1. Chapter 15 Chemical Equilibrium 1

2. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. 2

3. The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. 3

4. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. 4

5. Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow N 2 O 4 (g) 2 NO 2 (g) 5

6. The Equilibrium Constant N 2 O 4 (g) 2 NO 2 (g) 6

7. The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ] 7

8. The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2 8

9. The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = 9

10. The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = 10

11. The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD 11

12. What Are the Equilibrium Expressions for These Equilibria? 12

13. SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: PRACTICE EXERCISE Write the equilibrium-constant expression, K c for 13

14. The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b 14

15. Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV 15

16. Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant) 16

17. SAMPLE EXERCISE 15.2 Converting Between K c and K p In the synthesis of ammonia from nitrogen and hydrogen, K c = 9.60 at 300°C. Calculate K p for this reaction at this temperature. PRACTICE EXERCISE For the equilibrium is 4.08  10 –3 at 1000 K. Calculate the value for K p. 17

18. Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are. 18

19. Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide. 19

20. Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium. 20

21. What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. 21

22. What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium. 22

23. SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant The reaction of N 2 with O 2 to form NO might be considered a means of “fixing” nitrogen: The value for the equilibrium constant for this reaction at 25°C is K c = 1  10 –30. Describe the feasibility of fixing nitrogen by forming NO at 25°C. PRACTICE EXERCISE For the reaction at 298 K and K p = 54 at 700 K. Is the formation of HI favored more at the higher or lower temperature? 23

24. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 1 0.212 = K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g) 24

25. Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g) 25

26. Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. 26

27. SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the following information, determine the value of K c for the reaction 27

28. SAMPLE EXERCISE 15.5 continued PRACTICE EXERCISE Given that, at 700 K, K p = 54.0 for the reaction for the reaction determine the value of K p for the reaction 28

29. Heterogeneous Equilibrium 29

30. The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature. 30

31. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) 31

32. As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s) CO 2 (g) + CaO (s) 32

33. SAMPLE EXERCISE 15.6 Writing Equilibrium-Constant Expressions PRACTICE EXERCISE Write the following equilibrium-constant expressions: Write the equilibrium-constant expression for K c for each of the following reactions: for Heterogeneous Reactions 33

34. PRACTICE EXERCISE When added to Fe 3 O 4 (s) in a closed container, which one of the following substances—H 2 (g), H 2 O(g), O 2 (g) —will allow equilibrium to be established in the reaction SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium (a) pure CaCO 3, (b) CaO and a CO 2 pressure greater than the value of K p, (c) some CaCO 3 and a CO 2 pressure greater than the value of K p, (d) CaCO 3 and CaO? 34

35. The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. 35

36. If Q = K, the system is at equilibrium. 36

37. If Q > K, there is too much product and the equilibrium shifts to the left. 37

38. If Q < K, there is too much reactant, and the equilibrium shifts to the right. 38

39. SAMPLE EXERCISE 15.10 Predicting the Direction of Approach to Equilibrium At 448°C the equilibrium constant K c for the reaction is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0  10 –2 mol of HI, 1.0  10 –2 mol of H 2, and 3.0  10 –2 mol of I 2 in a 2.00-L container. 39

40. PRACTICE EXERCISE At 1000 K the value of K p for the reaction is 0.338. Calculate the value for Q p, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are 40

41. Equilibrium Calculations “ICE” 41

42. Equilibrium Calculations A closed system initially containing 1.000 x 10 −3 M H 2 and 2.000 x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g) 42

43. What Do We Know? ICE [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium 1.87 x 10 -3 43

44. [HI] Increases by 1.87 x 10 -3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3 44

45. Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3 45

46. We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3 46

47. …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 ) 47

48. SAMPLE EXERCISE 15.9 Calculating K from Initial and Equilibrium Concentrations A closed system initially containing 1.000  10 –3 M H 2 and 2.000T10 –3 M I 2 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87  10 –3 M. Calculate K c at 448°C for the reaction taking place, which is 48

49. PRACTICE EXERCISE Sulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel is charged at 1000 K with SO 3 (g) at a partial pressure of 0.500 atm. At equilibrium the SO 3 partial pressure is 0.200 atm. Calculate the value of K p at 1000 K. 49

50. SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrations A 1.000-L flask is filled with 1.000 mol of H 2 and 2.000 mol of I 2 at 448°C. The value of the equilibrium constant K c for the reaction at 448°C is 50.5. What are the equilibrium concentrations of H 2, I 2, and HI in moles per liter? from Initial Concentrations 50

51. PRACTICE EXERCISE For the equilibrium the equilibrium constant K p has the value 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl 5 (g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl 5, PCl 3, and Cl 2 at this temperature? 51

52. Le Châtelier’s Principle 52

53. Le Châtelier’s Principle “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.” 53

54. What Happens When More of a Reactant Is Added to a System? Play insert CD Play insert CD 54

55. The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH 3 ) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. 55

56. The Haber Process If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3. 56

57. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid. 57

58. The Effect of Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 (aq) + 6 H 2 O (l) 58

59. The Effect of Changes in Temperature Play insert CD Play insert CD 59

60. Catalysts increase the rate of both the forward and reverse reactions. 60

61. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. 61

62. SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium Consider the equilibrium In which direction will the equilibrium shift when (a) N 2 O 4 is added, (b) NO 2 is removed, (c) the total pressure is increased by addition of N 2 (g), (d) the volume is increased, (e) the temperature is decreased? 62

63. PRACTICE EXERCISE For the reaction in which direction will the equilibrium shift when (a) Cl 2 (g) is removed, (b) the temperature is decreased, (c) the volume of the reaction system is increased, (d) PCl 3 (g) is added? 63

64. Solution Analyze: We are asked to determine the standard enthalpy change of a reaction and how the equilibrium constant for the reaction varies with temperature. Plan: (a) We can use standard enthalpies of formation to calculate  H° for the reaction. (b) We can then use Le Châtelier’s principle to determine what effect temperature will have on the equilibrium constant. (a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction SAMPLE EXERCISE 15.14 Predicting the Effect of Temperature on K (b) Determine how the equilibrium constant for this reaction should change with temperature. Solve: (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation, less the same quantities for the reactants. At 25°C, for NH 3 (g) is The values for H 2 (g) and N 2 (g) are zero by definition because the enthalpies of formation of the elements in their normal states at 25°C are defined as zero (Section 5.7). Because 2 mol of NH 3 is formed, the total enthalpy change is 64

65. SAMPLE EXERCISE 15.14 continued (b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less NH 3 and more N 2 and H 2. This effect is seen in the values for K p presented in Table 15.2. Notice that K p changes markedly with changes in temperature and that it is larger at lower temperatures. Comment: The fact that K p for the formation of NH 3 from N 2 and H 2 decreases with increasing temperature is a matter of great practical importance. To form NH 3 at a reasonable rate requires higher temperatures. At higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH 3 is smaller. To compensate for this, higher pressures are needed because high pressure favors NH 3 formation. 65

66. PRACTICE EXERCISE Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction Use this result to determine how the equilibrium constant for the reaction should change with temperature. 66

67. SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together At temperatures near 800°C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H 2 : The mixture of gases that results is an important industrial fuel called water gas. (a) At 800°C the equilibrium constant for this reaction is K p = 14.1. What are the equilibrium partial pressures of H 2 O, CO, and H 2 in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H 2 O in a 1.00-L vessel? (b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25°C the value of K p for this reaction is 1.7  10 –21. Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H 2 at equilibrium, should the pressure of the system be increased or decreased? 67