It isn’t that they can’t see the solution. It is that they can’t see the problem.
—G. K. Chesterton
Catalysts have been used by humankind for over 2000 years.^{1} The first observed uses of catalysts were in the making of wine, cheese, and bread. For example, it was found that it was always necessary to add small amounts of bread from the previous batch to make the current batch. However, it wasn’t until 1835 that Berzelius began to tie together observations of earlier chemists by suggesting that small amounts of a foreign substance could greatly affect the course of chemical reactions. This mysterious force attributed to the substance was called catalytic (from the ancient Greek katálusis, “to dissolve away the inertness”). In 1894, Ostwald expanded Berzelius’s explanation by stating that catalysts were substances that accelerate the rate of chemical reactions without being consumed during the reaction. During the 180 years since Berzelius’s work, catalysts have come to play a major economic role in the world market. In the United States alone, sales of process catalysts will reach over $20 billion by 2020, the major uses being in petroleum refining and in chemical production.
^{1} S. T. Oyama and G. A. Somorjai, J. Chem. Educ., 65, 765.
What is catalysis?
A catalyst is a substance that affects the rate of a reaction but emerges from the process unchanged. A catalyst usually changes a reaction rate by promoting a different molecular path (“mechanism”) for the reaction. For example, gaseous hydrogen and oxygen are virtually inert at room temperature, but react rapidly when exposed to platinum. The reaction coordinate (cf. Figure 32) shown in Figure 101 is a measure of the progress along the reaction path as H_{2} and O_{2} approach each other and pass over the energy barrier to form H_{2}O. Catalysis is the occurrence, study, and use of catalysts and catalytic processes. Commercial chemical catalysts are immensely important. Approximately onethird of the material gross national product of the United States involves a catalytic process somewhere between raw material and finished product.^{2} The development and use of catalysts is a major part of the constant search for new ways of increasing product yield and selectivity in chemical reactions. Because a catalyst makes it possible to obtain an end product by a different pathway with a lower energy barrier, it can affect both the yield and the selectivity.
^{2} V. Haensel and R. L. Burwell, Jr., Sci. Am., 225(10), 46.
Catalysts can accelerate the reaction rate but cannot change the equilibrium.
Normally when we talk about a catalyst, we mean one that speeds up a reaction, although strictly speaking, a catalyst can either accelerate or slow the formation of a particular product species. A catalyst changes only the rate of a reaction; it does not affect the equilibrium.
The 2007 Nobel Prize for Chemistry was awarded to Gerhard Ertl for his pioneering work on heterogeneous catalytic reactions. A heterogeneous catalytic reaction involves more than one phase; usually the catalyst is a solid and the reactants and products are in liquid or gaseous form. One example is the production of benzene, which is mostly manufactured today from the dehydrogenation of cyclohexane (obtained from the distillation of petroleum crude oil) using platinumonalumina as the catalyst:
The simple and complete separation of the fluid product mixture from the solid catalyst makes heterogeneous catalysis economically attractive, especially because many catalysts are quite valuable and their reuse is demanded.
A heterogeneous catalytic reaction occurs at or very near the fluid–solid interface. The principles that govern heterogeneous catalytic reactions can be applied to both catalytic and noncatalytic fluid–solid reactions. The two other types of heterogeneous reactions involve gas–liquid and gas–liquid–solid systems. Reactions between a gas and a liquid are usually mass transfer–limited as will be discussed in Chapter 14.
Ten grams of this catalyst possess more surface area than a U.S. football field.
Because a catalytic reaction occurs at the fluid–solid interface, a large interfacial area is almost always essential in attaining a significant reaction rate. In many catalysts, this area is provided by an inner porous structure (i.e., the solid contains many fine pores, and the surface of these pores supplies the area needed for the high rate of reaction), see Figures 104(b) and 109. The area possessed by some porous catalysis materials is surprisingly large. A typical silicaalumina cracking catalyst has a pore volume of 0.6 cm^{3}/g and an average pore radius of 4 nm. The corresponding surface area can be the order of 300 m^{2}/g of these porous catalysts. Examples include the Raney nickel used in the hydrogenation of vegetable and animal oils, platinumonalumina used in the reforming of petroleum naphthas to obtain higher octane ratings, and promoted iron used in ammonia synthesis. Sometimes pores are so small that they will admit only small molecules and prevent large ones from entering and being catalyzed. Materials with this type of pore are called molecular sieves, and they may be derived from natural substances such as certain clays and zeolites, or they may be totally synthetic, such as some crystalline aluminosilicates (see Figure 102). These sieves can form the basis for quite selective catalysts; the pores can control the residence time of various molecules near the catalytically active surface to a degree that essentially allows only the desired molecules to react. One example of the high selectivity of zeolite catalysts is the formation of paraxylene from toluene and methane shown in Figure 102(b).^{3} Here, methane and toluene enter through the zeolite pore and react on the interior surface to form a mixture of ortho, meta, and paraxylenes. However, the size of the pore mouth is such that only paraxylene can exit through the pore mouth, as meta and orthoxylene with their methyl group on the side cannot fit through the pore mouth. There are interior sites that can isomerize orthoxylene and metaxylene to paraxylene. Hence, we have a very high selectivity to form paraxylene.
^{3} R. I. Masel, Chemical Kinetics and Catalysis, New York: Wiley Interscience, 2001, p. 741.
Catalyst types:
Porous
Molecular sieves
Monolithic
Supported
Unsupported
Typical zeolite catalyst
High selectivity to paraxylene
A very thorough and wellwritten book.
In some cases, a catalyst consists of minute particles of an active material dispersed over a lessactive substance called a support. The active material is frequently a pure metal or metal alloy. Such catalysts are called supported catalysts, as distinguished from unsupported catalysts. Catalysts can also have small amounts of active ingredients added called promoters, which increase their activity. Examples of supported catalysts are the packedbed catalytic converter in an automobile, the platinumonalumina catalyst used in petroleum reforming, and the vanadium pentoxide on silica used to oxidize sulfur dioxide in manufacturing sulfuric acid. Typical unsupported catalysts include platinum gauze for ammonia oxidation, promoted iron for ammonia synthesis, and the silicaalumina dehydrogenation catalyst used in butadiene manufacture.
In this section, we focus our attention on gasphase reactions catalyzed by solid surfaces. For a catalytic reaction to occur, at least one and frequently all of the reactants must become attached to the surface. This attachment is known as adsorption and takes place by two different processes: physical adsorption and chemisorption. Physical adsorption is similar to condensation. The process is exothermic, and the heat of adsorption is relatively small, being on the order of 1–15 kcal/mol. Here, the forces of attraction between the gas molecules and the solid surface are weak. These van der Waals forces consist of interaction between permanent dipoles, between a permanent dipole and an induced dipole, and/or between neutral atoms and molecules. The amount of gas physically adsorbed decreases rapidly with increasing temperature, and above its critical temperature only very small amounts of a substance are physically adsorbed.
The type of adsorption that affects the rate of a chemical reaction is chemisorption. Here, the adsorbed atoms or molecules are held to the surface by valence forces of the same type as those that occur between bonded atoms in molecules. As a result, the electronic structure of the chemisorbed molecule is perturbed significantly, causing it to be extremely reactive. Interaction with the catalyst causes bonds of the adsorbed reactant to be stretched, making them easier to break.
Figure 103 shows the bonding from the adsorption of ethylene on a platinum surface to form chemisorbed ethylidyne. Like physical adsorption, chemisorption is an exothermic process, but the heats of adsorption are generally of the same magnitude as the heat of a chemical reaction (i.e., 40–400 kJ/mol). If a catalytic reaction involves chemisorption, it must be carried out within the temperature range where chemisorption of the reactants is appreciable.
Chemisorption on active sites is what catalyzes the reaction.
In a landmark contribution to catalytic theory, H. S. Taylor suggested that a reaction is not catalyzed over the entire solid surface but only at certain active sites or centers.^{4} He visualized these sites as unsaturated atoms in the solids that resulted from surface irregularities, dislocations, edges of crystals, and cracks along grain boundaries. Other investigators have taken exception to this definition, pointing out that other properties of the solid surface are also important. The active sites can also be thought of as places where highly reactive intermediates (i.e., chemisorbed species) are stabilized long enough to react. This stabilization of a reactive intermediate is key in the design of any catalyst. Consequently, for our purposes we will define an active site as a point on the catalyst surface that can form strong chemical bonds with an adsorbed atom or molecule.
^{4} H. S. Taylor, Proc. R. Soc. London, A108, 105 (1928).
TOF
One parameter used to quantify the activity of a catalyst is the turnover frequency (TOF), f. It is the number of molecules reacting per active site per second at the conditions of the experiment. When a metal catalyst such as platinum is deposited on a support, the metal atoms are considered active sites. The dispersion, D, of the catalyst is the fraction of the metal atoms deposited on a catalyst that are on the surface. The TOFs for a number of reactions are shown in the Additional Material for Chapter 10 on the Web site (http://www.umich.edu/~elements/6e/10chap/expanded_ch10_TOF.pdf).
One common way to classify catalysts is in terms of the type of reaction they catalyze.
Table 101 gives a list of representative reactions and their corresponding catalysts. Further discussion of each of these reaction classes and the materials that catalyze them can be found on the CRE Web site’s Professional Reference Shelf R10.1 (http://www.umich.edu/~elements/6e/10chap/prof.html).
TABLE 101 TYPES OF REACTIONS AND REPRESENTATIVE CATALYSTS
Reaction 
Catalysts 
1. Halogenation–dehalogenation 
CuCl_{2}, AgCl, Pd 
2. Hydration–dehydration 
Al_{2}O_{3}, MgO 
3. Alkylation–dealkylation 
AlCl_{3}, Pd, Zeolites 
4. Hydrogenation–dehydrogenation 
Co, Pt, Cr_{2}O_{3}, Ni 
5. Oxidation 
Cu, Ag, Ni, V_{2}O_{5} 
6. Isomerization 
AlCl_{3}, Pt/Al_{2}O_{3}, Zeolites 
If, for example, we were to form styrene from an equimolar mixture of ethylene and benzene, we could first carry out an alkylation reaction to form ethyl benzene, which is then dehydrogenated to form styrene. We need both an alkylation catalyst and a dehydrogenation catalyst:
${{{\text{C}}_{2}\text{H}}_{4}+{\text{C}}_{6}\text{H}}_{6}\text{}\underset{\text{traceHCl}}{\overset{{\text{AlCl}}_{3}}{\to}}{{{{\text{C}}_{\text{6}}\text{H}}_{5}\text{C}}_{2}\text{H}}_{\text{5}}\text{}\stackrel{\text{Ni}}{\to}{\text{C}}_{\text{6}}{\text{}\text{H}}_{5}\text{CH\u2550}{\text{CH}}_{2}+{\text{H}}_{2}$
A photograph of different types and sizes of catalysts is shown in Figure 104(a). A schematic diagram of a tubular reactor packed with catalytic pellets is shown in Figure 104(b). The overall process by which heterogeneous catalytic reactions proceed can be broken down into the sequence of individual steps shown in Table 102 and pictured in Figure 105 for an isomerization reaction.
Each step in Table 102 is shown schematically in Figure 105.
A reaction takes place on the surface, but the species involved in the reaction must get to and from the surface.
The overall rate of reaction is limited by the rate of the slowest step in the sequence. When the diffusion steps (1, 2, 6, and 7 in Table 102) are very fast compared with the surface reactionrate steps (3, 4, and 5), the concentrations in the immediate vicinity of the active sites are indistinguishable from those in the bulk fluid. In this situation, the transport or diffusion steps do not affect the overall rate of the reaction. In other situations, if the reaction steps are very fast compared with the diffusion steps, mass transport does affect the reaction rate. In systems where diffusion from the bulk gas or liquid to the external catalyst surface or to the mouths of catalyst pores affects the rate, that is, Steps 1 and 7, changing the flow conditions past the catalyst should change the overall reaction rate (see Chapter 14). Once inside the porous catalysts, on the other hand, diffusion within the catalyst pores, that is, Steps 2 and 6, may limit the rate of reaction and, as a result, the overall rate will be unaffected by external flow conditions even though diffusion affects the overall reaction rate (see Chapter 15).
TABLE 102 STEPS IN A CATALYTIC REACTION

There are many variations of the situation described in Table 102. Sometimes, of course, two reactants are necessary for a reaction to occur, and both of these may undergo the steps listed above. Other reactions between two substances may have only one of them adsorbed.
In this chapter we focus on:
3. Adsorption
4. Surface reaction
5. Desorption
With this introduction, we are ready to treat individually the steps involved in catalytic reactions. In this chapter, only Steps 3, 4, and 5—that is, adsorption, surface reaction, and desorption—are considered as we assume that the diffusion steps (1, 2, 6, and 7) are very fast so that the overall reaction rate is not affected by mass transfer in any fashion. Further treatment of the effects involving diffusion limitations is provided in Figure 145 in Chapter 14 and Figure 155 in Chapter 15.
Where Are We Heading?^{†} As we saw in Chapter 7, one of the tasks of a chemical reaction engineer is to analyze rate data and to develop a rate law that can be used in reactor design. Rate laws in heterogeneous catalysis seldom follow powerlaw models (Eq. 33) and hence are inherently more difficult to formulate from the data. To develop an indepth understanding and insight as to how the rate laws are formed from heterogeneous catalytic data, we are going to proceed in somewhat of a reverse manner than what is normally done in industry when one is asked to develop a rate law. That is, we will first postulate catalytic mechanisms and then derive rate laws for the various mechanisms. The mechanism will typically have an adsorption step, a surface reaction step, and a desorption step, one of which is usually ratelimiting. Suggesting mechanisms and ratelimiting steps is not the first thing we normally do when presented with data. However, by deriving equations for different mechanisms, we will observe the various forms of the rate law one can have in heterogeneous catalysis. Knowing the different forms that catalytic rate equations can take, it will be easier to view the trends in the data and deduce the appropriate rate law. This deduction is usually what is done first in industry before a mechanism is proposed. Knowing the form of the rate law, one can then numerically evaluate the ratelaw parameters and postulate a reaction mechanism and ratelimiting step that are consistent with the experimental data. Finally, we use the rate law to design catalytic reactors. This procedure is shown in Figure 106. The dashed lines represent feedback to obtain new data in specific regions (e.g., concentrations, temperature) to evaluate the ratelaw parameters more precisely or to differentiate between reaction mechanisms.
^{†} “If you don’t know where you are going, you’ll probably wind up someplace else.” Yogi Berra, New York Yankees.
An algorithm
Knowing the various forms a catalytic rate law can take will help us interpret the data.
An overview
#ChanceCard: Do not pass go. Proceed directly to Section 10.2.3.
We will now discuss each of the steps shown in Figure 105 and Table 102. As mentioned earlier, this chapter focuses on Steps 3, 4, and 5 (the adsorption, surface reaction, and desorption steps) by assuming that Steps 1, 2, 6, and 7 are very rapid. Consequently, to understand when this assumption is valid, we shall give a quick overview of Steps 1, 2, 6, and 7. These steps involve diffusion of the reactants to and within the catalyst pellet. While these diffusion steps are covered in detail in Figure 145 in Chapter 14 and Figure 155 in Chapter 15, it’s worthwhile to give a brief description of these two masstransfer steps to better understand the entire sequence of steps. If you have had the core course in Mass Transfer or Transport Phenomena you can skip Sections 10.2.1 and 10.2.2, and go directly to Section 10.2.3.
For the moment, let’s assume that the transport of A from the bulk fluid to the external surface of the catalyst is the slowest step in the sequence shown in Figure 105. We lump all the resistance to transfer of reactant species A from the bulk fluid to the surface, in the mass transfer boundary layer surrounding the pellet. In this step the reactant A, which is at a bulk concentration C_{A}_{b} must travel (diffuse) through the boundary layer of thickness δ to the external surface of the pellet where the concentration is C_{A}_{s}, as shown in Figure 107. The rate of transfer (and hence rate of reaction, ${r}_{\text{A}}^{\prime}$) for this slowest step is
Rate = k_{C} (C_{A}_{b} – C_{A}_{s})
where the mass transfer coefficient, k_{C}, is a function of the hydrodynamic conditions, namely the fluid velocity, U, and the particle diameter, D_{p}.
External and internal mass transfer in catalysis are covered in detail in Chapters 14 and 15.
As we will see (Chapter 14), the mass transfer coefficient is inversely proportional to the boundary layer thickness, δ, and directly proportional to the diffusion coefficient (i.e., the diffusivity D_{AB}).
${k}_{\text{C}}=\frac{{\text{D}}_{\text{AB}}}{\delta}$
At low velocities of fluid flow over the pellet, the boundary layer across which A and B must diffuse is thick, and it takes a long time for A to travel to the surface, resulting in a small mass transfer coefficient k_{C}. As a result, mass transfer across the boundary layer is slow and limits the rate of the overall reaction. As the velocity over the pellet is increased, the boundary layer becomes thinner and the mass transfer rate is increased. At very high velocities, the boundary layer thickness, δ, is so small that it no longer offers any significant resistance to the diffusion across the boundary layer. As a result, external mass transfer no longer limits the rate of reaction. This external resistance also decreases as the particle size is decreased. As the fluid velocity increases and/or the particle diameter decreases, the mass transfer coefficient increases until a plateau is reached, as shown in Figure 108. On this plateau, C_{A}_{b} ≈ C_{A}_{s}, and one of the other steps in the sequence, that is, Steps 2–6, is the slowest step and limits the overall reaction rate. Further details on external mass transfer are discussed in Chapter 14.
High external fluid velocities decrease the resistance to external mass transfer.
Now consider that we are operating at a fluid velocity where external diffusion is no longer the ratelimiting step and that internal diffusion is the slowest step. In Step 2 of Figure 105, the reactant A diffuses from the external pellet surface at a concentration is C_{A}. As A diffuses into the interior of the pellet, it reacts with catalyst deposited on the sides of the catalyst pellet’s inner pore walls.
The smaller the size of the catalyst pellet, the smaller the resistance to internal mass transfer (i.e., diffusion).
For large pellets, it takes a long time for the reactant A to diffuse into the interior, compared to the time that it takes for the reaction to occur on the interior pore surface. Under these circumstances, the reactant is only consumed near the exterior surface of the pellet and the catalyst near the center of the pellet is wasted catalyst. On the other hand, for very small pellets it takes very little time to diffuse into and out of the pellet interior and, as a result, internal diffusion no longer limits the rate of reaction. When internal mass transfer does not limit the rate of reaction, the concentration inside the pores of the catalyst pellet is equal to the concentration at the external pellet surface, C_{A}_{s}. As a result the corresponding rate law is
Rate = k_{r} C_{A}_{s}
where C_{A}_{s} is the concentration at the external pellet surface and k_{r} is an overall rate constant, which is a function of particle size. The overall rate constant, k_{r}, increases as the pellet diameter decreases. In Chapter 15, we show that Figure 155 can be combined with Equation (1534) to arrive at the plot of k_{r} as a function of D_{P}, shown in Figure 109(b).
We see in Figure 109 that for small particle sizes, internal diffusion is no longer the slow step and that the surface reaction sequence of adsorption, surface reaction, and desorption (Steps 3, 4, and 5 in Figure 105) limit the overall rate of reaction. Consider now one more important point about internal diffusion and surface reaction. Steps (2–6) are not at all affected by flow conditions external to the pellet.
In the material that follows, we are going to choose our pellet size and external fluid velocity such that neither external diffusion (Chapter 14), nor internal diffusion (Chapter 15), limits the reaction. Instead, we assume that either Step 3 (adsorption), Step 4 (surface reaction), or Step 5 (desorption), or a combination of these steps, limits the overall rate of reaction.
Because chemisorption is a necessary part of a catalytic process, we shall discuss it before treating catalytic reaction rates. The letter S will represent an active site; alone, it will denote a vacant site, with no atom, molecule, or complex adsorbed on it. The combination of S with another letter (e.g., A⋅S) will mean that one unit of species A has been chemically adsorbed on the site S. Species A can be an atom, molecule, or some other atomic combination, depending on the circumstances. Consequently, the adsorption of A on a site S is represented by
$\text{A}+\text{S}\overrightarrow{\leftarrow}\text{A}\cdot \text{S}$
The total molar concentration of active sites per unit mass of catalyst is equal to the number of active sites per unit mass divided by Avogadro’s number and will be labeled C_{t} (mol/gcat). The molar concentration of vacant sites, C_{υ}(mol/gcat), is the number of vacant sites per unit mass of catalyst divided by Avogadro’s number. In the absence of catalyst deactivation, we assume that the total concentration of active sites, C_{t}, remains constant. Some further definitions include
P_{i} = partial pressure of species i in the gas phase (atm, bar, or kPa)
C_{i·S} = surface concentration of sites occupied by species i (mol/gcat)
A conceptual model depicting species A and B adsorbed on two different sites is shown in Figure 1010.
For the system shown in Figure 1010, the total concentration of sites is
Site balance
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{t}={\text{C}}_{\upsilon}+{\text{C}}_{\text{A}\cdot \text{S}}+{\text{C}}_{\text{B}\cdot \text{S}}\\ \hline\end{array}& \text{(101)}\end{array}$
This equation is referred to as a site balance. A typical value for the total concentration of sites could be the order of 10^{22} sites/gcat. To express the site concentration in terms of (moles/gcat) we divide by Avogadro’s number to get “moles of sites/gcat.”
Now consider the adsorption of a nonreacting gas onto the surface of a catalyst. Adsorption data are frequently reported in the form of adsorption isotherms. Isotherms portray the amount of a gas adsorbed on a solid at different pressures at a given temperature.
Postulate models; then see which one(s) fit(s) the data.
First, an adsorption mechanism is proposed, and then the isotherm (see Figure 1011, page 455) obtained from the mechanism is compared with the experimental data. If the isotherm predicted by the model agrees with the experimental data, the model may reasonably describe what is occurring physically in the real system. If the predicted curve does not agree with the experimental data, the model fails to match the physical situation in at least one important characteristic and perhaps more.
We will consider two types of adsorption: molecular adsorption and dissociative adsorption. To illustrate the difference between molecular adsorption and dissociative adsorption, we will postulate two models for the adsorption of carbon monoxide on a metal surface active site. In the molecular adsorption model, carbon monoxide is adsorbed as molecules, CO,
$\text{CO}+\text{S}\overrightarrow{\leftarrow}\text{CO}\cdot \text{S}$
as is the case on nickel
In the dissociative adsorption model, carbon monoxide is adsorbed as oxygen and carbon atoms instead of molecular CO
$\text{CO}+\text{2}\text{S}\overrightarrow{\leftarrow}\text{C}\cdot \text{S}+\text{O}\cdot \text{S}$
as is the case on iron^{5}
$\begin{array}{cccc}\text{CO}& \text{}& \text{}& \text{}\\ +& \text{}& \text{}& \text{}\\ \text{}& \text{Fe}\text{Fe}\text{Fe}& \overrightarrow{\leftarrow}& \stackrel{\underset{\underset{\cdot}{\cdot}}{\text{C}}}{\text{Fe}}\stackrel{\underset{\underset{\cdot}{\cdot}}{\text{O}}}{\text{Fe}}\text{Fe}\end{array}$
^{5} R. I. Masel, Principles of Adsorption and Reaction on Solid Surfaces, New York: Wiley, 1996.
Two models:
1. Adsorption as CO
2. Adsorption as C and O
The former is called molecular or nondissociated adsorption (e.g., CO) and the latter is called dissociative adsorption (e.g., C and O). Whether a molecule adsorbs nondissociatively or dissociatively depends on the surface.
Molecular Adsorption: The adsorption of carbon monoxide molecules will be considered first. Because the carbon monoxide does not react further after being adsorbed, we need only to consider the molecular adsorption process:
$\begin{array}{cc}\text{CO}+\text{S}\overrightarrow{\leftarrow}\text{CO}\cdot \text{S}& \text{(102)}\end{array}$
Molecular adsorption
In obtaining a rate law for the rate of adsorption, the reaction in Equation (102) can be treated as an elementary reaction. The rate of attachment of the carbon monoxide molecules to the active site on the surface is proportional to the number of collisions that CO molecules make with a surface active site per second. In other words, a specific fraction of the molecules that strike the surface become adsorbed. The collision rate is, in turn, directly proportional to the carbon monoxide partial pressure, P_{CO}. Because carbon monoxide molecules adsorb only on vacant sites and not on sites already occupied by other carbon monoxide molecules, the rate of attachment is also directly proportional to the concentration of vacant sites, C_{ν}. Combining these two facts along with the law of mass action means that the rate of attachment of carbon monoxide molecules to the surface is directly proportional to the product of the partial pressure of CO (P_{CO}) and the concentration of vacant sites (C_{ν}); that is,
$\text{Rate of attachment}={k}_{\text{A}}{\text{P}}_{\text{CO}}{\text{C}}_{\upsilon}$
P_{CO} = C_{CO}RT
The rate of detachment of molecules from the surface can be a firstorder process; that is, the detachment of carbon monoxide molecules from the surface is usually directly proportional to the concentration of sites occupied by the adsorbed molecules (e.g., C_{CO·S}):
$\text{Rate of detachment}={k}_{\text{\u2013A}}{\text{C}}_{\text{CO}\xb7S}$
The net rate of adsorption is equal to the rate of molecular attachment to the surface minus the rate of detachment from the surface. If k_{A} and k_{A} are the constants of proportionality for the attachment and detachment processes, then
$\begin{array}{cc}{\text{r}}_{\text{AD}}={k}_{\text{A}}{P}_{\text{CO}}{\text{C}}_{\upsilon}{k}_{\text{A}}{\text{C}}_{\text{CO}\cdot \text{S}}& \text{(103)}\end{array}$
The ratio K_{A} = k_{A}/k_{A} is the adsorption equilibrium constant. Using K_{A} to rearrange Equation (103) gives
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{AD}}={k}_{\text{A}}({P}_{\text{CO}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{CO}\cdot \text{S}}}{{K}_{\text{A}}})\\ \hline\end{array}& \text{(104)}\end{array}$
$\begin{array}{c}\text{A}+\text{S}\stackrel{\text{Adsorption}}{\stackrel{\text{}}{\overrightarrow{\leftarrow}}\text{A}\cdot \text{S}}\\ {\text{r}}_{\text{AD}}={k}_{\text{A}}({P}_{\text{A}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{A}\cdot \text{S}}}{{K}_{\text{A}}})\end{array}$
The adsorption rate constant, k_{A}, for molecular adsorption is virtually independent of temperature, while the desorption constant, k·_{A}, increases exponentially with increasing temperature. Consequently, the equilibrium adsorption constant K_{A} decreases exponentially with increasing temperature.
Site balance
Because carbon monoxide is the only material adsorbed on the catalyst, the site balance gives
$\begin{array}{cc}{\text{C}}_{t}={\text{C}}_{\text{}\upsilon}+{\text{C}}_{\text{CO}\cdot \text{S}}& \text{(105)}\end{array}$
$\begin{array}{c}\hline \hfill {\text{r}}_{\text{A}}^{\prime}={\text{r}}_{\text{AD}}=\left(\frac{\text{mol}}{\text{gcat}\cdot \text{s}}\right)\\ \hfill {k}_{\text{A}}=\left(\frac{1}{\text{atm}\cdot \text{s}}\right)\\ \hfill {P}_{\text{A}}=(\text{atm)}\text{}\\ \hfill {\text{C}}_{\upsilon}=\left(\frac{\text{mol}}{\text{gcat}}\right)\\ \hfill {K}_{\text{A}}=\left(\frac{1}{\text{atm}}\right)\\ \hfill {\text{C}}_{\text{A}\cdot \text{S}}=\left(\frac{\text{mol}}{\text{gcat}}\right)\\ \hline\end{array}$
At equilibrium, the net rate of adsorption equals zero, that is, r_{AD} ≡ 0. Setting the lefthand side of Equation (104) equal to zero and solving for the concentration of CO adsorbed on the surface, we get
$\begin{array}{cc}{\text{C}}_{\text{CO}\cdot \text{S}}={K}_{\text{A}}{\text{C}}_{\upsilon}{P}_{\text{CO}}& \text{(106)}\end{array}$
Using Equation (105) to give in terms of C_{CO·S} and the total number of sites C_{t}, we can solve for the equilibrium value of C_{CO·S} in terms of constants and the pressure of carbon monoxide
$\begin{array}{cc}{\text{C}}_{\text{CO}\cdot \text{S}}={K}_{\text{A}}{\text{C}}_{\upsilon}{P}_{\text{CO}}={K}_{\text{A}}{P}_{\text{CO}}({\text{C}}_{t}{\text{C}}_{\text{CO}\text{}\cdot \text{S}})& \text{}\end{array}$
Rearranging gives us a Langmuir isotherm^{6}
^{6} Named after Irving Langmuir (1881–1957), who first proposed it. He received the Nobel Prize in 1932 for his discoveries in surface chemistry (https://www.nobelprize.org/nobel_prizes/chemistry/laureates/1932/langmuirbio.html).
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{\text{CO}\cdot \text{S}}=\frac{{K}_{\text{A}}{P}_{\text{CO}}{\text{C}}_{t}}{1+{K}_{\text{A}}{P}_{\text{CO}}}\\ \hline\end{array}& \text{(107)}\end{array}$
Langmuir adsorption isotherm
This equation thus gives the equilibrium concentration of carbon monoxide adsorbed on the surface, C_{CO·S}, as a function of the partial pressure of carbon monoxide, and is an equation for the adsorption isotherm. This particular type of isotherm equation is called a Langmuir isotherm.^{7} Figure 1011(a) shows a singlesite Langmuir isotherm for molecular adsorption in terms of the adsorbed concentration (mass of CO adsorbed per unit mass of catalyst) as a function of the partial pressure of CO, that is, Equation (107).
^{7} Ibid.
Now, for the case of dissociative adsorption, Equation (1011), Figure 1011(b) shows the concentration of the dissociated atoms of C and O adsorbed per unit mass of catalyst of a function of partial pressure of CO, that is, Equation (1011).
One method of checking whether or not a model (e.g., molecular adsorption vs. dissociative adsorption) predicts the behavior of the experimental data is to linearize the model’s equation and then plot the indicated variables against one another. For example, the molecular adsorption isotherm, Equation (107), may be arranged in the form
$\begin{array}{cc}\frac{{P}_{\text{CO}}}{{\text{C}}_{\text{CO}\cdot \text{S}}}=\frac{1}{{K}_{\text{A}}{\text{C}}_{t}}+\frac{{P}_{\text{CO}}}{{\text{C}}_{t}}& \text{(108)}\end{array}$
Molecular adsorption
and the linearity of a plot of P_{CO}/C_{CO·S} as a function of P_{CO} will determine whether the data conform to molecular adsorption, that is, a Langmuir singlesite isotherm.
Dissociative Adsorption: We next derive the isotherm for carbon monoxide disassociating into separate atoms as it adsorbs on the surface, that is,
$\text{CO}+2\text{S}\overrightarrow{\leftarrow}\text{C}\cdot \text{S}+\text{O}\cdot \text{S}$
Dissociative adsorption
When the carbon monoxide molecule dissociates upon adsorption, it is referred to as the dissociative adsorption of carbon monoxide. As in the case of molecular adsorption, the rate of adsorption is proportional to the partial pressure of carbon monoxide in the system because this rate is governed by the number of gaseous CO collisions with the surface. For a molecule to dissociate as it adsorbs, however, two adjacent vacant active sites are required, rather than the single site needed when a substance adsorbs in its molecular form. The probability of two vacant sites occurring adjacent to one another is proportional to the square of the concentration of vacant sites, cf. the law of mass action. These two observations mean that the rate of adsorption is proportional to the product of the carbon monoxide partial pressure and the square of the vacantsite concentration, ${P}_{\text{CO}}{\text{C}}_{\upsilon}^{\text{2}}$.
For desorption to occur, two occupied sites must be adjacent, meaning that the rate of desorption is proportional to the product of the occupiedsite concentration, (C·S) × (O·S). The net rate of adsorption can then be expressed as
$\begin{array}{cc}{\text{r}}_{\text{AD}}={k}_{\text{A}}{P}_{\text{CO}}{\text{C}}_{\upsilon}^{\text{2}}{k}_{\text{A}}{\text{C}}_{\text{O}\cdot \text{S}}{\text{C}}_{\text{C}\cdot \text{S}}& \text{(109)}\end{array}$
Factoring out k_{A}, the equation for dissociative adsorption is
${\text{r}}_{\text{AD}}={k}_{\text{A}}({P}_{\text{CO}}{\text{C}}_{\upsilon}^{2}\frac{{\text{C}}_{\text{C}\cdot \text{S}}{\text{C}}_{\text{O}\cdot \text{S}}}{{K}_{\text{A}}})$
where
${K}_{\text{A}}=\frac{{k}_{\text{A}}}{{k}_{\text{A}}}$
Rate of dissociative adsorption
For dissociative adsorption, both k_{A} and k·_{A} increase exponentially with increasing temperature, while the adsorption equilibrium constant K_{A} decreases with increasing temperature because adsorption is an exothermic reaction step.
At equilibrium, r_{AD} ≡ 0, and
${k}_{\text{A}\text{}}{P}_{\text{CO}}{\text{C}}_{\upsilon}^{\text{2}}={k}_{\text{A}}{\text{C}}_{\text{C}\cdot \text{S}}{\text{C}}_{\text{O}\cdot \text{S}}$
For C_{C·S} = C_{O·S}
$\begin{array}{cc}{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}}{\text{C}}_{\upsilon}={\text{C}}_{\text{O}\cdot \text{S}}& \text{(1010)}\end{array}$
Substituting for C_{C·S} and C_{O·S} in a site balance equation (101),
$\begin{array}{cc}{\text{C}}_{t}& ={\text{C}}_{\upsilon}+{\text{C}}_{\text{O}\cdot \text{S}}+{\text{C}}_{\text{C}\cdot \text{S}}\hfill \\ \text{}& ={\text{C}}_{\upsilon}+{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}}{\text{C}}_{\upsilon}={\text{C}}_{\upsilon}(1+2{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}})\hfill \end{array}$
Site balance:
Solving for C_{υ}
${\text{C}}_{\upsilon}={\text{C}}_{\text{t}}/(1+2{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}})$
This value may be substituted into Equation (1010) to give an expression that can be solved for the equilibrium value of C_{O·S} as a function of the partial pressure of CO. The resulting equation for the isotherm shown in Figure 1011(b) is
$\begin{array}{cc}\begin{array}{c}\hline \text{}{\text{C}}_{\text{O}\cdot \text{S}}=\frac{{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}}{\text{C}}_{t}}{\text{1+2}{\left({K}_{\text{A}}{P}_{\text{CO}}\right)}^{\text{1/2}}}\\ \hline\end{array}& \text{(1011)}\end{array}$
Langmuir isotherm for carbon monoxide dissociative adsorption as atomic carbon and oxygen
Taking the inverse of both sides of the equation, then multiplying through by (P_{CO})^{1/2}, yields
$\begin{array}{cc}\frac{{\left({P}_{\text{CO}}\right)}^{\text{1/2}}}{{\text{C}}_{\text{O}\cdot \text{S}}}=\frac{1}{{\text{C}}_{t}{\left({K}_{\text{A}}\right)}^{\text{1/2}}}+\frac{2{\left({P}_{\text{CO}}\right)}^{\text{1/2}}}{{\text{C}}_{t}}& \text{(1012)}\end{array}$
If dissociative adsorption is the correct model, a plot of $\text{}({P}_{\text{CO}}^{1/2}/{\text{C}}_{\text{O}\cdot \text{S}})$ versus $\text{}{P}_{\text{CO}}^{1/2}$ should be linear with slope (2/C_{t}).
When more than one substance is present, the adsorption isotherm equations are somewhat more complex. The principles are the same, though, and the isotherm equations are easily derived. It is left as an exercise to show that the adsorption isotherm of A in the presence of another adsorbate B is given by the relationship
$\begin{array}{cc}{\text{C}}_{\text{A}\cdot \text{S}}=\frac{{K}_{\text{A}}{P}_{\text{A}}{\text{C}}_{\text{t}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}}& \text{(1013)}\end{array}$
When the adsorption of both A and B are firstorder processes, the desorptions are also first order, and both A and B are adsorbed as molecules. The derivations of other Langmuir isotherms are relatively easy.
In obtaining the Langmuir isotherm equations, several aspects of the adsorption system were presupposed in the derivations. The most important of these, and the one that has been subject to the greatest doubt, is that a uniform surface is assumed. In other words, any active site has the same attraction for an impinging molecule as does any other active site. Isotherms different from the Langmuir isotherm, such as the Freundlich isotherm, may be derived based on various assumptions concerning the adsorption system, including different types of nonuniform surfaces.
Note assumptions in the model and check their validity.
The rate of adsorption of species A onto a solid surface site
$\text{A}+\text{S}\overrightarrow{\leftarrow}\text{A}\cdot \text{S}$
is given by
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{AD}}={k}_{\text{A}}({P}_{\text{A}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{A}\cdot \text{S}}}{{K}_{\text{A}}})\\ \hline\end{array}& \text{(1014)}\end{array}$
After a reactant has been adsorbed onto the surface, that is, A · S, it is capable of reacting in a number of ways to form the reaction product. Three of these ways are:
Single site. The surface reaction may be a singlesite mechanism in which only the site on which the reactant is adsorbed is involved in the reaction. For example, an adsorbed molecule of A may isomerize (or perhaps decompose) directly on the site to which it is attached, such as pentene isomerization
The pentane isomerization can be written in generic form as
$\text{A}\cdot \text{S}\overrightarrow{\leftarrow}\text{B}\cdot \text{S}$
Each step in the reaction mechanism is elementary, so the surface reaction rate law is
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{S}}={k}_{\text{S}}{\text{C}}_{\text{A}\cdot \text{S}}{k}_{\text{S}}{\text{C}}_{\text{B}\cdot \text{S}}={k}_{\text{S}}({\text{C}}_{\text{A}\cdot \text{S}}\frac{{\text{C}}_{\text{B}\cdot \text{S}}}{{K}_{\text{S}}})\\ \hline\end{array}& \text{(1015)}\end{array}$
Single Site
${k}_{\text{s}}=\left(\frac{1}{\text{s}}\right)$
where K_{S} is the surfacereaction equilibrium constant K_{S} = k_{S}/k_{–S}
K_{s} = (dimensionless)
Dual site. The surface reaction may be a dualsite mechanism in which the adsorbed reactant interacts with another site (either unoccupied or occupied) to form the product.
First type of dualsite mechanism
For example, adsorbed A may need to react with an adjacent vacant site to further become destabilized to yield a vacant site and a site on which the product is adsorbed. In case of the dehydration of butanol, the reaction products adsorb on two adjacent sites.
For the generic reaction
$\text{A}\cdot \text{S}+\text{S}\overrightarrow{\leftarrow}\text{B}\cdot \text{S}+\text{S}$
the corresponding surfacereaction rate law is
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{S}}={k}_{\text{S}}[{\text{C}}_{\text{A}\cdot \text{S}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{B}\cdot \text{S}}{\text{C}}_{\upsilon}}{{K}_{\text{S}}}]\\ \hline\end{array}& \text{(1016)}\end{array}$
Dual Site
$\begin{array}{c}{\text{r}}_{\text{S}}=\left(\frac{\text{mol}}{\text{gcat}\cdot \text{S}}\right)\hfill \\ {k}_{\text{S}}=\left(\frac{\text{gcat}}{\text{mol}\cdot \text{S}}\right)\hfill \\ {K}_{\text{S}}=\left(\text{dimensionless}\right)\hfill \end{array}$
A second dualsite mechanism is the reaction between two adsorbed species, such as the reaction of CO with O.
For the generic reaction
$\text{A}\cdot \text{S}+\text{B}\cdot \text{S}\overrightarrow{\leftarrow}\text{C}\cdot \text{S}+\text{D}\cdot \text{S}$
the corresponding surfacereaction rate law is
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\text{A}\text{}\cdot \text{S}}{\text{C}}_{\text{B}\cdot \text{S}}\frac{{\text{C}}_{\text{C}\cdot \text{S}}{\text{C}}_{\text{D}\cdot \text{S}}}{{K}_{\text{S}}})\\ \hline\end{array}& \text{(1017)}\end{array}$
A third dualsite mechanism is the reaction of two species adsorbed on different types of sites S and S′, such as the reaction of CO with O.
For the generic reaction
$\text{A}\cdot \text{S}+\text{B}\cdot \text{S'}\overrightarrow{\leftarrow}\text{C}\cdot \text{S'}+\text{D}\cdot \text{S}$
the corresponding surfacereaction rate law is
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\text{A}\cdot \text{S}}{\text{C}}_{\text{B}\cdot \text{S'}}\frac{{\text{C}}_{\text{C}\cdot \text{S'}}{\text{C}}_{\text{D}\cdot \text{S}}}{{K}_{\text{S}}})\\ \hline\end{array}& \text{(1018)}\end{array}$
LangmuirHinshelwood kinetics
Reactions involving either single or dualsite mechanisms, which were described earlier, are sometimes referred to as following LangmuirHinshelwood kinetics.
Eley–Rideal. A third mechanism is the reaction between an adsorbed molecule and a molecule in the gas phase, such as the reaction of propylene and benzene (cf. the reverse reaction in Figure 1013 on page 463.)
For the generic reaction
$\text{A}\cdot \text{S}+\text{B}\left(g\right)\overrightarrow{\leftarrow}\text{C}\cdot \text{S}$
the corresponding surfacereaction rate law is
${k}_{s}=\left(\frac{1}{\text{atm}\cdot \text{s}}\right)$
${K}_{S}=\left(\frac{1}{\text{atm}}\right)$
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{S}}={k}_{\text{S}}[{\text{C}}_{\text{A}\cdot \text{S}}{P}_{\text{B}}\frac{{\text{C}}_{\text{C}\cdot \text{S}}}{{K}_{\text{S}}}]\\ \hline\end{array}& \text{(1019)}\end{array}$
Surface reaction models
This type of mechanism is referred to as an Eley–Rideal mechanism.
K_{DC} = (atm)
${k}_{\text{D}}=\left(\frac{1}{\text{S}}\right)$
In each of the preceding cases, the products of the surface reaction adsorbed on the surface are subsequently desorbed into the gas phase. For the desorption of a species (e.g., C)
$\text{C}\cdot \text{S}\overrightarrow{\leftarrow}\text{C}+\text{S}$
the rate of desorption of C is
$\begin{array}{cc}\begin{array}{c}\hline \text{}{\text{r}}_{\text{DC}}={k}_{\text{D}}({\text{C}}_{\text{C}\cdot \text{S}}\frac{{P}_{\text{C}}{\text{C}}_{\upsilon}}{{K}_{\text{DC}}})\\ \hline\end{array}& \text{(1020)}\end{array}$
where K_{DC} is the desorption equilibrium constant with units of atm. Now let’s look at the above desorption step of C•S from right to left. We note that the desorption step for C•S is just the reverse of the adsorption step. Consequently, the rate of desorption of C, r_{DC}, is just opposite in sign to the rate of adsorption of C, r_{ADC}
r_{DC} = –r_{ADC}
In addition, we see that the desorption equilibrium constant K_{DC} is just the reciprocal of the adsorption equilibrium constant for C, K_{C}
$\begin{array}{c}\hline {K}_{\text{DC}}=\frac{1}{{K}_{\text{C}}}\\ \hline\end{array}$
in which case the rate of desorption of C can be written
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{DC}}={k}_{\text{D}}({\text{C}}_{\text{C}\cdot \text{S}}{K}_{\text{C}}{P}_{\text{C}}{\text{C}}_{\upsilon})\\ \hline\end{array}& \text{(1021)}\end{array}$
$\begin{array}{c}{K}_{\text{DC}}=\left(\text{atm}\right)\\ {K}_{\text{C}}=\left(\frac{1}{\text{atm}}\right)\end{array}$
In the material that follows, the form of the equation for the desorption step that we will use to develop our rate laws will be similar to Equation (1021).
When heterogeneous reactions are carried out at steady state, the rates of each of the three reaction steps in series (adsorption, surface reaction, and desorption) are equal to one another
$\begin{array}{c}\hline {r}_{\text{A}}^{\prime}={r}_{\text{AD}}={r}_{\text{S}}={r}_{\text{D}}\\ \hline\end{array}$
However, one particular step in the series is usually found to be ratelimiting or ratecontrolling. That is, if we could make that particular step go faster, the entire reaction would proceed at an accelerated rate. Consider the analogy to the electrical circuit shown in Figure 1012. A given concentration of reactants is analogous to a given driving force or electromotive force (EMF). The current I (with units of coulombs/s) is analogous to the rate of reaction, ${\text{r}}_{\text{A}}^{\prime}$ (mol/s·gcat), and a resistance R_{i} (with units of Ohms, Ω) is associated with each step in the series. Because the resistances are in series, the total resistance R_{tot} is just the sum of the individual resistances, for adsorption (R_{AD}), surface reaction (R_{S}), and desorption (R_{D}). The current, I, for a given voltage, E, is
$\text{I}=\frac{E}{{R}_{\text{tot}}}=\frac{E}{{R}_{\text{AD}}+{R}_{\text{S}}+{R}_{\text{D}}}$
The concept of a ratelimiting step
Who is slowing us down?
Because we observe only the total resistance, R_{tot}, it is our task to find which resistance is much larger (say, 100 Ω) than the other two resistances (say, 0.1 Ω). Thus, if we could lower the largest resistance, the current I (i.e., ${\text{r}}_{\text{A}}^{\prime}$), would be larger for a given voltage, E. Analogously, we want to know which step in the adsorption–reaction–desorption series is limiting the overall rate of reaction.
An algorithm to determine the ratelimiting step
The approach in determining catalytic and heterogeneous mechanisms is usually termed the LangmuirHinshelwood approach, since it is derived from ideas proposed by Hinshelwood based on Langmuir’s principles for adsorption.^{8} The LangmuirHinshelwood approach was popularized by Hougen and Watson and occasionally includes their names.^{9} It consists of first assuming a sequence of steps in the reaction. In writing this sequence, one must choose among such mechanisms as molecular or atomic adsorption, and single or dualsite reaction. Next, rate laws are written for the individual steps as shown in the preceding section, assuming that all steps are reversible. Finally, a ratelimiting step is postulated, and steps that are not ratelimiting are used to eliminate all coveragedependent terms. The most questionable assumption in using this technique to obtain a rate law is the hypothesis that the activity of the surface is essentially uniform as far as the various steps in the reaction are concerned.
^{8} C. N. Hinshelwood, The Kinetics of Chemical Change, Oxford: Clarendon Press, 1940.
^{9} O. A. Hougen and K. M. Watson, Ind. Eng. Chem., 35, 529 (1943).
Industrial Example of AdsorptionLimited Reaction
An example of an adsorptionlimited reaction is the synthesis of ammonia from hydrogen and nitrogen
${\text{3H}}_{\text{2}}+{\text{N}}_{\text{2}}\overrightarrow{\leftarrow}{\text{2NH}}_{3}$
over an iron catalyst that proceeds by the following mechanism:^{10}
$\begin{array}{cc}\begin{array}{c}{\text{H}}_{2}+2\text{S}\overrightarrow{\leftarrow}\text{2H}\cdot \text{S}\hfill \end{array}\}\hfill & \text{Rapid}\hfill \\ \begin{array}{c}\text{}{\text{N}}_{2}+2\text{S}\overrightarrow{\leftarrow}2\text{N}\cdot \text{S}\hfill \end{array}\}\hfill & \text{Ratelimiting}\hfill \\ \begin{array}{c}\begin{array}{c}\begin{array}{c}\text{N}\cdot \text{S}+\text{H}\cdot \text{S}\overrightarrow{\leftarrow}\text{HN}\cdot \text{S}+\text{S}\hfill \end{array}\hfill \\ \begin{array}{c}\begin{array}{c}\text{NH}\cdot \text{S}+\text{H}\cdot \text{S}\overrightarrow{\leftarrow}{\text{H}}_{2}\text{N}\cdot \text{S}+\text{S}\hfill \end{array}\hfill \\ \begin{array}{c}\begin{array}{c}{\text{H}}_{\text{2}}\text{N}\cdot \text{S}+\text{H}\cdot \text{S}\overrightarrow{\leftarrow}{\text{NH}}_{3}\text{}\cdot \text{S}+\text{S}\hfill \end{array}\hfill \\ {\text{NH}}_{3}\text{}\cdot \text{S}\overrightarrow{\leftarrow}{\text{NH}}_{3}+\text{S}\hfill \end{array}\hfill \end{array}\hfill \end{array}\hfill \end{array}\}\hfill & \text{Rapid}\hfill \end{array}$
^{10} From the literature cited in G. A. Somorjai, Introduction to Surface Chemistry and Catalysis, New York: Wiley, 1994, p. 482.
Dissociative adsorption of N_{2} is ratelimiting.
The ratelimiting step is believed to be the adsorption of the N_{2} molecule as an N atom on an iron active site.
Industrial Example of SurfaceLimited Reaction
An example of a surfacelimited reaction is the reaction of two noxious automobile exhaust products, CO and NO
$\text{CO}+\text{NO}\to {\text{CO}}_{2}+\frac{1}{2}{\text{N}}_{2}$
carried out over a copper catalyst to form more environmentally acceptable products, N_{2} and CO_{2}
$\text{}\begin{array}{cc}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{}\text{CO}+\text{S}\overrightarrow{\leftarrow}\text{CO}\cdot \text{S}\end{array}\\ \text{NO}+\text{S}\overrightarrow{\leftarrow}\text{NO}\cdot \text{S}\end{array}\end{array}\}& \text{Rapid}\\ \begin{array}{c}\text{NO}\cdot \text{S}+\text{CO}\cdot \text{S}\overrightarrow{\leftarrow}{\text{CO}}_{2}+\text{N}\cdot \text{S}+\text{S}\end{array}\}& \text{Ratelimiting}\\ \begin{array}{c}\begin{array}{c}\begin{array}{c}\text{N}\cdot \text{S}+\text{N}\cdot \text{S}\overrightarrow{\leftarrow}{\text{N}}_{2}\text{}\cdot \text{S}+\text{S}\end{array}\\ {\text{N}}_{2}\text{}\cdot \text{S}\to {\text{N}}_{2}+\text{S}\end{array}\end{array}\}& \text{Rapid}\end{array}$
Surface reaction is ratelimiting.
Analysis of the rate law suggests that CO_{2} and N_{2} are weakly adsorbed, that is, have infinitesimally small adsorption constants (see Problem P109_{B}).
What If Two Steps Are Equally Rate Limiting?
If two steps, say surface reaction and desorption, are equally slow then one must resort to applying the PseudoSteadyState Hypothesis (PSSH) to the adsorbed species as shown in the Chapter 10 expanded material (http://www.umich.edu/~elements/6e/10chap/expanded_ch10_PSSH.pdf).
We now wish to develop rate laws for catalytic reactions that are not diffusionlimited. In developing the procedure of how to obtain a rate law, a mechanism and a ratelimiting step, consistent with experimental observation, we shall discuss the catalytic decomposition of cumene to form benzene and propylene. The overall reaction is
$\begin{array}{cc}\begin{array}{cc}{\text{C}}_{6}{\text{H}}_{5}\text{CH}{\left({\text{CH}}_{3}\right)}_{2}\text{}& \text{}\end{array}& \begin{array}{cc}\begin{array}{c}\to \end{array}& {\text{C}}_{6}{\text{H}}_{6}+{\text{C}}_{3}{\text{H}}_{6}\end{array}\end{array}$
Adsorption
Surface reaction
Desorption
A conceptual model depicting the sequence of steps in this platinumcatalyzed reaction is shown in Figure 1013. Figure 1013 is only a schematic representation of the adsorption of cumene; a more realistic model is the formation of a complex of the π orbitals of benzene with the catalytic surface, as shown in Figure 1014.
The nomenclature in Table 103 will be used to denote the various species in this reaction: C = cumene, B = benzene, and P = propylene. The reaction sequence for this decomposition is shown in Table 103.
TABLE 103 STEPS IN A LANGMUIRHINSHELWOOD KINETIC MECHANISM
$\begin{array}{cc}\hline \begin{array}{cc}\text{C}+\text{S}\underset{{k}_{\text{A}}}{\overset{{k}_{\text{A}}}{\overrightarrow{\leftarrow}\text{C}\cdot \text{S}}}& \text{Adsorption of cumene on the surface}\end{array}& \text{(1022)}\\ \begin{array}{cc}\text{C}\cdot \text{S}\underset{{k}_{\text{S}}}{\overset{{k}_{\text{S}}}{\overrightarrow{\leftarrow}\text{B}\cdot \text{S}}}+P& \begin{array}{c}\begin{array}{c}\text{Surface reaction to form adsorbed}\end{array}\\ \begin{array}{c}\begin{array}{c}\text{benzene and propylene in the}\end{array}\\ \text{gas phase}\end{array}\end{array}\end{array}& \text{(1023)}\\ \begin{array}{cc}\text{B}\cdot \text{S}\underset{{k}_{\text{D}}}{\overset{{k}_{\text{D}}}{\overrightarrow{\leftarrow}}}\text{B+S}& \text{Desorption of benzene from surface}\end{array}& \text{(1024)}\\ \hline\end{array}$ 
These three steps represent the mechanism for cumene decomposition.
Equations (1022)–(1024) represent the mechanism proposed for this reaction.
Ideal gas law P_{C} = C_{C}RT
When writing rate laws for these steps, we treat each step as an elementary reaction; the only difference is that the species concentrations in the gas phase are replaced by their respective partial pressures
C_{C} → P_{C}
There is no theoretical reason for this replacement of the concentration, C_{C}, with the partial pressure, P_{C}; it is just the convention initiated in the 1930s and used ever since. Fortunately, P_{C} can be calculated easily and directly from C_{C} using the ideal gas law (i.e., P_{C} = C_{C}RT ).
The rate expression for the adsorption of cumene as given in Equation (1022) is
${\text{r}}_{\text{AD}}={k}_{\text{A}}{P}_{\text{C}}{\text{C}}_{\upsilon}{k}_{\text{A}}{\text{C}}_{\text{C}\cdot \text{S}}$
$\begin{array}{cc}\begin{array}{cc}\hline \begin{array}{c}\text{Adsorption:}\end{array}& {\text{r}}_{\text{AD}}={k}_{\text{A}}({P}_{\text{C}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{C}\cdot \text{S}}}{{K}_{\text{C}}})\\ \hline\end{array}& \text{(1025)}\end{array}$
$\text{C}+\text{S}\underset{{k}_{\text{A}}}{\overset{{k}_{\text{A}}}{\overrightarrow{\leftarrow}}}\text{C}\cdot \text{S}$
If r_{AD} has units of (mol/gcat·s) and C_{C·S} has units of (mol cumene adsorbed/gcat) and P_{C} has units of kPa, bar or atm, then typical units of k_{A}, k·_{A}, and K_{C} would be
$\begin{array}{ccc}\text{}\left[{k}_{\text{A}}\right]& \equiv & {\begin{array}{c}(\text{kPa}\cdot \text{s})\end{array}}^{1}\text{or}{(\text{atm}\cdot \text{h})}^{1}\text{or}{(\text{bar}\cdot \text{h})}^{1}\hfill \\ \left[{k}_{\text{A}}\right]& \equiv & {\text{h}}^{1}\text{or}{\text{s}}^{1}\hfill \\ \left[{k}_{\text{C}}\right]& \equiv & \left[\frac{{k}_{\text{A}}}{{k}_{\text{A}}}\right]\equiv {\text{kpa}}^{1}\text{or}\left({\text{bar}}^{1}\right)\text{or}\left({\text{atm}}^{1}\right)\hfill \end{array}$
The rate law for the surfacereaction step producing adsorbed benzene and propylene in the gas phase
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{C}\cdot \text{S}& \underset{{k}_{\text{S}}}{\overset{{k}_{\text{S}}}{\begin{array}{c}\overrightarrow{\begin{array}{c}\leftarrow \end{array}}\end{array}}}\end{array}& \text{B}\cdot \text{S}\end{array}+\text{P}\left(\text{g}\right)& \text{(1023)}\end{array}$
is
${\text{r}}_{\text{S}}={k}_{\text{S}}{\text{C}}_{\text{C}\text{}\cdot \text{S}}{k}_{\text{S}}{P}_{{\text{}P}_{\text{}}}{\text{C}}_{\text{B}\cdot \text{S}}$
$\begin{array}{cc}\begin{array}{cc}\hline \begin{array}{c}\text{Surface reaction:}\end{array}& {\text{r}}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\text{C}\cdot \text{S}}\frac{{\text{P}}_{\text{P}}{\text{C}}_{\text{B}\cdot \text{S}}}{{\text{K}}_{\text{S}}})\\ \hline\end{array}& \text{(1026)}\end{array}$
with the surface reaction equilibrium constant being
${K}_{\text{S}}=\frac{{k}_{\text{S}}}{{k}_{\text{S}}}$
Typical units for k_{S} and K_{S} are s^{–1} and kPa (or atm or bar) respectively.
Propylene is not adsorbed on the surface. Consequently, its concentration on the surface is zero, that is,
C_{P·S} = 0
The rate of benzene desorption (see Equation (1024)) is
$\begin{array}{cc}{\text{r}}_{\text{D}}={k}_{\text{D}}{\text{C}}_{\text{B}\cdot \text{S}}{k}_{\text{D}}{P}_{\text{B}}{\text{C}}_{\upsilon}& \text{(1027)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\hline \begin{array}{c}\text{Desorption:}\end{array}& {\text{r}}_{\text{D}}={k}_{\text{D}}({\text{C}}_{\text{B}\cdot \text{S}}\frac{{P}_{\text{B}}{\text{C}}_{\upsilon}}{{K}_{\text{DB}}})\\ \hline\end{array}& \text{(1028)}\end{array}$
Typical units of k_{D} and K_{DB} are s^{–1} and kPa or bar, respectively. By viewing the desorption of benzene
$\begin{array}{cc}\begin{array}{cc}\text{B}\cdot \text{S}& \overrightarrow{\begin{array}{c}\leftarrow \end{array}}\end{array}& \text{B}+\text{S}\end{array}$
from right to left, we see that desorption is just the reverse of the adsorption of benzene. Consequently, as mentioned earlier, it is easily shown that the benzene adsorption equilibrium constant K_{B} is just the reciprocal of the benzene desorption constant K_{DB}
${K}_{\text{B}}=\frac{1}{{K}_{\text{DB}}}$
and Equation (1028) can be written as
$\begin{array}{ccc}\begin{array}{ccc}\hline \text{Desorption:}& {r}_{\text{D}}={k}_{\text{D}}({\text{C}}_{\text{B}\cdot \text{S}}{\text{K}}_{\text{B}}{\text{P}}_{\text{B}}{\text{C}}_{\upsilon})& \\ \hline\end{array}& & \text{(1029)}\end{array}$
Because there is no accumulation of reacting species on the surface, the rates of each step in the sequence are all equal as discussed in Figure 1012:
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}={r}_{\text{AD}}={r}_{\text{S}}={r}_{\text{D}}\\ \hline\end{array}& \text{(1030)}\end{array}$
For the mechanism postulated in the sequence given by Equations (1022)–(1024), we wish to determine which step is ratelimiting. We first assume one of the steps to be ratelimiting (ratecontrolling) and then formulate the reactionrate law in terms of the partial pressures of the species present. From this expression, we can determine the variation of the initial reaction rate with the initial partial pressures and the initial total pressure. If the predicted rate varies with pressure in the same manner as the rate observed experimentally, the implication is that the assumed mechanism and ratelimiting step are correct.
We will first start our development of the rate laws with the assumption that the adsorption step is ratelimiting and derive the rate law, and then proceed in the following reactions to assume that each of the other two steps’ surface reaction and desorption limit the overall rate and then derive the rate law for each of these other two limiting cases.
To answer this question, we shall first assume that the adsorption of cumene is indeed the ratelimiting step (RLS)
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{C}+\text{S}& \overrightarrow{\leftarrow}\end{array}& \text{}\end{array}& \text{C}\cdot \text{S}\end{array}$
derive the corresponding rate law, and then check to see whether it is consistent with experimental observation. By postulating that this (or any other) step is ratelimiting, we are assuming that the adsorption rate constant of this step (in this case k_{A}) is small with respect to the specific rates of the other steps (in this case k_{S} and k_{D}).^{11} The rate of adsorption is
^{11} Strictly speaking, one should compare the product k_{A}P_{C} with k_{S} and k_{D}.
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}={r}_{\text{AD}}={k}_{\text{A}}({P}_{\text{C}}{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{C}\cdot \text{S}}}{{K}_{\text{C}}})\\ \hline\end{array}& \text{(1025)}\end{array}$
Need to express C_{υ} and C_{C·S} in terms of P_{C}, P_{B}, and P_{P}.
$\begin{array}{c}{r}_{\text{AD}}={k}_{\text{A}}{P}_{\text{C}}[{\text{C}}_{\upsilon}\frac{{\text{C}}_{\text{C}\cdot \text{S}}}{{K}_{\text{C}}{P}_{\text{C}}}]\\ \frac{\text{mol}}{\text{s}\cdot \text{kgcat}}=\left(\frac{1}{\text{s atm}}\right)\cdot \left(\text{atm}\right)\cdot \left[\frac{\text{mol}}{\text{kgcat}}\right]=\left[\frac{1}{\text{S}}\right]\frac{\text{mol}}{\text{kgcat}}\end{array}$
Dividing r_{AD} by k_{A}P_{C}, we note $\frac{{r}_{\text{AD}}}{{k}_{\text{A}}{P}_{\text{C}}}=\frac{\text{mol}}{\text{kgcat}}$. The reason we do this is that in order to compare terms, the ratios $(\frac{{r}_{\text{AD}}}{{k}_{\text{A}}{P}_{\text{C}}}),\left(\frac{{r}_{\text{s}}}{{k}_{\text{s}}}\right)$, and $\left(\frac{{r}_{\text{D}}}{{k}_{\text{D}}}\right)$ must all have the same units $\left[\frac{\text{mol}}{\text{kgcat}}\right]$. Luckily for us, the end result is the same, however.
Because we cannot measure either C_{υ} or C_{C · }_{S} , we must replace these variables in the rate law with measurable quantities in order for the equation to be meaningful.
For steadystate operation we have
$\begin{array}{cc}{r}_{\text{C}}^{\prime}={r}_{\text{AD}}={r}_{\text{S}}={r}_{\text{D}}& \text{(1030)}\end{array}$
For adsorptionlimited reactions, k_{A} is very small and k_{S} and k_{D} are very, very large by comparison. Consequently, the ratios r_{S} / k_{S} and r_{D} / k_{D} are very small (approximately zero), whereas the ratio r_{AD} / k_{A} is relatively large.
The surface reactionrate law is
$\begin{array}{cc}{r}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\text{C}\cdot \text{S}}\frac{{\text{C}}_{\text{B}\cdot \text{S}}{P}_{P}}{{K}_{\text{S}}})& \text{(1031)}\end{array}$
Again, for adsorptionlimited reactions, the surfacespecific reaction rate k_{S} is large by comparison, and we can set
$\begin{array}{cc}\frac{{r}_{\text{S}}}{{k}_{\text{S}}}\simeq 0& \text{(1032)}\end{array}$
and solve Equation (1031) for C_{C · }_{S}
$\begin{array}{cc}{\text{C}}_{\text{C}\cdot \text{S}}=\frac{{\text{C}}_{\text{B}\cdot \text{S}}{P}_{P}}{{\text{k}}_{\text{S}}}& \text{(1033)}\end{array}$
From Equation (1033) we see that in order to be able to express C_{C · S} solely in terms of the partial pressures of the species present, we must evaluate C_{B · }_{S} . The rate of desorption of benzene is
$\begin{array}{cc}{r}_{\text{D}}={k}_{\text{D}}({\text{C}}_{\text{B}\cdot \text{S}}{K}_{\text{B}}{P}_{\text{B}}{\text{C}}_{\upsilon})& \text{(1029)}\end{array}$
Using $\frac{{r}_{\text{S}}}{{k}_{\text{S}}}\text{\u22430\u2243}\frac{{r}_{\text{D}}}{{k}_{\text{D}}}$ find C_{B · }_{S} and C_{C · }_{S} in terms of partial pressures.
However, for adsorptionlimited reactions, k_{D} is large by comparison, and we can set
$\begin{array}{cc}\frac{{r}_{\text{D}}}{{k}_{\text{D}}}\text{\u2243}0& \text{(1034)}\end{array}$
and then solve Equation (1029) for C_{B · }_{S}
$\begin{array}{cc}{\text{C}}_{\text{B}\cdot \text{S}}={K}_{\text{B}}{P}_{\text{B}}{\text{C}}_{\upsilon}& \text{(1035)}\end{array}$
After combining Equations (1033) and (1035), we have
$\begin{array}{cc}{\text{C}}_{\text{C}\cdot \text{S}}={K}_{\text{B}}\frac{{P}_{\text{B}}{P}_{\text{p}}}{{K}_{\text{S}}}{\text{C}}_{\upsilon}& \text{(1036)}\end{array}$
Replacing C_{C·S} in the rate equation by Equation (1036) and then factoring C_{υ}, we obtain
$\begin{array}{cc}{r}_{\text{AD}}={k}_{\text{A}}({P}_{\text{C}}\frac{{K}_{\text{B}}{P}_{\text{B}}{P}_{P}}{{K}_{\text{S}}{K}_{\text{C}}}){\text{C}}_{\upsilon}={k}_{\text{A}}({P}_{\text{C}}\frac{{P}_{\text{B}}{P}_{P}}{{K}_{P}}){\text{C}}_{\text{}\upsilon}& \text{(1037)}\end{array}$
Thermodynamics: Let’s look at how the thermodynamic constantpressure equilibrium constant, K_{P}, found its way into Equation (1037) and how we can find its value for any reaction. First we observe that at equilibrium r_{AD} = 0, Equation (1037) rearranges to
$\frac{{P}_{\text{B}e}{P}_{Pe}}{{P}_{\text{C}e}}=\frac{{K}_{\text{C}}{K}_{\text{S}}}{{K}_{\text{B}}}$
We also know from thermodynamics (Appendix C) that for the reaction
$\begin{array}{cc}\text{C}& \overrightarrow{\leftarrow}\text{B}+P\end{array}$
also at equilibrium $({r}_{\text{C}}^{\prime}\equiv 0)$, we have the following relationship for partial pressure equilibrium constant K_{P}
${K}_{P}=\frac{{P}_{\text{B}\text{e}}{P}_{P\text{e}}}{{P}_{\text{C}\text{e}}}$
Consequently, the following relationship must hold
$\begin{array}{cc}\begin{array}{c}\hline \frac{{K}_{S}{K}_{C}}{{K}_{\text{B}}}={K}_{P}\\ \hline\end{array}& \text{(1038)}\end{array}$
The thermodynamic equilibrium constant, K_{p}
The equilibrium constant can be determined from thermodynamic data and is related to the change in the Gibbs free energy, ΔG∘, by the equation (see Equation (C2) in Appendix C)
$\begin{array}{cc}\begin{array}{c}\hline RT\text{ln}\text{}K=\mathrm{\Delta}{\text{G}}^{\circ}\\ \hline\end{array}& \text{(1039)}\end{array}$
where R is the ideal gas constant and T is the absolute temperature.
Back to the rate law task at hand. The concentration of vacant sites, C_{υ}, can now be eliminated from Equation (1037) by utilizing the site balance to give the total concentration of sites, C_{t}, which is assumed constant^{12}
Total sites = Vacant sites + Occupied sites
^{12} #SurfaceReactionRate, Some (I won’t mention any names, but they know who they are) prefer to write the surface reaction rate in terms of the fraction of the surface of sites covered (i.e., f_{A} rather than the number of sites C_{A·S} covered, the difference being the multiplication factor of the total site concentration, C_{t}. In any event, the final form of the rate law is the same because C_{t}, K_{A}, k_{S}, and so on, are all lumped into the reactionrate constant, k.
Because cumene and benzene are adsorbed on the surface, the concentration of occupied sites is (C_{C·S} + C_{B·S}), and the total concentration of sites is
$\begin{array}{cc}{\text{C}}_{t}={\text{C}}_{\upsilon}+{\text{C}}_{\text{C}\cdot \text{S}}+{\text{C}}_{\text{B}\cdot \text{S}}& \text{(1040)}\end{array}$
Site balance
Substituting Equations (1035) and (1036) into Equation (1040), we have
${\text{C}}_{t}={\text{C}}_{\upsilon}+\frac{{K}_{\text{B}}}{{K}_{\text{S}}}{P}_{\text{B}}{P}_{P}{\text{C}}_{\upsilon}+{K}_{\text{B}}{P}_{\text{B}}{\text{C}}_{\upsilon}$
Solving for C_{υ}, we have
$\begin{array}{cc}{\text{C}}_{\upsilon}=\frac{{\text{C}}_{t}}{1+{P}_{\text{B}}{P}_{P}{K}_{\text{B}}/{K}_{\text{S}}+{K}_{\text{B}}{P}_{\text{B}}}& \text{(1041)}\end{array}$
Combining Equations (1041) and (1037), we find that the rate law for the catalytic decomposition of cumene, assuming that the adsorption of cumene is the ratelimiting step, is
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}={r}_{\text{AD}}=\frac{{\text{C}}_{\text{t}}{k}_{\text{A}}({P}_{\text{C}}{P}_{P}{P}_{\text{B}}/{K}_{P})}{1+{K}_{\text{B}}{P}_{P}{P}_{\text{B}}/{K}_{\text{S}}+{K}_{\text{B}}{P}_{\text{B}}}\\ \hline\end{array}& \text{(1042)}\end{array}$
Cumene reaction rate law iff adsorption were the limiting step
We now, as we usually do, sketch a plot of the initial rate of reaction as a function of the initial partial pressure, in this case cumene, P_{C0}. Initially, no products are present; consequently, P_{P} = P_{B} = 0. The initial rate is given by
$\begin{array}{cc}{r}_{\text{C0}}^{\prime}={\text{C}}_{\text{t}}{k}_{\text{A}}{P}_{\text{C0}}=k{P}_{\text{C0}}& \text{(1043)}\end{array}$
If the cumene decomposition is adsorption rate limited, then from Equation (1043) we see that the initial rate will be linear with the initial partial pressure of cumene, as shown in Figure 1015.
Iff adsorption were ratelimiting, the data should show $\begin{array}{cc}{r}_{\text{CO}}^{\prime}& \text{}\end{array}$ increasing linearly with P_{CO}.
Before checking to see whether Figure 1015 is consistent with experimental observation, we shall derive the corresponding rate laws for the other possible ratelimiting steps and then develop corresponding initial rate plots for the case when the surface reaction is ratelimiting and then for the case when the desorption of benzene is ratelimiting.
We next assume the surface reaction step
$\begin{array}{cc}\text{C}\cdot \text{S}& \overrightarrow{\to}\text{B}\cdot \text{S}+P\end{array}$
is the ratelimiting step (RLS) and derive a rate law in terms of the partial pressures of C and B. The rate law for the surface reaction is
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\text{C}\cdot \text{S}}\frac{{P}_{P}{\text{C}}_{\text{B}\cdot \text{S}}}{{K}_{\text{S}}})\\ \hline\end{array}& \text{(1026)}\end{array}$
Singlesite mechanism
Since we cannot readily measure the concentrations of the adsorbed species, we must utilize the adsorption and desorption steps to eliminate C_{C·S} and C_{B·S} from this equation.
From the adsorption rate expression in Equation (1025) and the condition that k_{A} and k_{D} are very large by comparison with k_{S} when the surface reaction is limiting (i.e., r_{AD}/k_{A}⋍ 0),^{13} we obtain a relationship for the surface concentration for adsorbed cumene
^{13} See footnote 11 on page 466.
C_{C·S} = K_{C}P_{C}C_{υ}
In a similar manner, the surface concentration of adsorbed benzene can be evaluated from the desorption rate expression, Equation (1029), together with the approximation
$\text{when}\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u2245}\frac{{r}_{\text{D}}}{{k}_{\text{D}}}$
Using $\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u22450}\text{\u2245}\frac{{r}_{\text{D}}}{{k}_{\text{D}}}$ find C_{B · }_{S} and C_{C · }_{S} in terms of partial pressures.
then we get the same result for C_{B•S} as before when we had adsorption limitation, that is,
C_{B·S} = K_{B}P_{B}C_{υ}
Substituting for C_{B·S} and C_{C·S} in Equation (1026) gives us
$\begin{array}{cc}{r}_{\text{S}}={k}_{\text{S}}({P}_{\text{C}}{K}_{\text{C}}\frac{{K}_{\text{B}}{P}_{\text{B}}{P}_{P}}{{K}_{\text{S}}}){\text{C}}_{\upsilon}={k}_{\text{S}}{K}_{\text{C}}({P}_{\text{C}}\frac{{P}_{\text{B}}{P}_{P}}{{K}_{P}}){\text{C}}_{\upsilon}& \text{(1026a)}\end{array}$
where the thermodynamic equilibrium constant was used to replace the ratio of surface reaction and adsorption constants, that is,
$\begin{array}{cc}{K}_{P}=\frac{{K}_{\text{C}}{K}_{\text{S}}}{{K}_{\text{B}}}& \text{(1026b)}\end{array}$
The only variable left to eliminate is and we use a site balance to accomplish this, that is,
$\begin{array}{cc}{\text{C}}_{\text{t}}={\text{C}}_{\upsilon}+{\text{C}}_{\text{B}\cdot \text{S}}+{\text{C}}_{\text{C}\cdot \text{S}}& \text{(1040)}\end{array}$
Site balance
Substituting for concentrations of the adsorbed species, C_{B·S}, and C_{C·S}, factoring out C_{υ}, and rearranging yields
${\text{C}}_{\upsilon}=\frac{{\text{C}}_{\text{t}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}{P}_{C}}$
Substituting for C_{υ} in Equation (1026a)
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}={r}_{\text{S}}=\frac{\stackrel{k}{\stackrel{}{\overbrace{{k}_{\text{S}}{\text{C}}_{\text{t}}{\text{K}}_{\text{C}}}}}({\text{P}}_{C}{\text{P}}_{\text{P}}{\text{P}}_{\text{B}}/{\text{K}}_{\text{P}})}{1+{\text{P}}_{\text{B}}{\text{K}}_{\text{B}}+{\text{K}}_{\text{C}}{\text{P}}_{\text{C}}}\\ \hline\end{array}& \text{(1044)}\end{array}$
Cumene rate law iff surfacereactionlimiting
The initial rate of reaction is
$\begin{array}{cc}{r}_{\text{C0}}^{\prime}=\frac{\stackrel{k}{\overbrace{\stackrel{}{{k}_{\text{S}}{\text{C}}_{\text{t}}{\text{K}}_{\text{C}}\text{}}}\text{}}\text{\hspace{0.17em}}\text{}{\text{P}}_{\text{C0}}\text{}}{1+{\text{K}}_{\text{C}}{\text{P}}_{\text{C0}}}=\frac{k{\text{P}}_{\text{C0}}}{1+{\text{K}}_{\text{C}}{\text{P}}_{\text{C0}}}& \text{(1045)}\end{array}$
Using Equation (1045), we make the sketch of the initial rate of reaction, $\begin{array}{cc}{r}_{\text{C0}}^{\prime}& \text{}\end{array}$, as a function of the initial partial pressure of cumene, P_{C0}, as shown in Figure 1016 for the case when surfacereaction limit is the limiting step.
At low partial pressures of cumene
${\text{1\u226bK}}_{\text{C}}{P}_{\text{C0}}$
and we observe that the initial rate will increase linearly with the initial partial pressure of cumene:
Iff surface reaction were ratelimiting, the data would show this behavior.
${r}_{\text{C0}}^{\prime}\approx k{P}_{\text{C}0}$
At high partial pressures
${K}_{\text{C}}{P}_{\text{C0}}\text{\u226b1}$
and Equation (1045) becomes
${r}_{\text{C0}}^{\prime}\text{\u2245}\frac{k{P}_{\text{C0}}}{{K}_{\text{C}}{P}_{\text{C0}}}=\frac{k}{{K}_{\text{C}}}$
and the initial rate is independent of the initial partial pressure of cumene.
To answer this question, we now want to determine the initial rate, ${r}_{C0}^{\prime}$, and a function of partial pressure of cumene, P_{C}, for the desorption step
$\text{B}\cdot \text{S}\overrightarrow{\leftarrow}\text{B}+\text{S}$
The rate expression for the desorption of benzene is
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{D}}={k}_{\text{D}}({\text{C}}_{\text{B}\cdot \text{S}}{K}_{\text{B}}{P}_{\text{B}}{\text{C}}_{\upsilon})\\ \hline\end{array}& \text{(1029)}\end{array}$
From the rate expression for surface reaction, Equation (1026), we set
For desorptionlimited reactions, both k_{A} and k_{S} are very large compared with k_{D}, which is small.
$\text{}\frac{{r}_{\text{S}}}{{k}_{\text{S}}}\text{\u2243}0$
to obtain
$\begin{array}{cc}{\text{C}}_{\text{B}\cdot \text{S}}={K}_{\text{S}}\left(\frac{{\text{C}}_{\text{C}\cdot \text{S}}}{{P}_{P}}\right)& \text{(1046)}\end{array}$
Similarly, for the adsorption step, Equation (1025), we set
$\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u2243}0$
then substitute for C_{C·S} in Equation (1046) to obtain
$\begin{array}{cc}{\text{C}}_{\text{B}\cdot \text{S}}=\frac{{K}_{\text{C}}{K}_{\text{S}}{P}_{\text{C}}{\text{C}}_{\upsilon}}{{P}_{P}}& \text{(1047)}\end{array}$
Combining Equations (1026b), (1029), and (1047) gives us
$\begin{array}{cc}{r}_{\text{D}}={k}_{\text{D}}{K}_{\text{C}}{K}_{\text{S}}(\frac{{P}_{\text{C}}}{{P}_{P}}\frac{{P}_{\text{B}}}{{K}_{P}}){\text{C}}_{\upsilon}& \text{(1048)}\end{array}$
where K_{C} is the cumene adsorption constant, K_{S} is the surface reaction equilibrium constant, and K_{P} is the thermodynamic gasphase equilibrium constant, Equation (1038), for the reaction. The expression for C_{υ} is obtained from a site balance:
$\begin{array}{ccc}\text{Site balance}:& {\text{C}}_{\text{t}}={\text{C}}_{\text{C}\cdot \text{S}}+{\text{C}}_{\text{B}\cdot \text{S}}+{\text{C}}_{\upsilon}& \text{(1040)}\end{array}$
After substituting for the respective surface concentrations, we solve the site balance for C_{υ}
$\begin{array}{cc}{\text{C}}_{\upsilon}=\frac{{\text{C}}_{\text{t}}}{1+{K}_{\text{C}}{K}_{\text{S}}{P}_{\text{C}}/{P}_{P}+{K}_{\text{C}}{P}_{\text{C}}}& \text{(1049)}\end{array}$
Replacing C_{υ} in Equation (1048) by Equation (1049) and multiplying the numerator and denominator by P_{P}, we obtain the rate expression for desorption control
Cumene decomposition rate law iff desorption were limiting
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}={r}_{\text{D}}=\frac{\stackrel{k}{\stackrel{}{\overbrace{{k}_{\text{D}}{\text{C}}_{\text{t}}{\text{K}}_{\text{S}}{\text{K}}_{\text{C}}}}}\text{}({\text{P}}_{\text{C}}{\text{P}}_{\text{B}}{\text{P}}_{\text{P}}/{\text{K}}_{\text{P}})\text{}}{{\text{P}}_{\text{P}}+{\text{P}}_{\text{C}}{\text{K}}_{\text{C}}{\text{K}}_{\text{S}}+{\text{K}}_{\text{C}}{\text{P}}_{\text{P}}{\text{P}}_{\text{C}}}\\ \hline\end{array}& \text{(1050)}\end{array}$
To determine the dependence of the initial rate of reaction on the initial partial pressure of cumene, we again set P_{P} = P_{B} = 0, and the rate law reduces to
${r}_{\text{C0}}^{\prime}={k}_{\text{D}}{\text{C}}_{\text{t}}$
If desorption limits, the initial rate is independent of the initial partial pressure of cumene.
with the corresponding plot of ${r}_{\text{C0}}^{\prime}$ shown in Figure 1017. If desorption were rate limiting, we would see that the initial rate of reaction would be independent of the initial partial pressure of cumene.
Now that we know what the functionality of initial rate versus partial pressure looks like when we assume each of the step is the RLS, it gives us a foot up in interpreting reaction rate data. The experimental observations of ${r}_{\text{C0}}^{\prime}$ as a function of P_{C0} are shown in Figure 1018. From the plot in Figure 1018, we can clearly see that neither adsorption nor desorption is ratelimiting. For the reaction and mechanism given by
$\begin{array}{cc}\text{C}+\text{S}\overrightarrow{\leftarrow}\text{C}\cdot \text{S}& \text{(1022)}\end{array}$
$\begin{array}{cc}\mathit{RateLimiting\; Step}\mathit{\left(}\mathit{RLS}\mathit{\right)}:\text{C}\cdot \text{S}\overrightarrow{\leftarrow}\text{B}\cdot \text{S+P}& \text{(1023)}\end{array}$
$\begin{array}{cc}\text{B}\cdot \text{S}\overrightarrow{\leftarrow}\text{B}+\text{S}& \text{(1024)}\end{array}$
the rate law derived by assuming that the surface reaction is the ratelimiting step (RLS) agrees with the data.
Cumene decomposition is surfacereactionlimited.
Surfacereactionlimited mechanism is consistent with experimental data.
The rate law for the case of no inerts adsorbing on the surface is
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{C}}^{\prime}=\frac{k({P}_{\text{C}}{P}_{\text{B}}{P}_{P}/{K}_{\text{p}})}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}{P}_{\text{C}}}\\ \hline\end{array}& \text{(1044)}\end{array}$
The forward cumene decomposition reaction is a singlesite mechanism involving only adsorbed cumene, while the reverse reaction of propylene in the gas phase reacting with adsorbed benzene is an Eley–Rideal mechanism.
If we were to have an adsorbing inert in the feed, the inert would not participate in the reaction but would occupy active sites on the catalyst surface:
$\text{I}+\text{S}\overrightarrow{\leftarrow}\text{I}\cdot \text{S}$
Our site balance is now
$\begin{array}{cc}{\text{C}}_{\text{t}}={\text{C}}_{\upsilon}+{\text{C}}_{\text{C}\cdot \text{S}}+{\text{C}}_{\text{B}\cdot \text{S}}+{\text{C}}_{\text{I}\cdot \text{S}}& \text{(1051)}\end{array}$
Because the adsorption of the inert is at equilibrium, the concentration of sites occupied by the inert is
$\begin{array}{cc}{\text{C}}_{\text{I}\cdot \text{S}}={K}_{\text{I}}{P}_{\text{I}}{\text{C}}_{\upsilon}& \text{(1052)}\end{array}$
Substituting for the inert sites in the site balance, the rate law for surface reaction control when an adsorbing inert is present is
$\begin{array}{cc}{r}_{\text{C}}^{\prime}=\frac{k({P}_{\text{C}}{P}_{\text{B}}{P}_{P}/{K}_{P})}{1+{K}_{\text{C}}{P}_{\text{C}}+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{I}}{P}_{\text{I}}\text{}}& \text{(1053)}\end{array}$
Adsorbing inerts
One observes that the rate decreases as the partial pressure of adsorbing inerts, P_{I}, increases.
We now consider a dualsite mechanism, which is a reforming reaction found in petroleum refining to upgrade the octane number of gasoline.
The more compact the molecule, the greater the octane number.
Catalyst manufacture
One common reforming catalyst is platinum on alumina. Platinum on alumina (Al_{2}O_{3}) (see SEM photo in Figure 1019) is a bifunctional catalyst that can be prepared by exposing alumina pellets to a chloroplatinic acid solution, drying, and then heating in air at 775–875 K for several hours. Next, the material is exposed to hydrogen at temperatures around 725–775 K to produce very small clusters of Pt on alumina. These clusters have sizes on the order of 10 Å, while the alumina pore sizes on which the Pt is deposited are on the order of 100–10,000 Å (i.e., 10–1000 nm).
Gasoline 

C_{5} 
10% 
C_{6} 
10% 
C_{7} 
20% 
C_{8} 
25% 
C_{9} 
20% 
C_{10} 
10% 
C_{1112} 
5% 
As an example of catalytic reforming, we shall consider the isomerization of npentane to ipentane
$\text{npentane}\underset{{\text{Al}}_{2}{\text{O}}_{3}}{\overset{0.75\text{wt\%}\text{Pt}}{\overrightarrow{\leftarrow}}}i\text{pentance}$
Normal pentane has an octane number of 62, while isopentane, which is more compact, has an octane number of 90! The npentane adsorbs onto the platinum, where it is dehydrogenated to form npentene. The npentene desorbs from the platinum and adsorbs onto the alumina, where it is isomerized to ipentene, which then desorbs and subsequently adsorbs onto platinum, where it is hydrogenated to form ipentane. That is,
$\text{npentane}\underset{\text{Pt}}{\overset{{\text{H}}_{2}}{\overrightarrow{\leftarrow}}}\text{npentane}\stackrel{{\text{Al}}_{2}{\text{O}}_{3}}{\overrightarrow{\leftarrow}}i\text{pentene}\underset{\text{Pt}}{\overset{+{\text{H}}_{2}}{\overrightarrow{\leftarrow}}}i\text{pentancce}$
We shall focus on the isomerization step to develop the mechanism and the rate law
$\begin{array}{c}\text{npentane}\stackrel{{\text{Al}}_{2}{\text{O}}_{3}}{\overrightarrow{\leftarrow}}i\text{pentancce}\\ \begin{array}{cc}\begin{array}{c}\text{N}\end{array}& \overrightarrow{\leftarrow \text{I}}\end{array}\end{array}$
The procedure for formulating a mechanism, ratelimiting step, and corresponding rate law is given in Table 104.
TABLE 104 ALGORITHM FOR DETERMINING THE REACTION MECHANISM AND RATELIMITING STEP
Isomerization of npentene (N) to ipentene (I) over alumina $\begin{array}{cc}\begin{array}{cc}\text{N}& \stackrel{{\text{Al}}_{2}{\text{O}}_{3}}{\begin{array}{c}\overrightarrow{\begin{array}{c}\leftarrow \end{array}}\end{array}}\end{array}& \text{I}\end{array}$ Reformingreaction to increaseoctane number of gasoline Step 1. Select a mechanism. (Let’s choose a DualSite Mechanism) $\begin{array}{cc}\text{Adsorption:}& \text{N}+\text{S}\overrightarrow{\leftarrow}\text{N}\cdot \text{S}\\ \text{Surface reaction:}& \text{N}\cdot \text{S}+\text{S}\overrightarrow{\leftarrow}\text{I}\cdot \text{S}+\text{S}\\ \text{Desorption:}& \text{I}\cdot \text{S}\overrightarrow{\leftarrow}\text{I}+\text{S}\end{array}$ Treat each reaction step as an elementary reaction when writing rate laws. Step 2. Assume a ratelimiting step. We choose the surface reaction first, because more than 75% of all heterogeneous reactions that are not diffusionlimited are surfacereactionlimited. We note that the PSSH must be used when more than one step is rate limiting (see Section 10.3.6). The rate law for the surface reaction step is $\begin{array}{cc}{r}_{\text{N}}^{\prime}={r}_{\text{S}}={k}_{\text{S}}({\text{C}}_{\upsilon}{\text{C}}_{\text{N}\cdot \text{S}}\frac{{\text{C}}_{\text{I}\cdot \text{S}}{\text{C}}_{\upsilon}}{{K}_{\text{S}}})& \text{(1054)}\end{array}$ Step 3. Find the expression for concentration of the adsorbed species C_{i·S}. Use the other steps that are not limiting to solve for C_{i·S} (e.g., C_{N·S} and C_{I·S}). For this reaction $\begin{array}{ccc}\text{From}& \frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u2243}0:& \begin{array}{cc}\begin{array}{c}{\text{C}}_{\text{N}\cdot \text{S}}={P}_{\text{N}}{K}_{\text{N}}{\text{C}}_{\upsilon}\end{array}& \text{(1055)}\end{array}\\ \text{From}& \frac{{r}_{\text{D}}}{{k}_{\text{D}}}\simeq \text{0:}& {\text{C}}_{\text{I}\cdot \text{S}}=\frac{{P}_{\text{I}}{\text{C}}_{\upsilon}}{{K}_{\text{D}}}={K}_{\text{I}}{P}_{\text{I}}{\text{C}}_{\upsilon}\end{array}$ Step 4. Write a site balance. $\begin{array}{cc}{\text{C}}_{\text{t}}={\text{C}}_{\upsilon}+{\text{C}}_{\text{N}\cdot \text{S}}+{\text{C}}_{\text{I}\cdot \text{S}}& \text{(1056)}\end{array}$ Step 5. Derive the rate law. Combine Steps 2, 3, and 4 to arrive at the rate law $\begin{array}{cc}{r}_{\text{N}}^{\prime}={r}_{\text{S}}=\frac{\stackrel{k}{\stackrel{}{\overbrace{{k}_{\text{S}}{\text{C}}_{\text{t}}^{2}{\text{K}}_{\text{N}}}}}({\text{P}}_{\text{N}}{\text{K}}_{\text{P}})}{{(1+{\text{K}}_{\text{N}}{\text{P}}_{\text{N}}+{\text{K}}_{\text{I}}{\text{P}}_{\text{I}})}^{2}}& \text{(1057)}\end{array}$ 
Step 6. Compare with data. Compare the rate law derived in Step 5 with experimental data. If they agree, there is a good chance that you have found the correct mechanism and ratelimiting step. If your derived rate law (i.e., model) does not agree with the data

An observation: We note that in Table 104 for the dualsite mechanism, the denominator of the rate law for $\begin{array}{c}{r}_{\text{N}}^{\prime}\end{array}$ is squared (i.e., in Step 5 [1/( )^{2})]), while for a singlesite mechanism, it is not squared (i.e., Step 6 [1/( )]). This fact is useful when analyzing catalytic reactor data.
TABLE 105 IRREVERSIBLE SURFACEREACTIONLIMITED RATE LAWS
Single site $\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{A}\cdot \text{S}\to \text{B}\cdot \text{S}& \text{}\end{array}& {r}_{\text{A}}^{\prime}=\frac{k{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}}\end{array}& \text{(1059)}\end{array}$
Dual site $\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{A}\cdot \text{S}+\text{S}\to \text{B}\cdot \text{S}+\text{S}& \text{}\end{array}& {r}_{\text{A}}^{\prime}=\frac{k{P}_{\text{A}}}{{(1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}})}^{2}}\end{array}& \text{(1060)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{A}\cdot \text{S}+\text{B}\cdot \text{S}\to \text{C}\cdot \text{S}+\text{S}& \text{}\end{array}& {r}_{\text{A}}^{\prime}=\frac{k{P}_{\text{A}}{P}_{\text{B}}}{{(1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}{P}_{\text{C}})}^{2}}\end{array}& \text{(1061)}\end{array}$
Eley–Rideal $\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{A}\cdot \text{S}+\text{B}\left(\text{g}\right)\to \text{C}\cdot \text{S}& \text{}\end{array}& {r}_{\text{A}}^{\prime}=\frac{k{P}_{\text{A}}{P}_{\text{B}}}{{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{C}}{P}_{\text{C}}}^{\text{}}}\end{array}& \text{(1062)}\end{array}$ 
We need a word of caution at this point. Just because the mechanism and ratelimiting step may fit the rate data does not imply that the mechanism is correct.^{14} Usually, spectroscopic measurements are needed to confirm a mechanism absolutely. However, the development of various mechanisms and ratelimiting steps can provide insight into the best way to correlate the data and develop a rate law.
^{14} R. I. Masel, Principles of Adsorption and Reaction on Solid Surfaces, New York: Wiley, 1996, p. 506, www.masel.com. This is a terrific book.
In Section 9.1 we discussed the PSSH, where the net rate of formation of reactive intermediates was assumed to be zero. An alternative way to derive a catalytic rate law rather than setting
$\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u22450}$
is to assume that each species adsorbed on the surface is a reactive intermediate. Consequently, the net rate of formation of species i adsorbed on the surface will be zero
$\begin{array}{cc}{r}_{i\cdot \text{S}}^{*}=0& \text{(1063)}\end{array}$
The PSSH is primarily used when more than one step is ratelimiting. The isomerization example shown in Table 104 is reworked using the PSSH in the Chapter 10 Expanded Material on the CRE Web site (http://www.umich.edu/~elements/6e/10chap/expanded_ch10_PSSH.pdf).
Consider a surfacereactionlimited irreversible isomerization
A → B
in which both A and B are adsorbed on the surface, and the rate law is
$\begin{array}{cc}{r}_{\text{A}}^{\prime}=\frac{k{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}}& \text{(1064)}\end{array}$
The specific reaction rate, k, will usually follow an Arrhenius temperature dependence and increase exponentially with temperature. However, the adsorption of all species on the surface is exothermic. Consequently, the higher the temperature, the smaller the adsorption equilibrium constant. That is, as the temperature increases, K_{A} and K_{B} decrease resulting in less coverage of the surface by A and B. Therefore, at high temperatures, the denominator of catalytic rate laws approaches 1. That is, at high temperatures (low coverage)
The rate law could then be approximated as
$\begin{array}{cc}{r}_{\text{A}}^{\prime}\text{\u2243}k{P}_{\text{A}}& \text{(1065)}\end{array}$
$\text{1}\text{\u226b}({P}_{\text{A}}{K}_{\text{A}}+{P}_{\text{B}}{K}_{\text{B}})$
Neglecting the adsorbed species at high temperatures
or for a reversible isomerization we would have
$\begin{array}{cc}{r}_{\text{A}}^{\prime}\text{\u2243}\text{}k({P}_{\text{A}}\frac{{P}_{\text{B}}}{{K}_{\text{P}}})& \text{(1066)}\end{array}$
Algorithm
Deduce
Rate law
Find
Mechanism
Evaluate
Ratelaw parameters
Design
PBR
CSTR
The algorithm we can use as a start in postulating a reaction mechanism and ratelimiting step is shown in Table 104. Again, we can never really prove a mechanism to be correct by comparing the derived rate law with experimental data. Independent spectroscopic experiments are usually needed to confirm the mechanism. We can, however, prove that a proposed mechanism is inconsistent with the experimental data by following the algorithm in Table 104. Rather than taking all the experimental data and then trying to build a model from the data, Box et al. describe techniques of sequential data collection and model building.^{15}
^{15} G. E. P. Box, J. S. Hunter, and W. G. Hunter, Statistics for Experimenters: Design, Innovation, and Discovery, 2nd ed., Hoboken, New Jersey: Wiley, 2005.
In this section, we focus on four operations that chemical reaction engineers need to be able to accomplish:
Developing an algebraic rate law consistent with experimental observations,
Analyzing the rate law in such a manner that the ratelaw parameters (e.g., k, K_{A}) can readily be determined from the experimental data,
Finding a mechanism and ratelimiting step consistent with the experimental data, and
Designing a catalytic reactor to achieve a specified conversion
We shall use the hydrodemethylation of toluene to illustrate these four operations.
Hydrogen and toluene are reacted over a solid mineral catalyst containing clinoptilolite (a crystalline silicaalumina) to form methane and benzene^{16}
^{16} J. Papp, D. Kallo, and G. Schay, J. Catal., 23, 168.
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{c}{\text{C}}_{\text{}6}{\text{H}}_{\text{5}}{\text{CH}}_{3}+{\text{H}}_{2}\text{}\end{array}& \underset{\text{catalyst}}{\begin{array}{c}\to \end{array}}\end{array}& \text{}\end{array}& \text{}\end{array}& {\text{C}}_{6}\end{array}{\text{H}}_{6}+{\text{CH}}_{\text{4}}$
We wish to design a packedbed reactor and a “fluidized bed” CSTR to process a feed consisting of 30% toluene, 45% hydrogen, and 25% inerts. Toluene is fed at a rate of 50 mol/min at a temperature of 640=C and a pressure of 40 atm (4052 kPa). To design the PBR, we must first determine the rate law from the differential reactor data presented in Table 106. In this table, we are given the rate of reaction of toluene as a function of the partial pressures of hydrogen (H_{2}), toluene (T), benzene (B), and methane (M). In the first two runs, methane was introduced into the feed together with hydrogen and toluene, while the other product, benzene, was fed to the reactor together with the reactants only in runs 3, 4, and 6. In runs 5 and 16, both methane and benzene were introduced in the feed. Because the conversion was less than 1% in the differential reactor, the partial pressures of the products, methane and benzene, formed in these runs were essentially zero, and the reaction rates were equivalent to initial rates of reaction.
TABLE 106 DATA FROM A DIFFERENTIAL REACTOR
${r}_{\text{T}}^{\prime}\times {\text{10}}^{10}$ 
Partial Pressure (atm) 

Run 
$\left(\frac{\text{g mol toluene}}{\text{gcat}\cdot \text{s}}\right)$ 
Toluene (T), P_{T} 
Hydrogen (H_{2}), P_{H}_{2} 
Methane (M), P_{M} 
Benzene (B), P_{B} 
Set A 





1 
71.0 
1 
1 
1 
0 
2 
71.3 
1 
1 
4 
0 
Set B 





3 
41.6 
1 
1 
0 
1 
4 
19.7 
1 
1 
0 
4 
5 
42.0 
1 
1 
1 
1 
6 
17.1 
1 
1 
0 
5 
Set C 





7 
71.8 
1 
1 
0 
0 
8 
142.0 
1 
2 
0 
0 
9 
284.0 
1 
4 
0 
0 
Set D 





10 
47.0 
0.5 
1 
0 
0 
11 
71.3 
1 
1 
0 
0 
12 
117.0 
5 
1 
0 
0 
13 
127.0 
10 
1 
0 
0 
14 
131.0 
15 
1 
0 
0 
15 
133.0 
20 
1 
0 
0 
16 
41.8 
1 
1 
1 
1 
Unscramble the data to find the rate law.
To unscramble the data, let’s first look at run 3. In run 3, there is no possibility of the reverse reaction taking place because the concentration of methane is zero, that is, P_{M} = 0, whereas in run 5 the reverse reaction could take place because all products are present. Comparing runs 3 and 5, we see that the initial rate is essentially the same for both runs, and we can assume that the reaction is virtually irreversible.
$\text{T}+{\text{H}}_{2}\text{}\overrightarrow{\text{catayst}}\text{}\text{M}+\text{B}$
We now ask what qualitative conclusions can be drawn from the data about the dependence of the rate of disappearance of toluene, ${r}_{\text{T}}^{\prime}$, on the partial pressures of toluene, hydrogen, methane, and benzene.
Dependence on the product methane. If methane were adsorbed on the surface, the partial pressure of methane would appear in the denominator of the rate expression and the rate would vary inversely with methane concentration
$\begin{array}{cc}{r}_{\text{T}}^{\prime}\sim \frac{[\text{}\cdot ]}{1+{K}_{\text{M}}{P}_{\text{M}}+\cdot \cdot \cdot}& \text{(1067)}\end{array}$
However, comparing runs 1 and 2 we observe that a fourfold increase in the pressure of methane has little effect on ${r}_{\text{T}}^{\prime}$. Consequently, we assume that methane is either very weakly adsorbed (i.e., K_{M}P_{M} ≪ 1) or goes directly into the gas phase in a manner similar to propylene in the cumene decomposition previously discussed.
Dependence on the product benzene. In runs 3 and 4, we observe that, for fixed concentrations (partial pressures) of hydrogen and toluene, the rate decreases with increasing concentration of benzene. A rate expression in which the benzene partial pressure appears in the denominator could explain this dependency
$\begin{array}{cc}{r}_{\text{T}}^{\prime}\sim \frac{1}{1+{K}_{\text{B}}{P}_{\text{B}}+\cdot \cdot \cdot}& \text{(1068)}\end{array}$
The type of dependence of on P_{B} given by Equation (1068) suggests that benzene is adsorbed on the clinoptilolite surface.
If it is in the denominator, it is probably on the surface.
Dependence on toluene. At low concentrations of toluene (runs 10 and 11), the rate increases with increasing partial pressure of toluene, while at high toluene concentrations (runs 14 and 15), the rate is virtually independent of the toluene partial pressure. A form of the rate expression that would describe this behavior is
$\begin{array}{cc}{r}_{\text{T}}^{\prime}\sim \frac{{P}_{\text{T}}}{1+{K}_{\text{T}}{P}_{\text{T}}+\cdot \cdot \cdot}& \text{(1069)}\end{array}$
A combination of Equations (1068) and (1069) suggests that the rate law may be of the form
$\begin{array}{cc}{r}_{\text{T}}^{\prime}\sim \frac{{P}_{\text{T}}}{1+{K}_{\text{T}}{P}_{\text{T}}+{K}_{\text{B}}{P}_{\text{B}}+\cdot \cdot \cdot}& \text{(1070)}\end{array}$
Dependence on hydrogen. When we compare runs 7, 8, and 9 in Table 106, we see that the rate increases linearly with increasing hydrogen concentration, and we conclude that the reaction is first order in H_{2}. In light of this fact, hydrogen is either not adsorbed on the surface or its coverage of the surface is extremely low $\left(1{\text{\u226bK}}_{{\text{H}}_{2}}{P}_{{\text{H}}_{2}}\right)$ for the pressures used. If H_{2} were adsorbed, ${r}_{\text{T}}^{\prime}$ would have a dependence on P_{H}_{2} analogous to the dependence of ${r}_{\text{T}}^{\prime}$ on the partial pressure of toluene, P_{T} (see Equation (1069)). For firstorder dependence on H_{2},
$\begin{array}{cc}{r}_{\text{T}}^{\prime}\text{}\sim {P}_{{\text{H}}_{2}}& \text{(1071)}\end{array}$
Deduced the rate law from data using our knowledge of catalysis
Combining Equations (1067)–(1071), we find that the rate law
$\begin{array}{cc}\hline {r}_{\text{T}}^{\prime}=\frac{k{P}_{{\text{H}}_{2}}{P}_{\text{T}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{T}}{P}_{\text{T}}}& \text{}\\ \hline\end{array}$
is in qualitative agreement with the data shown in Table 106.
We now propose a mechanism for the hydrodemethylation of toluene. Because the partial pressure of H_{2} only appears in the numerator, we assume that the reaction follows an Eley–Rideal mechanism. Applying the Eley–Rideal mechanism, toluene is adsorbed on the surface and then reacts with hydrogen in the gas phase to produce benzene adsorbed on the surface and methane in the gas phase. Benzene is then desorbed from the surface. Because approximately 75%–80% of all heterogeneous reaction mechanisms are surfacereactionlimited rather than adsorption or desorptionlimited, we begin by assuming the reaction between adsorbed toluene and gaseous hydrogen to be reactionratelimited. Symbolically, this mechanism and associated rate laws for each elementary step are
Approximately 75% of all heterogeneous reaction mechanisms are surfacereactionlimited.
Proposed Mechanism
Eley–Rideal mechanism
The rate law for the surface reaction step is
$\begin{array}{cc}\text{}{r}_{\text{S}}={k}_{\text{S}}({P}_{{\text{H}}_{2}}{\text{C}}_{\text{T}\cdot \text{S}}\frac{{\text{C}}_{\text{B}\cdot \text{S}}{P}_{\text{M}}}{{K}_{\text{S}}})& \text{(1073)}\end{array}$
For surfacereactionlimited mechanisms we see that we need to replace C_{T·S} and C_{B·S} in Equation (1073) by quantities that we can measure, for example, concentration or partial pressure.
For surfacereactionlimited mechanisms, we use the adsorption rate Equation (1072) for toluene to obtain C_{T·S}^{17}, that is,
^{17} See footnote 11 on page 466.
$\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\approx 0$
Then
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{\text{T}\cdot \text{S}}={K}_{\text{T}}{P}_{\text{T}}{\text{C}}_{\upsilon}\\ \hline\end{array}& \text{(1075)}\end{array}$
and we use the desorption rate Equation (1074) for benzene to obtain C_{B·S}:
$\frac{{r}_{\text{D}}}{{k}_{\text{D}}}\approx 0$
Then
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{\text{B}\cdot \text{S}}={K}_{\text{B}}{P}_{\text{B}}{\text{C}}_{\upsilon}\\ \hline\end{array}& \text{(1076)}\end{array}$
The total concentration of sites is
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{\text{t}}={\text{C}}_{\upsilon}+{\text{C}}_{\text{T}\cdot \text{S}}+{\text{C}}_{\text{B}\cdot \text{S}}\\ \hline\end{array}& \text{(1077)}\end{array}$
Substituting Equations (1075) and (1076) into Equation (1077) and rearranging, we obtain
$\begin{array}{cc}{\text{C}}_{\upsilon}=\frac{{\text{C}}_{\text{t}}}{1+{K}_{\text{T}}{P}_{\text{T}}+{K}_{\text{B}}{P}_{\text{B}}}& \text{(1078)}\end{array}$
Perform a site balance to obtain C_{υ}.
Next, substitute for C_{T·S} and C_{B·S}, in terms of partial pressures and then substitute for in Equation (1073) to obtain the rate law for the case when the reaction is surfacereactionratelimited
$\begin{array}{cc}{r}_{\text{T}}^{\prime}=\frac{\stackrel{k}{\stackrel{}{\overbrace{{\text{C}}_{\text{t}}{k}_{\text{S}}{\text{K}}_{\text{T}}}}}({\text{P}}_{{\text{H}}_{2}}{\text{P}}_{\text{T}}{\text{P}}_{\text{B}}{\text{P}}_{\text{M}}/{\text{K}}_{\text{P}})}{1+{\text{K}}_{\text{T}}{\text{P}}_{\text{T}}+{\text{K}}_{\text{B}}{\text{P}}_{\text{B}}}& \text{(1079)}\end{array}$
We have shown by comparing runs 3 and 5 that we can neglect the reverse reaction, that is, the thermodynamic equilibrium constant K_{P} is very, very large. Consequently, we obtain
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{T}}^{\prime}=\frac{k{P}_{{\text{H}}_{2}}{P}_{\text{T}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{T}}{P}_{\text{T}}}\\ \hline\end{array}& \text{(1080)}\end{array}$
Rate law for Eley–Rideal surfacereactionlimited mechanism
Again we note that the adsorption equilibrium constant of a given species is exactly the reciprocal of the desorption equilibrium constant of that species.
In the original work on this reaction by Papp et al.,^{18} over 25 models were tested against experimental data, and it was concluded that the preceding mechanism and ratelimiting step (i.e., the surface reaction between adsorbed toluene and H_{2} gas) is the correct one. Assuming that the reaction is essentially irreversible, the rate law for the reaction on clinoptilolite is
^{18} Ibid.
$\begin{array}{cc}{r}_{\text{T}}^{\prime}=k\frac{{P}_{{\text{H}}_{2}}{P}_{\text{T}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{T}}{P}_{\text{T}}}& \text{(1080)}\end{array}$
We now wish to determine how best to analyze the data to evaluate the ratelaw parameters, k, K_{T}, and K_{B}. This analysis is referred to as parameter estimation.^{19} We will use nonlinear least squares which require initial estimates of the parameters. These estimates can be obtained from a linear leastsquares analysis. To use linear least square, we now rearrange our rate law to obtain a linear relationship between our measured variables. For the rate law given by Equation (1080), we see that if both sides of Equation (1080) are divided by P_{H2}P_{T} and the equation is then inverted
^{19} See the Supplementary Reading for Chapter 10 on the CRE Web site (page 539) for a variety of techniques for estimating the ratelaw parameters.
$\begin{array}{cc}\begin{array}{c}\hline \frac{{P}_{{\text{H}}_{2}}{P}_{\text{T}}}{{r}_{\text{T}}^{\prime}}=\frac{1}{k}+\frac{{K}_{\text{B}}{P}_{\text{B}}}{k}+\frac{{K}_{\text{T}}{P}_{\text{T}}}{k}\\ \hline\end{array}& (1081)\end{array}$
Linearize the rate equation to extract the ratelaw parameters.
The regression techniques described in Chapter 7 could be used to determine the ratelaw parameters by using the equation
$\begin{array}{c}\hline {Y}_{j}={a}_{0}+{a}_{1}{X}_{\text{1}j}+{a}_{2}{X}_{2}j\\ \hline\end{array}$
A linear leastsquares analysis of the data shown in Table 106 is presented on the CRE Web site.
One can use the linearized leastsquares analysis (PRS 7.3) to obtain initial estimates of the parameters k, K_{T}, K_{B}, in order to obtain convergence in nonlinear regression. However, in many cases it is possible to use a nonlinear regression analysis directly, as described in Sections 7.5 and 7.6, and in Example 101.
Use nonlinear regression, as discussed in Section 7.5, along with the data in Table 106, to find the best estimates of the ratelaw parameters k, K_{B}, and K_{T} in Equation (1080).
Write the rate law solely as a function of the partial pressures.
Find the ratio of the sites occupied by toluene, C_{T•S}, to those occupied by benzene, C_{B•S}, at 40% conversion of toluene.
Solution
The data from Table 106 were entered into the Polymath nonlinear leastsquares program with the following modification. The rates of reaction in column 1 were multiplied by 10^{10}, so that each of the numbers in column 1 was entered directly (i.e., 71.0, 71.3, …). The model equation was
$\begin{array}{cc}\text{Rate}=\frac{k{P}_{\text{T}}{P}_{{\text{H}}_{2}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{T}}{P}_{\text{T}}}& \text{(E101.1)}\end{array}$
Following the stepbystep regression procedure in Chapter 7 and on the CRE Web site Summary Notes, we arrive at the following parameter values shown in Table E101.1 for this Living Example Problem (LEP). A tutorial is also given for LEP 101.
The best estimates are shown in the upperrighthand box of Table E101.1.
Converting the rate law to kilograms of catalyst and minutes,
$\begin{array}{cc}{r}_{\text{T}}^{\prime}=\frac{1.45\times {10}^{8}{P}_{\text{T}}{P}_{{\text{H}}_{2}}}{1+1.39{P}_{\text{B}}+1.038{P}_{\text{T}}}\frac{\text{mol T}}{\text{gcat}\cdot \text{S}}\times \frac{1000\text{g}}{1\text{kg}}\times \frac{60\text{s}}{\text{min}}& \text{(E101.2)}\end{array}$
we have
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{T}}^{\prime}=\frac{8.7\times {10}^{4}{P}_{\text{T}}{P}_{{\text{H}}_{2}}}{1+1.39{P}_{\text{B}}+1.038{P}_{\text{T}}}\left(\frac{\text{mol T}}{\text{gcat.}\cdot \text{min}}\right)\\ \hline\end{array}& \text{(E101.3)}\end{array}$
Ratio of sites occupied by toluene to those occupied by benzene
After we have the adsorption constants, K_{T} and K_{B}, we can calculate the ratio of sites occupied by the various adsorbed species. For example, taking the ratio of Equation (1075) to Equation (1076), the ratio of tolueneoccupied sites to benzeneoccupied sites at 40% conversion is
$\begin{array}{ccc}\frac{{\text{C}}_{\text{T}\cdot \text{S}}}{{\text{C}}_{\text{B}\cdot \text{S}}}& =\frac{{\text{C}}_{\upsilon}{K}_{\text{T}}{P}_{\text{T}}}{{\text{C}}_{\upsilon}{K}_{\text{B}}{P}_{\text{B}}}=\frac{{K}_{\text{T}}{P}_{\text{T}}}{{K}_{\text{B}}{P}_{\text{B}}}=\frac{{K}_{\text{T}}{P}_{\text{A0}}(1\text{X})}{{K}_{\text{B}}{P}_{\text{A0}}\text{X}}& \text{(E101.4)}\\ \text{}& =\frac{{K}_{\text{T}}(1\text{X})}{{K}_{\text{B}}\text{X}}=\frac{1.038(10.4)}{\text{1.39}\left(0.4\right)}=1.12& \text{}\end{array}$
We see that at 40% conversion there are approximately 12% more sites occupied by toluene than by benzene. This fact is common knowledge to every chemical engineering student at Jofostan University, Riça, Jofostan.
Analysis: This example shows once again how to determine the values of ratelaw parameters from experimental data using Polymath nonlinear regression. It also shows how to calculate the different fraction of sites, both vacant and occupied, as a function of conversion.
Our next step is to express the partial pressures P_{T}, P_{B}, and P_{H2} as a function of X, combine the partial pressures with the rate law, ${r}_{\text{A}}^{\prime}$, as a function of conversion, and carry out the integration of the packedbed design equation
$\begin{array}{cc}\frac{d\text{X}}{d\text{W}}=\frac{{r}_{\text{A}}^{\prime}}{{\text{F}}_{\text{A0}}}& \text{(217)}\end{array}$
The hydrodemethylation of toluene is to be carried out in a PBR catalytic reactor.
${\text{C}}_{6}{\text{H}}_{5}{\text{CH}}_{3}+{\text{H}}_{2}\text{}\underset{\text{catalyst}}{\to}\text{}{{\text{C}}_{6}\text{H}}_{\text{6}}+{\text{CH}}_{4}$
The molar feed rate of toluene to the reactor is 50 mol/min, and the reactor inlet is at 40 atm and 640°C. The feed consists of 30% toluene, 45% hydrogen, and 25% inerts. Hydrogen is used in excess to help prevent coking. The pressuredrop parameter, α, is 9.8 × 10^{–5} kg^{–1}.
Plot and analyze the conversion, the pressure ratio, p, and the partial pressures of toluene, hydrogen, and benzene as a function of PBR catalyst weight.
Determine the catalyst weight in a fluidized CSTR with a bulk density of 400 kg/m^{3} (0.4 g/cm^{3}) to achieve 65% conversion.
Solution
PBR with pressure drop
Mole Balance:
$\frac{{d\text{F}}_{T}}{dW}={r}_{\text{T}}^{\prime}$
$\begin{array}{cc}\begin{array}{c}\hline \frac{d\text{X}}{d\text{W}}=\frac{{r}_{\text{T}}^{\prime}}{{\text{F}}_{\text{T0}}}\\ \hline\end{array}& \text{(E102.1)}\end{array}$
Balance on toluene (T), the limiting reactant
Rate Law: From Equation (E101.1) we have
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{T}}^{\prime}=\frac{k{P}_{{\text{H}}_{2}}{P}_{\text{T}}}{1+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{T}}{P}_{\text{T}}}\\ \hline\end{array}& \text{(E102.2)}\end{array}$
with k = 0.00087 mol/atm^{2}/kgcat/min, K_{B} = 1.39 atm^{–1}, and K_{T} = 1.038 atm^{–1}.
Stoichiometry:
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\hline {P}_{\text{T}}={C}_{\text{T}}RT={C}_{\text{T0}}R{T}_{0}\left(\frac{1X}{\text{1+}\text{\u025b}X}\right)P={P}_{\text{T0}}\left(\frac{1X}{\text{1+\u025b}X}\right)P\\ \text{\epsilon}={y}_{\text{T0}}\delta =0.3\left(0\right)=0\\ \hline\end{array}\\ P=\frac{P}{{P}_{\text{0}}}\end{array}\\ {P}_{\text{T}}={P}_{\text{T0}}(1X)\text{p}\end{array}\\ {P}_{{\text{H}}_{2}}={P}_{\text{T0}}({\mathrm{\Theta}}_{{\text{H}}_{2}}X)\text{p}\end{array}& \text{(E102.3)}\end{array}$
Relating Toluene (T) Benzene (B) Hydrogen (H_{2})
$\begin{array}{cc}{\mathrm{\Theta}}_{{\text{H}}_{2}}=\frac{0.425}{0.30}=1.5& \text{}\\ {P}_{{\text{H}}_{\text{2}}}={P}_{\text{T0}}(1.5\text{X})P& \text{(E102.4)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline {P}_{\text{B}}={P}_{\text{T0}}{\text{X}}_{P}\\ \hline\end{array}& \text{(E102.5)}\end{array}$
Because ε = 0, we can use the integrated form of the pressuredrop term.
$\begin{array}{cc}\begin{array}{c}\hline P=\frac{P}{{P}_{0}}={(1\alpha \text{W})}^{\text{1/2}}\\ \hline\end{array}& \text{(533)}\end{array}$
P_{0} = total pressure at the entrance
α = 9.8 = 10^{–5} kg^{–1}
Note that P_{T0} designates the inlet partial pressure of toluene. In this example, the inlet total pressure is designated P_{0} to avoid any confusion. The inlet mole fraction of toluene is 0.3 (i.e., y_{T0} = 0.3), so that the inlet partial pressure of toluene is
P_{T0} = (0.3)(40) = 12 atm
Pressure drop in PBRs is discussed in Section 5.5.
We now calculate the maximum catalyst weight we can have, such that the exiting pressure will not fall below atmospheric pressure (i.e., 1.0 atm) for the specified feed rate. This weight is calculated by substituting the entering pressure of 40 atm and the exiting pressure of 1 atm into Equation (533), that is,
$\begin{array}{c}\begin{array}{c}\begin{array}{cc}p=\frac{P}{{P}_{0}}={(1\alpha \text{W})}^{\text{1/2}}& \text{}\end{array}\\ \text{}\frac{1}{40}={(1\text{9.8}\times {10}^{5}\text{W})}^{\text{1/2}}\end{array}\\ \text{W}=10197\text{kg}\end{array}$
Maximum catalyst weight for conditions given.
Conversion profile down the packed bed
Evaluate: Consequently, we will set our final weight at 10000 kg and determine the conversion profile as a function of catalyst weight up to this value. Equations (E102.1)–(E102.5) are shown in the Polymath program in Table E102.1. The conversion is shown as a function of catalyst weight in Figure E102.1, and profiles of the partial pressures of toluene, hydrogen, and benzene are shown in Figure E102.2. We note that the pressure drop causes (cf. Equation E102.5) the partial pressure of benzene to go through a maximum as one traverses the reactor.
Note the partial pressure of benzene goes through a maximum. Why?
If one had neglected ΔP it could have been very embarrassing.
See YouTube video, “Reaction Engineering Gone Wrong,” accessible from the CRE Web site home page.
PBR Analysis: For the case of no pressure drop, the conversion that would have been achieved with 10000 kg of catalyst would have been 79%, compared with 68.2% when there is pressure drop in the reactor. To carry out this calculation, for the case of no pressure drop, we use the Polymath Living Example Program (LEP) on the CRE Web site and simply multiply the pressuredrop parameter by zero, that is, line (5) would read α = 0.000098*0. For the feed rate given, eliminating or minimizing pressure drop would increase the production of benzene by up to 61 million pounds per year! Finally, we note in Figure E102.2 that the partial pressure of benzene (P_{B}) goes through a maximum. This maximum can be explained by recalling that P_{B} is just the product of the benzene mole fraction (y_{B}) times the total pressure (P) (i.e., P_{B} = y_{B}P_{T}). Near the middle to end of the bed, benzene is no longer being formed so that y_{B} stops increasing. However, because of the pressure drop, the total pressure decreases and, as a result, so does P_{B}.
“Fluidized” CSTR
We will now calculate the fluidized CSTR catalyst weight necessary to achieve the same (ca.) conversion as in the packedbed reactor at the same operating conditions. The bulk density in the fluidized reactor is 0.4 g/cm^{3} with p = 1. The design equation is
Mole Balance:
In 
– 
Out 
+ 
Gen 
= 
Accum 
F_{T0} 
– 
F_{T} 
+ 
${r}_{\text{T}}^{\prime}\text{W}$ 
= 
0 
Rearranging
$\begin{array}{cc}\text{W}=\frac{{\text{F}}_{\text{T0}}{\text{F}}_{\text{T}}}{{r}_{\text{T}}^{\prime}}=\frac{{\text{F}}_{\text{T0}}\text{X}}{{r}_{\text{T}}^{\prime}}& \text{(E102.6)}\end{array}$
Rate Law and
Stoichiometry same as in part (a) PBR calculation.
Combine and Evaluate: Writing Equation (E102.2) in terms of conversion (E102.3) and then substituting X = 0.65 and P_{T0} = 12 atm, we have
$\begin{array}{c}\begin{array}{c}{r}_{\text{T}}^{\prime}=\frac{8.7\times {10}^{4}{P}_{\text{T}\text{}}{P}_{{\text{H}}_{2}}}{1+1.39{P}_{\text{B}}+1.038{P}_{\text{T}}}=\frac{8.7\times {10}^{4}{{P}^{\text{}}}_{\text{T0}}^{2}(1\text{X})(1.5\text{X})}{1+1.39{P}_{\text{T0}}\text{X}+1.038{P}_{\text{T0}}(1\text{X})}=2.3\times {10}^{3}\frac{\text{mol}}{\text{kgcat}\cdot \text{min}}\\ \text{W}=\frac{{\text{F}}_{\text{T0}}\text{X}}{{r}_{\text{T}}^{\prime}}=\frac{\left(50\text{mol T/min}\right)\left(0.65\right)}{2.3\times {10}^{3}\text{mol}\text{T/kg cat}\cdot \text{min}}\end{array}\end{array}$
$\begin{array}{c}\hline \text{W}=1.41\times {10}^{\text{4}}\text{kg of catalyst}\\ \hline\end{array}$
The corresponding reactor volume is
$\text{}\text{V}=\frac{\text{W}}{{\rho}_{b}}=\frac{1.41\times {\text{10}}^{4}\text{}\text{\hspace{0.17em}}\text{kg}}{400{\text{kg/m}}^{3}}=35.25{\text{m}}^{3}\text{}$
This volume would correspond to a cylindrical reactor 2 m in diameter and 10 m high.
Analysis: This example used real data and the CRE algorithm to design a PBR and CSTR. An important point is that it showed how one could be embarrassed by not including pressure drop in the design of a packedbed reactor. We also note that for both the PBR and fluidized CSTR, the values of the catalyst weight and reactor volume are quite high, especially for the low feed rates given. Consequently, the temperature of the reacting mixture should be increased to reduce the catalyst weight, provided that side reactions and catalyst decay do not become a problem at higher temperatures.
How can the weight of catalyst be reduced?
Raise the temperature?
Example 102 illustrated the major activities pertinent to catalytic reactor design described earlier in Figure 106. In this example, the rate law was extracted directly from the data and then a mechanism was found that was consistent with experimental observation. Conversely, developing a feasible mechanism may guide one in the synthesis of the rate law.
We now extend the principles of the preceding sections to one of the newer technologies in chemical engineering. Chemical engineers are now playing an important role in the electronics industry. Specifically, they are becoming more involved in the manufacture of electronic and photonic devices, recording materials, and especially microfluidics and medical labonachip devices.
Surface reactions play an important role in the manufacture of microelectronic devices. One of the single most important developments of the twentieth century was the invention of the integrated circuit. Advances in the development of integrated circuitry have led to the production of circuits that can be placed on a single semiconductor chip the size of a pinhead and perform a wide variety of tasks by controlling the electron flow through a vast network of channels. These channels, which are made from semiconductors such as silicon, gallium arsenide, indium phosphide, and germanium, have led to the development of a multitude of novel microelectronic devices. Examples of microelectronic sensing devices manufactured using chemical reaction engineering principles are shown in the lefthand margin.
The manufacture of an integrated circuit requires the fabrication of a network of pathways for electrons. The principal reaction engineering steps of the fabrication process include depositing material on the surface of a material called a substrate (e.g., by chemical vapor deposition, abbreviated as CVD), changing the conductivity of regions of the surface (e.g., by boron doping or ion inplantation), and removing unwanted material (e.g., by etching). By applying these steps systematically, miniature electronic circuits can be fabricated on very small semiconductor chips. The fabrication of microelectronic devices may include as few as 30 or as many as 200 individual steps to produce chips with up to 10^{9} elements per chip. An abbreviated discussion of the steps involved to produce a typical metaloxide semiconductor fieldeffect transition (MOSFET) is given in the Additional Material for this chapter on the Web site (http://umich.edu/~elements/6e/10chap/pdf/Steps_in_Microchip_Fabrication.pdf).
The mechanisms by which CVD occurs are very similar to those of heterogeneous catalysis discussed earlier in this chapter. The reactant(s) adsorbs on the surface and then reacts on the surface to form a new surface. This process may be followed by a desorption step, depending on the particular reaction.
Ge used in solar cells
The growth of a germanium epitaxial film as an interlayer between a gallium arsenide layer and a silicon layer has received attention in the microelectronics industry.^{20} Epitaxial germanium is also an important material in the fabrication of tandem solar cells. The growth of germanium films can be accomplished by CVD. A proposed mechanism is
^{20} H. Ishii and Y. Takahashi, J. Electrochem. Soc., 135, p. 1539.
$\begin{array}{cc}\text{GasPhase dissociation:}& {\text{GeCl}}_{4}\left(\text{g}\right)\overrightarrow{\leftarrow}{\text{GeCl}}_{2}\left(\text{g}\right)+{\text{Cl}}_{2}\left(\text{g}\right)\end{array}$
$\begin{array}{ll}\text{Adsorption:}& {\text{GeCl}}_{2}\left(\text{g}\right)+\text{S}\text{}\stackrel{{k}_{\text{A}}}{\leftrightarrows}\text{}{\text{GeCl}}_{2}\text{}\cdot \text{S}\\ \text{Adsorption:}& {\text{H}}_{2}+2\text{S}\text{}\stackrel{{k}_{\text{H}}}{\leftrightarrows}\text{}2\text{H}\cdot \text{S}\\ \text{Surface reaction:}& \begin{array}{cc}\begin{array}{c}{\text{GeCl}}_{2}\text{}\cdot \text{S}+2\text{H}\cdot \text{S}\end{array}& \stackrel{{k}_{\text{S}}}{\begin{array}{c}\to \end{array}}\end{array}\text{Ge}\left(\text{S}\right)+2\text{HCl}\left(\text{g}\right)+2\text{S}\end{array}$
Mechanism
At first it may appear that a site has been lost when comparing the right and lefthand sides of the surface reaction step. However, the newly formed germanium atom on the righthand side is a site for the future adsorption of H_{2}(g) or GeCl_{2}(g), and there are three sites on both the right and lefthand sides of the surface reaction step. These sites are shown schematically in Figure 1020.
The surface reaction between adsorbed molecular hydrogen and germanium dichloride is believed to be ratelimiting. The reaction follows an elementary rate law with the rate being proportional to the fraction of the surface covered by GeCl_{2} times the square of the fraction of the surface covered by molecular hydrogen.
Rate law for ratelimiting step
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{Dep}}^{"}={k}_{\text{S}}f{\text{GeCl}}_{2}\text{}{f}_{\text{H}}^{2}\\ \hline\end{array}& \text{(1082)}\end{array}$
where
$\begin{array}{rr}{\text{r}}_{\text{Dep}}^{"}& =\text{deposition rate per unit surface area, nm/s}\hfill \\ {k}_{\text{S}}& =\text{surface specific reaction rate, nm/s}\hfill \\ {f}_{{\text{GeCl}}_{2}}& =\text{fraction of the surface occupied by germanium dichloride}\hfill \\ {f}_{\text{H}}& =\text{fraction of the surface covered by molecular hydrogen}\hfill \end{array}$
The deposition rate (film growth rate) is usually expressed in nanometers per second and is easily converted to a molar rate (mol/m^{2} · s) by multiplying by the molar density of solid germanium (mol/m^{3}).
The difference between developing CVD rate laws and rate laws for catalysis is that the site concentration (e.g., C_{υ}) is replaced by the fractional surface area coverage (e.g., the fraction of the surface that is vacant, f_{v}). The total fraction of surface available for adsorption should, of course, add up to 1.0.
Area balance
$\begin{array}{c}\text{Fractional area balance:}\end{array}\text{}\begin{array}{cc}\begin{array}{c}\hline \begin{array}{cc}& {f}_{\text{v}}+{f}_{{\text{GeCl}}_{2}}+{f}_{\text{H}}=1\end{array}\\ \hline\end{array}& \text{(1083)}\end{array}$
We will first focus our attention on the adsorption of GeCl_{2}. The rate of jumping on to the surface is proportional to the partial pressure of GeCl_{2}, P_{GeCl}_{2}, and the fraction of the surface that is vacant, f_{v} and the rate of jumping off is proportional to fraction area covered with GeCl_{2}. The net rate of GeCl_{2} adsorption is
$\begin{array}{cc}{\text{r}}_{\text{AD}}={k}_{\text{A}}({f}_{\text{v}}{\text{P}}_{{\text{GeCl}}_{2}}\frac{{f}_{{\text{GeCl}}_{2}}}{{\text{K}}_{\text{A}}})& \text{(1084)}\end{array}$
Since the surface reaction is ratelimiting, in a manner analogous to catalysis reactions, we have for the adsorption of GeCl_{2}
Adsorption of GeCl_{2} not ratelimiting
$\frac{{\text{r}}_{\text{AD}}}{{k}_{\text{A}}}\approx 0$
Solving Equation (1084) for the fractional surface coverage of GeCl_{2} gives
$\begin{array}{cc}\begin{array}{c}\hline {f}_{{\text{GeCl}}_{2}}={f}_{\text{v}}{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}\\ \hline\end{array}& (\text{1085)}\end{array}$
For the dissociative adsorption of hydrogen on the Ge surface, the equation analogous to Equation (1084) is
$\begin{array}{cc}{\text{r}}_{{\text{ADH}}_{2}}={k}_{\text{H}}({\text{P}}_{{\text{H}}_{2}}{f}_{\text{v}}^{2}\frac{{f}_{\text{H}}^{2}}{{\text{K}}_{\text{H}}})& \text{(1086)}\end{array}$
Since the surface reaction is ratelimiting
Adsorption of H_{2} is not ratelimiting.
$\frac{{\text{r}}_{{\text{ADH}}_{2}}}{{k}_{\text{H}}}\approx 0$
Then
$\begin{array}{cc}\begin{array}{c}\hline {f}_{\text{H}}={f}_{\text{v}}\sqrt{{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}}\\ \hline\end{array}& \left(\mathrm{1087}\right)\end{array}$
Recalling the rate of deposition of germanium, we substitute for f_{GeCl2} and f_{H} in Equation (1082) to obtain
$\begin{array}{cc}{\text{r}}_{\text{Dep}}^{"}={f}_{\text{v}}^{3}{k}_{\text{S}}{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}& \left(\mathrm{1088}\right)\end{array}$
We solve for f_{v} in an identical manner to that for C_{v} in heterogeneous catalysis. Substituting Equations (1085) and (1087) into Equation (1083) gives
${f}_{\text{v}}+{f}_{\text{v}}\sqrt{{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}}+{f}_{\text{v}}{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}=1$
Rearranging yields
$\begin{array}{cc}{f}_{\text{V}}=\frac{1}{1+{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}+\sqrt{{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}}}& \text{(1089)}\end{array}$
Finally, substituting for ${f}_{\text{v}}$ in Equation (1088), we find that
$\begin{array}{cc}{\text{r}}_{\text{Dep}}^{\u2033}=\frac{{k}_{\text{S}}{\text{K}}_{\text{H}}{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}{\text{P}}_{{\text{H}}_{2}}}{(1+{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}+{\sqrt{{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}})}^{3}}& \text{}\end{array}$
and lumping K_{A}, K_{H}, and k_{S} into a specific reaction rate k′ yields
Rate of deposition of Ge
$\begin{array}{cc}\begin{array}{cc}\hline \begin{array}{c}{\text{r}}_{\text{Dep}}^{\u2033}=\frac{{k}^{\prime}{\text{P}}_{{\text{GeCl}}_{2}}{\text{P}}_{{\text{H}}_{2}}}{(1+{\text{K}}_{\text{A}}{\text{P}}_{{\text{GeCl}}_{2}}+{\sqrt{{\text{K}}_{\text{H}}{\text{P}}_{{\text{H}}_{2}}})}^{3}}\end{array}& \text{}\\ \hline\end{array}& \text{(1090)}\end{array}$
We now need to relate the partial pressure of GeCl_{2} to the partial pressure of GeCl_{4} in order to calculate the conversion of GeCl_{4}. If we assume that the gasphase reaction
Equilibrium in gas phase
${\text{GeCl}}_{4}\left(g\right)\text{}\overrightarrow{\leftarrow}{\text{GeCl}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)$
is in equilibrium, we have
$\begin{array}{c}{\text{K}}_{\text{P}}=\frac{{\text{P}}_{{\text{GeCl}}_{2}}{\text{P}}_{{\text{Cl}}_{2}}}{{\text{P}}_{{\text{GeCl}}_{4}}}\\ {\text{P}}_{{\text{GeCl}}_{2}}=\frac{{\text{P}}_{{\text{GeCl}}_{4}}}{{\text{P}}_{{\text{Cl}}_{2}}}\text{}\cdot {\text{K}}_{\text{P}}\end{array}$
and if hydrogen is weakly adsorbed $(\sqrt{{K}_{H}{P}_{{H}_{2}}}<1)$, and letting k = k′ K_{P}, we obtain the rate of deposition as
$\begin{array}{cc}{\text{r}}_{\text{Dep}}^{\u2033}=\frac{k{\text{P}}_{{\text{GeCl}}_{4}}{\text{P}}_{{\text{H}}_{2}}{\text{P}}_{{\text{Cl}}_{2}}^{2}}{{({\text{P}}_{{\text{Cl}}_{2}}+{\text{K}}_{\text{A}}{\text{K}}_{\text{P}}{\text{P}}_{{\text{GeCl}}_{4}})}^{3}}& \text{(1091)}\end{array}$
Rate of deposition of Ge when H_{2} is weakly adsorbed
We now can use stoichiometry to express each of the species’ partial pressures in terms of conversion and the entering partial pressure of GeCl_{4}, P_{GeCl4,0}, and then proceed to calculate the conversion.
It should be noted that it is possible that GeCl_{2} may also be formed by the reaction of GeCl_{4} and a Ge atom on the surface, in which case a different rate law would result.
We have seen that for each mechanism and each ratelimiting step we can derive a rate law. Consequently, if we had three possible mechanisms and three ratelimiting steps for each mechanism, we would have nine possible rate laws to compare with the experimental data. We will use the regression techniques discussed in Chapter 7 to identify which model equation best fits the data by choosing the one with the smaller sums of squares and/or carrying out an Ftest. We could also compare the nonlinear regression residual plots for each model, which not only show the error associated with each data point but also show if the error is randomly distributed or if there is a trend in the error. If the error is randomly distributed, this result is an additional indication that the correct rate law has been chosen.
Regression
We need to raise a caution here about choosing the model with the smallest sums of squares. The caution is that the model parameter values that give the smallest sum must be realistic. In the case of heterogeneous catalysis, all values of the adsorption equilibrium constants must be positive. In addition, if the temperature dependence is given, because adsorption is exothermic, the adsorption equilibrium constant must decrease with increasing temperature. To illustrate these principles, let’s look at the following example.
The hydrogenation (H) of ethylene (E) to form ethane (EA),
H_{2} + C_{2}H_{4} → C_{2}H_{6}
is carried out over a cobalt molybdenum catalyst (Collect. Czech. Chem. Commun., 51, 2760). Carry out a nonlinear regression analysis to determine which of the following rate laws best describes the data in Table E103.1.
$\begin{array}{cccc}\left(\text{a}\right)& {\text{r}}_{\text{E}}^{\prime}=\frac{k{P}_{\text{E}}{P}_{\text{H}}}{1+{K}_{\text{EA}}\text{}\text{}\text{}{P}_{\text{EA}}+{K}_{\text{E}}{P}_{\text{E}}}\hfill & \left(\text{C}\right)& {\text{r}}_{\text{E}}^{\prime}=\frac{k{P}_{\text{E}}{P}_{E}\text{}}{{(1+{K}_{\text{E}}{P}_{\text{E}})}^{2}}\hfill \\ \left(\text{b}\right)& {\text{r}}_{\text{E}}^{\prime}=\frac{k{P}_{\text{E}}\text{}{P}_{\text{H}}}{1+{K}_{\text{E}}{P}_{\text{E}}}\hfill & \left(\text{d}\right)& {\text{r}}_{\text{E}}^{\prime}=k{P}_{\text{E}}^{a}{P}_{\text{H}}^{b}\hfill \end{array}$
Procedure
Enter data
Enter model
Make initial estimates of parameters
Run regression
Examine parameters and variance
Observe error distribution
Choose model
TABLE E103.1 DIFFERENTIAL REACTOR DATA
Run Number 
Reaction Rate (mol/kgcat. · s) 
P_{E} (atm) 
P_{EA} (atm) 
P_{H} (atm) 
1 
1.04 
1 
1 
1 
2 
3.13 
1 
1 
3 
3 
5.21 
1 
1 
5 
4 
3.82 
3 
1 
3 
5 
4.19 
5 
1 
3 
6 
2.391 
0.5 
1 
3 
7 
3.867 
0.5 
0.5 
5 
8 
2.199 
0.5 
3 
3 
9 
0.75 
0.5 
5 
1 
Solution
Polymath was chosen as the software package to solve this problem. The data in Table E103.1 were entered into the program. A screenshotbyscreenshot tutorial and set of instructions on how to carry out the regression is given on the CRE Web site, at the end of the Summary Notes for Chapter 7 (http://www.umich.edu/~elements/6e/07chap/summary.html). After entering the data and following the stepbystep procedures, the results shown in Table E103.2 were obtained.
Model (a) Single site, surfacereaction, ratelimited with hydrogen weakly adsorbed
From Table E103.2 data, we can obtain
$\begin{array}{cc}{\text{r}}_{\text{E}}^{\prime}=\frac{3.348{P}_{\text{E}}{P}_{\text{H}}}{1+0.043{P}_{\text{EA}}+2.21{P}_{\text{E}}}& \text{(E103.1)}\end{array}$
We now examine the sums of squares (variance) and range of variables themselves. The sums of squares is reasonable and, in fact, the smallest of all the models at 0.0049. However, let’s look at K_{EA}. We note that the value for the 95% confidence limit of ±0.0636 is greater than the nominal value of K_{EA} = 0.043 atm^{–1} itself (i.e., K_{EA} = 0.043 ± 0.0636). The 95% confidence limit means that if the experiment were run 100 times and then 95 times it would fall within the range (–0.021) < K_{EA} < (0.1066). Because K_{EA} can never be negative, we are going to reject this model. Consequently, we set K_{EA} = 0 and proceed to model (b).
Model (b) Single site, surfacereaction, ratelimited with ethane and hydrogen weakly adsorbed
From Table E103.2 we can obtain
The value of the adsorption constant K_{E} = 2.1 atm^{–1} is reasonable and is not negative within the 95% confidence limit. Also, the variance is small at ${\sigma}_{\text{B}}^{2}=0.0061$.
Model (c) Dual site, surfacereaction, ratelimited with hydrogen and ethane weakly adsorbed From Table E103.2 we can obtain
$\begin{array}{cc}{\text{r}}_{\text{E}}^{\prime}=\frac{2.0{P}_{\text{E}}{P}_{\text{H}}}{{(1+0.36{P}_{\text{E}})}^{2}}& \text{(E103.3)}\end{array}$
While K_{E} is small, it never goes negative within the 95% confidence interval. The variance of this model at ${\sigma}_{\text{B}}^{2}=0.0623$ is much larger than the other models. Comparing the variance of model (c) with model (b)
$\frac{{\sigma}_{\text{C}}^{2}}{{\sigma}_{\text{B}}^{2}}=\frac{0.0623}{0.0061}=10.2$
We see that the ${\sigma}_{\text{C}}^{2}$ is an order of magnitude greater than ${\sigma}_{\text{B}}^{2}$, and therefore we eliminate model (c).^{21}
^{21} See G. F. Froment and K. B. Bishoff, Chemical Reaction Analysis and Design, 2nd ed. New York: Wiley, 1990, p. 96.
Model (d) Empirical
Similarly for the powerlaw model, we obtain from Table E103.2
$\begin{array}{cc}{\text{r}}_{\text{E}}^{\prime}=0.894{P}_{\text{E}}^{0.26}{P}_{\text{H}}^{1.06}& \text{(E103.4)}\end{array}$
As with model (c), the variance is quite large compared to model (b)
$\frac{{\sigma}_{\text{D}}^{2}}{{\sigma}_{\text{B}}^{2}}=\frac{0.049}{0.0061}=8.03$
So we also eliminate model (d). For heterogeneous reactions, LangmuirHinshelwood rate laws are preferred over powerlaw models.
Analysis: Choose the Best Model. In this example, we were presented with four rate laws and were asked which law best fits the data. Because all the parameter values are realistic for model (b) and the sums of squares are significantly smaller for model (b) than for the other models, we choose model (b). We note again that there is a caution we need to point out regarding the use of regression! One cannot simply carry out a regression and then choose the model with the lowest value of the sums of squares. If this were the case, we would have chosen model (a), which had the smallest sums of squares of all the models with σ^{2} = 0.0049. However, one must consider the physical realism of the parameters in the model. In model (a), the 95% confidence interval was greater than the parameter itself, thereby yielding negative values of the parameter, K_{AE}, which is physically impossible.
In designing fixed and ideal fluidized bed catalytic reactors, we have assumed up to now that the activity of the catalyst remains constant throughout the catalyst’s life. That is, the total concentration of active sites, C_{t}, accessible to the reaction does not change with time. Unfortunately, Mother Nature is not so kind as to allow this behavior to be the case in most industrially significant catalytic reactions. One of the most insidious problems in catalysis is the loss of catalytic activity that occurs as the reaction takes place on the catalyst. A wide variety of mechanisms have been proposed by Butt and Petersen, to explain and model catalyst deactivation.^{22,23,24}
^{22} J. B. Butt and E. E. Petersen, Activation, Deactivation and Poisoning of Catalysts, New York: Academic Press, 1988.
^{23} D. T. Lynch and G. Emig, Chem. Eng. Sci., 44(6), 1275–1280 (1989).
^{24} R. Hughes, Deactivation of Catalysts, San Diego: Academic Press, 1984.
Catalytic deactivation adds another level of complexity to sorting out the reactionrate law parameters and pathways. In addition, we need to make adjustments for the decay of the catalysts in the design of catalytic reactors. However, please don’t worry, this adjustment is usually made by a quantitative specification of the catalyst’s activity, a(t). In analyzing reactions over decaying catalysts, we divide the reactions into two categories: separable kinetics and nonseparable kinetics. In separable kinetics, we separate the rate law and activity
$\text{Separable kinetics:}{\text{r}}_{\text{A}}^{\prime}=a\left(\text{Past history}\right)\times {\text{r}}_{\text{A}}^{\prime}\left(\text{Fresh catalyst}\right)$
When the kinetics and activity are separable, it is possible to study catalyst decay and reaction kinetics independently. However, nonseparability
$\text{Nonsearable kinetics:}{\text{r}}_{\text{A}}^{\prime}={\text{r}}_{\text{A}}^{\prime}\left(\text{Past history, fresh catalyst}\right)$
must be accounted for by assuming the existence of a nonideal surface or by describing deactivation by a mechanism composed of several elementary steps.^{23,24}
We shall consider only separable kinetics and define the activity of the catalyst at time t, a(t), as the ratio of the rate of reaction on a catalyst that has been used for a time t to the rate of reaction on a fresh catalyst (t = 0):
a(t): catalyst activity
$\begin{array}{cc}\begin{array}{c}\hline \text{}a\left(\text{t}\right)=\frac{{\text{r}}_{\text{A}}^{\prime}\left(\text{t}\right)}{{\text{r}}_{\text{A}}^{\prime}\left(\text{t=0}\right)}\\ \hline\end{array}& \text{(1092)}\end{array}$
Because of the catalyst decay, the activity decreases with time and a typical curve of the activity as a function of time is shown in Figure 1021.
Combining Equations (1092) and (32), the rate of disappearance of reactant A on a catalyst that has been utilized for a time t is
Reactionrate law accounting for catalyst activity
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{\text{A}}^{\prime}=a\left(\text{t}\right)k\left(\text{T}\right)\text{}\text{fn}({\text{C}}_{\text{A}},{\text{C}}_{\text{B}},\mathrm{...},{\text{C}}_{\text{P}})\\ \hline\end{array}& \text{(1093)}\end{array}$
$\begin{array}{ccc}\hfill \text{where}a\left(t\right)& =& \text{catalyticactivity,timedependent}\hfill \\ \hfill k\left(T\right)& =& \text{specificreactionrate,temperaturedependent}\hfill \\ \hfill {C}_{i}& =& \text{gasphaseconcentrationofreactants,products,orcontaminant}\hfill \end{array}$
The rate of catalyst decay, r_{d}, can be expressed in a rate law analogous to Equation (1093)
$\begin{array}{cc}\begin{array}{c}\hline {\text{r}}_{d}=\frac{da}{d\text{t}}=p\left[a\left(\text{t}\right)\right]{k}_{d}\left(\text{T}\right)h({\text{C}}_{\text{A}},\mathrm{...},{\text{C}}_{\text{P}})\\ \hline\end{array}& \text{(1094)}\end{array}$
Catalyst decay rate law
For example, p[a(t)] = [a(t)]^{2}
where p[a(t)] is some function of the catalyst activity, k_{d} is the specific decay constant, and h(C_{i}) is the functionality of rate of decay, r_{d}, on the species concentrations in the reacting mixture. For the cases presented in this chapter, this functionality either will be independent of concentration (i.e., h = 1) or as in the case of catalytic poisoning it could be a linear function of species concentration (i.e., h = C_{i}).
The functionality of the activity term, p[a(t)], in the decay law can take a variety of forms. For example, for a firstorder decay
$\begin{array}{cc}p\left(a\right)=a& \text{(1095)}\end{array}$
and for a secondorder decay
$\begin{array}{cc}p\left(a\right)={a}^{2}& \text{(1096)}\end{array}$
The particular function, p(a), will vary with the gas catalytic system being used and the reason or mechanism for catalytic decay.
There are three categories into which the loss of catalytic activity can traditionally be divided: sintering or aging, fouling or coking, and poisoning.
Sintering
Fouling
Poisoning
^{25} See G. C. Kuczynski, ed., Sintering and Catalysis, vol. 10 of Materials Science Research, New York: Plenum Press, 1975.
Sintering, also referred to as aging, is the loss of catalytic activity due to a loss of active surface area resulting from the prolonged exposure to high gasphase temperatures. The active surface area may be lost either by growth of the metals deposited on the support or by narrowing or closing of the pores inside the catalyst pellet. A change in the surface structure may also result from either surface recrystallization or the formation or elimination of surface defects (active sites). The reforming of heptane over platinum on alumina is an example of catalyst deactivation as a result of sintering.
Figure 1022 shows the loss of surface area resulting from the flow of the solid porous catalyst support at high temperatures to cause pore closure. Figure 1023 shows growth of small metal crystallites into larger particles where the interior atoms are not accessible to reactions thereby leading to a loss of metal surface area of the catalyst. Professor Abhaya Datye has recently identified that the dominant mechanism for this sintering of industrial catalysts involves the transport of mobile species, a process called Ostwald ripening.^{26}
^{26} T. W. Hansen, A. T. Delariva, S. R. Challa, and A. K. Datye, “Sintering of Catalytic Nanoparticles: Particle Migration or Ostwald Ripening?” Acc. Chem. Res., 46(8), 1720–1730 (2013).
Sintering is usually negligible at temperatures below 40% of the melting temperature of the solid.^{27} However, sintering can occur at lower temperatures when mobile species are generated through a reaction of the solid with the gas phase, for example the formation of volatile PtO_{2} due to reaction of Pt and O_{2}.
^{27} R. Hughes, Deactivation of Catalysts, San Diego: Academic Press, 1984.
The catalyst support becomes soft and flows, resulting in pore closure.
The catalyst support becomes soft and flows, resulting in pore closure.
The atoms move along the surface and agglomerate.
Deactivation by sintering may in some cases be a function of the mainstream gas concentration. Although other forms of the sintering decay rate laws exist, one of the most commonly used sintering decay laws is second order with respect to the present activity
$\begin{array}{cc}{r}_{d}={k}_{d}{a}^{2}=\frac{da}{dt}& \text{(1097)}\end{array}$
Integrating, with a = 1 at time t = 0, for constant k_{d} during the time t yields
Sintering: secondorder decay
$\begin{array}{cc}\begin{array}{c}\hline a\left(t\right)=\frac{1}{1+{k}_{d}t}\\ \hline\end{array}& \left(\mathrm{1098}\right)\end{array}$
The amount of sintering is usually measured in terms of the active surface area of the catalyst S_{a}
$\begin{array}{cc}{\text{S}}_{a}=\frac{{\text{S}}_{a0}}{1+{k}_{d}t}& \text{(1099)}\end{array}$
The sintering decay constant, k_{d}, follows the Arrhenius equation
$\begin{array}{cc}\begin{array}{c}\hline {k}_{d}={k}_{d}\left({\text{T}}_{0}\right)\text{exp}[\frac{{\text{E}}_{d}}{\text{R}}(\frac{1}{{\text{T}}_{0}}\frac{1}{\text{T}})]\\ \hline\end{array}& \text{(10100)}\end{array}$
Minimizing sintering
The decay activation energy, E_{d}, for the reforming of heptane on Pt/Al_{2}O_{3} is on the order of 70 kcal/mol, which is rather high. As mentioned earlier, sintering can be reduced by keeping the temperature below 0.3–0.4 times the metal’s melting point temperature, that is, (T < 0.3 Tmelting).
For the case where the temperature varies during the reaction time t, Equations (1093), (1094), along with the energy balance must be solved simultaneously.
We will now stop and consider reactor design for a fluid–solid system with decaying catalyst. To analyze these reactors, we only add one step to our algorithm; that is, determine the catalyst decay law. The sequence is shown here.
The algorithm
$\begin{array}{c}\hline \begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{Mole balance}& \to \end{array}& \text{Reactionrate law}\text{}\end{array}& \to \end{array}& \text{D}ecay\text{\hspace{0.17em}}\text{\hspace{0.17em}}rate\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}law& \to \end{array}\\ \begin{array}{cc}\begin{array}{c}\begin{array}{cc}\begin{array}{c}\begin{array}{cc}\begin{array}{c}\begin{array}{cc}\begin{array}{c}\text{Stoichiometry}\end{array}& \to \end{array}\end{array}& \text{Combine and sole}\end{array}\end{array}& \to \end{array}\end{array}& \text{Numerical techniques}\end{array}\\ \hline\end{array}$
The firstorder isomerization
A → B
is being carried out isothermally in a batch reactor on a catalyst that is decaying as a result of aging. Derive an equation for conversion as a function of time.
Solution
Mol Balance:
$\begin{array}{cc}{N}_{\text{A0}}\frac{{d\text{X}}_{d}}{dt}={r}_{\text{A}}^{\prime}W& \text{(E104.1)}\end{array}$
where X_{d} is the conversion of A when the catalyst is decaying.
ReactionRate Law:
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={k}^{\prime}a\left(\text{t}\right){\text{C}}_{\text{A}}& \text{(E104.2)}\end{array}$
Decay Law: For secondorder decay by sintering:
$\begin{array}{cc}\begin{array}{c}\hline a\left(t\right)=\frac{1}{1+{k}_{a}t}\\ \hline\end{array}& \text{(1098)}\end{array}$
One extra step (number 3) is added to the algorithm.
Stoichiometry:
$\begin{array}{cc}{C}_{\text{A}}={C}_{\text{A0}}(1{X}_{d})=\frac{{N}_{\text{A0}}}{\text{V}}(\text{1}{X}_{d})& \text{(E104.3)}\end{array}$
Combining gives us
$\begin{array}{cc}\frac{{dX}_{d}}{dt}={k}^{\prime}(1{X}_{d})a\frac{W}{V}& \text{(E104.4)}\end{array}$
Let k = k′W/V. Substituting for catalyst activity a, we have
$\begin{array}{cc}\begin{array}{c}\hline \frac{{dX}_{d}}{dt}=k(1{X}_{d})a=k\frac{(1{X}_{d})}{(1+{k}_{d}t)}\\ \hline\end{array}& \text{(E104.5)}\end{array}$
where X_{d} is the conversion when there is decay. We want to compare the conversion with and without catalyst decay.
For no decay k_{d} = 0
$\begin{array}{c}\hline \frac{dX}{dt}=k(1X)\\ \hline\end{array}$
The Polymath program and a comparison of the conversion with decay X_{d} and without decay X are shown below.
Analytical Solution
One can also obtain an analytical solution for this reaction order with catalyst decay. Separating variables and integrating yields
$\begin{array}{cc}{\int}_{0}^{\text{x}}\frac{{dX}_{d}}{1{X}_{d}}=k{\int}_{0}^{t}\frac{dt}{1+{k}_{d}t}& \text{(E104.6)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline \mathrm{ln}\frac{1}{1{X}_{d}}=\frac{k}{{k}_{d}}=\text{ln}(1+{k}_{d}t)\\ \hline\end{array}& \text{(E104.7)}\end{array}$
Solving for the conversion X_{d} at any time, t, in a decaying catalyst we find that
$\begin{array}{cc}{X}_{d}=1\frac{1}{{(1+{k}_{d}t)}^{k/{k}_{d}}}& \text{(E104.8)}\end{array}$
the analytical solution without decay
$\begin{array}{cc}\begin{array}{c}\hline \text{ln}X=1{e}^{kt}=kt\\ \hline\end{array}& \text{(E104.9)}\end{array}$
Parameters
$\begin{array}{ccc}k={k}_{1}\text{exp}\text{\hspace{0.17em}}\left[\frac{E}{R}(\frac{1}{{T}_{1}}\frac{1}{T})\right]& {k}_{d}={k}_{1d}\text{exp}\text{\hspace{0.17em}}\left[\frac{{E}_{d}}{R}(\frac{1}{{T}_{2}}\frac{1}{T})\right]& \text{}\end{array}$
$\begin{array}{lll}E=20000\text{cal/mol}& \text{}& {k}_{1d}=0.1{\text{s}}^{1}\\ {k}_{1}=0.01{\text{s}}^{1}& \text{}& \text{}{T}_{2}=300\text{K}\\ {T}_{1}=300\text{}\text{K}& \text{}& {E}_{d}=75000\text{cal/mol}\\ \text{}& \text{}& R\text{=1.9787}\end{array}$
Analysis: One observes that for long times the conversion in reactors with catalyst decay approaches a rather flat plateau and reaches conversion of about 30%. This is the conversion that will be achieved in a batch reactor for a firstorder reaction when the catalyst decay law is second order. By comparison, we obtain virtually complete conversion in 500 seconds when there is no decay. The purpose of this example was to demonstrate the algorithm for isothermal catalytic reactor design for a decaying catalyst. In problem P101_{B}(d), you are asked to sketch the temperature–time trajectories for various values of k and k_{d}.
This mechanism of decay (see Figures 1024, 1025, and 1026) is common to reactions involving hydrocarbons. It results from a carbonaceous (coke) material being deposited on the surface of a catalyst. The amount of coke on the surface after a time t has been found to obey the following empirical relationship
$\begin{array}{cc}\begin{array}{c}\hline {\text{C}}_{\text{C}}={\text{At}}^{\text{n}}\\ \hline\end{array}& \text{(10101)}\end{array}$
where C_{C} is the concentration of carbon on the surface (g/m^{2}) and n and A are empirical fouling parameters, which can be functions of the feed rate. This expression was originally developed by Voorhies and has been found to hold for a wide variety of catalysts and feed streams. Representative values for Equation (10101) for the cracking of a crude oil in fixedbed of catalyst given
C_{C} = 0.52 t^{0.38}
Another relationship for the weight percent coke for the cracking of East Texas light gas oil is^{28,29}
^{28} A. Voorhies, Ind. Eng. Chem., 37, 318.
^{29} C. O. Prater and R. M. Lago, Adv. Catal., 8, 293.
$\%\text{coke}=0.47\sqrt{\text{t}\left(\text{min}\right)}$
Different functionalities between the activity and amount of coke on the surface have been observed. One commonly used form is
$\begin{array}{cc}a=\frac{1}{{k}_{\text{Ck}}{\text{C}}_{\text{C}}^{p}+1}& \text{(10102)}\end{array}$
or, in terms of time, we combine Equations (10101) and (10102)
$\begin{array}{cc}a=\frac{1}{{k}_{\text{Ck}}{\text{A}pt}^{\text{np}}+1}=\frac{1}{1+{k}^{\prime}{\text{t}}^{m}}& \text{(10103)}\end{array}$
For light Texas gas oil being cracked at 750=F over a synthetic catalyst for short times, the decay law is
$\begin{array}{cc}\begin{array}{c}\hline a=\frac{1}{1+7.6{t}^{1/2}}\\ \hline\end{array}& \text{(10104)}\end{array}$
where t is in seconds.
Activity for deactivation by coking
Other commonly used forms are
$\begin{array}{cc}a={e}^{{\alpha}_{1}{\text{C}}_{\text{C}}}& \text{(10105)}\end{array}$
and
$\begin{array}{cc}\begin{array}{c}\hline a=\frac{1}{1+{\alpha}_{2}{\text{C}}_{\text{C}}}\\ \hline\end{array}& \text{(10106)}\end{array}$
A dimensionless fouling correlation has been developed by Pacheco and Petersen.^{30}
^{30} M. A. Pacheco and E. E. Petersen, J. Catal., 86, 75.
Minimizing coking
When possible, coking can be reduced by running at elevated pressures (2000–3000 kPa) and hydrogenrich streams. A number of other techniques for minimizing fouling are discussed by Bartholomew.^{31} Catalysts deactivated by coking can usually be regenerated by burning off the carbon.
^{31} R. J. Farrauto and C. H. Bartholomew, Fundamentals of Industrial Catalytic Processes, 2nd ed. New York: Blackie Academic and Professional, 2006. This book is one of the most definitive resources on catalyst decay.
Deactivation by this mechanism occurs when the poisoning molecules become irreversibly chemisorbed to active sites, thereby reducing the number of sites available for the main reaction. The poisoning molecule, P, may be a reactant and/or a product in the main reaction, or it may be an impurity in the feed stream.
It’s going to cost you. #Really?
Many petroleum feed stocks contain trace impurities such as sulfur, lead, and other components that are too costly to remove, yet poison the catalyst slowly over time. For the case of an impurity, P, in the feed stream, such as sulfur, for example, in the reaction sequence
the surface sites would change with time as shown in Figure 1027.
Progression of sites being poisoned
If we assume the rate of depletion of the poison from the reactant gas stream onto the catalyst sites, r_{P}, is proportional to the number of sites that are unpoisoned (C_{t}_{0} – C_{P·S}) and the concentration of poison in the gas phase is C_{P} then
Total site balance at time t C_{t} = C_{t0} – C_{P}·_{S}
r_{P·S} = k_{d}(C_{t0} – C_{P·S})C_{P}
where C_{P}·_{S} is the concentration of poisoned sites and C_{t}_{0} is the total number of sites initially available. Because every molecule that is adsorbed from the gas phase onto a site is assumed to poison the site, this rate is also equal to the rate of removal of total active sites (C_{t}) from the surface
$\frac{d{C}_{t}}{dt}=\frac{d{C}_{\text{P}\cdot \text{S}}}{dt}={r}_{\text{p}\cdot \text{s}}={k}_{d}({C}_{t0}{C}_{\text{p}\cdot \text{s}}){C}_{\text{p}}$
Dividing through by C_{t}_{0} and letting f be the fraction of the total number of sites that have been poisoned yields
$\begin{array}{cc}\frac{df}{dt}={k}_{d}(1f){C}_{\text{p}}& \text{(10108)}\end{array}$
The fraction of sites available for adsorption (1 = f ) is essentially the activity a(t). Consequently, Equation (10108) becomes
$\begin{array}{cc}\begin{array}{c}\hline \frac{da}{dt}=a\left(t\right){k}_{d}{C}_{\text{p}}\\ \hline\end{array}& \text{(10109)}\end{array}$
A number of examples of catalysts with their corresponding catalyst poisons are given by Farrauto and Bartholomew.^{32}
^{32} Ibid.
In packedbed reactors where the poison is removed from the gas phase by being adsorbed on the specific catalytic sites, the deactivation process can move through the packed bed as a wave front. Here, at the start of the operation, only those sites near the entrance to the reactor will be deactivated because the poison (which is usually present in trace amounts) is removed from the gas phase by the adsorption; consequently, the catalyst sites farther down the reactor will not be affected. However, as time continues, the sites near the entrance of the reactor become saturated, and the poison must travel farther downstream before being adsorbed (removed) from the gas phase and attaching to a site to deactivate it. Figure 1028 shows the corresponding activity profile for this type of poisoning process. We see in Figure 1028 that by time t4 the entire bed has become deactivated. The corresponding overall conversion at the exit of the reactor might vary with time as shown in Figure 1029. The partial differential equations that describe the movement of the reaction front shown in Figure 1028 are derived and solved in an example on the CRE Web site, at the very end of the Summary Notes for Chapter 10 (http://www.umich.edu/~elements/6e/10chap/summaryexample3.html).
For the case where the main reactant also acts as a poison, the rate laws are
$\begin{array}{lll}\text{M}\text{ain reaction}:& \begin{array}{cc}\begin{array}{c}\text{A}+\text{S}\end{array}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{B}+\text{S}\end{array}\end{array}& {r}_{\text{A}}^{\prime}={k}_{\text{A}}{\text{C}}_{\text{A}}^{\text{n}}\\ \text{Poisoning reaction:}& \begin{array}{cc}\begin{array}{c}\text{A+S}\end{array}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \underset{\xaf}{\begin{array}{c}\text{A}\cdot \text{S}\end{array}}\end{array}\end{array}& {r}_{d}={k}_{d}^{\prime}{\text{C}}_{\text{A}}^{m}{a}^{q}\end{array}$
An example where one of the reactants acts as a poison is in the reaction of CO and H_{2} over ruthenium to form methane, with
$\text{CO}+3{\text{H}}_{2}\to \text{C}{\text{H}}_{4}+{\text{H}}_{2}\text{O}$
$\begin{array}{c}{r}_{\text{CO}}=ka\left(t\right){\text{C}}_{\text{CO}}\\ \frac{da}{dt}={r}_{d}={k}_{d}^{\prime}a\left(t\right){\text{C}}_{\text{CO}}\end{array}$
Similar rate laws can be written for the case when the product B acts as a poison.
For separable deactivation kinetics resulting from contacting a poison at a constant concentration and no spatial variation
Separable deactivation kinetics
$\begin{array}{cc}\begin{array}{c}\hline \frac{da}{dt}={r}_{d}={k}_{d}^{\prime}{C}_{{\text{p}}_{\text{0}}}^{n}{a}^{n}\left(t\right)={k}_{d}{a}^{n}\\ \hline\end{array}& \text{(}\text{10110)}\end{array}$
where ${k}_{d}={k}_{d}^{\prime}{C}_{{\text{p}}_{\text{0}}}^{n}$. The solution to this equation for the case of firstorder decay, n = 1
$\begin{array}{cc}\begin{array}{cc}\frac{da}{dt}={k}_{d}a& \text{}\end{array}& \text{(10111)}\end{array}$
is
$\begin{array}{cc}a={e}^{{k}_{d}t}& \text{(10112)}\end{array}$
Table 107 gives a number of empirical decay laws along with the reaction systems to which they apply.
Key resource for catalyst deactivation
One should also see Fundamentals of Industrial Catalytic Processes, by Farrauto and Bartholomew, which contains rate laws similar to those in Table 107, and also gives a comprehensive treatment of catalyst deactivation.^{33}
^{33} Ibid.
Examples of reactions with decaying catalysts and their decay laws
TABLE 107 DECAY RATE LAWS
Functional Form of Activity 
Decay Reaction Order 
Differential Form 
Integral 
Form Examples 
Linear 
0 
$\frac{da}{dt}={\beta}_{0}$ 
a = 1 – β_{0}t 
Conversion of parahydrogen on tungsten when poisoned with oxygen^{a} 
Exponential 
1 
$\frac{da}{dt}={\beta}_{1}a$ 
a = e^{–β1t} 
Ethylene hydrogenation on Cu poisoned with CO^{b, c} 
Paraffin dehydrogenation on Cr? Al_{2}O_{3}c Cracking of gas oil^{d} Vinyl chloride monomer formation^{e} 

Hyperbolic 
2 
$\frac{da}{dt}={\beta}_{1}{a}^{2}$ 
$\frac{1}{a}=1+{\beta}_{2}t$ 
Vinyl chloride monomer formation^{f, g} Cyclohexane dehydrogenation on Pt/Al_{2}O_{3}g Isobutylene hydrogenation on Ni^{h} 
Reciprocal power 
$\frac{{\beta}_{3}+1}{{\beta}_{3}}=\gamma $ 
$\frac{da}{dt}={\beta}_{3}{a}^{n}{A}_{0}^{1/5}$ 
a = A_{0}t^{–β3} 
Cracking of gas oil and gasoline on clay^{i} 
$\frac{{\beta}_{4}+1}{{\beta}_{4}}=n$ 
$\frac{da}{dt}={\beta}_{4}an{A}_{0}^{1/5}$ 
a = A_{0}t^{–β4} 
Cyclohexane aromatization on NiAl^{j} 
^{a}D. D. Eley and E. J. Rideal, Proc. R. Soc. London, A178, 429 (1941).
^{b}R. N. Pease and L. Y. Steward, J. Am. Chem. Soc., 47, 1235 (1925).
^{c}E. F. K. Herington and E. J. Rideal, Proc. R. Soc. London, A184, 434 (1945).
^{d}V. W. Weekman, Ind. Eng. Chem. Process Des. Dev., 7, 90 (1968).
^{e}A. F. Ogunye and W. H. Ray, Ind. Eng. Chem. Process Des. Dev., 9, 619 (1970).
^{f}A. F. Ogunye and W. H. Ray, Ind. Eng. Chem. Process Des. Dev., 10, 410 (1971).
^{g}H. V. Maat and L. Moscou, Proc. 3rd lnt. Congr. Catal. Amsterdam: NorthHolland, 1965, p. 1277.
^{h}A. L. Pozzi and H. F. Rase, Ind. Eng. Chem., 50, 1075 (1958).
^{i}A. Voorhies, Jr., Ind. Eng. Chem., 37, 318 (1945); E. B. Maxted, Adv. Catal., 3, 129 (1951).
^{j}C. G. Ruderhausen and C. C. Watson, Chem. Eng. Sci., 3, 110 (1954).
Source: J. B. Butt, Chemical Reactor Engineering–Washington, Advances in Chemistry Series 109, Washington, D.C.: American Chemical Society, 1972, p. 259. Also see CES 23, 881(1968).
We will now consider three reaction systems that can be used to help offset systems with decaying catalyst. We will classify these systems as those having slow, moderate, and rapid losses of catalytic activity. To offset the decline in chemical reactivity of decaying catalysts in continuousflow reactors, the following three methods are commonly used:
Matching the reactor type with speed of catalyst decay
Slow decay – Temperature–Time Trajectories (10.8.1)
Moderate decay – MovingBed Reactors (10.8.2)
Rapid decay – StraightThrough Transport Reactors (10.8.3)
In many largescale reactors, such as those used for hydrotreating, and reaction systems where deactivation by poisoning occurs, the catalyst decay is relatively slow. In these continuousflow systems, constant conversion is usually necessary in order that subsequent processing steps (e.g., separation) are not upset. One way to maintain a constant conversion with a decaying catalyst in a packed or fluidized bed is to increase the reaction rate by steadily increasing the feed temperature to the reactor. Operation of a “fluidized” bed in this manner is shown in Figure 1030.
We are going to increase the feed temperature, T, in such a manner that the reaction rate remains constant with time:
${r}_{\text{A}}^{\prime}(t=0,{\text{T}}_{0})={r}_{\text{A}}^{\prime}(t,\text{T})=a(t,\text{T})[{r}_{\text{A}}^{\prime}(t=0,\text{T})]$
We will use a firstorder reaction to illustrate the salient points. For a firstorder reaction with decay we have
k(T_{0})C_{A} = a(t, T)k(T)C_{A}
Slow rate of catalyst decay
We will neglect any variations in concentration so that the product of the activity (a) and specific reaction rate (k) is constant and equal to the specific reaction rate, k_{0} at time t = 0 and temperature T_{0}; that is
$\begin{array}{cc}\begin{array}{c}\hline k\left(T\right)a(t,T)={k}_{0}\\ \hline\end{array}& \text{(10113)}\end{array}$
Note that for any powerlaw model, the reactant concentration dependence cancels out.
The goal is to find how the temperature should be increased with time (i.e., the temperature–time trajectory) to maintain constant conversion by keeping the rate ${r}_{\text{A}}^{\prime}$, constant. Using the Arrhenius equation to substitute for k in terms of the activation energy, E_{A}, gives
$\begin{array}{cc}{k}_{0}{e}^{({E}_{\text{A}}/R)(1/{T}_{0}1/T)}a={k}_{0}& \text{(10114)}\end{array}$
Solving for 1/T yields
Gradually raising the temperature can help offset effects of catalyst decay.
$\begin{array}{cc}\begin{array}{c}\hline \frac{1}{T}=\frac{R}{{E}_{\text{A}}}\text{ln}\text{\hspace{0.17em}}a+\frac{1}{{T}_{0}}\\ \hline\end{array}& \text{(10115)}\end{array}$
The decay law also follows an Arrheniustype temperature dependence
$\begin{array}{cc}\frac{da}{dt}={k}_{d0}{e}^{({E}_{d}/R)(1/{T}_{0}1/T)}{a}^{n}& \text{(10116)}\end{array}$
$\begin{array}{ccc}\hfill \text{where}{k}_{d0}& =& \text{decayconstantattemperature}{T}_{0,}\text{}{S}^{1}\hfill \\ \hfill {E}_{\text{A}}& =& \text{activationenergyforthemainreaction}(\text{e.g.,A}\to \text{B}),\text{kJ/mol}\hfill \\ \hfill {E}_{d}& =& \text{activationenergyforcatalystdecay,kJ/mol}\hfill \end{array}$
Substituting Equation (10115) into (10116) and rearranging yields
$\begin{array}{cc}\begin{array}{c}\hline \frac{da}{dt}={k}_{d0}\text{exp}\text{}(\frac{{\text{E}}_{d}}{{\text{E}}_{\text{A}}}\mathrm{ln}a){a}^{n}={k}_{d0}{a}^{(\text{n}{\text{E}}_{d}/{\text{E}}_{\text{A}})}\\ \hline\end{array}& \text{(10117)}\end{array}$
Integrating with a = 1 at t = 0 for the case n = (1 = E_{d}/E_{A}), we obtain
$\begin{array}{cc}t=\frac{1{a}^{1\text{n}+{\text{E}}_{d}/{\text{E}}_{\text{A}}}}{{k}_{d0}(1\text{n}+{\text{E}}_{d}/{\text{E}}_{\text{A}})}& \text{(10118)}\end{array}$
Solving Equation (10114) for a and substituting in (10118) gives
$\begin{array}{cc}\begin{array}{c}\hline t=\frac{1\mathrm{exp}\left[\frac{{E}_{\text{A}}\text{n}{E}_{\text{A}}+{E}_{d}}{\text{R}}(\frac{1}{T}\frac{1}{{T}_{0}})\right]}{{k}_{d0}(1\text{n}+{E}_{d}/{E}_{\text{A}})}\\ \hline\end{array}& \text{(10119)}\end{array}$
Equation (10119) tells us how the temperature of the catalytic reactor should be increased with time in order for the reaction rate and conversion to remain constant. However one would first want to solve this equation for temperature, T, as a function of time, t, in order to correctly operate the preheater shown in Figure 1030.
In many industrial reactions, the decay rate law changes as temperature increases. In hydrocracking, the temperature–time trajectories are divided into three regimes. Initially, there is fouling of the acidic sites of the catalyst followed by a linear regime due to slow coking and, finally, accelerated coking characterized by an exponential increase in temperature. The temperature–time trajectory for a deactivating hydrocracking catalyst is shown in Figure 1031.
For a firstorder decay, Krishnaswamy and Kittrell’s expression, Equation (10119), for the temperature–time trajectory reduces to
$\begin{array}{cc}t=\frac{{E}_{\text{A}}}{{k}_{d0}{E}_{d}}[1{e}^{({E}_{d}/R)(1/T1/{T}_{0})}]& \text{(10120)}\end{array}$
Comparing theory and experiment
One observes the model compares favorably with the experimental data.
Reaction systems with relatively rapid catalyst decay require the continual regeneration and/or replacement of the catalyst. Two types of reactors currently in commercial use that accommodate production with decaying catalysts are the movingbed for moderate decay and straightthrough transport reactor for rapid decay. A schematic diagram of a movingbed reactor (used for catalytic cracking) is shown in Figure 1032.
The freshly regenerated catalyst enters the top of the reactor and then moves through the reactor as a compact packed bed. The catalyst is coked continually as it moves through the reactor until it exits the reactor into the kiln, where air is used to burn off the carbon. The regenerated catalyst is lifted from the kiln by an airstream and then fed into a separator before it is returned to the reactor. The catalyst pellets are typically between onequarter and oneeighth inch in diameter.
The Chemical Safety Board video on the Process Safety Across the Chemical Engineering Curriculum Web site (http://umich.edu/~safeche/assets/pdf/courses/Problems/CRE/344ReactionEngrModule(3)PSExxon.pdf) shows an animation of the movingbed reactor at the ExxonMobil Torrance, California refinery.
The reactant feed stream enters at the top of the reactor and flows rapidly through the reactor relative to the flow of the catalyst through the reactor (Figure 1033). If the feed rates of the catalyst and the reactants do not vary with
Movingbed reactor, used for reactions with moderate rate of catalyst decay
time, the reactor operates at steady state; that is, conditions at any point in the reactor do not change with time. The mole balance on reactant A over ΔW is
$\begin{array}{ccccccc}\left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Molar}\end{array}\\ \text{flow}\end{array}\end{array}\\ \text{rate of A in}\end{array}\end{array}\right]& & \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Molar}\end{array}\\ \text{flow}\end{array}\end{array}\\ \text{rate of A out}\end{array}\end{array}\right]& +& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Molar}\end{array}\\ \text{rate of}\end{array}\end{array}\\ \text{generation of}\text{A}\end{array}\end{array}\right]& =& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Molar}\end{array}\\ \text{rate of}\end{array}\end{array}\\ \text{accumulation}\text{of A}\end{array}\end{array}\right]\\ {\text{F}}_{\text{A}}\left(\text{W}\right)& & {\text{F}}_{\text{A}}\left(\text{W+}\mathrm{\Delta}\text{W}\right)& +& {r}_{\text{A}}^{\prime}\mathrm{\Delta}\text{W}& =& \begin{array}{cc}0& \text{(10121)}\end{array}\end{array}$
Dividing by ΔW, letting ΔW approach zero, and expressing the flow rate in terms of conversion gives
Mole Balance
$\begin{array}{cc}\begin{array}{c}\hline {F}_{\text{A0}}\frac{dX}{dW}={r}_{\text{A}}^{\prime}\\ \hline\end{array}& \text{(217)}\end{array}$
The rate of reaction at any time t is
$\begin{array}{cc}{r}_{\text{A}}^{\prime}=a\left(t\right)[{r}_{\text{A}}^{\prime}(t=0)]=a(t=0)=a\left(t\right)\left[k\text{}\text{fn}({C}_{\text{A,}}{C}_{\text{B}},\mathrm{...},{C}_{\text{P}})\right]& \text{(1093)}\end{array}$
The activity, as before, is a function of the time the catalyst has been in contact with the reacting gas stream. The decay rate law is
Decay Law
$\begin{array}{cc}\begin{array}{c}\hline \frac{da}{dt}={k}_{d}{a}^{n}\\ \hline\end{array}& \text{(10110)}\end{array}$
We now need to relate the contact time to the weight of the catalyst. Consider a point z in the reactor, where the reactant gas has passed cocurrently through a catalyst weight W. Because the solid catalyst is moving through the bed at a rate U_{s} (mass per unit time), the time t that the catalyst has been in contact with the gas when the catalyst reaches a point z is
$\begin{array}{cc}t=\frac{W}{{U}_{\text{s}}}& \text{(10122)}\end{array}$
If we now differentiate Equation (10122)
$\begin{array}{cc}dt=\frac{dW}{{U}_{\text{s}}}& \text{(10123)}\end{array}$
and combine it with the decay rate law, we obtain
$\begin{array}{cc}\begin{array}{c}\hline \frac{da}{dW}=\frac{kd}{{U}_{\text{s}}}{a}^{\text{n}}\\ \hline\end{array}& \text{(10124)}\end{array}$
The activity equation is combined with the mole balance
The design equation for movingbed reactors
$\begin{array}{cc}\begin{array}{c}\hline \frac{dX}{dW}=\frac{a[{r}_{\text{A}}^{\prime}(t=0)]}{{F}_{\text{A0}}}\\ \hline\end{array}& \text{(10125)}\end{array}$
these two coupled differential equations (i.e., Equations (10124) and (10125)) are numerically solved simultaneously with an ODE solver, for example, Polymath.
The catalytic cracking of a gasoil charge, A, to form C_{5}= (B) and to form coke and dry gas (C) is to be carried out in a screwtype conveyor, movingbed reactor at 900°F:
This cracking of gas oil can also be written as
$\begin{array}{cc}\text{A}& \begin{array}{cc}\stackrel{{k}_{1}}{\begin{array}{c}\begin{array}{c}\to \end{array}\end{array}}& \text{Products}\end{array}\end{array}$
While pure hydrocarbons are known to crack according to a firstorder rate law, the fact that the gasoil exhibits a wide spectrum of cracking rates gives rise to the fact that the lumped cracking rate is well represented by a secondorder rate law with the following specific reaction rate:^{34}
^{34} Estimated from V. W. Weekman and D. M. Nace, AIChE J., 16, 397 (1970).
${r}_{\text{A}}^{\prime}=600\frac{{\left(\text{dm}\right)}^{6}}{\left(\text{kg cat}\right)\left(\text{mol}\right)\left(\text{min}\right)}{\text{C}}_{\text{A}}^{2}$
The catalytic deactivation is independent of gasphase concentration and follows a firstorder decay rate law, with a decay constant, k_{d} of 0.72 reciprocal minutes. The feed stream is diluted with nitrogen so that as a first approximation, volume changes can be neglected with reaction. The reactor contains 22 kg of catalyst that moves through the reactor at a rate of 10 kg/min. The gasoil is fed at a rate of 30 mol/min at a concentration of 0.075 mol/dm^{3}.
Determine the conversion that can be achieved in this reactor.^{†}
^{†} AWFOS–S. Be sure to see the animation of a movingbed reactor and description of the ExxonMobil Torrance, California accident (http://umich.edu/~safeche/assets/pdf/courses/Problems/CRE/344ReactionEngrModule(3)PSExxon.pdf).
Solution
Mole Balance:
$\begin{array}{cc}{F}_{\text{A0}}\frac{dX}{dW}=a({r}_{\text{A}}^{\prime})& \text{(E105.1)}\end{array}$
Rate Law:
$\begin{array}{cc}{r}_{\text{A}}^{\prime}=k{C}_{\text{A}}^{2}& \text{(E105.2)}\end{array}$
Decay Law. Firstorder decay:
$\frac{da}{dt}={k}_{d}a$
Moving beds: moderate rate of catalyst decay
Using Equation (10124), we obtain
$\begin{array}{cc}\frac{da}{dW}=\frac{{k}_{d}}{{U}_{\text{s}}}a& \text{(E105.3)}\end{array}$
Integrating
$\begin{array}{cc}a={e}^{({k}_{d}/{U}_{\text{S}})W}& \text{(E105.4)}\end{array}$
Stoichiometry. If υ ≈ υ_{0}, then
$\begin{array}{cc}{C}_{\text{A}}={C}_{\text{A0}}(1X)& \text{(E105.5)}\end{array}$
Combining we have
$\begin{array}{cc}\begin{array}{c}\frac{dX}{dW}=a\frac{k{C}_{\text{A}0}^{2}{(1X)}^{2}}{{F}_{\text{A0}}}\end{array}& \text{(E105.6)}\end{array}$
$\begin{array}{c}a={e}^{({k}_{d}/{U}_{\text{S}})W}\end{array}$
The polymath program is shown below along with a conversion profile.
Explore this problem using Wolfram or Python and then use Polymath to change the decay law and make two observations.
For the simple rate law and activity law given here, we also could have solved this problem analytically as shown below.
Separating and integrating yields (E105.7)
$\begin{array}{c}{F}_{\text{A0}}\frac{dX}{dW}={e}^{({k}_{d}/{U}_{\text{S}})W}k{C}_{\text{A0}}^{2}{(1X)}^{2}\end{array}$
$\begin{array}{cc}\frac{{F}_{\text{A0}}}{k{C}_{\text{A0}}^{2}}{\int}_{0}^{\text{x}}\frac{dX}{{(1X)}^{2}}={\int}_{0}^{W}{e}^{({k}_{d}/{U}_{\text{S}})W}\text{}dW& \text{(E105.7)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline \frac{X}{1X}=\frac{k{C}_{\text{A0}}^{2}{U}_{\text{s}}}{{F}_{\text{A0}}{k}_{d}}(1{e}^{{k}_{d}{\text{W/U}}_{\text{s}}})\\ \hline\end{array}& \text{(E105.8)}\end{array}$
Numerical evaluation:
$\begin{array}{ccc}\frac{X}{1X}& =& \frac{0.6\text{\hspace{0.17em}}{\text{dm}}^{6}}{\text{mol}\cdot \text{g}\cdot \text{cat.}\cdot \mathrm{min}}\times \frac{(0.075\text{\hspace{0.17em}}{\text{mol / dm}}^{3}{)}^{2}}{30\text{}\text{mol / min}}\frac{10000\text{\hspace{0.17em}}\text{g cat / min}}{0.72\text{}{\mathrm{min}}^{1}}\\ & & \times (1\text{exp}\left[\frac{(0.72{\text{min}}^{1})\left(22\text{g}\right)}{10\text{kg}\text{/ min}}\right])\\ & & \frac{X}{1X}=1.24\end{array}$
$\begin{array}{c}\begin{array}{c}\hline X=55\%\\ \hline\end{array}\end{array}$
If there were no catalyst decay, the conversion would be
$\begin{array}{lll}\frac{X}{1X}& =& \frac{{kc}_{\text{A0}}^{2}}{{F}_{\text{}\text{A0}}}W\\ & & 600(\frac{{\text{dm}}^{6}}{\left(\text{kg cat}\right)\left(\text{mol}\right)\left(\text{min}\right)}\times \frac{({\text{0.075 mol / dm}}^{3}{)}^{2}}{\left(\text{30 mol / min}\right)}(22\text{kg cat}\left)\right)\\ & & =2.47\end{array}$
$\begin{array}{c}\begin{array}{c}\hline X=71\%\\ \hline\end{array}\end{array}$
Analysis: The purpose of this example was to show stepbystep how to apply the algorithm to a movingbed reactor that has been used to reduce the effects of catalyst decay that would occur in a PBR. #Splaining. You would have some explaining to do if your company was expecting 71% conversion and you only observed 55%.
We will now rearrange Equation (E105.8) to a form more commonly found in the literature. Let λ be a dimensionless decay time
$\begin{array}{cc}\begin{array}{c}\hline \lambda ={k}_{d}t=\frac{{k}_{d}W}{{U}_{\text{s}}}\\ \hline\end{array}& \text{(10126)}\end{array}$
and Da_{2} be the Damköhler number for a secondorder reaction (a reaction rate divided by a transport rate) for a packedbed reactor
$\begin{array}{cc}D{a}_{2}=\frac{\left(k{C}_{\text{A0}}^{2}\right)\left(W\right)}{{F}_{\text{A0}}}=\frac{k{C}_{\text{A}0}W}{{\upsilon}_{0}}& \text{(10127)}\end{array}$
Through a series of manipulations we arrive at the equation for the conversion in a moving bed where a secondorder reaction is taking place^{35}
^{35} Ibid.
Secondorder reaction in a movingbed reactor
$\begin{array}{cc}\begin{array}{c}\hline X=\frac{D{a}_{2}(1{e}^{\lambda})}{\lambda +D{a}_{2}(1{e}^{\lambda})}\\ \hline\end{array}& \text{(10128)}\end{array}$
Similar equations are given or can easily be obtained for other reaction orders or decay laws.
This reactor is used for reaction systems in which the catalyst deactivates very rapidly. Commercially, the STTR is used in the production of gasoline from the cracking of heavier petroleum fractions where coking of the catalyst pellets occurs very rapidly. In the STTR, the catalyst pellets and the reactant feed enter together and are transported very rapidly through the reactor. The bulk density of the catalyst particle in the STTR is significantly smaller than in movingbed reactors, and the particles are carried through at the same velocity as the gas velocity. In some places the STTR is also referred to as a circulating fluidized bed (CFB). A schematic diagram is shown in Figure 1034.
A mole balance on the reactant A over the differential reactor volume
ΔV = A_{C} Δz
is
${F}_{\text{A}}{}_{z}{F}_{\text{A}}{}_{z+\mathrm{\Delta}z}+{r}_{\text{A}}{A}_{C}\text{}\mathrm{\Delta}z=0$
STTR: Used when catalyst decay (usually coking) is very rapid
Dividing by Δz and taking the limit as Δz → 0, letting ρ_{b} be the bulk density of the suspended catalyst and recalling that ${r}_{\text{A}}={\rho}_{\text{B}}{r}_{\text{A}}^{\prime}$, we obtain
$\begin{array}{cc}\frac{d{F}_{\text{A}}}{dz}={r}_{\text{A}}{\text{A}}_{\text{C}}={r}_{\text{A}}^{\prime}{\rho}_{\text{B}}{\text{A}}_{\text{C}}& \text{(10129)}\end{array}$
In terms of conversion [F_{A} = F_{A0} (– X)] and catalyst activity $[{r}_{\text{A}}^{\prime}={r}_{\text{A}}^{\prime}(\text{t}=0)a\left(\text{t}\right)]$, the reactor mole balance is written as
$\begin{array}{cc}\frac{d\text{X}}{dz}=\left(\frac{{\rho}_{\text{B}}{A}_{\text{C}}}{{F}_{\text{A0}}}\right)[{r}_{\text{A}}^{\prime}(\text{t}=0)]a\left(\text{t}\right)& \text{(10130)}\end{array}$
For a catalyst particle traveling through the reactor with a particle velocity U_{P}, the time the catalyst pellet has been in the reactor when it reaches a height z is just
$\begin{array}{cc}t=\frac{z}{{U}_{\text{P}}}& \text{(10131)}\end{array}$
Substituting for time t in terms of distance z (i.e., a(t) = a(z/U_{P})), the mole balance now becomes
$\frac{dX}{dz}=\frac{{\rho}_{\text{B}}{\text{A}}_{\text{C}}[{r}_{\text{A}}^{\prime}(t=0)]a\left(\mathrm{z/}{U}_{\text{p}}\right)}{{F}_{\text{A}0}}$
The entering molar flow rate, F_{A0}, can be expressed in terms of the gas velocity U_{0}, C_{A0}, and A_{C}
F_{A0} = U_{0}A_{C}C_{A0}
Substituting for F_{A0}, we have
$\begin{array}{cc}\begin{array}{c}\hline \frac{dX}{dz}=\frac{{\rho}_{\text{B}}a(z/{U}_{\text{p}})[{\text{r}}_{\text{A}}^{\prime}(t=0)]}{{C}_{\text{A0}}{U}_{0}}\\ \hline\end{array}& \begin{array}{c}\left(\mathrm{10132}\right)\end{array}\end{array}$
Equation (10132) describes how the conversion varies as we move up the reactor.
The vaporphase cracking of a gasoil is to be carried out in a straightthrough transport reactor (STTR) that is 10m high and 1.5 m in diameter. Gasoil is a mixture of normal and branched paraffins (C_{12}–C_{40}), naphthenes, and aromatics, all of which will be lumped as a single species, A. We shall lump the primary hydrocarbon products according to distillate temperature into two respective groups, dry gas (C_{1}–C_{4}) species B and gasoline (C_{5}–C_{14}) species C. The reaction
A typical cost of the catalyst in the reactor system is $1 million.
$\begin{array}{cc}\text{Gasoil}\left(\text{g}\right)& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{Products}\left(\text{g}\right)+\text{Coke}\end{array}\end{array}$
can be written symbolically as
$\begin{array}{cc}\text{A}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{B}+\text{C}+\text{Coke}\end{array}\end{array}$
Both B and C are adsorbed on the surface. The rate law for a gasoil cracking reaction on fresh catalyst can be approximated by
${r}_{\text{A}}^{\prime}=\frac{{k}^{\prime}{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}{P}_{\text{C}}}$
with k′ = 0.0014 kmol/kgcat·s·atm, K_{A} = 0.05 atm^{–1}, K_{B} = 0.15 atm^{–1}, and K_{C} = 0.1 atm^{–1}. The catalyst decays by the deposition of coke, which is produced in most cracking reactions along with the reaction products. The decay law is
$\begin{array}{cc}a=\frac{1}{1+{\text{At}}^{1/2}}& \text{with}\text{A}=7.6\text{}{\text{s}}^{1/2}\end{array}$
Pure gasoil enters at a pressure of 12 atm and a temperature of 400=C. The bulk density of catalyst in the STTR is 80 kgcat/m^{3}.
Plot the activity, a(z), and conversion, X(z), of gasoil up the 10m reactor for entering gas velocity U_{0} = 2.5 m/s.
Solution
Mole Balance:
$\begin{array}{c}{\text{F}}_{\text{A0}}\frac{d\text{X}}{d\text{z}}={\text{r}}_{\text{A}}{\text{A}}_{\text{C}}\end{array}$
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}\hline \frac{dX}{dz}=\frac{{r}_{\text{A}}}{{U}_{0}{C}_{\text{A0}}}\\ \hline\end{array}\end{array}& \left(\text{E106.1}\right)\end{array}$
The height of the catalyst particle at time “t” after entering the STTR is
$z={\int}_{0}^{t}{U}_{\text{p}}d\text{t}$
Differentiating, we can find a relation between the time the catalyst particle has been in the STTR and reached a height z, which we can use to find the activity a.
$\begin{array}{c}\hline \frac{dt}{dz}=\frac{1}{U}\\ \hline\end{array}$
Rate Law:
$\begin{array}{cc}{r}_{\text{A}}={\rho}_{\text{B}}({r}_{\text{A}}^{\prime})& \text{(E106.2)}\end{array}$
$\begin{array}{cc}{r}_{\text{A}}=a[{r}_{\text{A}}^{\prime}(t=0)]& \text{(E106.3)}\end{array}$
On fresh catalyst
$\begin{array}{cc}{r}_{\text{A}}^{\prime}(t=0)={k}^{\prime}\frac{{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}+{P}_{C}}& \text{(E106.4)}\end{array}$
We assume the solid particles travel up the reactor at the same velocity as the gas, that is, U_{P} = U_{g} = =/A_{c}
Combining Equations (E106.2)–(E106.4) gives
$\begin{array}{cc}\begin{array}{c}\hline {r}_{\text{A}}^{\prime}=a\left({\rho}_{B}{k}^{\prime}\frac{{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}+{K}_{\text{C}}+{P}_{C}}\right)\\ \hline\end{array}& \text{(E106.5)}\end{array}$
Decay Law. Assuming that the catalyst particle and gas travel up the reactor at the velocity U_{P} = U_{g}, we obtain the contact time with the catalyst at a height z to be
$\begin{array}{cc}t=\frac{z}{{U}_{\text{g}}}& \text{(E106.6)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline a=\frac{1}{1+A{\left({\mathrm{z/U}}_{g}\right)}^{1/2}}\\ \hline\end{array}& \text{(E106.7)}\end{array}$
where ${\text{U}}_{\text{g}}=\upsilon /{\text{A}}_{\text{C}}={\upsilon}_{0}(1+\text{\epsilon X})/{\text{A}}_{\text{C}}\text{and}{\text{A}}_{\text{C}}=\pi {\text{D}}^{2/4}$
Stoichiometry (gasphase isothermal and no pressure drop):
$\begin{array}{cc}\begin{array}{c}\hline {P}_{\text{A}}={P}_{\text{A0}}\frac{1X}{1+\epsilon X}\\ \hline\end{array}& \text{(E106.8)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline {P}_{\text{B}}=\frac{{P}_{\text{A0}}X}{1+\epsilon X}\\ \hline\end{array}& \text{(E106.9)}\end{array}$
$\begin{array}{cc}\begin{array}{c}\hline {P}_{\text{C}}={P}_{\text{B}}\\ \hline\end{array}& \text{(E106.10)}\end{array}$
Parameter Evaluation:
$\begin{array}{c}\begin{array}{c}\text{\epsilon}={y}_{\text{A0}}\delta =(1+11)=1\\ {U}_{\text{g}}={U}_{0}(1+\text{\epsilon X})\end{array}\\ {\text{C}}_{\text{A0}}=\frac{{P}_{\text{A0}}}{{RT}_{0}}=\frac{12\text{atm}}{(0.082{\text{m}}^{3}\cdot \text{atm/kmol}\cdot \text{K})\left(673\text{K}\right)}=0.22\frac{\text{kmol}}{{\text{m}}^{3}}\end{array}$
Equations (E106.1), (E106.5), (E106.7), and (E106.8)–(E106.10) are now combined and solved using an ODE solver. The Polymath program is shown in Table E106.1, and the computer output is shown in Figure E106.1.
Use Wolfram to learn how the conversion and activity profiles shown above in Figure E106.1 change as you vary the parameters.
Analysis: In this example we applied the algorithm to a STTR in which the gas velocity and, hence, particle velocity increases as we move through the reactor. The decay is quite rapid and the activity is only 15% of its initial value at z = 3 m into the reactor and the conversion begins to plateau at z = 6 m at 60% conversion. If there were no catalyst decay (a = 1), the conversion would have been 97% at z = 6 m.
This incident and video can be found on the Process Safety Across the Chemical Engineering Curriculum Web site and it is highly recommended to view the video first. On Monday February 18, 2015 an explosion occurred at ExxonMobil’s Torrance California refinery’s electrostatic precipitator unit, a pollution control device associated with the fluid catalytic cracking (FCC) unit. While the explosion occurred in the precipitator, the preinitiating events took place in the catalyst regeneration unit attached to a STTR, as shown in Figure 1035. In this incident, the spent catalyst from the STTR is covered with carbon compounds that are burned off in a regenerator. On a routine maintenance shut down, catalyst particles got lodged in the regenerator door allowing flammable vapor to flow out of the regeneration unit and proceed downstream to an electrostatic precipitator (ESP) where it found a spark and exploded.
Watch the Video: (https://www.csb.gov/exxonmobilrefineryexplosion/)
Incident Report Available At:: (https://www.csb.gov/assets/1/20/exxonmobil_report_for_public_release.pdf?15813)
CSB Incident Report No. 201502I CA ExxonMobil Torrance Refinery, February 18, 2015. If you need more detailed information that can be found in the video, review pages 1–22 of this report.
Figure 148 in Chapter 14 shows a schematic for the removal of coke from a catalyst pellet. The time necessary to completely burn the coke off the pellet is
$\begin{array}{c}\hline t=\frac{{\rho}_{\text{C}}{R}_{0}^{2}{\mathrm{\varphi}}_{\text{C}}}{{6D}_{e}{C}_{\text{A0}}}\\ \hline\end{array}$
as described by the shrinking core model in Section 14.6, read to learn about Coke Regeneration.
It is important that chemical engineers have an understanding of what the accident was, why it happened, and how it could have been prevented in order ensure similar accidents may be prevented. Applying a safety algorithm to the incident will help achieve this goal. In order to become familiar with a strategy for accident awareness and prevention, view the Chemical Safety Board video on the explosion and fill out the incident report as shown below.
Activity: The activity in this incident is the maintenance operation of a fluid catalytic cracking (FCC) unit at a refinery.
Hazard: The hazard in this incident is the flammability of light hydrocarbons.
Incident: The incident here is leakage of light hydrocarbons through the reactor into airside of the unit and finally into the electrostatic precipitator where they found a source of ignition and exploded.
Initiating Event: The initiating event was the lowering of pressure of steam in the reactor which allowed light hydrocarbons to flow back into the reactor and move toward the airside through the regenerator, as well as the failure of the Spent Catalyst Slide Valve to completely seal the regenerator from the reactor.
Preventative Actions and Safeguards: Some preventive actions or safeguards include strict inspection to control use of old equipment that may have undergone erosion due to use beyond their life, revaluation of operating procedure, and safe operating limits during Safe Park, extensive consideration of all possible failure scenarios. Other considerations include use of an alternative technology that avoids a possible ignition source (use of scrubber instead of ESP) when flammable substances are involved, regular inspection of plant equipment to determine their fitness, installation of gas detectors in the regenerator to detect presence of flammable gases that may flow to the ESP.
Contingency Plan/Mitigating Actions: Immediate shutdown of electrostatic precipitator or the FCC unit when a part of the plant is being opened, replaced, and so on, or when a safety critical safeguard, such as the barrier of catalyst, could not be established.
Lessons Learned: The lessons learned from this incident are that safe operating limits must be defined for all operations and process conditions must be verified to ensure that the process is safe. The equipment critical for safety must be maintained regularly to ensure that it performs its function and an assessment of various possible hazards must be made while changing practices, and they must be done only in consultation with experts from different fields.
A BowTie diagram for this incident is shown in Figure 1036.
Review the ExxonMobil Safety Module (http://umich.edu/~safeche/assets/pdf/courses/Problems/CRE/344ReactionEngrModule(3)PS20419Exxon.pdf) and the Catalyst Regeneration Web site.
In order to get the above BowTie diagram on one page, we had to divide it at the initiating event.
Types of adsorption:
Chemisorption
Physical adsorption
The Langmuir isotherm relating the concentration of species A on the surface to the partial pressure of A in the gas phase is
$\begin{array}{cc}{C}_{\text{A}\cdot \text{S}}=\frac{{K}_{\text{A}}{C}_{t}{P}_{\text{A}}}{1+{K}_{\text{A}}{P}_{\text{A}}}& \text{(S101)}\end{array}$
The sequence of steps for the solidcatalyzed isomerization
$\begin{array}{cc}\begin{array}{cc}\text{A}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{B}\end{array}\end{array}& \text{(S102)}\end{array}$
is:
Mass transfer of A from the bulk fluid to the external surface of the pellet
Diffusion of A into the interior of the pellet
Adsorption of A onto the catalytic surface
Surface reaction of A to form B
Desorption of B from the surface
Diffusion of B from the pellet interior to the external surface
Mass transfer of B away from the solid surface to the bulk fluid
Assuming that mass transfer is not ratelimiting, the rate of adsorption is
$\begin{array}{cc}\text{}{r}_{\text{AD}}={k}_{\text{A}}({C}_{\upsilon}{P}_{\text{A}}\frac{{C}_{\text{A}\cdot \text{S}}}{{K}_{\text{A}}})& \text{(S103)}\end{array}$
The rate of surface reaction is
$\begin{array}{cc}{r}_{\text{s}}={k}_{\text{s}}({C}_{\text{A}\cdot \text{S}}\frac{{C}_{\text{B}\cdot \text{S}}}{{K}_{\text{S}}})\text{}& \text{(S104)}\end{array}$
The rate of desorption is
$\begin{array}{cc}{r}_{\text{D}}={k}_{\text{D}}({C}_{\text{B}\cdot \text{S}}{K}_{\text{B}}{P}_{\text{B}}{C}_{\upsilon})& \text{(}\text{S105)}\end{array}$
At steady state
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={r}_{\text{AD}}={{r}_{\text{}}}_{\text{s}}={r}_{\text{D}}& \text{(S106)}\end{array}$
If there are no inhibitors present, the total concentration of sites is (S107)
$\begin{array}{cc}{C}_{\text{t}}={C}_{\upsilon}+{C}_{\text{A}\cdot \text{S}}+{C}_{\text{B}\cdot S}& \text{(S107)}\end{array}$
If we assume that the surface reaction is ratelimiting, we set
$\begin{array}{cc}\frac{{r}_{\text{AD}}}{{k}_{\text{A}}}\text{\u22cd}0& \frac{{r}_{\text{D}}}{{k}_{\text{D}}}\text{\u22cd0}\end{array}$
and solve for C_{A}·_{S} and C_{B}·_{S} in terms of P_{A} and P_{B}. After substitution of these quantities in Equation (S104), the concentration of vacant sites is eliminated with the aid of Equation (S107)
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={r}_{\text{s}}=\frac{\stackrel{k}{\stackrel{}{\overbrace{{C}_{1}{k}_{\text{}\text{s}}{K}_{\text{A}}}}}({P}_{\text{A}}{P}_{\text{B}}/{K}_{\text{p}})}{1+{K}_{\text{A}}{P}_{\text{A}}+{K}_{\text{B}}{P}_{\text{B}}}& \text{(S108)}\end{array}$
Recall that the equilibrium constant for desorption of species B is the reciprocal of the equilibrium constant for the adsorption of species B
$\begin{array}{cc}{K}_{\text{B}}=\frac{1}{{K}_{\text{DB}}}& \text{(S109)}\end{array}$
and the thermodynamic equilibrium constant, K_{P}, is
$\begin{array}{cc}{K}_{\text{P}}={K}_{\text{A}}{K}_{\text{S}}/{K}_{\text{B}}& \text{(S1010)}\end{array}$
Chemical vapor deposition (CVD)
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}{\text{SiH}}_{\text{4}}\left(g\right)& \overrightarrow{\leftarrow}\end{array}& {\text{SiH}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\end{array}& \text{(S1011)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}{\text{SiH}}_{2}\left(g\right)+\text{S}& \to \end{array}& {\text{SiH}}_{2}\cdot \text{S}\text{}\end{array}& \text{(S1012)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}{\text{SiH}}_{2}\text{}\cdot \text{S}& \to \end{array}& \text{Si(s)}+{\text{H}}_{2}\left(g\right)\end{array}& \text{(S1013)}\end{array}$
$\begin{array}{cc}{r}_{\text{Dep}}=\frac{{kP}_{{\text{SiH}}_{\text{4}}}}{{P}_{{\text{H}}_{2}}+{K}_{\text{p}}{P}_{{\text{SiH}}_{4}}}& \text{(S1014)}\end{array}$
Catalyst deactivation. The catalyst activity is defined as
$\begin{array}{cc}a\left(t\right)=\frac{{r}_{\text{A}}^{\prime}\left(t\right)}{{r}_{\text{A}}^{\prime}(t=0)}& \text{(S1015)}\end{array}$
The rate of reaction at any time t is
$\begin{array}{cc}{r}_{\text{A}}^{\prime}=a\left(t\right)k\left(T\right)\text{fn}({C}_{\text{A,}}{C}_{\text{B}},\mathrm{...},{C}_{\text{P}})& \text{(S1016)}\end{array}$
The rate of catalyst decay is
$\begin{array}{cc}{r}_{d}=\frac{da}{dt}=p\left[a\left(t\right)\right]{k}_{d}\left(T\right)g({C}_{\text{}\text{A,}}{C}_{\text{B}},\mathrm{...},{C}_{\text{P}})& \text{(S1017)}\end{array}$
For firstorder decay
$\begin{array}{cc}p\left(a\right)=a& \text{(S1018)}\end{array}$
For secondorder decay
$\begin{array}{cc}p\left(a\right)={a}^{2}& \text{(S1019)}\end{array}$
For slow catalyst decay, the idea of a temperature–time trajectory is to increase the temperature in such a way that the rate of reaction remains constant.
The coupled differential equations to be solved for a movingbed reactor are
$\begin{array}{cc}{F}_{\text{A0}}\frac{dX}{dW}=a({r}_{\text{A}}^{\prime})& \text{(S1020)}\end{array}$
For nthorder activity decay and m order in a gasphase concentration of species i in a moving bed where the solids move up with the velocity U_{S}
$\begin{array}{cc}\frac{da}{dW}=\frac{{k}_{d}{a}^{n}{C}_{i}^{m}}{{U}_{\text{s}}}& \text{(S1021)}\end{array}$
$\begin{array}{cc}t=\frac{W}{{U}_{\text{s}}}& \text{(S1022)}\end{array}$
The coupled equations to be solved in a straightthrough transport reactor are
$\begin{array}{cc}\frac{\mathit{d}X}{dz}=\frac{a\left(t\right)[{r}_{\text{A}}^{\prime}(t=0)]}{{U}_{0}}\frac{{\rho}_{bc}}{{C}_{\text{A0}}}& \text{(S1023)}\end{array}$
$\begin{array}{cc}U={U}_{0}(1+\epsilon X)\frac{T}{{T}_{0}}& \text{(S1024)}\end{array}$
$\begin{array}{cc}\mathit{\text{t}}=\frac{z}{{U}_{p}}& \text{(S1025)}\end{array}$
where U_{p} is the particle velocity up the STTR. For coking
$\begin{array}{cc}a\left(t\right)=\frac{1}{1+\text{A}{t}^{1/2}}& \text{(S1026)}\end{array}$
For the case when there is no slip between the catalyst particles and the gas velocity (S1027)
$\begin{array}{cc}{\text{U}}_{P}=\text{U}& \text{(S1027)}\end{array}$
The isomerization A → B is carried out over a decaying catalyst in a movingbed reactor with pressure drop. Pure A enters the reactor and the catalyst flows through the reactor at a rate of 2.0 kg/s.
$\begin{array}{cc}\frac{d\text{X}}{d\text{W}}=\frac{{r}_{\text{A}}^{\prime}}{{\text{F}}_{\text{A0}}}\hfill & k=0.1\text{mol/}(\text{kgcat}\cdot \text{s}\cdot \text{atm})\hfill \\ {r}_{\text{A}}^{\prime}=\frac{ak{P}_{\text{A}}}{1+\text{}{\text{K}}_{\text{A}}{P}_{\text{A}}}\hfill & {\text{K}}_{\text{A}}=1.5{\text{atm}}^{1}\hfill \\ \frac{da}{d\text{W}}=\frac{{k}_{d}2{P}_{\text{B}}}{{\text{U}}_{\text{s}}}\hfill & {k}_{d}=\frac{0.75}{\text{s}\cdot \text{atm}}\hfill \\ {P}_{\text{A}}={P}_{\text{A0}}(1\text{X})P\hfill & {\text{F}}_{\text{A0}}=10\text{mol/s}\hfill \\ {P}_{\text{B}}={P}_{\text{A0}}{\text{X}}_{p}\hfill & {P}_{\text{A0}}=20\text{atm}\hfill \\ \frac{dp}{d\text{W}}=\frac{\alpha}{2\text{p}}\hfill & {\text{U}}_{\text{s}}=2.0\text{kgcat/s}\hfill \\ \alpha =0.0019{\text{kg}}^{1}\hfill & {\text{W}}_{f}=500\text{kgcat}\hfill \end{array}$
LEP Explore this problem using Python and/or Wolfram on the CRE Web site.
Interactive Computer Games
Heterogeneous Catalysis (http://umich.edu/%7Eelements/6e/icm/hetcat.html)
After Reading Each Page in This Book, Ask Yourself a Question About What You Read
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
Questions
Q101_{A} QBR (Question Before Reading). The rate laws for heterogeneous reactions usually have concentration terms in both the numerator and denominator. How can this function come about?
Q102_{A} i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/10chap/iclicker_ch10_q1.html) and view at least 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.
Q103_{A} Read over the problems at the end of this chapter. Make up an original problem that uses the concepts presented in this chapter. See Problem P51_{A} for guidelines. To obtain a solution:
Create your data and reaction.
Use a real reaction and real data.
The journals listed at the end of Chapter 1 may be useful for part (b).
Choose an FAQ from Chapter 10 and explain why it was most helpful.
Listen to the audios on the Summary Notes, pick one, and explain why it was most helpful.
Q104_{B} For the decomposition of cumene discussed in this chapter, if an adsorbing inert is present, how would you compare the initial rate as a function of total pressure when desorption is the RLS, as shown in Figure 1018?
Q105_{A} Choose five Chapter 10 i>clicker questions, pick the one that was the most challenging, and explain why.
Q106_{A} Revisit Example 104. What if you were asked to sketch or explain the difference in the temperature–time trajectories and to find the catalyst lifetimes (say a = 0.1) for first and for secondorder decay when E_{A} = 35 kcal/mol, E_{d} = 10 kcal/mol, k_{d}_{0} = 0.01 day^{–1}, and T_{0} = 400 K? What would you say? Describe how the trajectory of the catalyst lifetime would change if E_{A} = 10 kcal/mol and E_{d} = 35 kcal/mol? At what values of k_{d}_{0} and ratios of E_{d} to E_{A} would temperature–time trajectories not be effective? Describe what your temperature–time trajectory would look like if n = 1 + E_{d}/E_{A}?
Q107_{A} Write a question for this problem that involves critical thinking and explain why it involves critical thinking.
Q108_{A} Go to the LearnChemE screencast link for Chapter 10 (http://www.umich.edu/~elements/6e/10chap/learnchemevideos.html). View one or more of the screencast 5 to 6minute videos.
List two items you learned from Professor Dave’s Lecture.
List four similarities between the Ely–Rideal mechanisms and the LangmuirHinshelwood screencast.
Write a twosentence evaluation of the videos you viewed.
Q109_{A} AWFOS–S10 View the CSB video (http://umich.edu/~safeche/assets/pdf/courses/Problems/CRE/344ReactionEngrModule(3)PSExxon.pdf). After watching the Chemical Safety Board video, what points do you feel should have been emphasized more in the Safety Analysis of the Incident algorithm and the BowTie Diagram? Write three or so sentences describing your takeaway lesson from this incident.
P101_{B}
Example 101: Nonlinear Regression to Determine Model Parameters and Site Concentration Ratio = C_{T·S}/C_{B·S}
Wolfram and Python
Find the critical value of K_{B} at which fraction of vacant sites starts to increase with conversion for the initial settings. Repeat for K_{T}.Vary only one parameter at a time.
Find the value K_{B} where the curve for $\left(\frac{{\text{C}}_{\text{B}\cdot \text{S}}}{{\text{C}}_{t}}\right)$ versus conversion X changes from convex to concave and explain why this shape change occurs.
Write a set of conclusions for your experiments (i) and (ii).
Example 102: Catalytic Reactor Design
Wolfram and Python
Describe how the partial pressure profiles change as you vary the sliders for α, K_{T}, and k.
What if the molar flow rate were reduced by 50%; how would X and p change?
After reviewing Generating Ideas and Solutions on the Web site (http://www.umich.edu/~elements/6e/toc/SCPS,3rdEdBook(Ch07).pdf), choose one of the brainstorming techniques (e.g., lateral thinking) to suggest two questions that should be included in this problem.
Write two conclusions from your experiments with the sliders in this example.
Polymath
What catalyst weight would be required for 60% conversion?
Which parameter will you vary so that P_{B} = P_{H2} at the middle of the reactor (i.e., W = 5000 kg).
Example 103: Hydrogenation Ethylene to Ethane
Use Polymath to learn how your answers would change if the following data for run 10 were incorporated in your regression table.
${r}_{\text{E}}^{\prime}=0.8\text{mol/kgcat}\cdot {\text{s, P}}_{\text{E}}=0.5\text{atm,}{P}_{\text{EA}}={\text{15 atm, P}}_{\text{H}}=2\text{atm}$
How do the rate laws (e) and (f)
$\begin{array}{cc}\left(\text{e}\right){r}_{\text{E}}^{\prime}=\frac{k{P}_{\text{E}}{P}_{\text{H}}}{{(1+{K}_{\text{A}}{P}_{\text{EA}}+{K}_{\text{E}}{P}_{\text{E}})}^{2}}& \left(\text{f}\right)\end{array}{r}_{\text{E}}^{\prime}=\frac{k{P}_{\text{H}}{P}_{\text{H}}}{1+{K}_{\text{A}}{P}_{\text{EA}}}$
compare with the nonlinear analysis with the other rate laws used to model the data?
Example 104: Calculating Conversion with Catalyst Decay in Batch Reactors
Wolfram and Python
What is the maximum conversion that can be achieved if there is no catalyst decay?
Vary k and k_{d} and describe what you find. Can you explain why there is no effect of catalyst decay on conversion at a high value of k?
Vary E and k_{d} and then write a few sentences and three conclusions describing the results of your experiments.
Explain why conversion with catalyst decay increases with increasing E_{d} and decreases with increasing E.
Polymath
Vary the ratio of (k/k_{d}) and describe what you find.
Repeat this example (i.e., the plotting of X vs. t) for a secondorder reaction with (C_{A0} = 1 mol/dm^{3}) and firstorder decay, and describe the differences from the base case.
Repeat this example for a firstorder reaction and firstorder decay and describe the differences from the base case.
Repeat this example for a secondorder reaction (C_{A0} = 1 mol/dm^{3}) and a secondorder decay and describe the differences from the base case.
Example 105: Catalytic Cracking in a MovingBed Reactor
Wolfram and Python
Suppose k_{d} and U_{S} are at their maximum values. What could you do to increase conversion?
Vary the parameters and write a set of conclusions.
Polymath
Use Polymath to learn what the conversion would be if there is no catalyst decay.
What if the solids and reactants entered from opposite ends of the reactor? How would your answers change?
What if the decay in the moving bed were second order? By how much must the catalyst charge, U_{S}, be increased to obtain the same conversion?
What if ε = 2 (e.g., A → 3B) instead of zero. How would the results be affected?
Example 106: Decay in a StraightThrough Transport Reactor
Wolfram and Python
What happens to a and X when T is varied?
What happens to a and X when U_{0} is varied?
Vary the parameters and write a set of conclusions.
Polymath
What if you varied the parameters P_{A0}, U_{g}, A, and k′ in the STTR? What parameter has the greatest effect on either increasing or decreasing the conversion?
Ask questions such as: What is the effect of varying the ratio of k to U_{g} or of k to A on the conversion? Make a plot of conversion versus distance as U_{g} is varied between 0.5 and 50 m/s.
Sketch the activity and conversion profiles for U_{g} = 0.025, 0.25, 2.5, and 25 m/s.
What gas velocity do you suggest operating at?
What is the corresponding entering volumetric flow rate?
What concerns do you have operating at the velocity you selected? Would you like to choose another velocity? If so, what is it? Which parameter will you vary so that conversion increases but activity decreases? Explain, if you can, this unusual behavior.
P102_{A} Download the Interactive Computer Games (ICG) from the CRE Web site (http://www.umich.edu/~elements/6e/icm/install.html). Play the game and then record your performance number for the game, which indicates your mastery of the material. Your professor has the key to decode your performance number. (This ICG is a little longer than the other ICGs.)
ICG Heterogeneous Catalysis Performance # _____________
P103_{A} tButyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline (Ind. Eng. Chem. Res., 27, 2224 (1988)). TBA was produced by the liquidphase hydration (W) of isobutene (I) over an Amberlyst15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility.
The reaction mechanism is believed to be
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{I}+\text{S}& \overrightarrow{\leftarrow}\end{array}& \text{I}\cdot \text{S}\end{array}& \text{(P103.1)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{W}+\text{S}& \overrightarrow{\leftarrow}\end{array}& \text{W}\cdot \text{S}\end{array}& \text{(P103.2)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{W}+\text{S}+\text{I}\cdot \text{S}& \overrightarrow{\leftarrow}\end{array}& \text{TBA}\cdot \text{S}+\text{S}\end{array}& \text{(P103.3)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\text{TBA}\cdot \text{S}& \overrightarrow{\leftarrow}\end{array}& \text{TBA}+\text{S}\end{array}& \text{(P103.4)}\end{array}$
Derive a rate law assuming:
The surface reaction is ratelimiting.
The adsorption of isobutene is limiting.
The reaction follows Eley–Rideal kinetics
$\begin{array}{cc}\begin{array}{cc}\text{I}\cdot \text{S}+\text{W}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{TBA}\cdot \text{S}\end{array}\end{array}& \text{(P103.5)}\end{array}$
and the surface reaction is limiting.
Isobutene (I) and water (W) are adsorbed on different sites.
$\begin{array}{cc}\text{I}+{\text{S}}_{1}\overrightarrow{\leftarrow}\text{}\text{I}\cdot {\text{S}}_{1}& \text{(P103.6)}\end{array}$
$\begin{array}{cc}\text{W}+{\text{S}}_{\text{2}}\overrightarrow{\leftarrow}\text{W}\cdot {\text{S}}_{2}& (\text{P103.7)}\end{array}$
TBA is not on the surface, and the surface reaction is ratelimiting.
$(\mathit{A}\mathit{n}\mathit{s}:{r}_{\text{TBA}}^{\prime}={r}_{1}^{\prime}=\frac{k[{C}_{1}{C}_{\text{W}}{C}_{\text{TBA}}/{K}_{c}]}{(1+{K}_{\text{W}}{C}_{\text{W}})(1+{K}_{1}{C}_{1})})$
What generalization can you make by comparing the rate laws derived in parts (a) through (d)?
P104_{B} OEQ (Old Exam Question). Consider the catalytic reaction as a function of the initial partial pressures
$2\text{A}\overrightarrow{\leftarrow}\text{B}+\text{C}$
The rate of disappearance of species A was obtained in a differential reactor and is shown below.
What species are on the surface?
What does Figure B tell you about the reversibility and what’s adsorbed on the surface?
Derive the rate law and suggest a rateliming step consistent with the above figures.
How would you plot your data to linearize the initial rate data in Figure A?
Assuming pure A is fed, and the adsorption constants for A and C are K_{A} = 0.5 atm^{–1} and K_{C} = 0.25 atm^{–1} respectively, at what conversion are the number of sites with A adsorbed on the surface and C adsorbed on the surface equal? (Ans: X = 0.66)
P105_{A} The rate law for the hydrogenation (H) of ethylene (E) to form ethane (A) over a cobaltmolybdenum catalyst (Collection Czech. Chem. Commun., 51, 2760 (1988)) is
${r}_{\text{E}}^{\prime}=\frac{k{P}_{\text{E}}{P}_{\text{H}}}{1+{K}_{\text{E}}{P}_{\text{E}}}$
Suggest a mechanism and ratelimiting step consistent with the above rate law.
What was the most difficult part in finding the mechanism?
P106_{B} The formation of propenol on a catalytic surface is believed to proceed by the following mechanism
$\begin{array}{c}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\begin{array}{c}{\text{O}}_{2}+2\text{S}\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\underleftarrow{\stackrel{\text{cat}}{\begin{array}{c}\to \end{array}}}\end{array}& \text{2}\text{O}\cdot \text{S}\end{array}\end{array}& \text{}\end{array}\\ {\text{C}}_{3}{\text{H}}_{6}+\text{O}\cdot \text{S}\to \text{\hspace{0.17em}}\text{}{\text{C}}_{3}\text{HOH}\cdot \text{S}\end{array}\\ \begin{array}{cc}\begin{array}{c}{\text{C}}_{3}\text{HOH}\cdot \text{S}\end{array}& \overrightarrow{\begin{array}{c}\leftarrow \end{array}}\end{array}\text{}{\text{C}}_{3}\text{HOH+S}\end{array}$
Suggest a ratelimiting step and derive a rate law.
P107_{B} OEQ (Old Exam Question). The dehydration of nbutyl alcohol (butanol) over an aluminasilica catalyst was investigated by J. F. Maurer (Ph.D. thesis, University of Michigan). The data in Figure P107_{B} were obtained at 750°F in a modified differential reactor. The feed consisted of pure butanol.
Suggest a mechanism and ratelimiting step that are consistent with the experimental data.
Evaluate the ratelaw parameters.
At the point where the initial rate is a maximum, what is the fraction of vacant sites? What is the fraction of occupied sites by both A and B? % vacant = 0.41
What generalizations can you make from studying this problem?
Write a question that requires critical thinking and then explain why your question requires critical thinking. Hint: See Preface Section G.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P108_{B} OEQ (Old Exam Question). The catalytic dehydration of methanol (ME) to form dimethyl ether (DME) and water was carried out over an ion exchange catalyst (K. Klusacek, Collection Czech. Chem. Commun., 49, 170 (1984)). The packed bed was initially filled with nitrogen, and at t = 0. The N_{2} feed is switched to pure methanol vapor entering the reactor at 413 K, 100 kPa, and 0.2 cm^{3}/s. The following partial pressures were recorded at the exit to the differential reactor containing 1.0 g of catalyst in 4.5 cm^{3} of reactor volume.
$\begin{array}{cc}\begin{array}{cc}\text{2}{\text{CH}}_{3}& \to \text{}\end{array}& {\text{CH}}_{3}{\text{OCH}}_{3}\end{array}+{\text{H}}_{2}\text{O}$
TABLE P108_{B} PARTIAL PRESSURE OF H_{2}, ME, AND DME EXITING THE REACTOR AFTER TIME
t(s) 

0 
10 
50 
100 
150 
200 
300 

P_{N2} (kPa) 
100 
50 
10 
2 
0 
0 
0 
P_{ME} (kPa) 
0 
2 
15 
23 
25 
26 
26 
P_{H2O} (kPa) 
0 
10 
15 
30 
35 
37 
37 
P_{DME} (kPa) 
0 
38 
60 
45 
40 
37 
37 
Use parts (a) through (f) to lead you to suggest a mechanism, ratelimiting step, and rate law consistent with this data.
Using the data above, sketch the exit concentrations as a function of time.
Which species took longer than others to exit the reactor in the gas phase? What could have caused this difference in exit times?
What species are adsorbed on the surface?
Are any species not adsorbed on the surface? If so, which ones?
Which set of figures in Figure P108_{B} correctly describes the functionality of the chemical reaction rate with the partial pressure P_{W}, P_{DME}, and P_{ME}?
Derive a rate law for the catalytic dehydration of methanol. Dimethyl Ether 2ME → DME + W.
P109_{B} In 1981, the U.S. government put forth the following plan for automobile manufacturers to reduce emissions from automobiles over the next few years.
Year 

1981 
1993 
2010 

Hydrocarbons 
0.41 
0.25 
0.125 
CO 
3.4 
3.4 
1.7 
NO 
1.0 
0.4 
0.2 
All values are in grams per mile. An automobile emitting 3.74 lb_{m} of CO and 0.37 lb _{m} of NO on a journey of 1000 miles would meet the current government requirements.
To remove oxides of nitrogen (assumed to be NO) from automobile exhaust, a scheme has been proposed that uses unburned carbon monoxide (CO) in the exhaust to reduce the NO over a solid catalyst, according to the reaction
$\begin{array}{cc}\begin{array}{cc}\text{CO}+\text{NO}& \begin{array}{cc}\begin{array}{c}\underset{\text{catalyst}}{\begin{array}{c}\to \end{array}}\end{array}& \text{Products}\left({\text{N}}_{2,}{\text{CO}}_{2}\right)\end{array}\end{array}& \text{}\end{array}$
Experimental data for a particular solid catalyst indicate that the reaction rate can be well represented over a large range of temperatures by
$\begin{array}{cc}{r}_{\text{N}}^{\prime}=\frac{k{P}_{\text{N}}{P}_{\text{C}}}{{(1+{K}_{1}{P}_{\text{N}}+{K}_{2}{P}_{\text{C}})}^{2}}& \text{(}\text{P109.1)}\end{array}$
$\begin{array}{ccc}\begin{array}{cc}\text{where}& {P}_{N}\end{array}& =& \text{gasphasepartialpressureofNO}\hfill \\ \hfill {P}_{\text{C}}& =& \text{gasphasepartialpressureofCO}\hfill \\ \hfill k,{K}_{1},{K}_{2}& =& \text{coefficientsdependingonlyontemperature}\hfill \end{array}$
Propose an adsorption–surface reaction–desorption mechanism and ratelimiting step that are consistent with the experimentally observed rate law. Do you need to assume any species are weakly adsorbed to get agreement with Equation (P109.1)?
A certain engineer thinks that it would be desirable to operate with a very large stoichiometric excess of CO to minimize catalytic reactor volume. Do you agree or disagree? Explain.
What would be the relevance of the problem if everyone were driving a hybrid by 2018? A driverless car by 2020?
P1010_{B} OEQ (Old Exam Question). Methyl ethyl ketone (MEK) is an important industrial solvent that can be produced from the dehydrogenation of butan2ol (Bu) over a zinc oxide catalyst (Ind. Eng. Chem. Res., 27, 2050 (1988)):
$\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\begin{array}{cc}\text{Bu}& \underset{\text{catalyst}}{\begin{array}{c}\to \end{array}}\end{array}\end{array}& \text{MEK}+{\text{H}}_{2}\end{array}& \text{}\end{array}$
The following data giving the reaction rate for MEK were obtained in a differential reactor at 490°C.
P_{Bu} (atm) 
2 
0.1 
0.5 
1 
2 
1 
P_{MEK} (atm) 
5 
0 
2 
1 
0 
0 
P_{H2} (atm) 
0 
0 
1 
1 
0 
10 
${r}_{\text{MEK}}^{\prime}(\text{mol/h}\cdot g\text{cat})$ 
0.044 
0.040 
0.069 
0.060 
0.043 
0.059 
Suggest a rate law consistent with the experimental data.
Suggest a reaction mechanism and ratelimiting step consistent with the rate law. Hint: Some species might be weakly adsorbed.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete prefaceintroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
Plot conversion (up to 90%) and reaction rate as a function of catalyst weight for an entering molar flow rate of pure butan2ol of 10 mol/min at an entering pressure P_{0} = 10 atm up to a catalyst weight W_{max} = 23 kg.
Write a question that requires critical thinking and then explain why your question requires critical thinking. Hint: See Preface Section G.
Repeat part (d), accounting for pressure drop and α = 0.03 kg^{–1}. Plot p and X as a function of catalyst weight down the reactor.
P1011_{B} Cyclohexanol was passed over a catalyst to form water and cyclohexene:
$\begin{array}{cc}\begin{array}{cc}\text{Cyclohexanol}& \underset{\text{}}{\underset{\text{catalyst}}{\begin{array}{c}\to \end{array}}}\end{array}& \text{Water}+\text{Cyclohexence}\end{array}$
It is suspected that the reaction may involve a dualsite mechanism, but it is not known for certain. It is believed that the adsorption equilibrium constant for cyclohexanol is around 1.0 and is roughly one or two orders of magnitude greater than the adsorption equilibrium constants for the other compounds. Using the data in Table P1011_{B}
TABLE P1011_{B} DATA FOR CATALYTIC FORMATION OF CYCLOHEXENE
Run 
Reaction Rate (mol/dm^{3} · s) × 10^{5} 
Partial Pressure of Cyclohexanol (atm) 
Partial Pressure of Cyclohexene (atm) 
Partial Pressure of Steam (H_{2}O) (atm) 
1 
3.3 
1 
1 
1 
2 
1.05 
5 
1 
1 
3 
0.565 
10 
1 
1 
4 
1.826 
2 
5 
1 
5 
1.49 
2 
10 
1 
6 
1.36 
3 
0 
5 
7 
1.08 
3 
0 
10 
8 
0.862 
1 
10 
10 
9 
0 
0 
5 
8 
10 
1.37 
3 
3 
3 
Suggest a rate law and mechanism consistent with the data given here.
Determine the values of the ratelaw parameters. (Ind. Eng. Chem. Res., 32, 2626–2632.)
Why do you think estimates of the ratelaw parameters were given?
For an entering molar flow rate of cychlohexanol of 10 mol/s at a partial pressure of 15 atm, what catalyst weight is necessary to achieve 85% conversion when the bulk density is 1500 gm/dm^{3}?
P1012_{B} OEQ (Old Exam Question). Experimental data for the gasphase catalytic reaction
A + B → C
is shown below. The limiting step in the reaction is known to be irreversible, so that the overall reaction is irreversible. The reaction was carried out in a differential reactor to which A, B, and C were all fed.
Run Number 
P_{A} (atm) 
P_{B} (atm) 
P_{C} (atm) 
Reaction rate (mol)/(gcat · s) 
1 
1 
1 
2 
0.114 
2 
1 
10 
2 
1.140 
3 
10 
1 
2 
0.180 
4 
1 
20 
2 
2.273 
5 
1 
20 
10 
0.926 
6 
20 
1 
2 
0.186 
7 
0.1 
1 
2 
0.0243 
Suggest a rate law consistent with the experimental data. Hint: Sketch $({r}_{\text{A}}^{\prime})$ as a function of P_{A}, as a function of P_{B}, and as a function of P_{C}.
From your rate expression, which species can you conclude are adsorbed on the surface?
Suggest a rate law and then show that your mechanism is constant with the rate law in part (a).
For an entering partial pressure of A of 2 atm in a PBR, what is the ratio of A to C sites at 80% conversion of A?
At what conversion are the number of A and C sites equal? (Ans: X = 0.235)
What reactor volume is necessary to achieve 90% conversion of A for a stoichiometric feed and flow of A 2 mol/s? (Ans: W = 8.9 gcat)
If necessary, feel free to use none, any, or all of the following parameter values:
$k=2.5\frac{\text{mol}}{{\text{atm}}^{2}\text{gcat}\cdot \text{S}},{K}_{\text{A}}=4{\text{atm}}^{1},{K}_{\text{C}}=13{\text{atm}}^{1},{K}_{\text{I}}=10\text{}{\text{atm}}^{1}$
P1013_{B} Solar Energy Capture: Water Splitting. Hydrogen and O_{2} can be combined in fuel cells to generate electricity. Solar energy can be used to split water to generate the raw reactants H_{2} and O_{2} for fuel cells. One method of solar thermal reduction is with NiFe_{2}O_{4} in the sequence
$\begin{array}{c}\text{Step}\left(1\right)\text{SolarEnergy}\text{}+\text{}\stackrel{\text{}}{\stackrel{\text{Surface}\left(\text{s}\right)}{\stackrel{}{\overbrace{{\text{NiFe}}_{2}{\text{O}}_{4}}}}\to \stackrel{\text{SolidSolution}\left(\text{s'}\right)}{\stackrel{}{\overbrace{1.2\text{FeO}+0.4{\text{Fe}}_{2}{\text{O}}_{3}+\text{NiO}}}}+{0.30}_{2}\uparrow}\text{}\\ \text{Step}\left(2\right)\text{}\stackrel{\text{}}{\stackrel{\text{SolidSolution}(s\prime )}{\overbrace{1.2\text{FeO}+0.4{\text{Fe}}_{2}{\text{O}}_{3}+\text{NiO}}}\text{}+0.6{\text{H}}_{2}\text{O}\to \text{}\stackrel{\text{Surface}}{\stackrel{}{\overbrace{\text{Ni}{\text{Fe}}_{2}{\text{O}}_{4}}}}\text{}+{\text{0.6H}}_{2}\text{}}\uparrow \text{}\end{array}$
We note NiFe_{2}O_{4} is regenerated in this process.^{36}
^{36} J. R. Scheffe, J. Li, and A. W. Weimer, “A spinel ferrite/hercynite watersplitting redox cycle,” Int. J. Hydrogen Energy, 35, 3333–3340 (2010).
Derive a rate law for Step (2), assuming that water adsorbs on the solid solution as a singlesite mechanism and that the reaction is irreversible.
Repeat (a) when the reaction is reversible and the solid solution adsorption site for water (S′) is different than the NiFe_{2}O_{4} site for the adsorption of H_{2}, (S).
$\begin{array}{c}\begin{array}{c}{\text{H}}_{2}\text{O}+{\text{S}}^{\prime}\overrightarrow{\leftarrow}{\text{S}}^{\prime}\cdot {\text{H}}_{2}\text{O}\\ {\text{S}}^{\prime}\cdot {\text{H}}_{2}\text{O}\overrightarrow{\leftarrow}\text{}\text{S}\cdot {\text{H}}_{2}+\frac{1}{2}{\text{O}}_{2}\end{array}\\ {\text{H}}_{2}\cdot \text{S}\overrightarrow{\leftarrow \text{}}\text{S}+{\text{H}}_{2}\end{array}$
How would your rate law change if we included the effect of hυ in Step 1?
$\begin{array}{c}\text{S}+\text{h}\upsilon \overrightarrow{\leftarrow}{\text{S}}^{\prime}\cdot {\text{O}}_{2}\\ {\text{S}}^{\prime}\cdot {\text{O}}_{2}\overrightarrow{\leftarrow}{\text{S}}^{\prime}+{\text{O}}_{2}\end{array}$
P1014_{A} Vanadium oxides are of interest for various sensor applications, owing to the sharp metal–insulator transitions they undergo as a function of temperature, pressure, or stress. Vanadium triisopropoxide (VTIPO) was used to grow vanadium oxide films by chemical vapor deposition (J. Electrochem. Soc., 136, 897 (1989)). The deposition rate as a function of VTIPO pressure for two different temperatures follows.
T = 120°C:
Growth Rate (μm/h) 
0.004 
0.015 
0.025 
0.04 
0.068 
0.08 
0.095 
0.1 
VTIPO Pressure (torr) 
0.1 
0.2 
0.3 
0.5 
0.8 
1.0 
1.5 
2.0 
T = 200°C:
Growth Rate (μm/h) 
0.028 
0.45 
1.8 
2.8 
7.2 
VTIPO Pressure (torr) 
0.05 
0.2 
0.4 
0.5 
0.8 
In light of the material presented in this chapter, analyze the data and describe your results. Specify where additional data should be taken.
P1015_{A} OEQ (Old Exam Question). Titanium dioxide is a widebandgap semiconductor that is showing promise as an insulating dielectric in VLSI capacitors and for use in solar cells. Thin films of TiO_{2} are to be prepared by chemical vapor deposition from gaseous titanium tetraisopropoxide (TTIP). The overall reaction is
$\begin{array}{ccc}\text{Ti}{\left(\text{O}{\text{C}}_{3}{\text{H}}_{7}\right)}_{4}& \to & {\text{TiO}}_{2}+4{\text{C}}_{3}{\text{H}}_{6}+2{\text{H}}_{2}\text{O}\\ \text{}& \text{}& \text{}\text{}\\ \text{}& \text{}& \text{}\end{array}$
The reaction mechanism in a CVD reactor is believed to be (K. L. Siefering and G. L. Griffin, J. Electrochem. Soc., 137, 814 (1990))
$\begin{array}{c}\text{}\\ \text{}\end{array}\begin{array}{ccc}\text{TTIP}\left(\text{g}\right)+\text{TTIP}\left(\text{g}\right)& \overrightarrow{\leftarrow}& \text{I}+P\\ \text{I}+\text{S}& \overrightarrow{\leftarrow}& \text{I}\cdot \text{S}\\ \text{I}\cdot \text{S}& \to & \text{}{\text{TiO}}_{2}+{P}_{2}\end{array}$
where I is an active intermediate and P_{1} is one set of reaction products (e.g., H_{2}O, C_{3}H_{6}), and P_{2} is another set. Assuming the homogeneous gasphase reaction for TTIP is in equilibrium, derive a rate law for the deposition of TiO_{2}. The experimental results show that at 200=C the reaction is second order at low partial pressures of TTIP and zero order at high partial pressures, while at 300=C the reaction is second order in TTIP over the entire pressure range. Discuss these results in light of the rate law you derived.
P1016_{B} The dehydrogenation of methylcyclohexane (M) to produce toluene (T) was carried out over a 0.3% Pt/ catalyst in a differential catalytic reactor. The reaction is carried out in the presence of hydrogen (H_{2}) to avoid coking (J. Phys. Chem., 64, 1559 (1960)).
Describe how you would determine the model parameters for each of the following rate laws.
$\begin{array}{cc}\left(1\right){r}_{\text{M}}^{\prime}=k{P}_{\text{M}}^{\alpha}{P}_{{\text{H}}_{2}}^{\beta}& \left(3\right){r}_{\text{M}}^{\prime}=\frac{k{P}_{\text{M}}{P}_{{\text{H}}_{2}}}{{(1+{K}_{\text{M}}{P}_{\text{M}})}^{2}}\\ \left(2\right){r}_{\text{M}}^{\prime}=\frac{k{P}_{\text{M}}}{1+{K}_{\text{M}}{P}_{\text{M}}}& \left(4\right){r}_{\text{M}}^{\prime}=\frac{k{P}_{\text{M}}{P}_{{\text{H}}_{2}}}{1+{K}_{\text{M}}{P}_{\text{M}}+{K}_{{\text{H}}_{2}}{P}_{{\text{H}}_{2}}}\end{array}$
Use the data in Table P1016_{B} below.
Which rate law best describes the data? Hint: Neither ${K}_{{\text{H}}_{2}}$ or K_{M} can take on negative values.
Where would you place additional data points?
Suggest a mechanism and ratelimiting step consistent with the rate law you have chosen.
TABLE P1016_{B} DEHYDROGENATION OF METHYLCYCLOHEXANE
P_{H2} (atm) 
P_{M} (atm) 
${r}_{\text{T}}^{\prime}\left(\frac{\text{moltoluene}}{\text{s}\xb7\text{kgcat}}\right)$ 
1 
1 
1.2 
1.5 
1 
1.25 
0.5 
1 
1.30 
0.5 
0.5 
1.1 
1 
0.25 
0.92 
0.5 
0.1 
0.64 
3 
3 
1.27 
1 
4 
1.28 
3 
2 
1.25 
4 
1 
1.30 
0.5 
0.25 
0.94 
2 
0.05 
0.41 
P1017_{A} OEQ (Old Exam Question). Sketch qualitatively the reactant, product, and activity profiles as a function of length at various times for a packedbed reactor for each of the following cases. In addition, sketch the effluent concentration of A as a function of time. The reaction is a simple isomerization:
A → B
Rate law: ${r}_{\text{A}}^{\prime}=ka{\text{C}}_{\text{A}}$
Decay law: ${r}_{d}={k}_{d}a{\text{C}}_{\text{A}}$
Case I: ${k}_{d}\text{\u226a}k,\text{CaseII:}{k}_{d}=k,\text{CaseIII:}{k}_{d}\text{\u226b}k$
${r}_{\text{A}}^{\prime}=ka{\text{C}}_{\text{A}}\text{and}\text{}{r}_{d}={k}_{d}{a}^{2}$
${r}_{\text{A}}^{\prime}=ka{\text{C}}_{\text{A}}\text{and}\text{}{r}_{d}={k}_{d}a{\text{C}}_{\text{B}}$
Sketch similar profiles for the rate laws in parts (a) and (c) in a movingbed reactor with the solids entering at the same end of the reactor as the reactant.
Repeat part (d) for the case where the solids and the reactant enter at opposite ends.
P1018_{B} The elementary irreversible gasphase catalytic reaction
$\begin{array}{cc}\begin{array}{cc}\text{A}+\text{B}& \stackrel{k}{\begin{array}{c}\to \end{array}}\end{array}& \text{C}+\text{D}\end{array}$
is to be carried out in a movingbed reactor at constant temperature. The reactor contains 5 kg of catalyst. The feed is stoichiometric in A and B. The entering concentration of A is 0.2 mol/dm^{3}. The catalyst decay law is zero order with k_{D} = 0.2 s^{–1} and k = 1.0 dm^{6}/(mol · kgcat ·s) and the volumetric flow rate is ν_{0} = 1 dm^{3}/s.
What conversion will be achieved for a catalyst feed rate of 0.5 kg/s? (Ans: X = 0.2)
Sketch the catalyst activity as a function of catalyst weight (i.e., distance) down the reactor length for a catalyst feed rate of 0.5 kg/s.
What is the maximum conversion that could be achieved (i.e., at an infinite catalyst loading rate)?
What catalyst loading rate is necessary to achieve 40% conversion? (Ans: U_{S} = 1.5 kg/s)
At what catalyst loading rate (kg/s) will the catalyst activity be exactly zero at the exit of the reactor?
What does an activity of zero mean? Can catalyst activity be less than zero?
How would your answer in part (a) change if the catalyst and reactant were fed at opposite ends? Compare with part (a).
Now consider the reaction to be zero order with
k = 0.2 mol/kgcat · min.
The economics:
The product sells for $160 per gram mole.
The cost of operating the bed is $10 per kilogram of catalyst exiting the bed.
What is the feed rate of solids (kg/min) that will give the maximum profit? (Ans: U_{s} = 4 kg/min.) Note: For the purpose of this calculation, ignore all other costs, such as the cost of the reactant, the cost to the company of providing free lunches to workers, and so on.
P1019_{B} OEQ (Old Exam Question). With the increasing demand for xylene in the petrochemical industry, the production of xylene from toluene disproportionation has gained attention in recent years (Ind. Eng. Chem. Res., 26, 1854 (1987)). This reaction,
$\begin{array}{c}\begin{array}{cc}\begin{array}{cc}2\text{Toluene}& \to \end{array}& \text{Benzene}\end{array}+\text{Xylene}\\ \begin{array}{cc}\begin{array}{c}\begin{array}{cc}\begin{array}{c}2\text{T}\end{array}& \stackrel{\text{catalyst}}{\begin{array}{c}\to \end{array}}\end{array}\end{array}& \text{B}+\text{X}\end{array}\end{array}$
was studied over a hydrogen mordenite catalyst that decays with time. As a first approximation, assume that the catalyst follows secondorder decay
${r}_{d}={k}_{d}{a}^{2}$
and the rate law for low conversions is
${r}_{\text{T}}^{\prime}={k}_{\text{T}}{P}_{\text{T}}a$
with k_{T} = 20 g mol/h·kgcat·atm and k_{d} = 1.6 h^{–1} at 735 K.
Compare the conversion–time curves in a batch reactor containing 5 kgcat at different initial partial pressures (1 atm, 10 atm, etc.). The reaction volume containing pure toluene initially is 1 dm^{3} and the temperature is 735 K.
What conversion can be achieved in a movingbed reactor containing 50 kg of catalyst with a catalyst feed rate of 2 kg/h? Toluene is fed at a pressure of 2 atm and a rate of 10 mol/min.
Explore the effect of catalyst feed rate on conversion.
Suppose that E_{T} = 25 kcal/mol and E_{d} = 10 kcal/mol. What would the temperature–time trajectory look like for a CSTR? What if E_{T} = 10 kcal/mol and E_{d} = 25 kcal/mol?
The decay law more closely follows the equation
${r}_{d}={k}_{d}{P}_{\text{T}}^{2}{a}^{2}$
with k_{d} = 0.2 atm^{–1}. Redo parts (b) and (c) for these conditions.
P1020_{A} The vaporphase cracking of gasoil in Example 106 is carried out over a different catalyst, for which the rate law is
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={k}^{\prime}{P}_{\text{A}}^{2}& \text{with}\text{}{k}^{\prime}\end{array}=5\times {10}^{5}\frac{\text{kmol}}{\text{kgcat}\cdot \text{s}\cdot {\text{atm}}^{2}}$
Assuming that you can vary the entering pressure and gas velocity, what operating conditions would you recommend?
What could go wrong with the conditions you chose?
Now assume the decay law is
$\begin{array}{cc}\frac{da}{dt}={k}_{\text{D}}a{\text{C}}_{\text{coke}}& \text{with}{k}_{\text{D}}\end{array}=100\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{S}}\text{at}\text{}{\text{40}\text{0}}^{\circ}\text{C}$
where the concentration, C_{coke}, in mol/dm^{3}, can be determined from a stoichiometric table.
For a temperature of 400=C and a reactor height of 15 m, what gas velocity do you recommend? Explain. What is the corresponding conversion?
The reaction is now to be carried in an STTR 15 m high and 1.5 m in diameter. The gas velocity is 2.5 m/s. You can operate in the temperature range between 100°C and 500°C. What temperature do you choose, and what is the corresponding conversion?
What would the temperature–time trajectory look like for a CSTR?
Additional information:
E_{R} = 3000 cal/mol
E_{D} = 15000 cal/mol
P1021_{C} When the impurity cumene hydroperoxide is present in trace amounts in a cumene feed stream, it can deactivate the silicaalumina catalyst over which cumene is being cracked to form benzene and propylene. The following data were taken at 1 atm and 420°C in a differential reactor. The feed consists of cumene and a trace (0.08 mol %) of cumene hydroperoxide (CHP).
Benzene in Exit Stream (mol %) 
2 
1.62 
1.32 
1.06 
0.85 
0.56 
0.37 
0.24 
t (s) 
0 
50 
100 
150 
200 
300 
400 
500 
Determine the order of decay and the decay constant. (Ans: k_{d} = 4.27 × 10^{–3} s^{–1})
As a first approximation (actually a rather good one), we shall neglect the denominator of the catalytic rate law and consider the reaction to be first order in cumene. Given that the specific reaction rate with respect to cumene is k = 3.8 = 10^{3} mol/kg fresh cat ·s·atm, the molar flow rate of cumene (99.92% cumene, 0.08% CHP) is 200 mol/min, the entering concentration is 0.06 kmol/m^{3}, the catalyst weight is 100 kg, and the velocity of solids is 1.0 kg/min, what conversion of cumene will be achieved in a movingbed reactor?
P1022_{C} The decomposition of spartanol to wulfrene and CO_{2} is often carried out at high temperatures (J. Theor. Exp., 15, 15 (2014)). Consequently, the denominator of the catalytic rate law is easily approximated as unity, and the reaction is first order with an activation energy of 150 kJ/mol. Fortunately, the reaction is irreversible. Unfortunately, the catalyst over which the reaction occurs decays with time on stream. The following conversion–time data were obtained in a differential reactor:
For T = 500 K:
t (days) 
0 
20 
40 
60 
80 
120 
X % 
1 
0.7 
0.56 
0.45 
0.38 
0.29 
For T = 550 K:
t (days) 
0 
5 
10 
15 
20 
30 
40 
X % 
2 
1.2 
0.89 
0.69 
0.57 
0.42 
0.33 
If the initial temperature of the catalyst is 480 K, determine the temperature–time trajectory to maintain a constant conversion.
What is the catalyst lifetime?
P1023_{B} The hydrogenation of ethylbenzene to ethylcyclohexane over a nickel–mordenite catalyst is zero order in both reactants up to an ethylbenzene conversion of 75% (Ind. Eng. Chem. Res., 28 (3), 260 (1989)). At 553 K, k = 5.8 mol ethylbenzene/(dm^{3} of catalyst · h). When a 100ppm thiophene concentration entered the system, the ethylbenzene conversion began to drop.
Time (h) 
0 
1 
2 
4 
6 
8 
12 
Conversion 
0.92 
0.82 
0.75 
0.50 
0.30 
0.21 
0.10 
The reaction was carried out at 3 MPa and a molar ratio of H_{2}/ETB = 10. Discuss the catalyst decay. Be quantitative where possible.
1. A terrific discussion of heterogeneous catalytic mechanisms and ratecontrolling steps may or may not be found in
THORNTON W. BURGESS, The Adventures of Grandfather Frog. New York: Dover Publications, Inc., 1915.
R. I. MASEL, Principles of Adsorption and Reaction on Solid Surfaces. New York: Wiley, 1996. A great reference.
G. A. SOMORJAI, Introduction to Surface Chemistry and Catalysis. New York: Wiley, 1994.
2. A truly excellent discussion of the types and rates of adsorption together with techniques used in measuring catalytic surface areas is presented in
R. I. MASEL, Principles of Adsorption and Reaction on Solid Surfaces. New York: Wiley, 1996.
3. Techniques for discriminating between mechanisms and models can be found in
G. E. P. BOX, J. S. HUNTER, and W. G. HUNTER, Statistics for Experimenters: Design, Innovation, and Discovery, 2nd ed. Hoboken, NJ: Wiley, 2005.
4. Examples of applications of catalytic principles to microelectronic manufacturing can be found in
JOHN B. BUTT, Reaction Kinetics and Reactor Design. Second Edition, Revised and Expanded. New York: Marcel Dekker, Inc., 1999.
D. M. DOBKIN AND M. K. ZURAW, Principles of Chemical Vapor Deposition. The Netherlands: Kluwer Academic Publishers, 2003.