Chemistry - Chemical Kinetics Question with Solution | TestHub

For first order reaction, $$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$$ Given : $\mathrm{k}=2.303\times 10^{-3}{\mathrm{s}}^{-1},\mathrm{a}=40\mathrm{g},\mathrm{a}-\mathrm{x}=10\mathrm{g}$ On substituting the given values in Eq. (i), we get $$\begin{array}{rl}t& =\frac{2.303}{2.303\times 10^{-3}}\log \frac{40}{10}\\ & =10^3\log 2^2=2\times 10^3\times \log 2\\ & =2\times 10^3\times 0.3010=602\mathrm{s}\end{array}$$ Alternative method For first order reaction, ${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$ $${\mathrm{t}}_{1/2}\left({\mathrm{t}}_{50\%}\right)=\frac{0.693}{2.303\times 10^{-3}}=301\mathrm{s}$$ Also, ${\mathrm{t}}_{75\%}=2{\mathrm{t}}_{50\%}$ $$\therefore {\mathrm{t}}_{75\%}=2\times 301=602\mathrm{s}$$