Calculate the enthalpy change for the combustion of 1 mole of methane (CH4) gas given that the standard enthalpy of formation of methane is -74.8 kJ/mol, and the standard enthalpies of formation for carbon dioxide (CO2) and water (H2O) are -393.5 kJ/mol and -241.8 kJ/mol respectively. The balanced chemical equation for the reaction is:CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

Question

Calculate the enthalpy change for the combustion of 1 mole of methane (CH4) gas given that the standard enthalpy of formation of methane is -74.8 kJ/mol, and the standard enthalpies of formation for carbon dioxide (CO2) and water (H2O) are -393.5 kJ/mol and -241.8 kJ/mol respectively. The balanced chemical equation for the reaction is:CH4 (g) + 2O2 (g) → CO

Answer 1

To calculate the enthalpy change for the combustion of 1 mole of methane, we can use the following equation:H_combustion =  H_f products  -  H_f reactants where H_combustion is the enthalpy change for the combustion, H_f products  is the sum of the standard enthalpies of formation for the products, and H_f reactants  is the sum of the standard enthalpies of formation for the reactants.For the given reaction:CH4  g  + 2O2  g   CO2  g  + 2H2O  l The products are CO2 and 2H2O, and the reactants are CH4 and 2O2. The standard enthalpy of formation for O2 is 0 kJ/mol since it is in its elemental form.Now, we can plug in the given values into the equation:H_combustion = [ -393.5 kJ/mol  + 2 -241.8 kJ/mol ] - [ -74.8 kJ/mol  + 2 0 kJ/mol ]H_combustion =  -393.5 - 483.6  -  -74.8 H_combustion = -877.1 + 74.8H_combustion = -802.3 kJ/molTherefore, the enthalpy change for the combustion of 1 mole of methane is -802.3 kJ/mol.