Question

Two-component solid–liquid system of naphthalene–benzene forms a simple eutectic mixture. Assuming an ideal solution, the mole fraction of naphthalene in benzene at \(300~\text{K}\) and \(1~\text{bar}\) is ........... (rounded off to two decimal places).
(Given: freezing point \(T_{\mathrm{fp}}\) and enthalpy of fusion \(\Delta H_{\mathrm{fus}}\) of naphthalene are \(353~\text{K}\) and \(19.28~\text{kJ mol}^{-1}\), respectively; \(R=8.31~\text{J K}^{-1}\text{mol}^{-1}\))

Hint:

Ideal solubility of a solid: \(\ln x = -\dfrac{\Delta H_{\mathrm{fus}}}{R}\left(\dfrac{1}{T}-\dfrac{1}{T_{\mathrm{m}}}\right)\). Works well for simple eutectic systems far from solid–solid interactions and at pressures near 1 bar.

Answer 1

For a simple eutectic mixture, the mole fraction of naphthalene in benzene at the eutectic temperature can be calculated using the following equation based on the freezing point depression formula for ideal solutions: 
ΔT_f = \(\frac{RT_f^2}{\Delta H_{\mathrm{fus}}}\) ln X_napht 
Where: 
- ΔT_f = T_fp - T_sol (Freezing point depression), 
- R is the gas constant, 
- T_f is the freezing point of the pure substance (naphthalene), 
- ΔH_fus is the enthalpy of fusion of naphthalene, 
- X_napht is the mole fraction of naphthalene in benzene. 
We assume that the temperature at which the mixture is being measured (T = 300 K) is lower than the freezing point of naphthalene, so we can calculate the freezing point depression and then solve for the mole fraction. 

Step 1: Freezing point depression: 
Given: 
- T_fp = 353 K, 
- ΔH_fus = 19.28 kJ/mol = 19280 J/mol, 
- R = 8.31 J/mol·K. 
We calculate ΔT_f = 353 K - 300 K = 53 K. 

Step 2: Plug into the formula: 
Rearrange to solve for the mole fraction X_napht: 
X_napht = exp \(\left( \frac{ \Delta T_f \Delta H_{\mathrm{fus}} }{ R T_f^2 } \right)\) 
Substituting the known values: 
X_napht = exp \(\left( \frac{ 53 \times 19280 }{ 8.31 \times 353^2 } \right)\) 

Step 3: Simplifying the expression: 
After performing the calculation: 
X_napht ≈ 0.35 
Therefore, the mole fraction of naphthalene in benzene at 300 K and 1 bar is approximately 0.35 (rounded to two decimal places).