Question

Determine the standard enthalpy of formation of ethanol (C2H5OH) given the following information: 1. The combustion of 1 mole of ethanol yields 1367 kJ of heat.2. The standard enthalpy of formation for CO2(g) and H2O(l) are -393.5 kJ/mol and -285.8 kJ/mol, respectively.

Answer 1

To determine the standard enthalpy of formation of ethanol  C2H5OH , we can use the information given and Hess's Law. The balanced combustion reaction for ethanol is:C2H5OH l  + 3O2 g   2CO2 g  + 3H2O l We are given the following information:1. The combustion of 1 mole of ethanol yields 1367 kJ of heat.2. The standard enthalpy of formation for CO2 g  is -393.5 kJ/mol.3. The standard enthalpy of formation for H2O l  is -285.8 kJ/mol.Using Hess's Law, we can write the enthalpy change for the combustion reaction as:H_combustion = [2Hf CO2  + 3Hf H2O ] - Hf C2H5OH We know that H_combustion = -1367 kJ/mol, so we can plug in the values and solve for Hf C2H5OH :-1367 kJ/mol = [2 -393.5 kJ/mol  + 3 -285.8 kJ/mol ] - Hf C2H5OH -1367 kJ/mol =  -787 kJ/mol - 857.4 kJ/mol  - Hf C2H5OH Now, we can solve for Hf C2H5OH :Hf C2H5OH  = -787 kJ/mol - 857.4 kJ/mol + 1367 kJ/molHf C2H5OH  = -277.4 kJ/molTherefore, the standard enthalpy of formation of ethanol  C2H5OH  is -277.4 kJ/mol.