Gibbs Free Energy.ppt

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., DG
r< 0)
…so the reaction won’t proceed if the reaction produces
an increase in Gibbs free energy
Gibbs Free Energy

3. formation ofenergy free D
o
fG 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) + O
2(gas) = SiO
2 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. ThermodynamicsKRTG
o
R lnD
K=equilibrium constant at standard T
T in kelvin 298.18K
R=gas constant=1.987 cal/mol
oKG
o
R log364.1D 364.1
10
o
R
G
K
D


7. Example: What is the DG
o
Rof calcite dissociation?
Use data in appendix B for DG
o
f
DG
o
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
CaCO
3 Ca
2+
+ CO
3
2-

8. What if T 25
o
C? Use the Van’t Hoff Equation






D

15.298
11
3025.2
loglog
TR
H
KK
o
Ro
TT o
RHD
Enthalpy of reaction
R=1.987 cal/mol°
T in KelvinKRTG
o
R lnD
DG°
r= DH°
r-TDS°
r
and
lnK
T-lnK
T°= (-DH°
r/R)(1/T-1/T°)
We can derive:

9. Example: What is K
Tof calcite dissociation at T=38°C?






D

15.298
11
3025.2
loglog
TR
H
KK
o
Ro
TT o
RHD
= [(-129.74)+(-161.8)] -[(-288.46)] = -3.0810
9
1085.8
0532.9
15.298
1
311
1
)987.1(3025.2
08.3
)1071.4log(log












xK
xK
T
T
When T increases, K decreases
(K
T°= 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
●[email protected]