JEE Main 2025 Apr 2 Shift 2, Chemistry Q17: Factors Affecting Equilibria

JEE Main2025Apr 2, Shift 2Chemistry

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature :

$A (g) ⇌ B (g) + C (g)$

If p being the total pressure, $K_{p}$ is the pressure equilibrium constant and $α$ is the degree of dissociation, then which of the following is true at equilibrium ?

A
If p value is extremely high compared to $K_{p}$ , $α \approx 1$
B
When p increases $α$ decreases
C
If $k_{p}$ value is extremely high compared to p, $α$ becomes much less than unity
D
When p increases $α$ increases

Solution

Starting with 1 mole of A and a total of $(1 + α)$ moles at equilibrium, the mole fractions are $\frac{1 - α}{1 + α}, \frac{α}{1 + α}, \frac{α}{1 + α}$ for A, B, C respectively.

Multiplying by total pressure $p$ and substituting into the expression for $K_{p}$ :

$K_{p} = \frac{P _{B} \cdot P _{C}}{P _{A}} = \frac{α ^{2} p}{1 - α ^{2}}$ .

At fixed temperature $K_{p}$ is constant, so increasing $p$ must be compensated by decreasing $\frac{α ^{2}}{1 - α ^{2}}$ , i.e. $α$ must decrease.

Physically this is Le Chatelier's principle: the forward step increases moles of gas ( $Δ n = + 1$ ), so higher pressure shifts equilibrium backward, lowering $α$ .

$∴$ When $p$ increases, $α$ decreases.

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JEE Main 2025 Apr 2 Shift 2, Chemistry Q17: Factors Affecting Equilibria

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