Conisder the reaction represented by the equation:
`CH_(3)Cl(g) + H_(2)O(g) rarr CH_(3)OH(g) + HCl(g)`
These kinetics data were obtained for the given reaction concentrations:
`{:("Initial conc (M)",,"Initial rate of disappearance"),([CH_(3)Cl],[H_(2)O],"of "CH_(3)Cl(Ms^(-1))),(0.2,0.2,1),(0.4,0.2,2),(0.4,0.4,8):}`
Order with respect to `[CH_(3)Cl]` will be

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To determine the order of the reaction with respect to \([CH_3Cl]\), we will analyze the provided kinetic data using the rate law expression.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [CH_3Cl]^x [H_2O]^y \]
where \(x\) is the order with respect to \([CH_3Cl]\) and \(y\) is the order with respect to \([H_2O]\).

2. **Select Two Experiments**:
We will use the data from the first two experiments to find \(x\). The data is:
- Experiment 1: \([CH_3Cl] = 0.2\, M\), \([H_2O] = 0.2\, M\), Rate = 1 \(Ms^{-1}\)
- Experiment 2: \([CH_3Cl] = 0.4\, M\), \([H_2O] = 0.2\, M\), Rate = 2 \(Ms^{-1}\)

3. **Set Up the Rate Ratio**:
We can set up the ratio of the rates from the two experiments:
\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{k [CH_3Cl]_1^x [H_2O]_1^y}{k [CH_3Cl]_2^x [H_2O]_2^y} \]
Substituting the values:
\[ \frac{1}{2} = \frac{(0.2)^x (0.2)^y}{(0.4)^x (0.2)^y} \]

4. **Simplify the Equation**:
The \([H_2O]^y\) terms will cancel out:
\[ \frac{1}{2} = \frac{(0.2)^x}{(0.4)^x} \]
This can be rewritten as:
\[ \frac{1}{2} = \left(\frac{0.2}{0.4}\right)^x \]
Since \(\frac{0.2}{0.4} = \frac{1}{2}\), we have:
\[ \frac{1}{2} = \left(\frac{1}{2}\right)^x \]

5. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ 1 = x \]

6. **Conclusion**:
Therefore, the order with respect to \([CH_3Cl]\) is:
\[ x = 1 \]

### Final Answer:
The order with respect to \([CH_3Cl]\) is **1**.