The reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ was initiated with the amount ' a ' of $\mathrm{A}(\mathrm{g})$. At equilibrium it is found that the amount of $\mathrm{A}(\mathrm{g})$ remaining is $(\mathrm{a}-\mathrm{x})$ at a total pressure of p. The equilibrium constant $\mathrm{K}_{\mathrm{p}}$ of the reaction can be calculated from the expression :

(B) $\frac{x^{2}}{a^{2}-x^{2}} \times p$

(A) $\frac{x^{2}}{a^{2}+x^{2}} \times p$

(C) $\frac{a+x^{2}}{x^{2}} \times p$

(D) $\frac{a-x^{2}}{x^{2}} \times p$

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Equilibrium

The reaction

A (g) ⇌ B (g) + C (g)

was initiated with the amount ' a ' of

A (g)

. At equilibrium it is found that the amount of

A (g)

remaining is

(a - x)

at a total pressure of p.
The equilibrium constant

K_{p}

of the reaction can be calculated from the expression :

Answer: (B)

Starting from amount

a

of A and losing

x

to reaction, equilibrium holds

a - x

of A with

x

each of B and C, a total of

a + x

. The partial pressures are

P_{A} = \frac{a - x}{a + x} p

and

P_{B} = P_{C} = \frac{x}{a + x} p

. Substituting into

K_{p} = \frac{P _{B} P _{C}}{P _{A}}

gives

K_{p} = \frac{( \frac{x}{a + x} p ) ^{2}}{\frac{a - x}{a + x} p} = \frac{x ^{2} p}{( a + x ) ( a - x )}

. The equilibrium constant is

K_{p} = \frac{x ^{2}}{a ^{2} - x ^{2}} \times p

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option