When a mixture of and in the volume ratio of 1:5 is allowed to react at 700 K and 10^3 atm pressure, — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ Initial: 1 mol N$_2$, 5 mol H$_2$. Mole fraction of NH$_3$ at equilibrium = 0.4; total pressure $P = 10^3$ atm. \textbf{Step 1 — Find $\alpha$ (mol N$_2$ that react):} Let $\alpha$ mol N$_2$ react. Equilibrium moles: N$_2 = 1-\alpha$, H$_2 = 5-3\alpha$, NH$_3 = 2\alpha$. Total = $6-2\alpha$. \[ x_{\text{NH}_3} = \frac{2\alpha}{6-2\alpha} = 0.4 \implies 2\alpha = 2.4 - 0.8\alpha \implies \alpha = \frac{6}{7} \] \textbf{Step 2 — Equilibrium moles:} \[ n_{\text{N}_2} = \frac{1}{7},\quad n_{\text{H}_2} = \frac{17}{7},\quad n_{\text{NH}_3} = \frac{12}{7},\quad n_{\text{total}} = \frac{30}{7} \] \textbf{Step 3 — Partial pressures:} \[ p_{\text{N}_2} = \frac{1/7}{30/7} \times 10^3 = \frac{10^3}{30}\ \text{atm}, \quad p_{\text{H}_2} = \frac{17 \times 10^3}{30}\ \text{atm}, \quad p_{\text{NH}_3} = \frac{12 \times 10^3}{30} = 400\ \text{atm} \] \textbf{Step 4 — $K_p$:} \[ K_p = \frac{p_{\text{NH}_3}^2}{p_{\text{N}_2} \cdot p_{\text{H}_2}^3} = \frac{(400)^2}{\dfrac{10^3}{30} \cdot \left(\dfrac{17000}{30}\right)^3} \] \[ = \frac{1.6\times10^5}{\dfrac{10^3}{30} \cdot \dfrac{4.913\times10^{12}}{2.7\times10^4}} = \frac{1.6\times10^5 \times 30 \times 2.7\times10^4}{10^3 \times 4.913\times10^{12}} \approx 2.6\times10^{-5}\ \text{atm}^{-2} \] \textbf{Answer: A} — $K_p = 2.6\times10^{-5}\ \text{atm}^{-2}$