For a reaction at equilibrium $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\frac{1}{2} \mathrm{C}(\mathrm{g})$ the relation between dissociation constant (K), degree of dissociation ($\alpha$) and equilibrium pressure (p) is given by :

(B) $\mathrm{K}=\frac{\alpha^{\frac{3}{2}} \mathrm{p}^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$

(A) $\mathrm{K}=\frac{\alpha^{\frac{1}{2}} \mathrm{p}^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}$

(C) $\mathrm{K}=\frac{(\alpha \mathrm{p})^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}$

(D) $K=\frac{(\alpha p)^{\frac{3}{2}}}{(1+\alpha)(1-\alpha)^{\frac{1}{2}}}$

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Equilibrium

For a reaction at equilibrium

A (g) ⇌ B (g) + \frac{1}{2} C (g)

the relation between dissociation constant (K), degree of dissociation (

α

) and equilibrium pressure (p) is given by :

(A) $K = \frac{α ^{\frac{1}{2}} p ^{\frac{3}{2}}}{( 1 + \frac{3}{2} α ) ^{\frac{1}{2}} ( 1 - α )}$
(B) $K = \frac{α ^{\frac{3}{2}} p ^{\frac{1}{2}}}{( 2 + α ) ^{\frac{1}{2}} ( 1 - α )}$ ✓ Correct
(C) $K = \frac{( α p ) ^{\frac{3}{2}}}{( 1 + \frac{3}{2} α ) ^{\frac{1}{2}} ( 1 - α )}$
(D) $K = \frac{( α p ) ^{\frac{3}{2}}}{( 1 + α ) ( 1 - α ) ^{\frac{1}{2}}}$

Answer: (B)

Starting with initial pressure

P_{i}

, at equilibrium:

P_{A} = P_{i} (1 - α)

P_{B} = P_{i} α

P_{C} = P_{i} \frac{α}{2}

. Total pressure

p = P_{i} (1 + \frac{α}{2})

, so

P_{i} = \frac{p}{1 + \frac{α}{2}} = \frac{2 p}{2 + α}

. Partial pressures:

P_{A} = \frac{2 p ( 1 - α )}{2 + α}

P_{B} = \frac{2 p α}{2 + α}

P_{C} = \frac{p α}{2 + α}

. The equilibrium constant

K_{p} = \frac{P _{B} \cdot P _{C}^{1/2}}{P _{A}} = \frac{\frac{2 p α}{2 + α} \cdot ( \frac{p α}{2 + α} ) ^{1/2}}{\frac{2 p ( 1 - α )}{2 + α}} = \frac{α ^{3/2} p ^{1/2}}{( 2 + α ) ^{1/2} ( 1 - α )}

. Answer: (B).

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option