Mass Transfer Heterogeneous and Homogeneous Chemical … · Mechanism of irreversible heterogeneous reactions -Heterogeneous reactions of unusual stoichiometries Overview. Mass Transfer

Heterogeneous and Homogeneous Chemical Reactions

Mass Transfer

Lecture 11, 29.11.2017, Dr. K. Wegner

Mass Transfer – Chemical reactions 11-2

- DefinitionsReaction rateReaction orderEquilibrium constant

- Heterogeneous & homogeneous reactions

- Diffusion-controlled reactions- Diffusion and 1st order heterogeneous reactions- Mechanism of irreversible heterogeneous reactions - Heterogeneous reactions of unusual stoichiometries

Overview


Review “Chemical reactions”A chemical reaction is a conversion of one set of chemical substances (reactants A, B) into another (products C, D).

The reaction rate is the decrease in reactant concentration or the increase in product concentration with time:

(1)

(Valid for a single reaction in a closed system of constant volume.)

a, b, c, d: stoichiometric coefficients

dtdc

d1

dtdc

c1

dtdc

b1

dtdc

a1r DCBA ==−=−=→

aA + bB → cC + dD


For single step (elementary) reactions, the reaction rate depends on the reactant concentration as follows:

(2)

κ is the rate constant.

orbB

aA cc~r ⋅→

bB

aA ccr ⋅⋅κ= →→

2HI

HI cdt

dc21r ⋅κ=−= →→2 HI → I2 + H2

For instance:

The rate (velocity) of a reaction can depend on many factors such as the concentrations (partial pressures) of the species, the temperature, catalysts, mass transfer,...


Many chemical reactions are reversible and forward and reverse reactions finally lead to an equilibrium:

We can define (1) and (2) also for the reverse single-step reaction:

dtdc

d1

dtdc

c1

dtdc

b1

dtdc

a1r DCBA −=−=+=+=←

dD

CC ccr ⋅⋅κ= ←←

2 HI ⇆ H2 + I2 22

22IH

IH ccdt

dcdt

dcr ⋅⋅κ=−=−= ←←

aA + bB ⇆ cC + dD


Now, the rate of the total reaction is the difference in the rates of the forward and reverse reactions:

the reaction has reached the chemical equilibrium and the ratio of the rate constants is the equilibrium constant of the chemical reaction:

If and

For an irreversible reaction

→← =→ rr :0r →← κ=κ→κ and 0 bB

aA ccr ⋅⋅κ=so, e.g.

bB

aA

dD

cC

1 ccccK⋅⋅

=κκ

=←

→ e.g.2HI

IH1 c

ccK 22

⋅=


Reaction orderA reaction rate that is proportional to the concentration of one reactant is a first order reaction, e.g.:

One proportional to the concentrations of two reactants is of second order, e.g.:

1Acr ⋅κ= →→

1B

1A ccr ⋅⋅κ= →→

2Acr ⋅κ= →→Note, that also is a second order reaction.

A reaction rate not depending on any concentration is called a zero order reaction.


Typically this dependence has to be determined by experiments.

2HI

HI cdt

dc21r ⋅κ=−= →

2 HI → I2 + H2but

2 N2O → O2 + 2 N2

ONON

2

2 cdt

dc21r ⋅κ=−= →

Note: Reactions typically occur in a series of multiple reaction steps. In general, the contributions of all steps result in an overall reaction rate. In this case the exponents of the rate equation (2) are not equal to the stoichiometric factors a, b in (1) !!

For instance,

Frequently, one reaction step is much slower than the others and thus is the rate-determining step. If this step is an elementary reaction, eq. (2) can be applied.


Reactions and mass transferChemical reactions are always coupled with mass transfer since the reactants have to travel to the location where the conversion takes place (“sink” for the reactants) while the products have to travel away (“source” for the products).

Comparing the speed of mass transfer with that of reaction allows to determine the rate-limiting step and if rxns have to be accounted for:

reaction rate >> mass transfer rate → neglect reactionsreaction rate ≈ mass transfer rate → include reactionsreaction rate << mass transfer rate → neglect mass transfer

When reaction rate and mass transfer are comparable, we need to incorporate reaction into the mass transfer models!


Activation energyActivation energy is required for the reaction to start and to carry on. The activation energy is typically supplied as heat (higher temperature) but sometimes it can be other electromagnetic radiation, such as UV light.

A catalyst lowers the activation energy and increases the speed of the reaction at a certain temperature. The catalyst (typically a solid, but could also be a liquid) is not consumed after the reaction.

Examples of chemical reactions• Combustion (gaseous, liquid or solid fuels)• Production of chemicals (catalysts)• Fuel cells, batteries (catalysts)• Gas cleaning (e.g. removal of SO2 by CaO)• (Nano-) particle synthesis


Heterogeneous & homogeneous reactions

Heterogeneous reactions take place AT an interface (e.g. solid catalyst surface, coal combustion) and diffusion and reaction occur by steps IN SERIES.

Homogeneous reactions take place THROUGHOUT the volume in the same phase (fuel combustion in engine). Diffusion and reaction occur by steps partially in PARALLEL.

For simplification of mass transfer problems some heterogeneous reactions can be treated as homogeneous ones, depending on the definition of the observation space.


For example, in combustion of coal particles in fluidized beds reaction initially takes place AT the coal surface – so heterogeneous reaction.

CO2

O2

O2

CO2

O2

O2

O2

Coal

O2O2

CO2 O2

CO2O2

Later on, however, a highly porous inorganic structure (e.g. fly ash) develops and combustion takes place THROUGHOUT the particle.

Even though microscopically the reaction still is a surface reaction, macroscopic observation of the entire particle gives the picture of reactions occurring homogeneously throughout the particle.


Thus, this case can be described as a homogeneous reaction and the “homogeneous reaction model” can be applied:

The above example shows that the assignment of the reaction type can be subjective!

Combustion is an irreversible reaction, so


In summary, coal particle combustion can be described as:

a) Heterogeneous reaction and transport:0

1121 ν⋅∇−∇=

∂∂ ccD

tc

The reaction is introduced at the boundary conditions!

b) Homogeneous (homogeneous-like) reaction model and transport:

10

1121 rccD

tc

+ν⋅∇−∇=∂∂

(see: Generalized Mass Balances)


Examples of reaction assignment

Oxidation of lead sulfide (semiconductor) particles:

PbS particles → porous PbOheat

Removal of ammonia from off-gas:

NH3 / air + H2O → NH4OHscrubbing

Adsorption of sulfur dioxide (from combustion off-gas):

SO2 + Ca(OH)2/H2O CaSO4scrubbing

using Ca(OH)2 slurry


1. Diffusion-controlled reaction

Every diffusion-controlled process involves multiple steps. E.g. in dehydrogenation of C2H6 on Pt crystals

(1) C2H6 diffuses to Pt surface(2) C2H6 reacts on the surface(3) Product diffuses away from the

surface

(a)

When step 1 takes much longer than step 2 then we have a diffusion-controlled reaction.


In (a) we have a diffusion-controlled heterogeneous reaction where the reactant diffuses to a (e.g. hot or catalytic) surface, reacts and the product diffuses away.

(a)


In (b) we have a porous particle and the reactant diffuses to catalytically active sites where it reacts and the product diffuses away while reaction continues. Very important in catalysis, scrubbing and extraction (homogeneous reaction model).

(b)


In (c) slow-moving molecules react with more mobile ones facilitating the transport of the former across a membrane (example: oxygen uptake by hemoglobin in the lungs).

In case (d), molecules are well mixed and react upon contact. The reaction rate depends on the Brownian motion of the molecules (acid/base reactions, CH4 combustion etc.)

(d)

(c)

Goal: To find the overall rate of the process


2. Diffusion and 1st order heterogeneous reactions(dilute solutions)

A first-order reaction with respect to a reactant is one where the reaction rate is proportional to the reactant concentration:

i ir = cκ

At steady state, the overall reaction rate per area equals the diffusion fluxes

2

2232

1111

r )(

)(

=

−=−=

−=

cckncckn

i

i


The surface reaction is

Species 1 ↔ Species 2κ2

κ-2

And the reaction rate is given by

i22i122 ccr −κ−κ=

Note that the equilibrium constant is

2

22K

−κκ

=

To find the overall reaction we must eliminate the surface (or interfacial) concentrations that are hard to measure.


Following the analysis of the mass transfer across interfaces (“overall MTC”) we obtain:

−==

2

2112 K

ccKnr

Where the overall MTC or “resistance” is:

( )2321 Kk11k11K+κ+

=


Similarities/differences with interfacial mass transfer:

1. The resistances 1 / k1, 1 / κ2 and 1 / (k3K2) add to the total resistance 1/K

⇔

2 species of ionconcentrat bulkexisting the withmequilibriu in

1 species of ionconcentrat/ 22 Kc

2. Henry’s law constant (partition coefficient) characterizes distribution between phases: K2 ↔ HK2 and H vary over a wide range. Here, similarly to H, the K2determines the relative impact of k1 and k3.


Example 1: Limiting cases of 1st order reactions

(a) Fast stirringk1 and k3 are large, so K → κ2. Then,

−κ=

2

2122 K

ccr

(b) High temperatureUsually at high T the reactions are fast so 1/κ2 is very small and can be neglected

( )

−

+=

2

21

2312 K

ccKk1k1

1r

(c) Irreversible reaction κ-2 = 0 so K2 → ∞ 121

2 c1k1

1rκ+

=

Note that ONLY in this case the resistances are simply additive.

Determine the overall rate of the process when:

( )2321 Kk11k11K+κ+

=


Example 2: Cholesterol solubilization in bile

Bile* is the body’s "detergent" for fat digestion through cholesterol excretion. Failure of bile leads to cholesterol gallstones. Typically, gallstones are removed by surgery. However, lab data show that gallstones can be dissolved by administering certain components of the bile.

*Bile = "Galle" in German

Goal: Find the dissolution (“reaction”) rate of a 1 cm gallstoneAs a cholesterol gallstone is a solid, mass transfer/chemical reaction at a solid/fluid interface needs to be understood: Use the rotating (spinning) disk apparatus for determining cholesterol dissolution rates.


cholesterol dissolution rate = 5.39 · 10-9 g/(cm2 s) = r2solubility = 1.48 · 10-3 g/cm3 = c1D = 2·10-6 cm2/sdisk diameter, d = 1.59 cmRe = 11200bile kinematic viscosity, ν = 0.036 cm2/scholesterol solution, ∆ρ density over the bile = 1·10-5 g/cm3

If the reaction is irreversible find:(a) The surface reaction rate constant (spinning disk experiment)(b) The overall mass transfer coefficient K ("effective dissolution

rate") of a cholesterol gallstone with 1cm diameter.


(a) For this case21 /1k1

1Kκ+

=

Find k1 from the appropriate mass transfer coefficient using the correlation for the spinning disk:

s/cm1016.2D

ddD62.0k 3

31212

1−⋅=

ν

νω

=

So κ2 can be obtained from k1 and K as κ2 = 3.6·10-6 cm/s

scm1064.3

cmg1048.1)scm(g1039.5

crK cKr 6

33

29

1

212

−−

−

⋅=⋅⋅

==→⋅=


(b) dissolution rate of a cholesterol gallstone 1 cm in diameter

The 1 cm cholesterol gallstone is assumed to fixed in place, but the density gradient between cholesterol and bile results in flow by free convection so the mass transfer coefficient is:

1/4 1/331

2k d d g2 0.6D D

∆ρ ν = + ρν

3 5 3 21 31

6 2 3 2 2k 1cm (1cm) 1 10 g / cm 980cm / s2 0.6 (18000)

2 10 cm / s 1g / cm (0.036cm / s)

−

−

⋅ ⋅ ⋅= + ⋅ ⋅

51k 5.6 10 cm / s−= ⋅


As we are still dealing with irreversible reactions:

s/cm104.3

s/cm106.3/1s/cm106.5/11

/1k11K

6

65

21

−

−−

⋅=

⋅+⋅=

κ+=

In a unstirred bile, the rate is also controlled by the reaction.


3. Finding the mechanism of irreversibleheterogeneous reactions

Typically, the reaction mechanism is not known so it has to be inferred.


Physical situationRate-controlling step

Size R = ƒ(time, reagent)

Size = ƒ(temperature) Size = ƒ(flow) Remarks

A Shrinking Reaction R ∝ c1t strong temperature Independant of flow Other reactionparticle variation stoichomerties can

be found easilyB Shrinking External diffusion R2 ∝ (c1t) small particles Weak temperature Independent for small The exact variation

particle R3/2 ∝ (c1t)larger particles variation particles only with flow dependson the mass transfercoefficient

C Shrinking Reaction R ∝ c1t strong temperature Independent of flow This is the same ascorea case A, except for

ash formationD Shrinking External diffusion R ∝ c1t Weak Usually about square This case is

corea root of flow uncommonE Shrinking Ash diffusion R ∝ (c1t)1/2 Weak Independant of flow This case is common,

corea an interesting contrast with theprevious one

Notes: aThis is often called the topochemical model. The size R refers to the cone

The surface reaction takes place as ( )products2species

solid1species

gaseous→

+


Cases A and C: Surface reaction controls

2122 ccr κ−=

As the solid concentration c2 is constant the reaction is 1st order with respect to c1.

Mass balance per particle:

12

22122

2

222

3

cdtdr

r4ccdtdrcr4

r4rcr34

dtd

κ−=

πκ−=π

π⋅=

π

Be careful, distinguish between radius r and reaction rate r2!

If the gas phase concentration c1 is constant and the particle has an initial radius R0 tcRr 120 ⋅κ−=


Case B: Diffusion outside of a shrinking particle controlsIdentify correlation (forced convection around solid sphere)

νν

1/2 1/31k d d v= 2 + 0.6D D

B.1) For very small particles Re 0, so rDk

Ddk

=→= 2

Mass balance:

tcDcRr

ccD

dtdrr

crDr

dtdrcr

ccrkcrdtd

i

⋅

−=

−=

π−=π

−π⋅−=

π

2

120

2

12

12

22

112

23

2

44

)(434


B.2) For large particles

2/16/1

3/22/1

1

3/12/11

rDv42.0k

Dvd6.0

Ddk

−⋅

ν≅

ν

ν

≅

Similarly,

⋅ = ⋅ ν ×

= × ν

1/ 2 2 / 31/ 2

2 11/ 6

1/ 2 2 / 33 / 2 3 / 2 1

0 1/ 62

dr v Dc r -0.42 cdt

c0.64 v Dr R - tc


Cases C, D, and E: Shrinking core model

Two diffusional resistances in series

Bulk fluid↓

Particle surface↓

Unreacted core

R0

Physical situation Rate-controlling step

A Shrinking Reactionparticle

B Shrinking External diffusionparticle

C Shrinking Reactioncorea

D Shrinking External diffusioncorea

E Shrinking Ash diffusioncorea


Case D: Diffusion in the surrounding bulk fluid controls

Mass balance:

tcckRr

kcr4dtdrcr4

kcr4cr34

dtd

2

10

12

22

12

23

⋅

−=

π−=π

⋅π−=

π

Similar dependence of r with t as with cases A and C. Dependence on flow (through k) but weak dependence on temperature.


Case E: Diffusion in the ash (or shell) controls

Here the thickness of the shell is important (film model):

rRDk0 −

=

where D is the effective diffusivity through the shell

Mass balance:

2/1

2

10

0

122

3

tc

cD2Rr

rRcDr4cr

34

dtd

⋅

⋅−=

−⋅

⋅π−=

π


4. Heterogen. reactions of unusual stoichiometries4.1 Irreversible second-order heterogeneous reaction

(additivity of resistances)


Steady-state mass transfer to the surface: )cc(knr i11111 −==

Surface reaction: 2i121 cr ⋅κ=

Find the c1i by equating the above

−

κ+

κ=

=⋅−⋅+⋅κ

1k

c412kc

0ckckc

1

12

2

1i1

11i112i12

So the overall reaction rate is: κ

= − + − κ

1 2 11 1 1

1 2 1

k 4 cr k c 1 1 12c k


Compare this with the corresponding first order reaction:

κ= − + − κ

1 2 11 1 1

1 2 1

k 4 cr k c 1 1 12c k

121

1 /1/11 c

kr ⋅

κ+=

first order reactionsecond order reaction


4.2 Heterogeneous Reactions in Concentrated SolutionsUntil now we assumed that k1 ≠ f(r2). This is good for dilute solutions.

Cracking of hydrocarbons:

1 mol of species 1 → n moles of species 2 n > 1

Reforming

1 mol of species 1 → n moles of species 2 n < 1

κ2 κ2

In both cases we must account for CONVECTION in the calculation of mass transfer.


In cracking, convection is AWAY from the surface: The reacting species must diffuse AGAINST the current of product species.

In reforming, convection is TOWARDS the surface as reaction "pulls" reactants. The product must diffuse away.

ν > 1 crackingν < 1 reformingν = 1 treat as before

Goal: To calculate the overall rate!


Overall rate of reaction across this film is:

= = − = κ ⋅ν2

1 1 2 1inr n c

We must calculate the flux n1 for concentrated solutions:

01

11 vc

dzdcDn +−=

Now the flux from convection is

1210 n)1(nncv ν−=+=

So the equation for flux n1 can be written as: 0Dncn

Dc1

dzdc 1

111 =+

−ν+


B.C.’s: z = 0: c1 = c10 and z = l: c1 = c1i

Solving for n1 gives: + ν − κ ⋅= = − ν − + ν −

1 1 21 1

10

k c 1 ( 1)n /( c)r n ln( 1) 1 ( 1)c / c

where lDk1 =

For diffusion control (κ2·c10/n1 is large).

The diagram shows the rate relative to that for dilute solutions:


Consider studying cracking of oil by flowing it over a hot plate.If the molecular weight of the product is only 25% of that of the oil, by how much the convection introduced mass transfer changes the cracking reaction rate?

1 molecule oil = ν x 0.25 molecules of the product. So ν ≈ 4

As only oil is flowing (c10/c) = 1. So 1 + (ν - 1) c10/c ≈ 4.From the above diagram:

%40ratereaction1:1

rateactual=

The convection REDUCES the reaction rate when the effect of reaction on the mass transfer coefficient is neglected.

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Mass Transfer Heterogeneous and Homogeneous Chemical … · Mechanism of irreversible heterogeneous reactions -Heterogeneous reactions of unusual stoichiometries Overview. Mass Transfer