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Diffusion: A Brief Review

Theresa Julia ZielinskiMonmouth UniversityDepartment of Chemistry, Medical Technology, and PhysicsWest Long Branch, NJ [email protected]

© Copyright Theresa Julia Zielinski, 2006. All rights reserved. You are welcome to use thisdocument in your own classes but commercial use is not allowed without the permission of theauthor.

Goal: To provide the user with the opportunity to explore some of the basic diffusion concepts via mathematicalmodels.

Objectives :Explain the diffusion process using molecular concepts.•Derive Fick's Second Law using a simple one dimensional model.•Sketch plots of diffusion from a thin concentrated segment of solution out into pure solvent as a•function of time and dustance from the starting point of the diffusion process.Describe how changing the diffusion coefficient affects the plots drawn in the previous bullet.•Relate several examples of how Fick's laws apply to cellular processes.•Explain the meaning of 'diffusion controlled' when used to describe biochemical reactions.•Compute the concentration of solute at any time and distance in the solution relative to the•sourse of the solute at time t=0s.Use units effectively in preparing plots and in calculations of diffusion conditions.•

Introduction

Diffusion is the process where by a solute moves from a region of high concentration to one of lowconcentration. The flow of solute is called the flux, J. This flow is examined using a plane in the solution that isperpendicular to the flow of solute. In the schematic shown here the flux, J, will be toward the right.

dx

x

low concentrationhigh concentration

J

dx

x


dx

x


J

Fick's First Law of Diffusion sums up this process in the following mathematical expression.

J D−x

cd

d

=Equation 1

where dc/dx is the concentration gradient per unit length and D is the diffusion constant. Note the negative signin this expression.

Remember that while J is predominantly from left to right in this image there is flow from right to left. Diffusionoccurs in both directions. The flow to the right in this case is greater than that to the left.

The central region in our diagram experiences the following change. Material flows in from the left side andflows out from the right side. The net change as a function of time is

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[ ]1( ) ( )

c J J x J x dx

t dx x

∂ ∂= − + = −∂ ∂

Using Fick's First Law we obtain Fick's Second Law of diffusion.

c c D

t x x

∂ ∂ ∂ = ∂ ∂ ∂ Equation 2

or

2

2

c c D

t x∂ ∂=∂ ∂

Equation 3Fick's Second Law

Exercise 1. Given Fick's First and Second Laws, determine the proper units for D.

Consider the following schematic

x

0

Sample placed at origin at t=0

x

0

Sample placed at origin at t=0

Here a sample is placed at the origin at time zero. What would be the concentration of the sample asa function of x and t?

For this example of diffusion we can write the solution to Fick's Second Law as

c x t,( ) At

exp x2

4 D t−

= Equation 4

Exercise 2: Varify that Equation 4 is a solution to Fick's Second Law, Equation 3 .

Exercise 3. Examine Equation 4 and determine what values of c are permitted when t=0.

Exercise 4. Given A = 10 and D = 50, prepare plots of c(x,t) at times above 0.01. Thefirst plot is given below.

A 10:=D 50:= x 10− 9.9−, 10..:=

c x t,( )A

texp

x2

4 D t−

:=

Plot of c(x) at at t=0.01

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10 5 0 5 100

50

100

c x .01,( )

x

Exercise 5. The parameter A is given by the following expression. Use this expression to examinethe diffusion plots with this more precise definition of A. How does the diffusion depend on c 0? How

does the diffusion depend on the width of the solute containing layer?

0

2

c x A Dπ

∆=

Exerciswe 6. Ribonuclease, molecular weight 13683 g/mol, has a diffusion coefficient

D=0.107*10 -5 cm 2 /s. Prepare appropriate plots of the concentration as a function of position for several times. Choose the various parameters so that you can tell how the ribonucleaseconcentration profile appears after 1 day, 1 week, two weeks. (Hint: one can avoidconsideration of the value of c 0 by evaluating the ratio c/c 0 .)

The Step-Function Situation.

An interesting type of diffusion situation is the one where a solvent is layered over a solution to create a

concentration step-function. How the step-function evolves over time is governed by the following equation. Thefigure shows a horizontal view of the solvent on the right and the solution on the left.

X=0-1 +1

c/c 0

X=0-1 +1

c/c 02

40 0

1 1( , )

2

x y Dt c x t c e dy

π

= − ∫

c0 1.0:=

D 2.5 105− cm

2

s:=

xmax 10 mm:= xinc .1 mm:=

t 15 min:= x xmax− xmax− xinc+, xmax..:=

c x t, D,( ) c01

2

1

π 0

x

4 D t

yexp y2−( )

⌠

⌡ d−

:=

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10 5 0 5 100

0.5

1

c x t, D,( )

x

mmOne can find the value of c at any distance along the x axis and at any time by typing the following.

c 1 cm 15 hr , D,( ) 0.271=

Exercise 7. Add several more curves to the plot above by specifying specific times (include units). Howdoes the concentration profile as shown in the plot change as the time increases?

Exercise 8. Tabulate the concentration at 1cm as a function of time. Discuss the trend you observe in thedata.

Exercise 9. Vary the value of D and explain how the curves in the figure above change. The currentvalue of D used at first is one that is typical for small molecules in noviscous solvent. The values for biomolecules run in the range of 6*10^-8 to 1*10^-6 or so. Given this information explore the timefor a biomolecule to diffuse across a cell if the cell plasma is nonviscous. How would a viscous cellplasma alter your conclusion. (Hint: Be careful with units.)

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