1. Chemical Equilibrium

2. [H2][I2] = K (equilibrium constant)
H2 + I HI
At equilibrium, the rate of the forward and reverse reactions are equal.
The concentration of reactants and products at equilibrium are related.
[HI]2
[H2][I2] = K (equilibrium constant)

3. H2 + I HI
Suppose the concentrations of H2 and I2 are each initially M at 425C and no HI is present. Over time, the concentrations of H2 and I2 will decrease and the concentration of HI will increase until equilibrium is reached. At equilibrium it was found the concentrations of [H2] = [I2] = M and [HI] = M.
This data can be presented in an I.C.E. table.

5. I.C.E. I - initial concentration C - change in concentration
E - equilibrium concentration

6. Equation H2 + I2 2 HI Initial conc.(I) 0.00175 0.00175 0
Change (C)
Equilibrium (E)

7. The change in concentrations of reactants and products are always equal to the difference between the equilibrium and initial concentrations.
Putting the equilibrium concentrations from the ICE table into the K expression gives:
[HI] (0.0276)2
[H2][I2] = (0.0037)(0.0037) = 56

8. As long as the temperature is kept constant, you can change the concentrations of reactants and the value of K = 56.

9. equilibrium constant expression
General Equation for Equilibrium:
aA + bB cC + dD
K = C c D d A a B b
Where K = equilibrium constant
equilibrium constant expression

10. Equilibrium Constant Expression
When a reaction is at equilibrium:
All concentrations are at equilibrium
Products – numerator / reactants – denominator
Powers are the coefficients from the balanced equation
K depends on reaction temperature
K has no units

11. Reactions involving solids:
The concentrations of solids are not included in the equilibrium constant expression:
Reactions in solution:
Since water concentration is very high, the concentration of water is not included in the equilibrium constant expression.

12. Sometimes you will see K written as:
Keq reaction is at equilibrium
Ka reaction is at equilibrium for acids
Kb reaction is at equilibrium for bases
Kc reaction is at equilibrium for molar concentrations (M)

13. Write the equilibrium constant expression for:
Problem 16.1 (page 725)
Write the equilibrium constant expression for:
N2 (gas) + 3H2 (gas) NH3 (gas)
H2 CO3 (aq) + H2O (liquid) HCO3- (aq) + H3O+ (aq)

16. Determining an Equilibrium Constant
If the experimental values of concentrations and products are known, an equilibrium constant can be obtained by substituting the data into the equation:
K = C c D d A a B b

17. Determining an Equilibrium Constant:
2SO2 (g) + O2 (g) SO3 (g)
At equilibrium:
[SO2] = 3.61 x 10-3 M
[O2] = 6.11 x 10-4 M
[SO3] = 1.01 x 10-2 M

18. K = SO3 2 SO2 2 O2
K = x 10− x 10− x 10−4
K = 1.28 x 104

19. Determining an Equilibrium Constant
More commonly, the concentrations of all reactants and products at equilibrium are not known.
You must use an ICE table for these problems.

20. Determining an Equilibrium Constant:
Suppose that 1.00 mol of SO2 and 1.00 mol of O2 are placed in a 1 L flask at 1000 K. When equilibrium is reached, mol of SO3 has been formed. Calculate the equilibrium constant.
2SO2 (g) + O2 (g) SO3 (g)
K = SO3 2 SO2 2 O2

21. Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0 Change (C)
Equilibrium (E)

22. Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0
Change (C) x -X +2X
Equilibrium (E)

23. Equation 2SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0
Change (C) x -X +2X
Equilibrium (E) x x x = 0.925
Solve for X
X = 0.463

24. Equation SO2 + O2 2 SO3 Initial conc.(I) 1.0 1.0 0
Change (C) (0.463) X
Equilibrium (E)

25. K = SO3 2 SO2 2 O2
K =
K = 2.8 x 102

26. 2 Fe+3 (aq) + 3 I- (aq) 2 Fe+2 (aq) + I3- (aq)
Example 16.3, page 731
In aqueous solution iron (III) ions react with iodine ions to give iron (II) ions and triiodide ions, I3-. Suppose the initial concentration of Fe+3 is 0.20 M, the initial I- ion concentration is 0.30 M, and the equilibrium concentration of I3- ions is M. What is the value of K?
2 Fe+3 (aq) + 3 I- (aq) Fe+2 (aq) + I3- (aq)

27. Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3-
Initial conc.(I)
Change (C)
Equilibrium (E)

28. Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3-
Initial conc.(I)
Change (C) x x +2x x
Equilibrium (E)

29. Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3-
Initial conc.(I)
Change (C) x x +2x x
Equilibrium (E) x x x
X =

30. Equation 2 Fe+3 + 3 I- 2 Fe+2 + I3-
Initial conc.(I)
Change (C) x x +2x x
Equilibrium (E)
X =

31. [Fe+2]2 [I3-]
[Fe+3]2 [I-]3
K =
(0.173)2 (0.0866)
(0.0270)2 (0.040)3
K =
K = 5.6 x 104

32. Example 16.4, page 732
Suppose a tank initially contains H2S at a pressure of 10.0 atm and a temperature of 800 K. When the reaction has come to equilibrium, the partial pressure of S2 vapor is atm. Calculate K.
2 H2S (g) H2 (aq) + S2(g)
Use equilibrium expression substituting gas pressure for concentrations

33. Equation 2 H2S 2 H2 + S2 Initial conc.(I) 10.0 0 0 Change (C)
Equilibrium (E)

34. In many cases the value of K and the initial concentrations of reactants are known, and you want to know the concentrations at equilibrium.
Use ICE tables to summarize the information given, and what you need to calculate.

35. Example 16.5, page 733:
Suppose that 1.00 mol of H2 and 1.00 mol of I2 are placed in a 0.5 L flask at 425 K. What are the concentrations of H2, I2 and HI when equilibrium is reached? K = 55.64
H2 (g) + I2 (g) HI (g)
K = HI 2 H2 I2 = 55.64

36. Equation H2 + I2 2HI Initial conc.(I) 2.0 M 2.0 M 0 Change (C)
Equilibrium (E)

37. Equation H2 + I2 2HI Initial conc.(I) 2.0 M 2.0 M 0
Change (C) x x x
Equilibrium (E) 2.0 – X – x X

38. [HI]2
[H2] [I2]
K =
(2x)2
(2.0 – x) (2.0 – x)
55.64 =
(2x)2
(2.0 – x)2
55.64 =

39. Take the square root of both sides:
(2x)2
(2.0 – x)2
55.64 = 2x
(2.0 – x)
7.459 =
7.459 (2.0 –x) = 2x
14.59 – 7.459x = 2x
14.9 = x
X = 1.58

40. Equation H2 + I2 2HI [H2] = 0.42 M [I2] = 0.42 M [HI] = 3.16 M
With x = 1.58, substitute in ICE Table:
Equation H I HI
Equilibrium (E) 2.0 – X – x X
[H2] = 0.42 M
[I2] = 0.42 M
[HI] = 3.16 M

41. Determining an Equilibrium Constant which involves a quadratic equation.
Assumptions can be used to avoid this problem.

42. Consider the dissociation of I2 molecules to I atoms
Consider the dissociation of I2 molecules to I atoms. The initial concentration of I2 is 0.45 M and the equilibrium constant K = 5.6 x at 500C. Calculate the concentrations at equilibrium.
I2 (g) I (g)
K = I 2 I2 = 5.6 x 10-12

43. Equation I2 2 I Initial conc.(I) 0.45 M 0 Change (C) -x +2x
Equilibrium (E) x x

44. [ I ]2
[ I2 ]
K = 5.6 x =
(2x)2
(0.45 – x)
5.6 x =
To avoid using the quadratic equation, assume x in the denominator is very small compared to 0.45.

45. (2x)2
(0.45)
5.6 x =
X = 7.9 x 10-7
From the value of x, we can determine
[ I2 ] = 0.45 – x, or 0.45 M
[ I ] = 2x = 2(7.9 x 10-7) = x 10-6 M

46. Example 16.6, page 736:
N2 (g) + O2 (g) NO (g)
The above reaction contributes to air pollution when gasoline is burned in air at a high temperature, such as a car engine. At 1500 K, K = 1.0 x Suppose the sample of air has an initial [N2] = 0.80 M and [O2] = 0.20 M. Calculate the equilibrium concentrations.

47. Equation N2 + O2 2 NO K = NO 2 N2 O2 = 1.0 x 10-5
Initial conc.(I)
Change (C) x x x
Equilibrium (E) 0.8 – X – x X

48. (2x)2
(0.80 – x)(0.20 – x)
1.0 x 10-5 =
To avoid using the quadratic equation, assume x in the denominator is very small.
(2x)2
(0.80)(0.20)
1.0 x 10-5 =
Solving for x: x = 6.3 x 10-4

49. Equation N2 + O2 2 NO Equilibrium (E) 0.8 – X 0.2 – x 2X
[ N2 ] = 0.8 M
[ O2 ] = 0.2 M
[NO] = 1.3 x -3 M

50. Problem 16.7, Page 739 Equation 1: N2 + 3H2 2 NH3 K1 = 3.5 x 108
Equation 2 is ½ Equation 1
So K2 = 𝐾1
K2 = 𝑥 = x 104

51. Problem 16.7, Page 739 Equation 1: N2 + 3H2 2 NH3 K1 = 3.5 x 108
Equation 3 is the reverse of Equation 1
so K3 = 1 / K1
K3 = 1 / 3.5 x = x 10-9

52. Disturbing Chemical Equilibrium
The equilibrium between reactants and products can be disturbed in three ways:
Change of temperature
Change of concentrations of reactants and/or products
Change in pressure (gases only)

53. LeChatelier’s Principle – a change in any of the factors that determine equilibrium conditions of a system (concentration, temperature, pressure) will cause the system to change to maintain equilibrium.

54. Change of concentrations of reactants and/or products
N H NH3 + heat N2 H2
NH3
heat

55. If you add more of a reactant (N2 or H2)
N H NH3 + heat
NH3
heat N2 H2

56. To re-establish equilibrium, there is a shift to the right, more NH3 is formed
N H NH3 + heat
NH3
heat N2 H2

57. If some reactant (N2 or H2) is removed N2 + 3H2 2 NH3 + heat

58. To re-establish equilibrium, there is a shift to the left, more N2 and H2 is formed
N H NH3 + heat H2 N2
NH3
heat

59. If you NH3 is removed form the reaction N2 + 3H2 2 NH3 + heat

60. To re-establish equilibrium, there is a shift to the right to from more NH3.
N H NH3 + heat
heat
NH3 N2 H2

61. 2) Change of temperature (exothermic)
N H NH3 + heat N2 H2
NH3
heat

62. If heat is increased for an exothermic reaction:
N H NH3 + heat N2 H2
heat
NH3

63. If heat is increased, equilibrium shifts to the left:
N H NH3 + heat H2 N2
heat
NH3

64. 2) Change of temperature (exothermic)
2 NO N2O4 + heat
NO2
N2O4
heat

65. If heat is decreased, equilibrium shifts to the right to form more product:
2 NO N2O4 + heat
N2O4
heat
NO2

66. 2) Change of temperature (endothermic)
N O2 + heat NO
heat N2 O2 NO

67. If heat is increased, equilibrium shifts to the right to form more product:
N O2 + heat NO NO
heat N2 O2

68. concentration and temperature changes only
General Rule
concentration and temperature changes only
TT AA
Take away Addition
Toward that side Away from that side