For the reaction `A + B to` products, what will be the order of reaction with respect to A and B?
`{:("Exp",[A](mol L^(-1)),[B](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,2.5 xx 10^(-4), 3 xx 10^(-5), 5 xx 10^(-4)),(2,5 xx 10^(-4),6 xx 10^(-5), 4 xx 10^(-3)),(3, 1xx 10^(-3), 6 xx 10^(-5), 1.6 xx 10^(-2)):}`

Options

A. 1 with respect to A and 2 with respect to B
B. 2 with respect to A and 1 with respect to B
C. 1 with respect to A and 1 with respect to B
D. 2 with respect to A and 2 with respect to B

Explanation

To determine the order of the reaction with respect to A and B for the reaction \( A + B \rightarrow \text{products} \), we will analyze the provided experimental data step by step.

### Step 1: Write the rate law expression
The rate of the reaction can be expressed as:
\[ \text{Rate} = k [A]^x [B]^y \]
where \( x \) is the order with respect to A and \( y \) is the order with respect to B.

### Step 2: Analyze the data from the experiments
We have the following experimental data:

| Experiment | [A] (mol L\(^{-1}\)) | [B] (mol L\(^{-1}\)) | Initial Rate (mol L\(^{-1}\) s\(^{-1}\)) |
|------------|----------------------|----------------------|-------------------------------------------|
| 1 | \(2.5 \times 10^{-4}\) | \(3 \times 10^{-5}\) | \(5 \times 10^{-4}\) |
| 2 | \(5 \times 10^{-4}\) | \(6 \times 10^{-5}\) | \(4 \times 10^{-3}\) |
| 3 | \(1 \times 10^{-3}\) | \(6 \times 10^{-5}\) | \(1.6 \times 10^{-2}\) |

### Step 3: Determine the order with respect to A (x)
To find \( x \), we can compare experiments 2 and 3, where the concentration of B remains constant.

From Experiment 2:
\[ \text{Rate}_2 = k [A]_2^x [B]_2^y \]
\[ 4 \times 10^{-3} = k (5 \times 10^{-4})^x (6 \times 10^{-5})^y \quad \text{(1)} \]

From Experiment 3:
\[ \text{Rate}_3 = k [A]_3^x [B]_3^y \]
\[ 1.6 \times 10^{-2} = k (1 \times 10^{-3})^x (6 \times 10^{-5})^y \quad \text{(2)} \]

Now, we divide equation (1) by equation (2):
\[ \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}} = \frac{k (5 \times 10^{-4})^x (6 \times 10^{-5})^y}{k (1 \times 10^{-3})^x (6 \times 10^{-5})^y} \]
The \( k \) and \( [B]^y \) terms cancel out:
\[ \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}} = \frac{(5 \times 10^{-4})^x}{(1 \times 10^{-3})^x} \]
Calculating the left side:
\[ \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}} = \frac{4}{16} = \frac{1}{4} \]
Thus, we have:
\[ \frac{1}{4} = \left(\frac{5 \times 10^{-4}}{1 \times 10^{-3}}\right)^x = (5)^x \cdot (10^{-4 + 3})^x = (5)^x \]
This implies:
\[ (5)^x = \frac{1}{4} \]
Taking logarithms:
\[ x \log(5) = \log(0.25) \implies x = \frac{\log(0.25)}{\log(5)} \]
Calculating gives \( x = 2 \).

### Step 4: Determine the order with respect to B (y)
Next, we will find \( y \) using experiments 1 and 2, where the concentration of A is constant.

From Experiment 1:
\[ 5 \times 10^{-4} = k (2.5 \times 10^{-4})^x (3 \times 10^{-5})^y \quad \text{(3)} \]

From Experiment 2:
\[ 4 \times 10^{-3} = k (5 \times 10^{-4})^x (6 \times 10^{-5})^y \quad \text{(4)} \]

Now we divide equation (4) by equation (3):
\[ \frac{4 \times 10^{-3}}{5 \times 10^{-4}} = \frac{k (5 \times 10^{-4})^x (6 \times 10^{-5})^y}{k (2.5 \times 10^{-4})^x (3 \times 10^{-5})^y} \]
The \( k \) terms cancel out:
\[ \frac{4 \times 10^{-3}}{5 \times 10^{-4}} = \frac{(5 \times 10^{-4})^x (6 \times 10^{-5})^y}{(2.5 \times 10^{-4})^x (3 \times 10^{-5})^y} \]
Calculating the left side:
\[ \frac{4 \times 10^{-3}}{5 \times 10^{-4}} = \frac{4}{5} \cdot 10 = 8 \]
Thus:
\[ 8 = \frac{(5)^x (6)^y}{(2.5)^x (3)^y} \]
Substituting \( x = 2 \):
\[ 8 = \frac{(5)^2 (6)^y}{(2.5)^2 (3)^y} \]
Calculating gives:
\[ 8 = \frac{25 \cdot (6)^y}{6.25 \cdot (3)^y} \]
This simplifies to:
\[ 8 = 4 \cdot \frac{(6)^y}{(3)^y} \implies 2 = \left(\frac{6}{3}\right)^y \implies 2 = 2^y \]
Thus, \( y = 1 \).

### Conclusion
The order of the reaction with respect to A is \( 2 \) and with respect to B is \( 1 \). Therefore, the overall order of the reaction is:
- Order with respect to A: 2
- Order with respect to B: 1

**Final Answer:** The correct option is \( 2 \) with respect to A and \( 1 \) with respect to B.