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 relationship between the time required to decompose half the substance (T_half) for an nth order reaction and the initial concentration (A), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Life in Reactions**:
The half-life (T_half) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Zero-Order Reaction**:
For a zero-order reaction, the half-life is given by:
\[ T_{1/2} = \frac{A}{2k} \]
Here, T_half is directly proportional to the initial concentration (A). Thus, we can say:
\[ T_{1/2} \propto A \]
Inversely, this means:
\[ T_{1/2} \propto \frac{1}{A} \]

3. **First-Order Reaction**:
For a first-order reaction, the half-life is independent of the initial concentration:
\[ T_{1/2} = \frac{0.693}{k} \]
Therefore, we conclude:
\[ T_{1/2} \propto 1 \quad (\text{independent of } A) \]

4. **Second-Order Reaction**:
For a second-order reaction, the half-life is given by:
\[ T_{1/2} = \frac{1}{kA} \]
Hence, we can say:
\[ T_{1/2} \propto \frac{1}{A} \]

5. **Nth-Order Reaction**:
For an nth-order reaction, the half-life can be expressed as:
\[ T_{1/2} = \frac{(n-1)}{kA^{n-1}} \]
This implies that:
\[ T_{1/2} \propto \frac{1}{A^{n-1}} \]

6. **Conclusion**:
From the analysis above, we can conclude that for an nth-order reaction, the time required to decompose half the substance (T_half) is inversely proportional to the initial concentration raised to the power of (n-1):
\[ T_{1/2} \propto \frac{1}{A^{n-1}} \]

### Final Answer:
The time required to decompose half the substance for an nth order reaction is inversely proportional to \( A^{(n-1)} \).