# What is the derivation for the expression of depression in freezing point?

In my textbook, it is written that the expression for depression in freezing point is

$$\Delta T_\mathrm f=K_\mathrm f\cdot m$$

where $$\Delta T_\mathrm f$$ is the freezing point depression (defined as positive value), $$K_\mathrm f$$ is a constant of the solvent, and $$m$$ is the molality of the solute.

But how did this expression come about? There is no derivation given in my textbook. Can someone please rigorously prove this expression?

• That's likely because you are using a basic introductory textbook. Getting into the derivation is usually considered beyond the scope of an introductory chemistry/biochemistry class. Most physical chemistry textbooks will show it. If you do the derivation, you'll end up with $\Delta T_f \approx \frac{R T_f^2}{\Delta H_f} m$ And since $\frac{R T_f^2}{\Delta H_f}$ is a constant, it's replaced with $K_f$, giving $\Delta T_f \approx m K_f$ May 7, 2021 at 2:09
• If you want to see what a derivation looks like, you can find one here: chem.libretexts.org/Bookshelves/… But this derivation won't make sense to you unless you understand the starting expression for chemical potential as a function of mole fraction, which is something you will learn if and when you take physical chemistry. May 7, 2021 at 2:10
• May 7, 2021 at 4:12
• The relationship can be determined experimentally. Take any substance, dissolve it in a liquid with some known concentration and record the new freezing point. The new freezing point is linearly correlated with the concentration of the solute regardless of the identity of the solute. See Raoult's 1882 study determining Kf for benzene: gallica.bnf.fr/ark:/12148/bpt6k30518.image.f187.langEN Jul 9, 2021 at 17:15

The most rigorous treatment comes from Edward Guggenheim's book "Thermodynamics: An Advanced Treatment for Chemists and Physicists" (ISBN 978-0444869517).

Let's assume an equilibrium between a liquid solution and the pure solid solvent 1. Let's also assume pressure is constant. At equilibrium between liquid mixture at its freezing point and the pure solid at temperatur $$T$$ we assume the chemical potential $$\mu$$ is equal:

$$\mu^{\rm mixture}_1(T) = \mu^{\rm solid}_1(T)$$

Let's say $$T^{\rm liquid}$$ is the equilibrium temperature of the pure liquid (the freezing point of the pure liquid), so we obtain:

$$\mu^{\rm liquid}_1(T^{\rm liquid}) = \mu^{\rm solid}_1(T^{\rm liquid})$$

If we divide equation $$(2)$$ by $$(1)$$, we get:

$$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \frac{\mu^{\rm solid}_1(T^{\rm liquid})}{\mu^{\rm solid}_1(T)}$$

This can be rewritten to:

$$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \left( \large \frac{\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}}{\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}} \right)$$

Taking the logarithms:

$$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) - \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right)$$

We now apply the following equation (which we can obtain from the Guggenheim square):

$$\displaystyle \frac{\partial \mu_1}{\partial T} = \frac{- H_1}{RT^2}$$

We get

$$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm solid}}{RT^2} dT$$

and

$$\displaystyle \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm liquid}}{RT^2} dT$$

Combining those equations results in:

$$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = - \int_{T^{\rm liquid}}^{T} = - \int_{T^{\rm liquid}}^{T} \frac{{\Delta}_f H_1^{\rm \ominus}}{RT^2} dT$$

In this equation, $${\Delta}_f H_1^{\rm \ominus}$$ is the proper enthalpy of fusion of the pure solvent.

We now apply the definition of the chemical potential $$\mu$$:

$$\mu_1 = \mu_1^{\ominus} + RT \phi {\sum}_s r_s$$

Here, $$\phi$$ denotes the osmotic coefficient of the solution at its freezing point. We can rewrite the equation to:

$$\displaystyle \phi {\sum}_s r_s = \left< {\Delta}_f H_1^{\rm \ominus} \right> \left( \frac{\frac{1}{RT}}{\frac{1}{RT^{\rm liquid}}}\right)$$

In this equation, $$\left< {\Delta}_f H_1^{\rm \ominus} \right>$$ is the average value of $${\Delta}_f H_1^{\rm \ominus}$$ over the reciprocal temperature interval $$1/T^{\rm liquid}$$ to $$1/T$$. Because $${\Delta}_f H_1^{\rm \ominus}$$ is always positive, it follows $$T (The freezing point of the solution is always below that of the pure solvent if the solid phase is pure solvent).

Approximation: For dilute solutions, $$\Delta T = T^{\rm liquid}-T << T^{\rm liquid}$$, so we get

$$\displaystyle \phi {\sum}_s r_s = \frac{{\Delta}_f H_1^{\ominus}(T^{\rm liquid}-T)}{RT^2}$$

We rearrange this equation to:

$$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s r_s \frac{RT^2}{{\Delta}_f H_1^{\ominus}}$$

For calculations, we prefer the molalities $$m_s$$ over solute-solvent mole ratios $$r_s$$:

$$m_s = r_s/r^{\ominus}$$

We obtain:

$$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s m_s \left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$$

The factor $$\left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$$ is the cryoscopic constant of the solvent.

To derive it, you have to know how an equilibrium depends on concentration and on temperature.

At the freezing point, pure liquid and the solid are at equilibrium. If you lower to concentration of the liquid by adding solute, you disturb the equilibrium. If you lower the temperature by the right amount, you are back at equilibrium.

The relationship you get is:

$$x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\ \Delta T$$

This is from a General Chemistry text that also does not derive it. With some algebra, you can express $$K_f$$ in terms of the enthalpy of fusion comparing this and the equation given by the OP.

Can someone please rigorously prove this expression?

Without knowing what established results of thermodynamics could be used in the derivation, one would have to develop multiple chapters of a thermodynamics textbook. Calling it a proof would probably be overstating it.