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

The correct answer is Option (4) → Double

For a first order reaction the time needed to reach a given fraction of completion does not depend on the starting concentration:

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

When 90% has reacted, 10% is left, so [R]₀/[R] = 10:

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

When 99% has reacted, 1% is left, so [R]₀/[R] = 100:

    \( \displaystyle t_{99\%} = \frac{2.303}{k}\log 100 = \frac{2 \times 2.303}{k} = 2\,t_{90\%} \)

So t_99% is double t_90%.

Shortcut: each extra “9” after the decimal point (90% → 99% → 99.9%) adds one more t_90%, because log 10, log 100 and log 1000 are 1, 2 and 3. So t_99% = 2 t_90% and t_99.9% = 3 t_90%.