In a first-order reaction `A rarr B`, if `K` is the rate constant and initial concentration of the reactant is `0.5 M`, then half-life is

Options

A. `(0.693)/(0.5K)`
B. `(log2)/(k)`
C. `(log2)/(ksqrt(0.5))`
D. `("ln"2)/(k)`

Explanation

To find the half-life of a first-order reaction \( A \rightarrow B \) with a given rate constant \( K \) and initial concentration of \( 0.5 \, M \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Half-Life**:
The half-life (\( t_{1/2} \)) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{K} \]
This formula indicates that the half-life is independent of the initial concentration of the reactant.

2. **Identify the Given Information**:
- Rate constant \( K \) (value not specified, but it is a variable).
- Initial concentration of the reactant \( [A]_0 = 0.5 \, M \) (not needed for the calculation).

3. **Substitute the Values into the Formula**:
Since we have the formula for half-life, we can express it as:
\[ t_{1/2} = \frac{0.693}{K} \]
This is the expression for half-life in terms of the rate constant \( K \).

4. **Rearranging the Expression (Optional)**:
If needed, we can express \( 0.693 \) in terms of logarithms:
\[ 0.693 = 2.303 \times \log(2) \]
Therefore, we can rewrite the half-life as:
\[ t_{1/2} = \frac{2.303 \times \log(2)}{K} \]
or equivalently,
\[ t_{1/2} = \frac{\ln(2)}{K} \]
This shows another form of the half-life expression.

5. **Conclusion**:
The half-life of the reaction is expressed as:
\[ t_{1/2} = \frac{0.693}{K} \quad \text{or} \quad t_{1/2} = \frac{\ln(2)}{K} \]
Since the question does not provide specific options, we conclude that the half-life is dependent on the rate constant \( K \) and is independent of the initial concentration.