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The bond enthalpy associated with a $\ce{O=O}$ double bond is equal to 495 kJ/mol. Does that mean that adding enough kinetic energy in the form of heat will eventually cause the bonds to break and create monoatomic oxygen?

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Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen.

$$\ce{O2 ->C[energy]\ 2O}$$

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Yes. For quantitative information see Gas Phase Reaction Kinetics of Neutral Oxygen Species, particularly section 2.1, especially table 4, which gives the equillibrium constant $K_D$ as a function of temperature.


$K_D = \frac{[O]^2}{[O_2]}$

$K_D$ exceeds 1 above ~1000K, in units of particles per cubic centimeter.

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It is important to point out that heat is a bulk (statistical mechanical) concept while bond strength is largely quantum mechanical. If I am not mistaken, heat/temperature can be defined using pictures like molecules striking the edges of a container, or perhaps from energy/entropy relationships that involve Avagadro's number of molecules.

To your question, this means that given a large enough thermal energy, other molecules of $\ce{O2}$ will likely collide with an $\ce{O_2}$ molecule in question -- thereby bringing their nuclear kinetic energy to bear, and perhaps causing a dissociation.

I don't think that an isolated molecule of $\ce{O_2}$ (or anything for that matter) will spontaneously rupture because of heat, because here heat is not a well-defined quantum mechanical concept, as it is mediated by many moving atoms.

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Yes. Adding enough thermal energy to anything will disrupt the bonds.

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In principle yes, but it will need much more than 495 kJ/mol for sure.

The bond enthalpy is based on (experimental) enthalpy of formation. It does not take into account the enthalpy reaction barrier of the $\ce{O2}$ dissociation.

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Ah, interesting! I knew it would need more but not much more Approximately how much more would you say? – VOKBY May 29 '14 at 19:41
"much more than 495 kJ/mol for sure" Hmmm, remember that the molecules have a Boltzmann distribution of energies to begin with. – ron May 29 '14 at 21:41
Did you mean Boltzmann's kT contribution? At room temperature it will be less than 1 kcal/mol, very insignificant to the enthalpy barrier. – nawesita May 30 '14 at 0:35
I mean there is a large distribution of kinetic energy in the molecules in the sample, such that even at room temperature some will occasionally dissociate. – ron May 30 '14 at 13:01
By enthalpy reaction barrier, do you mean an activation energy? There is no activation energy whatsoever for bond formation; atoms at infinity have zero potential energy, and at any value below that up to a certain equilibrium distance (the bond distance), it's a downhill slope. The curve can be modeled coarsely by a Lennard-Jones potential. Thus, the inverse processes of bond breaking only requires 495 kJ/mol to separate the atoms infinitely. – Nicolau Saker Neto May 31 '14 at 22:56

Heating it even further will rip all the electrons away from the nuclei themselves to create a plasma state of free floating positive nuclei and electrons whereby, if enough energy was put into the system, the oxygen nuclei would fuse

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It sure will. But as soon as the gas cools down it will just re-form into O2 unless you give it something else to react with.

It's possible to get it to react with fluorine at around 700 degrees by heating O2 and F2 through it. You now have monatomic O and F ready to party, the flourine in particular "losing it's gentle and forgiving nature" when reduced to single life. After cooling (to 90K), you will have a flask full of FOOF. That's assuming you still have an intact lab (hint: please do not ever try this, for your sake or the sake of others in your vicinity --Ed.).

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Thank you for your username homage, but it was not appreciated. Continued misbehavior will result in suspension. – jonsca Jun 1 '14 at 22:28

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