/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 In a general second-order chemic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 31

In a general second-order chemical reaction, chemicals \(A\) and \(B\) (the reactants) combine to form chemical \(C\) (the product). If the initial concentrations of the reactants \(A\) and \(B\) are \(a\) and \(b,\) respectively, then the concentration \(x(t)\) of the product satisfies the equation \(x^{\prime}(t)=[a-x(t)][b-x(t)] .\) What is the rate of change of the product when \(x(t)=a\) ? At this value, is the concentration of the product increasing, decreasing or staying the same? Assuming that \(a

Short Answer

Expert verified
The rate of change of the product is zero when \(x(t)=a\), implying the concentration stays constant at this point. Further, given conditions \(a<b\) and absence of product at start of the reaction, the maximum concentration of the product is \(x(t)=a\).

Step by step solution

01

Interpret the Differential Equation

We are given the differential equation \(x^{\prime}(t)=[a-x(t)][b-x(t)]\). This represents the rate of change of the product \(C\) with respect to time. When \(x(t)=a\), the rate of change of the product \(C\) becomes \(x^{\prime}(t)=[a-a][b-a]=0\).
02

Analyze the Rate of Change

Since the derivative \(x^{\prime}(t)\) is zero when \(x(t)=a\), this signifies that there is no change in concentration at \(x(t)=a\). That's because a zero rate of change implies the function is neither increasing nor decreasing at that particular point. Hence, the concentration of the product remains constant when \(x(t)=a\).
03

Find the Maximum Concentration of Product

Given that \(a<b\) and there is no product present when the reaction starts, the maximum concentration \(x(t)\) that can be achieved is \(a\). This is because \(a\) represents the initial concentration of the reactant \(A\), and since this reaction is second-order, the reactants \(A\) and \(B\) are consumed at the same rate. Therefore, once all of \(A\) has been used up (and turned into the product \(C\)), the product production will stop, hence why the maximum concentration of \(C\) is \(a\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Second-Order Reaction
A second-order reaction often involves two reactant molecules coming together to form a product. In general, this reaction can be simplified to a scenario where reactants \(A\) and \(B\) interact to produce a product \(C\). The rate at which the reaction proceeds depends on the concentration of both reactants.

In our case, the differential equation given is \(x^{\prime}(t)=[a-x(t)][b-x(t)]\). It tells us how the concentration of product \(C\) changes over time. The equation shows that the rate of change \(x^{\prime}(t)\) is influenced by how much of \(A\) and \(B\) has already reacted to form \(C\).
  • If both \(A\) and \(B\) are completely consumed, the reaction stops, and the concentration of \(C\) remains constant.
  • Second-order reactions ensure that both reactants have equal opportunity to participate in the reaction, which indicates a dependency on both initial concentrations \(a\) and \(b\).
Differential Equations in Chemistry
Differential equations are vital tools in chemistry, especially when modeling how concentrations of substances change over time. In the context of the given equation, \(x^{\prime}(t)=[a-x(t)][b-x(t)]\), it describes the rate at which the product \(C\) forms from reactants \(A\) and \(B\).

These equations are important because they enable chemists to predict how quickly a reaction will proceed, and they help identify points of interest in a reaction, like when it reaches equilibrium. In our example, when \(x(t) = a\), the derivative \(x^{\prime}(t) = 0\). This means that at this point, no further product is being formed.
  • The concentration metric described by the differential equation is dynamic, changing as reactants convert to products.
  • Differential equations help elucidate the point at which all reactants have been used up, indicating a maximum product concentration.
Reaction Rate Analysis
Analyzing reaction rates is a fundamental aspect of understanding chemical kinetics. The rate of a chemical reaction reveals how quickly or slowly a product is formed from reactants.

In our case, we are given the rate of change of product \(C\) as \(x^{\prime}(t)=[a-x(t)][b-x(t)]\). When \(x(t) = a\), the rate is zero since \(x^{\prime}(t) = [a-a][b-a] = 0\), meaning the formation of the product has stopped.
  • Conducting a rate analysis helps determine the point at which the concentration reaches its maximum; in this example, it's when \(x(t) = a\).
  • Understanding reaction rates allows for a deeper insight into the efficiency and speed of a reaction, which is critical in industrial and laboratory settings.

By knowing the reaction rate, chemists can adjust conditions such as temperature and pressure to optimize the speed and yield of product formation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Newton's theory of gravitation states that the weight of a person at elevation \(x\) feet above sea level is \(W(x)=P R^{2} /(R+x)^{2}\) where \(P\) is the person's weight at sea level and \(R\) is the radius of the earth (approximately 20,900,000 feet). Find the linear approximation of \(W(x)\) at \(x=0 .\) Use the linear approximation to estimate the elevation required to reduce the weight of a 120 -pound person by \(1 \%\)

State the domain for \(\sin ^{-1} x\) and determine where it is increasing and decreasing.

In the figure, a beam is clamped at its left end and simply supported at its right end. \(A\) force \(P\) (called an axial load) is applied at the right end. GRAPH CANT COPY If enough force is applied, the beam will buckle. Obviously, it is important for engineers to be able to compute this buckling load. For certain types of beams, the buckling load is the smallest positive solution of the equation \(\tan \sqrt{x}=\sqrt{x} .\) The shape of the buckled beam is given by \(y=\sqrt{L}-\sqrt{L} x-\sqrt{L} \cos \sqrt{L} x+\sin \sqrt{L} x,\) where \(L\) is the buckling load. Find \(L\) and the shape of the buckled beam.

The human cough is intended to increase the flow of air to the lungs, by dislodging any particles blocking the windpipe and changing the radius of the pipe. Suppose a windpipe under no pressure has radius \(r_{0} .\) The velocity of air through the windpipe at radius \(r\) is approximately \(V(r)=c r^{2}\left(r_{0}-r\right)\) for some constant \(c .\) Find the radius that maximizes the velocity of air through the windpipe. Does this mean the windpipe expands or contracts?

Use a CAS to determine the range of \(x\) 's in exercise 54 for which the approximation is accurate to within \(0.01 .\) That is, find \(x\) such that \(\left|e^{x}-(1+x)\right|<0.01\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.