Acetic acid `(CH_(3)COOH)` is partially dimerised to `(CH_(3)COOH_(2))` in the vapour phase. At a total pressure of 0.200 atm, acetic acid is `92.0%` dimerized at 298 K. The value of equilibrium constant of dimerization under these conditions is

Options

A. `57.5atm^(-1)`
B. `9.7atm^(-1)`
C. `97atm^(-1)`
D. `194atm^(-1)`

Answer

To solve the problem of finding the equilibrium constant of dimerization of acetic acid `(CH₃COOH)` at a total pressure of 0.200 atm, where it is 92.0% dimerized at 298 K, we can follow these steps:

### Step 1: Write the Dimerization Reaction
The dimerization of acetic acid can be represented as:
\[ 2 \text{CH}_3\text{COOH} \rightleftharpoons \text{(CH}_3\text{COOH)}_2 \]

### Step 2: Define Initial and Equilibrium Concentrations
Let the initial amount of acetic acid be 1 mole. At equilibrium, if `α` is the degree of dimerization, then:
- Moles of acetic acid remaining = \( 1 - \alpha \)
- Moles of dimer formed = \( \frac{\alpha}{2} \)

Given that the dimerization is 92% complete, we have:
\[ \alpha = 0.92 \]

### Step 3: Calculate Moles at Equilibrium
At equilibrium:
- Moles of acetic acid = \( 1 - 0.92 = 0.08 \)
- Moles of dimer = \( \frac{0.92}{2} = 0.46 \)

### Step 4: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 0.08 + 0.46 = 0.54 \]

### Step 5: Calculate Partial Pressures
Using the total pressure \( P = 0.200 \, \text{atm} \):
- Partial pressure of dimer \( P_d \):
\[ P_d = \frac{0.46}{0.54} \times 0.200 \]
- Partial pressure of acetic acid \( P_a \):
\[ P_a = \frac{0.08}{0.54} \times 0.200 \]

### Step 6: Substitute Values into the Equilibrium Constant Expression
The equilibrium constant \( K \) for the dimerization reaction is given by:
\[ K = \frac{P_d}{(P_a)^2} \]

### Step 7: Calculate the Values
1. Calculate \( P_d \):
\[ P_d = \frac{0.46}{0.54} \times 0.200 = 0.17037 \, \text{atm} \]

2. Calculate \( P_a \):
\[ P_a = \frac{0.08}{0.54} \times 0.200 = 0.02963 \, \text{atm} \]

3. Substitute into the equilibrium constant expression:
\[ K = \frac{0.17037}{(0.02963)^2} \]

### Step 8: Final Calculation
Calculate \( K \):
\[ K = \frac{0.17037}{0.000876} \approx 194.062 \]

### Conclusion
The value of the equilibrium constant \( K \) for the dimerization of acetic acid under the given conditions is approximately:
\[ K \approx 194.06 \, \text{atm}^{-1} \]