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Free Energy and Equilibria
DG < 0 process is spontaneous
DG > 0 reverse process is spontaneous
DG = 0 no change is spontaneous: Equilibrium
Le Châtelier’s principle gives us some understanding of how the
equilibrium of a system changes when we perturb it.
"Any change in one of the variables that determines the state of a system in
equilibrium causes a shift in the position of equilibrium in a direction that
tends to counteract the change in the variable under consideration."
We are interested in how systems tend towards equilibrium and what
obstacles keep them out of equilibrium.
The free energy, or more correctly, the chemical potential is our measure
of roughly how far we have to go to come to equilibrium. It is a
potential energy of sorts.
Chemical Potential
The chemical potential can be used to give quantitative meaning to Le
Châtelier’s principle.
Chemical potential of component A, µA, is defined as the partial molar
Gibbs free energy:
 G 
 A   
n A T, P, n n
j
A
This is the change in G with respect to a infinitesimal change in the
amount of component A with all other parameters held constant.
It is essentially the free energy increase (or decrease) associated with
adding a little of A to the system.
Directionality of a Chemical Reaction
Consider a closed system of four components (A, B, C, and D) undergoing
a reversible chemical reaction:
aA  bB  cC  dD
dG  SdT  VdP   AdnA   BdnB  C dnC   DdnD
In a closed system,
At constant T and P,
dn A dn B
dn
dn

  c   d  d
a
b
c
d
dG    i dni
i
 (c C  d D  a A  b B )d
 a A  b B  c C  d D d
At equilibrium:
a A  b B  cC  d D
dn A  ad
dnB  bd
Example
H2O (l)
100 °C
H2O (g)
This is at constant T and P. Its reversible. So,
DG°vap= 0 = DH°vap-T DS°vap
Note that this implies at equilibrium:
H 2 O(l)   H 2 O(g)
Chemical Potential and Partial Pressure
We found last time that:
P 
GP2   GP1   nRT ln  2 
P1 
Defining the standard state:
Then,
and,
GP1 1atm  G
0
 P 
0
G P   G  nRT ln
1atm 
 G 
 A   
n A T, P, n n
j
so,
(T = constant)
   0  RT ln
A
 P 
1atm 
In a mixture, the chemical potential of A can thus be expressed in terms of
its partial pressure and its standard (pure) chemical potential:
 A   0A  RT ln
 PA 
1atm 
DG of Mixing
Consider the isobaric, isothermal mixing of two gases:
Gas A
at
P Atm
Gas B
at
P Atm
Gas A+B
at
P Atm (total)
DGmix  nA RT ln X A  nB RT ln XB
DGmix  TDSmix
DSmix  nA Rln X A  nB Rln X B
Xi 
ni
 nj
Mole Fraction component i
j
XA 
nA
nB
and X B 
n A  nB
nA  nB
Equilibrium Constant
The Haber process of nitrogen fixation:
N2 + 3H2
2NH3
DG  2  NH3   N 2  3 H 2
 0
PNH3   0
PN 2   0
PH 2 





  N  RT ln
 3H  RT ln 
 2 NH3  RT ln
2
2
1atm  
1atm  
1atm 

Rewrite letting Pi 
Pi
unitless pressure ratio. No units inside “ln”.
1atm

0
DG  2  0NH3   0N 2  3 H
 RT 2 ln PNH3  ln PN 2  3ln PH2
2
  P 2

NH3
0
 DG  RT ln 
3 
P
P
  N 2 H 2 
a
We have used the identity: aln x  ln x

Equilibrium Constant
At any temperature and pressure there exists and equilibrium state for
this reaction.
A combination of
PN2 , PH 2 , PNH3 such that DG = 0.
  P 2

NH3
0
DG  DG  RT ln 
3 
P
P
  N 2 H2 
  P 2

NH3
0
0  DG  RT ln 
3 
P
P
  N 2 H2 
  P 2

NH3
0
DG  RT ln 
3 
P
P
  N 2 H 2 
This equation holds for equilibrium values of PN2 , PH 2 , PNH3
Equilibrium Constant
  
   
 
eq 2
PNH
0
3
DG  RT ln 
  eq
eq
P
P
  N 2 H 2
3
Define equilibrium constant (at constant T and P):
K


PNeq
PHeq
2
eq
PNH
3
  
2
More generally:
0
DG  RT ln K
3
2




K
aA + bB
cD + dD
  
n
eq n
P
P
 i 
 i 
products
products
eq c
PC
eq a
PA
i
Most generally:
K
 Pi
reac tan ts

eq n i
i
Q
 Pi
reac tan ts

ni
eq d
PD
eq b
PB
Chemical Potential Example
Activation barrier

Reactants
Reactants
D
Products
D
Products
Chemical potential is a measure of the thermodynamic free energy.
It tells us what the equilibrium distribution of reactants and products must be
It does not tell us the kinetic rate.
Equilibrium Example
Oxidation of CO:
2 CO (g) + O2 (g)
2 CO2 (g)
The free energy change for this reaction is simply:
DG ° rxn= 2 DG°289(CO2) - 2 DG°289(CO) - DG°289(O2)
We can calculate this using numbers from the appendix:
DG ° rxn= 2 (-394.36) - 2( -137.17) - 0 (kJ/mol) = -514.38 kJ/mol
A relatively large negative number.
DG ° rxn= -RTlnK
K=
e514,380/RT=
e207
=
 
eq
eq 2
PO PCO 
eq
PCO
2
2
2
The equilibrium for this reaction lies far in favor of the products.
Large negative DG means the reaction goes forward with high probability.
Equilibrium Example
The conversion of CO and O2 to CO2 is
energetically favorable, but the reaction is slow
Reactants
(CO,O2) D, DG
Products (CO2)
1) We have to first break an O-O double bond
2) The resultant atoms of oxygen must
then react with CO.
Reaction coordinate
For other reactions like
2 H2(g) + O2(g)
2 H2O (l)
We need a catalyst (e.g. platinum surface), and ideally we wish to convert the
released energy to work.
Catalysts enable a reaction to take an alternative path, with lower activation barrier.
DH0 for some Noncovalent Interactions
Reaction
Interaction
DH0(kJ mol-1)
Na+(g) + Cl-(g) NaCl(s)
Ionic
NaCl(s)
Na+(aq) + Cl-(aq)
Ionic +ion-dipole
Ar(g)
Ar(s)
London (fluctuation dipole)
Acetone(g)
Acetone(l)
London-van der Waals
H3 C
H3 C
(permanent dipole)
2
O
O
H O
Hydrogen bond
H
H
CH3
H
H2NON
H
H OH2
H2NON
Hydrogen bond (aq)
H2NONH2
H
O(NH2)2
HOH (Urea (aq))
C3H6(l) + H2O(l)
(TSWP p. 98)
C3H6(aq)
Hydrophobic
-785
+4
-8
-30
-20
-5
-10
Protein Unfolding
Proteins have a native state. (Really, they tend to have a tight cluster of native
states.)
Denaturation occurs when heat or denaturants such as guanidine, urea or detergent
are added to solution. Also, the pH can affect folding.
When performing a denaturation process non-covalent interactions are broken.
Ionic, van der-Waals, dipolar, hydrogen bonding, etc.
Solvent is reorganized.
Protein Unfolding
heat
Let’s consider denaturation with heat. We can determine a great deal about the
nature of the protein from such a consideration.
The experimental technique we use for measuring thermodynamic changes here is
the differential scanning calorimeter.
Basic experiment: Add heat to sample, measure its temperature change.
Protein Unfolding
T1
T2
Heat
Protein
+
Solvent
Solvent
T1-T2
In differential scanning calorimetry you have two samples:
Your material of interest
Control
You put in an amount of heat to raise the temperature of the control at a constant
rate, then measure the rate of change in temperature of the other sample as a
function of the input heat.
This is a measure of the heat capacity!
Protein Unfolding
pH8.0
Bacillus stearothermophilus
Rabbit
E. coli
pH6.0
Data for glyceraldehyde-3-phosphate dehydrogenase.
Is the protein more stable at pH 8 or 6? Why is B. stear. more stable?
Protein Unfolding
We are given the following data for the denaturation of lysozyme:
DG° kJ/mol
DH° kJ/mol
DS° J/ K mol
TDS° kJ/mol
10
25
60
100
67.4
137
297
69.9
60.7
236
586
175
27.8
469
1318
439
-41.4
732
2067
771
°C
Where is the denaturation temperature?
What then is special about the temperature at which the denaturation is spontaneous?