1. The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. => dynamic state : rate forward = rate reverse

2. The Equilibrium State 2NO 2 (g)N2O4(g)N2O4(g)

4. The Equilibrium Constant K c c C + dDaA + bB For a general reversible reaction: Kc =Kc = [A] a [B] b [C] c [D] d Equilibrium constant expression Equilibrium constant Reactants Products Equilibrium equation: Double arrows show reversible reaction

5. The Equilibrium Constant K c 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 c =

6. Examples Write the equilibrium constant expression, K c, for each of the following reactions: a.CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) b.2 NO(g) N 2 (g) + O 2 (g)

7. The Equilibrium Constant K c

8. Learning Check = 4.63 x 10 -3 0.0429 (0.0141) 2 [N 2 O 4 ] [NO 2 ] 2 K c == 4.64 x 10 -3 0.0337 (0.0125) 2 Experiment 2 Experiment 5 =

9. The Equilibrium Constant K c K c = 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 = 1 KcKc 2NH 3 (g)N 2 (g) + 3H 2 (g) [N 2 ][H 2 ] 3 [NH 3 ] 2 K c = ´

10. Equilibrium Constant, K c 4NH 3 (g)2N 2 (g) + 6H 2 (g) [N 2 ] 2 [H 2 ] 6 [NH 3 ] 4 = K c ´´ K c = 2

11. Examples Consider the following equilibria 2CO 2 (g) + 2 CF 4 (g) 4COF 2 (g) Write the equilibrium constant K’ c expression

12. 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

13. Examples In the industrial synthesis of hydrogen, mixtures of CO and H 2 are enriched in H 2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) What is the value of K p at 700K if the partial pressures in an equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H 2 O, 6.12 atm of CO 2, and 20.3 atm of H 2 ?

14. The Equilibrium Constant K p n K p = K c (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.

15. Examples Consider the following reaction and corresponding value of K c : H 2 (g) + I 2 (s) 2HI(g) K c = 6.2 x 10 2 at 25 o C What is the value of K p at this temperature?

16. Heterogeneous Equilibria

17. 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 c = K c = [CO 2 ]K p = P CO 2

18. Examples Consider the following unbalanced reaction (NH 4 ) 2 S(s) NH 3 (g) + H 2 S(g) An equilibrium mixture of this mixture at a certain temperature was found to have [NH 3 ] = 0.278 M and [H 2 S] = 0.355 M. What is the value of the equilibrium constant (K c ) at this temperature?

19. Using the Equilibrium Constant Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 13/19 K c = 4.2 x 10 -48 2H 2 (g) + O 2 (g)2H 2 O(g) (at 500 K) K c = 2.4 x 10 47 2H 2 O(g)2H 2 (g) + O 2 (g) (at 500 K) K c = 57.0 2H I (g)H 2 (g) + I 2 (g) (at 500 K)

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

21. Using the Equilibrium Constant

22. If Q c = K c no net reaction occurs. If Q c < K c net reaction goes from left to right (reactants to products). If Q c > K c net reaction goes from right to left (products to reactants).

23. Finding Equilibrium concentrations from Initial Concentrations Steps to follow in calculating equilibirium concentrations from initial concentration Write a balance equation for the reaction Make an ICE (Initial, Change, Equilibrium) table, involves The initial concentrations The change in concentration on going to equilibrium, defined as x The equilibrium concentration Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x Calculate the equilibrium concentrations form the calculated value of x Check your answers

24. Using the Equilibrium Constant Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense Quadratic equation ax2 + bx + c = 0

25. Finding Equilibrium concentrations from Initial Concentrations For Homogeneous Mixture aA(g) bB(g) + cC(g) Initial concentration (M) [A] initial [B] initial [C] initial Change (M) - ax + bx + cx Equilibrium (M) [A] initial – ax [B] initial + bx [C] initial + cx K c = [products] [reactants]

26. Finding Equilibrium concentrations from Initial Concentrations For Heterogenous Mixture aA(g) bB(g) + cC(s) Initial concentration (M) [A] initial [B] initial …….. Change (M) - ax + bx …….. Equilibrium (M) [A] initial – ax [B] initial + bx ………. K c = [products] [reactants]

27. 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 c is 57.0 at 700 K. 2H I (g)H 2 (g) + I 2 (g)

28. Examples The value of K c for the reaction is 3.0 x 10 -2. Determine the equilibrium concentration if the initial concentration of water is 8.75 M C(s) + H 2 O(g) CO(g) + H 2 (g)

29. Examples Consider the following reaction I 2 (g) + Cl 2 (g) 2ICl(g) K p = 81.9 (at 25 o C) A reaction mixture at 25 o C intially contains P I2 = 0.100 atm, P Cl2 = 0.100 atm, and P ICl = 0.100 atm. Find the equilibrium pressure of I 2, Cl 2 and ICl at this temperature. In which direction does the reaction favored?

30. Le Châtelier’s Principle Chapter 13/30 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.

31. Altering an Equilibrium Mixture: Concentration 2NH 3 (g)N 2 (g) + 3H 2 (g)

32. Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 13 Chemical Equilibrium John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition

33. Altering an Equilibrium Mixture: Concentration the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance. the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance. In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that

34. Altering an Equilibrium Mixture: Concentration Add reactant – denominator in Qc expression becomes larger Q c < K c To return to equilibrium, Qc must be increases More product must be made => reaction shifts to the right

35. Altering an Equilibrium Mixture: Concentration Remove reactant – denominator in Qc expression becomes smaller Q c > K c To return to equilibrium, Qc must be decreases Less product must be made => reaction shifts to the left

36. 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.

37. Examples Consider the following reaction at equilibrium CO(g) + Cl 2 (g) COCl 2 (g) Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following reaction mixture a.COCl 2 is added to the reaction mixture b.Cl 2 is added to the reaction mixture c.COCl 2 is removed from the reaction mixture: also account for the change using the reaction quotient Q c

38. Examples The reaction of iron (III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: Fe 2 O 3 (s) + 3 CO(g) 2 Fe(l) + 3CO 2 (g) Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by: a.adding Fe 2 O 3 b.Removing CO 2 c.Removing CO; also account for the change using the reaction quotient Q c

39. Altering an Equilibrium Mixture: Pressure and Volume 2NH 3 (g)N 2 (g) + 3H 2 (g)

40. Altering an Equilibrium Mixture: Changes in Pressure and Volume an increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas. a decrease in pressure by expanding the volume will bring about net reaction in the direction that increases the number of moles of gas. In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that

41. Altering an Equilibrium Mixture: Pressure and Volume If reactant side has more moles of gas Denominator will be larger Q c < K c To return to equilibrium, Qc must be increased Reaction shifts toward fewer moles of gas (to the product) If product side has more moles of gas Numerator will be larger Q c > K c To return to equilibrium, Q c must be decreases Reaction shifts toward fewer moles of gas (to the reactant)

42. Altering an Equilibrium Mixture: Pressure and Volume Reaction involves no change in the number moles of gas No effect on composition of equilibrium mixture For heterogenous equilibrium mixture Effect of pressure changes on solids and liquids can be ignored Volume is nearly independent of pressure Change in pressure due to addition of inert gas No change in the molar concentration of reactants or products No effect on composition

43. Altering an Equilibrium Mixture: Changes in Pressure and Volume 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. (1.00)(6.00) 3 (3.96) 2 = 0.0726 < 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 reducing the volume by a factor of 2.

44. Examples Consider the following reaction at chemical equilibrium 2 KClO 3 (s) 2 KCl(s) + 3O 2 (g) a.What is the effect of decreasing the volume of the reaction mixture? b.Increasing the volume of the reaction mixture? c.Adding inert gas at constant volume?

45. Examples Does the number moles of products increases, decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume? PCl 5 (g) PCl 3 (g) + Cl 2 (g) CaO(s) + CO 2 (g) CaCO 3 (s) 3 Fe(s) + 4H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g)

46. Altering an Equilibrium Mixture: Temperature 2NH 3 (g)N 2 (g) + 3H 2 (g)  H° = -2043 kJ As the temperature increases, the equilibrium shifts from products to reactants.

47. Altering an Equilibrium Mixture: Changes in Temperature

48. Altering an Equilibrium Mixture: Temperature Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 13/48 the equilibrium constant for an exothermic reaction (negative  H°) decreases as the temperature increases. Contains more reactant than product Kc decreases with increasing temperature the equilibrium constant for an endothermic reaction (positive  H°) increases as the temperature increases. Contains more product than reactant Kc increases with increasing temperature In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that

49. Examples The following reaction is endothermic CaCO 3 (s) CaO(s) + CO 2 (g) a.What is the effect of increasing the temperature of the reaction mixture? b.Decreasing the temperature?

50. Examples In the first step of Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 2 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(l) ΔH o = -905 kJ How does the temperature amount of NO vary with an increases in temperature?

51. The Effect of a Catalyst on Equilibrium

53. Catalyst increases the rate of a chemical reaction Provide a new, lower energy pathway Forward and reverse reactions pass through the same transition state Rate for forward and reverse reactions increase by the same factor Does not affect the composition of the equilibrium mixture Does not appear in the balance chemical equation Can influence choice of optimum condition for a reaction