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JEE Main 201325 AprilOnline
Exam
JEE Main
Year
2013
Date
25 April
Mode
Online
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2013 · 25 April · Online · Chemistry · Chemical Equilibrium

The ratio (K_p)/(K_c) for the reaction CO(g)+(1)/(2) O_2(g)...

The ratio KpKc\frac{K_p}{K_c} for the reaction CO(g)+12O2(g)⇌CO2(g)\mathrm{CO}(g)+\frac{1}{2} O_2(g) \rightleftharpoons C O_2(g) is:

Options

  1. A1RT\frac{1}{\sqrt{R T}}
  2. B(RT)1/2(R T)^{1 / 2}
  3. CRT\mathrm{RT}
  4. D11

Correct answer

A. 1RT\frac{1}{\sqrt{R T}}

Step-by-step solution

KP=KC(RT)Δng\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} For the reaction CO(g)+12O2(g)⇋CO2( g)Δng=1−(1+12)=−12\begin{aligned} & \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(g) \leftrightharpoons \mathrm{CO}_2(\mathrm{~g}) \\ & \Delta n_g=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2} \end{aligned}\therefore K_{\mathrm{P}}=\frac{\mathrm{K}_{\mathrm{C}}}{\sqrt{\mathrm{RT}}} ; \quad \frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{C}}}=\frac{1}{\sqrt{R T}}$
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