Chemistry 433
Lecture 14
Gibbs Free Energy
and Chemical Potential
NC State University
The internal energy expressed
in terms of its natural variables
We can use the combination of the first and second laws to
derive an expression for the internal energy in terms of its
natural variables. If we consider a reversible process:
dU = δq + δw
δw = -PdV (definition of work)
δq = TdS
S (second
(
law rearranged))
Therefore, dU = TdS - PdV
This expression expresses the fact that the internal energy
U has a T when the entropy changes and a slope -P when
the volume changes.
changes We will use this expression to derive
the P and T dependence of the free energy functions.
The Gibbs energy expressed
in terms of its natural variables
To find the natural variables for the Gibbs energy we begin
with the internal energy:
dU = TdS - PdV
and
d substitute
b i
iinto:
dH = dU + PdV + VdP
to find:
dH = TdS + VdP (S and P are natural variables of enthalpy)
and using:
dG = dH - TdS - SdT
we find:
dG = -SdT + VdP (T and P are natural variables of G)
O
Once
again
i we see why
h G is
i so useful.
f l It
Its natural
t l variables
i bl
are ones that we commonly experience: T and P.
The variation of the Gibbs
energy with pressure
We shown that dG = VdP - SdT. This differential can be
used to determine both the pressure and temperature
dependence
p
of the free energy.
gy At constant temperature:
p
SdT = 0 and dG = VdP.
The integrated form of this equation is:
ΔG =
P2
VdP
P1
For one mole of an ideal gas we have:
ΔGm = RT
P2
P1
dP = RT ln P2
P1
P
Note that we have expressed G as a molar quantity Gm = G/n.
The variation of the Gibbs
energy with pressure
We can use the above expression
p
to indicate the
free energy at some pressure P relative to the pressure
of the standard state P = 1 bar.
Gm T = G 0 T + RT ln
P
1 bar
G0(T) is the standard molar Gibb’s free energy for a gas.
As discussed above the standard molar Gibb’s
G ’ ffree energy
is the free energy of one mole of the gas at 1 bar of pressure.
The Gibb’s free energy
gy increases logarithmically
g
y with p
pressure.
This is entirely an entropic effect.
Note that the 1 bar can be omitted since we can write:
RT ln
P
= RT ln P – RT ln 1 = RT ln P
1 bar
The pressure dependence of G for
liquids and solids
If we are dealing
g with a liquid
q
or a solid the molar volume
is more or less a constant as a function of pressure.
Actually, it depends on the isothermal compressibility,
κ = -1/V(∂V/∂P)
1/V(∂V/∂P)T, but κ is very small
small. It is a number of the
order 10-4 atm-1 for liquids and 10-6 atm-1 for solids. We
have discussed the fact that the density of liquids is
nott strongly
t
l affected
ff t d by
b pressure. The
Th smallll value
l off κ
is another way of saying the same thing.
For our p
purposes
p
we can treat the volume as a constant
and we obtain
Gm T = G 0 T + Vm P – 1
Question
What is the isothermal compressibility κ = -1/V(∂V/∂P)T for an
ideal gas?
A. κ = -1/V
B. κ = 1/P
C. κ = 1/T
D. κ = nRT/P
Question
What is the isothermal compressibility κ = -1/V(∂V/∂P)T for an
ideal gas?
A. κ = -1/V
B. κ = 1/P
C. κ = 1/T
D. κ = nRT/P
Question
Compare the following two equations for the pressure
dependence of the Gibbs free energy. Which one applies
t the
to
th formation
f
ti off diamond
di
d from
f
graphite?
hit ?
P
1 bar
A.
Gm T = G 0 T + RT ln
B.
Gm T = G 0 T + Vm P – 1
Question
Compare the following two equations for the pressure
dependence of the Gibbs free energy. Which one applies
t the
to
th formation
f
ti off diamond
di
d from
f
graphite?
hit ?
P
1 bar
A.
Gm T = G 0 T + RT ln
B.
Gm T = G 0 T + Vm P – 1
Systems with more than one
component
Up
p to this p
point we have derived state functions for p
pure
systems. (The one exception is the entropy of mixing).
However, in order for a chemical change to occur we must
have more than one component present
present. We need generalize
the methods to account for the presence of more than one
type of molecule. In the introduction we stated that we would
d thi
do
this using
i a quantity
tit called
ll d th
the chemical
h i l potential.
t ti l Th
The
chemical potential is nothing more than the molar Gibbs
free energy
gy of a p
particular component.
p
Formally
y we write it
this way:
Rate of change of G as number of
moles of i changes with all other
μ i = ∂G
∂n i T,P, j ≠ i
variables held constant.
Example:
p ag
gas p
phase reaction
Let’s consider a gas phase reaction as an example. We will
use a textbook
t tb k example:
l
N2O4 (g) = 2 NO2 (g)
We know how to write the equilibrium
q
constant for this reaction.
2
NO 2
P
K=
PN 2O4
At constant T and P we will write the total Gibbs energy as:
dG = μ NO 2dn NO 2 + μ N 2O 4dn N 2O4
dG = 2μ NO 2dn – μ N 2O 4dn
We use the reaction stoichiometry to obtain the factor 2 for NO2.
Definition of the Gibbs free energy
change for chemical reaction
We now define ΔrxnG:
Δ rxnG = dG
dn
T,P
This is ΔrxnG but it is not ΔrxnGo! Note that we will use ΔrxnG
and ΔG interchangably.
If we now apply the pressure dependence for one component
component,
Gm T = G 0 T + RT ln
P
1 bar
t a multicomponent
to
lti
t system:
t
μ i T = μ 0i T + RT ln
Pi
1 bar
These two expressions are essentially identical. The
chemical potential, μi, is nothing more than a molar free energy.
Question
How should one think of the chemical potential, μ
of component j?
A. It is the potential energy of that component
B It is
B.
i a molar
l free
f
energy off that
th t componentt
C. It is potential entropy of that component
D. It is a molar entropy of that component
Question
How should one think of the chemical potential, μ
of component j?
A. It is the potential energy of that component
B It is
B.
i a molar
l free
f
energy off that
th t componentt
C. It is potential entropy of that component
D. It is a molar entropy of that component
Single component:
Gm T = G 0 T + RT ln
P
1 bar
Multiple components each have a μj:
μ i T = μ 0i T + RT ln
Pi
1 bar
Application of definitions to the
chemical reaction
We can write the Gibbs energy as:
ΔG = 2μ NO 2 – μ N 2O4
and use the chemical p
potentials:
μ NO 2= μ 0NO 2 + RT ln PNO 2
μ N 2O4= μ 0N 2O4 + RT ln PN 2O4
to obtain the following:
ΔG = 2μ 0NO 2 – μ 0N 2O4 + 2RT ln PNO 2 – RT ln PN 2O4
2
P
NO 2
ΔG = ΔG 0 + RT ln
PN 2O4
, ΔG 0 = 2μ
2 0NO 2 – μ 0N 2O4
Note the significance
g
of ΔG and ΔGo
The change ΔG is the change in the Gibbs energy function.
It h
has th
three possible
ibl ranges off value:
l
ΔG < 0 (process is spontaneous)
ΔG = 0 ((system
y
is at equilibrium)
q
)
ΔG > 0 (reverse process is spontaneous)
On the other hand ΔGo is the standard molar Gibbs energy
change for the reaction. It is a constant for a given chemical
reaction. We will develop these ideas for a general reaction
l t iin th
later
the course. F
For now, let’s
l t’ consider
id th
the system
t
att
equilibrium. Equilibrium means ΔG = 0 so:
2
P
NO 2
0
0
0 = ΔG + RT ln K , ΔG = –RT ln K , K =
PN 2O4
Question
Which statement is true at equilibrium if K = 2?
A. ΔG = -RT
A
RT ln(2),
l (2) ΔGo = 0
B. ΔG = 0, ΔGo = -RT ln(2)
C. ΔG = ΔGo = -RT ln(2)
B. ΔG = 0, ΔGo = 0
Question
Which statement is true at equilibrium if K = 2?
A. ΔG = -RT
A
RT ln(2),
l (2) ΔGo = 0
B. ΔG = 0, ΔGo = -RT ln(2)
C. ΔG = ΔGo = -RT ln(2)
B. ΔG = 0, ΔGo = 0
Temperature dependence of ΔGo
The van
van’tt Hoff equation
We use two facts that we have derived to determine the
t
temperature
t
dependence
d
d
off the
th free
f
energy. Here
H
are we
are skipping some derivations that are not often used and
deriving a very useful expression in biology.
ΔG 0 = –RT ln(K)
ΔG 0 = ΔH 0 – TΔS 0
0
0
ΔH
ΔS
ln(K) = –
+
RT
R
0
ΔH
1 – 1
ln(K 2) – ln(K 1) =
R T1 T2
If we plot ln(K) vs 1/T the slope is -ΔHo/R. This is a useful
expression for determining the standard enthalpy change.
Example:
p the formation of diamond
Graphite and diamond are two forms of carbon. Given that
th ffree energy off fformation
the
ti off diamond
di
d is:
i
C(s, graphite)
C(s, diamond)
ΔrGo = + 2.90 kJ/mol
and the densities are:
ρ(graphite) = 2.26 and ρ(diamond) = 3.51
calculate the pressure required to transform carbon into
diamond.
diamond
Solution: Graphite will be in equilibrium with diamond when:
0 = ΔG + ΔVm P – 1
0
P = 1 – ΔG = 1 – ΔG
M –M
ΔVm
ρ d ρ gr
0
0
Example:
p the formation of diamond
Plugging the values we find:
0 = ΔG 0 + ΔVm P – 1
0
0
ΔG
ΔG
P=1–
=1–
M –M
ΔVm
ρ d ρ gr
2900 J/mol
J/ l
=1–
0.012 kg/mol 1 – 1
3510 2260
9
= 1.5 x 10 Pa = 15000 bar
Protein folding example:
T
Two
state
t t model
d l
kf
U
F
ku
Unfolded
Folded
K = [F]/[U]
K = ff/(1-ff)
( )
Fraction folded ff Fraction unfolded 1-ff
Thermodynamic
y
model
ff = 1/(1+K)
o/RT
-ΔG
ΔG
K=e
o/RT
-ΔG
ff = 1/(1 + e
)
o/RT ΔSo/R
-ΔH
ff = 1/(1 + e
e
)
The temperature at which the protein
is 50% folded can be defined as Tm
the melt temperature.
o
o
o
At Tm , ΔG = 0 or Tm = ΔH /ΔS .
Equilibrium
q
melt curves
o
Mostly folded
o
Mostly unfolded
Tm
In this case: Tm = 300 K = ΔH /ΔS
o
o
Van’t Hoff plots
p
Slope = -ΔH /R
o
The standard method for obtaining the
reaction enthalpy is a plot of ln K vs. 1/T
Question
What determines the steepness of the equilibrium melt
curve?
A. The melt temperature
B. The equilibrium constant
C. The ratio of the enthalpy to the entropy
D. The magnitude of the enthalpy and entropy
Question
What determines the steepness of the equilibrium melt
curve?
A. The melt temperature
B. The equilibrium constant
C. The ratio of the enthalpy to the entropy
D. The magnitude of the enthalpy and entropy
Question
What can you say about the enthalpy of reaction for the
process shown in the figure
p
g
below?
A. The reaction is exothermic.
B. The reaction is endothermic.
C. The reaction is spontaneous.
D. None of the above.
Question
What can you say about the enthalpy of reaction for the
process shown in the figure
p
g
below?
A. The reaction is exothermic.
B. The reaction is endothermic.
C. The reaction is spontaneous.
D. None of the above.