If the standard enthalpy of combustion of octane, C8H18(l), at 298 K is -5471 kJ

Question

If the standard enthalpy of combustion of octane, C8H18(l), at 298 K is -5471 kJ x mol-1, calculate the standard enthalpy of formation of octane. The standard enthalpies of formation of carbon dioxide and liquid water are -393.51 and -285.83 kJ x mol-1 respectively.

Explanation / Answer

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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