Question

What is the standard enthalpy change for the combustion of methane (CH4) at 298 K and 1 atm pressure, given that the reactants and products are in their standard states?

Answer 1

The standard enthalpy change for the combustion of methane  CH4  can be found using the standard enthalpies of formation for the reactants and products involved in the reaction. The balanced chemical equation for the combustion of methane is:CH4 g  + 2 O2 g   CO2 g  + 2 H2O l The standard enthalpies of formation  Hf  for the substances involved 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  = -285.8 kJ/molUsing Hess's Law, the standard enthalpy change for the combustion of methane  Hc  can be calculated as:Hc = [Hf CO2  + 2  Hf H2O ] - [Hf CH4  + 2  Hf O2 ]Hc = [-393.5 + 2   -285.8 ] - [-74.8 + 2  0]Hc =  -393.5 - 571.6  -  -74.8 Hc = -965.1 + 74.8Hc = -890.3 kJ/molTherefore, the standard enthalpy change for the combustion of methane  CH4  at 298 K and 1 atm pressure is -890.3 kJ/mol.