Answer the following in detail.

Question

Derive an expression for the relation between half-life and rate constant for first-order reaction.
The half-life period for the first order reaction is $1.7$ hrs. How long will it take for $20\%$ of the reactant to disappear?

Accepted Answer

1. 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 } $

2. Given: Half life $\left( t _{1 / 2}\right)=1.7$ hours, $[ A ]_0=100 \%,[ A ]_{ t }=100-20=80 \%$
To find: Time for $20 \%$ of reactant to react $=t$
Formulae:
1. $t _{\frac{1}{2}}=\frac{0.693}{ k }$
2. $t =\frac{2.303}{ K } \log _{10} \frac{[ A ]_0}{[ A ]_{ t }}$

Calculation: $t _{\frac{1}{2}}=\frac{0.693}{ k }$
$k =\frac{0.693}{ t _{\frac{1}{2}}}=\frac{0.693}{1.7 h }=0.4076 h ^{-1}$
$t =\frac{2.303}{ k } \log _{10} \frac{[ A ]_0}{[ A ]_{ t }}$
$=\frac{2.303}{0.4076 h ^{-1}} \log \frac{100}{80}$
$t =\frac{2.303}{0.4076 h ^{-1}} \times 0.0969$
$=0.5475 h \times \frac{60 min }{1 h }$
$= 32.9$ min
The time required for $20\%$ of reaction to react is $0.5475$ h or $32.9$ min.

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