At $60^{\circ}$ and $1 \mathrm{~atm}, \mathrm{~N}_2 \mathrm{O}_4$ is $50 \%$ dissociated into…

At $60^{\circ}$ and $1 \mathrm{~atm}, \mathrm{~N}_2 \mathrm{O}_4$ is $50 \%$ dissociated into $\mathrm{NO}_2$ then $K_p$ is
  1. $1.33 \mathrm{~atm}$
  2. $2 \mathrm{~atm}$
  3. $2.67 \mathrm{~atm}$
  4. $3 \mathrm{~atm}$

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 $\alpha=\frac{1}{2}$ $K_p=\frac{p_{\mathrm{NO}_2}^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{\left(2 \times \frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)}=2 \mathrm{~atm}$

Asked in: NEET 2005

Practice more Chemical Equilibrium questions on Aicharya