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 introduces into a 20L reaction vessel at 500K. At this temperature, the equilibrium constant, $K_C$ for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ is $1.7\times {10}^2$. Is the reaction mixture at equilibrium? If not what is the direction of the net reaction?

• A. Not at equilibrium, forward shift
• B. Not at equilibrium, backward shift
• C. Cannot be predicted
• D. In equilibrium

Answer 1

The correct option is B Not at equilibrium, backward shift

The reaction for the formation of ammonia is $N_2+3H_2\leftrightharpoons 2N{H}_3$. The expression for the reaction quotient is $Q=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$.

Substitute values in the above expression.

$Q=\frac{(\frac{8.13}{20}{)}^2}{\left(\frac{1.57}{20}\right)(\frac{1.92}{20}{)}^3}=2.38\times {10}^3$

But the value of $K_c$ is $1.7\times {10}^2$. Thus, the value of the reaction quotient is greater than the value of the equilibrium constant $(Q>K_c)$.

Thus, the reaction is not at equilibrium. To attain an equilibrium, the reaction will shift backward.