The rate of a first order reaction is `1.5xx10^(-2) mol L^(-1) "min"^(-1)` at 0.8 M concentration of the reactant. The half - life of the reaction is

Options

A. 0.383 min
B. 36.97 min
C. 8.73 min
D. 7.53 min

Answer

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

### Step-by-Step Solution:

1. **Identify the given data:**
- Rate of reaction (R) = \(1.5 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}\)
- Concentration of reactant (A) = \(0.8 \, \text{M}\)

2. **Use the first-order rate equation:**
The rate of a first-order reaction can be expressed as:
\[ R = k[A] \]
where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.

3. **Rearrange to find the rate constant (k):**
\[ k = \frac{R}{[A]} \]
Substituting the values:
\[ k = \frac{1.5 \times 10^{-2}}{0.8} \]

4. **Calculate the rate constant (k):**
\[ k = \frac{1.5 \times 10^{-2}}{0.8} = 1.875 \times 10^{-2} \, \text{min}^{-1} = 0.01875 \, \text{min}^{-1} \]

5. **Use the half-life formula for first-order reactions:**
The half-life (\(t_{1/2}\)) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]

6. **Substitute the value of k into the half-life formula:**
\[ t_{1/2} = \frac{0.693}{0.01875} \]

7. **Calculate the half-life:**
\[ t_{1/2} \approx 36.96 \, \text{minutes} \]

8. **Final result:**
The half-life of the reaction is approximately \(36.97 \, \text{minutes}\).