The rate for the `1^(st)` order reaction is `0.69 xx10^(-2) mol L^(-1) min ^(-1)` and the initial concentration is 0.2 mol `L^(-1)` . The half life period is ____.

Options

A. 1200 s
B. `0.33s`
C. 600s
D. 1s

Answer

To find the half-life period of the first-order reaction, we can follow these steps:

### Step 1: Identify the given data
- Rate of reaction (r) = \(0.69 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}\)
- Initial concentration \([A]_0\) = \(0.2 \, \text{mol L}^{-1}\)

### Step 2: Write the rate equation for a first-order reaction
For a first-order reaction, the rate of reaction is given by:
\[ r = k \cdot [A] \]
where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 3: Rearrange the equation to solve for \(k\)
Substituting the given values into the equation:
\[ 0.69 \times 10^{-2} = k \cdot 0.2 \]
To find \(k\), we rearrange the equation:
\[ k = \frac{0.69 \times 10^{-2}}{0.2} \]

### Step 4: Calculate the value of \(k\)
Now, performing the calculation:
\[ k = \frac{0.69 \times 10^{-2}}{0.2} = 0.0345 \, \text{min}^{-1} \]

### Step 5: Use the half-life formula for a first-order reaction
The half-life (\(T_{1/2}\)) for a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]

### Step 6: Substitute the value of \(k\) into the half-life formula
Substituting the calculated value of \(k\):
\[ T_{1/2} = \frac{0.693}{0.0345} \]

### Step 7: Calculate the half-life
Now, performing the calculation:
\[ T_{1/2} \approx 20.08 \, \text{minutes} \]

### Step 8: Convert the half-life from minutes to seconds
Since the options are in seconds, we convert minutes to seconds:
\[ T_{1/2} = 20.08 \times 60 \approx 1204.8 \, \text{seconds} \approx 1200 \, \text{seconds} \]

### Final Answer
Thus, the half-life period is approximately **1200 seconds**.

---