Chemistry · Chemical Equilibrium

JEE Main 2026 — 28 January, Evening Shift — Question 53

Observe the following equilibrium in a 1 L flask.

A(g) \rightleftharpoons B(g)

At $\mathrm{T}(\mathrm{K})$ , the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to $\mathrm{T}(\mathrm{K})$ to establish the equilibrium again. The new equilibrium concentrations (in M ) of A and B are respectively.

Option A:
$0.367,0.275$
Option B:
$0.53,0.4$
Option C:
$0.742,0.557$
Option D:
$0.557,0.418$
Correct

Answer: D

Step-by-step solution

A ⇌ B

0.5 \mathrm{M} \quad 0.375 \mathrm{M} \quad \text { (At equilibrium) }

$\mathrm{K}_{\mathrm{eq}}=\frac{[\mathrm{B}]_{\mathrm{eq}}}{[\mathrm{A}]_{\mathrm{eq}}}=\frac{0.375}{0.5}=0.75$ Now 0.1 mole of A is added so reaction will move in forward direction. A ⇌ B $0.6-\mathrm{x} \quad 0.375+\mathrm{x}$ $\mathrm{K}_{\mathrm{eq}}=0.75=\frac{0.375+\mathrm{x}}{0.6-\mathrm{x}}$ $0.45-0.75 \mathrm{x}=0.375+\mathrm{x}$ $1.75 \mathrm{x}=0.075$ $\mathrm{X}=\frac{0.075}{1.75}=\frac{3}{70}=0.043$ Moles of $\mathrm{A}=0.043=0.557$ Moles of $\mathrm{B}=0.418$ Ans. (4) is correct.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient