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 rate law for the reaction \(2A + B \rightarrow C + D\) based on the provided data, we can follow these steps:

### Step 1: Write the general rate law expression
The general rate law expression for the reaction can be written 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 from the table
We have the following data from the experiment:

\[ \begin{array}{|c|c|c|c|} \hline \text{Run} & [A] \, (\text{mol L}^{-1}) & [B] \, (\text{mol L}^{-1}) & \text{Initial Rate of formation of } D \, (\text{mol L}^{-1} \text{ min}^{-1}) \\ \hline I & 0.1 & 0.1 & 6.0 \times 10^{-3} \\ II & 0.3 & 0.2 & 7.2 \times 10^{-2} \\ III & 0.3 & 0.4 & 2.88 \times 10^{-1} \\ IV & 0.4 & 0.1 & 2.40 \times 10^{-2} \\ \hline \end{array} \]

### Step 3: Determine the order with respect to \(B\)
To find the order with respect to \(B\) (\(y\)), we can compare runs II and III where the concentration of \(A\) is constant:

- Run II: \([A] = 0.3\), \([B] = 0.2\), Rate = \(7.2 \times 10^{-2}\)
- Run III: \([A] = 0.3\), \([B] = 0.4\), Rate = \(2.88 \times 10^{-1}\)

Calculating the ratio of the rates:
\[ \frac{\text{Rate III}}{\text{Rate II}} = \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = 4 \]
Now, the ratio of the concentrations of \(B\):
\[ \frac{[B]_{III}}{[B]_{II}} = \frac{0.4}{0.2} = 2 \]
Since the rate increases by a factor of 4 when the concentration of \(B\) doubles, we can conclude that:
\[ 2^y = 4 \implies y = 2 \]

### Step 4: Determine the order with respect to \(A\)
Next, we can find the order with respect to \(A\) (\(x\)) by comparing runs I and IV where the concentration of \(B\) is constant:

- Run I: \([A] = 0.1\), \([B] = 0.1\), Rate = \(6.0 \times 10^{-3}\)
- Run IV: \([A] = 0.4\), \([B] = 0.1\), Rate = \(2.40 \times 10^{-2}\)

Calculating the ratio of the rates:
\[ \frac{\text{Rate IV}}{\text{Rate I}} = \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4 \]
Now, the ratio of the concentrations of \(A\):
\[ \frac{[A]_{IV}}{[A]_{I}} = \frac{0.4}{0.1} = 4 \]
Since the rate increases by a factor of 4 when the concentration of \(A\) increases by a factor of 4, we can conclude that:
\[ 4^x = 4 \implies x = 1 \]

### Step 5: Write the final rate law expression
Now that we have determined the orders, we can write the rate law expression:
\[ \text{Rate} = k [A]^1 [B]^2 \]
or simply:
\[ \text{Rate} = k [A] [B]^2 \]

### Conclusion
The correct answer is that the rate law for the reaction is:
\[ \text{Rate} = k [A] [B]^2 \]

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