Talk:Rate equation/Archive 1

Archive 1

Requested Move

It seems that there are a few pages that already explain reaction rates in chemical reactions. This being one of them, another being the page called Chemical Reactions and the third being reaction rate. Both this article and the article discussing reaction rates can be easily trimmed down and merged into the Chemistry article. I do not have the understanding or ability to move so much information so I'd appreciate it if someone considered doing so. Aznph8playa 00:26, 30 May 2005 (UTC)

this page will get extented to include pseudo first order reactions and steady state etc (much more out there). The chemical reaction page should be as broad as possible (which it is not at the moment) and is now in need of attention (missing: biochemistry, electrochemisty, organometallic chemisty, inorganic chemistry). Reaction rates and rate laws are also not the same. Please consult the Wiki guidelines, this is the encyclopedia part, what you have in mind really belongs in wikibooks rikXL 15:44, 30 May 2005 (UTC)

Ah my mistake. It seems I haven't taken enough chemistry. I had thought there wasn't much more past the three rate laws: zero, first and second. Thanks for the clearification. Aznph8playa 19:16, 30 May 2005 (UTC)

Older comments

I've updated the page to include zero-order rate laws so there is an explanation of what they are instead of just being implied. I've also rearranged some of the formula's, most notably the second order AB integrated rate law, which is now in an exponential form. I've also removed the subscript t's associated with the concentrations in the integrated laws. Many textbooks leave out the subscripts, and it looks neater without them, however, if it becomes confusing, perhaps they should be reincluded?--Artorius 11:17, 3 June 2005 (UTC)

Closed system assumption

I think it is unfortunate that most texts (chemical textbooks as well as wiki texts) that try to explain reaction rates, rate laws etc. confuse rate law with the result of a mass balance for a closed system. I've tried bringing some clarity to this issue under mass balance but the more I look through wikipedia the more problems I find.

Consider for example a bathtub in which we have some bathing salt dissolved. If we fill in more water, the concentration will decrease. With the definition of rate law commonly used by non-chemical engineers many people will (trust me on this) misunderstand the system and think there is a reaction going on since ${\frac {dC}{dt}}\neq 0$. But ${\frac {dC}{dt}}$ does not define reaction rate! One must first write down a mass balance before a link between ${\frac {dC}{dt}}$ and the reaction rate can be found! Saittam 15:52, 12 August 2005

I guess it is up to us to strike a balance between chemists and chemical engineers view on reaction rates. Wiki should not respect the classical boundaries and there should not be two separate chemical kinetics articles for each group . The way you have included the mass balance link in rate law is the best way to do this. I may also help to use the bathtub example in the text itself and not just in talk because it really brings home your point. V8rik 20:59, 15 August 2005 (UTC)

Explanation?

Shouldn't there be an explanation of why the rate law works? I understand that a lot of scientific articles don't have an explanation section, but in those cases the explanation is either unknown, obvious, or exceedingly complex; the explanation for this law isn't any of those. Comments before I add it? — Preceding unsigned comment added by 72.140.134.165 (talk) 18:50, 14 December 2005 (UTC)

The reason that a rate law "works" is that it is simply a description of what one "sees"; it can be used to test hypotheses of a reaction mechanism, that is true, but not all rate laws are completely explained. Physchim62 (talk) 16:20, 15 December 2005 (UTC)

or do you mean how to derive the rate laws? just make the edit you have in mind, I am intrigued V8rik 20:22, 15 December 2005 (UTC)

Second-Order Example Not Possible

The example second-order reaction $\left(2{\mbox{NO}}_{2}\rightarrow {}2{\mbox{NO}}_{2}+{\mbox{O}}_{2}\right)$ isn't possible; perhaps $\left(2{\mbox{NO}}_{3}\rightarrow {}2{\mbox{NO}}_{2}+{\mbox{O}}_{2}\right)$ was intended? --Shadypalm88 06:34, 6 February 2006 (UTC)

According to these lecture notes, it seems that $2{\mbox{NO}}_{2}\rightarrow {}2{\mbox{NO}}+{\mbox{O}}_{2}$ was the intended equation. I have made this change and also cleaned up the LaTeX a little. --Shadypalm88 06:34, 6 February 2006 (UTC)

Sorry, I've made a mistake in writing up the example for 2nd order reaction.

What I wanted to write should be $2{\mbox{NO}}_{2}\rightarrow {}2{\mbox{NO}}+{\mbox{O}}_{2}$

Thank you for your correction. Ray 07:09, 6 February 2006 (UTC)

units for rate constants k

I guess the units for rate constants k in the table Summary for reaction orders 0, 1, 2 and n should be liter/(mol s) and not 1/(mol s). --193.174.169.1 12:31, 30 August 2007 (UTC)

first-order

$\ \ln {[A]}=-akt+\ln {[A]_{0}}$
What is "a"? --Saippuakauppias ⇄ 15:50, 14 December 2007 (UTC)

There is no a in that equation. The equation should be $ln{[A]_{T}}=-kt+ln{[A]_{0}}$ Redtails (talk) 12:58, 1 July 2009 (UTC)

Measuring Rate Laws

This article gives a good account of the equations that describe rate laws, but would probably be more complete if it explained how rate laws are determined in the real world. I know one common method is measuring the change in absorption with a photospectrometer as a reaction proceeds. Perhaps a chemistry buff could add a section about experimentally determining rate laws? fpgojkf;bldf —Preceding unsigned comment added by 125.209.115.133 (talk) 10:32, 30 January 2008 (UTC)

Its not a bad idea.

There are some somewhat more sophisticated methods as well that involve solving an inverse problem (e.g. assuming a model for a reaction, a performing an experiment and curve-fitting to the optimum value of the order.)

In my grad kinetics book it outlined some of the super-fancy (and super-expensive ways) of determining fundamental reaction parameters, e.g. order and activation energy. If I find the time I'll look into it.

Bradridder (talk) 00:17, 9 February 2010 (UTC)

2nd order rate equation

I was reading through my chemistry textbook, and i think we have a typo here under the second order rate equation. The 2nd order rate equation should be 1/[A] = 1/[A]0 + kt, and its respective half-life should be t = 1/k[A]0. Does anyone know how to fix this?Bcperson89 (talk) 10:44, 17 March 2010 (UTC)

this change was a recent one, it has been reverted V8rik (talk) 20:02, 17 March 2010 (UTC)

2nd order rate equation #2

Since my changes were removed by another user (he didn't like an edit from 147.155.200.124, which is my ameslab.gov computer, where I work as a PhD chemist, and my expertise is in chemical kinetics), I'll post one important point here. The proper way to write a 2nd-order rate law is this: 1/[A]=1/[A0] + 2kt.

The emphasis is on "2". This stems from the fact that for an elementary reaction A + A = B, the proper rate equation is -dA/dt = 2kA^2.

This is how the rate constant of a 2nd-order reaction is properly defined, and this is how you will find all 2nd-order rate constants in standard kinetics tables and databases. For example, NIST tables or RCDC at Notre Dame. Also, the half-life is therefore t1/2 = 1/2kA0.

EDIT: Sorry for multiple edits. I am only now learning how to properly edit wikipedia. I figured I'd give a reference to support the above claim.

Quote: Rate constants for second-order decay of the radicals are given as k (not 2k) where –dR/dt = 2k[R]2 has been determined.

http://www.rcdc.nd.edu/compilations/Ali/Ali.htm —Preceding unsigned comment added by 173.26.203.133 (talk) 13:40, 25 March 2010 (UTC)

that user would be me! . Problem is anonymous users with an IP are usually vandals and editors tend not to be patient with their edits (even when connecting from the Ames lab), so please consider getting a user account. Secondly many readers like myself will be confused when the equation in question differs from their particular textbook so perhaps if you change it in the way you suggest you add a comment that textbooks tend to differ on this issue and that the multiplication of two constants still is a constant (PS if you use 4 tildes in a row you auto sign your comment) V8rik (talk) 21:54, 25 March 2010 (UTC)

Good point. I'll make sure to register properly tonight. As for the content itself, it's a little confusing to say the least. I just checked my 3 general chemical kinetics books (Connors; Espenson; and Capellos and Bielski). The first 2 books give the final definition without the stoichiometric factor of 2. Capellos and Bielski do it right, the way I suggested. Both first 2 books mention the existence of the stoichiometric factor, but in the end obtain the wrong result. The reason why it is ultimately wrong is threefold:

1) Kinetics tables and actual peer reviewed papers from good kinetics groups give rate constants, for the lack of a better word, in the "proper" form (that is, not like in the first 2 books.)

2) It is counter intuitive to include the factor of 2 into the rate constant. Consider this hypothetical reaction: 2 A + B = C. The proper rate laws with respect to A and B are this: -dA/dt = 2 k A^2 B, and -dB/dt = k A^2 B. The stoichiometric coefficients should not be combined with the rate constant to avoid confusion.

3) Modern research in chemical kinetics pretty much requires using simulation software. Second-order rate constants used in simulations must be entered in the "proper" form. Otherwise, one would have the divide the rate constant by 2 before using it in simulations. —Preceding unsigned comment added by 147.155.200.124 (talk) 22:54, 25 March 2010 (UTC)

I registered, edited the article, and added a reference. Now, if only someone could help me figure out how to become a regular contributor to wikipedia and join some kind of a project related to chemistry and kinetics, so that my own credentials here would be different from just a random person. I think I'd then be motivated to go through kinetics articles and correct any mistakes or typos :) OP (talk) 01:23, 26 March 2010 (UTC)

Thanks for your input and if you have any questions on editing do not hesitate to ask V8rik (talk) 20:55, 26 March 2010 (UTC)

Hi OP , you have noticed your edits get reverted. Too many visitors to this page (many of them students preparing for exams) are unfamiliar with your particular interpretation of the second order rate equation. Consider leaving it as it is and insert your interpretation as a comment instead of the other way around. V8rik (talk) 22:22, 4 April 2010 (UTC)

Yes, thanks. I just noticed. I'll try to figure out how best to approach this problem and write it up. As a side note, I'd like point out that it's not just my interpretation (I'm sure you didn't mean it that way.) There are only a few (count on two hands) true chemical kinetics groups in this country (I'm not counting more biology and biochemistry oriented people, as I'm not that familiar with their work, but I'm sure there must be a lot of fine kinetics groups in that field.) And when someone writes a popular and really easy to comprehend kinetics text, like Connors, and then makes up a different definition, a lot of people not intimately familiar with the field will follow the trend. Even my former PhD adviser, Jim Espenson, only mentions this issue in passing in his book. I never asked him about this, but I imagine that the way he wrote it was to comply with other texts, like Connors's. While, all the peer reviewed literature I've seen used the proper definition, including Espenson's own papers. OP (talk) 01:29, 13 April 2010 (UTC)

old habits die hard, we have the same discussion over at lone pair: the general textbooks that all the students use insist water has rabbit ears while the more advanced textbooks insist that is not the case. V8rik (talk) 21:40, 13 April 2010 (UTC)

The problem arises because textbook authors tend to not use the fundamental definition of the rate when deriving the empirical rate laws. For example, consider the elementary reaction $2A\to B$. The rate for this reaction is defined as $r=-{\frac {1}{2}}{\frac {d\left[A\right]}{dt}}={\frac {d\left[B\right]}{dt}}$. The rate law is $r=k\left[A\right]^{2}$. Combining these two expressions gives $-{\frac {1}{2}}{\frac {d\left[A\right]}{dt}}=k\left[A\right]^{2}$, which integrates to the proper form. Not deriving the integrated rate law from the fundamental definition can lead to problems when analyzing complex systems such as consecutive and/or reversible reactions. ZPOT (talk) 16:51, 18 April 2010 (UTC)

Thank you, you nailed it. It's kind of sad to see that proper definitions can still be misunderstood, even after dozens of excellent kinetics texts are available. Sometimes, it almost seems like kinetics is viewed as some kind of voodoo magic, while it just happens to be one of the most useful, if not the most useful, methods in elucidating reaction mechanisms.OP (talk) 21:23, 4 May 2010 (UTC)

Alternative View of First Order Kinetics

I'm afraid I find this section to be confusing and inaccurate. The author is suggesting that the usual equation for 1st order kinetics is inaccurate and can give large errors. However, they assume that for 5% decay, 5% of the initial population decays every time interval. This is wrong, because 5% of the instantaneous population decays every time interval. Their analysis is only accurate for cases where the decay rate is very small compared to the time interval (otherwise the calculus involved in integrating the original rate equation wouldn't apply). This is also the difference between compounding interest daily and continuously. Should this section be removed/edited? Futurechemist1 (talk) 01:01, 13 May 2010 (UTC)

Confusion does not warrant deletion, use the tag. If the section is inaccurate then edit it, again there is no reason to delete, the difference between initial and instantaneous appears to me a subtle one V8rik (talk) 20:40, 13 May 2010 (UTC)

Sorry about that, this is my first time editing Wikipedia. I reread the section to figure out where I'm getting confused. I think it's how the author defines the rate constant. The rate constant represents the "speed" of the reaction, but it shouldn't be true that a rate of X% means a rate constant of X. If that was the case, a reaction consuming 50% of reactant in 1s would have k=0.5 and a half life of 1s by definition. But half life is also defined as ln(2)/k = ln(2)/0.5 = 1.38, which isn't the half life. In this case, the author's argument about the equation being inaccurate falls apart. I'll have a go at editing this. Futurechemist1 (talk) 22:09, 14 May 2010 (UTC)

As a follow up, I've been in contact with ESkeptic, who originally created this section, and I've changed the section based on our discussion that should correct any of the original confusion. Futurechemist1 (talk) 11:22, 30 May 2010 (UTC)

There is still a problem with the derived relationship. Given the fundamental definition of the rate of a first-order reaction is $r=-{\frac {dA}{dt}}=kA=kA_{0}e^{-kt}$, it would imply that the final derived relationship would be $A=A_{0}\left(1-{\frac {r}{A_{0}}}\right)^{t}=A_{0}\left(1-ke^{-kt}\right)^{t}$. This equation cannot be valid since as it would suggest that as $t\to \infty ,A\to A_{0}$ instead of the correct value of 0 (note: the rate of a first order reaction approaches zero as t approaches infinity). It appears that defining %BD as equivalent to ${\frac {r}{A_{0}}}$ is invalid. From the abstract of the reference to this approach, the authors claim that they "... use equations based on steady state approximations to directly model time course plots" for "single substrate enzyme-catalyzed reactions". These approximations do not appear to be a valid for all t associated with the analysis of traditional first-order kinetics. Enzyme catalyzed reactions are usually analyzed using only the initial rate data. Therefore, this analysis approach is probably only valid for enzyme systems and should be moved to a page discussing enzyme kinetics. ZPOT (talk) 18:28, 30 May 2010 (UTC)

You're right about that. But I think everything is OK up until defining r=%BD, so the equation $A=A_{0}\left(1-\%BD\right)^{t}$ should be fine. No assumptions about enzymes would have been made by this point, so it should be general for all 1st order behavior. Futurechemist1 (talk) 20:23, 30 May 2010 (UTC)

I still disagree with defining $e^{-k}$ as the fraction of the population that remains each time period (unless that time period is restricted to $\Delta$t = 1). After all, the fraction of the population that will react depends on the length of the associated time period. If the time period is restricted to 1, then a valid representation of the equation would be $A=A_{0}\left(1-\%BD\right)$. This equation is essentially a rearranged form of the equation for the average rate over the restricted time period: $A=A_{0}\left(1-{\frac {r_{avg}}{A_{0}}}\right)$. For any time period, the equation will be: $A=A_{0}-r_{avg}\Delta t$. ZPOT (talk) 22:11, 31 May 2010 (UTC)

As I understood the original text, I think the way %BD is used is ok. BD is the amount consumed every 1 time interval. Consider a 1st order reaction where 10% of the remaining reactant is consumed every 1s, therefore %BD=0.1. After 1s, 90% remains, after 2s, 81% remains, etc. Mathematically, this is $A=A_{0}\left(0.9\right)^{t}=A_{0}\left(1-0.1\right)^{t}=A_{0}\left(1-\%BD\right)^{t}$. Futurechemist1 (talk) 16:42, 1 June 2010 (UTC)

Yes, if the time period corresponding to the fraction of the population remaining is restricted to $\Delta t=1$, the approach is fine. It probably should be clarified that the time period refers to 1s, 1min, 1h, etc (depending on the units of the rate constant). ZPOT (talk) 22:09, 1 June 2010 (UTC)

The first rate equation in introduction

the very first rate equation is given as:

$r\;=\;k[\mathrm {A} ]^{x}[\mathrm {B} ]^{y}$

This is for an elementary reaction: aA + bB → C
In this case, due to what an elementary reaction is, would it not be that x=a and y=b? Why change it then? I presume it would be to show that, generally, x and y are not equal to a and b respectively (when it's not an elementary reaction). However, if this is the case, maybe it's better to show a general rate equation first, then specify the case for an elementary reaction (where x=a and y=b).
Or am I missing something here?

There is a major conceptual error in the introduction to this article in regards to the orders of elementary reactions and the observed reaction orders. Please consult any elementary general chemistry textbook for an explanation. — Preceding unsigned comment added by 75.141.106.196 (talk) 12:06, 2 August 2012 (UTC)

I agree, if the reaction is elementary then the raised powers are related to the stoichiometric coefficients. We know when a reaction is not elementary when the order of the reaction doesn't correspond to the values of the stoichiometric coefficients. The first paragraph is therefore in error. Rhodydog (talk) 18:06, 16 August 2013 (UTC)

I agree, for an elementary reaction there are no intermediate reaction steps possible (otherwise it would not be an elementary reaction), hence the reaction must follow the mass action law and therefore x = a and y = b. I am pretty confident about this (PhD in Chemical Engineering). After reading more in the article, I think several things are blatantly wrong or have underlying assumptions that are simply not mentioned therefore implying general validity. Necmon (talk) 00:25, 22 August 2013 (UTC)

Note that this error has been fixed by the rewrite of May 2015. The second paragraph now treats elementary reactions correctly. Dirac66 (talk) 01:58, 16 September 2015 (UTC)