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, we can follow these steps:

### Step-by-Step Solution

1. **Identify the formula for half-life of a first-order reaction**:
The half-life (t₁/₂) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Substitute the given value of the specific reaction rate**:
In this case, the specific reaction rate \( k \) is given as:
\[ k = 0.347 \times 10^{-3} \, \text{sec}^{-1} \]

3. **Plug the value of \( k \) into the half-life formula**:
Now, substituting the value of \( k \) into the half-life formula:
\[ t_{1/2} = \frac{0.693}{0.347 \times 10^{-3}} \]

4. **Perform the calculation**:
First, calculate the denominator:
\[ 0.347 \times 10^{-3} = 0.000347 \]
Now, calculate \( t_{1/2} \):
\[ t_{1/2} = \frac{0.693}{0.000347} \approx 1993.05 \, \text{seconds} \]

5. **Convert the result into scientific notation**:
The result can be expressed in scientific notation:
\[ t_{1/2} \approx 1.99 \times 10^{3} \, \text{seconds} \]

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