Verified 30 May 2026.
For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, if $K_p = 4 \text{ atm}$ at $300\text{K}$, the degree of dissociation ($\alpha$) at a total pressure of $1 \text{ atm}$ is:
$\sqrt{\frac{4}{5}}$
$\frac{2}{\sqrt{5}}$
$\frac{1}{2}$
$\sqrt{\frac{2}{3}}$
For $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$:
$$K_p = \frac{4\alpha^2 P}{1 - \alpha^2}$$
Substituting $K_p = 4$ and $P = 1$:
$$4 = \frac{4\alpha^2}{1 - \alpha^2}$$
$$1 - \alpha^2 = \alpha^2 \implies 2\alpha^2 = 1 \implies \alpha = \frac{1}{\sqrt{2}}$$
But $\frac{1}{\sqrt{2}} = \sqrt{\frac{1}{2}}$. Checking options: $\sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$. The answer is $\alpha = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
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