An amount of 16 moles and 4 moles of is confined in a vessel of volume one litre. The vessel is heat — Chemical Equilibrium Chemistry Question
Question
An amount of 16 moles $H_2$ and 4 moles of $N_2$ is confined in a vessel of volume one litre. The vessel is heated to a constant temperature until the equilibrium is established. At equilibrium, the pressure was found to be 9/10^th of the initial pressure. The value of $K_c$ for the reaction: $N_2$(g) + 3$H_2$(g) ⇌ 2$NH_3$(g) is:
💡 Solution & Explanation
Reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$; \quad V = 1\ \text{L} \textbf{Step 1 — Find moles at equilibrium using pressure ratio:} At constant V and T: $P \propto n$, so $P_2/P_1 = n_2/n_1$. \[ \frac{P_2}{P_1} = 0.9 \implies n_2 = 0.9 \times n_1 = 0.9 \times 20 = 18\ \text{mol} \] \textbf{Step 2 — Find extent of reaction $x$:} Let $x$ mol N$_2$ react. Total moles change by $-2x$ (since $\Delta n = 2 - 1 - 3 = -2$ per mol N$_2$). \[ 20 - 2x = 18 \implies x = 1 \] \textbf{Step 3 — Equilibrium concentrations} (V = 1 L): \[ [\text{N}_2] = 4-1 = 3\ \text{M}, \quad [\text{H}_2] = 16-3 = 13\ \text{M}, \quad [\text{NH}_3] = 2(1) = 2\ \text{M} \] \textbf{Step 4 — $K_c$:} \[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{4}{3 \times 13^3} = \frac{4}{3 \times 2197} = \frac{4}{6591} \approx 6.07\times10^{-4}\ \text{M}^{-2} \] \textbf{Answer: B} — $K_c = 6.07\times10^{-4}\ \text{M}^{-2}$