Question

A mixture of $1.57mol$ of $N_2$, $1.92mol$ of $H_2$ and $8.13mol$ of $N{H}_3$ is introduced into a $20L$ reaction vessel at $500K$. At this temperature, the equilibrium constant, $K_c$ for the reaction, $N_{2(g)}+3H_{2(g)}\rightleftharpoons 2N{H}_{3(g)}$ is $1.7\times 10^2$. What is the direction of the net reaction?

• A. Forward
• B. Backward
• C. At equilibrium
• D. Data is insufficient

Answer 1

Answer: (B)

The reaction is :
$N_{2(g)}+3H_{2(g)}\rightleftharpoons 2N{H}_{3(g)}$
$Q_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$
$=\frac{{\left(\frac{8.13}{20}mol{L}^{-1}\right)}^2}{\left(\frac{1.57}{20}mol{L}^{-1}\right){\left(\frac{1.92}{20}mol{L}^{-1}\right)}^3}$
$=2.38\times 10^3$
As $Q_c\ne K_c$, the reaction mixture is not in equilibrium.
As $Q_c>K_c$, the net reaction will be in the backward direction.