Several  problems  have  been  covered  in  various  lectures.  A  few  of  these  with  problem  statements  with  solutions  are  below.  

Lect. No.: 15 Problem : 15A Time : 12:26 The first-order reaction A B→ was carried out and the following experimental data were obtained (Table 1). All other conditions for these experiments were same. Assuming negligible external mass transfer resistance, (a) estimate the Thiele modulus and effectiveness factor for each pellet and (b) how small should the pellets be made to eliminate nearly all internal diffusion resistance?

Solution: Part (a)

( )( )

' 221 1 13 coth 1A c

e As

r obs RD C

ρηφ φ φ

−= = −               [1]

Suppose 11φ and 12φ are the Thiele Moduli at Run 1 and Run 2 with 1Arʹ′− and 2Arʹ′− being the corresponding observed reaction rates, R1 and R2 being the corresponding radii. Using Eq. (1), we obtain

' 22 2 12 12' 2

11 111 1

coth 1coth 1

A

A

r Rr R

φ φφ φ

− −=

−− [2]

Taking the ratio of the Thiele module for runs 1 and 2, we obtain

'

111 1 1

11 12 12 12'12 2 2

2

0.01 100.001

As c

e As

As c

e As

rRD C R R m

R R mrRD C

ρφ

φ φ φ φφ ρ

−

= = ⇒ = = =−

[3]

Using Eqs. (2) & (3) and introducing the information in Table 1, we obtain

( )12 12

12 12

coth 10.05

10 coth 10 1φ φφ φ

−=

− [4]

Solving which gives 12 1.65φ = and 11 1210 16.5φ φ= = . The corresponding effectiveness factors

obtained using Eq. (1) are   2 10.856; 0.182η η= =  

Part (b)

Suppose that operating at an effectiveness factor of 0.95 is sufficient to eliminate most of internal diffusion resistance.

Using Eq. (1), that is, ( )21 1 13 coth 1ηφ φ φ= − , 13 0.9φ = , where subscript 3 refers to the radius R3 at

which 0.95η = . Using Eq. (2), ( ) 4133 1

11

0.90.01 5.5*10 0.5516.5

R R m mmφφ

−⎛ ⎞= = = =⎜ ⎟⎝ ⎠

.

Table 1: Experimental data Measured Rate (obs)

(mol/g cat s) x 105 Pellet Radius

(m) Run 1 3.0 0.01 Run 2 15.0 0.001

Lect. No.: 16 Problem : 16A Time : 05:40 For the reaction 2 2C CO CO+ → conducted in a catalytic reactor containing particles of radius

0.7R cm= with bulk concentration being 5 31.22*10 /AsC mol cm−= , the observed reaction rate is 9 3( ) 4.67*10 / seccr obs mol cmρ −ʹ′− = . After the reaction was conducted, the particles were cut

open and the reacted carbon profiles were measured. These profiles suggested strong diffusional effects to be present. Verify this observation.

The rate law, in concentration units is 2 31

AA

D A

kCrK C K C

− =+ +

where, CA is the concentration of CO2

(species A) and CD is the concentration of CO at the surface. The constants 9 3

2 4.15*10 /K cm mol= and 5 33 3.38*10 /K cm mol= . k is the rate constant. Diffusivity of the

species in the catalyst is given by 20.1 / seceAD cm= .

Solution

Weisz-Prater parameter (CWP) under the given conditions is

( )' 2 9 23

5

4.67*10 *0.7 1.88*10 10.1*1.22*10

A cWP

eA As

r obs RC

D Cρ −

−−

−= = = << [1]

indicating no internal diffusion limitations present. However the experimental observations suggest otherwise. Poor prediction by the Weisz-Prater method is due to the fact that CWP in Eq. (1) uses Thiele modulus expression for a first order reaction when the actual reaction is not first-order. Therefore, this problem warrants the use of Generalized Thiele Modulus.

Assuming equimolar counter diffusion i.e.; eA eDD D= and that concentration of CO at surface

0DsC ≈ , the rate expression can be rewritten as,

( ) ( )'

2 3 21 2 2A

AAs A

kCrK C K K C

− =+ + −

              [2]

Assuming the pellet was infinitely long with , 0A eqC = , the modified parameter

( ) ( )

( )

( )( )

' 2 '2

'

01' 2

3 2 3

3 2 3 2 2

2

1 1 2 11 ln2 2 2 1 2

2.5 1

As

A c AsC

eA A A

A c As As As

eA As As

r obs R r

D r dC

r obs R K C K C K CD K K C K K K C

ρηφ

ρ−

− −Φ = =

−

⎧ ⎫⎡ ⎤− ⎡ ⎤+ + +⎪ ⎪= −⎨ ⎬⎢ ⎥⎢ ⎥− − +⎣ ⎦⎪ ⎪⎣ ⎦⎩ ⎭= >

∫

       [3]  

So, as observed experimentally, there is a strong internal diffusion limitation.

Lect. No.: 18 Problem : 18A Time : 00:00 Design a packed bed reactor in which the reaction 2A B C→ + is being conducted under internal diffusional limiting conditions and the exit conversion is 0.81. The fluid is being pumped into the reactor at a superficial velocity of 4 / secU m= . The reaction is being conducted at temperature

260 533T C K= ° = and at inlet pressure of 4.94P atm= . Assume 8 22.68 10 / seceAD m−= × , 6 251 / . .seck m m molʹ′ʹ′ = , 6 32.1 10 /b g mρ = × , 2410 /aS m g= , 0.38pd cm= . Assume rate law 2

A Abr k Cʹ′ʹ′ ʹ′ʹ′− =

Solution

The inlet concentration 04.94 0.113 /

0.082 533AbPC gmol lRT

= = =×

Mole balance for the reactor is given by

2" 2

2 0Ab AbeA a b Abd C dC

D U k S Cdzdz

ρ− −Ω = [1]

where Ω is the overall effectiveness factor. It should be noted that in general, for a second order reaction explicit expression for Ω is usually not available and will be a function of the local concentration of species A and as a result will be a function of position as well. Assuming the flow rate through the bed is very large and the axial diffusion can be neglected, that is,

2

2Ab Ab

eAd C dC

D Udzdz

<< , Eq (1) can be simplified to

2" 0Ab a b AbdC S Ck

dz Uρ

−Ω = [2]

along with the condition at the entrance of the reactor 0Ab AbC C= @ 0z = . Analytical solution for

Eq. (2) is usually unavailable due to the dependence of the overall effectiveness factor Ω whose explicit dependence on the concentration is a priori unknown. However, the reaction under the specified conditions is internal diffusion controlling. In this regime, the overall effectiveness factor may be approximated to the effectiveness factor η and assumed constant. Under this approximation, Eq. (2) can integrated to obtain the length required to achieve the desired conversion as

"0

1 11b a Ab

ULXk S Cρ

⎛ ⎞= −⎜ ⎟−Ω ⎝ ⎠ [3]

Using the expression for 2φ for a second order reaction, the effectiveness factor 1 2 1 2 1 2

87

2

2 3 2 3 2 3 9.47 101 2 1 2 1 2.59 10nn

ηφ φ

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Note that the Thiele Modulus will be a function of position. For the chosen parameters, as the variation with respect to position is negligible, the Thiele Modulus is evaluated at the inlet concentration and is assumed constant. 1η << implies strongly internal diffusion limited, therefore

approximating 89.47 10η −Ω ≈ = ×

2" 8 6

0

4 0.81 3.62 101 9.47 10 2.1 10 51 410 0.113 (1 0.81)b a Ab

U XL mk S C Xρ

−−

= = = ×Ω − × × × × × × −

Lect. No.: 22 Problem : 22A Time : 13:25 Ref.: - It is required to determine the value of kL and â for a batch absorber using the reaction

( ) 2 ( ) ( )A g B l C l+ →  which is first order in A. kL and â are expected to be about 10-4 m/s and 200 m2/m3 respectively. Da = 2.5x10-9 m2/s. A choice of liquid phase reactants is available with different rate constants. Determine what value of k will suit the purpose. Solution:

Given: kL = 1 x 10-4 m/s; â = 200 m2/m3; Da = 2.5x10-9 m2/s

To find k value at which given condition will satisfies Thickness of the film:

95

4

2.5*10 2.5*101*10

A

L

Dm

kδ

−−

−= ≅ ≅  

 δ . â = 200 * 2.5 * 10-5 = 5 * 10-3

 by assuming slow reaction regime,

2 2 101 1

19

2.5 *10 *0.25

2.5*10A

k kM k

Dδ −

−≈ = ≈  

 

We know that, P = M / (δ. â) ( )31 10.25 / 5*10 50k k−= =  

 Chosen a value of 1

1 0.2seck −≈ gives M = 0.05 and P=10, which satisfies the conditions such

as 1M << and 1P >> , also In general, rate of mass transfer for slow reaction regime is

*

1A L APR k aCP

⎛ ⎞= ⎜ ⎟+⎝ ⎠  

                                  * *1011L A L Ak aC k aC⎛ ⎞= ≈⎜ ⎟⎝ ⎠

 

Hence, the value of 11 0.2seck −≈ satisfies the condition for slow reaction regime.

Lect. No.: 23 Problem : 23A Time : 06:20

Ref.: - Part:1 Rate constant of an unknown reaction An oxidation reaction A B Pν+ → , which is first order in oxygen(A) is carried out in a stirred cell with a flat gas-liquid interface of 132 cm2 at atmospheric pressure with pure oxygen. Over a stirrer speed range of 60-200 RPM, the rate of absorption was measured to be nearly constant at 1.23x10-5 mol/s, as measured by the difference in the flow rates of gas at inlet and outlet; it was also

independent of the volume of liquid in the vessel. The solubility of A in the liquid phase follows Henry’s law with H = 5.8x10-7 mol/cm3/atm. Find the rate constant of the reaction. (DAB = 2.1x10-5 cm2/s, concentration of B = 0.01 mol/cm3). Solution:

Given: âVL = 132 cm2

51.23*10 / secA LR V mol−=  

2

* 7 3* 5.8*10 /A OC H p mol cm−= = 5 22.1*10 / secAD cm−=  

30.01 /BbC mol cm= To find the rate constant of the reaction We consider fast reaction regime, for an given information which suggest that, kL various with RPM leads to RAVL independent of RPM, kL and VL

So, rate reaction expressed as,

*1 * *A L A AR V D k C=  âVL

7 5 7

11.23*10 2.1*10 * *5.8*10 *k− − −= 132

1

1 1.229.13seck −=

Lect. No.: 23 Problem : 23B Time : 19:30

Ref.: - Part:2 Interfacial area by the chemical method. The same reaction is now conducted in an agitated, bubbling stirred tank, with air instead of oxygen. From a measurement of the oxygen content in the gas leaving, a rate of absorption of 3.95x10-5 mol/s was determined with a total dispersion volume of 1700 cm3. Determine the specific interfacial area per unit volume of the dispersion. Mass transfer co-efficient in such equipment usually varies in the range of 2-4x10-2 cm/s. Solution:

Given: Assume:

20.21Op atm=

2

* 7* 5.8*10 *0.21A OC H p −= =  

            7 31.218*10 /mol cm−= Total dispersion volume = 1700 cm3

22 4*10 / secLk cm−= −

To find the specific interfacial area per unit volume of the dispersion

Assumed 24*10 / secLk cm−= for an fast reaction regime ( )3M >

We know that, 5

12

2.1*10 *1229.134*10

A

L

D kM

k

−

−= =

4.02 3M = >    

The rate of absorption in fast reaction regime is,

*1 * *A L A AR V D k C=  âVL

5 5 73.95*10 2.1*10 *1229.13*1.218*10 *− − −= âVL

Total interfacial area (âVL) = 2018.56 cm2

Interfacial area per unit volume of dispersion (âVL) 2 32018.56 1.19 /1700

cm cm= =  

Lect. No.: 26 Problem : 26A Time : 24:10

Ref.: - Maximum and actual enhancement factors CO2 is being absorbed from a gas into a solution of NaOH at 20ºC, in a packed tower. At a certain point in the tower, the partial pressure of CO2 is 1 bar, and the concentration of NaOH 0.5 kmol/m3. Other data are as follows: kL = 10-4 m/s; interfacial area per unit volume of packed space is 100 m-1; *AC = 0.04 kmol/m3; second order rate constant of the reaction k = 104 m3/kmol s, DA = 1.8 x 10-9 m2/s

and DB = 3.06x10-9m2/s. Find the maximum enhancement possible and the actual enhancement. Find also the actual absorption rate, in units of kmol per sec per unit volume of packed space. The reaction is:

2 2 3 22CO NaOH Na CO H O+ → +  

 Solution:

Given: 30.5 /BbC kmol m= ;   * 30.04 /AC kmol m= ;   410 / secLk m−=  

4 31 10 / seck m kmol= ;   9 21.8*10 / secAD m−= ;   9 23.06*10 / secBD m−=  

1.7B AD D =  

(a) To find the maximum enhancement possible

We know that, *

1.7*0.5 10.6252*0.04

B Bb

A A

D Cq

D Cν= = =

Maximum enhancement factor,

( )1 1.7*11.625A

B

DE q

D∞ ≅ + =  

 

8.91E∞ ≅  

 (b) To find actual enhancement

We know that, 9 4

14

1.8*10 *10 *0.51*10

A Bb

L

D k CM

k

−

−= =

( )30 10.625M q= > =    

 Actual enhancement factor,

tanh1 1

E E E EE M M

E E∞ ∞

∞ ∞

⎛ ⎞− −= ⎜ ⎟⎜ ⎟− −⎝ ⎠

 

First approximation:

For a larger value of M and E∞, tanh 11

E EME∞

∞

⎛ ⎞−≅⎜ ⎟⎜ ⎟−⎝ ⎠

which lead to

301 7.91

E EE M E E

E∞

∞∞

− ⎛ ⎞≅ = −⎜ ⎟− ⎝ ⎠

 

 

8.30E = (by trial and error) Second approximation:

8.91 8.30tanh tanh 30 11 8.91 1

E EM

E∞

∞

⎛ ⎞ ⎛ ⎞− −= ≅⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

 

   

tanh 8.301 1

E E E EE M M

E E∞ ∞

∞ ∞

⎛ ⎞− −= ≅⎜ ⎟⎜ ⎟− −⎝ ⎠

(c) To find actual absorption rate, in units of kmol per sec per unit volume of packed space

 The Rate of absorption is

* 4 5 21*10 *0.04*8.3 3.32*10 / secA L AR k C E kmol m− −= = =  

 3 33.32*10 / secAR a kmol m−=

Lect. No.: 12 Problem : 12A Time : 38:35 Ref.: Scott Fogler, pg.: 858 A first-order heterogeneous irreversible reaction is taking place within a spherical catalyst pellet which is plated with platinum throughout the pellet. The reactant concentration halfway between the external surface and the centre of the pellet (i.e., r = R/2) is equal to one-tenth the concentration of pellet’s external surface. The concentration at the external surface is 0.001 g mol/dm3, the diameter (2R) is 2 x 10-3 cm, and the diffusion coefficient is 0.1cm2/s.

A B→ (a) What is the concentration of reactant at a distance of 3 x 10-4 cm in from the external pellet

surface?

(b) To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8?

Solution:

Given:

CA / CAS = 0.1; CAS = 0.001 g mol/dm3; dp = 2 x 10-3 cm; De = 0.1 cm2/s;

(a) To find the concentration of reactant at a distance of 3 x 10-4 cm in from the external pellet surface

We know that,

1

1

sinh1sinh

A

AS

CC

φ λψ

λ φ⎛ ⎞

= = ⎜ ⎟⎝ ⎠

     -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐>    (1)  

11

1

sinh 0.510.1 60.5 sinh

φφ

φ⎛ ⎞

= ⇒ =⎜ ⎟⎝ ⎠

(by trial & error method)

Dimensionless radius of the catalyst expressed in the form of  

4 3 4

3

3*10 1*10 3*10 0.71*10

r RR R

λ− − −

−

− −= = = =  

Substituting value of λ and φ1 in eq.(1), we get

1

1

sinh1 1 sinh(6*0.7)sinh 0.001 0.7 sinh6

A A

AS

C CC

φ λψ

λ φ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠  

4 32.36*10 /AC mol dm−=  

Lect. No.: 13 Problem : 12A (Cont.) Time : 00:00 Ref.: Scott Fogler, pg.: 858

(b) To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8

The Thiele modulus is,

1 1c a r

e e

k s kR R

D Dρ

φ = =  

3 1116 1*10 3600000sec

0.1r

rk

k− −= ⇒ =  

Calculating Thiele modulus for an effectiveness factor 0.8 is

[ ]1 1 121

30.8 coth 1 2η φ φ φφ

= = − ⇒ =  

The corresponding Thiele modulus expression to calculate diameter of the catalyst particle is,

41 36000002 3.4*100.1

r

e

kR R R cmD

φ −= = = ⇒ =  

46.8*10pd cm−=  

 

Lect. No.: 38 Problem : 38A Time : 28:40 Ref.: Scott Fogler, pg.: 971 Conversion using Dispersion and Tank-in-Series Models:

The first-order reaction

A B→

is carried out in a 10 cm diameter tubular reactor 6.36 m in length. The specific reaction rate is 0.25 min-1. The results of a tracer test carried out on this reactor are shown in Table T38A-1.

Table T38A-1. Effluent tracer concentration as a function of time

time(min) 0 1 2 3 4 5 6 7 8 9 10 12 14 C (mg/L) 0 1 5 8 10 8 6 4 3 2.2 1.5 0.6 0

Calculate conversion using (a) the closed vessel dispersion model, (b) PFR, (C) the tank-in-series model, and (d) a single CSTR.

Solution:

Given:

d = 10 cm, k = 0.25 min-1

time 0 1 2 3 4 5 6 7 8 9 10 12 14 C(t) 0 1 5 8 10 8 6 4 3 2.2 1.5 0.6 0

(a) To calculate conversion using the closed vessel dispersion model

Table T38A-2. Calculation to determine tm and σ2

time 0 1 2 3 4 5 6 7 8 9 10 12 14 C(t) 0 1 5 8 10 8 6 4 3 2.2 1.5 0.6 0 E(t) 0 0.02 0.1 0.16 0.2 0.16 0.12 0.08 0.06 0.044 0.03 0.012 0 tE(t) 0 0.02 0.2 0.48 0.8 0.80 0.72 0.56 0.48 0.40 0.3 0.14 0 t2E(t) 0 0.02 0.4 1.44 3.2 4.0 4.32 3.92 3.84 3.60 3.0 1.68 0

To find E(t) and then tm, we first find the area under the C curve, which is

( )0

50 minC t dt g∞

=∫  

Then ( )0

5.15minmt tE t dtτ∞

= = =∫  

Using Simpson rule, we find,

( ) ( ) ( ) ( )2

0

1 0.0 3.0 2 0.4 3.2 4.32 3.84 4 0.02 1.44 4.0 3.92 3.63

t E t dt∞

⎛ ⎞= + + + + + + + + + +⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠∫

                                              ( ) ( )2 3.0 0.0 4 1.683⎛ ⎞+ + +⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠

 

                                          232.63min=  

To obtain the variance, we substituting these values

( ) ( ) ( )22 2 2

0 0

t E t dt t E t dtσ τ τ∞ ∞

= − = −∫ ∫  

( )22 232.63 5.15 6.10minσ = − =  

Dispersion in a closed vessel is represented by

( )( )2

2 2

2 1 expPe PePe

στ

= − + −  

( )( )( )2 2

6.1 20.23 1 exp5.15

Pe PePe

= = = − + −  

Solving for Pe by trial and error, we obtained 7.5Pe =  

Next we need to calculate Da, ( )( )15.15min 0.25min 1.29Da kτ −= = =  

Using the equation for q and X gives

( )4 1.2941 1 1.307.5

DaqPe

= + = + =  

Then, substitute q and Pe value in conversion expressed for a dispersion model

( )( ) ( ) ( ) ( )2 2

4 exp 21

1 exp 2 1 exp 2

q PeX

q qPe q qPe= −

+ − − −  

( ) ( )( ) ( )( ) ( ) ( )( )2 2

4 1.30 exp 7.5 21

1 1.30 exp 1.30*7.2 2 1 1.30 exp 1.30*7.2 2X = −

+ − − −  

0.68X =  

When dispersion effects are present in this tubular reactor, 68% conversion is achieved.

(b) Conversion for Plug flow reactor:

If the reactor were operating ideally as a plug-flow reactor, the conversion would be

( ) ( ) ( )1 exp 1 exp 1 exp 1.29X k Daτ= − − = − − = − −  

0.725X =  

72.5% conversion would be achieved in an ideal plug-flow reactor.

(c) Conversion for tank-in-series:

First calculate the number of tanks in series,

( )22

2

5.154.35

6.1n τ

σ= = =  

To calculate the conversion for first-order for n tanks in series is

( ) ( )( ) ( )( )4.351 1 11 1 1

1 1 1 5.15 / 4.35 0.25n n

i

Xk n kτ τ

= − = − = −+ + +

 

0.677X =

67.7% conversion achieved for the tanks-in-series model

(d) Conversion for CSTR:

For a single CSTR,

1.291 2.29kXk

ττ

= =+

 

0.563X =

56.3% conversion achieved for the single CSTR.

ADDITIONAL PROBLEMS WITH SOLUTIONS

1. Consider the first order decomposition of A. The following data is given: mxL 4104 −=       KhrmkJke ///6.1=     hrmxDe /105 25−=  

KmhrkJhT ///160 2=     hrmkm /300=     molAkJH /160−=Δ  

3/20 mmolCAb =       hrmmolrobs //10 35−=  

 

Answer  the  following  questions:  

Is  external  mass  transfer  important  to  consider?  

Are  there  significant  limitations  due  to  pore  diffusion?  

Do  we  expect  significant  temperature  gradients  within  the  pellet  &  outside?    

SOLUTION:

2. The irreversible gas-phase reaction A B is carried out isothermally over a packed bed of solid catalyst particles. The reaction is first order in the concentration of A on the catalyst surface. The feed consists of 50% (mole) A and 50% inerts and enters the bed at a temperature 300 K. The entering volumetric flow rate is 10 lit/sec The relation between Sh and Re is Sh= 100 (Re)0.5

 

As  a  first  approximation  one  may  neglect  pressure  drop.  The  entering  concentration  of  A  is  1.0M.  Calculate  the  catalyst  weight  necessary  to  achieve  60%  conversion?  

 

Kinematic  viscosity:  0.02  cm2/sec;  Particle  diameter:  0.1  cm  

Superficial  velocity  10  cm/s;  Catalyst  surface  area  /mass  of  the  catalyst  bed:  60  cm2/g.  cat  

Diffusivity  of  A  10-­‐2  cm2/sec.  

Specific  rate  constant  (k)  is  0.01  cm3/sec  g  cat  with  E=  4000  cal/mol  

SOLUTION  

 

3. (a) Following is the observed reaction rate in an isothermal reactor as a function of particle size for an elementary first order liquid phase reaction. The bulk concentration (1 mol/lit) is same in each case. Find the approximate value of effective intra-particle diffusivity. Catalyst density is 1 gm/cc.

 

(b)  The  above  reaction   is  performed   in  a  fluidized  bed  reactor  which  received  the  feed  at  100  kmol/hr  and  a  conversion  of  10%  is  realized.  Predict  the  conversion  if  the  original  particle  radius  of  1.8cm  of  the  same  catalyst   is  reduced  by  half  under  otherwise  similar  conditions.    Fluidized  bed  reactor      can  be  considered  to  be  a  perfectly  back-­‐mixed  reactor  for  all  practical  purposes.  

SOLUTION:  

 

4. A first order irreversible cracking reaction A = B is performed in a fixed bed reactor on a catalyst particle size of 0.15 cm. Pure A enters the reactor at a superficial velocity of 2m/s, a temperature of 2000C and pressure of 1 atm. Under these conditions, the reaction is severely affected by internal diffusion effects. Calculate the length of bed necessary to achieve 60% conversion.

Data  given:    

The   intrinsic   reaction   rate   constant   calculated   by   performing   experiments   with   very   small  particle  size  of  the  same  catalyst  is  0.0003  m3/g  cat.  sec.    

Effective  diffusivity:  1.5  x  10-­‐8  m2/s  

Catalyst  density:  2  gm/cm3  

SOLUTION: