The equilibrium constant for the reaction SO 3 ( g ) rightleftharpoons SO 2 ( g )+1/2 O 2 ( g ) is K C =4.9 × 10 -2 . The value of K C for the reaction given below is 2 SO 2 ( g )+ O 2 ( g ) rightleftharpoons 2 SO 3 ( g ) is

Chemistry · Chemical Equilibrium

JEE Main 2024 — 4 April, Shift 2 — Question 60

The equilibrium constant for the reaction

$\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$

is $\mathrm{K}_{\mathrm{C}}=4.9 \times 10^{-2}$ .

The value of $\mathrm{K}_{\mathrm{C}}$ for the reaction given below is $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$ is

Option A:
4.9
Option B:
41.6
Option C:
49
Option D:
416
Correct

Answer: D

Step-by-step solution

$\quad \mathrm{K}_{\mathrm{C}}^{\prime}=\left(\frac{1}{\mathrm{~K}_{\mathrm{C}}}\right)^{2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^{2}$

$\mathrm{K}_{\mathrm{C}}^{\prime}=416.49$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient