1. Chapter 19 Chemical Thermodynamics © 2012 Pearson Education, Inc.

2. Chemical Thermodynamics First Law of Thermodynamics You will recall from Chapter 5 that energy cannot be created or 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. © 2012 Pearson Education, Inc.

3. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

4. Chemical Thermodynamics A.Yes, as the eggs moves downward. B.Yes, because the eggs undergo a change in velocity upon release. C.No, because the egg breaks upon hitting the surface. D.No, the yellow yolk is a liquid and changes locations within the egg.

5. Chemical Thermodynamics Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. © 2012 Pearson Education, Inc.

6. Chemical Thermodynamics A.Greater than 0.5 atm B.Equal to 0.5 atm C.Less than 0.5 atm

7. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

8. Chemical Thermodynamics A.Freezing of liquid water to ice B.Melting of ice to liquid water

9. Chemical Thermodynamics A.Yes. Non spontaneous processes can never occur under any circumstances. B.No. Non spontaneous processes can occur with some continuous external assistance.

10. Chemical Thermodynamics Sample Exercise 19.1 Identifying Spontaneous Processes Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) Water at 40 °C gets hotter when a piece of metal heated to 150 °C is added. (b) Water at room temperature decomposes into H 2 (g) and O 2 (g). (c) Benzene vapor, C 6 H 6 (g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C. Practice Exercise At 1 atm pressure, CO2(s) sublimes at –78 °C. Is this process spontaneous at –100 °C and 1 atm pressure?

11. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

12. Chemical Thermodynamics A.δT must be very large. B.δT must be close to zero. C.δT must be zero. D.δT must be infinitesimally small.

13. Chemical Thermodynamics A.Yes, because the system is restored to its original state. B.Yes, because the surroundings are also restored to their original state. C.No, because there is no evidence that the surroundings are also restored to their original state. D.No, because energy had to be used to restore the system.

14. Chemical Thermodynamics Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Spontaneous processes are irreversible. © 2012 Pearson Education, Inc.

15. Chemical Thermodynamics Entropy Entropy (S) is a term coined by Rudolph Clausius in the nineteenth century. Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered,. 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. qTqT © 2012 Pearson Education, Inc.

16. Chemical Thermodynamics Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial 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: © 2012 Pearson Education, Inc.  S = q rev T

17. Chemical Thermodynamics A.The value of  S is a state function because q is constant for a specified T irrespective of the path chosen. B.  S depends not merely on q but on q rev. There is only one reversible isothermal path between two states regardless of the number of possible paths. C.  S has negligible dependence on q and thus q does not affect the state function properties of S or  S. D.  H = q p.  H is directly related to q.  H is a state function; thus,  S is also a state function.

18. Chemical Thermodynamics Sample Exercise 19.2 Calculating ΔS for a Phase Change Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ΔH 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? Practice Exercise The normal boiling point of ethanol, C 2 H 5 OH, is 78.3 °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?

19. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

20. Chemical Thermodynamics 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 These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases. © 2012 Pearson Education, Inc.

21. Chemical Thermodynamics A.We need to know if the change involves a closed or open system to make a conclusion. B. The entropy of the surroundings must increase by the same amount as the entropy decrease of the system. C.The entropy of the surroundings must increase by a greater amount than the entropy decrease of the system. D.The entropy of the surroundings must decrease by a smaller amount than the entropy decrease of the system.

22. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

23. Chemical Thermodynamics A.Molecules and single atoms experience the same types of motion. B.A molecule can vibrate and rotate; a single atom undergoes neither. C.A molecule can undergo translational motion and rotate; a single atom undergoes neither. D.A molecule can vibrate and undergo translational motion; a single atom undergoes neither.

24. Chemical Thermodynamics 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 about an axis or rotation about  bonds. 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. Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy is S = k ln W where k is the Boltzmann constant, 1.38  10  23 J/K. © 2012 Pearson Education, Inc.

25. Chemical Thermodynamics A. S = 0 B.S = 1 C.S > 0 D.S < 0

26. Chemical Thermodynamics Entropy on the Molecular Scale The change in entropy for a process, then, is  S = k ln W final  k ln W initial W final W initial  S = k ln Entropy increases with the number of microstates in the system. The number of microstates and, therefore, the entropy, tends to increase with increases in −Temperature −Volume −The number of independently moving molecules. © 2012 Pearson Education, Inc.

27. Chemical Thermodynamics Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) © 2012 Pearson Education, Inc.

28. Chemical Thermodynamics A.VaporB. LiquidC. Ice

29. Chemical Thermodynamics Solutions Generally, when a solid is dissolved in a solvent, entropy increases. © 2012 Pearson Education, Inc.

30. Chemical Thermodynamics A.Increase in number of molecules during change B.Decrease in number of molecules during change C.Change in pressure of system D.Change in internal energy of system

31. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

32. Chemical Thermodynamics Predict whether is positive or negative for each process, assuming each occurs at constant ΔS 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) Sample Exercise 19.3 Predicting the Sign of ΔS Practice Exercise Indicate whether each process 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)

33. Chemical Thermodynamics In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Sample Exercise 19.4 Predicting Relative Entropies Practice Exercise Choose the system 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 °C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H 2 O(l) at 25 °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.

34. Chemical Thermodynamics A.The system is a substance at the triple point. B.The system is pure liquid at 0 K (absolute zero). C.The contents of the system are at their standard states. D.They system must be a perfect crystal at 0 K (absolute zero).

35. Chemical Thermodynamics Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. © 2012 Pearson Education, Inc.

36. Chemical Thermodynamics A.During melting or boiling at constant temperature entropy dramatically increases because energy is removed from the system during the change. B.During melting or boiling at constant temperature entropy changes significantly because more microstates exist for the final phase compared to the initial phase. C.During melting or boiling at constant temperature entropy changes significantly because fewer microstates exist for the final phase compared to the initial phase. D.During melting or boiling at constant temperature entropy changes significantly because no change in the number of microstates occurs during the phase change.

37. Chemical Thermodynamics Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. © 2012 Pearson Education, Inc.

38. Chemical Thermodynamics Standard Entropies Larger and more complex molecules have greater entropies. © 2012 Pearson Education, Inc.

39. Chemical Thermodynamics A.270 J/mol-K to 300 J/mole-K B.280 J/mol-K to 305 J/mole-K C.310 J/mol-K to 315 J/mole-K D.320 J/mol-K to 340 J/mole-K

40. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

41. Chemical Thermodynamics Sample Exercise 19.5 Calculating ΔS° from Tabulated Entropies Calculate the change in the standard entropy of the system, ΔS°, for the synthesis of ammonia from N 2 (g) and H 2 (g) at 298 K:N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Practice Exercise Using the standard molar entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K: Al 2 O 3 (s) + 3 H 2 (g)  2 Al(s) + 3 H 2 O(g)

42. Chemical Thermodynamics Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: At constant pressure, q sys is simply  H  for the system. The universe is composed of the system and the surroundings. Therefore,  S universe =  S system +  S surroundings For spontaneous processes  S universe > 0  S surr =  q sys T © 2012 Pearson Education, Inc.

43. Chemical Thermodynamics 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 © 2012 Pearson Education, Inc.

44. Chemical Thermodynamics A.Always increase B.Always decrease C.Sometimes increases and sometimes decreases, depending on the process

45. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

46. Chemical Thermodynamics A.Spontaneous B.Nonspontaneous

47. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

48. Chemical Thermodynamics A.The “downhill” in free energy leads to a transition state. B.An equilibrium is always achieved when both enthalpy and entropy decrease in magnitude, that is “downhill”. C.Free-energy difference between a product and a reactant is always negative, that is “downhill”. D.An equilibrium reaches a state of minimum free energy, which is lower than the free energies of the reactants and products, in a “downhill” direction.

49. Chemical Thermodynamics A.(a) The entropy of the universe increases and (b) the free energy of a system decreases at constant temperature. B.(a) The entropy of the system decreases and (b) the free energy of the universe increases at constant temperature. C.(a) The entropy of the system increases and (b) the free energy of the universe decreases at constant temperature. D.(a) The entropy of the universe decreases and (b) the free energy of a system increases at constant temperature.

50. Chemical Thermodynamics Sample Exercise 19.6 Calculating Free–Energy Change from ΔH°, T, and ΔS° 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 ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these conditions? Practice Exercise Calculate ΔG° for a reaction for which ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Is the reaction spontaneous under these conditions?

51. Chemical Thermodynamics 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. At temperatures other than 25  C,  G  =  H   T  S  How does  G  change with temperature? © 2012 Pearson Education, Inc.

52. Chemical Thermodynamics A.It indicates the process is spontaneous under standard conditions. B.It indicates the process has taken place under standard conditions. C.It indicates the process has taken place at 273 K and 1 torr. D.It indicates the process has taken place at 1 atm and 0 K.

53. Chemical Thermodynamics Sample Exercise 19.7Calculating Standard Free–Energy Change from Free Energies of Formation (a) Use data from Appendix C to calculate the standard free–energy change for the reaction P 4 (g) + 6 Cl 2 (g)  4 PCl 3 (g) run at 298 K. (b) What is ΔG° for the reverse of this reaction? Practice Exercise Use data from Appendix C to calculate ΔG° at 298 K for the combustion of methane: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g).

54. Chemical Thermodynamics Sample Exercise 19.8 Estimating and Calculating ΔG° In Section 5.7 we used Hess’s law to calculate ΔH° 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) ΔH° = –2220 kJ (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate ΔG° for the reaction at 298 K. Is your prediction from part (a) correct? Practice Exercise For the combustion of propane at 298 K, C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g), do you expect ΔG° to be more negative or less negative than ΔH°?

55. Chemical Thermodynamics 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. © 2012 Pearson Education, Inc.

56. Chemical Thermodynamics A.  H = T  S B.  H < T  S C.  H > T  S D.Cannot determine without additional information

57. Chemical Thermodynamics Sample Exercise 19.9Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for the reaction changes with increasing temperature. (b) Calculate ΔG° at 25 °C and 500 °C. Practice Exercise (a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g). (b) Use your values from part (a) to estimate ΔG° at 400 K.

58. Chemical Thermodynamics Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT ln Q (Under standard conditions, all concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.) © 2012 Pearson Education, Inc.

59. Chemical Thermodynamics Sample Exercise 19.10Relating ΔG to a Phase Change at Equilibrium (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl 4 (l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use data from Appendix C and Equation 19.12 to estimate the normal boiling point of CCl 4. Practice Exercise 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 Figure 11.5.)

60. Chemical Thermodynamics Sample Exercise 19.11Calculating the Free–Energy Change under Nonstandard Conditions Calculate at 298 K for a mixture of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3 being used in the Haber process: Practice Exercise Calculate at 298 K for the Haber reaction if the reaction mixture consists of 0.50 atm N 2, 0.75 atm H 2, and 2.0 atm NH 3.

61. Chemical Thermodynamics Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT ln K Rearranging, this becomes  G  =  RT ln K or K = e © 2012 Pearson Education, Inc.  G  /RT

62. Chemical Thermodynamics Sample Exercise 19.12 Calculating an Equilibrium Constant from ΔG° The standard free–energy change for the Haber process at was obtained in Sample Exercise 19.9 for the Haber reaction: Use this value of ΔG° to calculate the equilibrium constant for the process at 25 °C. Practice Exercise Use data from Appendix C to calculate ΔG° and K at 298 K for the reaction.