# What is the mechanism of endothermic dissolution?

My real question is in fact "How can dissolution happen if it's endothermic?" However, there have been many questions regarding this exact topic and the answers about Gibbs free energy determining spontaneity of the reaction doesn't help my intuition at all. The majority of answers list the increase of entropy as the cause of endothermic dissolution, but in my opinion, entropy doesn't make sense as the cause.

Let's take a similar problem, namely the melting of a solid. If we place a solid with lower temperature into surroundings with higher temperature (enough for the solid to reach its melting point), it's not that liquification would cause the temperature of the surroundings to drop, but rather that the transfer of heat from the surroundings to the solid would cause liquification. The acquired heat allows the solid to increase its entropy, not the other way around.

So, if increased entropy is the result of endothermic dissolution, what is the cause? Or, what is the mechanism that allows entropy to be increased?

Let's take the dissolution of NaCl in water as an example. This process is endothermic by around 4 kJ/mol. My guess is that, since some ions in the solid have higher kinetic energy than average, this energy will be sufficient to overcome the 4 kJ/mol barrier and leave the solid, taking some of its thermal energy away from it. As a result, the temperature of the remaining solid will drop a little. Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal. This will result in lowering the temperature of water and in the increase of entropy. Is my thinking correct? If not, why?

• quote: So, if increased entropy is the result of endothermic dissolution, what is the cause? The external temperature difference, set up using an external mechanism - work done in making this system is equal to the heat content it has, which increases the entropy after dissolution. Nov 7, 2020 at 14:40
• The main problem is to think on a part of the system, or taking the system as composed by the subject you are looking at. A piece of ice normally does cool down the normal environment. At least nearby.... That is heat flowing in. Nov 7, 2020 at 21:02
• Have you heard about entropic explosion ? Nov 8, 2020 at 9:23

Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal

It is fine to think that way as a rough draft in rationalizing this process. The only issue is that temperature is a macroscopic quantity, and mechanism tends to be a particular (molecular) concept. So you could say that the individual solid particles (e.g. ions) will have slightly lower kinetic energy than average, and collisions with individual solvent molecules will balance this out.

Or you could start with solvent molecules near the solid-liquid interface that already have a higher kinetic energy than the average become part of the solid, thus lowering the average kinetic energy of the remaining solute molecules.

Or you could say that the water molecules solvating the ions use parts of their kinetic energy in the desolvation process, directly lowering the temperature of the solvent.

Because all the species have a distribution of kinetic energies around the average, all these scenarios are correct.

As a result, the temperature of the remaining solid will drop a little.

I don't think the bulk of the solid will have a lower average kinetic energy, so I don't think the temperature of the solid will drop faster than that of the solution. I would think that there is a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand. This is maintained as long as there is a net transfer from solution to solid. Once equilibrium is reached, of course, this will include thermal equilibrium.

this energy will be sufficient to overcome the 4 kJ/mol barrier

"Barrier" sounds like an activation energy, something you can step over. Here, I would rather talk of "energetic cost" or "step up", something that you don't get back. The solvation energy is higher than the lattice energy, so there is a cost for taking the ions out of solution and adding them to the ionic solid.

• Thank you for the answer! If I may, I have a few points to clarify. Firstly, if an ion with kinetic energy equal to the average kinetic energy got separated from the rest of the solid, the average KE of the bulk solid would stay the same. But if it was an ion with more than the average KE, shouldn't its separation take away energy that was, so to say, meant for other ions, leaving the bulk with less KE on average? Furthermore, could you please clarify what you meant by "a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand"? Nov 7, 2020 at 19:02
• @HeatherB With the temperature gradient, I meant that the surface of the solid will be a bit colder than the inside of the solid. Also, the solution near the solid will be a bit colder than the rest of the solution. That's because nothing is happening anywhere except at the surface of the solid.
– Karsten
Nov 7, 2020 at 19:50
• @HeatherB "KE meant for other ions". Let's take a simpler example: a gas in two containers connected by a tube. By chance, a particle with high kinetic energy goes from the right to the left, and one with low kinetic energy goes from the left to the right. This will result in a small temperature difference. These temperature fluctuations are normal, and tiny given the large number of particles in most systems.
– Karsten
Nov 7, 2020 at 19:54