Chemistry 431
Lecture 27
The Ensemble Partition Function
Statistical Thermodynamics
NC State University
Representation of an Ensemble
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
N,V,T
An ensemble is a collection of systems that have certain
Thermodynamic variables fixed. The most common Ensemble is the canonical ensemble, which has fixed
N = number of particles, V = volume, T = temperature
Ensemble Partition Function
We distinguish here between the partition function of the ensemble, Q and that of an individual molecule, q. Since Q represents a sum over all states accessible to the system it can written as
Q(N,V,T) =
e
Σ
i,j,k,...
– β(ε i + ε j + ε k + ...)
where the indices i,j,k, represent energy levels of different particles. Proof of the Boltzmann
distribution
Consider a collection of systems at constant N, V, and T. We call this collection an ensemble. The energy and quantum states of the systems can vary. There are A total systems and a1 systems with energy E1, a2
systems with energy E2 etc. We are interested in knowing the probability that the system has energies E1, E2, etc. The relative number must depend on the energy so we can write: a2/a1 = f(E1,E2)
which says that the ratio of populations is some function of E1 and E2. Proof of the Boltzmann
distribution
However, since there is always a zero of energy (i.e. the ground state) the energy function can be written in terms of the energy difference between E1 and E2.
f(E1,E2) = f(E1 – E2)
Hence,
a2/a1 = f(E1 – E2)
This must be true for any pair of states.
a3/a2 = f(E2 – E3) and a3/a1 = f(E1 – E3)
and since
a3/a1 = (a3/a2) (a2/a1)
Proof of the Boltzmann
distribution
we have
f(E1 – E3) = f(E1 – E2) f(E2 – E3)
f must therefore be an exponential function.
ex+y = ex ey
so in general f(E) = eβE where β is an arbitrary constant.
eβ(E1 – E3) = eβ(E1 – E2) eβ(E2 – E3)
Thus, a2/a1 = eβ(E1 – E2) or in general an/am = eβ(Em – En)
Proof of the Boltzmann
distribution
This implies that both am and an are given by
aj = Ce‐βEj
(I)
where C is an constant.
Σj a j = CΣj e – βE
and since Σjaj = A we have
A = CΣ e – βE j
j
Since C=
A
Σ e – βE j
j
j
Proof of the Boltzmann
distribution
We can substitute into equation I above to obtain
– βE j
aj
e
=
A Σ e – βE j
j
Using the definition for the probability pj = aj/A we can calculate pj as
– βE j
e
pj =
Σ e – βE j
j
• The molecular partition function, q represents the energy levels of one individual molecule. We can rewrite the above sum as
• Q = qiqjqk… or Q = qN for N particles. Note that qi means a sum over states or energy levels accessible to molecule i and qj means the same for molecule j. q(V,T) = Σ e – βε i
i
• The molecular partition function counts the energy levels accessible to molecule i only.
• Q counts not only the states of all of the molecules, but all of the possible combinations of occupations of those states. However, if the particles are not distinguishable then we will have counted N! states too many (N! = N(N‐1)(N‐2)….). This factor is exactly how many times we can swap the indices in Q(N,V,T) and get the same value (again provided that the particles are not distinguishable). If we consider 3 particles we have i,j,k j,i,k, k,i,j k,j,i j,k,i i,k,j or 6 = 3!. • Thus we write the partition function as
Q = q N for distinguishable particles
qN
for indistinguishable particles
Q=
N!
Translational Partition Function
The translational partition function is the most important one for statistical thermodynamics. Pressure is caused by translational motion, i.e., momentum exchange with the walls of a container. For this reason it is important to understand the origin of the translational partition function.
Translational energy levels are so closely spaced as that they are essentially a continuous distribution. The quantum mechanical description of the energy levels is obtained from the quantum mechanical particle in a box. Stirling’s Approximation
Note that the logarithm of the partition function is important in many calculations. Since the partition function of indistinguishable particles is qN/N! we need to be able to calculate ln(N!). Stirling’s approximation states that lnN! = NlnN. The derivation is given in the attached supplementary informtion.
Particle‐in‐a‐box energy levels
• The energy levels are
2
h
2
2
2
εnx,ny,nz =
n
+n
+n
y
z
2 x
8ma
nx,ny,nz =1, 2,...
• The box is a cube of length a. The average quantum numbers will be very large for a typical molecule. This is very different than what we find for vibration and electronic levels where the quantum numbers are small (i.e. only one or a few levels are populated). Many translational levels are populated thermally.
The translational partition function is
∞
trans
q = Σ e–εnx,ny,nz/kT
n ,n ,n
x y z =1
∞
∞
∞
2
h
2
2
2
= Σ Σ Σ exp –
n
+n
+n
8ma2kT x y z
nx =1 ny =1 nz =1
The three summations are identical and so they can be
written as the cube of one summation
3
q
trans
∞
2 2
n
h
= Σ exp –
n=1
8ma 2kT
The fact that the energy levels are essentially continuous
and that the average quantum number is very large allows
us to rewrite the sum as an integral.
3
∞
q trans =
0
2 2
n
h
exp –
dn
8ma 2kT
The translational partition function is proportional to volume
The sum started at 1 and the integral at 0. This difference
is not important if the average value of n is ca. 109! If we
have the substitution a = h2/8ma2kT we can rewrite the
integral as
3
∞
q trans =
e
– αn 2
dn =
0
π
4α
1/2 3
This is a Gaussian integral. The solution of Gaussian
integrals is discussed the math section of the Website.
If we now plug in for a and recognize that the volume of
the box is V = a3 we have
q trans = 2πmkT
2
h
3/2
V
The molecular partition function
The total energy of a molecule is a sum of the energies of
the individual motions and electronic states.
ε tot = ε trans + ε rot + ε vib + ε elec
Therefore the partition function that includes all of the
quantum states of the molecule is:
∞
q =Σe
tot
i=1
∞
=Σe
– ε tot
i /kT
–
=q
trans
=Σe
ε trans
/kT
i
i=1
∞
∞
–
i=1
Σe
–
ε rot
j /kT
j=1
rot
vib
elec /kT
ε trans
+ ε rot
i
i + εi + εi
vib
q q q
∞
Σe
k=1
elec
–
ε vib
k /kT
∞
Σe
l=1
–
ε elec
/kT
l
Probability in the ensemble
The ensemble partition function is:
∞
Q
= iΣ= 0 e – Ei /kT
Where the ensemble energy is EJ. The population of a
particular state J with energy EJ is given by
– E J / kT
e
e
pJ = ∞
=
– E J / kT
Q
Σ
e
J=0
– E J / kT
This known as the Boltzmann distribution. The
normalization constant of the above probability is 1/Q.
The sum of all of the probabilities must equal 1.
Calculation of average properties
The importance of the canonical ensemble is evident once
we begin to calculate average thermodynamic quantities.
The basic approach is to sum over the probability of a state
being occupied times the value of the property in a given state.
In general, for an average property M we can write
< M >=
Σ PM
j
j
j
M could be energy or pressure etc. Pj is the Boltzmann
probability given by Pj = e−βεj/Q.
Average energy
If we denote the average energy <E> then
∞
∞
E =
Σ
J=0
PJ E J
=
Σ
J=0
E J e – βE J
∞
– βE J
Σ
e
J =0
This can be written compactly as
E =–
∂ln Q
∂β
Consistency check with kinetic
theory of gases
The average energy per molecule is given by
ε trans
∂lnqtrans
=–
∂β
= kT
2
= kT
V
2
∂lnqtrans
∂T
V
∂ 3 lnT + terms independent of T
∂T 2
V
= 3 kT 2 1 ∂ T
T ∂T
2
= 3 kT
2
which agrees with the kinetic theory of gases. The
second step follows from the fact that
ln(abc) = ln(a) + ln(b) + ln(c).
We can rewrite the logarithm as a sum. The terms that
do not depend on temperature will vanish.
Statistical entropy in the
canonical ensemble
The canonical ensemble can also be called the (N,V,T)
ensemble. Note that energy is not a constant in this
ensemble. The entropy has two contributions in this
ensemble. One is from the internal energy:
U – U0
S=
T
and the second is statistical:
S =k ln Q
Thus, when we put them together we have:
U – U0
S=
+ k lnQ
T
Microcanonical Ensemble
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,E
N,V,e
N,V,E
N,V,E
N,V,E
The microcanonical ensemble is an ensemble of closed
systems. Therefore, the fixed thermodynamic variables are N = number of particles, V = volume, E = energy.
If there are Ω microstates then the probability of being
In a given microstate is 1/Ω.
Statistical entropy
The entropy does not depend at all on the energy in the
microcanonical ensemble. One can understand this since
the energy is the same in each ensemble. Therefore, in an
ensemble of Ω systems the energy purely statistical,
S = R ln Ω
We will use this entropy in solids at low temperature and for
problems protein folding.
Conformational entropy
The entropy of a polymer or a protein depends on the number
of possible conformations. This concept was realized first more
than 100 years ago by Boltzmann. The entropy is proportional
to the natural logarithm of the number of possible
conformations, W.
S = R ln W
For a polymer W = MN where M is the number of possible
conformations per monomer and N is the number of monomers.
For a typical polypeptide chain in the unfolded state M could be
a number like 6 where the conformations include different φ,ψ
angles and side chain angles. On the other hand, when the
protein is folded the conformational entropy is reduced to W = 1
in the theoretical limit of a uniquely folded structure. Thus,
we can use statistical considerations to estimate the entropy
barrier to protein folding.
Multinomial distribution
The total number of ways that a group of N objects can be
arranged in M different categories is MN.
The multinomial coefficient applies to a situation where there
are more than two groups. In general, if there are M
different groups that we can place N objects in we have
as the total number of ways
W = N!
Πn
M
m=1
m
!
where Π is the product operator indicating that there are
M terms multiplied together in the denominator. The values
of the nm are the numbers of objects in each of the groups.
Clearly, the nm must sum up to the total number of objects.
Σ
M
m=1
nm = N
Permutation of Letters
We have already seen that the permutations of the
indices i,j,k, etc. gives rise to N! different combinations.
The permutation of indices in the sum over Boltzmann
factors lead to the factor of N! in the partition function
Q = qN/N! for indistinguishable particles. In this application
we are assuming that all of the indices are unique and
therefore the number of ways of arranging them is given by
W=
N!
1!1!1!1!....
i.e. there is only one index in each group.
Suppose we asked how many ways there are of arranging
the letters in the word MINIMUM. There are three Ms and
two Is with a total of seven letters. In this case then there
are four groups, the group of M, N, I, and U. The total
number of ways is
4⋅5⋅6⋅7
7!
W=
3!2!1!1
=
2
= 420
Application to a game of chance
This is discussion is based on a game called YAHTZEE.
On each turn you role five dice. You obtain points based
on the configuration of the dice. The following point scheme
is made in the instructions.
50 = all five dice have the same number (YAHTZEE)
30 = full house (two of and kind with three of kind)
40 = long straight (five sequential numbers)
12 = three of a kind (with any two not paired)
We define a configuration as {[1],[2],[3],[4],[5],[6]} where
[1] is number of ones etc. rolled on a particular cast of the dice.
Consider how many ways there are to get a YAHTZEE of all
ones. In this case each of the five die must have the same
number. The configuration is {5,0,0,0,0,0} and the number
of ways is given by 5!/5!0!0!0!0! = 1.
Application to a game of chance
It is obvious that there is only one way to get all ones.
But what about a full house? (three of one kind and two of another)
We must be specific since there are a number of full houses
possible (how many?). Let us take two twos and three sixes.
The configuration is {0,2,0,0,0,3} and the number of ways this
can be achieved is 5!/0!2!0!0!0!3! = 120/(2x6) = 10.
There are two possible long straights. The configuration
is {1,1,1,1,1,0} and the number of ways this can be
achieved is 5!/1!1!1!1!0! = 120.
The overall probability of obtaining any configuration is given
by W for that configuration compared to the total number of
configurations MN. For YAHTZEE is 65 = 7776. Therefore the
Probability of getting a long straight is 2*120/7776 = 0.03.
Protein folding example:
Two state model
kf
U
F
ku
Unfolded
Folded
K = [F]/[U]
K = ff/(1-ff)
Fraction folded ff Fraction unfolded 1-ff
The Levinthal Paradox
Energy
The Levinthal paradox
assumes that all of the
possible conformations
will be sampled with
equal probability until
the proper one (N = native)
is found. Thus, the funnel
surface looks like a hole
in a golf course. The paradox
Conformation
states that if a protein samples
all 6M conformations it will take a time longer than the age of
the universe to find the native fold (N). Note that the statistical
entropy is S = RlnW = R ln(6M) = Mln(6)R in this example.
M = number of residues, 6 = number of conformations per residue
Two-state folding mechanism
The Levinthal view reflects the idea that protein folding would
occur as a result of a two state mechanism:
U ⇔N
This view suggests that the cooperativity of protein folding,
i. e. the sharp transitions between unfolded and native, leads to
an all-or-none mechanism. This means that there are no
intermediates on the folding pathway. Because folding studies
were rooted in chemistry, there has always been a strong
background notion of analogy between a chemical reaction with
its intermediates and transition states. However, this viewpoint
suggests that the protein must hit a “hole-in-one” as indicated
by Levinthal-type funnel.
The Pathway Model
Energy
Imagine that the a unique
pathway winds through
the surface to the hole.
The path starts at A and
the folding goes through
a unique and well-defined
set of conformational
changes. Here the entropy
must decrease rapidly since
the number of degrees of
Conformation
freedom in the folding pathway is quite
small compared to 6M. The funnel diagram is an energy diagram.
The configurational entropy is implied by the width of the funnel.
In a free energy diagram the folding pathway would have a high
barrier due to the large entropy change required for a unique path.
BPTI folding pathways
BPTI has three disulfides (30-51, 5-55 and 14-38). Reduction of
the DS's thus gives 6 CysSH. Reoxidation is performed with a
mixture of oxidized and reduced glutathione, GSSG and GSH,
under alkaline conditions. Creighton used rapid quenching with
acid or excess iodoacetate to trap the remaining free SH groups,
and separate the resulting labeled protein by chromatographic
methods. The major findings were that there was one dominant
one-disulfide intermediate, out of the 15 possible ones, which
contained the native 30-51 DS. All subsequent intermediates
contained this DS. What was striking about Creighton's results
was that the subsequent two-disulfide species favored two
non-native second disulfide bonds. This led to idea of trapped
misfolded states. However, this view has been challenged in
BPTI by Kim et al.
BPTI: evidence for folding pathways
One piece of evidence for folding pathways comes from trapping
disulfide intermediates. This method was pioneered by Creighton
using BPTI (and has only been used on a couple of other proteins)
Creighton et al. Prog. Biophys. Mol. Biol. 33 231, 1978
BPTI folding pathways
Kim et al. showed that some of the
previous data and interpretations
were wrong. The major 2-DS
species contains the two native DS's,
30-51 and 5-55. The third disulfide
is formed quite slowly because it
is quite buried. It is possible to
isolate a stable species with only the
first two disulfides formed and the
third remaining in the reduced form.
Studies with a mutant in which the
third DS was replaced by 2 Ala, and
which folded at a similar rate to the
wild type, support the idea that the
trapped disulfide species have partial
native-like structure.
Kim (1993) Nature 365 185
The molten globule
Compact intermediate states: Investigations in the 80's on
a-lactalbumin mostly by Kuwajima and coworkers, and also by
Ptitsyn and coworkers, revealed that under certain denaturing
conditions, in particular low pH, or low denaturant concentrations,
that a stable intermediate state was formed which could be
distinguished form the native and unfolded states by virtue of
having a nearly native-like far-UV CD, but an unfolded-like near-UV
CD spectrum, i.e. secondary structure but no tertiary structure.
This was also manifested as non-coincident denaturation transitions
when monitored by near- and far-UV CD. Subsequent studies also
showed that the a-lactalbumin intermediate was quite compact.
This state was given the name molten globule.
Kuwajima (1989) Proteins 6, 87; Ptitsyn, (1987) J. Prot. Chem. 6, 273.
α‐lactalbumin
Is hydrophobic collapse the same as the molten globule state?
There is considerable controversy, stemming from a lack of
experimental data, as to whether the initial collapse involves a
non-specific hydrophobic collapse, followed by subsequent gain of
secondary structure, or whether the initial collapse involves
formation of secondary structure, followed subsequently by
hydrophobic collapse. Both may occur simultaneously. One
to suspect this is that formation of secondary structure is one of
the few ways in which the polypeptide can become compact.
Hydrophobic collapse is usually thought to precede the formation
of a molten globule. A further collapse and formation of additional
secondary structure takes place in a subsequent fast step (ms). This
leads to a compact intermediate with much of the secondary structure
in place but few tertiary interactions. Such species are sometimes
called molten globules
Formation of tertiary structure
Molten globules rearrange to a native-like conformation on a time
scale of 100's of msec. There can be slower transitions that involve
proline isomerization. This picture is for relatively small proteins
which are monomeric or long biopolymers such as collagen. For
oligomers the situation is slightly different in that the intermediate
states may (often) interact to form dimers (or higher oligomers),
prior to the final rearrangements to a native-like conformation.
There are now a few cases of proteins, which fold in less than
50 msec, with little evidence of populated intermediates. (e. g.
cytochrome c and CspA). It is likely that in these cases, there may
indeed be intermediates but their lifetimes are sufficiently short
that they have not been detected. For these proteins we can talk
of two state behavior.
The Folding Funnel
Energy
The classic folding funnel
shown here represents the
change in energy for a large
number of folding paths
that lead to the native
configuration. There are
no energy barriers. This
implies that all paths have
an equal probability leading
Conformation
to the folded state. The funnel
shown here has no energy barriers and all paths lead directly to
the native state. Thus, this funnel is consistent with two state
folding behavior.
Dill and Chan, Nature Struct. Biol. (1997), 4:10-19
Helmholtz free energy
The Helmholtz free energy is A = U – TS. Using the symbols
for the internal energy in the canonical ensemble we can
also write this as A = E – TS.
A = E − E0 −T
A = −kT ln Q
E − E0
−Tk ln Q
T
Note that the there is a reference energy used in this
derivation. Entropy is absolute, but the energy must always
be measured or calculated relative to an arbitrarily defined
standard.
Gibbs free energy
For distinguishable particles we have
N
q
Q=
N!
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
q
G − G(0) = −nRT ln m ⎟⎟
N A ⎟⎠
where qm is a molar partition function q/n.
For indistinguishable particles we have
Q = qN
G − G(0) = −nRT ln q
The equilibrium constant
For a hypothetical reaction
aA +bB ⇔cC +dD
The free energy of reaction is
ΔGrxn = dGD + cGC – bGB – aG A
o
o
o
o
o
The naught symbol signifies the molar free energy in its
standard state. The Gibb’s free energy is related to the
partition function by:
o
q
o
o
Gi = Gi (0) – RTln i
Ni
The equilibrium constant
Including all of the terms we have:
ΔG
o
rxn
o c
C
o b
B
o a
A
q
q
q
q
= –RT ln
+ ln
– ln
– ln
ND
NC
NB
NA
ΔGrxn = –RTln
o
o d
D
q
o d
D
q
/ND
o b
B
/NB
q
q
o c
C
o a
A
/NC
= – RT ln K
/N A
where K is the equilibrium constant. Thus, we can calculate
the equilibrium constant from first principles using quantum
states as the starting point.