The half-life of a substance in a first-order reaction is 15 minutes.…

The correct answer is Option (2) → 4.62 × 10^–2 min^–1

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 \( t=t_{1/2} \), \( [R]=\frac{[R]_0}{2} \), so the ratio inside the log is 2:

    \( \displaystyle k=\frac{2.303}{t_{1/2}}\log 2=\frac{2.303\times 0.3010}{t_{1/2}} \)

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

The half-life of a first-order reaction does not depend on the initial concentration, so k follows from \( t_{1/2} \) alone. With \( t_{1/2}=15\,\mathrm{min} \):

    \( \displaystyle k=\frac{0.693}{15\,\mathrm{min}} \)

    \( \displaystyle k=0.0462\,\mathrm{min^{-1}}=4.62\times10^{-2}\,\mathrm{min^{-1}} \)