Adsorption of Benzene on Activated Carbon - Langmuir and Freundlich Isotherms, BET Equation

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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 ) =log k+ 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