Chemistry - Chemical Kinetics Question with Solution | TestHub

$$\begin{array}{rl}& \because \lambda =\frac{0.6932}{t_{1/2}}=\frac{0.6932}{6.93}\stackrel{-1}{\min }\\ & \text{ Also }t=\frac{2.303}{\lambda }\log \frac{\left[A_{\circ }\right]}{[A]}\\ & \left[A_{\circ }\right]=\text{ initial concentration (amount) }\\ & [A]=\text{ final concentration (amount) }\\ & \therefore t=\frac{2.303\times 6.93}{0.6932}\log \frac{100}{1}\\ & =46.06\text{ minutes }\end{array}$$