Chemistry - Chemical Kinetics Question with Solution | TestHub

For first order reaction, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{20}=0.0346\min$ Here, if $[A{]}_0=1,[A{]}_1-\frac{1}{10},t_1/10=?$ $\begin{array}{l}k=\frac{2.303}{t}\log \frac{[\mathrm{A}{]}_0}{[\mathrm{A}{]}_1}\\ \therefore t_{1/10}=\frac{2.303}{0.0346}\log 10=66.56\min \end{array}$