# Entropy and Gibbs Free Energy

I've been struggling with the notion of entropy and gibbs free energy for almost three days now. Different sources on and off the internet say different things about entropy.

Gibbs Free Energy is said to be both a measure of spontaneity and maximum work extractable from a reaction, and somehow I am unable to reconcile the two ideas. While the laws of probability favour increases in entropy as a system can take on microstates, what is the cause of the energetics associated with it. How can it be both a "measure of spontaneity"and "maximum work".

You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here.

The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe $\Delta S_\text{univ}$ is either zero or strictly positive for any physical change that occurs in it. I specify physical to stress that not all possible changes necessarily meet this constraint. In particular, when it comes to chemical reactions, they tend to happen in one way and not the other.

It would be fine to stick to this definition: a chemical reaction is physically favoured if it leads to an increase of the entropy of the universe.

However, it is not very convenient because most of the time we care about a particular system and not the universe as a whole.

It is therefore common to partition the universe in two parts: the system of interest and the environment.

We then apply the property of additivity go entropy that says that $\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{env} \geq 0$

This is the most general statement we can make although it is not yet very useful.

It is now time to become more specific about the conditions under which the change or the reaction or the transformation in the system will occur.

For chemical reaction, it is often the case that they are carried at constant pressure $P$, temperature $T$ and mass $M$ or amount of matter.

The key is then to express $\Delta S_\text{env}$ in terms of the system thermodynamic properties to end up with a closed condition to be satisfied by the system only to fulfil the second law for the entire universe.

• Since the temperature is fixed it means that the environment acts as a thermostat and we can write, by definition, that $\Delta S_\text{env} = \frac{Q_\text{env}}{T}$ where $Q_\text{env}$ is the heat received by the environment during the change in the system. Then, since the only exchanges of heat occur between the environment and the system, it has to be the case that $Q_\text{env} = - Q_\text{sys}$ where $Q_\text{sys}$ is the heat received by the system during the transformation. We now apply the first law of thermodynamics that tells us that $\Delta U_\text{sys} = W_\text{sys} + Q_\text{sys}$ from which we deduce that $Q_\text{env} = W_\text{sys}-\Delta U_\text{sys}.$

• We now use the fact that the pressure is constant during the process. If $\Delta V_\text{sys}$ is the change of volume of the system during the transformation then we can write $W_\text{sys} = -P\Delta V_\text{sys} = \Delta (P_\text{sys}V_\text{sys}).$

Putting all this together we get that

$$\Delta S_\text{sys} + \frac{-\Delta (PV_\text{sys})-\Delta U_\text{sys}}{T} \geq 0$$

Upon multiplying this last equation by $-T$ and using that fact $T$ is constant during the reaction, we get that the second law of thermodynamics is satisfied (for the whole universe) iff the system satisfies:

$$\Delta (U_\text{sys}+PV_\text{sys}-TS_\text{sys}) = \Delta G_\text{sys}(P,T) \leq 0$$

That's for the spontaneity aspect.

For the maximum work extractable aspect, it comes from the assumption above that the work is only due to the imposed pressure and changes in the system volume. You can write more generally $W_\text{sys} = W_\text{other} -P\Delta V_\text{sys}$.

By redoing the same type of calculation as before, you then end up with the following relation for the second law of thermodynamic to hold:

$$\Delta G_\text{sys}(P,T) \leq W_\text{other}$$

In particular, this other work is also related to the work to provide to reverse a chemical reaction.

• my bad, I hadn't seen the date of the question... – gatsu Jun 30 '15 at 10:05
• This is the clearest explanation I've seen so far. But it assumes that the initial and final system state have the same P and T. What if they don't? Does G no longer apply, or do we just need a better answer to the question in that case? – JoelG Feb 5 '16 at 12:01
• @JoelG: indeed if it is not the same P and T, then this is not $G$ that matters anymore. That is because $G$ is the good thermodynamic potential, i.e. the function from which derives the driving forces in your system, only when P and T are fixed. If it is V and T that are fixed for instance, then the good potential, and the possible work available, is given by the Helmholtz free energy $A \equiv U-TS$. – gatsu Feb 6 '16 at 12:33

In chemical thermodynamics, it is derived from 2nd law of thermodynamics that in case the final temperature and pressure are the same as the initial ones, one equilibrium state can change into another (say, by removal of a partition) only if the final value of the Gibbs energy is not greater than the initial value.

When the reaction is done in a reversible way, the change in Gibbs energy quantifies work that is extracted from the system +reservoir. In practice it is impossible to have reversible reaction, so some potentially extractible work gets wasted into internal energy of the supersystem.

My advice is: do not try to reconcile different points of view on Gibbs energy and entropy at this time. The connections between thermodynamics, mechanics and theory of probability are so interesting that teachers forget they will make chaos of them and continue to make a great mistake by trying to teach them to students at the same time (say in a year-long course) even when nobody understands them as well as average practitioners of thermodynamics, mechanics and probability understand these subjects separately.

If you study chemistry, focus mainly on how the Gibbs energy $G$ is defined in thermodynamics, what is $\Delta G^°$ and how it is used, and why - learn what 2nd law of thermodynamics implies for it. You will find the other views of Gibbs energy and entropy from mechanics and probability are not that important.

If you study theoretical physics, study thermodynamics first, then when you get the idea what thermodynamic entropy is switch to statistical physics and confront the viewpoints. Some things match, some do not and there is a lot of nonsense floating in the literature caused by confusion of the two different concepts.

If you study theory of information, study that and take theoretical physics use of entropy later as an interesting area where it is possible to reformulate some parts in terms of information theory.

Just do not try to understand it all at the same time - people from chemistry, statistical physics and mathematicians mean different things with the same words like energy, entropy all the time. You need to learn one meaning of the words used in one subject before you redefine them and learn the other.