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.

Explanation

`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`