1.

The molar heat of formation of `NH_(4)NO_(3(s))` is `-367.54kJ` and those of `N_(2)O_((g)), H_(2)O_((l))` are `81.46` and `-285.8kJ` respectively at `25^(@)C` and 1 atmosphere pressure. Calculate `DeltaH` and `DeltaU` of the reaction `NH_(4)NO_(3(s))rarr N_(2)O_((g))+2H_(2)O_((l))`.

Answer» `NH_(4)NO_(3(s))rarr N_(2)O_((g))+2H_(2)O_((l)), DeltaH=?`
`DeltaH_(Reaction)=DeltaH_(Products)-DeltaH_(Rea c t a nts)`
`=DeltaH_(N_(2)O)+DeltaH_(H_(2)O)xx2-DeltaH_(NH_(4)NO_(3))`
Given, `DeltaH_(N_(2)O)=+81.46kJ` ltBRgt `DeltaH_(H_(2)O)=-285.8kJ`
`DeltaH_(NH_(4)NO_(3))=-367.54kJ`
`:. DeltaH_(Reaction)=+81.46+2(-285.8)-(-367.54)`
`DeltaH=-122.6kJ`
Further
`DeltaH=DeltaU+DeltanRT `
`(Deltan=1-0=1,R=8.314J,T=298K)`
`:. -122.6xx10^(3)=DeltaU+1xx8.314xx298`
`:. DeltaU=-125077j o u l e=-125.077kJ`


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