For the reaction `:`
`2A+B rarr A_(2)B`
the rate `=k[A][B]^(2)` with `k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1)`. Calculate the initial rate of the reaction when `[A]=0.1 mol L^(c-),[B]=0.2 mol L^(-1)`. Calculate the rate of reaction after `[A]` is reduced to `0.06 mol L^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial rate of the reaction

Given:
- Rate equation: \( \text{Rate} = k[A][B]^2 \)
- Rate constant \( k = 2.0 \times 10^{-6} \, \text{mol}^{-2} \, \text{L}^2 \, \text{s}^{-1} \)
- Initial concentrations: \( [A] = 0.1 \, \text{mol L}^{-1} \) and \( [B] = 0.2 \, \text{mol L}^{-1} \)

Substituting the values into the rate equation:

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.1) \times (0.2)^2 \]

Calculating \( (0.2)^2 = 0.04 \):

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.1) \times (0.04) \]

Calculating the product:

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.004) = 8.0 \times 10^{-9} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 2: Calculate the rate of the reaction after [A] is reduced to 0.06 mol L⁻¹

New concentration:
- \( [A] = 0.06 \, \text{mol L}^{-1} \)

We need to determine how much of \( A \) reacted. The initial concentration of \( A \) was \( 0.1 \, \text{mol L}^{-1} \), and it is now \( 0.06 \, \text{mol L}^{-1} \). Therefore, the change in concentration of \( A \) is:

\[ \Delta [A] = 0.1 - 0.06 = 0.04 \, \text{mol L}^{-1} \]

From the stoichiometry of the reaction \( 2A + B \rightarrow A_2B \), for every 2 moles of \( A \) that react, 1 mole of \( B \) reacts. Thus, the change in concentration of \( B \) is:

\[ \Delta [B] = \frac{1}{2} \Delta [A] = \frac{1}{2} \times 0.04 = 0.02 \, \text{mol L}^{-1} \]

The initial concentration of \( B \) was \( 0.2 \, \text{mol L}^{-1} \), so the new concentration of \( B \) is:

\[ [B] = 0.2 - 0.02 = 0.18 \, \text{mol L}^{-1} \]

Now we can calculate the new rate using the updated concentrations:

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

Substituting the new values:

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^2 \]

Calculating \( (0.18)^2 = 0.0324 \):

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.0324) \]

Calculating the product:

\[ \text{Rate} = (2.0 \times 10^{-6}) \times (0.001944) = 3.888 \times 10^{-9} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answers:
- Initial rate: \( 8.0 \times 10^{-9} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Rate after reduction of \( [A] \): \( 3.888 \times 10^{-9} \, \text{mol L}^{-1} \text{s}^{-1} \)