A reaction, which is second order,has a rate constant of `0.002 L mol^(-1)s^(-1)`. If the initial cond. Of the reactant is `0.2M`. How long will it take for the concentration to become `0.0400M` ?

Options

A. `1000s `
B. `400 s`
C. `200 s`
D. `10,000s`

Explanation

To solve the problem, we will use the integrated rate law for a second-order reaction. The formula for a second-order reaction is given by:

\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]

Where:
- \([A]\) is the final concentration of the reactant,
- \([A_0]\) is the initial concentration of the reactant,
- \(k\) is the rate constant,
- \(t\) is the time.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial concentration \([A_0] = 0.2 \, M\)
- Final concentration \([A] = 0.0400 \, M\)
- Rate constant \(k = 0.002 \, L \, mol^{-1} \, s^{-1}\)

2. **Substitute the Values into the Integrated Rate Law:**
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
Substituting the values:
\[ \frac{1}{0.0400} - \frac{1}{0.2} = 0.002 \cdot t \]

3. **Calculate the Left Side:**
\[ \frac{1}{0.0400} = 25 \, L/mol \]
\[ \frac{1}{0.2} = 5 \, L/mol \]
Therefore,
\[ 25 - 5 = 0.002 \cdot t \]
Simplifying gives:
\[ 20 = 0.002 \cdot t \]

4. **Solve for \(t\):**
\[ t = \frac{20}{0.002} \]
\[ t = 10000 \, seconds \]

### Final Answer:
The time taken for the concentration to decrease from \(0.2M\) to \(0.0400M\) is **10000 seconds**.