IF the `t_(1//2)` for a first order reaction `0.4` min, the time of or `99.9%` completion of the reaction is ............min.

Options

A. 16 min
B. 8 min.
C. 32 min
D. 40 min.

Explanation

To solve the problem of finding the time required for 99.9% completion of a first-order reaction given that the half-life (t₁/₂) is 0.4 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between half-life and rate constant (k)**:
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Calculate the rate constant (k)**:
Rearranging the formula to solve for \( k \):
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting \( t_{1/2} = 0.4 \) minutes:
\[ k = \frac{0.693}{0.4} \approx 1.7325 \, \text{min}^{-1} \]

3. **Determine the time for 99.9% completion**:
The formula to calculate the time \( t \) for a first-order reaction is:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A_t} \right) \]
For 99.9% completion, \( A_t \) (the concentration at time \( t \)) is 0.1% of \( A_0 \). Thus, \( A_t = 0.001 A_0 \):
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{0.001 A_0} \right) = \frac{2.303}{k} \log(1000) \]
Since \( \log(1000) = 3 \):
\[ t = \frac{2.303 \times 3}{k} \]

4. **Substituting the value of k**:
Now substituting \( k \approx 1.7325 \):
\[ t = \frac{2.303 \times 3}{1.7325} \approx \frac{6.909}{1.7325} \approx 3.99 \, \text{minutes} \]
Rounding this gives approximately 4 minutes.

### Final Answer:
The time required for 99.9% completion of the reaction is approximately **4 minutes**.