Gibbs Free Energy, G
∆Suniv = ∆Ssurr + ∆Ssys
Suniv =
Hsys
T
+ Ssys
Multiply through by -T
-T∆Suniv = ∆Hsys - T∆Ssys
-T∆Suniv = change in Gibbs free energy
for the system = ∆Gsystem
Under standard conditions —
o
∆G
sys
=
o
∆H
sys
-
o
T∆S
sys
1
o
∆G
Gibbs free
=
o
∆H
-
o
T∆S
2
energy change =
total energy change for system
- energy lost in disordering the system
If reaction is
• exothermic (negative ∆ Ho) (energy dispersed)
• and entropy increases (positive ∆So)
(matter dispersed)
• then ∆Go must be NEGATIVE
• reaction is spontaneous (and productfavored).
o
∆G
=
o
∆H
-
o
T∆S
3
Gibbs free energy change =
total energy change for system
- energy lost in disordering the system
If reaction is
• endothermic (positive ∆Ho)
• and entropy decreases (negative ∆So)
• then ∆Go must be POSITIVE
• reaction is not spontaneous (and is reactant-
favored).
Gibbs Free Energy, G
o
∆G
=
o
∆H
-
o
T∆S
∆Ho
∆So
∆Go Reaction
exo(–)
increase(+)
–
Prod-favored
endo(+)
decrease(-)
+
React-favored
exo(–)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
4
Gibbs Free Energy, G
o
∆G
=
o
∆H
-
5
o
T∆S
Two methods of calculating ∆Go
a)
Determine ∆Horxn and ∆Sorxn and use
GIbbs equation.
b)
Use tabulated values of
free energies
of formation, ∆Gfo.
∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)
Free Energies of Formation
Note that ∆G˚f for an element = 0
6
Calculating ∆Gorxn
7
Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Horxn = -1238 kJ
Use standard molar entropies to calculate
∆Sorxn = -97.4 J/K or -0.0974 kJ/K
∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
Reaction is product-favored in spite of negative
∆Sorxn.
Reaction is “enthalpy driven”
Calculating ∆Gorxn
NH4NO3(s) + heat ---> NH4NO3(aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
8
Calculating ∆Gorxn
9
NH4NO3(s) + heat ---> NH4NO3(aq)
From tables of thermodynamic data we find
∆Horxn = +25.7 kJ
∆Sorxn = +108.7 J/K or +0.1087 kJ/K
∆Gorxn = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ
Reaction is product-favored in spite of negative
∆Horxn.
Reaction is “entropy driven”
Gibbs Free Energy, G
o
∆G
=
o
∆H
-
10
o
T∆S
Two methods of calculating ∆Go
a)
Determine ∆Horxn and ∆Sorxn and use
GIbbs equation.
b)
Use tabulated values of
free energies
of formation, ∆Gfo.
∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)
Calculating ∆Gorxn
∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)
11
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
∆Gorxn = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]
∆Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
∆Gorxn = -394.4 kJ
Reaction is product-favored as expected.
Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
∆Horxn = +467.9 kJ
∆Sorxn = +560.3 J/K
∆Gorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does ∆Gorxn just change from
being (+) to being (-)?
When ∆Gorxn = 0 = ∆Horxn - T∆Sorxn
Hrxn
467.9 kJ
T =
=
= 835.1 K
Srxn
0.5603 kJ/K
12
13
More thermo?
You betcha!
Thermodynamics and Keq

FACT: ∆Gorxn is the change in free energy
when pure reactants convert COMPLETELY
to pure products.

FACT: Product-favored systems have
Keq > 1.

Therefore, both ∆G˚rxn and Keq are related
to reaction favorability.
14
Thermodynamics and Keq
Keq is related to reaction favorability and so
to ∆Gorxn.
The larger the value of K the more negative
the value of ∆Gorxn
o
∆G rxn
= - RT lnK
where R = 8.31 J/K•mol
15
Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
∆Gorxn = +4.8 kJ
∆Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
When ∆Gorxn > 0, then K < 1
16
∆G, ∆G˚, and Keq
• ∆G is change in free energy at nonstandard conditions.
• ∆G is related to ∆G˚
• ∆G = ∆G˚ + RT ln Q
where Q = reaction quotient
• When Q < K or Q > K, reaction is
spontaneous.
• When Q = K reaction is at equilibrium
• When ∆G = 0 reaction is at equilibrium
• Therefore, ∆G˚ = - RT ln K
17
∆G, ∆G˚, and Keq
Figure 19.10
18
∆G, ∆G˚, and Keq
• Product favored
reaction
• –∆Go and K > 1
• In this case ∆Grxn is
< ∆Gorxn , so state with
both reactants and
products present is
MORE STABLE than
complete conversion.
19
∆G, ∆G˚, and Keq
20
Product-favored
reaction.
2 NO2 ---> N2O4
∆Gorxn = – 4.8 kJ
Here ∆Grxn is less than
∆Gorxn , so the state
with both reactants
and products
present is more
stable than complete
conversion.
∆G, ∆G˚, and Keq
21
Reactant-favored
reaction.
N2O4 --->2 NO2
∆Gorxn = +4.8 kJ
Here ∆Gorxn is greater
than ∆Grxn , so the
state with both
reactants and
products present is
more stable than
complete conversion.
Thermodynamics and Keq

Keq is related to reaction favorability.

When ∆Gorxn < 0, reaction moves
energetically “downhill”

∆Gorxn is the change in free energy when
reactants convert COMPLETELY to
products.
22