5.5: Stokes-Einstein-Sutherland Equation
- Page ID
- 546093
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A useful approximate description of the diffusion constant in a liquid is the Stokes-Einstein-Sutherland equation:
\[D =\frac{k_B T}{6 \pi \eta R_{eff}},\label{SES}\]
where \(k_B\) is the Boltzmann constant, T the temperature in Kelvin, \(\eta\) the viscosity of the solvent and \(R_{eff}\) the effective radius of the diffusing particle including any solvent being carried along with the particle. This equation is most often used to estimate D or using measured D and \(\eta\) to calculate \(R_{eff}\). In water this effective radius is often referred to as the hydrodynamic radius.
Derivation
The diffusion constant D is the proportionality constant between the flux (J) and the slope of the concentration (c):
\[J_z = -D\frac{dc}{dz}.\]
Alternatively, the flux can be thought of as the product of the concentration (c) and a drift speed (s) in the direction of the flux:
\[J_z = sc.\]
Thus,
\[sc = -D\frac{dc}{dz}.\label{eq1}\]
The drift speed is fixed by the force pushing the particles (the effective potential created by the concentration gradient) and the frictional force resisting movement of the particles through the solvent.
The force created by a potential \(\mu\) along direction z is:
\[F_{pot} = -\frac{d\mu}{dz}.\]
In the case of a concentration gradient the appropriate potential is the chemical potential. Assuming an ideal solution the potential is:
\[\mu = \mu_o + RTlnc.\]
So,
\[F_{pot} = -\frac{d\mu}{dz} = -\frac{d}{dz}\left(\mu_o + RTlnc\right) = \frac{-RT}{c}\frac{dc}{dz}.\label{F_pot}\]
This expression is on a per mole basis, but we want to do our calculation on a per particle basis. Dividing the above equation by \(N_A\) to acheive this:
\[F_{pot} = \frac{-k_B T}{c}\frac{dc}{dz}.\label{F_pot_molec}\]
If we assume sperical particles and that the Stoke's macroscopic rule for the friction experienced by a spherical particle holds for molecule sized particles we have the balancing force as:
\[f = 6 \pi\eta R_{eff} s.\label{stokes}\]
Setting the forces in equations \(\ref{F_pot_molec}\) and \(\ref{stokes}\) equal and solving for \(sc = J_z\):
\[\frac{-k_B T}{c}\frac{dc}{dz} = 6 \pi\eta R_{eff} s \implies sc = \frac{-k_B T}{6 \pi\eta R_{eff}}\frac{dc}{dz}.\]
Comparing the final expression with equation \(\ref{eq1}\) we see that:
\[D =\frac{k_B T}{6 \pi \eta R_{eff}}.\label{SES2}\]