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Unformatted text preview: Summary of last lecture Chemical Equilibrium, examples of chemical reactions. Chemistry 161: Statistical Thermodynamics
Lecture 17 Quantum Statistics
Reading: Reif, Ch 9 McQuarrie Ch10 Key Concepts and Lessons: a) Binding  proteinprotein, proteinligand etc b) Quantum statistics  a general consideration Reactions involving polyatomic molecules.
Our derivations so far included mono and diatomic molecules and we did not consider multiatomic molecules. Briefly, the changes that are introduced when we consider multiatomic molecules are as follows: a) Vibrational motions. Now we consider not one vibration as in diatomic molecules but several vibrations. In fact the number of vibrational degrees can be estimated easily. For molecule of M atoms the total possible number of degrees of freedom (if atoms were allowed to move independently) is 3M. The molecule as whole has 6 degrees of freedom 3 rotations and 3 translations. Therefore the remaining 3M6 degrees of freedom are vibrations. Now one can introduce a mechanical concept of normal modes saying that each vibration is essentially a linear oscillator i=1,...3M6 . Then vibrational CPF generalizes to: move with some frequency i
hvi 3M i =1 6 qvib = e 1 2 kT hvi e kT Reactions involving polyatomic molecules.
b) Rotational partition function. A molecule can be presented as a rigid body with three major inertia moments I A , I B , I C The QM treatment is relatively straightforward for spherical top when I A = I B = IC it is the same as for the diatomic case but the degeneracy of each rotational level 2 now is . In general case the QM consideration is very involved but classical ( 2J + 1) mechanics treatment is still possible and valid (see and solve problem 8.16 of the McQuarrie book) . As before we can introduce three rotational temperatures: r , A, B,C h2 = 8 2 I A, B,C k and when all three rotational temperatures are low (always the case with complex molecules)
qrot =
1/ 2 8 2 I A kT h2 1/ 2 8 2 I B kT h2 1/ 2 8 2 I C kT h2 1/ 2 = 1/ 2 T3
r, A r,B r ,C 1/ 2 Polyatomic molecules reactions: Example of water
1 H 2 + O2 H 2O 2 qH O / V 2 K p (T ) = qH / V qO / V ( ( 2 )(
T ) 2 ) 1/ 2 3 qH O
2 V = 2 mH O kT
2 3/ 2 1/ 2 H2 O A 3 H2 O C 1/ 2
j=1 j v 3 j v 1 De kT h2
2 mH kT
2 H2 O B e 2kT j =1 1 e kT e qH V qo V 3/ 2 1 2 = T 2
3/ 2 r , H2 v ,H 2 v ,H 2 De h2 2 mo kT
2 e 2kT 1 e kT e
1 kT 2 =3 T 2
r ,O2 v ,O2 v ,O2 DeO 2 h2 e 2kT 1 e kT e kT The CPF for oxygen has an extra factor 3 in front because its electronic wave function has total spin 1 in the ground state I,e, it is 2S+1=3 degenerate. Reaction of water formation (Cont)
K p (T ) = AT
5/ 4 exp H kT where H=D H2 O el D H2 el 1 O2 1 D + 2 el 2 3 h
j =1 H2 o j 1 h 2 H2 1 h 4 O2 Change in enthalpy per product molecule upon reaction. At 1500K the equilibrium constant for water formation is 4.77*105 (atm)1/2 In general: when reactions are endothermic (I.e. enthalpy of the product is higher than enthalpy of reactants higher T shifts equilibrium towards products. In the opposite case when a reaction is exothermic higher T shifts equilibrium towards reactants. This is consistent with Les Chatellier Principle. Binding
Now consider two large molecules associating. (e.g. proteins). In this case equilibrium constant is:
Kc =
2 m1,2 kT qmonomer = V h2
3/ 2 dim ers 1 2
1/ 2 1,2 C 1/ 2 1,2 A T3
1,2 B s qdim er V = 2 (m1 + m2 )kT h2 3/ 2 1/ 2 dim er A T 3 dim er C 1/ 2 j v j=1 s j v 1 dim er B e 2kT j =1 1 e kT e U kT where we summed over all relative motions (s degrees of freedom). U is energy of interactions between proteins. If we assume that associating molecules exhibit classical (nonfrozen) vibrations in the complex for all 3 relative coordinates then
K = Ae
U kT In the opposite case of frozen vibrations of the molecules the binding constant will be:
K = AT 3e
U kT Binding: A simple example of rigid rotor binding model
j U12 qmonomer 2 mkT = V h2 3/ 2 1/ 2 mono A T 3 mono C 1/ 2 1/ 2 mono B = 2 mkT h2
1/ 2 3/ 2 1/ 2 4 (
144 T 3 mono A
j U12 ) 3 qdim er V 4 mkT = h2 3/ 2 1/ 2 dim er A T3
dim er B dim er C 1/ 2 qint ernal 4 mkT h2 3/ 2 1/ 2 T
dim er A 4
j U12 1 2 T
dim er B,C e
j =1 kT K = AT 144 3 j =1 e kT Binding: Flexible proteinprotein binding (i.e. proteins are flexible)
In the previous example we considered molecules (represented as cubes) to be rigid. However, we Know that there are internal vibrational motions in large molecules like proteins. How does that affect their binding? Consider binding two proteins or two fragments of a molecular crystal (nano objects) N1 + N2 N1+N2 Now imagine that first fragment (or protein) has N1 atomic groups (amino acid residues or molecules) Inside while second fragment has N2 of them. When these fragments join together they form a complex having N1+N2 groups. Now assume that all vibrational modes in both fragments are classical, I.e. that Debye T is low. Now we can evaluate all relevant q's:
2 m1kT q1 monomer = V h2
2 2 m2 kT qmonomer = V h2 3/ 2 1/ 2 mono A 3/ 2 1/ 2 mono A T3
mono B mono C 1/ 2 3N1 6 T
i v = A1T 3N1 3 i=1 1/ 2 3N 2 6 T3
mono B mono C T
i v = A2T 3N 2 3 i=1 Binding: Flexible proteinprotein binding (i.e. proteins are flexible) (Cont)
Now we consider that two fragments interact with binding energy Uint which `'glues'' them together; Upon binding they form a new fragment which has N1+N2 units and N1+N26 vibrational models all classical
qdim er V 2 m1 + m2 kT h
2 = ( ) 3/ 2 1/ 2 dim er A T3
dim er B dim er C 1/ 2 3 N + N 1 2 ( ) 6 T
i v U int e kT = Adim er T 3 N1 + N 2 ( ) 3 U int e kT i=1 Now we consider the binding constant: qdim er V K c (T ) = 2 qmonomer q1 monomer V V A = dim er T 3e A1 A2 U int kT Now note an extremely peculiar T3 dependence in that result. It suggests that under certain conditions (i.e at T high enough to be above Debye T association may be ENTROPICALLY favorable. Why? Because association creates 3 extra VIBRATIONAL degrees of freedom which contribute lnT Into entropy rather than 1/2 ln T for pure kinetic degrees of freedom  entropy of mutual positioning of two parts of the complex, which depends on T  higher T means more uncertainty In that quantity. Q: plot the dependence of K(T) above and discuss. (10pts) Quantum Statistics of Ideal gases
Now we descend to the world of low temperatures where multiple particles occupy single quantum levels. Particles are indistinguishable but now the simple trick of dividing Z by N! does not work  we need to be more careful here. For the ideal gas of noninteracting particles we can introduce singleparticle quantum levels  I.e. levels which are allowed for a single particle without a regard of any other particles in the box. We denote these quantum levels as 1....k and their energies are Now we denote occupancies of each state 1 ...... k n1 .......nk as  telling us how many particles occupy quantum level 1... and how many particles occupy QL k. Now we have obvious relations for total energy and total number of particles. E = N = alllevels nk
k alllevels k nk
k Now we want to calculate the CPF: Q=
n1 ,n2 ..n M n + n +...n e ( 1 1 2 2 M M) = n1 e n1 1 e
n2 n2 1 ..... =
k 1 1 e
k Looks nice and simple but with one small caveat: it is wrong!. Why? Because particles are not fully independent: their total number is fixed at N. Therefore we cannot sum over n's independently. Help!!!! Quantum Statistics of Ideal Gases (Cont) The way out is to relax the particle conservation condition, i.e. consider the Grand Canonical Ensemble! e pV = =
N v E e ( v Nv ) Now totalN constraint is removed and we can handily sum over all occupancy numbers, I.e. n's: =
n1 ,n2 ...n j exp
j ( j nj ) Now we should distinguish between BoseEinstein statistics (for particles with integer spin) which allows ANY occupancy and FermiDirac Statistics (for particles with semiinteger spin, I.e. electrons) Which does not allow more than one particle in a state I.e. for FD n can be either 0 or 1. With that we have: BE =e =e pV =
j nj =0 1 e ( ( j nj ) =
j 1 1 e ( j ) ) pV FD =
j e
nj =0 j nj ) =
j { 1+ e ( j } Quantum Statistics of Ideal Gases (Cont)
pV = ln pV = ln
BE =
j ln 1 ln 1 + e
j e (
j j ) FD = ( ) Where summation is taken over all quantum states. Average occupation numbers in each quantum state are determined in a straightforward way: nj nj BE = = ln ( ln (
j j ) ) = e = e ( ( 1
j ) 1 ) +1 1
j FD These relations contain an unknown quantity  chemical potential. It can be determined from the condition that total number of particles is conserved as N: nj
j = N Quantum Statistics of Ideal Gases: high T limit
At high T all occupancy numbers are very small:
n j << 1 which means that
e ( j ) 1 In order to satisfy this condition for each quantum level j we need to have:
nj = e 1 ( j ) = e e j i.e. we omitted 1 in the expression for occupancies. In this case the expression for chemical potential and occupancy numers simplifies: N e = j e
j nj = N e e
j j j Next, we use a standard Thermodynamic relation: ln Z(N ,V ,T ) =
Which in highT case becomes N + ln
( ) =
j ln Q( N ,V ,T ) =
(we expanded logs in full formulas for N +
j e N + N
j because at high T all e ( ) << 1 and therefore ln 1 e ( ( j ) ) e ( j ) Quantum Statistics of Ideal Gases: high T limit
We note that: = ln N
And therefore ln
j e j ln Q = N ln N + N + N ln
j e j So that finally we get the classical expression of CPF which is valid only at high T when all occupancy numbers are small:
N 1 Q = N! e
j j qN = N! Back to low T Introducing a convenient notation = e We get final relations for both BE ( sign) and FD (+ sign) cases: ( V ,T , )
1 e =
k (
k 1 e k ) 1 N =
k e e
k k nk = E = 1 e e k e
k k k k 1 k pV = kT ln 1 ( e k ) How to sum over quantum levels? `'Particle in a Box'' is a solution: k x ,k y ,k z h2 2 = k x2 + k y + k z2 8mV 2/ 3 ( )
3/ 2 1/ 2 Where k are integers. We derived the conversion in Lecture 4, slide #8 (included below) kx k y kz = () 8m d = V 2 4 h d And we get general closed form expressions for BE and FD ideal gases: N =2 2m h2 3/ 2 V
0 e d 1 e
3/ 2 pV = 2 kT 2m h2 V
0 ln 1 ( e )d We will next study a very rich behavior that these relations predict! Now let us try to evaluate density of states just for N=1 case In this case it is more convenient first to evaluate the function which gives the total number of states below certain energy. In other words we need to evaluate how many integer numbers satisfy the relation: ( ) ny h2 2 2 nx + ny + nz2 8ma 2 { } ; nx 0;ny 0;nz 0 It is clear from the picture that integer points are located equidistantly on the axes (with increment 1). Therefore their number is prportional to the volume of the octant (because they are all nonnegative). The radius of the sphere is:
nx
R = 8ma 2 h2 4 3 and the volume of the octant is R
3 ( 1 )= 8 = 6 8ma 2 h2 3/ 2 This expression gives the number of states with energy at or below . The number of states in the range ( , 4 +d ) is then:
3/ 2 1/ 2 ( +d ) ( )=d d d ( )=d 8ma 2 h2 +O d ( 2 ) Next lecture:
Quantum Statistics of Ideal gases: Bose Einstein case and BE condensation. (Reif Ch 9) ...
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This note was uploaded on 05/04/2010 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.
 Spring '10
 Shaklovich

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