Calculate the half life of the reaction `ArarrB`, when the initial concentration of A is `0.01 "mol" L^(-1)` and initial rate is `0.00352 "mol" L^(-1)"min"^(-1)`. The reaction is of the first order

Options

A. 19.69 min
B. 1.969 min
C. 7.75 min
D. 77.5 min

Explanation

To calculate the half-life of the reaction \( A \rightarrow B \), given that the reaction is first order, we can follow these steps:

### Step 1: Write the rate law for a first-order reaction.
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of reactant A.

### Step 2: Substitute the given values into the rate law.
We are given:
- Initial concentration of A, \( [A] = 0.01 \, \text{mol L}^{-1} \)
- Initial rate of the reaction, \( \text{Rate} = 0.00352 \, \text{mol L}^{-1} \text{min}^{-1} \)

Substituting these values into the rate law:
\[ 0.00352 = k \times 0.01 \]

### Step 3: Solve for the rate constant \( k \).
To find \( k \), rearrange the equation:
\[ k = \frac{0.00352}{0.01} = 0.352 \, \text{min}^{-1} \]

### Step 4: Use the rate constant to calculate the half-life.
For a first-order reaction, the half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.352} \]

### Step 5: Calculate the half-life.
Now perform the calculation:
\[ t_{1/2} \approx 1.969 \, \text{minutes} \]

### Final Answer:
The half-life of the reaction is approximately \( 1.969 \, \text{minutes} \).

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