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

$$ΔT_{f}=K_{f}\cdot m$$

where $ΔT_{f}$ is the freezing point depression (defined as positive value), $K_{f}$ 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?

  • $\begingroup$ 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 $\Delta T_f \approx m K_f$ $\endgroup$ – theorist May 7 at 2:09
  • $\begingroup$ 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. $\endgroup$ – theorist May 7 at 2:10
  • $\begingroup$ See these answers: chemistry.stackexchange.com/questions/61783/… ... chemistry.stackexchange.com/questions/61750/… $\endgroup$ – Nilay Ghosh May 7 at 4:12

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.


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Simplified description of the quantitative relationship between $\Delta T_f$ and $m$:

Assuming $T_{tr}=T_f$ (for $\ce{H2O}$, $T_{tr}=\pu{0.01^{\circ}C}$ and $T_f=\pu{0.00^{\circ}C}$ at $\pu{1atm}$) and small values of $\Delta T_f$ and $\Delta p$ (which mean $VP$ decreases on adding nonvolatile solute), $OAB$ can considered a straight line. Then $\triangle OAC$ and $\triangle OBD$ are similar, so $\frac{OC}{CA}=\frac{OD}{BD}$.

  • 3
    $\begingroup$ You should only include the actual figure as an image, with the rest being converted to text/equations. Images aren't able to be found in search and assistive reading software can't extract information from images. $\endgroup$ – Tyberius Jun 9 at 16:29

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