JEE Main 2025 Apr 8 Shift 2, Chemistry Q1: Reactions of Various Orders

JEE Main2025Apr 8, Shift 2Chemistry

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_{1}$ and $t_{2}$ (s), respectively. The ratio $t_{1} / t_{2}$ will be:

A
$\frac{4}{3}$
B
$\frac{3}{2}$
C
$\frac{3}{4}$
D
$\frac{2}{3}$

Solution

For a first order reaction the half-life $t_{1/2}$ is independent of the initial concentration, so it can be used as a natural clock.

Reaching $C_{0} /4$ is equivalent to completing two successive halvings of concentration, so $t_{1} = 2 t_{1/2}$ .

Reaching $C_{0} /8$ requires three successive halvings, so $t_{2} = 3 t_{1/2}$ .

Taking the ratio: $\frac{t _{1}}{t _{2}} = \frac{2 t _{1/2}}{3 t _{1/2}} = \frac{2}{3}$ .

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q1: Reactions of Various Orders

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