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: \(log 2 = 0.3010\))

3.48 months

The correct answer is option 2. 3.48 months.

Given,

\(t_{75\%} = 6.96\, \ months\)

For a first-order reaction,

\(k = \frac{2.303}{t}log\frac{a}{a - x}\)

Since the reactant is reduced to \(25\%\), hence the reaction is \(75\%\) complete

Let the assume the amount of the initial reaction, \(a = 100\)

After \(75\%\) completion, \(a - x = 100 - 75 = 25\)

Thus, applying all the values, we get

\(k = \frac{2.303}{t_{75\%}}log\frac{100}{25}\)

or, \(k = \frac{2.303}{6.96}log\frac{100}{25}\)

or, \(k = \frac{2.303}{6.96} log 4\)

or, \(k = 0.331 × 0.602\)

or, \(k = 0.19922\)

The half-life of a first-order reaction is given by

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

or, \(t_{1/2} = \frac{0.693}{0.19922}\)

or, \(t_{1/2} = 3.478\)

or, \(t_{1/2} \approx 3.48\, \ months\)