1. CHAPTER FIFTEEN Copyright © Tyna L. Gaylord 2002 - 2009 All Rights Reserved 1

2. 2 At equilibrium, the rate at which products form is EQUAL to the rate at which products decompose. Forward reaction: A --> B rate = k f [A] Reverse reaction: B --> A rate = k r [B] At equilibrium, k f [A] = k r [B] [B] = k f = a constant (K eq ) [A] k r Chem 15.1 Equilibrium

3. 3 Once equilibrium is established, the concentrations of A and B do not change.  The fact that the composition remains constant with time does not mean that A and B stop reacting  Compound A is still converted into compound B, but both processes occur at the same rate  Indicated by double arrow  Chem 15.1 Equilibrium

4. 3 Chem 15.1 Equilibrium

5. 3 Chem 15.1 Equilibrium One of the most important chemical systems is the synthesis of ammonia from nitrogen and hydrogen (HABER process)

6. 3 Chem 15.1 Equilibrium HABER process: N 2 + 3H 2  2NH 3(g) 3put N 2 + 3H 2 in a high pressure tank at a total pressure of several hundred atmospheres, in the presence of a catalyst, and at a temperature of several hundred degrees Celsius. 3Two gases react, but does not lead to complete consumption of gases

7. 3 Chem 15.1 Equilibrium HABER process: N 2 + 3H 2  2NH 3(g) at equilibrium, the relative concentrations of H 2, N 2 and NH 3 are the same, regardless of the starting mixture, and note that the equilibrium is reached from either direction !

8. 3 Chem 15.1 Equilibrium The equilibrium constant expression depends only on the stoichiometry of the reaction, not on its mechanism. The only thing can alter the proportionality constant of a balanced chemical equation is TEMPERATURE, altering anything else will cause the reaction to shift in order to get back to the same proportionality...

9. 3 Chem 15.1 Equilibrium …which is why the rates of the forward and reverse reactions are always EQUAL and the concentration of each chemical always remains CONSTANT

10. Chem 116: Prof. T.L. Heise 11 Chem 15.2 Sample exercise: Write the equilibrium constant expression for H 2(g) + I 2(g)  2HI (g) Equilibrium

11. Sample exercise: Write the equilibrium constant expression for H 2(g) + I 2(g)  2HI (g) K eq = [HI] 2 [H 2 ][I 2 ] Chem 116: Prof. T.L. Heise 12 Chem 15.2 Equilibrium

12. When the reactants and products of an equilibrium are gases, the partial pressures can be used… K p = (P P ) p (P Q ) q (P A ) a (P B ) b K p = K eq (RT)  n Chem 116: Prof. T.L. Heise 13 Chem 15.2 Equilibrium

13. Sample exercise: For the equilibrium 2SO 3(g)  2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3. Calculate the value for K p. Chem 116: Prof. T.L. Heise 14 Chem 15.2 Equilibrium

14. Sample exercise: For the equilibrium 2SO 3(g)  2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3. Calculate the value for K p. K p = K eq (RT)  n Chem 116: Prof. T.L. Heise 15 Chem 15.2 Equilibrium

15. Sample exercise: For the equilibrium 2SO 3(g)  2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3. Calculate the value for K p. K p = K eq (RT)  n = 4.08 x 10 -3 ((0.0821)(1000)) 1 Chem 116: Prof. T.L. Heise 16 Chem 15.2 Equilibrium

16. Sample exercise: For the equilibrium 2SO 3(g)  2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3. Calculate the value for K p. K p = K eq (RT)  n = 4.08 x 10 -3 ((0.0821)(1000)) 1 = 4.08 x 10 -3 (82.1) Chem 116: Prof. T.L. Heise 17 Chem 15.2 Equilibrium

17. Sample exercise: For the equilibrium 2SO 3(g)  2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3. Calculate the value for K p. K p = K eq (RT)  n = 4.08 x 10 -3 ((0.0821)(1000)) 1 = 4.08 x 10 -3 (82.1) = 0.335 Chem 116: Prof. T.L. Heise 18 Chem 15.2 Equilibrium

18. Equilibrium Constants can be very large or very small. The magnitude of the equilibrium constant provides us with important information about the equilibrium mixture. Chem 116: Prof. T.L. Heise 19 Chem 15.2 Equilibrium

19. Sample Exercise: The equilibrium constant for the reaction H 2(g) + I 2(g)  2HI (g) varies with temperature in the following way: K eq = 794 at 298 K K eq = 54 at 700 K Is the formation of products favored more at the higher or lower temperature? Chem 116: Prof. T.L. Heise 20 Chem 15.2 Equilibrium

20. Sample Exercise: The equilibrium constant for the reaction H 2(g) + I 2(g)  2HI (g) varies with temperature in the following way: K eq = 794 at 298 K K eq = 54 at 700 K Is the formation of products favored more at the higher or lower temperature? Lower because K eq is larger Chem 116: Prof. T.L. Heise 21 Chem 15.2 Equilibrium

21. The substances in the equilibrium are of different phases: the concentration of a pure solid or liquid equals its density divided by its molar mass D = g/cm 3 = mol M g/mol cm 3 the density is constant at any given temperature so... Chem 116: Prof. T.L. Heise 22 Chem 15.3 Equilibrium

22. CaCO 3 (s)  CaO(s) + CO 2 (g) K eq = [CaO][CO 2 ] [CaCO 3 ] K eq = constant 1[CO 2 ] constant 2 K eq ’ = K eq constant 1 = [CO 2 ] constant 2 Chem 116: Prof. T.L. Heise 23 Chem 15.3 If a pure solid or liquid is used it is NOT included in the equilibrium constant Equilibrium

23. Chem 116: Prof. T.L. Heise 24 Chem 15.3 Sample Exercise: Write the equilibrium expressions for K eq and K p for the reaction 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) Equilibrium

24. Sample Exercise: Write the equilibrium expressions for K eq and K p for the reaction 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) K eq = products reactants Chem 116: Prof. T.L. Heise 25 Chem 15.3 Equilibrium

25. Sample Exercise: Write the equilibrium expressions for K eq and K p for the reaction 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) K eq = products = [H 2 ] 4 reactants [H 2 O] 4 Chem 116: Prof. T.L. Heise 26 Chem 15.3 Equilibrium

26. Sample Exercise: Write the equilibrium expressions for K eq and K p for the reaction 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) K eq = products = [H 2 ] 4 reactants [H 2 O] 4 K p = P (H 2 ) 4 P (H 2 O) 4 Chem 116: Prof. T.L. Heise 27 Chem 15.3 Equilibrium

27. Sample Exercise: Which of the following substances - H 2 (g), H 2 O(g), O 2 (g) - when added to Fe 3 O 4 (s) in a closed container at high temperature, permits attainment of equilibrium in the rxn 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) Chem 116: Prof. T.L. Heise 28 Chem 15.3 Equilibrium

28. Sample Exercise: Which of the following substances - H 2 (g), H 2 O(g), O 2 (g) - when added to Fe 3 O 4 (s) in a closed container at high temperature, permits attainment of equilibrium in the rxn 3Fe(s) + 4H 2 O(g)  Fe 3 O 4 (s) + 4H 2 (g) Only hydrogen Chem 116: Prof. T.L. Heise 29 Chem 15.3 Equilibrium

29. Sample Exercise: Nitryl chloride, NO 2 Cl, is in equilibrium with NO 2 and Cl 2 : 2NO 2 Cl(g)  2NO 2 (g) + Cl 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq. Chem 116: Prof. T.L. Heise 30 Chem 15.4 Equilibrium

30. 2NO 2 Cl(g)  2NO 2 (g) + Cl 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq. Keq = products reactants Chem 116: Prof. T.L. Heise 31 Chem 15.4 Equilibrium

31. 2NO 2 Cl(g)  2NO 2 (g) + Cl 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq. Keq = products = [NO 2 ] 2 [Cl 2 ] reactants [NO 2 Cl] 2 Chem 116: Prof. T.L. Heise 32 Chem 15.4 Equilibrium

32. 2NO2Cl(g)  2NO2(g) + Cl2(g) At equilibrium the concentrations of the substances are [NO2Cl] = 0.00106 M, [NO2] = 0.0108 M, and [Cl2] = 0.00538 M. From these data calculate the equilibrium constant, Keq. Keq = [NO 2 ] 2 [Cl 2 ] = (0.0108) 2 (0.00538) [NO 2 Cl] 2 (0.00106) 2 Chem 116: Prof. T.L. Heise 33 Chem 15.4 Equilibrium

33. 2NO2Cl(g)  2NO2(g) + Cl2(g) At equilibrium the concentrations of the substances are [NO2Cl] = 0.00106 M, [NO2] = 0.0108 M, and [Cl2] = 0.00538 M. From these data calculate the equilibrium constant, Keq. Keq = [NO 2 ] 2 [Cl 2 ] = (0.0108) 2 (0.00538) [NO 2 Cl] 2 (0.00106) 2 = 0.558 Chem 116: Prof. T.L. Heise 34 Chem 15.4 Equilibrium

34. Rarely are we given all the information as in the first example problem, normally, one concentration will be given and stoichiometric calculations will give you the others. Sample Exercise pg 572 on board. Chem 116: Prof. T.L. Heise 35 Chem 15.4 Equilibrium

35. If K is very large, the reaction will tend to proceed far to the right creating a large amount of products. If K is very small, the reaction will not proceed, staying far to the left, leaving a large amount of reactants unused. Chem 116: Prof. T.L. Heise 36 Chem 15.5 Equilibrium

36. Predicting the Direction 3use initial concentrations to determine original K eq 3obtain K eq at temperature you want 3compare calculated K eq to actual K eq and it will tell you how the reaction is going Chem 116: Prof. T.L. Heise 37 Chem 15.5 Equilibrium

37. Predicting the Direction 32.00 mol H 2, 1.00 mol of N 2, and 2.00 mol of NH 3 at 472°C 3Q = (2) 2 /(1)(2) 3 = 0.500 3according to S.E. 15.7, K eq = 0.105 3our actual is much smaller than the original which telling me there is quite a bit less product and more reactant, so reaction is proceeding to the left Chem 116: Prof. T.L. Heise 38 Chem 15.5 Equilibrium

38. Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)  2SO 2 (g) + O 2 (g) is 4.08 x 10 -3. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 M, [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 -2 M. Chem 116: Prof. T.L. Heise 39 Chem 15.5 Equilibrium

39. Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)  2SO 2 (g) + O 2 (g) is 4.08 x 10 -3. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 M, [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 -2 M. Q = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 Chem 116: Prof. T.L. Heise 40 Chem 15.5 Equilibrium

40. Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)  2SO 2 (g) + O 2 (g) is 4.08 x 10 -3. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 M, [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 -2 M. Q = [SO 2 ] 2 [O 2 ] = (5 x 10 -3 ) 2 (3 x 10 -2 ) [SO 3 ] 2 (2 x 10 -3 ) 2 Chem 116: Prof. T.L. Heise 41 Chem 15.5 Equilibrium

41. Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)  2SO 2 (g) + O 2 (g) is 4.08 x 10 -3. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 M, [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 -2 M. Q = [SO 2 ] 2 [O 2 ] = (5 x 10 -3 ) 2 (3 x 10 -2 ) = 0.187 [SO 3 ] 2 (2 x 10 -3 ) 2 Chem 116: Prof. T.L. Heise 42 Chem 15.5 Equilibrium

42. Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)  2SO 2 (g) + O 2 (g) is 4.08 x 10 -3. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 M, [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 -2 M. Q = 0.2 ; 4.08 x 10 -3 is much smaller, so I need less products to form and more reactants which indicates a shift to the left Chem 116: Prof. T.L. Heise 43 Chem 15.5 Equilibrium

43. It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration. Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ? Chem 116: Prof. T.L. Heise 44 Chem 15.5 Equilibrium

44. It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration. Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ? K p = 1.45 x 10 -5 = [NH 3 ] 2 [H 2 ] 3 [N 2 ] 45 Chem 15.5 Equilibrium

45. It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration. Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ? K p = 1.45 x 10 -5 = x 2 (0.928) 3 (0.432) Chem 116: Prof. T.L. Heise 46 Chem 15.5 Equilibrium

46. It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration. Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ? x = 5.01 x 10 -6 Chem 116: Prof. T.L. Heise 46 Chem 15.5 Equilibrium

47. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? Chem 116: Prof. T.L. Heise 47 Chem 15.5 Equilibrium

48. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 116: Prof. T.L. Heise 48 Chem 15.5 Equilibrium

49. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 116: Prof. T.L. Heise 49 Chem 15.5 0.497 = (.350)(x) (0.860) Equilibrium

50. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 116: Prof. T.L. Heise 50 Chem 15.5 0.497(0.860) = (.350)(x) Equilibrium

51. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 116: Prof. T.L. Heise 51 Chem 15.5 0.497(0.860) = (x) (0.350) Equilibrium

52. Sample exercise:At 500 K the reaction PCl 5 (g)  PCl 3 (g) + Cl 2 (g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 116: Prof. T.L. Heise 52 Chem 15.5 1.22 = (x) Equilibrium

53. Chem 116: Prof. T.L. Heise 53 Chem 15.6 If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. 3 ways to change an equilibrium 4add or remove a reactant or product 4change the pressure 4change the temperature Equilibrium

54. Chem 116: Prof. T.L. Heise 54 Chem 15.6 Change in reactant or product concentrations 3adding a reactant or product forces a shift that will use up what has been added 3removing a reactant or product forces a shift that will create more of what was taken Equilibrium

55. Chem 116: Prof. T.L. Heise 55 Chem 15.6 Change in volume or pressure 3decreasing volume causes increasing pressure and vice versa 3increasing pressure forces a shift in equilibrium towards the production of fewer moles of gas 3decreasing pressure forces a shift in equilibrium towards the production of more moles of gas Equilibrium

56. Chem 116: Prof. T.L. Heise 56 Chem 15.6 Change in temperature 3increasing temperature forces a shift in the endothermic direction so increased energy can be used up 3decreasing temperature forces a shift in the exothermic direction so more energy can be produced to replace lost energy Equilibrium

57. Chem 116: Prof. T.L. Heise 57 Chem 15.6 Concentration of H 2 is altered and affects are graphed Equilibrium

58. Chem 116: Prof. T.L. Heise 58 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added (b) the temperature is increased (c ) the volume is decreased (d) PCl 5 (g) is added Equilibrium

59. Chem 116: Prof. T.L. Heise 59 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

60. Chem 116: Prof. T.L. Heise 60 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

61. Chem 116: Prof. T.L. Heise 61 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (b) the temperature is increased PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

62. Chem 116: Prof. T.L. Heise 62 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (b) the temperature is increased PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

63. Chem 116: Prof. T.L. Heise 63 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decreased Equilibrium

64. Chem 116: Prof. T.L. Heise 64 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decreased...so pressure is increased, favors less moles PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) 1 mole 1 mole + 1 mole Equilibrium

65. Chem 116: Prof. T.L. Heise 65 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decrease...so pressure is increased, favors less moles PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) 1 mole 1 mole + 1 mole Equilibrium

66. Chem 116: Prof. T.L. Heise 66 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl 5 (g) is added Equilibrium

67. Chem 116: Prof. T.L. Heise 67 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl5(g) is added PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

68. Chem 116: Prof. T.L. Heise 68 Chem 15.6 Sample Exercise: For the reaction PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl5(g) is added PCl 5 (g) + energy  PCl 3 (g) + Cl 2 (g) Equilibrium

69. Chem 116: Prof. T.L. Heise 69 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)  2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature. Equilibrium

70. Chem 116: Prof. T.L. Heise 70 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)  2PCl 3 (g) + O 2 (g) -542.2 -288.07 + 0  H = products - reactants 2(-288.07) - 2(-542.2) 508.26 kJ Equilibrium

71. Chem 116: Prof. T.L. Heise 71 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)  2PCl 3 (g) + O 2 (g) -542.2 -288.07 + 0  H = products - reactants 2(-288.07) - 2(-542.2) 508.26 kJ Endothermic Equilibrium

72. Chem 116: Prof. T.L. Heise 72 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy  2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature. Equilibrium

73. Chem 116: Prof. T.L. Heise 73 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy  2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature. K eq = [PCl 3 ] 2 [O 2 ] [POCl 3 ] 2 Equilibrium

74. Chem 116: Prof. T.L. Heise 74 Chem 15.6 Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy  2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature. K eq = [PCl 3 ] 2 [O 2 ]if energy inc., [POCl 3 ] 2 equilibrium shifts right, products are bigger, K eq is larger Equilibrium

75. Chem 116: Prof. T.L. Heise 75 Chem 15.6 Effects of a Catalyst: a catalyst changes the rate of a reaction only, not the amounts of the compounds!! Remember - Rates are Equal Concentrations are Constant Only temp. can change a constant Equilibrium