Conversion X Conversion X
CSTR 1 CSTR 1
3.1 Basic Definitions
3.1.4 Reversible Reactions
.All rate laws for reversible reactions must reduce to the thermodynamic relationship relating the reacting species concentrations at equilibrium. At lequi- librium, the rate of reaction is identically zero for all species (i.e., -rA E 0).
That is, for the general reaction
(2- 1 ) the concentrations at equilibrium are related by the thermodynamic relation- ship (see Appentdix C).
a A + b B
e
c C + d D( 3 - 1 0) The units of K , are ( m ~ l / d r n ~ ) ~ + ~ - ~ - ~ .
To illustrate how to write rate laws for reversible reactions we will use the combination of two benzene molecules to form one molecule of hydrogen
Ibid.
70 Rate Laws and Stoichiometry Chap. 3
and one of diphenyl. In this discussion we shall consider this gas-phase reac- tion to be elementary and reversible:
kB
k-B
'
C12HlO +H22C6H6 <
or symbolically,
The specific reaction defined w.r.t.
a particular species
The forward and reverse specific reaction rate constants, k, and k-, , Benzene (B) is being depleted by the forward reaction
k17 be respectively, will be dejined with respect to benzene.
' c,,
HI0 + H2kB
2c6 H6
in which the rate of disappearance of benzene is
rB,forward = k B C i -
If we multiply both sides of this equation by -1; we obtain the expression'for the rate of formation of benzene for the forward reaction:
rB,forward = - k B c i (3-11)
For the reverse reaction between diphenyl (D) and hydrogen ( H 2 ) , Cl2HlO + H2 k-B > 2C6H6
the rate of formation of benzene is given as
YB, reverse = ~-BCDCH, (3-12)
The net rate of formation of benzene is the sum of the rates of formation from the forward reaction [i.e., Equation (3-1 l)] and the reverse reaction [i.e., Equa- tion (3-12)]:
rB rB, net = rB, forward + 'B, reverse
(3-13) Multiplying both sides of Equation (3-13) by -1, we obtain the rate law for the rate of disappearance of benzene, -rB :
Elementary reversible A.B
- r , = k
[ 2)
C A - - (3-14)Sec. 3.1 Basic Definitions 79
where
kB
k-B - = K , = concentration equilibrium constant
The equilibrium constant decreases with increasing temperature for exotlhermic reactions and increases with increasing temperature for endothermic reactions.
We need to check to see if the rate law given by Equation (3-14) is ther- modynamically consistent at equilibrium. Using Equation (3- 10) and substitut- ing the appropriate species concentration and exponents, thermodynamics tells us that
(3-15)
At equilibrium, -rB
=
0, and the rate law given by Equation (3-14) becomes At equilibrium therate law mwt equation consistent with
thermodynamic Rearranging, we obtain equilibrium
reduce to an - r B ~ O = k C
which is identical to Equation (3-15).
A further discussion of the equilibrium constant and its thermodynamic relationship is given in Appendix C.
Finally, we want to rewrite the rate of formation of diphenyl and hydro- gen in terms of concentration. The rate of formation of these species must have the same functional dependence on concentrations as does the rate of disap- pearance of blenzene. The rate of formation of diphenyl is
L J
Using the relationship given by Equation (2-20) for the general reaction (3-16)
(2-20) we can obtain the relationship between the various specific reaction rates, k , , k , :
(3-17)
80 Rate Laws and Stoichiometry Chap. 3
These criteria must be satisfied
Comparing Equations (3-16) and (3-17), we see the relationship between the specific reaction rate with respect to diphenyl and the specific reaction rate with respect to benzene is
Example 3-3 Formulating a Reversible Rate Law
The exothermic reaction
A f 2 B ___) 2D (E3-3.1)
is virtually irreversible at low temperatures and the rate law is
-rA = k,Cy2CB (E3-3.2)
Suggest a rate law that is valid at high temperatures, where the reaction is revers- ible:
A + 2 B
e
2D (E3-3.3)Solution
The rate law for the reversible reaction must
1. satisfy thermodynamic relationships at equilibrium, and
2. reduce to the irreversible rate law when the concentration of one or more of the reaction products is zero.
We know from thermodynamics that the equilibrium relationship for Reaction (E3-3.1) as written is
with units [K,] = dm3 K c = -
c2,
cAecie
Rearranging Equation (E3-3.4) in the form\
suggests that we try a reversible rate law of the form -r, = k ,
[
CACi -g]
(E3-3.4)
(E3-3.5) Equation (E3-3.5) satisfies the equilibrium conditions but does not simplify to the initial, irreversible rate when C, = 0. Substituting C, = 0 into the equation being tested yields
Sec. :3.1 Basic Definitions 81
- IAO = kAc.40c& (E3-3.6)
Equation (E3-3.6) does not agree with Equation (E3-3.2) and therefore the rate law given by Equation (E3-3.5) is not valid.
The one-half power in the rate law suggests that we might take the square root of Equation (E3-3.4):
[ K C 2 ] is in units of - (E3-3.7)
C D e
Rearranging gives
112 CDe
c,, c,,
- Kc2 - = 0 (E3-3.8) Using this new equilibrium constant, Kc, , we can formulate another suggestion for thle reaction rate expression:(E3-3.9) Note that thiis expression satisfies both the thermodynamic relationship (see ithe def- inition of K c 2 ) and reduces to the irreversible rate law when C, = 0. The Form of thie irreversible rate law provides a big clue as to the form of the reversible reaction rate expression.
3.1.5 Rlonelementary Rate Laws and Reactions
It is interesting to note that although the reaction orders correspond to the stoichiometric coefficients for the reaction between hydrogen and iodine, the rate expression for the reaction between hydrogen and another halogen, bro- mine, is quite complex. This nonelementary reaction
H 2 + B r 2
--+
2HIBrproceeds by a free-radical mechanism, and its reaction rate law is
(3-18) Another reaction involving free radicals is the vapor-phase decomposition of acetaldehyde:
CH3CHO CH,+CO
At a temperature of about 500"C, the order of the reaction is three-halves with respect to acetaldehyde.
82 Rate Laws and Stoichiometry Chap. 3
3/2
- ~ C H , C H O = ~CCH,CHO (3-19)
In many gas-solid catalyzed reactions it is sometimes preferable to write the rate law in terms of partial pressures rather than concentrations. One such example is the reversible catalytic decomposition of cumene, C, to form ben- zene, B, and propylene, P:
c6 H&H(CH3)2
e
c6 H6+
c3 H6The reaction can be written symbolically as
It was found experimentally that the reaction follows Langmuir-Hinshelwood kinetics and the rate law is (see Chapter 10)
(3-20) where
K,
is the pressure equilibrium constant with units of atm (or kPa);K,
and
K,
are the adsorption constants with units of atm-' (orWa');
and the specific reaction rate, k, has units ofmol cumene [kl = kg cat s atm
We see that at equilibrium ( - r & = 0) the rate law for the reversible reaction is indeed thermodynamically consistent:
Solving for K , yields
pBepPe P C e
K p =
-
which is identical to the expression obtained from thermodynamics.
To express the rate of decomposition of cumene -rk as a function of conversion, replace the partial pressure with concentration, using the ideal gas law:
P, = C,RT (3-21)
and then express concentration in terms of conversion.
The rate of reaction per unit weight of catalyst, - r a , and the r$e of reaction per unit volume, - r A , are related through the bulk density ph of the catalyst particles in the fluid media:
Sec. 3.i! Present Status of Our Approach to Reactor Sizing and Design 83
- r A = P b ( - r A )
(3-22) moles - - -
[
mass1 [ --
moles1
time e volume volume time a mass
In fluidized catalytic beds the bulk density is normally a function of the Bow rate thirough the bed.
3.2 Present (Status
of
Our Approach to Reactor Sizing and DesignIn Chapter 2 we showed how it was possible to size CSTRs, PFRs, and PBRs using the design equations in Table 3-1
if
the,.rate of disappearance of '4 is known as a function of conversion, X :- r A = g
TABLE 3-1. DESIGN EQUATIONS
Differential Algebraic Integral
Form Form Form
d X
Batch NAo .- = -rAV (2-6) dt
Backmix.
The design (CSTR) equations
-rA. = f(C,)
c,
= h, (XII +
-'A = g ( x ) and then we can design isothermal reactors
Packed bed d X
(PBN FAO dW -- = -ra (2-17)
(2-9) t = NAo i o - dX
-rAV
dX o -rA
V = F A 0
1
- (2-16)dX o -ra
w
= F A , / - (2- 18)In general, information in the form -rA = g ( X ) is not available. However, we have seen in Section 3.1 that the rate of disappearance of A, - r A , is normally expressed in terms of the concentration of the reacting species. This functionality, ( 3 - 1) is called a rate lmv. In Section 3.3 we show how the concentration of the react- ing species may be written in terms of the conversion X ,
-rA = k[fn(C,,C,,
...)I
c,
= h, ( X )With thlese additiional relationships, one observes that if the rate law is given and the concentrations can be expressed as a function of conversion, then in fact we have -rli as a function of X and this is all that is needed to evaluate the design equations. One can use either the numerical techniques described in Chapter 2, or, as we shall see in Chapter 4, a table of integrals.
84 Rate Laws and Stoichiometry Chap. 3
3.3
Stoichiometric TableNow that we have shown how the rate law can be expressed as a function of concentrations, we need only express concentration as a function of conversion in order to carry out calculations similar to those presented in Chapter 2 to size reactors. If the rate law depends on more than one species, we must relate the concentrations of the different species to each other. This relationship is most easily established with the aid of a stoichiometric table. This table presents the stoichiometric relationships between reacting molecules for a single reaction.
That is, it tells us how many molecules of one species will be formed during a chemical reaction when a given number of molecules of another species disap- pears. These relationships will be developed for the general reaction
a A + b B
e
c C + d D (2- 1) Recall that we have already used stoichiometry to relate the relative rates of reaction for Equation (2-1):4
rA ‘B ‘C - ‘D
_ -
- - - -’ a -b c d (2-20)
In formulating our stoichiometric table we shall take species A as our basis of calculation (i.e,, limiting reactant) and then divide through by the sto- ichiometric coefficient of A,
(2-2)
b c d
U u a
A + - B
+
- C + - D in order to put everything on a basis of “per mole of A.”give the change in the number of moles of each species (Le., A, B, C , and D).
Next, we develop the stoichiometric relationships for reacting species that
3.3.1 Batch Systems
Figure 3-1 shows a batch system in which we will carry out the reaction given by Equation (2-2). At time t = 0 we will open the reactor and place a number of moles of species A, B, C, D, and I ( N A o , N,, , N c o , N,,, and N , , respectively) into the reactor.
Species A is our basis of calculation and NAo is the number of moles of A initially present in the reactor. Of these, N A , X moles of A are consumed in the system as a result of the chemical reaction, leaving (NAo - N A o X ) moles of A in the system. That is, the number of moles of A remaining in the reactor after conversion X has been achieved is
N A = N A , - NAOX = N A o (1 - X )
The complete stoichiometric table for the reaction shown in Equation (2-2) taking place in a batch reactor is presented in Table 3-2.
Sec. 3.3 Stoichiometric Table a5
NOD
t=o
Figure 3-1 Batch reactor.
Tal determine the number of moles of each species remaining after NA,oX moles of A have reacted, we form the stoichiometric table (Table 3-2). This stoichiometric table presents the following information:
Column I : the particular species
Column 2: the number of moles of each species initially present
Column 3: the change in the number of moles brought about by reaction Column 4: the number of moles remaining in the system at time t Components of the
stoichiometric table
Totals NTO N T = N T o +
k : :
- + - - - - 1)
NAoX86 Rate Laws and Stoichiometry Chap. 3
To calculate the number of moles of species B remaining at time t we recall that at time t the number of moles of A that have reacted is NAOX. For every mole of A that reacts, b/a moles of B must react; therefore, the total number of moles of B that have reacted is
moles B reacted
moles A reacted moles A reacted moles B reacted =
Because B is disappearing from the system, the sign of the “change” is nega- tive. N,, is the number of moles initially in the system. Therefore, the number of moles of B remaining in the system, N , , is given in the last column of Table 3-2 as
The complete stoichiometric table delineated in Table 3-2 is for all species in the reaction
(2-2)
b c d
A + - B a
+
a- C + - D aThe stoichiometric coefficients in parentheses (dla
+
c/a - b/a - 1) repre-sent the increase in the total number of moles per mole of A reacted. Because this term occurs often in our calculations it is given the symbol 6:
I I
(3-23)
I I
The parameter 6 tells us the change io the total number of moles per mole of A reacted. The total number of moles can now be calculated from the equation
N T = NTo
+
SNAOXWe recall from Chapter 1 that the kinetic rate law (e.g., - r A = kC2) is a function solely of the intensive properties of the reacting materials (e.g., tem- perature, pressure, concentration, and catalysts, if any). The reaction rate,
-
r A ,usually depends on the concentration of the reacting sbecies raised to some power. Consequently, to determine the reaction rate as a function of conversion X , we need to know the concentrations of the reacting species as a function of conversion.
We want
=
The concentration of A is the number of moles of A per unit volume:
Batch concentration
After writing similar equations for B, C, and D, we use the stoichiometric table to express the concentration of each component in terms of the conver- sion X
Sec. 3.3 Stoichiometric Table
I
87(3-24)
We further simplify these equations by defining the parameter
O i
, which allows us to factor N,, in each of the expressions,for concentration:@ - - NBO NAO
B - (3-25)
We need V ( X ) to
obtain C, h , ( X ) species concentration as a function of conversion.
We now need only to finc volume as a function of conversion to c,tain the
3,,3.2 Constant-Volume Reaction Systems
Some significant simplifications in the reactor design equations are pos- sible when the reacting system undergoes no change in volume as the reaction progresses. These systems are called constant-volume, or constant-density, because of the invariance of either volume or density during the reaction pro- cess. This situation may arise from several causes. In gas-phase batch systems, the reactor is usually a sealed vessel with appropriate instruments to measure pressure and temperature within the reactor. The volume within this vessel is fixed and will not change, and is therefore a constant-volume system. The lab- oratory bomb reactor is a typical example of this type of reactor.
Another example of a constant-volume gas-phase isothermal reaction occurs when the number of moles of product equals the number of moles of reactant. The water-gas shift reaction, important in coal gasification and many other processes, is one of these:
CO
+
H 2 0e
C 0 2+
H,88 Rate Laws and Stoichiometry Chap. 3
Concentration as a function of conversion when no volume change occurs with reaction
In this reaction, 2 mol of reactant forms 2 mol of product. When the number of reactant molecules forms an equal number of product molecules at the same temperature and pressure, the volume of the reacting mixture will not change if the conditions are such that the ideal gas law is applicable, or if the com- pressibility factors of the products and reactants are approximately equal.
For liquid-phase reactions taking place in solution, the solvent usually dominates the situation. As a result, changes in the density of the solute do not affect the overall density of the solution significantly and therefore it is essen- tially a constant-volume reaction process. Most liquid-phase organic reactions do not change density during the reaction, and represent still another case to which the constant-volume simplifications apply. An important exception to this general rule exists for polymerization processes.
For the constant-volume systems described above, Equation (3-25) can be simplified to give the following expressions relating concentration and con- version:
v = vo
c,
=c,, ( a , + - x 3
Example 3-4 Expressing = hj(X) for a Liquid-Phase Reaction
(3-26)
Soap consists of the sodium add potassium salts of various fatty acids such as oleic, stearic, palmitic, lauric, and myristic acids. The saponification for the formation of soap from aqueous caustic soda and glyceryl stearate is
3NaOH(aq)
+
(C,7H3SCOO)3C3 H, --+ 3C,, H3,C0ONa + C3 %(OW3 Letting X represent the conversion of sodium hydroxide (the moles of sodium hydroxide reacted per mole of sodium hydroxide initially present), set up a stoichi- ometric table expressing the concentration of each species in terms of its initial con- centration and the conversion X.Solution
Because we are taking sodium hydroxide as our basis, we divide through by the sto- ichiometric coefficient of sodium hydroxide to put the reaction expression in the form
Choosing a basis of calculation
Stoichiometric . table (batch)
~
See. 3.3 Stoichiclmetric Table
NaOH
+
(C17 H,,COO),C, H, _ _ j C,,H,,COONa+
C, H,(OH),A + J B 3
*
C+
f DWe may then perfom the calculations shown in Table E3-4.1. Because this is a liq- uid-phase reaction, the density p is considered to be constant; therefore, V = V,>.
TABLE E3-4.1. STOICHIOMETRIC TABLE FOR LIQUID-PHASE SOAP REACTION
Species Symbol Initially Change Remaining Concentration NaOlH
Example 3-5 What Is the Limiting Reactant?
Having set up thle stoichiometric table in Example 3-4, one can now readily use it to calcullate the concentrations at a given conversion. If the initial mixture consists solely of sodium hydroxide at a concentration of 10 mol/L (Le., 10 mol/dm3 or 10 kmoil’m3 ) and of glyceryl stearate at a concentration of 2 g mol/L, what is the con- centration of glycerine when the conversion of sodium hydroxide is (a) 20% ,and (b) 90%?
Solution
Only the reactants NaOH and (CI7 H3,COO),C, H, are initially present; therefore, (a) For 20% conlversion:
0, = 0, = 0.
CD = CAo(:] = (10)k= 0.67 ) g moVL = 0.67 mol/dm3
= lO(0.133) = 1.33moVdm3
90 Rate Laws and Stoichiometry Chap. 3
The basis of calculation should be the limiting reactant
I
(b) For 90% conversion:Negative concentration-impossible!
Ninety percent conversion of NaOH is not possible, because glyceryl stearate is the limiting reactant. Consequently, all the glyceryl stearate is used up before 90% of the NaOH could be reacted. It is important to choose the limiting reactant as the basis of calculation.
Let US find CB:
C - 10 - -- = lO(O.2-0.3) = - 1 mol/dm3
B - [l?o O:)