CHAPTER 4. THERMODYNAMICS: THE FIRST LAW

orbitals
orbitals
4-17
o
H (C ) = 718.38 kJ/mol
f
(g)
o
H (H ) = 217.94 kJ/mol.
f
o
Then H = 718.38 + 4(217.94)+ 74.85 = 1665.0 kJ,
and therefore the average C!H bond energy is 1665.0/4 = 416.0 kJ/mol.
(g)
As an illustration of how bond enthalpies may be used to estimate enthalpies of formation,
consider the case of benzene;
6C + 3H ! C H .
(gr) 2(g) 6 6(g)
Using the tabulated C=C, C!C, and C!H bond enthalpies, the
formation of benzene from the gaseous atoms as follows:
o
H can be estimated for the
Form 6 C!H bonds = 6 x !413 = !2478 kJ (Note: When bonds
3 C!C bonds = 3 x !348 = !1044 kJ are formed, H is
3 C=C bonds = 3 x !615 = !1845 kJ negative).
Total bond formation = !5367 kJ.
Then, to obtain the enthalpy of formation, we must add the enthalpy changes for the additional two
steps;
o
o
6C 6 6C , H = 6 H (C ) = 6 x 718.38 = 4308 kJ/mol
(gr) (g) f (g)
o
o
3H 6 6H , H = 6 H (H ) = 6 x 217.94 = 1308 kJ/mol.
2(g) (g) f (g)
o
Therefore, fH (C6H 6(g) ) = !5367 + 4308 + 1308 = 249 kJ/mol. Actually, this turns out to be a rather
poor estimate (the experimental value is 83 kJ/mol) because the double bonds in benzene are best
2
described in terms of sp hybridized σ
and delocalized
π
that extend around the
whole molecule. The delocalization causes the bond enthalpy to be considerably larger than that
expected for single and double bonds. This reduction in the bond enthalpy is referred to as
resonance stabilization.
H. Temperature dependence of enthalpy
It is often necessary to know reaction enthalpies at temperatures other than 298 K.
Consider the cycle shown below involving the same reaction at two temperatures T and T .
1 2