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.