Talk:Rate equation

The contents of the Order of reaction page were merged into Rate equation on 27 April 2018. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see its talk page.

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What exactly does it mean to say what the order of a reaction is with respect to a reactant? Does it always mean "If you multiply the concentration of one reactant by a certain power and leave the others constant, what power of that factor does the rate of reaction get multiplied by?" How do half order reactions happen? Blackbombchu (talk) 23:53, 14 September 2015 (UTC)[reply]

For your first question (Does it always mean ...), the answer is yes, your proposed meaning is correct. This could be spelled out in the article, but I think it is not really necessary because it is an obvious consequence of the mathematical definition :${\displaystyle r\;=\;k[\mathrm {A} ]^{x}[\mathrm {B} ]^{y}}$.

As for half order reactions, note that in the last sentence of the second paragraph (The equation may involve a fractional order, and may depend on the concentration of an intermediate species.), the words fractional order are linked to a section of the Order of reaction article which explains the origin of order 3/2 for the reaction CH₃CHO → CH₄ + CO. For questions not answered in a Wikipedia article, it is often a good idea to follow the links for relevant words. Dirac66 (talk) 18:20, 16 September 2015 (UTC)[reply]

Added an explanation as to why the 2nd order equation includes a factor of two multiplying the rate constant. — Preceding unsigned comment added by 2001:1388:803:CDFC:7C1F:DA6F:70F5:E9F0 (talk) 08:19, 14 October 2015 (UTC)[reply]

During an assignment I found something I suspect to be a mistake in this section, I do not know how to edit these things myself. First it states that A_n/A_n-1=f_RP, then it states that (f_RP)ⁿ=(1-f_BP)ⁿ, and at last it states that f_BP=1-A_n/A_n-1, which would suggest that f_RP=1+f_BP. I hope someone can help me by correcting either of the equations, so 1 of the subtraction signs become an addition sign, Thank you in advance! A happy user — Preceding unsigned comment added by 87.59.207.115 (talk) 16:28, 14 December 2017 (UTC)[reply]

Never mind this, I couldn't get it to make sense at first, after a few more minutes it began to make sense.87.59.207.115 (talk) 16:34, 14 December 2017 (UTC)[reply]

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I propose that Order of reaction be merged into Rate equation. Based on what is in these articles, they seem to be the same subject and even have many of the same section headings. "Rate equation" is the more general term, so I recommend that it be the target; but Order of reaction is better written overall, so its content would mostly replace the corresponding sections in Rate equation. RockMagnetist (DCO visiting scholar) (talk) 01:30, 19 April 2018 (UTC)[reply]

Support. Yes, the articles discuss very similar subject matter with a lot of overlap. In fact I have often had trouble deciding in which of the two articles to search first for a given point. Combining them would solve that problem for everyone. I also agree that more material should be taken from Order of reaction which is now better written, but that Rate equation is a better title for the combined article because it is more often used in chemistry textbooks. Dirac66 (talk) 01:52, 20 April 2018 (UTC)[reply]

O.k., it has been a week. I'm going to do it. RockMagnetist (DCO visiting scholar) (talk) 00:42, 27 April 2018 (UTC)[reply]

My impression is that everything in Equilibrium reactions or opposed reactions and the sections below belong somewhere else because they are not really discussing the rate equation. I'm not sure where, though - Chemical kinetics, maybe? RockMagnetist (DCO visiting scholar) (talk) 17:23, 27 April 2018 (UTC)[reply]

First I assume that you mean Equilibrium reactions or opposed reactions and its subsections, i.e. 10.1 and 10.2 but not 11, 12, etc. I think however that section 10 mainly needs a title change; it is really about rate equations but for the case where forward and reverse reactions are at comparable speeds, close enough to equilibrium so that both must be considered. Some of the equations are rate equations, and those that are not are used to help define the rate constants in terms of the rate equation.

Some other possible names for this section used in textbooks: Reactions approaching equilibrium (Atkins and de Paula, Physical Chemistry, 8th edn p.804) or Reversible first-order reactions (Espenson, Chemical kinetics and reaction mechanisms, p.42) or Opposing reactions of higher order (Espenson p.45). I will change the section title to Opposing reactions without the misleading word Equilibrium, since the conditions of interest are only close to equilibrium but not exactly at equilibrium. Also I will remove equilibrium from the first sentence of the section, although the mention of the equilibrium constant in the 4th sentence does seem necessary. Dirac66 (talk) 19:06, 27 April 2018 (UTC)[reply]

This section is a bit messy :

• The reaction rates are introduced by ${\displaystyle k_{1}}$ and ${\displaystyle k_{2}}$, and turned into ${\displaystyle k_{f}}$, ${\displaystyle k_{b}}$, respectively, in the simple example subsection
• It is very unlucky to choose ${\displaystyle t}$ as stoichiometric coefficient for reactant ${\displaystyle A}$, and denote time by ${\displaystyle t}$ as well.
• The somehow concluding two lines in the simple example

${\displaystyle v=k_{1}[{\ce {A}}]^{s}[{\ce {B}}]^{t}-k_{2}[{\ce {X}}]^{u}[{\ce {Y}}]^{v}}$

${\displaystyle -{\frac {d[{\ce {A}}]}{dt}}=k_{f}[{\ce {A}}]_{t}-k_{b}[{\ce {B}}]_{t}}$

seem pretty unrelated to each other - not mentioning the confusion of time and stoichiometric notation and the different reaction rate notations - there is no obvious way the second line follows from the first. Even if one uncoupled the reactants ${\displaystyle A,B}$ from the products ${\displaystyle X,Y}$ as is suggested by the argumentation above, it feels strange to consider ${\displaystyle A}$ and ${\displaystyle B}$ now as a separate reaction with forward and backward rates ${\displaystyle k_{f}}$, ${\displaystyle k_{b}}$, respectively, and swallowing all stoichiometric coefficients. 2A02:2788:925:F83D:5CFE:78AD:EEE6:CEDA (talk) 19:59, 27 March 2020 (UTC)[reply]

It's probably no coincidence that the section is unsourced. Also, in my opinion there are too many derivations for an encyclopedic article; this isn't a textbook. But this is low on my list of priorities. If you'd like to clean it up, go for it! RockMagnetist(talk) 16:31, 28 March 2020 (UTC)[reply]

Thanks to 2A02:2788:925:F83D:5CFE:78AD:EEE6:CEDA for pointing out the notational inconsistencies in this section. My impression is that different authors over the years have used different notations, possibly from different uncited sources. So to reduce the confusion I have rewritten the equations with a uniform notation for the section.

For the rate constants I have written ${\displaystyle k_{1}}$ for the forward reaction and ${\displaystyle k_{-1}}$ for the reverse reaction throughout the section.

I have written the general reaction as aA + bB = pP (for product) + qQ, which makes it clear which stoechiometric coefficient goes with which reactant or product. I agree that t is an unacceptable choice for an article about kinetics since it will be confused with t for time. And I have eliminated X and x since x is also used for a concentration in this section.

As for the two equations described above as "pretty unrelated to each other", the problem is that B denotes a reactant in the first equation and a product in the second equation! This is also resolved by using a uniform notation for each species.

Finally, I agree with Rock Magnetist that the section should be sourced. However if several sources are cited, it is important that all the symbols and equations are transcribed into a uniform notation for the section. Dirac66 (talk) 16:14, 2 April 2020 (UTC)[reply]

@Smokefoot: I notice that today you changed the 3 (unsourced) examples in the section First order to 3 other (still unsourced) examples, with the comment that they are more appropriate examples. For the record, could you indicate why you consider the new examples more appropriate. My opinion is that there are many possible examples and that the new examples seem neither more or less appropriate than the old. However we really should use examples with sources, which are not that hard to find in chemical kinetics textbooks.

Also, the removed example 2 N₂O₅ perhaps had the advantage of showing that the order cannot be deduced from the coefficients. An A + B → reaction would be even better, especially if neither A nor B is H₂O. Dirac66 (talk) 00:16, 30 November 2021 (UTC)[reply]

Thanks for the note. You raise good points. I'm not sure that my answer is for any record, and am unsure that a great answer exists, but let me show the before and after with my remarks.

• ${\displaystyle {\ce {H2O2 -> H2O + 1/2O2}}}$ H2O + 1/2O2}}}"/>

comment: pretty complicated process, at least to me. no ref.

• ${\displaystyle {\ce {SO2Cl2 -> SO2 + Cl2}}}$ SO2 + Cl2}}}"/>

comment: obscure and messy reactant. no ref.

• ${\displaystyle {\ce {2N2O5 -> 4NO2 + O2}}}$ 4NO2 + O2}}}"/>

comment: maybe a keeper, dunno. no ref.

• ${\displaystyle {\ce {Cl2 -> 2Cl}}}$ 2Cl}}}"/>

comment: no option, really, other than unimolecular, so clearer. We start with the basics. no ref but not sure we need one.

• ${\displaystyle {\ce {cis-2-butene -> trans-2-butene}}}$ trans-2-butene}}}"/>

comment: Organic example. Most chemists are organic, so we provide them in with this (uncited) example. I guess we could include something like t-BuCl hydrolysis, which is SN1.

• ${\displaystyle {\ce {[CoCl(NH3)5]2+ + H2O -> [Co(H2O)(NH3)5]+3 + Cl-}}}$ [Co(H2O)(NH3)5]+3 + Cl-}}}"/>

comment: thought provoking and, I thought, informative because there are two reactants, but one has no effect on rate. I have a source for this one.

If and when we settle this matter, we can scout for sources. --Smokefoot (talk) 01:39, 30 November 2021 (UTC)[reply]

Thanks for the reply. I think we should settle the matter by choosing among the reactions that we can source. After a 10-minute search, I found that the N₂O₅ kinetics are explained in Atkins and de Paula (7th ed, p.886) or (8th ed, p.813-4). It is also explained at Lindemann mechanism#Decomposition of dinitrogen pentoxide with a 3-step mechanism and sourced to Laidler's book.

As for ${\displaystyle {\ce {Cl2 -> 2Cl}}}$ 2Cl}}}"/>, I suspect it is actually BImolecular by a Lindemann mechanism: 2 Cl₂ → Cl₂ + Cl₂^*(activated), Cl₂^* → 2 Cl. But I have no source now.

For ${\displaystyle {\ce {[CoCl(NH3)5]2+ + H2O}}}$ let's put in the source you have. With H₂O as a reagent, it is best to check whether it is a pseudo-first order reaction (see section of this article after second order) with H₂O in large excess.

So for now I think we should keep only N₂O₅ and ${\displaystyle {\ce {[CoCl(NH3)5]2+ + H2O}}}$, until we find sources for another reaction or two.Dirac66 (talk) 03:31, 30 November 2021 (UTC)[reply]

Will do, but semi-distracted now. A few comments,

Re "As for ${\displaystyle {\ce {Cl2 -> 2Cl}}}$ 2Cl}}}"/>, I suspect it is actually BImolecular by a Lindemann mechanism: 2 Cl₂ → Cl₂ + Cl₂^*(activated), Cl₂^* → 2 Cl."

Isnt such collisional activation true of almost all supposed first order reactions?

"For ${\displaystyle {\ce {[CoCl(NH3)5]2+ + H2O}}}$ let's put in the source you have. With H₂O as a reagent, it is best to check whether it is a pseudo-first order reaction (see section of this article after second order) with H₂O in large excess." got it about detecting solvent dependence. Maybe Mo(CO)6 + L. Need to look.--Smokefoot (talk) 04:01, 30 November 2021 (UTC)[reply]

I have found a reference for ${\displaystyle {\ce {H2O2 -> H2O + 1/2O2}}}$ H2O + 1/2O2}}}"/> which is given as a first-order example by Petrucci, Ralph H.; Harwood, William S.; Herring, F. Geoffrey (2002). General Chemistry (8th ed.). Prentice Hall. p. 588. ISBN 0-13-014329-4., so we can restore this example. Just put the cite book info between two reference tags to make it a footnote. No reaction mechanism in this source though. Dirac66 (talk) 02:53, 4 December 2021 (UTC)[reply]

And re your question "Isn't such collisional activation true of almost all supposed first order reactions?" I think you meant "almost all supposed unimolecular reactions". According to the Lindemann mechanism, any unimolecular mechanism requires collisional activation as a first step, and therefore will become second order at suffciently low pressure.

Should we change the First-order examples to the 3 sourced examples we have now: N₂O₅, ${\displaystyle {\ce {[CoCl(NH3)5]2+ + H2O}}}$ and ${\displaystyle {\ce {H2O2 -> H2O + 1/2O2}}}$ H2O + 1/2O2}}}"/>? Dirac66 (talk) 17:09, 5 December 2021 (U

@Smokefoot: OK. I have now gone ahead and changed the examples to ${\displaystyle {\ce {2N2O5 -> 4NO2 + O2}}}$ 4NO2 + O2}}}"/>, ${\displaystyle {\ce {[CoCl(NH3)5]^2+ + H2O}}}$ and ${\displaystyle {\ce {H2O2 -> H2O + 1/2O2}}}$ H2O + 1/2O2}}}"/> as discussed. I have also added the sources I have for the first and third reaction. For the second I realize that I don't know your source, so could you please add it? I did modify the "chem" format to change the reactant from ${\displaystyle {\ce {[CoCl(NH3)5]2+}}}$ to ${\displaystyle {\ce {[CoCl(NH3)5]^2+}}}$ which I'm pretty sure is what you meant, and similarly for the 3+ product. Dirac66 (talk) 21:02, 20 December 2021 (UTC)[reply]