// hyper simplistic terminology in use for the question - apologies.

Molecules are formed by atoms bonding together, and those bonds have an equivalent energy.

At what temperature does the thermal energy of an atom exceed the bond energy, preventing it from forming any molecule?

I think an equivalent question is "what is the temperature needed to spontaneously dissociate any molecule into its component atoms?"

  • $\begingroup$ Depends on the molecule. Some are not stable above -200 C; others seem to be hold it together even at 4,000 C. $\endgroup$ – matt_black Jul 5 '18 at 9:51
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    $\begingroup$ $10\,000^\circ\;\rm C$ is more than enough for everyone. $\endgroup$ – Ivan Neretin Jul 5 '18 at 9:58
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    $\begingroup$ What Ivan meant, I think, is that you should ask for "what is the minimum temperature needed..." $\endgroup$ – Gaurang Tandon Jul 5 '18 at 10:46
  • $\begingroup$ That depends what's the pressure and what you call molecule. $\endgroup$ – Mithoron Jul 5 '18 at 16:52

The bond will break if you can put more than its dissociation energy into it; $\ce{CO}$ has a huge dissociation energy of $\pu{1080 kJ/mol}$, $\ce{I2}$ less than most and is only $\pu{151 kJ/mol}$. This energy can be easily obtained using a laser of the correct wavelength. (You will need to convert units to electron volts to get wavelength needed. $\pu{1 kJ/mol} = \pu{1.036E-2 eV}$, $\pu{1eV} = \pu{8065 cm^-1}$).

However, when we excite using a laser we only use only a small range of energy, using heat at a given temperature is different as temperature is an average measure of a distribution of energies given by the Boltzmann expression $n_i \approx \exp(-E_i/k_\mathrm{B}T)$ where $n_i$ is the number of molecules with energy $E_i$ at temperature $T$, $k_\mathrm{B}$ is the Boltzmann constant. If dealing with molar units replace $k_\mathrm{B}$ by $R$, the gas constant.

Thus, even at room temperature ($\pu{300 K}$), there are, in principle, some molecules of $\ce{CO}$ that will be decomposed even though the dissociation energy is so large; the chance of being decomposed is $\exp(-1080000/(300\times8.314) \approx 10^{-190}$. It is utterly minute even for $6\cdot 10^{23}$ molecules in a mole, but for iodine $\approx 10^{-27}$, still minute so these molecules are stable at room temperature. At $\pu{1000 K}$, 1 in $\approx 10^8$ $\ce{I2}$ molecule decompose and $10^{-4}$ at $\pu{200 K}$ so this is appreciable over a period of time.

(Strictly we should integrate from $T$ to infinity to get the exact answer but the exponential decays so quickly just one value gives us a good idea)

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  • $\begingroup$ I've plugged in the "10,000C is more than enough for everyone" comment (as 10000K) into the formula exp(-DissociationEnergy/(T * 8.314)) and get 0.2728 for CO. So even at 10,000K, only 27% of CO molecules have dissociated. Is that correct? $\endgroup$ – Neil Moss Jul 6 '18 at 13:25
  • $\begingroup$ Not quite, you need to multiply this by the number of collisions that give molecules that energy , in other words calculate the rate constant and then the lifetime of the decomposition to work out how quickly they would decompose. The simplest estimate is an Arrhenius expression, much more sophisticated is called RRKM theory. $\endgroup$ – porphyrin Jul 7 '18 at 7:06

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