to enhance the knowledge about phase transformation, and clear the basics of energy of reaction .

1. Gibbs Free Energy
Gibbs free energy is a measure of chemical energy
All chemical systems tend naturally toward states
of minimum Gibbs free energy
G = H - TS
Where:
G = Gibbs Free Energy
H = Enthalpy (heat content)
T = Temperature in Kelvins
S = Entropy (can think of as randomness)

2. • Products and reactants are in equilibrium when their Gibbs
free energies are equal
• A chemical reaction will proceed in the direction of lower
Gibbs free energy (i.e., DGr < 0)
…so the reaction won’t proceed if the reaction produces
an increase in Gibbs free energy
Gibbs Free Energy

3. formation
of
energy
free

D o
f
G
Gibbs Free Energy
DG°r = SnG°f (products) - SnG°f (reactants)
DG°r > 0, backwards reaction with deficient energy
DG°r < 0, forwards reaction with excess energy
DG°r = 0, reaction is in equilibrium
DG°r is a measure of the driving force

4. Thermodynamics
For a phase we can determine V, T, P, etc., but not G or H
We can only determine changes in G or H as we change some other
parameters of the system
Example: measure DH for a reaction by calorimetry - the heat
given off or absorbed as a reaction proceeds
Arbitrary (based on random choice) reference state and assign an
equally arbitrary value of H to it:
Choose 298.15 K/25°C and 0.1 MPa/1 atm/1 bar (lab conditions)
...and assign H = 0 for pure elements (in their natural state - gas,
liquid, solid) at that reference

5. Thermodynamics
In our calorimeter we can then determine DH for the reaction:
Si (metal) + O2 (gas) = SiO2 DH = -910,648 J/mol
= molar enthalpy of formation of quartz (at 25°C, 1 atm)
It serves quite well for a standard value of H for the phase
Entropy has a more universal reference state:
entropy of every substance = 0 at 0K, so we use that
(and adjust for temperature)
Then we can use G = H - TS to determine G for
quartz = -856,288 J/mol

6. Thermodynamics
K
RT
Go
R ln


D
K=equilibrium constant at standard T
T in kelvin 298.18K
R=gas constant=1.987 cal/molo
K
Go
R log
364
.
1


D
364
.
1
10
o
R
G
K
D



7. Example: What is the DGo
R of calcite dissociation?
Use data in appendix B for DGo
f
DGo
R = [(-132.3)+(-126.17)] - [(-269.9)] = +11.43 kcal
(+) means that the reaction goes from right to left
so K must be small
What is the value of K?
364
.
1
10
o
R
G
K
D


K = 10(-11.43/1.364) = 10-8.3798 = 4.171 x 10-9
CaCO3 Ca2+ + CO3
2-

8. What if T  25oC? Use the Van’t Hoff Equation







D


15
.
298
1
1
3025
.
2
log
log
T
R
H
K
K
o
R
o
T
T
o
R
H
D Enthalpy of reaction
R=1.987 cal/mol°
T in Kelvin
K
RT
Go
R ln


D DG°r = DH°r-TDS°r
and
lnKT - lnKT° = (-DH°r/R)(1/T-1/T°)
We can derive:

9. Example: What is KT of calcite dissociation at T=38°C?







D


15
.
298
1
1
3025
.
2
log
log
T
R
H
K
K
o
R
o
T
T
o
R
H
D = [(-129.74)+(-161.8)] - [(-288.46)] = -3.08
10
9
10
85
.
8
0532
.
9
15
.
298
1
311
1
)
987
.
1
(
3025
.
2
08
.
3
)
10
71
.
4
log(
log















x
K
x
K
T
T
When T increases, K decreases
(KT° = 4.171 x 10-9)

10. Thermodynamics
Summary thus far:
– G is a measure of relative chemical stability for a phase
– We can determine G for any phase by measuring H and S for
the reaction creating the phase from the elements
– We can then determine G at any T and P mathematically
• Most accurate if know how V and S vary with P and T
– dV/dP is the coefficient of isothermal compressibility
– dS/dT is the heat capacity (Cp)
If we know G for various phases, we can determine which is
most stable
• Why is melt more stable than solids at high T?
• Is diamond or graphite stable at 150 km depth?
• What will be the effect of increased P on melting?

11. Thanks!
Any questions?
You can find me at
● @hamzaahmed
● hamzaahmed0696@mail.me