Problem 49 - Thomas Calculus in SI Units

Question

When metallic tin is kept below \(13.2^{\circ} \mathrm{C}\), it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate \(\boldsymbol{v}=d x / d t\) of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, \(v\) may be considered to be a function of \(x\) alone, and $$v=k x(a-x)=k a x-k x^{2}$$ where \(x=\) the amount of product \(a=\) the amount of substance at the beginning \(k=\) a positive constant. At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v ?\)

Step-by-Step Solution

Answer

The rate \( v \) reaches its maximum at \( x = \frac{a}{2} \), with a maximum value of \( \frac{ka^2}{4} \).

Step 1: Understand the rate equation

The rate of reaction, which depends on the amounts of substance and product, is given by \( v = kx(a-x) = ka x - kx^2 \). Here, \(ka\) and \(k\) are constants, \(a\) is the initial amount of substance, and \(x\) is the amount of product formed.

Step 2: Identify the task

We need to find the value of \(x\) that maximizes the rate \(v\) and determine this maximum rate. This requires finding the critical points of the function \(v\).

Step 3: Differentiate the rate equation with respect to x

Differentiate \( v = ka x - kx^2 \) with respect to \( x \) to find \( \frac{dv}{dx} \). The differentiation yields: \( \frac{dv}{dx} = ka - 2kx \).

Step 4: Find critical points

Set the derivative \( \frac{dv}{dx} = ka - 2kx \) equal to zero to find critical points: \( 0 = ka - 2kx \). Solve for \( x \), obtaining \( x = \frac{a}{2} \).

Step 5: Determine the nature of the critical point

Perform the second derivative test. Differentiate \( \frac{dv}{dx} = ka - 2kx \) to get \( \frac{d^2v}{dx^2} = -2k \), which is negative, indicating that \( x = \frac{a}{2} \) is a maximum point.

Step 6: Find the maximum rate

Substitute \( x = \frac{a}{2} \) into the original rate equation \( v = ka x - kx^2 \). Calculate \( v \) using \[ v = ka \left(\frac{a}{2}\right) - k \left(\frac{a}{2}\right)^2 = \frac{ka^2}{2} - \frac{ka^2}{4} = \frac{ka^2}{4}. \]

Key Concepts

Rate of Reaction

One of the fundamental concepts in chemistry is the rate of reaction. It tells us how fast or slow a reaction is happening. Imagine a busy street; the rate of reaction would be like counting how many cars pass by in a minute. In chemical terms, it's often expressed as a change in concentration over time.
In our case study, where tin turns into gray powder in cold weather, the speed of this transformation is what we call the rate of reaction. Mathematically, the rate of reaction can be given as \( v = \frac{dx}{dt} \). Here, \( x \) is the amount of product formed at a certain time \( t \), and \( v \) denotes the rate at which this product forms.
The example situation involves an autocatalytic reaction. This is where things get interesting: the catalyst of a reaction is created by the reaction itself. In simpler terms, the product being formed helps speed up the formation of even more product.

Autocatalytic Reaction

An autocatalytic reaction is a fascinating type of chemical reaction where the product helps catalyze or speed up its own formation. Imagine it as a snowball effect—a small event triggers a similar event that leads to a chain reaction.
In the context given, imagine tin transforming into gray powder. As this gray powder forms, it aids in the creation of even more gray powder, acting as its own catalyst. Hence, the transformation starts slowly but picks up speed as more product is made.
The rate at which this process unfolds is initially slow, then accelerates when both the original substance and its product are plenty. Eventually, it slows again when most of the original substance is used up. This dynamic nature makes autocatalytic reactions compelling to study and analyze.

Differentiation

The process of differentiation is like peeling off layers of an onion to understand its different properties. In mathematics, differentiation helps us determine how a function changes at any point. It's like finding the slope of a line or how steep it is.
For the equation \( v = ka x - kx^2 \), we find how \( v \) changes with respect to \( x \) by differentiating it. If we let \( \frac{dv}{dx} = ka - 2kx \), this helps us find the critical points, where the change is zero or extreme.
Differentiation helps us delve deeper into calculus optimization and find the specific amounts where the reaction rate is maximized or minimized. This skill is essential in optimization, allowing us to harness the power of calculus to uncover important insights in the behavior of chemical reactions.

Critical Points

Critical points are like check-points in a chemical reaction rate. They tell us where the reaction's pace hits a high or low. Identifying these points helps us understand where adjustments could be made for improved efficiency.
For the function \( v = ka x - kx^2 \), a critical point occurs where the derivative is zero. This means the rate of change is neither increasing nor decreasing in speed, representing a peak or trough. From the differentiation, \( \frac{dv}{dx} = ka - 2kx \) is set to zero, leading to \( x = \frac{a}{2} \).
After finding a critical point, we can use the second derivative test to figure out if it's a maxima or minima. Here, the second derivative \( \frac{d^2v}{dx^2} = -2k \) tells us it’s a maximum since it's negative. Hence, the reaction rate peaks when half the initial substance has transformed into the product.