∆Hf of C (g) = 715 kJ/molSolution: For each mole of CH 4 formed, 2 moles of H—H bonds are broken and 4 moles of C—H
bonds are formed (each molecule of CH 4 contains 4 C—H bonds). The one additional
thing we have to take into consideration is that bond dissociation energies are always
in reference to the gas phase, yet here we have carbon in its solid form. This is why we
need the enthalpy of formation of gaseous carbon: The first step in our hypothetical
scheme is to convert carbon from its solid to its gaseous form. Thus the enthalpy
change is:
Again, we subtract the bond dissociation energy to reflect the fact that the C—H bonds are being
formed rather than broken.
Heats of Combustion
One more type of standard enthalpy change that is often used is the standard heat of combustion,
∆H°comb. A combustion reaction is one in which the reactant reacts with (excess) oxygen to yield (in
most cases) carbon dioxide and water, producing a flame during the reaction. (Excess oxygen is
specified because inadequate oxygen leads to the generation of carbon monoxide rather than
carbon dioxide.) The burning of a log, for example, is a combustion reaction. These reactions are
exothermic (release energy, have a negative enthalpy change). The reactions used in the C 3 H 8 (g)
example earlier were combustion reactions, and the corresponding values ∆Ha, ∆Hb, and ∆Hc in that
example were thus heats of combustion.
QUICK QUIZ
At any given temperature, which will have a higher degree of entropy, a solid or a gas?Answer: Gas