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Why does heat/spark start a combustion reaction?

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Moogie
#1
Aug5-10, 12:07 PM
P: 168
Hi

I guess the title says it all.

Why does heat start a combustion reaction?

Why does a spark start a combustion reaction?

thanks
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DaveC426913
#2
Aug5-10, 12:13 PM
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Quote Quote by Moogie View Post
Hi

I guess the title says it all.

Why does heat start a combustion reaction?

Why does a spark start a combustion reaction?

thanks
What are your thoughts on the matter?

Here's something to ponder: Chemically, what is the defining property of a fuel? And why don't fuels simply spontaneously burn?
Moogie
#3
Aug5-10, 12:28 PM
P: 168
I don't know :) I guess that's why I am asking. I'm used to the concept that you need a spark/heat to start a reaction and then this will release heat and there is a chain reaction.

Perhaps fuels don't spontaneously react with oxygen because they don't have sufficient energy to overcome activation energy for the reaction but the spark provides this. Not sure.

Moogie
#4
Aug5-10, 12:34 PM
P: 168
Why does heat/spark start a combustion reaction?

I don't know the defining property of a fuel either.

Why does a reaction with oxygen release so much energy? I'm from a biochemistry background so i am used to thinking that you release energy when combusting glucose because the bonds in co2 and h20 are stronger than the C-C and C-H bonds in glucose so the 'excess' energy is released

But why are bonds with oxygen so strong/stable?
DaveC426913
#5
Aug5-10, 01:04 PM
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OK, you have much more knowledge than your first post let on.

Your posts are all correct. Why do you doubt your instincts?

Oxygen is strategically placed on the periodic table, in both row and column. With the exception of Fluorine, it the most electro-negative of all the elements, meaning greatest affinity.

You don't need much of a kick in the form of heat to jostle the molecules and give oxygen access to the chemically active parts.
Borek
#6
Aug5-10, 01:29 PM
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Quote Quote by Moogie View Post
sufficient energy to overcome activation energy
Bingo!
Moogie
#7
Aug5-10, 05:05 PM
P: 168
Are you saying that the high electronegativity of oxygen means bonds to oxygen are particularly strong?


Why specifically does that make bonds stronger? Bonding electrons will experience higher Zeff as oxygen is on far right?

Why don't things 'burn' in fluorine or nitrogen?
DaveC426913
#8
Aug5-10, 06:49 PM
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Quote Quote by Moogie View Post
Why don't things 'burn' in fluorine or nitrogen?
They do! Fluorine is way more reactive than oxygen. It is so reactive that it will happily burn hydrocarbons at room temperature (i.e. spontaneously). Luckily it's not a common as free oxygen.


As for N, it really likes to bond with itself. And many explosives have N as a principle component because it releases a lot of energy when it converts back to N2.
Moogie
#9
Aug9-10, 04:23 PM
P: 168
Of course things burn in fluorine. Bad example. Sorry.

Why is fluorine so much more reactive than oxygen? I know it is the most electronegative element but i'm trying to understand in terms of energetics. Why does the fact fluorine has high electronegativity mean the activation energy of the reaction [e.g. with hydrocarbons] is so low it will occur at room temperature

thanks
Borek
#10
Aug9-10, 04:44 PM
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It badly wants to find a lower energy state. That's the reason behind most of the reactions (although not all, sometimes there is a strong entropic component).
Moogie
#11
Aug12-10, 05:40 PM
P: 168
I presume it all comes down to Marcus theory as pointed out in recent post by alxm
alxm
#12
Aug12-10, 05:52 PM
Sci Advisor
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Quote Quote by Moogie View Post
I presume it all comes down to Marcus theory as pointed out in recent post by alxm
Well, Marcus theory is just for the kinetics of electron-transfer reactions (biochemical example: moving an electron to/from the pheophytin intermediate in photosystem II). For 'ordinary' reactions such as combustion you use the Arrhenius and Eyring equations.

But that just describes the relation between kinetics, activation energy and temperature. Doesn't explain why oxides tend to have lower heats of formation. Which is actually a quite difficult question to answer. You could answer it in terms of Pauling electronegativities (and many textbooks do.)

You can't actually do that though, it's cheating - because the electronegativities are calculated from bond strengths, so you're explaining something in terms of itself!
Moogie
#13
Aug12-10, 07:00 PM
P: 168
sorry, i've got lost in the answer here now. I wanted to know why fluurine is more reactive than oxygen. I know its more electronegative and forms more stable compounds, but that's the 'end result' and not why the reaction occurs more quickly. I'm assuming here that being more reactive means faster reaction. So i thought for the reaction to occur faster it must be due to a lower activation energy. So then i thought F must be more reactive than O because the energetics of electron gain must have a lower AE for F than O
alxm
#14
Aug12-10, 07:51 PM
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Quote Quote by Moogie View Post
I'm assuming here that being more reactive means faster reaction. So i thought for the reaction to occur faster it must be due to a lower activation energy. So then i thought F must be more reactive than O because the energetics of electron gain must have a lower AE for F than O
Well, it's not as simple as an electron jumping from one to another. You typically have a concerted reaction where one bond is being formed while the other is broken. Electrons, being all quantum-mechanical, have no problems 'being in two places at once'. At the transition-state you often have, in effect, a quantum-mechanical superposition of reactant and product states. Which is not usually 50-50% either. The nuclei move so much slower than electrons that the rate is usually limited by the motion of the nuclei, to which the electrons adapt instantly. (This is why electron transfer reactions, by which we mean those where no atoms move and the electron just 'jumps', are distinct.)

Anyway, to give a simple direct answer, one important reason fluorine is more reactive than oxygen, is that the F-F bond is weaker than the O=O bond, so that half of the reaction occurs more easily.
Moogie
#15
Aug13-10, 06:23 AM
P: 168
thanks very much.

one question always raises another though :)

Why does the reactivity of the halogens decrease down the group? The X-X bond strength decreases down the group with increase in atomic radii so you would think reactivity would increase? I know the intermolecular forces increase down the group as atoms become more polarizable but i would have thought the X-X bond would be the main factor affecting reactivity
alxm
#16
Aug13-10, 10:54 AM
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Quote Quote by Moogie View Post
Why does the reactivity of the halogens decrease down the group? The X-X bond strength decreases down the group with increase in atomic radii so you would think reactivity would increase?
Other way around, actually. The bond strength is essentially the dissociation energy:
X* + X* <--> X2
I.e. the difference in energy between the bonded and free atoms. Now, the bond in itself does not change much as you go down the halogens, as it's still essentially the same bond with more-or-less the same electronic structure. But with the increase in atomic radii and polarizability the relative energy of the free atoms goes down; an iodine radical is more stable and less reactive than a fluorine radical.

So the energy of the right side of the reaction above stays more-or-less the same, but the energy on the left side goes down, leading to a smaller bonding energy. So you have a weaker bond and less reactivity.
Moogie
#17
Aug19-10, 01:54 PM
P: 168
Sorry I don't understand your answer.

If iodine radicals were more stable than fluorine radicals then the energy difference between diatomic iodine and iodine atoms would be greater than the energy difference for fluorine so the bond in iodine would be stronger as delta E would be bigger. But H-F is the strongest bond?

I've obviously misunderstood something. I've attached a picture showing what I mean about the energy levels
Attached Thumbnails
halogens.jpg  
Moogie
#18
Aug19-10, 01:59 PM
P: 168
I can see that the strength of the bond decreases down the group and I can see why the stability of the halogen radicals would increase down the group making them less reactive but I didn't understand your description in terms of energy levels. I have drawn a sketch of my interpretation of what you said about energy levels.


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