$$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g})$$ The correct relationship between $$\mathrm{K}_{\mathrm{P}}$$, $$\alpha$$ and equilibrium pressure P is

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

(A) $$\mathrm{K}_{\mathrm{P}}=\frac{\alpha^{1 / 2} \mathrm{P}^{1 / 2}}{(2+\alpha)^{1 / 2}}$$

(C) $$\mathrm{K}_{\mathrm{P}}=\frac{\alpha^{1 / 2} \mathrm{P}^{3 / 2}}{(2+\alpha)^{3 / 2}}$$

(D) $$\mathrm{K}_{\mathrm{P}}=\frac{\alpha^{1 / 2} \mathrm{P}^{1 / 2}}{(2+\alpha)^{3 / 2}}$$

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Equilibrium

A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)

The correct relationship between

K_{P}

α

and equilibrium pressure P is

Answer: (B)

Let

α

be the degree of dissociation of A at total equilibrium pressure P.

At equilibrium: moles of

A = (1 - α)

B = α

C = α /2

; total moles

= 1 + \frac{α}{2} = \frac{2 + α}{2}

.

Partial pressures:

P_{A} = \frac{2 ( 1 - α )}{2 + α} P

P_{B} = \frac{2 α}{2 + α} P

P_{C} = \frac{α}{2 + α} P

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

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option