To calculate the enthalpy change for the combustion of one mole of methane gas, we need to use the balanced chemical equation and the standard enthalpies of formation for each compound involved.The balanced chemical equation for the combustion of methane is:CH g + 2O g CO g + 2HO g Now, we need the standard enthalpies of formation Hf for each compound:Hf CH = -74.8 kJ/molHf O = 0 kJ/mol since O is an element in its standard state Hf CO = -393.5 kJ/molHf HO = -241.8 kJ/molUsing Hess's Law, we can calculate the enthalpy change for the reaction Hrxn as follows:Hrxn = [Hf CO + 2Hf HO ] - [Hf CH + 2Hf O ]Hrxn = [ -393.5 kJ/mol + 2 -241.8 kJ/mol ] - [ -74.8 kJ/mol + 2 0 kJ/mol ]Hrxn = -393.5 - 483.6 + 74.8 kJ/molHrxn = -802.3 kJ/molSo, the enthalpy change for the combustion of one mole of methane gas to form carbon dioxide gas and water vapor at standard temperature and pressure STP is -802.3 kJ/mol.