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Chapter 6 Chemical Equilibrium 1. 1. Reaction Rates 2. The Equilibrium Condition 3. The Equilibrium Constant 4. Heterogeneous Equilibria 5. Equilibrium.

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Presentation on theme: "Chapter 6 Chemical Equilibrium 1. 1. Reaction Rates 2. The Equilibrium Condition 3. The Equilibrium Constant 4. Heterogeneous Equilibria 5. Equilibrium."— Presentation transcript:

1 Chapter 6 Chemical Equilibrium 1

2 1. Reaction Rates 2. The Equilibrium Condition 3. The Equilibrium Constant 4. Heterogeneous Equilibria 5. Equilibrium Expressions Involving Pressures 6. Applications of the Equilibrium Constant 7. Solving Equilibrium Problems 8. Le Chatelier’s Principle 2

3 3 Reaction Rates Some chemical reactions proceed rapidly.  The precipitation reactions, where the products form practically the instant the two solutions are mixed. Other reactions proceed slowly.  Like the decomposition of dye molecules of a sofa placed in front of a window. The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time. Chemists study ways of controlling reaction rates.

4 4 2 N 2 O 5 (g) → 4 NO 2 (g) + O 2 (g) Over time, the concentrations of reactants decrease as products increase.

5 5 2 N 2 O 5 (g) → 4 NO 2 (g) + O 2 (g): Rate vs. Time Because reactant concentrations decrease, the rates of reactions slow down over time.

6 6 The higher the concentration of reactant molecules, the faster the reaction will generally go. Increases the frequency of reactant molecule collisions. Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time. Factors Effecting Reaction Rate: Reactant Concentration

7 7 Effect of Concentration on Rate Low concentrations of reactant molecules leads to fewer effective collisions, therefore a slower reaction rate. High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate.

8 8 Increasing the temperature increases the number of molecules in the sample with enough energy so that their collisions can overcome the activation energy. Increasing the temperature also increases the frequency of collisions. So the rate increases because the frequency of effective collisions increases. Both these mean that increasing temperature increases the reaction rate. Factors Effecting Reaction Rate: Temperature

9 9 Reaction Dynamics, Continued The forward reaction slows down as the amounts of reactants decreases. At the same time, the reverse reaction speeds up as the concentration of the products increases. Eventually, the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. Dynamic equilibrium is reached when the rates of two opposite processes are the same.

10 Effect of Temperature on Rate Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster reaction rate. 10

11 The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. 2NO 2 (g)N2O4(g)N2O4(g) BrownColorless 11

12 12 Chemical Equilibrium When a reaction reaches equilibrium, the amounts of reactants and products in the system stay constant. The forward and reverse reactions still continue. Because they go at the same rate, the amounts of materials do not change.

13 Equilibrium Initially, we only have reactant molecules in the mixture. The reaction can only proceed in the forward direction, making products. As the reaction proceeds, the forward reaction slows down as the reactants get used up. At the same time, the reverse reaction speeds up as product concentration increases. Eventually, the forward and reverse rates are equal. At this time equilibrium is established. Once equilibrium is established, the concentrations of the reactants and products in the final mixture do not change, (unless conditions are changed). 13

14 14 Equilibrium, Continued Time Rate Rate forward Rate reverse Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

15 15 Hypothetical Reaction 2 Red  Blue Time[Red][Blue] 00.4000.000 100.2080.096 200.1900.105 300.1800.110 400.1740.113 500.1700.115 600.1680.116 700.1670.117 800.1660.117 900.1650.118 1000.1650.118 1100.1640.118 1200.1640.118 1300.1640.118 1400.1640.118 1500.1640.118 The reaction slows over time, but the red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the red and the blue molecules no longer change— equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of red molecules turning into blue is the same as the rate of blue molecules turning into red.

16 16 Hypothetical Reaction 2 Red  Blue, Continued

17 17 Equilibrium  Equal The rates of the forward and reverse reactions are equal at equilibrium. But that does not mean the concentrations of reactants and products are equal. Some reactions reach equilibrium only after almost all the reactant molecules are consumed— we say the position of equilibrium favors the products. Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed—we say the position of equilibrium favors the reactants.

18 18 Equilibrium Constant Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them. For the reaction H 2 (g) + I 2 (g)  2HI(g) at equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant.

19 The Equilibrium Constant K eq cC + dDaA + bB For a general reversible reaction: K eq = [A] a [B] b [C] c [D] d Equilibrium constant expression Equilibrium constant Reactants Products For the following reaction:2NO 2 (g)N2O4(g)N2O4(g) [N 2 O 4 ] [NO 2 ] 2 = 4.64 x 10 -3 (at 25 °C)K eq = Equilibrium equation:

20 20 Equilibrium Constant, Continued For the general equation aA + bB  cC + dD, the relationship is given below: The lowercase letters represent the coefficients of the balanced chemical equation. Always products over reactants. The constant is called the equilibrium constant, K eq.

21 21 Writing Equilibrium Constant Expressions For aA + bB  cC + dD, the equilibrium constant expression is: So for the reaction 2 N 2 O 5  4 NO 2 + O 2, the equilibrium constant expression is:

22 22 Equilibrium Constants for Heterogeneous Equilibria Pure substances in the solid and liquid state have constant concentrations. Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution. Therefore, these substances are not included in the equilibrium constant expression. For the reaction CaCO 3 (s) + 2 HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l):

23 23 Heterogeneous Equilibria CaO(s) + CO 2 (g)CaCO 3 (s) LimeLimestone (1) (1)[CO 2 ] = [CO 2 ] [CaCO 3 ] [CaO][CO 2 ] = Pure solids and pure liquids are not included. K eq = K eq = [CO 2 ]K p = P CO 2

24 24 Write the Equilibrium Constant Expressions, K eq, for Each of the Following: 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba +2 (aq) + SO 4 -2 (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)

25 25 Write the Equilibrium Constant Expressions, K eq, for Each of the Following, Continued: 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba +2 (aq) + SO 4 -2 (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) K eq = [CO] 2 [O 2 ] [CO 2 ] 2 K eq =[Ba +2 ][SO 4 -2 ] K eq = [CO 2 ] [H 2 O] 2 [CH 4 ][O 2 ] 2

26 26 What Does the Value of K eq Imply? When the value of K eq > > 1, we know that when the reaction reaches equilibrium, there will be many more product molecules present than reactant molecules. The position of equilibrium favors products. When the value of K eq < < 1, we know that when the reaction reaches equilibrium, there will be many more reactant molecules present than product molecules. The position of equilibrium favors reactants.

27 27 A Large Equilibrium Constant

28 28 A Small Equilibrium Constant

29 29 Write the Equilibrium Constant Expressions, K eq, and Predict the Position of Equilibrium for Each of the Following: 2 HF(g)  H 2 (g) + F 2 (g) K eq = 1 x 10 -95 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K eq = 8 x 10 25 N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K eq = 3 x 10 -17

30 Write the Equilibrium Constant Expressions, K eq, and Predict the Position of Equilibrium for Each of the Following, Continued: 2 HF(g)  H 2 (g) + F 2 (g) K eq = 1 x 10 -95 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K eq = 8 x 10 25 N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K eq = 3 x 10 -17 Favors reactants. Favors products. Favors reactants. 30

31 31 Calculating K eq The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for K eq. Although you may have different amounts of reactants and products in the equilibrium mixture, the value of K eq will always be the same. The value of K eq depends only on the temperature. The value of K eq does not depend on the amounts of reactants or products with which you start.

32 32 Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g)  2HI(g) Initial Equili brium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] 0.50 0.00.11 0.78 0.0 0.500.055 0.39 0.50 0.165 1.17 1.00.50.00.530.0330.934

33 33 Example 1—Find the Value of K eq for the Reaction from the Given Concentrations: 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g). K eq is unitless. Check: Solve: Solution Map: Relationships: [CH 4 ] = 0.0203 M, [C 2 H 2 ] = 0.0451 M, [H 2 ] = 0.112 M K eq Given: Find: [CH 4 ], [C 2 H 2 ], [H 2 ]K eq

34 34 Practice—Calculate K eq for the Reaction 2 NO 2 (g)  N 2 O 4 (g) at 100  C if the Equilibrium Concentrations Are [NO 2 ] = 0.0172 M and [N 2 O 4 ] = 0.0014 M.

35 Practice—Find the Value of K eq for the Reaction from the Given Concentrations: 2 NO 2 (g)  N 2 O 4 (g). K eq is unitless. Check: Solve: Solution Map: Relationships: [NO 2 ] = 0.0172 M, [N 2 O 4 ] = 0.0014 M K eq Given: Find: [NO 2 ], [N 2 O 4 ]K eq 35

36 36 Example 3 —Find the Value of [HI] for the Reaction at Equilibrium from the Given Concentrations and K eq : H 2 (g) + I 2 (g)  2HI(g) Solve: Solution Map: Relationships: [I 2 ] = 0.020 M, [H 2 ] = 0.020 M, K eq = 69 [HI] Given: Find: [I 2 ], [H 2 ], K eq [HI]

37 37 A Sample of PCl 5 (g) Is Placed in a 0.500 L Container and Heated to 160 °C. The PCl 5 Is Decomposed into PCl 3 (g) and Cl 2 (g). At Equilibrium, 0.203 Moles of PCl 3 and Cl 2 Are Formed. Determine the Equilibrium Concentration of PCl 5 if K eq = 0.0635.

38 38 A Sample of PCl 5 (g) Is Placed in a 0.500 L Container and Heated to 160 °C. The PCl 5 Is Decomposed into PCl 3 (g) and Cl 2 (g). At Equilibrium, 0.203 Moles of PCl 3 and Cl 2 Are Formed. Determine the Equilibrium Concentration of PCl 5 if K eq = 0.0635, Continued. PCl 5  PCl 3 + Cl 2 Equilibrium Concentration, M ?

39 39 The Equilibrium Constant K eq Why?

40 The Equilibrium Constant K eq cC + dDaA + bB For a general reversible reaction: K eq = [A] a [B] b [C] c [D] d Equilibrium constant expression Equilibrium constant Reactants Products For the following reaction:2NO 2 (g)N2O4(g)N2O4(g) [N 2 O 4 ] [NO 2 ] 2 = 4.64 x 10 -3 (at 25 °C)K eq = Equilibrium equation:

41 The Equilibrium Constant K eq = 4.63 x 10 -3 0.0429 (0.0141) 2 Experiment 5 [N 2 O 4 ] [NO 2 ] 2 Keq == 4.64 x 10 -3 0.0337 (0.0125) 2 Experiment 1 41

42 Chapter 13/42 K eq = The Equilibrium Constant K eq The equilibrium constant and the equilibrium constant expression are for the chemical equation as written. N 2 (g) + 3H 2 (g)2NH 3 (g) [NH 3 ] 2 [N 2 ][H 2 ] 3 = 4NH 3 (g)2N 2 (g) + 6H 2 (g) [N 2 ] 2 [H 2 ] 6 [NH 3 ] 4 = Keq ´´ 1 K eq 2NH 3 (g)N 2 (g) + 3H 2 (g) [N 2 ][H 2 ] 3 [NH 3 ] 2 K eq = ´ 2

43 Chapter 13/43 The Equilibrium Constant K p 2NO 2 (g)N2O4(g)N2O4(g) K p = N2O4N2O4 P NO 2 P 2 P is the partial pressure of that component

44 Chapter 13/44 The Equilibrium Constant K p ∆n K p = K eq (RT) ∆n 0.082058 K mol L atm R is the gas constant, T is the absolute temperature (Kelvin). is the number of moles of gaseous products minus the number of moles of gaseous reactants.

45 Chapter 13/45 Using the Equilibrium Constant 10 -3 < K eq < 10 3 Appreciable concentrations of both reactants and products are present. K eq > 10 3 Products predominate over reactants. K eq < 10 -3 Reactants predominate over products.

46 46 Using the Equilibrium Constant cC + dDaA + bB Q = [A] t a [B] t b [C] t c [D] t d Reaction quotient: The reaction quotient, Q, is defined in the same way as the equilibrium constant, K eq, except that the concentrations in Q are not necessarily equilibrium values.

47 Using the Equilibrium Constant 47

48 48 Using the Equilibrium Constant If Q = K eq no net reaction occurs. If Q < K eq net reaction goes from left to right (reactants to products). If Q > K eq net reaction goes from right to left (products to reactants).

49 49 Using the Equilibrium Constant At 700 K, 0.500 mol of H I is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H 2, I 2, and H I. K eq is 57.0 at 700 K. 2H I (g)H 2 (g) + I 2 (g)

50

51 Copyright © 2010 Pearson Prentice Hall, Inc. 51 Using the Equilibrium Constant 2H I (g)H 2 (g) + I 2 (g) K eq = [H 2 ][ I 2 ] [H I ] 2 I 000.250 C+x -2x Exx0.250 - 2x 57.0 = x2x2 (0.250 - 2x) 2 Substitute values into the equilibrium expression: Set up a table:

52 Copyright © 2010 Pearson Prentice Hall, Inc. 52 Using the Equilibrium Constant Determine the equilibrium concentrations: x = 0.0262 x2x2 (0.250 - 2x) 2 57.0 = Solve for “x”: H I : 0.250 - 2(0.0262) = 0.198 M H 2 : 0.0262 M I 2 : 0.0262 M

53 53 Disturbing and Re-Establishing Equilibrium Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same. However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established. The new concentrations will be different, but the equilibrium constant will be the same. Unless you change the temperature.

54 Copyright © 2010 Pearson Prentice Hall, Inc. 54 Le Châtelier’s Principle Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. The concentration of reactants or products can be changed. The pressure and volume can be changed. The temperature can be changed.

55 55 Le Châtelier’s Principle Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change. “When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.”

56 56 The Effect of Concentration Changes on Equilibrium Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found. That has the same K eq. Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. You can use to this to drive a reaction to completion! Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium.

57 Copyright © 2010 Pearson Prentice Hall, Inc. 57 Altering an Equilibrium Mixture: Concentration 2NH 3 (g)N 2 (g) + 3H 2 (g) at 700 K, K c = 0.291 Since Q c < K c, more reactants will be consumed and the net reaction will be from left to right. (1.50)(3.00) 3 (1.98) 2 = 0.0968 < K c Q c = [N 2 ][H 2 ] 3 [NH 3 ] 2 = An equilibrium mixture of 0.50 M N 2, 3.00 M H 2, and 1.98 M NH 3 is disturbed by increasing the N 2 concentration to 1.50 M.

58 58 The Effect of Concentration Changes on Equilibrium, Continued When NO 2 is added, some of it combines to make more N 2 O 4.

59 59 The Effect of Concentration Changes on Equilibrium, Continued When N 2 O 4 is added, some of it decomposes to make more NO 2.

60 60 Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems: 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)

61 61 Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued: 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) Shift right, removing some of the added CO 2 and increasing the concentrations of CO and O 2. Shift left, removing some of the added Ba 2+ and reducing the concentration of SO 4 2-. Shift right, removing some of the added CO 2 and decreasing the O 2, while increasing the concentration of CO 2.

62 62 Effect of Volume Change on Equilibrium For solids, liquids, or solutions, changing the size of the container has no effect on the concentration. Changing the volume of a container changes the concentration of a gas. Same number of moles, but different number of liters, resulting in a different molarity.

63 63 Effect of Volume Change on Equilibrium, Continued Decreasing the size of the container increases the concentration of all the gases in the container. This increases their partial pressures. If their partial pressures increase, then the total pressure in the container will increase. According to Le Châtelier’s principle, the equilibrium should shift to remove that pressure. The way to reduce the pressure is to reduce the number of molecules in the container. When the volume decreases, the equilibrium shifts to the side with fewer molecules.

64 64 Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules—the reactant side. The Effect of Volume Change on Equilibrium, Continued

65 65 Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced. 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)

66 66 Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced, Continued. 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) Shift left because there are fewer gas molecules on the reactant side than on the product side. No effect because none of the substances are gases. Shift right because there are fewer gas molecules on the product side than on the reactant side.

67 67 The Effect of Temperature Changes on Equilibrium Exothermic reactions release energy and endothermic reactions absorb energy. If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes. However, heat is not matter and not written in a proper equation.

68 68 The Effect of Temperature Changes on Equilibrium for Exothermic Reactions For an exothermic reaction, heat is a product. Increasing the temperature is like adding heat. According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. The concentrations of C and D will decrease and the concentrations of A and B will increase. The value of K eq will decrease. How will decreasing the temperature effect the system? aA + bB  cC + dD + heat

69 69 The Effect of Temperature Changes on Equilibrium for Endothermic Reactions For an endothermic reaction, heat is a reactant. Increasing the temperature is like adding heat. According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. The concentrations of C and D will increase and the concentrations of A and B will decrease. The value of K eq will increase. How will decreasing the temperature effect the system? Heat + aA + bB  cC + dD

70 70 The Effect of Temperature Changes on Equilibrium

71 71 Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced. Heat + 2 CO 2 (g)  2 CO(g) + O 2 (g) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq)(endothermic) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)(exothermic)

72 72 Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced, Continued. Heat + 2 CO 2 (g)  2 CO(g) + O 2 (g) Heat + BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) + Heat Shift left, reducing the value of K eq. Shift right, increasing the value of K eq.


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