Under the same reaction conditions, initial concentration of $${\text{1}}{\text{.386 }}mol{\text{ }}d{m^{ - 3}}$$   of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio $$\left( {{k_1}/{k_0}} \right)$$  of the rate constant for first order $$\left( {{k_1}} \right)$$  and zero order $$\left( {{k_2}} \right)$$  of the reaction is -

Solution :
The values of rate constants $${k_0},{k_1}$$ for zero order and first order reaction, respectively, are given by the following equation :
$${k_0} = \frac{{{A_0}}}{{2 \times {t_{\frac{1}{2}}}}}$$     [ where $${{A_0} = }$$  initial concentration, and $${{t_{\frac{1}{2}}} = }$$  half - life period ]
and $${k_1} = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$$
substituting various given values, we get
$$\eqalign{ & {k_0} = \frac{{{\text{1}}{\text{.386}}\,{\text{mol litr}}{{\text{e}}^{ - 1}}}}{{2 \times 20\,\sec }}\,\,...{\text{(i)}} \cr & {\text{and}}\,{k_1} = \frac{{0.639}}{{40\,\sec }}\,\,...({\text{ii)}} \cr & {\text{Dividing (ii) by (i), we get}} \cr & \frac{{{k_1}}}{{{k_0}}} = \frac{{0.639}}{{40}} \times \frac{{2 \times 20}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr & = \frac{{0.639}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr & = 0.5\,mo{l^{ - 1}}{\text{litre}} \cr & = 0.5\,mo{l^{ - 1}}d{m^3}\,\,\,\,\,\,\,\,\,\left[ {1\,{\text{litre}} = 1d{m^3}} \right] \cr & {\text{Thus the correct answer is (A)}}{\text{.}} \cr} $$