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^(oplus) = SigmaDelta_(f) H_(Product)^(oplus) - Sigma Delta_(f)H_(Reactant)^(oplus)`
`= [Delta_(f)H^(oplus)(NH_(3))] - [1/2 Delta H_(f)^(@)(N_(2))]+[3/2 DeltaH_(r)^(@)(H_(2))]`
`[-11.02 kcal mol^(-1) ]-[1/2 xx(0)] + [3/2 xx(0)]`
`= -11.02 kcal mol^(-1) -0`
`-11.02 kcal mol^(-1)`
The standard enthalpies of formation of element in its references state is taken as zero, therefore`Delta _(f)H^(oplus)` for `N_(2) and H_(2)` has been taken as zero.