1. Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition Copyright © 2008 Pearson Prentice Hall, Inc.

2. Chapter 16/2 Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

3. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/3 Spontaneous Processes

4. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/4 Enthalpy, Entropy, and Spontaneous Processes State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state. Enthalpy Change (  H): The heat change in a reaction or process at constant pressure;  H =  E + P  V. Entropy (S): The amount of molecular randomness in a system.

5. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/5 Enthalpy, Entropy, and Spontaneous Processes = +40.7 kJ  H vap CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g) H 2 O(l)H 2 O(s) H 2 O(g)H 2 O(l) 2NO 2 (g)N2O4(g)N2O4(g) = +3.88 kJ H°H° = +6.01 kJ  H fusion = -890.3 kJ H°H° = +57.1 kJ H°H° Endothermic: Exothermic: Na 1+ (aq) + Cl 1- (aq)NaCl(s) H2OH2O

6. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/6 Enthalpy, Entropy, and Spontaneous Processes  S = S final - S initial

7. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/7 Enthalpy, Entropy, and Spontaneous Processes

8. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/8 Enthalpy, Entropy, and Spontaneous Processes

9. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/9 Entropy and Probability S = k ln W k = Boltzmann’s constant = 1.38 x 10 -23 J/K W =The number of ways that the state can be achieved.

10. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/10 Entropy and Temperature Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero.

11. Entropy and Temperature

12. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/12 Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

13. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/13 Standard Molar Entropies and Standard Entropies of Reaction cC + dDaA + bB  S° = S°(products) - S°(reactants)  S° = [cS°(C) + dS°(D)] - [aS°(A) + bS°(B)] ReactantsProducts

14. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/14 Standard Molar Entropies and Standard Entropies of Reaction Using standard entropies, calculate the standard entropy change for the decomposition of N 2 O 4. 2NO 2 (g)N2O4(g)N2O4(g) - (1 mol)  S° = 2 S°(NO 2 (g)) - S°(N 2 O 4 (g)) 304.2 K mol J  S° = 175.8 J/K 240.0 K mol J  S° = (2 mol)

15. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/15 Entropy and the Second Law of Thermodynamics First Law of Thermodynamics: In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant. Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases.

16. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/16 Entropy and the Second Law of Thermodynamics  S total > 0The reaction is spontaneous.  S total < 0The reaction is nonspontaneous.  S total = 0The reaction mixture is at equilibrium.  S total =  S sys +  S surr or  S total =  S system +  S surroundings

17. Entropy and the Second Law of Thermodynamics  s surr  -  H  s surr  T 1  s surr = T -  H

18. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/18 Free Energy  G =  H - T  S = -T  S total Free Energy: G = H - TS  G =  H - T  S  S =  S sys Using:  s surr = T -  H  S total =  S sys +  S surr

19. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/19 Free Energy  G < 0The reaction is spontaneous.  G > 0The reaction is nonspontaneous.  G = 0The reaction mixture is at equilibrium. Using the second law and  G =  H - T  S = -T  S total

20. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/20 Free Energy

21. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/21 Standard Free-Energy Changes for Reactions Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution.  G° =  H° - T  S°

22. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/22 Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:  S° = -198.7 J/K 2NH 3 (g)N 2 (g) + 3H 2 (g) = -33.0 kJ  H° = -92.2 kJ  G° =  H° - T  S° 1000 J 1 kJ K -198.7 J xx= -92.2 kJ - 298 K

23. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/23 Standard Free Energies of Formation  G° =  G° f (products) -  G° f (reactants)  G° = [c  G° f (C) + d  G° f (D)] - [a  G° f (A) + b  G° f (B)] ReactantsProducts cC + dDaA + bB

24. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/24 Standard Free Energies of Formation

25. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/25 Standard Free Energies of Formation Using table values, calculate the standard free-energy change at 25 °C for the reduction of iron(III) oxide with carbon monoxide: 2Fe(s) + 3CO 2 (g)Fe 2 O 3 (s) + 3CO(g) - [(1 mol)(-742.2 kJ/mol) + (3 mol)(-137.2 kJ/mol)]  G° = -29.4 kJ  G° = [2  G° f (Fe(s)) + 3  G° f (CO 2 (g))] - [1  G° f (Fe 2 O 3 (s)) + 3  G° f (CO(g))]  G° = [(2 mol)(0 kJ/mol) + (3 mol)(-394.4 kJ/mol)]

26. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/26 Free Energy Changes and the Reaction Mixture  G =  G° + RT ln Q  G = Free-energy change under nonstandard conditions. For the Haber synthesis of ammonia: 2NH 3 (g)N 2 (g) + 3H 2 (g) Q p = NH 3 P 2 H2H2 P N2N2 P 3

27. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/27 Free Energy Changes and the Reaction Mixture C2H4(g)C2H4(g)2C(s) + 2H 2 (g) Q p = C2H4C2H4 P H2H2 P 2 Calculate ln Q: Calculate  G for the formation of ethylene (C 2 H 4 ) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H 2 and 0.10 atm C 2 H 4. = -11.51 100 2 0.10 ln

28. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/28 Free Energy Changes and the Reaction Mixture Calculate  G: = 68.1 kJ/mol +  G = 39.6 kJ/mol (298 K)(-11.51) 1000 J 1 kJ 8.314 K mol J  G =  G° + RT ln Q

29. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/29 Free Energy and Chemical Equilibrium  G =  G° + RT ln Q Q >> 1RT ln Q >> 0  G > 0 The total free energy decreases as the reaction proceeds spontaneously in the reverse direction. When the reaction mixture is mostly products: Q << 1RT ln Q << 0  G < 0 The total free energy decreases as the reaction proceeds spontaneously in the forward direction. When the reaction mixture is mostly reactants:

31. Free Energy and Chemical Equilibrium At equilibrium,  G = 0 and Q = K.  G° = -RT ln K  G =  G° + RT ln Q

32. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/32 Free Energy and Chemical Equilibrium Calculate K p at 25 °C for the following reaction: Calculate  G°: CaO(s) + CO 2 (g)CaCO 3 (s) - [(1 mol)(-1128.8.0 kJ/mol)]  G° = +130.4 kJ/mol = [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]  G° = [  G° f (CaO(s)) +  G° f (CO 2 (g))] - [  G° f (CaCO 3 (s))]

33. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 16/33 Free Energy and Chemical Equilibrium Calculate K: =  G° = -RT ln K -130.4 kJ/mol ln K = -52.63 (298 K) 1000 J 1 kJ 8.314 K mol J = 1.4 x 10 -23 Calculate ln K: RT -G°-G° ln K = -52.63 K = e