1. Chapter 19 Spontaneous Change: Entropy and Free Energy Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php

2. 2 Spontaneous processes We have a general idea of what we consider spontaneous to mean: A spontaneous process WILL OCCUR in a system WITHOUT any outside action being performed on the system.

3. 3 Spontaneous processes Ice melting above zero Celcius is spontatneous Object falling to earth is spontaneous

4. 4 Spontaneous processes Ice will melt above zero Celcius. We don’t have to DO anything! Objects will fall to earth. We don’t have to DO anything!

5. 5 Spontaneous processes Since we DON’T have to DO anything for these spontaneous processes to occur it APPEARS that an overall energy change from potential energy to kinetic energy IS SPONTANEOUS

6. 6 Non-spontaneous processes We have a general idea of what we consider non-spontaneous to mean: A non-spontaneous process WILL NOT OCCUR in a system UNTIL an outside action is performed on the system.

7. 7 Non-spontaneous processes Ice freezing above zero Celcius is non-spontatneous Object rising from earth is non-spontaneous

8. 8 Non-spontaneous processes We can make water freeze above zero Celcius by increasing the pressure. We can make an object rise from the earth by picking it up.

9. 9 Non-spontaneous processes Since we DO have to ACT for these non-spontaneous processes to occur it APPEARS that an overall energy change from kinetic energy to potential energy IS NON-SPONTANEOUS

10. 10 Chemistry and spontaneity We know there are chemical processes that are spontaneous because we can put the chemical system together and reactants become products without us having to do anything. H 3 O + (aq) + OH - (aq)  2 H 2 O (l)

11. 11 Chemistry and non-spontaneity We know there are chemical processes that are non-spontaneous because we can put the chemical system together and reactants DO NOT become products. The system we put together stays like it is UNTIL WE CHANGE SOMETHING! 2 H 2 O (l)  2 H 2 (g) + O 2 (g)

12. 12 Spontaneous vs. non-spontaneous It is obvious by the examples we’ve looked at that the opposite of every spontaneous process is a non-spontaneous process. In chemical systems we’ve seen that if we put a chemical system together a reaction occurs until the system reaches equilibrium. Whether the forward reaction or the reverse reaction dominates depends on which of the two reactions is spontaneous at those conditions!

13. 13 Equilibrium and spontaneity are related!

14. 14 Spontaneity and energy In our examples it APPEARED that the spontaneous process ALWAYS takes a system to a lower potential energy.

15. 15 Spontaneity and energy If this were true all exothermic processes would be spontaneous and all endothermic processes would be non- spontaneous. THIS ISN’T TRUE! H 2 O NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) is spontaneous even though  H = +25.7 kJ

16. 16 Recall the First Law The First Law of thermodynamics stated that the energy of an ISOLATED system is constant. What’s the largest ISOLATED system we can think of? It’s the UNIVERSE! The energy of the universe is constant!

17. 17 Recall the First Law On the universal scale, there is no overall change in energy, and so lower energy CANNOT be the only requirement for spontaneity. There must be something else as well!

18. 18 Further proof lower energy isn’t enough If an ideal gas expands into a vacuum at a constant temperature, then no work is done and no heat is transferred

19. 19 Further proof lower energy isn’t enough No work done and no heat transferred means NO OVERALL CHANGE in energy of the system

20. 20 Further proof lower energy isn’t enough This spontaneous process has no overall change in energy!

21. 21 Entropy Entropy (from Greek, meaning “in transformation”) is a thermodynamic property that relates the distribution of the total energy of the system to the available energy levels of the particles.

22. 22 Entropy A general way to envision entropy is “differing ways to move” Consider mountain climbers on a mountain. Two factors affect the distribution of mountain climbers on a mountain: Total energy of all the climbers and how many places can you stop on the mountain

23. 23 Consider hungry mountain climbers Hungry mountain climbers have little total energy amongst themselves to climb a mountain, so most of them are near the bottom, while some are distributed on the lower parts of the mountain. Few energy levels can be reached!

24. 24 Consider well fed mountain climbers Well-fed mountain climbers have more total energy amongst themselves to climb a mountain, so the climbers will be more spread out on the whole mountain. More energy levels can be reached!

25. 25 Temperature and total energy The total energy shared by molecules is related to the temperature. A given number of molecules at a low temperature (less total energy) have less “differing ways to move” than the same number of molecules at a high temperature (more total energy).

26. 26 Higher mountain means more places to stop If a well-fed mountain climber tries to climb Signal Hill, they will most likely reach the top. They have only a few places to stop (levels) because Signal Hill is a small mountain. The same well-fed mountain climber on Mount Everest has a greater number of places to stop (levels) because it is a larger mountain.

27. 27 Volume and energy levels The number of levels of energy distribution of molecules is related to the volume. A given number of molecules in a small volume have less “differing ways to move” than the same number of molecules in a larger volume.

28. 28 Entropy The greater the number of “differing ways to move” molecules can take amongst the available energy levels of a system of a given state (defined by temperature and volume, and number of molecules), the greater the entropy of the system.

29. 29 Expansion into vacuum A gas expands into a vacuum because the increased volume allows for a greater number of “differing ways to move” for the molecules, even if the temperature is the same.

30. 30 Expansion into vacuum That is, the entropy increases when the gas is allowed to expand into a vacuum. Entropy increase plays a role in spontaneity!

31. 31 Entropy is a state function Entropy, S, is a state function like enthalpy or internal energy. The entropy of a system DEPENDS ONLY on the current state (n, T, V, etc.) of the system, and NOT how the system GOT TO BE in that state.

32. 32 Boltzmann equation and entropy More available energy levels when the size of a box increases – like expanding a gas into a vacuum – ENTROPY INCREASES! More energy levels are accessible when the temperature increases – ENTROPY INCREASES!

33. 33 Change in entropy is a state function Because entropy is a state function, then change in entropy  S is ALSO a state function. The difference in entropy between two states ONLY depends on the entropy of the initial and final states, and NOT the path taken to get there.

34. 34 Hess’s Law Recall Hess’s Law – as long as we get from the same initial state to the same final state then  H will be the same regardless of the steps we add together. Change in entropy  S will work exactly the same way! As long as we get from the same initial state to the same final state then  S will be the same regardless of the steps we add together.

35. 35 Boltzmann equation and entropy n, T, V help define the number of states (number of available energy levels) the system can have. The many “different ways to move” of molecules in a particular state are called microstates. Hopefully it makes sense that more total states should automatically mean more total microstates. The number of microstates is often symbolized by W.

36. 36 Playing cards Say we have a deck of 52 playing cards. Choosing one playing card is a state. If we choose the first card out of the pack, there are 52 microstates for this first state. The second card (second state) we choose has 51 microstates, and so on.

37. 37 Overall there are W = 52!  8 x 10 67 possible distributions (total microstates) for 52 playing cards! Playing cards

38. 38 If we flip 52 coins (a coin is one state), with two possible microstates (heads or tails) each, there are W = 2 52 = 4.5 x 10 15 possible distributions. (total microstates) A deck of 52 playing cards has greater entropy than 52 coins! Playing cards and coin flips

39. 39 Boltzmann equation and entropy Ludwig Boltzmann formulated the relationship between the number of microstates (W) and the entropy (S). S = k ln W The constant k is the Boltzmann constant which has a value equal to the gas constant R divided by Avagadro’s number N A

40. 40 Boltzmann equation and entropy S = k ln W where k = R / N A k = (8.3145 J  K -1  mol -1 ) / (6.022 x 10 23 mol -1 ) k = 1.381 x 10 -23 J  K -1 We can see the units for entropy will be Joules per Kelvin (J  K -1 )

41. 41 Measuring entropy change From the units for entropy (J  K -1 ) we get an idea of how we might measure entropy change  S It must involve some sort of energy change relative to the temperature change!

42. 42 Measuring entropy change  S = q rev / T The change in entropy is the heat involved in a reversible process at a constant temperature.

43. 43 Heat IS NOT a state function Since heat IS NOT a state function we need a reversible process to make it ACT LIKE a state function.

44. 44 Reversible processes In a reversible process a change in one direction is exactly equal and opposite to the change we see if we do the change in the reverse direction. In reality it is impossible to make a reversible process without making an infinite number of infinitesimally small changes.

45. 45 Reversible processes We can however imagine the process is done reversibly and calculate the heat involved in it, so we can calculate the reversible entropy change that could be involved in a process.  S rev = q rev / T

46. 46 Endothermic increases in entropy In these three processes the molecules gain greater “differing ability to move.” The molecules occupy more available microstates at the given temperature, and so the entropy increases in all three processes!

47. 47 Generally entropy increases when… …we go from solid to liquid. …we go from solid or liquid to gas. …we increase the amount of gas in a reaction. …we increase the temperature. …we allow gas to expand against a vacuum. …we mix gases, liquids, or otherwise make solutions of most types.

48. 48 Problem Predict whether entropy increases, decreases, or we’re uncertain for the following processes or reactions: Answers: a) decreases b) increases c) uncertain d) increases

49. 49 Evaluating entropy and entropy changes Phase transitions – In phase transitions the heat change does occur reversibly, so we can use the formula  S rev = q rev / T to calculate the entropy change. In this case the heat is the enthalpy of the phase transition and the temperature is the transition temperature  S =  H tr / T tr

50. 50 Evaluating entropy and entropy changes Phase transitions – For water going from ice (solid) to liquid,  H  fus = 6.02 kJ  mol -1 at the melting point (transition temp.) of 273.15 K (0  C)  S  fus =  H  fus / T mp  S  fus = 6.02 kJ  mol -1 / 273.15 K  S  fus = 22.0 J  K -1  mol -1

51. 51 Evaluating entropy and entropy changes Phase transitions – For water going from liquid to gas,  H  vap = 40.7 kJ  mol -1 at the boiling point (transition temp.) of 373.15 K (100  C)  S  vap =  H  vap / T bp  S  vap = 40.7 kJ  mol -1 / 373.15 K  S  vap = 109 J  K -1  mol -1

52. 52 Problem What is the standard molar entropy of vapourisation  S  vap for CCl 2 F 2 if its boiling point is -29.79  C and  H  vap = 20.2 kJ  mol -1 ? Answer:  S  vap = 83.0 J  K -1  mol -1

53. 53 Problem The entropy change for the transition from solid rhombic sulphur to solid monoclinic sulphur at 95.5  C is  S  tr = 1.09 J  K -1  mol -1. What is the standard molar enthalpy change  H  tr for this transition? Answer:  H  tr = 402 J  mol -1

54. 54 Absolute entropies Say we imagine a system of molecules that has no total energy. At this zero- point energy there can ONLY be ONE possible distribution of microstates, as no molecule has the energy to occupy a higher energy level. The entropy CAN NEVER get smaller than its value in this situation, so we define the entropy S of this situation as ZERO.

55. 55 Absolute entropies This kind of imagining is the Third Law of Thermodynamics which states that The entropy of a pure perfect crystal at 0K is zero. At conditions other than at absolute zero, our entropy is that of the perfect system (zero) PLUS any entropy changes that come changing temperature and/or volume. These are absolute entropies!

56. 56 Methyl chloride entropy as a function of temperature

57. 57 Standard molar entropies One mole of a substance in its standard state will have an absolute entropy that we often call the standard molar entropy S . These are usually tabulated at 298.15 K In a chemical process we can then use these standard molar entropies to calculate the entropy change in the process.

58. 58 Standard molar entropies  S  = [  p S  (products) -  r S  (reactants)] Hopefully this looks somewhat familiar! We have seen a special treatment of Hess’s Law in Chem 1050 where  H  = [  p  H f  (products) -  r  H f  (reactants)] We can do something similar with ANY thermodynamic property that IS A STATE FUNCTION!

59. 59 Standard molar entropies  S  = [  p S  (products) –  r S  (reactants)]  H  = [  p  H f  (products) –  r  H f  (reactants)] Enthalpies of formation ARE NOT absolute!

60. 60 Problem Use the data given to calculate the standard molar entropy change for the synthesis of ammonia from its elements. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) S  298 for N 2 = 191.6 J  K -1  mol -1 S  298 for H 2 = 130.7 J  K -1  mol -1 S  298 for NH 3 = 192.5 J  K -1  mol -1 Answer: -198.7 J  K -1  mol -1 (per mole of rxn)

61. 61 Problem N 2 O 3 is an unstable oxide that readily decomposes. The decomposition of 1.00 mol N 2 O 3 to nitrogen monoxide and nitrogen dioxide at 25  C is accompanied by the entropy change  S  = 138.5 J  K -1  mol -1. What is the standard molar entropy of N 2 O 3 (g) at 25  C? S  298 for NO (g) = 210.8 J  K -1  mol -1 S  298 for NO 2 (g) = 240.1 J  K -1  mol -1 Answer: 312.4 J  K -1  mol -1

62. 62 The second law of thermodynamics We’ve seen that entropy MUST play a role in spontaneity, because the total energy of the universe doesn’t change. We could say that an entropy increase leads to spontaneity, but we have to be careful.

63. 63 The second law of thermodynamics Ice freezing below 0 Celcius is spontaneous, but the entropy of the water decreases in the process!  S  freeze = -  H  fus / T mp Since  H  fus and T mp are +ve, then  S  freeze is –ve! -ve since “freezing” is the reverse of “fusion” (like we do in Hess’s Law)

64. 64 The second law of thermodynamics The water is ONLY the system. The rest of the universe (the surroundings) must experience an opposite heat change as it takes the heat the freezing water gave off (a +ve  H for the surroundings), which means the entropy of the REST OF THE UNIVERSE INCREASES in the process of water freezing!

65. 65 The second law of thermodynamics The total entropy change in any process is the entropy change for the system PLUS the entropy change for the surroundings  S universe =  S total =  S sys +  S surr Now we can connect entropy and spontaneity!

66. 66 The second law of thermodynamics In any spontaneous process the entropy of the universe INCREASES.  S universe =  S sys +  S surr > 0 This is the Second Law of Thermodynamics!

67. 67 Water freezing So while water freezing below zero Celcius decreases the entropy of the system, the heat given off to the surroundings increases the entropy of the surroundings to a greater extent. The total entropy change of the universe is positive and the process of water freezing below 0 Celcius is spontaneous!

68. 68 Free energy We’ve seen entropy increases when molecules have more ways to distribute themselves amongst the energy levels.

69. 69 Free energy However, some of the energy a molecule uses to put itself at a higher energy level CAN NO LONGER be used to to do work because doing work would put the molecule back at a lower energy level, which would automatically decrease entropy. The energy is NOT free (or available) to be used!

70. 70  S =  H rev / T Reversible since the rest of the universe is SO BIG then T  S =  H Free energy  S universe =  S sys +  S surr T  S universe = T  S sys + T  S surr T  S universe = T  S sys +  H surr

71. 71 Free energy T  S universe = T  S sys +  H surr T  S universe = T  S sys -  H sys Since  H sys = -  H surr

72. 72 Free energy T  S universe = T  S sys –  H sys -T  S universe =  H sys - T  S sys  G =  H sys – T  S sys  G is the free energy (Gibbs free energy)

73. 73 Free energy  G = -T  S univ For a spontaneous process  S univ > 0, which means for a spontaneous process  G < 0!

74. 74 Free energy  G < 0 is spontaneous  G > 0 is non-spontaneous  G = 0 is at equilibrium

75. 75

76. 76 Problem Predict the spontaneity at low and high temperatures for: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H  = -92.22 kJ 2 C (s) + 2 H 2 (g)  C 2 H 4 (g)  H  = 52.26 kJ

77. 77 Problem N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H  = -92.22 kJ Spontaneous @ low T and nonspontaneous @ high T 2 C (s) + 2 H 2 (g)  C 2 H 4 (g)  H  = 52.26 kJ Nonspontaneous @ all T

78. 78 Standard free energy change  G  Just like we can have a standard enthalpy change  H  for chemicals, we can also define the standard free energy change…  G  =  H  - T  S 

79. 79 Standard free energy of formation  G f  Standard enthalpies of formation  H f  of elements in their standard states are zero:  H  rxn = [  p  H f  (products) –  r  H f  (reactants)] Standard free energies of formation  G f  of elements in their standard states are zero:  G  = [  p  G f  (products) –  r  G f  (reactants)]

80. 80 Problem Calculate  G  at 298.15 K for the reaction 4 Fe (s) + 3 O 2 (g)  2 Fe 2 O 3 (s) by using the two following sets of data. Compare your answers. a)  H  = -1648 kJ  mol -1 and  S  = -549.3 J  K -1  mol -1 b)  G f  (Fe) = 0 kJ  mol -1  G f  (O 2 ) = 0 kJ  mol -1  G f  (Fe 2 O 3 ) = -742.2 kJ  mol -1

81. 81 Problem answer a)  G  = -1484 kJ  mol -1 b)  G  = -1484.4 kJ  mol -1 The answers are the same because free energy is a state function.

82. 82 Free energy and equilibrium We’ve already seen that  G < 0 is spontaneous  G > 0 is non-spontaneous One process in a written reaction is spontaneous while the reverse is not, as long as  G ≠ 0. Therefore  G = 0 is at equilibrium

83. 83 Water and steam H 2 O (l, 1 atm)  H 2 O (g, 1 atm)  G  373.15 K = 0 kJ  mol -1 The system is at equilibrium at 1 atm (standard conditions) and at the boiling point temperature!

84. 84 Water and steam H 2 O (l, 1 atm)  H 2 O (g, 1 atm)  G  298.15 K = 8.590 kJ  mol -1 The system is not at equilibrium at 1 atm (standard conditions) and at the room temperature!

85. 85 Water and steam H 2 O (l, 1 atm)  H 2 O (g, 1 atm)  G  298.15 K = 8.590 kJ  mol -1 The forward process is non- spontaneous (  G  > 0) so the reverse process is spontaneous and condensation occurs.

86. 86 Water and steam H 2 O (l, 0.03126 atm)  H 2 O (g, 0.03126 atm)  G 298.15 K = 0 kJ The system is at equilibrium at 0.03126 atm (non-standard conditions) and at the room temperature!

87. 87 Water and steam H 2 O (l, 0.03126 atm)  H 2 O (g, 0.03126 atm)  G 298.15 K = 0 kJ  mol -1 Water CAN evaporate at room temperature, just not to give an equilibrium pressure of 1 atm!

88. 88 Non-standard conditions As we’ve seen with the previous water example, our interest in an equilibrium system is often at non-standard conditions, so knowing  G  is usually not as useful as knowing  G.

89. 89 Non-standard conditions For an ideal gas  H does not change if pressure changes, so at all non- standard conditions  H =  H . For an ideal gas  S does change if pressure changes (expansion into vacuum shows us this!), so at all non-standard conditions  S ≠  S .

90. 90 Non-standard free energy Because of these facts, the non- standard free energy change is  G =  H  - T  S But the standard free energy change is  G  =  H  - T  S 

91. 91 Non-standard free energy The difference between standard and non-standard free energy is totally due to the difference in entropy change between the standard and non-standard conditions  G -  G  = - T(  S-  S  )  G =  G  + T(  S  -  S)

92. 92 Boltzmann distribution By the Boltzmann distribution S = R ln W for one mole of particles  G =  G  + T(R ln W - R ln W  )  G =  G  + T(R ln W / W  ) We are comparing a real system with a standard one. We are dealing with activities!

93. 93 Boltzmann distribution  G =  G  + RT ln Q eq In the comparison, we are looking at a reaction quotient! In other words, we are looking at how our non-standard system is different from the system we have at standard conditions!

94. 94 Boltzmann distribution  G =  G  + RT ln Q eq If our system is at equilibrium then Q eq = K eq and  G = 0

95. 95 Boltzmann distribution  G =  G  + RT ln K eq = 0 which means  G  = -RT ln K eq

96. 96 Equilibrium constant The thermodynamic equilibrium constant for a reaction is directly related to the standard free energy change!  G  = -RT ln K eq K eq = e -  G  /RT

97. 97 Equilibrium constant  G   0 K eq  1  G  >> 0 then K eq is very small  G  << 0 then K eq is very large

98. 98 Equilibrium constant at 298 K  G  K eq Meaning K eq = e -  G  /RT

99. 99 Predicting reaction direction  G < 0 means the forward reaction is spontaneous at the given non-standard conditions  G  1

100. 100 Predicting reaction direction  G = 0 means the system is at equilibrium at the given non- standard conditions  G  = 0 means the system is at equilibrium at standard conditions and so K eq = 1 which only occurs at one specific T!

101. 101 Predicting reaction direction  G > 0 means the forward reaction is non-spontaneous at the given non-standard conditions  G  > 0 means the forward reaction is non-spontaneous at standard conditions and so K < 1

102. 102 Predicting reaction direction  G =  G  ONLY at standard conditions!

103. 103 Thermodynamic equilibrium constant K eq The thermodynamic equilibrium constant K eq is expressed in terms of activities, which are unitless quantities. Activities relate properties like concentration or pressure compared to a standard property value, like 1 M for concentration or 1 bar for pressure.

104. 104 Thermodynamic equilibrium constant K eq a A + b B  c C + d D where a x = [X] / c 0 (c 0 is a standard concentration of 1 M) or a x = P x / P 0 (P 0 is a standard pressure of 1 atm) Note that a x = 1 for pure solids and liquids

105. 105 Free energy and equilibrium constants  G  = -RT ln K eq IS ALWAYS TRUE  G  = -RT ln K c and  G  = -RT ln K p DO NOT have to be true!

106. 106 Problem Write thermodynamic equilibrium constant expressions for each of the following reactions and relate them to K c and K p where appropriate: a) Si (s) + 2 Cl 2 (g)  SiCl 4 (g) b) Cl 2 (g) + H 2 O (l)  HOCl (aq) + H + (aq) + Cl - (aq)

107. 107 Problem answer a) Si (s) + 2 Cl 2 (g)  SiCl 4 (g) b) Cl 2 (g) + H 2 O (l)  HOCl (aq) + H + (aq) + Cl - (aq)

108. 108 Problem Use the given data to determine if the following reaction is spontaneous at standard conditions at 298.15 K: N 2 O 4 (g)  2 NO 2 (g)  G  f (N 2 O 4 ) = 97.89 kJ  mol -1  G  f (NO 2 ) = 51.31 kJ  mol -1 Answer:  G  = 4.73 kJ  mol -1, not spontaneous at standard conditions.

109. 109 Problem Based on the problem of the previous slide, determine which direction the reaction will go in if 0.5 bar of each gas is placed in an evacuated container: N 2 O 4 (g)  2 NO 2 (g) Answer: Since  G  = 4.73 kJ  mol -1, then K eq = 0.15 = K p. Since Q p = 0.5 the reaction should go from right to left.

110. 110 Problem Determine the equilibrium constant at 298.15 K for the following reaction using the given data: AgI (s)  Ag + (aq) + I - (aq)  G  f (AgI) = -66.19 kJ  mol -1  G  f (Ag + ) = 77.11 kJ  mol -1  G  f (I - ) = -51.57 kJ  mol -1

111. 111 Problem answer Since  G  = 91.73 kJ  mol -1, then K eq = 8.5 x 10 -17 = K c = K sp. If we compare to the K sp value in Table 18.1 (8.5 x 10 -17 ) we see we are definitely in the right ballpark!

112. 112  G  and K eq are functions of T We’ve already seen that equilibrium constants change with temperature. Why? K eq = e -  G  /RT and  G  =  H  - T  S 

113. 113  G  and K eq are functions of T  G  = -RT ln K eq =  H  - T  S  ln K eq = -(  H  /RT) + (  S  /R)

114. 114 Problem At what temperature will the following reaction have K p = 1.50 x 10 2 ? 2 NO (g) + O 2 (g)  2 NO 2 (g)  H  = -114.1 kJ  mol -1 and  S  = -146.5 J  K -1  mol -1 Answer: T = 606 K

115. 115  G  and K eq are functions of T  G  = -RT ln K eq =  H  - T  S  ln K eq = -(  H  /RT) + (  S  /R) If  H and  S are constant over a temperature range then

116. 116  G  and K eq are functions of T ln K 1 = -(  H  /RT 1 ) + (  S  /R) minus ln K 2 = -(  H  /RT 2 ) + (  S  /R) ln K 1 - ln K 2 = -(  H  /RT 1 ) - (-  H  /RT 2 )

117. 117  G  and K eq are functions of T ln K 1 - ln K 2 = [-  H  /R] [(1  T 1 ) - (1  T 2 )] OR ln [K 1 /K 2 ] = [-  H  /R] [(1  T 1 ) - (1  T 2 )]

118. 118 van’t Hoff equation The van’t Hoff equation relates equilibrium constants to temperatures ln [K 1 /K 2 ] = [-  H  /R] [(1  T 1 ) - (1  T 2 )] It looks very similar to the Arrhenius equation and so a plot of ln K versus 1/T should give a straight line with a slope of [-  H  /R]

119. 119 Plot of ln K p versus 1/T for the reaction. slope = -  H  / R, so for this reaction  H  = -180 kJ  mol -1 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)

120. 120 Problem The following equilibrium constant data have been determined for the reaction H 2 (g) + I 2 (g)  2 HI (g) K p = 50.0 @ 448  C K p = 66.9 @ 350  C Estimate  H  for the reaction. Answer:  H  = -11.1 kJ  mol -1.

121. 121 Coupled reactions If we have a non-spontaneous (  G > 0) reaction with a product that appears as a reactant in a different reaction that is spontaneous (  G < 0) then we can couple the two reactions (do them in the same container at the same time) to drive the non-spontaneous reaction!

122. 122 Coupled reactions This should make sense because if we have two equilibria in one container, one with a small K (like K sp ) and one with a large K (like K f for complex formation), then the coupled reaction that is the sum of the two reactions has a K value that is larger than K sp ! We have made the non-spontaneous reaction occur!