1. Chemical Equilibrium The reversibility of reactions

2. Equilibrium Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

3. Equilibrium The forward reaction rate slows down since the concentration of reactants has decreased. Since the concentration of products is significant, the reverse reaction rate gets faster. Eventually, the forward reaction rate = reverse reaction rate

4. Equilibrium At this point, the reaction is in equilibrium. The equilibrium is dynamic. Both the forward and reverse reactions occur, but there is no net change in concentration.

5. Equilibrium: Equilibrium: 2NO 2 ↔ N 2 O 4 The concentrations of N 2 O 4 and NO 2 level off when the system reaches equilibrium.

6. Equilibrium: Equilibrium: 2NO 2 ↔ N 2 O 4 The system will reach equilibrium starting with either NO 2, N 2 O 4, or a mixture of the two.

7. Equilibrium Reactions that can reach equilibrium are indicated by double arrows (↔ or ). The extent to which a reaction proceeds in a particular direction depends upon the reaction, temperature, initial concentrations, pressure (if gases), etc.

8. Equilibrium Scientists studying many reactions at equilibrium determined that for a general reaction with coefficients a, b, c and d : a A + b B ↔ c C + d D K is the equilibrium constant for the reaction. [C] c [D] d [A] a [B] b K=

9. Equilibrium a A + b B ↔ c C + d D This is the equilibrium constant expression for the reaction. K is the equilibrium constant. The square brackets indicate the concentrations of products and reactants at equilibrium. [C] c [D] d [A] a [B] b K=

10. Equilibrium Constants The value of the equilibrium constant depends upon temperature, but does not depend upon the initial concentrations of reactants and products. It is determined experimentally.

11. Equilibrium Constants

12. The units of K are usually omitted. The square brackets in the equilibrium constant expression indicate concentration in moles/liter. Some texts will use the symbol K c or K eq for equilibrium constants that use concentrations or molarity.

13. Equilibrium Constants The size of the equilibrium constant gives an indication of whether a reaction proceeds to the right.

14. Equilibrium Constants If a reaction has a small value of K, the reverse reaction will have a large value of K. At a given temperature, K = 1.3 x 10 -2 for: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) If the reaction is reversed, 2 NH 3 (g) ↔ N 2 (g) + 3 H 2 (g) K = 1/(1.3 x 10 -2 ) = 7.7 x 10 1

15. Equilibrium Constants Equilibrium constants for gas phase reactions are sometimes determined using pressures rather than concentrations. The symbol used is K p rather than K (or K c ). For the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) K p = (P ammonia ) 2 (P N 2 )(P H2 ) 3

16. Equilibrium Constants The numerical value of K p is usually different than that for K. The values are related, since P= nRT = MRT, where M = mol/liter. For reactions involving gases: K C = K P (RT) Δn where n is the change in moles of gases in the balanced chemical reaction. V

17. Equilibrium Constants The method for solving equilibrium problems with K p or K is the same. With K p, you use pressure in atmospheres. With K or K c, you use concentration in moles/liter, or molarity.

18. Heterogeneous Equilibria Many equilibrium reactions involve reactants and products where more than one phase is present. These are called heterogeneous equilibria. An example is the equilibrium between solid calcium carbonate and calcium oxide and carbon dioxide. CaCO 3 (s) ↔ CaO(s) + CO 2 (g)

19. Heterogeneous Equilibria CaCO 3 (s) ↔ CaO(s) + CO 2 (g) Experiments show that the position of the equilibrium does not depend upon the amounts of calcium carbonate or calcium oxide. That is, adding or removing some of the CaCO 3 (s) or CaO(s) does not disrupt or alter the concentration of carbon dioxide.

20. Heterogeneous Equilibria CaCO 3 (s) ↔ CaO(s) + CO 2 (g)

21. Heterogeneous Equilibria CaCO 3 (s) ↔ CaO(s) + CO 2 (g) This is because the concentrations of pure solids (or liquids) cannot change. As a result, the equilibrium constant expression for the above reaction is: K=[CO 2 ] or K p = P CO2

22. Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

23. The Reaction Quotient If reactants and products of a reaction are mixed together, it is possible to determine whether the reaction will proceed to the right or left. This is accomplished by comparing the composition of the initial mixture to that at equilibrium.

24. The Reaction Quotient If the concentration of one of the reactants or products is zero, the reaction will proceed so as to make the missing component. If all components are present initially, the reaction quotient, Q, is compared to K to determine which way the reaction will go.

25. The Reaction Quotient Q has the same form as the equilibrium constant expression, but we use initial concentrations instead of equilibrium concentrations. Initial concentrations are usually indicated with a subscript zero.

26. The Reaction Quotient K = 2.2 x 10 2 for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?

27. The Reaction Quotient K = 2.2 x 10 2 for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? 1. Calculate Q and compare it to K.

28. The Reaction Quotient A comparison of Q with K will indicate the direction the reaction will go.

29. The Reaction Quotient K = 2.2 x 10 2 for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? Suppose 1.00 mol N 2, 2.0 mol H 2 and 1.5 mol of NH 3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? 1. Calculate Q and compare it to K.

30. ∆G for Non-Standard Conditions The thermodynamic tables are for standard conditions. This includes having all reactants and products present initially at a temperature of 25 o C. All gases are at a pressure of 1 atm, and all solutions are 1 M.

31. Equilibrium and Free Energy Thermodynamic tables can be used to calculate ΔG o, the standard Gibbs free energy change. Since the direction in which a reaction will proceed depends upon the reaction quotient, Q, the value of ΔG, will also depend upon the initial concentrations of reactants and products. ΔG = ΔG o + RT ln Q

32. ∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆G o + RTlnQ Where R = 8.314 J/K-mol T is temperature in Kelvins Q is the reaction quotient

33. ∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆G o + RT ln Q For Q, gas pressures are in atmospheres, and concentrations of solutions are in molarity, M.

34. ∆G o and Equilibrium A large negative value of ∆G o indicates that the forward reaction or process is spontaneous. That is, there is a large driving force for the forward reaction. This also means that the equilibrium constant for the reaction will be large.

35. ∆G o and Equilibrium A large positive value of ∆G o indicates that the reverse reaction or process is spontaneous. That is, there is a large driving force for the reverse reaction. This also means that the equilibrium constant for the reaction will be small. When a reaction or process is at equilibrium, ∆G o = zero.

36. ∆G o and Equilibrium

37. ∆G = ∆G o + RT lnQ At equilibrium, ∆G is equal to zero, and Q = K. 0 = ∆G o + RT lnK ∆G o = - RT lnK

38. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: H 2 O(l) ↔ H 2 (g) + ½ O 2 (g)

39. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: H 2 O(l) ↔ H 2 (g) + ½ O 2 (g) Since hydrogen and oxygen react explosively to form water, the reverse reaction should be unfavorable, and have a positive ∆G o, and a small equilibrium constant.

40. Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: H 2 O(l) ↔ H 2 (g) + ½ O 2 (g)

41. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: H 2 O(l) ↔ H 2 (g) + ½ O 2 (g) ∆G o = Σ ∆G f o products – Σ ∆G f o reactants ∆G o = [(1 mol) (∆G f o H 2 (g)) +(½ mol) (∆G f o H 2 (g))] –[(1 mol) (∆G f o H 2 O(l)] ∆G o =0 –[( 1 mol)(-237.1kJ/mol)] = +237.1 kJ

42. Free Energy and Equilibrium ∆G o = - RT lnK where R = 8.314 J/K-mol lnK = –∆G o /RT For H 2 O(l) ↔ H 2 (g) + ½ O 2 (g) : lnK = –237.1 kJ(10 3 J/kJ)/(8.314J/mol-K) (298K) ln K = –95.70 K = e –95.70 = 2.7 x 10 -42

43. Free Energy and Equilibrium In summary, an unfavorable reaction has a positive free energy change, and a small equilibrium constant.

44. Problem: Calculation of K At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g)

45. Problem: Calculation of K At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) 1. Write the equilibrium constant expression.

46. Problem: Calculation of K At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) 1. Write the equilibrium constant expression. K = [NH 3 ] 2 /([N 2 ][H 2 ] 3 )

47. Problem: Calculation of K At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) 2. Make a table of concentrations.

48. Problem: Calculation of K N 2 (g) + 3 H 2 (g) ↔ 2NH 3 (g) [N 2 ] [H 2 ] [NH 3 ] initial 3.00mol 2.00L change equili-librium 2.50mol 2.00 L

49. Problem: Calculation of K N 2 (g) + 3 H 2 (g) ↔ 2NH 3 (g) 3. Complete the table. [N 2 ] [H 2 ] [NH 3 ] initial 3.00mol 2.00L change equili-librium 2.50mol 2.00 L

50. Problem: Calculation of K N 2 (g) + 3 H 2 (g) ↔ 2NH 3 (g) 3. Complete the table. [N 2 ] [H 2 ] [NH 3 ] initial 3.00mol 2.00L change-.50mol 2.00 L equili-librium 2.50mol

51. Problem: Calculation of K N 2 (g) + 3 H 2 (g) ↔ 2NH 3 (g) 3. Complete the table. [N 2 ] [H 2 ] [NH 3 ] initial00 3.00mol 2.00L change-.50mol 2.00 L equili-librium 2.50mol

52. [N 2 ] [H 2 ] [NH 3 ] initial00 3.00mol 2.00L change+.25mol 2.00 L +.75mol2.00L-.50mol equili-librium2.50mol N 2 (g) + 3H 2 (g) ↔2NH 3 (g)

53. [N 2 ] [H 2 ] [NH 3 ] initial00 3.00mol 2.00L change+.25mol 2.00 L +.75mol2.00L-.50mol equili-librium2.50mol N 2 (g) + 3H 2 (g) ↔2NH 3 (g) The equilibrium values are the sum of the initial concentrations plus any changes that occur.

54. [N 2 ] [H 2 ] [NH 3 ] initial00 3.00mol 2.00L change+.25mol 2.00 L +.75mol2.00L-.50mol equili-librium+.13M+.38M+1.25M N 2 (g) + 3H 2 (g) ↔2NH 3 (g) The equilibrium values are the sum of the initial concentrations plus any changes that occur.

55. 4. Substitute and solve for K. K = [NH 3 ] 2 /([N 2 ][H 2 ] 3 ) [N 2 ] [H 2 ] [NH 3 ] equili-librium+.13M+.38M+1.25M N 2 (g) + 3H 2 (g) ↔2NH 3 (g) K=(1.25) 2 /[(.13)(.38) 3 ] = 2.2 x 10 2

56. Problem H 2 (g) + I 2 (g) ↔ 2HI(g) K p = 1.00 x 10 2 H 2 (g) + I 2 (g) ↔ 2HI(g) K p = 1.00 x 10 2 Initially, P hydrogen = P iodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

57. Problem

58. Problem

59. Problem H 2 (g) + I 2 (g) ↔ 2HI(g) K p = 1.00 x 10 2 H 2 (g) + I 2 (g) ↔ 2HI(g) K p = 1.00 x 10 2 Initially, P hydrogen = P iodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases. 2. Make a table of initial, change and equilibrium pressures.

60. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) H2H2H2H2 I2I2I2I2HI initial.500 atm 0 change equili-brium

61. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) H2H2H2H2 I2I2I2I2HI initial.500 atm 0 change-x-x+2x equili-brium

62. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) H2H2H2H2 I2I2I2I2HI initial.500 atm 0 change-x-x+2x equili-brium.500-x.500-x2x

63. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) H2H2H2H2 I2I2I2I2HI equili-brium.500-x.500-x2x 3. Substitute and solve.

64. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) H2H2H2H2 I2I2I2I2HI equili-brium.500-x.500-x2x 3.Substitute and solve. = 1.00 x 10 2 K p = (P HI ) 2 /(P H2 )(P I2 ) = 1.00 x 10 2 = 1.00 x 10 2 (2x) 2 = 1.00 x 10 2 (.500-x) (.500-x)

65. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) 3. Substitute and solve. = 1.00 x 10 2 Kp= (P HI ) 2 /(P H2 )(P I2 ) = 1.00 x 10 2 = 1.00 x 10 2 (2x) 2 = 1.00 x 10 2 (.500-x) (.500-x) Take the square root of both sides: = 10.0 (2x) = 10.0 (.500-x)

66. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) 3. Substitute and solve. = 10.0 (2x) = 10.0 (.500-x) 2x = 10.0 (.500-x) = 5.00 -10.0x 12.0 x = 5.00 x =.417

67. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) Calculate the equilibrium partial pressures of all three gases. 4. Answer the question: Calculate the equilibrium partial pressures of all three gases. x =.417 H2H2H2H2 I2I2I2I2HI equili-brium.500-x.500-x2x

68. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) Calculate the equilibrium partial pressures of all three gases. 4. Answer the question: Calculate the equilibrium partial pressures of all three gases. x =.417 H2H2H2H2 I2I2I2I2HI equili-brium.500-x 0.083 atm.500-x 2x.834 atm

69. Problem: H 2 (g) + I 2 (g) ↔ 2HI(g) = 1.00 x 10 2 ? 5. Check your answer, if possible. Does (P HI ) 2 /(P H2 )(P I2 ) = 1.00 x 10 2 ? (.834) 2 /(.083)(.083) = (.696)/(.0069) = 1.0 x 10 2 H2H2H2H2 I2I2I2I2HI equili-brium.500-x 0.083 atm.500-x 2x.834 atm

70. Problem: N 2 (g) + O 2 (g) ↔ 2NO(g) At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium. At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.

71. Problem: N 2 (g) + O 2 (g) ↔ 2NO(g) At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium. At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium. 1. Write the equilibrium constant expression. K = 0.050 = [NO] 2 /[N 2 ][O 2 ]

72. Problem: N 2 (g) + O 2 (g) ↔ 2NO(g) At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium. At 2200 o C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium. 2. Make a table of initial, change and equilibrium concentrations.

73. Problem: N 2 (g) + O 2 (g) ↔ 2NO(g) N2N2N2N2 O2O2O2O2NO initial 1.60mol 2.00 L.400 mol 2.00 L 0 change-x-x+2x equili-brium.800 - x.200 - x 2x

74. Problem: N 2 (g) + O 2 (g) ↔ 2NO(g) N2N2N2N2 O2O2O2O2NO equili-brium.800 - x.200 - x 2x 3.Substitute and solve. K = 0.050 = [NO] 2 /[N 2 ][O 2 ]

75. Approaches to Solving Problems Some equilibrium problems require use of the quadratic equation to obtain an accurate solution. In cases where the equilibrium constant is quite small (roughly 10 -5 or smaller), it is often possible to make assumptions that will simplify the mathematics.

76. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) At 35 o C, K = 1.6 x 10 -5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl 2 are placed in a 1.0 liter flask. At 35 o C, K = 1.6 x 10 -5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl 2 are placed in a 1.0 liter flask. 1. K = 1.6 x 10 -5 =[NO] 2 [Cl 2 ]/[NOCl] 2

77. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) At 35 o C, K = 1.6 x 10 -5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl 2 are placed in a 1.0 liter flask. At 35 o C, K = 1.6 x 10 -5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl 2 are placed in a 1.0 liter flask. 2. Make a table of concentrations.

78. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) NOClNO Cl 2 initial2.0mol 1.00 L 0 1.0 mol 1.00 L change-2x+2x+x equili-brium 2.0 - 2x 2x1.0+x

79. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) NOClNO Cl 2 equili-brium 2.0 - 2x 2x1.0+x 3.Substitute and solve. K = 1.6 x 10 -5 =[NO] 2 [Cl 2 ]/[NOCl] 2 1.6 x 10 -5 =[2x] 2 [1.0 + x]/[2.0-2x] 2 1.6 x 10 -5 =[2x] 2 [1.0 + x]/[2.0-2x] 2

80. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) 3.Substitute and solve. 1.6 x 10 -5 = [2x] 2 [1.0 + x] 1.6 x 10 -5 = [2x] 2 [1.0 + x] [2.0-2x] 2 [2.0-2x] 2 Since K is small, x, the amount of product formed, will be a small number. As a result, 1.0 + x ≈1.0, and 2.0-2x≈2.0

81. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) 3.Substitute and solve. 1.6 x 10 -5 = [2x] 2 [1.0 + x] 1.6 x 10 -5 = [2x] 2 [1.0 + x] [2.0-2x] 2 [2.0-2x] 2 The expression simplifies to: 1.6 x 10 -5 = [2x] 2 [1.0] [2.0] 2 [2.0] 2 0 0

82. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) 3.Solve for x. 1.6 x 10 -5 = [2x] 2 [1.0] [2.0] 2 [2.0] 2 4x 2 = 1.6 x 10 -5 4x 2 = 1.6 x 10 -5 (4)/1 = 6.4 x 10 -5 x 2 = 1.6 x 10 -5 ; x =0.0040

83. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) 4. Check your assumption. x =0.0040 Does 1.0 + x ≈1.0, and 2.0-2x≈2.0? 1.0 +.004 = 1.0 2.0 – 2(.004) = 2.0 -.008=2.0

84. Validity of the Assumption These assumptions are generally considered valid if they cause an insignificant error. In this case, due to significant figures, no error is introduced. Usually, if the difference is within 5%, the error is considered acceptable.

85. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) NOClNO Cl 2 equili-brium 2.0 - 2x 2x1.0+x 5.Answer the question. [NOCl] = 2.0M; [NO]= 0.0080M; [Cl 2 ] = 1.0M

86. Problem: 2NOCl(g)↔2NO(g) + Cl 2 (g) 5.Check your answer. [NOCl] = 2.0M; [NO]= 0.0080M; [NOCl] = 2.0M; [NO]= 0.0080M; [Cl 2 ] = 1.0M [Cl 2 ] = 1.0M Does [NO] 2 [Cl 2 ]/[NOCl] 2 = 1.6 x 10 -5 ? (.0080) 2 (1.0)/(2.0) 2 = 1.6 x 10 -5

87. Le Chatelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift so as to counteract the stress.

88. Le Chatelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift so as to counteract the stress. Types of stresses: Addition or removal of reactants or products Addition or removal of reactants or products Changes in pressure or volume (for gases) Changes in pressure or volume (for gases) Changes in temperature Changes in temperature

89. Fe 3+ (aq) + SCN - ↔ FeSCN 2+

90. Heat + N 2 O 4 (g) ↔ 2 NO 2 (g)

91. CoCl 4 2- + 6 H 2 O  Co(H 2 O) 6 2+ + 4Cl 1- + heat

92. Le Chatelier’s Principle For reactions involving gaseous products or reactants, changes in pressure of volume may shift the equilibrium. The equilibrium will shift only if there is an unequal number of moles of gas on either side of the reaction. Reactions involving liquids or solids are not significantly affected by changes in pressure or volume.

93. Le Chatelier’s Principle

94. Pressure Changes

95. Le Chatelier’s Principle For the reaction For the reaction C(s) + 2 H 2 (g) ↔ CH 4 (g) ∆H o =-75kJ Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes. Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes. a) add carbon b) remove methane c) increase the temperature

96. Le Chatelier’s Principle The value of K will change with temperature. The effect of a change in temperature on K depends on whether the reaction is exothermic or endothermic. The easiest way to predict the result when temperature is changed is to write in heat as a product (for exothermic reactions) or a reactant (for endothermic reactions).

97. Le Chatelier’s Principle For the reaction For the reaction C(s) + 2 H 2 (g) ↔ CH 4 (g) ∆H o =-75kJ Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes. Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes. d) add a catalyst e) decrease the volume f) remove some carbon g) add hydrogen