1. Thermodynamics and Equilibrium Chapter 19

2. 2 Copyright © by Houghton Mifflin Company. All rights reserved. Thermodynamics Thermodynamics is the study of the relationship between heat and other forms of energy in a chemical or physical process. 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.

3. Chapter 193 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

4. Chapter 194 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

5. Chapter 195 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

6. Chapter 196 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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

7. Chapter 197 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

8. Chapter 198 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

9. Chapter 199 Copyright © by Houghton Mifflin Company. All rights reserved. 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. In general, the first law of thermodynamics states that the change in internal energy,  U, equals heat plus work.

10. Chapter 1910 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

11. Chapter 1911 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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

12. Chapter 1912 Copyright © by Houghton Mifflin Company. All rights reserved. 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. For example, when 1.00 mol Zn reacts with excess HCl, 1.00 mol H 2 is produced. q p = -152.4 kJ

13. Chapter 1913 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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

14. Chapter 1914 Copyright © by Houghton Mifflin Company. All rights reserved. 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. The work done by the chemical system in pushing back the atmosphere is

15. Chapter 1915 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Relating the change in internal energy to the heat of reaction, you have

16. Chapter 1916 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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).

17. Chapter 1917 Copyright © by Houghton Mifflin Company. All rights reserved. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure. 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.

18. Chapter 1918 Copyright © by Houghton Mifflin Company. All rights reserved. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure. 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.

19. Chapter 1919 Copyright © by Houghton Mifflin Company. All rights reserved. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure. 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.

20. Chapter 1920 Copyright © by Houghton Mifflin Company. All rights reserved. Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms of the relationship of  H to the heat at constant pressure. The standard enthalpy change for a reaction is

21. Chapter 1921 Copyright © by Houghton Mifflin Company. All rights reserved. Spontaneous Processes and Entropy A spontaneous process is a physical or chemical change that occurs by itself. 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

22. Chapter 1922 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy and the Second Law of Thermodynamics The second law of thermodynamics addresses questions about spontaneity in terms of a quantity called entropy. 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.

23. Chapter 1923 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

24. Chapter 1924 Copyright © by Houghton Mifflin Company. All rights reserved. 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. In other words, (For a spontaneous process)

25. Chapter 1925 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Normally, the quantity of entropy created during a spontaneous process cannot be directly measured.

26. Chapter 1926 Copyright © by Houghton Mifflin Company. All rights reserved. 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. If we delete it from our equation, we can conclude (For a spontaneous process)

27. Chapter 1927 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

28. Chapter 1928 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Under these conditions, no significant amount of entropy is created. The entropy results entirely from the absorption of heat. Therefore, (For an equilibrium process)

29. Chapter 1929 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

30. Chapter 1930 Copyright © by Houghton Mifflin Company. All rights reserved. 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? 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

31. Chapter 1931 Copyright © by Houghton Mifflin Company. All rights reserved. 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? 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.

32. Chapter 1932 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy, Enthalpy, and Spontaneity Now you can see how thermodynamics is applied to the question of reaction spontaneity. 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)

33. Chapter 1933 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy, Enthalpy, and Spontaneity Now you can see how thermodynamics is applied to the question of reaction spontaneity. Rearranging this equation, we find This inequality implies that for a reaction to be spontaneous,  H-T  S must be negative. If  H-T  S is positive, the reverse reaction is spontaneous. If  H-T  S=0, the reaction is at equilibrium (Spontaneous reaction, constant T and P)

34. Chapter 1934 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

35. Chapter 1935 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Standard state implies 25 o C, 1 atm pressure, and 1 M for dissolved substances.

36. Chapter 1936 Copyright © by Houghton Mifflin Company. All rights reserved. Standard Entropies and the Third Law of Thermodynamics The standard entroy 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. Note that the elements have nonzero values, unlike standard enthalpies of formation,  H f o, which by convention, are zero.

37. Chapter 1937 Copyright © by Houghton Mifflin Company. All rights reserved. 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. The symbol S o, rather than  S o, is used for standard entropies to emphasize that they originate from the third law.

38. Chapter 1938 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. Even without knowing the values for the entropies of substances, you can sometimes predict the sign of  S o for a reaction.

39. Chapter 1939 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. 1.A reaction in which a molecule is broken into two or more smaller molecules. The entropy usually increases in the following situations:

40. Chapter 1940 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. The entropy usually increases in the following situations: 2.A reaction in which there is an increase in the moles of gases.

41. Chapter 1941 Copyright © by Houghton Mifflin Company. All rights reserved. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. 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.

42. Chapter 1942 Copyright © by Houghton Mifflin Company. All rights reserved. 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. The calculation is similar to that used to obtain  H o from standard enthalpies of formation.

43. Chapter 1943 Copyright © by Houghton Mifflin Company. All rights reserved. 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. It is convenient to put the standard entropies (multiplied by their stoichiometric coefficients) below the formulas. So:So:2 x 19321417470

44. Chapter 1944 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

45. Chapter 1945 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Chapter 1946 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

47. Chapter 1947 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

48. Chapter 1948 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

49. Chapter 1949 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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

50. Chapter 1950 Copyright © by Houghton Mifflin Company. All rights reserved. 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. You can calculate  H o and  S o using their respective summation laws.

51. Chapter 1951 Copyright © by Houghton Mifflin Company. All rights reserved. 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. You can calculate  H o and  S o using their respective summation laws.

52. Chapter 1952 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Now substitute into our equation for  G o. Note that  S o is converted to kJ/K.

53. Chapter 1953 Copyright © by Houghton Mifflin Company. All rights reserved. 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. 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.

54. Chapter 1954 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ Place below each formula the values of  G f o multiplied by stoichiometric coefficients.

55. Chapter 1955 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ You can calculate  G o using the summation law.

56. Chapter 1956 Copyright © by Houghton Mifflin Company. All rights reserved. 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. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ You can calculate  G o using the summation law.

57. Chapter 1957 Copyright © by Houghton Mifflin Company. All rights reserved.  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.

58. Chapter 1958 Copyright © by Houghton Mifflin Company. All rights reserved.  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.

59. Chapter 1959 Copyright © by Houghton Mifflin Company. All rights reserved.  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.

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

61. Chapter 1961 Copyright © by Houghton Mifflin Company. All rights reserved. Free Energy Change During Reaction As a system approaches equilibrium, the instantaneous change in free energy approaches zero. 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.

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

63. Chapter 1963 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

64. Chapter 1964 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

65. Chapter 1965 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

66. Chapter 1966 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

67. Chapter 1967 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

68. Chapter 1968 Copyright © by Houghton Mifflin Company. All rights reserved. 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

69. Chapter 1969 Copyright © by Houghton Mifflin Company. All rights reserved. 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,

70. Chapter 1970 Copyright © by Houghton Mifflin Company. All rights reserved. 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,

71. Chapter 1971 Copyright © by Houghton Mifflin Company. All rights reserved. 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

72. Chapter 1972 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

73. Chapter 1973 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

74. Chapter 1974 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

75. Chapter 1975 Copyright © by Houghton Mifflin Company. All rights reserved. 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

76. Chapter 1976 Copyright © by Houghton Mifflin Company. All rights reserved. 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

77. Chapter 1977 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

78. Chapter 1978 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

79. Chapter 1979 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

80. Chapter 1980 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

81. Chapter 1981 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

82. Chapter 1982 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

83. Chapter 1983 Copyright © by Houghton Mifflin Company. All rights reserved. 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.

84. Chapter 1984 Copyright © by Houghton Mifflin Company. All rights reserved. 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.