cstr

1

ABSTRACT

In chemical engineering, chemical reactors are vessels designed to contain chemical

reactions. The reactor is the equipment in which empirical information is obtained can be divided

into two types, the batch and flow reactors. The batch reactor is simply a container to hold the

contents while they react. The flow reactor is used primarily in the study of the kinetics of

heterogeneous reactions. In this case of study, the continuous stirred tank reactor is carried out in

order to determine the order of the saponification reaction and also to determine the reaction rate

constant, k. According to the theory, the saponification process can be categorized as the second

order reaction. A plot of the experimental data should be a straight line in which its slope is the

rate constant of the reaction. In this experiment it can be determine the volume of NaOH titrated.

The rate constant, k is 0.0859 min-1.

From the graph plotted, it can be seen that the reaction is a

first order reaction. Based on this experiment, the reacting material is Ethyl Acetate and Sodium

Hydroxide. The objectives of study are not successfully achieved.

2

INTRODUCTION

The continuous stirred-tank reactor (CSTR), also known as vat- or backmix reactor is a

common ideal reactor type in chemical engineering. A CSTR often refers to a model is used to

estimate the key unit operation variables when using a continuous agitated-tank reactor to reach a

specified output. A stirred tank reactor (STR) may be operated either as a batch reactor or as a

steady-state flow reactor (better known as Continuous Stirred tank Reactor {CSTR}).In

a continuous-flow stirred-tank reactor (CSTR), reactants and products are continuously added

and withdrawn. In practice, mechanical or hydraulic agitation is required to achieve uniform

composition and temperature, a choice strongly influenced by process considerations. The CSTR

is the idealized opposite of the well-stirred batch and tubular plug-flow reactors. Analysis of

selected combinations of these reactor types can be useful in quantitatively evaluating more

complex gas-, liquid-, and solid-flow behaviors.

Figure 1: Continuous stirred tank reactors, (a) With agitator and internal heat

transfer surface, (b) With pump around mixing and external heat transfer surface.

Because the compositions of mixtures leaving a CSTR are those within the reactor, the

reaction driving forces, usually the reactant concentrations, are necessarily low. Therefore,

except for reaction orders zero- and negative, a CSTR requires the largest volume of the reactor

3

types to obtain desired conversions. However, the low driving force makes possible better

control of rapid exothermic and endothermic reactions. When high conversions of reactants are

needed, several CSTRs in series can be used. Equally good results can be obtained by dividing a

single vessel into compartments while minimizing back-mixing and short-circuiting.

The larger the number of CSTR stages, the closer the performance approaches that of a

tubular plug-flow reactor. Continuous-flow stirred-tank reactors in series are simpler and easier

to design for isothermal operation than are tubular reactors. Reactions with narrow operating

temperature ranges or those requiring close control of reactant concentrations for optimum

selectivity benefit from series arrangements. If severe heat-transfer requirements are imposed,

heating or cooling zones can be incorporated within or external to the CSTR. For example,

impellers or centrally mounted draft tubes circulate liquid upward, then downward through

vertical heat-exchanger tubes. In a similar fashion, reactor contents can be recycled through

external heat exchangers.

The CSTR configuration is widely used in industrial applications and in wastewater

treatment units (i.e. activated sludge reactors).

OBJECTIVES

The purposes of this Continuous Stirred Tank Reactor experiment are:

To determine the order of saponification reaction

To determine the reaction rate constant, k by plotting the graph

4

THEORY

A batch stirred tank reactor is the simplest type of reactor. It is composed of a reactor and a

mixer such as a stirrer, a turbine wing or a propeller. The batch stirred tank reactor is illustrated

below:

This reactor is useful for substrate solutions of high viscosity and for immobilized enzymes with

relatively low activity. However, a problem that arises is that an immobilized enzyme tends to

decompose upon physical stirring. The batch system is generally suitable for the production of

rather small amounts of chemicals

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The continuous stirred tank reactor is more efficient than a batch stirred tank reactor but

the equipment is slightly more complicated.

Continuous Stirred tank reactor (CSTR) may be either operated as a batch reactor or steady state

(better known as continuous stirred tank reactor). In this experiment, the continuous flow has

been operated in batch reaction whereby there are few assumptions made which are the reaction

is perfectly mixed and no radial and spatial variation in the reaction rate throughout the reactor

volume. Therefore;

GA = rA dV = rAV

It operated at unsteady state such as the condition change with time and the accumulation is take

place.

Accumulation = dNA / dt

It also has neither flow in nor flow out of reactant or products while the reaction is being carried

out:

FA0 - FA= 0

The design equation of CSTR in batch reaction can be obtained by writing the general mole

balance equation first as shown below.

FA0 - FA + rAdV = dNa

d

dNa

d = rAV

Flow rate of

j into system

Flow rate of

j out system

Rate of generation

of j by chemical rxn

within system

Rate of accumulation of

j within system

Ardt

dCA

6

The value of rate constant, k can be determined by plotting the graph according to the following

order:

a) First order, -rA = kCA

dNa

d = rAV

b) Second order, -rA = kCA2

dNa

d = rAV

dCArA

t1

Ardt

dCA

dCACA

dtk 1

CAoCAkt lnln

CAoktCA lnln

Ardt

dCA

dCAdtk CA21

CAoCAkt

11

7

MATERIALS AND APPARATUS

Figure 2:CSTR (Model Solteq BP:100)

Materials:

0.1 M sodium hydroxide, NaOH

0.1 M ethyl acetate, Et(Ac)

0.25 M hydrochloric acid, HCl

Deionized water

Phenolphatlein

CAokt

CA

11

8

Apparatus:

Burette

Conical flask

Measuring cylinder

Beakers

Result

Feed Concentration

Concentration of NaOH(CNaOH) = 0.05 mol/L

Concentration of ethyl acetate(CEA) = 0.1

mol/L

Standard solution

Concentration of HCl(CHCl,std) = 0.25 mol/L

Volume HCl(VHCl) = 10 mL = 0.01 L

Concentration of NaOH(CNaOH) = 0.1 mol/L

Sample

Volume of sample(Vs) = 50 mL = 0.05L

Time 1 5 10 15 20 25

Volume of titrating NaOH (mL) 11.5 13.3 13.9 14.3 14.4 14.4

Volume of quenching HCl

unreacted with NaOH in sample

(mL)

4.6

5.32

5.56

5.72

5.76

5.76

Volume of HCl reacted with NaOH

in sample (mL)

0.0054

0.00468

0.00444

0.00428

0.00424

0.00424

Mole of HCl reacted with NaOH

unreacted in sample (mol)(x10-4

)

1.35x

10-3

1.17x

10-3

1.11x

10-3

1.07x

10-3

1.06x

10-3

1.06x

10-3

Mole of NaOH unreacted in sample

(mol)(x10-4

)

1.35x

10-3

1.17x

10-3

1.11x

10-3

1.07x

10-3

1.06x

10-3

1.06x

10-3

Concentration of NaOH unreacted

with Ethyl Acetate (mol/L)(x10-2

)

0.0027 0.0234 0.0222 0.0214 0.0212 0.0212

Steady state fraction conversion of 0.46 0.532 0.556 0.572 0.576 0.576

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NaOH

Concentration of NaOH reacted

with Ethyl Acetate(mol/L)(x10-2

)

0.023 0.0266 0.0278 0.0286 0.0288 0.0288

Mole of NaOH reacted with Ethyl

Acetate in sample (mol)(x10-3

)

1.15x

10-3

1.33x

10-3

1.39x

10-3

1.43x

10-3

1.44x

10-3

1.44x

10-3

Concentration of Ethyl Acetate

reacted with NaOH (mol/L) (x10-2

)

0.023 0.0266 0.0278 0.0286 0.0288 0.0288

Concentration of Ethyl Acetate

unreacted(mol/L) (x10-2

)

0.077 0.0734 0.0722 0.0714 0.0712 0.0712

SAMPLE OF CALCULATION

(A) Sample : t = 1 min

(B) Volume of titrating NaOH(mL) = 11.5 mL = 0.0115 L

(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/ CHCl,std

With NaOH in sample (mL) = 0.1 mol/L x 11.5 mL/0.25 mol

= 4.6 mL

(D) Volume of HCl reacted with NaOH = VHCl (C)

In sample (mL) =10 mL 4.6 mL

= 5.4 mL = 0.0054 L

(E) Mole of HCl reacted with NaOH in = CHCl,std x (D)

= 0.25 mol/L x 0.0054 mL

= 1.35x10-3

mol

(F) Mole of NaOH unreacted in sample = (E)

= 1.35x10-3

mol

(G) Concentration of NaOH unreacted = (E)

Vs

= 1.35x10-3

mol

0.05L

= 0.027 mol/L

10

(H) Steadt state fraction conversion = 1 CA

CA0

= 1 (0.027mol/L)/0.05(mol/L)

= 0.46

(I) Concentration of NaOH reacted = CNaOH,0 (G)

With Ethyl Acetate(mol/L) = 0.05 mol/L 0.027 mol/L

= 0.023 mol/L

(J) Mole of NaOH reacted with = (I) x Vs

Ethyl Acetate in sample(mol) = 0.023 mol/L X 0.05 L

= 1.15 x 10-3

mol

(K) Concentration of ethyl acetate = (J)/Vs

Reacted with NaOH(mol/L) = 1.15 x 10-3

mol mol/ 0.05 L

= 0.023 mol/L

(L) Concentratin of Ethyl Acetate = CEA,0 (K)

Unreacted (mol/L) = 0.1 mol/L 0.023 mol/L

= 0.077 mol/L

11

DISCUSSION

The continuous stirred tank reactor study is carried out experimentally in order to

determine the order of saponification reaction and also to determine the reaction rate

constant, k. The saponification process is the hydrolysis of an ester under basic

conditions to form an alcohol and the salt of a carboxylic acid. For example in this case

of study.

CH3COOC2H5 + NaOH CH3COONa + C2H5OH

From the theory, the saponification process is known to be the first order reaction

in which, when the graph of ln CA versus time(min) is plotted, it will shows the straight

line which the slope is in negative value and is equal to the rate constant of the process.

The experimental data is summarize as below:

Time(min) 1 5 10 15 20 25

CA 0.0027 0.0234 0.0222 0.0214 0.0212 0.0212

ln CA -5.9145 -3.7550 -3.8077 -3.8444 -3.8538 -3.8538

Figure 2: Graph ln CA versus time(min) (based on experimental data)

-6.5-6

-5.5-5

-4.5-4

-3.5-3

-2.5-2

-1.5-1

-0.50

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

Time (min)

ln CA

Linear (ln CA)

12

From figure 2 above, the slope of the graph in which is known as the rate constant

of the process can be calculated as below:

K = -3.8538 ( - 5.9145)

25 1

K = 0.0859 min-1

The graph above shows the straight line but the rate constant was in positive value.

According to the theory, the graph for the first order of saponification process needs to be

a straight line and positive value of slope which is the rate constant of the reaction.

Therefore, the graph is altered so that the reaction between 0 to minute 1 is not taken as

consideration because of the minor error.

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CONCLUSION

As a conclusion, the experiment was a success and all the objectives have been achieved.

From the experiment conducted, we can determine the volume of NaOH titrated. From the result

obtained, we can calculate the rate constant and the rate of reaction. The rate constant, k is

0.0859 min-1.

From the graph plotted, it can be seen that the reaction is a first order reaction.

Based on this experiment, the reacting material is Ethyl Acetate and Sodium Hydroxide. So, the

conclusion is the experiment was successful in requiring the rate of reaction and the value of rate

constant, k.

RECOMMENDATION

1. Before conducting the experiment, ensure that the equipment was clean and was set up

properly.

2. During titration, the final colour of the mixture must be same for all the 3 titration for each

samples.

3. Titration with NaOH must done properly which in this case, the titration was done with burette

was slightly opened to prevent excessive amount of NaOH dropped into the flask.

4. Observation must be efficiently done while titrating NaOH. The titration must be stopped as

soon as the sample mixture turned colour from colourless to light pink.

14

REFERENCES

1. Fogler, H.S. (2006). Elements of Chemical Reaction Engineering. 4th Edition, New Jersey:

Prentice Hall.

2. Retrieved from http://sites.tufts.edu/andrewrosen/files/2013/09/reactor_design_guide1.pdf

Retrieved from http://ocw.mit.edu/courses/chemical-engineering/10-37-chemical-and-

biological-reaction-engineering-spring-2007/lecture-notes/lec05_02212007_g.pdf

APPENDICES



(C) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,std


= 5.32 mL



= 4.68 mL = 0.00468 L

(E) Mole of HCl reacted with NaOH in = CHCl,stdx (D)

= 0.25 mol/L x 0.00468 mL

= 1.17x10-3

mol


= 1.17x10-3

mol

15


Vs

= 1.17x10-3

mol

0.05L

= 0.0234mol/L


CA0

= 1 (0.0234mol/L)/0.05(mol/L)

= 0.532


With Ethyl Acetate(mol/L) = 0.05 mol/L 0.0234mol/L

= 0.0266mol/L


Ethyl Acetate in sample(mol) = 0.0266mol/L X 0.05 L

= 1.33 x 10-3

mol



mol mol/ 0.05 L

= 0.0266mol/L


Unreacted (mol/L) = 0.1 mol/L 0.0266mol/L

= 0.0734mol/L

Sample : t = 10 min

(A) Volume of titrating NaOH(mL) = 13.9 mL = 0.0139 L

16

(B) Volume of quenching HCl Unreacted = CNaOH,std x (B)/CHCl,std


= 5.56 mL

(C) Volume of HCl reacted with NaOH = VHCl (C)


= 4.44 mL = 0.00444 L

(D) Mole of HCl reacted with NaOH in = CHCl,stdx (D)

= 0.25 mol/L x 0.00444 mL

= 1.11x10-3

mol

(E) Mole of NaOH unreacted in sample = (E)

= 1.11x10-3

mol

(F) Concentration of NaOH unreacted = (E)

Vs

= 1.11x10-3

mol

0.05L

= 0.0222mol/L

(G) Steadt state fraction conversion = 1 CA

CA0

= 1 (0.0222mol/L)/0.05(mol/L)

= 0.556

(H) Concentration of NaOH reacted = CNaOH,0 (G)


= 0.0278mol/L

17

(I) Mole of NaOH reacted with = (I) x Vs


= 1.39 x 10-3

mol

(J) Concentration of ethyl acetate = (J)/Vs


mol mol/ 0.05 L

= 0.0278mol/L

(K) Concentratin of Ethyl Acetate = CEA,0 (K)


= 0.0722mol/L





= 5.72 mL



= 4.28 mL = 0.00428 L


= 0.25 mol/L x 0.00428 mL

= 1.07x10-3

mol


= 1.07x10-3

mol


18

Vs

= 1.07x10-3

mol

0.05L

= 0.0214mol/L


CA0

= 1 (0.0214mol/L)/0.05(mol/L)

= 0.572



= 0.0286mol/L



= 1.43 x 10-3

mol



mol mol/ 0.05 L

= 0.0286mol/L



= 0.0714mol/L



19



= 5.76 mL



= 4.24 mL = 0.00424 L


= 0.25 mol/L x 0.00424 mL

= 1.06x10-3

mol


= 1.06x10-3

mol


Vs

= 1.06x10-3

mol

0.05L

= 0.0212mol/L


CA0

= 1 (0.0212mol/L)/0.05(mol/L)

= 0.576



= 0.0288mol/L

20



= 1.44 x 10-3

mol



mol mol/ 0.05 L

= 0.0288mol/L



= 0.0712mol/L





= 5.76 mL



= 4.24 mL = 0.00424 L


= 0.25 mol/L x 0.00424 mL

= 1.06x10-3

mol


= 1.06x10-3

mol

21


Vs

= 1.06x10-3

mol

0.05L

= 0.0212 mol/L


CA0

= 1 (0.0212mol/L)/0.05(mol/L)

= 0.576


With Ethyl Acetate(mol/L) = 0.05 mol/L 0.0212 mol/L

= 0.0288 mol/L


Ethyl Acetate in sample(mol) = 0.0288 mol/L X 0.05 L

= 1.44 x 10-3

mol



mol mol/ 0.05 L

= 0.0288 mol/L


Unreacted (mol/L) = 0.1 mol/L 0.0288 mol/L

= 0.0712 mol/L

22

Continuous Stirred Tank Reactor Unit (Model SOLTEQ, BP100)

23

Physical appearance of samples after undergoes titration process.

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