Now let us return to the situation with solids, liquids, and gases. How do we think about entropy in these systems? Doesn’t a substance become more ordered as we move it from gas to liquid to solid? Clearly the entropy of a solid is lower than that of a liquid (and the entropy of a liquid is lower than that of a gas). We can calculate how entropies change for materials as they go from gas to liquid to solid, and as we have predicted they decrease. How can it be that a change in which the entropy of the system decreases (for example freezing of ice) can occur? Are we forced to conclude that things we know to happen are impossible according to the Second Law of Thermodynamics?

5.1 Systems
5.2 Temperature
5.3 Vibrations
5.4 Phase changes
5.5 Thermodynamics
5.6 Phases, again

The Second Law of Thermodynamics tells us that for every change that occurs, the entropy of the universe must increase. The problem with this is that we are all well aware of changes where the entropy apparently decreases. How can we resolve this seeming paradox? The answer lies in the fact that for any system the entropy may indeed decrease - water freezing is an example of this phenomenon. However for the universe as a whole (or more easily defined, the system and its surroundings) total entropy must increase. Consider when water freezes, the water molecules form stable interactions (hydrogen bonds). As we have seen previously, the formation of stabilizing interactions lowers the potential energy of the system. Because energy is conserved, this energy must be released to the surroundings as thermal (kinetic) energy. That is, the freezing of water is an exothermic process.

Now we can see the solution to our thermodynamic problem. The reason that the freezing of water does not violate the second law is because even though the system (ice) becomes more ordered and has lower entropy, the energy that is released to the surroundings makes those molecules move faster, which leads to an increase in the entropy of the surroundings. The increase in entropy in the surroundings is greater than the decrease in entropy of the ice! When we consider both system and surroundings, the change in entropy (ΔS) is positive. The Second Law is preserved (yet again), but only when we actively embrace systems thinking.

Free Energy to the Rescue

This idea, that we must consider changes in entropy for both the system and its surroundings when we are predicting which way a change will occur (that is, in which direction a process is thermodynamically favorable) is an important one. However, it is almost always easier to look at the system than it is to look at the surroundings (after all we define the system as that part of the universe we are studying.) It would therefor be much more convenient to use criteria for change that refer only to the system. Fortunately there is a reasonably simple way to do this. Let us think about water freezing again: we can measure the enthalpy change for this process. The thermal energy change for the system, ΔHfreezing, is about -6 kJ/mole. That is 6 kJ of thermal energy are released into the surroundings for every mole of water that freezes. We can relate this thermal energy release to the entropy change of the surroundings. Entropy is measured in units of J/K – that is energy/temperature. Since we know how much energy is added to the surroundings we can calculate the entropy change that this energy produces.

Mathematically we can express this as ΔSsurroundings = ΔHsurroundings/T. And since we know that ΔHsystem = -ΔHsurroundings, (that is: the energy lost by the system = minus (-) the energy gained by the surroundings) we can express the entropy change of the surroundings in terms of measurable variables for the system. That is ΔSsurroundings = -ΔHsystem/T.

If you recall, we can express the total entropy change (the one the second law ”cares” about) as ΔStotal = ΔSsystem + ΔSsurroundings. Substitute for the ΔSsurroundings term, we get ΔStotal = ΔSsystem -ΔHsystem/T. At this point, we have an equation that involves only variables that relate to the system, these are much easier to measure and calculate. We can rearrange the equation by multiplying throughout by -T, to get -TΔStotal = ΔHsystem-TΔSsystem

The quantity -TΔStotal has units of energy, and is commonly known as the free energy change ΔG (or the Gibbs Free energy). The equation is normally written as: ΔG = ΔH – TΔS. The free energy change of a reaction is probably the most important thermodynamic term that you will need to learn about. In most biological and biochemical systems, it is ΔG that is commonly used to determine whether reactions are thermodynamically favorable. However, it is important to remember that ΔG is a proxy for the entropy change of the universe; if it is negative, universal entropy is increasing (and the reaction occurs), if it is positive, universal entropy would decrease if the reaction occurred (and so it does not). It is possible however for reactions with a positive ΔG to occur, but only if they are coupled with a reaction with an even greater negative ΔG (see chapters 8 and 9).

There are numerous tables of thermodynamic data in most texts and on many websites. Since we often want to use thermodynamic data such as ΔH, ΔS, and ΔG, it is useful to have some reference state. This is known as the standard state and is defined as 298 K temperature, 1 atmosphere pressure, 1M concentrations (room temperature and pressure). When thermodynamic data refer to the standard state they are given the superscript º (nought), so ΔHº, ΔSº, and ΔGº all refer to the standard state. However we often apply these data at temperatures other than 298 K - and while small errors might creep in the results are usually accurate enough.

What is “Free” about Gibbs Free Energy?

The reason we use ΔG or ΔGº to describe many systems (and especially biological ones), is that both the magnitude and sign tell us a lot about how that system will behave. We use ΔG (the Gibbs free energy change) rather than ΔH (the enthalpy change) because ΔG tells us how much energy is actually available to bring about further change (or to do work). It factors out for us the energy produced from the change from what that is lost to the surroundings as increased entropy (and so is not available for us to use to do work.) As an example, when wood is burnt, it is theoretically impossible for all of the heat released to be used to do work, some of the energy goes to increase the entropy of the system. For any change in the system, some of the energy is always lost in this way to the surroundings; this is why it is impossible to build a machine that is 100% efficient in converting energy from one kind to another (although many have tried - google perpetual motion machines.) The term “free energy” then doesn’t mean that it is literally free - but rather means what is theoretically available for further transformations.

When ΔG is negative, we know that the reaction will be thermodynamically favored.[Many people use the term spontaneous - but this is misleading since it infers that the reaction will happen right away. In fact ΔG tells us nothing about when the process will happen - only that it is favored. As we will see later the rate at which a process occurs is governed by other factors. ] The best case scenario is when ΔH is negative (an exothermic change - that is the system is losing energy to the surroundings and becoming more stable), and ΔS is positive - (that is the system is increasing in entropy). Because T is always greater than 0 (in degrees K), TΔS will be positive and this value will be subtracted from ΔH to given an even larger negative ΔG value. A good example of such a process is the reaction (combustion) of sugar (C₆H₁₂O₆) with molecular oxygen (O₂):

C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(g).

This is an exothermic process and, as you can see from the reaction equation, it results in the production of more molecules than we started with (often a sign that entropy has increased.)

A process such as this (-ΔH and + ΔS) is thermodynamically favored at all temperatures. On the other hand an endothermic process (+ΔH) and a decrease in entropy (-ΔS) will never occur as an isolated reaction (Ah, but so few reactions are actually isolated from the rest of the universe.) For example a reaction that combined CO₂ and H₂O to form sugar (the reverse reaction to the combustion above) is never thermodynamically favored since ΔH is positive and ΔS is negative, making ΔG always positive at all temperatures. Now you may find yourself shaking your head - everyone knows that the formation of sugars from carbon dioxide and water goes on all over the world right now (in plants). The key here is that plants use energy from the sun.

The reaction is actually:
“captured energy” + 6 CO₂(g) + 6 H₂O(g) → C₆H₁₂O₆(s) + 6 O₂(g) + “excess energy” .

Just because a process is thermodynamically unfavored doesn’t mean that it can never occur; what it does mean is that that process cannot occur in an isolation, it must be “coupled” to other reactions.

Free energy, and temperature

So we have two very clear cut cases that allow us to predict whether a process will occur - where the enthalpy and entropy predict the same outcome. But there are also two other situations where the enthalpy and entropy “point” in different directions. When this happens we need to use the fact that the free energy change is temperature dependent to predict the outcome. Recall the expression ΔG = ΔH – TΔS depends upon temperature. For a system where the entropy change is positive (+ΔS), an increase in temperature will lead to an increasingly negative contribution to ΔG. That is, as the temperature rises - a process that involves an increase in entropy will become more favorable. Conversely, if the system change involves a decrease in entropy, (that is ΔS is negative) then ΔG will become more positive (and less favorable) as the temperature increases.

This idea, that the temperature affects the direction of some processes is perhaps a little disconcerting - it goes against common sense that if you heat something up a reaction might actually stop and go backwards (and rest assured we will come back to this point later). But in fact there are a number of common processes where we can apply this kind of reasoning and find that they make perfect sense.

Up to this point, we have been considering physical changes to a system, that is populations of molecules going from solid to liquid or liquid to gaseous states (and vice versa). Not really what one commonly thinks of as chemistry, but the fact is that these transformations involve the making and breaking of interactions between molecules. We can therefore consider phase transitions as analogous to chemical reactions, and because they are somewhat simpler, develop logic that applies to both processes. So let us begin by considering a phase change/reaction system H₂O (liquid) ↔ H₂O (gas).

We use a double arrow ↔ to indicate that, depending upon the conditions, the reaction could go either to the right (boiling) or to the left (condensing). So, let us assume for the moment that we do not already know that water boils (changes from liquid to gas) at 100ºC, what factors would determine whether the reaction H₂O (liquid) ↔ H₂O (gas) favors the liquid or the gaseous state at a particular temperature? As we have seen the criterion for whether a process will “go” at a particular temperature is ΔG. We also know that the free energy change for a reaction going in one direction is the negative of the ΔG for the reaction going in the opposite direction. So that, the ΔG for the reaction:

H₂O (liquid) ↔ H₂O (gas) is –ΔG for the reaction H₂O (gas) ↔ H₂O (liquid).

When water boils all the intermolecular attractions between the water molecules must be overcome, allowing the water molecules to fly off into the gaseous phase. Therefore the process of water boiling is endothermic (ΔHvaporization = +40.65 kJ/mol); it requires an energy input from the surroundings (when you put a pot of water on the stove you have to turn on the burner for it to boil). When the water boils, the entropy change is quite large (ΔSvaporization = 109 J/mol K), as the molecules go from being relatively constrained in the liquid to gas molecules that can fly around. At temperatures lower than the boiling point the enthalpy term predominates and ΔG is positive, but as you increase the temperature in your pan of water, eventually it reaches a point where the contributions to ΔG of ΔH and TΔS are equal. That is ΔG goes from being positive to negative and the process becomes favorable. At the temperature where this “cross over” occurs ΔG = 0 and ΔH = TΔS. At this temperature (373 K, 100 ºC) water boils (at 1 atmosphere). At temperatures above the boiling point ΔG is always negative and water will exist predominantly in the gas phase. However if we let the temperature drop below the boiling point, the enthalpy term becomes predominant again and ΔG for boiling is positive - that is: water will not boil at temperatures below 100 ºC (at one atmosphere).

Let us now consider a different phase change, for example water freezing. In this case as the water molecules in the liquid start to aggregate and form hydrogen bonds with each other, energy is released to the surroundings (remember it is this energy that is responsible for increasing the entropy of the surroundings.) Therefore ΔH is negative - freezing is an exothermic process (ΔH_fusion = – 6 kJ/mol.) However freezing is also a process that reduces the system entropy. When water molecules are constrained, as in ice, their positional entropy is reduced. So water freezing is a process that is favored by the change in enthalpy and disfavored by the change in entropy. As the temperature falls, the entropy term contributes less to ΔG, and eventually (at the cross over point) ΔGfusion goes to zero and then becomes negative - that is the process becomes thermodynamically favored.

Water freezes at temperatures below 0 ºC. At temperatures where phase changes take place, (boiling point, melting point) ΔG = 0. Furthermore, if the temperature were kept constant - there would be no observable change in the system. We say that the system is at equilibrium; for any system at equilibrium, ΔG = 0.

5.1 Systems
5.2 Temperature
5.3 Vibrations
5.4 Phase changes
5.5 Thermodynamics
5.6 Phases, again

Question to answer:

• For each of these processes give the change in entropy of the system, the direction of thermal energy transfer (the sign of ΔH), the change in entropy of the surroundings and the change in entropy of the universe.
  Water freezing at -10 °C,
  Water boiling at 110 °C
• For each of these processes predict the sign of change in entropy (ΔS) of the system, the direction of thermal energy transfer (the sign of ΔH),and the sign of the Gibbs free energy change ΔG. What does the sign of ΔG tell you?
  Water freezing at -10 °C,
  Water boiling at 110 °C
  Water boiling at -10 °C,
  Water freezing at 110 °C
  

Questions to ponder and questions for later:

• Why do we put excess energy as a product of this equation?
• What other processes do you know that must be coupled to external energy sources to make them “go”?