Half life period of a first - order reaction is 1396 seconds. The specific rate constant of the reaction is

Options

A. `0.5xx10^(-2)s^(-1)`
B. `4.9xx10^(-4)s^(-1)`
C. `5.0xx10^(-2)s^(-1)`
D. `5.0xx10^(-3)s^(-1)`

Answer

To find the specific rate constant (k) of a first-order reaction given the half-life period (t₁/₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data**:
- Half-life (t₁/₂) = 1396 seconds.

2. **Recall the formula for half-life of a first-order reaction**:
- The half-life of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the specific rate constant.

3. **Rearrange the formula to solve for k**:
- We can rearrange the formula to find \( k \):
\[ k = \frac{0.693}{t_{1/2}} \]

4. **Substitute the value of t₁/₂ into the equation**:
- Now, substitute the given half-life into the rearranged formula:
\[ k = \frac{0.693}{1396 \text{ seconds}} \]

5. **Calculate the value of k**:
- Performing the calculation:
\[ k = \frac{0.693}{1396} \approx 4.96 \times 10^{-4} \text{ s}^{-1} \]

6. **Final answer**:
- The specific rate constant \( k \) is approximately:
\[ k \approx 4.96 \times 10^{-4} \text{ s}^{-1} \]