Compound 'A' and B react according to the following chemical equation.
`A_((g)) + 2B_((g)) to 2C_((g))`
Concentration of either A or B were changed keeping the concentrations of one of the reactants constant and rates were mesured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
`{:("Experiment","Initial concentration of"[A]//mol L^(-1),"Initial concentration of"[B]mol L^(-1),"Initial rate of formation of"[c]//mol L^(-1)s^(-1)),(1,0.30,0.30,0.10),(2,0.30,0.60,0.40),(3,0.60,0.30,0.20):}`

Options

A. Rate `= k [A]^(2) [B]`
B. Rate `= k [A] [B]^(2)`
C. Rate `= k [A] [B]`
D. Rate = `k [A]^(2) [B]^(0)`

Explanation

To determine the rate equation for the reaction \( A(g) + 2B(g) \rightarrow 2C(g) \) based on the given experimental data, we will analyze the initial concentrations of reactants and the corresponding initial rates of formation of product \( C \).

### Step 1: Write the general rate law expression
The rate of 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 the experiments:

| Experiment | [A] (mol L\(^{-1}\)) | [B] (mol L\(^{-1}\)) | Rate (mol L\(^{-1}\) s\(^{-1}\)) |
|------------|-----------------------|-----------------------|-----------------------------------|
| 1 | 0.30 | 0.30 | 0.10 |
| 2 | 0.30 | 0.60 | 0.40 |
| 3 | 0.60 | 0.30 | 0.20 |

### Step 3: Determine the order with respect to \( B \) using Experiments 1 and 2
In Experiment 1 and Experiment 2, the concentration of \( A \) is constant, while the concentration of \( B \) changes.

- For Experiment 1:
\[ \text{Rate}_1 = k [0.30]^x [0.30]^y = 0.10 \]
- For Experiment 2:
\[ \text{Rate}_2 = k [0.30]^x [0.60]^y = 0.40 \]

Dividing the second equation by the first:
\[ \frac{0.40}{0.10} = \frac{k [0.30]^x [0.60]^y}{k [0.30]^x [0.30]^y} \]
This simplifies to:
\[ 4 = \frac{[0.60]^y}{[0.30]^y} \]
\[ 4 = \left(\frac{0.60}{0.30}\right)^y = 2^y \]
Thus, \( y = 2 \).

### Step 4: Determine the order with respect to \( A \) using Experiments 1 and 3
Now, we will compare Experiment 1 and Experiment 3, where the concentration of \( B \) is constant, and the concentration of \( A \) changes.

- For Experiment 1:
\[ \text{Rate}_1 = k [0.30]^x [0.30]^2 = 0.10 \]
- For Experiment 3:
\[ \text{Rate}_3 = k [0.60]^x [0.30]^2 = 0.20 \]

Dividing the third equation by the first:
\[ \frac{0.20}{0.10} = \frac{k [0.60]^x [0.30]^2}{k [0.30]^x [0.30]^2} \]
This simplifies to:
\[ 2 = \frac{[0.60]^x}{[0.30]^x} \]
\[ 2 = \left(\frac{0.60}{0.30}\right)^x = 2^x \]
Thus, \( x = 1 \).

### Step 5: Write the final rate equation
Now that we have determined \( x = 1 \) and \( y = 2 \), we can write the rate equation as:
\[ \text{Rate} = k [A]^1 [B]^2 \]

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