Time required to decompose half the substance for a `n^(th)` order reaction is inversely proportional to : (Given that a : initial concentration):

Options

A. `a^(n+1)`
B. `a^(n-1)`
C. `a^(n-2)`
D. `a^n`

Answer

To solve the problem of determining the time required to decompose half the substance for an \( n^{th} \) order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life**:
The half-life (\( t_{1/2} \)) is the time required for the concentration of a reactant to decrease to half of its initial concentration. For an \( n^{th} \) order reaction, we need to derive the expression for half-life in terms of the initial concentration (\( a \)).

2. **Using the Rate Law**:
The rate law for an \( n^{th} \) order reaction can be expressed as:
\[ -\frac{d[A]}{dt} = k[A]^n \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

3. **Integrating for \( n^{th} \) Order Reaction**:
The integrated rate law for an \( n^{th} \) order reaction is given by:
\[ \frac{1}{[A]^{n-1}} - \frac{1}{[A_0]^{n-1}} = k t \]
where \( [A_0] \) is the initial concentration.

4. **Substituting for Half-Life**:
At half-life, the concentration \( [A] \) becomes \( \frac{a}{2} \). Thus, we substitute \( [A] = \frac{a}{2} \) and \( [A_0] = a \):
\[ \frac{1}{\left(\frac{a}{2}\right)^{n-1}} - \frac{1}{a^{n-1}} = k t_{1/2} \]

5. **Simplifying the Expression**:
This can be simplified as follows:
\[ \frac{2^{n-1}}{a^{n-1}} - \frac{1}{a^{n-1}} = k t_{1/2} \]
\[ \frac{2^{n-1} - 1}{a^{n-1}} = k t_{1/2} \]

6. **Finding the Relationship**:
Rearranging gives us:
\[ t_{1/2} = \frac{a^{n-1}}{k(2^{n-1} - 1)} \]
From this expression, we can see that \( t_{1/2} \) is directly proportional to \( a^{n-1} \).

7. **Inversely Proportional Relationship**:
Since we are looking for the relationship of \( t_{1/2} \) being inversely proportional, we can conclude:
\[ t_{1/2} \propto \frac{1}{a^{n-1}} \]

### Conclusion:
Thus, the time required to decompose half the substance for an \( n^{th} \) order reaction is inversely proportional to \( a^{n-1} \). Therefore, the correct answer is option 2: \( a^{n-1} \).