Diffusion Molecular View

When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.

To see how the experimental coefficients can be treated to properties of the system and particularly of the solute macromolecules we take a molecular view of the diffusion process. Consider across a distance interval dx over which the concentration changes from c to c-dc. The force that drives the molecules to the ore dilute region can be related to the difference in the, molar free energy of the solute at concentration c and at concentration c-dc. If deal behaviour is assumed, the free energy differences per molecule is

Gc - dc - Gc = RT/N In (c -dc)/c 


dG = RT/N In (1 - dc/c) - RT/N dc/c  where the relation In (1 - y) = -y for small y has been used.

This free energy difference corresponds to the mechanical energy needed to transfer one macromolecule across the distance dx. This energy can therefore be written as a force times the distance dx. Thud dG = driving force × dx, or

Driving force = dG/dx = RT/N 1/c dc/dx

A frictional force sets in and balances this diffusion force when some constant velocity is reached. The frictional force exerted by a viscous solvent fluid of viscosity η has been derived for a macroscopic sphere of radius r by G. G strokes as 

Frictional force = 6∏rη dx/dt

It appears suitable to apply this expression to the motion of reasonably spherical macromolecules. The diffusion velocity increases, therefore, until the force balances that equation. Then

6∏rη dx/dt = - RT/N 1/c dc/dx 


cdx/dt = - RT/(6∏rη) dc/dx

Since c implies a mass per unit volume measure of concentrations, the product c dx/dt can be interrupted as the rate with which the diffusing substance moves through a unit cross section at x. this follows suggests, from the fact that dx/dt, the average diffusion velocity in the x direction, is the distance the diffusing molecules travel per unit time. Thus all the molecules within a distance dx/dt of a cross section will pass cross section in unit time. These molecules are in a volume equal to dx/dt times the cross section area. The mass of these molecules is the product of this volume and the concentration expressed as mass per unit volume. Thus c dx/dt is the amount per unit time, i.e. the rate with which the solute passes through the cross section. We can write now

D ∂c/∂x = - RT/(6∏rη) ∂c/∂x

This leads to the identification

D = RT/(6∏rη) 

And 6∏rη = RT/DN

Measurements of D and η could therefore lead to a value of the radius r for the macromolecule. Such a procedure is a little unsatisfactory. Molecules do not necessarily obey Strokes' law, even if they are spherical. Furthermore, macromolecules will generally be solvated and in moving through the solution will to some extent vary along this salvation layer. Equation is important however, in that it provides a way of determining the effective value of the group of terms 6∏rη for a solute characterized by molecules with radius r and a solvent characterized by viscosity η

   Related Questions in Chemistry

  • Q : Significance of the organic chemistry

    Describe some of the significance of the organic chemistry in brief?

  • Q : Molecular Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2

  • Q : What is laser and explain its working?

    Laser action relies on a non-Boltzmann population inversion formed by the absorption of radiation and vibrational deactivation that forms a long lived excited electronic state. An excited state molecule can move to a lower energy state or return to the

  • Q : Alkaline medium The amount of KMnO 4

    The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox

  • Q : Number of moles present in water

    Provide solution of this question. How many moles of water are present in 180 of water: (a)1 mole (b)18 mole (c)10 mole (d)100 mole

  • Q : Equimolar solutions Select the right

    Select the right answer of the question. Equimolar solutions in the same solvent have : (a)Same boiling point but different freezing point (b) Same freezing point but different boiling poin (c)Same boiling and same freezing points (d) Different boiling and differe

  • Q : Product of HCl Zn Illustrate  the

    Illustrate  the product of HCl Zn?

  • Q : Dipole moment of chloro-octane Describe

    Describe the dipole moment of chloro-octane in brief?

  • Q : What is depression in freezing point?

    Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its solid and liquid phases have the same vapour pressure. The freezing point o

©TutorsGlobe All rights reserved 2022-2023.