction. We can set up the ICE table for this reaction with the initial partial pressures specified.

ANALYZE

SOL V E Let x be the increase in partial pressure of H2 as a result of the reaction. The change in P eo, will also be x, and the changes in P eo and P H,O will be -x. Inserting these numbers into the ICE table, we get


Pco (atm)

PH20 (atm)

PC02 (atm)

PH2 (atm)

2.00

2.00

0.00

0.15

-x

-x

+x

2.00 - x

2.00 - x

x

+x

0.15

+x

Inserting these equilibrium terms into the equilibrium constant expression for the reaction gives

(Peo,)(PH) K = ...,..-----,-,-------, p (Peo)(P H,O) (x)(0.15 + x) (2.00 - x)(2.00 - x)

= 10

Solving for x gives two values, x = 1.50 and 2.96. Of the two solutions, one of them (2.96) is not physically possible because inserting it into the equilibrium row of the ICE table gives negative partial pressures of CO and H20. Therefore, x = 1.50 and the partial pressure ofH2 at equilibrium is 1.50 + 0.15 = 1.65 atm.

I

Calculations

Based on K

I 771


I


From the ICE table, the equilibrium partial pressures of CO, H20, H2' and CO2 are 0.50, 0.50,1.65, and 1.50, respectively. Using these values, the calculated value of K p = (1.65)(1.50) I (0.5)(0.5) = 9.9, which is equal to 10 when we consider the rounding errors. Notice that 25% of the reactants initially present remain at equilibrium. This much unreacted reactant reflects the relatively small value of K p . THINK

ABOUT

IT

Practice Exercise The value of K c for the reaction

is 0.21 at 373 K. If a reaction vessel at that temperature initially contains 0.030 M concentrations of both N02 and N204, what will be the concentrations of the two gases when equilibrium is achieved?

mDI Changing K with Changing Temperature Throughout this discussion we have repeatedly stated that values of K depend on temperature. Now that we have explored concepts and relationships involving equilibria in some detail, let's look more closely at how temperature influences K. In the course of the discussion, we will develop several important equations that express useful ideas.

Temperature, K, and ~Go In Section 15.5 we derived Equation 15.14, which relates the change in standard free energy of a chemical reaction, !::.Go, and its equilibrium constant, K: lnK=

-!::.Go

RT In Chapter 13, we introduced the following equation relating the changes in standard free energy, enthalpy, and entropy of a chemical reaction:

We can combine these two equations to derive one that relates K to !::.Ho and !::.So.

(15.16) o

@Xi)

CONNECTION We discussed the relation between free energy, enthalpy, entropy, and temperature in Chapter 13.

Note how a negative value of !::.H and lor a positive value of !::.So contributes to a large value of K. These dependencies make sense because negative values of !::.Ho and positive values of !::.So are the two factors that combine to make reactions spontaneous. Because we are discussing the influence of temperature on the value of K, let's identify the factors on which Texerts its influence. Equation 15.16 indicates that the influence of !::.So on the magnitude of K is independent of temperature. (The term in the equation that contains !::.So does not include temperature as a factor.) On the other hand, the influence of the !::.Ho term does depend on temperature: the higher the temperature, the smaller the liT term and the smaller the influence of a negative !::.Ho.

15.9

This temperature dependence makes sense if we think of heat as a product in exothermic reactions and a reactant in endothermic reactions. Applying Le Chatelier's principle: Exothermic reaction: Reactants ~ products + heat Increase in T causes a decrease in K Endothermic reaction: Reactants + heat ~ products Increase in T causes an increase in K

If the values of t:..Ho and t:..So do not vary much with temperature, then Equation 15.16 predicts that In K is a linear function of liT. Furthermore, we expect the line on the graph ofln K versus 11Tto have a positive slope for an exothermic process and a negative slope for an endothermic process. We can determine t:..Ho from the slope of the line and t:..So from its intercept (the value on the vertical axis of the plot when liT is zero). This means that we can calculate fundamental thermodynamic values of a reaction at equilibrium by determining its equilibrium constant at different temperatures. As an example of this method, let's determine thermodynamic values for the system

+

2 CO2(g) ~ 2 CO(g) The equilibrium temperatures:

constant

a/g)

Kc has the values shown below at three different

Temperature

K,

(K)

1500

5.5

2500

4

10-9

X X

10-1

40.3

3000

We can see from the data that Kc increases as temperature increases, so we know heat is a reactant and the reaction is endothermic. We can use these data to establish how endothermic the reaction is by plotting In Kc versus liT and calculating t:..H:xn' We can also establish if the entropy change in the reaction is positive or negative and determine t:..S:xn' Equation 15.16, rewritten here in the form of the equation for a straight line, links Kc to temperature and to t:..H:xn:

y = m (x) + b We can supplement the table we have been given by calculating In K and liT. Temperature

(K)

In K,

K,

liT

1500

5.5 X 10-

9

-19.02

6.67 X 10-4

2500

4.0 X 10-1

-0.92

4.00 X 10-4

3000

40.3

3.70

3.33 X 10-4

I

Changing K with Changing Temperature

I 773


I


If we plot In K versus liT, we see that the plot is a straight line. 5.00

t

y

=

+ 26.353

-68,070x

0.00 -5.00

In K -10.00 -15.00 -20.00 -25.00 0.0002

.l..-

L!

----'.

0.0004

0.0006

...J

0.0008

l/T~

The slope of this line is -68,070 value of t:..Horxn as follows:

-t:..Ho

K

= -t:..H:jR,

from which we can calculate the

68,070 K X 8.314 rxn

J

mol·K

= 566,000 Jlmol = 566 kJlmol The y-intercept is 26.4

= t:..So IR rxn

from which we can calculate the value of

t:..S:xn =

t:..So :

26.4 X 8.314 mol. K

rxn

J = 219 Jlmol.

K

As anticipated, based on the increasing value of Kc as T increased, the reaction is endothermic, as shown by the positive value of t:..H:xn = 566 kJlmol. It is accompanied by a positive entropy change (t:..S:xn = 219 J I mol· K), which is understandable because the forward reaction converts 2 moles of gaseous CO2 to 3 moles of gaseous products. From the values of t:..H:xn and t:..S:xn we can predict the value of K c at room temperature (298 K). The result is K c = 2 X 10-88, which means that there is virtually no probability that the CO2 we exhale will be converted to even trace amounts of poisonous carbon monoxide. In addition to enabling us to calculate useful values like enthalpy and entropy changes for reactions, Equation 15.16 expresses mathematically a number of principles. One principle is that a large gain in entropy of the system (positive t:..SO) contributes to making In K positive. This principle means that entropy favors equilibria when there are more moles of products (with a corresponding larger number of microstates) than of reactants. In addition to using Equation 15.16 to determine thermodynamic values, we can also use it to examine the effect of temperature on a reaction. Let's revisit an important reaction in automobile engines: the decomposition of NO into N2 and 02' We can calculate the value of the equilibrium constant of this reaction at the temperature at which an automobile engine operates (500°C) if we know the value of K for a reference temperature. We are looking for K for the decomposition reaction of NO at 500°C, so we start by writing Equation 15.16 in the form

InK=

t:..HO(l)

t:..So

-R T + Ii:

15.9

I

Changing K with Changing Temperature

I 775

As we just saw, this equation is that of a straight line (y = mx + b) where the y variable is In K and the x variable is liT. The slope of the line is -t:..HoIR and the intercept is t:..so/R. If we select two points along such a line (temperatures ~ and I;) at which the values of the equilibrium constant are K1 and K2' respectively, then

The difference between In K1 and In K2 is related to the difference between and I;:

In K1 - In K2 Rearranging

~

1) t:..HO(~) = - t:..HO( R ~ + R I;

the terms, we have

or

(15.17) Equation 15.17 is called the Clausius-Clapeyron equation, which we used in Chapter 12 in our discussion of the fractional distillation of crude oil. We can use it in the context of this chapter to calculate K p for the decomposition of NO

in the catalytic converter of an automobile at 500°C. To use Equation 15.17, we need a reference value for K.p We also need the standard enthalpy change (t:..HO) of the reaction. For the formation of NO

K p = 6.6 X 10-16 at 298 K. This reaction is the reverse of the decomposition reaction, so K p = 6.6 X 10-16 is the reciprocal of the value we need. For the reaction

the equilibrium constant K = 11Kp = 1.5 X 1015. The equilibrium constant for the decomposition reaction at 298 K is a large number, so this reaction is highly favorable at that temperature. The value of t:..Ho can be found in Appendix 4 for the formation of NO from its elements in their standard states:

t:..H; =

90.3

kJ

Thus, t:..Ho for the decomposition of NO is the negative of this value or -90.3 per mole of NO.

kJ

@ K, the reaction proceeds in the reverse direction (~) to make more reactants.

15.8

o

o

Calculating K from f1G

Use the relation 6.G = - RT In K. Be sure to convert 6.G to ]Imol and use R = 8.314 ]Imol' K.

15.9

Writing mass action expressions for heterogeneous equilibria

Molar concentrations of pure liquids and pure solids are omitted from equilibrium constant expressions because such concentrations are constants.

15.10

Adding or removing reactants or products to stress on equilibrium

Decreasing the concentration of a substance involved in an equilibrium shifts the equilibrium toward the production of more of that substance. Increasing the concentration of a substance shifts the equilibrium so that some of that substance is consumed in the reaction.

15.11

Assessing pressure effects on gas phase equilibria

Equilibria involving different numbers of gaseous reactants and products shift in response to an increase (or decrease) in pressure toward the side with fewer (or more) moles of gases, respectively.

15.12

o

779

sures. Equilibrium concentrations or pressures of reactants and products can be calculated from initial concentrations or pressures, the reaction stoichiometry, and the value of the equilibrium constant.

According to Le Chfttelier's principle, systems (including chemical reactions) at equilibrium respond to stress by shifting position to relieve the stress. Adding or removing a reactant or product, or applying pressure to a reaction mixture that includes gases creates stress that shifts the position of a chemical equilibrium. A catalyst decreases the time it takes a system to achieve equilibrium, but it does not change the value of the equilibrium constant. S ECTI 0 N 15.7

TYPE OF PROBLEM

I

780

I

Chapter Fifteen

I


SAMPLE

CONCEPTS

Predicting changes in equilibria with temperature

The value of K of an endothermic reaction increases with increasing temperatures; the value of K of an exothermic reaction decreases.

15.13

Using an ICE table to calculate equilibrium concentrations

An ICE table has a column for each reactant and product concentration (or partial pressure) and three rows labeled Initial, Change, and Equilibrium. Calculate Q to see which wayan equilibrium shifts under the specified conditions. Call one of the shifts x and write the others as decreases or increases in terms of x, taking into account the reaction stoichiometry. Add the I and C rows to get the expression for each reactant and product concentration in the E row. Substitute these data into the expression for K and solve for x.

15.14 and 15.15

Calculating equilibrium at a specific temperature

Use the relation

15.16

constants

AND EQUATIONS

EXERCISES

TYPE OF PROBLEM

In(K2) KI

= _ 6.HO(~ _~) R T, ~

where ~ and T, are Kelvin temperatures and the value of KI or K2 at the respective temperature is known. Be sure to convert 6.H to ]/mol to match the units of R. o


Figure P15.1 shows the energy profiles of reactions A ~ Band C ~ D, respectively. Which reaction has the larger forward rate constant? Which reaction has the smaller reverse rate constant? Which reaction has the larger equilibrium constant K/

t

GO

t

k]/mol

k]/mol

GO

A

D

Reaction progress ~

Reaction progress ~ FIGURE P15.1

15.2.

The progress with time of a reaction system is depicted in Figure P15.2. Red spheres represent the molar concentration of substance A and blue spheres represent the molar concentration of substance B. (a) Does the system reach equilibrium? (b) In which direction (A ~ B or B ~ A) is equilibrium attained? (c) What is the value of the equilibrium constant K/

Time

FIGURE P15.2


15.3.

I 781

In Figure P15.3 the red spheres represent reactant A and the blue spheres represent product B in equilibrium with A. (a) Write a chemical equation that describes the equilibrium. (b) What is the value of the equilibrium constant K}

300 K

FIGURE P1S.3 15.4.

The equilibrium constant K, for the reaction A (red spheres)

+

B (blue spheres) ~ AB

is 3.0 at 300.0 K. Does the situation depicted in Figure P15.4 correspond to equilibrium? If not, in what direction (to the left or to the right) will the system shift to attain equilibrium?

400 K FIGURE P15.5 15.6.

Figure P15.6 shows a plot ofln K, versus liT for the reaction A + 2 B ~ AB2• Is the reaction endothermic or exothermic?

FIGURE P1S.4 15.5.

The top and bottom diagrams in Figure P15.5 represent equilibrium states of the reaction. A (red spheres)

+

B (blue spheres) ~ AB

3.10

3.20

at 300 K and 400 K, respectively. Is this reaction endothermic or exothermic? Explain.

3.30 3.40 3.50 103/T (K-l)

3.60

FIGURE P15.6

Questions and Problems Understanding Equilibrium: A Qualitative View CONCEPT

15.7. 15.8. 15.9.

REVIEW

How are forward and reverse reaction rates related in a system at chemical equilibrium? Describe an example of a dynamic equilibrium that you experienced today. Does the reaction A ~ 2 B represented in Figure P15.9 reach equilibrium in 20 µ,s? Explain your answer.

~ c:: o

.~ tJ

c::

'"c:: o U

u

10

20

30

Time (µ.,s)

FIGURE P15.9

40

50

782

I Chapter Fifteen

I


15.10. At equilibrium, is the sum of the concentrations of all the reactants always equal to the sum of the concentrations of the products? Explain. 15.11. Suppose the forward rate constant of the reaction A ~ B is greater than the rate constant of the reverse reaction at a given temperature. Is the value of the equilibrium constant less than, greater than, or equal to 1? 15.12. Explain how it is possible for a reaction to have a large equilibrium constant but small forward and reverse rate constants.

PROBLEMS

15.21. Use the graph in Figure P15.21 to estimate the value of the equilibrium constant K, for the reaction NP(g)

+ ~02(g)·

0.0004 ~

0.0003

c::

0

.§

PROBLEMS

"" N2(g)

0.0002

1:: Cl)

u

15.13. In a study of the reaction 2 NP(g)

c::

~2N2(g)

+

a/g)

quantities of all three gases were injected into a reaction vessel. The N20 consisted entirely of isotopically labeled 15N20. Analysis of the reaction mixture after 1 day revealed the presence of compounds with molar masses 28, 29, 30, 32, 44, 45, and 46 g/mol. Identify the compounds and account for their appearance. 15.14. A mixture of!3CO, 12C02, and 02 in a sealed reaction vessel was used to follow the reaction 2CO(g)

+

a/g)

~2CO/g)

Analysis of the reaction mixture after 1day revealed the presence of compounds with molar masses 28, 29, 32, 44, and 45 g/mol. Identify the compounds and account for their appearance.

cS

0.0001 0.00000

The Equilibrium Constant Expression and K CONCEPT

REVIEW

15.17. Under what conditions are the numerical values of K, and Kp equal? 15.18. At 298 K, is Kp greater than or less than K, if there is a net increase in the number of moles of gas in a reaction and if K, > 1. Explain your answer. 15.19. Nitrogen oxides play important roles in air pollution. Write expressions for K, and Kp for the following reactions involving nitrogen oxides. a. N2(g) + 2a/g) ~ Np .(g) b. 3 NO(g) ~ N02(g) + NP(g) c. 2 NP(g) ~2N/g) + O/g) 15.20. Write expressions for K, and Kp for the following reactions, which contribute to the destruction of stratospheric ozone. a. CI(g) + 03(g) ~ CIO(g) + 02(g) b. 2 CIO(g) "" CI2(g) + 02(g) c. 203(g) ~30/g)

2

3 Time(s)

4

5

6

FIGURE P1S.21 15.22. Estimate the value of the equilibrium reaction 2NO(g)

+

constant K, for the

02(g) ~2NO/g)

from the data in Figure P15.22. 0.010

$ c::

15.15. Suppose the reaction A ~ B in the forward direction is first order in A and the rate constant is 1.50 X 10-2 S-l. The reverse reaction is first order in B and the rate constant is 4.50 X 10-2 S-1 at the same temperature. What is the value of the equilibrium constant for the reaction A ~ B at this temperature? 15.16. At 700 K the equilibrium constant K, for the gas phase reaction between NO and 02 forming N02 is 8.7 X 106• The rate constant for the reverse reaction at this temperature is 0.54 M-1 S-I. What is the value of the rate constant for the forward reaction at 700 K?

1

.g cc::

0.008 0.006

oj

Cl)

u

0.004

c::

0

U

0.002 2000

4000 Time (s)

6000

8000

FIGURE P1S.22

15.23. At 1200 K the partial pressures of an equilibrium mixture ofH2S, H2, and S are 0.020, 0.045, and 0.030 atm, respectively. Calculate the value of the equilibrium constant Kp at 1200 K.

15.24. At 1045 K the partial pressures of an equilibrium mixture of H20, H2' and 02 are 0.040, 0.0045, and 0.0030 atm, respectively. Calculate the value of the equilibrium constant Kp at 1045 K.

15.25. At equilibrium, the concentrations of gaseous N2, 02' and NO in a sealed reaction vessel are [N2] = 3.3 X 10-3 M, [02] = 5.8 X 10-3 M, and [NO] = 3.1 X 10-3 M. What is the value of K ,for the reaction

at the temperature

of the reaction mixture?


15.26. Analyses of an equilibrium mixture of gaseous N204 and N02 gave the following results: [N02] = 4.2 X 10-3 M and [N204] = 2.9 X 10-3 M. What is the value of the equilibrium constant K, for the following reaction at the temperature of the mixture?

I

783

The value of K, for the reaction is 5.0 at 32TC. What is the value of K p at 325°C? 15.36. If the value of Kp for the following reaction

+

SO/g)

N02(g) ~ NO(g)

+

S03(g)

is 3.45 at 298 K, what is the value of K, for the reverse reaction? 15.27. A sealed reaction vessel initially contains 1.50 X 10-2 moles of water vapor and 1.50 X 10-2 moles of CO. After the following reaction

+

HP(g)

+

CO(g) ~ H/g)

CO/g)

has come to equilibrium, the vessel contains 8.3 X 10-3 moles of CO2, What is the value of the equilibrium constant K, of the reaction at the temperature of the vessel? *15.28. A 100 mL reaction vessel initially contains 2.60 X 10-2 moles of NO and 1.30 X 10-2 moles ofH2• At equilibrium, the concentration of NO in the vessel is 0.161 M. At equilibrium the vessel also contains N2, H20, and H2• What is the value of the equilibrium constant K, for the following reaction? 2H2(g)

+

2 NO (g) ~2HP(g)

+

N2(g)

15.29. The equilibrium constant Kp for the following equilibrium is 32 at 298 K. What is the value of K, for this same equilibrium at 298 K? A(g)

+

CD(g) ~ C(g) the equilibrium N/g)

+

+

D(g)

3H2(g) ~2NH/g)

+

REVIEW

15.37. How is the value of the equilibrium constant affected by scaling up or down the coefficients of the reactants and products in the chemical equation describing the reaction? 15.38. Why must a written form of an equilibrium and the temperatures be given when reporting the value of an equilibrium constant? PROBLEMS

15.39. The equilibrium

constant K, for the reaction

+

A/g)

B/g) ~2AB(g)

is 4 at 300 K. What is the value of K, for the equilibrium

+ ~B/g)

~A/g) at 300 K? 15.40. The equilibrium

~AB(g)


2X(g)

+

2Y(g) ~2XY(g)

is 3.6 X 103 at 405 K. What is the value of K, for the equilibrium X(g) + Y(g) ~ XY(g) at 405 K? 15.41. The following reaction is one of the elementary steps in the oxidation of NO:

constant Kp for the synthesis of

is 1.45 X 10-5. What is the value of K?, 15.32. If the value of the equilibrium constant, K" for the following reaction is 5 X 105 at 298 K, what is the value of Kp at 298 K? 2CO(g)

CONCEPT

B(g) ~ AB(g)

15.30. The equilibrium constant K, for the following equilibrium is 6.0 X 104 at 500 K. What is the value of Kp for this same equilibrium at 500 K?

15.31. At 500T, ammonia

Manipulating Equilibrium Constant Expressions

a/g)

~2CO/g)

15.33. For which of the following reactions are the values of K, and Kp equal? a. 2 S02(g) + O/g) ~ 2 S03(g) b. Fe(s) + CO2(g) ~ FeO(s) + CO(g) c. HP(g) + CO(g) ~ H/g) + CO2(g) 15.34. For which of the following reactions are the values of K, and Kp different? a. 2 NO/g) ~NP/g) b. 2NO(g) + a/g) ,,"2NO/g) c. 203(g) ~30/g) 15.35. Bulletproof Glass Phosgene (COCl2) is used in the manufacture of foam rubber and bulletproof glass. It is formed from carbon monoxide and chlorine in the following reaction:

Write an expression for equilibrium constant K, for this reaction and for the reverse reaction:

How are the two K, expressions related? 15.42. Making Ammonia The value of the equilibrium for the formation of ammonia, N2(g)

+

3 H/g)

constant Kp

~ 2 NH3(g)

is 4.5 X 10-5 at 450°C. What is the value of K p for the following reaction?

15.43. Air Pollutants Sulfur oxides are major air pollutants. The reaction between sulfur dioxide and oxygen can be written in two ways:

and

Write expressions for the equilibrium reactions. How are they related?

constants for both

784

I Chapter Fifteen

I


15.44. At a given temperature, reaction

15.54. The equilibrium

constant K, for the reaction 2C~D

is 0.11. What is the equilibrium reaction?

15.45. At a given temperature, reaction

constant K, for the

the equilibrium

constant for the following

constant K, for the

the equilibrium

is 3 X 10-3. At a particular time, the composition of the reaction mixture is [C] = [D] = [E] = 5 X 10-4 M. In which direction will the reaction proceed to reach equilibrium? 15.55. Suppose the value of the equilibrium following reaction A(g)

is 2.4 X 10-3. What is the value of the equilibrium constant for each of the following reactions at that temperature? a. SOlg) + ~Olg) ~ S03(g) b. 2S03(g) ~2S0lg) + 0lg) c. S03(g) ~ SOlg) + ~Olg) 15.46. If the equilibrium constant K, for the reaction 2NO(g)

+

0lg)

15.47. Calculate the value of the equilibrium reaction

A(g)

2D~A+

constant K for the

2B

from the following information: A

+

K, = 0.041

C~2D

15.48. Calculate the value of the equilibrium reaction

constant K for the

from the following information:

H~2E+2F

K, = 3.1 X 10-4 K, = 2.8 X 10-2

Equilibrium Constants and Reaction Quotients CONCEPT

B(g) "" C(g)

+

B(g) ~ C(g)

K p = 1.00 at 300 K

a. PA = Pc = 1.0 atm, PB = 0.50 atm b. [A] = [B] = [C] = 1.0 M 15.57. If the equilibrium


N2(g)

+

02(g) ~ 2 NO(g)

is 1.5 X 10-3, in which direction will the reaction proceed if the partial pressures of the three gases are all 1.00 X 10-3 atm? 15.58. At 650 K, the value of the equilibrium constant Kp for the ammonia synthesis reaction

K , = 3.3

2B ~C

2G~H

+

constant Kp of the

is 1.00 at 300 K. Are either of the following reaction mixtures at chemical equilibrium at 300 K? a. PA = PE = Pc = 1.0 atm b. [A] = [B] = [C] = 1.0 M 15.56. In which direction will the following reaction proceed to reach equilibrium under the conditions given?

,,"2NOlg)

is 5 X 1012, what is the value of the equilibrium constant of each of the following reactions at the same temperature? a. NO(g) + ~Olg) ~ NOlg) b. 2NOlg) ~2NO(g) + 0lg) c. NOlg) ~ NO(g) + ~Olg)

+ E

REVIEW

15.49. What is a reaction quotient? 15.50. For a given equilibrium system how, in most cases, do the equilibrium constant and the reaction quotient differ? 15.51. What does it mean when the reaction quotient Q is numerically equal to the equilibrium constant K? 15.52. Explain how knowing Q and K for an equilibrium system enables you to say whether it is at equilibrium or whether it will shift in one direction or another. PROBLEMS

15.53. If the equilibrium constant K, for the reaction A(aq) ~ B(aq) is 22 at a given temperature, and if [A] = 0.10 M and [B] = 2.0 M in a reaction mixture at that temperature, is the reaction at chemical equilibrium? If not, in which direction will the reaction proceed to reach equilibrium?

is 4.3 X 10-4• If a vessel contains a reaction mixture in which [N2] = 0.010 M, [H2] = 0.030 M, and [NH3] = 0.00020 M, will more ammonia form? 15.59. The equilibrium X + Y "" Z has K, = 1.00 at 350 K. If the initial molar concentrations of X, Y, and Z in a solution are all 0.2 M, in which direction will the reaction shift to reach equilibrium? a. To the left, making more X and Y b. To the right, making more Z c. The system is at equilibrium and the concentrations will not change 15.60. In Problem 15.59, when the equilibrium shifts, does the concentration of X increase or decrease?

Equilibrium and Thermodynamics CONCEPT

REVIEW

15.61. Do all reactions with equilibrium constants < 1 have values of ~Go > O? *15.62. The equation ~Go = -RT In K relates the value of Kp' not K" to the change in standard free energy for a reaction in the gas phase. Explain why. 15.63. Starting with pure reactants, in which direction will an equilibrium shift if ~Go < O? 15.64. Starting with pure products, in which direction will an equilibrium shift if ~Go < O?


Heterogeneous Equilibria

PROBLEMS

15.65. Which of the following reactions has the largest equilibrium constant at 25°C? f1Go = 115.4 kJ a. CI/g) + F/g) ~ 2 CIF(g) b. CI/g) + Br/g) ~2CIBr(g) f1Go = -2.0kJ c. CI/g) + 12(g) ~ 2 ICI(g) f1Go = -27.9 kJ 15.66. Glycolysis Problems 15.66-15.69 focus on glycolysis, the multi-step biochemical process by which sugar is metabolized to create energy in the body. Which of the following steps in glycolysis has the largest equilibrium constant? a. Fructose 1,6-diphosphate ~ !1Go = 24 kJ 2 glyceraldehyde- 3-phosphate f1Go = 4.4 kJ b. 3-Phosphoglycerate ~ 2-phosphoglycerate f1Go = 1.8 kJ c. 2-Phosphoglycerate ~ phosphoenolpyruvate 15.67. The value of f1Go for the phosphorylation of glucose in glycolysis is 113.8 kJ/mol. Using Equation 15.13 what is the value of the equilibrium constant for the reaction at 298 K? 15.68. In glycolysis, the hydrolysis of ATP to ADP drives the phosphorylation of glucose:

+ ATP

Glucose

I 785

+

~ ADP

6.Go = -17.7 kJ

glucose 6-phosphate

What is the value of K ,for this reaction at 298 K? 15.69. Sucrose enters the series of reactions in glycolysis after its hydrolysis into glucose and fructose:

+

Sucrose

H20 ~ glucose

+

fructose

K, = 5.3 X 1012 at 298 K What is the value of f1Go for this process? 15.70. The value of the equilibrium constant Kp for the reaction H2(g)

+

CO2(g) ~ HP(g)

+

CO(g)

is 0.534 at 700°C. a. Calculate the value of !1Go of the reaction using Equation 15.13. b. Using values from Appendix 4, calculate the value of f1Go of the reaction and compare the result with that obtained in part a. 15.71. Use the following data to calculate the value of Kp at 298 K for the reaction N/g) N/g)

+

+

202(g) ,,"2NO/g)

O/g) ~ 2 NO(g)

f1Go = 173.2 kJ

+

f1Go = -69.7 kJ

2NO(g)

O/g) ~2N02(g)

15.72. Under the appropriate conditions, NO forms N20 and N02: 3 NO(g) ~ NP(g)

+

NO/g)

CONCEPT

REVIEW

15.73. Write the K, expression for the following reaction: CuS(s) ~ Cu2+(aq) + S2-(aq) 15.74. Write the K, expression for the following reaction: AIP3(s)

+

~2AP+(aq)

3HP(C)

+

60H-(aq)

15.75. Why does the K, expression for the reaction CaC03(s)

~ CaO(s)

+

CO/g)

not contain terms for the concentrations of CaC03 and CaO? 15.76. Use the equations f1Go = !1Ho - Tf1So and !1G = - RT In K to explain why the value of Kp for the reaction o

increases with increasing temperature.

Le Chatelier's Principle CONCEPT

REVIEW

15.77. Does adding reactants to a system at equilibrium increase the value of the equilibrium constant? 15.78. Increasing the concentration of a reactant shifts the position of chemical equilibrium towards formation of more products. What effect does adding a reactant have on the rates of the forward and reverse reactions? 15.79. Carbon Monoxide Poisoning Patients suffering from carbon monoxide poisoning are treated with pure oxygen to remove CO from the hemoglobin (Hb) in their blood. The two relevant equilibria are

+

Hb Hb

4CO(g) ~Hb(CO)4

+

40/g)

~Hb(0)4

The value of the equilibrium constant for CO binding to Hb is greater than that for 02' How, then, does this treatment work? 15.80. Is the equilibrium constant Kp for the reaction 2N02(g) ~ Np.(g) in air the same in Los Angeles as in Denver if the atmospheric pressure in Denver is lower but the temperature is the same? 15.81. Henry's law (Chapter 10) predicts that the solubility of a gas in a liquid increases with its partial pressure. Explain Henry's law in relation to Le Chatelier's principle. *15.82. For the reaction

Use the values for !1Go for the following reactions to calculate the value of K p for the above reaction at 500°C. 2NO(g)

+

2 NP(g)

~2NO(g)

N/g)

+

02(g) ~2NO/g)

+

N/g)

O/g) ~ 2 NO(g)

!1Go = -69.7 kJ !1Go = -33.8 kJ f1Go = 173.2 kJ

why does adding an inert gas such as argon to an equilibrium mixture of CO, 02' and CO2 in a sealed vessel increase the total pressure of the system but not affect the position of the equilibrium?

786

I Chapter Fifteen

I


PROBLEMS

15.83. Which of the following equilibria will shift towards formation of more products if an equilibrium mixture is compressed into half its volume? a. 2 NP(g) ,,"2Nlg) + 0lg) b. 2CO(g) + 02(g) ~2C02(g) c. N2(g) + 0lg) ~2NO(g) d. 2 NO(g) + 0lg) ~ 2 NOlg) 15.84. Which of the following equilibria will shift towards formation of more products if the volume of a reaction mixture at equilibrium increases by a factor of2? a. 2S02(g) + 02(g) ~2S0/g) b. NO(g) + 03(g) ~ N02(g) + 02(g) c. 2 NPh) ~2NOlg) + 02(g) d. NP4(g) ~2NOlg) 15.85. What would be the effect of the changes listed on the equilibrium concentrations of reactants and products in the following reaction?

a. 03 is added to the system. b. 02 is added to the system. c. The mixture is compressed to one tenth its initial volume. 15.86. How will the changes listed affect the position of the following equilibrium?

Calculations Based on K CONCEPT

REVIEW

15.91. Explain how checking the value of Q for a specified equilibrium system such as A "" B enables you to decide whether the reactant or product concentration increases as the system shifts to attain equilibrium. 15.92. If Q < K in a reversible reaction, will the reactant concentration increase or decrease as the system shifts to attain equilibrium? PROBLEMS

15.93. For the reaction K p = 23.6 at 500 K a. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are P PCls = 0.560 atm and P PCl3 = 0.500 atm. b. If more chlorine is added after equilibrium is reached, how will the concentrations of PC 15and PCl3 change? 15.94. Enough N02 gas is injected into a cylindrical vessel to produce a partial pressure, PN02' of 0.900 atm at 298 K. Calculate the equilibrium partial pressures ofN02 and N204, given K p = 4 at 298 K 15.95. The value of K, for the reaction between water vapor and dichlorine monoxide

a. The concentration of NO is increased. b. The concentration ofN02 is increased. c. The volume of the system is allowed to expand to 5 times its initial value. 15.87. How would reducing the partial pressure of 0lg) affect the position of the equilibrium in the following reaction?

*15.88. Ammonia is added to a gaseous reaction mixture containing H2' C12, and HCI that is at chemical equilibrium. How will the addition of ammonia affect the relative concentrations of H2, C12, and HCI if the equilibrium constant of reaction 2 is much greater than the equilibrium constant of reaction 1? (1) Hlg) + Cllg) ~2HCI(g) (2) HCI(g) + NH/g) ~ NH4CI(s)

is 0.0900 at 25T. Determine the equilibrium concentrations of all three compounds if the starting concentrations of both reactants are 0.00432 M and no HOCI is present. 15.96. The value of K p for the reaction

is 4.3 X 10-4 at 648 K. Determine the equilibrium partial pressure ofNH3 in a reaction vessel that initially contained 0.900 atm N2 and 0.500 atm H2 at 648 K. 15.97. The value of K p for the reaction NO(g)

+ ~ 02(g)

~ N02(g)

6

15.89. In which of the following equilibria does the product yield increase with increasing temperature? a. A + 2 B ~ C f1H > 0 b. A + 2 B ~ C f1H = 0 c. A + 2 B ~ C f1H < 0 15.90. In which of the following equilibria does the product yield decrease with increasing temperature? a. 2X + y~Z f1H> 0 b. 2X+Y~Z f1H=O c. 2X + Y~Z f1H< 0

is 2 X 10 at 25°C. At equilibrium, what is the ratio of P N02 to P NO in air at 25°C? Assume that P O2 = 0.21 atm and does not change. *15.98. Water Gas The water-gas reaction is a source of hydrogen. Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen:

The value of K ,for the reaction at 1000°C is 3.0 X 10-2. a. Calculate the equilibrium partial pressures of the products and reactants if PH20 = 0.442 atm and Pco = 5.0 atm at the start of the reaction. Assume that the carbon is in excess. b. Determine the equilibrium partial pressures of the reactants and products after sufficient CO and H2 are added to the equilibrium mixture in part a to initially increase the partial pressures of both gases by 0.075 atm.


15.99. The value of Kp for the reaction

is 1.5 at 700°C. Calculate the equilibrium partial pressures of CO and CO2 if initially Pco, = 5.0 atm and Pco = 0.0. Pure graphite is present initially and when equilibrium is achieved. 15.100. Jupiter's Atmosphere Ammonium hydrogen sulfide (NH4SH) was detected in the atmosphere of Jupiter subsequent to its collision with the comet Shoemaker-Levy. The equilibrium between ammonia, hydrogen sulfide, and NH4SH is described by the following equation: NH4SH(g) ~ NH3(g)

+

H2S(g)

The value of Kp for the reaction at 24°C is 0.126. Suppose a sealed flask contains an equilibrium mixture of gaseous NH4SH, NH3 and H2S. At equilibrium, the partial pressure ofH2S is 0.355 atm. What is the partial pressure ofNH/ *15.101. A flask containing pure N02 was heated to 1000 K, a temperature at which the value of Kp for the decomposition ofN02 is 158. 2N02(g) ~2NO(g)

+

02(g)

The partial pressure of02 at equilibrium is 0.136 atm. a. Calculate the partial pressures of NO and N02. b. Calculate the total pressure in the flask at equilibrium. 15.102. The equilibrium constant Kp of the reaction 2S03(g) ~2S02(g)

+

02(g)

is 7.69 at 830°C. If a vessel at this temperature initially contains pure S03 and if the partial pressure ofS03 at equilibrium is 0.100 atm, what is the partial pressure of 02 in the flask at equilibrium? *15.103. NOx Pollution In a study of the formation of NO x air pollution, a chamber heated to 2200°C was filled with air (0.79 atm N2, 0.21 atm 0,). What are the equilibrium partial pressures ofN2, 02' and NO if Kp = 0.050 for the following reaction at 2200 ? 0

*15.104. The equilibrium constant Kp for the thermal decomposition ofN02 2N02(g) ~2NO(g)

+

15.105. The value of K, for the thermal decomposition of hydrogen sulfide

+

S2(g)

K, = 6.1 X 10-3 *15.107. Chemical Weapon Phosgene, COCl2 gained notoriety as a chemical weapon in World War 1. Phosgene is produced by the reaction of carbon monoxide with chlorine

The value of K,for this reaction is 5.0 at 600 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which Pco = PCl = 0.265 atm and PCOCI = 0.000 atm? *15.108. At 2000°C, the val~e of K, for the reaction' 2CO(g)

+

02(g) ~2C02(g)

is 1.0. What is the ratio of[CO] to [C02] in an atmosphere in which [02] = 0.0045 M? *15.109. The water-gas shift reaction is an important source of hydrogen. The value of K, for the reaction CO(g)

+

H,o(g)

+

~ CO2(g)

H2(g)

at 700 K is 5.1. Calculate the equilibrium concentrations of the four gases if the initial concentration of each of them is 0.050 M. *15.110. Sulfur dioxide reacts with N02, forming S03 and NO: SO,(g)

+

N02(g) ~ S03(g)

+

NO(g)

If the value of K, for the reaction is 2.50, what are the equilibrium concentrations of the products if the reaction mixture was initially 0.50 M S02' 0.50 M N02, 0.0050 M S03' and 0.0050 M NO?

Changing K with Changing Temperature CONCEPT

REVIEW

15.111. The value of the equilibrium constant of a reaction decreases with increasing temperature. Is this reaction endothermic or exothermic? 15.112. The reaction

is exothermic. Does the value of Kp increase or decrease with increasing temperature? 15.113. The value of Kp for the water-gas shift reaction CO(g)

+

H,o(g)

~ H2(g)

+

CO2(g)

increases as the temperature decreases. Is the reaction exothermic or endothermic? 15.114. Does the value of K p for the reaction f1H = 206 kJ O

is 2.2 X 10-4 at 1400 K. A sample of gas in which [H2S] = 6.00 M is heated to 1400 K in a sealed highpressure vessel. After chemical equilibrium has been achieved, what is the value of[H2S]? Assume that no H2 or S2was present in the original sample.

787

15.106. Urban Air On a very smoggy day, the equilibrium concentration ofN02 in the air over an urban area reaches 2.2 X 10-7 M. If the temperature of the air is 25°C, what is the concentration of the dimer N204 in the air? Given:

02(g)

is 6.5 X 10-6 at 450"C. If a reaction vessel at this temperature initially contains 0.500 atm N02, what will be the partial pressures ofN02, NO, and 02 in the vessel when equilibrium has been attained?

2H2S(g) ~2H2(g)

I

increase, decrease, or remain unchanged as temperature increases?

788

I

I

Chapter Fifteen


PROBLEMS

15.115.

Air Pollution Automobiles and trucks pollute the air with NO. At 2000°C, K, for the reaction N2(g)

+

O/g) ~ 2 NO(g)

is 4.10 X 10-" and 6.Ho = 180.6 kJ. What is the value of K ,at 25°C? 15.116. At 400 K the value of K p for the reaction

+

N2(g) is 41, and 6.H 700 K?

O

=

3 H/g)

-92.2

~ 2 NH3(g) kJ. What is the value of K at p

*15.121. Elements of group 16 form hydrides with the generic formula H2X, When gaseous H2X is bubbled through a solution containing 0.3 M hydrochloric acid, the solution becomes saturated and [H2X] = 0.1 M. The following equilibria exist in this solution: HzX(aq)

+

HP(£) ~ HX-(aq)

HX-(aq)

+

HP(£) ~X2-(aq)

NO(g)

+

02(g) ~2NO/g)

decreases from 1.5 X 105 at 430T to 23 at 1000°C. From these data, calculate the value of 6.Ho for the reaction. 15.118. The value of K ,for the reaction A ~ B is 0.455 at 50°C and 0.655 at lOOT. Calculate I1H for the reaction. O

Additional Problems *15.119. CO as a Fuel Is carbon dioxide a viable source of the fuel CO? Pure carbon dioxide (Pco = 1 atm) decomposes at 2 high temperatures. For the system

the percentage of decomposition temperature as follows: Temperature (K)

1500

ofC02(g)

Decomposition

changes with

17.6

3000

54.8

X

10-8

Kz = 1 X 10-14

Hp+(aq)

+

SOz(g) ~ NO(g)

+

S03(g)

An equilibrium mixture is analyzed at a certain temperature and found to contain [NOz] = 0.100 M, [S02] = 0.300 M, [NO] = 2.00 M, and [S03] = 0.600 M. At the same temperature, extra SOz(g) is added to make [SOz(g)] = 0.800 M. Calculate the composition of the mixture when equilibrium has been reestablished. *15.123. Carbon disulfide is a foul-smelling solvent that dissolves sulfur and other nonpolar substances. It can be made by heating sulfur in an atmosphere of methane. For the reaction 4CH.(g)

+

S/5) ~ 4CS/g)

+

8H2(g)

I1H:xn = 150.3 kJ per mole of CH4 and I1S:xn = 336 J per mole of CH4• Calculate the equilibrium constant at 25°C and 500°C. *15.124. Making Hydrogen Debate continues on the practicality of H2 gas as a fuel. The equilibrium constant K, for the reaction

(%)

0.048

2500

+

K1 = 8.3

Hp+(aq)

Calculate the concentration ofXz- in the solution. *15.122. Homogeneous Catalysis Nitrogen dioxide catalyzes the reaction of SOz with 0z to form S03: NOz(g)

15.117. The equilibrium constant for the reaction

+

Is the reaction endotherm ic? Calculate the value of Kp at each temperature and discuss the results. Is the decomposition of CO2 an antidote for global warming? 15.120. Ammonia decomposes at high temperatures. In an experiment to explore this behavior, 2.000 moles of gaseous NH3 are sealed in a rigid 1 liter vessel. The vessel is heated at 800 K and some of the NH3 decomposes in the following reaction:

The system eventually reaches equilibrium and is found to contain 0.004 mole ofNH3' What are the values of Kp and K, for the decomposition reaction at 800 K?

is 1.0 X 105 at 25T. Starting with this value, calculate the value of 6.G:xn at 25°C, and, without doing any calculations, guess the sign of 6.H:xn' *15.125. Smokestack Cleanup Calcium oxide is used to remove the pollutant SOz from smokestacks and exhaust gases. The 6.G of the overall reaction o

CaO(5)

+

S02(g)

+ 10/g)

~ CaSO.(5)

is -418.6 kJ. What is Pso, in equilibrium with air (Po = 0.21 atm) and solid CaO?

,

*15.126. Volcanic Eruptions During volcanic eruptions, gases as hot as 700°C and rich in SOz are released into the atmosphere. As air mixes with these gases, the following reaction converts some of this SOz into S03:

Calculate the value of Kp for this reaction at 700°C. What is the ratio of Pso, to PS03 in equilibrium with Po, = 0.21 atm?


A LOOK AHEAD: A Balancing Act

16.1 Acids and Bases 16.2 The Br0nsted-Lowry

Model

16.3 pH and the Autoionization of Water

16.4 Calculations with pH and K 16.5 Polyprotic Acids 16.6 Acid Strength and Molecular Structure

16.7 pH of Salt Solutions 16.8 The Common-Ion Effect 16.9 pH Buffers 16.10 Acid-Base Titrations and Indicators

SHADES OF PURPLE, PINK, AND BLUE FLOWERS 1hecolorofhydrangea flowers depends on the acid content of the soil in which they are grown.

790

I Chapter Sixteen

I


A LOOK AHEAD: A Balancing Act Relative acidity or alkalinity is a crucially important characteristic of many systems. Farmers adjust the acid content of soil to enhance the growth of crops, and gardeners recognize the need of plants like blueberries and azaleas for an acidic environment. If the soil in which these plants are grown is insufficiently acidic, the plants cannot make use of nutrients, even if they are abundant in the ground. Most plants grow best in soils that are neutral to very slightly acidic because the microorganisms that play key roles in making nutrients available to plants thrive in such soils. Hydrangeas vary in color with the acidity of the soil in which they are grown. The acid content of the soil affects the ability of aluminum ions to be taken up by the plant and to interact with pigment molecules called anthocyanins present in the flowers. Aluminum ions are soluble in acidic solutions, and hydrangeas grown in acid soil have blue flowers. In neutral to slightly alkaline soil, however, aluminum precipitates as aluminum hydroxide and is not available to plants. Hydrangeas grown in these soils produce pink blooms. The relative acidity or alkalinity of soil and ground water depends on mineral content and the characteristics of the rainwater that falls on it. Normal rainwater is slightly acidic because of the presence of dissolved CO2, which reacts with water and produces hydrogen ions in solution. Acid rain, on the other hand, is formed when oxides of nitrogen and sulfur enter the air, typically as byproducts of the combustion of fossil fuels. Compared to CO2, NOx and S02 react to a greater extent with water. Air containing NO x and S02 yields rainwater with a hydrogen ion content that is 10 to 1000 times more acidic than that of rainwater containing only dissolved CO2, The more acidic rainwater alters the chemical balance in soils and interferes with plants' abilities to process nutrients. As a result, plants don't grow well. Sometimes they die. Carbon dioxide and carbonate compounds play important roles in determining the relative acidity and alkalinity in biological systems. We produce CO2 during metabolism, and we eliminate it every time we exhale. Our breathing maintains a balance of relative acidity and alkalinity by controlling the amount of dissolved CO2 in our systems. We actually need more controlled CO2 levels than O2 levels. Carbon dioxide in the fluids surrounding cells and in particular in blood maintains the acid-base balance our bodies need. This balance is vital because hydrogen ions interact with amino acids and alter the charge on proteins. Changes in charge disturb protein structure and function. The body can tolerate only a small change in hydrogen ion concentration in the blood, and deviations outside the normal range lead to coma and death. Ocean water is naturally alkaline and typically has a hydrogen ion concentration nearly 10 times lower than in human plasma. At the oceans' natural level of alkalinity, mollusks such as clams and oysters are able to form solid calcium carbonate shells and corals can form reefs of calcium carbonate, which has limited solubility under these conditions. However, if ocean water becomes more acidic because more CO2 or other nonmetal oxides dissolve in it, the solubility ofCaC03 increases, which could have disastrous consequences for marine life. The balance between acidity and alkalinity in rainwater, groundwater, and seawater clearly has a direct or indirect impact on all life forms. In this chapter we examine processes that disturb this balance and others that help restore it.

16.1

~

I

Acids and Bases

I 791

Acids and Bases

Acids and bases were originally identified as unique classes of compounds because of their distinctive properties. You must never taste or even casually smell unknown materials, but early chemists frequently used taste and aroma as identifying characteristics of substances, and the practice sometimes had fatal consequences. Acids taste sour; bases taste bitter. Aqueous solutions of bases have a slippery, soapy feel. Acids and bases can change the color of certain vegetable dyes. Litmus, a substance extracted from lichen, turns red in the presence of an acid; a base turns red litmus blue. Acids and bases react with each other in a process called neutralization because the reaction causes the acidic and basic properties of the two substances to disappear. Some of the acids and bases that we frequently encounter are listed in Table 16.1. Acids and bases are compounds of profound importance to industry, every living system, and the environment. Their commercial uses range from the production of steel, fertilizer, fabrics, soaps, pharmaceuticals, and cleaner-burning gasoline to enhancing oil recovery from wells, flavoring foods and beverages, and cleaning metals. In our bodies, hydrolysis of CO2 produces a weak acid that plays a major role in maintaining the acid-base balance necessary to assure normal functioning. In Chapter 4 we described the transformation of early Earth by reactions between the acids-formed when volcanic gases such as oxides of nitrogen and sulfur dissolved in rainwater-and basic minerals in the Earth's crust. In Chapters 14 and 15 we learned how internal combustion and diesel engines produce NO and how combustion of fossil fuels containing sulfur produces S02' Once in the atmosphere, these nitrogen and sulfur oxides react with additional O2 and water vapor to produce nitric acid (HN03), sulfuric acid (H2S04), and other acids that fall to Earth in the form of acid rain. The usefulness as well as the potential hazards associated with acids and bases depend in large part on the behavior of these compounds in water. Some acids and bases are completely dissociated in water; the behavior of others is characterized by equilibria that are governed by the same principles we explored for reactions in the gas phase in Chapter 15. In this chapter we apply the lessons we learned in Chapter 15 to an examination of the vitally important chemistry of acids and bases in aqueous solutions.

TABLE 16.1

@ pKa' If the concentration of the basic component is less than that of the acid, then the logarithmic term is less than zero, and pH < pKa. Consider the case in which the concentration of the base is 10 times the concentration of the acid, or [base] = 10[acid]. Substituting this ratio into the Henderson-Hasselbalch equation, we have

pH

=

[base]

pK a

+ log-[ aCl'd]

The

10 [acid]

= pK + Iog--a

=

pKa

+ log

[acid]

10

=

pKa

+

1

Henderson-Hasselbalch equation is used to calculate the pH of a solution in which the concentrations of acid and conjugate base are known.

I 821

822

I Chapter Sixteen

I


Therefore, a 10-fold higher concentration of base than acid produces a pH one unit above the pKa value. Similarly, if the concentration of the acid component is 10 times that of the base, pH is one unit below pKa' To demonstrate how the Henderson-Hasselbalch equation simplifies pH calculations when we know the concentrations of a weak acid and its conjugate base, let's use that equation to recalculate the pH of our river-water sample. Our starting information includes [H2C03]

=

[HC03-]

= 1 X 10-4 M

1.2 X 10-5 M

and the equilibrium and equilibrium constant

K = 4.3 X 10-7 '1

First we take the negative log of Ka1: pKa,

= -log Ka, = -log(4.3 X 10-7) = 6.37

and then insert pK., and the concentration values of [H2COJ and [HC03-] the Henderson-Hasselbalch equation. Solving the equation gives pH

into

[base]

= pK a + log-[ and -.-] = 6.37 +

4

1.0 X 10g 1.2 X 10-5

10

= 7.29 This result matches the one we calculated. As we have seen, the Henderson-Hasselbalch equation can be used to calculate the pH of a solution containing a weak acid and its conjugate base. It can also be used to calculate the pH of a solution of a base and its conjugate acid. There is an extra step involved in that calculation because we may know the Kb of the base but not the K. of its conjugate acid. Fortunately, the conversion from Kb to K., or Ka to Kb' for a conjugate pair makes use of the simple relation between Kb and Ka (Equation 16.19): Ka X Kb

= Kw

Because we use pKa values in the Henderson-Hasselbalch equation, we can take the negative logarithms of both sides of Equation 16.19. We first rearrange Equation 16.19 and then insert the value of Kw at 25°C:

_ Kw _ 1.00

.

K ------b K -logKb

pKb

X 10-14

K

.

= -log (1.00 X 10-14)

=

-

(-logKJ

14.00 - pKa

or

pKb

+ pK" =

14.00

(16.22)

16.9

I

pH Buffers

I 823

Calculating the pH of a Solution of a Base and Its Conjugate Acid Calculate the pH of a solution that is 0.200 Min The Kb ofNH3 is 1.76 X 10-5. COLLECT

AND

ORGANIZE

taining a weak base (NH) Hasselbalch equation,

NH3 and 0.300 Min

NH4Cl.

We are asked to calculate the pH of a solution conand a salt ofits conjugate acid (NH4 +). The Hendersonp H - pK a

[base]

+ log-[ aCl.d]

is used to calculate the pH of such a solution, but first we need to calculate the pf\, of the acid in the NH4 + solution. The pKa ofNH/ is related to the pKb ofNH3 by the following equation, which applies to all acid-base conjugate pairs in aqueous solutions at 25°C:

ANALYZE

pKb

+ pKa =

14.00

We are given the value of Kb ofNH3: 1.76 X 10-5. Our approach involves taking the negative logarithm of Kb to obtain pKb and then converting pKb to a pK .. SOLVE

We calculate pKb first: pKb

After inserting

= -logKb = -log(1.76

X 10-5)

this value of pKb into Equation pKa

= 4.75

16.22, we then solve for pKa:

= 14.00 - pKb = 14.00 - 4.75 = 9.25

By inserting this value and the given concentrations Henderson-Hasselbalch equation, we have pH

of NH3 and NH4

+

into the

[base]

= pK + log-a

[acid] 0.200

= 9.25 + log 0.300 = 9.07 T The Kb value of NH3 is greater than the Ka we calculated for NH4 + This statement makes sense because ammonia is a stronger base than the ammonium ion is an acid. It is reasonable that a solution containing similar concentrations of both should be slightly basic.

T HI N K ABO UTI

Practice Exercise Calculate the pH of a solution that is 0.150 M in benzoic acid and 0.100 M in sodium benzoate, its conjugate base .

.:IDI pH Buffers A solution that has the capacity to resist pH changes by neutralizing small additions of acid (or base) is called a pH buffer. Buffers are an important component of natural water systems; indeed, they are also essential to life itself. The internal pH of most living cells is regulated by buffer systems. The carbonic acid-bicarbonate system just described is one of the buffer systems that acts in living organisms as well as in rivers. When pH stabilization provided by these buffers is disturbed, proteins do not function optimally, and the health of individual cells and whole organisms is in jeopardy.

A pH buffer is a solution of acidic and basic solutes that resists changes in its pH when acids or bases are added to it.

824

I

Chapter Sixteen

I


An Environmental Buffer

~II

CHEMTOUR

Buffers

The presence of bicarbonate ion in the river water in the preceding section gives the water a capacity to resist pH change when, for example, acid rain falls into it. We can use the Henderson-Hasselbalch equation and other acid-base relations to examine the effect of adding a small amount of a strong acid to river water. For purposes of comparison, we can measure the effect on the pH of the same volume of pure (pH = 7.00) water at 25°C. We start with 1.00 L each of river water and pure water and add 10.0 mL of a 1.0 X 10-3 M HN03 solution to both. Adding acid to pure water is an exercise in dilution, which is a concept we introduced in Section 4.3. In this case, 10.0 mL of a 1.0 X 10-3 M HN03 solution is diluted to a final volume of 1.01 L. To calculate the final concentration of HN03 (and H+ ions) we use Equation 4.5: V,nitial X

Ginitial

=

Vfinal X Gfinal

where Ginitial and Gfinal are the initial and final concentrations of the acid and V,nitial and ~nal are its initial and final volumes. Expressing 10.0 mL as 0.0100 L so that the initial and final volume units match and then substituting the given information into Equation 4.5 leads us to 0.0100 L

(1.0

X

Gfinal

=

10-3 M)

X

[H+]

=

= 1.01 L

X

Gfinal

9.9 X 10-6 M

Nitric acid is completely dissociated, so the final concentration of H+ must also be 9.9 X 10-6 M. The corresponding pH value is -log(9.9 X 10-6 M) = 5.00, or 2.00 pH units lower than the initial pH of the pure water sample. To calculate the change in pH when the same quantity of nitric acid is added to one liter of our pH 7.29 river water, we need to reexamine the carbonic acidbicarbonate equilibrium: H2CO/aq)

~ HC03-(aq)

+ H+(aq)

When strong acid (H+) is added to a mixture ofH2C03 and HC03 -, it reacts with HC03- to produce H2C03.The ionization reaction runs in the reverse direction: H+(aq)

+ HC03-(aq)

~ H2C03(aq)

until essentially all of the added H+ ions have been consumed. We need to determine the number of moles of H + that were added so that we can calculate how many moles ofHC03- were consumed and, with that number, how many moles are left. There are 1.0 X 10-5 mol of H+ in 0.010 L of a 1.00 X 10-3 M HN03 solution. Given the 1:1:1 stoichiometric ratio of the reactants and products, what was added is completely consumed. In the process 1.0 X 10-5 mol of HC03- are consumed. The number of moles of bicarbonate present initially was (1.0

X

10-4 M)

X

(1.00 L)

=

1.0

X

10-4 mol

After subtracting the number of moles consumed, we have 9.0 X 10-5 mol of HC03 - in a final volume of 1.01 L. The resulting pH can be predicted from the K a, for carbonic acid. In the calculation, we let x = [H+] and we assume that [H2C03] is controlled by the solubility of CO2 in water and does not change from its initial value of 1.2 X 10-5 M. Inserting these values into the mass action expression for

1\"

ofH2C03

gives

[HC03-][H+] Ka, = ~C03]

(

=

9.0 X 10-5 mOl) 1.01 L x 1.2 X 10-5 M

=

7

4.3 X 10-

16.9

Solving for x gives x

= [H+] = 5.7

X 10-8 M

After taking the negative logarithm of this value we have pH

=

-log(5.7

X 10-8)

=

7.24

Recall that the original pH was 7.29. Therefore, the addition of nitric acid lowered the pH of the river water by only 0.05 pH units because of the action of the carbonic acid-hydrogen carbonate buffer. We can compare this result to the reduction by 2.00 pH units when the same amount of nitric acid was added to the same volume of pure unbuffered water. Calculating the Response to Addition of Acid or Base to a Buffer Calculate the change in pH upon addition of1.0 mL of a 1.0 MHCI solution to 100 mL of a solution that is 0.10 M sodium acetate and 0.10 M acetic acid at 25T. COLLECT

AN DORGAN

IZE A solution of a weak acid and a salt of its conjugate

base functions as a pH buffer. We know the volume and composition of the buffer and of the strong acid added to it. We need to calculate the composition of the reaction mixture after the addition. Equation 16.16 can be used to calculate the pH of such a mixture once we know the new concentration ratio of base (acetate ion) to acid (acetic acid). The H+ ions in the added strong acid react with acetate ions, forming unionized acetic acid

ANALYZE

The product of the volume times the concentration of the strong acid can be used to calculate the number of moles of CH3 COO- that is converted into CH3COOH. SOLVE

The initial quantities of CH3COO-

and CH3COOH

in the buffer are

both: 100 mf, X 0.100 mol f,

=

10.0 mmol

The quantity ofH+(aq) that was added 1.0 mf, X 1.00 mol f,

=

1.00 mmol

represents the quantity of CH3COOconverted into CH3COOH. the quantities of both after the addition of acid are (10.0 - 1.0) = CH3COO- and (10.0 + 1.0) = 11.0 mmol CH3COOH. Both these are in the same total volume. Therefore we can use the ratio of the two lieu of concentration values in Equation 16.16 to calculate pH: pH

[CH3COO-]

= pK - log ---a

[CH3COOH]

= 4.75 - log-

9.0

11.0

Therefore 9.0 mmol quantities of them in

= 4.66

The result makes sense because adding strong acid should lower the pH from the initial value of the buffer in which CH3COO= CH3COOH. This value is the same as the pKa of acetic acid (4.75). T HI N K ABO UTI T

Practice Exercise Calculate the change in pH after addition of 10.0 mL of a 0.100 M solution of NaOH to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic acid.

I

pH Buffers

I

825

826

I Chapter Sixteen

I


CONCEPT

TEST

Suggest a specific situation in which it may be advisable to have more weak acid than conjugate base in a buffer.

A Physiological Buffer The carbonate buffer system illustrated by the sample of river water is also of great importance in living systems. The rates of most biochemical reactions and their equilibrium constants are sensitive to changes in pH. Many biochemical reactions involved in life-sustaining processes like metabolism, respiration, the transmission of nerve impulses, and muscle contraction and relaxation take place only within a narrow pH range. The carbonic acid/bicarbonate conjugate acid-base pair form an important buffering system in blood and cellular fluids. The pH of blood can be altered by ingestion of acidic or basic substances, and the carbonate/bicarbonate buffer system compensates for such additions and maintains pH within the required range. This buffering system is intimately tied to our respiration, and an exceptional feature of pH control by this system is the role of breathing to maintain pH balance. Let's first consider how this buffer system is established in the body. Carbon dioxide is a normal product of metabolism. It is transported to the lungs, where it is eliminated from the body every time we exhale. When CO2 dissolves in blood plasma (or any aqueous medium), the following equilibria are established:

If biochemical processes in the body increase [H+] in the blood, bicarbonate ions react with the added hydrogen ions and are transformed into carbonic acid, which increases the concentration of dissolved CO2 in the blood. Respiration increases, and more CO2 is expelled from the lungs. These shifts are summarized in the following diagram. H+(aq)

U

+ HC03-(aq)

CO/aq)

~

c:::===>

c==

increase in

+ H20(t')

[H+]

CO2 exhaled

If biochemical processes consume H + ions, then the following equilibrium shifts occur: H+(aq)

+ HC03-(aq)

~ decrease in [H+]

~

NH3 > H20. Chemists use the parameter field strength to describe the relative abilities of ligands to split the energies of d orbitals in metal ions with which they form complexes. Chemists rank-order ligands on the basis of their field strengths in what is called a spectrochemical series. Table 17.4 contains one such series. The ligands at the top of the chart create the strongest fields, those at the bottom create the weakest. As the field strength of the ligand increases, the splitting energy (~) increases. Consequently, the strong-field ligands form complexes that absorb shorter wavelengths of light, whereas complexes of weak-field ligands absorb longer wavelengths.

series is a list of

A spectrochemical

ligands rank-ordered on their abilities to split the energies of the d orbitals of transition metal ions.

Spectrochemical Series of Some Common Ligands

TABLE 17.4

CW N0

-

2

en py= NH3

~

EDTN-

b

HO 2

"I'"

""0

;E

OHF-

~

.!

I'~ 11

In addition to determining the color of transition metal ions, crystal field splitting can influence their magnetic properties because these properties depend on the number of unpaired electrons in the valence shell d orbitals; the larger this number, the more paramagnetic are the ions. For example, an Fe3+ ion has five 3d electrons. In an octahedral field there are two ways to distribute these electrons among the five 3d orbitals. One way, and the one that conforms to Hund's rule, places a single electron in each orbital, leaving them all unpaired. However, when the ion is in a strong octahedral field where the value of ~o is relatively large, then all five electrons may go into the three orbitals of lower energy, leaving the upper two orbitals empty. This pattern occurs when the energy needed to pair the electrons in the lower-energy orbitals is less that the energy needed to promote an electron to one of the two higher-energy orbitals. In this configuration, only one electron is left unpaired. The first option with all five 3d electrons unpaired is called high spin because the spin on all five electrons is in the same direction and the resulting magnetic field produced by their spins is maximal. When four of the five electrons are paired and only one is left unpaired, the electron configuration is called low spin (Figure 17.17). Both configurations are paramagnetic because both have at least one unpaired electron. However, the high-spin state is much more paramagnetica property that can be measured with an instrument called a magnetometer. Other transition metal ions can have high-spin and low-spin states, but not all do. Consider, for example, C2+ ions in an octahedral field. Each ion has only

1l

;.:0 0.:

'" 0;1

:§ 0

Magnetism and Spin States

'"

"toc:: ~ ~ ~

ClBr-

1-

FIGURE 17.17 (a) Low-spin and highspin complexes. The ground state of a free Fe3+ ion has a degenerate, half-filled set of 3d orbitals. (b) In a strong octahedral field the energy of the 3d orbitals are split enough (6. > electron pairing energy) to produce the low-spin state. (c) A weak octahedral field (6. < electron pairing energy) produces the high-spin state. The Fe3+ ions in crystals of magnetite and (d) aquamarine are high spin. 0

0

IT] dx2 - y2 dz2

DJIJ

~ "c:: ~

dx2 - y2 dz2

DITIIIJ ITIIIIJ dxy

ITIIllIIIIJ FeH ion (a)

dyz

dxz

dxy

dyz

dxz

Low spin

High spin

(b)

(c)

(d)

17.9

three electrons distributed among its 3d orbitals. Each will be unpaired whether the orbitals are split a lot or only a little by the field. Therefore, for C2+ ions there is only one spin state. Other metal ions with full, or nearly full, sets of d orbitals have only one spin state because there are no, or only one or two, unpaired electrons in these orbitals. The following Sample Exercise illustrates the spin state limitations of having only few or near the maximum number of electrons in a set of d orbitals.

Predicting Spin States Not all transition metal ions can have high-spin and low-spin configurations. Determine which of the following ions could have high-spin and low-spin configurations in an octahedral field: (a) Mn4+; (b) Mn2+; (c) C03+; (d) Cu2+. COLLECT AND ORGANIZE To determine whether high- and low-spin states are possible, we first need to determine the number of d electrons in each of the ions. Then we need to distribute them among sets of d orbitals split by an octahedral field to see if the ions can have different high- and low-spin states. Mn, Co, and Cu are in groups 7, 9, and 11, respectively, of the periodic table. Also, in an octahedral field, a set of five d orbitals splits into two subsets: a low-energy subset of three orbitals, and a high-energy subset of two orbitals.

A N A L Y Z E The electron configurations

of the atoms of the three elements are

When they form cations, the atoms of these transition metal ions retain as many of their 3d electrons as possible. Therefore, the numbers of d electrons in the four ions are as follows: Ion

Mn4+

f-

3d electrons

3

5

6

9

SOLVE

a. Mn4+: Putting the electrons into the lowest-energy orbitals available, and keeping them unpaired as much as possible means placing each of the three in its own lower-energy orbital:

There is no low-spin option for Mn4+ if Hund's rule is obeyed. b. Mn2+: There are two options for distributing orbitals: (1)

five electrons among the five

(2)

IT]

The pattern in (1) represents a high-spin state.

state; the one in (2) is a low-spin

IXI)

I

Magnetism and Spin States

CONNECTION Substances made of atoms, ions, or molecules that contain unpaired electrons are paramagnetic (See Section 9.6).

I 873

874

I Chapter Seventeen

I


c. C03+: Distributing the six electrons as evenly as possible (1) produces a highspin state (four unpaired electrons). Placing electrons preferentially in the lower-energy orbitals (2) produces a low-spin state (no unpaired electrons): (1)

(2)

IT]

d. Cu2+: The nine 3d electrons completely fill the lower-energy orbitals and nearly fill the higher-energy ones. There is no other arrangement possible than the one shown, and so Cu2+ has only one spin state:

ABOUT IT In an octahedral field, metal ions with 4,5,6, or 7 d electrons can exist in high-spin and low spin states. The magnitude of the crystal field splitting energy, ~o' determines which spin state the ion occupies. THINK

Practice Exercise Which, if any, of the spin configurations

in Sample Exer-

cise 17.6 are diamagnetic? Whether a transition metal ion is in a high-spin or a low-spin configuration depends on the relative energy cost of pairing two electrons in a lower-energy d orbital compared with the energy cost of promoting one of them to a higherenergy d orbital (~). Several factors affect the size of ~o. We have already discussed a major one in the context of the spectrochemical series (see Table 17.4): a ligand that is a stronger Lewis base produces a stronger field and larger ~ 0 than a ligand that is a weaker Lewis base. For this reason, hydrated metal ions are more likely to be in high-spin states, and metals surrounded by nitrogen-based ligands are more likely to be in low-spin states because the N atoms are stronger Lewis bases; they induce larger d orbital splitting, and so more electrons are paired up in the lower-energy subset of orbitals. Other factors affecting spin state include the oxidation state of the metal ion. The higher the oxidation number (and ionic charge), the stronger the attraction of the electron pairs on the ligands for the ion. Greater attraction leads to more d orbital-ligand interaction and a larger ~o. The size of the ion counts, too. For example, fifth-row transition metal ions are larger than fourth-row ions with the same charge. For the same ligand, the larger the ion, the greater the extension of its d orbitals into the space around its nucleus. This leads to greater interaction with the lone pairs of electrons on a given ligand and to larger ~ 0 values. Our discussion of high- and low-spin states has focused entirely on d orbitals split by octahedral fields. What about spin states in tetrahedral fields? It turns out that most tetrahedral complexes are high spin because tetrahedral fields are created by only four ligands instead of the six in an octahedral field, and so they are weaker. Therefore, d orbital splitting tends to be less in a tetrahedral field; it is usually not large enough to offset the energies associated with pairing electrons in the same orbitals.

17.10

CONCEPT

I

Naming Complex Ions and Coordination Compounds

TEST

Explain the following: (a) Mn(pyridine)/+ is a high-spin complex ion, but Mn(CN)64- is low spin; (b) Fe(NH3)/+ is high spin, but Ru(NH)/+ is low spin.

mElI

Naming Complex Ions and Coordination Compounds

It is important to be able to name complex ions and coordination compounds so that we can express the identity and oxidation state of the central metal ion, the names and numbers of ligands bound to the inner coordination sphere of the metal ion, the overall charge of the complex ion, and the identities of the counter ions surrounding the complex. To convey all of this information, we need to follow some simple naming rules.

Complexes with a Positive Charge 1. Start with the name(s) of the ligand(s). The names of common ligands appear in Table 17.3. If there is more than one kind ofligand, put their names in alphabetical order. 2. Use the usual prefixes to indicate the number of each type ofligand in the complex ion: Number of Ligands

Prefix

2

Di-

3

Tri-

4

Tetra-

S

Penta-

6

Hexa-

3. Next write the name of the central metal ion using a Roman numeral to indicate the oxidation state of the transition metal ion. Following these rules, the names of the following three six-coordinate ions are Ni(HP\2+

Hexaaquanickel(II)

Co(NH)63+

Hexaamminecobalt(III)

[Cu(NH3) 4(HP)2]2+

Tetraam minediaquacopper(II)

It may seem strange having two a's together in these names, but it is permitted under current naming rules. In the third name, ammine comes before the aqua because, alphabetically, am comes before aq. The prefixes are ignored in determining alphabetical order. In these three examples the ligands are all neutral molecules, and so the charge of the complex ion is the same as the charge on the central metal ion, which also matches the oxidation state expressed by the Roman numeral. When the ligands include anions, determining the oxidation state of the central metal ion requires a bit more work, as we shall see in the next examples.

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Complexes with a Negative Charge 1. Follow the steps we use to name positively charged complexes. 2. Use an -ate ending on the name of the central metal ion to indicate the negative charge of the complex (just as we use -ate to end the names of oxoanions). For some metals, the base name changes, too. The two most common examples are iron, which becomes ferrate, and copper, which becomes cuprate. Following these rules, the names of the following complex anions are as follows:

N

We do not need a Roman numeral in the name of the aluminum complex because the oxidation number of Al in all its compounds is +3 (see Section 2.8). CONNECTION

Fe(CN)63-

Hexacyanoferrate(III)

[Fe(Hp)(CNls]3-

Aquapentacyanoferrate(II)

[Al(HPMOH)4]-

Diaquatetrahydroxoaluminate

In the first two examples we must determine the oxidation state of Fe, that is, the charge on its ion. To do so we start with the overall charge on the complex ion and then take into account the charge on the ligand anions to calculate the charge on the metal ion. For example, the overall charge of the aquapentacyanoferrate(II) ion is 3-. It contains five CN- ions. To reduce the combined charge of 5- from these cyanide ions to an overall charge of 3-, the charge on Fe must be 2+.

Coordination Compounds 1. If the counter first, followed 2. If the counter complex goes

ion of the complex ion is a cation, the cation name goes by the name of the anionic complex ion. ion of the complex ion is an anion, the name of cationic first followed by the name of the anion.

Applying these rules, the names of the following coordination [Ni(NH)6]CI2

Hexaamminenickel(II)

K3Fe(CN\

Potassium hexacyanoferrate(III)

[Co(NH)s(H2O)]Br2

Pentaammineaquacobalt(II)

compounds

are:

chloride

bromide

A key to naming coordination compounds is to recognize from their formulas that they are coordination compounds. For help with this, look for formulas that have: 1. The atomic symbols of one or more metallic elements followed by the formula of one or more of the ligands in Table 17.3. 2. The symbol of a metallic element and one or more of the ligands in Table 17.3, all in brackets, followed by the symbol of an anion. The following Sample Exercise provides practice in applying these rules.

Naming Coordination Name the following coordination

Compounds

compounds:

AND ORGANIZE Our task is to write a name for each compound that unambiguously identifies the composition of its complex ion and its overall composition. The formulas of the complex ions appear in brackets in both compounds. COLLECT

17.11

Cobalt is the central metal ion in both. Cobalt is a transition metal and so we express its oxidation state using Roman numerals. The names of common ligands are given in Table 17.3. can be useful to take an inventory of the ligands and counter ions shown in this table.

A N A LY ZEIt

Compound

Counter Ion

I I

LIGAND Name

Formula

Number

Prefix

Na4[Co(CN)6]

Na+

CN-

Cyano

6

Hexa-

[Co(NH3lsCI](N02)2

N02-

NH3

Ammine

5

Penta-

Cl-

Chloro

1

-

The charges of the cobalt ions in the two compounds can be calculated by setting the sum of the charges on all the ions in both compounds equal to zero: a. Ions: Charges:

(4 Na+ ions)

+

4+

+

(1 Co ion) x

+

(6 CN- ions)

+

6-

=0 x

b. Ions: Charges:

(1 Co ion) x

+ (1 Cl- ion) + (2 N02

+

1-

+

-

= 2+

ions)

=0

2x

= 3+

SOLVE

a. The name of the compound is sodium hexacyanocobaltate(II). b. The name of the compound is pentaamminechlorocobalt(III)

nitrite.

THINK ABOUT IT Naming coordination compounds requires you to (1) distinguish between ligands and counter ions and (2) recall which ligands are neutral and which are anions. Naming a coordination compound correctly shows that you understand these features of their structures.

Practice Exercise Identify the ligands and counter ions in the following compounds and name each compound:

IDII Isomerism in Coordination Compounds We introduced the concept of structural and geometric isomerism in Chapter 12. To review: • Structural isomers are compounds that have the same chemical formula but different arrangements of the bonds in their molecules. For example, these two hydrocarbons are structural isomers because they have the same formula (C4H8) but the C=C double bonds are in different locations in their structures: H......... C=C

H/

I-Butene

/CH2CH3

H3C......... /C=C

......... H H

/CH3 .........

H

2-Butene

• Geometric isomers, also called stereoisomers, are compounds with the same formulas and the same bonding pattern, but the orientation of the

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Isomerism in Coordination Compounds

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groups connected by those bonds is different. For example, these two hydrocarbons are geometric isomers: H......... C=C HC/ 3

/CH3 ......... H

trans-2-Butene

N

CONNECTION The rigidity of double bonds is described in Section 12.4.

H3C......... /C=C

/CH3 .........

H

H

cis-2- Butene

In trans-2-butene the -CH3 groups are on opposite sides of the C=C double bonds, but in cis-2-butene they are on the same side. Remember that C=C bonds are rigid so that the ends of molecules cannot rotate around the C=C axis as they can around C-C single bonds. Geometric Isomers

(a)

(b)

FIGURE 17.18 There are two ways to orient the two Cl- ions and the two NH3 molecules in the square planar coordination compound

Pt(NH3)2CI2:

two pairs ofligands

(a) each of the

could be on the

same side of the square, as they are in cis-diamminedichloroplatinum(II),

or (b) the

pairs could be in opposite corners, as in trans-diamminedichloroplatinum(II).

Both structural and geometric isomers occur in complex ions, but geometric isomerism is the focus of our discussion here. Let's begin with the square planar Pt2+ coordination compound that has the formula Pt(NH3\CI2. Both molecules of ammonia and the two chloride ions are coordinately bonded to the inner coordination sphere of the Pt2+ ion, so the name of the compound is the name of the complex: diamminedichloroplatinum(II). There are no counter ions because there is no net charge on the complex. Two geometric isomers of diamminedichloroplatinum(II) are possible because there are two ways to orient the two pairs ofligands around the square plane of the complex: each pair could be at the same side of the structure, or they could be at opposite corners, as shown in Figure 17.18(a) and (b), respectively. The isomer with each pair on the same side is called cis-diamminedichloroplatinum(II), and the isomer with the pairs in opposite corners is trans-diamminedichloroplatinum(II). Note that cis and trans have the same meaning in coordination chemistry as in organic nomenclature. To illustrate the importance of geometric isomerism, consider this: cis-diamminedichloroplatinum(II) is a widely used anticancer drug with the common name cisplatin, but the trans- isomer is ineffective in fighting cancer. The therapeutic power of cisplatin comes from changes in its structure when it is administered to a cancer patient. The first change is replacement of a chloro ligand (with a 1- charge) with a neutral water molecule, forming the cis-diammineaquachloroplatinum(II) ion with a 1+ charge. This compound is very reactive and binds to a nitrogen atom on DNA. Once bound, it readily undergoes another ligand exchange reaction in which the second chloride ion (with a 1- charge) is replaced by another neutral water molecule, forming another ion that is now directly attached to the DNA strand. This reactive ion forms another bond to a nitrogen atom on the DNA strand and causes distortion in the structure of the molecule. This two-point attack on the cancer cell's DNA has a remarkable biological effect: it prevents the DNA from replicating, which causes the cells to die. The trans isomer can also undergo ligand replacement of a chloride ion for water molecules, but the resultant ion cannot form links between nitrogen atoms on a DNA strand and is inactive as an anticancer agent. CONCEPT

TEST

How many isomers does the square-planar have?

coordination compound Pt(NH3)3CI

17.11

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Isomerism in Coordination Compounds

I 879

Geometric isomerism is also possible in six-coordinate octahedral complexes containing more than one type ofligand. For example, there are two possible geometric isomers of the coordination compound [Co(NH)4Cl2]CI as shown in the structures in Figure 17.19. The two chloro ligands in the [Co(NH)4Cl2] + complex ion are either on the same side of the complex with a 90° CI-Co-Cl bond angle, or they are across from each other with the metal atom in between. When the two chloro ligands are on the same side, the structure is called cis. The isomer with the chloro ligands on opposite sides of the complex is called trans. The full names of the two isomeric compounds are cis-tetraamminedichlorocobalt(III) chloride, which is violet, and trans-tetraamminedichlorocobalt(III) chloride, which is a lovely shade of green.

Identifying Geometric Isomers of Coordination Compounds


Sketch the structures and name the geometric isomers ofNi(NH)4CI2' COLLECT AND ORGAN IZE The formula contains no brackets, so there are no counter ions. This means that all four ammonia molecules and two chloride ions must occupy inner coordination sphere bonding sites. Both ligands are monodentate.

1- charges on the two chloride ions must be balanced by a charge of 2+ on the nickel ion. A total of six ligands means that the complex must be octahedral.

A N A L Y Z E The

(a) cis-tetraamminedichlorocobalt(III)

chloride

There are two ways to orient the two chloride ions: opposite each other with a CI-Ni-Cl bond angle of 180° or on the same side of the octahedron with a CI-Ni-Cl bond angle of 90°: SOLVE

CI

I....

NH 3

HN-Ni:""-NH 3

/

I

3

H3N CI The first isomer is trans-tetraamminedichloronickel(II); cis-tetraamminedichloronickel(II).

the

second

is

T HI N K ABO UTI T Although we could draw other tetraamminedichloronickel(II)

complex ions that might not initially look like these two isomers, if we consider their three-dimensional structures, we would see that each of them is the same as one of these two.

Practice Exercise Sketch the geometric isomers of [CoBr/en)(NH)2]+

and

name them.

Enantiomers Another kind of geometric isomerism is possible in complex ions and coordination compounds. Consider the octahedral Co (III) complex ion containing two ethylenediamine (en) molecules and two chloride ions. There are two ways to arrange the chloride ions: on adjacent bonding sites in a cis isomer or on opposite sides of the octahedron in a trans isomer. The cis option is shown in the molecular models in Figure 17.20. Note that the name of the isomer includes a prefix, bis, that we have not encountered before. In naming complexes with polydentate

(b) trans-tetraamminedichlorocobalt(III) chloride FIGURE 17.19 What a Chemist There are two geometric isomers of the coordination compound with the formula [Co(NH3)4Cl2]Cl: (a) cis-tetraamminedichlorocobalt(III) chloride and (b) trans-tetraamminedichlorocobalt(III) chloride.

Sees

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Metals Mirror

FIGURE 17.20 The complex ion cisdichlorobis(ethylenediamine)cobalt(III)

is

chiral, which means that the mirror

image of

one of its chiral forms, or enantiomers, superimposable

on the original.

this point we have rotated 180

0

in the structure

is not

To illustrate

the mirror image

on the right so that it

looks much like the original.

However,

that the top ethylenediamine

ligand is located

behind

note

the plane of this page in the original,

but in front of the plane in the rotated

mirror

image. Thus, the mirror images are not superimposable.

ligands we use bis- in place of di- to indicate that two molecules of the multitoothed ligand are present in the complex. a looking glass" as it Now let's look more closely at the cis isomer-"through were. Figure 17.20 shows that the cis-Co(en)2CI/ ion has a mirror image that is not identical to the original. The difference is demonstrated by the fact that there is no way to rotate the mirror image so that the atoms in its structure align exactly with those in the original. In other words, the two structures are not superimposable. You may wish to build models of the complex and its mirror image to satisfy yourself that this is so. Two such nonsuperimposable mirror images are said to be enantiomers, and such an ion or molecule is said to be chiral, from the Greek word cheir, meaning "hand" (the reflection of a left hand in a mirror looks like a right hand-but they are not identical, as anyone knows who tries to put their left hand into a right-hand glove). We will return to this concept of chirality and its profound implications in biochemistry and the life sciences in Chapter 21.

CONCEPT

TEST

Would four different ligands arranged in a square planar geometry produce a chiral complex ion? What about four different ligands in a tetrahedral geometry?

mIm Metal Complexes in Biomolecules

Enantiomers are geometric isomers that are nonsuperimposable mirror images of each other. Enantiomeric ions and molecules are called chiral. Porphyrin is a type of tetradentate cyclic ligand.

macro-

A macrocyclic ligand is a ring containing multiple electron-pair donors that bind to a metal ion.

At the beginning of this chapter we noted some of the important roles metals play in human health, and the importance of the presence of these metals in foods in chemical forms that can be readily absorbed by the body during digestion. In this chapter we have discussed how polydentate ligands such as EDT A can surround metal ions and prevent them from reacting with other species in solution. In this section we explore some biological polydentate ligands that involve metal ions in processes essential to nutrition and good health. Let's begin with the base of our food chain and the process we know as photosynthesis. Green plants can harness solar energy because they contain large biomolecules we call collectively chlorophyll. All molecules of chlorophyll have within them ring-shaped polydentate ligands called chlorins (Figure 17.2la). The structure of chlorin is very similar to that of another type of polydentate ligand found in biological systems called a porphyrin (Figure 17.21b). Note that the structures of the two differ by the presence of a C = C double bond in porphyrin (highlighted in red in Figure 17.21b) that is a single bond in chlorin. Their structures make them members of a class of compounds known generally as macrocyclic ligands. (A macrocycle is, literally, a big ring.)

17.12

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Metal Complexes in Siomolecules

I 881

FIGURE 17.21 Chlorin (a) and porphyrin

(b) rings are biologically

ligands that have similar core

structures:

the principal difference is a C=C

double bond in the porphyrin [highlighted

structure

in red in (b)] that is a single bond

in chlorin rings. The innermost rings are four nitrogen Chlorin ring system

(a)

Porphyrin

ring system

Metal-porphyrin

(b)

complex

(c)

important

tetradentate

of electrons

atoms in the

atoms whose lone pairs

are pointed toward the center of

the rings. When the two NH groups ionize, all four N atoms are free to form coordinate bonds with a metal ion, as shown with the

Two of the four nitrogen atoms in the porphyrin and chlorin rings are Sp3 hybridized and are also bound to hydrogen atoms, whereas the other two are Sp2 hybridized with no hydrogen atoms (see Figure 17.21). When a ring coordinates with a metal ion, the two hydrogen atoms ionize, giving the ring a charge of 2-. In the ionized structure there are four lone pairs of electrons oriented toward the center of the ring. They can occupy the four coordination sites in the equatorial plane of a sixcoordinate central metal ion or all four coordination sites in a square planar complex. Octahedral (six-coordinate) metal ions surrounded by the ring still have their two axial sites available for bonding to other ligands. Depending on the charge of the coordinated metal ion, the metal-ring complex may be ionic or neutral. Porphyrin and chlorin rings are widespread in nature and play many different biochemical roles. Their particular chemical and physical properties depend on

porphyrin

ring in (c). Metal ions at the

centers of these rings occupy four sites in the equatorial

planes of octahedral

remaining

sites in the axial positions

fields. The two above

and below the plane can be occupied by other ligands.

1. The identity of the coordinated metal ion. 2. The groups in the axial coordination sites. 3. The number and identity of organic groups attached to the outside of the rings. The chlorin ring in chlorophyll has a Mg2+ ion at its center (Figure 17.22). Delocalized p electrons in the conjugated double bonds within and around the ring stabilize it and give chlorophyll the ability to absorb wavelengths of red and blue-violet radiation (Figure 17.23). Because chlorophyll absorbs these colors, the color of chlorophyll and that of plants that contain it corresponds to the portion

400

500

Chlorophyll

FIGURE 17.22 Chlorophyll sunlight

(electromagnetic

the bonds of carbohydrates. have a

Mt+

(nm)

FIGURE 17.23 This is the combined

a

absorbs sunlight

700

600

Wavelength

absorption and, through

a series of reactions,

converts

energy), carbon dioxide, and water into chemical energy stored in All forms of chlorophyll,

including

ion in a chlorin ring within their molecular

chlorophyll

structures.

a shown here,

spectrum

other pigments pigments

of chlorophyll

and

in a typical green leaf. These

absorb most of the visible radiation

emitted by the sun except the yellow-green color of the leaves themselves.

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FIGURE 17.24 The colors offall in southern

Vermont are characterized

by the

reds, yellows, and gold colors ofleaves that have lost their green pigments

and are about

to fall to Earth.

(a)

(b)

FIGURE 17.25 Heme. (a) In this heme complex the four nitrogen atoms in a porphyrin

ring occupy equatorial

an octahedral

positions in

complex with an Fe2+ ion. (b)

Below the ring a fifth bond is formed between the Fe2+ ion and a lone pair of electrons on another nitrogen atom in the protein hemoglobin. porphyrin

The sixth ligand above the

ring is a loosely bonded molecule

of O2 as oxygenated lungs.

of the visible spectrum that is not absorbed, namely, green. In temperate climates chlorophyll is lost from the leaves of trees such as maples and oaks at the end of each growing season, which makes visible the colors of other pigments in the beautiful leaves of autumn (Figure 17.24). An important porphyrin complex has an Fe2+ ion in its center. This type of porphyrin is called a heme group (Figure 17.25). This group of atoms enables the protein called hemoglobin to transport O2 in the blood and a related protein, myoglobin, to transport O2 in muscle tissues. The four nitrogen atoms in the porphyrin ring of a heme group occupy equatorial positions in an octahedral complex of an Fe2+ ion. Below the ring a fifth bond is formed between the Fe2+ ion and a lone pair of electrons on another nitrogen atom in the protein. The sixth ligand, located above the porphyrin ring, is typically a molecule of O2, as in the oxygenated forms of hemoglobin in blood leaving the lungs. Each atom in an oxygen molecule has a lone pair of electrons that it can donate to the iron in heme. This coordinate covalent bond is strong enough to enable oxygen to be transported from the lungs to the cells in our bodies, but weak enough that it can be easily broken when the oxygen is delivered to a cell. Other ligands of similar size can bind to the Fe2+ ion in heme, and problems arise when some of them do. Carbon monoxide is such a ligand. Similar in size to oxygen, it easily fits into the sixth binding site. Unfortunately, it binds about 200 times more strongly than O2, If a person breathes air containing carbon monoxide, CO prevents O2 from being taken up by blood flowing through the lungs. Another class of proteins, called cytochromes, also contain heme groups (Figure 17.26). Cytochromes mediate oxidation and reduction processes connected with energy production in living cells. The heme group in cytochrome proteins serves as a conveyer of electrons as the half-reaction

hemoglobin

leaves the

rapidly and reversibly consumes or releases electrons needed in the biochemical reactions that sustain life. There are many kinds of cytochrome proteins with different substituents on the porphyrin rings and different axial ligands. Each of these variables influences the function of the metal-porphyrin complex. The association of metal ions with ligands results in complexes that vary in properties, appearance, and consequently in their use. The ions and neutral mol-

Summary I 883

FIGU RE 17.26 The structures of cytochrome proteins, such as cytochrome c (shown here), include heme complexes that mediate energy production and redox reactions in living cells. Different cytochromes have a variety of different ligands occupying the sixth octahedral coordination site, and different groups are attached to the porphyrin ring.

ecules that interact with metal ions in forming these complexes make the difference between a metal ion fulfilling an essential need in a living system or exerting a toxic effect and killing the organism. The same can be said for metals that act as catalysts in inorganic systems; ligands play an important role in determining the properties of the metal complex.

Summary A Lewis base is a substance that donates pairs of electrons to a Lewis acid during a chemical reaction. This donated pair forms a new covalent bond. In some of these reactions, such as that between CaO and S02' other bonds must break to accommodate the new one. SECTION 17.1

SECTION 17.2 Transition metal ions form complex ions when ligands donate pairs of electrons to empty valence-shell orbitals on the metal ion, thereby forming coordinate bonds. The number of coordinate bonds in a complex defines the coordination number of the metal ion. The most common coordination numbers are 6, 4, and 2. The metal ion in a complex ion acts as a Lewis acid (electron-pair acceptor), and the ligands act as Lewis bases (electron-pair donors). Compounds that contain complex ions are called coordination compounds; the net charges on complex ions are balanced by charges from counter ions. Ligands occupy binding sites in the inner coordination sphere of a metal ion. Counter ions surround complex ions in their compounds and dissociate from the complex ions when the compounds dissolve.

The stability of complex ions is expressed mathematically by their formation constants (Kf), which can be used to calculate the equilibrium concentration of free metal ions, Mn+(aq), in solutions of their complex ions. SECTION 17.3

SECTION 17.4 Hydrated metal ions with charges of 2+ or greater

can act as Bronsted-Lowry acids, which is why solutions of their soluble salts are acidic. Most transition metal ions have limited solubility in strongly basic solutions. An exception is Cr(III), which forms an anionic complex with the formula Cr(OH) 4 -.

17.5 The solubility of slightly soluble compounds is described by their K,p or solubility product, which is the value of the equilibrium constant for their dissolution.

SECTION

A molecule or ion of a monodentate ligand donates only one pair of electrons in a complex ion; polydentate ligands donate more than one in a process called chelation. EDTA is a particularly effective chelating and sequestering agent. It prevents metal ions in solution from reacting with other substances.

SECTION 17.6

17.7 Polydentate ligands are particularly effective in forming complex ions. This phenomenon is called the chelate effect and can be explained by the increase in entropy that accompanies the chelation process.

SECTION

SECTION 17.8 The colors of transition metals in crystalline solids

and in solutions can be explained by the interactions between electrons in different d orbitals and the lone pairs of electrons on surrounding ligands. Different interactions create crystal field splitting of the energies (labeled ~) of the d orbitals. A spectrochemical series ranks ligands on the basis of their field strengths and the wavelengths of electromagnetic radiation absorbed by their complex ions; the stronger the field the ligand produces, the shorter the wavelength of radiation the complex absorbs. The color of a complex ion in solution or in a crystalline solid is the complement of the color(s) it absorbs. Strong repulsions and large values of ~ can lead to electron pairing in lower-energy orbitals and an electron configuration called a low-spin state. Metals and their ions are less paramagnetic in low-spin states than when their d electrons are evenly distributed across all the d orbitals in their valence shell-a con figuration called a high-spin state.

SECTION 17.9

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17.10 The names of complex ions and coordinate compounds provide information about the identities and numbers of ligands, the identity and oxidation state of the central metal ion, and the identity of any counter ions whose charges balance the net charge of the complex ion.

SECTION

17.11 Complex metal ions contammg more than one type of ligand may form geometric isomers. For example, when one type occupies two adjacent corners of a square planar complex, the complex is a cis isomer; when the same ligand occupies opposite corners, it is a trans isomer. The chemical and biochemical properties of coordination compounds that are geometric isomers can be very dif-

SECTION

ferent, as illustrated by the difference between the chemotherapeutic power of cisplatin and that of its trans isomer. 17.12 Metal ion complexes play key roles in many biochemical processes. Among these are photosynthesis, which is mediated by chlorophyll, in which tetradentate chlorin rings coordinately bond to central Mg2+ ions; oxygen transport in the body based on the reversible bonding of O2 molecules to heme groups of Fe2+ ions in porphyrin rings; and energy production in cells, which is mediated by molecules called cytochromes that contain metals in different oxidation states.

SECTION



SAMPLE

Identifying

Determine which reactant donates a pair of electrons (the base) and which one accepts them (the acid).

17.1

Calculating the free metal ion concentration in a solution of one of its complex ions

Set up an ICE table based on formation of the complex. Let x = the concentration of M"+ that does not form the complex. If the value of Kfis large (it usually is), you may assume x is much less than the other concentrations.

17.2

Calculating

Use the mass action expression describing the solubility of a slightly soluble compound to determine concentrations in solution.

17.3 and 17.4

Identifying the donor groups in a molecule or structure

Look for functional groups that have atoms with lone pairs of electrons, such as -NH2' -NH-, and -COO-. Multiple coordinate bonds are possible if the groups can form stable ring structures with metal ions.

17.5

Predicting spin states

Sketch a d-orbital diagram based on crystal field splitting. Fill the lowest-energy orbitals with valence electrons. If more of them are paired than if they had been distributed evenly over all five d orbitals, multiple spin states are possible.

17.6

Follow the naming rules in Section 17.10.

17.7

Isomers are possible only if there are at least two types ofligands. If ligands of one type are all on the same side of the complex ion, it is a cis isomer, if on opposite sides, it is a trans isomer.

17.8

Lewis acids and bases

solubility using K,p

Naming coordination

compounds

Identifying geometric isomers of coordination compounds

EXERCISES


The chlorides of two of the four highlighted Figure P17.1 are colored. Which ones?

elements in

17.2.

Which of the highlighted transition metals in Figure P17.2 form M2+ cations that cannot have high-spin and low-spin states? ,-

,-

I-- I--

I

I-- I--

~ I



17.3.

Which of the highlighted transition metals in Figure P17.3 have M2+ cations that form colorless tetrahedral complex ions?

IT

17.6.

I

I 885

Therapy The compound with the structure shown in Figure P17.6 has been used to treat people exposed to plutonium, americium, and other actinide metal ions. (a) How many donor groups does the sequestering agent have in this structure, and (b) how many does it have when the carboxylic acid groups are ionized?

Chelation

OH O~

I

OH

0

HO~~N~Y

~

~o

~o

OH

FIGURE P17.3

OH

FIGURE P17.6 17.4.

Smoky quartz has distinctive lavender and purple colors due to the presence of manganese impurities in crystals of silicon dioxide. Which of the orbital diagrams in Figure P17.4 best describes the Mn2+ ion in a tetrahedral field?

IT]

[IIJ

[[]]]]]

ITIIllJ

(a)

(b)

17.7.

The three beakers in Figure P17.7 contain solutions of polyatomic Co(III) ions, [CoF6)3-, [Co(NH3\]3+, and [CO(CN)6]3-. Based on the colors of the three solutions, which compound is present in each of the beakers?

~N~

U b;d bJ (b)

(a)

[IIJ (d)

(c)

(c)

FIGURE P17.7 17.8.

Figure P17.8 shows the absorption spectrum of a solution of Ti(H20\3+. What color is the solution?

FIGURE P17.4 17.5.

Chelation Therapy The compound with the structure shown in Figure P17.5 is widely used in chelation therapy to remove excessive lead or mercury in patients exposed to these metals. (a) How many electron-pair donor groups ("teeth") does the sequestering agent have in this structure, and (b) how many does it have when the carboxylic acid groups are ionized?

JL IH /OH HO"" 'I If SH 0

FIGURE P17.5

v 400

v

v

450 500 550 600 650 700 750 Wavelength (nm) FIGURE P17.8

Questions and Problems Lewis Acids and Bases CONCEPT

17.9.

REVIEW

Can a substance be a Lewis base and not a Bronsted-Lowry base? Explain why or why not.

17.10. Can a substance be a Bronsted-Lowry acid and not be a Lewis acid? Explain why or why not. 17.11. Why is BF3 a Lewis acid but not a Bronsted-Lowry acid? 17.12. Would you expect NH3 or H20 to be a stronger Lewis base? Explain your selection.

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Metals

PROBLEMS

17.13. Use Lewis structures to show how electron pairs move and bonds form in the following reaction, and identify the Lewis acid and Lewis base.

17.14. Use Lewis structures to show how electron pairs move and bonds form and break in this reaction, and identify the Lewis acid and Lewis base. MgO(s)

+

CO/g)

-7

MgC03(s)

17.15. Use Lewis structures to show how electron pairs move and bonds form and break in this reaction, and identify the Lewis acid and Lewis base.



*17.18. Use Lewis structures to show how electron pairs move and bonds form and break in this reaction, and identify the Lewis acid and Lewis base. NOTE: HSbF6 is an ionic compound and one of the strongest Bmnsted-Lowry acids known.

17.26. When a strong base is added to a solution ofCuS04, which is pale blue, a precipitate forms and the solution above the precipitate is colorless. When ammonia is added, the precipitate dissolves and the solution turns a deep navy blue. Use appropriate chemical equations to explain why the observed changes occurred. 17.27. A lab technician cleaning glassware that contains residues of AgCl washes the glassware with an aqueous solution of ammonia. The AgCl, which is insoluble in water, rapidly dissolves in the ammonia solution. Why? 17.28. The procedure used in the previous question dissolves AgCl, but not AgI. Why? PROBLEMS NOTE: Appendix 5 contains formation constant (Kf) values that may be useful in solving the following problems. 17.29. One millimole ofNi(NO)2 dissolves in 250 mL of a solution that is 0.500 M in ammonia. (a) What is the concentration ofNi(NO)2 in the solution? (b) What is the concentration ofNi2+(aq) in the solution? 17.30. A 1.00 L solution contains 3.00 X 10-4 M CU(N03)2 and 1.00 X 10-3 M ethylenediamine. What is the concentration of Cu2+(aq) in the solution?

17.31. Suppose a 500 mL solution contains 1.00 millimoles of CO(N03)2' 100 millimoles ofNH3, and 100 millimoles of ethylenediamine. What is the concentration of Co2+(aq) in the solution? *17.32. To a 250 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNOv 0.100 M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously. Are the contents of the flask cloudy or clear? Support your answer with the appropriate calculations. (Hint: the Ksp of AgBr is 5.0 X 10-13.)

Complex Ions CONCEPT

REVIEW

17.19. When NaCl dissolves in water, which molecules or ions occupy the inner coordination sphere around the Na+ ions? 17.20. When CrCl3 dissolves in water, which of the following species are nearest the Cr3+ ions: (a) other Cr3+ ions, (b) CI- ions, (c) molecules of H20 with their 0 atoms closest to the Cr3+ ions, (d) molecules of H20 with their H atoms closest to the Cr3+ ions? 17.21. When Ni(N03)2 dissolves in water, what molecules or ions occupy the inner coordination sphere around the Ni2+ ions? 17.22. When [Ni(NH3)6]Cl2 dissolves in water, which molecules or ions occupy the inner coordination sphere around the Ni2+ ions? 17.23. Which ion is the counter ion in the coordination compound NaJZn(CN)4]? 17.24. Which ion is the counter ion in the coordination compound [Co(NH)4Cl2]N03?

Complex-Ion Equilibria CONCEPT

REVIEW

17.25. Trace Metal Toxicity Dissolved concentrations ofCu2+ as low 10-6 M are toxic to phytoplankton (microscopic algae). However, a solution that is 10-3 Min Cu(N03) 2 and 10-3 M in EDTA is not toxic. Why?

Hydrated Metal Ions as Acids CONCEPT

REVIEW

17.33. Which, if any, aqueous solutions of the following compounds are acidic: (a) CaCI2; (b) CrCl3; (c) NaCl; (d) FeCl2. 17.34. If 0.1 M aqueous solutions of each of these compounds were prepared, which one has the lowest pH? (a) BaCl2; (b) NiCl2; (c) KCl; (d) TiCl4 17.35. When ozone is bubbled through a solution ofiron(II) nitrate, dissolved Fe(II) ions are oxidized to Fe(III) ions. How does the oxidation process affect the pH of the solution? 17.36. As an aqueous solution ofKOH is slowly added to a stirred solution of AlCl3, the mixture becomes cloudy, but then clears when more KOH is added. (a) Explain the chemical changes responsible for the changes in the appearance of the mixture. (b) Would you expect to observe the same changes ifKOH were added to a solution of FeCl/ Explain why or why not. 17.37. Chromium(III) hydroxide is amphoteric. Write chemical equations showing how an aqueous suspension of this compound reacts to the addition of a strong acid and a strong base. 17.38. Zinc hydroxide is amphoteric. Write chemical equations showing how an aqueous suspension of this compound reacts to the addition of a strong acid and a strong base.

Questions and Problems I 887 17.39. Refining Aluminum To remove impurities such as calcium and magnesium carbonates and Fe(III) oxides from aluminum ore (which is mostly AI20), the ore is treated with a strongly basic solution. In this treatment AI(III) dissolves but the other metal ions do not. Why? *17.40. The same quantity of FeCl3 dissolves in 1 M aqueous solutions of each of these acids: HN03, HN02, H2S03, and CH3COOH. Is the concentration of Fe 3+ (aq) the same in all four solutions? Explain why or why not.

17.41. What is the pH of 0.50 M Al(NO)/ 17.42. What is the pH of 0.25 M CrCl/ 17.43. What is the pH of 0.100 MFe(NO)/ 17.44. What is the pH of 1.00 M Cu(NO)/ the titration NaOH) for the titration NaOH to a

curve (pH versus volume of a 25 mL sample of 0.5 M FeCI3. curve that results from the addition of sample containing 0.5 M KFeS04•

Solubilities of Ionic Compounds and CONCEPT

Ksp

REVIEW

17.47. What is the difference between solubility and solubility product? 17.48. Describe how the common-ion effect limits the dissolution of a sparingly soluble ionic compound. 17.49. Which of the following cations will precipitate first as a carbonate mineral from an equimolar solution ofMg2+, Ca2+, and Sr2+? 17.50. If the solubility of a compound increases with increasing temperature, does K,p increase or decrease? 17.51. The K of strontium sulfate increases from 2.8 X 10-7 at 'p 3TC to 3.8 X 10-7 at 7TC. Is the dissolution of strontium sulfate endothermic or exothermic? 17.52. How will adding concentrated NaOH(aq) affect the solubility of an AI (III) salt? 17.53. Chemistry of Tooth Decay Tooth enamel is composed of a mineral known as hydroxyapatite with the formula CaS(P04)30H. Explain why tooth enamel can be eroded by acidic substances released by bacteria growing in the mouth. 17.54. Fluoride and Dental Hygiene Fluoride ions in drinking water and toothpaste convert hydroxyapatite in tooth enamel into fluorapatite: Cas(PO 4)PH(s)

+

r(aq)

;='

Cas(PO 4)3F(S)

+

OH-(aq)

Why is fluorapatite less susceptible than hydroxyapatite to erosion by acids? PROBLEMS

17.55. At a particular temperature the value of[Ba2+] in a saturated solution of barium sulfate is 1.04 X 10-5 M. Starting with this information, calculate the K,p value of barium sulfate at this temperature. 17.56. Suppose a saturated solution of barium fluoride contains 1.5 X 10-2 Mr. What is the Ksp value ofBaF/ 17.57. What are the equilibrium concentrations ofCu+ and Cl- in a saturated solution of copper(l) chloride if K = 1.02 X 1O-6? 'p

17.59. What is the solubility of calcite (CaCO) in grams per milliliter at a temperature at which its K,p = 9.9 X 1O-9? 17.60. What is the solubility of silver iodide in grams per milliliter at a temperature at which its Ksp = 1.50 X 10-16. 17.61. What is the pH at 25T of a saturated solution of silver hydroxide, given K,p = 1.52 X 1O-8? 17.62. pH of Milk of Magnesia What is the pH of a saturated solution of magnesium hydroxide (the active ingredient in the antacid milk of magnesia)?

PROBLEMS

17.45. Sketch 0.50 M 17.46. Sketch 0.50 M

17.58. What are the equilibrium concentrations ofPb2+ and r in a saturated solution oflead fluoride if the Ksp value of PbF 2 is 3.2 X 1O-8?

17.63. Suppose you have 100 mL of each of the following solutions. In which will the most CaC03 dissolve? (a) 0.1 M NaCl; (b) 0.1 M Na2C03; (c) 0.1 M NaOH; (d) 0.1 M HCl 17.64. In which of the following solutions will CaF2 be most soluble? (a) 0.010 M Ca(NO)2; (b) 0.01 M NaF; (c) 0.001 M NaF; (d) 0.10 M Ca(NO)2 17.65. Composition of Seawater The average concentration of sulfate in surface seawater is about 0.028 M. The average concentration of Sr2+is 9 X lO-s M. If the K sp value of strontium sulfate is 3.4 X 10-7, is the concentration of strontium in the sea probably controlled by the insolubility of its sulfate salt? 17.66. Fertilizing the Sea to Combat Global Warming Some scientists have proposed adding Fe(III) compounds to large expanses of the open ocean to promote the growth of phytoplankton that would in turn remove CO2 from the atmosphere through photosynthesis. The average pH of open ocean water is 8.1. a. What is the maximum value of[Fe3+] in pH 8.1 seawater if the Ksp value ofFe(OH\ is 1.1 X 1O-36? *b. If you were to use the result from part a. to predict the solubility (s) ofFe(N03)3 in seawater using the equation Ksp = (mmxx)s(m+x), would the prediction be accurate? Explain your answer.

Polydentate Ligands; Ligand Strength and the Chelation Effect CONCEPT

REVIEW

17.67. What is meant by the term sequestering agent? What properties makes a substance an effective sequestering agent? 17.68. The condensed molecular structures of two compounds that each contain two -NH2 groups are shown in Figure P17.68. The one on the left is ethylenediamine, a bidentate ligand. Does the molecule on the right have the same ability to donate two pairs of electrons to a metal ion? Explain why you think it does or does not.

FIGURE P17.68 17.69. How does the chelating ability of an aminocarboxylate vary with changing pH? *17.70. The EDTA that is widely used as a food preservative is added to food, not as the undissociated acid, but rather

888

I

Chapter Seventeen

I

The Colorful Chemistry of Transition

Metals

as the calcium disodium salt: NaJCaEDTAl This salt is actually a coordination compound with a Ca2+ ion at the center of a complex ion. Draw a Iine structure of this compound.

Crystal Field Theory CONCEPT

REVIEW

17.71. Explain why the compounds of most of the first-row transition metals are colored. 17.72. Unlike the compounds of most transition metal ions, those ofTi4+ are colorless. Why? 17.73. Why is the dxy orbital higher in energy than the dxz and ~.z orbitals in a square-planar crystal field? 17.74. On average, the dorbitals ofa transition metal ion in an octahedral field are higher in energy than they are when the ion is in the gas phase. Why? PROBLEMS

17.75. Aqueous solutions of one the following complex ions of Cr(Ill) are violet; solutions of the other are yellow. Which is which? (a) Cr(Hp\3+; (b) Cr(NH)63+ 17.76. Which of the following complex ions should absorb the shortest wavelengths of electromagnetic radiation? (a) Cu(Cl)/-; (b) Cu(F)/-; (c) Cu(I)/-; (d) Cu(Br)/17.77. The octahedral crystal field splitting energy ~o of Co(phen)/+ is 5.21 X 10-19 Jlion. What is the color of a solution of this complex ion? 17.78. The octahedral crystal field splitting energy ~o of CO(CN)63- is 6.74 X 10-19 J/ion. What is the color of a solution of this complex ion? 17.79. Solutions ofNiCV- and NiBr/- absorb light at 702 and 756 nm, respectively. In which ion is the split of d-orbital energies greater? 17.80. Chromium(llI) chloride forms six-coordinate complexes with bipyridine, including cis-Cr(bipY)2CI2' which reacts slowly with water to produce two products, cis-Cr(bipY)2(HP)Cl+ and cis-Cr(bipy)/HP)/+. In which of these complexes should ~o be the largest?

Magnetism and Spin States CONCEPT

REVIEW

17.81. What determines whether a transition metal ion is in a high-spin configuration or a low-spin configuration? *17.82. In Section 7.2 we noted that the bonding in Fe(H20)63+ involves sp3d2 hybrid orbitals, but that the bonding in Fe(CN)63- comes from coordinate bonds with d2Sp3 hybrid orbitals on Fe3+. What is the reason for this difference in hybridization?

17.85. Which of the following cations can, in principle, have either a high-spin or a low-spin electron configuration in a tetrahedral field? C02+, Cr3+, Ni2+, and Zn2+ 17.86. How many unpaired electrons are in the following transition metal ions in an octahedral crystal field? High-spin Fe3+, Rh+, and Vl+, and low-spin Mn3+ 17.87. The manganese minerals pyrolusite, Mn02, and hausmannite, Mn3 0 4' contain Mn ions in octahedral holes formed by oxide ions. a. What are the charges of the Mn ions in each mineral? b. In which of these compounds could there be high-spin and low-spin Mn ions? 17.88. Dietary Supplement Chromium picolinolate is an overthe-counter diet aid sold in many pharmacies. The Cr3+ ions in this coordination compound are in an octahedral field. Is the compound paramagnetic or diamagnetic? *17.89. One method for refining cobalt involves the formation of the complex ion CoCV-. This anion is tetrahedral. Is this complex paramagnetic or diamagnetic? *17.90. Why is it that Ni(CN)/- is diamagnetic, but NiCl/- is paramagnetic?

Naming Complex Ions and Coordination Compounds PROBLEMS

17.91. What are the names of the following complex ions? a. Cr(NH)63+ b. CO(HP)63+ c. [Fe(NH3)sCl]2+ 17.92. What are the names of the following complex ions? a. Cu(NH3)2 + b. Ti(HP)/OH)/+ C. Ni(NH3)/HP)/+ 17.93. What are the names of the following complex ions? a. CoBr/b. Zn(Hp)(OH)3c. Ni(CN)/17.94. What are the names of the following complex ions? a. CoI/b. CuCI/c. [Cr(en)(OH)4]17.95. What are the names of the following coordination compounds? a. [Zn(en)]S04 b. [Ni(NH3)s(HP)]Cl2 c. K4Fe(CN)6 17.96. What are the names of the following coordination compounds? a. (NH4)JCo(CN)6] b. [Co(en)2Cl](N03)2 c. [Fe(Hp)/OH)JCl

Isomerism in Coordination Compounds; Metal Complexes in Biomolecules CONCEPT

PROBLEMS

17.83. How many unpaired electrons are there in the following transition-metal ions in a tetrahedral field? High-spin Fe2+, Cu2+, C02+, and Mn3+. 17.84. Which of the following cations can have either a high-spin or a low-spin electron configuration in an octahedral field? Fe2+, Co3+, Mn2+, and Cr3+

REVIEW

17.97. What do the prefixes cis- and trans- mean in the context of an octahedral complex ion? 17.98. What do the prefixes cis- and trans- mean in the context of a square-planar complex? 17.99. How many different types of donor groups are required to have geometric isomers of a square-planar complex? 17.100. With respect to your answer to the previous question, do all square-planar complexes with this many different types of donor groups have geometric isomers?

Questions and Problems PROBLEMS

17.101. Does the complex ion [Co(en)(H20)2C1Y+ have geometric isomers? 17.102. Does the complex ion Fe(en)/+ have geometric isomers? *17.103. Sketch the geometric isomers of the square planar complex ion CuCI2Br/-. Are any of these isomers chiral? *17.104. Sketch the geometric isomers of the complex ion Ni(en)Cl/CN)/-. Are any of these isomers chiral?

Additional Problems 17.105.

Photographic Film Processing During the processing of black-and-white photographic film, excess silver(I) halides are removed by washing the film in a bath containing sodium thiosulfate. This treatment is based on the following complexation reaction:

Ag+(aq) + 2 Sp/-(aq)

p

Ag(Sp)}-(aq)

Kf = 5

X

What is the ratio of [Ag+] to [Ag(S20)}-] in a bath in = 0.233 M? which [SP/-] 17.106. Lead Poisoning Children used to be treated for lead poisoning with intravenous injections ofEDTA. If the concentration of EDT A in the blood of a patient is 2.5 X 10-8 M and the formation constant for the complex, [Pb(EDTA)]2-, is 2.0 X 1018, what is the concentration ratio of the free (and potentially toxic) Pb2+(aq) in the blood to the much less toxic Pb2+-EDTA complex?

1013

I 889

17.107. Dissolving cobalt(II) nitrate in water gives a beautiful purple solution. There are three unpaired electrons in this cobalt(II) complex. When cobalt(II) nitrate is dissolved in aqueous ammonia and oxidized with air, the resulting yellow complex has no unpaired electrons. Which cobalt complex has the larger crystal field splitting energy Ll} 17.108. A solid compound containing Fe(II) in an octahedral crystal field has four unpaired electrons at 298 K. When the compound is cooled to 80 K, the same sample appears to have no unpaired electrons. How do you explain this change in the compound's properties? 17.109. When Ag20 reacts with peroxodisulfate (S20/-) ion (a powerful oxidizing agent), AgO is produced. Crystallographic and magnetic analyses of AgO suggest that it is not simply Ag(II) oxide, but rather a blend of Ag(I) and Ag(III) in a square planar environment. The Ag2+ ion is paramagnetic but, like AgO, Ag+ and Ag1+ are diamagnetic. Explain why. 17.110. The iron(II) compound Fe(bipy)/SCN)2 is paramagnetic, but the corresponding cyanide compound Fe(bipy)/CN)2 is diamagnetic. Why do these two compounds have different magnetic properties? 17.111. Aqueous solutions of copper(II)-ammonia complexes are dark blue. Will the color of the series of complexes, shift toward shorter or longer Cu(H20\_)NH3)/ wavelengths as the value of x increases from 0 to 6?

Electrochemistry and Electric Vehicles

A LOOK AHEAD: 100 mpg: Charging Up Hybrid Cars 18.1

Redox Chemistry Revisited

18.2

Electrochemical

Cells

18.3 Chemical Energy and Electrical Work

18.4 Standard Potentials (P) 18.5 A Reference Point: The Standard Hydrogen Electrode

18.6 The Effect of Concentration on

Eeell

18.7 Relating Battery Capacity to Quantities of Reactants

18.8 Electrolytic Cells and Rechargeable Batteries

18.9 Fuel Cells

AN ELECTRIFYING SPORTS CAR A bank o/lithium-ion batteries and a 185 kilowatt

motor power the high-

performance 2008 Tesla Roadster.

18.1

I


I

A LOOK AHEAD: 100 mpg: Charging Up Hybrid Cars The first decade of the 21st century has seen remarkable swings in the prices of gasoline and other fuels derived from petroleum and natural gas. These oscillations, which have included record prices for crude oil, have reinvigorated the development of alternative systems for powering vehicles. One popular alternative has been hybrid propulsion that relies on a combination of a small gasoline engine and an electric motor powered by rechargeable batteries. In some cars these batteries and motors can be the sole power source for short trips at moderate speeds. Battery-powered electric motors convert chemical energy into mechanical energy more efficiently than gasoline engines and they draw no power when the car is stopped in traffic. They can even be made to recharge a car's batteries when its driver applies the brakes. As the car slows, its decrease in kinetic energy is first turned into electrical energy and then into chemical energy as electric motors become electric generators and battery rechargers. Ultimately, the driving range of cars under electric power alone is limited by the capacity of their batteries to hold and deliver electrical charge. Research into new battery technologies that make these repositories of chemical energy more compact and more powerful has led to the development of high-performance all-electric vehicles and to "plug-in" hybrids: cars that have larger-capacity batteries than those in conventional hybrids such as the Toyota Prius. Plug-in hybrids can make most business or pleasure trips without their gasoline engines ever turning on. In this chapter, we examine the chemistry of modern batteries and another source of electrical power called fuel cells. Batteries and fuel cells harness the chemical energy released by spontaneous redox reactions, so we begin this chapter by revisiting the principles of redox chemistry introduced in Chapter 4. Then we look at the way free energy changes accompanying spontaneous redox reactions can create electromotive forces that push electrons through devices like electric motors. Later we'll see how to restore a battery's chemical energy by using an external source of electrical energy to force its spontaneous redox reaction to run in reverse. We conclude with an examination of the design of fuel cells based on combining H2 and O2 to form H20, and the challenges of building hydrogen-based energy systems.

ma


In Chapters 14 through 16 we discussed some of the environmental problems associated with the combustion of fossil fuels. In this chapter, we examine alternative technologies for powering vehicles that are more efficient than the internal combustion engine at converting chemical energy to mechanical energy and have the potential to reduce air pollution dramatically. These technologies are based on electrochemistry, the branch of chemistry that links redox reactions to the production or consumption of electrical energy. At the heart of electrochemistry are chemical reactions in which electrons are gained and lost at electrode surfaces. In other words, electrochemistry is based on reduction and oxidation, or redox, chemistry. The principles of redox reactions were introduced in Section 4.8. Let's review some basic definitions and concepts: • A redox reaction is the sum of two half-reactions: a reduction halfreaction, in which a reactant gains electrons, and an oxidation halfreaction, in which a reactant loses electrons.

Electrochemistry is the branch of chemistry that examines the transformations between chemical and electrical energy.

891

892

I Chapter Eighteen

I

Electrochemistry


• Reduction and oxidation half-reactions happen simultaneously so that the number of electrons gained during reduction exactly matches the number lost during oxidation. • A substance that is easily oxidized is one that readily gives up electrons. This electron-donating power makes the substance an effective reducing agent because electrons donated by this substance are accepted by another reactant (the one that is reduced). • A substance that readily accepts electrons and is thereby reduced is an effective oxidizing agent. The tendencies of different elements and their ions to gain or lose electrons are illustrated by the chemical properties of two neighbors in the periodic table of the elements, copper (Z = 29) and zinc (Z = 30). Zinc metal has considerable electron-donating power, and Cu2+ ions tend to be willing acceptors of donated electrons. These complementary chemical properties are captured in the images in Figure 18.1. When a strip ofZn metal is placed in a solution ofCuS04, Zn atoms spontaneously donate electrons to Cu2+ ions, forming Zn2+ ions and Cu atoms. The progress of this reaction can be monitored visually: shiny metallic zinc turns dark brown as a textured layer of copper accumulates, and the distinctive blue color of Cu2+(aq) ions fades as these ions acquire electrons and become atoms of copper metal. Writing a chemical equation that describes the spontaneous reaction between Zn and Cu2+ is easily done because every mole of Zn atoms that is oxidized loses 2 moles of electrons in an oxidation half-reaction:

and every mole ofCu2+ ions that is reduced accepts 2 moles of electrons in a reduction half-reaction:

Because the number of moles of electrons lost and gained in the two half-reactions is the same, writing a chemical equation for the overall redox reaction is simply a matter of adding the two half-reactions together:

Cu2+(aq) Zn(s)

Zn(s)

-?

Zn2+(aq)

+ 2 e-

-?

Cu(s)

-?

Cu(s)

+ Cu2+(aq) + ~

+ 2e-

+ Zn2+(aq) + ~

Canceling out the equal numbers of electrons gained and lost, we obtain a balanced net ionic equation for the redox reaction: Zn(s)

FIGURE 18.1 A strip of zinc immersed in a solution encrusted

of copper(II)

sulfate becomes

with a dark layer of copper. The

blue color of the solution

fades as Cu2+ ions

(the source of the color) are reduced metal, and Zn metal is oxidized Zn2+ ions.

to Cu

to colorless

+ Cu2+(aq)

-?

Cu(s)

+ Zn2+(aq)

Combining half-reactions is a convenient way to write balanced chemical equations for redox reactions, as we discussed in Section 4.8. In this chapter we will use a valuable resource in this process, the table of half-reactions in Appendix 6. The table sorts half-reactions in order of a parameter called their standard reduction potentials (E) We use EO values, which are expressed in volts (V), extensively in later sections of this chapter. For now, just keep in mind that the EO value of a reduction half-reaction is an indication of how likely it is to

18.1

occur. The half-reactions at the top of the table with the most positive EO values, for example,

EO

= 2.87 V

are those with reactants that are the strongest oxidizing agents (and therefore readily reduced), whereas the very negative EO values at the bottom of the table mean that the reactants, which include the major cations in biological systems and environmental waters (Na+, K+, Mg2+, and Ca2+) are not easily reduced. Instead, the products of the half-reactions at the bottom of the table are powerful reducing agents. In the following Sample and Practice Exercises we use half-reactions from Appendix 6 to write net ionic equations for redox reactions that involve molecular O2, In Sample Exercise 18.1 we write a net ionic equation for the reaction between O2 and Fe2+ under acidic conditions. In the Practice Exercise your task will be to write a net ionic equation describing the oxidation of N02 - to N03by O2 under basic conditions. Note that Appendix 6 contains two half-reactions describing the reduction of O2. One of them contains H+ ions and so applies to reactions in acidic solutions:

In alkaline solutions reduction of O2 produces OH- ions:

Writing Net Ionic Equations for Redox Reactions by Combining Half-Reactions Write a net ionic equation describing the oxidation ofFe2+(aq) by molecular oxygen in an acidic solution. COLLECT AN D ORGAN I ZE Perusing the reduction half-reactions in Appendix 6, we find the following half-reactions for the reduction of O2:

02(g) O/g)

+ 4H+(aq) + 4e- ~ + 2H20(t')

+ 4e- ~

2H20(t') 40H-(aq)

All of the half-reactions in Appendix 6 are written as reduction half-reactions, so there is none for the oxidation ofFe2+. However, there is one in which Fe2+ is the product of the reduction of Fe3+: Fe3+(aq) We can reverse this half-reaction half-reaction:

+ e- ~

Fe2+(aq)

to make Fe2+ the reactant in an oxidation

A N A L Y Z E The problem specifies acidic conditions, so we should use the first of

the two O2 half-reactions. In redox reactions the electrons gained by the substances that are reduced must equal the electrons lost by the substances that are oxidized. There is a gain of 4 moles of electrons in the O2 half-reaction and a loss of 1 mole of electrons in the Fe2+ half-reaction. Therefore, we need to multiply the Fe2+ half-reaction by 4.

I


I

893

894

I

Chapter Eighteen

I

Electrochemistry


Multiply half-reactions:

the Fe2+ half-reaction

SOLVE

by 4 and then

4 Fe2+(aq) ~ 4Fe3+(aq) 02(g) 4Fe2+(aq)

+ 4H+(aq) + 4e- ~

+ 02(g) + 4H+(aq) + 4-@::.

~

combine

the two

+ 4e-

2H20(t') 4Fe3+(aq)

+ 2H20(t')

+ 4-@::.

Simplifying gives 4Fe2+(aq)

+ 02(g) + 4H+(aq)

~ 4Fe3+(aq)

+ 2H20(t')

We can verify that this is a balanced net ionic equation by confirming that the number of moles of Fe, 0, and H atoms match on both sides of the reaction arrow and that the total electrical charges on both sides are the same. (Both are 12+.) Note that the result of combining these two half-reactions is not a complete, molecular equation but rather a net ionic equation for the redox reaction.

THINK

ABOUT

IT

Practice Exercise Write a net ionic equation describing the oxidation ofN02to N03

-

by O2 in a basic solution.


mD

~II

CHEMTOUR

Zinc-Copper

Cell

An electrochemical cell is an apparatus that converts chemical energy into electrical work or electrical work into chemical energy.

Electrochemical Cells

Having shown how the Zn/Cu2+ redox reaction is the sum of two discrete halfreactions, one oxidation and one reduction, let's now physically separate the two half-reactions using a device called an electrochemical cell (Figure 18.2). Electrochemical cells can convert the chemical energy of a spontaneous redox reaction into electrical energy, as happens in batteries. They can also convert electrical energy into chemical energy, as happens when a battery is recharged. In the cell in Figure 18.2 a strip of zinc is immersed in aIM solution of ZnS04, and in a separate compartment a strip of copper metal is immersed in aIM solution of CuSO 4' The two metal strips serve as the cell's electrodes, providing pathways for the electrons gained and lost by the two half-reactions to reach an external circuit. The separation of the Cu and Zn half-reactions forces electrons lost when Zn atoms are oxidized to travel through the external circuit (blue arrow in Figure 18.2) before they can be accepted by Cu2+ ions. As the two half-reactions in the electrochemical cell proceed, Cu2+ ions are reduced to Cu atoms in the right-hand compartment in Figure 18.2, and Zn atoms are oxidized to Zn2+ ions on the left. You might think that production of Zn2+ ions would result in a buildup of positive charge on the Zn side and that conversion ofCu2+ ions to Cu metal would create excess negative charge on the Cu side. No such charging occurs because the compartments are connected by a permeable bridge made of porous glass or plastic that allows ions to migrate from one compartment to the other to balance any buildup of charge. In most electrochemical cells the source of these ions is a strong electrolyte added to the solutions in both compartments. It is important that such a "background" electrolyte not interfere with the redox reactions at the electrodes. A useful candidate for the Zn/Cu2+ cell is Na2S04 because neither Na+ ions nor S042- ions interfere with the oxidation of Zn or reduction ofCu2+ ions. Migration ofNa+ ions into the Cu compartment and SO /- ions into the Zn compartment completes the electrical circuit and allows electrons to flow through the part of the circuit outside the cell.

18.2

FIGURE 18.2

What a Chemist Sees

consists of two compartments:

This electrochemical

the one on the left contains

immersed

in a 1 M solution of ZnSO 4; the other contains

immersed

in a 1 M solution of CuSO

connection through

between

it. If the electrodes

are connected

Electrochemical

I

Cells

895

cell

a zinc anode a copper cathode

~

A porous glass plug provides an electrical

the two compartments

to the negative terminal terminal-the

4'

I

as ions (and their charges) to a voltmeter-the

of the meter and the Cu electrode

migrate

II

Voltmeter Anode

(-)

~ Cathode

(+)

I~

Zn electrode to the positive

meter will read 1.10 V at 25°C.

Cu2+

(1.00 M)

With the passage of time, the mass of the Zn strip decreases; the mass of the Cu strip increases, and the blue color ofCu2+(aq) ions fades in the solution in the right-hand compartment in Figure 18.2. The decrease in chemical energy that accompanies this spontaneous cell reaction does electrical work in an external circuit. For example, it could be used to light a small light bulb or drive the motor in a small fan. This capacity to do electrical work makes this electrochemical cell a voltaic cell, that is, a battery. In this or any electrochemical cell, the electrode at which the oxidation half-reaction (loss of electrons) takes place (the Zn electrode in this case) is called the anode, and the electrode at which reduction half-reaction (gain of electrons) takes place (the Cu electrode in this case) is called the cathode.

CONCEPT

TEST

Will the increase in mass of the copper strip match the decrease in mass of the zinc strip during the reaction just described? (Answers

to Concept Tests are in the back

of the

book.)

The illustration in Figure 18.2 provides a useful view of the physical reality of a Zn/Cu2+ electrochemical cell, but we need a more compact way to symbolize the components of such a cell. That's where cell diagrams come in. A cell diagram uses symbols to show how the components of an electrochemical cell are connected. A cell diagram does not convey the stoichiometry of the reaction, and so any coefficients in the balanced equation for the cell reaction do not appear in the cell diagram. Here are the steps we follow when writing a cell diagram: (1) Write the chemical symbol of the anode at the far left of the diagram and write the symbol of cathode at the far right. (2) Work from the electrodes toward the connecting bridge using vertical lines to indicate phase changes (like that between a solid metal electrode and an aqueous solution). Represent solutions or ionic solids using the symbols of the ions or compounds that are changed by the cell reaction. (3) Use a double vertical line to represent the bridge connecting the anode and cathode half-reactions. Following these steps for the Zn/Cu2+

electrochemical

(3) Zn(s)

An anode is an electrode at which an oxidation half-reaction (loss of electrons) takes place. A cathode is an electrode at which a reduction half-reaction (gain of electrons) takes place.

Cu(s)

(1) Zn(s) (2) Zn(s)

cell,

In a voltaic cell, chemical energy is transformed into electrical energy by a spontaneous redox reaction.

I Zn2+(aq) I Zn2+(aq) II

Cu2+(aq) Cu2+(aq)

I Cu(s) I Cu(s)

A cell diagram uses symbols to show how the components of an electrochemical cell are connected.

896

I

Chapter Eighteen

I

Electrochemistry


Diagramming

F

4

Vohm,""

Ag+

~

I

Cell

This figure represents an electrochemical cell in which a copper electrode immersed in a 1.00 M solution of Cu2+ ions is connected to a silver electrode immersed in a 1.00 M solution of Ag+ ions. Write a balanced chemical equation for the cell reaction and diagram the cell. We need to write a balanced chemical equation for the cell reaction and diagram a cell in which current spontaneously flows from a copper electrode immersed in a solution of Cu2+ ions through an external circuit to a silver electrode immersed in a solution of Ag+ ions.

COLLECT

(1.00 M)

an Electrochemical

AND ORGANIZE

A N A L Y ZEIn a voltaic cell electrons flow from the electrode at which they are produced in an oxidation half-reaction and toward the electrode where they are consumed in a reduction half-reaction. Therefore, in the cell in the figure Cu is the anode and Ag is the cathode. The half-reactions are

Oxidation

at the anode:

Cu(s) ~ Cu2+(aq)

Reduction

at the cathode:

Ag+(aq)

+ e- ~

+ 2 e-

Ag(s)

Two moles of electrons are produced in the anode half-reaction, but only 1 mole is consumed in the cathode half-reaction. We need to multiply the cathode halfreaction by 2 before combining the two equations. In a cell diagram the anode and the species involved in the oxidation half-reaction are on the left and the species involved in the reduction half-reaction and the cathode are on the right. We use single lines to separate phases and a double line to separate the two half-reactions. Multiplying half-reaction,

SOLVE

the Ag+ half-reaction

2 [Ag+(aq)

+ e- ~

by two and adding

Ag(s)]

Cu(s) ~ Cu2+(aq) 2Ag+(aq)

+ Cu(s)

it to the Cu

~ 2 Ag(s)

+ 2 e-

+ Cu2+(aq)

Applying the rules for writing a cell diagram: Cu(s) T H INK

I Cu2+(aq) II Ag+(aq) I Ag(s)

The appearance of this diagram is much like the one in the text electrochemical cell reaction and so is reasonable.

ABO UTI T

for the Zn/Cu2+

Practice Exercise A cell like that shown in Figure 18.2 has a copper cathode immersed in a solution ofCu2+ ions and an aluminum anode immersed in a solution of AP+ ions. Write a balanced chemical equation for the cell reaction, and write the cell diagram.

18.3

mill

I

Chemical Energy and Electrical Work

I

897


When we connect the negative terminal of a voltmeter to the Zn electrode in Figure 18.2 and the positive terminal to the Cu electrode, the meter reads 1.10 V This voltage is a measure of the electromotive force (emf) generated by the cell reaction, which reflects how forcefully it pumps electrons through the external circuit from the negative terminal and toward the positive terminal. This voltage is also called cell potential (Eeen). It is directly proportional to the chemical (potential) energy released by simultaneously oxidizing Zn atoms and reducing Cu2+ ions. When the voltage produced by a voltaic cell pushes electrons through an external circuit, that flow of current can do electrical work, like lighting a light bulb or a light-emitting diode (LED), or turning an electric motor. The connection between the change in chemical free energy of a voltaic cell (~Gcen) and the electrical work (wei) that may result is simply ~Geen

=

(18.1)

welee

The sign of ~G is negative because the internal energy of a voltaic cell decreases as it is discharged, and the sign on We leeis negative because it represents work the cell does on its surroundings, which has a negative value from the perspective of the cell. The quantity of electrical work (wele) done by an electrochemical cell is the product of the electrical charge that the cell pushes through an external circuit times the electromotive force (cell potential Eeen) with which the cell pushes that charge. In equation form this definition is W

elec

=

CE cell

(18.2)

The quantity of charge is proportional to the number of electrons flowing through a circuit. As we noted in Chapter 2, the magnitude of the charge on a single electron is 1.602 X 10-19 coulombs (C). The magnitude of electrical charge in 1 mole of electrons (e-) is 1.602 X 10-19 C X 6.022 X 1023 e= mol e-

~

~II

CHEMTOUR

Free Energy

= 9.65 X 104 C mol e-

This quantity of charge, 9.65 X 104 C/mol, is called Faraday's constant (F) after Michael Faraday (1791-1867), the English chemist and physicist, who discovered that redox reactions take place when electrons are transferred between reacting species. The quantity of charge (C) flowing through an electrical circuit is the product of the number of moles of electrons (n) times the Faraday constant:

C= nF

@)@) CONNECTION The sign conventions used for work done on a thermodynamic system (+) and the work done by the system (-) are explained in Section 5.2.

(18.3)

Let's develop an equation relating welee and Eeen by combining Equations 18.2 and 18.3: (18.4) The negative sign on the right side of Equation 18.4 reflects the fact that work done by a voltaic cell on its surroundings (the external circuit) corresponds to energy lost by the cell.

Electromotive force (emf), sometimes called voltage, is the force that pushes electrons through an electrical circuit. Cell potential (£,,11) is the electromotive force expressed in volts (V) with which an electrochemical cell can push electrons through an external circuit connected to its terminals. Faraday's constant (F) is the quantity of electrical charge (in coulombs) in 1 mol of electrons. Its value to three significant figures is 9.65 X 104 C.

898

I

Chapter Eighteen

I

Electrochemistry


If we combine Equations 18.1 and 18.4, we connect the quantity of electrical work (wei) that a voltaic cell does on its surroundings energy that occurs inside the cell:

and the change in free

(18.5) The units on the quantities on the right side of Equation (mel) X (coulomb/mel)

18.5 are

X (volt)

or coulomb· volt Conveniently,

a coulomb· volt represents the same quantity of energy as a joule: 1 coulomb· volt

= 1 joule

or 1C·V=lJ The cell reaction in a voltaic cell must be spontaneous if it does work in an external circuit. Therefore, the sign of ~Gcell must be negative. In Equation 18.3, both nand F have positive values. Therefore, the value of Ecell for a voltaic cell with a spontaneous cell reaction must be positive. Let's use Equation 18.5 to calculate ~Gcell for the Zn/Cu2+ cell reaction in Figure 18.2 and see how its cell potential of1.10 V translates into a value of ~Gcell' The two half-reactions involve the loss and gain of2 moles of electrons. Therefore, the value of n in Equation 18.5 is 2, and the calculation proceeds as follows:

4

-2 mel X 9.65 X 10 C X 1.10 V mel -2.12 At full throttle, 12 million

horsepower

burns liquid hydrogen L (38,000

produces

X 105 V· C

X 105 J

-212 kJ

each of the three main

engines of a U.S. space shuttle

= -2.12

about

(9 X 106 kJ/s) as it at a rate of about 1,100

mol) per second.

As expected, the spontaneous cell reaction has a sizable negative ~G value. To put this value in perspective, on a mole-for-mole basis, the Zn/Cu2+ cell reaction produces slightly less useful energy than the combustion of hydrogen gas used to power the main engines of the U.S. space shuttles:

~GO


=

-228.6

kJ

Relating the Value of AGcell to the Value of Ecell

Silver oxide "button" batteries are used to power electric watches. These batteries, which generate a potential of 1.55 V, consist of a Zn anode and a Ag20 cathode immersed in a concentrated solution of KOH. The cathode reaction is based on the reduction of Ag20 to Ag metal. At the anode, Zn is oxidized to solid Zn(OH\. Write the net ionic equation for the electrochemical cell reaction, and calculate the value of ~Gcell'

Button batteries electronic

are used to power miniature

devices such as wristwatches.

COLLECT AND ORGANIZE We need to write the net ionic equation for the cell reaction and calculate the value of ~Gcell' We know that (1) Ecell = 1.55 V; (2) the cell reaction occurs in a strongly alkaline electrolyte; (3) the half-reaction at the cathode is based on the reduction of Ag20 to Ag metal; and (4) the half-reaction

18.3

at the anode is based on the oxidation of Zn metal to Zn(OH)2' We also know that the value of ~Gcell is related to the value of Ecell by Equation 18.5:

where F is 9.65 X 104 C/mol and n is the number of moles of electrons gained and lost in the cell reaction. A logical first step is to use half-reactions to write the net ionic equation for the cell reaction and to determine the value of n. The half-reaction at the cathode is based on the reduction of Ag2 0 to Ag metal. Appendix 6 contains the standard reduction potential at 25°C of the following half-reaction: A N A LY Z E

g = 0.342V In searching Appendix

6 for a half-reaction that describes the oxidation that occurs at the anode of a button battery, we need to reverse the reactant and product of the oxidation process and find an entry in which Zn(OH)2 is the reactant and metallic Zn is the product. There is such an entry in Appendix 6:

EO

-1.249V

=

Before we can combine the two half-reactions we need to first reverse the Zn half-reaction to make it an oxidation half-reaction. Doing so changes the sign of EO:

SOLVE

Zn(s)

+ 20H-(aq)

~ Zn(OH)/s)

+ 2e-

EO = 1.249V

The Ag and Zn half-reactions involve the gain and loss of the same number of electrons (two). Therefore, they can be added together without further modification: Ag20(s)

+ H20(f) Zn(s)

Ag20(s)

+ 2e-

+ 20H-(aq)

+ H20(f)

+

Zn(s)

~ 2Ag(s)

+ 20H-(aq)

~ Zn(OH)2(s)

EO

+ 2e-

=

0.342V

g = 1.249V

+ J OH-~a~ + ~ ~ 2Ag(s) + J OH-~6!~ +

Zn(OH)/s)

EO

=

+~ 1.591 V

Simplifying, we have the net ionic equation of the cell reaction and its standard cell potential:

EO = 1.591 V To calculate give

~Gcell

=

we insert the values of n, F, and

Ecell

into Equation

18.5 to

-2 +He! X 9.65 X 104 C/+He! X 1.591 V

= -3.07

X 105 C· V

=

-3.07

X 105

J=

-307

kJ

ABOUT IT The negative value of ~Gcell indicates that the cell reaction is spontaneous, which it must be to make a useful battery. The value is somewhat more negative than the ~Gcell we calculated for the Cu2+/Zn cell. This also makes sense because the button battery produces a somewhat larger cell potential and the same number of electrons are gained and lost in both cells. The calculated value of ~G is based on the reaction of one mole of Ag20 and one mole of Zn. The energy stored in a button battery that weighs only a gram or two would be a tiny fraction of the calculated value. Also, you may have noticed that there are no ions in the net ionic equation: they all cancel out when the two half-reactions THINK

I


I

899

900

I

I

Chapter Eighteen


are added together. This makes sense because all the reactants and products in the silver oxide battery reaction are solids, and so the net ionic and molecular equations are the same.

Practice Exercise The alkaline batteries used in flashlights and portable electronic devices produce a cell potential of 1.50 V. They have zinc anodes, an electrolyte of concentrated KOH, and a cathode half-reaction in which solid Mn02 is reduced to solid Mn203. What is the value of ~Gcell?

.':1.

Standard Potentials (£0)

In this section, we examine how the potentials produced by batteries are related to the chemical reactions occurring at their anodes and cathodes. We have defined electrical work as the product of the cell potential of the battery and the quantity of charge that it can deliver at that voltage: (18.4) Italian

physicist Alessandro

is credited

with building

Volta (1745-1827)

the first battery

in

1798. It consisted of a stack of alternating layers of zinc, blotter paper soaked in salt water, and silver.

The values of cell potentials depend on the half-reactions that take place inside the cells and on the standard potentials (EO) of those half-reactions. The superscript has its usual meaning-all reactants and products are in their standard states. For cell potentials, the standard state means that the concentrations of all dissolved substances are 1 M and the partial pressures of all gases are 1 bar

n

(105 Pa). By convention, tables of standard potentials such as those in Appendix 6 are for half-reactions written as reductions. Therefore, the standard potentials in There is no need for a sepaAppendix 6 are standard reduction potentials rate table of standard oxidation potentials, because reduction half-reactions can always be reversed, making them oxidation half-reactions. Reversing the reduction half-reaction does not affect the magnitude of its standard potential, but it does change the sign. In this respect, standard potentials are like other thermodynamic parameters of reactions, such as ~Go, ~Ho, and ~SO. Recall that reversing a chemical reaction changes the signs of these parameters, but their magnitudes stay the same. In equation form, this relation between the standard reduction Potential (E"re d) of a half-reaction and the standard oxidation potential (E"~ ) of the half-reaction running in reverse is

(E:cJ

~II

CHEMTOUR Cell Potential

EO ox

=

_EO red

(18.6)

The standard cell potential (E:cu) of an electrochemical cell may be calculated by combining the standard potentials of its cathode and anode half-reactions: A standard potential (EO) is the electromotive force of a half-reaction written as a reduction in which all reactants and products are in their standard states. This means that the concentrations of all dissolved substances are 1 M, and the partial pressures of all gases are 1 bar. It is also called the standard reduction potential

E:

(18.7)

For example, the value of ell based on the Zn/Cu2+ reaction we discussed in Section 18.1 can be calculated from the standard potentials of the cathode and anode half-reactions (see Appendix 6). The standard potential of the cathode halfreaction is

(E:,d)'

The standard cell potential (E;o11) is the electromotive force produced by an electrochemical cell when all reactants and products are in their standard states.

We can derive the standard potential for the oxidation half-reaction at the anode from the standard reduction potential in Appendix 6:

18.4

I

Standard Potentials (£O)

If we reverse this half-reaction and change the sign on the potential, we have the oxidation reaction of Zn and its EOoxvalue:

EOox= 0.762 V To calculate E:ell we combine this E:x value for the oxidation of Zn with the E~ed for the reduction ofCu2+:

EOcell

= EOred (Cu2+ICu) + EOox (Zn2+IZn) = 0.342 + 0.762 = 1.104 V

This is the voltage we measure when we connect a voltmeter to the cell shown in Figure 18.2. Equation 18.7 is a special case of Equation 18.8, the general equation that applies to the potential of any cell with any concentrations of reactants and products: Ecell= Erjcathode)

+ Eox(anode)

(18.8)

We will learn how to calculate Ecellvalues from E:ell values in Section 18.6. Before we can calculate Ecellfrom Erjcathode) and E)anode), we need to know which half-reaction occurs at the cathode and which occurs at the anode. In a Zn/Cu2+ cell, Cu2+ ions are reduced at the cathode and Zn atoms are oxidized at the anode because Cu2+ ions readily accept electrons and Zn atoms readily donate them. These redox properties can be predicted from the standard reduction potentials in Appendix 6. The value of E~ed for reducing Cu2+ ions to Cu atoms (0.342 V) is more positive than the value of E~ed for reducing Zn2+ ions to Zn atoms (-0.762 V). To have a positive standard cell potential, we must reverse the Zn half-reaction, making it an oxidation half-reaction, which means that it is the half-reaction that occurs at the anode. As a rule, a half-reaction that occurs above another in the table in Appendix 6 is the cathode half-reaction, and the half-reaction lower in the table runs in reverse at the anode.

Identifying Anode and Cathode HalfReactions Using Standard Potentials The standard reduction potentials batteries are

of the half-reactions

in single-use alkaline

and

Which half-reaction occurs at the cathode? Which occurs at the anode? What is the overall cell reaction? What is the value of el

E: ?

at the cathode and anode. To do so we need to identify the spontaneous cell reaction (~G:ell < 0, E:ell > 0) under standard conditions. We can calculate E:ell using Equation 18.7:

COL L EC TAN

D 0 R G A N I Z E We need to identify the chemical half-reactions

E:ell

=

E~e/cathode)

+ E:)anode)

For a spontaneous cell reaction E:ell must be greater than zero.

E cellis positive, we must reverse the half-reaction with the more negative E~ed value, turning it into an E:x value with the opposite sign.

A N A L Y Z E To ensure that

~II

CHEMTOUR

Alkaline Battery

I

901

902

I

Chapter Eighteen

I

Electrochemistry


SOL V E The ZnO reduction half-reaction

has the more negative E~ed value and so is the one that runs in reverse. Therefore, the oxidation half-reaction at the anode is

+ 20H-(aq)

Zn(s)

~ ZnO(s)

+ H20(t') + 2e-

gox = 1.25 V

and at the cathode

EOre d Combining

these half-reactions

=

0.15 V

to obtain the overall cell reaction and simplifying,

The result makes sense because the voltage of most alkaline batteries is nominally 1.5 V. During most of their useful lives, the potentials of alkaline batteries vary between 1.2 and 1.3 V. T HI N K ABO UTI

T

Practice Exercise The standard reduction potentials of the half-reactions nicad (nickel-cadmium) Cd(OH)/s)

in

batteries are

+ 2 e- ~

Cd(s)

+ 20H-(aq)

and 2NiO(OH)(s)

+ 2H20(t')

+ 2e-

~ 2 Ni(OH)2(s)

+ 20H-(aq) E~ed =

1.32 V

Which half-reaction occurs at the cathode? Which occurs at the anode? What is the overall cell reaction? What is the value of el

E: ?

Before closing this discussion of standard cell potentials, let's combine two halfreactions in which different numbers of electrons are gained and lost. This combination occurs in a type of battery that has a limitless supply of one of its reactants. It is called the zinc-air battery, and it powers devices where small size and light weight are high priorities, for example, hearing aids. Most of the internal volume of one of these button-shaped batteries is occupied by an anode consisting of a slurry of zinc particles packed in an aqueous solution of KOH. As in alkaline batteries (Sample Exercise 18.4), the anode half-reaction is

The cathode consists of porous carbon supported by a metal screen. Air diffuses through small holes in the battery and across a layer of Teflon that lets gases pass through but keeps electrolyte from leaking out. As air passes through the cathode, oxygen in the pores of the electrode is reduced to hydroxide ions:

Porous carbon/metal screen Most of the internal

volume of a zinc-air

E:jcathode) = 0.401 V

battery is occupied by the anode: a slurry of Zn particles surrounded negative

in an aqueous solution ofKOH by a metal cup that serves as the

pole of the battery.

air is the reactant through

Oxygen

at the cathode.

holes in an inverted

from the

Air enters

metal cup that

serves as the positive pole of the battery. inside the battery, of gas-permeable keep electrolyte

air diffuses through

Once layers

plastic film that let air in but from leaking out. Oxygen

in

To write the overall cell reaction, first we need to balance the number of electrons lost in the zinc oxidation half-reaction with the number gained by the reduction half-reaction of O2, There are 2 moles of electrons in the oxidation halfreaction and 4 moles in the reduction half-reaction. To balance them, let's multiply the oxidation half-reaction by 2 before combining the two half-reactions: 2 [Zn(s)

+ 20H-(aq)

the air is reduced at the porous carbon/metal cathode

to OH-

ions that migrate

anode, where they are consumed oxidation

half-reaction.

toward the in the Zn

02(g)

~ ZnO(s)

EOox (anode)

=

1.25 V

40H-(aq)

E:jcathode)

=

0.401 V

~ 2 ZnO(s)

E:

=

1.65 V

+ 2H20(t') + 4e- ~ 2 Zn(s)

+ O/g)

+ H20(t') + 2e-]

ell

18.5

I

A Reference Point: The Standard Hydrogen Electrode

I

903

When we multiplied the anode half-reaction by 2 and added it to the cathode half-reaction we did not multiply the EO of the anode half-reaction by 2. The reason we did not is that EO is an intensive property of a half-reaction or a complete cell reaction. It does not change when the quantities of reactants and products change. Thus, a zinc-air battery the size of a pea has the same E:ell as one the size of a book (like those being developed for electric vehicles). On the other hand, the capacity of a zinc-air battery to do electrical work does depend on how much zinc it has inside it because work is the product of cell potential times the quantity of charge it can deliver at that potential.

.:IIII

A Reference Point: The Standard Hydrogen Electrode

We can measure the value of Ecell using a voltmeter. How do we determine the individual values of E re d(cathode) and E ox (anode)? The way we do it is to arbitrarily assign a value to the potential produced by a particular half-reaction and then reference the potentials of all other half-reactions to it. The accepted reference point is the standard potential for the reduction of hydrogen ions to hydrogen gas:

E~ed

= 0.000 V

(18.9)

The EO for this half-reaction is defined as 0.000 V. An electrode that generates this reference potential, called the standard hydrogen electrode (SHE), consists of a platinum electrode in contact with a solution of a strong acid ([H+] = 1.00 M) and hydrogen gas (Figure 18.3). Platinum is not changed by the electrode reaction. Rather, it serves as a chemically inert conveyor of electrons. Electrons are conveyed to the electrode surface ifH+ ions are being reduced to hydrogen gas, or away from it if hydrogen gas is being oxidized to hydrogen ions. The potential of the SHE in either process is the same, 0.000 V. Because the electrode potential of the SHE is 0.000 V, the value of Ecell for any cell in which the SHE is one of the two components is the potential produced by the other component. Suppose, for example, that a cell consists of a strip of zinc metal immersed in a 1.00 M solution of Zn2+ ions in one compartment and an SHE in the other. Also suppose that a meter measures the electromotive force pushing electrons from the zinc electrode to the SHE, as shown in Figure 18.4(a). This direction of current flow means that the zinc electrode is the anode because Zn atoms are being oxidized to Zn2+ ions. In the process they donate electrons to the external circuit, making the anode the negative pole of the electrochemical cell. The SHE is the cathode and positive electrode of the cell. At 25°C, the meter reads 0.762 V. All the reactants and products are in their standard states, so the measured cell potential represents E:ell' The value of E:athodeis that of the SHE (ESHE = 0.000 V), and Eo)anode) is the standard potential of the oxidation of Zn to Zn2+:

Inserting

these values and symbols into Equation

E:ell = E;jcathode)

= E~HE + E:x 0.762 V = 0.000 V + E:ell

E:JZn)

=

0.762 V

CONNECTION Keep in mind that hydrogen ions in aqueous solutions are really hydronium ions, H30+(aq), as described in Chapters 4 and 16.

Pt electrode

Bubbles ofH2 FIGURE 18.3 standard

The potential

of the

hydrogen electrode (SHE) is

the reference point (0.000 V) for all other electrochemical

potentials.

The SHE consists

of a wire leading to a platinum immersed

electrode

in a 1.00 M solution ofH+(aq)

and bathed in a stream of pure H2 gas. Its potential

is the same whether H+(aq) ions are

reduced or H2 gas is oxidized.

18.7,

+ E:x(anode) (Zn)

E:

x

(Zn)

The potential of the standard hydrogen electrode (SHE) is defined as 0.000 V, and so the SHE serves as a reference electrode against which the potentials of other electrode half-reactions can be measured.

904

I Chapter Eighteen

I

Electrochemistry


FIGURE 18.4 The standard hydrogen electrode when used as a reference electrode allows the determination half-reaction. under standard

of the electrode potential

of another

(a) When coupled to a Zn electrode conditions,

~n1621l14 J +

~oltme:er

-

~H2 H2-,> (1 bar) (1 bar)

the SHE is the cathode

(+), which is connected to the positive terminal of the voltmeter. The Zn electrode is the anode (-), which is connected

to the negative terminal

of the voltmeter. With this setup, the voltmeter measures an E,dl value of 0.762 V, which is the same as E:JZn2+/Zn).

(b) When coupled to a Cu

electrode under standard

conditions,

the SHE is

the anode( -), the Cu electrode is the cathode (+), and E,," = E:jCu2+/Cu)

= 0.342

V.

Anode (-)

Cathode

Anode (-)

(+) (a)

Cathode

(+) (b)

This value is equal in magnitude but opposite in sign to the standard reduction potential of Zn2+:

Let's now consider the voltaic cell in Figure 18.4(b). In this case, the SHE is coupled to a copper electrode immersed in a 1.00 M solution ofCu2+ ions. In this cell, current flows from the SHE through an external circuit to the copper electrode at a cell potential of 0.342 Vat 25°C. The SHE is the anode, which means that it supplies electrons to the external circuit and so is the negative pole of the cell. Copper is the cathode and positive pole of the cell. The value of E;ed (Cu2+ I Cu) is calculated as follows:

E:ell

= E:e/cathode) =

0.342 V

E:e/Cu2+ ICu)

E:jCu2+ICu)

+ E:)anode) + E~HE

= E;e/Cu2+ICu) + 0.000 =

0.342 V

This half-reaction potential matches the value of Cu metal in Appendix 6.

E:ed

V

for the reduction ofCu2+ to

M':!W The Effect of Concentration on

feeu

Reactions stop when one of the reactants is completely consumed. This concept was the basis for our discussion of limiting reactants in Chapter 4. However, a commercial battery usually stops supplying current at its rated voltage before its reactants are completely consumed. Why does the cell potential of a battery decrease when it is nearly discharged? The answer is that the electromotive force produced by a voltaic cell is determined by the concentrations of the reactants and products.

The Nernst Equation The Nernst equation relates the potential of a cell (or half-cell) reaction to its standard potential (EO) and the concentrations of its reactants and products.

In 1889, a young German chemist named Walther Nernst (1864-1941) derived an expression, now called the Nernst equation, that describes the dependence of cell potentials on reactant and product concentrations. We can reconstruct his derivation starting with Equation 15.11, which relates the change in free

18.6

energy (~G) of any reaction to its change in free energy under standard tions (~GO): ~G

=

~Go

=

+ RTln Q

-nFE:ell

=g

cell

cell-

RTlnQ

(18.10)

nF

This is the equation Nernst developed in 1889. We can obtain a very useful form of the Nernst equation if we replace the symbols of the constants with their actual values. If we also assume that T = 25°C (298 K) and switch from natural logarithms to base 10 logarithms using the conversion: In Q = 2.303 log Q, then the coefficient of the logarithmic term has the value 0.0592 and the Nernst equation becomes Ecell

0.0592 log Q

o

= E cell- --n--

(18.11)

This is the form of the Nernst equation used in most calculations of Ecell' including the one in the following Sample Exercise.

Calculating Ecell from E:ell and the Concentrations of Reactants and Products If the cell reaction in Figure 18.2 proceeded until [Cu2+] 1.90 M, what would be the value of Ecell at 25°C? Given: Cu2+(aq)

+

Zn(s) ~ Cu(s)

+

Zn2+(aq)

= 0.10 M and [Zn2+]

gcell

=

= 1.104 V

AND ORGANIZE We are given the standard cell potential (1.104 V) and are asked to determine the value of Ecell when [Cu2+] = 0.10 M and [Zn2+] = 1.90 M. Solid copper and zinc are also part of the reaction system, but no terms for pure solids appear in reaction quotients. The Nernst equation (Equation 18.11) can be used to calculate the value of Ecell for different concentrations of reactants and products. COLLECT

The reaction quotient for the cell reaction has [Zn2+] as the only product term in its numerator and [Cu2+] as the only reactant term in its denominator. Each mole of Cu2+ ions acquires 2 moles of electrons, and each mole of Zn atoms donates 2 moles of electrons, so the value of n in the cell reaction is 2. Therefore, the Nernst equation for the Zn/Cu2+ cell reaction is

ANALYZE

Ecell

=

Ecell

I

905

+ RTlnQ

Dividing all terms by - nF,

E

The Effect of Concentration on

condi-

As we noted in Section 15.4, as a spontaneous reaction proceeds, the free energy changes from the negative ~Go values we associate with spontaneous reactions to ~G values that increase as reactants form products and the value of reaction quotient Q increases. When ~G reaches zero, the reaction has reached chemical equilibrium and no further conversion of reactants to products takes place. Now let's write an expression analogous to Equation 15.11 to relate Ecell to E:ell' We can do so by combining Equations 15.11 and 18.5 (~Gcell = -nFEcell) to give -nFEcell

I

1.104 V - 0.059210 [Zn2+] 2 g [Cu2+]

@

CONNECTION In Chapter 15 we discussed the relationship between free energy changes and the reaction quotient, Q.

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Chapter Eighteen

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Electrochemistry


SOLVE

Substituting the values of[Zn2+] and [Cu2+] in the Nernst equation gives E cell'

Solving for

Ecell'

= 1104 V - 0.0592 Iog-'-1 90 -2

0.10

we get E

II ce

= 1.104 V - 0.0592 (1.28) = 1.066 V 2

ABOUT IT The calculated Ecell value is 0.038 Vless thanE:ell' This decrease makes sense because the cell is 90% discharged and so a drop in voltage is to be expected. The drop is only 0.038 V because the logarithmic dependence of cell potential on the concentrations of reactants and products minimizes the impact of changing concentrations. How can we have a calculated Ecell value with four significant figures when we know the value of [Cu2+] to only two? The concentration values were used to calculate the change in cell potential, which was rounded off to only two significant figures: -0.038 V. That number was then combined with 1.104 V to obtain an answer with four significant figures. THINK

Practice Exercise What is the cell potential of the zinc-air battery if the partial pressure of oxygen in the air diffusing through its cathode is 0.21 bar? Given:

2 Zn(s)

+ 02(g)

~ 2 ZnO(s)

Equilibrium and Dead Batteries As a battery's cell reaction proceeds, the concentrations of the reactants decrease and the concentrations of the products increase. Therefore, the value of Q (and the logarithmic term in Equation 18.11) increases. Because of the negative sign in front of the logarithmic term, the value of Ecell decreases. Thus, the voltage produced by a battery decreases as the concentrations of its reactants decrease. Let's consider a common battery design that illustrates this point. In virtually every automobile, a lead-acid battery is used to drive an electric motor that starts the engine. The lead-acid battery (Figure 18.5) that provides power to start most motor vehicles contains six cells with anodes made oflead and cathodes made of Pb02

Cathode (+)

FIGURE 18.5 The lead-acid battery that provides power to start most motor vehicles contains six cells. Each has an anode made of lead and a cathode made ofPb02 in an electrolyte of 4.5 M H2SO

immersed 4'

The

electrodes are formed into plates and held in place by grids made of a lead alloy. The grids connect the cells together in series so that the operating voltage of the battery (12.0 V) is the sum of the six cell potentials

(each 2.0 V).

Cathode plate (Pb02)

Anode plate (Pb)

18.6

I

The Effect of Concentration on

I

Ecell

907

immersed in the electrolyte 4.5 M H2SO 4' The electrodes are formed into plates and held in place by grids made of a lead alloy. The grids connect the cells (each providing 2.0 V) together in series so that the operating voltage of the battery is the sum of the six cell potentials, or 12.0 V. As a lead-acid battery is discharged, Pb02 is reduced to solid PbSO 4 and Pb is oxidized to solid PbS04• The anode half-reaction involves the loss of2 moles of electrons for each mole oflead. Writing this half-reaction as an oxidation, Pb(s)

+

is reduced to PbSO

At the cathodes, Pb02 Pb02(s)

+ 2e-

S042-(aq) ~ PbSO/s)

+ 4H+(aq) +

+ 2e-

SO/-(aq)

EOox

=

0.356 V

4:

~ PbSO/s)

+ 2H20(t')

E~ed

=

1.685 V

in a half-reaction that consumes 2 moles of electrons. Combining these two halfreactions gives the following overall cell reaction with a standard cell potential of 2.041 V:

As the battery is discharged, [H2S04] decreases, and so does the value of Ecell calculated from the Nernst equation for the cell reaction:

= 2.04 V - 0.0592 10 __ 1

E

2

cell

g [H2SO

l

The decrease in Ecell is very gradual, not falling below 2.00 V until the battery is 97% discharged as shown in Figure 18.6. When the reactants are nearly consumed, the cell potential should approach zero as the cell reaction achieves chemical equilibrium. At this point the value of the reaction quotient (Q) is equal to the equilibrium constant (K) for the cell reaction. At equilibrium the cell potential drops to zero. Therefore, Equation 18.11 becomes

0= EO

cell

- 0.059210 K n g

or

10gK

=

2.1

nE:ell

(18.12)

0.0592

2.0

Equation 18.12 makes a mathematical connection between the equilibrium constant of a cell reaction and the standard cell potential. We can use this connection to calculate equilibrium constants for any redox reaction, not just those in electrochemical cells. For the more general case, we substitute E:xn for E:ell in Equation 18.12 to give

10gK

=

nE:xn 0.0592

(18.13)

as in Sample Exercise 18.6. The potential of every voltaic cell decreases as reactants are converted into products. In the case described here, the value of E:ell of the lead-acid battery cell reaction decreases as its sulfuric acid is consumed, but this decrease is only gradual until the battery is nearly completely discharged.

'".1l~ 0 0.-

1.9

;::;

" u 1.8

1.7' 0%

I

20%

40%

60%

80% Level of discharge (%)

FIGURE 18.6

The potential

lead-acid

decreases

converted potential completely

battery

into products,

ofa cell in a

as reactants

but the changes

are small until the battery discharged.

100%

are in

is nearly

908

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Chapter Eighteen

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Electrochemistry


Calculating the Equilibrium Constant of a Redox Reaction from the Standard Potentials of Its Half-Reactions Many of the methods used to determine the levels of mercury contamination in environmental samples begin by oxidizing all the mercury in the samples to Hg2+ and then reducing the Hg2+ to elemental Hg with Sn 2+. Use the appropriate standard potentials in Appendix 6 to calculate the equilibrium constant at 25°C for the reaction:

AND ORGANIZE We are asked to calculate the equilibrium constant for a redox reaction starting with the standard potentials of its component halfreactions. Equation 18.12 relates the value to the value of the equilibrium ell constant of the cell reaction. As we have seen, we can use Equation 18.13 to calculate the equilibrium constant for any redox reaction, using the standard potential EOrxn .

COLLECT

E:

To calculate E,oxn we need to combine the appropriate standard potentials. Appendix 6 lists the following half-reactions involving the reactants and products: Hg2+(aq)

+ 2 e-

-?

Hg(t')

E~ed

=

0.851 V

Sn4+(aq)

+ 2e-

-?

Sn2+(aq)

E~ed

=

0.154 V

Sn2+ is the reducing agent in the reaction, so it must be oxidized. Therefore, the second reaction runs in reverse, as an oxidation: ANALYZE

EOox

=

-0.154

V

Combining the half-reactions for the oxidation ofSn2+ and the reduction ofHg2+ and summing their standard potentials,

SOLVE

Hg2+(aq)

+ 2e-

Sn2+(aq)

E~ed = 0.851 V

-?

Hg(t')

-?

Sn4+(aq)

+ 2 e-

EOox = -0.154

V

g,xn = 0.697 V Using this reaction potential for the cell potential in Equation a value of 2 for n,

10 K = nE:xn g

=

0.0592

K

= 10235 = 3

2(0.697) 0.0592

=

18.13 and inserting

23.5

X 1023

E:

ABOUT IT Note how a positive value of xn corresponds to a huge equilibrium constant, indicating that the reaction essentially goes to completion even though E;xn is only about 0.7 V. That the reaction goes to completion is one reason why it can be reliably used to determine the concentrations of mercury in aqueous samples containing Hg2+ ions. THINK

Practice Exercise Using the standard reduction potentials in Appendix calculate the value of the equilibrium constant at 25°C for the reaction

6,

18.7

EIII

I

Relating Battery Capacity to Quantities of Reactants

I

909


An important performance characteristic of a battery is its capacity to do electrical work (wei)' which means its capacity to deliver electrical charge at its designed cell potential. This capacity was previously defined by Equation 18.2 w elec

=

CE cell

In our description of electrical work in Section 18.3 we noted that 1 joule of electrical energy is equivalent to 1 coulomb· volt of electrical work. Another important unit in electricity is the ampere (A), which is the SI base unit of electrical current. An ampere is defined as a current flow rate of 1 coulomb per second: 1ampere

=

The electrical energy rating of these rechargeable

AA batteries

is 1.6A· hr at 1.2 V.

1coulomb/second

or 1 coulomb

=

1 ampere' second

Multiplying both sides of the equation by volts, we get 1 coulomb· volt

= 1 joule

=

volt· ampere' second

Joules are rather small units, and so the capacities of batteries such as those used to power portable electronic devices and flashlights are usually expressed in units with longer time intervals. For example, the energy ratings of rechargeable AA batteries are often expressed in ampere-hours, or milliampere-hours, at a particular voltage. The energy stored in much larger batteries, such as the rechargeable battery packs used in hybrid vehicles, is usually expressed in kilowatt-hours (kWh) where 1 watt

=

1 volt· ampere

Note that a watt (W) is equivalent to a joule per second and so represents the rate of energy production or consumption. In the physical sciences we call this power; the watt is the SI unit of power. The power of the electric motors used in hybrid vehicles is sometimes expressed in horsepower (a unit widely used in the United States) and at other times in kilowatts (the SI-derived unit used in much of the rest of the world). Most hybrids get their best fuel mileage in low-speed city driving where most of the power is supplied by the more fuel-efficient electrical power system. For example, the U.S. EPA estimates that the 2008 Ford Escape gets 34 miles per gallon (mpg) of gasoline in city driving and 30 mpg (7.9 liters per 100 km) in highway driving.

The 2008 Ford Escape is powered by a 133 horsepower horsepower

gasoline engine and a 94

(70 kW) electric motor. As in

most of the first-generation from a pack of nickel-metal batteries under the floor.

Nickel-Metal

Hydride Batteries

The electrochemical cells in most rechargeable batteries and in the battery packs in hybrid cars are called nickel-metal hydride (NiMH) cells (Figure 18.7). In these cells NiO(OH) is reduced to Ni(OH)2 at the cathode and hydrogen is oxidized at an anode made of one or more transition metals. The two electrodes are separated by a porous spacer containing an aqueous solution of KOH. The cathode half-reaction is

EO

=

1.32 V

At the anode hydrogen is present not as H2' but rather as a metal hydride. To write the anode half-reaction we use the generic formula MH where M stands for a

hybrid vehicles,

power for the Escape's electric motor comes hydride (NiMH)

910

I Chapter Eighteen

I

Electrochemistry


transition metal or metal alloy that forms a hydride and serves as the anode. alkaline electrolyte the anode half-reaction can be written

In an

The standard potential of this half-reaction depends on the chemical properties of MH, but generally the value is near that of the SHE, 0.0 V. The cathode and anode half-reactions involve the gain and loss of 1 mole of electrons, respectively, so we can simply sum them to obtain the overall cell reaction. Canceling out the species on both sides of the reaction arrow, we get

MH(s)

Cathode Ni(OH)2

~~

NiO(OH)---------u NiO(OH)(s) Ni(OHMs)

""-

+ H20(C) + e~ + OH-(aq)

KOH FIGURE 18.7

Sees

What a Chemist

In the nickel-metal

hydride (NiMH)

battery, hydrogen atoms are stored in the interstices of a metal matrix. During the anode half-reaction

they are oxidized to H+

ions that combine with OH- ions, forming H20

and releasing electrons. At the cathode,

NiO(OH) electrolyte.

~ M(s)

+ Ni(OH\(s)

The value of E:ell for the NiMH battery cannot be calculated precisely because we have only an approximate value of E:x(anode). Most NiMH Anode cells deliver about 1.2 V. Now let's relate the electrical energy stored in a battery to the quantity of reactants that are needed to create that energy. Consider a rechargeMH(s) + OH-(aq) able AA battery. It is rated to deliver 2.5 ampere~ hours (A· hr) of electrical charge at 1.2 V. How M + H20(C) + emuch NiO(OH) has to be converted to Ni(OH)2 Electrolyte to deliver this much charge? To answer this question we need to relate the quantity of charge to a number of moles of electrons and then convert that to an equivalent number of moles of reactant and finally to a mass of reactant. Let's begin by noting that an ampere is the same as a coulomb per second (Cis), so the quantity of electrical charge delivered is

2.5 A·lH:

X

is reduced to Ni(OH)2' consuming

electrons and releasing OH- ions to the

+ NiO(OH)(s)

1C A· I>

X

60 min lH:

X

6~ I> = 9.0

X

103 C

f'fH*

A mole of charge is equivalent to 9.65 X 104 coulombs, so the number of moles of charge, which is equal to the number of moles of electrons that flow from the battery, is

9.0

X 103

e (~~) 9.65 X 10 e 4

= 0.0933 mol e-

The stoichiometry of the anode half-reaction (1 mole of electrons per mole of NiO(OH)) leads to the following approach to calculating the mass of NiO(OH) consumed:

0.0933 ~

_ (1 mel NiO(OH) 1~

)(91.701 g) = 8.6 g NiO(OH) ±-mol

The mass of an AA battery is 30 g so the result of the calculation is reasonable if we allow for the mass of the anode material, the electrolyte, and the battery's exterior shell.

Lithium-Ion Batteries Many electronic devices, including laptop computers, cell phones, and digital cameras, are powered by lithium-ion batteries (Figure 18.8). When these batteries are fully charged, Li+ ions are stored in anodes made of highly pure graphite. During discharge, the Li+ ions migrate toward highly porous cathodes made of transition-

18.7

Cathode

I


I

911

Anode (

Discharge

What a Chemist

FIGURE 18.8

Sees

This molecular

view of a lithium-

ion battery shows the migration stored within

graphite

toward a cathode is discharged. cathode

of Li+ ions

layers of the anode

made ofCo02

as the battery

The crystal structure

of the

is based on a cubic closest packed

array of oxide ions in which Co ions occupy half the octahedral

holes. Li+ ions can move

in and out of the other half of the holes. The migration

pattern

is reversed when the battery

is recharged. Cobalt ion

Lithium

ion

Oxide ion

Carbon atom

metal oxides or phosphates that form stable complexes with Li+ ions. One popular cathode material is cobalt oxide. Lithium-ion batteries with this cathode material have a cell potential of3.6 V (three times that of a NiMH battery) and are used to power most laptop computers, cell phones, and digital cameras. We can write the cell reaction for a lithium-ion battery with this cathode as follows: Lil-x Co02(s)

+ LixC6(s)

-?

6 C(s) + LiCo02(s)

In a fully charged cell the value of x in this equation is 1, which makes the formula of the cathode Co02.This formula means that Co is in its +4 oxidation state. As the cell discharges and Li+ ions migrate from the carbon anode toward the cobalt oxide cathode, the value of x falls from one toward zero. To balance this flow of positive charges inside the cell, electrons and their negative charges flow from the anode toward the cathode through an external circuit. If fully discharged, the formula of the cathode would be LiCo02, and Co would be in its +3 oxidation state. The electrodes in lithium-ion batteries may react with oxygen and water, so these batteries use electrolytes (for example, LiPF6) that dissolve in polar organic solvents such as tetrahydrofuran, ethylene carbonate, or propylene carbonate.

SAMPLE EXERCISE 2.8.7

Relating the Mass of Reactant in an Electrochemical Reaction to a Quantity of Electrical Charge

The energy capacity of the lithium-ion battery in a digital camera is 3.4 watthours at 3.6 V. How many grams ofLi+ ions must migrate from the anode to the cathode of the battery to produce this much electricity? We are asked to convert the electrical energy generated by an electrochemical cell into the mass of the ions carrying that current. We know the cell potential in volts and its energy capacity in watt-hours. Given these COLLECT

AND

ORGANIZE

Q Tetrahydrofuran

/\ 0yO o Ethylene carbonate

Propylene carbonate

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I

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I

Electrochemistry


starting points and the need eventually to calculate moles and grams of Li+ ions, the following equivalencies are likely to be useful: 1 watt

=

1 ampere (A) . volt (V)

1 coulomb (C)

=

1 ampere (A) . second (s)

We may also need to use Faraday's constant, 9.65 X 104 C/mol. A logical sequence of steps would be (1) to combine the starting potential and energy values to calculate the quantity of charge delivered by the battery and then (2) relate that quantity to the moles ofLi+ ions. In setting up our conversion factors we need to convert watts to ampere-volts and divide by the cell potential to cancel out volts. A N A LY Z E

SOLVE Using the equivalencies provided, we can calculate the amount of charge from the starting values:

3.4 W·IH: X 1 :A·V X _1_ W 3.6V

X 60 R'lia X 6017 X 1 C

IH:

mffi

= 3.4

X 103 C

A,'17

Dividing coulombs by Faraday's constant, 3.4 X 103

e (~~) 9.65 X 10 e4

= 0.0352 mol e-

For every mole of electrons that flows through an external circuit, one mole of singly charged Li+ ions migrates from the anode to the cathode inside the Li-ion battery that powers the circuit. This equality allows us to convert 0.0352 moles of electrons into an equivalent mass ofLi+ ions: 0.0352 ~

1

(

:::1/ .~

U,;+)(6941gLi+)

= 0.245 g Li

T HI N K ABO UTI T The battery that is the subject of this exercise weighs about 22 grams, so Li+ ions make up only about 1% of the mass of this lithium-ion battery. This small percentage is reasonable given the masses of the electrode materials that donate and accept the ions. In the anode each Li+ ion is surrounded by a hexagon of 6 carbon atoms (Figure 18.8) with about 10 times the mass of the Li+ ion. At the cobalt oxide cathode the corresponding mass ratio is about 13:1.

Practice Exercise Magnesium metal is produced by passing an electrical current through molten MgCI2• The reaction at the cathode is Mg2+(t')

+ 2e-

~ Mg(t')

How many grams of magnesium metal are produced if an average current of 63.7 amperes flows for 4.50 hours? Assume all of the current is consumed by the half-reaction shown.

mDI

Electrolytic Cells and Rechargeable Batteries

NiMH and lithium-ion batteries are rechargeable batteries, which means that the cell reactions in which chemical energy is converted into electrical energy can be reversed. They can be recharged because the products of the discharge reaction are substances that adhere to, or are embedded within, the electrodes. These substances are therefore available to be converted back to their original chemical forms when an opposing electrical potential is applied that is greater than the cell potential.

18.8

Voltaic Cell

I

I


913

Electrolytic Cell

Spontaneous cell reaction does work on its surroundings

External source of electrical power does work on system

-,r rI l1 ,Power

supply

\

Load

Anode (oxidation)

~

Cathode

Cathode

Anode

(reduction)

("d"";""i

(""d";""i

FIGURE 18.9 (a) In a voltaic cell, a spontaneous reaction

•

Cations

EZI:)

•

•

Anions

Anions

Electrolyte

EZI:)

•

Cations

electrical

and does electrical surroundings,

energy

work on its

such as lighting

a

light bulb. (b) In an electrolytic cell, an external

supply of electrical

energy does work on the chemical

Electrolyte

(a)

chemical

produces

system in the cell, driving a nonspontaneous

(b)

reaction.

Applying an opposing potential forces the cell reaction to run backward, converting electrical energy back into chemical energy. Recharging a battery is an example of electrolysis, that is, a chemical reaction driven by electricity. During recharging, a battery is transformed from a voltaic cell into an electrolytic cell, a process in which electrical energy is converted into chemical energy (Figure 18.9). The chemical changes that take place in a lead-acid battery in the spontaneous discharge process and during nonspontaneous recharging are shown in Figure 18.10. During discharging and recharging, current flows in opposite directions through the battery. Thus, the electrodes that serve as anodes (the sites of oxidation) during the discharge reaction turn into cathodes (the sites of reduction) during recharging. Any PbSO 4 that formed on them during discharge is reduced to Pb during recharging: E~ed

= -0.356 V

Similarly, the Pb02 electrodes that serve as cathodes during discharge become anodes during recharging, as any PbSO 4 that formed on the cathodes during discharge is oxidized to Pb02: PbSO /s)

+ 2 HP(t')

-?

Pb02(s)

+ 4H+(aq) +

SO/-(aq)

+ 2e-

EOox

=

-1.685 V

Recharging a lead-acid battery or any battery with an aqueous electrolyte can result in some undesirable and even dangerous side reactions. If too high a voltage forces too great a current through a lead-acid battery, the reduction oflead sulfate at the lead electrode may not keep up with the flow of electrons to the electrode surface. In that event, some other reducible species, such as the hydrogen ions in a concentrated solution of sulfuric acid, may accept the influx of electrons:

EO

= 0.000 V

Ifboth H+ ions and PbS04 are present at the surface of a lead electrode, which species is preferentially reduced? Note that the EO for reducing PbSO 4 (-0.356 V) is more negative than the standard potential for reducing H+ ions (0.000 V). In an electrolytic cell a species that is reduced at a less negative potential should

Electrolysis is a process in which electrical energy is used to drive a nonspontaneous chemical reaction. An electrolytic cell is a device in which an external source of electrical energy does work on a chemical system, turning reactant(s) into higher-energy product(s).

914

I

Chapter Eighteen

I

Electrochemistry


---------------------------~~Switch ~,-------------------------~~

{

0 (

~\/{

0(

Anode Pb(s)

+ SOi-(aq)

Cathode

----j

PbSOis)

+ 2 e-

4

PbS0 (s)

+ 2 e- ~(s)

+

SOi-(aq)

(-)

Recharge

Lead-acid battery

FIGURE 18.10

The lead-acid

batteries used in many vehicles are based on the oxidation of Pb and the reduction discharges

ofPb02•

PowerT~V~ supply~~ (alternator)

As the battery

Cathode Pb02(s)

+ 4H+(aq) + SOi-(aq) + 2 e----j PbSOis) + 2H20(f)

Anode PbS04(s) Pb02(s)

+ 2H20(f) ----j + 4H+(aq) + SOi-(aq)

+ 2 e-

(green circuit), Pb is

oxidized, and Pb02is

reduced. When

the engine is running,

an alternator

generates electrical energy that flows into the battery, recharging

it as

both electrode reactions are reversed: is oxidized to Pb02,

PbS04 PbSO

4

and

is reduced to Pb.

be preferentially reduced compared with one requiring a more negative potential. On this basis, we might conclude that H+ ions would be reduced to hydrogen gas before lead sulfate is reduced to lead metal. However, another factor affects the potential at which an electrode reaction begins during electrolysis. It has to do with the rates of the electron transfer processes at the electrode surfaces. Just as the rates of some spontaneous chemical reactions are slow, so too some half-reactions, particularly those in which a gas evolves, may be extremely slow. In homogeneous reactions we associate slow reaction rates with high activation energies. One way to overcome this barrier is to raise the temperature of the reaction mixture. In electrolysis, we can overcome a slow electrode reaction rate by applying an overpotential, that is, an even greater potential than predicted by calculated electrode potentials. The reduction ofH+ ions to H2 is such a reaction. Although there is little overpotential for the reduction ofH+ ions at the platinum electrodes used in the standard hydrogen electrodes, there is a large overpotential of1.0 V or more for reducing H+ ions at a lead electrode. This overpotential means that PbS04 is reduced before H+ ions when a lead-acid battery is recharged. Overpotentials also playa role in controlling the oxidation process that occurs at the anode of the lead-acid battery during recharging. There the desired oxidation process is the conversion of PbS04 into Pb02. However, another species is present in abundance that is, theoretically, more easily oxidized than PbSO 4' namely H20.

EOox An overpotential is the additional potential above the theoretically calculated value required to promote an electrode reaction in electrolysis.

= -1.229 V

The standard potential for the oxidation of water in acid solutions is more than 0.4 V less negative than the lead reaction. However, the water half-reaction involves the generation of the gas O2 and has a significant overpotential that inhibits water from oxidizing before PbSO 4 oxidizes.

18.8

I

I


915

Calculating the Time Required to Electrolyze a Quantity of Reactant A battery charger for AA NiMH batteries supplies a charging current of 1.00 A. How long will it take to oxidize 0.649 g of Ni(OH\ to NiO(OH)? COLLECT AND ORGANIZE We are asked to calculate the time required for a constant current to electrolyze a given mass of reactant. During recharging, the spontaneous half-reaction runs in reverse and Ni(OH)2 is oxidized to NiO(OH) at the anode:

EO One mole of electrons is produced for each mole ofNi(OH)2

= -1.32 V

oxidized.

We need to convert the given mass ofNi(OH)2 into moles ofNi(OH)2' Since the half-reaction involves the transfer of 1 mol of electron, that is also the number of moles of electrons pulled away from the electrode. Faraday's constant can be used to convert moles of electrons to coulombs of charge, and finally coulombs can be converted to an electrolysis time by dividing them by the amount of current flowing through the cell. ANALYZE

First we need to convert the mass ofNi(OH\ of moles of electrons:

into an equivalent number

SOLVE

n _ = 0.649 g Ni(OH)2 X

1 ffisl Ni(OH)2

71

Q')

e

•

1 mol eX ---ffisl Ni(OH)_

go

= 0.00700 mol ePower source

Converting moles of electrons to an equivalent positive electrical charge expressed in ampere-seconds gives:

0.00700 ~

X

9.65

X

~

4

10

eX

f

'II

1 A· s = 676 A· s

e

Therefore, the charger must deliver 1.00 A of current for 676 s or 1 min (6766 ) X --

60 ..

. = 11.3 mIll

ABOUT IT For those experienced with the time it takes to recharge batteries for portable devices, an interval of 11.3 min may seem short, but if we compare the quantity of the Ni(OH)2 to be oxidized with the quantity ofNiO(OH) in a fully charged AA size NiMH battery (see page 910), we realize that this battery was only slightly discharged. THINK

Practice Exercise Suppose that a car's starter motor draws 230 A of current for 6.0 s to start the car's engine on a cold winter morning. What mass oflead would have been oxidized to PbSO 4 in the car's battery to supply this much electricity?

Anode FIGURE 18.11

Cathode Silverware

typically

has only a thin coating of silver, which is

Electrolysis is used in many processes other than the recharging of batteries. Electrolytic cells are used to electroplate thin layers of silver (Figure 18.11), gold, and other metals onto objects, giving these objects the appearance, resistance to corrosion, and other properties of the electroplated metal, but at a fraction of the cost of fabricating the entire object out of the metal. In the chemical industry, the electrolysis of molten salts is used to produce high-energy, highly reactive substances, such as chlorine and fluorine, alkali and alkaline earth metals, and aluminum. When a salt such as NaCI is heated just

applied by a type of electrolysis electroplating.

known as

In the process, the positive

pole of an electric power supply is attached

to

a piece of pure silver, and the negative pole is connected

to the cutlery, such as a spoon, to

be electroplated.

The oxidation

the silver electrode

half-reaction

(anode), Ag -> Ag+

runs in reverse at the cathode

+ e-,

(spoon) as Ag+

ion is reduced to the free metal at the surfaces of the cutlery.

at

916

I Chapter Eighteen

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Electrochemistry


above its melting mous capacity to carbon electrodes negative electrode

point (above 800°C), it becomes an ionic liquid with an enorconduct electricity. If a sufficiently large voltage is applied to immersed in molten NaCI, sodium ions are attracted to the and are reduced, producing sodium metal:

Chloride ions will be attracted to the positive electrode and oxidized: 2 Cl-(t') ~ CI2(g)

(2)

Multiplying reaction (1) by two and combining overall electrolytic cell reaction: 2 Na+(t')

+ 2 CI-(t')

+ 2 e-

it with reaction (2), we have the

~ 2 Na(t')

+

CI/g)

A final note in summary about the polarities of the anodes and cathodes in electrolytic and voltaic cells is in order. The reactions in voltaic cells are spontaneous. These cells produce electric current. The anode in a voltaic cell is negative, and the oxidation reaction at the anode spontaneously releases electrons which flow toward a positive cathode. Reduction takes place at the cathode. The reactions in electrolytic cells are nonspontaneous. They require electric current (electrons) from an external power supply. These electrons are pushed into the cathode, which therefore has a negative charge, and reduction takes place at the cathode. Electrons flow from the negative cathode to the positive anode, where oxidation takes place. CONCEPT

TEST

The electrolysis of molten NaCI produces Na metal at the cathode. However, the electrolysis of an aqueous solution of NaCI results in the formation of gases at both the cathode and the anode. On the basis of the standard potentials in Appendix 6, predict which gases are formed at each electrode .

• !:it. Fuel Cells ~II

CHEMTOUR

Fuel Cell

Fuel cells are promising energy sources for many applications, from powering office buildings to cruise ships to electric vehicles. They are voltaic cells like batteries, and they rely on oxidizing a fuel such as H2 or CH4 to generate electrical energy.

In acidic electrolytes, the two half-reactions

are

(1) At the anode:

EOox = 0.000 V (2) At the cathode: O/g) A fuel cell is a voltaic cell in which the reaction is the oxidation of a fuel. The reaction is the equivalent of combustion, but the chemical energy is converted into electrical energy.

+ 4H+(aq) + 4e-

~ 2H20(t')

In fuel cells with alkaline material between the electrodes, such as molten sodium carbonate, the alkaline versions of these half-reactions

occur:

(1) At the anode: H2(g)

+ 20H-(aq)

~ 2H20(t')

+ 2e-

gox = 0.828V

18.9

I

Fuel Cells

I

917

(2) At the cathode: 02(g)

+ 2H20(t') + 4e-

~ 40H-(aq)

Note that, if we combine these two pairs of standard potentials by combining the standard potentials of each pair of half-reactions, we get the same ell value (1.229 V). This equality is logical because the energy released under standard conditions (~GO) by the oxidation of hydrogen gas to form liquid water should have but one value, and so ell should have only one value that does not depend on electrolyte pH. The same free-energy change could be used to power a vehicle by burning hydrogen in an internal combustion engine. However, typically less than a third of the energy from combustion is converted into mechanical energy in the internal combustion engines of automobiles. Most of the energy of combustion is lost to the surroundings as heat. In contrast, fuel-cell technologies can convert most of the energy released in a fuel-cell reaction to electrical energy. In addition, H2 -fueled vehicles emit only water vapor; they produce no oxides of nitrogen that can contribute to smog formation, and no carbon monoxide or carbon dioxide is released. Fuel cells are like batteries in that their chemisElectricmotor

E:

E:

try is based on two half-reactions occuring in environments separated from each other by a porous barrier through which electrolyte ions can pass. In one design, O2 is supplied to a cathode made of porous graphite containing atoms ofNi, and H2 gas is supplied to a graphite anode containing nickel(II) oxide. Both electrodes are immersed in an alkaline electrolyte. Hydroxide ions form during the reduction of O2 at the cathode: 02(g)

+ 2H20(t') + 4e-

Fuel cells used in electric vehicles incorporate acidic electrolytes and a proton-exchange membrane (PEM) to separate the anode and cathode half-reactions (Figure 18.12). On the anode side of the cell H2 fuel diffuses into a porous graphite electrode, where it is oxidized to H+ ions:

+ 2 e-

The electrons are pushed through an external circuit by an Ecell value of about 0.7 volts, which is about half the standard potential for the cell reaction 2H/g)

+ O/g)

H2 fuel

O2 from air

Anode

Cathode

~ 40H-(aq)

These migrate through the cell to the anode, where they combine with H2 as it is oxidized to water:

H2(g) ~ 2 H+(aq)

.=r-------fl=

~ 2H20(t')

As electrons are pumped out of the cell through the anode, H+ ions migrate through the protonexchange membrane to the cathode, where they participate in the reduction of O2:

Recirculate unusedH2

Air + wateroutput Proton-exchange membrane Catalytic electrodes Gas-permeable backing

FIGURE 18.12 Most of the fuel cellsused in vehiclesincorporatea protonexchangemembrane(PEM) between the two halvesof the cell. Hydrogen diffuses to the anode and oxygendiffusesto the cathode.Theseelectrodesare made of porous material,such as carbon nanofibers,that havehigh surfaceareas for a givenmass of material.The electrodesalso contain transition metals that serveas catalyststo speedup the electrodehalf-reactionsby increasingthe rate at which the bonds that hold moleculesof H2 and O2 together are broken. Platinum catalystsare effective at promoting H-H bond breaking, and a platinum-nickel alloywith the formula Pt3Ni is particularlyeffectivein catalyzingthe formationof free 0 atomsfrom O2 molecules.Thesecatalystsincreasethe rate of the overallcell reaction, 2 H/g) + 02(g) ---7 2 H20(C), and the poweroutput of the fuel cell.

Power, Pigments, and Phosphors: The Group 12 Elements The last group in the transition metals (group 12, Zn, Cd, Hg) contains elements used in many batteries. None of these three metals is particularly abundant in Earth's crust. Zinc is the most abundant of the three (78 mg/kg), and mercury is the least (67 µ,g/kg). All three are found in sulfide ores, including sphalerite (ZnS), greenockite (CdS), and cinnabar (HgS). Zinc is also found in carbonates and silicates. Most zinc ores contain cadmium impurities. The cathode ray tubes we discussed in Chapter 2 used by scientists like Ernest Rutherford and J. J. Thomson to detect cathode rays and X-rays were coated with ZnS because it emits light when excited by X-rays and electron beams. If another metal called an activator is added to zinc sulfide, the material phosphoresces. The color of the phosphorescence depends on the activator: a few ppm of silver produces a blue color; manganese yields orange-red; and copper produces a long-lasting dark green that is useful in electroluminescent panels. Zinc metal is recovered from zinc sulfide ore by heating the ore in air to produce zinc oxide: 2 ZnS(s)

+ 3 02(g) ~

2 ZnO(s)

(because it absorbs both UVA and UVB ranges of ultraviolet light), and on the insides of fluorescent light bulbs. A special grade of zinc oxide called Chinese white is used in artists' pigments. ZnO is reduced to molten metal with hot carbon:

+ C(s) ~

2 ZnO(s)

2 Zn(t')

+ CO/g)

Along with zinc, cadmium is produced from impurities in the ore and is separated by distillation of the molten metals. Cadmium is used in making nickel-cadmium (nicad) batteries, but large quantities are also used in producing pigments ranging from lemon yellow (CdS) through red (Cd with both Sand Se) to deep maroon (CdSe). The cadmium pigments are popular with artists because of their bright colors and their stability when exposed to light. Cinnabar (HgS) has been mined in Spain since Roman times. Also called vermillion or Chinese red, cinnabar has been used because of its unique red color in Chinese decorative art and lacquer ware.

+ 2 SO/g)

Zinc oxide is the most common ingredient in antibiotic creams for the treatment of diaper rash, in sunscreens

Cadmium medium

yellow

Cadmium

orange

Cadmium

hue

medium

15

o Zn

Cd

Hg

Element Zinc oxide is used to coat the insides of fluorescent

light bulbs and to make topical creams

that are also effective sunscreens.

The density of the group 12 elements increases with increasing

atomic number.

red

919

Mercury is produced directly from reaction ofHgS with oxygen:

2, tend to lose two valence electrons and form M2+ cations. Mercury also forms an Hg22+ cation, which contains a covalent Hg-Hg bond. The group 12 elements are more likely than the alkaline earth (group 2) metals to form covalent bonds. Zinc is an essential element in living organisms, but cadmium and mercury have no known health benefits and are quite toxic. The toxicity of these metals is related to their ability to attach to sulfur groups in some amino acids. These interactions allow the metals to be transported throughout the body attached to the peptides and proteins in blood. If they reach the brain, interactions with the proteins there can lead to mental retardation and other neurological disorders. The toxicity of mercury is increased by the formation of covalent alkyl mercury compounds such as methyl mercuric chloride, CH3HgCI, and dimethyl mercury, (CH3)2Hg. Mercury in these chemical forms can penetrate hydrophobic cell membranes more easily than inorganic Hg2+ ions and so can more severely disrupt internal cell functions. Microbial methylation of inorganic mercury wastes dumped in Minamata Bay, Japan, in the 1950s led to 52 sudden deaths and severe neurological damage to more than 3000 people. This incident sparked recognition of the environmental hazards of improperly disposing of mercury and other metals.

Mercury is the only metal, and one of only two elements (bromine is the other), that is a liquid at room temperature. Zinc and cadmium are low-melting, silvery solids that react rapidly with moist air to form the corresponding oxides. The oxides and sulfides of zinc and cadmium are semiconductors (see Chapter 11). The group 12 metals have (n - 1)dlOns2 electron configurations in their valence shells and, like the metals in group • Melting point 1000

•

Boiling point

Cd

Hg

o -273 Zn

Element The group 12 elements are lowmelting metals (Hg is a liquid at room temperature)

and are relatively low boiling

compared with other metals.

.Atomic 200

2

c

:§ os

bJl

g

g

1

u

~"

III

o

a

0..

;g

100

'i3

~

radius

.Ionic

radius

Cd

lLI.

Element The electronegativities

of the group 12

elements increase slightly as their atomic numbers increase.

Second ionization energy

~

E

;::; :::, 1000

~ c " ~

o Zn

Hg

First ionization energy

• 2000

o Zn

•

Cd

Zn

Hg

Element

Cd

Hg

Element

The radii of the 2+ cations of the group

The first and second ionization

12 elements increase as their atomic

of the group 12 elements show little

number increases. The atomic radii are less

dependence

dependent

on atomic number.

on atomic number.

energies

920

I Chapter Eighteen

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FIGURE 18.13 The 2008 Honda FCX Clarity is powered by a 100 kW fuel cell stack and 100kW (134 horsepower) electric motor that give the car a top speed of 100 miles per hour (160 km/hr). Hydrogen fuel is stored in a 171L tank at a maximum pressure of 350 atm, which gives the car a driving range of about 270 miles (430 km). A Li-ion battery pack provides supplemental energy storage including that produced by regenerative braking.

individual PEM fuel cell has a cell potential of about 1.0 V. That is not much, when hundreds of these cells are assembled into fuel cell stacks, such as the in the Honda FCX Clarity (Figure 18.13), they are capable of producing 100 of electrical power. That is enough to give the Clarity a top speed of100 miles hour (160 km/hr). As fuel cells become more efficient and less expensive, the principal limit on their use in passenger cars is the availability, cost, and safety of using hydrogen fuel. Hydrogen is flammable, but it has a higher ignition temperature than gasoline and spreads through the air more quickly, reducing the risk of fire. Unfortunately from the point of view of safety, hydrogen in air burns over a much wider range of concentrations than gasoline, and its flame is particularly hazardous because it is almost invisible. It is likely that most fuel-cell-powered vehicles will be buses and fleet vehicles operating from a central location, where hydrogen gas is available. For example, buses (with very large fuel tanks) powered by fuel cells can carry 60 passengers and have an operating range of 400 km. In 2008 the Honda Clarity was available in only a few communities in Southern California, where hydrogen gas was available at some filling stations. Longer ranges would be possible if better methods for storing hydrogen gas were available. Currently under development are mobile chemical-processing plants that can extract hydrogen from liquid fuels such as gasoline or methanol as needed by the vehicle. Making hydrogen from fossil fuels also generates carbon dioxide as a byproduct, so such processes do not improve the situation with greenhouse gases in the atmosphere. The most common method in use today is the production of hydrogen from natural gas, which results in about a twofold improvement in total emissions compared to those from a conventional gasolinepowered vehicle. Scientists are also developing new materials that can be used to store large quantities of hydrogen safely. In one promising technology, a carbon material, related in its structure to bucky balls, has been used to store many times the mass of hydrogen that could be compressed into the same volume without the carbon. Other scientists are developing new metal hydrides for hydrogen storage. There is an important safety feature to hydrogen sequestration in carbon structures or as metal hydrides: it would probably not be released and explode in a collision, as is likely if a high-pressure tank filled with H2 gas ruptured. The drawback to the metal hydride systems is that they tend to be heavy and add considerable weight to an automobile. A major focus of hydrogen research worldwide is finding a better storage method for hydrogen fuel. An but one kW per

@@ CONNECTION Processes for producing hydrogen gas by reacting methane and other fuels with steam are discussed in Chapter 15.

Problem-SolvingSummary

I

921

Summary The standard potentials in Appendix 6 are all reduction potentials, ed. The corresponding x values for the half-reactions in reverse have the same magnitudes but opposite signs.

SECTION 18.1 Electrochemistry is the branch of chemistry that links redox reactions to the production or consumption of electrical energy. Any redox reaction can be broken down into oxidation and reduction half-reactions. The gain of electrons in a reduction halfreaction must be balanced by the loss of electrons in the oxidation half-reaction coupled to it.

E:

18.5 All standard potentials are referenced to the potential of the standard hydrogen electrode (ESHE = 0.000 V).

SECTION

18.6 The potential of a voltaic cell decreases as reactants turn into products. The Nernst equation predicts how the potential of a cell changes with the concentrations of the cell reactants and products. The potential of a voltaic cell approaches zero as the cell reaction approaches chemical equilibrium. At equilibrium Ecell = 0 and Q = K.

SECTION

A voltaic cell is an electrochemical cell that produces an electrical current in an external circuit as a result of a spontaneous redox reaction in which oxidation occurs at the anode and reduction occurs at the cathode of the cell. A cell diagram uses symbols including single and double vertical lines to show how the components of the cathodic and anodic compartments of the cell are connected.

SECTION

18.2

18.7 The quantities of reactants consumed in a voltaic cell reaction are directly proportional to the coulombs of electrical charge that the cell delivers. Faraday's constant relates the quantity of electrical charge to the number of moles of electrons, and indirectly to the moles of reactants involved in the cell reaction. Most portable electronic devices are powered by nickel-metal hydride (NiMH) or lithium-ion batteries.

SECTION

18.3 The cell potential (Eccll) of a voltaic cell is the electromotive force that the cell generates. Ecell is related to the change in free energy (LlGcell) of the cell reaction by the equation:

SECTION

LlGcell

=

-nPE,dl

where Pis Faraday's constant (9.65 X 104 C/mol) and n is the number of electrons transferred in the balanced chemical equation describing the cell reaction. This change in free energy is available to do work in an external electrical circuit.

18.8 Reversing a spontaneous reaction by passing electric current through an electrochemical cell turns the cell into an electrolytic cell. Reactions in electrolytic cells that produce gases often require application of additional electromotive force, or overpotential. The polarities of cathodes and anodes in electrolytic cells are opposite the polarities of these electrodes in voltaic cells.

SECTION

18.4 The value Ecell under standard conditions is called standard cell potential (E;,II)' It is calculated by combining the standard potentials (EO) of a cell's half-reactions, E:ed of the cathode half-reaction and x of the anode half-reaction:

SECTION

E:

E;ell

=

E:,/cathode)

Problem-Solving

E:

18.9 Fuel cells directly convert the free energy released during the reaction 2H2 + O2 ~ 2H20 into electrical energy. Fuel cells with proton-exchange membranes have been developed as power supplies for electric vehicles.

SECTION

+ E:Janode)

Summary SAMPLE

TYPE OF PROBLEM

CONCEPTS

Writing net ionic equations for redox reactions

Combine the reduction and oxidation half-reactions after balancing the gain and loss of electrons.

18.1

Diagramming an electrochemical cell

Start with the anode half-reaction; use single lines to separate phases and a double line to separate the two half-reactions.

18.2

Relating the value of ..1G"II to the value of E,,"

AND EQUATIONS

18.3 where n is the number of moles of electrons transferred in the cell reaction and F is Faraday's constant.

Identifying anode and cathode half-reactions from their standard potentials

The half-reaction with the more positive standard reduction potential is the cathode half-reaction.

18.4

Calculating E,," from E:,II and the concentrations of reactants and products

E"II at 298 K is related to E:dl and the reaction quotient by the Nernst

18.5

equation: E

,oil

=

EO

,,11 -

0.0592 --n-logQ

EXERCISES

922

I

Chapter Eighteen

I

Electrochemistry


TYPE OF PROBLEM

CONCEPTS

AND EQUATIONS

SAMPLE

Calculating the equilibrium constant of a redox reaction from the standard potentials of its half-reactions

E:dl is the sum of E:jcathode) 298K by

Relating the mass of a reactant in an electrochemical reaction to a quantity of electrical charge

Determi ne the ratio of moles of reactants to moles of electrons transferred; relate coulombs of charge to moles of electrons using Faraday's constant.

18.7

Calculating the time to electrolyze a quantity of reactant

Use Faraday's constant, electrical current as charge per unit time, and the relation between moles of electrons and moles of reactants to describe an electrolytic process.

18.8

+ E:Janode)

and is related to Kat

EXERCISES

18.6

nE:

e11

10gK= 0.0592


Consider the four batteries in Figure P18.1. From top to bottom the sizes are AAA, AA, C, and D. The performance of batteries like these is often expressed in units such as (a) volts, (b) watt-hours, (c) milliampere-hours. Which of the values that go with each of these units are likely to differ significantly among the four batteries?

18.3.

The apparatus in Figure P18.3 is used for the electrolysis of water. Hydrogen and oxygen gas are collected in the two inverted burettes. An inert electrode at the bottom of the left burette is connected to the negative terminal of a 6-volt battery; the electrode in the burette on the right is connected to the positive terminal. A small quantity of sulfuric acid is added to speed up the electrolytic reaction. a. What are the half-reactions at the left and right electrodes and their standard potentials? b. Why does sulfuric acid make the electrolysis reaction go more rapidly?

FIGURE P18.1

18.2.

Which of the four curves in Figure P18.2 best represents the dependence of the potential of a lead-acid battery on the concentration of sulfuric acid? Note that the scale of the x-axis is logarithmic.

Overall cell reaction H20(C) ~ Hh) + ~02(g) FIGURE P18.3

18.4.

100 10-2 10-4 10-6 10-8 10-10 [H2S04] FIGURE P18.2

An electrolytic apparatus identical to the one shown in Problem 18.3 is used to electrolyze water, but the reaction is speeded up by the addition of sodium carbonate instead of sulfuric acid. a. What are the half-reactions and the standard potentials for the electrodes on the left and right? b. Why does sodium carbonate makes the electrolysis reaction go more rapidly?


18.5.

In many electrochemical cells the electrodes are metals that carry electrons to and from the cell but are not chemically changed by the cell reaction. Each of the highlighted clusters in the periodic table in Figure P18.S consists of three metals. Which of the highlighted clusters is best suited to form inert electrodes?

18.7.

I

923

In the voltaic cell shown in Figure P18.7, the concentrations ofCu2+ and Cd2+ are 1.00 M. On the basis ofthe standard potentials in Appendix 6, identify which electrode is the anode and which is the cathode. Indicate the direction of electron flow.


18.6.

In the voltaic cell shown in Figure P18.6, the greater density of a concentrated solution of CuSO 4 allows a less concentrated solution of ZnSO 4 solution to be (carefully) layered on top of it. Why is a porous separator not needed in this cell?

18.8.

In the voltaic cell shown in Figure P18.8, [Ag+] = [H+] = 1.00 M. On the basis of the standard potentials in Appendix 6, identify which electrode is the anode and which is the cathode. Indicate the direction of electron flow.

FIGURE P18.8

FIGURE P18.6

Questions and Problems Redox Chemistry Revisited; Electrochemical Cells CONCEPT

18.9.

REVIEW

What is the role of the porous separator in an electrochemical ceIl? 18.10. The chemical reactions in Figures 18.1 and 18.2 are the same. However the reaction in the cell in Figure 18.2 generates electricity; the reaction in the beaker in Figure 18.1 does not. Why? 18.11. Why can't a wire perform the same function as a porous separator in an electrochemical cell? 18.12. In a voltaic cell, why is the cathode labeled the positive terminal and the anode the negative terminal?

PROBLEMS

18.13. A voltaic cell with an aqueous electrolyte is based on the reaction between Pb2+(aq) and Zn(s), producing Pb(s) and Zn2+(aq). a. Write half-reactions for the anode and cathode. b. Write a balanced cell reaction. c. Diagram the cell. 18.14. A voltaic cell is based on the reaction between Ag+(aq) and Ni(s), producing Ag(s) and Ni2+(aq). a. Write the anode and cathode half-reactions. b. Write a balanced cell reaction. c. Diagram the cell.

924

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Chapter Eighteen

I


18.15. A voltaic cell with an alkaline aqueous electrolyte is based and the reduction on the oxidation of Cd(s) to Cd(OH)/s) ofMn04-(aq) to Mn02(s). a. Write half-reactions for the cell's anode and cathode. b. Write a balanced cell reaction. c. Diagram the cell. 18.16. A voltaic cell is based on the reduction of Ag+(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aq). a. Write half-reactions for the cell's anode and cathode. b. Write a balanced cell reaction. c. Diagram the cell. 18.17. Super Iron Batteries In 1999, scientists in Israel developed a battery based on the following cell reaction with iron(VI), nicknamed "super iron": 2 K2FeO .(aq)

+

3 Zn(s)

--7

+

Fe203(s)

ZnO(s)

+

2 K2ZnO/aq)

a. Determine the number of electrons transferred in the cell reaction. b. What are the oxidation states of the transition metals in the reaction? c. Diagram the cell. 18.18. Aluminum Air Batteries In recent years engineers have been working on an aluminum-air battery as an alternative energy source for electric vehicles. The battery consists of an aluminum anode, which is oxidized to solid aluminum hydroxide, immersed in an electrolyte of aqueous KOH. At the cathode oxygen from the air is reduced to hydroxide ions on an inert metal surface. Write the two half-reactions for the battery and diagram the cell. Use the generic M(s) symbol for the metallic cathode material.

Chemical Energy and Electrical Work CONCEPT REVIEW 18.19. How is it that a voltaic cell with a positive cell potential does negative work? *18.20. In physics, work (w) is done by exerting a force (F) to move an object through a distance (d) according to the equation w = F X d. Explain how this definition of work relates to electrical work (welec= C X E). PROBLEMS 18.21. For many years the 1.50 V batteries used to power flashlights were based on the following cell reaction: Zn(s)

+

+

2NH4Cl(s)

2 Mn02(s) --7 Zn(NH)2Cl/s)

+ Mnp/s)

+

HP(£)

What is the value of LlG"n? 18.22. The first generation oflaptop computers was powered by nickel-cadmium ("nicad") batteries, which generated 1.20 V based on the following cell reaction: Cd(s)

+

2 NiO(OH)(s)

+

2 HP(£)

--7

Cd(OH)2(s)

+

2 Ni(OH)/s)

What is the value of LlG"n? 18.23. The cells in the nickel-metal hydride battery packs used in many hybrid vehicles produce 1.20 V based on the following cell reaction: MH(s)

+

NiO(OH)(s)

--7

What is the value of LlG"n?

M(s)

+

Ni(OH)/s)

18.24. A cell in a lead-acid battery delivers exactly 2.00 V of cell potential based on the following cell reaction: Pb(s)

+

Pb02(s)

+

2 H2SO 4(aq)

2 PbSO .(s)

--7

+

2 HP(£)

What is the value of LlG"n?

Standard Potentials {EO} CONCEPT REVIEW 18.25. What is the function of platinum in the standard hydrogen electrode? 18.26. Is it possible to build a battery in which the anode chemistry is based on a half-reaction in which none of the species is a solid conductor, for example: Fe2+(aq)

--7

Fe3+(aq)

+

e-

18.27. In some textbooks the formula used to calculate standard cell potentials from the standard reduction potentials of the half-reactions occurring at the cathode and anode is given as

E;en

=

E:ed(cathode) - E:ed(anode)

Show how this equation is equivalent to Equation 18.7. 18.28. Suppose there were a scale for expressing electrode potentials in which the standard potential for the reduction of water in base

+

2HP(£)

2e-

--7

H2(g)

+

20H-(aq)

is assigned an EO value of 0.000 V. How would the standard potential values on this new scale differ from those in Appendix 6? 18.29. Is O2 a stronger oxidizing agent in acid or in base? Use standard reduction potentials in Appendix 6 to support your answer. *18.30. To inhibit corrosion of steel or aluminum structures in contact with seawater, pieces of other metals (often zinc) are attached to the structures to serve as "sacrificial anodes." Explain how these attached pieces of metal might protect the structures and describe which properties of zinc make it a good selection. PROBLEMS 18.31. Starting with the appropriate standard free energies of formation in Appendix 4, calculate the value of LlGo and E;en of the followi ng reactions: a. 2 Cu+(aq) --7 Cu2+(aq) + Cu(s) b. Ag(s) + Fe3+(aq) --7 Ag+(aq) + Fe2+(aq) 18.32. Starting with the appropriate standard free energies of formation in Appendix 4, calculate the values of LlGo and E;en of the followi ng reactions: a. FeO(s) + H/g) --7 Fe(s) + HP(£) b. 2Pb(s) + 02(g) + 2H2SO.(aq) --7 2PbSO.(s) + 2HP(£) 18.33. If a piece of silver is placed in a solution in which [Ag+] = [Cu2+] = 1.00 M, will the following reaction proceed spontaneously? 2Ag(s)

+

Cu2+(aq)

--7

2Ag+(aq)

+

Cu(s)

18.34. A piece of cadmium is placed in a solution in which [Cd2+] = [Sn2+] = 1.00 M. Will the following reaction proceed spontaneously? Cd(s)

+

Sn2+(aq)

--7

Cd2+(aq)

+

Sn(s)

Questions and Problems I 925 18.35. Sometimes the anode half-reaction in the zinc-air battery is written with the zincate ion, Zn(OH)/-, as the product. Write a balanced equation for the cell reaction based on this product. 18.36. Sometimes the cell reaction of nickel-cadmium batteries is written with Cd metal as the anode and solid Ni02 as the cathode. Assuming that the products of the electrode reactions are solid hydroxides ofCd(II) and Ni(II), respectively, write balanced chemical equations for the cathode and anode half-reactions and the overall cell reaction. 18.37. In a voltaic cell similar to the Cu-Zn cell in Figure 18.2, the Cu electrode is replaced with one made ofNi immersed in a solution of NiSO 4' Will the standard potential of this Ni-Zn cell be greater than, the same as, or less than 1.10 V? 18.38. Suppose the copper half of the Cu-Zn cell in Figure 18.2 were replaced with a silver wire in contact with 1 M Ag+(aq). a. What would be the value of E"n? b. Which electrode would be the anode? 18.39. Starting with standard potentials listed in Appendix 6, calculate the values of E;en and !lGo of the following reactions. a. Cu(s) + Sn2+(aq) --7 Cu2+(aq) + Sn(s) b. Zn(s) + NiH(aq) --7 Zn2+(aq) + Ni(s) 18.40. Starting with the standard potentials listed in Appendix 6, calculate the values of E;en and !lGo of the following reactions. a. Fe(s) + Cu2+(aq) --7 Fe2+(aq) + Cu(s) b. Ag(s) + Fe3+(aq) --7 Ag+(aq) + Fe2+(aq) 18.41. Voltaic cells based on the following pairs of half-reactions are prepared so that all reactants and products are in their standard states. For each pair, write a balanced equation for the cell reaction, and identify which half-reaction takes place at the anode and which at the cathode. a. HgH(aq) + 2e- --7 Hg(£) ZnH(aq) + 2e- --7 Zn(s) b. ZnO(s) + HP(£) + 2e- --7 Zn(s) + 20H-(aq) AgP(s) + HP(£) + 2e- --7 2Ag(s) + 20H-(aq) c. NiOH2 -(aq) + 2 e- --7 Ni(s) + 20H-(aq) O/g) + 2HP(£) + 4e- --740H-(aq) 18.42. Voltaic cells based on the following pairs of half-reactions are constructed. For each pair, write a balanced equation for the cell reaction, and identify which half-reaction takes place at each anode and cathode. a. Cd2+(aq) + 2e- --7 Cd(s) Ag+(aq) + e- --7 Ag(s) b. AgBr(s) + e- --7 Ag(s) + Br-(aq) Mn02(s) + 4H+(aq) + 2e- --7 MnH(aq) + 2 HP(£) c. PtCl/-(aq) + 2e- --7 Pt(s) + 4CI-(aq) AgCI(s) + e- --7 Ag(s) + CI-(aq) 18.43. The half-reactions and standard potentials for a nickelmetal hydride battery with a titanium-zirconium anode are as follows: Cathode:

NiO(OH)(s) + HP(£) + e- --7 Ni(OH)2(s) + OH-(aq)

EO

=

1.32 V

Anode:

TiZr2H(s)

+ OH-(aq) --7 TiZr2(s) + HP(£)

+ e-

EO = 0.00 V

a. Write the overall cell reaction for this battery. b. Calculate the standard cell potential. *18.44. New Lithium-Ion Batteries Scientists at the University of Texas, Austin, and at MIT have developed a new cathode material for lithium-ion batteries based on LiFePO 4' which is the composition of the cathode when the battery is fully discharged. Batteries with this cathode are more powerful than those of the same mass with LiCo02 cathodes. They are also more stable at high temperatures. a. What is the formula of the LiFePO 4 cathode when the battery is fully charged? b. Is Fe oxidized or reduced as the battery discharges? c. Is the cell potential of a lithium-ion battery with an iron phosphate cathode likely to differ from one with a cobalt oxide cathode? Explain your answer.

A Reference Point: The Standard Hydrogen Electrode; The Effect of Concentration on Ecell CONCEPT

REVIEW

18.45. Why does the operating cell potential of most batteries change little until the battery is nearly discharged? 18.46. The standard potential of the Cu-Zn cell reaction Zn(s)

+

Cu2+(aq)

--7

Zn2+(aq)

+

Cu(s)

is 1.10 V. Would the potential of the Cu-Zn cell differ from 1.10 V if the concentrations of both CuH and Zn2+were 0.25 M? PROBLEMS

18.47. Calculate the E"n value at 298 K for the cell based on the reaction Fe3+(aq)

+

Cr+(aq)

--7

Fe2+(aq)

+

Cr3+(aq)

when [Fe +] = [Cr2+] = 1.50 X 10- M and [Fe2+] = [Cr+] = 2.5 X 10-4 M. 18.48. Calculate the E"n value at 298 K for the cell based on the reaction 3

Cu(s)

3

+

2Ag+(aq)

when [Ag+] = 2.56

X

--7

Cu2+(aq)

+

2Ag(s)

10-3 M and [Cu2+] = 8.25

X

10-4 M.

18.49. Using the appropriate standard potentials in Appendix 6, determine the equilibrium constant for the following reaction at 298 K: Fe3+(aq)

+

Cr+(aq)

--7

Fe2+(aq)

+

Cr3+(aq)

18.50. Using the appropriate standard potentials in Appendix 6, determine the equilibrium constant at 298 K for the following reaction between Mn02 and FeH in acid solution: 4H+(aq)

+

MnO/s)

+

2 Fe2+(aq) --7 Mn2+(aq) + 2 Fe3+(aq)

+

2HP(£)

926

I Electrochemistry

I Chapter Eighteen


a. Determine EO for the reaction. For the half-reaction

18.51. If the potential of a hydrogen electrode based on the half-reaction 2 H+(aq)

+ 2 e-

+

+

2H+(aq)

+ 2 H+(aq) + 2 e-

Oxaloacetate-(aq)

+

e- --7 Cl02(g)

*18.57. The oxidation ofNH/ to N03 by the following equation:

+

2e- --7 NADH(aq)

2 H+(aq)

+

HP(£)

EO = 1.152V b. The reaction produces an atmosphere in the reaction vessel in which PClO, = 2.0 atm; PCl, = 1.00 atm. Calculate [Cl03 -] if, at equilibrium (T = 298 K), [H+] = [CI-] = 10.0 M.

is 0.000 V at pH = 0.00, what is the potential of the same electrode at pH = 7.00? 18.52. Glucose Metabolism The standard potentials for the reduction of nicotinamide adenine dinucleotide (NAD+) and oxaloacetate (reactants in the multistep metabolism of glucose) are as follows: NAD+(aq)

+

CI03-(aq)

--7 H/g)

H+(aq) EO = -0.320

V

--7 malate-(aq) g = -0.166

V

+ 202(g) +

NH/(aq)

in acid solution is described

-

HP(£)

--7 N03-(aq)

+

+

3 Hp

For the half-reaction

+

N03 -(aq)

10 H+(aq)

+

8 e- --7 NH/(aq)

a. Calculate the standard potential for the following reaction:

+

Oxaloacetate-(aq)

NADH(aq)

+

H+(aq) --7 malate-(aq)

+ NAD+(aq)

b. Calculate the equilibrium constant for the reaction at 298 K. 18.53. Permanganate ion can oxidize sulfite to sulfate in basic solution as follows:

+

2Mn04-(aq)

3S0/-(aq) + HP(£)--7 2MnO/s) + 3S0/-(aq)

+

20H-(aq)

Determine the potential for the reaction at 298 K when the concentrations of the reactants and products are as follows: [Mn04-] = 0.150 M, [SO/-] = 0.256 M, [SO/-] = 0.178 M, and [OH-] = 0.0100 M. Will the value of Ecxo increase or decrease as the reaction proceeds? *18.54. Manganese dioxide is reduced by iodide ion in acid solution as follows:

+

Mn02(s)

2l-(aq)

+

4H+(aq) --7 Mn2+(aq)

+

I2(aq)

+ 2 HP(£)

Determine the electrical potential of the reaction at 298 K when the initial concentrations of the components are as follows: [1-] = 0.225 M, [H+] = 0.900 M, [Mn2+] = 0.100 M, and [12] = 0.00114 M. If the solubility of iodine in water is approximately 0.114 M, will the value of Em increase or decrease as the reaction proceeds? 18.55. A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides. The following reaction describes the formation of NO, one of the products: 3Cu(s)

+

8H+(aq)

+

2 N03-(aq) --7 2NO(g) + 3Cu2+(aq)

+

4HP(£)

a. Starting with the appropriate standard potentials in Appendix 6, calculate Em for this reaction. b. Calculate E;xo at 298 K when [H+] = 0.100 M, [N03-] = 0.0250 M, [Cu2+] = 0.0375 M, and the partial pressure of NO = 0.00150 atm. 18.56. Chlorine dioxide (CIO) is produced by the following reaction of chlorate (CI03 -) with CI- in acid solution: 2Cl03-(aq)

+ 2CI-(aq)

+

4H+(aq)--7 2 Cl02(g)

+

Cl/g)

+ 2 HP(£)

2H+(aq)

EO a. Calculate EO for the overall reaction. b. If the reaction is in equilibrium with air (Po

(£) =

1.50 V

, = 0.21

atm)

at pH 5.60, what is the ratio of [N03-] to [NH/] at 298 K? 18.58. Recalculate the answers to Problem 18.57 for the oxidation ofNH/ at an altitude where Po, = 0.18 atm.

Relating Battery Capacity to Quantities of Reactants CONCEPT

REVIEW

18.59. One 12-volt lead-acid battery has a higher ampere-hour rating than another. Which ofthe following parameters are likely to be different for the two batteries? a. Individual cell potentials b. Anode half-reactions c. Total masses of electrode materials d. Number of cells e. Electrolyte composition f. Combined surface areas of their electrodes 18.60. In a voltaic cell based on the Cu-Zn cell reaction Zn(s)

+

Cu2+(aq) --7 Cu(s)

+

Zn2+(aq)

there is exactly 1 mole of each reactant and product. A second cell based on the Cd-Cu cell reaction Cd(s)

+

Cu2+(aq) --7 Cu(s)

+

Cd2+(aq)

also has exactly 1 mole of each reactant and product. Which of the following statements about these two cells is true? a. Their cell potentials are the same. b. The masses of their electrodes are the same. c. The quantities of electrical charge that they can produce are the same. d. The quantities of electrical energy that they can produce are the same. PROBLEMS

18.61. Which of the following voltaic cells will produce the greater quantity of electrical charge per gram of anode material? Cd(s) or 4AI(s)

+ 2 NiO(OH)(s) + 2 HP(£)

--7 2 Ni(OH)2(s)

+

30/g)

+

6HP(£)

+

40H-(aq)

+

Cd(OH)2(s)

--7 4Al(OH)4-(aq)


18.62. Which of the following voltaic cells will produce the greater quantity of electrical charge per gram of anode material? Zn(s)

+ Mn02(s) +

HP(£)

--7 ZnO(s)

+ Mn(OH)/s)

or

*18.63. Which of the following voltaic cell reactions delivers more electrical energy per gram of anode material at 298 K? Zn(s)

+

2NiO(OH)(s) + 2HP(£)--7 2 Ni(OH)/s) + Zn(OH)2(s)

or E"n

=

3.15V

*18.64. Which of the following voltaic cell reactions delivers more electrical energy per gram of anode material at 298 K? Zn(s)

+

Ni(OH)2(s) --7 Zn(OH)/s)

+

Ni(s)

E"n = 1.50V

or 2 Zn(s)

+

O/g) --7 2 ZnO(s)

E"n = 2.08V

Electrolytic Cells and Rechargeable Batteries CONCEPT

REVIEW

18.65. The positive terminal of a voltaic cell is the cathode. However, the cathode of an electrolytic cell is connected to the negative terminal of a power supply. Explain this difference in polarity. 18.66. The anode in an electrochemical cell is defined as the electrode where oxidation takes place. Why is the anode in an electrolytic cell connected to the positive (+) terminal of an external supply, whereas the anode in a voltaic cell battery is connected to the negative (-) terminal? 18.67. The salts obtained from the evaporation of seawater can act as a source of halogens, principally Cl2 and Br2, through the electrolysis of the molten alkali metal halides. As the potential of the anode in an electrolytic cell is increased, which of these two halogens forms first? 18.68. In the electrolysis described in Problem 18.67, why is it necessary to use molten salts rather than seawater itself? 18.69. In the electrolysis of H20 to H2 and O2, a strong electrolyte, such as Na2C03, is added to speed up the electrolysis process. How does Na2C03 speed up the process? 18.70. In the electrolysis of H20 to H2 and O2, why should Na2C03, but not NaCl, be added to speed up electrolysis? 18.71. Quantitative Analysis Electrolysis can be used to determine the concentration ofCu2+ in a given volume of solution by electrolyzing the solution in a cell equipped with a platinum cathode. If all of the CuH is reduced to Cu metal at the cathode, the increase in mass of the electrode provides a measure of the concentration of Cu2+ in the original solution. To ensure the complete (99.99%) removal of the CuH from a solution in which [CuH] is initially about 1.0 M, will the potential of the cathode (versus SHE) have to be more negative or less negative than 0.34 V (the standard potential for CuH + 2 e- --7 Cu)?

I

927

18.72. A high school chemistry student wishes to demonstrate how water can be separated into hydrogen and oxygen by electrolysis. She knows that the reaction will proceed more rapidly if an electrolyte is added to the water, and has these reagents available: 2.00 MH2S04, 2.00 M HCl, NaCI, Na2S04, Na2COv and CaC03• Which one(s) should she use? Explain your selection(s). PROBLEMS

18.73. Suppose the current flowing from a battery is used to electroplate an object with silver. Calculate the mass of silver that would be deposited by a battery that delivers 1.7 A· hr of charge. 18.74. A battery charger used to recharge the NiMH batteries used in a digital camera can deliver as much as 0.50 A of current to each battery. If it takes 100 min to recharge one battery, how much Ni(OH), (in grams) is oxidized to NiO(OH)? 18.75. A NiMH battery containing 4.10 g ofNiO(OH) was 50% discharged when it was connected to a charger with an output of2.00 A at 1.3 V. How long does it take to recharge the battery? *18.76. How long does it take to deposit a coating of gold 1.00 µm thick on a disk-shaped medallion 4.0 cm in diameter and 2.0 mm thick at a constant current of85 A? The density of gold is 19.3 glcm3• The gold solution contains Au(III). *18.77. Oxygen Supply in Submarines Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxygen by the electrolysis of water. a. How many liters of O2 at 298 K and 1.00 bar are produced in 1 hr in an electrolytic cell operating at a current of 0.025 A? b. Could seawater be used as the source of oxygen in this electrolysis? Explain why or why not. 18.78. In the electrolysis of water, how long will it take to produce 1.00 X 102 L ofH2 at STP (273 K and 1.00 bar) using an electrolytic cell through which a current of 52 mA flows? 18.79. Calculate the minimum (least negative) cathode potential (versus SHE) needed to begin electroplating nickel from 0.35 M NiH onto a piece of iron. Assume that the overpotential for the reduction ofNF+ on an iron electrode is negligible. *18.80. What is the minimum (least negative) cathode potential (versus SHE) needed to electroplate silver onto cutlery in a solution of Ag+ and NH3 in which most of the silver ions are present as the complex, Ag(NH),+, and the concentration of Ag+(aq) is only 3.50 X 10-5 M?

Fuel Cells CONCEPT

REVIEW

18.81. What are the advantages of hybrid (gasoline engine-electric motor) power systems over all-electric systems based on fuel cells? What are the disadvantages? 18.82. Small fleets of fuel-cell powered automobiles are operating in several U. S. cities. Describe three factors limiting the more widespread use of cars powered by fuel cells.

928

I

Chapter Eighteen

I

Electrochemistry


18.83. Methane can serve as the fuel for electric cars powered by fuel cells. Carbon dioxide is a product of the fuel cell reaction. All cars powered by internal combustion engines burning natural gas (mostly methane) produce CO2, Why are electric vehicles powered by fuel cells likely to produce less CO2 per mile? 18.84. To make the refueling offuel cells easier, several manufacturers offer converters that turn readily available fuels-such as natural gas, propane, and methanol-into H2 for the fuel cells and CO,. Although vehicles with such power systems are not truly "zero emission," they still offer significant environmental benefits over vehicles powered by internal combustion engines. Describe a few of them. PROBLEMS

18.85. Fuel cells with molten alkali metal carbonates as electrolytes can use methane as a fuel. The methane is first converted into hydrogen in a two-step process: CH.(g)

+

HP(g)

--7

CO(g)

CO(g)

+

HP(g)

--7

H2(g)

+ +

3 H/g)

CO/g)

a. Assign oxidation numbers to carbon and hydrogen in the reactants and products. b. Using the standard free energy of formation values in Appendix 4, calculate the standard free-energy changes in the two reactions and the overall LlGo for the formation ofH2 + CO2 from methane and steam. *18.86. Molten carbonate fuel cells fueled with H2 convert as much as 60% of the free energy released by the formation of water from H2 and O2 into electrical energy. Determine the quantity of electrical energy obtained from converting 1 mole ofH2 into H20(£) in such a fuel cell.

combining the standard potentials for the reduction of Ag+ and AP+ in Appendix 6 with the K,p expressions and values of Ag2S and AI(OH\ in Appendix 5.] *18.89. A magnesium battery can be constructed from an anode of magnesium metal and a cathode of molybdenum sulfide, M03S4. The half-reactions are Anode:

Mg(s)

Cathode:

Mg2+(aq)

--7

Mg2+(aq)

+

+ 2 e-

M03S .(s)

EOox = 2.37 V

+

2 e- --7 MgMo3S.(s)

a. If the standard cell potential for the battery is 1.50 V, what is the value of E;ed for the reduction ofM03S/ *b. The electrolyte in the battery contains a complex Why is it necessary to magnesium salt, Mg(AlCI3CH),. include Mg2+ ions in the electrolyte? c. What are the apparent oxidation states and electron configurations ofMo in M03S4 and in MgMo3S/ 18.90. Clinical Chemistry The concentration ofNa+ ions in red blood cells (11 mM) and in the surrounding plasma (140 mM) are quite different. Calculate the electrochemical potential (emf) across the cell membrane as a result of this concentration gradient. *18.91. The element fluorine, F2, was first produced in 1886 by electrolysis ofHF. Chemical syntheses ofF2 did not happen until 1986 when Karl O. Christe successfully prepared F2 by the reaction

a. Assign oxidation numbers to each compound and determine the number of electrons involved in the process. o b. Using the following values for LlH;, calculate LlH for the reaction.

Additional Problems

LlHr.SbF5(C) = -1324 kJlmol

LlHr•K,MoF6(;)

*18.87. Electrolysis of Seawater Magnesium metal is obtained by the electrolysis of molten Mg2+ salts obtained from evaporated seawater. a. Would elemental Mg form at the cathode or anode? b. Do you think the principal ingredient in sea salt (NaCI) would need to be separated from the Mg2+ salts before electrolysis? Explain your answer c. Would electrolysis of an aqueous solution ofMgCl2 also produce elemental Mg? d. If your answer to part c was no, what would be products of electrolysis? e. How does the phenomenon of overpotentials influence your answers to parts c and d? *18.88. Silverware Tarnish Low concentrations of hydrogen sulfide in air react with silver to form Ag2S, more familiar to us as tarnish. Silver polish contains aluminum metal powder in an alkaline suspension. a. Write a balanced net ionic equation for the redox reaction between Ag2S and AI metal that produces Ag metal and AI(OH\. b. Calculate EO for the reaction. [Hint: Derive EO values for the half-reactions in which Ag2S is reduced to Ag metal and Al(OH\ is reduced to AI metal. You can do this by

LlHr.MnF3(;) = -1579 kJlmol

LlHr.KsbF6(;) = -2080

= -2435 kJlmol kJlmol

c. If we assume that LlS is relatively small, such that LlG = LlH, estimate EO for this reaction. d. If LlS for the reaction is greater than zero, is our value for EO in part c too high or too low? e. The electrochemical synthesis ofF2 is described by the electrolytic cell reaction

Assign oxidation numbers and determine the number of electrons involves in this process. 18.92. Corrosion of Copper Pipes Copper is frequently the material of choice for household plumbing. In some communities heavily chlorinated water corrodes copper pipes. The corrosion reaction is believed to involve the formation of copper(l) chloride: 2 Cu(s)

+

CI2(aq)

--7

2 CuCl(s)

a. Write balanced equations for the half-reactions redox reaction. b. Calculate and LlG;xn for the reaction.

E;xo

in this


Power, Pigments, and Phosphors: The Group 12 Elements 18.93. Write balanced chemical equations for the following processes. a. Zinc(II) sulfide reacts with oxygen. b. Zinc(II) oxide reacts with carbon. c. Mercury(II) sulfide reacts with oxygen. 18.94. Write balanced chemical equations for the following processes: a. Zinc reacts with oxygen. b. Cadmium reacts with sulfur. c. Zinc metal reacts with aqueous copper(II) nitrate. 18.95. Zinc(II) sulfide and cadmium(II) sulfide crystallize in two different structures, an hcp (hexagonal closest-packed) and a ccp (cubic closest-packed) arrangement of sulfide ions with the cations in tetrahedral holes. a. Do these structures have the same packing efficiency? b. Are the tetrahedral holes the same size in both structures? Explain your answer. 18.96. Zinc(II) oxide and zinc(II) sulfide (wurtzite) have the same structure, an fcc (face-centered cubic) arrangement of anions with zinc cations in tetrahedral holes. a. Are the tetrahedral holes in ZnO the same size as in ZnS? b. How many tetrahedral holes are in the unit cell?

E:

I

929

18.97. Calculate the en value of the following reactions. a. Zn(s) + Cd2+(aq) --7 Cd(s) + Zn2+(aq) b. Hg(£) + Zn2+(aq) --7 Zn(s) + Hg2+(aq) c. Cd(s) + Hg22+(aq) --7 Cd2+(aq) + 2 Hg(£) 18.98. Which of the reactions in Problem 18.97 are spontaneous? 18.99. Which of these elements is the best reducing agent: Zn, Cd, or Hg? 18.100. Which cation, Hg2+ or Hg,'+, is the better oxidizing agent? 18.101. Why is the chemistry of zinc similar to the chemistry of magnesium? 18.102. Predict the products of the following reactions: a. Zn(g) + CI2(g) --7 b. ZnC03(s) + heat--7 c. Zn(OH)/s) + heat --7 18.103. The aqueous solutions of many transition metal compounds are colored, but solutions of zinc compounds are colorless. Why are solutions of Zn2+(aq) colorless?


A LOOK AHEAD: The Long and the Short of It

19.1 Biomolecules: Building Blocks of Life

19.2 Chirality 19.3 Protein Composition 19.4 Protein Structure and Function

19.5 Carbohydrates 19.6 Lipids 19.7 Nucleotides and Nucleic Acids

DNA AND

RNA

Scanning electron

micrograph showing strands o/DNA and RNA

(red).

(blue)

A Look Ahead: The Long and the Short of It

A LOOK AHEAD: The Long and the Short of It Perhaps the most fundamental questions asked by humans involve the origin of life. How did life begin on Earth? And if we do find out how, will we ever understand why? We may never have a completely satisfying answer to these questions. In searching for answers, however, we have learned much about the chemical reactions among the molecules found in living systems. As variable as life is, ranging from relatively simple one-celled organisms to green plants and the animals that consume them, several general patterns emerge when we consider the processes that constitute life. For all their stunning diversity, most living systems consist of substances made from only about 40 or 50 small molecules. Huge variations become possible when these few starting materials combine to form larger molecules and polymers. Seemingly subtle differences in the structure of the monomer units or in the orientation of bonds between units have vast and far-reaching consequences when large numbers of small molecules combine. The variation in characteristics of the final products is as wide-ranging as life itself Problems in the chemical machinery that makes biomolecules and unnatural variations in structures may result in catastrophic diseases. Many life-threatening illnesses are linked to the body's producing proteins of shorter than normal lengths. It is estimated that up to 70% of inherited disorders, including cystic fibrosis, are caused by the production of proteins that are too short. For example, if the synthesis of a protein called dystrophin is affected, Duchenne's muscular dystrophy may result. Because the shorter proteins are malformed, they are unable to build muscle fibers of sufficient strength to function normally. Muscle tissue wastes away and death follows, all because a biopolymer was too short to provide the strength needed in muscle tissue. Knowledge of the molecular mechanism that causes the shortened proteins to be produced has led to a promising new drug treatment for muscular dystrophy. The drug enables the system of reactions producing dystrophin to bypass the faulty step in the process and produce proteins of normal length. The relationship between chemical structure and the function of molecules in living organisms is important for health and normal function. The ramifications of recognizing the connection between structure and function range from understanding how life works at the molecular level, to the discovery of new drugs to treat disease, and even to the search for new sources of energy to power the vehicles used in modern transportation systems. The four major classes of organic compounds in living organisms-proteins, carbohydrates, lipids, and nucleic acids-carry out similar functions in all of the forms of life in which they are found. In this chapter, we discuss the small molecules that build membranes, provide energy, and signal biological responses to stimuli, and the large molecules and biopolymers that provide structure to living organisms, transmit genetic information from generation to generation, store energy, and break down food to provide energy for building the molecules required by and for the machinery of life. Life is not immune or exempt from obeying the laws of thermodynamics, and we conclude with a brief discussion of the molecular system that enables the storage and use of energy by living cells.

I

931

932

I Chapter Nineteen

I

Biochemistry:

The Compounds

A biomolecule is an organic molecule present naturally in a living system. Amino acids are molecules that contain one amine group and one carboxylic acid group. In an a-amino acid, the two groups are attached to the same a carbon atom. Proteins are polymers acids.

made of amino

A monosaccharide is a simple sugar and the simplest carbohydrate.

A polysaccharide (complex carbohydrate) is a polymer of monosaccharides. Lipids are water insoluble, oily organic compounds that, along with carbohydrates and proteins, are common structural materials in cells. Triglycerides are lipids consisting of esters of glycerol and long-chain fatty acids. A nucleotide is a monomer unit from which DNA and RNA are made.

l/a-carbon

H2N-E:-'f7-COOH H General

structure

of an a-amino

@Xi)

acid.

CONNECTION In ChapterlO we defined esters as molecules with the general formula

R_!O

\

O-R'

of Life

.u:. Biomolecules:

Building Blocks of Life

The molecules responsible for the architecture and the machinery of a living system are all synthesized from simple small-molecule precursors: N2, 02' H20, and CO2, The actual reactions that create the diversity of structure and function in living systems are well beyond the scope of this book. In this chapter, however, we discuss the end results of the reactions of these small molecules in terms of four families of biomolecules, organic molecules found naturally in living systems. These four groups-proteins, carbohydrates, lipids, and nucleic acids-are the building blocks of cells and entire organisms. Considering the huge number of different molecules that characterize even the simplest cell, it is amazing that all are primarily built from a relatively small number of precursors. Only two factors are responsible for all the diversity of form and function among biomolecules: the identity of the small molecules from which the biomolecules are built, and the shape of the resultant biomolecules. First we look at a brief summary of common small molecules from which larger biomolecules are built. Then we discuss each class ofbiomolecule in terms of its structure and function.

Small-Molecule Subunits Experiments conducted by Harold Urey and Stanley Miller in the 1950s marked a milestone in experimental research on the origin of life. Research on this topic continues, and the question of how the small molecules we now recognize as the building blocks of biomolecules actually arose on prebiotic Earth is still of great interest. The first experiments of Urey and Miller demonstrated that simple amino acids, molecules containing one amine group and one carboxylic acid group, could form in the absence of life. Although the conditions the two scientists chose are now thought to be different than those present on the early Earth, the result still stands: combinations of simple inorganic molecules can react to produce organic molecules found in living systems. Amino acids are fundamental building blocks of proteins, which are polymers of amino acids. The amino acid shown in the marginal figure is called an a-amino acid, because the -NH2 and the -COOH groups are attached to the same carbon atom. This carbon atom is referred to as the a carbon. The second class of small molecules from which an array of larger molecules and polymers are made is sugars. About 16 monosaccharides-the name means "one sugar" and identifies the simplest carbohydrates-are the common monomer units found in nature, from which the majority of polysaccharides or complex carbohydrates are made. Glucose, which we have seen many times in this book as a major product of photosynthesis, is a monosaccharide, and cellulose (the major structural component in plants) is a very common polysaccharide. The third class of molecules, the lipids, differs from carbohydrates and proteins because its members are not polymers. Lipids are best described by their physical properties, as opposed to any common structural subunit: lipids are insoluble in water, soluble in nonpolar solvents, and oily to the touch. They are common structural materials in cells. Because of their insolubility in water, they are ideal components for cell membranes, which separate the aqueous solutions within cells from the aqueous environment outside cells. An important class of lipids called triglycerides consists of fatty acid esters formed between glycerol and long-chain fatty acids, as shown on the next page. Nucleotides are the monomeric units from which deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are constructed. This fourth class of small

19.1

I

Biomolecules:

Building Blocks of Life

I

Glucose

o n

Cellulose

o

CH-CH-CH

I

I

2

OH

I

OH

2

OH

HO

Glycerol

0 II

CH-O

I

Fatty acid

0

2

CH-O 0

I

CH2-O These models show the glucose units that make up cellulose (a polysaccharide) polarity solubility

and a type of lipid called a triglyceride.

of glucose molecules

accounts

in water. Some of this polarity

polymerizes

to form cellulose,

long nonpolar the triglyceride

The

Triglyceride

for the compound's is lost when glucose

which is insoluble

in water. The

chains of fatty acids make their esters, including shown here, insoluble

in water.

~

fNH2

) molecules is of great importance to life, even though it makes up only about 1% of an organism by weight and is much less abundant than proteins, carbohydrates, and lipids. Whereas the other three classes of molecules play many roles in cellular structure, function, and energy storage, nucleotides have more-defined roles: they direct the metabolic activity of all cells. DNA carries the genetic blueprint of an organism and a variety ofRNAs use the DNA blueprint to build proteins. Five different nucleotide units make up DNA and RNA. Each nucleotide unit is composed of a simple sugar that is bound to a phosphate group and an organic base.

P

-0- !-O-CH]

I

0-

'yN

OHH1

o

H

I

H H

°

SCH3

o H N

I

H

Enkephalin FIGURE P19.7

~H

OH ~

0

972 I Chapter Nineteen

19.8.

I


Regulating Blood Pressure Angiotensin II is a polypeptide that regulates blood pressure. The structure of angiotensin II is shown in Figure P19.8. Which amino acids are in the structure?

O~W=p N~NHµH '"

OH

~

1.&

o

OH

(a)

R

= ~~./"'-..

1\ /14

-

.

(b)R=\~

~\

(c)R=

~

\

/14

FIGURE P19.9 19.10. Cocoa Butter Cocoa butter (Figure P19.10) is a key ingredient in chocolate. Cocoa butter is a triglyceride that results from esterification of glycerol with three different fatty acids. Identify the fatty acids produced by hydrolysis of cocoa butter.

Angiotensin II FIGURE P19.8

19.9.

Trans Fats The role of "trans fats" in human health has

been extensively debated both in the scientific community and in the popular press. What type of isomerism does the word "trans fat" refer to? Which of the molecules in Figure P19.9 are considered trans fats?

Cocoa butter FIGURE P19.10

Questions and Problems Biomolecules: CONCEPT

19.11.

Building

Blocks of Life

REVIEW

19.17. Are racemic mixtures considered homogeneous

*19.18.

In living cells, small molecules combine to make much

larger molecules. Are these processes accompanied by increases or decreases in entropy of the molecules? 19.12. What types of intermolecular forces might act to keep several strands ofbiomolecules together in supra molecular assemblies? 19.13. Give an example of a biomolecule that is considered to be a small molecule. 19.14. Match the following building blocks in the left-hand column with the appropriate biopolymer in the right-hand column: Amino acid Carbohydrate DNA Fatty acid Monosaccharide Protein Nucleotide Lipid

19.19.

19.20.

19.21.

*19.22. 19.23.

19.24.

or heterogeneous mixtures? Could a racemic mixture be distinguished from an achiral compound based on optical activity? Explain your answer. Why is the amino acid glycine achiral? Can geometric isomers of molecules such as cis and trans RCH=CH2 also have optical isomers? (R may be any of the functional groups we have encountered in this textbook.) Explain your answer. Which type of hybrid orbitals on a carbon atom, sp, Sp2, or sp" can give rise to enantiomers? Could an oxygen atom in an alcohol, ketone, or ether ever be a chiral center in the molecule? Which of the following objects are chiral? (a) a golf club; (b) a tennis racket; (c) a glove; (d) a shoe Which of the following objects are chiral? (a) a key; (b) a screwdriver; (c) a light bulb; (d) a baseball

PROBLEMS

19.25. Which

Chirality CONCEPT

o

o

+OH OH

~OH

(a)

(b)

REVIEW

19.15. Can all of the terms enantiomer,

of the molecules in Figure P19.25 are chiral?

achiral, and optically active

be used to describe a single compound? Explain. 19.16. Two compounds have the same structure and the same physical properties but also have the same optical activity. Are they enantiomers or the same molecule?

(e)

FIGURE P19.25


1

973

19.26. Which, if any, of the molecules shown in Figure P19.26 contains a chiral center?

~

~

~

(a)

(b)

(c)

Terpinol

FIGURE P19.26

FIGURE P19.30

19.27. Artificial Sweeteners Artificial sweeteners are fundamental to the diet food industry. Figure P19.27 shows three artificial sweeteners that have been used in food. Saccharin is the oldest, dating to 1879. Cyclamates were banned in 1969 following research suggesting they led to tumors. Aspartame may be more familiar to you under the name Nutrasweet. Each of these sweeteners contain between zero and two chiral carbon atoms. Circle the chiral center in each compound.

o

Protein Composition; Protein Structure and Function CONCEPT

REVIEW

19.31. Meteorites contain more L-amino acids than D-amino acids, which are the forms that make up the proteins in our bodies. What is meant by D-amino acids? 19.32. Of the 20 amino acids necessary to sustain life, which one is not chira!? 19.33. Which of the compounds in Figure P19.33 is not an a-amino acid?

~NH ~S)

o

0 Cyclamate

Aspartame

(a)

0

~OH H2N

19.34. Which of the compounds in Figure P19.34 are a-amino acids?

o

~OH 0

(b)

(c)

FIGURE P19.33

19.28. Identify the chiral centers in each of the molecules in Figure P19.28. )--- 83) are radioactive. Because there is no stable reference point, it is hard to say whether one of these nuclides is neutron rich or neutron poor. The heaviest nuclides tend to undergo radioactive decay by either lY decay, emitting an lY particle (~He) and producing nuclides with two fewer protons and two fewer neutrons, or f3 decay. Figure 20.5 summarizes

FIGURE 20.5 predictable

Radioactive

decay results in

changes in the number of protons

and neutrons

in a nucleus. Alpha decay leads

to the release of 2 protons and 2 neutrons,

and

ll'

1015

4-.

o I-
500°C). When a wire of NiTi is bent into a shape at room temperature and then heated, the crystal structure changes to the high-temperature form, removing the bend.

STENTS

Tantalum is used to coat artificial hip joints.

HpUn

~ ZnO(s) +

21.5

(b)

(a)

illustrates

I 1055

how they work: (a) These wires are originally

two have been deformed. When the middle wire (b) and the wire on the left (c) are heated gently, they restore themselves

shape. Similarly, stents have one shape at room temperature

and assume a different shape when they are heated to body temperature.

The unusual shape memory property of different selected alloys leads to their use in stents to prop open weakened and clogged arteries. The stent is formed into a coil or tube that can be surgically inserted into a partially blocked artery. Once the stent is in place it changes back to its original shape as it is warmed by body heat. The stent in its "body temperature" shape expands and props open the artery. At least 15 shape memory alloys have been identified, ranging from the most commonly used nickel-titanium alloy to alloys containing hafnium, titanium, and nickel. Perhaps hafnium-containing shape memory alloys will join the other group 4 elements as useful materials for medical applications. If so, such alloys would represent one of the very few medical applications of hafnium.

_

Toxic Elements

(c)

Shape memory alloys are used to prepare stents for heart patients. This set of photographs S-shaped;

I

Toxic Elements

Our survey of the naturally occurring elements has left us with relatively few elements that have no biological function and that have limited, if any, applications to medicine. These elements are scattered throughout the periodic table. Some of them are radioactive: they include the four naturally occurring actinide elements, actinium (Ac) , thorium (Th), protactinium (Pa), and uranium (U) as well as francium (Fr), radium (Ra), polonium (Po), astatine (At), and radon (Rn). Only radium and radon are found in appreciable amounts in nature. They have no known biological function; neither have they found application in medicine. In fact, as described in Chapter 20, radon gas and radium in granite and soil derived from granite pose serious health hazards from (Y decay. As noted in Chapter 20, (Y particles may have a short penetration distance, but their large relative biological effectiveness means they can have a devastating effect, especially on lung tissue. In this section we will focus on the compounds of beryllium, cadmium, mercury, and lead. Thallium could have been included among the quintessential toxic elements; however, we have seen that thallium isotopes are also useful imaging agents. Often it is the form of an element and its concentration that determine its utility in therapy or diagnosis as opposed to its inherent toxic effects. For these reasons we will bypass the toxicity issues associated with TI and take a closer look at the toxicities of Be, Cd, Hg, and Pb.

Beryllium Beryllium, the first member of the alkaline earth metals in group 2, is present in Earth's crust at concentrations comparable to those of tin and arsenic. Exposure to beryllium compounds in our lives fortunately is quite limited. Beryllium

to their original

1056

I Chapter Twenty-One

I


toxicity is most often encountered in industrial settings where berylliumcontaminated dust is inhaled. The Be2+ ion replaces M~+ in the body where it inhibits M~+ -catalyzed RNA and DNA synthesis in cells. On the other hand, we value beryllium when we find it in the gemstones beryl and emerald.

Lead

Beryl is a gemstone that contains beryllium.

The history of lead toxicity dates back thousands of years. Lead compounds like PbS, PbC03, and PbCI(OH) were used by the ancient Egyptians in cosmetics. The Romans used lead pipes for plumbing and wine carafes and suffered from lead poisoning as a result. In the Middle Ages, lead acetate, Pb(CH3COO)2' was added to wine to sweeten its taste, a practice with dire consequences. In the 20th century, lead in the form of tetraethyllead, [(CH3 CH)4Pb], was used as a gasoline additive to help automobile engines run smoothly. The effects of lead on arresting the development of children has been well documented, and the use of lead in motor fuels in the United States was phased out by 1986. Similarly, paints containing lead(II) carbonate ("white lead") and lead(II) chromate ("chrome yellow") have been banned from residential use in the U.S. since 1978. The toxicity oflead is traced to its strong binding to the oxygen groups of many enzymes. This action affects protein synthesis, hemoglobin synthesis, and many other physiological processes.

Mercury

~ CONNECTION Thevaporpressures ofliquids are discussed in Chapter 10. Pressure units (Pa) are defined in Chapter 6.

The history of mercury poisoning is nearly as old as that of lead. In the classic story Alice in Wonderland, Lewis Carroll's "mad-hatter" character exhibits the symptoms of mercury poisoning among hat makers. Mercury(II) nitrate, Hg(NO)2' was used to make felt easier to work. Mercury(II) ion binds easily to sulfur-containing amino acids like cysteine and is readily transported throughout the body. Some of the effects of mercury poisoning include memory loss, tremors, and impaired coordination. Although we no longer use mercury compounds in hat manufacture, mercury was used in other ways throughout the 20th century. Dental amalgams and older thermometers are two familiar applications of mercury. Mercury spills from broken thermometers are difficult to clean up. The dense liquid finds its way into cracks and corners. Mercury has a small but significant vapor pressure, about 0.1 Pa at 20°e. This volatility ofHg means that it can enter the body by inhalation. In our bodies, mercury can be transformed into methylmercury(II), CH3Hg+, a potent neurotoxin. The methyl group increases the solubility of mercury in the nonpolar portion of cell membranes and allows the CH3Hg+ to diffuse more readily through the lipid bilayer shown in Figure 21.2. In particular, the membranes separating blood from the brain are vulnerable to penetration by CH3Hg+, which leads to the symptoms of mercury poisoning already cited. The conversion of mercury to methylmercury in the environment (biomethylation) is responsible for the health problems encountered by Japanese fishermen and their families around Minamata Bay in the 1950s. Mercury-containing waste from industrial plants was routinely dumped into Minamata Bay where the mercury was converted to methylmercury. Fish accumulated the methylmercury in their tissues. The local residents ate the fish and suffered the effects of mercury poisoning. Sample Exercise 21.6 examines the role of equilibrium in the treatment of mercury and other heavy-metal poisoning.

21.5

I

Toxic Elements

I 1057

Calculating the Equilibrium Constant of a Ligand Exchange Reaction Mercury(II) ions (H~+) in the aqueous environment can be concentrated in fish, so governments have issued warnings about the health effects of excessive consumption of tuna and other fish species. Methylmercury (CH3Hg+) binds to sulfurcontaining amino acids in proteins. For example, CH3Hg+ combines with cysteine as described by the following equation and its associated formation constant:

+

CH3Hg+

cysteine ~

K

CH3Hg(cysteine)+

N

CONNECTION In Chapter 17 we discussed formation constants for metal complexes. Formation constants are equilibrium constants that describe the formation of a metal-ligand complex.

[CH3Hg(cysteine)+ ] f,Hg-cyst [CH3Hg+ ] [cysteine]

= -::-::-:=-=-=---_-=---_=_

K f,Hg-cyst = 3.98 X 1016 Mercury poisoning can be treated by injecting a compound, such as penicillamine, that has a larger formation constant for reaction with mercury(II) than cysteine. CH3Hg+

+ penicillamine K

~

CH3Hg(penicillamine)+

[CH3Hg(penicillamine)+ ] f,Hg-pen [CH3Hg+ ][penicillamine] = :----'---:-::-----___=_

K f,Hg-pen = 6.31 X 1016 The reaction in which penicillamine CH3Hg(cysteine)+

replaces cysteine is

+ penicillamine

~


+ cysteine

What is the equilibrium constant for the replacement reaction? Hint: You may wish to refer back to Chapter 17 for calculations involving formation constants of complexes, as well as Chapters 15 and 16 where we manipulated equilibrium constants, as you work through this exercise. COLLECT AND ORGANIZE We are given the formation constants (Kf,Hg-cyst and Kf ,Hg-pen ) of two reactions involving methylmercury cations. These data allow us to calculate the equilibrium constant for the replacement reaction.

A N A L Y Z E The

formation constant for the replacement reaction CH3Hg(penicillamine)+ and cysteine is [cysteine] [CH3Hg(penicillamine) +]

K

that

yields

= .,,----,-----'--.,.--::-::------_=_

f,Hg-pen+cyst [CH Hg(cysteine)+] 3

[penicillamine]

The equilibrium constant expression for KfH +cyst has [cysteine] in the numerator • g-pcn and [CH3Hg(cysteine)+] in the denominator, whereas the expression for Kf(H -cyst)has [cysteine] in the denominator and [CH3Hg(cysteine)+] in the numerator. Cysteine appears as a product in the replacement reaction, so let's write an equilibrium constant expression for the reverse reaction of the first reaction so that cysteine is a product: CH3Hg(cysteine)+

~

CH3Hg+

+

cysteine

Kf,Hg+cyst= [CH3Hg+] [cysteine ]/[ CH3Hg(cysteine)

= l/K f,Hg-cyst= 1/(3.98 X 10

16

)

+]

= 2.51 X 10-17

The replacement equation is the summation of the reaction to form CH3Hg (penicillamine)+ and the reaction to dissociate CH3Hg(cysteine)+: CH,Hg+

+ penicillamine

~


+

CH3Hg(cysteine)+

~

CH3Hg+

+

cysteine

1058


I


+ penicillamine

CH3Hg(cysteine)+ and the equilibrium Kf ,rep 1acement

=

~

+ cysteine


constant for this reaction is

[CH3Hg(penicillamine)+]

[cysteine ]/[ CH3Hg(cysteine)

+] [penicillamine]

The equilibrium constant for the replacement reaction is the product of the equilibrium constants for the two reactions that are added together to get the replacement reaction. Thus the equilibrium constant for the replacement reaction is

K f,replacement

= K f,Hg-pcn =

X

(l/K f,Hg + cyst )

[CH3Hg(penicillamine)

+]

[CH3Hg+ ] [penicillamine]

[CH3Hg+ ] [cysteine]

X--------------[CH3Hg(cysteine)

+]

We can calculate the equilibrium constant for the replacement reaction that breaks down CH3Hg(cysteine)+ by substituting the values of Kf,Hg-pen and

SOLVE

Kf,Hg+ cyst'.

K [,replacement = (6.31 X 1016)(2.51 X 10-17) = 1.58 T In this exercise, we see that the equilibrium between a methylmercury(cysteine) complex and penicillamine favors the binding of mercury to the latter ligand, although the equilibrium constant is small. Treatment of a patient suffering from mercury poisoning with penicillamine will eventually lead to reduction in the amount of mercury in the patient's body because the CH3Hg(penicillamine)+ is eliminated in urine, thereby reducing the concentration of one of the products and shifting the equilibrium of the replacement reaction to the right. The solution to this exercise applies concepts from Chapters 15 and 17. Specifically, we have revisited how to write equilibrium constant expressions for forward and reverse reactions as well as how to write equilibrium constant expressions for sequential reactions. We also have reviewed the concept of formation constants for metal ion reactions with chelating Lewis bases.

T HI N K ABO UTI

Practice Exercise The value of Kf for the complexation with methionine,

another sulfur-containing

CH3Hg+

+

methionine

~

of methylmercury amino acid, is 2.5 X 107•

CH3Hg(methionine)+

By using the data for penicillamine-mercury complexes given Exercise 21.6, calculate the equilibrium constant for the reaction CH3Hg(methionine)+

+ penicillamine

~

ion


in Sample

+ methionine

Cadmium

N

CONNECTION The applications of the group 12 elements in electrochemistry and battery technology are discussed in Chapter 18.

Until the latter part of the 20th century, the toxicity of cadmium compounds was of minimal concern to humans. Cadmium is not particularly abundant; it is found as 0.2%-0.4% by mass of zinc-containing ores. The demand for zinc exposed workers at zinc smelters to cadmium. However, the development of nickel/cadmium batteries as well as the use of cadmium sulfide and cadmium selenide in yellow, orange, and red pigments exposed a much larger population to cadmium compounds. For example, nicad batteries thrown into a landfill have the potential of introducing Cd2+ ions into groundwater. Cadmium(II) ions have an ionic radius (95 pm) that is close to that of calcium ions (Ca2+, 100 pm). The similarity in size leads to cadmium accumulation and subsequent embrittlement of bones. Like mercury and lead, cadmium ions bind strongly to sulfur-containing proteins and can displace Zn2+ from zinc enzymes.

Problem-Solving Summary I 1059

Summary 21.1 Essential elements have a physiological function in the body. Some nonessential elements may promote growth or have other stimulatory effects. Essential elements can be further categorized as major, trace, or ultra trace elements depending on their concentrations in living organisms.

SECTION

21.2 Sodium and potassium ions are important in generating and transmitting nerve impulses and maintaining proper osmotic pressure in cells. Transport of Na+ and K+ across cell membranes involves diffusion, Na+ -K+ ion pumps, or selective transport through ion channels formed by groups of helical proteins. Magnesium is important in the transfer of phosphate groups between ATP and ADP and is also found in chlorophyll, the green pigment in plants that mediates photosynthesis. Calcium is a major component of teeth and bones. A prolonged deficiency of calcium can lead to osteoporosis, whereas high concentrations of calcium in the intracellular fluid contribute to muscle cramps. Mammalian bones are composite materials that consist of mixtures of compounds with different compositions and structures. Bones are primarily elastic protein fibers and calcium salts including the mineral hydroxyapatite, Cas(PO 4)3(OH), which is also the principal component of teeth. Chloride ion is the most abundant anion in the human body and is involved in many functions. To maintain electrical neutrality in a cell, transport of alkali metal cations is accompanied by transport of chloride anions. Chloride ions also playa role in the elimination of chloride channels are an underlying cause of CO2, Malfunctioning cystic fibrosis. SECTION

Rubidium is probably nonessential, yet it is retained by the body because of its similarity in size and chemistry to K+. Cesium has no known function although it can also substitute for K+ and so can interfere with potassium-dependent functions. Strontium and barium ions are incorporated into human bones by replacing SECTIO N 21.3

Ca2+ ions but have no known function in the human body. However, Sr2+ and Ba2+ are found in the exoskeletons of certain single-cell organisms. Amorphous silica, Si02, makes up the exoskeletons of diatoms. Silicon scavenges aluminum in mammals, and the absence of silicon in their diets stunts growth. Germanium is a nonessential element and barely detectable in the human body. Trace quantities of arsenic may promote growth in some mammals. Selenium is an essential ultratrace element found in the body; the amino acid selenocysteine is a constituent of proteins that act as catalysts for biochemical reactions. In therapeutic treatments, fluoride ion replaces hydroxide in hydroxyapatite in teeth and inhibits the formation of cavities. Bromine does not appear to be an essential element. The body concentrates iodide (1-) in the thyroid gland where the iodide is incorporated into the hormones that regulate energy production and use. Many of the enzymes and coenzymes in our bodies contain transition metal ions including those of V, Mo, W, Fe, Co, Ni, Cu, and Zn. Manganese is important in photosynthesis. Radioactive isotopes with short half-lives that emit low-energy 'Y rays are used in the diagnosis of disease. The selection of a radioisotope is also governed by the toxicity of the element and its daughter isotopes, the speed at which the imaging agent is eliminated from the body, and the cost and availability. The availability of many isotopes has allowed medical researchers to investigate potential imaging agents across much of the periodic table. Therapeutic agents contain metallic elements, including lithium, transition elements, and heavier main group elements. A useful diagnostic or therapeutic agent must be sufficiently soluble in blood to be delivered to the target. Radioisotopes are used to both diagnose disease and treat cancer.

SECTION 21.4

21.5 Some elements pose health risks because they are reactive, whereas others, such as Be, Pb, Cd, TI, and Hg, are highly toxic because of their chemical properties.

SECTION



SAMPLE

Calculating an acid

Relate the pH of a solution to the [H +] by the equation

21.1

the pH and [H30+] of

Writing half-reactions oxoanions

pH = -log[H+] involving

Identifying particles in nuclear reactions

Use oxidation numbers to determine the number of electrons gained or lost; balance charges with H+ ions in acids or OH- ions in base; add water molecules to balance the numbers ofH and 0 atoms. (Another approach, based on balancing the number of 0 and H atoms with water molecules and H+ ions and then balancing charges with electrons, is described in Chapter 4.)

21.2

Sum the subscripts and superscripts on both sides of the radiochemical equation. The difference in the superscripts equals the mass of the missing particle, and the difference in the subscripts equals the charge on the particle.

21.3

EXERCISES

1060

TYPE

I Life and the Periodic Table


OF

Calculating

PROBLEM

CONCEPTS

a remaining

SAMPLE

EXERCISES

21.4

In A, = -0.693t Ao tll2 where Ao and A, are the amounts of material present initially and at time t, respectively.

the spontaneity

E:"

of a

redox reaction

Calculating

EQUATIONS

Use the equation

quantity

of a radioisotope

Predicting

AND

Calculate the by summing the oxidation potentials of the oxidation and reduction half-reactions. If is (+), the reaction is spontaneous.

21.5

Combine the equilibrium constant expressions for the formation of the two complexes, after reversing the one that dissociates in the overall reaction, to generate an equivalent equilibrium constant expression for the overall reaction.

21.6

E:

" X

the equilibrium

constant of a ligand exchange reaction

Visual Problems For Problems 21.1 and 21.2 refer to the following blank periodic tables with arrows showing trends. 21.1.

Which figure best describes the periodic trend in monoatomic cation radii moving up or down a group or across a period in the periodic table? (Arrows point in the direction of increasing radii.)

21.2.

---

r--

~ ~

---

~ ,

---

Which figure best describes the periodic trend in monoatomic anion radii moving up or down a group or across a period in the periodic table? (Arrows point in the direction of increasing radii.)

-

r--

~ ~

-

,~ ,. I

I

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(b)

(a) ---

r--

~ --~ ---

---

r--

~ --~

~ ~

,~

,

(c)

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(d)

Visual Problems

21.3.

21. 7.

Which group highlighted in the periodic table in Figure P21.3 typically forms ions that have larger radii than the corresponding neutral atoms?

I 1061

Describe the molecular geometry around each germanium atom in the structure of the germanium compound shown in Figure P21.7.

o

0

II

/CH2, HOOC

II

/Ge,

/Ge,

CH2

0

/CH2, CH2

COOH

FIGURE P21.7 21. 8.

Selenocysteine can exist as two enantiomers (stereoisomers). Identify the atom in Figure P21.8 responsible for the two enantiomers.

21.9.

The austenite structure for NiTi is similar to the CsCl structure. Which crystal structure shown in Figure P21.9 is correct for the austenite form of the shape memory alloy NiTi?

FIGURE P21.3 21.4.

Which group highlighted in the periodic table in Figure P21.4 typically forms ions that have smaller radii than the corresponding neutral atoms?

FIGURE P21.4 21.5.

As we saw in Chapter 18, the free energy (ilG) of a reaction is related to the cell potential by the equation ilG = -nFE. In Figure P21.5, two solutions ofNa+ of different concentration are separated by a semipermeable membrane. Calculate ilG for the transport ofNa+ from the side with higher concentration to the side with lower concentration. Membrane

(b)

(a)

(c)

Ni Ti

i/

FIGURE P21.9 *21.10. The unit cell for the martensite form of the shape memory alloy NiTi is shown in Figure P21.10. How many equivalent Ti and Ni atoms are in the unit cell?

[Na+] = 150 mM [Na+]

=

10 mM

FIGURE P21.S 21.6.

Two solutions ofK+ are separated by a semipermeable membrane in Figure P21.6. Calculate ilG for the transport ofK+ from the side with lower concentration to the side with higher concentration. Membrane

i/

Ni Ti FIGURE P21.10

[K+] = 100 mM [K+] = 12 mM

FIGURE P21.6

1062

I

Chapter Twenty-One

I


Questions and Problems The Periodic Table of Life CONCEPT

REVIEW

21.11. What is the difference between an essential element and a nonessential element? 21.12. Are all essential elements also major elements? 21.13. What distinguishes (a) bulk, (b) trace, and (c) ultratrace elements? 21.14. Should essential trace elements also be considered to be stimulatory? PROBLEMS

21.15. Which ion is larger: K+ or Na+? 21.16. Which ion is larger: Cl- or I-? 21.17. The concentrations of very dilute solutions are sometimes expressed as parts per million (mass of solute/total mass of solution) X 106. Express the concentration of each of the following essential trace elements in ppm: a. copper, 110 mg in 70 kg b. zinc, 3.3 X 10-2 g/kg c. iodine, 0.043 gin 100 kg 21.18. In the human body, the concentrations of essential ultratrace elements are even lower than those of essential trace elements and therefore are sometimes expressed in parts per billion (ppb). Units of ppb are defined as (mass of solute/ total mass of solution) X 109• Express the concentrations of each of the following essential ultratrace elements in ppb: a. cobalt, 4.3 X 10-5 g/kg b. boron, 0.014 g/100 kg c. chromium, 5.0 mg170 kg 21.19. In the following pairs, which element is more abundant in the human body: (a) silicon or oxygen; (b) iron or oxygen; (c) carbon or aluminum? 21.20. In the following pairs, which element is more abundant in the human body: (a) H or Si; (b) Ca or Fe; (c) N or Cr?

Essential Major Elements CONCEPT

REVIEW

21.21. Ion Transport in Cells Describe three ways in which ions of essential major elements (such as Na+ and K+) enter and exit cells. 21.22. Which transport mechanism for ions requires ATP: diffusion, ion channels, or ion pumps? 21.23. Why is it difficult for ions to diffuse across cell membranes? 21.24. Why does Sr2+substitute for Ca2+ in bones? 21.25. Which alkali metal ion is Rb+ most likely to substitute for? 21.26. Why don't alkaline earth metal cations substitute for alkali metal cations in cases where the ionic radii are similar? *21.27. Why might nature have chosen calcium carbonate over calcium sulfate as the major exoskeleton material in shells? 21.28. Bromide (Br-) and fluoride (F") are found in the body as nonessential elements. Would you expect their concentrations to be more similar to essential bulk or essential ultratrace elements?

PROBLEMS

21.29. Osmotic Pressure of Red Blood Cells One of the functions of the alkali metal cations Na+ and K+ in cells is to maintain the cells' osmotic pressure. The concentration ofNaCI in red blood cells is approximately 11 mM. Calculate the osmotic pressure of this solution at body temperature (37°C). (Hint: see Equation 10.10.) 21.30. Calculate the osmotic pressure exerted by a 92 mM solution ofKCI in a red blood cell at body temperature (37T). (Hint: see Equation 10.10.) *21.31. Electrochemical Potentials across Cell Membranes Very different concentrations ofNa+ ion exist in red blood cells (11 mM) and the blood plasma (160 mM) surrounding those cells. Solutions with two different concentrations separated by a membrane constitute a concentration cell of the type described in Chapter 18. E = EO (-0.0592/n)log{[Na+tn/[Na+]pb,mJ Calculate the electrochemical potential created by the unequal concentrations ofNa+. 21.32. The concentration ofK+ in red blood cells is 92 mM, and the concentration ofK+ in plasma is 10 mM. Calculate the electrochemical potential created by the two concentrations ofK+.

21.33. If the transport ofK+ across a cell membrane requires 5 kJ/mol, how many moles of ATP must be hydrolyzed to provide the necessary energy? The hydrolysis of ATP is described by the equation t::..GO = -34.5 kJ *21.34. Removing excess Na+ from a cell by an ion pump requires energy. How many moles of ATP must be hydrolyzed to overcome a cell potential of -0.07 V? The hydrolysis of 1 mol of ATP provides 34.5 kJ of energy. 21.35. Plankton Exoskeletons Exoskeletons of planktonic Acantharia contain strontium sulfate. Calculate the solubility in mol/L ofSrS04 in water at 25T given that Ksp = 3.4 X 10-7• 21.36. The algae Closterium contain structures built from barium sulfate (barite). Calculate the solubility in mol/L of BaSO 4 in water at 25°C given that K,p = 1.1 X 10-10.

Essential Trace and Ultratrace Elements CONCEPT

REVIEW

21.37. What danger to human health is posed by 137CS(t1l2 = 30 yr)? 21.38. Why is 137CS(t1l2 = 30 yr) considered to be dangerous to human health when naturally occurring 4°K (t1/2 = 1.28 X 106 yr) is benign? 21.39. What is the origin of most of the 137CSand 90Srin our environment? 21.40. Why does fluorapatite resist acid better than hydroxyapatite if both are insoluble in water?

Questions and Problems I 1063 21.41. What is the function of enzymes? 21.42. Enzymes are large proteins. Are all proteins enzymes? *21.43. What effect does an enzyme have on the activation energy of a biochemical reaction? *21.44. Why might reductases also be described as reducing agents? *21.45. When a transition metal ion like Cu2+ is incorporated into a metalloenzyme, is the formation constant likely to be much greater than one (K > > 1) or much less than one (K < < I)? Cu2+

+

[metalloenzyme] K = -[--]-[--.-]Cu2+ proteIn

protein"'" metalloenzyme

*21.46. When transition metals bind to proteins to form enzymes, is the process likely to be endergonic (tJ.G > 0) or exergonic (tJ.G < OJ? PROBLEMS

21.47. What are the products of radioactive decay of 137CS?Write a balanced equation for the nuclear decay reaction. 21.48. Potassium-40 decays by three pathways: f3 decay, positron emission, and electron capture. Write balanced equations for each of these processes. 21.49. Calculate the pH of a 1.00 X 10-3 M solution of selenocysteine (pK 31 = 2.21, pK a2 = 5.43). 21.50. Calculate the pH of a 1.00 X 10-3 M solution of cysteine (pK = 1.7, pK = 8.3) Is selenocysteine a stronger acid than cysteine? .1}

*21.57. Some sources give the formula of hydroxyapatite as Ca!O(PO 4\(OH)2' If the Ksp of Ca5(PO 4)30H is 2.3 X 10-59,what is the Ksp ofCa1o(P04)6(OH)/ *21.58. The same sources mentioned in the previous exercise cite the formula of fluorapatite as Ca!O(P04\(F)2' If the Ksp or Ca5(P04)3F is 3.2 X 10-6 what is the Ksp ofCa1o(P04)6(F)/

°,

*21.59. The activation energy for the uncatalyzed decomposition of hydrogen peroxide at 20°C is 75.3 kJlmol. In the presence of the enzyme catalase, the activation energy is reduced to 29.3 kJlmol. By using the following form of the Arrhenius equation, RT In (k/k2) = E" - E", how much faster is the catalyzed reaction? *21.60. Enzymatic Activity of Urease Urease catalyzes the decomposition of urea to ammonia and carbon dioxide.

o II

+

r(aq)

"'"Ca5(PO 4)3(F)(s)

+

OH-(aq)

*21.52. Effects of Excess Fluoridation on Teeth Too much fluoride might lead to the formation of calcium fluoride according to the reaction

Write the equilibrium constant expression for the reaction. Given the Ksp values for the following two reactions, calculate K for the reaction between Ca5(PO 4)30H and fluoride ion that forms CaF2. Ca5(PO 4)3(OH)(s) "'"5 Ca2+(aq)

+

3 PO /-(aq) + OH-(aq) K sp = 2.3 X 10-59 Ksp = 3.9 X 10-11

21.53. Tooth enamel is actually a composite material containing both hydroxyapatite and a calcium phosphate, CaS(HP04)/PO)4' 6Hp (Ksp = 1.1 X 10-47). Is this calcium mineral more or less soluble than hydroxyapatite (Ksp = 2.3 X 1O-59)? 21.54. The Ksp of actual tooth enamel is reported to be 1 X 10-58. Does this mean that tooth enamel is more soluble than pure hydroxyapatite (Ksp = 2.3 X 10-6°) ? Does the measured value of Ksp for tooth enamel support the idea that tooth enamel is a mixture of hydroxyapatite, Ca5(P04\OH, and a calcium phosphate Cas(HPO 4)/PO) 4'6 Hp (Ksp = 1.1 X 1O-47)?

(aq)

,/C,

.12

21.51. Composition of Tooth Enamel Tooth enamel contains the mineral hydroxyapatite. Hydroxyapatite reacts with fluoride ion in toothpaste to form fluorapatite. The equilibrium constant for the reaction between hydroxyapatite and fluoride ion is K = 8.48. Write the equilibrium constant expression for the following reaction. In which direction does the equilibrium lie? Ca5(PO 4)3(OH)(s)

21.55. Calculate the solubility in mol/L of hydroxyapatite, Ca5(PO 4)30H, Ksp = 2.3 X 10-59,and fluorapatite, Ca5(P04\F, Ksp = 3.2 X 10-60, in water at 25°C and pH = 7.0. 21.56. Calculate the solubility in mol/L of hydroxyapatite, Ca5(P04)30H, Ksp = 2.3 X 10-59, in water at 25°C and pH = 5.0.

H2N

+ H20(C)

-7 2 NH3(aq)

+ COig)

NH2

The rate constant for the uncatalyzed reaction at 20°C and pH 8 is k = 3 X 10-10 S-I. A urease isolated from the jack bean increases the rate constant to k = 3 X 104 S-I. By using the RT In (k/k) = Eo, - Eo, form of the Arrhenius equation, calculate the ratio of the activation energies.

Diagnosis and Therapy CONCEPT

REVIEW

21.61. List some of the considerations in choosing a radioisotope for imaging. 21.62. What advantages does an isotope like 99mTchave over 201TI for imaging the circulatory system? 21.63. Why might an a emitter be a good choice for chemotherapy? *21.64. What advantage might a f3 emitter have over an a emitter for imaging? 21.65. Gadolinium-153 decays by electron capture. What type of radiation does 153Gdproduce that makes it useful for imaging? 21.66. Gadolinium-153 and samarium-153 both have the same mass number. Why might !53Gd decay by electron capture whereas 153Smdecays by emitting f3 particles? 21.67. How do platinum- and ruthenium-containing drugs fight cancer? *21.68. Many transition metal complexes are brightly colored. Why might the titanium(IV) compound budotitane be colorless? (Hint: see Chapter 17.) *21.69. Is the glucose tolerance factor that contains chromium(III) paramagnetic or diamagnetic? (Hint: see Chapter 17.) 21.70. Why might lithium be an attractive anode material in batteries used in pacemakers? 21.71. Mercury compounds are generally toxic to humans. Why can we use mercury in dental amalgams? 21. 72. Tantalum is used in artificial joints. What property ofTa makes it attractive for this purpose?

1064

I

I

Chapter Twenty-One


Toxic Elements

PROBLEMS

21.73. The lanthanide isotopes cerium-141, terbium-160, thulium-170, and lutetium-177 all undergo f3 decay. Write a balanced nuclear equation for each decay reaction. Are all of the product isotopes in the same group of the periodic table as the reactants? 21.74. The lanthanide isotopes gadolinium-153 and ytterbium-169 decay by electron capture. Write a balanced nuclear equation for each decay reaction. Are both product isotopes in the same group of the periodic table? 21. 75. PET Imaging with Gallium A patient is injected with a 5 µM solution of gallium citrate containing 68Ga (tl12 = 9.4 hr) for a PET study. How long is it before the activity of the 68Ga drops to 5% of its initial value? 21.76. Indium-ll1 (t1/2 = 2.81 d) has been used in imaging. How long is it before the activity of lllIn drops to 15% ofits initial value? 21. 77. The bismuth in Pepto-Bismol™ and Kaopectate™ is found as BiO+. Draw the Lewis structure for the BiO+ cation. 21.78. Camcolit™ used in treating depression contains lithium carbonate. Draw the Lewis structure for Li2COJ. 21.79. Draw the Lewis structure for the citrate ion by using the skeletal drawing in Figure P21.79 as a guide. -OOC"

/OH C

H C/ 2

FIGURE P21.79

I

COO-

REVIEW

21.85. Why is methylmercury, CHJHg+, more toxic than mercury metal? 21.86. Why is Cd2+ more likely than Cr2+to replace Zn2+ in an enzyme like carbonic anhydrase in glucose tolerance factor? 21.87. Why is Be2+ more likely than Ca2+to displace Mg2+ in biomolecules? 21.88. PbS, PbCOJ, and PbCI(OH) have limited solubility in water. Which of them is/are more likely to dissolve in acidic solutions? PROBLEMS

21.89. Draw the Lewis structure for tetraethyllead, Pb(CH2CH)4' and determine its molecular geometry. 21.90. Draw the Lewis structure for the ionic compound mercury(II) nitrate. 21.91. The complexation of mercury(II) ion with methionine Hg2+ + methionine;=, Hg(methionine)2+ has a formation constant oflogK = 14.2, whereas the formation constant for the Hg2+ complex with penicillamine Hg2+ + penicillamine;=, Hg(penicillamine)2+ is 10gK = 16.3. Calculate the equilibrium constant for the reaction Hg(methionine)2+ + penicillamine ;=' Hg(penicillamine)2+ + methionine

"CH

I

CONCEPT

2

COO-

21. 80. Draw the Lewis structure for the thiosulfate ion, S2032-, which is found in the gold-bearing arthritis drug Sanochrysin. 21.81. Aluminum hydroxide is used in some antacids. Write a balanced net ionic equation for the reaction of aluminum hydroxide with HCl. 21.82. Aluminum carbonate is used in some antacids. Write a balanced net ionic equation for the reaction of aluminum carbonate with hydrochloric acid.

21.92. The complexation of mercury(II) ion with cysteine in aqueous solution Hg2+ + cysteine;=, Hg(cysteine)2+ has a formation constant oflogK = 14.2, whereas the formation constant for the Hg2+ complex with glycine Hg2+ + glycine;=, Hg(glycine)2+ is logK = 10.3. Calculate the equilibrium constant for the reaction Hg(cysteine)2+ + glycine;=, Hg(glycine)2+ + cysteine

21.83. A silverlzinc (Ag/Zn) battery has the advantage of being mercury-free. The half-reactions are as follows: ZnO(s) + HP(f)

+ 2 e-

-?

Zn(s) + 20H-(aq)

E = +0.342 V

AgP(s) + HP(f)

+ 2 e-

-?

2 Ag(s) + 20H-(aq)

E = -1.258 V O

What is the overall reaction of an electrochemical cell based on these materials? Using the EO values provided, determine the standard cell potential of the Ag/ZnO battery. 21.84. Power Sources for Pacemakers Another power source considered for pacemakers is the lithiumlcopper sulfide battery. Use the half-reactions shown here to determine the overall reaction for an electrochemical cell based on these materials. With the EO values provided, calculate the cell potential of the Li/CuS battery. Li+(s) CuS(s)

+ +

e-

-?

Li(s)

2 e-

-?

Cu(s)

r = +

S2-(aq)

21. 93. The equilibrium constant of the reaction

O

-3.05 V

EO = -0.851 V

CHJHg(penicillamine)+(aq) + cysteine(aq) ;=' CHJHg(cysteine)+(aq) + penicillamine(aq) is K = 0.633. Calculate the equilibrium concentrations of cysteine and penicillamine if we start with a 1.00 mM solution of cysteine and a 1.00 mM solution of CHJHg(penicillamine) +. 21.94. The equilibrium constant of the reaction in aqueous solution CHJHg(glutathione)+(aq) + cysteine(aq) CHJHg(cysteine)+(aq)

;='

+ glutathione(aq)

is K = 5.0. Calculate the equilibrium concentrations of cysteine and glutathione if we start with a 1.20 mM solution of cysteine and a 1.20 mM solution of CHJHg(glutathione)+ .

Mathematical

Procedures

WORKING

NOTATION

Qyantities

WITH SCIENTIFIC

that scientists work with often are very large, such as Earth's

mass of an electron. It is easier to work with these numbers The general

form of standard

raised to an integral standard

scientific

rameters

in standard

1. To convert

scientific

power. According

mass, or very small, such as the

if they are expressed in scientific notation.

is a value between

to this definition,

scientific

1 and 10 multipled

598 X 1022 kg (Earth's

but 5.98 X 1024 kg is. It is good practice

notation,

an "ordinary"

to use and report

number

to standard

scientific

notation,

(5,517 kg/m3) in scientific

requires

moving

the decimal

number

by 1000, or 103. To keep the value the same we add an exponent

problems

average density

then the exponent

is a negative

is 0.08206

2. To add or subtract

L . atom/(mol

Exercise

notation:

1 Calculate

in scientific

point two places to the right . K).

X 10-2 L . atom/(mol

notation,

their exponents

(and thus the decimal

when using a scientific

though

it could be writ-

must be the same. This

point position)

calculator

of some val-

because the calculator

the sum of the masses of the subatomic

particles

in an atom of

masses in grams: Particle

Mass (g)

proton

1.67263 X 10-24

neutron

1.67494 X 10-24

electron

9.10939 X 10-28

An atom of7Li has three protons,

total mass of the subatomic and an electron

the decimal

8.206

conversions.)

given the following

Solution

. K). Moving

the exponents

ues. (NOTE: This step is not necessary will make all necessary

of places you moved

the value of R used in solving ideal gas law

K).

numbers

may require you to change

the it by

is 5.517 X 103 kg/m3.

integer equal to the number

point to the right. For example,

ten 8.314 X 10° JI(mol·

Sample

notation

value of R, 8.314 JI(mol . K), does not need an exponent,

C. Another

lithium-7,

scientific

to multiply

point of a value less than one to the right to express it in scientific

the value of R to scientific

converts

notation

three places to the left. Doing so is the same as dividing

density in standard

B. If you move the decimal the decimal

point

point is located

digit.

to express Earth's

notation,

data or pa-

move the decimal

or to right for a small one, so that the decimal

A. For example,

103. So, Earth's

by 10

mass) is not in

notation.

to the left for a large number, after the first nonzero

notation

Appendix 1

particles

three electrons,

and four neutrons.

Therefore,

the

in an atom is the sum of three times the masses of a proton

and four times the mass of a neutron.

Doing

the multiplication

steps first yields

the following: mass of three protons mass of four protons

= 3 (1.67263

=

X 10-24 g) = 5.01789

4 (1.67494 X 10-24 g)

=

6.69976

X 10-24 g X 10-24 g

mass of three electrons = 3 (9.10939 X 10-28 g) = 2.732817 X 10-27 g Before adding convenient shifting

these masses together,

we must express them all with the same exponent.

is 10-24• To express the combined

the decimal

before the exponent

point in 2.732817 1000 times smaller

mass of three electrons

The most

using this exponent

requires

X 10-27 g three places to the left, which makes the number and compensates

for the exponent

becoming

1000 times

A-1

A-2

I

Appendix

larger. Adding

I

1

Mathematical

the resulting

Procedures

0.002732817 X

mass of the three electrons,

5.01789

X 10-24

+ 6.69976

g

Sample Solution

X 10-24 g

right of the decimal place, because that is the smallest number of digits to the right in two of the three values being summed. Therefore, the final answer is

to the base 10 of 100 (log 100).

Find the logarithm

values with exponents, together,

Sample Solution

Sample

the values in front of the but the exponents

automatically

Enter 100 into your calculator and press the "LOG"

button.

because 10 (the

Exercise

of 4.5.

6 Find the natural logarithm

Enter 4.5 into your calculator and press the "LN" button.

The answer should be 1.504.

11.72038 X 10-24 g

(these steps happen

5

base) raised to the power of the log (2) = 102 = 100.

We need to round off this sum so that we have only five digits to the

are multiplied

Exercise

button.'

figures).

The answer should be 2. This answer is as expected,

11.720382817 X 10-24 g

exponents

Enter 4.5 into your calculator and press the "LOG"

X 10-24 g

+ 0.002732817

3. To multiply

Solution

The answer should be 0.6532 (to four significant

10-24 g, to the other two masses, we have:

with scientific

7 Find the natural logarithm of 100.

Exercise

are added

calculators).

Solution

Enter 100 into your calculator and press the "LN" button.

The answer should be 4.61. Sample

Exercise

2 American

Steve Fossett circumnavigated

of Freedom

in early summer 2002 in the Spirit

the globe

balloon, which was par-

tially filled with 5.5 X 105 ft3 of helium. What is this volume in liters?

Let's compare the results of the four previous exercises. In both pairs of In and log values, the In value is 2.303 times the log value. These examples fit the general equation:

Given: 1 ft3 = 28.3 L Solution

In x = 2.30 log x

We convert the starting value by multiplying

(1

adding the exponents

(5.5

that the In of a value is greater that the log of the same

smaller base ofln units means that there are more of them than log units

1

=

4. To divide values with exponents,

15.5

X

106 L or 1.6

X

10-7 L

in a given value.

the values in front of the expo-

nents are divided, but the exponents steps happen automatically

It is reasonable

value because In is based on e (2.718), whereas log is to the base 10. The

105 w) (2.83 ~ 10 L)

X

5.5 by 2.83 and

+ 5):

are subtracted

Sample

8 Calculate

Exercise

In 1.2 X 10-3.

(again, these

with scientific calculators).

Solution

Enter 1.2 into your calculator, and then press the "EXP"

button, and then enter 3 and "+1-". Finally press "LN." The correSample

3 The speed oflight

Exercise

is 2.998 X 108 m/s. What is the

speed in miles per second? Given: 1 mile = 1.609 X 103 m

equivalent

sponding

keystrokes with a graphing

"(- )," 3, "ENTER."

calculator are "LN," 1.2, "x," "f\,"

The result should be -6.725,

which is negative

because the original number is less than 1. Keep in mind that values Solution dividing

Expressing

the speed of light in miles per second requires

the speed oflight

have negative logarithm

in meters per second by the conversion

factor given in the exercise. We divide the values in front of the exponents (2.998/1.609)

and subtract

to get their exponents

2.998 X 108 ffl/s 1.609 X 103 ffllmi

WORKING A logarithm

greater than zero have positive logarithm

(8 - 3)

5. Combining

logs: The following equations

rithms of the products or quotients to the individual

= 1.863 X 105 mils

values; those less than zero

values. summarize

how loga-

of two or more values are related

logs of those values.

logarithm

ab = logarithm

a

+ logarithm

b

logarithm

alb = logarithm

a - logarithm

b

and

WITH LOGARITHMS to the base 10 has the following form: log1Qx = log x = p, where x = lOP

We usually abbreviate the logarithm

function "log" if the logarithm

is to

If we know

the base 10, which means the scale in which the log 10 = 1. A logarithm

to the base e, called a natural logarithm,

has the fol-

LOGARITHMS

the value of log x, what

tion frequently

arises when working

is the negative log of the concentration

lowing form: log, x = In x = Scientific calculators

q, where

x = eq

have "log" and "In" buttons,

a number into its log or In form. The directions graphing

CONVERTING

so it is easy to convert

below apply to most non-

calculators.

INTO NUMBERS is the value of x? This ques-

with pH (see Chapter

16), which

of hydrogen ions, [H+], in solution:

pH = -log[W] Suppose the pH of a solution of a weak acid is 2.50. The concentration H+ is related to this pH value as follows:

2.50 = -log[H+] Sample

Exercise

4 Find the logarithm to the base 10 of 4.5 (log 4.5).

or

-2.50 alf you have a graphing press the "ENTER"

= log[W]

calculator such as a TI 83, press the "LOG"button,

button.

enter 4.5, and then

of

I

Appendix 1

To find the value of [H+], enter 2.5 into your calculator

and press the

where A is a constant for a particular

"+1-" button to change the value's sign to -2.5. The next step depends on the type of calculator you have. If yours has a "lOx" button, push it

tor), E, is the activation

to find the value of 10-2.5, which is the number we are looking for. The

equation gives

corresponding "ENTER."

Taking

Procedures

I

A-3

reaction (called the frequency facand R is the ideal gas

energy of the reaction,

the natural logarithms

of both sides of the Arrhenius

keystrokes with a graphing calculator are "10'','' "(- )," 2.5, there is no "lOx" key, but there is an

On some calculators

inverse function, Hitting

constant.

Mathematical

or "INV" key, that is used to invert other function keys.

the "INV" key followed by the "log" key takes the inverse of a

log, called an antilog, which is the same as raising 10 to the power (-2.5 in this case) that was entered. Some calculators, including the virtual one in many Windows

operating systems, have an "xl''' key. To use it you enter

10, push the "xy" key, enter 2.5, and then push the "+1-" key followed by the equals sign. All of these approaches 10 to the -2.50

do the same calculation,

taking

power, and give the same answer, [H+] = 3.2 X 10-

3•

In k = InA - (:~) This equation fits the general equation (In k) is the y-variable (liT)

Calculate the hydrogen ion concentration

(y

of a straight line

is the x-variable.

Plotting

should give a straight line with a slope equal to -E/R.

b) if

= mx +

(In k) versus The slopes

of these plots are negative because the activation energies, Ex' of chemical reactions are positive. The data for a reaction given in columns 2 and 4 of Table A1.1 are plotted in Figure Al.l.

The slope of the straight line

K) is used to calculate the value of E,,:

(-1281

-1281 Sample Exercise 9

and (liT)

in rain-

K= _Ex R

water in which pH = 5.62.

Ex = -(-1281 If you use one of the methods described above, you should find that the value of10-562 is 2.4 X 10-6•

(*)[8.31 ]/(moH~)]

Solution

SOLVING QUADRATIC

EQUATIONS

If the terms in an equation can be rearranged

so that they take the form

ax

they have the form of a quadratic

equation.

The value(s) of x can be

from the values of the coefficients a, b, and c by using the

equation

-b ± -vb2 x=-----2a

4ac

-

For example, if the solution to a problem yields the following expression where x is the concentration

of a solute:

+ 0.112x

X2

- 1.2

Then the value of x can be determined

X 10-3 =

0

-b ± -vb2 - 4ac -2a ± ~(0.112)2 - 4(1)(-1.2

= +0.010

500

0.0020

0.030

-3.5

0.0018

0.38

-0.97

600

0.0017

650

0.0015

17

2.8

700

0.0014

75

4.3

1

values; there-

fore we use only the +0.010 value.

DATA IN GRAPHICAL

curves to plots of experimental

ena obey exponential

(T). This relationship

(see Chapter

tool in de-

Many natural

For example, the rate constant

chemical reaction increases exponentially perature

-1

FORM

data is a powerful

between variables.

functions.

--

y

2

or -0.122

with increasing

is described

phenom-

by the Arrhenius

-2

(k) of a

absolute tem-

-3

equation

-4

14): k = Ae-E/RT

In k

550

3

no meaning because we cannot have negative concentration

the relationships

Rate Constant k

2.9

-

--

4

In this example, the negative value for x satisfies the equation, but it has

termining

(K-1)

X 10-3)

± -V0.01254 + 0.0048 2

--2 ± 0.132

Fitting

l/T

In k

-0.112

EXPRESSING

T (K)

T

5 2(1)

-0.112

Temperature

Rate Constant k as a Function of Temperature

as follows:

x= --

-0.112

TABLE Al.l

+ bx + c = 0

2

determined

= 10,645 ]/mol = 10.6 kJ/mol

FIGURE

Al.1

=

-1281x + 22.279

1.1

-

Appendix 2

SI Units and Conversion Factors SI Base Units

TABLE A2.1

I

SI Base Quantity

Unit

Symbol

length

meter

m

mass

kilogram

kg

second

s

mole

mol

kelvin

K

ampere

A

f--

time amount of substance r-temperature electric current

I I

, I

I SI-Derived Quantity r-electric charge

Unit

Symbol

--

Dimensions

coulomb

C

A-s

volt

V

llC

newton

N

kg'm/s2

hertz

Hz

S-I

momentum

newton-second

-

kg'm/s

power

watt

W

lis

pascal

Pa

N/m

Bq

S-I

meter per second

-

m/s

joule (newton-meter)

]

kg'm2/s2

electric potential r-force

--

frequency f--

radioactivity r-speed or velocity

becquerel

--

---

-

f--

pressure

-

-

f--

work, energy, or heat

TABLE A2.3 Prefix

SI Prefixes Multiplier

Symbol

f--

Multiplier

Symbol -

deci

d

10-1

deka

da

101

centi

c

10-2

hecto

h

102

m

10-

kilo

k

103

f--

micro r-nano

-

/-t

---

3

10-6

n

10-9

pico

P

femto at to

-

mega

--

M

--

106

giga

G

109

10-12

tera

T

1012

f

10-

peta

P

lOIS

a

10-18

exa

E

1018

f--

A-4

Prefix

-

milli

-

15

-

-

Appendix 2

TABLE A2.4 Quantity

Unit

Symbol

Conversion

-

electron-volt

eV

mass

pound

lb

heat

calorie

cal

length

angstrom

A

length

inch

in

mass

atomic mass unit

amu

atmosphere

atm

-

-

pressure

torr

temperature

Celsius scale

temperature

Fahrenheit scale

-

= 1 lb = 1 cal = 1A = 1 eV

453.592 g 4.184] 10-8 cm

=

10-10 m

= 2.54cm 1 amu = 1.6605402 -

torr

-

°C

min

time

hour

hr

time

day

d

1 torr

of

-

X 10-27 kg

1 atm = 1.01325 X 105 Pa

°C

OF

minute

-

= 1 atm/760 = K - 273.15 = 2(0C) + 32 5

= 60 s 1 hr = 60 min = 3600 s 1 d = 24 hr = 86,400 s 1 yr = 365.25 d 1 L = 1 dm3 = 10-3 m3 1 cm3 = 1 mL = 10-3 L 1 gal = 3.7854 L 1 min

time

year

yr

volume

liter

L

volume

cubic centi meter

em3, cc

volume

gallon

gal

TABLE A2.5

1.60217733 X 10-19]

1 in

time

-

-

Physical Constants

Quantity

r

-

energy

pressure

I

Special Units and Conversion Factors

Symbol

Value

acceleration due to gravity (Earth)

g

Avogadro's number

NA

Bohr radius

ao

5.29177249 X 10-11 m

Boltzmann's constant

kn

1.380658 X 1O-23]/K

electronic charge-to-mass ratio

-elm

elementary charge

e

1.60217733 X 10-19 C

Faraday's constant

F

9.6485309 X 104 C/mol

mass of an electron

m,

9.10939 X 10-28 g

mass of a neutron

m

1.67494 X 10-24 g

mass of a proton

m

permittivity of vacuum

9.80665 m/s2 (exact) 6.0221367 X 1023mol-I

-

1.75881962 X lOll C/kg

,

-

1.67263 X 10-24 g 22.41410 Llmol 8.854187817 X 10-12 C2/(N' m2)

h

speed oflight in vacuum

I universal gas constant

-

"

-L

molar volume of ideal gas at STP

Planck's constant

-

6.6260755 X 10-34]. s 2.99792458 X 108 m/s (exactJ

R

8.3145 ]I(mol' K) 0.082058 L· atm/(mol' K)

I

SI Units and Conversion Factors

I

A-5

Appendix 3

The Elements and Their Properties Ground-State Electron Configurations, Atomic Radii, and first Ionization Energies of the Elements

TABLE A3.1 I

Element

Atomic Number Z

Symbol

f--

hydrogen r---:helium lithium

beryllium boron

37

1312.0

He

2

1s2

32

2372.3

3

[He]2s' [He]2s2

---

4 5

carbon

C

nitrogen

N

oxygen

0

fluorine

F

neon

Ne

--

f--

-

152

--

112

-

520.2 899.5

88

6

[He]2s22pJ [He]2s22p2

77

1086.5

7

[He]2s22p3

75

1402.3

8

[He]2s22p4

73

1313.9

9

[He]2s22p5

71

1681.0

10

[He]2s22p6

----

69

-

2080.7

sodium

Na

11

magnesium

Mg

12

[Ne]3J2

160

aluminum

Al

13

577.5

Si

14

[Ne]3s 3pJ [Ne]3s23p2

143

silicon

117

786.5

15

[Ne]3s23p3

16

[Ne]3s23p4

phosphorus

-

f-"--

sulfur f--

--

P S

--

2

chlorine

Cl

17

[Ne]3s23p5

argon

Ar

18

[Ne]3s23p6

potassium

K

19

r-=--

calcium

Ca

scandium

Sc

-

418.8

20

197

589.8

21

[Ar]4s23d'

162

633.1

Cr

24

Mn Fe

iron cobalt

f--

nickel f--

--

--

Co Ni

-

---

147

-

658.8

135

650.9

[Ar]4s'3d5

128

652.9

25

[Ar]4s23d5

127

717.3

26

[Ar]4s23d6

126

762.5

27

[Ar]4s23d7

28

[Ar]4s23d8

copper

Cu

29

[Ar]4s'3dlO

zinc

Zn

30

gallium

Ga

31

[Ar]4s23dlO [Ar]4s23dI04pl

c....=:.

999.6

227

chromium

~nganese

-

1011.8

[Ar]4s'

[Ar]4s23d3

f--

-

[Ar]4s2

23

-

110 103

1520.6

22

Ti

737.7

97

V

-

495.3

1251.2

vanadium

titanium

--

186

99

[Ar]4s23d2

f--

--

---

125 124

-

-

800.6

[Ne]3s'

f--

A-6

lsI

B

f--

Ionization Energy (kJ/mol)

1

Be

-

f--

Atomic Radius (pm)

H

Li

-

f--

Ground-State Configuration

-

760.4 737.1

128

745.5

134

906.4

135

578.8

-

-

-

-

-

-

Appendix 3

I

Ground-State Electron Configurations, Atomic Radii, and first Ionization Energies of the Elements (Continued)

TABLE A3.1

Atomic Number Element

Z

Symbol

Atomic Radius (pm)


Ionization Energy (kJjmol)

germanium

Ge

32

[Ar]4s23d104p2

122

762.2

arsenic

As

33

[Ar]4s23d104p3

121

947.0

34

[Ar]4s23d104p4

35

[Ar]4s23d104p5

selenium

--

bromine

--

Se Br

-

krypton

Kr

36

[Ar]4s23d104p6

rubidium

Rb

37

403.0

215

549.5

180

599.8

160

640.1

Sr

38 39

[Kr]5s24dl

zirconium

Zr

40

[Kr]5s24d2

niobium

Nb

41

[Kr]5s14d4

molybdenum

Mo

42

[Kr]5s14d5

technetium

Tc

43

ruthenium

Ru

136

702.4

[Kr]5s14d7

134

710.2

134

719.7

137

804.4

Rh

45 46

[Kr]4dIO

47

[Kr]5s14d1o

--

Cd

48

[Kr]5s24dlO

indium

In

49

tin

Sn

50

[Kr]5s24d105pI [Kr]5s24d105p2

antimony

Sb

51

[Kr]5s24dI05p3

52 53

[Kr]5s24d105p4 [Kr]5s24d105p5

iodine

--

Te I

-

xenon

Xe

54

[Kr]5s24dI05p6

cesium

Cs

55

barium

Ba

144

--

cadmium

tellurium

652.1

44

Pd

-

-

[Kr]5s24d5

rhodium

-

146

--

684.3

palladium

Ag

--

139

[Kr]5s14d8

-

-

941.0 1139.9

247

Y

-

-

[Kr]5s1

strontium

silver

114

1350.8

yttrium

-

-

119

110

[Kr]5s2

-

-

-

-

-

731.0

151

867.8

167

558.3

140

708.6

141

833.6

143 133

-

869.3 1008.4

130

1170.4

[Xe]6s1

265

375.7

56

[Xe]6s2

222

502.9

187

538.1

182

534.4

lanthanum

La

57

[Xe]6s25dl

cerium

Ce

58

[Xe]6s2~r5dl

praseodymium

Pr

59

[Xe]6s2'if3

neodymium

Nd

60

promethium

Pm

61

[Xe]6s2'if4 [Xe]6s2'if5

183

535.5

samarium

Sm

62

[Xe]6s2'if6

180

544.5

europium

Eu

63

[Xe]6s2'if7

208

547.1

gadolinium

Gd

64

180

593.4

65

[Xe]6s2'if75dl [Xe]6s2'if9

terbium

-

Tb

-

-

dysprosium

Dy

66

[Xe]6s2'if

holmium

Ho

67

[Xe]6s2'if1l

-

--

Er

68

thulium

Tm

69

[Xe]6s2'if

70

[Xe]6s2'if14

71

[Xe]6s2'if145dl

-

Yb Lu

-

--

10

erbium

lutetium

-

181

[Xe]6s2'if 12

ytterbium

182

13

-

177

527.2

....

....

.... ....

.... ....

.... ....

533.1

-

565.8

178

573.0

176

581.0

176

589.3

176

596.7

193

....

-

174

.... ....

603.4 523.5 ~

Continued

on next page

I

The Elements and Their Properties

I

A-7

A-8

I

Appendix 3

I

The Elements and Their Properties

TABLE

Ground-State Electron Configurations, ~ Atomic Radii, and first Ionization Energies of the Elements (Continued)

A ...

Atomic Number

Atomic Radius (pm)


Ionization Energy (kJ/mol)

Element

Symbol

Z

hafnium

Hf

72

[Xe]6s2'if145J2

159

658.5

tantalum

Ta

73

[Xe]6r'if

146

761.3

74

[Xe]6s2'if145d4

75

[Xe]6s2'if145d5

f--

~gsten rhenium f--

W

--

Re

--

-

osmium

Os

76

[Xe]6s2'if

iridium

lr

77

f--

145J3

-

145d6

139 137

-

770.0 760.3

135

839.4

[Xe]6s2'if145J7

136

878.0

139

868.4

platinum

Pt

78

[Xe]6sl'if145d9

gold

Au

79

[Xe]6sl'if 145dJO

144

890.1

mercury

Hg

80

151

1007.1

r-=--

81

[Xe]6r'if145dJO [Xe]6s2'if 145d J06pl

lead

Pb

82

[Xe]6r'if

145d 106p2

154

bismuth

Bi

83

[Xe]6s2'if

145d J06p3

150

703.3

polonium

Po

84

[Xe]6r'if145dJ06p4

167

812.1

astatine

At

85

[Xe]6s2'if

145d 106p5

140

924.6

86

[Xe]6s2'if 145d J06p6

145

87

[Rn]7s1

270' 223'

509.3

thallium f--

Tl

--

f--

r-=--

radon

f--

francium f--

Rn

-

Fr

-

-

-

-

170

-

-

589.4 715.6

1037.1 380

radium

Ra

88

[Rn]7s2

actinium

Ac

89

[Rn]7s26dl

188

499

thorium

Th

90

[Rn]7s 6d

179

587

protactinium

Pa

91

[Rn]7s25j26dl

163

568

92

[Rn]7J25f36dl

93

f-f--

uranium

f--

-

U

-

2

2

plutonium

Pu

94

[Rn]7s25j-(aq) HP042-(aq);=' H+(aq) + PO/-(aq)

f--

perchloric

1

periodic

1

phenol

1

f--

phosphoric

propanoic pyruvic

1 2 3 1

1 2

sulfurous

1 2

HNO,(aq)

2.9

X X

10-11

-

8.55

-

7.54 10.5

-

1.7

X

10-

0.77

7.6

X

10-4

3.12

1.4

X

10-4

3.85

1.2

X

10-2

1.92

4.7

X

10-7

6.33

1.5

X

10-3

2.82

2.0

X

10-6

5.70

1

(strong)

-

-

pressure

A-33

Measure

of the energy released when nucleons

combine to form a nucleus. It is calculated using the equation E = (b.mk', where b.m is the mass defect of the nucleus.

calorie (cal)

The amount of heat necessary to raise the temperature

of

1 g of water by IT. calorimeter

A device used to measure the absorption

or release of heat

by a physical change or chemical process. calorimeter

constant

(Cca10,;m",,)

The heat capacity of a calorimeter.

A-34

I

Glossary

calorimetry

The measurement

of the change in heat that occurs dur-

capacity of a buffer

The concentration

of components

needed in solution

to consume the acid or base added without a significant change in pH. capillary

action

The rise of a liquid in a narrow tube due to adhesive

forces between the liquid and the tube and cohesive forces within the liquid. A carbon atom with a double bond to an oxygen atom.

carboxylic

An organic compound

acid

functional

containing

the -COOH

in the process.

An electrode at which a reduction

half-reaction

(gain of

Streams of electrons emitted by the cathode (negative

trochemical

force expressed in volts (V)

cell can push electrons through

compound

split-react

A self-sustaining

example, neutrons released when an atom is

(l!J.H)

The heat gained or lost by a thermodynamic

The volume of a fixed quantity

sure is directly proportional chelate effect compared

monodentate

of a metal with a polydentate

ligands

ligands. ligand

agent); one molecule of the ligand occupies two or more

coordination chemical

The attractive

force that holds two atoms or ions in a

equation

and quantities

A balanced equation that describes the identities of reactants

(substances that are consumed

chemical reaction) and products (substances chemical equilibrium

during a

that are formed).

A dynamic process in which the concentrations

of reactants and products remain constant over time and the rate of a reaction in the forward direction matches its rate in the reverse direction. chemical

formula

compound.

A representation

The representation

the subscripts

(and sometimes

of the constituent

elements in a

comprises the elemental symbols and coefficients) necessary to identify the

kinetics

of substances chemical

The study of the rates of change of concentrations A characteristic

of a substance that can be ob-

served only by reacting it chemically to form another substance. chemical

reaction

The transformation

of one or more substances into

The science of matter and the study of its composition,

structure, chiral

A molecule that is chiral is not superimposable

on its mirror

analysis

not

Based on the

variables change.

A method of analysis in which a substance is

masses are used to determine combustion

reactions

the composition

whose

of the original material,

Reactions between oxygen and another elewhen the other compound

products of complete combustion effect

is a hydrocarbon;

the

are carbon dioxide and water vapor.

The shift in the position of an equilibrium

caused

by the addition of an ion taking part in the reaction. complex carbohydrate

(also called polysaccharide)

A polymer of

monosaccharides. complex ion composite

An ionic species consisting

materials

compound

of a metal ion bonded to one

Contain

a mixture of substances with different

and structures.

A substance composed of two or more elements linked

together infixed condensation

proportions.

reaction

Two molecules combine to form a larger mol-

ecule and a small molecule (typically water). bands

Unoccupied

higher-energy

bands in which elec-

trons are free to migrate in a material, ofchiral

compounds

The actual positions in space

occupied by the groups surrounding conjugate

acid

conjugate

acid-base

a chiral center.

Formed when a Bmnsted-Lowry pairs

A Bremsted-Lowry

base accepts a proton. acid and base that dif-

fer from each other only by the presence or absence of a proton. conjugate

base

Formed when a Bmnsted-Lowry

acid donates an H+

ion to another species. conversion

factor

A fraction in which the numerator

the denominator coordinate

is equivalent

to

but is expressed in different units.

bond

coordination

Forms when one anion or molecule donates a pair of

compounds

coordination

number

pairs surrounding copolymer

Compounds

made up of at least one com-

of a metal ion

Identi fies the number of electron

it in a complex.

A heteropolymer

that contains two different monomers.

Those electrons in the filled, inner shells in an atom or

ion that are not involved in chemical reactions. The study of the physical nature and the form of the

universe as a whole. Coulomb's

A molecule has chirality ifits mirror image does not coin-

cide with itself.

Depend on the concentration,

gas law (also called general gas equation)

combustion

cosmology

and properties.

image. chirality

combined

core electrons

di fferent substances. chemistry

of solutions

plex ion.

involved in chemical reactions.

property

reactions.

properties

electrons to another ion or molecule to form a covalent bond.

number of atoms of each element in one molecule. chemical

colligative

configuration

substance together. chemical

that encodes for a specific

Like enzymes, are organic molecules that accelerate the

conduction

sites on the central metal,

bond

coenzymes

compositions

The greater affinity of metal ions for polydentate

The interaction

(chelating

of gas at constant pres-

to its absolute temperature.

to that for the corresponding

chelation

sequence of nucleotides

or more Lewis bases.

system during a reaction at constant pressure. law

A three-base

common-ion

material by heating.

series of reactions in which the prod-

further with other materials and continue the reaction.

change in enthalpy

system that exchanges energy but

not matter with the surroundings.

ment or compound;

or mixture that has been trans-

formed into a harder, more heat-resistant ucts of the reaction-for

an

to its terminals.

A solid inorganic

chain reaction

to its heat of vaporization.

A thermodynamic

burned completely in oxygen to produce known compounds

The electromotive

external circuit connected ceramic

of an elec-

cell are connected. (Ecd1)

with which an electrochemical

Charles's

stance at different temperatures

held constant while the remaining

Uses symbols to show how the components

cell potential

(together).

Relates the vapor pressure of a sub-

ideal gas law; it is used when one or more of the four gas variables are

Ions with positive charges.

cell diagram

word zusammen

equation

on the identity of particles dissolved in the solvent,

electrode) in a partially evacuated tube. cations

Z comes from the German Clausius-Clapeyron

rate of biochemical

electrons) takes place. rays

atoms) on the same side of a

amino acid.

A substance added to a reaction that increases the rate of the

cathode

An isomer that has two like groups

line drawn through a double bond; cis is Latin for "on this side";

codon

group.

reaction but is not consumed

cathode

(also called Z isomer)

closed system

carbonyl group

catalyst

cis isomer

(such as two R groups or two hydrogen

ing a physical change or chemical process.

law

The energy (E) of the interaction between two ions is di-

to the product of the charges of the two ions (QI and Q2) and inversely proportional to the distance (d) between their nuclei.

rectly proportional

I

Glossary

counter

ions

The ions that balance the electrical charges of complex

ions in coordination covalent

bond

A chemical bond formed between two atoms by shar-

pi (1T) bond

Electron

density is greatest above and below or

in front of and behind the bonding covalent

network

solid

Consists of atoms held together by extended

The minimum

amount of fissionable material needed to

guishable

That specific temperature

and pressure where the liquid

(T)

The temperature

A combustible

below which a material

dilution

The process oflowering

dipole-dipole

interactions

dispersion

forces

the concentration

Attractive

of hydrocarbons

and other organic molecules formed under Earth's

surface by the and pressure.

The separation

A separation

easily vaporized)

duet

technique

components

between electrons

in those orbitals and pairs of electrons in ligands surrounding

the

orbitals. The difference in energy among

subsets of d orbitals split by interactions crystal lattice

The three-dimensional

or molecules) in a crystalline crystalline

solid

them from the less volatile components.

Results when two atoms share two pairs of electrons.

The maximum

of two electrons contained

(ccp)

A crystal structure

in which the layers of

cubic unit cells have an

abcabcabc ... stacking pattern.

(also called trans isomer)

E isomer

from a

sample; one Ci equals 3.7 X 1010 Bq. Ring-containing

nuclear charge (Z,/f)

experienced

The attractive

alkanes with the general formula

A circular particle accelerator using the combined

of a constant magnetic field with an oscillating make particles move faster in an increasingly

atom shield it from the nucleus. effusion

The process by which a gas escapes from its container

through

a tiny hole into a region of lower pressure. A flow of electrons.

electrochemical

cell

An apparatus

electrochemistry

The six equivalent

electrostatic

action field to

larger spiral path.

hybrid orbitals that point

formed by mixing one (n)s orbital,

toward the apices of an octahedron

three (n)p orbitals, and two (n - l)d orbitals. law of partial

pressures

The branch of chemistry

electrolysis

The total pressure of any mixture

electrolyte

chemical reaction.

A substance that dissociates into ions when it dissolves,

electrolytic

cell

(also called atomic absorption

spectra)

Charac-

teristic series of dark lines produced when free, gaseous atoms are illuminated degenerate

magnetic

degree of ionization

The ratio of the quantity

ionized to the original concentration expressed as a percentage, delocalized

radiation

The pattern of alternating 1T

single and double bonds gives

called conjugation,

which means that

bonds are said to be delocalized

over the three

or more carbon atoms and the oxygen atom. density (d)

spectrum

The ratio of the mass (m) of an object to its volume (V); the

radiation,

force (emf)

electron electron

visible light, ultraviolet

Also called voltage, the force that pushes

an electrical circuit.

A negatively charged subatomic particle. affinity

electron

capture

rounding

The energy change occurring

when 1 mol of elec-

nucleus draws in one of its sur-

resulting in the nucleus with one more neutron

and one less proton. electron

configuration

Describes

the distribution

of electrons among

the orbitals of an atom or ion. electron-deficient

The direct conversion of a gas (vapor) into a solid without

A neutron-poor

electrons,

electron

liquid phase.

range of radiant energy

X-rays, and gamma rays.

substance. an intermediate

A continuous

that includes radio waves, infrared radiation,

density of a pure substance is an intensive physical property of that deposition

Any form of radiant energy in the electro-

trons combines with 1 mol of atoms or ions in the gas phase.

it is called percent ionization.

rise to a property sometimes the electrons in the

of a substance that is

of the substance present; when

into

spectrum.

electrons through

Have the exact same energy level.

reactant(s)

product(s).

electromagnetic

electromotive

by external sources of radiation.

orbitals

of the solvent.

A device in which an external source of electri-

cal energy does work on a chemical system, turning

electromagnetic

mixture.

that examines the trans-

A process in which electrical energy is used to drive a

nonspontaneous

higher-energy

of gases equals the sum of the partial pressures of each gas in the spectra

that converts chemical energy into

between chemical and electrical energy.

and enhances the conductivity d2 Sp3 hybrid orbitals

force toward the nucleus

by an electron in an atom; its value is the positive charge

formations

cyclotron

the double bond; trans is Latin

for "across"; E comes from the German word entgegen (opposite).

electrical work or electrical work into chemical energy.

C2H2n•

dark-line

An isomer in an alkene that has

two like groups (such as two R groups or two hydrogen atoms) on

electric current

A unit used to express the level of radioactivity

cycloalkanes

shell

on the nucleus reduced by the extent to which other electrons in the

solid.

atoms, ions, or molecules in face-centered curie (Ci)

in the outermost

of a hydrogen atom (n = 1).

effective

in a crystal field. array of particles (atoms, ions,

An ordered array of atoms, ions, or molecules.

cubic closest-packed

Dalton's

in which the more volatile (more

opposite sides of a line drawn through energy (.6,)

forces

of a mixture are vaporized and then

thereby separating

double bond

Intermolecular

dipoles in molecules.

of a set of d orbitals into subsets

with different energies resulting from interactions

crystal field splitting

of solutes in a

forces between polar mol-

(also called London forces)

caused by temporary

condensed,

liquid mixture of hundreds

processing of plant matter under high temperature crystal field splitting

as it passes

The spread of one substance (usually a gas or liquid) through

distillation

from each other.

becomes a superconductor. crude oil

radiation

narrow openings.

ecules.

and gas phases of a substance have the same density and are indistincritical temperature

The bending of electromagnetic

diffusion

electrons

solution by adding more solvent.

sustain a nuclear chain reaction. critical point

diffraction

Have no unpaired field.

another.

axis.

arrays of covalent bonds. critical mass

atoms, ions, and molecules

around the edge of an object or through

ing pairs of electrons. covalent

diamagnetic

and are weakly repelled by a magnetic

compounds.

A-35

transitions

Movements compounds

than H) in Lewis structures

of electrons between energy levels. Substances whose central atoms (other

have fewer than four electron pairs (that

is, less than an octet of electrons).

A-36

I

Glossary

e1ectronegativity

extensive

A relative measure of the ability of an atom in a

electron-pair

geometry

Describes

the arrangement

in 1T' bonds

face-centered

In a system with alternating

bonds, electrons in

7T

bonds can be delocalized

single and double

over several atoms or

in antibonding

orbitals

Destabilize

the molecule because

they do not increase the electron density between nuclear centers and therefore do not participate electrons

in bonding

increasing

Hold atoms in molecules together by

the electron density between nuclear centers.

electrostatic

energy (E)

potential

cubic (fcc)

(also called a group)

Elements in the same column of the

periodic table; elements in the same family have similar chemical properties. (F)

The quantity of electrical charge (in coulombs) in

1 mol of electrons; its value to three significant figures is 9.65 X 104 C. fats

The energy of a particle because of

A unit cell that has the same kind of particle

center of each face.

Faraday's constant

in holding the molecule together.

orbitals

of a substance that varies with the

(atom, ion, or molecule) at the eight corners of a cube and at the family

even across an entire molecule. electrons

A characteristic

of atoms and lone

pairs of electrons about a central atom. electrons

property

quantity of the substance present.

bond to attract electrons to itself.

Solid triglycerides

filtration

containing

A process for separating

primarily

saturated

fatty acids.

particles suspended

in a liquid or a

its position with respect to another particle; it is directly proportional

gas from that liquid or gas by passing the mixture through

to the product of the charges of the particles and inversely propor-

that retains the particles.

tional to the distance between them. element

first law of thermodynamics

A pure substance that cannot be separated

into simpler sub-

stances by chemical means. elementary

step formula

view of a single process taking

food value

imposable; end point

reactions

that is reached when just enough

endothermic

Nonspontaneous

process

roundings

reactions.

formation

formation

reaction

standard formula

mass

energy state

An allowed value of energy.

formula

unit

The sum of the internal energy and the pressure-volume

product of a thermodynamic of reaction

(also called heat of reaction)

The heat

(S)

A measure of the distribution

of energy in a system at a

constant

(K)

The value of the ratio of concentrations

partial pressures) in the equilibrium

(or

constant expression at a specific

temperature. equilibrium

expression

the relationship

(also called mass action expression)

between the concentration

(or partial pres-

sure) terms of reactants and products when a system is at equilibrium. equivalence

point

The point in a titration

of titrant added is stoichiometrically

where the number of moles

equal to the number of moles of

elements

Those elements in tissue, blood, or other body

fluids that have a physiological element ultimately ester

The -OH

function;

a deficiency in an essential

leads to damage to the organism.

Has the general formula R-O-R, (also called vaporization)

where

where R is any alkyl group or

may be different. The transformation

of mol-

Any energy state of an electron in an atom above the

ground state. exergonic exothermic

lines

reactions process

a mixture of com-

A set of dark lines in the otherwise An indication

continuous

solar

of the energy available to do useful

work; free energy is a thermodynamic criterion for spontaneous Molecules

and hence unpaired

that have an odd number of valence electrons

electrons in their Lewis structures.

change (I1G)

free-energy

for spontaneous

state function that provides a

change.

The change in the free energy of a process;

processes at a constant temperature

and pressure,

t::.G CaCl2 > NaCI 10.2. The melting point ofTi02 is highest, and the melting point of CaCl2 is higher than that ofPbCI,. 10.3. U= -3793 k]/mol 10.4. Ethylene glycol has two -OH groups to form hydrogen bonds, compared to isopropanol with only one. 10.5. Substance (a) has the largest dipole-dipole interactions. Substance (d) has the largest dispersion forces. Substance (c) has the lowest boiling point. 10.6. O2 < CO < Hp 10.7. Helium is less polarizable than nitrogen. 10.8. 9.5 X 10-5 mol/L 10.9. P,oln = 0.845 atm 10.10. At a pressure of25 atm, CO2 is probably a solid at -lOOT. CO2 melts between - 50°C and - 60°C, and it vaporizes between lOT and 20°C. 10.11. 0.841 m 10.12. 4.4 m 10.13. !::.T;, = 9.3T; boiling point = 109.3°C 10.14. Boiling point = 102.TC 10.15. !::.~ = -0.12°C 10.16. 7T = 20.3 atm 10.17. 7T = 8.40 atm 10.18 . .M = 180 g/mol 10.19. .M = 6.4 X 104 g/mol

A-50

I

Answers to Concept Tests and Practice Exercises

Chapter 11 CONCEPT

TESTS

p.527.

The 3s bands ofMg are full, so the conductivity ofMg is best explained by overlapping valence and conduction bands. p. 536 (top). Yes. Describing bronze as a solution is correct as long as the Sn is uniformly (homogeneously) distributed. p. 536 (bottom). No. Sn is randomly located in the unit cell. p.543. Region a is diamond.

PRACTICE

EXERCISES

11.1. Tin-doped GaAs is p-type; selenium-doped GaAs is n-type. r = 128 pm 11.3. Density of silver = 10.57 g/cm3; density of gold = 19.41 g/cm3 11.4. Gold (r = 144 pm) will form substitutional alloys with both silver (r = 144 pm) and copper (128 pm; 11%difference in radius). 11.5. Radius ofNa+ = 101 pm 11.6. Density ofNaCI = 2.16 g/cm3 11.7. 412 nm; 2fJ = 30.00° for n = 3; 2fJ = 40.44° for n = 4

p.605.

p.608.

Methanol has a higher percentage of oxygen in it than ethanol. This oxygen has no fuel value. Therefore, methanol has less fuel value than ethanol. MTBE > diethyl ether> ethanol> methanol

PRACTICE

EXERCISES

12.1. Yes. One mole of compound A reacts with 2 moles ofH2; one mole of compound B reacts with 3 moles ofH2. Because A and B have similar molar masses, a given mass ofB will react with close to 1.5 times as much ofH2 as the same mass of A. The structure of the product of the reaction with A is the same as that with B: CH3 -CH2 -CH2 -CH2 -CH2 -CH2 -CH3' 12.2. The carbon-skeleton structure of n-hexane is

11.2.

CH3CH2CH2CH2CH2CH2CH3 [or CH3(CH2\CH3]. 12.3. Molecules in sets a and c are isomers; those in set b have different formulas. 12.4. 28.4 k]/mol 12.5. 100

Chapter 12 CONCEPT

G e B

TESTS

No. A triple bond or two double bonds represent the same degree of unsaturation (two). p.581. The estimates based on bond energies are the same because the molecules of the two compounds contain the same number ofC-C bonds and C-H bonds. However, they are different compounds with different chemical and physical properties. It is reasonable that their actual heats of combustion would be a little different. p.583. IfIUPAC rules are followed, the name of a compound provides an unambiguous description of its molecular structure. p.584. The solubility of a gas is proportional to its partial pressure, which is in turn proportional to the product of its mole fraction in a mixture times total pressure. As crude oil comes up from deep underground, there is a decrease in total pressure, which reduces the solubility of the CI to C 4 alkanes. p.586. Gasoline has the lower average molar mass and so should have the higher vapor pressure. p. 593. The reaction is endothermic, It requires tremendous amounts of energy, which is impossible to control during production. Therefore this approach is impractical. p.597. (a) No. The two hydrogen atoms on the terminal carbon atom make both "sides" of the double bond equivalent. (b) No. The bond angles are 180° so there are no "sides" to C=C bonds. p.598. LDPE and HDPE have different properties and uses. If they are to be recycled for uses similar to their original ones, they need to be kept separate. p.600. Chlorine atoms are so bulky that the polymer chains in Saran cannot fit tightly together. The resulting loose fit means that the Saran is less rigid and more elastic than PVC. p.603. These alcohols and ethers are polar solutes that tend to be soluble in polar solvents such as water; nonpolar solutes such as hydrocarbons have little solubility in water. The -OH groups in alcohols can hydrogen-bond with water; water can hydrogenbond with -0- in ethers.

90

2...-

e

'"

70

~

60

0-

p.575.

80

S

1 0

50 Volume of distillate (mL)

•

12.6. The mole ratio is 1.4. 12.7. The five isomers are:

trans

12.8.

CIS

JyOCH' o

12.9. E-85 has 33% less energy per gram than gasoline. 12.10. Dipole-induced dipole interactions 12.11.

n

12.12. The polar materials from which the gloves are made allow moisture to pass through but repel nonpolar oil and grease. 12.13.

HO~OH

o

0

Chapte CONCEPl p.635.

(1) I' (2) 1 is ad

occu p.639.

FoUl

gree: p. 645. (a).A coml

p.647.

is ne phot Part to re tran: of g~ micr

p. 648 (top). vibn

p. 648 (botto "spo: proc p.660. It ca

PRACTIC 13.1. All t four.

t.S 13.2.

'Y'

than Onl~ reacl air i~

13.3.

13.4.

Pred CO2 The (a) 1 of g~ (b) ~ 10-3

t.so

u

A-52 I Answers to Concept Tests and Practice Exercises 14.6.

If we use a spreadsheet (like Excel) to calculate In [H202l values, we can then plot them and fit a linear trend line to them: -0.5,

14.11. We can use a spreadsheet (like Excel) to calculate In k and 1lTvalues, plot them, and fit a linear trend line to them:

, -27.4,

~ -1.0

i

-27.6

6'

~

.s

...

'"'< .s

-1.5

-27.8 -28.0

1000

-28.2 0.0032

2000

I

Time (s)

I

I

0.0034

0.0036

I

I

0.0038

0.0040

r..

0.0042

I

0.0044

liT The plot is linear, so the reaction is first order with a rate constant equal to minus the slope. Therefore, k = 8.3 X 1O-4s-l• 14.7. k = 0.693/28 d = 0.025 d-I 14.8. After calculating 1/[N02l values in a spreadsheet (like Excel), we can plot them versus time and fit a linear trend line to them: 6.

.,

I

Multiplying the negative of the slope by R gives E, = 6.9 k]lmol. No. The mechanism is not valid because none of the proposed elementary steps have a stoichiometry that fits the rate law for the overall reaction. 14.13. Yes. N02 acts as a homogeneous catalyst that is consumed in one step and regenerated in the next.

14.12.

Chapter 15 CONCEPT

2000

4000

6000

8000

10000

p.742. p.758. p.761.

Time (s) The plot is linear, so the reaction is second order with a rate constant equal to the slope. Therefore, k = 2,09 X 1O-4M-Is-I and the rate = 2.09 X 1O-4M-Is-I[N02F 14.9. tll2 = 9.23 X 1O-3s 14.10. By using a worksheet (like Excel) to calculate In [CI] values, we can plot them and fit a linear trend line:

p.765.

PRACTICE

-30.4 " ----------------,

S .s

TESTS

b. [C02l = [H2l > [COl = [H,Gl K should be small and is: 6.5 X 10-7 Pure liquid water is assigned an activity of "1"; therefore, the K, expression contains only a term for the concentration of H,G(g). Adding reactant shifts the equilibrium to the right forming more product and consuming some of the additional reactant. Removing reactant shifts the equilibrium to the left, consuming product and replacing some of the lost reactant.

EXERCISES [CO][H,P [CH ][H,Gl

-30.6

15.1. Kc

-30.8

K, = 2.9 X 102 15.3. K p = 2.7 X 104 15.4 . K p = 4.5 X 10-12 15.5. K= 3.6 X lOS 15.6. K, = 0.13 15.7. Kovc"lI = 1.7 X 102 15.8. Q < K; therefore, the mixture is not at chemical equilibrium and the reaction proceeds in the forward direction. 15.9. K p = 6.1 X 105

=

4

15.2.

-31.0 -31.2 -31.4 -31.6 -31.8

0

1000 Time(s)

2000

The plot is linear, so the reaction is first order with a rate constant equal to minus the slope. Therefore, k' = 616 S-I and k = k'/(8.5 X lO-lIM) = 7.3 X 1012 M-Is-I.

15.10.

(PCO)2 (a) Kp = (Pco,)

15.11. (a) Loss of ammonia shifts the equilibrium toward the formation of more ammonia; (b) some of the ammonia decomposes into nitrogen and hydrogen as the reaction runs in reverse; (c) removing these two reactants shifts the equilibrium toward forming more of them.


15.12. 15.13. 15.14. 15.15. 15.16.

Increasing pressure shifts the equilibrium to the right. The value of the equilibrium constant increases. Partial pressure of HI is 0.156 atm. [NP4] = 0.016 M; [NO,] = 0.058 M At 298 K, Kp = 3.2 X 10-37;at 2000 K, Kp = 3.3 X 10-12 p.862.

Chapter 16 CONCEPT p.800. p.805. p.813.

p.826. p.833.

p.868.

TESTS

The pH of the weak acid is higher because it is not completely ionized. In terms of acid strength: C > B > A No influence on the pH of phosphoric acid, but significant influence in citric acid because the K al and K a2 values are much closer together. If the desired buffer pH is below the pK, of the acid, more of the acidic component is needed. During the second step in the titration all of the bicarbonate produced by the titration of carbonate in the first step plus the bicarbonate initially in the sample are titrated.

p.875.

p.878. p.880.

16.1. 16.2. 16.3. 16.4. 16.5. 16.6. 16.7. 16.8. 16.9. 16.10. 16.11. 16.12.

16.13.

EXERCISES

CH3COOH, CH3COO-; Hp, Hp+ pH = 2.160 [H+] = 1.3 X 10-12 M; [OH-] = 8 X 10-3 M K, = 7.9 X 10-4 pH = 3.3 [OH-] = 9.4 X 10-3; pOH = 2.03; pH = 11.97 pH = 5.63 NH/(aq) + HP(C) H NH3(aq) + Hp+(aq) pH = 9.08 pH = 4.01 pH change is less than 0.01 units. Blood pH would rise because the mouse would be able to exhale CO2 and so shift the equilibrium to the left, thereby reducing [H +]. (a) Ammonia concentration = 0.4846 M; (b) possible indicators are methyl red, bromcresol green

EXERCISES

17.1. CO2 is the Lewis acid, and CaO (or its 02- ion) is the Lewis base. 17.2. [Ag+] = 9.2 X 10-9 M 17.3. Solubility = 2.6 X 10-3 M 17.4. The pH in the problem (10.33) matches the pK" of carbonic acid in Appendix 5, which means, according to the Henderson Hasselbalch equation, that ratio of CO/- to HC03 - ions will be 1:1. Therefore, half the CO/- ions that form when ZnC03 dissolves will remain as C032-. Thus, K,p = [Zn2+][CO/-] = (x) (0.5 x) = 1.2 X 10-10 and x = 1.5 X 10-5• 17.5. Assuming the hydroxyl group does not ionize, but the three carboxylic acid groups do, the molecule has three "teeth." 17.6. Low-spin COHo 17.7. (a) Ligand-ammonia; counter ion-chloride ion; nametetraamminezinc(II) chloride; (b) Ligands-ammonia and water; counter ion-nitrate; name-tetraamminediaquacobalt(III) nitrate + + 17.8.

Chapter 17 CONCEPT

TESTS

p.851. H-N:

I I

H

p.854.

:F:

H and

(a)

(b) +

I .. I ..

B-F: :F:

A coordinate bond is formed when the N atom in NH3 donates its lone pair of electrons to an empty Sp3 orbital of the B atom in BF3. Hybridization is sp3d2• Each F atom has 7 valence electrons and gains one more by sharing a single electron in a half-filled

A-53

orbital ofXe. To create 4 half-filled orbitals in the valence shell ofXe, two electrons from two of the three filled 6p orbitals are promoted to two of the empty 6d orbitals. Then those four halffilled orbitals are mixed with the filled 6s and third 6p orbital to form six equivalent hybrid orbitals. Since all six orbitals are in the n = 6 shell, the hybridization scheme is sp3d2• Solubility would have decreased because more of the fluoride ion that dissolved would stay as P-. Spontaneity must be due to a favorable (negative) change in enthalpy because there are no changes in the number of particles and they are all liquids or solutes. Therefore the change in entropy should be small. (a) Cyanide is a stronger field strength ligand than pyridine; greater splitting in the cyanide complex leads to population of the lower energy orbitals and low spin. (b) The Ru ion is a much larger ion than the Fe ion and its larger valence shell d orbitals interact more strongly with ligand lone pairs, producing more splitting and low spin. There is only one structure possible. The square planar complex is not chiral; the tetrahedral complex is chiral.

PRACTICE PRACTICE

I

(c) The names of (a), (b), and (c) are: (a) cis-diammine-trans-dibromoethylenediaminecobalt(III); (b) trans-diammine-cis-dibromoethylenediaminecobalt(III); (c) cis-diammine-cis-dibromoethylenediaminecobalt(III).

A-54 I Answers to Concept Tests and Practice Exercises

Chapter 18

PRACTICE

EXERCISES

19.1.

CONCEPT p.895.

p.916.

TESTS

No, the number of moles ofCu deposited match the number of moles of Zn that dissolve, but the molar mass of Zn is slightly larger than that ofCu so the changes in mass are not the same. H2(g) is produced at the anode; CI2(g) at the cathode

PRACTICE

0)

Chiral (c)

(b)

(a)

EXERCISES

18.1. 2N02-(aq) + 02(g) --72N03-(aq) 18.2. 2AI(s) + 3 Cu2+(aq)--7 2 AP+(aq) + 3 Cu(s) Al(s) I AP+(aq) II Cu2+(aq)I Cu(s) 18.3. LlGcdl = -290 kJ 18.4. The cadmium half-reaction occurs at the anode, the nickel halfreaction occurs at the cathode. Overall: NiO(OH)(s) + Cd(s) + 2HP(f)--7 2 Ni(OH)2(s) + Cd(OH)2(s) E:,II = 1.72 V 18.5. E"II = 1.64 V 18.6. K = 1.8 X 1062

~O~

~

NH2 HQ,J) •• COOH Chiral

I~

o

OH

OH

Chiral

(d)

(e)

19.2.

18.7. 130 g Mg 18.8. 1.48 g Pb

Chapter 19 CONCEPT p.939.

TESTS

A fragrant molecule is like the white hand and the receptor protein is like the brown one. When the fragrant molecule and receptor fit together, as when the two right hands shake, the protein signals the fragrance of the molecule to the brain. When the intermolecular fit is not as good, as illustrated by the bottom pair of hands, then no fragrance, or a different one, is sensed. p.942. Muscarine is a chiral molecule and a plant produces only one enantiomer, so a solution of muscarine from a natural source would rotate the plane of polarized light. Because the fluid in the person's stomach contained muscarine but did not rotate the plane of polarized light, it must be a racemic mixture resulting from a laboratory synthesis. Therefore the coroner concluded that the person did not die from eating toxic mushrooms but may have been murdered by someone who had access to synthetic muscarine. p. 945. All of the amino acids without an ionizable functional group on their side chains, e.g., alanine, can have only two pK, values. Those with an ionizable side chain group, e.g., glutamic acid, have three. p.949. Leucine or isoleucine p.955. Two isomers are possible because the -OH group on the C-5 oflinear fructose can approach the C=O bond on the C-2 carbon from either above or below the plane of the ring as the cyclic form of the sugar forms. p.959. As sugar is oxidized in the body it forms pyruvate, which, according to Figure 19.32, is a precursor of cholesterol. p.962. The unsaturated fats are liquids at low temperatures whereas saturated fats might solidify stopping the flow of blood to and from the reindeer's hooves. p.963. Olestra is a much larger molecule with many more C-H and C-C bonds. Therefore combusting a mole of it yields much more energy. p. 964. Fluoroacetic acid, the fluoroacetate ion, and acetamide have similar structures and shapes, so perhaps the acetamide competes with the other two for the enzymatic binding site.

+

+ OH-(aq)'= + Hp(aq) 1\'2:HiICH(COO-)~H(COOH)(aq) + OH-(aq)'= H3NCH(COO-)CH(COO-)(aq) + Hp(aq) + 1\'3:H3NCH(COO-)CH(COO-)(aq) + OH-(aq) ,= H2NCH(COO-)CH(COO-)(aq) + Hp(aq) For amino acids "A," "B," and "c" there are six different

19.3. K'I: H3NCH(COOH)CH(COOH)(aq)

HiICH(COO-)CH(COOH)(aq)

19.4.

sequences: ABC, ACB, BCA, BAC, CAB, and CBA. 19.5. Because cellobiose is indigestible, it must have a J3-1,4Iinkage, which means it has the structure in (a). 19.6. There are 3 possibilities for linking 1 molecule of each of the fatty acids "A," "B," and "c" to glycerol: ABC, CAB, BCA. Each has two enantionmers, so there are a total of 6 isomers. 19.7. CGGTATCCGAT TTAAGCCGCTAG 11111111111

111111111111

(a) GCCATAGGCTA

(b) AATTCGGCGATC

Chapter 20 CONCEPT

TESTS

p. 1008. Electron capture does not result in the emission of any subatomic particles. p.l012. The limestone is too old and nearly all the carbon-14 in it would have decayed leaving too little to detect and measure quantitatively.

PRACTICE 20.1. 20.2. 20.3. 20.4. 20.5.

EXERCISES

1.4% 1.25 X 10-12 J 1;!Xe Beta decay; ~! Si ~;Cl+ ~n--7~~Ar+

~_e


20.6. 20.7. 20.8.

3.65 x 109Bq = 9.86 x 10-2 Ci 1.4 X 1011X-rays 4.95 X 103 years

Chapter 21 CONCEPT p.1030.

TESTS

The electron configuration ofMgH is ls22s22p6. It has no electrons in a partially filled set of d orbitals that could undergo transitions resulting in absorption of photons of visible light.

PRACTICE 21.1. 21.2. 21.3. 21.4. 21.5. 21.6.

I A-55

EXERCISES

pH = 1.19 N2(g) + 8H+(aq) + 8e- --7 2NH/aq) + H2(g) 2~:TI+~_e--72~~Hg+y There are 20.7 mg ofl86Re, but only 9.4 mg ofl88Re left. SnsHg(s) + 02(g) + 4H+(aq) --7 Sn6Hg(s) + 2SnH(aq) + 2H,0(C) E:x" = 1.36 V K"" = 2.5 X 109

Answers to Selected End-ot-Chapter Questions and Problems Chapter 1 1. (a) A pure compound

in the gas phase

(b) A mixture of elements: red in the liquid phase and blue in the gas phase 3. (b) A mixture of two gaseous elements undergoes

a chemical reaction to form a solid

compound. 5. Both are correct because on the sun, matter is being turned into energy through

nuclear

fusion.

7. One chemical property of gold is its resistance to corrosion (oxidation). Gold's physical properties

include its density, its color, its melting temperature,

and its electrical and thermal

conductivity.

9. Add water to the salt-sand mixture to dissolve the salt. Filter this mixture to isolate the sand and then recover the salt by evaporating

the water from the solution that passed through

the

filter.

11. (b) combustion 13. A Snickers bar, an uncooked hamburger,

and a hot dog

15. Orange juice (with pulp) 17. They can be distinguished

by their physical state (sugar is a solid, water is a liquid, and oxy-

gen is a gas), by their melting and boiling points, and by their density.

19. Density, melting point, thermal and electrical conductivity, physical properties

whereas tarnishing

and softness (a-d) are all

and reaction with water (e and f) are both chemical

properties. 21. Distillation

will separate the water from the dissolved protein that has formed a homoge-

neous solution with the water. This process would have to be accomplished, temperature,

however, at low

or the protein will be denatured.

23. Fe is a solid, O2 is a gas, and Hg is a liquid.

25. Extensive properties change with the size of the sample and so cannot be used to identify a substance.

27. We need at least one observation,

experiment,

or idea (from examining

nature).

29. Yes 31. Theory as used in normal conversation

is an idea, an opinion, or a speculation

that can be

changed. 33. SI units can be easily converted into a larger or smaller unit by multiplying

or dividing by

multiples of 10. English units are sometimes based on other number multiples and thus are more complicated

A-56

to manipulate.

Answers to Selected End-of-Chapter Questions and Problems

87. Both mixtures (a) and (b) would react to make sodium chloride with no reactants left over.

35. 6.70 X 108 mi/h 37. 93.2% 39. 4.1

X

I A-57

89. 75 loaves of bread and 50 oz of mayonnaise

1013km

91. 17 bicycles

41. 1330 Cal

93. Day 11

43. 2.5 mi 45. 9.0 mi/h

Chapter 2

47. 23 g 49. 19.0 mL

1. (c) A mixture ofN02

51. 26.5 g; 2.65 X 10-2 kg

3. The element shaded dark blue-helium

53. 58.0 mL

5. (a) Chlorine (CI2, yellow) is a reactive nonmetal. (b) Neon (Ne, red) is a chemically inert gas. (c) Sodium (Na, dark blue) is a reactive metal.

55. 33.7 mL 57. 5.1 g/cm3 59. Yes, the cube has a density of 0.64 g/cm3• 61. 0.29 cm3 63. (a) Accuracy means that the experimental value agrees with the true value of the measurement. Precision means that several repeated measurements agree with each other with little variability. (b) No, the lawyer confuses the two. He is saying that ifhis weight is not known exactly to the ounce compared to the true value, then even the pound value is in error. (c) Yes, to be precisely accurate, a series of measurements would be each very close to each other and also in agreement with the true value. (d) The sign means that the weight that is measured by the scale is within ± the smallest division on the scale. 65. (a) Manufacturer #1 has a range of 0.516 - 0.504 = 0.012 j.Lm. Manufacturer #2 has a range of 0.514 - 0.512 = 0.002 j.Lm. Manufacturer #3 has a range of 0.502 - 0.500 = 0.002 j.Lm. (b) Yes, Manufacturers #2 and #3 can justify the claim. (c) Yes, in the case of Manufacturer #2, the lines are printed at wider widths than the widths specified. 67. (b) 0.08206, (c) 8.314, (f) 3.752 X 10-5, and maybe (d) 5.420 103, depending on whether the 0 is significant 69. (a) (b) (c) (d)

17.4 1 X 10-13 5.70 X 10-23 3.58 X 10-3

71. Yes, -40°C is equal to -40°F.

75. 172°F, 351 K

Mg (green) will form MgO. K (red) will form Kp. Ti (yellow) will form Ti02• Al (dark blue) will form AIPr

9. When looking at a single compound, the law of constant composition applies. In a compound, the ratio of the constituent atoms is fixed. When comparing two or more compounds made up of the same elements, the law of multiple proportions applies if the compounds can combine in different ratios to give distinct compounds. 11. Dalton's theory states that because atoms are indivisible, the ratio

of the atoms (elements) in a compound is a ratio of whole numbers. Thus, in water, the ratio of volumes of hydrogen to oxygen is the whole number ratio 2:1, because the ratio of hydrogen and oxygen atoms in water is 2:1. 13. 1.5 15. 30.5% 17. Rutherford concluded that the positive charge in the atom could not be spread out (to form the pudding) in the atom, but must result from a concentration of charge in the center of the atom (the nucleus). Most of the alpha particles were deflected only slightly or passed directly through the gold foil, so Rutherford reasoned that the nucleus must be small compared to the size of the entire atom. The negatively charged electrons do not deflect the alpha particles, and Rutherford reasoned that the electrons took up the remai nder of the space of the atom outside of the nucleus. 19. The fact that cathode rays were deflected by a magnetic field indicated that they were streams of charged particles.

23. Greater than 1

77. 39.2T 183.9 K

83. HgBa2CaCu206

25. (a) and (b) 27. 35.46 amu

81. -452°F

85. 0.12 mg/L

X

7. (a) (b) (c) (d)

21. A weighted average takes into account the proportion of each value in the group of values to be averaged.

73. -270T

79. -89.2°C,

and NO

has the highest ~

29. Yes, the average atomic mass of magnesium on Mars is the same as on Earth. 31. 47.95 amu

A-58 I Answers to Selected End-of-Chapter Questions and Problems 33. Mendeleev only knew the masses of the elements at the time he arranged the elements into his periodic table. 35. (a) (b) (c) (d) 37.

I

14C,6 protons, 8 neutrons, 6 electrons s9Fe,26 protons, 33 neutrons, 26 electrons 90Sr,38 protons, 52 neutrons, 38 electrons 2JOPb,82 protons, 128 neutrons, 82 electrons 23Na

8'Y

118Sn

Number of protons

11

39

50

Number of neutrons

12

50

68

Symbol

197Au -

79

Number of electrons

11

39

50

23

89

118

197

81Br-

""Ra2+

37CI-

23Na+

79

-

Number of protons

17

11

35

88

Number of neutrons

20

12

46

138

Number of electrons

18

10

36

86

Mass number

37

23

81

226

-

1

-

41. NaC!, MgCI2, CaCI2, SrCI2; Na2S04, MgS04' CaS04, SrS04 43. Compounds (a) and (d) are molecules. Compounds (b) and (c) consist of ions. 45. (c) Be 47. (c) S249. (a) S2-, (b) P3-, and (d) Ca2+ 51. (b) Br

55. The Roman numerals indicate the charge on the transition metal cation. 57. (a) (b) (c) (d) (e) (f) (g) (h)

N03, nitrogen trioxide N20S' dinitrogen pentoxide N204, dinitrogen tetroxide N02, nitrogen dioxide N203, dinitrogen trioxide NO, nitrogen monoxide N20, dinitrogen monoxide N40, tetranitrogen monoxide

59. (a) (b) (c) (d)

Na2S, sodium sulfide SrCI2, strontium chloride Al20y aluminum oxide LiH, lithium hydride

61. (a) Cobalt(II) oxide

(b) Cobalt(III) oxide (c) Cobalt(IV) oxide 63. (a) BrO(b) S042(c) 103(d) N0265. (a) (b) (c) (d)

Nickel(II) carbonate Sodium cyanide Lithium hydrogen carbonate Calcium hypochlorite

Hydrofluoric acid Bromic acid Hl04 HN02

69. (a) (b) (c) (d) (e)

Sodium Sodium Sodium Sodium Sodium

71. (a) (b) (c) (d) (e)

K2S K2Se Rb2S04 RbN02 MgS04

73. (a) (b) (c) (d) (e)

Manganese(II) sulfide Vanadium(II) nitride Chromium(III) sulfate Cobalt(II) nitrate lron(III) oxide

oxide sulfide sulfate nitrate nitrite

-

118

Mass number 39. [ Symbol

67. (a) (b) (c) (d)

75. (b) Na2S03 77. (b) Cl2 79. (a) Na 81. Chemistry is the study of the composition, structure, properties, and reactivity of matter. Cosmology is the study of the history, structure, and dynamics of the universe. These two sciences are related because (1) the universe is composed of matter, the study of which is chemistry, and therefore, the study of the universe is really chemistry; (2) the changing universe is driven by chemical and nuclear reactions that are also studied in chemistry; and (3) cosmology often asks what the universe is made of at the atomic level. 83. Because quarks combine to make up the three particles that are important to the properties and reactivity of atoms: protons, neutrons, and electrons. 85. The density of the universe is decreasing. 87. The higher the charge (number of protons), the more repulsion the nuclei feel for each other and the higher the temperature needed to overcome that repulsion. 89. The electrons in He are held more tightly by the nucleus so they are harder to remove. 91. The expanding universe was cooling and therefore could not support the high temperatures needed for fusion. Also, the expanding universe was not dense enough for nuclei to fuse. 93. Both 95. (a) 160 (b) 24Mg (c) 36Ar 97. (a) S9CO (b) 121Sb (c) lI°Hg

Answers to Selected End-of-Chapter Questions and Problems I A-59 99. (a) Electrons (b) The negatively charged electron was attracted to the positively charged plate as the electron passed through the electric field. (c) The light spot would be at the bottom. The electron would still be deflected toward the positively charged plate, which would now be at the bottom of the tube. (d) The position of the light spot on the fluorescent screen would be half way between the position where it was before the voltage was reduced and the "zero" spot location when there is no voltage between the plates. 101. (a) Two (b) Both spots would be seen below the center of the screen. The spot lowest on the screen (farthest from the center) would be the ll' particle and the one closest to the center would be the proton. 103. 1:2:4 105. 17Cu:1Sn 107. (a) Sc, Ga, Ge (b) Ekaaluminum is gallium, ekaboron is scandium, and ekasilicon is germanium. (c) Scandium was discovered in 1879, gallium was discovered in 1875, and germanium was discovered in 1886. 109. (a) 52.93% (b) 0.599 cm3 111. 60.11% 113. (a) 79BrJ9Br = 157.8366 amu, 79Br-81Br= 159.8346 amu, 81Br81Br = 161.8326 amu (b) 79BrJ9Br, 25.41%; 79Br-8IBr,50.00%; 8IBr-8IBr,24.59%

15. (a) (b) (c) (d)

7.53 X 1022 atoms 7.53 X 1022 atoms 1.51 X 1023 atoms 2.26 X 1023 atoms

17. (a) Both contain the same (b) NP4 (c) CO2 19. (a) 3.00 mol (b) 4.50 mol (c) 1.50 mol 21. 41.63 mol 23. 0.25 mol; 10 g 25. (a) (b) (c) (d)

1 mol 2 mol 1 mol 3 mol

27. (a) (b) (c) (d)

64.06 glmol 48.00 glmol 44.01 glmol 108.01 glmol

29. (a) (b) (c) (d)

152.15 164.20 148.21 132.16

glmol glmol glmol glmol

31. (a) NO (b) CO2 (c) 02 33. 0.752 mol

Chapter 3 1. (a) 4X(g) (b) 4X(g) (c) 4X(g) (d) 4X2(g)

+ 4 Y(g) -7 4 XY(g) + 4Y(g) -74XY(s) + 4Y(g) -7 2XY2(g) + 2X(g) + 4Y2(g) -7 8XY(g)

35. 10.3 g 37. Diamond 39. No 41. No

3. Fe(C)

43. (a) CH4(g) + HP(g) -7 CO(g) (b) 2 NH3(g) -7 N2(g) + 3 H/g) (c) CO(g) + HP(g) -7 CO/g)

5. It is thought that most of the water present on Earth arrived with colliding asteroids and comets when the Earth was young. A small amount may have been derived from water vapor outgassing from Earth's interior.

45. (a) 3 FeSi03(s) + 4HP(C) -7 Fe3Sips(OH),(s) + H4SiO,(aq) (b) Fe2Si04(s) +2C02(g) + 2HP(C) -7 2 FeC03(s) + H4Si04(aq) (c) Fe3Sips(OH),(s) + 3C02(g) + 2HP(C)-7 3FeC03(s) + 2 H4SiO ,(aq)

7. A dozen is too small a unit to express the very large number of atoms, ions, or molecules present in a mole.

47. (a) (b) (c) (d)

9. No, the molar mass of a substance does not directly correlate to the number of atoms in a molecular compound. The statement posed by the question would only be true if the two compounds were composed of the same element. 11. (a) (b) (c) (d)

7.3 X 10-10 mol Ne 7.0 X 10-11 mol CH4 4.2 X 10-12 mol 03 8.1 X lO-IS mol N02

13. (a) 2.5 X 10-12 mol bytes (b) 3.3 X lO-IS mol bytes

+ 3 H2(g) + H/g)

N2(g) + a/g) -72NO(g) 2NO(g) + 02(g) -72NO/g) NO(g) + N03(g) -72N02(g) 2N2(g) + a/g) -72NP(g)

49. (a) NPS 60 40 20

83. No. Each gas has a different molar mass; therefore the densities of different gases will not necessarily be the same for a particular temperature and pressure.

E

y

=

18603x - 0.0237

~ ~

+

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Science in the Context of Application (Boston Studies in the Philosophy of Science)

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