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. 400sec
B. 600sec
C. 800sec
D. 1200sec

Answer

To find the half-life period of a first-order reaction given the rate and initial concentration, 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: Calculate the rate constant (k)
For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k [A_0] \]
Where \(k\) is the rate constant. Rearranging this gives:
\[ k = \frac{\text{Rate}}{[A_0]} \]
Substituting the values:
\[ k = \frac{0.69 \times 10^{-2}}{0.2} \]
Calculating this:
\[ k = \frac{0.69 \times 10^{-2}}{0.2} = 3.45 \times 10^{-2} \, \text{min}^{-1} \]

### Step 3: Use the half-life formula for first-order reactions
The half-life (\(t_{1/2}\)) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{3.45 \times 10^{-2}} \]

### Step 4: Calculate the half-life
Calculating this gives:
\[ t_{1/2} = \frac{0.693}{3.45 \times 10^{-2}} \approx 20.0 \, \text{minutes} \]

### Step 5: Convert minutes to seconds
Since the options are given in seconds, we convert minutes to seconds:
\[ 20 \, \text{minutes} = 20 \times 60 = 1200 \, \text{seconds} \]

### Final Answer
The half-life period is \(1200 \, \text{seconds}\).

---