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

Correct Answer: 10

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