Problem 25 - Thomas Calculus

Question

First-order chemical reactions In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of \(\delta\) -glucono lactone into gluconic acid, for example, $$\frac{d y}{d t}=-0.6 y$$ when \(t\) is measured in hours. If there are 100 grams of \(\delta\) -glucono lactone present when \(t=0,\) how many grams will be left after the first hour?

Step-by-Step Solution

Answer

Approximately 54.88 grams remain after one hour.

Step 1: Identify the Differential Equation

We are given a differential equation \(\frac{dy}{dt} = -0.6y\), which models the rate of change of the amount of \(\delta\)-glucono lactone, \(y\), with respect to time, \(t\).

Step 2: Solve the Differential Equation

This is a separable differential equation. We can separate the variables as follows: \(\frac{dy}{y} = -0.6 \, dt\). Integrating both sides, we have \(\int \frac{1}{y} \, dy = \int -0.6 \, dt\), leading to \(\ln|y| = -0.6t + C\), where \(C\) is the constant of integration.

Step 3: Solve for the Constant of Integration

To find \(C\), we use the initial condition: when \(t = 0\), \(y = 100\). Substituting these values into \(\ln|y| = -0.6t + C\), we get \(\ln|100| = C\), so \(C = \ln 100\).

Step 4: Exponentiate to Solve for y(t)

We exponentiate both sides of the equation \(\ln|y| = -0.6t + \ln 100\) to solve for \(y(t)\). This gives us \(|y| = e^{\ln 100 - 0.6t}\). Simplifying, we find \(y = 100e^{-0.6t}\).

Step 5: Calculate Amount After One Hour

Substitute \(t = 1\) into the equation \(y = 100e^{-0.6t}\): \[ y(1) = 100e^{-0.6 \cdot 1} = 100e^{-0.6} \approx 100 \times 0.5488 = 54.88 \]Therefore, approximately 54.88 grams are left after one hour.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They describe how a particular quantity changes with respect to another. In the case of first-order chemical reactions, differential equations are often used to model the rate of change of concentrations of reactants over time.

For example, if a chemical reaction involves the transformation of a substance like \(\delta\)-glucono lactone, the rate at which this substance changes its concentration can be described by a differential equation. This model helps predict how much of the substance will remain at any given time. Understanding the behavior of differential equations is crucial in many scientific fields, such as chemistry, physics, and biology.

• A differential equation can often be recognized by the presence of derivatives, such as \( \frac{dy}{dt} \) in the given equation, representing "the rate of change" of \( y \) with respect to time \( t \).
• Solving differential equations helps in forecasting how systems evolve, which is essential for experimental and quantitative analysis.

Separable Differential Equations

A separable differential equation is a specific type of differential equation in which variables can be separated on opposite sides of the equation. This property makes them relatively straightforward to solve.

In our case, the equation \(\frac{dy}{dt} = -0.6y\) is a prime example of a separable differential equation. We can "separate" the variables by rearranging the terms:

• Move all \(y\)-terms to one side: \(\frac{dy}{y}\)
• And all the \(t\)-terms to the other side: \(-0.6 \, dt\)

By doing this, we create two simpler integrals that can be solved individually:

\[\int \frac{1}{y} \, dy = \int -0.6 \, dt\]
Solving these integrals ultimately gives us the solution to the original problem. This method is very helpful in biology, chemistry, and physics, as it can simplify modeling various natural processes.

Rate of Change

The rate of change is a fundamental concept that describes how one quantity changes in relation to another. In the context of chemical reactions, the rate of change refers to how fast the concentration of a reactant decreases or increases over time.

In first-order reactions, like the decay of \(\delta\)-glucono lactone into gluconic acid, the rate at which the substance amount changes is directly proportional to the amount present. This can be mathematically expressed as:
\[\frac{dy}{dt} = -0.6y\]
Here, \(\frac{dy}{dt}\) is the rate of change, indicating the instantaneous rate at which the concentration \(y\) decreases over time \(t\). The negative sign suggests the substance is decreasing.

• A higher absolute value of the rate constant (like 0.6) implies a faster rate of transformation.
• Understanding the rate of change is vital for controlling and optimizing reactions in industrial processes.

Exponential Decay

Exponential decay refers to a specific pattern of decrease over time, commonly observed in processes where quantities diminish at rates proportional to their current value. This is a hallmark of first-order chemical reactions.

In the given problem, the amount of \(\delta\)-glucono lactone decreases exponentially according to the function \(y = 100e^{-0.6t}\). This expression explains how the quantity declines at a consistent percentage rate.

• "Exponential decay" leads to a rapid decrease initially, which then slows down as time progresses.
• In the function \(100e^{-0.6t}\), 100 represents the initial amount of substance, while \(e^{-0.6t}\) captures the decay effect as time \(t\) advances.

Exponential decay is prevalent not only in chemical reactions but also in other phenomena like radioactive decay, population decrease, and cooling of objects. By comprehending such patterns, we better understand and predict the behavior of various natural systems.