1. Chapter 19 Chemical Thermodynamics Lecture Presentation Dr. Subhash C. Goel South GA State College Douglas, GA © 2012 Pearson Education, Inc.

2. Review Units of Energy The SI unit of energy is the joule (J): An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = 4.184 J 1 J = 1  kg m 2 s2s2

3. © 2012 Pearson Education, Inc. Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.

4. © 2012 Pearson Education, Inc. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

5. © 2012 Pearson Education, Inc. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial

6. © 2012 Pearson Education, Inc.  E, q, w, and Their Signs

7. © 2012 Pearson Education, Inc. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −P  V

8. © 2012 Pearson Education, Inc. Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

9. © 2012 Pearson Education, Inc. Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

10. © 2012 Pearson Education, Inc. Enthalpy Since  E = q + w and w = −P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q + w) − w  H = q So, at constant pressure, the change in enthalpy is the heat gained or lost.

11. © 2012 Pearson Education, Inc. Enthalpy of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants  H = enthaly of reaction or heat of reaction

12. © 2012 Pearson Education, Inc. Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

13. © 2012 Pearson Education, Inc. Enthalpies of Formation An enthalpy of formation,  H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

14. © 2012 Pearson Education, Inc. Calculation of  H We can use Hess’s law in this way:  H =  n  H f,products –  m  H f °,reactants where n and m are the stoichiometric coefficients.

15. 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.

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

17. 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.

18. Chemical Thermodynamics Exercise 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.

19. 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.

20. 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.

21. 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,. qTqT © 2012 Pearson Education, Inc.

22. Chemical Thermodynamics Entropy Entropy can be described as a measure of how spread out or dispersed the energy is among the different possible ways that system can contain energy. The greater is the dispersal, greater is the entropy. It is related to the various modes of motion in molecules. © 2012 Pearson Education, Inc.

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

24. Chemical Thermodynamics 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 © 2012 Pearson Education, Inc.

25. Chemical Thermodynamics Exercise 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?

26. 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.

27. 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 © 2012 Pearson Education, Inc.

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

29. 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.

30. 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. © 2012 Pearson Education, Inc.

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

32. Chemical Thermodynamics Entropy on the Molecular Scale 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.

33. 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. © 2012 Pearson Education, Inc.

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

35. 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.

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

37. 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) Exercise 3 Predicting the Sign of ΔS

38. 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. Exercise 4 Predicting Relative Entropies

39. 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.

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

41. 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.

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

43. 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.

44. Chemical Thermodynamics Exercise 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)

45. 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:  S surr =  q sys T At constant pressure, q sys is simply  H  for the system. © 2012 Pearson Education, Inc.

46. Chemical Thermodynamics 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 © 2012 Pearson Education, Inc.

47. 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.

48. 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.

49. 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.

50. Chemical Thermodynamics Exercise 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?

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. © 2012 Pearson Education, Inc.

52. Chemical Thermodynamics Exercise 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?

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

54. 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.

55. Chemical Thermodynamics Free Energy and Temperature © 2012 Pearson Education, Inc.

56. Chemical Thermodynamics Exercise 8Determining 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.

57. 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.

58. Chemical Thermodynamics Exercise 9Calculating 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:

59. 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

60. Chemical Thermodynamics Exercise 10 Calculating an Equilibrium Constant from ΔG° The standard free–energy change for the Haber process at was obtained in Exercise 8 for the Haber reaction: Use this value of ΔG ° to calculate the equilibrium constant for the process at 25 ° C.