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 provided kinetic data from three experiments. The goal is to find the order of the reaction with respect to A and B.

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

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

| Experiment | [A] (M) | [B] (M) | Rate of formation of C (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 to find \( y \)
First, we will compare experiments 1 and 2 to find the value of \( y \):
- From experiment 1:
\[ 1.2 \times 10^{-3} = k (0.01)^x (0.1)^y \]
- From experiment 2:
\[ 1.2 \times 10^{-3} = k (0.1)^x (0.2)^y \]

Now, we can divide the two equations:
\[ \frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{k (0.01)^x (0.1)^y}{k (0.1)^x (0.2)^y} \]
This simplifies to:
\[ 1 = \frac{(0.01)^x (0.1)^y}{(0.1)^x (0.2)^y} \]
Rearranging gives:
\[ 1 = \frac{(0.01)^x}{(0.1)^x} \cdot \frac{(0.1)^y}{(0.2)^y} \]
This simplifies to:
\[ 1 = 10^{-x} \cdot \left(\frac{1}{2}\right)^y \]
Thus, we have:
\[ 10^{-x} \cdot 2^{-y} = 1 \]

### Step 4: Solve for \( y \)
From the equation \( 10^{-x} \cdot 2^{-y} = 1 \), we can deduce that:
- If \( y = 0 \), then \( 10^{-x} = 1 \) implies \( x = 0 \).
- If \( y \neq 0 \), we need to find a specific value.

Since we see that the rates are the same for experiments 1 and 2, we conclude that \( y = 0 \).

### Step 5: Compare experiments to find \( x \)
Now, we will compare experiments 1 and 3 to find the value of \( x \):
- From experiment 1:
\[ 1.2 \times 10^{-3} = k (0.01)^x (0.1)^0 \]
- From experiment 3:
\[ 2.4 \times 10^{-3} = k (0.2)^x (0.1)^0 \]

Dividing these two equations gives:
\[ \frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}} = \frac{k (0.01)^x}{k (0.2)^x} \]
This simplifies to:
\[ \frac{1}{2} = \left(\frac{0.01}{0.2}\right)^x \]
This can be rewritten as:
\[ \frac{1}{2} = (0.05)^x \]
Taking logarithm on both sides:
\[ \log\left(\frac{1}{2}\right) = x \log(0.05) \]
Solving for \( x \):
\[ x = \frac{\log\left(\frac{1}{2}\right)}{\log(0.05)} \]
Calculating gives \( x = 1 \).

### Step 6: Write the final rate law
Now that we have determined \( x = 1 \) and \( y = 0 \), we can write the rate law for the formation of C:
\[ \text{Rate} = k [A]^1 [B]^0 = k [A] \]

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