The experiment data for the reaction `2A + B_(2) rarr 2AB` is
`|{:("Experiment",[A] M,[B_(2)] M,"Initial rate" (mol L^(-1) s^(-1)),),(I,0.50,0.5,1.6 xx 10^(-4),),(II,0.50,1.0,3.2 xx 10^(-4),),(III,1.00,1.0,3.2 xx 10^(-4),):}|`
Write the most probable rate equation for the reaction giving reason for your answer.

Options

A. rate = `K[A]^2[B_2]^2`
B. rate = `K[A]^2[B_2]`
C. rate = `k[B_2]`
D. rate = `k[B_2]^2`

Answer

To determine the most probable rate equation for the reaction \( 2A + B_2 \rightarrow 2AB \), we will analyze the provided experimental 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:
\[ \text{Rate} = k [A]^x [B_2]^y \]
where \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, \( [B_2] \) is the concentration of reactant \( B_2 \), and \( x \) and \( y \) are the orders of the reaction with respect to \( A \) and \( B_2 \), respectively.

### Step 2: Analyze the data from Experiment I
From Experiment I:
- \( [A] = 0.50 \, \text{M} \)
- \( [B_2] = 0.50 \, \text{M} \)
- Initial Rate = \( 1.6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)

Substituting into the rate law:
\[ 1.6 \times 10^{-4} = k (0.50)^x (0.50)^y \tag{1} \]

### Step 3: Analyze the data from Experiment II
From Experiment II:
- \( [A] = 0.50 \, \text{M} \)
- \( [B_2] = 1.00 \, \text{M} \)
- Initial Rate = \( 3.2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)

Substituting into the rate law:
\[ 3.2 \times 10^{-4} = k (0.50)^x (1.00)^y \tag{2} \]

### Step 4: Divide equations (1) and (2)
Dividing equation (2) by equation (1):
\[ \frac{3.2 \times 10^{-4}}{1.6 \times 10^{-4}} = \frac{k (0.50)^x (1.00)^y}{k (0.50)^x (0.50)^y} \]
This simplifies to:
\[ 2 = \frac{(1.00)^y}{(0.50)^y} \]
\[ 2 = 2^y \]
Thus, we find:
\[ y = 1 \]

### Step 5: Analyze the data from Experiment III
From Experiment III:
- \( [A] = 1.00 \, \text{M} \)
- \( [B_2] = 1.00 \, \text{M} \)
- Initial Rate = \( 3.2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)

Substituting into the rate law:
\[ 3.2 \times 10^{-4} = k (1.00)^x (1.00)^y \tag{3} \]

### Step 6: Divide equations (2) and (3)
Dividing equation (3) by equation (2):
\[ \frac{3.2 \times 10^{-4}}{3.2 \times 10^{-4}} = \frac{k (1.00)^x (1.00)^y}{k (0.50)^x (1.00)^y} \]
This simplifies to:
\[ 1 = \frac{(1.00)^x}{(0.50)^x} \]
\[ 1 = 2^x \]
Thus, we find:
\[ x = 0 \]

### Step 7: Write the final rate equation
Now substituting the values of \( x \) and \( y \) back into the rate law:
\[ \text{Rate} = k [A]^0 [B_2]^1 = k [B_2] \]

### Conclusion
The most probable rate equation for the reaction is:
\[ \text{Rate} = k [B_2] \]