During the kinetic study of the reaction, `2A +B rarrC+D` , following results were obtained
`{:("Run",[A]//molL^(-1),[B]molL^(-1),"Initial rate formation of "D//molL^(-1)"min"^(-1)),(I,0.1,0.1,6.0xx10^(-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)):}`
Based on the above data which one of the following is correct ?

Options

A. Rate = `K[A]^2[B]`
B. Rate = `K[A][B]`
C. Rate = `K[A]^2[B]^2`
D. Rate = `K[A][B]^2`

Answer

To determine the order of the reaction \(2A + B \rightarrow C + D\) 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-by-Step Solution:

**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 the Data**
We have the following data from the experiments:

| Run | [A] (mol/L) | [B] (mol/L) | Initial Rate of Formation of D (mol/L/min) |
|-----|-------------|--------------|----------------------------------------------|
| I | 0.1 | 0.1 | \(6.0 \times 10^{-3}\) |
| II | 0.3 | 0.2 | \(7.2 \times 10^{-2}\) |
| III | 0.3 | 0.4 | \(2.88 \times 10^{-1}\) |
| IV | 0.4 | 0.1 | \(2.40 \times 10^{-2}\) |

**Step 3: Determine the Order with Respect to A (x)**
To find \(x\), we can compare runs I and IV where the concentration of \(B\) is constant.

From Run I:
\[ 6.0 \times 10^{-3} = k (0.1)^x (0.1)^y \quad \text{(Equation 1)} \]

From Run IV:
\[ 2.40 \times 10^{-2} = k (0.4)^x (0.1)^y \quad \text{(Equation 2)} \]

Dividing Equation 2 by Equation 1:
\[ \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k (0.4)^x (0.1)^y}{k (0.1)^x (0.1)^y} \]
\[ 4 = \frac{(0.4)^x}{(0.1)^x} \]
\[ 4 = \left(\frac{0.4}{0.1}\right)^x = (4)^x \]
Thus, \(x = 1\).

**Step 4: Determine the Order with Respect to B (y)**
Next, we can find \(y\) by comparing runs II and III where the concentration of \(A\) is constant.

From Run II:
\[ 7.2 \times 10^{-2} = k (0.3)^x (0.2)^y \quad \text{(Equation 3)} \]

From Run III:
\[ 2.88 \times 10^{-1} = k (0.3)^x (0.4)^y \quad \text{(Equation 4)} \]

Dividing Equation 4 by Equation 3:
\[ \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k (0.3)^x (0.4)^y}{k (0.3)^x (0.2)^y} \]
\[ 4 = \frac{(0.4)^y}{(0.2)^y} \]
\[ 4 = \left(\frac{0.4}{0.2}\right)^y = (2)^y \]
Thus, \(y = 2\).

**Step 5: Write the Overall Rate Law**
Now, substituting the values of \(x\) and \(y\) into the rate law:
\[ \text{Rate} = k [A]^1 [B]^2 \]

**Step 6: Conclusion**
The overall order of the reaction is:
\[ n = x + y = 1 + 2 = 3 \]

### Final Answer
The correct option based on the analysis is that the order of the reaction with respect to \(A\) is 1 and with respect to \(B\) is 2.

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