Chemical Kinetics Question 7
Question 7 - 2024 (30 Jan Shift 1)
The rate of first order reaction is $0.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 10 minutes and $0.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 20 minutes after initiation. Half life of the reaction is ___________ minutes. (Given $\log 2=0.3010, \log 3=0.4771$ )
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Answer (24)
Solution
$$ \begin{align*} & 0.04=k[A]{0} \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \tag{1}\ & 0.03=\mathrm{k}[\mathrm{A}]{0} \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \tag{2} \end{align*} $$
$(1) /(2)$
Solutions
$$ \frac{4}{3}=\mathrm{e}^{600 \mathrm{k}(2-1)} $$
$$ \frac{4}{3}=\mathrm{e}^{600 \mathrm{k}} $$
$\ln \frac{4}{3}=600 \mathrm{k}$
$\ln \frac{4}{3}=600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}}$
$$ \mathrm{t}_{1 / 2}=600 \frac{\ln 2}{\ln \frac{4}{3}} \mathrm{sec} $$
$\mathrm{t}_{1 / 2}=600 \times \frac{\log 2}{\log 4-\log 3} \mathrm{sec} .=10 \times \frac{0.3010}{0.6020-0.477} \min$
$\mathrm{t}_{1 / 2}=24.08 \mathrm{~min}$
Ans. 24
