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)`

The molar heat of formation of `NH_(4)NO_(3)(s)` is `-360.0kJ` and tho

1.	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)`
Answer» `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`

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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)`

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