For courses in Thermodynamics.A visual, conceptual and contemporary approach to Physical ChemistryEngel and Reid'sT
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Table of contents :
Cover......Page 1
Title Page......Page 2
Copyright Page......Page 3
Brief Contents......Page 5
Detailed Contents......Page 6
About the Authors......Page 10
Preface......Page 11
Acknowledgments......Page 12
Math Essential 1: Units, Significant Figures, and Solving End of Chapter Problems......Page 18
1.1. What Is Thermodynamics and Why Is It Useful?......Page 22
1.2. The Macroscopic Variables Volume, Pressure, and Temperature......Page 23
1.3. Basic Definitions Needed to Describe Thermodynamic Systems......Page 27
1.4. Equations of State and the Ideal Gas Law......Page 29
1.5. A Brief Introduction to Real Gases......Page 31
Math Essential 2: Differentiation and Integration......Page 38
2.1. Internal Energy and the First Law of Thermodynamics......Page 46
2.2. Heat......Page 47
2.3. Work......Page 48
2.4. Equilibrium, Change, and Reversibility......Page 50
2.5. The Work of Reversible Compression or Expansion of an Ideal Gas......Page 51
2.6. The Work of Irreversible Compression or Expansion of an Ideal Gas......Page 53
2.7. Other Examples of Work......Page 54
2.8. State Functions and Path Functions......Page 56
2.9. Comparing Work for Reversible and Irreversible Processes......Page 58
2.10. Changing the System Energy from a Molecular Level Perspective......Page 62
2.11. Heat Capacity......Page 64
2.12. Determining U and Introducing the State Function Enthalpy......Page 67
2.13. Calculating q, w, U, and H for Processes Involving Ideal Gases......Page 68
2.14. Reversible Adiabatic Expansion and Compression of an Ideal Gas......Page 72
Math Essential 3: Partial Derivatives......Page 80
3.1. Mathematical Properties of State Functions......Page 82
3.2. Dependence of U on V and T......Page 85
3.3. Does the Internal Energy Depend More Strongly on V or T?......Page 87
3.4. Variation of Enthalpy with Temperature at Constant Pressure......Page 91
3.5. How are CP and CV Related?......Page 93
3.6. Variation of Enthalpy with Pressure at Constant Temperature......Page 94
3.7. The Joule–Thomson Experiment......Page 96
3.8. Liquefying Gases Using an Isenthalpic Expansion......Page 98
4.1. Energy Stored in Chemical Bonds Is Released or Absorbed in Chemical Reactions......Page 104
4.2. Internal Energy and Enthalpy Changes Associated with Chemical Reactions......Page 105
4.3. Hess’s Law Is Based on Enthalpy Being a State Function......Page 108
4.4. Temperature Dependence of Reaction Enthalpies......Page 110
4.5. Experimental Determination of U and H for Chemical Reactions......Page 112
4.6. Differential Scanning Calorimetry......Page 114
5.1. What Determines the Direction of Spontaneous Change in a Process?......Page 124
5.2. The Second Law of Thermodynamics, Spontaneity, and the Sign of S......Page 126
5.3. Calculating Changes in Entropy as T, P, or V Change......Page 127
5.4. Understanding Changes in Entropy at the Molecular Level......Page 131
5.5. The Clausius Inequality......Page 133
5.6. The Change of Entropy in the Surroundings and Stot = S + Ssur......Page 134
5.7. Absolute Entropies and the Third Law of Thermodynamics......Page 136
5.9. Entropy Changes in Chemical Reactions......Page 140
5.10. Heat Engines and the Carnot Cycle......Page 142
5.11. How Does S Depend on V and T ?......Page 147
5.12. Dependence of S on T and P......Page 148
5.13. Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines......Page 149
6.1. Gibbs Energy and Helmholtz Energy......Page 164
6.2. Differential Forms of U, H, A, and G......Page 168
6.3. Dependence of Gibbs and Helmholtz Energies on P, V, and T......Page 170
6.4. Gibbs Energy of a Reaction Mixture......Page 172
6.5. Calculating the Gibbs Energy of Mixing for Ideal Gases......Page 174
6.6. Calculating the Equilibrium Position for a GasPhase Chemical Reaction......Page 176
6.7. Introducing the Equilibrium Constant for a Mixture of Ideal Gases......Page 179
6.8. Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gases......Page 183
6.9. Variation of KP with Temperature......Page 184
6.10. Equilibria Involving Ideal Gases and Solid or Liquid Phases......Page 186
6.11. Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarity......Page 188
6.12. Expressing U, H, and Heat Capacities Solely in Terms of Measurable Quantities......Page 189
6.13. A Case Study: The Synthesis of Ammonia......Page 193
6.14. Measuring G for the Unfolding of Single RNA Molecules......Page 197
7.1. Real Gases and Ideal Gases......Page 206
7.2. Equations of State for Real Gases and Their Range of Applicability......Page 207
7.3. The Compression Factor......Page 211
7.4. The Law of Corresponding States......Page 214
7.5. Fugacity and the Equilibrium Constant for Real Gases......Page 217
8.1. What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?......Page 224
8.2. The Pressure–Temperature Phase Diagram......Page 227
8.4. Pressure–Volume and Pressure–Volume– Temperature Phase Diagrams......Page 234
8.5. Providing a Theoretical Basis for the P–T Phase Diagram......Page 236
8.6. Using the Clausius–Clapeyron Equation to Calculate Vapor Pressure as a Function of T......Page 238
8.7. Dependence of Vapor Pressure of a Pure Substance on Applied Pressure......Page 240
8.8. Surface Tension......Page 241
8.9. Chemistry in Supercritical Fluids......Page 244
8.10. Liquid Crystal Displays......Page 245
9.1. Defining the Ideal Solution......Page 254
9.2. The Chemical Potential of a Component in the Gas and Solution Phases......Page 256
9.3. Applying the Ideal Solution Model to Binary Solutions......Page 257
9.4. The Temperature–Composition Diagram and Fractional Distillation......Page 261
9.5. The Gibbs–Duhem Equation......Page 263
9.6. Colligative Properties......Page 264
9.7. Freezing Point Depression and Boiling Point Elevation......Page 265
9.8. Osmotic Pressure......Page 267
9.9. Deviations from Raoult’s Law in Real Solutions......Page 269
9.10. The Ideal Dilute Solution......Page 271
9.11. Activities are Defined with Respect to Standard States......Page 273
9.12. Henry’s Law and the Solubility of Gases in a Solvent......Page 277
9.13. Chemical Equilibrium in Solutions......Page 278
9.14. Solutions Formed from Partially Miscible Liquids......Page 281
9.15. Solid–Solution Equilibrium......Page 283
10.1. Enthalpy, Entropy, and Gibbs Energy of Ion Formation in Solutions......Page 290
10.2. Understanding the Thermodynamics of Ion Formation and Solvation......Page 292
10.3. Activities and Activity Coefficients for Electrolyte Solutions......Page 295
10.4. Calculating g+ Using the Debye–Hückel Theory......Page 297
10.5. Chemical Equilibrium in Electrolyte Solutions......Page 301
11.1. The Effect of an Electrical Potential on the Chemical Potential of Charged Species......Page 308
11.2. Conventions and Standard States in Electrochemistry......Page 310
11.4. Chemical Reactions in Electrochemical Cells and the Nernst Equation......Page 313
11.5. Combining Standard Electrode Potentials to Determine the Cell Potential......Page 315
11.7. Relationship Between the Cell EMF and the Equilibrium Constant......Page 317
11.8. Determination of E ~ and Activity Coefficients Using an Electrochemical Cell......Page 319
11.9. Cell Nomenclature and Types of Electrochemical Cells......Page 320
11.10. The Electrochemical Series......Page 321
11.11. Thermodynamics of Batteries and Fuel Cells......Page 322
11.12. Electrochemistry of Commonly Used Batteries......Page 323
11.13. Fuel Cells......Page 327
11.14. Electrochemistry at the Atomic Scale......Page 329
11.15. Using Electrochemistry for Nanoscale Machining......Page 332
12.1. Why Probability?......Page 338
12.2. Basic Probability Theory......Page 339
12.3. Stirling’s Approximation......Page 347
12.4. Probability Distribution Functions......Page 348
12.5. Probability Distributions Involving Discrete and Continuous Variables......Page 350
12.6. Characterizing Distribution Functions......Page 353
Math Essential 4: Lagrange Multipliers......Page 364
13.1. Microstates and Configurations......Page 366
13.2. Derivation of the Boltzmann Distribution......Page 372
13.3. Dominance of the Boltzmann Distribution......Page 377
13.4. Physical Meaning of the Boltzmann Distribution Law......Page 379
13.5. The Definition of β......Page 380
14.1. The Canonical Ensemble......Page 390
14.2. Relating Q to q for an Ideal Gas......Page 392
14.3. Molecular Energy Levels......Page 394
14.4. Translational Partition Function......Page 395
14.5. Rotational Partition Function: Diatomic Molecules......Page 397
14.6. Rotational Partition Function: Polyatomic Molecules......Page 405
14.7. Vibrational Partition Function......Page 407
14.8. The Equipartition Theorem......Page 412
14.9. Electronic Partition Function......Page 413
14.10. Review......Page 417
15.1. Energy......Page 424
15.2. Energy and Molecular Energetic Degrees of Freedom......Page 428
15.3. Heat Capacity......Page 433
15.4. Entropy......Page 438
15.5. Residual Entropy......Page 443
15.6. Other Thermodynamic Functions......Page 444
15.7. Chemical Equilibrium......Page 448
16.1. Kinetic Theory of Gas Motion and Pressure......Page 458
16.2. Velocity Distribution in One Dimension......Page 459
16.3. The Maxwell Distribution of Molecular Speeds......Page 463
16.4. Comparative Values for Speed Distributions......Page 466
16.5. Gas Effusion......Page 468
16.6. Molecular Collisions......Page 470
16.7. The Mean Free Path......Page 474
17.1. What Is Transport?......Page 480
17.2. Mass Transport: Diffusion......Page 482
17.3. Time Evolution of a Concentration Gradient......Page 486
17.4. Statistical View of Diffusion......Page 488
17.5. Thermal Conduction......Page 490
17.6. Viscosity of Gases......Page 493
17.7. Measuring Viscosity......Page 496
17.8. Diffusion and Viscosity of Liquids......Page 497
17.9. Ionic Conduction......Page 499
18.1. Introduction to Kinetics......Page 510
18.2. Reaction Rates......Page 511
18.3. Rate Laws......Page 513
18.4. Reaction Mechanisms......Page 518
18.5. Integrated Rate Law Expressions......Page 519
18.6. Numerical Approaches......Page 524
18.7. Sequential FirstOrder Reactions......Page 525
18.8. Parallel Reactions......Page 530
18.9. Temperature Dependence of Rate Constants......Page 532
18.10. Reversible Reactions and Equilibrium......Page 534
18.11. PerturbationRelaxation Methods......Page 538
18.12. The Autoionization of Water: A Temperature Jump Example......Page 540
18.13. Potential Energy Surfaces......Page 541
18.14. Activated Complex Theory......Page 543
18.15. DiffusionControlled Reactions......Page 547
19.1. Reaction Mechanisms and Rate Laws......Page 558
19.2. The Preequilibrium Approximation......Page 560
19.3. The Lindemann Mechanism......Page 562
19.4. Catalysis......Page 564
19.5. RadicalChain Reactions......Page 575
19.6. RadicalChain Polymerization......Page 578
19.7. Explosions......Page 579
19.8. Feedback, Nonlinearity, and Oscillating Reactions......Page 581
19.9. Photochemistry......Page 584
19.10. Electron Transfer......Page 596
20.1. What Are Macromolecules?......Page 610
20.2. Macromolecular Structure......Page 611
20.3. RandomCoil Model......Page 613
20.4. Biological Polymers......Page 616
20.5. Synthetic Polymers......Page 624
20.6. Characterizing Macromolecules......Page 627
20.7. SelfAssembly, Micelles, and Biological Membranes......Page 634
Appendix A: Data Tables......Page 642
Credits......Page 660
Index......Page 661
Answers to Selected Numerical Problems......Page 678
Back Cover......Page 689
ENGEL REID PHYSICAL CHEMISTRY
Thermodynamics, Statistical Thermodynamics, and Kinetics 4e
A visual, conceptual and contemporary approach to the fascinating field of Physical Chemistry guides students through core concepts with visual narratives and connections to cuttingedge applications and research. The fourth edition of Thermodynamics, Statistical Thermodynamics, & Kinetics includes many changes to the presentation and content at both a global and chapter level. These updates have been made to enhance the student learning experience and update the discussion of research areas. MasteringTM Chemistry, with a new enhanced Pearson eText, has been significantly expanded to include a wealth of new endofchapter problems from the 4th edition, new selfguided, adaptive Dynamic Study Modules with wrong answer feedback and remediation, and the new Pearson eText which is mobile friendly.
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Thermodynamics, Statistical Thermodynamics, and Kinetics 4e
Thomas Engel Philip Reid
PHYSICAL CHEMISTRY
Thermodynamics, Statistical Thermodynamics, and Kinetics FOURTH EDITION
Thomas Engel University of Washington
Philip Reid University of Washington
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ISBN 10: 0134804589; ISBN 13: 9780134804583 (Student edition) ISBN 10: 0134814614; ISBN 13: 9780134814612 (Books A La Carte edition)
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To Walter and Juliane, my first teachers, and to Gloria, Alex, Gabrielle, and Amelie. THOMAS ENGEL
To my family. PHILIP REID
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Brief Contents
THERMODYNAMICS, STATISTICAL THERMODYNAMICS, AND KINETICS
1 Fundamental Concepts of Thermodynamics 5
12 Probability 321
2 Heat, Work, Internal Energy, Enthalpy, and the
13 The Boltzmann Distribution 349
First Law of Thermodynamics 29
3 The Importance of State Functions: Internal Energy and Enthalpy 65
4 Thermochemistry 87
5 Entropy and the Second and Third Laws of Thermodynamics 107
6 Chemical Equilibrium 147
7 The Properties of Real Gases 189
8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases 207
9 Ideal and Real Solutions 237
10 Electrolyte Solutions 273
14 Ensemble and Molecular Partition Functions 373
15 Statistical Thermodynamics
407
16 Kinetic Theory of Gases 441 17 Transport Phenomena
463
18 Elementary Chemical Kinetics 493 19 Complex Reaction Mechanisms 541 20 Macromolecules APPENDIX
593
A Data Tables 625
Credits 643 Index 644
11 Electrochemical Cells, Batteries, and Fuel Cells 291
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Detailed Contents
THERMODYNAMICS, STATISTICAL THERMODYNAMICS, AND KINETICS Preface x
Math Essential 3 Partial Derivatives
Math Essential 1 Units, Significant Figures, and Solving End of Chapter Problems
3 The Importance of State Functions: Internal Energy and Enthalpy 65
1 Fundamental Concepts of Thermodynamics 5 1.1 What Is Thermodynamics and Why Is It Useful? 5 1.2 The Macroscopic Variables Volume, Pressure, and Temperature 6 1.3 Basic Definitions Needed to Describe Thermodynamic Systems 10 1.4 Equations of State and the Ideal Gas Law 12 1.5 A Brief Introduction to Real Gases 14 Math Essential 2 Differentiation and Integration
2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics 29 2.1 Internal Energy and the First Law of Thermodynamics 29 2.2 Heat 30 2.3 Work 31 2.4 Equilibrium, Change, and Reversibility 33 2.5 The Work of Reversible Compression or Expansion of an Ideal Gas 34 2.6 The Work of Irreversible Compression or Expansion of an Ideal Gas 36 2.7 Other Examples of Work 37 2.8 State Functions and Path Functions 39 2.9 Comparing Work for Reversible and Irreversible Processes 41 2.10 Changing the System Energy from a MolecularLevel Perspective 45 2.11 Heat Capacity 47 2.12 Determining ∆U and Introducing the State Function Enthalpy 50 2.13 Calculating q, w, ∆U, and ∆H for Processes Involving Ideal Gases 51 2.14 Reversible Adiabatic Expansion and Compression of an Ideal Gas 55
3.1 Mathematical Properties of State Functions 65 3.2 Dependence of U on V and T 68 3.3 Does the Internal Energy Depend More Strongly on V or T? 70 3.4 Variation of Enthalpy with Temperature at Constant Pressure 74 3.5 How are CP and CV Related? 76 3.6 Variation of Enthalpy with Pressure at Constant Temperature 77 3.7 The Joule–Thomson Experiment 79 3.8 Liquefying Gases Using an Isenthalpic Expansion 81
4 Thermochemistry 87 4.1 Energy Stored in Chemical Bonds Is Released or Absorbed in Chemical Reactions 87 4.2 Internal Energy and Enthalpy Changes Associated with Chemical Reactions 88 4.3 Hess’s Law Is Based on Enthalpy Being a State Function 91 4.4 Temperature Dependence of Reaction Enthalpies 93 4.5 Experimental Determination of ∆U and ∆H for Chemical Reactions 95 4.6 Differential Scanning Calorimetry 97
5 Entropy and the Second and Third Laws of Thermodynamics 107 5.1 What Determines the Direction of Spontaneous Change in a Process? 107 5.2 The Second Law of Thermodynamics, Spontaneity, and the Sign of ∆S 109 5.3 Calculating Changes in Entropy as T, P, or V Change 110 5.4 Understanding Changes in Entropy at the Molecular Level 114 5.5 The Clausius Inequality 116 v
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5.6 The Change of Entropy in the Surroundings and ∆Stot = ∆S + ∆Ssur 117 5.7 Absolute Entropies and the Third Law of Thermodynamics 119 5.8 Standard States in Entropy Calculations 123 5.9 Entropy Changes in Chemical Reactions 123 5.10 Heat Engines and the Carnot Cycle 125 5.11 How Does S Depend on V and T ? 130 5.12 Dependence of S on T and P 131 5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines 132
6 Chemical Equilibrium 147 6.1 Gibbs Energy and Helmholtz Energy 147 6.2 Differential Forms of U, H, A, and G 151 6.3 Dependence of Gibbs and Helmholtz Energies on P, V, and T 153 6.4 Gibbs Energy of a Reaction Mixture 155 6.5 Calculating the Gibbs Energy of Mixing for Ideal Gases 157 6.6 Calculating the Equilibrium Position for a GasPhase Chemical Reaction 159 6.7 Introducing the Equilibrium Constant for a Mixture of Ideal Gases 162 6.8 Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gases 166 6.9 Variation of KP with Temperature 167 6.10 Equilibria Involving Ideal Gases and Solid or Liquid Phases 169 6.11 Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarity 171 6.12 Expressing U, H, and Heat Capacities Solely in Terms of Measurable Quantities 172 6.13 A Case Study: The Synthesis of Ammonia 176 6.14 Measuring ∆G for the Unfolding of Single RNA Molecules 180
7 The Properties of Real Gases 189 7.1 Real Gases and Ideal Gases 189 7.2 Equations of State for Real Gases and Their Range of Applicability 190 7.3 The Compression Factor 194 7.4 The Law of Corresponding States 197 7.5 Fugacity and the Equilibrium Constant for Real Gases 200
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8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases 207 8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases? 207 8.2 The Pressure–Temperature Phase Diagram 210 8.3 The Phase Rule 217 8.4 Pressure–Volume and Pressure–Volume– Temperature Phase Diagrams 217 8.5 Providing a Theoretical Basis for the P–T Phase Diagram 219 8.6 Using the Clausius–Clapeyron Equation to Calculate Vapor Pressure as a Function of T 221 8.7 Dependence of Vapor Pressure of a Pure Substance on Applied Pressure 223 8.8 Surface Tension 224 8.9 Chemistry in Supercritical Fluids 227 8.10 Liquid Crystal Displays 228
9 Ideal and Real Solutions 237 9.1 Defining the Ideal Solution 237 9.2 The Chemical Potential of a Component in the Gas and Solution Phases 239 9.3 Applying the Ideal Solution Model to Binary Solutions 240 9.4 The Temperature–Composition Diagram and Fractional Distillation 244 9.5 The Gibbs–Duhem Equation 246 9.6 Colligative Properties 247 9.7 Freezing Point Depression and Boiling Point Elevation 248 9.8 Osmotic Pressure 250 9.9 Deviations from Raoult’s Law in Real Solutions 252 9.10 The Ideal Dilute Solution 254 9.11 Activities are Defined with Respect to Standard States 256 9.12 Henry’s Law and the Solubility of Gases in a Solvent 260 9.13 Chemical Equilibrium in Solutions 261 9.14 Solutions Formed from Partially Miscible Liquids 264 9.15 Solid–Solution Equilibrium 266
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CONTENTS
10 Electrolyte Solutions 273 10.1 Enthalpy, Entropy, and Gibbs Energy of Ion Formation in Solutions 273 10.2 Understanding the Thermodynamics of Ion Formation and Solvation 275 10.3 Activities and Activity Coefficients for Electrolyte Solutions 278 10.4 Calculating g{ Using the Debye–Hückel Theory 280 10.5 Chemical Equilibrium in Electrolyte Solutions 284
11 Electrochemical Cells, Batteries, and Fuel Cells 291 11.1 The Effect of an Electrical Potential on the Chemical Potential of Charged Species 291 11.2 Conventions and Standard States in Electrochemistry 293 11.3 Measurement of the Reversible Cell Potential 296 11.4 Chemical Reactions in Electrochemical Cells and the Nernst Equation 296 11.5 Combining Standard Electrode Potentials to Determine the Cell Potential 298 11.6 Obtaining Reaction Gibbs Energies and Reaction Entropies from Cell Potentials 300 11.7 Relationship Between the Cell EMF and the Equilibrium Constant 300 11.8 Determination of E ~ and Activity Coefficients Using an Electrochemical Cell 302 11.9 Cell Nomenclature and Types of Electrochemical Cells 303 11.10 The Electrochemical Series 304 11.11 Thermodynamics of Batteries and Fuel Cells 305 11.12 Electrochemistry of Commonly Used Batteries 306 11.13 Fuel Cells 310 11.14 Electrochemistry at the Atomic Scale 312 11.15 Using Electrochemistry for Nanoscale Machining 315
12 Probability 321 12.1 Why Probability? 321 12.2 Basic Probability Theory 322 12.3 Stirling’s Approximation 330 12.4 Probability Distribution Functions 331 12.5 Probability Distributions Involving Discrete and Continuous Variables 333 12.6 Characterizing Distribution Functions 336
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Math Essential 4 Lagrange Multipliers
13 The Boltzmann Distribution 349 13.1 Microstates and Configurations 349 13.2 Derivation of the Boltzmann Distribution 355 13.3 Dominance of the Boltzmann Distribution 360 13.4 Physical Meaning of the Boltzmann Distribution Law 362 13.5 The Definition of b 363
14 Ensemble and Molecular Partition Functions 373 14.1 14.2 14.3 14.4 14.5
The Canonical Ensemble 373 Relating Q to q for an Ideal Gas 375 Molecular Energy Levels 377 Translational Partition Function 378 Rotational Partition Function: Diatomic Molecules 380 14.6 Rotational Partition Function: Polyatomic Molecules 388 14.7 Vibrational Partition Function 390 14.8 The Equipartition Theorem 395 14.9 Electronic Partition Function 396 14.10 Review 400
15 Statistical Thermodynamics 407 15.1 Energy 407 15.2 Energy and Molecular Energetic Degrees of Freedom 411 15.3 Heat Capacity 416 15.4 Entropy 421 15.5 Residual Entropy 426 15.6 Other Thermodynamic Functions 427 15.7 Chemical Equilibrium 431
16 Kinetic Theory of Gases 441 16.1 Kinetic Theory of Gas Motion and Pressure 441 16.2 Velocity Distribution in One Dimension 442 16.3 The Maxwell Distribution of Molecular Speeds 446 16.4 Comparative Values for Speed Distributions 449 16.5 Gas Effusion 451 16.6 Molecular Collisions 453 16.7 The Mean Free Path 457
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CONTENTS
17 Transport Phenomena 463 17.1 What Is Transport? 463 17.2 Mass Transport: Diffusion 465 17.3 Time Evolution of a Concentration Gradient 469 17.4 Statistical View of Diffusion 471 17.5 Thermal Conduction 473 17.6 Viscosity of Gases 476 17.7 Measuring Viscosity 479 17.8 Diffusion and Viscosity of Liquids 480 17.9 Ionic Conduction 482
18 Elementary Chemical Kinetics 493 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9
Introduction to Kinetics 493 Reaction Rates 494 Rate Laws 496 Reaction Mechanisms 501 Integrated Rate Law Expressions 502 Numerical Approaches 507 Sequential FirstOrder Reactions 508 Parallel Reactions 513 Temperature Dependence of Rate Constants 515 18.10 Reversible Reactions and Equilibrium 517 18.11 PerturbationRelaxation Methods 521 18.12 The Autoionization of Water: A TemperatureJump Example 523 18.13 Potential Energy Surfaces 524 18.14 Activated Complex Theory 526 18.15 DiffusionControlled Reactions 530
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19 Complex Reaction Mechanisms 541 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8
Reaction Mechanisms and Rate Laws 541 The Preequilibrium Approximation 543 The Lindemann Mechanism 545 Catalysis 547 RadicalChain Reactions 558 RadicalChain Polymerization 561 Explosions 562 Feedback, Nonlinearity, and Oscillating Reactions 564 19.9 Photochemistry 567 19.10 Electron Transfer 579
20 Macromolecules 593 20.1 What Are Macromolecules? 593 20.2 Macromolecular Structure 594 20.3 RandomCoil Model 596 20.4 Biological Polymers 599 20.5 Synthetic Polymers 607 20.6 Characterizing Macromolecules 610 20.7 SelfAssembly, Micelles, and Biological Membranes 617 APPENDIX
A Data Tables 625
Credits 643 Index 644
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About the Authors THOMAS ENGEL taught chemistry at the University of Washington for more than 20 years, where he is currently professor emeritus of chemistry. Professor Engel received his bachelor’s and master’s degrees in chemistry from the Johns Hopkins University and his Ph.D. in chemistry from the University of Chicago. He then spent 11 years as a researcher in Germany and Switzerland, during which time he received the Dr. rer. nat. habil. degree from the Ludwig Maximilians University in Munich. In 1980, he left the IBM research laboratory in Zurich to become a faculty member at the University of Washington. Professor Engel has published more than 80 articles and book chapters in the area of surface chemistry. He has received the Surface Chemistry or Colloids Award from the American Chemical Society and a Senior Humboldt Research Award from the Alexander von Humboldt Foundation. Other than this textbook, his current primary science interests are in energy policy and energy conservation. He serves on the citizen’s advisory board of his local electrical utility, and his energyefficient house could be heated in winter using only a handheld hair dryer. He currently drives a hybrid vehicle and plans to transition to an electric vehicle soon to further reduce his carbon footprint. PHILIP REID has taught chemistry at the University of Washington since 1995. Professor Reid received his bachelor’s degree from the University of Puget Sound in 1986 and his Ph.D. from the University of California, Berkeley in 1992. He performed postdoctoral research at the University of MinnesotaTwin Cities before moving to Washington. Professor Reid’s research interests are in the areas of atmospheric chemistry, ultrafast condensedphase reaction dynamics, and organic electronics. He has published more than 140 articles in these fields. Professor Reid is the recipient of a CAREER Award from the National Science Foundation, is a Cottrell Scholar of the Research Corporation, and is a Sloan Fellow. He received the University of Washington Distinguished Teaching Award in 2005.
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Preface The fourth edition of Thermodynamics, Statistical Thermodynamics, and Kinetics includes many changes to the presentation and content at both a global and chapter level. These updates have been made to enhance the student learning experience and update the discussion of research areas. At the global level, changes that readers will see throughout the textbook include:
• Review of relevant mathematics skills. One of the
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primary reasons that students experience physical chemistry as a challenging course is that they find it difficult to transfer skills previously acquired in a mathematics course to their physical chemistry course. To address this issue, contents of the third edition Math Supplement have been expanded and split into 11 two to fivepage Math Essentials, which are inserted at appropriate places throughout this book, as well as in the companion volume Quantum Chemistry & Spectroscopy, just before the math skills are required. Our intent in doing so is to provide “justintime” math help and to enable students to refresh math skills specifically needed in the following chapter. Concept and Connection. A new Concept and Connection feature has been added to each chapter to present students with a quick visual summary of the most important ideas within the chapter. In each chapter, approximately 10–15 of the most important concepts and/or connections are highlighted in the margins. EndofChapter Problems. Numerical Problems are now organized by section number within chapters to make it easier for instructors to create assignments for specific parts of each chapter. Furthermore, a number of new Conceptual Questions and Numerical Problems have been added to the book. Numerical Problems from the previous edition have been revised. Introductory chapter materials. Introductory paragraphs of all chapters have been replaced by a set of three questions plus responses to those questions. This new feature makes the importance of the chapter clear to students at the outset. Figures. All figures have been revised to improve clarity. Also, for many figures, additional annotation has been included to help tie concepts to the visual program. Key Equations. An endofchapter table that summarizes Key Equations has been added to allow students to focus on the most important of the many equations in each chapter. Equations in this table are set in red type where they appear in the body of the chapter. Further Reading. A section on Further Reading has been added to each chapter to provide references for students and instructors who would like a deeper understanding of various aspects of the chapter material. Guided Practice and Interactivity TM ° Mastering Chemistry with a new enhanced eBook,
x
has been significantly expanded to include a wealth of
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new endofchapter problems from the fourth edition, new selfguided, adaptive Dynamic Study Modules with wrong answer feedback and remediation, and the new Pearson eBook which is mobile friendly. Students who solve homework problems using MasteringTM Chemistry obtain immediate feedback, which greatly enhances learning associated with solving homework problems. This platform can also be used for preclass reading quizzes linked directly to the eText that are useful in ensuring students remain current in their studies and in flipping the classroom. NEW! Pearson eText, optimized for mobile, gives students access to their textbook anytime, anywhere. Pearson eText mobile app offers offline access and can be downloaded for most iOS and Android phones/tablets from the Apple App Store or Google Play. Configurable reading settings, including resizable type and nightreading mode Instructor and student notetaking, highlighting, bookmarking, and search functionalities
■
■
■
° NEW! 66 Dynamic Study Modules help students °
°
study effectively on their own by continuously assessing their activity and performance in real time. Students complete a set of questions with a unique answer format that also asks them to indicate their confidence level. Questions repeat until the student can answer them all correctly and confidently. These are available as graded assignments prior to class and are accessible on smartphones, tablets, and computers. Topics include key math skills, as well as a refresher of general chemistry concepts, such as understanding matter, chemical reactions, and the periodic table and atomic structure. Topics can be added or removed to match your coverage.
In terms of chapter and section content, many changes were made. The most significant of these changes are as follows:
• A new chapter entitled Macromolecules (Chapter 20) has
•
been added. The motivation for this chapter is that assemblies of smaller molecules form large molecules, such as proteins or polymers. The resulting macromolecules can exhibit new structures and functions that are not reflected by the individual molecular components. Understanding the factors that influence macromolecular structure is critical in understanding the chemical behavior of these important molecules. A more detailed discussion of systembased and surroundingsbased work has been added in Chapter 2 to help clarify the confusion that has appeared in the chemical education literature about using the system or surroundings pressure in calculating work. Section 6.6
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•
•
PREFACE
has been extensively revised to take advances in quantum computing into account. The discussion on entropy and the second law of thermodynamics in Chapter 5 has been substantially revised. As a result, calculations of entropy changes now appear earlier in the chapter, and the material on the reversible Carnot cycle has been shifted to a later section. The approach to chemical equilibrium in Chapter 6 has been substantially revised to present a formulation in terms of the extent of reaction. This change has been made to focus more clearly on changes in chemical potential as the driving force in reaching equilibrium.
For those not familiar with the third edition of Thermodynamics, Statistical Thermodynamics, and Kinetics, our approach to teaching physical chemistry begins with our target audience— undergraduate students majoring in chemistry, biochemistry, and chemical engineering, as well as many students majoring in the atmospheric sciences and the biological sciences. The following objectives outline our approach to teaching physical chemistry.
• Focus on teaching core concepts. The central principles
•
•
•
•
of physical chemistry are explored by focusing on core ideas and then extending these ideas to a variety of problems. The goal is to build a solid foundation of student understanding in a limited number of areas rather than to provide a condensed encyclopedia of physical chemistry. We believe this approach teaches students how to learn and enables them to apply their newly acquired skills to master related fields. Illustrate the relevance of physical chemistry to the world around us. Physical chemistry becomes more relevant to a student if it is connected to the world around us. Therefore, example problems and specific topics are tied together to help the student develop this connection. For example, fuel cells, refrigerators, heat pumps, and real engines are discussed in connection with the second law of thermodynamics. Every attempt is made to connect fundamental ideas to applications that could be of interest to the student. Link the macroscopic and atomiclevel worlds. One of the strengths of thermodynamics is that it is not dependent on a microscopic description of matter. However, students benefit from a discussion of issues such as how pressure originates from the random motion of molecules. Present exciting new science in the field of physical chemistry. Physical chemistry lies at the forefront of many emerging areas of modern chemical research. Heterogeneous catalysis has benefited greatly from mechanistic studies carried out using the techniques of modern surface science. Atomicscale electrochemistry has become possible through scanning tunneling microscopy. The role of physical chemistry in these and other emerging areas is highlighted throughout the text. Provide a versatile online homework program with tutorials. Students who submit homework problems using MasteringTM Chemistry obtain immediate feedback,
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xi
a feature that greatly enhances learning. Also, tutorials with wrong answer feedback offer students a selfpaced learning environment. Use webbased simulations to illustrate the concepts being explored and avoid math overload. Mathematics is central to physical chemistry; however, the mathematics can distract the student from “seeing” the underlying concepts. To circumvent this problem, webbased simulations have been incorporated as endofchapter problems in several chapters so that the student can focus on the science and avoid a math overload. These webbased simulations can also be used by instructors during lecture. An important feature of the simulations is that each problem has been designed as an assignable exercise with a printable answer sheet that the student can submit to the instructor. Simulations, animations, and homework problem worksheets can be accessed at www.pearsonhighered.com/ advchemistry.
Effective use of Thermodynamics, Statistical Thermodynamics, and Kinetics does not require proceeding sequentially through the chapters or including all sections. Some topics are discussed in supplemental sections, which can be omitted if they are not viewed as essential to the course. Also, many sections are sufficiently selfcontained that they can be readily omitted if they do not serve the needs of the instructor and students. This textbook is constructed to be flexible to your needs. We welcome the comments of both students and instructors on how the material was used and how the presentation can be improved. Thomas Engel and Philip Reid University of Washington
ACKNOWLEDGMENTS Many individuals have helped us to bring the text into its current form. Students have provided us with feedback directly and through the questions they have asked, which has helped us to understand how they learn. Many of our colleagues, including Peter Armentrout, Doug Doren, Gary Drobny, Eric Gislason, Graeme Henkelman, Lewis Johnson, Tom Pratum, Bill Reinhardt, Peter Rosky, George Schatz, Michael Schick, Gabrielle Varani, and especially Bruce Robinson, have been invaluable in advising us. We are also fortunate to have access to some endofchapter problems that were originally presented in Physical Chemistry, 3rd edition, by Joseph H. Noggle and in Physical Chemistry, 3rd edition, by Gilbert W. Castellan. The reviewers, who are listed separately, have made many suggestions for improvement, for which we are very grateful. All those involved in the production process have helped to make this book a reality through their efforts. Special thanks are due to Jim Smith, who guided us through the first edition, to our current editor Jeanne Zalesky, to our developmental editor Spencer Cotkin, and to Jennifer Hart and Beth Sweeten at Pearson, who have led the production process.
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xii
PREFACE
4TH EDITION MANUSCRIPT REVIEWERS David Coker, Boston University Yingbin Ge, Central Washington University Eric Gislason, University of Illinois, Chicago
Nathan Hammer, University of Mississippi George Papadantonakis, University of Illinois, Chicago
Stefan Stoll, University of Washington Liliya Yatsunyk, Swarthmore College
4TH EDITION ACCURACY REVIEWERS Garry Crosson, University of Dayton
Benjamin Huddle, Roanoke College
Andrea Munro, Pacific Lutheran University
4TH EDITION PRESCRIPTIVE REVIEWERS Joseph Alia, University of Minnesota, Morris Herbert Axelrod, California State University, Fullerton Timothy Brewer, Eastern Michigan University Paul Cooper, George Mason University Bridget DePrince, Florida State University
Patrick Fleming, California State University, East Bay Richard Mabbs, Washington University, St. Louis Vicki Moravec, Trine University Andrew Petit, California State University, Fullerton Richard Schwenz, University of Northern Colorado
Ronald Terry, Western Illinois University Dunwei Wang, Boston College Gerard Harbison, University of Nebraska, Lincoln
Sophya Garashchuk, University of South Carolina Leon Gerber, St. John’s University Nathan Hammer, The University of Mississippi Cynthia Hartzell, Northern Arizona University Geoffrey Hutchinson, University of Pittsburgh John M. Jean, Regis University Martina Kaledin, Kennesaw State University George Kaminski, Central Michigan University Daniel Lawson, University of Michigan, Dearborn
William Lester, University of California, Berkeley Dmitrii E. Makarov, University of Texas at Austin Herve Marand, Virginia Polytechnic Institute and State University Thomas Mason, University of California, Los Angeles Jennifer Mihalik, University of Wisconsin, Oshkosh Enrique PeacockLópez, Williams College
PREVIOUS EDITION REVIEWERS Alexander Angerhofer, University of Florida Clayton Baum, Florida Institute of Technology Martha Bruch, State University of New York at Oswego David L. Cedeño, Illinois State University Rosemarie Chinni, Alvernia College Allen Clabo, Francis Marion University Lorrie Comeford, Salem State College Stephen Cooke, University of North Texas Douglas English, University of Maryland, College Park
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A Visual, Conceptual, and Contemporary Approach to Physical Chemistry
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A Visual, Conceptual, and Contemporary Approach to Physical Chemistry NEW! Math Essentials provide a review of relevant math skills, offer “just in time” math help, and enable students to refresh math skills specifically needed in the chapter that follows.
UPDATED! Introductory paragraphs of all chapters have been replaced by a set of three questions plus responses to those questions making the relevance of the chapter clear at the outset.
C H A P T E R
6
Commuting and Noncommuting Operators and the Surprising Consequences of Entanglement WHY is this material important?
6.1
Commutation Relations
6.2
The Stern–Gerlach Experiment
6.3
The Heisenberg Uncertainty Principle
6.4
(Supplemental Section) The Heisenberg Uncertainty Principle Expressed in Terms of Standard Deviations
6.5
(Supplemental Section) A Thought Experiment Using a Particle in a ThreeDimensional Box
6.6
(Supplemental Section) Entangled States, Teleportation, and Quantum Computers
The measurement process is different for a quantummechanical system than for a classical system. For a classical system, all observables can be measured simultaneously, and the precision and accuracy of the measurement is limited only by the instruments used to make the measurement. For a quantummechanical system, some observables can be measured simultaneously and exactly, whereas an uncertainty relation limits the degree to which other observables can be known simultaneously and exactly.
WHAT are the most important concepts and results? Measurements carried out on a system in a superposition state change the state of the system. Two observables can be measured simultaneously and exactly only if their corresponding operators commute. Two particles can be entangled, after which their properties are no longer independent of one another. Entanglement is the basis of both teleportation and quantum computing.
WHAT would be helpful for you to review for this chapter?
NEW! Concept and Connection features in each chapter present students with quick visual summaries of the core concepts within the chapter, highlighting key take aways and providing students with an easy way to review the material.
It would be helpful to review the material on operators in Chapter 2.
6.1 COMMUTATION RELATIONS Concept For a quantum mechanical system, it is not generally the case that the values of all observables can be known simultaneously.
Solid Critical point
Solid
Critical point
T
Gas
m l
q
id
b
Solid–Liqu
Pressure
Solid
lid
e
–G
as
f
Liq Ga uid e s
le
So
lum
Critical point
Gas
d
Tri p
p Vo
Liquid
c
a
Liq Ga uid s So lid –G as
0
k
g Triple point
A P–V–T phase diagram for a substance that expands upon melting. The indicated processes are discussed in the text.
Solid–Liquid Liquid
f
a
V
G
h
Lin
as
e
g
o n
T2 T1
T3Tc
T4
e tur era
mp
Te
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Figure 8.15
P m
P
uid
UPDATED! All figures have been revised to improve clarity and for many figures, additional annotation has been included to help tie concepts to the visual program.
105
Liq
In classical mechanics, a system can in principle be described completely. For instance, the position, momentum, kinetic energy, and potential energy of a mass falling in a gravitational field can be determined simultaneously at any point on its trajectory. The uncertainty in the measurements is only limited by the capabilities of the measurement technique. The values of all of these observables (and many more) can be known simultaneously. This is not generally true for a quantummechanical system. In the quantum world, in some cases two observables can be known simultaneously with high accuracy. However, in other cases, two observables have a fundamental uncertainty that cannot be eliminated through any measurement techniques. Nevertheless, as will be shown later,
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Continuous Learning Before, During, and After Class Mastering™ Chemistry
NEW! 66 Dynamic Study Modules help students study effectively on their own by continuously assessing their activity and performance in real time. Students complete a set of questions with a unique answer format that also asks them to indicate their confidence level. Questions repeat until the student can answer them all correctly and confidently. These are available as graded assignments prior to class and are accessible on smartphones, tablets, and computers. Topics include key math skills as well as a refresher of general chemistry concepts such as understanding matter, chemical reactions, and understanding the periodic table & atomic structure. Topics can be added or removed to match your coverage.
NEW! Enhanced EndofChapter and Tutorial Problems offer students the chance to practice what they have learned while receiving answerspecific feedback and guidance.
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Pearson eText NEW! Pearson eText, optimized for mobile gives students access to their textbook anytime,
anywhere.
Pearson eText is a mobile app which offers offline access and can be downloaded for most iOS and Android phones/tablets from the Apple App Store or Google Play: • Configurable reading settings, including resizable type and nightreading mode • Instructor and student notetaking, highlighting, bookmarking, and search functionalities
178
178
CHAPTER 6 Chemical Equilibrium
Figure 6.8
Enthalpy diagram for the reaction mechanism in the synthesis of ammonia. See Equations (6.91) through (6.95). The successive steps in the reaction proceed from left to right in the diagram.
CHAPTER 6 Chemical Equilibrium
Figure 6.8
314 3
N(g) 1 3H(g)
Enthalpy diagram for the reaction mechanism in the synthesis of ammonia. See Equations (6.91) through (6.95). The successive steps in the reaction proceed from left to right in the diagram.
N(g) 1 3H(g) 103
J
314 3 103 J NH(g) 1 2H(g)
1124 3 103 J
390 3 103 J NH2(g) 1 H(g)
NH(g) 1 2H(g)
466 3 103 J
1124 3 103 J
390 3 103 J
1
2
N2(g) 1
3
2
NH2(g) 1 H(g)
N2(g) 1
3
2
N21g2 + □ S N21a2
N21a2 + □ S 2N1a2
245.9 3 103 J
H2(g)
NH3(g) Progress of reaction
synthesis reaction, such a route is a heterogeneous catalytic reaction, using iron as a catalyst. The mechanism for this path between reactants and products is
466 3 103 J
1 2
245.9 3 103 J
H2(g)
H21g2 + 2□ S 2H1a2
NH3(g) Progress of reaction
N21g2 + □ S N21a2
(6.96)
H21g2 + 2□ S 2H1a2
N21a2 + □ S 2N1a2
(6.98)
N1a2 + H1a2 S NH1a2 + □
(6.99)
NH1a2 + H1a2 S NH21a2 + □
(6.100)
NH21a2 + H1a2 S NH31a2 + □ NH31a2 S NH31g2 + □
(6.97) (6.99) (6.100)
NH21a2 + H1a2 S NH31a2 + □ NH31a2 S NH31g2 + □
synthesis reaction, such a route is a heterogeneous catalytic reaction, using iron as a catalyst. The mechanism for this path between reactants and products is
(6.96) (6.98)
N1a2 + H1a2 S NH1a2 + □ NH1a2 + H1a2 S NH21a2 + □
(6.101) (6.102)
The symbol □ denotes an ensemble of neighboring Fe atoms, also called surface sites, which are capable of forming a chemical bond with the indicated entities. The designation (a) indicates that the chemical species is adsorbed (chemically bonded) to a surface site. The enthalpy change for the overall reaction N21g2 + 3>2 H21g2 S NH31g2 is the same for the mechanisms in Equations (6.91) through (6.95) and (6.96) through (6.102) because H is a state function. This is a characteristic of a catalytic reaction. A catalyst can affect the rate of the forward and backward reaction but not the position of equilibrium in a reaction system. The enthalpy diagram in Figure 6.9 shows that the
(6.97)
Homogeneous gasphase reactions N(g) 1 3H(g) 314 3 103 J NH(g) 1 2H(g)
(6.101)
Figure 6.9
1124 3 103 J
390 3 103 J NH (g) 1 H(g)
2 Enthalpy diagram for the homogeneous gasphase and heterogeneous catalytic NH3(g) 1 3 2 N2(g) 1 2 H2(g) 466 3 103 J reactions for the ammonia synthesis reaction. The activation barriers for the 0 245.9 3 103 J individual steps in the surface reaction Progress of reaction are shown. The successive steps in the Heterogeneous catalytic reactions reaction proceed from left to right in the diagram. See the reference to G. Ertl in NH2(a) 1 H(a) 245.9 3 103 J 1 3 Further Reading for more details. 2 N2(a) 1 2 H2(a) NH3(g) NH3(a) Adapted from G. Ertl, Catalysis NH(a) 1 2H(a) Ratelimiting step Reviews—Science and Engineering N(a) 1 3H(a) 21 (1980): 201–223.
(6.102)
The symbol □ denotes an ensemble of neighboring Fe atoms, also called surface sites, which are capable of forming a chemical bond with the indicated entities. The designation (a) indicates that the chemical species is adsorbed (chemically bonded) to a surface site. The enthalpy change for the overall reaction N21g2 + 3>2 H21g2 S NH31g2 is the same for the mechanisms in Equations (6.91) through (6.95) and (6.96) through (6.102) because H is a state function. This is a characteristic of a catalytic reaction. A catalyst can affect the rate of the forward and backward reaction but not the position of equilibrium in a reaction system. The enthalpy diagram in Figure 6.9 shows that the
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Homogeneous gasphase reactions N(g) 1 3H(g) 314 3 103 J NH(g) 1 2H(g)
Figure 6.9
1124 3 10
3J
390 3 103 J NH (g) 1 H(g)
2 Enthalpy diagram for the homogeneous gasphase and heterogeneous catalytic NH3(g) 1 3 2 N2(g) 1 2 H2(g) 466 3 103 J reactions for the ammonia synthesis reaction. The activation barriers for the 0 245.9 3 103 J individual steps in the surface reaction Progress of reaction are shown. The successive steps in the Heterogeneous catalytic reactions reaction proceed from left to right in the diagram. See the reference to G. Ertl in NH2(a) 1 H(a) 245.9 3 103 J 1 3 Further Reading for more details. 2 N2(a) 1 2 H2(a) NH3(g) NH3(a) Adapted from G. Ertl, Catalysis NH(a) 1 2H(a) Ratelimiting step Reviews—Science and Engineering N(a) 1 3H(a) 21 (1980): 201–223.
178
CHAPTER 6 Chemical Equilibrium
Figure 6.8
N(g) 1 3H(g)
Enthalpy diagram for the reaction mechanism in the synthesis of ammonia. See Equations (6.91) through (6.95). The successive steps in the reaction proceed from left to right in the diagram.
314 3 103 J NH(g) 1 2H(g)
1124 3 103 J
390 3 103 J NH2(g) 1 H(g)
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466 3 103 J
1
2
N2(g) 1
3
2
245.9 3 103 J
H2(g)
NH3(g) Progress of reaction
synthesis reaction, such a route is a heterogeneous catalytic reaction, using iron as a catalyst. The mechanism for this path between reactants and products is N21g2 + □ S N21a2
(6.96)
N21a2 + □ S 2N1a2
(6.98)
N1a2 + H1a2 S NH1a2 + □
(6.99)
NH1a2 + H1a2 S NH21a2 + □
(6.100)
H21g2 + 2□ S 2H1a2
NH21a2 + H1a2 S NH31a2 + □ NH31a2 S NH31g2 + □
(6.97)
(6.101) (6.102)
The symbol □ denotes an ensemble of neighboring Fe atoms, also called surface sites, which are capable of forming a chemical bond with the indicated entities. The designation (a) indicates that the chemical species is adsorbed (chemically bonded) to a surface site. The enthalpy change for the overall reaction N21g2 + 3>2 H21g2 S NH31g2 is the same for the mechanisms in Equations (6.91) through (6.95) and (6.96) through (6.102) because H is a state function. This is a characteristic of a catalytic reaction. A catalyst can affect the rate of the forward and backward reaction but not the position of equilibrium in a reaction system. The enthalpy diagram in Figure 6.9 shows that the Homogeneous gasphase reactions N(g) 1 3H(g) 314 3 103 J NH(g) 1 2H(g)
Figure 6.9
1124 3 103 J
390 3 103 J NH (g) 1 H(g)
2 Enthalpy diagram for the homogeneous gasphase and heterogeneous catalytic NH3(g) 1 3 2 N2(g) 1 2 H2(g) 466 3 103 J reactions for the ammonia synthesis reaction. The activation barriers for the 0 245.9 3 103 J individual steps in the surface reaction Progress of reaction are shown. The successive steps in the Heterogeneous catalytic reactions reaction proceed from left to right in the diagram. See the reference to G. Ertl in NH2(a) 1 H(a) 245.9 3 103 J 1 3 Further Reading for more details. 2 N2(a) 1 2 H2(a) NH3(g) NH3(a) Adapted from G. Ertl, Catalysis NH(a) 1 2H(a) Ratelimiting step Reviews—Science and Engineering N(a) 1 3H(a) 21 (1980): 201–223.
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MATH ESSENTIAL 1:
Units, Significant Figures, and Solving End of Chapter Problems ME1.1 UNITS
ME1.1 Units
Quantities of interest in physical chemistry such as pressure, volume, or temperature are characterized by their magnitude and their units. In this textbook, we use the SI (from the French Le Système international d'unités) system of units. All physical quantities can be defined in terms of the seven base units listed in Table ME1.1. For more details, see http://physics.nist.gov/cuu/Units/units.html. The definition of temperature is based on the coexistence of the solid, gaseous, and liquid phases of water at a pressure of 1 bar.
ME1.2 Uncertainty and Significant Figures ME1.3 Solving EndofChapter Problems
TABLE ME1.1 Base SI Units Base Unit
Unit
Definition of Unit
Unit of length
meter (m)
The meter is the length of the path traveled by light in vacuum during a time interval of 1>299,792,458 of a second.
Unit of mass
kilogram (kg)
The kilogram is the unit of mass; it is equal to the mass of the platinum iridium international prototype of the kilogram kept at the International Bureau of Weights and Measures.
Unit of time
second (s)
The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.
Unit of electric current
ampere (A)
The ampere is the constant current that, if maintained in two straight parallel conductors of infinite length, is of negligible circular cross section, and if placed 1 meter apart in a vacuum would produce between these conductors a force equal to 2 * 107 kg m s2 per meter of length. In this definition, 2 is an exact number.
Unit of thermodynamic temperature
kelvin (K)
The Kelvin is the unit of thermodynamic temperature. It is the fraction 1>273.16 of the thermodynamic temperature of the triple point of water.
Unit of amount of substance
mole (mol)
The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12 where 0.012 is an exact number. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
Unit of luminous intensity
candela (cd)
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540. * 1012 hertz and that has a radiant intensity in that direction of 1>683 watt per steradian.
Quantities of interest other than the seven base quantities can be expressed in terms of the units meter, kilogram, second, ampere, kelvin, mole, and candela. The most important of these derived units, some of which have special names as indicated, are listed in Table ME1.2. A more inclusive list of derived units can be found at http://physics .nist.gov/cuu/Units/units.html.
1
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MATH ESSENTIAL 1 Units, Significant Figures, and Solving End of Chapter Problems
TABLE ME1.2 Derived Units Unit
Definition
Relation to Base Units 2
Special Name
Abbreviation
Area
Size of a surface
m
m2
Volume
Amount of threedimensional space an object occupies
m3
m3
Velocity
Measure of the rate of motion
m s1
m s1
Acceleration
Rate of change of velocity
m s2
m s2 1
kg m s1
Linear momentum
Product of mass and linear velocity of an object
kg m s
Angular momentum
Product of the moment of inertia of a body about an axis and its angular velocity with respect to the same axis
kg m2 s1
Force
Any interaction that, when unopposed, will change the motion of an object
kg m s2
newton
N
Pressure
Force acting per unit area
kg m1 s2 N m2
pascal
Pa
Work
Product of force on an object and movement along the direction of the force
kg m2 s2
joule
J
Kinetic energy
Energy an object possesses because of its motion
kg m2 s2
joule
J
Potential energy
Energy an object possesses because of its position or condition
kg m2 s2
joule
J
Power
Rate at which energy is produced or consumed
kg m2 s3
watt
W
Mass density
Mass per unit volume
kg m3
kg m3
Radian
Angle at the center of a circle whose arc is equal in length to the radius
m>m = 1
m>m = 1
Steradian
Angle at the center of a sphere subtended by a part of the surface equal in area to the square of the radius
m2 >m2 = 1
m2 >m2 = 1
Frequency
Number of repeat units of a wave per unit time
s1
hertz
Hz
Electrical charge
Physical property of matter that causes it to experience an electrostatic force
As
coulomb
C
Electrical potential
Work done in moving a unit positive charge from infinity to that point
volt
V
Electrical resistance
Ratio of the voltage to the electric current that flows through a conductive material
kg m2 s3 >A W>A
ohm
Ω
kg m2 s3 >A2 W>A2
kg m2 s1
If SI units are used throughout the calculation of a quantity, the result will have SI units. For example, consider a unit analysis of the electrostatic force between two charges: F = =
q1q2 8pe0r 2
=
C2 A2 s2 = 8p * kg 1s4A2 m3 * m2 8p * kg 1s4A2 m3 * m2
1 1 kg m s2 = N 8p 8p
Therefore, in carrying out a calculation, it is only necessary to make sure that all quantities are expressed in SI units rather than carrying out a detailed unit analysis of the entire calculation.
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ME1.3 Solving EndofChapter Problems
3
ME1.2 UNCERTAINTY AND SIGNIFICANT FIGURES
In carrying out a calculation, it is important to take into account the uncertainty of the individual quantities that go into the calculation. The uncertainty is indicated by the number of significant figures. For example, the mass 1.356 g has four significant figures. The mass 0.003 g has one significant figure, and the mass 0.01200 g has four significant figures. By convention, the uncertainty of a number is {1 in the rightmost digit. A zero at the end of a number that is not to the right of a decimal point is not significant. For example, 150 has two significant figures, but 150. has three significant figures. Some numbers are exact and have no uncertainty. For example, 1.00 * 106 has three significant figures because the 10 and 6 are exact numbers. By definition, the mass of one atom of 12C is exactly 12 atomic mass units. If a calculation involves quantities with a different number of significant figures, the following rules regarding the number of significant figures in the result apply:
• In addition and subtraction, the result has the number of digits to the right of the decimal point corresponding to the number that has the smallest number of digits to the right of the decimal point. For example 101 + 24.56 = 126 and 0.523 + 0.10 = 0.62. In multiplication or division, the result has the number of significant figures corresponding to the number with the smallest number of significant figures. For example, 3.0 * 16.00 = 48 and 0.05 * 100. = 5.
•
It is good practice to carry forward a sufficiently large number of significant figures in different parts of the calculation and to round off to the appropriate number of significant figures at the end.
ME1.3 SOLVING ENDOFCHAPTER PROBLEMS Because calculations in physical chemistry often involve multiple inputs, it is useful to carry out calculations in a manner that they can be reviewed and easily corrected. For example, the input and output for the calculation of the pressure exerted by gaseous benzene with a molar volume of 2.00 L at a temperature of 595 K using the Redlich– RT a 1 Kwong equation of state P = in Mathematica is shown Vm  b 2T Vm1Vm + b2 below. The statement in the first line clears the previous values of all listed quantities, and the semicolon after each input value suppresses its appearance in the output. In[36]:= Clear[r, t, vm, a, b, prk] r = 8.314 * 10^ 2; t = 595; vm = 2.00; a = 452; b = .08271; rt a 1 prk = vm  b vm(vm + b) 2t out[42]= 21.3526 Invoking the rules for significant figures, the final answer is P = 21.4 bar. The same problem can be solved using Microsoft Excel as shown in the following table. A
B
C
D
E
F
1
R
T
Vm
a
b
=((A2*B2)/(C2E2))(D2/SQRT(B2))*(1/(C2*(C2+E2)))
2
0.08314
595
2
452
0.08271
21.35257941
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C H A P T E R
1
Fundamental Concepts of Thermodynamics
1.1 What Is Thermodynamics and Why Is It Useful?
1.2 The Macroscopic Variables Volume, Pressure, and Temperature
1.3 Basic Definitions Needed to
WHY is this material important? Thermodynamics is a powerful science that allows predictions to be made about chemical reactions, the efficiency of engines, and the potential of new energy sources. It is a macroscopic science and does not depend on a description of matter at the molecular scale. In this chapter, we introduce basic concepts such as the system variables pressure, temperature, and volume, and equations of state that relate these variables with one another.
Describe Thermodynamic Systems
1.4 Equations of State and the Ideal Gas Law
1.5 A Brief Introduction to Real Gases
WHAT are the most important concepts and results? Processes such as chemical reactions occur in an apparatus whose contents we call the system. The rest of the universe is the surroundings. The exchange of energy and matter between the system and surroundings is central to thermodynamics. We will show that the macroscopic gas property pressure arises through the random thermal motion of atoms and molecules. Equations of state such as the ideal gas law allow us to calculate how one system variable changes when another variable is increased or decreased.
WHAT would be helpful for you to review for this chapter? It would be useful to review the material on units and problem solving discussed in Math Essential 1.
1.1 WHAT IS THERMODYNAMICS AND WHY IS IT USEFUL?
Thermodynamics is the branch of science that describes the behavior of matter and the transformation between different forms of energy on a macroscopic scale, which is the scale of phenomena experienced by humans, as well as largerscale phenomena (e.g., astronomical scale). Thermodynamics describes a system of interest in terms of its bulk properties. Only a few variables are needed to describe such a system, and the variables are generally directly accessible through measurements. A thermodynamic description of matter does not make reference to its structure and behavior at the microscopic level. For example, 1 mol of gaseous water at a sufficiently low density is completely described by two of the three macroscopic variables of pressure, volume, and temperature. By contrast, the microscopic scale refers to dimensions on the order of the size of molecules. At the microscopic level, water is described as a dipolar triatomic molecule, H2O, with a bond angle of 104.5° that forms a network of hydrogen bonds. In the first part of this book (Chapters 1–11), we will discuss thermodynamics. Later in the book, we will turn to statistical thermodynamics. Statistical thermodynamics
Concept Because thermodynamics does not make reference to a description of matter at the microscopic level, it is equally applicable to a liter of garbage and a liter of pure water.
5
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CHAPTER 1 Fundamental Concepts of Thermodynamics
uses atomic and molecular properties to calculate the macroscopic properties of matter. For example, statistical thermodynamic analysis shows that liquid water is the stable form of aggregation at a pressure of 1 bar and a temperature of 90°C, whereas gaseous water is the stable form at 1 bar and 110°C. Using statistical thermodynamics, we can calculate the macroscopic properties of matter from underlying molecular properties. Given that the microscopic nature of matter is becoming increasingly well understood using theories such as quantum mechanics, why is the macroscopic science thermodynamics relevant today? The usefulness of thermodynamics can be illustrated by describing four applications of thermodynamics that you will have mastered after working through this book:
• You have built an industrial plant to synthesize NH3(g) gas from N2(g) and H2(g).
•
•
•
You find that the yield is insufficient to make the process profitable, and you decide to try to improve the NH3 output by changing either the temperature or pressure of synthesis, or both. However, you do not know whether to increase or decrease the values of these variables. As will be shown in Chapter 6, the ammonia yield will be higher at equilibrium if the temperature is decreased and the pressure is increased. You wish to use methanol to power a car. One engineer provides a design for an internal combustion engine that will burn methanol efficiently according to the reaction CH3OH(l) + 3>2O2(g) S CO2(g) + 2H2O(l). A second engineer designs an electrochemical fuel cell that carries out the same reaction. He claims that the vehicle will travel much farther if it is powered by the fuel cell rather than by the internal combustion engine. As will be shown in Chapter 5, this assertion is correct, and an estimate of the relative efficiencies of the two propulsion systems can be made. You are asked to design a new battery that will be used to power a hybrid car. Because the voltage required by the driving motors is much higher than can be generated in a single electrochemical cell, many cells must be connected in series. Because the space for the battery is limited, as few cells as possible should be used. You are given a list of possible cell reactions and told to determine the number of cells needed to generate the required voltage. As you will learn in Chapter 11, this problem can be solved using tabulated values of thermodynamic functions. Your attempts to synthesize a new and potentially very marketable compound have consistently led to yields that make it unprofitable to begin production. A supervisor suggests a major effort to make the compound by first synthesizing a catalyst that promotes the reaction. How can you decide if this effort is worth the required investment? As will be shown in Chapter 6, the maximum yield expected under equilibrium conditions should be calculated first. If this yield is insufficient, a catalyst is useless.
1.2 THE MACROSCOPIC VARIABLES VOLUME, PRESSURE, AND TEMPERATURE
Concept The origin of pressure in a gas is the random thermally induced motion of individual molecules.
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We begin our discussion of thermodynamics by considering a bottle of a gas such as He or CH4. At a macroscopic level, the sample of known chemical composition is completely described by the measurable quantities volume, pressure, and temperature for which we use the symbols V, P, and T. The volume V is just that of the bottle. What physical association do we have with P and T? Pressure is the force exerted by the gas per unit area of the container. It is most easily understood by considering a microscopic model of the gas known as the kinetic theory of gases. The gas is described by two assumptions: first, the atoms or molecules of an ideal gas do not interact with one another, and second, the atoms or molecules can be treated as point masses. The pressure exerted by a gas on the container that confines the gas arises from collisions of randomly moving gas molecules with the container walls. Because the number of molecules in a small volume of the gas is on the order of Avogadro’s number NA, the number of collisions between molecules is also
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z
large. To describe pressure, a molecule is envisioned as traveling through space with a velocity vector v that can be resolved into three Cartesian components: vx, vy, and vz, as illustrated in Figure 1.1. The square of the magnitude of the velocity v2 in terms of the three velocity components is v2 = v # v = v2x + v2y + v2z
vz
(1.1)
v
The particle kinetic energy is 1>2 mv2 such that
etr =
1 2 1 1 1 mv = mv2x + mv2y + mv2z = etrx + etry + etrz 2 2 2 2
mai F m dvi 1 dmvi 1 dpi P = = = a b = a b = a b A A A dt A dt A dt
∆ptotal =
∆p * (number of molecules) molecule
y
x
Figure 1.1
Cartesian components of velocity. The particle velocity v can be resolved into three velocity components: vx, vy, and vz.
2mvx mvx
(1.3)
In Equation (1.3), F is the force of the collision, A is the area of the wall with which the particle has collided, m is the mass of the particle, vi is the velocity component along the i direction (i = x, y, or z), and pi is the particle linear momentum in the i direction. Equation (1.3) illustrates that pressure is related to the change in linear momentum with respect to time that occurs during a collision. Due to conservation of momentum, any change in particle linear momentum must result in an equal and opposite change in momentum of the container wall. A single collision is depicted in Figure 1.2. This figure illustrates that the particle linear momentum change in the x direction is 2mvx (note that there is no change in momentum in the y or z direction). Accordingly, a corresponding momentum change of 2mvx must occur for the wall. The pressure measured at the container wall corresponds to the sum of collisions involving a large number of particles that occur per unit time. Therefore, the total momentum change that gives rise to the pressure is equal to the product of the momentum change from a singleparticle collision and the total number of particles that collide with the wall:
vy
vx
(1.2)
where e is kinetic energy and the subscript tr indicates that the energy corresponds to translational motion of the particle. Furthermore, this equation states that the total translational energy is the sum of translational energy along each Cartesian dimension. Pressure arises from the collisions of gas particles with the walls of the container; therefore, to describe pressure, we must consider what occurs when a gas particle collides with the wall. First, we assume that the collisions with the wall are elastic collisions, meaning that translational energy of the particle is conserved. Although the collision is elastic, this does not mean that nothing happens. As a result of the collision, linear momentum is imparted to the wall, which results in pressure. The definition of pressure is force per unit area, and, by Newton’s second law, force is equal to the product of mass and acceleration. Using these two definitions, we find that the pressure arising from the collision of a single molecule with the wall is expressed as
7
1.2 The Macroscopic Variables Volume, Pressure, and Temperature
x
Figure 1.2
Collision between a gas particle and a wall. Before the collision, the particle has a momentum of mvx in the x direction, whereas after the collision the momentum is mvx. Therefore, the change in particle momentum resulting from the collision is 2mvx. By conservation of momentum, the change in momentum of the wall must be 2mvx. The incoming and outgoing trajectories are offset to show the individual momentum components.
(1.4)
How many molecules strike the side of the container in a given period of time? To answer this question, the time over which collisions are counted must be considered. Consider a volume element defined by the area of the wall A multiplied by length ∆x, as illustrated in Figure 1.3. The collisional volume element depicted in Figure 1.3 is given by
V = A∆x
(1.5)
The length of the box ∆x is related to the time period over which collisions will be counted ∆t and the component of particle velocity parallel to the side of the box (taken to be the x direction):
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∆x = vx ∆t
(1.6)
Area 5 A Dx 5 vxDt
Figure 1.3
Volume element used to determine the number of collisions with a wall per unit time.
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8
CHAPTER 1 Fundamental Concepts of Thermodynamics
In this expression, vx is for a single particle; however, an average of this quantity will be used when describing the collisions from a collection of particles. Finally, the number of particles that will collide with the container wall Ncoll in the time interval ∼ ∆t is equal to the number density N. This quantity is equal to the number of particles in the container N divided by the container volume V and multiplied by the collisional volume element depicted in Figure 1.3:
nNA 1 1 ∼ Ncoll = N * 1Avx ∆t2 a b = 1Avx ∆t2 a b 2 V 2
(1.7)
We have used the equality N = n NA where NA is Avogadro’s number and n is the number of moles of gas in the second part of Equation (1.7). Because particles travel in either the +x or x direction with equal probability, only those molecules traveling in the +x direction will strike the area of interest. Therefore, the total number of collisions is divided by two to take the direction of particle motion into account. Employing Equation (1.7), we see that the total change in linear momentum of the container wall imparted by particle collisions is given by ∆ptotal = 12mvx21Ncoll2 = 12mvx2 a
=
nNA Avx ∆t b V 2
nNA A∆t m 8 v2x 9 V
(1.8)
In Equation (1.8), angle brackets appear around v2x to indicate that this quantity represents an average value, given that the particles will demonstrate a distribution of velocities. This distribution is considered in detail later in Chapter 13. With the total change in linear momentum provided in Equation (1.8), the force and corresponding pressure exerted by the gas on the container wall [Equation (1.3)] are as follows:
F =
P =
∆ptotal nNA = Am 8 v2x 9 ∆t V
nNA F = m 8 v2x 9 A V
(1.9)
Equation (1.9) can be converted into a more familiar expression once 1>2 m 8v2x 9 is recognized as the translational energy in the x direction. In Chapter 14, it will be shown that the average translational energy for an individual particle in one dimension is m 8v2x 9 k BT = 2 2
(1.10)
where T is the gas temperature. Substituting this result into Equation (1.9) results in the following expression for pressure:
Concept The ideal gas law assumes that individual molecules are point masses that do not interact.
Concept Temperature can only be measured indirectly through a physical property such as the volume of a gas or the voltage across a thermocouple.
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P =
nNA nNA nRT m 8 v2x 9 = k BT = V V V
(1.11)
We have used the equality NA kB = R where kB is the Boltzmann constant and R is the ideal gas constant in the last part of Equation (1.11). The Boltzmann constant relates the average kinetic energy of molecules to the temperature of the gas, whereas the ideal gas constant relates average kinetic energy per mole to temperature. Equation (1.11) is the ideal gas law. Although this relationship is familiar, we have derived it by employing a classical description of a single molecular collision with the container wall and then scaling this result up to macroscopic proportions. We see that the origin of the pressure exerted by a gas on its container is the momentum exchange of the randomly moving gas molecules with the container walls. What physical association can we make with the temperature T? At the microscopic level, temperature is related to the mean kinetic energy of molecules as shown by
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Equation (1.10). We defer the discussion of temperature at the microscopic level until Chapter 13 and focus on a macroscopic level discussion here. Although each of us has a sense of a “temperature scale” based on the qualitative descriptors hot and cold, a more quantitative and transferable measure of temperature that is not grounded in individual experience is needed. The quantitative measurement of temperature is accomplished using a thermometer. For any useful thermometer, the measured temperature, T, must be a singlevalued, continuous, and monotonic function of some thermometric system property such as the volume of mercury confined to a narrow capillary, the electromotive force generated at the junction of two dissimilar metals, or the electrical resistance of a platinum wire. The simplest case that one can imagine is when T is linearly related to the value of the thermometric property x:
T1x2 = a + bx
(1.12)
Equation (1.12) defines a temperature scale in terms of a specific thermometric property, once the constants a and b are determined. The constant a determines the zero of the temperature scale because T102 = a and the constant b determines the size of a unit of temperature, called a degree. One of the first practical thermometers was the mercuryinglass thermometer. This thermometer utilizes the thermometric property that the volume of mercury increases monotonically over the temperature range 38.8°C to 356.7°C in which Hg is a liquid. In 1745, Carolus Linnaeus gave this thermometer a standardized scale by arbitrarily assigning the values 0 and 100 to the freezing and boiling points of water, respectively. Because there are 100 degrees between the two calibration points, this scale is known as the centigrade scale. The centigrade scale has been superseded by the Celsius scale. The Celsius scale (denoted in units of °C) is similar to the centigrade scale. However, rather than being determined by two fixed points, the Celsius scale is determined by one fixed reference point at which ice, liquid water, and gaseous water are in equilibrium. This point is called the triple point (see Section 8.2) and is assigned the value 0.01°C. On the Celsius scale, the boiling point of water at a pressure of 1 atmosphere is 99.975°C. The size of the degree is chosen to be the same as on the centigrade scale. Although the Celsius scale is used widely throughout the world today, the numerical values for this temperature scale are completely arbitrary because a liquid other than water could have been chosen as a reference. It would be preferable to have a temperature scale derived directly from physical principles. There is such a scale, called the thermodynamic temperature scale or absolute temperature scale. For such a scale, the temperature is independent of the substance used in the thermometer, and the constant a in Equation (1.12) is zero. The gas thermometer is a practical thermometer with which the absolute temperature can be measured. A gas thermometer contains a dilute gas under conditions in which the ideal gas law of Equation (1.11) describes the relationship among P, T, and the molar density rm = n>V with sufficient accuracy:
9
1.2 The Macroscopic Variables Volume, Pressure, and Temperature
P = rm RT
(1.13)
P rm R
(1.14)
Ambient temperature 1024 s after Big Bang
1012 K
1010 K Core of red giant star 108 K
Core of sun Solar corona
106 K
Surface of sun
104 K
Mercury boils H2O is liquid
102 K
Oxygen boils Helium boils
1K
Average temperature of universe
1022 K 3He
1024 K
superfluid
Figure 1.4
Absolute temperature displayed on a logarithmic scale. The temperature of a number of physical phenomena is also shown. 0.1 L 2.5
2.0 Pressure (bar)
1.5
0.2 L
1.0
0.3 L 0.4 L 0.5 L 0.6 L
0.5
Equation (1.13) can be rewritten as
T =
showing that for a gas thermometer, the thermometric property is the temperature dependence of P for a dilute gas at constant V. The gas thermometer provides the international standard for thermometry at very low temperatures. At intermediate temperatures, the electrical resistance of platinum wire is the standard, and at higher temperatures the radiated energy emitted from glowing silver is the standard. The absolute temperature is shown in Figure 1.4 on a logarithmic scale together with associated physical phenomena. Equation (1.14) implies that as T S 0, P S 0. Measurements carried out by Guillaume Amontons in the 17th century demonstrated that the pressure exerted by a fixed amount of gas at constant V varies linearly with temperature as shown in Figure 1.5.
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2200 2100 0
100 200 300 400
Temperature (Celsius)
Figure 1.5
Relationship between temperature and pressure for a dilute gas. The pressure exerted by 5.00 * 103 mol of a dilute gas is shown as a function of the temperature measured on the Celsius scale for different fixed volumes. The dashed segments of lines indicate that the data are extrapolated to lower temperatures than could be achieved experimentally by early investigators.
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10
CHAPTER 1 Fundamental Concepts of Thermodynamics
At the time of these experiments, temperatures below 30°C were not attainable in the laboratory. However, the P versus TC (temperature on the Celsius scale) data can be extrapolated to the limiting TC value at which P S 0. It is found that these straight lines obtained for different values of V intersect at a common point on the TC axis that lies near 273°C. The data in Figure 1.5 show that at constant V the thermometric property P varies with temperature as
Nucleus
Plasma membrane Animal cell
(a) Nucleus
Vacuole
Cell wall Chloroplast
Mitochondrion
Plant cell (b)
Figure 1.6
Animal and plant cells are open systems. Sketch of (a) an animal cell and (b) a plant cell shown with selected organelles. The contents of the animal cell include the cytosol fluid and the numerous organelles (e.g., nucleus, mitochondria) that are separated from the surroundings by a lipidrich plasma membrane. The plasma membrane acts as a boundary layer that can transmit energy and is selectively permeable to ions and various metabolites. A plant cell is surrounded by a cell wall that similarly encases the cytosol and organelles, including chloroplasts (in which photosynthesis occurs).
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(1.15)
where a and b are experimentally obtained proportionality constants. Figure 1.5 shows that all lines intersect at a single point, even for different gases. This suggests a unique reference point for temperature, rather than the two reference points used in constructing the centigrade scale. The value zero is given to the temperature at which P S 0, so that a = 0. However, this choice is not sufficient to define the temperature scale because the size of the degree is undefined. By convention, the size of the degree on the absolute temperature scale is set equal to the size of the degree on the Celsius scale. With these two choices, the absolute and Celsius temperature scales are related by Equation (1.16). The scale measured by the ideal gas thermometer is the absolute temperature scale used in thermodynamics. The unit of temperature on this scale is called the kelvin, abbreviated K (without a degree sign):
Mitochondrion
P = a + bTC
T>K = TC >°C + 273.15
(1.16)
1.3 BASIC DEFINITIONS NEEDED TO DESCRIBE THERMODYNAMIC SYSTEMS
In the first two sections of this chapter, we discussed the macroscopic variables pressure, volume, and temperature. With this background information, we are now ready to introduce some important concepts used in thermodynamics. A thermodynamic system consists of all the materials involved in the process under study. This material could be the contents of an open beaker containing reagents, the electrolyte solution within an electrochemical cell, or the contents of a cylinder and movable piston assembly in an engine. In thermodynamics, the rest of the universe is referred to as the surroundings. If a system can exchange matter with the surroundings, it is called an open system; if not, it is a closed system. Living cells are open systems (see Figure 1.6). Both open and closed systems can exchange energy with the surroundings. Systems that can exchange neither matter nor energy with the surroundings are called isolated systems. The interface between the system and its surroundings is called the boundary. Boundaries determine if energy and mass can be transferred between the system and the surroundings and lead to the distinction between open, closed, and isolated systems. Consider Earth’s oceans as a system, with the rest of the universe being the surroundings. The system–surroundings boundary consists of the solid–liquid interface between the continents and the ocean floor and the water–air interface at the ocean surface. For an open beaker in which the system is the contents, the boundary surface is just inside the inner wall of the beaker, and it passes across the open top of the beaker. In this case, energy can be exchanged freely between the system and surroundings through the side and bottom walls, and both matter and energy can be exchanged between the system and surroundings through the open top boundary. The portion of the boundary formed by the beaker in the previous example is called a wall. Walls can be rigid or movable and permeable or nonpermeable. An example of a movable wall is the surface of a balloon. An example of a selectively permeable wall is the fabric used in raingear, which is permeable to water vapor but not to liquid water. The exchange of energy and matter across the boundary between system and surroundings is central to the important concept of equilibrium. The system and surroundings can be in equilibrium with respect to one or more of several different system variables such as pressure (P), temperature (T ), and concentration. Thermodynamic equilibrium refers to a condition in which equilibrium exists with respect to P, T, and
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1.3 Basic Definitions Needed to Describe Thermodynamic Systems
concentration. What conditions are necessary for a system to come to equilibrium with its surroundings? Equilibrium is established with respect to a given variable only if that variable does not change with time, and if it has the same value in all parts of the system and surroundings. For example, the interior of a soap bubble1 (the system) and the surroundings (the room) are in equilibrium with respect to P because the movable wall (the bubble) can reach a position where P on both sides of the wall is the same, and because P has the same value throughout the system and surroundings. Equilibrium with respect to concentration exists only if transport of all species across the boundary in both directions is possible. If the boundary is a movable wall that is not permeable to all species, equilibrium can exist with respect to P but not to concentration. Because N21g2 and O21g2 cannot diffuse through the (idealized) bubble, the system and surroundings are in equilibrium with respect to P but not to concentration. Equilibrium with respect to temperature is a special case that is discussed next. Two systems that have the same temperature are in thermal equilibrium. We use the concepts of temperature and thermal equilibrium to characterize the walls between a system and its surroundings. Consider the two systems with rigid walls shown in Figure 1.7a. Each system has the same molar density and is equipped with a pressure gauge. If we bring the two systems into direct contact, two limiting behaviors are observed. If neither pressure gauge changes, as in Figure 1.7b, the wall has not allowed the transfer of energy, and we refer to the wall as being adiabatic. Because P1 ≠ P2, the systems are not in thermal equilibrium and, therefore, have different temperatures. An example of a system surrounded by adiabatic walls is coffee in a Styrofoam cup with a Styrofoam lid.2 Experience shows that it is not possible to bring two systems enclosed by adiabatic walls into thermal equilibrium by bringing them into contact because adiabatic walls insulate against the transfer of “heat.” If we push a Styrofoam cup containing hot coffee against one containing ice water, they will not reach the same temperature. Rely on experience at this point regarding the meaning of heat; a thermodynamic definition will be given in Chapter 2. The second limiting case is shown in Figure 1.7c. In bringing the systems into intimate contact, both pressures change and reach the same value after some time. We conclude that the systems have the same temperature, T1 = T2, and say that they are in thermal equilibrium. We refer to the walls as being diathermal. Two systems in contact separated by diathermal walls reach thermal equilibrium because diathermal walls conduct heat. Hot coffee stored in a copper cup is an example of a system surrounded by diathermal walls. Because the walls are diathermal, the coffee will quickly reach room temperature. The zeroth law of thermodynamics generalizes the experiments illustrated in Figure 1.7 and asserts the existence of an objective temperature that can be used to define the condition of thermal equilibrium. The formal statement of this law is as follows:
Two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium with one another.
The unfortunate name was assigned to the “zeroth” law because it was formulated after the first law of thermodynamics, but logically it precedes it. The zeroth law tells us that we can determine if two systems are in thermal equilibrium without bringing them into contact. Imagine the third system to be a thermometer, which is defined more precisely in the next section. The third system can be used to compare the temperatures of the other two systems; if they have the same temperature, they will be in thermal equilibrium if placed in contact. 1
For this example, the surface tension of the bubble is assumed to be so small that it can be set equal to zero. This is in keeping with the thermodynamic tradition of weightless pistons and frictionless pulleys. 2
In this discussion, Styrofoam is assumed to be a perfect insulator.
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Concept Two systems are in thermodynamic equilibrium if pressure, temperature, and all concentrations are the same in both systems.
(a)
Adiabatic wall does not allow thermal equilibrium (b)
(c)
Diathermal wall allows thermal equilibrium
Figure 1.7
Comparison of adiabatic and diathermal walls. (a) Two separated systems with rigid walls and the same molar density have different temperatures. (b) Two systems are brought together so that their adiabatic walls are in intimate contact. The pressure in each system will not change unless heat transfer is possible. (c) As in part (b), two systems are brought together so that their diathermal walls are in intimate contact. The pressures become equal.
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CHAPTER 1 Fundamental Concepts of Thermodynamics
1.4 EQUATIONS OF STATE
AND THE IDEAL GAS LAW
Concept An equation of state links the variables that define the system.
4
Pressure ( 106 Pa)
Macroscopic models in which the system is described by a set of variables are based on experience. It is particularly useful to formulate an equation of state that relates the state variables. A dilute gas can be modeled as consisting of point masses that do not interact with one another; we call this an ideal gas. The equation of state for an ideal gas was first determined from experiments by the English chemist Robert Boyle in the 17th century. If the pressure of He is measured as a function of the volume for different values of temperature, the set of nonintersecting hyperbolas as shown in Figure 1.8 is obtained. The curves in this figure can be quantitatively fit by the functional form PV = aT
(1.17)
3
where T is the absolute temperature as defined by Equation (1.16), allowing a to be determined. The constant a is found to be directly proportional to the mass of gas used. It is useful to separate this dependence by writing a = nR, where n is the number of moles of the gas and R is a constant that is independent of the size of the system. The result is the ideal gas equation of state
2
700 K
1 200 K 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Volume (1022 m3)
Figure 1.8
Plot showing the relationship between pressure and volume. Red curves represent fixed values of temperature for 1.00 mol of He. The interval of curves is 100 K.
PV = NkBT = nRT
(1.18)
as derived in Equation (1.11). The equation of state given in Equation (1.18) is familiar as the ideal gas law. Because the four variables P, V, T, and n are related through the equation of state, any three of these variables are sufficient to completely describe the ideal gas. Of these four variables, P and T are independent of the amount of gas, whereas V and n are proportional to the amount of gas. A variable that is independent of the size of the system (for example, P and T ) is referred to as an intensive variable, and one that is proportional to the size of the system (for example, V ) is referred to as an extensive variable. Equation (1.18) can be written in terms of intensive variables exclusively:
P = rmRT
(1.13)
For a fixed number of moles, the ideal gas equation of state has only two independent intensive variables: any two of P, T, and rm. For an ideal gas mixture PV = a ni RT
i
(1.19)
because ideal gas molecules do not interact with one another. Equation (1.19) can be rewritten in the form
P = a i
ni RT = a Pi = P1 + P2 + P3 + c V i
(1.20)
In Equation (1.20), Pi is the partial pressure of each gas. This equation states that each ideal gas exerts a pressure that is independent of the other gases in the mixture. We also have ni RT ni RT Pi ni V V = xi = = = (1.21) n RT n P nRT i a V V i
Concept In an ideal gas mixture, the total pressure is the sum of the partial pressures.
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which relates the partial pressure of a component in the mixture Pi with its mole fraction, xi = ni >n, and the total pressure P. In the SI system of units, pressure is measured in Pascal (Pa) units, where 1 Pa = 1 N>m2. Volume is measured in cubic meters, and temperature is measured in kelvin. However, other units of pressure are frequently used; these other units are related to the Pascal, as indicated in Table 1.1. In this table, numbers that are not exact have been given to five significant figures. The other commonly used unit of volume is the liter (L), where 1 m3 = 103 L and 1 L = 1 dm3 = 103 m3.
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1.4 Equations of State and the Ideal Gas Law
13
TABLE 1.1 Units of Pressure and Conversion Factors Unit of Pressure
Symbol
Numerical Value
Pascal
Pa
1 N m2 = 1 kg m1 s2
Atmosphere
atm
1 atm = 101,325 Pa (exactly)
Bar
bar
1 bar = 105 Pa
Torr or millimeters of Hg
Torr
1 Torr = 101,325>760 = 133.32 Pa
Pounds per square inch
psi
1 psi = 6,894.8 Pa
EXAMPLE PROBLEM 1.1 Before you begin a day trip, you inflate the tires on your automobile to a recommended pressure of 3.21 * 105 Pa on a day when the temperature is 5.00°C. On your trip you drive to the beach, where the temperature is 28.0°C. (a) What is the final pressure in the tires, assuming constant volume? (b) Derive a formula for the final pressure, assuming more realistically that the volume of the tires increases with increasing pressure as Vƒ = Vi11 + g 3 Pƒ  Pi 4 2 where g is an experimentally determined constant.
Solution a. Because the number of moles is constant, PƒVƒ Pi Vi Tƒ Pi Vi = ; Pƒ = ; Ti Tƒ VƒTi Pƒ =
b.
Pi Vi Tƒ VƒTi
= 3.21 * 105 Pa *
PƒVi11 + g 3 Pƒ  Pi 4 2 Pi Vi = ; Ti Tƒ
1273.15 + 28.02 Vi * = 3.61 * 105 Pa Vi 1273.15  5.002
Pi Tƒ = PƒTi11 + g 3 Pƒ  Pi 4 2
P 2ƒ Ti g + PƒTi11  Pi g2  Pi Tƒ = 0 Pƒ =
Ti11  Pi g2 { 2T 2i 11  Pi g22 + 4Ti Tƒ gPi 2Ti g
We leave it to the endofchapter problems to show that this expression for Pƒ has the correct limit as g S 0.
In the SI system, the constant R that appears in the ideal gas law has the value 8.314 J K1 mol1, where the joule (J) is the unit of energy in the SI system. To simplify calculations for other units of pressure and volume, the values of the constant R with different combinations of units are given in Table 1.2. TABLE 1.2 The Ideal Gas Constant, R, in Various Units R = 8.314 J K1 mol1 R = 8.314 Pa m3 K1 mol1 R = 8.314 * 102 L bar K1 mol1 R = 8.206 * 102 L atm K1 mol1 R = 62.36 L Torr K1 mol1
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CHAPTER 1 Fundamental Concepts of Thermodynamics
EXAMPLE PROBLEM 1.2 Consider the composite system, which is held at 298 K, shown in the following figure. Assuming ideal gas behavior, calculate the total pressure and the partial pressure of each component if the barriers separating the compartments are removed. Assume that the volume of the barriers is negligible.
He 2.00 L 1.50 bar
Ne 3.00 L 2.50 bar
Xe 1.00 L 1.00 bar
Solution The number of moles of He, Ne, and Xe is given by PV RT PV nNe = RT PV nXe = RT n = nHe
nHe =
= =
8.314 8.314
=
8.314 + nNe +
1.50 bar * 2.00 L = 0.121 mol * 102 L bar K1 mol1 * 298 K 2.50 bar * 3.00 L = 0.303 mol * 102 L bar K1 mol1 * 298 K 1.00 bar * 1.00 L = 0.0403 mol * 102 L bar K1 mol1 * 298 K nXe = 0.464
The mole fractions are nHe 0.121 = = 0.261 n 0.464 nNe 0.303 = = = 0.653 n 0.464 nXe 0.0403 = = = 0.0860 n 0.464
xHe = xNe xXe The total pressure is given by P =
1nHe + nNe + nXe2RT V
0.464 mol * 8.314 * 102 L bar K1 mol1 * 298 K 6.00 L = 1.92 bar =
The partial pressures are given by PHe = xHeP = 0.261 * 1.92 bar = 0.501 bar PNe = xNeP = 0.653 * 1.92 bar = 1.25 bar PXe = xXeP = 0.0860 * 1.92 bar = 0.165 bar
1.5 A BRIEF INTRODUCTION TO REAL GASES The ideal gas law provides a useful means of describing a system in terms of macroscopic parameters. However, we should also emphasize the downside of not taking the microscopic nature of the system into account. For example, the ideal gas law only holds for gases at low densities. In practice, deviations from the ideal gas law that occur for real gases must be taken into account in such applications as a gas thermometer.
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15
1.5 A Brief Introduction to Real Gases
For most applications, calculations based on the ideal gas law are valid to pressures on the order of 1 bar. Real gases will be discussed in detail in Chapter 7. However, because we need to take nonideal gas behavior into account in Chapters 2 through 6, we introduce an equation of state that is valid to greater densities in this section. The assumptions of the ideal gas law—that atoms or molecules do not interact with one another and can be treated as point masses—limit the range of validity of the model. Consider the potential energy function typical for a real gas, as shown in Figure 1.9. This figure shows the potential energy of interaction of two gas molecules as a function of the distance between them. The intermolecular potential can be divided into regions in which the potential energy is essentially zero 1r 7 rtransition2, negative (attractive interaction) 1rtransition 7 r 7 rV = 02, and positive (repulsive interaction) 1r 6 rV = 02. The distance rtransition is not uniquely defined and depends on the energy of the molecule. The value of rtransition is on the order of the molecular size. As density is increased from very low values, molecules approach one another to within a few molecular diameters and experience a longrange attractive van der Waals force due to timefluctuating dipole moments in each molecule. This strength of the attractive interaction is proportional to the polarizability of the electron charge in a molecule and is, therefore, substance dependent. In the attractive region, P is lower than that calculated using the ideal gas law. This is the case because the attractive interaction brings the atoms or molecules closer than they would be if they did not interact. At sufficiently high densities, the atoms or molecules experience a shortrange repulsive interaction due to the overlap of the electron charge distributions. Because of this interaction, P is higher than that calculated using the ideal gas law. We see that for a real gas P can be either greater or less than the ideal gas value. Note that the potential becomes repulsive for a value of r greater than zero. As a consequence, the volume of a gas even well above its boiling temperature approaches a finite limiting value as P increases. By contrast, the ideal gas law predicts that V S 0 as P S ∞ . Given the potential energy function depicted in Figure 1.9 under what conditions is the ideal gas equation of state valid? A real gas behaves ideally only at low densities for which r 7 rtransition, and the value of rtransition is substance dependent. The van der Waals equation of state takes both the finite size of molecules and the attractive potential into account. It has the form
P =
nRT n2a  2 V  nb V
Real gas
Ideal gas
rtransition rV50
0 V(r)
r
0
Figure 1.9
The potential energy of interaction of two molecules or atoms as a function of their separation r. The red curve shows the potential energy function for an ideal gas. At values of r less than that indicated by the dashed line, an equation of state more accurate than the ideal gas law should be used. V1r2 = 0 at r = rV = 0 and as r S ∞ .
(1.22)
This equation of state has two parameters that are substance dependent and must be experimentally determined. The parameters b and a take the finite size of the molecules and the strength of the attractive interaction into account, respectively. (Values of a and b for selected gases are listed in Table 7.4.) The van der Waals equation of state is more accurate in calculating the relationship between P, V, and T for gases than the ideal gas law because a and b have been optimized using experimental results. However, there are other more accurate equations of state that are valid over a wider range than the van der Waals equation, as will be discussed in Chapter 7.
EXAMPLE PROBLEM 1.3 Van der Waals parameters are generally tabulated with either of two sets of units: a. Pa m6 mol2 or bar dm6 mol2 b. m3 mol1 or dm3 mol1 Determine the conversion factor to convert one system of units to the other. Note that 1 dm3 = 103 m3 = 1 L.
Solution Pa m6 mol2 * m3 mol1 *
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bar 105 Pa
*
106 dm6 = 10 bar dm6 mol2 m6
103 dm3 = 103 dm3 mol1 m3
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CHAPTER 1 Fundamental Concepts of Thermodynamics
In Example Problem 1.4, a comparison is made of the molar volume for N2 calculated at low and high pressures, using the ideal gas and van der Waals equations of state.
EXAMPLE PROBLEM 1.4
Connection The pressure exerted by a real gas can be larger or smaller than the pressure of an ideal gas under the same conditions.
a. Calculate the pressure exerted by N2 at 300. K for molar volumes of 250. L mol1 and 0.100 L mol1 using the ideal gas and the van der Waals equations of state. The values of parameters a and b for N2 are 1.370 bar dm6 mol2 and 0.0387 dm3 mol1, respectively. b. Compare the results of your calculations at the two pressures. If P calculated using the van der Waals equation of state is greater than those calculated with the ideal gas law, we can conclude that the repulsive interaction of the N2 molecules outweighs the attractive interaction for the calculated value of the density. A similar statement can be made regarding the attractive interaction. Is the attractive or repulsive interaction greater for N2 at 300. K and Vm = 0.100 L?
Solution a. The pressures calculated from the ideal gas equation of state are P =
nRT 1 mol * 8.314 * 102 L bar mol1K1 * 300. K = = 9.98 * 102 bar V 250. L
P =
nRT 1 mol * 8.314 * 102 L bar mol1K1 * 300. K = = 249 bar V 0.100 L
The pressures calculated from the van der Waals equation of state are P = =
nRT n2a  2 V  nb V 1 mol * 8.314 * 102 L bar mol1 K1 * 300. K 250. L  1 mol * 0.0387 dm3 mol1

11 mol22 * 1.370 bar dm6 mol2 1250. L22
= 9.98 * 102 bar P =
1 mol * 8.314 * 102 L bar mol1 K1 * 300 K 0.100 L  1 mol * 0.0387 dm3 mol1

11 mol22 * 1.370 bar dm6 mol2
= 270. bar
10.100 L22
b. Note that the result is identical to the result for the ideal gas law for Vm = 250. L and that the result calculated for Vm = 0.100 L deviates from the ideal gas law result. Because Preal 7 Pideal, we conclude that the repulsive interaction is more important than the attractive interaction for this specific value of molar volume and temperature.
VOCABULARY absolute temperature scale adiabatic Boltzmann constant boundary Celsius scale centigrade scale
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closed system diathermal elastic collision equation of state equilibrium extensive variable
gas thermometer ideal gas ideal gas constant ideal gas law intensive variable isolated system
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17
Conceptual Problems
kelvin macroscopic scale macroscopic variables mole fraction open system partial pressure
surroundings system system variables temperature temperature scale thermal equilibrium
thermodynamic equilibrium thermodynamic temperature scale thermometer van der Waals equation of state zeroth law of thermodynamics
KEY EQUATIONS Equation P = P =
nNA nNA m 8v2x 9 = kT V V nRT V
T>K = TC >°C + 273.15 P = a i
ni RT = a Pi = P1 + P2 + P3 + c V i
Significance of Equation
Equation Number
Molecular level origin of pressure
1.11
Ideal gas equation of state
1.11
Definition of absolute temperature scale relative to Celsius scale
1.16
Relation between total and partial pressure
1.20
Pi = P
niRT niRT ni V V = = = xi niRT n nRT a V V i
Relation between partial pressure, total pressure, and mole fraction
1.21
P =
nRT n2a  2 V  nb V
Van der Waals equation of state
1.22
CONCEPTUAL PROBLEMS Q1.1 The location of the boundary between the system and the surroundings is a choice that must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight room. Is the system open or closed if you place the boundary just outside the liquid water? Is the system open or closed if you place the boundary just inside the walls of the room? Q1.2 Real walls are never totally adiabatic. Order the following walls in increasing order with respect to being diathermal: 1cmthick concrete, 1cmthick vacuum, 1cmthick copper, 1cmthick cork. Q1.3 Why is the possibility of exchange of matter or energy a necessary condition for equilibrium between two systems? Q1.4 At sufficiently high temperatures, the van der Waals RT equation has the form P ≈ . Note that the attractive Vm  b part of the potential has no influence in this expression. Justify this behavior using the potential energy diagram in Figure 1.9. Q1.5 The parameter a in the van der Waals equation is greater for H2O than for He. What does this say about the form of the potential function in Figure 1.9 for the two gases?
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Q1.6 Can temperature be measured directly? Explain your answer. Q1.7 Give an example of two systems that are in equilibrium with respect to only one of two state variables. Q1.8 Explain how the ideal gas law can be deduced for the measurements shown in Figures 1.5 and 1.8. Q1.9 Give an example based on molecule–molecule interactions illustrating how the total pressure upon mixing two real gases could be different from the sum of the partial pressures. Q1.10 Which of the following systems are open? (a) a dog, (b) an incandescent light bulb, (c) a tomato plant, (d) a can of tomatoes. Explain your answers. Q1.11 Which of the following systems are isolated? (a) a bottle of wine, (b) a tightly sealed, perfectly insulated thermos bottle, (c) a closed tube of toothpaste, (d) our solar system. Explain your answers. Q1.12 Why do the z and y components of the velocity vector not change in the collision depicted in Figure 1.2?
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CHAPTER 1 Fundamental Concepts of Thermodynamics
Q1.13 If the wall depicted in Figure 1.2 were a movable piston, under what conditions would it move as a result of the molecular collisions? Q1.14 The mass of an He atom is less than that of an Ar atom. Does that mean that because of its larger mass, Argon exerts a higher pressure on the container walls than He at the same molar density, volume, and temperature? Explain your answer. Q1.15 Explain why attractive interactions between molecules in gas make the pressure less than that predicted by the ideal gas equation of state.
Q1.16 State whether the following are intensive or extensive variables. (a) temperature, (b) pressure, (c) mass. Explain your answers. Q1.17 State whether the following are intensive or extensive variables. (a) density, (b) mean square speed 8v2x 9 , (c) volume. Explain your answers.
NUMERICAL PROBLEMS Section 1.2 P1.1 Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas constant is 7.41J G1 mol1. P1.2 Suppose that you measured the product PV of 1 mol of a dilute gas and found that PV = 23.60 L atm at 0.00°C and 32.35 L atm at 100.°C. Assume that the ideal gas law is valid, with T = t1°C2 + a, and that the values of R and a are not known. Determine R and a from the measurements provided. Section 1.4 P1.3 A sample of propane 1C3H82 is placed in a closed vessel together with an amount of O2 that is 3.05 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation assuming that the H2O is present as a gas. P1.4 A gas sample is known to be a mixture of ethane and butane. A bulb having a 195.0cm3 capacity is filled with the gas to a pressure of 103 * 103 Pa at 17.5°C. If the weight of the gas in the bulb is 0.2988 g, what is the mole percent of butane in the mixture? P1.5 One liter of fully oxygenated blood can contain and transport 0.210 liters of O2 measured at T = 298 K and P = 1.00 atm. Calculate the number of moles of O2 carried per liter of blood. Hemoglobin, the oxygen transport protein in blood, has four oxygenbinding sites. How many hemoglobin molecules are required to transport the O2 in 2.50 L of fully oxygenated blood? P1.6 Yeast and other organisms can convert glucose 1C6H12O62 to ethanol 1CH3CH2OH2 by a process called alcoholic fermentation. The net reaction is C6H12O61s2 S 2C2H5OH1l2 + 2CO21g2
Calculate the mass of glucose required to produce 3.10 L of CO2 measured at P = 1.00 atm and T = 305 K. P1.7 A vessel contains 2.25 g H2O1l2 in equilibrium with water vapor at 30.°C. At this temperature, the vapor pressure of H2O is 31.82 Torr. What volume increase is necessary for all the water to evaporate?
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P1.8 Consider a 25.5L sample of moist air at 60.°C and 1.00 atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent N2, 21.0 mole percent O2, and 1.00 mole percent Ar. a. What are the mole percentages of each of the gases in the sample? b. The percent relative humidity is defined as %RH = PH2O >P *H2O, where PH2O is the partial pressure of water in the sample and P *H2O = 0.197 atm is the equilibrium vapor pressure of water at 60.°C. The gas is compressed at 60.°C until the relative humidity is 100.%. What volume does the mixture contain now? c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 75.0 atm? P1.9 A typical diver inhales 0.450 L of air per breath and carries a 25 L breathing tank containing air at a pressure of 250. bar. As she dives deeper, the pressure increases by 1 bar for every 10.08 m. How many breaths can the diver take from this tank at a depth of 50. m? Assume that the temperature remains constant. P1.10 Consider two connected rigid vessels separated by a valve. The first rigid vessel has a volume of 0.400 m3 and contains H2 at 19.5°C and a pressure of 500. * 103 Pa. The second rigid vessel has a volume of 0.750 m3 and contains Ar at 28.5°C at a pressure of 200. * 103 Pa. The valve separating the two vessels is opened, and both are cooled to a temperature of 15.0°C. What is the final pressure in the vessels? P1.11 A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the reaction CuO1s2 + H21g2 S Cu1s2 + H2O1l2, and oxygen reoxidizes the copper formed according to Cu1s2 + 1>2 O21g2 S CuO1s2. At 25°C and 750. Torr, 155.0 cm3 of the mixture yields 68.0 cm3 of dry oxygen measured at 25°C and 750. Torr after passage over CuO and the drying agent. What is the original composition of the mixture? P1.12 An athlete at high performance inhales ∼4.00 L of air at 1.00 atm and 298 K. The inhaled air and the exhaled air contain 0.45 and 5.9% by volume of water, respectively.
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Numerical Problems
For a respiration rate of 29 breaths per minute, how many moles of water per minute are expelled from the body through the lungs? P1.13 Aerobic cells metabolize glucose in the respiratory system. This reaction proceeds according to the overall reaction 6O21g2 + C6H12O61s2 S 6CO21g2 + 6H2O1l2
Calculate the volume of oxygen required at STP to metabolize 0.0375 kg of glucose 1C6H12O62. STP refers to standard temperature and pressure; that is, T = 273 K and P = 1.00 atm. Assume oxygen behaves ideally at STP. P1.14 An athlete at high performance inhales ∼4.00 L of air at 1.0 atm and 298 K at a respiration rate of 35 breaths per minute. If the exhaled and inhaled air contain 14.0 and 20.9% by volume of oxygen, respectively, how many mol of oxygen per minute are absorbed by the athlete’s body? P1.15 A mixture of 1.75 * 103 g of O2, 4.10 * 103 mol of N2, and 6.50 * 1020 molecules of CO are placed into a vessel of volume 4.75 L at 15.0°C. a. Calculate the total pressure in the vessel. b. Calculate the mole fractions and partial pressures of each gas. P1.16 In the absence of turbulent mixing, the partial pressure of each constituent of air would fall off with height above sea level in the Earth’s atmosphere as Pi = P 0i eMigz>RT where Pi is the partial pressure at the height z, P 0i is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, T is the absolute temperature, and Mi is the molecular mass of the gas. As a result of turbulent mixing, the composition of the Earth’s atmosphere is constant below an altitude of 100 km, but the total pressure decreases with altitude as P = P 0eMavegz>RT , where Mave is the mean molecular weight of air. At sea level, xN2 = 0.78084, xHe = 0.00000524, and T = 300 K. a. Calculate the total pressure at 10.5 km assuming a mean molecular mass of 28.9 g mol1 and that T = 300. K throughout this altitude range. b. Calculate the value that xN2 >xHe would have at 10.5 km in the absence of turbulent mixing. Compare your answer with the correct value. P1.17 An initial step in the biosynthesis of glucose C6H12O6 is the carboxylation of pyruvic acid CH3COCOOH to form oxaloacetic acid HOOCCOCH2COOH CH3COCOOH1s2 + CO21g2 S HOOCCOCH2COOH1s2
If you knew nothing else about the intervening reactions involved in glucose biosynthesis other than no further carboxylations occur, what volume of CO2 is required to produce 2.50 g of glucose? Assume P = 1 atm and T = 298. K. P1.18 Consider the oxidation of the amino acid glycine NH2CH2COOH to produce water, carbon dioxide, and urea NH2CONH2: NH2CH2COOH1s2 + 3O21g2 S NH2CONH21s2 + 3CO21g2 + 3H2O1l2
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19
Calculate the volume of carbon dioxide evolved at P = 1.00 atm and T = 305. K from the oxidation of 0.050 g of glycine. P1.19 Assume that air has a mean molar mass of 28.9 g mol1 and that the atmosphere has a uniform temperature of 25.0°C. Calculate the barometric pressure in Pa in Boulder, for which z = 5430. ft. Use the information contained in Problem P1.16. P1.20 When Julius Caesar expired, his last exhalation had a volume of 375 cm3 and contained 1.00 mole percent argon. Assume that T = 300. K and P = 1.00 atm at the location of his demise. Assume further that T has the same value throughout the Earth’s atmosphere. If all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere, how many inhalations of 375 cm3 must we make to inhale one of the Ar atoms exhaled in Caesar’s last breath? Assume the radius of the Earth to be 6.37 * 106 m. [Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of the Earth’s surface. See Problem P1.16 for the dependence of the barometric pressure on the height above the Earth’s surface.] P1.21 Calculate the number of molecules per m3 in an ideal gas at the standard temperature and pressure conditions of 0.00°C and 1.00 atm. P1.22 Consider a gas mixture in a [email protected] flask at 15.0°C. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent: a. 4.50 g H2 and 3.25 g O2 b. 3.25 g N2 and 2.00 g O2 c. 4.50 g CH4 and 2.25 g NH3 P1.23 A mixture of H21g2 and NH31g2 has a volume of 125 cm3 at 0.00°C and one atm. The mixture is cooled to the temperature of liquid nitrogen, at which point ammonia freezes out and the remaining gas is removed from the vessel. Upon warming the vessel to 0.00°C and one atm, the volume is 82.5 cm3. Calculate the mole fraction of NH3 in the original mixture. P1.24 A sealed flask with a capacity of 1.35 dm3 contains 10.5 g of carbon dioxide. The flask is so weak that it will burst if the pressure exceeds 1.00 * 106 Pa. At what temperature will the pressure of the gas exceed the bursting pressure? P1.25 A balloon filled with 55.5 L of He at 15.0°C and one atm rises to a height in the atmosphere where the pressure is 207 Torr and the temperature is 32.4°C. What is the final volume of the balloon? Assume that the pressure inside and outside the balloon have the same value. P1.26 Carbon monoxide competes with oxygen for binding sites on the transport protein hemoglobin. CO can be poisonous if inhaled in large quantities. A safe level of CO in air is 50. parts per million (ppm). When the CO level increases to 800. ppm, dizziness, nausea, and unconsciousness occur, followed by death. Assuming the partial pressure of oxygen in air at sea level is 0.20 atm, what proportion of CO to O2 is fatal?
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CHAPTER 1 Fundamental Concepts of Thermodynamics
P1.27 The total pressure of a mixture of oxygen and hydrogen is 2.00 atm. The mixture is ignited, and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.155 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole percent? P1.28 Liquid N2 has a density of 875.4 kg m3 at its normal boiling point. What volume does a balloon occupy at 298 K and a pressure of 1.00 atm if 5.25 * 103 L of liquid N2 is injected into it? Assume that there is no pressure difference between the inside and outside of the balloon. P1.29 Calculate the volume of all gases evolved by the complete oxidation of 0.500 g of the amino acid alanine NH2CHCH3COOH if the products are liquid water, nitrogen gas, and carbon dioxide gas, the total pressure is 1.00 atm, and T = 305 K. P1.30 As a result of photosynthesis, an acre of forest (1 acre = 4047 square meter) can take up 1000. kg of CO2. Assuming air is 0.0412% CO2 by volume, what volume of air is required to provide 500. kg of CO2? Assume T = 298 K and P = 1.00 atm. P1.31 A gas thermometer of volume 0.205 L contains 0.325 g of gas at 759.0 Torr and 15.0°C. What is the molar mass of the gas? P1.32 A 625 cm3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 3.025 g at 25.0°C and the pressure is 760. Torr, calculate the mole fraction of Xe in the mixture. P1.33 Use L’Hôpital’s rule, lim 3 ƒ1x2>g1x24 x S 0 = dƒ1x2>dx lim c d to show that the expression derived for dg1x2>dx x S 0 Pƒ in part (b) of Example Problem 1.1 has the correct limit as g S 0. P1.34 Many processes such as the fabrication of integrated circuits are carried out in a vacuum chamber to avoid reaction of the material with oxygen in the atmosphere. It is difficult to routinely lower the pressure in a vacuum chamber below 8.0 * 1011 Torr. Calculate the molar density at this pressure at 300. K. What fraction of the gasphase molecules initially present for 1.0 atm in the chamber is present at 1.0 * 1010 Torr? P1.35 Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person with
a mass of 71 kg? The following data are available: Oxygen consumption is approximately 38 mL of O2 per minute per kilogram of body weight, measured at T = 300. K and P = 1.00 atm. In an adult there are about 1.6 * 1010 cells per kg body mass. Each cell contains approximately 800. mitochondria. P1.36 A cylinder of compressed gas contains 3.21 * 103 g of N2 gas at a pressure of 3.75 * 107 Pa and a temperature of 15.0°C. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 2.35 * 105 Pa? Assume ideal behavior and that the gas temperature is unchanged. Section 1.5 P1.37 A real gas is in a regime in which it is accurately described by the ideal gas equation of state. 19.0 g of the gas occupies 50.0 L at a pressure of 0.500 atm and a temperature of 449 K. Identify the gas. P1.38 Calculate the pressure exerted by Ar for a molar volume of 1.54 L mol1 at 375 K using the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar dm6 mol2 and 0.0320 dm3 mol1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions? P1.39 Use the ideal gas and van der Waals equations to calculate the pressure when 1.75 mol H2 are confined to a volume of 2.05 L at 298 K. Is the gas in the repulsive or attractive region of the molecule–molecule potential? P1.40 Calculate the pressure exerted by benzene for a molar volume of 1.75 L at 605 K using the Redlich–Kwong equation of state: P = =
RT a 1 Vm  b V 1V 2T m m + b2 nRT n2a 1 V  nb V1V + nb2 2T
The Redlich–Kwong parameters a and b for benzene are 452.0 bar dm6 mol2 K1>2 and 0.08271 dm3 mol1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions? P1.41 Rewrite the van der Waals equation using the molar volume rather than V and n.
FURTHER READING Laugier, A., and Garai, J. “Derivation of the Ideal Gas Law.” Journal of Chemical Education (2007): 1832–1833. Eberhart, J. “The Many Faces of van der Waals’s Equation of State.” Journal of Chemical Education 66 (1989): 906–909.
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Rigby, M. “A van der Waals Equation for Nonspherical Molecules.” Journal of Physical Chemistry 76 (1972): 2014–2016. Soave, G. “Improvement of the van der Waals Equation of State.” Chemical Engineering Science 39 (1984): 357–369.
19/09/17 5:03 PM
MATH ESSENTIAL 2:
Differentiation and Integration Differential and integral calculus is used extensively in physical chemistry. In this unit we review the most relevant aspects of calculus needed to understand the chapter discussions and to solve the endofchapter problems.
ME2.1 The Definition and Properties of a Function
ME2.1 THE DEFINITION AND PROPERTIES
ME2.3 The Chain Rule ME2.4 The Sum and Product Rules
OF A FUNCTION
A function ƒ is a rule that generates a value y from the value of a variable x. Mathematically, we write this as y = ƒ1x2. The set of values x over which ƒ is defined is the domain of the function. Singlevalued functions have a single value of y for a given value of x. Most functions that we will deal with in physical chemistry are single valued. However, inverse trigonometric functions and 1 are examples of common functions that are multivalued. A function is continuous if it satisfies these three conditions:
ME2.2 The First Derivative of a Function
ME2.5 The Reciprocal Rule and the Quotient Rule ME2.6 HigherOrder Derivatives: Maxima, Minima, and Inflection Points ME2.7 Definite and Indefinite Integrals
ƒ1x2 is defined at a
lim ƒ1x2 exists
xSa
lim ƒ1x2 = ƒ1a2
xSa
(ME2.1)
ME2.2 THE FIRST DERIVATIVE OF A FUNCTION The first derivative of a function has as its physical interpretation the slope of the function evaluated at the point of interest. In order for the first derivative to exist at a point a, the function must be continuous at x = a, and the slope of the function at x = a must be the same when approaching a from x 6 a and x 7 a. For example, the slope of the function y = x 2 at the point x = 1.5 is indicated by the line tangent to the curve shown in Figure ME2.1. Mathematically, the first derivative of a function ƒ1x2 is denoted dƒ1x2>dx. It is defined by dƒ1x2 ƒ1x + h2  ƒ1x2 = lim (ME2.2) hS0 dx h
f(x) f(x)5x2
20
10
24
2
22
4 x
210
The symbol ƒ′1x2 is often used in place of dƒ1x2>dx. For the function of interest,
dƒ1x2 1x + h22  1x22 2hx + h2 = lim = lim = lim 2x + h = 2x (ME2.3) hS0 hS0 hS0 dx h h
In order for dƒ1x2>dx to be defined over an interval in x, ƒ1x2 must be continuous over the interval. Next, we present rules for differentiating simple functions. Some of these functions and their derivatives are as follows: d1ax n2 = anx n  1, where a is a constant and n is any real number dx d1aex2 = aex, where a is a constant dx d ln x 1 = x dx
M02_ENGE4583_04_SE_C02A_021028.indd 21
Figure ME2.1
The function y = x2 plotted as a function of x. The dashed line is the tangent to the curve at x = 1.5.
(ME2.4) (ME2.5) (ME2.6) 21
25/07/17 11:33 AM
22
MATH ESSENTIAL 2 Differentiation and Integration
d1a sin x2 = a cos x, where a is a constant dx
(ME2.7)
d1a cos x2 = a sin x, where a is a constant dx
(ME2.8)
ME2.3 THE CHAIN RULE In this section, we deal with the differentiation of more complicated functions. Suppose that y = ƒ1u2 and u = g1x2. From the previous section, we know how to calculate dƒ1u2>du. But how do we calculate dƒ1u2>dx? The answer to this question is stated as the chain rule: dƒ1u2 dƒ1u2 du = dx du dx
(ME2.9)
Several examples illustrating the chain rule follow: d sin13x2 d sin13x2 d13x2 = = 3 cos13x2 dx d13x2 dx
d ln1x 22 d ln1x 22 d1x 22 2x 2 = = 2 = x dx dx d1x 22 x
d ax +
dx
1 4 b x
1 4 1 b d ax + b x x 1 5 1 = = 4 ax + b a1  2 b x dx 1 x d ax + b x d ax +
(ME2.10)
(ME2.11)
(ME2.12)
d exp1ax 22 d exp1ax 22 d1ax 22 = = 2ax exp1ax 22, where a is a constant (ME2.13) dx dx d1ax 22
Additional examples of use of the chain rule include: d 1 23x 4>3 2 >dx = 14>3223x 1>3 d 1 5e322x 2 >dx = 1522e322x
(ME2.14) (ME2.15)
d14 sin kx2 = 4k cos x, where k is a constant dx
d 1 23 cos 2px 2 dx
= 223p sin 2px
(ME2.16)
ME2.4 THE SUM AND PRODUCT RULES Two useful rules in evaluating the derivative of a function that is itself the sum or product of two functions are as follows:
d3 ƒ1x2 + g1x24 dƒ1x2 dg1x2 = + dx dx dx
(ME2.17)
d1x 3 + sin x2 dx 3 d sin x = + = 3x 2 + cos x dx dx dx
(ME2.18)
For example,
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ME2.6 HigherOrder Derivatives: Maxima, Minima, and Inflection Points
d3 ƒ1x2g1x24 dƒ1x2 dg1x2 = g1x2 + ƒ1x2 dx dx dx
23
(ME2.19)
For example, d3 sin1x2 cos1x2 4 d sin1x2 d cos1x2 = cos1x2 + sin1x2 dx dx dx = cos2 x  sin2 x
(ME2.20)
ME2.5 THE RECIPROCAL RULE
AND THE QUOTIENT RULE
How is the first derivative calculated if the function to be differentiated does not have a simple form such as those listed in the preceding section? In many cases, the derivative is found by using the product rule and the quotient rule given by da
For example,
da
dc
For example,
1 b sin x 1 d sin x cos x =  2 = dx sin x dx sin2 x ƒ1x2 dƒ1x2 dg1x2 d g1x2  ƒ1x2 g1x2 dx dx = 2 dx 3 g1x24 da
1 b dƒ1x2 ƒ1x2 1 = 2 dx dx 3 ƒ1x24
x2 b sin x 2x sin x  x 2 cos x = dx sin2 x
(ME2.21)
(ME2.22)
(ME2.23)
(ME2.24)
ME2.6 HIGHERORDER DERIVATIVES: MAXIMA, MINIMA, AND INFLECTION POINTS
A function ƒ1x2 can have higherorder derivatives in addition to the first derivative. The second derivative of a function is the slope of a graph of the slope of the function versus the variable. In order for the second derivative to exist, the first derivative must be continuous at the point of interest. Mathematically, d 2ƒ1x2
dx
For example, d 2 exp1ax 22 dx 2
=
2
=
d dƒ1x2 a b dx dx
(ME2.25)
d3 2ax exp1ax 224 d d exp1ax 22 c d = dx dx dx
= 2a exp1ax 22 + 4a2x 2 exp1ax 22, where a is a constant (ME2.26)
The symbol ƒ″1x2 is often used in place of d 2ƒ1x2>dx 2. If a function ƒ1x2 has a concave upward shape 1∪2 at the point of interest, its first derivative is increasing with x and therefore ƒ″1x2 7 0. If a function ƒ1x2 has a concave downward shape 1¨2 at the point of interest, ƒ″1x2 6 0.
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24
MATH ESSENTIAL 2 Differentiation and Integration 4
f(x)5x225x 22
f(x)
2
1
21
x
2
22
The second derivative is useful in identifying where a function has its minimum or maximum value within a range of the variable, as shown next. Because the first derivative is zero at a local maximum or minimum, dƒ1x2>dx = 0 at the values xmax and xmin. Consider the function ƒ1x2 = x 3  5x shown in Figure ME2.2 over the range 2.5 … x … 2.5. By taking the derivative of this function and setting it equal to zero, we find the minima and maxima of this function in the range d1x 3  5x2 5 = 3x 2  5 = 0, which has the solutions x = { = 1.291 dx A3
24
Figure ME2.2
The maxima and minima can also be determined by graphing the derivative and finding the zero crossings, as shown in Figure ME2.3. Graphing the function clearly shows that the function has one maximum and one minimum in the range specified. Which criterion can be used to distinguish between these extrema if the function is not graphed? The sign of the second derivative, evaluated at the point for which the first derivative is zero, can be used to distinguish between a maximum and a minimum:
ƒ1 x2 = x3 − 5x plotted as a function of x. Note that it has a maximum and a minimum in the range shown.
f(x)
f(x)53x225 10
d 2ƒ1x2
dx
5
d 2ƒ1x2
22
1
21
dx
x
2
2
2
= =
d dƒ1x2 c d 6 0 for a maximum dx dx
d dƒ1x2 c d 7 0 for a minimum dx dx
(ME2.27)
We return to the function graphed earlier and calculate the second derivative: 25
Figure ME2.3
The first derivative of the function shown in the previous figure as a function of x.
f(x)
2.5
x 1
21
2
22.5 f(x)5x3
25.0 27.5
Figure ME2.4
d 2ƒ1x2 dx
2
d13x 2  52 d d1x 3  5x2 c d = = 6x dx dx dx at x = {
5 = {1.291 A3
(ME2.28)
(ME2.29)
we see that x = 1.291 corresponds to the minimum, and x = 1.291 corresponds to the maximum. If a function has an inflection point in the interval of interest, then
5.0
22
dx 2
=
By evaluating
7.5
d 21x 3  5x2
ƒ1 x2 = x3 plotted as a function of x. The value of x at which the tangent to the curve is horizontal is called an inflection point.
dƒ1x2 = 0 and dx
d 2ƒ1x2
= 0
dx 2
(ME2.30)
An example for an inflection point is x = 0 for ƒ1x2 = x 3. A graph of this function in the interval 2 … x … 2 is shown in Figure ME2.4. In this case,
dx 3 = 3x 2 = 0 at x = 0 and dx
d 21x 32
= 6x = 0 at x = 0
dx 2
(ME2.31)
ME2.7 DEFINITE AND INDEFINITE INTEGRALS In many areas of physical chemistry, the property of interest is the integral of a function over an interval in the variable of interest. For example, the work done in expanding an ideal gas from the initial volume Vi to the final volume Vƒ is the integral of the external pressure Pext over the volume xƒ
w = 
Lxi
Vƒ
Pexternal Adx = 
LVi
Pexternal dV
(ME2.32)
Equation ME2.13 defines a definite integral in which the lower and upper limits of integration are given. Geometrically, the integral of a function over an interval is the area
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25
ME2.7 Definite and Indefinite Integrals
under the curve describing the function. For example, the integral 12.31x 3  5x2dx is the sum of the areas of the individual rectangles in Figure ME2.5 in the limit within which the width of the rectangles approaches zero. If the rectangles lie below the zero line, the incremental area is negative; if the rectangles lie above the zero line, the incremental area is positive. In this case, the total area is zero because the total negative area equals the total positive area. The integral can also be understood as an antiderivative. From this point of view, the integral symbol is defined by the relation 2.3
ƒ1x2 =
dƒ1x2 dx L dx
L
1x 3  5x2dx =
x4 5x 2 + C 4 2
f(x)5x325x
22
f(x)
2
1
21
2
x
22 24
(ME2.33)
and the function that appears under the integral sign is called the integrand. Interpreting the integral in terms of area, we evaluate a definite integral, and the interval over which the integration occurs is specified. The interval is not specified for an indefinite integral. The geometrical interpretation is often useful in obtaining the value of a definite integral from experimental data when the functional form of the integrand is not known. For our purposes, the interpretation of the integral as an antiderivative is more useful. The value of the indefinite integral 1 1x 3  5x2dx is that function which, when differentiated, gives the integrand. Using the rules for differentiation discussed earlier, you can verify that
4
Figure ME2.5
The integral of a function over a given range corresponds to the area under the curve. The area under the curve is shown approximately by the green rectangles.
(ME2.34)
Note the constant that appears in the evaluation of every indefinite integral. By differentiating the function obtained upon integration, you should convince yourself that any constant will lead to the same integrand. In contrast, a definite integral has no constant of integration. If we evaluate the definite integral 2.3
L
2.3
1x 3  5x2dx = a
x4 5x 2 x4 5x 2 + Cb  a + Cb (ME2.35) 4 2 4 2 x = 2.3 x = 2.3
we see that the constant of integration cancels. The function obtained upon integration 2.3 is an even function of x, and 12.31x 3  5x2dx = 0, just as we saw in the geometric interpretation of the integral. Some indefinite integrals are encountered so often by students of physical chemistry that they become second nature and are recalled at will. These integrals are directly related to the derivatives discussed in Sections ME2.2–ME2.5 and include the following: L
dƒ1x2 = ƒ1x2 + C
L
x ndx =
xn + 1 + C n + 1
dx = ln x + C L x
M02_ENGE4583_04_SE_C02A_021028.indd 25
eax + C, where a is a constant a
(ME2.36)
L
eax =
L
sin x dx = cos x + C
(ME2.38)
L
cos x dx = sin x + C
(ME2.39)
(ME2.37)
25/07/17 11:33 AM
26
MATH ESSENTIAL 2 Differentiation and Integration
Although students will no doubt be able to recall the most commonly used integrals, the primary tool for the physical chemist in evaluating integrals is a good set of integral tables. Some commonly encountered integrals are listed below. The first group presents indefinite integrals. L L L L L L L L L L L L
1sin ax2dx = 
(ME2.40)
1 sin ax + C a
1cos ax2dx = 1sin 2 ax2dx = 1cos2 ax2dx =
1 cos ax + C a
(ME2.41)
1 1 x sin 2ax + C 2 4a
(ME2.42)
1 1 x + sin 2ax + C 2 4a
(ME2.43)
x sin2 bxdx =
x2 cos 2bx x sin 2bx + C 4 4b 8b2
(ME2.44)
x cos2 bxdx =
x2 cos 2bx x sin 2bx + + + C 4 4b 8b2
(ME2.45)
1x 2 sin2 ax2dx =
1 3 1 1 1 x  a x2 b sin 2ax x cos 2ax + C 6 4a 4a2 8a3
1x 2 cos2 ax2dx =
1 3 1 1 1 x + a x2 b sin 2ax + x cos 2ax + C 3 6 4a 4a2 8a
1sin3 ax2dx = 1x 2 sin ax2dx = 1x 2 cos ax2dx = x meaxdx =
(ME2.48)
1a2x 2  22cos ax
+
2x sin ax + C a2
(ME2.49)
1a2x 2  22sin ax
+
2x cos ax + C a2
(ME2.50)
a3
a
3
1 eax a eax eax dx = + dx + C m m  1 xm  1 m  1 L xm  1 Lx L
(ME2.47)
3 cos ax cos 3ax + + C 4a 12a
x meax m x m  1eaxdx + C a aL
r 2eardr = 
(ME2.46)
ear 2 2 1a r + 2ar + 22 + C a3
(ME2.51) (ME2.52) (ME2.53)
The following group includes definite integrals. a
a
0
0
npx mpx npx mpx a sin a b * sin a b dx = cos a b * cos a b dx = dmn a a a a 2 L L where dmn is one if m = n and 0 if m ≠ n
a
L 0
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c sin a
npx npx b d * c cos a b d dx = 0 a a
(ME2.54) (ME2.55)
23/10/17 2:30 PM
ME2.7 Definite and Indefinite Integrals p
L
p
2
sin mx dx =
0
∞
sin x
∞
dx =
∞
L 2x 0
x neaxdx =
0
2
cos x
n! a
∞
L
L
cos2 mx dx =
0
L 2x 0 L
x 2neax dx =
n+1
dx =
1#3#5#
∞
2
x 2n + 1eax dx =
0
∞
L 0
2
eax dx = a
p 2
(ME2.56)
p A2
(ME2.57)
1a 7 0, n positive integer2
# # 12n  12
n+1 n
2
0
L
27
a
p 1a 7 0, n positive integer2 Aa
n! 1a 7 0, n positive integer2 2 an + 1
p 1>2 b 4a
(ME2.58)
(ME2.59)
(ME2.60)
(ME2.61)
Commercially available software such as Mathematica™, Maple™, Matlab™, and MathCad™ can evaluate both definite and indefinite integrals.
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C H A P T E R
2
Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
2.1 Internal Energy and the First Law of Thermodynamics
2.2 Heat 2.3 Work 2.4 Equilibrium, Change, and Reversibility
2.5 The Work of Reversible Compression or Expansion of an Ideal Gas
WHY is this material important? Energy transformations are an important feature of many processes such as chemical reactions and phase transformations. In this chapter, you will learn how to calculate changes in the internal energy and enthalpy of a system and its surroundings as system variables are changed. You will also learn how macroscopic processes such as heat and work flow can be understood at the molecular level.
WHAT are the most important concepts and results? Changes in the internal energy and enthalpy depend only on the initial and final states of a system. By contrast, work and heat flow, which are responsible for changing the energy of a system, depend on the path between the initial and final states. The heat capacity of gases, liquids, and solids depends strongly on temperature, which can be understood by considering the translational, rotational, and vibrational degrees of freedom of molecules.
WHAT would be helpful for you to review for this chapter? Because differential and integral calculus is used extensively in this chapter, it would be helpful to review the material on calculus in Math Essential 2.
2.6 The Work of Irreversible Compression or Expansion of an Ideal Gas
2.7 Other Examples of Work 2.8 State Functions and Path Functions
2.9 Comparing Work for Reversible and Irreversible Processes
2.10 Changing the System Energy from a MolecularLevel Perspective
2.11 Heat Capacity 2.12 Determining ∆U and Introducing the Sate Function Enthalpy
2.13 Calculating q, w, ∆U, and ∆H for Processes Involving Ideal Gases
2.14 Reversible Adiabatic Expansion and Compression of an Ideal Gas
2.1 INTERNAL ENERGY AND THE FIRST LAW OF THERMODYNAMICS
This section focuses on the change in energy of the system and surroundings during a thermodynamic process such as an expansion or compression of a gas. In thermodynamics, we are interested in the internal energy of a system, as opposed to the energy associated with a system relative to a particular frame of reference. For example, a container of gas in an airplane has a kinetic energy relative to an observer on the ground. However, the internal energy of the gas is defined relative to a coordinate system fixed on the container that moves as the airplane moves. Viewed at a molecular level, the internal energy can take on a number of forms such as
• the translational energy of the molecules. • the potential energy of the constituents of the system. For example, a crystal con•
sisting of polarizable molecules will experience a change in its potential energy as an electric field is applied to the system. the internal energy stored in the form of molecular vibrations and rotations.
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30
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
• the internal energy stored in the form of chemical bonds that can be released •
through a chemical reaction. the potential energy of interaction between molecules.
The total of all these forms of energy for the system of interest is given the symbol U and is called the internal energy.1 The first law of thermodynamics is based on the results of many experiments showing that energy can be neither created nor destroyed, if the energies of both the system and the surroundings are considered. This law can be formulated in a number of equivalent forms. Our initial formulation of this law is as follows: The internal energy U of an isolated system is constant. This form of the first law looks uninteresting because it suggests that nothing happens in an isolated system when viewed from outside the system. How can the first law tell us anything about thermodynamic processes such as chemical reactions? Consider separating an isolated system into two subsystems, the system and the surroundings. When changes in U occur in a system in contact with its surroundings, the change in total energy∆Utot is given by ∆Utot = ∆Usys + ∆Usur = 0 (2.1) Therefore, the first law becomes
Concept The energy of a system can only be changed by transfer of heat or work between system and surroundings.
∆Usys =  ∆Usur
(2.2)
For any decrease of Usys, Usur must increase by exactly the same amount. For example, if a gas (the system) is cooled, the temperature of the surroundings must increase. How can the energy of a system be changed? There are many ways to alter U, several of which are discussed in this chapter. All changes in a closed system in which no chemical reactions or phase changes occur can be classified only as heat, work, or a combination of both. Therefore, the internal energy of such a system can only be changed by the flow of heat or work across the boundary between the system and surroundings. For example, U for a gas can be increased by putting it in an oven or by compressing it. In both cases, the temperature of the system increases. This important recognition leads to a second and more useful formulation of the first law:
∆U = q + w
(2.3)
where q and w designate heat and work, respectively. Throughout this textbook, quantities without a subscript such as ∆U, q, and w refer to the system. What do we mean by heat and work? In the following two sections, we define these important concepts and discuss how they differ. The symbol ∆ is used to indicate a change that occurs as a result of an arbitrary process. The simplest processes are those in which one of P, V, or T remains constant. A constant temperature process is referred to as isothermal, and the corresponding terms for constants P and V are isobaric and isochoric, respectively.
2.2 HEAT In this and the next section, we discuss the two ways in which the internal energy of a system can be changed. Heat,2 in the context of thermodynamics, is defined as the quantity of energy that flows across the boundary between the system and surroundings 1
We could include other terms such as the binding energy of the atomic nuclei, but we choose to include only the forms of energy that are likely to change in simple chemical reactions. 2 Heat is perhaps the most misused term in thermodynamics, as discussed in the article by Robert Romer (2001) cited in Further Reading. In common usage, it is incorrectly referred to as a substance as in the phrase “Close the door; you’re letting the heat out!” An equally inappropriate term is heat capacity (discussed in Section 2.11), because it implies that materials have the capacity to hold heat rather than to store energy. We use the terms heat flow or heat transfer to emphasize the transitory nature of heat. However, you should not think of heat as a fluid or a substance.
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because of a temperature difference between the system and the surroundings. Heat always flows spontaneously from regions of high temperature to regions of low temperature. Heat has several important characteristics, notably:
• Heat is transitory. That is, it only appears during a change in state of the system and •
•
31
2.3 Work
surroundings. Heat is not associated with the initial and final states of the system and the surroundings. The net effect of heat is to change the internal energy of the system and surroundings in accordance with the first law. If the only change in the surroundings is a change in temperature of a reservoir, heat has flowed between the system and the surroundings. The quantity of heat that has flowed is directly proportional to the change in temperature of the reservoir. The sign convention for heat is as follows. If the temperature of the system is raised, q is positive; if it is lowered, q is negative. It is common usage to state that if q is positive, heat is withdrawn from the surroundings and deposited in the system. If q is negative, heat is withdrawn from the system and deposited in the surroundings.
Defining the surroundings as the rest of the universe is impractical because it is not realistic to search through the whole universe to see if a mass has been raised or lowered and if the temperature of a reservoir has changed. Experience shows that, in general, only those parts of the universe close to the system interact with it. Experiments can be constructed to ensure that this is the case, as shown in Figure 2.1. Imagine that we are interested in an exothermic chemical reaction that is carried out in a rigid sealed container with diathermal walls. We define the system as consisting solely of the reactant and product mixture. The vessel containing the system is immersed in an inner water bath separated from an outer water bath by a container with rigid diathermal walls. During the reaction, heat flows out of the system 1q 6 02, and the temperature of the inner water bath increases to Tƒ. Using an electrical heater, we can increase the temperature of the outer water bath so that at all times, Touter = Tinner . Therefore, no heat flows across the boundary between the two water baths, and because the container enclosing the inner water bath is rigid, no work flows across this boundary. Therefore, ∆U = q + w = 0 + 0 = 0 for the composite system made up of the inner water bath and everything within it. This composite system is an isolated system that does not interact with the rest of the universe. To determine q and w for the reactant and product mixture, we need to examine only the composite system and we can disregard the rest of the universe.
Concept If heat is withdrawn from the surroundings and deposited in the system, q is positive.
Rest of universe
Thermometers
Outer water bath Inner water bath
Sealed reaction vessel Tinner Touter 5 Tinner Heating coil
Figure 2.1
Isolating a system of interest. A sealed reaction vessel with rigid walls is immersed in an inner water bath, which in turn is immersed in an outer water bath. A heater in the outer water bath is used to always keep the temperatures of the outer and inner water baths equal. This allows an isolated composite system to be created in which the surroundings to the system of interest are limited in extent.
2.3 WORK Work, in the context of thermodynamics, is defined as any action that transfers energy across the boundary between the system and surroundings; this transfer results when a force in the surroundings causes a displacement in the system along the direction of the force. For example, adding a mass to a spring (the system) tethered to the ceiling causes the spring to increase in length. This process involves work. Other examples of work are charging a battery using a current flow across an electrical potential or compressing a gas. In each of these examples, there is a motion along the direction of the force. The discussion throughout this chapter is limited to processes in which the system and surroundings have no kinetic energy at the beginning and end of the process. We also assume that there are no frictional forces between moving components in the system or surroundings. Just as for heat, work has several important characteristics:
• Work is transitory: it only appears during a change in state of the system and sur•
roundings. Only energy, and not work, is associated with the initial and final states of the system. The net effect of work is to change U of the system and surroundings in accordance with the first law, according to Equation (2.3).
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32
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Concept
• In principle, work can always be carried out so that a mass in the surroundings is
If in doing work, a mass in the surroundings is lowered, w is positive.
• •
raised or lowered in the Earth’s gravitational field; this is the signature of work as opposed to heat. If the mass in the surroundings has been raised, w is defined to be negative. In contrast, if the mass in the surroundings has been lowered, w is defined to be positive. If w 7 0, it is common usage to say that work has been carried out on the system by the surroundings; if w 6 0, we say that work has been carried out by the system on the surroundings. Work can be defined from a reference point in either the system or the surroundings. In the absence of friction, wsur = w. Work can be measured and calculated by following changes either in the system or in the surroundings.3 Work by or on the system results in the change of height of a mass in the surroundings, and the amount of work done is given by w = mg∆h
(2.4)
where ∆h is the change in height of a mass in the surroundings and g is the gravitational constant. Alternately, the work for the same process can be calculated from changes in the system using the following equation xƒ
•
w =
Lxi
F # dx.
(2.5)
where F is the force in the surroundings acting on the system and dx is the displacement. Note that because of the scalar product in the integral, the work will be zero unless the force has a component along the displacement direction. The sign of the work follows from evaluating the integral in Equation (2.5). If the vectors F and dx point in the same direction, w is positive. In contrast, if the vectors point in the opposite direction, w is negative. Example Problem 2.2 illustrates both ways to calculate work. At a molecular level, work is associated with a coordinated motion. For example, in doing electrical work, the electrons move from a lower to a higher electrical potential. In compressing a gas, there is a concerted motion of the molecules. Common forms of work are listed in Table 2.1 and shown in Figure 2.2.
Next, we will distinguish between two types of processes: reversible and irreversible. After introducing these two types of processes, we will discuss how to calculate the work of compressing or expanding an ideal gas. TABLE 2.1 Types of Work Equation for SystemBased Work
Types of Work
Variables, System definition
Gas expansion and compression
Pressure in the surroundings at the system– surroundings boundary 1Pext2, volume 1V2 The gas is the system.

Force 1F2, distance 1x2 The spring is the system.
w =
Surface tension 1g2, surface area 1s2 The content of the bubble is the system.
w = 
Electrical potential difference 1f2, electrical charge 1Q2 The conductor is the system.
w =
Spring stretching and compression Bubble expansion and contraction Current passes through conductor
Vƒ
LVi
Pext dV xƒ
Lxi
F # dx sƒ
Lsi
L0
gds
Q
f dQ′
SI Units Pa m3 = J Nm = J 1N m12 1m22 = J VC = J
3
The relative merits of calculating system and surroundingsbased work are discussed in the articles by Gislason and Craig (2005, 2007) and SchmidtRohr (2014) cited in Further Reading.
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2.4 Equilibrium, Change, and Reversibility
33
Mass
Mass Mechanical stops
Piston
Potential energy of a spring increases in both compression and expansion.
Mass
Pi ,Vi
Mass
Initial state
Mass
Piston
Mass
Mass
Force vector
Pf ,Vf
Mass Final state Increase in energy of the system: Temperature increase (a) Compression work done on a gas (the system). The initial and final volumes are defined by the mechanical stops indicated in the figure.
Increase in the surface area of water (b) The surface of water (the system) is deformed when a cylindrical mass with a hydrophobic coating is placed on it.
Increase in potential energy stored in the spring (c) A spring (the system), anchored at the top is stretched in response to attachment of a spherical mass to its lower end.
Figure 2.2
(d) A spring (the system), anchored at the bottom is compressed as a mass is attached to its upper end. In parts (c) and (d), the mass is moved horizontally and then attached to the spring in the inital state.
Four processes that involve work. The initial state is shown in the top row, and the final state is shown in the bottom row. (a) Compression work done on a gas (the system). (b) The surface of water (the system) is deformed by a mass placed on it. (c) A stretched spring (the system). (d) A compressed spring (the system). In each process, the mass in the surroundings is lowered, showing that work has been done on the system; thus, w 7 0. The increase in energy of the system manifests as a temperature increase in the gas (a), an increase in the surface area of the water (b), and an increase in potential energy stored in the spring (c and d). Note that unlike a gas, the potential energy of a spring increases in both compression and expansion. The darker red color in the bottom diagram in (a) relative to that in the top diagram implies a higher temperature.
2.4 EQUILIBRIUM, CHANGE, AND REVERSIBILITY Thermodynamics can only be applied to systems in internal equilibrium, and a requirement for equilibrium is that the overall rate of change of all processes such as diffusion or chemical reaction be zero. How do we reconcile these statements with our calculations of q, w, and ∆U associated with processes in which there is a macroscopic change in the system? To answer this question, it is important to distinguish between the system and surroundings each being in internal equilibrium, and between the system and surroundings each being in equilibrium with one another.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
First, there is the issue of internal equilibrium. Consider a system composed of an ideal gas, which satisfies the equation of state, P = nRT>V. All combinations of P, V, and T consistent with this B: Pressure and equation of state form a surface in P–V–T space, as shown in 3 temperature consistent Figure 2.3. All points on the surface correspond to states of interwith a V = 4.0 L nal equilibrium; that is, the system is uniform on all length scales 2 i and is characterized by single values of T, P, and concentration. Nonequilibrium situations cannot be represented on such a plot 1 f because T, P, and concentration do not have unique values. An C: Neither constant temperature nor 0 example of a system that is not in internal equilibrium is a gas constant volume 12.5 10 7.5 5.0 2.5 that is expanding so rapidly that the exchange of energy between V (L) molecules through collisions is slow compared with the rate of expansion. In this case, different regions of the gas may have difFigure 2.3 ferent values for the density, pressure, and temperature. A pressure–volume–temperature diagram for an ideal gas. Next, consider a process in which the system changes from an The surface shown satisfies the equation PV = RT, and therefore all possible states for one mole of an ideal gas lie on this initial state characterized by Pi, Vi, and Ti to a final state charactersurface. All combinations of pressure and volume consistent ized by Pƒ, Vƒ, and Tƒ, as shown in Figure 2.3. If the rate of change with T = 800 K (curve A) and all combinations of pressure of the macroscopic variables is small, the system passes through and temperature consistent with a volume of 4.0 L (curve B) a succession of states that are close to and essentially indistinare shown as black curves that lie in the P–V–T surface. The guishable from equilibrium states as it goes from the initial to the third curve corresponds to a path between an initial state i and final state. Such a process is called a quasistatic process. We a final state ƒ that is neither a constant temperature nor a now visualize a process in which the system undergoes a major constant volume path (curve C). change in terms of a directed path consisting of a sequence of quasistatic processes, and we distinguish between two important classes of quasistatic processes, namely, reversible and irreversible processes. The term path as used Concept in thermodynamics refers to a curve in the space defined by the changing variables In a quasistatic process, the system that represents a process. passes through a continuous Reversibility in a chemical system can be illustrated by a system consisting of liqsuccession of equilibrium states. uid water in equilibrium with gaseous water that is surrounded by a thermal reservoir. The system and surroundings are both at temperature T. An infinitesimally small increase in T results in a small increase of the amount of water in the gaseous phase and a small decrease in the liquid phase. An equally small decrease in the temperature has the opposite effect. Therefore, fluctuations in T give rise to corresponding fluctuations in the composition of the system. Because an infinitesimal change in the direction of the variable that drives the process (T in this case) causes a reversal in the direction of the process, we say that the process is reversible. If an infinitesimal change in the direction of the driving variable does not change the direction of the process, we say that the process is irreversible. For example, if a stepwise temperature increase is induced in the system discussed here using a heat pulse, the amount of water in the gas phase increases rapidly. In this case, the composition of the system cannot be returned to its initial value by an infinitesimal temperature decrease. Although any process that takes place at a rapid rate in the real world is irreversible, real processes can approach reversibility in the appropriate limit. For example, a slow increase in the electrical potential in an electrochemical cell can convert Piston reactants to products in a nearly reversible process. 1000 800 600 400 200
P (10
26
Pa)
T (K)
A: Pressure and volume consistent with T = 800 K
Piston
Ti ,Vi
Ti ,1/2 Vi Initial state
Final state
Figure 2.4
An ideal gas confined in a piston and cylinder assembly. Reversible isothermal compression is shown in which Vƒ = Vi >2. Reversibility is achieved by adjusting the rate at which the beaker on top of the piston is filled with water relative to the evaporation rate.
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2.5 THE WORK OF REVERSIBLE COMPRESSION OR EXPANSION OF AN IDEAL GAS
Imagine reversibly compressing an ideal gas using the setup shown in Figure 2.4. In this reversible process, the upward force on the inner face of the piston is given by PA, where P is the gas pressure and A is the area of the piston. The downward force in the surroundings is given by mg = Pext A, where m is the mass of the piston and water beaker and Pext is the pressure in the surroundings at the system–surroundings boundary. P and Pext differ only by an infinitesimal amount, and both pressures change continuously in the course of this quasistatic reversible process. The work done on the
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2.5 The Work of Reversible Compression or Expansion of an Ideal Gas
gas in moving the piston and mass from an initial height hi to a final height hƒ where hƒ 6 hi is hƒ
wrev compression =
Lhi
F # dx =
hƒ
Lhi
hƒ
mgdx =
Vƒ
= 
LVi
Lhi
Pext Adx
35
Concept In a reversible process, an infinitesimal change in the driving force can reverse the direction of change.
Vƒ
Pext dV = 
LVi
PdV
(2.6)
In the third integral, we have converted force to pressure, and in the fourth integral, we have changed the integration variable to volume rather than distance. The minus sign in the fourth integral arises because, for a compression, dV = Adx. Because P = Pext in a reversible process, either of these variables can be used in Equation (2.6); however, it is convenient to use P because it can be related to other measurable variables by the equation of state for the gas. In order to evaluate the integral, we need to know the functional dependence of P on V. For the example of an isothermal compression of an ideal gas, Vƒ
wisothermal rev compression = 
LVi
Vƒ
PdV = nRT
Vƒ dV = nRT ln 7 0 Vi LVi V
(2.7)
Because compression and expansion by the interchanges of the limits of integration and the sign of dV, wcompression = wexpansion. Note that because a compression process in the system is an expansion process in the surroundings, w = wsur. We can also calculate the work done in this process by observing changes in the surroundings rather than the system. In this case w = mg∆h, where m is the mass of the piston and water beaker and ∆h is the distance that the mass moves in the process. Both expressions for w give the same answer, as is shown in Example Problem 2.1.
EXAMPLE PROBLEM 2.1 1.00 mole of an ideal gas (the system) is confined in a piston and cylinder assembly similar to that shown in Figure 2.2. The initial temperature is 300. K, and the initial pressure is 0.200 bar. The surroundings are a water bath at 300. K. a. Write an expression in terms of the system variables for the work in a reversible process in which the volume of the gas is doubled. Can you obtain a numerical value for w? b. Assume that the reversible process is isothermal. Calculate w by evaluating the integral in part (a).
Solution Vƒ
a. wrev expansion = 
LVi
Vƒ
Pext dV = 
LVi
P1V2dV because P = Pext.
The integral can be evaluated only if the functional dependence P1V2 is known. nRT b. P1V2 = For an isothermal expansion, the product nRT can be taken outside V the integral. Vƒ
wrev expansion = 
LVi
Vƒ
PdV = nRT
Vƒ dV = nRT ln Vi LVi V
= 8.314 J mol1 K1 * 1.00 mol * 300. K ln 2 = 1.73 kJ In endofchapter Problem P2.18, you will be asked to calculate the work for this process by considering the movement of masses in the surroundings.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
2.6 THE WORK OF IRREVERSIBLE
COMPRESSION OR EXPANSION OF AN IDEAL GAS
Imagine the irreversible compression of an ideal gas in the apparatus shown in Figure 2.2 in which both stops are removed abruptly. The frictionless piston falls quickly and oscillates around its final height until coming to rest.4 The height of the mass on the piston decreases from hi to hƒ. In this case, the gas pressure will not be uniform or well defined if the piston falls quickly enough. In addition, we cannot calculate the work from the system variables because they do not have the same values throughout the system. However, if we wait until the piston has come to rest, we can calculate the work in the system from the change in potential energy in the surroundings using w = wsur = mg∆h
(2.8)
as shown in Example Problem 2.2. The change in potential energy in the surroundings has two components: the height of the mass on the piston has changed, and part of the surroundings at constant pressure Psur has been displaced. hƒ
wsur =
Lhi
hƒ
mgdx +
Lhi
hƒ
Psur Adx =
Lhi
a
Vƒ mg + Psur b Adx = Pext dVsur A LVi
(2.9)
Note that Pext also has two components: one from the gas pressure in the surroundings and the second from the force exerted by the piston. Because dVsur = dV, V w = wsur =  1Vi ƒPext dV for an irreversible expansion or compression. Note that we obtained the same formula for the reversible process in which P = Pext.
EXAMPLE PROBLEM 2.2
Concept Work for both reversible and irreversible processes can always be V calculated from w =  1V ƒPext dV . i
1.00 mole of an ideal gas (the system) is confined in a piston and cylinder assembly similar to that shown in Figure 2.2. The piston radius is 10.0 cm, and the total mass on the weightless piston is 64.0 kg. The initial temperature is 300. K. The gas is irreversibly expanded at constant temperature by removing half of the mass. The surroundings are evacuated. a. Describe in words the evolution of the system when the mass on the piston is reduced. b. Calculate w for this irreversible process using Equation (2.6). c. Calculate w for this process using Equation (2.8). Compare your result with your results for part (b).
Solution a. The piston moves rapidly upward and because of its inertia overshoots its equilibrium position. It oscillates back and forth around the equilibrium position until the oscillations are damped out and the piston comes to rest with P = Pext = 1.00 bar. In this case, Psur = 0. Because the expansion may be too rapid to allow internal equilibrium to be maintained, the systembased work cannot in general be calculated for an irreversible expansion.
4
Even for an ideal gas in a frictionless piston and cylinder assembly, the oscillations are damped out. See the references to Gislason and Craig (2005, 2007) in Further Reading for an explanation.
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2.7 Other Examples of Work
b. To use Equation (2.6), we must calculate the external pressure and the initial and final volumes. Pi =
64.0 kg * 9.81 m s2 mg Force = = = 2.00 * 104 Pa Area pr 2 p * 110.0 * 102 m22
Pƒ = 0.500Pi Vi =
nRT 8.314 J mol1 K1 * 1.00 mol * 300. K = = 0.125 m3 Pi 2.00 * 104 Pa
Vƒ =
PiVi = 2Vi = 0.250 m3 Pƒ
Mass
Pext = Pƒ = 1.00 * 104 Pa Pi ,Vi
Vƒ
LVi
w = 
Pext dV = Pext1Vƒ  Vi2
O2
H2 1
H2O
2
= 1.00 * 104 Pa * 10.250 m3  0.125 m32 = 1.25 * 103 J
c. To calculate w using Equation (2.8), we must first calculate the distance by which the piston is raised in the expansion. The initial height is hi =
Vi pr 2
=
0.125 m3 = 3.97 m p * 110.0 * 102 m22
Electrical generator
Mass
Because the volume is doubled, ∆h = hi
Initial state
w = mg∆h = 32.0 kg * 9.81 m s2 * 0.125 m = 1.25 * 103 J The results to parts (b) and (c) are the same as they must be if friction is neglected. Note that although the initial and final states of the system are the same in Example Problems 2.1 and 2.2, w is not the same. We will discuss the origin of this difference in Section 2.9.
We summarize this discussion of the reversible or irreversible compression or expansion of an ideal gas in a frictionless apparatus by stating that the work for both processes V can be calculated from the equation w =  1Vi ƒPext dV. Evaluation of the integral requires that the functional dependence of Pext on V be known. Note that unlike the case of the reversible expansion discussed in the previous section, only the surroundingsbased calculation can be carried out for an irreversible expansion or compression.
Mass
Pi,Vf
H2
1
O2
H2O
2
Electrical generator
2.7 OTHER EXAMPLES OF WORK An example of another important kind of work, electrical work, is shown in Figure 2.5 in which the contents of the cylinder is the system. Electrical current flows through a conductive aqueous solution, and water undergoes electrolysis to produce H2 and O2 gas. The current is produced by a generator. As current flows, the mass that drives the generator is lowered. In this case, the surroundings do electrical work on the system. As a result, some of the liquid water is transformed to H2 and O2. From electrostatics, the work done in transporting a charge, Q, through an electrical potential difference, f, is
welectrical = Qf
(2.10)
For a constant current, I, that flows for a time, t, Q = It. Therefore,
M02_ENGE4583_04_SE_C02_029062.indd 37
welectrical = Ift
(2.11)
Mass Final state
Figure 2.5
A piston and cylinder assembly that contains liquid water. The water is electrolyzed to produce H21g2 and O21g2 by passing current between the two electrodes shown in the figure. In doing so, work is done on the system, as shown by the lowered mass linked to the generator.
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38
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
The system also does work on the surroundings through the increase in the volume of the gas phase at the constant external pressure Pi , as shown by the raised mass on the piston. The total work done is Vƒ
w = wPV + welectrical = Ift 
LVi
Pext dV = Ift  Pext1Vƒ  Vi2
(2.12)
Several additional examples of work calculations are given in Example Problems 2.3 and 2.4.
EXAMPLE PROBLEM 2.3 a. Calculate the work to expand 20.0 L of an ideal gas to a final volume of 85.0 L against a constant external pressure of 2.50 bar. The system is the gas. b. An air bubble in liquid water is expanded from a radius of 0.100 cm to a radius of 0.325 cm. The surface tension of water is 71.99 N m1. Calculate the work for this process. The system is the contents of the bubble. c. A current of 3.20 A is passed through a heating coil for 30.0 s. The electrical potential across the resistor is 14.5 V. Calculate the work for this process. The system is the coil. d. The force that a H2 molecule exerts on whatever is stretching or compressing it through a distance x is given by F = kx with k = 575 N m1. The minus sign indicates that the force opposes the motion. Calculate the work to compress the bond 1.0 pm. The system is the molecule.
Solution Vƒ
a. w =
LVi
Pext dV = Pext1Vƒ  Vi2
= 2.50 bar * sƒ
b. w =
Lsi
105 Pa 103 m3 * 185.0 L  20.0 L2 * = 16.3 kJ bar L
gds = g4p1r 2ƒ  r 2i 2
= 4p * 71.99 N m110.3252 cm2  0.1002 cm22 *
104 m2 cm2
= 8.65 * 103 J
c. w =
L0
Q
f dQ′ = fQ = fIt = 14.5 V * 3.20 A * 30.0 s = 1.39 kJ
d. In calculating the work done on the molecule, we must consider the force exerted by the surroundings on the molecule, which is the negative of the force exerted by the molecule on the surroundings. w =
L
F # dx =
xƒ
L x0
= 2.9 * 1022 J
kxdx = c
kx 2 xƒ 575 N m1 * x 2 1.0 * 10 d = c d 2 x0 2 0
12
EXAMPLE PROBLEM 2.4 A heating coil is immersed in a 100. g sample of H2O liquid at the standard boiling temperature of 99.61 °C in an open insulated beaker on a laboratory bench at 1 bar pressure. In this process, 10.0% of the liquid is converted to the gaseous form at a pressure of 1 bar. A current of 2.00 A flows through the heater from a 12.0V battery
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39
2.8 State Functions and Path Functions
for 1.011 * 103 s to effect the transformation. The densities of liquid and gaseous water under these conditions are 997 and 0.590 kg m3, respectively. a. It is commonly useful to replace a real process by a model that exhibits the important features of the process. Design a model system and its surroundings, like those shown in Figures 2.1 and 2.2, that would allow you to measure the heat and work associated with this transformation. For the model system, define the system and surroundings as well as the boundary between them. b. Calculate q and w for the process. c. How can you define the system for the open insulated beaker on the laboratory bench such that the work is properly described?
Solution a. The following figure shows the system that consists only of liquid initially and gas and liquid in the final state. The cylinder walls and the piston form adiabatic walls. The external pressure is held constant by a suitable weight. b. In the system shown, the heat input to the liquid water can be equated with the work done on the heating coil. Therefore, q = Ift = 2.00 A * 12.0 V * 1.011 * 103 s = 24.3 kJ
105 Pa * a
mgas,ƒ rgas
10.0 * 103 kg 0.590 kg m3
+
+
mliq,ƒ rliq

mliq,i rliq
90.0 * 103 kg 997 kg m3
b

Mass
Heating coil
As the liquid is vaporized, the volume of the system increases at a constant external pressure. Therefore, the work done by the system on the surroundings is w = Pext1Vƒ  Vi2 = Pext a
Pext 5 1.00 atm
100.0 * 103 kg 997 kg m3
= 1.70 kJ
H2O(g)
A
12 V
H2O(l)
b
Note that the electrical work done on the heating coil is much larger than the P–V work done in the expansion. c. Define the system as the liquid in the beaker and the volume containing only molecules of H2O in the gas phase. This volume will consist of disconnected volume elements dispersed in the air above the laboratory bench.
2.8 STATE FUNCTIONS AND PATH FUNCTIONS An alternate statement of the first law is that ∆U depends only on the initial and final states and is independent of the path connecting the initial and final states. In the following, this assertion is verified for kinetic energy, and the argument can be extended to other forms of energy, such as potential energy and internal energy. Consider a single molecule in the system. Imagine that the molecule of mass m initially has the speed v1. We now change its speed incrementally following the sequence v1 S v2 S v3 S v4. The change in the kinetic energy along this sequence is given by
Concept The change in value of a state function depends only on the initial and final states and not on the path between them.
1 1 1 1 ∆Ekinetic = a m1v222  m1v122 b + a m 1v322  m1v222 b 2 2 2 2 1 1 + a m1v422  m1v322 b 2 2
1 1 = a m1v422  m1v122 b 2 2
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(2.13)
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Mass Mechanical stops
Piston
P1,V1,T1
Even though v2 and v3 can take on any arbitrary values, they still do not influence the result. We conclude that the change in the kinetic energy depends only on the initial and final speed and that it is independent of the path between these values. Because this conclusion holds for all molecules in the system, it also holds for ∆U. This example supports the assertion that ∆U depends only on the final and initial states and not on the path connecting these states. Any function that satisfies this condition is called a state function because it depends only on the state of the system and not on the path taken to reach the state. This property can be expressed in a mathematical form. Any state function, for example, U, must satisfy the equation ƒ
T3
∆U =
L
dU = Uƒ  Ui
(2.14)
i
Initial state
Mass Piston P2,V2,T2
T3 Intermediate state
Mass Piston P3,V2,T3
where i and ƒ denote the initial and final states. This equation states that in order for ∆U to depend only on the initial and final states characterized here by i and ƒ, the value of the integral must be independent of the path. If this is the case, U can be expressed as an infinitesimal quantity, dU, that, when integrated, depends only on the initial and final states. The quantity dU is called an exact differential. We defer a more detailed discussion of exact differentials to Chapter 3. It is useful to define a cyclic integral, denoted by the symbol A , as applying to a cyclic path such that the initial and final states are identical. For U or any other state function, (2.15)
V2
Final state
A system consisting of an ideal gas contained in a piston and cylinder assembly. The gas in the initial state V1,T1 is compressed to an intermediate state, whereby the temperature increases to the value T2. It is then brought into contact with a thermal reservoir at T3, leading to a further rise in temperature. The final state is V2,T3. The mechanical stops allow the system volume to be only V1 or V2. The color of the gas signifies temperature, with the redder colors indicating higher temperatures.
dU = Uƒ  Uƒ = 0
because the initial and final states are the same in a cyclic process. Next, we will show that q and w are not state functions. The state of a singlephase system at fixed composition is characterized by any two of the three variables P, T, and V. The same is true of U. Therefore, for a system of fixed mass, U can be written in any of the three forms U1V,T2, U1P,T2, or U1P,V2. Imagine that a gas of fixed mass characterized by V1 and T1 is confined in a piston and cylinder system that is isolated from the surroundings. There is a thermal reservoir in the surroundings at a temperature T3 7 T1. As shown in Figure 2.6, we do compression work on the system starting from an initial state in which the volume is V1 to an intermediate state in which the volume is V2 where V2 6 V1 using a constant external pressure Pext . The work is given by
T3
Figure 2.6
C
w = 
L V1
Pext dV = Pext1V2  V12 = Pext ∆V
(2.16)
Because work is done on the system in the compression, w is positive and U increases. Because the system consists of a uniform single phase, U is a monotonic function of T, and T also increases. The change in volume ∆V has been chosen such that the temperature of the system T2 in the intermediate state after the compression satisfies the inequality T1 6 T2 6 T3 . We next lock the piston in place and let an amount of heat q flow between the system and surroundings at constant by bringing the system into contact with the reservoir at temperature T3. The final state values of T and V after these two steps are T3 and V2. This twostep process is repeated for different values of the mass. In each case, the system is in the same final state characterized by the variables V2 and T3. The sequence of steps that takes the system from the initial state V1,T1 to the final state V2,T3 is referred to as a path. By changing the mass, a set of different paths is generated, all of which originate from the state V1,T1, and end in the state V2,T3. According to the first law, ∆U for this twostep process is
∆U = U1T3,V22  U1T1,V12 = q + w
(2.17)
Because ∆U is a state function, its value for the twostep process just described is the same for each of the different values of the mass.
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2.9 Comparing Work for Reversible and Irreversible Processes
Are q and w also state functions? For this twostep process, w = Pext ∆V
(2.18)
and Pext is different for each value of the mass, or for each path, whereas ∆V is constant. Therefore, w is also different for each path; we can choose one path from V1,T1 to V2,T3 and a different path from V2,T3 back to V1,T1. Because the work is different along these paths, the cyclic integral of work is not equal to zero. Therefore, w is not a state function. Using the first law to calculate q for each of the paths, we obtain the result q = ∆U  w = ∆U + Pext ∆V
(2.19)
Because ∆U is the same for each path and w is different for each path, we conclude that q is also different for each path. Just as for work, the cyclic integral of heat is not equal to zero. Therefore, neither q nor w is a state function, and they are called path functions. Because both q and w are path functions, there are no exact differentials for work and heat. Incremental amounts of these quantities are denoted by dq and dw, rather than dq and dw, to emphasize the fact that incremental amounts of work and heat are not exact differentials. Because dq and dw are not exact differentials, there are no such quantities as ∆q, qƒ, and qi nor ∆w, wƒ, and wi. One cannot refer to the work or heat possessed by a system or to the change in work or heat associated with a process. After a transfer of heat and/or work between the system and surroundings is completed, the system and surroundings possess internal energy but not heat or work. The preceding discussion emphasizes that it is important to use the terms work and heat in a way that reflects the fact that they are not state functions. Examples of systems of interest to us are batteries, fuel cells, refrigerators, and internal combustion engines. In each case, the utility of these systems is that work and/or heat flows between the system and surroundings. For example, in a refrigerator, electrical energy is used to extract heat from the inside of the device and to release it in the surroundings. One can speak of the refrigerator as having the capacity to extract heat, but it would be wrong to speak of it as having heat. In the internal combustion engine, chemical energy contained in the bonds of the hydrocarbon fuel molecules and in O2 is released in forming CO2 and H2O. This change in internal energy can be used to rotate the wheels of the vehicle, thereby inducing a flow of work between the vehicle and the surroundings. One can refer to the capability of the engine to do work, but it would be incorrect to refer to the vehicle or the engine as P1 containing or having work.
We next compare the work associated with the reversible and irreversible isothermal expansion and compression of an ideal gas (see Figure 2.7). Consider the following process. A quantity of an ideal gas is confined in a cylinder with a weightless movable piston. The walls of the system are diathermal, allowing heat to flow between the system and surroundings. Therefore, the process is isothermal at the temperature of the surroundings, T. The system is initially defined by the variables T, P1, and V1. The position of the piston is determined by Pext = P1, which can be changed by adding or removing weights from the piston. Because the weights are moved horizontally, no work is done in adding or removing them. The gas is first expanded at constant temperature by decreasing Pext to the value P2 (weights are removed), where P2 6 P1. A sufficient amount of heat flows into the system through the diathermal walls to keep the temperature at the constant value T. The system is now in the state defined by T, P2, and V2, where V2 7 V1. The system is then returned to its original state in an
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Heat and work are path functions; they are not state functions.
Total w for cycle
2.9 COMPARING WORK FOR
P2
w for compression step
Pext
REVERSIBLE AND IRREVERSIBLE PROCESSES
Concept
w for expansion step
0
V1
V
V2
Figure 2.7
A twostep process involving compression followed by expansion. The process is shown on a P–V diagram. This type of figure is also known as an indicator diagram. The work for each step and the total work can be obtained from the diagram. The arrows indicate the direction of change in V in the two steps. The sign of w is opposite for these two processes.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
isothermal process by increasing Pext to its original value P1 (weights are added). Heat flows out of the system into the surroundings in this step. The system has been restored to its original state and, because this is a cyclic process, ∆U = 0. The total work associated with this cyclic process is given by the sum of the work for each individual step: wtot = a  Pext,i ∆Vi = wexpansion + wcompression = P21V2  V12  P11V1  V22 i
= 1P2  P12 * 1V2  V12 7 0 because P2 6 P1 and V2 7 V1
(2.20)
The relationship between P and V for the process under consideration is shown graphically in Figure 2.7, in what is called an indicator diagram. An indicator diagram is useful because the work done in the expansion and contraction steps can be evaluated from the appropriate area in the figure, which is equivalent to evaluating the integral w =  1 Pext dV. Note that the work done in the expansion is negative because ∆V 7 0, and that done in the compression is positive because ∆V 6 0. Because P2 6 P1, the magnitude of the work done in the compression process is greater than that done in the expansion process and wtot 7 0. What can one say about qtot? The first law states that because ∆U = qtot + wtot = 0, qtot 6 0. The same cyclical process is carried out in a reversible cycle. A necessary condition for reversibility is that P = Pext at every step of the cycle. This means that P changes during the expansion and compression steps. The work associated with the expansion is
25
wisothermal expansion = 
Pi
L
P dV = nRT
V2 dV = nRT ln (2.21) V1 L V
This work is shown schematically as the red area in the indicator diagram of Figure 2.8. If this process is reversed and the compression work is calculated, the following result is obtained:
20
P (bar)
L
Pext dV = 
wisothermal compression = nRT ln
V1 V2
(2.22)
15
10
It is seen that the magnitudes of the work in the forward and reverse processes are equal. The total work done in this cyclical process is given by
5
Pf
0
5
10
15 V (L)
20
25
Figure 2.8
An indicator diagram for a reversible process. Unlike Figure 2.7, the areas under the P–V curves are the same in the forward and reverse directions.
w = wexpansion + wcompression = nRT ln = nRT ln
V2 V2 + nRT ln = 0 V1 V1
V2 V1  nRT ln V1 V2
(2.23)
Therefore, the work done in a reversible isothermal cycle is zero. Because ∆U = q + w is a state function, q = w = 0 for this reversible isothermal process. Looking at the heights of the weights in the surroundings at the end of the process, we find that they are the same as at the beginning of the process. To compare the work for reversible and irreversible processes, the state variables need to be given specific values, as is done in Example Problem 2.5.
EXAMPLE PROBLEM 2.5 In this example, 2.00 mol of an ideal gas undergoes isothermal expansion along three different paths: (1) reversible expansion from Pi = 25.0 bar and Vi = 4.50 L to Pƒ = 4.50 bar; (2) a singlestep irreversible expansion against a constant external pressure of 4.50 bar, and (3) a twostep irreversible expansion consisting initially of an expansion against a constant external pressure of 11.0 bar until P = Pext, followed by an expansion against a constant external pressure of 4.50 bar until P = Pext. Calculate the work for each of these processes. For which of the irreversible processes is the magnitude of the work greater?
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2.9 Comparing Work for Reversible and Irreversible Processes
43
Solution The processes are depicted in the following indicator diagram:
25
Pi
P (bar)
20
15
10
5
Pf
0
5
10
15
20
25
V (L)
We first calculate the constant temperature at which the process is carried out, the final volume, and the intermediate volume in the twostep expansion: T =
PiVi 25.0 bar * 4.50 L = = 677 K 2 nR 8.314 * 10 L bar mol1 K1 * 2.00 mol
Vƒ =
nRT 8.314 * 102 L bar mol1 K1 * 2.00 mol * 677 K = = 25.0 L Pƒ 4.50 bar
Vint =
nRT 8.314 * 102 L bar mol1 K1 * 2.00 mol * 677 K = = 10.2 L Pint 11.0 bar
The work of the reversible process is given by w = nRT1 ln
Vƒ Vi
= 2.00 mol * 8.314 J mol1 K1 * 677 K * ln
25.0 L = 19.3 * 103 J 4.50 L
We next calculate the work of the singlestep and twostep irreversible processes: wsingle = Pext ∆V = 4.50 bar * = 9.23 * 103 J wtwostep = Pext ∆V = 11.0 bar * 4.50 bar *
105 Pa 103 m3 * 125.00 L  4.50 L2 * bar L 105 Pa 103 m3 * 110.2 L  4.50 L2 * bar L
105 Pa 103 m3 * 125.00 L  10.2 L2 * bar L
= 12.9 * 103 J
The magnitude of the work is greater for the twostep process than for the singlestep process, but less than that for the reversible process.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Pressure
Pressure
Comparison of work performed under irreversible and reversible conditions. The graphs show (a) irreversible expansion (green areas) and (b) reversible compression (orange and yellow) work. Note that the total work done in the irreversible expansion and compression processes approaches that of the reversible process as the number of steps becomes large.
Pressure
Figure 2.9
Pressure
The amount of work that can be extracted by a process is a maximum if the process is reversible.
Pressure
Concept
Pressure
Volume Volume Volume (a) Work done in reversible expansion (red + yellow area) compared with work done in a multistep series of irreversible expansion processes at constant pressure (yellow area)
Volume Volume Volume (b) Area under black curve is reversible compression work. Green area is work done in a multistep series of irreversible compression processes at constant pressure
Example Problem 2.5 shows that the magnitude of w for the irreversible expansion is less than that for the reversible expansion, but it also suggests that the magnitude of w for a multistep expansion process increases with the number of steps. This is indeed the case, as shown in Figure 2.9. Imagine that the number of steps n increases indefinitely. As n increases, the pressure difference Pext  P for each individual step decreases. In the limit that n S ∞, the pressure difference Pext  P S 0, and the total area of the rectangles in the indicator diagram approaches the area under the reversible curve. In this limit, the irreversible process becomes reversible, and the value of the work equals that of the reversible process. By contrast, the magnitude of the irreversible compression work exceeds that of the reversible process for finite values of n and becomes equal to that of the reversible process as n S ∞. The difference between the expansion and compression processes results from the requirement that Pext 6 P at the beginning of each expansion step, whereas Pext 7 P at the beginning of each compression step. On the basis of these calculations for the reversible and irreversible cycles, we introduce another criterion to distinguish between reversible and irreversible processes. Suppose that a system undergoes a change through one or more individual steps, and that the system is restored to its initial state by following the same steps in reverse order. The system is restored to its initial state because the process is cyclical. If the surroundings are also returned to their original state (all masses at the same height and all reservoirs at their original temperatures), the process is reversible. If the surroundings are not restored to their original state, the process is irreversible.
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2.10 Changing the System Energy from a MolecularLevel Perspective
45
We are often interested in extracting work from a system. For example, it is the expansion of the fuel–air mixture in an automobile engine upon ignition that provides the torque that eventually drives the wheels. Is the capacity to do work similar for reversible and irreversible processes? This question is answered using the indicator diagrams of Figures 2.7–2.9 for the specific case of isothermal expansion work, noting that the work can be calculated from the area under the P–V curve. We compare the work for expansion from V1 to V2 in a single stage at constant pressure to that for the reversible case. For the singlestage expansion, the constant external pressure is given by Pext = nRT>V2. However, if the expansion is carried out reversibly, the system pressure is always greater than this value. By comparing the areas in the indicator diagrams of Figure 2.9, we see that 0 wrev 0 Ú 0 wirr 0 (2.24) By contrast, for the compression step,
0 wrev 0 … 0 wirr 0
(2.25)
The reversible work is the lower bound for the compression work and the upper bound for the expansion work. This result for the expansion work can be generalized to an important statement that holds for all forms of work: The maximum work that can be extracted from a process between the same initial and final states is that obtained under reversible conditions. Although the preceding statement is true, it suggests that it would be optimal to operate an automobile engine under conditions in which the pressure inside the cylinders differs only infinitesimally from the external atmospheric pressure. This is clearly not possible. A practical engine must generate torque on the drive shaft, and this can only occur if the cylinder pressure is appreciably greater than the external pressure. Similarly, a battery is only useful if one can extract a sizable rather than an infinitesimal current. To operate such devices under useful irreversible conditions, the work output is less than the theoretically possible limit set by the reversible process.
2.10 CHANGING THE SYSTEM ENERGY FROM A MOLECULARLEVEL PERSPECTIVE
Our discussion so far has involved changes in energy for macroscopic systems. But what happens at the molecular level if energy is added to the system? In shifting to a molecular perspective, we move from a classical to a quantummechanical description of matter. For this discussion, we need two results that will be discussed in the companion textbook Quantum Chemistry & Spectroscopy by Thomas Engel. First, in general the energy levels of quantummechanical systems are discrete, and molecules can only possess amounts of energy that correspond to these values. By contrast, in classical mechanics, energy is a continuous variable. Second, we use a result from statistical mechanics showing that the relative probability of a molecule being in a state corresponding to the allowed energy values e1 and e2 at temperature T is given by e13e2  e14>kBT2. This result will be discussed in Chapter 13. To keep the mathematics simple, consider an idealized model system: a gas consisting of a single He atom confined in a onedimensional container with a length of 8.0 nm. The principles of quantum mechanics inform us that the translational energy of the He atom confined in this box can only have the discrete values shown in Figure 2.10b. We lower the temperature of this onedimensional He gas to 0.20 K. The calculated probability of the atom being in a given energy level is shown in Figure 2.10b. If we now do work on the surroundings by expanding the box at constant temperature, the energy levels will change to those shown in Figure 2.10a. Note that all the energy levels are shifted to lower values as the container is made larger. If we keep the temperature constant at 0.20 K during this expansion, the distribution of the atoms among the energy levels changes to that shown in Figure 2.10a. This redistribution occurs because the total energy of the He atom remains constant in the expansion because the temperature is kept constant.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Figure 2.10
Arrows show how energy of each level decreases as length of box increases 1.4 3 10223 1.2 3 10223
Energy (joules)
Energy levels for a He atom confined to a onedimensional box. Boxes are of length (a) 10.0 nm and (b) 8.00 nm. Circles indicate the probability that the He atom has an energy corresponding to each of the energy levels at 0.20 K. Each circle indicates a probability of 0.010. For example, the probability that the energy of the He atom corresponds to the lowest energy level in the 10.0 nm box is 0.21.
1 3 10223
8 3 10224 6 3 10224 4 3 10224 2 3 10224
(a) a 5 10.0 nm
Concept At the molecular level, work is associated with changes in the allowed energy levels, whereas heat is associated with changes in the population of the energy levels.
(b) a 5 8.00 nm
What happens if we keep the container at the longer length and increase the system energy by heat flow so that T increases to 0.30 K? In this case, the energy levels are unchanged because they depend on the container length but not on the temperature of the gas. However, the energy of the system increases, and, as shown in Figure 2.11, this increase occurs by a redistribution of the probability of finding the He atom in the energy levels. The increase in system energy comes from an increase in the probability of the He atom populating higherenergy levels and a corresponding decrease in the probability of populating lowerenergy levels. Just as for a macroscopic system, the quantummechanical system energy increases through heat or work flow. However, in doing work, the energy levels change, whereas in transferring heat, the energy level remains constant and their population changes. Quantummechanical effects become clearer in the next section when we consider how molecules can take up energy through rotation and vibration.
1.4 3 10223
Energy (joules)
1.2 3 10223
Figure 2.11
Energy levels for the 10.0 nm box described in Figure 2.10 caption. Circles indicate the probability that a He atom has an energy corresponding to each of the energy levels at (a) 0.200 K and (b) 0.300 K. Each circle indicates a probability of 0.010.
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1 3 10223
8 3 10224 6 3 10224 4 3 10224 2 3 10224
(a) 0.200 K
(b) 0.300 K
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2.11 Heat Capacity
47
2.11 HEAT CAPACITY We can quantify heat flow in terms of the easily measured electrical work done on a heating coil immersed in the system, with the equation w = Ift. The response of a singlephase system of constant composition to heat input is an increase in T, as long as the system does not undergo a phase change such as the vaporization of a liquid. The thermal response of a system to heat flow is described by a very important thermodynamic property called the heat capacity, which is a measure of energy needed to change the temperature of a substance by a given amount. The symbol for heat capacity is C. The name heat capacity is unfortunate because it implies that a substance has the capacity to absorb heat. A much better name would be energy capacity. Heat capacity is a material and temperaturedependent property, as will be discussed later. Mathematically, heat capacity is defined by the relation
C1T2 = lim
∆T S 0 Tƒ
q dq =  Ti dT
(2.26)
where C is in the SI unit of J K1. It is an extensive quantity that doubles as the number of atoms or molecules in the system is doubled. Commonly, the molar heat capacity Cm is used in calculations. It is an intensive quantity with the units of J K1 mol1. Experimentally, the heat capacity of fluids is measured by immersing a heating coil in the gas or liquid and equating the electrical work done on the coil with the heat flow into the sample. For solids, the heating coil is wrapped around the solid. The value of the heat capacity depends on the experimental conditions under which it is determined. The most common conditions are constant V or P, for which the heat capacity is denoted CV and CP, respectively. Values of CP,m at 298.15 K for pure substances are tabulated in Table 2.3 (see Appendix A, Data Tables), and formulas for calculating CP,m at other temperatures for selected gases, liquids, and solids are listed in Tables 2.4–2.6. Next, let us consider heat capacity using a molecularlevel model, beginning with gases. Figure 2.12 illustrates the energylevel structure for a molecular gas. Molecules can absorb energy by moving faster, by rotating in threedimensional space, by periodic oscillations (known as vibrations) of the atoms around their equilibrium structure, and by electronic excitations. These energetic degrees of freedom are referred to as translation, rotation, vibration, and electronic excitation. Each of the degrees of freedom has its own set of energy levels, and the probability of an individual molecule occupying a higherenergy level increases as it gains energy. The energy levels for atoms and molecules associated with rotation, vibration, and electronic excitation are independent of the container size. In contrast, the energy levels for translation are dependent on container size.
Concept The heat capacity of a substance depends on the values of its rotational and vibrational energy levels, the temperature, and whether it is a gas, liquid, or solid.
DEelectronic DEvibration
DEtranslation
DErotation
Figure 2.12 Vibrational energy levels
Electronic energy levels
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Rotational energy levels
Translational energy levels
Schematic depiction of energy levels for each degree of freedom of a molecule. The dashed line between electronic energy levels on the left indicates what appear to be a continuous range of energies. However, as the energy scale is magnified stepwise, discrete energy levels for vibration, rotation, and translation can be resolved.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
TABLE 2.2 EnergyLevel Spacings for Degrees of Freedom Degree of Freedom
Typical EnergyLevel Spacing
Electronic
∼ 5 * 1019 J
Vibration
∼ 2 * 1020 J
Rotation
∼ 2 * 1023 J
Translation
∼ 2 * 1041 J
The amount of energy needed to move up the ladder of energy levels is very different for the different degrees of freedom: ∆Eelectronic W ∆Evibration W ∆Erotation W ∆Etranslation. Values for these ∆E are molecule dependent, but orderofmagnitude numbers are shown in Table 2.2. A molecule gains or loses energy through collisions with other molecules. An orderofmagnitude estimate of the energy that can be gained or lost by a molecule in a collision is kBT, where kB = R>NA is the Boltzmann constant and T is the absolute temperature. A given degree of freedom in a molecule can only take up energy through molecular collisions if the spacing between adjacent energy levels and the temperature satisfies the relationship ∆E ≈ kBT, which has the value 4.1 * 1021 J at 300. K. At 300. K, ∆E ≈ kBT is always satisfied for translation and rotation, but not for vibration and electronic excitation. We formulate the following general rule that relates the heat capacity CV,m and the degrees of freedom in a molecule, which will be discussed in more detail in Chapter 15.
The heat capacity CV,m for a gas at temperature T not much lower than 300. K is R>2 for each translation and rotational degree of freedom, where R is the ideal gas constant. Each vibrational degree of freedom for which the relation ∆E>kBT 6 0.1 is obeyed contributes R to CV,m. If ∆E>kBT 7 10, the degree of freedom does not contribute to CV,m. For 10 7 ∆E>kBT 7 0.1, the degree of freedom contributes partially to CV,m.
80
C2H4
CV, m ( J K21mol21 )
70 60 50 CO2
40 30 20
CO
10
He
0
200 400 600 800 1000 Temperature (K)
Figure 2.13
Molar heat capacity CV,m for selected gases as a function of temperature. Compared with molecules, atoms have only translational degrees of freedom and, therefore, have comparatively low values for CV,m that are independent of temperature. Molecules with vibrational degrees of freedom have higher values of CV,m at temperatures sufficiently high to activate the vibrations.
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Figure 2.13 shows the variation of CV,m for a monatomic gas and several molecular gases. Atoms only have 3 translational degrees of freedom. Linear molecules have an additional 2 rotational degrees of freedom and [email protected] vibrational degrees of freedom where n is the number of atoms in the molecule. Nonlinear molecules have 3 translational degrees of freedom, 3 rotational degrees of freedom, and [email protected] vibrational degrees of freedom. A He atom has only 3 translational degrees of freedom, and all electronic transitions are of high energy compared to kT. Therefore, CV,m = 3R>2 over the entire temperature range, as shown in the figure. CO is a linear diatomic molecule that has two rotational degrees of freedom for which ∆E>kBT 6 0.1 at 200. K. Therefore, CV,m = 5R>2 at 200. K. The single vibrational degree of freedom begins to contribute to CV,m above 200. K but does not contribute fully for T 6 1000. K because 10 7 ∆E>kBT below 1000. K. CO2 has 4 vibrational degrees of freedom, some of which contribute to CV,m near 200. K. However, CV,m does not reach its maximum value of 13R>2 below 1000. K. Similarly, CV,m for C2H4, which has 12 vibrational degrees of freedom, does not reach its maximum value of 15 R below 1000. K because 10 7 ∆E>kBT for some vibrational degrees of freedom. Electronic energy levels are too far apart for any of the molecular gases to contribute to CV,m. To this point, we have only discussed CV,m for gases. It is easier to measure CP,m than CV,m for liquids and solids because liquids and solids generally expand with increasing temperature and exert enormous pressure on a container at constant volume (see Example Problem 3.2). An example of how CP,m depends on T for solids and liquids is presented in Figure 2.14 for Cl2. To make the functional form of CP,m1T2 understandable, we briefly discuss the relative magnitudes of CP,m in the solid, liquid, and gaseous phases using an energylevel model. A solid can be thought of as a set of interconnected harmonic oscillators, and heat uptake leads to the excitations of the collective vibrations of the solid, which have much closer spaced energy levels than those for individual molecules. At very low temperatures, these vibrations cannot be activated because the spacing of the vibrational energy levels is large compared to kBT. Therefore, CP,m approaches zero as T approaches zero. For the solid, CP,m rises rapidly with T because the thermal energy available as T increases is sufficient to activate the vibrations of the solid. The heat capacity increases discontinuously as the solid melts to form a liquid. This is the case because the liquid
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49
2.11 Heat Capacity
Tsys,ƒ
L
qP =
Tsys,i
Tsur,ƒ
CP1system,T2dT = 
L
Tsur,i
CP1surroundings,T2dT
(2.27)
By measuring the temperature change of a thermal reservoir in the surroundings at constant pressure, qP can be determined as illustrated in Example Problem 2.6. In Equation (2.27), the heat flow at constant pressure has been expressed from the perspective of both the system and the surroundings. A similar equation can be written for a constant volume process. Water is a convenient choice of material for a heat bath in experiments because CP is nearly constant at the value 4.18 J g 1 K1 or 75.3 J mol1 K1 over the range from 0°C to 100.°C.
60 50 Cp,m ( J K21 mol21)
retains all the local vibrational modes of the solid, and more lowenergy modes become available upon melting. Therefore, the heat capacity of the liquid is greater than that of the solid. As the liquid vaporizes, the local vibrational modes present in the liquid are converted to translations that cannot take up as much energy as vibrations. Thus, CP,m decreases discontinuously at the vaporization temperature. The heat capacity in the gaseous state increases slowly with temperature as the vibrational modes of the individual molecules are activated, as discussed previously. These changes in CP,m can be calculated for a specific substance using a microscopic model and statistical thermodynamics, as will be discussed in detail in Chapter 15. Once the heat capacity of a variety of different substances has been determined, we have a convenient way to quantify heat flow. For example, at constant pressure, the heat flow between the system and surroundings can be written as
40 30 20
Solid
Liquid
Gas
10
100 200 300 Temperature (K)
400
Figure 2.14
Variation of CP,m with temperature for phases of Cl2. The colored fields correspond to the solid, liquid, and gas phases.
EXAMPLE PROBLEM 2.6 The volume of a system consisting of an ideal gas decreases at constant pressure. As a result, the temperature of a 1.50 kg water bath in the surroundings increases by 14.2°C. Calculate qP for the system.
Solution Tsys,ƒ
qP =
L
Tsys,i
Tsur,ƒ
CP1system,T2dT = 
L
Tsur,i
CP1surroundings,T2dT
= 1.50 kg * 4.18 J g 1 K1 * 14.2 K = 89.1 kJ
How are CP and CV related for a gas? Consider the processes shown in Figure 2.15 in which a fixed amount of heat flows from the surroundings into a gas. In the constant pressure process, the gas expands as its temperature increases. Therefore, the system does work on the surroundings. As a consequence, not all the heat flow into the system can be used to increase ∆U. No such work occurs for the corresponding constant volume process, and all the heat flow into the system can be used to increase ∆U. Therefore, dTP 6 dTV for the same heat flow dq. For this reason, CP 7 CV for gases. The same argument applies to liquids and solids as long as V increases with T. Nearly all substances follow this behavior, although notable exceptions occur, such as liquid water between 0°C and 4°C, for which the volume increases as T decreases. However, because ∆Vm upon heating is much smaller than for a gas, the difference between CP,m and CV,m for a liquid or solid is much smaller than for a gas. The preceding remarks about the difference between CP and CV have been qualitative in nature. However, the following quantitative relationship, which will be proved in Chapter 3, holds for an ideal gas:
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CP  CV = nR or CP,m  CV,m = R
(2.28)
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics w
Figure 2.15
Comparison of constant pressure and constant volume heating of a gas in a piston and cylinder assembly. (a) Constant pressure heating. Note that not all the heat flow into the system can be used to increase ∆U in a constant pressure process, because the system does work on the surroundings as it expands. (b) Constant volume heating. Note that no work is done for constant volume heating. A redder color of the gas corresponds to a higher temperature.
Mass Piston
Mass Piston Pi,Ti
Pi,Tf
Initial state
Final state
q
(a) Constant pressure heating
Concept For a given substance, for the gaseous phase, CP 7 CV. For the liquid or solid state, CP ≈ CV . q Vi,Ti
Vi,Tf
Initial state
Final state
(b) Constant volume heating
2.12 DETERMINING 𝚫U AND INTRODUCING THE STATE FUNCTION ENTHALPY
Concept ∆U can be measured by carrying out a process at constant volume and determining q, ∆U = qV .
Measuring the energy taken up or released in a chemical reaction is of particular interest to chemists. How can ∆U for a thermodynamic process be measured? Although these measurements will be the topic of Chapter 4, we briefly discuss such measurements here in order to enable you to carry out calculations on ideal gas systems in the endofchapter problems. The first law states that ∆U = q + w. Imagine that the process is carried out under constant volume conditions and that nonexpansion work is not possible. Because under these conditions w =  1 Pext dV = 0, ∆U = qV
(2.29)
Equation (2.29) states that ∆U can be experimentally determined by measuring the heat flow between the system and surroundings in a constant volume process. Chemical reactions are generally carried out under constant pressure rather than constant volume conditions. It would be useful to have an energy state function that has a relationship analogous to Equation (2.29), but at constant pressure conditions. Under constant pressure conditions, we can write dU = dqP  Pext dV = dqP  P dV
(2.30)
Integrating this expression between the initial and final states: ƒ
L i
dU = Uƒ  Ui =
L
dqP 
L
P dV = qP  P1Vƒ  Vi2
= qP  1PƒVƒ  PiVi2
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(2.31)
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2.13 CALCULATING q, w, ∆U, AND ∆H FOR PROCESSES INVOLVING IDEAL GASES
51
Note that in order to evaluate the integral involving P, we must know the functional relationship P1V2, which in this case is Pi = Pƒ = P where P is constant. Rearranging the last equation, we obtain
1Uƒ + PƒVƒ2  1Ui + PiVi2 = qP
(2.32)
H K U + PV
(2.33)
Because P, V, and U are all state functions, U + PV is a state function. This new state function is called enthalpy and is given the symbol H.
As is the case for U, H has the units of energy, and it is an extensive property. As shown in Equation (2.34), ∆H for a process involving only P–V work can be determined by measuring the heat flow between the system and surroundings at constant pressure: ∆H = qP
(2.34)
Concept ∆H can be measured by carrying out a process at constant pressure and determining q, ∆H = qP .
This equation is the constant pressure analogue of Equation (2.29). Because chemical reactions are conducted at constant P much more commonly than at constant V, the energy change measured experimentally by monitoring the heat flow is ∆H rather than ∆U. We classify a chemical reaction as being exothermic if heat is liberated to the surroundings 1∆H 6 02 or endothermic if heat flows from the surroundings into the system 1∆H 7 02. Combustion is an example of an exothermic reaction, and melting of a solid is an example of an endothermic process.
2.13 CALCULATING q, w, 𝚫U, AND 𝚫H FOR PROCESSES INVOLVING IDEAL GASES
In this section, we discuss how ∆U and ∆H, as well as q and w, can be calculated from the initial and final state variables if the path between the initial and final state is known. The problems at the end of this chapter ask you to calculate q, w, ∆U, and ∆H for simple and multistep processes. Because an equation of state is often needed to complete such calculations, the system will generally be an ideal gas. Using an ideal gas as a surrogate for more complex systems has the significant advantage that the mathematics is simplified, allowing one to concentrate on the process rather than the manipulation of equations and the evaluation of integrals. What does one need to know to calculate ∆U? The following discussion is restricted to processes that do not involve chemical reactions or changes in phase. Because U is a state function, ∆U is independent of the path between the initial and final states. To describe a fixed amount of an ideal gas (i.e., n is constant), the values of two of the variables P, V, and T must be known. Is this also true for ∆U for processes involving ideal gases? To answer this question, consider the expansion of an ideal gas from an initial state V1,T1 to a final state V2,T2. We first assume that U is a function of both V and T. Is this assumption correct? Because ideal gas atoms or molecules do not interact with one another, U will not depend on the distance between the atoms or molecules. Therefore, U is not a function of V, and we conclude that ∆U must be a function of T only for an ideal gas, ∆U = ∆U1T2. We also know that for a temperature range over which CV is constant, ∆U = qV = CV 1Tƒ  Ti2
(2.35)
∆H = ∆U1T2 + ∆1PV2 = ∆U1T2 + ∆1nRT2 = ∆H1T2
(2.36)
Is this equation only valid for constant V? Because U is a function of only T for an ideal gas, Equation (2.35) is also valid for processes involving ideal gases in which V is not constant. Therefore, if one knows CV, T1, and T2, ∆U can be calculated, regardless of the path between the initial and final states. How many variables are required to define ∆H for an ideal gas? We write
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
We see that ∆H is also a function of only T for an ideal gas. In analogy to Equation (2.35),
∆H = qP = CP1Tƒ  Ti2
(2.37)
Because ∆H is a function of only T for an ideal gas, Equation (2.37) holds for all processes involving ideal gases, whether or not P is constant, as long as it is reasonable to assume that CP is constant. Therefore, if the initial and final temperatures are known or can be calculated, and if CV and CP are known, ∆U and ∆H can be calculated regardless of the path for processes involving ideal gases using Equations (2.35) and (2.37), as long as no chemical reactions or phase changes occur. Because U and H are state functions, the previous statement is true for both reversible and irreversible processes. Recall that for an ideal gas CP  CV = nR, so that if one of CV and CP is known, the other can be readily determined. We next note that the first law links q, w, and ∆U. If any two of these quantities are known, the first law can be used to calculate the third. For example, if only expansion work takes place, one always proceeds from the equation
w = 
L
Pext dV
(2.38)
This integral can only be evaluated if the functional relationship between Pext and V is known. A commonly encountered case is Pext = constant, such that
w = Pext1Vƒ  Vi2
(2.39)
Because Pext ≠ P, the work considered in Equation (2.39) is for an irreversible process. A second commonly encountered case is that the system and external pressure differ only by an infinitesimal amount. In this case, it is sufficiently accurate to write Pext = P, and the process is reversible:
w = 
nRT dV L V
(2.40)
This integral can only be evaluated if T is known as a function of V. The most commonly encountered case is an isothermal process, in which T is constant. As was seen in Section 2.3, for this case
w = nRT
W2.1 Reversible Cyclic Processes
Vf dV = nRT ln Vi L V
(2.41)
In solving thermodynamic problems, it is very helpful to understand the process thoroughly before starting the calculation because it may be possible to obtain the value of one or more of q, w, ∆U, and ∆H without a calculation for an ideal gas. For example, ∆U = ∆H = 0 for an isothermal process because ∆U and ∆H depend only on T. For an adiabatic process, q = 0 by definition. If only P–V work is possible, w = 0 for a constant volume process. These guidelines are illustrated in Example Problems 2.7 and 2.8.
EXAMPLE PROBLEM 2.7
1
2 16.6 bar, 25 L
P (bar)
16.6 bar, 1.00 L
3 V (L)
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Consider a system that contains 2.50 mol of an ideal gas for which CV,m = 20.79 J mol1 K1. The system is taken through the cycle shown in the following diagram in the direction indicated by the arrows. The curved path corresponds to PV = nRT, where T = T1 = T3. a. Calculate q, w, ∆U, and ∆H for each segment and for the cycle, assuming that the heat capacity is independent of temperature. b. Calculate q, w, ∆U, and ∆H for each segment and for the cycle in which the direction of each process is reversed.
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2.13 CALCULATING q, w, ∆U, AND ∆H FOR PROCESSES INVOLVING IDEAL GASES
Solution We begin by asking whether we can evaluate q, w, ∆U, or ∆H for any of the segments without performing calculations. Because the path between states 1 and 3 is isothermal, ∆U and ∆H are zero for this segment. Therefore, from the first law, q3 S 1 = w3 S 1. For this reason, we only need to calculate one of these two quantities. Because ∆V = 0 along the path between states 2 and 3, w2 S 3 = 0. Therefore, ∆U2 S 3 = q2 S 3. Again, we only need to calculate one of these two quantities. Furthermore, because the total process is cyclic, the change in any state function is zero. Therefore, ∆U = ∆H = 0 for the cycle, no matter which direction is chosen. We now deal with each segment individually.
53
Concept For a process involving ideal gases, some of q, w, ∆U, and ∆H can usually be determined without calculation, based on the path between initial and final states.
Segment 1 S 2 The values of n, P1 and V1, and P2 and V2 are known. Therefore, T1 and T2 can be calculated using the ideal gas law. We use these temperatures to calculate ∆U, as follows: ∆U1 S 2 = nCV,m1T2  T12 = =
nCV,m nR
20.79 J mol1 K1 0.08314 L bar K1 mol1
1P2V2  P1V12
* 116.6 bar * 25.0 L  16.6 bar * 1.00 L2
= 99.6 kJ
The process takes place at constant pressure, so w = Pext1V2  V12 = 16.6 bar *
105 N m2 bar
* 125.0 * 103 m3  1.00 * 103 m32
= 39.8 kJ Using the first law,
q = ∆U  w = 99.6 kJ + 39.8 kJ = 139.4 kJ We next calculate T2: T2 =
P2V2 16.6 bar * 25.0 L = = 2.00 * 103 K nR 2.50 mol * 0.08314 L bar K1 mol1
We next calculate T3 = T1 and then ∆H1 S 2: T1 =
P1V1 16.6 bar * 1.00 L = = 79.9 K nR 2.50 mol * 0.08314 L bar K1 mol1
∆H1 S 2 = ∆U1 S 2 + ∆1PV2 = ∆U1 S 2 + nR1T2  T12
= 99.6 * 103 J + 2.5 mol * 8.314 J mol1 K1 *12.00 * 103 K  79.9 K2 = 139.4 kJ
Segment 2 S 3 As previously noted, w = 0, and ∆U2 S 3 = q2 S 3 = CV1T3  T22
= 2.50 mol * 20.79 J mol1 K1179.9 K  2000 K2 = 99.6 kJ
The numerical result is equal in magnitude, but opposite in sign to ∆U1 S 2, because T3 = T1. For the same reason, ∆H2 S 3 =  ∆H1 S 2.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Segment 3 S 1 For this segment, ∆U3 S 1 = 0 and ∆H3 S 1 = 0 as noted earlier, and w3 S 1 = q3 S 1. Because this is a reversible isothermal compression, w3 S 1 = nRT ln * ln
V1 = 2.50 mol * 8.314 J mol1 K1 * 79.9 K V3
1.00 * 103 m3 25.0 * 103 m3
= 5.35 kJ The results for the individual segments and for the cycle in the indicated direction are given in the following table. If the cycle is traversed in the reverse fashion, the magnitudes of all quantities in the table remain the same, but all signs change. q 1 kJ2
Path 1S2 2S3 3S1
w 1 kJ2
39.8 0 5.35 34.5
139.4 99.6 5.35 34.5
Cycle
∆U 1 kJ2 99.6 99.6 0 0
∆H 1 kJ2 139.4 139.4 0 0
EXAMPLE PROBLEM 2.8 In this example, 2.50 mol of an ideal gas with CV,m = 12.47 J mol1 K1 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure of the gas are 325 K and 2.50 bar, respectively. The final pressure is 1.25 bar. Calculate the final temperature, q, w, ∆U, and ∆H.
Solution Because the process is adiabatic, q = 0, and ∆U = w. Therefore, ∆U = nCV,m1Tƒ  Ti2 = Pext1Vƒ  Vi2
Using the ideal gas law,
nCV,m1Tƒ  Ti2 = nRPext a Tƒ anCV,m +
Tƒ Pƒ

Ti b Pi
nRPext nRPext b = Ti anCV,m + b Pƒ Pi RPext Pi ≤ RPext + Pƒ
CV,m + Tƒ = Ti ±
CV,m
8.314 J mol1 K1 * 1.00 bar 2.50 bar ≤ = 268 K 8.314 J mol1 K1 * 1.00 bar + 1.25 bar
12.47 J mol1 K1 + = 325 K * ±
12.47 J mol1 K1
We calculate ∆U = w from ∆U = nCV,m 1Tƒ  Ti2 = 2.5 mol * 12.47 J mol1 K1 * 1268 K  325 K2 = 1.78 kJ
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2.14 Reversible Adiabatic Expansion and Compression of an Ideal Gas
55
Because the temperature falls in the expansion, the internal energy and enthalpy decreases: ∆H = ∆U + ∆1PV2 = ∆U + nR1T2  T12
= 1.78 * 103 J + 2.5 mol * 8.314 J mol1 K1 * 1268 K  325 K2 = 2.96 kJ
2.14 REVERSIBLE ADIABATIC EXPANSION
AND COMPRESSION OF AN IDEAL GAS
The adiabatic expansion and compression of gases is an important meteorological process. For example, the cooling of a cloud as it moves upward in the atmosphere can be modeled as an adiabatic process because the heat transfer between the cloud and the rest of the atmosphere is slow on the timescale of its upward motion. Consider the adiabatic expansion of an ideal gas. Because q = 0, the first law takes the form ∆U = w
or CV dT = Pext dV
Connection The adiabatic expansion and compression of gases is an important meteorological process.
(2.42)
For a reversible adiabatic process, P = Pext, and CV dT = nRT
dV V
or, equivalently, CV
dT dV = nR T V
(2.43)
Integrating both sides of this equation between the initial and final states, Tƒ
Vƒ
dT dV CV = nR T L L V
Ti
(2.44)
Vi
If CV is constant over the temperature interval Tƒ  Ti, then
CV ln
Tƒ Ti
= nR ln
Vƒ Vi
(2.45)
3 Adiabatic
ln a
Tƒ Ti
b = 1g  12 ln a
Vƒ Vi
b
or, equivalently,
Tƒ Ti
= a
Vƒ Vi
b
1g
(2.46)
where g = CP,m >CV,m. Substituting Tƒ >Ti = PƒVƒ >PiVi in the previous equation, we obtain
PiV gi = PƒV gƒ
2
1
(2.47)
for the adiabatic reversible expansion or compression of an ideal gas. Note that our derivation is only applicable to a reversible process because we have assumed that P = Pext. Reversible adiabatic compression of a gas leads to heating, and reversible adiabatic expansion leads to cooling as illustrated in Example Problem 2.9. Adiabatic and isothermal expansion and compression are compared in Figure 2.16, in which two systems containing 1 mol of an ideal gas have the same volume at P = 1 atm. One system undergoes adiabatic compression or expansion, and the other undergoes isothermal compression or expansion. Under isothermal conditions, heat flows out of the system as it is compressed to P 7 1 atm, and heat flows into the system as it is expanded to P 6 1 atm to keep T constant. Because no heat flows into or out of the system under adiabatic conditions, its temperature increases in compression and decreases in expansion. Because Tadiabatic 7 Tisothermal for a compression starting at 1 atm, Padiabatic 7 Pisothermal for a given volume of the gas. Similarly, in a reversible adiabatic expansion originating at 1 atm, Padiabatic 6 Pisothermal for a given volume of the gas.
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P (atm)
Because CP  CV = nR for an ideal gas, Equation (2.45) can be written in the form
Isothermal 0
10
20
30
40 V (L)
50
60
Figure 2.16
Dependence of P on V for reversible adiabatic and isothermal compression or expansion of an ideal gas. Two systems containing 1 mol of N2 have the same P and V values at 1 atm. The red curve corresponds to reversible expansion and compression under adiabatic conditions. The blue curve corresponds to reversible expansion and compression under isothermal conditions.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
EXAMPLE PROBLEM 2.9 A cloud mass moving across the ocean at an altitude of 2000. m encounters a coastal mountain range. As it rises to a height of 3500. m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000. m and 3500. m is 0.802 and 0.602 atm, respectively. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that CP,m for air is 28.86 J K1 mol1 and that air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?
Solution Tƒ Vƒ ln a b = 1g  12 ln a b Ti Vi = 1g  12 ln a
Energy
= 
Tƒ Pi Tƒ Pi b = 1g  12 ln a b  1g  12 ln a b Ti Pƒ Ti Pƒ a
CP,m
 1b CP,m  R 1g  12 Pi Pi ln a b = ln a b CP,m g Pƒ Pƒ CP,m  R
28.86 J K1 mol1  1b 0.802 atm 28.86 J K1 mol1  8.314 J K1 mol1 = * ln a b 1 1 0.602 atm 28.86 J K mol 28.86 J K1 mol1  8.314 J K1 mol1 a
(a)
(b)
(c)
= 0.0826
Figure 2.17
Effects of adiabatic compression on occupation of energy levels. Blue dots signify relative populations of different energy levels. (a) Translational energy levels for a system before undergoing adiabatic compression. (b) After compression, energy levels are shifted upward, but occupation probability of the levels on the horizontal axis is unchanged. (c) Subsequent cooling to the original temperature at constant V restores energy to its original value by decreasing the probability of occupying higherenergy states. Each circle corresponds to a probability of 0.10.
Tƒ = 0.9207 Ti = 265 K You can expect snow.
It is instructive to consider an adiabatic compression or expansion from a microscopic point of view. In an adiabatic compression, the energy levels are all raised, but the probability that a given level is accessed is unchanged, as shown in Figure 2.17. This behavior, in contrast to that shown in Figure 2.11, is observed because in an adiabatic compression T and therefore U increase. For more details, see the reference to Dickerson (1969) in Further Reading. If the gas is subsequently cooled at constant V, the energy levels remain unchanged, but the probability that higherenergy states are populated decreases, as shown in Figure 2.17c.
VOCABULARY cyclic path degrees of freedom endothermic enthalpy exact differential exothermic
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first law of thermodynamics heat heat capacity indicator diagram internal energy irreversible
isobaric isochoric isothermal path path function quasistatic process
reversible state function work
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Conceptual Problems
KEY EQUATIONS Equation
Significance of Equation
∆U = q + w
First law of thermodynamics
2.3
w = mg∆h
Surroundingsbased work
2.4
Systembased work
2.5
Work of isothemal reversible compression of ideal gas
2.7
xƒ
w =
Lxi
F # dx
wisothermal rev compression =  nRT ln
Vƒ Vi
Equation Number
Vƒ
w =  wsur = 
LVi
Pext dV
General equation for reversible or irreversible P–V work
Following 2.9
Equation for electrical work
2.11
Definition of heat capacity
2.26
CP  CV = nR or CP,m  CV,m = R
Relation between CP and CV for ideal gas
2.28
∆U = qV , ∆H = qP
Experimental determination of ∆H, ∆U by measuring heat flow
H K U + PV
Definition of enthalpy
welectrical = Ift C1T2 = lim
∆T S 0 Tƒ
Tƒ Ti
= a
Vƒ Vi
b
1g
q dq =  Ti dT
, PiV gi = PƒV gƒ
2.29, 2.34 2.33
Relationship between T and V or P and V for a reversible adiabatic expansion or contraction
2.46, 2.47
CONCEPTUAL PROBLEMS Q2.1 Electrical current is passed through a resistor immersed in a liquid set in an adiabatic container. The temperature of the liquid is varied by 1°C. The system consists solely of the liquid. Does heat or work flow across the boundary between the system and surroundings? Justify your answer. Q2.2 Two ideal gas systems undergo reversible expansion under different conditions starting from the same P and V. At the end of the expansion, the two systems have the same volume. The pressure in the system that has undergone adiabatic expansion is lower than the pressure in the system that has undergone isothermal expansion. Explain this result without using equations. Q2.3 You have a liquid and its gaseous form in equilibrium in a piston and cylinder assembly in a constant temperature bath. Give an example of a reversible process. Q2.4 Describe how reversible and irreversible expansions differ by discussing the degree to which equilibrium is maintained between the system and the surroundings. Q2.5 For a constant pressure process, ∆H = qP. Does it follow that q is a state function? Explain. Q2.6 A cup filled with water (the system) at 278K is placed in a microwave oven, and the oven is turned on for 1 minute
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during which it begins to boil. Which of q, w, and ∆U are positive, negative, or zero? Q2.7 What is wrong with the following statement? Burns caused by steam at 100°C can be more severe than those caused by water at 100°C because steam contains more heat than water. Rewrite the sentence to convey the same information in a correct way. Q2.8 Why is it incorrect to speak of the heat or work associated with a system? Q2.9 You have a liquid and its gaseous form in equilibrium in a piston and cylinder assembly in a constant temperature bath. Give an example of an irreversible process. Q2.10 What is wrong with the following statement? Because the wellinsulated house stored a lot of heat, the temperature did not decrease much when the furnace failed. Rewrite the sentence to convey the same information in a correct way. Q2.11 Explain how a mass of water in the surroundings can be used to determine q for a process. Calculate q if the temperature of a 1.00kg water bath in the surroundings increases by 1.25°C. Assume that the surroundings are at a constant pressure.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Q2.12 A chemical reaction occurs in a constant volume enclosure separated from the surroundings by diathermal walls. Can you say whether the temperature of the surroundings increases, decreases, or remains the same in this process? Explain. Q2.13 Explain the relationship between the terms exact differential and state function. Q2.14 Imagine a horizontal cylinder with a frictionless piston midway between the ends held by a mechanical stop. To the left of the piston is an ideal gas at 1 bar pressure. To the right of the piston is a vacuum. The system consists only of the gas. Assume that the expansion is adiabatic. Discuss what happened when the stop is released. Assign a sign to w, q, and ∆U after all kinetic energy has dispersed. Q2.15 Discuss the following statement: If the temperature of the system increased, heat must have been added to it. Q2.16 Discuss the following statement: Letting heat flow into an object causes its temperature to increase. Q2.17 An ideal gas is expanded reversibly and adiabatically. Decide which of q, w, ∆U, and ∆H are positive, negative, or zero. Q2.18 An ideal gas is expanded reversibly and isothermally. Decide which of q, w, ∆U, and ∆H are positive, negative, or zero. Q2.19 An ideal gas is expanded adiabatically into a vacuum. Decide which of q, w, ∆U, and ∆H are positive, negative, or zero. Q2.20 A bowling ball (a) rolls across a horizontal table and (b) falls on the floor. Is the work associated with each part of this process positive, negative, or zero? Assume that friction can be neglected.
Q2.21 A perfectly insulating box is partially filled with water in which an electrical resistor is immersed. An external electrical generator converts the change in potential energy of a mass m that falls by a vertical distance h into electrical energy that in turn is dissipated in the resistor. Assign positive, negative, or zero values to q, w, and ∆U if the system is defined as the resistor and the water. Everything else is in the surroundings. Q2.22 Explain why ethene has a higher value for CV,m at 800. K than CO. Q2.23 Explain why CP,m is a function of temperature for ethane but not for argon in a temperature range in which electronic excitations do not occur. Q2.24 What is the difference between a quasistatic process and a reversible process? Q2.25 A dishwasher is opened immediately before the drying cycle starts. Both ceramic mugs and plastic storage containers are beaded with water. After five minutes, the water beads on the mugs are gone and those on the plastic container are still there. Explain. Q2.26 Arrange the following molecules or atoms in order of increasing heat capacity at 100. K. (a) He (b) HCl (c) C2H4 (d) C6H6 Q2.27 Arrange the following molecules or atoms in order of increasing heat capacity at 1000. K. (a) He (b) HCl (c) C2H4 (d) C6H6 Q2.28 Liquid water one degree above its freezing point is put into a glass bottle, and and a rubber stopper is put into the neck. The bottle is put into a freezer, and after equilibrium is reached, ice has formed and the bottle has broken. Determine the signs of q, w, and ∆U for the process if the water is the system and everything else is in the surroundings.
NUMERICAL PROBLEMS Section 2.3 P2.1 A muscle fiber contracts by 2.7 cm and in doing so lifts a weight. Calculate the work performed by the fiber. Assume the muscle fiber obeys Hooke’s law F = kx with a force constant, k, of 825 N m1. P2.2 The formalism of Young’s modulus is sometimes used to calculate the reversible work involved in extending or compressing an elastic material. Assume a force F is applied to an elastic rod of crosssectional area A0 and length L 0. As a result of this force, the rod changes in length by ∆L. Young’s modulus E is defined as tensile stress E = = tensile strain
F
A0
∆L
L0
=
FL 0 A0 ∆L
a. Relate k in Hooke’s law to Young’s modulus expression given above.
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b. Using your result in part (a), show that the magnitude of the reversible work involved in changing the length L 0 of an elastic cylinder of crosssectional area A0 by ∆L is 1 ∆L 2 a b EA0 L 0. 2 L0 P2.3 Young’s modulus (see P2.2) of muscle fiber is approximately 2.80 * 107 Pa. A muscle fiber 3.25 cm in length and 0.230 cm in diameter is suspended, with a mass M hanging at its end. Calculate the mass required to extend the length of the fiber by 4.5%. P2.4 DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of length L is bent so that it lies on the arc of a circle of radius Rc. The reversible work involved in bending DNA without twisting BL is wbend = , where B is the bending force constant. 2R2c The DNA in a nucleosome particle is about 680. Å in length. w =
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Numerical Problems
59
P2.8 As the volume of an ideal gas confined in a piston and cylinder assembly increases, 78.3 J of heat is absorbed from the surroundings. The external pressure is 1.00 bar, and the final volume of the gas is 65.6 L. If the ∆U for the process is 215 J, calculate the initial volume of the gas. Assume ideal Sections 2.5 and 2.6 gas behavior. P2.5 Assume the following simplified dependence of the P2.9 ∆U for a van der Waals gas increases by 450. J in pressure in a ventricle of the human heart as a function of the an expansion process, and the magnitude of w is 98.0 J. volume of blood pumped. Calculate the sign of w and the magnitude and sign of q for the process. P2.10 2.25 moles of an ideal gas at 32.0°C expands isotherPs mally from an initial volume of 19.0 dm3 to a final volume of 60.8 dm3. Calculate w for this process (a) for expansion Pd against a constant external pressure of 9.39 * 104 Pa and (b) for a reversible expansion. 0 75 150 P2.11 Consider the isothermal expansion of 1.75 mol of an V (cm3) ideal gas at 375 K from an initial pressure of 15.0 bar to a final pressure of 1.50 bar. Describe the process that will result The systolic pressure, Ps, is 130. mm Hg, corresponding to in the greatest amount of work being done by the system with 1.73 * 104 Pa. The diastolic pressure, Pd, is 90.0 mm Hg, Pext Ú 1.50 bar and calculate w. Describe the process that will corresponding to 1.20 * 104 Pa atm. If the volume of blood result in the least amount of work being done by the system pumped in one heartbeat is 75.0 cm3, calculate the work done with Pext Ú 1.50 bar and calculate w. What is the least amount in a heartbeat. of work done without restrictions on the external pressure? P2.6 A piston and cylinder assembly with a piston radius of P2.12 An ideal gas undergoes a singlestage expansion against 0.250 m contains 12.5 mol of an ideal gas with CV,m = 3R>2 a constant external pressure Pext at constant temperature from at 298 K. The assembly is surrounded by air at a pressure of T, P , V , to T, P , V 1.00 bar. The cylinder axis is vertical, and the piston is orii i ƒ ƒ. ented upward as in Figure 2.1. The system is at equilibrium a. What is the largest mass m that can be lifted through the when the length of the cylinder filled with gas is 0.750 m. height h in this expansion? (a) Calculate the mass of the piston. (b) Calculate the distance b. The system is restored to its initial state in a singlestate the piston moves if the temperature of the gas is increased to compression. What is the smallest mass m′ that must fall 750. K. (c) Calculate the work done by the piston and cylinder through the height h to restore the system to its initial state? assembly on the surroundings. c. If h = 10.0 cm, Pi = 1.00 * 106 Pa, Pƒ = 1.25 * 105 Pa, P2.7 By evaluating the integral w =  1 Pext dV, calculate the T = 280. K, and n = 1.75 mol, calculate the values of the work done in each of the cycles shown below in units of Joules. masses in parts (a) and (b). P2.13 An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to a final state described by Pƒ, Vƒ, T 4 4 4 in (a) a process at the constant external pressure Pƒ and (b) in 3 3 3 a reversible process. Derive expressions for the largest mass that can be lifted through a height h in the surroundings in 2 2 2 these processes. 1 1 1 P2.14 1.500 moles of N in a state defined by T = 250.0 K 2 i and Vi = 0.7500 L undergoes an isothermal reversible expansion until L. Calculate w assuming (a) that the 1 4 1 4 1 4 2 3 2 3 2Vƒ = 3 50.00 gas is described by the ideal gas law and (b) that the gas is V (L) V (L) V (L) described by the van der Waals equation of state. What is the (a) (b) (c) percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for N2 are listed in Table 7.4. Carry out your calculations of w to four 4 significant figures. 3 P2.15 2.50 mol of an ideal gas with CV,m = 3R>2 initially at 2 a temperature Ti = 298 K and Pi = 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed 1 by placing a 650. kg mass on the piston of diameter 19.8 cm. Calculate the work done in this process and the distance that the 1 4 4 2 3 2 3 piston travels. Assume that the mass of the piston is negligible. V (L) V (L) First calculate the external pressure and the initial volume.
3 2 1
3 (L)
4
1 (b)
P (bar)
P (bar)
P (bar)
4
P (bar)
P (bar)
Pressure
Nucleosomal DNA is bent around a protein complex called the histone octamer into a circle of radius 55 Å. Calculate the reversible work involved in bending the DNA around the histone octamer if the force constant B = 1.65 * 1028 J m1.
(c)
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.16 An ideal gas described by Ti = 275 K, Pi = 1.10 bar, and Vi = 10.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.10 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closedcycle process in a P–V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction? P2.17 A hotair balloon is expanded by using a propane burner to heat the gas inside the balloon. The amount of heat released in the process is 1.25 * 108 J, and the balloon expands from an initial volume to 2.25 * 106 L to a final volume of 3.75 * 106 L. Assume that the external pressure is constant at 1.00 bar. Calculate w and ∆U for the process assuming ideal gas behavior. P2.18 10.0 moles of an ideal gas (the system) is confined in a piston and cylinder assembly similar to that shown in Figure 2.2. The radius of the piston is 5.00 cm. The initial temperature is 300. K, and the initial pressure is 2.00 bar. The gas is expanded to twice its initial volume in an isothermal reversible process. Assume that the surroundings are a vacuum. a. Calculate w for this process considering only the movement of masses in the surroundings. b. Make a plot of the mass on the piston as a function of x, the height of the piston. Is m a linear function of x? Section 2.11 P2.19 A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packing emergency rations that, if completely metabolized, will release 35 kJ of heat per gram of rations consumed. How many grams of ration must the hiker consume to avoid a reduction in body temperature of 3.0 K as a result of heat loss? Assume the heat capacity of the body equals that of water and that the mass of the hiker is 62 kg. State any additional assumptions. P2.20 Count Rumford used a horse to turn cannon boring machinery. He showed that a single horse could heat 11.6 kg of ice water 1T = 273 K2 to T = 355 K in 2.5 hours. Assuming the same rate of work, how high could a horse raise a 195 kg weight in 3.0 minutes? Assume the heat capacity of water is 4.18 J K1 g 1. P2.21 At 298 K and 1 bar pressure, the density of water is 0.9970 g cm3, the coefficient of thermal expansion is 2.07 * 104 K1, and CP,m = 75.30 J K1 mol1. If the temperature of 275 g of water is increased by 32.0 K, calculate w, q, ∆H, and ∆U. P2.22 A system consisting of 115 g of liquid water at 300. K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 2.25 A passes through the 25.0ohm resistor for 100. s, what is the final temperature of the water? P2.23 The heat capacity of solid lead oxide is given by CP,m = 44.35 + 1.47 * 103
T in units of J K1 mol1 K
Calculate the change in enthalpy of 2.25 mol of PbO1s2 if it is cooled from 790. to 300. K at constant pressure.
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P2.24 A major league pitcher throws a baseball with a speed of 150. kilometers per hour. If the baseball weighs 195 grams and its heat capacity is 1.7 J g 1 K1, calculate the temperature increase of the ball when it is stopped by the catcher’s mitt. Assume that no heat is transferred to the catcher’s mitt and that the catcher’s arm does not recoil when he catches the ball. P2.25 Suppose that an adult is encased in a thermally insulating barrier and that the body retains all the heat evolved by metabolism of foodstuffs. Calculate the increase in body temperature after 2.5 hours. Assume that the heat capacity of the body is 4.18 J g 1 K1 and that the heat produced by metabolism is 9.4 kJ kg 1 hr 1. P2.26 A vessel containing 2.00 mol of an ideal gas with Pi = 1.00 bar and CP,m = 5R>2 is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at 310. K, and separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of CP = 2800. J K1. The gas is compressed reversibly to Pƒ = 15.0 bar. What is the temperature of the system after thermal equilibrium has been established? P2.27 An airflow rate of 250. cfm at a pressure of 1.00 bar is used to heat a house. Calculate the power in kW required to heat the air from the outside temperature of 41.0 °F to a final temperature of 68.0 °F. Assume that the heat capacity of air is constant at a value of 1.30 * 103 J cm3 K1. P2.28 River water at a temperature of 290. K is used for cooling in an industrial process. The water returned to the river is at a temperature of 320. K. What percentage of the river water can be used for cooling if the river temperature is required by law to increase by no more than 10.0 K? Section 2.13 P2.29 1.75 moles of an ideal gas, for which CV,m = 3R>2, is subjected to two successive changes in state: (1) From 15.0°C and 125. * 103 Pa, the gas is expanded isothermally against a constant pressure of 15.2 * 103 Pa to twice the initial volume. (2) At the end of the previous process, the gas is cooled at constant volume from 15.0°C to 35.0°C. Calculate q, w, ∆U, and ∆H for each of the stages. Also calculate q, w, ∆U, and ∆H for the complete process. P2.30 A 1.75 mol sample of an ideal gas for which CV,m = 3R>2 undergoes the following twostep process: (1) From an initial state of the gas described by T = 15.0°C and P = 5.00 * 104 Pa, the gas undergoes an isothermal expansion against a constant external pressure of 2.50 * 104 Pa until the volume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature falls to 19.0°C. Calculate q, w, ∆U, and ∆H for each step and for the overall process. P2.31 1.50 mol of an ideal gas, for which CV,m = 3R>2, initially at 15.0°C and 1.20 * 106 Pa undergoes a twostage transformation. For each of the stages described in the following list, calculate the final pressure, as well as q, w, ∆U, and ∆H. Also calculate q, w, ∆U, and ∆H for the complete process.
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a. The gas is expanded isothermally and reversibly until the volume triples. b. Beginning at the end of the first stage, the temperature is raised to 130°C at constant volume. P2.32 The temperature of 2.75 mol of an ideal gas is changed from 95.0°C to 15.0°C at constant pressure. CV,m = 3R>2. Calculate q, w, ∆U, and ∆H. P2.33 2.25 moles of an ideal gas are subjected to the changes below. Calculate the change in temperature for each case if CV,m = 3R>2. a. q = 375 J, w = 225 J b. q = 275. J, w = 275 J c. q = 0, w = 375 J P2.34 Calculate ∆H and ∆U for the transformation of 1.75 mol of an ideal gas from 15.0°C and 1.00 atm to 475.°C and T 25.0 atm if CP,m = 20.9 + 0.042 in units of J K1mol1. K P2.35 2.50 moles of an ideal gas for which CV,m = 20.8 J K1 mol1 is heated from an initial temperature of 15.0°C to a final temperature of 450.°C at constant volume. Calculate q, w, ∆U, and ∆H for this process. P2.36 2.50 moles of an ideal gas are compressed isothermally from 84.0 to 22.0 L using a constant external pressure of 2.80 atm. Calculate q, w, ∆U, and ∆H. P2.37 A pellet of Zn of mass 25.0 g is dropped into a flask containing dilute H2SO4 at a pressure of P = 1.00 bar and a temperature of T = 310. K. What reaction occurs? Calculate w for the process. Section 2.14 P2.38 Calculate q, w, ∆U, and ∆H if 1.50 mol of an ideal gas with CV,m = 3R>2 undergoes a reversible adiabatic expansion from an initial volume Vi = 19.0 dm3 to a final volume Vƒ = 60.8 dm3. The initial temperature is 32.0°C. P2.39 Calculate w for the adiabatic expansion of 1.75 mol of an ideal gas at an initial pressure of 1.50 bar from an initial temperature of 450. K to a final temperature of 300. K. Write an expression for the work done in the isothermal reversible expansion of the gas at 300. K from an initial pressure of 1.50 bar. What value of the final pressure would give the same value of w as the first part of this problem? Assume that CP,m = 5R>2. P2.40 In the adiabatic expansion of 2.50 mol of an ideal gas from an initial temperature of 15.0°C, the work done on the surroundings is 950. J. If CV,m = 3R>2, calculate q, w, ∆U, and ∆H. P2.41 A bottle at 300. K contains an ideal gas at a pressure of 145 * 103 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against Pext = 1.00 * 105 Pa, and some gas is expelled from the bottle in the process. When P = Pext, the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 300. K. What is the final pressure in the bottle for a monatomic gas, for which CV,m = 3R>2, and a diatomic gas, for which CV,m = 5R>2?
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Numerical Problems
61
P2.42 A 1.75 mole sample of an ideal gas with CV,m = 3R>2 initially at 310. K and 1.00 * 105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 1.95 * 106 Pa. Calculate the final temperature of the gas. Calculate q, w, ∆U, and ∆H for this process. P2.43 In an adiabatic compression of 1.75 mol of an ideal gas with CV,m = 5R>2, the temperature rises from 300. K to 500. K. Calculate q, w, ∆H, and ∆U. P2.44 A 2.50 mol sample of an ideal gas for which P = 3.25 bar and T = 298 K is expanded adiabatically against an external pressure of 0.225 bar until the final pressure is 0.225 bar. Calculate the final temperature, q, w, ∆H, and ∆U for (a) CV,m = 3R>2 and (b) CV,m = 5R>2. P2.45 2.25 mole of an ideal gas with CV,m = 3R>2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure are Ti = 310. K and Pi = 15.2 bar. The final pressure is Pƒ = 1.00 bar. Calculate q, w, ∆U, and ∆H for the process. P2.46 A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 300. K. The final pressure is Pƒ = 3.00 bar. Calculate the final temperature of the air in the tire. Assume that CV,m = 5R>2. Does your answer seem plausible? If not, explain why. P2.47 An automobile tire contains air at 2.00 * 105 Pa at 15.0°C. The stem valve is removed, and the air is allowed to expand adiabatically against the constant external pressure of one bar until P = Pexternal. For air, CV,m = 5R>2. Calculate the final temperature. Assume ideal gas behavior. The stem is reinserted, and the tire is allowed to warm up to 15.0°C. What is the final pressure? P2.48 Consider the adiabatic expansion of an ideal monatomic gas with CV,m = 3R>2. The initial state is described by P = 9.50 bar and T = 273 K. a. Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to a final pressure of P = 1.25 bar. b. Calculate the final temperature if the same gas undergoes an adiabatic expansion against an external pressure of P = 1.25 bar to a final pressure P = 1.25 bar. Explain the difference in your results for parts (a) and (b). P2.49 The temperature of 2.25 moles of an ideal gas increases from 15.1° to 39.5°C as the gas is compressed adiabatically. Calculate q, w, ∆U, and ∆H for this process assuming that CV,m = 3R>2. P2.50 1.50 mole of an ideal gas with CV,m = 3R>2 is compressed adiabatically and reversibly from the initial conditions P = 1.00 bar and 300. K to a final pressure of 95.0 bar. Calculate the final temperature and the work done in the process.
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CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.51 A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless adiabatic piston. Each part contains 55.0 L of an ideal monatomic gas with CV,m = 3R>2. Initially, Ti = 300. K and Pi = 2.00 * 105 bar in each part. Heat is slowly introduced into the left part using an electrical heater until the piston has moved sufficiently to the right to produce a final pressure Pƒ = 3.75 bar in the right part. Consider the compression of the gas in the right part to be a reversible process. a. Calculate the work done on the right part in this process and the final temperature in the right part. b. Calculate the final temperature in the left part and the amount of heat that flowed into this part.
P2.52 2.75 moles of an ideal gas is expanded from 375 K and an initial pressure of 4.50 bar to a final pressure of 1.00 bar, and CP,m = 5R>2. Calculate w for the following two cases: a. The expansion is isothermal and reversible. b. The expansion is adiabatic and reversible. Without resorting to equations, explain why the result for part (b) is greater than or less than the result for part (a). P2.53 A 1.75 mole sample of carbon dioxide, for which CP,m = 37.1 J K1 mol1 at 298 K, is expanded reversibly and adiabatically from a volume of 3.25 L and a temperature of 298 K to a final volume of 40.0 L. Calculate the final temperature, q, w, ∆H, and ∆U. Assume that CP,m is constant over the temperature interval.
WEBBASED SIMULATIONS, ANIMATIONS, AND PROBLEMS Simulations, animations, and homework problem worksheets can be accessed at www.pearsonhighered.com/advchemistry W2.1 Reversible cyclic processes are simulated in which the cycle is either rectangular or triangular on a P–V plot. For each segment and for the cycle, q, and w are determined. For
a given cycle type, the ratio of work done on the surroundings to the heat absorbed from the surroundings is determined for different P and V values.
FURTHER READING Dickerson, R. E. Molecular Thermodynamics. Menlo Park, CA: W.A. Benjamin, 1969, Chapter 4. Gislason, E. A., and Craig, N. C. “Cementing the Foundations of Thermodynamics: Comparison of SystemBased and SurroundingsBased Definitions of Work and Heat.” Journal of Chemical Thermodynamics 37 (2005): 954. Gislason, E. A., and Craig, N. C. “PressureVolume Integral Expressions for Work in Irreversible Processes.” Journal of Chemical Education. 84 (2007): 499.
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Romer, R. “Heat Is Not a Noun.” American Journal of Physics 69 (2001): 107–109. SchmidtRohr, K. “Expansion Work without the External Pressure and Thermodynamics in Terms of Quasistatic Irreversible Processes.” Journal of Chemical Education 91 (2014): 402.
19/09/17 5:29 PM
MATH ESSENTIAL 3:
Partial Derivatives
Many quantities that we will encounter in physical chemistry are functions of several variables. In that case, we have to reformulate differential calculus to take several variables into account. We define the partial derivative with respect to a specific variable just as we did in Section ME2.2 by treating all other variables indicated by subscripts as constants. For example, consider 1 mol of an ideal gas for which
P = ƒ1V, T2 =
RT V
(ME3.1)
Note that P can be written as a function of the two variables V and T. The change in P resulting from a change in V or T is proportional to the following partial derivatives: a
P1V + ∆V, T2  P1V, T2 0P R 1 1 b = lim ∆V S 0 = lim ∆V S 0 a  b 0V T ∆V ∆V V + ∆V V = lim ∆V S 0
a
R  ∆V RT b =  2 a 2 ∆V V + V∆V V
P1V, T + ∆T2  P1V, T2 0P 1 R1T + ∆T2 RT b = lim ∆T S 0 = lim ∆T S 0 c d 0T V ∆T ∆T V V =
R V
(ME3.2)
The subscript y in 10ƒ>0x2y indicates that y is being held constant in the differentiation of the function ƒ with respect to x. The partial derivatives in Equation (ME3.2) allow one to determine how a function changes when all of the variables change. For example, what is the change in P if the values of T and V both change? In this case, P changes to P + dP where
0P 0P R RT dP = a b dT + a b dV = dT  2 dV 0T V 0V T V V
y 10
(ME3.3)
z
0z 0z b dx + a b dy 0x y 0y x
(ME3.4)
The first term is the slope of the hill in the x direction, and the second term is the slope in the y direction. These changes in the height z as you move first along the x direction and then along the y direction are illustrated in Figure ME3.1. Because the slope of the hill is a function of x and y, this expression for dz is only valid for small changes dx and dy. Otherwise, higherorder derivatives need to be considered.
30 Able Hill 40
( )y dx 1 (−z −y )x dy ( )y dx
−z z 1 −x −z z 1 −x
x
Consider the following practical illustration of Equation (ME3.3). You are on a hill and have determined your altitude above sea level. How much will the altitude (denoted z) change if you move a small distance east (denoted by x) and north (denoted by y)? The change in z as you move east is the slope of the hill in that direction, 10z>0x2y, multiplied by the distance dx that you move. A similar expression can be written for the change in altitude as you move north. Therefore, the total change in altitude is the sum of these two changes or dz = a
20
Able Hill
50 m 40 m 30 m 20 m 10 m Sea level 50
Figure ME3.1
Able Hill contour plot and cross section. The cross section (bottom) is constructed from the contour map (top). Starting at the point labeled z on the hill, you first move in the positive x direction and then along the y direction. If dx and dy are sufficiently small, the change in height dz is 0z 0z given by dz = a b dx + a b dy. 0x y 0y x
63
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Math Essential 3 Partial Derivatives
Second or higher derivatives with respect to either variable can also be taken. The mixed second partial derivatives are of particular interest. Consider the two mixed second partial derivatives of P: RT 0c d 2 V ¢ Æ ¢ 0 0P 0P ° ° RT R a a b b = = 0 0T = a0 c  2 d n 0T b =  2 0T 0V T V 0T0V 0V V V T V V
a
2
0 0P 0P ° ° a b b = = 0 0V 0T V T 0T0V
0c
RT d V ¢ Æ ¢ R R 0V = a0 c d n 0Vb =  2 0T V V V T T (ME3.5)
For all state functions ƒ and for the specific case of P, the order in which the function is doubly differentiated does not affect the outcome, and we conclude that a
0 0ƒ1V, T2 0 0ƒ1V, T2 a b b = a a b b 0T 0V 0V 0T T V V T
(ME3.6)
Because Equation (ME3.5) is only satisfied by state functions ƒ, it can be used to determine if a function ƒ is a state function. If ƒ is a state function, one can write ƒ ∆ƒ = 1i dƒ = ƒfinal  ƒinitial. This equation states that ƒ can be expressed as an infinitesimal quantity, dƒ, that, when integrated, depends only on the initial and final states; dƒ is called an exact differential. We can illustrate these concepts with the following calculation.
PROBLEM a. Calculate 2
a
2
0ƒ 0ƒ 0ƒ 0ƒ ° b ,a b ,a b ,a b , 0x y 0y x 0x 2 y 0y 2 x
0a
x
0ƒ b 0x y ¢ 0y
, and
x
°
0a
0ƒ b 0y x ¢ 0x y
for the function ƒ1x, y2 = ye + xy + x ln y. b. Determine if ƒ1x, y2 is a state function of the variables x and y. c. If ƒ1x, y2 is a state function of the variables x and y, what is the total differential dƒ?
Solution 0ƒ a. a b = yex + y + ln y, 0x y a
02ƒ 0x
2
a
b = yex,
a
y
0ƒ 0a b ° 0x y ¢ 0y
x
b. Because we have shown that °
02ƒ 0y 2
°
1 = ex + 1 + , y
0a
0ƒ x b = ex + x + y 0y x
0ƒ b 0x y ¢ 0y
x
=
0a
°
b = x
x y2
0ƒ b 0y x ¢ 1 = ex + 1 + y 0x y
0a
0ƒ b 0y x ¢ 0x y
ƒ1x, y2 is a state function of the variables x and y. Generalizing this result, any wellbehaved function that can be expressed in analytical form is a state function. c. The total differential is given by dƒ = a
0ƒ 0ƒ b dx + a b dy 0x y 0y x
= 1yex + y + ln y2dx + aex + x +
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x b dy y
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C H A P T E R
3
The Importance of State Functions: Internal Energy and Enthalpy
3.1 Mathematical Properties of State Functions
3.2 Dependence of U on V and T 3.3 Does the Internal Energy Depend More Strongly on V or T ?
3.4 Variation of Enthalpy with Temperature at Constant Pressure
3.5 How Are CP and CV Related?
WHY is this material important? Expressing state functions in terms of their variables using partial derivatives is a powerful tool. It allows us to write equations linking state functions such as U and H with quantities that can be determined experimentally.
3.6 Variation of Enthalpy with Pressure at Constant Temperature
3.7 The Joule–Thomson Experiment 3.8 Liquefying Gases Using
WHAT are the most important concepts and results?
an Isenthalpic Expansion
The natural variables for U are T and V, but for most cases U can be regarded as a function of T alone. The natural variables for H are T and P, but for most cases H can be regarded as a function of T alone. The expansion of real gases can lead to cooling or heating, depending on the value of the Joule–Thomson coefficient of the gas at the given temperature and pressure.
WHAT would be helpful for you to review for this chapter? Because partial derivatives are used extensively in this chapter, it would be helpful to review the material in Math Essential 3.
3.1 MATHEMATICAL PROPERTIES OF STATE FUNCTIONS
In Chapter 2 we demonstrated that U and H are state functions and that w and q are path functions. We also discussed how to calculate changes in these quantities for an ideal gas. In this chapter, the path independence of state functions is exploited to derive relationships with which ∆U and ∆H can be calculated as functions of P, V, and T for real gases, liquids, and solids. In doing so, we develop the formal aspects of thermodynamics. We will show that the formal structure of thermodynamics provides a powerful aid in linking theory and experiment. The thermodynamic state functions of interest here are defined by two variables from the set P, V, and T. In formulating changes in state functions, we will make extensive use of partial derivatives, which are reviewed in Math Essential 3. In the following discussion, two important results from differential calculus will be used frequently. First consider a function z = ƒ1x, y2 that can be rearranged to x = g1y, z2 or y = h1x, z2. For example, if P = nRT>V, then V = nRT>P and T = PV>nR. In this case
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a
0P b = 0V T
1 0V a b 0P T
(3.1) 65
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Second, the cyclic rule will also be used: a
0P 0V 0T b a b a b = 1 0V T 0T P 0P V
(3.2)
It is called the cyclic rule because the variables x, y, and z in the three terms follow the orders x, y, z; y, z, x; and z, x, y. As we show next, Equations (3.1) and (3.2) can be used to reformulate Equation (3.3): dP = a
0P 0P b dT + a b dV 0T V 0V T
(3.3)
Suppose this expression needs to be evaluated for a specific substance, such as N2 gas. What quantities must be measured in the laboratory in order to obtain numerical values for 10P>0T2V and 10P>0V2T ? Using Equations (3.1) and (3.2),
Concept The easily measured thermal expansion and compressibility coefficients are directly related to partial derivatives.
a
0P 0V 0T b a b a b = 1 0V T 0T P 0P V
0V a b 0T P 0P 0P 0V a a b = a b a b = = kT 0T V 0V T 0T P 0V a b 0P T a
and
0P 1 b = 0V T kTV
(3.4)
where a and kT are the readily measured isobaric volumetric thermal expansion coefficient and the isothermal compressibility, respectively, defined by
a =
1 0V 1 0V a b and kT =  a b V 0T P V 0P T
(3.5)
Both 10V>0T2P and 10V>0P2T can be measured by determining the change in volume of the system when the pressure or temperature is varied, while keeping the second variable constant. The minus sign in the equation for kT is chosen so that values of the isothermal compressibility are positive. For small changes in T and P, Equations (3.5) can be written in the form with differences rather than derivatives: V1T22 = V1T1211 + a3 T2  T1 4 2 and V1P22 = V1P1211  kT 3 P2  P1 4 2. Values for a and kT for selected solids and liquids are shown in Tables 3.1 and 3.2, respectively. Equation (3.5) is an example of how seemingly abstract partial derivatives can be directly linked to experimentally determined quantities using the mathematical properties of state functions. Using the definitions of a and kT , we can write Equation (3.3) in the form
dP =
a 1 dT dV kT kTV
(3.6)
which can be integrated to give Tf
Vf
Vf a 1 a 1 ∆P = dT dV ≈ 1Tf  Ti2 ln kT kT Vi L kT L kTV Ti
(3.7)
Vi
The second expression in Equation (3.7) holds if ∆T and ∆V are small enough that a and kT are constant over the range of integration. Example Problem 3.1 shows a useful application of this equation.
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3.1 Mathematical Properties of State Functions
67
TABLE 3.1 Selected Isobaric Volumetric Thermal Expansion Coefficients at 298 K
Ag(s)
106 a>1K12 57.6
Hg1l2
Al(s)
69.3
11.4
Au(s)
42.6
CCl41l2
Cu(s)
Element
Element or Compound
104 a>1K12 1.81 14.6
49.5
CH3COCH31l2 CH3OH1l2
14.9
Fe(s)
36.9
C2H5OH1l2
11.2
Mg(s)
78.3
10.5
Si(s)
7.5
C6H5CH31l2
W(s) Zn(s)
13.8
C6H61l2 H2O1l2
2.04
90.6
H2O1s2
1.66
11.4
Sources: Based on Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York: Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch für Chemiker und Physiker. 4th ed. Berlin: Springer, 1998.
TABLE 3.2 Selected Isothermal Compressibility Coefficients at 298 K Substance
106 kT >bar 1
Substance
106 kT >bar 1
SiO2(s)
2.57
Br21l2
C2H5OH1l2
110
Ni(s)
0.513
C6H5OH1l2
61
TiO2(s)
0.56
C6H61l2
94
Al(s)
1.33
Cu(s)
0.702
CCl41l2
C(graphite)
0.156
CH3OH1l2
Mn(s)
0.716
Co(s)
0.525
CS21l2
Au(s)
0.563
Hg1l2
Pb(s)
2.37
Fe(s)
0.56
SiCl41l2
Ge(s)
1.38
Na(s)
13.4
CH3COCH31l2 H2O1l2
TiCl41l2
64
103 120
125 92.7 45.9 3.91 165 89
Sources: Based on Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York: Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton FL: CRC Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch für Chemiker und Physiker. 4th ed. Berlin: Springer, 1998.
EXAMPLE PROBLEM 3.1 You have accidentally arrived at the end of the range of an ethanolinglass thermometer so that the entire volume of the glass capillary is filled. By how much will the pressure in the capillary increase if the temperature is increased by another 10.0°C? aglass = 2.00 * 1051°C21, aethanol = 11.2 * 1041°C21, and kT, ethanol = 11.0 * 1051bar21. Will the thermometer survive your experiment?
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Solution Using Equation (3.7), ∆P = =
Vf aethanol aethanol 1 1 dT dV ≈ ∆T ln kT, ethanol kT, ethanol Vi L kT, ethanol L kT, ethanolV Vi11 + aglass ∆T2 aethanol 1 ∆T ln kT, ethanol kT, ethanol Vi
≈
Viaglass ∆T aethanol 1 ∆T kT, ethanol kT, ethanol Vi
=
1aethanol  aglass2
=
111.2  0.2002 * 1041°C21
kT, ethanol
∆T
11.0 * 1051bar21
* 10.0°C = 100. bar
In this calculation, we have used the relations V1T22 = V1T1211 + a3 T2  T1 4 2 and ln11 + x2 ≈ x if x V 1. The glass is unlikely to withstand such a large increase in pressure.
3.2 DEPENDENCE OF U ON V AND T In this section, we will investigate how the state function U varies with T and V. To do so, we will use an important property of state functions. If ƒ is a state function, one can ƒ write ∆ƒ = 1i dƒ = ƒfinal  ƒinitial. This equation states that ƒ can be expressed as an infinitesimal quantity dƒ that when integrated depends only on the initial and final states; df is called an exact differential. An example of a state function and its exact differential is U and dU = dq  Pext dV. For a given amount of a pure substance or a mixture of fixed composition, U is determined by any two of the three variables P, V, and T. The following discussion will demonstrate that it is particularly convenient to choose the variables T and V. Because U is a state function, an infinitesimal change in U can be written as
dU = a
0U 0U b dT + a b dV 0T V 0V T
(3.8)
This expression indicates that if the state variables change from T, V to T + dT, V + dV, the change in U, dU, can be determined as follows: First, determine the slopes of U1T,V2 with respect to T and V and evaluate them at T, V. Next, multiply the values of these slopes by the increments dT and dV, respectively, Finally, add the two terms. As long as dT and dV are infinitesimal quantities, higherorder derivatives can be neglected. How can numerical values for 10U>0T2V and 10U>0V2T be obtained? In the following, we only consider PV work. Combining Equation (3.8) and the differential expression of the first law,
dq  Pext dV = a
0U 0U b dT + a b dV 0T V 0V T
(3.9)
The symbol dq is used for an infinitesimal amount of heat as a reminder that heat is not a state function. We first consider processes at constant volume for which dV = 0, so that Equation (3.9) becomes
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dqV = a
0U b dT 0T V
(3.10)
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3.2 DEPENDENCE OF U ON V AND T
Note that in the previous equation, dqV is the product of a state function and an exact differential. Therefore, dqV behaves like an exact differential, but only because the path (constant V ) is specified. The quantity dq is not an exact differential. Although the quantity 10U>0T2V looks very abstract, it can be readily measured. For example, imagine immersing a container with rigid diathermal walls in a water bath, where the contents of the container are the system. A process such as a chemical reaction is carried out in the container, and the heat flow to the surroundings is measured. If heat flow dqV occurs, a temperature increase or decrease dT is observed in the system and the water bath surroundings. Both of these quantities can be measured. Their ratio, dqV >dT, is a special form of the heat capacity discussed in Section 2.5: dqV 0U = a b = CV dT 0T V
69
Concept The heat capacity at constant volume is a direct measure of the partial derivative of U with respect to T at constant V.
(3.11)
where dqV >dT corresponds to a constant volume path and is called the heat capacity at constant volume. The quantity CV is extensive (that is, it depends on the size of the system), whereas CV, m is an intensive quantity (that is, it does not depend on the size of the system). As discussed in Section 2.5, CV, m is different for different substances under the same conditions. Observations show that CV, m is always positive for a singlephase, pure substance or for a mixture of fixed composition, as long as no chemical reactions or phase changes take place in the system. For processes subject to these constraints, U increases monotonically with T. Now that we have an expression for CV , we have a way to experimentally determine changes in U with T at constant V for systems of pure substances or for mixtures of constant composition in the absence of chemical reactions or phase changes. After CV has been determined as a function of T as discussed in Section 2.11, the following integral is numerically evaluated: T2
∆UV =
L T1
T2
CV dT = n CV, m dT L
(3.12)
T1
Over a limited temperature range, CV, m can often be regarded as a constant. If this is the case, Equation (3.12) simplifies to T2
∆UV =
L
CV dT = CV ∆T = nCV, m ∆T
(3.13)
T1
Equation 3.11 can be written in a different form to explicitly relate qV and ∆U: ƒ
L
ƒ
dqV =
i
0U b dT or qV = ∆U L 0T V i
a
(3.14)
Although dq is not an exact differential, the integral has a unique value if the path is defined, as it is in this case (constant volume). Equation (3.14) shows that ∆U for an arbitrary process in a closed system in which only PV work is carried out can be determined by measuring q under constant volume conditions. As will be discussed in Chapter 4, the technique of bomb calorimetry uses this approach to determine ∆U for chemical reactions. Next consider the dependence of U on V at constant T, or 10U>0V2T . This quantity has the units of J>m3 = 1J>m2>m2 = kg ms2 >m2 = force>area = pressure and is called the internal pressure. To explicitly evaluate the internal pressure for different substances, a result will be used that is derived in the discussion of the second law of thermodynamics in Section 5.12:
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a
0U 0P b = T a b  P 0V T 0T V
(3.15)
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Using this equation, we can write the total differential of the internal energy as Vf , Tf
Vi , Tf A
T
B
C Vi , T i
Vf , Ti V
dU = dUV + dUT = CV dT + c T a
(3.16)
In this equation, the symbols dUV and dUT have been used, where the subscript indicates which variable is constant. Equation (3.16) is an important result that applies to systems containing gases, liquids, or solids in a single phase (or mixed phases at a constant composition) if no chemical reactions or phase changes occur. The advantage of writing dU in the form given by Equation (3.16) over that in Equation (3.8) is that 10U>0V2T can be evaluated in terms of the system variables P, V, and T and their derivatives, all of which are experimentally accessible. Once 10U>0V2T and 10U>0T2V are known, these quantities can be used to determine dU. Because U is a state function, the path taken between the initial and final states is unimportant. Three different paths are shown in Figure 3.1, and dU is the same for these and any other paths connecting Vi, Ti and Vƒ, Tƒ. To simplify the calculation, the path chosen consists of two segments in which only one of the variables changes in a given path segment. An example of such a path is Vi, Ti S Vf, Ti S Vf, Tf . Because T is constant in the first segment,
Figure 3.1
Volume–temperature diagram with three paths connecting initial and final states. Paths A, B, and C connect Vi, Ti and Vƒ, Tƒ. Because U is a state function, all paths are equally valid in calculating ∆U. Therefore, a specification of the path is irrelevant.
0P b  P d dV 0T V
dU = dUT = c T a
0P b  P d dV 0T V
Because V is constant in the second segment, dU = dUV = CV dT. Finally, the total change in U is the sum of the changes in the two segments, dUtotal = dUV + dUT .
3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T?
Chapter 2 demonstrated that U is a function of T alone for an ideal gas. However, this statement is not true for real gases, liquids, and solids for which the change in U with V must be considered. In this section, we investigate if the temperature or the volume dependence of U is most important in determining ∆U for a process of interest. To evaluate the relative importance of temperature and volume, we will consider separately systems that consist of an ideal gas, a real gas, a liquid, and a solid. Example Problem 3.2 shows that Equation (3.15) leads to a simple result for a system consisting of an ideal gas.
EXAMPLE PROBLEM 3.2 Evaluate 10U>0V2T for an ideal gas and modify Equation (3.16) accordingly for the specific case of an ideal gas.
Solution a
03 nRT>V4 0U 0P nRT b = Ta b  P = Ta b  P =  P = 0 0V T 0T V 0T V V
Therefore, dU = CVdT, which shows that for an ideal gas, U is a function of T only. Example Problem 3.2 shows that U is only a function of T for an ideal gas. Specifically, U is not a function of V. This result is understandable in terms of the potential function of Figure 1.9. Because ideal gas molecules do not attract or repel one another, no energy is required to change their average distance of separation (increase or decrease V ): Tƒ
∆U =
L Ti
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CV1T2 dT
(3.17)
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3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T?
Recall that because U is only a function of T, Equation (3.17) holds for an ideal gas even if V is not constant. Next consider the variation of U with T and V for a real gas. The experimental determination of 10U>0V2T was carried out in the mid19th century by James Joule using an apparatus consisting of two glass flasks separated by a stopcock, all of which were immersed in a water bath. An idealized view of the experiment is shown in Figure 3.2. As a valve between the volumes is opened, a gas initially in volume A expands to completely fill the volume A + B. In interpreting the results of this experiment, it is important to understand where the boundary between the system and surroundings lies. Here, the decision was made to place the system boundary so that it includes all the gas. Initially, the boundary lies totally within VA, but it moves during the expansion so that it continues to include all gas molecules. With this choice, the volume of the system changes from VA before the expansion to VA + VB after the expansion has taken place. The first law of thermodynamics [Equation (3.9)] states that dq  Pext dV = a
0U 0U b dT + a b dV 0T V 0V T
However, all the gas is contained in the system; therefore, Pext = 0 because a vacuum cannot exert a pressure. Therefore, Equation (3.9) becomes dq = a
0U 0U b dT + a b dV 0T V 0V T
71
Thermometer
A P5Pi
B P50 Water bath
Figure 3.2
Schematic depiction of the Joule experiment. In this experiment, which is used to determine 10U>0V2T , two spherical vessels, A and B, are separated by a valve. Both vessels are immersed in a water bath, the temperature of which is monitored. The initial pressure in each vessel is indicated.
(3.18)
To within experimental accuracy, Joule found that dTsur = 0. Because the water bath and the system are in thermal equilibrium, dT = dTsur = 0. With this observation, Joule concluded that dq = 0. Therefore, Equation (3.18) becomes a
0U b dV = 0 0V T
(3.19)
Because dV ≠ 0, Joule concluded that 10U>0V2T = 0. Joule’s experiment was not definitive because the experimental sensitivity was limited, as shown in Example Problem 3.3.
EXAMPLE PROBLEM 3.3 In Joule’s experiment to determine 10U>0V2T , the heat capacities of the gas and the water bath surroundings were related by Csur >Csys ≈ 1000. If the precision with which the temperature of the surroundings could be measured is {0.006°C, what is the minimum detectable change in the temperature of the gas?
Solution View the experimental apparatus as two interacting systems in a rigid adiabatic enclosure. The first is the volume within vessels A and B, and the second is the water bath and the vessels. Because the two interacting systems are isolated from the rest of the universe, q = Cwater ∆Twater + Cgas ∆Tgas = 0 ∆Tgas = 
Cwater ∆Twater = 1000 * 1{0.006°C2 = 6°C Cgas
In this calculation, ∆Tgas is the temperature change that the expanded gas undergoes to reach thermal equilibrium with the water bath, which is the negative of the temperature change during the expansion.
Because the minimum detectable value of ∆Tgas is rather large, this apparatus is clearly not suited for measuring small changes in the temperature of the gas in an expansion.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Concept U does not depend on V for an ideal gas.
Joule later carried out more sensitive experiments in collaboration with William Thomson (Lord Kelvin). These experiments, which are discussed in Section 3.8, demonstrate that 10U>0V2T is small but nonzero for real gases. Example Problem 3.2 has shown that 10U>0V2T = 0 for an ideal gas. We next calV culate 10U>0V2T and ∆UT = 1Vm,m,i f 10U>0V2T dVm for a real gas, in which the van der Waals equation of state is used to describe the gas, as illustrated in Example Problem 3.4.
EXAMPLE PROBLEM 3.4 For a gas described by the van der Waals equation of state, P = nRT>1V  nb2 an2 >V 2. Use this equation to complete these tasks: a. Calculate 10U>0V2T using 10U>0V2T = T10P>0T2V  P. V b. Derive an expression for the change in internal energy, ∆UT = 1Vi ƒ 10U>0V2T dV, in compressing a van der Waals gas from an initial molar volume Vi to a final molar volume Vƒ at constant temperature.
Solution
a. T a
0P ° b  P = T 0T V =
0c
nRT n2a  2d V  nb nRT V ¢  P =  P 0T V  nb V
nRT nRT n2a n2a + 2 = 2 V  nb V  nb V V
Vƒ
Vƒ
Vi
Vi
0U n2a 1 1 b. ∆UT = a b dV = dV = n2a a b 2 0V V V L T LV i ƒ Note that ∆UT is zero if the attractive part of the intermolecular potential is zero.
Example Problem 3.4 demonstrates that, in general, 10U>0V2T ≠ 0 and that ∆UT can be calculated if the equation of state of the real gas is known. This allows the T relative importance and value of ∆UV = 1Ti ƒ CV dT to be determined in a process in which both T and V change, as shown in Example Problem 3.5.
EXAMPLE PROBLEM 3.5 One mole of N2 gas undergoes a change from an initial state described by T = 200. K and Pi = 5.00 bar to a final state described by T = 400. K and Pƒ = 20.0 bar. Treat N2 as a van der Waals gas with the parameters a = 0.137 Pa m6 mol2 and b = 3.87 * 105 m3 mol1. We use the path N21g, T = 200. K, P = 5.00 bar2 S N21g, T = 200. K, P = 20.0 bar2 S N21g, T = 400. K, P = 20.0 bar2, keeping in mind that all paths will give the same answer for ∆U of the overall process. V a. Calculate ∆UT = 1Vi ƒ 10U>0V2T dV using the result of Example Problem 3.4. Note that Vi = 3.28 * 103 m3 and Vƒ = 7.88 * 104 m3 at 200. K, as calculated using the van der Waals equation of state. T b. Calculate ∆UV = n 1Ti ƒ CV, m dT using the following relationship for CV, m in this temperature range: CV, m J K1 mol1
= 22.50  1.187 * 102
T T2 T3 + 2.3968 * 105 2  1.0176 * 108 3 K K K
The ratios T n >Kn ensure that CV, m has the correct units. c. Compare the two contributions to ∆U. Can ∆UT be neglected relative to ∆UV?
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73
3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T?
Solution a. Using the result of Example Problem 3.4, ∆UT = n2a a * a
1 1 b = 0.137 Pa m6 Vm, i Vm, ƒ
1 1 b = 132 J 3 3 3.28 * 10 m 7.88 * 104 m3
Tƒ
b. ∆UV = n
L
CV, m dT
Ti
22.50  1.187 * 102
400. K
=
L
200. K
±
1.0176 * 108
T T2 + 2.3968 * 105 2 K K
T3 K3
≤ dT J
= 14.50  0.712 + 0.447  0.06102kJ = 4.17 kJ
c. ∆UT is 3.2% of ∆UV for this case. In this example, and for most processes, ∆UT can be neglected relative to ∆UV for real gases.
The calculations in Example Problems 3.4 and 3.5 show that to a good approximaV tion ∆UT = 1Vi ƒ 10U>0V2T dV ≈ 0 for real gases under most conditions. Therefore, it is sufficiently accurate to consider U as a function of T only 3 U = U1T24 for real gases in processes that do not involve unusually high gas densities. Having discussed ideal and real gases, what can be said about the relative magV T nitude of ∆UT = 1Vi ƒ 10U>0V2T dV and ∆UV = 1Ti ƒ CV dT for processes involving liquids and solids? From experiments, it is known that the density of liquids and solids varies only slightly with the external pressure over the range in which these two forms of matter are stable. This conclusion is not valid for extremely highpressure conditions such as those in the interior of planets and stars. However, it is safe to say that dV for a solid or liquid is very small in most processes. Therefore,
Concept For real gases, liquids, and solids, the dependence of U on T is much stronger than on V.
V2
liq ∆U solid, T
=
0U 0U b dV ≈ a b ∆V ≈ 0 0V T L 0V T V1
a
(3.20)
because ∆V ≈ 0. This result is valid even if 10U>0V2T is large. The conclusion that can be drawn from this section is as follows: Under most conditions encountered by chemists in the laboratory, U can be regarded as a function of T alone for all substances. The following equations give a good approximation even if V is not constant in the process under consideration: Tƒ
U1Tƒ, Vƒ2  U1Ti, Vi2 = ∆U =
L Ti
Tƒ
CV dT = n CV, m dT L
(3.21)
Ti
Note that Equation (3.21) is only applicable to a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no chemical reactions. Changes in U that arise from these processes will be discussed in Chapters 4 and 8.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
3.4 VARIATION OF ENTHALPY WITH
TEMPERATURE AT CONSTANT PRESSURE
Pext 5 P Mass Piston
Mass Piston
P, Vf ,Tf
P, Vi ,Ti
Initial state
Final state
As for U, H can be defined as a function of any two of the three variables P, V, and T. It was convenient to choose U to be a function of T and V because this choice led to the identity ∆U = qV . Using a similar reasoning, we choose H to be a function of T and P. How does H vary with P and T? The variation of H with T at constant P is discussed next, and a discussion of the variation of H with P at constant T is deferred to Section 3.6. Consider the constant pressure process shown schematically in Figure 3.3. For this process defined by P = Pext, (3.22)
Although the integral of dq is in general path dependent, it has a unique value in this case because the path is specified, namely, P = Pext = constant. Integrating both sides of Equation (3.22),
Figure 3.3
Constant pressure process in a piston and cylinder assembly. The initial and final states are shown for an undefined process that takes place at constant pressure.
dU = dqP  P dV
ƒ
L i
ƒ
dU =
L i
ƒ
dqP 
L i
P dV or Uƒ  Ui = qP  P1Vƒ  Vi2
(3.23)
Because P = Pƒ = Pi, this equation can be rewritten as
1Uƒ + PƒVƒ2  1Ui + PiVi2 = qP or
∆H = qP
(3.24)
The preceding equation shows that the value of ∆H can be determined for an arbitrary process at constant P in a closed system in which only PV work occurs by simply measuring qP, the heat transferred between the system and surroundings in a constant pressure process. Note the similarity between Equation (3.24) and Equation (3.14). For an arbitrary process in a closed system in which there is no work other than PV work, ∆U = qV if the process takes place at constant V and ∆H = qP if the process takes place at constant P. These two equations are the basis for the fundamental experimental techniques of bomb calorimetry and constant pressure calorimetry discussed in Chapter 4. A useful application of Equation (3.24) is in experimentally determining ∆H and ∆U of fusion and vaporization for a given substance. Fusion 1solid S liquid2 and vaporization 1liquid S gas2 occur at a constant temperature if the system is held at a constant pressure and heat flows across the system–surroundings boundary. In both of these phase transitions, attractive interactions between the molecules of the system must be overcome. Therefore, q 7 0 in both cases and CP S ∞ . Because ∆H = qP, ∆ fus H and ∆ vap H can be determined by measuring the heat needed to affect the transition at constant pressure. Because ∆H = ∆U + ∆1PV2, at constant P,
∆ vap H  ∆ vap U = P∆ vap V 7 0
(3.25)
The change in volume upon vaporization is ∆ vap V = Vgas  Vliq W 0; therefore, ∆ vap U 6 ∆ vap H. An analogous expression to Equation (3.25) can be written relating ∆ fus U and ∆ fus H. Note that ∆Vfus is much smaller than ∆Vvap and can be either positive or negative. Therefore, ∆ fus U ≈ ∆ fus H. The thermodynamics of fusion and vaporization will be discussed in more detail in Chapter 8. Because H is a state function, dH is an exact differential, allowing us to link 10H>0T2P to a measurable quantity. In analogy to the preceding discussion for dU, dH is written in the form
dH = a
0H 0H b dT + a b dP 0T P 0P T
(3.26)
Because dP = 0 at constant P, and dH = dqP from Equation (3.24), Equation (3.26) becomes
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dqP = a
0H b dT 0T P
(3.27)
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3.4 Variation of Enthalpy with Temperature at Constant Pressure
Equation (3.27) allows the heat capacity at constant pressure CP to be defined in a fashion analogous to CV in Equation (3.11): dqP 0H CP = = a b dT 0T P
(3.28)
Although this equation looks abstract, CP is a readily measurable quantity. To measure it, one need only measure the heat flow to or from the surroundings for a constant pressure process, together with the resulting temperature change in the limit in which dT and dq approach zero and form the ratio lim 1dq>dT2P . dT S 0 As was the case for CV , CP is an extensive property of the system and varies from substance to substance. The temperature dependence of CP must be known in order to calculate the change in H with T. For a constant pressure process in which there is no change in the phase of the system and no chemical reactions, Tƒ
∆HP =
L Ti
Concept The heat capacity at constant pressure is a direct measure of the partial derivative of H with respect to T at constant P.
Tƒ
CP1T2dT = n CP, m1T2dT L
(3.29)
Ti
If the temperature interval is small enough, it can usually be assumed that CP is constant. In that case, ∆HP = CP ∆T = nCP,m ∆T
(3.30)
Example Problem 3.6 shows a case in which the temperature dependence of CP cannot be neglected. The calculation of ∆H for chemical reactions and changes in phase will be discussed in Chapters 4 and 8.
EXAMPLE PROBLEM 3.6 A 143.0 g sample of C1s2 in the form of graphite is heated from 300. to 600. K at a constant pressure. Over this temperature range, CP, m has been determined to be CP, m J K1mol1
= 12.19 + 0.1126 7.800 * 1011
T T2 T3  1.947 * 104 2 + 1.919 * 107 3 K K K
T4 K4
Calculate ∆H and qP. How large is the relative error in ∆H if we neglect the temperaturedependent terms in CP, m and assume that CP, m maintains its value at 300. K throughout the temperature interval?
Solution Tƒ
m ∆H = C 1T2dT M L P, m Ti
=
T T2  1.947 * 104 2 + 1.919 K J K ± ≤ dT 3 4 T T mol L 7  7.800 * 1011 4 300. K * 10 K K3 600. K
143.0 g 12.00 g mol1
12.19 + 0.1126
600. K T T2 T3 + 0.0563 2  6.49 * 105 3 + 4.798 143.0 K K K = * ≥ ¥ J = 46.9 kJ 4 5 12.00 T T * 108 4  1.56 * 1011 5 K K 300. K
12.19
From Equation (3.24), ∆H = qP.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
If we had assumed CP, m = 8.617 J mol1 K1, which is the calculated value at 300. K, ∆H = 143.0 g>12.00 g mol1 * 8.617 J K1 mol1 * 3 600. K  300. K4 = 30.8 kJ. The relative error is 100 * 130.8 kJ  46.9 kJ2>46.9 kJ = 34.3%. In this case, it is not reasonable to assume that CP, m is independent of temperature.
3.5 HOW ARE CP AND CV RELATED? To this point, two distinct heat capacities, CP and CV, have been defined. How are these quantities related? To answer this question, the differential form of the first law is written as dq = CV dT + a
0U b dV + Pext dV 0V T
(3.31)
Consider a process that proceeds at constant pressure for which P = Pext . In this case, Equation (3.31) becomes Because dqP = CPdT,
CP = CV + a
dqP = CV dT + a
0U b dV + P dV 0V T
(3.32)
0U 0V 0V 0U 0V b a b + P a b = CV + c a b + P d a b 0V T 0T P 0T P 0V T 0T P
= CV + T a
0P 0V b a b 0T V 0T P
(3.33)
To obtain Equation (3.33), both sides of Equation (3.32) have been divided by dT, and the ratio dV>dT has been converted to a partial derivative at constant P. Equation (3.15) has been used in the last step. Using Equation (3.5) and the cyclic rule, one can simplify Equation (3.33) to 0V 2 a b 0T P 0P 0V CP = CV + T a b a b = CV  T 0T V 0T P 0V a b 0P T
CP = CV + TV
a2 a2 or CP, m = CV, m + TVm kT kT
(3.34)
Equation (3.34) provides another example of the usefulness of the formal theory of thermodynamics in linking seemingly abstract partial derivatives with experimentally available data. The difference between CP, m and CV, m can be determined at a given temperature knowing only the molar volume, the isobaric volumetric thermal expansion coefficient, and the isothermal compressibility. Equation (3.34) is next applied to ideal and real gases, as well as liquids and solids, in the absence of phase changes and chemical reactions. Because a and kT are always positive for real and ideal gases, CP  CV 7 0 for these substances. For an 0P 0V ideal gas, 10U>0V2T = 0 as shown in Example Problem 3.2, and T a b a b = 0T V 0T P nR nR T a b a b = nR, so that Equation (3.33) becomes V P CP  CV = nR (3.35) This result was stated without derivation in Section 2.4. Turning to liquids and solids, we note that the partial derivative 10V>0T2P ≈ Va is much smaller for liquids and solids than for gases. Therefore, generally
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CV W c a
0U 0V b + Pd a b 0V T 0T P
(3.36)
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3.6 Variation of Enthalpy with Pressure at Constant Temperature
77
so that CP ≈ CV for a liquid or a solid. As shown earlier in Example Problem 3.1, it is not feasible to carry out heating experiments for liquids and solids at constant volume because of the large pressure increase that occurs. Therefore, tabulated heat capacities for liquids and solids list CP, m rather than CV, m.
3.6 VARIATION OF ENTHALPY WITH PRESSURE AT CONSTANT TEMPERATURE
In the previous section, we learned how H changes with T at constant P. To calculate how H changes as both P and T change, 10H>0P2T must be calculated. The partial derivative 10H>0P2T is less straightforward to determine in an experiment than 10H>0T2P. As we will show, for many processes involving changes in both P and T, 10H>0T2P dT W 10H>0P2T dP, and the pressure dependence of H can be neglected relative to its temperature dependence. However, the knowledge that 10H>0P2T is not zero is essential for understanding the operation of a refrigerator and the liquefaction of gases. The following discussion is applicable to gases, liquids, and solids. Given the definition H = U + PV, we begin by writing dH as dH = dU + P dV + V dP
(3.37)
Substituting the differential forms of dU and dH, CPdT + a
0H 0U b dP = CV dT + a b dV + P dV + V dP 0P T 0V T = CV dT + c a
0U b + P d dV + V dP 0V T
(3.38)
For isothermal processes, dT = 0, and Equation (3.38) can be rearranged to a
0H 0U 0V b = c a b + P d a b + V 0P T 0V T 0P T
(3.39)
Using Equation (3.15) for 10U>0V2T , a
0H 0P 0V b = Ta b a b + V 0P T 0T V 0P T = V  Ta
0V b ≈ V11  Ta2 0T P
(3.40)
The second formulation of Equation (3.40) is obtained through application of the cyclic rule [Equation (3.3)]. This equation is applicable to all systems containing pure substances or mixtures at a fixed composition, provided that no phase changes or chemical reactions take place. The quantity 10H>0P2T is evaluated for an ideal gas in Example Problem 3.7.
EXAMPLE PROBLEM 3.7 Evaluate 10H>0P2T for an ideal gas.
Solution 10P>0T2V = 103 nRT>V4 >0T2V = nR>V and 10V>0P2T = 1d3 nRT>P4 >dP2T = nRT>P 2 for an ideal gas. Therefore, a
0H 0P 0V nR nRT nRT nRT + V = 0 b = T a b a b + V = T a 2 b + V = 0P T 0T V 0P T V P nRT P
This result could have also been derived directly from the definition H = U + PV. Thus, for an ideal gas U = U1T2 only and PV = nRT. Therefore, H = H1T2 for an ideal gas and 10H>0P2T = 0.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Because Example Problem 3.7 shows that H is only a function of T for an ideal gas,
Concept H does not depend on P for an ideal gas.
Tƒ
∆H =
L Ti
Tƒ
CP1T2dT = n CP,m1T2dT L
(3.41)
Ti
for an ideal gas. Because H is only a function of T, Equation (3.41) holds for an ideal gas even if P is not constant. This result is also understandable in terms of the potential function of Figure 1.9. Because ideal gas molecules do not attract or repel one another, no energy is required to change their average distance of separation (increase or decrease P). Equation (3.40) is next applied to several types of systems. We have seen that 10H>0P2T = 0 for an ideal gas. For liquids and solids, 1 W Ta for T 6 1000 K, as can be seen in Table 3.1. Therefore, for liquids and solids, 10H>0P2T ≈ V to a good approximation, and dH can be written as dH ≈ CP dT + V dP
(3.42)
for systems that consist only of liquids or solids.
EXAMPLE PROBLEM 3.8 Calculate the change in enthalpy when 124 g of liquid methanol initially at 1.00 bar and 298 K undergoes a change of state to 2.50 bar and 425 K. The density of liquid methanol under these conditions is 0.791 g cm3, and CP, m for liquid methanol is 81.1 J K1 mol1.
Solution Because H is a state function, any path between the initial and final states will give the same ∆H. We choose the path methanol (l, 1.00 bar, 298 K) S methanol (l, 1.00 bar, 425 K) S methanol (l, 2.50 bar, 425 K). The first step is isothermal, and the second step is isobaric. The total change in H is Tƒ
∆H = n
L
Pƒ
CP,mdT +
Ti
L Pi
V dP ≈ nCP,m 1Tƒ  Ti2 + V1Pƒ  Pi2
= 81.1 J K1 mol1 * +
124 g 0.791 g cm
3
*
124 g 32.04 g mol1
* 1425 K  298 K2
106 m3 105 Pa * 12.50 bar  1.00 bar2 * 3 bar cm
= 39.9 * 103 J + 23.5 J ≈ 39.9 kJ
Note that the contribution to ∆H from the change in T is far greater than that from the change in P.
Concept Except for large pressure changes, H does not depend on P for liquids and solids.
Example Problem 3.8 shows that because molar volumes of liquids and solids are small, H changes much more rapidly with T than with P. Under most conditions, H can be assumed to be a function of T only for solids and liquids. Exceptions to this rule are encountered in geophysical or astrophysical applications, for which extremely large pressure changes can occur. The following conclusion can be drawn from this section: under most conditions encountered by chemists in the laboratory, H can be regarded as a function of T alone for liquids and solids. It is a good approximation to write T2
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H1Tƒ, Pƒ2  H1Ti, Pi2 = ∆H =
L T1
T2
CP dT = n CP, m dT L
(3.43)
T1
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3.7 The Joule–Thomson Experiment
79
even if P is not constant in the process under consideration. The dependence of H on P for real gases is discussed in Section 3.8 and Section 3.9 in the context of the Joule– Thomson experiment. Note that Equation (3.43) is only applicable to a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no chemical reactions. Changes in H that arise from chemical reactions and changes in phase will be discussed in Chapters 4 and 8. Having dealt with solids, liquids, and ideal gases, we are left with real gases. For real gases, 10H>0P2T and 10U>0V2T are small, but the values of these partial derivatives still have a considerable effect on the properties of the gases upon expansion or compression. Conventional technology for the liquefaction of gases and for the operation of refrigerators is based on the fact that 10H>0P2T and 10U>0V2T are not zero for real gases. To derive a useful formula for calculating 10H>0P2T for a real gas, the Joule– Thomson experiment is discussed in the next section.
3.7 THE JOULE–THOMSON EXPERIMENT If the valve on a cylinder of compressed N2 at 298 K is opened fully, it will become covered with frost, demonstrating that the temperature of the valve is lowered below the freezing point of H2O. A similar experiment, using a cylinder that contains H2, leads to a considerable increase in temperature and, potentially, an explosion. How can these effects be understood? To explain them, we discuss the Joule–Thomson experiment. The Joule–Thomson experiment, shown in Figure 3.4, can be viewed as an improved version of the Joule experiment because it allows 10U>0V2T to be measured with a much higher sensitivity than in the Joule experiment. In this experiment, gas flows from the highpressure cylinder on the left to the lowpressure cylinder on the right through a porous plug in an insulated pipe. The pistons move to keep the pressure unchanged in each region until all the gas has been transferred to the region to the right of the porous plug. If N2 is used in the expansion process 1P1 7 P22, it is found that T2 6 T1; in other words, the gas is cooled as it expands. What is the origin of this effect? Consider an amount of gas equal to the initial volume V1 as it passes through the apparatus from left to right. The total work in this expansion process is the sum of the work performed on each side of the plug separately by the moving pistons: 0
w = wleft + wright = 
L V1
V2
P1 dV 
L
P2 dV = P1V1  P2V2
(3.44)
0
Pressure gauges
Figure 3.4
P1V1T1
Initial state Porous plug
P2V2T2
Final state
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The Joule–Thomson experiment. In the experiment, a gas is forced through a porous plug using a piston and cylinder mechanism. The pistons move to maintain a constant pressure in each region. There is an appreciable pressure drop across the plug, and the temperature change of the gas is measured. The upper and lower figures show the initial and final states, respectively. As shown in the text, if the piston and cylinder assembly forms an adiabatic wall between the system (the gases on both sides of the plug) and the surroundings, the expansion is isenthalpic.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Because the pipe is insulated, q = 0, and ∆U = U2  U1 = w = P1V1  P2V2
(3.45)
This equation can be rearranged to U2 + P2V2 = U1 + P1V1 or H2 = H1
Concept The Joule Thomson experiment provides a way to measure the partial derivative of U with respect to V at constant T and the partial derivative of H with respect to P at constant T for real gases.
TABLE 3.3 Joule–Thomson Coefficients for Selected Substances at 273 K and 1 atm Gas Ar
mJ  T 1K>MPa2
Note that the enthalpy is constant in the expansion; the expansion is isenthalpic. For the conditions of the experiment that uses N2, both dT and dP are negative, so 10T>0P2H 7 0. The experimentally determined limiting ratio of ∆T to ∆P at constant enthalpy is known as the Joule–Thomson coefficient: mJ  T = lim a
∆P S 0
4.38
CO2
10.9
H2
 0.34
He
 0.62
N2
2.15
Ne
 0.30
NH3
28.2
O2
2.69
Source: Data from Linstrom, P. J., and Mallard, W. G., eds. NIST Chemistry Webbook: NIST Standard Reference Database Number 69. Gaithersburg, MD: National Institute of Standards and Technology. Retrieved from http://webbook.nist.gov.
∆T 0T b = a b ∆P H 0P H
(3.47)
If mJ  T is positive, the conditions are such that the attractive part of the potential dominates, and if mJ  T is negative, the repulsive part of the potential dominates. Using experimentally determined values of mJ  T , 10H>0P2T can be calculated. For an isenthalpic process, 0H dH = CP dT + a b dP = 0 (3.48) 0P T Dividing through by dP and making the condition dH = 0 explicit,
3.66
CH4
(3.46)
CP a
0T 0H b + a b = 0 0P H 0P T
giving a
0H b = CPmJ  T 0P T
(3.49)
Equation (3.49) states that 10H>0P2T can be calculated using the measurement of materialdependent properties CP and mJ  T . Because mJ  T is not zero for a real gas, the pressure dependence of H for an expansion or compression process for which the pressure change is large cannot be neglected. Note that 10H>0P2T can be positive or negative, depending on the sign of mJ  T at the P and T of interest. If mJ  T is known from experiment, 10U>0V2T can be calculated as shown in Example Problem 3.9. This has the advantage that a calculation of 10U>0V2T based on measurements of CP, mJ  T and the isothermal compressibility kT is much more accurate than a measurement based on the Joule experiment. Values of mJ  T are shown for selected gases in Table 3.3. Keep in mind that mJ  T is a function of T, P and ∆P, so the values listed in the table are only valid for a small pressure decrease originating at 1 atm pressure.
EXAMPLE PROBLEM 3.9 Using Equation (3.39), 10H>0P2T = 3 10U>0V2T + P4 10V>0P2T + V, derive an expression giving 10U>0V2T entirely in terms of measurable quantities for a gas.
Solution
a
0H 0U 0V b = ca b + Pda b + V 0P T 0V T 0P T
a
0U b = 0V T =
a
0H b  V 0P T  P 0V a b 0P T
CPmJ  T + V  P kTV
In this equation, kT is the isothermal compressibility defined in Equation (3.5).
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3.8 Liquefying Gases Using an Isenthalpic Expansion 700
EXAMPLE PROBLEM 3.10 Using Equation (3.39),
600
Temperature (K)
0H 0U 0V a b = ca b + Pd a b + V 0P T 0V T 0P T
show that mJ  T = 0 for an ideal gas.
Solution mJ  T = 
1 0H 1 0U 0V 0V a b = ca b a b + P a b + Vd CP 0P T CP 0V T 0P T 0P T
N2
500 400 300 200
H2
100
1 0V = c0 + P a b + Vd CP 0P T
100
03 nRT>P4 1 1 nRT = cP a b + Vd = c+ Vd = 0 CP 0P C P T P
Example Problem 3.10 shows that for an ideal gas, mJ  T is zero. It can be shown that for a van der Waals gas in the limit of zero pressure mJ  T =
1 2a a  bb CP,m RT
(3.50)
3.8 LIQUEFYING GASES USING
AN ISENTHALPIC EXPANSION
For real gases, the Joule–Thomson coefficient mJ  T can take on either negative or positive values in different regions of PT space. If mJ  T is positive, a decrease in pressure leads to a cooling of the gas; if it is negative, the expansion of the gas leads to a heating. Figure 3.5 shows the variation of mJ  T with T and P for N2 and H2. All along the solid curve, mJ  T = 0. To the left of each curve, mJ  T is positive, and to the right, it is negative. The temperature for which mJ  T = 0 is referred to as the inversion temperature. If the expansion conditions are kept in the region in which mJ  T is positive, ∆T can be made sufficiently large as ∆P decreases in the expansion to liquefy the gas. Note that Equation (3.50) predicts that the inversion temperature for a van der Waals gas is independent of P, which is not in agreement with experiment. The Joule–Thomson coefficients for N2 and H2 are shown as a function of temperature and pressure in Figure 3.5. The information in Figure 3.5 is in accord with the observation that a highpressure 1100 6 P 6 500 atm2 expansion of N2 at 300 K leads to cooling, whereas similar conditions for H2 lead to heating. To cool H2 in an expansion, it must first be precooled below 200 K, and the pressure must be less than 160 atm. He and H2 are heated in an isenthalpic expansion at 300 K for P 6 200 atm. The Joule–Thomson effect can be used to liquefy gases such as N2. As shown in Figure 3.6, the gas at atmospheric pressure is first compressed to a value of 50 atm to 200 atm, which leads to a substantial increase in its temperature. It is cooled and subsequently passed through a heat exchanger in which the gas temperature decreases to a value within ∼50 K of the boiling point. At the exit of the heat exchanger, the gas expands through a nozzle to a final pressure of 1 atm in an isenthalpic expansion. The cooling that occurs because mJ  T 7 0 results in liquefaction. The gas that boils away passes back through the heat exchanger in the opposite direction in which the gas to be liquefied is passing. The two gas streams are separated, but they are in good thermal contact. In this process, the gas to be liquefied is effectively precooled, enabling a singlestage expansion to achieve liquefaction.
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200 300 400 Pressure (atm)
500
Figure 3.5
In this calculation, we have used the result that 10U>0V2T = 0 for an ideal gas.
81
Plot of Joule–Thomson coefficients for N2 and H2. Along the curves in the figure mJ  T = 0. The value of mJ  T is positive to the left of the curves and negative to the right. To experience cooling upon expansion at 100. atm, T must lie between 50. K and 150. K for H2. The corresponding temperatures for N2 are 100. K and 650. K.
Connection The dependence of H on P at constant T for real gases allows gases to be liquified.
Cooler
Gas feed
Compressor
Liquid out
Figure 3.6
Schematic depiction of the liquefaction of a gas using an isenthalpic Joule– Thomson expansion. Heat is extracted from the gas exiting from the compressor. It is further cooled in the countercurrent heat exchanger before expanding through a nozzle. Because its temperature is sufficiently low at the exit to the countercurrent heat exchanger, liquefaction occurs.
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CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
VOCABULARY internal pressure isenthalpic isobaric volumetric thermal expansion coefficient isothermal compressibility
cyclic rule exact differential heat capacity at constant pressure heat capacity at constant volume
Joule–Thomson coefficient Joule–Thomson experiment partial derivatives
KEY EQUATIONS Equation a
Significance of Equation
0P 0V 0T b a b a b = 1 0V T 0T P 0P V
a =
1 0V 1 0V a b and kT =  a b V 0T P V 0P T
dqV 0U = a b = CV dT 0T V ƒ
L
ƒ
dqV =
i
0U b dT or qV = ∆U L 0T V i
a
Tƒ
U1Tƒ, Vƒ2  U1Ti, Vi2 = ∆U =
L Ti
LTi
CV, m dT
∆H = qP
dqP 0H = a b dT 0T P
CP = CV + TV
a2 a2 or CP, m = CV, m + TVm kT kT T2
H1Tƒ, Pƒ2  H1Ti, Pi2 = ∆H = mJ  T = lim a ∆P S 0
L
Cyclic rule
3.2
Definition of volumetric thermal expansion coefficient and the isothermal compressibility
3.5
Definition of heat capacity at constant volume
3.11
∆U can be measured by determining heat flow at constant volume
3.14
Under most conditions, U = U1T2 alone
3.21
Changes in enthalpy can be measured by determining heat flow at constant pressure.
3.24
Definition of heat capacity at constant pressure
3.28
Relation between CP and CV for any substance
3.34
Under most conditions, H = H1T2 alone
3.43
Definition of Joule–Thomson coefficient
3.47
Tƒ
CV dT = n
1Uƒ + PƒVƒ2  1Ui + PiVi2 = qP or CP =
Equation Number
T2
CP dT = n
T1
L
CP, m dT
T1
∆T 0T b = a b ∆P H 0P H
CONCEPTUAL PROBLEMS Q3.1 The heat capacity CP,m is less than CV,m for H2O1l2 near 4°C. Explain this result. Q3.2 What is the physical basis for the experimental result that U is a function of V at constant T for a real gas? Under what conditions will U decrease as V increases? Q3.3 Explain why ethene has a higher value for CV,m at 800. K than CO.
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Q3.4 Why does the relation CP 7 CV always hold for a gas? Can CP 6 CV be valid for a liquid? Q3.5 Why is CP,m a function of temperature at 300. K for ethane but not for argon? Q3.6 Explain, without using equations, why 10H>0P2T is generally small for a real gas.
19/09/17 5:57 PM
Numerical Problems
Q3.7 Why is it reasonable to write dH ≈ CPdT + VdP for a liquid or solid sample? Q3.8 Refer to Figure 1.9 and explain why 10U>0V2T is generally small for a real gas. Q3.9 Can a gas be liquefied through an isenthalpic expansion if mJ  T = 0? Q3.10 Why is qv = ∆U only for a constant volume process? Is this formula valid if work other than P–V work is possible? Q3.11 Classify the following variables and functions as intensive or extensive: T, P, V, q, w, U, H. Q3.12 Why are q and w not state functions? T Q3.13 Why is the equation ∆H = 1Ti ƒCP1T2 dT = T n 1Ti ƒCP,m1T2 dT valid for an ideal gas even if P is not constant in the process? Is this equation also valid for a real gas? Why or why not? Q3.14 What is the relationship between a state function and an exact differential?
83
Q3.15 Is the following statement always, never, or sometimes valid? Explain your reasoning: ∆H is only defined for a constant pressure process. Q3.16 Is the following statement always, never, or sometimes valid? Explain your reasoning: A thermodynamic process is completely defined by the initial and final states of the system. Q3.17 Is the following statement always, never, or sometimes valid? Explain your reasoning: q = 0 for a cyclic process. Q3.18 The molar volume of H2O1l2 decreases with increasing temperature near 4°C. Can you explain this behavior using a molecularlevel model? Q3.19 Why was the following qualification made in Section 3.6? Note that Equation (3.43) is only applicable to a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no chemical reactions. T T Q3.20 Is the expression ∆UV = 1T12CV dT = n 1T12CV,m dT only valid for an ideal gas if V is constant?
NUMERICAL PROBLEMS Section 3.1 P3.1 One consequence of climate change is the rise in sea level as the oceans warm and as land based ice melts and increases the volume of the oceans. The volume of the Earth’s oceans is 1.35 * 109 km3, and the average depth is 3668 m. The volume of the Greenland ice sheet is 2.85 * 106 km3. Using the data tables, calculate (a) the rise in sea level expected if the ocean temperature increases by 2.0°C, the current maximum increase targeted by the UN intergovernmental panel on climate change and (b) the rise in sea level expected if the Greenland ice sheet melts. State any assumptions made. P3.2 Obtain an expression for the isothermal compressibility kT = 1>V10V>0P2T for a van der Waals gas. P3.3 A vessel is filled completely with liquid water and sealed at 15.0°C and a pressure of 1.00 bar. What is the pressure if the temperature of the system is raised to 75.5°C? Under these conditions, awater = 2.04 * 104 K1, avessel = 1.42 * 104 K1, and kT = 4.59 * 105 bar 1. P3.4 Integrate the expression a = 1>V10V>0T2P assuming that a is independent of pressure. By doing so, obtain an expression for V as a function of T and a at constant P. P3.5 The function ƒ1x, y2 is given by ƒ1x, y2 = xy sin 5x + 2 x 2 2y ln y + 3e2x cos y. Determine a
0ƒ 0ƒ 02ƒ 02ƒ 0 0ƒ 0 0ƒ b , a b , a 2 b , a 2 b , a a b b and a a b b 0x y 0y x 0x y 0y x 0y 0x y x 0x 0y x y
Obtain an expression for the total differential dƒ. P3.6 Using the result of Equation (3.4), 10P>0T2V = a>kT , express a as a function of kT and Vm for an ideal gas, and a as a function of b, kT , and Vm for a van der Waals gas.
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P3.7 Starting with the van der Waals equation of state, find an expression for the total differential dP in terms of dV and dT. By calculating the mixed partial derivatives 1010P>0V2T >0T2V and 1010P>0T2V >0V2T , determine if dP is an exact differential. P3.8 Use 10U>0V2T = 1aT  kTP2>kT to calculate 10U>0V2T for an ideal gas. P3.9 A differential dz = ƒ1x, y2dx + g1x, y2dy is exact if the integral 1 ƒ1x, y2dx + 1 g1x, y2dy is independent of the path. Demonstrate that the differential dz = 2xydx + x 2dy is exact by integrating dz along the paths (1,1) S (1, 8) S (6, 8) and (1, 1) S (1, 3) S (4, 3) S (4, 8) S (6, 8). The first number in each set of parentheses is the x coordinate, and the second number is the y coordinate. P3.10 Show that dr>r = adT + kTdP where r is the density r = m>V. Assume that the mass, m, is constant. P3.11 Because V is a state function, 1010V>0T2P >0P2T = 1010V>0P2T >0T2P. Using this relationship, show that the isothermal compressibility and isobaric expansion coefficient are related by 10a>0P2T = 10kT >0T2P. P3.12 Starting with a = 11>V210V>0T2P, show that a = 11>r210r>0T2P, where r is density. P3.13 This problem will give you practice in using the cyclic rule. Use the ideal gas law to obtain the three functions P = ƒ1V, T2, V = g1P, T2, and T = h1P, V2. Show that the cyclic rule 10P>0V2T 10V>0T2P 10T>0P2V = 1 is obeyed. P3.14 Regard the enthalpy as a function of T and P. Use the cyclic rule to obtain the expression CP =  a
0H 0T b na b 0P T 0P H
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P3.15 Derive the equation 10P>0V2T = 1>1kTV2 from basic equations and definitions. P3.16 Derive the equation 10H>0T2V = CV + Va>kT from basic equations and definitions. 0CV 02P P3.17 Show that a b = T a 2b 0V T 0T V
0CV P3.18 Show that a b = 0 for both an ideal and a van der 0V T Waals gas. P3.19 Calculate the isobaric volumetric thermal expansion coefficient and the isothermal compressibility, respectively, defined by 1 0V 1 0V a = a b and kT =  a b V 0T P V 0P T for an ideal gas at 298 K and 1.00 bar L.
Section 3.2 P3.20 Show that the expression 10U>0V2T = T10P>0T2V  P can be written in the form a
0U P P 1 b = T 2 a0 c d n 0T b = a0 c d n 0 c d b 0V T T T T V V
P3.21 Derive an expression for the internal pressure of a gas RT a that obeys the Berthelot equation of state, P = Vm  b TV 2m P3.22 Because U is a state function, 10>0V10U>0T2V2T = 10>0T10U>0V2T2V . Using this relationship, show that 10CV >0V2T = 0 for an ideal gas. P3.23 Derive the following relation, a
0U 3a b = 0Vm T 22TVm1Vm + b2
for the internal pressure of a gas that obeys the Redlich–Kwong equation of state, RT a 1 P = . Vm  b 2T Vm1Vm + b2
P3.24 A mass of 25.0 g of H2O(s) at 273 K is dropped into 130. g of H2O(l) at 298 K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP, m for H2O is constant at its values for 298 K throughout the temperature range of interest.
Section 3.3 P3.25 A mass of 29.5 g of H2O(g) at 373 K is allowed to flow into 280. g of H2O(l) at 298 K and 1 atm. Calculate the final temperature of the system once equilibrium has been reached. Assume that CP, m for H2O is constant at its values for 298 K throughout the temperature range of interest. P3.26 A 62.5g piece of gold at 650. K is dropped into 165 g of H2O(l) at 298 K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP, m for Au and H2O are constant at their values for 298 K throughout the temperature range of interest.
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P3.27 Calculate w, q, ∆H, and ∆U for the process in which 2.50 mol of water undergoes the transition H2O(l, 373 K) S H2O(g, 610. K) at 1 bar of pressure. The volume of liquid water at 373 K is 1.89 * 105 m3 mol1, and the volume of steam at 373 and 610. K is 3.03 and 5.06 * 102 m3 mol1, respectively. For steam, CP,m can be considered constant over the temperature interval of interest at 33.58 J mol1 K1. P3.28 The Joule coefficient is defined by 10T>0V2U = 11>CV23 P  T10P>0T2V 4 . Calculate the Joule coefficient for an ideal gas and for a van der Waals gas.
Section 3.4 0U 0H P3.29 For an ideal gas, a b and a b = 0. Prove that 0V T 0P T CV is independent of volume and CP is independent of pressure.
Section 3.5 P3.30 Derive the following expression for calculating the isothermal change in the constant volume heat capacity: 10CV >0V2T = T102P>0T 22V . P3.31 Equation (3.34), CP = CV + TV1a2 >kT2, links CP and CV with a and kT . Use this equation to evaluate CP  CV for an ideal gas. P3.32 Use the result of Problem P3.30 to derive a formula for 10CV >0V2T for a gas that obeys the Redlich–Kwong equation RT a 1 of state, P = Vm  b V 1V 2T m m + b2 P3.33 Use the relation
CP,m  CV,m = T a
0Vm 0P b a b 0T P 0T V
the cyclic rule, and the van der Waals equation of state to derive an equation for CP, m  CV, m in terms of Vm, T, and the gas constants R, a, and b. P3.34 For the equation of state Vm = RT>P + B1T2, show that a
0CP, m 0P
b = T T
d 2B1T2 dT 2
[Hint: Use Equation 3.40 and the property of state functions with respect to the order of differentiation in mixed second derivatives.] 0U 0V P3.35 Prove that CV =  a b a b 0V T 0T U
Section 3.6 P3.36 The molar heat capacity CP, m of SO21g2 is described by the following equation over the range 300 K 6 T 6 1700 K: CP, m R
= 3.093 + 6.967 * 103
+ 1.035 * 109
T3 K3
T T2  45.81 * 107 2 K K
In this equation, T is the absolute temperature in kelvins. The ratios T n >Kn ensure that CP, m has the correct dimension. Assuming ideal gas behavior, calculate q, w, ∆U, and ∆H if 2.25 moles of SO21g2 are heated from 10.0° to 1250°C at a constant pressure of 1 bar. Explain the sign of w.
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Further Reading
P3.37 For a gas that obeys the equation of state Vm =
RT + B1T2 P
derive the result a
dB1T2 0H b = B1T2  T 0P T dT
P3.38 Use the relation 10U>0V2T = T10P>0T2V  P and the cyclic rule to obtain an expression for the internal pressure, 10U>0V2T , in terms of P, a, T, and kT .
Section 3.7 P3.39 Because 10H>0P2T = CPmJ  T , the change in enthalpy of a gas expanded at constant temperature can be calculated. To do so, the functional dependence of mJ  T on P must be known. Treating Ar as a van der Waals gas, calculate ∆H when 1 mol of Ar is expanded from 325 bar to 1.75 bar at 375 K. Assume that mJ  T is independent of pressure and is given by mJ  T = 3 12a>RT2  b4 >CP,m and CP,m = 5>2R for Ar. What value would ∆H have if the gas exhibited ideal gas behavior?
FURTHER READING Apfelbaum, E. M., and Vorb’ev, V. S. “The Similarity Law for the Joule–Thomson Inversion Curve.” Journal of Physical Chemistry B 118 (2014): 12239–12242. Barrante, J. R. “Applied Mathematics for Physical Chemistry.” Chapter 4. Differential Calculus, 3rd Edition. New York: Pearson, 2004. Goussard, J.O., and Roulet, B. “Free Expansion for Real Gases.” American Journal of Physics 61 (1993): 845–848.
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Halpern, A. M., and Gozashti, S. “An Improved Apparatus for the Measurement of the Joule–Thomson Coefficient of Gases.” Journal of Chemical Education 63 (1986): 1001–1002. Hecht, C. E., and Zimmerman, G. “Measurement of the Joule–Thomson Coefficient.” Journal of Chemical Education 31 (1954): 530–533.
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C H A P T E R
4
Thermochemistry
4.1 Energy Stored in Chemical Bonds Is Released or Absorbed in Chemical Reactions
4.2 Internal Energy and Enthalpy
WHY is this material important? Thermochemistry is the branch of thermodynamics that investigates the heat flow into or out of a reaction system. As reactants are converted into products, energy can either be absorbed by the system or released to the surroundings. By measuring this energy transfer, chemical bond energies and enthalpies can be determined.
WHAT are the most important concepts and results? For a reaction that occurs at constant volume, the heat that flows into or out of the system is equal to ∆U for the reaction. For a reaction that takes place at constant pressure, the heat that flows into or out of the system is equal to ∆H for the reaction. Because H and U are state function, the reaction energy or enthalpy can be written as the energies or enthalpies of formation of the products minus those of the reactants. This property allows ∆H and ∆U for a reaction to be calculated for many reactions without carrying out an experiment.
Changes Associated with Chemical Reactions
4.3 Hess’s Law Is Based on Enthalpy Being a State Function
4.4 Temperature Dependence of Reaction Enthalpies
4.5 Experimental Determination of ∆U and ∆H for Chemical Reactions
4.6 (Supplemental Section) Differential Scanning Calorimetry
WHAT would be helpful for you to review for this chapter? You should be familiar with state functions and in particular with U and H.
4.1 ENERGY STORED IN CHEMICAL BONDS
IS RELEASED OR ABSORBED IN CHEMICAL REACTIONS
Most of the internal energy or enthalpy of a molecule is stored in the form of potential energy in chemical bonds. As reactants are transformed to products in a chemical reaction, energy can be released or taken up as bonds are made or broken. For example, consider a reaction in which N21g2 and H21g2 dissociate into atoms, and the atoms recombine to form NH31g2. The enthalpy changes associated with individual steps and with the overall reaction 1>2 N21g2 + 3>2 H21g2 ¡ NH31g2 are shown in Figure 4.1. Note that large enthalpy changes are associated with the individual steps but the enthalpy change in the overall reaction is much smaller. The change in enthalpy or internal energy resulting from chemical reactions appears in the surroundings in the form of a temperature increase or decrease resulting from heat flow and/or in the form of expansion or nonexpansion work. For example, nonexpansion electrical work is possible if the oxidation of a hydrocarbon or H21g2 is carried out in an electrochemical cell. In Chapters 6 and 11, the extraction of nonexpansion work from chemical reactions will be discussed. In this chapter, the focus is on using measurements of heat flow to determine changes in U and H due to chemical reactions. 87
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CHAPTER 4 Thermochemistry
Figure 4.1
Enthalpy changes for individual steps in the overall reaction 1>2 N21g2 + 3>2 H21g2 ¡ NH31g2.
N(g) 1 3H(g)
314 3 103 J NH(g) 1 2H(g)
1124 3 103 J
390 3 103 J NH2(g) 1 H(g)
466 3 103 J
1
2
N2 (g) 1 32 H2(g)
245.9 3 103 J NH3(g)
4.2 INTERNAL ENERGY AND ENTHALPY
CHANGES ASSOCIATED WITH CHEMICAL REACTIONS
Concept In an exothermic process, heat is transferred from the system to the surroundings. In an endothermic process, heat is transferred from the surroundings to the system.
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In the previous chapters, we discussed how ∆U and ∆H are calculated from work and heat flow between the system and the surroundings for processes that do not involve phase changes or chemical reactions. In this section, this discussion is extended to reaction systems. Imagine that a reaction involving a stoichiometric mixture of reactants (the system) is carried out in a constant pressure reaction vessel with diathermal walls that is immersed in a water bath (the surroundings). If the temperature of the water bath increases, heat has flowed from the system (the contents of the reaction vessel) to the surroundings (the water bath and the vessel). In this case, we say that the reaction is exothermic. If the temperature of the water bath decreases, the heat has flowed from the surroundings to the system, and we say that the reaction is endothermic. Consider the following reaction:
Fe 3O41s2 + 4 H21g2 ¡ 3 Fe1s2 + 4 H2O1l2
(4.1)
Note that the phase (solid, liquid, or gas) for each reactant and product has been specified because U and H are different for each phase. This reaction will only proceed at a measurable rate at elevated temperatures. However, as we show later, it is useful to tabulate values for ∆H for reactions at a pressure of 1 bar and a specified temperature, generally 298.15 K. The pressure value of 1 bar defines a standard state, and changes in H and U at the standard pressure of 1 bar are indicated by a superscript ~ as in ∆H ~ and ∆U ~ . The standard state for gases is a hypothetical state in which the gas behaves ideally at a pressure of 1 bar. For most gases, deviations from ideal behavior are very small. The enthalpy of reaction, ∆ r H, at specific values of T and P is defined as the
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4.2 Internal Energy and Enthalpy Changes Associated with Chemical Reactions
89
heat exchanged between the system and the surroundings as the reactants are transformed into products at conditions of constant T and P. The subscript r indicates that the process is a chemical reaction. By convention, heat flowing into the system is given a positive sign. Therefore, ∆ r H is a negative quantity for an exothermic reaction and a positive quantity for an endothermic reaction. The standard enthalpy of reaction, ∆ r H ~ , refers to one mole of the specified reaction at a pressure of 1 bar and, unless indicated otherwise, to T = 298.15 K. How can the reaction enthalpy and internal energy be determined? We proceed in the following way. The reaction is carried out at 1 bar pressure, and the temperature change ∆T that occurs in a finite size water bath, initially at 298.15 K, is measured. The water bath is large enough that ∆T is small. If ∆T is negative as a result of the reaction, the bath is heated to return it, the reaction vessel, and the system to 298.15 K using an electrical heater. By doing so, we ensure that the initial and final states are the same and therefore the measured ∆H is equal to ∆ r H ~ . The electrical work done on the heater that restores the temperature of the water bath and the system to 298.15 K is equal to ∆ r H ~ . If the temperature of the water bath increases as a result of the reaction, a separate experiment must be carried out. In this case, the electrical work done on a heater in the water bath at 298.15 K that increases the system and water bath temperature by ∆T is measured. For this scenario, ∆ r H ~ is equal to the negative of the electrical work done on the heater. The experimental method for determining ∆ r H ~ was described in the previous paragraph. Because the number of reactions among all possible reactants and products is very large, tabulation of the reaction enthalpies for all possible chemical reactions would be a monumental undertaking. Thus, another approach, in which ∆ r H ~ is calculated from tabulated enthalpy values for individual reactants and products, is generally preferred. The advantage of this approach is that there are far fewer reactants and products than there are reactions among them. As an example of how to calculate ∆ r H ~ from enthalpies of individual reactants and products, consider the determination of ∆ r H ~ for the reaction of Equation (4.1) at T = 298.15 K and P = 1 bar. These values for P and T are chosen because, by convention, thermodynamic values are tabulated at T = 298.15 K and P = 1 bar. However, ∆ r H can be calculated for other values of P and T using the methods discussed in Chapters 2 and 3. In principle, we could express ∆ r H ~ in terms of the individual enthalpies of reactants and products: ~ ~ ∆ r H ~ = H products  H reactants
= 3H m~ 1Fe, s2 + 4H m~ 1H2O, l2  H m~ 1Fe 3O4, s2  4H m~ 1H2, g2 (4.2)
The m subscripts refer to molar quantities. Although Equation (4.2) is correct, it does not provide a useful way to calculate ∆ r H ~ . That is, there is no experimental way to determine the absolute enthalpy for any element or compound because there is no unique reference point with a value of zero against which individual enthalpies can be measured. Only ∆H and ∆U, as opposed to H and U, can be determined experimentally. Equation (4.2) can be transformed into a more useful form by introducing the enthalpy of formation. The standard enthalpy of formation, ∆ f H ~ , is defined as the enthalpy change of a reaction in which the only reaction product is 1 mol of the species of interest, and only pure elements in their most stable state of aggregation under standard state conditions appear as reactants. We refer to these species as being in their standard reference state. For example, the standard reference state of water and carbon at 298.15 K and 1 bar pressure are H2O1l2 and solid carbon in the form of graphite. Note that with this definition ∆ f H ~ = 0 for an element in its standard reference state because the reactants and products are identical. Reaction enthalpies can be expressed in terms of formation enthalpies. The only compounds that are produced or consumed in the reaction Fe 3O41s2 + 4 H21g2 ¡ 3 Fe1s2 + 4 H2O1l2 are Fe 3O41s2 and H2O1l2. All elements that appear in the reaction
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Concept The standard reference state of a substance is the most stable state at 1 bar pressure and 298.15 K.
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CHAPTER 4 Thermochemistry
are in their standard reference states. The formation reactions for the compounds at 298.15 K and 1 bar are
∆r H ~
∆r H
~
1 O 1g2 ¡ H2O1l2 2 2 = ∆ f H ~ 1H2O, l2  0  0 = ∆ f H ~ 1H2O, l2 H21g2 +
3 Fe1s2 + 2 O21g2 ¡ Fe 3O41s2
= ∆ f H ~ 1Fe 3O4, s2  0  0 = ∆ f H ~ 1Fe 3O4, s2
(4.3)
(4.4)
If Equation (4.2) is rewritten in terms of the enthalpies of formation, a simple equation for the reaction enthalpy is obtained:
∆ r H ~ = 4∆ f H ~ 1H2O, l2  ∆ f H ~ 1Fe 3O4, s2
(4.5)
nAA + nBB + c ¡ nXX + nYY + c
(4.6)
Note that elements in their standard reference state do not appear in this equation because ∆ f H ~ = 0 for these species. This result can be generalized to any chemical transformation
which we write in the form
Concept The enthalpy change for a reaction can be expressed in terms of the standard formation enthalpies of reactants and products.
i
[
DrH
2nA DfH A 2nB DfH
[
nCC 1 nDD
[ B
(4.7)
The Xi refer to all species that appear in the overall equation. The unitless stoichiometric coefficients ni are positive for products and negative for reactants. The enthalpy change associated with this reaction is ∆ r H ~ = a ni ∆ f H i~
nAA 1 nBB
0 = a ni Xi
i
(4.8)
The rationale behind Equation (4.8) is depicted in Figure 4.2. Two paths are considered between the reactants A and B and the products C and D in the reaction vAA + vBB ¡ vCC + vDD. The first of these is a direct path for which ∆H ~ = ∆ r H ~ . In the second path, A and B are first broken down into their elements, each in its standard reference state. Subsequently, the elements are combined to form C and D. The enthalpy ~ ~ change for the second route is ∆ r H ~ = g i ni ∆ f H i,products  g i 0 ni 0 ∆ f H i,reactants = ~ g ni ∆ f H i . Because H is a state function, the enthalpy change is the same for both paths. i
[ [ nC DfH C 1nD DfH D
Elements in standard reference state
Figure 4.2
Two different pathways between the same reactants and products. The value of ∆H for both paths is the same because they both connect the same initial and final states. Equation (4.8) is a consequence of this conclusion.
This is stated in mathematical form in Equation (4.8). Writing ∆ r H ~ in terms of formation enthalpies is a great simplification over compiling measured values of reaction enthalpies. Standard formation enthalpies for atoms and inorganic compounds at 298.15 K are listed in Table 4.1, and standard formation enthalpies for organic compounds are listed in Table 4.2 (Appendix A, Data Tables). Another thermochemical convention involves the calculation of enthalpy changes of electrolyte solutions. The solution reaction that occurs when a salt such as NaCl is dissolved in water is NaCl1s2 ¡ Na+ 1aq2 + Cl 1aq2
Because it is not possible to form only positive or negative ions in solution, the measured enthalpy of solution of an electrolyte is the sum of the enthalpies of all anions and cations formed. To be able to tabulate values for enthalpies of formation of individual ions, the enthalpy for the following reaction is set equal to zero at P = 1 bar for all temperatures: 1>2 H21g2 ¡ H + 1aq2 + e 1metal electrode2
In other words, solution enthalpies of formation of ions are measured relative to that for H + 1aq2. The thermodynamics of electrolyte solutions will be discussed in detail in Chapter 10.
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4.3 Hess’s Law Is Based on Enthalpy Being a State Function
91
As the previous discussion shows, only the ∆ f H ~ of each reactant and product is needed to calculate ∆ r H ~ . Each ∆ f H ~ is a difference in enthalpy between the compound and its constituent elements, rather than an absolute enthalpy. However, there is a convention that allows absolute enthalpies to be specified using the experimentally determined values of the ∆ f H ~ of compounds. In this convention, the absolute enthalpy of each pure element in its standard reference state is set equal to zero. With this convention, the absolute molar enthalpy of any chemical species in its standard reference state H m~ is equal to ∆ f H ~ for that species. To demonstrate this convention, recall the reaction in Equation (4.4):
∆ r H ~ = ∆ f H ~ 1Fe 3O4, s2 = H m~ 1Fe 3O4, s2  3H m~ 1Fe, s2  2H m~ 1O2, g2 (4.4)
Setting H m~ = 0 for each element in its standard reference state,
∆ f H ~ 1Fe 3O4, s2 = H m~ 1Fe 3O4, s2  3 * 0  2 * 0 = H m~ 1Fe 3O4, s2
(4.9)
The value of ∆ r H ~ for any reaction involving compounds and elements is unchanged by this convention. In fact, one could choose a different number for the absolute enthalpy of each pure element in its standard reference state, and it would still not change the value of ∆ r H ~ . However, it is much more convenient (and easier to remember) if one sets H m~ = 0 for all elements in their standard reference state. This convention will be used again in Chapter 6 when chemical potential is discussed.
4.3 HESS’S LAW IS BASED ON ENTHALPY BEING A STATE FUNCTION
As discussed in the previous section, it is extremely useful to tabulate values of ∆ f H ~ for chemical compounds at one fixed combination of P and T. Tables 4.1 and 4.2 list this data for 1 bar and 298.15 K. With access to these values of ∆ f H ~ , ∆ r H ~ can be calculated for all reactions among these elements and compounds at 1 bar and 298.15 K. But how is ∆ f H ~ determined? Consider the formation reaction for C2H61g2: 2 C1graphite2 + 3 H21g2 ¡ C2H61g2
(4.10)
Graphite is the standard reference state for carbon at 298.15 K and 1 bar because it is slightly more stable than diamond under these conditions. However, it is unlikely that one would obtain only ethane if the reaction were carried out as written. Given this experimental hindrance, how can ∆ f H ~ for ethane be determined? To determine ∆ f H ~ for ethane, we take advantage of the fact that ∆H is path independent. In this context, path independence means that the enthalpy change for any sequence of reactions that sum to the same overall reaction is identical. This statement is known as Hess’s law. Therefore, one is free to choose any sequence of reactions that leads to the desired outcome. Combustion reactions are well suited for these purposes because in general they proceed rapidly, go to completion, and produce only a few products. To determine ∆ f H ~ for ethane, one can carry out the following combustion reactions: C2H61g2 + 7>2 O21g2 ¡ 2 CO21g2 + 3 H2O1l2
∆ r H I~
(4.11)
∆ r H II~
(4.12)
H21g2 + 1>2 O21g2 ¡ H2O1l2
~ ∆ r H III
(4.13)
2 * 3 C1graphite2 + O21g2 ¡ CO21g24
2∆ r H II~
(4.14)
2 CO21g2 + 3 H2O1l2 ¡ C2H61g2 + 7>2 O21g2
 ∆ r H I~
(4.15)
~ 3∆ r H III
(4.16)
2 C1graphite2 + 3 H21g2 ¡ C2H61g2
~ 2∆ r H II~  ∆ r H I~ + 3∆ r H III
C1graphite2 + O21g2 ¡ CO21g2
Concept Hess’s law states that the enthalpy change for any sequence of reactions that sum to the same overall reaction has the same value.
These reactions are combined in the following way to obtain the desired reaction:
3 * 3 H21g2 + 1>2 O21g2 ¡ H2O1l24
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CHAPTER 4 Thermochemistry
We emphasize again that it is not necessary for these reactions to be carried out at 298.15 K. The reaction vessel is immersed in a water bath at 298.15 K, and the combustion reaction is initiated. If the temperature in the vessel rises during the course of the reaction, the heat flow that restores the system and surroundings to 298.15 K after completion of the reaction is measured, allowing ∆ r H ~ to be determined at 298.15 K. Several points should be made about enthalpy changes in relation to balanced overall equations describing chemical reactions. First, because H is an extensive function, multiplying all stoichiometric coefficients with any number changes ∆ r H ~ by the same factor. Therefore, it is important to know which set of stoichiometric coefficients has been assumed if a numerical value of ∆ r H ~ is given. Second, because the units of ∆ f H ~ for all compounds in the reaction are kJ mol1, the units of the reaction enthalpy ∆ r H ~ are also kJ mol1. One might pose the question “per mole of what?” given that all the stoichiometric coefficients may differ from each other and from the value of one. The answer to this question is per mole of the reaction as written. Doubling all stoichiometric coefficients doubles ∆ r H ~ .
EXAMPLE PROBLEM 4.1 The average bond enthalpy of the O —H bond in water is defined as onehalf of the enthalpy change for the reaction H2O1g2 ¡ 2 H1g2 + O1g2. The formation enthalpies, ∆ f H ~ , for H1g2 and O1g2 are 218.0 and 249.2 kJ mol1, respectively, at 298.15 K, and ∆ f H ~ for H2O1g2 is 241.8 kJ mol1 at the same temperature. a. Use this information to determine the average bond enthalpy of the O —H bond in water at 298.15 K. b. Determine the average bond energy ∆U of the O —H bond in water at 298.15 K. Assume ideal gas behavior.
Solution a. We consider the sequence H2O1g2 ¡ H21g2+1>2 O21g2
∆ r H ~ = 241.8 kJ mol1
1>2 O21g2 ¡ O1g2
∆ r H ~ = 249.2 kJ mol1
H21g2 ¡ 2 H1g2
∆ r H ~ = 2 * 218.0 kJ mol1
H2O1g2 ¡ 2 H1g2 + O1g2
∆ r H ~ = 927.0 kJ mol1
This is the enthalpy change associated with breaking both O —H bonds under standard conditions. We conclude that the average bond enthalpy of the O —H bond in water is 1 * 927.0 kJ mol1 = 463.5 kJ mol1. We emphasize that this is the average value 2 because the values of ∆ r H ~ for the reactions H2O1g2 ¡ H1g2 + OH1g2 and OH1g2 ¡ O1g2 + H1g2 differ. b. ∆U ~ = ∆H ~  ∆1PV2 = ∆H ~  ∆nRT = 927.0 kJ mol1  2 * 8.314 J mol1K1 * 298.15 K = 922.0 kJ mol1
1 * 922.0 kJ mol1 = 2 461.0 kJ mol1. The bond energy and the bond enthalpy are nearly identical.
The average value for ∆U ~ for the O —H bond in water is
Example Problem 4.1 shows how bond energies can be calculated from reaction enthalpies. The value of a bond energy is of particular importance for chemists in estimating the thermal stability of a compound as well as its stability with respect to reactions with other molecules. Values of bond energies tabulated in the format of the periodic table together with the electronegativities are shown in Table 4.3. (For more details, see the article by N. K. Kildahl cited in Further Reading.) The value of the
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4.4 Temperature Dependence of Reaction Enthalpies
93
TABLE 4.3 Mean Bond Energies 1
2
H,2.20 432 432 459 565
Selected Bond Energies (kJ/mol)
14
15
16
17
18 He
B,2.04
293 389 536,636 613
C,2.55 346 602,835 411 358,799 485
N,3.04 167 418,942 386 201,607 283
O,3.44 142 494 459 142,494 190
F,3.98
Ne
Na,0.93 Mg,1.31 72 197 477
129 ,377 513
Al,1.61 272 583
Si,1.90 222 318 318 452,640 565
P,2.19
S,2.58
Cl,3.16 240 428 218 249
Ar
K,0.82
Sr,1.00 105 ,460 550
Ga,1.81 Ge,2.01 As,2.18 Se,2.55 Br,2.96 Kr
Li,0.98 105 243 573
49 180 490
Be,1.57
13
208 ,444 632
2O21g2 S CaO1s2 635.1 CaO1s2 + H2O1l2 S Ca1OH221s2 65.2
The standard enthalpies of combustion of graphite and C2H21g2 are 393.51 and 1299.58 kJ mol1, respectively. Calculate the standard enthalpy of formation of CaC21s2 at 25°C. P4.14 From the following data at 298.15 K, as well as data in Table 4.1, calculate the standard enthalpy of formation of H2S1g2 and of FeS21s2:
∆ r H ~ 1 kJ mol −1 2
Fe1s2 + 2H2S1g2 S FeS21s2 + 2H21g2 H2S1g2 + 3>2O21g2 S H2O1l2 + SO21g2
137.0 562.0
P4.15 From the following data at 298.15 K, calculate the standard enthalpy of formation of FeO1s2 and of Fe 2O31s2:
∆ r H ~ 1 kJ mol −1 2
Fe 2O31s2 + 3C1graphite2 S 2Fe1s2 + 3CO1g2 492.6 FeO1s2 + C1graphite2 S Fe1s2 + CO1g2 155.8 C1graphite2 + O21g2 S CO21g2 393.51 CO1g2 + 1>2O21g2 S CO21g2 282.98
Calculate the standard enthalpy of formation of FeO1s2 and of Fe 2O31s2. P4.16 Use the average bond energies in Table 4.3 to estimate ∆ rU ~ for the reaction C2H41g2 + H21g2 S C2H61g2. Also calculate ∆ rU ~ from the tabulated values of ∆ f H ~ for reactant and products (Appendix A, Data Tables). Calculate the percent error in estimating ∆U R~ from the average bond energies for this reaction. P4.17 Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of formation of CO21g2 and H2O1l2 to determine ∆ f H ~ for benzene. P4.18 Compare the heat evolved at constant pressure per mole of oxygen in the combustion of sucrose 1C12H22O112 and palmitic acid 1C16H32O22 with the combustion of a typical protein, for which the empirical formula is C4.3H6.6NO. Assume for the protein that the combustion yields N21g2, CO21g2, and H2O1l2. Assume that the enthalpies for combustion of sucrose, palmitic acid, and a typical protein are 5647 kJ mol1, 10,035 kJ mol1, and 22.0 kJ g 1, respectively. Based on these calculations, determine the average heat evolved per mole of oxygen consumed, assuming combustion of equal moles of sucrose, palmitic acid, and protein. Section 4.4 P4.19 At 1000. K, ∆ r H ~ = 123.77 kJ mol1 for the reaction N21g2 + 3H21g2 S 2NH31g2, with CP, m = 3.502R, 3.466R,
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105
Numerical Problems
and 4.217R for N21g2, H21g2, and NH31g2, respectively. From this information, calculate ∆ f H ~ of NH31g2 at 375 K. Assume that the heat capacities are independent of temperature. P4.20 Calculate ∆ f H ~ for NO1g2 at 590. K assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at 298.15 K. P4.21 Derive a formula for ∆ r H ~ for the reaction CO1g2 + 1>2O21g2 S CO21g2 assuming that the heat capacities of reactants and products do not change with temperature. P4.22 Calculate the standard enthalpy of formation of FeS21s2 at 550.°C from the data tables and the following data at 298.15 K. Assume that the heat capacities are independent of temperature. Substance 1
∆ f H 1kJ mol 2 ~
Fe1 s 2 FeS2 1 s 2 Fe 2O3 1s 2 S(rhombic) SO2 1g2  824.2
CP, m >R
3.02
7.48
296.81
2.72
You are also given that for the reaction 2FeS21s2 + 11>2O21g2 S Fe 2O31s2 + 4SO21g2, ∆ r H ~ = 1655 kJ mol1.
P4.23 At 298 K, ∆ r H ~ = 131.28 kJ mol1 for the reaction C1graphite2 + H2O1g2 S CO1g2 + H21g2, with CP, m = 8.53, 33.58, 29.12, and 28.82 J K1 mol1 for graphite, H2O1g2, CO1g2, and H21g2, respectively. From this information, calculate ∆ r H ~ at 195°C. Assume that the heat capacities are independent of temperature. P4.24 Consider the reaction TiO21s2 + 2 C1graphite2 + 2 Cl21g2 S 2 CO1g2 + TiCl41l2 for which ∆ r H ~ = 80.0 kJ mol1. Given the following data at 25°C, a. Calculate ∆ r H ~ at 135.8°C, the boiling point of TiCl4. b. Calculate ∆ f H ~ for TiCl41l2 at 25°C. TiO2 1 s 2 Cl2 1 g2 C(graphite) CO1 g2 TiCl4 1l2
Substance
∆ f H ~ 1kJ mol12 1
CP, m 1J K
 944
1
mol 2 55.06
110.5
33.91
8.53
29.12
145.2
Assume that the heat capacities are independent of temperature. P4.25 From the following data, calculate ∆ r H ~ 1391.4 K2 for the reaction CH3COOH1g2 + 2O21g2 S 2H2O1g2 + 2CO21g2: ∆ r H ~ 1 kJ mol −1 2 CH3COOH1l2 + 2O21g2 S 2H2O1l2 + 2CO21g2 871.5 H2O1l2 S H2O1g2 40.656 CH3COOH1l2 S CH3COOH1g2 24.4
Values for ∆ r H ~ for the first two reactions are at 298.15 K, and for the third reaction at 391.4 K. Substance CP, m >R
CH3COOH1 l2 O2 1 g2 CO2 1g2 H2O1l2 14.9
3.53
4.46
9.055
H2O1g2 4.038
P4.26 Calculate ∆ r H ~ at 520. K for the reaction 4 NH31g2 + 6 NO1g2 S 5 N21g2 + 6 H2O1g2 using the temperature dependence of the heat capacities from the data tables. Compare your result with ∆ r H ~ at 298.15 K. Is the difference large or small? Why?
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P4.27 Given the following heat capacity data, calculate ∆ f H ~ of CO21g2 at 685 K. Substance
CP, m >J mol1 K1
C(graphite) 8.52
O2 1 g2 28.8
CO2 1g2
37.1
P4.28 Calculate ∆H for the process in which Cl21g2 initially at 298.15 K at 1 bar is heated to 585 K at 1 bar. Use the temperaturedependent heat capacities in the data tables. How large is the relative error if the molar heat capacity is assumed to be constant at its value of 298.15 K over the temperature interval? Section 4.5 P4.29 A sample of K1s2 of mass 3.041 g undergoes combustion in a constant volume calorimeter. The calorimeter constant is 1849 J K1, and the measured temperature rise in the inner water bath containing 1450. g of water is 1.776 K. Calculate ∆ f U ~ and ∆ f H ~ for K2O. P4.30 Calculate ∆ c H ~ and ∆ cU ~ for the oxidation of benzene (g). Also calculate ∆ c H ~  ∆ cU ~ ∆c H ~ P4.31 Benzoic acid of mass 1.35 g is reacted with oxygen in a constant volume calorimeter to form H2O1l2 and CO21g2 at 298 K. The mass of the water in the inner bath is 1.55 * 103 g. The temperature of the calorimeter and its contents rises 2.76 K as a result of this reaction. Calculate the calorimeter constant. P4.32 A sample of Na 2SO41s2 is dissolved in 225 g of water at 298 K such that the solution is 0.288 molar in Na 2SO4. A temperature rise of 0.129°C is observed. The calorimeter constant is 330. J K1. Calculate the enthalpy of solution of Na 2SO4 in water at this concentration. Compare your result with that calculated using the data in Table 4.1 (Appendix A, Data Tables). P4.33 If 4.206 g of ethanol, C2H5OH1l2 is burned completely in a bomb calorimeter at 298.15 K, the heat produced is 124.34 kJ. a. Calculate ∆ c H ~ for ethanol at 298.15 K. b. Calculate ∆ f H ~ of ethanol at 298.15K. P4.34 A 0.1567 g sample of sucrose, C12H22O11, is burned in a bomb calorimeter at 298 K. In order to produce the same temperature rise in the calorimeter as the reaction, 2582 J must be expended. a. Calculate ∆ cU ~ and ∆ c H ~ for the combustion of 1 mole of sucrose. b. Using the data tables and your answer to (a), calculate ∆ f H ~ for sucrose. c. The rise in temperature of the calorimeter and its contents as a result of the reaction is 1.743 K. Calculate the heat capacity of the calorimeter and its contents. P4.35 Calcium chloride is soluble in water according to the reaction CaCl21s2 S Ca2+1aq2 + 2Cl1aq2. Assuming that no heat is transferred to the surroundings, calculate the change in temperature when 12.5 g of CaCl2 are dissolved in 150. g of water.
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CHAPTER 4 Thermochemistry
CP(T)
Section 4.6 P4.36 The following data are a DSC scan of a solution of a T4 lysozyme mutant. From the data, determine Tm. Determine also the excess heat capacity ∆CP at T = 308 K. In addition, trs determine the intrinsic dC int P and transition ∆C P excess heat capacities at T = 308 K. In your calculations use the extrapolated curves, shown as dotted lines in the DSC scan, where the y axis shows CP.
P4.37 Using the protein DSC data in Problem 4.36, calculate the enthalpy change between T = 288 K and T = 318 K. Give your answer in units of kJ per mole. Assume the molar mass of the protein is 14000. grams per mole. Hint: You can perform the integration of the heat capacity by estimating the area under the DSC curve and above the baseline in Problem 4.36. This can be done by dividing the area into small rectangles and summing the areas of the rectangles. Comment on the accuracy of this method.
0.418 J K21g21
288
298 308 Temperature (K)
318
FURTHER READING Höhne, G. W., Hemminger, W. F., and Flammersheim, H.J. Differential Scanning Calorimetry, 2nd Edition. Berlin: Springer, 2003. Jacobson, N. “Use of Tabulated Thermochemical Data for Pure Compounds.” Journal of Chemical Education 78 (2001): 814–819.
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Kildahl, N. K. “Bond Energy Data Summarized.” Journal of Chemical Education 72 (1995): 423– 424.
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C H A P T E R
5
Entropy and the Second and Third Laws of Thermodynamics
5.1 What Determines the Direction of Spontaneous Change in a Process?
5.2 The Second Law of Thermodynamics, Spontaneity, and the Sign of ∆S
5.3 Calculating Changes in Entropy as T, P, or V Change
5.4 Understanding Changes in Entropy at the Molecular Level
WHY is this material important?
5.5 The Clausius Inequality
Mixtures of chemically reactive species can evolve toward reactants or toward products, defining the direction of spontaneous change. How do we know which of these reactions is spontaneous? Entropy, designated by S, is the state function that predicts the direction of natural, or spontaneous, change. In this chapter, you will learn how to carry out entropy calculations.
5.6 The Change of Entropy in the Surroundings and ∆Stot = ∆S + ∆Ssur
5.7 Absolute Entropies and the Third Law of Thermodynamics
5.8 Standard States in Entropy
WHAT are the most important concepts and results? Heat flows from hotter bodies to colder bodies, and gases mix rather than separate. Entropy always increases for a spontaneous change in an isolated system. For a spontaneous change in a system interacting with its environment, the sum of the entropy of the system and that of the surroundings always increases. Thermodynamics can be used to devise ways to reduce our energy use and to transition away from fossil fuels.
WHAT would be helpful for you to review for this chapter? Because differential and integral calculus and partial derivatives are used extensively in this chapter, it would be helpful to review the material on calculus in Math Essential 2 and 3.
Calculations
5.9 Entropy Changes in Chemical Reactions
5.10 Heat Engines and the Carnot Cycle
5.11 (Supplemental Section) How Does S Depend on V and T ?
5.12 (Supplemental Section) Dependence of S on T and P
5.13 (Supplemental Section) Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
5.1 WHAT DETERMINES THE DIRECTION OF
SPONTANEOUS CHANGE IN A PROCESS?
The first law of thermodynamics constrains the range of possible processes to those that conserve energy and rules out machines that work indefinitely without an energy source (perpetual motion machines) that have been the dream of inventors over centuries. Any system that is not at equilibrium will evolve with time. We refer to the direction of evolution as the spontaneous direction. Spontaneous does not mean that the process occurs immediately, but rather that it will occur with high probability if any barrier to the change is overcome. For example, the transformation of a piece of wood to CO2 and H2O in the presence of oxygen is spontaneous, but it only occurs at elevated temperatures because an activation energy barrier must be overcome for the reaction to proceed. In this chapter, we discuss the second law of thermodynamics, which allows us to predict if a process is spontaneous. Before stating the second law, we discuss two processes that satisfy the first law because they conserve energy and attempt to identify criteria that can predict the spontaneous direction of evolution.
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Concept A process is defined to be spontaneous even if it occurs only for some conditions because of an energy barrier.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Energy uptake by floor
Figure 5.1
Path of a bouncing rubber ball. The bouncing ball rebounds each time with a smaller amplitude. The green semicircles in the floor indicate energy uptake by the floor.
qP T1
T2
Figure 5.2
Adjacent metal rods of differing temperatures. The rods, for which T1 7 T2, differ in temperature by ∆T, are placed in thermal contact at constant pressure. This composite system is contained in a rigid adiabatic enclosure (not shown) and is, therefore, an isolated system.
Consider the rubber ball shown in Figure 5.1 that falls in a gravitational field and bounces upward after colliding with the floor. We assume for simplicity that the ball and the floor consist of the same material. After the initial drop, the ball falls again and bounces back a number of times. We know from experience that each successive bounce will be lower than the previous one. How do we explain this observation? Because of its kinetic energy, the falling ball has excess energy relative to the floor. In an inelastic collision with the floor, the net energy transfer is to the lowerenergy floor and not to the higherenergy ball. Because the ball has lost energy, each successive bounce does not attain the height of the previous bounce. What would it take for each successive bounce to be higher? This would require that there be a decrease of floor energy in the region where the ball hits and a corresponding increase in the energy of the ball in such a way that the first law is not violated. In this case, the collision partner with higherenergy gains rather than loses energy. Experience tells us that this process is not spontaneous. We generalize this result and conclude that in a collision, excess energy is redistributed to the lowerenergy collision partner, and not to the higherenergy collision partner. This conclusion leads us to the following insight: As a system evolves toward equilibrium in the spontaneous direction, energy becomes more evenly distributed in space rather than being concentrated in a small region. In Figure 5.2, we show a second process in which a hot metal rod is put in contact with a cold metal rod, giving rise to a temperature difference between the two ends. What happens to the temperature difference as the system evolves toward equilibrium? Our experience is that the hot end becomes colder and the cold end becomes warmer until the temperature difference vanishes. We never observe that the hot end continues to become hotter while the cold end becomes colder, even though such a process could still obey the first law. As for the previous example, in the spontaneous direction, energy flows from the hot to the cold rod until the energy is distributed uniformly in space. These two examples suggest that the driving force in a spontaneous process is spreading out energy that occurs through heat flow rather than through work. We would like to identify a new thermodynamic function that allows us to predict the direction of spontaneous change. It should be a state function, so that the prediction depends only on the initial and final states associated with the process and not on the path between them. Because heat flow seems to be a key factor, we write the first law in the following form for an ideal gas undergoing changes in V and T in a reversible process. dqrev = dU  dwrev
= CV1T2dT + PdV nRT = CV1T2dT + dV V
(5.1)
Recall from Math Essential 3 that the criterion for an exact differential is that the mixed second partial derivatives are equal. Because a
0CV1T2 0V
b ≠ a T
01nRT>V2 0T
b V
(5.2)
dqrev is not an exact differential. However, if we divide both sides of Equation (5.1) by T, CV1T2 dqrev nR = dT + dV T T V
(5.3)
Tƒ Vƒ CV1T2 dqrev nR ∆S1V, T2 = dS = = dT + dV T T L L LTi LVi V
(5.4)
Concept dqrev is the differential form of a state T function that we call entropy.
dqrev is an exact differential1 because 103 CV1T2>T 4 >0V2T = T 101nR>V2>0T2V = 0. This means that there is a state function that we call entropy, dqrev which we denote by S for which the differential form is . For our example, the T change in S is given by we see that
1
Although we have assumed ideal gas behavior, the conclusion that dqrev >T is an exact differential is more general. See Lee et al. (2015) in Further Reading for details.
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5.2 THE SECOND LAW OF THERMODYNAMICS, SPONTANEITY, AND THE SIGN OF ∆S
109
If Tƒ  Ti is not large, we can assume that CV is constant over the temperature interval and can remove it from the integral to obtain ∆S1V, T2 = CV ln
Tƒ Ti
+ nR ln
Vƒ Vi
(5.5)
for the change in entropy of an ideal gas undergoing a change in volume and temperature.
5.2 THE SECOND LAW OF THERMODYNAMICS, SPONTANEITY, AND THE SIGN OF 𝚫S
We have shown that dqrev >T is an exact differential, and we have defined a state function S related to the exact differential. However, we have not yet made a link between ∆S and spontaneity. We now do so by considering two processes—heat flow in the metal rod shown in Figure 5.2 and the collapse of an ideal gas to onehalf its volume, in a quantitative manner. The first process concerns the spontaneous direction of change in a metal rod with a temperature gradient that we considered in the previous section. To model this process, consider the isolated composite system considered in Section 5.1. Two systems, in the form of identical metal rods with different temperatures T1 7 T2, are brought into thermal contact. In the following discussion, we assume a very small amount of heat dqP is withdrawn from one rod and deposited in the other rod under constant pressure conditions. Because dqP is very small, we can assume that T1 and T2 remain constant. To calculate dS for this irreversible process, we must imagine a reversible process in which the initial and final states are the same as for the irreversible process because dS = dqrev >T. For example, in an imaginary reversible process, the hotter rod is coupled to a reservoir whose temperature is lowered very slowly. The temperatures of the rod and the reservoir differ only infinitesimally throughout the process in which an amount of heat, dqP, is withdrawn from the rod. The decrease in temperature of the rod, dT, is related to qP by
dqP = CPdT
(5.6)
It has been assumed that dT = T1  T2 V T2 and T1. Because the path is defined (constant pressure), dqP is independent of how rapidly the heat is withdrawn (the path). More formally, because dqP = dH and because H is a state function, dqP is independent of the path between the initial and final states. Therefore, dqP = dqrev if the temperature increment dT is identical for the reversible and irreversible processes. We next couple the second rod at the lower temperature T2 to a heat reservoir and raise the reservoir temperature slowly by dT. We deposit the same amount of heat in the second rod that was withdrawn from the first rod. Next, we calculate the entropy change for this process in which heat flows from one rod to the other. Keep in mind that although we are considering a reversible process, S is a state function. Therefore, if the initial and final states are the same, ∆S has the same value for any process, whether reversible or irreversible. Because the composite system is isolated, dq1 + dq2 = 0, and dq1 = dq2 = dqP . The entropy change of the composite system is the sum of the entropy changes in each rod: dS =
dqrev, 1 T1
+
dqrev, 2 T2
=
Concept System entropy increases for the direction of spontaneous change in an isolated system.
dq1 dq2 T2  T1 1 1 + = dqP a  b = dqP a b (5.7) T1 T2 T1 T2 T2
Because T1 7 T2, the quantity in the parentheses is negative. We calculate ∆S for the two possible directions:
• If heat flows from the hotter to the colder rod, the temperature difference becomes •
smaller. dqP 6 0 and dS 7 0. If heat flows from the colder to the hotter rod, the temperature difference becomes larger. dqP 7 0 and dS 6 0.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Ti,Vi Ti,1/2 Vi Initial state
Final state
(a) Irreversible process
Piston
Ti,Vi
Piston Ti,1/2 Vi
Initial state Final state (b) Reversible isothermal compression
Figure 5.3
Two processes in which a confined gas is reduced to onehalf of its initial volume. (a) An irreversible process is shown in which an ideal gas confined in a container with rigid adiabatic walls is spontaneously reduced to onehalf its initial volume. (b) A reversible isothermal compression in a piston and cylinder assembly is shown for the same initial and final states as in the irreversible process in part (a). Reversibility is achieved by adjusting the rate at which the beaker on top of the piston is filled with water relative to the evaporation rate.
Note that dS and dqP have the same magnitude, but different signs, for the two directions of change. This shows that S is a useful function for measuring the direction of spontaneous change in an isolated system because it gives us different values for the two possible directions. All experimental evidence tells us that heat flows from the hotter to the colder body and the temperature gradient becomes smaller with time. In the absence of constraints, systems evolve to distribute energy uniformly throughout space. It can be concluded from this example that the process in which ∆S 7 0 is the direction of spontaneous change in an isolated system. Next, imagine a second process in which an ideal gas collapses to onehalf its initial volume without a force acting on it. This process and its reversible analog are shown in Figure 5.3. Recall that U is independent of V for an ideal gas. Because U does not change as V decreases, and U is a function of T only for an ideal gas, the temperature remains constant in the irreversible process. Therefore, the irreversible process shown in Figure 5.3a is both adiabatic and isothermal and is described by Vi, Ti S 11>2Vi2, Ti. From the ideal gas law, we know that the final pressure is twice the initial pressure. The imaginary reversible process that we use to carry out the calculation of ∆S is shown in Figure 5.3b. In this process, which must have the same initial and final states as the irreversible process, water is slowly and continuously added to the beaker on the piston to ensure that P = Pext throughout the process. The ideal gas undergoes a reversible isothermal transformation described by Vi, Ti S 11>2Vi2, Ti. Because ∆U = 0, q = w. We calculate ∆S for this process:
∆S =
1 qrev dqrev wrev 2 Vi = = = nR ln = nR ln 2 6 0 Ti Ti Vi L T
(5.8)
Experience tells us that the opposite process, in which the gas spontaneously expands so that it occupies twice the volume, is the spontaneous direction of change. For this case
∆S = nR ln
2Vi = nR ln 2 7 0 Vi
(5.9)
Again, we find that the process with ∆S 7 0 is the direction of spontaneous change in this isolated system. We can also understand this process in terms of a redistribution of energy; the kinetic energy of the gas molecules is spatially spread out by the gas uniformly filling the volume. The results obtained for the two isolated systems that we have discussed are generalized in the following statement of the second law of thermodynamics: For any process that proceeds in an isolated system, there is a unique direction of spontaneous change and ∆S 7 0 for the spontaneous process. ∆S 6 0 for the opposite or nonspontaneous direction of change, and ∆S = 0 only for a reversible process. In a reversible process, there is no direction of spontaneous change because the system is always in equilibrium. We cannot emphasize too strongly that ∆S 7 0 is a criterion for spontaneous change only if the system does not exchange energy in the form of heat or work with its surroundings. Note that if any process occurs in the isolated system, it is by definition spontaneous and the entropy increases. Whereas U can neither be created nor destroyed, S for a process occurring in an isolated system is created 1∆S 7 02 but never destroyed 1∆S 6 02.
5.3 CALCULATING CHANGES IN ENTROPY AS T, P, OR V CHANGE
The most important thing to remember in carrying out entropy calculations is that ∆S must be calculated along a reversible path because it is defined by 1 1dqrev >T2. For an irreversible process, ∆S must be calculated for a reversible process that proceeds
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5.3 CALCULATING CHANGES IN ENTROPY AS T, P, OR V CHANGE
111
between the same initial and final states as the irreversible process. Because S is a state function, ∆S is path independent, provided that the transformation is between the same initial and final states in both processes. Therefore, your calculation for the reversible path also holds for the irreversible process. We first consider two processes that require no calculation. First, for any reversible adiabatic process, qrev = 0, so that ∆S = 1 1dqrev >T2 = 0. Second, for any cyclic process, ∆S = A 1dqrev >T2 = 0, because the change in any state function for a cyclic process is zero. Next, consider ∆S for the reversible isothermal expansion or compression of an ideal gas, described by Vi, Ti S Vƒ, Ti. Because ∆U = 0 for this process, as was discussed in Section 2.5,
qrev = wrev = nRT ln
∆S =
Vƒ
(5.10)
Vƒ dqrev 1 = qrev = nR ln T Vi L T
(5.11)
Vi
and
Note that ∆S 7 0 for an expansion 1Vƒ 7 Vi2 and ∆S 6 0 for a compression 1Vƒ 6 Vi2. Next, consider ∆S for an ideal gas that undergoes a reversible change in T at constant V or P. For a reversible process described by Vi, Ti S Vi, Tƒ, it also holds that dqrev = CV dT, and furthermore that
∆S =
Tƒ nCV,m dT dqrev = ≈ nCV,m ln T Ti L T L
(5.12)
For a constant pressure process described by Pi, Ti S Pi, Tƒ, it also holds that dqrev = CP dT, and furthermore that
∆S =
Tƒ nCP,m dT dqrev = ≈ nCP,m ln T Ti L T L
(5.13)
The last expressions in Equations (5.12) and (5.13) are valid if the temperature interval is small enough that the temperature dependence of CV,m and CP,m can be neglected. Again, although expressions for ∆S have been derived for a reversible process, Equations (5.12) and (5.13) hold for any reversible or irreversible process between the same initial and final states for an ideal gas. The results of the last three derived expressions can be combined in the following way. Because the macroscopic variables V, T or P, T completely define the state of an ideal gas, any change Vi, Ti S Vƒ, Tƒ can be separated into two segments, Vi, Ti S Vƒ, Ti and Vƒ, Ti S Vƒ, Tƒ. A similar statement about P and T is also valid. Because ∆S is independent of the path, any reversible or irreversible process for an ideal gas described by Vi, Ti S Vƒ, Tƒ can be treated as consisting of two segments, one of which occurs at constant volume and the other at constant temperature. For this twostep process, ∆S is given by
∆S = nR ln
Vƒ
Tƒ + nCV,m ln Vi Ti
(5.14)
This result was previously obtained in Equation (5.5). Similarly, for any reversible or irreversible process for an ideal gas described by Pi, Ti S Pƒ, Tƒ as shown in Example Problem 5.1.
∆S = nR ln
Pƒ
Tƒ + nCP,m ln Pi Ti
Concept Because entropy is a state function, ∆S is the same for a reversible and irreversible process between the same initial and final states.
(5.15)
In writing Equations (5.14) and (5.15), it has been assumed that the temperature dependence of CV,m and CP,m can be neglected over the temperature range of interest.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
EXAMPLE PROBLEM 5.1 Using the equation of state and the relationship between CP,m and CV,m for an ideal gas, show that Equation (5.14) can be transformed into Equation (5.15).
Solution ∆S = nR ln
Vƒ Vi
= nR ln
Tƒ
+ nCV,m ln
Pƒ Pi
Ti
= nR ln
+ n1CV,m + R2ln
Tƒ Ti
Tƒ Pi Ti Pƒ
+ nCV,m ln
= nR ln
Pƒ Pi
Tƒ Ti
+ nCP,m ln
Tƒ Ti
Next consider ∆S for phase changes. Experience shows that a liquid is converted to a gas at a constant boiling temperature through heat input if the process is carried out at constant pressure. Because qP = ∆H, ∆S for the reversible process is given by ∆ vap S =
∆ vap H dqrev qrev = = Tvap Tvap L T
(5.16)
Similarly, for the phase change solid S liquid, ∆ fus S =
Concept Even if the equation of state is not known, ∆S can be calculated if a and kT are known.
dqrev qrev ∆ fus H = = Tfus Tfus L T
(5.17)
Finally, consider ∆S for an arbitrary process involving real gases, solids, and liquids for which the volumetric thermal expansion coefficient a and the isothermal compressibility kT are known, but the equation of state is not known. The calculation of ∆S for such processes is described in Supplemental Section 5.11 and 5.12, in which the properties of S as a state function are fully exploited. The results are stated here. For the system undergoing the change Vi, Ti S Vƒ, Tƒ, Tƒ
Vƒ
Tƒ CV a a ∆S = dT + dV = CV ln + 1Vƒ  Vi2 k k T T T T L L i
Ti
(5.18)
Vi
The last step in Equation (5.18) is only valid if CV, kT , and a are constant over the temperature and volume intervals of interest, which is only the case for a liquid or solid. For the system undergoing a change Pi, Ti S Pƒ, Tƒ, Tƒ
Pƒ
CP ∆S = dT Va dP L T L
Ti
(5.19)
Pi
For a solid or liquid, the last equation can be simplified to ∆S = CP ln
Tƒ Ti
 Va1Pƒ  Pi2
(5.20)
if CP, V, and a are assumed constant over the temperature and pressure intervals of interest. The integral forms of Equations (5.18) and (5.20) are valid for ideal and real gases, liquids, and solids. Calculations using these equations are presented in Example Problems 5.2 through 5.4.
EXAMPLE PROBLEM 5.2 One mole of CO gas is transformed from an initial state characterized by Ti = 320. K and Vi = 80.0 L to a final state characterized by Tƒ = 650. K and Vƒ = 120.0 L. Assuming ideal gas behavior, calculate ∆S for this process. For CO, CV,m J mol1 K1
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= 31.08  0.01452
T T2 T3 + 3.1415 * 105 2  1.4973 * 108 3 K K K
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5.3 CALCULATING CHANGES IN ENTROPY AS T, P, OR V CHANGE
113
Solution Consider the following reversible process in order to calculate ∆S. The gas is first heated reversibly from 320. to 650. K at a constant volume of 80.0 L. Subsequently, the gas is reversibly expanded at a constant temperature of 650. K from a volume of 80.0 L to a volume of 120.0 L. The entropy change for this process is obtained using Equation (5.14). Tƒ
∆S =
Vƒ CV dT + nR ln T Vi L Ti
∆S1J K12 =
650. K
1.00 mol *
L
320. K
a31.08  0.01452
T T2 T3 + 3.1415 * 105 2  1.4973 * 108 3 b K K K dT T
+ 1.00 mol * 8.314 J K1 mol1 * ln
120.0 L 80.0 L
= 22.024 J K1  4.792 J K1 + 5.027 J K1  1.207 J K1 + 3.371 J K1
= 24.4 J K1
EXAMPLE PROBLEM 5.3 In this problem, 2.50 mol of CO2 gas is transformed from an initial state characterized by Ti = 450. K and Pi = 1.35 bar to a final state characterized by Tƒ = 800. K and Pƒ = 3.45 bar. Calculate ∆S for this process using Equation (5.19). Assume ideal gas behavior and use the ideal gas value for a. For CO2, CP,m J mol1 K1
= 18.86 + 7.937 * 102
T T2 T3  6.7834 * 105 2 + 2.4426 * 108 3 K K K
Solution Consider the following reversible process in order to calculate ∆S. The gas is first heated reversibly from 450. to 800. K at a constant pressure of 1.35 bar. Subsequently, the gas is reversibly compressed at a constant temperature of 800. K from a pressure of 1.35 bar to a pressure of 3.45 bar. We first derive an expression for a. a = Tƒ
1 0V 1 03 nRT>P4 1 a b = a b = V 0T P V 0T T P Pƒ
Tƒ
Pƒ
Tƒ
Pi
Ti
Pi
Ti
Pƒ CP,m CP,m CP dP ∆S1J K 2 = dT Va dP = n dT  nR = n dT  nR ln Pi L T L L T L P L T 1
Ti
= 2.50 mol
*
800. K a18.86
L
+ 7.937 * 102
450. K
T T2 T3  6.7834 * 105 2 + 2.4426 * 108 3 b K K K dT T
2.50 mol * 8.314 * ln
3.45 bar J K1 mol1 1.35 bar
= 27.13 J K1 + 69.45 J K1  37.10 J K1 + 8.57 J K1 19.50 J K1 = 48.6 J K1
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EXAMPLE PROBLEM 5.4 In this problem, 3.00 mol of liquid mercury is transformed from an initial state characterized by Ti = 300. K and Pi = 1.00 bar to a final state characterized by Tƒ = 600. K and Pƒ = 3.00 bar. a. Calculate ∆S for this process; a = 1.81 * 104 K1, r = 13.54 g cm3, and CP,m for Hg1l2 = 27.98 J mol1 K1. b. What is the ratio of the pressuredependent term to the temperaturedependent term in ∆S? Explain your result.
Solution a. Because the volume changes only slightly with temperature and pressure over the range indicated, Tƒ
Pƒ
Ti
Pi
Tƒ CP ∆S = dT Va dP ≈ nCP,m ln  nVm,ia1Pƒ  Pi2 Ti L T L = 3.00 mol * 27.98 J mol1 K1 * ln 3.00 mol *
600. K 300. K
200.59 g mol1 13.54 g cm3
106 cm3 * m3
* 1.81 * 104 K1
* 2.00 bar * 105 Pa bar 1 = 58.2 J K1  1.61 * 103 J K1 = 58.2 J K1 b. The ratio of the pressuredependent to the temperaturedependent term is 3 * 105. Because the volume change with pressure is very small, the contribution of the pressuredependent term is negligible in comparison with the temperaturedependent term.
As Example Problem 5.4 shows, ∆S for a liquid or solid for a process in which both P and T change is dominated by the temperature dependence of S. Unless the change in pressure is very large, ∆S for liquids and solids can be considered to be a function of temperature only.
5.4 UNDERSTANDING CHANGES IN ENTROPY AT THE MOLECULAR LEVEL
In the first part of this chapter, we used the behavior of a bouncing ball, two rods placed in thermal contact, and an ideal gas expanding to fill its accessible volume to show that there is a natural tendency to spread out energy in spontaneous processes. At the molecular level, what is the basis for this tendency? This tendency can be explained in terms of probability as will be discussed in Chapter 12. At the molecular level, systems have discrete energy levels rather than a continuous energy spectrum. Consider the example of a He atom confined to a onedimensional box. The allowed energy levels for He atoms in boxes of length 8.00 and 10.00 nm are shown in Figure 5.4. Think of the probabilities depicted in Figure 5.4 by circles as individual He atoms. The total energy of the system is the sum of the individual energies. Assigning the number of He atoms to the various energy levels specifies a configuration, and the weight of a configuration is the number of ways (microstates)
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5.4 Understanding Changes in Entropy at the Molecular Level
Figure 5.4
Arrows show how energy levels change as box length increases from 8.00 nm to 10.0 nm.
Allowed energy levels for a He atom in a onedimensional box. Box length is 10.0 nm in (a) and 8.00 nm in (b). The length of the chain of circles on each energy level gives the probability that each energy level is occupied. One circle corresponds to a probability of 0.01. The connecting lines between the boxes show how the energy levels change as the box length increases from 8.00 nm to 10.0 nm.
1.4 3 10223 1.2 3 10223
Energy (joules)
115
1.0 3 10223
8.0 3 10224 6.0 3 10224 4.0 3 10224 2.0 3 10224
(a) 10.0 nm
(b) 8.00 nm
that a configuration can be generated by placing He atoms in the energy levels. The configurations depicted in Figure 5.4 for the two boxes are associated with the greatest number of microstates and are therefore the most probable configurations. Ludwig Boltzmann (1844–1906) derived the following relationship between the entropy and the number of microstates consistent with a given total energy.
S = k ln W
(5.21)
In this equation, W is the number of microstates and k is the Boltzmann constant, which is equal to the ideal gas constant divided by Avogadro’s number. The important conclusion to be drawn from this equation is that entropy increases as the number of microstates available to the system increases. The increase can occur either through lowering the energy levels or increasing the temperature, which allows more energy levels to be occupied. We use the length of the box in analogy to the volume of a threedimensional system to explain why S increases as V increases. For a given total energy, there are more ways to distribute the atoms in energy levels as the box length increases because the energy levels become more closely spaced and lower in energy. Therefore, more energy levels and more microstates are accessible for a given total energy as the system volume increases. This explanation is the molecularlevel basis of the increase in S as V increases. Keeping the box length constant but increasing the temperature to increase the system energy also leads to a redistribution of the atoms in the most probable configuration, as shown in Figure 5.5. We see that more energy levels are accessible to the system as the temperature increases. Employing the same reasoning as above, we see that more microstates will be available to the system as the temperature increases, which explains why S increases as T increases. We will revisit the molecular level definition of entropy and the dependence of S on V and T at a molecular level in a more rigorously quantitative manner in Chapter 15.
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Concept At the microscopic level, entropy is associated with the number of microstates consistent with a given total energy.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Figure 5.5 1.4 3 10223 1.2 3 10223
Energy (joules)
Probabilities for a He atom of occupying the allowed energy levels in a onedimensional box. Energy levels are for (a) 0.200 K and (b) 0.300 K. Box lengths for both diagrams are 10.0 nm. The length of the chain of circles on each energy level is proportional to the probability that each energy level is occupied. One circle corresponds to a probability of 0.01.
1.0 3 10223
8.0 3 10224 6.0 3 10224 4.0 3 10224 2.0 3 10224
(a) 0.200 K
(b) 0.300 K
5.5 THE CLAUSIUS INEQUALITY In Section 5.2, it was shown, by using two examples, that ∆S 7 0 provides a criterion to predict the natural direction of change in an isolated system. This result can also be obtained without considering a specific process. Consider the differential form of the first law for a process in which only P–V work is possible:
dU = dq  Pext dV
(5.22)
Equation (5.22) is valid for both reversible and irreversible processes. If the process is reversible, we can write Equation (5.22) in the following form:
dU = dqrev  P dV = T dS  P dV
(5.23)
Because U is a state function, dU is independent of the path, and Equation (5.23) holds for both reversible and irreversible processes, as long as there are no phase transitions or chemical reactions, and only P–V work occurs. To derive the Clausius inequality, we equate the expressions for dU in Equations (5.22) and (5.23):
Concept For any irreversible process in an isolated system, ∆S 7 0.
dqrev  dq = 1P  Pext2dV
(5.24)
dqrev  dq = TdS  dq Ú 0 or TdS Ú dq
(5.25)
If P  Pext 7 0, the system will spontaneously expand, and dV 7 0. If P  Pext 6 0, the system will spontaneously contract, and dV 6 0. For both possibilities, 1P  Pext2dV 7 0. Therefore, we conclude that
The equality holds only for a reversible process. We rewrite the Clausius inequality in Equation (5.25) for an irreversible process in the form
dS 7
dq T
(5.26)
However, for an irreversible process in an isolated system, dq = 0. Therefore, we have proved that for any irreversible process in an isolated system, ∆S 7 0.
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5.6 THE CHANGE OF ENTROPY IN THE SURROUNDINGS AND ∆Stot = ∆S + ∆Ssur
117
How can the result from Equations (5.22) and (5.23) that dU = dq  Pext dV = TdS  PdV be reconciled with the fact that work and heat are path functions? The answer is that dw Ú PdV and dq … TdS, where the equalities hold only for a reversible process. The result dq + dw = TdS  PdV states that the amount by which the work is greater than PdV and the amount by which the heat is less than TdS in an irreversible process involving only PV work are exactly equal. Therefore, the differential expression for dU in Equation (5.23) is obeyed for both reversible and irreversible processes. In Chapter 6, the Clausius inequality is used to generate two new state functions, the Gibbs energy and the Helmholtz energy. These functions allow predictions to be made about the spontaneous direction of change in processes for which the system interacts with its environment, using only information about the system.
5.6 THE CHANGE OF ENTROPY
IN THE SURROUNDINGS AND 𝚫S tot = 𝚫S + 𝚫Ssur
As shown in Section 5.2, the entropy of an isolated system increases in a spontaneous process. Is it always true that a process is spontaneous if ∆S for the system is positive? In this section, a criterion for spontaneity is developed that takes into account the entropy change in both the system and the surroundings. In general, a system interacts only with the part of the universe that is very close. Therefore, one can think of the system and the interacting part of the surroundings as forming an interacting composite system that is isolated from the rest of the universe, as shown in Figures 2.1 and 5.6. The part of the surroundings that is relevant for entropy calculations is a thermal reservoir at a fixed temperature, T. The mass of the reservoir is sufficiently large that its temperature is only changed by an infinitesimal amount dT when heat is transferred between the system and the surroundings. Therefore, the surroundings always remain in internal equilibrium during heat transfer. Next consider the entropy change of the surroundings, whereby the surroundings are at either constant V or constant P. We assume that the system and surroundings are at the same temperature. If this were not the case, heat would flow across the boundary until T is the same for system and surroundings, unless the system is surrounded by adiabatic walls, in which case qsur = 0. The amount of heat absorbed by the surroundings, qsur, depends on the process occurring in the system. If the surroundings are at constant V, qsur = ∆Usur, and if the surroundings are at constant P, qsur = ∆Hsur . Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly. Therefore,
dSsur =
dqsur qsur , or for a macroscopic change, ∆Ssur = T T
(5.27)
Note that the heat that appears in Equation (5.27) is the actual heat transferred. By contrast, in calculating ∆S for the system, dqrev for a reversible process that connects the initial and final states of the system must be used, not the actual dq for the process because dS is defined by dqrev >T and not dq>T. It is essential to understand this reasoning in order to carry out calculations for ∆S and ∆Ssur . Using the correct value for heat so as to correctly calculate the entropy change of the system, as opposed to that of the surroundings, is further explored in Figure 5.6. In this figure, a gas (the system) is enclosed in a piston and cylinder assembly with diathermal walls. The gas is reversibly compressed by the external pressure generated by water slowly accumulating on the top of the piston. The piston and cylinder assembly is in contact with a water bath thermal reservoir that keeps the temperature of the gas fixed at the value T. In Example Problem 5.5, ∆S and ∆Ssur is calculated for this reversible compression.
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Concept For a system that interacts with its surroundings, the spontaneity condition is ∆S tot = ∆S + ∆Ssur 7 0.
w
T
Piston P,V
q
Figure 5.6
Sample of an ideal gas (the system) confined in a piston and cylinder assembly with diathermal walls. The assembly is in contact with a thermal reservoir that holds the temperature at a value of 300. K. Water dripping into the beaker on the piston increases the external pressure slowly enough to ensure a reversible compression. The directions of work and heat flow are indicated.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
EXAMPLE PROBLEM 5.5 One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath is very large, the temperature of this thermal reservoir in the surroundings remains essentially constant at 300. K during the process. Calculate ∆S, ∆Ssur, and ∆Stot.
Solution Because this is an isothermal reversible process, ∆U = 0, and q = qrev = w. From Section 2.5, Vƒ
qrev = w = nRT
Vƒ dV = nRT ln Vi L V Vi
= 1.00 mol * 8.314 J mol1 K1 * 300. K * ln
10.0 L = 2.29 * 103 J 25.0 L
The entropy change of the system is given by ∆S =
dqrev qrev 2.29 * 103 J = = = 7.62 J K1 T T 300. K L
The entropy change of the surroundings is given by ∆Ssur =
qsur q 2.29 * 103 J =  = = 7.62 J K1 T T 300. K
The total change in the entropy is given by ∆Stot = ∆S + ∆Ssur = 7.62 J K1 + 7.62 J K1 = 0 Because the process in Example Problem 5.5 is reversible, there is no direction of spontaneous change, and therefore ∆Stot = 0. In Example Problem 5.6, this calculation is repeated for an irreversible process that goes between the same initial and final states of the system.
EXAMPLE PROBLEM 5.6 One mole of an ideal gas at 300. K is isothermally compressed by a constant external pressure equal to the final pressure in Example Problem 5.5. Because P ≠ Pext, this process is irreversible. The initial volume is 25.0 L, and the final volume is 10.0 L. The temperature of the surroundings is 300. K. Calculate ∆S, ∆Ssur , and ∆Stot .
Solution The initial and final states of the system in this problem are the same as those in Example Problem 5.5. Because S is a state function, we know that ∆S for this irreversible process must be the same as for the reversible process, therefore ∆S = 7.62 J>K. To calculate ∆Ssur, we first calculate the external pressure and the initial pressure in the system: Pext =
Pi =
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nRT 1 mol * 8.314 J mol1 K1 * 300. K = = 2.49 * 105 Pa V 1 m3 10.0 L * 3 10 L nRT 1 mol * 8.314 J mol1 K1 * 300. K = = 9.98 * 104 Pa V 1 m3 25.0 L * 3 10 L
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5.7 Absolute Entropies and the Third Law of Thermodynamics
119
Because Pext 7 Pi, we know that the direction of spontaneous change will be the compression of the gas to a smaller volume. Because ∆U = 0, q = w = Pext1Vƒ  Vi2 = 2.49 * 105 Pa * 110.0 * 103 m3  25.0 * 103 m32 = 3.74 * 103 J
The entropy change of the surroundings is given by ∆Ssur =
qsur q 3.74 * 103 J =  = = 12.45 J K 1 T T 300. K
It is seen that ∆S 6 0 and ∆Ssur 7 0. The total change in the entropy is given by ∆Stot = ∆S + ∆Ssur = 7.62 J K 1 + 12.45 J K 1 = 4.83 J K 1 The entropy of the system can decrease in a spontaneous process, as it does in this case, as long as ∆Stot 7 0.
5.7 ABSOLUTE ENTROPIES AND THE
THIRD LAW OF THERMODYNAMICS
All elements and many compounds exist in three different states of aggregation. One or more solid phases are the most stable forms at low temperature, and when the temperature is increased to the melting point, a constant temperature transition to the liquid phase is observed. As the temperature is increased further, a constant temperature phase transition to a gas is observed at the boiling point. At temperatures greater than the boiling point, the gas is the stable form. The entropy of an element or a compound is experimentally determined using heat capacity data through the relationship dqrev,P = CP dT. Just as for the thermochemical data discussed in Chapter 4, entropy values are generally tabulated for a standard temperature of 298.15 K and a standard pressure of 1 bar. As an example, we will describe such a determination for the entropy of O2 at 298.15 K, first in a qualitative fashion and then quantitatively in Example Problem 5.7. The experimentally determined heat capacity of O2 is shown in Figure 5.7 as a function of temperature for a pressure of 1 bar. O2 has three solid phases, and transitions between them are observed at 23.66 and 43.76 K. The solid form that is stable above 43.76 K melts to form a liquid at 54.39 K. The liquid vaporizes to form a gas at 90.20 K. These phase transitions are indicated as dashed lines in Figure 5.7. Experimental measurements of CP,m are available above 12.97 K. Below this temperature, the data are extrapolated to zero kelvin by assuming that in this very low temperature range CP,m varies with temperature as T 3. This extrapolation is based on a model of the vibrational spectrum of a crystalline solid that will be discussed in Chapter 15. The explanation for the dependence of CP,m on T is the same as that presented for Cl2 in Section 2.11.
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Concept The entropy of the universe is always increasing.
50
Cp,m (JK21 mol21)
The previous calculations lead to the following conclusion: If the system and the part of the surroundings with which it interacts are viewed as an isolated composite system, the criterion for spontaneous change is ∆Stot = ∆S + ∆Ssur 7 0. We can generalize this result by expanding the size of the composite system to infinity and conclude that a decrease in the entropy of the universe will never be observed because ∆Stot Ú 0 for any change that actually occurs. Any process that occurs in the universe is by definition spontaneous and leads to an increase of Stot. Note that a spontaneous process in a system that interacts with its surroundings is not characterized by ∆S 7 0 but by ∆Stot 7 0. The entropy of the system 1∆S2 can decrease in a spontaneous process, as long as the entropy of the surroundings 1∆Ssur2 increases by a greater amount. In Chapter 6, the spontaneity criterion ∆Stot = ∆S + ∆Ssur 7 0 will be used to generate two state functions, the Gibbs energy and the Helmholtz energy. These functions allow one to predict the direction of change in systems that interact with their environment using only the changes in system state functions.
40
30
Solid I Solid II Solid III Liquid
20
Gas
10
0
20
40 60 80 100 120 Temperature (K)
Figure 5.7
Experimentally determined heat capacity for O2 as a function of temperature. Dots are data from Giauque and Johnston [J. American Chemical Society 51 (1929), 2300], and the red solid curve below 90 K are polynomial fits to these data. The red line above 90 K is a fit to data from the NIST Chemistry Webbook. The blue curve is an extrapolation from 12.97 to 0 K as described in the text. Vertical dashed lines indicate constant temperaturephase transitions, and the most stable phase at a given temperature is indicated in the figure.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Under constant pressure conditions, the molar entropy of the gas can be expressed in terms of the molar heat capacities of the solid, liquid, and gaseous forms and the enthalpies of fusion and vaporization as Tƒ
Sm1T2 = Sm10 K2 +
Concept The entropy of a pure, perfectly crystalline substance (element or compound) is zero at zero Kelvin.
+
∆ vap H Tb
L
C solid P,m dT′
0
T
L
+
Tb
+
T′
∆Hƒusion Tƒ
Tb
+
L Tƒ
C liquid P,m dT′ T′
gas
C P,m dT′ T′
(5.28)
If the substance has more than one solid phase, each will give rise to a separate integral. Note that the entropy changes associated with the phase transitions solid S liquid and liquid S gas, discussed in Section 5.4, must be included in the calculation as well as any solid S solid transitions. To obtain a numerical value for Sm1T2, the heat capacity must be known down to zero Kelvin, and Sm10 K2 must also be known. We first address the issue of the entropy of a solid at zero Kelvin. The third law of thermodynamics can be stated in the following form, attributed to Max Planck: The entropy of a pure, perfectly crystalline substance (element or compound) is zero at zero Kelvin. Different forms of the same substance, if pure and perfectly crystalline, such as graphite, diamond, and C60 all have zero entropy at absolute zero. Although a more detailed discussion of the third law using a microscopic model will be presented in Chapter 15, a brief overview is presented here. Recall that in a perfectly crystalline atomic (or molecular) solid, the position of each atom is known. Because the individual atoms are indistinguishable, exchanging the positions of two atoms does not lead to a new state. Therefore, a perfect crystalline solid has only one microstate at zero kelvin and S = k ln W = k ln 1 = 0. Some crystalline solids such as CO(s) are not perfectly ordered as they are cooled to 0 K because the energy difference between parallel and antiparallel CO molecules is very small. At the freezing temperature, the thermal energy of CO molecules is insufficient to surmount the energy barrier to rotation. In such a case, disorder is frozen in, and the entropy approaches a nonzero value as T S 0 K, which is referred to as a residual entropy. The importance of the third law is that it allows determinations of the absolute entropies of elements and compounds to be carried out for any value of T. To calculate S at a temperature T using Equation (5.28), the CP,m data of Figure 5.7 are graphed in the form CP,m >T as shown in Figure 5.8. Entropy can be obtained as a function of temperature by numerically integrating the area under the curve in Figure 5.8 and adding the entropy changes associated with 1
CP,m (JK22 mol21) T
Solid I
Figure 5.8
CP,m , T as a function of temperature for O2. The vertical dashed lines indicate constant temperaturephase transitions, and the most stable phase at a given temperature is indicated in the figure.
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0.8 0.6 0.4 0.2
Solid II
Liquid
Solid III 0
Gas 50
100
150 200 Temperature (K)
250
300
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5.7 Absolute Entropies and the Third Law of Thermodynamics
Figure 5.9
S
The molar entropy of O2 as a function of temperature. The vertical dashed lines indicate constant temperaturephase transitions, and the most stable phase at a given temperature is indicated in the figure.
150 Solid I
100
Gas
Solid II
[
m (JK
21 mol21)
200
50
Liquid
Solid III
0
50
100
150 Temperature (K)
200
250
300
phase changes at the transition temperatures. Calculations of Sm~ for O2 at 298.15 K are carried out in Example Problem 5.7, and Sm~ 1T2 is shown in Figure 5.9. From the previous discussion, several general statements, listed below as bullet points, regarding the relative magnitudes of the entropy of different substances and different phases of the same substance can now be presented. These statements will be justified on the basis of a microscopic model in Chapter 15.
• Because CP >T in a singlephase region and ∆S for melting and vaporization are •
•
•
always positive, Sm for a given substance is greatest for the gasphase species. The gas molar entropies follow the order Sm 7 Sliquid 7 Ssolid m m . The molar entropy increases with the size of a molecule because the number of degrees of freedom increases with the number of atoms. A nonlinear gasphase molecule has three translational degrees of freedom, three rotational degrees of freedom, and 3n  6 vibrational degrees of freedom. A linear molecule has three translational, two rotational, and 3n  5 vibrational degrees of freedom. For a molecule in a liquid, the three translational degrees of freedom are converted to local vibrational modes. A solid has only vibrational modes. It can be modeled as a threedimensional array of coupled harmonic oscillators, as shown in Figure 5.10. This solid has a wide spectrum of vibrational frequencies, and solids with a large binding energy have higher frequencies than more weakly bound solids. Because modes with high frequencies are not activated at low temperatures, Ssolid is larger for weakly bound solids than m for strongly bound solids at low and moderate temperatures. The entropy of all substances is a monotonically increasing function of temperature.
EXAMPLE PROBLEM 5.7 The heat capacity of O2 has been measured at 1 atm pressure over the interval 12.97 K 6 T 6 298.15 K. The data have been fit to the following polynomial series in T>K: 0 6 T 6 12.97 K: CP,m1T2
J mol1 K1
J mol
1
K
1
= 5.666 + 0.6927
Schematic model of a solid. A useful model of a solid is a threedimensional array of coupled springs. In solids with a high binding energy, the springs are very stiff.
Concept Molar entropies follow the order liquid S gas 7 S solid m 7 Sm m .
T3 = 2.11 * 103 3 K
Concept
12.97 K 6 T 6 23.66 K: CP,m1T2
Figure 5.10
2
3
T T T  5.191 * 103 2 + 9.943 * 104 3 K K K
The entropy of all substances is a monotonically increasing function of temperature.
23.66 K 6 T 6 43.76 K: CP,m1T2
J mol1 K1
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= 31.70  2.038
T T2 T3 + 0.08384 2  6.685 * 104 3 K K K
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
43.76 K 6 T 6 54.39 K:
CP,m1T2
J mol1 K 1 54.39 K 6 T 6 90.20 K: CP,m1T2
J mol
1
K
= 81.268  1.1467
1
= 46.094
T T2 T3 + 0.01516 2  6.407 * 105 3 K K K
90.20 K 6 T 6 298.15 K: CP,m1T2
J mol1 K 1
= 32.71  0.04093
T T2 T3 + 1.545 * 104 2  1.819 * 107 3 K K K
The transition temperatures and the enthalpies for the transitions indicated in Figure 5.8 are as follows: Solid III S solid II Solid II S solid I Solid I S liquid Liquid S gas
93.80 J mol1 743 J mol1 445.0 J mol1 6815 J mol1
23.66 K 43.76 K 54.39 K 90.20 K
a. Using these data, calculate Sm~ for O2 at 298.15 K. b. What are the three largest contributions to Sm~ ?
Solution
heating solid phase III to 12.97 K
T
0K
heating solid phase II
T
23.66 K
743 J mol1 + + 43.76 K
T
L
C solid,I P,m dT
43.76 K
T
µ
L
54.39 K
C liquid P,m dT
54.39 K
vaporization
e
µ
445.0 J mol1 + 54.39 K
90.20 K
93.80 J 23.36 K
heating solid phase I
transition II S I
heating liquid
fusion
+
T
12.97 K
+
e
L
C solid,II P,m dT
L
C solid,III dT P,m
µ
+
e
43.76 K
23.66 K
•
C solid,III dT P,m
transition III S II
e
L
a. Sm1298.15 K2 = Sm10 K2 + ~
e
12.97 K ~
heating solid phase III
+
6815 J mol1 90.20 K
heating gas
e
298.15 K
+
L
90.20 K
gas
C P,m dT T
= 0 + 1.534 J mol1 K1 + 3.964 J mol1 K1 + + 16.98 J mol1 K1 + + 8.181 J mol1 K1 + + 75.59 J mol1 K1 + = 204.9 J mol1 K1
+ 6.649 J mol1 K1 19.61 J mol1 K1 10.13 J mol1 K1 27.06 J mol1 K1 35.27 J mol1 K1
There is an additional small correction for nonideality of the gas at 1 bar. The currently accepted value is Sm~ 1298.15 K2 = 205.152 J mol1 K1 (Linstrom, P. J., and Mallard, W. G., Eds., NIST Chemistry Webbook: NIST Standard Reference Database Number 69, National Institute of Standards and Technology, Gaithersburg, MD, retrieved from http://webbook.nist.gov.) b. The three largest contributions to Sm are ∆S for the vaporization transition, ∆S for heating of the gas from the boiling temperature to 298.15 K, and ∆S for heating of the liquid from the melting temperature to the boiling point.
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5.9 Entropy Changes in Chemical Reactions
The preceding discussion and Example Problem 5.7 show how numerical values of the entropy for a specific substance can be determined at a standard pressure of 1 bar for different values of the temperature. These numerical values can then be used to calculate entropy changes in chemical reactions, as will be shown in Section 5.9.
123
Concept Tabulated values of entropy refer to a standard pressure of 1 bar.
5.8 STANDARD STATES IN ENTROPY
[
Sm – S
As discussed in Chapter 4, changes in U and H are calculated using the result that ∆Hƒ values for pure elements in their standard state at a pressure of 1 bar and a temperature of 298.15 K are zero. For S, the third law provides a natural definition of zero, namely, the crystalline state at zero kelvin. Therefore, the absolute entropy of a compound as a given temperature can be experimentally determined from heat capacity measurements, as described in the previous section. Entropy is also a function of pressure, and tabulated values of entropy refer to a standard pressure of 1 bar. The value of S varies most strongly with P for a gas. From Equations (5.14) and (5.15), for an ideal gas at constant T, Vƒ Pƒ ∆Sm = R ln = R ln (5.29) Vi Pi
m
CALCULATIONS
0 1
Choosing Pi = P ~ = 1 bar, Sm1P2 = Sm~  R ln
P1bar2 P~
2
4
3
P (bar)
(5.30)
Figure 5.11 shows a plot of the molar entropy of an ideal gas as a function of pressure. It is seen that as P S 0, Sm S ∞. This is a consequence of the fact that as P S 0, V S ∞. As Equation (5.30) shows, entropy becomes infinite in this limit. Equation (5.30) provides a way to calculate the entropy of a gas at any pressure. For solids and liquids, S varies so slowly with variations in P, as shown in Section 5.3 and Example Problem 5.4, that the pressure dependence of S can usually be neglected.
Figure 5.11
Molar entropy of an ideal gas as a function of gas pressure. By definition, at 1 bar, Sm = Sm~ , the standard state molar entropy. As discussed in the text, as P S 0, entropy becomes infinite.
5.9 ENTROPY CHANGES IN CHEMICAL REACTIONS
The entropy change in a chemical reaction is a major factor in determining the equilibrium concentration in a reaction mixture. In an analogous fashion to calculating ∆H R~ and ∆U R~ for chemical reactions, ∆SR~ is equal to the difference in the entropies of products and reactions, which can be written as ∆ rS ~ = a vi Si~
i
(5.31)
Concept The entropy change for a chemical reaction is equal to the difference in the entropies of products and reactions.
In Equation (5.31), the stoichiometric coefficients vi are positive for products and negative for reactants. For example, in the reaction Fe 3O41s2 + 4H21g2 S 3Fe1s2 + 4H2O1l2
(5.32)
the entropy change under standard state conditions of 1 bar and 298.15 K is given by ~ ~ ~ ~ ~ ∆ r S298.15 = 3S298.15 1Fe, s2 + 4S298.15 K1H2O, l2  S298.151Fe 3O4, s2  4S298.151H2, g2
= 3 * 27.28 J K1 mol1 + 4 * 69.61 J K1 mol1  146.4 J K1 mol1  4 * 130.684 J K1 mol1
= 308.9 J K1 mol1 Note that unlike formation enthalpies, S ~ for elements in their standard reference state are not zero. For this reaction, ∆SR is large and negative, primarily because gaseous species are consumed in the reaction, and none are generated. For reactions involving gases, ∆SR~ is generally positive for ∆n 7 0, and negative for ∆n 6 0 where ∆n is the change in the number of moles of gas in the overall reaction.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Tabulated values of S ~ are generally available at the standard temperature of 298.15 K, and values for selected elements and compounds are listed in Tables 4.1 and 4.2 (see Appendix A, Data Tables). However, it is often necessary to calculate ∆S ~ at other temperatures. Such calculations are carried out using the temperature dependence of S discussed in Section 5.4 and shown in Example Problem 5.8: T
∆ r S 1T2 = ∆ r S 1298.15 K2 +
~
~
∆CP dT′ L T′
(5.33)
298.15
This equation is valid only if no phase changes occur in the temperature interval between 298.15 K and T. If phase changes occur, the associated entropy changes must be included as they were in Equation (5.28).
EXAMPLE PROBLEM 5.8 The standard entropies of CO, CO2, and O2 at 298.15 K are ~ S298.15 1CO, g2 = 197.67 J K1 mol1
~ S298.15 1CO2, g2 = 213.74 J K1 mol1
~ S298.15 1O2, g2 = 205.138 J K1 mol1
The temperature dependence of constant pressure heat capacity for CO, CO2, and O2 is given by CP,m1CO, g2
= 31.08  1.452 * 102
T T2 T3 + 3.1415 * 105 2  1.4973 * 108 3 K K K
CP,m1CO2, g2
= 18.86 + 7.937 * 102
T T2 T3  6.7834 * 105 2 + 2.4426 * 108 3 K K K
CP,m1O2, g2
= 30.81  1.187 * 102
J K1 mol1
J K1 mol1
J K1 mol1
T T2 + 2.3968 * 105 2 K K
Calculate ∆ r S ~ for the reaction CO1g2 + 1>2 O21g2 S CO21g2 at 475 K.
Solution ∆CP,m 1
JK
mol
1
= a18.86  31.08 
1 * 30.81b 2
+ a7.937 + 1.452 +
1 T * 1.187b * 102 2 K
 a6.7834 + 3.1415 +
1 T2 * 2.3968b * 105 2 2 K
+ 12.4426 + 1.49732 * 108
= 27.63 + 9.983 * 102
T3 K3
T T2 T3  1.112 * 104 2 + 3.940 * 108 3 K K K
∆ r S ~ = S ~ 1CO2, g, 298.15 K2  S ~ 1CO, g, 298.15 K2 = 213.74 J K1 mol1  197.67 J K1 mol1 
1 * S ~ 1O2, g, 298.15 K2 2
1 * 205.138 J K1 mol1 2
= 86.50 J K1 mol1
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5.10 Heat Engines and the Carnot Cycle T
∆ r S 1T2 = ∆ r S 1298.15 K2 + ~
~
L
298.15 K
∆Cp T′
125
dT′
= 86.50 J K1 mol1
475 K
+
L
a 27.63 + 9.983 * 102
298.15 K
T T2 T3  1.112 * 104 2 + 3.940 * 108 3 b K K K dT J K1 mol1 T
= 86.50 J K1 mol1 + 112.866 + 17.654  7.605 + 1.05942J K1 mol1 = 86.50 J K1 mol1  1.757 J K1 mol1 = 88.3 J K1 mol1
The value of ∆ r S ~ is negative at both temperatures because the number of moles of gaseous species is reduced in the reaction.
5.10 HEAT ENGINES AND THE CARNOT CYCLE The concept of entropy arose as 19thcentury scientists attempted to maximize the work output of engines. In this section, we obtain an alternate formulation of the second law of thermodynamics based on a theoretical study of idealized engines. A real engine, such as an automobile engine, operates in a cyclical process of fuel intake, compression, ignition and expansion, and exhaust, which occurs several thousand times per minute and is used to perform work on the surroundings. Because the work produced by such an engine is a result of the heat released in a combustion process, it is referred to as a heat engine. An idealized version of a heat engine is depicted in Figure 5.12. The system consists of a working substance (in this case an ideal gas) confined in a piston and cylinder assembly with diathermal walls. This assembly can be brought into contact with a hot reservoir at Thot or a cold reservoir at Tcold. The expansion or contraction of the gas caused by changes in its temperature drives the piston in or out of the cylinder. This linear motion is converted to circular motion, and the rotary motion is used to do work in the surroundings. The efficiency of a heat engine is of particular interest in practical applications. Experience shows that work can be converted to heat with 100% efficiency. Consider an example from calorimetry, discussed in Chapter 4, in which electrical work is performed on a resistive heater immersed in a water bath. We observe that all of the electrical work done on the heater has been converted to heat, resulting in an increase in the temperature of the water and the heater. What is the maximum theoretical efficiency of the reverse process, the conversion of heat to work? As shown below, it is less than 100%. There is a natural asymmetry in the efficiency of converting work to heat and converting heat to work. Thermodynamics provides an explanation for this asymmetry.
Concept An analysis of the Carnot cycle shows that there is a natural asymmetry in the efficiency of converting work to heat and converting heat to work.
Figure 5.12 Thot
Tcold
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A schematic depiction of a heat engine. The working substance (in this example an ideal gas) is confined by a piston and cylinder assembly. As the assembly is brought into contact with hot or cold reservoirs, temperature changes cause the piston to move, which generates a linear motion of the connecting rod, that is in turn mechanically converted to a rotary motion, which is used to perform work.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Figure 5.13
a
Pa
Isothermal expansion Thot
Pressure
A reversible Carnot cycle for a sample of an ideal gas working substance projected onto an indicator diagram. The cycle consists of two adiabatic and two isothermal segments. The arrows indicate the direction in which the cycle is traversed. The insets show the volume of gas and the coupling to the reservoirs at the beginning of each successive segment of the cycle. The coloring of the contents of the cylinder indicates the presence of the gas and not its temperature. The volume of the cylinder shown is that at the beginning of the appropriate segment.
Adiabatic compression
Tcold
Thot
Pb
b
Thot
Tcold
Pd
d
Pc
Isothermal compression
Va
Tcold
Thot
c
Tcold
0
Adiabatic expansion
Vd
Vb
Vc
Volume
Concept It is not possible to construct a cyclic process in an indicator diagram that results in nonzero net work using only adiabatic or only isothermal segments.
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As discussed in Section 2.7, the maximum work output in an isothermal expansion occurs in a reversible process. For this reason, we will next calculate the efficiency of a reversible heat engine because the efficiency of a reversible engine is an upper bound to the efficiency of a real engine. This reversible engine converts heat into work by exploiting the spontaneous tendency of heat to flow from a hot reservoir to a cold reservoir. It performs work on the surroundings by operating in a cycle of reversible expansions and compressions of an ideal gas in a piston and cylinder assembly. (We discuss automotive engines in Section 5.13.) This reversible cycle is shown in Figure 5.13 in a P–V diagram. The expansion and compression steps are designed so that the engine returns to its initial state after four steps. Recall that the area within the cycle equals the work done by the engine. Beginning at point a, the first segment is a reversible isothermal expansion in which the gas absorbs heat from the reservoir at Thot, and does work on the surroundings. In the second segment, the gas expands further, this time adiabatically. Work is also done on the surroundings in this step. At the end of the second segment, the gas has cooled to the temperature Tcold. The third segment is an isothermal compression in which the surroundings do work on the system and heat is absorbed by the cold reservoir. In the final segment, the gas is compressed to its initial volume, this time adiabatically. Work is done on the system in this segment, and the temperature returns to its initial value, Thot. In summary, heat is taken up by the engine in the first segment at Thot and is released to the surroundings in the third segment at Tcold. Work is done on the surroundings in the first two segments and on the system in the last two segments. An engine is only useful if net work is done on the surroundings—that is, if the magnitude of the work done in the first two steps is greater than the magnitude of the work done in the last two steps. The efficiency of the engine can be calculated by comparing the net work per cycle with the heat taken up by the engine from the hot reservoir. Before carrying out this calculation, we will discuss the rationale for the design of this reversible cycle in more detail. To avoid losing heat to the surroundings at temperatures between Thot and Tcold, the adiabatic segments 2 1b S c2 and 4 1d S a2 are used to move the gas between these temperatures. To absorb heat only at Thot and release heat only at Tcold, segments 1 1a S b2 and 3 1c S d2 must be isothermal. The reason for using alternating isothermal and adiabatic segments is that no two isotherms at different temperatures intersect and no two adiabats starting from two different temperatures intersect. Therefore, it is impossible to create a closed cycle of nonzero area in an indicator diagram out of isothermal or adiabatic segments alone. However, net work can be done using alternating adiabatic and isothermal segments. The reversible cycle depicted in Figure 5.13 is called a Carnot cycle, after the French engineer, Nicolas Carnot, who first studied such cycles in the early 19th century.
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5.10 Heat Engines and the Carnot Cycle
127
TABLE 5.1 Heat, Work, and 𝚫U for the Reversible Carnot Cycle Segment Initial State
Final State
aSb
Pa, Va, Thot
Pb,Vb, Thot
bSc
q
∆U
w
Pc,Vc, Tcold
wab1 2
∆Uab = 0
Pb, Vb, Thot
qab1+2
cSd
Pc, Vc, Tcold Pd,Vd, Tcold
Pa,Va, Thot
0
wcd1+2
∆Ucd = 0
dSa
Pd, Vd, Tcold qcd12
wbc12
Cycle
Pa, Va, Thot
Pa,Va, Thot
wda1+ 2
∆Uda = wda1+2
qab + qcd1+ 2 wab + wbc + wcd + wda12 ∆Ucycle = 0
0
∆Ubc = wbc12
The efficiency of the Carnot cycle can be determined by calculating q, w, and ∆U for each segment of the cycle, assuming that the working substance is an ideal gas. The results are shown in a qualitative fashion in Table 5.1. The appropriate signs for q and w are indicated. If ∆V 7 0, w 6 0 for the segment, and the corresponding entry for work has a negative sign. For the isothermal segments, q and w have opposite signs because ∆U = q + w = 0. From Table 5.1, it is seen that
wcycle = wcd + wda + wab + wbc and qcycle = qab + qcd
(5.34)
Because ∆Ucycle = 0,
wcycle = 1qcd + qab2
(5.35)
wcycle 6 0, therefore 0 qab 0 7 0 qcd 0
(5.36)
By comparing the areas under the two expansion segments with those under the two compression segments in the indicator diagram in Figure 5.13, one can see that the total work as seen from the system is negative, meaning that work is done on the surroundings in each cycle. Using this result, we arrive at an important conclusion that relates the heat flow in the two isothermal segments:
More heat is withdrawn from the hot reservoir than is deposited in the cold reservoir. It is useful to make a model of this heat engine that indicates the relative magnitude and direction of the heat and work flow; such a model is shown in Figure 5.14a. The figure makes it clear that not all of the heat withdrawn from the higher temperature reservoir is converted to work done by the system (the engine) on the surroundings. Hot reservoir
Hot reservoir
Work
Figure 5.14 Work
Cold reservoir (a) Schematic model of the heat engine operating in a reversible Carnot cycle.
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(b) According to the second law of thermodynamics, the heat engine shown cannot be constructed.
Carnot heat engine cycle. (a) A schematic model of a heat engine operating in a reversible Carnot cycle. (b) The second law of thermodynamics asserts that it is impossible to construct a heat engine as shown that operates using a single heat reservoir and converts the heat withdrawn from the reservoir into work with 100% efficiency as shown.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
The efficiency, e, of the reversible Carnot engine is defined as the ratio of the work output to the heat withdrawn from the hot reservoir. Referring to Table 5.1,
Concept No heat engine can convert heat entirely into work.
e = 
wcycle qab
= 1 
=
qab + qcd qab
0 qcd 0 6 1 because 0 qab 0 7 0 qcd 0 , qab 7 0, and qcd 6 0 0 qab 0
(5.37)
Equation (5.37) shows that the efficiency of a heat engine operating in a reversible Carnot cycle is always less than one. Equivalently, not all of the heat withdrawn from the hot reservoir can be converted to work. This conclusion is valid for all engines and illustrates the asymmetry in converting heat to work and work to heat. These considerations on the efficiency of reversible heat engines led to the Kelvin– Planck formulation of the second law of thermodynamics:
It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.
An alternative, but equivalent, statement of the second law by Clausius is as follows:
Pressure
It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a hot reservoir and the flow of an equivalent amount of heat out of the system into a cold reservoir.
Volume
Figure 5.15
Approximation of a reversible cycle. A reversible cycle of arbitrary shape, indicated by the ellipse, can be approximated to any desired accuracy by a large number of Carnot cycles, each having a sequence of alternating adiabatic and isothermal segments.
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The second law asserts that the heat engine depicted in Figure 5.14b cannot be constructed. The second law has been put to the test many times by inventors who have claimed that they have invented an engine that has an efficiency of 100%. No such claim has ever been validated. To test the assertion made in this statement of the second law, imagine that such an engine has been invented. We mount it on a boat in Seattle and set off on a journey across the Pacific Ocean. Heat is extracted from the ocean, which is the single heat reservoir, and is converted entirely to work in the form of a rapidly rotating propeller. Because the ocean is huge, the decrease in its temperature as a result of withdrawing heat is negligible. By the time we arrive in Japan, not a gram of diesel fuel has been used because all the heat needed to power the boat has been extracted from the ocean. The money that was saved on fuel is used to set up an office and begin marketing this wonder engine. Does this scenario sound too good to be true? It is. Such an impossible engine is called a perpetual motion machine of the second kind because it violates the second law of thermodynamics. A perpetual motion machine of the first kind violates the first law. The first statement of the second law can be understood using an indicator diagram. For an engine to produce work, the area of the cycle in a P–V diagram must be greater than zero. However, this is impossible in a simple cycle using a single heat reservoir. If Thot = Tcold in Figure 5.13, the cycle a S b S c S d S a collapses to a line, and the area of the cycle is zero. An arbitrary reversible cycle can be constructed that does not consist of individual adiabatic and isothermal segments. However, as shown in Figure 5.15, any reversible cycle can be approximated by a succession of adiabatic and isothermal segments, an approximation that becomes exact as the length of each segment approaches zero. It can be shown that the efficiency of such a cycle is also given by Equation (5.37), so that the efficiency of all heat engines operating in any reversible cycle between the same two temperatures, Thot and Tcold, is identical. A more useful form than Equation (5.37) for the efficiency of a reversible heat engine can be derived if the working substance in the engine is an ideal gas. Calculating
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5.10 Heat Engines and the Carnot Cycle
129
the work flow in each of the four segments of the Carnot cycle using the results of Sections 2.7 and 2.9, Vb Va
wab = nRThot ln
wbc = nCV,m1Tcold  Thot2 wbc 6 0 because Tcold 6 Thot
wcd
wab 6 0 because Vb 7 Va
Vd = nRTcold ln Vc
(5.38)
wcd 7 0 because Vd 6 Vc
wda = nCV,m1Thot  Tcold2 wda 7 0 because Thot 7 Tcold
As derived in Section 2.14, the volume and temperature in the reversible adiabatic segments are related by
g1
ThotV b
g1
= TcoldV c
g1
and TcoldV d
g1
= ThotV a
(5.39)
You will show in the endofchapter problems that Vc and Vd can be eliminated from the set of Equations (5.38) to yield wcycle = nR1Thot  Tcold2 ln
Vb 6 0 Va
(5.40)
Because ∆Ua S b = 0, the heat withdrawn from the hot reservoir is qab = wab = nRThot ln
Vb Va
(5.41)
and the efficiency of the reversible Carnot heat engine with an ideal gas as the working substance can be expressed solely in terms of the reservoir temperatures
e =
0 wcycle 0 qab
Thot  Tcold Tcold = = 1 6 1 Thot Thot
(5.42)
The efficiency of this reversible heat engine can approach one only as Thot S ∞ or Tcold S 0, neither of which can be accomplished in practice. Therefore, heat can never be totally converted to work in a reversible cyclic process. Because wcycle for an engine operating in an irreversible cycle is less than the work attainable in a reversible cycle, eirreversible 6 ereversible 6 1.
Concept The efficiency of a reversible Carnot engine can be calculated from the temperatures of the hot and cold reservoirs.
EXAMPLE PROBLEM 5.9 Calculate the maximum work that can be done by a reversible heat engine operating between 500. and 200. K if 1000. J is absorbed at 500. K.
Solution The fraction of the heat that can be converted to work is the same as the fractional fall in the absolute temperature. This is a convenient way to link the efficiency of an engine with the properties of the absolute temperature. e = 1 
Tcold 200. K = 1 = 0.600 Thot 500. K
w = eqab = 0.600 * 1000. J = 600. J In this section, only the most important features of heat engines have been discussed. It can also be shown that the efficiency of a reversible heat engine is independent of the working substance. Therefore, the result in Equation (5.42) is not restricted to an ideal gas working substance; rather, it is valid for any working substance and depends only on Thot and Tcold. For a more indepth discussion of heat engines, the interested reader is referred to Heat and Thermodynamics, seventh edition, by M. W. Zemansky and R. H. Dittman (McGrawHill, 1997). We will return to a discussion of the efficiency of engines when we discuss refrigerators, heat pumps, and real engines in Section 5.13.
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics S U P P L E M E N TA L
S E C T I O N
5.11 HOW DOES S DEPEND ON V AND T? Section 5.4 showed how entropy varies with P, V, and T for an ideal gas. In this section, we derive general equations for the dependence of S on V and T that can be applied to solids, liquids, and real gases. We do so by using the property that dS is an exact differential. A similar analysis of S as a function of P and T is carried out in Section 5.12. Consider Equation (5.23), rewritten in the form
Concept Using the properties of partial derivatives and state functions, the dependence of S on V and T can be expressed in therms of measurable quantities.
dS =
1 P dU + dV T T
(5.43)
Because 1>T and P>T are greater than zero, the entropy of a system increases with the internal energy at constant volume and increases with the volume at constant internal energy. However, because internal energy is not generally a variable under experimental control, it is more useful to obtain equations for the dependence of dS on V and T. We first write the total differential dS in terms of the partial derivatives with respect to V and T: 0S 0S dS = a b dT + a b dV (5.44) 0T V 0V T To evaluate 10S>0T2V and 10S>0V2T , we start with dS = dU>T + 1P>T2dV, allowing Equation (5.44) for dS to be rewritten in the form
dS =
CV 1 0U P 1 0U c CV dT + a b dV d + dV = dT + c P + a b d dV T 0V T T T T 0V T
(5.45)
Equating the coefficients of dT and dV in Equations (5.44) and (5.45), a
CV 0S b = 0T V T
and a
0S 1 0U b = cP + a b d 0V T T 0V T
(5.46)
The temperature dependence of entropy at constant volume can be calculated straightforwardly using the first equality in Equation (5.46):
dS =
CV dT, constant V T
(5.47)
The expression for 10S>0V2T in Equation (5.46) is not in a form that allows for a direct comparison with experiment. The property that dS is an exact differential (see Section 3.1) allows a more useful relation to be derived. Because a
0 0S 0 0S a b b = a a b b 0T 0V T V 0V 0T V T
(5.48)
one can take the mixed second derivatives of the expressions in Equation (5.46), a
0 0S 1 0 0U a b b = a a b b 0V 0T V T T 0V 0T V T
0 0S 1 0P 0 0U 1 0U a a b b = ca b + a a b b d  2 cP + a b d 0T 0V T V T 0T V 0T 0V T V 0V T T
(5.49)
Substituting the expressions for the mixed second derivatives in Equation (5.49) into Equation (5.48), canceling the double mixed derivative of U that appears on both sides of the equation, and simplifying the result, we obtain the following equation:
P + a
0U 0P b = Ta b 0V T 0T V
(5.50)
This equation provides the expression for the internal pressure 10U>0V2T that was used without a derivation in Section 3.2. It provides a way to calculate the internal pressure of the system if the equation of state for the substance is known.
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5.12 DEPENDENCE OF S ON T AND P
131
Comparing the result in Equation (5.50) with the second equality in Equation (5.46), we obtain a practical equation for the dependence of entropy on volume under constant temperature conditions: a
10V>0T2P 0S 0P a b = a b = = kT 0V T 0T V 10V>0P2T
(5.51)
where a is the coefficient for thermal expansion at constant pressure and kT is the isothermal compressibility coefficient. Both of these quantities are readily obtained from experiments. In simplifying this expression, the cyclic rule for partial derivatives has been used. The result of these mathematical manipulations is that dS can be expressed in terms of dT and dV as
dS =
CV a dT + dV kT T
(5.52)
Integrating both sides of this equation, Tƒ
Vƒ
CV a ∆S = dT + dV k T L L T
Ti
(5.53)
Vi
This result applies to a singlephase system of a liquid, solid, or gas that undergoes a transformation from Ti, Vi to Tƒ, Vƒ, provided that no phase changes or chemical reactions occur in the system. S U PPL EM EN TAL
SEC TI O N
5.12 DEPENDENCE OF S ON T AND P Because chemical transformations are normally carried out at constant pressure rather than constant volume, we need to know how S varies with T and P. The total differential dS is written in the form dS = a
0S 0S b dT + a b dP 0T P 0P T
(5.54)
Concept
Starting from the relation U = H  PV, we write the total differential dU as dU = T dS  P dV = dH  P dV  V dP
(5.55)
This equation can be rearranged to give an expression for dS:
dS =
1 V dH  dP T T
(5.56)
Using the properties of partial derivatives and state functions, the dependence of S on P and T can be expressed in terms of measurable quantities.
The previous equation is analogous to Equation (5.43) but contains the variable P rather than V. dH = a
0H 0H 0H b dT + a b dP = CP dT + a b dP 0T P 0P T 0P T
(5.57)
Substituting this expression for dH into Equation (5.56),
dS =
CP 1 0H 0S 0S dT + c a b  V d dP = a b dT + a b dP T T 0P T 0T P 0P T
(5.58)
Because the coefficients of dT and dP must be the same on both sides of Equation (5.58),
a
CP 0S b = 0T P T
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and a
0S 1 0H b = ca b  Vd 0P T T 0P T
(5.59)
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The ratio CP >T is positive for all substances, allowing one to conclude that S is a monotonically increasing function of the temperature. Just as for 10S>0V2T in Section 5.12, the expression for 10S>0P2T is not in a form that allows a direct comparison with experimental measurements. Just as in our evaluation of 10S>0V2T , we equate the mixed second partial derivatives of 10S>0T2P and 10S>0P2T : a
0 0S 0 0S a b b = a a b b 0T 0P T P 0P 0T P T
(5.60)
These mixed partial derivatives can be evaluated using Equation (5.59): a
a
0 0S 1 0CP 1 0 0H a b b = a b = a a b b 0P 0T P T T 0P T T 0P 0T P T
0 0S 1 0 0H 0V 1 0H a b b = ca a b b  a b d  2 ca b  Vd 0T 0P T P T 0T 0P T P 0T P 0P T T
(5.61) (5.62)
Equating Equations (5.61) and (5.62),
1 0 0H 1 0 0H 0V 1 0H a a b b = ca a b b  a b d  2 ca b  Vd T 0T 0P T P T 0T 0P T P 0T P 0P T T
(5.63)
Simplifying this equation results in a
0H 0V b  V = T a b 0P T 0T P
(5.64)
Using this result and Equation (5.59), we can write the pressure dependence of the entropy at constant temperature in a form that easily allows an experimental determination of this quantity to be made: a
0S 0V b =  a b = Va 0P T 0T P
(5.65)
Using these results, we can write the total differential dS in terms of experimentally accessible parameters as
dS =
CP dT  Va dP T
(5.66)
Integrating both sides of this equation, Tƒ
Pƒ
CP ∆S = dT Va dP L T L
Ti
(5.67)
Pi
This result applies to a singlephase system of a pure liquid, solid, or gas that undergoes a transformation from Ti Pi to Tƒ, Pƒ, provided that no phase changes or chemical reactions occur in the system.
SUP P LE ME NTAL
SE CTION
5.13 ENERGY EFFICIENCY, HEAT PUMPS,
REFRIGERATORS, AND REAL ENGINES
Thermodynamics provides the tools to study the release of energy through chemical reactions or physical processes such as harvesting light to perform work or generate heat. As the Earth’s population continues to increase in the coming decades, the need for energy in various forms will rise rapidly. Electricity is a major component of this
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5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
6.0
Highemissions scenario Moderateemissions scenario
5.0
Lowemissions scenario
4.0
2020–2029
Year 2000 constant concentrations 20th century
3.0
2090–2099
Moderateemissions scenario
1.0 0 –1.0 2000 Year
2100
(a) Global surface warming for years 1900 to 2000 (black curve) and projected to 2100 (colored curves, various scenarios).
Moderateemissions scenario Highemissions scenario
2.0
1900
133
Highemissions scenario
Lowemissions scenario
Global surface warming (°C)
Lowemissions scenario
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 (°C)
(b) Uncertainty in predicted temperature increase for a given scenario is represented by the height of bars between panels.
(c) Increase in temperature on a regional scale for different scenarios of greenhouse gas emissions.
Figure 5.16
energy demand, and fossil fuels are expected to be the major source for electricity production for the foreseeable future. The increased combustion of fossil fuels will continue to contribute to the rapid increase in the CO2 concentration in the atmosphere that began with the industrial revolution. This increased CO2 concentration is expected to lead to a significant increase in the global average surface temperature, as shown for different scenarios of the rate of atmospheric CO2 increase in Figure 5.16. For details on the scenarios and the possible consequences of this temperature increase, which include flooding of land areas near sea level, acidification of the oceans, and increasingly severe droughts and storms, see the Intergovernmental Panel on Climate Change website at www.ipcc.ch/. In order to slow or to reverse the buildup of greenhouse gases in the atmosphere, we must capture significant portions of the greenhouses gases produced in the combustion of fossil fuels and move to energy sources, such as wind and solar, that do not generate greenhouse gases. We must also find ways to produce more industrial output and work with less energy. Table 5.2 shows that per capita energy use in the United States is substantially higher than that in other regions of the world as well as Europe which has a similar climate and living standard. The energy flow in the United States for 2013 through various sectors of the economy is shown in Figure 5.17. The parallel pathway for CO2 emissions into the atmosphere is shown in Figure 5.18. By comparing the useful and lost energy,
Historical and projected global surface temperatures. Source: UN Intergovernmental Panel on Climate Change
TABLE 5.2 Per Capita Energy Use, 2009 Nation or Region
Per Capita Energy Use (mega BTU) 1 J = 1054 BTU
Ratio to U.S. per Capita Use
United States
308
1
Eurasia
137
0.444
Europe
133
0.431
Africa
16
0.0519
World
71
0.231
Source: U.S. Energy Information Administration
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Figure 5.17
U.S. energy flow trends for 2013. The energy contributions from different sources to each economic sector, such as electricity generation, are shown, together with net useful and rejected (wasted) energy for each sector. Note that 59% of the total consumed energy in all sectors is wasted. Source: Lawrence Livermore National Laboratory, U.S. Department of Energy
Figure 5.18
2013 U.S. carbon dioxide emissions from energy consumption. The pathways of CO2 generation in economic sectors are shown, together with a breakdown in the three fossil fuel sources of natural gas, coal, and petroleum. Source: Lawrence Livermore National Laboratory, U.S. Department of Energy
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5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
it is seen that 59% of the energy is lost in the form of heat, including friction in engines and turbines, resistive losses in the distribution of electricity, and heat loss from poorly insulated buildings. Note in particular the inefficiency of the transportation sector, where 79% of expended energy is wasted. As discussed in Section 5.10, the conversion of heat to work cannot be achieved with an efficiency of 100%, even if dissipative processes such as friction are neglected. Therefore, significant losses are inevitable, but substantial increases in energy efficiency are possible, as will be discussed below. Electricity generation by wind turbines, photovoltaic panels, hydroelectric plants, and fuel cells (see Section 11.13) is of particular importance because it involves the conversion of one form of work to another. The efficiency for these methods is not subject to the limitations imposed on the conversion of heat to work by the second law. What can we learn from thermodynamics that will help us use less energy and produce more work with less energy? To address this question, consider how energy is used in the residential sector, as shown in Figure 5.19. Heating buildings and water accounts for 47% of residential energy usage, and electrical energy is most commonly used for these purposes. Both building and water heating can be provided with significantly less energy input using an electrically powered heat pump, as explained below. In an idealized reversible heat pump, that can be used either to heat or refrigerate, the Carnot cycle in Figure 5.13 is traversed in the opposite direction. Thus, the signs of w and q in the individual segments and the signs of the overall w and q are reversed. Heat is now withdrawn from the cold reservoir (the refrigerator) and deposited in the home, which is the hot reservoir (Figure 5.20a). Because this is not a spontaneous process, work must be done on the system to effect this direction of heat flow. The heat and work flow for room heating is shown in Figure 5.20b. A heat pump is used to heat a building by extracting heat from a colder thermal reservoir, such as a lake, the ground, or the ambient air. The coefficient of performance Hot reservoir
135
Concept Adoption of electrical vehicles, heat pumps, and electricity generation using wind and sunlight can greatly reduce the rate of climate change.
Building heating 34%
Water heating 13% Refrigerator 8%
Appliances and lighting 34%
Electric A/C 11%
Figure 5.19
U.S. household energy use. Source: U.S. Energy Information Agency
Hot reservoir
Work
Work
Figure 5.20
Cold reservoir
Cold reservoir (a)
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(b)
Application of the principle of a reverse Carnot heat engine to refrigeration and household heating. The reverse Carnot heat engine can be used to induce heat flow from a cold reservoir to a hot reservoir with the input of work. The hot reservoir in both (a) and (b) is a room. The cold reservoir can be configured as the inside of a refrigerator or outside ambient air, earth, or water for a heat pump.
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of a heat pump, hhp, is defined as the ratio of the heat pumped into the hot reservoir to the work input to the heat pump:
hhp =
qhot qhot Thot = = w qhot + qcold Thot  Tcold
(5.68)
Assume that Thot = 294 K and Tcold = 278 K, typical for a mild winter day. The maximum hhp value is calculated to be 18. Such high values cannot be attained for real heat pumps operating in an irreversible cycle with dissipative losses. Typical values for commercially available heat pumps lie in the range of 2 to 3. This means that for every joule of work done on the system, 2 to 3 J of heat is made available for building heating. Heat pumps become less effective as Tcold decreases, as shown by Equation (5.68). Therefore, geothermal heat pumps that use ∼55°F soil 6 to 10 feet below the Earth’s surface, rather than ambient air, as the cold reservoir are much more efficient in very cold climates than air source heat pumps. Note that a coefficient of performance of 3 for a heat pump means that a house can be heated using 33% of the electrical power consumption that would be required to heat the same house using electrical baseboard heaters. This is a significant argument for using heat pumps for residential heating. Heat pumps can also be used to heat water. In addition to improving the efficiency of heating space and water, houses can also be designed to significantly reduce total energy use. A combination of a wellinsulated house with rooftop photovoltaic panels can result in a house that generates as much energy as it consumes in a year in most regions of the United States. A refrigerator is also a heat engine operated in reverse (Figure 5.20a). The interior of the refrigerator is the system, and the room in which the refrigerator is situated is the hot reservoir. Because more heat is deposited in the room than is withdrawn from the interior of the refrigerator, the overall effect of such a device is to increase the temperature of the room. However, the usefulness of the device is that it provides a cold volume for food storage. The coefficient of performance, hr, of a reversible Carnot refrigerator is defined as the ratio of the heat withdrawn from the cold reservoir to the work supplied to the device:
hr =
qcold qcold Tcold = = w qhot + qcold Thot  Tcold
(5.69)
This formula shows that as Tcold decreases from 0.9 Thot to 0.1 Thot, hr decreases from 9 to 0.1, showing that in order to provide a lower temperature, more work is required to extract a given amount of heat. A household refrigerator typically operates at 255 K in the freezing compartment and 277 K in the refrigerator section. Using the lower of these temperatures for the cold reservoir and 294 K as the temperature of the hot reservoir (the room), we find that the maximum hr value is 6.5. This means that for every joule of work done on the system, 6.5 J of heat can be extracted from the contents of the refrigerator. This is the maximum coefficient of performance and is only applicable to a refrigerator operating in a reversible Carnot cycle with no dissipative losses. When we take losses into account, we find that it is difficult to achieve hr values greater than ∼1.5 in household refrigerators. This shows the significant loss of efficiency in an irreversible dissipative cycle. We have seen that a heat engine cannot convert heat into electricity with 100% efficiency because some of the heat is injected into the cold reservoir. The cold reservoir in fossilfuelfired electrical generation plants is the atmosphere. Generally, heat and electricity are produced separately. However, it is possible to increase the efficiency of the overall process by collecting the waste heat from electricity production and using it to heat buildings, as shown in Figure 5.21, in a process called cogeneration. An example of cogeneration is burning a fuel such as coal to generate steam used to drive a turbine that generates electricity. The steam exiting the turbine is still at a high temperature and can be used to heat buildings. In New York City, many buildings are heated using cogeneration. Because of losses in piping heat over long distances, cogeneration is most suitable for urban areas. The U.S. Department of Energy has set a goal of producing 20% of electricity by cogeneration by the year 2030.
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Figure 5.21
Exhaust gases
Illustration of cogeneration. Conventionally, electricity and heat for buildings are produced separately. Waste heat is produced in each process. Through cogeneration, most of the waste heat generated in electricity production can be used to heat buildings.
Air intake
Catalytic converter
Hot water to building
Water pipes
Natural gas fuel
Electricity to building Engine
Cold water in from building
Generator
In addition to improvements in heating and electrical generation, possible increases in energy efficiency in the household sector include the areas of cooking and lighting. Traditional electric cooktops use either natural gas or electricity to create heat, which is transferred to the cooking pot and its contents by conduction and convection. About 45% of the energy produced by the combustion of natural gas and 35% of the electrical energy are lost because the air rather than the pot is heated. The most efficient method for cooking is induction cooking in which an induction coil generates a magnetic field that induces heating in metal cookware placed on top of it. The efficiency of this method is 90%. Traditional lighting technology is very inefficient, with incandescent lights converting only 2% of the electrical work required to heat the tungsten filament to visible light. The remaining 98% of the radiated energy appears as heat. Fluorescent lights, which are much more efficient, contain a small amount of Hg vapor that emits UV light from an excited state created by an electrical discharge. The UV light is absorbed by a fluorescent coating on the surface of the bulb, which emits visible light. The fraction of the emitted light in the visible spectrum is much larger than that in incandescent lighting. For this reason, fluorescent lighting is a factor of 5 to 6 more efficient. Illumination with light emitting diodes (LED lighting) is a rapidly growing sector of lighting. LED lights have an efficiency similar to fluorescent lighting, but they are more compact and can be highly directed or diffuse sources of light. Because the transportation sector is a major user of energy, we will next discuss real engines. We will use the Otto engine and the diesel engine as examples. The Otto engine is the most widely used engine in automobiles. The engine cycle consists of four strokes, as shown in Figure 5.22. In the intake stage, the intake valve opens as the piston is moving downward, drawing a fuel–air mixture into the cylinder (see yellow arrow). At the compression stage, the intake valve is closed, and the mixture is compressed as the piston moves upward. Just after the piston has reached its highest point, the fuel–air mixture is ignited by a spark plug, and the rapid heating that results from the combustion process causes the gas to expand and the pressure to increase. The combustion process drives the piston down in a power stroke. This is shown as the power stage in Figure 5.22. Finally, the combustion products are forced out of the cylinder by the upwardmoving piston as the exhaust valve is opened (exhaust stage). To arrive at a
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CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics Spark plug Intake valve
Exhaust valve
Cylinder Piston Connecting rod Crankshaft
Stroke 1: Intake Intake valve opens as piston moves downward, drawing in fuelair mixture.
Stroke 2: Compression Intake valve closes and mixture is compressed as piston moves upward.
Stroke 3: Power Spark plug ignites compressed mixture and combustion moves piston downward.
Stroke 4: Exhaust Exhaust valve opens as piston moves upward, forcing combustion products out of cylinder.
Figure 5.22
Schematic illustration of the fourstroke cycle of an Otto engine. See text for detailed description and discussion. The left valve (see yellow arrow) is the intake valve, and the right valve (see red arrow) is the exhaust valve.
Otto cycle
Pressure
a
Adiabatic
qhot
b
d Adiabatic
qcold c
e Volume (a) Otto engine
qhot a
b
Diesel cycle
maximum theoretical efficiency, we will analyze the reversible Otto cycle, shown in Figure 5.23a. We conclude this section with a discussion of the diesel engine cycle, shown in Figure 5.23b. The reversible Otto cycle begins with the intake stroke e S c, which is assumed to take place at constant pressure. At this point, the intake valve is closed, and the piston compresses the fuel–air mixture along the adiabatic path c S d in the second step. This path can be assumed to be adiabatic because the compression occurs too rapidly to allow much heat to be transferred out of the cylinder. Ignition of the fuel–air mixture takes place at d. The rapid increase in pressure occurs at constant volume. In this reversible cycle, the combustion is modeled as a quasistatic heat transfer from a series of reservoirs at temperatures ranging from Td to Ta . The power stroke is modeled as the adiabatic expansion a S b. At this point, the exhaust valve opens and the gas is expelled. This step is modeled as the constant volume pressure decrease b S c. The upward movement of the piston expels the remainder of the gas along the line c S e, after which the cycle begins again. The efficiency of this reversible cyclic engine can be calculated as follows. Assuming CV to be constant along the segments d S a and b S c, we write
qhot = CV 1Ta  Td2 and qcold = CV 1Tb  Tc2
(5.70)
The efficiency is given by Pressure
Adiabatic c Adiabatic
e =
0 qcold 0 qhot + qcold Tb  Tc = 1 = 1  a b qhot qhot Ta  Td
(5.71)
The temperatures and volumes along the reversible adiabatic segments are related by
qcold d
e
Volume (b) Diesel engine
TcV gc  1 = TdV gd  1 and TbV gc  1 = TaV gd  1
(5.72)
because Vb = Vc and Vd = Ve . Recall that g = CP >CV . Ta and Tb can be eliminated from Equation (5.72) to give e = 1 
Tc Td
(5.73)
Figure 5.23
Pressure–volume diagrams for idealized reversible cycles of engines. (a) Otto engine and (b) diesel engine.
where Tc and Td are the temperatures at the beginning and end of the compression stroke c S d. Temperature Tc is fixed at ∼300 K, and the efficiency can be increased only
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139
by increasing Td. This is done by increasing the compression ratio, Vc >Vd . However, if the compression ratio is too high, Td will be sufficiently high that the fuel ignites before the end of the compression stroke. A reasonable upper limit for Td is 600 K, and that for Tc = 300 K, e = 0.50. This value is an upper limit because a real engine does not operate in a reversible cycle and because heat is lost along the c S d segment. Achievable efficiencies in Otto engines used in passenger cars lie in the range of 0.20 to 0.30. However, additional losses occur in the drive chain, tire deformation, and displacement of air as the vehicle moves. As shown in Figure 5.18, the overall efficiency of the transportation sector is only 21%. In the diesel engine depicted in Figure 5.23b, higher compression ratios are possible because only air is let into the cylinder in the intake stroke. The fuel is injected into the cylinder at the end of the compression stroke, thus avoiding spontaneous ignition of the fuel–air mixture during the compression stroke d S a. Because T ∼ 950 K for the compressed air, combustion occurs spontaneously without a spark plug along the constant pressure segment a S b after fuel injection. Because the fuel is injected over a time period in which the piston is moving out of the cylinder, this step can be modeled as a constant pressure heat intake. In this segment, it is assumed that heat is absorbed from a series of reservoirs at temperatures between Ta and Tb in a quasistatic process. In the other segments, the same processes occur as described for the reversible Otto cycle. The heat intake along the segment a S b is given by
qhot = Cp1Tb  Ta2
(5.74)
and qcold is given by Equation (5.70). Therefore, the efficiency is given by
e = 1 
1 Tc  Td a b g Tb  Ta
(5.75)
Similarly to the treatment of the Otto engine, Tb and Tc can be eliminated from Equation (5.75), and the expression
Vb g Va g b  a b Vd 1 Vd e = 1 g Vb Va a b  a b Vd Vd a
(5.76)
can be derived. For typical values Vb >Vd = 0.2, Va >Vd = 1>15, g = 1.5, and e = 0.64. The higher efficiency achievable with the diesel cycle in comparison with the Otto cycle is a result of the higher temperature attained in the compression cycle. Real diesel engines used in trucks and passenger cars have efficiencies in the range of 0.30 to 0.35. Just as for building heating, energy usage in the U.S. transportation sector can be drastically reduced. The U.S. personal vehicle fleet currently averages approximately 23 miles per gallon of gasoline. The corresponding figure for the European Union, where gasoline prices are more than twice as high as in the United States, is 43 miles per gallon. Hybrid gasoline–electric vehicles, which use electrical motors to power the vehicle at low speeds, switch automatically to an Otto engine at higher speeds, and use regenerative braking to capture the energy used to slow the vehicle, are currently the most efficient vehicles, averaging more than 50 miles per gallon. Purely electric vehicles (EVs) and plugin hybrids, which allow the batteries that power the electrical operation mode to be recharged at the owner’s residence, are now available, and their energy use in gasoline equivalent is more than 100 miles per gallon. An additional benefit of EVs is that if they are charged with electricity generated without burning fossil or biofuels, their operation does not generate greenhouse gases, Advances in battery technology are likely to make the use of electrical vehicles more widespread in the near future. Automobiles that use fuel cells (see Section 11.12) as a power source are of particular interest for future development.
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140
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
VOCABULARY entropy heat engine heat pump Otto engine perpetual motion machine of the first kind perpetual motion machine of the second kind
Carnot cycle Clausius inequality coefficient of performance cogeneration configuration diesel engine
refrigerator second law of thermodynamics spontaneous direction third law of thermodynamics weight of a configuration
KEY EQUATIONS Equation ∆S =
Significance of Equation
dqrev L T
∆S = nR ln
Vƒ Vi
∆S =  nR ln
∆ vap S =
Definition of entropy
+ nCV,m ln
Pƒ
Tƒ Ti
+ nCP,m ln
Pi
Tƒ Ti
∆ vap H qrev dqrev = = Tvap Tvap L T
Equation Number 5.4 and 5.8
Entropy change for ideal gas in process Vi, Ti S Vƒ, Tƒ
5.14
Entropy change for ideal gas in process Pi, Ti S Pƒ, Tƒ
5.15
Entropy change for vaporization and fusion
5.16 and 5.17
qrev ∆ fus H dqrev ∆ fus S = = = Tfus Tfus L T ∆S =
Tƒ
Vƒ
Ti
Vi
CV a dT + dV k T L L T
Entropy change for arbitrary process involving gases, liquids, and solids
5.18
Molecularlevel definition of entropy
5.21
dq T
Clausius inequality relates entropy change and dq
5.26
dqsur or for a macroscopic change, T qsur = T
How to calculate entropy change in surroundings
S = k ln W dS 7
dSsur = ∆Ssur
Tƒ
Sm1T2 = Sm10 K2 + Tb
+
L
L
C solid P,m dT′
liquid C P,m dT′
T′
Tƒ
Sm1P2 = Sm~  R ln ∆ r S ~ = a vi Si~
T′
0
+
∆ vap H Tb
+
∆Hfusion Tƒ T
+
5.27
L Tb
gas C P,m dT′
Calculating absolute entropy of gas from heat capacities
5.28
Dependence of entropy of ideal gas on pressure
5.30
Calculating entropy change in chemical reaction
5.31
Temperature dependence of reaction entropies
5.33
T′
P1bar2 P~
i
T
∆ r S 1T2 = ∆ r S 1298.15 K2 + ~
~
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∆CP dT′ L T′
298.15
14/08/17 9:53 AM
Conceptual Problems
e =
0 wcycle 0 qab
=
Thot  Tcold Tcold = 1 6 1 Thot Thot
Tƒ
Vƒ
Ti
Vi
Tƒ
Pƒ
Ti
Pi
CV a ∆S = dT + dV L T L kT CP ∆S = dT Va dP L T L
Efficiency of reversible Carnot heat engine
5.42
Entropy change for liquid, solid, or gas that undergoes a transformation from Ti, Vi to Tƒ, Vƒ.
5.53
Entropy change for liquid, solid, or gas that undergoes a transformation from Ti Pi to Tƒ, Pƒ .
5.67
141
CONCEPTUAL PROBLEMS Q5.1 Under what conditions is ∆S 6 0 for a spontaneous process? Q5.2 Why are ∆ fus S and ∆ vap S always positive? Q5.3 An ideal gas in thermal contact with the surroundings is cooled in an irreversible process at constant pressure. Are ∆S, ∆Ssur, and ∆Stot positive, negative, or zero? Explain your reasoning. Q5.4 The amplitude of a pendulum consisting of a mass on a long wire is initially adjusted to have a very small value. The amplitude is found to decrease slowly with time. Is this process reversible? Would the process be reversible if the amplitude did not decrease with time? Q5.5 A process involving an ideal gas is carried out in which the temperature changes at constant volume. For a fixed value of ∆T, the mass of the gas is doubled. The process is repeated with the same initial mass, and ∆T is doubled. For which of these processes is ∆S greater? Why? Q5.6 You are told that ∆S = 0 for a process in which the system is coupled to its surroundings. Can you conclude that the process is reversible? Justify your answer. Q5.7 Under what conditions does the equality ∆S = ∆H>T hold? Q5.8 Is the following statement true or false? If it is false, rephrase it so that it is true. The entropy of a system cannot increase in an adiabatic process. Q5.9 Which of the following processes is spontaneous? a. The reversible isothermal expansion of an ideal gas b. The vaporization of superheated water at 102°C and 1 bar c. The constant pressure melting of ice at its normal freezing point by the addition of an infinitesimal quantity of heat d. The adiabatic expansion of a gas into a vacuum Q5.10 One joule of work is done on a system, raising its temperature by one degree centigrade. Can this increase in temperature be harnessed to do one joule of work? Explain. Q5.11 Your roommate decides to cool the kitchen by opening the refrigerator. Will this strategy work? Explain your reasoning.
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Q5.12 An ideal gas undergoes an adiabatic expansion into a vacuum. Are ∆S, ∆Ssur, ∆Stot positive, negative, or zero? Explain your reasoning. Q5.13 When a saturated solution of a salt is cooled, a precipitate crystallizes out. Is the entropy of the crystalline precipitate greater or less than the dissolved solute? Explain why this process is spontaneous. Q5.14 A system undergoes a change from one state to another along two different pathways, one of which is reversible and the other of which is irreversible. What can you say about the relative magnitudes of qrev and qirr? Q5.15 An ideal gas in a piston and cylinder assembly with adiabatic walls undergoes an expansion against a constant external pressure. Are ∆S, ∆Ssur, and ∆Stot positive, negative, or zero? Explain your reasoning. Q5.16 Is the equation Tƒ
Vƒ
Ti
Vi
Tƒ CV a a ∆S = dT + dV = CV ln + 1Vƒ  Vi2 k k Ti T L T L T
valid for an ideal gas? Q5.17 Why is the efficiency of a Carnot heat engine the upper bound to the efficiency of an internal combustion engine? Q5.18 Two vessels of equal volume, pressure, and temperature both containing Ar are connected by a valve. Allowing mixing of the two volumes, what is the change in entropy when the valve is opened? Is ∆S the same if one of the volumes contained Ar and the other contained Ne? Q5.19 Without using equations, explain why ∆S for a liquid or solid is dominated by the temperature dependence of S as both P and T change. Q5.20 Solid methanol in thermal contact with the surroundings is reversibly melted at the normal melting point at a pressure of 1 atm. Are ∆S, ∆Ssur, and ∆Stot positive, negative, or zero? Explain your reasoning. Q5.21 Can incandescent lighting be regarded as an example of cogeneration during the heating season? In a season where air conditioning is required?
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NUMERICAL PROBLEMS Section 5.3 P5.1 One mole of H2O1l2 is compressed from a state described by P = 1.00 bar and T = 298 K to a state described by P = 605 bar and T = 575 K. In addition, a = 2.07 * 104 K1, and the density of liquid water can be assumed to be constant at the value 997 kg m3. Calculate ∆S for this transformation, assuming that kT = 0. P5.2 1.75 moles of an ideal gas with CV,m = 5R>2 are transformed from an initial state at T = 725 K and P = 1.50 bar to a final state at T = 273 K and P = 5.50 bar. Calculate ∆U, ∆H, and ∆S for this process. P5.3 2.50 mol of an ideal gas with CV,m = 3R>2 undergoes a process in which the initial state is described by Ti = 398 K and Pi = 1.00 bar and the final state is described by Tƒ = 798 K and Pƒ = 125 bar. Calculate ∆S for (a) a reversible process and (b) for an irreversible process at a constant external pressure of 125 bar. P5.4 The diagram below shows three paths for 2.75 mol of a monatomic ideal gas between the initial (0.500 bar, 250. K) and final (6.50 bar, 650. K) states. Calculate ∆S for paths a and c and describe how you could calculate ∆S for path b. 700
a
T (K)
600 500
b
400
c
300
1
2
3 4 P (bar)
5
6
P5.5 2.50 moles of N2 at 15.0°C and 6.20 bar undergo a transformation to the state described by 375°C and 2.00 bar. Calculate ∆S if CP,m J mol1 K1
= 30.81  11.87 * 103 1.0176 * 108
T T2 + 2.3968 * 105 2 K K
T3 K3
P5.6 Calculate ∆S for the isothermal compression of 2.25 mol of Cu(s) from 1.00 bar to 1400. bar at 298 K. a = 0.492 * 104 K1, kT = 0.78 * 106 bar 1, and the density is 8.92 g cm3. Repeat the calculation assuming that kT = 0. CP dT  VadP, calculate the T decrease in temperature that occurs if 1.75 moles of water at 298 K and 1475 bar are brought to a final pressure of 1.00 bar in a reversible adiabatic process. Assume that kT = 0.
P5.7 Using the expression dS =
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P5.8 2.50 moles of an ideal gas with CV,m = 3R>2 undergoes the transformations described in the following list from an initial state described by T = 310. K and P = 6.00 bar. Calculate q, w, ∆U, ∆H, and ∆S for each process. a. The gas undergoes a reversible adiabatic expansion until the final pressure is onefourth its initial value. b. The gas undergoes an adiabatic expansion against a constant external pressure of 1.50 bar until the final pressure is onefourth its initial value. c. The gas undergoes an expansion against a constant external pressure of zero bar until the final pressure is equal to onefourth of its initial value. P5.9 At the transition temperature of 95.4°C, the enthalpy of transition from rhombic to monoclinic sulfur is 0.38 kJ mol1. a. Calculate the entropy of transition under these conditions. b. At its melting point, 119°C, the enthalpy of fusion of monoclinic sulfur is 1.23 kJ mol1. Calculate the entropy of fusion. c. The values given in parts (a) and (b) are for 1 mol of sulfur; however, in crystalline and liquid sulfur, the molecule is present as S8. Convert the values of the enthalpy and entropy of fusion in parts (a) and (b) to those appropriate for S8. P5.10 One mole of a van der Waals gas at 25.0°C is expanded isothermally and reversibly from an initial volume of 0.015 m3 to a final volume of 0.100 m3. For the van der Waals gas, 10U>0V2T = a>V 2m. Assume that a = 0.556 Pa m6 mol2 and that b = 64.0 * 106 m3 mol1. Calculate q, w, ∆U, ∆H, and ∆S for the process. P5.11 Calculate ∆H and ∆S if the temperature of 2.50 moles of Hg1l2 is increased from 0.00°C to 125.0°C at 1 bar. Over this temperature range, CP,m = 30.093  4.944 * T 103 J mol  1 K1. K P5.12 Calculate ∆S if the temperature of 1.75 mol of an ideal gas with CV = 5R>2 is increased from 298 to 720. K under conditions of (a) constant pressure and (b) constant volume. P5.13 Between 0°C and 100°C, the heat capacity of Hg1l2 is given by CP,m1Hg, l2 1
JK
mol
1
= 30.093  4.944 * 103
T K
Calculate ∆H and ∆S if 1.75 moles of Hg1l2 are raised in temperature from 0.00° to 94.5°C at constant P. P5.14 The heat capacity of aquartz is given by CP,[email protected], s2 J K1 mol1
= 46.94 + 34.31 * 103  11.30 * 105
T K
T2 K2
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Numerical Problems
The coefficient of thermal expansion is given by a = 0.3530 * 104 K1 and Vm = 22.6 cm3 mol1. Calculate ∆Sm for the transformation aquartz (0.00°C, 1 atm) S aquartz (575.°C, 850. atm). P5.15 a. Calculate ∆S if 1.50 mol of liquid water is heated from 0.00° to 15.5°C under constant pressure if CP,m = 75.3 J K1 mol1. b. The melting point of water at the pressure of interest is 0.00°C, and the enthalpy of fusion is 6.010 kJ mol1. The boiling point is 100.°C, and the enthalpy of vaporization is 40.65 kJ mol1. Calculate ∆S for the transformation H2O1s, 0.00°C2 S H2O1g, 100.°C2. P5.16 An ideal gas sample containing 2.25 moles for which CV,m = 3R>2 undergoes the following reversible cyclical process from an initial state characterized by T = 298 K and P = 1.00 bar: a. It is expanded reversibly and adiabatically until the volume triples. b. It is reversibly heated at constant volume until T increases to 298 K. c. The pressure is increased in an isothermal reversible compression until P = 1.00 bar. Calculate q, w, ∆U, ∆H, and ∆S for each step in the cycle and for the total cycle. P5.17 The standard entropy of Pb(s) at 298.15 K is 64.80 J K1 mol1. Assume that the heat capacity of Pb(s) is given by CP,m1Pb, s2
= 22.13 + 0.01172
J mol1 K1
T T2 + 1.00 * 105 2 K K
The melting point is 327.4°C, and the heat of fusion under these conditions is 4770. J mol1. Assume that the heat capacity of Pb1l2 is given by CP,m1Pb, l2 1
JK
mol
1
= 32.51  0.00301
T K
a. Calculate the standard entropy of Pb1l2 at 850.°C. b. Calculate ∆H for the transformation Pb1s, 25.0°C2 S Pb1l, 850.°C2. P5.18 1.75 moles of an ideal gas with CV = 13>22 R undergoes the transformations described in the following list from an initial state described by T = 298 K and P = 1.00 bar. Calculate q, w, ∆U, ∆H, and ∆S for each process. a. The gas is heated to 595 K at a constant external pressure of 1.00 bar. b. The gas is heated to 595 K at a constant volume corresponding to the initial volume. c. The gas undergoes a reversible isothermal expansion at 298 K until the pressure is onethird of its initial value.
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Section 5.6 P5.19 Calculate ∆S, ∆Stot, and ∆Ssur when the volume of 225 g of CO initially at 273 K and 1.00 bar increases by a factor of three in (a) an adiabatic reversible expansion, (b) an expansion against Pext = 0, and (c) an isothermal reversible expansion. Take CP,m to be constant at the value 29.14 J mol1 K1 and assume ideal gas behavior. State whether each process is spontaneous. The temperature of the surroundings is 273 K. P5.20 A copper block of mass 10.0 kg initially at 370. K is immersed in 25.0 kg of water initially at 298 K. Calculate ∆S for the water, the Cu block, and the overall entropy change. Is the process spontaneous? Assume that the heat capacity of Cu remains constant at its 298.15 K value over the temperature interval of interest. P5.21 A 34.0 g mass of ice at 273 K is added to 145 g of H2O1l2 at 298 K at constant pressure. Is the final state of the system ice or liquid water? Calculate ∆S for the process. Is the process spontaneous? P5.22 One mole of H2O1l2 is supercooled to 4.12°C at 1 bar pressure. The equilibrium freezing temperature of water at this pressure is 0.00°C. The transformation H2O1l2 S H2O1s2 is suddenly observed to occur. By calculating ∆S,∆Ssur, and ∆Stot, verify that this transformation is spontaneous at 4.12°C. The heat capacities are given by CP,m1H2O, l2 = 75.3 J K1 mol1 and CP,m1H2O, s2 = 37.7 J K1 mol1, and ∆ fus H = 6.008 kJ mol1 at 0.00°C. Assume that the surroundings are at 4.12°C. [Hint: Consider the two pathways at 1 bar: (a) H2O1l, 4.12°C2 S H2O1s, 4.12°C2 and (b) H2O1l, 4.12°C2 S H2O1l, 0.00°C2 S H2O1s, 0.00°C2 S H2O1s, 4.12°C2. Because S is a state function, ∆S must be the same for both pathways.] P5.23 Calculate ∆Ssur and ∆Stot for the processes described in parts (a) and (b) of Problem P5.8. Which of the processes is a spontaneous process? The state of the surroundings for each part is 298 K, 1.50 bar. P5.24 Calculate ∆Ssur and ∆Stot for part (c) of Problem P5.8. Is the process spontaneous? The state of the surroundings is T = 298 K, P = 0.250 bar. P5.25 25.00 g of steam at 373 K is added to 390. g of H2O1l2 at 298 K at constant pressure of 1 bar. Is the final state of the system steam or liquid water? Calculate ∆S for the process. Assume that the final state is liquid water. If this is not the case, the calculated temperature will be greater than 373 K. P5.26 The average heat evolved by the oxidation of foodstuffs in an average adult per hour per kilogram of body weight is 7.20 kJ kg 1 hr 1. Assume the weight of an average adult is 57.5 kg. Suppose the total heat evolved by this oxidation is transferred into the surroundings during a period lasting one week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at T = 310. K.
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Section 5.7 P5.27 From the data shown in the accompanying table, derive the absolute entropy of crystalline glycine at T = 300. K. You can perform the integration numerically using either a spreadsheet program or a curvefitting routine and a graphing calculator (see Example Problem 5.7).
for ∆Sden given above, calculate the excess entropy of denaturation. In your calculations, use the dashed curve below the yellow area as the heat capacity baseline, which defines dC trs P as shown in Figure 4.8. Assume the molecular weight of the protein is 14000. g mol1. You can perform the integration by counting squares.
Heat Capacity ~ C P,m 1 J K−1 mol −1 2
Temperature (K)
0.3 2.4 7.0 13.0 25.1 35.2 43.2 50.0 56.0 61.6 67.0 72.2 77.4 82.8 88.4 94.0 99.7
CP(T)
10. 20. 30. 40. 60. 80. 100. 120. 140. 160. 180. 200. 220. 240. 260. 280. 300.
0.418 J K21g21
288
298 308 Temperature (K)
318
P5.28 The following heat capacity data have been reported for Lalanine: T 1K2
10.
20.
40.
60.
80.
100.
140.
180.
220.
260.
300.
~ C P,m 1J K1 mol12 0.49 3.85 17.45 30.99 42.59 52.50 68.93 83.14 96.14 109.6 122.7
By a graphical treatment, obtain the molar entropy of Lalanine at T = 300. K. You can perform the integration numerically using either a spreadsheet program or a curvefitting routine and a graphing calculator (see Example Problem 5.7). P5.29 Calculate the entropy of one mole of water vapor at 175°C and 0.625 bar using the information in the data tables in Appendix A. P5.30 Using your result from Problem 5.28, extrapolate the absolute entropy of Lalanine to physiological conditions, T = 310. K. Assume the heat capacity is constant between T = 300. K and T = 310. K. P5.31 For protein denaturation, the excess entropy of denaT2
turation is defined as ∆ denS =
dC trs P dT, where dC trs P , is L T T1
the transition excess heat capacity. The way in which dC trs P can be extracted from differential scanning calorimetry (DSC) data is discussed in Section 4.6 and shown in Figure 4.8. The DSC data below are for a protein mutant that denatures between T1 = 288 K and T2 = 318 K. Using the equation
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Section 5.9 P5.32 Consider the formation of glucose from carbon dioxide and water, that is, the reaction of the photosynthetic process: 6CO21g2 + 6H2O1l2 S C6H12O61s2 + 6O21g2. The following table of information will be useful in working this problem: T = 298 K ∆ f H >kJ mol ~
1
S ~ >J mol1 K1
CP,m >J mol
1
1
K
CO2 1 g2 H2O 1 l2 C6H12O6 1 s 2 O2 1 g2 393.5
285.8
1273.1
0.0
213.8
70.0
209.2
205.2
37.1
75.3
219.2
29.4
Calculate the entropy and enthalpy changes for this chemical system at T = 298.15 K and T = 365 K. Also calculate the entropy change of the surroundings and the universe at both temperatures. P5.33 Calculate ∆ r S ~ for the reaction 3 H21g2 + N21g2 S 2 NH31g2 at 670. K. Omit terms in the temperaturedependent heat capacities greater than T 2 >K2.
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Numerical Problems
P5.34 Calculate ∆ r S ~ for the reaction H21g2 + Cl21g2 S 2 HCl1g2 at 725 K. Omit terms in the temperaturedependent heat capacities greater than T 2 >K2. P5.35 Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction: C6H12O61s2 S 2CH3CHOHCOOH1s2. Thermodynamic data at T = 298.15 K for glucose and lactic acid are given below. Glucose(s) Lactic Acid(s)
𝚫 f H ~ 1kJ mol−1 2 CP,m 1J K−1 mol−1 2 S ~ 1J K−1 mol−1 2
1273.1 673.6
219.2 127.6
209.2 192.1
Calculate ∆S for the system, the surroundings, and ∆Stot at T = 310. K. Assume the heat capacities are constant between T = 298 K and T = 310. K. P5.36 The amino acid glycine dimerizes to form the dipeptide glycylglycine according to the the reaction 2 Glycine1s2 S Glycylglycine1s2 + H2O1l2 Calculate, ∆S, ∆Ssurr , and ∆Stotal at T = 298.15 K. Useful thermodynamic data follow: Glycine Glycylglycine Water ∆ f H ~ 1kJ mol12
S ~ 1J K1 mol12
537.2
746.0
285.8
103.5
190.0
70.0
Section 5.10 P5.37 The Chalk Point, Maryland, generating station supplies electrical power to the Washington, D.C., area. Units 1 and 2 have a gross generating capacity of 710. MW (megawatt). The steam pressure is 25 * 106 Pa, and the superheater outlet temperature 1Th2 is 540.°C. The condensate temperature 1Tc2 is 30.0°C. a. What is the efficiency of a reversible Carnot engine operating under these conditions? b. If the efficiency of the boiler is 91.2%, the overall efficiency of the turbine, which includes the Carnot efficiency and its mechanical efficiency, is 46.7%, and the efficiency of the generator is 98.4%, what is the efficiency of the total generating unit? (Another 5.0% needs to be subtracted for other plant losses.) c. One of the coal burning units produces 355 MW. How many metric tons 11 metric ton = 1 * 106 g2 of coal per hour are required to operate this unit at its peak output if the enthalpy of combustion of coal is 29.0 * 103 kJ kg 1? P5.38 Consider the reversible Carnot cycle shown in Figure 5.13 with 1.25 mol of an ideal gas with CV = 5R>2 as the working substance. The initial isothermal expansion occurs at the hot reservoir temperature of Thot = 825 K from an initial volume of 2.25 L 1Va2 to a volume of 10.5 L 1Vb2. The system then undergoes an adiabatic expansion until the temperature falls to Tcold = 298 K. The system then undergoes an isothermal compression and a subsequent adiabatic compression until the initial state described by Ta = 825 K and Va = 2.25 L is reached.
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145
a. Calculate Vc and Vd. b. Calculate w for each step in the cycle and for the total cycle. c. Calculate e and the amount of heat that is extracted from the hot reservoir to do 1.00 kJ of work in the surroundings. P5.39 Using your results from Problem P5.38, calculate q, ∆U, and ∆H for each step in the cycle and for the total cycle described in Figure 5.13. P5.40 Beginning with Equation (5.38), use Equation (5.39) to eliminate Vc and Vd to arrive at the result wcycle = nR1Thot  Tcold2 ln Vb >Va. P5.41 Using your results from Problems P5.38 and P5.39, calculate ∆S, ∆Ssur, and ∆Stot for each step in the cycle and for the total Carnot cycle described in Figure 5.13. Section 5.13 P5.42 An electrical motor is used to operate a Carnot refrigerator with an interior temperature of 0.00°C. Liquid water at 0.00°C is placed into the refrigerator and transformed to ice at 0.00°C. If the room temperature is 300. K, what mass of ice can be produced in one day by a 0.375hp motor that is running continuously? Assume that the refrigerator is perfectly insulated and operates at the maximum theoretical efficiency. P5.43 An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting as the hot reservoir. Assume that an air conditioner consumes 1.25 * 103 W of electrical power and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 1.75, how much heat can be extracted from the house in a day? P5.44 The maximum theoretical efficiency of an internal combustion engine is achieved in a reversible Carnot cycle. Assume that the engine is operating in the Otto cycle and that CV,m = 5R>2 for the fuel–air mixture initially at 298 K (the temperature of the cold reservoir). The mixture is compressed by a factor of 7.5 in the adiabatic compression step. What is the maximum theoretical efficiency of this engine? How much would the efficiency increase if the compression ratio could be increased to 17? Do you see a problem in doing so? P5.45 Calculate ∆S, ∆Ssur, and ∆Stot per day for the airconditioned house described in Problem 5.43. Assume that the interior temperature is 65.0°F and the exterior temperature is 99.0°F. The corresponding absolute temperature values for 65.0 and 99.0 degrees are 291.48 K and 310.37 K, respectively. P5.46 A refrigerator is operated by a 0.25hp 11 hp = 746 watts2 motor. If the interior is to be maintained at 2.75°C and the room temperature is 34.5°C, what is the maximum heat leak (in watts) that can be tolerated? Assume that the coefficient of performance is 50. % of the maximum theoretical value. What happens if the leak is greater than your calculated maximum value?
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P5.47 The mean solar flux at the Earth’s surface is ∼2.00 J cm2 min1. In a nonfocusing solar collector, the temperature can reach a value of 95.5°C. A heat engine is operated using the collector as the hot reservoir and a cold reservoir at 298 K. Calculate the area of the collector needed to produce 1000. watts. Assume that the engine operates at the maximum Carnot efficiency.
P5.48 The interior of a refrigerator is held at 34°F and the interior of a freezer is held at 1.00°F. If the room temperature is 68°F, by what factor is it more expensive to extract the same amount of heat from the freezer than from the refrigerator? Assume that the theoretical limit for the performance of a reversible refrigerator is valid in this case.
FURTHER READING BenNaim, A. Entropy and the Second Law. Singapore: World Scientific Publishing Co., 2012. Lambert, F. “Configurational Entropy Revisited.” Journal of Chemical Education 84 (2007): 1548–1550. Lambert, F. “Entropy Is Simple, Qualitatively.” Journal of Chemical Education 79 (2002): 1241–1246. Lambert, F. “Shuffled Cards, Messy Desks and Disorderly Dorm Rooms—Examples of Entropy Increase? Nonsense!” Journal of Chemical Education 76 (1999): 1385–1387.
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Lee, S., Lee, K., and Lee, J. “An Alternative Presentation of the Second Law of Thermodynamics.” Journal of Chemical Education 92 (2015): 771–773. Zemansky, M. W., and Dittman, R. H. Heat and Thermodynamics (7th ed.). New York: McGrawHill, 1997, Chapter 7.
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C H A P T E R
6
Chemical Equilibrium
6.1 Gibbs Energy and Helmholtz Energy
6.2 Differential Forms of U, H, A, and G
6.3 Dependence of Gibbs and Helmholtz Energies on P, V, and T
WHY is this material important? In this chapter, spontaneity is discussed in the context of the approach to equilibrium of a reactive mixture of gases. Two new state functions are introduced that express spontaneity in terms of the properties of the system only, so that the surroundings do not have to be considered explicitly.
6.4 Gibbs Energy of a Reaction Mixture
6.5 Calculating the Gibbs Energy of Mixing for Ideal Gases
6.6 Calculating the Equilibrium Position for a GasPhase Chemical Reaction
WHAT are the most important concepts and results? The Helmholtz energy provides a criterion for determining if a reaction mixture will evolve toward reactants or products if the system is at constant V and T. The Gibbs energy is the criterion for determining if a reaction mixture will evolve toward reactants or products if the system is at constant P and T. Using the Gibbs energy, we derive a thermodynamic equilibrium constant KP that predicts the equilibrium concentrations of reactants and products in a mixture of reactive ideal gases.
6.7 Introducing the Equilibrium Constant for a Mixture of Ideal Gases
6.8 Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gases
6.9 Variation of KP with Temperature
WHAT would be helpful for you to review for this chapter?
6.10 Equilibria Involving Ideal Gases
It would be helpful to review the material in Chapter 5 because the current chapter is based on the discussion of spontaneity in that chapter.
6.11 Expressing the Equilibrium
6.1 GIBBS ENERGY AND
Capacities Solely in Terms of Measurable Quantities
In Chapter 5, it was shown that the direction of spontaneous change for an arbitrary process is predicted by ∆S + ∆Ssur 7 0. In this section, this spontaneity criterion is used to derive two new state functions, the Gibbs and Helmholtz energies. These new state functions provide the basis for all further discussions of spontaneity. Formulating spontaneity in terms of the Gibbs and Helmholtz energies rather than ∆S + ∆Ssur 7 0 has the important advantage that spontaneity and equilibrium can be defined using only properties of the system rather than those of both the system and surroundings. We will also show that the Gibbs and Helmholtz energies allow us to calculate the maximum work that can be extracted from a process such as a chemical reaction. The fundamental expression governing spontaneity is the Clausius inequality [Equation (5.26)], written in the form TdS Ú dq
Constant in Terms of Mole Fraction or Molarity
6.12 Expressing U and H and Heat
HELMHOLTZ ENERGY
and Solid or Liquid Phases
6.13 (Supplemental Section) A Case Study: The Synthesis of Ammonia
6.14 (Supplemental Section) Measuring ∆G for the Unfolding of Single RNA Molecules
(6.1)
The equality is satisfied only for a reversible process. Because dq = dU  dw,
TdS Ú dU  dw or, equivalently, dU + dw + TdS Ú 0
(6.2)
As discussed in Section 2.2, a system can do different types of work on the surroundings. Keep in mind that the work done by the system on the surroundings is equal in magnitude and opposite in sign to the work done by the surroundings on the system.
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CHAPTER 6 Chemical Equilibrium
It is particularly useful to distinguish between expansion work, in which the work arises from a volume change in the system, and nonexpansion work (for example, electrical work). As we will see in Example Problem 6.2, nonexpansion work is the more useful of the two types of work. We rewrite Equation (6.2) in the form
dU  Pext dV + dwnonexp + TdS Ú 0
(6.3)
Equation (6.3) expresses the condition of spontaneity for an arbitrary process in terms of the changes in the state functions U, V, S, and T as well as the pathdependent functions Pext dV and dwnonexp. To make a connection with the discussion of spontaneity in Chapter 5, consider a special case of Equation (6.3). For an isolated system, w = 0 and dU = 0. Therefore, Equation (6.3) reduces to the familiar result derived in Section 5.5: dS Ú 0
(6.4)
Because chemists are interested in systems that interact with their environment, we define equilibrium and spontaneity for such systems. As was done in Chapters 1 through 5, it is useful to consider transformations at constant temperature and either constant volume or constant pressure. Keep in mind that constant T and P (or V) does not imply that these variables are constant throughout the process, but rather that they are the same for the initial and final states of the process. We first consider isothermal processes. For an isothermal process, d1TS2 = TdS, and Equation (6.3) can be written in the following form: dU + TdS Ú dwexp  dwnonexp or, equivalently,
d1U  TS2 … dwexp + dwnonexp
(6.5)
The combination of state functions U  TS, which has the units of energy, defines a new state function that we call the Helmholtz energy, abbreviated A. By using this definition, the general condition of spontaneity for isothermal processes becomes
dA  dwexp  dwnonexp … 0
(6.6)
Because the equality applies for a reversible transformation, Equation (6.6) provides a way to calculate the maximum work that a system can do on the surroundings in an isothermal process by considering a reversible process, in which case the inequality in Equation (6.6) becomes an equality.
dwmax = dwexp + dwnonexp = dA
(6.7)
Example Problem 6.1 illustrates the usefulness of A for calculating the maximum work available through a chemical reaction.
EXAMPLE PROBLEM 6.1
Concept The value of the change in Helmholtz energy is the maximum work that can be extracted from a process.
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You have made a scientific breakthrough and can construct a fuel cell based on the electrochemical oxidation of a hydrocarbon fuel. This fuel cell will allow you to convert electrical work to other work such as driving the wheels of a car without the second law losses that arise from combustion in a heat engine. Two possible choices for a fuel are methane and octane. (a) Calculate the maximum work available through the electrochemical oxidation of these two hydrocarbons on a per mole and a per gram basis at 298.15 K and 1 bar pressure. (b) Another option to do work on the surroundings is to burn the hydrocarbon fuel in a heat engine, as discussed in Section 5.10. Calculate the maximum work available in the combustion of these hydrocarbons in a heat engine with an efficiency of 50.0%. The standard enthalpies of combustion ~ are ∆ com H ~ 1CH4, g2 = 891 kJ mol1, ∆H com 1C8H18, l2 = 5471 kJ mol1, and 1 1 ~ Sm 1C8H18, l2 = 361.1 J mol K . Use tabulated values in Appendix A for other data needed for these calculations. Assume ideal gas behavior. Are there any factors not mentioned that should be taken into account in making a decision between these two fuels?
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149
Solution a. At constant T, ∆A and ∆H are related by ∆A = ∆U  T∆S = ∆H  ∆1PV2 T∆S = ∆H  ∆nRT  T∆S The oxidation reactions are Methane: CH41g2 + 2O21g2 S CO21g2 + 2H2O1l2
Octane: C8H181l2 + 25>2 O21g2 S 8CO21g2 + 9H2O1l2
∆ ox A~ 1CH4, g2 = ∆ oxU ~ 1CH4, g2  T ¢
Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2 ≤ Sm~ 1CH4, g2  2Sm~ 1O2, g2
= ∆ com H ~ 1CH4, g2  ∆nRT
T1Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2  Sm~ 1CH4, g2 2Sm~ 1O2, g22
= 891 * 103 J mol1 + 2 * 8.314 J mol1 K1 * 298.15 K 298.15 K * a
∆ ox A~ 1CH4, g2 = 814 kJ mol1
213.8 J mol1 K1 + 2 * 70.0 J mol1 K1 b 186.3 J mol1 K1  2 * 205.2 J mol1 K1
∆ ox A~ 1C8H18, l2 = ∆ comU ~ 1C8H18, l2
T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2  Sm~ 1C8H18, l2 
25 ~ S 1O , g2b 2 m 2
T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2  Sm~ 1C8H18, l2 
25 ~ S 1O , g2b 2 m 2
= ∆ com H ~ 1C8H18, l2  ∆nRT
= 5471 * 103 J mol1 +
9 * 8.314 J mol1 K1 * 298.15 K 2
8 * 213.8 J mol1 K1 + 9 * 70.0 J mol1 K1 298.15 K * ° ¢ 25 361.1 J mol1 K1 * 205.2 J mol1 K1 2
∆ ox A~ 1C8H18, l2 = 5285 kJ mol1
The maximum work for both reactions is negative, which is necessary to do work by the fuel cell on the surroundings. On a per mol basis, octane (molecular weight 114.25 g mol1) is capable of producing a factor of 6.5 more work than methane (molecular weight 16.04 g mol1). However, on a per gram basis, methane and octane are nearly equal in their ability to produce work (50.6 kJ g 1 versus 46.3 kJ g 1). You might want to choose octane because it can be stored as a liquid at atmospheric pressure. By contrast, a pressurized tank is needed to store methane as a liquid at 298.15 K. b. If these hydrocarbons are burned and used in a heat engine to generate work, the maximum work is given by wmax = qhot * efficiency = ∆ com H ~ * efficiency = 5471 kJ mol1 * 0.500 = 2736 kJ mol1 for C8H18, l = 891 kJ mol1 * 0.500 = 446 kJ mol1 for CH4, g The efficiency enters in the calculation in part (b) because, as we showed in Chapter 5, heat cannot be converted into work with 100% efficiency. This limitation does not apply to the fuel cell where electrical work is converted to another type of work. Ignoring friction, which appears in both methods of generating work, one type of work can in principle (but not practically) be converted to another type of work with 100% efficiency. Comparing the answers to parts (a) and (b), we see why optimizing the performance of fuel cells, which at present are not widely used, is an active area of research.
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CHAPTER 6 Chemical Equilibrium
Concept ∆A provides a criterion for spontaneity in terms of system variables only for a process at constant V and T.
Having demonstrated the usefulness of A in calculating the maximum amount of work that can be generated in an arbitrary process, we return to the issue of spontaneity. In discussing the Helmholtz energy, dT = 0 was the only constraint applied. We now apply the additional constraint dV = 0, in which case we are considering a process at constant T and V. Because V is constant, dwexp = 0. If nonexpansion work also is not possible in the transformation, dwnonexp = dwexp = 0, and the condition that defines spontaneity and equilibrium for a process at constant T and V becomes dA … 0
(6.8)
Equation (6.8) states that a process for which T and V have the same initial and final values is spontaneous if dA 6 0. The system is at equilibrium with its surroundings if dA = 0, and neither direction of change is spontaneous. Chemical reactions are more commonly studied under constant pressure than constant volume conditions. Therefore, the condition for spontaneity is considered next for an isothermal constant pressure process. At constant P and T, d1PV2 = Pext dV and d1TS2 = TdS. In this case, using the relation H = U + PV, Equation (6.3) can be written in the form d1U + PV  TS2 = d1H  TS2 … dwnonexp (6.9) The combination of state functions H  TS, which has the units of energy, defines a new state function called the Gibbs energy, abbreviated G. The condition for spontaneity for an isothermal process at constant pressure becomes dG  dwnonexp … 0
(6.10)
For a reversible process, the equality holds, and the change in the Gibbs energy is a measure of the maximum nonexpansion work that can be produced in the transformation. Expansion work is included in the definition of Gibbs energy. A particularly important application of Equation (6.10) is to calculate the electrical work produced by a reaction in an electrochemical cell or fuel cell, as shown in Example Problem 6.2. This topic will be discussed in more detail in Chapter 11.
EXAMPLE PROBLEM 6.2
Concept The value of the change in Gibbs energy is the maximum nonexpansion work that can be extracted from a process.
(a) Calculate the maximum nonexpansion work that can be produced by the fuel cell oxidation reactions in Example Problem 6.1. (b) Compare your results with those of Example Problem 6.1. What can you conclude about the relative amounts of expansion and nonexpansion work available by carrying out these reactions?
Solution a. ∆ oxG ~ 1CH4, g2 = ∆ com H ~ 1CH4, g2 T ¢
Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2 ≤ Sm~ 1CH4, g2  2Sm~ 1O2, g2
= 891 * 103 J mol1 298.15 K * ¢
213.8 J mol1 K1 + 2 * 70.0 J mol1 K1 ≤ 186.3 J mol1 K1  2 * 205.2 J mol1 K1
∆ oxG ~ 1CH4, g2 = 819 kJ mol1
∆ oxG ~ 1C8H18, l2 = ∆ com H ~ 1C8H18, l2
T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2  Sm~ 1C8H18, l2 = 5471 * 103 J mol1
25 ~ S 1O , g2b 2 m 2
8 * 213.8 J mol1 K1 + 9 * 70.0 J mol1 K1 ¢ 25 361.1 J mol1 K1 * 205.2 J mol1 K1 2 1 ~ ∆ oxG 1C8H18, l2 = 5296 kJ mol 298.15K * °
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6.2 DIFFERENTIAL FORMS OF U, H, A, AND G
b. For methane, wexp + wnon = 814 kJ mol1 and wnon = 819 kJ mol1. Therefore, wexp = 5 kJ mol1. For octane, wexp + wnon = 5285 kJ mol1 and wnon = 5296 kJ mol1. Therefore, wexp = 11 kJ mol1. This comparison tells us that the magnitude of expansion work is much smaller than that of nonexpansion work. The sign of the expansion work is positive (work done on the system rather than by the system) because the number of moles of gaseous species decreases in the reaction for both methane and octane. Note that we are calculating the expansion work at 298 K. In an automobile engine, the combustion temperature is approximately 2000 K, so that expansion work is generated by heating a fuel–air mixture using the enthalpy of combustion. The hot gas mixture exerts a force on the pistons that is converted to a rotary motion of the wheels. After having demonstrated the usefulness of G for determining the amount of nonexpansion work available from an arbitrary process, we consider the usefulness of G for determining the spontaneous direction of a process. We consider a transformation at constant P and T for which nonexpansion work is not possible. In this case, the condition for spontaneity becomes dG … 0
151
Concept ∆G provides a criterion for spontaneity in terms of system variables only for a process at constant P and T.
(6.11)
In summary, Gibbs and Helmholtz energies are complementary state functions in predicting spontaneity. If nonexpansion work does not occur, a process for which the initial and final V and T values are the same is spontaneous if dA 6 0 and a process for which the initial and final P and T values are the same is spontaneous if dG 6 0. We emphasize that these criteria are useful because they are based on system properties; importantly, we need not be concerned about what happens in the surroundings. However, you should keep in mind that the fundamental criterion for spontaneity is still ∆S + ∆Ssur 7 0. By expressing ∆Ssur as dU>T for a process at constant V and T and as dH>T for a process at constant P and T, we have been able to write ∆Ssur in terms of changes in state functions of the system. We can see this clearly by dividing ∆G = ∆H  T∆S by T:

∆G ∆H = + ∆S = ∆Ssurr + ∆Ssystem T T
(6.12)
6.2 DIFFERENTIAL FORMS OF U, H, A, AND G To this point, we have defined four different state functions U, H, A, and G, all of which have the units of energy. How do these functions differ from each other, and for what purpose are they most useful? The total energy U is a measure of all of the kinetic and potential energy in the system. The enthalpy H tells us how much heat flows out of or into the system if a process is carried out at constant pressure. For example, ∆H for the combustion of natural gas allows us to calculate the burn rate in a furnace used to heat a home in winter. The Helmholtz energy A tells us the maximum total work that the process can produce. It also provides a criterion for spontaneity for processes that occur at constant T and V. The Gibbs energy G tells us the maximum nonexpansion work that the process can produce. This is a critical piece of information in designing batteries and fuel cells. The Gibbs energy also provides a criterion for spontaneity for processes that occur at constant T and P. In this section, we discuss how these energy state functions depend on the macroscopic system variables. To do so, the differential forms dU, dH, dA, and dG are developed. As we will see, these differential forms are essential in calculating how U, H, A, and G vary with state variables such as P and T. Starting from the definitions
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U H A G
= = = =
q + w U + PV U  TS H  TS = U + PV  TS
(6.13)
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CHAPTER 6 Chemical Equilibrium
the following total differentials can be formed:
dU dH dA dG
= = = =
TdS TdS TdS TdS
+
PdV PdV + PdV + VdP = TdS + VdP PdV  TdS  SdT = SdT  PdV VdP  TdS  SdT = SdT + VdP
(6.14) (6.15) (6.16) (6.17)
These differential forms express the internal energy as U(S,V ), the enthalpy as H(S,P), the Helmholtz energy as A(T,V ), and the Gibbs energy as G(T,P). Although other combinations of variables can be used, these natural variables are used because the differential expressions are compact. What information can be obtained from the differential expressions in Equations (6.14) through (6.17)? Because U, H, A, and G are state functions, two different equivalent expressions, such as those written for dU here, can be formulated:
dU = TdS  PdV = a
0U 0U b dS + a b dV 0S V 0V S
(6.18)
For Equation (6.17) to be valid, the coefficients of dS and dV on both sides of the equation must be equal. Applying this reasoning to Equations (6.14) through (6.17), we obtain the following expressions: a
Concept
Equations (6.19)–(6.22) give information on how U, H, A, and G change with their natural variables.
a
0U 0U b = T and a b = P 0S V 0V S 0H 0H b = T and a b = V 0S P 0P S
a
a
0A 0A b = S and a b = P 0T V 0V T
0G 0G b = S and a b = V 0T P 0P T
(6.19) (6.20) (6.21) (6.22)
These expressions state how U, H, A, and G vary with their natural variables. For example, because T and V always have positive values, Equation (6.20) states that H increases if either the entropy or the pressure of the system increases. We discuss how to use these relations for macroscopic changes in the system variables in Section 6.3. The differential expressions in Equations (6.14) through (6.17) can be used in a second way. From Section 3.1, we know that because dU is an exact differential: ¢
0 0U 1S,V2 0 0U 1S,V2 a b ≤ = ¢ a b ≤ 0V 0S 0S 0V V S S V
Equating the mixed second partial derivative derived from Equations (6.14) through (6.17), we obtain the following four Maxwell relations:
Concept The Maxwell relations connect seemingly unrelated quantities.
a
0T 0P b = a b 0V S 0S V
a
0S 0P a b = a b = kT 0V T 0T V
a
a
0T 0V b = a b 0P S 0S P
(6.23)
(6.24)
0S 0V b = a b = Va 0P T 0T P
(6.25) (6.26)
A derivation for one of the Maxwell relations is shown in Example Problem 6.3. Equations (6.23) and (6.24) refer to a partial derivative at constant S. What conditions must a transformation at constant entropy satisfy? Because dS = dqrev >T, a transformation at constant entropy refers to a reversible adiabatic process.
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6.3 DEPENDENCE OF GIBBS AND HELMHOLTZ ENERGIES ON P, V, AND T
153
EXAMPLE PROBLEM 6.3 Derive the Maxwell relation a
0T 0P b = a b 0V S 0S V
Solution Because the variables S and V appear in the equation, we look for a differential of U, H, A, or G that includes these variables. Equation (6.18) shows that this is the case for U. We next form the mixed partial derivatives of U as follows: a
0 0U 1S,V2 0 0U 1S,V2 a b b = ¢ a b ≤ . 0V 0S 0S 0V V S S V
Substituting dU = TdS  PdV in the previous expression, ¢ a
0 03 TdS  PdV4 0 03 TdS  PdV4 a b ≤ = ¢ a b ≤ 0V 0S 0S 0V V S S V
0T 0P b = a b 0V S 0S V
The Maxwell relations have been derived using only the property that U, H, A, and G are state functions. These four relations are extremely useful in transforming seemingly obscure partial derivatives to other partial derivatives that can be directly measured. For example, these relations will be used to express U, H, and heat capacities solely in terms of measurable quantities such as kT, a, and the state variables P, V, and T in Section 6.12.
6.3 DEPENDENCE OF GIBBS AND HELMHOLTZ ENERGIES ON P, V, AND T
The state functions A and G are particularly important for chemists because of their roles in determining the direction of spontaneous change in a reaction mixture. For this reason, we need to know how A changes with T and V, and how G changes with T and P to be able to predict if the direction of spontaneous change will change with changes in P, T, and V. We begin by asking how A changes with T and V. From Section 6.2,
a
0A 0A b = S and a b = P 0T V 0V T
(6.21)
where S and P always take on positive values. Therefore, the general statement can be made that the Helmholtz energy of a pure substance decreases as either the temperature or the volume increases. Because most reactions of interest to chemists are carried out under constant pressure rather than constant volume conditions, we will devote more attention to the properties of G than to those of A. From Section 6.2,
a
0G 0G b = S and a b = V 0T P 0P T
Concept G decreases as T increases and increases as P increases.
(6.22)
Thus, Gibbs energy decreases as temperature increases, but it increases as pressure increases. How can Equation (6.22) be used to calculate changes in G with macroscopic changes in the variables T and P? We will consider each of these variables separately. Because G is a state function, the total change in G as both T and P are varied is the sum of the separate contributions. We first discuss the change in G with P.
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CHAPTER 6 Chemical Equilibrium
For a macroscopic change in P at constant T, the second expression in Equation (6.22) is integrated at constant T, yielding P
L~
P
P
dG = G1T, P2  G ~ 1T, P ~ 2 =
L~
(6.27)
Vd P′
P
where we have chosen the initial pressure to be the standard state pressure P ~ = 1 bar. This equation takes on different forms for liquids and solids, on the one hand, and for gases, on the other hand. For liquids and solids, the volume is, to a good approximation, independent of P over a limited range in P and P
G1T, P2 = G ~ 1T, P ~ 2 +
L~ P
Vd P′ ≈ G ~ 1T, P ~ 2 + V1P  P ~ 2
(6.28)
By contrast, the volume of a gaseous system changes appreciably with pressure. In calculating the change of G with P at constant T, any path connecting the same initial and final states gives the same result. Choosing the reversible path and assuming ideal gas behavior, we find that P
G1T, P2 = G ~ 1T2 +
L~ P
P
VdP′ = G ~ 1T2 +
nRT P dP′ = G ~ 1T2 + nRT ln ~ P′ P L~ P (6.29)
The functional dependence of the molar Gibbs energy Gm on P for an ideal gas is shown in Figure 6.1, where Gm approaches minus infinity as the pressure approaches zero. This is a result of the volume dependence of S that was discussed in Section 5.5, ∆S = nR ln1Vf > Vi2 at constant T. As P S 0, V S ∞ and S S ∞ . Therefore, G = H  TS S  ∞ in this limit. We next investigate the dependence of G on T. As we will see in the next section, the thermodynamic equilibrium constant K is related to G>T. Therefore, it is more useful to obtain an expression for the temperature dependence of G>T than for the temperature dependence of G. Using the chain rule, we see that this dependence is given by
a
03 G>T 4 0T
b
P
= =
d3 1>T 4 1 0G a b + G T 0T P dT
1 0G G S G G + TS H a b  2 =   2 = =  2 2 T 0T P T T T T T
(6.30)
Figure 6.1
Molar Gibbs energy of an ideal gas relative to its standard state value. The value of G1T, P2  G ~ 1T2 is shown as a function of pressure at 298.15 K. G1T, P2  G ~ 1T2 approaches minus infinity as P approaches 0.
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[Gm(298.15 K,P) 2 G
[ m(298.15
K)] (J)
In the second line of Equation (6.30), we have used Equation (6.22), 10G>0T2P = S, and the definition G = H  TS. Equation (6.30) is known as the Gibbs–Helmholtz equation. Because d11>T2 1 =  2 dT T 4000 2000 0 1 22000
2 P/P
[
3
4
5
24000
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6.4 Gibbs Energy of a Reaction Mixture
155
the Gibbs–Helmholtz equation can also be written in the form a
03 G>T 4 03 1>T 4
b
P
= a
03 G>T 4 0T
b a P
dT H b =  2 1T 22 = H d3 1>T 4 T
(6.31)
The preceding equation also applies to the change in G and H associated with a process such as a chemical reaction. Replacing G by ∆G and integrating Equation (6.31) at constant P, T2
LT1
T
da
2 ∆G 1 b = ∆Hd a b T T LT1
∆G1T22 ∆G1T12 1 1 = + ∆H1T12 a  b T2 T1 T2 T1
(6.32)
Concept The Gibbs–Helmholtz equation is used to calculate the variation of ∆G with temperature.
(6.33)
It has been assumed in the second equation that ∆H is independent of T over the temperature interval of interest. If this is not the case, the integral must be evaluated numerically, using tabulated values of ∆ f H ~ and temperaturedependent expressions of CP,m for reactants and products.
EXAMPLE PROBLEM 6.4 The value of ∆ f G ~ for Fe1g2 is 370.7 kJ mol1 at 298.15 K, and the value of ∆ f H ~ for Fe1g2 is 416.3 kJ mol1 at the same temperature. Assuming that ∆ f H ~ is constant in the interval 250–400 K, calculate ∆ f G ~ for Fe1g2 at 400. K.
Solution ∆ f G ~ 1T22 = T2 c
∆ f G ~ 1T12 1 1 + ∆ f H ~ 1T12 * a  b d T1 T2 T1
370.7 * 103 J mol1 + 416.3 * 103 J mol1 298.15 K = 400. K * ≥ ¥ 1 1 *a b 400. K 298.15 K
∆ f G ~ 1400. K2 = 355.1 kJ mol1
Analogies to Equations (6.27) and (6.32) for the dependence of A on V and T are left to the endofchapter problems.
6.4 GIBBS ENERGY OF A REACTION MIXTURE To this point, the discussion has been limited to systems at a fixed composition. The rest of the chapter focuses on using the Gibbs energy to understand equilibrium in a reaction mixture under constant pressure conditions, which correspond to typical laboratory experiments. Because reactants are consumed and products are generated in chemical reactions, the expressions derived for state functions such as U, H, S, A, and G must be revised to include changes in composition. We focus on G in the following discussion. For a reaction mixture containing species 1, 2, 3, c, G is not only a function of the variables T and P. Because it depends on the number of moles of each species, G is written in the form G = G1T, P, n1, n2, n3, c2. The total differential dG is
dG = a
0G 0G 0G b dT + a b dP + a b dn 0T P,n1,n2 c 0P T,n1,n2 c 0n1 T,P,n2 c 1
+a
0G 0G b dn + a b dn + c 0n2 T,P,n1 c 2 0n3 P,n1,n2 c 3
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(6.34)
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CHAPTER 6 Chemical Equilibrium
where the variables dni have the units of moles. Note that if the concentrations do not change, all of the dni = 0, and Equation (6.34) reduces to dG = a
0G 0G b dT + a b dP. 0T P 0P T
Equation (6.34) can be simplified by defining the chemical potential, mi, as mi = a
0G b 0ni P,T,nj ≠ ni
(6.35)
It is important to realize that although mi is defined mathematically in terms of an infinitesimal change in G as the amount dni of species i changes by an infinitesimal amount, mi is the change in the Gibbs energy of the mixture per mole of substance i added at constant concentration. These two requirements are not contradictory. To keep the concentration constant, we add one mole of substance i to a huge vat containing many moles of the various species. In this case, 10G>0ni2P,T,nj ≠ ni, where dni S 0, or the ratio 1∆G> ∆ni2P,T,nj ≠ ni, where ∆ni is 1 mol have the same value. Using the notation of Equation (6.35), we can write Equation (6.34) as follows:
Concept The chemical potential of a chemical species is an intensive quantity numerically equal to the Gibbs energy of the pure substance.
dG = a
0G 0G b dT + a b dP + a mi dni 0T P,n1,n2 c 0P T,n1,n2 c i
(6.36)
To ensure that you are familiar with the units for all quantities in Equation (6.36), we recommend that you carry out a unit analysis of this equation. Now imagine integrating Equation (6.36) at constant composition and at constant T and P from an infinitesimal size of the system where ni S 0 and therefore G S 0 to a macroscopic size where the Gibbs energy has the value G. Because T and P are constant, the first two terms in Equation (6.36) do not contribute to the integral. Furthermore, because the composition is constant, mi is constant, and G
L
0
dG′ = a mi i
ni
L
G = a nimi i
dn′i
0
(6.37)
Note that mi depends on the number of moles of each species present. If the system consists of a single pure substance A, G = nAGm,A because G is an extensive quantity.1 Applying Equation (6.35),
mA = a
03 nAGm,A 4 0G b = a b = Gm,A 0nA P,T 0nA P,T
(6.38)
showing that mA is an intensive quantity equal to the molar Gibbs energy of A, Gm,A, for a pure substance. As we will show in the next section, mA in a mixture depends on the concentration of species A. Why is mi called the chemical potential of species i? This can be understood by assuming that the chemical potential for species i has the values mIi in region I, and mIIi in region II of a given mixture with mIi 7 mIIi. If dni moles of species i are transported from region I to region II, at constant T and P, the change in G is given by
dG = mIi dni + mIIi dni = 1mIIi  mIi2dni 6 0
(6.39)
Because dG 6 0, this process is spontaneous. For a given species, transport will occur spontaneously from a region of high chemical potential to one of low chemical potential. The flow of material will continue until the chemical potential has the same value in 1
As discussed in Chapter 2, we can only measure changes in energy rather than absolute values of energy. However, in some cases it is useful to define a reference zero and report absolute values of energy functions such as H and G. As for formation enthalpies, we set H m~ = Gm~ = 0 for all elements in their standard reference state at 298.15 K. With this convention, Gm~ = ∆ f G ~ for all compounds in their standard reference state.
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6.5 Calculating the Gibbs Energy of Mixing for Ideal Gases
all regions of the mixture. Note the analogy between this process and the flow of mass in a gravitational potential field or the flow of charge in an electrostatic potential field. Therefore, the term chemical potential is appropriate. In this discussion, we have defined a new criterion for equilibrium in a multicomponent mixture: At equilibrium, the chemical potential of each species is the same throughout a mixture.
157
Xe
6.5 CALCULATING THE GIBBS ENERGY OF MIXING FOR IDEAL GASES
Ar
In this section, we describe how the chemical potential of a reactant or product species changes through mixing with other species. Using these results, we show in Section 6.7 that the partial pressures of all constituents of a gaseous reaction mixture are related by the thermodynamic equilibrium constant, KP. Recall from Equation (6.29) that the molar Gibbs energy of a pure ideal gas depends on its pressure as G1T, P2 = G ~ 1T2 + nRT ln1P>P ~ 2. Therefore, the chemical potential of component i in a mixture of ideal gases can be written in the form mi1T, P2 = mi~ 1T2 + RT ln
Pi P~
Ne
(6.40)
where mi~ 1T2 is the chemical potential of the pure gas at a pressure of 1 bar and temperature T and Pi is the partial pressure of component i. We see that the chemical potential of a gas in a mixture depends logarithmically on its partial pressure. We next obtain a quantitative relationship between ∆ mixG and the mole fractions of the individual constituents of the mixture. Consider the system shown in Figure 6.2. The four compartments contain the gases He, Ne, Ar, and Xe at the same temperature and pressure. The volumes of the four compartments differ. To calculate ∆ mixG, we must compare G for the initial state shown in Figure 6.2 and the final state in which all gases are uniformly distributed in the container. For the initial state, in which we have four pure separated substances,
He
Figure 6.2
An isolated system consists of four separate subsystems that contain He, Ne, Ar, and Xe, each at the same pressure. The barriers separating these subsystems can be removed, leading to mixing.
Gi = GHe + GNe + GAr + GXe = nHemHe + nNemNe + nArmAr + nXemXe
~ = nHe amHe + RT ln
P P P ~ ~ b + nNe amNe + RT ln ~ b + nAr amAr + RT ln ~ b P~ P P
~ + nXe amXe + RT ln
P b P~
(6.41)
For the final state in which all four components are dispersed in the mixture, ~ Gƒ = nHe amHe + RT ln
PNe PHe PAr ~ ~ b + nNe amNe + RT ln ~ b + nAr amAr + RT ln ~ b ~ P P P
~ + nXe amXe + RT ln
PXe b P~
(6.42)
The Gibbs energy of mixing is Gƒ  Gi or ∆ mixG = Gƒ  Gi = nHe RT ln
PNe PHe PXe PAr + nNe RT ln + nAr RT ln + nXe RT ln P P P P (6.43)
Because the individual gas molecules in the mixture do not interact, the final pressure and temperature of the mixture remain at their initial values. For each component i of the mixture, we can write Pi = xi P, where xi is the mole fraction of component i and P is the total pressure. Similarly, ni = xi n where n is the total number of moles in the gas mixture. Using this notation, ∆ mixG becomes
∆ mixG = nRTxHe ln xHe + nRTxNe ln xNe + nRTxAr ln xAr + nRTxXe ln xXe = nRT a xi ln xi
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i
(6.44)
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Note that because all the xi 6 1, ln xi 6 0, and each term in Equation (6.44) is negative. Therefore, ∆ mixG 6 0, showing that mixing is a spontaneous process. Equation (6.44) allows us to calculate ∆ mixG for any given set of the mole fractions xi. It is easiest to graphically visualize the results for a binary mixture of species A and B. For this case, xB = 1  xA and
0
D mixG (J mol21)
2250 2500 2750
(6.45)
A plot of ∆ mixG versus xA is shown in Figure 6.3 for a binary mixture. Note that ∆ mixG is zero for xA = 0 and xA = 1 because only pure substances are present in these limits. Also, ∆ mixG has a minimum for xA = 0.5, when A and B are present in equal amounts. The entropy of mixing can be calculated from Equation (6.44):
21000 21250 21500
21750 0.2 0.4 0.6 0.8 Mole fraction xA
1
∆ mixS =  a
0∆ mixG b = nR a xi ln xi 0T P i
(6.46)
∆ mixS is positive because all of the ln xi 6 0. For a twocomponent mixture,
Figure 6.3
Gibbs energy of mixing of ideal gases A and B as a function of xA, with nA + nB = 1 mole and T = 298.15 K.
5
D mixS (J K21 mol21)
∆ mixG = nRT 3 xA ln xA + 11  xA2ln11  xA24
∆ mixS = nR1xA ln xA + xB ln xB2
(6.47)
if both components and the mixture are at the same pressure. As shown in Figure 6.4, the entropy of mixing is greatest for xA = 0.5. What is the cause of the increase in entropy? Each component of the mixture expands from its initial volume to the same larger final volume. Therefore, ∆ mixS arises from the dependence of S on V at constant T, as shown in Equation (6.24). We verify this assertion in Example Problem 6.5.
EXAMPLE PROBLEM 6.5
4
Two gases at the same pressure and separated by a barrier have volumes ViA and ViB. The barrier is removed, allowing each of the gases to occupy the volume Vƒ = ViA + ViB in an isothermal process. (a) Calculate ∆S for this process. Compare your result with Equation (6.47). (b) Use calculus to calculate which value of xA maximizes ∆S.
3 2
Solution a. From Equation (5.14), ∆S for an isothermal expansion is ∆S = nR ln Vƒ >Vi . Therefore, ∆S for this process is
1
0.2
0.4 0.6 0.8 Mole fraction xA
Figure 6.4
Entropy of mixing of ideal gases A and B as a function of mole fraction of component A at 298.15 K, with nA + nB = 1.
1
∆S = R anA ln
Vƒ
ViA
+ nB ln
Vƒ
ViB
b
Using the relations nA = n xA and xA = ViA >Vƒ , we obtain ∆S = R anxA ln
1 1 + nxB ln b = nR1xA ln xA + xB ln xB2 xA xB
This is the same result that was obtained in Equation (6.47). b. We differentiate ∆S = nR1xA ln xA + 11  xA2ln11  xA22 with respect to xA, and set the result equal to, zero. d5 nR3 xA ln xA + 11  xA2ln11  xA24 6 >dxA =
nR3 ln xA + 1  ln11  xA2  14 = 0 xA xA ln = 0 therefore = 1 and xA = 0.5 1  xA 1  xA
EXAMPLE PROBLEM 6.6 Consider the system shown in Figure 6.2. Assume that the separate compartments contain 1.0 mol of He, 3.0 mol of Ne, 2.0 mol of Ar, and 2.5 mol of Xe at 298.15 K. The pressure in each compartment is 1 bar. a. Calculate ∆ mixG. b. Calculate ∆ mixS.
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159
Solution
a. ∆ mixG = nRT a xi ln xi i
= 8.5 mol * 8.314 J K1 mol1 * 298.15 K *a
1.0 1.0 3.0 3.0 2.0 2.0 2.5 2.5 ln + ln + ln + ln b 8.5 8.5 8.5 8.5 8.5 8.5 8.5 8.5
= 2.8 * 104 J
b. ∆ mixS = nR a xi ln xi i
= 8.5 mol * 8.314 J K1 mol1 * a
1.0 1.0 3.0 3.0 2.0 2.0 2.5 2.5 ln + ln + ln + ln b 8.5 8.5 8.5 8.5 8.5 8.5 8.5 8.5
= 93 J K1
Example Problem 6.6 illustrates how to calculate ∆ mixG and ∆ mixS. What is the driving force for the mixing of gases? We saw in Section 6.2 that there are two contributions to ∆G: an enthalpic contribution ∆H and an entropic contribution T∆S. By calculating ∆ mixH from ∆ mixH = ∆ mixG + T∆ mixS using Equations (6.44) and (6.46), you will see that for the mixing of ideal gases, ∆ mixH = 0. Because the molecules in an ideal gas do not interact, there is no enthalpy change associated with mixing. We conclude that the mixing of ideal gases is driven entirely by ∆ mixS as shown above. Although the mixing of gases is always spontaneous, the same is not true of liquids. A very strong attractive potential is essential for a gas to condense to a liquid phase. Liquids can be either miscible or immiscible. How can this fact be explained? If the AB attractions of the molecules of liquids A and B are greater than AA or BB attractions, mixing is exothermic and ∆ mixH 6 0. For gases or liquids, ∆ mixS 7 0, and therefore ∆ mixG = ∆ mixH  T∆ mixS 6 0. However, if the AB attractions are less than AA or BB attractions, mixing is endothermic and ∆ mixH 7 0. If ∆ mixH 7 0 and ∆ mixH 7 T∆ mixS, ∆ mixG 7 0 and the mixing of the two liquids is not spontaneous. However, mixing is still spontaneous if ∆ mixH 7 0 and ∆ mixH 6 T∆ mixS. In this case, the attractive interaction between A and B molecules is slightly less than the AA and BB interactions. The increase in the entropic component of the Gibbs energy outweighs the increase in the enthalpic component and mixing is spontaneous.
Concept Mixing of ideal gases is driven by the resulting increase in entropy.
6.6 CALCULATING THE EQUILIBRIUM POSITION 2 FOR A GASPHASE CHEMICAL REACTION
Given a set initial number of moles of all pure separated gaseous components in a reaction mixture, we will see that there is only one set of partial pressures, Pi, that corresponds to equilibrium when the gases are combined. In this section, we combine the results of the two previous sections to determine the equilibrium position in a reaction mixture. Consider the balanced chemical reaction
aA + bB + xC + c N dM + eN + gO + c
(6.48)
2
See the articles by L. M. Raff and by E. A. Gislason and N. C. Craig listed in Further Reading for a more detailed discussion of conditions for spontaneity and equilibrium.
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in which Greek letters represent stoichiometric coefficients and the uppercase Roman letters represent reactants and products. We can write an abbreviated expression for this reaction in the form a viXi = 0
i
(6.49)
where a, b, c, g are represented by the dimensionless stoichiometric coefficients vi and A, B, c, O are represented by Xi. In Equation (6.49), the stoichiometric coefficients of the products are positive and those of the reactants are negative. What determines the equilibrium partial pressures of the reactants and products? Imagine that the reaction proceeds in the direction indicated in Equation (6.48) by an infinitesimal amount. We describe the change in the Gibbs energy dr G for this process. The subscript r refers to the process being a chemical reaction. Starting at the initial value of the concentrations of all species, the reaction proceeds spontaneously to the right if and only if the Gibbs energy decreases as the reaction proceeds. If the reaction is spontaneous in this direction at the initial concentrations, it continues to proceed to the right, with all concentrations changing in accord with the stoichiometric coefficients until the Gibbs energy reaches a minimum value. For this unique combination of the concentrations of reactants and products, the system is at equilibrium. Any further movement of the reaction to the right leads to an increase in the Gibbs energy, which is no longer a spontaneous change. Now, let us formulate this general picture mathematically. Assuming that P and T are constant as the reaction advances, starting from Equation (6.36), we can write the differential change in the Gibbs energy, dG, as dr G = a midni
Concept Extent of reaction, j, is useful in describing the equilibrium position in a chemical reaction.
i
(6.50)
where the various dni are determined by the stoichiometric equation for the reaction. We let the reaction move slightly to the right by j moles of a particular product, which we choose arbitrarily. We call j the extent of reaction. As the reaction advances, the number of moles of each species changes according to ni = n*i + ni j
(6.51)
n*i
where is the number of moles of species i present before the reaction moved slightly to the right. Writing Equation (6.51) in a differential form gives
dni = ni dj
and Equation (6.50) becomes
dr G = a ni mi dj or i
(6.52)
0r G a b = a ni mi = ∆m 0j P,T i
(6.53)
where mi are the chemical potentials of the various reactants and products at the com0r G position of the reaction mixture. We see that the derivative a b , which is the 0j P,T slope of a plot of G versus j, determines the equilibrium position for the reaction. At 0r G the minimum in the Gibbs energy for the reaction mixture, a b = 0. 0j P,T To demonstrate this analysis quantitatively, let us consider the simple reaction A S B. As j increases from zero to one mole, one mole of A is converted to one mole of B. From Equation (6.44), we see that the Gibbs energy for the reaction mixture has two contributions for each reaction species. The first arises from the pure unmixed substances and the second from mixing. Gunmixed and Gmixture are defined by Equation (6.54). Gunmixed1T, P2 = xB mB~ 1T2 + 11  xB2mA~ 1T2
Gmixture1T, P2 = xB mB~ 1T2 + xB RT ln xB + xAmA~ 1T2 + xART ln xA
= xB mB~ 1T2 + xB RT ln xB + 11  xB2mA~ 1T2 + 11  xB2RT ln11  xB2 (6.54)
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Figure 6.5
4000
Plot of Gmixture , Gunmixed , and 𝚫 mixG as a function of the extent of reaction. As j increases from zero to one mole, one mol of A is converted to one mol of B in the reaction A S B. Gmixture has two contributions, Gunmixed from the pure components and ∆ mixG from the mixing of A and B, which we know from Section 6.5 has a minimum value at j = 0.5 mole.
3000 Gunmixed Gunmixed,Gmixture,DmixG (J mol21)
161
2000
1000
Gmixture
Extent of
0 0.2
0.4
0.6
0.8
1.0 reaction j (moles)
21000
DmixG
22000
For our simple reaction xA = 1  xB and j is numerically equal to xB, although it ~ has the units of moles. If we assume that mA~ = G m,A = ∆ f G A~ = 3.80 kJ>mole, ~ ~ ~ mB = G m,B = ∆ f G B = 1.00 kJ>mole and that T = 298 K, we can evaluate Gmixture, Gunmixed, and ∆ mixG as a function of j. The results are shown in Figure 6.5. 0r G We see that for j = 0.756 mole, a b = 0, which corresponds to the equi0j P,T 0r G librium position. For all values j 6 0.756 mole, a b is negative, and therefore 0j P,T the reaction proceeds spontaneously in the direction of increasing j. For all values of 0r G j 7 0.756 mole, a b is positive, and the reaction proceeds spontaneously in the 0j P,T direction of decreasing j. The position of the minimum in Gmixture is strongly dependent on the difference G A~  G B~ . Figure 6.6 shows that as G A~  G B~ increases, the minimum moves closer to the reaction species, which has the lower G ~ value. However, it is the mixing term that gives rise to a minimum in a plot of Gmixture versus j. If the mixing did not change the value of Gmixture, the minimum in the plot of Gmixture versus j would have a minimum value at j = 1 mole and the reaction system would consist of pure B at equilibrium.
Plots of free energy versus extent of progress for various values of Gm,A. The plots were constructed for the reaction A S B for Gm,B = 50.5 kJ mol1 and Gm,A values of (a) 55.0 kJ mol1, (b) 60.0 kJ mol1, and (c) 85.0 kJ mol1. Only the portion of the graph for j values near one is shown for (b) and (c), so that the minimum in Gmixture is near j = 0.86 in (a), 0.98 in (b), and 0.999999 in (c) can be seen. Note that jeq approaches one as ~ ~ G m,A  G m,B increases.
Gunmixed
51000 50800 50600 50400
Gmixture
50200
Gunmixed,Gmixture (J mol21)
51200
0.85 0.90 0.95 1.00 Extent of Reaction (moles) (a)
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50800
Gunmixed
50700 50600 50500
Gunmixed,Gmixture (J mol21)
50900
51400 Gunmixed,Gmixture (J mol21)
Figure 6.6
Gmixture
50500
0.97 0.98 0.99 1.00 Extent of Reaction (moles) (b)
50500.1
Gunmixed
Gmixture
0.996 0.997 0.998 0.999 1.0 Extent of Reaction (moles) (c)
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Concept If there were no entropy of mixing, a stoichiometric reaction mixture would consist entirely of reactants or products.
The most important points presented in this section on using G to predict spontaneity at constant P and T, and extending the discussion to include using A to predict spontaneity at constant V and T, can be generalized by the following statements:
• Reaction spontaneity can only be determined for specific values of the partial pressures of all reactants and products. In general, it is not possible to determine spontaneity over a range of partial pressures. For small changes of the partial pressures in a reaction mixture at constant P and T in which no nonexpansion work is possible, the condition for spontaneity is
•
drG 6 0 or a
0r G b = a ni mi 6 0 0j P,T i
(6.55)
0r G b = a nimi = 0, the reaction mixture is at equilibrium, and neither 0j P,T i direction of change is spontaneous.
• If a
For small changes of the partial pressures in a reaction mixture at constant V and T in which no nonexpansion work is possible, A rather than G is the function that determines spontaneity, and the following points hold:
• The condition for spontaneity is dr A 6 0 or a
0r A b 6 0 0j V,T
(6.56)
0r A b = 0, the reaction mixture is at equilibrium, and neither direction of 0j V,T change is spontaneous.
• If a
EXAMPLE PROBLEM 6.7
Is the reaction 2NOg1g2 N N2O41g2 spontaneous if the partial pressures of NO21g2 and N2O41g2 are 0.500 bar at T = 298.15 K?
Solution As the answer is independent of the total mass, we assume one mole of reactant and product. a
0r G ~ b = a ni mi = ∆ f G N~ 2O41T2 + RT ln xN2O4  2∆ f G NO 1T2  2RT ln xNO2 2 0j P,T i = 99.8 kJ mol1 + 8.314 J mol1 K1 * 298.15 K ln
1 2
 2 * 51.3 kJ mol1  2 * 8.314 J mol1 K1 * 298.15 K ln
1 2
= 1.08 kJ mol1 Because a
0r G b is negative, the reaction is spontaneous under these conditions. 0j P,T
6.7 INTRODUCING THE EQUILIBRIUM CONSTANT FOR A MIXTURE OF IDEAL GASES
In this section, we focus on the pressure dependence of mi in order to calculate the equilibrium position for a typical reaction mixture. Consider the following reaction that takes place between ideal gas species A, B, C, and D at a total pressure of 1 bar:
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aA + bB N gC + dD
(6.57)
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163
In a reaction mixture with a total pressure of 1 bar, the partial pressures of each species, Pi, is less than the standard state pressure of 1 bar. Because all four species are present 0r G in the reaction mixture, a b for the arbitrary set of partial pressures PA, PB, PC, 0j P,T and P is given by D
a
PC 0r G PD b = a ni mi = g amC~ + RT ln ~ b + d amD~ + RT ln ~ b 0j P,T P P i
PA PB a amA~ + RT ln ~ b  b amB~ + RT ln ~ b P P
At equilibrium, a
(6.58)
0r G b = 0 and 0j P,T
PC PD PA PB  dRT ln ~ + aRT ln ~ + bRT ln ~ P~ P P P (6.59) We can rearrange the previous equation to give gmC~ + dmD~  amA~  bmB~ = gRT ln
PC g PD d b a ~b P~ P ∆m ~ = RT ln PA a PB b a ~b a ~b P P a
where ∆m ~ is given by
(6.60)
∆m ~ 1T2 = gmC~ 1T2 + dmD~ 1T2  amA~ 1T2  bmB~ 1T2 = a vi ∆ f G i~ 1T2
i
(6.61)
Because the chemical potential m ~ 1T2 is the molar Gibbs energy for the pure component at 1 bar pressure and temperature T, ∆m ~ can be calculated from the ∆ f G ~ 1298.15 K2 tabulated in Appendix A. The combination of the partial pressures of reactants and products in Equation (6.60) is called the reaction quotient of pressures, which is abbreviated QP and defined as follows: PC g PD d a ~b a ~b P P QP = (6.62) PA a PB b a ~b a ~b P P
Concept The reaction quotient of pressures can be used to calculate the direction of spontaneous change in a reaction mixture of ideal gases.
With these definitions of QP and ∆m ~ , Equation (6.58) becomes a
0r G b = ∆ r G ~ 1T2 + RT ln QP 0j P,T
(6.63)
0r G b can be separated into two terms. QP depends 0j P,T only on the partial pressures of reactants and products, and ∆ r G ~ 1T2 depends only on T. 0r G At equilibrium, a b = 0 and ∆ r G ~ = RT ln QP.3 We denote this special 0j P,T system configuration by adding a superscript eq (for equilibrium) to each partial pressure and renaming QP as KP. It is important to distinguish between QP and KP. QP is the reaction quotient of pressures for any arbitrary state of the reaction mixture. The reaction quotient of pressures obtained by shifting the reactants toward products or Note this important result that a
0r G b and ∆ r G = ∆ r G ~ 1T2 + RT ln QP, which incorrectly 0j P,T implies a difference in the G values of products and reactants rather than a derivative of G with respect to the 0r G extent of reaction. A better notation is to write ∆ r m = a b and ∆ r m = ∆ r G ~ 1T2 + RT ln QP. 0j P,T
3
Some authors write the equality ∆ r G = a
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CHAPTER 6 Chemical Equilibrium
vice versa to reach the equilibrium state is KP. The quantity KP is called the thermodynamic equilibrium constant: g P eq d P eq C D b a ~b ~ P P 0 = ∆ r G ~ 1T2 + RT ln eq a eq b or, equivalently, PA PB a ~b a ~b P P ~ ∆ r G 1T2 = RT ln KP or ∆ r G ~ 1T2 ln KP = (6.64) RT Because ∆ r G ~ 1T2 is a function of T only, KP is also a function of T only. The thermodynamic equilibrium constant KP does not depend on P. Note also that KP is a dimensionless number because each partial pressure is divided by P ~ . Next, we will show how Equation (6.63) can be used to predict the direction of spontaneous change for a given set of partial pressures of the reactants and products. If the partial pressures of the products C and D are large and those of the reactants A and B are small compared to their values at equilibrium, QP will be large. As a result, the 0r G value of RT ln QP will be large and positive, and a b = ∆ r G ~ + RT ln QP 7 0. 0j P,T In this case, the reaction as written in Equation (6.57) from left to right is not spontaneous, but the reverse of the reaction is spontaneous. Next, consider the opposite extreme. If the partial pressures of the reactants A and B are large and those of the products C and D are small compared to their values at equilibrium, QP will be small. As a result, the value of RT ln QP will be large and negative. In this case, 0r G a b = ∆ r G ~ + RT ln QP 6 0 and the reaction is spontaneous as written from 0j P,T left to right, as some of the reactants combine to form products. These concepts are illustrated in Example Problems 6.8 and 6.9.
a
Concept The thermodynamic equilibrium constant, KP , is a dimensionless quantity that depends only on T.
EXAMPLE PROBLEM 6.8 0rG b for the reaction conditions in Example Problem 6.7 at 0j P,T 525 K. Is the reaction spontaneous at this temperature?
Calculate ∆ r m = a
Solution 0rG b = ∆ r m at the elevated temperature, we use Equation (6.33) 0j P,T assuming that ∆ r H is independent of T: To calculate a
∆ r m1T22 = T2 c
∆ r m1298.15K2 1 1 + ∆ r H1T12 a bd 298.15K T2 298.15K
~ ∆ r H1298.15K2 = ∆ f H N~ 2O41T2  2∆ f H NO 1T2 2
= 9.16 * 103 J mol1  2 * 33.2 * 103 J mol1
= 57.2 kJ mol1
∆ r m1525 K2 = 525 K * c
1.08 * 103 J mol  1 298.15 K
 57.2 * 103 J mol  1 a ∆ r m1525 K2 = 45.4 kJ mol1
1 1 bd 525 K 298.15 K
Because ∆ r m1525 K2 7 0, the reaction as written is not spontaneous, but the reverse reaction is spontaneous.
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6.7 Introducing the Equilibrium Constant for a Mixture of Ideal Gases
165
It is important to understand that the equilibrium condition of Equation (6.64) links the partial pressures of all reactants and products. Imagine that an additional amount of species D is added to the system initially at equilibrium. The system will reach a new eq eq eq equilibrium state in which P eq A , P B , P C , and P D all have new values because the value c P eq d a P eq b P eq P eq C D A b a b n a b a B~ b has not changed. P~ P~ P~ P Tabulated values of ∆ f G ~ for compounds are used to calculate ∆ rG ~ at P ~ = 1 bar and T = 298.15 K, as shown in Example Problem 6.9. Just as for enthalpy, ∆ f G ~ for a pure element in its standard reference state is equal to zero because the reactants and products in the formation reaction are identical.
of KP = a
∆ rG ~ = a vi ∆ f G i~
i
(6.65)
EXAMPLE PROBLEM 6.9 Calculate ∆ rG ~ for the reaction Fe 3O41s2 + 4H21g2 S 3Fe1s2 + 4H2O1l2 at 298.15 K.
Solution ∆ rG ~ = 3∆ f G ~ 1Fe, s2 + 4∆ f G ~ 1H2O, l2  ∆ f G ~ 1Fe 3O4, s2  4∆ f G ~ 1H2, g2
= 3 * 0 kJ mol1  4 * 237.1 kJ mol1 + 1015.4 kJ mol1  4 * 0 kJ mol1 = 67.00 kJ mol1
At other temperatures, ∆ rG ~ 1T2 is calculated using Equation (6.33) as shown in Example Problem 6.10.
EXAMPLE PROBLEM 6.10 Calculate ∆ rG ~ for the reaction in Example Problem 6.9 at 360. K.
Solution To calculate ∆ rG ~ at the elevated temperature, we use Equation (6.32) assuming that ∆ r H ~ is independent of T: ∆ rG ~ 1T22 = T2 c
∆ rG ~ 1298.15K2 1 1 + ∆ r H ~ 1T12 a bd 298.15K1 T2 298.15K
∆ r H ~ 1298.15K2 = 3∆ f H ~ 1Fe, s2 + 4∆ f H ~ 1H2O, l2  ∆ f H ~ 1Fe 3O4, s2  4∆ f H ~ 1H2, g2
= 3 * 0 J mol1  4 * 285.8 * 103 J mol1 + 1118.4 * 103 J mol1  4 * 0 J mol1 = 24.80 kJ mol1 ∆ rG ~ 1360. K2 = 360. K * c
67.00 * 103 J mol1  24.80 298.15 K
* 103 J mol1 a
∆ rG ~ 1360. K2 = 80.9 kJ mol1
1 1 bd 360. K 298.15 K
Example Problem 6.11 shows how KP is calculated using tabulated ∆ f G ~ values.
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CHAPTER 6 Chemical Equilibrium
EXAMPLE PROBLEM 6.11 a. Using data from Table 4.1 (see Appendix A, Data Tables), calculate KP at 298.15 K for the reaction CO1g2 + H2O1l2 S CO21g2 + H21g2. b. Equimolar amounts of reactants and products are placed in a reaction vessel at constant pressure. Based on the value that you obtained for part (a), do you expect the mixture to consist mainly of CO21g2 and H21g2 or mainly of CO1g2 + H2O1l2 at equilibrium?
Solution
a. ln KP = = 
1 1 ∆ f G ~ 1CO2, g2 + ∆ f G ~ 1H2, g2  ∆ f G ~ 1H2O, l2 ∆ rG ~ = a b RT RT  ∆ f G ~ 1CO, g2
1 394.4 * 103 J mol1 + 0 + 237.1 * a b * 103 J mol1 + 137.2 * 103 J mol1 8.314 J mol1 K1 * 298.15 K
= 8.1087
KP = 3.32 * 103 b. Because KP W 1, the mixture will consist mainly of the products CO21g2 + H21g2 at equilibrium.
6.8 CALCULATING THE EQUILIBRIUM PARTIAL
PRESSURES IN A MIXTURE OF IDEAL GASES
Concept Given a set of initial partial pressures, there is a unique combination of reactant and product partial pressures at equilibrium for a given value of T.
As shown in the previous section, the partial pressures of the reactants and products in a mixture of gases at equilibrium cannot take on arbitrary values because they are related through KP. In this section, we show how the equilibrium partial pressures can be calculated for ideal gases. Similar calculations for real gases will be discussed in the next chapter. In Example Problem 6.12, we consider the dissociation of chlorine: Cl21g2 N 2Cl1g2
(6.66)
It is useful to set up the calculation in a tabular form, as shown in Example Problem 6.12.
EXAMPLE PROBLEM 6.12 In this example, n0 moles of chlorine gas are placed in a reaction vessel whose temperature can be varied over a wide range so that molecular chlorine can partially dissociate to atomic chlorine. a. Define the degree of dissociation as b = deq >n0, where 2deq is the number of moles of Cl(g) present at equilibrium and n0 represents the number of moles of Cl21g2 that would be present in the system if no dissociation occurred. Derive an expression for KP in terms of n0, deq, and P. b. Derive an expression for b as a function of KP and P.
Solution a. We set up the following table: Initial number of moles Moles present at equilibrium Mole fraction present at equilibrium, xi Partial pressure at equilibrium, Pi = xi P
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Cl2 1 g2
n0 n0  deq n0  deq n0 + deq n0  deq a bP n0 + deq
N
2Cl 1 g2 0 2deq 2deq
n0 + deq 2deq a bP n0 + deq
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6.9 VARIATION OF KP WITH TEMPERATURE
167
We next express KP in terms of n0, deq , and P:
KP1T2 =
a
P eq Cl
2
b P~ = eq P Cl2 a ~ b P
ca
4d 2eq
2deq
b
P 2 d P~
n0 + deq n0  deq P a b n0 + deq P ~
4d 2eq P P = = 2 2 P~ 1n0 + deq21n0  deq2 P ~ 1n02  d eq
This expression is converted into one in terms of b : 4d 2eq
KP1T2 =
b. aKP1T2 + 4
1n022  d 2eq
P b b2 = KP1T2 P~ b = R
KP1T2
KP1T2 + 4
4b2 P P = P~ 1  b2 P ~
P P~
Because KP1T2 depends strongly on temperature, b will also be a strong function of temperature. Note that b also depends on both KP and P for this reaction. Whereas KP1T2 is independent of P, b as calculated in Example Problem 6.12 does depend on P. In the particular case considered, b decreases as P increases for constant T. As will be shown in Section 6.11, b depends on P whenever ∆v ≠ 0 for a reaction.
6.9 VARIATION OF KP WITH TEMPERATURE In Section 6.7, we showed that KP and ∆ rG ~ are related by ln KP =  ∆ rG ~ >RT =  ∆ r H ~ >RT + ∆ r S ~ . Here we illustrate how these individual functions vary with temperature for two reactions of industrial importance, beginning with ammonia synthesis. This reaction is used to produce almost all of the fertilizer used in growing food crops and will be discussed in detail in Section 6.13. The overall reaction is 3 H21g2 + N21g2 S 2NH31g2. First, we will derive an expression for the dependence of ln KP on T. Starting with Equation (6.64), we write
d1∆ rG ~ >RT2 d ln KP 1 d1∆ rG ~ >T2 = = dT dT R dT
Concept Variation of KP with T can be calculated using the Gibbs–Helmholtz equation.
(6.67)
Using the Gibbs–Helmholtz equation [Equation (6.30)], the preceding equation reduces to
d ln KP ∆rH ~ 1 d1∆ rG ~ >T2 = = dT R dT RT 2
(6.68)
Because tabulated values of ∆ f G ~ are available at 298.15 K, we can calculate ∆ rG ~ and KP at this temperature; KP can be calculated at the temperature Tf by integrating Equation (6.68) between the appropriate limits:
ln KP1Tƒ2
L
ln KP1298.15 K2
Tƒ
1 d ln KP = R L
298.15 K
∆rH ~ T2
dT
(6.69)
Figure 6.7a shows that ∆ r H ~ is large, negative, and weakly dependent on temperature. Because the number of moles of gaseous species decreases as the reaction proceeds, ∆ r S ~ is negative and is weakly dependent on temperature. The temperature dependence of
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168
CHAPTER 6 Chemical Equilibrium
[
DrS 200
[
D rH
100 300
0
[
2DrH /T
400 2100
T(K)
1000
1200
[
2DrG /T
100
2200
2DrH /T [
0 400
500
2100
600
900 800 T(K)
700
1000
2300
DrH
[
2400
DrS
[
2200
600 800 2DrG /T [
200
(a) Ammonia synthesis
(b) Coal gasification
Figure 6.7
Temperature dependence of thermodynamic quantities for two reactions of importance to industry. The plots are for reactions involved in (a) ammonia synthesis and (b) coal gasification. In both plots, the quantities  ∆ r H ~ >T, ∆ r S ~ , ∆ r H ~ , and  ∆ r G ~ >T = R ln KP are shown as a function of temperature. The units of ∆ r H ~ are kJ mol1, and the units of ∆ r S ~ are J mol1 K1. The units of  ∆ r H ~ >T and  ∆ r G ~ >T are kJ mol1 K1.
ln KP =  ∆ r G ~ >RT is determined by ∆ r H ~ . Because ∆ r H ~ is negative, KP decreases as T increases. We see that KP 7 1 for T 6 450 K, and it decreases with increasing T throughout the temperature range shown. Figure 6.7b shows analogous results for the coal gasification reaction H2O1g2 + C1s2 N CO1g2 + H21g2. This reaction produces a mixture of CO1g2 and H21g2, which is known as syngas. It can be used to produce diesel engine fuel in the form of alkanes through the Fischer–Tropsch process in the reaction 12n + 12H21g2 + nCO1g2 S CnH12n + 221l2 + nH2O1l2. For the coal gasification reaction, the number of moles of gaseous species increases as the reaction proceeds so that ∆ r S ~ is positive. ∆ r H ~ is large, positive, and not strongly dependent on temperature. Again, the temperature dependence of KP is determined by ∆ r H ~ . Because ∆ r H ~ is positive, KP increases as T increases. The following conclusions can be drawn from Figure 6.7 and Equations (6.67)–(6.69):
• The temperature dependence of KP is determined by ∆ r H ~ . • If ∆ r H ~ 7 0, KP increases with increasing T; if ∆ r H ~ 6 0, KP decreases with • •
increasing T. ∆ r S ~ has little effect on the dependence of KP on T. If ∆ r S ~ 7 0, ln KP is increased by approximately the same amount relative to  ∆ r H ~ >RT at all temperatures. If ∆ r S ~ 6 0, ln KP is decreased by approximately the same amount relative to  ∆ r H ~ >RT at all temperatures.
If the temperature Tƒ is not much different from 298.15 K, it can be assumed that ∆ r H ~ is constant over the temperature interval for most reactions. As was shown in Figure 6.7, this assumption is better than it might appear at first glance. Although H is strongly dependent on temperature, the temperature dependence of ∆ r H ~ is governed by the difference in heat capacities ∆CP between reactants and products (see Section 4.4). If the heat capacities of reactants and products are nearly the same, ∆ r H ~ is nearly independent of temperature. With the assumption that ∆ r H ~ is independent of temperature, Equation (6.69) becomes
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ln KP1Tƒ2 = ln KP1298.15 K2 
∆r H ~ 1 1 a b R Tƒ 298.15 K
(6.70)
16/11/17 10:36 AM
6.10 Equilibria Involving Ideal Gases and Solid or Liquid Phases
169
EXAMPLE PROBLEM 6.13 Using the result of Example Problem 6.12 and the data tables, consider the dissociation equilibrium Cl21g2 N 2Cl1g2. a. Calculate KP at 800. K, 1500. K, and 2000. K for P = 0.010 bar. b. Calculate the degree of dissociation b at 800. K, 1500. K, and 2000. K.
Solution a. ∆ r G ~ = 2∆ f G ~ 1Cl, g2  ∆ f G ~ 1Cl2, g2 = 2 * 105.7 * 103 J mol1  0 = 211.4 kJ mol1
∆ r H ~ = 2∆ f H ~ 1Cl, g2  ∆ f H ~ 1Cl2, g2 = 2 * 121.3 * 103 J mol1  0 = 242.6 kJ mol1
ln KP1Tƒ2 = = 
∆rG ~ ∆r H ~ 1 1 a b RT R Tƒ 298.15 K
211.4 * 103 J mol1 242.6 * 103 J mol1 8.314 J K1 mol1 * 298.15 K 8.314 J K1 mol1
* a
1 1 b Tƒ 298.15 K
ln KP1800. K2 = 
211.4 * 103 J mol1 242.6 * 103 J mol1 8.314 J K1 mol1 * 298.15 K 8.314 J K1 mol1
* a
1 1 b = 23.888 800. K 298.15 K
KP1800. K2 = 4.22 * 1011
The values for KP at 1500. and 2000. K are 1.03 * 103 and 0.134, respectively. b. The value of b at 2000. K is given by b = R
KP1T2
KP1T2 + 4
P P~
=
0.134 = 0.878 A 0.134 + 4 * 0.01
The values of b at 1500. and 800. K are 0.159 and 3.23 * 105, respectively.
The degree of dissociation of Cl21g2 increases with temperature as shown in Example Problem 6.13. This is always the case for an endothermic reaction.
6.10 EQUILIBRIA INVOLVING IDEAL GASES AND SOLID OR LIQUID PHASES
In the preceding sections, we discussed chemical equilibrium in a homogeneous system of ideal gases. However, many chemical reactions involve a gas phase in equilibrium with a solid or liquid phase. An example is the thermal decomposition of CaCO31s2 : CaCO31s2 N CaO1s2 + CO21g2
(6.71)
In this case, a pure gas is in equilibrium with two solid phases at the pressure P. At equilibrium, ∆m = a ni mi = 0 i
0 = meq1CaO, s, P2 + meq1CO2, g, P2  meq1CaCO3, s, P2
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(6.72)
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CHAPTER 6 Chemical Equilibrium
Because the equilibrium pressure is P ≠ P ~ , the pressure dependence of m for each species must be taken into account. From Section 6.3, we know that the pressure dependence of G for a solid or liquid is very small:
Concept KP for an equilibrium involving gases and liquids or solids contains contributions only from gaseous species.
meq1CaO, s, P2 ≈ m ~ 1CaO, s2 and meq1CaCO3, s, P2 ≈ m ~ 1CaCO3, s2
(6.73)
You will verify the validity of Equation (6.73) in the endofchapter problems. Using the dependence of m on P for an ideal gas, we see that Equation (6.72) becomes 0 = m ~ 1CaO, s2 + m ~ 1CO2, g2  m ~ 1CaCO3, s2 + RT ln
∆rG ~
PCO2
or P~ PCO2 = m ~ 1CaO, s2 + m ~ 1CO2, g2  m ~ 1CaCO3, s2 = RT ln ~ P
(6.74)
Rewriting this equation in terms of KP, we obtain
ln KP = ln
PCO2 P~
= 
∆rG ~ RT
(6.75)
We generalize this result as follows. KP for an equilibrium involving gases with liquids and/or solids contains only contributions from the gaseous species. However, ∆ r G ~ is calculated taking all species into account.
EXAMPLE PROBLEM 6.14 Given the preceding discussion and using the tabulated values of ∆ f G ~ and ∆ f H ~ in Appendix A, calculate the CO21g2 pressure in equilibrium with a mixture of CaCO31s2 and CaO(s) at 1000., 1100., and 1200. K.
Solution For the reaction CaCO31s2 N CaO1s2 + O21g2, ∆ r G ~ = 131.1 kJ mol1 and ∆ r H ~ = 178.5 kJ mol1 at 298.15 K. We use Equation (6.75) to calculate KP1298.15 K2 and Equation (6.70) to calculate KP at elevated temperatures. ln
PCO2 P~
11000. K2 = ln KP11000. K2
= ln KP1298.15 K2 
=
∆ r H ~ 1298.15 K2 1 1 a b R 1000. K 298.15 K
131.1 * 103 J mol1 178.5 * 103 J mol1 8.314 J K1 mol1 * 298.15 K 8.314 J K1 mol1
*a
1 1 b 1000. K 298.15 K
= 2.348; PCO211000. K2 = 0.0956 bar
The values for PCO2 at 1100. K and 1200. K are 0.673 bar and 3.42 bar, respectively.
If the reaction involves only liquids or solids, the pressure dependence of the chemical potential is generally small and can be neglected. However, it cannot be neglected if P W 1 bar, as shown in Example Problem 6.15.
EXAMPLE PROBLEM 6.15 At 298.15 K, ∆ f G ~ 1C, graphite2 = 0 and ∆ f G ~ 1C, diamond2 = 2.90 kJ mol1. Therefore, graphite is the more stable solid phase at this temperature at P = P ~ = 1 bar. Given that the densities of graphite and diamond are 2.25 and 3.52 kg>L, respectively, at what pressure will graphite and diamond be in equilibrium at 298.15 K?
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6.11 Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarity
171
Solution At equilibrium ∆G = G1C, graphite2  G1C, diamond2 = 0. Using the pressure dependence of G given in Equation (6.22), 10Gm >0P2T = Vm, we establish the condition for equilibrium: ∆G = ∆ f G ~ 1C, graphite2  ∆ f G ~ 1C, diamond2 graphite diamond + 1V m  Vm 21∆P2 = 0
graphite diamond 0 = 0  2.90 * 103 kJ mol1 + 1V m  Vm 21P  1 bar2
P = 1 bar +
= 1 bar +
12.00 * 103 kg mol1
2.90 * 103 kJ mol1 1 1 MC a b rgraphite rdiamond
2.90 * 103 kJ mol1 1 1 * a b 3 3 2.25 * 10 kg m 3.52 * 103 kg m3
= 105 Pa + 1.51 * 109 Pa = 1.51 * 104 bar
Fortunately for all those with diamond rings, although the conversion of diamond to graphite at 1 bar and 298 K is spontaneous, the rate of conversion is vanishingly small.
6.11 EXPRESSING THE EQUILIBRIUM CONSTANT IN TERMS OF MOLE FRACTION OR MOLARITY
Chemists often find it useful to express the concentrations of reactants and products in units other than partial pressures. Two examples of other units that we will consider in this section are mole fraction and molarity. Note, however, that this discussion is still limited to a mixture of ideal gases. The extension of chemical equilibrium to include neutral and ionic species in aqueous solutions will be made in Chapter 10, after the concept of activity has been introduced. We first express the equilibrium constant in terms of mole fractions. The mole fraction xi and the partial pressure Pi are related by Pi = xi P. Therefore, g P eq d g x eqP d P eq x eq C CP D b a b a b a D~ b g eq d 1x eq P~ P~ P~ P P g+d ab C 2 1x D 2 KP = = = eq a eq b eq eq eq a eq b a P ~ b a b PA PB xA P xB P 1x A 2 1x B 2 a ~b a ~b a ~ b a ~ b P P P P
a
= Kx a
P ∆v b P~
Kx = KP a
P ∆v b P~
Concept The equilibrium constant Kx depends on T and the difference between the stoichiometric coefficients of reactant and products.
(6.76)
Recall that ∆v is the difference in the stoichiometric coefficients of products and reactants. Note that just as for KP, Kx is a dimensionless number. Because the molarity ci is defined as ci = ni >V = Pi >RT, we can write Pi >P ~ = 1RT>P ~ 2ci. To work with dimensionless quantities, we introduce the ratio ci >c ~ , which is related to Pi >P ~ by
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Pi c ~ RT ci = ~ P P~ c~
(6.77)
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CHAPTER 6 Chemical Equilibrium
c ~ is the standard state concentration of 1 mole/L. Using this notation, we can express Kc in terms of KP: g P eq d g ceq d P eq ceq C C D b a b a b a D~ b P~ P~ c~ c KP = eq a eq b = eq a eq b PA PB cA cB a ~b a ~b a ~b a ~b P P c c
a
Kc = KP a
a
c ~ RT g + d  a  b c ~ RT ∆v b = K a b c P~ P~
c ~ RT ∆v b P~
(6.78)
Equations (6.76) and (6.78) show that Kx and Kc are in general different from KP. They are only equal in the special case that ∆v = 0. Suppose that we have a mixture of reactive gases at equilibrium. Does the equilibrium shift toward reactants or products as T or P is changed? Because jeq increases if KP or Kx increases and decreases if KP or Kx decreases, we need only consider the dependence of KP on T or Kx on T and P. We first consider the dependence of KP on T. We know from Section 6.9 that for an exothermic reaction, d ln KP >dT 6 0, and jeq will shift toward the reactants as T increases. For an endothermic reaction, d ln KP >dT 7 0, and jeq will shift toward the products as T increases. The dependence of jeq on pressure can be determined from the relationship between Kx and P:
Kx = KP a
P ∆v b P~
(6.79)
Because KP is independent of pressure, the pressure dependence of Kx arises solely from 1P>P ~ 2∆v. If ∆v 7 0, the number of moles of gaseous products increases as the reaction proceeds. In this case, Kx decreases as P increases, and jeq shifts back toward the reactants. If the number of moles of gaseous products decreases as the reaction proceeds, Kx increases as P increases, and jeq shifts forward toward the products. If ∆v = 0, jeq is independent of pressure. A combined change in T and P leads to a superposition of the effects just discussed. According to the French chemist Le Chatelier, reaction systems at chemical equilibrium respond to an outside stress, such as a change in T or P, by countering the stress. Consider the Cl21g2 N 2Cl1g2 reaction for which ∆ r H 7 0, discussed in Example Problems 6.12 and 6.13. There, jeq responds to an increase in T in such a way that heat is taken up by the system; in other words, more Cl21g2 dissociates. This counters the stress imposed on the system by an increase in T. Similarly, jeq responds to an increase in P in such a way that the volume of the system decreases. Specifically, jeq changes in the direction that ∆v 6 0, and for the reaction under consideration, Cl(g) is converted to Cl21g2. This shift in the position of equilibrium counters the stress brought about by an increase in P.
6.12 EXPRESSING U, H, AND HEAT CAPACITIES SOLELY IN TERMS OF MEASURABLE QUANTITIES
In this section, we use the Maxwell relations derived in Section 6.2 to express U, H, and heat capacities solely in terms of measurable quantities such as kT, a and the state variables P, V, and T. Often, it is not possible to determine an equation of state for a substance over a wide range of the macroscopic variables. This is the case for most solids and liquids. However, material constants such as heat capacities, the thermal expansion coefficient, and the isothermal compressibility may be known over a limited range of their variables. If this is the case, the enthalpy and internal energy can be expressed in terms of the material constants and the variables V and T or P and T, as shown later. The Maxwell relations play a central role in deriving these formulas.
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6.12 EXPRESSING U, H, AND HEAT CAPACITIES SOLELY IN TERMS OF MEASURABLE QUANTITIES
173
We begin with the following differential expression for dU that holds for solids, liquids, and gases, provided that no nonexpansion work, phase transitions, or chemical reactions occur in the system: dU = TdS  PdV
(6.80)
Invoking a transformation at constant temperature and dividing both sides of the equation by dV, 0U 0S 0P a b = T a b  P = T a b  P 0V T 0V T 0T V
(6.81)
To arrive at the second expression in Equation (6.81), we have used the third Maxwell relation (Equation (6.25)). We next express 10P>0T2V in terms of a, the isobaric volumetric thermal expansion coefficient, and kT , the isothermal compressibility: a
10V>0T2P 0P a b = = kT 0T V 10V>0P2T
Concept It is possible to express U, H, and heat capacities solely in terms of measurable quantities if the equation of state is not known.
(6.82)
Equation (6.82) allows us to write Equation (6.81) in the form a
aT  kT P 0U b = kT 0V T
(6.83)
Through this equation, we have linked 10U>0V2T , which as discussed in Section 3.2, is called the internal pressure, to easily measured material properties. This allows us to calculate the internal pressure of a liquid or a gas, as shown in Example Problem 6.16.
EXAMPLE PROBLEM 6.16 At 298 K, the thermal expansion coefficient and the isothermal compressibility of liquid water are a = 2.04 * 104 K1 and kT = 45.9 * 106 bar 1. a. Calculate 10U>0V2T for water at 320. K and P = 1.00 bar. b. If an external pressure equal to 10U>0V2T were applied to 1.00 m3 of liquid water at 320. K, how much would its volume change? What is the relative change in volume? c. As shown in Example Problem 3.4, 10U>0Vm2T = a>V 2m for a van der Waals gas. Calculate 10U>0Vm2T = a>V 2m for N21g2 at 320 K and P = 1.00 atm, given that a = 1.35 atm L2 mol2. Compare the value that you obtain with that of part (a) and discuss the difference.
Solution a. a
aT  kTP 0U b = kT 0V T =
2.04 * 104 K1 * 320. K  45.9 * 106 bar 1 * 1.00 bar 45.9 * 106 bar 1
= 1.4 * 103 bar b.
kT = 
1 0V a b V 0P T Pƒ
∆VT = 
L Pi
V1T2kT dP ≈ V1T2kT1Pƒ  Pi2
= 1.00 m3 * 45.9 * 106 bar 1 * 11.4 * 103 bar  1 bar2 = 6.4 * 102 m3
∆VT 6.4 * 102 m3 = = 6.4, V 1 m3
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CHAPTER 6 Chemical Equilibrium
c. At atmospheric pressure, we can calculate Vm using the ideal gas law and a
1.35 atm L2 mol2 * 11.00 atm22 0U a aP 2 b = 2 = 2 2 = 0Vm T Vm RT 18.206 * 102 atm L mol1 K1 * 320. K22 = 1.96 * 103 atm
The internal pressure is much smaller for N2 gas than for H2O liquid. This is the case because H2O molecules in a liquid are relatively strongly attracted to one another, whereas the interaction between gas phase N2 molecules is very weak. Gases, which have a small internal pressure, are more easily compressed by an external pressure than liquids or solids, which have a high internal pressure. Using the result in Equation (6.83) for the internal pressure, we can write dU in the form aT  kT P 0U 0U dU = a b dT + a b dV = CV dT + dV (6.84) kT 0T V 0V T
Equation (6.84) is useful because it contains only the material constants CV, a, kT, and the macroscopic variables T and V. In Example Problems 6.17 and 6.18, we show that ∆U, ∆H, and CP  CV can be calculated even if the equation of state for the material of interest is not known, as long as the thermal expansion coefficient and the isothermal compressibility are known.
EXAMPLE PROBLEM 6.17 A 1000. g sample of liquid Hg undergoes a reversible change from an initial state Pi = 1.00 bar, Ti = 300. K to a final state Pƒ = 300. bar, Tƒ = 600. K. The density of Hg is 13,534 kg m3, a = 1.81 * 104 K1, CP,m = 27.98 J mol1 K1, and kT = 3.91 * 106 bar 1, in the range of the variables in this problem. Recall from Section 3.5 that CP  CV = 1TV a2 >kT2, and that CP ≈ CV for liquids and solids. a. Calculate ∆U and ∆H for the transformation. Describe the path assumed for the calculation. b. Compare the relative contributions to ∆U and ∆H from the change in pressure and the change in temperature. Can the contribution from the change in pressure be neglected?
Solution Because U is a state function, ∆U is independent of the path. We choose the reversible path Pi = 1.00 bar, Ti = 300. K S Pi = 1.00 bar, Tƒ = 600. K S Pƒ = 300. bar, Tƒ = 600. K. Note that because we have assumed that all the material constants are independent of P and T in the range of interest, the numerical values calculated will depend slightly on the path chosen. However, if the P and T dependence of the material constant were properly accounted for, all paths would give the same value of ∆U. Tƒ
∆U =
L
Vƒ
CP dT +
Ti
aT  kT P dV kT L Vi
Using the definition kT = 1>V10V>0P2T , ln Vƒ >Vi = kT1Pƒ  Pi2, and V1P2 = Vi ekT 1P  Pi2. We can also express P1V2 as P1V2 = Pi  11>kT2ln1V>Vi2 Tƒ
∆U =
L Ti
Tƒ
Vƒ
CP dT +
aT 1 V  aPi ln b d dV k k V T T L i Vi
c
Vƒ
Vƒ
Vi
Vi
aT 1 V = CP dT + a  Pi b dV + ln dV kT kT L Vi L L Ti
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6.12 EXPRESSING U, H, AND HEAT CAPACITIES SOLELY IN TERMS OF MEASURABLE QUANTITIES
175
Using the standard integral 1 ln 1x>a2dx = x + x ln 1x>a2 Tƒ
∆U =
L Ti
Cp dT + a
aT 1 V Vƒ  Pi b 1Vƒ  Vi2 + c V + V ln d kT kT Vi Vi
= CP1Tƒ  Ti2 + a
Vi =
Vƒ aT 1  Pi b 1Vƒ  Vi2 + aVi  Vƒ + Vƒ ln b kT kT Vi
1.000 kg m = 7.389 * 105 m3 = r 13534 kg m3
Vƒ = Vi ekT 1Pƒ  Pi2 = 7.389 * 105 m3 exp13.91 * 106 bar 1 * 299 bar2 = 7.380 * 105 m3
Vƒ  Vi = 9 * 108 m3 ∆U =
1000. g 200.59 g mol1 a
* 27.98 J mol1 K1 * 1600. K  300. K2
1.81 * 104 K1 * 300. K 105 Pa  1.00 barb * * 9 * 108 m3 6 1 1 bar 3.91 * 10 bar
9 * 108 m3 + 7.380 * 105 m3 1 105 Pa + * * ± ≤ 7.380 * 105 m3 1 bar 3.91 * 106 bar 1 * ln 5 3 7.389 * 10 m = 41.8 * 103 J  100 J + 1 J ≈ 41.7 * 103 J As shown in Section 3.1, V1P, 600. K2 = 3 1 + a1600. K  300. K24 V1Pi, 300. K2ekT 1P  Pi2 = V′ ekT 1P  Pi2 Tƒ
∆H =
L
Pƒ
Cp dT +
Ti
L
Tƒ
VdP =
Pi
= CP1Tƒ  Ti2 + =
1000. g
200.59 g mol1
L
Pƒ
kT P i
Cp dT + V′ e
Ti
L
ekT P dP
Pi
V′i ekT Pi k P 1e T i  ekT Pƒ2 kT
* 27.98 J mol1 K1 * 1600. K  300. K2
+ 7.389 * 105 m3 * 11 + 300. K * 1.81 * 104 K12 *
exp13.91 * 106 bar 1 * 1 bar2
3.91 * 106 bar 1 * 11 bar>105 Pa2
* 3 exp1 3.91 * 106 bar 1 * 1 bar2
 exp13.91 * 106 bar 1 * 300. bar24 ∆H = 41.8 * 103 J + 2.29 * 103 J = 44.1 * 103 J The temperaturedependent contribution to ∆U is 98% of the total change in U, and the corresponding value for ∆H is ∼95%. The contribution from the change in pressure to ∆U and to ∆H is very small.
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CHAPTER 6 Chemical Equilibrium
EXAMPLE PROBLEM 6.18 In Example Problem 6.17, it was assumed that CP = CV . a2 , and the experimentally determined kT value CP,m = 27.98 J mol1 K1 to obtain a value for CV,m for Hg1l2 at 300. K. b. Did the assumption that CP,m ≈ CV,m in Example Problem 6.15 introduce an appreciable error in your calculation of ∆U and ∆H ? a. Use Equation (3.38), CP,m = CV,m + TVm
Solution a. CP,m  CV,m =
TVma2 kT 300. K *
0.20059 kg mol1 13534 kgm3
=
= 3.73 J K1 mol1
* 11.81 * 104 K122
3.91 * 106 bar 1 * 11 bar>105 Pa2
CV,m = CP,m 
TVma2 kT
= 27.98 J K1 mol1  3.73 J K1 mol1 = 24.25 J K1 mol1 b. ∆U is in error by ∼15%, and ∆H is unaffected by assuming that CP,m ≈ CV,m. It is a reasonable approximation to set CP = CV for a liquid or solid. S U P P L E M E N TA L
S E C T I O N
6.13 A CASE STUDY: THE SYNTHESIS OF AMMONIA
Concept Nearly all inorganic nitrogen input to agricultural soils is provided by the Haber–Bosch ammonia synthesis process.
In this section, we discuss a specific reaction and show both the power and the limitations of thermodynamic principles and analysis in developing a strategy to maximize the rate of a desired reaction. Ammonia synthesis is the primary route to the manufacture of fertilizers used in agriculture. The commercial process used today is essentially the same as that invented by the German chemists Robert Bosch and Fritz Haber in 1908. In terms of its impact on human life, the Haber–Bosch synthesis may be the most important chemical process ever invented because the increased food production possible with fertilizers based on NH3 allowed the large increase in Earth’s population in the 20th century to occur. In 2000, more than 2 million tons of ammonia were produced per week using the Haber–Bosch process, and more than 98% of the inorganic nitrogen input to soils used in agriculture worldwide as fertilizer is generated using the Haber–Bosch process. What useful information can the ammonia synthesis reaction from thermodynamics give us? The overall reaction is
1>2 N21g2 + 3>2 H21g2 N NH31g2
(6.85)
Because the reactants are pure elements in their standard states at 298.15 K, ∆ r H ~ = ∆ f H ~ 1NH3, g2 = 45.9 * 103 J mol1 at 298.15 K ∆ r G ~ = ∆ f G ~ 1NH3, g2 = 16.5 * 103 J mol1 at 298.15 K
Assuming ideal gas behavior, we see that the equilibrium constant KP is PNH3 a ~ b 1xNH32 P P 1 KP = = a b 3 3 1 1 ~ PN2 2 PH2 2 1xN2221xH222 P a ~b a ~b P P
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(6.86) (6.87)
(6.88)
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6.13 A Case Study: The Synthesis of Ammonia
177
We will revisit this equilibrium and include deviations from ideal behavior in Chapter 7. Because the goal is to maximize xNH3, Equation (6.88) is written in the form
1
3
xNH3 = 1xN222 1xH222 a
P bK P~ P
(6.89)
What conditions of temperature and pressure are most appropriate to maximizing the yield of ammonia? Because d ln KP ∆r H ~ = 6 0 (6.90) dT RT 2 KP and, therefore, xNH3 decrease as T increases. Because ∆v = 1, xNH3 increases with P. We conclude that the best conditions to carry out the reaction are high P and moderate T. In fact, industrial production is carried out for P 7 100 atm and T ∼ 700 K. Our discussion has focused on reaching equilibrium because this is the topic of inquiry in thermodynamics. However, note that many synthetic processes are deliberately carried out far from equilibrium in order to selectively generate a desired product. For example, ethylene oxide, which is produced by the partial oxidation of ethylene, is an industrially important chemical used to manufacture antifreeze, surfactants, and fungicides. Although the complete oxidation of ethylene to CO2 and H2O has a more negative value of ∆ r G ~ than the partial oxidation, it is undesirable because it does not produce useful products. These predictions provide information about the equilibrium state of the system but lack one important component. How long will it take to reach equilibrium? Unfortunately, thermodynamics gives us no information about the rate at which equilibrium is attained. Consider the following example. Using values of ∆ f G ~ from the data tables, convince yourself that the oxidation of potassium and coal (essentially carbon) are both spontaneous processes. For this reason, potassium has to be stored in an environment free of oxygen. However, coal is stable in air indefinitely. Thermodynamics tells us that both processes are spontaneous at 298.15 K, but experience shows us that only one proceeds at a measurable rate. However, by heating coal to an elevated temperature, the oxidation reaction becomes rapid. This result implies that there is an energetic activation barrier to the reaction that can be overcome with the aid of thermal energy. As we will see, the same is true of the ammonia synthesis reaction. Although Equations (6.86) and (6.87) predict that xNH3 is maximized as T approaches 0 K, the reaction rate is vanishingly small unless T 7 500 K. To this point, we have only considered the reaction system using state functions. However, a chemical reaction consists of a number of individual steps, called a reaction mechanism. How might the ammonia synthesis reaction proceed? One possibility is the gasphase reaction sequence: N21g2 S 2N1g2 (6.91)
H21g2 S 2H1g2
(6.92)
N1g2 + H1g2 S NH1g2
(6.93)
NH1g2 + H1g2 S NH21g2
(6.94)
NH21g2 + H1g2 S NH31g2
(6.95)
In each of the recombination reactions, other species that are not involved in the reaction can facilitate energy transfer. The enthalpy changes associated with each of the individual steps are known and are indicated in Figure 6.8. This diagram is useful because it gives much more information than the singlevalue ∆ r H ~ . Although the overall process is slightly exothermic, the initial step, which is the dissociation of N21g2 and H21g2, is highly endothermic. Heating a mixture of N21g2 and H21g2 to high enough temperatures to dissociate a sizable fraction of the N21g2 would require such a high temperature that all the NH31g2 formed would dissociate into reactants. This can be shown by carrying out a calculation such as that in Example Problem 6.11. The conclusion is that the gasphase reaction between N21g2 and H21g2 is not a practical route to ammonia synthesis. The way around this difficulty is to find another reaction sequence for which the enthalpy changes of the individual steps are not prohibitively large. For the ammonia
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CHAPTER 6 Chemical Equilibrium
Figure 6.8
N(g) 1 3H(g)
Enthalpy diagram for the reaction mechanism in the synthesis of ammonia. See Equations (6.91) through (6.95). The successive steps in the reaction proceed from left to right in the diagram.
314 3 103 J NH(g) 1 2H(g)
390 3 103 J
1124 3 103 J
NH2(g) 1 H(g)
466 3 103 J
1
2
N2(g) 1
3
2
245.9 3 103 J
H2(g)
NH3(g) Progress of reaction
synthesis reaction, such a route is a heterogeneous catalytic reaction, using iron as a catalyst. The mechanism for this path between reactants and products is N21g2 + □ S N21a2
(6.96)
N21a2 + □ S 2N1a2
(6.98)
H21g2 + 2□ S 2H1a2
N1a2 + H1a2 S NH1a2 + □
(6.99)
NH1a2 + H1a2 S NH21a2 + □
(6.100)
(6.97)
NH21a2 + H1a2 S NH31a2 + □
(6.101)
NH31a2 S NH31g2 + □
(6.102)
The symbol □ denotes an ensemble of neighboring Fe atoms, also called surface sites, which are capable of forming a chemical bond with the indicated entities. The designation (a) indicates that the chemical species is adsorbed (chemically bonded) to a surface site. The enthalpy change for the overall reaction N21g2 + 3>2 H21g2 S NH31g2 is the same for the mechanisms in Equations (6.91) through (6.95) and (6.96) through (6.102) because H is a state function. This is a characteristic of a catalytic reaction. A catalyst can affect the rate of the forward and backward reaction but not the position of equilibrium in a reaction system. The enthalpy diagram in Figure 6.9 shows that the Homogeneous gasphase reactions N(g) 1 3H(g) 314 3 103 J NH(g) 1 2H(g)
Figure 6.9
1124 3 103 J
Enthalpy diagram for the homogeneous gasphase and heterogeneous catalytic 1 3 2 N2(g) 1 2 H2(g) reactions for the ammonia synthesis reaction. The activation barriers for the 0 individual steps in the surface reaction are shown. The successive steps in the reaction proceed from left to right in the diagram. See the reference to G. Ertl in 1 3 Further Reading for more details. 2 N2(a) 1 2 H2(a) Source: Adapted from G. Ertl, Catalysis Reviews—Science and Engineering 21 (1980): 201–223.
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Ratelimiting step
390 3 103 J NH2(g) 1 H(g) 466 3 103 J
NH3(g) 245.9 3 103 J
Progress of reaction
Heterogeneous catalytic reactions NH2(a) 1 H(a) NH(a) 1 2H(a) N(a) 1 3H(a)
NH3(a)
NH3(g)
245.9 3 103 J
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6.13 A Case Study: The Synthesis of Ammonia
enthalpies of the individual steps in the gas phase and surface catalyzed reactions are very different. These enthalpy changes have been determined in experiments but are not generally listed in thermodynamic tables. In the catalytic mechanism, the enthalpy for the dissociation of the gasphase N2 and H2 molecules is greatly reduced by stabilization of the atoms through their bond to the surface sites. However, a small activation barrier to the dissociation of adsorbed N2 remains. This barrier is the ratelimiting step in the overall reaction. As a result of this barrier, only one in a million gasphase N2 molecules incident on the catalyst surface per unit time dissociates and is converted into ammonia. Such a small reaction probability is not uncommon for industrial processes based on heterogeneous catalytic reactions. In industrial reactors, on the order of ∼107 collisions of the reactants with the catalyst occur in the residence time of the reactants in the reactor. Therefore, even reactions with a reaction probability per collision of 106 can be carried out with a high yield. How was Fe chosen as the catalyst for ammonia synthesis? Even today, catalysts are usually optimized using a trialanderror approach. Haber and Bosch tested hundreds of catalyst materials, and more than 2500 substances have been tried since. On the basis of this extensive screening, it was concluded that a successful catalyst must fulfill two different requirements: (1) it must bind N2 strongly enough that the molecule can dissociate into N atoms, but (2) it must not bind N atoms too strongly. If that were the case, the N atoms could not react with H atoms to produce gasphase NH3. As Haber and Bosch knew, both osmium and ruthenium are better catalysts than iron for NH3 synthesis, but these elements are far too expensive for commercial production. Why are these catalysts better than Fe? Quantummechanical calculations by Jacobsen et al. (2001) [See Further Reading for the reference] explained the “volcano plot” of the activity versus the N2 binding energy shown in Figure 6.10. Jacobsen et al. (2001) concluded that molybdenum is a poor catalyst because N is bound too strongly, whereas N2 is bound too weakly. Nickel is a poor catalyst for the opposite reason: N is bound too weakly and N2 is bound too strongly. Osmium and ruthenium are the best elemental catalysts because they fulfill both requirements well. Jacobsen et al. suggested that a catalyst that combined the favorable properties of Mo and Co to create CoMo surface sites would be a good catalyst, and they concluded that the ternary compound Co3Mo3N best satisfied these requirements. On the basis of these calculations, the catalyst was synthesized with the results shown in Figure 6.11. The experiments show that Co3Mo3N is far more active than Fe and more active than Ru over most of the concentration range. These experiments also show that the turnover frequency for Co3Mo3N lies at the point shown on the volcano plot of Figure 6.10. This catalyst is one of the few that have been developed on the basis of theoretical calculations as opposed to the trialanderror method.
Turnover frequency (s21)
101
“CoMo”
Concept The ideal ammonia catalyst binds N 2 molecules relatively strongly but does not bind N atoms strongly.
Ru Os
Figure 6.10
100 Fe 1021 Co
1022 1023 1024
179
Mo Ni
1025 2100.0
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275.0
250.0
225.0 20.0 25.0 [DE2DE(Ru)] (kJ/mol N2)
50.0
75.0
Calculated turnover frequency for NH3 production as a function of the calculated binding energy of N2. NH3 production is the rate of NH3 molecules per surface site per second. Values are relative to those on ruthenium. Reaction conditions are 400°C, 50 bar total pressure, and the gas composition of H2 : N2 = 3 :1 contains 5% NH3. “CoMo” is the compound Co3Mo3N, which best satisfied the requirements of the catalyst. For more details, see the reference to Jacobsen in Further Reading. Source: Adapted from Jacobsen et al, Journal of the American Chemical Society 123 (2001): 8404–8405.
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CHAPTER 6 Chemical Equilibrium
Figure 6.11
Source: Adapted from Jacobsen et al, Journal of the American Chemical Society 123 (2001): 8404–8405.
8 Ru
7 NH3 concentration (%)
Experimental data for turnover frequency for NH3 production as a function of the NH3 concentration in the mixture. The Co3Mo3N catalyst is superior to both Fe and Ru as predicted by the calculations. The reaction conditions are those for Figure 6.10. The number of active surface sites is assumed to be 1% of the total number of surface sites. See the reference to Jacobsen in Further Reading for more details.
Co3Mo3N 6 5
Fe
4 3 2 0
2.5
5.0 7.5 10.0 12.5 15.0 Turnover frequency [s21]
17.5
20.0
An unresolved issue is the nature of the surface sites indicated by □ in Equations (6.96) through (6.102). Are all adsorption sites on an NH3 synthesis catalyst surface equally reactive? To answer this question, it is important to know that metal surfaces at equilibrium consist of large, flat terraces separated by monatomic steps. It was long suspected that the steps rather than the terraces play an important role in ammonia synthesis. This was convincingly shown (see reference to Dahl in Further Reading) in an elegant experiment. Dahl et al. evaporated a small amount of gold on a Ru crystal. As had been demonstrated (see reference to Hwang in Further Reading) the inert Au atoms migrate to the step sites on Ru and thereby block these sites, rendering them inactive in NH3 synthesis. Dahl et al. (2001) showed that the rate of ammonia synthesis was lowered by a factor of 109 after the step sites—which make up only 1% to 2% of all surface sites— were blocked by Au atoms! This result clearly shows that in this reaction only step sites are active. They are more active because they have fewer nearest neighbors than terrace sites. This enables the step sites to bind N2 more strongly, promoting dissociation into N atoms that react rapidly with H atoms to form NH3. This discussion of catalysis of ammonia synthesis has shown the predictive power of thermodynamic principles in analyzing chemical reactions. Calculating ∆ r G ~ allowed us to predict that the reaction of N2 and H2 to form NH3 is a spontaneous process at 298.15 K. Calculating ∆ r H ~ allowed us to predict that the yield of ammonia increases with increasing pressure and decreases with increasing temperature. However, thermodynamics cannot be used to determine how rapidly equilibrium is reached in a reaction mixture. If thermodynamic tables of species such as N21a2, N(a), NH(a), NH21a2, and NH31a2 were available, it would be easier for chemists to develop synthetic strategies that are likely to proceed at an appreciable rate. However, whether an activation barrier exists in the conversion of one species to another—and how large the barrier is—are not predictable by thermodynamic analysis. Therefore, synthetic strategies must rely on chemical intuition, experiments, and quantummechanical calculations to develop strategies that reach thermodynamic equilibrium at a reasonable rate.
S U P P L E M E N TA L
S E C T I O N
6.14 MEASURING 𝚫G FOR THE UNFOLDING OF SINGLE RNA MOLECULES
Concept ∆G for unfolding biomolecules can be measured by mechanically pulling the molecule apart.
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In general, ∆G, ∆H, and ∆U are measured using thermochemical methods as discussed in Chapter 4. However, many biochemical systems are irreversibly changed upon heating, so that calorimetric methods are unsuitable for such systems. With the advent of atomicscale manipulation of molecules, which began with the invention of scanning
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6.14 MEASURING ∆G FOR THE UNFOLDING OF SINGLE RNA MOLECULES
220
Extension (nm) 240 260
(a)
(b)
181
P5abcDA C A5’ AC A GA G C G C C GU P5a A U G C U A C G G U U A 3helix G C junction G C AAAGGGUAU C GUU C C G A G A P5c U G A U Hairpin G C A U loop G C P5b U A U G U G C G A G AA
RNA/DNA Handle A Biotin
Antidigoxigenin bead
probe microscopies (see Chapter 16), it has become possible to measure ∆G for a change using an individual molecule. An interesting illustration is the measurement of ∆G for the unfolding of single RNA molecules. RNA molecules must fold into very specific shapes in order to carry out their function as catalysts for biochemical reactions. However, the mechanisms for how the molecule achieves its equilibrium shape— which is important because the energetics of folding is a key parameter in modeling the folding process—are not well understood. Liphardt et al. (see Further Reading) obtained ∆G for this process by unzipping a single RNA molecule. The measurement relies on Equation (6.10), which states that Gibbs energy is a measure of the maximum nonexpansion work that can be produced in a transformation. Figure 6.12 shows how the force associated with the unfolding of RNA into two strands can be measured. Each of the two strands that make up the RNA is attached to a chemical “handle” that allows the individual strands to be firmly linked to 2mmdiameter polystyrene beads. By using techniques based on scanning probe microscopy, the distance between the beads is increased by distances in the nanometer range, and the force needed to do so is measured. The plot of force versus distance for the P5abc∆A RNA is shown in Figure 6.13 for a rate of 1 pN s1. The red and black curves are nearly superimposed, showing that the process is essentially reversible. (No process that occurs at a finite rate is truly reversible.) Finally, we will discuss the form of the force versus distance curve. The initial increase in distance between ∼210 and ∼225 nm is due to the stretching of the handles. However, a marked change in the shape of the curve is observed for F ≈ 13 pN, for which the distance increases abruptly. This behavior is the signature of the unfolding of the two strands. The negative slope of the curve between 230 and 260 nm is determined by the experimental technique used to stretch the RNA. In a “perfect” experiment, this portion of the curve would be a horizontal line, corresponding to an increase in length at constant force. The observed length increase is identical to the length of the molecule, showing that the molecule unfolds all at once in a process in which all bonds between the complementary base pairs are broken. The analysis of the results to obtain ∆G from the force curves is complex and will not be discussed here. The experimentally obtained value for ∆G is 193 kJ>mol, showing that a change in the thermodynamic state function G can be determined by a measurement of work on a single molecule. The unfolding can also be carried out by increasing the temperature of the solution containing the RNA to ∼80°C, in which case the process is referred to as melting. In the biochemical environment, unfolding is achieved through molecular forces exerted by enzymes called DNA polymerases, which act as molecular motors.
Streptavidin bead
RNA/DNA Handle B Digoxigenin
Figure 6.12
RNA molecule and measurement of the force needed to unfold strands. (a) Schematic diagram of one of the three types of RNA studied by Liphardt et al. are shown. The letters G, C, A, and U refer to the bases guanine, cytosine, adenine, and uracil. The strands are bound together by hydrogen bonding between the complementary base pairs A and U, and G and C. The pronounced bulge at the center of P5abc∆A is a hairpin loop. Such loops are believed to play a special role in the folding of RNA. (b) An individual RNA molecule is linked by handles attached to the end of each strand to functionalized polystyrene spheres that are moved relative to one another to induce unfolding of the RNA mechanically. See the reference to J. Liphardt in Further Reading for more details. Source: Adapted from Liphardt et al, Science, 292 (2001): 733–737.
280
16
Force/pN
15
14
Figure 6.13
Measured force versus distance curve for the P5abc𝚫A form of RNA. The black trace represents the stretching of the molecule, and the red curve represents the refolding of the molecule as the beads are brought back to their original separation. The rate of increase in force is 1 pN s1.
13
12 1 pN s21
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Source: Adapted from J. Liphardt et al, Science, 292 (2001): 733–737.
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CHAPTER 6 Chemical Equilibrium
VOCABULARY activation barrier chemical potential extent of reaction Gibbs energy Gibbs–Helmholtz equation
Helmholtz energy maximum nonexpansion work Maxwell relations molar Gibbs energy natural variables
reaction mechanism reaction quotient of pressures thermodynamic equilibrium constant
KEY EQUATIONS Equation
Significance of Equation
d1U  TS2 … dwexp + dwnonexp
Definition of Helmholtz energy A = U  TS
Equation Number 6.5
and identification with maximum total work dA … 0
Spontaneity criterion for process at constant T and V
d1U + PV  TS2 = d1H  TS2 … dwnonexp
Definition of Gibbs energy G = H  TS
6.8 6.9
and identification with maximum nonexpansion work dG … 0
Spontaneity criterion for process at constant T and P, no nonexpansion work
6.11
dU = TdS  PdV dH = TdS  PdV + PdV + VdP = TdS + VdP dA = TdS  PdV  TdS  SdT = SdT  PdV
Differential forms of important state functions
6.14–6.17
Maxwell relations provide relationships between different functions that are useful in linking seemingly unconnected state functions.
6.23–6.26
dG = TdS + VdP  TdS  SdT = SdT + VdP a a a
0T 0P b = a b 0V S 0S V 0T 0V b = a b 0P S 0S P
0S 0P a b = a b = kT 0V T 0T V
a
0S 0V b =  a b = Va 0P T 0T P P
G1T, P2 = G 1T2 + ~
a
03G>T4 0T
∆G1T22 T2
mi = a
b
=
P
= 
L~ P
VdP′ = G ~ 1T2 + nRT ln
H T2
∆G1T12 T1
0G b 0ni P,T,nj ≠ ni
+ ∆H1T12 a
G = a ni mi i
∆ mixG = nRT a xi ln xi i
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1 1  b T2 T1
P P~
Pressure dependence of the Gibbs energy for an ideal gas
6.29
Gibbs–Helmholtz equation allows KP to be calculated at different temperatures
6.30
Calculation of Gibbs energy at elevated temperatures assuming ∆H is constant over temperature range
6.33
Definition of chemical potential
6.35
Gibbs energy of gas mixture
6.37
Gibbs energy of mixing ideal gases
6.44
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183
Conceptual Problems
∆ mixS =  a a
0∆ mixG b =  nR a xi ln xi 0T P i
0r G b = a ni mi = ∆m 0j P,T i
drG 6 0 or a
0r G b = a ni mi 6 0 0j P,T i
PC g PD d b a ~b P~ P ∆m ~ = a vi ∆ f G i~ 1T2 =  RT ln a PA PB b i a ~b a ~b P P a
0 = ∆ r G ~ 1T2 + RT ln
a
a
P eq C
c
b a ~
P P eq A P
a
b a ~
P eq D
P P eq B ~
P
b
b ~
6.46
Slope of plot of G versus extent of reaction equals change of chemical potential of mixture
6.53
Criterion for spontaneity of reaction at constant T and P
6.55
Relates reaction quotient of pressures to ∆m ~ and Gibbs energies of formation of reactants and products
6.60 and 6.61
Defines thermodynamic equilibrium constant KP
6.64
Relates ∆ r G ~ to Gibbs energies of formation of reactants and products
6.65
Calculation of KP at elevated temperatures assuming ∆H is constant over temperature range
6.70
Defines equilibrium constant in terms of mole fractions
6.76
Defines equilibrium constant in terms of molarity
6.78
d
b
or, equivalently, ∆ r G 1T2 = RT ln KP ~
Entropy change of mixing ideal gases
or
ln KP = 
∆ r G ~ 1T2 RT
∆ r G ~ = a vi ∆ f G i~ i
ln KP1Tƒ2 = ln KP1298.15 K2 Kx = KP a Kc = KP a
P ∆v b P~
∆r H ~ 1 1 a b R Tƒ 298.15 K
c ~ RT ∆v b P~
CONCEPTUAL PROBLEMS Q6.1 It is found that KP is independent of T for a particular chemical reaction. What does this tell you about the reaction? Q6.2 The reaction A + B S C + D is at equilibrium for j = 0.1. What does this tell you about the ∆ f G ~ for reactants and products? Q6.3 Under what condition is KP = Kx? Q6.4 What is the relationship between the KP for the two reactions 3>2H21g2 + 1>2N21g2 S NH31g2 and 3H21g2 + N21g2 S 2NH31g2? Q6.5 Under what conditions is dA … 0 a condition that defines the spontaneity of a process? Q6.6 By invoking the pressure dependence of the chemical potential, show that if a valve separating a vessel of pure A from a vessel containing a mixture of A and B is opened, mixing will occur. Both A and B are ideal gases, and the initial pressure in both vessels is 1 bar. Q6.7 Under what conditions is dG … 0 a condition that defines the spontaneity of a process?
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Q6.8 Is the equation 10U>0V2T = 1aT  kTP2>kT valid for liquids, solids, and gases? Q6.9 Why is it reasonable to assume that the chemical potential of a pure liquid or solid substance is independent of the pressure in considering chemical equilibrium? Q6.10 The reaction A + B S C + D is at equilibrium for j = 0.5. What does this tell you about the ∆ f G ~ for reactants and products? Q6.11 Which thermodynamic state function gives a measure of the maximum electrical work that can be carried out in a fuel cell? Questions 6.12–6.17 refer to the reaction system CO 1 g2 + 1 , 2O2 1 g2 u CO2 1 g2 at equilibrium for which 𝚫 r H ~ = −283.0 kJ mol −1. Q6.12 Predict the change in the partial pressure of CO2 as temperature is increased at constant total pressure. Q6.13 Predict the change in the partial pressure of CO2 as pressure is increased at constant temperature.
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CHAPTER 6 Chemical Equilibrium
Q6.14 Predict the change in the partial pressure of CO2 as Xe gas is introduced into the reaction vessel at constant pressure and temperature. Q6.15 Predict the change in the partial pressure of CO2 as Xe gas is introduced into the reaction vessel at constant volume and temperature. Q6.16 Predict the change in the partial pressure of CO2 as a platinum catalyst is introduced into the reaction vessel at constant volume and temperature. Q6.17 Predict the change in the partial pressure of CO2 as O2 is removed from the reaction vessel at constant pressure and temperature. Q6.18 Is the equation ∆G = ∆H  T∆S applicable to all processes? Q6.19 Is the equation ∆A = ∆U  T∆S applicable to all processes? Q6.20 Under what conditions is Kx 7 KP? Q6.21 Under what conditions is the distribution of products in an ideal gas reactions system at equilibrium unaffected by an increase in the pressure? Q6.22 If KP is independent of pressure, why does the degree of dissociation in the reaction Cl21g2 N 2Cl1g2 depend on pressure? Questions 6.23–6.26 refer to the reaction system H2 1 g2 + Cl2 1 g2 N 2HCl1 g2 at equilibrium. Assume ideal gas behavior. Q6.23 How does the total number of moles in the reaction system change as T increases?
Q6.24 If the reaction is carried out at constant V, how does the total pressure change if T increases? Q6.25 Is the partial pressure of H21g2 dependent on T? If so, how will it change as T decreases? Q6.26 If the total pressure is increased at constant T, how will the relative amounts of H21g2 and HCl1g2 change? Q6.27 If additional Cl21g2 is added to the reaction system at constant V and T, how will the degree of dissociation of HCl1g2 change? Q6.28 If T is increased at constant total pressure, how will the degree of dissociation of HCl1g2 change? Q6.29 Which of the following data provide sufficient information to calculate KP at an arbitrary temperature T? Assume ideal gas behavior and that the heat capacities of reactants and products are independent of temperature. a. ∆ r H ~ 1298.15 K2 and ∆ r G ~ 1298.15 K2 b. ∆ r H ~ 1T2 and ∆ r G ~ 1298.15 K2 c. ∆ r G ~ 1T2 d. ∆ r H ~ 1298.15 K2 and KP1298.15 K2 e. KP1T12 and KP1T22 where T1, T2 ≠ T Q6.30 Consider a gasphase reaction for which ∆m1T2 = 15.5 kJ mol1 and ∆ r H ~ = 21.0 kJ mol1 at 350. K. If the following statements are true, explain why. If they are false, rewrite them to make them true. a. The reaction is spontaneous at 350. K. b. The reaction becomes spontaneous at temperatures well below 350. K.
NUMERICAL PROBLEMS Section 6.1 P6.1 Collagen is the most abundant protein in the mammalian body. It is a fibrous protein that serves to strengthen and support tissues. Suppose a collagen fiber can be stretched reversibly with a force constant of k = 15.0 N m1 and that the force, F (see Table 2.1) is given by F = k l. When a collagen fiber is contracted reversibly, it absorbs heat qrev = 0.015 J. Calculate the change in the Helmholtz energy, ∆A, as the fiber contracts isothermally from l = 0.030 to 0.025 m. Also calculate the reversible work performed wrev, ∆S, and ∆U. Assume that the temperature is constant at T = 310. K. P6.2 Calculate the maximum nonexpansion work that can be gained from the combustion of benzene(l) and of H21g2 on a per gram and a per mole basis under standard conditions. Is it apparent from this calculation why fuel cells based on H2 oxidation are under development for mobile applications? C6H61l2 + 15>2O21g2 4 6CO21g2 + 3H2O1l2
P6.3 A hardworking horse can lift a 300. lb. weight 110. ft. in one minute. Assume the horse generates energy to accomplish this work by metabolizing glucose: C6H12O61s2 + 6O21g2 S 6CO21g2 + 6H2O1l2
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Calculate how much glucose a horse must metabolize to sustain this rate of work for one hour. Assume T = 298.15 K. P6.4 Assume the internal energy of an elastic fiber under tension is given by dU = T dS  P dV  F d/. Obtain an expression for 10G>0/2P,T and calculate the maximum nonexpansion work obtainable when a collagen fiber with a force constant of k = 15.0 N m1 contracts from / = 20.0 to 10.0 cm at constant P and T. Sections 6.2 and 6.3 P6.5 Calculate ∆G for the isothermal expansion of 1.50 mol of an ideal gas at 325 K from an initial pressure of 15.5 bar to a final pressure of 1.0 bar. P6.6 As shown in Example Problem 3.5, 10Um >0V2T = a>V 2m for a van der Waals gas. In this problem, you will compare the change in energy with temperature and volume for N2, treating it as a van der Waals gas. a. Calculate ∆U per mole of N21g2 at 1 bar pressure and 298 K if the volume is increased by 5.00% at constant T. Approximate the molar volume with the ideal gas value. b. Calculate ∆U per mole of N21g2 at 1 bar pressure and 298 K if the temperature is increased by 5.00% at constant V.
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c. Calculate the ratio of your results in part (a) to the result in part (b). What can you conclude about the relative importance of changes in temperature and volume on ∆U? P6.7 Calculate ∆A for the isothermal compression of 1.70 mol of an ideal gas at 325 K from an initial volume of 50.0 L to a final volume of 10.0 L. Does it matter whether the path is reversible or irreversible? P6.8 Derive an expression for A1V,T2 analogous to that for G1T,P2 in Equation (6.29). P6.9 Show that 01A>T2 c d = U 011>T2 V
Write an expression analogous to Equation (6.32) that would allow you to relate ∆A at two temperatures. P6.10 A sample containing 2.50 mol of an ideal gas at 325 K is expanded from an initial volume of 10.5 L to a final volume of 60.0 L. Calculate ∆G and ∆A for this process for (a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure of 1.50 bar. Explain why ∆G and ∆A do or do not differ from one another. Sections 6.4 and 6.5 P6.11 Assume that a sealed vessel at constant pressure of 1 bar initially contains 2.00 mol of NO21g2. The system is allowed to equilibrate with respect to the reaction 2 NO21g2 N N2O41g2. The number of moles of NO21g2 and N2O41g2 at equilibrium is 2.00  2j and j, respectively, where j is the extent of reaction. a. Derive an expression for the entropy of mixing as a function of j. b. Graphically determine the value of j for which ∆ mixS has its maximum value. c. Write an expression for Gpure as a function of j. Use Equation (6.104) to obtain values of G m~ for NO2 and N2O4. d. Plot Gmixture = Gpure + ∆ mixG as a function of j for T = 298 K and graphically determine the value of j for which Gmixture has its minimum value. Is this value the same as for part (b)? P6.12 Calculate mmixture 1298.15 K, 1 bar2 for nitrogen in air, N2 assuming that the mole fraction of N2 in air is 0.790. P6.13 A sample containing 3.25 moles of He (1 bar, 350. K) is mixed with 2.25 mol of Ne (1bar, 350. K) and 1.75 mol of Ar (1 bar, 350. K). Calculate ∆ mixG and ∆ mixS. P6.14 A gas mixture with 3.00 mol of Ar, x moles of Ne, and y moles of Xe is prepared at a pressure of 1 bar and a temperature of 298 K. The total number of moles in the mixture is five times that of Ar. Write an expression for ∆ mixG in terms of x. At what value of x does the magnitude of ∆ mixG have its minimum value? Calculate ∆ mixG for this value of x. P6.15 You have containers of pure O2 and N2 at 298 K and 1 atm pressure. Calculate ∆ mixG relative to the unmixed gases of a. a mixture of 12 mol of O2 and 6 mol of N2
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Numerical Problems
185
b. a mixture of 12 mol of O2 and 12 mol of N2 c. Calculate ∆ mixG if 10 mol of pure N2 are added to the mixture of 12 mol of O2 and 12 mole of N2. Sections 6.6 and 6.7 P6.16 Nitrogen is a vital element for all living systems, but except for a few types of bacteria, bluegreen algae, and some soil fungi, most organisms cannot utilize N2 from the atmosphere. The formation of “fixed” nitrogen is therefore necessary to sustain life, and the simplest form of fixed nitrogen is ammonia NH3. A possible pathway for ammonia synthesis by a living system is: 1>2N21g2 + 3>2H2O1l2 S NH31aq2 + 3>2O21g2
where (aq) means the ammonia is dissolved in water. ∆ f G ~ 1NH3, aq2 = 80.3 kJ mol  1. a. Calculate ∆ r G ~ for the biological synthesis of ammonia at 298 K. b. Calculate the equilibrium constant for the biological synthesis of ammonia. c. Based on your answer to part (b), does the biological synthesis of ammonia occur spontaneously? P6.17 Consider the reaction system N21g2 + 3H21g2 N 2NH31g2. Assume that initially 3.00 mol of N21g2, 2.00 mol of H21g2, and 1.00 mol of NH31g2 are present in a reaction vessel at a constant pressure of 0.50 bar. a. If z is the extent of reaction, write expressions for the number of mols and the mole fractions of N21g2, H21g2, and NH31g2 present at equilibrium. b. Calculate the partial pressures of all species present at equilibrium if z = 0.30. c. What value of z corresponds to the reaction going to completion? P6.18 Calculate ∆ r A~ and ∆ r G ~ for the reaction C6H61l2 + 15>2O21g2 S 6CO21g2 + 3H2O1l2 at 298 K from the combustion enthalpy of benzene and the entropies of the reactants and products. All gaseous reactants and products are treated as ideal gases. P6.19 In Example Problem 6.11, KP for the reaction CO1g2 + H2O1l2 S CO21g2 + H21g2 was calculated to be 3.32 * 103 at 298.15 K. At what temperature does KP = 8.25 * 103? What is the highest value that KP can have by changing the temperature? Assume that ∆H R~ is independent of temperature. Sections 6.8 and 6.9 P6.20 A sample containing 3.50 moles of N2 and 7.50 mol of H2 is placed in a reaction vessel and brought to equilibrium at 52.0 bar and 555 K in the reaction 1>2 N21g2 + 3>2 H21g2 S NH31g2. a. Calculate KP at this temperature. b. Set up an equation relating KP and the extent of reaction as in Example Problem 6.12. c. Using a numerical equation solver, calculate the number of moles of each species present at equilibrium.
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CHAPTER 6 Chemical Equilibrium
P6.21 You place 1.75 mol of NOCl1g2 in a reaction vessel. Equilibrium at constant pressure is established with respect to the decomposition reaction NOCl1g2 N NO1g2 + 1>2 Cl21g2. a. Derive an expression for KP in terms of the extent of reaction j. b. Simplify your expression for part (a) in the limit that j is very small. c. Calculate j and the degree of dissociation of NOCl in the limit that j is very small at 350. K and a pressure of 1.50 bar. d. Solve the expression derived in part (a) using a numerical equation solver for the conditions stated in the previous part. What is the relative error in j made using the approximation of part (b)? P6.22 Calculate the degree of dissociation of N2O4 in the reaction N2O41g2 S 2NO21g2 at 325 K and a total pressure of 1.00 bar. Do you expect the degree of dissociation to increase or decrease as the temperature is increased to 550. K? Assume that ∆ r H ~ is independent of temperature. P6.23 Calculate KP at 298 and 595 K for the reaction NO1g2 + 1>2O21g2 S NO21g2 assuming that ∆ r H ~ is constant over the interval 298–600. K. Do you expect KP to increase or decrease as the temperature is increased to 700. K? P6.24 Consider the equilibrium NO21g2 L NO1g2 + 1>2O21g2. One mole of NO21g2 is placed in a vessel and allowed to come to equilibrium at a total pressure of 1 bar. An analysis of the contents of the vessel gives the following results: T PNO >PNO2
700. K 0.730
800. K 2.11
a. Calculate KP at 700. and 800. K. b. Calculate ∆ r G ~ and ∆ r H ~ for this reaction at 298.15 K using only the data in the problem. Assume that ∆ r H ~ is independent of temperature. c. Calculate ∆ r G ~ and ∆ r H ~ using the data tables and compare your answer with that obtained in part (b). P6.25 Consider the equilibrium C2H61g2 L C2H41g2 + H21g2. At 1025 K and a constant total pressure of 1.00 bar, C2H61g2 is introduced into a reaction vessel. The total pressure is held constant at 1.00 bar, and at equilibrium the composition of the mixture in mole percent is H21g2: 30.1%, C2H41g2: 30.1%, and C2H61g2: 39.8%. a. Calculate KP at 1025 K. b. If ∆ r H ~ = 136.4 kJ mol1, calculate the value of KP at 298.15K. c. Calculate ∆ r G ~ for this reaction at 298.15 K. P6.26 Many biological macromolecules undergo a transition called denaturation. Denaturation is a process whereby a structured, biological active molecule, called the native form, unfolds or becomes unstructured and biologically inactive. The equilibrium is native1folded2 N denatured1unfolded2
For a protein at pH = 2 the enthalpy change associated with denaturation is ∆H ~ = 418.0 kJ mol1, and the entropy change is ∆S ~ = 1.30 kJ K1 mol1.
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a. Calculate the Gibbs energy change for the denaturation of the protein at pH = 2 and T = 305 K. Assume the enthalpy and entropy are temperature independent between 298.15 K and 305 K. b. Calculate the equilibrium constant for the denaturation of protein at pH = 2 and T = 305 K. c. Based on your answer for parts (a) and (b), is the protein structurally stable at pH = 2 and T = 305 K? P6.27 Calculate KP at 525 K for the reaction N2O41l2 S 2NO21g2 assuming that ∆ r H ~ is constant over the interval 298–725 K. P6.28 Consider the equilibrium CO1g2 + H2O1g2 L CO21g2 + H21g2. At 1000. K, the composition of the reaction mixture is Substance
CO2 1 g2
Mole %
24.2
H2 1 g2 24.2
CO 1 g2 25.8
H2O 1 g2 25.8
a. Calculate KP and ∆ r G ~ at 1150. K. b. Given the answer to part (a), use the ∆ f H ~ of the reaction species to calculate ∆ r G ~ at 298.15 K. Assume that ∆ r H ~ is independent of temperature. P6.29 For the reaction C1graphite2 + H2O1g2 N CO1g2 + H21g2, ∆ r H ~ = 131.28 kJ mol1 at 298.15 K. Use the values of CP,m at 298.15 K in the data tables to calculate ∆ r H ~ at 150.0°C. P6.30 You wish to design an effusion source for Br atoms from Br21g2. If the source is to operate at a total pressure of 3.00 Torr, what temperature is required to produce a degree of dissociation of 0.150? What value of the pressure would increase the degree of dissociation to 0.500 at this temperature? P6.31 Consider the reaction FeO1s2 + CO1g2 L Fe1s2 + CO21g2 for which KP is found to have the following values: T KP
700.°C 0.688
1200.°C 0.310
a. Calculate ∆ r G ~ , ∆ r S ~ and ∆ r H ~ for this reaction at 700.°C. Assume that ∆ r H ~ is independent of temperature. b. Calculate the mole fraction of CO21g2 present in the gas phase at 700.°C. P6.32 Assuming that ∆ r H ~ is independent of temperature, calculate ∆ r G ~ for the process 1H2O, g, 298 K2 S 1H2O, g, 725.0 K2 . Calculate the relative change in the Gibbs energy over the temperature interval. P6.33 Oxygen reacts with solid glycylglycine C4H8N2O3 to form urea CH4N2O, carbon dioxide, and water: 3O21g2 + C4H8N2O31s2 S CH4N2O1s2 + 3CO21g2 + 2H2O1l2. At T = 298 K and 1.00 atm solid glycylglycine has the following thermodynamic properties: ∆ f G ~ = 491.5 kJ mol1, ∆ f H ~ = 746.0 kJ mol1,
S ~ = 190.0 J K1 mol1.
Calculate ∆ r G ~ at T = 298 K and at T = 310. K. State any assumptions that you make.
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Numerical Problems
P6.34 Calculate ∆ r G ~ for the reaction CO1g2 + 1>2O21g2 S CO21g2 at 298.15 K. Calculate ∆ r G ~ at 750. K, assuming that ∆ r H ~ is constant in the temperature interval of interest. P6.35 In this problem, you calculate the error in assuming that ∆ r H ~ is independent of T for the reaction 2CuO 1s2 L 2Cu1s2 + O21g2. The following data are given at 25°C: Compound
∆ f H ~ 1kJ mol12 1
∆ f G 1kJ mol 2 ~
1
CP,m1J K
CuO 1 s 2 157
O21g2
130.
1
mol 2
Cu 1 s 2
42.3
24.4
KP1Tƒ2
Tƒ
KP1Ti2
T0
29.4
1 ∆r H ~ a. From Equation (6.69) d ln KP = dT. R L T2 L To a good approximation, we can assume that the heat capacities are independent of temperature over a limited range in temperature, giving ∆ r H ~ 1T2 = ∆ r H ~ 1Ti2 +
∆CP1T  Ti2 where ∆CP = a ni CP,m1i2. By integrating i Equation (6.69), show that ln KP1T2 = ln KP1Ti2 +
∆ r H ~ 1Ti2 1 1 a  b R T Ti
Ti * ∆CP 1 ∆CP T 1 a  b + ln R T Ti R Ti
b. Using the result from part (a), calculate the equilibrium pressure of oxygen over copper and CuO1s2 at 1100. K. How is this value related to KP for the reaction 2CuO1s2 N 2Cu1s2 + O21g2? c. What value would you obtain if you assumed that ∆ r H ~ were constant at its value for 298.15 K up to 1100. K? P6.36 Consider the equilibrium in the reaction 3O21g2 N 2O31g2, with ∆ r H ~ = 285.4 * 103 J mol1 at 298 K. Assume that ∆ r H ~ is independent of temperature. a. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the pressure is increased. b. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the temperature is increased. c. Calculate KP at 800. and 900. K. d. Calculate Kx at 800. K and pressures of 1.00 and 2.50 bar. P6.37 N2O3 dissociates according to the equilibrium N2O31g2 N NO21g2 + NO1g2. At 275 K and one bar pressure, the degree of dissociation defined as the ratio of moles of NO1g2 or NO21g2 to the moles of the reactant assuming that no dissociation occurs is 0.486. Calculate ∆ r G ~ for this reaction at 275 K. P6.38 If the reaction Fe 2N1s2 + 3>2H21g2 L 2Fe1s2 + NH31g2 comes to equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700. and 800. K, PNH3 >PH2 = 1.971 and 0.9780, respectively, if only H21g2
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187
was initially present in the gas phase and Fe 2N1s2 was in excess. a. Calculate KP at 700. and 800. K. b. Calculate ∆ r S ~ at 700. K and 800. K and ∆ r H ~ , assuming that it is independent of temperature. c. Calculate ∆ r G ~ for this reaction at 298.15 K. P6.39 Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction C6H12O61s2 S 2CH3CHOHCOOH1s2. Thermodynamic data at T = 298 K for glucose and lactic acid are given in the table. Glucose(s) Lactic acid(s)
𝚫 r H ~ 1kJ mol −1 2 CP,m 1J K−1 mol −1 2 S ~ 1J K−1 mol −1 2
1273.1
219.2
209.2
673.6
127.6
192.1
Calculate ∆ r G ~ at T = 298 K and T = 325 K using the data in the table. Assume that all heat capacities are constant in this temperature interval and that ∆ r H ~ and ∆ r S ~ are constant in this temperature interval. P6.40 Consider the equilibrium 3O21g2 N 2O31g2. a. Using the data tables, calculate KP at 298 K. b. Set up a table as in Example Problem 6.10 describing the approach to equilibrium. At equilibrium, the number of moles of O21g2 and O31g2 are n0  3deq and 2deq, respectively. Assuming that the extent of reaction at equilibrium is much less than one, show that the degree of reaction defined as deq >n0 is given by 11>222KP * P>P ~ . c. Calculate the degree of reaction at 500. K and a pressure of 5.00 bar. d. Calculate Kx at 500. K and a pressure of 5.00 bar. d ln KP = P6.41 As shown in Equation (6.64), dT ~ ~ ∆r H 1 d1∆ r G >T2 = . For most cases, it is reasonable R dT RT 2 to assume that ∆ r H ~ is independent of temperature in calculating the temperature dependence of KP. However, a more accurate calculation takes the temperature dependence of ∆ r H ~ into account. As listed in the data tables, CP,m for individual reactants and products can be expressed in the form CP,m = A112 + A122T + A132T 2 + A142T 3, and the difference in the heat capacity between products can be expressed as ∆CP,m = ∆A112 + ∆A122T + ∆A132T 2 + ∆A142T 3. a. Show that ∆ r H ~ 1T2 = ∆A112T +
∆A122T 2 + 2
∆A132T 3 ∆A142T 4 + + B where B is a constant of 3 4 integration that can be evaluated as ∆ r H ~ can be calculated from standard enthalpies of formation at 298.15 K.
b. Show that ln KP1T2 =
∆A112 ∆A122T ln T + + R 2R
∆A132T 2 ∆A142T 3 + + C where C is a constant of 6R 12R integration that can be evaluated if KP is known at a given temperature.
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CHAPTER 6 Chemical Equilibrium
Section 6.10 P6.42 The shells of marine organisms are constructed of biologically secreted calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite are given below. Properties 1 T = 298 K, P = 1.00 bar2
Calcite
Aragonite
∆ f H ~ 1kJ mol12
1206.9
1207.0
1
1128.8
1127.7
S 1J K
1
92.9
88.7
81.9
81.3
2.710
2.930
∆ f G 1kJ mol 2 ~
~
1
mol 2
CP,m1J K1 mol12 1
Density 1g mL 2
a. Based on the thermodynamic data given, would you expect an isolated sample of calcite at T = 298 K and P = 1.00 bar to convert to aragonite, given sufficient time? Explain. b. Suppose the pressure applied to an isolated sample of calcite is increased. Can the pressure be increased to the point that isolated calcite will be converted to aragonite? Explain.
c. What pressure must be achieved to induce the conversion of calcite to aragonite at T = 298 K. Assume both calcite and aragonite are incompressible at T = 298 K. d. Can calcite be converted to aragonite at P = 1.00 bar if the temperature is increased? Explain. P6.43 The pressure dependence of G is quite different for gases and condensed phases. Calculate ∆Gm for the processes (C, solid, graphite, 1 bar, 298.15 K) S (C, solid, graphite, 400. bar, 298.15 K) and (He, g, 1 bar, 298.15 K) S (He, g, 400. bar, 298.15 K). By what factor is ∆Gm greater for He than for graphite? P6.44 Ca1HCO3221s2 decomposes at elevated temperatures according to the stoichiometric equation Ca1HCO3221s2 S CaCO31s2 + H2O1g2 + CO21g2. a. If pure Ca1HCO3221s2 is put into a sealed vessel, the air is pumped out, and the vessel and its contents are heated, the total pressure is 0.150 bar. Determine KP under these conditions. b. If the vessel also contains 0.150 bar H2O1g2 at the final temperature, what is the partial pressure of CO21g2 at equilibrium?
Section 6.12 P6.45 Use the equation CP,m  CV,m = TVma2 >kT and the data tables to determine CV,m for H2O1l2 at 325 K. Calculate 1CP,m  CV,m2>CP,m.
FURTHER READING Dahl. S., Logadottir, A., Egeberg, R. C., Larsen, J. H., Chorkendorff, I., Tornqvist, E., and Norskov, J. K. “Role of Steps in N2 Activation on Ru(0001).” Physical Review Letters 83 (1999): 1814–1817. Ertl, G. “Surface Science and Catalysis—Studies on the Mechanism of Ammonia Synthesis: The P.H. Emmett Award Address.” Catalysis Reviews—Science and Engineering 21 (1980): 201–223. Gislason, E. A., and Craig, N. C. “Criteria for Spontaneous Processes Derived from the Global Point of View.” Journal of Chemical Education 90 (2013): 584–590. Hwang, R. Q., Schroder, J., Gunther, C., and Behm, R. J. “Fractal Growth of TwoDimensional Islands: Au on Ru(0001).” Physical Review Letters 67 (1991): 3279–3283.
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Jacobsen, C. J. H., Dahl, S., Clausen, B. S., Bahn, S., Logadottir, A., and Norskov, J. K. “Catalyst Design by Interpolation in the Periodic Table: Bimetallic Ammonia Synthesis Catalysts.” Journal of the American Chemical Society 123 (2001): 8404–8405. Liphardt, J., Onoa, B., Smith, S., Tinoco, I., and Bustamante, C. “Reversible Unfolding of Single RNA Molecules by Mechanical Force.” Science, 292 (2001): 733–737. Raff, L. M. “Spontaneity and Equilibrium: Why ‘∆G 6 0 Denotes a Spontaneous Process’ and ∆G = 0 Means the System Is at Equilibrium Are Incorrect.” Journal of Chemical Education 91 (2014a): 386–395. Raff, L. M. “Spontaneity and Equilibrium III: A History of Misinformation.” Journal of Chemical Education 91 (2014b): 2128–2136.
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C H A P T E R
7
The Properties of Real Gases
7.1 Real Gases and Ideal Gases 7.2 Equations of State for Real Gases and Their Range of Applicability
7.3 The Compression Factor
WHY is this material important? The ideal gas law is only accurate for gases at low values of density. To design production plants that use real gases at high pressures, equations of state valid for gases at higher densities are needed. Such equations must take into account the finite volume of a molecule and the intermolecular potential.
7.4 The Law of Corresponding States
7.5 Fugacity and the Equilibrium Constant for Real Gases
WHAT are the most important concepts and results? Equations of state for real gases accurately describe the P–V relationship of a given gas at a fixed value of T within their range of validity. These equations use parameters specific to a given gas. An important consequence of nonideality is that the chemical potential of a real gas must be expressed in terms of its fugacity rather than its partial pressure. Fugacities, rather than pressures, must also be used in calculating the thermodynamic equilibrium constant Kp for a real gas at elevated pressures.
WHAT would be helpful for you to review for this chapter? It would be useful to review the material on ideal gases in Chapter 1.
7.1 REAL GASES AND IDEAL GASES To this point, the ideal gas equation of state has been assumed to be sufficiently accurate to describe the P–V–T relationship for a real gas. This assumption has allowed calculations of expansion work and of the equilibrium constant KP in terms of partial pressures using the ideal gas law. In fact, the ideal gas law provides an accurate description of the P–V–T relationship for many gases, such as He, for a wide range of P, V, and T values. However, it describes the P–V relationship for water for a wide range of P and V values within {10, only for T 7 1300 K, as shown in Section 7.4. What is the explanation for this different behavior of He and H2O? Is it possible to derive a “universal” equation of state that can be used to describe the P–V relationship for gases as different as He and H2O? In Section 1.5, the two main deficiencies in the microscopic model on which the ideal gas law is based were discussed. The first assumption is that gas molecules are point masses. However, molecules occupy a finite volume; therefore, a real gas cannot be compressed to a volume that is less than the total molecular volume. The second assumption that was made is that the molecules in the gas do not interact; but molecules in a real gas do interact with one another through a potential as depicted in Figure 1.9. Because the potential has a short range, its effect is negligible at low densities, which correspond to large distances between molecules. Additionally, at low densities, the molecular volume is negligible compared with the volume that the gas occupies. Therefore, the P–V–T relationship of a real gas is the same as that for an ideal gas at sufficiently low densities and high temperatures. At higher densities and low temperatures, molecular interactions cannot be neglected. Because of these interactions, the pressure
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Concept Molecules of real gases have a finite molecular volume and interact with one another.
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CHAPTER 7 The Properties of Real Gases
of a real gas can be higher or lower than that for an ideal gas at the same density and temperature. What determines which of these two cases applies? The questions raised in this section are the major themes of this chapter.
7.2 EQUATIONS OF STATE FOR REAL GASES AND THEIR RANGE OF APPLICABILITY
This section discusses several equations of state for real gases and the range of the variables P, V, and T over which they accurately describe a real gas. Such equations of state must exhibit a limiting P–V–T behavior identical to that for an ideal gas at low density. They must also correctly model the deviations for ideal gas behavior that real gases exhibit at moderate and high densities. The first two equations of state considered here include two parameters, a and b, that must be experimentally determined for a given gas. The parameter a is a measure of the strength of the attractive part of the intermolecular potential, and b is a measure of the minimum volume that a mole of molecules can occupy. Real gas equations of state are best viewed as empirical equations whose functional form has been chosen to fit experimentally determined P–V–T data. The most widely used equation for real gases is the van der Waals equation of state:
P =
RT a nRT n2a  2 =  2 Vm  b V m V  nb V
(7.1)
A second useful equation of state is the Redlich–Kwong equation of state:
Concept All equations of state for real gases have a limited range of P, V, and T over which they are applicable.
P =
RT a 1 nRT n2a 1 = Vm  b V 1V + b2 V nb V1V + nb2 2T m m 2T
(7.2)
Although the same symbols are used for parameters a and b in both equations of state, they have different values for a given gas. Figure 7.1 shows that the degree to which the ideal gas, van der Waals, and Redlich– Kwong equations of state correctly predict the P–V behavior of CO2 depends on P, V, and T. At 426 K, all three equations of state reproduce the correct P–V behavior reasonably well over the range shown, with the ideal gas law having the largest error. By contrast, the three equations of state give significantly different results at 310. K. The ideal gas law gives unacceptably large errors, and the Redlich–Kwong equation of state is more accurate than is the van der Waals equation. We will have more to say about the range over which the ideal gas law is reasonably accurate when discussing the compression factor in Section 7.3. A third widely used equation of state for real gases is the Beattie–Bridgeman equation of state. The form of the equation is P =
RT c A a1 b 1Vm + B2  2 2 3 Vm Vm VmT
A = A0 a1 
a b Vm
and B = B0 a1 
with b b Vm
(7.3)
This equation uses five experimentally determined parameters to fit P–V–T data. ecause of its complexity, it will not be discussed further. B A further important equation of state for real gases has a different form than any of the previous equations. The virial equation of state is written in the form of a power series in 1>Vm
P = RT c
B1T2 1 + + gd Vm V 2m
(7.4)
The power series does not converge at high pressures where Vm becomes small. The B1T2, C1T2, and so on are called the second, third, and so on virial coefficients. This equation is more firmly grounded in theory than the previously discussed three equations because a series expansion is always valid in its convergence range. In practical
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7.2 Equations of State for Real Gases and Their Range of Applicability
191
Figure 7.1
Comparison of real gas equations of state to ideal gas equation of state. Isotherms for CO21g2 are shown at (a) 426 K and (b) 310 K. The purple curve is calculated using the van der Waals equation of state, the blue curve is calculated using the Redlich–Kwong equation of state, and the red curve is calculated using the ideal gas equation of state. The black dots are accurate values taken from the NIST Chemistry Webbook, which are derived from experimental results. Note that in (a) the purple curve is mostly obscured by the blue curve.
140
Pressure (bar)
120 100
T 5 426 K
80 60 40 20
0
0.25
0.50
0.75 1.00 Molar volume (L)
1.25
1.50
1.75
(a)
Pressure (bar)
500 400 300 T 5 310. K 200 100
0
0.1
0.2
0.3 Molar volume (L)
0.4
0.5
0.6
(b)
use, the series is usually terminated after the second virial coefficient because values for the higher coefficients are not easily obtained from experiments. Table 7.1 (see Appendix A, Data Tables) lists values for the second virial coefficient for selected gases for different temperatures. If B1T2 is negative, the attractive part of the potential dominates at that value of T. In contrast, if B1T2 is positive, the repulsive part of the potential dominates at that value of T. Statistical thermodynamics can be used to relate the virial coefficients with the intermolecular potential function. As you will show in the endofchapter problems, B1T2 for a van der Waals gas is given by B1T2 = b  1a>RT2. The principal limitation of the ideal gas law is that it does not predict that a gas can be liquefied under appropriate conditions. Consider the approximate P–V diagram for CO2 shown in Figure 7.2. (It is approximate because it is based on the van der Waals equation of state rather than experimental data.) Each of the curves is an isotherm (that is, each corresponds to a fixed temperature). The behavior predicted by the ideal gas law is shown for T = 334 K. To understand the behavior indicated by diagram in Figure 7.2, consider the isotherm for T = 258 K. Starting at large values of Vm, at first as Vm decreases the pressure increases. Then, with a further decrease in Vm, P is constant over a range of values of Vm. The value of Vm at which P becomes constant depends on T. As Vm decreases further, the pressure suddenly increases rapidly. The reason for this unusual dependence of P on Vm becomes clear when the CO2 compression experiment is carried out in the piston and transparent cylinder assembly shown in Figure 7.3. The system consists of either a single phase or two phases separated by a sharp interface, depending on the values of T and Vm. For points a, b, c, and d on
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Figure 7.2 140
365 K
120 334 K Ideal gas
334 K Pressure (bar)
Calculated isotherms for CO2 modeled as a van der Waals gas. The gas (green) and liquid (blue) regions and the gas–liquid (yellow) coexistence region are shown. The dashed curve was calculated using the ideal gas law. The isotherm at T = 304.12 K is at the critical temperature and is called the critical isotherm.
Gas
100 80 60
304.12 K Liquid
274 K d
40
c
258 K b
243 K
20
0
Concept Ideal gases do not condense to form liquids or solids.
Concept Real gases have twophase coexistence regions.
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Gas 1 Liquid
0.1
0.2
0.3 0.4 Molar volume (L mol–1)
0.5
0.6
the 258 K isotherm of Figure 7.2, the system has the following composition: at point a: the system consists entirely of CO21g2. However, at points b and c, a sharp interface separates CO21g2 and CO21l2. Along the line linking points b and c, the system contains CO21g2 and CO21l2 in equilibrium with one another. The proportion of liquid to gas changes, but the pressure remains constant. The temperaturedependent equilibrium pressure is called the vapor pressure of the liquid. As the volume is decreased further, the system becomes a singlephase system again, consisting of CO21l2 only as at point d. The pressure changes slowly with increasing V if Vm 7 0.33 L because CO21g2 is quite compressible. However, the pressure changes rapidly with decreasing V if Vm 6 0.06 L because CO21l2 is nearly incompressible. The singlephase regions and the twophase gas–liquid coexistence region are shown in Figure 7.2. If the same experiment is carried out at successively higher temperatures, it is found that the range of Vm in which two phases are present becomes smaller, as seen in the 243, 258, and 274 K isotherms of Figure 7.2. The temperature at which the range of Vm has shrunk to a single value is called the critical temperature, Tc. For CO2, Tc = 304.12 K. At T = Tc, the isotherm exhibits an inflection point so that
W7.1 van der Waals equation of state isotherms
a
a
0P 02P b = 0 and a 2 b = 0 0Vm T = Tc 0V m T = Tc
(7.5)
What is the significance of the critical temperature? Although critical behavior will be discussed in Chapter 8, at this point it is sufficient to know that as the critical point is approached along the 304.12 K isotherm, the density of CO21l2 decreases and the density of CO21g2 increases, and at T = Tc the densities are equal. At temperatures greater than Tc, no interface is observed in the experiment depicted in Figure 7.3, and liquid and gas phases can no longer be distinguished. Instead, the term supercritical fluid is used to describe the state of CO2. As will be discussed in Chapter 8, Tc and the corresponding values Pc and Vc, which together are called the critical constants, take on particular significance in describing the phase diagram of a pure substance. The critical constants for a number of different substances are listed in Table 7.2 (see Appendix A, Data Tables). The parameters a and b for the van der Waals and Redlich–Kwong equations of state are chosen so that the equation of state best represents real gas data. This can be done by using the values of P, V, and T at the critical point Tc, Pc, and Vc, as shown in Example Problem 7.1. Parameters a and b calculated from critical constants in this way are listed in Table 7.4 (see Appendix A, Data Tables).
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7.2 Equations of State for Real Gases and Their Range of Applicability
193
Figure 7.3
Pa Pb Pc Pd
CO2(g)
CO2(g) CO2(g) CO2(l)
CO2(l)
Point c
Point d
CO2(l) Point a
Point b
Schematic diagram of CO2 volume and phases present in a transparent cylinder undergoing compression. The experiment is conducted at a constant temperature of 258 K, which corresponds to the brown isotherm in Figure 7.2. Also, cylinders with labeled points a, b, c, and d correspond to the same labeled points in Figure 7.2. The liquid and gas volumes are not shown to scale. Note that Pb = Pc.
EXAMPLE PROBLEM 7.1 At T = Tc, 10P>0Vm2T = Tc = 0 and 102P>0V 2m2T = Tc = 0. Use this information to determine a and b in the van der Waals equation of state in terms of the experimentally determined values Tc and Pc.
Solution ° °
0c
02 c
RT a  2d Vm  b Vm ¢ RTc 2a = + 3 = 0 2 0Vm 1Vmc  b2 V mc T = Tc
RTc 2a RT a 0c + 3d  2d Vm  b 1Vm  b22 Vm ¢ Vm ¢ ° = 0Vm 0V 2m T = Tc T = Tc =
2RTc 3
1Vmc  b2
+
6a = 0 V 4mc
Equating RTC from these two equations gives RTc =
2a 3a 3 1Vmc  b22 = 4 1Vmc  b23, which simplifies to 1V  b2 = 1 2Vmc mc V mc V 3mc
The solution to this equation is Vmc = 3b. Substituting this result into 10P>0Vm2T = Tc = 0 gives 
RTc
1Vmc  b22
+
RTc 8a 2a 2a 2a 12b22 = = + = 0 or T = c 27Rb 12b22 V 3mc 13b23 13b23 R
Substituting these results for Tc and Vmc in terms of a and b into the van der Waals equation gives the result Pc = a>27b2. We only need two of the critical constants Tc, Pc, and Vmc to determine a and b. Because the measurements for Pc and Tc are more accurate than those for Vmc, we use these constants to obtain expressions for a and b. These results for Pc and Tc can be used to express a and b in terms of Pc and Tc. The results are b =
RTc 8Pc
and a =
27R2T 2c 64Pc
A similar analysis for the Redlich–Kwong equation gives a =
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R2T 5>2 c 9Pc121>3  12
and b =
121>3  12RTc 3Pc
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Concept Real gases do not exhibit oscillations predicted by van der Waals equation of state.
140
Pressure (bar)
120 100 80 60
304 K 274 K 258 K
40 20
0
243 K
0.1 0.2 0.3 0.4 0.5 0.6 Molar volume (L mol21)
Figure 7.4
Van der Waals isotherms and their corresponding Maxwell constructions. The calculated relationship between pressure and molar volume for CO2 is shown at the indicated temperatures. The Maxwell constructions are shown in the oscillating portion of isotherms. Note that no oscillations in the calculated isotherms occur for T Ú Tc.
Concept The compression factor is a convenient way to quantify nonideality.
What is the range of validity of the van der Waals and Redlich–Kwong equations? No twoparameter equation of state is able to reproduce the isotherms shown in Figure 7.2 for T 6 Tc because it cannot reproduce either the range in which P is constant or the discontinuity in 10P>0V2T at the ends of this range. This failure is illustrated in Figure 7.4, in which isotherms calculated using the van der Waals equation of state are plotted for some of the values of T also used in Figure 7.2. Below Tc, all calculated isotherms have an oscillating region that is unphysical because V increases as P increases. In a Maxwell construction, the oscillating portion of an isotherm is replaced by the horizontal line for which the areas above and below the line are equal, as indicated in Figure 7.4. The Maxwell construction is used in generating the horizontal segments of isotherms shown in Figure 7.2. The Maxwell construction can be justified on theoretical grounds, but the equilibrium vapor pressure determined in this way for a given value of T is only in qualitative agreement with experiment. The van der Waals and Redlich–Kwong equations of state do a good job of reproducing P–V isotherms for real gases only in the singlephase gas region T 7 Tc and for densities well below the critical density, rc = M>Vmc, where Vmc is the molar volume at the critical point. The Beattie–Bridgeman equation, which has three more adjustable parameters, is accurate above Tc for higher densities.
7.3 THE COMPRESSION FACTOR How large is the error in P–V curves if the ideal gas law is used rather than the van der Waals or Redlich–Kwong equations of state? To address this question, it is useful to introduce the compression factor, z, defined by
z =
Vm V ideal m
=
PVm RT
(7.6)
For the ideal gas, z = 1 for all values of P and Vm. If z 7 1, the real gas exerts a greater pressure than the ideal gas, and if z 6 1, the real gas exerts a smaller pressure than the ideal gas for the same values of T and Vm. The compression factor for a given gas is a function of temperature, as shown in Figure 7.5. In this figure, z has been calculated for N2 at two values of T using the van der Waals and Redlich–Kwong equations of state. The dots are calculated from accurate values of Vm taken from the NIST Chemistry Workbook. Although the results calculated using these equations of state are not in quantitative agreement with accurate results for all P and T, the trends in the functional dependence of z on P for different T values are correct. Because it is inconvenient to rely on tabulated data for individual gases, we focus on z1P2 curves calculated from real gas equations of state. Both the 1.2
RK 400. K
Plot of calculated compression factors for N2 1 g2 as a function of pressure. The calculated compression factors, z, are shown for 200. and 400. K using the Redlich–Kwong (red lines) and van der Waals (purple lines) equations of state. For comparison, calculated values of z are also shown (colored dots and squares) for the accurate values for Vm taken from the NIST Chemistry Webbook. The Redlich– Kwong equation does not give physically meaningful solutions at P greater than 175 bar for 200. K, for which r = 2.5rc.
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Compression factor (z)
Figure 7.5
1.1 vdW 400. K 1.0 50
100
150
200
250 300 Pressure (bar)
0.9 RK 200. K
0.8
vdW 200. K
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7.3 The Compression Factor
van der Waals and Redlich–Kwong equations predict that for T = 200 K, z initially decreases with pressure. The compression factor only becomes greater than the ideal gas value of one for pressures in excess of 200 bar. For T = 400. K, z increases linearly with T. This functional dependence is also predicted by the Redlich–Kwong equation. The van der Waals equation predicts that the initial slope is zero, and z increases slowly with P. For both temperatures, z S 1 as P S 0. This result shows that the ideal gas law is obeyed if P is sufficiently small. To understand why the lowpressure value of the compression factor varies with temperature for a given gas, we use the van der Waals equation of state. Consider the variation of the compression factor with P at constant T, a
0z 0z 1 0z b = a b = a b 0P T 03 RT>Vm 4 T RT 03 1>Vm 4 T
for a van der Waals gas in the ideal gas limit as 1>Vm S 0. As shown in Example Problem 7.2, the result 10z>0P2T = b>RT  a>1RT22 is obtained.
W7.2 Quantitative comparison between van der Waals and ideal gas equations of state
EXAMPLE PROBLEM 7.2 Show that the slope of z as a function of P as P S 0 is related to the van der Waals parameters by lim a
pS0
0z 1 a b = ab b 0P T RT RT
Solution Rather than differentiating z with respect to P, we transform the partial derivative to one involving Vm: z =
Vm V ideal m
PVm = = RT
a
RT a  2 b Vm Vm  b Vm Vm a = RT Vm  b RTVm
Vm 0z 0 a c d 0Vm 0Vm Vm  b RTVm 0z a b = ± ≤ = ± ≤ 0P T 0P 0 RT a a  2b 0Vm T 0Vm Vm  b Vm T Vm 1 a + Vm  b 1Vm  b22 RTV 2m = ± ≤ RT a + 2 1Vm  b22 V 3m Vm  b Vm a + 2 2 1Vm  b2 1Vm  b2 RTV 2m = ± ≤ RT a + 2 1Vm  b22 V 3m b a b a + + 2 2 2 1Vm  b2 RTV m Vm RTV 2m = ± ≤ S± ≤ S RT a RT + 2 3  2 1Vm  b22 Vm Vm S
1 a ab b as Vm S ∞ RT RT
RT a RT + 2 3 S  2 as Vm S ∞ because the sec2 1Vm  b2 Vm Vm ond term approaches zero more rapidly than the first term and Vm W b.
In the second to last line, 
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CHAPTER 7 The Properties of Real Gases
TABLE 7.3 Boyle Temperatures of Selected Gases
He
TB1K2
H2
110.
Ne
122
N2
327
CO
352
O2
400.
CH4
510.
Kr
575
Ethene
735
H2O
1250
Gas
23
Source: Calculated from data in Lide, D. R., ed. CRC Handbook of Thermophysical and Thermochemical Data. Boca Raton, FL: CRC Press, 1994.
Concept Real gases behave similarly to ideal gases at the Boyle temperature.
From Example Problem 7.2, the van der Waals equation predicts that the initial slope of the z versus P curve is zero if b = a>RT. The corresponding temperature is known as the Boyle temperature TB
TB =
a Rb
(7.7)
Values for the Boyle temperature of several gases are shown in Table 7.3. Because the parameters a and b are substance dependent, TB is different for each gas. Using the van der Waals parameters for N2, TB = 425 K, whereas the experimentally determined value is 327 K. The agreement is qualitative rather than quantitative. At the Boyle temperature, both z S 1, and 10z>0P2T S 0 as P S 0, which is the behavior exhibited by an ideal gas. It is only at T = TB that a real gas exhibits ideal behavior as P S 0 with respect to lim 10z>0P2T . Above the Boyle temperature, 10z>0P2T 7 0 PS0 as P S 0, and below the Boyle temperature, 10z>0P2T 6 0 as P S 0. These inequalities provide a criterion to predict whether z increases or decreases with pressure at low pressures for a given value of T. Note that the initial slope of a z versus P plot is determined by the relative magnitudes of b and a>RT. Recall that the repulsive interaction is represented through b, and the attractive part of the potential is represented through a. From the previous discussion, lim 10z>0P2T is always positive at high temperatures because b  1a>RT2 S b 7 0 as PS0 T S ∞. This means that molecules primarily feel the repulsive part of the potential for T W TB. Conversely, for T V TB, lim 10z>0P2T, will always be negative because the PS0 molecules primarily feel the attractive part of the potential. For high enough values of P, z 7 1 for all gases, showing that the repulsive part of the potential dominates at high gas densities, regardless of the value of T. Next consider the functional dependence of z on P for different gases at a single temperature. Calculated values for oxygen, hydrogen, and ethene obtained using the van der Waals equation at T = 400. K are shown in Figure 7.6, together with accurate results for these gases. It is seen that lim 10z>0P2T at 400. K is positive for PS0 H2, negative for ethane, and approximately zero for O2. These trends are correctly predicted by the van der Waals equation. How can this behavior be explained? It is useful to compare the shape of the curves for H2 and ethene with those in Figure 7.5 for N2 at different temperatures. This comparison suggests that at 400. K the gas temperature is well above TB for H2 so that the repulsive part of the potential dominates.
Figure 7.6
Plot of compression factor as a function of pressure at T = 400. K for three different gases. The solid lines have been calculated using the van der Waals equation of state. The dots are calculated from accurate values for Vm taken from the NIST Chemistry Webbook. Red, hydrogen; purple, oxygen; and light blue, ethene.
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Compression factor (z)
1.2
Hydrogen
1.1
Oxygen 1.0 50
100
150
200
250
300
Pressure (bar) 0.9 Ethene
0.8
0.7
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7.4 The Law of Corresponding States
197
By contrast, it suggests that 400. K is well below TB for ethene, and the attractive part of the potential dominates. Because the initial slope is nearly zero, the data in Figure 7.6 suggest that 400. K is near TB for oxygen. As is seen in Table 7.3, TB = 735 K for ethene, 400. K for O2, and 110. K for H2. These values are consistent with the previous explanation. The results shown in Figure 7.6 can be generalized as follows. If lim 10z>0P2T 6 0 PS0 for a particular gas, T 6 TB, and the attractive part of the potential dominates. If lim 10z>0P2T 7 0 for a particular gas, T 7 TB, and the repulsive part of the PS0 potential dominates.
7.4 THE LAW OF CORRESPONDING STATES
W7.3 Compression factor for a van der Waals gas
As shown in Section 7.3, the compression factor is a convenient way to quantify deviations from the ideal gas law. In calculating z using the van der Waals and Redlich– Kwong equations of state, different parameters must be used for each gas. Is it possible to find an equation of state for real gases that does not explicitly contain materialdependent parameters? Real gases differ from one another primarily in the value of the molecular volume and in the magnitude of the attractive potential. Because molecules that have a stronger attractive interaction exist as liquids to higher temperatures, one might think that the critical temperature is a measure of the magnitude of the attractive potential. Similarly, one might think that the critical volume is a measure of the molecular volume. If this is the case, different gases should behave similarly if T, V, and P are measured relative to their critical values. These considerations suggest the following hypothesis. Different gases have the same equation of state if each gas is described by the dimensionless reduced variables Tr = T>Tc, Pr = P>Pc, and Vmr = Vm >Vmc, rather than by T, P, and Vm. The preceding statement is known as the law of corresponding states. If two gases have the same values of Tr, Pr, and Vmr, they are in corresponding states. The values of P, V, and T can be very different for two gases that are in corresponding states. For example, H2 at 12.93 bar and 32.98 K in a volume of 64.2 * 103 L and Br2 at 103 bar and 588 K in a volume of 127 * 103 L are in the same corresponding state. Next we justify the law of corresponding states and show that it is obeyed by many gases. The parameters a and b can be eliminated from the van der Waals equation of state by expressing the equation in terms of the reduced variables Tr, Pr, and Vmr. This can be seen by writing the van der Waals equation in the form
PrPc =
RTrTc a  2 2 VmrVmc  b V mrV mc
(7.8)
Next replace Tc, Vmc, and Pc by the relations derived in Example Problem 7.1:
Pc =
a 8a , Vmc = 3b, and Tc = 27Rb 27b2
(7.9)
Equation (7.8) becomes aPr
8aTr a  2 2 27b13bVmr  b2 27b 9b V mr 8Tr 3 Pr =  2 3Vmr  1 V mr 2
=
or
(7.10)
Equation (7.10) relates Tr, Pr and Vmr without reference to the parameters a and b. herefore, it has the character of a universal equation, like the ideal gas equation T of state. To a good approximation, the law of corresponding states is obeyed for a large number of different gases as shown in Figure 7.7, as long as Tr 7 1. However, Equation (7.10) is not really universal because the materialdependent quantities enter through values of Pc, Tc, and Vc rather than through a and b.
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CHAPTER 7 The Properties of Real Gases
Figure 7.7
Concept
1.0 Tr 5 2.00 Tr 5 1.50 Compression factor (z)
Values of compression factor plotted as a function of the reduced pressure Pr for seven gases. Compression factors calculated for the six values of reduced temperatures are indicated in the figure. The solid curves are drawn to guide the eye.
0.8 Tr 5 1.30 Nitrogen Tr 5 1.20
0.6
Methane Propane
Tr 5 1.10
Ethane Butane
0.4
Isopentane
The law of corresponding states allows the compression factors for very different gases to be estimated.
Ethene
Tr 5 1.00 0.2 1
4 5 3 Reduced pressure (Pr)
2
6
7
The law of corresponding states implicitly assumes that two parameters are sufficient to describe an intermolecular potential. This assumption is best for molecules that are nearly spherical because for such molecules the potential is independent of the molecular orientation. It is not nearly as good for dipolar molecules such as HF, for which the potential is orientation dependent. The results shown in Figure 7.7 demonstrate the validity of the law of corresponding states. How can this law be applied to a specific gas? The goal is to calculate z for specific values of Pr and Tr and to use these z values to estimate the error in using the ideal gas law. A convenient way to display these results is in the form of a graph. Calculated results for z using the van der Waals equation of state as a function of Pr for different values of Tr are shown in Figure 7.8. For a given gas and specific P and T values, Pr and Tr can be calculated. A value of z can then be read from the curves in Figure 7.8. From Figure 7.8, we see that for Pr 6 5.5, z 6 1, as long as Tr 6 2. This means that the real gas exerts a smaller pressure than an ideal gas in this range of Tr and Pr. We conclude that for these values of Tr and Pr, the molecules are more influenced by the attractive part of the potential than by the repulsive part that arises from the finite molecular volume. However, z 7 1 for Pr 7 7 for all values of Tr as well as for all
1.6
Compression factor (z)
1.4
Figure 7.8
The compression factor plotted as a function of Pr for selected values of Tr. Values of Tr are indicated adjacent to the curves. The curves were calculated using the van der Waals equation of state.
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1.2 4.0 1.0 2 0.8 1.3
0.6
2.0 1.7 1.5 1.4
4
6
8 10 Reduced pressure (Pr)
1.2 1.1
0.4 1.0
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199
values of Pr if Tr 7 4. Under these conditions, the real gas exerts a larger pressure than an ideal gas. The molecules are influenced more by the repulsive part of the potential than by the attractive part. Using the compression factor, we can define the error in assuming that the pressure can be calculated using the ideal gas law as Error = 100, *
z  1 z
(7.11)
Figure 7.8 shows that the ideal gas law is in error by less than 30% in the range of Tr and Pr where the repulsive part of the potential dominates if Pr 6 8. However, the error can be as great as 300% in the range of Tr and Pr where the attractive part of the potential dominates. The error is greatest near Tr = 1 because at this value of the reduced temperature, the liquid and gaseous phases can coexist. At or slightly above the critical temperature, a real gas is much more compressible than an ideal gas. These curves can be used to estimate the temperature range over which the molar volume for H2O1g2 predicted by the ideal gas law is within 10% of the result predicted by the van der Waals equation of state over a wide range of pressure. From Figure 7.8, this is true only if Tr 7 2.0 or if T 7 1300 K.
EXAMPLE PROBLEM 7.3 Using Figure 7.8 and the data in Table 7.2 (see Appendix A, Data Tables), calculate the volume occupied by 1.000 kg of CH4 gas at T = 230. K and P = 68.0 bar. Calculate V  Videal and the relative error in V if V were calculated from the ideal gas equation of state.
Solution From Table 7.2, Tr and Pr can be calculated: Tr =
230. K 68.0 bar = 1.21 and Pr = = 1.48 190.56 K 45.99 bar
From Figure 7.8, z = 0.63. znRT V = = P
0.63 *
1000. g 16.04 g mol1
V  Videal V  Videal V
* 0.08314 L bar K1mol1 * 230. K
68.0 bar 11.0 L = 11.0 L = 6.5 L 0.63 6.5 L = = 58, 11.0 L
= 11.0L
Because the critical variables can be expressed in terms of the parameters a and b as shown in Example Problem 7.1, the compression factor at the critical point can also be calculated. For the van der Waals equation of state,
zc =
PcVc 1 a 27Rb 3 = * * 3b * = 2 RTc R 8a 8 27b
(7.12)
Equation (7.12) predicts that the critical compressibility is independent of the parameters a and b and should, therefore, have the same value for all gases. A comparison of this prediction with the experimentally determined value of zc in Table 7.2 shows qualitative but not quantitative agreement. A similar analysis using the critical parameters obtained from the Redlich–Kwong equation also predicts that the critical compression factor should be independent of a and b. In this case, zc = 0.333. This value is in better agreement with the values listed in Table 7.2 than that calculated using the van der Waals equation.
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CHAPTER 7 The Properties of Real Gases
7.5 FUGACITY AND THE EQUILIBRIUM CONSTANT FOR REAL GASES
As shown in the previous section, the pressure exerted by a real gas can be greater or less than that for an ideal gas. We next discuss how this result affects the value of the equilibrium constant for a mixture of reactive gases. For a pure ideal gas, the chemical potential as a function of the pressure has the form (see Section 6.3) mideal1T, P2 = m ~ 1T2 + RT ln
P P~
(7.13)
To construct an analogous expression for a real gas, we write mreal1T, p2 = m ~ 1T2 + RT ln
Real gas
Gm– Gm (298.15 K)
Ideal gas
4 2 3 5 Relative Pressure, P/P \
\
1
f
P Repulsive part of potential dominates
Figure 7.9
The chemical potential for an ideal gas and a real gas as a function of pressure. For densities corresponding to the attractive range of the potential, P 6 Pideal. ideal Therefore, G real and ƒ 6 P. The m 6 Gm inequalities are reversed for densities corresponding to the repulsive range of the potential.
ƒ ƒ~
(7.14)
where the quantity ƒ is called the fugacity of the gas. The fugacity can be viewed as the effective pressure that a real gas exerts. For densities corresponding to the attracideal tive range of the intermolecular potential, G real and ƒ 6 P. For densities corm 6 Gm ideal responding to the repulsive range of the intermolecular potential, G real and m 6 Gm ƒ 7 P. These relationships are depicted in Figure 7.9. The fugacity has the limiting behavior that ƒ S P as P S 0. The standard state of fugacity, denoted ƒ ~ , is defined as the value that the fugacity would have if the gas behaved ideally at 1 bar pressure. This is equivalent to saying that ƒ ~ = P ~ . This standard state is a hypothetical standard state because a real gas will not exhibit ideal behavior at a pressure of one bar. However, the standard state defined in this way makes Equation (7.14) become identical to Equation (7.13) in the ideal gas limit ƒ S P as P S 0. The preceding discussion provides a method to calculate the fugacity for a given gas. How are the fugacity and the pressure related? For any gas, real or ideal, at constant T,
dGm = VmdP
(7.15)
Therefore, dmideal = V ideal m dP dmreal = V real m dP (7.16)
Equation (7.16) shows that because V ideal ≠ V real m m , the chemical potential of a real gas will change differently with pressure than the chemical potential of an ideal gas. We form the difference real dmreal  dmideal = 1V real m  V m 2dP
(7.17)
and integrate Equation (7.17) from an initial pressure Pi to a final pressure P: P
L
Concept
Pi
Fugacity is the effective pressure that a real gas exerts.
1dmreal  dmideal2 = 3 mreal1P2  mreal1Pi24  3 mideal1P2  mideal1Pi24 P
=
L Pi
ideal 1V real m  V m 2dP′
(7.18)
P′ is used as a dummy variable in the previous equation because P appears in the upper limit of integration. Equation (7.18) allows us to calculate the difference in chemical potential between a real and an ideal gas at pressure P. Now let Pi S 0. In this limit, mreal 1Pi2 = mideal 1Pi2 because all real gases approach ideal gas behavior at a sufficiently low pressure. Equation (7.18) becomes P
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mreal1P2  mideal1P2 =
L 0
ideal 1V real m  V m 2dP′
(7.19)
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7.5 Fugacity and the Equilibrium Constant for Real Gases
Equation (7.19) provides a way to calculate the fugacity of a real gas. Using Equations (7.13) and (7.14) for mreal1P2 and mideal1P2, P and ƒ are related at the final pressure by P
1 ln ƒ = ln P + 1V real  V ideal m 2dP′ RT L m
(7.20)
0
Because tabulated values of the compression factor z of real gases are widely available, ideal it is useful to rewrite Equation (7.20) in terms of z by substituting z = V real m >V m . The result is P
P
0
0
z  1 z  1 ln ƒ = ln P + dP′ or ƒ = P expc a b dP′ d or ƒ = g1P, T2P (7.21) L P′ L P′
W7.4 Calculating the fugacity coefficient from the compression factor
Equation (7.21) provides a way to calculate the fugacity if z is known as a function of pressure. Note that ƒ and P are related by the proportionality factor g, which is called the fugacity coefficient. However, g is not a constant; it depends on both P and T.
EXAMPLE PROBLEM 7.4 For T 7 TB, the equation of state P1Vm  b2 = RT is an improvement over the ideal gas law because it takes the finite volume of the molecules into account. Derive an expression for the fugacity coefficient for a gas that obeys this equation of state.
Solution Because z = PVm >RT = 1 + Pb>RT, the fugacity coefficient is given by P
P
0
0
f z  1 d bP ln g = ln = dP′ = dP′ = P RT L P′ L RT bP>RT
Equivalently, g = e . Note that g 7 1 for all values of P and T because the e quation of state does not take the attractive part of the intermolecular potential into account.
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1.5 H2 1.4 Fugacity coefficient (g)
Calculated values for g 1P, T2 for H2, N2, and NH3 are shown in Figure 7.10 for T = 700. K. It can be seen that g S 1 as P S 0. This must be the case because the fugacity and pressure are equal in the ideal gas limit. The following general statements can be made about the relationship between g and z: g 1P, T2 7 1 if the integrand of Equation (7.21) satisfies the condition 1z  12>P 7 0 for all pressures up to P. Similarly, g 1P, T2 6 1 if 1z  12>P 7 0 for all pressures up to P. These predictions about fugacity and the fugacity coefficient can be related to the Boyle temperature. If T 7 TB, then g1P, T2 7 1 for all pressures, the fugacity is greater than the pressure, and g 7 1. However, if T 6 TB, then g 1P, T2 6 1, and the fugacity is smaller than the pressure. The last statement does not hold for very high values of Pr because, as shown in Figure 7.8, z 7 1 for all values of T at such high relative pressures. Are these conclusions in accord with the curves in Figure 7.10? The Boyle temperatures for H2, N2, and NH3 are 110, 327, and 995 K, respectively. Because at 700. K, T 7 TB for H2 and N2, but T 6 TB for NH3, we conclude that g 7 1 at all pressures for H2 and N2, and g 6 1 at all but very high pressures for NH3. These conclusions are consistent with the results shown in Figure 7.10. The fugacity coefficient can also be graphed as a function of Tr and Pr. This is convenient because it allows g for any gas to be estimated once T and P have been expressed as reduced variables. Graphs of g as functions of Pr and Tr are shown in Figure 7.11. Note that for large values of Tr , g values are close to 1.0 even for large values of Pr because the attractive potential becomes unimportant in this limit. The curves have been calculated using Beattie–Bridgeman parameters for N2, and are known to
N2
1.3 1.2 1.1 Pressure (bar)
1.0
100 200 300 400 500 600 700 0.9
NH3
Figure 7.10
Fugacity coefficients for H2 1g2 , N2 1g2 , and NH3 1g2 plotted as a function of the partial pressure of the gases for T = 700. K. The calculations were carried out using the Beattie–Bridgeman equation of state.
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be accurate for N2 over the indicated range of reduced temperature and pressure. Their applicability to other gases assumes that the law of corresponding states holds. As Figure 7.7 shows, this is generally a good assumption. What are the consequences of the fact that except in the dilute gas limit ƒ ≠ P? Because the chemical potential of a gas in a reaction mixture is given by Equation (7.14), the thermodynamic equilibrium constant for a real gas Kƒ must be expressed in terms of the fugacities. Therefore Kƒ for the reaction 3>2H2 + 1>2N2 S NH3 is given by
Kƒ = a
ƒN2 ƒ~
a
b
ƒNH3 ƒ
1>2
~
a
b
ƒH2 ƒ~
b
3>2
= a
a
gNH3PNH3
gN2 PN2
= KP
P~
P~ b
1>2
a
b
gH2PH2 P~
gNH3
1gN221>2 1gH223>2
b
3>2
(7.22)
We next calculate the error in using Kp rather than Kƒ to calculate an equilibrium constant, using the ammonia synthesis reaction at 700. K and a total pressure of 400. bar as an example. In the industrial synthesis of ammonia, the partial pressures of H2, N2, and NH3 are typically 270, 90, and 40 bar, respectively. The calculated fugacity coefficients under these conditions are gH2 = 1.11, gN2 = 1.04, and gNH3 = 0.968 using the Beattie–Bridgeman equation of state. Therefore,
Fugacities must be used to accurately calculate the composition of a reactive mixture of real gases at high pressures.
11.0421>2 11.1123>2
xNH3 = 1xN221>21xH223>2 a
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(7.23)
P b Kƒ P~
(7.24)
What conclusions can be drawn from this calculation? If KP was used rather than Kƒ, the calculated mole fraction of ammonia would be in error by 9%, which is a significant error in calculating the economic viability of a production plant. However, for pressures near 1 bar, activity coefficients for most gases are very close to unity, as can be seen in Figure 7.11. Therefore, under typical laboratory conditions, the fugacity of a gas can be set equal to its partial pressure if P, V, and T are not close to their critical values.
3.0 2.4 2.0 1.8 1.7 1.6
Fugacity coefficient (g)
1.2
Fugacity coefficient plotted as a function of the reduced pressure for the indicated values of the r educed temperature. To avoid overlap, curves for reduced temperature values of 1.0–3.0 are shown in (a) and curves for reduced temperature values of 3.0–24 are shown in (b).
= 0.917 KP
Note that Kƒ is smaller than Kp for the conditions of interest. The NH3 concentration at equilibrium is proportional to the equilibrium constant, as shown in Section 6.14.
1.4
Figure 7.11
10.9682
1.0
1.5 1.4
0.8
1.3
0.6
1.2
0.4
1.1
0.2
1.0
3.0
Fugacity coefficient (g)
Concept
Kƒ = KP
1.3
6.0
9.0
1.2
12 18
1.1
24
1.0 2 4 6 8 10 12 14 Reduced pressure (Pr)
(a) Reduced temperature values of 1.0–3.0
0
2 4 6 8 10 12 14 Reduced pressure (Pr)
(b) Reduced temperature values of 3.0–24
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Conceptual Problems
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VOCABULARY Beattie–Bridgeman equation of state Boyle temperature compression factor critical constants critical temperature
fugacity fugacity coefficient law of corresponding states Maxwell construction Redlich–Kwong equation of state
supercritical fluid van der Waals equation of state vapor pressure virial equation of state
KEY EQUATIONS Equation RT a nRT n2a  2 =  2 Vm  b V m V  nb V
P =
RT a 1 nRT n2a 1 = Vm  b V  nb 2T Vm1Vm + b2 2T V1V + nb2
P =
P = RT c z =
Significance of Equation
Vm
B1T2 1 + + gd Vm V 2m
V ideal m
TB =
=
PVm RT
a Rb
m1T, p2 = m ~ 1T2 + RT ln P
g1P, T2 = expc
L 0
a
ƒ ƒ~
z  1 bdP′ d P′
P
P
0
0
z  1 z  1 ln ƒ = ln P + dP′ or ƒ = P expc a bdP′ d L P′ L P′
Equation Number
van der Waals equation of state
7.1
Redlich–Kwong equation of state
7.2
Virial equation of state
7.4
Definition of compression factor
7.6
Boyle temperature for van der Waals gas
7.7
Chemical potential of real gas and definition of fugacity
7.14
Definition of fugacity coefficient
from 7.21
Calculation of fugacity from compressibility
7.21
CONCEPTUAL PROBLEMS Q7.1 Using the concept of the intermolecular potential, explain why two gases in corresponding states can be expected to have the same value for z. Q7.2 Consider the comparison made between accurate results and those based on calculations using the van der Waals and Redlich–Kwong equations of state in Figures 7.1 and 7.5. Is it clear that one of these equations of state is better than the other under all conditions? Q7.3 Why is the standard state of fugacity, ƒ ~ , equal to the standard state of pressure, P ~ ? Q7.4 Explain the significance of the Boyle temperature. Q7.5 A van der Waals gas undergoes an isothermal revers ible expansion under conditions such that z 7 1. Is the work
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performed more or less than if the gas followed the ideal gas law? Q7.6 For a given set of conditions, the fugacity of a gas is greater than the pressure. What does this tell you about the interaction between the molecules of the gas? Q7.7 What can you conclude about the ratio of fugacity to pressure for N2, H2, and NH3 at 500 bar using the data in Figure 7.10? Q7.8 Is the ratio of fugacity to pressure greater to or less than one if the attractive part of the interaction potential between gas molecules dominates? Q7.9 A gas is slightly above its Boyle temperature. Do you expect z to increase or decrease as P increases?
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Q7.10 Explain why the oscillations in the twophase coexistence region using the Redlich–Kwong and van der Waals equations of state (see Figure 7.4) do not correspond to reality. Q7.11 A van der Waals gas undergoes an isothermal reversible expansion under conditions such that z 6 1. Is the work done more or less than if the gas followed the ideal gas law? Q7.12 The value of the Boyle temperature increases with the strength of the attractive interactions between molecules. Arrange the Boyle temperatures of the gases Ar, CH4, and C6H6 in increasing order. Q7.13 A system containing argon gas is at pressure P1 and temperature T1. How would you go about estimating the fugacity coefficient of the gas? Q7.14 By looking at the a and b values for the van der Waals equation of state, decide whether 1 mole of O2 or H2O has the higher pressure at the same value of T and V.
Q7.15 Will the fugacity coefficient of a gas at a temperature greater than the Boyle temperature be less than one at low pressures? Q7.16 Show that the van der Waals and Redlich–Kwong equations of state reduce to the ideal gas law in the limit of low gas density. Q7.17 Which of Ne or Ar has the larger van der Waals parameter a? Explain your reasoning. Q7.18 Which of Ne or Ar has the larger van der Waals parameter b? Explain your reasoning. Q7.19 You have calculated the pressure exerted by ethane using the ideal gas law and the Redlich–Kwong equations of state. How do you decide if the repulsive or attractive part of the molecular potential dominates under the given conditions? Q7.20 Equation 1.20 states that the total pressure in a mixture of gases is equal to the sum of the partial pressures. Is this equation valid for real gases? If so, under what conditions is it valid?
NUMERICAL PROBLEMS Section 7.2 P7.1 A van der Waals gas has a value of z = 1.00032 at 298 K and 1 bar, and the Boyle temperature of the gas is 195 K. Because the density is low, you can calculate Vm from the ideal gas law. Use this information and the result of Problem 7.21, z ≈ 1 + 1b  a>RT211>Vm2, to estimate a and b. P7.2 Assume that the equation of state for a gas can be written in the form P1Vm  b1T22 = RT. Derive an expression for a = 1>V10V>0T2P and kT = 1>V10V>0P2T for such a gas in terms of b1T2, db1T2>dT, P, and Vm. P7.3 One mole of Ar initially at 298 K undergoes an adiabatic expansion against a pressure Pexternal = 0 from a volume of 4.0 L to a volume of 50.0 L. Calculate the final temperature using the ideal gas and van der Waals equations of state. Assume CV,m = 3>2 R. P7.4 At 725 K and 280. bar, the experimentally determined density of N2 is 4.13 mol L 1. Compare this with values calculated from the ideal and Redlich–Kwong equations of state. Use a numerical equation solver to solve the Redlich–Kwong equation for Vm or use an iterative approach starting with Vm equal to the ideal gas result. Discuss your results. P7.5 2.50 moles of Ar undergoes an isothermal reversible expansion from an initial volume of 1.75 L to a final volume of 70.00 L at 310. K. Calculate the work done in this process using the ideal gas and van der Waals equations of state. What percentage of the work done by the van der Waals gas arises from the attractive potential? P7.6 The volume of a spherical molecule can be estimated as V = b>4NA where b is the van der Waals parameter for the excluded molar volume and NA is Avogadro’s number. Justify this relationship by considering a spherical molecule of radius
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r, with volume V = 4>3pr 3. What is the volume centered at the molecule that is excluded for the center of mass of a second molecule in terms of V? Multiply this volume by NA and set it equal to b. Apportion this volume equally among the molecules to arrive at V = b>4NA. Calculate the radius of a tetrachloromethane molecule from the value of its van der Waals parameter b. P7.7 Show that the van der Waals and Redlich–Kwong equations of state reduce to the ideal gas equation of state in the limit of low density. P7.8 Another equation of state is the Berthelot equation, Vm = RT>P + b  a>1RT 22. Derive expressions for a = 1>V10V>0T2P and kT = 1>V10V>0P2T from the Berthelot equation in terms of V, T, and P. 0 ln z b for a real gas where P7.9 Show that PkT = 1  P a 0P T k is the isothermal compressibility. P7.10 Calculate the density of O21g2 at 480. K and 280. bar using the ideal gas and the van der Waals equations of state. Use a numerical equation solver to solve the van der Waals equation for Vm or use an iterative approach starting with Vm equal to the ideal gas result. Based on your result, does the attractive or repulsive contribution to the interaction potential dominate under these conditions? P7.11 A sample containing 50.0 g of Ar is enclosed in a container of volume 0.0925 L at 375 K. Calculate P using the ideal gas, van der Waals, and Redlich–Kwong equations of state. Based on your results, does the attractive or repulsive contribution to the interaction potential dominate under these conditions?
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WebBased Simulations, Animations, and Problems
Section 7.3 P7.12 For values of z near one, it is a good approximation to write z1P2 = 1 + 10z>0P2TP. Assume that z = 1.00221 at 298 K and 1 bar, and the Boyle temperature of a gas is 140. K. Using an equation solver, the relation for z1P2 above, the van a der Waals equation of state, and the relation TB = , calcuRb late the values of a, b, and Vm for the van der Waals gas. P7.13 For a gas at a given temperature, the compression factor z is described by the empirical equation P P 2 z = 1  8.50 * 103 ~ + 3.50 * 105 a ~ b P P ~ where P = 1 bar. Calculate the fugacity coefficient for P = 100., 200., 300., 400., 500., and 600. bar. For which of these values is the fugacity coefficient greater than one? P7.14 The experimentally determined density of O2 at 140. bar and 298 K is 192 g L 1. Calculate z and Vm from this information. Compare this result with what you would have estimated from Figure 7.8. What is the relative error in using Figure 7.8 for this case? P7.15 Show that the second virial coefficient for a van der Waals gas is given by B1T2 =
1 0z a = b RT ° 1 ¢ RT 0 Vm T
0 ln z b for a real gas where 0T P a is the volumetric thermal expansion coefficient.
P7.16 Show that Ta = 1 + T a
P7.17 At what temperature does the slope of the z versus P curve as P S 0 have its maximum value for a van der Waals gas? What is the value of the maximum slope? P7.18 Show that Ta = 1 + T10 ln z>0T2P and that PkT = 1  P10 ln z>0P2T . P7.19 The observed Boyle temperatures of H2, N2, and CH4 are 110., 327, and 510. K, respectively. Compare these values with those calculated for a van der Waals gas with the appropriate parameters.
205
P7.20 For the Berthelot equation, Vm = 1RT>P2 + b 1a>RT 22, find an expression for the Boyle temperature in terms of a, b, and R. P7.21 For a van der Waals gas, z = Vm >1Vm  b2  a>RTVm. Expand the first term of this expression in a Taylor series in the limit Vm W b to obtain z ≈ 1 + 1b  a>RT211>Vm2.
Section 7.4 P7.22 a. Using the relationships derived in Example Problem 7.1 and the values of the critical constants for ethanol from Table 7.2, calculate values for the van der Waals parameters a, b, and R from zc, Tc, Pc, and Vc. Do your results agree with those in Tables 1.2 and 7.4 in Appendix A? b. Calculate the van der Waals parameters a and b using the critical constants for ethanol and the correct value for R. Do these results agree with those in Tables 1.2 and 7.4? P7.23 Calculate the P and T values for which Br21g2 is in a corresponding state to F21g2 at 300. K and 75.0 bar. P7.24 Use the law of corresponding states and Figure 7.8 to estimate the molar volume of propane at T = 500. K and P = 75.0 bar. The experimentally determined value is 0.438 mol L 1. What is the relative error of your estimate? P7.25 Calculate the van der Waals parameters of pyridine from the values of the critical constants. P7.26 Calculate the Redlich–Kwong parameters of hydrogen from the values of the critical constants. P7.27 Calculate the critical volume for ethane using the data for Tc and Pc in Table 7.2 (see Appendix A, Data Tables) assuming (a) the ideal gas equation of state and (b) the van der Waals equation of state. Use an iterative approach to obtain Vc from the van der Waals equation, starting with the ideal gas result. How well do the calculations agree with the tabulated values for Vc? P7.28 The experimental critical constants of Ar are listed in Table 7.2. Use the values of Pc and Tc to calculate Vc. Assume that at the critical point Ar behaves as (a) an ideal gas, (b) a van der Waals gas, and (c) a Redlich–Kwong gas. For parts (b) and (c), use the formulas for the critical compression factor. Compare your answers with the experimental value. Are any of your calculated results close to the experimental value in Table 7.2?
WEBBASED SIMULATIONS, ANIMATIONS, AND PROBLEMS Simulations, animations, and homework problem worksheets can be accessed at www.pearsonhighered.com/advchemistry W7.1 In this problem, the student gains facility in using the van der Waals equation of state. A set of isotherms will be generated by varying the temperature and initial volume using sliders for a given substance. Buttons allow a choice among more than 20 gases. a. The student generates a number of P–V curves at and above the critical temperature for the particular gas, and explains trends in the ratio PvdW >Pideal as a function of V for a given T and as a function of T for a given V.
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b. The compression factor z is calculated for two gases at the same value of Tr and is graphed versus P and Pr. The degree to which the law of corresponding states is valid is assessed. W7.2 A quantitative comparison is made between the ideal gas law and the van der Waals equation of state for 1 of more than 20 different gases. The temperature is varied using sliders, and Pideal, PvdW, the relative error PvdW >PvdW  Pideal, and the density of gas relative to that at the critical point are calculated. The student is asked to determine the range of pressures and temperatures in which the ideal gas law gives reasonably accurate results.
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W7.3 The compression factor and molar volume are calculated for an ideal gas and a van der Waals gas as a function of pressure and temperature. These variables can be varied using sliders. Buttons allow a choice among more than 20 gases. The relative error VvdW >VvdW  Videal and the density of gas relative to that at the critical point are calculated. The student is asked to determine the range of pressures and temperatures in which the ideal gas law gives reasonably accurate results for the molar volume.
W7.4 The fugacity and fugacity coefficient are determined as a function of pressure and temperature for a model gas. These variables can be varied using sliders. The student is asked to determine the Boyle temperature and also the pressure range in which the fugacity is either more or less than the ideal gas pressure for the temperature selected.
FURTHER READING BenAmotz, D. Ben, and Levine, R. “Updated Principle of Corresponding States.” Journal of Chemical Education 81 (2004): 142–146. David, C. “Fugacity Examples.” Journal of Chemical Education 81 (2004): 1653–1654. Eberhart, J. “The Many Faces of van der Waals’ Equation of State.” Journal of Chemical Education 66 (1989): 906–909.
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Rigby, M. “A van der Waals Equation for Nonspherical Molecules.” Journal of Physical Chemistry 76 (1972): 2014–2016. Soave, G. “Improvement of the van der Waals Equation of State.” Chemical Engineering Science 39 (1984): 357–369.
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C H A P T E R
8
Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
8.2 The Pressure–Temperature Phase Diagram
8.3 The Phase Rule 8.4 Pressure–Volume and Pressure– Volume–Temperature Phase Diagrams
WHY is this material important? At room temperature, the most stable form of water is the liquid phase. At temperatures greater than 100°C, the most stable form of H2O is gaseous water vapor. Boiling water is the manifestation of the transition from liquid water to gaseous water. The process of boiling occurs at constant temperature. In the elevated pressure of a pressure cooker, food cooks much more quickly than in a pot on the stove at a pressure of 1 bar. In this chapter, you will learn how all of this information can be represented in a phase diagram.
WHAT are the most important concepts and results? The occurrence of a single stable phase or the coexistence of several phases is predictable with the application of the principles of thermodynamics. For example, coexistence of two phases of a pure substance occurs if and only if the chemical potential is equal in both phases. It turns out that no more than three phases of a pure substance can coexist. Furthermore, the vapor pressure of a liquid increases exponentially with temperature. Moreover, the pressure inside a small liquid droplet can be significantly greater than that inside a large droplet because of the effects of surface tension. This observation explains how condensation occurs in clouds.
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
8.6 Using the Clausius–Clapeyron Equation to Calculate Vapor Pressure as a Function of T
8.7 Dependence of Vapor Pressure of a Pure Substance on Applied Pressure
8.8 Surface Tension 8.9 (Supplemental Section) Chemistry in Supercritical Fluids
8.10 (Supplemental Section) Liquid Crystal Displays
WHAT would be helpful for you to review for this chapter? This chapter builds on the concept of chemical potential, introduced in Chapter 6. Thus, it would be useful to have a firm understanding of this concept.
8.1 WHAT DETERMINES THE RELATIVE STABILITY OF THE SOLID, LIQUID, AND GAS PHASES?
Substances occur in solid, liquid, and gaseous phases. Phase refers to a form of matter that is uniform with respect to chemical composition and the state of aggregation on both microscopic and macroscopic length scales. For example, liquid water in a beaker is a singlephase system, but a mixture of ice and liquid water consists of two distinct phases, each of which is uniform on microscopic and macroscopic length scales. A substance may exist in only one gaseous phase. Similarly, most substances have a 207
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Concept The chemical potential determines which of several phases is the most stable.
single liquid state, although there are exceptions. For example, helium can exist as a normal liquid or as a superfluid. In contrast, many substances may exist as several different solid phases. In this section, the conditions under which a pure substance spontaneously forms a solid, liquid, or gas are discussed. Experience demonstrates that as T is lowered from 300. to 250. K at atmospheric pressure, liquid water is converted to a solid phase. Similarly, as liquid water is heated to 400. K at atmospheric pressure, it vaporizes to form a gas. Experience also shows that if a solid block of carbon dioxide is placed in an open container at 1 bar, it sublimes over time without passing through the liquid phase. Because of this property, solid CO2 is known as dry ice. These observations can be generalized as follows. Whereas the solid phase is the most stable state of a substance at sufficiently low temperatures, the gas phase is the most stable state of a substance at sufficiently high temperatures. At intermediate temperatures, the liquid state is stable, if it exists at the pressure of interest. What determines which of the solid, liquid, or gas phases exists and which phase is most stable at a given temperature and pressure? The answers to these questions are the subject of this chapter. As discussed in Chapter 6, the criterion for stability at constant temperature and pressure is that the Gibbs energy, G1T, P, n2, is minimized. For a pure substance, the chemical potential m is defined as m = a
03 nGm 4 0G b = a b = Gm 0n T, P 0n T, P
where n designates the number of moles of substance in the system, dm = dGm. We can therefore express the differential dm as dm = Sm dT + Vm dP
(8.1)
Chemical potential,
From this equation, it is possible to determine how m varies with changes in P and T:
Solid
Liquid
Gas Tm
Tb
Temperature
Figure 8.1
Chemical potential plotted as a function of temperature and the relative stability of solid, liquid, and gas phases at a given pressure. The curves for the solid (green curve), liquid (blue curve), and gaseous (brown curve) states are shown for a given value of pressure. The stable state of the system at any given temperature is that phase with the lowest value of m, and the temperature range of the most stable phase is indicated by colored areas. The substance melts at the temperature Tm, corresponding to the intersection of the solid and liquid curves. It boils at the temperature Tb, corresponding to the intersection of the liquid and gas curves. The three curves are actually curved slightly downward but have been approximated as straight lines.
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a
0m 0m b = Sm and a b = Vm 0T P 0P T
(8.2)
Because Sm and Vm are always positive, m decreases as the temperature increases and increases as the pressure increases. Section 5.4 demonstrated that S varies slowly with T (approximately as ln T ). Therefore, over a limited range in T, a plot of m versus T at constant P is a curve with a slowly increasing negative slope, approximating a straight line. It is also known from experience that heat is absorbed as a solid melts to form a liquid and as a liquid vaporizes to form a gas. Both processes are endothermic with ∆H 7 0. Because at the melting and boiling temperatures the entropy increases in a stepwise fashion by ∆S = ∆ transH > T, the entropy of the three phases follows this order:
gas Sm 7 Sliquid 7 Ssolid m m
(8.3)
Figure 8.1 graphs the functional relation between m and T for the solid, liquid, and gas phases at a given value of P. The molar entropy of a phase is the negative of the slope of the m versus T curve, and the relative entropies of the three phases are given by Equation (8.3). The stable state of the system at any given temperature is that phase that has the lowest m. Assume that the initial state of the system is described by the dot in Figure 8.1. The most stable phase is the solid phase because m is much larger for the liquid and gas phases than for the solid. As the temperature is increased, the chemical potential decreases as m remains on the solid curve. However, because the slopes of the liquid and gas curves are greater than the slope of the solid, each of these m versus T curves will intersect the solid curve for some value of T. In Figure 8.1, the liquid curve intersects the solid curve at Tm, which is called the melting temperature. At this temperature, the solid and liquid phases coexist and are in thermodynamic equilibrium. However, if the temperature is raised by an infinitesimal amount dT, the solid will melt completely because the liquid phase has the lower chemical potential at Tm + dT. Similarly, the liquid and gas phases are in thermodynamic equilibrium
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209
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
at the boiling temperature Tb. For T 7 Tb, the system is entirely in the gas phase. Note that the progression of solid S liquid S gas as T increases at this value of P can be explained with no other information than that 10m > 0T2P = Sm and that gas Sm 7 Sliquid 7 Ssolid m m . If the temperature is changed too quickly, the equilibrium state of the system may not be reached and a metastable phase may persist. For example, it is possible to form a superheated liquid in which the liquid phase is metastable at temperatures greater than Tb. Superheated liquids are dangerous because of the large volume increase that occurs if the system suddenly converts to the stable vapor phase. Boiling chips are often used in chemical laboratories to avoid the formation of superheated liquids. Similarly, it is possible to form a supercooled liquid, in which case the liquid is metastable at temperatures less than Tm. Glass is made by cooling a viscous liquid fast enough to avoid crystallization. These disordered materials lack the periodicity of crystals but behave mechanically like solids. Seed crystals can be used to increase the rate of crystallization if the viscosity of a liquid is not too high and the cooling rate is sufficiently slow. Liquid crystals, which can be viewed as a state of matter intermediate between a solid and a liquid, are discussed in Section 8.10. In Figure 8.1, we considered changes of m with T at constant P. How is the relative stability of the three phases affected if P is changed at constant T ? From Equation gas (8.2), 10m > 0P2T = Vm and V m W V liquid . For most substances, V liquid 7 V solid m m m . Therefore, the m versus T curve for the gas changes much more rapidly with pressure (by a factor of ∼1000) than the liquid and solid curves. This behavior is illustrated in Figure 8.2, where it can be seen that the temperature at which the solid and the liquid gas curves intersect shifts as the pressure is increased. Because V m W V liquid 7 0, an m increase in P always leads to a boiling point elevation. An increase in P leads to a freezing point elevation if V liquid 7 V solid and to a freezing point depression if m m liquid solid Vm 6 V m . Few substances obey the relation V liquid 6 V solid m m , with water being an important example. The consequences of this unusual behavior for water will be discussed in Section 8.2. The m versus T curve for a gas shifts along the T axis much more rapidly with changes in P than the liquid and solid curves. As a consequence, changes in P can alter the way in which a system progresses through the phases with increasing T from the “normal” order solid S liquid S gas shown in Figure 8.1. For example, the sublimation of dry ice at 298 K and 1 bar can be explained using Figure 8.3a. For CO2 at a given pressure, the m versus T curve for the gas intersects the corresponding curve for the solid at a lower temperature than the curve for the gas intersects the liquid curve. Therefore, the solid S liquid transition is energetically unfavorable with respect to the solid S gas transition at this pressure. Under these conditions, the solid
Solid
Either liquid or gas, depending on P.
Liquid DTm
Gas
Either solid or liquid, depending on P.
Chemical potential, M
Chemical potential, M
Either solid or liquid, depending on P.
Solid
DTm
Either liquid or gas, depending on P.
Liquid
DTb
DTb
Tm T'm
Tb
T'm Tm
T'b
Temperature (a)
M08_ENGE4583_04_SE_C08_207236.indd 209
Temperature (b)
Gas
Tb
T'b
Concept Metastable phases do not correspond to equilibrium states.
Concept Boiling and freezing temperatures vary with pressure.
Figure 8.2
The effect of pressure on the relative stability of solid, liquid, and gas phases. Chemical potential is plotted as a function of temperature. The diagrams are for cases in which (a) V liquid 7 V solid m m solid and (b) V liquid 6 V . The solid curves m m show m as a function of temperature for all three phases at P = P1. The dashed curves show the same information for P = P2,where P2 7 P1. The unprimed temperatures refer to P = P1 and the primed temperatures refer to P = P2. The shifts in the solid and liquid curves are greatly exaggerated. The colored areas indicate the temperature range in which the phases are most stable.
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Figure 8.3
Solid
Gas
Chemical potential, M
At Ttp, all three phases coexist in equilibrium.
Chemical potential, M
Sublimation and triple point behavior. The chemical potential of a substance in the solid (green curve), liquid (blue curve), and gaseous (brown curve) states is plotted as a function of temperature for a fixed value of pressure. (a) Pressure less than triple point pressure. (b) Triple point pressure. The colored areas indicate the temperature range in which the phases are the most stable. The liquid phase is not stable at any temperature in part (a), and is only stable at the single temperature Ttp in part (b).
Solid
Ts Temperature (a) The pressure lies below the triple point pressure, and the solid sublimes.
Gas
Tt p Temperature (b) The pressure corresponds to the triple point pressure.
sublimes, and the transition temperature Ts is called the sublimation temperature. There is also a pressure at which the m versus T curves for all three phases intersect. The P, Vm, T values for this point specify the triple point, so named because all three phases coexist in equilibrium at this point. This case is shown in Figure 8.3b. Triple point temperatures for a number of substances are listed in Table 8.1 (see Appendix A, Data Tables).
8.2 THE PRESSURE–TEMPERATURE PHASE DIAGRAM
Concept A P–T phase diagram summarizes information on the stability of different phases as the values of P and T change.
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As shown in the previous section, at a given value of pressure and temperature a system containing a pure substance may consist of a single phase, two phases in equilibrium, or three phases in equilibrium. The usefulness of a phase diagram is that it displays this information graphically. Although any two of the macroscopic system variables P, V, and T can be used to construct a phase diagram, the P9T diagram is particularly useful. In this section, the features of a P9T phase diagram that are common to pure substances are discussed. Phase diagrams must generally be determined experimentally because materialspecific forces between atoms determine the temperatures and pressures at which different phases are stable. Increasingly, calculations have become sufficiently accurate that major features of phase diagrams can be obtained using microscopic theoretical models. However, as shown in Section 8.4, thermodynamics can say a great deal about the phase diagram without considering the microscopic properties of the system. The P–T phase diagram displays stability regions for a pure substance as a function of pressure and temperature. An example is shown in Figure 8.4. Most P, T points correspond to a single solid, liquid, or gas phase. At the triple point, all three phases coexist. The triple point of water is 273.16 K and 611 Pa. All P, T points for which the same two phases coexist at equilibrium fall on a curve. Such a curve is called a c oexistence curve. Three separate coexistence curves are shown in Figure 8.4, corresponding to solid–gas, solid–liquid, and gas–solid coexistence. As shown in Section 8.5, the slopes of the solid–gas and liquid–gas curves are always positive. The slope of the solid– liquid curve can be either positive or negative. The boiling point of a substance is defined as the temperature at which the vapor pressure of the substance is equal to the external pressure. The standard boiling temperature is the temperature at which the vapor pressure of the substance is 1 bar.
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8.2 The Pressure–Temperature Phase Diagram Constant temperature (isothermal) path greater than triple point
Figure 8.4
Pressure, P (bar)
Critical point Solid
1
Liquid
d
Constant pressure (isobaric) path for P > Ptp
Gasliquid coexistence curve is crossed in the constant pressure process moving to lower temperatures.
a Triple point
c Tm
Tb Temperature
The normal boiling temperature is the temperature at which the vapor pressure of the substance is 1 atm. Values of the normal boiling and freezing temperatures for a number of substances are shown in Table 8.2. Because 1 bar is slightly less than 1 atm, the standard boiling temperature is slightly less than the normal boiling temperature. Along twophase curves in which one of the coexisting phases is the gas, P refers to the vapor pressure of the substance. In all regions, P refers to the external pressure that would be exerted on the pure substance if it were confined in a piston and cylinder assembly. The solid–liquid coexistence curve traces out the melting point as a function of pressure. The magnitude of the slope of this curve is large, as shown in Section 8.5. Therefore, Tm changes very little with pressure. If the solid is more dense than the liquid, the slope of this curve is positive, and the melting temperature increases with pressure. This is the case for most substances. If the solid is less dense than the liquid, the slope is negative and the melting temperature decreases with pressure. Water is one of the few substances that exhibits this behavior. Imagine the fate of aquatic plants and animals in climate zones where the temperature routinely falls below 0°C in the winter if water behaved “normally.” Lakes would begin to freeze over at the water–air interface, and the ice formed would fall to the bottom of the lake. This would lead to more ice formation until the whole lake was full of ice. In cool climate zones, it is likely that ice at the bottom of the lakes would remain throughout the summer. The aquatic life that we are familiar with would not survive under such a scenario. The slope of the liquid–gas coexistence curve is much smaller than that of the solid–liquid coexistence curve, as will be shown in Section 8.5. Therefore, the boiling point is a much stronger function of the pressure than the freezing point. The boiling point always increases with pressure. This property is utilized in a pressure cooker, where increasing the pressure by 1 bar increases the boiling temperature of water by approximately 20°C. The rate of the chemical processes involved in cooking increase exponentially with T. Therefore, a pressure cooker operating at P = 2 bar can cook food in 20% to 40% of the time required for cooking at atmospheric pressure. By contrast, a mountain climber in the Himalayas would find that the boiling temperature of water is reduced by approximately 25°C relative to sea level. Cooking takes significantly longer under these conditions.
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A P9T phase diagram. Shown are singlephase regions, coexistence curves for which two phases coexist at equilibrium, and a triple point. The processes corresponding to paths a, b, c, and the two processes starting at point d are described in the text. Two solid–liquid coexistence curves are shown. For most substances, the solid curve, which has a positive slope, is observed. For water or any other substance for which volume increases in a transition from the liquid to the solid phase, the red dashed curve corresponding to a negative slope is observed.
Gas
b
Constant pressure (isobaric) path for P < Ptp
211
Concept A P–T phase diagram generally shows gaseous, liquid, and solid single phase regions, two phase coexistence curves, and the triple and critical points.
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
TABLE 8.2 Melting and Normal Boiling Temperatures and Enthalpies of Transition at 1 atm Pressure
Density, r/(g/cm3)
1
Liquid
rc Gas 0
0
100
200
300
Tc
Temperature, T (°C)
Figure 8.5
Density of the liquid and gaseous phases of water as a function of temperature. At the critical point, the densities are equal.
Tm1K2
∆ fusH 1kJ mol12 at Tm 1.12
Tb1K2
6.41
239.18
∆ vapH 1kJ mol12 at Tb
Substance
Name
Ar
Argon
83.8
Cl2
Chlorine
171.6
Fe
Iron
1811
H2
Hydrogen
13.81
0.12
20.4
0.90
H2O
Water
273.15
6.010
373.15
40.65
13.81
He
Helium
0.95
I2
Iodine
386.8
14.73
0.021
N2
Nitrogen
63.5
0.71
Na
Sodium
370.87
2.60
87.3
3023
4.22 457.5 77.5 1156
6.43
20.41 349.5
0.083 41.57 5.57 98.0
NO
Nitric oxide
109.5
2.3
121.41
13.83
O2
Oxygen
54.36
0.44
90.7
6.82
SO2
Sulfur dioxide
197.6
7.40
263.1
24.94
Si
Silicon
1687
50.21
2628
359
W
Tungsten
3695
52.31
5933
422.6
Xe
Xenon
161.4
1.81
165.11
12.62
CCl4
Carbon tetrachloride
250
3.28
349.8
29.82
CH4
Methane
90.68
0.94
111.65
8.19
CH3OH
Methanol
175.47
3.18
337.7
35.21
CO
Carbon monoxide
68
0.83
81.6
6.04
C2H4
Ethene
103.95
3.35
169.38
13.53
C2H6
Ethane
90.3
2.86
184.5
14.69
C2H5OH
Ethanol
159.0
5.02
351.44
38.56
C3H8
Propane
85.46
3.53
231.08
19.04
C5H5N
Pyridine
231.65
8.28
388.38
35.09
C6H6
Benzene
278.68
9.95
353.24
30.72
C6H5OH
Phenol
314.0
455.02
45.69
C6H5CH3
Toluene
178.16
6.85
383.78
33.18
C10H8
Naphthalene
353.3
17.87
491.14
43.18
11.3
Sources: Data from Lide, D. R., ed. Handbook of Chemistry and Physics; 83rd ed. Boca Raton, FL: CRC Press, 2002; Lide, D. R., ed. CRC Handbook of Thermophysical and Thermochemical Data. Boca Raton, FL: CRC Press, 1994; and Blachnik, R., ed. D'Ans Lax Taschenbuch für Chemiker und Physiker, 4th ed. Berlin: Springer, 1998.
Concept Above the critical pressure and temperature, liquid and gas phases are not distinguishable.
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The termination of the coexistence curves is different for each curve. Specifically, the solid–gas coexistence curve ends at the triple point, the liquid–solid coexistence curve extends indefinitely, and the liquid–gas curve ends at the critical point. The critical point is characterized by T = Tc and P = Pc. For T 7 Tc and P 7 Pc, the liquid and gas phases have the same density as shown in Figure 8.5, so it is not meaningful to refer to distinct phases. Because the liquid and gas phases are indistinguishable at the critical point, ∆ vapH approaches zero as the critical point is reached, as shown in Figure 8.6.
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50
40
DsubH
30
20
10
0 0
100
200
Tc
300
Temperature, T (°C)
Figure 8.6
The enthalpy of vaporization for water plotted as a function of temperature. At the critical point, ∆ vapH approaches zero.
Solid Temperature
Substances for which T 7 Tc and P 7 Pc are called supercritical fluids. As discussed in Section 8.9, supercritical fluids have unusual properties that make them useful in chemical technologies. Each of the paths labeled a, b, c, and d in Figure 8.4 corresponds to a process that demonstrates the usefulness of the P9T phase diagram. In the following, each process is considered individually. Process a follows a constant pressure (isobaric) path. An example of this path is heating one mole of ice. The system is initially in the solid singlephase region. Assume that heat is added to the system at a constant rate using current flow through a resistive heater. Because the pressure is constant, qP = ∆H. Furthermore, as discussed in Section 2.4, ∆H ≈ CP ∆T in a singlephase region. Combining these equations, ∆T = qP > C solid P . Along path a, the temperature increases linearly with qp in the singlephase solid region, as shown in Figure 8.7. At the melting temperature Tm, heat continues to be absorbed by the system as the solid is transformed into a liquid. During melting, the system consists of two distinct phases, solid and liquid. As heat continues to flow into the system, the temperature will not increase until the system consists entirely of liquid. The heat taken up per mole of the system at the constant temperature Tm is ∆ fusH. As soon as the last amount of solid melts, the temperature again increases linearly with qp and ∆T = qP > C liquid until the boiling point is P reached. At this temperature, the system consists of two phases, liquid and gas. During boiling, the temperature remains constant until all the liquid has been converted into gas. The heat taken up per mol of the system at the constant temperature Tb is ∆ vapH. Finally, the system enters the singlephase gas region. Along path b, the pressure is less than the triple point pressure. Therefore, the liquid phase is not stable, and the solid is converted directly into the gaseous form. As Figure 8.4 shows, there is only one twophase interval along this path. A diagram illustrating the relationship between temperature and heat flow is shown in Figure 8.8. Note that the initial and final states in process b can be reached by an alternative route described by the vertical dashed arrows in Figure 8.4. The pressure of the system in process b is first increased to the value of the initial state of process a at constant temperature. The process then follows that described for process a, after which the pressure is returned to the final pressure of process b. Invoking Hess’s law, the enthalpy change for this pathway and for pathway b are equal. Now imagine that the constant pressures for processes a and b differ only by an infinitesimal amount, a lthough that for a is higher than the triple point pressure and that for b is lower than the triple point pressure. We can express this mathematically by setting the pressure for process a equal to Ptp + dP and that for process b equal to Ptp  dP. We examine the limit dP S 0. In this limit, ∆H S 0 for the two steps in the process indicated by
Temperature
213
8.2 The Pressure–Temperature Phase Diagram
DvapH /(kJ mol–1)
Solid 1 Gas
Gas
Tm
Tb
DsubH
Tm DfusH
DvapH
Heat flow at constant pressure, qP
Figure 8.8 Solid
Liquid 1 Solid
Liquid
Liquid 1 Gas
Gas
Heat flow at constant pressure, qP
Figure 8.7
Trends in a temperature versus heat curve for solid u liquid u gas. This schematic diagram illustrates the process corresponding to path a in Figure 8.4. Temperature rises linearly with qp in singlephase regions and remains constant along the twophase curves as the relative amounts of the two phases in equilibrium change (not to scale).
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Trends in a temperature versus heat curve for solid u gas. This schematic diagram illustrates the process corresponding to path b in Figure 8.4. Temperature rises linearly with qp in singlephase regions and remains constant along the twophase curves as the relative amounts of the two phases in equilibrium change (not to scale).
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the dashed arrows because dP S 0. Therefore, ∆H for the transformation solid S liquid S gas in process a and for the transformation solid S gas in process b must be identical. We conclude that
∆H = ∆ subH = ∆ fusH + ∆ vapH
(8.4)
This statement is strictly true at the triple point. Far from the triple point, the temperaturedependent heat capacities of the different phases and the temperature dependence of the transition enthalpies must be taken into account. Path c indicates an isothermal process in which the pressure is increased. The initial state of the system is the singlephase gas region. As the gas is compressed, it is liquefied as it crosses the gas–liquid coexistence curve. As the pressure is increased further, the sample freezes as it crosses the liquid–solid coexistence curve. Freezing is exothermic, and heat must flow to the surroundings as the liquid solidifies. If the process is reversed, heat must flow into the system to keep T constant as the solid melts. If T is less than the triple point temperature, the liquid exists at equilibrium only if the slope of the liquid–solid coexistence curve is negative, as is the case for water. Solid water below the triple point temperature can melt at constant T if the pressure is increased sufficiently to cross the liquid–solid coexistence curve. In an example of such a process, a thin wire to which a heavy weight is attached on each end is stretched over a block of ice. With time, it is observed that the wire lies within the ice block and eventually passes through the block. There is no visible evidence of the passage of the wire in the form of a narrow trench. What happens in this process? Because the wire is thin, the force on the wire results in a high pressure in the area of the ice block immediately below the wire. This high pressure causes local melting of the ice below the wire. Melting allows the wire to displace the liquid water, which flows to occupy the volume immediately above the wire. Because water in this region no longer experiences a high pressure, it freezes again and hides the passage of the wire. The consequences of having a critical point in the gas–liquid coexistence curve are illustrated in Figure 8.4. The gas–liquid coexistence curve is crossed in the constant pressure process starting at point d and moving to lower temperatures. A clearly visible interface will be observed as the twophase gas–liquid coexistence curve is crossed. However, the same overall process can be carried out in four steps indicated by the blue arrows. In this case, a twophase coexistence is not observed because the gas–liquid coexistence curve ends at Pcritical. The overall transition is the same along both paths; namely, gas is transformed into liquid. However, no interface will be observed if P for the higher pressure constant pressure process is greater than Pcritical.
EXAMPLE PROBLEM 8.1 Draw a generic P9T phase diagram like that shown in Figure 8.4. Draw pathways in the diagram that correspond to the processes described here: a. You hang washed clothing out to dry for a temperature below the triple point value. Initially, the water in the wet clothing is frozen. However, after a few hours in the sun, the clothing is warmer, dry, and soft. b. A small amount of ethanol is contained in a thermos bottle. A test tube is inserted into the neck of the thermos bottle through a rubber stopper. A few minutes after filling the test tube with liquid nitrogen, the ethanol is no longer visible at the bottom of the bottle. c. A transparent cylinder and piston assembly contains only a pure liquid in equilibrium with its vapor phase. An interface is clearly visible between the two phases. When you increase the temperature by a small amount, the interface disappears.
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8.2 The Pressure–Temperature Phase Diagram
215
Solution The phase diagram with the paths is shown here: c Critical point Liquid
Pressure
Solid
b
Gas
Triple point
a
Temperature
Paths a, b, and c are not unique. Path a must occur at a pressure lower than the triple point pressure. Process b must occur at a pressure greater than the triple point pressure and originate on the liquid–gas coexistence curve. Path c will lie on the liquid–gas coexistence curve up to the critical point, but it can deviate once T 7 Tc and P 7 Pc. A P9T phase diagram for water at high P values is shown in Figure 8.9. Because the logarithm of pressure is plotted on the y axis, the curvature of the solid–liquid and liquid–gas coexistence curves decreases with increasing temperature rather than increases as shown in Figure 8.4. Water has a number of solid phases that are stable in different pressure ranges because they have different densities. Eleven different crystalline forms of ice have been identified up to a pressure of 1012 Pa. Note, for example, in Figure 8.9 that ice VI does not melt until the temperature is raised to ∼300 K for P ≈ 1000 MPa X VII
VIII 109
VI II
V
Pressure, P (Pa)
IX
Liquid
III
Critical point
106 I
XI
103
Gas
Triple point
1 0
50
100
150
200
250
300 350 Temperature, T (K)
400
450
500
550
600
Figure 8.9
The P–T phase diagram for H2O for pressures to P = 1012 Pa and T = Tc. Source: Adapted from “Water Structure and Science,” http://www.lsbu.ac.uk/water/phase.html, courtesy of Martin Chaplin.
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(a) Hexagonal ice is the stable form of ice under atmospheric conditions. It has a relatively open structure and a density less than that of liquid water.
and that ice VII and ice X are stable to very high temperatures (though at very high pressures). For a comprehensive collection of material on the phase diagram of water, see http://www.lsbu.ac.uk/water/phase.html. Hexagonal ice (ice I) is the normal form of ice and snow. The structure shown in Figure 8.10 may be thought of as consisting of a set of parallel sheets connected to one another through hydrogen bonding. Hexagonal ice has a moderately open structure with a density of 0.931 g cm3 near the triple point. Figure 8.10 also shows the crystal structure of ice VI. All water molecules in this structure are hydrogen bonded to four other molecules. Ice VII is much more closely packed than hexagonal ice and does not float on liquid water. As shown in Figure 8.9, phase diagrams can be quite complex for simple substances because a number of solid phases can exist at different values of P and T. Another example is sulfur, which can also exist in several different solid phases. A portion of the phase diagram for sulfur is shown in Figure 8.11, and the solid phases are described by the symmetry of their unit cells. Note that several points correspond to threephase equilibria. By contrast, the CO2 phase diagram shown in Figure 8.12 is simpler. It is similar in structure to that of water, but the solid–liquid coexistence curve has a positive slope. Several of the endofchapter problems and questions refer to the phase diagrams in Figures 8.11, 8.12, and 8.13.
Liquid Pressure, P (atm) 1
Monoclinic solid
153°C, 1420 atm 115.21°C, 1 atm
95.39°C, 1 atm
444.6°C, 1 atm Rhombic solid
(b) Ice VI has a much more compact arrangement of water molecules and thus a density much greater than ice I (~1.3 g cm23).
95.31°C, 5.1 3 1026 atm
Figure 8.10
Crystal structures for two different forms of ice. (a) Hexagonal ice, (b) Ice VI. Ice VI is only stable at elevated pressures, as shown in the phase diagram of Figure 8.9. The dotted lines indicate hydrogen bonds.
115.18°C, 3.2 3 1025 atm
Gas
Temperature
Figure 8.11
The P–T phase diagram for sulfur. The diagram is not to scale.
10000 Solid
Pressure, P (atm)
1000
Solid
Liquid
100 10
uid
1 Liq
Sublimation point 278.5°C at 1 atm
as
Critical point
d1G
Liqui
Triple point 256.6°C at 5.11 atm
1
Figure 8.12
The P–T phase diagram for CO2. The diagram is not to scale.
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So
0.01
lid
1
Ga s
0.1
Gas
0.001 2140 2120 2100 280
260 240 220 0 20 Temperature, T (°C)
40
60
80
100
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8.4 Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
8.3 THE PHASE RULE As we have seen in Section 8.1, the coexistence of two phases, a and b, of a substance requires that the chemical potentials are equal:
ma1T, P2 = mb1T, P2
(8.5)
ma1T,P2 = mb1T,P2 = mg1T,P2
(8.6)
The state variables T and P, which are independent in a singlephase region, are linked through Equation (8.5) for a twophase coexistence region. Consequently, only one of T and P is needed to describe the system, and the other variable is determined by Equation (8.5). For this reason, the twophase coexistence regions in Figures 8.4, 8.9, 8.11, and 8.12 are curves that have the functional form P1T2 or T1P2. If three phases, a, b, and g, coexist in equilibrium, then
Equation (8.6) imposes another requirement on the variables T and P. Now only one combination of T and P satisfies the requirement posed by Equation (8.6), and threephase coexistence for a pure substance occurs only at a single point in the P, T phase diagram called the triple point. These restrictions on the number of independent variables accessible to a system can be summarized as follows. Because T and P can be varied independently in a singlephase region, a system of a pure substance has two degrees of freedom. The degrees of freedom are the number of independent variables needed to describe the physical state of the system. Because only one of T and P can be varied independently along a twophase coexistence curve, the system has one degree of freedom. Finally, because neither T nor P can be varied at a triple point, such a system has zero degrees of freedom. The American chemist J. W. Gibbs (1839–1903) derived the phase rule, which links the number of degrees of freedom to the number of phases in a system at equilibrium. For a pure substance, the phase rule takes the form F = 3  p
(8.7)
Concept The phase rule for a pure system relates the number of phases that can coexist at equilibrium and the number of degrees of freedom accessible to the system.
where F is the number of degrees of freedom and p is the number of phases. Gibbs proved that no more than three phases of a pure substance can be in equilibrium as F cannot be a negative number. As we will see in Chapter 9, the number of degrees of freedom increases if a system contains several chemically independent species, for example, in a system consisting of ethanol and water.
8.4 PRESSURE–VOLUME AND PRESSURE–
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Liquid
Solid + Liquid
Solid
In Section 8.2, the regions of stability and equilibrium among the solid, liquid, and gas phases were described using a P9T phase diagram. Any phase diagram that includes only two of the three state variables is limited because it does not contain information on the third variable. We first complement the information contained in the P9T phase diagram with a P–V phase diagram, and then we combine these two representations into a P9V9T phase diagram. Figure 8.13 shows a P9V phase diagram for a substance for which V liquid 7 V solid m m . Significant differences are seen in the way that two and threephase coexistences are represented in the twophase diagrams. The twophase coexistence curves of the P9T phase diagram become twophase regions in the P9V phase diagram because the volume of a system in which two phases coexist varies with the relative amounts of the material in each phase. For pressures well below the critical point, the range in V over which the gas and liquid coexist is large compared to the range in V over which gas the solid and liquid coexist because V solid 6 V liquid V Vm . Therefore, the gas–liquid m m coexistence region is broader than the solid–liquid coexistence region. Note that the triple point in the P9T phase diagram becomes a triple line in the P9V diagram. Although P and T have unique values at the triple point, V can range between a maximum value
Pressure
VOLUME–TEMPERATURE PHASE DIAGRAMS Critical point
Liquid+Gas
Gas
a
Triple line Solid + Gas
b
Volume
Figure 8.13
A P–V phase diagram. In this type of diagram, singlephase regions, twophase coexistence regions, a critical point, and a triple line are displayed.
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Concept A P–V phase diagram displays the information contained in a P–T phase diagram in a different manner.
Concept A P–V–T phase diagram is three dimensional and displays all information contained in P–T and P–V phase diagrams.
for which the system consists almost entirely of gas with trace amounts of the liquid and solid phases and a minimum value for which the system consists almost entirely of solid with trace amounts of the liquid and gas phases. The usefulness of the P9V diagram is illustrated by tracing several processes in Figure 8.13. In process a, a solid is converted to a gas by increasing the temperature in an isobaric process for which P is greater than the triple point pressure. This same process was depicted as path a in Figure 8.4. In the P9V phase diagram, it is clear that this process involves large changes in volume in the twophase coexistence region over which the temperature remains constant, which was not obvious in Figure 8.4. Process b shows an isobaric transition from solid to gas for P below the triple point pressure, for which the system has only one twophase coexistence region (see path b in Figure 8.4). Process b in Figure 8.13 is known as freeze drying. Assume that the food to be freeze dried is placed in a vessel at 10°C, and the system is allowed to equilibrate at this temperature. The partial pressure of H2O1g2 in equilibrium with the ice crystals in the food under these conditions is 260 Pa. A vacuum pump is now started, and the gas phase is pumped away. The temperature reaches a steadystate value determined by heat conduction into the sample and heat loss through sublimation. The solid has an equilibrium vapor pressure determined by the steadystate temperature. As the pump removes water from the gas phase, ice spontaneously sublimes in order to keep the gasphase water partial pressure at its equilibrium value. After the sublimation is complete, the food has been freeze dried. The advantage of freeze drying over boiling off water at 1 atm is that food can be dehydrated at low temperatures, and thus the food is not cooked. All the information on the values of P, V, and T corresponding to singlephase regions, twophase regions, and the triple point displayed in the P–T and P–V phase diagrams is best displayed in a threedimensional P–V–T phase diagram. Such a diagram is shown in Figure 8.14 for an ideal gas, which does not exist in the form of condensed phases. It is easy to obtain the P–T and P–V phase diagrams from the P–V–T phase diagram. The P–T phase diagram is a projection of the threedimensional surface on the P–T plane, and the P–V phase diagram is a projection of the threedimensional surface on the P–V plane. Figure 8.15 shows a P–V–T diagram for a substance that expands upon melting. The usefulness of the P–V–T phase diagram can be illustrated by revisiting the isobaric conversion of a solid to a gas at a temperature above the triple point shown as process a in Figure 8.4. This process is shown as the path a S b S c S d S e S f in Figure 8.15. We can see now that the temperature increases along the segments a S b, c S d, and V1,V2,V3 P
V1
P
V2 V3
T1,T2,T3,T4 T1 T2 T3 T4
Pressure
T
V
Figure 8.14
A P–V–T diagram for an ideal gas. Constant pressure (gray curves), constant volume (red curves), and constant temperature (blue curves) paths are shown.
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Vo
lum
e
ure
rat
e mp
Te
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8.5 PROVIDING A THEORETICAL BASIS FOR THE P–T PHASE DIAGRAM
Figure 8.15
P m
P
Solid Critical point
Liquid
Critical point
T m l
q
a g
Pressure
Solid
b a
c
lid
lum
Critical point
f
Gas
d
Li
qu i Gas id e le
So
Vo
Liquid
Tri p
p
Liq Ga uid s So lid –G as
0
k
Solid–Liqu id
Gas
Triple point
A P–V–T phase diagram for a substance that expands upon melting. The indicated processes are discussed in the text.
Solid–Liquid
f
Liq uid
Solid
219
–G
as
as
e
g
o n
e
V
G
h
Lin
T2 T1
T3Tc
T4
e
tur
ra pe
m
Te
e S f , all of which lie within singlephase regions, and it remains constant along the segments b S c and d S e, which lie within twophase regions. Similarly, process c in Figure 8.4 is shown as the path g S h S i S k S l S m in Figure 8.15.
8.5 PROVIDING A THEORETICAL BASIS FOR THE P–T PHASE DIAGRAM
In this section, a theoretical basis is provided for the coexistence curves that separate different singlephase regions in the P–T phase diagram. Along the coexistence curves, two phases are in equilibrium. From Section 5.6, we know that if two phases, a and b, are in equilibrium at a pressure P and temperature T, their chemical potentials must be equal:
ma1P, T2 = mb1P, T2
(8.8)
ma1P, T2 + dma = mb1P, T2 + dmb
(8.9)
dma = dmb
(8.10)
If the macroscopic variables are changed by a small amount, P, T S P + dP, T + dT, such that the system pressure and temperature still lie on the coexistence curve, then
In order for the two phases to remain in equilibrium,
Because dm can be expressed in terms of dT and dP,
dma = Sma dT + Vma dP and dmb = Smb dT + Vmb dP
(8.11)
The expressions for m can be equated, giving
Sma dT + Vma dP = Smb dT + Vmb dP or 1Smb  Sma2 dT = 1Vmb  Vma2 dP
(8.12)
∆Sm dP = dT ∆Vm
(8.13)
Assume that as P, T S P + dP, T + dT an incremental amount of phase a is transformed to phase b. In this case, ∆Sm = Smb  Sma and ∆Vm = Vmb  Vma. Rearranging Equation (8.12) gives the Clapeyron equation:
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
The Clapeyron equation allows us to calculate the slope of the coexistence curves in a P–T phase diagram if ∆Sm and ∆Vm for the transition are known. This information is necessary when constructing a phase diagram. We next use the Clapeyron equation to estimate the slope of the solid–liquid coexistence curve. At the melting temperature, ∆ fusG = ∆ fus H  T∆ fus S = 0
(8.14)
Therefore, the ∆Sm values for the fusion transition can be calculated from the enthalpy of fusion and the fusion temperature. Values of the normal fusion and vaporization temperatures, as well as ∆Hm for fusion and vaporization at 1 atm pressure, are shown in Table 8.2 for a number of different elements and compounds. Although there is a significant variation in these values, for our purposes it is sufficient to use the average value of ∆ fusS = 22 J mol1 K1 calculated from the data in Table 8.2 in order to estimate the slope of the solid–liquid coexistence curve. For the fusion transition, ∆V is small because the densities of the solid and liquid states are quite similar. The average ∆ fusV for Ag, AgCl, Ca, CaCl2 K, KCl, Na, NaCl, and H2O is +4.0 * 106 m3 mol1. Of these substances, only H2O has a negative value for ∆ fusV. We next use the average values of ∆ fusS and ∆ fusV to estimate the slope of the solid–liquid coexistence curve: a
∆ fusS dP 22 J mol1 K1 b = ≈ dT fus ∆ fusV {4.0 * 106 m3mol1
= {5.5 * 106 Pa K1 = {55 bar K1
(8.15)
Inverting this result, we see that 1dT > dP2fus ≈ {0.018 K bar . An increase of P by ∼50 bar is required to change the melting temperature by one degree. This result explains the very steep solid–liquid coexistence curve shown in Figure 8.4. The same analysis applies to the liquid–gas coexistence curve. Because ∆ vapH gas and ∆ vapV = V m  V liquid are always positive, 1dP > dT2vap is always positive. The m average of ∆ vapS for the substances shown in Table 8.2 is 95 J mol1 K1. This value is in accord with Trouton’s rule, which states that ∆ vapS ≈ 90 J mol1 K1 for liquids. The rule fails for liquids in which there are strong interactions between molecules such as OH or NH2 groups capable of forming hydrogen bonds. For these substances, ∆ vapS 7 90 J mol1 K1. The molar volume of an ideal gas is approximately 20. L mol1 in the temperature gas range in which many liquids boil. Because V m W V liquid , ∆ vapV ≈ 20.*103 m3mol1, m the slope of the liquid–gas coexistence curve is given by 1
Concept Trouton’s rule is useful in predicting ∆ vapS for liquids.
a
∆ vapS dP 95 J mol1 K1 b = ≈ ≈ 4.8 * 103 Pa K1 dT vap ∆ vapV 2.0 * 102 m3 mol1 = 4.8 * 102 bar K1
(8.16)
3
This slope is a factor of 10 smaller than the slope for the liquid–solid coexistence curve. Inverting this result, we find that 1dT > dP2vap ≈ 21 K bar 1. This result shows that it takes only a modest increase in the pressure to increase the boiling point of a liquid by a significant amount. For this reason, a pressure cooker or an automobile radiator does not need to be able to withstand high pressures. Note that the slope of the liquid–gas coexistence curve in Figure 8.4 is much less than that of the solid–liquid coexistence curve. The slope of both curves increases with T because ∆S increases with increasing T and ∆V decreases with increasing P. The solid–gas coexistence curve can also be analyzed using the Clapeyron equation. Because entropy is a state function, the entropy change for the processes solid 1P, T2 S gas1P, T2 and solid 1P, T2 S liquid 1P, T2 S gas 1P, T2 must be the same. Therefore, ∆ subS = ∆ fusS + ∆ vapS 7 ∆Svap. However, because the molar volume of the gas is so much larger than that of the solid or liquid, ∆ subV ≈ ∆ vapV. We conclude that 1dP > dT2sub 7 1dP > dT2vap. Therefore, the slope of the solid–gas coexistence curve will be greater than that of the liquid–gas coexistence curve. Because this comparison applies to a common value of the temperature, it is best made for temperatures just above and just below the triple point temperature. This difference in slope of these two coexistence curves is exaggerated in Figure 8.4.
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8.6 USING THE CLAUSIUS–CLAPEYRON EQUATION TO CALCULATE VAPOR PRESSURE AS A FUNCTION OF T
221
8.6 USING THE CLAUSIUS–CLAPEYRON EQUATION TO CALCULATE VAPOR PRESSURE AS A FUNCTION OF T
From watching a pot of water as it is heated on a stove, it is clear that the vapor pressure of a liquid increases rapidly with increasing temperature. The same conclusion holds for a solid below the triple point. To calculate the vapor pressure at different temperatures, the Clapeyron equation must be integrated. Consider the solid–liquid coexistence curve: Pf
Tf
Tf
Tf
∆ fusH dT ∆ fusS ∆ fusH dT dP = dT = ≈ ∆ fusV L T L L ∆ fusV L ∆ fusV T Pi
Ti
Ti
(8.17)
Ti
where the integration is along the solid–liquid coexistence curve. In the last step, it has been assumed that ∆ fusH and ∆ fusV are independent of T over the temperature range of interest. Assuming that 1Tf  Ti2>Ti is small, we can simplify the previous equation to give
Pf  Pi =
∆ fusH Tf ∆ fusH Ti + ∆T ∆ fusH ∆T ln = ln ≈ ∆ fusV Ti ∆ fusV Ti ∆ fusV Ti
(8.18)
The last step uses the result ln11 + x2 ≈ x for x V 1, obtained by expanding ln11 + x2 in a Taylor series about x = 0. We see that the vapor pressure of a solid varies linearly with ∆T in this limit. (The value of the slope dP>dT was discussed in the previous section.) For the liquid–gas coexistence curve, we have a different result because ∆V ≈ V gas. Assuming that the ideal gas law holds, we obtain the Clausius–Clapeyron equation, ∆ vapS ∆ vapH P∆ vapH dP = ≈ gas = dT ∆ vapV TV m RT 2
∆ vapH dT dP = P R T2
(8.19)
Assuming that ∆ vapH remains constant over the range of temperature of interest, the variation of the vapor pressure of the liquid with temperature is given by Pf
Tf
Pi
Ti
∆ vapH dP dT = * 2 P R L LT
ln
Pf Pi
= 
∆ vapH R
* a
1 1  b Tf Ti
(8.20)
We see that the vapor pressure of a liquid rises exponentially with temperature. The same procedure is followed for the solid–gas coexistence curve. The result is the same as Equation (8.20), with ∆ subH substituted for ∆ vapH. Equation (8.20) provides a way to determine the enthalpy of vaporization for a liquid by measuring its vapor pressure as a function of temperature, as shown in Example Problem 8.2. In this discussion, it has been assumed that ∆ vapH is independent of temperature. More accurate values of the vapor pressure as a function of temperature can be obtained by fitting experimental data. This leads to an expression for the vapor pressure as a function of temperature. These functions for selected liquids and solids are listed in Tables 8.3 and 8.4 (see Appendix A, Data Tables).
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Concept The vapor pressure of a liquid increases exponentially with temperature.
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EXAMPLE PROBLEM 8.2 The normal boiling temperature of benzene is 353.24 K, and the vapor pressure of liquid benzene is 1.19 * 104 Pa at 20.0°C. The enthalpy of fusion is 9.95 kJ mol1, and the vapor pressure of solid benzene is 137 Pa at 44.3°C. Calculate the following: a. ∆ vapH b. ∆ vapS c. Triple point temperature and pressure
Solution a. We can calculate ∆ vapH using the Clausius–Clapeyron equation because we know the vapor pressure at two different temperatures: ln
Pf Pi
∆ vapH 1 1 a  b R Tf Ti
= 
Pf
101,325 Pa Pi 1.19 * 104 Pa ∆ vapH = = 1 1 1 1 a  b a b Tf Ti 353.24 K 273.15 K + 20.0 K R ln
8.314 J mol1 K1 * ln
= 30.7 kJ mol1
b. ∆ vapS =
∆ vapH Tb
30.7 * 103 J mol1 = 86.9 J mol1 K1 353.24 K
=
c. At the triple point, the vapor pressures of the solid and liquid are equal: ln
P liquid tp P~
ln ln
R aln
8.314 J K1 mol1
∆ vapH 1 P liquid 1 i a  liquid b ~ P R Ttp Ti
P solid ∆ subH 1 P solid 1 tb i = ln a  solid b ~ P P~ R Ttp Ti
∆ vapH ∆ vapH  ∆ subH ∆ subH P liquid P solid i i ln + = ~ ~ solid liquid P P RTtp RT i RT i ∆ vapH  ∆ subH
Ttp =
=
= ln
P liquid i
 ln
P~
∆ vapH ∆ subH P solid i + b ~ solid P RT i RT liquid i
9.95 * 103J mol1 130.7 * 103 + 9.95 * 1032 J mol1 1.19 * 104 Pa 137 Pa ln ln 8.314 J K1 mol1 * 228.9 K 105 Pa 105 Pa * ± ≤ 3 1 30.7 * 10 mol + 8.314 J K1mol1 * 293.15 K
= 277 K We calculate the triple point pressure using the Clapeyron equation: ln ln ln
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Pf Pi
Ptp P
~
Ptp
= = 
∆ vapH 1 1 a  b R Tf Ti
30.7 * 103 J mol1 1 1 * a b 278 K 353.24 K 8.314 J mol1 K1
= 8.70483 P~ Ptp = 6.03 * 103 Pa
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8.7 Dependence of Vapor Pressure of a Pure Substance on Applied Pressure
223
8.7 DEPENDENCE OF VAPOR PRESSURE OF A
PURE SUBSTANCE ON APPLIED PRESSURE
To evaluate the effects of applied pressure on vapor pressure, consider the piston and cylinder assembly shown in Figure 8.16. If the cylinder contains only water at 25°C, the equilibrium vapor pressure of water at this temperature is P * = 3.16 * 103 Pa, or 0.0316 bar. Therefore, if the weightless piston is loaded with a mass sufficient to generate a pressure of 1 bar, the system is in the singlephase liquid region of the phase diagram in Figure 8.4. This state of the system is shown in Figure 8.16a. Next, the mass is reduced so that the pressure is exactly equal to the vapor pressure of water. The system now lies in the twophase liquid–gas region of the phase diagram described by the liquid–gas coexistence curve. The piston can be pulled outward or pushed inward while maintaining this pressure. This action leads to a larger or smaller volume for the gas phase, but the pressure will remain constant at 3.16 * 103 Pa as long as the temperature of the system remains constant. This state of the system is shown in Figure 8.16b. Finally, keeping the temperature constant, enough argon gas is introduced into the cylinder such that the sum of the argon and H2O partial pressures is 1 bar. This state of the system is shown in Figure 8.16c. What is the vapor pressure of water in this case, and does it differ from that for the system shown in Figure 8.16b? The vapor pressure P is used to denote the partial pressure of water in the gas phase, and P is used to denote the sum of the argon and water partial pressures. To calculate the partial pressure of water in the argon–water mixture, the following equilibrium condition holds: mliquid1T, P2 = mgas1T, P2
(8.21)
Mass Piston H2O(l) Pext 5 1.00 bar (a) Pressure is greater than the vapor pressure, and the contents consists of only liquid water.
Mass Piston H2O(g) H2O(l) Pext 5 0.0316 bar (b) Pressure is equal to the vapor pressure, and the contents consist of liquid and gaseous water.
Differentiating this expression with respect to P, we obtain a
0mliquid1T, P2 0P
b = a T
0mgas1T, P2 0P
b a T
0P b 0P T
Mass
(8.22)
Piston H2O(g)1Ar(g)
Because dm = Sm dT + Vm dP, 1dm>dP2T = Vm, the previous equation becomes gas V liquid = Vm a m
V liquid 0P 0P m b or a b = 0P T 0P T V gas m
(8.23)
This equation shows that the vapor pressure P increases if the total pressure P increases. gas However, the rate of increase is small because V liquid >V m V 1. It is reasonable to rem gas place V m with the ideal gas value RT>P in Equation (8.23). This leads to the equation P
P
RT dP′ dP = V liquid d P or RT = V liquid d P′ m m P L* P′ L*
P
(8.24)
P
H2O(l) Pext 5 1.00 bar (c) Pressure is 1 bar, and the contents consist of a mixture of argon and liquid and gaseous water.
Figure 8.16
A piston and cylinder assembly containing water at 298 K.
Integrating Equation (8.24) gives RT ln a
P P*
b = V liquid 1P  P *2 m
(8.25)
For the specific case of water, P * = 0.0316 bar, P = 1 bar, and V liquid = m 1.81 * 105 m3 mol1: ln a
P P
b = * =
V liquid 1P  P *2 m RT
1.81 * 105m3 mol1 * 11  0.03162 * 105 Pa 8.314 J mol1 K1 * 298 K
= 7.04 * 104
P = 1.0007 P * ≈ 0.0316 bar
For an external pressure of 1 bar, the effect is negligible. However, for P = 100. bar, P = 0.0339 bar, amounting to an increase in the vapor pressure of 7%.
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Concept The vapor pressure of a pure substance depends on the applied pressure.
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
8.8 SURFACE TENSION
Concept Surface tension arises at an atomic level because atoms at the surface have fewer nearest neighbors than those in the interior of a liquid or solid.
In discussing the liquid phase, the effect of the boundary surface on the properties of the liquid has thus far been neglected. In the absence of a gravitational field, a liquid droplet will assume a spherical shape because in this geometry the maximum number of molecules is surrounded by neighboring molecules. Because the interaction between molecules in a liquid is attractive, minimizing the surfacetovolume ratio minimizes the energy. Therefore, increasing the area of a liquid droplet requires work to be done on the droplet. How does the energy of the droplet depend on its surface area? Starting with the equilibrium spherical shape, assume that the droplet is distorted to create more area while keeping the volume constant. The work associated with the creation of additional surface area ds at constant V and T is
(8.26)
dw = dA = gds
The proportionality constant, g, linking dw and ds in Equation 8.26 is called the s urface tension. It has the units of energy/area or J m2, which is equivalent to N m1. Equation (8.26) predicts that a liquid, a bubble, or a liquid film suspended in a wire frame will tend to minimize its surface area because dA 6 0 for a spontaneous process at constant V and T. Consider the spherical droplet depicted in Figure 8.17. There must be a force acting on the droplet in the radially inward direction for the liquid to assume a spherical shape. An expression for the force can be generated as follows. If the radius of the droplet is increased from r to r + dr, the area increases by ds. s = 4pr 2 so ds = 8pr dr
(8.27)
From Equation (8.26), the work done in the expansion of the droplet is 8pgr dr. The force, which is normal to the surface of the droplet, is the work divided by the distance or
Figure 8.17
The forces acting on a spherical droplet that arise from surface tension.
(8.28)
F = 8pgr
The net effect of this force is to generate a pressure differential across the droplet surface. At equilibrium, there is a balance between the inward and outward acting forces. The inward acting force is the sum of the force exerted by the external pressure and the force arising from the surface tension, whereas the outward acting force arises solely from the pressure in the liquid: 4pr 2Pouter + 8pgr = 4pr 2Pinner or 2g Pinner = Pouter + (8.29) r Note that Pinner  Pouter S 0 as r S ∞ . Therefore, the pressure differential exists only for a curved surface. From the geometry in Figure 8.17, it is apparent that the larger pressure is always on the concave side of the interface. Values for the surface tension for a number of liquids are listed in Table 8.5. TABLE 8.5 Surface Tension of Selected Liquids at 298 K g 1mN m12
Formula
Name
40.95
CS2
Carbon disulfide
Water
71.99
C2H5OH
Ethanol
21.97
Hg
Mercury
485.5
C6H5N
Pyridine
36.56
CCl4
Carbon tetrachloride
26.43
C6H6
Benzene
28.22
CH3OH
Methanol
22.07
C8H18
Octane
21.14
Formula
Name
Br2
Bromine
H2O
g 1mN m12 31.58
Source: Data from Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
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8.8 Surface Tension
Equation (8.29) has interesting implications for the relative stabilities of bubbles with different curvatures 1>r. Consider two airfilled bubbles of a liquid with the same surface tension g, one with a large radius R1 and one with a smaller radius R2. Assume there is a uniform pressure outside both bubbles. From Equation (8.29) we obtain the difference between the pressures P1 and P2 within each bubble:
P1  P2 =
2g 2g 1 1 = 2g a b R1 R2 R1 R2
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h =
2g rgr
r2 r1
(8.30)
Now suppose the two bubbles come into contact as shown at the top of Figure 8.18. Because R1 7 R2, from Equation (8.30) the pressure P2 inside bubble 2 (with respect to some common exterior pressure) is greater than the pressure P1 in bubble 1. Thus, air will flow from bubble 2 to bubble 1 until the smaller bubble disappears entirely, as shown at the bottom of Figure 8.18. In foams this process is called coarsening, and in crystals it is called Ostwald ripening. A biologically relevant issue that follows from Equations (8.29) and (8.30) is the stability of lung tissue. Lung tissue is composed of small waterlined, airfilled chambers called alveoli. According to Equation (8.29), the pressure difference across the surface of a spherical airfilled cavity is proportional to the surface tension and inversely proportional to the radius of the cavity. Therefore, if two airfilled alveoli, approximated as spheres with radii r and R and for which r 6 R are interconnected, the smaller alveolus will collapse and the larger alveolus will expand because the pressure within the smaller alveolus is greater than the pressure within the larger alveolus. In theory, because of the surface tension of water that lines the alveoli, these cavities will coarsen (as defined above) as smaller alveoli collapse, leaving an ever diminishing number of alveoli of increasing size. Coarsening of the alveoli according to Equations (8.29) will result eventually in collapse of the lungs. In the alveoli, a lipid surfactant called phosphatidylcholine lowers the surface tension of the water lining the alveoli, thus stabilizing the lungs. But in persons lacking sufficient surfactant in the lining of their lungs, such as prematurely born infants and heavy smokers, the surface tension of the water in the alveoli is not lowered, and as a consequence lung tissue is destabilized by coarsening. Another consequence of the pressure differential across a curved surface is that the vapor pressure of a droplet depends on its radius. By substituting numbers in Equation (8.29) and using Equation (8.25) to calculate the vapor pressure, we find that the vapor pressure of a 107 m water droplet is increased by 1%, that of a 108 m droplet is increased by 11%, and that of a 109 m droplet is increased by 270% relative to a droplet with zero surface tension. [At such a small diameter, the application of Equation (8.29) is questionable because the size of an individual water molecule is comparable to the droplet diameter. Therefore, a microscopic theory is needed to describe the forces within the droplet.] This effect plays a role in the formation of liquid droplets in a condensing gas such as fog. Small droplets evaporate more rapidly than large droplets, and the vapor condenses on the larger droplets, allowing them to grow at the expense of small droplets. Capillary rise and capillary depression are other consequences of the pressure differential across a curved surface. Assume that a capillary of radius r is partially immersed in a liquid. When the liquid comes in contact with a solid surface, there is a natural tendency to minimize the energy of the system. If the surface tension of the liquid is lower than that of the solid, the liquid will wet the surface, as shown in Figure 8.19a. However, if the surface tension of the liquid is higher than that of the solid, the liquid will avoid the surface, as shown in Figure 8.19b. In either case, there is a pressure differential in the capillary across the gas–liquid interface because the interface is curved. If we assume that the liquid–gas interface is tangent to the interior wall of the capillary at the solid– liquid interface, the radius of curvature of the interface is equal to the capillary radius. The difference in the pressure across the curved interface 2g>r is balanced by the weight of the column in the gravitational field rgh. Therefore, 2g>r = rgh, and the capillary rise or depression is given by
225
Airflow
Figure 8.18
Interaction of two bubbles with unequal radii. In the top diagram, bubbles with radii R1 and R2 make contact and interact. Because the pressure within a bubble varies inversely with the bubble radius, air flows from the smaller into the larger bubble (middle diagram) until a single large bubble remains (bottom).
(8.31)
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CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Capillary rise Pouter
Pouter
Pinner 1 rgh
h
In the preceding discussion, it was assumed that either (1) the liquid completely wets the interior surface of the capillary, in which case the liquid coats the capillary walls but does not fill the core, or (2) the liquid is completely nonwetting, in which case the liquid does not coat the capillary walls but fills the core. In a more realistic model, the interaction is intermediate between these two extremes. In this case, the liquid–surface is characterized by the contact angle u, as shown in Figure 8.20. Complete wetting corresponds to u = 0°, and complete nonwetting corresponds to u = 180°. For intermediate cases, Pinner = Pouter +
2g cos u 2g cos u and h = r rgr
(8.32)
Measurement of the contact angle is one of the main experimental methods used to measure the difference in surface tension at the solid–liquid interface.
EXAMPLE PROBLEM 8.3 The sixlegged water strider supports itself on the surface of a pond on four of its legs. Each of these legs causes a depression to be formed in the pond surface. Assume that each depression can be approximated as a hemisphere of radius 1.2 * 104 m and that u (as in Figure 8.20) is 0°. Calculate the force that one of the insect’s legs exerts on the pond. (a) If the liquid wets the interior wall of the capillary, a capillary rise is observed. The combination Pyrex–water exhibits this behavior.
Solution 2g cos u 2 * 71.99 * 103 N m1 ∆P = = = 1.20 * 103 Pa r 1.2 * 104 m F = PA = P * pr 2 = 1.20 * 103 Pa * p11.2 * 104 m22 = 5.4 * 105 N
EXAMPLE PROBLEM 8.4 Water is transported upward in trees through channels in the trunk called xylem. Although the diameter of the xylem channels varies from species to species, a typical value is 2.0 * 105 m. Is capillary rise sufficient to transport water to the top of a redwood tree that is 100 m high? Assume complete wetting of the xylem channels.
Pouter
Solution From Equation (8.31),
Pouter h Capillary depression Pouter 1rgh
(b) If the liquid does not wet the capillary surface, a capillary depression is observed. The combination Pyrex–mercury exhibits this behavior.
Figure 8.19
Capillary rise and depression in a capillary partially immersed in a liquid. (a) Capillary rise, (b) capillary depression.
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h =
2g 2 * 71.99 * 103 N m1 = = 0.74 m rgr cos u 997 kg m3 * 9.81 m s2 * 2.0 * 105 m * 1
No, capillary rise is not sufficient to account for water supply to the top of a redwood tree. As Example Problem 8.4 shows, capillary rise is insufficient to account for water transport to the leaves in all but the smallest plants. The property of water that accounts for water supply to the top of a redwood is its large tensile strength. Imagine pulling on a piston in a cylinder containing only liquid water to create a negative pressure. How hard can a person pull on the water without “breaking” the water column? The answer to this question depends on whether bubbles are nucleated in the liquid. This phenomenon is called cavitation. If cavitation occurs, the bubbles will grow rapidly as the piston is pulled outward. The bubble pressure is given by Equations (8.31), where Pext is the vapor pressure of water. The height of the water column in this case is limited to about 9.7 m. However, bubble nucleation is a kinetic phenomenon initiated at certain sites at the wall surrounding the water, and under the conditions present in xylem tubes it is largely suppressed. In the absence of bubble nucleation, theoretical calculations predict that negative pressure in excess of 1000 atm can be generated.
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8.9 Chemistry in Supercritical Fluids
The pressure is negative because the water is under tension rather than compression. Experiments on water inclusions in very small cracks in natural rocks have verified these estimates. However, bubble nucleation occurs at much lower negative pressures in capillaries similar in d iameter to xylem tubes. Even in these capillaries, negative pressures of more than 50 atm have been observed. How does the high tensile strength of water explain the transport of water to the top of a redwood? If one cuts into a tree near its base, the sap oozes rather than spurts out, showing that the pressure in the xylem tubes is ∼1 atm at the base of a tree, whereas the pressure at the base of a tall water column would be much greater than 1 atm. Imagine the redwood in its infancy as a seedling. Capillary rise is sufficient to fill the xylem tubes to the top of the plant. As the tree grows taller, the water is pulled upward because of its large tensile strength. As the height of the tree increases, the pressure at the top becomes increasingly negative. As long as cavitation does not occur, the water column remains intact. As water evaporates from the leaves, it is resupplied from the roots through the pressure gradient in the xylem tubes that arises from the weight of the column. If the tree (and each xylem tube) grows to a height of ∼100 m, and P = 1 atm at the base, the pressure at the top must be ∼ 9 atm, from ∆P = rgh. We again encounter a negative pressure because the water is under tension. If water did not have a sufficiently high tensile strength, gas bubbles would form in the xylem tubes. This would disrupt the flow of sap, and tall trees could not exist.
S U PPL EM EN TAL
227
u
Figure 8.20
The shape of the liquid–air interface in a capillary. For cases intermediate between wetting and nonwetting, the contact angle u lies in the range 0° 6 u 6 180°.
SEC TI O N
8.9 CHEMISTRY IN SUPERCRITICAL FLUIDS Chemical reactions can take place in the gas phase, in solution (homogeneous reactions), or on the surfaces of solids (heterogeneous reactions). Solvents with suitable properties can influence both the yield and the selectivity of reactions in solution. It has been found that the use of supercritical fluids as solvents increases the number of parameters that chemists have at their disposal to tune a reaction system to meet their needs. Supercritical fluids (SCFs) near the critical point with reduced temperatures and pressures (see Section 7.4) in the range Tr ∼ 1.091.1 and Pr ∼ 192 have a density that is an appreciable fraction of the liquidphase density. They are unique in that they exhibit favorable properties of liquids and gases. Because the density is high, the solubility of solid substances is quite high, but the diffusion of solutes in the fluid is greater than that in the normal liquid. This is the case because the density of the SCF is lower than that in normal liquids. For similar reasons, the viscosity of SCFs is lower than that in normal liquids. As a result, mass transfer is faster, and the overall reaction rate can be increased. Because an SCF is more gaslike than a liquid, the solubility of gases can be much higher in the SCF. This property is of particular usefulness in enhancing the reactivity of reactions in which one of the reactants is a gas, such as oxidation or hydrogenation. For example, hydrogen generally has a low solubility in organic solvents but is quite soluble in organic SCFs. Carbon dioxide and water exhibit the unusual properties of SCFs. Because the values for the critical pressure and temperature of CO2 1Pc = 73.74 bar and Tc = 304 K2 are easily attainable, it is possible to use chemically inert supercritical CO2 to replace toxic organic solvents in the dry cleaning industry. Although supercritical CO2 is a good solvent for nonpolar molecules, it must be mixed with other substances to achieve a required minimum solubility for polar substances. This can be done without increasing the toxicity of the process greatly. Supercritical CO2 is also used in the decaffeination of coffee and tea. The critical constants of H2O are considerably higher 1Pc = 220.64 bar and Tc = 647 K2, so the demands on the vessel used to contain the supercritical fluid are greater than for CO2 However, as supercritical H2O is formed, many of the hydrogen bonds in the normal liquid are broken. As a result, the dielectric constant can be varied between the normal value for the liquid of 80 and a value of 5. At the lower end of
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Concept Supercritical fluids have unique properties than can be used to carry out reactions that would not be possible under “normal” conditions.
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this range, supercritical water acts like a nonpolar solvent and is effective in dissolving organic materials. Supercritical water has considerable potential for use at the high temperatures at which organic solvents begin to decompose. For example, it can be used to destroy toxic substances through oxidation in the remediation of contaminated groundwater. One challenge in using supercritical water is that it is highly corrosive, placing demands on the materials used to construct a practical facility.
SU P P LE ME NTAL
SE CTION
8.10 LIQUID CRYSTAL DISPLAYS CN
O2 N1 H3CO
OCH3
N
Figure 8.21
Examples of molecules that form liquid crystals. Liquid crystals are generally formed from polar organic molecules with a rodlike shape.
Liquid crystals are an exception to the general statement (superfluid He is another) that there are three equilibrium states of matter, namely, solids, liquids, and gases. Glass is a liquid of such high viscosity that it cannot achieve equilibrium on the timescale of a human lifetime. Glasses are a commonly encountered nonequilibrium state of matter. An example is SiO2 in the form of window glass. The properties of liquid crystals are intermediate between liquids and solids. Molecules that form liquid crystals typically contain rodshaped structural elements. Several such molecules are shown in Figure 8.21. How do liquid crystals differ from the other states of matter? The ordering in solid, liquid, and liquid crystal phases of such molecules is shown schematically in Figure 8.22. Whereas the crystalline solid phase is perfectly ordered and the liquid phase has no residual order, the liquid crystal phase retains some order. The long axes of all molecules deviate only by a small amount from their value in the liquid crystal structure (Figure 8.22c). The twisted nematic phase shown in Figure 8.23 is of particular importance because liquid crystals with this structure are the basis of the multibilliondollar liquid crystal display (LCD) industry and also the basis for sensor strips that change color with temperature. The twisted nematic phase consists of parallel planes in which the angle of the preferred orientation direction varies in a welldefined manner as the number of layers increases. If light is incident on such a crystal, it is partially reflected and partially transmitted from a number of layers. A maximum in the reflection occurs if the angle of incidence and the spacing between layers are such that constructive interference occurs in the light reflected from successive layers. Because this occurs only in a narrow range of wavelengths, the film appears colored, with the color determined by the wavelength. As temperature increases or decreases, the layer spacing and, therefore, the color of the reflected light change. For this reason, the color of the liquid crystal film changes with temperature. The way in which an LCD display functions is illustrated in Figure 8.24. A twisted nematic liquid crystal film is sandwiched between two transparent conducting
Figure 8.23
Schematic molecular diagram of a twisted nematic phase. The direction of preferential orientation of the rodlike molecules varies in a welldefined manner from one plane to the next.
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(a) Solid
(b) Liquid
(c) Liquid crystal
Figure 8.22
Schematic molecular structures of (a) solid, (b) liquid, and (c) liquid crystal phases.
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Vocabulary
Figure 8.24
Unpolarized light
Schematic diagram of an LCD display. The display consists of a twisted nematic liquid film enclosed between parallel transparent conducting electrodes. Polarizers whose transmission direction is rotated 90° with respect to one another are mounted on the electrodes. (a) Polarized light transmitted, (b) no light transmitted, (c) resulting display.
Unpolarized light Polarizer Polarizer E50
Electrodes
229
Electrodes (voltage on)
E
Analyzer Analyzer Polarized light (a) Light passing through the first polarizer is transmitted by the second polarizer because the plane of polarization of the light is rotated by the crystal.
No light (b) The orientational ordering of the twisted nematic phase is destroyed by application of the electric field. No light is passed.
(c) Arrangement of electrodes in an LCD alphanumeric display.
electrodes. The thickness of the film is such that the orientational direction rotates by 90° from the bottom to the top of the film. Ambient light is incident on an upper polarizing filter that allows only one direction of polarization to pass. The rodlike molecules act as a waveguide and rotate the plane of polarization as the light passes through the liquid crystal film. Therefore, light passes through the lower polarizer, which is rotated by 90° with respect to the upper polarizer. If an electric field is applied to the electrodes, the rotation of the orientational direction from plane to plane no longer occurs because the polar molecules align along the electric field. Therefore, the plane of polarization of the light transmitted by the upper polarizer is not rotated as it passes through the film. As a result, the light is not able to pass through the second polarizer. Now imagine the lower electrode to be in the form of a mirror. In the absence of the electric field (Figure 8.24a), the display will appear bright because the plane of polarization of the light reflected from the mirror is again rotated 90° as it passes back through the film. Consequently, it passes through the upper polarizer. By contrast, no light is reflected if the field is on (Figure 8.24b), and the display appears dark. Now imagine each of the electrodes to be patterned as shown in Figure 8.24c, and it will be clear why, in a liquid crystal watch display, dark numbers are observed on a light background.
Concept Liquid crystal displays are possible because of the ways that liquid crystals transmit polarized light.
VOCABULARY boiling point elevation capillary depression capillary rise Clapeyron equation Clausius–Clapeyron equation coexistence curve contact angle critical point degrees of freedom freeze drying freezing point depression freezing point elevation
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glass LCD display liquid crystals metastable phase nonwetting normal boiling temperature phase phase diagram phase rule P–T phase diagram P–V phase diagram P–V–T phase diagram
standard boiling temperature sublimation temperature supercritical fluids surface tension tensile strength triple point Trouton’s rule twisted nematic phase vapor pressure wetting
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KEY EQUATIONS Equation
Significance of Equation
liquid Sgas 7 Ssolid m 7 Sm m
Order of states of matter based on entropy values
8.3
∆H = ∆ sub H = ∆ fus H + ∆ vap H
Relation between enthalpies of fusion, vaporization, and sublimation
8.4
ma1T,P2 = mb1T,P2 = mg1T,P2
Condition for coexistence of three phases
8.6
F = 3  p
Phase rule for a pure substance
8.7
∆Sm dP = dT ∆Vm
Clapeyron equation allows calculation of slope of coexistence curves
8.13
Change in vapor pressure of liquid with temperature
8.20
Change in vapor pressure of liquid with applied pressure
8.25
Difference in pressure on liquid across a curved interface due to surface tension
8.29
Capillary rise as a function of contact angle and curvature of the interface
8.32
ln
Pf Pi
= 
RT ln a
P P*
∆ vap H R
1 1  b Tf Ti
b = V liquid 1P  P *2 m
Pinner = Pouter + h =
* a
2g r
2g cos u rgr
Equation Number
CONCEPTUAL PROBLEMS Q8.1 Why is it reasonable to show the m versus T segments for the three phases as straight lines as is done in Figure 8.1? More realistic curves would have some curvature. Is the curvature upward or downward on a m versus T plot? Explain your answer. Q8.2 Why do the temperature versus heat curves in the solid, liquid, and gas regions of Figure 8.7 have different slopes? Q8.3 Figure 8.7 is not drawn to scale. What would be the relative lengths on the qP axis of the liquid + solid, liquid, and liquid + gas segments for water if the drawing were to scale and the system consisted of H2O? Q8.4 Show the paths n S o S p S q and a S b S c S d S e S f of the P–V–T phase diagram of Figure 8.15 in the P–T phase diagram of Figure 8.4. Q8.5 At a given temperature, a liquid can coexist with its vapor at a single value of pressure. However, you can measure the presence of H2O1g2 above the surface of a lake, and H2O1g2 is still there if the barometric pressure rises or falls at constant temperature. How is this possible? Q8.6 Why are the triple point temperature and the normal freezing point very close in temperature for most substances? Q8.7 Give a molecularlevel explanation as to why the surface tension of Hg(l) is not zero. Q8.8 A vessel containing a liquid is opened inside an evacuated chamber. Will you see a liquid–gas interface if the volume of the initially evacuated chamber is (a) less than the
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critical volume, (b) much larger than the critical volume, and (c) slightly larger than the critical volume. Q8.9 Why are there no points in the phase diagram for sulfur in Figure 8.11 that show rhombic and monoclinic solid phases in equilibrium with liquid and gaseous sulfur? Q8.10 What is the physical origin of the pressure difference across a curved liquid–gas interface? Q8.11 A triple point refers to a point in a P–T phase diagram for which three phases are in equilibrium. Do all triple points correspond to a gas–liquid–solid equilibrium? Q8.12 Why does the liquid–gas coexistence curve in a P–T phase diagram end at the critical point? Q8.13 How can you get a P–T phase diagram from a P–V–T phase diagram? Q8.14 Why does the triple point in a P–T phase diagram become a triple line in a P–V phase diagram? Q8.15 Why does water have several different solid phases but only one liquid and one gaseous phase? Q8.16 As the pressure is increased at 45°C, ice I is converted to ice II. Which of these phases has the lower density? Q8.17 Why is ∆ subHm = ∆ fusHm + ∆ fusHm? Q8.18 What can you say about the density of liquid and gaseous water as the pressure approaches the critical value from lower (gas) and higher (liquid) values? Assume that the temperature is constant at the critical value.
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Numerical Problems
Q8.19 What can you say about ∆ vapH of a liquid as the temperature approaches the critical temperature? Q8.20 Is the following statement correct? Because dry ice sublimes, carbon dioxide has no liquid phase. Explain your answer. Q8.21 Make a drawing indicating the threestep process d of Figure 8.4 in Figure 8.13. Q8.22 Consider the phase diagram of Figure 8.4 and answer the following questions. a. Which phase is present if P is increased at constant T starting at the triple point?
231
b. Which phase is present if P is increased at constant T starting at the normal boiling point? c. Which phase is present if T is decreased at constant P starting at the critical point? d. Are any of your answers different if the substance under consideration is water? Q8.23 Use Trouton’s rule to predict the enthalpy of vaporization for H2O1l2, CCl41l2, CH3OH1l2, C3H81l2, and C6H61l2 using boiling temperatures from Table 8.2. Are your estimated enthalpies high or low? Explain the trends in your results.
NUMERICAL PROBLEMS Section 8.2 P8.1 Within what range can you restrict the values of P and T if the following information is known about CO2? Use Figure 8.12 to answer this question. a. As temperature is increased, the solid is first converted to liquid and subsequently to the gaseous state. b. As pressure on a cylinder containing pure CO2 is increased from 65 to 80. atm, no interface delineating liquid and gaseous phases is observed. c. Solid, liquid, and gas phases coexist at equilibrium. d. An increase in pressure from 10. to 50. atm converts the liquid to the solid. e. An increase in temperature from 80.° to 20.°C converts a solid to a gas with no intermediate liquid phase. P8.2 A P–T phase diagram for potassium is shown in the following diagram. POTASSIUM fcc (II)
12
8
Diamond
Liquid bcc (I)
4
Liquid
Pressure
Pressure, P (GPa)
16
P8.3 It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. At 1 atm pressure, ice melts at 273.15 K, ∆ fus H = 6010 J mol1, the density of ice is 920. kg m3, and the density of liquid water is 997 kg m3. a. What pressure is required to lower the melting temperature by 3.50°C? b. Assume that the width of the skate in contact with the ice has been reduced by sharpening to 18.0 * 103 cm and that the length of the contact area is 17.5 cm. If a skater of mass 78 kg is balanced on one skate, what pressure is exerted at the interface of the skate and the ice? c. What is the melting point of ice under this pressure? d. If the temperature of the ice is 3.50°C, do you expect melting of the ice at the iceskate interface to occur? P8.4 Answer the following questions using the P–T phase diagram for carbon sketched in the following diagram.
Graphite 0
Vapor 0
200 400 600 Temperature, T (K)
800
a. Which phase has a greater density, the fcc or the bcc phase? Explain your answer. b. Indicate the range of P and T in the phase diagram for which fcc and liquid potassium are in equilibrium. Does fcc potassium float on or sink in liquid potassium? Explain your answer. c. Extend P and T on this diagram and indicate where you might expect to find the vapor phase. Explain how you choose the slope of your liquid–vapor coexistence curve.
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Temperature
a. Which substance is denser, graphite or diamond? Explain your answer. b. Which phase is denser, graphite or liquid carbon? Explain your answer. c. Why does the phase diagram have two triple points? Explain your answer.
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Liquid Solid
Pressure
Pressure
P8.5 Are the two P–T phase diagrams shown in this problem likely to be observed for a pure substance? If not, explain all features of the diagram that will not be observed.
Vapor
Liquid
Solid
Vapor
Temperature
Temperature (a)
(b)
P8.6 The phase diagram of NH3 can be characterized by the following information. The normal melting and boiling temperatures are 195.2 and 239.82 K, respectively, whereas the triple point pressure and temperature are 6077 Pa and 195.41 K, respectively. The critical point parameters are 112.8 * 105 Pa and 405.5 K. Make a sketch of the P–T phase diagram (not necessarily to scale) for NH3. Place a point in the phase diagram for the following conditions. State which and how many phases are present. a. 195.41 K, 1050 Pa b. 195.41 K, 6077 Pa c. 237.51 K, 101325 Pa d. 420. K, 130 * 105 Pa e. 190. K, 6077 Pa P8.7 Prove that a substance for which the solid–liquid coexistence curve has a negative slope contracts upon melting. P8.8 Are the two P–T phase diagrams shown here likely to be observed for a pure substance? If not, explain all features of the diagram that will not be observed. Solid II
Vapor Solid
Pressure
Pressure
Liquid Solid I
Vapor
Temperature
Temperature (a)
Liquid
(b)
P8.9 Within what range can you restrict the values of P and/or T if the following information is known about sulfur? Use Figure 8.11 to answer this problem. a. Only the monoclinic solid phase is observed for P = 1 atm. b. When the pressure on the vapor is increased, the liquid phase is formed. c. Solid, liquid, and gas phases coexist at equilibrium. d. As temperature is increased, the rhombic solid phase is converted to the liquid directly. e. As temperature is increased at 1 atm, the monoclinic solid phase is converted to the liquid directly.
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P8.10 The normal melting point of H2O is 273.15 K, and ∆ fusH = 6010 J mol1. Calculate the change in the normal freezing point at 200. and 400. bar, assuming that the density of the liquid and solid phases remains constant at 997 and 917 kg m3, respectively. Explain why your answer is positive or negative. P8.11 The densities of a given solid and liquid of molar mass 122.5 g mol1 at its normal melting temperature of 432.5 K are 1075 and 1012 kg m3, respectively. If the pressure is increased to 150. bar, the melting temperature increases to 437.0 K. Calculate ∆ fusH and ∆ fusS for this substance. P8.12 In this problem, you will calculate the differences in the chemical potentials of ice and supercooled water and of steam and superheated water, all at 1 atm pressure, shown schematically for an arbitrary substance in Figure 8.1. For this problem, S ~ 1H2O, s2 = 48.0 J mol1 K1, S ~ 1H2O, l2 = 70.0 J mol1 K1, and S ~ 1H2O, g2 = 188.8 J mol1 K1. a. By what amount does the chemical potential of water exceed that of ice at 1.95°C? b. By what amount does the chemical potential of water exceed that of steam at 101.95°C? c. Which is the most stable phase in parts (a) and (b)? P8.13 24.5 g of water is in a container of 30.0 L at 298.15 K. The vapor pressure of water at this temperature is 23.76 Torr. a. What phases are present? b. At what volume would only the gas phase be present? c. At what volume would only the liquid phase be present? P8.14 Consider the transition between two forms of solid tin, Sn1s,gray2 N Sn1s,white2. The two phases are in equilibrium at 1 bar and 18°C. The densities for gray and white tin are 5750 and 7280 kg m3, respectively, and the molar entropies for gray and white tin are 44.14 and 51.18 J K1 mol1, respectively. Calculate the temperature at which the two phases are in equilibrium at 425 bar. Section 8.6 P8.15 Use the vapor pressures of CCl4(l) given in the following table to calculate the enthalpy of vaporization using a graphical method or a least squares fitting routine. T 1K2 280. 290. 300. 310.
P1 Pa2 6440. 10540 16580 25190
T 1K2 320. 330. 340
P1 Pa2 37130. 53250. 74520.
P8.16 The vapor pressure of benzene(l) is given by ln a
P 2.7738 * 103 b = 20.767 Pa T  53.08 K
a. Calculate the standard boiling temperature. b. Calculate ∆ vap H at 298 K and at the standard boiling temperature.
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Numerical Problems
P8.17 Use the vapor pressures for tetrachloromethane given in the following table to estimate the temperature and pressure of the triple point as well as the enthalpies of fusion, vaporization, and sublimation. T 1K2
Phase
T 1K2
Phase
P1 Pa2
Solid 160 0.0185 Solid 170 0.204 Liquid 275 667 Liquid 305 2690 P8.20 Autoclaves that are used to sterilize surgical tools require a temperature of 135°C to kill some bacteria. If water is used for this purpose, at what pressure must the autoclave operate? P8.21 You have collected a tissue specimen that you would like to preserve by freeze drying. To ensure the integrity of the specimen, the temperature should not exceed 6.10°C. The vapor pressure of ice at 273.16 K is 624 Pa; ∆ fusH = 6.008 kJ mol1 and ∆ vap H = 40.65 kJ mol1. What is the maximum pressure at which the freeze drying can be carried out? P8.22 Use the vapor pressures of ice given here to calculate the enthalpy of sublimation using a graphical method or a least squares fitting routine. T 1K2
P1 Torr2
200. 0.1676 210. 0.7233 220. 2.732 230. 9.195 240. 27.97 250. 77.82 260. 200.2 P8.23 The vapor pressure of a liquid can be written in the empirical form known as the Antoine equation, where A112, A122, and A132 are constants determined from measurements: P1T2 A122 ln = A112 Pa T + A132 K Starting with this equation, derive an equation giving ∆ vap H as a function of temperature.
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P8.24 Use the following vapor pressures of propane to calculate the enthalpy of vaporization using a graphical method or a least squares fitting routine. T 1K2
P1 Pa2
Solid 230. 225.7 Solid 250. 905 Liquid 280. 6440 Liquid 340. 62501 P8.18 You have a compound dissolved in chloroform and need to remove the solvent by distillation. Because the compound is heat sensitive, you hesitate to raise the temperature above 7.50°C and decide on vacuum distillation. What is the maximum pressure at which the distillation takes place? P8.19 Use the vapor pressures for hexane given in the following table to estimate the temperature and pressure of the triple point as well as the enthalpies of fusion, vaporization, and sublimation.
233
100. 120. 140. 160. 180. 200. 220.
P1 Pa2 0.01114 2.317 73.91 838.0 5054 2.016 * 104 6.046 * 104
P8.25 The vapor pressure of liquid benzene is 19846 Pa at 308.22 K and ∆ vap H = 30.72 kJ mol1. Calculate the normal and standard boiling points. Does your result for the normal boiling point agree with that in Table 8.3? If not, suggest a possible cause. P8.26 Benzene(l) has a vapor pressure of 0.48423 bar at 331.06 K and an enthalpy of vaporization of 30.72 kJ mol1. Assume that CP,m of the vapor and liquid phases at that temperature have their 298.15 K values of 82.4 and 136.0 J K1 mol1, respectively. Calculate the vapor pressure of C6H61l2 at 349.0 K assuming that a. the enthalpy of vaporization does not change with temperature. b. the enthalpy of vaporization at temperature T can be calculated from the equation ∆ subH1T2 = ∆ subH1T02 + ∆CP1T  T02, assuming that ∆CP does not change with temperature. c. the experimentally determined value is 0.88717 bar. If your calculated values differ from this value, provide a possible explanation. P8.27 Use the values for ∆ fG ~ 1H2O, g2 and ∆ fG ~ 1H2O, l2 from Appendix B to calculate the vapor pressure of H2O at 298.15 K. P8.28 The vapor pressure of an unknown solid is approximately given by ln1P>Torr2 = 22.6  2200. 1K>T2, and the vapor pressure of the liquid phase of the same substance is approximately given by ln1P>Torr2 = 18.4  1700. 1K>T2. a. Calculate ∆ vap H and ∆ sub H. b. Calculate ∆ fus H. c. Calculate the triple point temperature and pressure. P8.29 For water, ∆ vap H is 40.65 kJ mol1, and the normal boiling point is 373.15 K. Calculate the boiling point for water at the summit of Mount Kilimanjaro (elevation 5895 m), where the normal barometric pressure is 51530 Pa. P8.30 Use the vapor pressures of SO21l2 given in the following table to calculate the enthalpy of vaporization using a graphical method or a least squares fitting routine. T 1K2 190. 200. 210. 220.
P1 Pa2 824.1 2050. 4591 9421
T 1K2 230. 240. 250. 260.
P1 Pa2 17950 32110 54410 87930
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P8.31 The vapor pressure of ethanol(l) is 7.87 * 103 Pa at 298.15 K. Use this value to calculate ∆ f G ~ 1CH3OH, g2 ∆ f G ~ 1CH3OH, l2. Compare your result with those in Table 4.2. P8.32 Carbon tetrachloride melts at 250. K. The vapor pressure of the liquid is 10,539 Pa at 290. K and 74,518 Pa at 340. K. The vapor pressure of the solid is 270. Pa at 232 K and 1092 Pa at 250. K. a. Calculate ∆ vap H and ∆ sub H. b. Calculate ∆ fusH. c. Calculate the normal boiling point and ∆ vap S at the boiling point. d. Calculate the triple point pressure and temperature. P8.33 In Equation (8.16), 1dP>dT2vap was calculated by gas assuming that V m W V liquid . In this problem, you will test m the validity of this approximation. For water at its normal boiling point of 373.13 K, ∆ vap H = 40.65 * 103 J mol1, rliquid = 958.66 kg m3, and rgas = 0.58958 kg m3. Compare the calculated values for 1dP>dT2vap with and without the approximation of Equation (8.16). What is the relative error in making the approximation? P8.34 The variations of the vapor pressure of the liquid and solid forms of a pure substance near the triple Psolid K point are given by ln = 8500. + 33.75 and Pa T Pliquid K ln = 4000. + 20.00. Calculate the temperature Pa T and pressure at the triple point. P8.35 Use the vapor pressures of nbutane given in the following table to calculate the enthalpy of vaporization using a graphical method or a least squares fitting routine. T 1 °C 2
134.3 121 104
P1 Pa2 1.00 10.00 100.00
T 1 °C 2
81.1 49.1 0.800
P1 Pa2 1000. 10000. 100000.
P8.36 At 298.15 K, ∆ f G ~ 1CH3OH, g2 = 162.3 kJ mol1 and ∆ f G ~ 1CH3OH, l2 = 166.6 kJ mol1. Calculate the vapor pressure of methanol at this temperature. P8.37 Solid iodine, I21s2, at 25.0°C has an enthalpy of sublimation of 56.30 kJ mol1. The CP, m of the vapor and solid phases at that temperature are 36.9 and 54.4 J K1 mol1, respectively. The sublimation pressure at 25.00°C is 0.30844 Torr. Calculate the sublimation pressure of the solid at the melting point 1113.6°C2 assuming that a. the enthalpy of sublimation and the heat capacities do not change with temperature. b. the enthalpy of sublimation at temperature T can be calculated from the equation ∆ sub H1T2 ∆ sub H1T02 + ∆CP1T  T02 P8.38 A reasonable approximation to the vapor pressure of krypton is given by log101P>Torr2 = b  0.052231a>T2
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For solid krypton, a = 10065 and b = 7.1770. For liquid krypton, a = 9377.0 and b = 6.92387. Use this formula and these constants to estimate the triple point temperature and pressure as well as the enthalpies of vaporization, fusion, and sublimation of krypton. Section 8.7 P8.39 Calculate the vapor pressure of CH3OH1l2 at 310. K if He is added to the gas phase at a partial pressure of 200. bar. By what factor does the vapor pressure change? (Hint: You need to calculate the vapor pressure of methanol at this temperature and 1 bar using the data in Table 8.3.) P8.40 32.5 g of H2O1s2 at 260. K are transformed to H2O1l2 with an increase in entropy of 47.0 J K1. Use this information and the heat capacities in Table 2.4 to calculate ∆ fusH for water. P8.41 At temperatures below the triple point, H2O1s2 can sublime to form H2O1g2. Calculate ∆S if 24.0 g of H2O1s2 sublimes at 273 K and is subsequently heated to 298 K. Assume that the heat capacity of H2O1g2 is constant at its 298 K value over the temperature interval of interest. What fraction of ∆S comes from heating the gas? (Hint: Use Equation (8.4) and an analogy of Equation (5.16) to calculate the entropy of sublimation.) Section 8.8 P8.42 In Section 8.8, it is stated that the maximum height of a water column in which cavitation does not occur is ∼9.7 m. Show that this is the case at 298 K. P8.43 A cell is roughly spherical with a radius of 15.0 * 106 m. If the cell grows to double its volume, calculate the magnitude of the work required to expand the cell surface. Assume the cell is surrounded by pure water and that T = 298.15 K. P8.44 Calculate the vapor pressure for a mist of spherical water droplets of radius (a) 1.70 * 108 m and (b) 2.00 * 106 m surrounded by water vapor at 298 K. The vapor pressure of water at this temperature is 25.2 Torr. P8.45 Calculate the vapor pressure of water droplets of radius 1.65 * 108 m at 360. K in equilibrium with water vapor. Use the tabulated value of the density and the surface tension at 298 K from Appendix B for this problem. (Hint: You need to calculate the vapor pressure of water at this temperature using the data in Table 8.3.) P8.46 Calculate the difference in pressure across the liquid– air interface for a droplet of radius 100. nm for (a) mercury and (b) methanol. P8.47 Calculate the vapor pressure of a droplet of benzene of radius 1.00 * 108 m at 15.0°C in equilibrium with its vapor. Use the tabulated value of the density and the surface tension at 298 K from Appendix B for this problem. (Hint: You need to calculate the vapor pressure of benzene at this temperature using the data in Table 8.3.)
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Further Reading
FURTHER READING Alper, J. S. “The Phase Rule Revisited: Interrelationships between Components and Phases.” Journal of Chemical Education 76 (1999): 1567–1569. Brooks, S. D., Gonzales, M., and Farias, R. “Using Surface Tension Measurements to Understand How Pollution Can Influence Cloud Formation, Fog, and Precipitation.” Journal of Chemical Education 86 (2009): 838–841.
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Jensen, W. B. “Generalizing the Phase Rule.” Journal of Chemical Education 78 (2001): 1369–1370. Lin, F., Liu, D., Das, S., Prempeh, N., Hua, Y., and Lu, J. “Recent Progress in Heavy Metal Extraction by Supercritical CO2 Fluids.” Industrial and Chemical Engineering Research 53 (2014): 1866–1877. Prasad, R. “On Capillary Rise and Nucleation.” Journal of Chemical Education 85 (2008): 1389–1391.
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C H A P T E R
9
Ideal and Real Solutions
9.1 Defining the Ideal Solution 9.2 The Chemical Potential of a Component in the Gas and Solution Phases
9.3 Applying the Ideal Solution Model to Binary Solutions
WHY is this material important? Solutions are commonly encountered in chemistry. Because the components of a solution interact strongly, there is no equivalent to the ideal gas equation of state for liquid solutions. The ideal solution model and ideal dilute solution model discussed in this chapter allow calculations to be carried out for real solutions.
9.4 The Temperature–Composition
WHAT are the most important concepts and results?
9.7 Freezing Point Depression
For ideal solutions, the partial pressure of a component in the vapor phase above a solution is simply related to the mole fraction in the liquid phase by Raoult’s law. For solutions characterized by the ideal dilute solution model, the solvent is best described by Raoult’s law and the solute is best described by Henry’s law. Osmotic pressure, freezing point depression, and boiling point elevation can all be described in terms of the ideal and ideal dilute solution models.
9.5 The Gibbs–Duhem Equation 9.6 Colligative Properties and Boiling Point Elevation
9.8 Osmotic Pressure 9.9 Deviations from Raoult’s Law in Real Solutions
9.10 The Ideal Dilute Solution 9.11 Activities Are Defined with Respect to Standard States
9.12 Henry’s Law and the Solubility
WHAT would be helpful for you to review for this chapter?
of Gases in a Solvent
It would be helpful to review chemical equilibrium discussed in Chapter 6.
9.13 Chemical Equilibrium in Solutions
9.1 DEFINING THE IDEAL SOLUTION
9.14 Solutions Formed from Partially Miscible Liquids
In an ideal gas, the atoms or molecules do not interact with one another. Clearly, this is not a good model for a liquid, in which the atoms and molecules clearly do interact. In fact, without attractive interactions, a liquid would transform into a gas. The attractive interaction in liquids varies greatly, as shown by the large variation in boiling points among the elements; for instance, helium has a normal boiling point of 4.2 K, whereas hafnium has a boiling point of 5400 K. In developing a model for solutions, the vapor phase in equilibrium with the solution must be taken into account. Consider pure liquid benzene in a beaker placed in a closed room. Because the liquid is in equilibrium with the vapor phase above it, there is a nonzero partial pressure of benzene in the air surrounding the beaker. This pressure is called the vapor pressure of benzene at the temperature of the liquid. What happens when toluene is added to the beaker? It is observed that the partial pressure of benzene is reduced, and the vapor phase now contains both benzene and toluene. For this particular mixture, the partial pressure of each component (i) above the liquid is given by
Diagram and Fractional Distillation
Pi = xiP *i i = 1, 2
(9.1)
9.15 Solid–Solution Equilibrium
Concept The ideal solution is defined by Raoult’s law.
P *i
where xi is the mole fraction of that component in the liquid and is the vapor pressure of the pure component. This equation states that the partial pressure of each of the two components is directly proportional to P *i and that the proportionality constant is xi. Equation (9.1) is known as Raoult’s law and is the definition of an ideal solution. 237
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CHAPTER 9 Ideal and Real Solutions
Solvent
Solute
Figure 9.1
Schematic model of a solution. The white and black spheres represent solvent and solute molecules, respectively.
Raoult’s law holds for each substance in an ideal solution over the range from 0 … xi … 1. In binary solutions, one refers to the component that has the higher value of xi as the solvent and the component that has the lower value of xi as the solute. Few solutions satisfy Raoult’s law. However, it is useful to study the thermodynamics of ideal solutions and to introduce departures from ideal behavior later. Why is Raoult’s law not generally obeyed over the whole concentration range of a binary solution consisting of molecules A and B? Equation (9.1) is only obeyed if the A–A, B–B, and A–B interactions are all equally strong. This criterion is satisfied for a mixture of benzene and toluene because the two molecules are similar in size, shape, and chemical properties. However, it is not satisfied for arbitrary molecules. Raoult’s law is an example of a limiting law; the solvent in a real solution obeys Raoult’s law as the solution becomes highly dilute. Raoult’s law is derived in Example Problem 9.1 and can be rationalized using the model depicted in Figure 9.1. In the solution, molecules of solute are distributed in the solvent. The solution is in equilibrium with the gas phase, and the gasphase composition is determined by a dynamic balance between evaporation from the solution and condensation from the gas phase, as indicated for one solvent and one solute molecule in Figure 9.1.
EXAMPLE PROBLEM 9.1 Assume that the rates of evaporation Revap and condensation Rcond of the solvent from the surface of pure liquid solvent are given by the expressions Revap = Akevap Rcond = Akcond P *solvent where A is the surface area of the liquid and kevap and kcond are the rate constants for evaporation and condensation, respectively. Derive a relationship between the vapor pressure of the solvent above the pure solvent and above a solution.
Solution For the pure solvent, the equilibrium vapor pressure is found by setting the rates of evaporation and condensation equal: Revap = Rcond Akevap = Akcond P *solvent P *solvent =
kevap kcond
The superscript * is used to indicate a pure substance. Next, consider the ideal solution. In this case, the rate of evaporation is reduced by the factor xsolvent because only that fraction of the surface is available for evaporation of the solvent. Revap = Akevap xsolvent Rcond = Akcond Psolvent and at equilibrium Revap = Rcond Akevap xsolvent = Akcond Psolvent Psolvent = xsolvent
kevap kcond
= xsolvent P *solvent
The derived relationship is Raoult’s law.
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9.2 The Chemical Potential of a Component in the Gas and Solution Phases
239
9.2 THE CHEMICAL POTENTIAL OF A COMPONENT IN THE GAS AND SOLUTION PHASES
If the liquid and vapor phases are in equilibrium, the following equation holds for each component of the solution, where mi is the chemical potential of species i: msolution = mivapor i
(9.2)
Recall from Section 6.3 that the chemical potential of a substance in the gas phase is related to its partial pressure Pi by Pi mivapor = mi~ + RT ln ~ (9.3) P where mi~ is the chemical potential of pure component i in the gas phase at the standard state pressure P ~ = 1 bar. Because at equilibrium msolution = mivapor = mi, i Equation (9.3) can be written in the form mi = mi~ + RT ln
Pi P~
(9.4)
For pure liquid i in equilibrium with its vapor, m*i 1liquid2 = m*i 1vapor2 = m*i . Therefore, the chemical potential of the pure liquid is given by m*i = mi~ + RT ln
P *i P~
(9.5)
Subtracting Equation (9.5) from (9.4) gives mi = m*i + RT ln
Pi P *i
(9.6)
For an ideal solution, Pi = xiP *i . Combining Equations (9.6) and (9.1), the central equation describing ideal solutions is obtained: mi = m*i + RT ln xi
(9.7)
This equation relates the chemical potential of a component in an ideal solution to the chemical potential of the pure liquid form of component i and the mole fraction of that component in the solution. This equation is most useful in describing the thermodynamics of solutions in which all components are volatile and miscible in all proportions. Keeping in mind that mi = ∆ f Gi, the form of Equation (9.7) is identical to that derived for the Gibbs energy of a mixture of gases in Section 6.5. Therefore, one can derive relations for the thermodynamics of mixing to form ideal solutions as was done for ideal gases in Section 6.5. These relations are shown in Equation (9.8). Note in particular that ∆ mixH = ∆ mixV = 0 for an ideal solution:
Concept In an ideal solution, the chemical potential of a component is determined by the chemical potential of the pure component and the mole fraction of the component.
∆ mixG = nRT a xi ln xi i
∆ mixS = a ∆ mixV = a
0∆ mixG b = nR a xi ln xi 0T P, n1, n2 i 0∆ mixG b = 0 and 0P T, n1, n2
∆ mixH = ∆ mixG + T∆ mixS
= nRT a xi ln xi  T anR a xi ln xi b = 0 i
i
(9.8)
A calculation for ∆ mixG and ∆ mixS is shown in Example Problem 9.2.
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EXAMPLE PROBLEM 9.2 An ideal solution is made from 5.00 mol of benzene and 3.25 mol of toluene. Calculate ∆ mixG and ∆ mixS at 298 K and 1 bar pressure. Is mixing a spontaneous process?
Solution The mole fractions of the components in the solution are xbenzene = 0.606 and xtoluene = 0.394. ∆ mixG = nRT a xi ln xi i
= 8.25 mol * 8.314 J mol1 K1 * 298 K * 10.606 ln 0.606 + 0.394 ln 0.3942
= 13.7 * 103 J
∆ mixS = nR a xi ln xi i
= 8.25 mol * 8.314 J mol1 K1 * 10.606 ln 0.606 + 0.394 ln 0.3942
= 46.0 J K1
Mixing is spontaneous because ∆ mixG 6 0 for an ideal solution. If two liquids are miscible, it is always true that ∆ mixG 6 0.
9.3 APPLYING THE IDEAL SOLUTION MODEL TO BINARY SOLUTIONS
Although the ideal solution model can be applied to any number of components, to simplify the mathematics, the focus in this chapter is on binary solutions, which consist of only two components. Because Raoult’s law holds for both components of the mixture, P1 = x1P *1 and P2 = x2P *2 = 11  x12P *2. The total pressure in the gas phase varies linearly with the mole fraction of each of its components in the liquid: Ptot = P1 + P2 = x1P *1 + 11  x12P *2 = P *2 + 1P *1  P *22x1
(9.9)
The individual partial pressures as well as Ptot above a benzene–1,2 dichloroethane (C2H4Cl22 solution are shown in Figure 9.2. Small deviations from Raoult’s law are seen. Such deviations are typical because few solutions obey Raoult’s law exactly. 250 Total gas pressure 200
Comparison of gas pressures calculated from Raoult’s law with experimental results for a binary solution. The vapor pressure of benzene (blue), 1,2 dichloroethane (red), and the total pressure (purple) above the solution is shown as a function of the mole fraction of 1,2 dichloroethane. The triangles, squares, and circles are experimental data points [Zawidski, J. V. Zeitschrift für Physikalische Chemie 35 (1900): 129]. The solid curves are polynomial fits to the data. The dashed lines are calculated using Raoult’s law.
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Pressure, P (Torr)
Figure 9.2
Benzene 150
Dichloroethane
100
50
0
0.2
0.4 0.6 Mole fraction of 1,2 dichloroethane
0.8
1
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9.3 Applying the Ideal Solution Model to Binary Solutions
Nonideal solutions, which generally exhibit large deviations from Raoult’s law, are discussed in Section 9.9. The concentration unit used in Equation (9.9) is the mole fraction of each component in the liquid phase. The mole fraction of each component in the gas phase can also be calculated. Using the definition of the partial pressure and the symbols y1 and y2 to denote the gasphase mole fractions, we can write
y1 =
P1 = * Ptot P2 +
x1P *1 (P *1 
P *22x1
(9.10)
To obtain the pressure in the vapor phase as a function of y1, we first solve Equation (9.10) for x1:
x1 =
y1P *2 P *1 + (P *2  P *12y1
Ptot =
P *1P *2
P *1
+
(P *2

P *12y1
80
(9.12)
P *1Ptot
 P *1P *2 Ptot(P *1  P *22
(9.13)
The variation of the total pressure with x1 and y1 is not the same, as is seen in Figure 9.3. In Figure 9.3a, the system consists of a singlephase liquid for pressures above the curve and of a twophase vapor–liquid mixture for points lying on the curve in the P  x1 diagram. Only points lying on or above the curve are meaningful because points lying below the curve do not correspond to equilibrium states at which liquid is present. In Figure 9.3b, the system consists of a singlephase vapor for pressures below the curve and of a twophase vapor–liquid mixture for points lying on the curve in the P  y1 diagram. Points lying above the curve do not correspond to equilibrium states at which vapor is present. The excess vapor would condense to form liquid. Note that pressure is plotted as a function of different variables in the two parts of Figure 9.3. To compare the gas phase and liquid composition at a given total pressure, both are graphed as a function of the average composition of benzene, denoted by Zbenzene, in the whole system in Figure 9.4. The average composition Z is defined by Zbenzene =
vapor nliquid benzene + nbenzene vapor liquid vapor nliquid toluene + ntoluene + nbenzene + nbenzene
nbenzene = ntotal
In the region labeled Liquid in Figure 9.4, the system consists entirely of a liquid phase and Zbenzene = xbenzene. In the region labeled Vapor, the system consists entirely of a gaseous phase and Zbenzene = ybenzene. The area separating the singlephase liquid and vapor regions corresponds to the twophase liquid–vapor coexistence region. We next apply the phase rule to the benzene–toluene ideal solution for which there are two components (benzene and toluene) and three independent state variables, namely, P, T, and Zbenzene (or Ztoluene). Because Zbenzene = 1  Ztoluene, these two variables are not independent. For a system containing C components, the phase rule discussed in Section 8.3 takes the form
F = C  p + 2
40
0
Liquid + vapor
0.2
0.4 0.6 xbenzene
0.8
1
(a)
100
Pressure, P (Torr)
y1 =
Liquid
20
Equation (9.12) can be rearranged to give an equation for y1 in terms of the vapor pressures of the pure components and the total pressure:
60
(9.11)
and obtain Ptot from Ptot = P *2 + (P *1  P *22x1
100
Pressure, P (Torr)
80
Liquid + vapor
60 40 Vapor 20
0
0.2
0.4 0.6 ybenzene
0.8
1
(b)
Figure 9.3
Total pressure above a solution as a function of gas and liquid composition. For a benzene–toluene ideal solution, the total gasphase pressure is shown for different values of (a) the mole fraction of benzene in the solution and (b) the mole fraction of benzene in the vapor. Points on the curves correspond to vapor–liquid coexistence. Only the curves and the darkly shaded areas are of physical significance, as explained in the text.
(9.14)
For the benzene–toluene ideal solution, C = 2 and F = 4  p. Therefore, in the singlephase regions labeled Liquid and Vapor in Figure 9.4, the system has three degrees of freedom. One of these degrees of freedom is already fixed because the temperature is held constant in Figure 9.4. Therefore, the two remaining degrees of freedom are P and Zbenzene. In the twophase region labeled Liquid + Vapor, the system has only one degree of freedom. Once P is fixed, the Z values at the ends of the horizontal line determine xbenzene and ybenzene and therefore Zbenzene. We note that the number of components
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CHAPTER 9 Ideal and Real Solutions
Figure 9.4 100
a Liquid
80 Pressure, P (Torr)
A pressure–composition phase diagram for a benzene–toluene ideal solution. Z is the average composition of a component of interest, in this case benzene. The upper curve shows the vapor pressure as a function of xbenzene. The lower curve shows the vapor pressure as a function of ybenzene. Above the two curves, the system is totally in the liquid phase, whereas below the two curves, the system is totally in the vapor phase. The elongateshaped area between the two curves is the liquid– vapor coexistence region. The horizontal lines connecting the curves are tie lines.
60
b
Vapor pressure as a function of xbenzene
c Liquid + vapor
40
Vapor
d 20
Vapor pressure as a function of ybenzene
0
0.2
0.4
0.6
0.8
1
Zbenzene
Concept The vapor phase above a solution is enriched in the most volatile component relative to the solution.
can be less than the number of chemical species. For example, a solution containing the three species A, B, and C linked by the chemical equilibrium A + B N C contains only two independent components because only two of the three concentrations are independent variables. To demonstrate the usefulness of this pressure–average composition (P–Z) diagram, consider a constant temperature process in which the pressure of the system is decreased from the value corresponding to point a in Figure 9.4. Because P 7 Ptot, initially the system is entirely in the liquid phase. As the pressure is decreased at constant composition, the system remains entirely in the liquid phase until the constant composition line intersects the P versus xbenzene curve. At this point, the system enters the twophase vapor–liquid coexistence region. As point b is reached, what are the values for xbenzene and ybenzene, the mole fractions of benzene in the liquid and vapor phases? These values can be determined by constructing a tie line in the twophase coexistence region. The values for xbenzene and ybenzene are given by the Z values at the ends of the tie line. A tie line graphically connects two coexisting phases on a phase diagram. In Figure 9.4, tie lines are horizontal lines (two are shown) at the pressure of interest that connects the P versus xbenzene and P versus ybenzene curves. Only the values at the ends of the tie line have a physical meaning. Note that for all values of the pressure, ybenzene is greater than xbenzene, showing that the vapor phase is always enriched in the more volatile or higher vapor pressure component in comparison with the liquid phase. Calculations illustrating these concepts are shown in Example Problems 9.3 and 9.4.
EXAMPLE PROBLEM 9.3 An ideal solution of 5.00 mol of benzene and 3.25 mol of toluene is placed in a piston and cylinder assembly. At 298 K, the vapor pressure of the pure substances are P *benzene = 96.4 Torr and P *toluene = 28.9 Torr. a. The pressure above this solution is reduced from 760. Torr. At what pressure does the vapor phase first appear? b. What is the composition of the vapor under these conditions?
Solution a. The mole fractions of the components in the solution are xbenzene = 0.606 and xtoluene = 0.394. The vapor pressure above this solution is Ptot = xbenzene P *benzene + xtoluene P *toluene = 0.606 * 96.4 Torr + 0.394 * 28.9 Torr = 69.8 Torr No vapor will be formed until the pressure has been reduced to this value.
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9.3 Applying the Ideal Solution Model to Binary Solutions
243
b. The composition of the vapor at a total pressure of 69.8 Torr is given by ybenzene =
Pbenzene Ptot
=
xbenzeneP *benzene Ptot
=
0.606 * 96.4 Torr = 0.837 69.8 Torr
Liquid
l
ytoluene = 1  ybenzene = 0.163 Note that the vapor is enriched relative to the liquid in the more volatile component, which has the higher vapor pressure and lower boiling temperature.
To calculate the relative amount of material in each of the two phases in a coexistence region, we will derive the lever rule for a binary solution of the components A and B. Figure 9.5 shows a magnified portion of Figure 9.4 centered at the tie line that passes through point b. We derive the lever rule using the following geometrical argument. The lengths of the line segments lb and bn are given by
lb = ZB  xB =
ntot B tot
(9.15)
bv = yB  ZB =
nvapor ntot B B  tot tot nvapor n
(9.16)
n

nliq B ntot liq
n b Liquid + vapor
Vapor
Figure 9.5
An enlarged portion of the twophase coexistence region of Figure 9.4. The vertical line through point b indicates a constant system composition. The lever rule (see text) is used to determine the fraction of the liquid and vapor phases present in the system.
The superscripts on n indicate whether we are referring to the moles of component B in the vapor or liquid phases or the total number of moles of B in the system. If Equatot tion (9.15) is multiplied by ntot liq, Equation (9.16) is multiplied by nvapor, and after the modified Equation (9.16) is subtracted from the modified Equation (9.15), we find that tot lb ntot liq  bv nvapor =
We conclude that
ntot B liq vapor tot 1ntot + ntot 2 = ntot vapor2  1nB + nB B  nB = 0 ntot liq ntot liq
ntot vap
=
bv lb
(9.17)
It is convenient to restate this result as tot ntot liq1ZB  xB2 = nvapor1yB  ZB2
(9.18)
Concept The lever rule allows the proportions of a component between different phases to be calculated.
Equation (9.18) is called the lever rule by analogy with the torques acting on a lever of length lb + bv with the fulcrum positioned at point b. For the specific case shown tot in Figure 9.5, ntot liq >nvapor = 2.36. Therefore, 70.1% of the total number of moles in the system is in the liquid phase, and 29.9% is in the vapor phase.
EXAMPLE PROBLEM 9.4 For the benzene–toluene solution of Example Problem 9.3, calculate a. the total pressure b. the liquid composition c. the vapor composition when 1.50 mol of the solution has been converted to vapor.
Solution The lever rule relates the average composition, Zbenzene = 0.606, and the liquid and vapor compositions: tot ntot vapor 1ybenzene  Zbenzene2 = nliq 1Zbenzene  xbenzene2
Entering the parameters of the problem, this equation simplifies to 6.75xbenzene + 1.50ybenzene = 5.00
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CHAPTER 9 Ideal and Real Solutions
The total pressure is given by Ptot = xbenzene P *benzene + 11  xbenzene2P *toluene
= 3 96.4xbenzene + 28.911  xbenzene24 Torr
and the vapor composition is given by
ybenzene =
P *benzenePtotal  P *benzeneP *toluene Ptotal(P *benzene  P *toluene2
Ptot  2786 Torr ¥ Ptotal 67.5 Torr
96.4 = ≥
These three equations in three unknowns can be solved by using an equation solver or by eliminating the variables by combining equations. For example, the first equation can be used to express ybenzene in terms of xbenzene. This result can be substituted into the second and third equations to give two equations in terms of xbenzene and Ptot. The solution for xbenzene obtained from these two equations can be substituted in the first equation to give ybenzene. The relevant answers are xbenzene = 0.561, ybenzene = 0.810, and Ptot = 66.8 Torr.
9.4 THE TEMPERATURE–COMPOSITION
DIAGRAM AND FRACTIONAL DISTILLATION
The enrichment of the vapor phase above a solution in the more volatile component is the basis for fractional distillation, an important separation technique in chemistry and in the chemical industry. It is more convenient to discuss fractional distillation using a temperature–composition diagram than using the pressure–composition diagram discussed in the previous section. The temperature–composition diagram gives the temperature of the solution as a function of the average system composition for a predetermined total vapor pressure, Ptot. It is convenient to use the value Ptot = 1 atm so the vertical axis is the normal boiling point of the solution. Figure 9.6 shows a boiling temperature–composition diagram for a benzene–toluene solution. Neither the Tb versus xbenzene nor the Tb versus ybenzene curves are linear in a temperature–composition diagram. Note that because the more volatile component has the lower boiling point, the vapor and liquid regions are inverted when compared with the pressure–composition diagram.
Figure 9.6
Temperature–composition diagram for the benzene–toluene system. The boiling temperature for an ideal solution of benzene and toluene is plotted against the average system composition, Zbenzene. The upper curve shows the boiling temperature as a function of ybenzene, and the lower curve shows the boiling temperature as a function of xbenzene. The area intermediate between the two curves shows the vapor–liquid coexistence region.
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Boiling Temperature, T (K)
380
b
a
Vapor
375
c
d
Boiling temperature as a function of ybenzene
370 Liquid
365
Boiling temperature as a function of xbenzene
360
f
e
h
g i
355
j k
0
0.2
0.4
0.6 Zbenzene
0.8
1
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9.4 The Temperature–Composition Diagram and Fractional Distillation
The principle of fractional distillation can be illustrated using the sequence of lines labeled a through k in Figure 9.6. The vapor above the solution at point a is enriched in benzene and has a composition given by point b. If this vapor is separated from the original solution and condensed by lowering the temperature, the resulting liquid, indicated by point c, will have a composition whose mole fraction of benzene is greater than the original solution, as indicated by point a. As for the original solution, the vapor above this separately collected liquid is enriched in benzene. As this process is repeated, the successively collected vapor samples become more enriched in benzene. In the limit of a very large number of steps, the last condensed samples are essentially pure benzene. The multistep procedure described earlier is very cumbersome because it requires the collection and evaporation of many different samples. In practice, the separation into pure benzene and toluene is accomplished using a distillation column, shown schematically in Figure 9.7. Rather than have the whole system at a uniform temperature, a distillation column operates with a temperature gradient so that the top part of the column is at a lower temperature than the boiling solution. For such a distillation column, the horizontal segments ab, cd, ef, and so on in Figure 9.6 correspond to successively higher (cooler) portions of the column. Only the most volatile component of the solution initially passes through the column and is collected as a liquid. As the most volatile component is depleted from the solution, the boiling temperature increases, and the next most volatile component can be separated and collected, and so on. Distillation plays a major role in the separation of crude oil into its various components. The lowest boiling liquid is gasoline, followed by kerosene, diesel fuel and heating oil, and gas oil, which is further broken down into lowermolecularweight components by catalytic cracking. It is important in distillation to keep the temperature of the boiling liquid as low as possible to avoid the occurrence of thermally induced reactions. A useful way to reduce the boiling temperature is to distill the liquid under a partial vacuum, which is done by pumping away the gas phase. The boiling point of a typical liquid mixture can be reduced by approximately 100°C if the pressure is reduced from 760 to 20 Torr. Although the principle of fractional distillation is the same for real solutions, it is not possible to separate a binary solution into its pure components if the nonideality is strong enough. If the A–B interactions are more attractive than the A–A and B–B interactions, the boiling point of the solution will go through a maximum at a concentration intermediate between xA = 0 and xA = 1. An example of such a case is an acetone–chloroform mixture. The hydrogen on the chloroform forms a hydrogen bond with the oxygen in acetone, as shown in Figure 9.8, leading to stronger A–B than A–A and B–B interactions. A boiling point diagram for this case is shown in Figure 9.9a. At the maximum boiling temperature, the liquid and vapor composition lines coincide and are tangent to one another. Fractional distillation of a solution with a maximum boiling temperature is shown schematically in Figure 9.9a. Beginning with an initial value of xA greater than that corresponding to the maximum boiling point, the component with the lowest boiling point will initially emerge at the top of the distillation column. In this case, it will be pure component A. However, the liquid left in the heated flask will not be pure component B, but rather the solution corresponding to the concentration at which the maximum boiling point is reached. Continued boiling of the solution at this composition will lead to evaporation at constant composition. Such a mixture is called an azeotrope, and because Tb, azeotrope 7 Tb, A, Tb, B, it is called a maximum boiling azeotrope. An example for a maximum boiling azeotrope is a mixture of H2O 1Tb = 100°C2 and HCOOH 1Tb = 100°C2 at xH2O = 0.427, which boils at 107.2°C. Other commonly occurring azeotropic mixtures are listed in Table 9.1. If the A–B interactions are less attractive than the A–A and B–B interactions, a minimum boiling azeotrope is formed (see Figure 9.9b). Fractional distillation of such a solution beginning with an initial value of xA less than that corresponding to the minimum boiling point leads to a liquid with the azeotropic composition initially emerging at the top of the distillation column. An example for a minimum boiling azeotrope is a mixture of CS2 (Tb = 46.3°C) and acetone (Tb = 56.2°C) at xCS2 = 0.608, which boils at 39.3°C.
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245
Resistive heater
Initial solution
Figure 9.7
Schematic diagram of a fractional distillation column. The initial solution is introduced at the bottom of the column. The resistive heater provides the energy needed to vaporize the liquid. It is assumed that the liquid and vapor are at equilibrium at each level of the column. The equilibrium temperature decreases from the bottom to the top of the column.
H3C C
O
H
CCI3
H3C
Figure 9.8
Hydrogen bond formation between acetone and chloroform. The relatively strong bonding between species leads to the formation of a maximum boiling azeotrope.
Concept The different components in an ideal solution can be separated by fractional distillation.
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TABLE 9.1 Composition and Boiling Temperatures of Selected Azeotropes
Boiling Temperature
Vapor
Water–ethanol
Boiling Temperature of Components 1°C2 100. and 78.5
0.096
Azeotrope Boiling Point 1°C2
Water–trichloromethane
100. and 61.2
0.160
56.1
Water–benzene
100. and 80.2
0.295
69.3
Water–toluene
100. and 111
0.444
84.1
Ethanol–hexane
78.5 and 68.8
0.332
58.7
Ethanol–benzene
78.5 and 80.2
0.440
67.9
Ethyl acetate–hexane
78.5 and 68.8
0.394
65.2
Carbon disulfide–acetone
46.3 and 56.2
0.608
39.3
Tolueneacetic acid
111 and 118
0.625
100.7
Azeotropic Mixture
Liquid xA (initial) 1
0 xA (a) Maximum boiling point azeotrope
78.2
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
Vapor
Boiling Temperature
Mole Fraction of First Component
It is still possible to collect one component of an azeotropic mixture using the property that the azeotropic composition is a function of the total pressure, as shown in Figure 9.10. The mixture is first distilled at atmospheric pressure, and the volatile distillate is collected. This vapor is condensed and subsequently distilled at a reduced pressure for which the azeotrope contains less A (more B). If this mixture is distilled, the azeotrope evaporates, leaving some pure B behind.
Liquid xA (initial) 1
0 xA (b) Minimum boiling point azeotrope
Figure 9.9
Boiling behavior of azeotroic binary solutions. (a) Maximum boiling point azeotrope and (b) minimum boiling point azeotrope. The dashed lines indicate the initial composition of the solution and the composition of the azeotrope. The sequence of horizontal segments with arrows corresponds to successively higher (cooler) portions of the column.
9.5 THE GIBBS–DUHEM EQUATION In this section, we will show that the chemical potentials of the two components in a binary solution are not independent. This is an important result because it allows the chemical potential of a nonvolatile solute such as sucrose in a volatile solvent such as water to be determined. Such a solute has no measurable vapor pressure; therefore, its chemical potential cannot be measured directly. As shown here, its chemical potential can be determined if only the chemical potential of the solvent as a function of concentration is known. From Chapter 6, the differential form of the Gibbs energy is given by
dG = S dT + V dP + a mi dni
so that at constant T and P,
Concept The Gibbs–Duhem equation states that not all chemical potentials can be varied independently.
i
dG = a mi dni i
(9.19)
(9.20)
However, from Equation (6.37) we have G = a nimi and i
dG = a ni dmi + a mi dni i
i
(9.21)
Because both of the preceding equations for dG are correct, we conclude that
a ni dmi = 0 i
(9.22)
which is known as the Gibbs–Duhem equation. This equation states that not all chemical potentials can be varied independently.
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9.6 Colligative Properties
For a binary solution at constant T and P, the Gibbs–Duhem equation reduces to
n1dm1 + n2dm2 = 0 or x1dm1 + x2dm2 = 0
Atmospheric pressure
(9.23)
dm2 = 
n1dm1 n2
(9.24)
The use of the Gibbs–Duhem equation is illustrated in Example Problem 9.5.
EXAMPLE PROBLEM 9.5
Boiling Temperature
If the change in the chemical potential of the first component is dm1, the change of the chemical potential of the second component is given by
Reduced pressure
One component in a solution follows Raoult’s law, m1 = m*1 + RT ln x1, over the entire composition range 0 … x1 … 1. Using the Gibbs–Duhem equation, show that the second component must also follow Raoult’s law.
Solution From Equation (9.24), x1dm1 n1 x1 dx1 dm2 = =  d1m*1 + RT ln x12 = RT x2 n2 x2 x1
Because x1 + x2 = 1, then dx2 = dx1 and dm2 = RT dx2 >x2. Integrating this equation, one obtains m2 = RT ln x2 + C, where C is a constant of integration. This constant can be evaluated by examining the limit x2 S 1. This limit corresponds to the pure substance 2 for which m2 = m*2 = RT ln 1 + C. We conclude that C = m*2 and, therefore, m2 = m*2 + RT ln x2.
9.6 COLLIGATIVE PROPERTIES Many solutions consist of nonvolatile solutes that have limited solubility in a volatile solvent. Examples are solutions of sucrose or sodium chloride in water. Important properties of these solutions, including boiling point elevation, freezing point depression, and osmotic pressure only depend on the solute concentration—not on the nature of the solute. These properties are called colligative properties. In this section, colligative properties are discussed using the ideal solution model. Corrections for nonideality will be made in Section 9.13. As discussed in Section 9.1, the vapor pressure above a solution containing a solute is lowered with respect to the pure solvent. For the case of dissolution of a nonvolatile solute in which the solute does not crystallize out with the solvent during freezing (because the solute cannot be easily integrated into the solvent crystal structure), the effect on the solvent chemical potential is shown in Figure 9.11a. The chemical potential of the solid is unaffected because of the assumption that the solute does not crystallize out with the solvent. Furthermore, although the gas pressure above the solution is lowered by the addition of the solute, the chemical potential of the gas is unaffected. Only the chemical potential of the liquid is affected through formation of the solution. As shown in Figure 9.11a, the melting temperature Tm, defined as the intersection of the solid and liquid m versus T curves, is lowered. Similarly, the boiling temperature Tb, defined as the intersection of the gas and liquid m versus T curves, is raised by dissolution of a nonvolatile solute in the solvent. The same information is shown in a P–T phase diagram in Figure 9.11b. Because the vapor pressure above the solution is lowered by the addition of the solute, the liquid– gas coexistence curve intersects the solid–gas coexistence curve at a lower temperature than for the pure solvent. This intersection defines the triple point, and it must also be the origin of the solid–liquid coexistence curve. Therefore, the solid–liquid coexistence curve is shifted to lower temperatures through the dissolution of the nonvolatile solute. The overall effect of the shifts in the solid–gas and solid–liquid coexistence curves is a
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0
1
xA
Figure 9.10
A minimum boiling point diagram at two different pressures. Because the azeotropic composition depends on the total pressure, pure B can be recovered from the A–B mixture by first distilling the mixture under atmospheric pressure and, subsequently, under a reduced pressure. Note that the boiling temperature is lowered as pressure is reduced.
Concept Colligative properties depend on the solute concentration and not on the identity of the solute.
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Pure solid Chemical Potential, M
Pure liquid
Solution
freezing point depression and a boiling point elevation, both of which arise through lowering the vapor pressure of the solvent. This effect depends only on the concentration, and not on the identity, of the nonvolatile solute. The preceding discussion is qualitative in nature. In the next two sections, quantitative relationships are developed between the colligative properties and the concentration of the nonvolatile solute.
9.7 FREEZING POINT DEPRESSION
AND BOILING POINT ELEVATION
DTm
If a solution is in equilibrium with a pure solid solvent, the following relation must be satisfied: msolution = m*solid (9.25)
Pure gas
Recall that msolution refers to the chemical potential of the solvent in the solution, and m*solid refers to the chemical potential of the pure solvent in solid form. From Equation (9.7), we can express msolution in terms of the chemical potential of the pure solvent and its concentration and rewrite Equation (9.25) in the form
DTb T¿m T¿m
Tb T¿b
Temperature (a)
m*solvent + RT ln xsolvent = m*solid
(9.26)
This equation can be solved for ln xsolvent: Pure substance
m*solid  m*solvent RT
1 bar
(9.27)
m*solid  m*solvent = ∆ fG solid  ∆ fG solvent =
The difference in chemical potentials ∆ fusG so that
Pressure
Solution
ln xsolvent =
ln xsolvent =
 ∆ fusG RT
(9.28)
Because we are interested in how the freezing temperature is related to xsolvent at constant pressure, we need the partial derivative 10T>0xsolvent2P. This quantity can be obtained by differentiating Equation (9.28) with respect to xsolvent: T¿m Tm
Tb T¿b
Temperature (b)
0
∆ fusG T ¢ 0T a b 0T P 0xsolvent P
(9.29)
The first partial derivative on the right side of Equation (9.29) can be simplified using the Gibbs–Helmholtz equation (see Section 6.3), giving
Figure 9.11
Illustration of boiling point elevation and freezing point depression. Addition of a nonvolatile solute affects the chemical potential of a solution but has essentially no effect on the chemical potentials of the solid and gas phases. This difference in behavior leads to an expansion of the liquid stability field, which in turn produces elevation of the boiling point, T S T′b, and depression of the freezing point, T S T′m. These effects are illustrated on plots of (a) chemical potential versus temperature and (b) pressure versus temperature.
0ln xsolvent 1 1 ° a b = = xsolvent 0xsolvent P R
1 xsolvent
=
∆ fusH RT
2
a
0T 0xsolvent
b
P
or
dxsolvent ∆ fusH dT = d ln xsolvent = (constant P) xsolvent R T2
(9.30)
This equation can be integrated between the limits given by the pure solvent (xsolvent = 1), for which the fusion temperature is Tfus, and an arbitrary small concentration of solute, for which the fusion temperature is T: xsolvent
T
∆ fusH dT′ dx = 2 L R T′ L x 1
(9.31)
Tfus
For xsolvent, not very different from 1, ∆ fusH is assumed to be independent of T, and Equation (9.31) simplifies to
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R ln xsolvent 1 1 = T Tfus ∆ fusH
(9.32)
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9.7 Freezing Point Depression and Boiling Point Elevation
In discussing solutions, it is more convenient to use molality of the solute expressed in moles of solute per kg of solvent as the concentration unit, rather than the mole fraction of the solvent. For dilute solutions, ln xsolvent = ln1nsolvent >1nsolvent + msolute + Msolvent nsolvent22 = ln11 + Msolvent msolute2, and Equation (9.32) can be rewritten in terms of the molality (m) rather than the mole fraction. The result is ∆Tf = 
RMsolventT 2fus msolute = Kf msolute ∆ fusH
249
Concept Freezing point depression and boiling point elevation arise because of the lowering of the vapor pressure of the solvent above the solution.
(9.33)
where Msolvent is the molar mass of the solvent. In recasting concentration units from Equation (9.32) to Equation (9.33), we have made the approximation ln11 + Msolvent msolute2 ≈ Msolvent msolute and 1>T  1>Tfus ≈  ∆Tf >T 2fus. Note that Kf depends only on the properties of the solvent and is primarily determined by the molecular mass and the enthalpy of fusion. For most solvents, the magnitude of Kf lies between 1.5 and 10, as shown in Table 9.2. However, Kf can reach unusually high values—for example, 30. 1K kg2>mol for carbon tetrachloride and 40. 1K kg2>mol for camphor. The boiling point elevation can be calculated using the same argument with ∆ vapG and ∆ vapH substituted for ∆ fusG and ∆ fusH. The result is and
RMsolvent T 2vap 0T a b = 0msolute P, m S 0 ∆ vapH ∆Tb =
RMsolventT 2vap ∆ vapH
(9.34)
(9.35)
msolute = Kbmsolute
Because ∆ vapH 7 ∆ fusH, it follows that Kf 7 Kb. Typically, Kb values range between 0.5 and 5.0 K > 1mol kg 12, as shown in Table 9.2. Note also that, by convention, both Kf and Kb are positive; hence, the negative sign in Equation (9.33) is replaced by a positive sign in Equation (9.35). Example Problem 9.6 illustrates calculations for the freezing point depression and determination of the molecular mass.
TABLE 9.2 Freezing Point Depression and Boiling Point Elevation Constants Standard Boiling Point (K)
Substance
Standard Freezing Point (K)
Acetic acid
289.6
Kf 1K kg mol12 3.59
391.2
Kb1K kg mol12
Benzene
278.6
5.12
353.3
2.53
Camphor
449
482.3
5.95
Carbon disulfide
161
319.2
2.40
Carbon tetrachloride
250.3
30.
349.8
4.95
Cyclohexane
279.6
20.0
353.9
2.79
Ethanol
158.8
2.0
351.5
1.07
Phenol
314
7.27
455.0
3.04
Water
273.15
1.86
373.15
0.51
40. 3.8
3.08
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
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EXAMPLE PROBLEM 9.6 In this example, 4.50 g of a substance dissolved in 125 g of CCl4 leads to an elevation of the boiling point of 0.650 K. Calculate the freezing point depression, the molar mass of the substance, and the factor by which the vapor pressure of CCl4 is lowered.
Solution ∆Tf = a
Kf Kb
b ∆Tb = 
30. K>1mol kg 12
4.95 K>1mol kg 12
* 0.650 K = 3.9 K
To avoid confusion, we use the symbol m for molality and m for mass. We solve for the molar mass Msolute using Equation (9.35): ∆Tb = Kb msolute = Kb * a
msolute >Msolute b msolvent
Msolute =
Kbmsolute msolvent ∆Tb
Msolute =
4.95 K kg mol1 * 4.50 g = 274 g mol1 0.125 kg * 0.650 K
We solve for the factor by which the vapor pressure of the solvent is reduced by using Raoult’s law: Psolvent P *solvent
= xsolvent = 1  xsolute = 1 
nsolute nsolute + nsolvent
4.50 g 274 g mol1
= 1 a
4.50 g 274 g mol
b + a 1
125 g 153.8 g mol
= 0.980 b 1
9.8 OSMOTIC PRESSURE P
P1P
Solution
Pure solvent
Some membranes allow the passage of small molecules such as water, yet do not allow larger molecules such as sucrose to pass through them. Such a semipermeable membrane is an essential component in medical technologies, including kidney dialysis, which is described below. Consider a sac of such a membrane that contains a solute that cannot pass through the membrane. If the sac is immersed in a beaker filled with pure solvent, then initially the solvent diffuses into the sac. Diffusion ceases when equilibrium is attained, and at equilibrium the pressure is higher in the sac than in the surrounding solvent. This result is shown schematically in Figure 9.12. The process in which solvent diffuses through a membrane and dilutes a solution is known as osmosis. The increase in pressure in the solution is known as the osmotic pressure. To understand the origin of the osmotic pressure, denoted by p, the equilibrium condition is applied to the contents of the sac and the surrounding solvent:
* msolution solvent 1T, P + p, xsolvent2 = msolvent 1T, P2
(9.36)
* msolution solvent 1T, P + p, xsolvent2 = msolvent 1T, P + p2 + RT ln xsolvent
(9.37)
Using Raoult’s law to express the concentration dependence of msolvent, Semipermeable membrane
Figure 9.12
Osmotic pressure experiment. An osmotic pressure arises if a solution containing a solute that cannot pass through the membrane boundary is immersed in the pure solvent.
M09_ENGE4583_04_SE_C09_237272.indd 250
Because m for the solvent is lower in the solution than in the pure solvent, only an increased pressure in the solution can raise its m sufficiently to achieve equilibrium with the pure solvent. The dependence of m on pressure and temperature is given by dm = dGm = VmdP  SmdT. At constant T we can write P+p
m*solvent1T,
P + p, xsolvent2 
m*solvent1T,
P2 =
L
V *mdP′
(9.38)
P
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9.8 Osmotic Pressure
251
where V *m is the molar volume of the pure solvent and P is the pressure in the solvent outside the sac. Because a liquid is nearly incompressible, it is reasonable to assume that V *m is independent of P to evaluate the integral in the previous equation. Therefore, m*solvent 1T, P + p, xsolvent2  m*solvent 1T, P2 = V *m p, and Equation (9.37) reduces to pV *m + RT ln xsolvent = 0
(9.39)
For a dilute solution, nsolvent W nsolute, and ln xsolvent = ln(1  xsolute2 ≈ xsolute = 
nsolute nsolute ≈ (9.40) nsolvent nsolute + nsolvent
Equation (9.39) can be simplified further by recognizing that for a dilute solution, V ≈ nsolventV *m. With this substitution, Equation (9.39) becomes
p =
nsolute RT V
(9.41)
Note the similarity in form between this equation and the ideal gas law. An important application of the selective diffusion of the components of a solution through a membrane is dialysis. In healthy individuals, the kidneys remove waste products from the bloodstream, whereas individuals with damaged kidneys use a dialysis machine for this purpose. Blood from the patient is shunted through tubes made of a selectively porous membrane surrounded by a flowing sterile solution made up of water, sugars, and other components. Blood cells and other vital components of blood are too large to fit through the pores in the membranes, but urea and salt flow out of the bloodstream through membranes into the sterile solution and are removed as waste. Example Problem 9.7 illustrates a calculation of the osmotic pressure.
Concept Osmotic pressure arises as a consequence of the selective diffusion of solution components through a membrane.
EXAMPLE PROBLEM 9.7 Calculate the osmotic pressure generated at 298 K if a cell with a total solute concentration of 0.500 mol L1 is immersed in pure water. The cell wall is permeable to water molecules but not to the solute molecules.
Solution nsolute RT = 0.500 mol L1 * 8.206 * 102 L atm K1 mol1 * 298 K V = 12.2 atm
p =
As this calculation shows, the osmotic pressure generated for moderate solute concentrations can be quite high. Hospital patients have died after pure water has accidentally been injected into their blood vessels because the osmotic pressure is sufficient to burst the walls of blood cells.
Plants use osmotic pressure to achieve mechanical stability in the following way. A plant cell is bounded by a cellulose cell wall that is permeable to most components of the aqueous solutions that it encounters. A semipermeable cell membrane through which water, but not solute molecules, can pass is located just inside the cell wall. When the plant has sufficient water, the cell membrane expands, pushing against the cell wall, which gives the plant stalk a high rigidity. However, if there is a drought, the cell membrane is not totally filled with water and it moves away from the cell wall. As a result, only the cell wall contributes to the rigidity, and plants droop under these conditions. Another important application involving osmotic pressure is the desalination of seawater using reverse osmosis. Seawater is typically 1.1 molar in NaCl. Equation (9.41) shows that an osmotic pressure of 27 bar is needed to separate the solvated Na+ and Cl ions from the water. If the seawater side of the membrane is subjected to a pressure greater than 27 bar, H2O from the seawater will flow through the membrane, resulting in a separation of pure water from the seawater. This process is called reverse osmosis. The challenge in carrying out reverse osmosis on the industrial scale needed to provide
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CHAPTER 9 Ideal and Real Solutions
a coastal city with potable water is to produce robust membranes that accommodate the necessary flow rates without getting fouled by algae and also effectively separate the ions from the water. The mechanism that prevents passage of the ions through suitable membranes is not fully understood. It is not based on the size of pores within the membrane alone, and it involves charged surfaces within the hydrated pores. These membraneanchored ions repel the mobile Na+ and Cl ions, while allowing the passage of the neutral water molecule.
9.9 DEVIATIONS FROM RAOULT’S LAW IN REAL SOLUTIONS
In Sections 9.1 through 9.8, the discussion was limited to ideal solutions. However, in general, if two volatile and miscible liquids are combined to form a solution, Raoult’s law is not obeyed. This is the case because the A–A, B–B, and A–B interactions in a binary solution of A and B are in general not equal. If the A–B interactions are less attractive than the A–A and B–B interactions, positive deviations from Raoult’s law will be observed. In contrast, if the A–B interactions are more attractive than the A–A and B–B interactions, negative deviations from Raoult’s law will be observed. An example of a binary solution with positive deviations from Raoult’s law is CS2 9acetone. Experimental data for this system are shown in Table 9.3, and the data are plotted in Figure 9.13. How can a thermodynamic framework analogous to that presented for the TABLE 9.3 Partial and Total Pressures above a CS2–Acetone Solution xCS2 0
PCS2 (Torr) 0
Pacetone (Torr)
Ptot (Torr)
xCS2
PCS2 (Torr)
Pacetone (Torr)
Ptot (Torr)
343.8
343.8
0.4974
404.1
242.1
646.2
0.0624
110.7
331.0
441.7
0.5702
419.4
232.6
652.0
0.0670
119.7
327.8
447.5
0.5730
420.3
232.2
652.5
0.0711
123.1
328.8
451.9
0.6124
426.9
227.0
653.9
0.1212
191.7
313.5
505.2
0.6146
427.7
225.9
653.6
0.1330
206.5
308.3
514.8
0.6161
428.1
225.5
653.6
0.1857
258.4
295.4
553.8
0.6713
438.0
217.0
655.0
0.1991
271.9
290.6
562.5
0.6713
437.3
217.6
654.9
0.2085
283.9
283.4
567.3
0.7220
446.9
207.7
654.6
0.2761
323.3
275.2
598.5
0.7197
447.5
207.1
654.6
0.2869
328.7
274.2
602.9
0.8280
464.9
180.2
645.1
0.3502
358.3
263.9
622.2
0.9191
490.7
123.4
614.1
0.3551
361.3
262.1
623.4
0.9242
490.0
120.3
610.3
0.4058
379.6
254.5
634.1
0.9350
491.9
109.4
601.3
0.4141
382.1
253.0
635.1
0.9407
492.0
103.5
595.5
0.4474
390.4
250.2
640.6
0.9549
496.2
85.9
582.1
0.4530
394.2
247.6
641.8
0.9620
500.8
73.4
574.2
0.4933
403.2
242.8
646.0
0.9692
502.0
62.0
564.0
1
512.3
0
512.3
Source: Zawidski, J. V. Zeitschrift für Physikalische Chemie 35 (1900): 129.
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9.9 Deviations from Raoult’s Law in Real Solutions
Figure 9.13
Deviations from Raoult’s law. The data in Table 9.3 are plotted versus xCS2. Dashed lines show the expected behavior if Raoult’s law were obeyed.
Ptotal
600
Pressure, P (Torr)
500 400 300 200
0
0.2
0.4
xCS2
0.6
0.8
1
ideal solution in Section 9.2 be developed for real solutions? This issue is addressed throughout the rest of this chapter. Figure 9.13 shows that the partial and total pressures above a real solution can differ substantially from the behavior predicted by Raoult’s law. Another way that ideal and real solutions differ is that the set of equations denoted in Equation (9.8), which describes the change in volume, entropy, enthalpy, and Gibbs energy that results from mixing, are not applicable to real solutions. For real solutions, these equations can only be written in a much less explicit form. Assuming that A and B are miscible,
∆ mixG ∆ mixS ∆ mixV ∆ mixH
6 0 7 0 ≠ 0 ≠ 0
V ideal = xAV *m, A + 11  xA2V *m, B m
(9.43)
as expected for 1 mol of an ideal solution, where V *m, A and V *m, B are the molar volideal umes of the pure substances A and B. Figure 9.14 shows ∆Vm = V real for an m  Vm acetone–chloroform solution as a function of xchloroform. Note that ∆Vm is positive for some values of xchloroform but negative for other values of xchloroform. The deviations from ideality are small but clearly evident. The deviation of the volume from ideal behavior can best be understood by defining the concept of partial molar quantities. One can form partial molar quantities for any extensive property of a system (for example, U, H, G, A, and S). Partial molar quantities (other than the chemical potential, which is the partial molar Gibbs energy) are usually denoted by the appropriate symbol topped by a horizontal bar, such as Vi. The concept of partial molar quantities can be illustrated by discussing the partial molar volume. The volume of 1 mol of pure water at 25°C is 18.1 cm3. However, if 1 mol of water is added to a large volume of an ethanol–water solution with xH2O = 0.75, the volume of the solution increases by only 16 cm3. This is the case because the local structure around a water molecule in the solution is more compact than that in pure water. The partial molar volume of a component in a solution is defined as the volume by which the solution changes if 1 mol of the component is added to such a
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Real solutions can show appreciable deviations from Raoult’s law.
(9.42)
Whereas ∆ mixG 6 0 and ∆ mixS 7 0 always hold for miscible liquids, ∆ mixV and ∆ mixH can be positive or negative, depending on the nature of the A–B interaction in the solution. As indicated in Equation (9.42), the volume change upon mixing is not generally 0. Therefore, the volume of a solution will not be given by
Concept
0.2
0.4
0.6
0.8
1
0 Volume Change DVsol, (cm3)
100
Pacetone
PCS2
20.025
x chloroform
20.050 20.075 20.100 20.125 20.150 20.175
Figure 9.14
Deviations of measured molar volume for the acetone–chloroform system relative to that for ideal solution behavior. Ideal solution volume calculated from Equation 9.43. Molar volume deviations plotted as a function of the mole fraction of chloroform.
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Figure 9.15
82.0
75.0 Acetone
74.5
81.5
74.0 81.0
73.5
80.5
73.0 72.5
80.0
Chloroform 72.0
79.5
71.5
0.2
0.4
xchloroform
0.6
0.8
Partial molar volume of acetone (cm3 mol21)
The partial molar volumes in a binary system. Chloroform (blue curve) and acetone (red curve) partial molar volumes in a chloroform–acetone binary solution are shown as a function of xchloroform. The arrows indicate the appropriate yaxis labels for the given curve.
Partial molar volume of chloroform (cm3 mol21)
254
1
large volume that the solution composition can be assumed constant. This statement is expressed mathematically in the following form:
V11P, T, n1, n22 = a
0V b 0n1 P, T, n2
(9.44)
With this definition, the volume of a binary solution is given
Concept The Gibbs–Duhem equation also applies to partial molar volumes in real solutions.
V = n1V11P, T, n1, n22 + n2V21P, T, n1, n22
(9.45)
Note that because the partial molar volumes depend on the concentration of all components, the same is true of the total volume. Partial molar volume is a function of P, T, n1, and n2, and Vi can be greater than or less than the molar volume of the pure component. Therefore, the volume of a solution of two miscible liquids can be greater than or less than the sum of the volumes of the pure components of the solution. Figure 9.15 shows data for the partial volumes of acetone and chloroform in an acetone–chloroform binary solution at 298 K. Note that the changes in the partial molar volumes with concentration are small, but not negligible. In Figure 9.15, we can see that V1 increases if V2 decreases and vice versa. This is the case because partial molar volumes are related in the same way as the chemical potentials are related in the Gibbs–Duhem equation [Equation (9.23)]. In terms of the partial molar volumes, the Gibbs–Duhem equation takes the form x2 x1 dV1 + x2 dV2 = 0 or dV1 =  dV2 (9.46) x1 Therefore, as seen in Figure 9.15, if V2 changes by an amount dV2 over a small concentration interval, V1 will change in the opposite direction. The Gibbs–Duhem equation is applicable to both ideal and real solutions.
9.10 THE IDEAL DILUTE SOLUTION Although no simple model exists that describes all real solutions, there is a limiting law that merits further discussion. In this section, we describe the model of the ideal dilute solution, which provides a useful basis for describing the properties of real solutions if they are dilute. Just as for an ideal solution, at equilibrium the chemical potentials of a component in the gas and solution phase of a real solution are equal. As for an ideal solution, the chemical potential of a component in a real solution can be separated into
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255
two terms, a standard state chemical potential and a term that depends on the partial pressure. mi = m*i + RT ln
Pi P *i
(9.47)
Recall that for a pure substance m*i (vapor) = m*i (liquid) = m*i . Because the solution is not ideal, Pi ≠ xi P *i . First consider only the solvent in a dilute binary solution. To arrive at an equation for mi that is similar to Equation (9.7) for an ideal solution, we define the dimensionless activity, asolvent, of the solvent by
asolvent =
Psolvent P *solvent
(9.48)
Note that asolvent = xsolvent for an ideal solution. For a nonideal solution, the activity and the mole fraction are related through the activity coefficient gsolvent, defined by
gsolvent =
asolvent xsolvent
(9.49)
The activity coefficient quantifies the degree to which the solution is nonideal. The activity plays the same role for a component of a solution that the fugacity plays for a real gas in expressing deviations from ideal behavior. In both cases, ideal behavior is observed in the appropriate limit, namely, P S 0 for the gas, and xsolute S 0 for the solution. To the extent that there is no atomicscale model that tells us how to calculate g, it should be regarded as a correction factor that exposes the inadequacy of our model, rather than as a fundamental quantity. As will be discussed in Chapter 10, there is such an atomicscale model for dilute electrolyte solutions. How is the chemical potential of a component related to its activity? Combining Equations (9.47) and (9.48), one obtains a relation that holds for all components of a real solution: mi = m*i + RT ln ai
(9.50)
Equation (9.50) is a central equation in describing real solutions. It is the starting point for the discussion in the rest of this chapter. To this point in the chapter, we have focused on the solvent in a dilute solution. However, the ideal dilute solution is defined by the conditions xsolute S 0 and xsolvent S 1. Because the solvent and solute are considered in different limits, we use different expressions to relate the mole fraction of a component and the partial pressure of the component above the solution. Consider the partial pressure of acetone as a function of xCS2 shown in Figure 9.16. Although Raoult’s law is not obeyed over the whole concentration range, it is obeyed in the limit that xacetone S 1 and xCS2 S 0. In this limit, the average acetone molecule at the surface of the solution is surrounded by acetone molecules. Therefore, to a good approximation, Pacetone = xacetone P *acetone as xacetone S 1. Because the majority species
Pressure, P (Torr)
Henry’s law Pacetone
300 200 Raoult’s law 100
Figure 9.16 0
0.2
0.4
0.6 xCS2
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0.8
1
Partial pressure of acetone plotted as a function of xCS2. Note that Pacetone follows Raoult’s law as xCS2 S 0 and Henry’s law as xCS2 S 1. Data from Table 9.3.
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TABLE 9.4 Henry’s Law Constants for Aqueous Solutions Near 298 K Substance
kH (Torr)
kH (bar)
Ar
2.80 * 107
3.72 * 104
C2H6
2.30 * 107
3.06 * 104
CH4
3.07 * 107
4.08 * 104
CO
4.40 * 106
5.84 * 103
CO2
1.24 * 106
1.65 * 103
H2S
4.27 * 105
5.68 * 102
He
1.12 * 108
1.49 * 106
N2
6.80 * 107
9.04 * 104
O2
3.27 * 107
4.95 * 104
Source: Alberty, R. A., and Silbey, R. S. Physical Chemistry. New York: John Wiley & Sons, 1992.
Concept In an ideal dilute solution, the solvent follows Raoult’s law and the solute follows Henry’s law at low solute concentrations.
is defined to be the solvent, we see that Raoult’s law is obeyed for the solvent in a dilute solution. This limiting behavior is also observed for CS2 in the limit in which it is the solvent, as seen in Figure 9.13. Consider the opposite limit in which xacetone S 0. In this case, the average acetone molecule at the surface of the solution is surrounded by CS2 molecules. Therefore, the molecule experiences very different interactions with its neighbors than if it were surrounded by acetone molecules. For this reason, Pacetone ≠ xacetone P *acetone as xacetone S 0. However, it is apparent from Figure 9.16 that Pacetone also varies linearly with xacetone in this limit. This behavior is described by the following equation:
Pacetone = xacetone k H acetone as xacetone S 0
(9.51)
This relationship is known as Henry’s law, and the constant kH is known as the Henry’s law constant. The value of the constant depends on the nature of the solute and solvent and quantifies the degree to which deviations from Raoult’s law occur. As * the solution approaches ideal behavior, k H i S P i . For the data shown in Figure 9.13, H H k CS2 = 1750 Torr and k acetone = 1950 Torr. Note that these values are substantially greater than the vapor pressures of the pure substances, which are 512.3 and 343.8 Torr, respectively. The Henry’s law constants are less than the vapor pressures of the pure substances if the system exhibits negative deviations from Raoult’s law. Henry’s law constants for aqueous solutions are listed for a number of solutes in Table 9.4. Based on these results, the ideal dilute solution is defined as a solution in which the solvent is described using Raoult’s law and the solute is described using Henry’s law. As shown by the data in Figure 9.13, the partial pressures above the CS2acetone mixture are consistent with this model in either of two limits, xacetone S 1 or xCS2 S 1. In the first of these limits, we consider acetone to be the solvent and CS2 to be the solute. In the second limit, acetone is the solute and CS2 is the solvent.
9.11 ACTIVITIES ARE DEFINED WITH RESPECT TO STANDARD STATES
Predictions based on the ideal dilute solution model that Raoult’s law is obeyed for the solvent and Henry’s law is obeyed for the solute are not valid over a wide range of concentration. The concept of the activity coefficient introduced in Section 9.10 is used to quantify these deviations. In doing so, it is useful to define the activities in such a way that the solution approaches ideal behavior in the limit of interest, which is generally
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xA S 0, or xA S 1. With this choice, the activity approaches the concentration, and it is reasonable to set the activity coefficient equal to 1. Specifically, ai S xi as xi S 1 for the solvent, and ai S xi as xi S 0 for the solute. The reason for this choice is that numerical values for activity coefficients are generally not known. Choosing the standard state in this way ensures that the concentration (divided by the unit concentration to make it dimensionless), which is easily measured, is a good approximation of the activity. In Section 9.10, the activity and activity coefficient for the solvent in a dilute solution (xsolvent S 1) were defined by the relations
ai =
Pi P *i
and gi =
ai xi
257
Concept Standard states for real solutions are chosen to make the value of the activity coefficient approach one if possible.
(9.52)
As shown in Figure 9.13, the activity approaches unity as xsolvent S 1. We refer to an activity calculated using Equation (9.52) as being based on a Raoult’s law standard state. The standard state chemical potential based on Raoult’s law is m*solvent, which is the chemical potential of the pure solvent. However, this definition of the activity and choice of a standard state is not optimal for the solute at a low concentration because the solute obeys Henry’s law rather than Raoult’s law; therefore, the value of the activity coefficient will differ appreciably from one. In this case, * msolution solute = msolute + RT ln
kH solute xsolute P *solute
= m*H solute + RT ln xsolute as xsolute S 0 (9.53)
The standard state chemical potential is the value of the chemical potential when xi = 1. We see that the Henry’s law standard state chemical potential is given by
* m*H solute = msolute + RT ln
kH solute P *solute
(9.54)
The activity and activity coefficient based on Henry’s law are defined, respectively, by
ai =
Pi kH i
and gi =
ai xi
(9.55)
Note that Henry’s law is still obeyed for a solute that has such a small vapor pressure that we refer to it as a nonvolatile solute. The Henry’s law standard state is a state in which the pure solute would have a vapor pressure kH,solute rather than its actual value P *solute. It is a hypothetical state that does not exist. Recall that the value k H solute is obtained by extrapolation from the low coverage range in which Henry’s law is obeyed. Although this definition may seem peculiar, only in this way can we ensure that asolute S xsolute and gsolute S 1 as xsolute S 0. We reiterate the reason for this choice. If the preceding conditions are satisfied, the concentration (divided by the unit concentration to make it dimensionless) is, to a good approximation, equal to the activity. Therefore, equilibrium constants for reactions in solution can be calculated without knowing the values of the activity coefficients. We emphasize the difference in the Raoult’s law and Henry’s law standard states because it is the standard state chemical potentials m*solute and m*H solute that are used to calculate ∆G ~ and the thermodynamic equilibrium constant K. Different choices for standard states will result in different numerical values for K. The standard chemical potential m*H solute refers to the hypothetical standard state in which xsolute = 1, and each solute species is in an environment characteristic of the infinitely dilute solution. We now consider a less welldefined situation. For solutions in which the components are miscible in all proportions, such as the CS2 9acetone system, either a Raoult’s law or a Henry’s law standard state can be defined, as we show with sample calculations in Example Problems 9.8 and 9.9. In this case, there is no unique choice for the standard state over the entire concentration range in such a system. Numerical values for the activities and activity coefficients will differ, depending on whether the Raoult’s law or the Henry’s law standard state is used.
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1.0
EXAMPLE PROBLEM 9.8 Calculate the activity and activity coefficient for CS2 at xCS2 = 0.3502 using data from Table 9.3. Assume a Raoult’s law standard state.
CS2 activity (RL)
0.8
Solution
0.6
aRCS2 = 0.4
g RCS2 =
0.2
0
0.2
0.4
0.6 xCS
0.8
1
2
Activity coefficient CS2 (RL)
4.0 3.5 3.0 2.5
PCS2 P *CS2 aRCS2 xCS2
=
358.3 Torr = 0.6994 512.3 Torr
=
0.6994 = 1.997 0.3502
The activity and activity coefficients for CS2 are concentration dependent. Calculations of activity and activity coefficients using a Raoult’s law standard state and the same methods as in this Example Problem are shown in Figure 9.17 as a function of xCS2. For this solution, g RCS2 7 1 for all values of the concentration for which xCS2 6 1. Note that g RCS2 S 1 as xCS2 S 1, as the model requires. The activity and activity coefficients for CS2 using a Henry’s law standard state are shown in Figure 9.18 as a function of xCS2. For this solution, g H CS2 6 1 for all values of the concentration for which xCS2 7 0. Note that g H CS2 S 1 as xCS2 S 0 as the model requires. Which of these two possible standard states should be chosen? There is a good answer to this question only in the limits xCS2 S 0 or xCS2 S 1. For intermediate concentrations, either standard state can be used.
2.0
EXAMPLE PROBLEM 9.9
1.5 1.0
Calculate the activity and activity coefficient for CS2 at xCS2 = 0.3502 using data from Table 9.3. Assume a Henry’s law standard state.
0.5
Solution
0
0.2
0.4
0.6
0.8
aH CS2 =
1
xCS2
gH CS2
Figure 9.17
Activity and activity coefficient for CS2 in a CS2 9acetone solution based on a Raoult’s law standard state. Activity and activity coefficient are plotted as a function of xCS2. RL = Raoult’s law.
=
PCS2 kH CS2 aH CS2 xCS2
=
358.3 Torr = 0.204 1750 Torr
=
0.204 = 0.584 0.3502
The Henry’s law standard state is defined with respect to concentration measured in units of mole fraction. This is not a particularly convenient scale, and either the molarity or molality concentration scales are generally used in laboratory experiments. The mole fraction of the solute can be converted to the molality scale by dividing the numerator and denominator of the first expression in Equation (9.56) by nsolvent Msolvent:
xsolute =
nsolute = nsolvent + nsolute
msolute 1 Msolvent
(9.56)
+ msolute
where msolute is the molality of the solute, and Msolvent is the molar mass of the solvent in kg mol1. We see that msolute S xsolute >Msolvent as xsolute S 0. Using molality as the concentration unit, we define the activity and activity coefficient of the solute by
amolality solute =
g molality solute =
Psolute
k molality H amolality solute msolute
with amolality solute S msolute as msolute S 0 and
(9.57)
with g molality solute S 1 as msolute S 0
(9.58)
The Henry’s law constants and activity coefficients determined on the mole fraction scale must be recalculated to describe a solution characterized by its molality or
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9.11 Activities Are Defined with Respect to Standard States
molarity. Example Problem 9.10 shows the conversion from mole fraction to molarity; similar conversions can be made if the concentration of the solution is expressed in terms of molality. What is the standard state if concentrations rather than mole fractions are used? The standard state in this case is the hypothetical state in which Henry’s law is obeyed by a solution that is 1.0 molar (or 1.0 molal) in the solute concentration. It is a hypothetical state because at this concentration, substantial deviations from Henry’s law will be observed. Although this definition may seem peculiar at first reading, only in this way can we ensure that the activity becomes equal to the molarity (or molality), and the activity coefficient approaches 1, as the solute concentration approaches 0.
0.25
0.20 CS2 activity (HL)
0.15
0.10 0.05
EXAMPLE PROBLEM 9.10 a. D erive a general equation, valid for dilute solutions, relating the Henry’s law constants for the solute on the mole fraction and molarity scales. b. Determine the Henry’s law constants for acetone on the molarity scale. Use the results for the Henry’s law constants on the mole fraction scale cited in Section 9.10. The density of acetone is 789.9 g L1.
a. We use the symbol csolute to designate the solute molarity, and c ~ to indicate a 1 molar concentration. dxsolute dP dP = csolute csolute dxsolute da ~ b da ~ b c c
To evaluate this equation, we must determine
xsolute =
dxsolute csolute da ~ b c
0.8
1
0.2
0.4 0.6 xCS2
0.8
1
0.6 0.4 0.2
Figure 9.18
Activity and activity coefficient for CS2 in a CS2 9acetone solution based on a Henry’s law standard state. Activity and activity coefficient are plotted as a function of xCS2. HL = Henry’s law.
Therefore, c ~ Msolvent dP dP = rsolution dxsolute csolute da ~ b c
b.
0.4 0.6 xCS2
0.8
0
nsolute Msolvent Msolvent nsolute nsolute ≈ = = csolute rsolution nsolvent nsolute + nsolvent Vsolution rsolution
0.2
1.0 Activity coefficent CS2 (HL)
Solution
0
1 mol L1 * 58.08 g mol1 dP = * 1950 Torr = 143.4 Torr csolute 789.9 g L1 da ~ b c
The colligative properties for an ideal solution discussed in Sections 9.7 and 9.8 refer to the properties of the solvent in a dilute solution. The Raoult’s law standard state applies to this case and in an ideal dilute solution, Equations (9.33), (9.35), and (9.41) can be used with activities replacing concentrations:
∆Tf = Kf gmsolute ∆Tb = Kbgmsolute p = gcsolute RT
(9.59)
The activity coefficients are defined with respect to molality for the boiling point elevation ∆Tb and freezing point depression ∆Tf, and with respect to molarity for the osmotic pressure p. Equations (9.59) provide a useful way to determine activity coefficients, as shown in Example Problem 9.11.
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EXAMPLE PROBLEM 9.11 In 500. g of water, 24.0 g of a nonvolatile solute of molecular weight 241 g mol1 is dissolved. The observed freezing point depression is 0.359°C. Calculate the activity coefficient of the solute.
Solution ∆Tƒ = Kƒ gmsolute; g = g =
1.86 K kg mol1
∆Tƒ Kƒ msolute
0.359 K = 0.969 24.0 * mol kg 1 241 * 0.500
9.12 HENRY’S LAW AND THE SOLUBILITY OF GASES IN A SOLVENT
The ideal dilute solution model can be applied to the solubility of gases in liquid solvents. An example for this type of solution equilibrium is the amount of N2 absorbed by water at sea level, which is considered in Example Problem 9.12. In this case, one of the components of the solution is a liquid and the other is a gas. The equilibrium of interest between the solution and the vapor phase is
Concept Henry’s law is widely used to describe the solubility of gases in liquids.
N21aqueous2 N N21vapor2
(9.60)
The chemical potential of the dissolved N2 is given by
msolution = m*H N2 N2 1vapor2 + RT ln asolute
(9.61)
In this case, a Henry’s law standard state is the appropriate choice because the nitrogen is sparingly soluble in water. The mole fraction of N2 in solution, xN2, is given by
xN2 =
nN2 nN2 + nH2O
≈
nN2
(9.62)
(9.63)
n H 2O
The amount of dissolved gas is given by
nN2 = nH2O xN2 = nH2O
PN2 kH N2
Example Problem 9.12 shows how Equation (9.63) is used to model the dissolution of a gas in a liquid.
EXAMPLE PROBLEM 9.12 The average human with a body weight of 70. kg has a blood volume of 5.00 L. The Henry’s law constant for the solubility of N2 in H2O is 9.04 * 104 bar at 298 K. Assume that this is also the value of the Henry’s law constant for blood and that the density of blood is 1.00 kg L1. a. Calculate the number of moles of nitrogen absorbed in this amount of blood in air of composition 80.% N2 at sea level, where the pressure is 1 bar, and at a pressure of 50. bar. b. Assume that a diver accustomed to breathing compressed air at a pressure of 50. bar is suddenly brought to sea level. What volume of N2 gas is released as bubbles in the diver’s bloodstream?
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261
Solution nN2 = nH2O a. =
PN2 kH N2
5.0 * 103g 18.02 g mol
1
*
0.80 bar 9.04 * 104 bar
= 2.5 * 103 mol at 1 bar total pressure At 50. bar, nN2 = 50.*2.5 * 103 mol = 0.13 mol. b. V =
nRT P
(0.13 mol  2.5 * 103 mol) * 8.314 * 102 L bar mol1 K1 * 300. K 1.00 bar = 3.2 L =
The symptoms induced by the release of air into the bloodstream are known to divers as the bends. The volume of N2 just calculated is far more than is needed to cause the formation of arterial blocks due to gasbubble embolisms.
9.13 CHEMICAL EQUILIBRIUM IN SOLUTIONS The concept of activity can be used to express the thermodynamic equilibrium constant in terms of activities for real solutions. Consider a reaction between solutes in a solution. At equilibrium, the following relation must hold: a a vjmj1solution2b
j
equilibrium
= 0
(9.64)
where the subscript states that the individual chemical potentials must be evaluated under equilibrium conditions. Each of the chemical potentials in Equation (9.64) can be expressed in terms of a standard state chemical potential and a concentrationdependent term. Assume a Henry’s law standard state for each solute. Equation (9.64) then takes the form
*H eq vj a vjmj 1solution2 + RT a ln 1ai 2 = 0 j
j
(9.65)
Using the relation between the Gibbs energy and the chemical potential, we can write the previous equation in the form
vj ∆ rG ~ = RT a ln (aeq i 2 = RT ln K j
Concept Chemical equilibrium in solutions is defined in terms of activities rather than concentrations.
(9.66)
The equilibrium constant in terms of activities is given by
eq vj vj eq vj ci K = q 1aeq i 2 = q 1g i 2 a ~ b c i
i
(9.67)
where the symbol Π indicates that the terms following the symbol are multiplied with one another. It can be viewed as a generalization of KP, defined in Equation (6.60), and it can be applied to equilibria involving gases, liquids, dissolved species, and solids. For gases, the fugacities divided by ƒ ~ (see Section 7.5) are the activities. To obtain a numerical value for K, the standard state Gibbs reaction energy ∆ rG ~ must be known. In general, the ∆ rG ~ must be determined experimentally. This can be done by measuring the individual activities of the species in solution and calculating K from these results. After a series of ∆ rG ~ for different reactions has been determined, they can be combined to calculate the ∆ rG ~ for other reactions, as discussed for reaction enthalpies in Chapter 4. Because of the significant interactions between the solutes and the solvent, K values depend on the nature of the solvent. Furthermore, for electrolyte solutions K values depend on the ionic strength, as will be discussed in Chapter 10.
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An equilibrium constant in terms of molarities or molalities can also be defined starting from Equation (9.67) and setting all activity coefficients equal to 1. This is most appropriate for a dilute solution of a nonelectrolyte, using a Henry’s law standard state: eq vj vj ceq i vj ci K = q (g eq i 2 a ~ b ≈ qa ~ b c c
i
i
(9.68)
Example Problem 9.13 illustrates an application of Equation (9.68).
EXAMPLE PROBLEM 9.13
a. W rite the equilibrium constant for the reaction N21aq, m2 N N21g, P2 in terms of activities at 25°C, where m is the molality of N21aq2. b. By making suitable approximations, convert the equilibrium constant of part (a) into one in terms of pressure and molality only.
Solution nj eq nj a. K = q 1aeq i 2 = q 1g i 2 a ~ b c
ceq i
i
i
nj
=
a
a
P gN2, g a P ~ b = gN2, aq m a ~b m
gN2, g P P~
b
gN2, aq m m~
b
b. Using a Henry’s law standard state for dissolved N2, gN2, aq ≈ 1, because the concentration is very low. Similarly, because N2 behaves like an ideal gas to quite high pressures at 25°C, gN2, g ≈ 1. Therefore, P b P~ K ≈ m a ~b m Note that in this case, the equilibrium constant is simply the Henry’s law constant in terms of molality. a
The numerical values for the dimensionless thermodynamic equilibrium constant depend on the choice of the standard states for the components involved in the reaction. The same is true for ∆ rG ~ . Therefore, it is essential to know which standard state has been assumed before an equilibrium constant is used. The activity coefficients of most neutral solutes are close to 1 with the appropriate choice of standard state. Therefore, example calculations of chemical equilibria using activities will be deferred until electrolyte solutions are discussed in Chapter 10. For such solutions, gsolute differs substantially from 1, even for dilute solutions. An important example of equilibrium processes in solution is the physical interaction of a molecule with a binding site. Such binding equilibria processes are ubiquitous in chemistry and biochemistry, being central to the function of oxygen transport proteins such as hemoglobin or myoglobin, the action of enzymes, and the adsorption of molecules to surfaces. All of these processes involve equilibrium between the free molecule (R) and the molecule absorbed or bound to another species (protein, solid, etc.) that contains a binding site (M). In the limit where there is a single binding site per species, the equilibrium can be written as
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R + M N RM
(9.69)
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9.13 Chemical Equilibrium in Solutions
The equilibrium constant for this reaction in terms of the concentrations of the species of interest is
(9.71)
Expressing this fraction in terms of the equilibrium constant yields
v =
1>1cRcM2 cRM KcR K a b = = cM + cRM 1>1cRcM2 1>cR + K 1 + KcR
(9.72)
This expression demonstrates that v depends on both the molecular concentration and the value of the equilibrium constant. The variations of v with c R for different values of K are presented in Figure 9.19. With an increase in K (consistent with the bound form of the molecule being lower in Gibbs energy relative to the free form), the average number of bound molecules per species approaches unity (that is, the binding sites are completely occupied) even at low values of cR. To this point in this section, the development of a theory for chemical equilibrium in a solution was limited to a single binding site for a given molecule. To consider molecules with multiple binding sites, two issues must be considered. First, binding to one site may have an effect on binding at other sites. This effect is important, for example, in the binding of oxygen to hemoglobin. However, in the current theory development, we will restrict ourselves to independent site binding, where the binding at one site has no effect on binding at other sites. In this case, the average number of molecules bound to a specific site (i) is
vi =
KicR 1 + KicR
K5100
0.8
0.6 K55 0.4
0 0
Figure 9.19
Average number of bound molecules per species plotted against the concentration of free molecules for different equilibrium constant values. The average number of bound molecules per species, v; concentration of free species, cR; equilibrium constant, K. Values for K of 5, 10, 30, and 100 are shown.
(9.73) 500
400
(9.74)
The second issue to address is the equivalency of the binding sites. Assuming that all of the binding sites are equivalent, v becomes
N KicR NKcR v = a = 1 + K c 1 + KcR i R i=1
(9.75)
This expression is almost identical to our previous expression for v for a single binding site, but with the addition of a factor of N. Therefore, at large values of cR, the average number of bound molecules per species will approach N instead of 1. To determine K and N for specific binding equilibrium, the expression for v is recast in a linear form referred to as the Scatchard equation: v =
NKcR 1 + KcR
v = K v + NK cR
300
200
100
0 1 2 3 4 5 0 Average number of bound molecules per species, n
Figure 9.20
(9.76)
The Scatchard equation demonstrates that a plot of v>cR versus v should yield a straight line having a slope of K and a yintercept of NK from which N can be readily determined. A Scatchard plot for K = 100 and N = 4 is presented in Figure 9.20.
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0.1 0.2 0.3 0.4 0.5 0.6 0.7 Concentration of free species, CR
n CR
N N KicR v = a vi = a 1 + KicR i=1 i=1
K510
0.2
If there are N binding sites, then the average number of molecules adsorbed per species is simply the sum of the average values for each specific binding site:
K530
1
(9.70)
(In biochemistry, K is commonly expressed in terms of concentrations and therefore has units.) Assuming that there is a single binding site for the molecule, then the average number of bound molecules per species (v2 is given by cRM v = cM + cRM
1.2
Average number of bound – molecules per species, n
cRM K = cR * cM
263
Scatchard plot of independent binding for K = 100 and N = 4. K, equilibrium constant; N, number of binding sites for the molecule of interest. Note that the y intercept is K * N = 400, the slope is K =  100, and the x intercept is 4.
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Deviations from linear behavior in a Scatchard plot can occur if binding is not independent or if the binding sites are not equivalent. Example Problem 9.14 illustrates an application of Equation 9.76.
EXAMPLE PROBLEM 9.14 A variety of molecules physically bind to DNA by intercalation or insertion between the base pairs. The intercalation of chlorobenzylidine (CB) to calf thymus DNA was studied by circular dichroism spectroscopy [Zhong, W., Yu, J.S., Liang, Y., Fan, K., and Lai, L., Spectrochimica Acta, Part A, 60A (2004): 2985]. Using the following representative binding data, determine the binding constant (K) and the number of binding sites per base (N). cCB 1 :106 M 2 v
1.0
2.0
3.0
5.0
10.0
0.0063
0.0123
0.0180
0.0285
0.0515
Solution Plot of v>cCB versus v. Using the provided binding data allows for a determination of K and N. Using the data, we can construct the following Scatchard plot:
– 21 n/C CB (M )
6500
6000
5500
5000 Tuc 5 65.85°C
Temperature
a9
L1 T1
l1
L2 l2
a L1 + L2
0
x1
x2 xphenol
x3 1
Figure 9.21
A temperature–composition (T–x) phase diagram with an upper consolute temperature for the partially miscible liquids water and phenol. Pressure is held constant at 1 bar. The beigecolored region labeled L 1 + L 2 is a twophase region. The bluecolored balance of the diagram is a onephase region.
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0
0.02 0.04 – Average number of bound molecules per species, n
0.06
A best fit to a straight line results in a slope of 26,000 M1 and a y intercept of 6500. Comparison to the Scatchard equation reveals that the slope is equal to K, so that K = 26,000 M 1. In addition, the y intercept is equal to the product of N and K; therefore, N = 0.25. This value of N is consistent with a binding site composed of four bases, or two base pairs, which is consistent with intercalation between base pairs.
9.14 SOLUTIONS FORMED FROM PARTIALLY MISCIBLE LIQUIDS
Two liquids need not be miscible in all proportions. Over a limited range of relative concentration, a homogeneous solution can be formed, and outside of this range, two liquid phases separated by a phase boundary can be observed. An example is the water–phenol system for which a temperature–composition (T–x) phase diagram is shown in Figure 9.21. The pressure is assumed to be sufficiently high that no gas phase is present. As the phenol concentration increases from 0 at the temperature T1, a single, waterrich liquid phase of variable composition is formed. However, at x1, the appearance of a second, phenolrich liquid phase is observed, separated from the first phase by a
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9.14 Solutions Formed from Partially Miscible Liquids
210
l2
a0
Temperature, T (°C)
Temperature, T (°C)
l1
L1 + L2 L1
18.5
a9
265
L2
L1
L1 1 L2
L2
61
a 0
1
0
1 xnicotine
xtriethylamine
Figure 9.22
Figure 9.23
A T–x phase diagram with a lower consolute temperature for the partially miscible liquids water and triethylamine. Pressure is held constant at 1 bar. The beigecolored region labeled L 1 + L 2 is a twophase region. The bluecolored balance of the diagram is a onephase region.
A T–x phase diagram with an upper and lower consolute temperature for the partially miscible liquids water and nicotine. Pressure is held constant at 1 bar. The beigecolored region labeled L 1 + L 2 is a twophase region. The bluecolored balance of the diagram is a onephase region.
horizontal boundary with the denser phase below the less dense phase. If we start with pure phenol and add water at T1, separation into two phases is observed at x3. In the phenol–water system, solubility increases with temperature so that the extent in x of the twophase region decreases with increasing T, and phase separation is observed only at a single composition at the upper consolute temperature Tuc. For T 7 Tuc, phenol and water are completely miscible. The composition of the two phases in equilibrium for the phenol concentration given by x2 at a given temperature is given by the x values at the ends of the tie line l1l2, and the relative amount of the two phases present is given by the lever rule:
x3  x2 moles L 1 al2 = = x2  x1 moles L 2 al1
Concept Phase diagrams are useful in describing partially miscible liquids.
(9.77)
Although less common, liquid–liquid systems are known in which the solubility decreases as the temperature increases (at least for part of the temperature range). In this case, a lower consolute temperature is observed, below which the two liquids are completely miscible. An example is the triethylamine–water system shown in Figure 9.22. The water–nicotine system has both a lower and an upper consolute temperature, and phase separation is only observed between 61°C and 210°C, as shown in Figure 9.23. We next consider solutions of partially miscible liquids at pressures for which a vapor phase is in equilibrium with one or more liquid phases. The composition of the vapor for a given value of the average composition of the liquid depends on the relative values of the upper consolute temperature and the minimum boiling temperature for the azeotrope. For the system shown in Figure 9.24a, the vapor phase is only present at temperatures greater than the upper consolute temperature. In this case, the vapor and liquid composition is the same as that described in Section 9.4 for minimum boiling point azeotropes. However, if the upper consolute temperature is greater than the boiling temperature of the minimum boiling azeotrope, the T–x diagram shows new features, as illustrated for the water–butanol system in Figure 9.24b. First, consider the system at point a in Figure 9.24b. At this temperature, the system consists of a single liquid phase enriched in butanol. If the temperature is increased to Tb at constant pressure, the gasphase composition is given by point b. If the vapor of this composition is removed from the system and cooled to the original temperature Ta, it is at point c in the phase diagram, which corresponds to two liquid phases of composition d and e. This system consists of two components and, therefore, F = 3  p = 1.
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Figure 9.24
T–x composition phase diagrams for partially miscible liquids. (a) A generic phase diagram in which the upper consolute temperature is at a temperature lower than the azeotrope boiling temperature. (b) The system water–butanol. The upper consolute temperature is at a temperature higher than the minimum azeotrope boiling temperature. For both diagrams, pressure is held constant at 1 bar and the beigecolored regions labeled L 1 + L 2 and L + V are twophase regions, whereas the bluecolored regions are onephase regions.
L1V
Temperature
T0a
c
a
b Solid
0
1
xB
(a)
Temperature
Liquid T0a
g
Te
h
f e
T0b
d
L1 1 L2
Z1
xB
100
l k
Tj
L2 1 V
b
Tb
j
L1 1 V
94 L1
g
d
i f
h c
L2 Ta
e a
L1 1 L2 0
ZB
1
0
1 Z butanol
(b)
Next, consider heating the liquid described by point ƒ in the twoliquidphase region at constant pressure. Vapor first appears when the temperature is increased to 94°C and has the composition given by point i, corresponding to a minimum boiling point azeotrope. The vapor is in equilibrium with two liquid phases that have the composition given by points g and h. Because the pressure is constant in Figure 9.24b and p = 3 at point i, the system has zero degrees of freedom. Distillation of the solution under these conditions results in two separate liquid phases of fixed composition g and h. As more heat is put into the system, liquid is converted to vapor, but the composition of the two liquid phases and the vapor phase remains constant as described by points g, h, and i. Because the vapor is enriched in butanol with respect to the average composition of the system, the butanolrich phase L 2 is converted to vapor before the waterrich phase L 1. At the point when the L 2 phase disappears, a degree of freedom is gained, the temperature increases beyond 94°C, and the vapor composition changes along the i–j curve. As the vapor approaches the composition corresponding to j, the entire system is converted to vapor and the last drop of L 1, which disappears at Tj, has the composition given by point k. As the last drop of liquid is converted to vapor, an additional degree of freedom is gained, and further heat input results in an increase in the temperature of the vapor, with no change in its composition. To summarize, continued input of heat into a system of composition ƒ at 94°C increases the amount of material in the vapor phase. As long as the two phases of composition g and h are present, the temperature is constant and the amount of vapor increases, but its composition is fixed at i. After the last drop of L 2 of composition h boils off, the vapor composition continuously changes along the curve i–j. As the last drop of L 1 of composition k evaporates, the vapor composition approaches the average composition of the system given by ƒ, j, and l.
9.15 SOLID–SOLUTION EQUILIBRIUM
Solid
0
L1V Liquid
(a)
Liquid
118
Vapor Temperature, T (°C)
Temperature
Vapor
ZB
1
(b)
Figure 9.25
T–x phase diagrams showing solid– liquid equilibria in a twocomponent system. Pressure is held constant at 1 bar. (a) Component B is assumed to be the solute. (b) Fullphase diagram, including eutectic point for all possible concentrations of A and B.
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Consider the binary system A–B. At low temperatures, the components separate into pure solid phases A and B. At higher temperatures, this system consists of a single liquid phase, as A and B are miscible for all proportions of A and B. The transition between two immiscible solid phases and a single liquid phase exhibits interesting behavior, which we will now discuss. If a liquid solution of B in A (A is the solvent and therefore xB 6 xA2 is cooled sufficiently, pure solvent A will crystallize out in a solid phase, as discussed in Section 9.7 in connection with the freezing point depression. For example, point a in Figure 9.25a represents a liquid solution of composition b in equilibrium with pure solid A (point c). The relative number of moles of solution and solid can be found using the lever rule. The temperature T at which freezing occurs is given by
R ln xA 1 1 = T Tfus1A2 ∆ fusH1A2
(9.78)
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Vocabulary
Concept A minimum in the melting temperature with concentration can be observed for solidsolid solutions.
1500 1300 Temperature, T (°C)
where Tfus1A2and ∆ fus H1A2 are, respectively, the freezing temperature and the enthalpy of fusion of pure solvent A. The temperature T at which freezing occurs is shown as a function of xB by the dark curve in Figure 9.25a, and the dashed range indicates the region beyond which Equation (9.78) is unlikely to be valid. The same argument holds for a solution of A in B whereby we have interchanged the solvent and solute roles. Therefore, the appropriate phase diagram for all values of xB is that shown in Figure 9.25b. For the solid, we use the variables ZA and ZB rather than xA and xB because the solid consists of an inhomogeneous mixture of crystallites of pure A and pure B. In Figure 9.25b, point d represents pure solid B in equilibrium with a solution of composition ƒ. Point g represents pure solid A in equilibrium with the solution of composition h. Point e is a special point called the eutectic point that is the minimum melting point observed in the A–B system. The temperature Te is called the eutectic temperature (from the Greek word meaning easily melted). Because the tie line at Te extends across the entire phase diagram, a solution at Te is in equilibrium with both pure A (point i) and pure B (point j). As more heat is put into the system, the volume of the solution phase increases, but its composition remains constant at Ze until one of the pure solids is completely melted because, while the three phases are present, F = 0. Eutectic materials are widely used in making plumbing and electrical connections with a solder. The benefit of such materials is that the components to be joined do not need to be heated to high temperatures. Until the 1980s, copper plumbing components were connected using Pb–Sn solders. Now that it is recognized that even small amounts of lead present a health hazard, Pb–Sn solders have been replaced by leadfree solders, which may contain tin, copper, silver, indium, zinc, antimony, or bismuth. Another important application of eutectic materials is in the manufacture of electrical connections to silicon chips in the microelectronics industry. Gold is deposited onto the Si surface using masks to establish an appropriate pattern. Subsequently, the chip is heated to nearly 400°C. As the temperature increases, gold atoms diffuse rapidly into the silicon, and a very thin liquid film forms at the interface once enough gold has diffused to reach the eutectic composition. As more heat is put into the chip, a larger volume of eutectic alloy is produced, which is ultimately the electrical connection after the chip is cooled rapidly to room temperature, freezing in the hightemperature equilibrium state. The Au–Si phase diagram is shown in Figure 9.26; it is seen that the eutectic temperature is approximately 700°C below the melting point of pure gold and 950°C below the melting point of pure Si. The eutectic mixture is useful in bonding a gold wire to a silicon chip to form an electrical connection. Imagine that a gold wire is placed in contact with a Si chip on which Au has been deposited. If the chip is heated, the eutectic liquid will form before the Si or the Au melts, allowing the eutectic mixture to form a connection between the gold wire and the chip without melting the Au or the Si.
1100 Liquid
900 700
Liquid + Solid
500
363°C
300
18.6%
100 0
Solid
20 40 60 80 100 Atomic percent silicon
Figure 9.26
The T–x phase diagram for the gold– silicon system. Note that the eutectic melting point is much lower than the melting points of the pure substances Au and Si. Pressure is constant at 1 bar.
VOCABULARY activity activity coefficient average composition binding equilibria boiling point elevation colligative properties eutectic temperature fractional distillation freezing point depression Gibbs–Duhem equation Henry’s law
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Henry’s law constant Henry’s law standard state ideal dilute solution ideal solution independent site binding lever rule limiting law lower and upper consolute temperature osmosis osmotic pressure partial molar volume
pressure–average composition (P–Z) diagram Raoult’s law Raoult’s law standard state Scatchard equation Scatchard plot semipermeable membrane solute solvent temperature–composition diagram tie line
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KEY EQUATIONS Equation
Significance of Equation
Equation Number
Pi = xiP *i i = 1, 2
Definition of ideal solution
9.1
msolution = mvapor i i
Definition of equilibrium between liquid and vapor phases
9.2
Chemical potential of pure liquid in equilibrium with its vapor
9.6
Central equation describing ideal solutions
9.7
Pressure in vapor above ideal binary solution in terms of mole fraction in liquid phase
9.9
Mole fraction of component in vapor phase above binary solution
9.10
Pressure in vapor above ideal binary solution in terms of mole fraction in vapor phase
9.12
Phase rule
9.14
Lever rule for calculating amounts of vapor and liquid in coexistence region
9.18
Gibbs–Duhem equation for binary solution relates chemical potential of solute and solvent
9.23
Freezing point depression
9.33
Boiling point elevation
9.35
van’t Hoff equation for osmotic pressure
9.41
Definition of partial volume
9.44
Volume of binary solution in terms of partial molar volumes
9.45
Gibbs–Duhem equation relating partial molar volumes for binary solution
9.46
mi = m*i + RT ln
Pi
P *i mi = m*i + RT ln xi Ptot = P *2 + 1P *1  P *22x1 y1 =
P1 x1P *1 = * Ptot P 2 + 1P *1  P *22x1 P *1P *2
Ptot =
P *1
+
1P *2

F = C  p + 2
P *12y1
tot ntot liq1ZB  xB2 = nvapor1yB  ZB2
n1dm1 + n2dm2 = 0 or x1dm1 + x2dm2 = 0 ∆Tƒ = ∆Tb = p =
RMsolventT 2fus msolute =  Kƒ msolute ∆ fusH
RMsolventT 2vap ∆ vapH
msolute = Kb msolute
nsolute RT V
V1(P, T, n1, n22 = a
0V b 0n1 P, T, n2
V = n1V1(P, T, n1, n22 + n2V21P, T, n1, n22 x1 dV1 + x2 dV2 = 0 or dV1 = 
asolvent =
ai =
Pi kH i
Psolvent P *solvent
; gsolvent =
and gi =
x2 dV x1 2
asolvent xsolvent
ai xi
Definition of activity and activity coefficient for ideal dilute solution
9.48, 9.49
Definition of activity and activity coefficient for ideal dilute solution based on Henry’s law
9.55
CONCEPTUAL PROBLEMS Q9.1 Why is the magnitude of the boiling point elevation less than that of the freezing point depression? Q9.2 Fractional distillation of a particular binary liquid mixture leaves a residual liquid that consists of both components, the composition of which does not change as the liquid is boiled off. Is this behavior characteristic of a maximum or a minimum boiling point azeotrope?
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Q9.3 In the description of Figure 9.24, the following sentence appears: “At the point when the L 2 phase disappears, the temperature increases beyond 94°C and the vapor composition changes along the i–j curve.” Why does the vapor composition change along the i–j curve?
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Q9.4 Explain why chemists conducting quantitative work using liquid solutions prefer to express concentration in terms of molality rather than molarity. Q9.5 Explain the usefulness of a tie line on a P–Z phase diagram such as that of Figure 9.4. Q9.6 For a pure substance, the liquid and gaseous phases can only coexist for a single value of pressure at a given temperature. Is this also the case for an ideal solution of two volatile liquids? Q9.7 Using the differential form of G, dG = VdP  SdT, show that if ∆ mixG = nRT a i xi ln xi, then ∆ mixH = ∆ mixV = 0. Q9.8 What information can be obtained from a tie line in a P–Z phase diagram? Q9.9 You boil an ethanol–benzene mixture with xethanol = 0.35. What is the composition of the vapor phase that first appears? What is the composition of the last liquid left in the vessel? Q9.10 What can you say about the composition of the solid at temperatures less than the eutectic temperature in Figure 9.26 on a microscopic scale?
Numerical Problems
269
Q9.11 Is a whale likely to get the bends when it dives deep into the ocean and resurfaces? Answer this question by considering the likelihood of a diver getting the bends if he or she dives and resurfaces on one lung full of air as opposed to breathing air for a long time at the deepest point of the dive. Q9.12 Why is the preferred standard state for the solvent in an ideal dilute solution the Raoult’s law standard state? Why is the preferred standard state for the solute in an ideal dilute solution the Henry’s law standard state? Is there a preferred standard state for the solution in which xsolvent = xsolute = 0.5? Q9.13 The entropy of two liquids is lowered if they mix. How can immiscibility be explained in terms of thermodynamic state functions? Q9.14 Explain why colligative properties depend only on the concentration, and not on the identity of the solute molecules. Q9.15 The statement “The boiling point of a typical liquid mixture can be reduced by approximately 100°C if the pressure is reduced from 760 to 20 Torr” is found in Section 9.4. What figure or figures in Chapter 9 can you identify to support this statement in a qualitative sense?
NUMERICAL PROBLEMS Section 9.3 P9.1 At 325 K, the vapor pressure of benzene is 295 Torr and that of hexane is 439 Torr. Calculate the vapor pressure of a solution for which xbenzene = 0.28. Assume ideal behavior. P9.2 An ideal solution is made up of the volatile liquids A and B, for which P *A = 120. Torr and P *B = 70.5 Torr. Initially, the pressure is high enough that only the liquid phase is present. As pressure is reduced, the first vapor is observed at a total pressure of 98.5. Torr. Calculate xA. P9.3 At 280. K, the vapor pressure of ethyl bromide is 0.295 bar, and that of ethyl chloride is 0.823 bar. Assume that the solution is ideal. For the case in which there is only a trace of liquid present and the mole fraction of ethyl chloride in the vapor is 0.750, answer these questions: a. What are the total pressure and the mole fraction of ethyl chloride in the liquid? b. If there are 4.25 mol of liquid and 2.75 mol of vapor present at the same pressure as in part (a), what is the overall composition of the system? P9.4 A and B form an ideal solution at 298 K. With xA = 0.275, P *A = 95.0 Torr, and P *B = 44.5 Torr. a. Calculate the partial pressures of A and B in the gas phase. b. A portion of the gas phase is removed and condensed in a separate container. Calculate the partial pressures of A and B in equilibrium with this liquid sample at 298 K. P9.5 Given the vapor pressures of the pure liquids and the overall composition of the system, what are the upper and lower limits of pressure between which liquid and vapor coexist in an ideal solution?
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P9.6 A and B form an ideal solution. At a total pressure of 0.650 bar, yA = 0.485 and xA = 0.390. Use this information to calculate the vapor pressure of pure A and pure B. P9.7 An ideal solution is formed by mixing liquids A and B at 298 K. The vapor pressure of pure A is 175 Torr, and that of pure B is 104 Torr. If the mole fraction of A in the vapor is 0.550, what is the mole fraction of A in the solution? P9.8 A solution is prepared by dissolving 54.0 g of a nonvolatile solute in 150 g of water. The vapor pressure above the solution is 22.97 Torr, and the vapor pressure of pure water is 23.76 Torr at this temperature. What is the molecular mass of the solute? P9.9 A volume of 7.32 L of air is bubbled through liquid toluene at 360. K, thus reducing the mass of toluene in the beaker by 11.2 g. Assuming that the air emerging from the beaker is saturated with toluene, determine the vapor pressure of toluene at this temperature. P9.10 The vapor pressures of 1bromobutane and 1chlorobutane can be expressed in the form ln
Pbromo 1584.8 = 17.076 Pa T  111.88 K
and ln
Pchloro 2688.1 = 20.612 Pa T  55.725 K
Assuming ideal solution behavior, calculate xbromo and ybromo at 280. K and a total pressure of 4000. Pa.
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P9.11 In an ideal solution of A and B, 2.50 mol are in the liquid phase and 4.75 mol are in the gaseous phase. The overall composition of the system is ZA = 0.250 and xA = 0.175. Calculate yA. P9.12 Assume that 1bromobutane and 1chlorobutane form an ideal solution. At 285 K, P *chloro = 7242 Pa and P *bromo = 2756 Pa. When only a trace of liquid is present at 285 K, ychloro = 0.800. a. Calculate the total pressure above the solution. b. Calculate the mole fraction of 1chlorobutane in the s olution. c. Calculate Zchloro and Zbromo. P9.13 An ideal dilute solution is formed by dissolving the solute A in the solvent B. Write expressions equivalent to Equations (9.9) through (9.13) for this case. P9.14 At 325 K, pure toluene and hexane have vapor pressures of 1.42 * 104 Pa and 5.77 * 104 Pa, respectively. a. Calculate the mole fraction of hexane in the liquid mixture that boils at a pressure of 0.400 atm. b. Calculate the mole fraction of hexane in the vapor that is in equilibrium with the liquid of part (a). P9.15 You wish to remediate soil that has been contaminated by 1chlorobutane by passing air through it at a total pressure of 1.00 bar. (a) Assuming that the air leaving the soil is saturated with 1chlorobutane, calculate how many liters of air are required to remove 10.0 g of 1chlorobutane at 298 K. (b) Would raising the temperature of the soil make the process more efficient? Answer this question by calculating how many liters of air are required to remove 10.0 g of 1chlorobutane at 320. K using the data in Table 8.3. Section 9.7 P9.16 The dissolution of 14.5 g of a substance in 450. g of benzene at 298 K raises the boiling point by 0.855°C. Note that Kƒ = 5.12 K kg mol1, Kb = 2.53 K kg mol1, and the density of benzene is 876.6 kg m3. Calculate the freezing point depression, the ratio of the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, and the molar mass of the solute. P *benzene = 103 Torr at 298 K. P9.17 The heat of fusion of ice is 6.008 * 103 J mol1 at its normal melting point of 273.15 K. Calculate the freezing point depression constant Kƒ. Section 9.8 P9.18 The osmotic pressure of an unknown substance is measured at 298 K. Determine the molar mass if the concentration of this substance is 28.5 kg m3 and the osmotic pressure is 4.00 * 104 Pa. The density of the solution is 997 kg m3. P9.19 A sample of glucose (C6H12O62 of mass 9.75 g is placed in a test tube of radius 2.00 cm. The bottom of the test tube is a membrane that is semipermeable to water. The tube is partially immersed in a beaker of water at 298 K, so that the bottom of the test tube is only slightly below the level of the water in the beaker. The density of water at this temperature is 997 kg m3. After equilibrium is reached, how high is the level of the water in the tube above that in the beaker? What is the value of the osmotic pressure? You may find the approximation ln 11>11 + x22 ≈ x useful.
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Section 9.9 P9.20 The partial molar volumes of water and ethanol in a solution with xH2O = 0.45 at 25°C are 17.0 and 57.0 cm3 mol1, respectively. Calculate the volume change upon mixing sufficient ethanol with 4.50 mol of water to give this concentration. The densities of water and ethanol are 0.997 and 0.7893 g cm3, respectively, at this temperature. P9.21 A solution is made up of 199.0 g of ethanol and 116.7 g of H2O. If the volume of the solution is 360.1 cm3 and the partial molar volume of H2O is 17.0 cm3 mol1, what is the partial molar volume of ethanol under these conditions? P9.22 The densities of pure water and ethanol are 997 and 789 kg m3, respectively. The partial molar volumes of ethanol and water are 55.2 and 17.8 * 103 L mol1, respectively. Calculate the change in volume relative to the pure components when 3.25 L of a solution with xethanol = 0.540 is prepared. Section 9.10 P9.23 The partial pressures of Br2 above a solution containing CCl4 as the solvent at 25°C have the values listed, as a function of the mole fraction of Br2 in solution, in the following table [G. N. Lewis and H. Storch, J. American Chemical Society 39 (1917), 2544]. Use these data and a graphical method to determine the Henry’s law constant for Br2 in CCl4 at 25°C. xBr2
0.00394 0.00420 0.00599 0.0102
PBr2 (Torr)
1.52 1.60 2.39 4.27
xBr2
PBr2 (Torr)
0.0130 0.0236 0.0238 0.0250
5.43 9.57 9.83 10.27
P9.24 The data from Problem P9.23 can be expressed in terms of the molality rather than the mole fraction of Br2. Use the data from the following table and a graphical method to determine the Henry’s law constant for Br2 in CCl4 at 25°C in terms of molality. mBr2 1mol kg −1 2
0.026 0.028 0.039 0.067
PBr2 (Torr)
1.52 1.60 2.39 4.27
mBr2 1mol kg −1 2
0.086 0.157 0.158 0.167
PBr2 (Torr)
5.43 9.57 9.83 10.27
P9.25 (a) How many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at 298 K if the pressure used in the carbonation process was 2.4 bar? (b) What volume would the dissolved CO2 occupy as a gas at 298 K and 1.00 bar? The density of water at this temperature is 998 kg m3. P9.26 Calculate the concentration of O2 dissolved in water in equilibrium with air at 298 K and a total pressure of 1.00 bar. The vapor pressure of water at this temperature is 3.17 * 103 Pa, and the mol fraction of O2 in air is 0.2095.
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Numerical Problems
Section 9.11 P9.27 G. Ratcliffe and K. Chao [Canadian Journal of Chemical Engineering 47 (1969), 148] obtained the following tabulated results for the variation of the total pressure above a solution of isopropanol (P *1 = 1008 Torr) and ndecane (P *2 = 48.3 Torr) as a function of the mole fraction of the ndecane in the solution and vapor phases. Using these data, calculate the activity coefficients for both components using a Raoult’s law standard state. P (Torr)
x2
y2
942.6 909.6 883.3 868.4 830.2 786.8 758.7
0.1312 0.2040 0.2714 0.3360 0.4425 0.5578 0.6036
0.0243 0.0300 0.0342 0.0362 0.0411 0.0451 0.0489
P9.28 At a given temperature, a nonideal solution of the volatile components A and B has a vapor pressure of 320. Torr. For this solution, yA = 0.375. In addition, xA = 0.250, P *A = 610. Torr, and P *B = 495 Torr. Calculate the activity and activity coefficient of A and B. P9.29 At 39.9°C, a solution of ethanol (x1 = 0.9006, P *1 = 130.4 Torr) and isooctane (P *2 = 43.9 Torr) forms a vapor phase with y1 = 0.6667 at a total pressure of 185.9 Torr. a. Calculate the activity and activity coefficient of each component. b. Calculate the total pressure that the solution would have if it were ideal. P9.30 Calculate the activity and activity coefficient for CS2 at xCS2 = 0.6124 using the data in Table 9.3 for both a Raoult’s law and a Henry’s law standard state. Section 9.12 P9.31 At high altitudes, mountain climbers are unable to absorb a sufficient amount of O2 into their bloodstreams to maintain a high activity level. At a pressure of 1 bar, blood is typically 95% saturated with O2, but near 12,000 feet where the pressure is 0.660 bar, the corresponding degree of saturation is 60.%. Assuming that the Henry’s law constant for blood is the same as that for water, calculate the amount of O2 dissolved in 1.00 L of blood for pressures of 1 bar and 0.660 bar. Air contains 20.99% O2 by volume. Assume that the density of blood is 998 kg m3. P9.32 a. Calculate the solubility of N2 in 1 L of water at 298 K if its pressure above the solution is 12.75 bar. The density of water is 997 kg m3. b. If the pressure were suddenly reduced to 1.00 bar, as could happen in a diving accident, what volume of N2 is released per liter of blood, assuming that the solubility of N2 is the same as that in water?
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Section 9.13 P9.33 The binding of NADH to human liver mitochondrial isozyme was studied [Biochemistry 28 (1989) 5367], and it was determined that only a single binding site is present with K = 2.0 * 107 M1. What concentration of NADH is required to occupy 10% of the binding sites? P9.34 DNA is capable of forming complex helical structures. An unusual triplehelix structure of poly(dA).2poly(dT) DNA was studied by P. V. Scaria and R. H. Shafer [Journal of Biological Chemistry, 266 (1991) 5417] where the intercalation of ethidium bromide was studied using UV absorption and circular dichroism spectroscopy. The following representative data were obtained using the results of this study: cEth 1 MM 2
N
0.63
1.7
5.0
10.0
14.7
0.007
0.017
0.039
0.059
0.070
Using the above data, determine K and N for the binding of ethidium bromide to the DNA triplehelical structure. Section 9.14 P9.35 Two liquids, A and B, are immiscible for T 6 75.0°C and for T 7 45.0°C and are completely miscible outside of this temperature range. Sketch the phase diagram, showing as much information as you can from these observations. P9.36 Describe the changes you would observe as the temperature of a mixture of phenol and water at point a in Figure 9.21 is increased until the system is at point a′. How does the relative amount of separate phases of phenolrich liquid L 2 and waterrich liquid L 1 change along this path? P9.37 Describe what you would observe if you heated the liquid mixture at the composition corresponding to point i in Figure 9.24b from a temperature below Ta to 118°C. P9.38 Two liquids, A and B, are immiscible for xA = xB = 0.5, for T = 75.0°C, and completely miscible for T 7 75.0°C. Sketch the phase diagram, showing as much information as you can from these observations. P9.39 Describe the changes you would observe as the temperature of a mixture of triethylamine and water at point a in Figure 9.22 is increased until the system is at point a″. How does the relative amount of separate phases of triethylamine and water change along this path? P9.40 Describe the changes in a beaker containing water and butanol that you would observe along the path a S b S c in Figure 9.24b. How would you calculate the relative amounts of different phases present along the path? P9.41 Describe the changes in a beaker containing water and butanol that you would observe along the path ƒ S j S l in Figure 9.24b. How would you calculate the relative amounts of different phases present along the path? P9.42 Describe the changes in a beaker containing water and butanol that you would observe along the path ƒ S j S k in Figure 9.24b. How would you calculate the relative amounts of different phases present along the path?
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b. The vertical dashed line shows a cooling process, starting at point a which corresponds to a single phase liquid region. Describe which phases and components, and the relative amounts of the components are present at the points labeled a–e. a b c
Temperature
Section 9.15 P9.43 Describe what you would observe if you heated the solid at the composition 40. atomic percent Si in Figure 9.26 from 300.°C to 1300.°C. P9.44 Describe the system at points d and g in Figure 19.25b. How would you calculate the relative amounts of different phases present at these points? P9.45 A phase diagram for an alloy, the two components of which are completely soluble in the liquid phase and completely insoluble in the solid phase, is shown below. a. Label the four regions of the phase diagram indicating what components (A and/or B, where A and B denote solids) and phases are present. b. The vertical dashed line shows a cooling process, starting at point a. Describe which components and the relative amounts of the components that are present at the points labeled a–e.
d e
a 0
b
P9.47 A phase diagram for an alloy for which an intermetallic compound, AB, is formed is shown here. a. Label the different regions of the phase diagram indicating what components (A and/or B and/or AB where A, B, and AB denote solids) and phases are present. b. The vertical dashed line shows a cooling process, s tarting at point a. Describe which components and the r elative amounts of the components are present at the points labeled a–d.
c
Temperature
1
xA
d e
a
xA
1
P9.46 A phase diagram for an alloy, the two components of which are completely soluble in the liquid phase and partially soluble in the solid phase, is shown here. a. Label the different regions of the phase diagram, indicating what components (A and/or B where A and B denote components) and phases are present.
b
Temperature
0
c d
AB
FURTHER READING Castellan, G. “Phase Equilibria and the Chemical Potential.” Journal of Chemical Education 32 (1955): 424–427. Chinard, F. P. “The Definition of Osmotic Pressure.” Journal of Chemical Education 31 (1954): 66–69. McInnes, D. M., and Kampbell, D. “Bubble Stripping to Determine Hydrogen Concentrations in Ground Water: A Practical Application of Henry’s Law.” Journal of Chemical Education 80 (2003): 516–519.
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0
xa
xA
1
Nikitas, P. “Applications of the Gibbs–Duhem Equation.” Journal of Chemical Education, 78 (2001): 1070–1075. Rosenberg, R. M., and Peticolas, W. L. “Henry’s Law: A Retrospective.” Journal of Chemical Education 81 (2004): 1647–1652.
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C H A P T E R
10
Electrolyte Solutions
10.1 Enthalpy, Entropy, and Gibbs Energy of Ion Formation in Solutions
10.2 Understanding the Thermodynamics of Ion Formation and Solvation
WHY is this material important? Electrolyte solutions occur all around us, with seawater being the most obvious example. These solutions are quite different from the ideal and real solutions of neutral solutes discussed in Chapter 9. The fundamental reason for this difference is that solutes in electrolyte solutions exist as solvated positive and negative ions.
10.3 Activities and Activity Coefficients for Electrolyte Solutions
10.4 Calculating g { Using the Debye–Hückel Theory
10.5 Chemical Equilibrium in Electrolyte Solutions
WHAT are the most important concepts and results? Electrolyte and nonelectrolyte solutions are so different because the Coulomb interactions between ions in an electrolyte solution are of much longer range than the interactions between neutral solutes. For this reason, electrolyte solutions deviate from ideal behavior at much lower concentrations than do solutions of neutral solutes. The Debye–Hückel limiting law provides a useful way to calculate activity coefficients for dilute electrolyte solutions.
WHAT would be helpful for you to review for this chapter? It would be helpful to review the material on solutions in Chapter 9.
10.1 ENTHALPY, ENTROPY, AND GIBBS ENERGY OF ION FORMATION IN SOLUTIONS
This chapter discusses substances called electrolytes, which dissociate into positively and negatively charged mobile solvated ions when dissolved in an appropriate solvent. Consider the following overall reaction in water:
1 > 2 H21g2 + 1 > 2 Cl21g2 S H +1aq2 + Cl1aq2
(10.1)
∆ r H ~ = ∆ f H ~ 1H +, aq2 + ∆ f H ~ 1Cl, aq2
(10.2)
in which H +1aq2 and Cl1aq2 represent mobile solvated ions. Although similar in structure, Equation (10.1) represents a reaction that is quite different from the gasphase dissociation of an HCl molecule to give H +1g2 and Cl1g2. For the reaction in solution, ∆ r H ~ is 167.2 kJ mol1. The shorthand notation H +1aq2 and Cl1aq2 refers to positive and negative ions together with their associated solvation shell. A solvation shell around an ion consists of those solvent molecules that are bound to the ion by electrostatic forces and do not diffuse freely in the solution. The solvation shell is essential in lowering the energy of the ions, thereby making the previous reaction spontaneous. Although energy flow into the system is required to dissociate and ionize hydrogen and chlorine, even more energy is gained in the reorientation of the dipolar water molecules around the ions in the solvation shell. Therefore, the reaction is exothermic. The standard state enthalpy for this reaction can be written in terms of formation enthalpies:
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Concept The reference zero of energy for electrolyte solutions is set by the equation ∆ f G ~ 1H +, aq2 = 0 for all T.
There is no contribution of H21g2 and Cl21g2 to ∆ r H ~ in Equation (10.2) because ∆ f H ~ for a pure elem