Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, $\begin{array}{rl}& k=\frac{0.693}{t_{1/2}}\\ & t_{1/2}=\frac{0.693}{0.02303}=30\min \end{array}$ Percent of reactant that remains after $t_{1/2}=50\%$. $2\times {\mathrm{t}}_{1/2}=60\min$ Therefore, percent of reactant that remains after $2{\mathrm{t}}_{1/2}=25\%$