Question

The equilibrium constant for the reaction: $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2{NH}_{3\left(g\right)}$ at $715K$, is $6.0\times {10}^{-2}$. If, in a particular reaction, there are $0.25mol{L}^{-1}$ of $H_2$ and $0.06mol{L}^{-1}$ of ${NH}_3$ present, calculate the concentration of $N_2$ at equilibrium.

Answer 1

Given: Concentration of $\left[H_2\right]=0.25mol{L}^{-1}$
Concentration of $\left[{NH}_3\right]=0.06mol{L}^{-1}$
now, for the reaction $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2{NH}_{3\left(g\right)}$
$K=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$
or
$\left[N_2\right]=\frac{{\left[{NH}_3\right]}^2}{K\times {\left[H_2\right]}^3}$
$\left[N_2\right]=\frac{{\left[{NH}_3\right]}^2}{K\times {\left[H_2\right]}^3}$
$=\frac{{\left(0.06\right)}^2}{6.0\times {10}^{-2}\times {\left(0.25\right)}^3}mol{L}^{-1}=3.84mol{L}^{-1}$
Thus, the concentration of $\left[N_2\right]$ at equilibrium is $3.84mol{L}^{-1}$.