Value of `t_(1//2)` for first order reaction is

Options

A. `(0.693)/(K)`
B. `(0.2303)/(K)`
C. `[R]/(2)`
D. `(0.301)/(k)`

Explanation

To find the half-life (\(t_{1/2}\)) for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Half-Life**:
- The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Write the General Equation for a First-Order Reaction**:
- The equation for a first-order reaction can be expressed as:
\[ [A] = [A_0] e^{-kt} \]
- Where:
- \([A]\) = concentration of the reactant at time \(t\)
- \([A_0]\) = initial concentration of the reactant
- \(k\) = rate constant
- \(t\) = time

3. **Set Up the Equation for Half-Life**:
- At half-life, the concentration of the reactant is half of its initial concentration:
\[ [A] = \frac{[A_0]}{2} \]
- Substitute this into the first-order equation:
\[ \frac{[A_0]}{2} = [A_0] e^{-kt_{1/2}} \]

4. **Simplify the Equation**:
- Divide both sides by \([A_0]\):
\[ \frac{1}{2} = e^{-kt_{1/2}} \]

5. **Take the Natural Logarithm of Both Sides**:
- Taking the natural logarithm gives:
\[ \ln\left(\frac{1}{2}\right) = -kt_{1/2} \]

6. **Use the Property of Logarithms**:
- We know that \(\ln\left(\frac{1}{2}\right) = -\ln(2)\):
\[ -\ln(2) = -kt_{1/2} \]

7. **Solve for Half-Life**:
- Rearranging gives:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
- Since \(\ln(2) \approx 0.693\), we can express the half-life as:
\[ t_{1/2} = \frac{0.693}{k} \]

8. **Conclusion**:
- The value of \(t_{1/2}\) for a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]

### Final Answer:
The correct option for the half-life of a first-order reaction is \( \frac{0.693}{k} \).

---