If the standard enthalpy of combustion of ethanol, C₂H₅OH(l),At 298 K is -1368 kJ.mol^-¹,What is the standard enthalpy of formation of ethanol? The standard enthalpies of formation of carbon dioxide and liquid water are -393.51 and -285.83 kJ.mol^-¹,Respectively. A) +688.7 kJ.mol^-¹ B) -344.3 kJ.mol^-¹ C) +276.5 kJ.mol^-¹ D) -276.5 kJ.mol^-¹ E) -688.7 kJ.mol^-¹

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The standard enthalpy of formation of ethanol can be calculated using the standard enthalpy of combustion and the standard enthalpies of formation of the combustion products. The balanced chemical equation for the combustion of ethanol is: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l). Using Hess's law, the standard enthalpy of formation of ethanol is the sum of the standard enthalpies of formation of the products minus the standard enthalpy of combustion of ethanol. Calculation: (2*(-393.51) + 3*(-285.83)) - (-1368) = -276.5 kJ/mol.

If the Standard Enthalpy of Combustion of Ethanol, C2H5OH(l),At 298

If the Standard Enthalpy of Combustion of Ethanol, C₂H₅OH(l),At 298