Chemistry - Chemical Kinetics Question with Solution | TestHub

For first order reaction, $$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{\frac{1}{2}}}=\frac{0.693}{6.93}=0.1{\text{ hour }}^{-1}$$ Here, $[\mathrm{A}{]}_0=100,[\mathrm{A}{]}_{\mathrm{t}}=100-80=20$ $$\begin{array}{rl}& \text{ Now, }\mathrm{k}=\frac{2.303}{\mathrm{t}}{\log }_{10}\frac{[\mathrm{A}{]}_0}{[\mathrm{A}{]}_{\mathrm{t}}}\\ & \therefore \mathrm{t}=\frac{2.303}{0.1}{\log }_{10}\frac{100}{20}=23.03\times {\log }_{10}5=23.03\times 0.699\\ & =16.10\text{ hours. }\end{array}$$