Question:
If $75 \%$ of a first order reaction was completed in 90 minutes, $60 \%$ of the same reaction would be completed in approximately (in minutes)________.
(Take : $\log 2=0.30 ; \log 2.5=0.40$ )
Solution:
$\mathrm{t}_{0.75}=2 \times \frac{\ln 2}{\mathrm{k}}=90$
$\mathrm{k}=\frac{\ln 2}{45} \mathrm{~min}^{-1}$
$\mathrm{kt}=\ln \frac{1}{1-0.6}=\ln 2.5$
$\frac{\ln 2}{45} \times \mathrm{t}=\ln 2.5$
$\mathrm{t}=45 \times \frac{\log 2.5}{\log 2}=45 \times \frac{0.4}{0.3}=60 \mathrm{~min}$
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