Decomposition of A is a first order reaction at T ( K ) and is given by A ( g ) rightarrow B ( g )+ C ( g ) . In a closed 1 L vessel, 1 bar A ( g ) is allowed to decompose at T ( K ) . After 100 minutes, the total pressure was 1.5 bar . What is the rate constant (in min -1 ) of the reaction ? ( log 2=0.3)

Chemistry · Chemical Kinetics

JEE Main 2026 — 21 January, Evening Shift — Question 53

Decomposition of A is a first order reaction at $\mathrm{T}(\mathrm{K})$ and is given by $\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ . In a closed 1 L vessel, 1 bar $\mathrm{A}(\mathrm{g})$ is allowed to decompose at $\mathrm{T}(\mathrm{K})$ . After 100 minutes, the total pressure was 1.5 bar . What is the rate constant (in $\min ^{-1}$ ) of the reaction ? ( $\log 2=0.3)$

Option A:
$6.9 \times 10^{-1}$
Option B:
$6.9 \times 10^{-3}$
Correct
Option C:
$6.9 \times 10^{-2}$
Option D:
$6.9 \times 10^{-4}$

Answer: B

Step-by-step solution

$\mathrm{A}_{\mathrm{g}} \rightarrow \mathrm{B}_{\mathrm{g}}+\mathrm{C}_{\mathrm{g}}$ 1 - - 1-P P P $\mathrm{P}_{\text {total }}=1+\mathrm{P}$ $1.5=1+\mathrm{P}$ $\mathrm{P}=0.5$ $\mathrm{K}=\frac{1}{100} \ln \frac{1}{0.5}$ $=\frac{0.693}{100}$ $=6.9 \times 10^{-3} \mathrm{~min}^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws