For a first order reaction with rate constant 'k' and initial concentration 'a', the half-life period is given by

Options

A. `("In 2")/(k)`
B. `(1)/(ka)`
C. `(3)/(2k.a^(2))`
D. none of these

Explanation

To find the half-life period for a first-order reaction with rate constant 'k' and initial concentration 'a', we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law for First-Order Reactions**:
The rate of a first-order reaction can be expressed as:
\[ \frac{d[A]}{dt} = -k[A] \]
where \([A]\) is the concentration of the reactant at time \(t\), and \(k\) is the rate constant.

2. **Separate Variables**:
Rearranging the equation gives:
\[ \frac{d[A]}{[A]} = -k \, dt \]

3. **Integrate**:
Integrate both sides. The left side is integrated from the initial concentration \(A_0\) to the concentration at time \(t\) (\(A_t\)), and the right side is integrated from 0 to \(t\):
\[ \int_{A_0}^{A_t} \frac{d[A]}{[A]} = -k \int_0^t dt \]
This results in:
\[ \ln[A_t] - \ln[A_0] = -kt \]

4. **Simplify the Equation**:
Using the properties of logarithms, we can rewrite the equation:
\[ \ln\left(\frac{A_t}{A_0}\right) = -kt \]
Rearranging gives:
\[ \ln[A_0] - \ln[A_t] = kt \]

5. **Define Half-Life**:
The half-life (\(t_{1/2}\)) is the time required for the concentration to decrease to half of its initial value. Therefore, at half-life:
\[ A_t = \frac{A_0}{2} \]

6. **Substitute into the Equation**:
Substitute \(A_t = \frac{A_0}{2}\) into the equation:
\[ \ln\left(\frac{A_0/2}{A_0}\right) = -kt_{1/2} \]
This simplifies to:
\[ \ln\left(\frac{1}{2}\right) = -kt_{1/2} \]

7. **Use Logarithmic Identity**:
We know that \(\ln\left(\frac{1}{2}\right) = -\ln(2)\):
\[ -\ln(2) = -kt_{1/2} \]
Thus, we can write:
\[ t_{1/2} = \frac{\ln(2)}{k} \]

8. **Conclusion**:
The half-life period for a first-order reaction is given by:
\[ t_{1/2} = \frac{\ln(2)}{k} \]

### Final Answer:
The half-life period for a first-order reaction with rate constant 'k' is:
\[ t_{1/2} = \frac{\ln(2)}{k} \]