For the reaction 2X+Y+Z`to X_2 YZ`, the rate equation is : Rate `=k [X] [Y]^2` with k `= 3.0 xx 10^(-6) mol^(-2) L^(2) s^(-)` If[X]=0.1 mol `L^(-)` ,[Y]=0.2 mol L-and [Z]=0.7 mol` L^(-)`, determine
the rate after 0.02 mole of X has been reacted.

Explanation

To determine the rate of the reaction after 0.02 moles of X has been reacted, we will follow these steps:

### Step 1: Determine the initial concentrations of X, Y, and Z.
Given:
- Initial concentration of X, [X] = 0.1 mol/L
- Initial concentration of Y, [Y] = 0.2 mol/L
- Initial concentration of Z, [Z] = 0.7 mol/L

### Step 2: Calculate the change in concentrations after 0.02 moles of X has reacted.
Since the reaction is:
\[ 2X + Y + Z \rightarrow X_2 YZ \]

For every 2 moles of X that react, 1 mole of Y reacts. Therefore, if 0.02 moles of X react, the amount of Y that reacts is:
\[ \text{Moles of Y reacted} = \frac{0.02}{2} = 0.01 \text{ moles} \]

Now, we can find the new concentrations:
- New concentration of X:
\[ [X] = 0.1 - 0.02 = 0.08 \text{ mol/L} \]
- New concentration of Y:
\[ [Y] = 0.2 - 0.01 = 0.19 \text{ mol/L} \]
- The concentration of Z does not need to be calculated for the rate equation.

### Step 3: Write the rate equation.
The rate equation is given as:
\[ \text{Rate} = k [X] [Y]^2 \]

Where:
- \( k = 3.0 \times 10^{-6} \text{ mol}^{-2} \text{ L}^2 \text{ s}^{-1} \)

### Step 4: Substitute the values into the rate equation.
Now substituting the values into the rate equation:
\[ \text{Rate} = 3.0 \times 10^{-6} \times (0.08) \times (0.19)^2 \]

Calculating \( (0.19)^2 \):
\[ (0.19)^2 = 0.0361 \]

Now substituting this back into the rate equation:
\[ \text{Rate} = 3.0 \times 10^{-6} \times 0.08 \times 0.0361 \]

### Step 5: Calculate the rate.
Calculating the product:
\[ \text{Rate} = 3.0 \times 10^{-6} \times 0.08 \times 0.0361 \]
\[ = 3.0 \times 10^{-6} \times 0.002888 \]
\[ = 8.664 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate of the reaction after 0.02 moles of X has been reacted is:
\[ \text{Rate} \approx 8.664 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1} \]