The standard enthalpy change for the combustion of 1 mole of graphite solid in oxygen gas to form 1 mole of carbon dioxide gas at standard conditions can be calculated using the standard enthalpy of formation of CO2 g .The balanced chemical equation for the combustion of graphite is:C graphite + O2 g CO2 g Since the standard enthalpy of formation of graphite C and oxygen gas O2 are both 0 kJ/mol as they are in their standard states , we can directly use the standard enthalpy of formation of CO2 g to find the standard enthalpy change for the reaction.H = [standard enthalpy of formation of products] - [standard enthalpy of formation of reactants]H = [1 mol -393.5 kJ/mol ] - [1 mol 0 kJ/mol + 1 mol 0 kJ/mol]H = -393.5 kJ/molTherefore, the standard enthalpy change for the combustion of 1 mole of graphite in oxygen gas to form 1 mole of carbon dioxide gas at standard conditions is -393.5 kJ/mol.