RD Lab - Exp-07 - G-A1.pdf

Koya University
Faculty of Engineering
Chemical Engineering Department
3rd Stage (2021-2022)
Reactor Laboratory
Lab Report
Number of Experiment:7
Experiment Name: Constant Rate of Reaction (40 o
C)
Experiment Date: 17/11/2021
Submitted on: 10/12/2021
Instructor: Mr. Ahmed Abdulkareem Ahmed
Group: A1
Prepared by:
Safeen Yaseen Jafar
Ibrahim Ali
Ahmed Mamand Aziz
Rokan Mohammad Omer
Rivan Dler Ali
Ramazan Shkur Kakl

Table of Content
1. Aim of Experiment..................................................................................................................................1
2. Procedure.................................................................................................................................................2
3. Tools and Apparatus...............................................................................................................................3
4. Table of Reading .....................................................................................................................................5
5. Calculation and Results..........................................................................................................................7
6. Discussion...............................................................................................................................................13

1
1. Aim of Experiment
➢ To determine the reaction order.
➢ To find out the reaction constant (K) at 40 o
C.
➢ Finding Activation Energy (Ea).

2
2. Procedure
1. Prepare 1 L and 0.05 M of NaOH (solution). after it, we need to prepare second reactant
for reaction occur prepare 1 L and 0.05 M of CH3COOHC2H5 (solution) .
2. First of all, in the service units close all valve if open. After that put the bottles in specific
places in service unit. And be careful to that the pipes and valves are connected as well.
3. Turn on the switch control box (power supply).
4. This experiment operates in the room temperature (at 40 oC in this experiment).
5. Set the limited flow rate of the reagents before run the steps.
6. Switch on the valves and pumps of reactants
7. Take the reactants from their containers to inside the reactor.
8. We fill the reactor by both reactant liquids and the flow rate would be limited at both
flow meters in the control box.
9. Switch on the stirrer from the main control box.
10. Turn on the conductivity meter (which connected initially to the reactor).
11. Feeds enter the reactor and out the reactor (when reach overflow valve).
12. The conductivity measurements (Conductivity sensor) must be noted while change of
the conductivity reach the constant value. The readings should be taken at every 10
second.
13. At the end of the experiment, turn off the pumps, stirrer.
14. Turn off the power of control interface box.
15. Reactants should be removed from both (1 and 2) reactant bottle container. Then, the
liquids must be kept for following test.

3
3. Tools and Apparatus
Figure 2: Service Unit - Back
Part
Figure 1: Service Unit - Front Part
1
2
4
3
5
6
7
Figure 3: Control Unit
8
10
11
12
Figure 4: Batch Reactor
13
9
14

4
Service Unit, control unit and its parts:
1. Water Bath: is the tank which contain water it used for control temperature of
reactants.
2. Water Bath temperature switch button and controllers.
3. Reactant Container 1: For storage reactant 1.
4. Reactant Container 2: For storage reactant 2.
5. Water Pump AB-1: it used for pumping water.
6. Pump AB-2: It used for pump the reactant of reactor 1.
7. Pump AB-3: It used for pump the reactant of reactor 2.
8. Pump AB-1 on/off button: It used for switch on or switch of pump AB-1.
9. Power Button: Used to turn on control unit.
10. Temperature Display: For displaying the temperatures
11. Temperature Sensor: for record or estimate temperature.
12. Volumetric Switch Button and Controller: For control the flowrate of reactant.
13. Sensor Selector: it used for select the temperature sensor that you want.
14. Temperature Controller: It can be used when we want to select specified
temperature or control temperature in water bath that we want.
Batch reactor parts:
1. Coil: for control the temperature of the reaction.
2. Conductivity Sensor: for record the conductivity.
3. Temperature Sensor: for record temperature.
4. Stirrer: for mix the reactants and make a collision for reactants.
5. Vessel: it is the closed tank that contain reactants and the reaction happen in it.
6. Stirrer: for mix the reactants and make a collision for reactants.
7. Overflow valve: to out the product at the end of reaction.
8. Drain valve: to empty the vessel at the end of experiment.

5
4. Table of Reading
Time (sec) Conductivity (mS) or (λ) λo λ∞
0 6.82 6.82 2.19
10 6.76 6.82 2.19
20 6.7 6.82 2.19
30 6.64 6.82 2.19
40 6.43 6.82 2.19
50 6.24 6.82 2.19
60 5.84 6.82 2.19
70 5.6 6.82 2.19
80 5.27 6.82 2.19
90 5.02 6.82 2.19
100 4.8 6.82 2.19
110 4.6 6.82 2.19
120 4.44 6.82 2.19
130 4.24 6.82 2.19
140 4.11 6.82 2.19
150 3.87 6.82 2.19
160 3.71 6.82 2.19
170 3.56 6.82 2.19
180 3.41 6.82 2.19
190 3.33 6.82 2.19
200 3.21 6.82 2.19
210 3.1 6.82 2.19
220 3.01 6.82 2.19
230 2.87 6.82 2.19
240 2.82 6.82 2.19
250 2.81 6.82 2.19
260 2.79 6.82 2.19
270 2.73 6.82 2.19
280 2.71 6.82 2.19
290 2.7 6.82 2.19
300 2.65 6.82 2.19
310 2.63 6.82 2.19
320 2.57 6.82 2.19
330 2.53 6.82 2.19
340 2.52 6.82 2.19
350 2.48 6.82 2.19

6
360 2.47 6.82 2.19
370 2.43 6.82 2.19
380 2.39 6.82 2.19
390 2.35 6.82 2.19
400 2.33 6.82 2.19
410 2.32 6.82 2.19
420 2.31 6.82 2.19
430 2.28 6.82 2.19
440 2.27 6.82 2.19
450 2.26 6.82 2.19
460 2.22 6.82 2.19
470 2.19 6.82 2.19
480 2.19 6.82 2.19
*T = 40 oC (313 K)

7
5. Calculation and Results
We obtained conductivity value from recording by conductivity meter in lab
Concentration for each 10 sec. can calculated by this equation:
CA
CAo
=
λo− λ
λo− λ∞
Conversion calculated by this equation: X =
CAo−CA
CAo
We can find out the order of the reaction and constant (K) of the reaction by
know linear or non-linear below equations:
ln
CA
CAo
= −Kt
1
CA
−
1
CAo
= Kt
Then we will calculate Ea by graph and by slope as follow
Draw a line between Ln(k) and 1/T by using constant rates (K) in previous
experiments (exp. 6, 2 and this exp. 7) and T = 21, 30 and 40 o
C, respectively.
Time (sec) Conductivity (mS) Concentration Conversion Ln CA/CAo 1/CA
0 6.82 0.05 0 0 20
10 6.76 0.049352052 0.012958963 -0.013043663 20.26258206
20 6.7 0.048704104 0.025917927 -0.026259715 20.53215078
30 6.64 0.048056156 0.03887689 -0.039652772 20.80898876
40 6.43 0.045788337 0.084233261 -0.087993599 21.83962264
50 6.24 0.043736501 0.125269978 -0.133839987 22.86419753
60 5.84 0.039416847 0.211663067 -0.237829701 25.36986301
70 5.6 0.036825054 0.26349892 -0.305844577 27.15542522
80 5.27 0.033261339 0.334773218 -0.407627271 30.06493506
90 5.02 0.030561555 0.388768898 -0.492280156 32.72084806
100 4.8 0.028185745 0.436285097 -0.573206647 35.4789272
110 4.6 0.026025918 0.479481641 -0.652930121 38.42323651
120 4.44 0.024298056 0.514038877 -0.721626652 41.15555556
130 4.24 0.022138229 0.557235421 -0.814717075 45.17073171
140 4.11 0.020734341 0.585313175 -0.880231682 48.22916667
150 3.87 0.018142549 0.637149028 -1.013763075 55.11904762
160 3.71 0.016414687 0.671706263 -1.113846533 60.92105263
170 3.56 0.014794816 0.704103672 -1.217746128 67.59124088
180 3.41 0.013174946 0.73650108 -1.333706009 75.90163934

8
190 3.33 0.012311015 0.753779698 -1.401528606 81.22807018
200 3.21 0.011015119 0.779697624 -1.512754241 90.78431373
210 3.1 0.009827214 0.803455724 -1.626867548 101.7582418
220 3.01 0.008855292 0.822894168 -1.731007807 112.9268293
230 2.87 0.007343413 0.853131749 -1.918219349 136.1764706
240 2.82 0.006803456 0.863930886 -1.994592328 146.984127
250 2.81 0.006695464 0.866090713 -2.010592669 149.3548387
260 2.79 0.006479482 0.870410367 -2.043382492 154.3333333
270 2.73 0.005831533 0.88336933 -2.148743008 171.4814815
280 2.71 0.005615551 0.887688985 -2.186483336 178.0769231
290 2.7 0.005507559 0.889848812 -2.205901421 181.5686275
300 2.65 0.004967603 0.900647948 -2.309085658 201.3043478
310 2.63 0.00475162 0.904967603 -2.35353742 210.4545455
320 2.57 0.004103672 0.917926566 -2.500140894 243.6842105
330 2.53 0.003671706 0.926565875 -2.611366529 272.3529412
340 2.52 0.003563715 0.928725702 -2.641219493 280.6060606
350 2.48 0.003131749 0.937365011 -2.770431224 319.3103448
360 2.47 0.003023758 0.939524838 -2.805522544 330.7142857
370 2.43 0.002591793 0.948164147 -2.959673224 385.8333333
380 2.39 0.002159827 0.956803456 -3.141994781 463
390 2.35 0.001727862 0.965442765 -3.365138332 578.75
400 2.33 0.001511879 0.969762419 -3.498669724 661.4285714
410 2.32 0.001403888 0.971922246 -3.572777697 712.3076923
420 2.31 0.001295896 0.974082073 -3.652820404 771.6666667
430 2.28 0.000971922 0.980561555 -3.940502477 1028.888889
440 2.27 0.000863931 0.982721382 -4.058285512 1157.5
450 2.26 0.00075594 0.98488121 -4.191816905 1322.857143
460 2.22 0.000323974 0.993520518 -5.039114765 3086.666667
470 2.19 0 1 #NUM! #DIV/0!
480 2.19 0 1 #NUM! #DIV/0!
T = 40 o
C (313 K)

9
1. Draw: Ln (CA/CAo) vs time
The curve is nonlinear so this is second order reaction.
So, k calculated by:
ln
CA
CAo
= −Kt
At t = 380s → Concentration (CA) = 0.002159827
At t = 380s → ln
0.002159827
0.05
= −K × 380
−3.14199488 = −K × 380
K = 8.268 × 10-3 at 40 oC (313 K)
Then, we can find out Rate of reaction (r) from reaction rate law equation:
−ra = K × [CA]α
ra = 8.268 × 10-3 × [0.002713704]2
ra = 3.8571 × 10-8
M/sec
-6
-5
-4
-3
-2
-1
0
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
300
320
340
360
380
400
420
440
460
Ln(CA/CAo)
vs
t
Time (sec)
Ln(CA/CAo) vs t

10
2. 1/CA vs time
This is the first order because is linear and K can calculate by slope = K:
1
CA
−
1
CAo
= Kt
𝐊 = 𝐒𝐥𝐨𝐩𝐞 =
Δy
Δx
= 0.177166013
−ra = K × [CA]α
At (t=380 s) CA = 0.002159827
ra = 0.177166013 × [0.002159827]1
ra = 3.826479 × 10-4
M/sec
0
500
1000
1500
2000
2500
3000
3500
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
300
320
340
360
380
400
420
440
460
1/CA
Time (sec)
1/CA

11
Exp. Constant Rate (K) Temperature (oC) Temperature (K) 1/T (in Kelvin) Ln(K)
1 1.291 × 10-3 21 294 0.003401361 -6.652338167
2 5.946 × 10-3 30 303 0.00330033 -5.125036554
3 8.268 × 10-3 40 313 0.003194888 -4.795362637
R = 8.314 J/Kmol
Slope = −8951.149905 =
−Ea
R
Ea = 74419.86 J/Kmol = 74.42 KJ/Kmol
-7
-6
-5
-4
-3
-2
-1
0
0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345
Ln(K)
1/T
Lnk versus 1/T

13
6. Discussion
Discussion – Safeen Yaseen Ja’far
From this experiment we learnt determine order of reaction and reaction rate
constant and also Activation Energy (Ea) by graph and relation Ln(K) with (1/T).
As in the previous experiments (exp. 5 and 6) and this exp. 7. We should
record and calculated the conductivity (λ) and concentration with time (CA) by (λ).
We recorded the value of (λ) until we reach a constant value of it in the lab.
After that, by the equation or graph determine/find out the value of 𝛂 by know
the linearity of the graph. If graph was linear means order of reaction equal to 1
then calculate reaction rate constant (K) must be calculate by Slope. But if the
graph was nonlinear means the reaction order equal to 2 then calculate reaction
rate constant (K) by equation.
But in this experiment, we want to determine Ea and it can be determined by
calculate Ln of all value of K and inverse T (means 1/T) for this three exp. (5, 6
and 7), Then, draw a chart between them and we can find -slope which equal to (-
Ea/R) and know Ea have been determined.

14
Discussion – Rivan Dler Ali

15
Discussion – Ibrahim Ali
In this experiment we select the temperature at 40 C and then we want to
determine the reaction order by the line if the curve is straight the reaction order
the first and if the curve is not straight the we told the second order reaction and we
draw the curve by the lnCA/CA0 with time and 1/CA , and then we want to
determine reaction constan (K) by this to law ln CA/CAo = −Kt ; 1/CA − 1/CAo=
Kt and then we can find the Activation Energy (Ea) by the slope of the curve and R
constant gas.

16
Discussion – Ahmed Mamand Aziz
In this experiment we will determine reaction rate constant and the reaction
orders, the solution should be prepared that between {Sodium hydroxide and Ethyl
acetate}, should be known that our experiment done in [40 °C]. Each reactant
prepared in 1 L contain of 0.05M. Our first reading that recorded is conductivity by
conductivity meter that λ∞=2.19, λo=6.82. By using the little low we could find
the Conversion. After that to create the relation table we should find the value of
(Ln CA/Cao) and (1/CA) for know that the relation linear or nonlinear. we know
that the relation between (cond. and time) is opposite, and also known the
relationship between (conv. and time) that is proportional. and we know that the
relationship between (Ln (CA/Cao) and time) is opposite, and the relationship
between (1/CA and time) is also proportional.
During creating the first curve (second order) known that it is nonlinear because of
this the value of K can be found just by law and that is found K=8.268 × 10-3, but
in the first order because is linear so the value of K can be found by slop [slop=K]
and found that K=0.177166013.During the calculation in second order the curve is
nonlinear because of this the value of K can be found just by law, but in the first
order because is linear so the value of K can be found by slop [slop=K]

17
Discussion – Rokan Mohammad
In Constant Rate of Reaction experiment we recorded the conductivity value
by conductivity meter of the (NaOH and CH3COOC2H5)¬ reaction both about
0.05M at (40 0C), and after that we found conversion by using conversion
equation. As we know we should estimate the value of Ln CA/Cao and 1/CA for
creating the table to find the K and the order. So, about curves we have two order
(first order & second order). We should find value of 𝛂 either by graph or equation
if the graph is linear means order of reaction equal to 1 and K can be found by
slope, but if the graph is not linear the order of reaction equal to 2 and K will find
by equation. If we see the first curve these is second order and nonlinear curve so
we can find (K) by just low {Ln CA/CAo=−Kt}. However, in second curve these is
first order and linear so we can find the (K) value by slop that slop=K.

18
Discussion – Ramazan Shkur Kakl

RD Lab - Exp-07 - G-A1.pdf

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