Question
Derive the relation between half$-$life period and rate constant for first order reaction.
✓
Answer
The relation between half$-$life and rate constant for first$-$order reaction:
The integrated rate law for the first$-$order reaction is $k=\frac{2.303}{t} \log _{10} \frac{[A]_0}{[A]_t}$
where, $[A]_0$ is the initial concentration of reactant at $t = 0.$ It falls to $[A]_t$ at time $t$ after the start of the reaction.
The time required for $[ A ]_0$ to become $\frac{[ A ]_0}{2}$ is denoted as $t _{1 / 2}$ or $[ A ]_{ t }=\frac{[ A ]_0}{2}$ at $t = t_{1/2}$
Putting this condition in the integrated rate law we write
$k =\frac{2.303}{ t _{1 / 2}} \log _{10} \frac{[ A ]_{ t }}{[ A ]_0 / 2}=\frac{2.303}{ t _{1 / 2}} \log _{10} 2$
Substituting value of $log_{10}2,$
$k =\frac{2.303}{ t _{1 / 2}} \times 0.3010$
$\therefore k =\frac{0.693}{ t _{1 / 2}}$
$\therefore t _{1 / 2}=\frac{0.693}{ k }$
The integrated rate law for the first$-$order reaction is $k=\frac{2.303}{t} \log _{10} \frac{[A]_0}{[A]_t}$
where, $[A]_0$ is the initial concentration of reactant at $t = 0.$ It falls to $[A]_t$ at time $t$ after the start of the reaction.
The time required for $[ A ]_0$ to become $\frac{[ A ]_0}{2}$ is denoted as $t _{1 / 2}$ or $[ A ]_{ t }=\frac{[ A ]_0}{2}$ at $t = t_{1/2}$
Putting this condition in the integrated rate law we write
$k =\frac{2.303}{ t _{1 / 2}} \log _{10} \frac{[ A ]_{ t }}{[ A ]_0 / 2}=\frac{2.303}{ t _{1 / 2}} \log _{10} 2$
Substituting value of $log_{10}2,$
$k =\frac{2.303}{ t _{1 / 2}} \times 0.3010$
$\therefore k =\frac{0.693}{ t _{1 / 2}}$
$\therefore t _{1 / 2}=\frac{0.693}{ k }$
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