JEE Main · 2019 · Shift-IhardCEQ-052

Consider the reaction N2(g) + 3H2(g) 2NH3(g). The equilibrium constant is KP. If pure ammonia is left to dissociate,…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Consider the reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ . The equilibrium constant is $K_P$ . If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume $P_{\mathrm{NH_3}} \ll P_{total}$ at equilibrium)

Options

a
$\frac{3^{3/2} K_P^{1/2} P^2}{16}$
✓
b
$\frac{K_P^{1/2} P^2}{16}$
c
$\frac{K_P^{1/2} P^2}{4}$
d
$\frac{3^{3/2} K_P^{1/2} P^2}{4}$

Correct Answera

$\frac{3^{3/2} K_P^{1/2} P^2}{16}$

Detailed Solution

Step 1 — Reverse reaction (dissociation of $\mathrm{NH_3}$ ): $\mathrm{2NH_3 \rightleftharpoons N_2 + 3H_2}, \quad K_P' = \frac{1}{K_P}$

Step 2 — Let $P_{\mathrm{NH_3}} = p$ (small), total pressure $\approx P$ : From stoichiometry: $P_{\mathrm{N_2}} = \frac{P}{4}$ , $P_{\mathrm{H_2}} = \frac{3P}{4}$ (since $\mathrm{N_2}:H_2 = 1:3$ )

Step 3 — Write $K_P$ for forward reaction: $K_P = \frac{P_{\mathrm{NH_3}}^2}{P_{\mathrm{N_2}} \cdot P_{\mathrm{H_2}}^3} = \frac{p^2}{\frac{P}{4} \cdot \left(\frac{3P}{4}\right)^3} = \frac{p^2}{\frac{P}{4} \cdot \frac{27P^3}{64}} = \frac{p^2 \cdot 256}{27P^4}$

Step 4 — Solve for p: $p^2 = \frac{27 K_P P^4}{256} \Rightarrow p = \frac{3^{3/2} K_P^{1/2} P^2}{16}$

Answer: Option (1)

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