9. Reaction Mechanisms, Pathways, Bioreactions, and Bioreactors

The next best thing to knowing something is knowing where to find it.

—Samuel Johnson (1709–1784)

were presented, where n was an integer of 0, 1, or 2 corresponding to a zero-, first-, or second-order reaction. However, for a large number of reactions, the orders are noninteger, such as the decomposition of acetaldehyde at 500°C

CH₃CHO → CH₄ + CO

where the rate law developed in problem P9-5_B(b) is

The rate law could also have concentration terms in both the numerator and denominator such as the formation of HBr from hydrogen and bromine

H₂ + Br₂ → 2HBr

where the rate law developed in problem P9-5_B(c) is

Rate laws of this form usually involve a number of elementary reactions and at least one active intermediate. An active intermediate is a high-energy molecule that reacts virtually as fast as it is formed. As a result, it is present in very small concentrations. Active intermediates (e.g., A*) can be formed by collision or interaction with other molecules.

A + M → A* + M

Properties of an active intermediate A*

Here, the activation occurs when translational kinetic energy is transferred into internal energy, i.e., vibrational and rotational energy.¹ An unstable molecule (i.e., active intermediate) is not formed solely as a consequence of the molecule moving at a high velocity (high translational kinetic energy). The energy must be absorbed into the chemical bonds, where high-amplitude oscillations will lead to bond ruptures, molecular rearrangement, and decomposition. In the absence of photochemical effects or similar phenomena, the transfer of translational energy to vibrational energy to produce an active intermediate can occur only as a consequence of molecular collision or interaction. Collision theory is discussed in the Professional Reference Shelf in Chapter 3 on the CRE Web site. Other types of active intermediates that can be formed are free radicals (one or more unpaired electrons; e.g., CH₃•), ionic intermediates (e.g., carbonium ion), and enzyme-substrate complexes, to mention a few.

¹ W. J. Moore, Physical Chemistry, (Reading, MA: Longman Publishing Group, 1998).

The idea of an active intermediate was first postulated in 1922 by F. A. Lindemann, who used it to explain changes in the reaction order with changes in reactant concentrations.² Because the active intermediates were so short-lived and present in such low concentrations, their existence was not really definitively confirmed until the work of Ahmed Zewail, who received the Nobel Prize in Chemistry in 1999 for femtosecond spectroscopy.³ His work on cyclobutane showed that the reaction to form two ethylene molecules did not proceed directly, as shown in Figure 9-1(a), but formed the active intermediate shown in the small trough at the top of the energy barrier on the reaction-coordinate diagram in Figure 9-1(b). As discussed in Chapter 3, an estimation of the barrier height, E, can be obtained using computational software packages such as Spartan, Cerius², or Gaussian as discussed in the Molecular Modeling Web Module in Chapter 3 on the CRE Web site, www.umich.edu/~elements/5e/index.html.

² F. A. Lindemann, Trans. Faraday. Soc., 17, 598 (1922).

³ J. Peterson, Science News, 156, 247 (1999).

This condition is also referred to as the Pseudo-Steady-State Hypothesis (PSSH). If the active intermediate appears in n reactions, then

PSSH

To illustrate how rate laws of this type are formed, we shall first consider the gas-phase decomposition of azomethane, AZO, to give ethane and nitrogen

Experimental observations show that the rate of formation of ethane is first order with respect to AZO at pressures greater than 1 atm (relatively high concentrations)⁴

⁴ H. C. Ramsperger, J. Am. Chem. Soc., 49, 912 (1927).

r_C2H6 ∝ C_AZO

and second order at pressures below 50 mmHg (low concentrations):

Why the reaction order changes

We could combine these two observations to postulate a rate law of the form

To find a mechanism that is consistent with the experimental observations, we use the steps shown in Table 9-1.

TABLE 9-1 STEPS TO DEDUCE A RATE LAW

We will now follow the steps in Table 9-1 to develop the rate law for azomethane (AZO) decomposition, –r_AZO.

Step 1. Propose an active intermediate. We will choose as an active intermediate an azomethane molecule that has been excited through molecular collisions, to form AZO*, i.e., [(CH₃)₂N₂]*.

Step 2. Propose a mechanism.

In reaction 1, two AZO molecules collide and the kinetic energy of one AZO molecule is transferred to the internal rotational and vibrational energies of the other AZO molecule, and it becomes activated and highly reactive (i.e., AZO*). In reaction 2, the activated molecule (AZO*) is deactivated through collision with another AZO by transferring its internal energy to increase the kinetic energy of the molecules with which AZO* collides. In reaction 3, this highly activated AZO* molecule, which is wildly vibrating, spontaneously decomposes into ethane and nitrogen.

Step 3. Write rate laws.

Because each of the reaction steps is elementary, the corresponding rate laws for the active intermediate AZO* in reactions (1), (2), and (3) are

Note: The specific reaction rates, k, are all defined wrt the active intermediate AZO*.

(Let k₁ = k_1AZO*, k₂ = k_2AZO*, and k₃ = k_3AZO*)

The rate laws shown in Equations (9-3) through (9-5) are pretty much useless in the design of any reaction system because the concentration of the active intermediate AZO* is not readily measurable. Consequently, we will use the Pseudo-Steady-State-Hypothesis (PSSH) to obtain a rate law in terms of measurable concentrations.

Step 4. Write rate of formation of product.

We first write the rate of formation of product

Step 5. Write net rate of formation of the active intermediate and use the PSSH.

To find the concentration of the active intermediate AZO*, we set the net rate of formation of AZO* equal to zero,⁵ r_AZO* ≡ 0.

⁵ For further elaboration on this section, see R. Aris, Am. Sci., 58, 419 (1970).

Solving for C_AZO*

Step 6. Eliminate the concentration of the active intermediate species in the rate laws by solving the simultaneous equations developed in Steps 4 and 5. Substituting Equation (9-8) into Equation (9-6)

Step 7. Compare with experimental data.

At low AZO concentrations,

for which case we obtain the following second-order rate law:

At high concentrations

in which case the rate expression follows first-order kinetics

Apparent reaction orders

In describing reaction orders for this equation, one would say that the reaction is apparent first order at high azomethane concentrations and apparent second order at low azomethane concentrations.

(CH₃)₂O → CH₄ + H₂ + CO

Symbolically, this reaction will be represented as A going to product P; that is,

A → P

with

–r_A = kC_A

The reaction is first order but the reaction is not elementary. The reaction proceeds by first forming an active intermediate, A*, from the collision of the reactant molecule and an inert molecule of M. Either this wildly oscillating active intermediate, A^*, is deactivated by collision with inert M, or it decomposes to form product. See Figure 9-2.

The mechanism consists of the three elementary reactions:

Writing the rate of formation of product

r_P = k₃C_A*

⁶ The reaction pathway for the reaction in Figure 9-2 is shown in the margin. We start at the top of the pathway with A and M, and move down to show the pathway of A and M touching k₁ (collision) where the curved lines come together and then separate to form A^* and M. The reverse of this reaction is shown starting with A^* and M and moving upward along the pathway where arrows A^* and M touch, k₂, and then separate to form A and M. At the bottom, we see the arrow from A^* to form P with k₃.

Reaction pathways⁶

and using the PSSH to find the concentrations of A* in a manner similar to the azomethane decomposition described earlier, the rate law can be shown to be

Because the concentration of the inert M is constant, we let

to obtain the first-order rate law

–r_A = kC_A

First-order rate law for a nonelementary reaction

Consequently, we see the reaction

A → P

follows an elementary rate law but is not an elementary reaction.

General Considerations. Developing a mechanism is a difficult and time-consuming task. The rules of thumb listed in Table 9-2 are by no means inclusive but may be of some help in the development of a simple mechanism that is consistent with the experimental rate law.

TABLE 9-2 RULES OF THUMB FOR DEVELOPMENT OF A MECHANISM

Upon application of Table 9-2 to the azomethane example just discussed, we see the following from rate equation (9-12):

1. The active intermediate, AZO^*, collides with azomethane, AZO [Reaction 2], resulting in the concentration of AZO in the denominator.

2. AZO^* decomposes spontaneously [Reaction 3], resulting in a constant in the denominator of the rate expression.

3. The appearance of AZO in the numerator suggests that the active intermediate AZO^* is formed from AZO. Referring to [Reaction 1], we see that this case is indeed true.

Glow sticks
Web Module

A discussion of luminescence in the Web Module, Glow Sticks on the CRE Web site (www.umich.edu/~elements/5e/index.html). Here, the PSSH is applied to glow sticks. First, a mechanism for the reactions and luminescence is developed. Next, mole balance equations are written on each species and coupled with the rate law obtained using the PSSH; the resulting equations are solved and compared with experimental data.

1. Initiation: formation of an active intermediate

2. Propagation or chain transfer: interaction of an active intermediate with the reactant or product to produce another active intermediate

3. Termination: deactivation of the active intermediate to form products

Steps in a chain reaction

An example comparing the application of the PSSH with the Polymath solution to the full set of equations is given in the Professional Reference Shelf R9.1, Chain Reactions, on the CRE Web site for the cracking of ethane. Also included in Professional Reference Shelf R9.2 is a discussion of Reaction Pathways and the chemistry of smog formation.

S → P

The figure also shows the catalyzed reaction pathway that proceeds through an active intermediate (E · S), called the enzyme–substrate complex, that is,

Because enzymatic pathways have lower activation energies, enhancements in reaction rates can be enormous, as in the degradation of urea by urease, where the degradation rate is on the order of 10¹⁴ higher than without the enzyme urease.

An important property of enzymes is that they are specific; that is, one enzyme can usually catalyze only one type of reaction. For example, a protease hydrolyzes only bonds between specific amino acids in proteins, an amylase works on bonds between glucose molecules in starch, and lipase attacks fats, degrading them to fatty acids and glycerol. Consequently, unwanted products are easily controlled in enzyme-catalyzed reactions. Enzymes are produced only by living organisms, and commercial enzymes are generally produced by bacteria. Enzymes usually work (i.e., catalyze reactions) under mild conditions: pH 4 to 9 and temperatures 75°F to 160°F. Most enzymes are named in terms of the reactions they catalyze. It is a customary practice to add the suffix -ase to a major part of the name of the substrate on which the enzyme acts. For example, the enzyme that catalyzes the decomposition of urea is urease and the enzyme that attacks tyrosine is tyrosinase. However, there are exceptions to the naming convention, such as α-amylase. The enzyme α-amylase catalyzes the transformation of starch in the first step in the production of the controversial soft drink (e.g., Red Pop) sweetener high-fructose corn syrup (HFCS) from corn starch, which is a $4 billion per year business.

⁷ M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River, NJ: Prentice Hall, 2002).

Two models for enzyme–substrate interactions are the lock-and-key model and the induced fit model, both of which are shown in Figure 9-5. For many years the lock-and-key model was preferred because of the sterospecific effects of one enzyme acting on one substrate. However, the induced fit model is the more useful model. In the induced fit model, both the enzyme molecule and the substrate molecules are distorted. These changes in conformation distort one or more of the substrate bonds, thereby stressing and weakening the bond to make the molecule more susceptible to rearrangement or attachment.

There are six classes of enzymes and only six:

More information about enzymes can be found on the following two Web sites: http://us.expasy.org/enzyme/ and www.chem.qmw.ac.uk/iubmb/enzyme. These sites also give information about enzymatic reactions in general.

⁸ N. Levine and W. C. LaCourse, J. Biomed. Mater. Res., 1, 275.

1. The unbound enzyme urease (E) reacts with the substrate urea (S) to form an enzyme–substrate complex (E · S):

The reaction mechanism

2. This complex (E · S) can decompose back to urea (S) and urease (E):

3. Or, it can react with water (W) to give the products (P) ammonia and carbon dioxide, and recover the enzyme urease (E):

Symbolically, the overall reaction is written as

We see that some of the enzyme added to the solution binds to the urea, and some of the enzyme remains unbound. Although we can easily measure the total concentration of enzyme, (E_t), it is difficult to measure either the concentration of free enzyme, (E), or the concentration of the bound enzyme (E · S).

Letting E, S, W, E · S, and P represent the enzyme, substrate, water, the enzyme–substrate complex, and the reaction products, respectively, we can write Reactions (9-13), (9-14), and (9-15) symbolically in the forms

Here, P = 2NH₃ + CO₂.

The corresponding rate laws for Reactions (9-16), (9-17), and (9-18) are

where all the specific reaction rates are defined with respect to (E · S). The net rate of formation of product, r_P, is

For the overall reaction

we know that the rate of consumption of the urea substrate equals the rate of formation of product CO₂, i.e., –r_S = r_P.

This rate law, Equation (9-19), is of not much use to us in making reaction engineering calculations because we cannot measure the concentration of the enzyme substrate complex (E · S). We will use the PSSH to express (E · S) in terms of measured variables.

The net rate of formation of the enzyme–substrate complex is

r_E⋅S = r_1E⋅S + r_{2 E⋅S} + r_3E⋅S

Substituting the rate laws, we obtain

Using the PSSH, r_E·S = 0, we can now solve Equation (9-20) for (E · S)

and substitute for (E · S) into Equation (9-19)

We need to replace unbound enzyme concentration (E) in the rate law.

We still cannot use this rate law because we cannot measure the unbound enzyme concentration (E); however, we can measure the total enzyme concentration, E_t.

In the absence of enzyme denaturation, the total concentration of the enzyme in the system, (E_t), is constant and equal to the sum of the concentrations of the free or unbounded enzyme, (E), and the enzyme–substrate complex, (E · S):

Total enzyme concentration = Bound + Free enzyme concentration.

Substituting for (E · S)

solving for (E)

and substituting for (E) in Equation (9-22), the rate law for substrate consumption is

Note: Throughout the following text, E_t ≡ (E_t) = total concentration of enzyme with typical units such as (kmol/m³) or (g/dm³).

Dividing the numerator and denominator of Equation (9-24) by k₁, we obtain a form of the Michaelis–Menten equation:

The final form of the rate law

Turnover number k_cat

Michaelis constant K_M

The parameter k_cat is also referred to as the turnover number. It is the number of substrate molecules converted to product in a given time on a single-enzyme molecule when the enzyme is saturated with substrate (i.e., all the active sites on the enzyme are occupied, (S)>>K_M). For example, the turnover number for the decomposition of hydrogen-peroxide, H₂O₂, by the enzyme catalase is 40 × 10⁶ s^–1. That is, 40 million molecules of H₂O₂ are decomposed every second on a single-enzyme molecule saturated with H₂O₂. The constant K_M (mol/dm³) is called the Michaelis constant and for simple systems is a measure of the attraction of the enzyme for its substrate, so it’s also called the affinity constant. The Michaelis constant, K_M, for the decomposition of H₂O₂ discussed earlier is 1.1 M, while that for chymotrypsin is 0.1 M.⁹

⁹ D. L. Nelson and M. M. Cox, Lehninger Principles of Biochemistry, 3rd ed. (New York: Worth Publishers, 2000).

If, in addition, we let V_max represent the maximum rate of reaction for a given total enzyme concentration

V_max = k_cat(E_t)

the Michaelis–Menten equation takes the familiar form

Michaelis–Menten equation

For a given enzyme concentration, a sketch of the rate of disappearance of the substrate is shown as a function of the substrate concentration in Figure 9-6.

Michaelis–Menten plot

A plot of this type is sometimes called a Michaelis–Menten plot. At low substrate concentration, , Equation (9-26) reduces to

and the reaction is apparent first order in the substrate concentration. At high substrate concentrations,

Equation (9-26) reduces to

–r_S ≅ V_max

Interpretation of Michaelis constant

and we see the reaction is apparent zero order.

What does K_M represent? Consider the case when the substrate concentration is such that the reaction rate is equal to one-half the maximum rate

then

Solving Equation (9-27) for the Michaelis constant yields

K_M = (S_1/2)

The Michaelis constant is equal to the substrate concentration at which the rate of reaction is equal to one-half the maximum rate, i.e., –r_A = V_max/2. The larger the value of K_M, the higher the substrate concentration necessary for the reaction rate to reach half of its maximum value.

The parameters V_max and K_M characterize the enzymatic reactions that are described by Michaelis–Menten kinetics. V_max is dependent on total enzyme concentration, whereas K_M is not.

Two enzymes may have the same values for k_cat but have different reaction rates because of different values of K_M. One way to compare the catalytic efficiencies of different enzymes is to compare their ratios k_cat/K_M. When this ratio approaches 10⁸ to 10⁹ (dm³/mol/s), the reaction rate approaches becoming diffusion limited. That is, it takes a long time for the enzyme and substrate to find each other, but once they do they react immediately. We will discuss diffusion-limited reactions in Chapters 14 and 15.

Briggs–Haldane Rate Law

Applying the PSSH, after much effort and some approximations we obtain

which is often referred to as the Briggs–Haldane Equation [see P9-8_B (a)] and the application of the PSSH to enzyme kinetics, often called the Briggs–Haldane approximation.

Mole balance

Because this reaction is liquid phase, V = V₀, the mole balance can be put in the following form:

The rate law for urea decomposition is

Rate law

Substituting Equation (9-31) into Equation (9-30) and then rearranging and integrating, we get

Combine

Integrate

We can write Equation (9-32) in terms of conversion as

Time to achieve a conversion X in a batch enzymatic reaction

The parameters K_M and V_max can readily be determined from batch-reactor data by using the integral method of analysis. Dividing both sides of Equation (9-32) by (tK_M/V_max) and rearranging yields

We see that K_M and V_max can be determined from the slope and intercept of a plot of (1/t) ln [1/(1 – X)] versus X/t. We could also express the Michaelis–Menten equation in terms of the substrate concentration S

where S₀ is the initial concentration of substrate. In cases similar to Equation (9-33) where there is no possibility of confusion, we shall not bother to enclose the substrate or other species in parentheses to represent concentration [i.e., C_S ≡ (S) ≡ S]. The corresponding plot in terms of substrate concentration is shown in Figure 9-7.

¹⁰ M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River, NJ: Prentice Hall, 2002), p. 77.

¹¹ S. Aiba, A. E. Humphrey, and N. F. Mills, Biochemical Engineering (New York: Academic Press, 1973), p. 47.

The three most common types of reversible inhibition occurring in enzymatic reactions are competitive, uncompetitive, and noncompetitive. The enzyme molecule is analogous to a heterogeneous catalytic surface in that it contains active sites. When competitive inhibition occurs, the substrate and inhibitor are usually similar molecules that compete for the same site on the enzyme. Uncompetitive inhibition occurs when the inhibitor deactivates the enzyme–substrate complex, sometimes by attaching itself to both the substrate and enzyme molecules of the complex. Noncompetitive inhibition occurs with enzymes containing at least two different types of sites. The substrate attaches only to one type of site, and the inhibitor attaches only to the other to render the enzyme inactive.

In competitive inhibition, another substance, i.e., I, competes with the substrate for the enzyme molecules to form an inhibitor–enzyme complex, as shown in Figure 9-9.

In addition to the three Michaelis–Menten reaction steps, there are two additional steps as the inhibitor (I) reversely ties up the enzyme, as shown in Reaction Steps 4 and 5.

The rate law for the formation of product is the same [cf. Equations (9-18A) and (9-19)] as it was before in the absence of inhibitor

Applying the PSSH, the net rate of reaction of the enzyme–substrate complex is

The net rate of formation of the inhibitor–substrate complex (E · I) is also zero

Reaction Steps

Competitive Inhibition Pathway

The total enzyme concentration is the sum of the bound and unbound enzyme concentrations

Combining Equations (9-35), (9-36), and (9-37), solving for (E · S) then substituting in Equation (9-34) and simplifying

Rate law for competitive inhibition

V_max and K_M are the same as before when no inhibitor is present; that is,

and the inhibition constant K_I (mol/dm³) is

By letting , we can see that the effect of adding a competitive inhibitor is to increase the “apparent” Michaelis constant, . A consequence of the larger “apparent” Michaelis constant is that a larger substrate concentration is needed for the rate of substrate decomposition, –r_S, to reach half its maximum rate.

Rearranging Equation (9-38) in order to generate a Lineweaver–Burk plot

From the Lineweaver–Burk plot (Figure 9-10), we see that as the inhibitor (I) concentration is increased, the slope increases (i.e., the rate decreases), while the intercept remains fixed.

As with competitive inhibition, two additional reaction steps are added to the Michaelis–Menten kinetics for uncompetitive inhibition, as shown in Reaction Steps 4 and 5 in Figure 9-11.

Starting with the equation for the rate of formation of product, Equation (9-34), and then applying the pseudo-steady-state hypothesis to the intermediate (I · E · S), we arrive at the rate law for uncompetitive inhibition

Rate law for uncompetitive inhibition

The intermediate steps are shown in the Chapter 9 Summary Notes in Learning Resources on the CRE Web site. Rearranging Equation (9-40)

Reaction Steps

Uncompetitive inhibition pathway

The Lineweaver–Burk plot is shown in Figure 9-12 for different inhibitor concentrations. The slope (K_M/V_max) remains the same as the inhibitor (I) concentration is increased, while the intercept [(1/V_max)(1 + (I)/K_I)] increases.

In noncompetitive inhibition, also sometimes called mixed inhibition, the substrate and inhibitor molecules react with different types of sites on the enzyme molecule. Whenever the inhibitor is attached to the enzyme, it is inactive and cannot form products. Consequently, the deactivating complex (I · E · S) can be formed by two reversible reaction paths.

1. After a substrate molecule attaches to the enzyme at the substrate site, then the inhibitor molecule attaches to the enzyme at the inhibitor site. (E · S + I I · E · S)

2. After an inhibitor molecule attaches to the enzyme molecule at the inhibitor site, then the substrate molecule attaches to the enzyme at the substrate site. (E · I + S I · E · S)

These paths, along with the formation of the product, P, are shown in Figure 9-13. In noncompetitive inhibition, the enzyme can be tied up in its inactive form either before or after forming the enzyme–substrate complex as shown in Steps 2, 3, and 4.

Again, starting with the rate law for the rate of formation of product and then applying the PSSH to the complexes (I · E) and (I · E · S), we arrive at the rate law for the noncompetitive inhibition

Rate law for noncompetitive inhibition

The derivation of the rate law is given in the Summary Notes on the CRE Web site. Equation (9-42) is in the form of the rate law that is given for an enzymatic reaction exhibiting noncompetitive inhibition. Heavy metal ions such as Pb²⁺, Ag⁺, and Hg²⁺, as well as inhibitors that react with the enzyme to form chemical derivatives, are typical examples of noncompetitive inhibitors.

Rearranging

Reaction Steps

Mixed inhibition

For noncompetitive inhibition, we see in Figure 9-14 that both the slope and intercept increase with increasing inhibitor concentration. In practice, uncompetitive inhibition and mixed inhibition are generally observed only for enzymes with two or more substrates, S₁ and S₂.

The three types of inhibition are compared with a reaction without inhibitors and are summarized on the Lineweaver–Burk plot shown in Figure 9-15.

Summary plot of types of inhibition

In summary, we observe the following trends and relationships:

1. In competitive inhibition, the slope increases with increasing inhibitor concentration, while the intercept remains fixed.

2. In uncompetitive inhibition, the y-intercept increases with increasing inhibitor concentration, while the slope remains fixed.

3. In noncompetitive inhibition (mixed inhibition), both the y–intercept and slope will increase with increasing inhibitor concentration.

Problem P9-12_B asks you to find the type of inhibition for the enzyme catalyzed reaction of starch.

Consequently, we see that by replacing (I) by (S) in Equation (9-40), the rate law for –r_S is

We see that at low substrate concentrations

then

and the rate increases linearly with increasing substrate concentration.

At high substrate concentrations ((S)2/K_I) >>(K_M + (S)), then

Substrate inhibition

and we see that the rate decreases as the substrate concentration increases. Consequently, the rate of reaction goes through a maximum in the substrate concentration, as shown in Figure 9-16. We also see that there is an optimum substrate concentration at which to operate. This maximum is found by setting the derivative of –r_S wrt S in Equation (9-44) equal to 0, to obtain

When substrate inhibition is possible, the substrate is fed to a semibatch reactor called a fed batch to maximize the reaction rate and conversion.

Our discussion of enzymes is continued in the Professional Reference Shelf on the CRE Web site where we describe multiple enzyme and substrate systems, enzyme regeneration, and enzyme co-factors (see R9.6).

The growth of biotechnology

$200 billion

The use of living cells to produce marketable chemical products is becoming increasingly important. The number of chemicals, agricultural products, and food products produced by biosynthesis has risen dramatically. In 2016, companies in this sector raised over $200 billion of new financing.¹³ Both microorganisms and mammalian cells are being used to produce a variety of products, such as insulin, most antibiotics, and polymers. It is expected that in the future a number of organic chemicals currently derived from petroleum will be produced by living cells. The advantages of bioconversions are mild reaction conditions; high yields (e.g., 100% conversion of glucose to gluconic acid with Aspergillus niger); and the fact that organisms contain several enzymes that can catalyze successive steps in a reaction and, most importantly, act as stereospecific catalysts. A common example of specificity in bioconversion production of a single desired isomer that, when produced chemically, yields a mixture of isomers is the conversion of cis-proenylphosphonic acid to the antibiotic (–) cis-1,2-epoxypropyl-phosphonic acid. Bacteria can also be modified and turned into living chemical factories. For example, using recombinant DNA, Biotechnic International engineered a bacteria to produce fertilizer by turning nitrogen into nitrates.¹⁴

¹³ http://www.statista.com/topics/1634/biotechnology-industry/

¹⁴ Chem. Eng. Progr., August 1988, p. 18

More recently, the synthesis of biomass (i.e., cell/organisms) has become an important alternative energy source. Sapphire energy, the world’s first integrated algae-oil production facility, has pilot growth ponds in New Mexico. In this process, they grow the algae, flocculate and concentrate it, extract it, and then refine it and convert it to fuel oil in a liquid-phase flow reactor (see Problems P9-20_B and P9-21_B). In 2009, ExxonMobil invested over 600 million dollars to develop algae growth and harvest it in waste ponds. It is estimated that one acre of algae can provide 2,000 gallons of gasoline per year.

In biosynthesis, the cells, also referred to as the biomass, consume nutrients to grow and produce more cells and important products. Internally, a cell uses its nutrients to produce energy and more cells. This transformation of nutrients to energy and bioproducts is accomplished through a cell’s use of a number of different enzymes in a series of reactions to produce metabolic products. These products can either remain in the cell (intracellular) or be secreted from the cells (extracellular). In the former case, the cells must be lysed (ruptured) and the product filtered and purified from the whole broth (reaction mixture). A schematic of a cell is shown in Figure 9-17.

The cell consists of a cell wall and an outer membrane that encloses the cytoplasm containing a nuclear region and ribosomes. The cell wall protects the cell from external influences. The cell membrane provides for selective transport of materials into and out of the cell. Other substances can attach to the cell membrane to carry out important cell functions. The cytoplasm contains the ribosomes that contain ribonucleic acid (RNA), which are important in the synthesis of proteins. The nuclear region contains deoxyribonucleic acid (DNA), which provides the genetic information for the production of proteins and other cellular substances and structures.¹⁵

¹⁵ M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River, NJ: Prentice Hall, 2002).

The reactions in the cell all take place simultaneously and are classified as either class (I) nutrient degradation (fueling reactions), class (II) synthesis of small molecules (amino acids), or class (III) synthesis of large molecules (polymerization, e.g., RNA, DNA). A rough overview with only a fraction of the reactions and metabolic pathways is shown in Figure 9-18. A more detailed model is given in Figures 5.1 and 6.14 of Shuler and Kargi.¹⁶ In the Class I reactions, adenosine triphosphate (ATP) participates in the degradation of nutrients to form products to be used in the biosynthesis reactions (Class II) of small molecules (e.g., amino acids), which are then polymerized to form RNA and DNA (Class III). ATP also transfers the energy it releases when it loses a phosphonate group to form adenosine diphosphate (ADP).

¹⁶ M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River, NJ: Prentice Hall, 2002), pp.135, 185.

In general, the growth of an aerobic organism follows the equation

Cell multiplication

A more abbreviated form of Equation (9-49) generally used is that a substrate in the presence of cells produces more cells plus product, i.e.,

The products in Equation (9-50) include carbon dioxide, water, proteins, and other species specific to the particular reaction. An excellent discussion of the stoichiometry (atom and mole balances) of Equation (9-49) can be found in Shuler and Kargi,¹⁷ Bailey and Ollis,¹⁸ and Blanch and Clark.¹⁹ The substrate culture medium contains all the nutrients (carbon, nitrogen, etc.) along with other chemicals necessary for growth. Because, as we will soon see, the rate of this reaction is proportional to the cell concentration, the reaction is autocatalytic. A rough schematic of a simple batch biochemical reactor and the growth of two types of microorganisms, cocci (i.e., spherical) bacteria and yeast, is shown in Figure 9-20.

¹⁷ M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River, NJ: Prentice Hall, 2002).

¹⁸ J. E. Bailey and D. F. Ollis, Biochemical Engineering, 2nd ed. (New York: McGraw-Hill, 1987).

¹⁹ H. W. Blanch and D. S. Clark, Biochemical Engineering (New York: Marcel Dekker, Inc. 1996).

Lag phase

Phase I, shown in Figure 9-22, is called the lag phase. There is little increase in cell concentration in this phase. In the lag phase, the cells are adjusting to their new environment, carrying out such functions as synthesizing transport proteins for moving the substrate into the cell, synthesizing enzymes for utilizing the new substrate, and beginning the work for replicating the cells’ genetic material. The duration of the lag phase depends upon many things, one of which is the growth medium from which the inoculum was taken relative to the reaction medium in which it is placed. If the inoculum is similar to the medium of the batch reactor, the lag phase can be almost nonexistent. If, however, the inoculum were placed in a medium with a different nutrient or other contents, or if the inoculum culture were in the stationary or death phase, the cells would have to readjust their metabolic path to allow them to consume the nutrients in their new environment.²⁰

²⁰ B. Wolf and H. S. Fogler, “Alteration of the Growth Rate and Lag Time of Leuconostoc mesenteroides NRRL-B523,” Biotechnology and Bioengineering, 72 (6), 603 (2001). B. Wolf and H. S. Fogler, “Growth of Leuconostoc mesenteroides NRRL-B523, in Alkaline Medium,” Biotechnology and Bioengineering, 89 (1), 96 (2005).

Exponential growth phase

Phase II is called the exponential growth phase, owing to the fact that the cells’ growth rate is proportional to the cell concentration. In this phase, the cells are dividing at the maximum rate because all of the enzyme’s pathways for metabolizing the substrate are now in place (as a result of the lag phase) and the cells are able to use the nutrients most efficiently.

Antibiotics produced during the stationary phase

Phase III is the stationary phase, during which the cells reach a minimum biological space where the lack of one or more nutrients limits cell growth. During the stationary phase, the net cell growth rate is zero as a result of the depletion of nutrients and essential metabolites. Many important fermentation products, including many antibiotics, are produced in the stationary phase. For example, penicillin produced commercially using the fungus Penicillium chrysogenum is formed only after cell growth has ceased. Cell growth is also slowed by the buildup of organic acids and toxic materials generated during the growth phase.

Death phase

The final phase, Phase IV, is the death phase, where a decrease in live cell concentration occurs. This decline is a result of the toxic by-products, harsh environments, and/or depletion of nutrient supply.

the most commonly used expression is the Monod equation for exponential growth:

The cell concentration is often given in terms of weight (g) of dry cells per liquid volume and is specified “grams dry weight per dm³,” i.e., (gdw/dm³).

The specific cell growth rate can be expressed as

Representative values of μ_max and K_s are 1.3 h^–1 and 2.2 × 10^–5 g/dm³, respectively, which are the parameter values for the E. coli growth on glucose. Combining Equations (9-51) and (9-52), we arrive at the Monod equation for bacterial cell growth rate

Monod equation

For a number of different bacteria, the constant K_s is very small, with regard to typical substrate concentrations, in which case the rate law reduces to

The growth rate, r_g, often depends on more than one nutrient concentration; however, the nutrient that is limiting is usually the one used in Equation (9-53).

In many systems the product inhibits the rate of growth. A classic example of this inhibition is in winemaking, where the fermentation of glucose to produce ethanol is inhibited by the product ethanol. There are a number of different equations to account for inhibition; one such rate law takes the empirical form

where

Empirical form of Monod equation for product inhibition

with

For the glucose-to-ethanol fermentation, typical inhibition parameters are

In addition to the Monod equation, two other equations are also commonly used to describe the cell growth rate; they are the Tessier equation

and the Moser equation,

where λ and k are empirical constants determined by a best fit of the data. The Moser and Tessier growth laws are often used because they have been found to better fit experimental data at the beginning or end of fermentation. Other growth equations can be found in Dean.²¹

²¹ A. R. C. Dean, Growth, Function, and Regulation in Bacterial Cells (London: Oxford University Press, 1964).

The cell death rate is a result of harsh environments, mixing shear forces, local depletion of nutrients, and the presence of toxic substances. The rate law is

where C_t is the concentration of a substance toxic to the cell. The specific death rate constants k_d and k_t refer to the natural death and death due to a toxic substance, respectively. Representative values of k_d range from 0.1 h^–1 to less than 0.0005 h^–1. The value of k_t depends on the nature of the toxin.

Doubling times

Microbial growth rates are measured in terms of doubling times. Doubling time is the time required for a mass of an organism to double. Typical doubling times for bacteria range from 45 minutes to 1 hour but can be as fast as 15 minutes. Doubling times for simple eukaryotes, such as yeast, range from 1.5 to 2 hours but may be as fast as 45 minutes.

Effect of Temperature. As with enzymes (cf. Figure 9-8), there is an optimum in growth rate with temperature, owing to the competition of increased rates with increasing temperature and enzyme denaturation at high temperatures. An empirical law that describes this functionality is given in Aiba et al.²² and is of the form

²² S. Aiba, A. E. Humphrey, and N. F. Millis, Biochemical Engineering (New York: Academic Press, 1973), p. 407.

where I′ is the fraction of the maximum growth rate, T_m is the temperature at which the maximum growth occurs, and μ(T_m) is the growth rate at this temperature. For the rate of oxygen uptake of Rhizobium trifollic, the equation takes the form

The maximum growth of Rhizobium trifollic occurs at 310°K. However, experiments by Prof. Dr. Sven Köttlov of Jofostan University in Riça, Jofostan, show that this temperature should be 312°K, not 310°K.

In order to relate the substrate consumed, new cells formed, and product generated, we introduce the yield coefficients. The yield coefficient for cells and substrate is

The yield coefficient Y_c/s is the ratio of the increase in the mass concentration of cells, ΔC_C, to the decrease in the substrate concentration (–ΔC_s), (–ΔC_s = C_s0 – C_s), to bring about this increase in cell mass concentration. A representative value of Y_c/s might be 0.4 (g/g).

The reciprocal of Y_c/s, i.e., Y_s/c

gives the ratio of –ΔC_s, the substrate that must be consumed, to the increase in cell mass concentration ΔC_c.

Product formation can take place during different phases of the cell growth cycle. When product formation only occurs during the exponential growth phase, the rate of product formation is

Growth associated product formation

where

The product of Y_p/c and μ, that is, (q_P = Y_p/c μ), is often called the specific rate of product formation, q_P, (mass product/volume/time). When the product is formed during the stationary phase where no cell growth occurs, we can relate the rate of product formation to substrate consumption by

Nongrowth associated product formation

The substrate in this case is usually a secondary nutrient, which we discuss in more detail later when the stationary phase is discussed.

The stoichiometric yield coefficient that relates the amount of product formed per mass of substrate consumed is

In addition to consuming substrate to produce new cells, part of the substrate must be used just to maintain a cell’s daily activities. The corresponding maintenance utilization term is

Cell maintenance

A typical value is

The rate of substrate consumption for maintenance, r_sm, whether or not the cells are growing is

When maintenance can be neglected, we can relate the concentration of new cells formed to the amount of substrate consumed by the equation

Neglecting cell maintenance

This equation can be used for both batch and continuous flow reactors.

If it were possible to sort out the substrate (S) that is consumed in the presence of cells to form new cells (C) from the substrate that is consumed to form product (P), that is

the yield coefficients would be written as

These yield coefficients will be discussed further in the substrate utilization section.

Substrate Utilization. We now come to the task of relating the rate of nutrient (i..e., substrate) consumption, –r_s, to the rates of cell growth, product generation, and cell maintenance. In general, we can write

Substrate accounting

In a number of cases, extra attention must be paid to the substrate balance. If product is produced during the growth phase, it may not be possible to separate out the amount of substrate consumed for cell growth (i.e., produce more cells) from that consumed to produce the product. Under these circumstances, all the substrate consumed for growth and for product formation is lumped into a single stoichiometric yield coefficient, Y_s/c, and the rate of substrate disappearance is

The corresponding rate of product formation is

Growth-associated product formation in the growth phase

The Stationary Phase. Because there is no growth during the stationary phase, it is clear that Equation (9-70) cannot be used to account for substrate consumption, nor can the rate of product formation be related to the growth rate [e.g., Equation (9-63)]. Many antibiotics, such as penicillin, are produced in the stationary phase. In this phase, the nutrient required for growth becomes virtually exhausted, and a different nutrient, called the secondary nutrient, is used for cell maintenance and to produce the desired product. Usually, the rate law for product formation during the stationary phase is similar in form to the Monod equation, that is

Nongrowth-associated product formation in the stationary phase

The net rate of secondary nutrient consumption, r_sn, during the stationary phase is

In the stationary phase, the concentration of live cells is constant.

Because the desired product can be produced when there is no cell growth, it is always best to relate the product concentration to the change in secondary nutrient concentration. For a batch system, the concentration of product, C_p, formed after a time t in the stationary phase can be related to the secondary nutrient concentration, C_sn, at that time.

Neglects cell maintenance

We have considered two limiting situations for relating substrate consumption to cell growth and product formation: product formation only during the growth phase and product formation only during the stationary phase. An example where neither of these situations applies is fermentation using lactobacillus, where lactic acid is produced during both the logarithmic growth and stationary phase.

The specific rate of product formation is often given in terms of the Luedeking–Piret equation, which has two parameters, α (growth) and β (nongrowth)

with

r_p = q_pC_c

Luedeking–Piret equation for the rate of product formation

The assumption here in using the β-parameter is that the secondary nutrient is in excess.

Cell mass balance

The corresponding substrate balance is

Substrate balance

In most systems, the entering microorganism concentration, C_c0, is zero for a flow reactor.

The mass balances

Dividing by V yields the substrate balance for the growth phase

Growth phase

For cells in the stationary phase, where there is no growth in cell concentration, cell maintenance and product formation are the only reactions to consume the secondary substrate. Under these conditions the substrate balance, Equation (9-76), reduces to

Stationary phase

Typically, r_p will have the same Monod form of the rate law as r_g [e.g., Equation (9-71)]. Of course, Equation (9-79) only applies for substrate concentrations greater than zero.

Batch stationary growth phase

During the growth phase, we could also relate the rate of formation of product, r_p, to the cell growth rate, r_g, Equation (9-63), i.e., r_p = Y_p/cr_g. The coupled first-order ordinary differential equations above can be solved by a variety of numerical techniques.

and is simply the reciprocal of the space time τ, i.e., . Dividing Equations (9-75) and (9-76) by V and using the definition of the dilution rate, we have

CSTR mass balances

Using the Monod equation, the growth rate is determined to be

Rate law

For steady-state operation we have

Steady state

and

We now neglect the death rate, rd, and combine Equations (9-51) and (9-84) for steady-state operation to obtain the mass flow rate of cells out of the chemostat, , and the rate of generation of cells, r_gV. Equating and r_gV, and then substituting for r_g = μC_c, we obtain

Dividing by C_cV we see the cell concentration cancels to give the dilution rate D

Dilution rate

How to control cell growth

An inspection of Equation (9-87) reveals that the specific growth rate of the cells can be controlled by the operator by controlling the dilution rate D, i.e., . Using Equation (9-52)

to substitute for μ in terms of the substrate concentration and then solving for the steady-state substrate concentration yields

Assuming that a single nutrient is limiting, cell growth is the only process contributing to substrate utilization, and that cell maintenance can be neglected, the stoichiometry is

Substituting for C_s using Equation (9-87) and rearranging, we obtain

We see that if D > μ, then (dC_c/dt) will be negative, and the cell concentration will continue to decrease until we reach a point where all cells will be washed out:

C_c = 0

The dilution rate at which wash-out will occur is obtained from Equation (9-90) by setting C_c = 0.

Flow rate at which wash-out occurs

We next want to determine the other extreme for the dilution rate, which is the rate of maximum cell production. The cell production rate per unit volume of reactor is the mass flow rate of cells out of the reactor (i.e., ) divided by the volume V, or

Maximum rate of cell production (DC_c)

Using Equation (9-90) to substitute for C_c yields

Figure 9-25 shows production rate, cell concentration, and substrate concentration as functions of dilution rate.

We observe a maximum in the production rate, and this maximum can be found by differentiating the production rate, Equation (9-94), with respect to the dilution rate D:

Then

Maximum rate of cell production

The organism Streptomyces aureofaciens was studied in a 10-dm³ chemostat using sucrose as a substrate. The cell concentration, C_c (mg/ml), the substrate concentration, C_s (mg/ml), and the production rate, DC_c (mg/ml/h), were measured at steady state for different dilution rates. The data are shown in Figure 9-26.²³ Note that the data follow the same trends as those discussed in Figure 9-25.

²³ B. Sikyta, J. Slezak, and M. Herold, Appl. Microbiol., 9, 233 (1961).

This approximation is justified when the active intermediate is highly reactive and present in low concentrations.

2. The azomethane (AZO) decomposition mechanism is

By applying the PSSH to AZO*, we show the rate law, which exhibits first-order dependence with respect to AZO at high AZO concentrations and second-order dependence with respect to AZO at low AZO concentrations.

3. Enzyme kinetics: enzymatic reactions follow the sequence

Using the PSSH for (E · S) and a balance on the total enzyme, E_t, which includes both the bound (E · S) and unbound enzyme (E) concentrations

E_t = (E) + (E · S)

we arrive at the Michaelis–Menten equation

where V_max is the maximum reaction rate at large substrate concentrations (S >> K_M) and K_M is the Michaelis constant. K_M is the substrate concentration at which the rate is half the maximum rate (S_1/2 = K_M).

4. The three different types of inhibition—competitive, uncompetitive, and noncompetitive (mixed)—are shown on the Lineweaver–Burk plot:

5. Bioreactors:

(a) Phases of bacteria growth:

I. Lag    II. Exponential    III. Stationary    IV. Death

(b) Unsteady-state mass balance on a chemostat

(c) Monod growth rate law

(d) Stoichiometry

Substrate consumption

1. Puzzle Problem “What’s Wrong with this Solution?”

2. Physiologically Based Pharmacokinetic Model for Alcohol Metabolism

• Learning Resources

1. Summary Notes

2. Web Modules

A. Ozone Layer

Earth Probe TOMs Total Ozone September 8, 2000

Ozone (Dotson Units)

B. Glow Sticks

Photo courtesy of Goddard Space Flight Center (NASA). See the Web Modules on the CRE Web site for color pictures of the ozone layer and the glow sticks.

3. Interactive Computer Games

Enzyme Man

• Living Example Problems

1. Example 9-5 Bacteria Growth in a Batch Reactor

2. Example Chapter 9 CRE Web site: PSSH Applied to Thermal Cracking of Ethane

3. Example Chapter 9 CRE Web site: Alcohol Metabolism

4. Example Web Module: Ozone

5. Example Web Module: Glowsticks

• Professional Reference Shelf

R9-1. Chain Reactions Example Problem

R9-2. Reaction Pathways

R9-3. Polymerization

A. Step Polymerization

Example R9-3.1 Determining the Concentration of Polymers for Step Polymerization

B. Chain Polymerizations

Example R9-3.2 Parameters of MW Distribution

C. Anionic Polymerization

Example R9-3.3 Calculating the Distribution Parameters from Analytic Expressions for Anionic Polymerization

Example R9-3.4 Determination of Dead Polymer Distribution When Transfer to Monomer Is the Primary Termination Step

R9-4. Oxygen-Limited Fermentation Scale-Up

R9-5. Receptor Kinetics

A. Kinetics of Signaling

B. Endocytosis

R9-6. Multiple Enzyme and Substrate Systems

A. Enzyme Regeneration

Example PRS9-6.1 Construct a Lineweaver–Burk Plot for Different Oxygen Concentration

B. Enzyme Cofactors

(1) Example PRS9-6.2 Derive a Rate Law for Alcohol Dehydrogenase

(2) Example PRS9-6.3 Derive a Rate Law for a Multiple Substrate System

(3) Example PRS9-6.4 Calculate the Initial Rate of Formation of Ethanol in the Presence of Propanediol

R9-7. Physiologically Based Pharmacokinetic (PBPK) Models. Case Study: Alcohol Metabolism in Humans

²⁴ P. K. Wilkinson, et al., “Pharmacokinetics of Ethanol After Oral Administration in the Fasting State,” J. Pharmacoket. Biopharm., 5(3): 207–224 (1977).

R9-8. Pharmacokinetics in Drug Delivery

Pharmacokinetic models of drug delivery for medication administered either orally or intravenously are developed and analyzed.

You may wish to refer to W. Strunk and E. B. White, The Elements of Style, 4th ed. (New York: Macmillan, 2000) and Joseph M. Williams, Style: Ten Lessons in Clarity & Grace, 6th ed. (Glenview, IL: Scott, Foresman, 1999) to enhance the quality of your sentences.

P9-1_A ICG Enzyme Man.

(a) Load the ICG on your computer and carry out the exercise. Performance number = _______________________________.

(b) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-2_B

(a) Example 9-1. How would the results change if the concentration of CS₂ and M were increased?

(b) Example 9-2. (1) The following additional runs were carried out when an inhibitor was present:

What type of inhibition is taking place? (2) Sketch the curves for no inhibition as well as competitive, uncompetitive, noncompetitive (mixed), and substrate inhibition on a Woolf–Hanes plot and on an Eadie–Hofstee plot.

(c) Example 9-3. (1) What would the conversion be after 10 minutes if the initial concentration of urea were decreased by a factor of 100? (2) What would be the conversion in a CSTR with the same residence time, τ, as the batch reactor time t? (3) In a PFR?

(d) Example 9-4. What is the total mass of substrate consumed in grams per mass of cells plus what is consumed to form product? Is there disparity here?

(e) Example 9-5. Download the Living Example Problem. (1) Modify the code to carry out the fermentation in a fed-batch (e.g., semibatch) reactor in which the substrate is fed at a rate of 0.5 dm³/h and a concentration of 5 g/dm³ to an initial liquid volume of 1.0 dm³ containing a cell mass with an initial concentration of C_ci = 0.2 mg/dm³ and an initial substrate concentration of C_ci = 0.5 mg/dm³. Plot and analyze the concentration of cells, substrate, and product as a function of time, along with the mass of product up to 24 hours. (2) Repeat (1) when the growth is uncompetitively inhibited by the substrate with K_I = 0.7 g/dm³. (3) Set , and compare your results with the base case.

(f) Chain Reaction Example discussed in Professional Reference Shelf R9.1 on the CRE Web site. Over what range of time is the PSSH not valid? Download the Living Example Problem. Vary the temperature (800 < T < 1600). What temperature gives the greatest disparity with the PSSH results? Specifically compare the PSSH solution with the full numerical solution.

(g) Example on Alcohol Metabolism on the CRE Web site. This problem is a gold mine for things to be learned about the effect of alcohol on the human body. Download the Polymath Living Example Program from the CRE Web site. (1) Start by varying the initial doses of alcohol. (2) Next consider individuals who are ALDH enzyme deficient, which includes about 40% to 50% of Asians and Native Americans. Set V_max for acetaldehydes between 10% and 50% of its normal value and compare the concentration-time trajectories with the base cases. Hint: Read the journal article in the Summary Notes [Alcohol 35, p.1 (2005)].

P9-3_C (Flame retardants) Hydrogen radicals are important to sustaining combustion reactions. Consequently, if chemical compounds that can scavenge the hydrogen radicals are introduced, the flames can be extinguished. While many reactions occur during the combustion process, we shall choose CO flames as a model system to illustrate the process (S. Senkan et al., Combustion and Flame, 69, 113). In the absence of inhibitors

The last two reactions are rapid compared to the first two. When HCl is introduced to the flame, the following additional reactions occur:

Assume that all reactions are elementary and that the PSSH holds for the O·, OH ·, and Cl · radicals.

(a) Derive a rate law for the consumption of CO when no retardant is present.

(b) Derive an equation for the concentration of H · as a function of time, assuming constant concentration of O₂, CO, and H₂ O for both uninhibited combustion and combustion with HCl present. Sketch H · versus time for both cases.

P9-4_A The pyrolysis of acetaldehyde is believed to take place according to the following sequence:

(a) Derive the rate expression for the rate of disappearance of acetaldehyde, –r_Ac .

(b) Under what conditions does it reduce to the equation at the beginning of Section 9.1 on page 334?

(c) Sketch a reaction pathway diagram for this reaction. Hint: See margin note on page 338.

P9-5_B For each of the reactions in parts (a), (b), and (c), suggest a mechanism and apply the PSSH to learn if the mechanism is consistent with the rate law.

(a) The gas-phase homogeneous oxidation of nitrogen monoxide (NO) to dioxide (NO₂)

is known to have a form of third-order kinetics, which suggests that the reaction is elementary as written, at least for low partial pressures of the nitrogen oxides. However, the rate constant k actually decreases with increasing absolute temperature, indicating an apparently negative activation energy. Because the activation energy of any elementary reaction must be positive, some explanation is in order.

Provide an explanation, starting from the fact that an active intermediate species, NO₃, is a participant in some other known reactions that involve oxides of nitrogen. Draw the reaction pathway. Hint: See margin in Section 9.1.2.

(b) The rate law for formation of phosgene, COCl₂, from chlorine, Cl₂, and carbon monoxide, CO, has the rate law

Suggest a mechanism for this reaction that is consistent with this rate law and draw the reaction pathway. Hint: Cl formed from the dissociation of Cl₂ is one of the two active intermediates.

(c) Suggest an active intermediate(s) and mechanism for the reaction H₂ + Br₂ → 2HBr. Use the PSSH to show whether or not your mechanism is consistent with the rate law

P9-6_C (Tribology) Why you change your motor oil? One of the major reasons for engine-oil degradation is the oxidation of the motor oil. To retard the degradation process, most oils contain an antioxidant [see Ind. Eng. Chem. 26, 902 (1987)]. Without an inhibitor to oxidation present, the suggested mechanism at low temperatures is

where I₂ is an initiator and RH is the hydrocarbon in the oil.

When an antioxidant is added to retard degradation at low temperatures, the following additional termination steps occur:

(a) Derive a rate law for the degradation of the motor oil in the absence of an antioxidant at low temperatures.

(b) Derive a rate law for the rate of degradation of the motor oil in the presence of an antioxidant for low temperatures.

(c) How would your answer to part (a) change if the radicals I · were produced at a constant rate in the engine and then found their way into the oil?

(d) Sketch a reaction pathway diagram for both high and low temperatures, with and without antioxidant.

(e) See the open-ended problem G.2 in Appendix G and on the CRE Web site for more on this problem.

P9-7_A Epidemiology. Consider the application of the PSSH to epidemiology. We shall treat each of the following steps as elementary, in that the rate will be proportional to the number of people in a particular state of health. A healthy person, H, can become ill, I, spontaneously, such as by contracting smallpox spores:

or the person may become ill through contact with another ill person:

The ill person may become healthy:

or the ill person may expire:

The reaction given in Equation (P9-7.4) is normally considered completely irreversible, although the reverse reaction has been reported to occur.

(a) Derive an equation for the death rate.

(b) At what concentration of healthy people does the death rate become critical? (Ans.: When [H] = (k₃ + k₄)/k₂.)

(c) Comment on the validity of the PSSH under the conditions of Part (b).

(d) If k₁ = 10^-8 h^-1, k₂ = 10^-16 (people ·h)^-1, k₃ = 5 × 10^-10 h^-1, k₄ = 10^-11 h^-1, and H_o = 10⁹ people, use Polymath to plot H, I, and D versus time. Vary k_i and describe what you find. Check with your local disease control center or search online to modify the model and/or substitute appropriate values of k_i . Extend the model, taking into account what you learn from other sources (e.g., the Internet).

(e) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-8_B Derive the rate laws for the following enzymatic reactions and sketch and compare, where possible, with the plots shown in Figure E9-2.1.

(a)

(b)

(c)

(d)

(e) Which of the reactions (a) through (d), if any, lend themselves to analysis by a Lineweaver–Burk plot?

P9-9_B Beef catalase has been used to accelerate the decomposition of hydrogen peroxide to yield water and oxygen [Chem. Eng. Educ., 5, 141 (1971)]. The concentration of hydrogen peroxide is given as a function of time for a reaction mixture with a pH of 6.76 maintained at 30°C.

(a) Determine the Michaelis–Menten parameters V_max and K_M. (Ans.: K_M = 0.031 mol/dm³)

(b) If the total enzyme concentration is tripled, what will the substrate concentration be after 20 minutes?

(c) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

(d) List ways you can work this problem incorrectly.

P9-10_B It has been observed that substrate inhibition occurs in the following enzymatic reaction:

(a) Show that the rate law for substrate inhibition is consistent with the plot in Figure P9-10_B of –r_s (mmol/L ·min) versus the substrate concentration S (mmol/L).

(b) If this reaction is carried out in a CSTR that has a volume of 1000 dm³, to which the volumetric flow rate is 3.2 dm³/min, determine the three possible steady states, noting, if possible, which are stable. The entrance concentration of the substrate is 50 mmol/dm³. What is the highest conversion?

(c) What would be the effluent substrate concentration if the total enzyme concentration is reduced by 33%?

(d) List ways you can work this problem incorrectly.

(e) How could you make this problem more difficult?

P9-11_B The following data on baker’s yeast in a particular medium at 23.4°C were obtained in the presence and in the absence of an inhibitor, sulfanilamide. The reaction rate (–r_S) was measured in terms of the oxygen uptake rate Q_O2, obtained as a function of oxygen partial pressure.

(a) Assume the rate Q_O2 follows Michaelis–Menten kinetics with respect to oxygen. Calculate the Q_O2 maximum (i.e., V_max), and the Michaelis–Menten constant K_M.

(Ans.: V_max = 52.63 μL O₂/h · mg cells.)

(b) Using the Lineweaver–Burk plot, determine the type of inhibition sulfanilamide that causes the O₂ uptake to change.

(c) List ways you can work this problem incorrectly.

(d) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-12_B The enzymatic hydrolysis of starch was carried out with and without maltose and α-dextrin added. [Adapted from S. Aiba, A. E. Humphrey, and N.F. Mills, Biochemical Engineering (New York: Academic Press, 1973).]

Starch → α-dextrin → Limit dextrin → Maltose

Determine the types of inhibition for maltose and for α-dextrin.

P9-13_B The hydrogen ion, H⁺, binds with the enzyme (E^–) to activate it in the form EH. The hydrogen ion, H⁺, also binds with EH to deactivate it by forming

where E^– and are inactive.

(a) Determine if the preceding sequence can explain the optimum in enzyme activity with pH shown in Figure P9-13_B.

(b) List ways you can work this problem incorrectly.

(c) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-14_B An Eadie-Hofstee plot is shown below for the different types of enzyme inhibition. Match the line with the type of inhibition.

(a) Line A Inhibition Mechanism. Ans: __________

(b) Line B Inhibition Mechanism. Ans: __________

(c) Line C Inhibition Mechanism. Ans: __________

P9-15_B The biomass reaction

takes place in a 12-dm³ CSTR (chemostat) where the entering concentration of substrate is 200 g/dm³. The rate law follows the Monod equation with μ_max = 0.5s^–1 and K_S = 50 g/dm³. What is the volumetric flow rate, v₀ (dm³/h), that will give the maximum production rate of cells (g/h)?

P9-16_B The production of a product P from a particular gram-negative bacteria follows the Monod growth law

with μ_max = 1 h^–1, K_S = 0.25 g/dm³, and Y_c/s = 0.5 g/g.

(a) The reaction is to be carried out in a batch reactor with the initial cell concentration of C_c0 = 0.1 g/dm³ and substrate concentration of C_s0 = 20 g/dm³.

C_c = C_c0 + Y_{c/ s}(C_s0 – C_s)

Plot r_g, –r_s, –r_c, C_s, and C_c as a function of time.

(b) The reaction is now to be carried out in a CSTR with C_s0 = 20 g/dm³ and C_c0 = 0. What is the dilution rate at which wash-out occurs?

(c) For the conditions in part (b), what is the dilution rate that will give the maximum product rate (g/h) if Y_p/c = 0.15 g/g? What are the concentrations C_c, C_s, C_p, and –r_s at this value of D?

(d) How would your answers to (b) and (c) change if cell death could not be neglected with k_d = 0.02 h^–1?

(e) How would your answers to (b) and (c) change if maintenance could not be neglected with m = 0.2 g/h/dm³?

(f) Redo part (a) and use a logistic growth law

and plot C_c and r_c as a function of time. The term C_∞ is the maximum cell mass concentration and is called the carrying capacity, and is equal to C_∞ = 1.0 g/dm³. Can you find an analytical solution for the batch reactor? Compare with part (a) for C_∞ = Y_c/s C_s0 + C_c0.

(g) List ways you can work this problem incorrectly.

(h) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-17_B Redo Problem P9-16_B (a), (c), and (d) using the Tessier equation

with μ_max = 1.0 h^–1 and k = 8 g/dm³.

(a) List ways you can work this problem incorrectly.

(b) How could you make this problem more difficult?

P9-18_B The bacteria X-II can be described by a simple Monod equation with μ_max = 0.8 h^–1 and K_S = 4 g/dm³, Y_p/c = 0.2 g/g, and Y_s/c = 2 g/g. The process is carried out in a CSTR in which the feed rate is 1000 dm³/h at a substrate concentration of 10 g/dm³.

(a) What size fermentor is needed to achieve 90% conversion of the substrate? What is the exiting cell concentration?

(b) How would your answer to (a) change if all the cells were filtered out and returned to the feed stream?

(c) Consider now two 5000-dm³ CSTRs connected in series. What are the exiting concentrations C_s, C_c, and C_p from each of the reactors?

(d) Determine, if possible, the volumetric flow rate at which wash-out occurs and also the flow rate at which the cell production rate (C_c v₀) in grams per day is a maximum.

(e) Suppose you could use the two 5000-dm³ reactors as batch reactors that take two hours to empty, clean, and fill. What would your production rate be in (grams per day) if your initial cell concentration is 0.5 g/dm³? How many 500-dm³ batch reactors would you need to match the CSTR production rate?

(f) List ways you can work this problem incorrectly.

(g) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P9-19_A A CSTR is being operated at steady state. The cell growth follows the Monod growth law without inhibition. The exiting substrate and cell concentrations are measured as a function of the volumetric flow rate (represented as the dilution rate), and the results are shown below. Of course, measurements are not taken until steady state is achieved after each change in the flow rate. Neglect substrate consumption for maintenance and the death rate, and assume that Y_p/c is zero. For run 4, the entering substrate concentration was 50 g/dm³ and the volumetric flow rate of the substrate was 2 dm³/h.

(a) Determine the Monod growth parameters μ_max and K_S .

(b) Estimate the stoichiometric coefficients, Y_c/s and Y_s/c.

(c) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

(d) How could you make this problem more difficult?

P9-20_B Alternative Energy Source.²⁵ In the summer of 2009, ExxonMobil decided to invest 600 million dollars on developing algae as an alternative fuel. Algae would be grown and their oil extracted to provide an energy source. It is estimated that one acre of a biomass pond can provide 6,000 gallons of gasoline per year, which would require the capture of a CO₂ source more concentrated than air (e.g., fuel gas from a refinery) and also contribute to the sequestration of CO₂. The biomass biosynthesis during the day is

²⁵ The contributions of John Benemann to this problem are appreciated.

Sunlight + CO₂ + H₂O + Algae → More Algae + O₂

Consider a 5,000-gallon pond with perforated pipes into which CO₂ is injected and slowly bubbled into the solution to keep the water saturated with CO₂.

The doubling time during the day is 12 h at high-noon sunlight and zero during the night. As a first approximation, the growth during the 12 hours of daylight law is

r_g = fμC_C

with f = sunlight = sin (π t/12) between 6 a.m. and 6 p.m., otherwise f = 0, C_C is the algae concentration (g/dm³) and μ = 0.9 day^–1 (assumes constant CO₂ saturation at 1 atm is 1.69g/kg water). The pond is 30-cm deep and for effective sunlight penetration, the algae concentration cannot exceed 200 mg/dm³.

(a) Derive an equation for the ratio of the algae cell concentration C_C at time t to initial cell concentration C_C0, i.e., (C_C/C_C0). Plot and analyze (C_C/C_C0) versus time up to 48 hours.

(b) If the pond is initially seeded with 0.5 mg/dm³ of algae, how long will it take the algae to reach a cell density (i.e., concentration) of 200 mg/dm³, which is the concentration at which sunlight can no longer effectively penetrate the depth of the pond? Plot and analyze r_g and C_C as a function of time. As a first approximation, assume the pond is well mixed.

(c) Suppose the algae limit the sun’s penetration significantly even before the concentration reaches 200 mg/dm³ with e.g., μ = μ₀ (1 – C_C/200). Plot and analyze r_g and C_C as a function of time. How long would it take to completely stop growth at 200 mg/dm³?

(d) Now, let’s consider continuous operation. Once the cell density reaches 200 mg/dm, one-half of the pond is harvested and the remaining broth is mixed with fresh nutrient. What is the steady-state algae productivity in gm/year, again assuming the pond is well mixed?

(e) Now consider a constant feed of waste water and removal of algae at a dilution rate of one reciprocal day. What is the mass flow rate of algae out of the 5,000 gallon pond (g/d)? Assume the pond is well mixed.

(f) Now consider that the reaction is to be carried out in an enclosed, transparent reactor. The reactor can be pressurized with CO₂ up to 10 atm with K_S = 2 g/dm³. Assume that after the initial pressurization, no more CO₂ can be injected. Plot and analyze the algae concentration as a function of time.

(g) An invading algae can double twice as fast as the strain you are cultivating. Assume that it initially is at 0.1 mg/l concentration. How long until it is the dominant species (over 50% of the cell density)?

P9-21_A Short calculations on the algae ponds.

(a) If the pond is initially seeded with 0.5 mg/dm³ of algae, how long will it take the algae to reach a cell density (i.e., concentration) of 200 mg/dm³? Sketch a rough plot of r_g and C_C over time.

(b) Suppose the algae limit the sun’s penetration significantly even before the concentration reaches 200 mg/dm³ by using μ = μ₀ (1-C_C/200). Assume μ₀ = 0.9 day^–1. Qualitatively, what happens to the growth rate as the concentration of cells increases? Approximately how long would it take for the concentration to reach 200 mg/dm³? Why?

(c) An invading algae species can double twice as fast as the strain you are cultivating. Assume that it is initially at a concentration of 0.01 mg/dm³. How long until it becomes the dominant species in the pond (over 50% of the cell density)?

• Additonal Homework Problems

A number of homework problems that can be used for exams or supplementary problems or examples are found on the CRE Web site, www.umich.edu/~elements/5e/index.html.

www.enzymes.com

www.pharmacokinetics.com

FROST, A. A., and R. G. PEARSON, Kinetics and Mechanism, 2nd ed. New York: Wiley, 1961, Chapter 10. Old but great examples.

LAIDLER, K. J., Chemical Kinetics, 3rd ed. New York: HarperCollins, 1987.

PILLING, M. J., Reaction Kinetics, New York: Oxford University Press, 1995.

2. Further discussion of enzymatic reactions:

Just about everything you want to know about basic enzyme kinetics can be found in SEGEL, I. H., Enzyme Kinetics. New York: Wiley-Interscience, 1975.

An excellent description of parameter estimation, biological feedback, and reaction pathways can be found in VOIT, E. O., Computational Analysis of Biochemical Systems. Cambridge, UK: Cambridge University Press, 2000.

CORNISH-BOWDEN, A., Analysis of Enzyme Kinetic Data. New York: Oxford University Press, 1995.

NELSON, D. L., and M. M. COX, Lehninger Principles of Biochemistry, 3rd ed. New York: Worth Publishers, 2000.

SHULER, M. L., and F. KARGI, Bioprocess Engineering Principles, 2nd ed. Upper Saddle River, NJ: Prentice Hall, 2002.

3. Material on bioreactors can be found in

BAILEY, T. J., and D. OLLIS, Biochemical Engineering, 2nd ed. New York: McGraw-Hill, 1987.

BLANCH, H. W., and D. S. CLARK, Biochemical Engineering. New York: Marcel Dekker, 1996.

4. Also see

BURGESS, THORNTON W., The Adventures of Old Mr. Toad. New York: Dover Publications, Inc., 1916.

KEILLOR, GARRISON, Pretty Good Joke Book: A Prairie Home Companion. St. Paul, MN: HighBridge Co., 2000.

MASKILL, HOWARD, The Investigation of Organic Reactions and Their Mechanisms. Oxford UK: Blackwell Publishing Ltd, 2006.