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Chemistry

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  • $\begingroup$ 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