CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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CHEMICAL REACTION EQUILIBRIA
Raw materials are transformed into products by chemical reaction.
A chemical equilibrium has been established when a chemical reaction
reaches a state where the concentrations of reactants and products
remain constant.
At equilibrium, the rate of the forward reaction is equal to the rate of the
reverse reaction.
Both the rate and the equilibrium conversion of a chemical reaction
depend on the temperature, pressure and composition of reactants.
The study of thermodynamics give the equilibrium values. The study of
rate gives speed of reaching those values.
The purpose of this chapter is to determine the effect of temperature,
pressure and initial composition on the equilibrium conversion of chemical
reactions.
2
• THE REACTION COORDINATE
• APPLICATION OF EQUILIBRIUM CRITERIA
TO CHEMICAL REACTIONS
• THE STANDARD GIBBS-ENERGY
CHANGE AND THE EQUILIBRIUM
CONSTANT
• EFFECT OF TEMPERATURE ON THE
EQUILIBRIUM CONSTANT
• EVALUATION OF EQUILIBRIUM
CONSTANTS
3
For general chemical reaction
where is stoichiometric coefficient
Ai is chemical formula.
i is stoichiometric number, and it is
positive (+) for product negative (-) for reactant
Example:
the stoichiometric numbers are
The stoichiometric number for inert species is zero.
As the reaction in eq. (13.1) progresses, the changes in the numbers of moles of
species present are in direct proportion to the stoichiometric numbers. E.g. If 0.5
mol CH4 disappears, 0.5 mol H2O also disappears; simultaneously 0.5 mol CO and
1.5 mol H2 are formed.
4
4 2 23CH H O CO H
4 2 21 1 1 3CH H O CO H
1 1 2 2 3 3 4 4A A A A
i
(13.1)
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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Applied to differential amount of reaction, this principle provides the equations:
The list continues to include all species. Comparison yields
All terms being equal, they can be identified collectively by a single quantity
representing an amount of reaction. A definition of d is given by the equation:
The general relation connecting the differential change dni with d is therefore
is called the reaction coordinate (also known as degree of reaction or extent of
reaction), characterizes the extent or degree to which a reaction has taken place.
31 2 4
1 2 3 4
...dndn dn dn
d
5
1, 2, ..., Ni idn d i
(13.2)
(13.3)
32 1 1
2 1 3 1
etc.dndn dn dn
31 2 4
1 2 3 4
...dndn dn dn
Integration of eq. (13.3) from an initial unreacted state where = 0 and ni = nio to a
state reached after an amount of reaction gives
or
Summation over all species yields
or
where
Thus the mole fractions yi of the species present are related to by
6
0i i i
i i i
n n n
0n n
0 0i i i
i i i
n n n n
0
0
i iii
nny
n n
(13.5)
0 0
i
i
n
i in
dn d
0 1, 2, ..., i i in n i N (13.4)
For a system in which the following reaction occurs,
assume there are present initially 2 mol CH4, 1 mol H2O, 1 mol CO and 4 mol H2.
Determine expressions for the mole fractions yi as functions of .
4 2 23CH H O CO H
7
Consider a vessel which initially contains only n0 mol of water vapor. If
decomposition occurs according to the reaction,
find expressions which relate the number of moles and the mole fraction of each
chemical species to the reaction coordinate .
Solution:
For the reaction,
Application of eq. (13.5) yields
The fractional decomposition of water vapor is
When n0 = 1, is directly related to the fractional decomposition of the water vapor.
8
0
0
i iii
nny
n n
12 2 2 2H O H O
1 12 2
1 1
2 2 2
120
1 1 12 2 20 0 0
H O H O
ny y y
n n n
20 0 0
0 0 0
H On n n n
n n n
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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For multireaction, i,j designates the stoichiometric number of species i in reaction j.
The general relation, eq. (13.3) becomes
Integration from ni = ni0 and j = 0 to ni and j gives
Summing over all species yields
The definition of a total stoichiometric number for a single reaction has
its counterpart here in the definition:
Combination of this last equation with eq. (13.6) gives the mole fraction
9
, 1, 2, ..., i i j j
j
dn d i N
0 , 1, 2, ..., i i i j j
j
n n i N (13.6)
0 , 0 ,i i j j i j j
i i j j i
n n n
ii
, 0thusj i j j j
i j
n n
0 ,
0
1, 2, ..., i i j jj
i
j jj
ny i N
n
(13.7)
i = CH4 H2O CO CO2 H2
j j
1 -1 -1 1 0 3 2
2 -1 -2 0 1 4 2
10
Consider a system in which the following reactions occur:
where the numbers (1) and (2) indicate the value of j, the reaction index. If there
are present initially 2 mol CH4 and 3 mol H2O, determine expressions for the yi as
functions of 1 and 2.
Solution:
The stoichiometric numbers i,j can be arrayed as follows:
4 2 2
4 2 2 2
3 1
2 4 2
CH H O CO H
CH H O CO H
Application of eq. (13.7) gives
11
0 ,
0
1, 2, ..., i i j jj
i
j jj
ny i N
n
4
2 2
2
1 2 1
1 2 1 2
1 2 2
1 2 1 2
1 2
1 2
2
5 2 2 5 2 2
3 2
5 2 2 5 2 2
3 4
5 2 2
CH CO
H O CO
H
y y
y y
y
The Gibbs function serves as variable that determines whether a given chemical
change is thermodynamically possible.
In a spontaneous change, Gibbs energy always decreases and never increases. The
physical meaning of G is that it tells how far the free energy of the system has
changed from Go of the pure reactants.
Composition of a chemical reaction system will tend to change in a
direction that brings it closer to its equilibrium composition. Once this
equilibrium composition has been attained, no further change in the quantities of
the components will occur.
12
G<0 Reaction can spontaneously proceed to the right: AB
G>0 Reaction can spontaneously proceed to the left: AB
G=0 The reaction is at equilibrium; the quantities of A and B will not change
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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For a reaction A B, the standard free energy of
the products (point 2) is smaller than that of the
reactants (point 1), so the reaction will take place
spontaneously.
This does not mean that each mole of pure A will
be converted into one mole of pure B. For
reactions in which products and reactants occupy
a single phase (gas or solution), the meaning of
spontaneous is that the equilibrium composition
will correspond to an extent of reaction greater
than 0.5 but smaller than unity.
As the reaction proceeds to the right, the
composition changes, and G begins to fall. When
the composition reaches point 3, G reaches its
minimum value and further reaction would
cause it to rise. But because free energy can only
decrease but never decrease, this does not
happen. The composition of the system
remains permanently at its equilibrium
value.
13
Fig. 1 Spontaneous reaction (G<0)
Fig. 2 Non-spontaneous reaction (G>0)
Fig. 13.1 shows the relation of Gt and . The arrows along the curve indicate the
directions of changes in (Gt)T,P. The reaction coordinate, has its equilibrium value e
at the minimum of the curve.
This figure (also the two previous figure) indicates two distinctive features of
the equilibrium state for given T and P,
◦ Total Gibbs energy Gt is a minimum.
◦ Its differential is zero
14
(For a single chemical reaction)
,
0t
T PdG
reactants
products
Forward reaction
Equation (11.2) provides an expression for the total differential of the Gibbs energy:
If changes in the mole numbers ni occur as the result of a single chemical reaction in
a closed system, then by eq. (13.3) each dni may be replaced by the product id. Eq.
(11.2) then becomes
Because nG is a state function, the right side of this equation is an exact differential
expression,
The quantity represents, in general, the rate of change of total Gibbs energy
of the system with respect to the reaction coordinate at constant T and P.
( ) ( ) i i
i
d nG nV dP nS dT d
15
( ) ( ) i i
i
d nG nV dP nS dT dn (11.2)
, ,
t
i i
i T P T P
GnG
i i
i
Fig. 13.1 shows that this quantity is zero at the equilibrium state. A criterion of
chemical reaction equilibrium is therefore
Recall the definition of the fugacity of a species in solution:
In addition, eq. (11.31) may be written for pure species i in its standard state at the
same temperature:
The difference between these two equations is
16
0i i
i
(13.8)
ˆlni i iT RT f (11.46)
lno o
i i iG T RT f
ˆlno i
i i o
i
fG RT
f (13.9)
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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Combining eq. (13.8) with eq. (13.9) to eliminate i gives for the equilibrium state
of a chemical reaction:
or
or
where i signifies the product over all species i.
In exponential form, this equation becomes
where the definition of K and its logarithm are given by
ˆln 0io o
i i i i
i i
G RT f f
17
ˆln 0o o
i i i i
i
G RT f f
ˆlno
i i iiGo
RTi i
i
f f
ˆ io
i i
i
f f K
(13.10)
exp
ln
o
o
GK
RT
GK
RT
(13.11a)
(13.11b)
Also by definition,
Gio, Go and K are functions of temperature.
The function Go iiGio in eq. (13.12) is the difference between the Gibbs energies
of the products and reactants when each is in its standard state as a pure substance at the standard state pressure, but at the system temperature.
Other standard property changes of reaction are similarly defined. For general property M:
The relation between the standard heat of reaction and the standard Gibbs energy change of reaction is
18
o o
i iiG G (13.12)
K is called the equilibrium constant for the reaction; iiGio,
represented by Go, is called the standard Gibbs-energy change of
reaction.
o o
i i
i
M M
2
o
od G RT
H RTdT
(13.13)
Because the standard state temperature is that of the equilibrium mixture, the standard property changes of reaction, such as Go and Ho, vary with the equilibrium temperature.
The dependence of Go on T is given by eq. (13.13), which may be rewritten as
In view of eq. (13.11b), this becomes
If Ho is negative, i.e., if the reaction is exothermic, the equilibrium constant decreases as the temperature increases. K increases with T for endothermic reaction.
If Ho is assumed independent of T, integration of eq. (13.14) from T’ to T leads to the simple result:
A plot of ln K vs. the reciprocal of absolute temperature is a straight line.
19
2
o od G RT H
dT RT
2
ln od K H
dT RT
(13.14)
' '
1 1ln
oK H
K R T T
(13.15)
20
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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Effect of temperature on equilibrium constant is based on the definition of the
Gibbs energy at standard state:
Multiply by i and summation over all species gives
As a result of the definition of a standard property change of reaction, this reduces
to
The standard heat of reaction is related to temperature:
The temperature dependence of the standard entropy change of reaction is:
Multiply by i and summation over all species gives
21
o o oG H T S (13.16)
o o o
i i iG H TS
o o o
i i i i i i
i i i
G H T S
00
oT po o
T
CH H R dT
R
(4.18)
i
o o
i P
dTdS C
T
o o
P
dTd S C
T
Integration gives
where So and S0o are the standard entropy changes of reaction at temperature T
and at reference temperature T0.
Eqs. (13.16), (4.18) and (13.17) are combined to yield
However,
Hence,
Finally, division by RT yields
0 0
0 0 0
0
o oT T
o o o o P P
T T
T C CG H H G R dT RT dT
T R R
22
00
oT
o o P
T
C dTS S R
R T
(13.17)
0
0 0
0 0
0
1oo oo o o
T TP P
T T
G H HG C C dTdT
RT RT RT T R R T
0 00 0
o oT T
o o oP P
T T
C CG H R dT T S RT dT
R R
0 00
0
o oo H G
ST
(13.18)
0
2
0 0 2 2
0
1ln 1
2
oT
P
T
C dT DA BT CT
R T T
23
The first integral of eq. (13.18) is
where
OR
The second integral of eq. (13.18) is
OR
0
2 2 3 3
0 0 0
0
11 1 1
2 3
oT
P
T
C B C DdT A T T T
R T
(13.19)
0
T
T
(4.19)
0
2 2
0 0 2 2
0
1 1ln
2 2
T o
P
oT
C dT T C DA B T T T T
R T T T T
0
2 2 3 3
0 0
1 1
2 3
oT
Po
To
C B CdT A T T T T T T D
R T T
Go/RT(= - ln K) as given in eq. (13.18) is readily calculated at any temperature from
the standard heat of reaction and the standard Gibbs energy change of reaction at a
reference temperature (usually 298.15 K).
Factor K may be organized into three terms:
24
0 1 2K K KK
00
0
expoG
KRT
(13.21)
0 01
0
exp 1oH T
KRT T
0 02
1exp
o oT T
P P
T T
C C dTK dT
T R R T
(13.22)
(13.23)
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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EXAMPLE 13.4 Calculate the equilibrium constant for the vapor phase hydration of ethylene at
145oC (418.15 K) and at 320oC (593.15 K) from data given in App. C.
Solution:
The chemical reaction:
From Table C.1 and C.4:
25
2 4 2 2 5C H g H O g C H OH g
vi A B (x10-3) C (x10-6) D (x105) Ho298
(J/mol)
Go298
(J/mol)
C2H5OH 1 3.518 21.001 -6.002 - -235100 -168490
C2H4 -1 1.424 14.394 -4.392 - 52510 68460
H2O -1 3.470 1.450 - 0.121 -241818 -228572
For T = 418.15 K, values of integrals in eq. (13.18) are
Substitute into eq. (13.18) for a reference temperature of 298.15 K gives:
-1
298 ,298 8378 Jmolo o
i i
i
G G
26
-1
298 ,298 45792 Jmolo o
i i
i
H H
0 0
23.121 . 4.19 0.0692 . 13.19o o
T TP P
T T
C C dTdT eqn eqn
R R T
418.15 8378 45792 45792 23.121
0.0692 1.93568.314 298.15 8.314 418.15 418.15
oG
RT
-3
-6
5
-1.376
4.157 10
1.610 10
- 0.121 10
i i
i
i i
i
i i
i
i i
i
A A
B B
C C
D D
0
0 0
0 0
0
1oo oo o o
T TP P
T T
G H HG C C dTdT
RT RT RT T R R T
(13.18)
For T = 593.15 K,
Finally, by eq. (13.11b)
At 418.15 K, ln K = -1.9356 and K = 1.443 x 10-1
At 593.15 K, ln K = -5.8286 and K = 2.942 x 10-3
Alternative solution by using eq. (13.21), (13.22) and (13.23).
0
0
22.632
0.0173
oT
P
T
oT
P
T
CdT
R
C dT
R T
593.15 8378 45792 45792 22.632
0.0173 5.82868.314 298.15 8.314 593.15 593.15
oG
RT
27
lnoG
KRT
•RELATION OF EQUILIBRIUM
CONSTANTS TO COMPOSITION
• EQUILIBRIUM CONVERSIONS FOR
SINGLE REACTIONS
28
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Gas phase reactions
The standard state for a gas is the ideal gas state of the pure gas at the standard
state pressure Po of 1 bar.
Because the fugacity of ideal gas is equal to its pressure, fio = Po for each species i.
Therefore for gas phase reactions and eq. (13.10) becomes
This eq. relates K to fugacities of the reacting species as they exist in the real
equilibrium mixture. These fugacities reflect the non-idealities of the equilibrium
mixture and are functions of temperature, pressure and composition.
The fugacity is related to the fugacity coefficient by eq. (11.52):
29
ˆ ˆo o
i i if f f P
ˆ i
i
oi
fK
P
(13.25)
ˆ ˆi i if y P
ˆ io
i i
i
f f K
(13.10)
Substitute into eq. (13.25) provides an equilibrium expression displaying the
pressure and the composition:
where and Po is the standard state pressure of 1 bar.
The yi’s may be eliminated in favor of the equilibrium value of the reaction
coordinate e. For fixed temperature eq. (13.26) relates e to P. If P is specify, e can
be solved. can be evaluated by using eq. (11.63 or 11.64).
If the equilibrium mixture is an ideal solution, then each becomes i , then eq.
(13.26) becomes
Each i for a pure species can be evaluated from a generalized correlation.
For pressure sufficiently low or temperatures sufficiently high, the equilibrium
mixture behaves essentially as an ideal gas. Each and eq. (13.26) reduces to:
i
i i oi
Py K
P
30
ˆi
i i oi
Py K
P
(13.26)
ˆi
(13.27)
ˆ 1i
i
i oi
Py K
P
(13.28)
ii
ˆi
• Some conclusions based on eq. (13.28) that are true in general:
▫ According to eq. (13.14), the effect of temperature on the equilibrium constant K
is determined by the sign of Ho.
When Ho is positive, i.e., when the standard reaction is endothermic, an increase
in T results in an increase in K. Eq. (13.28) shows that an increase in K at constant
P results in an increase in i(yi)vi; this implies a shift of the reaction to the right
and an increase in e.
When Ho is negative, i.e., when the standard reaction is exothermic, an increase
in T causes a decrease in K and a decrease in i(yi)vi at constant P. This implies a
shift of reaction to the left and a decrease in e.
▫ If the total stoichiometric number v (≡ivi) is negative, eq. (13.28) shows that an
increase in P at constant T causes an increase in i(yi)vi , implying a shift
of the reaction to the right and an increse in e .
▫ If v is positive, an increase in P at constant T causes a decrease in i(yi)vi ,
implying a shift of the reaction to the left and a decrease in e .
31
2
ln od K H
dT RT
(13.14)
i
i oi
Py K
P
(13.28)
Liquid phase reactions
For a reaction occurring in the liquid phase,
Standard state for liquid fio is the fugacity of pure liquid i at the temperature of the
system and at 1 bar.
According to eq. (11.90),
The fugacity ratio can be expressed as
Because the fugacities of liquids are weak function of pressure, the ratio fi/fio is often
taken as unity.
For pure liquid i, eq. (11.31) is written for temperature T and pressure P, and for the
same temperature T but for standard state pressure Po.
32
ˆ io
i i
i
f f K
(13.10)
ˆi i i if x f
ˆi i i i i
i io o o
i i i
f x f fx
f f f
(13.29)
ln
ln
i i i
o o
i i i
G T RT f
G T RT f
ˆi
i
i i
f
x f (11.90)
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The difference between these two equations is
Integration of eq. (6.10) at constant T for the change of state of pure liquid i from Po
to P yields
As a result,
Because Vi changes little with pressure for liquids (and solids), integration from Po to
P gives
With eq. (13.29) and (13.30), eq. (13.10) may be written as
33
lno ii i o
i
fG G RT
f
o
Po
i i iP
G G VdP
lno
Pi
io Pi
fRT VdP
f
ln
o
ii
o
i
V P Pf
f RT
(13.30)
expi
o
i i i i
ii
P Px K V
RT
(13.31)
Except for high pressure, the exponential term is close to unity and may be omitted.
Then,
If the equilibrium mixture is an ideal solution, then i is unity and eq. (13.32)
becomes
34
i
i i
i
x K
(13.32)
i
i
i
x K (13.33)
EXAMPLE 13.5
A water gas shift reaction,
is carried out under the different sets of conditions described below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas.
35
2 2 2CO g H O g CO g H g
(a) The reactants consist of 1 mol of H2O vapor and 1 mol of CO. The
temperature is 1100 K and the pressure is 1 bar.
For the given reaction at 1100 K, 104/T = 9.05, and from Figure 13.2, ln K = 0 and K = 1.
Because the reaction mixture is an ideal gas, eq. (13.28) applies
By eq. (13.5),
Substitute into eq. (A) gives
Therefore the fraction of the steam that reacts is
36
1 1 1 1 0ii
2 2 2
2 2
2
0
1 1 11 11
1
1
CO H O CO H
H CO
CO H O
bary y y y
bar
y y
y y
(A)
2 2 2
1 1
2 2 2 2
e e e eCO H O CO Hy y y y
2
e21 or 0.5
1
e
e
i
i oi
Py K
P
0
0
i iii
nny
n n
2 2
2
1 10.5
1
oH O H O
oH O
n n
n
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(b) Same as (a) except that the pressure is 10 bar.
(c) Same as (a) except that 2 mol of N2 is included in the reactants.
(d) The reactants are 2 mol of H2O and 1 mol of CO. Other conditions are the same as in (a).
(e) The reactants are 1 mol of H2O and 2 mol of CO. Other conditions are the same as in (a).
(f) The initial mixture consists of 1 mol of H2O, 1 mol CO and 1 mol of CO2. Other conditions are the same as in (a).
(g) Same as (a) except that the temperature is 1650 K.
37
Estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration
at 250oC (523.15 K) and 35 bar for an initial steam to ethylene ratio of 5. Assume
the reaction mixture is an ideal solution.
Solution:
The calculation of K for this reaction is treated in Example 13.4.
At T = 523.15 K, K = 10.02 x 10-3
For ideal solution,
Because this equation becomes
38
i
i i oi
Py K
P
1 1 1 1ii
(13.27)
2 4 2 4 2 2
310.02 10EtOH EtOH
o
C H C H H O H O
y PA
y y P
Fugacity coefficients are evaluated by eq. (11.68).
The results are summarized in the following table:
Substitute values of i and P = 35 bar into eqn. (A) gives:
39
Tc/K Pc/bar i Tri Pri B0 B1 i
C2H4 282.3 50.40 0.087 1.853 0.694 -0.074 0.126 0.977
H2O 647.1 220.55 0.345 0.808 0.159 -0.511 -0.281 0.887
EtOH 513.9 61.48 0.645 1.018 0.569 -0.327 -0.021 0.827
0 1 0 1ln or expr r
r r
P PB B B B
T T
0 1
1.6 4.2
0.422 0.1720.083 and 0.139
r r
B BT T
2 4 2
30.997 0.887 3510.02 10 0.367
0.827 1
EtOH
C H H O
yB
y y
2 4 2 4 2 2
310.02 10EtOH EtOH
o
C H C H H O H O
y PA
y y P
By eqn. (13.5),
Substitute these into eqn. (B) yields:
The solution for the smaller root is e = 0.233. Therefore the maximum conversion
of ethylene to ethanol is
40
2 4 2
1 5
6 6 6
e e eC H H O EtOH
e e e
y y y
2
60.367 6 1.342 0
5 1
e e
e e
e e
or
2 4 2 4
2 4
1 10.233 23.3%
1
oC H C H
oC H
n n
n
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
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In a laboratory investigation, acetylene is catalytically hydrogenated to ethylene at
1120oC (1393.15 K) and 1 bar. If the feed is an equimolar mixture of acetylene and
hydrogen, what is the composition of the product stream at equilibrium?
Solution:
The required reaction is obtained by addition of the two formation reactions
written as follows:
The sum of reactions (I) and (II) is the hydrogenation reaction:
Also,
By eq. (13.11b),
41
2 2 2
2 2 4
2 I
2 2 II
C H C H
C H C H
2 2 2 2 4C H H C H
o oo
I IIG G G
ln ln lnI II I IIRT K RT K RT K or K K K
Data for both reactions (I) and (II) are given by Fig. 13.2. For 1120oC (1393.15 K),
104/T = 7.18, the following values are read from the graph:
Therefore, K = KIKII = 1.0
At this temperature and pressure of 1 bar, assume ideal gases.
By eq. (13.28),
On the basis of one mole initially of each reactant, eq. (13.5) gives
Therefore,
The smaller root of this quadratic expression is e = 0.293.
42
5
6
ln 12.9 4.0 10
ln 12.9 2.5 10
I I
II II
K K
K K
2 4
2 2 2
1C H
H C H
y
y y
2 2 2 2 4
1and
2 2
e eH C H C H
e e
y y y
2
21
1
e e
e
i
i oi
Py K
P
The equilibrium composition of the product gas is then
43
2 2 2 2 4
1 0.293 0.2930.414 and 0.172
2 0.293 2 0.293H C H C Hy y y
Acetic acid is esterified in the liquid phase with ethanol at 100oC (373.15 K) and
atmospheric pressure to produce ethyl acetate and water according to the
reaction:
If initially there is one mole each of acetic acid and ethanol, estimate the mole
fraction of ethyl acetate in the reacting mixture at equilibrium.
Solution:
44
3 2 5 3 2 5 2CH COOH l C H OH l CH COOC H l H O l
Ho298 (J) Go
298 (J)
CH3COOC2H5 -480 000 -332 000
H2O -285 830 -237 130
CH3COOH -484 500 -389 900
C2H5OH -277 690 -174 780
CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013
12
By eq. (13.11b),
For small temperature change from 298.15 K to 373.15 K, eq. (13.15) is adequate
for estimation of K. Therefore,
or
and
45
298
298
1 480000 1 285830 1 484500 1 277690 3640 J
1 332200 1 237130 1 389900 1 174780 4650 J
o o
i ii
o o
i ii
H H
G G
298
298 298
4650ln 1.8759 or 6.5266
8.314 298.15
oGK K
RT
373 298
298
1 1ln
373.15 298.15
oK H
K R
373 3640 1 1ln 0.2951
6.5266 8.314 373.15 298.15
K
373 6.5266 0.7444 4.8586K
For the given reaction, eq. (13.5), with x replacing y, yields
Because the pressure is low, eq. (13.32) is applicable. In the absence of data for
activity coefficients, assume the reacting species form an ideal solution. In this case
eq. (13.33) is employed, giving
Solution yields
46
2
1
2 2
e eAcH EtOH EtAc H Ox x x x
2
2
4.85861
EtAc H O
AcH EtOH
e
e
x xK
x x
0.6879 and 0.6879 /2 0.344e EtAcx
i
i
i
x K (13.33)
Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to Chemical
Engineering Thermodynamics. Seventh Edition. Mc Graw-Hill.
http://www.chem1.com/acad/webtext/chemeq/Eq-01.html#SEC1
http://www.chem1.com/acad/webtext/thermeq/TE4.html
http://www.chem1.com/acad/webtext/thermeq/TE5.html
47
PREPARED BY:
MDM. NORASMAH MOHAMMED MANSHOR
FACULTY OF CHEMICAL ENGINEERING,
UiTM SHAH ALAM.
03-55436333/019-2368303
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