For the first order reaction, the time required for the completion of…

The correct answer is Option (1) → t₉₀ = 3.3 t_1/2

For a first order reaction, the time needed for the concentration to fall from [R]₀ to [R] is:

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

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

Half-life (50% completion): [R] = [R]₀/2

    \( \displaystyle t_{1/2}=\frac{2.303}{k}\log 2=\frac{2.303\times0.3010}{k}=\frac{0.693}{k} \)

90% completion: 10% of the reactant is left, so [R] = [R]₀/10

    \( \displaystyle t_{90}=\frac{2.303}{k}\log 10=\frac{2.303}{k} \)

Dividing the two:

    \( \displaystyle \frac{t_{90}}{t_{1/2}}=\frac{2.303}{0.693}=3.32 \)

    \( \displaystyle t_{90}\approx3.3\,t_{1/2} \)

Shortcut: The factor 2.303/k cancels, so the ratio is log 10 / log 2 = 1/0.3010 = 3.32.