# Talk:Enthalpy

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 To-do list for Enthalpy: CREATE Enthalpy of Denaturation page to go with the link. Definitely requires REFERENCING for all formulae. Section about enthalpy changes needs cleaning up verification/referencing. 88.217.26.61 (talk) 18:53, 28 April 2008 (UTC) Standard enthalpy table -> Standard enthalpy ? Done 88.217.26.61 (talk) 18:53, 28 April 2008 (UTC)

## Enthalpy is Not Heat

There's a contradiction here, and I don't have the grounding in this subject to be able to fix it.

   Adding or removing energy through heat is the only way to change the enthalpy, and
The amount of change in enthalpy is equal to the amount of energy added through heat.


... However, heat is not the only way to change enthalpy.

Maybe I'm missing something? As I'm re-reading, I'm getting that my interpretation of the section is incorrect, but I can't figure out why. Could use clarification. — Preceding unsigned comment added by 108.57.35.49 (talk) 15:27, 31 May 2011 (UTC)

No, enthalpy H is not heat, H = U + PV. 86.168.55.23 (talk) 02:08, 20 October 2012 (UTC)

H is a state function, q is a path function, however at constant pressure (and where only expansion/contraction work is allowed) the mathematics show that dq = dH, therefore dq is proved to be a state function under the special case of constant pressure, whereas generally speaking q is a path function, and also in this special case change in heat (dq) is equal to the change in enthalpy (dH). 86.168.55.23 (talk) 02:13, 20 October 2012 (UTC)

## Query

I would really like to know what enthalpy means: help anyone? —Preceding unsigned comment added by 62.252.0.5 (talk) 00:42, January 9, 2005 (UTC)

I myself am just an undergraduate student at the University of Illinois. It appears the word comes from the greek language and means "to warm", or "to heat". A possible meaning of the word could then be expressed as the amount of warmth (which is amount of energy) in a room at a certain temperature and pressure.

For example: In a house each room could be assumed to be at the same pressure (atmospheric pressure). However, the garage would be at a lower temperature than the living room. The amount of enthalpy would be less in the garage (lower temp = less warmth,energy) than in the living room (higher temp = more warmth,energy). —Preceding unsigned comment added by 12.203.115.91 (talk) 16:52, January 31, 2005 (UTC)

I've believed for many years that enthalpy is an antonym of entropy - to have a tendency towards a more ordered situation (as opposed to entropy), opposite that of moving towards disorder (entropy). Checking with a dictionary, this appears to be a consistent view with the meanings of the words in one particular sense - that of enthalpy being defined as a measure of the ability to perform mechanical work, which will only increase as a function of order. Likewise, the definition of entropy in one context is as a measure denoting the disorder of the system being considered. The mathematical discussion of entropy / enthalpy reflect that in their formulas, but it is left to infer those definitions as a consequence to the understanding of the reader. Perhaps a short discussion of the meaning of the words in a more abstract sense would be beneficial to assist other readers? —Preceding unsigned comment added by 24.255.211.121 (talk) 08:02, July 5, 2005 (UTC)

The following was moved here from a section called "Agreed....." by Riick (talk) 05:01, 2 August 2010 (UTC)

== Agreed..... ==

I strongly agree. I'm not at all trying to divert this entry from hard science, but I think that if enthalpy is not a tendency towards order, there must be another term for the tendency towards order. I've been wondering about this for quite some time, maybe someone has an answer?

I have this feeling that as hard as entropy is to quantify, so must be enthalpy...... How on earth does a tree grow in the forest when the entropic nature of the universe, as we've been taught to believe, creates an inevitable pull towards a state of unified disorder, the decay and rot of natural matter.

There is a certain duality in forces here, and it just seems to me that enthalpy is the natural opposite..... I just don't know. And I can't seem to find anything defining or discussing enthalpy as a tendency towards order, save for a few sparse references in a few early-mid 20th century physics volumes. And in such cases, there was barely enough to go on.

Perhaps there is a nice Professor of Physic roaming the wikii that might enlighten us. —Preceding unsigned comment added by 69.221.174.207 (talk) 07:15, July 12, 2005 (UTC)

deolijn: umm just a question. entthalpy is a move towards order but when you mention a tree please understand it is a solid and thus different rules apply —Preceding unsigned comment added by 58.162.26.205 (talk) 12:35, July 21, 2006 (UTC)

Newbie: I'm no scholar, but i've always used "orthotropy" to be the antonym of entropy. If entropy means "turning in" or "in-turning" (like the way food turns?), then "orthotropy" would mean "right-" or "correct-turning". —Preceding unsigned comment added by 66.236.159.171 (talk) 21:47, July 30, 2005 (UTC)

Unregistered Roger: If you're trying to relate entropy to enthalpy then I'm afraid you're going nowhere. Entropy is a statistical measure and indicated what is the most likely state(ie.the one with the highest entropy) and the direction of the change in state. Enthalpy is a measure of energy and is not a fundamental property, but a convenient compound of 'internal energy' and 'pressure x volume'. It turns out that in a system where pressure is constant or where we are looking at the steady flow of a fluid through a system, then the change in enthalpy gives us a measure of the amount of energy that must flow into that system. —Preceding unsigned comment added by 137.205.146.206 (talk) 15:44, June 22, 2006 (UTC)

Freshman: I'd like to know whose P in the enthalpy equation refer to. The system's or the surrounding's pressure. Some articles listed it as surrounding's. This wikipedia article listed it as the system's. Please clarify the confusion. —Preceding unsigned comment added by 128.253.197.151 (talk) 05:08, October 23, 2006 (UTC)

Not trying to question hard science but seriously, a young and up and comcming scientist faced with this as the opening sentance. "In thermodynamics and molecular chemistry, the enthalpy (denoted as H, h, or rarely as χ) is a quotient or description of thermodynamic potential of a system, which can be used to calculate the heat transfer during a quasistatic process taking place in a closed thermodynamic system under constant pressure. " As one of my old teachers said to me. "If I hear another bloody quadrillion-syllabic word out of your mouth, detention!!" how about a nice simple intro?? Then you can lose the young whippersnapper later in the page. —Preceding unsigned comment added by 92.238.94.146 (talk) 02:39, 19 April 2009 (UTC)

(In reply to the one below above). Guys, enthalpy is NOT the antonym of entropy. If you have taken a class on thermodynamics, it is obvious that enthalpy is the sum of U and PV (i.e. H = U + PV). There is a great possibility that you are mistaking the concept of enthalpy with Gibbs Free Energy (G), which is G = H - TS. Remember that the equality sign here represents the system at equilibrium, otherwise you are going to have a <= sign there. Thank you and please refer back to your thermodynamics notes before writing on any science entry.

Guys, the tendency towards order is obviously the negative of entropy (-S). Thanks!

Guys, we can use Maxwell's Equation for it. Ouch man! —Preceding unsigned comment added by 168.122.13.29 (talk) 22:48, 27 March 2012‎

## Redirect

1. ) "Standard enthalpy change of denaturation" should link to a page titled "Standard enthalpy change of denaturation," not, as it now does, to the page on "Denaturation." This does not define ∆H(denat). Or, if "denaturation" is the preferred link, only that word should be highlighted. (12sep09) —Preceding unsigned comment added by Schmin (talkcontribs) 20:56, September 12, 2009 (UTC)

2.) Would it be reasonable to redirect heat of reaction here? I don't know enough about the topic, but it's a request article. A redirect here would seem sufficient to me, but can someone more knowledge inform? Would be greatly appreciated! Agentsoo 18:18, 14 August 2005 (UTC)

RE: redirect

redirect from enthalpy to heat of reaction? ... enthalpy has uses and meaning outside chemistry like the thermodynamic one treated here so while a link might be appropriate, it shouldn't redirect automatically —Preceding unsigned comment added by 137.99.21.199 (talk) 18:57, October 25, 2005 (UTC)

No. This would not be reasonable. The redirect to Standard enthalpy change of reaction is appropriate for now, but ideally there would be a separate Heat of reaction article with the alternate name Enthalpy of reaction to reflect usage under non-standard conditions. I am removing the redirect tag. Flying Jazz 03:08, 13 February 2006 (UTC)

## contradiction? - is dQ = dH in isobaric chemical reactions?

On the one hand it is written here:

It is seen that, not only must the Vdp  term be set to zero by requiring the pressures of the initial and final states to be the same, but the ${\displaystyle \mu _{i}dN_{i}}$ terms must be zero as well, by requiring that the particle numbers remain unchanged.

But on the other hand, the situation in which the use of dH=dQ is the most important, is in chemical reactions - where the sign of dH decides whether the reaction is exothermic or endothermic. But on chemical reacions ${\displaystyle \mu _{i}dN_{i}\neq 0}$, and usually (under constant T and P) ${\displaystyle \ \sum _{i}\mu _{i}dN_{i}=dG}$, which in spontanous reactions is smaller then 0.

So what do I miss? eman 15:56, 28 June 2006 (UTC)

The last few paragraphs of the "some useful relationships" section may be confusing you. I think the writer has the goal of showing the situations where the first and second law combine to result in the requirement that dH must be less than or equal to zero. Next, the writer includes the posibility of chemical reactions, but only to once again show the situations where the first and second law combine to result in the requirement that dH must be less than or equal to zero. I think that what the writer means by "interpretation of the enthalpy" when writing "any further generalization will add even more terms whose extensive differential term must be set to zero in order for the interpretation of the enthalpy to hold" is an interpretation of enthalpy as a potential whose derivative must be less than or equal to zero.
I think that the writer is trying to say is "The second law means that dH must be less than or equal to zero if entropy, pressure, and the number and types of particles remain unchanged." In other words, you're not missing anything because when there's a chemical reaction, the writer's "interpretation of the enthalpy" no longer applies. By the way, I think you meant "exo-" and "endothermic." I'll try to edit this in a couple days. Flying Jazz 03:29, 29 June 2006 (UTC)
Yes, of course I meant exothermic. I corrected myself above.
My main problem is that I know that chemists use the citeria of the sign of ΔH to determine whether a reaction is exothermic or endothermic, and now I can't see how this can be justified? So could it be that all the chemists, and all the chemistry books are wrong?!
eman 15:21, 29 June 2006 (UTC)
Enthalpy is just heat content, so the chemists are correct by definition. I guess I don't understand why you think they are incorrect. If you take some of the equations you wrote above and combine them with some of the equations in the text, you'll get the equation ΔG = ΔH - TΔS which should be familiar and might explain some of what's going on. Believe the textbooks. The section of the article that you're confused about should be ignored and will be removed soon. Flying Jazz 03:45, 1 July 2006 (UTC)
What bothers me is that for cheical reactions ΔG is usually not zero (for spontanous reactions it should be negative). So ${\displaystyle \Delta G\neq 0}$ means that ${\displaystyle \Delta H-T\Delta S\neq 0}$, thus ${\displaystyle \Delta H\neq T\Delta S}$, . eman 10:41, 1 July 2006 (UTC)
Yes. The equations that use the equality only apply for reversible processes (where delta G = 0). This is the case at equilibrium. A spontaneous chemical reaction isn't at equilibrium. After the concentrations change or other reaction conditions change so that equilibrium is achieved, the equalities hold. Flying Jazz 12:25, 1 July 2006 (UTC)
Still it doesn't solve my original problem about the ${\displaystyle \ \mu _{i}dN_{i}}$ in dH. The fact that the Enthalpy is also named "heat content" doesn't help. Nomeclature can be wrong, and has only symbolic importance. eman 11:17, 2 July 2006 (UTC)

OK, I think I understand what you're asking now, and it's similar to the question you asked on Talk:Internal energy with a similar answer, and it's not a simple answer and I might not explain it fully because I don't understand it 100% myself, but I'll try. Just as there are many types of internal energies depending on whether no ${\displaystyle \ \mu _{i}}$, some ${\displaystyle \ \mu _{i}}$ or all ${\displaystyle \ \mu _{i}}$ are natural variables, there are also many enthalpies for the same reason. For a "black box with some unknown amount of unknown reactions" inside:

${\displaystyle dH=TdS+VdP+\sum _{i}\mu _{i}dN_{i}\,}$

and all the natural variables (S, P, and all Ni) are independant of each other. But when you have a "heat of reaction" for a defined chemical reaction, all of the Ni are no longer independant of each other. They are related by stoichiometry, so one of the ${\displaystyle \ \mu _{i}}$ must be treated as a natural variable instead of a conjugate variable and subtracted out as discussed here and here. The nomenclature has been ambiguous for a long time, and maybe Alberty's IUPAC paper will change that in the future. Flying Jazz 16:51, 2 July 2006 (UTC)

I want to know the enthalpy —Preceding unsigned comment added by 80.191.144.10 (talk) 07:57, November 11, 2006 (UTC)

## Article has become messy

The article needs to be re-invigorated by explaining alongside definitions of thermodynamics. I was actually slowed down by downloading an image, and the article looked horrendous when the MathTeX markup wasn't on the screen. There needs to be some definate walkthroughs of the actual processes, rather than just explaining different aspects of enthalpy. This is the wikipedia, not a textbook! J O R D A N [talk ] 16:33, 20 February 2007 (UTC)

## ΔH

Im not sure if this has been mentioned, but in the intro but it describes enthalpy as either H or ΔH, Δ actually denotes the change in a value, so ΔH is actually the change of enthalpy, not the enthalpy itself. —Preceding unsigned comment added by 81.178.106.177 (talk) 22:34, April 22, 2007 (UTC)

yeah, so change it. —Preceding unsigned comment added by 86.27.184.240 (talk) 12:57, 15 February 2008 (UTC)

## The product of an exothermic reaction will be warm to the touch.

Is this necessarily true? Can an exothermic reaction not occur at -200 degrees, and be very very cold to the touch?

Ariel Hoffman —Preceding unsigned comment added by 87.69.68.65 (talk) 17:36, 31 December 2007 (UTC)

You are right, this is not always true. The statement about "being warm to the touch" is only meaningful if we assume that the reaction started at room temperature and the reaction is fast enough so that the products increase the temperature noticeably before losing the heat to their environment. Of course, if the reaction starts at a high temperature, it can feel warm to the touch whether the reaction is endothermic, exothermic, or not even happening at all! --Itub (talk) 11:06, 7 January 2008 (UTC)

## Molar enthalpy

The requested natural sciences page included a request for "molar enthalpy." I redirected here, as per "standard enthalpy" - should we add a short definition of molar enthalpy to the "definitions" section? Coanda-1910 (talk) 02:27, 6 April 2008 (UTC)

## Useless for the layman

There's an old quote that goes "you're so high-minded that you're of no earthly good." That's what I'm seeing out of a lot of the scientific/technical articles on Wikipedia, including this one. The articles are written on a level that is not appropriate for the layman to quickly grasp the concept. Instead, they drift into lofty and even high-theoretical issues without ever giving a firm grounding.

In this case, I used to know what enthalpy was. It was a term I used a lot back in my nuclear power days. However, I've been out of that industry for a while, and found my knowledge was getting a bit rusty. I took a look at this article, and now I still don't know what it is. That's pretty bad when someone who knows the subject can't figure it out.

I have no problem with the hard-science approach. But that should be only one part of it. We also need an overview that gives the average reader a good understanding of what it is (the article on internal energy does a decent job at that). Then follow it up with the hard science.

The overview in this one borders on the estoeric and really doesn't help much.

I would be happy to re-write it... but first I have to re-discover just what the hell enthalpy is. Izuko (talk) 12:19, 6 May 2008 (UTC)

I agree that the article is more inscrutable than necessary. At least it should have a more legible lead that, instead of bombarding the reader with jargon such as "thermodynamic potential" from the first sentence, tries to put the better-known and simpler aspects of enthalpy first. For example, the definition given in IUPAC's glossary is more readable: "Internal energy of a system plus the product of pressure and volume. Its change in a system is equal to the heat brought to the system at constant pressure."[1] For many people, "heat at constant pressure" is all they need to know about enthalpy, and yet this is hidden in various places throughout the article but not mentioned in the lead. --Itub (talk) 13:19, 6 May 2008 (UTC)

I have to agree with Itub. It is likely that most readers of this article will be students who have come across the term "enthalpy" in a chemistry class. The article is written by physicists for, it seems, physicists. It is important that the chemistry reader comes across the sentence from IUPAC, "Internal energy of a system plus the product of pressure and volume. Its change in a system is equal to the heat brought to the system at constant pressure.". Can we add that to the lead? --Bduke (talk) 23:31, 20 May 2008 (UTC)

Actually despite the jargon, this article contained a lot of mistakes. I corrected most of them. Removing the jargon would be a good idea. I really doubt whether the article was writen by real experts, given the stupid mistakes it contained. I've just rewritten Helmholtz free energy. That article contained so many false statements, it was really abominable. Of course, the article now looks much more technical, but then there is really no shortcut to explaining the things from fundamental concepts. Count Iblis (talk) 23:59, 20 May 2008 (UTC)
I am aware of the errors you fixed and I have no problem with that. We should not make errors. Your comment "there is really no shortcut to explaining the things from fundamental concepts" seems to me to be a typical physicist response. There is a way. It starts from recognising who our readers are and where they are coming from. As I said, I think a lot of our readers are chemistry students. Some might well be people with even less background than that. Just think where people are going to meet the term "Enthalpy" and come to WP to find out about it. It is more than likely to be something chemical, such as heats of reaction. We should start from a simple lead, even if it is skating close to oversimplification, in order to lead people into the content. Perhaps the article should be reorganised so the chemistry student knows when to stop leaving the rest for those who want the physicists detail and absolute rigor. I have spent a career trying to teach physical chemistry to students who every year seemed to have less background in physics and mathematics than those the year before. I am not going to wade in and edit the article unless there is some kind of consensus. I spent a lot of time on Entropy maybe two years ago. It just kept reverting back to being quite unintelligible to people with a poor background in physics and mathematics. This is a major problem in dozens of articles on topics that chemistry students meet in physical chemistry courses. --Bduke (talk) 01:14, 21 May 2008 (UTC)
I do agree with rewriting this article. Thing is that the mathematical formalism and/or jargon is often just a shortcut to a correct explanation. But that correct explanation must be given. If we look at what I corrected, it was the flawed inequalities. But the reason why the mistake was made points to a fundamental misunderstanding about the basic physics. The physics that is behind this can be explained to lay persons as well using only simple examples. Take e.g. the free expansion of a gas. One can explain that as follows:
There the entropy increased despite the fact that no heat is supplied. But the gas does no work either. So, pressure times volume change (assume moving the piston extremely fast by a small amount and then lock it in place so that the gas crashes into the piston at the new locked position). One can then heuristically reason as follows: The P dv term led to the gas being accelerated, increasing the kinetic energy of the gas as it shoots out toward the piston. The gas could have performed work, but all that energy will go to waste as thermal energy in the gas.
For an ideal gas, then end result is that the temperature does not change. If you imagine measuring the temperature during the free expansion, you would actually at first see that the temperature went down as just like in case of adiabatic expansion. But then all that kinetic energy get's dissipated into heat and then the temperature get's back to where it was. That is the cause of the entropy increase you now get that you wouldn't have gotten had the gas performed work.
One can then go a step further and point out that if you make the process more violent the concept of temperature would break down. One can argue that the velocity distribution is not like that of a gas at any temperature so one needs more variables to describe the macroscopic state of the gas as it expands than just pressure, volume and temperature. But you can apply thermodynamics to the initial and final states when things have settled down. From this perspective the equations of thermodynamics becomes much more interesting because it allows you to compute the outcome of complicated processes.
The problem is that the conventions used in wikipedia is to explain some subject comprehensively in a few pages. Then you cannot spend a page to discuss the free expansion experiment in detail to make some point about thermodynamics. But but perhaps we need to change this and do what is necessary to let people really understand the topic. Count Iblis (talk) 02:24, 21 May 2008 (UTC)
Unfortunately, it seems the article has been written and re-written by laymen or by chemistry students with a very poor knowledge/ understanding of Thermodynamics and the mathematics that go with Thermodynamics. I'll try and fix it when I have the time, but in the meantime, if any professors of thermodynamics wish to edit this article, please feel free. JC 86.178.174.199 (talk) 23:08, 27 May 2014 (UTC)
Or even worse still, written by biologists!!! 86.178.174.199 (talk) 22:58, 28 May 2014 (UTC)

See the discusion here. You can also check my rewrite of the derivations in Helmholtz free energy and of Fundamental thermodynamic relation. I'll rewrite the relevant derivations later in the week if no one else has the time to correct them. It should be understood that keeping these errors in the page does enormous damage to the reputation of wikipedia. Count Iblis (talk) 02:25, 19 May 2008 (UTC)

I have corrected the mistakes. Count Iblis (talk) 19:53, 20 May 2008 (UTC)

## Constant Pressure in the Definition

The "original definition" section has a question "?? Except that the pressure in a combustion cylinder is far from constant! Maybe the example is referring to a steam engine with 100% duty cycle?"

That is true, the pressure in a practical cylinder is far from constant if the internal reaction is a typical explosive combustion. If one wished to determine the work done by the gas during combustion by considering the pressure on the interior face of the piston, one would certainly have to know the detailed history of pressure vs. distance; a constant pressure could not usually be assumed. However, it is simpler to observe that a long time after the combustion is complete, when the system has returned to a quiescent state, the work done by the gas on the interior face of the piston must equal the work received by the environment at the exterior face. If the exterior face is in contact with a very large reservoir, e.g. the Earth's atmosphere, and the combustion isn't too fast, the pressure there is so close to constant that constant may be safely assumed.

Of course, a practical cylinder has a massy piston, and if the combustion is explosive, the piston may move quickly enough for the gas on its exterior face to be out of equilibrium with its surroundings. There may be sudden or oscillatory motion, producing either sound or shock waves in the environment subsequent to combustion. This real occurrence is inconsistent with assuming a constant exterior pressure. Nonetheless, for the purposes of a definition, the custom is to assume an ideal cylinder with a massless piston, and ideal combustion in which heat is produced slowly enough to maintain equilibrium of the exterior gas. The difference between real and ideal situations must be taken into account for experimental work, but is conventionally neglected for theoretical work.

The difference in sign between work done and work received was neglected in the text. This resulted from neglecting: a complete definition of work dW = F dot-product dX; whether the force under discussion is that exerted on the piston by the exterior gas or on the exterior gas by the piston; and a definition of the direction of increasing x. The positive signum shown in the text applies if one chooses these definitions: the force is that exerted by the piston on the exterior gas, and x increases on the exterior side away from the piston.

128.149.22.239 (talk) 18:55, 8 September 2009 (UTC)Jay Breidenthal jbreid@integrity.com

## Fluid

This entry's basic definition uses the word "Fluid" but i believe enthalpy is not specific to fluids;just a system. Fluids are simply commonly used when measuring enthalpy. Could someone weigh in on this? - tnx JScribner (talk) 22:40, 14 September 2009 (UTC)

I agree. Internal energy is defined in terms of a thermodynamic system. Enthalpy should be defined the same way, and thermodynamic system is more accurate than fluids. I will make the change. Dolphin51 (talk) 01:45, 15 September 2009 (UTC)

## Standard condidtions

I think the link Standard conditions redirects to Standard conditions for temperature and pressure. From what I understand enthalpy is calculated at a constant pressure not pressure and temperature. Shouldn't it redirect to standard state? --kupirijo (talk) 10:36, 25 October 2009 (UTC)

## Changed a sentence

There was a sentence that said "the change in enthalpy at constant pressure is equal to the heat added plus the work done". That is false. I changed it to "the change in enthalpy at constant pressure is equal to the heat added".

After all, that is why the concept of enthalpy even exists, it was because for the scientists in previous centuries with bad equipment, it let them run their experiments with pistons at constant pressure. Enthalpy eliminated a variable "work". This made life easier for them, so they created the concept of enthalpy. —Preceding unsigned comment added by Apc3161 (talkcontribs) 17:10, 5 January 2010 (UTC)

## Sudden change of notation

in the 'Difference between H and U: the additional term pV' the writer suddenly introduces a variable uppercase P to represent pressure [as far as i can tell]. I'm almost certain that lowercase 'p' is intended, but the switch is confusing. Perhaps someone with a more complete knowledge of the subject would have the confidence to make the appropiate modifications. Hai2410 (talk) —Preceding undated comment added 01:00, 24 January 2010 (UTC).

It's always been P in my experience. JIMp talk·cont 22:55, 6 June 2011 (UTC)

## Non-mechanical = heat??

In the beginning of this wiki, the term "non-mechanical" is used 6 times. So, I assume this means heat energy or something kind of like the type of energy when a photon of energy enters an atom system? Does this mean heat is going into the system, or out of? Clarification would be helpful. —Preceding unsigned comment added by 71.193.147.120 (talk) 20:40, 7 June 2010 (UTC)

The statement "ΔH of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it." is a bit baffling. It is not clear what the writer meant by "non-mechanical work". From the definition of H = U + PV, it follows that dH = dU + PdV + VdP. The heat flow is δQ = dU + δW where δW is the work done BY the system. For reversible processes, δW = PdV so dH = (dU + PdV) + VdP = δQrev + VdP. This means that ΔH = Qrev + ∫VdP. So, ΔH is equal to the sum of the integral of VdP and the reversible heat flow. So it seems to me that the statement as it stands is wrong, as well as being unclear. AMSask (talk) 00:50, 4 June 2014 (UTC)

## External fields

This version of the article says, "In the absence of an external field, the enthalpy may be defined, as it is generally known, by: H = U + pV". I have removed the bit about the absence of the external field because it seems the definition should still hold even in the presence of an external field. Either the external field affects the internal energy, pressure, and/or volume; or it does not. In either case I cannot envision a situation where the presence of an external field would make this definition incorrect. Anyone else want to weigh in on this? Riick (talk) 19:34, 16 August 2010 (UTC)

## Changes by Adwaele on June 21, 2012

In the Section “Formal definition” it is stated that p is the pressure at the boundary of the system and its environment. This is wrong/unclear for the following reasons:

1. The pressure need not be the same all over the boundary of the system. E.g. in the case of throttling usually an open system is chosen that contains the plug. In this case the pressures at the inlet and outlet differ. Which pressure should be taken in this case?
2. This formulation leaves unclear which pressure (and temperature?) must be chosen in calculating U. Is this also the value at the boundary?
3. In an inhomogeneous system the value of the enthalpy of the total system depends on the pressure and temperature distribution inside the system and not only on the pressure at the boundary.

The Article refers to Zumdahl, but Zumdahl talks about “the” pressure of the system (suggesting a homogeneous value) and not about the pressure at the boundary of the system and its environment.

The solution is simple: first define the enthalpy for homogeneous systems. This has also the advantage that δQ=TdS, since no irreversible processes can take place in a homogeneous system. Also the pressure is well-defined so δW=pdV. Next use the property that enthalpy is an extensive parameter. So the enthalpy of an inhomogeneous system is the sum of the enthalpies of the subsystems. In case of continuous variation of p and T composition etc. the summation becomes an integration.

The Subsection “Difference between enthalpy and internal energy” is not part of the formal definition and should be moved to a separate Section.

I have modified the text accordingly.

I like to add that there is a lot of overlap between the various Sections that needs to be cleaned up.

(Adwaele (talk) 21:05, 21 July 2012 (UTC)).

## The article is wrong

This article is wrong starting from the beginning: "Enthalpy is a measure of the total energy of a thermodynamic system." Enthalpy is not a measure of the total energy of a thermodynamic system. It is an artificial mathematical function H where H = U + PV, where U is the total energy of a thermodynamic system, and P is the pressure and V is the volume. The beauty of this artificial mathematical function is that thermodynamic experiments are usually carried out to measure changes in a quantity against one variable whilst holding other variables constant, and useful properties of the function themselves can be proved through the mathematics. For example, dq can be shown to be a state function and not a path function as might be expected from the property of q, under constant pressure. JC86.168.55.23 (talk) 02:04, 20 October 2012 (UTC)

Agreed: The first sentence of the present article "Enthalpy is a measure of the total energy of a thermodynamic system." is not correct. This should be changed. I can't believe that this error has remained for so long. No cite is provided. Where did this statement come from?AMSask (talk) 13:52, 1 April 2014 (UTC)
I would suggest a change to the first paragraph:
Enthalpy is a defined thermodynamic potential of a thermodynamic system consisting of the internal energy of that system plus the product of pressure and volume of the system:
${\displaystyle H=U+PV}$
The unit of measurement for enthalpy in the International System of Units (SI) is the joule, but other historical, conventional units are still in use, such as the British thermal unit and the calorie:
Since enthalpy, H, consists of internal energy, U, plus the product of pressure, P and the volume of the system, V, which are all functions of the state of the thermodynamic system, enthalpy is a state function.
While the units of enthalpy are units of energy, the enthalpy of a system should not be confused with internal energy, which is the total energy of a system.AMSask (talk) 19:23, 1 April 2014 (UTC)

The historical significance of putting H = U + PV by definition as an artificial mathematical function is as follows: During the Industrial Revolution, engines powered by burning coal to do work were invented and built. People were building better engines over time, better in the sense that more efficient engines, ie more work for the same amount of coal consumed, were built. So people wanted to know whether there was a limit on how much work you can get out from burning a fixed amount of coal. Obviously you cannot prove this one way or another by building more and more engines, because say in the first year, a train engine moved 1 mile by burning 1 tonnes of coal; in the second year a more efficient engine was built and it gave a mileage of 2; in the third year a better engine was built giving a mileage of 3, etc. So using this projection an engine built a thousand years later should give a mileage of 1000. What was needed was a mathematical way of making a proof one way or the other. By setting H = U + PV (a state function), it was proved that dq = dH at constant pressure. Therefore it was proved that dq was in fact a state function. That is to say, the heat change by a system moving from one state to another state is a fixed amount. ie Only a fixed amount of heat (which can only be turned into a fixed amount of work) can be obtained by burning a tonne of coal. JC 86.178.174.199 (talk) 00:41, 27 May 2014 (UTC)

Someone has made the opening even worse: "Enthalpy is a defined thermodynamic potential". Enthalpy is definitely not a defined thermodynamic potential. Enthalpy H is simply defined as U + PV. It may look simple, but H is a function of three variables, U, P and V, so is mathematically sophisticated. An example of a thermodynamic potential would be dH/dT (where T is temperature) or dH/dM (where M is mole quantity), (derivatives and partial derivatives). Please would only people who understand Thermodynamics and its associated mathematics edit the article. I am going to change this opening sentence. JC.86.178.174.199 (talk) 00:55, 27 May 2014 (UTC)

JC: I am not sure what your statement "Enthalpy is definitely not a defined thermodynamic potential." is based on. The 4 functions U, H, G, F are commonly referred to as thermodynamic potentials. See, for example: http://en.wikipedia.org/wiki/Thermodynamic_square AND http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thepot.html#c1 AMSask (talk) 01:17, 31 May 2014 (UTC)
Nope. Enthalpy is not thermodynamic potential. It is a thermodynamic state function. A potential in physics is the energy change per unit change of something else, and its units are joules per something, eg, joules per metre, joules per coulomb, joules per degree, joules per mole, etc. The unit of enthalpy is joules. AMSask may have confused enthalpy with the chemical potential, mu, (I apologise but I don't have Greek fonts), and the chemical potential (units of joules per mole) = the following (again I apologise for not having the partial derivative font, so I'll just use d with conditions being kept constant explicitly stated, where n indicates a given substance:
chemical potential for substance 1, mu(1) = (dG/dn1)at constant pressure, temperature, all other substances eg n2, n3 etc

                                          = (dU/dn1) at constant volume, entropy, n2,n3,n....

                                          = (dH/dn1) at constant pressure, entropy, n2, n3, n....

                                          = (dA/dn1) at constant volume, temperature, n2, n3, n....


As you can see one of the definition of chemical potential is dH/dn with conditions.

Ref: Peter Atkins, Physical Chemistry second edition reprinted 1984, ISBN 0-19-855150-9, pages 175 &176

109.158.28.28 (talk) 00:19, 17 November 2015 (UTC) JC

Just to point out to AMSask, the article in Wiki he/she cites titled thermodynamic potential does not cite any reference on where the title came from. 109.158.28.28 (talk) 00:26, 17 November 2015 (UTC)

I've changed the opening 2 paragraphs, only to discover the remainder of the opening is also gibberish. I'll try and fix it when I have the time. The Laws of Thermodynamics are general and not just for chemical reactions or changes. JC86.178.174.199 (talk) 01:14, 27 May 2014 (UTC)