# Rate of Diffusion through a Solution

Diffusion in a gas is the random motion of particles involved in the net movement of a substance from an area of high concentration to an area of low concentration. Each particle in a given gas with kinetic energy will continue to collide with other particles. In regions of the gas where the particle density is the highest, the particles will bounce off each other and the boundary of their container at a greater rate in comparison to regions of the gas that contains less particle density.

### Introduction

Upon perforation of the membrane separating gas A and gas B, the later image will result.

For a gas (a simplified case), the rate at which diffusion occurs is proportional to the square root of the density of the gas. The density of a gas is equal to the mass of the gas divided by the volume of the gas. If the volume is held constant and we want to compare one gas with another:

$$\dfrac{R_2}{R_1} = \sqrt{\dfrac{M_1}{M_2}}$$

Known as Graham's Law of Diffusion, where $$R$$ is the rate of diffusion in mol/s and $$M$$ is the molar mass in g/mol.

### Fick's First Law of Diffusion

For a volume of solution that does not change:

$J = -D\dfrac{dc}{dx}$

• J is the flux, or movement, of the molecules in a given time interval denoted by units of moles/(time*area)
• D is the diffusivity constant which describes the speed at which at object diffuses and is denoted by units of area/time
• Delta c refers to the change in concentration from a point in time where no diffusion has taken place to a point in time where the diffusion has been completed, denoted by units of (moles/volume)
• Delta x refers to the change in distance that a given particle undergoes during diffusion and is denoted by units of length such as (meters)

In the above diagram there is a cylinder with a region of high concentration and a region of low concentration. The volume of the region with no molecules can be designated as Area(of a circle) * L(length of the gap).

• The high concentration point on the left may be treated as $$C(X-1/2L)$$
• The low concentration point on the right may be treated as $$C(X+1/2L)$$

If the time interval is designated as Delta T, then:

• Average concentration coming from the left $$\propto C(X-1/2L) * L * Area * \Delta{T}$$
• Average concentration coming from the right $$\propto C(X+1/2L) * L * Area * \Delta{T}$$

The Total Flux may be represented as:

$J \propto \dfrac{[C(X-1/2L) * L * Area * \Delta{T}] [C(X+1/2L) * L * Area * \Delta{T}]}{(Area)(\Delta{T})}$

If we set the following:

$$C(X+1/2L) = C(X)+1/2L*\dfrac{dc}{dx}C(X-1/2L) = C(X)-1/2L * \dfrac{dc}{dx}$$

Then:

$J \propto -L^2\dfrac{dc}{dx}$

$$L_2$$ may be incorporated into the diffusivity constant D and thus the equation for the flux may be rewritten as:

$J = -D\dfrac{dc}{dx}$

### Fick's Second Law of Diffusion

An incremental volume may have diffusion which takes place both at the rear and front surface of the volume. There is a correlation between the rate at which the concentration in the volume of question changes with respect to the curvature of the concentration gradient and may be given as the following formula:

$$\dfrac{\partial{C}}{\partial{T}} = x\dfrac{\partial ^2C}{\partial x^2}$$

If the front gradient allows for (and thus drives) an uptake of solution into the incremental volume more quickly than the rear gradient allows for its dissipation, then the curvature is positive. If the rear gradient allows for (and thus drives) the concentration out of the incremental volume more quickly than the front gradient allows for its uptake, then the curvature would be negative.

### Diffusion and Activation

When two different particles end up near each other in solution, they may be trapped as a result of the particles surrounding them, which is known as the Cage Effect or Solvent Cage. In the diagram below, the two yellow particles may be unable to diffuse through the solution to get away from each other.

We can represent the two different particles colliding as a 2nd order reaction: $$A + B \rightarrow AB$$

$AB = K_d[A][B]$

Notice the use of Kd to denote the diffusion rate constant.

If circumstances change and either of the particles is able to diffuse out of the Solvent Cage then the following 1st order Reaction is possible, $$AB \rightarrow A + B$$:

$AB = K_d'[AB]$

There now exists a reaction for the formation of the AB complex as well as the breakdown of the AB complex into what may constitute products.

$k = \dfrac{K_aK_d}{K_a + K_d'}$

$$V = K[A][B]$$

$A + B \rightarrow AB \rightarrow Products$

The net rate of formation for AB can now be seen:

$$\dfrac{d[AB]}{dt} = (A+B \rightarrow AB) - (AB \rightarrow A + B) - (AB \rightarrow Products)$$

$\dfrac{d[AB]}{dt} = K_d[A][B] - K_d'[AB] - K_a[AB]$

$\dfrac{d[AB]}{dt} = K_d[A][B] - K_d'[AB] - K_a[AB] = 0$

$$[AB] = \dfrac{K_d[A][B]}{K_a + K_d'}$$

The final rate of product formation taking into account both Diffusion and Activation:

$$V = \dfrac{K_d[A][B]}{K_a + K_d'}$$

### Diffusion-controlled limit

If the rate at which particle A encounters particle B is much slower than the rate at AB? dissociates then we can safely ignore Kd' in the denominator of the equation.

$$V = K_d[A][B]$$

### Activation-controlled limit

The rate at which particle A encounters and reacts with particle B may exceed the rate at which the AB complex breaks apart into a product by a significant quantity. If the rate of AB? --> roducts is slow enough we may ignore Ka in the denominator:

$$V = \dfrac{K_d[A][B]}{K_d'}$$

### The Rate Constant Kd

A correlation between viscosity and rate of diffusion may be related by the following formula:

$$K_d = \dfrac{8RT}{3n}$$

Where n is the viscosity of the solution.

### Practice Problems

1. Compare the rate of diffusion between fluorine and chlorine gases.  F2, Fluorine gas has a molecular mass of 32 grams.  Cl2, Chlorine has a molecular mass of 70.90 grams
2. Gas A is 0.75 times as fast as Gas B.  The mass of Gas B is 32 grams.  What is the mass of Gas A?
3. Determine the rate of diffusion (flux) for aspirin dissolving through the stomach lining.  C1= 50 mg/L  and C2 = 290 mg/L.  The diffusivity constant of aspirin is 0.29 * 10-9 cm2/s and the thickness of the stomach lining is approximately 0.5 cm.
4. How long will it take oxygen to diffuse 0.5 cm below the surface of a still lake if D = 1*10-5cm2/s?
5. If it takes 5 seconds for oxygen to diffuse to the center of a bacterial cell that is 0.02 cm in diameter, determine the diffusivity constant.

### Solutions to Practice Problems

1.  Using Graham's Law of Diffusion:

(Rate1/Rate2) = (Mass2/Mass1)1/2

(RateF2/RateCl2) = (70.9g/32g)1/2 = 1.49

This tells us Fluorine gas is 1.49 times as fast as Chlorine gas.

2.  Using Graham's Law of Diffusion:

(Rate1/Rate2) = (Mass2/Mass1)1/2

(RateA/RateB) = (MassB/MassA)1/2

0.75 = (32g/MassA)1/2

0.752=(32g/MassA)

MassA = (32g/0.5625)

MassA = 56.8888g

3.  Using Fick's First Law:

J = -D(dc/dx)

Where:

J = our unknown (flux)

D = 0.29*10-9 cm2/s

dc = (C1 - C2) = 50mg/L - 290mg/L = -240mg/L, which is equivalent to -240mg/1000cm3 =-0.24mg/cm3

dx = 0.5cm

J = (0.29*10-9cm2/s)[(-0.24mg/cm3)/(0.5cm)] = 1.39*10-10mg/s*cm2

J = 1.39*10-10mg/s*cm2

4.  Using Fick's Second Law:

T = x2/2D

Where:

T = our unknown (time)

x = 0.5 cm

D = 1*10-5cm2/s

T = (0.5cm)2/[2(1*10-5cm2/s)]

T = 1.25*104 seconds

5.  Using Fick's Second Law:

First, rearrange the equation T = x2/2D to solve for D --> D = x2/2T

Where:

D = our unknown (diffusivity constant)

x = 0.01 cm (distance from the outside to the center of the cell)

T = 5s

D = (0.01cm)2/[2(5s)]

D = 1*10-5cm2/s

### References

1. Chang, Raymond. Physical Chemistry for the Biosciences. University Science Books, California 2005
2. Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. W.H. Freeman and Company, New York 2005

### Contributors

• Eric Stouffer (UCD)
• Anne Slisz (UCD)

21:03, 24 Mar 2014

## Tags

### Textbook Maps

An NSF funded Project