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 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.

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. 1.3806504*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

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= 125.2 days

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. 1.3806504*10-23 J/K D: Diffusion Coefficient I.S. S.I. m2/s T: Absolute Temperature K 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) 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:

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

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=dX2 dC1=dC2 C1≠C2 ≠C3 J = -D dC/dx J= -D (DC/Dx)

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 = 3.1416*(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 x1 x2 x3 J1 J2 J3 J1≠J2 ≠ J3 Particles Flux dX1=dX2 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

2.4.- Diffusion Process through Membranes 2.4.1. 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

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:0-02-321020-6 Physical Chemistry of Foods. Chapter 5. Pieter Walstra. Marcel Decker Inc. New York.2003.ISBN:0-8247-9355-2