Robert Curl, Richard Smalley, and Harold Kroto were awarded the Nobel Prize in Chemistry in 1996 for the discovery of the soccer ball shaped molecule C₆₀. This molecule was the first of a new series of molecular allotropes of carbon. The enthalpy of combustion of C₆₀ is 225 937 kJ · mol^–1, and its enthalpy of sublimation is 1233 kJ · mol^–1.

There are 90 bonds in C₆₀, of which 60 are single bonds and 30 are double bonds. Like benzene, C₆₀ has a set of multiple bonds for which resonance structures may be drawn.

(a) Determine the enthalpy of formation of C₆₀ from its enthalpy of combustion.

(b) Calculate the expected enthalpy of formation of C₆₀ from bond enthalpies, assuming the bonds to be isolated double and single bonds.

(c) Is C₆₀ more or less stable than predicted on the basis of the isolated-bond model?

(d) Quantify the answer to part (c) by dividing the difference between the enthalpy of formation calculated from the combustion data and that obtained from the bond enthalpy calculation by 60 to obtain a per-carbon value.

(e) How does the number in part (d) compare with the per-carbon resonance stabilization energy of benzene (the total resonance stabilization energy of benzene is approximately 150 kJ · mol^–1)?

(f) Why might these values differ? The enthalpy of atomization of C(gr) is 1717 kJ · mol^–1.