To calculate the change in enthalpy for the combustion of 2 moles of methane gas CH4 , we first need to write the balanced chemical equation for the combustion reaction:CH4 g + 2O2 g CO2 g + 2H2O l Next, we need the standard enthalpies of formation Hf for all the reactants and products involved in the reaction. The given standard enthalpy of formation for methane is -74.81 kJ/mol. The standard enthalpies of formation for the other substances are:O2 g : 0 kJ/mol since it's an element in its standard state CO2 g : -393.5 kJ/molH2O l : -285.8 kJ/molNow we can use Hess's Law to calculate the change in enthalpy H for the reaction:H = [Hf products ] - [Hf reactants ]H = [1 mol -393.5 kJ/mol + 2 mol -285.8 kJ/mol ] - [1 mol -74.81 kJ/mol + 2 mol 0 kJ/mol]H = -393.5 kJ + 2 -285.8 kJ - -74.81 kJ H = -393.5 kJ - 571.6 kJ + 74.81 kJH = -890.29 kJHowever, this is the enthalpy change for the combustion of 1 mole of methane. Since we are asked to calculate the enthalpy change for the combustion of 2 moles of methane, we need to multiply the result by 2:H 2 moles of CH4 = 2 -890.29 kJ = -1780.58 kJSo, the change in enthalpy for the combustion of 2 moles of methane gas CH4 at a constant pressure of 1 atmosphere and a temperature of 298 K is -1780.58 kJ.