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):}`
Overall order of the reaction will be

Options

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

Explanation

To determine the overall order of the reaction represented by the equation:

\[ \text{CH}_3\text{Cl}(g) + \text{H}_2\text{O}(g) \rightarrow \text{CH}_3\text{OH}(g) + \text{HCl}(g) \]

we will analyze the provided kinetics data and apply the method of initial rates.

### 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 \(x\) 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
Using the data from the first two experiments:

- 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 \(Ms^{-1}\)

We can set up the following equation using the rate law:

\[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{k [0.2]^x [0.2]^y}{k [0.4]^x [0.2]^y} \]

This simplifies to:

\[ \frac{1}{2} = \frac{(0.2)^x}{(0.4)^x} \]

### Step 3: Simplify the equation
Since \((0.4)^x = (2 \times 0.2)^x = 2^x (0.2)^x\), we can rewrite the equation as:

\[ \frac{1}{2} = \frac{(0.2)^x}{2^x (0.2)^x} \implies \frac{1}{2} = \frac{1}{2^x} \]

### Step 4: Solve for \(x\)
From the equation, we can equate:

\[ 2^x = 2 \implies x = 1 \]

Thus, the order with respect to \(\text{CH}_3\text{Cl}\) is \(1\).

### Step 5: Analyze the second and third experiments
Now, we will use the second and third experiments:

- Experiment 2: \([CH_3Cl] = 0.4 \, M\), \([H_2O] = 0.2 \, M\), Rate = 2 \(Ms^{-1}\)
- Experiment 3: \([CH_3Cl] = 0.4 \, M\), \([H_2O] = 0.4 \, M\), Rate = 8 \(Ms^{-1}\)

Setting up the equation:

\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \frac{k [0.4]^x [0.2]^y}{k [0.4]^x [0.4]^y} \]

This simplifies to:

\[ \frac{2}{8} = \frac{(0.2)^y}{(0.4)^y} \]

### Step 6: Simplify the equation
This can be rewritten as:

\[ \frac{1}{4} = \frac{(0.2)^y}{(0.4)^y} = \frac{(0.2)^y}{(2 \times 0.2)^y} = \frac{1}{2^y} \]

### Step 7: Solve for \(y\)
From the equation, we can equate:

\[ 2^y = 4 \implies y = 2 \]

Thus, the order with respect to \(\text{H}_2\text{O}\) is \(2\).

### Step 8: Calculate the overall order
The overall order of the reaction is given by:

\[ \text{Overall order} = x + y = 1 + 2 = 3 \]

### Final Answer
The overall order of the reaction is **3**.

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