1. Chapter 19 Chemical Thermodynamics

2. No Review Quiz No Lab

3. Chemical Thermodynamics The study of the energy transformations that accompany chemical and physical changes.

4. The Driving Forces All reactions (changes) in nature occur because of the interplay of two driving forces: (1) The drive toward lower energy. Decrease in enthalpy (2) The drive toward increased disorder. Increase in entropy

5. Enthalpy Enthalpy is the internal energy of a system at constant pressure. We cannot measure enthalpy directly but we can measure changes in the enthalpy of a system. Changes in enthalpy normally are in the form of heat.

6. Hidden Slide Analogy of Why we cannot measure enthalpy. You climb to the top of a building that is 203 feet high. You don't know how far away the center of the earth is, but you know it's 203 feet further away than it was before you climbed the building.

7. Enthalpy (Energy) Change Exothermic reactions –∆H Endothermic reactions +∆H

8. EnergyEntropy

9. Table 19.1 page 613

10. Hess’s Law If a process can be written as a sum of several steps, the enthalpy change of the total process is equal to the sum of the enthalpy changes for the various steps.

11. P 4 O 10 + 6H 2 O (l) → 4H 3 PO 4 Use the information below to determine the ∆H. 4P + 5O 2 → P 4 O 10 ∆H = -2984 kJ/mol H 2 + ½ O 2 → H 2 O (l) ∆H = -285.83 kJ/mol 3/2 H 2 + P + 2O 2 → H 3 PO 4 ∆H = -1267kJ/mol ∆H = -369 kJ/mol

12. 2NH 3 + 3O 2 + 2CH 4 → 2HCN + 6H 2 O Use the information below to determine the ∆H. ½ N 2 + 3/2 H 2 → NH 3 ∆H = - 46 kJ/mol C + 2H 2 → CH 4 ∆H = -75 kJ/mol ½ H 2 + C + ½ N 2 → HCN∆H = +135.1 kJ/mol H 2 + ½ O 2 → H 2 O∆H = -242 kJ/mol ∆H = -940 kJ/mol

13. Another method to determine ∆H for a reaction

14. ∆H = ∑∆H f (products) ─ ∑∆H f (reactants) Appendix I in your notebook has ∆H f values

15. ∆H = ∑∆H f (products) ─ ∑∆H f (reactants) Determine ∆H for the reaction using the above formula. Na(s) + O 2 (g) + CO 2 (g) → Na 2 CO 3 (s) ∆H f for Na 2 CO 3 (s) = -1130.8kJ/mol ∆H = -737.3 kJ/mol

16. ∆H = ∑∆H f (products) ─ ∑∆H f (reactants) Determine ∆H for the reaction using the above formula. C 2 H 5 OH(g) + 3O 2 (g)→ 2CO 2 (g) + 3H 2 O(g) ∆H = -1277.4 kJ/mol

17. Calculate the amount of energy released when 100.0g of C 2 H 5 OH is burned. C 2 H 5 OH(g) + 3O 2 (g)→ 2CO 2 (g) + 3H 2 O(g) ∆H = -1277.4 kJ -2772 kJ = 2772 kJ released

18. Change in Entropy (∆S) (+∆S) = increase in disorder (entropy) (-∆S) = decrease in disorder (entropy)

19. An Increase in Entropy (+∆S) Is a Force that Drives Physical and Chemical Changes.

20. Second Law of Thermodynamics Every time a change occurs it increases the entropy of the universe.

21. Second Law of Thermodynamics The second law implies that whenever a change occurs, some of the energy will be wasted which will lead to an increase in the disorder of the universe.

22. Entropy Entropy is a complicated concept to truly understand. It may be best to think of entropy as the degree of dispersion. –As matter or energy disperses (becomes more free to move or becomes more spread out) entropy will increase.

23. Entropy 1 Video ≈ 2:25

24. Entropy increases when matter is dispersed. Production of liquid or gas from a solid “or” production of gas from a liquid results in the dispersal of matter. The individual particles become more free to move and generally occupy a larger volume which causes entropy to increase.

25. Increase in Entropy (+∆S) Expansion of a gas.

26. Increase in Entropy (+∆S) Formation of a mixture.

27. Increase in Entropy (+∆S) Dissolution of a crystalline solid in water.

28. Increase in Entropy (+∆S) More particles are created.

29. Increase in Entropy (+∆S) More particles are created.

30. Decrease in Entropy (-∆S) Is simply a reverse of the previous processes. A gas dissolves in a liquid A precipitate forms

31. Third Law of Thermodynamics The entropy of any pure substance at 0 K is zero. We can interpret this to mean that as temperature increases entropy increases.

32. A more complex molecule has a higher entropy.

33. Calculating Entropy Change

34. ∆S = ∑S (products) ─ ∑S (reactants) Determine ∆S for the reaction using the above formula. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆S = +95.64 J/mol K

35. “Gibbs” Free Energy Allows use to determine whether a reaction is spontaneous or not spontaneous. We will say a process is or is not “thermodynamically favored” instead of using the terms spontaneous or not spontaneous. This avoids common confusion with associating the term spontaneous with the idea that something must occur immediately or without cause.

36. “Gibbs” Free Energy If a process is “thermodynamically favored” means the products are favored at equilibrium. The term “not thermodynamically favored” means the reactants are favored at equilibrium.

37. Change in Free Energy (ΔG) It is the change in free energy (ΔG) that determines whether a reaction is thermodynamically favorable or not.

38. ∆G = ∑∆G f (products) ─ ∑∆G f (reactants) Determine ∆G for the reaction using the above formula. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆G = -1306 kJ/mol

39. (-∆G) vs (+∆G) If -∆G the process is “thermodynamically favored” and the products are favored at equilibrium. If +∆G the process is “not thermodynamically favored” and the reactants are favored at equilibrium.

40. Understanding ΔG Just because a process is spontaneous or “thermodynamically favored” (-∆G) does not mean that it will proceed at any measurable rate. A reaction that is thermodynamically favored, (spontaneous), may occur so slowly that in practice it does not occur at all.

41. Understanding ΔG Thermodynamically favored reactions that do not occur at any measurable rate are said to be under “kinetic control”. These reactions often have a very high activation energy. The fact that a process does not proceed at a noticeable rate does not mean that the reaction is at equilibrium. We would conclude that such a reaction is simply under kinetic control.

42. Example The reaction of diamond (pure carbon) with oxygen in the air to form carbon dioxide is thermodynamically favored (spontaneous) but has an extremely slow rate.

43. Understanding ΔG A reaction with a with a +ΔG may be forced to occur with the application of energy or by coupling it to thermodynamically favorable reactions.

44. +ΔG+ΔG Energy can be used to cause a process to occur that is not thermodynamically favored. Using electricity to charge a battery.

45. +ΔG+ΔG Energy can be used to cause a process to occur that is not thermodynamically favored. Photoionization of an atom by light.

46. +ΔG+ΔG A thermodynamically unfavorable reaction may be made favorable by coupling it to a favorable reaction or series of favorable reactions. This process involves a series of reactions with common intermediates, such that the reactions add up to produce an overall reaction that is thermodynamically favorable (–ΔG).

47. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol

48. ∆G = 0 Reactants ↔ Products -∆G = thermodynamically favored = products favored +∆G = not thermodynamically favored = reactants favored ∆G is 0 = equilibrium

49. ∆G = ∆H – T∆S

50. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) Determine ∆G given: ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol or -1,306,000J/mol

51. ∆G = ∑∆G f (products) ─ ∑∆G f (reactants) Determine ∆G for the reaction using the above formula. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆G = -1306 kJ/mol ∆G = ∆H – T∆S C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) Determine ∆G given: ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol

52. Solve for ∆H: ∆G = ∆H – T∆S ∆H = ∆G + T∆S

53. Solve for ∆S: ∆G = ∆H – T∆S or

54. Solve for T: ∆G = ∆H – T∆S or

55. CaO(s) + SO 3 (g) → CaSO 4 (s) Calculate ∆S at 298K using the data given below. ∆H = -401.5 kJ/mol ∆G = -345.0 kJ/mol ∆S = -0.1896 kJ/mol K or -189.6 J/mol K

56. ∆G = ∆H – T∆S Can be used to explain the driving forces and thermodynamic favorability.

57. C 2 H 5 OH(g) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) ∆H = -1277.3 kJ/mol (exothermic) ∆S = +95.64 J/mol K (increased disorder) ∆G = ∆H – T∆S ∆G = (-) – [+(+)] negative – positive = always negative ∆G = -1305.2 kJ/mol (thermodynamically favored)

58. ∆G = ∆H – T∆S What are the other possibilities?

59. ∆G = ∆H – T∆S +∆H = endothermic -∆S = decreased entropy ∆G = ∆H – T∆S ∆G = (+) – [+(-)] positive – negative = always positive +∆G = not thermodynamically favored

60. ∆G = ∆H – T∆S +∆H = endothermic +∆S = increased entropy ∆G = ∆H – T∆S ∆G = (+) – [+(+)] positive – positive = ? + or - ∆G = may be thermodynamically favored

61. ∆G = ∆H – T∆S -∆H = exothermic -∆S = decreased entropy ∆G = ∆H – T∆S ∆G = (-) – [+(-)] negative – negative = ? + or - ∆G = may be thermodynamically favored

62. Hidden Slide Table 19.3 Page 621 gives some specific examples that support the previous slides.

63. What is the determining factor for thermodynamic favorability when the factors ∆H and T∆S contradict? ∆G = ∆H – T∆S ∆G = (+) – [+(+)] Temperature ∆G = ∆H – T∆S ∆G = (-) – [+(-)]

64. A process may be thermodynamically favored at one temperature and not at another. Is it thermodynamically favored for ice to melt or water to freeze?

65. CaO(s) + SO 3 (g) → CaSO 4 (s) The data given below are for 298K. Calculate ∆G using the ∆H and ∆S values below at 2463K. ∆H = -401.5 kJ/mol ∆G = -345.0 kJ/mol ∆S = -189.6 J/mol K ∆G = +65.5 kJ/mol at 2463K Conclusion: ∆S becomes a larger factor as temperature increases

66. CaO(s) + SO 3 (g) → CaSO 4 (s) Estimate the temperature at which the reaction becomes spontaneous. ∆H = -401.5 kJ/mol ∆S = -189.6 J/mol K T = 2118K = 1845°C

67. CaO(s) + SO 3 (g) → CaSO 4 (s) Estimate the temperature at which the reaction becomes spontaneous. ∆H = -401.5 kJ/mol ∆S = -189.6 J/mol K T = 2118K = 1845°C Why is this temperature only an estimate of the temperature?

68. CaO(s) + SO 3 (g) ↔ CaSO 4 (s) Calculate ∆S at 298K using the data given below. ∆H = -401.5 kJ/mol ∆G = -345.0 kJ/mol ∆S = -0.1896 kJ/mol K As the temperature increases above ≈ 2118K the reaction becomes nonspontaneous as ∆G becomes positive. Why?

69. Boiling & Equilibrium Boiling is a reversible reaction: H 2 O (l) ↔ H 2 O (g) What is the value of ∆G for the boiling water?

70. Estimate the boiling point of ethanol (C 2 H 5 OH). BP = 350K = 77°C

71. Estimate the boiling point of ethanol (C 2 H 5 OH). Estimated BP = 350K = 77°C FormulaNamesBoiling Point C 2 H 5 OHEthanol Ethyl alcohol Hydroxyethane Grain alcohol 78.4°C Why is there a discrepancy in the boiling point?

72. Thermodynamic Favorability (∆G) and Equilibrium (K) ∆GKReaction Negative>1Thermodynamically favored (products favored) 0=1At equilibrium Positive<1Not thermodynamically favored (reactants favored)

73. ∆G° = -R T lnK R = 8.314 J/K Used to determine: (1) K when given ∆G° (2) ∆ G° when a reaction is at equilibrium.

74. ∆G° = -R T lnK Calculate ∆G° for a reaction at equilibrium where K = 1.

75. ∆G° = -R T lnK If K ≠ 1 then ∆G° ≠ 0. This means the reactions is moving to the right towards products or moving to the left towards reactants and is not at equilibrium. LeChatlier’s Principle should help you to understand this. What does changing the temperature do to an equilibrium system.

76. What does changing the temperature do to an equilibrium system? The reaction shifts right or left favoring the products or reactants and raising or lowering K.

77. Every reaction has a temperature at which K = 1 and the reaction is neither favoring products nor reactants Reaction Kelvin Temperature K

78. The equilibrium constant for a reaction at 0°C is 1.657 x 10 -5. What is ∆G°? ∆G° = 24980J/mol = 24.98 kJ/mol

79. Calculate the equilibrium constant for a reaction at 298K if ∆Gº = -228.59 kJ. K = 1.17 x 10 40

80. K vs ∆G Compare K and ∆G from the last two problems. ln K = -11.0079 and K = 1.657 x 10 -5 and ∆G° = 24.98 kJ/mol ln K = 92.264 and K = 1.17 x 10 40 and ∆G° = -228.59 kJ/mol

81. 5 Formulas 1.∆H = ∑∆H f (products) ─ ∑∆H f (reactants) 2.∆S = ∑S (products) ─ ∑S (reactants) 3.∆G = ∑∆G (products) ─ ∑∆G (reactants) 4.∆G = ∆H – T∆S 5.∆G° = -RTlnK

82. Entropy 2 Video ≈ 8:21

83. Hidden Slides All the slides that follow are hidden however can be used if we are not running short on time.

84. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Calculate ∆Hº at 298K ∆H = 32.5 kJ/mol

85. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Calculate ∆Sº at 298K ∆S = 41.8 J/mol K

86. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Calculate ∆Gº at 298K ∆G = 20.0 kJ/mol

87. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Calculate K at 298K. K = 3.12 x 10 - 4

88. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Estimate ∆Gº at 798K. ∆G = -0.856 kJ/mol

89. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Estimate K at 798K Don’t Try It Yet.

90. Determination of K at different temperatures

91. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Estimate ∆Gº at 798K. Remember we just calculated this. ∆G = -0.856 kJ/mol

92. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Estimate K at 798K K = 1.14

94. CuS(s) + H 2 (g) ↔ Cu(s) + H 2 S(g) K at 798K = 1.14 K at 298K = 3.13 x 10 - 4

95. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Estimate the temperature at which this reaction becomes spontaneous. T = 778 K = 505°C

96. CuS(s) + H 2 (g)→ Cu(s) + H 2 S(g) Calculate the equilibrium constant at 778 K. K = 1