Question

The degree of dissociation \(\alpha\) of the reaction \(N_2O_4(g) \rightleftharpoons 2NO_2(g)\) can be related to \(K_p\) as

• A. \(\displaystyle \alpha = \frac{K_p}{4 + \dfrac{K_p}{P}}\)
• B. \(\displaystyle \frac{K_p}{4 + K_p}\)
• C. \(\displaystyle \left[\frac{\dfrac{K_p}{P}}{4 + \dfrac{K_p}{P}}\right]^{1/2}\)
• D. \(\displaystyle \alpha = \left(\frac{K_p}{4 + K_p}\right)^{1/2}\)

Hint:

For dissociation reactions of the type: \[ A \rightleftharpoons 2B \] the relation between \(K_p\), pressure, and degree of dissociation usually involves a square-root term.

Answer 1

The Correct Option is C

Step 1: Consider 1 mole of \(N_2O_4\) initially. Let the degree of dissociation be \(\alpha\). \[ \begin{array}{c|c|c|c} \text{Species} & N_2O_4 & NO_2
\hline \text{Initial (mol)} & 1 & 0
\text{Change (mol)} & -\alpha & +2\alpha
\text{Equilibrium (mol)} & 1-\alpha & 2\alpha \end{array} \]
Step 2: Total moles at equilibrium: \[ n_{\text{total}} = (1-\alpha) + 2\alpha = 1+\alpha \]
Step 3: Partial pressures: \[ P_{NO_2} = \frac{2\alpha}{1+\alpha}P, \quad P_{N_2O_4} = \frac{1-\alpha}{1+\alpha}P \]
Step 4: Expression for equilibrium constant: \[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \] \[ K_p = \frac{\left(\dfrac{2\alpha P}{1+\alpha}\right)^2}{\dfrac{(1-\alpha)P}{1+\alpha}} \] \[ K_p = \frac{4\alpha^2 P}{1-\alpha^2} \]
Step 5: Rearranging: \[ \alpha^2 = \frac{\dfrac{K_p}{P}}{4 + \dfrac{K_p}{P}} \] \[ \alpha = \left[\frac{\dfrac{K_p}{P}}{4 + \dfrac{K_p}{P}}\right]^{1/2} \]