The rate of first order reaction is `1.5 xx 10^(-2)` mol` L^(-1) min ^(-1)` at 0.5 M concentration of the reactant. The half-life of the reaction is

Options

A. `7.53` min
B. `0.383 `min
C. `23.1 ` min
D. `8.73 ` min

Explanation

To find the half-life of a first-order reaction given the rate and concentration of the reactant, we can follow these steps:

### Step 1: Understand the relationship between rate, rate constant (k), and concentration.
For a first-order reaction, the rate of the reaction can be expressed as:
\[ R = k[A] \]
where:
- \( R \) is the rate of the reaction,
- \( k \) is the rate constant,
- \( [A] \) is the concentration of the reactant.

### Step 2: Substitute the given values into the equation.
We are given:
- Rate \( R = 1.5 \times 10^{-2} \) mol L\(^{-1}\) min\(^{-1}\)
- Concentration \( [A] = 0.5 \) M

Substituting these values into the equation:
\[ 1.5 \times 10^{-2} = k \times 0.5 \]

### Step 3: Solve for the rate constant (k).
Rearranging the equation to solve for \( k \):
\[ k = \frac{1.5 \times 10^{-2}}{0.5} \]

Calculating \( k \):
\[ k = 1.5 \times 10^{-2} \div 0.5 = 3.0 \times 10^{-2} \, \text{min}^{-1} \]

### Step 4: Use the rate constant to find the half-life (t½).
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{k} \]

### Step 5: Substitute the value of k into the half-life formula.
Using \( \ln(2) \approx 0.693 \):
\[ t_{1/2} = \frac{0.693}{3.0 \times 10^{-2}} \]

Calculating \( t_{1/2} \):
\[ t_{1/2} = \frac{0.693}{0.03} = 23.1 \, \text{minutes} \]

### Final Answer:
The half-life of the reaction is approximately **23.1 minutes**.

---