Chemistry - Chemical Kinetics Question with Solution | TestHub

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

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

Given $t_{1/2}=20\text{ min}$,

$k=\frac{0.693}{20}{\text{ min}}^{-1}$

The time $t$ required for the concentration to reduce to $(1/10{)}^{\text{th}}$ of its initial value is:

$t=\frac{2.303}{k}{\log }_{10}\left(\frac{[A{]}_0}{[A{]}_t}\right)$

Substituting $[A{]}_t=\frac{[A{]}_0}{10}$ and $k=\frac{0.693}{20}$:

$t=\frac{2.303\times 20}{0.693}{\log }_{10}(10)$

$t=\frac{46.06}{0.693}\times 1$

$t=66.46\text{ min}$

Answer: 66.46 min