If the half life of a first order reaction is 6.93 minutes then the...

For a first-order reaction, the rate constant $k$ is given by:

$k = \dfrac{2.303 \log 2}{t_{1/2}}$

Using the standard approximation $2.303 \log 2 \approx 0.693$:

$k = \dfrac{0.693}{6.93} = 0.1 \text{ min}^{-1}$

The time required for $99\%$ completion of the reaction is given by the integrated rate law:

$t = \dfrac{2.303}{k} \log \left( \dfrac{a}{a - x} \right)$

Here, the initial concentration $a = 100$ and the amount reacted $x = 99$, so $a - x = 1$.

$t = \dfrac{2.303}{0.1} \log \left( \dfrac{100}{1} \right)$

$t = 23.03 \times \log(10^2)$

$t = 23.03 \times 2 = 46.06 \text{ minutes}$

Answer: $46.06$