Chemistry 331

1
Chemistry 331
Lecture 18 Gibbs Free Energy and Chemical
Potential
NC State University
2
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 dq dw dw -PdV (definition of
work) dq TdS (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. We will use this expression to
derive the P and T dependence of the free energy
functions.
3
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 substitute into 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) Once again we see why G is so
useful. Its natural variables are ones that we
commonly experience T and P.
4
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 of the free energy. At
constant temperature SdT 0 and dG VdP.
The integrated form of this equation is For
one mole of an ideal gas we have Note that we
have expressed G as a molar quantity Gm G/n.
5
The variation of the Gibbs energy with pressure
We can use the above expression to indicate the
free energy at some pressure P relative to the
pressure of the standard state P 1
bar. G0(T) is the standard molar Gibbs free
energy for a gas. As discussed above the
standard molar Gibbs free energy is the free
energy of one mole of the gas at 1 bar of
pressure. The Gibbs free energy increases
logarithmically with pressure. This is entirely
an entropic effect. Note that the 1 bar can be
omitted since we can write
6
The pressure dependence of G for liquids and
solids
If we are dealing with a liquid or a solid the
molar volume is more or less a constant as a
function of pressure. Actually, it depends on
the isothermal compressibility, k
-1/V(V/P)T, but k is very 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 not strongly
affected by pressure. The small value of k is
another way of saying the same thing. For our
purposes we can treat the volume as a constant
and we obtain
7
Question
What is the isothermal compressibility k
-1/V(V/P)T for an ideal gas? A. k -1/V B. k
1/P C. k 1/T D. k nRT/P
8
Question
What is the isothermal compressibility k
-1/V(V/P)T for an ideal gas? A. k -1/V B. k
1/P C. k 1/T D. k nRT/P
9
Question
Compare the following two equations for the
pressure dependence of the Gibbs free energy.
Which one applies to the formation of diamond
from graphite? A. B.
10
Question
Compare the following two equations for the
pressure dependence of the Gibbs free energy.
Which one applies to the formation of diamond
from graphite? A. B.
11
Systems with more than one component
Up to this point we have derived state functions
for 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. 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 do this using a quantity
called the chemical potential. The chemical
potential is nothing more than the molar
Gibbs free energy of a particular component.
Formally we write it this way
Rate of change of G as number of moles of i
changes with all other variables held constant.
12
Example a gas phase reaction
Lets consider a gas phase reaction as an
example. We will use a textbook example
N2O4 (g) 2 NO2 (g) We
know how to write the equilibrium constant for
this reaction. At constant T and P we will
write the total Gibbs energy as
We use the reaction stoichiometry to obtain the
factor 2 for NO2.
13
Definition of the Gibbs free energy change for
chemical reaction
We now define DrxnG This is DrxnG but it is
not DrxnGo! Note that we will use DrxnG and DG
interchangably. If we now apply the pressure
dependence for one component, to a
multicomponent system These two expressions
are essentially identical. The chemical
potential, mi, is nothing more than a molar free
energy.
14
Question
How should one think of the chemical potential, m
of component j? A. It is the potential energy
of that component B. It is a molar free energy of
that component C. It is potential entropy of that
component D. It is a molar entropy of that
component
15
Question
How should one think of the chemical potential, m
of component j? A. It is the potential energy
of that component B. It is a molar free energy of
that component C. It is potential entropy of that
component D. It is a molar entropy of that
component Single component Multiple
components each have a mj
16
Application of definitions to the chemical
reaction
We can write the Gibbs energy as and use the
chemical potentials to obtain the following
17
Note the significance of DG and DGo
The change DG is the change in the Gibbs energy
function. It has three possible ranges of
value DG lt 0 (process is spontaneous) DG 0
(system is at equilibrium) DG gt 0 (reverse
process is spontaneous) On the other hand DGo 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 later in the course. For
now, lets consider the system at equilibrium.
Equilibrium means DG 0 so
18
Question
Which statement is true at equilibrium if K
2? A. DG -RT ln(2), DGo 0 B. DG 0, DGo
-RT ln(2) C. DG DGo -RT ln(2) B. DG 0, DGo
0
19
Question
Which statement is true at equilibrium if K
2? A. DG -RT ln(2), DGo 0 B. DG 0, DGo
-RT ln(2) C. DG DGo -RT ln(2) B. DG 0, DGo
0
20
Temperature dependence of DGoThe vant Hoff
equation
We use two facts that we have derived to
determine the temperature dependence of the free
energy. Here are we are skipping some
derivations that are not often used and deriving
a very useful expression in biology. If
we plot ln(K) vs 1/T the slope is -DHo/R. This is
a useful expression for determining the standard
enthalpy change.
21
Example the formation of diamond
Graphite and diamond are two forms of carbon.
Given that the free energy of formation of
diamond is C(s, graphite) C(s, diamond)
DrGo 2.90 kJ/mol and the densities
are r(graphite) 2.26 and r(diamond)
3.51 calculate the pressure required to transform
carbon into diamond. Solution Graphite will be
in equilibrium with diamond when
22
Example the formation of diamond
Plugging the values we find
23
Protein folding exampleTwo state model
kf

• U F
• Unfolded Folded
• K F/U
• K ff/(1-ff)
• Fraction folded ff Fraction unfolded 1-ff

ku
24
Thermodynamic model

• ff 1/(1K)
• K e-DGo/RT
• ff 1/(1 e-DGo/RT)
• ff 1/(1 e-DHo/RTeDSo/R)
• The temperature at which the protein is 50
  folded can be defined as Tm the melt temperature.
  
• At Tm , DGo 0 or Tm DHo/DSo.

25
Equilibrium melt curves
o
o
Mostly folded
Mostly unfolded
Tm

• In this case Tm 300 K DHo/DSo

26
Vant Hoff plots
Slope -DHo/R

• The standard method for obtaining the reaction
  enthalpy is a plot of ln K vs. 1/T

27
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

28
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

29
Question

• What can you say about the enthalpy of reaction
  for the
• process shown in the figure below?
• A. The reaction is exothermic.
• B. The reaction is endothermic.
• C. The reaction is spontaneous.
• D. None of the above.

30
Question

• What can you say about the enthalpy of reaction
  for the
• process shown in the figure below?
• A. The reaction is exothermic.
• B. The reaction is endothermic.
• C. The reaction is spontaneous.
• D. None of the above.