Question

The rate of first-order reaction is $0.04{\mathrm{mol}}^{-1}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ at $10\mathrm{s}$ and $0.03{\mathrm{mol}}^{-1}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ at $20\mathrm{s}$ after initiation of the reaction. What is the half-life period of the reaction?

Answer 1

Step 1: Calculating the Rate constant

The rate law expression for the first-order reaction is,

$\mathrm{r}=\mathrm{K}\left[\mathrm{A}\right]$

The rate of a first-order reaction is $0.04{\mathrm{mol}}^{-1}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ at $10\mathrm{s}$

$0.04=\mathrm{K}{\left[\mathrm{A}\right]}_{10}$ ….(1)

The rate of a first-order reaction is $0.03{\mathrm{mol}}^{-1}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ at $20\mathrm{s}$

$0.03=\mathrm{K}{\left[\mathrm{A}\right]}_{20}$ …..(2)

Divide equation (1) by equation (2).

$$\begin{gathered} \frac{0.04}{0.03}=\frac{K[\mathrm{A}{]}_{10}}{K[\mathrm{A}{]}_{20}} \\ \frac{4}{3}=\frac{[\mathrm{A}{]}_{10}}{[\mathrm{A}{]}_{20}}.....\left(3\right) \end{gathered}$$

At 10 min, the expression for the rate constant will be $\mathrm{t}=\frac{2.303}{\mathrm{K}}\log \frac{[\mathrm{A}{]}_{10}}{[\mathrm{A}{]}_{20}}$ ……(4)

Substituting equation (3) in equation (4) we get,

$$\begin{gathered} 10=\frac{2.303}{\mathrm{K}}\log \frac{4}{3} \\ \mathrm{K}=\frac{2.303}{10}\log \frac{4}{3} \\ \mathrm{K}=0.0288{\min }^{-1} \end{gathered}$$

Step 2: Calculating the half-life period of the reaction

The half-life period is represented by ${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$$\begin{gathered} {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\mathrm{K}} \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{0.0288{\min }^{-1}} \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=24.06\min \end{gathered}$$

Hence the half-life period of the reaction is $24.06\min$.