Chem 402 Lecture 3: L3 1:23:17

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Washington University in St. Louis
University College - Chemistry
University College - Chemistry Chem 402
Barnes Alexander

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, βˆ’πœ€ , if non- 𝑖 covalently bonded subunits in neighboring β€œlattice” sites a. There is 0 energy for the 3 degenerate unbound states b. Molecular partition function = q conf βˆ‘π‘– π‘’βˆ’πœ€π‘–/π‘˜π‘‡ = 𝑒 πœ€π‘–π‘›π‘‘/π‘˜π‘‡ + 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 β€œg”i a. Molecular energy degeneracy for distinct energy levels, q = βˆ‘ 𝑖𝑔 𝑖 βˆ’πœ€π‘–/π‘˜π‘‡ β€’ β€œg” is the degeneracy of molecular energy levels i β€’ β€œΞ© 𝑖 is the degeneracy of microstates/system energies b. System/state degeneracy levels, Q = βˆ‘ π‘’π‘›π‘’π‘Ÿπ‘”π‘–π‘’π‘  𝐸𝑖Ω 𝑖 βˆ’πΈ π‘–π‘˜π‘‡ 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 βˆ’πœ€ /π‘˜π‘‡ 3. Using P =i 𝑒 𝑖 , where Q is the summation of all system states we introduce Q into 𝑄 the equation a. Note:  = 1/kT b. Remember to be explicit that we are holding V and N constant 4. The βˆ‘ 𝐸 𝑒 βˆ’ο’πΈπ‘– term looks like the derivative of ( ) 𝑖 𝑖 πœ•ο’ V,N 5. After substituting in, (1/Q) is the chain rule product of (lnQ/)V,N 6. Then, we want an expression in terms of T, so we must multiply by (T/B) N,V a. Trick: however, evaluation /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‘ˆ = π‘˜π‘‡ ( 2 πœ•π‘™π‘›π‘„) πœ•π‘‡ 𝑉.𝑁 B. Helmholtz Free Energy 1. Every thermodynamic quantity has a set of β€œnatural” variables a. Helmholtz Free Energy (A) has temp, volume, and # of particles, same natural variables as U b. βˆ†G = βˆ†H – Tβˆ†S (for changes, predict if process will be spontaneous) c. G = H – 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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