Find the value of `Delta_(f) H^(@)` for the reaction
`N_(2) O_(4) (g) + 3 CO(g) rarr N_(2) O (g) + 3 CO_(2) (g)`
Standard enthalpies of formation of `CO(g), CO_(2) (g), N_(2) O (g)`, and `N_(2) O_(4) (g)` are `- 110, - 393, 81`, and `9.7 kJ mol^(-1)`, respectively.
Strategy : The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpie of formation of the products each multiplied by its stiochiometric coefficient in the balanced equation, minus the corresponding sum of the standard molar enthalpies of formation of the reactants

Answer

`Delta_(r)H=-778 kJ`

Explanation

`Delta_(r)H=` for a reaction is defined as the difference between `Delta_(f)H` value of products and `Delta_(f)H` value of reacteants.
`Delta_(r)H=sumDelta_(f)H("products")-sumDelta_(f)H("reactants")`
For the given reaction,
`N_(2)O_(4(g))+_3CO_((g))rarrN_(2)O_((g))+3CO_(2(g))`
`Delta_(r)H=[{Delta_(f)H(N_(2)O)+3Delta_(f)H(CO_(2))}-{Delta_(f)H(N_(2)O_(4))+3Delta_(f)H(CO)}]`
Substituting the values of `Delta_(f)H "for"N_(2)O_(2),CO_(2)m_(2)O_(4)and CO` from the question, we get :
`Delta_(r)H=[{81 K "mol"^(-1)+3(-393)kJ"mol"^(-1)}-{9.7 kJ "mol"^(-1)+3(-110)kJ "mol"^(-1)}]`
`Delta_(r)H=-777.7 kJ "mol"^(-1)`
Hence, the value of `Delta_(r)H` for the reactio is `-777.7 kJ "mol"^(-1)`