Chemistry 431. Lecture 27 The Ensemble Partition Function Statistical Thermodynamics. NC State University

Transcription

1 Chemistry 431 Lecture 27 The Ensemble Partition Function Statistical Thermodynamics NC State University

2 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

3 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 β(ε i + ε j + ε k +...) Σ i,j,k,... where the indices i,j,k, represent energy levels of different particles.

4 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 a 1 systems with energy E 1, a 2 systems with energy E 2 etc. We are interested in knowing the probability that the system has energies E 1, E 2, etc. The relative number must depend on the energy so we can write: a 2 /a 1 = f(e 1,E 2 ) which says that the ratio of populations is some function of E 1 and E 2.

5 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 E 1 and E 2. f(e 1,E 2 ) = f(e 1 E 2 ) Hence, a 2 /a 1 = f(e 1 E 2 ) This must be true for any pair of states. a 3 /a 2 = f(e 2 E 3 ) and a 3 /a 1 = f(e 1 E 3 ) and since a 3 /a 1 = (a 3 /a 2 ) (a 2 /a 1 )

6 Proof of the Boltzmann distribution we have f(e 1 E 3 ) = f(e 1 E 2 ) f(e 2 E 3 ) f must therefore be an exponential function. e x+y = e x e y so in general f(e) = e βe where β is an arbitrary constant. e β(e1 E3) = e β(e1 E2) eβ(e2 E3) Thus, a 2 /a 1 = e β(e1 E2) or in general a n /a m = eβ(em En)

7 Proof of the Boltzmann distribution This implies that both am and an are given by a j = Ce βej (I) where C is an constant. and since Σ j a j = A we have Since Σ a j = C e βe j j Σ j A = C C = Σ j Σ j e βe j A e βe j

8 Proof of the Boltzmann distribution We can substitute into equation I above to obtain a j A = e βe j e βe j Σ j Using the definition for the probability p j = a j /A we can calculate p j as p j = e βe j e βe j Σ j

9 The molecular partition function, q represents the energy levels of one individual molecule. We can rewrite the above sum as Q = q i q j q k or Q = q N for N particles. Note that q i means a sum over states or energy levels accessible to molecule i and q j 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.

10 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 Q = q N N! for indistinguishable particles

11 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.

12 Stirling s Approximation Note that the logarithm of the partition function is important in many calculations. Since the partition function of indistinguishable particles is q N /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.

13 Particle in a box energy levels The energy levels are ε nx,n y,n z = h2 8ma n x +n y2 +n z n x,n y,n z =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.

14 The translational partition function is q trans = Σ e ε n x,n y,n /kt z n x,n y,n z =1 Σ Σ Σ = exp 8ma 2 kt n 2 2 x +n y2 +n z n x =1 n y =1 n z =1 The three summations are identical and so they can be written as the cube of one summation Σ q trans = exp n =1 h 2 n 2 8ma 2 kt 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. q trans = exp 0 h 2 3 h 2 n 2 8ma 2 kt dn 3

15 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. 10 9! If we have the substitution a = h 2 /8ma 2 kt we can rewrite the integral as q trans = 0 e αn2 dn 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 = a 3 we have q trans = 2πmkT h 2 3 = π 4α 1/2 3/2 V 3

16 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 tot = Σ i =1 e ε i tot /kt = Σ e εtrans i /kt i =1 = Σ e εtrans i + εi rot + εi vib + εi elec /kt Σ j =1 i =1 e ε j rot /kt = q trans q rot q vib q elec Σ k =1 e ε k vib /kt Σ l =1 e ε l elec /kt

17 Probability in the ensemble The ensemble partition function is: Q = e E i /kt Σ i =0 Where the ensemble energy is E J. The population of a particular state J with energy E J is given by p J = Σ J = 0 e E J / kt e E J / kt = e E J / kt Q 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.

18 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 P M j j j M could be energy or pressure etc. P j is the Boltzmann probability given by P j = e βε j /Q.

19 Average energy If we denote the average energy <E> then E = ΣJ =0 P J This can be written compactly as E J = ΣJ =0 Σ J = 0 E = ln Q β E J e βe J e βe J

20 Consistency check with kinetic theory of gases The average energy per molecule is given by ε trans = lnqtrans β V = kt 2 lnq trans T V = kt 2 T 3 2 lnt + terms independentof T V = 3 2 kt2 1 T T T = 3 2 kt 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.

21 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: S = U U 0 T and the second is statistical: S = klnq Thus, when we put them together we have: S = U U 0 T + k lnq

22 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.

23 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.

24 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 = M N 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.

25 Multinomial distribution The total number of ways that a group of N objects can be arranged in M different categories is M N. 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! M n m! Πm =1 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. M Σ m =1 n m = N

26 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 = q N /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 W = 7! 3!2!1!1 = = 420

27 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.

28 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 M N. For YAHTZEE is 6 5 = Therefore the Probability of getting a long straight is 2*120/7776 = 0.03.

29 Protein folding example: Two state model k f U F k u Unfolded Folded K = [F]/[U] K = ff/(1-ff) Fraction folded ff Fraction unfolded 1-ff

30 The Levinthal Paradox The Levinthal paradox assumes that all of the possible conformations will be sampled with equal probability until Energy the proper one (N = native) is found. Thus, the funnel surface looks like a hole in a golf course. The paradox states that if a protein samples Conformation all 6 M 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(6 M ) = Mln(6)R in this example. M = number of residues, 6 = number of conformations per residue

31 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.

32 The Pathway Model 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 6 M. 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. Energy

33 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 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.

34 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 , 1978

35 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, and 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

36 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.

37 α lactalbumin

38 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

39 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.

40 The Folding Funnel The classic folding funnel shown here represents the change in energy for a large number of folding paths Energy that lead to the native configuration. There are no energy barriers. This implies that all paths have an equal probability leading to the folded state. The funnel Conformation 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

41 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 E T 0 A= kt lnq E E T 0 Tk lnq 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.

42 Gibbs free energy For distinguishable particles we have Q = q N N! G G(0) = nrt ln q N m A where q m is a molar partition function q/n. For indistinguishable particles we have Q=q N G G( 0) = nrt lnq

43 The equilibrium constant For a hypothetical reaction The free energy of reaction is o ΔG rxn aa + bb cc + d D = dg D o + cg C o bg B o ag A 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: G i o = G i o (0) RTln q i o N i

44 The equilibrium constant Including all of the terms we have: o ΔG rxn d = RT ln q o D +ln q o C ln q o B ln q o A N D N C N B c b N A a o ΔG rxn = RTln q D o d /N D q C o c /N C q B o b /N B q A o a /N A = RT ln K where K is the equilibrium constant. Thus, we can calculate the equilibrium constant from first principles using quantum states as the starting point.