Half life period of 2 nd order reaction is

Options

A. Proportional to the initial concentration of reactants
B. Independent of the initial concentration of reactants
C. Inversely proportional to initial concentration of reactants
D. Inversely proportional to square of initial concentration of reactants

Answer

To determine the half-life period of a second-order reaction, we can derive the expression step by step.

### Step-by-Step Solution:

1. **Understanding Reaction Order**:
- A second-order reaction is one where the rate of reaction is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants.

2. **General Expression for Half-Life**:
- The general expression for the half-life (\(t_{1/2}\)) of an nth order reaction is given by:
\[ t_{1/2} = \frac{2^{n-1} - 1}{(n-1) [A_0]^{n-1} k} \]
- Here, \(k\) is the rate constant, \([A_0]\) is the initial concentration, and \(n\) is the order of the reaction.

3. **Substituting for Second Order**:
- For a second-order reaction, \(n = 2\). Substituting \(n = 2\) into the general expression:
\[ t_{1/2} = \frac{2^{2-1} - 1}{(2-1) [A_0]^{2-1} k} \]
- This simplifies to:
\[ t_{1/2} = \frac{2^1 - 1}{1 \cdot [A_0]^{1} k} \]

4. **Simplifying Further**:
- Continuing the simplification:
\[ t_{1/2} = \frac{2 - 1}{[A_0] k} = \frac{1}{[A_0] k} \]

5. **Final Expression**:
- Thus, the half-life period of a second-order reaction is given by:
\[ t_{1/2} = \frac{1}{k [A_0]} \]

### Conclusion:
The half-life of a second-order reaction is inversely proportional to the initial concentration of the reactant.