The value of $$K_p$$ for the equilibrium reaction N$$_2$$O$$_4(g) \rightleftharpoons$$ 2NO$$_2(g)$$ is $$2$$. The percentage dissociation of N$$_2$$O$$_4(g)$$ at a pressure of $$0.5$$ atm is

For the dissociation equilibrium $$N_2O_4(g) \rightleftharpoons 2\,NO_2(g)$$ the equilibrium constant in terms of pressure is

$$K_p=\dfrac{(P_{NO_2})^{\,2}}{P_{N_2O_4}} \qquad -(1)$$

Assume we start with 1 mol of pure $$N_2O_4$$ in a vessel whose total pressure is kept at $$0.5\;\text{atm}$$. If the fraction dissociated is $$\alpha$$, then at equilibrium

  • moles of $$N_2O_4 = 1-\alpha$$
  • moles of $$NO_2 = 2\alpha$$
  • total moles $$n_{\text{tot}} = 1-\alpha+2\alpha = 1+\alpha$$

The partial pressures are obtained from mole-fraction × total pressure:

$$P_{NO_2}= \dfrac{2\alpha}{1+\alpha}\times 0.5 = \dfrac{\alpha}{1+\alpha}$$
$$P_{N_2O_4}= \dfrac{1-\alpha}{1+\alpha}\times 0.5 = \dfrac{0.5\,(1-\alpha)}{1+\alpha}$$

Substitute these in equation $$-(1)$$ and set $$K_p=2$$:

$$2 = \dfrac{\left(\dfrac{\alpha}{1+\alpha}\right)^{2}} {\dfrac{0.5\,(1-\alpha)}{1+\alpha}} = \dfrac{\alpha^{2}}{(1+\alpha)^{2}}\, \dfrac{1+\alpha}{0.5\,(1-\alpha)} = \dfrac{2\alpha^{2}}{1-\alpha^{2}}$$

Cancel the factor 2 on both sides:

$$\dfrac{\alpha^{2}}{1-\alpha^{2}} = 1 \;\;\Longrightarrow\;\; \alpha^{2}=1-\alpha^{2} \;\;\Longrightarrow\;\; 2\alpha^{2}=1 \;\;\Longrightarrow\;\; \alpha=\sqrt{\dfrac{1}{2}}\approx 0.707$$

Percentage dissociation = $$\alpha\times100\% \approx 70.7\% \approx 71\%$$.

Hence the correct choice is
Option D which is: $$71$$.