Problem 53 - Thomas Calculus

Question

Tin pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C},\) it slowly becomes britle 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 \(v=d x / d t\) 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 \begin{equation}v=k x(a-x)=k a x-k x^{2}\end{equation} where \begin{equation} \begin{aligned} x &=\text { the amount of product } \\ a &=\text { the amount of substance at the beginning } \\ k &=\text { a positive constant. } \end{aligned} \end{equation} 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 \( v \) is maximum at \( x = \frac{a}{2} \), and the maximum rate is \( \frac{ka^2}{4} \).

Step 1: Understand the rate equation

The rate of the reaction is given by the equation \( v = kx(a-x) \). This equation shows that the rate depends on both the amount of product \( x \) and the initial amount of the substance \( a \).

Step 2: Rewrite the rate equation

The rate equation can be expanded as \( v = ka x - kx^2 \). This form of the equation is quadratic in \( x \), which will help us find the maximum rate.

Step 3: Find the critical points

To find the maximum value of a function, we need to find the critical points by setting the first derivative equal to zero. The first derivative of \( v \) with respect to \( x \) is \( \frac{dv}{dx} = ka - 2kx \). Set \( \frac{dv}{dx} = 0 \) to find critical points:\[ ka - 2kx = 0 \]

Step 4: Solve for x

Solve the equation \( ka - 2kx = 0 \) for \( x \):\[ ka = 2kx \]Divide both sides by \( k \) (assuming \( k eq 0 \)):\[ a = 2x \]Thus, \( x = \frac{a}{2} \).

Step 5: Evaluate the second derivative

To confirm that \( x = \frac{a}{2} \) is a maximum, evaluate the second derivative \( \frac{d^2v}{dx^2} = -2k \). The second derivative is negative, indicating a maximum at \( x = \frac{a}{2} \).

Step 6: Calculate the maximum rate

Substitute \( x = \frac{a}{2} \) into the original rate equation \( v = ka x - kx^2 \) to find the maximum rate:\[ v = ka\left(\frac{a}{2}\right) - k\left(\frac{a}{2}\right)^2 \]\[ v = \frac{ka^2}{2} - \frac{ka^2}{4} \]\[ v = \frac{ka^2}{4} \]

Key Concepts

Chemical Kinetics

Chemical kinetics is a fascinating area of chemistry that studies the rates at which chemical reactions occur and the factors influencing them. It helps us understand how quickly a reaction proceeds from reactants to products. This is crucial not only for theoretical comprehension but also for practical applications such as designing chemical processes.

In many chemical reactions, the rate at which the substances react can vary. Some reactions happen instantly, like the combustion of gasoline, while others, such as the rusting of iron or the tin pest scenario, proceed much more slowly.

Understanding how and why these rates vary involves looking at the concentrations of the reactants and products, temperature, and presence of catalysts. In our example with tin, the transformation to a gray powder is an autocatalytic process, where the product itself accelerates the ongoing reaction.

Rate Equations

Rate equations are mathematical expressions that describe the speed of a chemical reaction. They usually show how the rate of change of concentration of a substance relates to the concentration itself. The rate equation is directly aligned with the concept of chemical kinetics, providing a quantitative way to predict reaction behavior.

In the exercise, we dealt with the rate equation \( v = kx(a-x) \). This is a specific type of rate equation that involves both the initial substance and the product.

Here, \( x \) represents the amount of product formed, and \( a \) is the initial quantity of the substance. The equation \( v = ka x - kx^2 \) allows us to calculate how quickly the reaction progresses for different values of \( x \). It shows that at different stages of the reaction, the rate can increase or decrease based on the values of \( x \) and \( a \). This type of behavior is typical in autocatalytic reactions.

Maximum Rate of Reaction

Determining the maximum rate of reaction is crucial for optimizing chemical processes, where efficiency and time are of the essence. In an autocatalytic reaction like ours, the rate reaches a maximum under certain conditions which can be systematically uncovered.

The exercise guides us to find the maximum using a specific rate equation. To find the peak, one assesses where the product and the original substance are present in balanced amounts.

As calculated, the maximum rate occurs when \( x = \frac{a}{2} \), meaning that the amount of product is half of the original amount of the substance. At this point, both substance and its product are abundantly available to sustain a high rate, yielding the equation \( v = \frac{ka^2}{4} \), marking the maximum velocity of the reaction. This understanding helps in controlling and predicting reaction dynamics in practical settings.

First Derivative Test

The first derivative test is a powerful tool in calculus used to identify critical points of functions, helping determine where these points are maximum, minimum, or neither. In chemical kinetics, it is applied to assess the change in reaction rates.

For the tin pest problem, finding the maximum reaction rate involves taking the derivative \( \frac{dv}{dx} \) of the rate equation \( v = kx(a-x) \). The derivative \( \frac{dv}{dx} = ka - 2kx \) is set to zero to find critical points: \( ka - 2kx = 0 \).

Solving this gives \( x = \frac{a}{2} \), which signifies a potential maximum. To confirm, the second derivative test evaluates \( \frac{d^2v}{dx^2} \), and its negativity verifies the presence of a maximum point at \( x = \frac{a}{2} \). The first derivative test thus allows us to systematically predict reaction dynamics, crucial for efficient chemical management.