Chapter 13. Chemical Equilibrium

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1 Chapter 13 Chemical Equilibrium

2 Section 13.1 The Equilibrium Condition

3 Section 13.1 The Equilibrium Condition

4 Section 13.1 The Equilibrium Condition

5 Section 13.1 The Equilibrium Condition

6 Section 13.1 The Equilibrium Condition

7 Section 13.5 Applications of the Equilibrium Constant

8 Section 13.5 Applications of the Equilibrium Constant c

9 Section 13.1 The Equilibrium Condition Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Copyright Cengage Learning. All rights reserved 9

10 Section 12.1 Reaction Rates Average Kinetic Energy

11 Section 13.1 The Equilibrium Condition Equilibrium Is: Macroscopically static Microscopically dynamic Copyright Cengage Learning. All rights reserved 11

12 Section 13.1 The Equilibrium Condition Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. Copyright Cengage Learning. All rights reserved 12

13 Section 13.1 The Equilibrium Condition Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Copyright Cengage Learning. All rights reserved 13

14 Section 13.1 The Equilibrium Condition

15 Section 13.1 The Equilibrium Condition

16 Section 13.1 The Equilibrium Condition The Changes with Time in the Rates of Forward and Reverse Reactions Copyright Cengage Learning. All rights reserved 16

17 Section 13.5 Applications of the Equilibrium Constant Law of mass action Q = K Law of mass action (Q=K) Chemical equilibrium systems obey the law of mass action. This law states that the reaction quotient, Q, will be equal to a constant when a system reaches chemical equilibrium.

18 Section 13.5 Applications of the Equilibrium Constant

19 Section 13.5 Applications of the Equilibrium Constant

20 Section 13.5 Applications of the Equilibrium Constant

21 Section 13.5 Applications of the Equilibrium Constant

22 Section 13.1 The Equilibrium Condition CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. Copyright Cengage Learning. All rights reserved 22

23 Section 13.1 The Equilibrium Condition CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: The concentrations H 2 O(g) + of CO(g) each product H 2 (g) will + CO increase, 2 (g) the concentration of CO will decrease, and the concentration of You water add will more be higher H 2 O(g) than to the the flask. original How equilibrium does the concentration, but of lower each chemical than the initial compare total to amount. its original concentration after equilibrium is reestablished? Justify your answer. Copyright Cengage Learning. All rights reserved 23

24 Section 13.1 The Equilibrium Condition CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. Copyright Cengage Learning. All rights reserved 24

25 Section 13.1 The Equilibrium Condition This is the opposite scenario of the previous slide. The concentrations of water and CO will increase. The concentration of carbon dioxide decreases and the concentration of hydrogen will be higher than the original equilibrium concentration, but lower than the initial total amount.

26 Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q Used when all of the initial concentrations are nonzero. Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Law of mass action (Q=K) Chemical equilibrium systems obey the law of mass action. This law states that the reaction quotient, Q, will be equal to a constant when a system reaches chemical equilibrium. Copyright Cengage Learning. All rights reserved 26

27 Section 13.2 The Equilibrium Constant Consider the following reaction at equilibrium: ja + kb lc + md K = c [C] l [A] j [D] [B] m k A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). Copyright Cengage Learning. All rights reserved 27

28 Section 13.2 The Equilibrium Constant Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n. K values are usually written with no units. Copyright Cengage Learning. All rights reserved 28

29 Section 13.2 The Equilibrium Constant K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. Copyright Cengage Learning. All rights reserved 29

30 Section 13.3 Equilibrium Expressions Involving Pressures K involves concentrations. K p involves pressures. Copyright Cengage Learning. All rights reserved 30

31 Section 13.3 Example Equilibrium Expressions Involving Pressures Write the equilibrium constant for this reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) K = 2 PNH P N PH p K NH 3 = N 2 H Copyright Cengage Learning. All rights reserved 31

32 Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) K K c = NH = 3 N 3 H 2 PNH P N PH p Copyright Cengage Learning. All rights reserved 32

33 Section 13.3 Example Equilibrium Expressions Involving Pressures Write the equilibrium constant involving gas pressures for this reaction. N 2 (g) + 3H 2 (g) 2NH 3 (g) K = 2 PNH P N PH p K NH 3 = N 2 H Copyright Cengage Learning. All rights reserved 33

34 Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) K = 2 PNH P N PH p Copyright Cengage Learning. All rights reserved 34

35 Section 13.3 Example Equilibrium Expressions Involving Pressures Calculate the equilibrium constant for the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature: P P P NH N H = atm 1 = atm 3 = atm Copyright Cengage Learning. All rights reserved 35

36 Section 13.3 Equilibrium Expressions Involving Pressures K = N 2 (g) + 3H 2 (g) 2NH 3 (g) 2 PNH P N PH p K 2 = p K p = Copyright Cengage Learning. All rights reserved 36

37 Section 13.3 Equilibrium Expressions Involving Pressures

38 Section 13.3 Equilibrium Expressions Involving Pressures

39 Section 13.3 Equilibrium Expressions Involving Pressures

40 Section 13.3 Equilibrium Expressions Involving Pressures

41 Section 13.3 Equilibrium Expressions Involving Pressures The Relationship Between K and K p K p = K(RT) Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = L atm/mol K T = temperature (in Kelvin) Copyright Cengage Learning. All rights reserved 41

42 Section Equilibrium Expressions Involving Pressures

43 Section 13.3 Equilibrium Expressions Involving Pressures Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p ( ) from the previous example, calculate the value of K at 35 C. K p = = K L atm/mol K 308K K K RT = n 7 Copyright Cengage Learning. All rights reserved 43

44 Section 13.3 Equilibrium Expressions Involving Pressures

45 Section Homogeneous 13.3 Equilibria Equilibrium Expressions Involving Pressures Heterogeneous Equilibria

46 Section 13.4 Heterogeneous Equilibria Homogeneous Equilibria Homogeneous equilibria involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq) Copyright Cengage Learning. All rights reserved 46

47 Section 13.3 Equilibrium Expressions Involving Pressures

48 Section 13.3 Equilibrium Expressions Involving Pressures

49 Section 13.3 Equilibrium Expressions Involving Pressures

50 Section 13.4 Heterogeneous Equilibria Heterogeneous Equilibria Heterogeneous equilibria involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g) Copyright Cengage Learning. All rights reserved 50

51 Section 13.4 Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) K = O 3 2 Copyright Cengage Learning. All rights reserved 51

52 Section 13.4 Heterogeneous Equilibria

53 Section 13.4 Heterogeneous Equilibria

54 Section 13.4 Heterogeneous Equilibria Additional Concepts

55 Section 13.4 Heterogeneous Equilibria

56 Section 13.3 Equilibrium Expressions Involving Pressures

57 Section 13.3 Equilibrium Expressions Involving Pressures

58 Section 13.4 Heterogeneous Equilibria

59 Section 13.4 Heterogeneous Equilibria

60 Section 13.3 Equilibrium Expressions Involving Pressures

61 Section 13.4 Heterogeneous Equilibria

62 Section 13.4 Heterogeneous Equilibria

63 Section 13.4 Heterogeneous Equilibria

64 Section The Extent 13.4 of a Reaction Heterogeneous Equilibria

65 Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products the equilibrium lies to the right. Reaction goes essentially to completion. Copyright Cengage Learning. All rights reserved 65

66 Section 13.5 Applications of the Equilibrium Constant The Extent of a Reaction A very small value of K means that the system at equilibrium will consist of mostly reactants the equilibrium position is far to the left. Reaction does not occur to any significant extent. Copyright Cengage Learning. All rights reserved 66

67 Section 13.5 Applications of the Equilibrium Constant CONCEPT CHECK! If the equilibrium lies to the right, the value for K is. large (or >1) If the equilibrium lies to the left, the value for K is. small (or <1) Copyright Cengage Learning. All rights reserved 67

68 Section 13.5 Applications of the Equilibrium Constant

69 Section 13.5 Applications of the Equilibrium Constant

70 Section 13.5 Applications of the Equilibrium Constant

71 Section 13.5 Applications of the Equilibrium Constant 2014 AP Exam

72 Section 13.5 Applications of the Equilibrium Constant 2014 AP Exam

73 Section 13.5 Applications of the Equilibrium Constant 2014 AP Exam

74 Section 13.5 Applications of the Equilibrium Constant 2014 AP Exam

75 ICE Section 13.5 Applications of the Equilibrium Constant

76 Section 13.5 Applications of the Equilibrium Constant

77 Section 13.5 Applications of the Equilibrium Constant

78 Section 13.5 Applications of the Equilibrium Constant

79 Section 13.5 Applications of the Equilibrium Constant

80 Section 13.5 Applications of the Equilibrium Constant

81 Section 13.5 Applications of the Equilibrium Constant

82 Section 13.5 Applications of the Equilibrium Constant

83 Section 13.5 Applications of the Equilibrium Constant

84 Section 13.5 Applications of the Equilibrium Constant

85 Section 13.5 Applications of the Equilibrium Constant

86 Section 13.5 Applications of the Equilibrium Constant

87 Section 13.5 Applications of the Equilibrium Constant

88 Section 13.5 Applications of the Equilibrium Constant

89 Section Le Chatelier's 13.5 principle Applications of the Equilibrium Constant Reaction Quotient Q In order to determine the direction of the equilibrium shift, the reaction quotient Q is compared to the equilibrium constant K.

90 Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q Used when all of the initial concentrations are nonzero. Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Copyright Cengage Learning. All rights reserved 90

91 Section 13.5 Applications of the Equilibrium Constant

92 Section 13.5 Applications of the Equilibrium Constant

93 Section 13.5 Applications of the Equilibrium Constant Reaction Quotient, Q Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium. Copyright Cengage Learning. All rights reserved 93

94 Section 13.5 Applications of the Equilibrium Constant

95 Section 13.3 Equilibrium Expressions Involving Pressures

96 Section 13.3 Equilibrium Expressions Involving Pressures

97 Section 13.3 Equilibrium Expressions Involving Pressures

98 Section 13.3 Equilibrium Expressions Involving Pressures

99 Section 13.5 Applications of the Equilibrium Constant

100 Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? Copyright Cengage Learning. All rights reserved 100

101 Section 13.5 Applications of the Equilibrium Constant K = Copyright Cengage Learning. All rights reserved 101 Set up ICE Table Fe 3+ (aq) + SCN (aq) FeSCN 2+ (aq) Initial Change Equilibrium FeSCN 4.00 = M K = Fe 3 SCN 2.00 M 6.00 M

102 Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN (aq) (same temperature as Trial #1) Equilibrium:? M FeSCN 2+ (aq) 5.00 M FeSCN 2+ Copyright Cengage Learning. All rights reserved 102

103 Section 13.5 Applications of the Equilibrium Constant EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN (aq) Equilibrium:? M FeSCN 2+ (aq) 3.00 M FeSCN 2+ Copyright Cengage Learning. All rights reserved 103

104 Section 13.6 Solving Equilibrium Problems Solving Equilibrium Problems 1) Write the balanced equation for the reaction. 2) Write the equilibrium expression using the law of mass action. 3) List the initial concentrations. 4) Calculate Q, and determine the direction of the shift to equilibrium. Copyright Cengage Learning. All rights reserved 104

105 Section 13.6 Solving Equilibrium Problems Solving Equilibrium Problems 5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6) Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7) Check your calculated equilibrium concentrations by making sure they give the correct value of K. Copyright Cengage Learning. All rights reserved 105

106 Section 13.6 Solving Equilibrium Problems EXERCISE! Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial # M 5.00 M 1.00 M Trial # M 2.00 M 5.00 M Trial # M 9.00 M 6.00 M Find the equilibrium concentrations for all species. Copyright Cengage Learning. All rights reserved 106

107 Section 13.6 Solving Equilibrium Problems EXERCISE! Answer Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M Copyright Cengage Learning. All rights reserved 107

108 Section 13.6 Solving Equilibrium Problems CONCEPT CHECK! A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH 3 (g) N 2 (g) + H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69 Copyright Cengage Learning. All rights reserved 108

109 Section 13.6 Solving Equilibrium Problems CONCEPT CHECK! A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = M Concentration of NO 2 = M Copyright Cengage Learning. All rights reserved 109

110 Section 13.7 Le Châtelier s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Copyright Cengage Learning. All rights reserved 110

111 Section 13.7 Le Châtelier s Principle Effects of Changes on the System 1. Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2. Temperature: K will change depending upon the temperature (endothermic energy is a reactant; exothermic energy is a product). Copyright Cengage Learning. All rights reserved 111

112 Section 13.7 Le Châtelier s Principle Effects of Changes on the System 3. Pressure: a) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b) Addition of inert gas does not affect the equilibrium position. c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas. Copyright Cengage Learning. All rights reserved 112

113 Section 13.7 Le Châtelier s Principle

114 Section 13.7 Le Châtelier s Principle To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 114

115 Section 13.7 Le Châtelier s Principle Equilibrium Decomposition of N 2 O 4 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 115

116 Section 13.7 Le Châtelier s Principle

117 Section 13.7 Le Châtelier s Principle

118 Section 13.7 Le Châtelier s Principle

119 Section 13.7 Le Châtelier s Principle 2014 AP Exam

120 Section 13.7 Le Châtelier s Principle 2014 AP Exam

121 Section 13.7 Le Châtelier s Principle 2014 AP Exam

122 Section 13.7 Le Châtelier s Principle 2014 AP Exam

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