Ratio of Rate Constants for First and Zero Order Reactions

To determine the ratio $\left(\frac{k_1}{k_0}\right)$ of the rate constants, we use the respective half-life expressions for first-order and zero-order kinetics:

1. For the first-order reaction:
   The half-life $t_{1/2}^{(1)}$ is independent of the initial concentration and is given by:
   $t_{1/2}^{(1)} = \frac{\ln 2}{k_1}$
   Given $t_{1/2}^{(1)} = 40\text{ s}$ and $\ln 2 \approx 0.693$:
   $k_1 = \frac{\ln 2}{40\text{ s}} = \frac{0.693}{40}\text{ s}^{-1}$

2. For the zero-order reaction:
   The half-life $t_{1/2}^{(0)}$ depends on the initial concentration $[A]_0$ and is given by:
   $t_{1/2}^{(0)} = \frac{[A]_0}{2k_0}$
   Given $[A]_0 = 1.386\text{ mol dm}^{-3}$ and $t_{1/2}^{(0)} = 20\text{ s}$:
   $k_0 = \frac{[A]_0}{2 \times t_{1/2}^{(0)}} = \frac{1.386\text{ mol dm}^{-3}}{2 \times 20\text{ s}} = \frac{1.386}{40}\text{ mol dm}^{-3}\text{ s}^{-1}$

3. Ratio of the rate constants $\left(\frac{k_1}{k_0}\right)$:
   $\frac{k_1}{k_0} = \frac{\frac{0.693}{40}\text{ s}^{-1}}{\frac{1.386}{40}\text{ mol dm}^{-3}\text{ s}^{-1}} = \frac{0.693}{1.386}\text{ mol}^{-1}\text{ dm}^{3} = 0.5\text{ mol}^{-1}\text{ dm}^{3}$

Thus, the correct option is A.