If the rate constant for the first order reaction is 2 ×10^-5 s^-1 ,…

The correct answer is Option (1) → 577.5 min

For a first order reaction the integrated rate law gives the half-life, the time for the concentration to fall to half its initial value:

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

With \( [\mathrm{R}] = [\mathrm{R}]_0/2 \):

    \( \displaystyle t_{1/2} = \frac{2.303 \log 2}{k} = \frac{0.693}{k} \)

The half-life of a first order reaction does not depend on the initial concentration. Substituting \( k = 2 \times 10^{-5}\,\mathrm{s^{-1}} \):

    \( \displaystyle t_{1/2} = \frac{0.693}{2 \times 10^{-5}\,\mathrm{s^{-1}}} = 34650\,\mathrm{s} \)

    \( \displaystyle t_{1/2} = \frac{34650}{60}\,\mathrm{min} = 577.5\,\mathrm{min} \)

So the half-life is 577.5 min.