Question

A second-order chemical reaction involves the interaction (collision) of one molecule of a substance \(P\) with one molecule of a substance \(Q\) to produce one molecule of a new substance \(X ;\) this is denoted by \(P+Q \rightarrow X .\) Suppose that \(p\) and \(q,\) where \(p \neq q,\) are the initial concentrations of \(P\) and \(Q,\) respectively, and let \(x(t)\) be the concentration of \(X\) at time \(t .\) Then \(p-x(t)\) and \(q-x(t)\) are the concentrations of \(P\) and \(Q\) at time \(t\) and the rate at which the reaction occurs is given by the equation \(\frac{d x}{d t}=\alpha(p-x)(q-x)\) where \(\alpha\) is a positive constant. a. If \(x(0)=0,\) determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and find \(x(t)\) for any \(t\). b. If the substances \(P\) and \(Q\) are the same, then \(p=q\) and equation ( 32 ) is replaced by \(\frac{d x}{d t}=\alpha(p-x)^{2}\) If \(x(0)=0,\) determine the limiting value of \(x(t)\) as \(t \rightarrow \infty\) without solving the differential equation. Then solve the initial value problem and determine \(x(t)\) for any \(t\)

Short answer

Answer: The limiting value of the reaction as t approaches infinity when substances P and Q are different is the minimum of p and q, i.e., \(lim_{t \to \infty} x(t) = \min(p, q)\).

Step-by-step solution

Finding the limiting value of x(t) as t -> infinity

As time progresses, the rate at which the reaction occurs is expected to decrease since the reactants will be getting consumed. Thus, when \(\frac{dx}{dt}\) approaches zero, the reaction is effectively stopping. Therefore, when \(t \rightarrow \infty\), we can set \(\frac{dx}{dt} = 0\) and solve for \(x\): $$0 = \alpha(p-x)(q-x)$$ From this equation, the possible limiting values of \(x\) are \(x = p\) or \(x = q\). Since \(x\) represents the concentration of \(X\) and \(p \neq q\), we know that \(x\) cannot equal both \(p\) and \(q\) simultaneously. We have \(x(0) = 0\), which means that the initial concentration of \(X\) is 0. As the reaction proceeds, \(x\) will increase but always stay less than \(p\) and \(q\). Hence, the limiting value of \(x(t)\) as \(t \rightarrow \infty\) is the minimum of \(p\) and \(q\), i.e., \(\lim_{t \to \infty} x(t) = \min(p, q)\). Next, we will solve the IVP to find \(x(t)\) for any \(t\).

Solving the IVP

We have the differential equation and the initial condition: $$\frac{dx}{dt} = \alpha(p-x)(q-x), \quad x(0) = 0$$ To solve this, we can use separation of variables. Rearrange the equation and integrate both sides: $$\frac{dx}{(p-x)(q-x)} = \alpha dt$$ Now integrate: $$\int_{0}^{x} \frac{d\xi}{(p-\xi)(q-\xi)} = \int_{0}^{t} \alpha d\tau$$ To solve the integration on the left side, apply the partial fraction decomposition: $$\frac{1}{(p-\xi)(q-\xi)} = \frac{A}{p-\xi} + \frac{B}{q-\xi}$$ Subtract the fractions and set the numerators equal: $$1 = A(q-\xi) + B(p-\xi)$$ To find A and B, set \(\xi=p\) and \(\xi=q\), respectively: $$1 = A(q-p) \Rightarrow A = \frac{1}{q-p}$$ $$1 = B(p-q) \Rightarrow B = -\frac{1}{q-p}$$ Now, perform the integration: $$\int_{0}^{x} \left(\frac{1}{\xi-p} - \frac{1}{\xi-q}\right) d\xi = \alpha \int_{0}^{t} d\tau$$ $$\left[-\ln|\xi-p| + \ln|\xi-q|\right]_{0}^{x} = \alpha t$$ Note that \(\alpha\) is positive. Now take the natural exponential of both sides and simplify: $$\frac{q-x}{p-x} = \text{e}^{\alpha t}$$ Now, isolate \(x\) on one side: $$x = \frac{qp - p(q - \text{e}^{\alpha t})}{1 + \text{e}^{\alpha t}}$$ Hence, we have found the concentration of substance \(X\) at any time \(t\), given by \(x(t) = \frac{qp - p(q - \text{e}^{\alpha t})}{1 + \text{e}^{\alpha t}}\). #Part b# Now we need to handle the special case when \(P\) and \(Q\) are the same substances, making \(p = q\). Our given differential equation in this case is \(\frac{dx}{dt} = \alpha(p-x)^2\). We will first find the limiting value of \(x(t)\) as \(t \rightarrow \infty\) and then solve the IVP.

Finding the limiting value of x(t) as t -> infinity, when p = q

As in Step 1, we can set \(\frac{dx}{dt} = 0\) as \(t \rightarrow \infty\): $$0 = \alpha(p-x)^2$$ Since \(\alpha\) is positive, the equation implies \(x = p\) as the limiting value of \(x(t)\) when \(t \rightarrow \infty\). Now we will proceed to solve the IVP.

Solving the IVP when p = q

We have the differential equation and the initial condition: $$\frac{dx}{dt} = \alpha(p-x)^2, \quad x(0) = 0$$ Separate variables and integrate: $$\int_{0}^{x} \frac{d\xi}{(p-\xi)^2} = \int_{0}^{t} \alpha d\tau$$ Now perform the integration: $$\left[-\frac{1}{p-x}\right]_{0}^{x} = \alpha t$$ Solve for \(x\): $$x = \frac{pt}{1+\alpha pt}$$ Hence, we have found the concentration of substance \(X\) at any time \(t\) for the case \(p = q\), given by \(x(t) = \frac{pt}{1+\alpha pt}\).

Key concepts

Second-order Chemical Reaction

In the realm of chemical kinetics, a second-order reaction involves two reactant molecules colliding to form a product. Specifically, this exercise discusses a reaction involving substances \( P \) and \( Q \) colliding to produce a new substance \( X \). Such reactions are crucial in understanding chemical processes where the rate of reaction depends on the concentration of both reactants.

This type of reaction can be represented by the formula \( P + Q \rightarrow X \), indicating that one molecule each of \( P \) and \( Q \) yields one molecule of \( X \). Importantly, the rate of reaction is governed by the concentrations of \( P \) and \( Q \) at any time \( t \), articulated mathematically as \( \frac{d x}{d t} = \alpha (p-x)(q-x) \), where \( \alpha \) is a positive constant. Here, \( p \) and \( q \) are initial concentrations of \( P \) and \( Q \), and \( x(t) \) denotes the concentration of \( X \) at time \( t \).

Understanding these dynamics is fundamental in chemical engineering and reaction design, as control and prediction of product formation is key to efficiency and economic feasibility.

Initial Value Problem

An initial value problem (IVP) in differential equations addresses finding a specific solution given a set of initial conditions. Our task involves solving the differential equation: \[ \frac{dx}{dt} = \alpha (p-x)(q-x), \quad x(0) = 0 \] where \( x(0) = 0 \) specifies that initially, there is no \( X \) present.

This initial condition allows the formulation of a particular solution applicable to this situation.

• The differential equation is separable, which means it can be rearranged into a form that enables direct integration.
• Solving this kind of IVP is essential for modeling situations in various fields, allowing for precise predictions based on initial circumstances.
• The successfully derived solution, as per the steps detailed in the exercise, provides a model of how \( X \) changes over time based on the initial conditions set by \( p \) and \( q \).

Note that such modeling is a critical concept not only in chemistry but also in physics, biology, and engineering.

Separation of Variables

Separation of Variables is a powerful mathematical technique used to solve differential equations. In this exercise, we use it to handle the rates of reaction. The goal is to rearrange the equation such that all terms involving the dependent variable \( x \) are on one side, and all terms involving \( t \) are on the other.

In our case the equation \( \frac{dx}{dt} = \alpha (p-x)(q-x) \) is manipulated into \[ \frac{dx}{(p-x)(q-x)} = \alpha dt \] which separates \( x \) and \( t \). Through integration of both sides, the solution emerges, revealing how to express \( x \) as a function of \( t \).

• This ultimately leads to a formula that provides \( x(t) \), the concentration of the product \( X \) at any given time.


• This technique is a staple in solving not only chemical reaction equations but also in fluid dynamics and other time-evolution equations of physical systems.

With mastery of this method, one can solve vast classes of differential equations.

Limiting Values

Limiting values in the context of this reaction refer to the concentrations that reactants and products approach as time progresses toward infinity. As \( t \rightarrow \infty \), reactions are expected to slow down due to the decreased availability of reactants. Essentially, the reaction rate diminishes as \( \frac{dx}{dt} \rightarrow 0 \).

In this specific scenario, the examination reveals that the limiting value of \( x(t) \)—the concentration of \( X \)—reaches the smaller of \( p \) or \( q \). This is determined by setting \( \frac{dx}{dt} = 0 \) and solving for \( x \). Because \( x \) cannot exceed the initial concentrations of \( P \) or \( Q \), it converges to \( \min(p,q) \).

Understanding limiting values is not only essential in theoretical analysis but also practical in determining the maximum yield of a reaction.

• In chemistry, this helps calculate the efficiency and feasibility of reactions.
• In broader scientific contexts, it assists in predicting long-term behavior of dynamic systems.
• Such insights are invaluable in fields requiring precise control over process outcomes.