Given, CH4+2O2⟶CO2+2H2O ΔH=−210kcal/mol...(i) C2H6+27O2⟶2CO2+3H2O ΔH=−368kcal/mol...(ii)
On subtracting Eq (i) from Eq (ii), we get CH2+23O2⟶CO2+H2O ΔH=−158kcal/mol ∴ Enthalpy of combustion of one CH2 unit =−158kcal/mol ΔHcomb (C10H22)=ΔHcomb (CH4)+9×ΔHcomb (CH2) =−210+(9×−158) =−1632kcal