75% of a first-order reaction was completed in 32 min, in how much time was 50% of the reaction completed?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 16 min
For a first order reaction the half-life depends only on the rate constant k:
\( \displaystyle t_{1/2}=\frac{0.693}{k} \)
So for a first order reaction the half-life is the same whatever the starting concentration. Each half-life removes half of whatever reactant is left.
When 75% of the reaction is complete, 25% of the reactant is left:
\( \displaystyle 100\% \xrightarrow{t_{1/2}} 50\% \xrightarrow{t_{1/2}} 25\% \)
So 75% completion takes two half-lives.
Given t75% = 32 min:
\( \displaystyle 2\,t_{1/2}=32\,\mathrm{min} \)
\( \displaystyle t_{1/2}=\frac{32}{2}=16\,\mathrm{min} \)
50% of the reaction is completed in 16 min.
NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2. … It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa.”