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Chemical Kinetics — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics5 Marks
Question
Rate constant for a first order reaction has been found to be $2.54 \times 10^{-3} \mathrm{~s}^{-1}$. Calculate its three-fourth life.
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Answer

$\text{|R|}=\text{|R|}_0-\frac{3}{4}\text{|R}|_0=\frac{\text{|R|}_0}{4}$
Substituting $\text{|R|}=\frac{\text{|R|}_0}{4},\text{k}=2.54\times10^{-3}8^{-1}$ in the expression $\text{t}=\frac{2.303}{\text{k}}\log \frac{\text{|R|}_0}{\text{|R|}},$ we get
$\text{t=}\frac{2.303}{2.54\times10^{-3}}\log\frac{\text{|R|}_0}{\text{|R|}4}=\frac{2.303\times10^3}{2.54}\log4$
$\text{t}=\frac{2.303\times10^3}{2.54}\times06021=545.92\text{ s}$
Alternative methode:
For a first order reaction, $\text{t}_{3/4}=2\times\text{t}_{1/2}$
$\therefore \text{t}_{3/4}=2\times\frac{0.693}{\text{k}}$
$\text{t}_{3/4}=\frac{2\times.963}{2.54\times10^{-3}\text{s}}=545.67\text{s}$

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