Q27 / 90 · JEE Main 2023 (29 Jan Shift 2) · Chemistry

At $298 K$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g), K_{1} = 4 \times 10^{5}$
$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g), K_{2} = 1.6 \times 10^{12}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g), K_{3} = 1.0 \times 10^{- 13}$
Based on above equilibria, the equilibrium constant of the reaction,
$2 {NH}_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 NO (g) + 3 H_{2} O (g)$
is _____ $\times 10^{- 33}$ (Nearest integer)

Correct answer: 4

Solution

Given,
$N_{2} (g) 3 H_{2} (g) ⇌ 2 {NH}_{3} (g), K_{1} = 4 \times 10^{5} \dots (1)$

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g), K_{2} = 1.6 \times 10^{12} \dots (2)$

$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g), K_{3} = 1.0 \times 10^{- 13} \dots (3)$
Using properties of equilibrium constant:

$(2) + 3 \times (3) - (1)$

$2 {NH}_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 NO (g) + 3 H_{2} O (g)$

$K_{eq} = \frac{K_{2} \times K_{3}^{3}}{K_{1}} = \frac{1.6 \times 10^{12} \times {(10^{- 13})}^{3}}{4 \times 10^{5}}$

$= \frac{1.6}{4} \times 10^{- 32} = 4 \times 10^{- 33}$

Hence, the correct answer is 4.