1. Micelles as Drug Carriers for Controlled Release
Margarita Valero Juan
Physical Chemistry Department
Pharmacy Faculty
Salamanca University
ATHENS 2014

2. SALAMANCA, SPAIN
Margarita Valero

3. SALAMANCA MAIN SQUARE

4. SALAMANCA CATHEDRAL

6. PHARMACY FACULTY

7. Physical Chemistry Department

8. TRANSPORT PHENOMENA 2.1.- Concept of Transport 2.2.- Diffusion
2.3.- Diffusion of Matter
First Fick´s Law
Second Fick´s Law
2.4.- Diffusion through Membranes
Permeable Membranes
Semi-Permeables Membranes
2.5.- Bibliography

9. 2.1.- Transport J = f (X) Transport: FLUX (J):
Transference of “some amount” of a physical property between two regions of a system.
DRIVING FORCE (X)  SOME EFFECT: FLUX (J)
FLUX (J):
AMOUNT OF PHYSICAL MAGNITUD TRANSFERRED BY UNIT OF AREA AND TIME
J = f (X)
Physical Magnitud:
* Energy: Heat: X: Difference of Temperature
* Matter: X: Difference in the Concentration.
* Electric Charge: Electric Potential Diference.

10. 2.2.- Diffusion <x>=0 <x>2 = 2Dt D = kT/f D = kT/6pr
Definition: movement of molecules due to the thermal or kinetic energy.
Brownian Movement: in the absence of concentration gradient
“random walk”: by collision among particles
D: Diffusion Coefficient I.S. m2/s
t: time: seconds (s)
<x>2: mean square distance: I.S.: m2
<x>=0
<x>2 = 2Dt
f: frictional coefficient
k: Boltzman´s Constant I.S *10-23 J/K
D: Diffusion Coefficient I.S. S.I. m2/s
T: Temperature K
D = kT/f
Einstein´s Law:
Stokes-Einstein´s Law :
D = kT/6pr
h: solvent viscosity I.S.: Pa*s ((N/m2)*s)
r: particle radius (spherical particles) (rH= hydrodynamic radius): length

11. 2.2.- Diffusion <x>2 = 2Dt t = <x>2 / 2D
EXAMPLE 1: The diffusion coefficient of glucose is 4.62*10-2m2s-1. Calculate the time required for a glucose molecule to diffuse through: a) 10000Å b) 0.1 m
<x>2 = 2Dt
t = <x>2 / 2D
D: Diffusion Coefficient I.S. m2/s
t: time: s
<x>2: mean square distance: I.S. m2
<x>2 =(10000 Å*10-8m/Å)2=10-4m2
t=10-4m2/(2* 4.62*10-2m2s-1)=1.08*10-3s
b) <x>2 =(0.1m)2=10-2m2
t=10-2m2/(2* 4.62*10-2 m2s-1)=10.82*106s= days

12. 2.2.- Diffusion D = kT/6pr r = kT/6pD Stokes-Einstein´s Law :
EXAMPLE 2: Calculate the hydrodynamic radius of a sucrose molecule in water knowing that at 25ºC, Dsucrose= 69*10-9m2s-1 and H2O.=1.0*10-9 Ns/m2.
Stokes-Einstein´s Law :
D = kT/6pr
r = kT/6pD
h: solvent viscosity I.S.: Pa*s ((N/m2)*s)
r: particle radius (spherical particles) (rH= hydrodynamic radius): length
k: Boltzman´s Constant I.S *10-23 J/K
D: Diffusion Coefficient I.S. S.I. m2/s
T: Absolute Temperature K
r =( *10-23 J/K)(25+273)K/ (6*3.1416*1.0*10-9 Ns/m2.* 69*10-9m2s-1)=
= 3.16*10-10m = 3.16Å
J=N*m

13. 2.3.- Diffusion of Matter J = f (X) J = dn/A dt dC/dx: J = f (X) Flux:
J : particles/ length 2 time
Speed:
v = dn/dt
v: particles/ time
dC/dx:
Concentration Gradient:
particles/ length 4
J = f (X)
Leyes de Fick
Cuantificación del Proceso de Difusión:

14. 2.3.1- First Fick´s Law J = f (X) J =-D dC/dx J = dn/A dt = -D dC/dx
Flux of particles
J =-D dC/dx
D: Diffusion Coefficient
dC/dx: Concentration Gradient
UNITS:
* dC/dx: particles/length4 (c=particles/length3)
* dn/dt: particles/ time
* D: length2/time
A: length2
I.S: length: m; time: seconds
J = dn/A dt = -D dC/dx
v = dn/dt = -D A dC/dx

15. 2.3.1- First Fick´s Law Steady State Conditions:
J =cte and dC/dx= cte along x x1 x2 x3 J1 J2 J3
J1=J2=J3
J = dn/A dt = -D dC/dx
v = dn/dt = -D A dC/dx
dX1=dX dC1=dC2
C1≠C2 ≠C3
J = -D dC/dx
J= -D (DC/Dx)

16. 2.3.1- First Fick´s Law Steady State Conditions: J =cte and dC/dx= cte
EXAMPLE 3: In one container there is a wall that separates two regions through a circular disc of 6 mm of diameter and 5 mm in thickness. In the compatmet 1, there is an 0.2m aqueous urea solution; whereas compartment 2 has only water. How many grams of urea passes from compartment 1 to 2 in 1s?, Durea= 9.37*10-10m2s-1 and Murea=60g/mol.
Steady State Conditions: J =cte and dC/dx= cte
0.2M
Urea
H2O
J = Dn/A t
H2O
J = -D (DC/Dx)
Dn/t= -DA (DC/Dx)
5mm
D = 9.37*10-10m2s-1
A= pr2 = *(3 mm*10-3m/mm)2=2.83*10-6 m2
DC=-0.2M
DX=5 mm*10-3m/mm=5*10-3m
Dn/t=-9.37*10-10m2s-1*2.83*10-6 m2 *(-0.2M/5*10-3m)= 91.69*10-6 mol/s
91.69*10-6 mol*60g/mol/s= 5.5*10-3g= 5.5 mg

17. 2.3.2- Second Fick´s Law Non Steady State Flux:
J ≠ cte and dC/dx ≠ cte along x x1 x2 x3 J1 J2 J3
J1≠J2 ≠ J3
Particles Flux
dX1=dX dC1 ≠ dC2
C1≠C2 ≠C3
J = f (X)
∂C/∂t = D (∂/∂x(∂C/∂x))= D(∂2C/∂x2)
J =-D dn/dx
D:Diffusion Coefficient
dC/dx: Concentration Gradient

18. 2.4.- Diffusion Process through Membranes
Permeable Membranes
Steady State Conditions:J=cte and dC/dx =cte along X C1 C2 x1 x2 l
C1*P
C2*P C1 C2 x1 x2 l
C1*P
C2*P
J= -D (DC/Dx) C1 C2 x1 x2 l
P= Cm/C
P= Cm/C
J= -D P (DC/Dx)
PERMEABILITY: DP

19. 2.4.- Diffusion Process through Membranes
Semi-Permeable Membranes
DIALYSIS: diffusion of a permeable solute
OSMOSIS: diffusion of solvent molecules

20. 2.5.- Bibliography
Physical Chemistry with Applications to Biological Systems. Chapter 5. Raymond Chang. Collier Macmillan Canadá, Ltd ISBN:
Physical Chemistry of Foods. Chapter 5. Pieter Walstra. Marcel Decker Inc. New York.2003.ISBN: