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Chemistry · Chemical Kinetics

JEE Main 2026 — 5 April, Morning Shift — Question 55

First order gas phase reaction A→B+C\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C} pi=p_{i}= initial pressure of gas A, pt=p_{t}= total pressure of the reaction mixture at time t

Expression of rate constant ( k ) is

  1. Option A:

    1tln⁡pi2pi−pt\frac{1}{t} \ln \frac{p_{i}}{2 p_{i}-p_{t}}

    Correct
  2. Option B:

    1tln⁡2pipi−pt\frac{1}{t} \ln \frac{2 p_{i}}{p_{i}-p_{t}}

  3. Option C:

    1tln⁡pi3pi−2pt\frac{1}{t} \ln \frac{p_{i}}{3 p_{i}-2 p_{t}}

  4. Option D:

    1tln⁡3pi4pi−pt\frac{1}{t} \ln \frac{3 p_{i}}{4 p_{i}-p_{t}}

Answer: A

Step-by-step solution

A(g)→B(g)+C(g)t=0Pi−−t=tPi−xxx\begin{array}{c c c c} \mathrm{A(g)} & \rightarrow & \mathrm{B(g)} & +\mathrm{C(g)} \\ \hline t=0 & P_i & - & - \\ t=t & P_i-x & x & x \end{array}

As given, (Pi−x)+x+x=Pt\left(P_i-x\right)+x+x=P_t

x=Pt−Pix=P_t-P_i

K=1tln⁡PiPi−xK=\frac{1}{t}\ln\frac{P_i}{P_i-x}

=1tln⁡PiPi−(Pt−Pi)=\frac{1}{t}\ln\frac{P_i}{P_i-\left(P_t-P_i\right)}

K=1tln⁡Pi2Pi−PtK=\frac{1}{t}\ln\frac{P_i}{2P_i-P_t}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws

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