Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively.
The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction is

Options

A. `0.5 mol^(-1) dm^(3)`
B. `1.0 mol dm^(-3)`
C. `1.5 mol dm^(-3)`
D. `2.0 mol^(-1) dm^(3)`

Explanation

To solve the problem, we need to determine the ratio of the rate constants \( \frac{k_1}{k_0} \) for a reaction that follows first-order and zero-order kinetics.

### Step-by-Step Solution:

1. **Identify the given data:**
- Initial concentration, \( [A_0] = 1.386 \, \text{mol/dm}^3 \)
- Half-life for first-order kinetics, \( t_{1/2}^{(1)} = 40 \, \text{s} \)
- Half-life for zero-order kinetics, \( t_{1/2}^{(0)} = 20 \, \text{s} \)

2. **Use the half-life formula for first-order kinetics:**
The half-life for a first-order reaction is given by:
\[ t_{1/2}^{(1)} = \frac{0.693}{k_1} \]
Rearranging this gives:
\[ k_1 = \frac{0.693}{t_{1/2}^{(1)}} = \frac{0.693}{40} \]

3. **Use the half-life formula for zero-order kinetics:**
The half-life for a zero-order reaction is given by:
\[ t_{1/2}^{(0)} = \frac{[A_0]}{2k_0} \]
Rearranging this gives:
\[ k_0 = \frac{[A_0]}{2 \cdot t_{1/2}^{(0)}} = \frac{1.386}{2 \cdot 20} \]

4. **Calculate \( k_1 \) and \( k_0 \):**
- For \( k_1 \):
\[ k_1 = \frac{0.693}{40} = 0.017325 \, \text{mol/dm}^3/\text{s} \]
- For \( k_0 \):
\[ k_0 = \frac{1.386}{40} = 0.03465 \, \text{mol/dm}^3/\text{s} \]

5. **Calculate the ratio \( \frac{k_1}{k_0} \):**
\[ \frac{k_1}{k_0} = \frac{0.017325}{0.03465} \]
Simplifying this gives:
\[ \frac{k_1}{k_0} = \frac{0.017325}{0.03465} = \frac{1}{2} = 0.5 \]

### Final Answer:
The ratio \( \frac{k_1}{k_0} = 0.5 \).

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