Problem 46 - Applied Calculus

Question

A chemical reaction converts substance \(A\) to substance \(Y\); the presence of \(Y\) catalyzes the reaction. At the start of the reaction, the quantity of \(A\) present is \(a\) grams. At time \(t\) seconds later, the quantity of \(Y\) present is \(y\) grams. The rate of the reaction, in grams/sec, is given by Rate \(=k y(a-y), \quad k\) is a positive constant. (a) For what values of \(y\) is the rate nonnegative? Graph the rate against \(y\). (b) For what values of \(y\) is the rate a maximum?

Step-by-Step Solution

Answer

(a) Rate is nonnegative for \(0 \leq y \leq a\). (b) Rate is maximum at \(y = \frac{a}{2}\).

Step 1: Understanding Nonnegative Rate Condition

The rate of reaction is given by \( R = k y (a-y) \), where \( k \) is a constant, \( y \) is the quantity of substance \( Y \), and \( a \) is the initial quantity of \( A \). The rate is nonnegative if \( R \geq 0 \). Since \( k \) is a positive constant, \( R \geq 0 \) if \( y(a-y) \geq 0 \). Therefore, the values of \( y \) for which the rate is nonnegative are those satisfying \( 0 \leq y \leq a \).

Step 2: Graphing the Rate Function

To graph the rate against \( y \), plot \( R = k y (a-y) \) as a function of \( y \). This is a downward-opening parabola (inverted quadratic function) with zeros at \( y = 0 \) and \( y = a \). The vertex, which represents the maximum rate, is located exactly halfway between these zeros, at \( y = \frac{a}{2} \). Thus, the graph has the shape of a parabola with its maximum at \( y = \frac{a}{2} \).

Step 3: Finding Maximum Rate Values

To find the value of \( y \) for which the rate is maximum, look for the vertex of the parabola \( R = k y (a-y) \), which is symmetric around \( y = \frac{a}{2} \). This is the maximum point of the parabola because the parabola opens downward. Thus, the rate of reaction is at its maximum when \( y = \frac{a}{2} \).

Key Concepts

Understanding Catalysis

Catalysis is a fascinating concept in chemical kinetics, where a substance, known as a catalyst, increases the rate of a chemical reaction without being consumed in the process. In our exercise, substance \( Y \) acts as the catalyst. As \( Y \) is formed, it enhances the reaction of substance \( A \) converting into more \( Y \). This creates a feedback loop that accelerates the reaction, making it proceed faster than it would without \( Y \).

Catalysts are vital in the world of chemistry for several reasons:

• They lower the activation energy needed for a reaction to occur, allowing reactions to proceed more quickly.
• Catalysts are not used up, so they can participate in multiple reaction cycles.
• They can increase the selectivity of a reaction, targeting specific pathways over others.

In our specific scenario, since the reaction requires \( Y \) to catalyze, its presence directly influences how quickly \( A \) converts into \( Y \). Catalysts do not alter the thermodynamics of the reaction; they merely provide a different pathway that has a lower energy barrier. This is key to understanding why reactions with catalysts can move faster under the same conditions.

Exploring Reaction Rates

In chemical kinetics, the reaction rate measures how fast or slow a reaction occurs. It's a pivotal concept, especially in our exercise, where the reaction rate is dependent on both the amount of substance \( Y \) and the remaining \( A \). We can see that the rate equation is given by:

\[R = k y (a-y)\]where:

• \( R \) represents the reaction rate.
• \( k \) is a constant that scales the reaction rate.
• \( y \) is the amount of substance \( Y \).
• \( a-y \) is the remaining amount of \( A \).

Reaction rates are crucial for predicting how quickly products are formed in a chemical process. The reaction rate is nonnegative when \( 0 \leq y \leq a \), ensuring physically meaningful values. When \( y = 0 \) or \( y = a \), the reaction rate drops to zero, indicating no ongoing reaction. For practical purposes, understanding the conditions that maximize reaction rates can guide decisions in industrial and laboratory settings to optimize production and efficiency.

Analyzing Parabolic Graphs

In this exercise, visualizing the reaction rate function involves graphing it against \( y \). The given rate function \( R = k y (a-y) \) represents a parabolic graph, specifically an inverted parabola, meaning it opens downward. This type of graph is crucial to identify because it indicates how the reaction rate changes as \( y \) varies.

The parabola's key characteristics include:

• The graph touches the x-axis (zeros) at \( y = 0 \) and \( y = a \), where the reaction rate is zero.
• The vertex of the parabola, located at \( y = \frac{a}{2} \), signifies the maximum reaction rate.
• The symmetry of the parabola around this vertex provides insight into how the rate changes uniformly as \( y \) approaches \( 0 \) and \( a \).

Recognizing the position of the vertex is vital because it tells us the point where the reaction is fastest. Graphs like these are intuitive tools, offering a visual snapshot of the relationship between reaction conditions and rates. For students and chemists alike, mastering the ability to interpret parabolic graphs provides a solid foundation for analyzing more complex kinetic models.