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 calculating the equilibrium constant for the dimerization of acetic acid, 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 Conditions
Let’s assume we start with 1 mole of acetic acid (CH₃COOH) and no dimer (D). At equilibrium, if α is the degree of dimerization, then:
- Moles of CH₃COOH at equilibrium = \( 1 - \alpha \)
- Moles of dimer (D) at equilibrium = \( \frac{\alpha}{2} \)

### Step 3: Total Moles at Equilibrium
The total number of moles at equilibrium (N) can be calculated as:
\[ N = (1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2} \]

### Step 4: Calculate Mole Fractions
The mole fraction of acetic acid (A) and dimer (B) can be defined as:
- Mole fraction of A:
\[ x_A = \frac{1 - \alpha}{1 - \frac{\alpha}{2}} \]
- Mole fraction of B:
\[ x_B = \frac{\frac{\alpha}{2}}{1 - \frac{\alpha}{2}} \]

### Step 5: Calculate Partial Pressures
Using the total pressure \( P_t = 0.200 \, \text{atm} \):
- Partial pressure of A (\( P_A \)):
\[ P_A = x_A \cdot P_t = \frac{1 - \alpha}{1 - \frac{\alpha}{2}} \cdot P_t \]
- Partial pressure of B (\( P_B \)):
\[ P_B = x_B \cdot P_t = \frac{\frac{\alpha}{2}}{1 - \frac{\alpha}{2}} \cdot P_t \]

### Step 6: Substitute Values
Given that \( \alpha = 0.92 \) (92% dimerized) and \( P_t = 0.200 \, \text{atm} \):
1. Calculate \( P_A \):
\[ P_A = \frac{1 - 0.92}{1 - \frac{0.92}{2}} \cdot 0.200 = \frac{0.08}{1 - 0.46} \cdot 0.200 = \frac{0.08}{0.54} \cdot 0.200 \]
\[ P_A \approx 0.02963 \, \text{atm} \]

2. Calculate \( P_B \):
\[ P_B = \frac{\frac{0.92}{2}}{1 - \frac{0.92}{2}} \cdot 0.200 = \frac{0.46}{0.54} \cdot 0.200 \]
\[ P_B \approx 0.17037 \, \text{atm} \]

### Step 7: Calculate the Equilibrium Constant (K)
The equilibrium constant \( K \) for the dimerization can be expressed as:
\[ K = \frac{P_B}{(P_A)^2} \]
Substituting the values:
\[ K = \frac{0.17037}{(0.02963)^2} \]
\[ K \approx \frac{0.17037}{0.000876} \approx 194.06 \]

### Final Answer
The value of the equilibrium constant \( K \) for the dimerization of acetic acid under the given conditions is approximately:
\[ K \approx 194.06 \]