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 question regarding the relationship between the half-life of a reaction and the initial concentration of the reactant for an nth order reaction, we can follow these steps:

### Step-by-Step Solution:

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

2. **Setting Up the Reaction**:
For the reaction \( A \rightarrow \text{Products} \), let the initial concentration of reactant \( A \) be \( a \) mol/L.

3. **Using the Formula for Half-Life**:
The half-life for an nth order reaction is given by the formula:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{1}{(a - x)^{n-1}} - \frac{1}{a^{n-1}} \right) \]
where \( k \) is the rate constant, \( n \) is the order of the reaction, and \( x \) is the amount of reactant that has reacted.

4. **Substituting for Half-Life**:
At \( t = t_{1/2} \), the concentration of \( A \) will be \( \frac{a}{2} \). Therefore, \( x = \frac{a}{2} \). Substitute \( x \) into the half-life formula:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{1}{(a - \frac{a}{2})^{n-1}} - \frac{1}{a^{n-1}} \right) \]
Simplifying gives:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{1}{(\frac{a}{2})^{n-1}} - \frac{1}{a^{n-1}} \right) \]

5. **Simplifying the Expression**:
The term \( \frac{1}{(\frac{a}{2})^{n-1}} \) can be rewritten as \( \frac{2^{n-1}}{a^{n-1}} \). Thus, the expression becomes:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{2^{n-1}}{a^{n-1}} - \frac{1}{a^{n-1}} \right) \]
This simplifies to:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{2^{n-1} - 1}{a^{n-1}} \right) \]

6. **Final Expression**:
From the above expression, we can see that:
\[ t_{1/2} \propto \frac{1}{a^{n-1}} \]
This indicates that the half-life is inversely proportional to \( a^{n-1} \).

7. **Conclusion**:
Therefore, the correct proportionality for the half-life of the reaction is:
\[ t_{1/2} \propto \frac{1}{a^{n-1}} \]