In a first order reaction, concentration of reactant is reduced to $(1 / 8)^{ ext {th }}$ of concentration in 23.03 minutes. What is half-life period of reaction?

Question

In a first order reaction, concentration of reactant is reduced to $(1 / 8)^{	ext {th }}$ of concentration in 23.03 minutes. What is half-life period of reaction?

Aicharya · Accepted Answer

Initial concentration $\mathrm{a}_0=\mathrm{a}$ (let) Final concentration $a_t=\frac{a}{8}$ $1^{	ext {st }}$ order reaction $\begin{aligned} & \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}}=\frac{2.303}{23.03} \log \frac{\mathrm{a}}{\mathrm{a} / 8} \ & \mathrm{k}=\frac{1}{10} \log 8=\frac{3}{10} \log 2=0.3 \log 2 \end{aligned}$ Half -life $\mathrm{t}_{1 / 2}=\frac{2.303 \log 2}{\mathrm{k}}$ $\begin{aligned} & =\frac{2.303 	imes \log }{0.3 \log 2} \ & =7.7 \mathrm{~min} \end{aligned}$

In a first order reaction, concentration of reactant is reduced to $(1 / 8)^{\text {th }}$ of concentration…

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