Chemistry - Chemical Kinetics Question with Solution | TestHub

Given that, 

Under the same reaction conditions, initial concentration of $1.386 \mathrm{mol} {\mathrm{dm}}^{-3}$ of a substance becomes half in $40 \mathrm{s}$ and $20 \mathrm{s}$for first order and zero order kinetics, respectively.

For first order reaction,

${\mathrm{t}}_{1/2}=\frac{\ln 2}{{\mathrm{k}}_1}=40 \mathrm{s}...\left(\mathrm{i}\right)$

For zero order reaction, ${\mathrm{t}}_{1/2}=\frac{{\left[\mathrm{A}\right]}_0}{2{\mathrm{k}}_0}=20 \mathrm{s}....\left(\mathrm{ii}\right)$

Divide eqn. (i) by (ii), we get

$\frac{1}{2}=\frac{[\mathrm{A}{]}_0}{2{\mathrm{k}}_0}\times \frac{{\mathrm{k}}_1}{\ln 2}$

$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_0}=\frac{\ln 2}{[\mathrm{A}{]}_0}=\frac{0.693}{1.386}=0.5 {\mathrm{mol}}^{-1}{\mathrm{dm}}^3$