1. AP Notes Chapter 16 Equilibrium
Dynamic chemical system in which two reactions, equal and opposite, occur simultaneously

2. Properties 1. Appear from outside to be inert or not functioning
2. Can be initiated in both directions

3. Pink to blue
Co(H2O)6Cl2  Co(H2O)4Cl H2O
Blue to pink
Co(H2O)4Cl H2O  Co(H2O)6Cl2

4. Product conc. increases and then becomes constant at equilibrium
Equilibrium achieved
Product conc. increases and then becomes constant at equilibrium
Reactant conc. declines and then becomes constant at equilibrium

5. At any point in the reaction
H2 + I2 2 HI

6. In the equilibrium region
Equilibrium achieved
In the equilibrium region

7. Kinetics Definition
Rf = Rr

8. At equilibrium, the rates of the forward and reverse reactions are equal.

9. aA + bB  cC Rf(eq) = kf [A]a [B]b Rr(eq) = kr [C]c Rf = Rr
kf [A]a [B]b = kr [C]c

11. By convention

12. K is a concentration quotient for a system at equilibrium. K = Q

13. For a system NOT at equilibrium
Q ≠ K

14. The reverse reaction will occur until equilibrium is achieved.
if Q > K
The reverse reaction will occur until equilibrium is achieved.

15. The forward reaction will occur until equilibrium is achieved.
if Q < K
The forward reaction will occur until equilibrium is achieved.

16. Achieving equilibrium is a driving force in chemical systems and will occur when possible. It cannot be stopped (spontaneous)

17. 1. For the equilibrium system, 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?

18. 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m =
[reactant]n

19. 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L)
[reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2 =

20. 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C
2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?
2NO2 (g)  N2O4 (g)
KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L)
[reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2
= 8888
KC = 8.8 so… Q = 8888
Is K = Q? Concentrations and orders must be used!!
NO Q > K so [product] is BIGGER than is should be to be at Equilibrium with given K value so products will convert to reactants to reach equilibrium.

21. Types of Reactions 1. one way (goes to completion)
NaOH(s)  Na+(aq) + OH-(aq)

22. Types of Reactions 2. Equilibrium
(two opposite reactions at same time)
a. dimerization
2NO2(g)  N2O4(g)

23. b. dissociation of a weak electrolyte
CH3COOH + H2O 
CH3COO- + H3O+

24. c. saturated aqueous solutions
AgCl(s)  Ag+(aq) + Cl-(aq)
C6H12O6(s)  C6H12O6(aq)

25. EQUILIBRIUM CONSTANT Keq
By convention

26. [products]coeff > [reactants]coeff
if Keq > 1
[products]coeff > [reactants]coeff
the forward reaction proceeded to a greater extent than the reverse reaction to achieve equilibrium (i.e. the products predominate at equilibrium)

28. [products]coeff < [reactants]coeff
if Keq < 1
[products]coeff < [reactants]coeff
the forward reaction proceeded to a lesser extent than the reverse reaction to achieve equilibrium (i.e. the reactants predominate at equilibrium)

30. How are kf and kr related to temperature?
kf and kr are temperature dependent
thus, Keq is temperature dependent

31. N2O4 + heat 2 NO2 (colorless) (brown) ∆Ho = + 57.2 kJ
Kc (273 K) =
Kc (298 K) =

32. Examples of Equilibrium
Expressions
N2O4(g)  2NO2(g)

33. CH3COOH + H2O CH3COO- +H3O+

34. AgCl(s)  Ag+(aq) + Cl-(aq)

35. Concentrations of pure liquids and solids are NOT included in equilibrium expressions, as their concentrations are themselves constants.

36. The value of Keq may appear to change based on way equation is balanced.

38. A value that is mathematically related to another (eg
A value that is mathematically related to another (eg. temp) is NOT considered a new value

39. Multiple Equilibria H3PO4 + 3 H2O  PO43- + 3 H3O+
H3PO4 + H2O  H2PO4- + H3O+
H2PO4- + H2O  HPO42- + H3O+
HPO42- + H2O  PO43- + H3O+

40. for the complete dissociation of
Keq = K1. K2. K3
for the complete dissociation of
phosphoric acid

41. So far, Keq has been studied as a function of concentration, or expressed with appropriate notation, Kc

42. But, what about equilibrium systems where all components are gases?
Partial pressures  mole distribution

44. where
V = a container
parameter
(constant for all gases)
T = constant for given
values of K
R = constant

45. aA(g) + bB(g)  cC(g)

46. Substituting for a gas

48. Let
c - (a + b) = n
where n is the change in # of moles of gas (product - reactant) for the forward reaction.

49. If we express the equilibrium constant as a function of partial pressures

50. Thus
KC = KP(RT)-n or
KP = Kc(RT)Dn

51. 2. When 2. 0 moles of HI(g) are placed in a 1
2. When 2.0 moles of HI(g) are placed in a 1.0 L container and allowed to come to equilibrium with it’s elements, it is found that 20% of the HI decomposes. What is KC and KP?

52. Applications of the
Equilibrium Constant
&
LeChatelier’s
Principle

53. 3. 017 mol of n-butane is placed in a 0
mol of n-butane is placed in a 0.50 L container and allowed to come to equilibrium with its isomer isobutane. KC at 250C is What are the equilibrium concentrations of the two isomers?

54. Set up an ICE table Initial [ ] of components Change in [ ]
Equilibrium [ ]

55. n-butane  isobutane I
C x x
E x x

56. solve

57. mols Br2 are placed in a 2.0 L flask at 1756 K, which is of sufficient energy to split apart some of the molecules. If KC = 4.0 x 10-4 at 1756 K, what are the equilibrium concentrations of the bromine molecules and atoms?

58. Br2(g)  2 Br(g) I
C x x
E x x

59. solve

60. if K <<< [A]0, then can assume amount that dissociated to reach equilibrium is VERY small, thus

61. solve

62. 5. Calculate [OH-] at equilibrium of a solution that is initially 0
5. Calculate [OH-] at equilibrium of a solution that is initially M nicotine.

63. 2H2O + Nic  NicH OH- I
C x x x
E x x x

64. KC = K1 . K2
KC = (7.0 x 10-7)(1.1 x 10-10)
KC = 7.7 x 10-17

66. LeChatlier’s Principle
When a stress is placed on a system at equilibrium, the system will adjust so as to relieve that stress.

67. Stress Factors 1. Change in concentration of reactants or products
2. Change in volume or pressure (for gases)
3. Change in temperature

68. Responses to Stress 1. Change concentration a. add either H2 or N2
3 H2(g) + N2(g)  2 NH3(g)
1. Change concentration
a. add either H2 or N2
b. remove NH3

69. Responses to Stress 2. Change in volume or pressure a. increase volume
3H2(g) + N2(g)  2 NH3(g)
2. Change in volume or pressure
a. increase volume
b. Increase pressure
c. add He

70. Responses to Stress 3. Change in temperature a. increase temperature
3H2(g) + N2(g)  2 NH3(g)
H = -92 kJ
3. Change in temperature
a. increase temperature

71. AgCl(s)  Ag+(aq) + Cl-(aq)
a. add AgCl
b. add H2O
c. add NaCl
d. add NH3(aq)

72. slope of line is different
Keq is a temperature dependent constant, similar to the rate constant, kf or kr
slope of line is different

73. m = -HR/R
ln Keq
1/T

76. 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”