Question

What is the standard enthalpy change for the combustion of methane (CH4) to produce carbon dioxide (CO2) and water (H2O) if the reactants and products are at standard conditions (298 K and 1 atm)?

Answer 1

To calculate the standard enthalpy change for the combustion of methane, we need to use the standard enthalpies of formation for the reactants and products involved in the reaction. The balanced chemical equation for the combustion of methane is:CH g  + 2 O g   CO g  + 2 HO l The standard enthalpies of formation  Hf  for the substances involved are:Hf [CH g ] = -74.8 kJ/molHf [O g ] = 0 kJ/mol  since O is in its standard state Hf [CO g ] = -393.5 kJ/molHf [HO l ] = -285.8 kJ/molNow, we can use Hess's Law to calculate the standard enthalpy change for the reaction  Hrxn :Hrxn = [Hf products  - Hf reactants ]Hrxn = [1 mol   -393.5 kJ/mol  + 2 mol   -285.8 kJ/mol ] - [1 mol   -74.8 kJ/mol  + 2 mol  0 kJ/mol]Hrxn =  -393.5 kJ + 2  -285.8 kJ  -  -74.8 kJ Hrxn =  -393.5 kJ - 571.6 kJ  + 74.8 kJHrxn = -890.3 kJ + 74.8 kJHrxn = -815.5 kJTherefore, the standard enthalpy change for the combustion of methane to produce carbon dioxide and water at standard conditions  298 K and 1 atm  is -815.5 kJ/mol.