91Ó°ÊÓ

Chemical reactions: In a second-order reaction, one molecule of a substance \(A\) collides with one molecule of a substance \(B\) to produce a new substance, the product. If \(t\) denotes time and \(x=x(t)\) denotes the concentration of the product, then its rate of change \(\frac{d x}{d t}\) is called the rate of reaction. Suppose the initial concentration of \(A\) is \(a\) and the initial concentration of \(B\) is \(b\). Then, assuming a constant temperature, \(x\) satisfies the equation of change $$ \frac{d x}{d t}=k(a-x)(b-x) $$ for some constant \(k\). This is because the rate of reaction is proportional both to the amount of \(A\) that remains untransformed and to the amount of \(B\) that remains untransformed. Here we study a reaction between isobutyl bromide and sodium ethoxide in which \(k=0.0055, a=51\), and \(b=76\). The concentrations are in moles per cubic meter, and time is in seconds. 10 a. Write the equation of change for the reaction between isobutyl bromide and sodium ethoxide. b. Make a graph of \(\frac{d x}{d t}\) versus \(x\). Include a span of \(x=0\) to \(x=100\). c. Explain what can be expected to happen to the concentration of the product if the initial concentration of the product is 0 .

Step by step solution

01

Write the Equation of Change

Using the given values of \( k = 0.0055 \), \( a = 51 \), and \( b = 76 \), we substitute these into the equation of change for the reaction: \( \frac{d x}{d t} = 0.0055(51-x)(76-x) \). This equation describes how the concentration of the product \( x \) changes over time based on the concentrations of reactants \( A \) and \( B \).

02

Explore the Behavior of the Reaction Rate as a Function of Concentration

To understand the relationship, we plot \( \frac{d x}{d t} \) versus \( x \) for values of \( x \) ranging from 0 to 100, using the equation from Step 1. This graph will be part of a more complex nonlinear curve, showing how the rate of change depends intricately on the current concentration \( x \).

03

Analyze the Graph to Predict Product Concentration Behavior

The graph of \( \frac{d x}{d t} \) versus \( x \) shows that initially, when \( x = 0 \), the rate of reaction is calculated as \( \frac{d x}{d t} = 0.0055 \times 51 \times 76 = 21.39 \). This means the concentration of the product will initially increase rapidly. As \( x \) increases, the growth rate \( \frac{d x}{d t} \) decreases, indicating that the creation of the product will slow down as the reactants are consumed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second-Order Reaction

In chemistry, a second-order reaction is one where the rate of reaction depends on the product of the concentrations of two reactants. In our exercise, this means that the formation of the new product occurs with the collision of molecules from two different substances, namely substance \( A \) and substance \( B \).
This type of reaction is described by the equation \( \frac{d x}{d t} = k(a-x)(b-x) \), where:

• \( k \) is the rate constant, a measure of how quickly the reaction proceeds.
• \( a \) and \( b \) are the initial concentrations of substances \( A \) and \( B \), respectively.
• \( x \) represents the concentration of the product over time \( t \).

By understanding this equation, we can predict how fast or slow a reaction will occur based on the initial amounts and ongoing changes in concentrations of the reactants.

Concentration

Concentration refers to the amount of a substance present in a certain volume of solution, often measured in moles per liter (mol/L). In our problem, the concentrations given for isobutyl bromide and sodium ethoxide are 51 and 76 moles per cubic meter respectively. These initial values directly affect the rate and outcome of the reaction.
As the reaction takes place, the concentrations of \( A \) and \( B \) decrease because they are transformed into the product \( x \). This change in concentration is central to how the reaction progresses. Lower concentrations of reactants over time lead to a decreased rate of reaction, hence why the speed of product formation slows as the reaction proceeds.

Rate of Reaction

The rate of reaction describes how quickly or slowly a reaction occurs. It is mathematically represented as \( \frac{d x}{d t} \), which is the change in concentration of the product \( x \) with respect to time \( t \).
This rate depends on the ongoing concentrations of \( A \) and \( B \). Initially, when both \( A \) and \( B \) are abundant, the rate of reaction is higher. As the reaction continues and these concentrations decrease, the rate decreases as well. This is illustrated in the exercise by plotting \( \frac{d x}{d t} \) against \( x \), showing an initially steep rate that tapers off over time as reactants are consumed.

Equation of Change

The equation of change for a chemical reaction mathematically describes how the concentration of a substance alters over time due to the reaction. For the second-order reaction in our exercise, the equation \( \frac{d x}{d t} = 0.0055(51-x)(76-x) \) tells us precisely how fast the concentration of the product \( x \) changes as the reaction progresses.
This change is an interplay of the remaining reactants; as more of \( A \) and \( B \) are transformed into the product, their decreasing concentrations mean the rate at which \( x \) increases also slows. The rate constant \( k = 0.0055 \) highlights the specific reaction's inherent speed under the given conditions. Understanding this equation allows predictions about how concentrations evolve, letting us foresee how fast the product will form and when the reactants will be depleted.