Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ If p being the total pressure, $\mathrm{K}_{\mathrm{p}}$ is the pressure equilibrium constant and $\alpha$ is the degree of dissociation, then which of the following is true at equilibrium?

(B) When $p$ increases $\alpha$ decreases

(A) If $p$ value is extremely high compared to $\mathrm{K}_{\mathrm{p}}, \alpha \approx 1$

(C) If $\mathrm{k}_{\mathrm{p}}$ value is extremely high compared to $\mathrm{p}, \alpha$ becomes much less than unity

(D) When $p$ increases $\alpha$ increases

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Equilibrium

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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✓ Correct
(C)If $k_{p}$ value is extremely high compared to $p, α$ becomes much less than unity
(D)When $p$ increases $α$ increases

Answer: (B)

Solution

For

A (g) B (g) + C (g)

starting from one mole of A, the equilibrium amounts are

1 - α

α

and

α

in a total of

1 + α

moles, which gives

K_{p} = \frac{α ^{2}}{1 - α ^{2}} p

. Rearranging,

α^{2} = \frac{K _{p}}{K _{p} + p}

. Since

K_{p}

is fixed at constant temperature, raising the total pressure

p

lowers

α

, consistent with Le Chatelier's principle that increased pressure favours the side with fewer gas moles. An increase in pressure decreases the degree of dissociation.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Equilibrium

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