At `25^(@)C` the standard heat of formation of liquid `H_(2)O` is -286.0 `kJ*mol^(-1)`. Calculate the change in standard internal energy for this formation reaction.

Explanation

Formation reaction of `H_(2)O(l):H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l)`
For this reaction, `Deltan=0-(1+(1)/(2))=-(3)/(2)`
We know, `DeltaH^(0)=DeltaU^(0)+DeltanRT`
`therefore -286.0kJ=DeltaU^(0)+(-(3)/(2))xx8.314xx10^(-3)xx(273+25)kJ`
or, `DeltaU^(0)=-282.28kJ`
Therefore, the change in internal energy for the formation reaction of `H_(2)O(l)` is -282.28kJ.