Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction $k=\frac{2.303}{t}\log \frac{a}{a-x}$ When $t=60$ and $x=60\%$ $\begin{array}{rl}k& =\frac{2.303}{60}\log \frac{100}{100-60}\\ & =\frac{2.303}{60}\log \frac{100}{40}\\ & =0.0153\end{array}$ Now, $\begin{array}{rl}{t}^{1/2}& =\frac{2.303}{0.0153}\log \frac{100}{100-50}\\ & =\frac{2.303}{0.0153}\times \log 2\\ & =\frac{2.303}{0.0153}\times 0.3010\\ & =45.31\min \end{array}$