1. 1 © 2009 Brooks/Cole - Cengage CHEMICAL EQUILIBRIUM Chapter 16 Pb 2+ (aq) + 2 Cl – (aq)  PbCl 2 (s) PLAY MOVIE

2. 2 © 2009 Brooks/Cole - Cengage Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Pink to blue Co(H 2 O) 6 Cl 2  Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O  Co(H 2 O) 6 Cl 2 PLAY MOVIE

3. 3 © 2009 Brooks/Cole - Cengage Chemical Equilibrium Fe 3+ + SCN - e FeSCN 2+ + Fe(H 2 O) 6 3+ Fe(SCN)(H 2 O) 5 2+ + SCN - + H 2 O

4. 4 © 2009 Brooks/Cole - Cengage Chemical Equilibrium Fe 3+ + SCN - e FeSCN 2+ After a period of time, the concentrations of reactants and products are constant.After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.The forward and reverse reactions continue after equilibrium is attained. PLAY MOVIE

5. 5 © 2009 Brooks/Cole - Cengage Examples of Chemical Equilibria Phase changes such as H 2 O(s)  H 2 O(liq) PLAY MOVIE

6. 6 © 2009 Brooks/Cole - Cengage Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g)  Ca 2+ (aq) + 2 HCO 3 - (aq) CaCO 3 (s) + H 2 O(liq) + CO 2 (g)  Ca 2+ (aq) + 2 HCO 3 - (aq)

7. 7 © 2009 Brooks/Cole - Cengage Chemical Equilibria CaCO 3 (s) + H 2 O(liq) + CO 2 (g)  Ca 2+ (aq) + 2 HCO 3 - (aq) CaCO 3 (s) + H 2 O(liq) + CO 2 (g)  Ca 2+ (aq) + 2 HCO 3 - (aq) At a given T and P of CO 2, [Ca 2+ ] and [HCO 3 - ] can be found from the EQUILIBRIUM CONSTANT.

8. 8 © 2009 Brooks/Cole - Cengage Reaction Quotient & Equilibrium Constant Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium Equilibrium achieved See Active Figure 16.2

9. 9 © 2009 Brooks/Cole - Cengage Reaction Quotient & Equilibrium Constant At any point in the reaction  H 2 + I 2  2 HI

10. 10 © 2009 Brooks/Cole - Cengage Reaction Quotient & Equilibrium Constant Equilibrium achieved In the equilibrium region

11. 11 © 2009 Brooks/Cole - Cengage The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. a A + b B  c C + d D If Q = K, then system is at equilibrium.

12. 12 © 2009 Brooks/Cole - Cengage THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B  c C + d D a A + b B  c C + d D the following is a CONSTANT (at a given T) If K is known, then we can predict concs. of products or reactants.

13. 13 © 2009 Brooks/Cole - Cengage Determining K 2 NOCl(g)  2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ICE” table of concentrations [NOCl][NO][Cl 2 ] [NOCl][NO][Cl 2 ] Initial2.0000 Change Equilibrium0.66

14. 14 © 2009 Brooks/Cole - Cengage Determining K 2 NOCl(g)  2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ICE” table of concentrations [NOCl][NO][Cl 2 ] [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

15. 15 © 2009 Brooks/Cole - Cengage Determining K 2 NOCl(g)  2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

16. 16 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g)  SO 2 (g) S(s) + O 2 (g)  SO 2 (g)

17. 17 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq)

18. 18 © 2009 Brooks/Cole - Cengage The Meaning of K 1.Can tell if a reaction is product- favored or reactant-favored. For N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

19. 19 © 2009 Brooks/Cole - Cengage The Meaning of K For AgCl(s)  Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 Conc. of products is much less than that of reactants at equilibrium. The reaction with small K is strongly reactant-favored. Ag + (aq) + Cl - (aq) e AgCl(s) e AgCl(s) is product-favored. Ag + (aq) + Cl - (aq) e AgCl(s) e AgCl(s) is product-favored.

20. 20 © 2009 Brooks/Cole - Cengage Product- or Reactant Favored Product-favored K > 1 Reactant-favored K < 1

21. 21 © 2009 Brooks/Cole - Cengage The Meaning of K K comes from thermodynamics. ( See Chapter 19) ∆G˚ < 0: reaction is product favored ∆G˚ > 0: reaction is reactant-favored If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive

22. 22 © 2009 Brooks/Cole - Cengage The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. PLAY MOVIE

23. 23 © 2009 Brooks/Cole - Cengage The Meaning of K If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction “shift” to approach equilibrium? See Chemistry Now

24. 24 © 2009 Brooks/Cole - Cengage The Meaning of K All reacting chemical systems are characterized by their REACTION QUOTIENT, Q. Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become ________ and [n] must ____________. If Q = K, then system is at equilibrium.

25. 25 © 2009 Brooks/Cole - Cengage Æ Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g)  2 HI(g)

26. 26 © 2009 Brooks/Cole - Cengage H 2 (g) + I 2 (g)  2 HI(g) K c = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 ChangeEquilib

27. 27 © 2009 Brooks/Cole - Cengage Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 Change-x-x+2x Equilib1.00-x1.00-x2x where x is defined as am’t of H 2 and I 2 consumed on approaching equilibrium. H 2 (g) + I 2 (g)  2 HI(g) K c = 55.3

28. 28 © 2009 Brooks/Cole - Cengage H 2 (g) + I 2 (g)  2 HI(g) K c = 55.3 Step 2. Put equilibrium concentrations into K c expression.

29. 29 © 2009 Brooks/Cole - Cengage [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium H 2 (g) + I 2 (g)  2 HI(g) K c = 55.3

30. 30 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g) e 2 NO 2 (g) ee

31. 31 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 ChangeEquilib

32. 32 © 2009 Brooks/Cole - Cengage If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilib0.50 - x2x Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g)

33. 33 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) Step 2. Substitute into K c expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029 - 0.0059x = 4x 2 0.0029 - 0.0059x = 4x 2 4x 2 + 0.0059x - 0.0029 = 0 4x 2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

34. 34 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

35. 35 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion: x = 0.026 M Conclusion: x = 0.026 M [N 2 O 4 ] = 0.050 - x = 0.47 M [N 2 O 4 ] = 0.050 - x = 0.47 M [NO 2 ] = 2x = 0.052 M [NO 2 ] = 2x = 0.052 M x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

36. 36 © 2009 Brooks/Cole - Cengage Solving Quadratic Equations Recommend you solve the equation exactly on a calculator or use the “method of successive approximations” See Appendix A.

37. 37 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Adding equations for reactions S(s) + O 2 (g)  SO 2 (g) SO 2 (g) + 1/2 O 2 (g)  SO 3 (g) Net equation S(s) + 3/2 O 2 (g)  SO 3 (g) S(s) + 3/2 O 2 (g)  SO 3 (g)

38. 38 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O 2 (g)  SO 3 (g) 2 S(s) + 3 O 2 (g)  2 SO 3 (g)

39. 39 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Changing direction S(s) + O 2 (g)  SO 2 (g) S(s) + O 2 (g)  SO 2 (g) SO 2 (g)  S(s) + O 2 (g) SO 2 (g)  S(s) + O 2 (g)

40. 40 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions Concentration Units We have been writing K in terms of mol/L. These are designated by K c But with gases, P = (n/V)·RT = conc · RT P is proportional to concentration, so we can write K in terms of P. These are designated by K p. K c and K p may or may not be the same.

41. 41 © 2009 Brooks/Cole - Cengage Writing and Manipulating K Expressions K using concentration and pressure units K p = K c (RT) ∆n For S(s) + O 2 (g)  SO 2 (g) ∆n = 0 and K p = K c For SO 2 (g) + 1/2 O 2 (g)  SO 3 (g) ∆n = –1/2 and K p = K c (RT) –1/2

42. 42 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Temperature, catalysts, and changes in concentration affect equilibria.Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE CHATELIER’S PRINCIPLEThe outcome is governed by LE CHATELIER’S PRINCIPLE “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

43. 43 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics.

44. 44 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Temperature change  change in KTemperature change  change in K Consider the fizz in a soft drink CO 2 (aq) + HEAT  CO 2 (g) + H 2 O(liq)Consider the fizz in a soft drink CO 2 (aq) + HEAT  CO 2 (g) + H 2 O(liq) K = P (CO 2 ) / [CO 2 ]K = P (CO 2 ) / [CO 2 ] Increase T. What happens to equilibrium position? To value of K?Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(CO 2 ) increases and [CO 2 ] decreases.K increases as T goes up because P(CO 2 ) increases and [CO 2 ] decreases. Decrease T. Now what?Decrease T. Now what? Equilibrium shifts left and K decreases.Equilibrium shifts left and K decreases.

45. 45 © 2009 Brooks/Cole - Cengage Temperature Effects on Equilibrium N 2 O 4 (colorless) + heat  2 NO 2 (brown) ∆H o = + 57.2 kJ K (273 K) = 0.00077 K c (273 K) = 0.00077 K (298 K) = 0.0059 K c (298 K) = 0.0059 PLAY MOVIE

46. 46 © 2009 Brooks/Cole - Cengage Temperature Effects on Equilibrium See Figure 16.8

47. 47 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Add catalyst  no change in KAdd catalyst  no change in K A catalyst only affects the RATE of approach to equilibrium.A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

48. 48 © 2009 Brooks/Cole - Cengage N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heatN 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat K = 3.5 x 10 8 at 298 KK = 3.5 x 10 8 at 298 K Haber-Bosch Process for NH 3

49. 49 © 2009 Brooks/Cole - Cengage Haber-Bosch Ammonia Synthesis Fritz Haber 1868-1934 Nobel Prize, 1918 Carl Bosch 1874-1940 Nobel Prize, 1931

50. 50 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Concentration changesConcentration changes –no change in K –only the equilibrium composition changes.

51. 51 © 2009 Brooks/Cole - Cengage Le Chatelier’s Principle Adding a “reactant” to a chemical system. PLAY MOVIE

52. 52 © 2009 Brooks/Cole - Cengage Le Chatelier’s Principle Removing a “reactant” from a chemical system. PLAY MOVIE

53. 53 © 2009 Brooks/Cole - Cengage Le Chatelier’s Principle Adding a “product” to a chemical system. PLAY MOVIE

54. 54 © 2009 Brooks/Cole - Cengage Le Chatelier’s Principle Removing a “product” from a chemical system. PLAY MOVIE

55. 55 © 2009 Brooks/Cole - Cengage EQUILIBRIUM AND EXTERNAL EFFECTS Temperature, catalysts, and changes in concentration affect equilibria.Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE CHATELIER’S PRINCIPLEThe outcome is governed by LE CHATELIER’S PRINCIPLE “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

56. 56 © 2009 Brooks/Cole - Cengage Butane- Isobutane Equilibrium butane isobutane

57. 57 © 2009 Brooks/Cole - Cengage Butane e Isobutane At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane.Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]?When the system comes to equilibrium again, what are [iso] and [butane]?

58. 58 © 2009 Brooks/Cole - Cengage Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. Q is LESS THAN K. Therefore, the reaction will shift to the ____________. Butane e Isobutane

59. 59 © 2009 Brooks/Cole - Cengage You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right — away from butane and toward isobutane. Set up ICE table [butane][isobutane] [butane][isobutane]InitialChangeEquilibrium 0.50 + 1.50 1.25 - x + x 2.00 - x 1.25 + x Butane e Isobutane

60. 60 © 2009 Brooks/Cole - Cengage You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. Butane e Isobutane

61. 61 © 2009 Brooks/Cole - Cengage Le Chatelier’s Principle Change T – change in K – therefore change in P or concentrations at equilibrium Use a catalyst: reaction comes more quickly to equilibrium. K not changed. Add or take away reactant or product: –K does not change –Reaction adjusts to new equilibrium “position”

62. 62 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) Increase P in the system by reducing the volume (at constant T). e e PLAY MOVIE

63. 63 © 2009 Brooks/Cole - Cengage Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO 2 decreases and P of N 2 O 4 increases.