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Файл:Chemical Engineering of Natural Fuels and Carbon Materials. Study Guide
.pdf
аА + bВ + ... ↔ сС + dD + ...
C2H4 + C4H
8
(I)
2C3H
6
(II)
C2H6 + C4H
6
(III)
∆G = [c·∆G°С + d·∆ G°D + …] – [а·∆ G°А + b·∆ G°В + …].
Free energy change of the reaction can be calculated as a sum of
free energies of the products minus the sum of free energies of the reactants.
The value of ΔG varies with temperature change (there is reference
data). In general, the temperature dependence of Gibbs free energy ΔG
looks like
∆G
= А + В·Т,
T
where А and В are coefficients that are constant for any particular reaction.
These coefficients can be calculated by solving a system of
equations:
∆G
= А + В·Т1
T1
∆GТ2 = А + В·Т2.
The temperature at which ΔG = 0 is called the temperature limit of
thermodynamic feasibility (T
) of the reaction.
lim
Exothermic reactions are thermodynamically feasible at
temperatures below this temperature limit Т < T
lim
.
Endothermic reactions are thermodynamically feasible at
temperatures higher than this temperature limit Т > T
lim
.
Example 3.1.
At a temperature of 527 °C, cyclohexane thermal decomposition is
possible in three ways: 1) resulting in the formation of butene; 2) resulting
in the formation of propylene; 3) resulting in the formation of divinyl.
What direction is the most thermodynamically possible at this temperature?
Solution:
21
First of all, we need to calculate the values of ΔG
Hydrocarbon
∆G
800
, kJ/mol
С6Н
12
С2Н
4
С2Н
6
С3Н
6
С4Н
6
(divinyl)
С4Н
8
(1- butene)
+318.272
+102.613
+66.705
+145.854
+228.271
+207.195
for each of the reactions.
800
Using reference tables (Appendix 2) we find the values of Gibbs
free energy for substances involved in the reaction.
Table 3.1 – Gibbs free energies for substances
∆G
(I) = 102.613 + 207.195 – 318.272 = – 8.464 kJ/mol;
800
∆G
(II) = 2·145.854 – 318.272 = – 26.564 kJ/mol;
800
∆G
(III) = 66.705 + 228.271– 318.272 = – 23.296 kJ/mol.
800
Based on the calculation results, we can conclude that the second
reaction has the greatest negative value of ΔG
, and therefore, the second
800
reaction is the most thermodynamically probable at this temperature
compared to the reactions (I) and (II).
Task 3.1.
What direction of cyclopentane conversion is thermodynamically
more probable at the temperature 800K: 1) decomposition into methane and
butylene or 2) dehydrogenation to cyclopentene?
Determine the temperature limit of the thermodynamic feasibility of
the second reaction.
Task 3.2.
Normal alkanes can decompose according to the next scheme:
С
n+mH2(n+m)+2
→ C
nH2n
+ СmH
2m+2
.
Herewith, the molecule of alkane may decompose in the center of
the molecule, or closer to its end. n-octane (C8H18) decomposition being an
example, calculate the following: a) What direction is thermodynamically
22
more probable at temperature 427 оС? b) In the case of molecule
3H2
+
CH
3
CH
3
decomposition into unequal parts (according to the number of carbon
atoms) at the temperature 527 оС alkane or alkene will obtain the higher
molecular weight?
Task 3.3.
For the dehydrogenation reaction of methylcyclohexane into
toluene, find the temperatures at which the reaction is thermodynamically
possible.
Task 3.4.
For the alkylation reaction of benzene with ethylene, determine the
temperature limit of its thermodynamic feasibility.
23
4. THE EQUILIBRIUM CONSTANT OF A CHEMICAL REACTION
...PP
...PP
K
b
B
a
A
e
E
d
D
p
i
i
i
n
n
N
In case of equality of forward and reverse reaction rates the
chemical equilibrium occurs. Since the chemical equilibrium in the reacting
system is set, the concentrations and the partial pressures of the substances
become constant or become unchanged.
Fractional conversion (conversion coefficient, conversion degree)
of the substance in a chemical reaction is the ratio of the moles of the
substance reacted to the initial number of the moles of the substance.
The theoretical degree of the reaction completeness corresponds to
the establishment of equilibrium in the system, in other words, corresponds
to the most possible degree of conversion, which can be easily determined
based on the expression for the equilibrium constant of the process.
Therefore, for the processes occurring at constant pressure (P =
const) and temperature (T = const) the expression for the equilibrium
constant looks like:
аА + bВ … ↔ dD + еЕ ...
,
where Pi is the partial pressures of the substances participating in the
chemical reaction. Kp is called chemical equilibrium constant.
In this case, Dalton's law is valid:
Pi = Р
total
∙ Ni,
where Ni, the mole fraction, is equal to the ratio of moles of i-component to
the sum of the moles of all components in the system.
.
The higher the value of the equilibrium constant is, the higher the
quantity of final products in reaction mixture is and vice versa.
The equilibrium constant and Gibbs energy are related to each other
through the equation:
24
RT
G
Р
lnК
Reaction
Н
2
= 2Н
Initial state of the system
1 моль
0 моль
Equilibrium state of the system
(1 – α) моль
2α моль
The total quantity of reagents
(1 + α) моль
The partial pressures of the reactants
at equilibrium
1
1
P
1
2
P
...PP
...PP
K
b
B
a
A
e
E
d
D
p
2
2
2
H
2
H
p
1
4
P
1
1
P
1
2
P
P
P
K
2
From this equation, it can be concluded that a large negative value
of –ΔG corresponds to a larger value of the equilibrium constant, and vice
versa. If ΔG = 0, the equilibrium constant is equal to K = 1.
Example 4.1.
Derive an equation for the calculation of the equilibrium constant
for the transformation reaction of molecular hydrogen into atomic hydrogen
through the conversion degree of the starting reagent.
Solution:
Let us look at the various states of this reaction from its beginning
till achieving the chemical equilibrium, denoting the degree of reaction
completeness with the letter α.
Table 4.1 – States of the reaction
According to the reaction:
,
Example 4.2.
Find the theoretically possible output of isobutane:
at the temperature of 327 oC and atmospheric pressure.
n-С
4Н10
↔ iso-С
25
4Н10
.
Solution:
RT
G
Р
lnК
1
1
1
104
НСn
P
1
1
104
НСiso
P
Using the reference tables, we find Gibbs free energies for both
hydrocarbons at given temperature:
n-С
iso-С
4Н10
4Н10
– ∆G
– ∆G
= 101.73 kJ/mol;
600
= 102.57 kJ/mol.
600
Gibbs free energy change for the reaction:
∆G
= 102.57 – 101.73 = 0.84 kJ/mol = 840 J/mol.
600
,
Кр = е
– 0,1685
= 0.85
After denoting the output of the product through the α we can
calculate the partial pressures of substances in a state of equilibrium, using
Dalton's law:
,
α ≈ 0.46 or 46 % mol. is theoretically possible output of isobutane.
Task 4.1.
Determine the value of Gibbs energy and the equilibrium constant
for the dehydrogenation reaction of cyclohexane at a temperature of 427 oC
26
and atmospheric pressure, if these conditions lead to the formation of 20 %
CH
3
of benzene.
Task 4.2.
Determine the value of Gibbs energy and the equilibrium constant
for the alkylation reaction of isobutane with butene-1 at a temperature of 7
°C and atmospheric pressure, if 30 mol %. of iso-octane is formed. The
molar ratio of iso-C4H
: C4H8 = 5:1.
10
Task 4.3.
Find the value of Gibbs free energy change, the equilibrium
constant and the theoretically possible output of cyclohexane for
methylcyclopentane isomerization reaction at 327 oC and atmospheric
pressure.
27
5. KINETICS OF CHEMICAL REACTIONS
...,CCKV
BA
n
B
n
A
AeK
RT/E
o
Chemical reaction rate and chemical mechanisms are studied by
chemical kinetics.
The reaction rate depends on temperature. The temperature
dependence of the reaction rate constant is described by the Arrhenius
equation:
К = А∙е
-Е/RT
,
where A is the pre-exponential factor and E is the activation energy of the
reaction.
Formally, the order of the reaction for the specific substance is a
measure of the concentration degree of this substance in the kinetic equation
(it is also called the special order of the reaction).
аА + bВ … ↔ dD + еЕ ...
where V is the reaction rate, nА is the reaction order for substance A; K is
the reaction rate constant.
The reaction order is determined experimentally, generally by the
reaction rate dependence on the concentration of the specific substance at
constant concentrations of all other substances.
The activation energy of the reaction is the average excess energy E,
which should the reacting particles acquire to overcome the potential barrier
separating the initial and final state of the system.
Both non-catalytic and catalytic reaction rate are associated with the
activation energy:
– for monomolecular reaction.
, it could not be less than this value;
Ко(term.) ≈ 1013 sec
Ко(cat.) ≈ 105 sec
-1
-1
, it could not be more than this value.
E
cat.
< E
term.
28
The relationship between the reaction rate and the reaction time
xa
a
ln
1
K
K
693,0
K
2ln
2/1
)xb(a
)xa(b
ln
)ba(
1
K
)xa(a
x
K
aK
1
2/1
For a first order reaction, the rate constant of the reaction is
determined by the equation:
,
where τ is the reaction time, seconds; a is the initial concentration of the
substance; x is the decrease in the concentration of the substance from the
beginning of the reaction up to this point of time (τ); (a – x) is the
concentration of the substance at a given moment time.
Half-life time for a first order reaction can be expressed by the
equation:
For a second order reaction rate, the constant of the reaction is
determined by the equation:
In the case when the initial concentrations of substances are equal to
each other а = b:
Half-life time for a second order reaction can be expressed by the equation:
Temperature dependence of the reaction rate
Temperature dependence of the reaction rate can be described by
the following equation if the reaction rate constants at two given
temperatures are known:
.
.
.
.
29
12
12
T
T
TTR
)TT(E
K
K
ln
1
2
12
12
T
T
TTR303,2
)TT(E
K
K
lg
1
2
1
2
T
T
12
12
a
K
K
ln
)TT(
TTR
E
1
2
T
T
12
12
a
K
K
lg
)TT(
TTR303,2
E
44ln9,2
75,01
1
ln
3465,0
1
%75
Using this equation, the reaction activation energy can be
determined:
.
.
The reaction temperature coefficient shows how many times the
reaction rate constant will change at 10 ° temperature change (γ).
The reaction temperature gradient shows how many degrees of
temperature are needed to change the reaction rate twofold (α).
Example 5.1.
For the first order reaction, half-life time is equal to 2 seconds.
Determine the reaction rate constant under the same conditions. Determine
the necessary time for the reaction to proceed up to 75 percent under these
conditions?
Solution
Half-life time for a first order reaction is expressed by the equation:
τ
= 0,693 / К → К = 0,693/2 = 0,3465 sec-1,
1/2
The necessary time for the reaction to proceed up to 75 percent is 4
seconds.
.
30
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