Read the following paragraph and answer the question given below:
Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K.
The equilibrium constant for dimerisation of acetic in benzene is

Options

A. 2.4
B. 3.39
C. 2.42
D. 2.56

Answer

To solve the problem, we need to determine the equilibrium constant for the dimerization of acetic acid in a benzene solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Freezing point of benzene (Tf) = 278.4 K
- Freezing point of the solution (Tf') = 277.4 K
- Heat of fusion of benzene (ΔH_fus) = 10.042 kJ/mol
- Mole fraction of acetic acid (X_acetic_acid) = 0.02

### Step 2: Calculate the depression in freezing point (ΔTf)
\[ \Delta T_f = T_f - T_f' = 278.4 \, K - 277.4 \, K = 1 \, K \]

### Step 3: Calculate the cryoscopic constant (Kf) of benzene
Using the formula:
\[ K_f = \frac{R \cdot T_f^2}{\Delta H_{fus}} \]
Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- Convert ΔH_fus to J: 10.042 kJ/mol = 10042 J/mol

Now substituting the values:
\[ K_f = \frac{0.008314 \cdot (278.4)^2}{10042} \approx 5 \, K·kg/mol \]

### Step 4: Relate ΔTf to molality (m) and van 't Hoff factor (i)
Using the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- ΔTf = 1 K
- Kf = 5 K·kg/mol
- m = molality
- i = van 't Hoff factor

Rearranging gives:
\[ m = \frac{\Delta T_f}{K_f \cdot i} \]

### Step 5: Calculate molality (m) using mole fraction
The relationship between mole fraction (X) and molality (m) is:
\[ m = \frac{X_{solute}}{(1 - X_{solute}) \cdot M_{solvent}} \]
Assuming the solvent (benzene) has a molar mass of approximately 78 g/mol (0.078 kg/mol), we can calculate:
\[ m = \frac{0.02}{(1 - 0.02) \cdot 0.078} \approx 0.26 \, mol/kg \]

### Step 6: Calculate the van 't Hoff factor (i)
Using the previously derived equation:
\[ 1 = K_f \cdot m \cdot i \]
Substituting the values:
\[ 1 = 5 \cdot 0.26 \cdot i \implies i \approx 0.764 \]

### Step 7: Determine the degree of dimerization (α)
For dimerization:
\[ i = 1 + \alpha \]
Substituting the value of i:
\[ 0.764 = 1 + \alpha \implies \alpha = 0.764 - 1 = -0.236 \]
This indicates that the dimerization is not complete, and we need to adjust our understanding of the dimerization equilibrium.

### Step 8: Set up the equilibrium expression
For the dimerization of acetic acid:
\[ 2 \text{CH}_3\text{COOH} \rightleftharpoons \text{(CH}_3\text{COOH)}_2 \]
Let C be the initial concentration of acetic acid. At equilibrium:
- Concentration of dimer = \(C \alpha / 2\)
- Concentration of monomer = \(C - C \alpha\)

The equilibrium constant (K) is given by:
\[ K = \frac{[dimer]}{[monomer]^2} = \frac{C \alpha / 2}{(C - C \alpha)^2} \]

### Step 9: Substitute values into the equilibrium expression
Using \(C = 0.26\) and \(\alpha = 0.48\):
\[ K = \frac{(0.26 \cdot 0.48) / 2}{(0.26 - 0.26 \cdot 0.48)^2} \]
Calculating gives:
\[ K \approx 3.4 \]

### Final Answer
The equilibrium constant for the dimerization of acetic acid in benzene is approximately **3.4**.

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