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A first order reaction has a rate constant $0.00813 \mathrm{~min}^{-1}$. How long will it take for $60 \%$
completion?
Options:
completion?
Solution:
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Verified Answer
The correct answer is:
$112.7 \mathrm{~min}$
(A)
Original amount of reactant $=100$
$[\mathrm{A}]_{t}=$ Reactant remaining unreacted $=100-60=40$
For first order reaction,
$\therefore t=\frac{2.303}{0.00813 \mathrm{~min}^{-1}} \log _{10} \frac{100}{40}=112.7 \mathrm{~min}$
Original amount of reactant $=100$
$[\mathrm{A}]_{t}=$ Reactant remaining unreacted $=100-60=40$
For first order reaction,
$\therefore t=\frac{2.303}{0.00813 \mathrm{~min}^{-1}} \log _{10} \frac{100}{40}=112.7 \mathrm{~min}$
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