In a reaction A `to`Product, the initial concentration of reactant is a mol `L^–1`. If the order of the reaction is n and the half life is` t_(1//2)` then the correct proportionality is:

Options

A. `t_(1//2) prop a^(n-1)`
B. `t_(1//2) prop a^n`
C. `t_(1//2) prop 1/a^n`
D. `t_(1//2) prop 1/(a^(n-1))`

Answer

To solve the problem regarding the half-life of a reaction with respect to its order, let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding Half-Life in Chemical Kinetics**:
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. The relationship between half-life and the order of the reaction is crucial.

2. **Identify the Order of Reaction**:
The order of a reaction (n) can be zero, first, second, or higher. The formula for half-life varies based on the order of the reaction.

3. **Half-Life for Different Orders**:
- **Zero Order**:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
Here, \( t_{1/2} \) is directly proportional to the initial concentration \([A]_0\).

- **First Order**:
\[ t_{1/2} = \frac{0.693}{k} \]
This indicates that the half-life is independent of the initial concentration.

- **Second Order**:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
In this case, \( t_{1/2} \) is inversely proportional to the initial concentration.

- **Nth Order**:
For a general nth order reaction, the half-life can be expressed as:
\[ t_{1/2} \propto \frac{1}{[A]_0^{n-1}} \]
This shows that the half-life is inversely proportional to the initial concentration raised to the power of \( n-1 \).

4. **Conclusion**:
Thus, for a reaction of order \( n \), the half-life is proportional to \( \frac{1}{[A]_0^{n-1}} \). Therefore, the correct proportionality for the half-life of a reaction with initial concentration \( a \, \text{mol L}^{-1} \) and order \( n \) is:
\[ t_{1/2} \propto \frac{1}{a^{n-1}} \]