For the non-stoichiometric reaction `2A+BrarrC+D`
The following kinetic data were obtained in theee separate experiment, all at `98 K`
`|{:("Initial concentration (A)","Initial concentration (B)","Initial rate of formation of C" (molL^(-1) s^(-1))),(0.01 M,0.1 M,1.2 xx 10^(-3)),(0.1 M,0.2 M,1.2 xx 10^(-3)),(0.2 M,0.1 M,2.4 xx 10^(-3)):}|`
The rate law for the formation of `C` is:

Options

A. `(dc)/(dt)=K[A][B]^(2)`
B. `(dc)/(dt)=k[A]`
C. `(dc)/(dt)=k[A][B]`
D. `(dc)/(dt)=k[A]^(2)[B]`

Explanation

To determine the rate law for the formation of C in the reaction \(2A + B \rightarrow C + D\), we will analyze the given kinetic data and derive the rate law step by step.

### Step 1: Write the general form of the rate law
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 of the reaction with respect to A, and \(y\) is the order of the reaction with respect to B.

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

| Experiment | [A] (M) | [B] (M) | Rate (mol L\(^{-1}\) s\(^{-1}\)) |
|------------|---------|---------|-----------------------------------|
| 1 | 0.01 | 0.1 | \(1.2 \times 10^{-3}\) |
| 2 | 0.1 | 0.2 | \(1.2 \times 10^{-3}\) |
| 3 | 0.2 | 0.1 | \(2.4 \times 10^{-3}\) |

### Step 3: Compare experiments 1 and 2 to find \(y\)
We will compare the rates of experiments 1 and 2 to determine the order with respect to B (\(y\)).

\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = 1 \]

Substituting the concentrations into the rate law:
\[ 1 = \frac{k [0.01]^x [0.1]^y}{k [0.1]^x [0.2]^y} \]

This simplifies to:
\[ 1 = \frac{[0.01]^x [0.1]^y}{[0.1]^x [0.2]^y} \]
\[ 1 = \frac{0.01^x}{0.1^x} \cdot \frac{0.1^y}{0.2^y} \]
\[ 1 = 0.1^{x-y} \cdot 2^{-y} \]

Since \(0.1^{x-y} \cdot 2^{-y} = 1\), we can conclude that \(y = 0\). Thus, the order with respect to B is 0.

### Step 4: Compare experiments 1 and 3 to find \(x\)
Next, we will compare the rates of experiments 1 and 3 to determine the order with respect to A (\(x\)).

\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}} = \frac{1}{2} \]

Substituting the concentrations into the rate law:
\[ \frac{1}{2} = \frac{k [0.01]^x [0.1]^y}{k [0.2]^x [0.1]^y} \]

This simplifies to:
\[ \frac{1}{2} = \frac{[0.01]^x}{[0.2]^x} \]
\[ \frac{1}{2} = \left(\frac{0.01}{0.2}\right)^x \]
\[ \frac{1}{2} = \left(\frac{1}{20}\right)^x \]

Taking logarithms on both sides gives:
\[ x \log(20) = \log(2) \]
Thus,
\[ x = 1 \]

### Step 5: Write the final rate law
Now that we have determined the orders with respect to A and B:
- \(x = 1\)
- \(y = 0\)

The rate law for the formation of C is:
\[ \text{Rate} = k [A]^1 [B]^0 = k [A] \]

### Final Answer
The rate law for the formation of C is:
\[ \text{Rate} = k [A] \]