CENG 4710 Environmental Control: Benzene Adsorption and BET Surface

This assignment solution delves into the adsorption of benzene on activated carbon at varying conditions. It begins by fitting the Langmuir isotherm to experimental data at 298 K, estimating the parameters, followed by fitting the Freundlich isotherm and extracting its parameters, ultimately comparing the models. The analysis reveals the limitations of the Freundlich isotherm in this context. Further, the assignment explores the determination of the surface area of activated carbon using the BET equation based on nitrogen adsorption data at 77 K, discussing the limitations of the assumed isotherm model and the accuracy of the BET method under specific pressure conditions. The student provides detailed calculations and explanations for each part, highlighting the importance of understanding adsorption phenomena in environmental control.

CHEMICAL ENGINEERING
STUDENT NAME/ID
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Question 1
Adsorption of benzene on activated carbon (298 K)
(a) The Langmuir isotherm
0 200 400 600 800 1000 1200
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4 Langmuir Isotherm
Benzene partial pressure (Pa)
Adsorbed phase concentration of benzene (g
benzene/g solid)
Now,
θ= kp
1+ kp
0.124= 15.33∗k
1+ 15.33 k
k =9.2∗10−3 Pa−1
(b) Freundlich isotherm
1

q=K P
1
n (n>1)
log ( q ) =logk + 1
n log P
Where,
Slope = 1/n
Intercept = log K
Now,
0 0.5 1 1.5 2 2.5 3 3.5
-2.5
-2
-1.5
-1
-0.5
0
f(x) = − 0.344908112943666 x − 0.533977835217286
R² = 0.231528210480036
Freundlich Isotherm
log (p)
log (q)
Here,
Intercept = log K = -0.3475
K=0.449Pa^-1
Further,
Slope = 1/n = -0.0985
n=−10.1528
It can be seen that n value comes out to be negative which implies that Freundlich isotherm is
invalid.
Further,
2

Special case scenarios when the pressure is too low then n would tend to 1. (Graph would be
same as Langmuir isotherm)
n=1
q=K P
1
n
K= ( q )
P = 0.124
15.33 =8∗10−3 0.348Pa ^-1
Now,
When the pressure is too high then n would tend to infinity. (Graph would be same as Langmuir
isotherm)
n=∞
q=K P1/ ∞
K=(q)=0.348Pa ^-1
(c)The Langmuir isotherm model is considered to be a better model as compared with the
Freundlich isotherm to describe the experimental data.
Question 2
Adsorption of benzene on activated carbon (77 K)
Brunauer Emmett Teller BET Equation
3

Z
1−Z ∗1
V = ( C−1 ) Z
C V ' + 1
C V '
Where,
V=Volume of gas absorbed
V’ = Amount of gas corresponding to monolayer capacity
Z= P
P 0
C=Constant
The simpler form would be used when the coefficient is large.
V
V ' = 1
1−z
From the graph,
Slope=C−1
C V ' =0.7399
Intercept= 1
C V ' =0.165
Now,
( C−1 )∗0.165=0.7399
4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
f(x) = 0.739946131698709 x + 0.164958205628309
R² = 0.884037470719331
P/P0
q (g N2 /g solid)

C=5.48(Constant)
Further,
1
C V ' =0.165
1
5.48V ' =0.165
V ' =1.1051 (m^3)
Avogadro constant = 6.022 *10^23 mol ^-1
Contact area of one N2 molecule = 16.2*10^-20 m^2
Surface area S = (1.1051 /22400)* 6.022 *10^23 *16.2*10^-20
Surface area =4.8129 m^2/g
Limitations
The peaks of the graphs (isotherm) depend on the system characteristics and temperature. This
isotherm shows stepwise multilayer adsorption on uniform non-porous surface which indicates
that main assumption of isotherm has not been satisfied which is monolayers adsorption of the
gas on solid surface. It is true only for the first layer of the adsorption. BET isotherm would be
accurate only when the P/P0 falls between 0.05 and 0.3. This isotherm is not valid for very low
pressure and it has been assumed that at very low pressure, only few gas molecules have
absorbed.
5