Chemistry - Chemical Kinetics Question with Solution | TestHub

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

$k=\frac{0.693}{t_{1/2}}$

Substituting the given half-life $t_{1/2}=13.86$ min:

$k=\frac{0.693}{13.86}=0.05{\text{ min}}^{-1}$

The time $t$ required for the reactant to be reduced to $20\%$ of its initial concentration is calculated using the first order integrated rate equation:

$t=\frac{2.303}{k}\log \left(\frac{[A{]}_0}{[A]}\right)$

Given that the reactant is reduced to $20\%$, $[A]=0.20[A{]}_0$.

$t=\frac{2.303}{0.05}\log \left(\frac{100}{20}\right)$

$t=\frac{2.303}{0.05}\log 5$

Using $\log 5=0.699$:

$t=46.06\times 0.699=32.19\text{ min}\approx 32.2\text{ min}$

Answer: $32.2$ min.