JEE Main 2026 — Apr 5, Shift 1 — Chemistry solution

JEE Main2026Apr 5, Shift 1Chemistry

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

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
$\frac{1}{t} ln \frac{3 p _{i}}{4 p _{i} - p _{t}}$

Solution

Let $x$ be the partial pressure of A consumed at time $t$ . Then $p_{A} = p_{i} - x$ , $p_{B} = p_{C} = x$ , and the total is $p_{t} = (p_{i} - x) + 2 x = p_{i} + x$ , so $x = p_{t} - p_{i}$ .

Partial pressure of A remaining: $p_{i} - x = p_{i} - (p_{t} - p_{i}) = 2 p_{i} - p_{t}$ .

First-order integrated law in pressure form: $k = \frac{1}{t} ln \frac{p _{i}}{2 p _{i} - p _{t}}$ .

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