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2. 2 17 Chemical Equilibrium

3. 3 Chapter Goals 1.Basic Concepts 2.The Equilibrium Constant 3.Variation of K c with the Form of the Balanced Equation 4.The Reaction Quotient 5.Uses of the Equilibrium Constant, K c 6.Disturbing a System at Equilibrium: Predictions

4. 4 Chapter Goals 7.The Haber Process: A Commercial Application of Equilibrium 8.Disturbing a System at Equilibrium: Calculations 9.Partial Pressures and the Equilibrium Constant 10.Relationship between K p and K c 11.Heterogeneous Equilibria

5. 5 Chapter Goals 12.Relationship between  G o rxn and the Equilibrium Constant 13.Evaluation of Equilibrium Constants at Different Temperatures

6. 6 Basic Concepts Reversible reactions do not go to completion. –They can occur in either direction –Symbolically, this is represented as:

7. 7 Basic Concepts Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. –A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate. Chemical equilibria are dynamic equilibria. –Molecules are continually reacting, even though the overall composition of the reaction mixture does not change.

8. 8 Basic Concepts One example of a dynamic equilibrium can be shown using radioactive 131 I as a tracer in a saturated PbI 2 solution.

9. 9 Basic Concepts This movie depicts a dynamic equilibrium.

10. 10 Basic Concepts Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.

11. 11 Basic Concepts One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.

12. 12 Basic Concepts

13. 13 Basic Concepts

14. 14 The Equilibrium Constant For a simple one-step mechanism reversible reaction such as: The rates of the forward and reverse reactions can be represented as:

15. 15 The Equilibrium Constant When system is at equilibrium: Rate f = Rate r

16. 16 The Equilibrium Constant Because the ratio of two constants is a constant we can define a new constant as follows :

17. 17 The Equilibrium Constant Similarly, for the general reaction: we can define a constant

18. 18 The Equilibrium Constant Kc is the equilibrium constant. Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

19. 19 The Equilibrium Constant Example 17-1: Write equilibrium constant expressions for the following reactions at 500 o C. All reactants and products are gases at 500 o C.

20. 20 The Equilibrium Constant

21. 21 The Equilibrium Constant

22. 22 The Equilibrium Constant Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity. –Activities are directly related to molarity

23. 23 The Equilibrium Constant Example 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate K c for the reaction. Equil []’s 0.028 M 0.172 M 0.086 M You do it!

24. 24 The Equilibrium Constant Example 17-3: The decomposition of PCl 5 was studied at another temperature. One mole of PCl 5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl 3 was present in the container. Calculate the equilibrium constant at this temperature.

25. 25 The Equilibrium Constant Example 17-4: At a given temperature 0.80 mole of N 2 and 0.90 mole of H 2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH 3 was present. Calculate K c for the reaction. You do it!

26. 26 Variation of K c with the Form of the Balanced Equation The value of K c depends upon how the balanced equation is written. From example 17-2 we have this reaction: This reaction has a K c =[PCl 3 ][Cl 2 ]/[PCl 5 ]=0.53

27. 27 Variation of K c with the Form of the Balanced Equation Example 17-5: Calculate the equilibrium constant for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction. Equil. []’s 0.172 M 0.086 M 0.028 M The concentrations are from Example 17-2.

28. 28 The Reaction Quotient

29. 29 The Reaction Quotient

30. 30 The Reaction Quotient The mass action expression or reaction quotient has the symbol Q. –Q has the same form as Kc The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.

31. 31 The Reaction Quotient Why do we need another “equilibrium constant” that does not use equilibrium concentrations? Q will help us predict how the equilibrium will respond to an applied stress. To make this prediction we compare Q with K c.

32. 32 The Reaction Quotient

33. 33 The Reaction Quotient Example 17-6: The equilibrium constant for the following reaction is 49 at 450 o C. If 0.22 mole of I 2, 0.22 mole of H 2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

34. 34 Uses of the Equilibrium Constant, K c Example 17-7: The equilibrium constant, K c, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO 2 and 1.00 mole of NO 2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

35. 35 Uses of the Equilibrium Constant, K c Example 17-8: The equilibrium constant is 49 for the following reaction at 450 o C. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

36. 36 Disturbing a System at Equilibrium : Predictions LeChatelier’s Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium. –We first encountered LeChatelier’s Principle in Chapter 14. Some possible stresses to a system at equilibrium are: 1.Changes in concentration of reactants or products. 2.Changes in pressure or volume (for gaseous reactions) 3.Changes in temperature.

37. 37 Disturbing a System at Equilibrium : Predictions 1Changes in Concentration of Reactants and/or Products Also true for changes in pressure for reactions involving gases. –Look at the following system at equilibrium at 450 o C.

38. 38 Disturbing a System at Equilibrium : Predictions 1Changes in Concentration of Reactants and/or Products Also true for changes in pressure for reactions involving gases. –Look at the following system at equilibrium at 450 o C.

39. 39 Disturbing a System at Equilibrium : Predictions 1Changes in Concentration of Reactants and/or Products Also true for changes in pressure for reactions involving gases. –Look at the following system at equilibrium at 450 o C.

40. 40 Disturbing a System at Equilibrium : Predictions 2Changes in Volume (and pressure for reactions involving gases) –Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:

41. 41 Disturbing a System at Equilibrium : Predictions

42. 42 Disturbing a System at Equilibrium : Predictions

43. 43 Disturbing a System at Equilibrium : Predictions

44. 44 Disturbing a System at Equilibrium : Predictions 3Changing the Temperature

45. 45 Disturbing a System at Equilibrium : Predictions 3Changing the Reaction Temperature –Consider the following reaction at equilibrium:

46. 46 Disturbing a System at Equilibrium : Predictions Introduction of a Catalyst –Catalysts decrease the activation energy of both the forward and reverse reaction equally. Catalysts do not affect the position of equilibrium. –The concentrations of the products and reactants will be the same whether a catalyst is introduced or not. –Equilibrium will be established faster with a catalyst.

47. 47 Disturbing a System at Equilibrium : Predictions Example 17-9: Given the reaction below at equilibrium in a closed container at 500 o C. How would the equilibrium be influenced by the following?

48. 48 Disturbing a System at Equilibrium : Predictions Example 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

49. 49 Disturbing a System at Equilibrium : Predictions Example 17-11: How will an increase in temperature affect each of the following reactions?

50. 50 The Haber Process: A Practical Application of Equilibrium The Haber process is used for the commercial production of ammonia. –This is an enormous industrial process in the US and many other countries. –Ammonia is the starting material for fertilizer production. Look at Example 17-9. What conditions did we predict would be most favorable for the production of ammonia?

51. 51 The Haber Process: A Practical Application of Equilibrium

52. 52 The Haber Process: A Practical Application of Equilibrium

53. 53 The Haber Process: A Practical Application of Equilibrium This diagram illustrates the commercial system devised for the Haber process.

54. 54 Disturbing a System at Equilibrium: Calculations To help with the calculations, we must determine the direction that the equilibrium will shift by comparing Q with K c. Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C. What is the value of K c for this reaction?

55. 55 Disturbing a System at Equilibrium: Calculations

56. 56 Disturbing a System at Equilibrium: Calculations If the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations? 1 Calculate Q, after the volume has been doubled

57. 57 Disturbing a System at Equilibrium: Calculations Since Q<K c the reaction will shift to the right to re-establish the equilibrium. 2 Use algebra to represent the new concentrations.

58. 58 Disturbing a System at Equilibrium: Calculations Since Q<K c the reaction will shift to the right to re-establish the equilibrium. 2 Use algebra to represent the new concentrations.

59. 59 Disturbing a System at Equilibrium: Calculations Since Q<K c the reaction will shift to the right to re-establish the equilibrium. 2 Use algebra to represent the new concentrations.

60. 60 Disturbing a System at Equilibrium: Calculations Since Q<K c the reaction will shift to the right to re-establish the equilibrium. 2 Use algebra to represent the new concentrations.

61. 61 Disturbing a System at Equilibrium: Calculations

62. 62 Disturbing a System at Equilibrium: Calculations

63. 63 Disturbing a System at Equilibrium: Calculations

64. 64 Disturbing a System at Equilibrium: Calculations Example 17-13: Refer to example 17-12. If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M. You do it!

65. 65 Disturbing a System at Equilibrium: Calculations Example 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl 2, 0.60 moles of CO and 0.20 mole of Cl 2. Calculate the equilibrium constant. You do it!

66. 66 Disturbing a System at Equilibrium: Calculations An additional 0.80 mole of Cl 2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl 2, and COCl 2 when the new equilibrium is established. You do it!

67. 67 Partial Pressures and the Equilibrium Constant For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. For gases, the pressure is proportional to the concentration. We can see this by looking at the ideal gas law. –PV = nRT –P = nRT/V –n/V = M –P= MRT and M = P/RT

68. 68 Partial Pressures and the Equilibrium Constant For convenience we may express the amount of a gas in terms of its partial pressure rather than its concentration. To derive this relationship, we must solve the ideal gas equation.

69. 69 Partial Pressures and the Equilibrium Constant Consider this system at equilibrium at 500 0 C.

70. 70 Partial Pressures and the Equilibrium Constant

71. 71 Relationship Between K p and K c From the previous slide we can see that the relationship between K p and K c is:

72. 72 Relationship Between K p and K c Example 17-15: Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25 o C, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of K p ?

73. 73 Relationship Between K p and K c The numerical value of K c for this reaction can be determined from the relationship of K p and K c.

74. 74 Relationship Between K p and K c Example 17-16: K c is 49 for the following reaction at 450 o C. If 1.0 mole of H 2 and 1.0 mole of I 2 are allowed to reach equilibrium in a 3.0-liter vessel, (a) How many moles of I 2 remain unreacted at equilibrium? You do it!

75. 75 Relationship Between K p and K c (b) What are the equilibrium partial pressures of H 2, I 2 and HI? You do it!

76. 76 Relationship Between K p and K c (c) What is the total pressure in the reaction vessel? You do it!

77. 77 Heterogeneous Equlibria Heterogeneous equilibria have more than one phase present. –For example, a gas and a solid or a liquid and a gas. How does the equilibrium constant differ for heterogeneous equilibria? –Pure solids and liquids have activities of unity. –Solvents in very dilute solutions have activities that are essentially unity. –The Kc and Kp for the reaction shown above are:

78. 78 Heterogeneous Equlibria

79. 79 Heterogeneous Equlibria What are K c and K p for this reaction? You do it!

80. 80 Heterogeneous Equlibria What are K c and K p for this reaction?

81. 81 Relationship Between  G o rxn and the Equilibrium Constant  G o rxn is the standard free energy change. –  G o rxn is defined for the complete conversion of all reactants to all products.  G is the free energy change at nonstandard conditions For example, concentrations other than 1 M or pressures other than 1 atm.  G is related to  G o by the following relationship.

82. 82 Relationship Between  G o rxn and the Equilibrium Constant At equilibrium,  G=0 and Q=K c. Then we can derive this relationship:

83. 83 Relationship Between  G o rxn and the Equilibrium Constant For the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:

84. 84 Relationship Between  G o rxn and the Equilibrium Constant The relationships among  G o rxn, K, and the spontaneity of a reaction are:  G o rxn KSpontaneity at unit concentration < 0> 1Forward reaction spontaneous = 0= 1System at equilibrium > 0< 1Reverse reaction spontaneous

85. 85 Relationship Between  G o rxn and the Equilibrium Constant

86. 86 Relationship Between  G o rxn and the Equilibrium Constant Example 17-17: Calculate the equilibrium constant, K p, for the following reaction at 25 o C from thermodynamic data in Appendix K. Note: this is a gas phase reaction.

87. 87 Relationship Between  G o rxn and the Equilibrium Constant K p for the reverse reaction at 25 o C can be calculated easily, it is the reciprocal of the above reaction.

88. 88 Relationship Between  G o rxn and the Equilibrium Constant Example 17-18: At 25 o C and 1.00 atmosphere, K p = 4.3 x 10 -13 for the decomposition of NO 2. Calculate  G o rxn at 25 o C. You do it.

89. 89 Relationship Between  G o rxn and the Equilibrium Constant The relationship for K at conditions other than thermodynamic standard state conditions is derived from this equation.

90. 90 Evaluation of Equilibrium Constants at Different Temperatures From the value of  H o and K at one temperature, T 1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T 2.

91. 91 Evaluation of Equilibrium Constants at Different Temperatures Example 17-19: For the reaction in example 17-18,  H o = 114 kJ/mol and K p = 4.3 x 10 -13 at 25 o C. Estimate K p at 250 o C.

92. 92 Synthesis Question Mars is a reddish colored planet because it has numerous iron oxides in its soil. Mars also has a very thin atmosphere, although it is believed that quite some time ago its atmosphere was considerably thicker. The thin atmosphere does not retain heat well, thus at night on Mars the surface temperatures are 145 K and in the daytime the temperature rises to 300 K. Does Mars get redder in the daytime or at night?

93. 93 Synthesis Question The formation of iron oxides from iron and oxygen is an exothermic process. Thus the equilibrium that is established on Mars shifts to the iron oxide (red) side when the planet is cooler - at night. Mars gets redder at night by a small amount.

94. 94 Group Question If you are having trouble getting a fire started in the barbecue grill, a common response is to blow on the coals until the fire begins to burn better. However, this has the side effect of dizziness. This is because you have disturbed an equilibrium in your body. What equilibrium have you affected?

95. 95 17 Chemical Equilibrium