If I understand correctly, $\ce{H2}$ in the presence of $\ce{Pd}$ readily dissociates as it dissolves into the metal. With the dissociation energy for the $\ce{H—H}$ bond being so large, how is this possible?

At first I thought that the $\ce{H}$ atoms were falling to a lower energy level in the $\ce{Pd}$ that was somehow only available if they were monatomic, but if that were true wouldn't some heat be generated by the absorption?

  • $\begingroup$ The one-line answer would be that the two Pd-H bonds that form must be lower in energy, but right now I don't have anything to back that up. $\endgroup$
    – Ben Norris
    Commented Dec 7, 2012 at 12:32
  • $\begingroup$ That seems a reasonable conclusion and is what I assumed until I read that Palladium hydride readily releases H₂ by the reverse process of hydrogen absorption. $\endgroup$
    – Cargo
    Commented Dec 7, 2012 at 14:42
  • $\begingroup$ There is an equilibrium process, then, that can be manipulated by changing the conditions (for example low pressure vs. high pressure). I just don't know how it is done. $\endgroup$
    – Ben Norris
    Commented Dec 7, 2012 at 16:51
  • $\begingroup$ Metallic (Interstitial) hydrides are quite well-knows, and proximity of Delta G (Free Energy) of formation for some hydride to zero is not that strange. What is much more interesting, is easiness, with which protons travels through solid palladium. This is a thing that hard to understand. $\endgroup$
    – permeakra
    Commented Dec 7, 2012 at 16:56
  • $\begingroup$ @BenNorris: And presumably that the transition between the Kubas-bound state $\ce{Pd(\mu -H2)}$ and the dissociated hydride must have a very low barrier and not involve much of an energy change. $\endgroup$
    – Aesin
    Commented Dec 9, 2012 at 10:56

1 Answer 1


$\ce{Pd}$ can dissociate $\ce{H2}$ because the resulting $\ce{Pd-H}$ bonds are more stable than the starting $\ce{H2}$. But the reason why $\ce{Pd}$ is so good at dissociating $\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of $\ce{H2}$ on a $\ce{Pd}$ surface (and on $\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the $\ce{H2}$ in an excited state to go over a barrier and create $\ce{Pd-H}$ bonds. On $\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite $\ce{H2}$ to dissociate it.

Here is a good theoretical article (with a great title):

It shows the different barriers to dissociation of $\ce{H2}$ on $\ce{Pt}$, $\ce{Ni}$, $\ce{Cu}$ and $\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).

  • 1
    $\begingroup$ May you type the title of the article, please ? because the link has an error .@Guillaume $\endgroup$
    – M.ghorab
    Commented Jan 10, 2016 at 20:18

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