Question

A mixture of $1.57\,mol$ of $N_2$, $1.92\, mol$ of $H_2$ and $8.13\,mol$ of $NH_3$ is introduced into a $20\,L$ reaction vessel at $500\, K$. At this temperature, the equilibrium constant, $K_c$ for the reaction, ${N_{2(g)} + 3H_{2(g)} <=> 2NH_{3(g)}}$ 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)

Backward

Answer 2

The reaction is : ${N2_{(g)} +3H2_{(g)} <=> 2NH3_{(g)}}$ $Q_{c}=\frac{\left[NH_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}$ $=\frac{\left(\frac{8.13}{20}mol\,L^{-1}\right)^{2}}{\left(\frac{1.57}{20}\,mol\,L^{-1}\right)\left(\frac{1.92}{20}\,mol\,L^{-1}\right)^{3}}$ $=2.38\times10^{3}$ 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.