Problem 49 - Calculus for Biology and Medicine

Question

The reaction rate \(R(x)\) of the irreversible reaction $$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}$$ is a function of the concentration \(x\) of the product \(\mathrm{AB}\) and is given by $$R(x)=k(a-x)(b-x)$$ where \(k\) is a constant, \(a\) is the concentration of \(\mathrm{A}\) at the beginning of the reaction, and \(b\) is the concentration of \(\mathrm{B}\) at the beginning of the reaction. Express the percentage error of the reaction rate. \(100 \frac{\Delta R}{R}\), as a function of the percentage error of the concentration \(x, 100 \frac{\Delta x}{x}\)

Step-by-Step Solution

Answer

The percentage error of reaction rate is \( -100 \frac{(a+b-2x)}{(a-x)(b-x)} \frac{\Delta x}{x} \).

Step 1: Understand the given formula

The reaction rate is given by the expression \( R(x) = k(a-x)(b-x) \). Here, \(a\) and \(b\) are initial concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) respectively, and \( k \) is a constant. \( x \) is the concentration of the product \( \mathrm{AB} \).

Step 2: Express rate change in terms of concentration change

The change in reaction rate \( \Delta R \) can be expressed in terms of the change in concentration \( \Delta x \) using the derivative: \( \Delta R \approx \frac{dR}{dx}\Delta x \).

Step 3: Compute the derivative \( \frac{dR}{dx} \)

First, differentiate \( R(x) = k(a-x)(b-x) \) with respect to \( x \): \[ \frac{dR}{dx} = k \left[ (b-x)(-1) + (a-x)(-1) \right] = k \left[ -a - b + 2x \right] = -k(a+b-2x). \]

Step 4: Relate \( \Delta R \) and \( \Delta x \) using the derivative

Substitute the expression of \( \frac{dR}{dx} \) from Step 3: \( \Delta R \approx -k(a+b-2x) \Delta x \).

Step 5: Express percentage error of reaction rate

The percentage error of the reaction rate is \( 100 \frac{\Delta R}{R} \). Substitute \( \Delta R \) from Step 4: \[ 100 \frac{\Delta R}{R} \approx 100 \frac{-k(a+b-2x) \Delta x}{k(a-x)(b-x)}. \]

Step 6: Simplify and relate to percentage error of concentration

Simplify the expression from Step 5: \[ 100 \frac{\Delta R}{R} = 100 \frac{-(a+b-2x) \Delta x}{(a-x)(b-x)}. \] Now, introduce the percentage error of concentration \( 100 \frac{\Delta x}{x} \), and the relation becomes: \[ 100 \frac{\Delta R}{R} = -100 \frac{(a+b-2x)}{x} \frac{\Delta x}{(a-x)(b-x)} \].

Step 7: Final expression for percentage error

Combine steps to express the percentage error in terms of the errors: \[ 100 \frac{\Delta R}{R} \approx -100 \frac{(a+b-2x)}{(a-x)(b-x)} \times 100 \frac{\Delta x}{x} \]. This relates the percentage error in reaction rate to the percentage error in concentration \( x \).

Key Concepts

Differentiation

Differentiation is a core concept in calculus used to find the rate at which a function changes at any given point. When applied to the reaction rate, differentiation helps in determining how a small change in concentration affects the reaction rate. In our exercise, you might wonder why it’s needed: it’s because calculating the derivative of the reaction rate, \( R(x) \), allows us to express how the rate changes with respect to changes in concentration, \( x \).
To differentiate the equation \( R(x) = k(a-x)(b-x) \) with respect to \( x \), we apply the product rule. This rule is pivotal when dealing with products of functions in differentiation. The result of this differentiation gives us the derivative \( \frac{dR}{dx} = -k(a+b-2x) \).
Understanding this derivative is important because it indicates whether the reaction rate increases or decreases with changes in concentration. By connecting these ideas, you're using differentiation to predict the dynamic behavior of chemical reactions under different conditions.

Percentage Error

When working with measurements and calculations, understanding the concept of percentage error is critical. Percentage error quantifies the difference between an approximate or measured value and an exact or known value, expressed as a percentage. In chemistry and our problem, it helps evaluate the accuracy of the concentration measurements and the calculated reaction rates.
In our exercise, we're particularly interested in the percentage error of the reaction rate \( 100 \frac{\Delta R}{R} \), and how it relates to the percentage error in concentration \( x \), which is \( 100 \frac{\Delta x}{x} \). By deriving an expression linking these two percentage errors, you gain insight into how much the uncertainty in concentration affects the perceived accuracy of the reaction rate.
The expression \[ 100 \frac{\Delta R}{R} = -100 \frac{(a+b-2x)}{(a-x)(b-x)} \times 100 \frac{\Delta x}{x} \] accurately provides this relationship. This equation illustrates a direct proportionality between errors, highlighting that understanding and reducing errors in measurements can lead to significantly more reliable and accurate reaction rate data.

Concentration

Concentration is a fundamental concept in chemistry, referring to the amount of a substance in a given volume. It’s essential for predicting and understanding the behavior and rate of chemical reactions. In the given reaction, \( \mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB} \), the concentrations \( a \), \( b \), and \( x \) represent the initial concentrations of reactants and the concentration of the product, respectively.
In this exercise, the rate function depends on these concentrations, making it crucial to understand how varying \( x \) – the concentration of the product \( \mathrm{AB} \) – directly influences the reaction rate.
Concentration plays a significant role not just in affecting reaction rates but also in determining the system's dynamic state at any moment. Increasing or decreasing the concentration, especially of the reactants or products, can shift the equilibrium position of the reaction, affecting the yield and efficiency.
Thus, maintaining accurate concentration measurements ensures the precision of predictions for the reaction behavior, highlighted in the relationship between percentage errors for reaction rate and concentration changes, which is derived in the solution.