Equilibrium Calculations – Calculating a Missing Equilibrium Concentration

When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated by solving the K expression algebraically.

A computation of this sort is illustrated in the next example exercise:

Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the $\mathit K_c$ for the reaction, $N_2\;(g)+O_2\;(g)⇌2NO\;(g)$ is $4.1×10^{−4}$. Calculate the equilibrium concentration of $NO\;(g)$ in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of $N_2$ and $O_2$ at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

$$K_c=\frac{[NO]^2}{[N_2][O_2]}\\ [NO]^2=K_c[N_2][O_2]\\ [NO]=\sqrt{K_c[N_2][O_2]}\\ [NO]=\sqrt{(4.1×10^{−4})(0.036)(0.0089)}\\ [NO]=\sqrt{1.31×10^{−7}}\\ \mathbf{[NO]=3.6×10^{-4}\, \text{M}}$$

Thus [NO] is $3.6×10^{−4}$ mol/L at equilibrium under these conditions.

To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for $K$: $$K_c=\frac{[NO]^2}{[N_2][O_2]}\\ K_c=\frac{(3.6×10^{-4})^2}{(0.036)(0.0089)}\\ K_c=4.0×10^{-4}$$

This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.

Check Your Learning

The equilibrium constant $K_c$ for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is $6.00×10^{−2}$. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.


First, write the reaction described in the question: $$ N_2 + 3\,H_2 \rightleftharpoons 2\, NH_3 $$ Use this reaction to determine the formula for KC and solve as above: $$ K_C = \frac{[NH_3]^2}{[N_2][H_2]^3} \\ 6.00×10^{−2} = \frac{[NH_3]^2}{[4.26\, \text{M}][2.09\, \text{M}]^3} \\ [NH_3]^2 = 2.33_{3456} \\ [NH_3] = 1.52_{7566}\\ [NH_3] = 1.53\, \text{M}$$