Free Energy
◆  One
Free Energy and Spontaneity
CHEM 107
T. Hughbanks
more state function ....
◆  We know ∆Suniverse > 0 for a spontaneous
change, but ...
◆  We are still looking for a state function of
the system that will predict spontaneity.
◆  Define a new function that satisfies this
need. Call it “free energy.” (sometimes
“Gibbs free energy”)
Free Energy: Definition
Define the free energy by:
G = H – TS
◆  G
is a state function, since H, T, & S are. If
T & P are variables we control, G is the
function that predicts spontaneity.
◆  Consider a process that occurs at constant
temperature.
∆G =∆H – T∆S
◆  This
is the central equation in chemical
thermodynamics!
∆G & Spontaneity
∆G = ∆H – T∆S
T∆Suniv = T∆S – ∆H
So
∆G = – T∆Suniv
◆  From this, we see that ∆G and ∆Suniv
will always have opposite signs. (T > 0)
◆  Spontaneous process → ∆Suniv > 0, so
◆  Spontaneous process → ∆G < 0.
∆G & Spontaneity
∆G = ∆H - T∆S
(for any X, when not shown, ∆X = ∆Xsys)
◆  Compare this with ∆Suniv .
∆Suniv = ∆Ssys + ∆Ssurr = ∆S + ∆Ssurr
∆Ssurr = qsurr / T = –qsys / T
=∆H/T (const P)
◆  So:
∆Suniv = ∆S – ∆H/T
◆  Or:
T∆Suniv = T∆S – ∆H
∆G & Spontaneity
is thus the function we have been
seeking:
◆  ∆G
◆  a
state function of the system
tells us whether a process (reaction or
phase change) is spontaneous
◆  sign
is generally the most useful
thermodynamic function for a chemist.
◆  ∆G
Spontaneity: Role of ∆H & ∆S
∆G - Change in Free Energy
◆  Predictor
of spontaneity. A spontaneous
reaction has ∆G < 0.
◆  Also tells us the maximum amount of
energy which can be produced and used
to do work. So ∆G is useful in
determining amounts of fuel needed, etc.
◆  The
form of ∆G shows us the role of ∆H
and ∆S in determining spontaneity.
∆G = ∆H – T∆S
< 0 → exothermic
→ favors spontaneity
◆  ∆H
◆  ∆S
> 0 → entropy increases → favors spontaneity
Spontaneity: Role of ∆H, ∆S, T
Spontaneity: Role of T
tells us whether or not a reaction
will occur spontaneously.
∆G = ∆H – T∆S
◆  Usually, we assume that ∆H and ∆S do
not depend on T. This means that only
the T∆S term varies with T.
◆  Effect of temperature depends on signs.
◆  ∆G
2 NO2 ⇌ N2O4, ∆S < 0, ∆H < 0
low T: N2O4 favored; high T: NO2 favored
N2O4: colorless
ΔH ΔS
− +
+
−
+
+
−
−
∆G: Using Tabulated Data
◆  Thermodynamic
tables (Appendix E)
usually include ∆Gf° values.
◆  These are defined and used like ∆Hf° ’s.
◆  refer
◆  same
120 pm
O
N
134˚
O
NO2: brown
Always
Spontaneous
Never
Spontaneous
Spontaneous at
sufficiently high T
Spontaneous at
sufficiently low T
◆  ∆G°rxn
◆  Can
to formation reactions
standard state convention
= ∑ n ∆Gf° products – ∑ n ∆Gf° reactants
also use ∆H° and ∆S° to find ∆G°
Problem - Mond Process
Tabulated Data
◆  For
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
N2(g):
∆Hf° = 0, ∆Gf° = 0, S° = 191.61 J K-1 mol-1
◆  Note
that the data give S° and NOT ∆Sf°
note that ∆Gf° IS NOT EQUAL
TO ∆Hf° – TS°
◆  Also,
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
◆  ∆H° =
–160.8 kJ ; ∆S°= –409.5 J/K
◆  ∆G° =
∆H° – T∆S°
+ ½ O2(g) → CO(g)
earlier problem, ∆H° = –160.8 kJ
and ∆S° = –409.5 J/K )
◆  Given
∆Gf° = –137.17 kJ/mol for CO,
find ∆Gf° for Ni(CO)4(g).
◆  Use
∆G° and ∆S° to find the range of
temperatures at which the reaction is
spontaneous.
(1)
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
◆  ∆H° =
= –160.8 kJ – (298 K)(–0.4095 kJ/K) = –38.77 kJ/mol
◆  C(graphite)
◆  (From
(2)
–160.8 kJ ; ∆S°= –409.5 J/K
◆  Use
∆G° and ∆S° to find the range of
temperatures at which the reaction is
spontaneous.
∆Gf° = –137.17 kJ/mol (given)
◆
Find ∆Gf° :
Ni(s) + 4 C + 2 O2(g) → Ni(CO)4(g) (3)
Mond Process
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
◆
∆H° and ∆S° are both negative, so at
low T the reaction is spontaneous.
◆  At
high T, ∆G becomes positive, so
reaction proceeds spontaneously to the
left. At high T, Ni(CO)4 decomposes.
Mond Process
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
◆  For Ni purification:
First react impure Ni with pure CO to
form Ni(CO)4. (This works only for Ni,
other metals are not as reactive with CO.)
◆  Need
“low” T for reaction to go to right.
Run at 50 ˚C. (Tboil = 42˚C for Ni(CO)4.)
Mond Process
Mond Process
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
step in purification:
◆  Remove gas phase from reaction
chamber. This is a mixture of CO and
Ni(CO)4.
◆  Run reaction “backwards” to produce
pure nickel as a precipitate.
◆  2nd
Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)
◆  To produce Ni, we want the reaction to
go from right to left. This means high T
needed.
◆  Typically run at 230 ˚C, where
∆G = ∆H – T∆S = + 45.18 kJ
◆  Finally, remove pure Ni from vessel.
∆G & phase changes:
Example: H2O(l) ⇌ H2O(g)
∆G & phase changes
∆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
◆  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
◆  We
saw that at 298 K, ∆G > 0
= (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
◆  (∆H298/∆S298)
Generalization
P-T Phase diagram for H2O
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
Phase diagram for CO2
Free Energies and Phase Diagrams
note: boiling and
freezing points are
not defined for CO2.
after Figure 13-16,
not to scale
Example: recent exam
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.
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 °
(J mol–1 K–1) (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
temperature ranges (if any)
Snwhite → Sngray
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?