# Solubility of Organic Compound confusion

My teacher wrote the following on the board today:

A solute will only dissolve if the energy required to break the bonds in the solute and solvent is less than the energy released from the bonds formed between the solute and solvent.

I am aware that "like dissolves like" but not entirely sure what the sentence above means.

From the bond enthalpy chapter I learnt that, in an exothermic reaction the bonds formed are stronger due to lower in energy and thus more stable. Although I am not sure whether this is related to solubility, it seems to me that they are contradictory.

So can someone explain to my as to why bonds formed has less energy released than the broken bonds?

• Where exactly do you see a contradiction? – Ivan Neretin Jun 1 at 12:18
• suggestion: re read the quoted para again; you are confusing bond formation with bond breaking (in the solution) – Aditya Jun 1 at 13:28
• @IvanNeretin Sorry for not being specific enough, it says bond broken is less energy, whereas in the context of bond enthalpy, bonds forms are less energy. – Negrawh Jun 2 at 1:57
• Got it. Well, then you oversimplify things down to complete nonsense. Indeed, it says that bonds broken mean less energy than something (what?), while bonds formed mean less energy than another something. – Ivan Neretin Jun 2 at 5:25
• In addition to enthalpy, beware of entropic effects when molecules are very flexible. – Alchimista Jun 2 at 12:13

There are two regions of bond energy to look at here:

1. The bonds IN a molecule: a carbon-hydrogen bond has about $$\pu{100 kcal/mol}$$ of stabilization. An oxygen-hydrogen bond is a little stronger, about $$\pu{110 kcal/mol}$$. Stronger means HIGHER in stabilization energy; but we draw diagrams where the stronger bonds are designated by lines that are LOWER on the diagram. We hope no confusion results!

2. The bonds BETWEEN molecules of a pure material or a mixture. These (much weaker) bonds between molecules of a substance can be estimated by heats of fusion or vaporization. The heat/enthalpy of fusion of $$\ce{H2O}$$ is $$\pu{1.435 kcal/mol}$$, and for $$\ce{CH4}$$, $$\pu{0.223 kcal/mol}$$. Hydrogen bonding (not the regular bond to hydrogen, but the polar bonding between molecules) falls in this region of energy and is frequently encountered when examining solutions.

The bonding between water molecules is strong enough to resist separation by methane molecules because the gain could be no more than the stabilization in methane ($$\pu{0.223 kcal/mol}$$), but the loss would be as much as $$\pu{1.435 kcal/mol}$$. The first sentence in the question refers to this region of bond energies, not to the much higher energy region (~ $$\pu{100 kcal/mol}$$) that would be extensively covered in a chapter on bond enthalpy.

You can think of the process of solubilization of a solute A as a reaction:

$$\ce{A(\text{solid})-> A(\text{soln})}$$

If the reaction results in a net formation of bonds - by which it is meant that the interactions in the product are stronger than in the reagent - then there must be an accompanying release of excess thermal energy to the solvent to balance the reduction in potential energy. If the reaction is performed at constant temperature and pressure (in contact with a thermostat) then that thermal energy will be transferred as heat to the surroundings, and we call the process exothermic.

Unfortunately however, the statement that your teacher wrote, while at times true, is not general. It is in fact common to dissolve a solute with accompanying net breaking of bonds, that is, with bonds among product being higher in energy (weaker) than in the reagent. As you might imagine, this requires an input of thermal energy, and if the temperature is kept constant the result is a transfer of heat into the system.

While the exothermic case can occur (up to a point) even when the system loses entropy, the later endothermic scenario requires some way of compensating the surroundings for the loss of thermal energy, since that loss results in a decrease in the entropy of the surroundings. This compensation takes the form of an increase in the entropy of the system.

A good example is a solution of $$\ce{NaCl}$$ (at normal TP). The process is endothermic ($$\Delta_{\text{solv}} H>0$$) but the entropy change compensates for the heat gained from the surroundings.