Problem 19 - Calculus for Biology and Medicine

Question

Consider the chemical reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} $$ If \(x(t)\) denotes the concentration of \(\mathrm{AB}\) at time \(t\), and \(k\) is the rate constant for of the reaction, explain why: $$ \frac{d x}{d t}=k(a-x)(b-x) $$ where \(k\) is a positive constant and \(a\) and \(b\) denote the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), respectively, at time 0 .

Step-by-Step Solution

Answer

The rate equation \( \frac{d x}{d t} = k(a-x)(b-x) \) describes how the concentration of \( \mathrm{AB} \) increases by using available concentrations of \( \mathrm{A} \) and \( \mathrm{B} \), with \( k \) as a positive constant.

Step 1: Understand the Reaction

The reaction equation is \( \mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB} \). This reaction is a simple combination reaction where substances \( \mathrm{A} \) and \( \mathrm{B} \) combine to form the product \( \mathrm{AB} \). Let \( x(t) \) be the concentration of \( \mathrm{AB} \) at time \( t \).

Step 2: Initial Concentrations

At time \( t = 0 \), \( a \) and \( b \) represent the initial concentrations of reactants \( \mathrm{A} \) and \( \mathrm{B} \), respectively. Since no reaction has occurred at \( t = 0 \), the concentration of \( \mathrm{AB} \) is zero, i.e., \( x(0) = 0 \).

Step 3: Concentration Changes Over Time

As the reaction proceeds, the concentration of \( \mathrm{A} \) decreases by \( x(t) \), and the concentration of \( \mathrm{B} \) also decreases by \( x(t) \). Therefore, the concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) at any time \( t \) are \( a-x(t) \) and \( b-x(t) \), respectively.

Step 4: Derive the Rate Law

The rate of formation of \( \mathrm{AB} \) is proportional to the concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) that are available to react. This is expressed as \( \frac{d x}{d t} = k (a-x)(b-x) \), where \( k \) is the rate constant and it is positive, reflecting that the reaction proceeds forward.

Step 5: Confirm the Relationship

In chemical kinetics, the rate of a reaction is often expressed as a differential equation representing changes in reactant or product concentrations over time. The equation \( \frac{d x}{d t} = k(a-x)(b-x) \) follows this pattern, indicating how the concentration of \( \mathrm{AB} \) changes based on the remaining concentrations of \( \mathrm{A} \) and \( \mathrm{B} \).

Key Concepts

Differential Equations in Chemistry

In the realm of chemical kinetics, differential equations are essential in modeling the rate of reaction processes. Specifically, they describe how the concentration of a reactant or product changes over time. Let's consider the reaction \( \mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB} \). The differential equation utilized here is \( \frac{d x}{d t} = k(a-x)(b-x) \), representing how the concentration of \( \mathrm{AB} \), denoted as \( x(t) \), changes relative to time.

The equation can be broken down as follows:

• \( \frac{d x}{d t} \) is the rate at which \( \mathrm{AB} \) forms.
• The term \( (a-x) \) indicates the concentration of \( \mathrm{A} \) left unreacted at time \( t \).
• Similarly, \( (b-x) \) signifies the concentration of \( \mathrm{B} \) that remains at time \( t \).

Thus, the rate at which \( \mathrm{AB} \) forms is proportionate to the product of the remaining concentrations of \( \mathrm{A} \) and \( \mathrm{B} \). In this differential equation, the negative sign appears due to the consumption of reactants as the reaction transpires.

Differential equations, like the one above, provide a systematic method to predict how a reaction will progress over time by defining relationships between the rate of change of concentrations and the present concentrations of reactants. Indeed, these equations serve as pivotal tools in translating the dynamics of a chemical reaction into a quantifiable form, allowing for a deeper understanding and analysis.

Rate Constant

The rate constant, denoted as \( k \), plays a critical role in chemical kinetics. It provides a measure of the speed of a reaction at a given temperature. In our differential equation, \( \frac{d x}{d t} = k(a-x)(b-x) \), the value of \( k \) influences how quickly \( \mathrm{AB} \) is formed.

Here are some insights about \( k \):

• **Magnitude and Units**: The rate constant's units depend on the reaction's order. For a second order reaction like this one, where two reactants are involved, the units for \( k \) are \( \, \mathrm{M}^{-1} \mathrm{s}^{-1} \).
• **Factors Affecting \( k \)**: Primarily, temperature greatly influences the rate constant. Typically, as temperature increases, so does \( k \), correlating with the Arrhenius equation \( k = A \exp\left(-\frac{E_a}{RT}\right) \) where \( A \) is the pre-exponential factor and \( E_a \) is the activation energy.
• **Role in Rate Law**: The constant \( k \) determines the proportionality between the rate of reaction and the product of the reactant concentrations, effectively scaling the rate equation.

With \( k \) being just as integral as the concentration terms in the rate equation, a thorough understanding of this parameter helps in controlling and optimizing reaction rates in practical applications.

Concentration Changes Over Time

Understanding how concentrations change during a chemical reaction is crucial. The equation \( \frac{d x}{d t} = k(a-x)(b-x) \) models how the concentration of product \( \mathrm{AB} \) evolves as reactants \( \mathrm{A} \) and \( \mathrm{B} \) are consumed over time.

Key points to consider include:

• **Initial Concentrations**: At \( t = 0 \), the concentration of \( \mathrm{AB} \) is zero as the reaction has just begun. Therefore, \( x(0) = 0 \). As time progresses, \( x(t) \) increases, tracking the formation of product as \( \mathrm{A} \) and \( \mathrm{B} \) react.
• **Reactant Depletion**: The concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) decrease as products form, described by \( a-x(t) \) and \( b-x(t) \) respectively. These expressions track how much of each reactant remains unreacted at any point in time.
• **Dynamic Equilibrium**: If the reaction were reversible, changing concentrations would affect equilibrium positions. However, in our scenario, we focus solely on the forward reaction.

By using a differential equation to map these changes, we can predict the time-dependent pathway of product accumulation. This information is vital in understanding reaction dynamics and rates, and it aids chemists and engineers in designing systems where precise reactant and product management is essential.