Question

Calculate the enthalpy change for the oxidation of methane gas (CH4) to carbon dioxide (CO2) and water (H2O) using the following balanced chemical equation:CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (l)The standard enthalpy of formation of methane, carbon dioxide, and water are -74.81 kJ/mol, -393.5 kJ/mol, and -285.83 kJ/mol respectively.

Answer 1

To calculate the enthalpy change for the reaction, we can use the formula:H_reaction =  H_f_products -  H_f_reactantswhere H_reaction is the enthalpy change for the reaction, H_f_products are the standard enthalpies of formation of the products, and H_f_reactants are the standard enthalpies of formation of the reactants.For the given reaction:CH4  g  + 2O2  g  -> CO2  g  + 2H2O  l The standard enthalpy of formation for the reactants are:H_f  CH4  = -74.81 kJ/molH_f  O2  = 0 kJ/mol  since O2 is in its standard state The standard enthalpy of formation for the products are:H_f  CO2  = -393.5 kJ/molH_f  H2O  = -285.83 kJ/molNow, we can plug these values into the formula:H_reaction = [ -393.5 kJ/mol  + 2 -285.83 kJ/mol ] - [ -74.81 kJ/mol  + 0]H_reaction =  -393.5 - 571.66  -  -74.81 H_reaction =  -965.16  + 74.81H_reaction = -890.35 kJ/molThe enthalpy change for the oxidation of methane gas to carbon dioxide and water is -890.35 kJ/mol.