Question

Calculate the enthalpy change for the oxidation of 1 mole of methane gas to form carbon dioxide and water vapor, given the following balanced chemical equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)(enthalpy of formation: ΔHf(CH4)=-74.81 kJ/mol, ΔHf(CO2)=-393.51 kJ/mol, ΔHf(H2O)= -241.83 kJ/mol)

Answer 1

To calculate the enthalpy change for the oxidation of 1 mole of methane gas to form carbon dioxide and water vapor, we can use the following equation:H_reaction =  Hf products  -  Hf reactants For the given balanced chemical equation:CH4 g  + 2O2 g   CO2 g  + 2H2O g The enthalpy of formation values are given as:Hf CH4  = -74.81 kJ/molHf CO2  = -393.51 kJ/molHf H2O  = -241.83 kJ/molSince O2 is an element in its standard state, its enthalpy of formation is 0 kJ/mol.Now, we can plug these values into the equation: