ADVANCE OF CHEMICAL REACTION ENGINEERING
4.
Gambarkan langkah-langkah reaksi katalitik! (hubungkan dengan mass transfer)
Jawab:
7.
Cari metode untuk mengkarakterisasi luas surface area katalis padatan! (buat flow
chart)
Jawab: Sudah lengkap di “Teknik Karakterisasi Katalis”
Characterization
of Catalyst
Surface area
Particle
characteristic
Porosity
Density
Size of particle
Mechanical &
diffusivity
Structure &
Surface
characteristic
morphology
Chemisorptions
Acidity
Element composition
Bulk characteristic
Phase structure
Molecule structure
Bulk reactivity
Experimental method of estimating surface areas solid catalyst Sg (m2/g)
Volumetric Method
Dynamic Method
Gravimetric Method
Capable for Isotherm I,
Continuous flow of
Capable for recording
II, IV
adsorable gas through
complete isotherms
Langmuir Eq.
𝑷
𝑷
𝟏
=
+
𝑽 𝑽𝑴 𝒃𝑽𝑴
Point B method
adsorbent column
Beam
Balance
Easier to assemble
Gravity operated
Sophisticated
Plotting V vs P
Spring
to find ‘point B’
Balance
Inexpensive
BET method
Appropriate for physical
𝑷
𝑽(𝑷𝟎 − 𝑷)
=
(𝒄 − 𝟏) 𝑷
𝟏
+
( )
𝑽𝑴 𝑪
𝑽𝑴 𝑪 𝑷 𝟎
ads. using inert gas
𝒍𝒓𝟐
(𝟐𝒎 + 𝒍𝝁)
𝒛=
𝝅𝒚𝒂𝟒
= 𝒌𝒎 + 𝝀
8.
Kerjakan exercise 1!
Dinitrogen adsorption data:
Volume adsorbed (cm3/g)
Sample 1
Sample 2
0.02
23.0
0.15
0.03
25.0
0.23
0.04
26.5
0.32
0.05
27.7
0.38
0.10
31.7
0.56
0.15
34.2
0.65
0.20
36.1
0.73
0.25
37.6
0.81
0.30
39.1
0.89
(a) Calculate the BET surface area per gram of solid for Sample 1 using the full BET
P/P0
equation and the one-point BET equation. Are the values the same? What is the BET
constant?
(b) Calculate the BET surface area per gram of solid for Sample 2 using the full BET
equation and the one-point BET equation. Are the values the same? What is the BET
constant and how does it compare to the value obtained in (a)?
Jawab:
(a) BET surface area for Sample 1 using the one-point BET equation:
Normal boiling point of dinitrogen is 77 K and the saturated vapour pressure P0 = 1.05
bar = 101.3 kPa. Assuming mass of each sample is 1 gram.
P (kPa)
2.026
3.039
4.052
5.065
10.13
15.195
20.26
25.325
30.39
Volume adsorbed (cm3/g)
Sample 1
Sample 2
23.0
0.15
25.0
0.23
26.5
0.32
27.7
0.38
31.7
0.56
34.2
0.65
36.1
0.73
37.6
0.81
39.1
0.89
Plotting V against P is shown in Figure 1 below, which yields an ordinate value of
27.7 cm3/g at point B. This plot is identified by nothing the ordinate value of the
volume (when V is plotted against P) as the isotherm bends over sharply or by
applying the BET theory. The former method is euphemistically referred to as the
‘point B’ method and is only reliable when there is a well-defined and sharp change
of curvature in the isotherm as shown in graph below.
45
y = 0,5423x + 24,253
40
35
30
25
20
0
5
10
15
20
25
30
35
Hence, VM = 27.7 cm3/g = 2.77 x 10-8 m3/g. So, value of the specific surface area is:
𝑆𝑔 =
𝑉𝑀
× 6.023 × 1023 × 𝐴
0.0224
If the area occupied by each adsorbate molecule is A m2 (sometimes expressed as
Angstrom squared, which is equivalent to 10-20 m2). From equation above, value of
the specific area of solid for Sample 1 is 7.45 x 10-3 m2/g.
BET surface area for Sample 1 using the full BET equation:
𝑷
To apply the full BET equation, plotting data in the form 𝑽(𝑷
𝟎 −𝑷)
against P/P0 as
shown in Table and Figure below:
P/P0
P (kPa)
(𝑷𝟎 − 𝑷)
0.02
0.03
0.04
0.05
0.10
0.15
0.20
0.25
0.30
2.026
3.039
4.052
5.065
10.13
15.195
20.26
25.325
30.39
99.274
98.261
97.248
96.235
91.17
86.105
81.04
75.975
70.91
Sample 1
Volume
𝑷
adsorbed
𝑽(𝑷𝟎 − 𝑷)
(cm3/g)
23.0
0.000887
25.0
0.00124
26.5
0.00157
27.7
0.0019
31.7
0.0035
34.2
0.00515
36.1
0.00693
37.6
0.00887
39.1
0.011
Sample 2
Volume
𝑷
adsorbed
𝑽(𝑷𝟎 − 𝑷)
(cm3/g)
0.15
0.136
0.23
0.134
0.32
0.130
0.38
0.139
0.56
0.198
0.65
0.271
0.73
0.342
0.81
0.412
0.89
0.482
0,012
y = 0,0353x + 9E-05
0,01
0,008
0,006
0,004
0,002
0
0
0,05
0,1
0,15
0,2
0,25
0,3
0,35
From graphic above, if the slope of such a plot is (s) and its intercept is (i), then it is
clear from Equation:
(𝒄 − 𝟏) 𝑷
𝑷
𝟏
=
+
( )
𝑽(𝑷𝟎 − 𝑷) 𝑽𝑴𝑪
𝑽𝑴 𝑪 𝑷 𝟎
That 1/(s + i) is equal to VM. From the graph displayed, s = 3.5 x 10-2 and i = 9 x 10-5,
then VM = 28.49 cm3/g = 2.849 x 10-8 m3/g. Hence, value of the specific surface area
is:
𝑆𝑔 =
𝑉𝑀
× 6.023 × 1023 × 𝐴
0.0224
From equation above, value of the specific area of solid for Sample 1 is 7.66 x 10-3
m2/g. The value between those two method is not same (has a difference about
0.21 x 10-3 point).
BET constant:
The equation for BET constant (c) is:
C
s
1
i
Therefore, BET constant of solid for Sample 1 is = 389.889
(b) BET surface area for Sample 2 using the one-point BET equation:
Using same technique and equation in point (a), then plotting graph V against P as
below:
1,2
1
y = 0,0244x + 0,211
0,8
0,6
0,4
0,2
0
0
5
10
15
20
25
30
35
which yields an ordinate value of 0.38 cm3/g as point B. Hence, VM = 0.38 cm3/g =
3.8 x 10-10 m3/g. So, value of the specific surface area using same equation is 1.02 x
10-4 m2/g.
BET surface area for Sample 2 using the full BET equation:
Plotting data in the form
𝑷
𝑽(𝑷𝟎 −𝑷)
against P/P0 for Sample 2, then obtained graph
below:
0,6
0,5
y = 1,2903x + 0,086
0,4
0,3
0,2
0,1
0
0
0,05
0,1
0,15
0,2
0,25
0,3
0,35
From the graph displayed, s = 1.29 and i = 0.086, then VM = 0.72 cm3/g = 7.2 x 10-10
m3/g. Hence, value of the specific surface area is 1.95 x 10-4 m2/g. The value
between those two method is not same (has a difference about 0.93 x 10-4 point).
BET constant:
The equation for BET constant (c) is:
C
s
1
i
Therefore, BET constant of solid for Sample 2 is = 16