Under the same reaction conditions, intial concentration of `1.386 mol dm^(-1)` of a substance becomes half in 40 s and 20 s through first order and zero order kinetics respectively. 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. `1.5 mol dm^(-3)`
C. `1.5 mol dm^(-3)`
D. `2.0 mol^(-1) dm^(-1)`

Explanation

To solve the problem, we need to find the ratio of the rate constants for a first-order reaction \( k_1 \) and a zero-order reaction \( k_0 \) given the half-lives for both reactions.

### Step-by-Step Solution:

1. **Identify the half-lives**:
- The half-life for the first-order reaction is given as \( t_{1/2} = 40 \, \text{s} \).
- The half-life for the zero-order reaction is given as \( t_{1/2} = 20 \, \text{s} \).

2. **Calculate the rate constant for the first-order reaction \( k_1 \)**:
- The formula for the rate constant \( k_1 \) for a first-order reaction is:
\[ k_1 = \frac{0.693}{t_{1/2}} \]
- Substituting the half-life:
\[ k_1 = \frac{0.693}{40 \, \text{s}} = 0.017325 \, \text{dm}^{-3} \text{s}^{-1} \]

3. **Calculate the rate constant for the zero-order reaction \( k_0 \)**:
- The formula for the rate constant \( k_0 \) for a zero-order reaction is:
\[ k_0 = \frac{[A]_0}{2 \cdot t_{1/2}} \]
- Given the initial concentration \( [A]_0 = 1.386 \, \text{mol dm}^{-1} \) and substituting the half-life:
\[ k_0 = \frac{1.386}{2 \cdot 20 \, \text{s}} = \frac{1.386}{40 \, \text{s}} = 0.03465 \, \text{mol dm}^{-3} \text{s}^{-1} \]

4. **Calculate the ratio \( \frac{k_1}{k_0} \)**:
- Now, we can find the ratio of the rate constants:
\[ \frac{k_1}{k_0} = \frac{0.017325}{0.03465} \]
- Simplifying this gives:
\[ \frac{k_1}{k_0} = 0.5 \]

### Final Answer:
The ratio \( \frac{k_1}{k_0} \) is \( 0.5 \) or \( \frac{1}{2} \).

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