Calculate the enthalpy change where the standard heat of formation for gaseous `NH_(3)` is -11.02 kcal/mol at 298 K. The reaction given is
`(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) rarr NH_(3)(g)`

Explanation

`Delta_(r)H=sumDeltaunderset(f"(product)")(H^(0))-sumDeltaunderset(f"(reactant)")(H^(0))`
`=Deltaunderset(f(NH_(3)))(H^(0))-(1)/(2)Deltaunderset(f(N_(2)))(H^(0))+(3)/(2)Deltaunderset(f(H_(2)))(H^(0))`
`=-11.02k.calmol^(-1)-[(1)/(2)xx(0)]+[(3)/(2)xx(0)]`
`=-11.02kcalmol^(-1)-0=-11.02kcalmol^(-1)`
The standard enthalpies of formation of elements in their reference state is taken as zero, therefore `DeltaH_(f)^(0)` for `H_(2)` and K has been taken as zero