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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
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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).