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):}`
The rate law for the reaction will be

Options

A. `r = k[CH_(3)Cl][H_(2)O]`
B. `r = k[CH_(3)Cl]^(2)[H_(2)O]`
C. `r = k[CH_(3)Cl][H_(2)O]^(2)`
D. `r = k[CH_(3)Cl]^(2)[H_(2)O]^(4)`

Explanation

To determine the rate law for the reaction \( \text{CH}_3\text{Cl}(g) + \text{H}_2\text{O}(g) \rightarrow \text{CH}_3\text{OH}(g) + \text{HCl}(g) \), we will analyze the provided kinetic data and use it to find the reaction orders with respect to each reactant.

### Step 1: Write the rate law expression
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}]^x [\text{H}_2\text{O}]^y \]
where \( k \) is the rate constant, \( x \) is the order with respect to \( \text{CH}_3\text{Cl} \), and \( y \) is the order with respect to \( \text{H}_2\text{O} \).

### Step 2: Analyze the first two experiments
From the data provided:
1. Experiment 1: \( [\text{CH}_3\text{Cl}] = 0.2 \, M, [\text{H}_2\text{O}] = 0.2 \, M, \text{Rate} = 1 \, M/s \)
2. Experiment 2: \( [\text{CH}_3\text{Cl}] = 0.4 \, M, [\text{H}_2\text{O}] = 0.2 \, M, \text{Rate} = 2 \, M/s \)

We can set up the following ratio:
\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{1}{2} = \frac{(0.2)^x (0.2)^y}{(0.4)^x (0.2)^y} \]
The \( (0.2)^y \) terms cancel out, leading to:
\[ \frac{1}{2} = \frac{(0.2)^x}{(0.4)^x} = \left(\frac{0.2}{0.4}\right)^x = \left(\frac{1}{2}\right)^x \]
From this, we find:
\[ 1 = 2^{1-x} \implies x = 1 \]

### Step 3: Analyze the second and third experiments
Next, we compare the second and third experiments:
1. Experiment 2: \( [\text{CH}_3\text{Cl}] = 0.4 \, M, [\text{H}_2\text{O}] = 0.2 \, M, \text{Rate} = 2 \, M/s \)
2. Experiment 3: \( [\text{CH}_3\text{Cl}] = 0.4 \, M, [\text{H}_2\text{O}] = 0.4 \, M, \text{Rate} = 8 \, M/s \)

Setting up the ratio:
\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \frac{2}{8} = \frac{(0.4)^x (0.2)^y}{(0.4)^x (0.4)^y} \]
The \( (0.4)^x \) terms cancel out, leading to:
\[ \frac{1}{4} = \frac{(0.2)^y}{(0.4)^y} = \left(\frac{0.2}{0.4}\right)^y = \left(\frac{1}{2}\right)^y \]
From this, we find:
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^y \implies 2^{-2} = 2^{-y} \implies y = 2 \]

### Step 4: Write the final rate law
Now that we have determined \( x = 1 \) and \( y = 2 \), we can write the final rate law:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}]^1 [\text{H}_2\text{O}]^2 \]
or simply:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}] [\text{H}_2\text{O}]^2 \]

### Summary
The rate law for the reaction is:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}] [\text{H}_2\text{O}]^2 \]