Problem 61 - Calculus: Early Transcendental Functions

Question

In a second-order chemical reaction, one molecule each of substances \(A\) and \(B\) combine to produce one molecule of substance X. If \(a\) and \(b\) are the initial concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), respectively, the concentration \(x\) of the substance X satisfies the differential equation \(x^{\prime}=r(a-x)(b-x)\) for some positive rate constant \(r .(\text { a) If } r=0.4, a=6, b=8 \text { and } x(0)=0,\) find \(x(t)\) and \(\lim _{t \rightarrow \infty} x(t) .\) Explain this answer in terms of the chemical process. (b) Repeat part (a) with \(r=0.6 .\) Graph the solutions and discuss differences and similarities.

Step-by-Step Solution

Answer

The solution involves solving a separable differential equation to find \(x(t)\), the concentration of substance X over time, and calculating \(\lim_{t \rightarrow \infty}{x(t)}\) for the given rate constants of the reactions. The graphs will show the behavior of chemical reactions over time for different rate constants.

Step 1 - Identifying the differential equation

Identify the differential equation \(x^{\prime}=r(a-x)(b-x)\) and the initial condition \(x(0)=0\). Given constants: \(r=0.4, a=6, b=8\) for part a.

Step 2 - Solving the differential equation

This equation can be solved using separable differential equation techniques. The idea is to bring one variable on one side with its differential, and the rest on the other side. Divide both side by \((a - x)(b - x)\) and multiply with \(dt\) on both sides. Now integrating both sides gives us the solution.

Step 3 - Calculating concentration of X over time

Calculate the integral on both sides, using the initial conditions to solve for the constants of integration. This gives the concentration of substance X at any time t, \(x(t)\).

Step 4 - Calculate the limit of \(x(t)\) at infinity

Evaluate the limit of \(x(t)\) as \(t\) approaches infinity, \(\lim_{t \rightarrow \infty}{x(t)}\). This gives the equilibrium concentration of substance X.

Step 5 - Interpret the result

Explanation of the answer in the terms of chemical process. As time progresses, the reaction will reach a state of equilibrium. The limit gives us the final concentration of substance X at this equilibrium state.

Step 6 - Repeat for \(r=0.6\)

Repeat the same process for \(r=0.6\) to see the effect of rate constant on substance X's concentration.

Step 7 - Graph and compare

Plot both equations over time to visualize the quantity of substance X with the different rate constants. Discuss the differences and similarities.

Key Concepts

Second-order Reactions

A second-order reaction in chemical kinetics involves two reactant molecules colliding and reacting. Here, one molecule each of substances A and B combines to form one molecule of the product X. Given initial concentrations of A and B, the concentration of X evolves over time as described by a specific differential equation. This type of reaction is second-order because the rate depends on the concentration of both reactants involved.

• Second-order reactions are sensitive to changes in the concentration of reactants.
• The rate of formation of product X is proportional to the product of the concentrations of A and B.

This sensitivity means even small changes in concentration or rate constant can significantly affect the speed of the reaction and the equilibrium state.

Chemical Kinetics

Chemical kinetics studies the speed of reactions and factors affecting these rates. In our problem, it is represented by the differential equation. The rate equation for the reaction described is:\[ x^{\prime} = r(a-x)(b-x) \]where:- \( x^{\prime} \) is the rate of change of concentration of X with respect to time.- \( r \) is the rate constant, showing how quickly A and B react.- \( a-x \) and \( b-x \) represent current concentrations of A and B as the reaction progresses.

• The rate constant \( r \) changes the speed at which the reaction approaches equilibrium.
• A higher \( r \) means the reaction reaches equilibrium faster.

Understanding these principles helps predict how quickly products will form in any given chemical process.

Separable Equations

The differential equation in this problem is a classic example of a separable equation, a type where variables can be arranged to isolate each on either side of the equal sign with their corresponding differentials. We rewrite it as:\[ \frac{dx}{(a-x)(b-x)} = r dt \]This allows us to solve it through integration, a fundamental calculus technique. Breaking the equation apart into separable parts enables mathematicians and scientists to find solutions that describe how concentrations change over time.Integrating both sides of the equation helps determine the function \( x(t) \), giving insight into the concentration of X at any given time. By applying initial conditions, such as \( x(0) = 0 \), we solve for any constants of integration needed.

Equilibrium Concentration

As time progresses, chemical reactions often reach a state of balance known as equilibrium, where the concentrations of reactants and products remain constant. In our exercise, finding \( \lim_{t \rightarrow \infty} x(t) \) helps determine this state for product X.When the reaction rate decreases and stabilizes, the process closely approaches equilibrium:- At equilibrium, the rate of transformation of reactants into products equals the rate at which products revert to reactants (in reversible reactions).- In our case, the reaction between A and B progresses until the reactants are consumed optimally, leading to a stable concentration of product X.The equilibrium concentration reflects the maximum amount of X achievable for given initial conditions and rate constants, essential for understanding the system's final state.