In the first order reaction the concentration of the reactant is reduced \(\frac{1}{4}\)th in 60 minutes, what will be its half-life?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 30 minutes
For a first order reaction the half-life is constant: in each half-life the concentration falls to half, whatever its starting value. Falling to one fourth means two half-lives:
\( \displaystyle [R]_0 \xrightarrow{t_{1/2}} \frac{[R]_0}{2} \xrightarrow{t_{1/2}} \frac{[R]_0}{4} \)
\( \displaystyle 2\,t_{1/2} = 60\ \mathrm{min} \;\Rightarrow\; t_{1/2} = 30\ \mathrm{min} \)
The same result from the integrated rate equation:
\( \displaystyle k = \frac{2.303}{60}\log 4 = \frac{2.303 \times 0.602}{60} = 0.0231\ \mathrm{min^{-1}} \)
\( \displaystyle t_{1/2} = \frac{0.693}{0.0231} = 30\ \mathrm{min} \)