Foundations of Chemical Kinetics Lecture 27: Further developments of diffusion theory Marc R. Roussel Department of Chemistry and Biochemistry
Stokes-Einstein theory ◮ In the last lecture, we obtained the equation D i = k B T / f i where f i is a frictional coefficient. ◮ Equations for frictional coefficients exist for objects of various shapes immersed in a fluid. In particular, for a sphere, f i = 6 π r i η where r i is the radius of the sphere, and η is the viscosity of the solvent. ◮ By combining the two, we get the Stokes-Einstein equation: D i = k B T 6 π r i η
Stokes-Einstein theory: example ◮ C 60 has a diameter of 10.18 ˚ A. ◮ Benzonitrile has a viscosity of 1.24 mPa s at 25 ◦ C. ◮ According to the Stokes-Einstein theory, C 60 in benzonitrile should have a diffusion coefficient at 25 ◦ C of D = k B T 6 π r η = (1 . 380 6488 × 10 − 23 J K − 1 )(298 . 15 K) 6 π (5 . 09 × 10 − 10 m)(1 . 24 × 10 − 3 Pa s) = 3 . 46 × 10 − 10 m 2 s − 1 ◮ Experimental value: (4 . 1 ± 0 . 3) × 10 − 10 m 2 s − 1
Diffusive motion of ions in solution ◮ In addition to diffusion, ions in solution experience forces due to their charges. ◮ The force on an ion of charge z i (in elementary units) in an electric field E is F i = z i e E Warning: The textbook uses E to represent the electrostatic potential, which is very nonstandard. I use E to represent the electric field vector. The electrostatic potential V and electric field E are related by E = −∇ V ∇ E in the textbook corresponds roughly to ∇ V , give or take some funny stuff with signs.
Diffusive motion of ions in solution (continued) ◮ The electric force is balanced by the drag force F ( d ) = − f i v i , i so f i v i = z i e E or v i = z i e E / f i ◮ Each component of the velocity is proportional to the corresponding component of the electric field. Define the mobility of an ion as µ i = v i / E = | z i | e / f i so that v i = sgn( z i ) µ i E where sgn( z i ) is the sign of the charge.
Diffusive motion of ions in solution (continued) ◮ The flux of ion i due solely to the electric field is J ( E ) = c i v i = sgn( z i ) c i µ i E i ◮ Combining the flux due to the electric field with the flux due to diffusion, we get an overall flux J i = − D i ∇ c i + sgn( z i ) c i µ i E ◮ Applying the transport equation, ∂ c i /∂ t = −∇ · J i , we get the diffusion-conduction equation: ∂ c i ∂ t = D i ∇ 2 c i + sgn( z i ) µ i ∇ · ( c i E )
Diffusion coefficient and mobility ◮ Recall D i = k B T / f i and µ i = | z i | e / f i ◮ Therefore, µ i = | z i | e k B T D i
Mathematical interlude: Gradient in spherical polar coordinates In spherical polar coordinates, ˆ ˆ r ∂ θ ∂ φ ∂ ∇ = ˆ ∂ r + ∂θ + r r sin θ ∂φ r , ˆ θ and ˆ where ˆ φ are unit vectors in the corresponding directions. z ˆ φ ˆ r ˆ θ r θ y φ x
Equilibrium distribution in a spherically symmetric potential ◮ Suppose that an ion is placed in a spherically symmetric electrostatic potential (e.g. the potential due to another ion) with V ( r → ∞ ) = 0. For a spherically symmetric potential, V = V ( r ). ◮ The equilibrium solution of the diffusion-conduction equation will have the same symmetry, i.e. c i = c i ( r ). ◮ At equilibrium, J = 0 .
Equilibrium distribution in a spherically symmetric potential (continued) ◮ The flux is given by J i = − D i ∇ c i + sgn( z i ) c i µ i E = − D i ∇ c i − sgn( z i ) c i µ i ∇ V since E = −∇ V . ◮ If c i and V only depend on r , the equilibrium condition becomes dc i dV J i = 0 = − D i dr − sgn( z i ) c i µ i dr ◮ Separation of variables: dc i = − sgn( z i ) µ i dV c i D i
Equilibrium distribution in a spherically symmetric potential (continued) ◮ Substitute µ i / D i = | z i | e / k B T : dc i = − z i e k B T dV c i � c i ( r ) � V ( r ) dc i = − z i e dV ∴ c i k B T c ◦ 0 i � c i ( r ) � = − z i e k B T V ( r ) = − U ( r ) ∴ ln c ◦ k B T i where U ( r ) is the electrostatic potential energy of ion i in the potential V ( r ). � − U ( r ) � ∴ c i ( r ) = c ◦ i exp k B T which is a Boltzmann distribution!
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