The enthalpy change `(DeltaH)` for the reaction
`N_2(g) + 3H_2 (g) to 2NH_3(g)`
is -92.38 kJ at 298 K. What is `DeltaU` at 298 K ? `(R = 8.314 j K^(-1) mol^(-1))`

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)`