CP Chemistry
Equilibrium Worksheet Answer Key
1. Write the equilibrium expressions for each of the following reactions. Then, given the
equilibrium concentrations, solve for the equilibrium constant.
a.
N2 (g) +
Keq
Keq
Keq
b.
H2 (g) ↔
NH3 (g)
[N2] = 0.0200 M
[H2] = 0.0200 M
2
3
2
3
= [NH3] / [N2] [H2] = (0.60 M) / (0.30 M) [NH3] = 0.0100 M
= (0.0100 M)2 / (0.0200 M)(0.0200 M)3
= 625 M-2
KClO3 (s) ↔
KCl (s) +
O2 (g)
Keq = [O2]3 = (0.0500 M)3 = 0.000125 M3
c.
H2O (l) ↔
H+ (aq) +
OH- (aq)
[KClO3] = 0.0600 M
[KCl] = 0.0400 M
[O2] = 0.0500 M
[OH-] = 1.0x10-7
[H+] = 1.0x10-7
Keq = [H+] [OH-] = (1.0x10-7 M) / (1.0x10-7 M)
Keq = 1.0x10-14 M2
d.
CO (g) +
O2 (g) ↔
CO2 (g)
Keq = [CO2]2 / [CO]2 [O2]
Keq = (3.0 M)2 / (2.0 M)2(1.5 M)
Keq = 1.5 M-1
e.
Li2CO3 (s) ↔
Li+1 (aq) +
CO3-2 (aq)
[CO] = 2.0 M
[O2] = 1.5 M
[CO2] = 3.0 M
[Li+1] = 0.20 M
[CO3-2] = 0.10 M
Keq = [Li+]2 [CO3-] = (0.20 M)2 / (0.10 M)
Keq = 0.4 M
2. What is the molar concentration of each of the ions in a silver dichromate solution at
equilibrium?
Ag2Cr2O7 (s)  2 Ag+ + Cr2O7-2
2x
x
Ksp = [Ag+]2 [Cr2O7-2] = (2x)2 (x)
Ksp = 2.0x10-7 = 4x3
x = 3.68x10-3
[Cr2O7-2] = 3.68x10-3 M; [Ag+] = 7.36x10-3 M
3. Consider the synthesis of gaseous phosphorus trichloride. Assume this is a reversible
equation.
a. Write the balanced equation for this reaction. Include states of matter.
2 P (s) +
3 Cl2 (g) 
2 PCl3 (g)
b. At equilibrium, the concentrations of the substances in the reaction vessel are shown
below. Calculate the equilibrium constant.
[P] = 0.75 M
[Cl2] = 0.30 M
[PCl3] = 0.60 M
Keq = [PCl3]2 / [Cl2]3 = (0.60 M)2 / (0.30 M)3
=
c. Use the equilibrium constant for the reaction to calculate the molar concentration of
PCl3 present if [Cl2] is changed to 0.45 M.
Keq = [PCl3]2 / [Cl2]3 = (x)2 / (0.45 M)3
[PCl3] =
d. In which direction will the equilibrium shift if …
Cl2 is added? right
P is removed? No shift
e. How will [Cl2] change if …
P is added? No change
PCl3 is removed? 
4. Consider the placement of 0.48 grams aluminum hydroxide in a beaker containing 150
mL H2O.
a. Describe what you see in the beaker.
It appears to the naked eye that all of the solid drifts to the bottom of the beaker,
leaving a clear, colorless liquid above it.
b. Write the balanced equation for the dissociation of aluminum hydroxide.
Al(OH)3 (s)  Al3+ (aq) + 3 OH1- (aq)
c. If [OH-] = 2.4x10-6 M, calculate [Al+3].
[OH-1] = 3 [Al+3] from mole ratio so
[Al+3] = [OH-1]/3 = 8x10-7 M