If you are thinking about someone while reading this book, you are most definitely in love.
—Tim Newberger Undergraduate ChE Student W2013
Now 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 and sequence 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 single reactions. 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 disappears. These relationships will be developed for the general reaction
$\begin{array}{cc}aA+bB\leftrightarrows cC+dD& \left(\mathrm{21}\right)\end{array}$
Recall that we have already used stoichiometry to relate the relative rates of reaction for Equation (21):
This stoichiometric relationship relating reaction rates will be used in Chapters 6 and 8.
In formulating our stoichiometric table, we shall take species A as our basis of calculation (i.e., the limiting reactant) and then divide through by the stoichiometric coefficient of A
$\begin{array}{cc}\text{A}+\frac{b}{a}\text{B}\to \frac{c}{a}\text{C}+\frac{d}{a}\text{D}& \left(\mathrm{22}\right)\end{array}$
in order to put everything on a basis of “per mole of A.”
Next, we develop the stoichiometric relationships for reacting species that give the change in the number of moles of each species (i.e., A, B, C, and D).
Batch reactors (BRs) are primarily used for the production of specialty chemicals and to obtain reaction rate data in order to determine reactionrate laws and ratelaw parameters such as k, the specific reaction rate.
Figure 41 shows a starving artist’s rendition of a batch system in which we will carry out the reaction given by Equation (22). At time t = 0, we will open the reactor and place a number of moles of species A, B, C, and D, and inerts I (N_{A0}, N_{B0}, N_{C0}, N_{D0}, and N_{I0}, respectively) into the reactor and then of course close the port hole.
Species A is our basis of calculation, and N_{A0} is the number of moles of A initially present in the reactor. After a time t, N_{A0}X moles of A are consumed in the system as a result of the chemical reaction, leaving (N_{A0} – N_{A0}X) moles of A in the system. That is, the number of moles of A remaining in the reactor after a conversion X has been achieved is
N_{A} = N_{A0} – N_{A0}X = N_{A0}(1 – X)
We now will use conversion in a similar fashion to express the number of moles of B, C, and D in terms of conversion.
To determine the number of moles of each species remaining after N_{A0}X moles of A have reacted, we form the stoichiometric table (Table 41). This stoichiometric table presents the following information:
TABLE 41 STOICHIOMETRIC TABLE FOR A BATCH SYSTEM
Species 
Initially (mol) 
Change (mol) 
Remaining (mol) 
A 
N_{A0} 
−(N_{A0}X) 
N_{A} = N_{A} N_{A0} – N_{A0}X 
B 
N_{B0} 
$\frac{b}{a}\left({N}_{\text{A0}}X\right)$ 
${N}_{\text{B}}={N}_{\text{B}0}\frac{b}{a}{N}_{\text{A0}}X$ 
C 
N_{C0} 
$\frac{c}{a}\left({N}_{\text{A0}}X\right)$ 
${N}_{\text{C}}={N}_{\text{C}0}+\frac{c}{a}{N}_{\text{A0}}X$ 
D 
N_{D0} 
$\frac{d}{a}\left({N}_{\text{A0}}X\right)$ 
${N}_{\text{D}}={N}_{\text{D}0}+\frac{d}{a}{N}_{\text{A0}}X$ 
I (inerts) 
N_{I0} 
– 
N_{I} = N_{I0} 
Totals 
N_{T0} 
${N}_{\text{T}}={N}_{\text{T0}}+\underset{\delta}{\underbrace{\left(\frac{d}{a}+\frac{c}{a}\frac{b}{a}1\right)}}{N}_{\text{A0}}X$ 
Batch Reactor
Column 1: the species in the reaction
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
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 N_{A0}X. 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
$\begin{array}{c}\text{Moles}\text{Breacted}=\frac{\text{Moles}\text{Breacted}}{\text{Moles}\text{}\text{A}\text{reacted}}\cdot \text{Moles}\text{Areacted}\\ \hfill =\frac{b}{a}\left({N}_{\text{A0}}X\right)\hfill \end{array}$
Because B is disappearing from the system, the sign of the “change” is negative. N_{B0} is the number of moles of B initially in the system. Therefore, the number of moles of B remaining in the system, N_{B}, at a time t, is the number of moles of B initially minus the moles of B that have reacted and is given in the last column of Table 41 as
${N}_{\text{B}}={N}_{\text{B0}}\frac{b}{a}{N}_{\text{A0}}X$
The complete stoichiometric table delineated in Table 41 is for all species in the general reaction
Let’s take a look at the totals in the last column of Table 41. The stoichiometric coefficients in parentheses $\begin{array}{c}\hline (d/a+c/ab/a1)\\ \hline\end{array}$ represent the change in the total number of moles per mole of A reacted. Because this term occurs so often in our calculations, it is given the symbol δ:
The parameter δ
Definition of δ
$\overline{)\delta =\frac{\text{Changeinthetotalnumberofmoles}}{\text{MolesofAreacted}}}$
The total number of moles can now be calculated from the equation
N_{T} = N_{T0} + δ(N_{A0}X)
We recall from Chapters 1 and 3 that the kinetic rate law (e.g., ${r}_{\text{A}}=k{C}_{\text{A}}^{2}$) is a function solely of the intensive properties of the reacting system (e.g., temperature, pressure, concentration, and catalysts, if any). The reaction rate, –r_{A}, usually depends on the concentration of the reacting species 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, X. Let’s do it!
We want C_{j} = h_{j} (X)
The concentration of A is the number of moles of A per unit volume
$\text{}{C}_{A}=\frac{{N}_{\text{A}}}{V}$
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 conversion X:
$\begin{array}{cc}{C}_{\text{A}}=\frac{{N}_{\text{A}}}{V}=\frac{{N}_{\text{A0}}(1X)}{V}& \text{(42)}\end{array}$
$\begin{array}{cc}{C}_{B}=\frac{{N}_{B}}{V}=\frac{{N}_{B\text{0}}(b/a){N}_{\text{A0}}X}{V}& \text{(43)}\end{array}$
$\begin{array}{cc}{C}_{C}=\frac{{N}_{C}}{V}=\frac{{N}_{C\text{0}}+(c/a){N}_{\text{A0}}X}{V}& \text{(44)}\end{array}$
$\begin{array}{cc}{C}_{D}=\frac{{N}_{D}}{V}=\frac{{N}_{D\text{0}}+(d/a){N}_{\text{A0}}X}{V}& \text{(45)}\end{array}$
Because almost all batch reactors are solid vessels, the reactor volume is constant, so we can take V =V_{0}, then
$\begin{array}{c}{C}_{\text{A}}=\frac{{N}_{\text{A}}}{{V}_{0}}=\frac{{N}_{\text{A0}}(1X)}{{V}_{0}}\end{array}$
Constant Volume, V = V_{0}
We will soon see that Equation (46) for constantvolume batch reactors also applies to continuousflow liquidphase systems.
We further simplify these equations by defining the parameter Θ_{i}, which allows us to factor out N_{A0} in each of the expressions for concentration
$\begin{array}{c}\overline{){\text{\Theta}}_{i}=\frac{{N}_{i0}}{{N}_{\text{A0}}}=\frac{{C}_{i0}}{{C}_{\text{A0}}}=\frac{{y}_{i0}}{{y}_{\text{A0}}}},\\ \overline{){\text{\Theta}}_{i}=\frac{\text{Molesofspecies}\u201c\text{i}\u201d\text{initially}}{\text{MolesofspeciesAinitially}}}\end{array}$
${C}_{\text{B}}=\frac{{N}_{\text{A0}}[{N}_{\text{B}\text{0}}/{N}_{\text{A0}}(b/a)X]}{{V}_{0}}=\frac{{N}_{\text{A0}}[{\mathrm{\Theta}}_{\text{B}}(b/a)X]}{{V}_{0}}$
with ${\mathrm{\Theta}}_{\text{B}}=\frac{{N}_{\text{B0}}}{{N}_{\text{A}\text{0}}}$
Feed
Equalmolar:
Θ_{B} = 1
Stoichiometric:
${\mathrm{\Theta}}_{\text{B}}=\frac{b}{a}$
for an equalmolar feed Θ_{B} = 1 and for a stoichiometric feed Θ_{B} = b/a.
Continuing for species C and D
$\begin{array}{cc}{C}_{\text{C}}=\frac{{N}_{\text{A0}}[{\mathrm{\Theta}}_{C}+(c/a\left)X\right]}{{V}_{0}}& \\ {C}_{\text{C}}={C}_{\text{A0}}({\mathrm{\Theta}}_{\text{C}}+\frac{c}{a}X)& \text{(48)}\end{array}$
$\text{with}{\mathrm{\Theta}}_{C}=\frac{{N}_{\text{C0}}}{{N}_{\text{A0}}}$
Concentration for constantvolume batch reactors and for liquidphase continuousflow reactors
$\begin{array}{cc}{C}_{\text{D}}=\frac{{N}_{\text{A0}}[{\mathrm{\Theta}}_{\text{D}}+(d/a\left)X\right]}{{V}_{0}}& \\ {C}_{\text{D}}={C}_{\text{A0}}({\mathrm{\Theta}}_{\text{D}}+\frac{d}{a}X)& \text{(49)}\end{array}$
$\text{with}{\mathrm{\Theta}}_{\text{D}}=\frac{{N}_{\text{D}\text{0}}}{{N}_{\text{A0}}}$
For constantvolume batch reactors, for example, steel containers V = V_{0}, we now have concentration as a function of conversion. If we know the rate law, we can now obtain –r_{A} = f(X) to couple with the differential mole balance in terms of conversion in order to solve for the reaction time, t, to achieve a specified conversion, X.
For liquidphase reactions taking place in solution, the solvent usually dominates the situation. For example, most liquidphase organic reactions do not change density during the course of the reaction and represent still another case for which the constantvolume simplifications apply. As a result, changes in the density of the solute do not affect the overall density of the solution significantly and therefore it is essentially a constantvolume reaction process, V = V_{0} and υ = υ_{0}. Consequently, Equations (46)–(49) can be used for liquidphase reactions as well as constantvolume (V = V_{0}) gasphase reactions. An important exception to this general rule exists for some polymerization processes.
For liquids V = V_{0} and υ = υ_{0}
To summarize for constantvolume batch systems and for liquidphase reactions, we can use a rate law for Reaction (22) such as –r_{A} = k_{A}C_{A}C_{B} to obtain = f(X); that is,
${r}_{\text{A}}=k{C}_{\text{A}}{C}_{\text{B}}=k{C}_{\text{A0}}^{2}\text{(}1X\text{)}({\mathrm{\Theta}}_{\text{B}}\frac{b}{a}X)=f\text{(}X\text{)}$
Substituting for the given parameters k, C_{A0}, and Θ_{B}, we can now use the techniques in Chapter 2 to size the CSTRs and PFRs for liquidphase reactions.
Soap is necessary to clean such things as dirty softball and soccer players’ uniforms, dirty hands, and foul mouths. Soap consists of the sodium and potassium salts of various fatty acids, such as oleic, stearic, palmitic, lauric, and myristic acids. The saponification reaction for the formation of soap from aqueous caustic soda and glyceryl stearate is
3NaOH(aq) (C_{17} H_{35}COO)_{3}C_{3} H_{5} → 3C_{17} H_{35}COONa + C_{3} H_{5}(OH)_{3}
Letting X represent the conversion of sodium hydroxide (the moles of sodium hydroxide reacted per mole of sodium hydroxide initially present), set up a stoichiometric table expressing the concentration of each species in terms of its initial concentration and the conversion, X.
Solution
Because we have taken sodium hydroxide as our basis of calculation, we divide through by the stoichiometric coefficient of sodium hydroxide to put the reaction expression in the form
$\begin{array}{c}\text{NaOH}+\frac{1}{3}\text{(}{\text{C}}_{17}\text{}{\text{H}}_{35}{\text{COO)}}_{3}{\text{C}}_{3}{\text{H}}_{5}\to \text{}{\text{C}}_{17}{\text{H}}_{35}\text{COONa}+\frac{1}{3}{\text{C}}_{3}{\text{H}}_{5}\text{(OH}{\text{)}}_{3}\\ \end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{A}& +\end{array}& \frac{1}{3}\end{array}\text{B}& \to \end{array}& \text{C}\end{array}& +\end{array}& \frac{1}{3}\text{D}\end{array}$
Choosing a basis of calculation
We may next perform the calculations shown in Table E41.1. Because this reaction is a liquidphase reaction, the density ρ is considered to be constant; therefore, V = V_{0}
$\begin{array}{c}{C}_{\text{A}}=\frac{\text{}{N}_{\text{A}}}{V}=\frac{{N}_{\text{A}}}{{V}_{0}}=\frac{{N}_{\text{A0}}\text{(1}X)}{\text{}{V}_{0}}={C}_{\text{A0}}\text{(1}X\text{)}\\ {\mathrm{\Theta}}_{B}=\frac{{C}_{\text{B0}}}{{C}_{\text{A0}}}\text{\hspace{0.17em}}{\mathrm{\Theta}}_{C}=\frac{\text{}{C}_{C0}}{{C}_{\text{A0}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{\Theta}}_{D}=\frac{{C}_{\text{D0}}}{{C}_{\text{A0}}}\hfill \end{array}$
TABLE E41.1 STOICHIOMETRIC TABLE FOR LIQUIDPHASE SOAP REACTION
Species 
Symbol 
Initially 
Change 
Remaining 
Concentration 
NaOH 
A 
N_{A0} 
–N_{A0} X 
N_{A0} (1 – X) 
C_{A0} (1 – X) 
(C_{17} H_{35}COO)_{3}C_{3} H_{5} 
B 
N_{B0} 
$\frac{1}{3}{N}_{\text{A0}}X$ 
${N}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{X}{3})$ 
${C}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{X}{3})$ 
C_{17} H_{35}COONa 
C 
N_{C0} 
N_{A0} X 
N_{A0}(Θ_{C} + X) 
C_{A0}(Θ_{C} + X 
C_{3} H_{5}(OH)_{3} 
D 
N_{D0} 
$\frac{1}{3}{N}_{\text{A0}}X$ 
${N}_{\text{A0}}({\mathrm{\Theta}}_{\text{D}}+\frac{X}{3})$ 
${C}_{\text{A0}}({\mathrm{\Theta}}_{\text{D}}+\frac{X}{3})$ 
Water (inert) 
I 
N_{10} 
– 
N_{10} 
C_{I0} 
Totals 
N_{T0} 
0 
N_{T} = N_{T0} 
Stoichiometric table (batch)
Analysis: The purpose of this example was to show how the generic reaction in Table 41 is applied to a real reaction and how to develop the corresponding stoichiometric table.
Having set up the stoichiometric table in Example 41, one can now readily use it to calculate the concentrations at a given conversion. If the initial mixture consists of sodium hydroxide at a concentration of 10 mol/dm^{3} (i.e., 10 mol/L or 10 kmol/m^{3}) and glyceryl stearate at a concentration of 2 mol/dm^{3}, what are the concentrations of glycerol stearate, B, and of glycerine, D, when the conversion of sodium hydroxide is (a) 20% and (b) 90%?
Glyceryl Stearate (Chapter 2 AWFOS–S2)
Solution
Only the reactants NaOH and (C_{17} H_{35}COO)_{3}C_{3} H_{5} are initially present; therefore, Θ_{C} = Θ_{D} = 0.
For 20% conversion of NaOH
$\begin{array}{c}{C}_{\text{D}}={C}_{\text{A0}}\left(\frac{X}{3}\right)=\text{(10)}\left(\frac{0.2}{3}\right)=0.67\text{mol/L =}\text{}\text{0.67}\text{}\text{mol/d}{\text{m}}^{3}\hfill \\ {C}_{B}={C}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{X}{3})=10(\frac{2}{10}\frac{0.2}{3})=10\text{(0.133) = 1.33 mol/d}{\text{m}}^{3}\hfill \end{array}$
For 90% conversion of NaOH
${C}_{\text{D}}={C}_{\text{A0}}\left(\frac{X}{3}\right)=10\left(\frac{0.9}{3}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\text{mol/d}{\text{m}}^{3}$
Let us find C_{B}
${C}_{\text{B}}=10(\frac{2}{10}\frac{0.9}{3})=10\text{(0.2}\text{0.3) =}1\text{mol/d}{\text{m}}^{3}$
Oops!! Negative concentration—impossible! What went wrong?
Analysis: We purposely chose the wrong basis of calculation to make the point that we must choose the limiting reactant as our basis of calculation!! Ninetypercent conversion of NaOH is not possible because glyceryl stearate is the limiting reactant and is used up before 90% of the NaOH can be reacted. Glyceryl stearate should have been our basis of calculation and therefore we should not have divided the reaction as written by the stoichiometric coefficient of 3. It’s always important to make sure you learn something from a mistake, such as choosing the wrong basis of calculation. There is no such thing as failure, unless one fails to learn from it.
The basis of calculation must be the limiting reactant.
#LifeLesson!
The form of the stoichiometric table for a continuousflow system (see Figure 42) is virtually identical to that for a batch system (Table 41), except that we replace N_{j}_{0} by F_{j}_{0} and N_{j} by F_{j} (Table 42). Again taking A as the basis, we divide Equation (21) through by the stoichiometric coefficient of A to obtain
$\begin{array}{cc}A+\frac{b}{a}B\to \frac{c}{a}C+\frac{d}{a}D& \text{(22)}\end{array}$
TABLE 42 STOICHIOMETRIC TABLE FOR A FLOW SYSTEM
Species 
Feed Rate to Reactor (mol/time) 
Change within Reactor (mol/time) 
Effluent Rate from Reactor (mol/time) 
A 
F_{A0} 
–F_{A0}X 
F_{A} = F_{A0} (1 – X) 
B 
F_{B0} = Θ_{B}F_{A0} 
$\frac{b}{a}{F}_{\text{A0}}X$ 
${F}_{B}={F}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{b}{a}X)$ 
C 
F_{C0} = Θ_{C}F_{A0} 
$\frac{c}{a}{F}_{\text{A0}}X$ 
${F}_{C}={F}_{\text{A0}}({\mathrm{\Theta}}_{\text{C}}+\frac{c}{a}X)$ 
D 
F_{D0} = Θ_{D}F_{A0} 
$\frac{d}{a}{F}_{\text{A0}}X$ 
${F}_{\text{D}}={F}_{\text{A0}}({\mathrm{\Theta}}_{\text{D}}+\frac{d}{a}X)$ 
I 
F_{I0}= Θ_{I}F_{A0} 
– 
F_{I} = F_{A0}θ_{I} 
Totals 
F_{T0} 
${F}_{\text{T}}={F}_{\text{T0}}+(\frac{d}{a}+\frac{c}{a}\frac{b}{a}1){F}_{\text{A0}}X$ 

F_{T} = F_{T0} + δF_{A0} X 
Stoichiometric table for a flow system
where
$\overline{){\text{\Theta}}_{\text{B}}=\frac{{F}_{\text{B0}}}{{F}_{\text{A0}}}=\frac{{C}_{\text{B0}}{\upsilon}_{0}}{{C}_{\text{A0}}{\upsilon}_{0}}=\frac{{C}_{\text{B0}}}{{C}_{\text{A0}}}=\frac{{y}_{\text{B0}}}{{y}_{\text{A0}}}}$
and Θ_{C}, Θ_{D}, and Θ_{I} are defined similarly
and, as before,
For a flow system, the concentration C_{A} at a given point can be determined from the molar flow rate F_{A} and the volumetric flow rate υ at that point:
Definition of concentration for a flow system
Units of υ are typically given in terms of liters per second, cubic decimeters per second, or cubic feet per minute. We now can write the concentrations of A, B, C, and D for the general reaction given by Equation (22) in terms of their respective entering molar flow rates (F_{A0}, F_{B0}, F_{C0}, F_{D0}), the conversion, X, and the volumetric flow rate, υ.
$\begin{array}{ccc}{C}_{\text{A}}=\frac{{F}_{\text{A}}}{\upsilon}=\frac{{F}_{\text{A0}}}{\upsilon}\text{(1}X\text{)}\hfill & {C}_{\text{B}}=\frac{{F}_{B}}{\upsilon}=\frac{{F}_{\text{B0}}(b/a){F}_{\text{A0}}X}{\upsilon}& \\ {C}_{\text{C}}=\frac{{F}_{\text{C}}}{\upsilon}=\frac{{F}_{\text{C}\text{0}}+(c/a){F}_{\text{A0}}X}{\upsilon}& {C}_{D}=\frac{{F}_{\text{D}}}{\upsilon}=\frac{{F}_{\text{D}\text{0}}(d/a){F}_{\text{A0}}X}{\upsilon}& \text{(411)}\end{array}$
We now focus on determining the volumetric flow rate, υ.
For liquids, the fluid volume change with reaction is negligible when no phase changes are taking place. Consequently, we can take
υ = υ_{0}
Then
For liquids
C_{A} = C_{A0}(1 – X)
${C}_{B}={C}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{b}{a}X)$
Therefore, for a given rate law we have –r_{A} = g(X)
and so forth, for C_{C} and C_{D}.
Consequently, using any one of the rate laws in Chapter 3, we can now find –r_{A} = f(X) for liquidphase reactions. However, for gasphase reactions the volumetric flow rate most often changes during the course of the reaction because of a change in the total number of moles or a change in temperature or pressure. Hence, one cannot always use Equation (413) to express concentration as a function of conversion for gasphase reactions. No worries. In the next section we will sort this out.
In our previous discussions, we considered primarily systems in which the reaction volume or volumetric flow rate did not vary as the reaction progressed. Most batch and liquidphase and some gasphase systems fall into this category. There are other systems, though, in which either V or υ does vary, and these will now be considered.
A situation where one encounters a varying flow rate occurs quite frequently in gasphase reactions that do not have an equal number of product and reactant moles. For example, in the synthesis of ammonia
N_{2} + 3H_{2} ⇄ 2NH_{3}
4 mol of reactants gives 2 mol of product. In flow systems where this type of reaction occurs, the molar flow rate will be changing as the reaction progresses. Because equal numbers of moles occupy equal volumes in the gas phase at the same temperature and pressure, the volumetric flow rate will also change.
#We’reGood The same stoichiometric table holds for both liquidphase and gasphase reactions.
In the stoichiometric tables presented on the preceding pages, it was not necessary to make assumptions concerning a volume change in the first four columns of the table (i.e., the species, initial number of moles or molar feed rate, change within the reactor, and the remaining number of moles or the molar effluent rate). All of these columns of the stoichiometric table are independent of the volume or density, and they are identical for constantvolume (constantdensity) and varyingvolume (varyingdensity) situations. Only when concentration is expressed as a function of conversion does variable density enter the picture.
Flow Reactors with Variable Volumetric Flow Rate. To derive the concentrations of each species in terms of conversion for a gasphase flow system, we shall start by using the relationships for the total concentration. The total concentration, C_{T}, at any point in the reactor is the total molar flow rate, F_{T}, divided by the volumetric flow rate, υ (cf. Equation (410)). In the gas phase, the total concentration is also found from the gas law, C_{T} = P/ZRT. Equating these two relationships gives
$\begin{array}{cc}{C}_{\text{T}}=\frac{{F}_{\text{T}}}{\upsilon}=\frac{P}{ZRT}& \text{(414)}\end{array}$
At the entrance to the reactor
$\begin{array}{cc}{C}_{\text{T0}}=\frac{{F}_{\text{T0}}}{{\upsilon}_{0}}=\frac{{P}_{0}}{{Z}_{0}R{T}_{0}}& \text{(415)}\end{array}$
Taking the ratio of Equation (414) to Equation (415) and assuming negligible changes in the compressibility factor occur, that is, Z ≅ Z_{0}, during the course of the reaction, we have upon rearrangement
Gasphase reactions
We can now express the concentration of species j for a flow system in terms of its flow rate, F_{j}, the temperature, T, and total pressure, P.
${C}_{j}=\frac{{F}_{j}}{\upsilon}=\frac{{F}_{j}}{{\upsilon}_{0}\left(\frac{{F}_{\text{T}}}{{F}_{\text{T}}0}\frac{{P}_{0}}{P}\frac{T}{{T}_{0}}\right)}=\left(\frac{{F}_{\text{T0}}}{{\upsilon}_{0}}\right)\left(\frac{{F}_{j}}{{F}_{\text{T}}}\right)\left(\frac{P}{{P}_{0}}\right)\left(\frac{{T}_{0}}{T}\right)$
Use this gasphase concentration equation for membrane reactors (Chapter 6) and for multiple reactions (Chapter 8).
The total molar flow rate is just the sum of the molar flow rates of each of the species in the system and is
$\begin{array}{cc}\begin{array}{cc}{F}_{\text{T}}={F}_{\text{A}}+{F}_{\text{B}}+{F}_{\text{C}}+{F}_{\text{D}}+{F}_{\text{I}}+\mathrm{...}=\underset{j=1}{\overset{\text{n}}{\mathrm{\Sigma}}}& {F}_{j}\end{array}& \text{(418)}\end{array}$
We can also write Equation (417) in terms of the mole fraction of species j, y_{j}, and the pressure ratio, p, with respect to the initial or entering conditions, that is, sub “0”
$\begin{array}{c}{y}_{i}=\frac{{F}_{j}}{{F}_{T}}\\ p=\frac{P}{{P}_{0}}\end{array}$
The molar flow rates, F_{j}, are found by solving the mole balance equations. The concentration given by Equation (417) will be used for measures other than conversion when we discuss membrane reactors (Chapter 6) and multiple gasphase reactions (Chapter 8).
Now, let’s express the concentration in terms of conversion for gas flow systems. From Table 42, the total molar flow rate can be written in terms of conversion and is
F_{T} = F_{T0} + F_{A0} δX
We divide this equation through by F_{T0}
$\frac{{F}_{\text{T}}}{{F}_{\text{T0}}}=1+\frac{{F}_{\text{A0}}}{{F}_{\text{T0}}}\delta X=1+\stackrel{\epsilon}{\overbrace{{y}_{\text{A0}}\delta}}X$
Then
$\begin{array}{cc}\frac{{F}_{\text{T}}}{{F}_{\text{T0}}}=1+\epsilon X& \left(\mathrm{420}\right)\end{array}$
where y_{A0} is the mole fraction of A at the inlet (i.e., (F_{A0}/F_{T0})), and where δ is given by Equation (41) and ɛ is given by
$\epsilon =(\frac{d}{a}+\frac{c}{a}\frac{b}{a}1)\frac{{F}_{\text{A0}}}{{F}_{\text{T0}}}={y}_{\text{A0}}\delta $
Relationship between δ and ɛ
Equation (421) holds for both batch and flow systems. To interpret ɛ, let’s rearrange Equation (420) at complete conversion (i.e., X = 1 and F_{T} = F_{T}_{f})
$\epsilon =\frac{{F}_{\text{T}f}{F}_{\text{T0}}}{{F}_{\text{T0}}}$
Interpretation of ɛ
Substituting for (F_{T}/F_{T0}) in Equation (416) for the volumetric flow rate, υ, we have
Gasphase volumetric flow rate
The concentration of species j in a flow system is
$\begin{array}{cc}{C}_{j}=\frac{{F}_{j}}{v}& \left(\mathrm{424}\right)\end{array}$
The molar flow rate of species j is
F_{j} = F_{j}_{0} + v_{j} (F_{A0}X) = F_{A0}= (Θ_{j} + v_{j}X)
where v_{j} is the stoichiometric coefficient for species j, which is negative for reactants and positive for products. For example, for the reaction
$\begin{array}{cc}A+\frac{b}{a}B\to \frac{c}{a}C+\frac{d}{a}D& \left(\mathrm{22}\right)\end{array}$
ν_{A} = –1, ν_{B} = –b/a, ν_{C} = c/a, ν_{D} = d/a, and Θ_{j} = F_{j}_{0}/F_{A0}.
Substituting for the gas volumetric flow rate, υ, using Equation (423) and for F_{j}, we have
${C}_{j}=\frac{{F}_{\text{A0}}({\mathrm{\Theta}}_{j}+{\nu}_{j}X)}{{\upsilon}_{0}\left(\right(1+\epsilon X\left)\frac{{P}_{0}}{P}\frac{T}{{T}_{0}}\right)}$
Rearranging
Gasphase concentration as a function of conversion
Recall that y_{A0} = F_{A0}/F_{T0}, C_{A0} = y_{A0}C_{T0}, and ɛ is given by Equation (421) (i.e., ɛ = y_{A0}δ).
The stoichiometric table for the gasphase reaction (22) is given in Table 43.
TABLE 43 CONCENTRATIONS IN A VARIABLEVOLUME GAS FLOW SYSTEM
${C}_{\text{A}}=\frac{{F}_{\text{A}}}{\upsilon}=\frac{{F}_{\text{A0}}(1X)}{\upsilon}$ 
$=\frac{{F}_{\text{A0}}(1X)}{{\upsilon}_{0}(1\epsilon X)}\left(\frac{{T}_{0}}{T}\right)\frac{P}{{P}_{0}}$ 
$={C}_{\text{A0}}\left(\frac{1X}{1+\epsilon X}\right)\frac{{T}_{0}}{T}\left(\frac{P}{{P}_{0}}\right)$ 
${C}_{\text{B}}=\frac{{F}_{\text{B}}}{\upsilon}=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{B}}(b/a)X]}{\upsilon}$ 
$=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{B}}(b/a)X]}{{\upsilon}_{0}(1+\epsilon X)}\left(\frac{{T}_{0}}{T}\right)\frac{P}{{P}_{0}}$ 
$={C}_{\text{A0}}\left(\frac{{\mathrm{\Theta}}_{\text{B}}(b/a)X}{1+\epsilon X}\right)\frac{{T}_{0}}{T}\left(\frac{P}{{P}_{0}}\right)$ 
${C}_{\text{C}}=\frac{{F}_{\text{C}}}{\upsilon}=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{C}}+(c/a)X]}{\upsilon}$ 
$=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{C}}+(c/a)X]}{{\upsilon}_{0}(1+\epsilon X)}\left(\frac{{T}_{0}}{T}\right)\frac{P}{{P}_{0}}$ 
$={C}_{\text{A0}}\left(\frac{{\mathrm{\Theta}}_{\text{C}}+(c/a)X}{1+\epsilon X}\right)\frac{{T}_{0}}{T}\left(\frac{P}{{P}_{0}}\right)$ 
${C}_{\text{D}}=\frac{{F}_{\text{D}}}{\upsilon}=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{D}}+(d/a)X]}{\upsilon}$ 
$=\frac{{F}_{\text{A0}}[{\mathrm{\Theta}}_{\text{D}}+(d/a)X]}{{\upsilon}_{0}(1+\epsilon X)}\left(\frac{{T}_{0}}{T}\right)\frac{P}{{P}_{0}}$ 
$={C}_{\text{A0}}\left(\frac{{\mathrm{\Theta}}_{\text{D}}+(d/a)X}{1+\epsilon X}\right)\frac{{T}_{0}}{T}\left(\frac{P}{{P}_{0}}\right)$ 
${C}_{\text{I}}=\frac{{F}_{\text{I}}}{\upsilon}=\frac{{F}_{\text{A0}}{\mathrm{\Theta}}_{\text{I}}}{\upsilon}$ 
$=\frac{{F}_{\text{A0}}{\mathrm{\Theta}}_{\text{I}}}{{\upsilon}_{0}(1+\epsilon X)}\left(\frac{{T}_{0}}{T}\right)\frac{P}{{P}_{0}}$ 
$=\frac{{C}_{\text{A0}}{\mathrm{\Theta}}_{\text{I}}}{1+\epsilon X}\frac{{T}_{0}}{T}\left(\frac{P}{{P}_{0}}\right)$ 
At last!
We now have
C_{j} = h_{j} (X)
and
–r_{A} = g(X)
for variablevolume gasphase reactions.
One of the major objectives of this chapter is to learn how to express any given rate law –r_{A} as a function of conversion. The schematic diagram in Figure 43 helps to summarize our discussion on this point. The concentration of the reactant species B in the generic reaction is expressed as a function of conversion in both flow and batch systems for various conditions of temperature, pressure, and volume.
At last, by coupling Table 43 with the rate law for single gasphase reactions, we obtain what we needed and used in Chapter 2 to size and sequence flow reactors, that is,
–r_{A} = f(X)
A mixture of 28% SO_{2} and 72% air is charged to a flow reactor in which is oxidized.
2SO_{2} + O_{2} → 2SO_{3}
First, set up a stoichiometric table using only the symbols (i.e., Θ_{i}, F_{i} ).
Next, prepare a second table evaluating the species concentrations as a function of conversion for the case when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227°C.
Evaluate the parameters and make a plot of each of the concentrations SO_{2}, SO_{3}, N_{2} as a function of conversion.
Solution
Stoichiometric table. Taking SO_{2} as the basis of calculation, we divide the reaction through by the stoichiometric coefficient of our chosen basis of calculation
$\text{SO}{\text{}}_{2}+\text{\hspace{0.17em}}\frac{1}{2}{\text{O}}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\to {\text{SO}}_{3}$
The stoichiometric table is given in Table E43.1.
TABLE E43.1 STOICHIOMETRIC TABLE FOR $\text{SO}{\text{}}_{2}+\text{\hspace{0.17em}}\frac{1}{2}{\text{O}}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\to {\text{SO}}_{3}$
Species 
Symbol 
Initially 
Change 
Remaining 
SO_{2} 
A 
F_{A0} 
–F_{A0}X 
F_{A} F_{A0} (1 – X) 
O_{2} 
B 
F_{B0} = Θ_{B}F_{A0} 
$\frac{{F}_{\text{A0}}X}{2}$ 
${F}_{\text{B}}={F}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{1}{2}X)$ 
SO_{3} 
C 
0 
+F_{A0}X 
F_{C} = F_{A0}X 
N_{2} 
I 
F_{I0} = Θ_{I}F_{A0} 
– 
F_{I} = F_{I0} = Θ_{I}F_{A0} 
Totals 
F_{T0} 
${F}_{\text{T}}={F}_{\text{T0}}\frac{{F}_{\text{A0}}X}{2}$ 
Expressing the concentration as a function of conversion. From the definition of conversion, we substitute not only for the molar flow rate of SO_{2} (A) in terms of conversion but also for the volumetric flow rate as a function of conversion
${C}_{\text{A}}=\frac{{F}_{\text{A}}}{\upsilon}=\frac{{F}_{\text{A0}}(1X)}{\upsilon}$
Recalling Equation (423), we have
Neglecting pressure drop in the reaction, P = P_{0}, yields
$\upsilon ={\upsilon}_{0}(1+\epsilon X)\frac{T}{{T}_{0}}$
Neglecting pressure drop, P = P_{0}, that is, $p=\frac{p}{{p}_{0}}\approx 1$
If the reaction is also carried out isothermally, T = T_{0}, we obtain
υ = υ_{0}(1 + ɛX)
Isothermal operation, T = T_{0}
Similarly for B also with T = T_{0} and P = P_{0} (i.e., p = 1)
Parameter evaluation and plots of concentrations as a function of conversion. The inlet concentration of A is equal to the inlet mole fraction of A multiplied by the total inlet molar concentration. The total concentration can be calculated from an equation of state such as the ideal gas law. Recall that y_{A0} = 0.28, T_{0} = 500 K, and P_{0} = 1485 kPa.
$\overline{){C}_{\text{A0}}={y}_{\text{A0}}{C}_{\text{T0}}={y}_{\text{A0}}\left(\frac{{P}_{0}}{R{T}_{0}}\right)}$
$\begin{array}{c}=0.28\left[\frac{1485\text{}\text{kPa}}{8.314\text{}\text{kPa}\cdot {\text{dm}}^{3}/(\text{mol}\cdot \text{K})\times 500\text{}\text{K}}\right]\\ =\begin{array}{cc}0.1{\text{}\text{mol/dm}}^{3}& \underset{\xaf}{\begin{array}{c}\text{Answer}\end{array}}\end{array}\text{\hspace{0.17em}}\hfill \end{array}$
The total concentration at constant temperature and pressure is
${C}_{\text{T0}}=\frac{{P}_{\text{o}}}{R{T}_{\text{o}}}=\frac{1485\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{kpa}}{[8.314\text{\hspace{0.17em}}\text{kPa}\cdot {\text{dm}}^{3}/(\text{mol}\cdot \text{K})](500\text{\hspace{0.17em}}\text{K})}=0.357\frac{\text{mol}}{{\text{dm}}^{3}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E43.4}\right)$
We now evaluate ɛ.
$\varepsilon ={y}_{A0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta =\left(0.28\right)(11\frac{1}{2})=0.14\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E43.5}\right)$
Initially, 72% of the total number of moles is air containing 21% O_{2} and 79% N_{2}, along with 28% SO_{2}.
$\begin{array}{cc}\begin{array}{l}{F}_{\text{A0}}=\left(0.28\right)\left({F}_{\text{T0}}\right)\hfill \\ {F}_{\text{B0}}=\left(0.72\right)\left(0.21\right)\left({F}_{\text{T0}}\right)\hfill \\ {\text{\Theta}}_{B}=\frac{{F}_{\text{B}0}}{{F}_{\text{A0}}}=\frac{\left(0.72\right)\left(0.21\right)}{0.28}=0.54\hfill \\ {\text{\Theta}}_{I}=\frac{{F}_{\text{I}0}}{{F}_{\text{A0}}}=\frac{\left(0.72\right)\left(0.79\right)}{0.28}\hfill \end{array}& \underset{\_}{\text{Answer}}\end{array}$
Substituting for C_{A0} and ɛ in the species concentrations:
${\mathbf{\text{SO}}}_{2:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{\text{A}}={C}_{\text{A0}}\left(\frac{1X}{1+\u03f5X}\right)=0.1\left(\frac{1X}{\mathrm{1\u20130.14}X}\right){\text{mol/dm}}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E43.6}\right)$
The concentrations of different species at various conversions are calculated in Table E43.2 and plotted in Figure E43.1. Note that the concentration of is changing even though it is an inert species in this reaction!! #Seriously?
TABLE E43.2 CONCENTRATION AS A FUNCTION OF CONVERSION
C_{i} (mol/dm^{3}) 

Species 
X = 0.0 
X = 0.25 
X = 0.5 
X = 0.75 
X = 1.0 

SO_{2} 
C_{A} = 
0.100 
0.078 
0.054 
0.028 
0.000 
O_{2} 
C_{B} = 
0.054 
0.043 
0.031 
0.018 
0.005 
SO_{3} 
C_{C} = 
0.000 
0.026 
0.054 
0.084 
0.116 
N_{2} 
C_{I} = 
0.203 
0.210 
0.218 
0.227 
0.236 
Total 
C_{T} = 
0.357 
0.357 
0.357 
0.357 
0.357 
Note: Because the volumetic flow rate varies with conversion, υ = υ_{0}(1 – 0.14X), the concentration of inerts (N_{2}) is not constant.
Now we could use techniques presented in Chapter 2 to size reactors.
We are now in a position to express –r_{A} as a function of X and use the techniques in Chapter 2 to size and sequence reactors. However, we will use a better method to solve CRE problems, namely a Table of Integrals (Appendix A) or the Polymath, MATLAB, Wolfram, or Python software, discussed in the next chapter.
Analysis: In this example, we first formed a stoichiometric table in terms of molar flow rates. We then showed how to express the concentrations of each species in a gasphase reaction in which there is a change in the total number of moles. Next, we plotted each species concentration as a function of conversion and noted that the concentration of the inert, N_{2}, was not constant but increased with increasing conversion because of the decrease in the total molar flow rate, F_{T}, with conversion.
As previously mentioned many, if not most, rate laws for catalyst reactions are given in terms of partial pressures. Luckily, partial pressures are easily related to conversion with the aid of the ideal gas law and Equation (417).
${P}_{i}={C}_{i}RT={C}_{\text{T0}}\frac{{F}_{i}}{{F}_{T}}\stackrel{p}{\overbrace{\left(\frac{P}{{P}_{0}}\right)}}\frac{{T}_{0}}{T}RT=\stackrel{{P}_{\text{T}0}}{\overbrace{{C}_{\text{T0}}R{T}_{0}}}\stackrel{{y}_{i}}{\overbrace{\frac{{F}_{i}}{{F}_{T}}}}p$
The following equation is used when the mole balance is written in terms of molar flow rates:
However, when the mole balance is written in terms of conversion, we use Equation (425):
$\begin{array}{c}{P}_{i}={C}_{i}RT={C}_{\text{A0}}\frac{({\mathrm{\Theta}}_{i}+{v}_{i}X)}{(1+\epsilon X)}\frac{P}{{P}_{0}}\frac{{T}_{0}}{T}RT\\ =\stackrel{{P}_{\text{A}0}}{\overbrace{{C}_{\text{A0}}{RT}_{0}}}\frac{({\mathrm{\Theta}}_{i}+{v}_{i}X)}{(1+\epsilon X)}p\end{array}$
For example, the rate law for the hydrodemethylation of toluene (T) to form methane (M) and benzene (B) given by Equation (1080) on page 483 can now be written in terms of conversion.
${r\prime}_{T}=\frac{k{\mathrm{P}}_{\mathrm{A}0}^{2}(1X)({\mathrm{\Theta}}_{{H}_{2}}X)}{1+{K}_{B}{P}_{A\text{\hspace{0.17em}}0}X+{K}_{B}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{A\text{\hspace{0.17em}}\text{\hspace{0.17em}}0}(1X)}(p{)}^{2}$
T + H_{2} → M + B
If you haven’t decided which computer to buy or borrow, or don’t have your integral tables, I suppose you could resort to the graphical techniques in Chapter 2 and use a Levenspiel plot, ${F}_{\text{A0}}/{r}_{\text{A}}^{\prime}$ versus X, to achieve a specified conversion of toluene. However, that may put you at a severe disadvantage as other chemical reaction engineers will be using software packages (e.g., Polymath) to solve chemical reaction engineering problems.
The SO_{2} oxidation discussed in Example 43 is to be carried out over a solid platinum catalyst. As with almost all gas–solid catalytic reactions, the catalytic rate law is expressed in terms of partial pressures instead of concentrations. The rate law for this SO_{2} oxidation was found experimentally to be^{1}
${\text{r}}_{\text{S}{0}_{2}}^{\prime}=\frac{k\left[{P}_{{\text{SO}}_{2}}\sqrt{{P}_{{\text{O}}_{2}}}\frac{{P}_{{\text{SO}}_{\text{3}}}}{{K}_{\text{P}}}\right]}{{\left(1+\sqrt{{P}_{{\text{O}}_{\text{2}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{K}_{{\text{O}}_{\text{2}}}}+{P}_{{\text{SO}}_{\text{3}}}{K}_{{\text{SO}}_{\text{3}}}\right)}^{2}}\text{\hspace{0.17em}},{\text{molSO}}_{2}\text{oxidized/}\left(\text{h}\right)/\left(\text{gcat}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E44}\text{.1}\right)$
^{1} O. A. Uychara and K. M. Watson, Ind. Engrg. Chem, 35, 541.
where P_{i} (kPa, bar, or atm) is the partial pressure of species i.
The reaction is to be carried out isothermally at 400°C. At this temperature, the rate constant k per gram of catalyst (gcat), the adsorption constants for O_{2} (${K}_{{\text{O}}_{2}}$) and SO_{2} (${K}_{{\text{SO}}_{2}}$), and the pressure equilibrium constant, K_{P}, were experimentally found to be
k = 9.7 mol SO_{2}/atm^{3/2}/h/gcat
${K}_{{O}_{2}}=38.5\text{\hspace{0.17em}}{\text{atm}}^{1},\text{\hspace{0.17em}}{K}_{{\text{SO}}_{3}}=42.5\text{\hspace{0.17em}}{\text{atm}}^{1},\text{\hspace{0.17em}and\hspace{0.17em}}{K}_{\text{P}}=930{\text{\hspace{0.17em}atm}}^{1/2}$
The total pressure and the feed composition (e.g., 28% SO_{2}) are the same as in Example 43. Consequently, the entering partial pressure of SO_{2} is 4.1 atm. There is virtually no pressure drop in this reactor.
Write the rate law as a function of conversion evaluating all other parameters.
Prepare a Levenspiel plot by sketching the reciprocal rate [1/(−r_{A})] as a function of X up to the equilibrium conversion X_{e}, and note where the rate goes to zero and [1/(–r_{A})] goes to infinity.
Solution
Part (a)
No Pressure Drop and Isothermal Operation
For SO_{2}
First we need to recall the relationship between partial pressure and concentration, followed by the relationship between concentration and conversion. Because we know how to express concentration as a function of conversion, we therefore know how to express partial pressure as a function of conversion.
${P}_{{\text{SO}}_{2}}={C}_{{\text{SO}}_{2}}\text{\hspace{0.17em}}RT\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{F}_{{\text{SO}}_{2}}}{\upsilon}\text{\hspace{0.17em}}RT\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{F}_{{\text{SO}}_{2,\text{\hspace{0.17em}}}0}\left(1X\right)RT}{{\upsilon}_{0}\left(1+\varepsilon X\right)\frac{T}{{T}_{0}}\text{\hspace{0.17em}}\frac{{P}_{0}}{P}}=\frac{{F}_{{\text{SO}}_{2},0}}{{\upsilon}_{0}}\frac{R{T}_{0}\left(1X\right)\frac{P}{{P}_{0}}}{1+\varepsilon X}$
${P}_{{\text{SO}}_{2}}=\text{\hspace{0.17em}}\frac{{P}_{{\text{SO}}_{2,\text{\hspace{0.17em}}}0}\left(1X\right)\frac{P}{{P}_{0}}}{\left(1+\varepsilon X\right)}=\frac{{P}_{{\text{SO}}_{2,\text{\hspace{0.17em}}}0}\left(1X\right)P}{\left(1+\varepsilon X\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E44}\text{.2}\right)$
${P}_{i}={P}_{\text{A0}}\frac{({\mathrm{\Theta}}_{i}+{v}_{i}X)}{(1+\epsilon X)}\frac{P}{{P}_{0}}={P}_{\text{A0}}\frac{({\mathrm{\Theta}}_{i}+{v}_{i}X)}{(1+\epsilon X)}p$
For no pressure drop P = P_{0}, that is, p = 1
$\overline{){P}_{{\text{SO}}_{2}}=\frac{{P}_{{\text{SO}}_{2},0}\left(1X\right)}{\left(1+\epsilon X\right)}}$
For SO_{3}
For O_{2}
From Example 43
Θ_{B} = 0.54
Factoring out $\frac{1}{2}$ in Equation (E44.5) gives
From Equation (E43.5)
Substitute for the partial pressure in the ratelaw equation (E44.1)
${{{r}^{\prime}}_{{\text{SO}}_{2}}}^{}=k\left[\frac{{P}_{{\text{SO}}_{2,0}}^{3/2}\left(\frac{1x}{\mathrm{1\u20130.14}X}\right)\sqrt{\frac{(1.08X)}{2\left(\mathrm{1\u20130.14}X\right)}}\frac{{P}_{{\text{SO}}_{2,0}}X}{\left(\mathrm{1\u20130.14}X\right)}\left(\frac{1}{930\text{\hspace{0.17em}}{\text{atm}}^{1/2}}\right)}{(1+\sqrt{\frac{38.5\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{}P}_{{\text{SO}}_{2,\text{0}}}(1.08X)}{2(10.14\text{\hspace{0.17em}}\text{\hspace{0.17em}}X)}}+\frac{42.5\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{{\text{SO}}_{2,0}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}X}{(10.14X)}{)}^{2}}\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E44.7}\right)$
with k = 9.7 mol SO_{2}/atm^{3/2}/h/gcat P_{SO2,0} = 4.1 atm, ${P}_{{\text{SO}}_{2,0}}^{3/2}$ = 8.3 atm ^{3/2}
${r}_{\text{A}}^{\prime}=f\left(x\right)$
Part (b)
We can now use Equation (E44.8) to plot the reciprocal rate as a function of X as shown in Figure E44.1 to design a packedbed catalytic reactor. We note at the equilibrium conversion, X_{e}, the reciprocal rate goes to infinity. We could now use a Levenspiel plot to find the catalyst weight W in a packedbed reactor (PBR) to achieve a specified conversion.
Levenspiel plots are useful for conceptually visualizing the differences in reactor sizes and the sequencing of reactors. However, we will see in Chapter 5 that using numerical software packages are so much better ways to solve for the catalysis weight, W, than using Levenspiel plots. For example, we would couple Equation (E44.8) with Equation (217) and use an ordinary differential equation (ODE) solver, such as Polymath or MATLAB, to find the conversion X as a function of catalyst weight W. So, be sure to buy, rent, or borrow a laptop computer before attempting to solve the problems in Chapter 5 and beyond.
Analysis: In most heterogeneous catalytic reactions, rate laws are expressed in terms of partial pressures instead of concentration. However, we see that through the use of the ideal gas law we could easily express the partial pressure as a function of concentration then use the stoichiometric table and conversion in order to express the rate law as a function of conversion. In addition, for most all heterogeneous reactions you will usually find a term like (1 + K_{A}P_{A} + K_{B}P_{B} + . . .) in the denominator of the rate law, as will be explained in Chapter 10.
Thus far in this chapter, we have focused mostly on irreversible reactions. The procedure one uses for the isothermal reactor design of reversible reactions is virtually the same as that for irreversible reactions, with one notable exception: the maximum conversion that can be achieved at the reaction temperature is the equilibrium conversion, X_{e}. In the following example, it will be shown how our algorithm for reactor design is easily extended to reversible reactions.
Need to first calculate X_{e}
The reversible gasphase decomposition of nitrogen tetroxide, N_{2}O_{4}, to nitrogen dioxide, NO_{2},
N_{2}O_{4} ⇄ 2NO_{2}
Nitrogen Tetroxide
is to be carried out at constant temperature. The feed consists of pure N_{2}O_{4} at 340 K and 202.6 kPa (2 atm). The concentration equilibrium constant, K_{C}, at 340 K is 0.1 mol / dm^{3} and the rate constant ${k}_{{\text{N}}_{2}{\text{O}}_{4}}$ and the rate constant is 0.5 min^{–1}.
Set up a stoichiometric table and then calculate the equilibrium conversion of N_{2}O_{4} in a constantvolume batch reactor.
Set up a stoichiometric table and then calculate the equilibrium conversion of N_{2}O_{4} in a flow reactor.
Assuming the reaction is elementary, express the rate of reaction solely as a function of conversion for a flow system and for a batch system.
Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.
Solution
N_{2}O_{4} ⇄ 2NO_{2}
A ⇄ 2B
At equilibrium, the concentrations of the reacting species are related by the relationship dictated by thermodynamics (cf. see Equation (310) and Appendix C).
${K}_{C}=\frac{{C}_{\text{Be}}^{2}}{{C}_{\text{Ae}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.1}\right)$
Batch system—constant volume, V = V_{0}.
TABLE E45.1 STOICHIOMETRIC TABLE FOR BATCH REACTOR
Species 
Symbol 
Initially 
Change 
Remaining 
N_{2}O_{4} 
A 
N_{A0} 
–N_{A0}X 
N_{A} = N_{A0}(1 – X) 
NO_{2} 
B 
0 
+2N_{A0}X 
N_{B} = 2N_{A0}X 
N_{T0} = N_{A0} 
N_{T} = N_{T0} + N_{A0}X 
Living Example Problem
For batch systems C_{i} = N_{i}/V
${C}_{\text{A}}=\frac{{N}_{A}}{V}=\frac{{N}_{A}}{{V}_{0}}=\frac{{N}_{\text{A0}}(1X)}{{V}_{0}}={C}_{\text{A0}}(1X)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.2}\right)$
${C}_{B}=\frac{{N}_{B}}{V}=\frac{{N}_{B}}{{V}_{0}}=\frac{2{N}_{\text{A0}}X}{{V}_{0}}=2{C}_{\text{A0}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}X\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.3}\right)$
$\begin{array}{cc}\hfill {C}_{A\text{\hspace{0.17em}}\text{\hspace{0.17em}}0}& =\frac{{y}_{A\text{\hspace{0.17em}}0}{P}_{0}}{R{T}_{0}}=\frac{\left(1\right)(2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{atm})}{(0.082\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{atm}\cdot {\text{dm}}^{3}/\text{mol}\cdot \text{K})\text{\hspace{0.17em}}\text{\hspace{0.17em}}(340\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{K})}\hfill \\ \hfill & \begin{array}{cc}\begin{array}{c}=0.07174\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{mol}/\text{}{\text{dm}}^{3}\hfill \end{array}& \text{Answer}\end{array}\end{array}$
At equilibrium in a batch reactor, X = X_{eb}, we substitute Equations (E45.2) and (E45.3) into Equation (E45.1)
${K}_{\text{c}}=\frac{{C}_{\text{Be}}^{2}}{{C}_{\text{Ae}}}=\frac{4{C}_{\text{A0}}^{2}{X}_{eb}^{2}}{{C}_{\text{A0}}(1{X}_{eb})}=\frac{4{C}_{\text{A0}}{X}_{eb}^{2}}{1{X}_{eb}}$
(mathmathmathmath) to get
${X}_{eb}=\sqrt{\frac{{K}_{c}(1X)}{\text{4}\text{}{C}_{\text{A0}}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.4}\right)$
We will use the software package Polymath to solve for the equilibrium conversion and let Xeb represent the equilibrium conversion in a constantvolume batch reactor. Equation (E45.4) written in Polymath format becomes
f(Xeb) = Xeb – [Kc* (1 – Xeb)/(4*Cao)] ^0.5
The Polymath program and solution are given in Table E45.2.
When looking at Equation (E45.4), you probably asked yourself, “Why not use the quadratic formula to solve for the equilibrium conversion in both batch and flow systems?” That is from Equation (E45.4) we would have
$\begin{array}{cc}\hline Batch:& {X}_{eb}=\frac{1}{8}\left[(1+\sqrt{1+16{C}_{\text{A0}}/{K}_{\text{C}}})/({C}_{\text{A0}}/{K}_{C})\right]\\ \hline\end{array}$
and from Equation (E45.8) we would have
$\begin{array}{cc}\hline Flow:& {X}_{ef}=\left[\frac{(\epsilon 1)+\sqrt{{(\epsilon +1)}^{2}+4(\epsilon +{4C}_{\text{A0}}/{K}_{\text{C}})}}{2(\epsilon +4{C}_{\text{A0}}/{K}_{\text{C}})}\right]\\ \hline\end{array}$
There is a Polymath tutorial in the Summary Notes for Chapter 1 on the CRE Web site.
The answer is that future problems will be nonlinear and require Polymath solutions; therefore, this simple exercise increases the reader’s ease in using Polymath.
Polymath Tutorial Chapter 1
As seen in Table E45.2, the equilibrium conversion in a constantvolume batch reactor (X_{eb}) is
$\overline{){X}_{eb}=0.44}$
Note: A tutorial on Polymath can be found in the summary notes for Chapter 1 and on the CRE Web site (www.umich.edu/~elements/6e/tutorials/Polymath_LEP_tutorial.pdf).
Flow system. The stoichiometric table is the same as that for a batch system except that the number of moles of each species, N_{i}, is replaced by the molar flow rate of that species, F_{i}.
For constant temperature and pressure, the volumetric flow rate υ = υ_{0}(1 + ɛX), is and the resulting concentrations of species A and B are
TABLE E45.3 STOICHIOMETRIC TABLE FOR CONTINUOUSFLOW REACTOR
Species 
Symbol 
Entering 
Change 
Leaving 
N_{2}O_{2} 
A 
F_{A0} 
–F_{A0}X 
F_{A} = F_{A0}(1–X) 
NO_{2} 
B 
0 
+2F_{A0}X 
F_{B} = 2F_{A0}X 
F_{T} = F_{A0} 
F_{T} = F_{T0} + F_{A0} X 
${C}_{A}=\frac{F\text{A}}{\upsilon}=\text{\hspace{0.17em}}\frac{{F}_{\text{A0}}(1X)}{\upsilon}=\frac{{F}_{\text{A0}}(1X)}{{\upsilon}_{0}(1+\varepsilon X)}=\frac{{C}_{\text{A0}}(1X)}{1+\varepsilon X}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.5}\right)$
${C}_{B}=\frac{{F}_{B}}{\upsilon}=\text{\hspace{0.17em}}\frac{2{F}_{\text{A0}}X}{{\upsilon}_{0}(1+\varepsilon X)}=\frac{2{C}_{\text{A0}}X}{(1+\varepsilon X)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.6}\right)$
At equilibrium, X = X_{ef}, we can substitute Equations (E45.5) and (E45.6) into Equation (E45.1) to obtain the expression
${K}_{C}=\frac{{{C}_{Be}}^{2}}{{C}_{Ae}}=\frac{[2{C}_{\text{A0}}{X}_{ef}/(1+\varepsilon {X}_{ef}){]}^{2}}{{C}_{\text{A0}}(1{X}_{ef})/(1+\varepsilon {X}_{ef})}$
Simplifying gives
${K}_{C}=\frac{4{C}_{\text{A0}}{{X}_{ef}}^{2}}{(1{X}_{ef})(1+\varepsilon {X}_{ef})}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.7}\right)$
Rearranging to use Polymath yields
${X}_{ef}=\sqrt{\frac{{K}_{C}(1{X}_{ef})(1+\varepsilon {X}_{ef})}{4{C}_{\text{A0}}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.8}\right)$
For a flow system with pure N_{2}O_{4} feed, ɛ = y_{A0} δ = 1(2–1)=1.
We shall let Xef represent the equilibrium conversion in a flow system. Equation (E45.8) written in the Polymath format becomes
$\overline{)\text{f}(\text{Xef})=\text{Xef}{[\text{kc}*(1\text{Xef})*(1+\text{eps}*\text{Xef})/4/\text{cao}]}^{^}0.5}$
This solution is also shown in Table E45.3 $\begin{array}{c}\hline {X}_{ef}=0.51\\ \hline\end{array}$.
Note that the equilibrium conversion in a flow reactor (i.e., X_{ef} = 0.51), with no pressure drop, is greater than the equilibrium conversion in a constantvolume batch reactor (X_{eb} = 0.44). Recalling Le Châtelier's principle, can you suggest an explanation for this difference in X_{e}? This example is continued in Problem P41_{A} (a). Be sure to use Wolfram and/or Python in the Living Example Problems (LEPs) to see how the ratio (X_{eb}/X_{ef} ) varies with ɛ and C_{T0}.
LEP Sliders
Rate laws. Assuming that the reaction follows an elementary rate law, then
$\begin{array}{cc}{r}_{\text{A}}={k}_{\text{A}}[{C}_{\text{A}}\frac{{C}_{\text{B}}^{2}}{{K}_{C}}]& \left(\text{E45.9}\right)\end{array}$
For a constant volume (V = V_{0}) batch system
Here, C_{A} = N_{A}/V_{0} and C_{B} = N_{B}/V_{0}. Substituting Equations (E45.2) and (E45.3) into the rate law, we obtain the rate of disappearance of A as a function of conversion
–r_{A} =f(X) for a batch reactor with V = V_{0}
For a flow system
Here, C_{A} = F_{A}/υ and C_{B} = F_{B}/υ with υ = υ_{0} (1 + ɛX). Consequently, we can substitute Equations (E45.5) and (E45.6) into Equation (E45.9) to obtain
–r_{A} = f (X) for a flow reactor
As expected, the dependence of reaction rate on conversion, that is, –r_{A} = f(X), for a constantvolume batch system (i.e., Equation (E45.10)) is different than that for a flow system (Equation (E45.11)) for gasphase reactions.
If we substitute the values for C_{A0}, K_{C}, ɛ, and k_{A} = 0.5 min^{–1} in Equation (E45.11), we obtain –r_{A} solely as a function of X for the flow system.
${r}_{\text{A}}=\frac{0.5}{\text{min}}\left[0.072\frac{\text{mol}(1X)}{{\text{dm}}^{3}(1+X)}\frac{4(0.075{\text{mol/dm}}^{3}{)}^{2}{X}^{2}}{0.01{\text{mol/dm}}^{3}(1+X{)}^{2}}\right]$
${r}_{\text{A}}=0.036[\frac{(1X)}{(1+X)}\frac{2.88\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{X}^{2}}{\text{\hspace{0.17em}}(1+X{)}^{2}}]\text{\hspace{0.17em}}\left(\frac{\text{mol}}{{\text{dm}}^{3}\cdot \text{\hspace{0.17em}}\text{min}}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E45.12}\right)$
We can now use Equation (E45.12) to prepare our Levenspiel plot to find either the CSTR or PFR volume.
Figure E45.1 Levenspiel plot for a flow system.
X_{e} = 0.51
We see as we approach equilibrium, X_{e} = 0.51, –r_{A} goes to zero and (1/–r_{A}) goes to infinity.
CSTR volume. Just for fun (and this really is fun), let’s fast forward and calculate the CSTR reactor volume necessary to achieve 80% of the equilibrium conversion of 51% (i.e., X = 0.8X_{e} = (0.8)(0.51) = 0.4) for a molar feed rate of A of 3 mol/min.
Plotting the reciprocal of Equation (E45.12) to obtain Levenspiel plot as we did in Chapter 2, it can be seen in Figure E45.1 when X = 0.4 then as
$(\frac{1}{{r}_{\text{A}}}{)}_{x=0.4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=143\frac{{\text{dm}}^{3}\text{\hspace{0.17em}}\mathrm{min}}{\text{mol}}$
then for F_{A0} = 3 mol/min
Keeping an eye on things to come
$\begin{array}{cc}\hfill V& ={F}_{\text{A0}}X(\frac{1}{{r}_{\text{A}}}{)}_{X=0.4}\hfill \\ \hfill & =\left(\frac{3\text{\hspace{0.17em}}\text{mol}}{\mathrm{min}}\right)\left(0.4\right)\left(143\right)\frac{{\text{dm}}^{3}\mathrm{min}}{\text{mol}}\hfill \\ \hfill \text{V}& {=171\text{\hspace{0.17em}dm}}^{3}\hfill \end{array}$
A quicker way to find the CSTR volume, rather than drawing a Levenspiel plot, is to simply evaluate Equation (E45.12) at X = 0.4
${r}_{A}=0.036[\frac{\left(\mathrm{1\u20130.4}\right)}{(1+0.4)}\frac{2.88(0.4{)}^{2}}{(1+\left(0.4\right){)}^{2}}]=0.0070\text{\hspace{0.17em}}{\text{mol/dm}}^{3}/\mathrm{min}$
Using Equation (213), we can now find the reactor volume
$\begin{array}{cc}\text{V}=& \frac{{F}_{\text{A}0}X}{{r}_{\text{A}}{}_{\text{X}}}=\frac{{F}_{\text{A0}}\left(0.4\right)}{{r}_{\text{A}}{}_{0.4}}=\frac{(3\text{)mol/min)(0.4)}}{0.0070\text{)}\frac{\text{mol}}{{\text{dm}}^{3}\cdot \text{min}}}\hfill \\ \text{V=}& {\begin{array}{c}171\text{\hspace{0.17em}}\text{dm}\end{array}}^{3}=0.171{\text{}\text{m}}^{3}\hfill \end{array}$
The CSTR volume required to achieve 40% conversion is 0.171 m^{3}. See Problem P41A (b) to calculate the PFR volume.
Analysis: The purpose of this example was to calculate the equilibrium conversion first for a constantvolume batch reactor in part (a), and then for a constant pressure flow reactor in part (b). One notes that there is a change in the total number of moles in this reaction and, as a result, these two equilibrium conversions (batch and flow) are not the same!! We next showed in part (c) how to express –r_{A} = f(X) for a reversible gasphase reaction. Here we noted that the equations for the reactionrate laws as a function of conversion, that is, –r_{A} = f(X), are different for batch and flow reactors. Finally, in Part (d) having –r_{A} = f(X), we specified a molar flow rate of A (i.e., 3.0 mol A/min) and calculated the CSTR volume necessary to achieve 40% conversion. We did this calculation to give insight to the types of analyses we, as chemical reaction engineers, will carry out as we move into similar but more complex calculations in Chapters 5 and 6.
The Swiss Cheese Model shown in Figure 44 is another risk assessment tool, one that offers a deeper understanding into the layers of protection for chemical processes. This section (i.e., 4.4) is presented to give an overview of the model while the Process Safety Across the Chemical Engineering Curriculum Web site (http://umich.edu/~safeche/swiss_cheese.html) gives more indepth explanation. A layer of protection is either a preventative action put in place to reduce the chance that an incident will occur, or a mitigating action put in place to lessen the severity of an accident. These layers of protection can include using inherently safer designs, following proper lab procedures, wearing adequate personal protective equipment (PPE), and having an emergency response plan.
Figure 44 Swiss Cheese Model.
Even with the different levels or layers of protection, represented in Figure 44 as Swiss cheese slices, there is a possibility that a hazard will result in an incident. For an incident to occur, there must be vulnerability in each layer of protection, represented by holes in the slices of cheese. A hole is a weakness in that layer of protection (e.g., a backup cooling system, a faulty pressure relief valve). The size and number of holes in the cheese represent the relative lack of reliability of that layer. This model illustrates the importance of having redundant and strong protective layers so that hazards do not pass through multiple protective actions undetected and unresolved.
Understanding and implementing each protective level is important when running a chemical process. Table 44 shows and discusses each of the three control “cheese” slices (i.e., Engineering, Administrative, and Behavioral) while Table 45 shows and discusses the mitigating barriers.
TABLE 44 PREVENTATIVE ACTIONS
Engineering Controls 
Administrative Controls 
Behavioral Controls 
Engineering controls involve preventative actions that are designed into the system, for example, pressure relief valves on heat exchangers. 
Administrative controls involve managerial actions taken to ensure everyone involved in the process possesses the same knowledge of the established safety procedures, for example, Management of Change (MoC) procedures: Before making a change to a process, employees are required to identify, review, and approve modifications to ensure changes are implemented safely. 
Behavioral controls focus on actions that individuals can take to prevent an accident from occurring by using proper tools, and following lab procedures, MoC and safety checklists as instructed. 
TABLE 45 MITIGATING ACTIONS
Mitigating Barriers 
These barriers focus on actions that lessen the severity or impact of an accident, for example, having an emergency response plan, wearing personal protective equipment (PPE), or the installation of a fire suppression system. 
Engineering Controls, Administrative Controls, and Behavioral Controls represent Preventative Actions designed to reduce the likelihood of an incident occurring, while Mitigating Barriers represent Mitigating Actions designed to lessen the impact or severity of an incident. It’s important to recognize that all of these layers of protection will have some weaknesses.
The main purpose of the Swiss Cheese Model is to visualize the vulnerability and how a hazard may be able to pass through the many different preventative actions in place.
Further Reading on the Swiss Cheese Model:
Board, U.S. Chemical Safety and Hazard Investigation. Fire at Praxair St. Louis: Dangers of Propylene Cylinders. Washington DC, 2006.
D. Hatch, P. McCulloch, and I. Travers. “Visual HAZOP,” Chem. Eng., (917), 27–32 (2017, November).
USCSB. “CSB Safety Video: Dangers of Propylene Cylinders.” Online video clip. YouTube, 10 October 2007. Web. 8 May 2018.
The stoichiometric table for the reaction given by Equation (S41) being carried out in a flow system is
$\begin{array}{cc}A+\frac{b}{a}B\to \frac{c}{a}C+\frac{d}{a}D& (\text{S41)}\end{array}$
In the case of ideal gases, Equation (S43) relates volumetric flow rate to conversion.
Batch reactor constant volume: V = V_{0}
$\begin{array}{cc}V={V}_{0}& \begin{array}{c}(\text{S42)}\end{array}\end{array}$
Flow systems: Gas:
$\begin{array}{cc}v={v}_{0}\left(\frac{{P}_{0}}{P}\right)\text{(1}\text{+}\text{}\epsilon X)\frac{T}{{T}_{0}}& \text{(S43)}\end{array}$
Liquid: ()
$\begin{array}{cc}v={v}_{0}& (\text{S4}\end{array}4)$
Species 
Entering 
Change 
Leaving 
A 
F_{A0} 
–F_{A0}X 
F_{A0}(1 = X) 
B 
F_{B0} 
$\left(\frac{b}{a}\right){F}_{\text{A0}}X$ 
${F}_{\text{A0}}({\mathrm{\Theta}}_{\text{B}}\frac{b}{a}X)$ 
C 
F_{C0} 
$\left(\frac{c}{a}\right){F}_{\text{A0}}X$ 
${F}_{\text{A0}}({\mathrm{\Theta}}_{\text{C}}+\frac{c}{a}X)$ 
D 
F_{D0} 
$\left(\frac{d}{a}\right){F}_{\text{A0}}X$ 
${F}_{\text{A0}}({\mathrm{\Theta}}_{\text{D}}+\frac{d}{a}X)$ 
I 
F_{I0} 
   
F_{I0} 
Totals 
F_{T0} 
δF_{A0}X 
F_{T} = F_{T0} +δF_{A0}X 
Definitions of δ and ɛ: For the general reaction given by Equation (S41), we have
$\delta =\frac{\text{Changeintotalnumberofmoles}}{\text{moleofAreacted}}$
and
$\begin{array}{cc}\overline{)\epsilon ={y}_{\text{A0}}\delta}& \left(\text{S46}\right)\end{array}$
$\epsilon =\frac{\text{Changeintotalnumberofmolesforcompleteconversion}}{\text{Totalnumberof}\text{molesfedtothereactor}}$
For incompressible liquids or for batch gasphase reactions taking place in a constant volume, V = V_{0}, the concentrations of species A and C in the reaction given by Equation (S41) can be written as
$\begin{array}{cc}{C}_{\text{A}}\text{}=\text{}\text{\hspace{0.17em}}\frac{{F}_{\text{A}}}{v}\text{=}\frac{{F}_{\text{A0}}}{{v}_{0}}(1\text{}X)\text{=}{\text{C}}_{\text{A0}}(1\text{}X)& \left(\text{S47}\text{}\right)\end{array}$
$\begin{array}{cc}{C}_{c}={C}_{\text{A0}}({\mathrm{\Theta}}_{\text{C}}+\frac{c}{a}X)& (\text{S48}\end{array}\text{}\text{})$
Equations (S47) and (S48) also hold for gasphase reactions carried out in constantvolume batch reactors.
For gasphase reactions in continuousflow reactors, we use the definition of concentration (C_{A} = F_{A}/υ) along with the stoichiometric table and Equation (S43) to write the concentration of A and C in terms of conversion.
$\begin{array}{cc}{C}_{\text{A}}=\frac{{F}_{\text{A}}}{v}=\frac{{F}_{\text{A0}}(1X)}{v}={C}_{\text{A0}}\left[\frac{1X}{1+\varepsilon X}\right]P\left(\frac{{T}_{\text{0}}}{T}\right)& (\text{S49}\end{array})$
$\begin{array}{cc}{C}_{\text{c}}=\frac{{F}_{c}}{v}={C}_{\text{A0}}\left[\frac{{\mathrm{\Theta}}_{\text{c}}+(c/a)X}{1+\epsilon X}\right]P\left(\frac{{T}_{0}}{T}\right)& \text{(S410)}\end{array}$
$\text{with}{\mathrm{\Theta}}_{c}=\frac{{F}_{\text{c0}}}{{F}_{\text{A0}}}=\frac{{C}_{\text{c0}}}{{C}_{\text{A0}}}=\frac{{\mathrm{y}}_{\mathrm{c}0}}{{\mathrm{y}}_{\mathrm{A}0}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}p=\frac{p}{{p}_{0}}$
In terms of gasphase molar flow rates, the concentration of species i is
$\begin{array}{cc}{C}_{i}={C}_{T0}\frac{{F}_{i}}{{F}_{T}}\frac{P}{{P}_{0}}\frac{{T}_{0}}{T}& \text{(S411)}\text{}\end{array}$
Equation (S411) must be used for membrane reactors (Chapter 6) and for multiple reactions (Chapter 8).
Many catalytic rate laws are given in terms of partial pressure, for example,
$\begin{array}{cc}{r}_{A}^{\prime}=\frac{{K}_{A}{P}_{A}}{1+{K}_{A}{P}_{A}}& \text{(S412)}\end{array}$
The partial pressure is related to conversion through the stoichiometric table. For any species “i” in the reaction
$\begin{array}{cc}{P}_{i}={P}_{A0}\frac{({\mathrm{\Theta}}_{i}+{v}_{i}X)}{(1+\varepsilon X)}& \text{(S413)}\end{array}$
For species A
${P}_{A}={P}_{A0}\frac{(1X)}{(1+\varepsilon X)}p$
Substituting in the rate law
$\begin{array}{cc}{r}_{A}^{\prime}=\frac{{K}_{A}{P}_{A0}\frac{(1X)}{(1+\varepsilon X)}P}{1+{K}_{A}{P}_{A0\frac{(1X)}{(1+\varepsilon X)}P}}=\frac{{kP}_{A0}(1X)p}{(1+\varepsilon X)+{K}_{A}{p}_{A0}(1X)P}& \text{(S414)}\end{array}$
Interactive Computer Games (http://umich.edu/~elements/6e/icm/index.html) Quiz Show II Jeopardy (http://umich.edu/~elements/6e/icm/kinchal2.html). A Jeopardy Game competition between AIChE student chapters is held at the annual AIChE meeting.
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
Q41_{A} QBR (Question Before Reading). What are the differences in writing the concentrations as a function of conversion for gasphase and liquidphase reactions?
Q42_{A} i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/04chap/iclicker_ch4_q1.html) and view at leave five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case, explaining why the question should not be on the next exam. In either case, explain your reasoning.
Q43_{A} Example 41. Would the example be correct if water were considered an inert? Explain.
Q44_{A} Example 42. How would the answer change if the initial concentration of glyceryl stearate were 3 mol/dm^{3}? Rework Example 42 correctly using the information given in the problem statement.
Q45_{A} Example 43. Under what conditions will the concentration of the inert nitrogen be constant? Plot Equation (E45.2) in terms of (1/–r_{A}) as a function of X up to value of X = 0.99. What did you find?
Q46_{A} Can you name the English rock group that was famous worldwide in the 1970s and 1980s and whose lead singer was Freddie Mercury? The group had a resurgence in 2019 when Adam Lambert took over for Freddie, who had died in 1991. Hint: See Problem P522_{A}.
Q47_{A} What did you learn from the Swiss Cheese Model (e.g., make a list)?
Q48_{A} Go to the LearnChemE screencast link for Chapter 4 (http://www.umich.edu/~elements/6e/04chap/learnchemevideos.html).
View one or more of the screencast 5 to 6minute videos and write a twosentence evaluation.
What did the screencast do well and what could be improved?
Q49_{A} AWFOS–S4 Swiss Cheese Model.
What is the main purpose of implementing the Swiss Cheese Model in a workplace?
Categorize engineering controls, administrative controls, behavioral controls, and mitigating barriers as preventative actions or mitigating actions. List an example of each.
What happens when all of the weaknesses/vulnerabilities in the controls/barriers line up?
What was the takeaway lesson from this safety section?
P41_{A} (a) Example 44: SO_{2} Oxidation
Wolfram and Python
Vary Θ_{B} and observe the change in reaction rate. Go to the extremes and explain what is causing the curve to change the way it does.
Vary the parameters and list the parameters that have the greatest and the least effect on the reaction rate.
Write a set of conclusions about the experiments you carried out by varying the sliders in parts (i) through (iii).
Polymath
Using the molar flow rate of SO_{2} (A) of 3 mol/h and Figure E44.1, calculate the fluidized bed (i.e., CSTR catalyst weight) necessary for 40% conversion.
Next, consider the entering flow rate of SO_{2} is 1000 mol/h. Plot (F_{A0}/–r_{A}) as a function of X to determine the fluidized bed catalyst weight necessary to achieve 30% conversion, and 99% of the equilibrium conversion, that is, X = 0.99 X_{e}.
(b) Example 45: Equilibrium Conversion
Using Figure E45.1 and the methods in Chapter 2, estimate the PFR volume for 40% conversion for a molar flow rate of A of 5 dm^{3}/min.
Using Figure E45.1 and the methods in Chapter 2, estimate the time to achieve 40% conversion in a batch reactor when the initial concentration is 0.05 mol/dm^{3}.
Wolfram and Python
What values of K_{C} and C_{A0} cause X_{ef} to be the farthest away from X_{eb}?
What values of K_{C} and C_{A0} will cause X_{eb} and X_{ef} to be the closest together?
Observe the plot of the ratio of $\left(\frac{{X}_{eb}}{{X}_{er}}\right)$ as a function of y_{A}_{0}. Vary the values of K_{C} and C_{T0}. What conclusions can you draw?
Write a set of conclusions from your experiment (i) through (vi).
P42_{A} Load the Interactive Computer Games (ICG) Kinetic Challenge from the CRE Web site. Play the game and then record your performance number for the module that indicates your mastering of the material. Your professor has the key to decode your performance number. ICG Kinetics Challenge Performance # ______________.
A Jeopardy Game competition between AIChE student chapters is held at the annual AIChE meeting. You might find one of these questions asked.
Kinetics Challenge II
Rate 
Law 
Stoich 
100 
100 
100 
200 
200 
200 
300 
300 
300 
P43_{A} The elementary reversible reaction
2A ⇄ B
is carried out in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm^{3}. If the equilibrium conversion is found to be 60%,
What is the equilibrium constant, K_{C} if the reaction is a gasphase reaction? (Ans: Gas: K_{C} = 0.328 dm^{3}/mol)
What is the K_{C} if the reaction is a liquidphase reaction? (Ans: Liquid: K_{C} = 0.469 dm^{3}/mol)
Write =r_{A} solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, reversible, gasphase, isothermal reaction with no pressure drop with k_{A} = 2 dm^{6}/mol•s and K_{C} = 0.5 all in proper units.
Repeat (c) for a constantvolume batch reactor.
P44_{B} Stoichiometry. The elementary gas reaction
2A + B → C
is carried out isothermally in a PFR with no pressure drop. The feed is equal molar in A and B, and the entering concentration of A is 0.1 mol/dm^{3}. Set up a stoichiometric table and then determine the following.
What is the entering concentration (mol/dm^{3}) of B?
What are the concentrations of A and C (mol/dm^{3}) at 25% conversion of A?
What is the concentration of B (mol/dm^{3}) at 25% conversion of A? (Ans: C_{B} = 0.1 mol/dm^{3})
What is the concentration of B (mol/dm^{3}) at 100% conversion of A?
If at a particular conversion the rate of formation of C is 2 mol/min/dm^{3}, what is the rate of formation of A at the same conversion?
Write =r_{A} solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, irreversible, gasphase, isothermal reaction with no pressure drop with an equal molar feed and with C_{A0} = 2.0 mol/dm^{3} at, k_{A} = 2 dm^{6}/mol•s.
What is the rate of reaction at X = 0.5?
P45_{A} Set up a stoichiometric table for each of the following reactions and express the concentration of each species in the reaction as a function of conversion, evaluating all constants (e.g., =, =). Next, assume the reaction follows an elementary rate law, and write the reaction rate solely as a function of conversion, that is, –r_{A} = f(X).
For the liquidphase reaction
the entering concentrations of ethylene oxide and water, after mixing the inlet streams, are 16.13 and 55.5 mol/dm^{3}, respectively. The specific reaction rate is k = 0.1 dm^{3}/mol · s at 300 K with E = 12500 cal/mol.
After finding –r_{A} = f(X), calculate the CSTR spacetime, τ, for 90% conversion at 300 K and also at 350 K.
If the volumetric flow rate is 200 liters per second, what are the corresponding reactor volumes? (Ans: At 300 K: V = 439 dm^{3} and at 350 K: V = 22 dm^{3})
For the isothermal gasphase pyrolysis
C_{2}H_{6} → C_{2}H_{4} = H_{2}
pure ethane enters a flow reactor at 6 atm and 1100 K, with ΔP = 0. Set up a stoichiometric table and then write –r_{A} = f(X). How would your equation for the concentration and reaction rate, that is, –r_{A} = f(X), change if the reaction were to be carried out in a constantvolume batch reactor?
For the isothermal, isobaric, catalytic gasphase oxidation
the feed enters a PBR at 6 atm and 260°C, and is a stoichiometric mixture of only oxygen and ethylene. Set up a stoichiometric table and then write ${r}_{\text{A}}^{\prime}$ as a function of partial pressures. Express the partial pressures and ${r}_{\text{A}}^{\prime}$ as a function of conversion for (1) a fluidized batch reactor and (2) a PBR. Finally, write ${r}_{\text{A}}^{\prime}$ solely as a function of the rate constant and conversion.
Set up a stoichiometric table for the isothermal, catalytic gasphase reaction carried out in a fluidized CSTR.
The feed is stoichiometric and enters at 6 atm and 170°C. What catalyst weight is required to reach 80% conversion in a fluidized CSTR at 170°C and at 270°C? The rate constant is defined with respect to benzene and υ_{0} = 50 dm^{3}/min.
${K}_{B}=\frac{53\text{\hspace{0.17em}}\text{mol}}{\text{Kgcat}\cdot \mathrm{min}\cdot {\text{atm}}^{3}}\text{at 300 k with}\text{\hspace{0.17em}}E=80\text{\hspace{0.17em}}\text{kJ/mol}$
First write the rate law in terms of partial pressures and then express the rate law as a function of conversion. Assume ΔP ≡ 0.
P46_{A} Orthonitroanaline (an important intermediate in dyes—called fast orange) is formed from the reaction of orthonitrochlorobenzene (ONCB) and aqueous ammonia (see explosion in Figure E132.1 in Example 132).
The liquidphase reaction is first order in both ONCB and ammonia with k = 0.0017 m^{3}/kmol · min at 188°C with E = 11273 cal/mol. The initial entering concentrations of ONCB and ammonia are 1.8 and 6.6 kmol/m^{3}, respectively (more on this reaction in Chapter 13).
Set up a stoichiometric table for this reaction for a flow system.
Write the rate law for the rate of disappearance of ONCB in terms of concentration.
Explain how parts (a) and (b) would be different for a batch system.
(d) Write –r_{A} solely as a function of conversion.
–r_{A} = ______ 
What is the initial rate of reaction (X = 0)
at 188°C?  –r_{A} = ______ 
at 25°C?  –r_{A} = ______ 
at 288°C?  –r_{A} = ______ 
What is the rate of reaction when X = 0.90
at 188°C?  –r_{A} = ______ 
at 25°C?  –r_{A} = ______ 
at 288°C?  –r_{A} = ______ 
What would be the corresponding CSTR reactor volume at 25°C to achieve 90% conversion and at 288°C for a feed rate of 2 dm^{3}/min
at 25°C?  V = ______ 
at 288°C?  V = ______ 
P47_{B} OEQ (Old Exam Question). Consider the following elementary gasphase reversible reaction to be carried out isothermally with no pressure drop and for an equal molar feed of A and B with C_{A0} = 2.0 mol/dm^{3}.
2A + B ⇄ C
What is the concentration of B initially? C_{B0} = ____ (mol/dm^{3})
What is the limiting reactant? ______
What is the exit concentration of B when the conversion of A is 25%? C_{B} = _____ (mol/dm^{3})
Write –r_{A} solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, reversible, gasphase, isothermal reaction with no pressure drop with an equal molar feed and with C_{A0} = 2.0 mol/dm^{3}, k_{A} = 2 dm^{6}/mol^{2}•s, and K_{C} = 0.5 all in proper units –r_{A} = ____.
What is the equilibrium conversion?
What is the rate when the conversion is
0%?
50%?
0.99 X_{e}?
P48_{B} OEQ (Old Exam Question). The gasphase reaction
$\frac{1}{2}{\text{N}}_{2}+\frac{3}{2}{\text{H}}_{2}\to {\text{NH}}_{3}$
is to be carried out isothermally first in a flow reactor. The molar feed is 50% H_{2} and 50% N_{2}, at a pressure of 16.4 atm and at a temperature of 227°C.
Construct a complete stoichiometric table.
Express the concentrations in mol/dm^{3} of each for the reacting species as a function of conversion. Evaluate C_{A0},δ, and ɛ, and then calculate the concentrations of ammonia and hydrogen when the conversion of H_{2} is 60%. (Ans: ${\text{C}}_{{\text{H}}_{2}}=0.1\text{\hspace{0.17em}}{\text{mol/dm}}^{3}$)
Suppose by chance the reaction is elementary with ${\text{k}}_{{\text{N}}_{2}}=40{\text{dm}}^{3}/\text{mol/s}$. Write the rate of reaction solely as a function of conversion for (1) a flow reactor and for (2) a constantvolume batch reactor.
P49_{B} OEQ (Old Exam Question). There were 820 million pounds of phthalic anhydride produced in the United States in 1995. One of the end uses of phthalic anhydride is in the fiberglass of sailboat hulls. Phthalic anhydride can be produced by the partial oxidation of naphthalene in either a fixed or a fluidized catalytic bed. A flow sheet for the commercial process is shown below. Here, the reaction is carried out in fixed bed reactor with a vanadium pentoxide catalyst packed in 25 mm diameter tubes. A production rate of 31,000 tons per year would require 15,000 tubes.
Assume the reaction is first order in oxygen and second order in naphthalene with k_{N} = 0.01 dm^{6}/mol^{2}/s.
Adapted from Process Technology and Flowsheet Volume II. Reprints from Chemical Engineering, McGraw Hill Pub. Co., p. 111.
Set up a stoichiometric table for this reaction for an initial mixture of 3.5% naphthalene and 96.5% air (mol %), and use this table to develop the relations listed in parts (b) and (c). The initial/entering pressure and temperature are P_{0} = 10 atm and T_{0} = 500 K.
Express the following solely as a function of the conversion of naphthalene, X for a constantvolume isothermal batch reactor V = V_{0}. (1) The concentration of O_{2}, C_{O2}. (2) The total pressure P. (3) The rate of reaction, –r_{N}. (4) Then calculate the time to achieve 90% conversion.
Repeat (b) for an isothermal flow reactor to find C_{O2}, volumetric flow rate and the reaction rate –r_{N} as a function of conversion.
What CSTR and PFR volumes are necessary to achieve 98% conversion of naphthalene for an entering volumetric flow rate of 10 dm^{3}/s.
Write a few sentences about the above flow sheet to manufacture phthalic anhydride.
P410_{B} OEQ (Old Exam Question). The elementary reversible reaction
2A ⇄ B
is carried out isothermally in a flow reactor with no pressure drop and where pure A is fed at a concentration of 4.0 mol/dm^{3}. If the equilibrium conversion is found to be 60%
What is the equilibrium constant, K_{C}, if the reaction occurs in the gas phase?
What is the K_{C} if the reaction is a liquidphase reaction?
P411_{B} Consider the elementary gasphase reversible reaction is carried out isothermically
A ⇄ 3C
Pure A enters at a temperature of 400 K and a pressure of 10 atm. At this temperature, K_{C} = 0.25(mol/dm^{3})^{2}. Write the rate law and then calculate the equilibrium conversion for each of the following situations:
The gasphase reaction is carried out in a constantvolume batch reactor.
The gasphase reaction is carried out in a constantpressure batch reactor.
Can you explain the reason why there would be a difference in the two values of the equilibrium conversion?
P412_{B} Repeat parts (a)–(c) of Problem P411_{B} for the reaction
3A ⇄ C
Pure A enters at 400 K, 10 atm and the equilibrium constant is K_{C} = 2.5 (dm^{3}/mol)^{2}.
Compare the equilibrium conversions in Problems P411_{B} and P412_{B}. Explain the trends, for example, is X_{e} for constant pressure always greater than X_{e} for constant volume? Explain the reasons for the similarities and differences in the equilibrium conversions in the two reactions.
P413_{C} OEQ (Old Exam Question). Consider a cylindrical batch reactor that has one end fitted with a frictionless piston attached to a spring (Figure P413_{C}). The reaction
A + B → 8C
with the rate law
${r}_{\text{A}}={k}_{1}{C}_{\text{A}}^{2}{C}_{\text{B}}$
Figure P413_{C}
is taking place in this type of reactor.
Write the rate law solely as a function of conversion, numerically evaluating all possible symbols. (Ans: –r_{A} = 5.03 × 10^{–9} [(1 − X)^{3}/(1 + 3X)^{3/2}] lb mol/ft^{3}·s.)
What is the conversion and rate of reaction when V = 0.2 ft^{3}? (Ans: X = 0.259, –r_{A} = 8.63 × 10^{–10} lb mol/ft^{3}·s.)
Additional information:
Equal moles of A and B are present at t = 0
Initial volume: 0.15 ft^{3}
Value of k_{1}: 1.0 (ft^{3}/lb mol)^{2}=s·^{–1}
The spring constant is such that the relationship between the volume of the reactor and pressure within the reactor is
V = (0.1)(P) (V in ft^{3}, P in atm)
Temperature of system (considered constant): 140°F
Gas constant: 0.73 ft^{3}·atm/lb mol·°R
Further elaboration of the development of the general balance equation may be found on the CRE Web site www.umich.edu/~elements/6e/index.html and also may or may not be found in
G. KEILLOR AND T. RUSSELL, Dusty and Lefty: The Lives of the Cowboys (Audio CD), St. Paul, MN: Highbridge Audio, 2006.
R. M. FELDER, R. W. ROUSSEAU, and L. G. Bullard, Elementary Principles of Chemical Processes, 4th ed. Global Edition, Singapore: John Wiley & Sons Singapore PTE Ltd., 2017, Chap. 4.