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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)` |
Answer» `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` |
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