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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?

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?

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?

------------------Update------------------

I just read that saturated palladium hydride, 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?

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, forming $\ce{H2O}$.

Am I hilariously wrong here?

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?

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?

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?

------------------Update------------------

I just read that saturated palladium hydride, 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?

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, forming $\ce{H2O}$.

Am I hilariously wrong here?