Freezing point of solution

Question

It's stated in my textbook that at the freezing point, the vapour pressure of the solid phase equals the vapour pressure of the liquid phase. In a mixture, since the vapour pressure of the solution is lower than the vapour pressure of pure solvent, the temperature has to be lowered in order for the vapour pressures to equalize and hence freezing to take place.

According to what I understand after reading my textbook, when a solution freezes, it's the solvent which converts into solid and leaves behind liquid phase consisting of both the solvent and solute. Thus, how can the pure solid solvent and liquid phase mixture have the same vapour pressure at the same temperature, since the vapour pressure of pure solvent at any temperature is greater than the vapour pressure of mixture?

Answer 1

The figure shows the relationship between the solid and liquid solvent and a solution.

Instead of freezing at the freezing point of the solvent $T_1$ a solution freezes at $T_2$, where the vapour pressure of the solution equals that of the pure solid solvent.

Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect the solid solvent curve. The vapour pressure of the solution follows the line $\displaystyle \frac{d\ln(p)}{dT}=\frac{\Delta H_{vap}}{RT}$ and the pure solid solvent $\displaystyle \frac{d\ln(p_{ss})}{dT}=\frac{\Delta H_{sub}}{RT}$ where $\Delta H_{sub}$ is the molar heat of sublimation. This in turn is given by $\Delta H_{sub}=\Delta H_{vap}+\Delta H_{fus}$ where 'fus' means heat of fusion. Thus $\Delta H_{sub} \gt \Delta H_{vap}$ which in turn means that the slope of the solid solvent curve is always greater than that of the solution $\displaystyle \frac{d\ln(p_{ss})}{dT} \gt \frac{d\ln(p)}{dT}$. As the freezing point of the pure solvent is higher than that of the solution the vapour pressure curve of the solid solvent and solution must always intersect.