The half life of a second order process, `2A rarr` products , is

Options

A. Independent of initial concentration
B. Directly proportional to initial concentration of A
C. Inversely proportional to initial concentration of A
D. Inversely proportional to square of initial concentration

Answer

To find the half-life of a second-order reaction of the form \( 2A \rightarrow \text{products} \), we can follow these steps:

### Step 1: Understand the Reaction Order
The reaction is second-order with respect to reactant A. This means that the rate of the reaction depends on the square of the concentration of A.

### Step 2: Write the Integrated Rate Law for a Second-Order Reaction
For a second-order reaction, the integrated rate law is given by:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
where:
- \([A]\) is the concentration of A at time \(t\),
- \([A_0]\) is the initial concentration of A,
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 3: Define Half-Life
The half-life (\(t_{1/2}\)) is the time required for the concentration of A to decrease to half of its initial value. Therefore, at half-life:
\[ [A] = \frac{[A_0]}{2} \]

### Step 4: Substitute into the Integrated Rate Law
Substituting \([A] = \frac{[A_0]}{2}\) into the integrated rate law gives:
\[ \frac{1}{\frac{[A_0]}{2}} - \frac{1}{[A_0]} = kt_{1/2} \]

### Step 5: Simplify the Equation
This simplifies to:
\[ \frac{2}{[A_0]} - \frac{1}{[A_0]} = kt_{1/2} \]
\[ \frac{2 - 1}{[A_0]} = kt_{1/2} \]
\[ \frac{1}{[A_0]} = kt_{1/2} \]

### Step 6: Solve for Half-Life
Rearranging the equation gives:
\[ t_{1/2} = \frac{1}{k[A_0]} \]

### Conclusion
Thus, the half-life of a second-order reaction is:
\[ t_{1/2} = \frac{1}{k[A_0]} \]