t_99.9% with respect to t_90% for a first order reaction is:

The correct answer is Option (3) → Three

For a first order reaction the time for a given fraction to react depends only on the ratio [R]₀/[R]:

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

    \( \displaystyle t_{90\%}:\ [R] = 0.1[R]_0,\quad t_{90\%} = \frac{2.303}{k}\log 10 = \frac{2.303}{k} \)

    \( \displaystyle t_{99.9\%}:\ [R] = 0.001[R]_0,\quad t_{99.9\%} = \frac{2.303}{k}\log 1000 = 3 \times \frac{2.303}{k} \)

    \( \displaystyle \frac{t_{99.9\%}}{t_{90\%}} = \frac{\log 1000}{\log 10} = 3 \)

So t_99.9% is three times t_90%.