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 $6.0\times {10}^{-2}{M}^{-2}$. 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

The given reaction is :-

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$\Rightarrow K_C=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

Now, $\left[H_2\right]=0.25M;\left[{NH}_3\right]=0.06M$

$K_C=6\times {10}^{-2}{M}^{-2}$

$\Rightarrow 6\times {10}^{-2}=\frac{0.06\times 0.06}{{\left(0.25\right)}^3\times \left[N_2\right]}$

$\Rightarrow \left[N_2\right]=\frac{0.06\times 0.06}{0.06\times 0.25\times 0.25\times 0.25}$

$=3.84M$