The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.5kJ ` and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ and -285.78kJ` respectively at `25^(@)C` and 1 atmospheric pressure. Calculate the `DeltaH` and `DeltaU` for the reaction,
`NH_(4)NO_(3)(s)toN_(2)O(g)+2H_(2)O(l)`

Explanation

`DeltaH^(@)=DeltaH_(f("products"))^(@)-DeltaH_(f("reactants"))^(@)`
`=[DeltaH_(f(N_(2)O))^(@)+2xxDeltaH_(f(H_(2)O))^(@)]-[DeltaH_(f(NH_(4)NO_(3)))^(@)]`
`=81.46+2xx(-285.78)-(-367.5)`
`=81.46-571.56+367.5`
`=-122.56kJ`
We know that `DeltaH=DeltaU+DeltanRT`
or `DeltaU=DeltaH-DeltanRT`
`Deltan=1,R=8.314xx10^(-3)kJ" "mol^(-1)K^(-1),T=298K`
`DeltaU=-122.56-(1)(8.314xx10^(-3))(298)`
`=-122.56-2.477`
`=-125.037kJ`