# Actual meaning of the energy in the relation $E=k_bT$

I found in tables that the experimental energy required to brake the bond of the hydrogen molecule H$$_2$$ corresponds to $$E=436$$ kJ/mol. So I wanted to convert this value to temperature using the relation $$E=k_bT$$. The temperature corresponding to that energy is $$T= 51~960$$ K. This is roughly 9 times the temperature of the sun. It doesn't make sense to me that we require such a high temperature (51,960 K) to dissociate the hydrogen molecule. At that temperature even nuclear reactions are occurring. My only explanation is that I am misunderstanding the actual meaning of the energy in $$E=k_bT$$. ¿What is the actual meaning?

• In general, keep in mind that temperature is a thermodynamic variable, which means it is astatistical distribution tag when one goes down to molecules and atoms.. For your particular problem see this jpl.nasa.gov/nmp/st5/SCIENCE/sun.html , which supports the answer by Stephen – anna v Sep 29 '20 at 4:15

## 2 Answers

The nuclear reactions in the sun are not happening at the relatively cool surface, but in the core. The core of the sun is ~15 million deg Celsius. Also, the sun's size means that there are lots of reactions, so even very low probability events can happen very often.

Recall also that when that reaction happens in some sort of medium (gas or solution), the energy dissipates into the medium very rapidly.

If a similar reaction (say some sort of oxidation, is this an oxidation?) were to occur in an explosive, and say 10% of the explosive were to react, the explosive would find itself at a few thousand degrees ... that sounds intuitively about right to me.

The quantity $$k_B T$$ represents the scale of random energy fluctuations per molecule, per degree of freedom (up to a scale factor on the order of unity). If the dissociation energy per mole is $$D$$ and a gas of $$H_2$$ molecules is in thermal equilibrium at temperature $$T$$, it means the fraction of $$H_2$$ molecules with enough energy to dissociate will be on the order of $$e^{-(D/N_A k_B T)}$$ ($$N_A$$ is Avogadro's number). So yes, if the temperature is far below $$5e4 K$$, only a small fraction will dissociate (but note that some atomic $$H$$ will exist at any temperature). At room temperature, less than 0.5% of the molecules will dissociate at equilibrium, consistent with our understanding that molecular hydrogen exists as $$H_2$$ at room temperature.

Note, however, that there are other ways to dissociate $$H_2$$ than just through thermal fluctuations. In a plasma, such as an arc discharge, there are electrons and ions bombarding the molecules that can rip them apart in collisions, for example.