Margarita Valero Juan

Physical Chemistry Department

Pharmacy Faculty

Salamanca University

Micelles as Drug Carriers for Controlled Release

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.

J = f (X)

FLUX (J):

Physical Magnitud: * Energy: Heat: X: Difference of Temperature* Matter: X: Difference in the Concentration.* Electric Charge: Electric Potential Diference.

DRIVING FORCE (X) SOME EFFECT: FLUX (J)

AMOUNT OF PHYSICAL MAGNITUD TRANSFERRED BY UNIT OF AREA AND TIME

2.2.- Diffusion

<x>2 = 2Dt<x>=0

Brownian Movement: in the absence of concentration gradient

“random walk”: by collision among particles

Definition: movement of molecules due to the thermal or kinetic energy.

Einstein´s Law: D = kT/f

Stokes-Einstein´s Law : D = kT/6r

f: frictional coefficientk: Boltzman´s Constant I.S. 1.3806504*10-23 J/KD: Diffusion Coefficient I.S. S.I. m2/sT: Temperature K

: solvent viscosity I.S.: Pa*s ((N/m2)*s)r: particle radius (spherical particles) (rH= hydrodynamic radius): length

D: Diffusion Coefficient I.S. m2/st: time: seconds (s)<x>2: mean square distance: I.S.: m2

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

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

<x>2 = 2Dt

D: Diffusion Coefficient I.S. m2/st: time: s<x>2: mean square distance: I.S. m2

t = <x>2 / 2D

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.

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

Stokes-Einstein´s Law :

D = kT/6r

: solvent viscosity I.S.: Pa*s ((N/m2)*s)r: particle radius (spherical particles) (rH= hydrodynamic radius): lengthk: Boltzman´s Constant I.S. 1.3806504*10-23 J/KD: Diffusion Coefficient I.S. S.I. m2/sT: Absolute Temperature K

r = kT/6D

J=N*m

2.3.- Diffusion of Matter

Speed:

J = dn/A dt

v = dn/dt

Flux:

v: particles/ time

J : particles/ length 2 time

J = f (X)

Concentration Gradient: dC/dx: particles/ length 4

Leyes de FickJ = f (X) Cuantificación del Proceso de Difusión:

2.3.1- First Fick´s Law

J = f (X)

J =-D dC/dxD: Diffusion CoefficientdC/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

Flux of particles

2.3.1- First Fick´s LawSteady State Conditions:

J =cte and dC/dx= cte along x

J = dn/A dt = -D dC/dx

v = dn/dt = -D A dC/dx

J = -D dC/dx J= -D (C/x)

x1 x2 x3

J1 J2 J3

J1=J2=J3

C1≠C2 ≠C3

dX1=dX2 dC1=dC2

2.3.1- First Fick´s Law

Steady State Conditions: J =cte and dC/dx= cte

J = n/A tJ = -D (C/x)

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.

D = 9.37*10-10m2s-1

A= r2 = 3.1416*(3 mm*10-3m/mm)2=2.83*10-6 m2

C=-0.2MX=5 mm*10-3m/mm=5*10-3mn/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

n/t= -DA (C/x) 5mm

0.2MUreaH2O

H2O

2.3.2- Second Fick´s Law

J = f (X)

J =-D dn/dxD:Diffusion Coefficient dC/dx: Concentration Gradient

∂C/∂t = D (∂/∂x(∂C/∂x))= D(∂2C/∂x2)

Particles Flux

Non Steady State Flux: J ≠ cte and dC/dx ≠ cte along x

x1 x2 x3

J1 J2 J3

J1≠J2 ≠ J3

C1≠C2 ≠C3

dX1=dX2 dC1 ≠ dC2

2.4.- Diffusion Process through Membranes

Steady State Conditions:J=cte and dC/dx =cte along X

J= -D (C/x)

C1

C2

x1 x2

l

2.4.1. Permeable Membranes

P= Cm/C

C1

C2

x1 x2

l

C1*P

C2*P

C1

C2

x1 x2

l

C1*P

C2*P

P= Cm/C

J= -D P (C/x)

PERMEABILITY: DP

2.4.- Diffusion Process through Membranes

2.4.2. Semi-Permeable Membranes

DIALYSIS: diffusion of a permeable soluteOSMOSIS: diffusion of solvent molecules

2.5.- Bibliography

-Physical Chemistry with Applications to Biological Systems. Chapter 5. Raymond Chang. Collier Macmillan Canadá, Ltd. 1977.ISBN:0-02-321020-6

-Physical Chemistry of Foods. Chapter 5. Pieter Walstra. Marcel Decker Inc. New York.2003.ISBN:0-8247-9355-2