Chemical EquilibriumHard

Question

For the reaction N2O4(g)$\rightleftharpoons$ 2NO2(g), the relation between the degree of dissociation of N2O4(g) at pressure, P with its equilibrium constant KP is

Options

A.$\alpha = \frac{K_{p}/P}{4 + K_{p}/P}$
B.$\alpha = \frac{K_{p}}{4 + K_{p}}$
C.$\alpha = \left\lbrack \frac{K_{p}/P}{4 + K_{p}/P} \right\rbrack^{1/2}$
D.$\alpha = \left\lbrack \frac{K_{p}}{4 + K_{p}} \right\rbrack^{1/2}$

Solution

N2O4 $\rightleftharpoons$ 2NO2

Equilibrium 1 – α 2α

$K_{p} = \frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}} = \frac{\left( \frac{2\alpha}{1 + \alpha}.P \right)^{2}}{\left( \frac{1 - \alpha}{1 + \alpha}.P \right)} = \frac{4\alpha^{2}.P}{1 - \alpha^{2}} $$$\therefore\alpha = \sqrt{\frac{K_{p}/4p}{1 + \frac{K_{p}}{4p}}}$$

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