For a first order reaction `A rarrB` , the reaction rate at reactant concentration of 0.01 M is found to be `2.0xx10^(-5) "mol L"^(-1) s^(-1)`. The half life period of the reaction is

Options

A. 220 s
B. 30 s
C. 300 s
D. 347 s

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 reaction can be expressed as:
\[ \text{Rate} = k [A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of reactant A.

### Step 2: Rearrange the equation to find the rate constant \( k \)
Given the rate of reaction and the concentration of A:
- Rate = \( 2.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Concentration of \( A \) = \( 0.01 \, \text{M} \)

We can rearrange the equation to solve for \( k \):
\[ k = \frac{\text{Rate}}{[A]} \]

### Step 3: Substitute the values to calculate \( k \)
Substituting the given values:
\[ k = \frac{2.0 \times 10^{-5}}{0.01} = \frac{2.0 \times 10^{-5}}{1.0 \times 10^{-2}} = 2.0 \times 10^{-3} \, \text{s}^{-1} \]

### Step 4: Use the rate constant to find the half-life \( t_{1/2} \)
The half-life 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
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} = 346.5 \, \text{s} \]

### Step 6: Round the answer to the nearest whole number
Rounding \( 346.5 \) gives us:
\[ t_{1/2} \approx 347 \, \text{s} \]

### Final Answer
The half-life period of the reaction is approximately \( 347 \, \text{s} \).

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