Question

\text{Time required for completion of 99.9\% of a first order reaction is \\_\\_\\_\\_\\_\\_ times of half life } (t_{1/2}) \text{ of the reaction.}

Answer 1

For a first-order reaction, the time required for a certain percentage of reaction completion can be calculated using the formula:

$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]},$

where $[A]_0$ is the initial concentration, $[A]$ is the concentration at time $t$, and $k$ is the rate constant.

For 99.9% completion, $\frac{[A]}{[A]_0} = 0.001$:

$t = \frac{2.303}{k} \log \frac{1}{0.001} = \frac{2.303}{k} \log(10^3) = \frac{2.303}{k} \times 3 = 10 \times t_{1/2}.$

Thus, the time required for 99.9% completion is 10 times the half-life.
The Correct answer is: 10