For a first order reaction A→B, the reaction rate at reactant concentration of 0.01 M is found to be `2.0×10^(−5)molL^(−1)s^(−1)`.The half-life period of the reaction is:

Answer

To find the half-life period of a first-order reaction \( A \rightarrow B \), we can follow these steps:

### Step 1: Understand the relationship between rate, concentration, and rate constant
For a first-order reaction, the rate of the reaction is given by the equation:
\[ \text{Rate} = k \cdot [A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 2: Substitute the known values to find the rate constant \( k \)
We are given:
- Rate = \( 2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Concentration \([A] = 0.01 \, \text{M}\)

Substituting these values into the rate equation:
\[ 2.0 \times 10^{-5} = k \cdot 0.01 \]

### Step 3: Solve for \( k \)
Rearranging the equation to solve for \( k \):
\[ k = \frac{2.0 \times 10^{-5}}{0.01} \]
Calculating this gives:
\[ k = 2.0 \times 10^{-3} \, \text{s}^{-1} \]

### Step 4: Use the rate constant to find the half-life \( t_{1/2} \)
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} \]

### Step 5: Calculate the half-life
Calculating this gives:
\[ t_{1/2} = 346.5 \, \text{s} \]
Rounding this to the nearest whole number, we get:
\[ t_{1/2} \approx 347 \, \text{s} \]

### Final Answer
The half-life period of the reaction is approximately **347 seconds**.

---