If `a` is the initial concentration of the rectant, the half life period of the reaction of `n^(th)` order is inversely proportional to :

Options

A. `a^(n-1)`
B. `a^(n)`
C. `a^(1-n)`
D. `a^(n+1)`

Answer

To solve the question regarding the half-life period of a reaction of nth order, we need to analyze the relationship between the half-life period (T_half) and the initial concentration of the reactant (a).

### Step-by-Step Solution:

1. **Understanding Half-Life in Chemical Reactions**:
The half-life period (T_half) is the time required for the concentration of a reactant to decrease to half of its initial value. The relationship between half-life and the order of the reaction is important.

2. **Formulas for Half-Life**:
- For a **zero-order reaction** (n = 0):
\[ T_{1/2} = \frac{a}{2k} \]
Here, T_half is directly proportional to the initial concentration (a).

- For a **first-order reaction** (n = 1):
\[ T_{1/2} = \frac{0.693}{k} \]
In this case, T_half does not depend on the initial concentration (a).

3. **General Formula for nth Order Reaction**:
For a reaction of order n, the half-life can be expressed as:
\[ T_{1/2} \propto \frac{1}{a^{n-1}} \]
This indicates that the half-life period is inversely proportional to the initial concentration raised to the power of (n-1).

4. **Conclusion**:
From the general formula, we can conclude that the half-life period (T_half) of a reaction of nth order is inversely proportional to \( a^{n-1} \). Therefore, the correct answer to the question is that T_half is inversely proportional to \( a^{n-1} \).

### Final Answer:
The half-life period of the reaction of nth order is inversely proportional to \( a^{n-1} \).