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Overview
Each chapter is self-contained and features an introduction that identifies its basic goals, their significance, and a general plan for their achievement. This text's important aims are to demonstrate that the basic kinetic principles are essential to the solution of modern chemical problems, and to show how the underlying question — "How do chemical reactions occur?" — leads to exciting, vibrant fields of modern research. The first aim is achieved by using relevant examples in presenting the basic material, and the second is attained by inclusion of chapters on surface processes, photochemistry, and reaction dynamics.
Product Details
| ISBN-13: | 9780486131696 |
|---|---|
| Publisher: | Dover Publications |
| Publication date: | 09/12/2012 |
| Series: | Dover Books on Chemistry |
| Sold by: | Barnes & Noble |
| Format: | eBook |
| Pages: | 352 |
| File size: | 12 MB |
| Note: | This product may take a few minutes to download. |
Read an Excerpt
Chemical Kinetics and Reaction Dynamics
By Paul L. Houston
DOVER PUBLICATION, INC.
Copyright © 2001 Paul L. HoustonAll rights reserved.
ISBN: 978-0-486-13169-6
CHAPTER 1
Kinetic Theory of Gases
Chapter Outline
1.1 Introduction
1.2 Pressure of an Ideal Gas
1.3 Temperature and Energy
1.4 Distributions, Mean Values, and Distribution Functions
1.5 The Maxwell Distribution of Speeds
1.6 Energy Distributions
1.7 Collisions: Mean Free Path and Collision Number
1.8 Summary
Appendix 1.1 The Functional Form of the Velocity Distribution
Appendix 1.2 Spherical Coordinates
Appendix 1.3 The Error Function and Co-Error Function
Appendix 1.4 The Center-of-Mass Frame
1.1 INTRODUCTION
The overall objective of this chapter is to understand macroscopic properties such as pressure and temperature on a microscopic level. We will find that the pressure of an ideal gas can be understood by applying Newton's law to the microscopic motion of the molecules making up the gas and that a comparison between the Newtonian prediction and the ideal gas law can provide a function that describes the distribution of molecular velocities. This distribution function can in turn be used to learn about the frequency of molecular collisions. Since molecules can react only as fast as they collide with one another, the collision frequency provides an upper limit on the reaction rate.
The outline of the discussion is as follows. By applying Newton's laws to the molecular motion we will find that the product of the pressure and the volume is proportional to the average of the square of the molecular velocity, <v2>, or equivalently to the average molecular translational energy [member of]. In order for this result to be consistent with the observed ideal gas law, the temperature T of the gas must also be proportional to <v2> or <[member of]>. We will then consider in detail how to determine the average of the square of the velocity from a distribution of velocities, and we will use the proportionality of T with <v2> to determine the Maxwell-Boltzmann distribution of speeds. This distribution, F(v) dv, tells us the number of molecules with speeds between v and v + dv. The speed distribution is closely related to the distribution of molecular energies, G([member of]) d[member of]. Finally, we will use the velocity distribution to calculate the number of collisions Z that a molecule makes with other molecules in the gas per unit time. Since in later chapters we will argue that a reaction between two molecules requires that they collide, the collision rate Z provides an upper limit to the rate of a reaction. A related quantity λ is the average distance a molecule travels between collisions or the mean free path.
The history of the kinetic theory of gases is a checkered one, and serves to dispel the impression that science always proceeds along a straight and logical path. In 1662 Boyle found that for a specified quantity of gas held at a fixed temperature the product of the pressure and the volume was a constant. Daniel Bernoulli derived this law in 1738 by applying Newton's equations of motion to the molecules comprising the gas, but his work appears to have been ignored for more than a century. A school teacher in Bombay, India, named John James Waterston submitted a paper to the Royal Society in 1845 outlining many of the concepts that underlie our current understanding of gases. His paper was rejected as "nothing but nonsense, unfit even for reading before the Society." Bernoulli's contribution was rediscovered in 1859, and several decades later in 1892, after Joule (1848) and Clausius (1857) had put forth similar ideas, Lord Rayleigh found Waterston's manuscript in the Royal Society archives. It was subsequently published in Philosophical Transactions. Maxwell (Illustrations of Dynamical Theory of Gases, 1859— 1860) and Boltzmann (Vor-lesungen über Gastheorie, 1896—1898) expanded the theory into its current form.
1.2 PRESSURE OF AN IDEAL GAS
We start with the basic premise that the pressure exerted by a gas on the wall of a container is due to collisions of molecules with the wall. Since the number of molecules in the container is large, the number colliding with the wall per unit time is large enough so that fluctuations in the pressure due to the individual collisions are immeasurably small in comparison to the total pressure. The first step in the calculation is to apply Newton's laws to the molecules to show that the product of the pressure and the volume is proportional to the average of the square of the molecular velocity, <v2>.
Consider molecules with a velocity component vx in the x direction and a mass m. Let the molecules strike a wall of area A located in the z-y plane, as shown in Figure 1.1. We would first like to know how many molecules strike the wall in a time Δt, where Δt is short compared to the time between molecular collisions. The distance along the x axis that a molecule travels in the time Δt is simply vxΔt, so that all molecules located in the volume AvxΔt and moving toward the wall will strike it. Let n* be the number of molecules per unit volume. Since one half of the molecules will be moving toward the wall in the +x direction while the other half will be moving in the — x direction, the number of molecules which will strike the wall in the time Δt is ½n*AυxΔt
The force on the wall due to the collision of a molecule with the wall is given by Newton's law: F = ma = m dv/dt = d(mv)/dt, and integration yields FΔt = Δ(mv). If a molecule rebounds elastically (without losing energy) when it hits the wall, its momentum is changed from +mvx to —mvx, so that the total momentum change is Δ (mv) = 2mvx. Consequently, FΔt = 2mvx for one molecular collision, and FΔt = (½n*Aυx Δt)(2mυx) for the total number of collisions. Canceling Δt from both sides and recognizing that the pressure is the force per unit area, p = F/A, we obtain p = n*mv2x
All the molecules in the box that are moving toward the z-y plane will strike the wall.
Of course, not all molecules will be traveling with the same velocity vx. We will learn below how to characterize the distribution of molecular velocities, but for now let us simply assume that the pressure will be proportional to the average of the square of the velocity in the x direction, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. The total velocity of an individual molecule most likely contains other components along y and z. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] where [??], [??], and k are unit vectors in the x, y, and z directions, respectively, <[υ2]> = <[υ2x> + <[υ 2x> + <[υ2x> and <[υ2]> = <[υsup.2.sub.x]> + <[υ 2x> + <[υ 2x>. In an isotropic gas the motion of the molecules is random, so there is no reason for the velocity in one particular direction to differ from that in any other direction. Consequently, <[υ.sup.2.sub.x]> = <[υ.sup.2.sub.y]> = <[υ.sup.2.sub.z]> = <[υ2]>/3 When we combine this result with the calculation above for the pressure, we obtain
p = 1/3n*m<υ2>
(1.1)
Of course, n* in equation 1.1 is the number of molecules per unit volume and can be rewritten as nNA/V, where NA is Avogadro's number and n is the number of moles. The result is
pV = 1/3nNAm<υ2
(1.2)
Since the average kinetic energy of the molecules is <[member of]> = ½m<υ2> another way to write equation 1.2 is
pV = 2/3nNA<[member of]>
(1.3)
Equations 1.2 and 1.3 bear a close resemblance to the ideal gas law, pV = nRT. The ideal gas law tells us that the product of p and V will be constant if the temperature is constant, while equations 1.2 and 1.3 tell us that the product will be constant if <v2> or <[member of]> is constant. The physical basis for the constancy of pV with <v2> or <[member of]> is clear from our previous discussion. If the volume is increased while the number, energy, and velocity of the molecules remain constant, then a longer time will be required for the molecules to reach the walls; there will thus be fewer collisions in a given time, and the pressure will decrease. To identify equation 1.3 with the ideal gas law, we need to consider in more detail the relationship between temperature and energy.
1.3 TEMPERATURE AND ENERGY
Consider two types of molecule in contact with one another. Let the average energy of the first type be [<[member of]>1] and that of the second type be [<[member of]>2] . If [<[member of]>1] is greater than [<[member of]>2], then when molecules of type 1 collide with those of type 2, energy will be transferred from the former to the latter. This energy transfer is a form of heat flow. From a macroscopic point of view, as heat flows the temperature of a system of the type 1 molecules will decrease, while that of the type 2 molecules will increase. Only when [<[member of]>1] = [<[member of]>2] will the temperatures of the two macroscopic systems be the same. In mathematical terms, we see that T1= T2 when [<[member of]>1] = [<[member of]>2] and that T1 > T2 when [<[member of]>1] > [<[member of]>2]. Consequently, there must be a correspondence between <[member of]> and T so that the latter is some function of the former: T = T(<[member of]>).
The functional form of the dependence of T on <[member of]> cannot be determined solely from kinetic theory, since the temperature scale can be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR). Experimentally, this corresponds to measuring the temperature either by measuring the volume of an ideal gas held at constant pressure or by measuring the pressure of an ideal gas held at constant volume. Division of both sides of equation 1.3 by nR and use of the ideal gas relation gives us the result
T = pV/nR = 2/3 NA/R<[member of]>.
(1.4)
<[member of]> = 3/2kT
(1.5)
where k, known as Boltzmann's constant, is defined as R/NA. Note that since <[member of]> = ½m<[υ2]>.
<[[upsilon.sup.2]> = 3kT/m.
(1.6)
example 1.1
Calculation of Average Energies and Squared Velocities
Objective
Calculate the average molecular energy, <[member of]>, and the average squared velocity, <v2>, for a nitrogen molecule at T = 300 K.
Method
Use equations 1.5 and 1.6 with m = (28 g/mole)(1 kg/1000 g)/ (NA molecule/mole) and k = 1.38 × 10—23 J/K.
Solution
<e> = 3kT/2 = 3(1.38 × 10—23 J/K)(300 K)/2 = 6.21 × 10—21 J. <[υ.sup.2] = 3kT/m
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
To summarize the discussion so far, we have seen from equation 1.2 that pV is proportional to <v2> and that the ideal gas law is obtained if we take the definition of temperature to be that embodied in equation 1.5. Since <[member of]> = ½m<[υ.sub.2]>, both temperature and pV are proportional to the average of the square of the velocity. The use of an average recognizes that not all the molecules will be moving with the same velocity. In the next few sections we consider the distribution] of molecular speeds. But first we must consider what we mean by a distribution.
1.4 DISTRIBUTIONS, MEAN VALUES, AND DISTRIBUTION FUNCTIONS
Suppose that five students take a chemistry examination for which the possible grades are integers in the range from 0 to 100. Let their scores be S1 = 68, S2 = 76, S3 = 83, S4 = 91, and S5 = 97. The average score for the examination is then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
where NT = 5 is the number of students. In this case, the average is easily calculated to be 83.
Now suppose that the class had 500 students rather than 5. Of course, the average grade could be calculated in a manner similar to that in equation 1.7 with an index i running from 1 to NT = 500. However, another method will be instructive. Clearly, if the examination is still graded to one-point accuracy, it is certain that more than one student will receive the same score. Suppose that, instead of summing over the students, represented by the index i in equation 1.7, we form the average by summing over the scores themselves, which range in integer possibilities from j = 0 to 100. In this case, to obtain the average, we must weight each score Sj by the number of students who obtained that score, Nj:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.8)
Note that the definition of Nj requires that ΣNj = NT. The factor 1/NT in equation 1.8 is included for normalization, since, for example, if all the students happened to get the same score Sj = S then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.9)
Now let us define the probability of obtaining score Sj as the fraction of students receiving that score:
Pj = Nj/NT
(1.10)
Then another way to write equation 1.8 is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.11)
where ΣjPj = 1 from normalization.
Equation 1.11 provides an alternative to equation 1.7 for finding the average score for the class. Furthermore, we can generalize equation 1.11 to provide a method for finding the average of any quantity,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.12)
where Pj is the probability of finding the jth result.
example 1.2
Calculating Averages from Probabilities
Objective
Find the average throw for a pair of dice.
Method
Each die is independent, so the average of the sum of the throws will be twice the average of the throw for one die. Use equation 1.12 to find the average throw for one die.
Solution
The probability for each of the six outcomes, 1—6, is the same, namely, 1/6. Factoring this out of the sum gives <T> = (1/6) ΣTi, where Ti = 1,2,3,4,5,6 for i = 1—6. The sum is 21, so that the average throw for one die is <T> = 21/6 = 3.5. For the sum of two dice, the average would thus be 7.
The method can be extended to calculate more complicated averages. Let f(Qj) be some arbitrary function of the observation Qj. Then the average value of the function f(Q) is given by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.13)
For example, if Q were the square of a score, then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.14)
Suppose now that the examination is a very good one, indeed, and that the talented instructor can grade it not just to one-point accuracy (a remarkable achievement in itself!) but to an accuracy of dS, where dS is a very small fraction of a point. Let P(S) dS be the probability that a score will fall in the range between S and S + dS, and let dS become infinitesimally small. The fundamental theorems of calculus tell us that we can convert the sum in equation 1.11 to the integral
= ∫ P(S)S dS,
(1.15)
or, more generally for any observable quantity,
= ∫ P(Q)Q dQ
(1.16)
Equation 1.16 will form the basis for much of our further work. The probability function P(Q) is sometimes called a distribution function, and the range of the integral is over all values of Q where the probability is nonzero. Note that normalization of the probability requires
∫ P(Q)dQ = 1.
(1.17)
The quantity |ψ(x)|2dx is simply a specific example of a distribution function. Although knowledge of quantum mechanics is not necessary to solve it, you may recognize a connection to the particle in the box in Problem 1.7, which like Example 1.3 is an exercise with distribution functions.
example 1.3
Determining Distribution Functions
Objective
Bees like honey. A sphere of radius r0 is coated with honey and hanging in a tree. Bees are attracted to the honey such that the average number of bees per unit volume is given by Kr—5, where K is a constant and r is the distance from the center of the sphere. Derive the normalized distribution function for the bees. They can be at any distance from the honey, but they cannot be inside the sphere. Using this distribution, calculate the average distance of a bee from the center of the sphere.
Method
First we need to find the normalization constant K by applying equation 1.17, recalling that we have a three-dimensional problem and that in spherical coordinates the volume element for a problem that does not depend on the angles is 4πr2 dr. Then, to evaluate the average, we apply equation 1.16.
Solution
Recall that, by hypothesis, there is no probability for the bees being at r < r0, so that the range of integration is from r0 to infinity. To determine K we require
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.18)
or
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.19)
so that
K = r20/2π
Having determined the normalization constant, we now calculate the average distance:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(1.20)
1.5 THE MAXWELL DISTRIBUTION OF SPEEDS
We turn now to the distribution of molecular speeds. We will denote the probability of finding vx in the range from vx to vx + dvx, vy in the range from vy to vy + dvy, and vz in the range from vz to vz + dvz by F(vx, vy, vz) dvx dvy dvz. The object of this section is to determine the function F(vx, vy, vz)]. There are four main points in the derivation:
1. In each direction, the velocity distribution must be an even function of v.
2. The velocity distribution in any particular direction is independent from and uncorrelated with the distributions in orthogonal directions.
3. The average of the square of the velocity <v2> obtained using the distribution function should agree with the value required by the ideal gas law: <v2> = 3kT/m.
4. The three-dimensional velocity distribution depends only on the magnitude of v (i.e., the speed) and not on the direction.
We now examine these four points in detail.
1.5.1 The Velocity Distribution Must Be an Even Function of v]
Consider the velocities vx of molecules contained in a box. The number of molecules moving in the positive x direction must be equal to the number of molecules moving in the negative x direction. This conclusion is easily seen by examining the consequences of the contrary assumption. If the number of molecules moving in each direction were not the same, then the pressure on one side of the box would be greater than on the other. Aside from violating experimental evidence that the pressure is the same wherever it is measured in a closed system, our common observation is that the box does not spontaneously move in either the positive or negative x direction, as would be likely if the pressures were substantially different. We conclude that the distribution function for the velocity in the x direction, or more generally in any arbitrary direction, must be symmetric; i.e., F(vx) = F(—vx). Functions possessing the property that f(x) = f(—x) are called even functions, while those having the property thai f(x) = —f(—x) are called odd functions. We can ensure that F(υx) be an even function by requiring that the distribution function depend on the square of the velocity: F(υx) = f(υ2x) As shown in Section 1.5.3, this condition is also in accord with the Boltzmann distribution law.
(Continues...)
Excerpted from Chemical Kinetics and Reaction Dynamics by Paul L. Houston. Copyright © 2001 Paul L. Houston. Excerpted by permission of DOVER PUBLICATION, INC..
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Table of Contents
Contents
Title Page,Bibliographical Note,
Copyright Page,
Dedication,
Preface,
Introduction - A User's Guide to Chemical Kinetics and Reaction Dynamics,
Errata,
1 - Kinetic Theory of Gases,
2 - The Rates of Chemical Reactions,
3 - Theories of Chemical Reactions,
4 - Transport Properties,
5 - Reactions in Liquid Solutions,
6 - Reactions at Solid Surfaces,
7 - Photochemistry,
8 - Molecular Reaction Dynamics,
Answers and Solutions to Selected Problems,
Index,