LAB 2 CSTR (1)

ABSTRACT

INTRODUCTION

Reactor is one of the most important parts in industrial sector. Reactor is equipment that changes

the raw materials to the product that we want. A good reactor will give a high production and

economical. Two types of commonly use industrial reactors are Continuous Stirred Tank Reactor

(CSTR) and Plug Flow Reactor (PFR).

CSTR also known as vat- or back mix reactor is a common ideal reactor types in chemical

engineering. CSTR operates at steady state by feeding the reactants into the reactor and

continuously withdrawing the products. The CSTR is the idealized opposite of the well-stirred

batch and tubular plug-flow reactors. Ideally, a CSTR is perfectly well-mixed and has no spatial

variation in concentration or temperature. Since concentration and temperature are identical at all

points in the reactor, they are also the same at the point of discharge from the tank. Other

advantages are a wide temperature and pressure range and the ability to control temperature. A

disadvantage of the CSTR is that the conversion of reactants per volume of the reactor is the

smallest among the flow reactors for most reactions.

The key or main feature of this reactor is that mixing is complete so that properties such as

temperature and concentration of the reaction mixture are uniform in all parts of the vessel.

Material balance of a general chemical reaction described below. The conservation principle

requires that the mass of species A in an element of reactor volume dV obeys the following

statement:

(Rate of A into volume element) - (rate of A out of volume element) + (rate of A produced within

volume element) = (rate of A accumulated within vol. element)

OBJECTIVES

EXPERIMENT A: BATCH STIRRED TANK REACTOR EXPERIMENT

1. To determine the order saponification reaction.

2. To determine the reaction rate constant.

EXPERIMENT B: EFFECT TEMPERATURE ON REACTION RATE CONSTANT

1. To determine the effect temperature on reaction rate constant, k for batch reaction.

2. To determine the activation energy of saponification.

THEORY

IDEAL STIRRED-TANK REACTOR

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)). The key or main feature of

this reactor is that mixing is complete so that properties such as temperature and concentration of

the reaction mixture are uniform in all parts of the vessel. Material balance of a general chemical

reaction is described below.

The conservation principle requires that the mass of species A in an element of reactor volume

∆V obey the following statement:

Rate of

A

Rate of

A Rate of A Rate of A

into - out of + produced =

Accumulate

d

volume volume

within

volume

within

volume

elemen elemen element Element

t t

BATCH STIRRED-TANK REACTOR (BSTR)

The reaction studied in this is the saponification of ethyl acetate. This reaction is elementary and

second-order. The reaction equation is

NaOH + EtOAc NaOAc + EtOH

A + B --> C + D

In batch reactions, there are no feeds or exit streams and therefore equation (1) can be simplified

into:

Rate of A Rate of A

produced = accumulated

within volume within volume

element element (2)

For a constant volume isothermal batch reactor, the design equation is:

For a bimolecular second order reaction, the rate equation is:

-rA = k CACB

The rate of reaction of component A is defined as:

r A=1V ( dN A

dt )by−reaction =

moles−A−appear−by ¿reaction ¿

volume⋅time (3)

By this definition, if A is a reaction product, the rate is positive; whereas if it is a reactant, which

is consumed, the rate is negative.

Rearranging equation (3);

(−r A )V=N AO

dX A

dt (4)

Integrating equation (4);

t=N AO∫

dX A

(−r A )V (5)

where t is the time required to achieve a conversion XA for either isothermal or non-isothermal

operation.

STEADY STATE MIXED FLOW REACTOR

The general material balance for this reactor is as equation (1) except no accumulation of the

material A in rector. As shown in figure 2 below, if FAO = vOCAO is the molar feed rate of the

component A to the reactor, then considering the reactor as a whole we have,

Input of A (moles/time) = FAO (1 – XAO) = FAO

Output of A (moles/time) = FA = FAO (1 – XAO)

Disappearance of A by reaction (moles/time) = (-rA) V

=

moles−A−reacting( time)( volume )

(volume−of −reactor )

Replacing equation (1) with mathematical formula above,

FAOXA = (-rA) V

Which on arranging, will form the performance equation for mixed flow reactors,

Figure 1: Graphical representation of the performance equations for batch reactor, isothermal or non-isothermal at constant density.

VF AO

= τC AO

=ΔX A

−r A or,τ=1

s= V

vO=

VC AO

F AO=

C AO X A

−r A (7)

In mixed flow reactors, XA = XAF and CA= CAF. In a constant density system,

XA = 1 – (CA/CAO)

The performance equation can be rewritten in terms of concentration, or

VF AO

=X A

−r A=

CAO−C A

C AO (−r A), or , τ=V

v=

C AO X A

(−r A) (8)

These expressions relate the four terms XA, -rA, V, FAO; thus, knowing any three allows the fourth

to be found directly. In design, the size of reactor needed for a given duty or the extent of

conversion in a reactor of given size is found directly. Each steady-state point in a mixed flow

reactor gives the reaction rate for the conditions within the reactor. The mixed flow reactor

provides easier interpretation of reaction rate data and makes it very attractive in kinetic studies.

Figure 2: Components of Mixed Flow Reactor

Graphical Representation of the Design Equations for Steady State Mixed Flow Reactor.

Or in constant systems,

Irreversible Second-order Reaction (Bimolecular Type)

Consider the reaction:

A + B Products

The rate equation can be written as:

−r A=−dC A

dt=−

dC A

dt=kC AC B (11)

Plot of ln (CB/CA) versus t (time) will produce a straight line with slope equals (CBO – CAO)k.

Figure 3: Plot of 1 / (-rA) versus XA

Figure 4: Plot of 1 / (-rA) versus CA

While for steady state mixed flow reactor, the plot of rA versus CACB will give a straight line (pass

through the origin) with the slope equal to k.

For second order reaction with equal initial concentrations of A and B, the rate equation can be

written based on only one component:

−r A=−dC A

dt=kC A

2 =kC AO2 (1−X A)2

(12)

A plot of 1 / CA versus t will produce a straight line with slope equals to k.

Figure 5: Plot of ln (CB/CA) versus t, CAO ¿ CBO (for equation 11)

Figure 6: Plot of –rA versus CACB, CAO ¿ CBO (for equation 11)

In the case of steady state mixed flow reactor, the plot –rA versus CA2 will give the value of k and

a straight-line pass through the origin.



EXPERIMENT B: Effect of Temperature on Reaction Rate Constant, k

In any single homogenous reaction, temperature, composition, and reaction rate are uniquely

related. They can be represented graphically in one of three ways as shown below:

Figure 9: Plot of C versus T, r versus T, and r versus C

The effect of temperature on reaction rate constant can be demonstrated by performing a batch

reaction run at different temperatures.

r3

r2

r1

C3

C2

C1

RESULTS

EXPERIMENT A

Table 1:

Time (min) (A)

Volume of Titrating

NaOH (ml) (B)

Volume of Quenching

HCL Unreacted with NaOH in Sample

(ml) (C)

Volume of HCL

Reacted with NaOH in Sample

(ml) (D)

Mole of HCL Reacted

with NaOH in Sample

(E)

Mole of NaOH

Unreacted in Sample

(F)

1 16.2 6.48 3.52 0.00099 0.000995 17.7 7.08 2.92 0.00073 0.00073

10 18.7 7.48 2.52 0.00063 0.0006315 20.4 8.16 1.84 0.00052 0.0005220 21.5 8.6 1.4 0.00039 0.0003925 22.5 9.0 1.0 0.00028 0.00028

Table 2 :

Steady State

Fraction Conversion

of NaOH XA (H)

Concentration of NaOH

Reacted with Ethyl Acetate

(moles/litre) (I)

Mole of NaOH

Reacted with Ethyl Acetate in

Sample (J)

Concentration of Ethyl

Acetate Reacted

with NaOH (moles/litre)

(K)

Concentration of Ethyl

Acetate Unreacted

(moles/litre) (L)

ln(CB/CA)

0.604 0.0302 0.00151 0.0302 0.0698 1.2530.708 0.0354 0.00177 0.0354 0.0646 1.4660.748 0.0374 0.00187 0.0374 0.0626 1.5780.792 0.0396 0.00198 0.0396 0.0604 1.7590.844 0.0402 0.00201 0.0402 0.0598 2.0370.888 0.0444 0.00222 0.0444 0.0556 2.295

The time required for 95% conversion is

m = (CBo /CAo) = 0.1/0.05 = 2

ln m – XA = (CBO – CAO) kt m(1-XA)

ln m – XA t = m(1-XA)

(CBO – CAO) k

ln 2 - 0.95 = 2(1-0.95)

(0.1 – 0.05)0.0173

t = 2764.71 min

EXPERIMENT B

Temperature = 35°C

Table 3 :

Time (min) (A)

Volume of Titrating

NaOH (ml) (B)

Volume of Quenching

HCl Unreacted with NaOH in

Sample (ml) (C)

Volume of HCl Reacted with

NaOH in Sample

(ml) (D)

Mole of HCL Reacted with

NaOH in Sample

(E)

Mole of NaOH Unreacted in

Sample (F)


Unreacted with Ethyl Acetate (moles/litre)

CA (G)

1 21.7 8.69 1.32 0.00033 0.00033 0.00665 22.1 8.84 1.16 0.00029 0.00029 0.005810 22.4 8.96 1.04 0.00026 0.00026 0.005215 23.1 9.24 0.76 0.00019 0.00019 0.003820 23.6 9.44 0.56 0.00014 0.00014 0.002825 24 9.6 0.4 0.0001 0.0001 0.0020

Table 4 :

Steady State Fraction

Conversion of NaOH

XA (H)


Reacted with Ethyl Acetate (moles/litre)

(I)

Mole of NaOH Reacted with Ethyl Acetate

in Sample (J)

Concentration of Ethyl Acetate

Reacted with NaOH

(moles/litre) (K) (D)


Unreacted (moles/litre)

(L)

ln(CB/CA) ln K

0.868 0.0434 0.00217 0.0434 0.0566 2.149 -3.1370.884 0.0442 0.00221 0.0442 0.0558 2.264 -3.1190.896 0.0448 0.00224 0.0448 0.0552 2.362 -3.1060.924 0.0462 0.00231 0.0462 0.0538 2.650 -3.0750.944 0.0472 0.00236 0.0472 0.0528 2.937 -3.0530.960 0.048 0.00240 0.0480 0.0520 3.258 -3.037

0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

3

3.5

ln Ca/Cb VS Time

Series2

Time

ln C

a/Cb

Figure 1 Graph ln Ca/Cb versus Time for Temperature 35 Degree

TEMPERATURE = 40°C

Table 5 :

Time (min) (A)

Volume of Titrating

NaOH (ml) (B)

Volume of Quenching


Sample (ml) (C)


NaOH in Sample

(ml) (D)


NaOH in Sample

(E)


Sample (F)



CA (G)

1 16.3 6.52 3.48 0.000874 0.000874 0.01755 17.4 6.96 3.04 0.000760 0.000760 0.015210 17.6 7.04 2.96 0.000740 0.000740 0.014815 18.3 7.32 2.68 0.000670 0.000670 0.013420 19.2 7.68 2.32 0.000580 0.000580 0.011625 20.4 8.16 1.84 0.000460 0.000460 0.0092

Table 6 :


Conversion of NaOH

XA (H)



(I)


in Sample (J)


Reacted with NaOH




(L)

ln(CB/CA) ln K

0.650 0.0325 0.00163 0.0325 0.0675 1.3508 -3.42590.696 0.0348 0.00174 0.0348 0.0652 4.2895 -3.35810.704 0.0352 0.00176 0.0352 0.0648 4.3784 -3.34670.732 0.0366 0.00183 0.0366 0.0634 4.7313 -3.30770.768 0.0384 0.00192 0.0384 0.0616 5.3103 -3.25970.816 0.0408 0.00204 0.0408 0.0592 6.4348 -3.1991

1 2 3 4 5 6 70

5

10

15

20

25

30

ln ( Cb/Ca) versus time

time

ln (C

a/Cb

)

Figure 2 Graph of ln Ca/Cb versus Time for Temperature 40 Degree

TEMPERATURE = 45°C

Table 7 :

Time (min) (A)

Volume of Titrating

NaOH (ml) (B)

Volume of Quenching


Sample (ml) (C)


NaOH in Sample

(ml) (D)


NaOH in Sample

(E)


Sample (F)



CA (G)

1510152025

Table 8 :


Conversion of NaOH

XA (H)



(I)


in Sample (J)


Reacted with NaOH




(L)

ln(CB/CA) ln K

-1.253-1.466-1.578-1.759-2.037-2.295

0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

ln Cb/Ca versus time

time(min)

ln C

a/Cb

Figure 3 Graph ln Ca/Cb versus Time for Temperature 45 Degree

Figure 4 Graph ln K versus 1/T

0 0.2 0.4 0.6 0.8 1 1.2

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

f(x) = 0.71934706169974 x − 1.90597881079634

f(x) = − 0.160386196504073 x − 3.27726090459209f(x) = − 0.0746016417846237 x − 3.06972129806872

ln K versus 1/T

T=35 Linear (T=35) T=40Linear (T=40) T=45 Linear (T=45)

1/T

ln K

Figure 5 Graph ln K versus 1/T

From graph ln k vs 1/T, we can find the activation energy from the slope.

Knowing that, slope = −E /RWhere, E is the activation energy

R is the gas law constantTherefore, E = − (slope X R)

E = − (0.16143X 8.314)= 1.3422 J / mole

DISCUSSION

In this Continuous Stirred-Tank Reactor (CSTR) experiment, the objective is to determine and examine the reaction of chemical process. There are three area of investigation of reaction, the stoichiometry, the kinetics and the mechanism. First by using the stoichiometry of the equation, the kinetics is then investigated. With empirical rate expression available, the mechanism is then looked into.

Let B be the component with highest concentration (Ethyl Acetate), while the most convenient way to follow reaction is by titration of A (NaOH). The reaction is elementary and since it is bimolecular, it is natural to first try second order kinetics. In this experiment, the volume of Ethyl Acetate must equal to volume of NaOH and completely fed into the reactor at beginning since batch reaction is considered. Stirrer is used because agitation is also necessary to equalize the temperature in the reactor and to keep it at the desired level.

In the integral method of analysis, the first step is to guess a particular form of rate equation and after appropriate integration and mathematical manipulation, predict that the plot of certain concentration function versus time should yield a straight line.

Rate constant, k for the n-order reaction of homogenous reaction is (time)-1 concentration)1-n where n is the order of the reaction. In this experiment, there are experiment A and experiment B which is respective to time and temperature respectively. In experiment A, the which is batch reactor experiment we use 0.1 M, 1.25 liter Ethyl acetate and 0.05 M, 1.25 liter sodium hydroxide as the reactants. After that, the stirrer is switched to mix the reactant well. The sample of the products produced in every 1, 5, 10, 15, 20 and 25 min is taken out. In each sample titration has been carried out with 0.1 M NaOH to

evaluate the amount of unreacted HCL. From Table 1 it shows that at 1 minute of sample it needs 16.2 ml of 0.1 M NaOH to react. At 5 minute of sample it needs 17.7 ml of 0.1 M NaOH to react, whereas in sample of 10, 15, 20 and 25 minute it needs 18.7, 20.4, 21.5 and 22.5 ml of 0.1 M of NaOH to react respectively. It shows that the volume of NaOH is increasing. When plotting the graph of ln CB/CA vs. time, it shows that it gives almost straight line. From this result it showed that it follows second order of reaction. The rate constant from experiment A is 1.4595

with the time taken of 2764.17 min

For the experiment B, the procedures using same as the previous but at different temperatures are varying. The range of the temperature is 35°C, 40oC and 45oC. The result shows the effect of temperature on reaction rate constant. The temperatures are not constant for the whole experiment as well as the pattern of the graph. The heater is using to raise the temperature and in order to get a constant temperature, the cooling system is using. The start up experiment is same as experiment A, it is just in experiment B, heat is supplied to the system. The temperature was set up at first but this system also cannot give a very constant temperature. Therefore, it affects the value of concentration for NaOH and ethyl acetate. The graph should be a straight line but it is not because since there are some errors due to human, environment and the equipment itself. The activation energy of saponification is 1.3422J/mole and from the plotted graph of each temperature shows that temperature did affect the reaction rate constant, k for batch reaction. The pH also is not constant at one specific pH because it is fluctuate. The range of the pH changes is ±0.01.

For the temperature 30 oC, the time taken is 1, 5, 10, 15, 20 and 25 minutes show reading of volume NaOH needed is 16.3, 17.4, 17.6, 18.3, 19.2 and 20.4 ml respectively. The volume needed is increasing as the time increasing. For the temperature 35 oC, the time taken is 1, 5, 10, 15, 20 and 25 minutes show reading of volume NaOH needed is 21.7, 22.1, 22.4, 23.1, 23.6 and 24 ml respectively. After that table 7 shown the reading of volume NaOH needed for temperature 45 0C. One of the common errors that occur during this experiment is the human errors which are during the titration process. This is because sometimes over titrate the solution make the concentration that being obtained is not accurate. In addition, during taking reading of amount HCL from the burette also contribute to the error which is parallax

errors. The environment also contributes errors since this experiment base on the effect of temperature. During taking a sample may cause heat loss as well as the cool condition in the laboratory because air – conditioner is on. The machine itself also causes some deviations to the result because of inconstant of temperature level and speed of stirrer.

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LAB 2 CSTR (1)