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Author Topic: O+O2 binding energy. Why isn't it released as a UV-Photon?  (Read 4311 times)

Offline Roju

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Hello, and thanks for clicking on my question

I've learnt in school that the energy required to break Ozone O3 into O2 +O is equivalent to a UV-Phtoton with a wavelength of 220-310 nm.

Why is it that when O+O2 bond to become O3, a photon with a wavelength of 220-310 nm isn't emited?

Everywhere i look i find that the energy is released in the form of heat and photons with less energy than UV, but I don't understand how that can happen. Can someone explain what is happening, and if there is a physics concept I need to read up on to understand this?

This has been bothering me for over a year now, so if you know the answere i would really appreciate it if you took the time to explain.
« Last Edit: 22/10/2014 16:24:08 by Roju »


 

Offline alancalverd

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Because the reaction 3O2 → 2O3 requires energy input, so it absorbs photons.

The reverse reaction 2O3 → 3O2 requires the presence of a "third body" to remove the heat of reaction. It is not spontaneous (ozone is stable in the absence of any oxidisable substances or catalysts) so won't emit or absorb a photon.

I have a feeling that your school textbook has got it wrong.
« Last Edit: 22/10/2014 17:11:12 by alancalverd »
 

Offline Roju

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Because the reaction 3O2 → 2O3 requires energy input, so it absorbs photons.

The reverse reaction 2O3 → 3O2 requires the presence of a "third body" to remove the heat of reaction. It is not spontaneous (ozone is stable in the absence of any oxidisable substances or catalysts) so won't emit or absorb a photon.

I have a feeling that your school textbook has got it wrong.
I don't think you understood my question. I know that when O and O2 bond to become O3 energy is released in the form of heat.
My question is: Why and how is it released in the form of heat instead of being emitted as a photon with a wavelenght of 220-310 nm (Which is equivalent to the energy required to break the bond between O2 and O). Maybe it is totally obvious to you and you are leaving that part out, try to be as detailed as possible.
« Last Edit: 22/10/2014 19:43:37 by Roju »
 

Offline Bored chemist

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You may have learned that "I've learnt in school that the energy required to break Ozone O3 into O2 +O is equivalent to a UV-Phtoton with a wavelength of 220-310 nm. "
but it's wrong.
Energy is required to get the reaction to goo the other way.
Energy is released by that reaction (as written)
There are 2 reasons why the energy released when ozone decomposes isn't set free as a photon.
Firstly the reaction only occurs in the presence of a 3rd body, and that takes some of the energy.
Also some of the energy is carried away by the molecules- partly as kinetic energy :partly as excitation of the O2 molecule and the O atom.
 

Offline Roju

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Erm, now i'm super confused.

Are you saying that these two sentences are wrong?
To break a bond, for example the single bond between O and O2 in Ozone you need to input energy.
And when bonds form, for example when O bonds with O2 to make Ozone, energy is released.

Also, do the kinds of reactions you two mentioned, which require a "Third body", have a specific name?

Sorry if this is super basic to you and i'm being annoying, im just trying to understand.
« Last Edit: 22/10/2014 22:13:12 by Roju »
 

Offline chiralSPO

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The reaction of O and O2 to give O3 does release some energy. The reaction forming O3 from O2 requires significant input of energy, but this is to break the O-O bond to make O.

Actually both the creation and destruction of ozone in the atmosphere require the absorbtion of UV light. The reason for this is the activation energy. So even though overall 2O3 --> 3O2 releases energy, you need to put energy in to get the reaction started. I believe that in both cases the UV light puts the starting molecule into an excited electronic state, in which that O-O band can be more easily broken than in just the resting state (O3 + hv --> O3*-->O2 + O). So the reverse reaction won't release the light because O + O2 --> O3, not O3*

If the excited ozone does not break down, it can re-emit that UV photon (O3 + hv --> O3*-->O3 + hv)

Also, alancalvard, ozone is not very stable, even in the absence of oxidizible species. Unless it is really cold (and it is possible to make ozone crystals, which are fairly stable at liquid nitrogen temperatures) ozone will react with itself (sometime explosively) to produce oxygen if the concentration of ozone is too high. One can pump oxygen through ozone-producing machines produce ozone, which comes out as a mixture of ozone and oxygen, but there is a maximum possible concentration of ozone (I think its about 200 mmHg), above which it will spontaneously revert to oxygen.
 

Offline alancalverd

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A matter of interpretation! I read "may decompose explosively" as fairly stable, but then I play with stuff like positronium which selfannihilates in femtoseconds or less, so I assume that if someone has actually described and photographed a blue liquid using ordinary garden photons and a test tube, it's "stable"! I withdraw my assertion. 
 

Offline PmbPhy

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Quote from: Roju
I've learnt in school that the energy required to break Ozone O3 into O2 +O is equivalent to a UV-Phtoton with a wavelength of 220-310 nm.

Why is it that when O+O2 bond to become O3, a photon with a wavelength of 220-310 nm isn't emited?
Hi Roju: Welcome to the forum! :)

This is a problem that's best answered by a chemist. However let me say this. Just because it takes a certain amount of energy E_a to cause energy to be released it doesn't mean that the energy released is E_a. For example; think of a high cliff where at the top there is a bowl shaped indentation next to the edge. The particle is initially at rest at the bottom of the bowl. It takes an amount of energy E_a to raise the particle to the edge of the indentation, the energy going into just enough work to raise the particle from the bottom of the bowl to the top edge of the cliff and push it off. The particle then falls off the cliff and falls a very large distance to the ground. When the particle hits the ground a large amount of energy is released.

Now think of the potential well and an electron in the potential well. Let the potential be defined as follows

V(x) = x2 + a, x < R
V(x) = 0, x > R

Let the particle be at rest at x = 0. If we input an amount of energy E_a = R^2 then the electron can transition to V(x) = 0 while releasing the amount of energy E_b = a + R^2 > E_a. Do you see how this works?

If the energy released is less than the energy put in then the potential energy function is different than this.
« Last Edit: 22/10/2014 23:55:53 by PmbPhy »
 

Offline evan_au

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If you imagine atomic bonds as little springs, if you release an atomic bond (or form a new one), you will get a mechanical vibration in the molecule. A 2-atom molecule like oxygen can wobble or spin in a number of ways - and the number of wobble modes increases rapidly with the number of atoms in the molecule.

These wobbles store energy - until the molecule bumps into another molecule and causes it to wobble and/or spin too, thus sharing the energy throughout the gas. We measure an increase of energy in a gas as an increase in its temperature. The more vibration modes in the molecule, the more energy it takes to raise the temperature by 1 degree.

If some of the energy from the chemical reaction goes into making the molecule wobble or spin mechanically (ie heat), then not so much energy is available for release in the form of photons. The Laws of Thermodynamics tell us that most energy will eventually end up as heat, rather than light.

Edit: Added spin. Thanks, chiralSPO.
« Last Edit: 25/10/2014 23:03:06 by evan_au »
 

Offline chiralSPO

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A two atom molecule can only wobble in one way... (there are two rotational degrees of freedom, but only one vibrational)

A three atom molecule can wobble in many 3 or 4 ways, and larger molecules get increasingly more (3N5 = vibrational degrees of freedom for a linear molecule with N atoms or 3N6 vibrational degrees of freedom for a nonlinear molecule)

Otherwise evan_au's post is spot on.
 

Offline chiralSPO

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Re: O+O2 binding energy. Why isn't it released as a UV-Photon?
« Reply #10 on: 23/10/2014 19:31:29 »
A matter of interpretation! I read "may decompose explosively" as fairly stable, but then I play with stuff like positronium which selfannihilates in femtoseconds or less, so I assume that if someone has actually described and photographed a blue liquid using ordinary garden photons and a test tube, it's "stable"! I withdraw my assertion.

Yes, stable and unstable are relative terms... Really we can only compare the stability of two systems that have the same composition (compare Gibbs free energy of each to see which form(s) will exist at the expense of the others). Otherwise we are just comparing half lives, or the amount of effort required to keep something in one form (stays the same for months if you put it in a jar on the shelf vs 'you need to store this in a 20C freezer in the dark under an atmosphere of carbon monoxide or it will decompose')
« Last Edit: 23/10/2014 19:34:49 by chiralSPO »
 

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Re: O+O2 binding energy. Why isn't it released as a UV-Photon?
« Reply #10 on: 23/10/2014 19:31:29 »

 

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