Chemistry · Chemical Kinetics

JEE Main 2025 — 7 April, Evening Shift — Question 5

$\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ is a first order reaction.

Time	t	$\infty$
$\mathrm{P}_{\text {system }}$	pt	$\mathrm{p}_{\infty}$

The reaction was started with reactant A only. Which of the following expression is correct for rate constant k?

Option A:
$k=\frac{1}{t} \ln \frac{p_{\infty}}{p_{t}}$
Option B:
$k=\frac{1}{t} \ln \frac{p_{\infty}}{\left(p_{\infty}-p_{t}\right)}$
Option C:
$k=\frac{1}{\mathrm{t}} \ln \frac{2\left(\mathrm{p}_{\infty}-\mathrm{p}_{\mathrm{t}}\right)}{\mathrm{p}_{\mathrm{t}}}$
Option D:
$k=\frac{1}{t} \ln \frac{p_{\infty}}{2\left(p_{\infty}-p_{t}\right)}$
Correct

Answer: D

A(g) \longrightarrow B(g) + C(g)

\begin{array}{lccc} t = 0 & p_0 & 0 & 0 \\ t = t & p_0 - p & p & p \\ t = \infty & 0 & p_0 & p_0 \end{array}

\text{At } t = t \quad p_t = p_0 + p \implies p = p_t - p_0

t = \infty \quad 2p_0 = p_\infty

\text{For first order reaction}

k = \frac{1}{t} \ln \frac{(p_0)_A}{(p_t)_A}

= \frac{1}{t} \ln \frac{p_0}{p_0 - p}

= \frac{1}{t} \ln \frac{p_0}{p_0 - (p_t - p_0)} = \frac{1}{t} \ln \frac{p_0}{2p_0 - p_t}

= \frac{1}{t} \ln \frac{\frac{1}{2} p_\infty}{p_\infty - p_t} = \frac{1}{t} \ln \frac{p_\infty}{2(p_\infty - p_t)}

p_0 - p = \frac{p_\infty}{2} - (p_t - p_0) = \frac{p_\infty}{2} - p_t + \frac{p_\infty}{2} = p_\infty - p_t

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Catalysis, Parallel and Sequential Reactions + Miscellaneous Problems