To calculate the enthalpy change for the combustion of 1 mole of methane gas CH4 at standard conditions, we need to use the standard enthalpies of formation for the reactants and products involved in the reaction.The balanced chemical equation for the combustion of methane is:CH g + 2O g CO g + 2HO l The standard enthalpies of formation Hf for the substances involved in the reaction are:Hf CH = -74.8 kJ/molHf O = 0 kJ/mol since it is an element in its standard state Hf CO = -393.5 kJ/molHf HO = -285.8 kJ/molUsing Hess's Law, the enthalpy change for the reaction Hrxn can be calculated as follows:Hrxn = [Hf products - Hf reactants ]Hrxn = [1 -393.5 kJ/mol + 2 -285.8 kJ/mol ] - [1 -74.8 kJ/mol + 2 0 kJ/mol]Hrxn = -393.5 - 571.6 + 74.8 kJ/molHrxn = -890.3 kJ/molSo, the enthalpy change for the combustion of 1 mole of methane gas at standard conditions is -890.3 kJ/mol.