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.

Solution

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

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.

## Answer

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 K_{C} and solve as above:
$$ K_C = \frac{[NH_3]^2}{[N_2][H_2]^3} \\
6.00×10^{−2} = \frac{[NH_3]^2}{[4.26\, \text{M}][2.09\, \text{M}]^3} \\
[NH_3]^2 = 2.33_{3456} \\
[NH_3] = 1.52_{7566}\\
[NH_3] = 1.53\, \text{M}$$