Chemistry - Chemical Kinetics Question with Solution | TestHub

The half life ${\mathrm{t}}_{1/2}=10$ days The decay constant $$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{10\text{ days }}=0.0693{\text{ days }}^{-1}$$ The time required for one fourth conversion $$\begin{array}{rl}& \mathrm{t}=\frac{2.303}{\mathrm{k}}{\log }_{10}\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ & =\frac{2.303}{0.0693{\text{ day }}^{-1}}{\log }_{10}\frac{1}{1-(1/4)}=4.1\text{ days }\end{array}$$