Advanced Chemistry Practice Problems

Chemical Equilibrium: Calculating the Equilibrium Constant

1. Question: What is the value of Kc for the decomposition of ammonium carbamate

if the equilibrium concentrations are [NH3] = 0.212 M and [CO2] = 0.106 M?

NH4CO2NH2(s) ⇌ 2NH3(g) + CO2(g)

Answer: To calculate Kc, the law of mass action must be known for the

equilibrium reaction. Only NH3 and CO2 are included in the expression because

the ammonium carbamate is a solid and is excluded from the expression.

0.00476

0.1060.212

CONH

c

2

c

2

2

3c

K

K

K

2. Question: Given [A]eq = 0.150 M and [B]eq = 0.225, what is [C]eq?

A(aq) + B(aq) ⇌ 2C(aq) Kc = 2.61

Answer: When given the equilibrium constant and the equilibrium concentrations

of all but one species, the concentration of the last one can be calculated. The

law of mass action is written based on the balanced chemical equation and the

known values are substituted to solve for the unknown.

M 0.297[C]

M 0.0881C

0.2250.150

C 2.61

BA

C

2

2

2

c

K

3. Question: A reaction vessel contains N2O5 with an initial concentration of 1.00 M.

If N2O5 decomposes to form NO2 and O2, what concentrations of NO2 and O2 are

present at equilibrium if [N2O5]eq = 0.268 M? Hint: Write a balanced chemical

equation first.

Answer: As suggested in the hint, write the balanced chemical equation for the

decomposition of N2O5 to NO2 and O2.

Advanced Chemistry Practice Problems

2N2O5 4NO2 + O2

Based on the information in the problem, an ICE (initial, change, equilibrium)

table can be constructed.

INITIAL: The problem states that the initial concentration of N2O5 is 1.00 M.

Since no initial concentrations of NO2 and O2 are given, they are assumed to be

zero.

CHANGE: The change row is determined by the stoichiometry of the balanced

chemical reaction. The coefficients in the balanced equation will also be

coefficients for ‘x’ because the amount of change for each substance is

proportional to the other substances. Since there are only reactants present

initially, N2O5 must be decreasing and NO2 and O2 will increase.

EQUILIBRIUM: The final row is simply the sum of the initial and change rows in

the table.

N2O5 (M) NO2 (M) O2 (M)

I 1.00 0 0

C –2x +4x +x

E 1.00 – 2x 4x x

The problem also gives the equilibrium concentration of N2O5 as 0.268.

Therefore

366.0

732.02

00.1268.02

268.0200.1

x

x

x

x

Now that the value of x is known, the value of 4x can be determined (1.46). The

terms in the equilibrium row can be replaced with numerical values giving the

equilibrium concentrations of all three substances.

N2O5 (M) NO2 (M) O2 (M)

E 0.268 1.46 0.366