For the second order reaction, concentration (x) of the product at time t starting with initial concentration `[A]_(o)` is :

Options

A. `(kt[A_(o)]^(2))/(1+kt[A_(o)])`
B. `(k[A_(o)]^(2))/(1+kt)`
C. `(1+kt[A_(o)]^(2))/(k+[A_(o)]^(2))`
D. None of these

Explanation

To derive the expression for the concentration (x) of the product at time t for a second-order reaction starting with an initial concentration \([A]_0\), we can follow these steps:

### Step 1: Write the rate law for a second-order reaction
For a second-order reaction, the rate of reaction can be expressed as:
\[ \text{Rate} = k[A]^2 \]
where \(k\) is the rate constant and \([A]\) is the concentration of reactant A.

### Step 2: Use the integrated rate law for a second-order reaction
The integrated rate law for a second-order reaction is given by:
\[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]
where \([A]_0\) is the initial concentration of A, and \([A]\) is the concentration at time \(t\).

### Step 3: Express \([A]\) in terms of \(x\)
Let \(x\) be the concentration of the product formed at time \(t\). Therefore, the concentration of A at time \(t\) can be expressed as:
\[ [A] = [A]_0 - x \]

### Step 4: Substitute \([A]\) into the integrated rate law
Substituting \([A] = [A]_0 - x\) into the integrated rate law gives:
\[ \frac{1}{[A]_0 - x} - \frac{1}{[A]_0} = kt \]

### Step 5: Simplify the equation
Rearranging the equation:
\[ \frac{1}{[A]_0 - x} = kt + \frac{1}{[A]_0} \]
Taking the reciprocal of both sides:
\[ [A]_0 - x = \frac{1}{kt + \frac{1}{[A]_0}} \]

### Step 6: Solve for \(x\)
Now, solving for \(x\):
\[ x = [A]_0 - \frac{1}{kt + \frac{1}{[A]_0}} \]
To combine the terms, we can rewrite it as:
\[ x = [A]_0 - \frac{[A]_0}{kt[A]_0 + 1} \]
This can be simplified to:
\[ x = \frac{[A]_0(kt[A]_0)}{kt[A]_0 + 1} \]

### Final Expression
Thus, the concentration of the product \(x\) at time \(t\) for a second-order reaction is:
\[ x = \frac{kt[A]_0^2}{1 + kt[A]_0} \]