A reaction is found to have a rate constant, k = 5.5 × 10–14 s–1. The…

The correct answer is Option (1) → 1.26 × 10¹³s

The unit of the rate constant tells the order of a reaction. For Rate = k[A]ⁿ, the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹, so a rate constant in s⁻¹ belongs to a first order reaction.

For a first order reaction the integrated rate equation is

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

At the half-life, [R] = [R]₀/2, so

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

    \( \displaystyle t_{1/2}=\frac{0.693}{k} \)

The half-life of a first order reaction does not depend on the initial concentration. Here k = 5.5 × 10⁻¹⁴ s⁻¹:

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

    \( \displaystyle t_{1/2}=0.126\times10^{14}\ \mathrm{s} \)

    \( \displaystyle t_{1/2}=1.26\times10^{13}\ \mathrm{s} \)

So the half-life of the reaction is 1.26 × 10¹³ s.

Shortcut: 0.693/5.5 ≈ 0.126, which fixes the digits 1.26. Dividing by 10⁻¹⁴ multiplies by 10¹⁴, and 0.126 × 10¹⁴ = 1.26 × 10¹³ s.

Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Half-Life of a Reaction – …

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: