1. Chapter 17 The Direction of Chemical Change (The Second Law of Thermodynamics) Life thrives on Earth because there is a constant supply of energy from the Sun. Some of this energy is stored in plants through photosynthesis. Although photosynthesis is only about 3% efficient, it supports nearly all plants on Earth and the animals that feed on them. The concepts in this chapter are critical for understanding the conversion of energy from one form to another and for research into the availability and deployment of energy.

2. Assignment for Chapter 17 Exercises: 17.3, 17.8, 17.14, 17.19, 17.28, 17.35, 17.45, 17.53 Supplementary Exercises: 17.64, 17.69, 17.78 Applied Exercises: 17.81 Integrated Exercises: 17.90

3. The Big Question Energy (loss or gain) is required—obvious! How energy is distributed also affects the tendency of a change—not that obvious! Both entropy and enthalpy are important to drive a change although some processes are primarily driven by enthalpy while others by entropy. Overall, the Gibbs free energy is the parameter that controls the possibility of a change. Why do some changes take place? What parameter(s) control or drive a change? We’ll find how this conclusion is drawn.

4. Spontaneous Change Fig.1 The direction of spontaneous change is for a hot block of metal (left) to cool to the temperature of its surroundings (right). A block at the same temperature as its surroundings does not spontaneously become hotter. A change that tends to occur without needing to be driven by an external influence.

5. Fig.2 The direction of spontaneous change for a gas is toward filling its container. A gas that already fills its container does not collect spontaneously in a small region of the container. A glass cylinder containing a brown gas (upper piece of glassware in the left illustration) is attached to an empty flask. When the stopcock between them is opened, the brown gas fills both upper and lower vessels (right illustration). The brown gas is nitrogen dioxide. NO 2

6. Fig.3 We can understand the natural direction of the migration of heat from a hot region to a cold region by thinking about the jostling between the vigorously moving atoms in the hot region. Molecules jostle their neighbors, and the thermal motion spreads. Spontaneity—the spreading of energy to more and more degrees of freedom. Entropy is the measure of the number of degrees of freedom affected by thermal motion.

7. Fig.4 We can understand the natural direction of the migration of matter by visualizing how the random motion of molecules results in their spreading throughout the available space. The second law of thermodynamics: The entropy of an isolated system tends to increase. Spontaneity—the spreading of energy to more and more degrees of freedom. Entropy is the measure of the number of degrees of freedom affected by thermal motion.

8. Fig.5 A representation of the arrangement of molecules in (a) a solid and (b) a liquid. When the solid melts, there is an increase in the disorder of the system and hence a rise in entropy. For a given sample and at the same temperature, the entropy of liquid state is larger than that of solid state.

9. Fig.6 The entropy of a solid increase as its temperature is raised. The entropy increase sharply when the solid melts to form the more disordered liquid and then gradually increases again up to the boiling point. A second, larger jump in entropy occurs when the liquid turns into a vapor. For a given sample, the entropy is larger at higher temperatures than that at lower temperatures.

10. Example: predicting the relative entropies of two samples Which has the greater entropy: (a) 1 g of pure solid NaCl or 1 g of NaCl dissolved in 100 mL of water; (b) 1 g of water at 25 o C or a 1 g of water at 50 o C? Ans: 1 g of NaCl dissolved in 100 mL of water. 1 g of water at 50 o C. For a given sample and at the same temperature, the entropy of liquid state is larger than that of solid state. For a given sample, the entropy is larger at higher temperatures than that at lower temperatures.

11. Exercise: predicting the relative entropies of two samples Which has the greater entropy: 1 mol of CO 2 (s) or 1 mol of CO 2 (g) at the same temperature? Ans: 1 mol of CO 2 (g) Which has the greater entropy: a sample of liquid mercury at -15 o C or the same sample at 0 o C? Ans: The sample of liquid mercury at 0 o C.

12. Entropy: Macroscopic Definition and Calculation The greater the energy transferred to the system as heat, the greater the increase in entropy. If the transfer is made to a hot system, the increase in entropy is smaller than when the same amount of energy is transferred to a cool system. A reversible process is one that can be reversed by an infinitesimal change in a variable. (when heat is transferred to a system, the source must have the same temperature as the system itself.)

13. Calculating the Change in Entropy Calculate the change in entropy of a large tank of water when a total of 100.0 J of energy is transferred to it reversibly as heat at 20.0 o C. Calculate the change in entropy of a large iron block at 20.0 o C when 1500.0 J of energy escapes as heat from the block to the surroundings.

14. Classroom Exercise: Calculating the Change in Entropy Calculate the change in entropy of a large swimming pool at 28.0 o C when 240.0 J of escapes from the pool as heat to the surroundings.

15. Fusion and vaporization entropies The discontinuities correspond to phase transition. The jump of entropy is the Signature of first order phase transition.

16. Calculating the Entropy of Vaporization Calculate the change in molar entropy when water vaporizes at its boiling point. Calculating the Entropy of Fusion Calculate the change in molar entropy of ice at its melting point. (Look Table 6.2 for fusion enthalpy)

17. Classroom Exercise: Calculating the Entropy of Vaporization Calculate the change in molar entropy when ammonia vaporizes at its boiling point (239.7 K). The vaporization enthalpy of ammonia is 23.4 kJ/mol.

18. Entropy: Microscopic Definition and Calculation k: Boltzmann constant Entropy is a measure of how the energy of a system is stored for a given macroscopic state.

20. Absolute Entropies The third law of thermodynamics: The entropy of a perfect crystal approaches 0 as the absolute temperature approaches 0. All absolute entropies are positive  Standard molar entropy S m o (298 K).

21. More complicated compounds have higher entropies

22. Entropies are higher at higher temperatures

23. Fig.7 The entropy change due to heat transfer depends on both the amount of heat transferred and the temperature of the system. A lot of heat transferred to a cold system (upper left) results in a large increase in the entropy of the system. A small quantity of heat transferred to a hot system (lower right) results in a small increase in entropy of the system. You take more entropy after you drink a cup of hot tea than a cup of iced tea. You take more entropy when you’re in fever than when you’re normal after you drink a cup of tea.

24. The entropy is higher for Higher temperature Larger volume More complex structures Larger sample size Heavier atoms (entropy is not disorder!) Vapor relative to liquid or solid Liquid relative to solid

25. Estimating the relative value of the molar entropy Which substance in each pair has the higher molar entropy: (a) CO 2 at 25 o C and 1 atm or CO 2 at 25 o C and 3 atm; (b) Br 2 (l) or Br 2 (g) at the same temperature and pressure; (c) methane gas, CH 4, or propane gas, CH 3 CH 2 CH 3 at the same temperature and pressure? (a) One mole of CO 2 at 25 o C and 1 atm occupies larger volume than one mole of CO 2 at 25 o C and 3 atm  One mole of CO 2 at 25 o C and 1 atm has the higher molar entropy. (b) Gas has the higher molar entropy than liquid. Therefore, Br 2 (g) has the higher molar entropy at the same temperature and pressure. (c) One mole of CH 4 is lighter than one mole of CH 3 CH 2 CH 3. Therefore, CH 3 CH 2 CH 3 has the higher molar entropy at the same temperature and pressure.

26. Exercise Which substance in each pair has the higher molar entropy: (a) He at 25 o C or He at 100 o C in a container of the same volume; (b) Br(g) or Br 2 (g) at the same temperature and pressure? The higher the temperature, the higher the molar entropy  He at 100 o C has the higher molar entropy than He at 25 o C in a container of the same volume. The heavier the molecule/atom, the higher the molar entropy  Br 2 (g) has the higher molar entropy than Br(g) at the same temperature and pressure.

27. Classroom Exercise Which substance in each pair has the higher molar entropy at the same temperature and pressure : (a) Pb(s) or Pb(l); (b) SbCl 3 (g) or SbCl 5 (g) ? Liquid has the higher molar entropy than solid. Therefore, Pb(l) has the higher molar entropy than Pb(s) at the same temperature and pressure. The heavier the molecule/atom, the higher the molar entropy  SbCl 5 (g) has the higher molar entropy than SbCl 3 (g) at the same temperature and pressure.

28. Reaction Entropy Because the molar entropy of a gas is so much greater than that of Solids and liquids, a change in the amount of gas normally dominates any other entropy change in a reaction. A net consumption of gas usually results in a negative reaction entropy. A net production of gas usually results in a positive reaction entropy. N 2 (g) + 3H 2 (g)  2NH 3 (g) Reactant: 4 mol, Product: 2 mol  decrease in entropy. The standard molar entropies of common compounds are listed in Appendix 2

29. Exercise N 2 O 4 (g)  2NO 2 (g) Reactant: 1 mol, Product: 2 mol  increase in entropy.

30. Classroom Exercise C 2 H 4 (g) +H 2 (g)  C 2 H 6 (g) Reactant: 2 mol, Product: 1 mol  decrease in entropy.

31. Why does ice freeze spontaneously? Why exothermic reactions occur spontaneously? A process is spontaneous as long as the total entropy change is positive. A spontaneous process does NOT require the increase of the entropy of the system.

32. Fig.8 (a) In an exothermic process, heat escapes into the surroundings and increases their entropy. (b) In an endothermic process, the entropy of the surroundings decreases. The blue-green arrows indicate the direction of entropy change in the surroundings.

33. Exothermic reactions occur spontaneously because the increase of the entropy of the surroundings is more than the decrease of the system. N 2 (g) + 3H 2 (g)  2NH 3 (g) Standard reaction enthalpy = -92.22 kJ/mol < 0  exothermic  The entropy of the surroundings increases.

34. Fig.9 In an exothermic reaction, (a) the overall entropy change is certainly positive when the entropy of the system increases. (b) The overall entropy change is positive even when the entropy of the system decreases, provided that the entropy increase in the surroundings is greater. The reaction is spontaneous in both cases.

35. Fig.10 An endothermic reaction is spontaneous only when the entropy of the system increases enough to overcome the decrease in entropy of the surroundings, as it does here.

36. A process is spontaneous if the change of total entropy is positive Is the dissolution of ammonium nitrate to form a dilute aqueous solution spontaneous at 25 o C? NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) Reaction entropy: (endothermic)

37. A process is spontaneous if the change of total entropy is positive A model for the combustion of wood: C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) +6H 2 O(g) Reaction entropy: (exothermic)

38. Gibbs free energy must decrease for a spontaneous change: ΔS tot >0  ΔG = -TΔS tot <0 Josiah Willard Gibbs (1839–1903). Free energy is effectively the same as the total entropy (except for the negative sign and coefficient T) ΔG < 0  the process is spontaneous ΔG > 0  the reverse of the process is spontaneous ΔG = 0  both the process and its reverse are spontaneous  equilibrium.

39. Free energy must decrease for a spontaneous change: ΔG<0

40. Free energy and equilibrium At equilibrium, both directions are equally spontaneous, then This condition applies to any phase change and any chemical reaction at equilibrium at constant temperature and pressure. Example: water-ice equilibrium----

41. Predicting the boiling point of a substance Liquid metals, such as mixtures of sodium and potassium, are used as coolants in some nuclear reactors. Predict the normal boiling point of liquid sodium, given that the standard entropy of vaporization of liquid sodium is 84.8 J/K/mol and that its standard enthalpy of vaporization is 98.0 kJ/mol At equilibrium, both directions are equally spontaneous, then

42. Predicting the melting point of a substance Predict the normal melting point of solid chlorine, given that the standard entropy of fusion is 837.3 J/K/mol and that its standard enthalpy of fusion is 6.41 kJ/mol. At equilibrium, both directions are equally spontaneous, then

43. Classroom Exercise: Predicting the boiling point of methanol Predict the normal boiling point of methanol, CH 3 OH, given that the standard entropy of vaporization is 104.7 J/K/mol and that its standard enthalpy of vaporization is 35.3 kJ/mol.

44. Case Study 17 (a) The bubbles on the leaves of this underwater plant are oxygen produced by photosynthesis. Molecules such as the chlorophyll that colors the leaves green capture sunlight to begin the transformation of carbon dioxide and water to glucose and oxygen.

45. Case Study 17 (b) A weight with a small mass can be lifted into the air by another weight of the same or greater mass. What would appear unnatural if we saw it by itself (a weight rising) is actually part of a spontaneous event overall. The “natural” fall of the heavier weight causes the “unnatural” rise of the smaller weight.

46. Standard Reaction Free Energies Standard free energies of formation

47. The most stable form of an element is the state with lowest free energy of formation.

48. The most stable form of a compound is the state with lowest free energy of formation.

49. Standard Reaction Free Energies Calculate the standard free energy of formation of HI(g) at 25 o C from Its standard entropy and standard enthalpy of formation.

50. More Exercise Calculate the standard free energy of formation of NH 3 (g) at 25 o C.

51. Classroom Exercise Calculate the standard free energy of formation of cyclopropane, C 3 H 6 (g) at 25 o C.

52. Figure 17.12 The standard free energies of formation of compounds are defined as the standard reaction free energy for their formation from the elements. They represent a thermodynamic “altitude” with respect to the elements at “sea level.” The numerical values are in kilojoules per mole.

53. If the standard free energy of formation of a compound is smaller than 0,, then it is thermodynamically stable. Standard Free Energy of Formation Decides Thermodynamic Stability Thermodynamically unstable Thermodynamically stable Example: If the standard free energy of formation of a compound is larger than 0, then it is thermodynamically unstable.

54. Standard Free Energy of Formation Decides Thermodynamic Stability Is glucose stable relative to its elements at 25 o C and under standard conditions? C 6 H 12 O 6 (s)  6C (s) + 6H 2 (g) + 3O 2 (g) Standard free energy of formation (From Appendix 2A—Page A13):  thermodynamically stable. Is methylamine, CH 3 NH 2, stable relative to its elements at 25 o C and under standard conditions?  thermodynamically unstable.

55. Standard Free Energy of Formation Decides Thermodynamic Stability Is methylamine, CH 3 NH 2, stable relative to its elements at 25 o C and under standard conditions?  thermodynamically unstable. Look up Appendix 2A, page A13, and we find the standard free energy of formation of methylamine is 32.16 kJ/mol.

56. The Chemical Reaction Proceeds So That The Free Energy of Reaction < 0.  spontaneous reaction.

57. Negative Free Energy of Reaction Means Spontaneous Reaction Very negative  spontaneous reaction. Look up Appendix 2A, we have

58. Negative Free Energy of Reaction Means Spontaneous Reaction Negative  spontaneous reaction. Look up Appendix 2A, we have

59. Negative Free Energy of Reaction Means Spontaneous Reaction Very positive  not spontaneous reaction (the reverse is). Look up Appendix 2A, we have In biological systems, glucose is synthesized by assistance of a special bioenzyme.

60. Reaction Free Energy Varies With Temperature Reaction free energy can be determined from reaction quotient.

61. Figure 17.13 At constant temperature and pressure, the direction of spontaneous change is toward lower free energy. The equilibrium composition of a reaction mixture corresponds to the lowest point on the curve. In this example, substantial quantities of both reactants and products are present at equilibrium, and K is close to 1.

62. Figure 17.14 In this reaction, the free energy is a minimum when products are much more abundant than reactants. The equilibrium lies in favor of the products, and K  1. This reaction effectively goes almost to completion.

63. Figure 17.15 In this reaction, the free energy is a minimum when the reactants are much more abundant than the products. The equilibrium lies in favor of the reactants, and K  1. This reaction “does not go.”

64. Free Energy of Reaction Gives The Temperature Dependence of The Equilibrium Constant

65. Equilibrium Constant Can Be Found From Free Energies Calculate K p at 25 o C for the equilibrium

66. Figure 17.16 A negative value of the standard reaction free energy corresponds to an equilibrium constant greater than 1 and to products (yellow) favored over reactants (purple) at equilibrium. A positive value of the standard reaction free energy corresponds to an equilibrium constant of less than 1 and to reactants favored over products at equilibrium.

67. Estimate the minimum temperature at which K>1 K>1 when the reaction free energy becomes negative 

68. Objectives (1) Skills You Should Have Mastered Conceptual 1. State and explain the implications of the second law of thermodynamics, Section 17.2. 2. Explain how temperature, volume, and state of matter affect the entropy of a substance, Sections 17.2 and 17.3. 3. Show how Δ S surr is related to Δ H for a change at constant temperature and pressure and justify the relationship, Section 17.5. 4. Show how the free energy change accompanying a process is related to the direction of spontaneous reaction and the position of equilibrium, Sections 17.7 and 17.10. Descriptive 1. Describe the criteria for spontaneity of a reaction, Sections 17.5 and 17.6. 2. Identify thermodynamically unstable compounds from their standard free energies of formation, Section 17.8.

69. Objectives (2) Problem-Solving 1. Predict which of two systems has the greater entropy, given their compositions and conditions, Toolbox 17.1 and Example 17.1. 2. Calculate the change in entropy of a system due to heat transfer and phase changes, Toolbox 17.1 and Examples 17.2 and 17.3. 3. Estimate the relative molar entropies of two substances, Toolbox 17.1 and Example 17.4. 4. Calculate the standard reaction entropy from standard molar entropies, Example 17.5. 5. Judge the spontaneity of a reaction from its standard reaction enthalpy and standard reaction entropy, Example 17.6. 6. Predict the boiling point and melting point of a substance from the changes in entropy and enthalpy of the substance, Example 17.7. 7. Calculate a standard free energy of formation from the standard enthalpy of formation and standard molar entropies, Example 17.8. 8. Calculate the standard reaction free energy from free energies of formation, Example 17.9. 9. Calculate the reaction free energy from Δ Gr° and the reaction quotient, Example 17.10. 10. Calculate an equilibrium constant from Δ Gr° at a given temperature, Toolbox 17.2 and Example 17.11. 11. Predict the temperature at which a process with known DH and DS becomes spontaneous, Example 17.12.

70. Assignment for Chapter 17 Exercises: 17.3, 17.8, 17.14, 17.19, 17.28, 17.35, 17.45, 17.53 Supplementary Exercises: 17.64, 17.69, 17.78 Applied Exercises: 17.81 Integrated Exercises: 17.90