Q70 / 180 · AIIMS 2012 · Chemistry
At $60^{\circ}$ and $1 \mathrm{~atm}, \mathrm{~N}_2 \mathrm{O}_4$ is $50 \%$ dissociated into $\mathrm{NO}_2$ then $K_p$ is
- A$1.33 \mathrm{~atm}$
- Correct answerB$2 \mathrm{~atm}$
- C$2.67 \mathrm{~atm}$
- D$3 \mathrm{~atm}$
Solution
$\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2$
$\begin{array}{lcc}\text { Initially } & 1 & 0 \\ \text { At equilibrium } & 1-\alpha & 2 \alpha\end{array}$
$\mathrm{N}_2 \mathrm{O}_4$ is $50 \%$ dissociated, so $\alpha=\frac{1}{2}$
$K_p=\frac{p_{\mathrm{NO}_2}^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{\left(2 \times \frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)}=2 \mathrm{~atm}$
Hence, the correct answer is option: 'B'.