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Washington University in St. Louis

University College - Chemistry

University College - Chemistry Chem 402

Barnes Alexander

Spring

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23 January 2017
L3: Deriving Thermodynamic Functions from Q
I. Statistical Mechanics
A. Partition Function + Drug Binding
1. A 4 polymer subunit may have some favorable interaction energy, βπ , if non-
π
covalently bonded subunits in neighboring βlatticeβ sites
a. There is 0 energy for the 3 degenerate unbound states
b. Molecular partition function = q conf βπ πβππ/ππ = π ππππ‘/ππ + 3
2. Thus we can write the partition function as a sum over energy levels instead of
individual states, if we account for their degeneracies βgβi
a. Molecular energy degeneracy for distinct energy levels, q = β ππ π βππ/ππ
β’ βgβ is the degeneracy of molecular energy levels
i
β’ βΞ© π is the degeneracy of microstates/system energies
b. System/state degeneracy levels, Q = β ππππππππ πΈπΞ© π βπΈ πππ
3. All thermodynamic functions can be calculated from Q
a. Many cases is easier to describe Q in terms of degeneracy of energy levels rather
than explicit states, important for molecular understanding of entropy, S
b. If we can come up with an expression for All of the possible configurations (what
we call Q = canonical partition function), then we can describe the system in
terms of its thermodynamic properties
4. Often common to have drugs that maintain flexibility because of larger degeneracy in
certain states
II. Deriving Thermodynamic Functions from Q
A. Internal Energy, U 1. U is average energy of the system
2. By the Ergodic Hypothesis, the average (sum of n E fi iall i) is equivalent to the
probability or snapshot in time
βπ /ππ
3. Using P =i π π , where Q is the summation of all system states we introduce Q into
π
the equation
a. Note: ο’ = 1/kT
b. Remember to be explicit that we are holding V and N constant
4. The β πΈ π βο’πΈπ term looks like the derivative of ( )
π π πο’ V,N
5. After substituting in, (1/Q) is the chain rule product of (ο€lnQ/ο€ο’)V,N
6. Then, we want an expression in terms of T, so we must multiply by (ο€T/ο€B) N,V
a. Trick: however, evaluation ο€ο’/ο€T is much easier, which is -1/kT 2
b. Therefore, what we are looking for = -kT 2
7. This is then substituted in to give use our expression for U
a. π = ππ ( 2 ππππ)
ππ π.π
B. Helmholtz Free Energy
1. Every thermodynamic quantity has a set of βnaturalβ variables
a. Helmholtz Free Energy (A) has temp, volume, and # of particles, same natural
variables as U
b. βG = βH β TβS (for changes, predict if process will be spontaneous)
c. G = H β TS (describes processes at constant pressure, temperature, and # of
particles)
2. Helmholtz Free Energy (A)
a. Helmholtz free energy is similar, but describes process at constant volume,

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