Question

Calculate the standard enthalpy change for the combustion of ethane (C2H6) gas to form carbon dioxide (CO2) gas and water (H2O) vapor at 298 K and 1 atm. The balanced chemical equation for the reaction is: C2H6(g) + 3.5 O2(g) → 2 CO2(g) + 3H2O(g) Given that the standard enthalpy of formation for C2H6(g) is -84.68 kJ/mol, CO2(g) is -393.51 kJ/mol, and H2O(g) is -241.82 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the combustion of ethane, we can use the following formula:H_combustion =  H_f products  -  H_f reactants where H_combustion is the standard enthalpy change for the combustion, H_f products  is the sum of the standard enthalpies of formation of the products, and H_f reactants  is the sum of the standard enthalpies of formation of the reactants.For the products, we have 2 moles of CO2 g  and 3 moles of H2O g : H_f products  =  2  -393.51 kJ/mol  +  3  -241.82 kJ/mol  = -787.02 kJ/mol - 725.46 kJ/mol = -1512.48 kJ/molFor the reactants, we have 1 mole of C2H6 g  and 3.5 moles of O2 g . The standard enthalpy of formation for O2 g  is 0 kJ/mol since it is in its elemental form: H_f reactants  =  1  -84.68 kJ/mol  +  3.5  0 kJ/mol  = -84.68 kJ/molNow, we can calculate the standard enthalpy change for the combustion:H_combustion =  H_f products  -  H_f reactants  = -1512.48 kJ/mol -  -84.68 kJ/mol  = -1427.80 kJ/molTherefore, the standard enthalpy change for the combustion of ethane  C2H6  gas to form carbon dioxide  CO2  gas and water  H2O  vapor at 298 K and 1 atm is -1427.80 kJ/mol.