If a first order reaction takes 15 minutes for 90% completion, then t…

The correct answer is Option (3) → 4.5 minutes

For a first order reaction, the time to reach any fraction of completion depends only on the rate constant, not on the starting concentration:

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

For 90% completion, 10% of the reactant is left, so [R]₀/[R] = 100/10 = 10:

    \( \displaystyle 15=\frac{2.303}{k}\log10=\frac{2.303}{k} \)

    \( \displaystyle k=\frac{2.303}{15}\ \mathrm{min^{-1}} \)

For 50% completion, [R]₀/[R] = 2, which is the half-life:

    \( \displaystyle t_{1/2}=\frac{2.303}{k}\log2=\frac{2.303\times0.3010}{k} \)

    \( \displaystyle t_{1/2}=15\times0.3010 \)

    \( \displaystyle t_{1/2}=4.5\ \mathrm{min} \)

So 50% completion takes 4.5 minutes.

Shortcut: Since t_90% = 2.303/k and t_50% = 2.303 log 2/k, the ratio is log 2: t_50% = 0.301 × t_90% = 0.301 × 15 ≈ 4.5 min.