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1. HOMOGENEOUS CHEMICAL REACTIONS

Recall NH3 absorption into water (EXERCISES):

How much is mass transfer altered by chemical reaction?

Chemical reactions can alter mass fluxes by orders of magnitude.

2

First order reactions are typical - even though two reactants may

participate as usually one of them is in excess. In the latter case we

have pseudo first-order reactions.

1.1 Mass Transfer with 1st Order Reactions

42 CH

2

O1

2224

c]c[)rate reaction(

OH2COO2CH

where ]c[ 2

O1 2 is the pseudo-first order rate constant.

As mass transfer and reaction are intimately coupled when homoge-

neous reactions are involved, detailed calculations of k are extremely

hard! Instead, the mass transfer correlations will be corrected to

account for reaction. This is rather easy for 1st order reactions.

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A liquid is in contact with well mixed gas:

Irreversible Reactions

Mass balance in the film (dilute

concentrations) without chemical

reaction (steady state):

21

2110

dz

cdD

dz

dj

dz

dn

z = 0: c1 = c1i and z = l: c1 = 0

Boundary conditions:

)l

z1(cc i11 )0c(

l

Dj i11 lDk0

4

Now including chemical reaction (conversion of species 1 in the film):

1121

2

0 cdz

cdD

Solving this equation and inserting the B.C.s gives

]/sinh[

)](/sinh[

1

1

1

1

lD

zlD

c

c

i

This profile results in the curvature in our schematic instead of the

standard straight line.

Now when the reaction becomes rather slow:

111

1 1

/ ( ) ...lim ( 0) 1

/ ...i

D l zc l z z

c l lD l

@ so You recover the

diffusion result!

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Now back to the full problem:

The flux in the presence of reaction is:

]/sinh[

)](/cosh[

1

111

11

lD

zlDcD

dz

dcDj i

At the interface z=0: iclDDj 1111 )/coth(

where )/coth( 11 lDD is k with chemical reaction.

Again as 01 then l

Dkk 0

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The problem now is to determine l in order to obtain k. However, we

don’t have to, we only need the correction to k0. For this reason we

only need to calculate k/k0:

1 1

2 200 0

D Dkcoth

k k k

When reaction is slow 01 1

0 0 21

3

Dk...

k (k )and

When the reaction is fast 1

so 1coth 1

Dk and

Note that now )( 0kfk but only of 1

D

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Is the simple film model adequate?

Predicted corrections for chemical reaction are almost the same,

independent of the model chosen!

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Coupling between mass transfer and reaction greatly affects the

temperature-dependence of k.

1If doubles every 10 oC, then k doubles every 20 oC

In the absence of reaction k doubles every 50 oC!

Example: Variation of mass transfer with fluid flow

A spinning disk of reagent 1 is immersed in a dilute solution of

reagent 2. We measure reagent 1 lost from the spinning disk as a

function of rotation speed of the disk. How can we distinguish

between the two possible reaction mechanisms:

a) reagent dissolution and irreversible homogeneous reaction

b) irreversible heterogeneous reaction at the solid surface

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ϴ

Remember from Chapter 4: «Generalized Mass Balances»

)sat(cScRed

D62.0j

)sat(cD

d

d

D62.0j

1

chaptersgminupcotheinMore)tcoefficientransfermass(k

3/12/1

1

1

3/12/12

1

312120

D

d62.0

D

dk

10

Key dependence (no chemical reactions):

)flow fluid(0 fk or (Re)fSh

Spinning disk:

2/13/12/10 dva

dv0.62

D

dk

D

1/20

d

vDak

or

a) For 1st order irreversible HOMOGENEOUS reaction (see above):

11

20 1 20

/

DDkcoth

k aD(v / d)k

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There is a clear dependence

of k with v. At low v, .constk

while at high v, 2/1vk

b) For a 1st order irreversible HETEROGENEOUS reaction:

0

2

1 1 1

k k

1 2

1 2

2

1

/

/

(d )

aDv

a

1

Dv

dcothDk

21

11

Here, when v is small, the second

RHS term becomes large so .vk 2

1

When now v is large, the second

RHS term is insignificant and .k 2

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1.2 Mass Transfer with 2nd Order Reactions

Mass balances:

Species 1:2112

12

10 ccdz

cdD

21122

2

20 ccdz

cdD Species 2:

How do second-order

chemical reactions

enhance mass transfer?

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With the B.C.’s:

z = 0: c1 = c1i dc2/dz = 0

z = l: c1 = 0 c2 = c2l

These are difficult to solve in the general case.

Of course, if one of the reactants is in excess then it reduces to the

first order reaction problem.

Another special case is when the reaction is FAST and irreversible:

The two reactants disappear forming a reaction FRONT between

TWO films.

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(species 1) + (species 2) → products

11 1

0 i

c

Dn (c )

z

where zc is the location of the reaction front

22 2

0

l

c

Dn ( c )

l z

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From a mass balance: 0nn 21

The unknown zc can be eliminated to give:

i1

i11

l2211 c

cD

cD1

l

Dn

As lDk 1

0 then: i11

l22

0 cD

cD1

k

k

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Example: SO2 absorption in a packed tower

By how much the mass transfer will be improved as a function of

cNaOH, if we change the absorbing medium from water to a dilute

NaOH solution?

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First of all recognize that you have an acid-base second order

reaction that is FAST as all acid-base reactions.

OHSONaNaOHSO 2322 2 Species 1: SO2

Species 2: NaOH

At steady state: i11

0

p1 ppkj

0ck i1L

i1

i11

l220

L ccD

cD1k

From gas-liquid equilibrium at the interface: i1i1 cHp

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The interfacial concentration can be calculated

𝑘𝑝0 ∙ 𝑝1 −𝐻 ∙ 𝑐1𝑖 = 𝑘𝐿

0 ∙ 𝑐1𝑖 +𝐷2 ∙ 𝑐2𝑙𝜈 ∙ 𝐷1

−𝑐1𝑖 𝑘𝑝0 ∙ 𝐻 + 𝑘𝐿

0 = −𝑘𝑝0 ∙ 𝑝1 + 𝑘𝐿

0𝐷2 ∙ 𝑐2𝑙𝜈 ∙ 𝐷1

𝑐1𝑖 =𝑘𝑝0 ∙ 𝑝1 − 𝑘𝐿

0 𝐷2 ∙ 𝑐2𝑙𝜈 ∙ 𝐷1

𝑘𝑝0 ∙ 𝐻 + 𝑘𝐿

0

Now following what we learned about the overall mass transfer

coefficient𝑗1 = 𝐾𝐿

′ ∙ 𝑐1∗ − 0 = 𝐾𝐿

′ ∙𝑝1𝐻

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Now as we have a packed tower 𝐾𝐿′ = 𝐾𝐿 ∙ 𝑎

with the specific surface area (surface/volume) 𝑎

Then in absence of reaction:

In the presence of reaction:

Then𝐾𝐿 ∙ 𝑎

𝐾𝐿0 ∙ 𝑎

= 1 +𝐷2 ∙ 𝑐2,𝑙 ∙ 𝐻

𝜈 ∙ 𝐷1 ∙ 𝑝1

Hak

1

ak

1

1aK

0

p

0

L

0

L

Hak

1

ak

1

pD

HcD1

aK

0

p

0

L

11

L,22

L

20

Calculate H from our equilibrium data

𝑝1 = 𝐻 ∙ 𝑐1

g64

SOmol1

cm1

g001.0H

mmHg760

atm1mmHg10 2

3

mol/atmcm840H 3

2

25

325

0

L

L c

atm1/mmHg760

mmHg10)s/cm109.1(2

)mol/atmcm840()s/cm101.2(1

aK

aK

)liter/mol(cmol/liter351 2

21

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2. Diffusion-controlled Reactions

Examples:

Acid – base reactions

H2 or CH4 combustion

These reactions are so fast that they are always diffusion-controlled.

Goal: To determine the overall reaction rate

As the reaction is fast, it is

determined by the rate at which

molecules collide with each other.

3species2species1species 1

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Consider now a single molecule 2 of perfect spherical shape in a

volume , where c2 is the concentration of species 2 and NAV

is the Avogadro number.

AV2 Nc1

Consider a single sphere of radius a1 at a fixed point. Molecules of

radius a2 are in Brownian motion and diffuse to the surface of a1:

A mass balance gives

𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑𝑏𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

=𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑

𝑏𝑦 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛

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2

1 1

2

14

AV

r r jc N

We would like to calculate the concentration profile c1 away from the

surface of a sphere of radius σ12 so we can calculate the flux of

molecules 1 to the surface of molecules 2. This will give the rate of

collisions of molecules 1 and 2 per unit area.

Now we know the flux of molecules 1 using the earlier result (e.g.

dissolution of sphere, Chapter 2):

1 1 21 12

D (a a )j c

r

Combining the equations gives: 1 1 1 2 1 24 AVr D (a a )N c c

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Now consider that the molecule of radius a2 is in Brownian motion.

Then we introduce the diffusion coefficient describing the relative

motion of the two molecules:

Einstein equation (lecture on Diffusivity):

2

1 2

122

x xD D

t

t2

x

t2

xx2

t2

xD

2

2

0

21

2

112

1 2 D D

Thus the reaction rate can be written as

1 1 2 1 2 1 24 AVr (D D )(a a )N c c

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As the reaction is: 1 1 1 2 r c c

The κ1 becomes 1 1 2 124 AV(D D ) N

Where σ12 is the intermolecular distance that is tabulated in the

Cussler book and in various handbooks as the molecular collision

diameter:

12 1 2

1

2 ( )

κ1 gives a good approximation of very fast rates (within a factor of 10).