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1/18 2/18 Transport phenomena co07 —macroscopic view co07

Transport (kinetic) phenomena: : ~J of compound  (units: mol m 2 s 1) First Fick Law  − − diffusion, electric conductivity, , conduction . . . For ~ J = D∇~c concentration Flux of mass, charge, , heat, ...... − in kg m 3, ~J = amount (of quantity) transported per unit area (perpendicular is proportional to the concentration gradient − the flux is in to the vector of flux) within time unit ∂ ∂ ∂ ∂c ∂c ∂c 2 1 2 1 2 kg m s Units: /heat flux: J m s = W m , ∇~c = grad c = , , c = , , − − − − − ∂ ∂y ∂z ∂ ∂y ∂z current : A m 2     − 2 1 D = diffusion coefficient (diffusivity) of , unit: m s Cause = (generalized, thermodynamic) − ~ = gradient of a potential Example. A U-shaped pipe of lenght  = 20 cm and cross section F − 0.3 cm2. One end is in Coca-Cola (11 wt.% of sugar), other (/concentration, electric potential, tempera- A =

ture) end in pure water. How much sugar is transported by diffusion in 6 2 1 mg 0.74 one day? Dsucrose(25 C) = 5.2 10 cm s .

Small —linearity ◦ × − −

× × × × × ×

~ ~ A

− − − − J const − = = =

kg 10 4 . 7 s 60 24 m 10 3 . 0 s m kg 10 56 . 2 t J = m

×

7 2 2 4 1 2

F 7

− − −

= · =

 grad

s m kg 10 56 . 2 c D J × ×

1 2 7

− − − −

= In gases we use the kinetic theory: molecules (simplest: hard spheres) fly through =

s m 10 2 . 5 s cm 10 2 . 5 D

1 2 10 1 2 6

  grad

5 gm kg 550 = / c = c

space and sometimes collide 4

− −

= = 

gm kg 110 dm g 110 c L: 1 in sucrose g 110

3 3

3/18 [blend -g che/sucrose] 4/18 Diffusion—microscopic view co07 Einstein–Stokes equation co07

Flux is given by the mean velocity of molecules ~: Colloid particles or large spherical molecules of radius R in a solvent of viscosity η it holds (Stokes formula) ƒ ~J = ~c Difference of chemical poten- Arrhenius law for vis- Thermodynamic force = grad ~ = 6πηR. ~ tials = reversible work needed F cosity (decreases with of the chemical potential:− to move a particle (mole) from Einstein–Stokes equation: increasing T), diffusiv- ⇒ μ kBT one state to another ity (increases with in- ~ = ∇~ = ∇~c kBT kBT F − NA − c D = D = creasing T   ƒ 6πηR st  ⇒  where formula μ = μ e + RT ln(c/c ) for infinity dillution was used.  Opposite reasoning—hydrodynamic (Stokes) radius defined as: Friction force acting against moving by velocity ~ through a medium is: kBT ~fr ~ R =  = ƒ 6πηD F −  where ƒ is the friction coeficient. Both forces are in equilibrium: effective molecule size (incl. solvation shell) ~ ≈ fr fr J kBT Example. Estimate the size of the sucrose molecule. Wa- ~ +  = 0 i.e. ~ = ƒ~ = ƒ =  = ∇~c

F F F c F c ter viscosity is 0.891 10 3 m 1 kg s 1 at 25 C.

−  −  − − − ◦

= nm 47 . 0 × R kBT On comparing with ~J = D∇~c we get the Einstein equation: D = − ƒ (also Einstein–Smoluchowski equation)

5/18 [plot/cukr.sh] 6/18 Second Fick Law co07 Second Fick Law co07

Non-stationary phenomenon (c changes with time). Example. Coca-Cola in a cylinder (height 10 cm) + pure water (10 cm). What time The amount of substance increases within is needed until the surface concentration = half of bottom concentration? months 4 time dτ in volume dV = ddydz: Fourier method: 2 [J() J( + d)] dydz ∂c ∂ c c0  < /2 = D c(, 0) = ,y,z − ∂τ ∂2 0  > /2 X § ∂J 2 J  {J  d} dydz c0 2c0 π π = [ ( ) ( ) + )] c , τ cos exp Dτ ,y,z − ∂ ( ) = +  2 X 2 π  −  ! ∂J  ‹ = ddydz = ∇~ ~J dV = ∇~ ( D∇~c) dV  − ,y,z ∂ − · − · − 2 2 2 2 X 1 3π 3 π 1 5π 5 π ∂2 ∂2 ∂2 cos exp Dτ + cos exp Dτ 2 3  2 5  2  = D∇~ c dV = D + + c dV −   − !   − ! ··· ∂2 ∂y2 ∂z2 !  This type of equation is called ∂c  2 “equation of heat conduction”. = D∇ c ∂τ It is a parrabolic partial differ- ential equation

[traj/brown.sh] 7/18 [show/galton.sh] 8/18 Diffusion and the co07 Brownian motion as a random walk + co07 Instead of for c(r,~ τ), let us solve the 2nd Fick (Smoluchowski, Einstein) law for the probability of finding a particle, start- 0.5 t=0 t=1 within time Δτ, a particle moves randomly ing from origin at τ = 0. We get the Gaussian 0.4 t=2 – by Δ with probability 1/2 with half-width ∝ distribution t=3 – by Δ with probability 1/2 2 0.3 t=4 − 1/2 

c(x,t) t=5 1D: c(, τ) = (4πDτ)− exp − 4Dτ! 0.2 Using the central limit theorem: 2 2 r2 in one step: Var  =  = Δ 3/2 0.1 〈 〉 3D: c(r,~ τ) = (4πDτ)− exp 2 − 4Dτ! in n steps (in time τ = nΔτ): Var  = nΔ 0.0 2 −3 −2 −1 0 1 2 3 Gaussian normal distribution with σ = nΔ = τ/ΔτΔ: x ⇒ p p 2 1 2 2 1 pΔτ  Δτ 2 200 e  /2σ exp 1D:  = 2Dτ − = − 2 〈 〉 p2πσ p2πτ Δ − 2τ Δ    Last example – order-of-magnintude which is for 2D Δ2/Δτ the same as c , τ 2 y y 0 = ( ) τ  /2D = 4 months ≈ (for  = 0.1 m) 0 def. 2 2 NB: Var  = (  ) , for  = 0, then Var  = 

〈 − 〈 〉 〉 〈 〉 〈 〉 2 1/6 3D: r = 6Dτ 0 0 Example. Calculate Var , where  is a random number from interval ( 1, 1) 〈 〉 x x − 9/18 10/18 Electric conductivity co07 Electric conductivity co07

Ohm Law (here: U = voltage, U = ϕ2 ϕ1): substance κ/ S m 1 − ( − ) U 1 graphene 1 108 R =  = U 1/R = conductivity, [1/R] = 1/Ω = S = Siemens ×  R silver 63 106 × (Specific) conductivity (conductance) κ is 1/resistance of a unit cube sea water 5 1 Ge 2.2 1 = κ A = area,  = layer thickness, [κ] = S m R  A − tap water 0.005 to 0.05 Si 1.6 10 3 − ~ ~ ~ × Vector notation: j = κ = κ∇ϕ distilled water (contains CO ) 7.5 10 5 E − 2 − × 6 ~j = el. current density, j = / , ~ = el. field intensity, = U/ deionized water 5.5 10 A E E × − glass 1 10 15–1 10 11 × − × − teflon 1 10 25–1 10 23 × − × −

11/18 12/18 Molar conductivity co07 Mobility and molar conductivity co07

Strong electrolytes: conductivity proportional to concentration. Mobility of an ion = averaged velocity in a unit electric field: Molar conductivity λ:   = = U/ = el. intensity, U = voltage κ E λ = E c Charges ze of velocity  and concentration c cause the current density “κ ” Units: κ S m 1, λ S m2 mol 1.  [ ] = − [ ] = − Watch units—best convert c to mol m 3! ! . − j = czF =  czF = λc λ = zF = molar conductivity of ion  E E ⇒

Example. Conductivity of a 0.1 M solution of HCl o is 4 S m 1. Calculate the molar Ions (in dilute solutions) migrate independently (Kohlrausch law), for electrolyte

− −

z zA mol .4Sm S 0.04 +

1 conductivity. 2 Cννν⊕ Aννν −: here we define zA > 0 ⊕ j j j λ c λ c λ ννν λ ννν c = A + C = ( A A + C C)E = ( A A + C C) E Mathematically κ λ = = νννλ c  X

[cd pic;mz grotthuss.gif]13/18 14/18 Nothing is ideal co07 Conductivity of weak electrolytes co07 Limiting molar conductivity = molar conductivity at infinite dilution We count ions only, not unionized acid ∞ λ = lim λ  c 0 In the limiting concentration: Departure from the limiting linear behavior→ (cf. Debye–Hückel theory): ∞ ∞ ∞ ∞ def. exptl 0.04 λ = λ(c) = λ constpc nebo λ = λ const c κ = λ cions = λ αc = λ c − − HCl Typical values: Æ λexptl cation λ∞/ S m2 mol 1 anion λ∞/ S m2 mol 1 α = 0.03 ( − ) ( − ) λ∞ +

H 0.035 OH 0.020 /mol

− 2 Na+ 0.0050 Cl 0.0076 Ostwald’s dilution law: NaOH − 2 exptl 2 0.02 Ca2+ 0.012 SO 2 0.016 c α c (λ ) / S m 4 λ − K = = cst 1 α cst λ∞ λ∞ λexptl Mobility and molar conductivity decreases with the ion size (Cl is slow), solva- ( ) − − − + tion (small Li + 4 H2O is slow) 0.01 AgNO3

H+, OH are fast movie credit: Matt K. Petersen, Wikipedia − CuSO4 CH3COOH 0 0 0.2 0.4 0.6 0.8 1 1.2 c1/2 / (mol/L)1/2 →

15/18 [pic/nernstovavrstva.sh]16/18 Conductivity and the diffusion coefficient co07 Transference numbers co07 Einstein (Nernst–Einstein) equation: microscopically: Transference number (transport number) of an ion is the fraction of the total z e k T k T k T k T RT  current that is carried by that ion during migration (electrolysis). B B B B   = D D = = = = = kBT    = velocity ƒ / ze /( ) ze/ zF t F E E here z is with sign = =   +  ννν = stechiom. coeff. 2 2 ⊕ z F z + z zFD = RT λ = zF = D Ions move at different speeds under the same field. For Kννν⊕ Aννν − (electroneutrality: RT ⇒ z c = z c ; here z > 0) ⊕ ⊕ ⊕ diffusion: caused by a gradient of concentration/chemical potential  c z   z D ννν λ t = = = = = D  c z +  c z  +   +  z D + z D ννν λ + ννν λ ~J = D∇~c = c ∇~μ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ − − RT t  ze zFD Properties: t + t = 1, =  =  ,  = D, λ = zF ~j = c ∇~μ = c∇~μ ⊕ t  E kBT − RT − ⊕ ⊕ Example. Electrolysis of CuSO : t 2+ 40 %, t 2 60 %. 4 Cu = SO4 = migration: caused by el. field − Nernst layer bulk ~j = κ∇~ϕ = cλ∇~ϕ = czF∇~ϕ cathode − − − total 100 % Cu2+ Let us define the electrochemical potential μ˜ μ zFϕ, then migration 40 % = + ←− ←− diffusion 60 % 2+ 2+ ··· Cu D z F λ Cu ←− ←−    ←− ←− ~j = c∇~μ˜ = c ∇~μ˜ = c ∇~μ˜ Cu diffusion 2 60 % migration 2 60 % RT z F SO4 SO4 − − −  − − ←− −→ −→ ←−

17/18 18/18 Examples co07 Conductivity measurements co07

Example. Calculate the specific conductivity of a uni-univalent electrolyte MA of Usage: determination of ion concentrations (usually small) concentration 0.01 mol dm 3, assuming that both M and A are of the same size as solubility, dissociation constants, conductometric titration. . . −

6 2 1 ⇒

the sucrose molecule (D = 5.2 10 cm s at 25 C)? ±

− −

− − ◦ −

= =

) mol m S 002 . 0 λ ( m S 04 . 0

× κ 1 2 1

Note: 0.01 M of KCl has κ 0.14 S m 1 > 0.04 S m 1, because the ions are smaller = − − than sucrose

Calculate the migration speeds of ions M+,A between electrodes 1 cm

Example. −

×

apart with applied voltage 2 V. −

− − −

A M = = =

+ mh mm 15 s m 10 4  

1 1 6

Resistance constant of conductance cell (probe) C ( = m 1): ze −  =   = D λ = zF E kBT 1  = κ A Rκ = = C R  · ⇒ A C determined from a solution of known conductivity (e.g., KCl), C = R κ .