JEE Main 2025 Jan 23 Shift 2, Chemistry Q5: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2025Jan 23, Shift 2Chemistry

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.

A
$x = 3 \frac{2 K _{p}}{p}$
B
$x = 3 \frac{2 K _{p}^{2}}{p}$
C
$x = 3 \frac{K _{p}}{2 p}$
D
$x = 3 \frac{K _{p}}{p}$

Solution

Start from 1 mol of $X_{2} Y$ at pressure $P$ . At equilibrium with dissociation $x$ :

$n (X_{2} Y) = 1 - x, n (X_{2}) = x, n (Y_{2}) = x /2$ . Total $n_{T} = 1 + x /2$ .

Partial pressures: $P_{X_{2} Y} = \frac{1 - x}{1 + x /2} P, P_{X_{2}} = \frac{x}{1 + x /2} P, P_{Y_{2}} = \frac{x /2}{1 + x /2} P$ .

$K_{p} = \frac{P _{X_{2}} P _{Y_{2}}^{1/2}}{P _{X_{2} Y}}$ . Substituting and simplifying for $x ≪ 1$ (so $1 + x /2 \approx 1$ and $1 - x \approx 1$ ):

$K_{p} \approx x \cdot (\frac{x}{2})^{1/2} P^{1/2} = \frac{x ^{3/2}}{2} P^{1/2}$

Squaring both sides: $K_{p}^{2} = \frac{x ^{3} P}{2}$ , hence $x^{3} = \frac{2 K _{p}^{2}}{P}$ and $x = 3 \frac{2 K _{p}^{2}}{P}$ .

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JEE Main 2025 Jan 23 Shift 2, Chemistry Q5: Law of Chemical Equilibrium And Equilibrium Constant

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