For a first-order reaction `A rarr B` the reaction rate at reactant concentration of `0.01 M` is found to be `2.0 xx 10^(-5) "mol" L^(-1) s^(-1)`. The half-life period of the reaction is

Options

A. 200s
B. 30s
C. 300s
D. 347s

Explanation

To solve the problem step-by-step, we will follow the outlined approach in the video transcript.

### Step 1: Understand the Reaction and Given Data
We have a first-order reaction:
\[ A \rightarrow B \]
The concentration of reactant \( A \) is given as \( [A] = 0.01 \, \text{M} \), and the reaction rate at this concentration is given as:
\[ \text{Rate} = 2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 2: Use the Rate Law for First-Order Reactions
For a first-order reaction, the rate law is expressed as:
\[ \text{Rate} = k [A] \]
Where \( k \) is the rate constant. We can rearrange this equation to solve for \( k \):
\[ k = \frac{\text{Rate}}{[A]} \]

### Step 3: Substitute the Given Values to Calculate \( k \)
Substituting the known values into the equation:
\[ k = \frac{2.0 \times 10^{-5}}{0.01} \]
\[ k = \frac{2.0 \times 10^{-5}}{1.0 \times 10^{-2}} \]
\[ k = 2.0 \times 10^{-3} \, \text{L mol}^{-1} \text{s}^{-1} \]

### Step 4: Calculate the Half-Life of the Reaction
For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{\ln 2}{k} \]
Where \( \ln 2 \approx 0.693 \).

### Step 5: Substitute \( k \) into the Half-Life Formula
Now, substituting the value of \( k \) into the half-life formula:
\[ t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} \]

### Step 6: Perform the Calculation
Calculating the half-life:
\[ t_{1/2} = \frac{0.693}{0.002} \]
\[ t_{1/2} = 346.5 \, \text{s} \]

### Step 7: Conclusion
The half-life period of the reaction is approximately:
\[ t_{1/2} \approx 347 \, \text{s} \]