The next best thing to knowing something is knowing where to find it.
—Samuel Johnson (1709–1784)
In Chapter 3, a number of simple powerlaw models, for example,
$\mathrm{}{\mathit{r}}_{\mathrm{A}}\mathrm{=}\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathrm{n}}$
were presented, where n was an integer of 0, 1, or 2 corresponding to a zero, first, or secondorder reaction. However, for a large number of reactions, the orders are noninteger, such as the decomposition of acetaldehyde at 500°C
$\mathrm{C}{\mathrm{H}}_{\mathrm{3}}\mathrm{C}\mathrm{H}\mathrm{O}\mathrm{\to}\mathrm{C}{\mathrm{H}}_{\mathrm{4}}\mathrm{+}\mathrm{C}\mathrm{O}$
where the rate law developed in Problem P95_{B}(b) is
$\mathrm{}{\mathit{r}}_{\mathrm{C}{\mathrm{H}}_{\mathrm{3}}\mathrm{C}\mathrm{H}\mathrm{O}}=\mathit{k}{\mathit{C}}_{\mathrm{C}{\mathrm{H}}_{\mathrm{3}}\mathrm{C}\mathrm{H}\mathrm{O}}^{\mathrm{3}\mathrm{/}\mathrm{2}}$
Many rate laws have concentration terms in both the numerator and denominator such as the formation of HBr from hydrogen and bromine
H_{2} + Br_{2} → 2HBr
where the rate law developed in Problem P95_{B}(c) is
${r}_{\text{HBr}}=\frac{{k}_{1}{C}_{{\text{H}}_{2}}{C}_{{\text{Br}}_{2}}^{3/2}}{{C}_{\text{HBr}}\text{}+\text{}{k}_{2}{C}_{{\text{Br}}_{2}}}$
Rate laws of this form usually involve a number of elementary reactions and at least one active intermediate. An active intermediate is a highenergy 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 collisions or interactions 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, that is, 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 highamplitude 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 (http://www.umich.edu/~elements/6e/03chap/prof.html) on the CRE Web site. Other types of active intermediates that can be formed are free radicals (one or more unpaired electrons; e.g., CH_{3}•), ionic intermediates (e.g., carbonium ion), and enzyme–substrate complexes, to mention a few (http://www.umich.edu/~elements/6e/09chap/prof.html).
^{1}W. J. Moore, Physical Chemistry, Reading, MA: Longman Publishing Group, 1998.
“Wild” oscillations
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 shortlived 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 91(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 91(b). As discussed in Chapter 3, an estimation of the barrier height, E, can be obtained using computational software packages such as Spartan, Cerius^{2}, or Gaussian as discussed in the Molecular Modeling Web Module in Chapter 3 on the CRE Web site, www.umich.edu/~elements/6e/index.html.
^{2}F. A. Lindemann, Trans. Faraday. Soc., 17, 598 (1922).
^{3}J. Peterson, Science News, 156, 247 (1999).
In the theory of active intermediates, decomposition of the intermediate does not occur instantaneously after internal activation of the molecule; rather, there is a time lag, although infinitesimally small, during which the species remains activated. Zewail’s work was the first definitive proof of a gasphase active intermediate that exists for an infinitesimally short time. Because a reactive intermediate reacts virtually as fast as it is formed, the net rate of formation of an active intermediate (e.g., A*) is zero, that is,
$\begin{array}{cc}{r}_{\mathrm{A}\mathrm{*}}\mathrm{\equiv}\mathrm{0}& \left(\mathrm{91}\right)\end{array}$
This condition is also referred to as the PseudoSteadyState Hypothesis (PSSH). If the active intermediate appears in n reactions, then
PSSH
To illustrate how noninteger power rate laws are formed, we shall first consider the gasphase decomposition of azomethane, AZO, to give ethane and nitrogen
${{\text{(CH}}_{3}\text{)}}_{2}{\text{N}}_{2}\to {\text{C}}_{2}{\text{H}}_{6}{+\text{N}}_{2}$
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}
r_{C2H2} ∝ C_{AZO}
^{4} H. C. Ramsperger, J. Am. Chem. Soc., 49, 912.
and second order at pressures below 50 mmHg (low concentrations):
${r}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{\alpha}\text{}{C}_{\text{AZO}}^{2}$
Why the reaction order changes
We could combine these two observations to postulate a rate law of the form
$\mathrm{}{\mathit{r}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}=\frac{{\mathit{k}}_{\mathrm{1}}{\mathrm{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}^{\mathrm{2}}}{\mathrm{1}\mathrm{+}{\mathit{k}}_{\mathrm{2}}{\mathrm{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}}$
To find a mechanism that is consistent with the experimental observations, we use the steps shown in Table 91.
TABLE 91 STEPS TO DEDUCE A RATE LAW

We will now follow the steps in Table 91 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*, that is, [(CH_{3})_{2}N_{2}]*.
Step 2. Propose a mechanism.
$\text{Mechanism}\{\begin{array}{ll}\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{ctio}\mathrm{n}\text{}\mathrm{1}:& \hfill \mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\mathrm{+}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\stackrel{{k}_{\text{1AZO*}}}{\to}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\mathrm{+}\mathrm{[}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}{\mathrm{]}}^{\mathrm{*}}\hfill \\ \mathrm{R}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\text{}\mathrm{2}:& \hfill \mathrm{[}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}{\mathrm{]}}^{\mathrm{*}}\mathrm{+}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\stackrel{{k}_{\text{2AZO*}}}{\to}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\mathrm{+}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}\hfill \\ \mathrm{R}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\text{}\mathrm{3}:& \hfill \mathrm{[}\mathrm{(}\mathrm{C}{\mathrm{H}}_{\mathrm{3}}{\mathrm{)}}_{\mathrm{2}}{\mathrm{N}}_{\mathrm{2}}{\mathrm{]}}^{\mathrm{*}}\stackrel{{k}_{\text{3AZO*}}}{\to}{\mathrm{C}}_{\mathrm{2}}{\mathrm{H}}_{\mathrm{6}}\mathrm{+}{\mathrm{N}}_{\mathrm{2}}\hfill \end{array}$
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
$\begin{array}{cc}(1)\text{}{r}_{1\text{AZO*}}={k}_{1\text{AZO*}}{C}_{\text{AZO}}^{\text{2}}& \left(\mathrm{93}\right)\end{array}$
$\begin{array}{cc}(2)\text{}{r}_{2\text{AZO*}}=\text{}{k}_{2\text{AZO*}}{C}_{2\text{AZO*}}{C}_{\text{AZO}}& \left(\mathrm{94}\right)\end{array}$
$\begin{array}{cc}(3)\text{}{r}_{\text{3AZO*}}=\text{}{k}_{\text{3AZO*}}{C}_{\text{AZO*}}& \left(95\right)\end{array}$
Note: The specific reaction rates, k, are all defined wrt the active intermediate AZO*.
(Let k_{1} = k_{1AZO*}, k_{2} = k_{2AZO*}, and k_{3} = k_{3AZO*})
The rate laws shown in Equations (93)–(95) 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 PseudoSteadyState 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).
$\begin{array}{cclc}{r}_{\text{AZO*}}& =& {r}_{1\text{AZO*}}+{r}_{2\text{AZO*}}+{r}_{3\text{AZO*}}=\text{0}& \\ & =& {k}_{\text{1}}{C}_{\text{AZO}}^{\text{2}}{k}_{\text{2}}{C}_{\text{AZO*}}{C}_{\text{AZO}}{k}_{\text{3}}{C}_{\text{AZO*}}=\text{0}& \left(\mathrm{97}\right)\end{array}$
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 (98) into Equation (96)
Step 7. Compare with experimental data.
At low AZO concentrations,
${\mathit{k}}_{\mathrm{2}}{\mathit{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}\ll {\mathit{k}}_{\mathrm{3}}$
for which case we obtain the following secondorder rate law:
${\mathit{r}}_{{\mathrm{C}}_{\mathrm{2}}{\mathrm{H}}_{\mathrm{6}}}={\mathit{k}}_{\mathrm{1}}{\mathit{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}^{\mathrm{2}}$
At high concentrations
${\mathit{k}}_{\mathrm{2}}{\mathit{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}\gg {\mathit{k}}_{\mathrm{3}}$
in which case the rate expression follows firstorder kinetics
${\mathit{r}}_{{\mathrm{C}}_{\mathrm{2}}{\mathrm{H}}_{\mathrm{6}}}=\frac{{\mathit{k}}_{\mathrm{1}}{\mathit{k}}_{\mathrm{3}}}{{\mathit{k}}_{\mathrm{2}}}{\mathit{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}=\mathit{k}{\mathit{C}}_{\mathrm{A}\mathrm{Z}\mathrm{O}}$
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.
Apparent reaction orders
The PSSH can also explain why one observes so many firstorder reactions such as
(CH_{3})_{2}O → CH_{4} + H_{2} + 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 A 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 92.
The mechanism consists of the three elementary reactions:
1. Activation $\begin{array}{cc}\mathrm{A}\mathrm{+}\mathrm{M}\stackrel{{\mathit{k}}_{\mathrm{1}}}{\to}{\mathrm{A}}^{\mathrm{*}}\mathrm{+}\mathrm{M}& {r}_{\text{1A*}}\end{array}={k}_{1}{C}_{\text{A}}{C}_{\text{M}}$
2. Activation $\begin{array}{cc}{\mathrm{A}}^{*}\mathrm{+}\mathrm{M}\stackrel{{\mathit{k}}_{2}}{\to}\mathrm{A}\mathrm{+}\mathrm{M}& {r}_{2\text{A*}}\end{array}={k}_{2}{C}_{\text{A}}*{C}_{\text{M}}$
3. Decomposition $\begin{array}{cc}{\mathrm{A}}^{\mathrm{*}}\stackrel{{\mathit{k}}_{3}}{\to}\mathrm{P}& {r}_{\text{3A*}}\end{array}={k}_{3}{C}_{\text{A}}*$
^{6} The reaction pathway for the reaction in Figure 92 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 k_{1} (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, k_{2}, and then separate to form A and M. At the bottom, we see the arrow from A^{*} to form P with k_{3}.
Writing the rate of formation of product
r_{P} = k_{3}C_{A*}
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
$\begin{array}{cc}{\mathit{r}}_{\mathrm{P}}=\mathrm{}{\mathit{r}}_{\mathrm{A}}=\frac{{\mathit{k}}_{\mathrm{3}}{\mathit{k}}_{\mathrm{1}}{\mathit{C}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{M}}}{{\mathit{k}}_{\mathrm{2}}{\mathit{C}}_{\mathrm{M}}\mathrm{+}{\mathit{k}}_{\mathrm{3}}}& \left(\mathrm{910}\right)\end{array}$
Because the concentration of the inert M is constant, we let
$\begin{array}{cc}\mathit{k}=\frac{{\mathit{k}}_{\mathrm{1}}{\mathit{k}}_{\mathrm{3}}{\mathit{C}}_{\mathrm{M}}}{{\mathit{k}}_{\mathrm{2}}{\mathit{C}}_{\mathrm{M}}\mathrm{+}{\mathit{k}}_{\mathrm{3}}}& \left(\mathrm{911}\right)\end{array}$
Reaction pathways^{6}
to obtain the firstorder rate law
–r_{A} = kC_{A}
Consequently, we see the reaction
A → P
follows an elementary rate law but is not an elementary reaction. Reactions similar to the AZO decomposition mechanism would also be apparent first order if the corresponding to the term (k_{2} C_{AZO}) in the denominator of Equation (99) were larger than the other term k_{3} (i.e., (k_{2} C_{AZO}) >> k_{3}). The reaction pathway for this reaction is shown in the margin here. Reaction pathways for other reactions are given on the Web site (http://www.umich.edu/~elements/6e/09chap/profrxnpath.html).
Firstorder rate law for a nonelementary reaction
In many instances the rate data are correlated before a mechanism is found. It is a normal procedure to reduce the additive constant in the denominator to 1.
We therefore divide the numerator and denominator of Equation (99) by k_{3} to obtain
General Considerations. Developing a mechanism is a difficult and timeconsuming task. The rules of thumb listed in Table 92 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 92 RULES OF THUMB FOR DEVELOPMENT OF A MECHANISM

Upon application of Table 92 to the azomethane example just discussed, we observe the following from rate equation (912):
The active intermediate, AZO*, collides with azomethane, AZO [Reaction 2], resulting in the concentration of AZO in the denominator.
AZO* decomposes spontaneously [Reaction 3], resulting in a constant in the denominator of the rate expression.
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.
Light is given off when a highintensity ultrasonic wave is applied to water.^{7} You should be able to see the glow if you turn out the lights and irradiate water with a focused ultrasonic tip. This light results from microsize gas bubbles (0.1 mm) being formed by the ultrasonic wave and then being compressed by it. During the compression stage of the wave, the contents of the bubble (e.g., water and whatever else is dissolved in the water, CS_{2}, O_{2}, N_{2}) are compressed adiabatically.
^{7} P. K. Chendke and H. S. Fogler, J. Phys. Chem., 87, 1362.
This compression gives rise to high temperatures and kinetic energies of the gas molecules, which through molecular collisions generate active intermediates and cause chemical reactions to occur in the bubble.
$\mathrm{M}\mathrm{+}{\mathrm{H}}_{\mathrm{2}}\mathrm{O}\mathrm{\to}{\mathrm{H}}_{\mathrm{2}}{\mathrm{O}}^{\mathrm{*}}\mathrm{+}\mathrm{M}$
Collapsing cavitation microbubble
The intensity of the light given off, I, is proportional to the rate of deactivation of an activated water molecule that has been formed in the microbubble.
$\begin{array}{c}{\text{H}}_{\text{2}}{\text{O}}^{\text{*}}\stackrel{k}{\to}{\text{H}}_{\text{2}}\text{O}+hv\\ \text{Lightintensity(I)}\text{}\text{\alpha}({r}_{{\text{H}}_{\text{2}}{\text{O}}^{\text{*}}}\text{)}=k\text{\hspace{0.17em}}{C}_{{\text{H}}_{\text{2}}{\text{O}}^{\text{*}}}\end{array}$
An orderofmagnitude increase in the intensity of sonoluminescence is observed when either carbon disulfide or carbon tetrachloride is added to the water. The intensity of luminescence, I, for the reaction
$\mathrm{C}{\mathrm{S}}_{\mathrm{2}}^{*}\stackrel{{\mathit{k}}_{\mathrm{4}}}{\to}{\mathrm{C}\mathrm{S}}_{\mathrm{2}}\mathrm{+}\mathit{h}\mathit{v}$
is
$I\text{}\text{\alpha}\text{}\text{(}{r}_{{\text{CS}}_{\text{2}}^{\text{*}}}\text{)}={k}_{\text{4}}{C}_{{\text{CS}}_{\text{2}}^{\text{*}}}$
A similar result exists for CCl_{4}.
However, when an aliphatic alcohol, X, is added to the solution, the intensity decreases with increasing concentration of alcohol. The data are usually reported in terms of a Stern–Volmer plot in which relative intensity is given as a function of alcohol concentration, C_{X}. (See Figure E91.1, where I_{0} is the sonoluminescence intensity in the absence of alcohol and I is the sonoluminescence intensity in the presence of alcohol.)
Suggest a mechanism consistent with experimental observation.
Derive a rate law consistent with Figure E91.1.
Stern–Volmer plot
Solution
(a) Mechanism
From the linear plot in Figure E91.1 we know that
$\begin{array}{cccc}\frac{{I}_{\mathrm{0}}}{I}=\mathrm{A}\mathrm{+}\mathrm{B}{\mathit{C}}_{\mathrm{X}}\mathrm{\equiv}\mathrm{A}\mathrm{+}\mathrm{B}\mathrm{(}\mathrm{X)}& & \left(\text{E}\mathrm{91.1}\right)& \end{array}$
where C_{X} ≡ (X) and A and B are numerical constants. Inverting Equation (E91.1) yields
$\begin{array}{cc}\frac{\mathit{I}}{{\mathit{I}}_{\mathrm{0}}}\mathrm{=}\frac{\mathrm{1}}{\mathrm{A}\mathrm{+}\mathrm{B}\mathrm{(}\mathrm{X}\mathrm{)}}& \left(\text{E}\mathrm{91.2}\right)\end{array}$
From Rule 1 of Table 92, the denominator suggests that alcohol (X) collides with the active intermediate:
$\begin{array}{cc}\begin{array}{c}\text{X}+\text{Intermediate}\to \text{Deactivationproducts}\end{array}& \left(\text{E}\mathrm{91.3}\right)\end{array}$
The alcohol acts as what is called a scavenger to deactivate the active intermediate. The fact that the addition of CCl_{4} or CS_{2} increases the intensity of the luminescence
$\begin{array}{cc}I\text{}\mathrm{\alpha}\text{}\left({\mathrm{CS}}_{2}\right)& \left(\text{E}\mathrm{91.4}\right)\end{array}$
leads us to postulate (Rule 3 of Table 92) that the active intermediate ${\mathrm{CS}}_{2}^{*}$ was probably formed from the collision of either CS_{2} or another gas M or both in the collapsing bubble
$\begin{array}{cc}\text{M}+{\text{CS}}_{2}\to {\text{CS}}_{2}^{*}+\text{M}& \left(\text{E}\mathrm{91.5}\right)\end{array}$
Reaction pathways
where M is a third body (CS_{2}, H_{2}O, N_{2}, etc.).
We also know that deactivation can occur by the reverse of reaction (E91.5). Combining this information, we have as our mechanism:
$\begin{array}{cc}\text{Activation:}\text{}\text{M}+{\text{CS}}_{2}\stackrel{{k}_{1}}{\to}{\text{CS}}_{2}^{*}+\text{M}& \left(\text{E}\mathrm{91.5}\right)\end{array}$
$\begin{array}{cc}\text{Deactivation:}\text{}\text{M}+{\text{CS}}_{2}^{*}\stackrel{{k}_{2}}{\to}{\text{CS}}_{2}+\text{M}& \left(\text{E}\mathrm{91.6}\right)\end{array}$
$\text{Deactivation:}\text{}\begin{array}{cc}\text{M}+{\text{CS}}_{2}^{*}\stackrel{{k}_{3}}{\to}{\text{CS}}_{2}+\text{X}& \left(\text{E}\mathrm{91.3}\right)\end{array}$
$\text{Luminescence:}\text{}\begin{array}{cc}{\text{CS}}_{2}^{*}\stackrel{{k}_{4}}{\to}{\text{CS}}_{2}+hv& \left(\text{E}\mathrm{91.7}\right)\end{array}$
The mechanism
(b) Rate Law
Using the PSSH on $\mathrm{C}{\mathrm{S}}_{\mathrm{2}}^{\mathrm{*}}$ in each of the above elementry reactions yields
${r}_{{\text{CS}}_{\text{2}}^{\text{*}}}=\text{0}={k}_{\text{1}}{\text{(CS}}_{\text{2}}\text{)(M)}{k}_{\text{2}}{\text{(CS}}_{\text{2}}^{*}\text{)(M)}{k}_{\text{3}}{\text{(X)(CS}}_{\text{2}}^{\text{*}}\text{)}{k}_{\text{4}}{\text{(CS}}_{\text{2}}^{\text{*}}\text{)}$
Solving for ($\mathrm{C}{\mathrm{S}}_{\mathrm{2}}^{\mathrm{*}}$) and substituting into Equation (E91.8) gives us
$\begin{array}{cc}\mathit{I}=\frac{{k}_{\mathrm{4}}{k}_{\mathrm{1}}\mathrm{(}\mathrm{C}{\mathrm{S}}_{\mathrm{2}}\mathrm{)}\mathrm{(}\mathrm{M}\mathrm{)}}{{k}_{\mathrm{2}}\mathrm{(}\mathrm{M}\mathrm{)}\mathrm{+}{k}_{\mathrm{3}}\mathrm{(}\mathrm{X}\mathrm{)}\mathrm{+}{k}_{\mathrm{4}}}& \left(\begin{array}{c}\text{E91.9}\end{array}\right)\end{array}$
LEP Sliders
In the absence of alcohol
$\begin{array}{cc}{I}_{0}\text{}\mathrm{=}\frac{{k}_{\mathrm{4}}{k}_{\mathrm{1}}\mathrm{(}\mathrm{C}{\mathrm{S}}_{\mathrm{2}}\mathrm{)}\mathrm{(}\mathrm{M}\mathrm{)}}{{k}_{\mathrm{2}}\mathrm{(}\mathrm{M}\mathrm{)}\mathrm{+}{k}_{\mathrm{4}}}& \left(\begin{array}{c}\text{E91.10}\end{array}\right)\end{array}$
For constant concentrations of CS_{2} and the third body, M, we take a ratio of Equation (E91.10) to (E91.9)
which is of the same form as that suggested by Figure E91.1. Equation (E91.11) and similar equations involving scavengers are called Stern–Volmer equations.
Analysis: This example showed how to use the Rules of Thumb (Table 92) to develop a mechanism. Each step in the mechanism is assumed to follow an elementary rate law. The PSSH was applied to the net rate of reaction for the active intermediate in order to find the concentration of the active intermediate. This concentration was then substituted into the rate law for the rate of formation of product to give the rate law. The rate law from the mechanism was found to be consistent with experimental data.
A discussion of luminescence is given in the Web Module, Glow Sticks on the CRE Web site (www.umich.edu/~elements/6e/09chap/web.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. See Problem P91_{A}.
A chain reaction consists of the following sequence:
Initiation: formation of an active intermediate
Propagation or chain transfer: interaction of an active intermediate with the reactant or product to produce another active intermediate
Termination: deactivation of the active intermediate to form products
Glow sticks Web Module
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 (http://www.umich.edu/~elements/6e/09chap/profchain.html). Also included in Professional Reference Shelf R9.2 is a discussion of Reaction Pathways and the chemistry of smog formation (http://www.umich.edu/~elements/6e/09chap/profrxnpath.html).
LEP Sliders
We now extend our discussion on active intermediates to enzymatic reactions. An enzyme^{†} is a highmolecularweight protein or proteinlike substance that acts on a substrate (reactant molecule) to transform it chemically at a greatly accelerated rate, usually 10^{3}–10^{17} times faster than the uncatalyzed rate. The pharmaceutical business that will synthesize therapeutic proteins using CHO has been estimated to reach $200 billion by 2020. A colleague, Prof. Greg Thurber, and his students are doing exciting research in drug development using enzyme kinetics. The group applies the principles of enzyme kinetics to describe where drugs distribute and degrade in the body to design new therapeutics and diagnostic imaging agents for diseases such as breast cancer.^{8,9} Without enzymes, essential biological reactions would not take place at a rate necessary to sustain life. Enzymes are usually present in small quantities and are not consumed during the course of the reaction, nor do they affect the chemical reaction equilibrium. Enzymes provide an alternate pathway for the reaction to occur, thereby requiring a lower activation energy. Figure 93 shows the reaction coordinate for the uncatalyzed reaction of a reactant molecule, called a substrate (S), to form a product (P)
^{†} See http://www.rsc.org/Education/Teachers/Resources/cfb/enzymes.htm and https://en.wikipedia.org/wiki/Enzyme.
^{8} S. Bhatnagar, et. al., “Oral administration and detection of a nearinfrared molecular imaging agent in an orthotopic mouse model for breast cancer screening,” Mol. Pharm., 15, 1746–1754 (2018).
^{9} C. Cilliers, B. Menezes, I. Nessler, J. Linderman, and G. M. Thurber, “Improved tumor penetration and singlecell targeting of antibodydrug conjugates increases anticancer efficacy and host survival.,” Cancer Res., 78, 758–768 (2018).
S → P
This figure also shows the catalyzed reaction pathway that proceeds through an active intermediate (E · S), called the enzyme–substrate complex, that is,
$\begin{array}{c}\text{S}+\text{E}\text{\hspace{0.17em}}\rightleftarrows \text{\hspace{0.17em}}\text{E}\cdot \text{S}\to \text{E}+\text{P}\end{array}$
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 10^{14} faster than would be without the enzyme urease.
#You’reTheOnlyOne ForMe!
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 enzymecatalyzed 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 4–9 and temperatures 75°F–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 a few exceptions to the naming convention, such as αamylase. The enzyme αamylase catalyzes the transformation of starch in the first step in the production of the controversial soft drink (e.g., Red Pop) sweetener highfructose corn syrup (HFCS) from corn starch, which is a $4 billion per year business.
$\begin{array}{c}\text{Cornstarch}\stackrel{\alpha \text{amylase}}{\to}\text{Thinnedstarch}\underset{\text{amylase}}{\overset{\text{gluco}}{\to}}\text{Glucose}\underset{\text{isomerase}}{\overset{\text{Glucose}}{\to \text{}}}\text{HFCS}\end{array}$
The key factor that sets enzymatic reactions apart from other catalyzed reactions is the formation of an enzyme–substrate complex, (E · S). Here, substrate, (S), binds with a specific active site of the enzyme, (E), to form this complex.^{10} Figure 94 shows a schematic of the enzyme chymotrypsin (MW = 25,000 Daltons), which catalyzes the hydrolytic cleavage of polypeptide bonds. In many cases, the enzyme’s active catalytic sites are found where the various folds or loops interact. For chymotrypsin, the catalytic sites are noted by the amino acid numbers 57, 102, and 195 in Figure 94. Much of the catalytic power is attributed to the binding energy of the substrate to the enzyme through multiple bonds with the specific functional groups on the enzyme (amino side chains, metal ions). The type of interactions that stabilize the enzyme–substrate complex include hydrogen bonding and hydrophobic, ionic, and London van der Waals forces. If the enzyme is exposed to extreme temperatures or pH environments (i.e., both high and low pH values), it may unfold, losing its active sites. When this occurs, the enzyme is said to be denatured (see Figure 9.8 and Problem P913_{B}).
^{10} M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. Upper Saddle River, NJ: Prentice Hall, 2002.
Schematic showing the loops and folds
Two often used models for enzyme–substrate interactions are the lockandkey model and the induced fit model, both of which are shown in Figure 95. For many years, the lockandkey 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.
Two models for the enzyme–substrate complex
There are six classes of enzymes and only six:
1. Oxidoreductases 
AH_{2} + B + E → A + BH_{2} + E 
2. Transferases 
AB + C + E → AC + B + E 
3. Hydrolases 
AB + H_{2}O + E → AH + BOH + E 
4. Isomerases 
A + E → isoA + E 
5. Lyases 
AB + E → A + B + E 
6. Ligases 
A + B + E → AB + E 
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.
In developing some of the elementary principles of the kinetics of enzyme reactions, we shall discuss an enzymatic reaction that has been suggested by Levine and LaCourse as part of a system that would reduce the size of an artificial kidney.^{11} The desired result is a prototype of an artificial kidney that could be worn by the patient and would incorporate a replaceable unit for the elimination of the body’s nitrogenous waste products, such as uric acid and creatinine. A schematic of this device is shown on the CRE Web site (http://www.umich.edu/~elements/6e/09chap/summary.html#sec2). In the microencapsulation scheme proposed by Levine and LaCourse, the enzyme urease would be used in the removal of urea from the bloodstream. Here, the catalytic action of urease would cause urea to decompose into ammonia and carbon dioxide. The mechanism of the reaction is believed to proceed by the following sequence of elementary reactions:
^{11} N. Levine and W. C. LaCourse, J. Biomed. Mater. Res., 1, 275.
The unbound enzyme urease (E) reacts with the substrate urea (S) to form an enzyme–substrate complex (E · S):
$\begin{array}{cc}{\text{NH}}_{\text{2}}{\text{CONH}}_{\text{2}}+\text{Urease}\stackrel{{k}_{1}}{\to}{{\text{[NH}}_{\text{2}}{\text{CONH}}_{\text{2}}\cdot \text{Urease]}}^{*}& \left(\text{E}\mathrm{91.13}\right)\end{array}$
This complex (E · S) can decompose back to urea (S) and urease (E):
$\begin{array}{cc}{[{\text{NH}}_{\text{2}}{\text{CONH}}_{\text{2}}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\text{Urease]}}^{*}\text{}\stackrel{{k}_{2}}{\to}\text{\hspace{0.17em}}\text{Urease}+{\text{NH}}_{\text{2}}{\text{CONH}}_{\text{2}}& \left(\mathrm{914}\right)\end{array}$
Or, it can react with water (W) to give the products (P) ammonia and carbon dioxide, and recover the enzyme urease (E):
$\begin{array}{cc}[\mathrm{N}{\mathrm{H}}_{\mathrm{2}}\mathrm{C}\mathrm{O}\mathrm{N}{\mathrm{H}}_{\mathrm{2}}\xb7{\text{Urease]}}^{*}+{\text{H}}_{2}\text{O}\text{}\stackrel{{k}_{3}}{\to}{\text{2NH}}_{3}+{\mathrm{CO}}_{2}\text{+}\text{}\text{Urease}& \left(\mathrm{915}\right)\end{array}$
The reaction mechanism
Symbolically, the overall reaction is written as
$\text{S}+\text{E}\text{\hspace{0.17em}}\rightleftarrows \text{\hspace{0.17em}}\text{E}\cdot \text{S}\stackrel{+{\text{H}}_{2}\text{O}}{\to}\text{P}+\text{E}$
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, which is the active intermediate, and the reaction products, respectively, we can write Reactions (913)–(915) symbolically in the forms
$\begin{array}{cc}\mathrm{S}\mathrm{+}\mathrm{E}\stackrel{{\mathit{k}}_{\mathrm{1}}}{\to}\mathrm{E}\mathrm{\cdot}\mathrm{S}& \left(\mathrm{916}\right)\end{array}$
$\begin{array}{cc}\mathrm{E}\mathrm{\xb7}\mathrm{\text{S}}\stackrel{{\mathit{k}}_{2}}{\to}\mathrm{E}\mathrm{+}\mathrm{S}& \left(\mathrm{917}\right)\end{array}$
$\begin{array}{cc}\mathrm{E}\mathrm{\xb7}\mathrm{\text{S}}\mathrm{\text{}}\mathrm{\text{+}}\text{W}\stackrel{{\mathit{k}}_{3}}{\to}\text{\hspace{0.17em}}\text{P}\mathrm{+}\text{E}& \left(\mathrm{918}\right)\end{array}$
Here, P = 2NH_{3} = CO_{2}.
The corresponding rate laws for Reactions (916)–(918) are
$\begin{array}{cc}{\mathit{r}}_{\mathrm{1}\mathrm{E}\mathrm{\cdot}\mathrm{S}}={\mathit{k}}_{\mathrm{1}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{S)}& \left(\mathrm{916}\text{A}\right)\end{array}$
$\begin{array}{cc}{\mathit{r}}_{2\mathrm{E}\mathrm{\cdot}\mathrm{S}}={\mathit{k}}_{2}\mathrm{(E}\mathrm{\xb7}\mathrm{S)}& \left(\mathrm{917}\text{A}\right)\end{array}$
$\begin{array}{cc}{\mathit{r}}_{3\mathrm{E}\mathrm{\cdot}\mathrm{S}}={\mathit{k}}_{3}\mathrm{(E}\mathrm{\xb7}\mathrm{S)}\left(\mathrm{W}\right)& \left(\mathrm{918}\text{A}\right)\end{array}$
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 CO_{2}, that is, –r_{S} = r_{P}.
This rate law, Equation (919), is of not much use to us in making reaction engineering calculations because we cannot measure the concentration of the active intermediate, which is the enzyme–substrate complex (E · S). We will use the PSSH to express (E · S) in terms of measured variables.
–r_{S} = r_{P}, that is,
–r_{NH2CONH2}
= rCO_{2}
The net rate of formation of the enzyme–substrate complex is
r_{E·S} = r_{1E·S} + r_{2E·S} + r_{3E·S}
Substituting the rate laws, we obtain
$\begin{array}{ccc}{\mathit{r}}_{\mathrm{E}\mathrm{\cdot}\mathrm{S}}={\mathit{k}}_{\mathrm{1}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{}{\mathit{k}}_{\mathrm{2}}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S}\mathrm{)}\mathrm{}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W}\mathrm{)}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S)}& & \left(\mathrm{920}\right)\end{array}$
Using the PSSH, r_{E·S} = 0, we can now solve Equation (920) for (E · S)
and substitute for (E · S) into Equation (919)
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{s}}={\mathit{r}}_{\mathrm{P}}=\frac{{\mathit{k}}_{\mathrm{1}}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{(}\mathrm{W}\mathrm{)}}{{\mathit{k}}_{\mathrm{2}}\mathrm{+}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W}\mathrm{)}}& \left(\mathrm{922}\right)\end{array}$
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}.
We need to replace unbound enzyme concentration (E) in the rate law.
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):
$\begin{array}{cc}\mathrm{(}{\mathrm{E}}_{\mathit{t}}\mathrm{)}\mathrm{=}\mathrm{(}\mathrm{E)}\mathrm{+}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S)}& \left(\mathrm{923}\right)\end{array}$
Total enzyme concentration = Bound + Free enzyme concentration.
Substituting for (E · S)
$\mathrm{(}{\mathrm{E}}_{\mathit{t}}\mathrm{)}\mathrm{=}\mathrm{(}\mathrm{E)}+\frac{{\mathit{k}}_{\mathrm{1}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{S}\mathrm{)}}{{\mathit{k}}_{\mathrm{2}}\mathrm{+}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W}\mathrm{)}}$
solving for (E)
$\left(\text{E}\right)=\frac{\mathrm{(}{\mathrm{E}}_{\mathit{t}}\mathrm{)}\mathrm{(}{\mathit{k}}_{\mathrm{2}}+{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W}\mathrm{)}\mathrm{)}}{{\mathit{k}}_{\mathrm{2}}\mathrm{+}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W)}+{\mathit{k}}_{\mathrm{1}}\mathrm{(}\mathrm{S}\mathrm{)}}$
and substituting for (E) in Equation (922), 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/m^{3}) or (g/dm^{3}).
Because the reaction of urea and urease is carried out in an aqueous solution, water is, of course, in excess, and the concentration of water (W) is therefore considered constant, ca. 55 mol/dm^{3}. Let
${\mathit{k}}_{\mathrm{c}\mathrm{a}\mathrm{t}}={\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{W}\mathrm{)}\text{and}\text{let}{\mathit{K}}_{\mathrm{M}}=\frac{{\mathit{k}}_{\mathrm{c}\mathrm{a}\mathrm{t}}\mathrm{+}{\mathit{k}}_{\mathrm{2}}}{{\mathit{k}}_{\mathrm{1}}}$
After dividing the numerator and denominator of Equation (924) by k_{1}, we obtain a form of the Michaelis–Menten equation:
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{S}}=\frac{{\mathit{k}}_{\mathrm{c}\mathrm{a}\mathrm{t}}\mathrm{(}{\mathrm{E}}_{\mathit{t}}\mathrm{)}\mathrm{(}\mathrm{S}\mathrm{)}}{\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{+}{\mathit{K}}_{\mathrm{M}}}& \left(\mathrm{925}\right)\end{array}$
The final form of the rate law
Turnover number k_{cat}
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 singleenzyme molecule when the enzyme is saturated with substrate (i.e., all the active sites on the singleenzyme molecule are occupied, (S)>>K_{M}). For example, the turnover number for the decomposition of hydrogen peroxide, H_{2}O_{2}, by the enzyme catalase is 40 × 10^{6} s^{–1}. That is, 40 million molecules of H_{2}O_{2} are decomposed every second on a singleenzyme molecule saturated with H_{2}O_{2}. The constant K_{M} (mol/dm^{3}) 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 H_{2}O_{2} discussed earlier is 1.1 M, while that for chymotrypsin is 0.1 M.^{12}
^{12} D. L. Nelson and M. M. Cox, Lehninger Principles of Biochemistry, 3rd ed. New York: Worth Publishers, 2000.
Michaelis constant K_{M}
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
A sketch of the rate of disappearance of the substrate, –r_{s}, at a given concentration of total enzyme, E_{T}, is shown as a function of the substrate concentration, S, in Figure 96.
A plot of this type is sometimes called a Michaelis–Menten plot.
At low substrate concentrations, K_{M} ≫ (S), Equation (926) reduces to
$\mathrm{}{\mathit{r}}_{\mathrm{s}}\cong \frac{{\mathit{V}}_{\mathrm{max}}\mathrm{(}\mathrm{S}\mathrm{)}}{{\mathit{K}}_{\mathrm{M}}}$
and the reaction is apparent first order in the substrate concentration.
At high substrate concentrations,
(S) ≫K_{M}
Michaelis–Menten plot
Equation (926) reduces to
–r_{S} ≅ V_{max}
and we see the reaction is apparent zero order.
Interpretation of Michaelis constant
What does K_{M} represent? Consider the case when the substrate concentration is such that the reaction rate is equal to onehalf the maximum rate
$\mathrm{}{\mathit{r}}_{\mathrm{s}}\mathrm{=}\frac{{\mathit{V}}_{\mathrm{max}}}{\mathrm{2}}$
then
$\begin{array}{cc}\frac{{\mathit{V}}_{\mathrm{max}}}{\mathrm{2}}\mathrm{=}\frac{{\mathit{V}}_{\mathrm{max}}\mathrm{(}{\mathrm{S}}_{\mathrm{1}\mathrm{/}\mathrm{2}}\mathrm{)}}{{\mathit{K}}_{\mathrm{M}}\mathrm{+}\mathrm{(}{\mathrm{S}}_{\mathrm{1}\mathrm{/}\mathrm{2}}\mathrm{)}}& \left(\mathrm{927}\right)\end{array}$
Solving Equation (927) for the Michaelis constant yields
The Michaelis constant is equal to the substrate concentration at which the rate of reaction is equal to onehalf the maximum rate, that is, −r_{S} = V_{max}/2. The larger the value of K_{M}, the higher the substrate concentration(s) necessary for the reaction rate to reach half of its maximum value.
K_{M} = (S_{1/2})
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 10^{8}–10^{9} (dm^{3}/mol/s), the reaction rate approaches becoming diffusion limited. That is, it takes a long time for the enzyme and substrate to diffuse around in the fluid and find each other, but once they do they react immediately. We will discuss diffusionlimited reactions in Chapters 14 and 15.
In this example, we illustrate how to determine the enzymatic reactionrate parameters V_{max} and K_{M} for the reaction rate.
$\begin{array}{c}\text{Urea}+\text{Urease}\underset{{k}_{2}}{\overset{{k}_{1}}{\rightleftarrows}}\text{[Urea}\xb7{\text{Urease]}}^{*}\text{}\underset{+{\text{H}}_{2}\text{O}}{\overset{{k}_{3}}{\to}}\text{\hspace{0.17em}}2{\text{NH}}_{3}+{\text{CO}}_{2}+\text{Urease}\\ \text{S}+\text{E}\rightleftarrows \text{\hspace{0.17em}}\text{E}\xb7\text{S}\stackrel{+{\text{H}}_{2}\text{O}}{\to}\text{P}+\text{E}\end{array}$
The rate of reaction is given as a function of urea concentration in Table E92.1, where (S) ≡ C_{urea}.
TABLE E92.1 RATE DATA^{†}
C_{urea} (kmol/m^{3}) 
0.2 
0.02 
0.01 
0.005 
0.002 
–r_{urea} (kmol/m^{3}·s) 
1.08 
0.55 
0.38 
0.2 
0.09 
^{†} Unfiltered real data.
Determine the Michaelis–Menten parameters V_{max} and K_{M} for the reaction.
In addition to forming a Lineweaver–Burk Plot, form Hanes–Woolf and Eadie–Hofstee plots, and discuss the attributes of each type of plot.
Use nonlinear regression to find V_{max} and K_{M}.
Solution
Part (a)
Inverting Equation (926) gives us the Lineweaver–Burk equation
$\begin{array}{cc}\frac{\mathrm{1}}{\mathrm{}{\mathit{r}}_{\mathrm{s}}}\mathrm{=}\frac{\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{+}{\mathit{K}}_{\mathrm{M}}}{{\mathit{V}}_{\mathrm{max}}\mathrm{(}\mathrm{S}\mathrm{)}}\mathrm{=}\frac{\mathrm{1}}{{V}_{\mathrm{max}}}\mathrm{+}\frac{{\mathit{K}}_{\mathrm{M}}}{{V}_{\mathrm{max}}}\frac{\mathrm{1}}{\mathrm{(}\mathrm{S}\mathrm{)}}& \begin{array}{ccc}& \left(\text{E}\mathrm{92.1}\right)& \end{array}\end{array}$
or in terms of urea
Lineweaver–Burk equation
A plot of the reciprocal reaction rate versus the reciprocal urea concentration should be a straight line with an intercept (1/V_{max}) and slope (K_{M}/V_{max}). This type of plot is called a Lineweaver–Burk plot. We shall use the data in Table E92.2 to make two plots: a plot of –r_{urea} as a function of C_{urea} using Equation (926), which is called a Michaelis– Menten plot and is shown in Figure E92.1(a); and a plot of (1/–r_{urea}) as a function (1/C_{urea}), which is called a Lineweaver–Burk plot and is shown in Figure 92.1(b).
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TABLE E92.2 RAW AND PROCESSED DATA
C_{urea} (kmol/m^{3}) 
–r_{urea} (kmol/m^{3}·s) 
1/C_{urea} (m^{3}/kmol) 
1/–r_{urea} (m^{3}·s/kmol) 
0.20 
1.08 
5.0 
0.93 
0.02 
0.55 
50.0 
1.82 
0.01 
0.38 
100.0 
2.63 
0.005 
0.20 
200.0 
5.00 
0.002 
0.09 
500.0 
11.11 
Michaelis–Menten Plot
Lineweaver–Burk Plot
The intercept on Figure E92.1(b) is 0.75, so
$\frac{\mathrm{1}}{{V}_{\mathrm{max}}}=\mathrm{0}\mathrm{.}\mathrm{7}\mathrm{5}{\mathrm{m}}^{\mathrm{3}}\mathrm{\cdot}\mathrm{s}\mathrm{/}\text{kmol}$
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Therefore, the maximum rate of reaction is
$\begin{array}{c}\hline {V}_{\text{max}}=1.33\text{}{\text{kmol/m}}^{3}\cdot \text{s}=1.33{\text{mol/dm}}^{3}\cdot \text{s}\\ \hline\end{array}$
From the slope, which is 0.02 s, we can calculate the Michaelis constant, K_{M}
$\frac{{K}_{\text{M}}}{{V}_{\text{max}}}=\text{slope}=0.02\text{s}$
For enzymatic reactions, the two key ratelaw parameters are V_{max} and K_{M}.
Solving for K_{M}:
$\begin{array}{c}\hline {\begin{array}{c}{K}_{\text{M}}\end{array}=0.0266\text{kmol/m}}^{3}\mathrm{.}\\ \hline\end{array}$
Substituting K_{M} and V_{max} into Equation (926) gives us
where C_{urea} has units of (kmol/m^{3}) and –r_{urea} has units of (kmol/m^{3}·s). Levine and LaCourse suggest that the total concentration of urease, (E_{t}), corresponding to the value of V_{max} above is approximately 5 g/dm^{3}.
Part (b)
In addition to the Lineweaver–Burk plot, one can also use a Hanes–Woolf plot or an Eadie–Hofstee plot. Here, S ≡ C_{urea}, and –r_{S} ≡ –r_{urea}.
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{s}}\mathrm{=}\frac{{\mathit{V}}_{\mathrm{max}}\mathrm{(}\mathrm{S}\mathrm{)}}{{\mathit{K}}_{\mathrm{M}}\mathrm{+}\mathrm{(}\mathrm{S}\mathrm{)}}& \left(\mathrm{926}\right)\end{array}$
Hanes–Woolf Plot
For the Hanes–Woolf form, we can rearrange Equation (926) to
Equation (926) can be rearranged in the following forms. For the Eadie–Hofstee form
and for the Hanes–Woolf model, we plot [(S)/–r_{S}] as a function of (S) and for the Eadie–Hofstee model, we plot –r_{S} as a function of [–r_{S}/(S)].
Eadie–Hofstee Plot
When to use the different models? The Eadie–Hofstee plot does not bias the data points at low substrate concentrations, while the Hanes–Woolf plot gives a more accurate evaluation of V_{max}. In Table E92.3, we add two columns to Table E92.2 to generate these plots (C_{urea} ≡ S).
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TABLE E92.3 RAW AND PROCESSED DATA
S (kmol/m^{3}) 
–r_{S} (kmol/m^{3} · s) 
1/S (m^{3}/kmol) 
1/–r_{S} (m^{3} · s/kmol) 
S/–r_{S} 
–r_{s}/S (1/s) 
0.20 
1.08 
5.0 
0.93 
0.185 
5.4 
0.02 
0.55 
50.0 
1.82 
0.0364 
27.5 
0.01 
0.38 
100.0 
2.63 
0.0263 
38 
0.005 
0.20 
200.0 
5.00 
0.0250 
40 
0.002 
0.09 
500.0 
11.11 
0.0222 
45 
The slope of the Hanes–Woolf plot in Figure E92.2 (i.e., (1/V_{max}) = 0.826 s · m^{3}/kmol), gives V_{max} = 1.2 kmol/m^{3} · s and from the intercept, K_{M}/V_{max} = 0.02 s, we obtain K_{M} to be 0.024 kmol/m^{3}.
Best plot to evaluate V_{max}
$\frac{\mathrm{(}\mathrm{S}\mathrm{)}}{\mathrm{}{\mathit{r}}_{\mathrm{S}}}\mathrm{=}\frac{{\mathit{K}}_{\mathrm{M}}}{{\mathit{V}}_{\mathrm{max}}}\mathrm{+}\frac{\mathrm{1}\mathrm{(}\mathrm{S}\mathrm{)}}{{\mathit{V}}_{\mathrm{max}}}$
Hanes–Woolf plot
We next generate an Eadie–Hofstee plot from the data in Table E92.3. From the slope (–0.0244 kmol/m^{3}) in the EadieHofstee plot in Figure E92.3, we find K_{M} = 0.024 kmol/m^{3}. Next, extrapolating the line (–r_{S}/S) to zero, we find the intercept to be 1.22, therefore V_{max} = 1.22 kmol/m^{3} · s.
$\mathrm{}{\mathit{r}}_{\mathrm{S}}={\mathit{V}}_{\mathrm{max}}\mathrm{}{\mathit{K}}_{\mathrm{M}}\mathrm{(}\frac{\mathrm{}{\mathit{r}}_{\mathrm{S}}}{\mathrm{(}\mathrm{S}\mathrm{)}}\mathrm{)}$
Eadie–Hofstee plot
Part (c) Regression
Equation (926) and Table E92.2 were used in the regression program of Polymath with the following results for V_{max} and K_{M}, as shown in Table E92.4.
TABLE E92.4 REGRESSION RESULTS
A comparison of the values V_{max} and K_{M} is shown in Table E92.5.
TABLE E92.5 COMPARISON OF MICHAELIS–MENTON PARAMETER CALCULATIONS

Lineweaver–Burk 
Eadie–Hofstee 
Hanes–Woolf 
Regression 
V_{max} (kmol/m^{3} • s) 
1.33 
1.22 
1.21 
1.2 
K_{M} (kmol/m^{3}) 
0.027 
0.024 
0.024 
0.023 
These values are within experimental error of those values of V_{max} and K_{m} determined graphically.
Analysis: This example demonstrated how to evaluate the parameters V_{max} and K_{M} in the Michaelis–Menten rate law from enzymatic reaction data. Four techniques were used to evaluate V_{max} and K_{M} from experimental data: Lineweaver–Burk plot, Hanes–Woolf plot, Eadie–Hofstee plot, and nonlinear regression. The advantage of each type of plot was discussed for evaluating the parameters V_{max} and K_{M}.
A mole balance on urea in a batch reactor gives
$\begin{array}{c}\frac{\mathit{d}{\mathit{N}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}{dt}=\mathrm{}{\mathit{r}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}\mathit{V}\end{array}\begin{array}{c}\left(\text{929}\right)\end{array}$
Mole balance
Because this reaction is liquid phase, V = V_{0}, the mole balance can be put in the following form:
$\begin{array}{cc}\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}{\mathit{d}\mathit{t}}\mathrm{=}\mathrm{}{\mathit{r}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}& \left(\mathrm{930}\right)\end{array}$
The rate law for urea decomposition is
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}\mathrm{=}\frac{{\mathit{V}}_{\mathrm{max}}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}{{\mathit{K}}_{\mathrm{M}}\mathrm{+}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}& \left(\mathrm{931}\right)\end{array}$
Rate law
Substituting Equation (931) into Equation (930) and then rearranging and integrating, we get
$\mathit{t}\text{}\mathrm{=}{{\displaystyle \int}}_{{C}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}^{{C}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{0}}}\frac{\mathit{d}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}{\mathrm{}{\mathit{r}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}\mathrm{=}{{\displaystyle \int}}_{{C}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}^{{C}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{0}}}\frac{{\mathit{K}}_{\mathrm{M}}\mathrm{+}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}{{\mathit{V}}_{\mathrm{max}}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}}\mathit{d}{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}}$
Combine
Integrate
We can write Equation (932) in terms of conversion as
C_{urea} = Curea0 (1 − X)
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. Here, as shown in Chapter 7 Figures 71, 72, and 73, we want to arrange our equation so that we can obtain a linear plot of our data. Letting S = C_{urea} and S_{0} = Curea0, we arrange Equation (932) in the form
The corresponding plot in terms of substrate concentration is shown in Figure 97.
From Equation (933), we see that the slope of a plot of $\mathrm{\left[}\frac{\mathrm{1}}{\mathit{t}}\mathrm{ln}\frac{{\mathrm{S}}_{\mathrm{0}}}{\mathrm{S}}\mathrm{\right]}$ as a function of $\mathrm{\left[}\frac{{\mathrm{S}}_{\mathrm{0}}\mathrm{}\mathrm{S}}{\mathit{t}}\mathrm{\right]}$ will be $\mathrm{(}\mathrm{}\frac{\mathrm{1}}{{\mathrm{K}}_{\mathit{m}}}\mathrm{)}$ and the y intercept will be $\mathrm{(}\frac{{\mathit{V}}_{\mathrm{max}}}{{\mathit{K}}_{\mathrm{m}}}\mathrm{)}$. In cases similar to Equation (933) 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). You will note you cannot use the substrate concentration at t = 0 to construct this plot.
Calculate the time needed to convert 99% of the urea to ammonia and carbon dioxide in a 0.5dm^{3} batch reactor. The initial concentration of urea is 0.1 mol/dm^{3}, and the urease concentration is 0.001 g/dm^{3}. The reaction is to be carried out isothermally at the same temperature at which the data in Table E92.2 were obtained.
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Solution
We can use Equation (932a)
$\begin{array}{cc}\mathit{t}\mathrm{=}\frac{{\mathit{K}}_{\mathrm{M}}}{{\mathit{V}}_{\mathrm{max}}}\mathrm{ln}\frac{\mathrm{1}}{\mathrm{1}\mathrm{}\mathit{X}}\mathrm{+}\frac{{\mathit{C}}_{\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{0}}\mathit{X}}{{\mathit{V}}_{\mathrm{max}}}& \left(\mathrm{932a}\right)\end{array}$
From Table E92.4 we know K_{M} = 0.0233 mol/dm^{3}, V_{max} = 1.2 mol/dm^{3}= s. The conditions given are X = 0.99 and Cureao = 0.1 mol / dm^{3} (i.e., 0.1 kmol/m^{3}). However, for the conditions in the batch reactor, the enzyme concentration is only 0.001 g/dm^{3}, compared with 5 g/dm^{3} in Example 92. Because V_{max} = E_{t}·k_{3}, V_{max} for the second enzyme concentration is
$\begin{array}{c}{\mathit{V}}_{\mathrm{max}\mathrm{2}}=\frac{{\mathrm{E}}_{\mathit{t}\mathrm{2}}}{{\mathrm{E}}_{\mathit{t}\mathrm{1}}}{\mathit{V}}_{\mathrm{max}\mathrm{1}}=\frac{\mathrm{0}\mathrm{.}\mathrm{0}\mathrm{0}\mathrm{1}}{\mathrm{5}}\mathrm{\times}\mathrm{1.2}\mathrm{=}\mathrm{2.4}\mathrm{\times}\mathrm{1}{\mathrm{0}}^{\mathrm{}\mathrm{4}}\text{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{/}\mathrm{s}\mathrm{\cdot}\mathrm{d}{\mathrm{m}}^{\mathrm{3}}\\ {\mathit{K}}_{\mathrm{M}}=\mathrm{0.0}\mathrm{2}\mathrm{3}\mathrm{3}\text{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{/}\mathrm{d}{\mathrm{m}}^{\mathrm{3}}\text{and}X=0.99\end{array}$
Substituting into Equation (932)
$\begin{array}{lll}t\hfill & =\hfill & \frac{2.33\times {10}^{2}{\text{mol/dm}}^{3}}{2.4\times {10}^{4}{\text{mol/dm}}^{3}/\text{s}}1\text{n}\left(\frac{1}{0.01}\right)+\frac{\left(0.1{\text{mol/dm}}^{3}\right)\left(0.99\right)}{2.4\times {10}^{4}{\text{mol/dm}}^{3}/\text{s}}\hfill \\ \hfill & =\hfill & 447\text{s+412s}\hfill \\ \hfill & =\hfill & 859\text{s}\left(14.325\text{min}\right)\text{\hspace{1em}}\underset{\xaf}{\text{Answer}}\hfill \end{array}$
Analysis: This example shows a straightforward Chapter 5type calculation of the batch reactor time to achieve a certain conversion X for an enzymatic reaction with a Michaelis–Menten rate law. This batch reaction time is very short; consequently, a continuousflow reactor might be better suited for this reaction.
The effect of temperature on enzymatic reactions is very complex. If the enzyme structure would remain unchanged as the temperature is increased, the rate would probably follow the Arrhenius temperature dependence. However, as the temperature increases, the enzyme can unfold and/or become denatured and lose its catalytic activity. Consequently, as the temperature increases, the reaction rate, –r_{S}, increases up to a maximum and then decreases as the temperature is increased further. The descending part of this curve is called temperature inactivation or thermal denaturizing.^{13}
^{13} M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. Upper Saddle River, NJ: Prentice Hall, 2002, p. 77.
Figure 98 shows an example of this optimum in enzyme activity.^{14}
^{14} S. Aiba, A. E. Humphrey, and N. F. Mills, Biochemical Engineering, New York: Academic Press, 1973, p. 47.
In addition to temperature and solution pH, another factor that greatly influences the rates of enzymecatalyzed reactions is the presence of an inhibitor. Inhibitors are species that interact with enzymes and render the enzyme either partially or totally ineffective to catalyze its specific reaction. The most dramatic consequences of enzyme inhibition are found in living organisms, where the inhibition of any particular enzyme involved in a primary metabolic pathway will render the entire pathway inoperative, resulting in either serious damage to or death of the organism. For example, the inhibition of a single enzyme, cytochrome oxidase, by cyanide will cause the aerobic oxidation process to stop; death occurs in a very few minutes. There are also beneficial inhibitors, such as the ones used in the treatment of leukemia and other neoplastic diseases. Aspirin inhibits the enzyme that catalyzes the synthesis of the module prostaglandin, which is involved in the painproducing process. Aspirin and other covalent drugs with enzyme inhibitors are discussed in Nature Reviews, 10, 307 (2011) “The resurgence of covalent drugs,” by Professor Adrian Whitty. Data for an inhibitor discussed in this article is given in Problem P915. As shown in TV advertisements, the recent discovery of DDP4 enzyme inhibitor Januvia has been approved for the treatment of Type 2 diabetes, a disease affecting 240 million people worldwide. Work Problem P920_{C} and then as the TV ad said, “Ask your doctor if Januvia is right for you.”
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.
Competitive inhibition is of particular importance in pharmacokinetics (drug therapy). If a patient were administered two or more drugs that react simultaneously within the body with a common enzyme, cofactor, or active species, this interaction could lead to competitive inhibition in the formation of the respective metabolites and produce serious consequences. An example of competitive inhibition that is important in blood pressure regulation is the inhibition of the enzyme angiotensin by MercaptopropanalLpraline.^{†}
^{†} Biochemistry, 16 (25), 5481.
In competitive inhibition, another substance, that is, I, competes with the substrate for the enzyme molecules to form an inhibitor–enzyme complex, as shown in Figure 99. That is, 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 in Reaction Step 3 is the same (cf. Equations (918A) and (919)) as it was before, that is, in the absence of inhibitor
$\begin{array}{cc}{\mathit{r}}_{\mathrm{P}}={\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S}\mathrm{)}& \left(\mathrm{934}\right)\end{array}$
Applying the PSSH, the net rate of reaction of the enzyme–substrate complex is
$\begin{array}{cc}\begin{array}{ccc}{\mathit{r}}_{\mathrm{E}\mathrm{\cdot}\mathrm{S}}=\mathrm{0}={\mathit{k}}_{\mathrm{1}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{}{\mathit{k}}_{\mathrm{2}}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S}\mathrm{)}\mathrm{}{\mathit{k}}_{\mathrm{3}}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S}\mathrm{)}& & \end{array}& \left(\mathrm{935}\right)\end{array}$
In a similar fashion, apply the PSSH to the net rate of formation of the inhibitor–substrate complex (E · I) is also zero
$\begin{array}{cc}\begin{array}{ccc}{\mathit{r}}_{\mathrm{E}\mathrm{\cdot}\mathrm{I}}=\mathrm{0}={\mathit{k}}_{\mathrm{4}}\mathrm{(}\mathrm{E}\mathrm{)}\mathrm{(}\mathrm{I}\mathrm{)}\mathrm{}{\mathit{k}}_{\mathrm{5}}\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{I}\mathrm{)}& & \end{array}& \left(\mathrm{936}\right)\end{array}$
Reaction Steps
$\begin{array}{cc}\left(1\right)& E+\mathit{S}\stackrel{{\mathrm{k}}_{\mathrm{1}}}{\to}\mathrm{E}\mathrm{\cdot}\mathrm{S}\hfill \\ \left(2\right)& \mathrm{E}\mathrm{\cdot}\mathrm{S}\stackrel{{\mathrm{k}}_{2}}{\to}E\mathrm{+}\mathit{S}\hfill \\ \left(3\right)& \mathrm{E}\mathrm{\cdot}\mathrm{S}\stackrel{{\mathrm{k}}_{3}}{\to}P\mathrm{+}E\hfill \\ \left(4\right)& I\mathrm{+}E\stackrel{{\mathrm{k}}_{4}}{\to}\mathrm{E}\mathrm{\cdot}\mathrm{I}\text{}\left(inactive\right)\hfill \\ \left(5\right)& \mathrm{E}\mathrm{\cdot}\mathrm{I}\stackrel{{\mathrm{k}}_{5}}{\to}E\mathrm{+}I\hfill \end{array}$
Competitive Inhibition Pathway
Competitive inhibition pathway
The total enzyme concentration is the sum of the bound and unbound enzyme concentrations
$\begin{array}{cc}\begin{array}{ccc}{\mathrm{E}}_{\mathit{t}}=\mathrm{(}\mathrm{E}\mathrm{)}+\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{S}\mathrm{)}+\mathrm{(}\mathrm{E}\mathrm{\cdot}\mathrm{I}\mathrm{)}& & \end{array}& \left(\mathrm{937}\right)\end{array}$
Combining Equations (935), (936), and (937), solving for (E · S) then substituting in Equation (934) and simplifying
Rate law for competitive inhibition
V_{max} and K_{M} are the same as before when no inhibitor is present; that is,
${\mathit{V}}_{\mathrm{max}}={k}_{3}{\text{E}}_{t}\text{and}{K}_{M}=\frac{{k}_{2}+{k}_{3}}{{k}_{1}}$
and the competitive inhibition constant K_{I} (mol/dm^{3}) is defined as
${K}_{\mathrm{I}}=\frac{{\mathit{k}}_{\mathrm{5}}}{{\mathit{k}}_{\mathrm{4}}}$
By letting ${K}_{\mathrm{M}}^{\prime}={K}_{\mathrm{M}}\mathrm{(}\mathrm{1}\mathrm{+}\mathrm{(}I\mathrm{)}\mathrm{/}{K}_{I}\mathrm{)}$, we can see that the effect of adding a competitive inhibitor is to increase the “apparent” Michaelis constant, ${K}_{\text{M}}^{\prime}$. A consequence of the larger “apparent” Michaelis constant is ${K}_{\text{M}}^{\prime}$ that a larger substrate concentration is needed for the rate of substrate decomposition, –r_{S}, to reach half its maximum rate.
Rearranging Equation (938) in order to generate a Lineweaver–Burk plot
From the Lineweaver–Burk plot (Figure 910), we see that as the inhibitor (I) concentration is increased, the slope increases (i.e., the rate decreases), while the intercept remains fixed.
An example of uncompetitive inhibition is the inhibitor epristeride that affects androgen testosterone, as enzyme that stimulates growth. In uncompetitive inhibition, the inhibitor has no affinity for the enzyme by itself and thus does not compete with the substrate for the enzyme; instead, it ties up the enzyme–substrate complex by forming an inhibitor–enzyme–substrate complex, (I · E · S), which is inactive. In uncompetitive inhibition, the inhibitor reversibly ties up enzyme–substrate complex after it has been formed.
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 911.
Reaction Steps
$\begin{array}{cc}\left(1\right)& E+\mathit{S}\stackrel{{\mathrm{k}}_{\mathrm{1}}}{\to}\mathrm{E}\mathrm{\cdot}\mathrm{S}\hfill \\ \left(2\right)& \mathrm{E}\mathrm{\cdot}\mathrm{S}\stackrel{{\mathrm{k}}_{2}}{\to}E\mathrm{+}\mathit{S}\hfill \\ \left(3\right)& \mathrm{E}\mathrm{\cdot}\mathrm{S}\stackrel{{\mathrm{k}}_{3}}{\to}P\mathrm{+}E\hfill \\ \left(4\right)& I\mathrm{+}\mathrm{E}\mathrm{\cdot}\mathrm{S}\stackrel{{\mathrm{k}}_{4}}{\to}\mathrm{I}\mathrm{\cdot}\mathrm{E}\mathrm{\cdot}\mathrm{S}\text{}\left(inactive\right)\hfill \\ \left(5\right)& \mathrm{I}\mathrm{\cdot}\mathrm{E}\mathrm{\cdot}\text{S}\stackrel{{\mathrm{k}}_{5}}{\to}\text{I+}\text{E}\mathrm{\cdot}\text{S}\hfill \end{array}$
Uncompetitive Pathway
Uncompetitive inhibition pathway
Starting with the equation for the rate of formation of product, Equation (934), and then applying the pseudosteadystate 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 (http://www.umich.edu/~elements/6e/09chap/summaryexample4.html).
Rearranging Equation (940)
The Lineweaver–Burk plot for uncompetitive enzyme inhibition is shown in Figure 912 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.
^{15} In some literature, mixed inhibition is a combination of competitive and uncompetitive inhibition.
In noncompetitive inhibition, also sometimes called mixed inhibition, the substrate and inhibitor molecules react with different types of sites on the enzyme molecule. Examples of noncompetitive inhibitors include heavy metal ions such as Pb^{2+}, Ag^{+}, and Hg^{2+} and the inhibitor FK866, which affects the enzyme nicotinamide phosphoribosyltransferase.^{†} 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.
^{†} Cancer Res., 63 (21), 7436.
After a substrate molecule (S) attaches to the enzyme molecule (E) at the substrate site, then the inhibitor molecule attaches to the enzyme at the inhibitor site. (E • S + I ⇄ I • E • S)
After an inhibitor molecule (I) attaches to the enzyme molecule (E) 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 913. 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 (http://www.umich.edu/~elements/6e/09chap/summaryderive3.html). Equation (942) is in the form of the rate law that is given for an enzymatic reaction exhibiting noncompetitive inhibition.
Reaction Steps
$\begin{array}{cc}\left(1\right)& \mathrm{E}+\mathrm{S}\text{}\rightleftarrows \text{}\text{E}\cdot \text{S}\hfill \\ \left(2\right)& \mathrm{E}+\text{I}\text{}\rightleftarrows \text{I}\cdot \text{E}\text{}\left(\text{inactive}\right)\hfill \\ \left(3\right)& \text{I}+\mathrm{E}\cdot \text{S}\text{}\rightleftarrows \text{I}\cdot \text{E}\cdot \text{S}\text{}\left(\text{inactive}\right)\hfill \\ \left(4\right)& \text{S}+\text{I}\cdot \text{E}\text{}\rightleftarrows \text{I}\cdot \text{E}\cdot \text{S}\text{}\left(\text{inactive}\right)\hfill \\ \left(5\right)& \text{E}\cdot \text{S}\text{}\to \text{P}+\text{E}\hfill \end{array}$
Mixed inhibition
Noncompetitive Pathway
For noncompetitive inhibition, we see on the Lineweaver–Burk plot in Figure 914 both the slope $\left(\frac{{K}_{\mathrm{M}}}{{\mathit{V}}_{\mathrm{max}}}\mathrm{\right[}\mathrm{1}\mathrm{+}\frac{\mathrm{(}I\mathrm{)}}{{K}_{\mathrm{I}}}\mathrm{\left]}\right)$ and intercept $\left(\frac{\mathrm{1}}{{\mathit{V}}_{\mathrm{max}}}\mathrm{\right[}\mathrm{1}\mathrm{+}\frac{\mathrm{(}I\mathrm{)}}{{K}_{\mathrm{I}}}\mathrm{\left]}\right)$ increase with increasing inhibitor concentration. In practice, uncompetitive inhibition and mixed inhibition are generally observed only for enzymes with two or more substrates, S_{1} and S_{2}.
The three types of inhibition are compared with a reaction without inhibitors and are summarized on the Lineweaver–Burk plot shown in Figure 915.
Summary plot of types of inhibition
In summary, we observe the following trends and relationships:
In competitive inhibition, the slope increases with increasing inhibitor concentration, while the intercept remains fixed.
In uncompetitive inhibition, the yintercept increases with increasing inhibitor concentration, while the slope remains fixed.
In noncompetitive inhibition (mixed inhibition), both the yintercept and slope will increase with increasing inhibitor concentration.
Problem P912_{B} asks you to use the experimental data to identify the type of inhibition for the enzyme catalyzed reaction of starch.
Differences on Lineweaver–Burk Plot
In a number of cases, the substrate itself can act as an inhibitor, especially at high substrate concentrations. In the case of uncompetitive inhibition, the inactive molecule (S · E · S) is formed by the reaction
$\begin{array}{cc}\text{S}+\text{E}\cdot \text{S}\to \text{S}\cdot \text{E}\cdot \text{S}& \left(\text{inactive}\right)\end{array}$
Consequently, we see that by replacing (I) by (S) in Equation (940), the rate law for –r_{S} is
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{s}}=\frac{{\mathit{V}}_{\mathrm{max}}\mathrm{(}\mathrm{S}\mathrm{)}}{{K}_{\mathrm{M}}\mathrm{+}\mathrm{(}\mathrm{S}\mathrm{)}\mathrm{+}\frac{\mathrm{(}S{\mathrm{)}}^{\mathrm{2}}}{{K}_{\mathrm{I}}}}& \left(\mathrm{944}\right)\end{array}$
We see that at low substrate concentrations
$\begin{array}{cc}{K}_{\mathrm{M}}\mathrm{>}\mathrm{>}\mathrm{(}\mathrm{(}S\mathrm{)}\mathrm{+}\frac{\mathrm{(}S{\mathrm{)}}^{\mathrm{2}}}{{K}_{\mathrm{I}}}\mathrm{)}& \left(\mathrm{945}\right)\end{array}$
then
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{s}}\u02dc\frac{{\mathit{V}}_{\mathrm{max}}\mathrm{(}S\mathrm{)}}{{K}_{\mathrm{M}}}& \left(\mathrm{946}\right)\end{array}$
and the rate increases linearly with increasing substrate concentration.
At high substrate concentrations ((S)^{2} / K_{I}) >>(K_{M} + (S)), then
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{s}}=\frac{{\mathit{V}}_{\mathrm{max}}\mathrm{}{K}_{I}\mathrm{}}{S}& \left(\mathrm{947}\right)\end{array}$
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 916. 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 (944) 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 cofactors (see R9.6; http://www.umich.edu/~elements/6e/09chap/prof07prof2.html).
C&E News
Because of the increased interest in chemical engineering of faculty and students in biochemical engineering, many chemical engineering departments have changed their names to chemical and biochemical engineering. Biochemical engineering can encompass biomedical applications involving enzymatic reactions as well as cellular dynamics. A bioreactor is a reactor that sustains and supports life for cells and tissue cultures. Virtually all cellular reactions necessary to maintain life are mediated by enzymes as they catalyze various aspects of cell metabolism such as the transformation of chemical energy and the construction, breakdown, and digestion of cellular components. Because enzymatic reactions are involved in the growth of microorganisms (biomass), we now proceed to study microbial growth and bioreactors. Not surprisingly, the Monod equation, which describes the growth law for a number of bacteria and microorganisms, is similar to the Michaelis–Menten equation. Consequently, even though bioreactors are not truly homogeneous because of the presence of living cells, we include them in this chapter as a logical progression from enzymatic reactions. This discussion is far from being even slightly inclusive on bioreactions, but is meant to at least introduce a vocabulary for bioreactions and bioreactors.
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. Both microorganisms and mammalian cells are being used to produce a variety of products, such as insulin, most antibiotics, and polymers. The January 2019 issue of Cell Press indicated that the monoclonal antibodies (mAb) market is expected to grow to $130–$200 billion by 2020. mAb have been successful in fighting a number of cancers such as lymphoma, melanoma, along with breast, lung, and ovarian cancers. Mammalia cells are typically used in the production of mAb and the fundamentals of cell growth presented in this chapter may be used to model their growth. The reactors used for the grow mAb are batch, batch fed (semibatch), and perfusion (https://cellculturedish.com/perfusionbioreactorswithsomuchtooffertheydeserveacloserlook/). Additionally, 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 is the production of a single desired isomer that, when produced chemically, yields a mixture of isomers.
The growth of biotechnology
C&E News
In the fifth edition, I focused on Sapphire Energy and its initiative to grow and use live algae to produce biomass as an alternative energy source. Because I visited the New Mexico plant and was privy to the process, two homework problems were created (cf. Homework Problem P920_{C} and P921_{B}). Unfortunately, on April 19, 2017, the Algae World News reported that Sapphire Energy no longer exists and was bought by a local farmer for pennies on the dollar (https://news.algaeworld.org/2017/04/happenedsapphireenergy/).
Pennies!
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 917 along with a photo showing cell division, which is also shown in Figure 919.
The Algorithm
Cell Balance
Substrate Balance
Rate Law
Stoichiometry
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.^{16}
^{16} 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 918. A more detailed model is given in Figures 5.1 and 6.14 of Shuler and Kargi.^{17} 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).
^{17} Ibid., pp. 135, 185.
ATP + H_{2}O → ADP + P + H_{2}O_{} + Energy
The cell growth and division typical of mammalian cells is shown schematically in Figure 919. The four phases of cell division are called G1, S, G2, and M, and are also described in Figure 919.
In general, the growth of an aerobic organism follows the equation
Cell multiplication
A more abbreviated form of Equation (949) generally used is that a substrate in the presence of cells produces more cells plus product, that is,
$\begin{array}{cc}\text{Substrate}\text{}\stackrel{\text{Cells}}{\to}\text{More}\text{cells}\text{+}\text{Product}& \left(\mathrm{950}\right)\end{array}$
The products in Equation (950) 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 (949) can be found in texts by Shuler and Kargi,^{18} Bailey and Ollis,^{19} and Blanch and Clark.^{20} 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 920.
^{18} M. L. Shuler and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. Upper Saddle River, NJ: Prentice Hall, 2002.
^{19} J. E. Bailey and D. F. Ollis, Biochemical Engineering, 2nd ed. New York: McGrawHill, 1987.
^{20} H. W. Blanch and D. S. Clark, Biochemical Engineering, New York: Marcel Dekker, Inc., 1996.
Autocatalytic
Stages of cell growth in a batch reactor are shown schematically in Figures 921 and 922. Initially, a small number of cells is inoculated into (i.e., added to) the batch reactor containing the nutrients and the growth process begins, as shown in Figure 921. In Figure 922, the number of living cells is shown as a function of time. Actual cell concentration time data taken in the author’s research laboratory by PhD student Barry Wolf is shown on the CRE Web site^{21} (http://www.umich.edu/~elements/6e/09chap/summary.html#sec3).
Lag phase
Phase I, shown in Figure 922, 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.^{21}
^{21} B. Wolf and H. S. Fogler, “Alteration of the growth rate and lag time of Leuconostoc mesenteroides NRRLB523,” Biotechnol. Bioeng., 72 (6), 603 (2001). B. Wolf and H. S. Fogler, “Growth of Leuconostoc mesenteroides NRRLB523, in alkaline medium,” Biotechnol. Bioeng., 89 (1), 96 (2005).
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.
Exponential growth 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 cells are functioning but 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.
Antibiotics produced during the stationary phase
Phase IV the final phase, is the death phase, where a decrease in live cell concentration occurs. This decline is a result of the toxic byproducts, harsh environments, and/or depletion of nutrient supply.
Death phase
While many laws exist for the cell growth rate of new cells, that is,
$\text{Cells}+\text{Substrate}\to \text{More cells}+\text{Product}$
the most commonly used expression is the Monod equation for exponential growth:
$\begin{array}{cc}{r}_{g}=\mathrm{\mu}{C}_{c}& \left(9\text{}51\right)\end{array}$
where
r_{g} = cell growth rate, g/dm^{3}·s
C_{c} = cell concentration, g/dm^{3}
μ = specific growth rate, s^{–1}
The cell concentration is often given in terms of weight (g) of dry cells per liquid volume and is specified “grams dry weight per dm^{3},” that is, (gdw/dm^{3}).
The specific cell growth rate can be expressed as
$\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\mathrm{\mu}={\mathrm{\mu}}_{\mathrm{max}}\frac{{\mathit{C}}_{\mathit{s}}}{{K}_{\mathit{s}}+{\mathit{C}}_{\mathit{s}}}\mathrm{}\end{array}& {\text{S}}^{1}\end{array}& \left(9\text{}52\right)\end{array}$
where
μ_{max} = maximum specific growth reaction rate, s^{–1}
K_{s} = Monod constant, g/dm^{3}
C_{s} = substrate (i.e., nutrient) concentration, g/dm^{3}
Representative values μ_{max} = 1.3 h^{–1} K_{S} = 2.2 × 10^{–5} (g/dm^{3})
and is also shown in Figure 923. Representative values of μ_{max} and K_{s} are 1.3 h^{–1} and 2.2 × 10^{–5} g/dm^{3}, respectively, which are the parameter values for the E. coli growth on glucose. Combining Equations (951) and (952), 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
$\begin{array}{cc}{r}_{g}={\mathrm{\mu}}_{\mathrm{max}}{C}_{c}& \left(9\text{}54\right)\end{array}$
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 (953).
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
$\begin{array}{cc}{\mathit{k}}_{\mathrm{o}\mathrm{b}\mathrm{s}}={(\mathrm{1}\mathrm{}\frac{{\mathit{C}}_{p}}{{\mathit{C}}_{p}^{*}})}^{n}& \left(9\text{56}\right)\end{array}$
Empirical form of Monod equation for product inhibition
with
C_{p}= product concentration (g/dm^{3})
${C}_{p}^{*}$ = product concentration at which all metabolism ceases, g/dm^{3}
n = empirical constant
For the glucosetoethanol fermentation, typical inhibition parameters are
$\begin{array}{ccc}n=0.5& \mathrm{and}& {C}_{p}^{*}=93\text{}g/\end{array}{\mathrm{dm}}^{3}$
In addition to the Monod equation, two other equations are also commonly used to describe the cell growth rate; they are the Tessier equation
$\begin{array}{cc}{\mathit{r}}_{\mathit{g}}\mathrm{=}{\mathrm{\mu}}_{\mathrm{max}}[\mathrm{1}\mathrm{}\mathrm{exp}(\mathrm{}\frac{{\mathit{C}}_{\mathit{s}}}{\mathit{k}})]{\mathit{C}}_{\mathit{c}}& \left(9\text{}57\right)\end{array}$
Tessier Equation
and the Moser equation,
$\begin{array}{cc}{\mathit{r}}_{\mathit{g}}\mathrm{=}\frac{{\mathrm{\mu}}_{\mathrm{max}}{\mathit{C}}_{\mathit{c}}}{(\mathrm{1}+\mathit{k}{\mathit{C}}_{\mathit{s}}^{\mathrm{}\mathit{\lambda}})}& \left(9\text{}58\right)\end{array}$
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.^{22}
^{22} 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
$\begin{array}{cc}{\mathit{r}}_{d}\mathrm{=}({k}_{d}+{k}_{t}{C}_{t}){C}_{c}& \left(9\text{}59\right)\end{array}$
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.
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.
Doubling times
Effect of Temperature. As with enzymes (cf. Figure 98), 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.^{23} and is of the form
^{23} S. Aiba, A. E. Humphrey, and N. F. Millis, Biochemical Engineering, New York: Academic Press, 1973, p. 407.
$\begin{array}{cc}\begin{array}{c}\mathrm{\mu}\left(T\right)=\mathrm{\mu}\left({T}_{\text{m}}\right)I\prime \hfill \\ I\prime =\frac{{\mathit{aTe}}^{{E}_{1}/\mathit{RT}}}{{1+\mathit{be}}^{{E}_{2}/\mathit{RT}}}\hfill \end{array}\hfill & \left(9\text{60}\right)\end{array}$
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 trifolii, the equation takes the form
$\begin{array}{cc}I\prime ={\displaystyle \frac{\mathrm{0}\mathrm{.}\mathrm{0}\mathrm{0}\mathrm{3}\mathrm{8}\mathrm{T}{\mathit{e}}^{\mathrm{[}\mathrm{2}\mathrm{1}\mathrm{.}\mathrm{6}\mathrm{}\mathrm{6}\mathrm{7}\mathrm{0}\mathrm{0}\mathrm{/}\mathrm{T}\mathrm{]}}}{\mathrm{1}+{\mathit{e}}^{\mathrm{[}\mathrm{1}\mathrm{5}\mathrm{3}\mathrm{}\mathrm{4}\mathrm{8}\mathrm{,}\mathrm{0}\mathrm{0}\mathrm{0}\mathrm{/}\mathrm{T}\mathrm{]}}}}\hfill & \left(9\text{61}\right)\end{array}$
The maximum growth of Rhizobium trifolii was thought to occur at 310°K. However, experiments by Prof. Dr. Sven Köttlov and his students at Jofostan University in Riça, Jofostan, show that this temperature should be 312°K, not 310°K. A major turn of events in the Rhizobium trifolii scientific community!!
The stoichiometry for cell growth is very complex and varies with the microorganism/nutrient system and environmental conditions such as pH, temperature, and redox potential. This complexity is especially true when more than one nutrient contributes to cell growth, as is usually the case. We shall focus our discussion on a simplified version for cell growth, one that is limited by only one nutrient in the medium. In general, we have
$\text{Cells}+\text{Substrate}\to \text{More cells}+\text{Product}$
Tells us how much substrate is needed to produce new cells.
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}, that is, Y_{s/c}
${Y}_{\text{s}/\text{c}}=\frac{1}{{Y}_{\text{c}/\text{s}}}$
gives the ratio of (–ΔC_{s}), the substrate that must be consumed, to the increase in cell mass concentration ΔC_{c} with a representative value of 2.5 (g/g).
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
$\begin{array}{cc}\begin{array}{c}{\mathit{r}}_{\mathit{p}}={\mathit{Y}}_{\mathit{p}\mathrm{/}\mathit{c}}{\mathit{r}}_{\mathit{g}}={\mathit{Y}}_{\mathit{p}\mathrm{/}\mathit{c}}\text{}\mathrm{\mu}{\mathit{C}}_{C}={\mathit{Y}}_{\mathit{p}\mathrm{/}\mathit{c}}\frac{{\mathrm{\mu}}_{\mathrm{max}}{\mathit{C}}_{c}{\mathit{C}}_{s}}{{K}_{s}\text{+}{\mathit{C}}_{s}}\end{array}& \left(9\text{}63\right)\end{array}$
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
$\begin{array}{cc}{\mathit{r}}_{p}={Y}_{p/s}({r}_{s})& \left(9\text{6}5\right)\end{array}$
Nongrowthassociated 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
Cell yield
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
$m=0.05\text{}\text{}{\displaystyle \frac{g\text{}\text{}\mathrm{substrate}}{g\text{}\mathrm{dry}\text{}\mathrm{weight}\text{}}}{\displaystyle \frac{1}{\text{h}}}=0.05\text{}{\text{h}}^{1}$
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 is often used for both batch and continuousflow 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
$S\stackrel{\mathrm{cells}}{\to}Y{\prime}_{c/s}C+{Y\prime}_{p/s}P$
the yield coefficients would be written as
$\begin{array}{cc}{Y\prime}_{s/c}={\displaystyle \frac{\mathrm{Mass}\text{}\mathrm{of}\text{}\mathrm{substrate}\text{}\mathrm{consumed}\text{}\mathrm{to}\text{}\mathrm{form}\text{}\mathrm{new}\text{}\mathrm{cells}}{\mathrm{Mass}\text{}\text{}\mathrm{of}\text{}\mathrm{new}\text{}\mathrm{cells}\text{}\mathrm{formed}}}& \left(9\text{}69\text{A}\right)\end{array}$
$\begin{array}{cc}{Y\prime}_{s/p}={\displaystyle \frac{\mathrm{Mass}\text{}\mathrm{of}\text{}\mathrm{substrate}\text{}\mathrm{consumed}\text{}\mathrm{to}\text{}\mathrm{form}\text{}\mathrm{product}}{\mathrm{Mass}\text{}\text{}\mathrm{of}\text{}\mathrm{product}\text{}\mathrm{formed}}}& \left(9\text{}69\text{B}\right)\end{array}$
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
$\begin{array}{ccccccc}\left[\begin{array}{l}\mathrm{Net}\text{}\mathrm{rate}\text{}\mathrm{of}\\ \mathrm{substrate}\\ \mathrm{consumption}\end{array}\right]& =& \left[\begin{array}{l}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{substrate}\hfill \\ \mathrm{consumed}\end{array}\hfill \\ \mathrm{by}\text{cells}\end{array}\right]& +& \left[\begin{array}{l}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{substrate}\hfill \\ \mathrm{consumed}\text{}\mathrm{to}\end{array}\hfill \\ \mathrm{from}\text{}\mathrm{product}\end{array}\right]& +& \left[\begin{array}{l}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{substrate}\hfill \\ \mathrm{consumed}\text{}\text{}\mathrm{for}\end{array}\hfill \\ \mathrm{maintenance}\end{array}\right]\\ {r}_{s}& =& {\begin{array}{c}Y\prime \end{array}}_{s/c}{r}_{g}& +& {\begin{array}{c}Y\prime \end{array}}_{s/p}{r}_{p}& +& m\text{}{C}_{c}\end{array}$
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
Equation (970) will be the rate law for the substrate consumption that we will use in our analyses of cell growth. The corresponding rate of product formation is
Growthassociated product formation in the growth phase
The Stationary Phase. Because there is no growth during the stationary phase, it is clear that Equation (970) cannot be used to account for substrate consumption, nor can the rate of product formation be related to the growth rate (e.g., Equation (963)). 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
Nongrowthassociated product formation in the stationary phase
where
k_{p} = specific rate constant with respect to product, (dm^{3}/g · s)
C_{sn} = concentration of the secondary nutrient, (g/dm^{3})
C_{c} = cell concentration, g/dm^{3} (g ≡ gdw = gram dry weight)
K_{sn} = Monod constant for secondary nutrient, (g/dm^{3})
r_{pn} = Y_{p}_{/}_{sn}(=r_{sn}), (g/dm^{3} · s)
The net rate of secondary nutrient consumption, r_{sn}, during the stationary phase is relates to the rate of product formation, r_{p}
–r_{sn} = mC_{c} + Y_{sn/P}r_{p}
$\begin{array}{cc}{r}_{\mathit{sn}}=m\text{}{C}_{c}+\frac{{Y}_{\mathit{sn}/p}{k}_{p}{C}_{\mathit{sn}}{C}_{c}}{{K}_{\mathit{sn}}+{C}_{\mathit{sn}}}& \left(9\text{72}\right)\end{array}$
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 change in the secondary nutrient concentration, C_{sn}, at that time, from the secondary nutrient’s initial concentration, at time t = 0 C_{sno}, that is,
$\begin{array}{cc}{C}_{p}={Y}_{p/\mathit{sn}}({C}_{\mathit{sn}0}{C}_{\mathit{sn}})& \left(9\text{7}3\right)\end{array}$
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)
$\begin{array}{cc}{q}_{p}=\alpha \text{}{\mathrm{\mu}}_{g}+\mathrm{\beta}& \left(9\text{7}4\right)\end{array}$
such that the rate of product formation is
r_{p} = q_{p}C_{c}
Luedeking–Piret equation for the rate of product formation
The assumption here in using the =parameter is that the secondary nutrient is in excess.
The following data (Table E94.1) was obtained from batch reactor experiments for the yeast Saccharomyces cerevisiae
TABLE E94.1 RAW DATA
$\text{Glucose}\stackrel{\text{cells}}{\to}\text{More cells + Ethanol}$ 

Time, t (hr) 
Cells, C_{c} (g/dm^{3}) 
Glucose, C_{s} (g/dm^{3}) 
Ethanol, C_{p} (g/dm^{3}) 
0 
1 
250 
0 
1 
1.5 
244 
2.14 
2 
2.2 
231 
5.03 
3 
3.29 
218 
8.96 
Add another data point?
Determine the yield coefficients Y_{s}_{/}_{c}, Y_{c}_{/}_{s}, Y_{s}_{/}_{p}, Y_{p}_{/}_{s}, and Y_{p}_{/}_{c}. Assume no lag and neglect maintenance at the start of the growth phase when there are just a few cells.
Find the ratelaw parameters μ_{max} and K_{s}.
Solution
Yield coefficients
Calculate the substrate and cell yield coefficients, Y_{s}_{/}_{c} and Y_{c}_{/}_{s}.
Between t = 0 and t = 1 h
$\begin{array}{cc}{Y}_{s/c}=\frac{\mathrm{\Delta}\mathrm{\text{}}{\mathit{C}}_{s}}{\mathrm{\Delta}\mathrm{\text{}}{\mathit{C}}_{c}}=\frac{244250}{1.51}=12\text{}g/g& \left(\text{E}9\text{}4.1\right)\end{array}$
Between t = 2 and t = 3 h
$\begin{array}{cc}{Y}_{s/c}={\displaystyle \frac{218231}{3.292.2}}={\displaystyle \frac{13}{1.09}}=11.93\text{}g/g& \left(\text{E}9\text{}4.2\right)\end{array}$
Taking an average
Equation (E94.3) tells us that 11.96 grams of substrate are consumed to produce 1.0 grams of cells. We could also have used Polymath regression to obtain
that is 0.084 grams of cell are produced per gram of substrate consumed. Similarly, using the data at 1 and 2 hours, the substrate/product yield coefficient is
$\begin{array}{cc}{Y}_{s/p}=\frac{\mathrm{\Delta}\text{}{C}_{s}}{\mathrm{\Delta}\text{}{C}_{P}}=\frac{231244}{5.032.14}=\frac{13}{2.89}=4.5\text{}g/g& \left(\text{E}9\text{}4.5\right)\end{array}$
Here, approximately 0.22 grams of product are generated when 1.0 grams of substrate are consumed and the product/cell yield coefficient is
$\begin{array}{cc}{Y}_{p/c}=\frac{\mathrm{\Delta}\text{}{C}_{p}}{\mathrm{\Delta}\text{}{C}_{c}}=\frac{5.032.14}{2.21.5}=4.13\text{}g/g& \left(\text{E}9\text{}4.7\right)\end{array}$
We see 4.13 grams of product are generated with every gram of cells grown
Ratelaw parameters
We now need to determine the ratelaw parameters μ_{max} and K_{s} in the Monod equation
$\begin{array}{cc}{r}_{g}=\frac{{\mathrm{\mu}}_{\mathrm{max}}{C}_{c}{C}_{s}}{{K}_{S}+{C}_{s}}& \left(9\text{53}\right)\end{array}$
For a batch system
$\begin{array}{cc}{r}_{g}=\frac{d\text{}{C}_{c}}{d\text{}t}& \left(\text{E}9\text{}4.9\right)\end{array}$
To find the ratelaw parameters μ_{max} and K_{s}, we first apply the differential formulas in Chapter 7 to columns 1 and 2 of Table E94.1 to find r_{g} and add another column to Table E94.1. We will apply the differentiation formulas in Table 73 to the data in Table E94.1 in order to find r_{g} in Equation E94.9.
$\begin{array}{cc}{r}_{g0}=\frac{3{C}_{C0}+4{C}_{C1}{C}_{C2}}{2\mathrm{\Delta}\text{}t}=\frac{3\left(1\right)+4\left(1.5\right)2.2}{2\left(1\right)}=0.4& \left(\text{E}9\text{}4.10\right)\end{array}$
$\begin{array}{cc}{r}_{g1}=\frac{{C}_{C2}{C}_{C0}}{2\mathrm{\Delta}\text{}t}=\frac{2.21}{2\left(1\right)}=0.6& \left(\text{E}9\text{}4.11\right)\end{array}$
$\begin{array}{cc}{r}_{g1}=\frac{{C}_{C3}{C}_{C1}}{2\mathrm{\Delta}\text{}t}=\frac{3.291.5}{2\left(1\right)}=0.9& \left(\text{E}9\text{}4.12\right)\end{array}$
$\begin{array}{cc}{r}_{g3}=\frac{{C}_{C1}{4C}_{C2}+3{C}_{C3}}{2\mathrm{\Delta}\text{}t}=\frac{1.54\left(2.2\right)+3\left(3.29\right)}{2\left(1\right)}=1.29& \left(\text{E}9\text{}4.13\right)\end{array}$
We now form the processed data Table E94.2.
TABLE E94.2 RATE DATA
t 
C_{C} 
C_{S} 
r_{g} 
0 
1 
250 
0.4 
1 
1.5 
244 
0.6 
2 
2.2 
231 
0.9 
3 
3.29 
218 
1.29 
Because C_{s} >> K_{s} initially, it is best to regress the data using the Hanes–Woolf form of the Monod equation
$\begin{array}{cc}\frac{{C}_{c}{C}_{s}}{{r}_{g}}=\frac{{K}_{s}}{{\mathrm{\mu}}_{\mathrm{max}}}+\frac{{C}_{s}}{{\mathrm{\mu}}_{\mathrm{max}}}& \left(\text{E}9\text{}4.14\right)\end{array}$
How to regress the Monod equation for μ_{max} and K_{s}
We will use nonlinear least squares to find μ_{max} and K_{s} from the data in Table E94.2. We now use the newly calculated r_{g} along with C_{c} and C_{s} in Table E94.2 to prepare Table E94.3 given of $\left[\frac{{C}_{c}{C}_{s}}{{r}_{g}}\right]$ as a function of (C_{s}) to use in Polymath regression.
TABLE E94.3 PROCESSED DATA
t 
0 
1 
2 
3 
C_{c}C_{s}/r_{g} 
625 
610 
568 
558 
C_{s} 
250 
244 
231 
218 
From the output of Polymath’s nonlinear regression of Equation (E94.14), we find μ_{max} = 0.46 h^{–1} and K_{s} = 33.5 g/dm^{3}.
$\begin{array}{c}\hline {r}_{g}=\frac{0.46\left({\text{h}}^{1}\right){C}_{c}{C}_{s}}{33.5\left({\text{g/dm}}^{3}\right)+{C}_{s}}\\ \hline\end{array}$
Analysis: In this example we showed how to use the concentrations of cells, substrate and product given in Table E94.1 to calculate the yield coefficients Y_{s}_{/}_{c}, Y_{c}_{/}_{s}, Y_{s}_{/}_{p}, Y_{p}_{/}_{s}, Y_{c}_{/}_{p}, and Y_{p}_{/}_{c}. We used simplified calculations in this example in order to gain a quick understanding of how to obtain the parameters. In practice, we would most likely collect many more data points and use nonlinear regression to evaluate all the parameters. Next, we differentiated the cell concentration times data and then used nonlinear regression to find the Monod ratelaw parameters μ_{max} and K_{s}.
There are two ways that we could account for the growth of microorganisms. One is to account for the number of living cells, and the other is to account for the mass of the living cells. We shall use the latter, grams of cells. A mass balance on the microorganisms in a CSTR (chemostat; e.g., margin figure that follows and Figure 924) of constant volume is (975)
$\begin{array}{cc}\begin{array}{ccccccc}\left[\begin{array}{c}\begin{array}{l}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{accumulation}\hfill \\ \mathrm{of}\text{}\mathrm{cells},\hfill \end{array}\hfill \\ g/s\end{array}\right]\hfill & =\hfill & \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{cells}\hfill \end{array}\hfill \\ \mathrm{entering},\hfill \end{array}\hfill \\ g/s\end{array}\right]\hfill & \hfill & \left[\begin{array}{l}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{cells}\hfill \\ \mathrm{leaving},\hfill \end{array}\hfill \\ \hfill g/s\hfill \end{array}\right]\hfill & +\hfill & \left[\begin{array}{l}\begin{array}{c}\mathrm{Net}\text{rate}\mathrm{of}\hfill \\ \mathrm{generation}\hfill \\ \mathrm{of}\text{}\mathrm{live}\text{}\mathrm{cells},\hfill \end{array}\hfill \\ \hfill g/s\hfill \end{array}\right]\hfill \\ V\frac{d\text{}{C}_{c}}{d\text{}\text{}t}\hfill & =\hfill & {\upsilon}_{0}{C}_{c0}\hfill & \hfill & {\upsilon}_{0}{C}_{c}\hfill & +\hfill & ({r}_{g}{r}_{d})v\hfill \end{array}\hfill & \left(9\text{}75\right)\hfill \end{array}$
Cell mass balance
The corresponding substrate balance is
$\begin{array}{cc}\begin{array}{ccccccc}\left[\begin{array}{c}\begin{array}{l}\mathrm{Rate}\text{}\mathrm{of}\\ \mathrm{accumulation}\\ \mathrm{of}\text{}\mathrm{substrate},\end{array}\\ g/s\end{array}\right]& =& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\\ \mathrm{substrate}\hfill \end{array}\hfill \\ \mathrm{entering},\end{array}\hfill \\ g/s\end{array}\right]& & \left[\begin{array}{l}\begin{array}{c}\mathrm{Rate}\text{}\mathrm{of}\hfill \\ \mathrm{substrate}\hfill \\ \mathrm{leaving},\end{array}\hfill \\ \hfill g/s\hfill \end{array}\right]& +& \left[\begin{array}{l}\begin{array}{c}\text{Rate}\mathrm{of}\hfill \\ \mathrm{substrate}\hfill \\ \mathrm{generation},\end{array}\hfill \\ \hfill g/s\hfill \end{array}\right]\\ V\frac{d\text{}{C}_{s}}{d\text{}\text{}t}& =& {\upsilon}_{0}{C}_{s0}& & {\upsilon}_{0}{C}_{s}& +& {r}_{g}V\end{array}& \left(9\text{}76\right)\end{array}$
In most systems, the entering microorganism concentration, C_{c}_{0}, is zero for a flow reactor.
For a batch system υ = υ_{0} = 0, the mass balances are as follows:
Cell Mass Balance
$V\frac{d\text{}{C}_{c}}{d\text{}\text{}t}={r}_{g}V{r}_{d}V$
The mass balances
Dividing by the reactor volume V gives
Substrate Mass Balance
The rate of disappearance of substrate, =r_{s}, results from substrate used for cell growth and substrate used for cell maintenance
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 (976), reduces to
Stationary phase
Typically, the growth rate utilizing the secondary nutrients, r_{pn}, will have the same Monod form of the rate law as r_{g} (e.g., Equation (971)). Of course, Equation (979) only applies for substrate concentrations greater than zero.
The rate of product formation, r_{p}, can be related to the rate of substrate consumption, –r_{s}, through the following balance when maintenance is neglected, that is, m = 0:
$\begin{array}{cc}V\begin{array}{c}\frac{d\text{}{C}_{p}}{d\text{}t}={r}_{p}V={Y}_{p/s}\end{array}\text{}({r}_{s})V& \left(9\text{}81\right)\end{array}$
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 (963), that is, r_{p} = Y_{p/c}r_{g}. The coupled firstorder ordinary differential equations above can be solved by a variety of numerical techniques.
Glucosetoethanol fermentation is to be carried out in a batch reactor using an organism such as Saccharomyces cerevisiae. Plot the concentrations of cells, substrate, and product and the rates for growth, death, and maintenance, that is, r_{g}, r_{d}, and r_{sm} as functions of time. The initial cell concentration is 1.0 g/dm^{3}, and the substrate (glucose) concentration is 250 g/dm^{3}.
Additional data (Partial source: R. Miller and M. Melick, Chem. Eng., 113 (February 16, 1987):
$\begin{array}{c}\begin{array}{cccccc}\hfill {C}_{p}^{*}& =& 93\text{}g/{\mathrm{dm}}^{3}\hfill & \hfill {Y}_{c/s}& =& 0.08\text{}g/g\hfill \\ \hfill n& =& 0.52\hfill & \hfill {Y}_{p/s}& =& 0.45\text{}g/g\hfill \\ \hfill {\begin{array}{c}\mathrm{\mu}\end{array}}_{\mathrm{max}}& =& 0.46{\text{h}}^{1}\hfill & \hfill {Y}_{p/c}& =& 5.6\text{}g/g\hfill \\ \hfill {K}_{s}& =& 33.5\text{}g/{\mathrm{dm}}^{3}\hfill & \hfill {k}_{d}& =& 0.01{\text{h}}^{1}\hfill \end{array}\\ m=0.03\text{}\left(g\text{}\mathrm{substrate}\right)/(g\text{}\mathrm{cells}\cdot \text{h})\end{array}$
Solution
Mass balances:
Cells: $\begin{array}{cc}V\frac{d\text{}{C}_{c}}{d\text{}t}=({r}_{g}{r}_{d})V& \left(\text{E}9\text{}5.1\right)\end{array}$
Substrate: $\begin{array}{cc}V\frac{d\text{}{C}_{s}}{d\text{}t}={Y}_{s/c}({r}_{g})V{r}_{\mathit{sm}}V& \left(\text{E}9\text{}5.2\right)\end{array}$
Product: $\begin{array}{cc}V\frac{d\text{}{C}_{p}}{d\text{}t}={r}_{p}V={Y}_{p/c}\left({r}_{g}V\right)& \left(\text{E}9\text{}5.3\right)\end{array}$
The algorithm
Rate laws:
Growth: $\begin{array}{cc}{r}_{g}={\mathrm{\mu}}_{\mathrm{max}}{(1\frac{{C}_{p}}{{C}_{p}^{*}})}^{0.52}\frac{{C}_{c}{C}_{s}}{{K}_{s}+{C}_{s}}& \left(\text{E}9\text{}5.4\right)\end{array}$
Death: $\begin{array}{cc}{r}_{d}={k}_{d}{C}_{c}& \left(\text{E}9\text{}5.5\right)\end{array}$
Maintenance: $\begin{array}{cc}{r}_{\mathit{sm}}=m{C}_{c}& \left(9\text{67}\right)\end{array}$
Stoichiometry:
$\begin{array}{cc}{r}_{p}={Y}_{p/c}{r}_{g}& \left(\text{E}9\text{5.6}\right)\end{array}$
Combining gives
$\begin{array}{c}\begin{array}{cc}\frac{d\text{}{C}_{c}}{d\text{}\text{}t}={\mathrm{\mu}}_{\mathrm{max}}{(1\frac{{C}_{p}}{{C}_{p}^{*}})}^{0.52}\frac{{C}_{c}{C}_{s}}{{K}_{s}+{C}_{s}}{k}_{d}{C}_{c}& \left(\text{E}9\text{}5.7\right)\end{array}\hfill \end{array}$
Cells Substrate Product
These equations were solved using an ODE equation solver (see Table E95.1). The results are shown in Figure E95.1 for the parameter values given in the problem statement.
TABLE E95.1 POLYMATH PROGRAM
Differential equations 

Explicit equations 

Calculated values of DEQ variables 


Variable 
Initial value 
Final value 
1 
Cc 
1. 
16.55544 
2 
Cp 
0 
92.96161 
3 
Cs 
250. 
39.36185 
4 
kobs 
0.46 
0.007997 
5 
Ks 
33.5 
33.5 
6 
m 
0.03 
0.03 
7 
rd 
0.01 
0.1655544 
8 
rg 
0.4056437 
0.071523 
9 
rsm 
0.03 
0.4966631 
10 
t 
0 
12. 
11 
umax 
0.46 
0.46 
12 
Ypc 
5.6 
5.6 
13 
Ysc 
12.5 
12.5 
(http://www.umich.edu/~elements/6e/09chap/live.html)
The substrate concentration C_{s} can never be less than zero. However, we note that when the substrate is completely consumed, the first term on the righthand side of Equation (E95.8) (and line 2 of the Polymath program) will be zero but the second term for maintenance, mC_{c}, will not. Consequently, if the integration is carried further in time, the integration program will predict a negative value of C_{s}!! This inconsistency can be addressed in a number of ways, such as including an if statement in the Polymath program (e.g., if C_{s} is less than or equal to zero, then m = 0). The Wolfram sliders and typical Wolfram results are shown in Figures E95.2 and E95.3, respectively.
Analysis: In this example, we applied a modified CRE algorithm to biomass formation and solved the resulting equations using the ODE solver Polymath. We note in Figure E95.1(d) the growth rate, r_{g}, goes through a maximum, increasing at the start of the reaction as the concentration of cells, C_{c}, increases then decreasing as the substrate concentration decreases, and at the end C_{p} approaches C_{p}* and the growth rate approaches zero(nutrient) as does k_{obs} in Figure E95.1(c). One also observes in (d) the death rate r_{d} and substrate consumption for maintenance are very small and only become important at large cell concentrations. We see from Figures E95.1(a) and (b) that the cell concentration increases dramatically with time while the product concentration does not. The reason for this difference is that part of the substrate is consumed for maintenance and part for cell growth, leaving only the remainder of the substrate to be transformed into product.
Chemostats are essentially CSTRs that contain microorganisms. A typical chemostat is shown in Figure 924, along with the associated monitoring equipment and pH controller. One of the most important features of the chemostat is that it allows the operator to control the cell growth rate. This control of the growth rate is achieved by adjusting the volumetric feed rate (dilution rate).
In this section, we return to the mass balance equations on the cells (Equation (975)) and substrate (Equation (976)), and consider the case where the volumetric flow rates in and out are the same, υ = υ_{0}, and that no live (i.e., viable) cells enter the chemostat C_{C0} = 0. We next define a parameter common to bioreactors called the dilution rate, D. The dilution rate is
$D=\frac{{\upsilon}_{0}}{V}$
and is simply the reciprocal of the space time τ, that is, $D=\frac{1}{\tau}$. Dividing Equations (975) and (976) by V and using the definition of the dilution rate, and neglecting maintenance, that is, m = 0, we have
$\begin{array}{cc}\text{Accumulation = In}& \u2013\text{Out+}\text{}\text{Generation}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\mathrm{Cell}:& \frac{d\text{}{C}_{c}}{d\text{}t}=0& D\text{}{C}_{\text{\hspace{0.17em}}c}+({r}_{g}{r}_{d})\end{array}& \left(9\text{}82\right)\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\mathrm{Substrate}:& \frac{d\text{}{C}_{s}}{d\text{}t}=D\text{}{C}_{s0}& D\text{}{C}_{\text{\hspace{0.17em}}s}+{r}_{s}\end{array}& \left(9\text{}83\right)\end{array}$
CSTR mass balances
Using the Monod equation, the growth rate is determined to be
$\begin{array}{cc}{r}_{g}=\mathrm{\mu}\mathrm{\text{}}{C}_{c}=\frac{{\mathrm{\mu}}_{\mathrm{max}}{C}_{s}{C}_{c}}{{K}_{s}+{C}_{s}}& \left(9\text{}53\right)\end{array}$
Rate law
For steadystate operation we have
$\begin{array}{cc}D\text{}{C}_{C}={r}_{g}{r}_{d}& \left(9\text{}84\right)\end{array}$
Steady state
and
$\begin{array}{cc}D\text{}({C}_{s0}{C}_{s})={r}_{s}& \left(9\text{85}\right)\end{array}$
We now neglect the death rate, r_{d}, and combine Equations (951) and (984) for steadystate operation to obtain the mass flow rate of cells out of the chemostat, ${\dot{m}}_{c}={C}_{c}{\upsilon}_{0}$, and the rate of generation of cells, r_{g}V. Equating and r_{g}V, and then substituting for r_{g} = =C_{c}, we obtain
$\begin{array}{cc}{\dot{m}}_{c}={C}_{c}{\upsilon}_{0}={r}_{g}V=\mathrm{\mu}\text{}{C}_{c}V& \left(9\text{}\text{86}\right)\end{array}$
Dividing by C_{c}V we see the cell concentration cancels to give the dilution rate D
Dilution rate
An inspection of Equation (987) reveals that the specific growth rate of the cells can be controlled by the operator by controlling the dilution rate D, that is, $D=\frac{{\upsilon}_{0}}{V}$. That is, simply by increasing the volumetric flow rate we can increase the specific growth rate μ. Using Equation (952)
How to control cell growth
$\begin{array}{cc}\begin{array}{cc}\mathrm{\mu}={\mathrm{\mu}}_{\mathrm{max}}\frac{{C}_{s}}{{K}_{s}+{C}_{s}}& {\text{s}}^{1}\end{array}& \left(9\text{}\text{52}\right)\end{array}$
to substitute for μ in terms of the substrate concentration and then solving for the steadystate substrate concentration yields
$\begin{array}{cc}{C}_{s}=\frac{D\text{}{K}_{s}}{{\mathrm{\mu}}_{\mathrm{max}}D}& \left(9\text{}\text{}88\right)\end{array}$
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
$\begin{array}{cc}{r}_{s}={r}_{g}{Y}_{s/c}& \left(9\text{}\text{}89\right)\end{array}$
$\begin{array}{cc}{C}_{c}={Y}_{s/c}({C}_{s0}{C}_{s})& \left(9\text{}\text{}68\right)\end{array}$
Substituting for C_{s} using Equation (968) and rearranging, we obtain
To learn the effect of increasing the dilution rate, we combine Equations (982) and (954), and set m and r_{d} to zero to get
$\begin{array}{cc}\frac{d{C}_{c}}{d\text{}\text{}t}=(\mathrm{\mu}D){C}_{c}& \left(9\text{}\text{}91\right)\end{array}$
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
That is, as we continually increase we reach a point at which the concentration of live cells exiting the reactor will be zero and the point is called cell washout. The dilution rate at which washout will occur is obtained from Equation (990) by setting C_{c} = 0.
Flow rate at which washout 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., ${\dot{m}}_{c}={C}_{c}{\upsilon}_{0}$) divided by the volume V, or
$\frac{{\dot{M}}_{c}}{V}=\frac{{\upsilon}_{0}{C}_{c}}{V}=D{C}_{c}\phantom{000000000}\left(\mathrm{993}\right)$
Maximum rate of cell production (DC_{c})
Using Equation (990) to substitute for C_{c} in Equation (993) yields
$\begin{array}{cc}D\text{}{C}_{c}=D\text{}{Y}_{c/s}({C}_{s0}\frac{D\text{}{K}_{s}}{{\mathrm{\mu}}_{\mathrm{max}}D})& \left(9\text{}\text{}94\right)\end{array}$
Figure 925 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 (994), with respect to the dilution rate D:
$\begin{array}{cc}\text{}\frac{d\left(D\text{}{C}_{c}\right)}{d\text{}D}=0& \left(9\text{}\text{}95\right)\end{array}$
Then
Maximum rate of cell production
The organism Streptomyces aureofaciens was studied in a 10dm^{3} 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 926.^{24} Note that the data follow the same trends as those discussed in Figure 925.
^{24} B. Sikyta, J. Slezak, and M. Herold, Appl. Microbiol., 9, 233.
^{†} http://umich.edu/~safeche/processtriangle.html
To ensure that chemical processes run safely, we must pay attention to safety indicators—those practices and situations that tell us if we are likely to have an accident. Process Safety Triangles are used to illustrate the different indicators and actions that can lead to an accident. This tool highlights how the smallest unsafe act can lead to a major accident. The triangle is applied from the bottom up, where each layer can be thought of as a preventative measure to the layer above it. The purpose is to show how an unsafe mindset can grow and produce tragic consequences.
Let’s go to page 4 of the CCPS Process Safety document that discusses three metrics: Lagging Indicators, Leading Indicators and Near Misses. (https://www.aiche.org/sites/default/files/docs/pages/CCPS_ProcessSafety_Lagging_2011_224.pdf). Starting at the top of the pyramid and working down we have the following levels shown in Figure 927. A Google search on “process safety pyramid” will show you variations of the triangle/pyramid (https://www.google.com/search?safe=strict&q=process+safety+pyramid&tbm=isch&source=univ&sa=X&ved=2ahUKEwiD2OuX5uvhAhVNY6wKHbYDDPgQsAR6BAgHEAE&biw=1266&bih=524&dpr=1.25).
Fatality: Loss of life by an accident.^{†}
^{†} Google Search. Google. Web. 20 June 2018.
Lost Time/Serious Injury: The classification for an occupational injury which includes: (a) all disabling work injuries and (b) nondisabling work injuries. These injuries include: (1) eye injuries requiring treatment by a physician, (2) fractures, (3) injuries requiring hospitalization, (4) loss of consciousness, (5) injuries requiring treatment by a doctor, and (6) injuries requiring restriction of motion or work, or assignment to another job.^{‡}
^{‡} Center for Chemical Process Safety. “CCPS Process Safety Glossary.” American Institute of Chemical Engineers, www.aiche.org/ccps/resources/glossary.
Minor Injuries: Any injury sustained that does not meet one of the requirements of a serious injury listed above. For example, any injury that can be treated quickly on site, such as a bruise, a minor cut, or scratch.
Near Misses: An event in which an accident (i.e., property damage, environmental impact, or human loss) or an operational interruption could have plausibly resulted if circumstances had been slightly different.^{§}
^{§} Near Miss Reporting Systems by the National Safety Council and Alliance (an OSHA Cooperative Program). Document can be found at https://www.nsc.org/Portals/0/Documents/WorkplaceTrainingDocuments/NearMissReportingSystems.pdf.
Unsafe Acts: Any act that deviates from a generally recognized safe way or specified method of doing a job and which increases the probabilities for an accident.^{¶}
^{¶} US Legal, Inc. “USLegal.” Unsafe Act Law and Legal Definition (definitions.uslegal.com/u/unsafeact/).
One of the main purposes of the process safety triangle is the illustration of how unsafe acts can lead to a major incident. The process safety triangle is also used to visualize the different layers of protection and help redesign systems to ensure better preventative practices. The bottom level, unsafe acts, is categorized as a leading metric. Leading metrics are preventative actions taken to avoid an incident from occurring. Unsafe behaviors of employees, such as not wearing personal protective equipment or the lack of protection layers for a system, are the fundamental causes of process safety incidents. The wider the base of your process safety triangle is the wider the top level is. This triangle illustrates how the more unsafe behaviors you have, the more likely a fatality from a process safety incident is to occur.
Near misses and the levels above it are categorized as lagging metrics. In many companies, near misses are analyzed at the same level as an accident. Lagging metrics are events that happen in a process that are reported for the improvement of the safety of the process. Analyzing past incidents has proven that if near misses are treated like injuries and incidents by reporting them, then the likelihood of a fatality from a process safety incident decreases. All past accidents have had warning signs (near misses) that indicated something needed to change and could have prevented the accident from occurring.^{§} Reporting near misses makes others aware of a potential issue and allows for measures to be implemented to prevent the escalation to a process safety incident. Most, if not all, companies require that all unsafe acts–be logged in safety report whether or not an injury occurs.
The key takeaways from the process safety triangle are that eliminating unsafe acts drastically reduces the risk of a process safety incident occurring. Reporting near misses allows for the process to be reviewed and improve to prevent injuries and fatalities from occurring.
Below is an example of a process safety triangle with an estimation of each reported type of action on the triangle (Figure 928).
In the PSSH, we set the rate of formation of the active intermediates equal to zero. If the active intermediate A* is involved in m different reactions, we set it to
$\begin{array}{cc}\text{}{r}_{\text{A}*,\mathrm{net}}\equiv \underset{i=1}{\overset{m}{\mathrm{\Sigma}}}{r}_{\text{A}*i}=0& \left(\text{S}9\text{}\text{}1\right)\end{array}$
This approximation is justified when the active intermediate is highly reactive and present in low concentrations.
The azomethane (AZO) decomposition mechanism is
$\begin{array}{cc}{r}_{\text{}{\text{N}}_{2}}=\frac{\text{k}{\left(\mathrm{AZO}\right)}^{2}}{1+\text{k\u2032}\left(\mathrm{AZO}\right)}& \left(\text{S}9\text{}\text{}3\right)\end{array}$
By applying the PSSH to AZO*, we show the rate law, which exhibits firstorder dependence with respect to AZO at high AZO concentrations and secondorder dependence with respect to AZO at low AZO concentrations.
Enzyme kinetics: enzymatic reactions follow the sequence
$\text{E}+\text{S}\underset{{k}_{2}}{\overset{{k}_{1}}{\overrightarrow{\leftarrow}}}\text{E}\cdot \text{S}\stackrel{{k}_{3}}{\to}\text{E}+\text{P}$
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}).
The three different types of inhibition—competitive, uncompetitive, and noncompetitive (mixed)—are shown on the Lineweaver–Burk plot:
Bioreactors:
$\text{Cells}+\text{Substrate}\to \text{More cells}+\text{Product}$
Phases of bacteria growth:
I. Lag II. Exponential III. Stationary IV. Death
Unsteadystate mass balance on a chemostat
$\begin{array}{cc}\frac{d\text{}{C}_{c}}{d\text{}t}=D({C}_{c0}{C}_{c})+{r}_{g}{r}_{d}& \left(\text{S}9\text{}\text{}5\right)\end{array}$
$\begin{array}{cc}\frac{d\text{}{C}_{s}}{d\text{}t}=D({C}_{s0}{C}_{s})+{r}_{s}& \left(\text{S}9\text{}\text{}6\right)\end{array}$
Monod growth rate law
${r}_{g}={\mathrm{\mu}}_{\mathrm{max}}\begin{array}{cc}\frac{\text{}{C}_{c}{C}_{s}}{{K}_{s}+{C}_{s}}& \left(\text{S}9\text{}\text{}7\right)\end{array}$
Stoichiometry
$\begin{array}{cc}{Y}_{c/s}=\frac{\mathrm{Mass}\text{}\text{}\mathrm{of}\text{}\text{}\mathrm{new}\text{}\text{}\mathrm{cells}\text{}\mathrm{formed}}{\mathrm{Mass}\text{}\mathrm{of}\text{}\text{}\mathrm{substrate}\text{}\mathrm{consumed}}& \left(\text{S}9\text{}\text{}8\right)\end{array}$
$\begin{array}{cc}{Y}_{s/c}=\frac{1}{{Y}_{c/s}}& \left(\text{S}9\text{}\text{}9\right)\end{array}$
Substrate consumption
$\begin{array}{cc}{r}_{s/c}={Y}_{s/c}{r}_{g}+m\text{}{C}_{c}& \left(S9\text{}10\right)\end{array}$
Maximum cell concentration and cell washout
The dilution rate at which the cells become washed out is
${D}_{\text{max}}=\frac{{\mu}_{\text{max}}{C}_{\text{s}0}}{{K}_{\text{s}}+{C}_{\text{s}0}}$
The dilution rate that gives the maximum rate of production of the cells exiting the reactor, m_{c} is
${D}_{\text{maxprod}}={\mu}_{\text{max}}(1\u2122\frac{\sqrt{{K}_{s}}}{{K}_{s}+{C}_{s0}})$
(http://www.umich.edu/~elements/6e/09chap/obj.html#/)
Earth Probe TOMs Total Ozone September 8, 2000
Ozone Layer
(http://umich.edu/~elements/6e/web_mod/ozone/index.htm)
Glow Sticks,
(http://www.umich.edu/~elements/6e/web_mod/new/glowsticks/index.htm)
Interactive Computer Games (http://umich.edu/~elements/6e/icm/index.html)
Ecology (http://umich.edu/~elements/6e/icm/ecology.html)
Enzyme Man (http://umich.edu/~elements/6e/icm/enzyme.html)
R91 Pharmacokinetics in Drug Delivery (http://www.umich.edu/~elements/6e/09chap/prof07prof5.html)
Pharmacokinetic models of drug delivery for medication administered either orally or intravenously are developed and analyzed.
R92 Around 2005, the author of this book carried out a research project on ethane that was published in the medical journal Alcohol, 35 (1), 3–12 (2005). Abbreviated salient features from the article can be found in the Summary Notes (http://www.umich.edu/~elements/6e/09chap/summary.html#sec3) and in the Summary Notes (http://www.umich.edu/~elements/6e/09chap/profpharmacokinetics.html). Physiologically Based Pharmacokinetic (PBPK) Models. Case Study: Alcohol Metabolism in Humans (http://www.umich.edu/~elements/6e/09chap/profpharmacokinetics.html).
${{C}_{2}H}_{5}OH\text{}\underset{ADH}{\overrightarrow{\leftarrow}}{CH}_{3}CHO\text{}\overleftarrow{AIDH}\text{}{CH}_{3}COOH$
^{25} P. K. Wilkinson, et al., “Pharmacokinetics of ethanol after oral administration in the fasting state,” J. Pharmacoket. Biopharm., 5(3), 207–224 (1977).
Go to the LEP Wolfram sliders to vary some of the parameters and write two conclusions.
The author of this textbook and two students published an article on ethanol metabolics in the medical journal Alcohol that can be found at (http://www.umich.edu/~elements/6e/09chap/pdf/Alcohol.pdf).
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
In each of the following questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include.
To enhance the quality of your sentences, 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.
Q91_{A} QBR (Question Before Reading). What factors affect the enzymatic rate law and cell growth law both positive and negative ways.
Q92_{A} i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/09chap/iclicker_ch9_q1.html) and view at least five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.
Q93_{A} Go to the LearnChemE screencast link for Chapter 9 (http://www.umich.edu/~elements/6e/09chap/learnchemevideos.html).
View one of the screencast 5 to 6minute video (1) PSSH, (2) Cell Growth, (3) MichaelisMenten Kinetics in a CSTR, and (4) Competitive Inhibition tutorials and list two of the most important points that the presenter was making.
In the cell growth screencast, why is there a minus right before r_{s} and not before r_{g}?
Q94_{A} What key points does the Process Safety Triangle want to emphasize? Which point do you think some companies have recently been required to report? What is the overall takeaway lesson from the Process Safety Triangle?
P91_{A}
Example 91: The Stern–Volmer Equation
Wolfram and Python
Explain how I would change if the concentrations of CS_{2} and M were increased.
Explain how the light intensity varies with an increase in concentration of M.
Write two conclusions from your experiment in (i)–(ii).
Example 92: Evaluation of the Michaelis–Menten Parameters V_{max} and K_{M}
Wolfram and Python
At very low values of K_{M} in the Michaelis–Menton plot (e.g., K_{M} = 0.0001), how does the rate of substrate consumption, –r_{S}, vary with substrate concentration, C_{urea}?
At very low values of K_{M} in the Eadie–Hofstee plot, what are the values of K_{M} and V_{max} such that the substrate rate, –r_{S}, becomes zero?
Write two conclusions from your experiment in (i)–(ii).
Polymath
The following additional runs were carried out when an inhibitor was present:
C_{urea}(kmol/m^{3}) 
C_{inhibitor}(kmol/m^{3}) 
–r_{urea}(kmol/m^{3} · s) 
0.01 
0.1 
0.125 
0.005 
0.1 
0.065 
What type of inhibition is taking place?
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.
Example 93: Batch Enzymatic Reactors
Wolfram and Python
As the conversion approaches one, what does the reaction time, in seconds, approach?
What happens to X versus t when K_{M} is varied?
Vary each parameter and write a set of conclusions.
Polymath
What would the conversion be after 15 minutes if the initial concentration of urea were increased by a factor of 10?
What is the minimum value of K_{M} so that conversion is below 0.5 after half an hour?
Example 94: Estimate the Yield Coefficients 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?
Example 95: Bacteria Growth in a Batch Reactor
Wolfram and Python
Which parameter will you vary so that C_{C} goes through a maxima at t = 9 hr? Explain the results.
What is the minimum value of K_{S} such that C_{S} is always higher than C_{p}?
What is the maximum value of μ_{max} so that r_{g} continuously increases up to t = 12 hr?
What is the value of Y_{S/C} such that virtually 100% conversion is achieved at the end of the reaction?
Write a set of conclusions based on your experiment (i)–(iv) above.
Polymath
Modify the code to carry out the fermentation in a fedbatch (e.g., semibatch) reactor in which the substrate is fed at a rate of 0.5 dm^{3}/h and a concentration of 5 g/dm^{3} to an initial liquid volume of 1.0 dm^{3} containing a cell mass with an initial concentration of C_{ci} = 0.2 mg/dm^{3} and an initial substrate concentration of C_{ci} = 0.5 mg/dm^{3}. 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.
Repeat (i) when the growth is uncompetitively inhibited by the substrate with K_{I} = 0.7 g/dm^{3}.
Set ${C}_{p}^{*}=10000{\text{g/dm}}^{3}$, and compare your results with the base case.
LEP 9 Enzyme Inhibition.
Vary V_{max} and K_{m} between their minimum and maximum values and describe what you find.
Vary K_{I} and describe how each of the three types of inhibition is affected.
Write three conclusions based on your experiments in (i) and (ii).
LEP 97 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.
Hint: Read the journal article in the Summary Notes (Alcohol 35, 1 (2005)), available at http://www.umich.edu/~elements/6e/09chap/profpharmacokinetics.html.
Write a summary paragraph describing what you learned from reading the article.
Wolfram and Python
Vary the initial concentration for ethanol in the stomach and describe how the concentrations of ethanol and acetaldehyde in the blood are affected.
Set V_{max} for acetaldehydes between 10% and 50% of its normal value and compare the concentrationtime trajectories with the base cases.
Write two conclusions about alcohol metabolism.
P92_{A}
ICG Enzyme Man. Load the ICG on your computer and carry out the exercise. Performance number =________________________.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
Rederive Equation (99) assuming the inert gas M (e.g., N_{2}) involved is also the reaction with the added steps by
$\begin{array}{l}\text{AZO}+\text{M}\stackrel{{k}_{4}}{\to}{\text{AZO}}^{\cdot}+\text{M}\hfill \\ {\text{AZO}}^{\cdot}+\text{M}\stackrel{{k}_{5}}{\to}\text{AZO}+\text{M}\hfill \end{array}$
P93_{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 as a retardant with forward and reverse specific rate constants k_{5} and k_{6}, the following additional reactions occur:
$\begin{array}{c}\text{H}\cdot \text{HCl}\stackrel{{k}_{5}}{\to}{\text{H}}_{2}+\text{Cl}\cdot \\ \text{H}\cdot +\text{Cl}\cdot \stackrel{{k}_{6}}{\to}\text{HCl}\end{array}$
Assume that all reactions are elementary and that the PSSH holds for the O·, OH·, and Cl· radicals.
Derive a rate law for the consumption of CO when no retardant is present.
Derive an equation for the concentration of H· as a function of time, assuming constant concentration of O_{2}, CO, and H_{2}O for both uninhibited combustion and combustion with HCl present. Sketch H· versus time for both cases.
Use Polymath to find out what happens when k_{1} = 0.0001, k_{4} = 0.02, k_{5} = 0.05, and k_{6} = 0.005 appropriate units. Write one sentence conclusion.
P94_{A} The pyrolysis of acetaldehyde is believed to take place according to the following sequence:
$\begin{array}{ccc}\hfill {\mathrm{CH}}_{3}\mathrm{CHO}& \stackrel{{k}_{1}}{\begin{array}{c}\to \end{array}}& {\mathrm{CH}}_{3}\cdot +\mathrm{CHO}\hfill \\ \hfill {\mathrm{CH}}_{3}\cdot +{\mathrm{CH}}_{3}\mathrm{CHO}& \stackrel{{k}_{2}}{\to}& {\mathrm{CH}}_{3}\cdot +\mathrm{CO}+{\mathrm{CH}}_{4}\hfill \\ \hfill \mathrm{CHO}\cdot +{\mathrm{CH}}_{3}\mathrm{CHO}& \stackrel{{k}_{3}}{\to}& {\mathrm{CH}}_{3}+2\mathrm{CO}+{\text{H}}_{2}\hfill \\ \hfill 2{\mathrm{CH}}_{3}\cdot & \stackrel{{k}_{4}}{\to}& {\text{C}}_{2}{\text{H}}_{6}\hfill \end{array}$
Derive the rate expression for the rate of disappearance of acetaldehyde, –r_{Ac}.
Under what conditions does it reduce to the equation at the beginning of Section 9.1 on page 368?
Sketch a reaction pathway diagram for this reaction. Hint: See margin note on page 373.
P95_{B}
OEQ (Old Exam Question). 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.
The gasphase homogeneous oxidation of nitrogen monoxide (NO) to dioxide (NO_{2})
$2\mathrm{NO}+{\text{O}}_{2}\stackrel{k}{\to}2{\mathrm{NO}}_{2}$
is known to have a form of thirdorder 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, NO_{3}, 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.
The rate law for formation of phosgene, COCl_{2}, from chlorine, Cl_{2}, and carbon monoxide, CO, has the rate law
${r}_{{\mathrm{COCl}}_{2}}=k\text{}{\text{C}}_{\mathrm{CO}}{\text{C}}_{{\mathrm{Cl}}_{2}}^{\mathit{3}/\mathit{2}}$
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 Cl_{2} is one of the two active intermediates.
Suggest an active intermediate(s) and mechanism for the reaction H_{2} + Br_{2} → 2HBr. Use the PSSH to show whether or not your mechanism is consistent with the rate law
${r}_{\mathrm{HBr}}=\frac{{k}_{1}\text{}{\text{C}}_{{\text{H}}_{2}}{\text{C}}_{\text{Br}}^{\mathit{3}/\mathit{2}}}{{\text{C}}_{\mathrm{HBr}}+{k}_{2}{\text{C}}_{{\mathrm{Br}}_{2}}}$
P96_{C} (Tribology) Why do 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
$\begin{array}{ccc}\hfill {I}_{2}& \stackrel{{k}_{0}}{\to}& 2I\cdot \hfill \\ \hfill I\cdot +\mathrm{RH}& \stackrel{{k}_{1}}{\to}& \text{R}\cdot +\text{H}I\hfill \\ \hfill \text{R}\cdot +{\text{O}}_{2}& \stackrel{{k}_{p\text{}\text{l}}}{\to}& {\mathrm{RO}}_{2}^{\cdot}\hfill \\ \hfill {\mathrm{RO}}_{2}^{\cdot}+\mathrm{RH}& \stackrel{{k}_{p\text{2}}}{\to}& \mathrm{ROOH}+\text{R}\cdot \hfill \\ \hfill 2{\mathrm{RO}}_{2}^{\cdot}& \stackrel{{k}_{t}}{\to}& \mathrm{inactive}\hfill \end{array}$
where I_{2} 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:
Derive a rate law for the degradation of the motor oil in the absence of an antioxidant at low temperatures.
Derive a rate law for the rate of degradation of the motor oil in the presence of an antioxidant for low temperatures.
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?
Sketch a reaction pathway diagram for both high and low temperatures, with and without antioxidant.
See the openended problem G.2 in Appendix G and on the CRE Web site for more on this problem.
P97_{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:
$\begin{array}{cc}\text{H}\stackrel{{k}_{1}}{\to}\text{I}& \left(\text{P}9\text{}7.1\right)\end{array}$
or the person may become ill through contact with another ill person:
$\begin{array}{cc}\text{I}+\text{H}\stackrel{{k}_{2}}{\to}2\text{I}& \left(\text{P}9\text{}7.2\right)\end{array}$
The ill person may become healthy:
$\begin{array}{cc}\text{I}\stackrel{{k}_{3}}{\to}\text{H}& \left(\text{P}9\text{}7.3\right)\end{array}$
or the ill person may expire:
$\begin{array}{cc}\text{I}\stackrel{{k}_{4}}{\to}\text{D}& \left(\text{P}9\text{}7.4\right)\end{array}$
Application of the above sequence to the 2020 Coronavirus Pandemic can be found on page 737.
Derive an equation for the death rate.
At what concentration of healthy people does the death rate become critical? (Ans: When [H] =(k_{3} + k_{4})/k_{2}.)
Comment on the validity of the PSSH under the conditions of Part (b).
If k_{1} = 10^{–8}h^{–1}, k_{2} = 10^{–16} (people·h)^{–1}, k_{3} = 5 × 10^{–10} h^{–1}, k_{4} = 10^{–11} h^{–1}, and H_{o} = 10^{9} 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).
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P98_{B} Derive the rate laws for the following enzymatic reactions and sketch and compare, where possible, with the plots shown in Figure E92.1.
E + S ⇄ E · S ⇄ P + E
E + S ⇄ E · S ⇄ → P + E
E + S_{1} ⇄ E · S_{1}
E · S_{1} + S_{2} ⇄ E · S_{1}S_{2}
E · S_{1}S_{2} → P + E
E + S ⇄ E · S → P
P + E ⇄ E · P
Which of the reactions (a)–(d), if any, lend themselves to analysis by a Lineweaver–Burk plot?
P99_{B} OEQ (Old Exam Question). Beef catalase has been used to accelerate the decomposition of hydrogen peroxide to yield water and oxygen (Chem. Eng. Educ., 5, 141). 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.
t (min) 
0 
10 
20 
50 
100 
C_{H2O2} (mol/L) 
0.02 
0.01775 
0.0158 
0.0106 
0.005 
Determine the Michaelis–Menten parameters V_{max} and K_{M}.
If the total enzyme concentration is tripled, what will the substrate concentration be after 20 minutes?
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
List ways you can work this problem incorrectly.
P910_{B} OEQ (Old Exam Question). It has been observed that substrate inhibition occurs in the following enzymatic reaction:
E + S → P + E
Show that the rate law for substrate inhibition is consistent with the plot in Figure P910_{B} of –r_{s} (mmol/L·min) versus the substrate concentration S (mmol/L).
If this reaction is carried out in a CSTR that has a volume of 1000 dm^{3}, to which the volumetric flow rate is 3.2 dm^{3}/min, determine the three possible steady states, noting, if possible, which are stable. The entrance concentration of the substrate is 50 mmol/dm^{3}. What is the highest conversion?
What would be the effluent substrate concentration if the total enzyme concentration is reduced by 33%?
List ways you can work this problem incorrectly.
How could you make this problem more difficult?
P911_{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.
Assume the rate follows Michaelis–Menten kinetics with respect to oxygen. Calculate the maximum (i.e., V_{max}), and the Michaelis–Menten constant K_{M}.
(Ans: V_{max} = 52.63μL O_{2}/h · mg cells.)
Using the Lineweaver–Burk plot, determine the type of inhibition sulfanilamide that causes the O_{2} uptake to change.
P_{O2}* 
Q_{O2} (no sulfanilamide) 
Q_{O2} (20 mg sulfanilamide/mL added to medium) 
0.0 
00.0 
00.0 
0.5 
23.5 
17.4 
1.0 
33.0 
25.6 
1.5 
37.5 
30.8 
2.5 
42.0 
36.4 
3.5 
43.0 
39.6 
5.0 
43.0 
40.0 
*P_{O2} = oxygen partial pressure, mmHg; Q_{O2} = oxygen uptake rate, μL of O_{2} per hour per mg of cells.
List ways you can work this problem incorrectly.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P912_{B} OEQ (Old Exam Question). The enzymatic hydrolysis of starch was carried out with and without maltose and αdextrin added. (Adapted from S. Aiba, A. E. Humphrey, and N.F. Mills, Biochemical Engineering, New York: Academic Press, 1973.)
Starch → αdextrin → Limit dextrin → Maltose
No Inhibition 


C_{S} (g/dm^{3}) 
12.5 
9.0 
4.25 
1.0 

–r_{S} (relative) 
100 
92 
70 
29 
Maltose added (I = 12.7 mg/dm^{3}) 


C_{S} (g/dm^{3}) 
10 
5.25 
2.0 
1.67 

–r_{S} (relative) 
77 
62 
38 
34 
αdextrin added (I = 3.34 mg/dm^{3}) 


C_{S} (g/dm^{3}) 
33 
10 
3.6 
1.6 

–r_{S} (relative) 
116 
85 
55 
32 
Determine the types of inhibition for maltose and for αdextrin.
P913_{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 ${\mathit{EH}}_{2}^{+}$
$\begin{array}{l}\begin{array}{cccc}{\text{H}}^{+}+{\text{E}}^{}\hfill & \overrightarrow{\leftarrow}\hfill & \text{EH}\hfill & {K}_{1}=\frac{\left(\text{EH}\right)}{\left(\text{}{\text{H}}^{+}\right)\left({\text{E}}^{}\right)}\hfill \\ {\text{H}}^{+}+\text{EH}\hfill & \overrightarrow{\leftarrow}\hfill & {\begin{array}{c}\text{EH}\end{array}}_{2}^{+}\hfill & {K}_{2}=\frac{\left({\text{EH}}_{2}^{+}\right)}{\left({\text{H}}^{+}\right)\left(\text{EH}\right)}\hfill \end{array}\hfill \\ \begin{array}{cc}\begin{array}{c}\begin{array}{cc}\begin{array}{c}\text{EH}+S\end{array}& \overrightarrow{\leftarrow}\end{array}\end{array}& \mathrm{EHS}\to \mathrm{EH}+P,{K}_{\text{M}}=\frac{\left(\mathrm{EHS}\right)}{\left(\mathrm{EH}\right)\left(\text{S}\right)}\end{array}\hfill \end{array}$
where E^{–} and ${\mathit{EH}}_{2}^{+}$ are inactive.
Determine whether the preceding sequence can explain the optimum in enzyme activity with pH shown in Figure P913_{B}.
List ways you can work this problem incorrectly.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P914_{B} OEQ (Old Exam Question). An Eadie–Hofstee Plot is shown in Figure P914_{B} for different types of enzyme inhibition.
Match the line with the type of inhibition.
Line A Inhibition Mechanism. Ans: ________
Line B Inhibition Mechanism. Ans: ________
Line C Inhibition Mechanism. Ans: ________
Draw a Hanes–Woolf plot analogous to Figure P914_{B} for the three types of inhibition and the case of no inhibition.
P915_{B} An enzymatic reaction that follows, MichaelsMenten kinetics rate law with initial enzyme concentration C_{E0} is
The rate constant, k_{2}, was measured as a function of inhibitor concentration and shown in Table P915_{B}
TABLE P915_{B} SUBSTRATE CONCENTRATION AND K_{2} INHIBITION DATA^{†}
C_{s} (μ mol/dm^{3}) 
0.1 
0.04 
1.0 
2 
3 
4.0 
6.0 
8.0 
k_{2} (min^{–1}) 
0.015 
0.04 
0.062 
0.09 
0.0105 
0.0115 
0.0123 
0.0128 
^{†}Nature Reviews, Lit. cit.
Determine the type of inhibition and write the full rate law.
P916_{B} The biomass reaction
$\text{SubstrateS}+\text{Cells}\to \text{Morecells}+\text{Product}$
takes place in a 12dm^{3} CSTR (chemostat) where the entering concentration of substrate is 200 g/dm^{3}. The rate law follows the Monod equation with μ_{max} = 0.5s^{–1} and K_{S} = 50 g/dm^{3}. What is the volumetric flow rate, υ_{0} (dm^{3}/h), that will give the maximum production rate of cells (g/h)?
P917_{B} The production of a product P from a particular gramnegative bacteria follows the Monod growth law
${r}_{g}=\frac{{\mathrm{\mu}}_{\mathit{max}}{C}_{s}{C}_{c}}{{K}_{S}+{C}_{s}}$
with μ_{max} = 1 h^{–1}, K_{S} = 0.25 g/dm^{3}, and Y_{c/s} = 0.5 g/g.
The reaction is to be carried out in a batch reactor with the initial cell concentration of C_{c}_{0} = 0.1 g/dm^{3} and substrate concentration of C_{s}_{0} = 20 g/dm^{3}.
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.
The reaction is now to be carried out in a CSTR with C_{s}_{0} = 20 g/dm^{3} and C_{c}_{0} = 0. What is the dilution rate at which washout occurs?
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?
How would your answers to (b) and (c) change if cell death could not be neglected with k_{d} = 0.02 h^{–1}?
How would your answers to (b) and (c) change if maintenance could not be neglected with m = 0.2 g/h/dm^{3}?
Redo part (a) and use a logistic growth law
${r}_{g}={\mathrm{\mu}}_{\mathrm{max}}(1\frac{{C}_{c}}{{C}_{\infty}}){C}_{c}$
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/dm^{3}. Can you find an analytical solution for the batch reactor? Compare with part (a) for C_{∞} = Y_{c/s} C_{s}_{0} + C_{c}_{0}.
List ways you can work this problem incorrectly.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P918_{B} Redo Problem P917_{B} (a), (c), and (d) using the Tessier equation
r_{g} = μ_{max}[1 – e^{–Cs/k}]C_{c}
with μ_{max} = 1.0 h^{–1} and k = 8 g/dm^{3}.
List ways you can work this problem incorrectly.
How could you make this problem more difficult?
P919_{B} The bacteria XII can be described by a simple Monod equation with μ_{max} = 0.8 h^{–1} and K_{S} = 4 g/dm^{3}, 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 dm^{3}/h at a substrate concentration of 10 g/dm^{3}.
What size fermentor is needed to achieve 90% conversion of the substrate? What is the exiting cell concentration?
How would your answer to (a) change if all the cells were filtered out and returned to the feed stream?
Consider now two 5000dm^{3} CSTRs connected in series. What are the exiting concentrations C_{s}, C_{c}, and C_{p} from each of the reactors?
Determine, if possible, the volumetric flow rate at which washout occurs and also the flow rate at which the cell production rate (C_{c} υ_{0}) in grams per day is a maximum.
Suppose you could use the two 5000dm^{3} reactors as batch reactors that take 2 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/dm^{3}? How many 500dm^{3} batch reactors would you need to match the CSTR production rate?
List ways you can work this problem incorrectly.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
P920_{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/dm^{3} and the volumetric flow rate of the substrate was 2 dm^{3}/h.
Run 
C_{s} (g/dm^{3}) 
D (day^{21}) 
C_{c} (g/dm^{3}) 
1 
1 
1 
0.9 
2 
3 
1.5 
0.7 
3 
4 
1.6 
0.6 
4 
10 
1.8 
4 
Determine the Monod growth parameters μ_{max} and K_{S}.
Estimate the stoichiometric coefficients, Y_{c}_{/}_{s} and Y_{s}_{/}_{c}.
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
How could you make this problem more difficult?
P921_{C} Diabetes is a global epidemic affecting more than 240 million people worldwide. The vast majority of the cases are Type 2. Recently a drug, Januvia (J), was discovered to treat Type 2 diabetes. When food enters the stomach, a peptide, GLP1 (glucagonlike peptide 1) is released, which leads to glucosedependent insulin secretion and glucagon suppression. The halflife of GLP1 is very short because it is rapidly degraded by an enzyme dipeptidyl peptidaseIV (DPPIV), which cleaves the two terminal amino acids of the peptide thus deactivating it. DPIV rapidly cleaves the active form of GLP1 (GLP1[736] amide) to its inactive form (GLP1[936] amide) with a halflife of 1 minute, that is, t_{1/2} = 1 min., and it is thought to be the primary enzyme responsible for this hydrolysis.^{26}
^{26} Also see (a) J. J. Holst and D. F. Deacon, Diabetes, 47, 1663 (1998); (b) B. Balkan, et al., Diabetiologia, 42, 1324 (1999); (c) K. Augustyns, et al., Curr. Med. Chem., 6, 311 (1999).
$\begin{array}{c}\begin{array}{c}\mathrm{GLP}\text{}1\left(9\text{}36\right)\leftarrow \text{E}\mathrm{.}\mathrm{GLP}\text{}1\left(7\text{}36\right)\text{}\stackrel{\underset{\text{E}}{\mathrm{Bad}}}{\to}\text{}\mathrm{GLP}\text{}1\left(7\text{}\text{}36\right)\stackrel{\mathrm{Good}}{\to}\end{array}\\ \begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\\ \end{array}\\ \end{array}\\ \end{array}\\ \end{array}\end{array}\text{}\text{}\begin{array}{l}\mathrm{Acts}\text{}\mathrm{in}\text{}\mathrm{pancreas}\text{}\text{}\mathrm{to}\\ \begin{array}{c}\begin{array}{c}\mathrm{stimulate}\text{}\text{}\mathrm{insulin}\hfill \\ \mathrm{release}\text{}\mathrm{and}\text{}\mathrm{suppress}\end{array}\hfill \\ \mathrm{glucagon}\hfill \end{array}\end{array}$
Inhibition of the DPPIV Enzyme, (E), therefore, is expected to significantly reduce the extent of the inactivation of GLP1[736] and should lead to an increase in circulating levels of the active form of the hormone. Supporting evidence for this comes from DPPIVenzymedeficient mice, which have elevated levels of GLP1[736]^{27} amide. As a very rough approximation, let’s treat the reaction as follows: The new drug, an inhibitor of the DPPIV enzyme (E), is Januvia (J), which prevents the enzyme from deactivating GLP1.
^{27} D. Marguet, et al., Proc. Natl. Acad. Sci., 97, 6864 (2000).
E + GLP1 ⇄ E • GLP1 → Glucose Release
Inhibited
E + J ⇄ E • J (Inactive)
By delaying the degradation of GLP1, the inhibitor is able to extend the action of insulin and also suppress the release of glucagon.
Plot the ratio of reaction of –r_{GLP} (without inhibition) to the rate –r_{GLP} (with inhibition) as a function of DDP4 inhibitor constant for both competitive inhibition and uncompetitive inhibition.
Assuming the body is a wellmixed reactor, develop a model similar to Problems P85 and P86 for the dosage schedule for Januvia.
P922_{B} Alternative Energy Source.^{28} In the summer of 2009, ExxonMobil decided to invest 600 million dollars on developing algae as an alternative fuel and their TV commercials on this initiative were recently shown (e.g., December 2019). 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 6000 gal of gasoline per year, which would require the capture of a CO_{2} source more concentrated than air (e.g., fuel gas from a refinery) and also contribute to the sequestration of CO_{2}. The biomass biosynthesis during the day is
^{28} The contributions of John Benemann to this problem are appreciated.
Sunlight + CO_{2} + H_{2}O + Algae → More Algae + O_{2}
Consider a 5000gal pond with perforated pipes into which CO_{2} is injected and slowly bubbled into the solution to keep the water saturated with CO_{2}.
The doubling time during the day is 12 h at highnoon 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 (πt/12) between 6 A.M. and 6 P.M., otherwise f = 0, C_{C} is the algae concentration (g/dm^{3}) and μ = 0.9 day^{–1} (assumes constant CO_{2} saturation at 1 atm is 1.69 g/kg water). The pond is 30cm deep and for effective sunlight penetration, the algae concentration cannot exceed 200 mg/dm^{3}.
Derive an equation for the ratio of the algae cell concentration C_{C} at time t to initial cell concentration C_{C0}, that is, (C_{C}/C_{C0}). Plot and analyze (C_{C}/C_{C0}) versus time up to 48 hours.
If the pond is initially seeded with 0.5 mg/dm^{3} of algae, how long will it take the algae to reach a cell density (i.e., concentration) of 200 mg/dm^{3}, 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.
Suppose the algae limit the sun’s penetration significantly even before the concentration reaches 200 mg/dm^{3} with, for example, μ = μ_{0} (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/dm^{3}?
Now, let’s consider continuous operation. Once the cell density reaches 200 mg/dm, onehalf of the pond is harvested and the remaining broth is mixed with fresh nutrient. What is the steadystate algae productivity in gm/year, again assuming the pond is well mixed?
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 5000gal pond (g/d)? Assume the pond is well mixed.
Now consider that the reaction is to be carried out in an enclosed, transparent reactor. The reactor can be pressurized with CO_{2} up to 10 atm with K_{S} = 2 g/dm^{3}. Assume that after the initial pressurization, no more CO_{2} can be injected. Plot and analyze the algae concentration as a function of time.
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)?
P923_{A} Short calculations on the algae ponds.
If the pond is initially seeded with 0.5 mg/dm^{3} of algae, how long will it take the algae to reach a cell density (i.e., concentration) of 200 mg/dm^{3}? Sketch a rough plot of r_{g} and C_{C} over time.
Suppose the algae limit the sun’s penetration significantly even before the concentration reaches 200 mg/dm^{3} by using μ = μ_{0} (1 – C_{C}/200). Assume μ_{0} = 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/dm^{3}? Why?
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/dm^{3}. 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/6e/index.html.
Web
Review the following Web site:
www.pharmacokinetics.com
Text
A discussion of complex reactions involving active intermediates is given in
A. K. DATTA, Heat and Mass Transfer: A Biological Context, 2nd ed. Boca Raton, FL: CRC Press, 2017.
A. A. FROST and R. G. PEARSON, Kinetics and Mechanism, 2nd ed. New York: Wiley, 1961, Chap. 10. Old but great examples.
K. J. LAIDLER, Chemical Kinetics, 3rd ed. New York: HarperCollins, 1987.
M. J. PILLING, Reaction Kinetics, New York: Oxford University Press, 1995.
Further discussion of enzymatic reactions:
Just about everything you want to know about basic enzyme kinetics can be found in I. H. SEGEL, Enzyme Kinetics, New York: WileyInterscience, 1975.
An excellent description of parameter estimation, biological feedback, and reaction pathways can be found in E. O. VOIT, Computational Analysis of Biochemical Systems, Cambridge, UK: Cambridge University Press, 2000.
A. CORNISHBOWDEN, Analysis of Enzyme Kinetic Data, New York: Oxford University Press, 1995.
D. L. NELSON and M. M. COX, Lehninger Principles of Biochemistry, 3rd ed. New York: Worth Publishers, 2000.
M. L. SHULER and F. KARGI, Bioprocess Engineering Principles, 2nd ed. Upper Saddle River, NJ: Prentice Hall, 2002.
Material on bioreactors can be found in
T. J. BAILEY and D. OLLIS, Biochemical Engineering, 2nd ed. New York: McGrawHill, 1987.
H. W. BLANCH and D. S. CLARK, Biochemical Engineering, New York: Marcel Dekker, 1996.
Also see
THORNTON W. BURGESS, The Adventures of Old Mr. Toad, New York: Dover Publications, Inc., 1916.
GARRISON KEILLOR, Pretty Good Joke Book: A Prairie Home Companion, St. Paul, MN: HighBridge Co., 2000.
HOWARD MASKILL, The Investigation of Organic Reactions and Their Mechanisms, Oxford, UK: Blackwell Publishing Ltd, 2006.