The combustion of ethylene proceeds according to the equation C2H4 + 3O2 -> 2CO2 + 2H2O. The enthalpy change for the combustion of 1 mol of this hydrocarbon equals -1411.0 kJ/mol. Use the following from Hess's law and calculate the enthalpy of formation of 1 mol of ethylene, if AHf (CO2) = -393.5 kJ/mol, AHf (H2O) = -241.8 kJ/mol, AHf (O2) = 0.
AHf = sum AHf products / sum AHf reactants
AHf = 2AHf(CO2) + 2AHf(H2O) - AHf(C2H4) + 3AHf(O2) -1411
AHf = (2(-393.5) + 2(-241.8)) - (AHf(C2H4) + 3(0)) -1411
AHf = -1270.6 - AHf(C2H4) - 1411
AHf(C2H4) = -1270.6 + 1411
AHf(C2H4) = 140.4 kJ/mol