Energetics

While working in a chemistry laboratory, you may have noticed that your beaker gets hot or sometimes cold when you dissolve a compound in water. Have you wondered, what is the reason behind this observation? 

To understand this process, we must think about the individual steps that must be carried out when a solute is dissolved in a solvent:

Let’s break the process of solution formation into three small steps.

1. Particles of solute must separate –  In the solutes, which are in solid or liquid state, particles are held together through various inter molecular forces, such as ionic bonds,, hydrogen bonds, dipole dipole or London forces, as discussed in previous chapter.  When these compounds are allowed to mix with solvent, first the molecular units of solutes must be pulled apart.  To achieve that we need to overcome some intermolecular forces and therefore this step requires energy.

Fig 11.5

• Particles of Solvent separate – Water is a polar compound. It consists of an extended network of H₂O molecules linked together by dipole-dipole attractions that we call hydrogen bonds. These types of compounds are also known as “associated liquids”.

When a solute molecule is introduced into an associated liquid, a certain amount of energy is required to break the local hydrogen-bond structure and make space for the newly introduced solute molecule.  Hence the process will be endothermic.

Fig 11.6

• Solute particles will get surrounded by solvent particles and make a homogenous mixture.

Fig 11.7

Since water is polar, if the solute  is an ion or a polar molecule, new ion-dipole or dipole-dipole attractions come into play. In favorable cases, these may release sufficient potential energy to largely compensate for the energy required in first two steps of solution formation. As a result in some cases solution may even fill hot to touch or oven start boiling.

As we can see in this picture a polar ammonia molecule will be attracted towards polar water molecule and they will have strong force of attraction towards each other.  Each NH₃ molecule can form three hydrogen bonds with surrounding water molecules, so the resulting solution is even more hydrogen-bonded than is pure water — accounting for the considerable amount of heat released in the process and results in extraordinarily large solubility of ammonia in water.

Fig 11.8                                

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Fig. 11.9

For easy understanding, these three steps can be summarized in the diagram above. Here, the first two processes, separation of solute and separation of solvent, require energy, or you can say these two steps are endothermic. The ∆H value for these steps will be positive. However, when solute and solvent interact and mix into each other, this process will release energy, or it will be exothermic and ∆H_Mix value will be negative.

∆H_Solution = ∆H_Solute + ∆H_solvent + ∆H_Mix

Depending on the relative values of ∆H_Solute , ∆H_solvent , and ∆H_Mix , the overall process can be exothermic or endothermic.

The diagram above fits in perfectly for the solution of NaCl (Table Salt) in water.

First step:, solute particles separate. There is a strong force of attraction between Na⁺ and Cl^– ions. To separate these two ions in solution, a large amount of energy is needed.  If you recall from earlier chapters, we call it the lattice energy of NaCl. ∆H_Solute will be a large positive number in this case.

Second Step: Solvent particles separate.  Water molecules hold on to each other through strong intermolecular forces, hydrogen bonds.   To separate water molecules, we will also need a large amount of energy. So, ∆H_Solvent will also be a large positive number in this case.

In the third step of solution formation, these water molecules interact with sodium and chloride ions. Na⁺ and Cl^– ions will have a strong force of attraction towards polar water molecules, ion-dipole inter molecular force will be there. Hence ∆H_Mix will be a large negative number. Thus, a solution will form, and depending on the relative values of ∆H_Solute , ∆H_solvent,  ∆H_Mix, the ∆H_Solution can be negative, positive, or zero.

On the other hand, dissolving the oil in water has very different dynamics because oil is nonpolar and has weak London forces between its molecules.

Step 1: Solute particles separate. For the separation of the solute, the oil, in this case, you do not need as much energy. Oil is nonpolar and has weak London forces between its molecules.  To overcome these forces and separate oil molecules only a small amount of energy will be needed. So, ∆H_Solute will be a small positive number.

Step 2: Solvent particles separate. To separate water molecules from each other, which are holding on to each other through hydrogen bond, we again need a lot of energy.  Hence, ∆H_Solvent will be a large positive number like in the last example.

Step 3: solute and solvent particles interact. The most significant difference in this example will be that there will be negligible interactions between the nonpolar oil and polar water molecules. Hence ∆H_Mix will be a small number. Since ∆H_Solution = ∆H_Solute + ∆H_solvent + ∆H_Mix, the ∆H_Solution will turn out to be a large positive number. That’s why non polar solutes are insoluble or sparingly soluble in polar solvents such as water.                             

The hydrophobic effect:

It turns out, that this is not the only reason for small solubility of non polar solutes in water. It is now known that the H₂O molecules that surround a non polar intruder and find themselves unable to form energy-lowering polar or hydrogen-bonded interactions with the solute will rearrange themselves into a configuration that maximizes the hydrogen bonding with other water molecules. In doing so, this creates a cage-like shell around the solute molecule such as oil. In terms of the energetics of the process, these new H2O-H2O interactions largely compensate for the lack of solute-H2O interactions.  As you can see in the picture on the left, water molecules formed a cage like structure around the non polar solute particle.

Fig.11.10

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There is one more factor which is responsible for low solubility of nonpolar solutes in polar solvent.  It is the overall entropy of solution. Entropy is the measure of randomness, how disorderly particles are in a system determines the entropy of system. Things in the universe always favor more randomness or higher entropy.  Thus, an increase in entropy will also favor the solution formation.

Dissolution of a solute normally increases the entropy by spreading the solute molecules through the larger volume of the solvent molecules. However, the type of solute and solvent interation we saw above between non polar oil and polar water molecules, the cage of highly organized water molecules exerts its own toll on the solubility by reducing the entropy of the system. Here H₂O molecules are constrained to their location in this cage like structure, and their number is sufficiently great to reduce the overall entropy of this solution. Hence, further reduce the solubility of solute.

The small solubility of a non polar solute in an associated liquid such as water results more from the negative entropy change rather than from energetic considerations. This phenomenon is known as the hydrophobic effect.

The implications of the hydrophobic effect extend far beyond the topic of solubility. It governs the way that proteins fold, the formation of soap bubbles, and the formation of cell membranes.

The Effect of Entropy on Dissolution

When we put a small quantity of solid or liquid in a much larger volume of solvent, the solute particles spread randomly all over the solvent and the entropy of solute will increase.

If the energetics of dissolution are favorable, this increase in entropy means that the conditions for solubility will always be met. Even if the energetics are slightly endothermic the entropy effect can still allow the solution to form, although it may limit the maximum concentration that can be achieved.

In such a case, we may describe the solute as being slightly soluble in a certain solvent and a greater volume of solvent will be required to completely dissolve a given mass of solute.