Question

What is the enthalpy change for the combustion of 1 mole of methane gas (CH4) at standard conditions, given that the enthalpies of formation for methane, carbon dioxide, and water are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively?

Answer 1

To calculate the enthalpy change for the combustion of 1 mole of methane gas, we need to use the balanced chemical equation for the combustion reaction:CH g  + 2O g   CO g  + 2HO l Now, we can use Hess's Law, which states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the products minus the sum of the enthalpy changes for the reactants.H = [Hf CO  + 2Hf HO ] - [Hf CH  + 2Hf O ]Since O is an element in its standard state, its enthalpy of formation is zero. Therefore, the equation becomes:H = [Hf CO  + 2Hf HO ] - Hf CH Now, we can plug in the given enthalpy values:H = [ -393.5 kJ/mol  + 2 -285.8 kJ/mol ] -  -74.8 kJ/mol H =  -393.5 - 571.6 + 74.8  kJ/molH = -890.3 kJ/molThe enthalpy change for the combustion of 1 mole of methane gas at standard conditions is -890.3 kJ/mol.