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Chemistry 2e

16.4 Free Energy

Chemistry 2e16.4 Free Energy
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  1. Preface
  2. 1 Essential Ideas
    1. Introduction
    2. 1.1 Chemistry in Context
    3. 1.2 Phases and Classification of Matter
    4. 1.3 Physical and Chemical Properties
    5. 1.4 Measurements
    6. 1.5 Measurement Uncertainty, Accuracy, and Precision
    7. 1.6 Mathematical Treatment of Measurement Results
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  3. 2 Atoms, Molecules, and Ions
    1. Introduction
    2. 2.1 Early Ideas in Atomic Theory
    3. 2.2 Evolution of Atomic Theory
    4. 2.3 Atomic Structure and Symbolism
    5. 2.4 Chemical Formulas
    6. 2.5 The Periodic Table
    7. 2.6 Molecular and Ionic Compounds
    8. 2.7 Chemical Nomenclature
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  4. 3 Composition of Substances and Solutions
    1. Introduction
    2. 3.1 Formula Mass and the Mole Concept
    3. 3.2 Determining Empirical and Molecular Formulas
    4. 3.3 Molarity
    5. 3.4 Other Units for Solution Concentrations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  5. 4 Stoichiometry of Chemical Reactions
    1. Introduction
    2. 4.1 Writing and Balancing Chemical Equations
    3. 4.2 Classifying Chemical Reactions
    4. 4.3 Reaction Stoichiometry
    5. 4.4 Reaction Yields
    6. 4.5 Quantitative Chemical Analysis
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  6. 5 Thermochemistry
    1. Introduction
    2. 5.1 Energy Basics
    3. 5.2 Calorimetry
    4. 5.3 Enthalpy
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  7. 6 Electronic Structure and Periodic Properties of Elements
    1. Introduction
    2. 6.1 Electromagnetic Energy
    3. 6.2 The Bohr Model
    4. 6.3 Development of Quantum Theory
    5. 6.4 Electronic Structure of Atoms (Electron Configurations)
    6. 6.5 Periodic Variations in Element Properties
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  8. 7 Chemical Bonding and Molecular Geometry
    1. Introduction
    2. 7.1 Ionic Bonding
    3. 7.2 Covalent Bonding
    4. 7.3 Lewis Symbols and Structures
    5. 7.4 Formal Charges and Resonance
    6. 7.5 Strengths of Ionic and Covalent Bonds
    7. 7.6 Molecular Structure and Polarity
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  9. 8 Advanced Theories of Covalent Bonding
    1. Introduction
    2. 8.1 Valence Bond Theory
    3. 8.2 Hybrid Atomic Orbitals
    4. 8.3 Multiple Bonds
    5. 8.4 Molecular Orbital Theory
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  10. 9 Gases
    1. Introduction
    2. 9.1 Gas Pressure
    3. 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
    4. 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
    5. 9.4 Effusion and Diffusion of Gases
    6. 9.5 The Kinetic-Molecular Theory
    7. 9.6 Non-Ideal Gas Behavior
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  11. 10 Liquids and Solids
    1. Introduction
    2. 10.1 Intermolecular Forces
    3. 10.2 Properties of Liquids
    4. 10.3 Phase Transitions
    5. 10.4 Phase Diagrams
    6. 10.5 The Solid State of Matter
    7. 10.6 Lattice Structures in Crystalline Solids
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  12. 11 Solutions and Colloids
    1. Introduction
    2. 11.1 The Dissolution Process
    3. 11.2 Electrolytes
    4. 11.3 Solubility
    5. 11.4 Colligative Properties
    6. 11.5 Colloids
    7. Key Terms
    8. Key Equations
    9. Summary
    10. Exercises
  13. 12 Kinetics
    1. Introduction
    2. 12.1 Chemical Reaction Rates
    3. 12.2 Factors Affecting Reaction Rates
    4. 12.3 Rate Laws
    5. 12.4 Integrated Rate Laws
    6. 12.5 Collision Theory
    7. 12.6 Reaction Mechanisms
    8. 12.7 Catalysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  14. 13 Fundamental Equilibrium Concepts
    1. Introduction
    2. 13.1 Chemical Equilibria
    3. 13.2 Equilibrium Constants
    4. 13.3 Shifting Equilibria: Le Châtelier’s Principle
    5. 13.4 Equilibrium Calculations
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  15. 14 Acid-Base Equilibria
    1. Introduction
    2. 14.1 Brønsted-Lowry Acids and Bases
    3. 14.2 pH and pOH
    4. 14.3 Relative Strengths of Acids and Bases
    5. 14.4 Hydrolysis of Salts
    6. 14.5 Polyprotic Acids
    7. 14.6 Buffers
    8. 14.7 Acid-Base Titrations
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  16. 15 Equilibria of Other Reaction Classes
    1. Introduction
    2. 15.1 Precipitation and Dissolution
    3. 15.2 Lewis Acids and Bases
    4. 15.3 Coupled Equilibria
    5. Key Terms
    6. Key Equations
    7. Summary
    8. Exercises
  17. 16 Thermodynamics
    1. Introduction
    2. 16.1 Spontaneity
    3. 16.2 Entropy
    4. 16.3 The Second and Third Laws of Thermodynamics
    5. 16.4 Free Energy
    6. Key Terms
    7. Key Equations
    8. Summary
    9. Exercises
  18. 17 Electrochemistry
    1. Introduction
    2. 17.1 Review of Redox Chemistry
    3. 17.2 Galvanic Cells
    4. 17.3 Electrode and Cell Potentials
    5. 17.4 Potential, Free Energy, and Equilibrium
    6. 17.5 Batteries and Fuel Cells
    7. 17.6 Corrosion
    8. 17.7 Electrolysis
    9. Key Terms
    10. Key Equations
    11. Summary
    12. Exercises
  19. 18 Representative Metals, Metalloids, and Nonmetals
    1. Introduction
    2. 18.1 Periodicity
    3. 18.2 Occurrence and Preparation of the Representative Metals
    4. 18.3 Structure and General Properties of the Metalloids
    5. 18.4 Structure and General Properties of the Nonmetals
    6. 18.5 Occurrence, Preparation, and Compounds of Hydrogen
    7. 18.6 Occurrence, Preparation, and Properties of Carbonates
    8. 18.7 Occurrence, Preparation, and Properties of Nitrogen
    9. 18.8 Occurrence, Preparation, and Properties of Phosphorus
    10. 18.9 Occurrence, Preparation, and Compounds of Oxygen
    11. 18.10 Occurrence, Preparation, and Properties of Sulfur
    12. 18.11 Occurrence, Preparation, and Properties of Halogens
    13. 18.12 Occurrence, Preparation, and Properties of the Noble Gases
    14. Key Terms
    15. Summary
    16. Exercises
  20. 19 Transition Metals and Coordination Chemistry
    1. Introduction
    2. 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
    3. 19.2 Coordination Chemistry of Transition Metals
    4. 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
    5. Key Terms
    6. Summary
    7. Exercises
  21. 20 Organic Chemistry
    1. Introduction
    2. 20.1 Hydrocarbons
    3. 20.2 Alcohols and Ethers
    4. 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
    5. 20.4 Amines and Amides
    6. Key Terms
    7. Summary
    8. Exercises
  22. 21 Nuclear Chemistry
    1. Introduction
    2. 21.1 Nuclear Structure and Stability
    3. 21.2 Nuclear Equations
    4. 21.3 Radioactive Decay
    5. 21.4 Transmutation and Nuclear Energy
    6. 21.5 Uses of Radioisotopes
    7. 21.6 Biological Effects of Radiation
    8. Key Terms
    9. Key Equations
    10. Summary
    11. Exercises
  23. A | The Periodic Table
  24. B | Essential Mathematics
  25. C | Units and Conversion Factors
  26. D | Fundamental Physical Constants
  27. E | Water Properties
  28. F | Composition of Commercial Acids and Bases
  29. G | Standard Thermodynamic Properties for Selected Substances
  30. H | Ionization Constants of Weak Acids
  31. I | Ionization Constants of Weak Bases
  32. J | Solubility Products
  33. K | Formation Constants for Complex Ions
  34. L | Standard Electrode (Half-Cell) Potentials
  35. M | Half-Lives for Several Radioactive Isotopes
  36. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
  37. Index
By the end of this section, you will be able to:
  • Define Gibbs free energy, and describe its relation to spontaneity
  • Calculate free energy change for a process using free energies of formation for its reactants and products
  • Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
  • Explain how temperature affects the spontaneity of some processes
  • Relate standard free energy changes to equilibrium constants

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:

G=HTSG=HTS

Free energy is a state function, and at constant temperature and pressure, the free energy change (ΔG) may be expressed as the following:

ΔG=ΔHTΔSΔG=ΔHTΔS

(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)

The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:

ΔSuniv=ΔS+qsurrTΔSuniv=ΔS+qsurrT

The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, so this expression may be rewritten as:

ΔSuniv=ΔSΔHTΔSuniv=ΔSΔHT

Multiplying both sides of this equation by −T, and rearranging yields the following:

TΔSuniv=ΔHTΔSTΔSuniv=ΔHTΔS

Comparing this equation to the previous one for free energy change shows the following relation:

ΔG=TΔSunivΔG=TΔSuniv

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔSuniv. Table 16.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Relation between Process Spontaneity and Signs of Thermodynamic Properties
ΔSuniv > 0 ΔG < 0 spontaneous
ΔSuniv < 0 ΔG > 0 nonspontaneous
ΔSuniv = 0 ΔG = 0 at equilibrium
Table 16.3

What’s “Free” about ΔG?

In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work (w) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.

For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by

ΔG=ΔHTΔSΔG=ΔHTΔS

may be interpreted as representing the difference between the energy produced by the process, ΔH, and the energy lost to the surroundings, TΔS. The difference between the energy produced and the energy lost is the energy available (or “free”) to do useful work by the process, ΔG. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:

ΔG=wmaxΔG=wmax

However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.

Calculating Free Energy Change

Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, ΔG°, according to the following relation.

ΔG°=ΔH°TΔS°ΔG°=ΔH°TΔS°

Example 16.7

Using Standard Enthalpy and Entropy Changes to Calculate ΔG°Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process?

Solution The process of interest is the following:

H2O(l)H2O(g)H2O(l)H2O(g)

The standard change in free energy may be calculated using the following equation:

ΔG°=ΔH°TΔS°ΔG°=ΔH°TΔS°

From Appendix G:

Substance ΔHf°(kJ/mol)ΔHf°(kJ/mol) S°(J/K·mol)S°(J/K·mol)
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8

Using the appendix data to calculate the standard enthalpy and entropy changes yields:

ΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=[−241.82 kJ/mol(−286.83)]kJ/mol=45.01 kJΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=[−241.82 kJ/mol(−286.83)]kJ/mol=45.01 kJ
ΔS°=1mol×S°(H2O(g))1mol×S°(H2O(l))=(1 mol)188.8J/mol·K(1 mol)70.0J/mol K=118.8J/KΔS°=1mol×S°(H2O(g))1mol×S°(H2O(l))=(1 mol)188.8J/mol·K(1 mol)70.0J/mol K=118.8J/K
ΔG°=ΔH°TΔS°ΔG°=ΔH°TΔS°

Substitution into the standard free energy equation yields:

ΔG°=ΔH°TΔS°=45.01 kJ(298 K×118.8J/K)×1 kJ1000 JΔG°=ΔH°TΔS°=45.01 kJ(298 K×118.8J/K)×1 kJ1000 J
45.01 kJ35.4 kJ=9.6 kJ45.01 kJ35.4 kJ=9.6 kJ

At 298 K (25 °C) ΔG°>0,ΔG°>0, so boiling is nonspontaneous (not spontaneous).

Check Your Learning Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?

C2H6(g)H2(g)+C2H4(g)C2H6(g)H2(g)+C2H4(g)

Answer:

ΔG°=102.0 kJ/mol;ΔG°=102.0 kJ/mol; the reaction is nonspontaneous (not spontaneous) at 25 °C.

The standard free energy change for a reaction may also be calculated from standard free energy of formation ΔGf° values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, ΔGf°ΔGf° is by definition zero for elemental substances under standard state conditions. The approach used to calculate ΔG°ΔG° for a reaction from ΔGf°ΔGf° values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

mA+nBxC+yD,mA+nBxC+yD,

the standard free energy change at room temperature may be calculated as

ΔG°=νΔG°(products)νΔG°(reactants)=[xΔGf°(C)+yΔGf°(D)][mΔGf°(A)+nΔGf°(B)].ΔG°=νΔG°(products)νΔG°(reactants)=[xΔGf°(C)+yΔGf°(D)][mΔGf°(A)+nΔGf°(B)].

Example 16.8

Using Standard Free Energies of Formation to Calculate ΔG° Consider the decomposition of yellow mercury(II) oxide.

HgO(s,yellow)Hg(l)+12O2(g)HgO(s,yellow)Hg(l)+12O2(g)

Calculate the standard free energy change at room temperature, ΔG°,ΔG°, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?

Solution The required data are available in Appendix G and are shown here.

Compound ΔGf°(kJ/mol)ΔGf°(kJ/mol) ΔHf°(kJ/mol)ΔHf°(kJ/mol) S°(J/K·mol)S°(J/K·mol)
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2

(a) Using free energies of formation:

ΔG°=νGf°(products)νΔGf°(reactants)ΔG°=νGf°(products)νΔGf°(reactants)
=[1ΔGf°Hg(l)+12ΔGf°O2(g)]1ΔGf°HgO(s,yellow)=[1ΔGf°Hg(l)+12ΔGf°O2(g)]1ΔGf°HgO(s,yellow)
=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]1 mol(−58.43 kJ/mol)=58.43 kJ/mol=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]1 mol(−58.43 kJ/mol)=58.43 kJ/mol

(b) Using enthalpies and entropies of formation:

ΔH°=νΔHf°(products)νΔHf°(reactants)ΔH°=νΔHf°(products)νΔHf°(reactants)
=[1ΔHf°Hg(l)+12ΔHf°O2(g)]1ΔHf°HgO(s,yellow)=[1ΔHf°Hg(l)+12ΔHf°O2(g)]1ΔHf°HgO(s,yellow)
=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]1 mol(−90.46 kJ/mol)=90.46 kJ/mol=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]1 mol(−90.46 kJ/mol)=90.46 kJ/mol
ΔS°=νΔS°(products)νΔS°(reactants)ΔS°=νΔS°(products)νΔS°(reactants)
=[1ΔS°Hg(l)+12ΔS°O2(g)]1ΔS°HgO(s,yellow)=[1ΔS°Hg(l)+12ΔS°O2(g)]1ΔS°HgO(s,yellow)
=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]1 mol(71.13 J/mol K)=107.4 J/mol K=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]1 mol(71.13 J/mol K)=107.4 J/mol K
ΔG°=ΔH°TΔS°=90.46 kJ298.15 K×107.4 J/K·mol×1 kJ1000 JΔG°=ΔH°TΔS°=90.46 kJ298.15 K×107.4 J/K·mol×1 kJ1000 J
ΔG°=(90.4632.01)kJ/mol=58.45 kJ/molΔG°=(90.4632.01)kJ/mol=58.45 kJ/mol

Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.

Check Your Learning Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

C2H4(g)H2(g)+C2H2(g)C2H4(g)H2(g)+C2H2(g)

Answer:

(a) 140.8 kJ/mol, nonspontaneous

(b) 141.5 kJ/mol, nonspontaneous

Free Energy Changes for Coupled Reactions

The use of free energies of formation to compute free energy changes for reactions as described above is possible because ΔG is a state function, and the approach is analogous to the use of Hess’ Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:

H2O(l)H2O(g)H2O(l)H2O(g)

An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:

H2(g)+12O2(g)H2O(g)ΔGf°gas H 2O(l) H 2(g)+ 1 2 O 2(g)Δ G f °liquid ¯ H2O(l)H2O(g)ΔG°=ΔGf°gasΔGf°liquidH2(g)+12O2(g)H2O(g)ΔGf°gas H 2O(l) H 2(g)+ 1 2 O 2(g)Δ G f °liquid ¯ H2O(l)H2O(g)ΔG°=ΔGf°gasΔGf°liquid

This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for ΔG°:

ZnS(s)Zn(s)+S(s)ΔG1°=201.3kJZnS(s)Zn(s)+S(s)ΔG1°=201.3kJ

The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:

S(s)+O2(g)SO2(g)ΔG2°=300.1kJS(s)+O2(g)SO2(g)ΔG2°=300.1kJ

The coupled reaction exhibits a negative free energy change and is spontaneous:

ZnS(s)+O2(g)Zn(s)+SO2(g)ΔG°=201.3kJ+300.1kJ=98.8kJZnS(s)+O2(g)Zn(s)+SO2(g)ΔG°=201.3kJ+300.1kJ=98.8kJ

This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.

Example 16.9

Calculating Free Energy Change for a Coupled Reaction Is a reaction coupling the decomposition of ZnS to the formation of H2S expected to be spontaneous under standard conditions?

Solution Following the approach outlined above and using free energy values from Appendix G:

Decomposition of zinc sulfide:ZnS(s)Zn(s)+S(s)ΔG1°=201.3kJFormation of hydrogen sulfide:S(s)+H2(g)H2S(g)ΔG2°=33.4kJCoupled reaction:ZnS(s)+H2(g)Zn(s)+H2S(g)ΔG°=201.3kJ+33.4kJ=167.9kJDecomposition of zinc sulfide:ZnS(s)Zn(s)+S(s)ΔG1°=201.3kJFormation of hydrogen sulfide:S(s)+H2(g)H2S(g)ΔG2°=33.4kJCoupled reaction:ZnS(s)+H2(g)Zn(s)+H2S(g)ΔG°=201.3kJ+33.4kJ=167.9kJ

The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.

Check Your Learning What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?

FeS(s)+O2(g)Fe(s)+SO2(g)FeS(s)+O2(g)Fe(s)+SO2(g)

Answer:

−199.7 kJ; spontaneous

Temperature Dependence of Spontaneity

As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

ΔG=ΔHTΔSΔG=ΔHTΔS

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

  1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
  2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
  3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
  4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 16.12.

A table with three columns and four rows is shown. The first column has the phrase, “Delta S greater than zero ( increase in entropy ),” in the third row and the phrase, “Delta S less than zero ( decrease in entropy),” in the fourth row. The second and third columns have the phrase, “Summary of the Four Scenarios for Enthalpy and Entropy Changes,” written above them. The second column has, “delta H greater than zero ( endothermic ),” in the second row, “delta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,” in the third row, and “delta G greater than zero at any temperature, Process is nonspontaneous at any temperature,” in the fourth row. The third column has, “delta H less than zero ( exothermic ),” in the second row, “delta G less than zero at any temperature, Process is spontaneous at any temperature,” in the third row, and “delta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.”
Figure 16.12 There are four possibilities regarding the signs of enthalpy and entropy changes.

Example 16.10

Predicting the Temperature Dependence of Spontaneity The incomplete combustion of carbon is described by the following equation:

2C(s)+O2(g)2CO(g)2C(s)+O2(g)2CO(g)

How does the spontaneity of this process depend upon temperature?

Solution Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

Check Your Learning Popular chemical hand warmers generate heat by the air-oxidation of iron:

4Fe(s)+3O2(g)2Fe2O3(s)4Fe(s)+3O2(g)2Fe2O3(s)

How does the spontaneity of this process depend upon temperature?

Answer:

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

ΔG=ΔHTΔSΔG=ΔHTΔS
y=b+mxy=b+mx

Such a plot is shown in Figure 16.13. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

ΔG=0=ΔHTΔSΔG=0=ΔHTΔS
T=ΔHΔST=ΔHΔS

So, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

A graph is shown where the y-axis is labeled, “Free energy,” and the x-axis is labeled, “Increasing temperature ( K ).” The value of zero is written midway up the y-axis with the label, “delta G greater than 0,” written above this line and, “delta G less than 0,” written below it. The bottom half of the graph is labeled on the right as, “Spontaneous,” and the top half is labeled on the right as, “Nonspontaneous.” A green line labeled, “delta H less than 0, delta S greater than 0,” extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, “delta H less than 0, delta S less than 0,” extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, “delta H greater than 0, delta S greater than 0,” extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, “delta H greater than 0, delta S less than 0,” extends from three quarters of the way up the y-axis to the top right of the graph.
Figure 16.13 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example 16.11

Equilibrium Temperature for a Phase TransitionAs defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.

Solution The process of interest is the following phase change:

H2O(l)H2O(g)H2O(l)H2O(g)

When this process is at equilibrium, ΔG = 0, so the following is true:

0=ΔH°TΔS°orT=ΔH°ΔS°0=ΔH°TΔS°orT=ΔH°ΔS°

Using the standard thermodynamic data from Appendix G,

ΔH°=1mol×ΔHf°(H2O(g))1mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(286.83 kJ/mol)=44.01 kJΔH°=1mol×ΔHf°(H2O(g))1mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(286.83 kJ/mol)=44.01 kJ
ΔS°=1mol×ΔS°(H2O(g))1mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/KΔS°=1mol×ΔS°(H2O(g))1mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/K
T=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°CT=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°C

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Check Your Learning Use the information in Appendix G to estimate the boiling point of CS2.

Answer:

313 K (accepted value 319 K)

Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).

In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:

ΔG=ΔG°+RTlnQΔG=ΔG°+RTlnQ

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 16.12.

Example 16.12

Calculating ΔG under Nonstandard Conditions What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, PN2=0.870 atm,PN2=0.870 atm, PH2=0.250 atm,PH2=0.250 atm, and PNH3=12.9 atmPNH3=12.9 atm

2NH3(g)3H2(g)+N2(g)ΔG°=33.0 kJ/mol2NH3(g)3H2(g)+N2(g)ΔG°=33.0 kJ/mol

Solution The equation relating free energy change to standard free energy change and reaction quotient may be used directly:

ΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/molΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/mol

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.

Check Your Learning Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Answer:

ΔG = +45.1 kJ/mol; no

For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as

0=ΔG°+RTlnK(at equilibrium)0=ΔG°+RTlnK(at equilibrium)
ΔG°=RTlnKorK=eΔG°RTΔG°=RTlnKorK=eΔG°RT

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 16.4.

Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Composition of an Equilibrium Mixture
> 1 < 0 Products are more abundant
< 1 > 0 Reactants are more abundant
= 1 = 0 Reactants and products are comparably abundant
Table 16.4

Example 16.13

Calculating an Equilibrium Constant using Standard Free Energy Change Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.

Solution The reaction of interest is the following:

AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+][Cl]AgCl(s)Ag+(aq)+Cl(aq)Ksp=[Ag+][Cl]

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

ΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/molΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/mol

The equilibrium constant for the reaction may then be derived from its standard free energy change:

Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×10−10Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×10−10

This result is in reasonable agreement with the value provided in Appendix J.

Check Your Learning Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

2NO2(g)N2O4(g)2NO2(g)N2O4(g)

Answer:

K = 6.9

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure 16.14). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Three graphs, labeled, “a,” “b,” and “c” are shown where the y-axis is labeled, “Gibbs free energy ( G ),” and, “G superscript degree sign ( reactants ),” while the x-axis is labeled, “Reaction progress,” and “Reactants,” on the left and, “Products,” on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G less than 0,” while the lowest point on the graph is labeled, “Q equals K greater than 1.” In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G greater than 0,” while the lowest point on the graph is labeled, “Q equals K less than 1.” In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is equal to the starting point on the y-axis which is labeled, “G superscript degree sign ( reactants ).” The lowest point on the graph is labeled, “Q equals K equals 1.” At the top of the graph is the label, “Delta G superscript degree sign equals 0.”
Figure 16.14 These plots show the free energy versus reaction progress for systems whose standard free energy changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.
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