3. Calculate the standard molar enthalpy for the complete combustion of lquid et

Question

3. Calculate the standard molar enthalpy for the complete combustion of lquid ethanol (C2H5OH) using the standard enthalpies of formation of the reactants and products C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O answer in kJ

Explanation / Answer

C2H5OH (l) + 3O2 (g) -----> 2CO2 (g) + 3H2O (l) The heat of combustion of ethanol is -1368 kJ/mol. Heat of formation of CO2 (g) = -393.509 kJ/mol Heat of formation of H2O (l) = -285.83 kJ/mol Total enthalpy change = total heat of formation of CO2 reaction + total heat of formation of H2O in reaction Therefore, total enthalpy change = 2(-393.509) + 3(-285.83) = -787.018 + (-857.49) = -1644.508 kJ/mol Obviously, there is a difference between the total enthalpy change of the reaction and the heat of combustion of ethanol. This difference is the enthalpy of formation of the ethanol. The difference = -1644.508 - (-1368) kJ/mol = -276.51 kJ/mol Therefore, enthalpy change of formation of ethanol = -276.51 kJ/mol

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