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InterviewSolution
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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)` |
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Answer» `Delta_(r)H^(Theta)=sum Delta_(f)H_("Product")^(Theta) - sum Delta_(f)H_("Reactant")^(Theta)` `=[Delta_(f)H^(Theta)(NH_(3))]-[(1)/(2)Delta H_(f)^(@)(N_(2))]+[(3)/(2)Delta H_(f)^(@)(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 reference state is taken as zero, therefore `Delta_(f)H^(Theta)` for `N_(2) and H_(2)` has been taken as zero. |
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