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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?
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 $T<T^{\rm liquid}$ (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.
