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Q.
If rate constant of a reaction is $4.606 \times 10^{-3} \,s ^{-1}$ then the time required for the completion of $90 \%$ of the reaction is
Solution:
It is first order reaction since unit of the rate constant is $s ^{-1}$
$t=\frac{2.303}{k} \log \frac{a}{a-x}$
$=\frac{2.303}{4.606 \times 10^{-3}} \log \frac{100}{10}$
$=0.5 \times 10^{3}$
$=500\, s$