Q25 / 90 · JEE Main 2021 (22 Jul Shift 1) · Chemistry

Value of $K_{P}$ for the equilibrium reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2 (g)}$ at $288 K$ is $47.9 .$ The $K_{C}$ for this reaction at same temperature is (Nearest integer)
$(R = 0.083 L bar K^{- 1} {mol}^{- 1})$

Correct answer: 2

Solution

$K_{p} = K_{c} {(RT)}^{∆ n}$

$∆ n = 1$

$K_{C} = \frac{K_{p}}{RT} = \frac{47.9}{0.083 \times 288} = 2$

Hence, the correct answer is 2.

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