1. PROTEIN PHYSICS LECTURE 5

2. Energy: E (enthalpy: H=E+PV) PV=Nk B T for gases, PV << Nk B T for solids & liquids Entropy: S = k B ln(#STATES) If AB=A+B  E AB = E A +E B ; V AB = V A +V B ; S AB = S A +S B while # AB = # A # B while # AB = # A # B Free energy: G = H-TS  F = E-TS ------- ------- G: used when P=const for solids & liquids Chemical potential:   G (1) = G/N Probability ~ exp(-G/k B T)  max G  min G  min

3. G = H-TS; dG = (dH-TdS) - SdT At T=const (i.e., at dT=0): at equilibrium, G = min, i.e., dH-TdS=0. Thus: 1) T = dH/dS 2) S = -dG/dT 3) H = G+TS = G-T(dG/dT) Probability ~ exp(-G/k B T) k B = 2cal/(6  10 23 ) = 2cal/mol = R k B  300 o = R  300 o = 0.6 kcal/mol

4. int : “Free energy of interactions” int : “Free energy of interactions” Chemical potential: Chemical potential:  G (1) = G int - Tk B ln(V (1) )  G int + Tk B ln[C] EQUILIBRIUM for transition of molecule 1 from A to B: G A (1) = G B (1) chemical potentials in A and B are equal chemical potentials in A and B are equal  G int A  B = G int B – G int A  G int A  B = k B Tln([C inA ]/[C inB ]) ===================================================

5. Experiment:  G int A  B = k B T ln([C 1 in A ]/[C 1 in B ])  S int A  B = -d(  G int A  B )/dT  H int A  B =  G int A  B +T  S int A  B C 6 H 12 [C] of C 6 H 12 in H 2 O: 50 times less than in gas; 100000 times less than in liquid C 6 H 12 T=298 0 K=25 0 C

6. -2/3 +1/3 Loss: S (usual case) (usual case)-2/3Loss: LARGE E (rare case) (rare case) H-bond: directed

7. High heat capacity d(  H)/dT: Melting of “iceberg”

9. 20-25 cal/mol per Å 2 of accessible non-polar surface

10. ______largeeffect _______ small small______ large large