Talk:Arrhenius equation

Query1

The universal gas constant R has indeed the same units as the Boltzmann's constant (normally also written as "k", not to confuse with the "k" in Arrhenius equation!).

Let me explain:

[R] = E K^-1 mol^-1
[k] = E K^-1

and the potential confusion comes from the facty that a mol is indeed dimensionless (it is as if you say "a dozen", though a "mol" is much bigger (Avogadro's number)). So, RT has also dimensions of energy, and E/RT is indeed dimensionless as expected. jbc

Oh yes, that's right. But I guess the energy in the exponential would have to be per unit mole as well, as opposed to the energy of all molecules. Isn't that right? But is mol really dimensionless? I think I disagree about that part. If you have a molar mass, like 20 grams/mole. Actually scratch that, you are right it is "dimensionless", but not "unitless". And I think RT has dimensions of energy/mole, not energy as you say. Maybe we are saying the same thing though, I see what you mean. dave

Sorry, we can't have energy per unit mole on top, it's just the activation energy... But I still think there should be $n$ multiplied by R. Actually, I think the best form of the equation is exp(E/kT), then it is more general, and not specific only to chemistry where they deal with moles.

You are talking about different things in your comments. One thing is which is Arrhenius equation (as Arrhenius wrote it and it is used today), and another is whether it is right or not. For the first one, Arrhenius equation is without that $n$, as you can see in http://www.shodor.org/unchem/advanced/kin/arrhenius.html or any chemistry book you like. As for the second, I haven't worked the equation myself, but I would tend to thing it is right (wouldn't anyone have noticed by now?). If it turned out to be wrong, it would be worth to mention it in the article, but it would take some research before making such a statement, don't you think?

In particular, for what concerned you initially, the equation is dimensionally correct. A mol is indeed dimensionless, and you cannot disagree with it more than with a distance having dimensions of length, because it comes by its definition. You said "I think RT has dimensions of energy/mole, not energy as you say", which is contradictory, because the dimensions of energy/mol are the same as that for energy, mol being dimensionless. Of course it doesn't mean that "mol" is as if it didn't appear, the same way that "4 dozen" is not the same as "4".

I don't quite understand your second paragraph. Anyway, you mentioned earlier "molar mass", which is indeed a good example, because when you say "the mass of CO₂ is 44.01 g/mol", you are saying, "if you take 6.022 x 10²³ molecules of CO₂, they weight 44.1 g". See, you could play the same game with "dozens", or "pairs", or "hundreds", the numbers will of course vary, but the only dimension appearing is mass. jbc May 26 03:45 UTC 2003

I have a request - I know (or believe I know :) that there are reactions that don't follow the Arrhenius equation. Could someone knowledgeable update the page with the categories of reactions that don't obey the equation? I understand that the equation won't be accurate when, for example, the output energy of the equation drives other reactions that consume the reactants for the reaction you're interested in. I'm wondering if there are other examples.

--Could you give a more specific example of your output energy driving other reactions example?

Query2

Any fundamental reaction, which is any reaction where one or more reactants combine and turn into one or more products with no intermediate steps, will follow this equation. This is because the Arrhenius equation is identical to a Boltzmann distribution, which is derived from statistical mechanics. So the Arrhenius equation is not just a handy mathematical approximation. Physics dictates this must be true.

Is this really true? Doesn't the Boltzmann distribution just apply to gases? Olin

The Maxwell-Boltzmann distribution of molecular velocities is not restricted to gases - it applies to nearly all liquids (except at very low temperature) and, to a limited extent, even to solids. It is a rigorous result of classical statistical mechanics and, as such, applies whenever classical mechanics is valid. Nevertheless, you are right to be suspicious about the preceding poster's highly exaggerated claims about the Arrhenius equation. The Arrhenius equation is by no means "identical to a Boltzmann distribution". In order to get from the Boltzmann distribution to a formula for a chemical reaction rate, one must introduce a number of additional assumptions. These assumptions are reasonably well satified for a great many chemical reactions, so the Arrhenius equation is very useful. But it is nevertheless an approximation.--Rparson 02:30, 30 January 2006 (UTC)

The problem is when a reaction we're looking at is not fundamental. Then there are actually a bunch of sub-reactions going on. There might be steps that are not reactions at all - for example, if it's a catalytic reaction, some chemical needs to stick to a solid surface before it reacts. Adsorption (the chemical sticking of a molecule to a solid) goes *down* with temperature. The rate constant of the fundamental reaction steps increase with temperature, but at different, er, rates. There could also be reverse reactions, and if they increase with temperature, the overall rate for a reaction (forward minus backward) will go down.

The rate for a complicated reaction will (hopefully) consist of an algebraic expression of the reactants and the rate constants for many of the fundamental reactions that comprise the overall reaction. Each one of these fundamental rate constants will follow an Arrhenius law, but the overall reaction cannot follow an Arrhenius model because there is more than one rate constant. ---Alchemy3083 26 April 2005

Eyring equation

Really, there should be a different article for the Eyring equation than the Arrhenius equation, as they are derived from different theories.

See Eyring equation V8rik 19:33, 20 January 2006 (UTC)

The Arrhenius equation is not really "derived" at all - as far as I know there is no microscopic model that leads exactly to the Arrhenius equation. It's a quasi-empirical result. I think it is reasonable to keep some discussion of how microscopic models such as "collision theory" and transition state theory lead to Arrhenius-like expressions, but perhaps the present discussion is too long. What we really need is a much better, much more thorough article on transition state theory. --Rparson 02:30, 30 January 2006 (UTC)

Query3

How is the Arrhenius equation linked to what is called Kramers or Kramers-Bell theory for a two-state system (e.g. protein folding/unfolding). The equations look pretty much the same. —Preceding unsigned comment added by 128.214.3.229 (talk) 15:15, 2 October 2007 (UTC)

how to get the exponential factor

In all articles related to the arrhenius equation it is stated that the exponential factor stems from the Maxwell-Boltzmann distribution. Can anyone show how this is actually derived? —Preceding unsigned comment added by 84.191.237.160 (talk) 18:52, 16 December 2007 (UTC)

Doubling with 10 deg C increase

The tendency of chemical reaction rates to double with each 10 deg C increase in temperature is a fairly important characteristic of the natural world. It informs much of what we understand about pedogenesis, chemical weathering, and biogeochemical cycles. This article has been the only place in Wikipedia that frames this important concept in a way that is so easily and immediately recognizable, and I urge the watchers of this article to retain this popular treatment. I have restored and referenced a single sentence to this effect. I support expanding it. --Paleorthid (talk) 00:30, 8 August 2008 (UTC)

"Fudge factor"

Really? God Emperor^Talk 20:05, 12 June 2011 (UTC)

Context for quote re experiment & pre-exponential?

The article has a quote "it is not feasible to establish ... whether the predicted ... dependence ... is observed experimentally". I don't have access to the original source, but it appears to be so out of context as to be meaningless. The precision with which rates of different reactions can be measured, varies by many orders of magnitudes. What sort of system was the original talking about? — Preceding unsigned comment added by 131.203.251.94 (talk) 19:26, 14 October 2013 (UTC)

Prefactor

I am new to this page and to Wikipedia editing in general. Perhaps someone has already touched on this point somewhat in earlier discussion. If so, I apologize.

I attempted to work out the Arrhenius equation given in this article from the viewpoint of statistical mechanics and the canonical partition function. First, view each interaction (reacting or non-reacting) as individual subsystems of the system of all the collisions that occur in a given time interval, assumes to obey the assumptions of the canonical ensemble. Thus, finding the partition function of a system undergoing a reaction is simply the partition function of each pair, multiplied together (assuming no reactions occur with more than 2 reactants). The partition function

$Z=\sum _{s}\mathrm {e} ^{-\beta E_{s}}$

and its associated ensemble yields the probability of a certain collision resulting in a reaction

$P_{s}={\frac {1}{Z}}\mathrm {e} ^{-\beta E_{r}},$ where $E_{r}$ is the energy state of the products.

Now, in order to be completely general, we cannot pretend to know what the partition function is, as every interaction can have multiple ways for the reactants to interact, leading to multiple energies to sum over, with perhaps degenerate states. However, if we assume that the temperature remains approximately constant (and that the energies of the reaction and their corresponding degenerates do not depend on the extent of the reaction), we can say that ${\frac {1}{Z}}$ is approximately constant. Thus, when we multiply $P$ by the number of collisions per second, we have

$A{\frac {1}{Z}}\mathrm {e} ^{-\beta E_{s}},$

which is the average number of reacting collisions per second. However, this would yield

$k=A{\frac {1}{Z}}e^{-E_{a}/(k_{B}T)}=Ce^{-E_{a}/(k_{B}T)}$ as the Arrhenius equation. This of course would not affect measurements, but it would affect how we interpret the prefactor. Instead of interpreting $A$ as the total number of collisions per second, it would be interpreted as the number of collisions per second divided by the partition function of the reaction in question.

Essentially, my argument is that the probability of a collision resulting in a reaction needs to be normalized before it can be interpreted as a probability.

Physicsman19 (talk) 14:16, 17 July 2014 (UTC)physicsman19 7/17/2014

This is not a simple question as I'm sure you realize. For one thing, in chemical kinetics the partition function is part of the transition state theory, whereas the number of collisions is part of the collision theory. So you seem to be mixing two separate theories of the rate constant, each of which has its own article. This leads me to believe that your argument is what Wikipedia calls original research (see WP:OR), that is working things out your own way rather than using reliable sources such as textbooks or review articles. I would suggest you try to find one or more sources which address your points, and then try to summarize their arguments in the appropriate article. This article is really for the empirical Arrhenius law, so a better place would be either Transition state theory or Collision theory. Dirac66 (talk) 19:24, 17 July 2014 (UTC)