To calculate the enthalpy change for the combustion of 1 mole of butane CH gas under standard conditions, we first need to write the balanced chemical equation for the combustion reaction:CH g + 13/2 O g 4 CO g + 5 HO l Next, we need to find the standard enthalpies of formation Hf for each compound involved in the reaction. The standard enthalpies of formation are:Hf CH = -125.7 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/molNow, we can use Hess's Law to calculate the enthalpy change for the reaction:Hrxn = [Hf products ] - [Hf reactants ]Hrxn = [4 -393.5 kJ/mol + 5 -285.8 kJ/mol ] - [ -125.7 kJ/mol + 13/2 0 kJ/mol]Hrxn = -1574 kJ/mol - 1429 kJ/mol - -125.7 kJ/mol Hrxn = -3003.7 kJ/mol + 125.7 kJ/molHrxn = -2878 kJ/molTherefore, the enthalpy change for the combustion of 1 mole of butane CH gas under standard conditions is approximately -2878 kJ/mol.