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Q.
A first order reaction completes $60 \%$ in $20$ minutes. The time required for the completion of $90 \%$ of the reaction is approx
Chemical Kinetics
Solution:
$K =\frac{2.303}{ t } \log \frac{100}{100- x }$
$K =\frac{2.303}{20} \log \frac{100}{100-60}\,\,\,...(i)$
$K =\frac{2.303}{ t } \log \frac{100}{100-90}\,\,\,...(ii)$
$( i )/( ii )$
$\frac{ K =\frac{2.303}{20} \log \frac{100}{40}}{ K =\frac{2.303}{ t } \log \frac{100}{10}} \rightarrow \frac{1}{ t } \log 10=\frac{1}{20}[\log 10-\log 4]$
$\Rightarrow \frac{1}{t} \times 1=\frac{1}{20}[1-0.6020]$
$\therefore t =50$ minutes