Equilibrium Practice Problems
1. Write the equilibrium expression for each of the following reactions:
N2 (g) + 3 H2 (g)  2 NH3 (g)
K =
[NH3]2
[N2] [H2]3
I2 (s) + Cl2 (g)  2 ICl
K =
[ICl]2
[Cl2]
NO2 (g)  NO
K =
(g)
(g)
+ ½ O2 (g)
[NO] [O2]1/2
[NO2]
2. The dissociation of acetic acid, CH3COOH, has an equilibrium constant at
25C of 1.8 x 10-5. The reaction is
CH3COOH (aq)  CH3COO- (aq) + H+ (aq)
If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of
water and that of CH3COO- is 8.1 x 10-3 moles in the same 0.500 L,
calculate [H+] for the reaction.
K =
[CH3COO-] [H+]
[CH3COOH]
Plug in known values and solve.
1.8 x 10-5 = [8.1 x 10-3 moles/0.500 L] [H+]
[0.46 moles/0.500 L]
[H+] = 1.0 x 10-3 M
3. Indicate the effect of a catalyst, pressure, temperature and concentration
on each of the following on:
catalyst
pressure
temperature
concentration
a. speed presence
increase
direct
relationship
direct
relationship
direct
relationship
b. posn
based on moles
based on
enthalpy of rxn
inverse
relationship
nothing
4. Given the initial partial pressures of (PPCl5) = 0.0500 atm, (PPCl3) = 0.150
atm, and (PCl2) = 0.250 atm at 250C for the following reaction, what must
each equilibrium partial pressure be?
PCl5 (g)  PCl3 (g) + Cl2 (g)
Kp = 2.15
Determine reaction quotient, Q = 0.150 x 0.250 = 0.75
0.0500
Shift to products
Make a chart describing relationships in change.
PCl5 (g)

PCl3 (g) +
Cl2 (g)
Initial
0.0500
0.150
0.250
Change
-x
+x
+x
After
0.0500 -x
0.150 +x
0.250 + x
Plug into equilibrium expression and solve for x.
2.15 = (.150 +x)(0.250 + x)
so x = 0.0272
(0.0500 –x)
Determine Concentrations.
PCl5 (g)

After
0.0500 -x
Substitute
0.0228
PCl3 (g) +
0.150 +x
.177
Cl2 (g)
0.250 + x
.277
Confirm value of K with these values to check your answer.
5. Calculate Kp for the following equilibrium at 250C if K = 0.042.
PCl5 (g)  PCl3 (g) + Cl2 (g)
K = Kp(RT)- n
SO
0.042 = Kp (0.08206x523)-(2-1)
Therefore Kp = 1.8
6. The reaction
2 NO (g)  N2 (g) + O2 (g)
has a value of
K= 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L
vessel, calculate the equilibrium concentrations of NO, N2, and O2.
Make a chart describing relationships in change.
2 NO (g) 
N2 (g) +
O2 (g)
-3
Initial
6.8 x 10
0
0
Change
-2x
+x
+x
-3
After
6.8 x 10 -2x
x
x
Plug into equilibrium expression and solve for x.
2400 =
[x] [x]
so x = 0.0034
-3
2
[6.8 x 10 -2x]
Determine Concentrations.
2 NO (g) 
After
6.8 x 10-3- 2X
Substitute
0
N2 (g) +
X
0.0034
O2 (g)
X
0.0034
Does this make sense considering the value of K.
7. Using the same reaction as in #6, calculate the equilibrium
concentrations of NO, N2, and O2 if the initial concentrations of each
species are:
[NO] = 0 M, [N2] = 0.850 M, [O2] = 0.560 M.
Make a chart describing relationships in
2 NO (g) 
N2 (g) +
Initial
0
0.850 M
Change
+2x
-x
After
2x
0.850 M - x
Plug into equilibrium expression and solve for
2400 =
[0.850 M - x] [0.560 M - x]
[2x]2
Determine Concentrations.
2 NO (g) 
N2 (g) +
After
2x
0.850 M - x
Substitute
0.014
0.843
change.
O2 (g)
0.560 M
-x
0.560 M - x
x.
so x = 0.007
O2 (g)
0.560 M - x
0.553
Does this make sense considering the value of K.
8. The following reaction has an equilibrium constant of 620 at a certain
temperature. Calculate the equilibrium concentrations of all species if 4.5
mol of each component were added to a 3.0 L flask.
H2 (g) + F2 (g)  2 HF (g)
Determine molarity of solutions
[4.5 mol / 3.0L ] = 1.5 M of all 3 solutions
Make a chart describing relationships in change.
H2 (g) +
F2 (g)
 2 HF (g)
Initial
1.5 M
1.5 M
1.5 M
Change
-x
-x
+2x
After
1.5 –x
1.5 – x
1.5 +2x
Plug into equilibrium expression and solve for x.
620 = [1.5 +2x]2 so x = 1.33
[1.5 – x]2
Determine Concentrations.
After
1.5 –x
1.5 – x
1.5 +2x
Substitute
0.17 M
0.17 M
4.13 M
9. Ammonia undergoes hydrolysis according to the following reaction:
NH3 (aq) + H2O  NH4+ (aq) + OH- (aq)
K = 1.8 x 10-5
+
Calculate [NH3], [NH4 ] and [OH ] in a solution originally 0.200 M NH3.
NH3 (aq) + H2O  NH4+ (aq) + OH- (aq
Initial
0.200 M
0
0
Change
-x
+x
+x
After
0.2 – x
x
x
Plug into equilibrium expression and solve for x.
1.8 x 10-5 =
[x]2
so x = .0018
[0.2 – x]
Determine Concentrations.
After
0.2 – x
x
x
Substitute
0.2 M
.0018 M
.0018M
10. The equilibrium constant for the following reaction is 600C is 4.0.
Initially, two moles of CO and one mole of H2O were mixed in a 1.0 liter
container. Determine the concentration of all species at equilibrium.
CO(g) +
H2O(g)  CO2 (g) +
Initial
2.0 M
1.0 M
0
Change
-x
-x
+x
After
2.0 – x
1.0 - x
x
Plug into equilibrium expression and solve for x.
4.0
=
[x]2
so
x = 0.85
[1.0 – x][2.0 – x]
Determine Concentrations.
After
2.0 – 0.85 1.0 - 0.85
0.85
Substitute
1.15
0.15
0.85
H2 (g)
0
+x
x
0.85
0.85
11. If 5.0 moles of O2 and 4.0 moles of NO were entered into an empty 1.0
liter flask, calculate the equilibrium constant if the amount of NO2 found at
equilibrium was 1.5 moles.
2 NO (g) + O2 (g)  2 NO2 (g)
Initial
4.0 M
5.0 M
0
Change
-2x
-x
+2x
After
4.0 – 2x
5.0 - x
2x
Using 1.5 = 2x and therefore x = 0.75 M
After
4.0 – 1.5
5.0 – 0.75
1.5
Substitute
2.5
4.25
1.5
K=
(1.5)2
= 0.085
(2.5)2(4.25)
12. Two moles of NH3 were entered in a 1.0 liter container at 650C. At
equilibrium only 71% of the original NH3 was found. Determine the
equilibrium constant of the following reaction.
2 NH3 (g) 
N2 (g) +
3 H2 (g)
Initial
2.0 M
0
0
Change
-2x
+x
+3x
After
2.0 – 2x
+x
+3x
Using 2.0 – 2x = 2.0(71%) so x = 0.29
After
2.0 – 2(.29)
0.29
+3(0.29)
Substitute
1.42
0.29
0.87
K=
(0.29)(0.87)3
= 0.095
(1.42)2
13. If the equilibrium constant for the following reaction is 0.10, determine
the final concentration of ICl, if 4.0 moles of I2 and Cl2 were entered
initially into an empty 1.0 liter flask.
I2 (g) +
Cl2 (g) 
2 ICl (g)
Initial
4.0 M
4.0 M
0
Change
-x
-x
+2x
After
4.0 – x
4.0 - x
2x
Plug into equilibrium expression and solve for x.
0.10
=
[2x]2
so
x = 0.546
2
[4.0 – x]
Determine Concentrations.
After
4.0 – 0.546
4.0 - 0.546 2(0.546)
Substitute
3.5
3.5
1.1
14. The reaction of carbon disulfide with chlorine is as follows:
CS2 (g) + 3 Cl2 (g)  CCl4 (g) + S2Cl2 (g)
H = -238 kJ
Predict the effect of the following changes to the system on the direction
of equilibrium.
a. The pressure on the system is doubled by halving the volume.
b. CCl4 is removed as it is generated.
c. Heat is added to the system.
15. The reaction of nitrogen gas with hydloric acid is as follows:
N2 (g) + 6 HCl (g)  2 NH3 (g) + 3 Cl2 (g)
H = 461 kJ
Predict the effect of the following changes to the system on the direction
of equilibrium
a. Triple the volume of the system.
b. The amount of nitrogen is doubled.
c. Heat is added to the system.
Sometimes it’s our fault that things are out of balance.
(See below.)