Under the same reaction conditions, initial concentration of 1 ....

Given that,&

Under the same reaction conditions, initial concentration of $1. 386mol{dm}^{- 3}$ of a substance becomes half in $40s$ and $20s$for first order and zero order kinetics, respectively.

For first order reaction,

${t}_{1/2} = \dfrac{\ln 2}{k1} = 40s... \left(i\right)$

For zero order reaction, {t}_{1/2} = \dfrac{ \left[A\right]0}{2{k}_{0}} =$20s.... \left(ii\right)$

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

$\dfrac{1}{2} = \dfrac{[ A{]}_{0}}{2{k}_{0}} \times \dfrac{k1}{\ln 2}$

$\dfrac{k1}{k0} = \dfrac{\ln 2}{[ A{]}_{0}} = \dfrac{0. 693}{1. 386} =0. 5{mol}^{- 1}{dm}^{3}$&