75% of a first order reaction was completed in 32 minutes. How long w…

The correct answer is Option (3) → 16 min

For a first order reaction, the integrated rate equation is:

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

The time for 50% completion is the half-life. For a first order reaction it does not depend on the initial concentration:

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

When 75% of the reaction is complete, 25% of the reactant is left. So [R] = [R]₀/4 and [R]₀/[R] = 4, with t = 32 min:

    \( \displaystyle k=\frac{2.303}{32\ \mathrm{min}}\log 4=\frac{2.303\times 2\times 0.3010}{32\ \mathrm{min}} \)

    \( \displaystyle k=\frac{1.386}{32}\ \mathrm{min^{-1}}=0.0433\ \mathrm{min^{-1}} \)

    \( \displaystyle t_{1/2}=\frac{0.693}{0.0433\ \mathrm{min^{-1}}}=16\ \mathrm{min} \)

So 50% completion takes 16 min.

Shortcut: In a first order reaction, each half-life halves the amount of reactant left: 100% → 50% → 25%. So 75% completion takes two half-lives, and t_1/2 = 32 min / 2 = 16 min, with no logarithms needed.