For a first order reaction, `t_(0.75)` is 1386 seconds, then the specific rate constant in `sec^(-1)` is

Options

A. `10^(-3)`
B. `10^(-2)`
C. `10^(-9)`
D. `10^(-5)`

Explanation

To find the specific rate constant \( k \) for a first-order reaction given that \( t_{0.75} = 1386 \) seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand \( t_{0.75} \)**:
- The term \( t_{0.75} \) refers to the time taken for 75% of the reactant to be converted into products. This means that 25% of the reactant remains.

2. **Set up the relationship**:
- For a first-order reaction, the relationship between \( t_{0.75} \) and the rate constant \( k \) is given by the formula:
\[ t_{0.75} = \frac{2.303}{k} \log\left(\frac{[A_0]}{[A]}\right) \]
- Here, \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t_{0.75} \).

3. **Determine concentrations**:
- If we assume the initial concentration \( [A_0] = 100 \) (arbitrary units), then after 75% of the reaction, the remaining concentration \( [A] \) will be:
\[ [A] = 100 - 75 = 25 \]
- Thus, the ratio \( \frac{[A_0]}{[A]} \) becomes:
\[ \frac{100}{25} = 4 \]

4. **Substitute values into the formula**:
- Now, substituting the values into the equation:
\[ 1386 = \frac{2.303}{k} \log(4) \]

5. **Calculate \( \log(4) \)**:
- We know that \( \log(4) = 2 \log(2) \). Using \( \log(2) \approx 0.301 \):
\[ \log(4) \approx 2 \times 0.301 = 0.602 \]

6. **Rearranging the equation**:
- Now, we can rearrange the equation to solve for \( k \):
\[ k = \frac{2.303 \cdot \log(4)}{1386} \]
- Substituting \( \log(4) \):
\[ k = \frac{2.303 \cdot 0.602}{1386} \]

7. **Calculate \( k \)**:
- Performing the multiplication:
\[ k \approx \frac{1.384}{1386} \approx 0.000999 \, \text{sec}^{-1} \]
- This can be approximated to \( k \approx 10^{-3} \, \text{sec}^{-1} \).

### Final Answer:
The specific rate constant \( k \) is approximately \( 10^{-3} \, \text{sec}^{-1} \).