If the half-life (t1/2) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction
is closest to :

The correct option is 10 minutes.

Here is the step-by-step derivation to find the time required for 99.9% completion of a first-order reaction:

For a first-order reaction, the relation between the rate constant ($k$) and the half-life ($t_{1/2}$) is given by the formula:
$t_{1/2}=\frac{\ln \left(2\right)}{k}\approx \frac{0.693}{k}$

Given that the half-life ($t_{1/2}$) is 1 minute, we can write the rate constant as:
$k=\frac{0.693}{1\text{min}}=0.693{\text{min}}^{-1}$

The integrated rate equation for a first-order reaction is expressed as:
$t=\frac{2.303}{k}\log \left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)$
where:
• ${\left[A\right]}_0$ is the initial concentration of the reactant.
• ${\left[A\right]}_t$ is the concentration remaining at time $t$.

For 99.9% completion of the reaction:
• Let the initial concentration ${\left[A\right]}_0=100$.
• The remaining concentration at time $t$ will be: ${\left[A\right]}_t=100-99.9=0.1$.

Substituting these values into the first-order rate equation:
$t_{99.9\%}=\frac{2.303}{k}\log \left(\frac{100}{0.1}\right)$

Simplify the term inside the logarithm:
$\frac{100}{0.1}=1000={10}^3$

Thus:
$\log \left({10}^3\right)=3$

Substituting this back into the equation for $t$:
$t_{99.9\%}=\frac{2.303}{0.693}\times 3$

Calculate the final value:
$t_{99.9\%}\approx 3.32\times 3=9.96\text{minutes}$

Since 9.96 minutes is closest to 10 minutes, the time required for 99.9% completion is closest to 10 minutes.