91Ó°ÊÓ

In an enzyme-catalyzed reaction with stoichiometry \(\mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}\) is consumed at a rate given by an expression of the Michaelis-Menten form: $$r_{\mathrm{A}}[\operatorname{mol} /(\mathrm{L} \cdot \mathrm{s})]=\frac{k_{1} C_{\mathrm{A}}}{1+k_{2} C_{\mathrm{A}}}$$ where \(C_{\mathrm{A}}(\operatorname{mol} / \mathrm{L})\) is the reactant concentration, and \(k_{1}\) and \(k_{2}\) depend only on temperature. (a) The reaction is carried out in an isothermal batch reactor with constant reaction mixture volume \(V\) (liters), beginning with pure \(A\) at a concentration \(C_{\mathrm{A} 0}\). Derive an expression for \(d C_{\mathrm{A}} / d t\), and provide an initial condition. Sketch a plot of \(C_{\mathrm{A}}\) versus \(t,\) labeling the value of \(C_{\mathrm{A}}\) at \(t=0\) and the asymptotic value as \(t \rightarrow \infty\) (b) Solve the differential equation of Part (a) to obtain an expression for the time required to achieve a specified concentration \(C_{\mathrm{A}}\) (c) Use the expression of Part (b) to devise a graphical method of determining \(k_{1}\) and \(k_{2}\) from data for In versus the pour plot should involve fitting a straight line and determining the two parameters \(C_{\mathrm{A}}\) (int the parting of the partating and and the conting are a contation a conting from the slope and intercept of the line. (There are several possible solutions.) Then apply your method to determine \(k_{1}\) and \(k_{2}\) for the following data taken in a 2.00 -liter reactor, beginning with A at a concentration \(C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}\) $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$

Step by step solution

01

Develop the differential equation

The rate of change of concentration of reactant A, \( \frac{dC_A}{dt} \), is equal to the negative of the rate of reaction since reactant A is being consumed in the reaction. This gives us: \[\frac{dC_A}{dt} = - \frac{k_1 C_A}{1+k_2 C_A} \]At \( t = 0 \), the concentration of A, \( C_A \) is initially \( C_{A0} \). So, the initial condition is \( C_A = C_{A0} \) at \( t = 0 \).

02

Sketch a plot of \( C_A \) versus \( t \)

When you plot \( C_A \) against \( t \), you would observe that initially at \( t = 0 \), \( C_A = C_{A0} \), and as \( t \rightarrow \infty \), \( C_A \) approaches zero, indicating that all of the reactant has been consumed. The plot will start high and steadily decrease towards zero, representing the reaction's progress.

03

Solve the differential equation

Solving the differential equation to obtain an expression for the time required to achieve a particular concentration is a matter of integrating the rate expression. However, this requires knowledge of calculus and may be subject to numerical approximation methods since the rate equation is non-linear.

04

Graphical method to determine \( k_1 \) and \( k_2 \)

You can plot \( ln(C_A) \) versus \( t \) according to the data given. The intercept of this plot gives \( k_1 \) while the slope gives \( -k_2 \). By fitting a line to the data and determining the slope and intercept, the values of \( k_1 \) and \( k_2 \) can be approximated.

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Key Concepts

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

Michaelis-Menten Kinetics

Understanding Michaelis-Menten kinetics is essential for students delving into enzyme-catalyzed reactions. This concept outlines how the rate of reaction is influenced by the concentration of the substrate, which in our case is reactant A. The classic formula encapsulates this relationship as follows:
\[r_{\mathrm{A}} = \frac{k_1 C_{\mathrm{A}}}{1 + k_2 C_{\mathrm{A}}}\]
where \(r_{\mathrm{A}}\) denotes the reaction rate, \(k_1\) the maximum rate of reaction, \(k_2\) the Michaelis constant which is reflective of the affinity between the enzyme and the substrate, and \(C_{\mathrm{A}}\) represent the concentration of the substrate A. The kinetics indicate that as the substrate concentration increases, the reaction rate initially increases, but eventually it levels off, achieving a maximum velocity where the enzyme is saturated by the substrate. It's critical to comprehend this saturation effect, as it is pivotal in predicting the reaction behavior under varied conditions.

Reaction Rate Expression

The reaction rate expression is a mathematical representation that indicates how swiftly a chemical reaction occurs. It is governed by parameters like reactant concentration and rate constants, which depend on the nature of the reaction as well as external factors such as temperature.
For an enzyme-catalyzed reaction governed by Michaelis-Menten kinetics, the reaction rate expression is non-linear, meaning that it forms a hyperbolic relationship with substrate concentration. The rate at which substance A is used up can be described by the differential rate expression:
\[\frac{dC_{\mathrm{A}}}{dt} = - \frac{k_1 C_{\mathrm{A}}}{1 + k_2 C_{\mathrm{A}}}\]
This equation underlines a crucial detail: as A is consumed, its concentration decreases over time, which, in turn, affects the velocity of the reaction. Thus, understanding this expression is vital in predicting how long it takes for the reaction to reach a certain stage.

Batch Reactor Kinetics

Batch reactors are a common setup for carrying out chemical reactions on a limited scale and are frequently used for enzyme-catalyzed processes. Kinetics within a batch reactor requires consideration of how the reaction progresses over time, starting with initial concentrations and without the continuous addition or removal of reactants (or products).
In the given exercise, an isothermal batch reactor works with a constant volume. We use the initial condition, \(C_{\mathrm{A}0}\), which is the concentration of reactant A at the beginning (time \(t = 0\)). As the reaction advances, you can visualize the declining concentration of A using a plot of \(C_{\mathrm{A}}\) versus time \(t\). This graphical representation encapsulates the entirety of the batch reactor kinetics, showing how the concentration diminishes asymptotically as time progresses, eventually reaching an equilibrium where no further reaction takes place. Internalizing this behavior helps with predicting outcomes in batch reactions.

Differential Equation Solving

In the realm of enzyme-catalyzed reactions, solving differential equations is a fundamental skill required to glean insight into reaction dynamics. The rate equation we're dealing with is a differential equation that describes how the substrate concentration changes over time during the reaction.
To solve the rate differential equation given in the exercise: \[\frac{dC_{\mathrm{A}}}{dt} = - \frac{k_1 C_{\mathrm{A}}}{1 + k_2 C_{\mathrm{A}}}\] you would typically separate variables and integrate both sides. This involves calculus and sometimes requires numerical methods, especially if the integration cannot be done analytically due to the equation's complexity. Upon successful integration, you obtain a function that gives you the time required to reach a specific substrate concentration \(C_{\mathrm{A}}\).
Understanding and being able to apply these mathematical tools can crack the code of complex enzyme kinetics, providing pathways to the practical determination of rates and constants based on experimental data.