1. Thermodynamics and Equilibrium

2. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–2 –We introduced the thermodynamic property of enthalpy, H, in Chapter 6. –We noted that the change in enthalpy equals the heat of reaction at constant pressure. –In this chapter we will define enthalpy more precisely, in terms of the energy of the system. Thermodynamics Thermodynamics is the study of the relationship between heat and other forms of energy in a chemical or physical process.

3. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–3 –The internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system. –Internal energy is a state function. That is, a property of a system that depends only on its present state. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

4. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–4 –Thus, 1 mol of water at 0 o C and 1 atm pressure has a definite quantity of energy. –When a system changes from one state to another, its internal energy changes. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

5. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–5 –Changes in U manifest themselves as exchanges of energy between the system and surroundings. –These exchanges of energy are of two kinds; heat and work. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

6. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–6 –Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings. –Work, on the other hand, is the energy exchange that results when a force F moves an object through a distance d; work (w) = F  d (See Animation: Work Versus Energy Flow)(See Animation: Work Versus Energy Flow) First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

7. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–7 –Remembering our sign convention. Work done by the system is negative. Work done on the system is positive. Heat evolved by the system is negative. Heat absorbed by the system is positive. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

8. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–8 –Figure 19.2 illustrates the distinction between heat and work. –The system in Figure 19.2 gains internal energy from the heat absorbed and loses internal energy via the work done. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

9. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–9 –In general, the first law of thermodynamics states that the change in internal energy,  U, equals heat plus work. First Law of Thermodynamics To state the laws of thermodynamics, we must first understand the internal energy of a system and how you can change it.

10. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–10 –When gases are produced, they exert force on the surroundings as pressure. –If the reaction is run at constant pressure, then the gases produced represent a change in volume analogous to a distance over which a force is exerted. Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

11. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–11 –You can calculate the work done by a chemical reaction simply by multiplying the atmospheric pressure by the change in volume,  V. –It follows therefore, that Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

12. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–12 –For example, when 1.00 mol Zn reacts with excess HCl, 1.00 mol H 2 is produced. q p = -152.4 kJ Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

13. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–13 –At 25 o C and 1.00 atm (1.01 x 10 5 Pa), this amount of H 2 occupies 24.5 L (24.5 x 10 -3 m 3 ). q p = -152.4 kJ Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

14. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–14 –The work done by the chemical system in pushing back the atmosphere is Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

15. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–15 –Relating the change in internal energy to the heat of reaction, you have Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

16. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–16 –When the reaction occurs, the internal energy changes as the kinetic and potential energies change in going from reactants to products. –Energy leaves the system mostly as heat (q p = -152.4 kJ) but partly as expansion work (-2.47 kJ). Heat of Reaction and Internal Energy When a reaction is run in an open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

17. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–17 –We now define enthalpy, H, precisely as the quantity U + PV. –Because U, P, and V are state functions, H is also a state function. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure.

18. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–18 –This means that at a given temperature and pressure, a given amount of a substance has a definite enthalpy. –Therefore, if you know the enthalpies of substances, you can calculate the change in enthalpy,  H, for a reaction. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure.

19. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–19 –In practice, we measure certain heats of reactions and use them to tabulate enthalpies of formation,  H o f. –Standard enthalpies of formation for selected compounds are listed in Table 6.2 and in Appendix C. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure.

20. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–20 –The standard enthalpy change for a reaction is Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure.

21. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–21 –Examples include: A rock at the top of a hill rolls down. Heat flows from a hot object to a cold one. An iron object rusts in moist air. –These processes occur without requiring an outside force and continue until equilibrium is reached. (see Figure 19.4)(see Figure 19.4) Spontaneous Processes and Entropy A spontaneous process is a physical or chemical change that occurs by itself. (See Animation: Spontaneous Reaction) (See Animation: Spontaneous Reaction)

22. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–22 –Entropy, S, is a thermodynamic quantity that is a measure of the randomness or disorder of a system. –The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function. Entropy and the Second Law of Thermodynamics The second law of thermodynamics addresses questions about spontaneity in terms of a quantity called entropy.

23. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–23 –The net change in entropy of the system,  S, equals the sum of the entropy created during the spontaneous process and the change in energy associated with the heat flow. Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

24. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–24 –In other words, (For a spontaneous process) Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

25. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–25 –Normally, the quantity of entropy created during a spontaneous process cannot be directly measured. Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

26. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–26 –If we delete it from our equation, we can conclude (For a spontaneous process) Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

27. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–27 –The restatement of the second law is as follows: for a spontaneous process, the change in entropy of the system is greater than the heat divided by the absolute temperature. Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

28. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–28 –Under these conditions, no significant amount of entropy is created. –The entropy results entirely from the absorption of heat. Therefore, (For an equilibrium process) Entropy Change for a Phase Transition If during a phase transition, such as ice melting, heat is slowly absorbed by the system, it remains near equilibrium as the ice melts.

29. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–29 –Other phase changes, such as vaporization of a liquid, also occur under equilibrium conditions. –Therefore, you can use the previous equation to obtain the entropy change for a phase change. Entropy Change for a Phase Transition If during a phase transition, such as ice melting, heat is slowly absorbed by the system, it remains near equilibrium as the ice melts.

30. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–30 –When liquid CCl 4 evaporates, it absorbs heat:  H vap = 43.0 kJ/mol (43.0 x 10 3 J/mol) at 25 o C, or 298 K. The entropy change,  S, is A Problem To Consider The heat of vaporization,  H vap of carbon tetrachloride, CCl 4, at 25 o C is 43.0 kJ/mol. If 1 mol of liquid CCl 4 has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor at this temperature?

31. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–31 –In other words, 1 mol of CCl 4 increases in entropy by 144 J/K when it vaporizes. –The entropy of 1 mol of vapor equals the entropy of 1 mol of liquid (214 J/K) plus 144 J/K. A Problem To Consider The heat of vaporization,  H vap of carbon tetrachloride, CCl 4, at 25 o C is 43.0 kJ/mol. If 1 mol of liquid CCl 4 has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor at this temperature?

32. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–32 –Recall that the heat at constant pressure, q p, equals the enthalpy change,  H. –The second law for a spontaneous reaction at constant temperature and pressure becomes (Spontaneous reaction, constant T and P) Entropy, Enthalpy, and Spontaneity Now you can see how thermodynamics is applied to the question of reaction spontaneity.

33. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–33 –Rearranging this equation, we find –This inequality implies that for a reaction to be spontaneous,  H-T  S must be negative. –If  H-TDS is positive, the reverse reaction is spontaneous. If  H-T  S=0, the reaction is at equilibrium (Spontaneous reaction, constant T and P) Entropy, Enthalpy, and Spontaneity Now you can see how thermodynamics is applied to the question of reaction spontaneity.

34. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–34 –When temperature is raised, however, the substance becomes more disordered as it absorbs heat. –The entropy of a substance is determined by measuring how much heat is required to change its temperature per Kelvin degree. Standard Entropies and the Third Law of Thermodynamics The third law of thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy of zero.

35. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–35 –Standard state implies 25 o C, 1 atm pressure, and 1 M for dissolved substances. Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion (Table 19.1), also called its absolute entropy, S o, is the entropy value for the standard state of the species.

36. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–36 –Note that the elements have nonzero values, unlike standard enthalpies of formation,  H f o, which by convention, are zero. Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion (Table 19.1), also called its absolute entropy, S o, is the entropy value for the standard state of the species.

37. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–37 –The symbol S o, rather than  S o, is used for standard entropies to emphasize that they originate from the third law. Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion (Table 19.1), also called its absolute entropy, S o, is the entropy value for the standard state of the species.

38. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–38 –Even without knowing the values for the entropies of substances, you can sometimes predict the sign of  S o for a reaction. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  S o.

39. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–39 1.A reaction in which a molecule is broken into two or more smaller molecules. –The entropy usually increases in the following situations: Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  S o.

40. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–40 –The entropy usually increases in the following situations: 2.A reaction in which there is an increase in the moles of gases. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  S o.

41. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–41 –The entropy usually increases in the following situations: 3.A process in which a solid changes to liquid or gas or a liquid changes to gas. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  S o.

42. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–42 –The calculation is similar to that used to obtain  H o from standard enthalpies of formation. A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol.K). See Table 19.1 for other values.

43. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–43 –It is convenient to put the standard entropies (multiplied by their stoichiometric coefficients) below the formulas. So:So:2 x 19321417470 A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol.K). See Table 19.1 for other values.

44. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–44 –We can now use the summation law to calculate the entropy change. A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol.K). See Table 19.1 for other values.

45. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–45 A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol.K). See Table 19.1 for other values. –We can now use the summation law to calculate the entropy change.

46. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–46 –This quantity gives a direct criterion for spontaneity of reaction. 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.

47. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–47 –Thus, if you can show that  G is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous. Free Energy and Spontaneity Changes in H an S during a reaction result in a change in free energy,  G, given by the equation

48. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–48 –The next example illustrates the calculation of the standard free energy change,  G o, from  H o and  S o. 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.

49. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–49 –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 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 19.1.

50. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–50 –You can calculate  H o and  S o using their respective summation laws. 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 19.1.

51. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–51 –You can calculate  H o and  S o using their respective summation laws. 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 19.1.

52. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–52 –Now substitute into our equation for  G o. Note that  S o is converted to kJ/K. 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 19.1.

53. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–53 –By tabulating  G f o for substances, as in Table 19.2, you can calculate the  G o for a reaction by using a summation law. 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 stablest states at 1 atm pressure and 25 o C.

54. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–54 Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ –Place below each formula the values of  G f o multiplied by stoichiometric coefficients. 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 19.2.

55. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–55 Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ –You can calculate  G o using the summation law. 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 19.2.

56. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–56 –You can calculate  G o using the summation law. A Problem To Consider Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ 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 19.2.

57. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–57 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.  G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction.

58. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–58 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. The following rules are useful in judging the spontaneity of a reaction.  G o as a Criteria for Spontaneity

59. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–59 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. The following rules are useful in judging the spontaneity of a reaction.  G o as a Criteria for Spontaneity

60. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–60 –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. Maximum Work Often reactions are not carried out in a way that does useful work.

61. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–61 –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. Maximum Work Often reactions are not carried out in a way that does useful work.

62. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–62 –Figure 19.9 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. Free Energy Change During Reaction As a system approaches equilibrium, the instantaneous change in free energy approaches zero.

63. Figure 19.9 Free-energy change during a spontaneous reaction.

64. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–64 –Figure 19.10 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. Free Energy Change During Reaction As a system approaches equilibrium, the instantaneous change in free energy approaches zero.

65. Figure 19.10 Free-energy change during a nonspontaneous reaction.

66. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–66 –Here Q is the thermodynamic form of the reaction quotient. 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.

67. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–67 –  G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium. 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. Relating  G o to the Equilibrium Constant

68. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–68 –At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K. 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. Relating  G o to the Equilibrium Constant

69. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–69 –At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K. 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. Relating  G o to the Equilibrium Constant

70. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–70 –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. This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant. Relating  G o to the Equilibrium Constant

71. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–71 –Rearrange the equation  G o =-RTlnK to give 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.

72. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–72 –Substituting numerical values into the equation, 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.

73. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–73 –Hence, 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.

74. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–74 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 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.

75. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–75 –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. 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.

76. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–76 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. 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.

77. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–77 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. 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.

78. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–78 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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.

79. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–79 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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.

80. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–80 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o.

81. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–81 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o.

82. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–82 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –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.

83. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–83 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o.

84. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–84 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –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.

85. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–85 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –To determine the minimal temperature for spontaneity, we can set  G f º=0 and solve for T.

86. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–86 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –To determine the minimal temperature for spontaneity, we can set  G f º=0 and solve for T.

87. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–87 So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ 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. –Thus, CaCO 3 should be thermally stable until its heated to approximately 848 o C.

88. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–88 Operational Skills Calculating the entropy change for a phase transition Predicting the sign of the entropy change of a reaction Calculating  S o for a reaction 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

89. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–89 Animation: Work Versus Energy Flow Return to slide 6 (Click here to open QuickTime animation)

90. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–90 Animation: Spontaneous Reaction Return to slide 21 (Click here to open QuickTime animation)

91. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 19–91 Figure 19.4: Examples of a spontaneous and nonspontaneou s process. Return to slide 21