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 solve the problem regarding 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-by-Step Solution:

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

2. **Using the Integrated Rate Law**:
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 and \( [A] \) is the concentration at time \( t \).

3. **Expressing Concentration at Time \( t \)**:
Rearranging the integrated rate law gives:
\[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \]
Thus, we can express \( [A] \) as:
\[ [A] = \frac{1}{kt + \frac{1}{[A]_0}} \]

4. **Relating Concentration of Product \( x \)**:
The concentration of the product \( x \) formed at time \( t \) can be related to the change in concentration of \( A \):
\[ x = [A]_0 - [A] \]
Substituting the expression for \( [A] \):
\[ x = [A]_0 - \frac{1}{kt + \frac{1}{[A]_0}} \]

5. **Finding a Common Denominator**:
To simplify the expression, we can find a common denominator:
\[ x = [A]_0 - \frac{1}{kt[A]_0 + 1} \]
This can be rewritten as:
\[ x = \frac{[A]_0(kt[A]_0 + 1) - 1}{kt[A]_0 + 1} \]

6. **Final Expression for \( x \)**:
After simplifying, we arrive at the expression:
\[ x = \frac{[A]_0^2 kt}{kt[A]_0 + 1} \]

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