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Feb 25, 2015 at 15:40 history edited ron CC BY-SA 3.0
Jan 30, 2013 at 3:24 answer added Guillaume timeline score: 15
Jan 23, 2013 at 3:51 history edited Cargo CC BY-SA 3.0
Jan 20, 2013 at 4:42 history edited Cargo CC BY-SA 3.0
added to question
Jan 15, 2013 at 10:14 history edited F'x CC BY-SA 3.0
Bump unanswered question
Dec 9, 2012 at 10:56 comment added Aesin @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.
Dec 9, 2012 at 7:45 history tweeted twitter.com/#!/StackChemistry/status/277680298655879169
Dec 9, 2012 at 0:48 history edited jonsca CC BY-SA 3.0
edited body; edited title
Dec 7, 2012 at 16:56 comment added permeakra 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.
Dec 7, 2012 at 16:51 comment added Ben Norris 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.
Dec 7, 2012 at 14:42 comment added Cargo 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.
Dec 7, 2012 at 12:32 comment added Ben Norris 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.
Dec 7, 2012 at 7:39 history edited Mad Scientist CC BY-SA 3.0
Minor fixes, added catalysis tag
Dec 7, 2012 at 2:37 history edited Cargo CC BY-SA 3.0
revised title and removed pedantic cold fusion joke
Dec 6, 2012 at 16:54 history edited Cargo CC BY-SA 3.0
added 5 characters in body
Dec 6, 2012 at 16:41 history asked Cargo CC BY-SA 3.0