Half life time for second order reaction is proportional

Options

A. initial concentration of the reactant
B. independent of initial concentration
C. inversely proportional to initial concentration
D. `prop ( 1)/(["concentration"]^(2) )`

Answer

To find the half-life period for a second-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Half-Life**:
The half-life (t₁/₂) is the time required for the concentration of a reactant to decrease to half of its initial concentration.

2. **Identify the Order of Reaction**:
In this case, we are dealing with a second-order reaction. This means that the order (n) is 2.

3. **Use the Formula for Half-Life of a Second-Order Reaction**:
The formula for the half-life of a second-order reaction is given by:
\[ t_{1/2} = \frac{1}{k \cdot [A_0]} \]
where:
- \( t_{1/2} \) = half-life
- \( k \) = rate constant
- \( [A_0] \) = initial concentration of the reactant

4. **Substitute the Values**:
Since we know the reaction is second-order (n = 2), we can rewrite the half-life formula specifically for this order:
\[ t_{1/2} = \frac{1}{k \cdot [A_0]} \]

5. **Conclusion**:
The half-life for a second-order reaction is inversely proportional to the initial concentration of the reactant and directly proportional to the rate constant.

### Final Answer:
The half-life time for a second-order reaction is proportional to \( \frac{1}{[A_0]} \).