Free energy of dissolution | Applications of thermodynamics | AP Chemistry | Khan Academy

The term dissolution refers to the dissolving of one substance in a solvent. The dissolved substance is now called a solute, and the solute plus the solvent form a solution. If the standard change in free energy, delta G naught, is less than zero, the dissolution is thermodynamically favorable. So, if we were to put this substance in a solvent like water, the substance would dissolve, and it would form a solution.

However, if delta G naught is greater than zero, the dissolution is thermodynamically unfavorable. So, if we tried to dissolve the substance in water, it wouldn't dissolve, and therefore we would just see it on the bottom of the beaker. Since we're talking about potentially making a solution, sometimes you see a subscript S-O-L-N written next to delta G naught, so this would be delta G naught of solution. We could also call this the free energy of dissolution.

We can calculate delta G naught of solution by taking the standard change in enthalpy of the solution and from that subtracting the absolute temperature times the standard change in entropy of the solution. So, let's look in more detail at what determines the signs for delta H naught and delta S naught.

We can think about the dissolution of a solid in three hypothetical steps. The first step involves breaking up the solid. Let's think about the change in enthalpy, delta H1, for this first step. Solids are held together by attractive forces. For example, if we had an ionic solid, it would be ionic bonds or electrostatic interactions holding together the opposite charges. So, if our goal is to pull apart or break apart the solid, it would take energy to overcome these attractive forces. Therefore, delta H1 would be positive.

Next, let's think about the change in entropy, delta S1, for the first step. When the solid is broken up, the particles have a greater number of possible positions. An increased number of possible positions means an increase in the number of possible microstates. An increase in the number of microstates means an increase in entropy. Therefore, breaking up the solid means an increase in entropy, and delta S1 will be positive.

Step two is the preparation of the solvent to receive the solute. Let's think about the change in enthalpy for this second step. The solvent particles are held together by intermolecular forces. For water molecules, the most important intermolecular force is hydrogen bonding. The goal of this theoretical second step is to break apart the solvent particles and to move them far apart from each other so there's enough room to fit in a solute particle. Since it takes energy to overcome the intermolecular forces holding the solvent particles together, delta H2 is positive.

Next, let's think about the change in entropy, delta S2, for this second step. When the water molecules are pulled apart, there's an increase in the number of possible positions of the water molecules. Like the first step, an increase in the number of possible positions means an increase in the number of microstates, which means an increase in entropy. Therefore, delta S2 is positive.

The third step is called solvation, which refers to the interaction of the solute and the solvent. Imagine there's an attractive force between this white solute particle and these four blue solvent particles. If the solvent is water, this process is called hydration. If we think about, say, a positive cation interacting with water, since opposite charges attract, the positive cation is attracted to the negative end of the water molecule. Since we have a positively charged ion interacting with water, which is a polar molecule with a dipole moment, this type of interaction is called an ion-dipole interaction.

Let's think about the change in enthalpy for this third step. When the solute comes together with the solvent, energy is released. Therefore, delta H3 is negative. One way to think about why this is negative is to consider the ion-dipole interactions on the right. Since it would take energy to break these ion-dipole interactions, when those ion-dipole interactions form, energy must be given off.

Next, let's think about the change in entropy, delta S3, for the third step. Because the water molecules are attracted to the ions in solution, the water molecules have a decreased freedom of movement. Therefore, there are fewer positions possible for the water molecules. A decrease in the number of possible positions means a decrease in the number of microstates, which means a decrease in entropy. Therefore, for the third step, delta S3 is negative.

Now that we've talked about the signs for delta H and delta S for each of the three steps, let's think about how they influence the overall changes in delta H naught of solution and delta S naught of solution. Let's start with delta H naught of solution, which is equal to the sum of the changes in enthalpy for the three steps: delta H1 plus delta H2 plus delta H3. We've already seen that delta H1 is positive, delta H2 is positive, and delta H3 is negative.

So, adding delta H1 and delta H2 together gives us a positive value. Since we're adding a negative in delta H3, it's the magnitude of delta H3 that determines if the overall delta H naught of solution is positive or negative. If the magnitude of delta H1 plus delta H2 is greater than the magnitude of delta H3, delta H naught of solution will be positive. However, if the magnitude of delta H3 is greater than the magnitude of delta H1 plus delta H2, delta H naught of solution will be negative.

Therefore, it's possible for delta H naught of solution to be positive or negative depending on the substance being dissolved. An example of a substance that has a positive value for delta H naught of solution is sodium chloride. At 25 degrees Celsius, delta H naught of solution for sodium chloride is positive 3.9 kilojoules per mole of reaction. An example of a substance that has a negative value for delta H naught of solution is magnesium chloride. At 25 degrees Celsius, delta H naught of solution for magnesium chloride is equal to negative 160.0 kilojoules per mole of reaction.

The main reason why these two substances have such different enthalpies of solution has to do with the third step: the value of delta H3. So, let's take a look at some diagrams showing the ion-dipole interaction of the cation of both of these substances in aqueous solution. For sodium chloride, we would have a sodium cation in solution with a one plus charge. For magnesium chloride, it would be an Mg2 plus cation interacting with the water molecules.

Because the Mg2 plus cation has a higher charge than the sodium cation and because it's smaller, the Mg2 plus cation exerts a greater electric force on the water molecules. Because the magnesium two plus cation exerts a greater electric force on the water molecules, more energy is released when the magnesium two plus cation is hydrated compared to when the sodium one plus cation is hydrated. Since more energy is given off in the third step for the hydration of the magnesium two plus cation, that's enough to turn the overall delta H naught of solution negative. Therefore, the dissolution of magnesium chloride is an exothermic process.

For sodium chloride, when the sodium cation is hydrated, not as much energy is released. Therefore, the overall delta H of solution is positive, and the dissolution of sodium chloride is an endothermic process. Next, let's think about delta S naught of solution. Delta S naught of solution is equal to delta S1 plus delta S2 plus delta S3. Delta S1, the change in entropy for the first step, we saw was positive. Delta S2 was also positive, but delta S3 was negative.

Therefore, just like we saw for delta H naught, the sign for delta S naught depends on the magnitude of delta S1 plus delta S2 compared to delta S3. If delta S1 plus delta S2 is greater in magnitude than delta S3, delta S naught of solution will be positive. However, if the magnitude of delta S3 is greater than the magnitude of delta S1 plus delta S2, delta S naught of solution will be negative.

An example of a substance that has a positive value for delta S naught of solution is sodium chloride. At 25 degrees Celsius, delta S naught of solution is equal to positive 43.4 joules per kelvin mole of reaction. An example of a substance that has a negative value for delta S naught of solution is magnesium chloride. At 25 degrees Celsius, delta S naught of solution is equal to negative 114.7 joules per kelvin mole of reaction.

The main reason for the difference in entropies for these two substances has to do with the third step, so the magnitude of delta S3. Once again, let's take a look at the diagrams showing the hydration of the cations in aqueous solution. We've already talked about the fact that the smaller and more positive Mg2 plus cation has a greater electric force on its surrounding water molecules than the sodium one plus cation does on its surrounding water molecules.

The stronger electric force means a decreased freedom of movement of water molecules around the magnesium two plus cation. A decreased freedom of movement means a decreased number of possible microstates, which means a greater decrease in entropy. So, the hydration of the magnesium two plus cation leads to a greater decrease in entropy for step three, which outweighs the positive values for steps one and two, and that gives an overall negative value for delta S naught of solution.

Therefore, dissolving magnesium chloride in water results in a decrease in entropy. Since the sodium cation has a weaker electrostatic attraction for water molecules, the decrease in entropy is not as much, and therefore the magnitude of the third term does not overcome the magnitude of the first two terms. This means the overall delta S naught of solution is positive, and dissolving sodium chloride in water results in an increase in entropy.

We've just looked at delta H naught of solution and delta S naught of solution for two salts, sodium chloride and magnesium chloride. Let's calculate delta G naught of solution for both salts at 25 degrees Celsius. Let's start with the dissolution of solid sodium chloride and water to form sodium cations and chloride anions. 25 degrees Celsius is 298 Kelvin, so we can plug that into our equation. We can also plug in the values for delta H naught and delta S naught, so both of them were positive values for the dissolution of sodium chloride.

Kelvin cancels out and gives us delta G naught of solution is equal to negative 9.0 kilojoules per mole of reaction. Since delta G naught is negative, the dissolution of solid sodium chloride is a thermodynamically favorable process, which means at 25 degrees Celsius, sodium chloride is soluble in water. Notice how in this case the favorable positive entropy term outweighs the unfavorable positive enthalpy term to give a negative value for delta G naught. Therefore, for the dissolution of sodium chloride, the increase in entropy drives the dissolution.

Next, let's do the same calculation for the dissolution of solid magnesium chloride in water to form the magnesium two plus cation and two chloride anions. We've already seen how the increased positive charge of the Mg2 plus cation in solution gave us a negative value for delta H naught and a negative value for delta S naught. So, plugging in all our numbers at a temperature of 298 Kelvin, Kelvin cancels out and gives us delta G naught of solution is equal to negative 125.8 kilojoules per mole of reaction.

Since delta G naught of solution is negative, the dissolution of magnesium chloride is a thermodynamically favorable process. So, at room temperature of 25 degrees Celsius, magnesium chloride is soluble in water. However, this time it's the favorable negative value for the enthalpy term that outweighs the unfavorable negative value for the entropy term to give a negative value overall for delta G naught of solution. So, this time it's the decrease in enthalpy that drives the dissolution.

Delta G naught has been negative for both sodium chloride and for magnesium chloride. However, those are just the two examples that I chose for this video. It's certainly possible to get a positive value for delta G naught of solution, which would mean an insoluble salt. Because it can be very difficult to predict the signs for delta H naught and for delta S naught for a particular salt, it's even more difficult to predict if delta G naught is positive or negative for that particular salt. Therefore, it's often necessary to do a calculation to see if delta G naught is positive or negative.