Question

Calculate the enthalpy change for the combustion of 1 mole of methane gas (CH4) at standard conditions (25°C and 1 atm), given that the standard enthalpy of formation of methane is -74.87 kJ/mol and the standard enthalpy of formation of carbon dioxide (CO2) and water (H2O) are -393.51 kJ/mol and -285.83 kJ/mol, respectively.

Answer 1

To calculate the enthalpy change for the combustion of 1 mole of methane gas  CH4 , we need to consider the balanced chemical equation for the combustion reaction: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. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.H_combustion = [H_f CO2  + 2 * H_f H2O ] - [H_f CH4  + 2 * H_f O2 ]Since the standard enthalpy of formation of an element in its standard state is zero, the standard enthalpy of formation of O2 is 0 kJ/mol.H_combustion = [ -393.51 kJ/mol  + 2 *  -285.83 kJ/mol ] - [ -74.87 kJ/mol  + 2 *  0 kJ/mol ]H_combustion =  -393.51 + 2 * -285.83 + 74.87  kJ/molH_combustion =  -393.51 - 571.66 + 74.87  kJ/molH_combustion = -890.3 kJ/molTherefore, the enthalpy change for the combustion of 1 mole of methane gas  CH4  at standard conditions is -890.3 kJ/mol.