A first-order reaction requires 6.96 months for the concentration of reactant \(A\) to be reduced to \(25\%\) of its original value. The half-life of the reaction is:
(Given: \(log 2 = 0.3010\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3.48 months
For a first order reaction the half-life is constant, independent of the initial concentration. Reduction to 25% means the reaction has gone through two half-lives (100% → 50% → 25%), so t1/2 = 6.96/2 = 3.48 months.
The same result from the integrated rate equation:
\( k = \displaystyle\frac{2.303}{t}\log\displaystyle\frac{[R]_0}{[R]} = \displaystyle\frac{2.303}{6.96}\log\displaystyle\frac{100}{25} = \displaystyle\frac{2.303 \times 0.602}{6.96}\ \text{month}^{-1} \)
\( t_{1/2} = \displaystyle\frac{0.693}{k} = \displaystyle\frac{2.303 \times 0.3010 \times 6.96}{2.303 \times 0.602} = \displaystyle\frac{6.96}{2} = 3.48\ \text{months} \)
NCERT (2026-27 edition):
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.”
Half-Life of a Reaction – “… For the first order reaction, k = (2.303/t) log [R]0/[R]; at t1/2, [R] = [R]0/2, so t1/2 = (2.303/k) log 2 = (2.303/k) × 0.301, t1/2 = 0.693/k (3.17). 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.”