The process 1/2 `H_(2)(g)+e^(-) to H^(-)(g)` is endothermic `(DeltaH=+151kJ mol^(-))`, still ionic hydrides are known. How do you accound for this:

The process 1/2 `H_(2)(g)+e^(-) to H^(-)(g)` is endothermic `(DeltaH=+

1.	The process 1/2 `H_(2)(g)+e^(-) to H^(-)(g)` is endothermic `(DeltaH=+151kJ mol^(-))`, still ionic hydrides are known. How do you accound for this:
Answer» In the formation of solid ionic hydrides, energy is released as released as enthalpy `Delta_(f)H` from the corresponding ions in the gaseous state that are involved in the hydride. This compensates the energy which is needed for the formation of `H^(-)` ions from `H_(2)` gas

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The process 1/2 `H_(2)(g)+e^(-) to H^(-)(g)` is endothermic `(DeltaH=+151kJ mol^(-))`, still ionic hydrides are known. How do you accound for this:

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