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Wikipedia is an encyclopedia, which means it must place a stress (and this is codified by our most fundamental policies) on authoritative and reliable sources, as well as stay up-to-date. IUPAC, the most authoritative body for chemical nomenclature, recommends that the quantity that forms the subject of this article be called Gibbs energy or Gibbs function. This is not a particularly new recommendation, and all major textbooks on the market that have been revised recently follow this convention (including the standard reference, Atkins).

Of course more literature uses the previous term, but that's only because there has not yet been enough time for newer, post-recommendation literature to surpass the older literature in volume. I don't think an encyclopedia can afford to stay out-dated. The personal dislikes or preferences of contributors should not overwhelm the recommendation of IUPAC. I feel that the renaming of this article is required if we are to follow the policy of using reputable sources over original research. Loom91 09:27, 27 April 2007 (UTC)Reply[reply]

This has been debated to death more than once and the consensus was to keep the current name. I don't see anything new in your argument that hasn't been discussed before. Please read the archives, but to summarize, some of the objections were: free energy does not "belong" only to chemists; many new books still call it Gibbs free energy; many IUPAC recommendations are not followed by anyone, and we should be descriptive rather than normative; precedent in article naming for chemistry article titles (especially for chemical substances) is to prefer what is agreed as the most common name and not necessarily the IUPAC name; Wikipedia guideline for article titles favors the most common name. --Itub 09:59, 27 April 2007 (UTC)Reply[reply]

"Many new books still call it the Gibbs free energy". Do you have any references for that? Loom91 14:48, 28 April 2007 (UTC)Reply[reply]

Loom91, please read the archives. You will find the answers you seek there. This issue has already been debated and concensus has been reached. Thank-you: --Sadi Carnot 10:46, 30 April 2007 (UTC)Reply[reply]

lol pointless wiki edit war --Your Mom 10:46 21 November 2011 (UTC)

It's funny how people who prefer the status quo are happy to say that the issue has been decided already ;-p

—DIV (128.250.80.15 (talk) 04:32, 1 April 2008 (UTC))Reply[reply]

As a guy who is just trying to understand this stuff... The Term "Gibbs Energy" tells me nothing about the nature of the energy. At least "Gibbs Free Energy" is a little bit descriptive. The fact of the matter is I'm ticked off with you guys naming things after yourselves instead of giving things names that are descriptive and thus helpful to us who don't live this stuff 24/7. I think your behavior is inconsiderate to people like me and an unfortunate product of your egos. Do you not realize that you are making the subject less accessible to us? I'm as thankful to Mr. Gibbs as the next guy but ease of comprehension should not be sacrificed.

Dave3457 (talk) 05:03, 10 February 2009 (UTC)Reply[reply]

Your point is a good one, though I've always written 'Gibbs Energy', because the boundaries to geologic systems vary. You might consider running over to Gibbs' phase rule [sic], where Gibbs's equation F = (C + 2) - P, or (independent linear variables) = (total linear variables) - (independent linear relations among them) has always been rewritten F = C - P + 2 and attributed to Gibbs. If you've studied some linear algebra (one of Gibbs's specialties), you may recognize Gibbs's forgotten writing of his phase rule as the definition of the 'rank' of the 'basis' of a linear space of intensities.

i think that gibbs free energy just means that if the internal energy of the sysyem is greater than the energy needed to preform the reaction, it will be spontaeous. —Preceding unsigned comment added by 75.28.97.33 (talk) 11:10, 1 February 2008 (UTC)Reply[reply]

I think this article needs to be reviewed again. It might even qualify for featured article. 89.139.222.107 16:28, 13 August 2007 (UTC)Reply[reply]

How is the derivation in the derivation section a derivation? you use its definition in the second line. shouldn't you arrive at the definition from 1st and 2nd laws? Sorry but if you're taking shortcuts it doesn't really make sense.

I hope I don't come off as unappreciative of all the time you guys are putting into these pages but...

Is it just me or is the below, which is in bold, completely incomprehensible?

Gibbs defined what he called the “available energy” of a body as such:

The greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition.

For what it is worth Microsoft Word's grammar check didn't flag it

I appreciate that you can't be holding the hand of a beginner like me but Wikipedia should be a place for a guy like me to go to when he wants to understand something like Gibbs Free Energy. My inability to understand the way that they spoke in 1873 should not be what holds me back.

The year 1873 is Gibbs. No one, I believe, can read all of Gibb's reasoning. It's not the language. Try Pierre Duhem's 'Thermodynamics and Chemistry' on Google Books. The sections on equlibrium and stability are the clearest I've read. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

I suspect he meant something like "except in the situation where the system returns to its initial state"
Learning something shouldn't be more difficult than it needs to be.
Dave3457 (talk) 06:23, 10 February 2009 (UTC)Reply[reply]

The bold text from Gibbs is a long way from incomprehensible. The leap to 1870s usage seems to be no more than a shift to a less implicit style for the word "such." This feels like an attempt to have it both ways: expecting to access developments achieved by people who lived in a different time, but being annoyed if their work is delivered in the language they used! If the older usage seems awkward, that's a shame, but we can't very well resent Gibbs for having a better command of English than ours. What does it really mean, to say that difficulty with Gibbs' language "should not be" what holds us back? Do we mean that Gibbs "should" get his act together and write in a style that makes us more comfortable? When he talks about zero net change (from initial conditions), he refers (clearly) to heat transfer with external bodies. Arguably wordy to some modern tastes, but clear and unambiguous! Nowadays, we might say something like "external bodies, except those that at the end of the processes..." But it's "completely incomprehensible" to me, to suggest that the onus to understand what Gibbs is saying lies on anybody but ourselves. Of course the grammar check didn't flag it, because there's nothing wrong with Gibbs' grammar! Aboctok (talk) 16:10, 27 April 2014 (UTC)Reply[reply]

What do your extensive properties have to do with the integration-ability of the equation in the "derivation"? Do you mean to say "since these variables are usually constants?" Are you guys trying to use as few words as possible? Would it be too much trouble to elaborate on these steps? kthxBye. — Preceding unsigned comment added by 174.26.57.26 (talk) 20:47, 4 October 2013 (UTC)Reply[reply]

This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 09:52, 10 November 2007 (UTC)Reply[reply]

So, I removed this section relating entropy and life. It doesn't seem to fit in with the rest of the article. --HappyCamper (talk) 20:11, 22 December 2007 (UTC)Reply[reply]

The second sentence of this article does not make sense:

Technically, the Gibbs free energy is the maximum amount of non-expansion work which can be extracted from a closed system or this maximum can be attained only in a completely reversible process.

I'm not sure what the writer is trying to say. —Preceding unsigned comment added by 78.149.104.116 (talk) 23:23, 8 February 2008 (UTC) Yes, the terminology in thermodynamics is confusing and obfuscating, f.e. a reversible car is not reversible in thermodynamics. See reversibility to see other options to reverse. Additionally, 'heat' in thermodynamics means the total spectrum of electromagnetic radiation, but probably U know this already...Reply[reply]

Doesn't the reference to 'non-expansive work' refer to a different characteristic function, such as Helmholtz's? A system rolls along the Gibbs function's minimum as the system travels along an equilibrium path slow enough to keep the system uniform in temperature, pressure, and escaping tendencies of chemical substances. The Gibbs-Duhem equation and Duhem's Theorem allow one to calculate the thermodynamic properties of each substance in this open system. I never use the concept of 'free', but perhaps it fits in here. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

I have moved the section "What does the term ‘free’ mean?" to a more appropriate home at Thermodynamic free energy. People who have been following this article for years will know the 'reason' it was originally inserted by one particular 'contributor'. For the rest of you: I trust you will agree that it is common sense. For the moment I have only moved the content. The heading should disappear eventually too, but I don't want anyone to complain about 'broken links'.
—DIV (128.250.247.158 (talk) 07:18, 15 March 2009 (UTC))Reply[reply]

The section you speak of is still (or again?) in the article. As a newcomer to this page (but not the material), I was pleased to see so obvious an issue with terminology addressed right here; this is where such a discussion may be most relevant. Its being here is one reason I believe this article is so good as it is; it makes the material and thinking more accessible to non-specialists.

So please don't move or remove this section again! yoyo (talk) 17:04, 11 September 2009 (UTC)Reply[reply]

This is what hyperlinks are for! —DIV (138.194.11.244 (talk) 04:18, 13 May 2011 (UTC))Reply[reply]

It is the maximum amount of non-expansion (including contraction ie negative expansion)work extractable from a system changing from one state (an initial state) to another state (a final state) at constant pressure. In other words, it is the amount "free" to do this type of work. 86.136.201.235 (talk) 01:53, 25 December 2013 (UTC)Reply[reply]

I think it would be useful to put in the article dGo' (supposed to be a delta symbol, the letter G, a degrees sign, and an apostrophe), the definition of working in a biological system. I don't know how to format the text or where to insert it at. An example of such notation is found here http://www.sci.sdsu.edu/class/bio202/TFrey/ThermoEnzymes.html (section VIII, portion C, entry 7). AStudent (talk) 08:57, 28 May 2009 (UTC)Reply[reply]

The statement 'The equation can also be seen from the perspective of both the system and its surroundings (the universe).' is confusing, since it implies at first reading that the surroundings are the universe.

They should be the same. Being a geologist, I prefer 'system' and 'environment'. Heat and work shuffle back and forth. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

The last statement in the section 'The input of heat into an "endothermic" chemical reaction (e.g. the elimination of cyclohexanol to cyclohexene) can be seen as coupling an inherently unfavourable reaction (elimination) to a favourable one (burning of coal or the energy source of a heat source) such that the total entropy change of the universe is above or equal to zero, making the Gibbs free energy of the coupled reaction negative.' does not offer much information at all. It should be expanded and clarified with some description of what is meant by favourable and unfavourable, as well as providing additional information and insight regarding how this makes the "Gibbs free energy of the coupled reaction" negative.

I do not know enough about the matter at hand to suggest detailed meaningful alternatives, however I offer this feedback to you as the response from a reasonably intelligent reader who desires to obtain understanding of the topic. Suneecat (talk) 22:39, 27 November 2009 (UTC)Reply[reply]

This may not help, but thermodynamics predicts changes of the state of a system in an environment. There are two methods of change: reactions and diffusion. If it's reactions that change the state, the initial state undergoes perhaps many reactions to reach the final state. The change in state is determined by the initial state and the extents of all the reactions. One reaction can easily run backward, so long as the Gibbs function (say) of the final difference in state is less or equal to zero. Whether the reactions are 'coupled' or not seems irrelevant. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

I recently changed "endothermic" to "endergonic". I'm unfortunately not familiar with cyclochexanol/cyclohexane reactions, but if the point is that they are not spontaneous, then endergonic is the right word, not endothermic. In fact, claiming that reactions with Gibbs energies greater than zero are "endothermic" perpetuates the pernicious misconception that a reduction in the energy of a system is what determines spontaneity. I hope somebody else will check the reaction to confirm that it is indeed endergonic. Otherwise, I don't think that "coupled" reactions are irrelevant. I'm not sure if they appear anywhere else in this article but I believe that thinking in terms of "coupled reactions", in a thermodynamic sense, explains how metabolism and thus life are possible. 24 December 2013

My knowledge of thermodynamics is very limited so my questions refer only to the mathematics of the derivation.

In the section Derivation:

1. I don't see why it is necessary to integrate dG to obtain equation (8) for G. Simply comparing equations (2) and (3) leads to the same result.

The above point is well taken. The derivation is foreign to me, and switches between functions of state and their total derivatives, not partial (which should be there if something is held constant). Internal energy (which I avoid for methodological reasons) is converted into Gibbs Energy. These are just characteristic functions, functions whose extreme point is at equilibrium at constant V, S, and N(i) and that at a chemist's lab conditions: constant p, T, and mu(i). The Gibbs function (a characteristic function) varies with pressure and temperature, and this is not indicated by the final equation. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

2. Given the formula G=\sum \mu_i N_i should not dG=\sum \mu_i dN_i meaning that Vdp-SdT=0 identically, and if so, why is this fact not reflected in equation (6)? —Preceding unsigned comment added by Danielgenin (talk • contribs) 03:32, 3 March 2010 (UTC)Reply[reply]

Equation 6 is, I believe, wrong. It should read, I believe, dG = total derivative of -pV, TS, sum M(i) d(mu(i)) (Gibbs, eq 88). Each term has two differentials. From this, one step will subtract half and derive the Gibbs-Duhem equation, an identity which is very similar to your correctly derived identity. The latter has vast applications, and is valid for equilibrium, open systems. Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

should have units as joules? doesn't mention units of gibbs energy anywhere in the article... —Preceding unsigned comment added by 128.250.5.247 (talk) 13:21, 31 May 2010 (UTC)Reply[reply]

Yes, I think units deserve a section. Just keep units out of the equations, which would make the equations non-theoretical.

Please don't ignore DanielGenin's query, which draws attention (I believe) to a very serious typo. The section 'Definitions' also needs two typos fixed. One substitutes a(i) for mu(i) and the other has some confusion with N(i).

Could someone please change Gibbs's description as an 'American Engineer' to American Scientist? We think of Joseph Conrad as a writer, not a custom's agent; and Charles Pierce as a philosopher, not Coast Guard employee (though he was a scientist there).

(It may be hard to believe, but in the 19th Century faculty in different Departments actually spoke. Even geologists lunched with physicists: the discovery of eutectics was used immediately by Teall to explain the textures of some granites.) Gibbs was one of the greatest scientists of the 19th Century. You probably know him for his correct invention and extensive development theorems in chemical thermodynamics and for the notation you learned in advanced calculus. Few remember him for his dissertation on gears:-) Thanks!) Geologist (talk) 01:22, 10 December 2010 (UTC)Reply[reply]

There is an image at Wikimedia Commons if it's useful, not my area.ref1--Billymac00 (talk) 06:12, 26 December 2010 (UTC)Reply[reply]

Dear Colleague, this is a description of the formation of water from its beginning. It is presented in layman’s terms for a number of reasons. Your comments or your description of water formation are welcome.

The Universe and the universal laws of physics, thermodynamics, gravity and matter have all been well established for more than 13 billion yrs. Only 5 billion yrs. ago, in our Milky Way Galaxy of 400 billion stars, a benign cloud containing 90% of Hydrogen, combined with clouds containing 10% of other Nobel gases. The gases were interlaced according to their atomic weights as they, merged and began to drift. The formed gases drifted into an enormous amount of intrinsic electrically charged fissile atoms of naturally occurring galactic elements, integrating with the gases and combining. As the cloud drifted through an absolute zero region of the galaxy, its incorporated gases froze into gigantic, towering mountains of a combination of galactic mixed gases and elements. Embedded in the frozen gas cloud were also some extremely hot, cloud consuming ions. A fusion process was set into motion when friction from the charged galactic ions, ignited and began to consume the flammable gas cloud. The ignited gases, fused a hot spot which, then rendered liquid molten from the galactic elements. The impurities from the solid elements in the hot spot, formed into a slag cauldron which then began to heat up and thaw our frozen young star 4.9 billion yrs. ago. The frozen gas mountains thawed and breached the ruptured surface of our new Sun, flowing toward and finally reaching the extremely hot cauldron, with a mixture of liquefied homogenous gases.

When the thawed, liquefied gases made contact with the molten elements, a cataclysmic molten “splash”, gas atomic explosion resulted within the enclosed cauldron (Galvani’s Law).

The combination of ionic energy from fission and galvanic/ gas steam pressure produced from fusion caused the first internal solar eruption from atomic energy. The tremendous heat and pressure from the first powerful internal explosion caused the thawed, extremely hot neutrons of the fissionable atoms to melt. In that extreme environment the atoms were stripped bare of their neutrons, to a state of unstable ionized and un-ionized electrons (-) and protons (+). A Fission process began from the extreme heat, which caused the bare nuclei to collide and produce enough energy to split and separate the atoms into their sub-atomic state of quarks, neutrinos and leptons (Pauli’s Exclusion Principle). The tremendous internal atomic explosion with the force of an exploding star ejected the mountains of thawed unstable sub-atomic photosphere and gases upward with extreme pressure and force within a confined area of the cauldron. The tremendous pressure from the first internal atomic explosion, blasted the small lighter atoms of Hydrogen and other sub-atomic Noble gases into the larger, heavier Oxygen, Etc. atoms. The explosion pierced and impregnated the unstable Oxygen atoms, with the propelled Hydrogen atoms, combining and altering the sub-atomic Noble gases for the first time. This amalgamating and incorporating of sub-atomic ions under extreme heat and pressure began the natural process of the formation of water and other compounds. The process continued, when the impregnated Oxygen atoms were exploded into the absolute zero temperature of interstellar space under extreme atomic pressure from within the confined cauldron of the hot spot. When the encapsulated Hydrogen atoms were blasted into the absolute zero environment, they bonded forever with the Oxygen atoms and became one compound atom with a ratio of 2H:1O, for the first time ever. The extreme pressure and temperatures permanently bonded and compounded the atoms of H.,O.,C.,N.,and etc. into their present, compounded state of H2O (water) and other compounded atoms e.g. CO2, N2+O2,etc. The first eruption simultaneously forged the Kuiper Belt, extending it out 50 AU from the Sun. It also established The Oort (Hydrogen) Cloud which our Sun uses as its fuel source. The first eruption also and simultaneously defined the boundaries of our Heliosphere more than 4.8 billion yrs. ago. Thank you. Angelo Pettolino Author — Preceding unsigned comment added by 67.167.40.217 (talk) 17:08, 25 August 2011 (UTC)Reply[reply]

This has been said by so many people for such a long time, perhaps starting with Hamilton but probably even before him, that I believe asking for a citation there is unnecessary. What are your thoughts about this? Moreover, I don't see why there would be any need for the system to be in STP in order for this general rule of thumb to be true... 201.137.10.188 (talk) 06:21, 5 January 2012 (UTC)Reply[reply]

This is completely wrong. Changes can only occur when they result in the increase of universal entropy. (Ref: Peter Atkins' Physical Chemistry). A transfer of heat from a hot body to a cold body will always result in an increase in total entropy. Entropy can always be created by turning energy into heat. Since energy can neither be created nor destroyed a decrease of total energy (including work or heat) in one part of the universe will necessitate the increase in total energy in another part of the universe. Therefore if the argument that "Every system seeks to achieve a minimum free energy" is true, why then should this other part of the universe accept an energy transfer into it? The numerical value of a Gibbs function is such that if its value is positive, then the changes as written will give rise to a decrease in entropy, and therefore cannot happen. If the value is negative, then there is an overall increase in entropy and is therefore allowed. If the value is zero, then the process is in equilibrium. It also turns out from the mathematics that at a constant pressure, the numerical value of the Gibbs function is the maximum amount of non-expansion (including contraction) work (such as electrical work) that the system of changes can do. 81.129.179.74 (talk) 12:56, 5 February 2013 (UTC)Reply[reply]

The two-dimensional graph from Gibbs's 1873 paper carries a caption that says that the length AB is the Gibbs free energy. I don't think this can be right, since the graph is labelled as being a slice through constant volume (I also checked the original source). That must mean that AB is what modernly is called the Helmholtz free energy. I corrected this in the caption on the wikibio of Gibbs, which I've been editing extensively. I think someone should look after the other uses of that image in Wikipedia. - Eb.hoop (talk) 19:33, 25 February 2012 (UTC)Reply[reply]

I like wikipedia and have benefited from it many times. I don't want to direspect everyone's contributions to this page. But I've been teaching thermodynamics for twenty years and this page is *full* of serious logical and notational errors. In order to fix it I would have to substantially rewrite it, and I don't know how the community will respond to that. In the meantime, students using this page need to realize that it's got some serious errors at present. — Preceding unsigned comment added by 24.38.51.160 (talk) 16:46, 9 August 2012 (UTC)Reply[reply]

list the major errors here (you can summarize).

• Like this
• And this
• And this

Make simple fixes on the main article that involve a few words, insertions of sentences, etc. Delete incorrect statements. Do each one as a single edit, and give an edit summary. Don't do it all at once in a single sitting. Don't freak if somebody disagrees and reverts you-- probably that means you only have to come back here to talk that POINT over (one reason not to do the whole rewrite at once.)

Finally, you're new to WP or else you'd sign your TALK edits with ~~~~. Why don't you pick out a username? (it can be anything so long as not obscene, racist or distracting). I'll leave a "welcome" template on your page.SBHarris 19:07, 9 August 2012 (UTC)Reply[reply]

List of errors:

"Every system seeks to achieve a minimum free energy" 81.129.179.74 (talk) 13:01, 5 February 2013 (UTC)Reply[reply]

"is a thermodynamic potential that measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric)". No, the Gibbs function is simply a mathematical function G defined as G = H - TS. You have to work out the mathematics at specified conditions to obtain a physical interpretation for those specified conditions. 81.129.179.74 (talk) 13:07, 5 February 2013 (UTC)Reply[reply]

In section "Free Energy of Reactions", it is said that "to obtain equation (2) from equation (1) we must assume that T is constant", which seems to imply that equations (1) and (2) are not equivalent, and somehow one would be more valid than the other if we don't assume T to be constant. In reality either one is equally valid (as long as we don't concern ourselves on how to place the 'zero' of Gibbs Energy), and the hypothesis of T being constant is implicitly taken at the beggining of the section when the entropy change of the sorroundings is taken to be \Delta S = Q/T (which is only true if T=constant). Also, the phrase "Any adiabatic process that is also reversible is called an isentropic" seems unnecessary and, eventhough it is not strictly wrong, isentropic processes are a wider set of processes of which adiabatic+reversible processes are just a subset.--ChocolateMoccaFrappe (talk) 01:32, 11 February 2013 (UTC)Reply[reply]

The Gibbs free energy is in general

${\displaystyle G=U-TS+PV}$

The article seamlessly jumps into a special case of a homogeneous liquid or gas, giving

${\displaystyle G=\sum _{i}\mu _{i}N_{i))$

but not all systems are homogeneous. For a physics example, consider an electron system with electron-electron interactions [Brachman, M. K. (1954). "Fermi Level, Chemical Potential, and Gibbs Free Energy". The Journal of Chemical Physics. 22 (6): 1152–1151. doi:10.1063/1.1740312.]. In chemistry, inhomogeneity can come when you start to consider adhesion of chemicals to the walls of the beaker, walls whose surface area does not grow proportional to volume. --Nanite (talk) 13:12, 26 May 2013 (UTC)Reply[reply]

let me know if it's unnecessary or could use improvements. — Preceding unsigned comment added by 71.57.71.55 (talk) 07:00, 24 December 2013 (UTC)Reply[reply]

In the current article, it said (before my edit) that the Helmholtz free energy is for systems at constant volume and temperature.

In the article on Helmholtz free energy, it says the Helmholtz free energy is for systems at constant temperature.

So I changed the current article to agree with the Helmholtz article.

But is this correct? And why or why not?

178.38.83.46 (talk) 07:49, 9 March 2015 (UTC)Reply[reply]

The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Gibbs free energy G at T = constant, P = constant or Helmholtz free energy F at T = constant, V = constant), whilst the heat given out is usually a measure of the diminution of the total energy of the system (internal energy).

This can't be right. By conservation of energy, the change in the internal energy U must equal the amount of energy that leaves the system, which is heat plus work.

Why does the article say that the heat given out is "usually" a "measure" of the diminution of the internal energy? The only thing that prevents this from being outright wrong is the use of vague language. In any case, it clouds the issue.

178.38.83.46 (talk) 10:33, 9 March 2015 (UTC)Reply[reply]

Why does the intro show Gibbs represented by delta P when the rest of the article uses delta G, which seems widely used elsewhere? Gwideman (talk) 02:52, 16 July 2015 (UTC)Reply[reply]

The statement "Gibbs free energy is a thermodynamic potential that measures the maximum or reversible work that may be performed by a thermodynamic system at a constant temperature and pressure" is not correct! It is the negative of the change in Gibbs free energy that measures the reversible work. If there's not objection, I would make the necessary correction.--LaoChen (talk)06:41, 18 May 2016 (UTC)Reply[reply]

I would like to replace the "Gibbs free energy of reactions" section with a much simpler derivation. I would appreciate it if anyone could look at the derivation and critique it before I make this large change. It goes something like this;

The system under consideration is held at constant temperature and pressure, and is closed (no matter can come in or out). The Gibbs energy of any system is G=U+PV-TS and an infinitesimal change in G, at constant temperature and pressure yields:

${\displaystyle dG=dU-PdV-TdS}$

By the first law of thermodynamics, a change in the internal energy U is given by

${\displaystyle dU=\delta Q+\delta W}$

where δQ is energy added as heat, and δW is energy added as work. The work done on the system may be written as δW=-PdV+δW_x, where -PdV is the mechanical work of compression/expansion done on the system and δW_x is all other forms of work, which may include electrical, magnetic, etc. Assuming that only mechanical work is done,

${\displaystyle dU=\delta Q-PdV}$

and the infinitesimal change in G is:

${\displaystyle dG=\delta Q-TdS}$

The second law of thermodynamics states that for a closed system, ${\displaystyle TdS\geq \delta Q}$, and so it follows that:

${\displaystyle dG\leq 0}$

This means that for a system which is not in equilbrium, its Gibbs energy will always be decreasing, and when it is in equilibrium (i.e. no longer changing), the infinitesimal change dG will be zero. In particular, this will be true if the system is experiencing any number of internal chemical reactions on its path to equilibrium.

Please avoid using ‘Spanish masculine ordinal indicator’ in equations like ΔGº = ΔHº-TΔSº. I was completely confused what the hell it the underscored o is doing there. Not sure what actual mathematical symbol is supposed to be there, so cannot fix it. — Preceding unsigned comment added by 147.229.99.101 (talk) 11:09, 10 April 2018 (UTC)Reply[reply]

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Cheers.—InternetArchiveBot (Report bug) 04:50, 16 October 2017 (UTC)Reply[reply]

There is currently a tag at the top of this article that says it may be too technical. The tag is dated April 2018. As I look through the comments, the only ones that complain about it being too complicated are much, much older than that. This is a complex subject. I have had a year of college physics and that was forever ago. I came to this article because this topic came up recently in a discussion on information and I wanted to get the basic idea. The article seems clear and well written to me. Does anyone have any specifics as to why the article is too technical? If not then I’m going to remove the tag --MadScientistX11 (talk) 02:33, 7 July 2020 (UTC)Reply[reply]

I agree that the tag should be removed. This article is written quite clearly for anyone who has had a course in calculus. The only section that seems too technical is the one on Graphical interpretation, which could be moved further down as it is not really necessary to understand the rest of the article. Dirac66 (talk) 11:22, 7 July 2020 (UTC)Reply[reply]

Sounds good. Since no one has objected I'll take this as a consensus and am removing the tag. --MadScientistX11 (talk) 15:47, 8 July 2020 (UTC)Reply[reply]

OK. And I have now moved the Graphical interpretation section to the end of the article, since this is the only section which really does seem too technical. And not in any textbook which I have read. Dirac66 (talk) 20:06, 8 July 2020 (UTC)Reply[reply]

Assuming the superscripts in ${\displaystyle p^{\circ ))$, ${\displaystyle V^{\circ ))$ or ${\displaystyle T^{\circ ))$ indicate the standard states (1atm, 298K)

Given that

${\displaystyle pV=RT}$

and

${\displaystyle p^{\circ }V^{\circ }=RT^{\circ ))$

These two statements 1 and 2 below seem contradictory.

Is the ${\displaystyle {G^{\circ ))}$ in Statement 1 the same as the ${\displaystyle {G^{\circ ))}$ in the Statement 2 ?

This needs an explanation.

The sentence "If the volume is known rather than pressure" is not precise enough.

Statement 1

The temperature dependence of the Gibbs energy for an ideal gas is given by the Gibbs–Helmholtz equation, and its pressure dependence is given by

${\displaystyle {\frac {G}{N))={\frac {G^{\circ )){N))+kT\ln {\frac {p}{p^{\circ ))}.}$

Statement 2

If the volume is known rather than pressure, then it becomes

${\displaystyle {\frac {G}{N))={\frac {G^{\circ )){N))+kT\ln {\frac {V^{\circ )){V)),}$

Above comment by editor Maajdi on 7 March 2021.

Statement 1 is the usual textbook version. I will add Atkins and de Paula as a source.

Statement 2 is derived from statement 1 using :${\displaystyle p^{\circ }V^{\circ }=pV}$ for an ideal gas at constant temperature (Boyle's law). However statement 2 is not found in textbooks that I know of, so for Wikipedia it counts as unsourced and even original research. And as you say its meaning is not precise enough because it is not too clear exactly which volumes we are talking about. I will remove statement 2, which was added by a numbered editor in 2007 with no further comment. Dirac66 (talk) 21:01, 26 March 2021 (UTC)Reply[reply]