1. 11111 Chemistry 132 NT I wish I would have a real tragic love affair and get so bummed out that I’d just quit my job and become a bum for a few years, because I was thinking about doing that anyway. Jack Handey

2. 22222 Thermodynamics and Equilibrium Chapter 18 Module 2 Sections 18.4, 18.5, 18.6 and 18.7 Photomicrograph of urea crystals under polarized light.

3. 33333 Review Calculating the entropy change for a phase transition Predicting the sign of the entropy change of a reaction Calculating  S o for a reaction

4. 44444 Free Energy Concept The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS. This quantity gives a direct criterion for spontaneity of reaction.

5. 55555 Free Energy and Spontaneity Changes in H an S during a reaction result in a change in free energy,  G, given by the equation Thus, if you can show that  G is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous.

6. 66666 Standard Free-Energy Change The standard free energy change,  G o, is the free energy change that occurs when reactants and products are in their standard states. The next example illustrates the calculation of the standard free energy change,  G o, from  H o and  S o.

7. 77777 A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables 6.2 and 18.1. Place below each formula the values of  H f o and S o multiplied by stoichiometric coefficients. So:So: 3 x 130.6191.52 x 193 J/K Hfo:Hfo: 002 x (-45.9) kJ

8. 88888 A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables 6.2 and 18.1. You can calculate  H o and  S o using their respective summation laws.

9. 99999 A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables 6.2 and 18.1. You can calculate  H o and  S o using their respective summation laws.

10. 10 A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables 6.2 and 18.1. Now substitute into our equation for  G o. Note that  S o is converted to kJ/K. (see Exercise 18.6 and Problem 18.37)

11. 11 Standard Free Energies of Formation The standard free energy of formation,  G f o, of a substance is the free energy change that occurs when 1 mol of a substance is formed from its elements in their most stable states at 1 atm pressure and 25 o C. By tabulating  G f o for substances, as in Table 18.2, you can calculate the  G o for a reaction by using a summation law.

12. 12 A Problem To Consider Calculate  G o for the combustion of 1 mol of ethanol, C 2 H 5 OH, at 25 o C. Use the standard free energies of formation given in Table 18.2. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ Place below each formula the values of  G f o multiplied by stoichiometric coefficients.

13. 13 A Problem To Consider Calculate  G o for the combustion of 1 mol of ethanol, C 2 H 5 OH, at 25 o C. Use the standard free energies of formation given in Table 18.2. You can calculate  G o using the summation law. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ

14. 14 A Problem To Consider Calculate  G o for the combustion of 1 mol of ethanol, C 2 H 5 OH, at 25 o C. Use the standard free energies of formation given in Table 18.2. You can calculate  G o using the summation law. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ (see Exercise 18.7 and Problem 18.41)

15. 15  G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 1.When  G o is a large negative number (more negative than about –10 kJ), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.

16. 16  G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 2.When  G o is a large positive number (more positive than about +10 kJ), the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.

17. 17  G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 3.When  G o is a small negative or positive value (less than about 10 kJ), the reaction gives an equilibrium mixture with significant amounts of both reactants and products. (see Exercise 18.8 and Problem 18.45)

18. 18 Maximum Work Often reactions are not carried out in a way that does useful work. As a spontaneous precipitation reaction occurs, the free energy of the system decreases and entropy is produced, but no useful work is obtained. In principle, if a reaction is carried out to obtain the maximum useful work, no entropy is produced.

19. 19 Maximum Work Often reactions are not carried out in a way that does useful work. It can be shown that the maximum useful work, w max, for a spontaneous reaction is  G. The term free energy comes from this result.

20. 20 Free Energy Change During Reaction As a system approaches equilibrium, the instantaneous change in free energy approaches zero. Figure 18.9 illustrates the change in free energy during a spontaneous reaction. As the reaction proceeds, the free energy eventually reaches its minimum value. At that point,  G = 0, and the net reaction stops; it comes to equilibrium.

21. Figure 18.9 Free-energy change during a spontaneous reaction.

22. 22 Free Energy Change During Reaction As a system approaches equilibrium, the instantaneous change in free energy approaches zero. Figure 18.10 illustrates the change in free energy during a nonspontaneous reaction. Note that there is a small decrease in free energy as the system goes to equilibrium.

23. Figure 18.10 Free-energy change during a nonspontaneous reaction.

24. 24 Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation. Here Q is the thermodynamic form of the reaction quotient.

25. 25 Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation.  G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium.

26. 26 Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation. At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K.

27. 27 Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation. At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K.

28. 28 Relating  G o to the Equilibrium Constant This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant. When K > 1, the ln K is positive and  G o is negative. When K < 1, the ln K is negative and  G o is positive.

29. 29 A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Rearrange the equation  G o =-RTlnK to give

30. 30 A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Substituting numerical values into the equation,

31. 31 A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Hence, (see Exercises 18.10 and 18.11 and Problems 18.53, 18.55, and 18.57)

32. 32 Spontaneity and Temperature Change All of the four possible choices of signs for  H o and  S o give different temperature behaviors for  G o. HoHo SoSo GoGo Description –+– Spontaneous at all T +–+ Nonspontaneous at all T ––+ or – Spontaneous at low T; Nonspontaneous at high T +++ or – Nonspontaneous at low T; Spontaneous at high T

33. 33 Calculation of  G o at Various Temperatures In this method you assume that  H o and  S o are essentially constant with respect to temperature. You get the value of  G T o at any temperature T by substituting values of  H o and  S o at 25 o C into the following equation.

34. 34 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Place below each formula the values of  H f o and S o multiplied by stoichiometric coefficients.

35. 35 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ You can calculate  H o and  S o using their respective summation laws.

36. 36 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ

37. 37 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ

38. 38 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o.

39. 39 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o.

40. 40 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o. So the reaction is nonspontaneous at 25 o C.

41. 41 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o.

42. 42 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o. So the reaction is spontaneous at 1000 o C.

43. 43 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ To determine the minimum temperature for spontaneity, we can set  G f o =0 and solve for T.

44. 44 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ To determine the minimum temperature for spontaneity, we can set  G f o =0 and solve for T.

45. 45 A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Thus, CaCO 3 should be thermally stable until its heated to approximately 848 o C. (see Exercises 18.12 and 18.13 and Problems 18.59 and 18.61)

46. 46 Operational Skills Calculating  G o from  H o and  S o Calculating  G o from standard free energies of formation Interpreting the sign of  G o Writing the expression for a thermodynamic equilibrium constant Calculating K from the standard free energy change Calculating  G o and K at various temperatures

47. 47 Key Equations (For a spontaneous process) (For an equilibrium process)

48. 48 Key Equations or

49. 49 Homework Chapter 18 Homework: collected at the first exam. Review Questions: 7, 11, 16. Problems: 37, 41, 45, 49, 53, 57, 61, 83, 91. Time for a few review questions

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