Lecture 8: Gibbs Free Energy
• Reading: Zumdahl 10.7, 10.9
• Outline
– Defining the Gibbs Free Energy (DG)
– Calculating DG
– Pictorial Representation of DG
The Second Law
• The Second Law: there is always an
increase in the entropy of the universe.
• From our definitions of system and
surroundings:
DSuniverse = DSsystem + DSsurroundings
The Second Law (cont.)
• Three possibilities:
– If DSuniv > 0…..process is spontaneous
– If DSuniv < 0…..process is spontaneous in opposite
direction.
– If DSuniv = 0….equilibrium
• We need to know DS for both
the system and surroundings to predict if a
reaction will be spontaneous!
Defining DG
• Recall, the second law of thermodynamics:
DSuniv = DStotal = DSsystem + DSsurr
• Also recall:
DSsurr = -DHsys/T
• Then, DStotal = DSsystem +-DH
DSsurr
sys/T
-TDStotal = -TDSsystem+
DHsys
Defining DG (cont.)
• We then define:
DG = -TDStotal
• Then:
DG = -TDSsys+ DHsys
• Or:
DG = DH - TDS
DG = The Gibbs Free Energy w/ Pconst
DG and Spontaneous Processes
• Recall from the second law the condititions
of spontaneity:
• Three possibilities:
– If DSuniv > 0…..process is spontaneous
– If DSuniv < 0…..process is spontaneous in opposite
direction.
– If DSuniv = 0….equilibrium
• In our derivation of DG, we divided by -T;
therefore, the direction of the inequality
changes relative to entropy.
DG and Spontaneous Processes
(cont.)
• Three possibilities:
– If DSuniv > 0…..process is spontaneous
– If DSuniv < 0…..process is spontaneous in opposite
direction.
– If DSuniv = 0….equilibrium
• In terms of DG:
– If DG < 0…..process is spontaneous
– If DG > 0…..process is spontaneous in opposite
direction.
– If DG = 0….equilibrium
DG and Spontaneous Processes
(cont.)
• Note that DG is composite of both DH and DS
DG = DH - TDS
• A reaction is spontaneous if DG < 0. Such
that:
If DH < 0 and DS > 0….spontaneous at all T
If DH > 0 and DS < 0….not spontaneous at all T
If DH < 0 and DS < 0….spontaneous at low T
If DH > 0 and DS > 0….spontaneous at high T
Example
• At what T is the following reaction
spontaneous?
Br2(l)
Br2(g)
where DH° = 30.91 kJ/mol, DS° = 93.2 J/mol.K
• Ans:
DG° = DH° - TDS°
Example (cont.)
• Try 298 K just to see:
DG° = DH° - TDS°
DG° = 30.91 kJ/mol
- (298K)(93.2 J/mol.K)
DG° = (30.91 - 27.78) kJ/mol
= 3.13 kJ/mol > 0
Not spontaneous at 298 K
Example (cont.)
• At what T then?
DG° = DH° - TDS° = 0
T = DH/DS
T = (30.91 kJ/mol) /(93.2 J/mol.K)
T = 331.65 K
Just like our previous calculation
Calculating DG°
• In our previous example, we needed to
determine DH°rxn and DS°rxn to determine
DG°rxn
• Now, DG is a state function; therefore, we
can use known DG° to determine DG°rxn
using:
DGrxn =  DGprod. - DGreact.
Standard DG of Formation: DGf°
• Like DHf° and S°, DGf° 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.”
• Like DHf°, DGf° = 0 for an element in its
standard state:
Example: DGf° (O2(g)) = 0
Example
• Determine the DG°rxn for the following:
C2H4(g) + H2O(l)
C2H5OH(l)
• Tabulated DG°f from Appendix 4:
DG°f(C2H5OH(l)) = -175 kJ/mol
DG°f(C2H4(g)) = 68 kJ/mol
DG°f(H2O (l)) = -237 kJ/mol
Example (cont.)
• Using these values:
C2H4(g) + H2O(l)
C2H5OH(l)
DGrxn =  DGprod. - DGreact.
DG°rxn = DG°f(C2H5OH(l)) - DG°f(C2H4(g))
-DG°f(H2O (l))
DG°rxn = -175 kJ - 68 kJ -(-237 kJ)
DG°rxn = -6 kJ < 0 ; therefore, spontaneous
More DG° Calculations
• Similar to DH°, one can use the DG° for
various reactions to determine DG° for the
reaction of interest (a “Hess’ Law” for DG°)
• Example:
C(s, diamond) + O2(g)
CO2(g) DG° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) DG° = -394 kJ
More DG° Calculations (cont.)
C(s, diamond) + O2(g)
CO2(g) DG° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) DG° = -394 kJ
CO2(g)
C(s, graphite) + O2(g) DG° = +394 kJ
C(s, diamond)
C(s, graphite)
DG° = -3 kJ
DG°rxn < 0…..rxn is spontaneous
DG°rxn ≠ Reaction Rate
• Although DG°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)
DG°rxn = -397 kJ
<<0
But diamonds are forever…. DG°rxn ≠ rate