For the reaction N 2 O 4 ( g ) rightleftharpoons 2 NO 2 ( g ) , K p =0.492 atm at 300 K . K c for the reaction at same temperature is × 10 -2 . (Given : R =0.082 L atm -1 Kol -1 )

Chemistry · Chemical Equilibrium

JEE Main 2024 — 29 January, Shift 1 — Question 76

For the reaction $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$ , $\mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}$ at $300 \mathrm{~K} . \mathrm{K}_{\mathrm{c}}$

for the reaction at same temperature is $\qquad$ $\times 10^{-2}$ .

(Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm}^{-1} \mathrm{Kol}^{-1}$ )

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$

$\Delta \mathrm{n}_{\mathrm{g}}=1$

$\Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{0.492}{0.082 \times 300}=2 \times 10^{-2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient