The enthalpy change `(Delta H)` for the reaction
`N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)`
is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Explanation

`DeltaH=DeltaU+Deltan_(g)RT`
Therefore, `DeltaU=DeltaH-Deltan_(g)RT`
`Deltan=2-(1+3)=-2`
`Deltan_(g)RT=-2xx8.314xx10^(-3) kJ mol^(-1) K^(-1)xx298 K`
`=-4.955 kJ mol^(-1)`
`DeltaU=-92.38 kJ-(-4.955 kJ mol^(-1))`
`=-92.38+4.955=-87.425 kJ mol^(-1)`