Q12 / 90 · JEE Main 2012 (Offline) · Chemistry

The equilibrium constant $\left(\mathrm{K}_{\mathrm{c}}\right)$ for the reaction $\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature $\mathrm{T}$ is $4 \times 10^{-4}$. The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction, $\mathrm{NO}(\mathrm{g}) \rightarrow{1 / 2} \mathrm{~N}_2(\mathrm{g})+{1 / 2} \mathrm{~O}_2(\mathrm{g})$ at the same temperature is :

  • A
    $0.02$
  • B
    $2.5 \times 10^2$
  • C
    $4 \times 10^{-4}$
  • Correct answer
    D
    $50.0$

Solution

$\begin{array}{ll}\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_{\mathrm{C}}=4 \times 10^{-4} \\ \mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\frac{1}{2} \mathrm{O}_2 & \mathrm{~K}_{\mathrm{C}}^1=\sqrt{\frac{1}{\mathrm{~K}_{\mathrm{C}}}} \\ \mathrm{K}_{\mathrm{C}}^1=\frac{1}{\sqrt{4 \times 10^{-4}}}=50 & \end{array}$

Hence, the correct answer is option: 'D'.

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