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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?

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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 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).

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Welcome to Chemistry Stack Exchange! – ManishEarth Jan 30 '13 at 4:33
Thank you for the welcome and the formatting @manishearth – Guillaume Jan 30 '13 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 '13 at 9:02

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