Question

Calculate the enthalpy change for the combustion of methane, given the following balanced chemical equation:CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)The enthalpies of formation for methane, carbon dioxide, and water are -74.8 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol respectively.

Answer 1

To calculate the enthalpy change for the combustion of methane, we can use the following formula: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 enthalpies of formation of the products, and H_f reactants  is the sum of the enthalpies of formation of the reactants.From the balanced chemical equation, we have:1 mol of CH4 g   methane  and 2 mol of O2 g   oxygen  as reactants.1 mol of CO2 g   carbon dioxide  and 2 mol of H2O g   water  as products.Using the given enthalpies of formation:H_f CH4  = -74.8 kJ/molH_f CO2  = -393.5 kJ/molH_f H2O  = -241.8 kJ/molNow, we can calculate the enthalpy change for the combustion:H_combustion = [1  H_f CO2  + 2  H_f H2O ] - [1  H_f CH4  + 2  H_f O2 ]Since the enthalpy of formation for an element in its standard state is zero, H_f O2  = 0 kJ/mol.H_combustion = [1   -393.5  + 2   -241.8 ] - [1   -74.8  + 2  0]H_combustion =  -393.5 - 483.6  -  -74.8 H_combustion = -877.1 + 74.8H_combustion = -802.3 kJ/molThe enthalpy change for the combustion of methane is -802.3 kJ/mol.