To calculate the enthalpy change for the combustion of 2 moles of methane gas, we first need to write the balanced chemical equation for the combustion of methane:CH4 g + 2 O2 g CO2 g + 2 H2O l Next, we need the standard enthalpies of formation Hf for all the compounds involved in the reaction. We are given the Hf for methane gas CH4 as -74.8 kJ/mol. The standard enthalpies of formation for O2 g , CO2 g , and H2O l are 0 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.Now we can use Hess's Law to calculate the enthalpy change for the reaction:H = Hf products - Hf reactants H = [1 * -393.5 kJ/mol + 2 * -285.8 kJ/mol ] - [1 * -74.8 kJ/mol + 2 * 0 kJ/mol]H = -393.5 - 571.6 + 74.8 kJ/molH = -890.3 kJ/molSince we want to calculate the enthalpy change for the combustion of 2 moles of methane gas, we multiply the enthalpy change per mole by 2:H_total = 2 * -890.3 kJ/mol = -1780.6 kJSo, the enthalpy change for the combustion of 2 moles of methane gas at constant pressure is -1780.6 kJ.