Question

For the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), the correct relation between degree of dissociation (\(\alpha\)) of \( \text{N}_2\text{O}_4(g) \) and equilibrium constant, \( K_p \) is (P = total pressure of mixture)

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

Hint:

For a dissociation reaction of type \( A \rightleftharpoons nB \), the term \( 1-\alpha^2 \) often appears in the denominator. Rearranging \( K_p = \frac{4\alpha^2 P}{1-\alpha^2} \) is a standard procedure.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to derive the expression for the equilibrium constant \( K_p \) in terms of partial pressures, which depend on the mole fractions (determined by the degree of dissociation \(\alpha\)) and the total pressure \( P \).
Step 2: Key Formula or Approach:
1. Reaction: \( \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2 \). 2. Let initial moles be 1. At equilibrium: - \( n_{\text{N}_2\text{O}_4} = 1 - \alpha \) - \( n_{\text{NO}_2} = 2\alpha \) - Total moles \( n_{total} = 1 + \alpha \). 3. Partial Pressure \( p_i = \left( \frac{n_i}{n_{total}} \right) P \). 4. \( K_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} \).
Step 3: Detailed Explanation:
Calculate partial pressures: \[ p_{\text{NO}_2} = \left( \frac{2\alpha}{1+\alpha} \right) P \] \[ p_{\text{N}_2\text{O}_4} = \left( \frac{1-\alpha}{1+\alpha} \right) P \] Substitute these into the \( K_p \) expression: \[ K_p = \frac{\left[ \frac{2\alpha P}{1+\alpha} \right]^2}{\left[ \frac{(1-\alpha)P}{1+\alpha} \right]} \] \[ K_p = \frac{4\alpha^2 P^2}{(1+\alpha)^2} \times \frac{1+\alpha}{(1-\alpha)P} \] \[ K_p = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1-\alpha^2} \] Now, solve for \( \alpha \): \[ K_p (1 - \alpha^2) = 4\alpha^2 P \] \[ K_p - K_p \alpha^2 = 4\alpha^2 P \] \[ K_p = \alpha^2 (4P + K_p) \] \[ \alpha^2 = \frac{K_p}{4P + K_p} \] Divide numerator and denominator by \( P \): \[ \alpha^2 = \frac{K_p/P}{4 + K_p/P} \] Taking the square root: \[ \alpha = \left( \frac{K_p/P}{4 + K_p/P} \right)^{1/2} \]
Step 4: Final Answer:
The relation is \( \alpha = \left( \frac{K_p/P}{4 + K_p/P} \right)^{1/2} \).