Download Calculate dCA/dt for Reversible Reaction

Department of Chemical Engineering
Faculty of Engineering
Universitas Indonesia
2013
HOMOGENEOUS
HETEROGENEOUS
Form
Soluble metal complexes,
usually mononuclear
Metals, usually supported,
or metal oxides
Active site
Well-defined, discrete molecules
Poorly defined
Phase
Liquid
Gas/Solid
Temperature
Low (<250C)
High (250-500oC)
Activity
High
Variable
Selectivity
High
Variable
Diffusion
Facile
Can be very important
Heat transfer
Facile
Can be problematic
Product separation
Generally problematic
Facile
Catalyst recycle
expensive
simple
Catalyst modification
Easy
Difficult
Reaction mechanism
Reasonably well understood
Not obvious
Sensitivity to
deactivation
Low
High
Definition
HOMOGENEOUS REACTION
all reactants are in the same
phase
HETEROGENEOUS REACTION
more than one phase in
reactants
Equilibrium
Constant Rate (K)
Equal between forward and
reverse reaction
Difference between forward
and reverse reaction
Surface area affects
the reaction rate
No
Yes
Example
3H2(g) + N2(g) --> 2NH3(g)
Zn(s) + 2HCl(aq) --> H2(g) +
ZnCl2(aq)
Ag+(aq) + Cl-(aq) --> AgCl(s)
C(s) + O2(g) --> CO2(g)
CATALYST
Definition
Molecular
weight
Alternate terms
Reaction rate
BIOCATALYST
may be broadly defined as the use of
catalysts are substances that
enzymes or whole cells to increase
increases or decrease the rate of
speed in which a reaction takes place
a chemical reaction but remain
but do not affects the
unchanged
thermodynamics of reaction
low molecular weight
High molecular weight globular
compounds
protein or whole cells
Inorganic catalyst
Organic catalyst
Typically slower
Several times faster
Specificity
They are not specific and
therefore end up producing
residues with errors
Conditions
High temperature
Example
vanadium oxide
Biocatalyst are highly specific
producing large amount of good
residues
Mild conditions, physiological pH and
temperature
amylase, lipase
Steps in a Catalytic Reaction:
1. Mass transfer (diffusion) of the reactant(s) (e.g.,
species A) from the bulk fluid to the external surface
of the catalyst pellet
2. Diffusion of the reactant from the pore mouth
through the catalyst pores to the immediate vicinity
of the internal catalytic surface
3. Adsorption of reactant A onto the catalyst surface
4. Reaction on the surface of the catalyst (e.g., AB)
5. Desorption of the products (e.g., B) from the surface
6. Diffusion of the products from the interior of the
pellet to the pore mouth at the external surface
7. Mass transfer of the products from the external
pellet surface to the bulk fluid
 The
reaction given below:
 The
solution is:
(Pseudo Equilibrium)
 The
reaction given below:
 The
solution is:

Apparatus for the
volumetric method

Sensitive beam-type
balance used for the
gravimetric method

Equipment
arrangement for the
dynamic method
Dinitrogen adsorption data:
P/P0
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 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)?
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. Table modification for the answer:
Sample 1
P/P0
P (kPa)
Sample 2
Volume
Volume
adsorbed
adsorbed
(cm3/g)
(cm3/g)
0.02
2.026
99.274
23.0
0.000887
0.15
0.136
0.03
3.039
98.261
25.0
0.00124
0.23
0.134
0.04
4.052
97.248
26.5
0.00157
0.32
0.130
0.05
5.065
96.235
27.7
0.0019
0.38
0.139
0.10
10.13
91.17
31.7
0.0035
0.56
0.198
0.15
15.195
86.105
34.2
0.00515
0.65
0.271
0.20
20.26
81.04
36.1
0.00693
0.73
0.342
0.25
25.325
75.975
37.6
0.00887
0.81
0.412
0.30
30.39
70.91
39.1
0.011
0.89
0.482
Equation needed:
C 
s
1
i
BET surface area for Sample 1 using the one-point BET equation:
Plotting V against P to get the ‘Point B’ as VM
45
y = 0.5423x + 24.253
40
35
30
25
20
0
5
10
15
20
25
30
35
VM = 27.7 cm3/g = 2.77 x 10-8 m3/g, then using Eq. 2 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:
From Eq. 1, plotting data in the form P/[V(P0-P)] against P/P0 to get
slope (s) & intercept (i) that 1/(s + i) is equal to VM. From graphic,
VM = 28.49 cm3/g = 2.849 x 10-8 m3/g then using Eq. 2 specific area
of solid for Sample 1 is 7.66 x 10-3 m2/g
Graphic full BET method of solid for Sample 1:
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
BET constant for Sample 1:
Using Eq. 3, then BET constant of solid for Sample 1 is = 389.889
BET surface area for Sample 2 using the one-point BET equation:
Plotting V against P to get the ‘Point B’ as VM
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
VM = 0.38 cm3/g = 3.8 x 10-10 m3/g, then using Eq. 2 specific area of
solid for Sample 1 is 1.02 x 10-4 m2/g
BET surface area for Sample 2 using the full BET equation:
From Eq. 1, plotting data in the form P/[V(P0-P)] against P/P0 to
get slope (s) & intercept (i) that 1/(s + i) is equal to VM.
0.6
0.5
y = 1.2903x + 0.086
0.4
0.3
0.2
0.1
0
0
0.05
cm3/g
0.1
10-10
0.15
m3/g,
0.2
0.25
0.3
0.35
VM = 0.72
= 7.2 x
then using Eq. 2 specific area of
solid for Sample 1 is 1.95 x 10-4 m2/g
BET constant for Sample 2:
Using Eq. 3, then BET constant of solid for Sample 2 is = 16
 Difference
value of surface area using one-point
BET eq. and full BET eq. :
Discrepancy value between those method
illustrated the dangers in relying on the
estimation of a single point either by inspection
(point B method) therefore point B is not
particularly well defined and the BET full method
more empirical.
 Comparison of BET constant between Sample 1
and 2 :
Comparison of the BET constant obtained from
Sample 1 & 2 indicated its depends on the
difference on volume adsorbed of each sample
that showed by slope and intercept of line that
used to calculate the layer of adsorbed gas
quantity