Rate constant of a first order reaction is 2.5 × 10−3 s−1. Its half l…

The correct answer is Option (2) → 277.2 s

The half-life (t_1/2) of a reaction is the time in which the concentration of a reactant falls to half its initial value. For a first order reaction, the integrated rate equation is:

    \( \displaystyle k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} \)

At t = t_1/2, [R] = [R]₀/2, so the ratio [R]₀/[R] = 2:

    \( \displaystyle k=\frac{2.303}{t_{1/2}}\log 2 \)

    \( \displaystyle t_{1/2}=\frac{2.303\times0.3010}{k}=\frac{0.693}{k} \)

So the half-life of a first order reaction is independent of the initial concentration.

Here k = 2.5 × 10⁻³ s⁻¹:

    \( \displaystyle t_{1/2}=\frac{0.693}{2.5\times10^{-3}\ \mathrm{s^{-1}}} \)

    \( \displaystyle t_{1/2}=0.2772\times10^{3}\ \mathrm{s}=277.2\ \mathrm{s} \)

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Half-Life of a Reaction – “… A first order reaction is found to have a rate constant, k = 5.5 × 10⁻¹⁴ s⁻¹. Find the half-life of the reaction.

Half-life for a first order reaction is

    \( \displaystyle t_{1/2}=\frac{0.693}{k}=\frac{0.693}{5.5\times10^{-14}\ \mathrm{s^{-1}}}=1.26\times10^{13}\ \mathrm{s} \)”