The ratio frac K P K C for the reaction: CO ( g ) +1/2 O 2( g ) rightleftharpoons CO 2( g ) is:

Chemistry · Chemical Equilibrium

JEE Main 2024 — 6 April, Shift 2 — Question 76

The ratio $\frac{K_{P}}{K_{C}}$ for the reaction: $\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}$ is:

Option A:
$(\mathrm{RT})^{1 / 2}$
Option B:
RT
Option C:
1
Option D:
$\frac{1}{\sqrt{\mathrm{RT}}}$
Correct

Answer: D

Step-by-step solution

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

$\Delta \mathrm{n}_{\mathrm{g}}=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}$

$\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{C}}}=(\mathrm{RT})^{\Delta \mathrm{ng}_{\mathrm{g}}}=\frac{1}{\sqrt{\mathrm{RT}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient