Talk:Bond-dissociation energy

Latest comment: 5 years ago by 192.91.171.36 in topic Strongest bond in non-neutral molecule?

Bond dissociation energies not found in source edit

I pulled a copy of the paper in reference two and found that it does not include homonuclear diatomics, so cannot be the source for the first table. Would anyone care to post a better reference? 173.17.185.1 (talk) 01:59, 8 December 2011 (UTC)WillReply

Possible Reference: http://www.nist.gov/data/nsrds/NSRDS-NBS31.pdf Elleacampbell (talk) 20:49, 27 November 2013 (UTC)Reply

Bond dissociation energy vs. bond energy edit

Science world says this and bond energy are the same. We say different. Which is correct? - Taxman Talk 17:37, August 1, 2005 (UTC)

  • in many case the values will be similar but in other there are differences, see the water example , to state that they are the same would be an oversimplification V8rik 15:22, 2 August 2005 (UTC)Reply
This article currently states that bond energy is the sum of all the bond dissociation energies in a molecule, but the Bond energy article does not seem to go with this definition. How can this be clarified?--GregRM 13:58, 8 July 2006 (UTC)Reply
I think I understand it somewhat now...the bond energy is not the sum itself, but is calculated from the sum.--GregRM 14:02, 8 July 2006 (UTC)Reply
I agree with the OP - Wikipedia articles must come from prior published material, and every mention I can find of bond dissociation energy says it is equal to the bond energy. My main source is the Silberberg Chemistry 4th edition, but the OP mentions a Science World article as well. --169.233.62.35 19:00, 14 July 2006 (UTC)Reply
  • Fair enough, I provided a reference and another example explaining the difference from the same reference. I quess that in other publications the two are the same for practical purposes. and who is OP? V8rik 20:14, 14 July 2006 (UTC)Reply
    • I am guessing OP stands for "original poster" or something similar.--GregRM 00:21, 24 July 2006 (UTC)Reply

Clarification of bond energy method edit

The sentence mentioning that the bond energy is calculated from the sum seems potentially misleading...it says that the bond energy is calculated based on the sum of the bond dissociation energies of all bonds within the molecule. However, one might say for water, for example, that since the two OH bonds in the water molecule are equivalent, the average based on the sum of the two is the same as the first OH dissociation. Perhaps it could be worded more precisely to emphasize the distinction illustrated in the example; perhaps something along the lines of: "...bond energy, which is calculated based on the average of the bond dissociation energies associated with sequentially breaking all X-Y bonds in a molecule (where X-Y is the bond of interest)."; or perhaps "...bond energy, which is calculated based on the sum of the bond dissociation energies associated with sequentially breaking up a molecule into its constituent atoms." Note that I'm not sure which (if any) of these statements is correct and still holds for more complicated molecules...I would have to look into this some more. Does either of these agree with the general method discussed by the textbook reference?--GregRM 00:21, 24 July 2006 (UTC)Reply

Units edit

The units listed in the article are kcal/mol, should the units be SI (J/mol) as that is the proper scientific convention? later units in the article are also in J/mol

The Wikipedia:Manual of Style (dates and numbers) suggests using SI for all articles, where there are not compelling reasons to do otherwise. Based on this argument, I would use kJ/mol. However, the scientific literature (J. of Chemical Physics, J. of Physical Chemistry) are rife with both units. Therefore, I suggest sticking to the units of the source, with converted units in parenthesis, as suggested in the style manual, under the units of measurement heading. Az7997 18:28, 17 November 2006 (UTC)Reply

Use of Symbols edit

In the introduction, BDE is defined as abstandard enthalpy change. However, in the section Homolytic vs heterolytic dissociation, the following appears: H2 → 2 H· ΔG = 102 kcal/mol (see table below) H2 → H+ + H- ΔG = 66 kcal/mol (in water) Should not the symbols be ΔH instead of ΔG? --BingoChem (talk) 19:00, 26 July 2010 (UTC)Reply

In any case, ΔG is a symbol used without definition in this article. That should not be allowed. Please define all symbols where or before they are used. The Wikipedia article on Gibbs free energy explains that ΔG = ΔH - TΔS internal. A reaction that proceeds spontaneously has a negative ΔG. T is absolute temperature. So at absolute zero, or if ΔS is negligible, ΔG = ΔH. At the level of the present article, please consider mentioning the but minimizing this distinction, since the purpose of the article seems to be to enable the reader to compare different kinds of bonds without accounting for change of temperature, pressure, volume or agitation.Nhy67ygv (talk) 02:30, 11 March 2012 (UTC)Reply

Alkane bonds ".. readily broken by catalytic cracking" edit

Is this comment in the table really accurate? In the catalytic process, the bond length and energy is altered when a molecule bounds to the surface of catalyst. So this seems to be the case of considerably lower energies. Maybe it would be more correct to mention "broken during industrially used thermal cracking processes" ? --Esmu Igors (talk) 17:56, 13 December 2011 (UTC)Reply

I dared alter this point by myself. Please sorry if I am not right about this. --Esmu Igors (talk) 19:54, 15 December 2011 (UTC)Reply

Quantum Mechanical treatment of Bond dissociation energies edit

I think that this article is missing this section, and I'm not educated enough to add it. I actually came to wikipedia looking for the QM subject on dissociation. I hope that someone will work on this. Elleacampbell (talk) 20:44, 27 November 2013 (UTC)Reply

The simplest possible case is two protons linked by one electron, otherwise known as the dihydrogen cation, q.v. for the relevant Schroedinger equation which was first applied to this cation within a year of Schroedinger's 1926 paper. Diatomic molecules are relatively tractable, triatomic and up are much hairier. Vaughan Pratt (talk) 05:03, 2 September 2014 (UTC)Reply

How to compute the Dissociation Energy per bond? edit


I was intrigued by the way the Energy per bond computations were carried out.
For example,

CH3CH2–H → CH3CH2 + H
D0 = ΔH = 101.1 kcal/mol = 423.0 kJ/mol = 4.40 eV (per bond)

Could a practical by-hand computation be shown please?

Next, for example, from the table,
O–H Hydroxyl 110kcal/mol 460kJ/mol 4.77eV
The computation from kcal to kJ is fairly simple. The number 4.18 is used. But what about a single bond?
In the second table, the per bond computations are not given.
The file, http://www.nist.gov/data/nsrds/NSRDS-NBS31.pdf , courtesy Elleacampbell, have been perused. It doesn't have the per-bond dissociation energy, but an exhaustive list. A great reference! Thank you for the link!
The guidance provided here could be used to compute any given data in the said file.
Thanking you in anticipation.
Regards
Bkpsusmitaa (talk) 01:46, 24 March 2017 (UTC)Reply

What you need here is the energy e of an electron volt, namely e = 0.160218 attojoules, and the fact that a mole of anything (molecules, badgers, whatever) is Avogadro's number N_A of them. Simply multiply those two numbers together and you obtain 96.485 kJ/mole as the energy density corresponding to 1 ev per bond in the case of one broken bond per molecule.
The energy needed to convert a mole of ethane to a mole of ethyl radicals, 423.0 kJ/mol, is, per bond in that mole, 423.0/96.485 = 4.38 ev/bond. (I have no idea why they give 4.40 there.) Likewise 460 kJ/mol becomes 460/96.485 = 4.77 ev/bond.
Incidentally one way to think about electron volts is that a visible photon has an energy of around 2 or 3 electron volts, and that this is about the voltage needed to move an electron from where it's stuck in the valence band of a photovoltaic cell to the more freely moving conductance band. This is why photovoltaic cells work best in visible light, as opposed to ultraviolet or infrared light on either side. UV leaves a surplus of energy that heats the cell counterproductively while IR isn't quite enough to lift the electron out of the valence dungeon.
Oh, and also a plug for defining Avogrado's number as an integer, namely n quadrillion where n is roughly 602 million and more precisely 602214086. To memorize the latter write it as
602
214
086
The powers 1, 2, 4, 8 of 2 form the letter T below 602 in the most logical order, leaving two holes that you fill clockwise by repeating the top row, namely 60 (and then you bump into 2). Vaughan Pratt (talk) 06:26, 25 March 2017 (UTC)Reply
Thank you for your extended reply! In our programs we used 6.023 X 10^23 units. Personally, I am indebted.
I would rather have wanted to have a column added and then the per unit bond dissociation energy, inserted. But I don't have the expertise to add a column and then apply the formula in each of those columns. I will learn, if it is not done within the period.
I once again thank you for your prompt and kind reply.
Bkpsusmitaa (talk) 08:50, 25 March 2017 (UTC)Reply
Yes, converted and added a column. Bkpsusmitaa (talk) 10:42, 25 March 2017 (UTC)Reply

Visually Orphaned Ref. #6 edit

@GregRM: @V8rik: @Esmu Igors: @Vaughan Pratt:

In the article there are the editing marks "<  !- -" and "- ->" respectively before and after the statement below:

Following dissociation, if new bonds of greater bond-dissociation energy are formed, these products are at lower enthalpy, there is a net loss of energy, and thus the process overall is exothermic. In particular, the conversion of the weak double bonds in O2 to the stronger bonds in CO2 and H2O makes combustion exothermicwell that depends on the C and H sources!

Because of the editing mark the Ref.#6, "Schmidt-Rohr, K. (2015). "Why Combustions Are Always Exothermic, Yielding About 418 kJ per Mole of O2". J. Chem. Educ. 92: 2094–2099. doi:10.1021/acs.jchemed.5b00333." appears as an orphan in the article.
Could anything be done about this please? The author and the editor may like to look into this!
Bkpsusmitaa (talk) 04:12, 25 March 2017 (UTC)Reply

Solution to orphaned ref. No.6 edit

@Vaughan Pratt: Could you please check and suggest?
@Primefac: 'Vaughan_Pratt' is just one person who helped on an earlier query. Why did sir use 'they'?
@Chrissymad: Is sir a paid editor and suffering from a chronic shortage of time? It appears that sir has not even noticed that this is not a submission for a new article, but an addition for an orphaned reference in the article Bond-dissociation energy that was commented out way back!
@Timothyjosephwood: okay! BEBOLD followed. Thanks.

Explanatory portion added to illustrate the objection raised, commenting out the said portion.
Bkpsusmitaa (talk) 01:06, 28 March 2017 (UTC)Reply

@GregRM: @V8rik: @Esmu Igors: @Vaughan Pratt:

To be added:

As illustration, first, normal combustion in open air be considered. Such reactions usually take place under standard atmospheric pressure, without constraints in the volumes of the reactants and products.
The case of the combustion of butane in open air is one such case employed for efficient cooking using LPG. A combination of n-butane and isopropane (2, methyl propane), its melting-point varies upto 6 units and boiling-point varies upto two units. There are four carbon-carbon bonds and 10 carbon-hydrogen bonds. At a very crude level, looking at the table of bond-dissociation energy, the four C-C bonds have a total ~ 4 X 3.65 (average) eV/Bond ≈ 14.6 eV for the four bonds and the ten C-H bonds have a total ~ 10 X 4.37 (an average of methyl-, ethyl- and tertiary-bonds) ≈ 43.7 eV, a total of 58.3 eV for one molecule of n-Butane.
BDE of O-O bond is 5.15 eV, whereas for the products, CO2 and H2O, the BDEs are [11.16(C-O bond) + 5.51(CO-O bond)] 16.17 eV and [4.77 eV (O-H Bond) + 2.78 eV (OH-H bond)] 7.55 eV, respectively.
A balanced chemical equation is as follows:
2C4H10 + 13O2 + Initial Energy = 8CO2 + 10H2O
⇒ (2 X 58.3 eV) + (13 X 5.15 eV) + Initial Energy yields (8 X 16.17 eV) + (10 X 7.55 eV)
⇒ 116.6 eV + 66.95 eV + Initial Energy → 129.36 eV + 75.5 eV
⇒ 183.55 + Initial Energy → 204.86 eV
It is clearly apparent that the final BDE is higher than those of the reactants, and this energy is released when the products are formed. Since for all the reactants the BDE is positive, energy is to be provided externally to break up the bonds between C-C, C-H and O-O bonds to initiate the reaction. However, since the BDEs of the products are much higher than those of the reactants, the energy released during the initiation is utilised to break up the bonds of next set of reactants. But how?

Though the reaction takes place in open air, the burner hob, the cooking utensil surface exposed to the heat and the surrounding air keeps the local energy just about adequate to keep the bonds of the continually-added reactants breaking in a cascade and sustain the exothermic reaction.

Hence, the final chemical equation could be re-written as:
2C4H10 + 13O2 + Initial Energy = 8CO2 + 10H2O + Energy Released
This is a reasonably logical analysis based on the axiomatic fact that an initial match-stick or a spark-lighter is enough to keep the burner running as long as required.

Next, for an Internal Combustion Engine using methane and air mixture as reactants and running adiabatically, a similar balanced equation and logical analysis could be attempted, but in this case the enthalpy of the system would be conserved, slightly modifying the operational set-up to account for heat dissipation and cooling of the engine. However, such methodical set-ups are routine in the laboratories of engine manufacturers and research institutes to increase the efficiency of the engine, which is essentially and perpetually a work in progress.

The Applications Section edit

Can someone please explain why this long rambling section is currently relevant, especially with the strange discussion of "initial input" energy?

I moved this text there because I thought there might be something to salvage, but I just ended up getting frustrated by the editor's writing style and failure to follow the general style and tone of an encyclopedia article, as well as questionable scientific correctness.

I move to delete this section, unless someone understood the point and wants to do a massive rewrite of it. Alsosaid1987 (talk) 05:53, 21 September 2018 (UTC)Reply

Wikipedia is an evolving project and at some early phases in article development, editors go into teaching/preaching mode. I guess that stuff is better than a blank page, but eventually it should be wiped, IMHO. So that is what I did.
BTW, what about BDFE's? That seems to be the direction inorganic/organommetallics has shifted. DOI 10.1021/cr100085k.--Smokefoot (talk) 11:35, 21 September 2018 (UTC)Reply
Thanks -- I agree with the suggestion. These days, BDFE's are quoted more and more - in organic chemistry as well, esp. among workers in radical chemistry. It should at least be defined in the article. I don't know much about how it's actually determined though -- it seems like people have always dealt with Delta H nought of formation because that's what you measure in a calorimeter -- I don't know whether it is any harder to determine everything in terms of Delta Gs these days with modern spectrometric methods or photoacoustic calorimetry, for instance. Most of the numbers I've seen are computed ones from DFT. It's not clear to me whether that's because they are hard to determine experimentally, or the "infrastructure" hasn't been built up yet to have tables of values to be available. Alsosaid1987 (talk) 22:43, 21 September 2018 (UTC)Reply

Color of the halogens and BDE edit

The color of the halogens, as far as I recall, is due to pi* to sigma* excitation. Is that directly correlated with the BDE? If so, could someone provide a reference? In fact, the BDE of I2 and F2 are quite similar (both around 35 kcal/mol), but their colors are not. Alsosaid1987 (talk) 04:26, 23 October 2018 (UTC)Reply

Those comments were probably mine. I removed them today. --Smokefoot (talk) 08:27, 23 October 2018 (UTC)Reply
Think we've all made edits that seem less apt on second consideration. Thanks for the changes! Alsosaid1987 (talk) 12:23, 23 October 2018 (UTC)Reply

Strongest bond in non-neutral molecule? edit

The first table of bond energies says carbon monoxide has the "strongest bond in neutral molecule," then just leaves the obvious question hanging. The page on Chemical bond has a narrower table that also mentions the energy of the CO bond (with a slightly different value) and shows nothing stronger. So what is the actual strongest bond between atoms? 192.91.171.36 (talk) 19:18, 5 April 2019 (UTC)Reply