91Ó°ÊÓ

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 1 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 \(t \text { is } M-f(t) .\)]

Step by step solution

01

Define Variables

Let the function of time be denoted by two variables. Let \(y = f(t)\) be the amount of trypsin present at time \(t\) and \(M\) be the total amount of trypsin and trypsinogen.

02

Determine the Amount of Trypsinogen

The amount of trypsinogen present at time \(t\) is given by the total amount minus the amount of trypsin. So, \(\text{amount of trypsinogen} = M - y\).

03

State the Rate of Change

The rate of formation of trypsin is proportional to the product of the amounts of trypsin and trypsinogen present. Thus, \(\frac{dy}{dt} \propto \left(y(M - y)\right)\).

04

Write the Differential Equation

To make the proportionality an equation, introduce the proportionality constant \(k\). The differential equation becomes: \(\frac{dy}{dt} = ky(M - y)\).

05

Determine the Value of Fastest Reaction

To find the value of \(y\) that maximizes the rate of reaction, differentiate the rate equation \(R = y(M - y)\) with respect to \(y\) and find its critical points: \(\frac{dR}{dy} = M - 2y\).

06

Solve for the Critical Point

Set \(\frac{dR}{dy} = 0\), solving gives: \((M - 2y = 0)\Rightarrow y = \frac{M}{2}\). This is the point where the reaction proceeds the fastest.

07

Sketch the Solution

Sketch the graph of \(y(t)\). Initially, \(y\) will be increasing rapidly, but as \(y\) approaches \frac{M}{2}\for a while and then slows down as \(y\) approaches \(M\). Mark the point \frac{M}{2}\the curve as the point of maximum velocity.

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Key Concepts

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

autocatalytic reaction

In an autocatalytic reaction, a product of the reaction catalyzes the reaction itself. This means that the more product is formed, the faster the reaction goes. Consider trypsinogen converting into trypsin: the trypsin produced helps convert more trypsinogen into trypsin. This creates a situation where the reaction can speed itself up as long as there is more reactant to be converted.

rate of formation

The rate at which a product forms in a chemical reaction is a crucial aspect of understanding how fast a reaction proceeds. For our autocatalytic reaction, the rate of formation of trypsin is proportional to the amount of trypsin already present and the amount of trypsinogen available. Mathematically, this is expressed as \(\frac{dy}{dt} \propto \left(y(M - y)\right)\) where \(\frac{dy}{dt}\) is the rate of change of the amount of trypsin over time \(t\), \(y\) is the amount of trypsin, and \(M - y\) is the amount of trypsinogen.

proportionality constant

To turn the proportional relationship between the rate of formation and the reactant and product amounts into an equation, we introduce a proportionality constant \(\text{k}\). This gives us the differential equation \(\frac{dy}{dt} = ky(M - y)\). The constant \(\text{k}\) represents the rate at which the autocatalytic reaction proceeds per unit concentration of the reactants.

maximizing reaction rate

To find the maximum rate of the reaction, we need to determine the value of \(y\) (the amount of trypsin) that maximizes the rate of change. Differentiating the rate equation \(\text{R} = y(M - y)\) with respect to \(y\) and setting the derivative to zero, we get \(\frac{dR}{dy} = M - 2y = 0\). Solving for \(y\) gives us \(y = \frac{M}{2}\), meaning the reaction proceeds fastest at this point. This is a critical concept in optimizing reaction conditions in chemical processes.