I know that the freezing point depression and boiling point elevation are given by the formulas $\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}$ and $\Delta\mathrm{T}_{f}=ib\mathrm{K}_{F}.$ It seems that when this formula is applied for equimolar concentrations, the magnitude of freezing point depression is greater than boiling point elevation. Is there a qualitative reason for this? If so, what is it?
1 Answer
To find an answer, let us take a look at how the cryoscopic constant (for decrease in melting point) and the ebullioscopic constant (for increase of boiling point) are defined:
Please note that the Kelvin-scale is used for temperature here. The boiling point is always above the melting point (we keep pressure constant to keep the example simple).
For most substances, the enthalpy of vaporization is greater than the enthalpy of fusion. It follows that the denominator for K_eb is greater, resulting in a smaller K_eb.
This means that the effect of greater enthalpy of vaporization compared with enthalpy of fusion outweighs the effect of boiling temperature being greater than the melting temperature (even though it's the square of absolute temperature).