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JEE Main 202122 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 22 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2021 · 22 July · Morning Shift · Chemistry · Chemical Equilibrium

Value of K _ P for the equilibrium reaction N _ 2 O _ 4 ( g ) ⇌ 2...

Value of

{K}_{P}

for the equilibrium reaction

{N}_{2}{O}_{4} ( g ) ⇌ 2{NO}_{2 ( g )}

at

288K

is

47 . 9 . The {K}_{C}

for this reaction at same temperature is (Nearest integer)

\left(R = 0 . 083Lbar{K}^{- 1}{mol}^{- 1}\right)

Correct answer

0

Step-by-step solution

${K}_{p} = {K}_{c}{\left(RT\right)}^{∆ n}$

$∆ n = 1$

${K}_{C} = \dfrac{{K}_{p}}{RT} = \dfrac{47 . 9}{0 . 083 \times 288} = 2$

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