Chem Rxn Equil

CPE553 CHEMICAL ENGINEERING THERMODYNAMICS 5/16/2013

1

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)


2

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


3

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


4

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)


5

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


6

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)


7

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


8

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 ô o

i i if f f P

ˆ i

i

oi

fK

P

(13.25)

ˆ î 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

î

i i oi

Py K

P

(13.26)

î

(13.27)

ˆ 1i

i

i oi

Py K

P

(13.28)

ii

î

• 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)

î i i if x f

î 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

î

i

i i

f

x f (11.90)


9

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


10

(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


11

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


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.

[email protected]

03-55436333/019-2368303

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Chem Rxn Equil