∆G & phase changes
∆G & phase changes:
Example: H2O(l) ⇌ H2O(g)
◆  For
P = 1 atm; T < 373 K (100 ˚C) ∆G > 0
(we know water doesn’t boil below 100 ˚C)
◆  For P = 1 atm; T > 373 K
∆G < 0
◆  For P = 1 atm; T = 373 K
∆G = 0
At the boiling point, the liquid and vapor
are in equilibrium. The boiling point is
defined as the temp. at which the liquid and
vapor phases are at equilibrium at 1 atm.
∆G : H2O(l) ⇌ H2O(g) — more
◆  From
Appendix E:
H2O(g)
H2O(l )
∆Hf° (kJ/mol) –241.8
–285.8
S° (J/mol K) 188.8
69.9
◆  For the vaporization process at 298 K:
∆H298 = ∆Hf° (gas) - ∆Hf° (liq.)
= –241.8 –(–285.8) = +44.8 kJ
∆S298 = S° (gas) - S° (liq.)
= 188.8 – 69.9 = 118.9 J/K = 0.119 kJ/K
◆  ∆G298 = ∆H298 – (298 K)∆S298 = +8.9 kJ
∆G : H2O(l) ⇌ H2O(g) — more
saw that at 298 K, ∆G > 0
◆  (∆H298/∆S298) = (44.8/0.119) K = 376 K this
is close to 373 K, the boiling point! Why?
rearrange: ∆H298 = (376 K) ∆S298
∆H298 - (376 K) ∆S298 = 0
recall: ∆H373 -(373 K)∆S373 = ∆G373 = 0 (why?)
if ∆H373 ≈ ∆H298 and ∆S373 ≈ ∆S298
then ∆G373 = 0 ≈ ∆H298 - (373 K)∆S298
◆  We
Generalization
and ∆S often don’t change much for a
process or reaction from ∆H298 and ∆S298
◆  Often, can estimate for ∆G at other temps.:
∆G(T) ≈ ∆H298 –T∆S298
◆  ∆H
P-T Phase diagram for H2O
Phase diagram for CO2
note: boiling and
freezing points are
not defined for CO2.
after Figure 13-16,
not to scale
Free Energies and Phase Diagrams
Phase Diagrams
◆  All
phase changes are determined by ∆G,
phases are in equilibrium when ∆G = 0.
◆  P-T phase diagrams tell us what phases
are stable at various values of T and P.
◆  A line separating two regions of a phase
diagram represents the temperatures and
pressures where ∆G = 0 for the change
from one phase to another.
Example: recent exam
Aluminum metal has a melting point of 660.3 ˚C.
Use the table below to calculate an estimate for the
boiling point of aluminum in ˚C
Substa
nce
∆Hf °
(kJ mol–1)
∆Gf °
(kJ mol–1)
Al(s)
S°
Cp °
–1
–1
(J mol K ) (J mol–1 K–1)
28.33
24.35
Al(l)
10.56
7.20
39.55
24.21
Al(g)
326.4
285.7
164.54
21.38
Explain why this is just an estimate. A one- or twosentence explanation is all that is necessary.
Example - from old exam
Solid tin exists in two common (b) The standard molar entropy,
forms (allotropes), “white”
S˚, of gray tin is 44.14 J mol-1
and “gray” tin. The standard
K-1. What is the standard
molar enthalpy and molar free
molar entropy of white tin?
energy for the transformation (c) The melting point of tin is
between these two allotropes
505 K. For temperatures
are given below
below 505 K, give the
Snwhite → Sngray temperature ranges (if any)
over which white and gray tin
∆H˚ = -2.090 kJ/mol; ∆G˚ =
are the most stable allotropes.
0.130 kJ/mol
(a) Under standard conditions
(298 K, 1 atm), which
allotrope is
thermodynamically stable?