During the kinetic study of the reaction `2A +B rarr C + D` following results were obtained.
`{:(, Run[A],[B] i n M,"Initial rate of fo rmation of D in "m s^(-1),),(I,0.1,0.1,6.0 xx 10^(-3),),(II,0.3,0.2,7.2xx10^(-2),),(III,0.3,0.4,2.88xx10^(-1),),(IV,0.4,0.1,2.40xx10^(-2),):}`
On the basis of above data which one is correct ?

Options

A. `r = k[A]^(2)[B]`
B. `r = k[A][B]`
C. `r = k[A]^(2)[B]^(2)`
D. `r = k[A][B]^(2)`

Explanation

To determine the correct order of the reaction `2A + B → C + D` based on the provided data, we will analyze the initial rates of formation of D in relation to the concentrations of A and B.

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

### Step 2: Analyze data from runs I and II
Using runs I and II to find \( x \):
- Run I: \([A] = 0.1 \, M\), \([B] = 0.1 \, M\), Rate = \(6.0 \times 10^{-3} \, \text{m/s}\)
- Run II: \([A] = 0.3 \, M\), \([B] = 0.2 \, M\), Rate = \(7.2 \times 10^{-2} \, \text{m/s}\)

Set up the equations:
\[ 6.0 \times 10^{-3} = k (0.1)^x (0.1)^y \quad \text{(1)} \]
\[ 7.2 \times 10^{-2} = k (0.3)^x (0.2)^y \quad \text{(2)} \]

### Step 3: Divide equations (2) by (1)
Dividing equation (2) by equation (1):
\[ \frac{7.2 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{(0.3)^x (0.2)^y}{(0.1)^x (0.1)^y} \]
This simplifies to:
\[ 12 = \frac{(0.3)^x (0.2)^y}{(0.1)^x (0.1)^y} \]
\[ 12 = \frac{(0.3)^x (0.2)^y}{(0.1)^{x+y}} \]

### Step 4: Rearranging and solving for \( x \) and \( y \)
Rearranging gives:
\[ 12 \cdot (0.1)^{x+y} = (0.3)^x (0.2)^y \]

### Step 5: Analyze data from runs II and III
Using runs II and III to find \( y \):
- Run II: \([A] = 0.3 \, M\), \([B] = 0.2 \, M\), Rate = \(7.2 \times 10^{-2} \, \text{m/s}\)
- Run III: \([A] = 0.3 \, M\), \([B] = 0.4 \, M\), Rate = \(2.88 \times 10^{-1} \, \text{m/s}\)

Set up the equations:
\[ 7.2 \times 10^{-2} = k (0.3)^x (0.2)^y \quad \text{(3)} \]
\[ 2.88 \times 10^{-1} = k (0.3)^x (0.4)^y \quad \text{(4)} \]

### Step 6: Divide equations (4) by (3)
Dividing equation (4) by equation (3):
\[ \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{(0.3)^x (0.4)^y}{(0.3)^x (0.2)^y} \]
This simplifies to:
\[ 4 = \left(\frac{0.4}{0.2}\right)^y \]
\[ 4 = 2^y \]
Thus, \( y = 2 \).

### Step 7: Substitute \( y \) back to find \( x \)
Now substitute \( y = 2 \) back into the earlier equation to find \( x \):
Using the earlier derived equation:
\[ 12 \cdot (0.1)^{x+2} = (0.3)^x (0.2)^2 \]

### Step 8: Solve for \( x \)
This will yield \( x = 1 \).

### Conclusion
The overall order of the reaction is:
\[ x + y = 1 + 2 = 3 \]

### Final Answer
The order of the reaction is 3.