°sys < 0
S°sys
S°surr
Lecture 8 Summary
S°univ
Spontaneous, exothermic
°sys < 0
S°surr
S°univ
Spontaneous, exothermic
S°sys
°sys > 0
S°sys
S°univ
S°surr
Spontaneous, endothermic
Lecture 9: Gibbs Free Energy G
Reading: Zumdahl 10.7, 10.9
Outline
Defining the Gibbs Free Energy (G)
Calculating G (several ways to)
Pictorial Representation of G
Defining G
Recall, the second law of thermodynamics:
Suniv = Stotal = Ssys + Ssurr
Also recall:
Ssurr = -Hsys/T
Substituting,
Stotal = Ssys +-H
S
/T
syssurr
Multipling all by (-T) gives
-TStotal = -TSsys+ Hsys
We then define:
G = -TStotal
Substituting
G = -TSsys+ Hsys
Giving finally
G = H - TS
G = The Gibbs Free Energy @const P
G and Spontaneous Processes
Recall from the second law the conditions of spontaneity:
Three possibilities:
If Suniv > 0…..process is spontaneous
If Suniv < 0…..process is spontaneous in opposite
direction.
If Suniv = 0….equilibrium
In our derivation of G, we divided by -T;
therefore, the direction of the inequality
relative to entropy is now reversed.
Three possibilities in terms of S:
If Suniv > 0…..process is spontaneous.
If Suniv < 0…..process is spontaneous in opposite
direction.
If Suniv = 0…. system is in equilibrium.
Three possibilities in terms of G:
If G < 0…..process is spontaneous.
If G > 0…..process is spontaneous in opposite
direction.
If G = 0….system is in equilibrium
Spontaneous Processes: temperature dependence
Note that G is composed of both H and S terms
G = H - TS
A reaction is spontaneous if G < 0. So,
If H < 0 and S > 0….spontaneous at all T
If H > 0 and S < 0….not spontaneous at any T
If H < 0 and S < 0…. becomes spontaneous at low T
If H > 0 and S > 0….becomes spontaneous at high T
Example: At what T is the following reaction
spontaneous?
Br2(l)
Br2(g)
where H° = 30.91 kJ/mol, S° = 93.2 J/mol.K
G° = H° - TS°
Try 298 K just to see result at standard conditions
G° = H° - TS°
G° = 30.91 kJ/mol
- (298K)(93.2 J/mol.K)
G° = (30.91 - 27.78) kJ/mol
= 3.13 kJ/mol > 0
Not spontaneous at 298 K
At what T then does the process become spontaneous?
G° = H° - TS° = 0
T = /S
T = (30.91 kJ/mol) /(93.2 J/mol.K)
T = 331.65 K
Just like our previous calculation
Calculating G°
In our previous example, we needed to determine
H°rxn and S°rxn separately to determine G°rxn
But ∆ G is a state function; therefore, we can
use known G° to determine G°rxn using:
Grxn =  Gprod. - Greact.
Standard G of Formation: Gf°
Like Hf° and S°, the standard Gibbs free
energy of formation Gf° is defined as the
“change in free energy that accompanies the
formation of 1 mole of that substance for its
constituent elements with all reactants and
products in their standard state.”
As for Hf° , Gf° = 0 for an element in its standard
state:
Example: Gf° (O2(g)) = 0
Example
• Determine the G°rxn for the following:
C2H4(g) + H2O(l)
C2H5OH(l)
• Tabulated G°f from Appendix 4:
G°f(C2H5OH(l)) = -175 kJ/mol
G°f(C2H4(g)) = 68 kJ/mol
G°f(H2O (l)) = -237 kJ/mol
• Using these values:
C2H4(g) + H2O(l)
C2H5OH(l)
Grxn =  Gprod. - Greact.
G°rxn = G°f(C2H5OH(l)) - G°f(C2H4(g))
-G°f(H2O (l))
G°rxn = -175 kJ - 68 kJ -(-237 kJ)
G°rxn = -6 kJ < 0 ; therefore, spontaneous
More G° Calculations
• Similar to H°, one can use the G° for
various reactions to determine G° for the
reaction of interest (a “Hess’ Law” for G°)
• Example:
C(s, diamond) + O2(g)
CO2(g) G° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) G° = -394 kJ
C(s, diamond) + O2(g)
CO2(g) G° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) G° = -394 kJ
CO2(g)
C(s, graphite) + O2(g) G° = +394 kJ
C(s, diamond)
C(s, graphite)
G° = -3 kJ
G°rxn < 0…..rxn is spontaneous
G°rxn ≠ Reaction Rate
• Although G°rxn can be used to predict if a
reaction will be spontaneous as written, it
does not tell us how fast a reaction will
proceed.
• Example:
C(s, diamond) + O2(g)
CO2(g)
G°rxn = -397 kJ
<<0
But diamonds are forever…. G°rxn ≠ rate
Example Problem
• Is the following reaction spontaneous under standard conditions?
4KClO3 (s) 
 3KClO4 (s)  KCl(s)

H°f (kJ/mol)
S° (J/mol.K)
KClO3(s)
-397.7
143.1
KClO4(s)
-432.8
151.0
KCl (s)
-436.7
82.6
Example Problem Solution
• Calulating H°rxn
4KClO3 (s) 
 3KClO4 (s)  KCl(s)
Hrxn = 3H f (KClO4 )  H f (KCl) - 4H f (KClO3 )

= 3(-432.8kJ)  (-436.7kJ) - 4(397.7kJ)
= -144kJ
• Calulating S°rxn

Srxn = 3S(KClO4 )  S(KCl) - 4S(KClO3 )
= 3(151.0 J K )  (82.6 J K ) - 4(143.1 J K )
= -36.8 J K
Example Problem Solution
• Calulating G°rxn
Grxn = Hrxn - TSrxn
(
= -144kJ - (298K ) -38.6 J
 1kJ 
K 1000J 
)
= -133kJ

G°rxn < 0 ;therefore, reaction is spontaneous.
Example Problem Continued
For what temperatures will this reaction be spontaneous?
Answer: For T in which Grxn < 0.
Grxn = H rxn - TSrxn
0 = H rxn - TSrxn
H rxn
=
Srxn
-133kJ
= 3446K = T


1kJ
-38.6 J 
K 1000J 
(
)
Spontaneous as long as T < 3446 K.
