Question

A reaction has a half-life of $1$ min. The time required for $99.9\%$ completion of the reaction is _____ min. (Round off to the nearest integer) [ Use $\ln 2=0.69$, $\ln 10=2.3$]

Answer 1

Step 1: Integrated rate equation is given by

$\mathrm{t}=\frac{1}{\mathrm{k}}\ln \frac{{\mathrm{A}}_0}{A_t}$

Where,

$\mathrm{t}$ - time

$\mathrm{k}$ - rate constant

$A_0$ - initial concentration of the substrate

$A_t$ - concentration of the substrate after time $\mathrm{t}$

Step 2: From the question

${\mathrm{t}}_{50\%}=\frac{1}{\mathrm{k}}\ln \frac{100}{50}$ here ${\mathrm{t}}_{50\%}=1\min$

${\mathrm{t}}_{99.9\%}=\frac{1}{\mathrm{k}}\ln \frac{100}{0.1}$

Step 3: Required time is

$\frac{{\mathrm{t}}_{99.9\%}}{{\mathrm{t}}_{50\%}}=\frac{{\displaystyle \frac{1}{\mathrm{k}}}\ln {\displaystyle \frac{100}{0.1}}}{{\displaystyle \frac{1}{\mathrm{k}}}\ln {\displaystyle \frac{100}{50}}}$

$$\begin{gathered} {\mathrm{t}}_{99.9\%}=\frac{\ln 1000}{\ln 2}\times {\mathrm{t}}_{50\%} \\ {\mathrm{t}}_{99.9\%}=\frac{3\times \ln 10}{\ln 2}\times {\mathrm{t}}_{50\%} \\ {\mathrm{t}}_{99.9\%}=\frac{3\times 2.3}{0.69}\times 1 \\ =10\min \end{gathered}$$

Hence the time required is $\mathbf{10}\mathbf{}\mathbf{min}$.