Chemical Engineering DepartmentCDB 2043 REACTION ENGINEERINGCHAPTER 5: Collection and Analysis of Rate Data1

Failing to plan is planning to fail-Effie Jones-

At the end of the chapter, student should be able to apply the concept of CRE algorithm in solving design problems related to:1. Determine the reaction order and specific reaction rate from experimental data 2. Decide the most suitable type of analysis technique for a particular problem 2

LEARNING OUTCOMES

Recap from Chapter 4MOLE BALANCE & DESIGN EQUATIONRATE LAWSTOICHIOMETRYCOMBINE INFORMATION

GRAPHICAL METHODNUMERICAL METHODANALYTICAL METHODSOFTWARE

EVALUATE

GRAPHICAL METHOD

TopicsAnalysis of data to find rate law:1) Integral method of analysis2) Differential method of analysis

Analysis of Data to find Rate Law 2 common reactors used to obtain rate data:

Batch reactor

Differential reactor'L

FA0 FAe

Catalyst

9 For homogeneous reactions9 By measuring CA as f(t)

Analysis of Data to find Rate Law Methods used to analyze rate data:

1. Integral method (try and error - reaction order)2. Differential method (

) (Fogler: page 258-259)

Methods of finding

9 Graphical9 Finite Different Method (Numerical Method)9 Polynomial3. Half live method4. Initial rate method5. Linear regression6. Non-linear regression

1. Need to guess reaction order 2. Integrate the differential form of equation used to model the reactor used.3. If the right reaction order is assume, the plot of concentration time data should be linear

Integral method analysis

3 examples will be considered: Zero order First order Second order

Y

X

Step 1: Mole balance: = Step 2: Rate Law: = Step 3: Stoichiometry: V=V0, = 0

8

Analysis of Data to find Rate Law Consider the following reaction:

A Products

Step 4: Combine: =

Zero order reaction:

= rA = k Integration gives:

Integral method analysis

Slope = -k 0

=

0

= 0

y C m

x

CA

t

CA0

10

Integral method analysis First order reaction:

= =

Integration gives: 0

1 =

0

0=

y m

x

Slope = k

0

t

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Integral method analysis Second order reaction:

= = 2

Integration gives: 0

12 =

0

110

=

ym

x

Slope = k1

tC CA0

The reaction of trityl (A) and methanol (B)

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Example 5.2||O(C6H5)3CCl + CH3OH (C6H5)3CCH3 + HCl

A B C D

Time (min) 0 50 100 150 200 250 300

CA(mol/dm3) 0.0500 0.0380 0.00306 0.00256 0.00222 0.00195 0.00174It was reported that the reaction is second order w.r.t A.Given, CB0 = 0.5 mol/dm3Mission:1. Using the data, confirm that the reaction is second order.

010

20

30

40

50

60

70

0 50 100 150 200 250 300 350

13

Example 5.1= 2

0

12 =

0

Time (min)

CA(mol/dm3)

1/CA(dm3/mol)0 0.0500 20.0050 0.0380 26.32100 0.00306 326.80150 0.00256 390.63200 0.00222 450.45250 0.00195 512.82300 0.00174 574.71

1= +

10

1

t

Slope = k

10

y = 0.1248x + 20.118

k = 0.1248 dm3

mol.minCA0= 0.0497 mol/dm3

1. Guess reaction order, integrate mole balance equation.2. Calculate the resulting function in term of concentration data, and plot f(C) vs t3. If the plot is linear, then you have guessed the correct reaction order.4. Otherwise, guess another order and repeat the procedure.14

Summary of Topic 1

15

Differential method analysis Take natural log of:

= +

=

y m

x

C

Plot graph of

vs. give straight line graph with: Slope = Intercept = ln k

H ? ?

Graphical Method Finite Difference Method

Initial point : 0

= 30+4122

Interior point :

= 12

+1 1

Last point : 5

= 12

3 44 + 35

16

Finding

The following reaction takes place in an isothermal constant volume batch reaction system. The initial concentration of B is 0.5 mol/dm3. The concentration of A at different time was recorded as shown in the table.

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Example 5.2||O(C6H5)3CCl + CH3OH (C6H5)3CCH3 + HCl

A B C D

Time (min) 0 50 100 150 200 250 300

CA(mol/dm3) 0.0500 0.0380 0.00306 0.00256 0.00222 0.00195 0.00174Mission:

Determine reaction order w.r.t. A

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Example 5.2=

= +

Using graphical method:=

2 12 1

Time (min)

CA(mol/m3)x103

0 5050 38100 30.6150 25.6200 22.2250 19.5300 17.4= 38 5050 0 = 0.24 3.

0.240.1480.10.0680.0540.042

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Example 5.2

0.05

0.10

0.15

0.20

0.25

0.30

00 50 100 150 200 250 300

moldm3.min

t (min)

Time (min)

CA(mol/m3)x103

0 5050 38100 30.6150 25.6200 22.2250 19.5300 17.4

0.240.1480.10.0680.0540.042

0.30.1860.120.080.050.047

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Example 5.2t CA ln CA

ln [

]

0 50 -0.6931 0.3 -1.2039750 38 -0.9676 0.186 -1.68201100 30.6 -1.1842 0.12 -2.12026150 25.6 -1.3626 0.08 -2.52573200 22.2 -1.5051 0.05 -2.99573250 19.5 -1.6348 0.047 -3.05761300 17.4 -3.5-3-2.5

-2

-1.5

-1

-0.5

0-2.0000 -1.5000 -1.0000 -0.5000 0.0000

y = 2.0868x + 0.2912

= 2.0868

= +

21

Summary of Topic 21. Find

.

2. Plot

3. Slope is (reaction order).

END OF LECTURE

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