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):}`
Unit of rate constant will be

Options

A. `s^(-1)`
B. `L^(2) mol^(-2) s^(-1)`
C. `L mol^(-1) s^(-1)`
D. `mol L^(-1) s^(-1)`

Explanation

To solve the problem, we will follow these steps:

### 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, and \( x \) and \( y \) are the orders of the reaction with respect to each reactant.

### Step 2: Use the initial rate data to find \( x \) and \( y \)
We will use the provided experimental data to find the values of \( x \) and \( y \).

#### Experiment 1 and Experiment 2:
From the first two experiments:
- Experiment 1: \([\text{CH}_3\text{Cl}] = 0.2 \, M\), \([\text{H}_2\text{O}] = 0.2 \, M\), Rate = 1 \( \text{Ms}^{-1} \)
- Experiment 2: \([\text{CH}_3\text{Cl}] = 0.4 \, M\), \([\text{H}_2\text{O}] = 0.2 \, M\), Rate = 2 \( \text{Ms}^{-1} \)

Setting up the ratio of rates:
\[ \frac{1}{2} = \frac{0.2^x \cdot 0.2^y}{0.4^x \cdot 0.2^y} \]
This simplifies to:
\[ \frac{1}{2} = \frac{0.2^x}{0.4^x} = \left(\frac{0.2}{0.4}\right)^x = \left(\frac{1}{2}\right)^x \]
Thus, we have:
\[ 1 = 2^{1-x} \implies x = 1 \]

#### Experiment 2 and Experiment 3:
From the second and third experiments:
- Experiment 2: \([\text{CH}_3\text{Cl}] = 0.4 \, M\), \([\text{H}_2\text{O}] = 0.2 \, M\), Rate = 2 \( \text{Ms}^{-1} \)
- Experiment 3: \([\text{CH}_3\text{Cl}] = 0.4 \, M\), \([\text{H}_2\text{O}] = 0.4 \, M\), Rate = 8 \( \text{Ms}^{-1} \)

Setting up the ratio of rates:
\[ \frac{2}{8} = \frac{0.4^x \cdot 0.2^y}{0.4^x \cdot 0.4^y} \]
This simplifies to:
\[ \frac{1}{4} = \frac{0.2^y}{0.4^y} = \left(\frac{0.2}{0.4}\right)^y = \left(\frac{1}{2}\right)^y \]
Thus, we have:
\[ 1 = 2^{2-y} \implies y = 2 \]

### Step 3: Write the complete rate law
Now that we have \( x = 1 \) and \( y = 2 \), we can write the rate law as:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}]^1 [\text{H}_2\text{O}]^2 \]

### Step 4: Determine the overall order of the reaction
The overall order \( n \) is given by:
\[ n = x + y = 1 + 2 = 3 \]

### Step 5: Find the unit of the rate constant \( k \)
The unit of the rate constant \( k \) can be determined using the formula:
\[ \text{Unit of } k = \text{mol L}^{-1} \text{s}^{-1} \text{ (Rate)} \text{ divided by } [\text{Concentration}]^n \]
Substituting \( n = 3 \):
\[ \text{Unit of } k = \text{mol L}^{-1} \text{s}^{-1} \div (\text{mol L}^{-1})^3 = \text{mol L}^{-1} \text{s}^{-1} \div \text{mol}^3 \text{L}^{-3} = \text{L}^2 \text{mol}^{-2} \text{s}^{-1} \]

### Final Answer
The unit of the rate constant \( k \) is:
\[ \text{L}^2 \text{mol}^{-2} \text{s}^{-1} \]

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