The heat of combustion of `C_(2)H_(6)` is `-368.4 kcal`. Calculate the heat of combustion of `C_(2)H_(2)`, when the heat of combustion of `H_(2)` is `68.32 kcal mol^(-1)`.
`C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = 37.1 kcal`.

Explanation

`C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = -37.1 kcl`
`Delta H_(c )C_(2)H_(6) = -368.4 kcal, Delta H_(c )C_(2)H_(4) = ? , Delta H_(c )H_(2(g)) = -68.32 kcal`
`Delta H = SigmaDelta H_("c(reactants)") = -Sigma Delta H_("c(products)")`
`Delta H = Delta H_(c )C_(2)H_(4) + Delta H_(c )H_(2(g)) - Delta H_(c )C_(2)H_(6(c ))`
`-37.1 kcal = Delta H_(c )C_(2)H_(4) - 68.32 - (-368.4)`
`Delta H_(c )C_(2)H_(4) = -337.18 k cal`.