Micelles as Drug Carriers for Controlled
Release
Margarita Valero Juan
Physical Chemistry Department
Pharmacy Faculty
Salamanca University
ATHENS 2014
SALAMANCA, SPAIN
Margarita Valero
SALAMANCA MAIN
SQUARE
SALAMANCA
CATHEDRAL
PHARMACY FACULTY
Physical Chemistry Department
TRANSPORT PHENOMENA
2.1.- Concept of Transport
2.2.- Diffusion
2.3.- Diffusion of Matter
2.3.1.- First Fick´s Law
2.3.2.- Second Fick´s Law
2.4.- Diffusion through Membranes
2.4.1.-Permeable Membranes
2.4.2.- Semi-Permeables Membranes
2.5.- Bibliography
2.1.- Transport
Transport:
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.
2.2.- Diffusion
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
<x>=0
<x>2 = 2Dt
Einstein´s Law: D = kT/f
Stokes-Einstein´s Law :
D: Diffusion Coefficient I.S. m2/s
t: time: seconds (s)
<x>2: mean square distance: I.S.: m2
f: frictional coefficient
k: Boltzman´s Constant I.S. 1.3806504*10-23 J/K
D: Diffusion Coefficient I.S. S.I. m2/s
T: Temperature K
D = kT/6pr
: solvent viscosity I.S.: Pa*s ((N/m2)*s)
r: particle radius (spherical particles) (rH= hydrodynamic radius): length
2.2.- Diffusion
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
a) <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= 125.2 days
2.2.- Diffusion
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
: solvent viscosity I.S.: Pa*s ((N/m2)*s)
r: particle radius (spherical particles) (rH= hydrodynamic radius): length
k: Boltzman´s Constant I.S. 1.3806504*10-23 J/K
D: Diffusion Coefficient I.S. S.I. m2/s
T: Absolute Temperature K
a) r =(1.3806504*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
2.3.- Diffusion of Matter
J = f (X)
Flux:
Speed:
J = dn/A dt J : particles/ length 2 time
v = dn/dt v: particles/ time
Concentration Gradient:
J = f (X) Leyes de Fick
dC/dx: particles/ length 4
Cuantificación del Proceso de Difusión:
2.3.1- First Fick´s Law
Flux of particles
J = f (X)
J =-D dC/dx
D: Diffusion Coefficient
dC/dx: Concentration Gradient
J = dn/A dt = -D dC/dx
v = dn/dt = -D A dC/dx
UNITS:
* dC/dx: particles/length4
(c=particles/length3)
* dn/dt: particles/ time
* D: length2/time
•A: length2
I.S: length: m; time: seconds
2.3.1- First Fick´s Law
Steady State Conditions:
J =cte and dC/dx= cte along x
J1=J2=J3
J1 J2 J3
J = dn/A dt = -D dC/dx
v = dn/dt = -D A dC/dx
x1 x2 x3
C1≠C2 ≠C3
dX1=dX2
J = -D dC/dx
dC1=dC2
J= -D (DC/Dx)
2.3.1- First Fick´s Law
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
H2O
5mm
J = -D (DC/Dx)
J = Dn/A t
Dn/t= -DA (DC/Dx)
D = 9.37*10-10m2s-1
A= pr2 = 3.1416*(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
2.3.2- Second Fick´s Law
Non Steady State Flux:
J ≠ cte and dC/dx ≠ cte along x
J1≠J2 ≠ J3
J1 J2 J3
Particles Flux
x1 x2 x3
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
dX1=dX2
dC1 ≠ dC2
2.4.- Diffusion Process through
Membranes
2.4.1. Permeable Membranes
Steady State Conditions:J=cte and dC/dx =cte along X
J= -D (DC/Dx)
C1
x2
x1
x2
x1
C1*P
C2*P
C1
C1
C2
C1*P
C2
C2*P
C2
l
l
x2
x1
P= Cm/C
l
P= Cm/C
J= -D P (DC/Dx)
PERMEABILITY: DP
2.4.- Diffusion Process through
Membranes
2.4.2. Semi-Permeable Membranes
DIALYSIS: diffusion of a permeable solute
OSMOSIS: diffusion of solvent molecules
2.5.- Bibliography
-Physical Chemistry with Applications to Biological Systems. Chapter 5.
Raymond Chang. Collier Macmillan Canadá, Ltd. 1977.ISBN:0-02321020-6
-Physical Chemistry of Foods. Chapter 5. Pieter Walstra. Marcel Decker
Inc. New York.2003.ISBN:0-8247-9355-2