c ° 2011 Ismail Tosun Chapter 13 Chemical Reaction Equilibrium A chemical reactor is the heart of a chemical plant in which reactants are converted into prod- ucts. Design of a reactor should provide answers to questions like: Which products will be obtained from the reaction(s) of given reactants? What are the best conditions, i.e., tempera- ture, pressure, addition of inerts, etc., to convert more of the reactants into products? What quantities of products will be formed from the given amounts of reactants? This chapter starts with the fundamentals such as stoichiometry, molar extent of reaction, and the condition of chemical reaction equilibrium. Calculation of the equilibrium constant is then introduced. Estimation of the equilibrium extent for reactions taking place in a single phase, as well as factors affecting it, will be discussed. 13.1 STOICHIOMETRY OF A CHEMICAL REACTION Balancing of a chemical equation is based on the conservation of mass for a closed thermo- dynamic system. If a chemical reaction takes place in a closed container, the mass does not change even if there is an exchange of energy with the surroundings. Consider carbon dioxide reforming 1 of methane described by the reaction CH 4 (g)+ CO 2 (g) 2 H 2 (g)+2 CO (g) (13.1-1) If A 1 = CH 4 , A 2 = CO 2 , A 3 = H 2 , and A 4 = CO, Eq. (13.1-1) is expressed as A 1 + A 2 =2 A 3 +2 A 4 (13.1-2) It is convenient to write all the chemical species on one side of the equation and give a positive sign to the species that are regarded as the products of the reaction. Thus, 2 A 3 +2 A 4 − A 1 − A 2 =0 (13.1-3) In general, any chemical reaction is expressed as k X i=1 α i A i =0 (13.1-4) where α i is the stoichiometric coefficient of the i th chemical species (positive if species is a product, negative if species is a reactant), k is the total number of species in the reaction, and A i is the chemical symbol for the i th chemical species, representing the molecular weight of 1 The term "reforming" is used for the production of hydrogen from hydrocarbons. 435
Transcript
c° 2011 Ismail Tosun
Chapter 13
Chemical Reaction Equilibrium
A chemical reactor is the heart of a chemical plant in which reactants are converted into prod-ucts. Design of a reactor should provide answers to questions like: Which products will beobtained from the reaction(s) of given reactants? What are the best conditions, i.e., tempera-ture, pressure, addition of inerts, etc., to convert more of the reactants into products? Whatquantities of products will be formed from the given amounts of reactants?
This chapter starts with the fundamentals such as stoichiometry, molar extent of reaction,and the condition of chemical reaction equilibrium. Calculation of the equilibrium constant isthen introduced. Estimation of the equilibrium extent for reactions taking place in a singlephase, as well as factors affecting it, will be discussed.
13.1 STOICHIOMETRY OF A CHEMICAL REACTION
Balancing of a chemical equation is based on the conservation of mass for a closed thermo-dynamic system. If a chemical reaction takes place in a closed container, the mass does notchange even if there is an exchange of energy with the surroundings.
Consider carbon dioxide reforming1 of methane described by the reaction
CH4 (g) + CO2 (g) 2H2 (g) + 2CO (g) (13.1-1)
If A1 = CH4, A2 = CO2, A3 = H2, and A4 = CO, Eq. (13.1-1) is expressed as
A1 +A2 = 2A3 + 2A4 (13.1-2)
It is convenient to write all the chemical species on one side of the equation and give a positivesign to the species that are regarded as the products of the reaction. Thus,
2A3 + 2A4 −A1 −A2 = 0 (13.1-3)
In general, any chemical reaction is expressed as
kXi=1
αiAi = 0 (13.1-4)
where αi is the stoichiometric coefficient of the ith chemical species (positive if species is aproduct, negative if species is a reactant), k is the total number of species in the reaction, andAi is the chemical symbol for the ith chemical species, representing the molecular weight of
1The term "reforming" is used for the production of hydrogen from hydrocarbons.
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species. Equation (13.1-4) is simply a mathematical statement representing the conservationof mass2. In matrix notation, Eq. (13.1-4) is expressed in the form
£A1 A2 A3 ... Ak
¤·
⎡⎢⎢⎢⎢⎢⎣α1α2α3...αk
⎤⎥⎥⎥⎥⎥⎦ = [ 0 ] (13.1-5)
Each chemical species, Ai, is the sum of the chemical elements, Ej , such that
Ai =tX
j=1
βjiEj (13.1-6)
where βji represents the number of chemical elements Ej in the chemical species Ai, and t isthe total number of chemical elements. Substitution of Eq. (13.1-6) into Eq. (13.1-4) gives
kXi=1
αi
⎛⎝ tXj=1
βjiEj
⎞⎠ =tX
j=1
ÃkXi=1
βji αi
!Ej = 0 (13.1-7)
Since all the Ej are linearly independent3, then
kXi=1
βji αi = 0 j = 1, 2, ..., t (13.1-8)
In matrix notation, Eq. (13.1-8) is expressed in the form⎡⎢⎢⎢⎣β11 β12 β13 ... β1kβ21 β22 β23 ... β2k...
......
......
βt1 βt2 βt3 ... βtk
⎤⎥⎥⎥⎦| z
·
β
⎡⎢⎢⎢⎣α1α2...αt
⎤⎥⎥⎥⎦ = [0] (13.1-9)
where the element-by-species matrix, [β], is formed as
Species→ A1 A2 A3 Ak
[β] =
E1E2...Et
⎡⎢⎢⎢⎣β11 β12 β13 ... β1kβ21 β22 β23 ... β2k...
......
......
βt1 βt2 βt3 ... βtk
⎤⎥⎥⎥⎦ (13.1-10)
2For the reaction given by Eq. (13.1-1), Eq. (13.1-4) reads
MCH4 +MCO2 = 2MH2 + 2MCO
where M represents the molecular weight.3The expression
nXi=1
αixi = α1x1 + α2x2 + ...+ αnxn
where α1, α2, ..., αn is a set of scalars, is called a linear combination of the elements of the set S =x1, x2, ..., xn . The elements of the set S are said to be linearly dependent if there exists a set of scalars
α1, α2, ..., αn with elements αi not all equal to zero, such that the linear combinationnPi=1
αixi = 0 holds. If
nPi=1
αixi = 0 holds for all αi = 0, then the set S is linearly independent. See also Appendix F.
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Using the property of a "reduced row echelon form" of a matrix as explained in Appendix F,i.e., Eqs. (F.6-10) and (F.6-11), Eq. (13.1-9) can be rewritten in the form
rref[β]·
⎡⎢⎢⎢⎣α1α2...αt
⎤⎥⎥⎥⎦ = [0] (13.1-11)
where rref[β] represents the reduced row echelon form of the element-by-species matrix. Equa-tion (13.1-11) is used to balance chemical reactions.
Example 13.1 Consider carbon dioxide reforming of methane, i.e.,
α1CH4 + α2CO2 + α3H2 + α4CO = 0
Balance this equation by applying Eq. (13.1-11).
Solution
The element-by-species matrix is written as
Species→ CH4 CO2 H2 CO
[β] =CHO
⎡⎣ 1 1 0 14 0 2 00 2 0 1
⎤⎦ (1)
Reduced row echelon form of [β] is
rref [β] =
⎡⎢⎢⎣1 0 0 1
2
0 1 0 12
0 0 1 − 1
⎤⎥⎥⎦ (2)
Therefore, Eq. (13.1-11) becomes⎡⎢⎢⎣1 0 0 1
2
0 1 0 12
0 0 1 − 1
⎤⎥⎥⎦·⎡⎢⎢⎣α1α2α3α4
⎤⎥⎥⎦ =⎡⎣000
⎤⎦ (3)
Multiplication of the matrices yields
α1 +1
2α4 = 0 (4)
α2 +1
2α4 = 0 (5)
α3 − α4 = 0 (6)
Since there are 4 unknowns and 3 equations, expressing α1, α2, and α3 in terms of α4, a freevariable, gives
α1 = − 12 α4 α2 = − 1
2 α4 α3 = α4 (7)
If we take α4 = 2, then α1 = α2 = − 1, and α3 = 2, Hence, the reaction becomes
CH4 +CO2 2H2 + 2CO
Comment: Stoichiometric coefficients have units. For example, in the above equation thestoichiometric coefficient of H2 indicates that there are 2 moles of H2 per mole of CH4.
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13.2 THE LAW OF COMBINING PROPORTIONS
Stoichiometric coefficients have the units of moles of i per mole of basis species, where basisspecies is arbitrarily chosen. Consider, for example, a reaction between nitrogen and hydrogento form ammonia, i.e.,
N2 (g) + 3H2 (g) 2NH3 (g)
If N2 is chosen as the basis species, then the stoichiometric coefficients 3 and 2 represent the"moles of H2 per mole of N2" and "moles of NH3 per mole of N2", respectively. Suppose thatthe reactor is initially charged with (nN2)o moles of N2, (nH2)o moles of H2, and (nNH3)o molesof NH3. At any given instant during the reaction, let the number of moles of N2, H2, and NH3in the reacting mixture be nN2 , nH2 , and nNH3 , respectively. Therefore, the number of molesof N2 and H2 reacted as well as the number of moles of NH3 formed can be expressed as
Moles of N2 reacted = (nN2)o − nN2
Moles of H2 reacted = (nH2)o − nH2
Moles of NH3 formed = nNH3 − (nNH3)o
(13.2-1)
The moles of N2 reacted can also be expressed by using either the moles of H2 reacted or themoles of NH3 formed as
Moles of N2 reacted = (nN2)o − nN2 =(nH2)o − nH2
3=
nNH3 − (nNH3)o2
(13.2-2)
Rearrangement of Eq. (13.2-2) gives
Moles of N2 reacted =nN2 − (nN2)o
− 1 =nH2 − (nH2)o
− 3 =nNH3 − (nNH3)o
2(13.2-3)
Equation (13.2-3) can be generalized as
moles of basis species =moles of i reacted
(moles of i/mole of basis species)(13.2-4)
which is known as the law of combining proportions. The mathematical expression for the lawof combining proportions is given by
ni − nioαi
= (13.2-5)
where is called the molar extent of the reaction4. Rearrangement of Eq. (13.2-5) gives
ni = nio + αi (13.2-6)
Once has been determined, the number of moles of any chemical species participating in thereaction can be determined by using Eq. (13.2-6).
The molar extent of the reaction should not be confused with the fractional conversion,which can only take values between 0 and 1. The molar extent of the reaction is an extensiveproperty measured in moles and its value can be greater than unity.
4The term has been given various names in the literature, such as degree of advancement, reaction ofcoordinate, degree of reaction, and progress variable.
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It is also important to note that the fractional conversion may be different for each reactingspecies, i.e.,
Xi =nio − ninio
(13.2-7)
On the other hand, molar extent is unique for a given reaction. Comparison of Eqs. (13.2-6)and (13.2-7) indicates that
=nio(−αi)
Xi (13.2-8)
The total number of moles, nT , of a reacting mixture at any instant can be calculated bythe summation of Eq. (13.2-6) over all species, i.e.,
nT = nTo + α (13.2-9)
where nTo is the initial total number of moles and α is defined by
α =Piαi (13.2-10)
Example 13.2 A reactor is initially charged with 2 moles of C2H4, 5 moles of H2O, and 1mole of H2. After setting up the temperature and pressure, the following reaction takes place:
C2H4 (g) + 2H2O (g) CH4 (g) + CO2 (g) + 2H2 (g)
a) Determine expressions for the number of moles, ni, and the mole fractions, yi, of eachspecies as a function of the molar extent of the reaction.
b) Calculate the number of moles and mole fractions of each species when = 1.8.
Solution
a) The use of Eq. (13.2-6) expresses number of moles of each species as a function of the molarextent of the reaction in the form
nC2H4 = 2−nH2O = 5− 2nCH4 =nCO2 =nH2 = 1 + 2
nT = 8 +
The mole fractions can be calculated from yi = ni/nT as
yC2H4 =2−8 +
yH2O =5− 28 +
yCH4 = yCO2 = 8 +yH2 =
1 + 2
8 +
b) When = 1.8, number of moles of each species becomes
Comment: The value of that makes ni = 0 for a reactant gives the greatest possible valueof . For the given reaction
nC2H4 = 0 ⇒ = 2nH2O = 0 ⇒ = 2.5
Therefore, ethylene is the limiting reactant and cannot exceed 2.
The molar concentration of the ith species, ci, is defined by
ci =niV
(13.2-11)
Dividing Eq. (13.2-6) by the volume V givesniV=
nioV+ αi
³V
´(13.2-12)
orci = cio + αi ξ (13.2-13)
where cio is the initial molar concentration of the ith species and ξ is the intensive extent of
the reaction in moles per unit volume. Note that ξ is related to fractional conversion by
ξ =cio
(−αi)Xi (13.2-14)
The total molar concentration, c, of a reacting mixture at any instant can be calculated bythe summation of Eq. (13.2-13) over all species. The result is
c = co + α ξ (13.2-15)
where co is the initial total molar concentration.When more than one reaction takes place in a reactor, Eq. (13.2-6) takes the form
nij = nijo + αij j (13.2-16)
where
nij = number of moles of the ith species in the jth reaction
nijo = initial number of moles of the ith species in the jth reaction
αij = stoichiometric coefficient of the ith species in the jth reaction
j = extent of the jth reaction
Summation of Eq. (13.2-16) over all reactions taking place in a reactor givesXj
nij =Xj
nijo +Xj
αij j (13.2-17)
or
ni = nio +Xj
αij j (13.2-18)
Example 13.3 The following two reactions occur simultaneously in a batch reactor :
CH4 (g) +H2O (g) CO (g) + 3H2 (g)
CO (g) +H2O (g) CO2 (g) +H2 (g)
A mixture of 20 mol % CH4, 70% H2O, and 10% inerts is fed into a reactor and the reactionsproceed until 6% CO and 43% H2 are formed. Determine the percentage of each species in areacting mixture.
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Solution
Basis: 1 mole of a feed mixture
Let 1 and 2 be the extents of the first and second reactions, respectively. Then the number ofmoles of each species can be expressed as
nCH4 = 0.2− 1
nH2O = 0.7− 1 − 2
nCO = 1 − 2
nH2 = 3 1 + 2
nCO2 = 2
ninert = 0.1
nT = 1 + 2 1
The mole fractions of CO and H2 are given in the problem statement. These values are usedto determine the extent of the reactions as
yCO =1 − 2
1 + 2 1= 0.06 ⇒ 0.88 1 − 2 = 0.06
yH2 =3 1 + 2
1 + 2 1= 0.43 ⇒ 2.14 1 + 2 = 0.43
Simultaneous solution of these two equations gives
1 = 0.162 and 2 = 0.083
Therefore, the mole fractions of CH4, H2O, CO2, and the inerts are
yCH4 =0.2− 1
1 + 2 1=
0.2− 0.1621 + (2)(0.162)
= 0.029
yH2O =0.7− 1 − 2
1 + 2 1=0.7− 0.162− 0.0831 + (2)(0.162)
= 0.344
yCO2 =2
1 + 2 1=
0.083
1 + (2)(0.162)= 0.063
yinert =0.1
1 + 2 1=
0.1
1 + (2)(0.162)= 0.076
Comment: Although inert species do not take part in the reactions, they must be consideredin determining the total number of moles.
13.3 EQUILIBRIUM FOR A SINGLE REACTION
Consider a single chemical reaction taking place in a single phase. The change in the totalGibbs energy of the system is given by Eq. (7.1-13), i.e.,
dG = V dP − S dT +kXi=1
Gi dni (13.3-1)
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The changes in the number of moles occur as a result of a chemical reaction and fromEq. (13.2-6)
dni = αi d (13.3-2)
Substitution of Eq. (13.3-2) into Eq. (13.3-1) gives
dG = V dP − S dT +kXi=1
αiGi d (13.3-3)
Differentiation of Eq. (13.3-3) with respect to the molar extent of reaction, , by keepingtemperature and pressure constant givesµ
∂G
∂
¶T,P
=kXi=1
αiGi (13.3-4)
which indicates that the rate of change of the total Gibbs energy of the system with the molarextent of reaction is equal to
PαiGi. Since (dG)T,P = 0 at equilibrium as shown in Figure
13.1, then it follows that
kXi=1
αiGi = 0 Criterion for chemical reaction equilibrium (13.3-5)
In other words, when a chemical reaction reaches equilibrium, total Gibbs energy of a reactingmixture reaches its minimum value.
Equilibrium
0=εd
dG
Constant T & P
ε
G
Figure 13.1 Variation of Gibbs energy with molar extent of reaction.
13.3.1 The Equilibrium Constant
The solution of Eq. (13.3-5) requires an expression for the partial molar Gibbs energy. FromEq. (7.3-5)
The molar Gibbs energy of a pure component i at the temperature of the system, T , and atsome reference pressure, P o, is
eGoi (T, P
o) = λi(T ) +RT ln foi (T,Po) (13.3-7)
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The superscript o indicates the standard state conditions, i.e., T , P o, and the phase. Subtrac-tion of Eq. (13.3-7) from Eq. (13.3-6) gives
Gi(T, P, xi) = eGoi (T, P
o) +RT lnbai(T, P, xi) (13.3-8)
in which the activity, bai, is defined bybai(T,P, xi) = bfi(T, P, xi)
foi (T, Po)
(13.3-9)
The use of Eq. (13.3-8) in Eq. (13.3-5) leads to
kXi=1
αiGi = 0 =kXi=1
αi eGoi +RT
kXi=1
αi lnbai=
kXi=1
αi eGoi +RT
kXi=1
lnbaαii=
kXi=1
αi eGoi +RT ln
kYi=1
baαii (13.3-10)
The equilibrium constant, Ka, and the standard Gibbs energy change of reaction, ∆Gorxn, are
defined by
Ka =kYi=1
baαii (13.3-11)
∆Gorxn(T ) =
kXi=1
αi eGoi (T, P
o) (13.3-12)
so that Eq. (13.3-10) takes the form
Ka(T ) = exp
∙− ∆G
orxn(T )
RT
¸(13.3-13)
It is important to keep in mind that the equilibrium constant, Ka, is independent of pressure,and varies only with temperature. The presence of a catalyst does not affect the equilibriumconstant. A catalyst influences reaction rate.
13.4 EVALUATION OF THE EQUILIBRIUM CONSTANT
Calculation of the equilibrium constant, Ka, from Eq. (13.3-13) requires the standard Gibbsenergy change of reaction, ∆Go
rxn, to be known at the temperature of reaction, T . StandardGibbs energy change of reaction can be calculated from the relation
∆Gorxn = ∆H
orxn − T ∆So
rxn (13.4-1)
where ∆Horxn, standard enthalpy change of reaction (or standard heat of reaction), and ∆S
orxn,
standard entropy change of reaction, ∆Sorxn, are defined by
∆Horxn =
kXi=1
αi eHoi (13.4-2)
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∆Sorxn =
kXi=1
αi eSoi (13.4-3)
Standard heat of reaction is the difference between the enthalpies of pure products and reactantsin their standard states. Since it is not practical to list ∆Ho
rxn and ∆Sorxn for every reaction,
calculations of ∆Horxn and ∆G
orxn are based on the formation reactions of compounds involved
in the reaction.The difference between the enthalpy of one mole of a pure compound i and the total
enthalpy of the elements of which it is composed at the temperature of the system, T , and thestandard state pressure of 1 bar is called the standard enthalpy of formation of the compound,(∆ eHo
f )i. The enthalpies of formation of elements in their standard state are defined to be zeroat any temperature. As a result, the standard enthalpy of formation of a compound is just thestandard heat of reaction in which one mole of it is formed from elements. Thus,
eHoi =
³∆ eHo
f
´i
(13.4-4)
Substitution of Eq. (13.4-4) into Eq. (13.4-2) leads to
∆Horxn =
kXi=1
αi
³∆ eHo
f
´i
(13.4-5)
The difference between the Gibbs energy of one mole of a pure compound and the totalGibbs energy of the elements of which it is composed at the temperature of the system, T ,and the standard state pressure of 1 bar is called the standard Gibbs energy of formation of thecompound, (∆ eGo
f )i. Gibbs energies of formation of elements in their standard state are definedto be zero at any temperature. Thus,
eGoi =
³∆ eGo
f
´i
(13.4-6)
Substitution of Eq. (13.4-6) into Eq. (13.3-12) leads to
∆Gorxn =
kXi=1
αi
³∆ eGo
f
´i
(13.4-7)
In the literature, standard enthalpy and Gibbs energy of formation data are generallytabulated at 298K and 1 bar. The values of ∆ eHo
Once ∆Gorxn,298 is known, Ka at 298K can be easily calculated from Eq. (13.3-13). To
determine Ka at any other temperature, it is first necessary to differentiate Eq. (13.3-13) withrespect to temperature to get
d lnKa
dT= − 1
R
d
dT
µ∆Go
rxn
T
¶
= − 1R
⎛⎜⎜⎝ 1T d∆Gorxn
dT| z −∆Sorxn
− ∆Gorxn
T 2
⎞⎟⎟⎠ =
∆Horxnz |
T ∆Sorxn +∆G
orxn
RT 2(13.4-8)
ord lnKa
dT=∆Ho
rxn
RT 2(13.4-9)
which is known as the van’t Hoff equation. The van’t Hoff equation is valid for both homoge-neous and heterogeneous reactions5.
If heat is evolved in the reaction, the reaction is called exothermic. If heat is absorbed, thereaction is called endothermic. Therefore,
∆Horxn
½> 0 for an endothermic reaction< 0 for an exothermic reaction
(13.4-10)
When the reaction is exothermic, Ka decreases with increasing temperature. On the otherhand, if the reaction is endothermic, Ka increases with increasing temperature. Integration ofEq. (13.4-9) is dependent on whether ∆Ho
rxn remains constant or not.5A homogeneous reaction occurs throughout the given phase. A heterogeneous reaction occurs at an interface
between at least two different phases and is usually a catalytic reaction.
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13.4.1 ∆Horxn is Independent of Temperature
For small temperature ranges ∆Horxn can be considered independent of temperature. In this
case, integration of Eq. (13.4-9) from 298K to any T gives
ln
µKa
Ka,298
¶= − ∆H
orxn
R
µ1
T− 1
298
¶(13.4-11)
Rearrangement of Eq. (13.4-11) gives
lnKa = −∆Ho
rxn
R
1
T+∆So
rxn
R(13.4-12)
Equations (13.4-11) and (13.4-12) both indicate that a plot of lnKa versus the reciprocal ofthe absolute temperature, 1/T , is a straight line with a slope −∆Ho
rxn/R as shown in Figure13.2.
1/T
lnKa
Exothermic Reaction
Endothermic Reaction
Figure 13.2 lnKa versus 1/T when ∆Horxn is independent of temperature.
Example 13.5 Calculate the equilibrium constant for the water-gas shift reaction, i.e.,
CO (g) +H2O (g) CO2 (g) +H2 (g)
at 350K by assuming ∆Horxn to be independent of temperature.
Comment: Since ∆Horxn < 0, the reaction is exothermic.
13.4.3 An Alternative Way of Evaluating Ka
Once standard Gibbs energy change of reaction, ∆Gorxn, is known at the temperature of re-
action, then the equilibrium constant can be easily calculated from Eq. (13.3-13). For thispurpose, it is first necessary to obtain a relationship between (∆ eGo
f )i and (∆ eGof,298)i. Note
that
(∆ eHof )i = (∆
eHof,298)i +
Z T
298∆ eCo
Pi dT (13.4-21)
where∆ eCo
Pi =eCoPi −
Xe
αei eCoPe (13.4-22)
in which αei is the number of atoms of an element e in species i, and eCoPeis the molar heat
capacity of element e. For example, for ethanol (C2H5OH), αeC = 2, αeH2 = 3, and αeO2 = 0.5.The use of Eq. (13.4-21) in the Gibbs-Helmholz equation, Eq. (5.7-8), gives the standard Gibbsenergy of formation at the temperature of reaction as
(∆ eGof )i
T=(∆ eGo
f,298)i
298−Z T
298
"(∆ eHo
f,298)i +
Z T
298∆ eCo
PidT
#dT
T 2(13.4-23)
The use of Eq. (13.4-23) requires heat capacities of elements to be known. Heat capacitiesof elements in the gaseous form can be obtained from Appendix B. The molar heat capacity ofcarbon (graphite) is given as (Kubaschewski and Alcock, 1979)
The use of Eq. (13.3-3) gives the equilibrium constant as
Ka = exp
∙− 18, 289
(8.314)(500)
¸= 0.012
13.5 GAS PHASE REACTIONS
For gas phase reactions, the standard state is defined as pure components at the temperatureof the system and standard state pressure, P o. The standard state pressure is chosen in such away that a pure component behaves as an ideal gas6. Under these conditions, fugacity is equalto pressure, i.e., foi (T, P
o) = P o. Thus, the equilibrium constant is expressed as
Ka =kYi=1
" bfi(T, P, yi)foi (T, P
o)
#αi=
kYi=1
" bfi(T, P, yi)P o
#αi(13.5-1)
Substitution of bfi = bφiyiP (13.5-2)
into Eq. (13.5-1) gives
Ka =kYi=1
bφαii| z Kφ
kYi=1
yαii| z Ky
kYi=1
µP
P o
¶αi
(13.5-3)
6The standard state pressure is usually chosen as 1 bar.
450
The third term on the right-hand side of Eq. (13.5-3) is expressed as
kYi=1
µP
P o
¶αi
=
µP
P o
¶Σαi
=
µP
P o
¶α
(13.5-4)
so that Eq. (13.5-3) becomes
Ka = KφKy
µP
P o
¶α
(13.5-5)
The units of P and P o must be consistent so that the ratio P/P o is a dimensionless quantity.For many gas phase systems, the Lewis-Randall rule is applicable. In that case the fugacity
coefficient becomes
bφi(T, P, yi) = bfi(T, P, yi)yiP
=yi fi(T,P )
yiP=
fi(T, P )
P= φi(T,P ) (13.5-6)
where fi is the fugacity of pure i at the temperature and pressure of the reacting system. Theuse of Eq. (13.5-6) in Eq. (13.5-5) leads to
Ka = Kf/P Ky
µP
P o
¶α
Ideal mixture of nonideal gases (13.5-7)
where
Kf/P =kYi=1
µfiP
¶αi
=kYi=1
φαii (13.5-8)
If pure gases behave ideally, then Kf/P = 1 and Eq. (13.5-7) reduces to
Ka = Ky
µP
P o
¶α
Ideal gas mixture (13.5-9)
Example 13.9 Cyclohexane is mainly produced from the hydrogenation of benzene accordingto the reaction
C6H6 (g) + 3H2 (g) C6H12 (g)
The reactor temperature is 550K and hydrogen to benzene feed mole ratio is 4.5:1. Estimatethe gas composition under equilibrium conditions if the reactor pressure is
a) 1 barb) 15 bar
Solution
a) Since the pressure is low, we can assume ideal gas behavior. Choosing P o = 1bar, Eq.(13.5-9) reduces to
Ka = Ky Pα (1)
Note that
α =3X
i=1
αi = 1− 1− 3 = − 3 and P = 1bar
451
so that Eq. (1) reduces to
Ka = Ky =nC6H12/nT
(nC6H6/nT ) (nH2/nT )3=
nC6H12 n3T
nC6H6 n3H2
(2)
From Appendix E, enthalpies and Gibbs energies of formation of the species taking place in thereaction are as follows:
Substitution of the numerical values into Eq. (13.4-18) gives
lnKa = − 10.108 lnT − 1.083× 10−3 T + 7.788× 10−6 T 2 − 3.831× 10−9 T 3
+ 6.696× 10−13 T 4 + 22, 020T
+ 22.945 (3)
When T = 550KKa = 1.47
Choosing 1mol of benzene as a basis, the number of moles of each species present in the reactorcan be expressed as a function of the molar extent of the reaction from Eq. (13.2-6) as
nC6H6 = 1−nH2 = 4.5− 3
nC6H12 =
nT = 5.5− 3
452
Therefore, Eq. (2) can be expressed as
1.47 =
µ1−
¶µ5.5− 34.5− 3
¶3⇒ = 0.4
The mole fractions of each species are
yC6H6 =1− 0.4
5.5− (3)(0.4) = 0.140
yH2 =4.5− (3)(0.4)5.5− (3)(0.4) = 0.767
yC6H12 =0.4
5.5− (3)(0.4) = 0.093
b) Since the pressure is moderately high, it is plausible to assume that nonideal gases form anideal mixture, i.e., the Lewis-Randall rule applies. Taking P o = 1bar, Eq. (13.5-7) reduces to
Ka = Kf/P Ky P−3 ⇒ Ky =
Ka P3
Kf/P(4)
Fugacity coefficients of pure components, calculated from the Peng-Robinson equation of state,Eq. (5.3-12), are given in the following table:
Substitution of numerical values into Eq. (4) gives
Ky =(1.47)(15)3
0.969= 5120
The molar extent of the reaction is calculated from
Ky = 5120 =
µ1−
¶µ5.5− 34.5− 3
¶3⇒ = 0.999
which indicates that percent conversion of benzene to cyclohexane is almost 100% and the molefractions are
yC6H6 = 0.0004 yC6H12 = 0.3991 yH2 = 0.6005
Example 13.10 A mixture of methane and steam in the mole ratio 1:5 enters a reactoroperating at 800K and 2 bar. Estimate the composition of the product stream at equilibrium ifthe following reactions take place within the reactor :
Following the procedure outlined in Example 13.6, the equilibrium constants at 800K are cal-culated as
(Ka)1 = 0.031 and (Ka)2 = 0.096
Let 1 and 2 be the extents of the first and second reactions, respectively. The number of molesof each species present in the reactor can be expressed as a function of the molar extents of thereactions in the form
nCH4 = 1− 1 − 2
nH2O = 5− 1 − 2 2
nCO = 1
nCO2 = 2
nH2 = 3 1 + 4 2
nT = 6 + 2 1 + 2 2
Assuming an ideal gas mixture and taking P o = 1bar, Eq. (13.5-9) gives
(Ka)1 = (Ky)1 P2 (1)
and(Ka)2 = (Ky)2 P
2 (2)
Ky values are given by
(Ky)1 =(nCO/nT ) (nH2/nT )
3
(nCH4/nT ) (nH2O/nT )=
nCO n3H2
nCH4 nH2O n2T
=1 (3 1 + 4 2)
3
(1− 1 − 2) (5− 1 − 2 2) (6 + 2 1 + 2 2)2
(3)
454
(Ky)2 =(nCO2/nT ) (nH2/nT )
4
(nCH4/nT ) (nH2O/nT )2=
nCO2 n4H2
nCH4 n2H2O n
2T
=2 (3 1 + 4 2)
4
(1− 1 − 2) (5− 1 − 2 2)2 (6 + 2 1 + 2 2)
2(4)
Substitution of Eqs. (3) and (4) into Eqs. (1) and (2), respectively, yields
0.031 =
"1 (3 1 + 4 2)
3
(1− 1 − 2) (5− 1 − 2 2) (6 + 2 1 + 2 2)2
#(2)2 (5)
0.096 =
"2 (3 1 + 4 2)
4
(1− 1 − 2) (5− 1 − 2 2)2 (6 + 2 1 + 2 2)
2
#(2)2 (6)
Simultaneous solution of Eqs. (5) and (6) by MATHCADR°gives
1 = 0.079 and 2 = 0.463
Therefore, the mole fractions of the gas mixture under equilibrium conditions are given by
Example 13.11 You are given the task of producing methanol from hydrogenolysis of methylformate (HCOOCH3) over copper-containing catalyst, i.e.,
HCOOCH3 (g) + 2H2 (g) 2CH3OH (g) (1)
Your boss is suspicious about the possibility of the following decarbonylation reaction
HCOOCH3 (g) CH3OH (g) + CO (g) (2)
taking place in the reactor. If the hydrogen to methyl formate feed mole ratio is 4:1, and the re-actor temperature and pressure are 500K and 40 bar, respectively, estimate the gas compositionunder equilibrium conditions.
DATA: For methyl formate
Tc = 487.2K Pc = 60bar ω = 0.257
The gases can be represented by the Soave-Redlich-Kwong equation of state. Kim et al. (1990)reported the equilibrium constants as
indicating that while the first reaction is highly exothermic the second is highly endothermic.Calculation of the equilibrium constants gives
Ka1 = 4.233 and Ka2 = 0.012
Since Ka1 ÀKa2, it seems that at the reaction temperature of 500K the first reaction isdominant and the effect of the second reaction on the equilibrium gas composition can be safelyneglected7.
Taking P o = 1bar, Eq. (13.3-9) becomes
Ka = KφKy P−1 = 4.233 (1)
Since the reaction pressure is 40 bar, the assumption of ideal mixture behavior is doubtful. Inthis case, direct calculation of Kφ is impossible since determination of fugacity coefficientsrequires a priori knowledge of the equilibrium gas composition. For this reason, iterative calcu-lations will be carried out indirectly as follows.
Choosing 1mol of methyl formate as a basis, the number of moles of each species present inthe reactor can be expressed as a function of the molar extent of the reaction in the form
nHCOOCH3 = 1−nH2 = 4− 2
nCH3OH = 2
nT = 5−
Therefore, mole fractions of the species are expressed as a function of the molar extent ofreaction as
yHCOOCH3 =1−5− yH2 =
4− 25− yCH3OH =
2
5− (2)
The expression for Ky becomes
Ky =y2CH3OH
yHCOOCH3y2H2
=4 2(5− )
(1− )(4− 2 )2 (3)
7At higher temperatures, however, this assumption is certainly not valid. For example, Ka1 = 0.283 andKa2 = 0.145 at 600K, indicating that the two reactions are competing with each other. At higher temperatures,due to its endothermic behavior, the second reaction is expected to be the dominant one. Indeed, the equilibriumconstants at 800K are Ka1 = 7.477× 10−3 and Ka2 = 4.706.
456
The iterative procedure is given as follows:
1. Assume ,2. Calculate gas phase composition from Eq. (2),3. Calculate Ky from Eq. (3),4. Once fugacity coefficients are determined from Eq. (7.5-15), calculate Kφ from
Kφ =
bφ2CH3OHbφHCOOCH3bφ2H25. Check whether Ka −KφKy P
−1 = 0 is satisfied.
According to the stoichiometry of the reaction, 1 mole of HCOOCH3 reacts with 2 moles ofH2. Since H2 is in excess, HCOOCH3 is the limiting reactant. If all HCOOCH3 were depleted,the maximum value of would be 1. Thus, may take values less than unity. The results areshown in the table below :
Example 13.12 A 100 L constant-volume reactor is evacuated and then filled with 1 mol ofdinitrogen tetraoxide (N2O4) at 298K. It decomposes according to the reaction
N2O4 (g) 2NO2 (g)
Estimate the equilibrium gas composition if the temperature is kept constant at 298K.
Solution
From Appendix E
Species ∆ eGof ( kJ/mol)
N2O4 97.79NO2 51.26
Therefore, ∆Gorxn at 298K is
∆Gorxn,298 = (2)(51.26)− 97.79 = 4.73 kJ = 4730 J
457
The equilibrium constant at 298K is
Ka,298 = exp
µ−∆Go
rxn,298
RT
¶= exp
∙− 4730
(8.314)(298)
¸= 0.148
The initial pressure, Po, within the reactor is
Po =nToRT
V=(1)(8.314× 10−2)(298)
100= 0.248 bar
Assuming ideal gas behavior and choosing P o = 1bar, Eq. (13.5-9) reduces to
Ka = Ky P =
³nNO2/nT
´2nN2O4/nT
P =n2NO2
nN2O4nTP (1)
The number of moles of each species present in the reactor can be expressed as a function ofthe molar extent of the reaction as
nN2O4 = 1−nNO2 = 2
nT = 1 +
In this case, the pressure within the reactor is not constant and changes as a function of themolar extent of reaction in the form
P =
µnTnTo
¶Po = 0.248 (1 + ) bar (2)
Thus, Eq. (1) takes the form
0.148 =
∙4 2
(1− )(1 + )
¸0.248 (1 + ) ⇒
2
1− = 0.149
The solution of the above quadratic equation gives = 0.319. The mole fractions are
yN2O4 =1− 0.3191 + 0.319
= 0.516 and yNO2 =(2)(0.319)
1 + 0.319= 0.484
13.5.1 Variables Affecting the Extent of Reaction
For simplicity let us consider an ideal gas mixture. Taking the standard state pressure as 1 bar,Eq. (13.5-9) reduces to
Ky = Ka P−α (13.5-10)
Note that while Ka depends only on temperature Ky is a function of both temperature andpressure. Before investigating the effect of various factors, i.e., temperature, pressure, and theamount of inerts, on the extent of reaction, it is first necessary to determine how Ky varieswith the extent of reaction.
The mole fraction of each species is represented by
yi =ninT
=nio + αinTo + α
=yio + αi
∗
1 + α ∗ (13.5-11)
458
where∗ =
nTo(13.5-12)
By definition
Ky =kYi=1
yαii ⇒ lnKy =kXi=1
αi ln yi (13.5-13)
Differentiation of Eq. (13.5-13) with respect to ∗ yields
d lnKy
d ∗ =kXi=1
αid ln yid ∗ =
kXi=1
αiyi
dyid ∗ (13.5-14)
From Eq. (13.5-11)
dyid ∗ =
αi(1 + α ∗)− α(yio + αi∗)
(1 + α ∗)2=
αi1 + α ∗ −
αyi1 + α ∗ (13.5-15)
Substitution of Eq. (13.5-15) into Eq. (13.5-14) results in
d lnKy
d ∗ =1
1 + α ∗
⎡⎣ kXi=1
α2iyi−Ã
kXi=1
αi
!2⎤⎦ (13.5-16)
According to Cauchy’s inequalityÃkXi=1
a2i
!ÃkXi=1
b2i
!≥Ã
kXi=1
aibi
!2(13.5-17)
If we letai =
αi√yi
and bi =√yi (13.5-18)
then Eq. (13.5-17) gives ÃkXi=1
α2iyi
!ÃkXi=1
yi
!| z
1
≥Ã
kXi=1
αi
!2(13.5-19)
indicating that the terms in brackets in Eq. (13.5-16) are always greater than zero. Also notethat
1 + α ∗ =nTnTo
> 0 (13.5-20)
Therefore, one can conclude that
d lnKy
d ∗ > 0 ⇒ dKy
d ∗ > 0 ⇒ dKy
d> 0 (13.5-21)
which implies that the molar extent of reaction increases with increasing Ky or vice versa.Substitution of Eq. (13.3-13) into Eq. (13.5-10) results in
Ky = exp
µ− ∆G
orxn
RT
¶| z
Ka
P−α (13.5-22)
459
indicating that
Large negativevalues of ∆Go
rxn→ Large values
of Ka→ Large values
of Ky→ Large equilibrium
extent
Large positivevalues of ∆Go
rxn→ Small values
of Ka→ Small values
of Ky→ Small equilibrium
extent
• Effect of temperature at constant pressure
Differentiation of the molar extent of the reaction with respect to temperature, while keepingpressure constant, is given by the chain rule asµ
∂
∂T
¶P
=d
dKy
µ∂Ky
∂T
¶P
=d
dKy
µdKa
dTP−α
¶(13.5-23)
The van’t Hoff equation, Eq. (13.4-9), gives
d lnKa
dT=∆Ho
rxn
RT 2⇒ dKa
dT= Ka
µ∆Ho
rxn
RT 2
¶(13.5-24)
Substitution of Eq. (13.5-24) into Eq. (13.5-23) yieldsµ∂
∂T
¶P
=
µd
dKy
Ka
RT 2P−α
¶∆Ho
rxn (13.5-25)
In Eq. (13.5-25), the terms in parentheses are always greater than zero. Therefore, the valueof (∂ /∂T )P is dependent on the sign of ∆Ho
rxn, i.e.,
Exothermic rxn ⇒ ∆Horxn < 0 ⇒ (∂ /∂T )P < 0 ⇒ decreases with increasing T
Endothermic rxn ⇒ ∆Horxn > 0 ⇒ (∂ /∂T )P > 0 ⇒ increases with increasing T
For a highly exothermic reaction, while equilibrium conversion decreases at higher tempera-tures, the reaction rate increases. Thus, the optimum reactor temperature should be determinedas a compromise between reaction kinetics and thermodynamics8.
• Effect of pressure at constant temperature
Differentiation of the molar extent of the reaction with respect to pressure at constant temper-ature is given by the chain rule asµ
∂
∂P
¶T
=d
dKy
µ∂Ky
∂P
¶T
=
µd
dKy
Ky
P
¶(−α) (13.5-26)
Therefore, if the summation of the stoichiometric coefficients, α, is negative, increase in pressureat constant temperature results in an increase in the equilibrium extent of the reaction. Onthe other hand, if α is positive, then an increase in pressure at constant temperature causes adecrease in the equilibrium extent of the reaction.
It should be kept in mind that the value of Ka does not change with pressure. However,the position of equilibrium may change with a change in pressure.
8Thermodynamics, reaction kinetics, and economics must be considered together in the design and operationof chemical reactors.
460
• Effect of inerts addition at constant temperature and pressure
Since Ky is fixed at constant temperature and pressure, then
Ky = constant =kYi=1
yαii =kYi=1
µninT
¶αi
=
kQi=1
nαii
nαT
(13.5-27)
Since the total number of moles, nT , increases with the addition of inerts, the variation ofwill be dependent on the summation of the stoichiometric coefficients, α.
When α > 0, addition of inerts increases the denominator of Eq. (13.5-27). Thus, the
numerator,kQi=1
nαii , should increase so as to satisfy the equality, leading to an increase in .
When α < 0, rearrangement of Eq. (13.5-27) gives
Ky = constant = n−αT
kYi=1
nαii (13.5-28)
In this case, addition of inerts increases n−αT . Thus,
kQi=1
nαii should decrease to satisfy the
equality, leading to a decrease in .When α = 0, addition of inerts has no influence on the molar extent of the reaction.
Example 13.13 Consider the production of cyclohexane from the hydrogenation of benzeneas given in Example 13.9 according to the reaction
C6H6 (g) + 3H2 (g) C6H12 (g)
Estimate the equilibrium extent of the reaction under the different sets of conditions givenbelow :
a) The hydrogen to benzene feed mole ratio is 4.5:1, and the reaction takes place at 600K and1 bar.
b) The hydrogen to benzene feed mole ratio is 4.5:1, and the reaction takes place at 600K and2 bar.
c) The hydrogen to benzene feed mole ratio is 3:1, and the reaction takes place at 600K and1 bar.
d) The hydrogen to benzene feed mole ratio is 6:1, and the reaction takes place at 600K and1 bar.
Solution
In Example 13.9, the equilibrium constant is expressed as a function of temperature in the form
lnKa = − 10.108 lnT − 1.083× 10−3 T + 7.788× 10−6 T 2 − 3.831× 10−9 T 3
+ 6.696× 10−13 T 4 + 22, 020T
+ 22.945 (1)
When T = 600KKa = 0.027
461
Assuming ideal gas behavior and taking P o = 1bar, Eq. (13.5-9) becomes
Ka =Ky
P 3(2)
where
Ky =nC6H12nC6H6
µnTnH2
¶3(3)
a) In Example 13.9, the equilibrium constant at 550K was calculated as 1.47. Since the increasein temperature from 550K to 600K decreases Ka from 1.47 to 0.027, it is obvious that theequilibrium extent of the reaction will decrease. The same conclusion can also be reached by theapplication of Le Chatelier’s principle, which states that "if a stress (or disturbance) is appliedto a system at equilibrium, the system will react so as to relieve the stress." In this specificcase, the reaction is exothermic (∆Ho
rxn,298 = − 206, 180 J) and, as a result of the reaction,the heat produced can be regarded as a "product". If the reaction temperature is increased, thereaction will shift to the left so as to relieve the disturbance with a concomitant decrease in theyield of cyclohexane.
To quantify this conclusion, let us choose 1mol of benzene as a basis. Thus, the number ofmoles of each species present in the reactor can be expressed as a function of the molar extentof the reaction in the form
nC6H6 = 1−nH2 = 4.5− 3
nC6H12 =
nT = 5.5− 3Thus, Eq. (2) takes the form
0.027 =
µ1−
¶µ5.5− 34.5− 3
¶3⇒ = 0.014
b) In this case, the pressure is different from unity and Eq. (2) becomes
0.027 =
µ1−
¶µ5.5− 34.5− 3
¶3 1
(2)3⇒ = 0.102
indicating an increase in the molar extent of the reaction with an increase in pressure. Thesame conclusion can also be reached by the application of Le Chatelier’s principle. Since thereare 4 moles of gaseous reactants on the left-hand side yielding 1 mole of gaseous cyclohexaneon the right-hand side, an increase in pressure will shift the reaction to the right.
c) In this case, benzene and hydrogen are introduced to the reactor in stoichiometric proportions.The number of moles of each species present in the reactor can be expressed as
nC6H6 = 1−nH2 = 3− 3
nC6H12 =
nT = 4− 3
and Eq. (2) becomes
0.027 =(4− 3 )3
27 (1− )4⇒ = 0.012
462
In part (a), i.e., when the hydrogen to benzene mole ratio was 4.5:1, the molar extent of thereaction was calculated as 0.014. Thus, increasing the number of moles of hydrogen increasesthe molar extent of the reaction. This phenomenon is also explained by Le Chatelier’s principle.When a reactant is added to a system at equilibrium, part of it will be consumed by the reactionas it establishes a new equilibrium state. This offsets some of the stress of the increase inreactant.
d) The number of moles of each species present in the reactor can be expressed as
nC6H6 = 1−nH2 = 6− 3
nC6H12 =
nT = 7− 3
and Eq. (2) becomes
0.027 =
µ1−
¶µ7− 36− 3
¶3⇒ = 0.017
Therefore, as hydrogen is introduced in excess of the stoichiometric amount, the extent of thereaction slightly increases.
13.5.2 Exceptions to Le Chatelier’s Principle
Consider the reaction between nitrogen and hydrogen to form ammonia, i.e.,
N2 (g) + 3H2 (g) 2NH3 (g)
Once equilibrium is established at the given temperature and pressure, what happens if morenitrogen is added to the system?
According to Le Chatelier’s principle, one can immediately conclude that the reaction shiftsto the right, producing more NH3, so as to relieve the disturbance. This conclusion, however,is not generally true as will be shown in the following analysis.
Assuming an ideal gas mixture and taking the standard state pressure as 1 bar, Eq. (13.5-9)becomes
Ka =y2NH3
yN2 y3H2
P−2 (13.5-29)
Since temperature and pressure are kept constant, the use of yi = ni/nT and rearrangementreduce Eq. (13.5-29) to
J = KaP2 = constant =
n2NH3n2T
nN2 n3H2
(13.5-30)
The variation of J with respect to the moles of nitrogen is given by
dJ
dnN2=
n2NH3
n3H2
d
dnN2
"(nN2 + nH2 + nNH3)
2
nN2
#
=
Ãn2NH3nT
n3H2n
2N2
!| z Always positive
(2nN2 − nT ) (13.5-31)
463
Thus, the sign of dJ/dnN2 is dependent on the sign of (2nN2 − nT ).When
2nN2 − nT > 0 ⇒ yN2 > 0.5
addition of nitrogen leads to an increase in J . To re-establish the constancy of J , the reactionmust shift to the left, producing more nitrogen. This is in contradiction with Le Chatelier’sprinciple.
When2nN2 − nT < 0 ⇒ yN2 < 0.5
addition of nitrogen leads to a decrease in J . To re-establish the constancy of J , the reactionmust proceed to the right, producing more ammonia as predicted by Le Chatelier’s principle.
For a more detailed discussion on the subject, see de Heer (1957), Katz (1961), and Cortiand Franses (2003).
13.6 LIQUID (OR SOLID) PHASE REACTIONS
The equilibrium constant, Ka, is defined by Eq. (13.3-11), i.e.,
Ka =kYi=1
" bfi(T, P, xi)foi (T, P
o)
#αi(13.6-1)
Substitution of bfi(T,P, xi) = γi(T, P, xi)xi fi(T, P ) (13.6-2)
into Eq. (13.6-1) gives
Ka =kYi=1
γαii| z Kγ
kYi=1
xαii| z Kx
kYi=1
∙fi(T, P )
foi (T, Po)
¸αi(13.6-3)
Note that while fi is the fugacity of pure liquid (or solid) at the temperature and pressure ofthe system foi is the fugacity of pure liquid (or solid) at the temperature of the system but atstandard state pressure, P o. These two quantities are related by Eq. (5.4-11), i.e.,
fi(T, P )
foi (T, P
o)= exp
" eVi(P − P o)
RT
#(13.6-4)
Substitution of Eq. (13.6-4) into Eq. (13.6-3) gives
Ka = Kγ Kx
kYi=1
(exp
" eVi(P − P o)
RT
#)αi
= Kγ Kx exp
"(P − P o)
RT
kXi=1
αieVi# (13.6-5)
Except for high pressures, the exponential term in Eq. (13.6-5) will approach unity and maybe neglected. Hence,
Ka = Kγ Kx Pressure correction neglected (13.6-6)
For an ideal mixture, Kγ = 1 and Eq. (13.6-6) reduces to
Ka = Kx Ideal mixture (13.6-7)
Example 13.14 Methyl tert-butyl ether (MTBE) has been used in gasoline as an octaneenhancer since the late 1970’s. It is typically manufactured in petroleum refineries by reacting
464
isobutene with methanol over a catalyst. Since it can easily mix with water, leading to ground-water pollution, there have been restrictions on the use of MTBE over the last decade. As aresult, refineries have been seeking new uses for isobutene, one of which is to use it as a rawmaterial in the production of isooctene. For this purpose, isobutene is first dimerized, 2,4,4-trimethyl-1-pentene (TMP1) and 2,4,4-trimethyl-2-pentene (TMP2) being the main dimericproducts. The isomerization reaction between TMP1 and TMP2 is given by
TMP1 (l) TMP2 (l)
Karinen et al. (2001) studied this reaction and represented the equilibrium constant as
lnKa = −421.67
T− 0.056
Determine the equilibrium composition of the mixture if isomerization takes place at 333K andatmospheric pressure.
Solution
The equilibrium constant Ka is
Ka = exp
µ− 421.67
333− 0.056
¶= 0.267 (2)
Assuming that the reacting species form an ideal mixture, from Eq. (13.6-7)
Ka = 0.267 = Kx =xTMP2xTMP1
=xTMP2
1− xTMP2(1)
The solution givesxTMP2 = 0.211 and xTMP1 = 0.789
Example 13.15 The isomerization reaction
A (l) B (l)
takes place at 298K. The molar excess Gibbs energy of the liquid mixture is given by
eGex
RT= 0.8xAxB
Determine the equilibrium composition of the mixture if ∆Gorxn,298 = − 2500 J.
Solution
When pressure correction is neglected, the expression for Ka is given by Eq. (13.6-6), i.e.,
Ka = Kγ Kx (1)
The equilibrium constant Ka is
Ka = exp
µ− ∆G
orxn
RT
¶= exp
∙2500
(8.314)(298)
¸= 2.743 (2)
465
From the given expression for the excess Gibbs energy (two-suffix Margules equation), the ac-tivity coefficients are
ln γA = 0.8x2B and ln γB = 0.8x
2A (3)
Therefore, Kγ is
Kγ =γBγA
= exph0.8(x2A − x2B)
i= exp
h0.8 (2xA − 1)
i(4)
The expression for Kx is
Kx =xBxA
=1− xAxA
(5)
The use of Eqs. (2), (4), and (5) in Eq. (1) leads to
2.743 =
µ1− xAxA
¶exp
h0.8 (2xA − 1)
i(6)
The solution of this nonlinear equation gives xA = 0.179.
Comment: If we were to take Kγ = 1, the result would be xA = 0.267.
The presence of two species in Examples 13.14 and 13.15 certainly simplifies the calcula-tions. Liquid phase reactions usually involve more than two species and the resulting mixturecannot considered ideal in most cases. As a result, calculation of activity coefficients andtheir dependence on the mole fraction of species complicate the calculations. In this case, thefollowing procedure should be used:
1. Calculate Ka from Eq. (13.3-13),2. Express mole fractions of each species, xi, in terms of the extent of the reaction, ,3. Assume ,4. Calculate xi and Kx,5. Calculate activity coefficients of species involved in the reaction. Usually UNIFAC is usedto estimate γi values in a liquid mixture.6. Calculate Kγ,7. Check whether Ka −Kγ Kx = 0 is satisfied.
13.7 AN ALTERNATIVE WAY OF CALCULATING EQUILIBRIUM COMPO-SITION
In Section 3.5, equilibrium compositions of the gas phase reactions are determined by firstexpressing the number of moles of each species in terms of the molar extent of reaction andthen substituting these into the equilibrium constant expression(s).
An alternative way of calculating equilibrium composition is to first express the equilibriumconstant in terms of the mole fractions of each species. The constraint on the mole fractions,i.e.,
kXi=1
yi = 1 (13.7-1)
provides one additional equation. The conservation of elements is expressed in the form
Ej =kXi=1
βjini j = 1, 2, ..., t (13.7-2)
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where t represents the number of elements. Dividing Eq. (13.7-2) by the total number of moles,nT , gives
Ej
nT=
kXi=1
βjiyi (13.7-3)
In matrix notation, Eq. (13.7-3) is expressed as⎡⎢⎢⎢⎣E1/nTE2/nT...
where [β] is the element-by-species matrix defined by Eq. (13.1-10). Simultaneous solutionof Eqs. (13.7-1) and (13.7-4) with the equilibrium constant expression gives the compositionunder equilibrium conditions. This approach is generally preferred for heterogeneous reactions.
Example 13.16 Resolve part (a) of Example 13.9 using an alternative approach.
Solution
The equilibrium constant expression reduces to
Ka = Ky =yC6H12
yC6H6 y3H2
= 1.47 (1)
To determine 4 unknowns, i.e., yC6H12 , yC6H6 , yH2, and nT , 3 more equations are needed besidesEq. (1). One of these equations is Eq. (13.7-1), i.e.,
yC6H6 + yH2 + yC6H12 = 1 (2)
The remaining 2 equations come from the conservation of elements. The element-by-speciesmatrix is written as
Species→ C6H6 H2 C6H12
[β] =CH
∙6 0 66 2 12
¸(3)
Choosing 1 mole of C6H6 as a basis, 6 atoms of carbon and 15 atoms of hydrogen enter areactor. Hence, Eq. (13.7-4) takes the form∙
6/nT15/nT
¸=
∙6 0 66 2 12
¸·
⎡⎣yC6H6yH2yC6H12
⎤⎦ (4)
or6 yC6H6 + 6 yC6H12 =
6
nT(5)
6 yC6H6 + 2 yH2 + 12 yC6H12 =15
nT(6)
Simultaneous solution of Eqs. (1), (2), (5), and (6) by MATHCADR°yields
13.3 A mixture of 25mol % CO and 75% H2 is fed to a reactor in which the following reactiontakes place
CO (g) + 2H2 (g) CH3OH (g)
The reaction proceeds until the mole fraction of H2 in the gas mixture drops to 0.6. Estimatethe mole fractions of CO and CH3OH.
(Answer: yCO = 0.1, yCH3OH = 0.3)
Problem related to Section 13.3
13.4 Rearrange Eq. (13.3-13) in the form
lnKa = −∆Ho
rxn
RT+∆So
rxn
R(1)
In his Priestley Medal Address, Rossini (1971) stated that the equilibrium constant, which is ameasure of the reaction’s spontaneity, increases with either an increase in ∆So
rxn or a decreasein ∆Ho
rxn. An increase in ∆Sorxn leads to an increase in randomness and a decrease in ∆H
orxn
implies bond formation leading to a more stable system. Thus, by analogy, Rossini related theterms ∆So
rxn/R and −∆Horxn/RT to personal freedom and personal security, respectively. He
then concluded that one cannot have a maximum of freedom and a maximum of security at thesame time. In other words, an increase in security is accompanied by a decrease in freedom.Do you agree with Rossini’s statement? For further discussion on the subject, see Leonard(2006), Wojcik (2006), and Silverstein (2006).
Problems related to Section 13.4
13.5 Assume ∆Horxn to be independent of temperature and show that the equilibrium con-
stant of the reaction2CO (g) 2C (s) +O2 (g)
is given by
lnKa = −26, 606
T− 21.64
13.6 Ethyl tert-butyl ether (ETBE), a good octane booster, is obtained by the gas phasereaction of ethanol (EtOH) and isobutylene (IB) according to the reaction
EtOH (g) + IB (g) ETBE (g)
Iborra et al. (1989) reported enthalpy and Gibbs energy of formation data in the gas phase asfollows:
13.8 2, 6-Dimethylnapthalene (2, 6-DMN) is an intermediate for producing high-performancethermoplastic polyethylene naphthalate. It is obtained by the liquid phase isomerization of1, 5-DMN according to the following reactions:
1, 5-DMN 1, 6-DMN Rxn 1
1, 6-DMN 2, 6-DMN Rxn 2
The equilibrium constants for these reactions are reported by Kraikul et al. (2005) as
lnKa,1 = 3.872−1263
T
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lnKa,2 = 0.6746−319.5
T
For 2, 6-DMN∆ eHo
f = 15, 196 J/mol ∆ eGof = 184, 924 J/mol
Estimate ∆ eHof and ∆ eGo
f for 1, 5-DMN.
(Answer: 2039 J/mol and 183, 031 J/mol)
13.9 The standard Gibbs energy of formation values for PbO (s) are reported by Ganesan etal. (2003) as a function of temperature as follows:
∆ eGof ( kJ/mol) − 149.2 − 129.3 − 109.4
T (K) 700 900 1100
Calculate ∆ eHof for PbO (s) at 800K.
(Answer: − 218.8 kJ/mol)
Problems related to Section 13.5
13.10 Oxidation of HCl by air (oxygen) in the presence of a catalyst (CuCl2) to produce Cl2is known as the Deacon process and is represented by
2HCl (g) +1
2O2 (g) Cl2 (g) +H2O (g)
The feed consisting of 80 mol % HCl, 19.5% O2, and 0.5% N2 enters the reactor operating at700K and 1 bar. Estimate the composition of the equilibrium mixture.
13.11 Estimate the temperature at which the fractional conversion of propane in the followingreaction
C3H8 (g) C2H4 (g)+CH4 (g)
is 90%. The pressure is 1 bar. Assume ideal gases and∆Horxn to be independent of temperature.
(Answer: 655.7K)
13.12 One of the current commercially important processes to produce acetone (C3H6O) isthe catalytic dehydrogenation of isopropanol (C3H8O) in the vapor phase at 1 bar accordingto the reaction
C3H8O (g) C3H6O (g)+H2 (g)
a) Using the procedure outlined in Section 13.4.3, estimate ∆ eGof for isopropanol and acetone
at 400, 450, 500, 550, 600, and 650K.
b) Calculate Ka at these temperatures.
c) Plot lnKa versus 1/T and show that
lnKa = −6823.7
T+ 14.559
d) Neglecting other reactions, estimate the minimum temperature for 95% conversion of iso-propanol to acetone.
(Answer: d) 553.3K)
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13.13 Nitric oxide, NO, is a common air pollutant produced by automobile engines and powerplants according to the reaction
N2 (g) +O2 (g) 2NO (g)
A constant-volume reactor of 100 L is initially charged with 4 moles of N2 and 1 mole of O2and then temperature is increased to 800K. Estimate the equilibrium extent of the reaction.
(Answer: = 5.567× 10−6)
13.14 Ethylene is produced by the dehydrogenation of ethane by high temperature crackingaccording to the reaction
C2H6 (g) = C2H4 (g)+H2 (g)
In order to suppress side reactions, the ethane is diluted with steam (0.3 moles of steam permole of ethane) before it enters the reactor. A simplified flowsheet of the process is given below.
Pure H2
Pure C2H4
Pure H2O
Recycle (Pure C2H6)
New C2H6
Steam
REACTOR
SEPARATOR
a) If the reactor operates at 1100K and 2 bar, determine the composition of the stream exitingthe reactor. Assume ideal gas behavior.
b) Calculate the recycle ratio of C2H6, i.e., moles of recycle per mole of new C2H6.
c) When you report to work one morning, you notice that the recycle ratio of C2H6 hasdecreased. What conditions or situations might have caused such a response and why?
