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A set of questions about Unit Operations Adsorption
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Adsorption Adsorption process/kinetics Mechanisms of Adsorption Factors affecting adsorption process Adsorption Isotherms CE6301:
Home made charcoal from paper waste.Full description
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The aim of this project is to investigate the ascorbic acid content in different citrus fruits.
The solar adsorption Refrigeration system is very useful in rural area without grid. The Adsorber desorber are the important role in this system. Among the adsorbent refrigerant, the activated carbon and ethanol is applied. This paper focuses on the
Title : Adsorption of acetic acid on charcoal surface Objective : To find out the reaction rate of adsorption of acetic acid on charcoal surface . Balanced Equations: CH3COOH + NaOH CH3COONa + H2O KHP + NaOH KNaP + H2O Observations : When KHP was added to the water , it dissolved . When phenolphthalein was added into KPH or acetic acid , the color of the solution was colorless . When titrate the KPH or acetic acid with NaOH , the color of the solution was changed from colorless to pink . Experimental data: Weight of charcoal used for each flask Conical flask no. 1
2
3
4
[CH3COOH](M) 0.25
0.125
0.075
0.050
Charcoal (g)
1.002
1
1.001
1
Volume of NaOH used for titration with KHP Trial 1
Trial 2
Trial 3
Mass of KHP (g)
0.1
0.111
0.102
Initial volume (mL)
3
15.7
22.3
Final volume (mL)
9
22.3
28.8
Vol. of NaOH used (mL)
6
6.6
6.5
Volume of NaOH used for titration with acetic acid Trial 1 Flask No.
Vol. of Initial HAc used vol. of (mL) NaOH
Trial 2 Final vol. of NaOH used
Initial vol. of NaOH
Final vol. of NaOH used
Ave. vol. of NaOH used
used
(mL)
used
(mL)
(mL)
(mL)
(mL)
1
5.00
2.8
14.5
14.5
26.3
11.75
2
5.00
27.7
33.4
33.4
38.6
5.45
3
10.0
26.3
32.6
32.6
38.6
6.20
4
10.0
38.6
42.5
45.5
46.2
3.80
Data treatment : Molarity of NaOH : 0.082M M.W. of acid = 60 g/mol Flask No.
Ce
X
1
0.1927
0.1719
2
0.08938
0.10686
3
0.05084
0.07248
4
0.03116
0.05652
Ce = equilibrium [HAc](M) X = mass of acetic acid adsorbed (g)
Calculation : In flask 1 , Ce = (11.75/1000 x 0.082 ) / (50/1000) = 0.1927 X = (0.25-0.1927) x 60 x 50/1000 = 0.1719 In flask 2 , Ce = (5.45/1000 x 0.082) / (50/1000) = 0.08938 X = (0.125 – 0.08938) x 60 x 50/1000 = 0.10686 In flask 3, Ce = (6.2/1000 x 0.082) / (50/1000) = 0.05084 X = (0.075-0.05084) x 60 x 50/1000 = 0.07248 In flask 4, Ce = (3.8/1000 x 0.082) / (50/1000) = 0.03116 X = (0.05-0.03116 ) x 60 x 50/1000 = 0.05625 Question : 1. Considering log(x/m) = nlogc + logk log ( x/m)
log c
-0.76
-0.60
-0.97
-0.90
-1.14
-1.12
-1.25
-1.30
a graph of log(x/m) against log c 1
log (x/m)
0.5 0
-1.5
-1
-0.5
0
0.5
1
1.5
2
-0.5 y = 0.7056x - 0.3359 -1 -1.5
log c
Where n = 0.71 , logk = -0.34 The value of k = 10-0.34 = 0.46 s−1 2) 1. 2. 3. 4. 5.
Water treatment use as antidote for medication purify car exhaust gas make mask for workers like construction workers act as catalyst in chemical reaction
Conclusion : The reaction rate of adsorption of acetic acid on charcoal surface is 0.46 s−1 .
