South Pasadena • AP Chemistry
Name
3 ▪ Chemical Equilibrium
Period
Date
UNIT PREPARATION
PROBLEMS
1. Consider the reaction: 2 A  B
(a) What is “equal” at equilibrium? How can you tell the system reached equilibrium?
At equilibrium, the rate of the forward reaction (2A → B) is equal to the rate of the reverse reaction
(B →2A). Equilibrium is reached when there is no change in measureable properties and all
species are present in the reaction mixture.
(c) What is the value of the equilibrium constant for this reaction?
[B]
(0.5)
Keq =
= 12.5
2 =
[A]
(0.2)2
1.2
Concentration (M)
(b) The following graph was plotted when 1.2 M A is placed in a
container. At what time did the system appear to reach
equilibrium? Explain briefly.
Equilibrium is reached at t =4 s, because at that time, there is
no change in concentration of reactants and products.
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3 4 5
Time (s)
2. Consider the reversible reaction:
BaF2 (s) +2 H+ (aq)  Ba2+ (aq) + 2 HF (aq)
(a) Write the equilibrium expression for the reaction.
[Ba2+]eq[HF]2eq
Keq =
[H+]2eq
6
Keq = 15
(b) Is this process considered reactant-favored, product-favored, or neither? Justify your answer.
The process is product favored because Keq >> 1.
(c) Suppose a solution contained: [H+] = 0.010 M, [Ba2+] = 0.45 M, [HF] = 0.050 M
Is this reaction mixture at equilibrium? If not, in which direction will it shift? Justify your answer.
[Ba2+][HF]2 (0.45 M)(0.050 M)2
Q=
=
= 11.25
[H+]2
(0.010 M)2
Reaction will shift in the forward direction (to the right) because Q < K eq.
(d) When this reaction is heated up, the mass of BaF2 (s) increases. Is the reaction endothermic, exothermic,
or neither? Justify your answer.
Reaction is exothermic. As the temperature is increased, the reaction shifts in the endothermic
direction, to the left, forming more BaF2 (s).
3. Consider the reaction:
2 HI (g)  H2 (g) + I2 (g)
Keq = 2.9 × 10−5
Suppose 0.50 mol of HI (g) is placed in a 1.0 L container. What is the concentration of all species when the
system reaches equilibrium?
(a) Solve this by taking the square root of both sides of the Keq equation.
2 HI (g)  H2 (g) + I2 (g)
Initial
0.50
0
0
Change
− 2x
+x
+x
Equilibrium 0.50−2x
x
x
[H2]eq[I2]eq
Keq =
= 2.9 × 10−5
[HI]2eq
(x)(x)
= 2.9 × 10−5
(0.50−2x)2
x
= 5.4 × 10−3
0.50−2x
x = 0.0027 – 0.011x
x = 0.00266
[HI]eq = 0.495 M; [H2]eq = [I2]eq = 0.00266 M
(b) Solve this by assuming x << 0.50.
[H2]eq[I2]eq
Keq =
= 2.9 × 10−5
[HI]2eq
(x)(x)
= 2.9 × 10−5
Assume 2x << 0.50, so 0.50−2x = 0.50
(0.50−2x)2
x2
= 2.9 × 10−5
(0.50)2
x = 0.00269
[HI]eq = 0.495 M; [H2]eq = [I2]eq = 0.00269 M
4. Consider the reaction:
2 SO3 (g)  2 SO2 (g) + O2 (g)
A 1.5 M sample of SO3 was placed in a container and allowed to react. When equilibrium was established,
0.30 M of O2 was present in the container. Find the value of the equilibrium constant, Keq.
Initial
Change
Equilibrium
Keq =
2 SO3 (g)  2 SO2 (g) + O2 (g)
1.5
0
0
−0.60
+0.60
+0.30
0.90
0.60
0.30
[SO2]2eq[O2]eq (0.60)2(0.30)
=
= 0.133
[SO3]2eq
(0.90)2