The half-life period for a first order reaction is:

Options

A. Independent of concentration
B. Proportional of concentration
C. Inversely proportional to concentration
D. Inversely proportional to the square of the concentration

Explanation

To determine the half-life period for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life**: The half-life (t₁/₂) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **First-Order Reaction Rate Law**: For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k[A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

3. **Integrated Rate Equation**: The integrated rate equation for a first-order reaction is given by:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
where \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

4. **Substituting for Half-Life**: At half-life, the concentration \( [A] \) becomes \( \frac{[A_0]}{2} \). Substituting this into the integrated rate equation gives:
\[ \ln\left(\frac{[A_0]}{\frac{[A_0]}{2}}\right) = kt_{1/2} \]
Simplifying this, we have:
\[ \ln(2) = kt_{1/2} \]

5. **Solving for Half-Life**: Rearranging the equation to solve for \( t_{1/2} \):
\[ t_{1/2} = \frac{\ln(2)}{k} \]
Since \( \ln(2) \) is approximately 0.693, we can express the half-life as:
\[ t_{1/2} = \frac{0.693}{k} \]

6. **Conclusion**: The half-life of a first-order reaction is independent of the initial concentration of the reactant and depends only on the rate constant \( k \).

### Final Answer:
The half-life period for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]