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 when the initial concentrations of A and B are equal, i.e., \([A_0] = [B_0]\).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ A + B \rightarrow P \]
The rate of the reaction is given by:
\[ -\frac{d[A]}{dt} = -\frac{d[B]}{dt} = k[A][B] \]
This indicates that the reaction is second-order, as it depends on the concentrations of both A and B.

2. **Setting Up the Integrated Rate Law**:
Since \([A_0] = [B_0]\), we can denote both initial concentrations as \(C_0\):
\[ [A_0] = [B_0] = C_0 \]
Thus, we can express the rate law as:
\[ -\frac{d[A]}{dt} = k[A][B] = k[A]^2 \]

3. **Rearranging the Rate Equation**:
Rearranging the equation gives:
\[ -\frac{d[A]}{[A]^2} = k \, dt \]

4. **Integrating**:
We will integrate both sides. The left side will be integrated from \([A_0]\) to \([A]\) and the right side from \(0\) to \(t\):
\[ \int_{[A_0]}^{[A]} -\frac{d[A]}{[A]^2} = \int_0^t k \, dt \]
The left side integrates to:
\[ \left[ \frac{1}{[A]} \right]_{[A_0]}^{[A]} = \frac{1}{[A]} - \frac{1}{[A_0]} \]
The right side integrates to:
\[ kt \]

5. **Combining the Results**:
Combining the results from the integration gives:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
Rearranging this, we have:
\[ \frac{1}{[A]} = \frac{1}{[A_0]} + kt \]

6. **Substituting Back for B**:
Since \([A_0] = [B_0]\), we can also express this in terms of B:
\[ \frac{1}{[B]} = \frac{1}{[B_0]} + kt \]

### Final Result:
Thus, the integrated rate law when \([A_0] = [B_0]\) is:
\[ \frac{1}{[A]} = \frac{1}{[A_0]} + kt \quad \text{and} \quad \frac{1}{[B]} = \frac{1}{[B_0]} + kt \]