CH3OH + 1.5O2 -> CO2 + 2H2O. Using the information from Table 10 of the Data Booklet, determine the theoretical enthalpy of combustion of methanol.
By definition, ΔH°rxn = Σ ΔH°f (products) − Σ ΔH°f (reactants).The enthalpy of combustion is the same as the enthalpy of reaction in this case. From Table 10, we can get the standard enthalpy of formation (ΔH°f ) for each species in the reaction. These are as follows: CH3OH : -239 kJ·mol−1, O2 : 0 kJ·mol−1, CO2 : -393.5 kJ·mol−1, H2O : -285.8 kJ·mol−1.So ΔH°rxn = Σ ΔH°f (products) − Σ ΔH°f (reactants) = (-393.5 - 2 x 285.8) - (-239 - 1.5 x 0) = 726.1 kJ·mol−1, which is the final answer.
