Step-by-step solution:

1. Analyze the First-Order Reaction

For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and is related to the rate constant ($k_1$) by the formula: $t_{1/2} = \frac{\ln(2)}{k_1}$ Given that the half-life for the first-order reaction is 40 seconds: $40 \text{ s} = \frac{\ln(2)}{k_1}$ Solving for $k_1$: $k_1 = \frac{\ln(2)}{40} \text{ s}^{-1}$ We know that $\ln(2) \approx 0.693$. So, $k_1 = \frac{0.693}{40} \text{ s}^{-1}$

2. Analyze the Zero-Order Reaction

For a zero-order reaction, the half-life ($t_{1/2}$) depends on the initial concentration ($[A]_0$) and is related to the rate constant ($k_0$) by the formula: $t_{1/2} = \frac{[A]_0}{2k_0}$ Given that the initial concentration is $[A]_0 = 1.386 \text{ mol dm}^{-3}$ and the half-life for the zero-order reaction is 20 seconds: $20 \text{ s} = \frac{1.386 \text{ mol dm}^{-3}}{2k_0}$ Solving for $k_0$: $k_0 = \frac{1.386 \text{ mol dm}^{-3}}{2 \times 20 \text{ s}} = \frac{1.386}{40} \text{ mol dm}^{-3} \text{ s}^{-1}$

3. Calculate the Ratio ($k_1 / k_0$)

Now, we need to find the ratio of the rate constants, $k_1 / k_0$: $\frac{k_1}{k_0} = \frac{\frac{\ln(2)}{40} \text{ s}^{-1}}{\frac{1.386}{40} \text{ mol dm}^{-3} \text{ s}^{-1}}$ The term '40' in the denominator of both expressions cancels out: $\frac{k_1}{k_0} = \frac{\ln(2)}{1.386} \frac{\text{s}^{-1}}{\text{mol dm}^{-3} \text{s}^{-1}}$ Substitute the value of $\ln(2) \approx 0.693$: $\frac{k_1}{k_0} = \frac{0.693}{1.386} \text{ mol}^{-1} \text{ dm}^{3}$ Notice that $1.386 = 2 \times 0.693$. Therefore: $\frac{k_1}{k_0} = \frac{0.693}{2 \times 0.693} \text{ mol}^{-1} \text{ dm}^{3}$ $\frac{k_1}{k_0} = \frac{1}{2} \text{ mol}^{-1} \text{ dm}^{3} = 0.5 \text{ mol}^{-1} \text{ dm}^{3}$

4. Conclusion

The ratio of the rate constants $(k_1/k_0)$ is $0.5 \text{ mol}^{-1} \text{ dm}^{3}$. This matches option A.