Chemistry - Chemical Kinetics Question with Solution | TestHub

Rate constant $k=1.155\times 10^{-3}{\mathrm{s}}^{-1}$ $k=\frac{2.303}{t}\log \frac{a}{(a-x)}\because a=a,(a-x)=\frac{a}{2}$ $\begin{array}{rl}{t}_{1/2}& =\frac{2.303}{k}\log \frac{a}{a/2}\\ & =\frac{2.303}{1.155\times 10^{-3}}\log 2\\ & =\frac{2.303}{1.155\times 10^{-3}}\times 0.3010\end{array}$ $\begin{array}{rl}& =\frac{0.693\times 10^3}{1.155}\\ \text{ or }{t}_{1/2}& =\frac{0.693}{k}=\frac{0.693}{1.155\times 10^{-3}}\\ & =600\mathrm{s}\end{array}$