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 \( A + 3B \rightarrow AB_3 \) based on the provided initial rates and concentrations, we will follow these steps:

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

### Step 2: Analyze the provided data
We have the following data:

1. \( [A] = 0.1 \, M, [B] = 0.1 \, M, \text{Rate} = 0.002 \, M/s \)
2. \( [A] = 0.2 \, M, [B] = 0.1 \, M, \text{Rate} = 0.002 \, M/s \)
3. \( [A] = 0.3 \, M, [B] = 0.2 \, M, \text{Rate} = 0.008 \, M/s \)
4. \( [A] = 0.4 \, M, [B] = 0.3 \, M, \text{Rate} = 0.018 \, M/s \)

### Step 3: Determine the order with respect to \( A \) (x)
We will compare the first and second experiments where the concentration of \( B \) remains constant:

\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{0.002}{0.002} = 1 \]
\[ \frac{[A]_1^x [B]_1^y}{[A]_2^x [B]_2^y} = \frac{(0.1)^x (0.1)^y}{(0.2)^x (0.1)^y} \]
This simplifies to:
\[ 1 = \frac{(0.1)^x}{(0.2)^x} = \left(\frac{0.1}{0.2}\right)^x = \left(\frac{1}{2}\right)^x \]
Since \( 1 = \left(\frac{1}{2}\right)^x \), we find that:
\[ x = 0 \]
Thus, the order with respect to \( A \) is 0.

### Step 4: Determine the order with respect to \( B \) (y)
Now, we will compare the first and third experiments where the concentration of \( A \) is varied and \( B \) is also varied:

\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \frac{0.002}{0.008} = \frac{1}{4} \]
\[ \frac{[A]_1^x [B]_1^y}{[A]_3^x [B]_3^y} = \frac{(0.1)^x (0.1)^y}{(0.3)^x (0.2)^y} \]
Substituting \( x = 0 \):
\[ \frac{1}{4} = \frac{(0.1)^0 (0.1)^y}{(0.3)^0 (0.2)^y} = \frac{(0.1)^y}{(0.2)^y} = \left(\frac{0.1}{0.2}\right)^y = \left(\frac{1}{2}\right)^y \]
Thus, we have:
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^y \]
This implies:
\[ \left(\frac{1}{2}\right)^y = \frac{1}{2^2} \implies y = 2 \]
Thus, the order with respect to \( B \) is 2.

### Step 5: Write the final rate law
Now that we have determined \( x \) and \( y \):
\[ \text{Rate} = k [A]^0 [B]^2 = k [B]^2 \]

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