When 2mole of `C_(2)H_(6)` are completely burnt `-3129kJ` of heat is liberated. Calculate the heat of formation of `C_(2)H_(6).Delta_(f)H^(Theta)` for `CO_(2)` and `H_(2)O` are `-395` and `-286kJ`, respectively.
The equation for the combustion of `C_(2)H_(6)` is:
`2C_(2)H_(6) + 7O_(2) rarr 4CO_(2) + 6H_(2)O, DeltaH =- 3129 kJ`
`DeltaH^(Theta) = Delta_(f)H^(Theta) (“products”) – Delta_(f)H^(Theta)(“reactants”)`
`=[4xxDelta_(f)H^(Theta)underset((CO_(2))).+6xxDelta_(f)H^(Theta)underset((H_(2)O)).]`
`-[2xxDelta_(f)H^(Theta)underset((CO_(2)H_(6))).+7xxDelta_(f)H^(Theta)underset((O_(2))).]`
`-3129 = [4xx-395) + 6 xx (-286)]`
`-[2xxDelta_(f)H^(Theta)underset((C_(2)H_(6))).+7xx0]`
or `2 xx Delta_(f)H^(Theta)(C_(2)H_(6)) =- 167`
So `Delta_(f)H^(Theta)(C_(2)H_(6)) =- (167)/(2) =- 83.5kJ`
`2C_(2)H_(6)(g)+7O_(2)(g)rarr 4CO_(2)(g)+6H_(2)O(l) , Delta H=-3129 kJ`
`Delta H_(“reaction”)=Sigma Delta H_(“products”)-Sigma Delta H_(“reac tan ts”)`
`-3129 =[4xxDelta_(f)H_(CO_(2))+6xxDelta_(f)H_(H_(2)O)]-[2xxDelta_(f)H_(C_(2)H_(6)]`
`2Delta_(f)H_(C_(2)H_(6))=-167` kJ [for elements `Delta H = 0`]
`Delta_(f)H_(C_(2)H_(6))=-83.5 kJ`