The specific reaction rate of a first order reaction is `0.347 times 10^–3 sec^–1`. The half life period of the reaction will be :

Options

A. `0.347 times 10^-3 sec`
B. `0.694 times 10^3 sec`
C. `0.347 times 10^3 sec`
D. `2 times 10^3 sec`

Answer

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

### Step-by-Step Solution:

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

2. **Substitute the Given Rate Constant:**
We are given that the specific reaction rate \( k = 0.347 \times 10^{-3} \, \text{sec}^{-1} \). We will substitute this value into the half-life formula:
\[ t_{1/2} = \frac{0.693}{0.347 \times 10^{-3}} \]

3. **Calculate the Half-Life:**
Now, perform the calculation:
\[ t_{1/2} = \frac{0.693}{0.347 \times 10^{-3}} = \frac{0.693}{0.000347} \]
To simplify this, we can calculate:
\[ t_{1/2} \approx 2000 \, \text{seconds} \]
(Note: The exact calculation yields approximately 1995.4 seconds, which can be rounded to 2000 seconds.)

4. **Express the Result in Scientific Notation:**
The half-life can also be expressed in scientific notation:
\[ t_{1/2} \approx 2.0 \times 10^{3} \, \text{seconds} \]

5. **Final Answer:**
Thus, the half-life period of the reaction is:
\[ t_{1/2} \approx 2 \times 10^{3} \, \text{seconds} \]