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):}`
If `H_(2)O` is taken in large excess, the order of the reaction will be

Options

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

Explanation

To determine the order of the reaction when water (H₂O) is taken in large excess, we can analyze the given kinetics data step by step.

### Step 1: Write the Rate Law Expression
The rate of the reaction can be expressed in terms of the concentrations of the reactants as follows:
\[ \text{Rate} = k [\text{CH}_3\text{Cl}]^y [\text{H}_2\text{O}]^x \]
where \( k \) is the rate constant, \( y \) is the order with respect to CH₃Cl, and \( x \) is the order with respect to H₂O.

### Step 2: Analyze the Given Data
We have the following data:
1. For \([CH_3Cl] = 0.2\) M and \([H_2O] = 0.2\) M, Rate = 1
2. For \([CH_3Cl] = 0.4\) M and \([H_2O] = 0.2\) M, Rate = 2
3. For \([CH_3Cl] = 0.4\) M and \([H_2O] = 0.4\) M, Rate = 8

### Step 3: Determine the Order with Respect to CH₃Cl
Using the first two sets of data:
- From the first data point:
\[ 1 = k (0.2)^y (0.2)^x \] (Equation 1)
- From the second data point:
\[ 2 = k (0.4)^y (0.2)^x \] (Equation 2)

Now, divide Equation 2 by Equation 1:
\[ \frac{2}{1} = \frac{k (0.4)^y (0.2)^x}{k (0.2)^y (0.2)^x} \]
This simplifies to:
\[ 2 = \frac{(0.4)^y}{(0.2)^y} = \left(\frac{0.4}{0.2}\right)^y = 2^y \]
Thus, we find:
\[ y = 1 \]

### Step 4: Determine the Order with Respect to H₂O
Now we can use the second and third sets of data:
- From the second data point:
\[ 2 = k (0.4)^y (0.2)^x \] (Equation 2)
- From the third data point:
\[ 8 = k (0.4)^y (0.4)^x \] (Equation 3)

Now, divide Equation 3 by Equation 2:
\[ \frac{8}{2} = \frac{k (0.4)^y (0.4)^x}{k (0.4)^y (0.2)^x} \]
This simplifies to:
\[ 4 = \frac{(0.4)^x}{(0.2)^x} = \left(\frac{0.4}{0.2}\right)^x = 2^x \]
Thus, we find:
\[ x = 2 \]

### Step 5: Total Order of the Reaction
The total order of the reaction is the sum of the orders with respect to each reactant:
\[ \text{Total Order} = x + y = 2 + 1 = 3 \]

### Conclusion
When water is taken in large excess, the order of the reaction is **3**.

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