Department  of  Chemistry  

PbI2(s ) Pb(aq)2+ + 2I(aq)

−

K = [Pb2+ ][I− ]2

ΔGo = 46.1 kJ/mol ⇒ [Pb2+ ] = 1.14 ×10−3 M

measured: [Pb2+ ] = 1.37 ×10−3 M

K = e−ΔGo /RT

Explain  the  difference  

Department  of  Chemistry  

solute(water) solute(octanol)

log(Pwo)  =  3.37  

ΔGo = −19.2 kJ/mol

What  is  the  most  probable  sign  of  ΔSo  and  why?  

Department  of  Chemistry  

CH3COOH CH3COO− + H+

pKa = − log Ka( )

Ka =[CH3COO

− ][H+ ][CH3COOH]

the  pKa  in  water  is  4.76.    Is  the  pKa  higher  or  lower  DMSO  and  why?  

S  

ε  =  47  

Department  of  Chemistry  

P + L P ⋅L

Ligand  binding  to  the  enzyme  carbonic  anhydrase  (at  25  oC)  

ΔSo                                        9.7                                                                    67.4  J/molK  

DOI:  10.1002/anie.201301813  

Explain  the  difference  in  ΔSo    

Department  of  Chemistry  

Explain  why  ΔSo  is  posiUve  

hWp://youtu.be/zKNmBjqGijI  

ΔSo = 25.8 J/molK at 25 oC

(Video  of  benzamidine  binding  to  trypsin)