Talk:Standard enthalpy of formation

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equation incorrect

equation for standard enthalpy change of reaction should be changed to reflect that fact that the stoichiometric numbers of reactants are already negative, thus there is no need for to have separate reactant and product terms in the equation. It also needs general cleanup; words like "products" do not belong in equations, and "reaction" should be simply "r" according to convention. Use the equation on the standard enthalpy change of reaction as a model. 71.121.232.215 (talk) 05:13, 7 April 2009 (UTC)[reply]

Also, aren't the stoichiometric values wrong?

CH4 + 3 O2 → CO2 + 2 H2O

On left: 1C, 4H, 6O On Right 1C, 4H, 4O

I'm assuming the eqn should be: CH4 + 2 O2 → CO2 + 2 H2O but I'm still a little new to enthalpy so will leave it to somebody else! :-) (Chris Fletcher) —Preceding unsigned comment added by Chris Fletcher (talk • contribs) 13:14, 2 June 2010 (UTC)[reply]


This article states what it is and how to use it. But fails to define the origin of the tables or how the individual molecules enthalpy change of formation are derived. Does there exist a known mathematical way in which to come up with these numbers? 96.2.120.230 (talk) 18:32, 6 August 2010 (UTC)[reply]

NaOH data wrong

Standard heat of formation is listed in table here as 426 kJ/mol while wikipedia article on Sodium Hydroxide claims it is 735 kJ/mol. Both can not be right.71.31.146.16 (talk) 07:00, 20 June 2012 (UTC)[reply]

Atkins and de Paula list -425.61 for NaOH(s) and the NIST webbook gives -425.93, so I have corrected the NaOH article. The value in this article is close. However the table does need reformatting - the minus sign for this value (and a few others) is hidden on the preceding line because the columns are too narrow. Dirac66 (talk) 14:37, 20 June 2012 (UTC)[reply]
OK, formatting fixed. Dirac66 (talk) 19:21, 20 June 2012 (UTC)[reply]

Isotopic differences in elements

Are there some differences in the values of this enthalpy for elements in various isotopic mixtures similar to those that exist in different allotropic states of various elements (for example the diamond and graphite for carbon, where the diamond has a non-zero enthalpy of formation)? What isotopic mixture should have the zero value? I think this is a useful content addition to article--5.15.53.219 (talk) 12:26, 16 June 2015 (UTC)[reply]

I see that editor 5.2.200.163 added two sentences yesterday on isotopic differences, which imply that the standard state corresponds to a definite isotopic composition for each element. Although this does seem plausible in principle, it would be better to have a source which confirms the statement and which specifies how the standard-state isotopic composition is determined for each element.
Also, the second sentence refers to the enthalpy of atomization of hetero-isotopic diatomic molecules. I would question whether individual molecules actually have enthalpies, since enthalpy is a thermodynamic property of macroscopic amounts of matter. Again a source would help. Dirac66 (talk) 02:06, 9 October 2015 (UTC)[reply]
The second aspect could be easily fixed by a different wording, instead of molecules we can say coumpounds or hetero-isotopic diatomic molecules compounds which underlines the macroscopic view.--5.2.200.163 (talk) 12:00, 9 October 2015 (UTC)[reply]
Surely source(s) for these assertions would help. I wonder though what is the probability that the standard-state isotopic composition for each element enthalpy of formation issue could have gone unnoticed in sources until now, could it be higher than the default 50%-50% assigning?--5.2.200.163 (talk) 12:08, 9 October 2015 (UTC)[reply]
Compounds is a better word provided the example is actually a compound. The present example 35Cl-36Cl is actually an element and not a compound, so the enthalpy of formation is zero, at least without worrying about isotopes. A better example would be HDO or perhaps CH3D. Dirac66 (talk) 00:18, 10 October 2015 (UTC)[reply]
Perhaps element would be a better choice. But isn't this diatomic molecule from two different isotopes (of chlorine, oxigen, nitrogen, etc) like hydrogen deuteride technically a compound molecule/substance (isotopologue)?--5.2.200.163 (talk) 13:40, 13 October 2015 (UTC)[reply]
I have now found a source which contradicts the claim that Also isotopes of an element other than the standard state have non-zero standard enthalpies of formation, differing from the stardard state isotope or isotopic mixture by the energy of neutron capture for one or more neutrons. The NIST thermochemical table here gives the value of ΔHf°(HDO, g) = -245.37 kJ/mol, compared to -241.83 kJ/mol for ΔHf°(H2O, g). The difference of 3.54 kJ/mol is many orders of magnitude too small to be an energy of neutron capture which is a nuclear reaction. In fact the energy of neutron capture is not included because ΔHf°(HDO, g) is the enthalpy of the reaction 1/2 H2 + 1/2 D2 + 1/2 O2 → HDO, with D in both reactants and products so that no nuclear reaction is involved. Instead the 3.54 kJ/mol corresponds to a difference in zero-point vibrational energy between reactants and products, which is very much smaller.
This illustrates the dangers of adding unsourced material. I will remove the reference to isotopic differences in the Key Concepts section. It is true that many other statements in the article are also unsourced and should eventually be sourced, but I will leave them alone since they appear to be generally correct.Dirac66 (talk) 18:32, 14 October 2015 (UTC)[reply]
I don't see exactly what is the exact contradiction you claim, the issue is still needing further details to be solved. The difference in the heat of formation of water and semi-deuterated water is inconclusive about the differences heat of atomization and of formation of different isotopes as it is clear that transition from one isotope to another requires a nuclear reaction of at least one neutron capture or beta decay such as in the tritiation of heavy water moderator due to neutron flux or similarly to the deuteration of light water. It is clear that some paradoxical/anomalous situation is involved requiring further investigation. Of course statements should be backed by sources in order to be a verified information in wikiarticles.
Does the source you mention contain the heat of atomization of deuterium, tritium and hydrogen deuteride to be compared with that of hydrogen to see where the anomaly comes from? It is clear that there must be a difference of some sort between isotopes, the exact nature and numerical value has to be determined. So about the wording I'd leave out the statement about neutron capture difference but clearly the difference between isotopes should be mentioned with the specification that is of an unspecified nature.--5.2.200.163 (talk) 12:55, 15 October 2015 (UTC)[reply]
An important aspect of this discussion involves the measured energy of a neutron capture reaction. Are there some tables with such measurements to be checked? Surely the neutron capture is or must be included in some thermochemical cycle. Is then the heat of formation of molecular deuterium D2 considered zero like that of molecular protium H2? This seems to be a suggested answer, but if so something does not add up here. Where dissapears the heat of neutron capture involved in transforming H2 into D2?--5.2.200.163 (talk) 09:45, 19 October 2015 (UTC)[reply]
How about the heat of atomization of D2 (aka deuterium bond energy)? What is the difference from the case of H2?--5.2.200.163 (talk) 09:48, 19 October 2015 (UTC)[reply]
The NIST tables display a difference of 3.72 kj/mol between heat of atomizations.--5.2.200.163 (talk) 09:56, 19 October 2015 (UTC)[reply]
Perhaps the following thoughts may be useful. They concern the "minus-oneth" law of thermodynamics, a presupposition of the subject, that for a system, there exist states of internal thermodynamic equilibrium.
According to Münster:
... We shall now give this concept a precise form by means of the definition:
      An isolated system is in thermodynamic equilibrium when, in the system, no changes of state are occurring at a measurable rate.
...
      The proviso 'at a measurable rate' implies that we can consider an equilibrium only with respect to specified processes and defined experimental conditions.
...
... This can, however, be completely ignored since thermonuclear processes do not occur at a measurable rate under conditions usually considered in thermodynamics.
      The concept 'absolute equilibrium' or 'equilibrium with respect to all imaginable processes', has, therefore, no physical significance.[1]
According to Callen:
      In actuality, few systems are in absolute and true equilibrium. In absolute equilibrium, all radioactive materials would have decayed completely and nuclear reactions would have transmuted all nuclei to the most stable of isotopes. Such processes, which would take cosmic times to complete, generally can be ignored.[2]
  1. ^ Münster, A. (1970). Classical Thermodynamics, translated by E.S. Halberstadt, Wiley–Interscience, London, ISBN 0-471-62430-6, pp. 52–53.
  2. ^ Callen, H.B. (1960/1985). Thermodynamics and an Introduction to Thermostatistics, (1st edition 1960) 2nd edition 1985, Wiley, New York, ISBN 0-471-86256-8, p. 15.
I think it important that Münster implies that one must presuppose and explicitly prescribe the range of processes that one will admit as relevant.Chjoaygame (talk) 13:14, 19 October 2015 (UTC)[reply]
I agree with Chjoaygame's explanation of the reason for ignoring nuclear reactions, but I will add a note about the meaning of the NIST value. For HD, the value given for ΔHf is ΔH of the reaction 1/2 H2 + 1/2 D2 → HD, so that neutron capture is not included. (As for HDO explained above, but without the oxygen.) The neutron capture would be included if one considered the reaction H2 + n → HD, but as Chjoaygame has explained this is of no practical interest on any useful timescale so it is ignored. Dirac66 (talk) 14:39, 19 October 2015 (UTC)[reply]
I think the most important aspect when discussing about the synthesis of HD from H2 and D2 to mention explicitly what is the heat of formation of D2. Is it considered zero by default like in the case of H2? This assumption seems problematic. Another approach should be considered by applying Hess's law, similarly to the case of NaCl which is mentioned in the article and involves heats of atomization, sublimation, ionization, etc. The synthesis of HD from H2 and D2 involves only heats of atomization (and eventually neutron capture, depending on the absence or presence of initial deuterium). But the neutron capture cannot be omitted when considering the initial reactant only hydrogen and neutron and not deuterium which is not present.
Timescales are not to be taken into consideration when applying Hess law. The heat of a reaction does not depend on its reaction rate, so timescales, a kinetic factor, is relevant only in thermokinetic reasonings which are not the object of this discussion. The formation of deuterium from hydrogen could not ignor neutron capture which is esential for formation of an isotope from the lightest isotope of an element. The heat of formation of isotopes reference is the subject of this discussion, no reason to ignor neutron capture and to assume that the heat of formation of D2 is zero.--5.2.200.163 (talk) 13:17, 20 October 2015 (UTC)[reply]
In my limited reading of this section of the talk page, I see it as discussing a proposal to add information about isotopic mixtures. The question is asked "What isotopic mixture should have the zero value?" and the comment added "I think this is a useful content addition to article". Perhaps this topic should be added to the article, depending on factors that I do not pretend to comprehend. But if it is to be added, it needs far more than editorial chatter; it needs a fair survey of reliable sources. I suppose that in some quarters, such as perhaps nuclear physics, neutron capture will be part of the array of presupposed reactions, while in other quarters, such as perhaps ordinary chemistry, it will not? To bring it into the article, I suppose a fair survey of reliable sources would extend to perhaps five standard textbooks on nuclear physics with specific interest in such reactions; and a check on standard ordinary chemistry texts to see what they have to say on the subject. If those sources are in fair agreement, they should be cited for that in the relevant section of the article. The section should make it clear that its topic assumes interest in those reactions. This talk page is not a general discussion page. It is about editing the article.Chjoaygame (talk) 14:45, 20 October 2015 (UTC)[reply]
I agree about the survey of sources needed both in nuclear physical chemistry and engineering and ordinary physical chemistry focused on chemical thermodynamics. The problem is where should the search for sources be started? There is a possibility that some sources not contain the topic mentioned. In this case, when to end the search? This seemed at first a rather trivial chemical thermodynamics textbook/seminary problem which has turned out not so trivial as expected initially. Also some extra-wiki discussions with knowledgeable people in both domains could point to some sources to be checked.--5.2.200.163 (talk) 15:34, 20 October 2015 (UTC)[reply]
Yes.Chjoaygame (talk) 15:47, 20 October 2015 (UTC)[reply]
The NIST page for deuterium is here and shows a value for entropy but not enthalpy. I believe this is because, yes, it is considered zero by default. The NIST values only make sense if it is assumed that the standard state for each isotopic species is the elemental state for that isotopic species, so that H2 and D2 are both considered zero. Yes, this point could be mentioned in the article if someone can find an explicit source for it. Dirac66 (talk) 16:01, 20 October 2015 (UTC)[reply]
Can NIST data assumption be mentioned without a source to explicitly state the assumption on D2?--5.2.200.163 (talk) 15:44, 23 October 2015 (UTC)[reply]

Bond energy calculation

An equivalent method to calculate/check the heat effect of a reaction is that based on bond energy of products minus the bond energy of reactants. Applied to the reaction of interest in this case the formation of HD from H2 and D2 (whose heat is considered to be the heat of formation of HD considering D2 heat zero) involves the bond energy of HD. A key question is: Does the value of this bond energy appear in NIST tables?--5.2.200.163 (talk) 10:36, 22 October 2015 (UTC)[reply]

No, the NIST tables do not include bond energies. One can calculate them from the ΔHf values: E(H-D) = ΔHf(H) + ΔHf(D) - ΔHf(HD), although of course this will not serve as a check.
I think the three of us have taken this discussion as far as is useful without additional sources. The situation is that the NIST values we have found only make sense if it is assumed that NIST excludes nuclear reactions from the ΔHf values, either for the reason given by Chjoaygame above or for some other reason. So in this article we can either say that nuclear reactions are excluded or we can just omit the point (which I favor), but we cannot say that they are included. Dirac66 (talk) 01:12, 23 October 2015 (UTC)[reply]
I agree that this talk section has gone far enough. I am favour of omitting mention because so far no suitable survey of reliable sources has emerged.Chjoaygame (talk) 01:58, 23 October 2015 (UTC)[reply]
I also agree about the length of this discussion. For the moment, lacking sources, we can omit the aspects discussed here. Perhaps we could bring other editors knowledgeable in nuclear engineering or/and chemical engineering thermodynamics to help searching for relevant sources.--5.2.200.163 (talk) 15:33, 23 October 2015 (UTC)[reply]
There is one more aspect to be clarified at the present stage of the discussion: the one re NIST (default) assumption about D2 heat of formation. Could it or not be mentioned in article just by acces to NIST tables without other source to explicitly notice what NIST tables contain?--5.2.200.163 (talk) 15:37, 23 October 2015 (UTC)[reply]
Above are two sources saying it is not considered in ordinary thermodynamics. I think that since you think it important enough to justify mention of the less ordinary case, you should think it important enough to find sources for that.Chjoaygame (talk) 18:24, 23 October 2015 (UTC)[reply]

Sign wrong?

IANA physicist or chemist, but I believe that the values for acetylene(g) and benzene(l) should be negative, not positive, since that's their base phase at 25°C. I see that a lot of compound pages have had their numeric data swept off into a "data" page, presumably so it's easier for the cognoscenti to watch for malicious graffiti on the key data. Perhaps this page needs something/someone similar? Even (or maybe especially) if I'm wrong! :)


Oh, yeah... if the page is going to differentiate between subphases (C-graphite vs C-diamond and S vs S-monoclinic) perhaps the "phase" column could be renamed "allotrope".

--Chem Doormouse — Preceding unsigned comment added by 71.109.148.145 (talk) 06:59, 7 February 2016 (UTC)[reply]

The positive signs for acetylene and benzene are correct. They indicate that acetylene and benzene have higher enthalpies than the standard state; for example C2H2(g) has a higher enthalpy than 2 C(s) + H2(g). This is deduced from experimental values of the heat of combustion. C2H2(g) has been found to liberate more heat on burning to form 2 CO2(g) + H2O(l) than do 2 C(s) + H2(g), so it must start at a higher enthalpy.
As for naming the column allotrope, the problem is that the word is only used for elements and not for compounds. Most of the substances considered are compounds so phase seems the best word for the majority of entries in the table. Dirac66 (talk) 16:27, 7 February 2016 (UTC)[reply]

Standard enthalpy change of formation (data table listed at Redirects for discussion

An editor has asked for a discussion to address the redirect Standard enthalpy change of formation (data table. Please participate in the redirect discussion if you wish to do so. Headbomb {t · c · p · b} 01:01, 12 May 2019 (UTC)[reply]

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