First order gas phase reaction $\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$ $p_{i}=$ initial pressure of gas $A, p_{t}=$ total pressure of the reaction mixture at time t Expression of rate constant ( k ) is

(A) $\frac{1}{t} \ln \frac{p_{i}}{2 p_{i}-p_{t}}$

(B) $\frac{1}{t} \ln \frac{2 p_{i}}{p_{i}-p_{t}}$

(C) $\frac{1}{t} \ln \frac{p_{i}}{3 p_{i}-2 p_{t}}$

(D) $\frac{1}{t} \ln \frac{3 p_{i}}{4 p_{i}-p_{t}}$

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Chemical Kinetics

First order gas phase reaction

A \to B + C

p_{i} =

initial pressure of gas

A, p_{t} =

total pressure of the reaction mixture at time t
Expression of rate constant ( k ) is

Answer: (A)

Let

x

be the fall in the partial pressure of A after time

t

. For

A \to B + C

the partial pressures become

p_{i} - x

for A and

x

each for B and C, so the total pressure is

p_{t} = (p_{i} - x) + x + x = p_{i} + x

, which gives

x = p_{t} - p_{i}

and a remaining pressure of A equal to

p_{i} - x = 2 p_{i} - p_{t}

. Inserting this into the integrated first-order law gives

k = \frac{1}{t} ln \frac{p _{i}}{2 p _{i} - p _{t}}

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option