Question

Calculate the standard enthalpy of formation of methanol, CH3OH (l), given that the standard enthalpy of combustion of methanol is -726.4 kJ/mol and the standard enthalpy of formation of carbon dioxide and water are -393.5 kJ/mol and -285.8 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy of formation of methanol, we can use the following combustion reaction:CH3OH  l  + 3/2 O2  g   CO2  g  + 2 H2O  l The standard enthalpy change for this reaction  H_comb  is -726.4 kJ/mol.We can also express the standard enthalpy change for this reaction in terms of the standard enthalpies of formation  H_f  of the reactants and products:H_comb = [H_f CO2  + 2H_f H2O ] - [H_f CH3OH  + 3/2H_f O2 ]Since the standard enthalpy of formation of an element in its standard state is zero, H_f O2  = 0. We can now plug in the given values for the standard enthalpies of formation of CO2 and H2O:-726.4 kJ/mol = [ -393.5 kJ/mol  + 2 -285.8 kJ/mol ] - [H_f CH3OH ]Now, we can solve for H_f CH3OH :-726.4 kJ/mol =  -393.5 kJ/mol - 571.6 kJ/mol  - H_f CH3OH -726.4 kJ/mol = -965.1 kJ/mol - H_f CH3OH Adding 965.1 kJ/mol to both sides:H_f CH3OH  = 238.7 kJ/molTherefore, the standard enthalpy of formation of methanol, CH3OH  l , is 238.7 kJ/mol.