Equilibrium constant for the reaction $\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(g) ightleftharpoons…

Equilibrium constant for the reaction $\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(g) ightleftharpoons \mathrm{H}_2(g)+\mathrm{CO}_2(g)$ is 81 . If velocity constant of the forward reaction is $162 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$. What is the velocity constant (in $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$ ) for the backward reaction ?
  1. $13122$
  2. $2$
  3. $261$
  4. $243$

Solution

$K_c=\frac{k_f}{k_b}=\frac{162}{81}=2$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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