For a first-order reaction `A rarr B` the reaction rate at reactant concentration of `0.10 M` is found to be `2.0 xx 10^(-5) "mol" L^(-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:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 2: Substitute the given values into the rate equation
We are given:
- Rate = \( 2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Concentration \([A] = 0.10 \, \text{M}\)

Substituting these values into the rate equation:
\[ 2.0 \times 10^{-5} = k \times 0.10 \]

### Step 3: Solve for the rate constant \( k \)
Rearranging the equation to solve for \( k \):
\[ k = \frac{2.0 \times 10^{-5}}{0.10} = 2.0 \times 10^{-4} \, \text{s}^{-1} \]

### Step 4: Use the formula for half-life of 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 5: Substitute the value of \( k \) into the half-life formula
Now substituting \( k = 2.0 \times 10^{-4} \, \text{s}^{-1} \):
\[ t_{1/2} = \frac{0.693}{2.0 \times 10^{-4}} = 3465 \, \text{s} \]

### Step 6: Round the answer
Rounding \( 3465 \, \text{s} \) gives approximately \( 3465 \, \text{s} \).

### Final Answer
The half-life period of the reaction is approximately \( 3465 \, \text{s} \).

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