7.5. MOLECULAR DIFFUSION PLUS CONVECTION AND CHEMICAL REACTION

7.5A. Different Types of Fluxes and Fick's Law

In Section 6.2B the flux was defined as the molar flux of A in kg mol A/s · m2 relative to the molar average velocity νM of the whole or bulk stream. Also, NA was defined as the molar flux of A relative to stationary coordinates. Fluxes and velocities can also be defined in other ways. Table 7.5-1 lists the different types of fluxes and velocities often used in binary systems.

Table 7.5-1. Different Types of Fluxes and Velocities in Binary Systems
  Mass Flux (kg A/s · m2)Molar Flux (kg mol A/s · m2)
Relative to fixed coordinatesnA = ρAνANA = cAνA
Relative to molar average velocity νM
Relative to mass average velocity νjA =ρA(νAν)JA = cA(νAν)
Relations Between Fluxes Above
 NA + NB = MNA = nA/MANA = JA + cAν
 JA = jA/MAnA + nB = ρν
 jA + jB = 0nA = jA + ρAν
Different Forms of Fick's Law for Diffusion Flux

The velocity ν is the mass average velocity of the stream relative to stationary coordinates and can be obtained by actually weighing the flow for a timed increment. It is related to the velocity νA and νB by

Equation 7.5-1


where wA is ρA/ρ, the weight fraction of A; wB is the weight fraction of B; and νA is the velocity of A relative to stationary coordinates in m/s. The molar average velocity νM in m/s is relative to stationary coordinates:

Equation 7.5-2


The molar diffusion flux relative to the molar average velocity νM defined previously is

Equation 7.5-3


The molar diffusion flux JA relative to the mass average velocity ν is

Equation 7.5-4


Fick's law from Table 7.5-1 as given previously is relative to νM:

Equation 7.5-5


Fick's law can also be defined in terms of a mass flux relative to ν as given in Table 7.5-1:

Equation 7.5-6


EXAMPLE 7.5-1. Proof of Mass Flux Equation

Table 7.5-1 gives the following relation:

Equation 7.5-7


Prove this relationship using the definitions of the fluxes in terms of velocities.

Solution: From Table 7.5-1, substituting ρA(νAν) for jA and ρB(νBν) for jB, and rearranging,

Equation 7.5-8


Equation 7.5-9


Substituting Eq. (7.5-1) for ν and ρ for ρA + ρB, the identity is proved.


7.5B. Equation of Continuity for a Binary Mixture

A general equation can be derived for a binary mixture of A and B for diffusion and convection that also includes the terms for unsteady-state diffusion and chemical reaction. We shall make a mass balance on component A on an element Δx Δy Δz fixed in space as shown in Fig. 7.5-1. The general mass balance on A is

Equation 7.5-10


Figure 7.5-1. Mass balance for A in a binary mixture.


The rate of mass A entering in the direction relative to stationary coordinates is (nAx|xy Δz kg A/s and leaving is (nAx|x+Δxy Δz. Similar terms can be written for the y and z directions. The rate of chemical production of A is rA kg A generated/s · m3 volume and the total rate generated is rAx Δy Δz) kg A/s. The rate of accumulation of A is (∂ρA/∂tx Δy Δz. Substituting into Eq. (7.5-10) and letting Δx, Δy, and Δz approach zero,

Equation 7.5-11


In vector notation,

Equation 7.5-12


Dividing both sides of Eq. (7.5-11) by MA,

Equation 7.5-13


where RA is kg mol A generated/s · m3. Substituting NA and Fick's law from Table 7.5-1,

Equation 7.5-14


and writing the equation for all three directions, Eq. (7.5-13) becomes

Equation 7.5-15


This is the final general equation.

7.5C. Special Cases of the Equation of Continuity

1. Equation for constant c and DAB

In diffusion with gases the total pressure P is often constant. Then, since c = P/RT, c is constant for constant temperature T. Starting with the general equation (7.5-15) and substituting ∇xA = ∇cA/c, we obtain

Equation 7.5-16


2. Equimolar counterdiffusion for gases

For the special case of equimolar counterdiffusion of gases at constant pressure and no reaction, c = constant, νM = 0, DAB = constant, RA = 0, and Eq. (7.5-15) becomes

Equation 7.1-9


This equation is Eq. (7.1-9), derived previously, which is also used for unsteady-state diffusion of a dilute solute A in a solid or a liquid when DAB is constant.

3. Equation for constant ρ and DAB (liquids)

In dilute liquid solutions, the mass density ρ and DAB can often be considered constant. Starting with Eq. (7.5-12) we substitute nA = −ρDABwA + ρAν from Table 7.5-1 into this equation. Then using the fact that for constant ρ, ∇wA = ∇ρA/ρ and also that (▽ · ν) = 0, substituting these into the resulting equation, and dividing both sides by MA, we obtain

Equation 7.5-17


7.5D. Special Cases of the General Diffusion Equation at Steady State

1. Introduction and physical conditions at the boundaries

The general equation for diffusion and convection of a binary mixture in one direction with no chemical reaction has been given previously:

Equation 6.2-14


To integrate this equation at steady state, it is necessary to specify the boundary conditions at z1 and z2. In many mass-transfer problems the molar ratio NA/NB is determined by the physical conditions occurring at the two boundaries.

As an example, one boundary of the diffusion path may be impermeable to species B because B is insoluble in the phase at this boundary. Diffusion of ammonia (A) and nitrogen (B) through a gas phase to a water phase at the boundary is such a case, since nitrogen is essentially insoluble in water. Hence, NB = 0, since at steady state NB must have the same value at all points in the path of z2z1. In some cases, a heat balance in the adjacent phase at the boundary can determine the flux ratios. For example, if component A condenses at a boundary and releases its latent heat to component B, which vaporizes and diffuses back, the ratios of the latent heats determine the flux ratio.

In another example, the boundary concentration can be fixed by having a large volume of a phase flowing rapidly by with a given concentration xA1. In some cases the concentration xA1 may be set by an equilibrium condition, whereby xA1 is in equilibrium with some fixed composition at the boundary. Chemical reactions can also influence the rates of diffusion and the boundary conditions.

2. Equimolar counterdiffusion

For the special case of equimolar counterdiffusion, where NA = −NB, Eq. (6.2-14) becomes, as shown previously, for steady state and constant c,

Equation 7.5-18


3. Diffusion of A through stagnant, nondiffusing B

For gas A diffusing through stagnant nondiffusing gas B, NB = 0, and integration of Eq. (6.2-14) gives Eq. (6.2-22):

Equation 6.2-22


Several other more complicated cases of integration of Eq. (6.2-14) are considered next.

4. Diffusion and chemical reaction at a boundary

Often in catalytic reactions, where A and B are diffusing to and from a catalyst surface, the relation between the fluxes NA and NB at steady state is controlled by the stoichiometry of a reaction at a boundary. An example is gas A diffusing from the bulk gas phase to the catalyst surface, where it reacts instantaneously and irreversibly in a heterogeneous reaction as follows:

Equation 7.5-19


Gas B then diffuses back, as shown in Fig. 7.5-2.

Figure 7.5-2. Diffusion of A and heterogeneous reaction at a surface.


At steady state 1 mol of A diffuses to the catalyst for every 2 mol of B diffusing away, or NB = −2NA. The negative sign indicates that the fluxes are in opposite directions. Rewriting Eq. (6.2-14) in terms of mole fractions,

Equation 7.5-20


Next, substituting NB = −2NA into Eq. (7.5-20),

Equation 7.5-21


Rearranging and integrating with constant c (P = constant), we obtain the following:

1. Instantaneous surface reaction

Equation 7.5-22


Equation 7.5-23


Since the reaction is instantaneous, xA2 = 0, because no A can exist next to the catalyst surface. Equation (7.5-23) describes the overall rate of the process of diffusion plus instantaneous chemical reaction.

2. Slow surface reaction

If the heterogeneous reaction at the surface is not instantaneous but slow for the reaction A → 2B, and the reaction is first order,

Equation 7.5-24


where is the first-order heterogeneous reaction velocity constant in m/s. Equation (7.5-23) still holds for this case, but the boundary condition xA2 at z = δ is obtained by solving for xA in Eq. (7.5-24),

Equation 7.5-25


For steady state, NAz=δ = NA. Substituting Eq. (7.5-25) into (7.5-23),

Equation 7.5-26


The rate in Eq. (7.5-26) is less than in Eq. (7.5-23), since the denominator in the latter equation is 1 + xA2 = 1 + 0 and in the former is .

EXAMPLE 7.5-2. Diffusion and Chemical Reaction at a Boundary

Pure gas A diffuses from point 1 at a partial pressure of 101.32 kPa to point 2 a distance 2.00 mm away. At point 2 it undergoes a chemical reaction at the catalyst surface and A → 2B. Component B diffuses back at steady state. The total pressure is P = 101.32 kPa. The temperature is 300 K and DAB = 0.15 × 104 m2/s.

  1. For instantaneous rate of reaction, calculate xA2 and NA.

  2. For a slow reaction where m/s, calculate xA2 and NA.

Solution: For part (a), pA2 = xA2 = 0 since no A can exist next to the catalyst surface. Since NB = −2NA, Eq. (7.5-23) will be used as follows: δ = 2.00 × 103m, T = 300 K, c = P/RT = 101.32 × 103/(8314 × 300) = 4.062 × 102 kg mol/m3, xA1 = pA1/P = 101.32 × 103/101.32 × 103 = 1.00.


For part (b), from Eq. (7.5-25), . Substituting into Eq. (7.5-26),


Solving by trial and error, NA = 1.004 × 104 kg mol A/s · m2. Then xA2 = (1.004 × 104)/(5.63 × 103 × 4.062 × 102) = 0.4390.


Even though in part (a) of Example 7.5-2 the rate of reaction is instantaneous, the flux NA is diffusion-controlled. As the reaction rate slows, the flux NA is decreased also.

5. Diffusion and homogeneous reaction in a phase

Equation (7.5-23) was derived for the case of chemical reaction of A at the boundary on a catalyst surface. In some cases component A undergoes an irreversible chemical reaction in the homogeneous phase B while diffusing, as follows: AC. Assume that component A is very dilute in phase B, which can be a gas or a liquid. Then at steady state the equation for diffusion of A is as follows, where the bulk-flow term is dropped:

Equation 7.5-27


Writing a material balance on A shown in Fig. 7.5-3 for the Δz element for steady state,

Equation 7.5-28


Figure 7.5-3. Homogeneous chemical reaction and diffusion in a fluid.


The first-order reaction rate of A per m3 volume is

Equation 7.5-29


where k' is the reaction velocity constant in s1. Substituting into Eq. (7.5-28) for a cross-sectional area of 1 m2 with the rate of accumulation being 0 at steady state,

Equation 7.5-30


Next we divide through by Δz and let Δz approach zero:

Equation 7.5-31


Substituting Eq. (7.5-27) into (7.5-31),

Equation 7.5-32


The boundary conditions are cA = cA1 for z = 0 and cA = cA2 for z = L. Solving,

Equation 7.5-33


This equation can be used at steady state to calculate cA at any z and can be used for reactions in gases, liquids, or even solids, where the solute A is dilute.

As an alternative derivation of Eq. (7.5-32), we can use Eq. (7.5-17) for constant ρ and DAB:

Equation 7.5-17


We set the first term ∂cA/∂t = 0 for steady state. Since we are assuming dilute solutions and neglecting the bulk flow term, ν = 0, making the second term in Eq. (7.5-17) zero. For a first-order reaction of A where A disappears, kg mol A generated/s · m3. Writing the diffusion term for the z direction only, we obtain

Equation 7.5-34


which is, of course, identical to Eq. (7.5-32).

7.5E. Unsteady-State Diffusion and Reaction in a Semi-infinite Medium

Here we consider a case where dilute A is absorbed at the surface of a solid or stagnant fluid phase and then unsteady-state diffusion and reaction occur in the phase. The fluid or solid phase B is considered semi-infinite. At the surface, where z = 0, the concentration of cA is kept constant at cA0. The dilute solute A reacts by a first-order mechanism

Equation 7.5-35


and the rate of generation is . The same diagram as in Fig. 7.5-3 holds. Using Eq. (7.5-30) but substituting (∂cA/∂t)(Δz)(1) for the rate of accumulation,

Equation 7.5-36


This becomes

Equation 7.5-37


The initial and boundary conditions are

Equation 7.5-38


The solution by Danckwerts (D1) is

Equation 7.5-39


The total amount Q of A absorbed up to time t is

Equation 7.5-40


where Q is kg mol A absorbed/m2. Many actual cases are approximated by this case. The equation is useful where absorption occurs at the surface of a stagnant fluid or a solid and unsteady-state diffusion and reaction occur in the solid or fluid. The results can be used to measure the diffusivity of a gas in a solution, to determine reaction rate constants k' of dissolved gases, and to determine solubilities of gases in liquids with which they react. Details are given elsewhere (D3).

EXAMPLE 7.5-3. Reaction and Unsteady-State Diffusion

Pure CO2 gas at 101.32 kPa pressure is absorbed into a dilute alkaline buffer solution containing a catalyst. The dilute, absorbed solute CO2 undergoes a first-order reaction, with k' = 35 s−1 and DAB = 1.5 × 109 m2/s. The solubility of CO2 is 2.961 × 107 kg mol/m3 · Pa (D3). The surface is exposed to the gas for 0.010 s. Calculate the kg mol CO2 absorbed/m2 surface.

Solution: For use in Eq. (7.5-40), k't = 35(0.01) = 0.350. Also, cA0 = 2.961 × 107 (kg mol/m3 · Pa)(101.32 × 103 Pa) = 3.00 × 102 kg mol SO2/m3.



7.5F. Multicomponent Diffusion of Gases

The equations derived in this chapter have been for a binary system of A and B, which is probably the most important and most useful one. However, multicomponent diffusion sometimes occurs where three or more components A, B, C, . . . are present. The simplest case is for diffusion of A in a gas through a stagnant, nondiffusing mixture of B, C, D, . . . at constant total pressure. Hence, NB = 0, Nc = 0, . . . . The final equation derived using the Stefan–Maxwell method (G1) for steady-state diffusion is

Equation 7.5-41


where piM is the log mean of pi1 = PpA1 and pi2 = PpA2. Also,

Equation 7.5-42


where = mol B/mol inerts = xB/(1−xA), .

EXAMPLE 7.5-4. Diffusion of A Through Nondiffusing B and C

At 298 K and 1 atm total pressure, methane (A) is diffusing at steady state through nondiffusing argon (B) and helium (C). At z1 = 0, the partial pressures in atm are pA1 = 0.4, pB1 = 0.4, and pC1 = 0.2, and at z2 = 0.005 m, pA2 = 0.1, pB2 = 0.6, and pC2 = 0.3. The binary diffusivities from Table 6.2-1 are DAB = 2.02 × 105 m2/s, DAC = 6.75 × 105 m2/s, and DBC = 7.29 × 105 m2/s. Calculate NA.

Solution: At point 1, = xB/(1 − xA) = 0.4/(1 − 0.4) = 0.667. At point 2, = 0.6/(1 − 0.1) = 0.667. The value of is constant throughout the path. Also, .

Substituting into Eq. (7.5-42),


For calculating piM, pi1 = PpA1 = 1.0 − 0.4 = 0.6 atm, pi2 = PpA2 = 1.0 − 0.1 = 0.90. Then,


Substituting into Eq. (7.5-41),


Using atm pressure units,


A number of analytical solutions have been obtained for other cases, such as for equimolar diffusion of three components, diffusion of components A and B through stagnant C, and the general case of two or more components diffusing in a multicomponent mixture. These are discussed in detail with examples by Geankoplis (G1); the reader is referred there for further details.


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