Question

Calculate the standard enthalpy change for the complete combustion of 2 moles of methane gas under standard conditions, where all reactants and products are in their standard states. The balanced chemical equation for the combustion of methane is CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), and the standard enthalpy of formation ΔH°f 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 standard enthalpy change for the complete combustion of 2 moles of methane gas, we can use the following equation:Hrxn =  Hf products  -  Hf reactants For the given balanced chemical equation, CH4 g  + 2O2 g   CO2 g  + 2H2O l , we have:Hrxn = [1 * Hf CO2  + 2 * Hf H2O ] - [2 * Hf CH4  + 2 * Hf O2 ]Since the standard enthalpy of formation for elemental oxygen  O2  is zero, the equation becomes:Hrxn = [1 *  -393.5 kJ/mol  + 2 *  -285.8 kJ/mol ] - [2 *  -74.8 kJ/mol ]Hrxn =  -393.5 kJ/mol - 571.6 kJ/mol  -  -149.6 kJ/mol Hrxn = -965.1 kJ/mol + 149.6 kJ/molHrxn = -815.5 kJ/molSince we are calculating the standard enthalpy change for the complete combustion of 2 moles of methane gas, we need to multiply the enthalpy change for 1 mole by 2:Hrxn  2 moles  = -815.5 kJ/mol * 2Hrxn  2 moles  = -1631 kJTherefore, the standard enthalpy change for the complete combustion of 2 moles of methane gas under standard conditions is -1631 kJ.