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=+151

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» Solution :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