Talk:Activation energy
![]() | This ![]() It is of interest to the following WikiProjects: | |||||||||||||||||||||||
|
Untitled
editthe activation energy of the catalyst is not equivalent with the heat, in the process that potassium chlorate produces oxygen, can we increase the temperature insteady of the catalyst -- manganese dioxide in the experiment? no!thus there is a difference between the catalyst and the energy (for example heat). though the catalyst can decrease the activation energy.
Bond information needed?
editThe article does not supply enough information about how the activation energy breaks the first bonds in a reactant (or similar, i forget the correct term) and how this then continues the reaction. Instead the article talks more theoretically how the Activation Energy is a point where the needed energy is supplied and the reaction continues.
Although I'm just a student learning about this, I beleive that more information about chemical bonds etc would be good. If I'm wrong or confusing all this with something else then forget it. Thanks!
84.92.204.165 16:33, 14 January 2007 (UTC)Spike
"Etc" needed?
editIn the first paragraph, this sentence:
- This requires energy - activation energy - and comes from the heat of the system i.e., the translational, vibrational etc... energy of each molecule.
contains an "etc..." that I don't think is needed. First of all, "etc." and "..." is redundant (both mean "and so on"). Secondly, the heat is contained in the translations, vibrations, and rotations of the molecules. Couldn't the "etc." be replaced by "and rotational" without any loss of explanatory power. At room temperature, other degrees of freedom (such as electronic and nuclear transitions) make a negligible contribution to the heat capacity. So unless anyone objects, could the sentence instead read:
- This requires energy - activation energy - and comes from the heat of the system, i.e. the translational, vibrational, and rotational energy of each molecule.
- Hmm. Actually, that's wrong. There are a number of reactions where the activation energy is not supplied by heat. Specifically, there are electrochemcial reactions and photochemical reactions where the activation energy comes from electric fields or incident photons. Rare, sure, but certinally important. I'll edit the article to point out these cases. Syntax 15:03, 9 December 2005 (UTC)
Figure 1
editThe article refers to Figure 1 but there doesn't appear to be a figure 1. Was this text copied from somewhere? Perhaps it needs rewriting to remove the reference, but better if a figure 1 could be sourced and supplied showing a typical transition state energy plot? ++Lar: t/c 16:55, 1 January 2006 (UTC)
Removed a lot of material
editI've taken a lot of material out of this page. It wasn't vandalism (despite lots of people trying to revert me!) The info was simply wrong, or, at best, of limited relevance or misleading.
And I've taken the technical flag off.
129.16.97.227 (talk) 15:53, 11 December 2007 (UTC)
Aktivierungsenergie: Die Chemische Darstellung von Kleinsten Millimeter Planetenbodens
edit- "Die Gesteinsproben sollen unter anderem auf rund 1000 Grad Celsius erhitzt werden, um Aufschluss über den Wassergehalt und die in der Probe enthaltenen Minerale zu erhalten.
Die Forscher suchen insbesondere nach Spuren von Kohlenstoff, Wasserstoff und Stickstoff."
- http://www.sueddeutsche.de/wissen/artikel/410/179858/
- http://de.wiki.x.io/wiki/Thermogravimetrie
- http://de.wiki.x.io/wiki/Mikrowaage
- http://de.wiki.x.io/wiki/Aktivierungsenergie
- Der Mars besteht zu 95,3 % aus CO2
"Bond breaking is equivalent to bond making"
editThis article states that the transition state is equivalent to the point on the reaction coordinate at which bond breaking and bond making are balanced. Bond making by definition would reduce the total free energy of the system, and hence this statement is patently wrong. I think the key to the problem is the word "balanced"; there is no balance, and there isn't an intermediate of "partial bonds" or any of those myths. If my understanding is correct, the transition state, if we want to speak loosely in terms of bonds, is the point at which all the bonds are broken [input of energy], but no bonds are being made —Preceding unsigned comment added by N1r4v (talk • contribs) 19:46, 6 May 2009 (UTC)
- That's not quite true either, though, is it? Consider the SN2 transition state. There is some bonding interaction between the carbon undergoing inversion and both the nucleophile and leaving group. The system is not the same as three isolated species - for example, where does the negative charge reside - on the nucleophile or the leaving group? Answer: on both.
activation energy unit
editMy kinetics teacher (though not being a reliable source) stated that the unit for activation energy is simply kJ (or J, whichever order of magnitude you'd like to express it in), instead of kJ/mol. I'm starting to doubt what has been said. It's true each molecule needs to be brought to the transition state in order to react (hence why it'd be energy per unit of amount), but a closed system with a known amount of mass could be seen as having a constant amount of molecules.
According to the Arrhenius_equation , the activation energy has to be either kJ/mol or J/mol because of the gas constant... Redtails (talk) 12:48, 1 July 2009 (UTC)
- Your kinetics teacher is wrong. When we measure activation energy using the Arrhenius equation, we are measuring it on a bulk system where the differences in energy of individual molecules can be averaged out (or, more accurately, represented using the Boltzmann distribution). However energy is an extensive property so, to compare values between different systems, you have to divide by the amount of substance present. This gives units of kJ/mol (in real-life cases).
- You can use computational chemistry to calculate an "activation energy" for an individual molecule (or pair of molecules). We usually refer to this as the "energy barrier", as it is not exactly the same as the activation energy you measure on the laboratory scale. It is quoted in atomic scale energy units such as hartrees or electronvolts: you could quote it in attojoules, but nobody does because there would be an additional uncertainty in the conversion factor. Physchim62 (talk) 14:42, 1 July 2009 (UTC)
- "Amount of substance" is a dimensionless quantity, so "kJ/mol" are dimensionally identical to "kJ" - the dimensions are in both cases those of energy, which are Mass(Length^2)(Time^-2). Perhaps this is what your teacher was trying to express? 129.67.84.166 (talk) 11:44, 13 November 2009 (UTC)
- Your kinetics teacher was not wrong! just mislead you. Activation energy is energy/amount of substance but it does not have to be a mol. When they said it is expressed in kJ they meant kJ/atom or molecule. Then the Arrhenius equation is where k_B is the Boltzmann constant. Recall that the gas constant is just the Boltzmann constant multiplied by Avogadro's number. Thus, if you got your answer as 40 J for the activation energy, this implicitly means 40 J per atom or molecules or particle. multiplying in by Avogadros number will convert it to J/mol. Just as dividing the activation energy in J/mol by Avogadro's number will convert the units to J or rather Joules per atom.The Lamb of God (talk) 03:47, 24 November 2009 (UTC)
- For instance if you look at the wikipedia page for Avogadro's number it will tell you that it is in units of mol^-1. It is however implicitly molecules per mol. It is simply convention not to right it out formally for whatever reason. So, your teacher in not trying to confuse you by explaining this, just ended up confusing you anyway.The Lamb of God (talk) 03:51, 24 November 2009 (UTC)
catalysis
editI really feel that it would be more correct to say that the catalyst provides for an energetically more favorable pathway. Rather than just to say it lowers the activation energy. The activation energy is the difference between the energy at the initial state and the energy of the transition state. So, a catalyst would therefore induce, in a sense, an alternative transition state that is lower in energy than the original transition state. Consequently, it would provide for a more energetically favorable reaction path. Its presence does not lower the energy of the transition state, but rather it participates in the formation of an alternate transition state that has a lower energy. This is admittedly nit-picky however.The Lamb of God (talk) 02:52, 5 December 2009 (UTC)
I neither agree nor disagree with your proposed change in saying, "energetically more favorable", instead of the current, "lowers the activation energy". Both are correct in their own sense, perhaps instead offer one or the other as a secondary title. What I would propose adding to the catalysis section however is the topic of Binding Energy and how specific chemical interactions act to lower the activation energy. There is no mention of how the binding energy allows for lowering activation energy needed to achieve the transition state. Will be planning to make additions and edits regarding this topic in the near future. RockstheBlugold (talk) 10:41, 14 November 2019
I have concluded my editing of this section, mostly highlighting the specific reasons behind "why" catalysts reduce activation energy. I linked more of the confusing topics to other wiki-articles. I believe my explanation of binding energy in this section is straightforward yet in-depth enough for a common-person to grasp the topic's relation to catalytic activity.RockstheBlugold (talk) 22:34, 22 November 2019 (UTC)
Explanation under the first picture (with sparks and flame)
editQuote: "The blue flame will sustain itself after the sparks are extinguished because the continued combustion of the flame is now energetically favorable."
"Energetically favorable"? - Isn't it simply because the constant stream of incoming volatile gas runs into already burning gas...
- At the higher temperature it is favorable where at the lower temperature it is not. Favorable refers to a formula (a mathematical inequality). What you say is true, but omits the reference to a numerical test.
- Actually, I see that there is an ambiguity. "Energetically favorable" could refer either to thermodynamically favorable, or to the reaction rate (kinetics). Thermodynamically favorable means the sign of the Gibbs free energy difference of the two end states, which changes with temperature. The reaction rate depends on the supply of energetic particles, which increases at higher temperature, as well as the activation energy, which I'm reading this article to find out about. So both of these kinds of "favorable" can change with temperature. But I don't know which effect dominates in common situations.
Assessment comment
editThe comment(s) below were originally left at Talk:Activation energy/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
Rated "high" as high school/SAT biology content; important for enzyme catalysis. - tameeria 22:58, 18 February 2007 (UTC) |
Last edited at 22:58, 18 February 2007 (UTC). Substituted at 06:38, 29 April 2016 (UTC)
Generalization of Activation Energy
editIn physics[1], materials science,[2] and nuclear science[3] there are significant uses of the phrase "activation energy" each of which have a similar physical meaning as is used for the chemical reaction application. I believe that this page should be significantly modified to reflect these changes. — Preceding unsigned comment added by Antwan718 (talk • contribs) 01:21, 22 May 2018 (UTC)
The following text,
Activation energy may also be defined as the minimum energy required to start a chemical reaction or ignite a combustible vapor, gas or dust cloud.
is scientifically incorrect.
The use here would more adequately be expressed as the minimum ignition energy for a given mixture. [4] — Preceding unsigned comment added by Antwan718 (talk • contribs) 01:29, 22 May 2018 (UTC)
References
Activation energy is not Ignition energy
editThis entire statement is scientifically incorrect as ignition energy and activation energy describe two different quantities. [1].[2]
Explosives, hydrogen, unsaturated hydrocarbons and alkanes in oxygen have the lowest activation energy: in the range of 1 to 100 μJ. Alkanes in air, distillate fuels, hybrid mixtures and extremely sensitive dusts have a activation energyrange of 0.1 to 10 mJ. Combustible dusts have a activation energy range of 0.01 to 10 J.
For most materials the stoichiometric concentration can be calculated from the carbon and hydrogen content and is typically about twice the lower flammable limit. The lowest ignition energy value rarely occurs at the stoichiometric combustion concentration. For heavier gasses, the minimum usually occurs slightly above stoichiometric concentration and for lighter gasses such as Hydrogen, slightly below.[3]
References
- ^ https://www.sciencedirect.com/science/article/pii/S0010218001002486
- ^ An introduction to combustion, Turns
- ^ Magison, Ernest C. (1998). Electrical instruments in hazardous locations (4th ed.). Research Triangle Park, NC: ISA. ISBN 978-0979234316.
Section - "Relationship with Gibbs Energy of Activation"
editI have added a small concluded blurb at the bottom of this section to wrap things up. This blurb is meant to affirm activation's energy relation to overall reaction energy - them being independent of each other. Perhaps the wording can be altered slightly, though I think this piece wraps up the section nicely in understanding activation energy's role in a whole process.RockstheBlugold (talk) 22:37, 22 November 2019 (UTC)
Standard States
editThe entropy and enthalpy of activation are typically the standard enthalpy of activation and standard entropy of activation. The standard state symbol should also be there. This is not actually just a question of semantics, it is a common misconception that the tabulated values are free from standard states. Usually they are not free from standard states. http://goldbook.iupac.org/terms/view/E02142 — Preceding unsigned comment added by Ashi Starshade (talk • contribs) 21:52, 3 May 2020 (UTC)
Thanks. I have now included this point in the section Relationship with Gibbs energy of activation. Dirac66 (talk) 03:05, 11 May 2020 (UTC)
Activation energy in a 2D potential energy surface
editSee commits: Special:Diff/1000520950, Special:Diff/1182119303
I'm removing this section. Potential energy surfaces enable the estimation of activation enthalpy/entropy in transition state theory, the successor to the Arrhenius theory of reaction rates. The transition state theory article already includes a short discussion of the connection (PES saddle point energy determines enthalpy); it's less relevant to this article.
Moreover, this section appears to be an edited copy of the Fig. 3 images and text in Kristinsdóttir, Lilja; Skúlason, Egill (2012-09-01). "A systematic DFT study of hydrogen diffusion on transition metal surfaces". Surface Science. 606 (17): 1400–1404. doi:10.1016/j.susc.2012.04.028. ISSN 0039-6028.
It includes several errors which suggest the wiki author may not be one of the paper authors, contrary to the copyright statement in File:Comparative_between_1D_and_2D_PES_for_tungsten_and_hydrogen_gas_reaction.jpg:
- The 2D PES maps the energy of hydrogen adatoms (relative to gas-phase ) bound to a W(110) surface - it's not a PES for the surface binding/dissociation of .
- Likewise, the 1D reaction coordinate diagram is for the minimum-energy path of hydrogen adatoms between W(110) surface binding sites, not for "the reaction of hydrogen gas and tungsten".
- The non-word "depics" in the sentence "The depics correspond to the trajectories" appears to be a misphrasing of the paper's sentence beginning "Fig. 3 depicts that the trajectories corresponding to the one Cu has, have somewhat symmetrical shape".
So even if we conclude that a section on potential energy surfaces is appropriate in this article, I think we'll have to write one from scratch, rather than reworking this one. Preimage (talk) 07:16, 27 October 2023 (UTC)
- Preimage. I don't understand any of that, but I just want to say that I appreciate you using your subject matter expertise to improve the encyclopedia. I support your efforts and I hope you stick around! Happy editing :) –Novem Linguae (talk) 08:14, 27 October 2023 (UTC)