Question

The standard change in internal energy for the combustion of buckyball is -25,968 kJ/mol. What is the enthalpy of combustion? What is the enthalpy of formation of buckyball? What is the molar enthalpy associated with converting buckyball to graphite? How does this compare to the enthalpies associated with converting diamond to graphite?

Answer 1

C60 + 60 O2 ==> 60CO2
dE = q + w
Combustions are done at constant and W = -pdV. 60 mols CO2 come from 60 mols O2 so there no change in volume and no work. Thus dE = q = dH.
dH formation would be the reverse; i.e., 60C --> C60 and
dHrxn = dHformation C60. We don't know that but can calculate that from the above.
C60 + 60O2 ==> 60CO2
dHrxn = (n*dHf CO2)-(dHfC60)-(dHf O2)
-26988 kJ/mol = 60*-393.5) - 0
Calculate dHf C60 and use that to calculate dH formation. Post your work if you get stuck.