Chap 10 Chemreactions

7/29/2019 Chap 10 Chemreactions

1/12

Mass Transfer

Chapter 10

Hererogeneous and Homogeneous

Chemical reactions

r. . c e - ec ure . .

28. November 2012

Overview: Chemical reactions

-

Reaction rateorder reaction

equilibrium constant

- Hetero eneous & Homo eneous reactions

- -

- Diffusion and 1st Order Heterogeneous Reactions

-

- Heterogeneous Reactions of Unusual Stoichiometries

Mass Transfer Chemical reactions 10-2

Introduction to chemical reactions

A chemical reaction is a conversion of one set of chemical

substances into another.

AB+CD AD +BF

The reaction rate is the decrease in reactant concentration or the

increase in product concentration with time:

(1)


reactants:

With being the rate constant.

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 reaction:



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Now, the rate of the total reaction is the difference in the rates

of the forward and reverse reactions:

If and

the reaction has reached the chemical equilibrium and the ratio

reaction:


eac on or er Apart from temperature and pressure, the reaction rate can

depend on the concentration of the reactants.

reactant is a first order reaction, e.g.:

one depending on two concentrations is of second order:

or an rrevers e secon or er reac on


The forward reaction

is third order, since:

Note: Reactions typically occur in a series of multiple reactions. One step

is the rate determining one. Therefore, the overall reaction would be

of second or first order. (depending also on concentration)

A reaction rate not depending on any concentration is called a

.


Reactions and 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).

reaction allows to determine the rate-limiting step and whether

the reactions have to be accounted for:

mass transfer >> reaction rate include reactions



3/12

When reaction rate and mass transfer are comparable then we

need to incorporate reaction into the models.

iiii

rccDt

c

02

(see: Generalized Mass Balances)


Activation energyc va on energys requ re or e reac on o s ar an o carry

on. The activation energy is typically supplied as heat (higher

radiation, such as UV light.

the reaction at a certain temperature. The catalyst (typically a

solid but could also be a li uid is not consumed after the

reaction.

Exam les of chemical reactions

Combustion (gaseous, liquid or solid fuels)

Production of chemicals (catalysts) Fuel cells, batteries (catalysts)

Gas cleaning (e.g. removal ofSO2by CaO)


-

Heterogeneous & Homogeneous reactions

Heterogeneous reactions take place AT an interface (e.g. solid

catalyst surface, coal combustion) and diffusion and reaction

.

in the same phase (fuel combustion in engine). Diffusion and

reaction occur by steps partially in PARALLEL.

For sim lification of mass transfer roblems some

heterogeneous reactions can be treated as homogeneousones, depending on the definition of the observation space.


For example, in combustion of coalCO2O2

par c es n u ze e s n a y

reaction takes place AT the coal

O2

O2Coal

.

O2CO2

O2

O2O2

CO2

a er on, owever, a g y porous

inorganic structure (e.g. fly ash)

O2

O2

THROUGHOUT the particle.

CO2

Even though microscopically the reaction still is a surface reaction,


macroscop c o serva on o e en re par c e g ves e p c ure o

reactions occurring homogeneously throughout the particle.


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Thus, this case can be described as a homogeneous reaction

and the homogeneous reaction model can be applied:

assignment of the reaction type can be

sub ective!

Combustion is an irreversible reaction, so


In summary, coal particle combustion can be described as:

01121 ccDc

The reaction is introduced at the boundary conditions!

b) Homogeneous (homogeneous-like) reaction model and

transport:

1

0

11

21 rccDc


Examples of reaction assignment

PbS particles porous PbOheat

emova o ammon a rom o -gas:

scrubbing

3 2 4

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


2 2 2 4using Ca(OH)2 slurry

8.1 Diffusion-Controlled Reaction

Every diffusion-controlled process involves

multi le ste s. E. . in deh dro enation of

(a)

C2H6 on Pt crystals

(1) C2H6 diffuses to Pt surface

(2) C2H6 reacts on the surface

surface

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



5/12

In a) we have a diffusion-controlled

heterogeneous reaction where the(a)

reactant diffuses to a (e.g. hot or

catalytic) surface, reacts and the.

In b) we have a porous particle and

the reactant diffuses to catalytically

active sites where it reacts and the

pro uc uses away w e reac on

continues. Very important in catalysis,


(homogeneous reaction model).

In c) a slow-moving molecules

react with more mobile ones

(c)

facilitating the transport of the

former across a membraneexamp e: oxygen upta e y

hemoglobin in the lungs).

In case d), molecules are well

mixed and react u on contact. The

reaction rate depends on the

Brownian motion of the moleculesac ase reac ons, 4combustion etc.)


(a)

(b)


Goal: To find the overall rate of the process

. us on an r er e erogeneous

Reactions (dilute solutions)

A first-order reaction with respect

to a reactant is one where the

reac on ra e s propor ona o ereactant concentration:

i i

At the steady state the overall

reaction rate/area equals the

diffusion fluxes

2232

1111

)(

cckn

ccn

i

i


2r


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The surface reaction is

Species 1 Species 22

-2

And the reaction rate is given by

Note that the e uilibrium constant is

2

2

2K

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

2ccKnr 2

ere e overa or res s ance s:

1

2321 Kk11k1


Similarities/differences with interfacial mass transfer:

1. The resistances 1 / k1, 1 / 2 and 1 / (k3K2) add to the total

resistance 1/K

1speciesofionconcentrat

2speciesofionconcentratbulk

ex s ngewmequ r un22c

2. Henrys law constant (partition coefficient) characterizes

2

K2 and H vary over a wide range. Here, similarly to H, the K2determines the relative impact of k1 and k3.


Example 8.2.1: Limit ing cases of 1st order reactions

(a) Fast stirring

Determine overall rate of the process when:

k1 and k3 are large, so K 2. Then,

2122

Kcr

(b) High temperatureUsually at high Tthe reactions are fast

2c1

so 1/2 is very small and can be

neglected 2231 KKk1k1

(c) Irreversible reaction -2 = 0 so K2 1212 c

1k1

1r

Note that ONLY in this case the resistances are simplyadditive.



7/12

Example 8.2.2: Cholesterol so lubilization in bile

Bile* is the bodys "detergent" for fat digestion through cholesterol

excretion. Failure of bile leads to cholesterol gallstones. Typically,

. ,

gallstones can be dissolved by administering certain components

of the bile.

Goal: Find dissolution (reaction) rate of gallstone

As a cholesterol gallstone is a solid, mass transfer/chemical

(spinning) disk apparatus for cholesterol dissolution rates.


*Bile = Galle (in German)

cholesterol dissolution rate= 5.39 10-9 g/(cm2 s) = r2= -3 3 =. 1

D = 210-6 cm2/s

disk diameter, d = 1.59 cmRe = 11200

bile kinematic viscosity, = 0.036 cm2/sc o es ero so u on, ens y over e e = - g cm

If the reaction is irreversible find:

a surface reaction rate constantb) dissolution rate of a 1 cm radius cholesterol gallstone


(a) For this case21 /1k1

K

9 22

2 1 3 3

1

5.39 10 g cm srr K c K

c 1.48 10 g cm

Find k1 from the appropriate mass transfer coefficient using the

dD 331212

.Dd

.1

o 2 can e o a ne rom 1 an as 2 = . - cm s


(b) dissolution rate of a 1 cm radius cholesterol gallstone

,

density gradient between cholesterol and bile results in flow by

free convection so the mass transfer coefficient is:

1/4 1/331k d d g

2.

D D

3 5 3 21 31

6 2 3 2 2

k 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



8/12

As we are still dealing with irreversible reactions:

1K

1

/1k1

65

21

s/cm104.3

..

6

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


8.3 Finding the Mechanism of Irreversible Heterogeneous Reactions

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

inferred.

productssolidgaseous


spec esspec es

Physical situation Rate-controlling

step

Size

R = (time, reagent)

Size = (temperature) Size = (f low) Remarks

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

be found easily

2 1

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

coefficient

C Shrinkin Reaction R c stron tem erature Inde endent of flow This is the same as 1 corea case A, except for

ash formation

D Shrinking External diffusion R c1t Weak Usually about square This case iscorea root of flow uncommon

E Shrinking Ash diffusion R (c1t)1/2 Weak Independant of flow This case is common,corea an interesting

contrast with the

previous one

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


Cases A and C: surface reaction controls

2122ccr

As the solid concentration (c2) is constant the reaction is 1st

order with respect to c1.

22

dr

r4rcr3dt

2122

dr

rccdt

cr

Be careful, distinguish

12dt

reaction rate r2!

If the gas phase concentration c1 is constant and the particlehas an initial radius R0


120


9/12

Case B: Diffusion outside of a shrinking particle controls

Identify correlation (forced convection around solid sphere)

1/2 1/3

1 =2+0.6D D

. or very sma par c es e , sor

kD

2

Ddr

ccrkcrdt i

22

112 )(43

cDdr

r

crdt

c

12

tDc

Rr

cdt

122

2

2


c 2

B.2) For large particles 3/12/1

1 vd6.0dk

2/13/22/1 rDv42.0k

DD

Similarly,

1/ 2 2 / 31/ 2

2 11/ 6

dr v Dc r -0.42 c

dt

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

01/ 6

c0.64 v Dr R - t

c


Cases C, D, and E: shrinking core model:

Two diffusional resistances in seriesPhysical situation Rate-controlling step

Bulk fluid

R0 particle

ar c e sur aceUnreacted core

particle

corea

corea

E Shrinking Ash diffusiona


Case D: Diffusion in the surrounding bulk fluid controls

Mass balance: kcr4cr3dt

12

23

kcr4dtrcr4 1222

tc

ckRr

2

10

Similar dependence ofrwith tas with cases A and C.

Dependence on flow (through k) but weak dependence on

temperature.



10/12

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

Here the thickness of the shell is important Film model:

DrR0

123 cD4d

2/1

0

cD2

rR3dt

20 c


8.4 Heterogeneous Reactions of Unusual Stoichiometries

8.4.1 Irreversible second-order heterogeneous reaction

(additivity of resistances)


Steady-state mass transfer to the surface:

)cc(knr i11111

Surface reaction: 2i121 cr

Find the c1iby equating the above 0ckckc 11i112i12

1k

c1

2c

1

12

2

1i1

So the overall reaction rate is: 1 2 1

1 1 1

k 4 cr k c 1 1 1


ompare s w e correspon ng rs or er reac on:

1 2 11 1 1

1 2 1

k 4 cr k c 1 1 1

2c k 11 /1/1

1c

kr

first order reactionsecond order reaction



11/12

8.4.2 Heterogeneous Reactions in Concentrated Solutions

Until now we assumed that k1 f(r2). This is good for dilute

solutions.

Cracking of hydrocarbons:

xamp e:

Reforming

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

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


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" " . .

> 1 cracking

< 1 reforming

= 1 treat as be ore

Goal: To calculate the overall rate!


Overall rate of reaction across this film is:

2

1 1 2 1ir n c

We must calculate the flux n1 for concentrated solutions:

dc11 vc

dzn

ow e ux rom convec on s

0 n1nncv

So the e uation for flux n can be written as: n1dc


DDcdz

11

. . 1 10z = l: c1 = c1i

Solving forn1 gives:

1 1 21 1

10

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

whereD

k

(2c10/n1 is large) the

following diagram shows

the rate relative to that ofdilute solutions:



12/12

Consider studying cracking of oil by flowing it over a hot plate.

If the molecular wei ht of the roduct is onl 25% of that of the oil,

by how much the convection introduced mass transfer changes

the cracking reaction rate?

1 molecule oil = x 0.25 molecules of the product.

= - .

the above figure:%40

rateactual

The convection REDUCES the reaction rate when the effect of


reaction on the mass transfer coefficient is neglected.

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Chap 10 Chemreactions

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