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

The equation for the combustion of `C_(2)H_(6)` is
`2C_(2)H_(6)+7O_(2)to4CO_(2)+6H_(2)O,DeltaH=-3129kJ`
`DeltaH=DeltaH_(f("Products"))-DeltaH_(f("reactants"))`
`=[4xxDeltaH_(f(CO_(2)))+6xxDeltaH_(f(H_(2)O))]-[2xxDeltaH_(f(C_(2)H_(6))+7DeltaH_(f(O_(2)))]`
`-3129=[4xx(-395)+6xx(-286)]-[2xxDeltaH_(f(C_(2)H_(6)))+7xx0]`
or `2xxDeltaH_(f(C_(2)H_(6)))=-167`
So, `DeltaH_(f(C_(2)H_(6)))=-(167)/(2)=-83.5kJ`