Question

The Law of Mass Action in chemistry results in the differential equation $$ \frac{d x}{d t}=k(a-x)(b-x), \quad k>0, \quad a>0, \quad b>0 $$ where \(x\) is the amount of a substance at time \(t\) resulting from the reaction of two others. Assume that \(x=0\) when \(t=0\). (a) Solve this differential equation in the case \(b>a\). (b) Show that \(x \rightarrow a\) as \(t \rightarrow \infty\) (if \(b>a\) ). (c) Suppose that \(a=2\) and \(b=4\), and that 1 gram of the substance is formed in 20 minutes. How much will be present in 1 hour? (d) Solve the differential equation if \(a=b\).

Short answer

For \( b>a \), \( x(t) = \frac{ab(1-e^{(b-a)kt})}{b-ae^{(b-a)kt}} \). As \( t \to \infty \), \( x \to a \). For \( a=2, b=4 \), solve for \( k \) and calculate \( x(60) \). For \( a=b \), \( x(t) = \frac{a^2}{a+k(a^2)t} \).

Step-by-step solution

Understand the Differential Equation

We have the equation \( \frac{d x}{d t} = k(a-x)(b-x) \), where \( k, a, \) and \( b \) are constants. We need to solve this differential equation under different conditions.

Solve for the Case b > a

For part (a), separate variables to solve the equation. Integrating, we get \[ \int \frac{1}{(a-x)(b-x)} \, dx = \int k \, dt. \] Use partial fraction to decompose the left-hand side: \[ \frac{1}{(a-x)(b-x)} = \frac{1}{b-a} \left( \frac{1}{a-x} - \frac{1}{b-x} \right). \] Integrate to solve for \( x(t) \). The integration gives \( \frac{1}{b-a} \left( -\ln|a-x| + \ln|b-x| \right) = kt + C \). Solve for \( x \) in terms of \( t \).

Find Constant C Using Initial Condition

The initial condition \( x=0 \) when \( t=0 \) helps us find \( C \). Substitute \( x=0, t=0 \) into the integrated result to get \( C = \frac{1}{b-a} \ln\frac{b}{a} \).

Solve and Simplify for x(t)

Simplifying the integrated expression: \[ \ln\frac{b-x}{a-x} = (b-a)kt + \ln\frac{b}{a}. \] Therefore, we have \[ \frac{b-x}{a-x} = \left(\frac{b}{a}\right) e^{(b-a)kt}. \] Solve for \( x(t) \), giving \[ x = \frac{ab(1-e^{(b-a)kt})}{b-ae^{(b-a)kt}}. \]

Show x Approaches a as t Approaches Infinity

As \( t \to \infty \), \( e^{-(b-a)kt} \to 0 \), making: \( x(t) \to \frac{ab}{b} = a \).

Solve Part (c) with Given Values

Given \( a=2, b=4 \), and \( x(20) = 1 \). Substitute into \( \frac{b-x}{a-x} = \left(\frac{b}{a}\right) e^{2kt} \) to find \( k \). Use this result to calculate \( x(60) \).

Solve When a = b

When \( a=b \), the differential equation simplifies to \( \frac{d x}{d t} = k(a-x)^2 \). This is a separable differential equation. Integrate to find \( x(t) = \frac{a^2}{a+k(a^2)t} \).

Key concepts

Law of Mass Action

The Law of Mass Action is a fundamental principle in chemistry and governs the rates of reactions. It states that the rate of a chemical reaction is proportional to the product of the concentrations of the reacting substances, each raised to a power corresponding to the number of moles involved in the reaction. In the context of the given differential equation:\[\frac{d x}{d t} = k(a-x)(b-x), \quad k>0, \quad a>0, \quad b>0\]The law predicts how the concentration of a substance, represented by \(x\), changes over time as it is produced through a chemical reaction. Here, \(a\) and \(b\) represent the initial concentrations of two reactants, while \(k\) is a rate constant. The reaction's progression is dictated by how \(x\), the produced substance, evolves according to this law. By modeling this with a differential equation, one can analyze how quickly the reaction proceeds or what happens to the system as time goes on.

partial fractions

Partial fraction decomposition is a mathematical technique used to simplify the integration of rational functions. When we have the expression \[\frac{1}{(a-x)(b-x)}\]we can break it down into simpler fractions to make it easier to integrate. For example, this expression is decomposed as:\[\frac{1}{(a-x)(b-x)} = \frac{1}{b-a} \left( \frac{1}{a-x} - \frac{1}{b-x} \right)\]This allows us to integrate each term separately. Partial fractions are a powerful tool in calculus, particularly in solving differential equations. By converting a complex fraction into a sum of simpler ones, we use basic integration methods to find solutions. This step is vital when solving our differential equation as it simplifies the process, leading to the integrated expression needed to find the solution for \(x(t)\).

limit behavior

When analyzing the behavior of solutions as time progresses, it is essential to consider the concept of limit behavior. This refers to the long-term behavior of a system described by a differential equation as time approaches infinity. In the provided exercise, we're interested in determining what happens to \(x(t)\) as \(t\) tends towards infinity.For the equation:\[\frac{d x}{d t} = k(a-x)(b-x)\]it is shown that as \(t \to \infty\), \(x(t)\) approaches \(a\). This is concluded because the exponential term \(e^{-(b-a)kt}\) tends to zero over time, simplifying the expression for \(x(t)\). Understanding limit behavior helps predict the ultimate outcome of a reaction, indicating stability or equilibrium. It is crucial when determining if \(x(t)\) will plateau or continue changing indefinitely.

initial conditions

Initial conditions are specific values that help define the constants in the solution of a differential equation. They provide the necessary background to solve an equation uniquely. In our exercise, the initial condition is given as \(x=0\) when \(t=0\).Using initial conditions, we calculate constants like \(C\) in our integrated function to solve the differential equation:\[\ln\frac{b-x}{a-x} = (b-a)kt + C\]By substituting \(x=0\) and \(t=0\), we find \(C = \frac{1}{b-a} \ln\frac{b}{a}\). This step ensures that our solution \(x(t)\) respects the original conditions of the problem. Initial conditions are vital as they ground the solution in the context of the specific reaction being modeled, making it applicable to real-world scenarios like predicting the amount of substance formed after a given time.