For the reaction,
`A+BrarrP,-(d[A])/(dt)=-(d[B])/(dt)=k[A][B] and Rt=(1)/([A_(0)]-[B]_(0))ln.([A][B]_(0))/([B][A]_(0))` when `[A]_(0)ne[B]_(0)`
If `[A]_(0)=[B]_(0)` then the integrated rate law will be

Options

A. `kt=ln.([A])/([B])`
B. `(1)/([B])=(1)/([A]_(0))+kt`
C. `(1)/([A])=(1)/([A]_(0))+kt`
D. `(1)/([A])=(1)/([A]_(0))+kt or (1)/([B])=(1)/([B]_(0))+kt`

Answer

To solve the problem, we need to derive the integrated rate law for the reaction \( A + B \rightarrow P \) given that the initial concentrations of \( A \) and \( B \) are equal, i.e., \( [A]_0 = [B]_0 \).

### Step-by-Step Solution:

1. **Write the Rate Equation**:
The rate of the reaction is given by:
\[ -\frac{d[A]}{dt} = -\frac{d[B]}{dt} = k[A][B] \]

2. **Assume Initial Concentrations**:
Let \( [A]_0 = [B]_0 = C_0 \). At any time \( t \), let the concentration of \( A \) and \( B \) be \( [A] = C_0 - x \) and \( [B] = C_0 - x \), where \( x \) is the amount of \( A \) and \( B \) that has reacted.

3. **Substitute into the Rate Equation**:
Substitute \( [A] \) and \( [B] \) into the rate equation:
\[ -\frac{d[A]}{dt} = k[C_0 - x][C_0 - x] \]
This simplifies to:
\[ -\frac{d[A]}{dt} = k(C_0 - x)^2 \]

4. **Express in Terms of \( x \)**:
Since \( [A] = C_0 - x \), we can rewrite the equation:
\[ \frac{dx}{dt} = -k(C_0 - x)^2 \]

5. **Separate Variables**:
Rearranging gives:
\[ \frac{dx}{(C_0 - x)^2} = -k dt \]

6. **Integrate Both Sides**:
Integrate the left side from \( 0 \) to \( x \) and the right side from \( 0 \) to \( t \):
\[ \int_0^x \frac{dx'}{(C_0 - x')^2} = -k \int_0^t dt' \]
The left side integrates to:
\[ -\frac{1}{C_0 - x} \bigg|_0^x = -\frac{1}{C_0 - x} + \frac{1}{C_0} \]
The right side integrates to:
\[ -kt \]

7. **Combine Results**:
Thus, we have:
\[ -\frac{1}{C_0 - x} + \frac{1}{C_0} = -kt \]
Rearranging gives:
\[ \frac{1}{C_0 - x} = \frac{1}{C_0} + kt \]

8. **Final Form**:
Taking the reciprocal gives the integrated rate law:
\[ C_0 - x = \frac{C_0}{1 + C_0 kt} \]
Therefore, the integrated rate law when \( [A]_0 = [B]_0 \) is:
\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]