For the reaction :
`2A + B to A_2B` , the rate = `K[A][B]^2` with `K = 2.0 xx 10^(-6) mol^(-1) lit^(2) sec^(-1)`. Initial concentration of A and B are 0.2 mol/lt and 0.4 mol/lit respectively. Calculate the rate of reaction after [A] is reduced to 0.12 mol/lit

Options

A. ` 3.11 xx 10 ^( - 8 ) `
B. ` 8.6 xx 10 ^( -8) `
C. ` 5.21 xx 10 ^( -6 ) `
D. None of these

Explanation

To solve the problem, we need to calculate the rate of the reaction after the concentration of A is reduced to 0.12 mol/L. The reaction is given as:

\[ 2A + B \rightarrow A_2B \]

The rate law is provided as:

\[ \text{Rate} = K[A][B]^2 \]

where \( K = 2.0 \times 10^{-6} \, \text{mol}^{-1} \, \text{L}^2 \, \text{s}^{-1} \).

### Step 1: Determine the initial concentrations
The initial concentrations of A and B are given as:
- \([A]_0 = 0.2 \, \text{mol/L}\)
- \([B]_0 = 0.4 \, \text{mol/L}\)

### Step 2: Calculate the change in concentration of A
We know that the final concentration of A is given as:
- \([A] = 0.12 \, \text{mol/L}\)

The change in concentration of A can be calculated as:
\[ \Delta [A] = [A]_0 - [A] = 0.2 - 0.12 = 0.08 \, \text{mol/L} \]

### Step 3: Determine the change in concentration of B
From the stoichiometry of the reaction, 2 moles of A react with 1 mole of B. Therefore, for every 2 moles of A that react, 1 mole of B will react. Since 0.08 moles of A have reacted, the amount of B that reacts can be calculated as:
\[ \Delta [B] = \frac{1}{2} \times \Delta [A] = \frac{1}{2} \times 0.08 = 0.04 \, \text{mol/L} \]

### Step 4: Calculate the final concentration of B
Now, we can find the final concentration of B:
\[ [B] = [B]_0 - \Delta [B] = 0.4 - 0.04 = 0.36 \, \text{mol/L} \]

### Step 5: Substitute the final concentrations into the rate equation
Now we can substitute the final concentrations of A and B into the rate equation:
\[ \text{Rate} = K[A][B]^2 \]
Substituting the values:
\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.12) \times (0.36)^2 \]

### Step 6: Calculate the rate
Calculating \( (0.36)^2 \):
\[ (0.36)^2 = 0.1296 \]
Now substituting this back into the rate equation:
\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.12) \times (0.1296) \]
Calculating this gives:
\[ \text{Rate} = 2.0 \times 10^{-6} \times 0.12 \times 0.1296 = 3.1104 \times 10^{-8} \, \text{mol/L/s} \]

### Final Answer
Thus, the rate of the reaction after the concentration of A is reduced to 0.12 mol/L is approximately:
\[ \text{Rate} \approx 3.11 \times 10^{-8} \, \text{mol/L/s} \]