To calculate the change in enthalpy H for the given reaction, we first need to write the balanced chemical equation for the reaction:CH4 g + 2O2 g CO2 g + 2H2O g Next, we need to find the standard enthalpies of formation Hf for each substance involved in the reaction. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states at 298 K and 1 atm pressure. The values can be found in a standard thermodynamic table. The values are:Hf CH4 = -74.8 kJ/molHf O2 = 0 kJ/mol since O2 is in its standard state Hf CO2 = -393.5 kJ/molHf H2O = -241.8 kJ/molNow, we can calculate the change in enthalpy H for the reaction using the following equation:H = [ moles of products Hf of products ] - [ moles of reactants Hf of reactants ]H = [ 1 mol CO2 -393.5 kJ/mol + 2 mol H2O -241.8 kJ/mol ] - [ 1 mol CH4 -74.8 kJ/mol + 2 mol O2 0 kJ/mol ]H = -393.5 - 2 241.8 + 74.8 kJ/molH = -393.5 - 483.6 + 74.8 kJ/molH = -802.3 kJ/molNow, we need to find the moles of CH4 in 25 grams:moles of CH4 = mass / molar massmoles of CH4 = 25 g / 12.01 g/mol C + 4 1.01 g/mol H moles of CH4 = 25 g / 16.05 g/molmoles of CH4 = 1.556 molFinally, we can calculate the change in enthalpy for 25 grams of CH4:H = -802.3 kJ/mol 1.556 molH = -1248.3 kJThe change in enthalpy for the reaction of 25 grams of methane with excess oxygen is -1248.3 kJ.