Problem 12 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

In a chemical reaction of the form$$2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{A}_{2} \mathrm{~B} $$, where the reaction does not involve intermediate compounds, the reaction rate is proportional to [A] \(^{2}\) [B] where [A] and [B] denote, respectively, the concentrations of the components A and B. Let \(a\) and \(b\) denote \([\mathrm{A}]\) and \([\mathrm{B}]\), respectively, and assume that \([\mathrm{B}]\) is much greater than \([\mathrm{A}]\) so that \((b(t) \gg a(t) .\) The rate at which \(a\) changes may be written $$a^{\prime}(t)=-k(a(t))^{2} b(t)=-K(a(t))^{2}$$ We have assumed that \(b(t)\) is (almost) constant because [A] is the limiting concentration of the reaction. Let $$a(t)=\frac{a_{0}}{1-a_{0} K t}$$ Show that \(\quad a(0)=a_{0}\). Use the Definition of Derivative 3.22 , $$a^{\prime}(t)=\lim _{b \rightarrow t} \frac{a(b)-a(t)}{b-t}$$ to compute \(a^{\prime}(t)\). Then compute \((a(t))^{2}\) and show that $$a^{\prime}(t)=-K(a(t))^{2}$$

Step-by-Step Solution

Answer

The derivative matches the given reaction rate, confirming \(a'(t) = -K (a(t))^2\). The initial condition \(a(0) = a_0\) is satisfied.

Step 1: Verify Initial Condition

To show that \(a(0)=a_0\), substitute \(t = 0\) into the expression for \(a(t)\). We have:\[a(0) = \frac{a_0}{1 - a_0 K \cdot 0} = \frac{a_0}{1} = a_0.\] Thus, the condition \(a(0) = a_0\) is satisfied.

Step 2: Definition and Setup of Derivative

According to Definition 3.22, the derivative \(a'(t)\) is:\[a'(t) = \lim_{b \rightarrow t} \frac{a(b) - a(t)}{b - t}.\]Substitute \(a(t) = \frac{a_0}{1-a_0 K t}\) into the expression:\[a(b) = \frac{a_0}{1-a_0 K b}.\]Therefore, substitute into the derivative formula:\[a'(t) = \lim_{b \rightarrow t} \frac{\frac{a_0}{1-a_0 K b} - \frac{a_0}{1-a_0 K t}}{b-t}.\]

Step 3: Simplify the Derivative Expression

Combine fractions in the numerator:\[\frac{a_0(1-a_0 K t) - a_0(1-a_0 K b)}{(1-a_0 K b)(1-a_0 K t)}.\]Simplify the numerator:\[a_0 - a_0^2 K t - a_0 + a_0^2 K b = a_0^2 K(b - t).\]Thus, the expression becomes:\[a'(t) = \lim_{b \rightarrow t} \frac{a_0^2 K (b - t)}{(b - t)(1-a_0 K b)(1-a_0 K t)}.\]

Step 4: Evaluate the Limit

Cancel \((b - t)\) from numerator and denominator:\[a'(t) = \lim_{b \rightarrow t} \frac{a_0^2 K}{(1-a_0 K b)(1-a_0 K t)}.\]Take the limit as \(b \rightarrow t\):\[a'(t) = \frac{a_0^2 K}{(1-a_0 K t)^2}.\]

Step 5: Compute \\(a(t)^2\\)

Compute \(a(t)^2\):\[a(t)^2 = \left(\frac{a_0}{1-a_0 K t}\right)^2 = \frac{a_0^2}{(1-a_0 K t)^2}.\]

Step 6: Verification of Reaction Rate

Now verify whether \(a'(t) = -K (a(t))^2\):We previously found that:\[a'(t) = \frac{a_0^2 K}{(1-a_0 K t)^2},\]and \(a(t)^2 = \frac{a_0^2}{(1-a_0 K t)^2}.\)Therefore:\[-K(a(t))^2 = -K \frac{a_0^2}{(1-a_0 K t)^2}.\]These match, thus confirming:\[a'(t) = -K (a(t))^2.\]

Key Concepts

Chemical Kinetics

Chemical kinetics is an area of chemistry that looks at the speed of chemical reactions and the factors influencing this speed. For the reaction rate equation given by \[2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{A}_{2} \mathrm{~B}\], it's crucial to understand that the rate of reaction is determined by the concentrations of the reactants. The rate is proportional to \([A]^2[B]\), where \([A]\) and \([B]\) are the respective concentrations of species A and B.

In simpler terms, chemical kinetics tells us how fast a reaction proceeds based on how much of each substance we have. This helps us predict how different situations will change a reaction, like alterations in temperature or concentration.

• Higher concentrations generally lead to faster reactions.
• Temperature can also affect reaction rates.
• Catalysts are substances that can speed up or slow down reactions without being consumed.

Understanding these principles helps chemists control and optimize reactions for desired outcomes.

Differential Equations

Differential equations are mathematical expressions involving derivatives, which represent how a function changes. In chemistry, they are widely used to model the changes in concentrations of reactants over time during a reaction.

In the given exercise, the differential equation for a'(t) is:\[a^{\prime}(t)=-K(a(t))^{2}\]. This equation helps us understand how the concentration of reactant A decreases over time due to the reaction.

Differential equations can be complex, but they are powerful tools. They allow us to:

• Predict the behavior of reactants in a chemical reaction.
• Understand the relationship between concentrations and time.
• Solve for unknown functions given initial conditions like a(0) = a_0.

These equations are essential for making precise predictions in chemical systems.

Derivative Definition

The derivative of a function provides the rate at which the function's value changes as its input changes. In the context of the exercise, the derivative a'(t) represents how the concentration of reactant A changes over time.

Using the formal definition of the derivative:\[a'(t) = \lim_{b \rightarrow t} \frac{a(b) - a(t)}{b - t}\],we can compute a'(t) precisely. This step is crucial in verifying the proposed expression for \(a(t)\) given in the exercise.

Derivatives are foundational in calculus and play a significant role in various scientific and engineering disciplines. In chemical kinetics:

• They describe how quickly reactions proceed.
• They help in adjusting variables to control reaction rates.
• They provide insights into the stability and dynamics of chemical systems.

Limiting Reactant

The concept of a limiting reactant is fundamental in chemistry and plays a key role in understanding chemical reactions. A limiting reactant is the one that is entirely consumed first, dictating the maximum amount of product that can be formed.

In this exercise, A is considered the limiting reactant since we assume \(b(t) \gg a(t)\). This means B is in excess, and the reaction's progress is determined by how much A is available. As A is depleted, the reaction rate decreases.

Key features of limiting reactants include:

• They limit the extent of the reaction.
• Once used up, the reaction stops, and no more product can be formed.
• Knowing the limiting reactant helps in calculating theoretical yields of products.

Overall, understanding the limiting reactant concept is essential for predicting the outcomes and efficiency of chemical reactions.