The reaction quotient $(\mathrm{Q})$ for the reaction: $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)}…
$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$
is given by $\mathrm{Q}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}$ The reaction will proceed from right to left if:
- $\mathrm{Q}=\mathrm{K}_e$
- $\mathrm{Q} < \mathrm{K}_c$
- $\mathrm{Q} > \mathrm{K}_c$
- $Q=0$
Solution
$\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}: \\
& \Delta n=2-4=-2
\end{aligned}$
Thus, the reaction moves in forward direction, when $\mathrm{Q} > \mathrm{K}_c$.
Asked in: NEET 2003