# Enthalpy and it's use in Gibb's Free Energy

1. Nov 26, 2009

### cavalier

Lately I've been struggling with the idea of enthalpy and what it means conceptually, especially in its use in Gibb's free energy. There is nothing in the definition of change in enthalpy that would connect logically to spontaneity or free energy.

After thinking about it for a couple days, here are my ideas about spontaneity. I expect some or even all of them to be wrong, so it would help me very much if someone could correct my mistakes or show me where I managed to get things right.
1. Enthalpy is just a way of measuring the energy released by something at constant pressure. We could measure it using the change in internal energy at constant volume, but in that case pressure would not be constant.
2. Things tend to become disordered because disorder is statistically favored. Things also tend to want to lose potential energy. It takes a certain amount of the latter tendency to overcome the former tendency and vice versa.
3. $$\Delta G=\Delta H - T\Delta S$$ is just a way of apply (2) to systems of gases, most of the time involving chemical reactions.
4. $$\Delta G=-nF\Delta E$$ is an application of (2) to electrochemistry.
5. Any spontaneous changes where entropy decreases is accompanied and balanced out by the conversion of potential energy into other forms.

2. Nov 26, 2009

### Gerenuk

You ask good questions. I think that is the way to understand physics.
You seem to have used "applications" of enthalpy only, which isn't best to understand it's origins. I try to explain the view I'm comfortable with.

First I should say that all laws are originally expressed at differentials like
$$\mathrm{d}E=T\mathrm{d}S-p\mathrm{d}V$$
That is important. All other equations are special cases only. The definition of enthalpy is
$$H=E+pV$$
That has only mathematical reasons, as with this definition the differential now uses another variable (Legendre transformation)
$$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}p$$

Now in a constant pressure process ($\mathrm{d}p=0$, $\mathrm{d}Q=T\mathrm{d}S$) we have exactly
$$Q=\Delta H\qquad(\text{const }p)$$
That's why the heat in a chemical reaction is given by the change in enthalpy.

The free energy concept is something else. Maybe you can find a good statistical mechanics book (Reichl maybe?) and look up the section about "availability". Basically if you have a system in contact with an environment, then the total entropy of these two system can only increase. Consequently the work that you can extract from the system will be equal to
$$A=E-T_\text{envir}S+p_\text{envir}V$$
Incidently for processes with some defined values in the final variables this availability will be equal to the change in one of the 4 thermodynamic potentials (E, F, H, G). Have a look at the section.

Enthalpy is rather the change of plain heat Q in a processes where pressure is kept constant.
The work at constant pressure is of course $W=-p\Delta V$.

It is not a law that things want to lose energy! The only real law says that total entropy (system+environment) wants to increase. The statement that "free energy" should minimize derives from it under special circumstances. One can show that the increase of total entropy (system+environment) is equivalent to the statement that the Gibbs free energy of the system wants to decrease, if you consider only final states that have a defined pressure and temperature.
And btw, you can conclude that the normal internal energy of the system goes to a minimum, if we have defined value of entropy and volume of the final state.

The importance for chemistry stems from the fact that
a) if you want to know the heat at constant pressure you need the difference in enthalpy
b) since after the experiment you have a defined temperature and pressure, you want to minimize Gibbs free energy in order to find the final state

The total entropy never decreases. There is no statement about the entropy of the components. Saying the individual entropy is balanced out is very hand-wavy, but maybe some people find it useful to think this way.

I hope I haven't mixed up stuff, so it's up to you to read and ask questions

Last edited: Nov 26, 2009
3. Nov 29, 2009

### cavalier

This confuses me. I usually think of heat as energy being transferred from one thing to another. Can we really speak of a transfer of energy in all chemical reactions? Suppose we have a reaction of two solid chemicals that produces gas at constant temperature. The enthalpy for such a reaction will be positive, which makes sense because the gas produced has a lot of random molecular translational energy, but that energy simply arises from the chemical energy; I would argue there wasn't really a transfer.

Does E in this equation refer to something besides internal energy?
Would it make sense to apply enthalpy for changes occurring at changing pressure? Engines perform under nonconstant pressure. Can we determine the enthalpy of an engine cycle?

It was definitely hand-wavy. The idea came to me when I thought about a sealed container of muddy water. In space, the mud will never settle because a mixture is statistically favored. On Earth, the mud will settle. If I have the correct idea of entropy, it seems like the second law is being violated. It occurred to me that potential energy is being converted into some other form when the denser mud sinks, but I'm not sure how that fits in, and I don't know how to reconcile the sinking of mud in water with the second law and Gibb's free energy (or even if makes sense to apply Gibb's free energy to this change).

4. Nov 29, 2009

### Gerenuk

Hmm, I haven't understood what you meant in this paragraph.

Yes, you are probably right that one should rather use U.

Enthalpy is defined for all equilibrium states (i.e. homogeneous temperature and pressure) and you can calculate it. Here is basically the tools that you have:
$$\Delta U=Q-\int p\mathrm{d}V$$
$$H=U+pV$$
$$\Delta H=Q+\int V\mathrm{d}p$$
$$W=-\int p\mathrm{d}V$$
$$\Delta U=Q-p\Delta V\qquad\text{(const p or V)}$$
$$\Delta H=Q+V\Delta p\qquad\text{(const p or V)}$$
$$W=-p\Delta V\qquad\text{(const p or V)}$$
Now you can use any of these equations purely mathematically without even understanding what they stand for. But take care that some of the equation apply for certain contraints on the processes only - in doubt you have to apply the more general equations.
With these equations you see that only for constant pressure processes the change in enthalpy is equal to the heat. In general these quantities aren't equal but exist. They might be useful for other purposes.

The second law of entropy only works for (energy) isolated systems! However, the mud interacts with the gravitation of earth upon which both object move closer together.
The earth and the mud together could be considered as an isolated system. The entropy of earth plus the entropy of mud will increase as a total, but what can we say about a "subcomponent" like the mud? One can show that
(maximum entropy of earth+mud) is equivalent to (minimum free energy of the mud)
This way we can make statements about the mud.
[I have to think for a moment which free energy to take here. The problem is that mud isn't merely a gas with pressure and temperature only. It has macroscopic particle positions and also is subjected to gravity. This might require some special free energy.]

5. Mar 12, 2011

### Zeppos10

In the posts above no systematic distinction is made between the internal pressure and the environmental pressure. On Post #3 Cavalier defines the free enthalpy / Gibbs free energy as A=G=U + p(env)V -T(env)S, but he does not define enthalpy as H=U+p(env)V.
May be the attachment (post #7) at https://www.physicsforums.com/showthread.php?t=88987&referrerid=219693 can resolve some of the problems: the discussion above indicates many entangled problems.