For a first order reaction, the half-life period is independent of

Options

A. initial concentration
B. cube root of initial concentration
C. first power of final concentration
D. square root of final concentration

Explanation

To solve the question regarding the half-life period of a first-order reaction, 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:
\[ \text{Rate} = -\frac{d[A]}{dt} = k[A] \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

2. **Rearranging the Rate Law:**
We can rearrange the equation to separate variables:
\[ \frac{d[A]}{[A]} = -k \, dt \]

3. **Integrate Both Sides:**
We will integrate both sides. The left side will be integrated from the initial concentration \([A_0]\) to \([A_0/2]\) (since we are looking for the half-life), and the right side will be integrated from \(0\) to \(t_{1/2}\):
\[ \int_{[A_0]}^{[A_0/2]} \frac{d[A]}{[A]} = -k \int_0^{t_{1/2}} dt \]

4. **Performing the Integration:**
The left side integrates to:
\[ \ln\left(\frac{[A_0/2]}{[A_0]}\right) = \ln\left(\frac{1}{2}\right) = \ln 2 \]
The right side integrates to:
\[ -kt_{1/2} \]

5. **Setting the Integrals Equal:**
Equating both sides gives us:
\[ \ln 2 = -k t_{1/2} \]

6. **Solving for Half-Life:**
Rearranging this equation to solve for \(t_{1/2}\):
\[ t_{1/2} = \frac{\ln 2}{k} \]

7. **Conclusion:**
From the derived expression for half-life, we can conclude that the half-life period \(t_{1/2}\) of a first-order reaction is dependent on the rate constant \(k\) and is independent of the initial concentration \([A_0]\).

### Final Answer:
The half-life period of a first-order reaction is independent of the initial concentration of the reactant.