Problem 1 - Calculus with Concepts in Calculus

Question

In an autocatalytic chemical reaction a substance \(A\) is converted into a substance \(B\) in such a manner that $$ \frac{d x}{d t}=k x(a-x) $$ where \(x\) is the concentration of substance \(B\) at time \(t, a\) is the initial concentration of substance \(A\), and \(k\) is a positive constant. Determine the value of \(x\) at which the rate \(d x / d t\) of the reaction is maximum.

Step-by-Step Solution

Answer

The rate is maximum at \( x = \frac{a}{2} \).

Step 1: Understand the derivative function

Given the rate of reaction \( \frac{d x}{d t} = k x (a-x) \), we need to find the value of \( x \) where this rate is maximized.

Step 2: Differentiate the rate function

To find the maximum rate, we first differentiate the rate function \( \frac{d x}{d t} = k x (a-x) \) with respect to \( x \). Use the product rule:\[\frac{d}{dx}[kx(a-x)] = k[(a-x) + x] = k[a-2x].\]

Step 3: Set the derivative to zero

To find the critical points, set the derivative \( k[a-x(2)] = 0 \) equal to zero:\[a - 2x = 0.\]Thus, \( 2x = a \), solving for \( x \) gives \( x = \frac{a}{2} \).

Step 4: Analyze for maximum

The second derivative test or analyzing nearby values can determine if this is a maximum. Differentiating the first derivative gives a negative value, indicating a maximum. However, as the form is quadratic opening downwards, \( x = \frac{a}{2} \) will be at maximum.

Key Concepts

Differential Equations

In many fields, differential equations serve as a powerful tool to model change. They describe how a quantity changes over time, based on its current state and certain parameters. This is particularly useful in modeling chemical reactions, where concentrations of substances change as the reaction progresses.
In an autocatalytic reaction, we often encounter a differential equation that relates the rate of change of a substance to its current concentration. In the example given, the differential equation is \( \frac{dx}{dt} = kx(a-x) \). This tells us the rate of change of substance \( B \) based on its concentration \( x \) at any time \( t \).
Solving this type of equation involves differentiation, which helps us find critical points. These points tell us where significant changes occur, such as maximum rates or equilibrium points in the reaction.

Rate of Reaction

The rate of a chemical reaction is a measure of how quickly reactants are converted into products. In our equation \( \frac{dx}{dt} = kx(a-x) \), the rate depends on the concentration \( x \) as well as the initial concentration \( a \) and rate constant \( k \).
To find when this rate is at its peak, we must use calculus to differentiate the rate function. The product rule helps us expand and differentiate the original rate formula, yielding the derivative \( k(a-2x) \).
By setting this derivative to zero, we discover the concentration \( x \) at which the rate is maximized, which in this case is \( x = \frac{a}{2} \). This point represents the highest speed at which the substance \( A \) is converted into \( B \), optimizing the reaction dynamics.

Calculus in Chemistry

Calculus plays an instrumental role in understanding chemical reactions by providing the analytical tools necessary to explore reaction dynamics. Derivatives, like the one found in \( \frac{dx}{dt} = kx(a-x) \), allow chemists to understand how reaction rates vary with concentration, which is crucial for reaction efficiency and effectiveness.
Using calculus to differentiate and find critical points helps in predicting how the reaction progresses over time. With this information, chemists can adjust conditions to optimize reactions, whether that's by changing temperature, concentrations, or other reaction components.
Furthermore, the use of calculus allows us to apply mathematical techniques, such as second derivative tests, to determine whether calculated critical points indicate maximum or minimum rates. This makes calculus an essential part of predicting and controlling reaction outcomes.