Question

What is the enthalpy change for the combustion of 1 mole of methane (CH4) at constant pressure if the standard enthalpy of formation of methane is -74.8 kJ/mol? Write a balanced equation for the combustion reaction and show your calculations.

Answer 1

To calculate the enthalpy change for the combustion of 1 mole of methane, we need to use the balanced equation for the combustion reaction and the standard enthalpies of formation for all the reactants and products.The balanced equation for the combustion of methane is:CH g  + 2O g   CO g  + 2HO l Now, we need the standard enthalpies of formation  Hf  for all the reactants and products: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/molUsing Hess's Law, we can calculate the enthalpy change for the reaction  Hrxn  as follows:Hrxn =  [Hf products ] -  [Hf reactants ]Hrxn = [1   -393.5 kJ/mol  + 2   -285.8 kJ/mol ] - [1   -74.8 kJ/mol  + 2  0 kJ/mol]Hrxn =  -393.5 - 571.6  -  -74.8 Hrxn = -965.1 + 74.8Hrxn = -890.3 kJ/molSo, the enthalpy change for the combustion of 1 mole of methane at constant pressure is -890.3 kJ/mol.