# 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

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.