Sciencemadness Discussion Board - Energy Requirements for Breaking C=C bonds

Energy Requirements for Breaking C=C bonds

Hexavalent - 14-4-2012 at 11:55

Can anyone please tell me how to calculate the energy required (in kJ to break chemical bonds, specifically in this case C=C ones? Thanks in advance.

barley81 - 14-4-2012 at 14:20

Table of bond enthalpies: http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm

It takes 614kJ to break 1 mole of C=C bonds.

[Edited on 14-4-2012 by barley81]

Hexavalent - 14-4-2012 at 22:44

Thanks barley, but it's the actual calculation I need to know as opposed to just the answer

watson.fawkes - 15-4-2012 at 06:05

Quote: Originally posted by Hexavalent

Thanks barley, but it's the actual calculation I need to know as opposed to just the answer

The calculations are all done with software. There's no closed-form version of these calculations. The field is called computational quantum chemistry. If you want to understand how to do the calculations, learn the software. If you want to understand why the calculations work, learn quantum mechanics.

DJF90 - 15-4-2012 at 06:07

Quote: Originally posted by watson.fawkes

Quote: Originally posted by Hexavalent

Thanks barley, but it's the actual calculation I need to know as opposed to just the answer

Both are a little out of the range of a 13-year old, dont you think Watson?

watson.fawkes - 15-4-2012 at 06:16

Quote: Originally posted by DJF90

Both are a little out of the range of a 13-year old, dont you think Watson?

No one is 13 forever.

Nicodem - 15-4-2012 at 06:24

Actually, I believe the field of computational chemistry is very much within the grasp of teenagers. It does require learning the use of the pertaining software and at least some understanding of the theory, but it does not require years of experience or any lab facilities or chemicals. Unfortunately, there is little interest in the amateur community for this type of science even though it is the most suitable resource-wise. A lot of the required software is pirated and can easily be obtained. I would expect today's teenagers are more literate in computer usage than I was at their age. I'm sure many of them are familiar with the Linux environments needed to run the calculations, though the lack of a visual interface in most common software packages and the command line usage probably puts them off. Still, there is nothing that a 13 year old could not easily learn.

DJF90 - 15-4-2012 at 07:39

Sorry Nicodem, despite some pertinent points I have to disagree. I dont know how easy it is to interact with a quantum chemistry program, but learning the quantum mechanics behind it will be far from his grasp. Before he can even start on that he'll need extensive mathematical tuition, as I doubt he's capable of the calculus that is extensively involved. Thats not to say he's never going to be able to do it, just not right now (and at least not for a while).

After doing some breif reading, it appears as if software packages such as Spartan may be fairly intuitive to use. However I'm far from an expert in the field of QM or Computational chemistry, so I guess I'll keep my mouth shut.

Hexavalent - 15-4-2012 at 07:41

OK, I recently did a mock exam the other day as practice (I do chemistry with people older than myself) and this was a question;

8a. Ethene, C2H4, burns in air to give carbon dioxide and water. The following equation shows the chemical changes that occur as ethene burns.

[Here was a line diagram showing ethene and oxygen burning completely and stoichiometrically to give CO2 and water]

The table below gives the relative amounts of energy needed to break the bonds shown in the equation.

C--H, 413kJ
C=C, ?
O=O, 496kJ
O--H, 464kJ
C=O, 743

Note - the amount of energy released in making a bond is equal and opposite to that needed to break the bond.

The overall relative energy change during the reaction is -1076kJ, showing that the reaction is exothermic.

Calculate the energy needed to break the C=C bond.

[ANSWER SPACE FOR CALCULATIONS]

How would one answer this question?

watson.fawkes - 15-4-2012 at 09:33

Quote: Originally posted by Hexavalent

The overall relative energy change during the reaction is -1076kJ, showing that the reaction is exothermic.

First things first. This is one of those exam questions that has false assumptions behind it, namely, that every bond of the same kind requires the same energy to break. This is true only approximately. Bonds vary in energy depending on what molecule they are in. Do not infer from the way the question is worded that there's some universal and exact truth behind these numbers. I find such questions pedagogically damaging, frankly, though that is veering away from the issue.

To solve the question, treat it as a stoichiometric problem, with individual bonds as the molecular species. It's a bookkeeping problem at heart. Write down the combustion reaction, decompose into individual bonds, and tally.

arsphenamine - 15-4-2012 at 09:52

Hexavalent,

This is more precisely called thermochemistry, which studies heats and energies of reactions.

Classically, you burn something and measure how much heat it produces.
Today, you can do ab initio computations, but at the end of it,
you use thermochemical ideas to make your results useful.

I will walk you through it.

From Hess's Law, we know that:
 the sum of reactant energies equals the sum of product energies.

Your balanced reaction is:
H2C=CH2 + 3O2 = 2CO2 + 2H2O

Count the number of different bond types as you have listed them (CH, C=C, etc.)
For example, H2C=CH2 also has 4 CH bonds and H2O has 2 OH bonds.

The CH bond energy, we can symbolize as E[CH] until we need to use 413 kJ.

Then your reactant energies are 4*E[CH] + E[C=C] + 3*E[O=O]
and your product energies are 2*2*E[C=O] + 2*2*E[OH]

Product energies = reactant energies.
E[C=C] is your unknown, so rearrange the equation and plug in the numbers.

[Edited on 15-4-2012 by arsphenamine]

arsphenamine - 15-4-2012 at 10:18

As I know exactly how easy it is to interact with quantum chemical programs,
I mention that absent any guidance, it is a difficult and months long learning curve.

With some practical guidance, you can do a few instructive calculations in a normal lecture period.
Accordingly, the North Carolina High School system has their own <a href="http://chemistry.ncssm.edu/">online computational chemistry server</a>

There are many free (for the asking if not the downloading) ab initio software packages.
Check the chemistry menus on Open Science Project at http://openscience.org/links/ for goto's.

Personally, I have settled on Avogadro for editing molecules, GAMESS-US and
MoPac2009 for computations, MolDen for visualizing the results. Your needs will vary.

The student version of Spartan is an utter piece of shit and any $50 you might pay for it
would be better spent on getting drunk and puking on a dumpster, but perhaps that puts
too fine a point on it.

Hexavalent - 15-4-2012 at 10:20

Thankyou very much arsphenamine, I think now understand the concept better . . .is this correct;

Reactants = e(C--H)*4 = 413*4 = 1652kJ
= e(O=O)*3 = 496*3 = 1488kJ

SUM= 1652+1488=3140kJ

Products = e(C=O)*4 = 743*4 = 2972kJ
=e(O--H)*4=464*4=1856kJ

SUM= 2972+ 1856=4828

e(C=C)=4828-3140=1688kJ

Hmmm . . .not quite right, the last time I checked 614, the literature value, is different from what I got . . .what have I done wrong?

[Edited on 15-4-2012 by Hexavalent]

arsphenamine - 15-4-2012 at 10:27

Quote: Originally posted by Hexavalent

Thankyou very much arsphenamine, I now understand the concept better and can do the calculations with ease

You are very welcome.

Since you are in the UK, I recommend your native PC-GAMESS computational chemistry package if you every get interested in that.

arsphenamine - 15-4-2012 at 11:16

NIST cites ethylene heat of combustion at 1411 kJ/mol.
http://webbook.nist.gov/cgi/cbook.cgi?ID=C74851&Mask=1#T...

DJF90 - 15-4-2012 at 12:19

Quote: Originally posted by arsphenamine

Hexavalent,

This is more precisely called thermochemistry, which studies heats and energies of reactions.

Classically, you burn something and measure how much heat it produces.
Today, you can do ab initio computations, but at the end of it,
you use thermochemical ideas to make your results useful.

I will walk you through it.

From Hess's Law, we know that:
 the sum of reactant energies equals the sum of product energies.

Your balanced reaction is:
H2C=CH2 + 3O2 = 2CO2 + 2H2O

Count the number of different bond types as you have listed them (CH, C=C, etc.)
For example, H2C=CH2 also has 4 CH bonds and H2O has 2 OH bonds.

The CH bond energy, we can symbolize as E[CH] until we need to use 413 kJ.

Then your reactant energies are 4*E[CH] + E[C=C] + 3*E[O=O]
and your product energies are 2*2*E[C=O] + 2*2*E[OH]

Product energies = reactant energies.
E[C=C] is your unknown, so rearrange the equation and plug in the numbers.

[Edited on 15-4-2012 by arsphenamine]

Good advice on the whole, except for the whole "reactant energy = product energy" bullshit. If that were the case, then there'd be no enthalpic driving force for a reaction, and it wouldnt be exothermic or endothermic either. What you mean to say is that energy is conserved!

So here is a more appropriate walkthrough, which gives the a correct answer. The first thing (other than conservation of mass and energy) is that "Note - the amount of energy released in making a bond is equal and opposite to that needed to break the bond."

So, to MAKE the bonds in the reactants, we have:
E(reactants) = 4(-416)+3(-496)+ E(-C=C) = (-3152kJ - E(C=C))

And to MAKE the bonds in the products, we have:
E(products) = 4(-743)+4(-464) = -4828kJ

We know that the energy given out from combustion is -1076kJ. So...

(-3152 - E(C=C)) -1076 = -4828

Rearranging to get E(C=C) on its own:
E(C=C) = (-1076)+(-3152) +4828 = 600kJ

Which is at least in the right region, unlike arsphenamine's answer. I found that the answer never quite matched up with the data elsewhere, probably so the teacher could see if the student had actually worked it out or if they just copied it out of a databook.

[Edited on 15-4-2012 by DJF90]

Hexavalent - 15-4-2012 at 12:48

Thanks a lot, DJF! But why do you have the numbers as negatives when calculating the energies for the reactants and products?

DJF90 - 15-4-2012 at 18:34

Something along the lines of making a Hess Cycle, delHf and all that (being negative), if you assume you start with atomic elements (and these have a E = 0).