The reaction A ( g ) rightleftharpoons 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 :

Chemistry · Chemical Equilibrium

JEE Main 2026 — 5 April, Evening Shift — Question 54

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 :

Option A:
$\frac{x^{2}}{a^{2}+x^{2}} \times p$
Option B:
$\frac{x^{2}}{a^{2}-x^{2}} \times p$
Correct
Option C:
$\frac{a+x^{2}}{x^{2}} \times p$
Option D:
$\frac{a-x^{2}}{x^{2}} \times p$

Answer: B

Step-by-step solution

$\quad \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ $\begin{array}{lccl}t=0 & a & 0 & 0 t_{e q} & a-x & x & x\end{array}$ $\mathrm{K}_{\mathrm{p}}=\frac{(\mathrm{x})(\mathrm{x})}{(\mathrm{a}-\mathrm{x})} \times\left(\frac{\mathrm{P}}{\mathrm{a}+\mathrm{x}}\right)$ $K_{P}=\frac{x^{2} P}{a^{2}-x^{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient