The molar heat of formation of `NH_(4)NO_(3)(g)` is -367.54kJ and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ` and -285.78 kJ respectively at `25^(@)C` and 1.0 atmospheric pressure. Calculate `Delta H` and `DeltaU` for the reaction.
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Correct Answer – `DeltaH = – 122. 56 kJ mol^(-1) , DeltaU = – 125. 04kJ mol^(-1)`
`NH_(4)NO_(3)(s) rarrN_(2)O(g) + 2H_(2)O(l), DeltaH _(“reaction”) = SigmaDelta_(f) H (` Products `) -SigmaDelta_(f) H (` Reactants)
`Deltan_(g) =n_(p)- n_(r) = 1-0 = 1`
`DeltaU = DeltaH – Delta n_(g) RT`