Ch 17 – Free Energy:
Using entropy, thermodynamics can predict the direction in which a process will occur spontaneously:
ΔSuniv = ΔSsys + ΔSsurr
 For a spontaneous process, ΔSuniv must be positive
 For a process at constant temperature and pressure:
 ΔSsys is dominated by “positional” entropy = changes in number of gas molecules
 ΔSsurr determined by heat
 ΔSsurr positive for exothermic changes (ΔH is negative)
ΔSsurr = -
∆𝐻
𝑇

Because ΔSsurr depends inversely on T, exothermicity becomes a more important driving
force at low temperatures.
 Vaporization involves a much greater change in disorder than melting (gases are very
disordered compared to liquids, whereas liquids and solids vary less in disorder, distance
between molecules and number of possible positions for molecules); hence ΔSvap > ΔSfus
Third Law of Thermodynamics:
Entropy of a perfect crystal at 0 K is ΔS = 0 (lowest possible entropy)
Free Energy (G) is a state function:
ΔG = ΔH - TΔS
 A process occurring at a constant temperature and pressure is spontaneous in the direction in
which its free energy decreases (ΔG < 0)
 When a liquid boils at its boiling temperature or freezes at its freezing temperature no useful
work can be done (ΔG = 0)
 So, in melting and boiling, ΔS =
∆𝑯
𝑻
 Free Energy depends on temperature and pressure:
ΔG = ΔG⁰ + RT ln P
 Derive relationship between ΔG⁰ of a reaction and value of equilibrium constant (K)
ΔG⁰ = -RT ln K
Answer this question (use back if necessary):
1. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol,
respectively. The melting temperature of benzene is 5.5 ⁰C and it boils at 80.1 ⁰C.
a. Calculate the entropy changes for solid → liquid, and for liquid → vapor for benzene.
b. Would you expect the ΔS for these two changes to be about the same?
c. Comment on the physical significance of the difference in these two values.
d. Why are the values for heat of vaporization usually so much greater than the heats of
fusion?