# 9.5: Problems

• • R.L. Cerro, B. G. Higgins, S Whitaker
• Professors (Chemical Engineering) at University of Alabama at Huntsville & University of California at Davis

## Section 9.1

1. Apply a column/row partition to show how Eq. $$(9.1.6)$$ is obtained from Eq. $$(9.1.5)$$.

2. Illustrate how a row/row partition leads from Eq. $$(9.1.23)$$ to Eq. $$(9.1.24)$$.

3. Use Eqs. $$(9.1.35)$$ through $$(9.1.43)$$ to obtain a local chemical reaction rate equation for species $$D$$.

4. Develop the local chemical reaction rate equation for species $$C$$ on the basis of Eqs. $$(9.1.35)$$ through $$(9.1.43)$$.

5. Re-write Eqs. $$(9.1.41)$$ and $$(9.1.43)$$ using the stoichiometric coefficients, $$\nu_{AI}$$, $$\nu_{DII}$$, etc., in place of $$\alpha$$, $$\beta$$, $$\gamma$$, etc. Show that this change in the nomenclature leads to the form encountered in Eq. $$(6.2.28)-(6.2.31)$$.

6. Develop the representation for $$R_B$$ given in Eqs. $$(9.1.44)$$.

7. It is difficult to find a real system containing three species for which the reactions can be described by

$A \stackrel{k_1}{\longrightarrow} B \stackrel{k_2}{\longrightarrow} C \label{1.1}$

however, this represents a useful model for the exploration of the condition of local reaction equilibrium. The stoichiometric constraint for this series of first order reactions is given by

$R_A + R_B + R_C \label{1.2}$

since the three molecular species must all contain the same atomic species. For a constant volume batch reactor, and the initial conditions given by

IC.

$c_A = c^{\circ}_A, \quad c_B = 0, \quad c_C = 0, \quad t = 0 \label{1.3}$

one can determine the concentrations and reaction rates of all three species as a function of time. If one thinks of species $$B$$ as a reactive intermediate, the condition of local equilibrium takes the form Local reaction equilibrium:

$R_B = 0 \label{1.4}$

In reality, the reaction rate for species $$B$$ cannot be exactly zero; however, we can have a situation in which

$R_B << |R_A| \label{1.5}$

Often this condition is associated with a very large value of $$k_2$$, and in this problem you are asked to develop and use the exact solution for this process to determine how large is very large. You can also use the exact solution to see why the condition of local reaction equilibrium might be referred to as the steady-state assumption for batch reactors.

## Section 9.2

8. In our study of Michaelis-Menten kinetics we simplified the analysis by ignoring the influence of the substrate $$B$$, or any other substrate, on the enzyme catalyzed reaction of species $$A$$. When multiple substrates are considered, the analysis becomes very complex20; however, if we assume that the reversible binding steps are at equilibrium, the analysis of two substrates becomes quite tractable. The binding between enzyme $$E$$ and substrate $$A$$ is described by

$E+A \quad \overset{\stackrel{k_{I}}{\longrightarrow}}{\stackrel{\longleftarrow}{k_{II}}} \quad EA\label{2.1}$

and when $$A$$ is in equilibrium with $$EA$$ we can replace Eqs. $$(9.2.1)-(9.2.3)$$ and $$(9.2.4)-(9.2.6)$$ with the equilibrium relation 4/24/2013 given by

$c_{EA} = \frac{k_I}{K_{II}} c_E c_A = K^I_{eq} c_E c_A \label{2.2}$

The reversible binding between enzyme $$E$$ and substrate $$B$$ is similarly described by

$E + B \rightleftarrows EB \label{2.3}$

and we express the equilibrium relation as

$c_{EB} = K^{II}_{eq} c_E c_B \label{2.4}$

In this model we assume that once substrate $$B$$ has reacted with enzyme $$E$$ to form the complex $$EB$$, no additional reaction with substrate $$A$$ can take place. In this step substrate $$B$$ acts as an inhibitor since it removes some enzyme $$E$$ from the system. However, the complex $$EA$$ can further react with substrate $$B$$ to form the complex $$EAB$$ as indicated by

$EA + B \rightleftarrows EAB \label{2.5}$

The equilibrium relation associated with this process is given by

$c_{EAB} = K^{III}_{eq} c_{EA} c_B \label{2.6}$

In this step substrate $$B$$ acts as a reactant since it produces the complex $$EAB$$ that is the source of the product $$D$$.

Because of the assumed equilibrium relations indicated by Eqs. \ref{2.2}, \ref{2.4} and \ref{2.6}, we have only a single rate equation based on an irreversible reaction. This irreversible reaction is described by

Elementary chemical kinetic schema III:

$EAB \stackrel{k_{III}}{\longrightarrow} E + D \label{2.7a}$

Elementary stoichiometry III:

$R^{III}_{EAB} =- R^{III}_{E}, \quad R^{III}_{D} = R^{III}_{E} \label{2.7b}$

Elementary chemical kinetic rate equation III:

$R^{III}_{D} = k_{III} c_{EAB} \label{2.7c}$

Since the product $$D$$ is involved in only this single reaction, we express the rate of production as

$R_{D} = k_{III} c_{EAB} \label{2.8}$

In the absence of significant cell death, one can assume that the enzyme $$E$$ remains within the cells, thus the total concentration of enzyme is constant as indicated by

$c_E + c_{EA} + c_{EB} + c_{EAB} = c^{\circ}_{E} \label{2.9}$

In this problem you are asked to show that Eqs. \ref{2.2} through \ref{2.9} lead to an equation having the form of max

$R_{D}=\mu_{\max }\left(\frac{c_{A}}{K_{A}+c_{A}}\right)\left(\frac{c_{B}}{K_{B}+c_{B}}\right) \label{2.10}$

provided that you impose the special condition given by

$K^{III}_{eq} = K^{II}_{eq} \label{2.11}$

This restriction is based on the idea that $$B$$ binds with $$EA$$ in the same manner that $$B$$ binds with $$E$$.

9. In Problem 8 we considered a case in which the substrate $$B$$ acted both as a reactant in the production of species $$D$$ and as an inhibitor in that process. In some cases we can have an analog of the substrate that acts as a pure inhibitor and the Michaelis-Menten process takes the form

$E + A \quad \overset{\stackrel{k_{I}}{\rightharpoonup}}{\stackrel{\leftharpoondown}{k_{II}}} \quad EA \stackrel{k_{III}}{\longrightarrow} E + D \label{2.9.1}$

$E + H \quad \overset{\stackrel{k_{IV}}{\rightharpoonup}}{\stackrel{\leftharpoondown}{k_{V}}} \quad EH\label{2.9.2}$

in which we have used $$H$$ to represent the inhibitor.

In this problem we assume that $$k_{III} << k_{II}$$ in order to utilize (as an approximation) the equilibrium relation given by

$c_{EA} = K^I_{eq} c_E c_A \label{2.9.3}$

In addition we assume that the process illustrated by Equation \ref{2.9.2} can be approximated by the following equilibrium relation

$c_{EA} = K^{II}_{eq} c_E c_H \label{2.9.4}$

In this problem you are asked to make use of the condition given by

$c_E + c_{EA} + c_{EH} = c^{\circ}_E \label{2.9.5}$

in order to develop an expression for $$R_D$$ in which the concentration of the inhibitor, $$c_H$$, is an unknown. Consider the special cases that occur when $$K^{II}_{eq} c_H / K^{I}_{eq}$$ becomes both large and small relative to some appropriate parameter.

## Section 9.3

10. Indicate how Eqs. $$(9.3.22)-(9.3.26)$$ are obtained from Eqs. $$(9.3.17)-(9.3.21)$$.

11. Beginning with the second of Eqs. $$(9.3.29)$$ derive Eq. $$(9.3.36)$$ using elementary row operations.

12. In the analysis of the hydrogen bromide reaction described by Eqs. $$(9.3.43)-(9.3.47)$$, the concept of local reaction equilibrium was imposed on the reactive intermediates, H and Br, according to

Local reaction equilibrium:

$R_{\ce{H}} = 0, \quad R_{\ce{Br}} = 0 \label{12.1}$

Use of these simplifications, along with the chemical kinetic schemata and the associated mass action kinetics given by Eqs. $$(9.3.48)$$ though $$(9.3.67)$$, led to the net rate of production of hydrogen bromide given by Eq. $$(9.3.91)$$. In reality, $$R_{\ce{H}}$$ and $$R_{\ce{Br}}$$ will not be zero but they may be small enough to recover Eq. $$(9.3.91)$$. The concept that something is small enough so that it can be set equal to zero is explored by Eqs. $$(9.1.73)$$ though $$(9.1.77)$$. In this problem you are asked to develop an analysis indicating that Eqs. \ref{12.1} listed above are acceptable approximations when the following inequalities are satisfied:

$R_{\ce{Br}}<<2 k_{I} c_{\ce{Br2}} / k_{V}, \quad R_{\ce{H}}<<2 k_{I} c_{\ce{Br2}} / k_{V}, \quad R_{\ce{H}}<<k_{II} c_{\ce{H2}} \sqrt{2 k_{I} c_{\ce{Br2}} / k_{V}} \label{12.2}$

One should think of Eqs. \ref{12.1} as being assumptions concerning the rates of production of the reactive intermediates, while Eqs. \ref{12.2} should be thought of as restrictions on the magnitude of these rates.

13. Use Eq. $$(9.3.68)$$ to verify Eq. $$(9.3.73)$$.

14. The global stoichiometric schema associated with the decomposition of $$\ce{N2O5}$$ to produce $$\ce{NO2}$$ and $$\ce{O2}$$ can be expressed as

$\ce{N2O5} \to 2\ce{NO2} + \frac{1}{2}\ce{O2} \label{14.1}$

and experimental studies indicate that the reaction can be modeled as first order in $$\ce{N2O5}$$. Show why the following elementary chemical kinetic schemata give rise to a first order decomposition of $$\ce{N2O5}$$.

Elementary chemical kinetic schema I:

$\ce{N2O5} \stackrel{k_I}{\longrightarrow} \ce{NO2} + \ce{NO3} \label{14.2}$

Elementary chemical kinetic schema II:

$\ce{NO2} + \ce{NO3} \stackrel{k_{II}}{\longrightarrow} \ce{NO2} + \ce{O2} + \ce{NO} \label{14.3}$

Elementary chemical kinetic schema III:

$\ce{NO} + \ce{NO3} \stackrel{k_{III}}{\longrightarrow} \ce{2NO2} \label{14.4}$

Elementary chemical kinetic schema IV:

$\ce{NO2} + \ce{NO3} \stackrel{k_{IV}}{\longrightarrow} \ce{N2O5} \label{14.5}$

Since neither $$\ce{NO}$$ nor $$\ce{NO3}$$ appear in the global stoichiometric schema given by Equation \ref{14.1}, one can assume that these two compounds are reactive intermediates or Bodenstein products and their rates of production can be set equal to zero as an approximation

15. In Problem 14 we considered the decomposition of $$\ce{N2O5}$$to produce $$\ce{NO2}$$ and $$\ce{O2}$$ by means of the kinetic schemata illustrated by Eqs. \ref{14.2} through \ref{14.5} in Problem 14. The reactive intermediates were identified as $$\ce{NO}$$ and $$\ce{NO3}$$. The rate equations for these two reactive intermediates and for $$\ce{N2O5}$$are given by

$R_{\ce{NO3}}=k_{I} c_{\ce{N2} \ce{O5}}-k_{II} c_{\ce{NO3}} c_{\ce{NO2}}-k_{III} c_{\ce{NO}} c_{\ce{NO3}}-k_{\ce{IV}} c_{\ce{NO2}} c_{\ce{NO3}} \label{15.1}$

$R_{\ce{NO}}=k_{II} c_{\ce{NO3}} c_{\ce{NO2}}-k_{III} c_{\ce{NO}} c_{\ce{NO3}}\label{15.2}$

$R_{\ce{N2} \ce{O5}}=-k_{I} c_{\ce{N2} \ce{O5}}+k_{\ce{IV}} c_{\ce{NO2}} c_{\ce{NO3}}\label{15.3}$

If the condition of local reaction equilibrium is imposed according to

Local reaction equilibrium:

$R_{\ce{NO3}} = 0, \quad R_{\ce{NO}} = 0 \label{15.4}$

one can obtain a simple representation for $$R_{\ce{N2O5}}$$ in terms of only $$c_{\ce{N2O5}}$$. In this problem you are asked to replace the assumptions given by Eqs. \ref{15.4} with restrictions indicating that $$R_{\ce{NO3}}$$ and $$R_{\ce{NO}}$$ are small enough so that Eqs. \ref{15.4} become acceptable approximations.

## Section 9.4

16. If species $$C$$ in Eq. $$(9.3.27)$$ is a reactive intermediate or Bodenstein product, identify the stoichiometric matrix and the Bodenstein matrix associated with Eq. $$(9.3.27)$$. Impose the condition of local reaction equilibrium on the Bodenstein product in order to derive an expression for $$R_E$$ in terms of $$c_A$$, $$c_B$$ and $$c_D$$.

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