In a certain gaseous reaction between `A` and `B`, `A + 3B rarr AB_(3)`. The initial rate are reported as follows:
`{:([A],[B],Rate,,),(0.1 M,0.1 M,0.002 M s^(-1),,),(0.2 M,0.1 M,0.002 M s^(-1),,),(0.3 M,0.2 M,0.008 M s^(-1),,),(0.4 M,0.3 M,0.018 M s^(-1),,):}`
The rate law is

Options

A. `r = k[A][B]^(3)`
B. `r = k[A]^(0)[B]^(2)`
C. `r = k[A][B]`
D. `r = k[A]^(0)[B]^(3)`

Explanation

To determine the rate law for the reaction \( A + 3B \rightarrow AB_3 \), we will analyze the given data step by step.

### Step 1: Write the general form of the rate law
The rate law can be expressed as:
\[ \text{Rate} = k [A]^n [B]^m \]
where \( n \) is the order with respect to \( A \) and \( m \) is the order with respect to \( B \).

### Step 2: Analyze the effect of changing \( [A] \) while keeping \( [B] \) constant
From the data provided:
1. For \( [A] = 0.1 \, M \) and \( [B] = 0.1 \, M \), Rate = \( 0.002 \, M/s \)
2. For \( [A] = 0.2 \, M \) and \( [B] = 0.1 \, M \), Rate = \( 0.002 \, M/s \)

Here, when \( [A] \) is doubled from \( 0.1 \, M \) to \( 0.2 \, M \), the rate remains the same. This indicates that the rate does not depend on the concentration of \( A \).

Thus, we can conclude:
\[ n = 0 \]

### Step 3: Analyze the effect of changing \( [B] \) while keeping \( [A] \) constant
Next, we will compare two sets of data where \( [A] \) is constant and \( [B] \) changes:
1. For \( [A] = 0.1 \, M \) and \( [B] = 0.1 \, M \), Rate = \( 0.002 \, M/s \)
2. For \( [A] = 0.3 \, M \) and \( [B] = 0.2 \, M \), Rate = \( 0.008 \, M/s \)

Now, we can set up the equations based on the rate law:
- For the first case:
\[ 0.002 = k [0.1]^0 [0.1]^m \implies 0.002 = k \cdot 0.1^m \]
- For the second case:
\[ 0.008 = k [0.3]^0 [0.2]^m \implies 0.008 = k \cdot 0.2^m \]

### Step 4: Divide the two equations to eliminate \( k \)
Dividing the second equation by the first:
\[ \frac{0.008}{0.002} = \frac{k \cdot 0.2^m}{k \cdot 0.1^m} \]
This simplifies to:
\[ 4 = \left(\frac{0.2}{0.1}\right)^m = 2^m \]

### Step 5: Solve for \( m \)
Taking logarithm or recognizing that \( 2^m = 4 \) gives:
\[ m = 2 \]

### Step 6: Write the final rate law
Now that we have determined \( n \) and \( m \):
\[ \text{Rate} = k [A]^0 [B]^2 = k [B]^2 \]

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