Consider the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$ The equilibrium constant of the above reaction is $\mathrm{K}_{3}$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $\mathrm{P}_{\mathrm{NH}_{3}} \ll \mathrm{P}_{\text {total }}$ at equilibrium)

(A) $\frac{3^{3 / 2} \mathrm{~K}_{\mathrm{p}}^{1 / 2} \mathrm{P}^{2}}{16}$

(B) $\frac{\mathrm{K}_{\mathrm{p}}^{1 / 2} \mathrm{P}^{2}}{16}$

(C) $\frac{\mathrm{K}_{\mathrm{p}}^{1 / 2} \mathrm{P}^{2}}{4}$

(D) $\frac{3^{3 / 2} \mathrm{~K}_{\mathrm{p}}^{1 / 2} \mathrm{P}^{2}}{4}$

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Equilibrium

Consider the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)

The equilibrium constant of the above reaction is

K_{3}

. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that

P_{NH_{3}} ≪ P_{total}

at equilibrium)

Answer: (A)

Let ammonia dissociate as

2 NH X_{3} N X_{2} + 3 H X_{2}

. If the equilibrium ammonia pressure is

P_{1}

, the nitrogen and hydrogen appear in the ratio

1 : 3

; writing their partial pressures as

x

and

3 x

and using

P_{1} ≪ P

, the total pressure is

P \approx x + 3 x = 4 x

, so

x = \frac{P}{4}

. The given formation constant is

K_{p} = \frac{P _{1}^{2}}{P _{N X_{2}} P _{H X_{2}}^{3}} = \frac{P _{1}^{2}}{x ( 3 x ) ^{3}} = \frac{P _{1}^{2}}{27 x ^{4}}

. Solving for the ammonia pressure,

P_{1} = 27 K_{p} x^{2} = 3^{3/2} K_{p}^{1/2} (\frac{P}{4})^{2} = \frac{3 ^{3/2} K _{p}^{1/2} P ^{2}}{16}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option