1. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Chapter 19 Chemical Thermodynamics Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten John D. Bookstaver St. Charles Community College Cottleville, MO

2. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. First Law of Thermodynamics You will recall from Chapter 5 that energy cannot be created nor destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

3. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. HH  H = heat transferred between the system and its surroundings during a constant-pressure process  E = q + w  E =  in internal E of a system q = heat absorbed by the system from surroundings w = work done on the system by the surroundings (see Chapter 5 for details)

4. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.

5. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

6. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous.

7. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.

8. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Spontaneous processes are irreversible.

9. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Entropy (S) is a term coined by Rudolph Clausius in the 19th century. Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered,. qTqT

10. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Entropy can be thought of as a measure of the randomness of a system. It is related to the various modes of motion in molecules.

11. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial

12. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:  S = q rev T

13. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.2 (p. 807) The element, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 o C, and its molar enthalpy of fusion is  H fusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg (l) freezes at the normal freezing point? Hint: Calculate q, as you have done before, then reverse the sign and divide by T in K. (  S sys = -2.44 J/K)

14. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.2 The normal boiling point of ethanol, C 2 H 5 OH, is 78.3 o C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C 2 H 5 OH (g) at 1 atm condenses to liquid at the normal boiling point? (-163 J/K)

15. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.

16. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics In other words: For reversible processes:  S univ =  S system +  S surroundings = 0 For irreversible processes:  S univ =  S system +  S surroundings > 0

17. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.

18. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample.

19. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale Molecules exhibit several types of motion: –Translational: Movement of the entire molecule from one place to another. –Vibrational: Periodic motion of atoms within a molecule. –Rotational: Rotation of the molecule on about an axis or rotation about  bonds.

20. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. –This would be akin to taking a snapshot of all the molecules. He referred to this sampling as a microstate of the thermodynamic system.

21. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy is S = k lnW where k is the Boltzmann constant, 1.38  10  23 J/K.

22. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale The change in entropy for a process, then, is  S = k lnW final  k lnW initial lnW final lnW initial  S = k ln Entropy increases with the number of microstates in the system.

23. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tends to increase with increases in –Temperature. –Volume. –The number of independently moving molecules.

24. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S (g) > S (l) > S (s)

25. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Solutions Generally, when a solid is dissolved in a solvent, entropy increases.

26. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes In general, entropy increases when –Gases are formed from liquids and solids; –Liquids or solutions are formed from solids; –The number of gas molecules increases; –The number of moles increases.

27. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

28. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.3 (p. 814) Predict whether  S is positive or negative for each of the following processes, assuming each occurs at constant temperature: a) H 2 O (l)  H 2 O (g) b) Ag + (aq) + Cl - (aq)  AgCl (s) c) 4 Fe (s) + 3 O 2(g)  2 Fe 2 O 3(s) d) N 2(g) + O 2(g)  2 NO (g)

29. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.3 Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO 2(s)  CO 2(g) b) CaO (s) + CO 2(g)  CaCO 3(s) c) HCl (g) + NH 3(g)  NH 4 Cl (s) d) 2 SO 2(g) + O 2(g)  2 SO 3(g)

30. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.4 (p. 815) Choose the sample of matter that has greater entropy in each pair, and explain your choice: a) 1 mol of NaCl (s) or 1 mol of HCl (g) at 25 o C b) 2 mol of HCl (g) or 1 mol of HCl (g) at 25 o C c) 1 mol of HCl (g) or 1 mol or Ar (g) at 298 K

31. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.4 Choose the substance with the greater entropy in each case: a) 1 mol of H 2(g) at STP or 1 mol of H 2(g) at 100 o C and 0.5 atm b) 1 mol of H 2 O (s) at 0 o C or 1 mol of H 2 O (l) at 25 o C c) 1 mol of H 2(g) at STP or 1 mol of SO 2(g) at STP d) 1 mol of N 2 O 4(g) at STP or 2 mol of NO 2(g) at STP

32. Chemical Thermodynamics Where we’ve been:  H, including Hess’s Law – the heat component  S, including #microstates – the entropy component Where we’re going:  H and  S  Gibbs Free Energy – will a reaction occur? Application to equilibrium Application to electrochemistry (after Spring Break) © 2009, Prentice-Hall, Inc.

33. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass.

34. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Entropies Larger and more complex molecules have greater entropies.

35. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which  H is estimated:  S  =  n  S  (products) —  m  S  (reactants) where n and m are the coefficients in the balanced chemical equation.

36. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process:  S surr =  q sys T At constant pressure, q sys is simply  H  for the system.

37. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore,  S universe =  S system +  S surroundings For spontaneous processes  S universe > 0

38. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Entropy Change in the Universe Since  S surroundings = and q system =  H system This becomes:  S universe =  S system + Multiplying both sides by  T, we get  T  S universe =  H system  T  S system  H system T  q system T

39. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Gibbs Free Energy  T  S universe is defined as the Gibbs free energy,  G. When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous.

40. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Gibbs Free Energy 1.If  G is negative, the forward reaction is spontaneous. 2.If  G is 0, the system is at equilibrium. 3.If  G is positive, the reaction is spontaneous in the reverse direction.

41. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.6 (p. 821) Calculate the standard free energy change for the formation of NO (g) from N 2(g) and O 2(g) at 298 K: N 2(g) + O 2(g)  2 NO (g) given that  H o = 180.7 kJ and  S o = 24.7 J/K. Is the reaction spontaneous under these circumstances? (no,  G = 173.3 kJ)

42. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.6 A particular reaction has  H o = 24.6 kJ and  S o = 132 J/K at 298 K. Calculate  G o. Is the reaction spontaneous under these conditions? (Yes,  G o = -14.7 kJ)

43. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation,  G . f  G  =  n  G  (products)   m  G  (reactants) f f where n and m are the stoichiometric coefficients.

44. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Standard Conditions Solid pure solid Liquidpure liquid Gas1 atm pressure Solution1 M concentration Elementsstandard free energy of formation of an element in its free state = zero

45. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.7 (p. 823) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: a) P 4(g) + 6 Cl 2(g)  4 PCl 3(g) (-1102.8 kJ) b) What is  G o for the reverse of the above reaction? (+1102.8 kJ)

46. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.7 By using the data from Appendix C, calculate  G o at 298 K for the combustion of methane: CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (g) (-800.7 kJ)

47. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.8 (p. 823) In Section 5.7 we used Hess’s law to calculate  H o for the combustion of propane gas at 298 K: C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (l)  H o = -2220 kJ Without using data from Appendix C, predict whether  G o for this reaction is more negative or less negative than  H o. Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?(-2108 kJ)

48. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.8 Consider the combustion of propane to form CO 2(g) and H 2 O (g) at 298 K: C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  H o = -2220 kJ Would you expect  G o to be more negative or less negative than  H o ?

49. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy Changes At temperatures other than 25°C,  G° =  H   T  S  How does  G  change with temperature?

50. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Temperature There are two parts to the free energy equation: –  H  — the enthalpy term –T  S  — the entropy term The temperature dependence of free energy, then comes from the entropy term.

51. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Temperature

52. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.9 (p. 826) The Haber process for the production of ammonia involves the following equilibrium: N 2(g) + 3 H 2(g)  2 NH 3(g) Assume that  H o and  S o for this reaction do not change with temperature. a) Predict the direction in which  G o for this reaction changes with increasing temperature. b) Calculate the values of  G o for the reaction at 25 o C and 500 o C. (-33.3 kJ, 61 kJ)

53. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.9 a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate  H o and  S o at 298 K for the following reaction: 2 SO 2(g) + O 2(g)  2 SO 3(g) (  H o = -196.6 kJ,  S o = -189.6 J/K) b) Using the values obtained in part (a), estimate  G o at 400 K. (  G o = -120.8 kJ)

54. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium Reminders:  G o and K eq apply to standard conditions (pure solid or liquid, gases at 1 atm and 1M solutions).  G and Q (equilibrium quotient) apply to any conditions.

55. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.) R = 8.314 J/mol.K, T is in K

56. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.10 (p. 827) As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. a)Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl 4(l). b) What is the value of  G o for the equilibrium in (a)? c) Use thermodynamic data in Appendix C and  G o =  H o – T  S o to estimate the normal boiling point of CCl 4. (70 o C)

57. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.10 Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br 2(l). (The experimental value is given in Table 11.3). (330 K)

58. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK or, K = e -  G  RT

59. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium If  G o 1 when ln K is +  more negative  G o  larger K  products are favored If  G o = 0, then K = 1 If  G o > 0, then K <1 when ln K is -  more positive  G o  smaller K  reactants are favored

60. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.11 (p. 828) We will continue to explore the Haber process for the synthesis of ammonia: N 2(g) + 3 H 2(g)  2 NH 3(g) Calculate  G at 298 K for a reaction mixture that consists of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3. (-44.9 kJ/mol)

61. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.11 Calculate  G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N 2, 0.75 atm H 2, and 2.0 atm NH 3. (-26.0 kJ/mol)

62. Chemical Thermodynamics © 2009, Prentice-Hall, Inc.

63. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Exercise 19.12 (p. 829) Use standard free energies of formation to calculate the equilibrium constant K eq at 25 o C for the reaction involved in the Haber process: N 2(g) + 3 H 2(g)  2 NH 3(g) (7 x 10 5 )

64. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Practice Exercise 19.12 Use data from Appendix C to calculate the standard free-energy change,  G o, and the equilibrium constant, K, at 298 K for the following reaction: H 2(g) + Br 2(l)  2 HBr (g) (-106.4 kJ/mol; 4 x 10 18 )

65. Chemical Thermodynamics © 2009, Prentice-Hall, Inc. Sample Integrative Exercise 19 (p. 831) Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: NaCl (s)  Na + (aq) + Cl - (aq) AgCl (s)  Ag + (aq) + Cl - (aq) a) Calculate the value of  G o at 298 K for each of the preceding reactions. b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? c) Use the values of  G o to calculate K sp values for the two salts at 298 K. d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? e) How will  G o for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?