cathodic protection design

1. 1
Gibbs Free Energy, G
Multiply through by -T
-T∆Suniv = ∆Hsys - T∆Ssys
-T∆Suniv = change in Gibbs free energy
for the system = ∆Gsystem
Under standard conditions —
∆Go
sys = ∆Ho
sys - T∆So
sys
Suniv =
Hsys
T
+ Ssys
∆Suniv = ∆Ssurr + ∆Ssys

2. 2
∆Go = ∆Ho - T∆So
Gibbs free 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 product-
favored).

3. 3
∆Go = ∆Ho - T∆So
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).

4. 4
Gibbs Free Energy, G
∆Go = ∆Ho - T∆So
∆Ho ∆So ∆Go Reaction
exo(–) increase(+) – Prod-favored
endo(+) decrease(-) + React-favored
exo(–) decrease(-) ? T dependent
endo(+) increase(+) ? T dependent

5. 5
Gibbs Free Energy, G
∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆Ho
rxn and ∆So
rxn and use
GIbbs equation.
b) Use tabulated values of free energies
of formation, ∆Gf
o.
∆Go
rxn =  ∆Gf
o (products) -  ∆Gf
o (reactants)

6. 6
Free Energies of Formation
Note that ∆G˚f for an element = 0

7. 7
Calculating ∆Go
rxn
Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Ho
rxn = -1238 kJ
Use standard molar entropies to calculate
∆So
rxn = -97.4 J/K or -0.0974 kJ/K
∆Go
rxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
Reaction is product-favored in spite of negative
∆So
rxn.
Reaction is “enthalpy driven”

8. 8
Calculating ∆Go
rxn
Is the dissolution of ammonium nitrate product-
favored?
If so, is it enthalpy- or entropy-driven?
NH4NO3(s) + heat ---> NH4NO3(aq)

9. 9
Calculating ∆Go
rxn
From tables of thermodynamic data we find
∆Ho
rxn = +25.7 kJ
∆So
rxn = +108.7 J/K or +0.1087 kJ/K
∆Go
rxn = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ
Reaction is product-favored in spite of negative
∆Ho
rxn.
Reaction is “entropy driven”
NH4NO3(s) + heat ---> NH4NO3(aq)

10. 10
Gibbs Free Energy, G
∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆Ho
rxn and ∆So
rxn and use
GIbbs equation.
b) Use tabulated values of free energies
of formation, ∆Gf
o.
∆Go
rxn =  ∆Gf
o (products) -  ∆Gf
o (reactants)

11. 11
Calculating ∆Go
rxn
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
∆Go
rxn = ∆Gf
o(CO2) - [∆Gf
o(graph) + ∆Gf
o(O2)]
∆Go
rxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
∆Go
rxn = -394.4 kJ
Reaction is product-favored as expected.
∆Go
rxn =  ∆Gf
o (products) -  ∆Gf
o (reactants)

12. 12
Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
∆Ho
rxn = +467.9 kJ ∆So
rxn = +560.3 J/K
∆Go
rxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does ∆Go
rxn just change from
being (+) to being (-)?
When ∆Go
rxn = 0 = ∆Ho
rxn - T∆So
rxn
T =
Hrxn
Srxn
=
467.9 kJ
0.5603 kJ/K
= 835.1 K

13. 13
More thermo? You betcha!

14. 14
 FACT: ∆Go
rxn 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.
Thermodynamics and Keq

15. 15
Keq is related to reaction favorability and so
to ∆Go
rxn.
The larger the value of K the more negative
the value of ∆Go
rxn
∆Go
rxn = - RT lnK
where R = 8.31 J/K•mol
Thermodynamics and Keq

16. 16
Calculate K for the reaction
N2O4 --->2 NO2 ∆Go
rxn = +4.8 kJ
∆Go
rxn = +4800 J = - (8.31 J/K)(298 K) ln K
∆Go
rxn = - RT lnK
lnK = -
4800 J
(8.31 J/K)(298K)
= - 1.94
Thermodynamics and Keq
K = 0.14
When ∆Go
rxn > 0, then K < 1

17. 17
∆G, ∆G˚, and Keq
• ∆G is change in free energy at non-
standard 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

18. 18
∆G, ∆G˚, and Keq
Figure 19.10

19. 19
• Product favored
reaction
• –∆Go and K > 1
• In this case ∆Grxn is
< ∆Go
rxn , so state with
both reactants and
products present is
MORE STABLE than
complete conversion.
∆G, ∆G˚, and Keq

20. 20
Product-favored
reaction.
2 NO2 ---> N2O4
∆Go
rxn = – 4.8 kJ
Here ∆Grxn is less than
∆Go
rxn , so the state
with both reactants
and products
present is more
stable than complete
conversion.
∆G, ∆G˚, and Keq

21. 21
Reactant-favored
reaction.
N2O4 --->2 NO2
∆Go
rxn = +4.8 kJ
Here ∆Go
rxn is greater
than ∆Grxn , so the
state with both
reactants and
products present is
more stable than
complete conversion.
∆G, ∆G˚, and Keq

22. 22
 Keq is related to reaction favorability.
 When ∆Go
rxn < 0, reaction moves
energetically “downhill”
 ∆Go
rxn is the change in free energy when
reactants convert COMPLETELY to
products.
Thermodynamics and Keq