5  Reaction Mechanisms and Mechanistic Rate Expressions

During a non-elementary reaction, the involved bonds do not break and form simultaneously. As a result, intermediate species are formed that are not observed at the macroscopic level. Thus, what appears to be a single reaction event at the macroscopic level actually occurs as two or more elementary reactions at the microscopic level. The set of elementary reaction events that actually takes place at the molecular level is known as the reaction mechanism. Because it appears from a macroscopic perspective that only the non-elementary reaction is taking place, it is desirable to have an expression for the apparent rate of that non-elementary reaction. The focus of this chapter is the generation of apparent rate expressions for non-elementary reactions from proposed mechanisms.

5.1 Reaction Mechanisms

Understanding the mechanism of a non-elementary reaction can be useful in a number of ways. Often reaction mechanisms are studied with the intention of using that knowledge to modify the reactive process. Because the intermediate species in a reaction mechanism are not readily observable using typical analysis instruments, the experimental study of mechanisms is demanding. Reaction Engineering Basics does not touch upon the experimental study of reaction mechanisms. Its focus with respect to mechanisms is on generating apparent rate expressions for non-elementary reactions given either a proposed mechanism or a mechanism that is well-established by experimental study.

The species that participate in a reaction mechanism but do not appear in the reaction expression for the non-elementary reaction are called reactive intermediates. The concentrations of reactive intermediates are always very small. Otherwise they would be observed at the macroscopic level. The reason reactive intermediates have small concentrations is that they react quickly making their lifetimes very short. That’s why they are called reactive intermediates. The identification of reactive intermediates when solving problems involving mechanisms is illustrated in every example at the end of this chapter.

By definition, every reaction in a reaction mechanism is an elementary reaction. Consequently, the rate expression for each mechanistic step will have the functional form shown in Equation 4.16, or for gases, in Equation 4.17. Those equations include the equilibrium constant for the elementary step. Very often in the case of mechanistic steps, thermodynamic data are not available for the reactive intermediates. In those cases, the equilibrium constant for the mechanistic step won’t be known. For this reason, in this chapter the rate expressions for elementary mechanistic steps will be written as shown in Equation 5.1, and the reverse rate coefficients treated as kinetics parameters. The square brackets in that equation can represent concentrations or (if the reagents are gases) partial pressures. When writing the rate expressions in this way, it is important to remember that the forward and reverse rate coefficients must be consistent with thermodynamics. The use of Equation 5.1 when solving problems involving mechanisms also is illustrated in every example at the end of this chapter.

\[ r_j = k_{j,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j}} - k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j}} \tag{5.1}\]

\(\qquad\) where \(j\) denotes a step in a reaction mechanism,
\(\qquad \qquad\) \(i^\prime\) indexes the reactants in step \(j\), and
\(\qquad \qquad\) \(i^{\prime\prime}\) indexes the products in step \(j\).

There are a few rules that all mechanisms must obey. There must be some linear combination of the mechanistic steps that exactly equals the apparent, non-elementary reaction. Being elementary reactions, every step in a reaction mechanism must be reversible. This is required by the principle of microscopic reversibility. In addition, the mechanistic steps must be consistent with all available experimental data including (isotopic effects, isotope distributions in products, spectroscopic measurements, etc.) and not just the reaction kinetics.

Reaction mechanisms are sometimes differentiated as having either an open sequence of steps or a closed sequence. Mechanisms with a closed sequence of steps are commonly referred to as chain reaction mechanisms. The distinguishing feature of a chain reaction mechanism is that it includes elementary steps known as propagation steps. A propagation step is a step wherein one reactive intermediate participates as a reactant and another reactive intermediate is formed as a product. The full set of propagation steps in the mechanism forms a closed sequence where each reactive intermediate is produced in one propagation step and consumed in a different propagation step. In a chain reaction mechanism, the apparent, non-elementary reaction is, in fact, a linear combination of the full set of propagation steps. As a consequence, propagation steps occur many, many more times than the other steps in the mechanism.

HBr synthesis, Equation 5.2, is a non-elementary reaction. The HBr synthesis mechanism, Equations 5.3 through 5.6, is a classic example of a chain reaction mechanism. Notice that reactions 5.4 and 5.5 each consume one reactive intermediate and generate a different reactive intermediate. This makes reactions 5.4 and 5.5 propagation steps. More specifically, reaction 5.4 consumes a Br· reactive intermediate and generates an H· reactive intermediate. Then reaction 5.5 consumes an H· reactive intermediate and generates a Br· reactive intermediate. Adding just these two propagation steps together yields the apparent, non-elementary reaction 5.2.

\[ H_2 + Br_2 \rightleftarrows 2 HBr \tag{5.2}\]

\[ Br_2 \rightleftarrows 2 Br \cdot \tag{5.3}\]

\[ Br \cdot + H_2 \rightleftarrows HBr + H \cdot \tag{5.4}\]

\[ H \cdot + Br_2 \rightleftarrows HBr + Br \cdot \tag{5.5}\]

\[ 2 H \cdot \rightleftarrows H_2 \tag{5.6}\]

While the sum of the two propagation steps, 5.4 and 5.5, equals the apparent non-elementary reaction, the other steps are still needed. This can be seen by noting that if neither reaction 5.3 nor reaction 5.6 ever occurred, then there would not be any reactive intermediates present, and that would mean that the propagation steps could not occur. In theory, either reaction 5.3, in the forward direction, or reaction 5.6, in the reverse direction, would need to occur at least one time in order to get the propagation sequence going. Steps 5.4 and 5.5 could then occur over and over until all of the Br2 or all of the H2 was used up. At that point, reaction 5.3, in the reverse direction, or reaction 5.6, in the forward direction, would need to occur one time to use up the reactive intermediates.

Reactions 5.3 and 5.6 are examples of two additional kinds of steps found in chain reaction mechanisms. These kinds of steps are known as initiation steps and termination steps. An elementary initiation step does not have a reactive intermediate as a reactant, but it generates one or more reactive intermediates as products. Similarly, termination steps consume reactive intermediates without producing any new ones. As noted above, even though initiation and termination steps only need to occur a small number of times compared to the propagation steps they are still a necessary kind of elementary reaction step for a chain reaction mechanism.

It has already been noted that every elementary reaction must be reversible. As such, the definitions just given for initiation and termination steps don’t make much sense because the reverse of any initiation step will be a termination step by definition, and similarly, the reverse of any termination step will be an initiation step. It is perhaps better to refer to them as initiation/termination steps. Usually, when one says that a step is an initiation step, what is meant is that the step in the forward direction, as written, is an initiation step. Commonly, the mechanistic steps corresponding to an apparent, non-elementary reaction will be written so that the reactants in the initiation step are also reactants in the apparent, non-elementary reaction.

Chain transfer steps can also appear in chain reaction mechanisms. These steps are common in free radical polymerization mechanisms. In free radical polymerization, a very long hydrocarbon molecule (the so-called chain) has a free radical at one end. Monomer molecules react with the free radical end of the growing polymer chain, with the net result that the chain is longer by one monomer unit and still has a free radical at the end. In a chain transfer step, a monomer adds to the end of a growing polymer chain, but the free radical transfers to a different molecule. That is, a chain transfer step terminates one growing chain and starts a new one.

Yet another type of step in chain reactions is known as a chain branching step. Chain branching steps must be kept under control; if they are not controlled, they can lead to explosions. In an elementary chain branching step, one reactive intermediate is consumed, but two new reactive intermediates are generated. Clearly if these steps get out of control, the number of reactive intermediates will increase geometrically. This, coupled with the fact that reactive intermediates are so highly reactive, is why explosions can result from uncontrolled chain branching steps.

If it is not possible to identify propagation steps in a reaction mechanism, then the mechanism consists of an open sequence of reaction steps. Open sequence mechanisms are a little less common than closed sequences. The reason has to do with the energy required for a reaction to take place. If, for example, in a chain reaction mechanism, the initiation step requires a significant energy input, but the propagation steps do not, the apparent, non-elementary reaction is able to proceed with relative ease because the initiation step only needs to occur a small number of times. In contrast, if one step in an open sequence of steps requires a significant energy input, that amount of energy will be required every time the apparent, non-elementary reaction takes place.

Reagents participate in a chemical reaction according to a fixed stoichiometry. For example every time one H2 molecule participates in reaction 5.2, one Br2 molecules also participates and two HBr molecules are produced. The propagation steps in a chain reaction mechanism, and all of the steps in an open sequence reaction mechanism, must obey a similar kind of stoichiometry. For example, each time the apparent, non-elementary HBr synthesis reaction occurs, each of the propagation steps, 5.4 and 5.5, must occur one time. The stoichiometric number, \(\sigma_j\), of mechanistic step \(j\) (not to be confused with a stoichiometric coefficient, \(\nu_{i,j}\)) is defined as the number of times mechanistic step \(j\) must occur when the apparent, non-elementary reaction occurs one time. Put differently, in the linear combination of the mechanistic steps that equals the apparent, non-elementary reaction, the coefficient that multiplies each mechanistic step is the stoichiometric number of that step. Thus, the stoichiometric numbers of steps 5.4 and 5.5 are each equal to 1. For some steps in a reaction mechanism, it is not possible to assign a unique stoichiometric number. For example, the initiation and terminations steps above, reactions 5.3 and 5.6, do not have unique stoichiometric numbers.

Identifying the types of reactions in a chain reaction mechanism and ensuring that the mechanism satisfies the rule that there must be some linear combination of the mechanistic steps that exactly equals the apparent, non-elementary reaction is illustrated in Example 5.3.1.

5.2 Generating Rate Expressions from Mechanisms

Once a proposed mechanism has been checked to make sure it obeys the rules given above, it can be used to generate an expression for the apparent rate of generation of a reactant or product in the corresponding non-elementary reaction. Specifically, the apparent net rate of generation of species \(i\) in the overall macroscopically observed reaction, \(j\), will be the sum of its rates of generation in each of the mechanistic steps. That is, the apparent rate of generation of \(i\) via the non-elementary reaction, \(j\), is given by Equation 5.7. Because the steps are elementary, their rates are given by Equation 5.1, leading to Equation 5.8 as the expression for the apparent rate of generation of \(i\) via non-elementary reaction \(j\). In Reaction Engineering Basics, such rate expressions are referred to as mechanistic rate expressions. Examples 5.3.1, 5.3.2, 5.3.4, and 5.3.5 illustrate the use of Equations 5.7 and 5.8 in problems involving the generation of a rate expression from a mechanism.

\[ r_{i,j} =\sum_{j^\prime}\nu_{i,j^\prime}r_{j^\prime} \tag{5.7}\]

\[ r_{i,j} = \sum_{j^\prime}\nu_{i,j^\prime} \left(k_{j^\prime,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j^\prime}} - k_{j^\prime,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j^\prime}} \right) \tag{5.8}\]

\(\qquad\) where \(i\) is a reactant or product in the non-elementary reaction \(j\),
\(\qquad \qquad\) \(j^\prime\) indexes the steps in the mechanism for non-elementary
\(\qquad \qquad \qquad\) reaction \(j\),
\(\qquad \qquad\) \(i^\prime\) indexes the reactants in mechanistic step \(j^\prime\), and
\(\qquad \qquad\) \(i^{\prime\prime}\) indexes the products in mechanistic step \(j^\prime\).

It is important to note that \(i\) in Equations 5.7 and 5.8, can be any reactant or product in the apparent, non-elementary reaction, \(j\). The resulting rates of generation of reagent \(i\) via non-elementary reaction \(j\) will all be stoichiometrically equivalent.

Generally Equation 5.8 leads to a mechanistic rate expression that is not very useful for reaction engineering purposes. Specifically, the resulting mechanistic rate expression will contain several terms, and it will include concentrations of reactive intermediates. The latter are present in a real system in very small concentrations. These concentrations are typically unknown, and they are very difficult to measure accurately. That makes it difficult to use a mechanistic rate expression in its initial form and makes it desirable to simplify it and eliminate concentrations of reactive intermediates from it. Simplification is accomplished by making different assumptions. Of course, the resulting simplified rate expression will only be valid if the assumptions are valid and if the underlying mechanism is correct.

It is sometimes observed that the rate of one particular mechanistic step has virtually no effect upon the apparent rate of the macroscopically observed reaction. Steps of this kind can be referred to as kinetically insignificant steps. This leads to an assumption that can be used to simplify a mechanistic rate expression. If a mechanistic step is assumed to be kinetically insignificant, its rate can be set equal to zero as shown in Equation 5.9. Example 5.3.2 shows the use of Equation 5.9 during the generation of mechanistic rate expression.

\[ r_{j,\text{insig}} = 0 \tag{5.9}\]

Similarly, it is sometimes observed that the rate in the reverse direction of a particular step is very small compared to its forward rate or that the step is thermodynamically able to go essentially to completion. In other words, it is an effectively irreversible step. If a mechanistic step is assumed to be effectively irreversible, the term corresponding to its rate in the reverse direction in Equation 5.1, is set equal to zero as shown in Equation 5.10. Examples 5.3.1, 5.3.4, and 5.3.5 illustrate the use of Equation 5.10.

\[ r_{j,\text{irrev}} = k_{j,f} \prod_{i^\prime} \left[ i^\prime \right]^{-\nu_{i^\prime,j}} \tag{5.10}\]

\(\qquad\) where \(i^\prime\) indexes the reactants in mechanistic step \(j\).

5.2.1 The Bodenstein Steady State Approximation

Bodenstein, M. and Lind, S. C. (1907) suggested an assumption related to reactive intermediates appearing in a reaction mechanism. As noted previously, reactive intermediates will be present at very low concentrations. This is a consequence of their high reactivity. As the reaction begins, the concentration of these species increases, but because they are so reactive, those intermediates formed undergo subsequent reaction very rapidly. In this way a steady state is quickly established whereby the rate at which reactive intermediates are being formed equals the rate at which they are undergoing subsequent reaction. Hence their concentration becomes constant and the overall rate of generation of the reactive intermediates becomes equal to zero. When the concentration of a species does not change over time, that concentration is said to be at steady state. For reactive intermediates, one can assume that this steady state condition always exists, ignoring the brief time required for the concentrations of the reactive intermediates to build up to their steady state values. This is known as the Bodenstein steady state approximation. To apply the steady state approximation, the rate of generation of each reactive intermediate via the non-elementary reaction is set equal to zero, as shown in Equation 5.11. Examples 5.3.1, 5.3.2, 5.3.4, and 5.3.5 illustrate the use the Bodenstein steady state approximation in problems involving the generation of a rate expression from a mechanism.

\[ 0 = \sum_{j^\prime}\nu_{RI,j^\prime}r_{j^\prime} \tag{5.11}\]

\(\qquad\) where \(j^\prime\) indexes the steps in the reaction mechanism.

If there are \(N\) reactive intermediates in the mechanism, then Equation 5.11 can be used once for each intermediate. This results in a set of \(N\) algebraic equations. This set of \(N\) equations can be solved for the concentrations of the \(N\) reactive intermediates in terms of the rate coefficients in the mechanism and the concentrations of stable species such as the reactants and products of the apparent, non-elementary reaction. The resulting expressions for the concentrations of the reactive intermediates can be substituted back into the mechanistic rate expression from Equation 5.8. After doing so, the mechanistic rate expression for the apparent, non-elementary reaction will no longer contain concentrations of reactive intermediates, and it will be much more useful for reaction engineering purposes. This process is illustrated in Examples 5.3.4 and 5.3.5.

5.2.2 Rate Determining Step

Sometimes one step in the mechanism is much more difficult or demanding than any of the other steps. This step introduces a bottleneck in the reaction kinetics; if the rate of this one step were somehow increased, the net rates of all the other steps and the observed overall rate would increase proportionally. This kind of step is referred to as a rate-determining step or a rate-limiting step. In situations where there is a rate-determining step, the rate expression can often be simplified considerably by making the assumption that the apparent rate of the non-elementary reaction, \(r_j\), is equal to the rate of the rate-determining step, \(r_{rds}\), as shown in Equation 5.12. Note that \(r_j\) in Equation 5.12 is the apparent general rate of the non-elementary reaction, \(j\); it is not a species generation rate. It is important to recognize that not all mechanisms have a rate-determining step. The assumption of a rate-determining step during the generation of a rate expression from a mechanism is illustrated in Examples 5.3.3, 5.3.6, and 5.3.7.

\[ r_j = r_{rds} = k_{rds,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,rds}} - k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},rds}} \tag{5.12}\]

\(\qquad\) where \(i^\prime\) indexes the reactants in the rate-determining step and
\(\qquad \qquad\) \(i^{\prime\prime}\) indexes the products in the rate-determining step.

It is often stated that the rate determining step is the slowest step in the mechanism, but this is not an accurate statement. When the steps all have the same stoichiometric number (see above), then every step in the mechanism proceeds at the same net rate. If the rate determining step was much slower than the steps before it, then the concentrations of the reactants of the rate-determining step would continually build up. Eventually, the concentrations of the reactants of the rate-determining step would become sufficiently large to be observed macroscopically. At that point, it would no longer appear from a macroscopic perspective, that only the non-elementary reaction was taking place. It is more accurate to say that the rate-determining step is the kinetic bottleneck or that it is the most demanding step, than to incorrectly state that it is the slowest step. Its rate determines the rates of all the other mechanistic steps.

The concept of one step being more difficult or demanding can be put in more scientific terms. Recall that multiplying each mechanistic step by its stoichiometric number and summing yields the apparent, non-elementary reaction. Since free energy is a state function, the sum of the free energy change for each mechanistic step multiplied by the stoichiometric number of that step is equal to the free energy change for the apparent, non-elementary reaction, as expressed in Equation 5.13. If there is one mechanistic step wherein essentially all of the overall free energy change takes place, that one step is the rate-determining step, Equation 5.14. Since essentially all of the free energy change occurs in the rate-determining step, the free energy change for the other steps is essentially zero, Equation 5.15. Note, again, that there does not have to be a rate-determining step in a reaction mechanism.

\[ \Delta G_j = \sum_{j^\prime} \sigma_{j^\prime} \Delta G_{j^\prime} \tag{5.13}\]

\[ \Delta G_j = \sigma_{rds} \Delta G_{rds} \tag{5.14}\]

\[ \Delta G_{nrd} \approx 0 \tag{5.15}\]

\(\qquad\) where \(j^\prime\) indexes the steps in the mechanism for non-elementary
\(\qquad \qquad \qquad\) reaction \(j\)

If a rate-determining step exists for a particular reaction mechanism, there is another consequence. Since the free energy changes for all steps other than the rate-determining step are essentially zero, it may be assumed that all steps other than the rate-determining step reach a state of quasi-equilibrium. (Recall from thermodynamics that by definition, the free energy change for a process at thermodynamic equilibrium is equal to zero.) This is expressed in Equation 5.16.

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \tag{5.16}\]

Note that the rate expression that results from the application of Equation 5.12 is likely to include the concentration of one or more reactive intermediates, and if it does, it is not suitable for many reaction engineering purposes. In those cases, however, the quasi-equilibration assumption, Equation 5.16, can be applied to all other steps. The resulting equilibrium expressions can be solved for the concentrations of the reactive intermediates in terms of the concentrations of the reactants, products and the equilibrium constants for the steps other than the rate-determining step. Upon substitution into Equation 5.12, a rate expression more suitable for reaction engineering purposes results. The use of the assumption that steps other than the rate-determining step are quasi-equilibrated is illustrated in Examples 5.3.3, 5.3.6, and 5.3.7.

Rate expressions that are derived with the assumption of a rate-determining step will only apply for environmental conditions that are far from equilibrium. Recall that the free energy change due to the rate-determining step is equal to the overall free energy change. However, as the system approaches thermodynamic equilibrium the overall free energy change approaches zero (by definition) and the free energy change for every step approaches zero (according to the principle of microscopic reversibility). Thus, as the system approaches thermodynamic equilibrium it is no longer possible to identify a rate-determining step, and as a consequence, rate expressions that are derived with the assumption of a rate-determining step will only apply for environmental conditions that are far from equilibrium.

5.2.3 Homogeneous Catalytic and Enzymatic Reaction Mechanisms

Recall that a catalyst is a material that causes the rate of one or more chemical reactions to increase, but the catalyst itself is not a reactant or product of the apparent non-elementary reaction. In some cases, the catalyst is present within the same phase as the reacting fluid. In this case the catalyst is referred to as a homogeneous catalyst. For present purposes, an enzyme may be considered to be a homogeneous catalyst for biological reactions, and for the remainder of this sub-section any discussion concerning catalysts also applies to enzymes unless noted otherwise.

In mechanisms for catalytic reactions, the catalyst appears in several different chemical forms. That is, some of the catalyst will be free (not bonded to anything), while some catalyst will be complexed (chemically bound) with reactants, products, or other species. Each of the chemical forms of the catalyst can be treated as a reactive intermediate because it appears in the reaction mechanism, but not in the apparent, non-elementary reaction. The concentrations of the various chemical forms of the catalyst can be difficult to measure, making it desirable to eliminate them from mechanistic rate expressions using either the steady-state approximation or, when there is rate-determining step in the mechanism, quasi-equilibrium assumptions.

The presence of a catalyst leads to an additional complication when simplifying a mechanistic rate expression. When the Bodenstein steady state approximation is applied to each of the chemical forms of the catalyst, it is found that the resulting equations cannot be solved to obtain expressions for the concentrations of the reactive intermediates as is done for non-catalytic mechanisms. The reason is that when the free catalyst and all the complexes it forms are treated as a reactive intermediates, the equations generated using the Bodenstein steady state approximation are not mathematically independent. One of the Bodenstein steady state equations must be replaced.

The equation that replaces one of the Bodenstein steady state equations is an expression for the conservation of catalyst. While the concentrations of the various chemical forms of the catalyst aren’t known, the total amount of catalyst originally added to the system usually is known, and it is a constant. Since catalyst is not generated nor consumed by reaction, the sum of the concentrations of all forms of the catalyst (free, reactant-complexed, intermediate-complexed, and product-complexed) must equal the known total concentration of catalyst. This is expressed in Equation 5.17 where \(\kappa_{i^\prime}\) is the number of catalyst species in the form originally added to the system that are needed to create one complex of the catalyst with species \(i^\prime\). When one of the Bodenstein steady state equations is replaced by Equation 5.17, the resulting set of equations is mathematically independent. They can be solved to obtain expressions for the concentration of each reactive intermediate in terms of only rate coefficients, concentrations of stable species and the total concentration of the catalyst in its initial form, \(C_{0,cat}\). The amount of catalyst originally added to the system, \(C_{0,cat}\), is a known constant, so its presence in the rate expression is acceptable. The use of Equation 5.17 is illustrated in Examples 5.3.4 and 5.3.5.

\[ C_{0,cat} = C_{cat,free} + \sum_{i^\prime}\kappa_{i^\prime}C_{i^\prime} \tag{5.17}\]

\(\qquad\) where \(i^\prime\) indexes all complexed forms of the catalyst.

A similar complication can arise for either catalytic or non-catalytic reactions if the mechanism involves ionic species. In that case, similar to requiring the total amount of catalyst to be conserved, charge conservation must be enforced. If the reacting solution is uncharged, this means that the sum of the amounts of all positively charged species multiplied by their respective charges must equal the sum of the amounts of all negatively charged species multiplied by their respective charges, as expressed in Equation 5.18. That equation can then be used to replace one of the Bodenstein steady state equations, leading to a mathematically independent set of equations.

\[ 0 = \sum_{i+} C_{i+}q_{i+} + \sum_{i-} C_{i-}q_{i-} \tag{5.18}\]

\(\qquad\) where \(i+\) indexes the positively charged species in the system, and
\(\qquad \qquad\) \(i-\) indexes the negatively charged species in the system.

While enzyme catalysis and homogeneous chemical catalysis are the same in many respects, there are some differences. The term “substrate” is typically used instead of “reactant”. In catalytic systems, a reagent that binds with the catalyst and renders it catalytically inactive is called a catalyst poision. The equivalent in an enzymatic reaction is called an enzyme inhibitor. Typically an enzyme is a large molecule that has many twists and folds in its structure, and the catalysis associated with the enzyme takes place when the substrate binds to one particular location within the overall structure. Inhibitors are often molecules that also bind to the particular location where the catalysis takes place. When an inhibitor molecule is bound in this way, the enzyme becomes catalytically inactive until such time that the inhibitor releases from it. In addition to inhibition, some enzymes require cofactor molecules. A cofactor molecule is often a relatively small inorganic molecule that has the effect of activating an enzyme when it binds to it. That is, the enzyme alone is not active, but when a cofactor binds to it, it becomes catalytically active.

In terms of simplifying mechanistic rate expressions, the concentrations of poisons, inhibitors, and cofactors that are not complexed with a catalyst or enzyme is often easily measured. In this case, it is acceptable for reaction engineering purposes for the concentrations of uncomplexed poisons, inhibitors, and cofactors to appear in the rate expression for the apparent, non-elementary reaction. If, for some reason, the presence of these concentrations in the rate expression was undesirable, the poision, inhibitor, or cofactor could be treated as a conserved species. Doing so would lead to their total concentration appearing in the rate expression instead of the concentration of the uncomplexed form. In any case, it is not acceptable for the concentration of a poison, inhibitor or cofactor that is complexed to the catalyst to appear in the rate expression for the non-elementary reaction. Generation of a mechanistic rate expression when the mechanism includes an inhibitor is illustrated in Example 5.3.5.

5.2.4 Heterogeneous Catalytic Reaction Mechanisms

Many industrial processes utilize heterogeneous catalysts; they are a separate phase that is in contact with the reacting fluid. Most typically the catalyst is solid while the reagents are in a gaseous or liquid phase. The reaction actually takes place on the surface of the catalyst at specific locations called active sites. When writing a mechanistic step that involves an active site, it is common to use some type of star to represent the site, e. g. \(\ast\). Reactants, intermediates and products can bond to these sites in much the same way as they can form complexes with homogeneous catalysts or enzymes. In heterogeneous catalysis, the bonding of a species in the reacting fluid phase to an active site on the surface of the catalyst is a process referred to as adsorption, and the surface complexes that are generated are called adsorbed species. The reverse process, where a species leaves the surface and enters the fluid phase is called desorption.

In simple situations, it is usually safe to assume that the total surface area of the heterogeneous catalyst remains constant during the course of the reaction. It can also be safely assumed that the number (and therefore the concentration) of active sites, \(C_{\text{sites}}\), remains constant. Note that this is a two-dimensional concentration (sites per surface area) not a three-dimensional one (sites per volume). In heterogeneous catalytic kinetics, the concentration of an adsorbed species, \(k\) is usually expressed in terms of the fraction of the total sites that it covers, \(\theta_k\). This quantity is then referred to as the fractional coverage or the surface coverage of surface species \(k\). When this is done, the concentration, \(C_k\) of any surface species adsorbed on a single site is given by Equation 5.19.

\[ C_k = C_{\text{sites}} \theta_k \tag{5.19}\]

Transition state theory can be used to derive the mathematical form of rate expressions for an elementary surface reaction. When this is done, the concentration of sites, \(C_{\text{sites}}\), can be combined with the pre-exponential factor of the rate coefficient. After doing so, the resulting rate expression for an elementary surface reaction is given in Equation 5.20.

\[ \begin{align} r_j &= k_{j,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,j}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,j}} \\ &- k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},j} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},j}} \end{align} \tag{5.20}\]

\(\qquad\) where \(i^\prime\) indexes the fliud phase reactants in step \(j\),
\(\qquad \qquad\) \(k^\prime\) indexes the surface reactants in step \(j\),
\(\qquad \qquad\) \(i^{\prime\prime}\) indexes the fluid phase products in step \(j\), and
\(\qquad \qquad\) \(k^{\prime\prime}\) indexes the surface products in step \(j\).

Note that because the rate expression for a heterogeneous catalytic mechanistic step is different than those for non-catalytic, catalytic and enzymatic mechanistic steps, if there is a rate-determining step (rds) in a heterogeneous catalytic reaction mechanism, Equation 5.21 should be used in place of Equation 5.12 for the apparent rate of the non-elementary reaction \(j\). Also, when equilibrium expressions are written for surface reactions, the fractional coverage of surface species replaces their concentration.

\[ \begin{align} r_j &= r_{rds} \\ &= k_{rds,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,rds}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,rds}} \\ &- k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},rds} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},rds}} \end{align} \tag{5.21}\]

\(\qquad\) where \(i^\prime\) indexes the fliud phase reactants in the rds,
\(\qquad \qquad\) \(k^\prime\) indexes the surface reactants in the rds,
\(\qquad \qquad\) \(i^{\prime\prime}\) indexes the fluid phase products in the rds, and
\(\qquad \qquad\) \(k^{\prime\prime}\) indexes the surface products in the rds.

For reaction engineering purposes, it is not desirable to have a rate expression that contains surface coverages because they are typically unknown and difficult to measure. Therefore, the mechanistic rate expression is simplified using the Bodenstein steady state approximation or other valid assumptions. Similar to homogeneous and enzymatic catalysis, the heterogeneous catalytic sites are conserved, and one Bodenstein steady state equation must be replaced by an expression for the conservation of catalytic sites. In the case of heterogeneous catalysis, the conservation of catalytic sites takes the form shown in Equation 5.22.

\[ 1 = \theta_{\text{vacant}} + \sum_k \theta_k \tag{5.22}\]

The generation of a mechanistic rate expression from a heterogeneous catalytic reaction mechanism is illustrated in Examples 5.3.6 and 5.3.7.

5.2.5 Most Abundant Intermediate

Catalytic reaction mechanisms (homogeneous, enzymatic or heterogeneous) are subtly different from non-catalytic reaction mechanisms. During non-catalytic reactions, the amount of every reactive intermediate is small. This makes their concentration difficult to measure, leading to the necessity to eliminate those concentrations from rate expressions for the non-elementary reaction. During catalytic reactions the total number of reactive intermediates is not necessarily small, it is determined by the amount of catalyst present. As a result, some of the intermediates in catalytic reactions will be present in significant amounts. Nonetheless, measuring the concentrations of those intermediates is still very difficult, so it is common to eliminate their concentrations from mechanistic rate expressions. That is, even though their concentration may be greater than reactive intermediates in non-catalytic reactions, the non-complexed catalyst and each complexed form of a catalyst or enzyme is treated as a reactive intermediate. This also applies to vacant sites and each adsorbed surface species in a heterogeneous catalytic reaction mechanism.

As a consequence of the larger concentrations of catalytic intermediates, it is possible that during reaction, most of the catalyst, enzyme or catalyst sites are complexed with one reagent. If so, that complex is called the most abundant intermediate. Mathematically, when there is a most abundant intermediate, Equation 5.23 applies for a homogeneous catalytic or enzymatic mechanism and Equation 5.24 applies for a heterogeneous catalytic reaction mechanism. These inequalities can sometimes lead to additional simplification of the apparent rate expression for the non-elementary reaction. This happens when the apparent rate expression has a denominator with several terms. When there is a most-abundant intermediate, some of the terms in these denominators may be much smaller than others and so can be deleted without affecting the accuracy of the rate expression. The simplification of a mechanistic rate expression when there is a most abundant intermediate in the mechanism is illustrated in Example 5.3.7.

\[ C_{ma} \gg C_{nma} \tag{5.23}\]

\[ \theta_{ma} \gg \theta_{nma} \tag{5.24}\]

5.3 Examples

All of the examples included in Chapter 5 can be classified as “mechanism problems.” Each example begins with a problem statement. Immediately following the problem statement there is a green “expert thinking” callout. like the one at the end of this paragraph. By default the content of these callouts is hidden. Clicking on the callout opens it and shows its content. The intention of these callouts is to provide some insight into what an expert might be thinking after reading the problem statement as they ponder how to solve the problem. To reduce the amount of duplication, the callout below presents general expert thinking that applies to every example in Chapter 5. The individual expert thinking callouts in each of the examples then provide thinking specific to that problem. In other words, the reader should mentally insert the content of this callout at the start of all of the remaining expert thinking callouts in this chapter.

This is what I call a “mechanism problem.” Its characteristic features are that it identifies a non-elementary reaction, provides a mechanism for that reaction, and asks for an expression for the apparent rate of the non-elementary reaction. I’ve learned that solving this type of problem has two parts. The first part is to express the apparent rate of the non-elementary reaction in terms of the rates of the steps in the mechanism. If there is a rate-determining step in the mechanism and there is not a heterogeneous catalyst, Equation 5.12 is used to do this. If there is a rate-determining step in the mechanism and there is a heterogeneous catalyst, Equation 5.21 is used. If there is not a rate-determining step, Equation 5.7 is used.

The second part involves eliminating concentrations of reactive intermediates, concentrations of catalyst complexes or heterogeneous catalyst surface coverages from the apparent rate expression. (Note in the present context, “catalyst complexes” includes the free catalyst as well as each of its complexed forms, and “surface species” includes the vacant site as well as each adsorbed species adsorbed on the surface.)

If (a) there is a rate-determining step in the mechanism, this is done by writing Equation 5.16 (quasi-equilibrium expressions) for all the other steps. If there is a homogeneous catalyst or an enzyme in the mechanism, Equation 5.17 is added to the set of quasi-equilibrium expressions. Alternatively, if there is a heterogeneous catalyst in the mechanism, Equation 5.22 is added to the set of quasi-equilibrium expressions.

If (b) there is not a rate-determining step, Equation 5.11 (the Bodenstein steady state approximation) is used to eliminate the concentrations of reactive intermediates, catalyst complexes or surface species coverages from the apparent rate expression. If the mechanism does not involve any type of catalyst or enzyme, Equation 5.11 is written for each reactive intermediate appearing in the reaction mechanism. If there is a homogeneous catalyst or an enzyme in the mechanism, Equation 5.11 is written for all but one catalyst complex and Equation 5.17 (the catalyst conservation expression) is added to the set of Bodenstein steady state expressions. If there is a heterogeneous catalyst in the mechanism, Equation 5.11 is written for all but one surface species and Equation 5.22 (the surface site conservation expression) is added to the set of Bodenstein steady state approximations.

In either case (a) or case (b), the resulting set of equations is then solved to obtain expressions for the concentrations of the reactive intermediates, the concentrations of the catalyst complexes, or the heterogeneous catalyst surface coverages. Then the results can be used to eliminate the concentrations of the reactive intermediates, catalyst complexes or surface coverages from the apparent rate expression.

Following the expert thinking callout in each example, a concise formulation of the problem solution is presented. The formulations contain multiple instances of “where did that come from” callouts. By default, the contents of these callouts is also hidden. These callouts provide added detail about how the equations they follow were generated. The idea is that with the content of all of the callouts hidden, the reader can see the solution in a very concise format, but if the reader does not perceive where certain equations came from, they can open the corresponding callout to see more details.

5.3.1 Validity, Classification of Steps and Mechanistic Rate Expression for a Chain Reaction Mechanism

Suppose that nitrogen oxidation, equation (1), occurs via the chain reaction mechanism given in equations (2) through (5). Classify each of the mechanistic steps as initiation/termination, propagation, chain branching or chain transfer; show that there is a linear combination of equations (2) through (5) that is equal to equation (1); and write, but do not simplify, the expression for the apparent rate of non-elementary reaction (1) that is predicted by the mechanism. Discuss the suitability of the resulting rate expression for reaction engineering purposes.

\[ N_2 + O_2 \rightleftarrows 2 NO \tag{1} \]

\[ O_2 \rightleftarrows 2 O \cdot \tag{2} \]

\[ O \cdot + N_2 \rightleftarrows NO + N \cdot \tag{3} \]

\[ N \cdot + O_2 \rightleftarrows NO + O \cdot \tag{4} \]

\[ 2 N \cdot \rightleftarrows N_2\tag{5} \]

This is a mechanism problem. The problem asks me to classify each step in the mechanism. I can do that knowing that initiation steps have no reactive intermediates as reactants and one or more reactive intermediates as products, while termination steps have one or more reactive intermediates as reactants and no reactive intermediates as products. Propagation steps have one reactive intermediate among the reactants and a different reactive intermediate among the products. Chain branching steps have one reactive intermediate among the reactants and two reactive intermediates among the products. Chain transfer steps are usually found in polymerization reactions, so I don’t expect to find any chain transfer steps in this mechanism.

The problem next asks me to show that there is a linear combination of the steps that equals the apparent, non-elementary reaction. I know that in chain mechanisms that linear combination usually involves only the propagation steps. For simple mechanisms like those considered in Reaction Engineering Basics, the necessary linear combination can usually be found by inspection.

Finally, the problem asks me to write the generalized rate expression predicted by the mechanism. This mechanism does not have a rate-determining step, so I’ll set the rate of generation of a reactant or product in the non-elementary reaction equal to the sum of its rates of generation in the mechanistic steps. In order to do that I need to identify the reactive intermediates. The reactive intermediates are the reagents that appear in one or more mechanistic steps, but they do not appear in the apparent, non-elementary reaction. Here I can see that N2, O2, and NO appear in the apparent, non-elementary reaction, so they are not reactive intermediates. The only other reagents that I see in the mechanism (reactions (2) through (5)) are N∙ and O∙, so they are the reactive intermediates.

Initiation/Termination Steps: Steps (2) and (5)

Propagation Steps: Steps (3) and (4)

By inspection it can be seen that adding steps (3) and (4) yields the apparent, non-elementary reaction, reaction (1).

Reactive Intermediates: \(N \cdot\), \(O \cdot\)

Apparent Rate Expression

Setting the apparent rate of generation of NO in the non-elementary reaction equal to the sum of its rates of generation in each of the mechanistic steps yields equation (6).

\[ r_{NO,1} = r_3 + r_4 \tag{6} \]

Start from Equation 5.7.

\[ r_{i,j} =\sum_{j^\prime}\nu_{i,j^\prime}r_{j^\prime} \]

I’ve chosen to generate an expression for the apparent generation of NO in the non-elementary reaction (1), so \(i = NO\) and \(j = 1\). In Equation 5.7, \(j^\prime\) indexes all of the steps in the mechanism (reactions (2) through (5)), so here \(j^\prime\) includes 2, 3, 4, and 5.

\[ r_{NO,1} = \nu_{NO,2}r_2 + \nu_{NO,3}r_3 + \nu_{NO,4}r_4 + \nu_{NO,5}r_5 \]

\[ r_{NO,1} = \left(0\right)r_2 + \left(1\right)r_3 + \left(1\right)r_4 + \left(0\right)r_5 \]

\[ r_{NO,1} = r_3 + r_4 \]

The rate expressions for mechanistic steps (3) and (4) are given in equations (7) and (8).

\[ r_3 = k_{3,f} \left[ O \cdot \right] \left[ N_2 \right] - k_{3,r} \left[ NO \right] \left[ N \cdot \right] \tag{7} \]

\[ r_4 = k_{4,f} \left[ N \cdot \right] \left[ O_2 \right] - k_{4,r} \left[ NO \right] \left[ O \cdot \right] \tag{8} \]

Mechanism steps are elementary reactions, so start from Equation 5.1.

\[ r_j = k_{j,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j}} - k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j}} \]

To generate the rate expression for reaction (3), \(j = 3\). In this equation, \(i^\prime\) indexes all of the reactants in the reaction and \(i^{\prime\prime}\) indexes all of the products in the reaction, so here \(i\) includes O∙ and N2 and \(i^\prime\) includes NO and N∙.

\[ r_3 = k_{3,f} \left[ O \cdot \right]^{-\nu_{O \cdot,3}}\left[ N_2 \cdot \right]^{-\nu_{N_2,3}} - k_{3,r}\left[ NO \right]^{\nu_{NO,3}}\left[ N \cdot \right]^{\nu_{N \cdot,3}} \]

\[ r_3 = k_{3,f} \left[ O \cdot \right]^{-(-1)}\left[ N_2 \cdot \right]^{-(-1)} - k_{3,r}\left[ NO \right]^{1}\left[ N \cdot \right]^{1} \]

\[ r_3 = k_{3,f} \left[ O \cdot \right] \left[ N_2 \right] - k_{3,r} \left[ NO \right] \left[ N \cdot \right] \tag{7} \]

The rate expression for step (4) is generated analogously.

The problem asks for apparent rate of reaction (1), not the apparent rate of generation of NO via reaction (1). It can be found using equation (9).

\[ r_1 = \frac{r_{NO,1}}{2} \tag{9} \]

Start from Equation 4.2.

\[ r_{i,j} = \nu_{i,j} r_j \]

Here \(i = NO\) and \(j = 1\).

\[ r_{NO,1} = \nu_{NO,1} r_1 \]

\[ r_{NO,1} = (2) r_1 \qquad \Rightarrow \qquad r_1 = \frac{r_{NO,1}}{2} \]

Calculations

  1. Substitute equation (6) into equation (9).
  2. Substitute equations (7) and (8) into the resulting equation.

Final Answer and Discussion

\[ \begin{align} r_1 =&\, 0.5k_{3,f} \left[ O \cdot \right] \left[ N_2 \right] - 0.5k_{3,r} \left[ NO \right] \left[ N \cdot \right] \\ & + 0.5k_{4,f} \left[ N \cdot \right] \left[ O_2 \right] - 0.5k_{3,r} \left[ NO \right] \left[ O \cdot \right] \end{align} \tag{10} \]

It will be seen in later chapters of Reaction Engineering Basics, that a reaction engineer will commonly use the rate expression by substituting it into mole and energy balances for a reactor (see Appendix H). When writing those mole and energy balances, the reaction engineer will not know the reaction mechanism; the reaction engineer will only know that reaction (1) takes place and that equation (10) is the rate expression for reaction (1).

The mole and energy balances will include variables for the amounts of N2, O2, and NO, but they will not include variables for the amounts of N∙ and O∙. Furthermore, there is no way to calculate the concentrations of N∙ and O∙ because they are not stoichiometrically related to the concentrations of N2, O2, and NO. Without concentrations of N∙ and O∙, it is not possible to evaluate the reaction rate, and consequently the reaction engineer will not be able to solve the mole and energy balances on the reactor.

For this reason, the initial form of rate expressions generated from mechanisms is not useful for reaction engineering purposes. In order to perform reaction engineering tasks, the rate expression must be simplified so that it does not contain concentrations of reactive intermediates.

Note

The reaction engineer could use an alternative approach if the mechanism was known. Instead of assuming that only the apparent, non-elementary reaction takes place, the reaction engineer could assume that reactions (2) through (5) occur in the reactor. In that case, the mole and energy balances would include variables for the amounts of the reactive intermediates and it would be possible to evaluate the rates of the reactions.

However, that approach introduces a different problem. If the mechanistic reactions are used, the rate expressions for each of the mechanistic steps are needed. That means, for example, that experiments would need to be performed to find the values of \(k_{3,f}\), and \(k_{3,r}\) in equation (7). Performing those experiments would require setting and/or measuring the concentrations of N∙ and O∙. Because N∙ and O∙ are highly reactive, setting and measuring their concentrations is very difficult. In other words, the difficult and demanding task of estimating the values of the kinetics parameters in the mechanistic rate expressions would be necessary if the reaction engineer opted for this alternative approach.

5.3.2 Mechanistic Rate Expression for HBr Synthesis using the Bodenstein Steady State Approximation

In the mechanism for synthesis of HBr given below, assume step (4) is effectively irreversible and step (5) is kinetically insignificant. Use the Bodenstein steady-state approximation to derive a mechanistic rate expression for the apparent rate of generation of HBr via non-elementary reaction (1). Simplify the mechanism so that it contains only partial pressures of stable species and not those of reactive intermediates. Comment on estimating the kinetics parameters in the resulting rate expression.

Overall Reaction:

\[ H_2 + Br_2 \rightleftarrows 2 HBr \tag{1} \]

Proposed Mechanism:

\[ Br_2 \rightleftarrows 2 Br \cdot \tag{2} \]

\[ Br \cdot + H_2 \rightleftarrows HBr + H \cdot \tag{3} \]

\[ H \cdot + Br_2 \rightleftarrows HBr + Br \cdot \tag{4} \]

\[ 2 H \cdot \rightleftarrows H_2 \tag{5} \]

I always like to begin mechanism problems by identifying the reactive intermediates. I see that H2, Br2, and HBr all appear in the apparent, non-elementary reaction. Looking at the mechanism proposed in reactions (2) through (5), I see that H∙ and Br∙ also appear, so they are the reactive intermediates.

This mechanism does not include a rate-determining step, so I will use Equation 5.7 to write the rate of generation of a reactant or product in the non-elementary reaction in terms of the rates of the mechanistic steps. The problem asks for an expression for the rate of generation of HBr, so I’ll write the rate expression for HBr.

I know that when the mechanism does not have a rate-determining step, the Bodenstein steady state approximation, Equation 5.11, must be used to eliminate concentrations of reactive intermediates from the rate expression for the non-elementary reaction.

Reactive Intermediates: H∙ and Br∙

Necessary Equations

The apparent rate of generation of HBr via the non-elementary reaction (1) is the sum of its rates of generation in the mechanistic steps, equation (6)

\[ r_{HBr,1} = r_3 + r_4 \tag{6} \]

Since there isn’t a rate-determining step, start with Equation 5.7.

\[ r_{i,j} =\sum_{j^\prime}\nu_{i,j^\prime}r_{j^\prime} \]

I want an expression for the apparent rate of generation of HBr by reaction (1), so \(i = HBr\) and \(j = 1\). In Equation 5.7, \(j^\prime\) indexes all of the reaction steps in the mechanism, so here \(j^\prime\) includes 2, 3, 4, and 5.

\[ r_{HBr,1} = \nu_{HBr,2}r_2 + \nu_{HBr,3}r_3 + \nu_{HBr,4}r_4 + \nu_{HBr,5}r_5 \]

\[ r_{HBr,1} = \left(0\right)r_2 + \left(1\right)r_3 + \left(1\right)r_4 + \left(0\right)r_5 \]

\[ r_{HBr,1} = r_3 + r_4 \tag{10} \]

The mechanistic steps are elementary reactions, so their rate expressions, equations (7) through (10), are known from theory.

\[ r_2 = k_{2,f} \left[ Br_2 \right] - k_{2,r} \left[ Br \cdot \right]^2 \tag{7} \]

\[ r_3 = k_{3,f} \left[ Br \cdot \right] \left[ H_2 \right] - k_{3,r} \left[ HBr \right] \left[ H \cdot \right] \tag{8} \]

\[ r_4 = k_{4,f} \left[ H \cdot \right] \left[ Br_2 \right] \tag{9} \]

\[ r_5 = 0 \tag{10} \]

To generate an expression for the rate of step (2), start with Equation 5.1.

\[ r_j = k_{j,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j}} - k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j}} \]

In the case of step 2, \(j = 2\). In Equation 5.1, \(i^\prime\) indexes all of the reactants in the reaction and \(i^{\prime\prime}\) indexes all of the products in the reaction, so here \(i^\prime\) includes only Br2 and \(i^{\prime\prime}\) includes only Br∙.

\[ r_2 = k_{2,f} \left[ Br_2 \right]^{-\nu_{Br_2,2}} - k_{2,r}\left[ Br \cdot \right]^{\nu_{Br \cdot,2}} \]

\[ r_2 = k_{2,f} \left[ Br_2 \right]^{-(-1)} - k_{2,r}\left[ Br \cdot \right]^{2} \]

\[ r_2 = k_{2,f} \left[ Br_2 \right] - k_{2,r} \left[ Br \cdot \right]^2 \]

The rate expressions for steps (3) and (4) are generated analogously except the second term in the expression for the rate of step (4), which is the rate in the reverse direction, is set equal to zero because the problem says that step (4) is effectively irreversible.

The rate of step (5) is set equal to zero because the problem states that it is kinetically insignificant.

The Bodenstein steady state approximation can be written for each of the reactive intermediates.

\[ 0 = 2r_2 - r_3 + r_4 \tag{11} \]

\[ 0 = r_3 - r_4 - 2r_5 \tag{12} \]

In the case of Br∙, start with Equation 5.11.

\[ 0 = \sum_{j^\prime}\nu_{RI,j^\prime}r_{j^\prime} \]

In this equation, \(j^\prime\) indexes all steps in the mechanism and RI denotes the reactive intermediate, so here, \(j^\prime\) includes 2, 3, 4, and 5, and RI = Br∙.

\[ 0 = \nu_{Br \cdot,2}r_2 + \nu_{Br \cdot,3}r_3 + \nu_{Br \cdot,4}r_4 + \nu_{Br \cdot,5}r_5 \]

\[ 0 = \left( 2 \right) r_2 + \left( -1 \right)r_3 + \left( 1 \right)r_4 + \left( 0 \right)r_5 \]

\[ 0 = 2r_2 - r_3 + r_4 \tag{11} \]

The Bodenstein steady state approximation for H∙ is generated analogously.

After substitution of equations (6) through (9) into these equations, they can be solved to obtain expressions for the concentrations of H∙ and Br∙. Those expressions can then be used to eliminate the reactive intermediate concentrations from the rate expression for the non-elementary reaction.

Calculations

  1. Substitute equations (7) through (10) to generate new versions of equations (6), (11), and (12).
  2. Simultaneously solve equations (11) and (12) to obtain expressions for [H∙] and [Br∙].
  3. Substitute the results into equation (6).

Results, Final Answer and Comments

Solving the Bodenstein steady state equations yields equations (13) and (14) for the concentrations of H∙ and Br∙. Substitution of those results into the rate expression yields a mechanistic rate expression for the apparent rate of generation of HBr via non-elementary reaction (1), as shown in equation (15).

\[ \left[ Br \cdot \right] = \sqrt{\frac{k_{2,f}}{k_{2,r}}} \sqrt{\left[ Br_2 \right]} \tag{13} \]

\[ \left[ H \cdot \right] = \frac{k_{3,f}\sqrt{\frac{k_{2,f}}{k_{2,r}}} \sqrt{\left[ Br_2 \right]}\left[ H_2\right]}{k_{3,r}\left[ HBr\right] + k_{4,f}\left[ Br_2 \right]} \tag{14} \]

\[ r_{HBr,1} = 2\frac{k_{3,f}k_{4,f}\sqrt{\frac{k_{2,f}}{k_{2,r}}} \left[ Br_2 \right]^{3/2}\left[ H_2\right]}{k_{3,r}\left[ HBr\right] + k_{4,f}\left[ Br_2 \right]} \tag{15} \]

Before the rate expression in equation (15) could be used, it would be necessary to estimate the values of the kinetics parameters appearing in it. Using experimental data where the concentrations or partial pressures of H2, Br2 and HBr were varied and the corresponding values of the rate were measured, it would not be possible to obtain unique values for each of the five rate coefficients. This can be seen most easily by dividing the numerator and denominator of equation (15) by \(k_{4,f}\) as shown in equation (16).

\[ r_{HBr,1} = 2\frac{k_{3,f}\sqrt{\frac{k_{2,f}}{k_{2,r}}} \left[ Br_2 \right]^{3/2}\left[ H_2\right]}{\frac{k_{3,r}}{k_{4,f}}\left[ HBr\right] + \left[ Br_2 \right]} \tag{16} \]

Notice that \(k_{2,f}\), \(k_{2,r}\), and \(k_{3,f}\) do not appear individually in the rate expression. They only appear together as the term, \(k\), defined in equation (17). Similarly, \(k_{3,r}\) and \(k_{4,f}\) do not appear individually, but only together as the term, \(k^\prime\), defined in equation (18). Consequently, the rate expression can be written as shown in equation (19).

\[ k = k_{3,f}\sqrt{\frac{k_{2,f}}{k_{2,r}}} \tag{17} \]

\[ k^\prime = \frac{k_{3,r}}{k_{4,f}} \tag{18} \]

\[ r_{HBr,1} = 2\frac{k \left[ Br_2 \right]^{3/2}\left[ H_2\right]}{k^\prime\left[ HBr\right] + \left[ Br_2 \right]} \tag{19} \]

Statistical parameter estimation methods can be employed to find unique “best” values for \(k\) and \(k^\prime\). Looking at equation (17), it can be seen that arbitrary values can be selected for any two of the three rate coefficients, \(k_{2,f}\), \(k_{2,r}\), and \(k_{3,f}\). As long as the value of the third rate coefficient was chosen so that \(k\) was equal to its “best” value, the rate expression in equation (16) would be equally accurate. Thus, \(k_{2,f}\), \(k_{2,r}\), and \(k_{3,f}\) are coupled. An infinite number of sets of values could be used and the accuracy of the rate expression would not change. The rate coefficients, \(k_{3,r}\) and \(k_{4,f}\), are similarly coupled. Whenever possible, coupled kinetics parameters in mechanistic rate expressions should be combined into a single parameter for which a unique value can be estimated. Put differently, for reaction engineering purposes, rate expression (19) is preferred over rate expression (16).

Note

The coupled rate coefficients in this problem appeared in a term where they were multiplied, divided and raised to a constant power (here, the power was 1/2 corresponding to the square root). When rate coefficients are combined in this way, the combined parameters, \(k\) and \(k^\prime\) in this case, will display Arrhenius temperature dependence and long as the original rate coefficients did so.

If rate coefficients are added or subtracted, this will not be true. That is, if \(k = k_1 + k_2\), the combined rate coefficient \(k\) will not display Arrhenius temperature dependence, even if \(k_1\) and \(k_2\) each do display Arrhenius temperature dependence.

5.3.3 Mechanistic Rate Expression from a Mechanism with a Rate-Determining Step

Suppose that iodopropane disproportionates to produce iodine according to the apparent non-elementary reaction (1). By separately assuming each of the three mechanistic steps proposed in the mechanism, equations (2) through (4), to be rate-determining, generate three possible mechanistic rate expressions for reaction (1) that do not contain concentrations of reactive intermediates.

\[ 2 C_3H_5I \rightleftarrows C_6H_{10} + I_2 \tag{1} \]

\[ C_3H_5I \rightleftarrows C_3H_5 \cdot + I \cdot \tag{2} \]

\[ C_3H_5I + I \cdot \rightleftarrows C_3H_5 \cdot + I_2 \tag{3} \]

\[ 2C_3H_5 \cdot \rightleftarrows C_6H_{10} \tag{4} \]

This is a mechanism problem, and in order to begin I need to identify the reactive intermediates. The reactive intermediates are the reagents that appear in the mechanism, but not in the non-elementary reaction. Here I see that C3H5I, C6H10 and I2 appear in the apparent, non-elementary reaction (1). Looking at the mechanism, I see that it additionally includes C3H5∙ and I∙, so they are reactive intermediates.

I know that when there is a rate-determining step in the mechanism, the apparent rate of the non-elementary reaction is equal to the rate of the rate-determining step, Equation 5.12.

I also know that when one step is rate-determining, quasi-equilibrium expressions, Equation 5.16, can be written for the other steps and used to eliminate the concentrations of the reactive intermediates from the rate expression.

This problem asks me to derive three apparent rate expressions, so I’ll have to repeat the process three times.

Reactive Intermediates: \(C_3H_5 \cdot\) and \(I \cdot\)

Necessary Equations

For the case where reaction (2) is the rate-determining step, the apparent rate of the non-elementary reaction, equation (5), is equal to the rate of reaction (2), because reaction (2) is the rate-determining step.

\[ r_1 = r_2 = k_{2,f} \left[ C_3H_5I \right] - k_{2,r} \left[ C_3H_5 \cdot \right] \left[ I \cdot \right] \tag{5} \]

Since there is a rate-determining step, start with Equation 5.12.

\[ r_j = r_{rds} = k_{rds,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,rds}} - k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},rds}} \]

In this expression \(j\) is the non-elementary reaction (in this problem, reaction (1)), \(rds\) denotes the rate-determining step (here, step (2)), \(i^\prime\) indexes the reactants in the rate-determining step (here only C3H5I) and \(i^{\prime \prime}\) indexes the products in the rate determining step (here C3H5∙ and I∙).

\[ r_1 = r_{2} = k_{2,f} \left[ C_3H_5I \right]^{-\nu_{C_3H_5I,2}} - k_{2,r}\left[ C_3H_5 \cdot \right]^{\nu_{C_3H_5 \cdot,2}}\left[ I \cdot \right]^{\nu_{I \cdot,2}} \]

\[ r_1 = r_{2} = k_{2,f} \left[ C_3H_5I \right]^{-(-1)} - k_{2,r}\left[ C_3H_5 \cdot \right]^1\left[ I \cdot \right]^1 \]

\[ r_1 = r_2 = k_{2,f} \left[ C_3H_5I \right] - k_{2,r} \left[ C_3H_5 \cdot \right] \left[ I \cdot \right] \]

The expressions for the apparent rate of reaction for the other two cases (where step (3) is rate-determining and where step (4) is rate-determining) are generated analogously.

The mechanistic steps other than the rate-determining step can be assumed to be quasi-equilibrated. This gives the equilibrium expressions in equations (6) and (7).

\[ K_3 = \frac{\left[ C_3H_5 \cdot \right] \left[ I_2 \right]}{\left[ C_3H_5I \right] \left[ I \cdot \right]} \tag{6} \]

\[ K_4 = \frac{\left[ C_6H_{10} \right]}{\left[ C_3H_5 \cdot \right]^2} \tag{7} \]

For reaction (3), start with Equation 5.16.

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \]

In this problem, \(nrd\) denotes a step other than the rate-determining step, in this case, step (3), and \(i\) indexes all reagents present in the system, in this case, C3H5I, C6H10, I2, C3H5∙, and I∙.

\[ K_3 = \left[ C_3H_5I \right]^{\nu_{C_3H_5I,3}} \left[ C_6H_{10} \right]^{\nu_{C_6H_{10},3}} \left[ I_2 \right]^{\nu_{I_2,3}} \left[ C_3H_5 \cdot \right]^{\nu_{C_3H_5 \cdot,3}} \left[ I \cdot \right]^{\nu_{I \cdot,3}} \]

\[ K_3 = \left[ C_3H_5I \right]^{-1} \left[ C_6H_{10} \right]^0 \left[ I_2 \right]^1 \left[ C_3H_5 \cdot \right]^1 \left[ I \cdot \right]^{-1} \]

\[ K_3 = \frac{\left[ C_3H_5 \cdot \right] \left[ I_2 \right]}{\left[ C_3H_5I \right] \left[ I \cdot \right]} \]

The quasi-equilibrium expression for step (4) is generated analogously, as is the quasi-equilibrium expression for step (2) for the cases where step (3) is rate-determining and where step (4) is rate-determining.

Equations (6) and (7) can be solved to obtain expressions for the concentrations of the reactive intermediates, C3H5∙ and I∙. Those expressions can then be substituted into the equation (5) to produce an acceptable rate expression.

Calculations

  1. Solve equations (6) and (7) to obtain expressions for [C3H5∙] and [I∙].
  2. Substitute the results into equation (5).

For the case where reaction (3) is the rate-determining step, the apparent rate of the non-elementary reaction is equal to the rate of step (3) and steps (2) and (4) are quasi-equilibrated leading to equations (8) through (10).

\[ r_1 = r_3 = k_{3,f} \left[ C_3H_5I \right] \left[ I \cdot \right] - k_{3,r} \left[ C_3H_5 \cdot \right] \left[ I_2 \right] \tag{8} \]

\[ K_2 = \frac{\left[ C_3H_5 \cdot \right] \left[ I \cdot \right]}{\left[ C_3H_5I \right]} \tag{9} \]

\[ K_4 = \frac{\left[ C_6H_{10} \right]}{\left[ C_3H_5 \cdot \right]^2} \tag{10} \]

Equations (9) and (10) can be solved to obtain expressions for the concentrations of the reactive intermediates, C3H5∙ and I∙ which can be substituted into the equation (8) to produce an acceptable rate expression.

For the case where reaction (4) is the rate-determining step, the apparent rate of the non-elementary reaction is equal to the rate of step (4) and steps (2) and (3) are quasi-equilibrated leading to equations (11) through (13).

\[ r_1 = r_4 = k_{4,f} \left[ C_3H_5 \cdot \right]^2 - k_{4,r} \left[ C_6H_{10} \right] \tag{11} \]

\[ K_2 = \frac{\left[ C_3H_5 \cdot \right] \left[ I \cdot \right]}{\left[ C_3H_5I \right]} \tag{12} \]

\[ K_3 = \frac{\left[ C_3H_5 \cdot \right] \left[ I_2 \right]}{\left[ C_3H_5I \right] \left[ I \cdot \right]} \tag{13} \]

Equations (12) and (13) can be solved to obtain expressions for the concentrations of the reactive intermediates, C3H5∙ and I∙ which can be substituted into the equation (11) to produce an acceptable rate expression.

Final Answer

When step (2) is assumed to be rate-determining, the expressions for the concentrations of [C3H5∙] and [I∙] shown in equations (14) and (15) are obtained by solving equations (6) and (7). Substitution into equation (5) yields the expression for the apparent rate of non-elementary reaction (1) shown in equation (16).

\[ \left[ C_3H_5 \cdot \right] = \frac{\sqrt{\left[ C_6H_{10} \right]}}{\sqrt{K_4}} \tag{14} \]

\[ \left[ I \cdot \right] = \frac{\sqrt{\left[ C_6H_{10} \right]}\left[ I_2 \right]}{K_3 \sqrt{K_4}\left[ C_3H_5I \right]} \tag{15} \]

\[ r_1 = k_{2,f} \left[ C_3H_5I \right] - \frac{k_{2,r}}{K_3 K_4} \frac{\left[ C_6H_{10} \right] \left[ I_2 \right]}{\left[ C_3H_5I \right]} \tag{16} \]

When step (3) is assumed to be rate-determining, the expressions for the concentrations of [C3H5∙] and [I∙] shown in equations (17) and (18) are obtained by solving equations (9) and (10). Substitution into equation (8) yields the expression for the apparent rate of non-elementary reaction (1) shown in equation (19).

\[ \left[ C_3H_5 \cdot \right] = \frac{\sqrt{\left[ C_6H_{10} \right]}}{\sqrt{K_4}} \tag{17} \]

\[ \left[ I \cdot \right] = \frac{K_2 \sqrt{K_4} \left[ C_3H_5I \right]}{\sqrt{\left[ C_6H_{10} \right]}} \tag{18} \]

\[ r_1 = k_{3,f} K_2 \sqrt{K_4} \frac{\left[ C_3H_5I \right]^2}{\sqrt{\left[ C_6H_{10} \right]}} - \frac{k_{3,r}}{\sqrt{K_4}}\sqrt{\left[ C_6H_{10} \right]}\left[ I_2 \right] \tag{19} \]

When step (4) is assumed to be rate-determining, the expressions for the concentrations of [C3H5∙] and [I∙] shown in equations (20) and (21) are obtained by solving equations (12) and (13). Substitution into equation (11) yields the expression for the apparent rate of non-elementary reaction (1) shown in equation (22).

\[ \left[ C_3H_5 \cdot \right] = \sqrt{K_2K_3} \frac{\left[ C_3H_5I \right]}{\sqrt{\left[ I_2 \right]}} \tag{20} \]

\[ \left[ I \cdot \right] = \frac{\sqrt{K_2}}{\sqrt{K_3}} \sqrt{\left[ I_2 \right]} \tag{21} \]

\[ r_1 = \frac{k_{4,f}K_2K_3\left[ C_3H_5I \right]^2}{\left[ I_2 \right]} - k_{4,r} \left[ C_6H_{10} \right] \tag{22} \]

Note

If thermodynamic data for the mechanistic steps are not available and the equilibrium constants are treated as kinetics parameters, then the parameters in the second term of equation (16) are coupled as was the case in Example 5.3.2. That is, it would not be possible to estimate unique values for k2,r, K3 and K4 using experimental kinetics data. Grouping the parameters to define an apparent rate coefficient as shown in equation (23) allows the rate expression to be written as equation (24). The apparent rate coefficient, \(k_r^\prime\), will exhibit Arrhenius temperature dependence and it will be possible to estimate a unique value for it.

\[ k_r^\prime = \frac{k_{2,r}}{K_3 K_4} \tag{23} \]

\[ r_1 = k_{2,f} \left[ C_3H_5I \right] - k_r^\prime \frac{\left[ C_6H_{10} \right] \left[ I_2 \right]}{\left[ C_3H_5I \right]} \tag{24} \]

Similar grouping of the kinetics parameters in rate expressions (19) and (22) leads to equations (25) and (26)

\[ r_1 = k_f^\prime \frac{\left[ C_3H_5I \right]^2}{\sqrt{\left[ C_6H_{10} \right]}} - k_r^\prime \sqrt{\left[ C_6H_{10} \right]}\left[ I_2 \right] \tag{25} \]

\[ r_1 = k_f^\prime \frac{\left[ C_3H_5I \right]^2}{\left[ I_2 \right]} - k_{4,r} \left[ C_6H_{10} \right] \tag{22} \]

5.3.4 Michaelis-Menten Rate Expression for an Enzymatic Reaction

Suppose an enzyme, E, catalyzes the conversion of substrate, S, into product, P, as indicated in the apparent, non-elementary reaction (1). If the mechanism consists of reactions (2) and (3), derive a mechanistic rate expression for the apparent generation of P via reaction (1) assuming step (3) is effectively irreversible. The resulting rate expression should be suitable for reaction engineering purposes. That is, it should not contain concentrations of reactive intermediates.

\[ S \rightleftarrows P \tag{1} \]

\[ E + S \rightleftarrows E\!\!-\!\!S \tag{2} \]

\[ E\!\!-\!\!S \rightleftarrows E + P \tag{3} \]

This is a mechanism problem. There isn’t a rate-determining step in the mechanism, but it does involve an enzyme. In problems with enzymes (or with homogeneous catalysts), each form of the enzyme is treated the same way reactive intermediates are treated, so for this problem the reactive intermediates are the free (uncomplexed) enzyme and the enzyme-substrate complex.

Since there isn’t a rate-determining step, I will need to choose a reagent in the non-elementary reaction and set the apparent rate of generation of that reagent via the non-elementary reaction equal to the sum of the rates at which it is generated in each mechanistic step. In this problem, I’ll do this for the product, P, since it only appears in one step in the mechanism.

There are two reactive intermediates, but if I wrote the Bodenstein steady state approximation for both of them, the equations would not be mathematically independent because the mechanism includes an enzyme. I know that in this situation I need to write the Bodenstein steady state approximation for all but one of the reactive intermediates and then add an equation for the conservation of catalyts. Those equations can be solved to obtain expressions for the concentrations of the reactive intermediates that can be used to eliminate the reactive intermediate concentrations from the rate expression.

Reactive Intermediates: E and E-S

Necessary Equations

Setting the apparent rate of generation of P via reaction (1) equal to the sum of its rates of generation in each of the mechanistic steps yields equation (4).

\[ r_{P,1} = r_3 \tag{4} \]

Start with Equation 5.7. \[ r_{i,j} =\sum_{j^\prime}\nu_{i,j^\prime}r_{j^\prime} \]

Here I want an expression for the apparent generation of P via non-elementary reaction (1), so \(i = P\) and \(j = 1\). In Equation 5.7, \(j^\prime\) indexes all of the steps in the mechanism, so \(j^\prime\) includes 2, and 3.

\[ r_{P,1} = \nu_{P,2}r_2 + \nu_{P,3}r_3 \]

\[ r_{P,1} = (0)r_2 + (1)r_3 \]

\[ r_{P,1} = r_3 \]

The mechanistic steps are elementary reactions, so their rate expressions are known, equations (5) and (6).

\[ r_2 = k_{2,f} \left[ E \right] \left[ S \right] - k_{2,r} \left[ E\!-\!S \right] \tag{5} \]

\[ r_3 = k_{3,f} \left[ E\!-\!S \right] \tag{6} \]

The mechanistic steps are elementary reactions, so start with Equation 5.1 \[ r_j = k_{j,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j}} - k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j}} \]

In the case of step (2), \(j = 2\). In Equation 5.1, \(i^\prime\) indexes all of the reactants in the reaction and \(i^{\prime\prime}\) indexes all of the products in the reaction, so for reaction (2) \(i\) includes E and S and \(i^\prime\) includes only E-S.

\[ r_2 = k_{2,f} \left[ E \right]^{-\nu_{E,2}} \left[ S \right]^{-\nu_{S,2}} - k_{2,r}\left[ E\!-\!S \right]^{\nu_{E-S,2}} \]

\[ r_2 = k_{2,f} \left[ E \right]^{-(-1)} \left[ S \right]^{-(-1)} - k_{2,r}\left[ E\!-\!S \right]^1 \]

\[ r_2 = k_{2,f} \left[ E \right] \left[ S \right] - k_{2,r} \left[ E\!-\!S \right] \]

The expression for the rate of step (3) is generated analogously, except the second term is set equal to zero because the problem states that step (3) is effectively irreversible.

There are two reactive intermediates, but only one of the corresponsing Bodenstein steady state approximations can be used because the other will not be mathematically independent. The Bodenstein steady state approximation for E-S is shown in equation (7).

\[ 0 = r_2 - r_3 \tag{7} \]

Start with Equation 5.11. \[ 0 = \sum_{j^\prime}\nu_{RI,j^\prime}r_{j^\prime} \]

In this equation, \(RI\) is one of the reactive intermediates and \(j^\prime\) indexes all of the steps in the mechanism. Thus,in the caser of E-S, \(RI\) = E-S and \(j^\prime\) includes reactions (2) and (3).

\[ 0 = nu_{E\!-\!S,2}r_2 + nu_{E\!-\!S,3}r_3 \]

\[ 0 = \left( 1 \right) r_2 + \left( -1 \right)r_3 \]

\[ 0 = r_2 - r_3 \]

Since there is an enzyme in the mechanism, I can write an expression for the conservation of enzyme, equation (8).

\[ E_0 = \left[ E \right] + \left[ E\!-\!S \right] \tag{8} \]

Start with Equation 5.17.

\[ C_{0,cat} = C_{cat,free} + \sum_{i^\prime}\kappa_{i^\prime}C_{i^\prime} \]

Equation 5.17 was written for homogeneous catalysts, so here I’ve replace the C’s with E’s. E0 represents the total amount of enzyme originally added to the system, expressed as a concentration. It is known and constant, so its appearance in the rate expression is acceptable. The index, \(i^\prime\) includes all complexed forms of the enzyme. In this problem, the only complexed form of the enzyme is E-S. That complex contains one original enzyme unit, so \(\kappa_{E\!-\!S} = 1\).

\[ E_0 = \left[ E \right] + \left[ E\!-\!S \right] \]

Equations (9) and (10) can be solved to obtain expressions for the concentrations of E and E-S which can be used to eliminate the reactive intermediate concentrations from the rate expression.

Calculations

  1. Generate new versions of equations (4) and (7) by substitution of equations (5) and (6).
  2. Solve equations (7) and (8) to obtain expressions for [E] and [E-S].
  3. Substitute the resulting expressions in equation (4).

Final Answer

The expressions for the concentrations of E and E-S shown in equations (9) and (10) result from solving equations (7) and (8). Substitution of those expression into equation (4) yields the expression for the apparent rate of generation of P via non-elementary reaction (1) shown in equation (11).

\[ \left[ E \right] = \frac{E_0}{1 + \frac{k_{2,f}}{k_{2,r} + k_{3,f}} \left[ S \right]} \tag{9} \]

\[ \left[ E\!-\!S \right] = \frac{k_{2,f} E_0 \left[ S \right]}{k_{2,r}+ k_{3,f} + k_{2,f}\left[ S \right]} \tag{10} \]

\[ r_{P,1} = \frac{k_{2,f} k_{3,f} E_0 \left[ S \right]}{k_{2,r} + k_{3,f} + k_{2,f}\left[ S \right]} = \frac{k_{3,f} E_0 \left[ S \right]}{\frac{k_{2,r} + k_{3,f}}{k_{2,f}} + \left[ S \right]} \tag{11} \]

Note
  1. Michaelis and Menten (1913) studied enzymatic conversion of a substrate to product and were the first to propose a mechanistic rate epxression of the form of equation (11). They defined two kinetics parameters, \(V_{max}\) and \(K_m\), as shown in equations (12) and (13). Substitution of those definitions in equation (11) yields the so-called Michaelis-Menten rate expression, equation (14).

\[ V_{max} = k_{3,f} E_0 \tag{12} \]

\[ K_m = \frac{k_{2,r} + k_{3,f}}{k_{2,f}} \tag{13} \]

\[ r_{P,1} = \frac{V_{max} \left[ S \right]}{K_m + \left[ S \right]} \tag{14} \]

  1. The Michaelis-Menten rate expression above is non-linear with respect to the kinetics parameters. The parameters, \(V_{max}\) and \(K_m\), can be estimated using experimental kinetics data together with software for fitting non-linear models to data. Notice, however, that if the reciprocals of the two sides of the equation are set equal to each other as shown in equation (15), a linear model results. Specifically, defining \(y\) and \(x\) as shown in equations (16) and (17), it can be seen that the model has the linear form shown in equation (18) where the slope, \(m\), and intercept, \(b\), are given in equations (19) and (20).

\[ \frac{1}{r_{P,1}} = \left( \frac{K_m}{V_{max}} \right) \frac{1}{\left[ S \right]} + \frac{1}{V_{max}} \tag{15} \]

\[ y = \frac{1}{r_{P,1}} \tag{16} \]

\[ x = \frac{1}{\left[ S \right]} \tag{17} \]

\[ y = mx + b \tag{18} \]

\[ m = \frac{K_m}{V_{max}} \tag{19} \]

\[ b = \frac{1}{V_{max}} \tag{20} \]

As was the case when the Arrhenius expression was linearized (see the note at the end of Example 4.5.4), a spreadsheet program could be used to create a plot of \(\frac{1}{r_{P,1}}\) vs. \(\frac{1}{\left[ S \right]}\). A plot of this type was used by Lineweaver and Burk (1934), and today plots of the type are commonly called Lineweaver-Burk plots. By adding a trendline to the Lineweaver-Burk plot, the best estimates for \(m\) and \(b\) will be obtained. As equation (20) shows, the best estimate for \(V_{max}\) is equal to the reciprocal of the best estimate for \(b\). Then, as equation (19) shows, the best estimate for \(K_m\) is equal to the best estimate for \(m\) multiplied by \(V_{max}\).

It can be noted in equation (13) that \(K_m\) includes the sum of two rate coefficients. As such, \(K_m\) will not display Arrhenius temperature dependence if it is measured over a sufficiently broad range of temperatures. Fortunately, enzymes are often active only in a narrow range of temperatures, and this does not become a problem.

5.3.5 Mechanistic Rate Expression from an Enzymatic Mechanism with an Inhibitor

Suppose that enzyme E catalyzes the conversion of substrate S to product P, but if the reagent I is present in the system, it inhibits the enzyme. The apparent reaction is shown in equation (1), and the proposed mechanism consists of reactions (2) through (4).

\[ S \rightleftarrows P \tag{1} \]

\[ E + S \rightleftarrows E\!\!-\!\!S \tag{2} \]

\[ E + I \rightleftarrows E\!\!-\!\!I \tag{3} \]

\[ E\!\!-\!\!S \rightleftarrows E + P \tag{4} \]

Assume that step (4) is effectively irreversible and derive a Michaelis-Menten type of rate expression for this system that is suitable for reaction engineering purposes (i. e. that does not contain concentrations of reactive intermediates). You may assume that the concentration of the free inhibitor can be easily measured, so that its presence in the rate expression is acceptable.

This is again a mechanism problem that includes an enzyme and does not have a rate determining step. In mechanism problems that involve a homogeneous catalyst or an enzyme, the free and compelexed forms of the catalyst or enzyme are treated like reactive intermediates. Thus, in this problem the reactive intermediates are E, E-S and E-I.

Since there isn’t a rate-determining step, I will choose a reagent from the apparent reaction and set its apparent rate of generation in the non-elementary reaction equal to the sum of its rates of generation in the mechanistic steps. Here I will do this for P because it only appear in one mechanistic step.

To eliminate the concentrations of reactive intermediates from the rate expressions, I’ll write the Bodenstein steady state approximation for two of the three reactive intermediates. The Bodenstein steady state approximation for the third reactive intermediate would not be mathematically independent, so I will replace it with an expression for the conservation of catalyst. The resulting set of three equations can be solved to obtain expressions for the concentrations of the three reactive intermediates. Those expressions can then be used to eliminate the concentrations of reactive intermediates from the rate expression.

Reactive Intermediates: E, E-S, and E-I

Necessary Equations

Setting the apparent rate of generation of P equal to the sum of the rates at which P is generated in each of the mechanistic steps yields equation (5).

\[ r_{P,1} = r_4 \tag{5} \]

Start with Equation 5.7.

\[ r_{i,j} =\sum_{j^\prime}\nu_{i,j^\prime}r_{j^\prime} \]

Noting that \(j^\prime\) indexes all of the steps in the mechanism and setting \(i=P\) and \(j=1\) gives the expression for the apparent rate of generation of P via non-elementary reaction (1).

\[ r_{P,1} = \nu_{P,2}r_2 + \nu_{P,3}r_3 + \nu_{P,4}r_4 \]

\[ r_{P,1} = \left( 0 \right)r_2 + \left( 0 \right)r_3 + \left( 1 \right)r_4 \]

\[ r_{P,1} = r_4 \]

The mechanistic steps are elementary, so their rates are known, equations (6) through (8).

\[ r_2 = k_{2,f} \left[ E \right] \left[ S \right] - k_{2,r} \left[ E\!-\!S \right] \tag{6} \]

\[ r_3 = k_{3,f} \left[ E \right] \left[ I \right] - k_{3,r} \left[ E\!-\!I \right] \tag{7} \]

\[ r_4 = k_{4,f} \left[ E\!-\!S \right] \tag{8} \]

Start with Equation 5.1.

\[ r_j = k_{j,f} \prod_{i^\prime}\left[ i^\prime \right]^{-\nu_{i^\prime,j}} - k_{j,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{\nu_{i^{\prime\prime},j}} \]

To generate an expression for step (2), set \(j=2\) and note that \(i^\prime\) indexes all of the reactants in the reaction and \(i^{\prime\prime}\) indexes all of the products in the reaction. Thus, \(i^\prime\) includes E and S and \(i^{\prime\prime}\) includes only E-S.

\[ r_2 = k_{2,f} \left[ E \right]^{-\nu_{E,2}}\left[ S \right]^{-\nu_{S,2}} - k_{2,r}2\left[ E-S \right]^{\nu_{E-S,2}} \]

\[ r_2 = k_{2,f} \left[ E \right]^{-(-1)}\left[ S \right]^{-(-1)} - k_{2,r}2\left[ E-S \right]^1 \]

\[ r_2 = k_{2,f} \left[ E \right] \left[ S \right] - k_{2,r} \left[ E\!-\!S \right] \]

Expressions for \(r_3\) and \(r_4\) are generated in analogous manner except the second term in the rate expression for step (4) is set equal to zero because the problem states that reaction (4) is effectively irreverisble.

Applying the Bodenstein steady state approximation to E-S and E-I yields equations (9) and (10).

\[ 0 = r_2 - r_4 \tag{9} \]

\[ 0 = r_3 \tag{10} \]

Start with the Bodenstein steady state approximation, Equation 5.11.

\[ 0 = \sum_{j^\prime}\nu_{RI,j^\prime}r_{j^\prime} \]

In this equation \(j\) indexes all of the steps in the mechanism and \(RI\) denotes one of the reactive intermediates. Consequently, in the Bodenstein steady state approximation for E-S, \(j\) includes steps (2), (3), and (4), and \(RI\) equals E-S.

\[ 0 = \nu_{ES,2} r_2 + \nu_{ES,3} r_3 + \nu_{ES,4} r_4 \]

\[ 0 = \left( 1 \right) r_2 \left( 0 \right) r_3 + \left( -1 \right) r_4 \]

\[ 0 = r_2 - r_4 \]

The Bodenstein steady state approximation for E-I is generated analogously.

Equation (11) is an expression for the conservation of enzyme.

\[ E_0 = \left[ E \right] + \left[ E\!-\!S \right] + \left[ E\!-\!I \right] \tag{11} \]

Start with Equation 5.17.

\[ C_{0,cat} = C_{cat,free} + \sum_{i^\prime}\kappa_{i^\prime}C_{i^\prime} \]

Equation 5.17 was written for homogeneous catalysts, so here I’ve replace the C’s with E’s. E0 represents the total amount of enzyme originally added to the system, expressed as a concentration. It is known and constant, so its appearance in the rate expression is acceptable. The index, \(i^\prime\) includes all complexed forms of the enzyme. In this problem, the complexed forms of the enzyme are E-S and E-I. Both complexes contains one original enzyme unit, so \(\kappa_{E\!-\!S} = 1\) and \(\kappa_{E\!-\!I} = 1\).

\[ C_{0,cat} = C_{cat,free} + \sum_{i^\prime}\kappa_{i^\prime}C_{i^\prime} \]

\[ E_0 = \left[ E \right] + \left[ E\!-\!S \right] + \left[ E\!-\!I \right] \]

Equations (9) through (11) can be solved to obtain expressions for the concentrations of the reactive intermediates. The results can be used to eliminate the reactive intermediates from the rate expression.

Note that since the stoichiometric coefficient of P in the non-elementary reaction is 1, the apparent rate of reaction (1) is equal to the apparent rate of generation of P via reaction (1), equation (12).

\[ r_{P,1} = \nu_{P,1}r_1 \qquad \Rightarrow \qquad r_1 = r_{P,1} \tag{12} \]

Calculations

  1. Generate new versions of equations (5), (9), and (10) by substitution of equations (6) through (8).
  2. Solve equations (9) through (11) to obtain expressions for \(\left[ E \right]\), \(\left[ E\!-\!S \right]\), and \(\left[ E\!-\!I \right]\).
  3. Substitute the results from step 2 into equation (5).

Results and Final Answer

Solving equations (9) through (11) yields the expressions for the concentrations of E, E-S, and E-I shown in equations (13) through (15). Substitution of these expressions into equation (5) then gives the rate expression shown in equation (16).

\[ \left[ E \right] = \frac{k_{3,r} \left( k_{2,r} + k_{4,f} \right) E_0}{k_{2,f} k_{3,r} \left[ S \right] + \left( k_{2,r} k_{3,f} + k_{3,f} k_{4,f} \right)\left[ I \right] + \left( k_{2,r} k_{3,r} + k_{3,r} k_{4,f} \right)} \tag{13} \]

\[ \left[ E\!-\!I \right] = \frac{k_{3,f} \left( k_{2,r} + k_{4,f} \right) E_0 \left[ I \right]}{k_{2,f} k_{3,r} \left[ S \right] + \left( k_{2,r} k_{3,f} + k_{3,f} k_{4,f} \right)\left[ I \right] + \left( k_{2,r} k_{3,r} + k_{3,r} k_{4,f} \right)} \tag{14} \]

\[ \left[ E\!-\!S \right] = \frac{k_{2,f} k_{3,r} E_0 \left[ S \right]}{k_{2,f} k_{3,r} \left[ S \right] + \left( k_{2,r} k_{3,f} + k_{3,f} k_{4,f} \right)\left[ I \right] + \left( k_{2,r} k_{3,r} + k_{3,r} k_{4,f} \right)} \tag{15} \]

\[ r_{P,1} = \frac{k_{4,f} E_0 \left[ S \right]}{\left[ S \right] + \frac{k_{2,r} k_{3,f} + k_{3,f} k_{4,f}}{k_{2,f} k_{3,r}} \left[ I \right] + \frac{k_{2,r} k_{3,r} + k_{3,r} k_{4,f}}{k_{2,f} k_{3,r}}} \tag{16} \]

The problem asks for a “Michaelis-Menten type rate expression.” In a manner analogous to the note at the end of Example 5.3.4, The kinetics parameters, \(V_{max}\), \(K_m\), and \(K_I\), can be defined as shown in equations (17) through (19). Substitution into equation (15) then yields the Michaelis-Menten type rate expression shown in equation (20).

\[ V_{max} = k_{4,f} E_0 \tag{17} \]

\[ K_m = \frac{k_{2,r} k_{3,r} + k_{3,r} k_{4,f}}{k_{2,f} k_{3,r}} \tag{18} \]

\[ K_I = \frac{k_{2,r} k_{3,f} + k_{3,f} k_{4,f}}{k_{2,f} k_{3,r}} \tag{19} \]

\[ r_{P,1} = \frac{V_{max} \left[ S \right]}{K_m + K_I \left[ I \right] + \left[ S \right]} \tag{20} \]

Note

Because they contain sums of rate coefficients, \(K_m\) and \(K_I\) in equation (20) are not expected to exhibit Arrhenius temperature dependence over a wide range of temperatures. That said, enzymes often are stable only over a relatively narrow range of temperatures in which case this probably won’t cause problems.

5.3.6 Langmuir-Hinshelwood Rate Expression for a Heterogeneous Catalytic Reaction

Suppose the non-elementary reaction (1) is heterogeneously catalyzed, and the corresponding reaction mechanism is given by equations (2) through (6). Assume step (4) is rate-determining and derive an expression for the apparent rate of reaction (1) that does not include fractional coverages.

\[ A + B \rightleftarrows C + D \tag{1} \]

\[ A + \ast \rightleftarrows A\!-\!\ast \tag{2} \]

\[ B + \ast \rightleftarrows B\!-\!\ast \tag{3} \]

\[ A\!-\!\ast + B\!-\!\ast \rightarrow C\!-\!\ast + D\!-\!\ast \tag{4} \]

\[ C\!-\!\ast \rightleftarrows C + \ast \tag{5} \]

\[ D\!-\!\ast \rightleftarrows D + \ast \tag{6} \]

This mechanism involves a heterogeneous catalyst, and there is a rate-determining step in the mechanism. Consequently, the apparent rate of the non-elementary reaction is equal to the rate of the rate-determining step as given in Equation 5.21.

\[ \begin{align} r_j &= r_{rds} \\ &= k_{rds,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,rds}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,rds}} \\ &- k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},rds} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},rds}} \end{align} \]

To make the expression for the apparent rate of the heterogeneously catalyzed, non-elementary reaction useful for reaction engineering purposes, the fractional coverages must be eliminated from it. Because there is a rate determining step, quasi-equilibrium expressions, Equation 5.16, are written for all other steps. (When doing so, if \(i\) is a surface species, then \(\left[i\right] = \theta_i\).)

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \]

An expression for the conservation of sites, Equation 5.22, is added to to the quasi-equilibrium expressions.

\[ 1 = \theta_{\text{vacant}} + \sum_k \theta_k \]

The quasi-equilibrium expressions together with the site conservation equation are solved to obtain expressions for the surface coverages. The results are used to eliminate the surface coverages from the expression for the apparent rate of the non-elementary reaction.

Reactive Intermediates: \(\ast\), \(A\!-\!\ast\), \(B\!-\!\ast\), \(C\!-\!\ast\), and \(D\!-\!\ast\)

Necessary Equations

Setting the apparent rate of the non-elementary reaction equal to the rate of the rate-determining step gives equation (7).

\[ r_1 = r_4 = k_{4,f} \theta_A \theta_B - k_{4,r} \theta_C \theta_D \tag{7} \]

Start with Equation 5.21.

\[ \begin{align} r_j &= r_{rds} \\ &= k_{rds,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,rds}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,rds}} \\ &- k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},rds} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},rds}} \end{align} \]

For this problem, \(j = 1\), \(rds = 4\), there are no fluid phase reactants or products in the rds, so \(i^\prime\) and \(i^{\prime\prime}\) do not index anything, \(k^\prime\) indexes \(A\!-\!\ast\) and \(B\!-\!\ast\) and \(k^{\prime\prime}\) indexes \(C\!-\!\ast\) and \(D\!-\!\ast\).

\[ r_1 = k_{4,f} \theta_A^{-\nu_{A\!-\!\ast,4}} \theta_B^{-\nu_{B\!-\!\ast,4}} - k_{4,r} \theta_C^{\nu_{C\!-\!\ast,4}} \theta_D^{\nu_{D\!-\!\ast,4}} \]

\[ r_1 = k_{4,f} \theta_A^{-(-1)} \theta_B^{-(-1)} - k_{4,r} \theta_C^1 \theta_D^1 \]

\[ r_1 = k_{4,f} \theta_A \theta_B - k_{4,r} \theta_C \theta_D \]

Assuming steps (2), (3), (5), and (6) are quasi-equilibrated yields equations (8) through (11).

\[ K_2 = \frac{\theta_A}{\left[ A \right] \theta_{\text{vacant}}} \tag{8} \]

\[ K_3 = \frac{\theta_B}{\left[ B \right] \theta_{\text{vacant}}} \tag{9} \]

\[ K_5 = \frac{\left[ C \right] \theta_{\text{vacant}}}{\theta_C} \tag{10} \]

\[ K_6 = \frac{\left[ D \right] \theta_{\text{vacant}}}{\theta_D} \tag{11} \]

Start with Equation 5.16.

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \]

In the case of reaction (2), \(nrd = 2\), \(i\) indexes all reagents in the system (\(A\), \(B\), \(C\), \(D\), \(\ast\), \(A\!-\!\ast\), \(B\!-\!\ast\), \(C\!-\!\ast\), and \(D\!-\!\ast\)), and for the surface species \(\left[ i \right] = \theta_i\).

\[ K_2 = \left[A\right]^{\nu_{A,2}}\left[B\right]^{\nu_{B,2}}\left[C\right]^{\nu_{C,2}}\left[D\right]^{\nu_{D,2}}\theta_{\text{vacant}}^{\nu_{\ast,2}}\theta_A^{\nu_{A\!-\!\ast,2}}\theta_B^{\nu_{B\!-\!\ast,2}}\theta_C^{\nu_{C\!-\!\ast,2}}\theta_D^{\nu_{D\!-\!\ast,2}} \]

\[ K_2 = \left[A\right]^{-1}\left[B\right]^0\left[C\right]^0\left[D\right]^0\theta_{\text{vacant}}^{-1}\theta_A^1\theta_B^0\theta_C^0\theta_D^0 \]

\[ K_2 = \left[A\right]^{-1}\theta_{\text{vacant}}^{-1}\theta_A^1 = \frac{\theta_A}{\left[ A \right] \theta_{\text{vacant}}} \]

The generation of the expressions for K3, K5, and K6 is analogous.

An expression for the conservation of sites is given in equation (12).

\[ 1 = \theta_{\text{vacant}} + \theta_A +\theta_B + \theta_D + \theta_D \tag{12} \]

Start with Equation 5.22.

\[ 1 = \theta_{\text{vacant}} + \sum_k \theta_k \]

For this problem, \(k\) indexes all adsorbed species: \(A\!-\!\ast\), \(B\!-\!\ast\), \(C\!-\!\ast\), and \(D\!-\!\ast\)

\[ 1 = \theta_{\text{vacant}} + \theta_A +\theta_B + \theta_D + \theta_D \]

Calculations

  1. Solve the five equations, (8) through (12), to obtain expressions for the five surface coverages, \(\theta_{\text{vacant}}\), \(\theta_A\), \(\theta_B\), \(\theta_C\), and \(\theta_D\).
  2. Substitute the results into equation (7).

Results and Final Answer

Solving equations (8) through (12) for \(\theta_{\text{vacant}}\), \(\theta_A\), \(\theta_B\), \(\theta_C\), and \(\theta_D\) yields equations (13) through (17).

\[ \theta_{\text{vacant}} = \frac{1}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{13} \]

\[ \theta_A = \frac{K_2 \left[ A \right]}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{14} \]

\[ \theta_B = \frac{K_3 \left[ B \right]}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{15} \]

\[ \theta_C = \frac{\frac{1}{K_5} \left[ C \right]}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{16} \]

\[ \theta_D = \frac{\frac{1}{K_6} \left[ D \right]}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{17} \]

Substitution of equations (13) through (17) into equation (7) yields the requested rate expression, equation (18).

\[ r_1 = \frac{k_{4,f}K_2K_3\left[ A \right] \left[ B \right]}{1 + K_2 \left[ A \right] + K_3 \left[ B \right] + \frac{1}{K_5} \left[ C \right] + \frac{1}{K_6} \left[ D \right]} \tag{18} \]

Note

The rate expressions for heterogeneous catalytic reactions that result when it is assumed that the adsorption and desorption steps are all equilibrated and a surface reaction step is controlling are known as Langmuir-Hinshelwood rate expressions. They are named for Irving Langmuir, who developed models for adsorption equilibrium (Langmuir, I 1918), and Cyril Hinshelwood who used them to develop rate expressions for surface reactions.

Langmuir-Hinshelwood rate expressions for heterogeneous catalytic reactions, Michaelis-Menten rate expressions for enzymatic reactions and general rate expressions for homogeneous catalytic reactions often take the form of a fraction. The numerator is typically a single positive term when the apparent reaction is irreversible or the difference between two terms when the apparent reaction is reversible. The denominator is typically the sum of several terms. Example 5.3.7 shows how the summation in the denominator in these rate expressions often reduces to a single term when the amount of one of the reactive intermediates is much greater than all the other reactiven intermediates.

5.3.7 Mechanistic Rate Expression from a Mechanism with a Most Abundant Intermediate

The oxidation of carbon monoxide, reaction (1) is heterogeneously catalyzed. A proposed reaction mechanism is shown in reactions (2) through (6). Assume that step (6) is rate-limiting and derive an expression for the apparent rate of reaction (1) that contains only partial pressures, rate coefficients and equilibrium constants. How does the apparent rate expression change if \(O\!-\!\ast\) is the most abundant surface intermediate?

Overall Reaction:

\[ 2 CO + O_2 \rightleftarrows 2 CO_2 \tag{1} \]

Proposed Mechanism: \[ O_2 + \ast \rightleftarrows O_2\!-\!\ast \tag{2} \]

\[ CO + O_2\!-\!\ast \rightleftarrows CO_3\!-\!\ast \tag{3} \]

\[ CO_3\!-\!\ast \rightleftarrows CO_2 + O\!-\!\ast \tag{4} \]

\[ CO + O\!-\!\ast \rightleftarrows CO_2\!-\!\ast \tag{5} \]

\[ CO_2\!-\!\ast \rightleftarrows CO_2 + \ast \tag{6} \]

The mechanism has a rate-determining step and involves a heterogeneous catalyst. In that situation the apparent rate of the non-elementary reaction is equal to the rate of the rate-determining step as given in Equation 5.21.

\[ \begin{align} r_j &= r_{rds} \\ &= k_{rds,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,rds}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,rds}} \\ &- k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},rds} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},rds}} \end{align} \]

To make the apparent rate expression useful for reaction engineering purposes, all surface coverages must be eliminated from it. When there is a rate-determining step, quasi-equilibrium expressions for all other steps, Equation 5.16, are used.

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \]

In this problem there are 4 steps that are not rate-determining, and therefore 4 quasi-equilibrium expressions, but there are five surface species (\(\ast\), \(O_2\!-\!\ast\), \(CO_3\!-\!\ast\), \(O\!-\!\ast\), and \(CO_2\!-\!\ast\)). An expression for the conservation of catalyst sites, Equation 5.22, provides a fifth equation.

\[ 1 = \theta_{\text{vacant}} + \sum_k \theta_k \]

Those five equations can be solved to get expressions for the surface coverages. Substitution of the surface coverage expressions into the rate expression then eliminates the surface coverages from the rate expression.

In the second part of this problem there is a most abundant surface intermediate. This allows four inequalities of the form given in Equation 5.24 to be written.

\[ \theta_{ma} \gg \theta_{nma} \]

Substitution of the expressions for the surface coverages into these equalities followed by algebraic simplification may then show that some terms in the expression for the apparent rate of the non-elementary reaction can be eliminated because they are negligible.

Reactive Intermediates: \(\ast\), \(O_2\!-\!\ast\), \(CO_3\!-\!\ast\), \(O\!-\!\ast\), and \(CO_2\!-\!\ast\).

Necessary Equations

Setting the apparent rate of the non-elementary reaction equal to the rate of the rate-determining step, (6) yields equation (7).

\[ r_1 = r_6 = k_{6,f} \theta_{CO_2} - k_{6,r} \left[ CO_2 \right] \theta_{\text{vacant}} \tag{7} \]

Start with Equation 5.21.

\[ \begin{align} r_j &= r_{rds} \\ &= k_{rds,f}\prod_{i^\prime}\left[ i^\prime \right]^{- \nu_{i^\prime,rds}} \prod_{k^\prime}\theta_{k^\prime}^{- \nu_{k^\prime,rds}} \\ &- k_{rds,r}\prod_{i^{\prime\prime}}\left[ i^{\prime\prime} \right]^{ \nu_{i^{\prime\prime},rds} }\prod_{k^{\prime\prime}}\theta_{k^{\prime\prime}}^{ \nu_{k^{\prime\prime},rds}} \end{align} \]

For this problem, \(j = 1\), \(rds = 6\), there are no fluid phase reactants in the rds, so \(i^\prime\) does not index anything, \(i^{\prime\prime}\) indexes the only fluid phase product in the rds, CO2, \(k^\prime\) indexes the only surface reactant in the rds, \(CO_2\!-\!\ast\), and \(k^{\prime\prime}\) indexes the only surface product in the rds, \(\ast\).

\[ r_1 = k_{6,f} \theta_{CO_2}^{-\nu_{CO_2\!-\!\ast,6}} - k_{6,r} \left[ CO_2 \right]^{\nu_{CO_2,6}} \theta_{\text{vacant}}^{\nu_{\ast,6}} \]

\[ r_1 = k_{6,f} \theta_{CO_2}^{-(-1)} - k_{6,r} \left[ CO_2 \right]^1 \theta_{\text{vacant}}^1 \]

\[ r_1 = k_{6,f} \theta_{CO_2} - k_{6,r} \left[ CO_2 \right] \theta_{\text{vacant}} \]

Assuming the other steps, reactions (2) through (5), to be at quasi-equilibrium gives equations (8) through (11).

\[ K_2 = \frac{\theta_{O_2}}{ \left[ O_2 \right] \theta_{\text{vacant}}} \tag{8} \]

\[ K_3 = \frac{\theta_{CO_3}}{ \left[ CO \right] \theta_{O_2}} \tag{9} \]

\[ K_4 = \frac{ \left[ CO_2 \right] \theta_{O}}{ \theta_{CO_3}} \tag{10} \]

\[ K_5 = \frac{\theta_{CO_2}}{ \left[ CO \right] \theta_O} \tag{11} \]

Start with Equation 5.16.

\[ K_{nrd} = \prod_i \left[i\right]^{\nu_{i,nrd}} \]

In the case of step (2), \(nrd=2\), \(i\) indexes every species in the system, and when \(i\) is a surface species, \([i] = \theta_i\).

\[ K_2 = \left[ CO \right]^{\nu_{CO,2}} \left[ O_2 \right]^{\nu_{O_2,2}} \left[ CO_2 \right]^{\nu_{CO_2,2}} \theta_{\text{vacant}}^{\nu_{\ast ,2}} \theta_{O_2}^{\nu_{O_2\!-\!\ast ,2}} \theta_{CO_3}^{\nu_{CO_3\!-\!\ast ,2}} \theta_{O}^{\nu_{O\!-\!\ast ,2}} \theta_{CO_2}^{\nu_{CO_2\!-\!\ast ,2}} \]

\[ K_2 = \left[ CO \right]^0 \left[ O_2 \right]^{-1} \left[ CO_2 \right]^0 \theta_{\text{vacant}}^{-1} \theta_{O_2}^{1} \theta_{CO_3}^0 \theta_{O}^0 \theta_{CO_2}^0 \]

\[ K_2 = \left[ O_2 \right]^{-1} \theta_{\text{vacant}}^{-1} \theta_{O_2}^{1} = \frac{\theta_{O_2}}{ \left[ O_2 \right] \theta_{\text{vacant}}} \]

The equations for steps (3), (4), and (5) are generated analogously.

Requiring the conservation of catalyst sites yields equation (12).

\[ 1 = \theta_{\text{vacant}} + \theta_{O_2} + \theta_{CO_3} + \theta_O + \theta_{CO_2} \tag{12} \]

Start with Equation 5.22.

\[ 1 = \theta_{\text{vacant}} + \sum_k \theta_k \]

The index \(k\) includes the four adsorbed species: \(O_2\!-\!\ast\), \(CO_3\!-\!\ast\), \(O\!-\!\ast\), and \(CO_2\!-\!\ast\)

\[ 1 = \theta_{\text{vacant}} + \theta_{O_2} + \theta_{CO_3} + \theta_O + \theta_{CO_2} \]

The five equations, (8) through (12), can be solved to obtain expressions the five surface coverages, \(\theta_{\text{vacant}}\), \(\theta_{O_2}\), \(\theta_{CO_3}\), \(\theta_O\), and \(\theta_{CO_2}\). Those expressions can then be substituted into equation (7) to generate an expression for the apparent rate of the non-elementary reaction (1).

Assuming \(O\!-\!\ast\) to be the most abundant surface intermediate leads to inequalities (13) through (16).

\[ \theta_{O} \gg \theta_{\text{vacant}} \tag{13} \]

\[ \theta_{O} \gg \theta_{CO_2} \tag{14} \]

\[ \theta_{O} \gg \theta_{CO_3} \tag{15} \]

\[ \theta_{O} \gg \theta_{O_2} \tag{16} \]

Start with Equation 5.24.

\[ \theta_{ma} \gg \theta_{nma} \]

In all four cases, \(ma = O\!-\!\ast\); in each case \(nma\) is one of the other surface species: \(\ast\), \(O_2\!-\!\ast\), \(CO_3\!-\!\ast\), and \(CO_2\!-\!\ast\).

After substitution of the expressions for the surface coverages into these inequalities, they may show that some terms in the expression for the apparent rate of the non-elementary reaction are negligible and can be eliminated.

Calculations

  1. Solve equations (8) through (12) to obtain expressions for \(\theta_{\text{vacant}}\), \(\theta_{O_2}\), \(\theta_{CO_3}\), \(\theta_{O}\), and \(\theta_{CO_2}\).
  2. Substitute the results into equation (7) to eliminate the surface coverages from the expression for the apparent rate of the non-elementary reaction.
  3. Substitute the results from step 1 into equations (13) through (16) and simplify.
  4. See whether the inequalities resulting from step 3 allow simplification of the rate expression.
    1. Typically the rate expression will have a denominator that contains a sum of several terms.
    2. If one of the inequalities shows that one term in the denominator is much much greater than another, delete the smaller term from the denominator.
    3. Repeat step 4.b for each inequality.

Results and Final Answer

Solving equations (8) through (12) leads to the expressions for the surface coverages shown in equations (17) through (21).

\[ \theta_{\text{vacant}} = \frac{1}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \tag{17} \]

\[ \theta_{O_2} = \frac{K_2 \left[ O_2 \right]}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \tag{18} \]

\[ \theta_{CO_3} = \frac{K_2 K_3 \left[ CO \right] \left[ O_2 \right]}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \tag{19} \]

\[ \theta_O = \frac{\frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]}}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \tag{20} \]

\[ \theta_{CO_2} = \frac{\frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \tag{21} \]

Substitution of equations (17) through (21) then leads to the expression for the apparent rate of non-elementary reaction (1) given in equation (22).

\[ \begin{align} r_1 =&\, \frac{\frac{k_{6,f}K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]} - k_{6,r} \left[ CO_2 \right]}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \\ & \\ &= \frac{k_{6,f}K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right] - k_{6,r} \left[ CO_2 \right]^2}{\begin{pmatrix}\left[ CO_2 \right] + K_2 \left[ O_2 \right]\left[ CO_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right]\left[ CO_2 \right] \\ + K_2K_3K_4 \left[ CO \right] \left[O_2 \right] + K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right] \end{pmatrix}} \end{align} \tag{22} \]

When equations (17) through (21) are substituted into the inequality expressions in equations (13) through (16), simplification leads to the inequalities shown in expressions (23) through (26).

\[ K_2K_3K_4 \left[ CO \right] \left[O_2 \right] \gg \left[ CO_2 \right] \tag{23} \]

\[ K_2K_3K_4 \left[ CO \right] \left[O_2 \right] \gg K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right] \tag{24} \]

\[ K_2K_3K_4 \left[ CO \right] \left[O_2 \right] \gg K_2 K_3 \left[ CO \right] \left[ O_2 \right]\left[ CO_2 \right] \tag{25} \]

\[ K_2K_3K_4 \left[ CO \right] \left[O_2 \right] \gg K_2 \left[ O_2 \right]\left[ CO_2 \right] \tag{26} \]

To generate equation (23), start with equation (13).

\[ \theta_{O} \gg \theta_{\text{vacant}} \]

Substitute equations (17) and (21).

\[ \begin{align} &\frac{\frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]}}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \\ &\gg \frac{1}{\begin{pmatrix}1 + K_2 \left[ O_2 \right] + K_2 K_3 \left[ CO \right] \left[ O_2 \right] \\ + \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} + \frac{K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right]}{\left[ CO_2 \right]}\end{pmatrix}} \end{align} \]

The denominators on the two sides of the inequality are the same. Multiply each side by the denominator.

\[ \frac{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]}{\left[ CO_2 \right]} \gg 1 \]

Multiply both sides by \(\left[ CO_2 \right]\).

\[ K_2K_3K_4 \left[ CO \right] \left[O_2 \right] \gg \left[ CO_2 \right] \]

The procedure for generating equations (24), (25), and (26) is analogous, but the starting equations are (14), (15), and (16), respectively.

Expression (23) shows that the fourth term in the denominator of equation (22) is much, much greater than the first term. Put differently, expression (23) shows that the first term in the denominator of equation (22) is negligible compared to the fourth term, and so the first term can be set equal to zero. Expression (24) similarly shows that the last term in the denominator is negligible; expression (25) shows that the third term is negligible; and expression (26) shows that the second term is negligible. When all of the negligible terms are set equal to zero, equation (22) simplifies to equation (27).

\[ r_1 = \frac{k_{6,f}K_2K_3K_4K_5\left[ CO \right]^2 \left[ O_2 \right] - k_{6,r} \left[ CO_2 \right]^2}{K_2K_3K_4 \left[ CO \right] \left[O_2 \right]} \tag{27} \]

Note

As often occurs in mechanistic rate expressions, several of the rate coefficients and equilibrium constants in the rate expression, equation (27), are coupled. It would not be possible to estimate unique values for each of them by fitting equation (27) to experimental rate data. Assuming that thermodynamic data are not available for the surface species so that the equilibrium constants are being treated as kinetics parameters, the rate coefficients and equilbrium constants can be combined as shown in equations (28) and (29), leading to the rate expression shown in equation (30).

\[ k_{1,f}^\prime = k_{6,f}K_5 \tag{28} \]

\[ k_{1,r}^\prime = \frac{k_{6,r}}{K_2K_3K_4} \tag{29} \]

\[ r_1 = k_{1,f}^\prime \left[ CO \right] - k_{1,r}^\prime \frac{\left[ CO_2 \right]^2}{\left[ CO \right] \left[O_2 \right]} \tag{30} \]

It is possible to estimate unique values for \(k_{1,f}^\prime\) and \(k_{1,r}^\prime\) using experimental data for the apparent rate of reaction (1) at varying concentrations of CO, O2, and CO2. The estimated apparent rate coefficients, \(k_{1,f}^\prime\) and \(k_{1,r}^\prime\), would be expected to exhibit Arrhenius temperature dependence.

5.4 Symbols Used in Chapter 5

Symbol Meaning
\(i\) Subscript denoting a specific reagent that is present in the system.
\(j\) Subscript denoting a reaction taking place in the system.
\(k\) Subscript denoting a surface species or variable representing an apparent rate coefficient.
\(k_j\) Rate coefficient for reaction \(j\); an additional subscripted \(f\) denotes the forward reaction, an additional subscripted \(r\) denotes the reverse reaction.
\(ma\) Subscript denoting the most abundant intermediate.
\(nma\) Subscript denoting an intermediate that is not the most abundant intermediate.
\(nrd\) Subscript denoting a step in a mechanism other than the rate-determining step.
\(q_i\) Net charge on reagent \(i\).
\(rds\) Subscript denoting the rate-determining step in a mechanism.
\(r_{i,j}\) Net rate (or apparent net rate) of generation of reagent \(i\) via reaction \(j\)
\(r_j\) Net rate of reaction \(j\), an additional subscripted \(\text{insig}\) denotes a kinetically insignificant reaction; an additional subscripted \(\text{irrev}\) deontes an effectively irreversible reaction.
\(C_{0,cat}\) Equivalent concentration of catalyst initially added to the system.
\(C_{cat,free}\) Concentration of uncomplexed catalyst.
\(C_i\) Concentration of reagent \(i\).
\(C_{sites}\) Surface concentration of sites on a heterogeneous catalytic surface.
\(E_0\) Equivalent concentration of enzyme initially added to the system.
\(K_m\) Parameter in a Michaelis-Menten type rate expression.
\(K_{nrd}\) Equilibrium constant for step \(nrd\) in a mechanism.
\(K_I\) Parameter in a Michaelis-Menten type rate expression.
\(RI\) Subscript denoting a reactive intermediate.
\(V_{max}\) Parameter in a Michaelis-Menten type rate expression.
\(\theta_k\) Fractional coverage of the catalyst surface by reagent \(k\).
\(\theta_{\text{vacant}}\) Fraction of the catalyst surface that is not covered by any reagent.
\(\kappa_i\) Number of catalyst moieties as originally added to the system that are part of catalyst complex \(i\).
\(\nu_{i,j}\) Stoichiometric coefficient of reagent \(i\) in reaction \(j\).
\(\sigma_j\) Stoichiometric number of mechanistic step \(j\).
\(\Delta G_j\) Gibbs free energy change for reaction \(j\).
\(\left[\,\,\right]\) Square brackets indicate the concentration or partial pressure of the reagent they contain.