Question

Calculate the enthalpy of oxidation of methane gas (CH4) at standard conditions given the following information:- Enthalpy of formation of CH4 = -74.8 kJ/mol- Enthalpy of formation of H2O = -285.8 kJ/mol- Enthalpy of formation of CO2 = -393.5 kJ/mol

Answer 1

To calculate the enthalpy of oxidation of methane gas  CH4 , we need to consider the balanced chemical equation for the complete combustion of methane:CH4 g  + 2 O2 g   CO2 g  + 2 H2O l Now, we can use Hess's Law to calculate the enthalpy change for this reaction. The enthalpy change for the reaction  H  is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:H = [Hf CO2  + 2 * Hf H2O ] - [Hf CH4  + 2 * Hf O2 ]Since O2 is in its standard state, its enthalpy of formation is 0 kJ/mol. Now, we can plug in the given values:H = [ -393.5 kJ/mol  + 2 *  -285.8 kJ/mol ] - [ -74.8 kJ/mol  + 2 *  0 kJ/mol ]H =  -393.5 - 571.6 + 74.8  kJ/molH = -890.3 kJ/molSo, the enthalpy of oxidation of methane gas  CH4  at standard conditions is -890.3 kJ/mol.