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

Options

A. `0.383` min
B. `23.1` min
C. `8.73` min
D. `7.53` 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, and concentration.
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k \cdot [A] \]
where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of the reactant.

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

Using the rate law:
\[ 1.5 \times 10^{-2} = k \cdot 0.5 \]

### Step 3: Solve for the rate constant \( k \).
Rearranging the equation to find \( k \):
\[ k = \frac{1.5 \times 10^{-2}}{0.5} \]
\[ k = 3.0 \times 10^{-2} \, \text{min}^{-1} \]

### Step 4: Use the rate constant to find the half-life.
The half-life \( t_{1/2} \) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{3.0 \times 10^{-2}} \]

### Step 5: Calculate the half-life.
Calculating the above expression:
\[ t_{1/2} = \frac{0.693}{0.03} \]
\[ t_{1/2} = 23.1 \, \text{minutes} \]

### Final Answer:
The half-life of the reaction is \( 23.1 \, \text{minutes} \).

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