75% of a first order reaction was completed in 16 minutes. When was i…

The correct answer is Option (1) → 8 min

For a first order reaction the time needed for a given fraction to react does not depend on the starting concentration:

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

After 75% completion, 25% remains, so \( [\mathrm{R}] = 0.25\,[\mathrm{R}]_0 \) and

    \( \displaystyle \log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} = \log\frac{1}{0.25} = -\log 0.25 = 0.602 \)

    \( \displaystyle k = \frac{2.303 \times 0.602}{16\ \mathrm{min}} = 0.0866\ \mathrm{min^{-1}} \)

The half-life of a first order reaction is

    \( \displaystyle t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0866\ \mathrm{min^{-1}}} = 8\ \mathrm{min} \)

So the reaction was half completed after 8 minutes.

Shortcut: 75% completion leaves one-fourth, which is two half-lives (100% → 50% → 25%). So \( t_{1/2} = 16/2 = 8 \) min, without any logarithm.