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

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

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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. – Ben Norris Dec 7 '12 at 12:32
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. – Cargo Dec 7 '12 at 14:42
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. – Ben Norris Dec 7 '12 at 16:51
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. – permeakra Dec 7 '12 at 16:56
@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. – Aesin Dec 9 '12 at 10:56

1 Answer

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$\ce{Pd}$ 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).

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Welcome to Chemistry Stack Exchange! – ManishEarth♦ Jan 30 at 4:33
Thank you for the welcome and the formatting @manishearth – Guillaume Jan 30 at 5:31
No, thank you for answering this question (which somehow went unanswered for a while despite the fact that it isn't that hard). If you'd like, we've got quite a few more unanswered questions here (of which I suspect a few are easy, but forgotten) :) – ManishEarth♦ Jan 30 at 9:02

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