The half-life periof of a first order reaction is…………..of initial concentration.

Answer

**Step-by-Step Solution:**

1. **Understanding Half-Life in First-Order Reactions:**
- The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Rate Law for First-Order Reactions:**
- 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. **Expression for Concentration Over Time:**
- The integrated rate law for a first-order reaction is given by:
\[ \ln\left(\frac{[A_0]}{[A_t]}\right) = kt \]
where \([A_0]\) is the initial concentration and \([A_t]\) is the concentration at time \(t\).

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

5. **Simplifying the Equation:**
- This simplifies to:
\[ \ln(2) = kt_{1/2} \]

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

7. **Conclusion:**
- The expression for half-life (\(t_{1/2}\)) of a first-order reaction does not include the initial concentration \([A_0]\). Thus, it is independent of the initial concentration.

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

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