--%>

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 : 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

  • Q : P block why BiF3 is ionic whereas other

    why BiF3 is ionic whereas other trihalides are covalent in nature?

  • Q : Ionic radius of chloride ion The edge

    The edge length of the unit cell of Nacl crystal lattice is 552 pm. If ionic radius of sodium ion is 95. What is the ionic radius of chloride ion:(a) 190 pm  (b) 368 pm  (c) 181 pm  (d) 276 pm     <

  • 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

  • Q : Dipole attractions for london dispersion

    Illustrate how are dipole attractions London dispersion forces and hydrogen bonding similar?

  • Q : Explain the catalyst definition and

    Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. The phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.

  • Q : Mass percent Help me to go through this

    Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9

  • Q : Molar mass what is the equation for

    what is the equation for calculating molar mass of non volatile solute

  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen

  • Q : Problem on normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N