The molar heat of formation of `NH_(4)NO_(3)(s)` is `-360.0kJ` and those of `N_(2)O(g)`and `H_(2)O(l)` are `+80.00kJ` and `-285.00kJ`, respectively, at `25^(@)C` and `1atm`. Calculate `DeltaH` and `DeltaU` for the reaction.
`NH_(4)NO_(3)(s) rarr N_(2)O(g) +2H_(2)O(l)`

Explanation

`DeltaH^(Theta) = DeltaH = Delta_(f)H^(Theta) ("products") - Delta_(f)H^(Theta) ("reactants")`
`=[Delta_(f)H_((N_(2)O))^(Theta)+2xxDelta_(f)H_((H_(2)O))^(Theta)-[Delta_(f)H_((NH_(4)NO_(3)))^(Theta)]`
`= 80.00 + 2 xx (-280.00) - (-360.0)`
`= 80.0 - 560.0 + 360.0 =- 120.0kJ`
We know that `DeltaH = DeltaU + DeltanRT`
or `DeltaU = DeltaH - DeltanRT`
`Deltan = 1,R = 8.314 xx 10^(-3) kJ mol^(-1) K^(-1), T = 298K`
`DeltaU =- 120.0 -(1) (8.314xx10^(-3)) (298)`
`=- 120.0 - 2.477 =- 122.477 kJ`