JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Equilibrium

Consider the reaction

X_{2} Y (g) ⇌ X_{2} (g) + \frac{1}{2} Y_{2} (g)

The equation representing correct relationship between the degree of dissociation (x) of

X_{2} Y (g)

with its equilibrium constant

K_{p}

is ______.
Assume x to be very very small.

Answer: (B)

Solution

For

X X_{2} Y (g) X X_{2} (g) + \frac{1}{2} Y X_{2} (g)

, start with 1 mol of

X X_{2} Y

at total pressure

P

and degree of dissociation

x

. At equilibrium:

X X_{2} Y = 1 - x

X X_{2} = x

Y X_{2} = x /2

, total moles

= 1 + x /2

. The partial pressures are

P_{X_{2} Y} = \frac{( 1 - x ) P}{1 + x /2}

P_{X_{2}} = \frac{x P}{1 + x /2}

P_{Y_{2}} = \frac{( x /2 ) P}{1 + x /2}

, so

K_{p} = \frac{P _{X_{2}} P _{Y_{2}}^{1/2}}{P _{X_{2} Y}} = (\frac{x}{1 - x}) (\frac{x}{2 ( 1 + x /2 )})^{1/2} P^{1/2}

For

x ≪ 1

1 - x \approx 1

and

1 + x /2 \approx 1

, so

K_{p} \approx \frac{x ^{3/2}}{2 ^{1/2}} P^{1/2}

; squaring gives

K_{p}^{2} = \frac{x ^{3} P}{2}

, so

x = 3 \frac{2 K _{p}^{2}}{P}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option