--%>

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 

Or

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 

Or

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 : Procedure for separating the components

    Briefly describe the procedure for separating the components of the gun-powder?

  • Q : Vapour pressure of volatile substance

    Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc

  • Q : Calculating total number of moles

    Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20

  • Q : Explain equilibrium and molecular

    The equilibrium constant can be treated as a particular type of molecular distribution. Consider the simplest gas-phase reaction, one in which molecules of A are converted to molecules of B. the reaction, described by the equation

    Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio

  • Q : Molarity of the final mixture Can

    Can someone please help me in getting through this problem. Two solutions of a substance (that is, non electrolyte) are mixed in the given manner 480 ml of 1.5M first solution + 520 ml of 1.2M second solution. Determine the molarity of the final mixture

  • Q : Polyhalogen compounds we need 10

    we need 10 examples for the polyhalogen compounds....please help me....need it urgently...

  • Q : Effect of addition of mercuric iodide

    Give me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the:(a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

  • Q : Real vapour pressure Choose the right

    Choose the right answer from following. The pressure under which liquid and vapour can coexist at equilibrium is called the : (a) Limiting vapour pressure (b) Real vapour pressure (c) Normal vapour pressure (d) Saturated vapour pressure

  • Q : Molality of Sulfuric acid Choose the

    Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml]  (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6