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) for a first-order reaction given its half-life period (t_half), we can use the relationship between the half-life and the rate constant. The formula for the half-life of a first-order reaction is:

\[ t_{1/2} = \frac{0.693}{k} \]

### Step-by-Step Solution:

1. **Identify the given data**:
- Half-life period (t_half) = 1396 seconds.

2. **Write the formula for half-life of a first-order reaction**:
\[ t_{1/2} = \frac{0.693}{k} \]

3. **Rearrange the formula to solve for k**:
\[ k = \frac{0.693}{t_{1/2}} \]

4. **Substitute the value of t_half into the equation**:
\[ k = \frac{0.693}{1396} \]

5. **Calculate the value of k**:
- Performing the division:
\[ k = 0.000496 \, \text{s}^{-1} \]

6. **Convert to scientific notation**:
\[ k = 4.96 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer:
The specific rate constant (k) of the reaction is approximately:
\[ k \approx 4.96 \times 10^{-4} \, \text{s}^{-1} \]

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