t75% and t25% are the time required for 75% and 25% completion of the…

The correct answer is Option (4) → t_75% = 4.8 t_25%

For a first order reaction, the time needed for the reactant concentration to fall from [R]₀ to [R] is given by the integrated rate equation:

    \( \displaystyle t=\frac{2.303}{k}\log\frac{[R]_0}{[R]} \)

This time depends only on the fraction of reactant left, not on the starting concentration.

75% completion: 25% of the reactant is left, so [R] = [R]₀/4.

    \( \displaystyle t_{75\%}=\frac{2.303}{k}\log 4 \)

    \( \displaystyle \log 4=2\times0.3010=0.6020 \)

25% completion: 75% of the reactant is left, so [R] = 3[R]₀/4.

    \( \displaystyle t_{25\%}=\frac{2.303}{k}\log\frac{4}{3} \)

    \( \displaystyle \log\frac{4}{3}=0.6020-0.4771=0.1249 \)

Dividing the two times:

    \( \displaystyle \frac{t_{75\%}}{t_{25\%}}=\frac{0.6020}{0.1249}=4.82 \)

    \( \displaystyle t_{75\%}\approx4.8\,t_{25\%} \)

So the correct relationship is t_75% = 4.8 t_25%.

Shortcut: The factor 2.303/k is the same in both times and cancels, so only the ratio log 4 / log (4/3) is needed.