Chapter 14 Chemical Equilibrium

1. Chapter 14 Chemical Equilibrium

2. Contents and Concepts Describing Chemical Equilibrium • Chemical Equilibrium—A Dynamic Equilibrium • The Equilibrium Constant • Heterogeneous Equilibria; Solvents in Homogeneous Equilibria Using the Equilibrium Constant • Qualitatively Interpreting the Equilibrium Constant • Predicting the Direction of Reaction • Calculating Equilibrium Concentrations

3. Changing Reaction Conditions: Le Châtelier’s Principle • Removing Products or Adding Reactants • Changing the Pressure and Temperature • Effect of a Catalyst

4. Learning Objectives Describing Chemical Equilibrium • Chemical Equilibrium—A Dynamic Equilibrium • a. Define dynamic equilibrium and chemical equilibrium. • b. Apply stoichiometry to an equilibrium mixture.

5. 2. The Equilibrium Constant • a. Define equilibrium-constant expression and equilibrium constant. • b. State the law of mass action. • c. Write equilibrium-constant expressions. • d. Describe the kinetics argument for the approach to chemical equilibrium. • e. Obtain an equilibrium constant from reaction composition.

6. 2. The Equilibrium Constant (cont) • f. Describe the equilibrium constant Kp; indicate how Kpand Kcare related.State the law of mass action. • g. Obtain Kcfor a reaction that can be written as a sum of other reactions of known Kcvalues.

7. 3. Heterogeneous Equilibria; Solvents in Homogeneous Equilibria • a. Define homogeneous equilibrium and heterogeneous equilibrium. • b. Write Kcfor a reaction with pure solids or liquids. Using the Equilibrium Constant 4. Qualitatively Interpreting the Equilibrium Constant • a. Give a qualitative interpretation of the equilibrium constant based on its value.

8. 5. Predicting the Direction of Reaction • a. Define reaction quotient, Q. • b. Describe the direction of reaction after comparing Q with Kc. • c. Use the reaction quotient.

9. 6. Calculating Equilibrium Concentrations • a. Obtain one equilibrium concentration given the others. • b. Solve an equilibrium problem (involving a linear equation in x). • c. Solve an equilibrium problem (involving a quadratic equation in x).

10. Changing the Reaction Conditions; Le Châtelier’s Principle 7. Removing Products or Adding Reactants • a. State Le Châtelier’s principle. • b. State what happens to an equilibrium when a reactant or product is added or removed. • c. Apply Le Châtelier’s principle when a concentration is altered.

11. 8. Changing the Pressure and Temperature • a. Describe the effect of a pressure change on chemical equilibrium. • b. Apply Le Châtelier’s principle when the pressure is altered. • c. Describe the effect of a temperature change on chemical equilibrium. • d. Apply Le Châtelier’s principle when the temperature is altered. • e. Describe how the optimum conditions for a reaction are chosen.

12. 9. Effect of a Catalyst • a. Define catalyst. • b. Compare the effect of a catalyst on rate of reaction with its effect on equilibrium. • c. Describe how a catalyst can affect the product formed.

13. Chemical reactions often seem to stop before they are complete. • Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants. • When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists.

14. CO(g) + 3H2(g) CH4(g) + H2O(g) • The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium.

15. CO(g) + 3H2(g) CH4(g) + H2O(g) • This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium.

16. Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

17. We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.

18. When heated PCl5, phosphorus pentachloride, forms PCl3 and Cl2 as follows: • PCl5(g) PCl3(g) + Cl2(g) • When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain 0.135 mol PCl3. What is the molar composition of the mixture?

19. We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts. • Initially we had 1.00 mol PCl5 and no PCl3 or Cl2. • The change in each is stoichiometric: If x moles of PCl5 react, then x moles of PCl3 and x moles of Cl2 are produced. • For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.

20. We were told that the equilibrium amount of PCl3 is 0.135 mol. That means x = 0.135 mol. • We can now find the amounts of the other substances.

21. Moles PCl5 = 1.00 – 0.135 = 0.87 mol • (2 decimal places) • Moles PCl3 = 0.135 mol • (given with 3 significant figures) • Moles Cl2 = 0.135 mol

22. The Equilibrium Constant, Kc • The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation. • The equilibrium constant, Kc, is the value obtained for the Kc expression when equilibrium concentrations are substituted.

23. For the reaction • aA + bB cC + dD • The equilibrium constant expression is • Kc =

24. Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst: • 2H2(g) + CO(g) CH3OH(g) • What is the Kc expression for this reaction?

25. When we are given some information about equilibrium amounts, we are able to calculate the value of Kc. • We need to take care to remember that the Kc expression uses molar concentrations.

26. Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: • 2CO2(g) 2CO(g) + O2(g) • At 3000 K, 2.00 mol CO2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains. • What is the value of Kc at this temperature?

27. 2CO2(g) 2CO(g) + O2(g) Initial 2.00 mol 0 0 Change –2x +2x +x Equilibrium 2.00 – 2x 2x x 0.90 mol 1.10 mol 0.55 mol We can find the value of x. 2.00 – 2x = 0.90 1.10 = 2x x = 0.55 mol 2x = 2(0.55) = 1.10 mol

28. 2CO2(g) 2CO(g) + O2(g)

29. In a heterogeneous equilibrium, in the Kc expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density). • As a result, we incorporate those constants into the value of Kc, thereby making a new constant, Kc. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present. • More simply, we write the Kc expression by replacing the concentration of a solid or pure liquid with 1.

30. Write the Kc expression for the following reaction: • H2O(g) + C(s) CO(g) + H2(g)

31. Given: • aA + bB cC + dD; K1 • When the reaction is reversed: • cC + dD aA + bB; K2 • The equilibrium constant expression is inverted: • K2 =

32. Given: • aA + bB cC + dD; K1 • When the reaction is doubled: • 2aA + 2bB 2cC + 2dD; K2 • The equilibrium constant expression, K2 , is the square of the equilibrium constant expression, K1: • K2 =

33. For the reaction • aA(g) + bB(g) cC(g) + dD(g) • The equilibrium constant expressions are • Kc = and Kp =

34. How are these related? • We know • From the ideal gas law, we know that • So,

35. When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, Kp. • In general, the value of Kpis different from that of Kc. • We will explore this relationship on the next slides. Recall the ideal gas law and the relationship between pressure and molarity of a gas:

36. Kp = Kc (RT)Dn

37. For catalytic methanation, • CO(g) + 3H2(g) CH4(g) + H2O(g) • the equilibrium expression in terms of partial pressures becomes • and

38. The value of Kc at 227°C is 0.0952 for the following reaction: • CH3OH(g) CO(g) + 2H2(g) • What is Kp at this temperature? Kp = 0.0952(RT)Dn where T = 227 + 273 = 500. K R = 0.08206 L  atm/(mol  K) Dn = 2 Kp = 1.60 × 102

39. We can use the value of the equilibrium constant in several ways. • First, we can qualitatively describe the content of the reaction mixture by looking at the magnitude of Kc. • Second, we can determine the direction in which a reaction will proceed by comparing Kc to the value of the reaction quotient, Q, which has the same expression as Kc but uses nonequilibrium values.

40. Finally, we can determine equilibrium concentrations given the initial concentrations and the value of Kc.

41. When Kc is very large (>102), the equilibrium mixture is mostly products. • When Kc is very small (<10-2), the equilibrium mixture is mostly reactants. • When Kc approaches 1, the equilibrium mixture contains appreciable amounts of both reactants and products.

42. Kc = 0.82 for a reaction. Describe the composition of the equilibrium mixture. Because Kc < 100 and > 0.01, at equilibrium there will be substantial amounts of both reactants and products.

43. Reaction Quotient, Q • The reaction quotient has the same form as the equilibrium constant, but uses initial concentrations for its value. • When Kc > Q, the reaction proceeds to the right. • When Kc < Q, the reaction proceeds to the left. • When Kc = Q, the reaction is at equilibrium.

44. Qc must move toward Kc. Here the numerator must increase; more products must be produced. Here the denominator must increase; more reactants must be produced.

45. Calculating Equilibrium Concentrations • 1. When all but one equilibrium concentration and the value of Kc are known. • 2. When the value of Kc and the initial concentrations are known. • When the Kc expression is a perfect square: solving a linear equation. • When the Kc expression is not a perfect square: solving a quadratic equation.

46. Nickel(II) oxide can be reduced to the metal by treatment with carbon monoxide. • CO(g) + NiO(s) ⇌ CO2(g) + Ni(s) • If the partial pressure of CO is 100. mmHg and the total pressure of CO and CO2 does not exceed 1.0 atm, will this reaction occur at 1500 K at equilibrium? (Kp = 700. at 1500 K.)

48. Nitrogen and oxygen form nitric oxide. • N2(g) + O2(g) 2NO(g) • If an equilibrium mixture at 25°C contains 0.040 M N2 and 0.010 M O2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 × 10-30.

49. N2(g) + O2(g) 2NO(g)

50. When the initial concentration and the value of Kc are known, we return to the stoichiometric chart of initial, change, and equilibrium (ICE) amounts or concentrations to find the equilibrium concentrations.