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.

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    $\begingroup$ +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? $\endgroup$
    – S R Maiti
    Commented Jul 22, 2021 at 7:08
  • 1
    $\begingroup$ Thanks SR Maiti, I use Quantum Espresso which is a DFT based software. $\endgroup$
    – astha
    Commented Jul 22, 2021 at 10:34

1 Answer 1


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.

  • $\begingroup$ Let me answer myself now. For a reaction mechanism, there are two states, initial and final. It may be possible that the final state is more stable than the first one. If the system is in the state one, then we need to give some external forces that will push the reaction to happen and thus making the product. These external forces can be measured in the terms of kinetic energy barrier. Thanks @wcchem5 $\endgroup$
    – astha
    Commented Jul 23, 2021 at 2:06

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