Question

A mixture of 1.57 mol of $N_2$, 1.92 mol of $H_2$ and 8.13 mol of $N{H}_3$ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, $K_c$ for the reaction
$N_2\left(g\right)+3H_2\left(g\right)\leftrightarrow 2N{H}_3\left(g\right)is1.7\times {10}^2$
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer 1

The given reaction is:
$N_{2\left(g\right)}+3H_{2\left(g\right)}\leftrightarrow 2N{H}_{3\left(g\right)}$
The given concentration of various species is
$\left[N_2\right]=\frac{1.57}{20}mol{L}^{-1}\left[H_2\right]=\frac{1.92}{20}mol{L}^{-1}\left[N{H}_3\right]=\frac{80.13}{20}mol{L}^{-1}$

Now, reaction quotient $Q_c$ is:
$Q_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}=\frac{{\left(\frac{\left(8.13\right)}{20}\right)}^2}{\left(\frac{1.57}{20}\right){\left(\frac{1.92}{20}\right)}^3}=2.4\times {10}^3$
Since, $Q_c\ne K_c$, the reaction mixture is not at equilibrium.
Again, $Q_c>K_c$. Hence, the reaction will proceed in the reverse direction.