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Chapter 7: Problem 22

Consider the following hypothetical chemical system: $$ \begin{array}{ll} M_{0} \text { (a catalyst) }+\mathrm{X} \underset{k_{1}}{\stackrel{k_{1}}{\mathrm{M}}} \mathrm{M}_{\mathrm{t}} & \text { (active complex), } \\ \mathrm{M}_{1}+\mathrm{X} \stackrel{k_{0}}{k_{-2}} \mathrm{M}_{2} & \text { (inactive complex), } \\ \mathrm{M}_{1}+\mathrm{Y} \stackrel{k_{2}}{\longrightarrow} \mathrm{P}+\mathrm{Q}+\mathrm{M}_{0} & \text { (products plus catalyst). } \end{array} $$ This system is called a substrate-inhibited reaction since the chemical \(\mathrm{X}\) can deactivate the complex \(M_{1}\) which is required in forming the products. (a) Write equations for the chemical components. (b) Assume \(M_{0}+M_{1}+M_{2}=C\) (where \(m_{0}, m_{1}\), and \(m_{2}\) are concentrations of \(M_{0}, M_{1}\), and \(M_{2}\), and \(C\) is a constant). Make a quasi-steady-state assumption for \(M_{0}, M_{1}\), and \(M_{2}\) and show that $$ m_{0}=m_{1} \frac{k_{-1}+k_{3} y}{k_{1} x}, \quad m_{2}=k_{2} x \frac{m_{1}}{k-2} . $$ (c) Use these results to show that $$ m_{1}=\frac{C k_{1} x}{k_{-1}+k_{3} y+k_{1} x\left[1+\left(k_{2} / k-2\right) x\right]} $$ (d) Now show that \(x\) will satisfy an equation whose dimensionless form is $$ \frac{d x}{d t}=\gamma-x-\frac{\beta x y}{1+x+y+(\alpha / \delta) x^{2}} $$ Identify the various combinations of parameters. When can the term \(y\) in the denominator be neglected?

Short Answer

Expert verified
Initial steps to set up the equations and apply the quasi-steady-state assumption are crucial. Following these, simplify using conservation laws to find expressions and finally identify parameter combinations for the dimensionless form.

Step by step solution

01

Writing Equations for the Chemical Components

Given the reaction and its rate constants, the equations for the chemical components based on the provided system are as follows: 1. Formation of the active complex: \[ \frac{d[M_t]}{dt} = k_{1}[M_0][X] - k_{-1}[M_t] \] 2. Formation of the inactive complex: \[ \frac{d[M_2]}{dt} = k_{0}[M_1][X] - k_{-2}[M_2] \] 3. Product formation and catalyst regeneration: \[ \frac{d[P]}{dt} = k_{2}[M_1][Y] \] \[ \frac{d[Q]}{dt} = k_{2}[M_1][Y] \] 4. Conservation of concentrations: \[ M_0 + M_1 + M_2 = C \]
02

Applying the Quasi-Steady-State Assumption

Based on the quasi-steady-state assumption for \(M_0\), \(M_1\), and \(M_2\): 1. For \(M_t\), set \( \frac{d[M_t]}{dt} = 0 \): \[ 0 = k_{1}[M_0][X] - k_{-1}[M_t] \implies [M_t] = \frac{k_1[M_0][X]}{k_{-1}} \] 2. For \(M_2\), set \( \frac{d[M_2]}{dt} = 0 \): \[ 0 = k_{0}[M_1][X] - k_{-2}[M_2] \implies [M_2] = \frac{k_0[M_1][X]}{k_{-2}} \]
03

Solving for \(m_0\) and \(m_2\)

Substitute the equilibrium expressions into the conservation equation, \( M_0 + M_1 + M_2 = C \): \[ M_0 + [M_t] + [M_2] = C \] Replace \([M_t]\) and \([M_2]\) using results from Step 2: \[ M_0 + \frac{k_1 M_0 X}{k_{-1}} + \frac{k_0 M_1 X}{k_{-2}} = C \] Solving for \(M_0\) and \(M_2\):\[ M_0 = M_1 \frac{k_{-1} + k_{3}[Y]}{k_1[X]} \] \[ M_2 = \frac{k_0 M_1 [X]}{k_{-2}} \]
04

Expressing \(m_1\) in Terms of Known Quantities

Using the conservation equation and rearranging for \(M_1\): \[ M_1 = \frac{C k_1 x}{k_{-1} + k_{3} y + k_1 x \biggl(1+ \frac{k_2}{k_{-2}} x \biggl)} \]
05

Writing the Differential Equation

Using the results from the previous steps and expressing in dimensionless form: \[ \frac{dx}{dt} = \frac{d[x]}{dt} = \frac{d[x]}{dt} = \frac{\beta x y}{1+x+y+\frac{\beta}{u} x^2} \] The parameters are defined as: \( \beta = k_2/k_{-2} \), \( u = k_{-1} [k_3] \), \( x = [X] \), and \( y = [Y] \). In the case where \( y \) is much smaller than all other terms in the denominator, the term \( y \) can be neglected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Catalytic reaction kinetics
Chemical reactions involving catalysts are fundamental in chemistry. In these reactions, a catalyst speeds up the reaction without being consumed. The given problem describes a complex scenario where the catalyst exists in different states.

Active and inactive complexes dynamically form and break down. It's vital for understanding the conversion processes from reactants to products.

In substrate-inhibited reactions, the substrate itself can deactivate the catalyst, making the system more challenging. The rate of product formation depends on the concentration of different species and their interactions, governed by rate constants.

To fully grasp these kinetics, one must consider:
  • The interaction rates between the catalyst and substrates.
  • Formation and dissociation rates of complex species.
  • The conservation of total catalyst concentration.
By analyzing these factors, students can predict how the reaction proceeds and how efficient the catalyst will be under different conditions.
Quasi-steady-state assumption
The quasi-steady-state assumption (QSSA) is a crucial simplification in reaction kinetics. It assumes that the concentration of intermediate complexes remains relatively constant over time.

In the given problem, QSSA applies to complexes formed during the reaction. For example, the concentration of the active complex \( M_t \) is set to be steady. This simplification leads to easier differential equations that describe the system.

Applying QSSA involves setting the derivatives of intermediate species concentrations to zero. Consequently, this provides algebraic expressions instead of differential ones for these species:
  • The steady concentration of active and inactive complexes can be derived.
  • Complex simplifications emerge, allowing deeper insights into system behavior.
Mastering the QSSA application helps students simplify and solve complex kinetic problems more efficiently.
Differential equations in chemistry
Differential equations are mathematical tools used to describe the rate of change in chemical systems. In the context of catalytic reactions, these equations capture how concentrations of reactants, intermediates, and products change over time.

For the given system:
  • Equations describe the rate of formation and breakdown of active and inactive complexes.
  • Equations also capture the overall concentration changes of reactants and products.
Solving these equations requires techniques like applying the QSSA. This reduces the complexity and helps focus on the key dynamics.

Understanding the differential equations behind these systems is crucial. It provides insights into reaction rates, equilibrium points, and the effectiveness of catalysts under various conditions.
Chemical concentration equilibrium
Chemical equilibrium occurs when the concentration of reactants and products remains constant over time. This happens when the rates of the forward and reverse reactions are equal.

In the context of catalytic reactions, achieving equilibrium is critical for stability and efficiency. The given problem highlights:
  • Equilibrium concentrations of active and inactive catalyst complexes.
  • Equilibrium between reactants and products considering catalytic effects.
By understanding how equilibrium is reached and maintained, students can design better catalytic systems. They can predict the most efficient conditions for maximum product yield and minimum catalyst deactivation, ensuring a deeper grasp of real-world chemical reactions.

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Most popular questions from this chapter

The Brusselator. This hypothetical system was first proposed by a group working in Brussels [see Prigogine and Lefever \((1968)]\) in connection with spatially nonuniform chemical patterns. Because certain steps involve trimolecular reactions, it is not a model of any real chemical system but rather a prototype that has been studied extensively. The reaction steps are $$ \begin{aligned} & A & \rightarrow X \\ B+& X & \rightarrow Y+D \\ 2 X+Y & \rightarrow 3 X \\ X & \rightarrow E \end{aligned} $$ It is assumed that concentrations of \(A, B, D\), and \(E\) are kept artificially constant so that only \(\mathrm{X}\) and \(\mathrm{Y}\) vary with time. (a) Show that if all rate constants are chosen appropriately, the equations describing a Brusselator are: $$ \begin{aligned} &\frac{d x}{d t}=A-(B+1) x+x^{2} y \\ &\frac{d y}{d t}=B x-x^{2} y \end{aligned} $$ (b) Find the steady state. (c) Calculate the Jacobian and show that if \(\mathrm{B}>1\), the Brusselator is a positive-feedback system as described in Section \(7.8\).

Suppose \(\mathrm{A}\) and \(\mathrm{B}\) are monomers that undergo dimerization in a rapidly equilibrating reaction: $$ \mathrm{A}+\mathrm{B} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}[\mathrm{A}-\mathrm{B}] $$ Show that the concentration of dimers is proportional to the product of the monomer concentrations.

An activator-inhibitor system (Gierer-Meinhardt). A system also studied in connection with spatial pattems (see Chapter 10) consists of two substances. The activator enhances its own synthesis as well as that of the inhibitor. The inhibitor causes the formation of both substances to decline. Several versions have been studied, among them is the following system: $$ \begin{aligned} &\frac{d x}{d t}=\rho+\frac{x^{2}}{y}-x \\ &\frac{d y}{d t}=x^{2}-\gamma y \end{aligned} $$ (a) Find the steady state for this system. (b) Show that if the input of activator is sufficiently small compared to thedecay rate of inhibitor, the system is an activator-inhibitor system as described in Section \(7.8\). (c) In a modified version of these equations the term \(x^{2} / y\) is replaced by \(x^{2} / y\left(1+k x^{2}\right)\). Suggest what sort of chemical interactions may be occurring in this and in the original system. (d) Why is it not possible to solve for the steady state of the modified GiererMeinhardt equations? (e) Find the Jacobian of the modified system. When does the Jacobian have the sign pattern of an activator-inhibitor system?

A model for control of synthesis of a gene product that activates mitosis was suggested by Tyson and Sachsenmaier (1979), based on repression and derepression (the reversal of repression) of a genetic operon (a sequence of genes that are controlled as a unit by a single gene called the operator.) They assumed that the gene consists of two portions, one replicating earlier than the second, with the following control system. Protein \(\mathbf{R}\) (coded by 1) binds to the operator region \(\mathbf{O}\) of gene 2, repressing transcription of genes \(G_{P}\) and \(G_{A}\). Protein \(P\) (product of \(G_{P}\) ) inactivates the repressor and thus has a positive influence on its own synthesis, as well as on synthesis of A (see figure). Assume that \(\mathrm{R}, \mathrm{P}\), and \(\mathrm{A}\) have removal rates \(\ell_{1}, \ell_{2}\), and \(\ell_{3}\). Let \(G_{\mathrm{R}}, G_{\mathrm{P}}\), and \(G_{A}\) be the number of genes coding for \(\mathrm{R}, \mathrm{P}\), and \(\mathrm{A}\), at rates \(K_{1}, K_{2}\), and \(K_{3}\) (when actively transcribing). Let \(f\) be the fraction of operons of the late-replicating DNA that are active at a given time. (a) Give equations governing the concentrations of \(\mathrm{R}, \mathrm{P}\), and \(\mathrm{A}\) in terms of \(R_{i} \mathrm{~s}, G_{i} \mathrm{~s}\), and \(f\). (b) Following are the operator and repressor binding reactions: $$ \begin{aligned} &P+R \frac{k_{1}}{k_{-1}} X \\ &R+O \frac{k_{2}}{k_{-2}} O_{R} \end{aligned} $$ where \(X\) is an inducer-repressor complex and \(O_{R}\) is a repressed operator. Write down equations for P, R, and \(O\) based on these kinetics. (c) Now assume that these reactions are always in equilibrium and that the total number of operator molecules is \(O_{T}\), a constant. Let $$ K_{2}=\frac{k_{2}}{k_{-2}} \quad \text { and } \quad K_{1}=\frac{k_{1}}{k_{-1}} \text {. } $$ Find an expression relating the fraction \(f\) of repressed operators \(\left(f=O / O_{t}\right)(1)\) to these rate constants and to the concentration of \(\mathrm{R}\). (d) For the specific situation in which more than one repressor molecule can bind the operator, $$ n \mathrm{R}+\mathrm{O} \rightleftharpoons \mathrm{OR}_{n} \text {, } $$ (again where \(K_{2}=k_{2} / k_{-2}\) is the equilibrium constant), Tyson and Sachsenmaier showed that $$ \begin{aligned} f=(\epsilon+x)^{n}\left[1+(\epsilon+x)^{n}\right]^{-1}, & \text { for } \quad x=\frac{K_{1} P}{K_{2} X} \\ \epsilon &=\left(K_{2} R_{r}\right)^{-1} . \end{aligned} $$ provided \(\left(O_{T} \& X\right)\). Sketch \(f\) as a function of \(x\) for \(n=2\).

The following model was proposed by Othmer and Aldridge (1978): A cell can produce two chemical species \(x\) and \(y\) from a substrate according to the reaction substrate \(\rightarrow x \rightarrow y \rightarrow\) products. Species \(x\) can diffuse across the cell membrane at a rate that depends linearly on its concentration gradient. The ratio of the volume of cells to the volume of extemal medium is given by a parameter \(\epsilon ; x\) and \(y\) are intracellular concentrations of \(X\) and \(Y\) and \(x^{0}\) is the extracellular concentration of \(X .\) The equations they studied were $$ \begin{aligned} &\frac{d x}{d t}=\delta-F(x, y)+P\left(x^{0}-x\right) m \\ &\frac{d y}{d t}=\alpha[F(x, y)-G(y)] \\ &\frac{d x^{0}}{d t}=\epsilon P\left(x-x^{0}\right) \end{aligned} $$ (a) Explain the equations. Determine the values of \(\bar{x}, F(\bar{x}, \bar{y})\) and \(G(\bar{y})\) at the steady state \(\left(\bar{x}, \bar{y}, \bar{x}^{0}\right)\). (b) The matrix of linearization of these equations about this steady state is $$ \mathbf{J}=\left(\begin{array}{ccc} k_{11}-P & k_{12} & P \\ k_{21} & k_{22} & 0 \\ \epsilon P & 0 & -\epsilon P \end{array}\right) $$ What are the constants \(k_{j j} ?\) (c) For the characteristic equation $$ \lambda^{3}+a_{1} \lambda^{2}+a_{2} \lambda+a_{3}=0 $$ find \(a_{1}, a_{2}\), and \(a_{3}\) in terms of \(k_{i j}\) and in terms of partial derivatives of \(F\) and \(G\).
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