I just read that saturated palladium hydride, $\ce{PdH}$0.7, when exposed to oxygen will generate heat and water on it's surface, $\ce{4 H}$(Pd) $\ce{+O2}$(g) $\ce{\to 2 H2O}$(g, l). What is going on?


2 Answers 2


I'm not an expert, but typically there are a series of surface reactions where the reactive species are all adsorbed on the surface. So atomic hydrogen wouldn't "leave" the palladium before it reacts, it just diffuses from an internal crystal site to a surface site.

Likewise, the oxygen would adsorb onto the Pd surface, I'll assume dissociatively. So now there is atomic O and H diffusing around the surface sites of the Pd. These can react and form adsorbed OH, which can further react and form adsorbed H2O, which can then desorb and give you your water.

Your enthalpy calculation is off, the equation for water already includes the fact that an oxygen molecule must break apart to form it.

If you include the hydrogen gas reaction with Pd in your picture, all you're doing is turning hydrogen gas and oxygen gas into water, using Pd as a catalyst. So the total enthalpy change is just -285.83kJ/mol. You don't have enough information to determine how much heat is released (or absorbed possibly) in the hydrogen absorption reaction and how much is released in the water production reaction, but the total of the two will be -285.83kJ/mol.


I'm guessing that when the $\ce{H}$ atoms leave the palladium are they still monatomic for an instant, but without the $\Delta H^0_f = 436 kJ/mol$ extra energy that free $\ce{H}$ atoms would normally have, as bypassed by the $\ce Pd$ somehow. Most of the $\ce{H}$ atoms simply fall in step with one another, reforming $\ce{H2}$ without a peep. But some latch onto nearby $\ce{O2}$ molecules, disassociating them and forming $\ce{H2O}$.

The heat generated is the difference between heat required to dissociate the $\ce{O2}$ and the heat generated by the formation of $\ce{H2O}$.

$\ce{O2 \to 2 O}$ $\Delta H^o_f = 247.5 kJ/mol$

$\ce{ H2 + 1/2 O2 \to H2O}$ $\Delta H^o_f = -285.83 kJ/mol$

$∴ \ce{2H_{(Pd)} + 1/2O2_{(g)} -> H2O_{(g, l)}} \quad \Delta H^o_{\mathrm{f}} = -38.33 \mathrm{kJ/mol}$

Being as this is only$\ 5.68 kJ/mol$ short of the heat required to vaporise $\ce{H2O}$, it explains why some of the product would be gas and some liquid.

This is a complete guess, am I hilariously wrong here?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.