Question

Calculate the standard enthalpy change (ΔH°) for the reaction of combustion of 1 mole of ethanol (C2H5OH) given that the standard enthalpy of formation (ΔH°f) for ethanol and carbon dioxide (CO2) are -277.7 kJ/mol and -393.5 kJ/mol respectively, and the standard enthalpy of formation (ΔH°f) for water (H2O) is -285.8 kJ/mol. The balanced chemical equation for the reaction is as follows:C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

Answer 1

To calculate the standard enthalpy change  H  for the combustion of 1 mole of ethanol, we can use the following equation:H =  Hf  products  -  Hf  reactants For the balanced chemical equation:C2H5OH  l  + 3 O2  g   2 CO2  g  + 3 H2O  l The standard enthalpy of formation  Hf  for the products are:- CO2: -393.5 kJ/mol- H2O: -285.8 kJ/molThe standard enthalpy of formation  Hf  for the reactants are:- C2H5OH: -277.7 kJ/mol- O2: 0 kJ/mol  since O2 is in its standard state Now, we can plug these values into the equation:H = [2 -393.5  + 3 -285.8 ] - [-277.7 + 3 0 ]H = [-787 +  -857.4 ] - [-277.7]H = -1644.4 + 277.7H = -1366.7 kJ/molTherefore, the standard enthalpy change  H  for the combustion of 1 mole of ethanol is -1366.7 kJ/mol.