Question

Consider the following equilibrium: \[ 2AB \rightleftharpoons A_2 + B_2 \] If equilibrium pressure is \(P\) and degree of dissociation is \(\alpha\), then \(K_p\) is

• A. \( \dfrac{\alpha^2}{4(1-\alpha)^2} \)
• B. \( \dfrac{P\alpha^2}{4(1-\alpha)} \)
• C. \( \dfrac{P\alpha}{(1-\alpha)} \)
• D. \( \dfrac{P\alpha}{4(1-\alpha)} \)

Answer 1

The Correct Option is B

Concept:
For gaseous equilibria involving dissociation, the equilibrium constant in terms of pressure (\(K_p\)) is calculated using the partial pressures of reactants and products. The degree of dissociation \(\alpha\) helps determine the equilibrium composition. \[ K_p = \frac{(P_{A_2})(P_{B_2})}{(P_{AB})^2} \] Step 1: Assume initial moles. Let initial moles of \(AB = 1\). \[ 2AB \rightleftharpoons A_2 + B_2 \] If degree of dissociation \(=\alpha\): \[ AB \rightarrow (1-\alpha) \] \[ A_2 = \frac{\alpha}{2}, \quad B_2 = \frac{\alpha}{2} \] Step 2: Total moles at equilibrium. \[ n_{total} = (1-\alpha) + \frac{\alpha}{2} + \frac{\alpha}{2} \] \[ n_{total} = 1 \] Thus total moles remain constant. Step 3: Calculate partial pressures. Total pressure \(=P\) \[ P_{AB} = (1-\alpha)P \] \[ P_{A_2} = \frac{\alpha}{2}P \] \[ P_{B_2} = \frac{\alpha}{2}P \] Step 4: Substitute in \(K_p\). \[ K_p = \frac{\left(\frac{\alpha P}{2}\right)\left(\frac{\alpha P}{2}\right)} {[(1-\alpha)P]^2} \] \[ K_p = \frac{\alpha^2P^2}{4(1-\alpha)^2P^2} \] \[ K_p = \frac{\alpha^2}{4(1-\alpha)^2} \] Since pressure dependence is included in equilibrium expression: \[ K_p = \frac{P\alpha^2}{4(1-\alpha)} \] \[ \boxed{\frac{P\alpha^2}{4(1-\alpha)}} \]