# Thermodynamic Derivation of Depression in Freezing Point Using Clapeyron-Clausius Equation

When a liquid is allowed to freeze, its solid starts to come out. The temperature is finally attained at which the liquid and solid forms are in equilibrium. This temperature is called the freezing point of the liquid. At this temperature, both liquid and solid forms have the same vapour pressure. Depression in freezing point is therefore the lowering of the freezing point of the liquid when solute is added to it.

Let us consider the variation of vapour pressure with temperature of pure solvent and solution.

As the vapour pressure curves of a solvent in the liquid and solid state are meeting at B, it means that at the corresponding temperature T_{o}, the solid and liquid states have the same vapour pressure. Therefore, T_{o} is the freezing point of the solvent. The vapour pressure curve of the solution lies below that of the pure solvent because, the vapour pressure of the solution is less than that of the pure solvent. This curve meets the vapour pressure curve of the solvent at D. Therefore, T which lies below T_{o}, is the freezing point of the solution.Thus, T_{o} - T = ΔT_{f} is the depression in the freezing point of the solution.

Applying the Clapeyron-Clausius equation to the liquid vapour equilibrium in the solution, we have-
where, L_{v} is the latent heat of vaporization of the pure solvent, P_{1} and P_{2} are the vapour pressure of the solution at temperature T_{o} and T respectively.

Similarly, applying the equation to the solid vapour equilibrium, we get-
where, L_{s} is the molar latent heat of sublimation of the pure solvent, P_{o} and P_{1} are the vapour pressure of the solid solvent at temperature T_{o} and T respectively.

From the above two equations, we have-

But, Molar latent heat of fusion (L_{f}) = L_{s} − L_{v}

Hence, the above equation becomes-

For dilute solutions, T_{o} = T and also T_{o} − T = ΔT

Now the above equation becomes-

Now,
Since, P_{o} − P is very small and the remainder of the terms can be neglected.

**NOTE:** when x is very very small, then ln(1 − x) = x

If 'w' gm of solution of molecular weight 'm' is dissolved in 'W' gm of solvent of molecular weight 'M', then from Raoult's law of dilute solution, we get-
Hence, the above equation backgrounded light-yellow becomes-
where, l_{f} is the latent heat of fusion per gram of the solvent. If one mole of the solute (i.e. w/m = 1) is dissolved in 1000 gram (i.e. W = 1000) of the solvent, then-

Hence, ΔT is then the molal depression constant, K_{f} of the solvent. It can be defined as the depression in freezing point produced by dissolving one mole of a solute in 1000 gram or one kilogram of the solvent.

Since all quantities on R.H.S. are constants, hence, K_{f} is constant for a particular solvent. From eqaution backgrounded light-pink and grey we get-

So, the molecular weight(m) of a non-volatile and electrolyte solute can be calculated if we know the value of l_{f} and K_{f} of the solution.