If the half-life (t_1/2) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

The correct answer is 10 minutes.

Step-by-Step Derivation and Explanation:

1. Formula for First-Order Kinetics:
For a first-order chemical reaction, the rate constant $k$ is related to the half-life ($t_{1/2}$) by the following relation:
$k=\frac{\ln \left(2\right)}{t_{1/2}}\approx \frac{0.693}{t_{1/2}}$

Given that the half-life $t_{1/2}=1\text{minute}$:
$k=0.693{\text{min}}^{-1}$

2. Finding the Time Required for 99.9% Completion:
Let the initial concentration of the reactant be ${\mathrm{[A]}}_0=100$.
If the reaction is 99.9% complete, the amount of reactant consumed is 99.9.
Therefore, the remaining concentration of reactant at time $t$, denoted as ${\mathrm{[A]}}_t$, is:
${\mathrm{[A]}}_t=100-99.9=0.1$

The integrated rate equation for a first-order reaction is:
$t=\frac{2.303}{k}\log \left(\frac{{\mathrm{[A]}}_0}{{\mathrm{[A]}}_t}\right)$

Substitute the values into the equation:
$t=\frac{2.303}{0.693}\log \left(\frac{100}{0.1}\right)$
Since $\frac{100}{0.1}=1000={10}^3$ and $\log \left({10}^3\right)=3$:
$t=\frac{2.303}{0.693}\times 3$

3. Simplifying the Calculation:
Note that $2.303\times 3=6.909$.
Thus, the expression becomes:
$t=\frac{6.909}{0.693}\approx 9.97\text{minutes}$

This value is closest to 10 minutes.