Chemistry - Chemical Equilibrium Question with Solution | TestHub

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g) ; $\Delta G^{◦}=30kJ$

At equilibrium, $\Delta G^{◦}=-RT\ln K_p$.

$-30=-5.705\log K_p$

$\therefore \log K_p=\frac{30}{5.705}=5.26$

$K_p=10^{5.26}$

(a) $Q_p=\frac{(P_{N{H}_3}{)}^2}{(P_{N_2})(P_{H_2}{)}^3}=\frac{(1{)}^2}{(1)(1{)}^3}=1$

Since $Q_p<K_p$, the reaction will move in the forward direction.

(b) $Q_p=\frac{(0.01{)}^2}{(10{)}^3(10{)}^2}=10^{-8}$

The reaction moves in the forward direction.

(c) $Q_p=\frac{(0.001{)}^2}{(1{)}^3(1{)}^2}=10^{-6}$

The reaction moves in the forward direction.

(d) $Q_p=\frac{(0.01{)}^2}{(0.001{)}^3(0.01{)}^2}=10^7$

Since $Q_p>K_p$, the reaction moves in the backward direction.