1. Equilibrium © 2009, Prentice-Hall, Inc. Chapter 14 Chemical Equilibrium

2. Equilibrium © 2009, Prentice-Hall, Inc. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3. Equilibrium © 2009, Prentice-Hall, Inc. 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.

4. Equilibrium © 2009, Prentice-Hall, Inc. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

5. Equilibrium © 2009, Prentice-Hall, Inc. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

6. Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant

7. Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate Law: Rate = k f [N 2 O 4 ]

8. Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate Law: Rate = k r [NO 2 ] 2

9. Equilibrium © 2009, Prentice-Hall, Inc. 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 ] =

10. Equilibrium © 2009, Prentice-Hall, Inc. 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 ] =

11. Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

12. Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Since 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 )

13. Equilibrium © 2009, Prentice-Hall, Inc. Relationship Between K c and K p From the Ideal Gas Law we know that Rearranging it, we get PV = nRT P = RT nVnV

14. Equilibrium © 2009, Prentice-Hall, Inc. 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)

15. Equilibrium © 2009, Prentice-Hall, Inc. 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.

16. Equilibrium © 2009, Prentice-Hall, Inc. Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

17. Equilibrium © 2009, Prentice-Hall, Inc. Equilibrium Can Be Reached from Either Direction It doesn’t 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.

18. Equilibrium © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium.

19. Equilibrium © 2009, Prentice-Hall, Inc. 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.

20. Equilibrium © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 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)

21. Equilibrium © 2009, Prentice-Hall, Inc. 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 ] N 2 O 4(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)

22. Equilibrium © 2009, Prentice-Hall, Inc. 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.

23. Equilibrium © 2009, Prentice-Hall, Inc. Heterogeneous Equilibrium

24. Equilibrium © 2009, Prentice-Hall, Inc. The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by multiplying the density of the substance by its molar mass — and both of these are constants at constant temperature.

25. Equilibrium © 2009, Prentice-Hall, Inc. 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)

26. Equilibrium © 2009, Prentice-Hall, Inc. 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)

27. Equilibrium © 2009, Prentice-Hall, Inc. Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

28. Equilibrium © 2009, Prentice-Hall, Inc. Equilibrium Calculations

29. Equilibrium © 2009, Prentice-Hall, Inc. An Equilibrium Problem 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 (s) 2 HI (g)

30. Equilibrium © 2009, Prentice-Hall, Inc. What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium1.87 x 10 -3

31. Equilibrium © 2009, Prentice-Hall, Inc. [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 equilibrium1.87 x 10 -3

32. Equilibrium © 2009, Prentice-Hall, Inc. 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 equilibrium1.87 x 10 -3

33. Equilibrium © 2009, Prentice-Hall, Inc. 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 equilibrium6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

34. Equilibrium © 2009, Prentice-Hall, Inc. …and, therefore, the equilibrium constant. K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

35. Equilibrium © 2009, Prentice-Hall, Inc. Example Sulfuryl chloride, SO 2 Cl 2, reacts to produce sulfur dioxide, and chlorine gas. 3.509 grams of the compound decomposes in a 1.00 L flask, the temp is raised to 375K, determine the pressure in the flask. At equilibrium the total pressure is 1.43 atm, calculate the partial pressure of all the substances. Calculate the Kp value for the reaction.

36. Equilibrium © 2009, Prentice-Hall, Inc. The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

37. Equilibrium © 2009, Prentice-Hall, Inc. If Q = K, the system is at equilibrium.

38. Equilibrium © 2009, Prentice-Hall, Inc. If Q > K, there is too much product, and the equilibrium shifts to the left.

39. Equilibrium © 2009, Prentice-Hall, Inc. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

40. Equilibrium © 2009, Prentice-Hall, Inc. The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

41. Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle

42. Equilibrium © 2009, Prentice-Hall, Inc. 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.”

43. Equilibrium © 2009, Prentice-Hall, Inc. If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5

44. Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

45. Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

46. Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5 colder hotter

47. Equilibrium © 2009, Prentice-Hall, Inc. uncatalyzedcatalyzed 14.5 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

48. Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5

49. Equilibrium © 2009, Prentice-Hall, Inc. Catalysts

50. Equilibrium © 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of both the forward and reverse reactions.

51. Equilibrium © 2009, Prentice-Hall, Inc. Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.

52. Equilibrium © 2009, Prentice-Hall, Inc. Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK 18.6

53. Equilibrium © 2009, Prentice-Hall, Inc.  G 0 =  RT lnK 18.6

54. Equilibrium © 2009, Prentice-Hall, Inc. Ch 16: Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 3 3 - ] 2 Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form

55. Equilibrium © 2009, Prentice-Hall, Inc. 16.6

56. Equilibrium © 2009, Prentice-Hall, Inc. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. 16.6

57. Equilibrium © 2009, Prentice-Hall, Inc. What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10 16.6

58. Equilibrium © 2009, Prentice-Hall, Inc. 16.6

59. Equilibrium © 2009, Prentice-Hall, Inc. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? 16.6 The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form

60. Equilibrium © 2009, Prentice-Hall, Inc. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 16.7 AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.020 = = 3.9 x 10 -11 M [Ag + ] = K sp [Br - ] 1.6 x 10 -10 0.020 = = 8.0 x 10 -9 M 3.9 x 10 -11 M < [Ag + ] < 8.0 x 10 -9 M

61. Equilibrium © 2009, Prentice-Hall, Inc. The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 s 2 = K sp s = 8.8 x 10 -7 NaBr (s) Na + (aq) + Br - (aq) [Br - ] = 0.0010 M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = 0.0010 + s  0.0010 K sp = 0.0010 x s s = 7.7 x 10 -10 16.8