Can someone help me understand enthelpy and free energy?

[GravitatisVis]

I'm teaching myself thermodynamics and I'm having trouble understanding Helmholtz and Gibbs free energy.

My understanding of enthalpy (H) is pretty solid I think. I understand that enthalpy is a state variable, meaning its change tells you how much heat (Q) is absorbed or released by a system at constant pressure. (ΔH=Q)

I also understand that if the change in enthalpy is negative (ΔH<0) for example the process is exothermic, which means the system's temperature increases and this increase in temperature causes heat to spontaneously flow from the system to surroundings in accordance with the first law of thermodynamics. This also means that in accordance with the second law, the entropy (S) of the system decreases (since it loses heat) and the entropy of the surroundings increase, but the increase outweighs the decrease.

I'm really having trouble, however, understanding why Gibb's free energy is defined as G = H - TS and ΔG = ΔH - TΔS. What's the deal with this TS term? What wasn't enthalpy telling us before? In this case the temperature and pressure are held fixed. I know that if a process is quasistatic, Q=TΔS. And since ΔH=Q, wouldn't the Qs cancel out? I'm pretty sure sure this isn't the case. And since T is constant, how can heat be release/ absorbed at all? Heat only flows in the context of a temperature gradient.

I understand that Gibb's free energy tells you two things: 1) the spontaneity of the reaction and 2) the energy available to do useful work. What I'm having difficulty understanding is the nature of the TS term that's slapped on to enthalpy. Since T is constant is there heat flow between the system and surroundings at all?

Same with Helmholtz Free Energy: F = U - TS. The change in F is defined as ΔF = ΔU - TΔS. What is the motivation for introducing the TS term? In this case the temperature and volume are held fixed.

Thanks everyone.

 

[quietrain]

i don't really understand too but this might help

"When a system changes from a well-defined initial state to a well-defined final state, the Gibbs free energy ΔG equals the work exchanged by the system with its surroundings, minus the work of the pressure forces, during a reversible transformation of the system from the same initial state to the same final state."

from wiki. it seems to imply that the work exchanged by the system with its surroundings = U + PdV = H

and minus the work of the pressure forces = -TS

i am not very sure though ...

 

[Studiot]

Yes, quietrain you have spotted the omission from Gravitasvis' post.

When you are considering an energy balance you need also to consider work in or out as well as heat exchanged.
Note this work can present itself in a huge variety of ways in the full first law. We don't usually consider electrical/gravitational etc but sometimes have to add these in as well.

In the first ( and second) law the equality sign applies in a reversible process. That is why temperature is held constant,
An example of such a process is melting or boiling.

You didn't ask explicity but your question about entropy can be answered by the following.

The area under a curve on a PV diagram represents energy - the energy (work) difference between the initial and final states in a process.

Entropy was introduced as a function to act in a similar way with temperature so the area under a curve in a TS diagram also corresponds to an energy change.

There are other pairs of quantities that might also be included depending upon the application
see my post#11 in this thread

https://www.physicsforums.com/showthread.php?t=465492&highlight=magnetic+moment&page=2

 

[A. Neumaier]

I'm teaching myself thermodynamics and I'm having trouble understanding Helmholtz and Gibbs free energy.[...]
I'm really having trouble, however, understanding why Gibb's free energy is defined as G = H - TS and ΔG = ΔH - TΔS. What's the deal with this TS term? What wasn't enthalpy telling us before?

The reason for the different energies is that they describe extremality principles valid under different boundary conditions: Enthalpy at constant S, P, N, Helmholtz free energy at constant T, V, N, and Gibbs free energy at constant T, P, N. For the latter two, see, e.g., Section 4.4 of http://lanl.arxiv.org/abs/0810.1019 , where things are summarized in a concise form.

 

[GravitatisVis]

Let me try to state my current understanding. The reason to introduce new thermodynamic potentials is because they're required to explain what's going on in a system that has artificially imposed constraints.

While internal energy works great for an isolated system (const. U), new potentials are needed for closed and open systems. So for a closed system with constant S,P,N, Enthalpy is introduced. For a closed system with constant S,V,N, Helmholtz free energy is introduced. And for a closed system with constant T,P,N, it's Gibb's Free energy. And you derive these new potentials using the Legendre Transform to switch variables around to get the constraints you're looking for; i.e. in the case of internal energy, you subtract a TS to switch variables from S,V,N to T,V,N to get Helmholtz free energy!

I've read that closed system can exchange energy with its surroundings (or reservoir or environment) in the form of heat, work, or both. Do the constraints on the system impose what kind of energy can be transferred? It looks like a closed system with constant T,V,N, wouldn't be able to perform any mechanical work, since it's volume is held fixed. Is that right?

 

[A. Neumaier]

Let me try to state my current understanding. The reason to introduce new thermodynamic potentials is because they're required to explain what's going on in a system that has artificially imposed constraints.

While internal energy works great for an isolated system (const. U), new potentials are needed for closed and open systems. So for a closed system with constant S,P,N, Enthalpy is introduced. For a closed system with constant S,V,N, Helmholtz free energy is introduced. And for a closed system with constant T,P,N, it's Gibb's Free energy.

And you derive these new potentials using the Legendre Transform to switch variables around to get the constraints you're looking for; i.e. in the case of internal energy, you subtract a TS to switch variables from S,V,N to T,V,N to get Helmholtz free energy!

I've read that closed system can exchange energy with its surroundings (or reservoir or environment) in the form of heat, work, or both. Do the constraints on the system impose what kind of energy can be transferred? It looks like a closed system with constant T,V,N, wouldn't be able to perform any mechanical work, since it's volume is held fixed. Is that right?

Yes.

 

[Studiot]

It is my understanding that from a thermodynamic point of view:

A system is closed if no matter (mass) is exchanged across the system boundary.

A system is isolated if no energy or matter is exchanged across the system boundary.

Edit see posts 8 & 9.

 

[A. Neumaier]

It is my understanding that from a thermodynamic point of view:

A system is closed if no matter (mass) is exchanged across the system boundary.

Indeed, you are right; I just checked two other sources. Thus I need to change this in my lecture notes... Thanks for the correction.

A system is isolated if no energy is exchanged across the system boundary.

No. According to both wikipedis (http://en.wikipedia.org/wiki/Closed_system ) and
Reichl's book, isolated means that nothing at all is exchanged. If no energy is exchanged, one says thermally isolated (S=const), and if no work is exchanged (V=const), one says mechanically isolated.

 

[Studiot]

Yes indeed that should have been the transfer of both matter and energy banned for an isolated system.

Sorry.

 

[GravitatisVis]

I'd like to thank everyone for the previous clarifications. I've been able to clear a lot up because of them. But I still have some questions. In my book Thermal Physics by Schroeder, it says that for a system with constant T,V,N, the change in Helmholtz Free Energy places a limit on the total amount of work (mechanical and otherwise) the system can do. (ΔF≤W)

It also says that F approaches a minimum by maximizing entropy and minimizing internal energy. So how does the system do this exactly? I realize that since it's an open system, it must exchange energy with the environment in such a way to minimize F. By what mechanism does this process take place? I'm confused because I thought that since V is constant, no mechanical work could be done by the system at all. So how does the internal energy decrease? Also since T is constant, heat won't flow spontaneously between the system and reservoir, so where does the increase in entropy come from?

Thanks again.

 

[A. Neumaier]

I'd like to thank everyone for the previous clarifications. I've been able to clear a lot up because of them. But I still have some questions. In my book Thermal Physics by Schroeder, it says that for a system with constant T,V,N, the change in Helmholtz Free Energy places a limit on the total amount of work (mechanical and otherwise) the system can do. (ΔF≤W)

It must read W<=Delta F. One cannot use more energy than is there. This holds under arbitrary boundary conditions with the corresponding free energy potential. Thus also W<= Delta G at constant T,P,N, etc.

It also says that F approaches a minimum by maximizing entropy and minimizing internal energy.

This is not true.
The entropy is maximized at constant U,V,N.
The internal energy is minimized at constant S,V,N.
The Helmholtz free energy is minimized at constant T,V,N.
The Gibbs free energy is minimized at constant T,P,N.
The enthalpy is minimized at constant S,P,N.
The common reason for this is the second law of thermodynamics in its most general form; see the reference given in #4. (But the identification of the meaning of ''closed'' is wrong there. It should be constant N, not constant V.)

By what mechanism does this process take place? I'm confused because I thought that since V is constant, no mechanical work could be done by the system at all. So how does the internal energy decrease? Also since T is constant, heat won't flow spontaneously between the system and reservoir, so where does the increase in entropy come from?

It is your book that is confused, not you!

 

[GravitatisVis]

It also says that F approaches a minimum by maximizing entropy and minimizing internal energy.

It literally says, "in a constant temperature environment, saying F tends to decrease is the same as saying that U tends to decrease while S tends to increase." But it doesn't mention how this is done. I thought that since Helmholtz free energy only applies to an open system with const T,V,N, that it must exchange energy with its environment somehow to minimize F at equilibrium. Is this wrong? Since T is constant, the system and environment are in thermal equilibrium right? This rules out heat transfer right? How can the system and environment be at different pressures and not "exchange" volume?

 

[DrDu]

If no energy is exchanged, one says thermally isolated (S=const), and if no work is exchanged (V=const), one says mechanically isolated.

A thermally isolated system does not exchange heat with the surrounding. This does not exclude irreversible processes to take place in the system, e.g. chemical reactions or the conversion of mechanical work into internal energy. Another example is the free expansion of an ideal gas. So S=const is certainly not a condition for a thermally isolated system.

 

[A. Neumaier]

A thermally isolated system does not exchange heat with the surrounding. This does not exclude irreversible processes to take place in the system, e.g. chemical reactions or the conversion of mechanical work into internal energy. Another example is the free expansion of an ideal gas. So S=const is certainly not a condition for a thermally isolated system.

I thought we were discussing equilibrium thermodynamics.

Outside of equilibrium, all the stuff we discussed is not very well defined, and one has to specify the framework within which one considers the nonequilibrium concepts.

 

[DrDu]

I thought we were discussing equilibrium thermodynamics.

Outside of equilibrium, all the stuff we discussed is not very well defined, and one has to specify the framework within which one considers the nonequilibrium concepts.

Yes, but I am not talking about the detailed description of an irreversible process but only of the possible equilibrium starting and end points. Then the initial and final equilibrium states of a system may differ in entropy although the system was all the time thermally isolated. I would consider this as a typical problem of classical thermodynamics.

 

[A. Neumaier]

Yes, but I am not talking about the detailed description of an irreversible process but only of the possible equilibrium starting and end points. Then the initial and final equilibrium states of a system may differ in entropy although the system was all the time thermally isolated. I would consider this as a typical problem of classical thermodynamics.

If you start in equilibrium, nothing happens at all. Equilibrium states are stable in time.

Chemical reactions happen only when the systems starts in a nonequilibrium state, typically obtained by merging two subsystems that each were in equilibrium. You pour two things together, thereby creating a nonequilibrium situation.

What then happens depends on the conditions that you can keep constant, typically T, N, and either V or P - which explains the dominance of the Helmholtz and the Gibbs free energy in engineering work. The combined system attains the state of minimal free energy under the available constraints. Thus by solving the corresponding optimization problem (or, in simple cases, setting the gradient to zero), one can predict the outcome when equilibrium is restored.

 

[A. Neumaier]

It literally says, "in a constant temperature environment, saying F tends to decrease is the same as saying that U tends to decrease while S tends to increase." But it doesn't mention how this is done. I thought that since Helmholtz free energy only applies to an open system with const T,V,N, that it must exchange energy with its environment somehow to minimize F at equilibrium. Is this wrong? Since T is constant, the system and environment are in thermal equilibrium right? This rules out heat transfer right? How can the system and environment be at different pressures and not "exchange" volume?

As I already said, your understanding of the situation is here better than the description in the book.

Most books contain errors due to lack of care or oversight (even very good books tend to have a few such errors). You need to learn to recognize when an author makes or doesn't make sense, and find out when you can trust your own view more than that of a particular source. This is not always possible while you learn the basics of a field, but in these cases you should leave the case open until new evidence arrives to resolve the issue.

 

[DrDu]

I'd like to thank everyone for the previous clarifications. I've been able to clear a lot up because of them. But I still have some questions. In my book Thermal Physics by Schroeder, it says that for a system with constant T,V,N, the change in Helmholtz Free Energy places a limit on the total amount of work (mechanical and otherwise) the system can do. (ΔF≤W)

It also says that F approaches a minimum by maximizing entropy and minimizing internal energy. So how does the system do this exactly? I realize that since it's an open system, it must exchange energy with the environment in such a way to minimize F. By what mechanism does this process take place? I'm confused because I thought that since V is constant, no mechanical work could be done by the system at all. So how does the internal energy decrease? Also since T is constant, heat won't flow spontaneously between the system and reservoir, so where does the increase in entropy come from?

Thanks again.

Typically, the systems people have in mind when using this theorem is able to do other work than volume work; think of a battery.
Also, the temperature mustn't be constant during the whole process. It is sufficient that the system has the same temperature at the beginning and end of the process. So heat can be exchanged.
On the other hand, if the process is irreversible, it is not necessary to exchange heat for an increase of entropy.
In the case of a system with fixed pressure, one considers work other than volume work -pdV.

 

[Zeppos10]

I'm really having trouble, however, understanding why Gibb's free energy is defined as G = H - TS and ΔG = ΔH - TΔS. What's the deal with this TS term? What wasn't enthalpy telling us before?

As far as I am concerned, this is the crux of this discussion. I do not see a straightforward answer to this problem in the response sofar. I will add separate comments on separate responses.

 

[Zeppos10]

The reason to introduce new thermodynamic potentials is because they're required to explain what's going on in a system that has artificially imposed constraints.
While internal energy works great for an isolated system (const. U), new potentials are needed for closed and open systems. So for a closed system with constant S,P,N, Enthalpy is introduced. ?

However, the usefulness of internal energy is not at all limited to isolated systems (cf first law), and the usefulness of enthalpy is not at all limited to systems with constant S,P,N, as everybody knows who measures heat-of-reaction in a calorimeter. (Note that here it is not specified what N=constant means here: if these are the number of particles (molecules), N is not a conserved quantity in chemical systems. Also note that it is very difficult for any actual proces to keep S constant.)

And you derive these new potentials using the Legendre Transform to switch variables around.

Legendre transforms in thermodynamics = science fiction; see elsewhere in Physics Forums.

 

[A. Neumaier]

the usefulness of enthalpy is not at all limited to systems with constant S,P,N, as everybody knows who measures heat-of-reaction in a calorimeter. (Note that here it is not specified what N=constant means here: if these are the number of particles (molecules), N is not a conserved quantity in chemical systems. Also note that it is very difficult for any actual process to keep S constant.)

We are discussing a substance of identical particles without chemical reactions, so N is constant. And the equilibrium case, where a calorimeter keeps S constant.

Legendre transforms in thermodynamics = science fiction

It is you adding the fiction.

 

[DrDu]

We are discussing a substance of identical particles without chemical reactions, so N is constant. And the equilibrium case, where a calorimeter keeps S constant.

Who said so? Chemical reactions is where the concept of the thermodynamical potential is most usefull. And, as we already discussed, we are interested not in a single equilibrium state but in the transition from one equilibrium state to another with the intermediate states not necessarily being equilibrium states. The whole fun of potentials is that their difference depends only on the initial and final state and not on the path.

You also said in a previous post:
"Chemical reactions happen only when the systems starts in a nonequilibrium state, typically obtained by merging two subsystems that each were in equilibrium. You pour two things together, thereby creating a nonequilibrium situation. "
This isn't true. You can - at least in principle, and often also in practice - do a chemical reaction so that it is arbitrarily well approximated by equilibrium states, that is, as a quasistatic process. This is usually discussed assuming some machine where the reactants can be supplied separately by semipermeable membranes which are permeable each only for one of the components. This has been discussed especially by Wolfgang Pauli who discusses also alternative mechanisms. See his Lectures on Thermodynamics or
http://www.springerlink.com/content/b3xj217k4518667x/

 

[A. Neumaier]

Who said so?

I gathered it from the notation used. If there are chemical reactions, there are different N_i for each chemical species, and there are constraints given by conservation laws for the atoms - the whole set-up becomes much more complex. The original question was phrased in a quite specific context - about clarifying the nature of the potentials and why they matter.

Chemical reactions is where the concept of the thermodynamical potential is most usefull. And, as we already discussed, we are interested not in a single equilibrium state but in the transition from one equilibrium state to another with the intermediate states not necessarily being equilibrium states.

But equilibrium thermodynamics is only about reversible transformation, where equilibrium is preserved.

The whole fun of potentials is that their difference depends only on the initial and final state and not on the path.

Of course, the use of equilibrium thermodynamics is far wider. But if you want to discuss chemical reactions, I think you'd open a corresponding new thread, and say which aspect you want to discuss. If I sopt it, I'd see what I can contribute there.

You also said in a previous post:
"Chemical reactions happen only when the systems starts in a nonequilibrium state, typically obtained by merging two subsystems that each were in equilibrium. You pour two things together, thereby creating a nonequilibrium situation. "
This isn't true. You can - at least in principle, and often also in practice - do a chemical reaction so that it is arbitrarily well approximated by equilibrium states, that is, as a quasistatic process. This is usually discussed assuming some machine where the reactants can be supplied separately by semipermeable membranes which are permeable each only for one of the components.

Yes, the situation I described was indeed not representative for all processes;
slowly changing the chemical boundary conditions is indeed described by reversible transformations and preserves chemical equilibrium. But in this case, my remark that a calorimeter keeps S constant still applies, since a reversible ttransformation does not involve any entropy production.

In any case, chemical equilibrium is a far more complex subject than simple equilibrium, and the thermodynamic potentials are then functions of many parameters rather than only of three.

 

[DrDu]

But equilibrium thermodynamics is only about reversible transformation, where equilibrium is preserved.

Then your view of thermodynamics is at variance with common definitions, e.g. consider this paragraph from C. Caratheodory's axiomatization from 1909 (Annalen d. Mathematik, 3:355-386 (from a time when irreversible thermodynamics didn't exist):

"Eine zweite Voraussetzung, die fuer einfache Systeme gelten soll, ist
die, dass durch den Anfangszustand und die Endgestalt allein die waehrend
einer adiabatischen Zustandsaenderung geleistete aeussere Arbeit A noch
nicht eindeutig bestimmt sein soll. Es sollen I am Gegenteil adiabatische
Zustandsaenderungen moeglich sein, die yon einem gegebenen Anfangs-
zustand zu derselben vorgeschriebenen Endgestalt fuehren, und denen von-
einander verschiedene Arbeitsleistungen entsprechen. Wenn z. B. ein Gas
sich in einem adiabatischen Zylinder befindet, der durch einen beweg-
lichen Kolben verschlossen ist, so ist die durch den Kolben geleistete
Arbeit bei vorgeschriebener Ausdehnung des Gases eine andere je nach
der Geschwindigkeit, mit welcher man den Kolben herauszieht."

So he explicitly considers adiabatic processes where different amounts of volume work are delivered due to different speeds of the process.

 

[A. Neumaier]

Then your view of thermodynamics is at variance with common definitions, e.g. consider this paragraph from C. Caratheodory's axiomatization from 1909 (Annalen d. Mathematik, 3:355-386 (from a time when irreversible thermodynamics didn't exist): [...]
So he explicitly considers adiabatic processes where different amounts of volume work are delivered due to different speeds of the process.

I was careful to say that ''equilibrium thermodynamics is only about reversible transformation''; you didn't notice my qualification ''equilibrium''.

Thermodynamics has always been more general than equilibrium thermodynamics - steam engines didn't operate at equilibrium. Before the time when irreversible thermodynamics was developped, the only (but extremely important) statement thermodynamics could make about the nonequilibrium situation was the second law in its global form
(in contrast with the local form in irreversible thermodynamics):

The simple version of the global second law says:

Under constant and fixed equilibrium boundary conditions, a nonequilibrium system will ultimately reach (metastable) equilibrium, the equilibrium state reached will have the same intensive state variables as those corresponding to the boundary conditions, and the thermodynamic potential appropriate for the boundary conditions applied (-S, U, H, G, etc.) will attain a (local) minimum.

Thus, if one knows the analytic form of the potential, one can predict the final state by solving a corresponding optimization problem.

This scenario is used for a huge number of applications, including phase equilibrium and chemical equilibrium.

A more general version of the global second law allows for reversible processes (implemented through changing boundary conditions), and is formulated in Chapter 4 of my book from
http://lanl.arxiv.org/abs/0810.1019

The local second law says that the local mean entropy production is always nonnegative.
It implies both versions of the global form.

While the above might still be at variance with common definitions, it can be so only if the latter are too specialized or too inaccurately formulated.

 

[Zeppos10]

I'm really having trouble, however, understanding why Gibb's free energy is defined as G = H - TS and ΔG = ΔH - TΔS. What's the deal with this TS term? What wasn't enthalpy telling us before?

As far as I am concerned, this is the crux of this discussion. I do not see a straightforward answer to this problem in the response sofar.

This position (crux of discussion) stands:
2 aspects (questions) might be of interest at this poimt:
a) how many applications can be fouind for G, when no chemical reactions / phase transitions are involved ?
b) if we do look at chemical reactions (eg 2H2 + O2 = 2H2O) the ΔS term is the difference between the entropy of products and reactants: this quantity can be negative, which implies it is not the total entropy change ?.
I agree: this specifies the question, it does not answer it.

 

[DrDu]

Just want to remark that the "Delta" notation in thermodynamics does not (or not always) have the meaning of a difference but, e.g., $$\Delta U=\sum_i \nu_i \partial U/\partial \xi |_{T,p}$$ where the $$\nu_i$$ are the stochiometric coefficients of the reaction and $$\xi$$ is the coordinate of the reaction, so it refers to an infinitesimal change due to the reaction.

 

[A. Neumaier]

I'm really having trouble, however, understanding why Gibb's free energy is defined as G = H - TS and ΔG = ΔH - TΔS. What's the deal with this TS term? What wasn't enthalpy telling us before?

Did you read the whole thread? What precisely do you ask for?

G is much more useful under conditions where T,P, N are held constant - which is far more often the case than the conditions under which the enthalpy is best.

 

[Zeppos10]

We are discussing a substance of identical particles without chemical reactions, so N is constant. And the equilibrium case, where a calorimeter keeps S constant.

This disdussion is about the variable of state indicated by G (free enthalpy). Allmost all the applications I know of G involve systems with chemical reactions or phase transitions. Those systems seems to be excluded by Neumaier so I like to know which (real) systems he is thinking of, where G is a useful variable of state.
There is no calorimeter I know of where S is kept constant.

 

[Zeppos10]

Did you read the whole thread? What precisely do you ask for?
G is much more useful under conditions where T,P, N are held constant - which is far more often the case than the conditions under which the enthalpy is best.

Yes, I read the whole thread, but not all of it pertains to the interpretation of G=H-TS and this is what I emphasized.
I would like to know applications of G outside chemistry: is it used in mechanics ? In electrical systems ? I look for the domain of application, which might help us.
btw: If G=H-TS, then G only applies to systems for which H is defined: which cannot be more often then for systems for which H applies.

 

[A. Neumaier]

Yes, I read the whole thread, but not all of it pertains to the interpretation of G=H-TS and this is what I emphasized.
I would like to know applications of G outside chemistry: is it used in mechanics ? In electrical systems ? I look for the domain of application, which might help us.
btw: If G=H-TS, then G only applies to systems for which H is defined: which cannot be more often then for systems for which H applies.

G and H are defined for _every_ thermodynamic system; you can go between the two by a standard Legendre transform. depending on the application (and on the tabulated values for your system of interest, you choose one of the two (or another potential, such as that by Helmholtz). All these descriptions are equivalent, but calculating with one description may be far simpler than with another one.

 

[Zeppos10]

G and H are defined for _every_ thermodynamic system; you can go between the two by a standard Legendre transform. depending on the application (and on the tabulated values for your system of interest, you choose one of the two (or another potential, such as that by Helmholtz). All these descriptions are equivalent, but calculating with one description may be far simpler than with another one.

For the discussion on "Why (not) the Legendre Transform" see post #7 at
https://www.physicsforums.com/showthread.php?t=313535

 

[A. Neumaier]

For the discussion on "Why (not) the Legendre Transform" see post #7 at
https://www.physicsforums.com/showthread.php?t=313535

Callen, one of the first to convert to the LT-approach of thermodynamics, writes in 1987 (Thermodynamics etc, p138): "the introduction of the transformed representations is purely a matter of convenience".

True. And in complicated problems, there is a BIG difference between which formulation one uses - in one a particular problem may be tractable, in another one not.

1. First of all: there is nothing convenient about the Legendre transform.

Others find it _very_ convenient; that's why it made it into all textbooks.

2. In general (for an arbitrary, but well defined system), both U(V,S) and H(p,S) are unknown (unspecified) functions:

But in general (for any particular specific system), both U(V,S) and H(p,S) are convex functions, and one of the two (or, more often the Gibbs or Helmholtz free energy) is usually approximately known, and can be used for predictions.

 

[Studiot]

I do believe that equations and conditions involving G are of particular use in multicomponent systems.

As such you will find G used in alloy metallurgy and flow processes such as combustion engines, where the chemical reaction is secondary to the mechanical considerations.

Whether you consider this Chemistry, Chemical Engineering or Mechanical Engineering is moot.