Half-life period of a first-order reaction is `1386` seconds. The specific rate constant of the reaction is

Options

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

Explanation

To find the specific rate constant (K) of a first-order reaction given the half-life period, we can use the relationship between the half-life (t₁/₂) and the rate constant (K) for first-order reactions. The formula is:

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

### Step-by-Step Solution:

1. **Identify the given data**:
- Half-life period (t₁/₂) = 1386 seconds

2. **Use 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 t₁/₂ into the equation**:
\[ K = \frac{0.693}{1386} \]

5. **Calculate K**:
- Performing the division:
\[ K = 0.0005 \text{ (approximately)} \]
- This can also be expressed in scientific notation:
\[ K = 0.5 \times 10^{-3} \text{ per second} \]

6. **Conclusion**:
- The specific rate constant (K) is \( 0.5 \times 10^{-3} \) per second.

### Final Answer:
The specific rate constant of the reaction is \( 0.5 \times 10^{-3} \) per second.

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