Chemistry - Chemical Kinetics Question with Solution | TestHub

(A) $\begin{array}{l}{\mathrm{t}}_{1/2}=3\min \\ \therefore \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{3}=0.231{\min }^{-1}\end{array}$ $\begin{array}{l}[\mathrm{A}{]}_0=\text{ Original amount of reactant }=100\\ [\mathrm{A}{]}_t=\text{ Reactant remaining unreacted }=100-90=10\end{array}$ For first order reaction, $t=\frac{2.303}{k}{\log }_{10}\frac{[A{]}_0}{[A{]}_t}=\frac{2.303}{0.231{\min }^{-1}}{\log }_{10}\frac{100}{10}=9.969\min$