# Can the dehydrogenation energy be negative?

I am studying the dehydrogenation of a cyclohexane ring containing $$\ce{B-N}$$ on the adjacent position. or simply we can think about $$\ce{H3N-BH3}$$. If we release one $$\ce{H2}$$ molecule from it then the Gibbs free energy and the enthalpy of dehydrogenation are negative (See Scheme 1 under Fig. 1 here: https://pubs.acs.org/doi/10.1021/ja9106622 ). If we calculate simple dehydrogenation energy for $$\ce{H3N-BH3}$$ then it is also negative.

Negative dehydrogenation energy means it's a spontaneous reaction and we do not need anything to release the $$\ce{H2}$$ molecule from the system. My concern is "How the system is stable in its pristine condition if its dehydrogenation enthalpy/energy is negative"?

Under the scheme 3rd of Fig. 1 in the above-mentioned paper. If we release $$\ce{1H2}$$ from the ring then the dehydrogenation energy is negative. I do not understand how I can explain this negative value of dehydrogenation energy.

• +1 And please note that chemical formulas can be rendered with $\ce{..}$. Also, is your question about an completely experimental study, or is there a theoretical modelling somewhere? Jul 22 at 7:08
• Thanks SR Maiti, I use Quantum Espresso which is a DFT based software. Jul 22 at 10:34

Because the hydrogenation release has a barrier (activation energy) that is not too low. The fact that the dehydrogenation energy of $$\ce{H3N-BH3}$$ is negative does prove that it will eventually decompose if left at room temperature, but it does so extremely slowly so that you don't notice, and it's more useful to treat it as stable. It's like, your oxidation energy is negative too, yet you haven't disappeared into $$\ce{CO2}$$ and $$\ce{H2O}$$ yet.