Calculate the heat of formation for C2H5OH, given the standard enthalpies of for

Question

Calculate the heat of formation for C2H5OH, given the standard enthalpies of formation for CO2, H2O, and O2 are -393.5 kJ/mol, -285.83 kJ/mol, and 0 kJ/mol, respectively.

Explanation / Answer

Calculate the standard enthalp of formation of ethanol C2H5OH (l) from the heat of combustion of ethanol, which is -1368 kJ/mol, by using tabulated standard enthalpies of formation for CO2 (g) and H20 (l). The combustion of C2H5OH is as follows: 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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