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In an autocatalytic reaction, one substance is converted into a second substance in such a way that the second substance catalyzes its own formation. This is the process by which trypsinogen is converted into the enzyme trypsin. The reaction starts only in the presence of some trypsin, and each molecule of trypsinogen yields one molecule of trypsin. The rate of formation of trypsin is proportional to the product of the amounts of the two substances present. Set up the differential equation that is satisfied by \(y=f(t)\), the amount (number of molecules) of trypsin present at time \(t .\) Sketch the solution. For what value of \(y\) is the reaction proceeding the fastest? [Note: Letting \(M\) be the total amount of the two substances, the amount of trypsinogen present at time \(t\) is \(M-f(t) .]\)

Step by step solution

01

Identify the variables and relationships

Define the variables involved: Let \( y = f(t) \) be the amount of trypsin present at time \( t \). The total amount of trypsin and trypsinogen is \( M \). The amount of trypsinogen at time \( t \) is thus \( M - y \).

02

Set up the rate of formation expression

Given that the rate of formation of trypsin is proportional to the product of the amounts of trypsin and trypsinogen present, we write: \[ \frac{dy}{dt} = k y (M - y) \] where \( k \) is the rate constant.

03

Determine the critical points for the maximum rate

To find when the reaction is proceeding the fastest, take the derivative of the rate function \( y (M - y) \): \[ \frac{d}{dy}[y (M - y)] = M - 2y \]. Set this equal to zero to find the critical points: \[ M - 2y = 0 \Rightarrow y = \frac{M}{2} \].

04

Confirm maximum rate

To confirm this is a maximum, check the second derivative: \[ \frac{d^2}{dy^2}[y (M - y)] = -2 \]. Since \( -2 < 0 \), \( y = \frac{M}{2} \) is indeed a maximum point.

05

Plot the differential equation solution

Sketch the graph of \( y(t) \) as dictated by the differential equation: \[ \frac{dy}{dt} = k y (M - y) \]. The solution is a logistic curve starting from the initial presence of trypsin (since reaction starts only in its presence) and asymptotically approaching \( M \) as \( t \to \infty \). At \( y = \frac{M}{2} \), the rate is maximum.

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Key Concepts

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

Autocatalytic Reaction

An autocatalytic reaction is a type of chemical reaction where the product of the reaction also acts as a catalyst for the same reaction. This means that as the reaction progresses, it accelerates itself. A common example of an autocatalytic reaction is the conversion of trypsinogen to trypsin. Here, trypsin catalyzes the formation of more trypsin from trypsinogen.
In autocatalytic reactions, the presence of a small initial quantity of the catalyst is crucial. Without it, the reaction would not proceed. Therefore, an initial amount of trypsin is necessary to start converting trypsinogen into more trypsin.

Differential Equations

Differential equations are mathematical equations that describe the rate at which quantities change. They are fundamental in modeling the behavior of systems that change over time, such as chemical reactions.

In the context of an autocatalytic reaction, a differential equation can describe the rate of formation of a product. For instance, the differential equation
\( \frac{dy}{dt} = k y (M - y) \),
where
\( y \) is the amount of trypsin at time
\( t \), and
\( M \) is the total amount of trypsin and trypsinogen.

The equation expresses that the rate of formation of trypsin is proportional to the product of the amounts of trypsin and trypsinogen present.

Rate of Reaction

The rate of reaction in chemical kinetics refers to how quickly a reactant is converted into a product. For the autocatalytic reaction, the rate of formation of trypsin is given by the equation
\( \frac{dy}{dt} = k y (M - y) \)

This equation states that the rate of the reaction depends on both
\( y \) (the amount of trypsin present) and
\( M - y \) (the remaining amount of trypsinogen).
The reaction rate is fastest when
\( y = \frac{M}{2} \),
meaning that when half of the total amount is trypsin, the remaining half being trypsinogen, the rate of reaction reaches its peak.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that studies the speed or rate at which chemical reactions occur and the factors that influence these rates. It provides insights into reaction mechanisms and helps predict the progress of chemical reactions.

The autocatalytic reaction involving trypsin and trypsinogen is a classic example. The rate law for this reaction can be expressed as
\( \frac{dy}{dt} = k y (M - y)\),

where:

• \( y \): amount of trypsin.
• \( M - y \): amount of trypsinogen.
• \( k \): rate constant.

This equation captures how changes in the amounts of trypsin and trypsinogen affect the overall reaction rate.

Logistic Growth Model

The logistic growth model is often used to describe how a quantity grows in a limited environment. While it's typically applied to population growth, it can also describe the rate of chemical reactions with self-limiting behavior. For the autocatalytic reaction, the differential equation
\( \frac{dy}{dt} = k y (M - y) \)
describes a type of logistic growth.

Initially, when
\( y \) is small, the growth is nearly exponential, and as
\( y \) grows, the term
\( M - y \)
(the remaining reactant) reduces the rate of reaction. Eventually, the system approaches an equilibrium state where
\( y \) asymptotically approaches
\( M \). At this point, the amount of trypsinogen has been nearly exhausted, and the reaction rate slows significantly.