Question

The degree of dissociation $\alpha $ of the reaction
$N_{2}O_{4}\left(g\right)\rightleftharpoons 2NO_{2}\left(g\right)$
Can be related to $K_{p}$ as [Given: Total pressure at equilibrium $= P$]

• A. $\alpha =\frac{\frac{K_{P}}{P}}{4 + \frac{K_{P}}{P}}$
• B. $\alpha =\frac{K_{P}}{4 + K_{P}}$
• C. $\alpha =\left[\frac{K_{P} / P}{4 + K_{P} / P}\right]^{1 / 2}$
• D. $\alpha =\left[\frac{K_{P}}{4 + K_{P}}\right]^{1 / 2}$

Answer 1

Answer: (C)

$N_{2}O_{4}\left(g\right)\rightleftharpoons2NO_{2}\left(g\right)$

$K_{P}=\frac{\left(2 \alpha \right)^{2}}{\left(1 - \alpha \right)}\times \left(\frac{P}{1 + \alpha }\right)^{1}$
$K_{P}=\frac{4 \alpha ^{2}}{1 - \alpha ^{2}}P$
$\alpha =\left[\frac{K_{P} / P}{4 + K_{P} / P}\right]^{1 / 2}$