Find the values of A,B and C in the following table for the reaction `X + Y to Z`. The reaction is of first order w.r.t X and zero order w.r.t. Y.
`{:("Exp",[X](mol L^(-1)),[Y](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,0.1,0.1,2xx10^(-2)),(2,A,0.2,4 xx 10^(-2)),(3,0.4,0.4,B),(4,C,0.2,2 xx 10^(-2)):}`

Options

A. `A = 0.2 mol L^(-1), B = 8 xx 10^(-2) mol L^(-1) s^(-1), C = 0.1 mol L^(-1)`
B. `A = 0.4 mol L^(-1), B = 4 xx 10^(-2) mol L^(-1) s^(-1), C = 0.02 mol L^(-1)`
C. `A = 0.2 mol L^(-1), B = 2 xx 10^(-2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)`
D. `A = 0.4 mol L^(-1), B = 2 xx 10^(2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)`

Explanation

To solve for the values of A, B, and C in the given reaction table for the reaction \( X + Y \rightarrow Z \), we will use the information that the reaction is first order with respect to X and zero order with respect to Y.

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [X]^1 [Y]^0 = k [X] \]
This indicates that the rate depends only on the concentration of X.

2. **Using Experimental Data**:
We will analyze the data from the table provided.

- **Experiment 1**:
\[ [X] = 0.1 \, \text{mol L}^{-1}, \quad [Y] = 0.1 \, \text{mol L}^{-1}, \quad \text{Initial Rate} = 2 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]
From this, we can write:
\[ 2 \times 10^{-2} = k \times 0.1 \quad \text{(Equation 1)} \]

- **Experiment 2**:
\[ [X] = A, \quad [Y] = 0.2 \, \text{mol L}^{-1}, \quad \text{Initial Rate} = 4 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]
From this, we can write:
\[ 4 \times 10^{-2} = k \times A \quad \text{(Equation 2)} \]

3. **Finding A**:
Dividing Equation 2 by Equation 1:
\[ \frac{4 \times 10^{-2}}{2 \times 10^{-2}} = \frac{k \times A}{k \times 0.1} \]
Simplifying gives:
\[ 2 = \frac{A}{0.1} \implies A = 2 \times 0.1 = 0.2 \, \text{mol L}^{-1} \]

4. **Using Experiment 3 to Find B**:
- **Experiment 3**:
\[ [X] = 0.4 \, \text{mol L}^{-1}, \quad [Y] = 0.4 \, \text{mol L}^{-1}, \quad \text{Initial Rate} = B \]
From this, we can write:
\[ B = k \times 0.4 \quad \text{(Equation 3)} \]
Dividing Equation 3 by Equation 1:
\[ \frac{B}{2 \times 10^{-2}} = \frac{k \times 0.4}{k \times 0.1} \]
Simplifying gives:
\[ \frac{B}{2 \times 10^{-2}} = 4 \implies B = 4 \times 2 \times 10^{-2} = 8 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Using Experiment 4 to Find C**:
- **Experiment 4**:
\[ [X] = C, \quad [Y] = 0.2 \, \text{mol L}^{-1}, \quad \text{Initial Rate} = 2 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]
From this, we can write:
\[ 2 \times 10^{-2} = k \times C \quad \text{(Equation 4)} \]
Dividing Equation 4 by Equation 1:
\[ \frac{2 \times 10^{-2}}{2 \times 10^{-2}} = \frac{k \times C}{k \times 0.1} \]
Simplifying gives:
\[ 1 = \frac{C}{0.1} \implies C = 0.1 \, \text{mol L}^{-1} \]

### Final Values:
- \( A = 0.2 \, \text{mol L}^{-1} \)
- \( B = 8 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \)
- \( C = 0.1 \, \text{mol L}^{-1} \)