The half-life periods of two first-order reactions are in the ratio 3: 2. If t₁ is the time required for 25% completion of the first reaction and t₂ is the time required for 75% completion of the second reaction, then the ratio, t₁ : t₂, is (log 3 = 0.48, log 2 = 0.3)

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3 : 5

3 : 2

3 : 10

12 : 25

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{t_{1}}{t_{2}} = \frac{\frac{{(t_{1 / 2})}_{1}}{\log 2} \log (\frac{100}{75})}{\frac{{(t_{1 / 2})}_{2}}{\log_{2}} \cdot \log (\frac{100}{25})} = \frac{{(t_{1 / 2})}_{1} \log (\frac{4}{3})}{{(t_{1 / 2})}_{2} \log (4)} \\ = \frac{3}{2} \times \frac{0.6 - 0.48}{0.6} = \frac{3}{10} \end{array}$