I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check with. My book is also very confusing on this subject, and I don't have a world renowned memory (meaning I'd rather understand than memorize the equation), so I tried to think about it and came with the following conclusions:

-Freezing Point Depression: When a solvent is not pure and has particles dissolved in it, there are constituents (be it molecules, ions, etc.) that take up volume in the solution. For freezing to occur, these constituents need to aggregate to form a lattice, bonds, etc. Say we have a pure solvent X. This process occurs at a temperature Y. However, when this solvent is not pure and has dissolved solutes, there is "blockage" impeding these constituents from aggregating and bonding. Therefore, the temperature Y needs to be reduced further to slow down the kinetic energy of all constituents in solution (solvent and solute), which gives a greater statistical probability of intermolecular forces getting an opportunity to bind solvent-solvent constituents.

-Boiling Point Elevation: Boiling, if I'm not mistaken, occurs when the partial pressure of a vapor exceeds atmospheric pressure. For a gas to enter the vapor phase, molecules in solution need to have a certain kinetic energy (as specified by a Boltzmann distribution) to free themselves of the intermolecular forces in the solution phase and become a vapor. When a pure solvent is now diluted with other particles, the new solute-solvent bonding energy needs to be overcome to allow a solvent molecule to exist as a vapor. One way to overcome this energy is to increase the average kinetic energy of all molecules, and this is done through a temperature increase.

This is for an introductory chemistry course, and I'm not sure if my explanations are sound. I just want a yay or a nay in terms of my logic, because I don't have the most intimate background in chemistry and I'm trying to conceptually understand things.


2 Answers 2


I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.

So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. Boltzmann's entropy formula tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$.

Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by

\begin{equation} G = H - TS \end{equation}

where $H$ is the enthalpy and $T$ is the temperature, so that

\begin{equation} \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S}) \end{equation}

Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} < S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.

An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$. Requiring the Gibbs free energy of vapourization

\begin{equation} \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S}) \end{equation}

to be equal to zero gives

\begin{equation} \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S \end{equation}

Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} > S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} < \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.

  • 2
    $\begingroup$ "the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules” Can you explain why you’ve made this assumption? $\endgroup$ Apr 2, 2016 at 14:47
  • 2
    $\begingroup$ This is an excellent answer. I wish more people had seen and upvoted it. Dang it, I wonder why the O.P didn't accept it. @lightweaver: The solid phase only consists of the solvent molecules. Since we're looking at the situation from an entropic point of view, it is safe to say that in both cases, the entropy of the solid phase doesn't change due to this (correct) assumption that the solid phase is pure. $\endgroup$
    – user33789
    Dec 17, 2016 at 4:30
  • $\begingroup$ @Kaumudi.H But why is the solid phase pure? Where did the solute disappear to? Did it crystallize into its own phase? $\endgroup$ Dec 18, 2016 at 5:35
  • $\begingroup$ Philipp, in almost every textbook that I look into, and online notes indicates that colligative property is only and only dependent on a solute's concentration in a solution. And my doubt has been, is it temperature dependent as well? because concentration of a solution is highly temperature dependent. Colligative properties on their own, thrive on "points" (freezing point - a temperature, boiling point - a temperature, vapour pressure - is temperature dependent). Is colligative property temperature dependent or not? $\endgroup$
    – bonCodigo
    Jul 10, 2017 at 22:51
  • $\begingroup$ @bonCodigo Well, if you look at the derivation of the freezing point depression, e.g. here, you'll find that it is dependent on the freezing temperature of the pure solvent. In that respect it certainly is temperature dependent. Apart from that: most statements about colligative properties are made for ideal solutions and only for those they will hold. In real solutions you get interactions between the molecules and those lead to (sometimes sizeable, sometimes very small) deviations from the predicted behavior... $\endgroup$
    – Philipp
    Jul 11, 2017 at 1:55

The depression of freezing point and elevation of boiling point are, as has been described by @phillip, due to entropic effects.

The basic idea can be seen in the change in free energy with temperature as shown below. The phase change occurs where two solid lines cross. However, rather that use free energy G, which is extensive it is more general to use the molar free energy $G_m$ which is intensive (just like T and p) and is usually given the symbol $\mu$ and called the chemical potential.

Just as in mechanics where a ball will spontaneously roll down hill to the lowest potential so in a chemical system the direction of spontaneous change is that which lowers the chemical potential.

Free energy is defined as $\mu = G_m = H_m-TS_m$ where the subscripts m indicate molar quantities. The most stable state of a substance is that with the lowest chemical potential.

At low temperatures $\mu$ is determined by the enthalpy $H_m$ because the entropy changes only slowly on increasing the temperature. This is because a solid has strong intermolecular interactions and molecules remain in their relative places as the temperature increases.

At high temperature $\mu$ is dominated by $TS_m$ and so the chemical potential of a gas decreases rapidly with temperature. In the liquid the situation is such that neither term dominates entirely on its own. But the entropy does increase from that of the solid simply because molecules can now exchange places with one another, which is not possible in a solid, and have more freedom to move in other ways.

The chemical potential of a solution of an involatile solute is lower than that of a pure solution because it has a higher entropy due to the increased number of arrangements that the two types of molecules can take compared to a pure liquid. Entropy is $ S=R\ln(\Omega)$ for $\Omega$ different arranegments. The figure shows plots of chemical potential vs temperature. It can be seen that compared to the pure liquid the solution has a lower melting point and a higher boiling point.



The dotted lines show only that it is possible to super-heat and super-cool at a phase change.

The change in free energy vs temperature at constant pressure is calculated using $dG=Vdp-SdT$. As $dG$ is an exact differential ($dG$ depends only on starting and ending values, not the path taken) differentiating at constant pressure gives $$\left (\frac{\partial G}{\partial T}\right)_p=-S$$ or $$\left (\frac{\partial \mu}{\partial T}\right)_p=-S_m$$ which shows that the slope of the change of free energy (or chemical potential) with temperature is $-S$. The third law means that entropy $S$ is always a positive quantity, and thus we can conclude that at constant pressure the chemical potential always decreases with increase in temperature and the rate of decrease is greatest for systems with the greatest entropy.

  • $\begingroup$ In case of solutions we can say that $G=G(T, p, n_1, n_2)$ where $n_1$ is the number of solvent's particles and $n_2$ the number of solute's particles. Is it valid to divided by $n$, where $n=n_1 + n_2$ in order to get $G_m$? Is it because $G$ is an extensive quantity with respect to $n_1$ and $n_2$? $\endgroup$
    – ado sar
    Jan 2, 2022 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.