Talk:Chemical equilibrium

message: I've moved the recent comments here so as not to confuse them with comments on the original article  Petergans 20:33, 16 May 2007 (UTC)
replies: text: Despite the fact that I agreed that your complete rewrite was better, may I remind you that no one owns articles in Wikipedia? Asking people to respect your version is just as arbitrary as them asking you to respect the previous version. In that case I think the best course of action would be to revert to the old version of the article (before your rewrite) and resume the discussion from there. Also, discussing your credentials is usually far less effective at convincing people here than discussing the actual content of the article. --Itub 13:26, 17 May 2007 (UTC)

text: You have violated Wikipedia policies with that post. You did not wait for consensus before doing a complete rewrite of the article, I don't have to wait for your permission before editing your version. My (or your) qualifications are irrelevant in Wikipedia, only our submissions and sources are. Threatening a revert war is an offence under Wikipedia policies and you will be banned if you do it. If you want to do away with large parts of an existing article, it's your responsibility to cite sources and gather consensus to show they are incorrect/unnecessary. Please maintain civility and discuss which of my edits you find objectionable, like I did. Loom91 13:29, 17 May 2007 (UTC)
thread_title:  Comments 

message: This arises because an obsolete and unsound theory (law of mass-action) is apparently still being taught in schools, but not in Universities. I attempted to address this issue by means of an historical introduction. Perhaps a separate, historical, article on mass-action and kinetics would be better. In this way the details of kinetics could be handled separately from equilibrium.
I'm not sure how sound Mass action is. It appears to confuse the molecularity of a reaction with the rate expression, but I haven't looked at it carefully.  The text of the original paper by Guldberg and Waage can be seen in translation. 
Showing that the equilibrium state can be approached from either direction was important historically in establishing the nature of the state. This idea could be illustrated by a real isomerisation reaction which is both unimolecular and first order kinetics, or by one of G&W's original subjects of study. BTW, the reaction H2+I2 \to  2HI is a chain reaction, not second order, as was originally thought (Atkins).
In summary, put the historical stuff in a separate article?
Petergans 12:17, 22 May 2007 (UTC)
thread_title: The big dilemma

message: The middle paragraph could possibly be rephrased to better include the mentioning of reaction quotients. - BeardedPhysicist 15:17, 7 Mar 2005 (UTC)
thread_title: RXN quotient

message: A wikilink to ICE chart needs to placed in this article. s d 3 1 4 1 5 final exams! 17:25, 16 December 2006 (UTC)
thread_title: ICE

message: There is lots of duplicated material between here and at Equilibrium constant. I think we should decide to merge some things together, probably in the direction Equilibrium constant --> Chemical equilibrium. Also, would anyone object to a rewrite of this article? We have someone who is potentially interested in doing this. I think it would be a really great thing for this article. --HappyCamper 16:55, 24 April 2007 (UTC)
replies: text: Alright, that sounds good too. Sort of like what we have at quantum mechanics? Sections with brief paragraphs describing concepts, and then a link to a main article which goes into more detail. --HappyCamper 17:26, 24 April 2007 (UTC)
replies: text:  I'm not in favour of merging. These two topics have different scope. Equilibrium constant is only a specific mathematical idea related to chemical equilibrium. This is like the difference between the scope of the articles quantum mechanics and wavefunction. They are bound to have a lot of overlap, but they are distinctly different topics. Loom91 18:56, 24 April 2007 (UTC)
replies: text: So what does this mean? Leave this as is? Or go with V8rik's idea? Am I ever glad I dropped a note here first before being bold...the edit history doesn't betray the fact that there are multiple people watching this article! --HappyCamper 19:46, 24 April 2007 (UTC)
replies: text:  V8rik has not given a specific example of what he intends to do. But I don't see any problem with the overall idea. Loom91 15:57, 25 April 2007 (UTC)
replies: text: I am also not in favor of the merge.  I think that the differences between the topics are great enough that they should be treated seperately.  Then again, I don't have fancy credentials, so anyone that does, feel free to ignore me.  — Preceding unsigned comment added by 76.26.104.63 (talk) 04:24, 1 September 2011 (UTC)
thread_title:  Merging and rewriting 

message: OK, so it seems there's some contention over this paragraph: . I think the original edit is this one , but what is the author really trying to say? I don't see how the justification given connects with the first statement. I'd like to remove this entirely, or at least reword it somewhat. --HappyCamper 19:43, 24 April 2007 (UTC)
replies: text: There are some problems with those those paragraphs, but the general idea that equilibrium is due to the entropy of mixing is correct (I'm looking at Atkins and de Paula, 7th edition, p. 226 ["Molecular interpretation 9.1" in the chapter about chemical equilibrium]). However, the entropy increases, not decreases, due to mixing. Also, it would be more general to talk about free energy of mixing. The entropy [times T] and the free energy of mixing are the same only in the case of ideal gases, because in that case the enthalpy of mixing is zero. --Itub 08:09, 25 April 2007 (UTC)
replies: text:  Actually, it's more general to talk about entropy, because it is a more fundamental and universal quantity than Gibbs energy. The minimisation of the latter is a criterion that can be derived from the maximisation of the former under isobaric conditions. We generally talk about free energies because it's more convenient to look at affairs from the systems perspective rather than the universe's perspective. In standard state, the products of a spontaneous reaction are more favoured (actually, more probable) than the reactants. Yet in ordinary conditions the reaction does not go to completion because the mixing of reactants and products contributes an entropy to the equilibrium state. When an equilibrium is 'driven' by removing the products as they are formed, it's this contribution we are trying to remove. Loom91 16:06, 25 April 2007 (UTC)
replies: text: It's more general to talk about entropy only for isolated systems, or the entire universe... but for the mixing process in a non-isolated system, there is a huge difference between entropy and free energy that matters a lot for non-ideal mixtures. It is the enthalpy of mixing and it also contributes to the chemical equilibrium. I also note that the treatment in Atkins uses the free energy of mixing rather than the entropy of mixing when discussing the causes of chemical equilibrium. --Itub 08:18, 26 April 2007 (UTC)
replies: text: Let me get a hold of the Atkins book. This is very novel and I want to find out more :-). --HappyCamper 19:13, 26 April 2007 (UTC)

text: You miss the point. Get any introductory book on physical chemistry and read the part on Gibbs function. Enthalpy change of the system is simply a more convenient way of expressing the entropy change of the surroundings. The only deciding factor of spontaneuity is entropy change of the universe (or any isolated part of it). The Gibbs function is simply a way of incorporating the entropy change of the surroundings into a system property, the enthalpy change. Loom91 09:07, 27 April 2007 (UTC)
replies: text: I know that. The problem is that what people usually call the "entropy of mixing" is what results from the equation \Delta S_m =  -nR(x_1\ln x_1 + x_2\ln x_2)\, (at least in my experience). This equation does not include the change in entropy due to changing the temperature (I assume that's because of the interest in isolating the contribution due to "mixing"); this means that it only applies to an actual mixing process if the process isothermal. What I'm calling the "entropy of mixing" is therefore not equal in general to the total entropy change for the universe and cannot be used a spontaneity criterion by itself ("total entropy change" = "entropy of mixing" + "entropy of cooling/heating due to the mixing process"). A process can only be isothermal if the system is not isolated or if the mixture is ideal. And since people usually care about equilibrium at a given temperature, it is better to talk of the free energy of mixing. If I'm misunderstanding the common usage of the term "entropy of mixing", please give me some references, and feel free to correct the entropy of mixing article as well. --Itub 09:43, 27 April 2007 (UTC)

text: But the temperature change is not relevant here. The question is what stops reactions from completely transforming the unfavoured product to the more favoured product? The answer is not in cooling/heating, they will not stop the reaction from going to completion. It is the contribution of mixing that makes the equilibrium state more favoured than the pure product state. Loom91 14:47, 28 April 2007 (UTC)
thread_title:  Entropy of mixing 

message: The previous thermodynamics section in my view lacked an explanation exactly what was being derived (why this chapter exists in the first place). Before that it lacked the introduction of any relationship between Gibbs energy and chemical potential. I have modified this section, I hope it will more clearly demonstrate how you get from standard Gibbs energy equation and relationship chemical potential & activity to the fundamental delta G = lnK/RT. Bear in mind that Wiki is not written specifically for university students and that more steps are required to explain a concept. Also if we ever want this article the get the featured article status (and we all want that!) we really have to dumb down. V8rik 17:44, 31 May 2007 (UTC)
replies: text: I agree. Loom91 12:03, 1 June 2007 (UTC)
thread_title:  Thermodynamics 

message: Let's have consistency in the symbol used for thermodynamic standard. It should be the plimsoll line. I suggest the use of ..o  or  ..^\ominus  as in the following examples. text ΔGo; math \Delta G^\ominus Petergans 10:54, 1 June 2007 (UTC)
replies: text: I agree. Loom91 12:03, 1 June 2007 (UTC)
thread_title:  Notation 

message: The equation for ionic strenght differs from that in the ionic strength article, why are there the additional terms cX and cY? 
: I = \frac{1}{2}\left(c_X + c_Y + \sum_{i=1}^n c_i z_i^2\right) 
V8rik 19:01, 3 June 2007 (UTC)
replies: text: I = \frac{1}{2}\left(c_X z_X^2 + c_Y z_Y^2 + \sum_{i=1}^n c_i z_i^2\right) 
would be better, with the added comment that zX and zY are (to my knowledge) always equal to one in practice and that then the ionic strength is then effectively equal to the salt concentration.
The point about writing the terms separately is that together they will be much larger than the terms due to species in equilibrium, so that ionic strength will be nearly constant, whatever the position of the equilibrium, for example, throughout a  weak acid/strong base titration. It is then assumed that the variation in activity coefficients will be negligible and this assumption has been extensively validated by experimental results using the ionic medium method. If you want a reference to back this up you can use [IUPAC SC-Database A comprehensive database of published data on equilibrium constants of metal complexes and ligands] Petergans 08:06, 4 June 2007 (UTC)
thread_title:  Ionic strength equation 

message: The statement as it stands is incorrect. Consider a simple equilibrium reaction that can give two products.
:A+B \rightleftharpoons C; [C]=K1[A][B]
:A+B \rightleftharpoons D; [D]=K2[A][B]
If the system is at equilibrium at all times the product ratio is then given by
:\frac{[C]}{[D]}=\frac{K_1}{K_2}
regardless of any reaction intermediates. The Curtin-Hammett principle applies when the intermediates are not at equilibrium. Suppose the two products are formed from two intermediates X and Y which are formed slowly from A and B. if they are converted rapidly into the products; the concentrations [X] and [Y] will not be the equilibrium concentrations and the product ratio will be as per Curtin-Hammett. This principle is about control of product ratio by means of kinetics and does not belong in an article on equilibria. Petergans 09:21, 17 June 2007 (UTC)
replies: text: I agree with you there. The Curtin-Hammett principle is always used in kinetically-controlled reactions. Loom91 08:27, 18 June 2007 (UTC)

text: (Note: this was a response to V8rik, but Petergans and I posted almost at the same time and there was an edit conflict). The Curtin-Hammett principle article is correct; I know it because I wrote most of it! :P But the question here is whether it is worth mentioning it in the article about chemical equilibrium. Good arguments can be made either way. The situation considered by the Curtin-Hammett principle certainly involves an equilibrium between the two intermediates, but I wouldn't go so far as to say it is an "application" of equilibrium or a "type" of equilibrium; it is more of an application of kinetics. If I mentioned it here, it would be more as an "exception", by saying something like: "when two reactants that are an equilibrium can react irreversibly to give each a different product, the product ratio is not the same as the reactant ratio. This is explained by the Curtin-Hammett principle." This is more or less what you added, but the context is not the best. A similar argument applies to the case of oscillating reactions. I suggest that this article could have a short section near mentioning the existence of kinetics and non-equilibrium thermodynamics. These topics are not completely separable IMO, and at the very least they deserve some mention as related topics. --Itub 09:04, 19 June 2007 (UTC)
thread_title:  Curtin-Hammett 

message: In my opinion this topic is pure kinetics and does not belong here. It is true that the final product in time is an equilibrium mixture, but then that's true of all reactions if one accepts that no reaction ever goes 100% to completion. Petergans 09:53, 18 June 2007 (UTC)
replies: text: You are placing unnecessary emphasis on the distinction between equilibrium and kinetics. The phenomena of oscillating reactions certainly deserves at leastt passing mention in the article. Loom91 20:48, 19 June 2007 (UTC)
thread_title:  Oscillating reactions 

message: The article Mass action is currently the subject of an editorial dispute between me and Petergans. I request that you take a look at the revision history and the talk page and participate in the discussion. Loom91 08:21, 28 June 2007 (UTC)
thread_title:  Request for comment at [[Mass action]] 

message: In my opinion, the first sentence is very confusing and spoils an otherwise good article. "In a chemistry, chemical equilibrium is the distribution of interverting chemical compounds." What is that supposed to mean?? Opening sentences should be simple, and I suggest we return to the previous first sentence slightly modified to: "In a chemical reaction, chemical equilibrium is the state in which of the concentrations of the reactants and products do not change with time. (I would delete the mention of activity in the opening sentence, although it can certainly be mentioned later. Dirac66 (talk) 00:24, 7 February 2011 (UTC)
thread_title:  Confusing first sentence 

message: I just read this article after revisiting A-level (highschool) chemistry. There the accepted definition of equilibrium is as given in the opening paragraph: "rates of forwards and reverse reactions equal, concentrations don't change with time." As a theoretical chemist that got me worried, to me that describes steady state and not the very special (interesting) case. The article does go on to describe equilibrium as situtation in which the chemical potential is zero; my wonder is, is it worth making stronger the distinction between steady state and equilibrium?  — Preceding unsigned comment added by Bcbccouk (talk • contribs) 17:40, 5 January 2012 (UTC)
replies: text: As a start, I have added a "See also" link to Steady state (chemistry) which explains the distinction between the two concepts in the second paragraph. Yes, we can also add something in this article, but that will take more thought to do correctly. Dirac66 (talk) 02:36, 6 January 2012 (UTC)
thread_title:  Steady state 

message: (moved from User_talk:Petergans#Field_effects_on_equilibrium
Could you insert some formulae concerning electric and magnetic field effects influence on chemical equilibrium?--82.137.11.71 (talk) 19:38, 29 April 2012 (UTC)
replies: text: Unfortunately not. I've never heard of any such effects. Petergans (talk) 20:05, 29 April 2012 (UTC)
thread_title:  Field effects 

message: This article could benefit by having an entry level section.
Mis4930 (talk) 02:42, 20 September 2012 (UTC)
thread_title:  Entry level 

message: Please check Ref 12. The page cannot be found.  — Preceding unsigned comment added by DD dev (talk • contribs) 18:32, 27 November 2012 (UTC)
replies: text: I have now updated the URL for this Reference. Dirac66 (talk) 02:56, 29 November 2012 (UTC)
thread_title: Reference 12

message: Regarding the following excerpt from the article:
Solids also do not appear in the equilibrium equation. An example is the Boudouard reaction:[10]
2CO \rightleftharpoons CO_2 + C 
for which the equation (without solid carbon) is written as:
K_c=\frac{[CO_2]} {[CO]^2}
In fact, it is common in metallurgy to include the term aC, or activity of carbon, in the Keq expression.  The reason is that if we want to calculate the "carbon potential" of a gas containing CO and CO2, we include a term for the aC in the numerator, and knowing the value of Keq, and the partial pressures of CO and CO2, we can calculate the value of the aC.  A value of aC = 0.9 can carburize steel.  Also, it may be useful to determine the aC in a CO/CO2 mixture to ascertain if the mixture is metastable with reference to the deposition of carbon (i.e., if the aC is >1).  
Also, this section of the article uses the word "equation" when the words "equilibrium constant expression" are really what is meant.  Thermbal (talk) 05:10, 20 June 2013 (UTC)
replies: text: I will add the words "if they are considered to be pure". No phase is completely pure, but for many applications it is a reasonable approximation to consider that some phases are pure. For metallurgy, as you say, this is often not the case. Dirac66 (talk) 19:35, 20 June 2013 (UTC)
replies: text: Another aspect which can impose the consideration of the activity of solids is the solubility of gases in solids and the underlying adsorbtion equilibrium of gases on solids. In this case it is known that carbon can adsorb considerable amounts of gases.--188.26.22.131 (talk) 10:10, 31 July 2013 (UTC)

text: Activity of solids in equlibrium constant is required by dimensional considerations namely the condition that equilibrium constant be dimensionless.--188.26.22.131 (talk) 10:36, 31 July 2013 (UTC)

text: Another case of necessity (of metallurgical importance) to consider activities of solids is given by the activity of a metal in an alloy.--188.26.22.131 (talk) 12:58, 31 July 2013 (UTC)
thread_title:  Pure substances