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. `(ln 2)/(K)`
B. `(ln 2)/(Ksqrt(0.5))`
C. `(log_(10) 2)/(K)`
D. `(0.693)/(0.5 K)`

Explanation

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

### Step-by-Step Solution:

1. **Understand the half-life formula for first-order reactions**:
The half-life \( T_{1/2} \) for a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{K} \]
Here, \( 0.693 \) is the natural logarithm of 2 (\( \ln 2 \)).

2. **Identify the given values**:
- The initial concentration \( [A]_0 = 0.5 \, M \) (though it is not needed for calculating half-life in first-order kinetics).
- The rate constant \( K \) (this is provided in the question but not numerically specified).

3. **Substitute the values into the formula**:
Since we do not have a specific value for \( K \), we will express the half-life in terms of \( K \):
\[ T_{1/2} = \frac{0.693}{K} \]

4. **Conclusion**:
The half-life of the reaction is:
\[ T_{1/2} = \frac{0.693}{K} \]

### Final Answer:
The half-life of the first-order reaction \( A \rightarrow B \) is \( \frac{0.693}{K} \).

---