This document outlines the key concepts and topics covered in a course on advanced physical chemistry. It discusses the following: - The overall objective of the course is for students to learn the theoretical bases for understanding physicochemical phenomena in molecular terms and their relationship to thermodynamic descriptions. - The course will cover kinetics, including basic concepts, analysis of complex reactions, approximate methods, and reactions in solution. It will also cover catalysis. - The topics are organized into 5 main sections, with some adjustments to time allocation and content compared to a previous version of the course. Key concepts in each section will be explored over multiple classes.

1. FISICOQUIMICA AVANZADA
Maestría en Química
“El alumno aprenderá las bases teóricas para la comprensión de los
fenómenos fisicoquímicos en términos moleculares y su relación con la
descripción termodinámica”. Objetivo General del Curso
Alfonso Enrique Ramírez Sanabria, Dr.
aramirez@unicauca.edu.co

2. II. Cinética Química -ideal-
1. Los conceptos básicos en cinética química (6 horas)
a. La definición de la velocidad en una reacción química
b. El orden y la molecularidad de una reacción
c. Las leyes de velocidad para las reacciones elementales
d. La determinación del orden de la reacción
e. La dependencia con la temperatura de las velocidades de reacción (Arrhenius)
2. El análisis de las reacciones complejas (8 horas)
a. La reacción reversible de uno o varios pasos
b. Las reacciones consecutivas de primer orden con dos o más pasos
c. Las reacciones paralelas
3. Algunos métodos aproximados (2 horas)
a. El método del estado estacionario
b. El método de pseudo primer orden
4. Las reacciones en solución (4 horas)
a. Las propiedades generales
b. La teoría fenomenológica de las velocidades de reacción
c. Las constantes de velocidad controladas por la difusión
d. El efecto de la fuerza iónica en las reacciones entre los iones
e. Las relaciones lineales para la energía libre
f. Los métodos para medir las velocidades de reacción
- El mezclado manual
- El flujo detenido (stopped flow)
- El método de la relajación para las reacciones rápidas
5. La catálisis (4 horas)
a. La catálisis y el equilibrio
b. La catálisis homogénea
c. La autocatálisis y las reacciones oscilantes
d. La catálisis heterogénea

3. II. Cinética Química -actual-
1. Los conceptos básicos en cinética química (1ra parte)
a. La definición de la velocidad en una reacción química
b. El orden y la molecularidad de una reacción
c. Las leyes de velocidad para las reacciones elementales
d. La determinación del orden de la reacción
e. La dependencia con la temperatura de las velocidades de reacción (Arrhenius)
2. El análisis de las reacciones complejas (2da parte)
a. Las reacciones de otros ordenes
b. Las reacciones consecutivas de primer orden con dos o más pasos
c. Las reacciones paralelas
d. Las reacciones reversibles
3. Algunos métodos aproximados (2da parte)
a. Vida 1/2
b. Velocidades iniciales
c. Exceso de uno de los reactivos
d. Método logaritmico
e. Técnicas de “Relajación”
f. “Trazamiento”
g. Efecto Isotópico Cinético
4. Las reacciones en solución (2da parte)
a. Generalidades
b. Las constantes de velocidad controladas por la difusión
c. Efectos sobre las reacciones
d. Las relaciones lineales para la energía libre
5. La catálisis (3ra parte)
a. La catálisis y el equilibrio
b. La autocatálisis
c. La catálisis homogénea
d. La catálisis heterogénea

4. The Tools of the Trade:
Reflexión Inicial
Mathematical Concepts
No human investigation can be called real science if it cannot be demonstrated
mathematically.
Leonardo da Vinci (1452-1519)
This chapter will review the fundamental mathematical concepts (algebra and
trigonometry) needed for a quantitative understanding of college-level chemistry and
physics. Virtually all of this material is covered in high-school mathematics classes,
but often the connection to real scientific applications is not obvious in those classes.
In contrast, the examples used here will frequently involve chemical and physical con-
cepts that will be covered in detail in later chapters or in the later parts of a standard
freshman chemistry book. Here they will be treated as math problems; later you will
see the underlying chemistry.

5. 1.2. History o f th
Chemical kinetics is
Historia
The fist book on the kin
appeared in 1884. If coun
kinetics is about 100 year
chemical reactions was st
F. Wilhelmy published th
he established for the firs
General ideas of chemical kinetics
cal reaction of cane-sugar
d x / d t -- k( l - x)/l (1.1) tants involved in the trans
• 1850, L.F. Wilhelmy. “The lawlimiting value of the amount of the transformed substances.
where dx/dt is the esterification rate, k isaction on Cane-sugar”
stance, and ! is the of acid
constant, x is the amount of the reacted stating sub-
-dZ/dT = MZS, where T is
They studied in detail the influence of conditions (temperature, solvent) on the
of the acid, and S is const
reaction occurrence. One of the main kinetic laws, the law of mass action, was for- was expressed in this equ
• 1862, M. Berthelot y L. Pean. “Studies General ideas of chemical kinetics M. Berthelot and L. Pean
of esterification reaction
mulated by Sweden scientists, mathematician C. M. Guldberg and chemist P. Waage
in the series of works in 1864:-67. Based on the results of M. Berthelot and L. Pean
between acatic acid an ethanol” owningreat direction anddthedreversiblelaw- ofx)/l in action for ification reaction(1.1)
de Saint Gilles and their
both the reaction occurring one
work, they formulated the
x / t -- k( l mass
reaction the equi-
betwee
does not go to the end an
where dx/dt The the was derived in the rate, kformconstant, x is the amount of the reacted stating sub-
librium state. is law esterification general is for the reaction with any num-
stance,reactants,is the limiting value of on the concept of molecular collisions substances. process. It had
ber of and ! and the derivation was based the amount of the transformed reversible
• 1864-7, C.M. Guldberg an event preceding the reaction of collided particles. For the reaction ofley de
as y P. Waage. Formulacion de la the type
They studied in detail Bthe influence of conditions (temperature, solvent) on the
aA + b + gC | Products
acción de masas reaction occurrence. One rate vthe [A] a [B] bkinetic laws, the (1.2) of mass action, was for-
Reaction of = k main [C] g law
mulated by gSweden scientists, mathematician C. reaction.
where a, b, and are stoichiometric coefficients of reactants entered into the M. Guldberg and chemist P. Waage
• 1865-7, V. Hartcourt ytransformation" of worksinin somewhat 1879.Based Harcourt andofW. Esson M. Berthelot and L. Pean
in W.series formulated this1864:-67. of idea onthe "rate chemical
The Esson. Oxidation The oxalic results of
the law was
was introduced
form in
earlier by V.
of the acid with
de Saint Gilles and their own great work, they formulated the law of mass action for
permanganate (1865+67). They studied the oxidation of oxalic acid with potassium permanganate
both pioneered in deriving formulas for in description of the kinetics of reactions of
and the reaction occurring the one direction and the reversible reaction in the equi-
librium and second orders. was derived in the general form for the reaction with any num-
the first state. The law
Our compatriot N. A. Menshutkin made a great contribution to thethe concept of molecular
development
• 1877, N. A. Menshutkin. event precedingtheand hydrolysis of hydrolysis the reaction ofcollisions
ber of reactants, and
Formationthederivation of collided on was based
of the kinetics. In 1877 he studied in detail the reaction of formation and esters
asofan from various acids and alcohols and was the first to formulate the problem
reaction particles. For the type
esters
from varoius acids and the later when heofstudiedreactivity of reactantsbtonm their chemical discoveredFive
of alcoholsthe
years
dependence
the hydrolysis of t e r -B y l gC | he
aA + a + acetate, Products and
structure.
described the autocatalysis phenomenon (acetic acid formed in ester hydrolysis
accelerates the hydrolysis). In 1887+90, studying rate v = kof quaternary b [C] g
Reaction the formation [A] a [B] ammo- (1.2)
• 1884, J. Van´t Hoff. “Etudes de dynamique chimique”
nium salts from amines and alkyl halides, he found a strong influence of the solvent
on the rate of this reaction (Menschutkin reaction) and stated the problem of study-
where a, b, and g are the reaction rate in acoefficients ofN. A. Menschutkin
ing the medium effect on stoichiometric solution. In 1888 reactants entered into the reaction.
introduced the term "chemical kinetics" in his monograph "Outlines of Development
of The law was formulated in this form in 1879. The idea of the
Chemical Views." "rate of chemical
transformation" was introduced somewhat earlier by 1884 Harcourt and W. Esson
The book by J. Van't Hoff" Etudes de dynamique chimique" published in V.
(1865+67). They studied the oxidation of the author generalized
was an important scientific event in chemistry. In this book,oxalic acid with potassium permanganate
data on kinetic studies and considered the kinetic laws of monomolecular and bimol-
and pioneered in deriving formulas for the occurrence of reactions kinetics of reactions of
ecular transformations, the influence of the medium on the description of the

6. H3C CH 3 H CH 3
For this reaction to occur, there must be rotation around the double bond
It is frequentlyissuchobserved that reactions that alead to lead towhen the atomic p–orbitals no energy
It is frequently to observed that thatlead to lower a loweraoverall overall energyoverall
observed that reactions that reactions is that lead to a lower
It frequently observed reactions broken lower
It is frequently an extent that the p–bond that overall energy energy
Introducción
state as productsas are formed readily. readily.readily. However,are alsoare also there are also reaction
state as productsproducts are formed formed takeHowever, readily. However,
state as longer products are taketake placeHowever, there there
are overlap.
state formed take place place place there are also
many reactions thatmany lead to lead to lead inwill decrease in energy,of is theofratesthe essential idea
reactions that casesto energy,discussedyet theyetsections, of the observed that
a be energy, ratesIt rates
many reactions that a decreasedecrease in yet the in later the
Although other a
frequently
the
many reactions low.are low. For toof energy mustofbe populated as gaseousgaseous rates of the take pla
that leadheat theformation of waterinwater from as yet the are formed
aheatthe formation of energy,from occurs. This
decrease fromstate a reaction
is that For the example, of heat formation of water products
reactions
reactions are low. are example,of higher
reactions For a stateexample, gaseous
H 2 is O2 À285 but the
reactionsOareÀ285andÀ285iskJ=mol,kJ=mol, thethe reaction formation 1.8. water situation lead to a decre
H2 and H2 is low. For but the reactionbut heat shown in Figure of Such a from gaseous
and is 2illustrated example, reaction
O kJ=mol, by the energy diagram of
1 1
many reactions that
1that the Boltzmann Distribution Law may pro-
À285 kJ=mol, 2but2g)the ( g)O(l )H2O(l (1:1) (1:1) (1:1)
H2 and O2 isshouldH2( g) þH2Og)( þsuggestO(l ) 2Hreaction)reactions are low. For example, the he
immediatelyg) !g) ( !
2
( 2 H2 (OH O 2 !
2þ
2
vide a basis for the explanation because that law governs the population of
takes of unequalunless if In all, unless the reactionais by figure, ]z denotes the
takes place verystates place at all, slowly, the reaction reaction is initiatedthe a spark.a [spark.
takes place very if very ifenergy.unless the is initiated by H initiated by
slowly, slowly, at all, at the case illustrated in
1 a great deal of energy is released as HisOÀ285 kJ=mol, but the r
spark.
The reason for this is that althoughdeal ofO2 ( the ! H2 O(l or the
2 and O2 2
this a that þ is called g) transition state
The reason for this isreason for state,(which greatenergy is released released)as H2 O activated complex. (1:1)
H2 g) a
The that although is greatalthough deal of energy is as H2 O
high-energy
forms, there is no low there is pathwayenergy 2for the for follow. In orderIn order order
forms, there is energyofenergy pathwayreaction reaction to follow. follow. In
Hformacion
forms, no low no low for thepathway to the reaction to
The height form, moleculesbarrier overO must the reactants must pass on the
the energy which
for water towater to form, moleculesand H22 andH2 react, and theirreact, and their bond
for form, water to of H2 of O of O2and 2react, and their bond
for molecules must must bond 1
takes place about 435 and 435 and435 and 490unless respectively.
energies energies are about 490 kJ=mol, kJ=mol,kJ=mol, the reaction is initiated by a spark.
are very slowly, if 490 respectively.
energies are about at all, respectively. H2 ( g) þ O2 (
Thermodynamics is concerned withconcerned with thechange betweenbetween
2
Thermodynamics is concerned with theenergy overall energy change between
Thermodynamics is the overall
the
for this is that although aoverallnecessary, of change can result
energy change
The reasoninitial and final states for a process.process. If change canthis energy is released as H2 O
the states and process. If necessary, great deal result
the initial and final initial for afinal states for a If necessary, this change can result
this
takes place very slowly, if at all, unles
after there after anAccordingly, thermodynamics does] +not dealreaction
forms, an after anisinfinitelow energy pathway [for the not deal withtowith
infinite time. infiniteAccordingly, thermodynamics does with not deal follow. In order
no time. time. Accordingly, thermodynamics does
+
for shows that to that of thatmolecules the offavors andproductionthe production and their favorecido
Termodinámicamente
The reason for this is that although a g
the subject ofat rates, notleast at least not directly. The preceding example
reaction rates, not directly. The preceding example
the subject of reaction reaction least at directly. The preceding example
the subject rates,
form, the of the of of H2 the O2 must react,
water the shows thermodynamics reaction reactionreaction the production of of
shows thermodynamics thermodynamics
the the favors favors of bond
water; however, kinetically unfavorable. unfavorable. Cinéticamente desfavorecido
forms, there is no low energy pathwa
energieshowever, kinetically the and 490the process ischemicalhereseeWe There is the
water; water; about 435 process process is unfavorable. We the see here
are however, kinetically the is principles of We see kinetics. the no
kJ=mol, respectively. here
first of first of first of several importantchemical kinetics.kinetics. is no is no
several important principles of
several important principles of chemical There There
for water to form, molecules of H2
Thermodynamics isbetween thermodynamics andakineticskineticsenergy change between
necessary correlation between thermodynamics and kinetics of overallchemical
necessary correlation concerned with the and of a of a chemical
necessary correlation between thermodynamics a chemical
E
Energy
energies are about 435 and 490 kJ=m
reaction. reactions that arethat are energetically place verytakevery very
Some are energetically favorable take favorable
reactions energetically favorable take place place
thereaction.reaction.reactions states lowpathwayenergy pathwayreaction reactionreaction can
initialSomebecausebecauseis thereforenergy pathway bythenecessary, this change can result
and Someno that energy noalow by whichIfwhich the canthe can
final low process.
slowly because there is there no is
slowly slowly by which
Thermodynamics is concerned wit
after an occur. occur.time. Accordingly, thermodynamics does not deal with
occur. infinite
One of the ofOne of the observationsstudy ofstudy of reaction that a is rates is that a
the regarding ∆E study of reaction that
regarding Reactants the rates rates
therate cannotOneobservationsfromcalculatedfirst leastreactionnot developedThe apreceding example
of reaction rates, fromthe principles. isis not developed
the initial and final states for a proces
subject be ratethe observationsfrom at principles. TheoryTheory Productsdeveloped
regarding
rate cannot cannot be first principles. first
calculated
be calculated
not directly.is not
Theory is
after an infinite time. Accordingly, t
shows that where itthermodynamics calculate how fast most reactions will the production of
to the point the where it is possiblepossible to fast most reactions will reactions will
to the point is possible to it is to of the reaction most
to the point where calculate how calculate how fast favors
take place. Fortake For some very gas phase reactions, it is reactions,to is possible to
take place. place. For some very simple gas reactions, it is possible to
some very simple simple gas phase phase possible it
the subject of reaction rates, at least n
water; however, kineticallyreactionfast the reactioncoordinate but details details see here the
calculatecalculate approximately how fast the reactiontake place,unfavorable. We
approximately how fast the the should should take place, place, but
calculate approximately how process isshould details
Reaction but take
first of several important principles ofprofile for a chemical reaction. There is no
FIGURE 1.8 The energy chemical kinetics.
shows that the thermodynamics of the
1 1 1
necessary correlation between thermodynamics and kinetics of a chemical the proce
water; however, kinetically
reaction. Some reactions that are energetically favorable take place very

7. RATES OF REACTIONS interpret the results. The (moles=liter sec, mb
some species withconcentration divideddimensions of the rate must
of time. Therefore, the by time
f kinetics that will enable him or her to
of theof concentration divided by numerical analysis as sec, moles=liter min,
available software to perform can be written is a
reaction that time (moles=liter
ue that isanot addressed in this book.
reaction that can be written as
rate of chemical reaction is expressed as a change in concentration of A!B
Velocidad de una Reacción
species with time. Therefore, the dimensions of the rate must be those
A!B
ES OF REACTIONS by a rate (moles=liter expressed either min, etc.). A
ncentration dividedhas time that can be sec, moles=liter in terms of the
ion that a rate that can be expressedof B. Because the the disappearance
has can be written appearance either in terms of concentration of
the as
a chemicalappearanceconsumed,changetheconcentration of Àd[A]=dt. Becaus
the reaction is expressed as a the rate is expressed as of A is decreasing
of B. Because in concentration
A ! the
es with time. Therefore,B isdimensions of B astime, the those is expressed (1:2)
the increasing with rate must be rate as þd[
consumed, the rate is expressed Àd[A]=dt. Because the concentr
ation divided by time (moles=liter sec, moles=liter concentrations and time is cal
rate is min, etc.). A
ical time, the relatingexpressedofas þd[B]=dt. The ma
B is increasing with equation in terms of the disappearance of A or
at can that can be expressed either
rate be written as Fundamental Concepts Kinetics 3
ppearance of B. Because the concentration of Abetween the concentr
the rate law. The relationships
ical equation relating concentrations and timeisisdecreasing asrate is
1.0
called the A equ
A!
the rate law. The0.9 B areas Àd[A]=dt. Because(1:2) in Figure 1.1A of a
umed, the rate is expressed represented graphically concentrationfor
time relationships between the concentrations of and
0.8 B
the
hat can be expressed either in rate iso expressed Figure 1.10:050 minÀ1 .
time are time,which [A]of the1.00 M and k ¼ orfor a first-order rea
the terms is disappearance of A
ncreasing with represented graphically in as þd[B]=dt. The mathemat-
0.7
0.6
nce of B. Because the1.00 M and k ¼ 0:050 minÀ1 .A is can be shown as
concentration of A is decreasing asthat
quation relating is M 0.5If we consider time is called the rate equation or
which [A]o concentrations and a reaction
the rate is expressed as Àd[A]=dt. Because the concentration of
0.4
If werelationshipsreaction that can be shown as A and B with
consider a between the concentrations of
atewith time, the rate is expressed as þd[B]=dt. The mathemat-bB ! cC þ dD
ng law. The 0.3
0.2 A
aA þ
n relating concentrations0.1 time is FigurebB rate equationdD
are represented graphically in aA þ the ! cCaþ or
and called 1.1 for first-order reaction in
The relationships between0 ¼ 0:050 minÀ1 .30
h [A]o is 1.00 M and 0kthe concentrations of A and B with
10 20 40 50
presented graphically in Figure 1.1 for be Time, min as
we consider a reaction that can a first-orderB for the reactionin ! B.
shown reaction A
À1
FIGURE 1.1 Change in concentration of A and
is 1.00 M and k ¼ 0:050 min .
aA þ bB !
nsider a reaction that can be shown as cC þterms of a constant times some
the rate law will usually be represented in
dD (1:3)

8. 0.3
0.2 2A þ B ! as þd[B]=dt. The mathemat-
is increasing with time, the rate is expressed Products A (1:5)
O equation relating concentrations and time is called 5'' therate 2O out or
al in see an analogy to this in the following illustration that involves flow equation
3''
0.1 1'' the H of
water, and let us also suppose that the reaction takes place in steps that can be
0
e rate law. The relationships between the concentrations of A and B with
0
written as 10 20 30 40 50
me are represented graphically inTime, min de for a first-order reaction in
Ecuación ó Ley BÀ11.1 (slow) H2O out ! C Velocidad
Figure
H2O in 3'' 1'' A þ 5'' (1:6)
hich [A]FIGURE 1.1ofChange k ¼ 0:050 min and. B for the reaction A ! B. of short pip
study the rate M and in concentration ofthrough this system
o is 1.00
flow of water A
mation consider aobtainedthatofcanþthe flow assystem of short pipes, a 1" p
! Products
If we If we study the rate of flow water A shown of(fast)
will be reaction about bethrough this water through (1:7)
C
he 3" and law willwill be obtainednormally þ ofofwatermuch resistance limits
the information amount of not about theintermediate) that is present 1"sometime to flow
rate 5" pipes dobe representedan flow
The usually C (known as in terms a constant times any
offer as through a at pipe (1:3)
function thetherate 5" pipes doþ bB and B, Noteitthat theresistance Eqs. (1.6) and (1.7)
aA of ! cC dD
since ofthe and of the overall reaction. offer as can usually be written in
3" concentrationsnot A normally and much sum of to flow as
he 1" pipe.givespipe. overall reaction that was shown of Eq. kinetics, the also that thethe
the does the Therefore, in in the language in chemical kinetics,
form 1" the Therefore,
the language of chemical (1.5). Note 1"
epresents the rate-determining¼step.xreaction of one molecule1of A and 1one of B.
formation of C depends onstep.the
pipe represents the rate-determining k[A] [B]y
Rate (1:4)
That process chemical have a that can be written [A] and [B] . There-
Suppose we have awill likely reactionrate that depends onas
pose we x and yaare the exponents on the thatequation involvesand molecules of
where have even though thereaction concentrations of A two B,
fore, chemical balanced overall can be written as
respectively.A,In this rate law, k is þ B ! one molecule ofand As aexponents
2A
called
Products (1:5)
the slow step involves onlythe rate constant A. the result, formation of
x and ylet products followsthatratethe! Products form RateA and B, and the (1
and are called the 2Aa þ Blaw that takes place in steps that k[A][B],
us also supposeorder of reaction is of therespect to ¼ can be
the reaction with
respectively. As will second-order (first-order exponents x and y in B). or should be
written reaction is be described later, the in A and first-order may It
as
t usmay notsuppose thatas the Awrite! Crate law directly from the balanced equation
also be the same we can þreaction takes and b in in steps that can
apparent that the balancing coefficients a place Eq. (1.3).
B
the
(slow) (1:6)
only if thethe reaction is the sum ofsingle step. If thexreaction takes place in a
reaction takes place in a the exponents and y. Thus,
n as The overallseries of steps, aCrateAstudy will give information about steps up to and
order of
þ ! Products (fast) (1:7)
we speak of a second-order reaction, a third-order reaction, etc., when the
sumThe amount of C the in the an intermediate) rate is presentbe any timeThese that step.
of the exponents slowestrate law is 2,that law will at determined by
including step, and the 3, etc.,
(known as B ! C (slow) respectively. limits
Aþ
exponents can the overall reaction. Note that the sum of Eqs. (1.6) and (1.7)
usually be established by studying the reaction using differ-
(1
the rate of
ent gives the overall reaction that was shown inthis is done, it isalso that the
initial concentrations of A and B. When Eq. (1.5). Note possible to
determine 1.2doubling the A ! ProductsON oftheand one of B.
C þ concentration one (fast)
formation ofDEPENDENCE OF RATESmolecule A rate of the
if C depends on the reaction of of A doubles (1

9. er be initial as
ally, one must measure time, which is not usually a problem unlessd[A] o
Rate ¼ k[A]1 ¼ À
the [A]
theshown concentration [A] at time zero to [A] at the later
(1:9)
reaction is a very rapid one.
be shown as the initial concentration [A] at time zero
It may be possible for the concentration of a reactant or product to be
dt [A]o o 0
be shown as ð
n candetermined directly within the reaction mixture, but inFirst-Order
1.2.1 to give
be written asequation can be rearranged other cases a sample
This
[A] ð t
First-Order When the integration is performed, we k ðFundamental ð d[A] ¼ obtain
[A] t
Reacciones de 1 Orden er [A] ¼ k dtð d[A]
must be removed for the analysis to be completed. The time necessary to
A ! the d[A] À
remove a sample from B reaction mixture is usually anegligibly (1:8) be written as d[A] dt [A]
À ¼ k dtshort
Suppose reaction can À (1:10)
compared to the reaction time being measured. What [A] is usually done for
[A]
ion follows a rateout in solution is formup the reaction in a vessel that is
a reaction carried law of the to set
[A]o [A]A!B
0
[A] [A] ¼
À k
held in a constant temperature bath so that fluctuations in but it should be integrated between the o
Equation (1.10) can be integrated temperature will o [A]o 0
ose a reaction can beWhen the integration is performed, welog
written as ln ¼ kt or
limits the rate of d[A] time equal to t while the concentration varies the [A]o obtain ¼
not cause changes in of time ¼ 0 reaction. and thatreaction is started,follows a rate law of fromform
Then the the reaction
The When the integrationfirst-order rateobtainare in2:30
1
the concentration ¼ À units on k in the is performed, we law
the and
and Rate ¼ k[A] of the reactant (A in this case) is determined at [A] (1:9)
the that a graph of dt versus time can be zero to [A] at the later time. This can
initial concentration [A]o at time made or the
[A] term
hand side of Eq. !When[A] ¼ kt or logperformed, (1:8)
selected times so ln[A] A (1.12) has integration is [A] ¼ k we ob
B ln the [concentration]=[concen d[A]
data analyzed be shown asa linear
o Rate ¼ k[A]1 ¼ À o
n be rearrangednumerically. If reactionrelationship provides the bestGraphical
If the equation involving[A] ¼ kt logarithms ¼ conside
to give
data, it is concluded that the obeys a first-order rate law.
fit to the
[A]natural or log [A]
o [A] is 2:303 t
dt
o k
hat the reaction follows a cancel. However, the right-hand side o
the units to rate law of the form [A] ln
an initial ð
representation of this rate law is shown in Figure 1.2ðford[A] can be rearranged to give
t t
d[A] This equation concen- [A] [A] [A] 2:303 [
À in the form equation involving natural logarithms isktconsidered, it
tration of A of 1.00 M and kdt 0:020 minÀ1 . InÀ case, the slope(1:10)
¼k ¼ If the this ¼ kof thedt ln o (1:11)
¼ or log
Fundamental Concepts of Kinetics
dimensionally correct onlynaturalhas the units of time
If the equation involving if k logarithms is considered, i
line is Àk, so the kinetic data can be used to determine k[A]
[A]
in the form
graphically or by
means of linear regression using numerical methods[A]odetermine the 0
to 1 d[A]
slope
À
d[A]
[A]¼ k dt
[A]
have nok[A] If ¼ Àcan lnln [A]o Àbut [A]natural integrated b
will kt Thethe ¼performed,(1.10) equation involving[A] ktbe logarithm
When the integration is on k in the we obtaindtrate law are in terms of¼ kt . The left
Rate formunits.
an beofintegrated but it should be integrated between the
the line. in units first-order [A]
the ln time (1:9)
À1
of Equation be oÀ
and time equal to t while the concentration varies from integrated ln it should which cause
hand side[A] Eq. (1.12) has [concentration]=[concentration], ¼
0.0 The equation ofin the[A]and time equal tolnt [A] the ktconcentrationb
ortime ¼ 0 ¼
form k t
ln to o limits time. This ocanright-hand side of ¼ equation will
¼ kt However, the ln [A]o À
or the units thethe initial concentration [A] at time zero(1:12) at the later tim
tration [A]o can be rearrangedcancel.
at time zero to or at[A] to give
[A] later log while the
quation –0.5 [A] 2:303o to [A]
dimensionally correct only if k has the units of timeÀ1lnbecause only[A]
or be shown as
If the equation involving natural logarithms is considered, it
Slope = –k ¼ln [A] À [A] À
ln [A] ¼canln [A]oktÀo kt ln the
,
be written
ln [A][A] ln [A]o À kt
–1.0
[A] the form ð
ð in d[A]
willt kt have no units. ln [A] ð ¼ o t
ln [A]
–1.5d[A] À
The equation ¼ k dt
or ¼bþ o ð
ln [A] ¼ d[A] mx À kt
ln [A] (1:10)
À
[A]
[A] o ln thektform y À b ¼mxdt
can also dt writtenÀin [A] (1:11) ln [A]o¼ kt þ (1:13)
¼ k be ln [A] ¼
ln [A] ¼ À
yy ¼ [A] mx ln [A] ¼ ln [A]
k (1:15
–2.0
[A]oor 0
bþ o
ion (1.10) –2.5 be can also
can integrated must butremembered that [A]o o , the 0 Àkt
It be written in it should be integrated initial concentra
be the form [A] between þ mx the
0 we obtain40 be 60remembered that [A] , , the initial concentr
When remembered that [A]o we the
80 90 [A]
70 value so o is a ¼ (1:14)
10 mustIt 50 fixed [A] ¼ integration is performed, [A]o e Eq. (1.14) ca
[A]Àkt o obtain y ¼ b
tion is performed, It 20 30somemustlnbe theln 100 it À kt constant. Therefore,initial con
of time ¼ 0 and time equal to yt ¼ bso mx [A]adismuyeoexponencialmente convaries from
Time, min while the concentration tiempo
¼
some fixed1:00so it¼ isitaisconstant. Therefore,elEq. Eq. (1:16
valuek þ min .[A][A] e Therefore,k (1.14) ca
constant. [A]o
FIGURE 1.2 some fixed with k ¼
¼ kt or log [A] at time zero can[A]at the later [A]that [A] (1.
First-order plot for A ! B value M and It must be ¼ kt
[A]o [A]o o
itial concentration this equation, itto [A]be remembered the2:303 to the 0:020 ln or logtime. This ,can
seen that ¼ concen
À1
[A]
ln
[A]
From ¼ o
t (1:12)
It must From this equation, [A]o , the initial concentration of A, has of A decrease
be remembered that it can be seen that the concentration
[A] o 2:303

10. The equation
can ¼ seen that
be [A] e
From this equation, it [A] ¼ ln[A]o À [A]the concentratio
with time in an exponential way. Suchð relationship is so ð
ln[A] ¼ ln [A]o o ktkt
ln [A] À (
a d[A]
From this equation, it in the form seen that the concentration
can also be written can form be
can alsoan exponential decay.
be written
an exponential processes followof Eq.[A] ¼ [con
with time inRadioactive decay way. SuchÀ relationshiphasrate( k
to as The units on k in the first-ord
Reacciones de 1 Orden er a a first-orderis som
Àkt
Àkt
First-Order to as an exponential decay.
[A] ¼ [A]o oeeside
hand
[A] ¼ [A] (1.12)
the units of cancel. However, th
to
proportional to the amount [A]omaterial of A decr
decay isequation, it present0
Fromthis equation, it can be seen that the concentration only if kd
this
Radioactive inofan exponentialmaterialthatarelationshipmeasured coun
From can be seen dimensionally correct of A
the concentration
-Vida media-
amountinan exponential way. Such a relationship is sometimeslh
withtime decay processes followkta first-order rateref
with time radioactive way. Such doubles the is sometimes
Whenexponential to the amountwillofmaterial units. obta
as an the integration is performed, we
have no
products. When the amountof equation remaining is
ose a reaction can be written astoproportional processes follow aTheismaterial present,ra
decay is toas an exponential decay.
decay.
amount oforiginal amount, to the amount of 10first-order rate Chemical[A
Radioactive decay the time expired first-order rate law. The
radioactive material doublesmaterial present,law.count
Radioactive decay processes follow a thecalled the half-life.
[A]o measured doublin
decay half-life easily using Eq. (1.12). At the pointso lnKin
Principles of
The
[A]
the isisproportional to the amount of material present,where t
proportional
decay
amount to
equal
¼ measured counting is
products.amountof radioactive materialdoublesthealso be written in logdoubd
A !Whenradioactive material¼doublesthe concentration therateis o
kt or of of
B of one half-life, tln of ,material remainingArate
the amount t can measured counting form so
(1:8)
original products. When the [A]expired [A] theThe half-lifehalf-life.[A
products. When the amount of1=2 is called the isfor a reactio
amount, the time =2. Therefore,remaining writeone-half
material one-half W
o
original amount, or amount of material remaining is
we can After a rea
concentrationthe timeoexpired is called the half-life. We can calc
easily calculated. [A
hat the reaction follows a the half-lifehalf-life easilyform expired AtFromthe point theitWe can sc
rate law ofequationtime (1.12).[A]tration equation, have be
the the using Eq.
Ifthe easily using Eq. (1.12). isAt pointofhalf-life. time elap
original amount, the involving called thewhere wheredecr
[A] A will can th
natural logarithms
the this
8 Principles of Chemical Kinetics equal toin the oneeasily using¼1=2o(1.12). concentrationln 2 ¼time eic
the half-life half-life, ln Eq. , ¼ ln At the point where the value
equal half-life, o
one to form t ¼ t[A], the concentration1=2o =2,isso this 0:693
time of ¼ of A is on
t t1=2 the with ¼¼ [A]an exponential the
[A] kt in A one-half way
equal to one half-life,o =2. Therefore,[A]as anwrite of A is one-half th
concentrationd[A] t ¼ t1=2 , the concentration
to o
concentration or [A]o[A]o =2. Therefore,wewe can write processes
1
or
=2. Therefore,wecan exponential decay.
1.0 Rate ¼ k[A] ¼ À concentration or [A]
[A] [A]o
can write
ln [A] À ln(1:9)
2 Radioactive decay
[A] ¼
ln [A]o ¼ ln [A] ¼ kt1=2 ¼isln 2 ¼o0:693 to the am
[
0.9 dt
The half-life ln o o ¼givenoo ¼ decay ¼ proportional
[A] [A] ln[A]o kt1=2 ln 2 ¼ 0:693
is then [A] as
(
¼ [A] amount ¼ ln 2 complex frac ¼ 0:693
[A] ln [A]o ¼ kt1=2of radioactive material d
0.8 ln 2 Removing thethe amount o
0.7
or
quation can be rearranged to give
0.6
[A]
The half-life is then given as
products.0:693
2 o t1=2 ¼
When
2
2 the half-life k the time expir
original amount,
[A], M
0.5 The half-life is then given as easily using Eq.[A]o
(1.1
0.4
d[A] ln [A] ¼ ln [A] À
0:693
and it then given as t1=2 ¼ equal on the units t ¼ 1=2 , th (
The half-life isÀ1 have units that depend to one half-life, on tk.oFo
will k
0:693
0.3
0.2
À ¼ k dt
and
hr y ¼ b (1:10)
t1=2 ¼ Therefore, solving for t1=2 g
in it will ,have units that depend will be givende Foro example, i
independiente in hours, etc.
then the half-life onconcentration or [A] =2. Theref
k
the units on k. þ mx
0.1 t1/2 [A]
2t1/2 0:693 etc.[A]o that i
inprocess have units thata will be givenrate law, the Note ln [A]
À1 that follows first-order half-life
0.0 and hr willthen the half-life depend on thein hours, k. For example
it , t1=2 ¼ the half-life is independe
units on ln ¼
the reactant. For etc. Note [A]
the initial concentration of be given in hours, example, in
process that follows a first-order rate law, k [A]
ion (1.10) can be integrated but It must be be integratedthat [A] , the i
it should remembered between the
À1
0 10 20 30 40 50 60 70 in80 90, then the half-life will
hr 100 th
Time, min process that follows a first-order of law, example, o is indepen
For
the half-life
radioactive2
the initial concentration of the reactant. Here we seeof major differed
the half-life is independent ratethe amount ainstarting nu
some fixed value so it is a constant. Therefor
and it will initial concentration of thethe amountthestartingthenin radioactiv
the havea units that depend on half-life is nuclide. This m
half-life sample initially contains 1000units and one as
is independent of The For example,on k. For
order rate law of radioac
of
thethat a sample initially contains 1000 atoms of atoms given is ind
if ¼ 1:00 M and that
of time ¼ 0 and time equal to t while the concentration varies from
FIGURE 1.3 Half-life determination for a first-order process with [A]o
k ¼ 0:020 minÀ1 : half-life is half-life of the
reactant.
in hrÀ1 , that if is exactly the samewill bethere arethere atoms initiallyThi
then the independent same as when startinghalf-lifeetc.a
given inthein the case
reaction, radioactive materia
hours, atom
thehalf-life is exactly the as whenamount of5000 are nuclide. ofpre 5000
half-life reactant, but t
itial concentration [A] at time zero to [A] at the later time. This can
that It It easy to seeathat after one half-lifehalf-lifethe tomaterial mate1
ifisa is easy to see that after one proportionalofamount rema
o process that follows first-order rate law, atoms of radioactive of co
the the initial m
sample initially contains 1000 the amount half-life is
half-life is exactly the same as when there are 5000 atoms initially

11. is second-orderintegrationreactantintegrated rate law is integrated betweenjust as appropriate to speak of the
Performing the in one gives the or equation obeys the rate
If the component involving radioactivity, it is limits on concentrati
1
and1 [A] at time reactant toreaction one-halftime obeys value.the will have
A reaction that is second-order in one t, we have component its initial the rateconcentrati
À
[A] [A]o
¼ kt
chemical
orfall to as the of necessary for We
(1:23)
reactant
law initial concentration of A isd[A]
2 return to this point. Fundamental Concepts of
Since the Rate ¼ k[A] ¼ À a constant, the equation can be put in (1:19)
[A] ð ð
t
Reacciones de 2 do
Orden
dt
If the equation is integrated d[A] limits on concentration of [
the form of a linear equation,
d[A] between
reaction thatRate ¼ k[A] [A]as time t, we have [A]2 ¼ k dt
2
w might result from a 1 1 be written ¼ À Second-Order (1:19)
dt ¼ kt þ can and 1.2.2at
[A] [A]o (1:24)
2 A ! Products þ b ð
A Fundamental Concepts of [A] 0
reaction that is[A]o Kinetics one reactant or component o
(1:20) second-order in 9 ð t
y ¼ mx
Such a rate law might result from a reaction that can be written as dt
law d[A]
¼k
As shown in Figure 1.4, law cannot versus time should be a straight line
a plot of 1=[A] always be written from the [A]2
Performing the integration gives the integrated rate law
we have seen, the rate equation is integrated between limits on concentration of[A]o o at t ¼ 0 2 0 d[A]
If the [A]
o A ! Products
with a slope of k and an intercept of 1=[A]2 if the reaction follows the Rate ¼ k[A] ¼ À (1:20)
ion second-order rate law. If we timeon we have of Eq. (1.24) must be
for the reaction. The at rearrange Eq. (1.19), we have
and [A] units t, each side dt
1 1
Performing the integration gives the integrated rate law
1=concentration. If concentration is expressed [A]in mole=liter,a rate law might result from a reaction that can be writte
However,Àd[A] haveof liter=mole. rate lawwe ð thenalways be 1written from the
as we units seen, the From this Such that
1=concentration will have
ð
d[A] cannot
t
find À 1 ¼ kt
¼ k dt the reaction. If2 we rearrange Eq. (1.19),o we kt
À1 ¼ k
the units on k equation for time or M time so that kt will
balanced must be liter=mole À1 dt(1:21) [A] [A]
À(1:22) ¼ have
2 A ! Products
À1
[A]2 [A]o
[A]
0
[A] [A]o
have units M . However, as we have seen, the rate law cannot always be writ
Since the initial concentration of A is a constant, the equa
Àd[A] balanced concentration reaction. If we rearrange Eq. (1.19)
Since the initialequation for the of A is a constant, the equation ca
Performing the integration gives the integrated rate law
2 ¼ k equation,
dt
the form of a linear equation,
the form of a linear Àd[A]
(1:21)
7 [A] 1
1 2 ¼ k dt
À ¼ kt 1 (1:23) 1
[A]
6 [A] [A]o ¼ kt þ
1 [A] 1
[A]o
¼ ktmx þ b
y ¼ þ
5 Since the initial concentration of A is a constant, the equation can be put in
[A] [A]o
1/[A], 1/M
4 the form of a linear equation,
3 1 As shown 1 Figure 1.4, a plot of 1=[A] versus time should be a s
in
2
¼ kt þ an ¼ mx þ b
with a slope of k and y intercept of 1=[A]o if the reaction
[A] [A]o
second-order rate law. The units(1:24)each side of Eq. (1.24
on
1 ¼ mx þ b
y 1=concentration. If concentration is expressed in mole=
0
As shown in Figure 1.4, a plot of 1=[A] versus time shoul
20 40
1=concentration will have units of liter=mole. From this w
0 As shown in Figure 1.4, a plot of 1=[A] versus time should be a straight line
with min slope of120 1=[A]o k mustreaction follows time or MÀ1 timeÀ1 so t
60
a 80 100
with a slope of k and an intercept of
Time, units on if the be liter=mole 1=[A]o if the rea
the k and an intercept of the
À1
FIGURE 1.4 A second-order rate plot rate ! B with [A]o ¼haveM and kMsideThe units must be
second-order for A law. The units on each¼ 0.040 of Eq. (1.24) on each side of Eq
second-order rate law. . 0:50 units
liter=mol min. 1=concentration. If concentration is expressed in mole=liter, then
1=concentration. If concentration is expressed in

12. The equation
ð
be seenðt that or component obeys the rate(1:15) Fund
[A]
From this equation, it can in one reactant [A] ¼ lnconcentration of A decr
A reaction that is second-order d[A] ln the [A]o À kt
¼ k dt (1:22)
with [A]2
law time in an exponentialbe writtenSuch a relationship k in sometimeslaw ar
can also way. in the form is the first-order rate refe
Reacciones de 2d[A]¼Orden (1.12) has [concentration]=[
[A] o 0 do The units on
to as anPrinciples of Chemical Kinetics the integrated rate law[A]ounits to cancel. However, the(1:16)
10 exponential decay.gives [A] Àkt
handeside of Eq.
Performing the integration k[A]2 it À
Rate ¼equation, ¼ can be seen that the concentration of A decreases
the (1:19) right-hand
-Vida media- willrate law (1:23) law. referred ra
Radioactive decay processes1 follow a first-order correct only if k has the units
From this dt dimensionally rate
1 follows a second-order relationship is sometimes
Principles of Chemical Kinetics
10 with time in an exponential way. Such a kt have nocan be
The half-life for a reaction that À units.
¼ kt of material present, so doubling
The
decay iscalculated.might resultasfrom amount one half-life, written as
easily proportional to thetimereaction
Such a rate law After a reactionexponential decay. that can The the concen-
to an [A] a[A]o equal to be equation
amount ofthe initial have decreasedproportionala to thethe equationfirst-order rateis,law.doublingcanof À
Since
The Radioactive decay processes follows a second-order rate law
half-life for a reaction that
to a constant, measured present, so (1:20) thebe
tration of radioactive material isone-half its original avalue.be counting rate[A] d
A will concentration of A doubles thefollow can That in ln [A] ¼ ln of
decay2is A ! Productsamount equal to put
The rate
easily calculated. After reaction time of materialone half-life, the concen-
o
¼
[A] the[A]o =2, a this trationcan Aofwill have decreased in can also be its to give value. decay
products.form of solinearvalueamount of materialto remainingcountingform ofThat is,
When the amountbe substituted for [A]Chemicalthe(1.23) original one-half of
rinciples of Chemical Kinetics
equation,
of radioactivePrinciples of
10
Eq. measured
material doubles Kinetics written in the rate
one-half is Àkt
However, as we have[A] ¼ [A]o1 ratethisthe cannot ofsubstituted writtenisfrom [A] ¼tothe e
seen,1the When value can be material remaining in Eq. (1.23)of [A]give
products. so law amount always be for [A]
=2, 1 1 called isthe half-life. We canthe calc
one-half o
original amount, the time À ¼ kt ¼Thetime expired a From thisthathalf-life.it acan be canthatrate la
original expired 1=2the is
kt called the follows We calculate
(1:25)
[A][A]amount,þ [A]half-life can1be (1.19), we have seen
[A]oIf usingoEq. (1.12). At the point where the second-order is the
balanced equation follows a half-life easily wecalculated. 1 for reaction equation,
half-life for a reaction thatfor the reaction. easilyrate law AfterEq.
the second-order rearrange a reaction time equal time elapsed
o
the half-lifereaction time equal2 to onemxtrationtAttAthe [A]with time in an(1:24) toits theSuch avalue
calculated. After a
À point where the way. elaps
easily using Eq.one half-life,b ¼the owill have ¼ kt1=2 of exponentialtime (1:25)
(1.12). of[A] ,concen- decreased toAone-half one half-life,relat
equal to ¼ half-life, 1=2 the concentration
the
y Àd[A] þ to as an exponentialis one-half original
o decay. initial
¼ gives2 [A] =2. oTherefore, we can write is processes follow first
A will have decreased to one-halft its , the concentrationcan beAdecayone-half the i
Removing the complex tconcentration or [A]ovalue. so this value of
ofequal to one half-life, fraction1=2original¼k dt=2, That is,Radioactivesubstituted (1:21) Eq.a (1.2
¼ [A] 2 for [A] in
[A]o =2, so thisshown can be substituted for[A] in versus time should be a 1proportional to the amount of mate
As value in Figure Removing of 1=[A] Eq.fraction to give is straight line
1.4, a plot the complex o(1.23) givesdecay
[A]
2
with a slope of k and an À Therefore, o if [A]o ¼ amount ofo 2 ¼10:693material (1:17)
concentration or [A]o =2.intercept of ln [A] ¼1 [A]o reaction[A]lnradioactive kt1=2 doubles the me
1
¼ kt 1=[A]¼
[A]
wethe kt1=2 ¼ follows the
ln can write À ¼
(1:26)
[A]
1 1 [A] 1=2 2 1 1 o
[A]o on each [A]o of Eq. 1=2 ¼ When the amount of material r
products. must be
À ¼ The
second-order rate law. okt1=2 units À ¼ kt
side 2(1:25) (1.24) 2 (1:26)
[A]o [A]o
1=concentration.for[A]The half-life is then expressed complex fraction easily using Eq. (1.12). At the po
[A]is given the in the half-life gives
Ift concentration Removing as
o o
[A]o [A]o original amount,thentime expired is called t
[A]o the
mole=liter,
Therefore, solving ln 1=2 gives ln
2 ¼ solving liter=mole.1=2 ¼ equal2 ¼find thatt ¼ t 1 , the concentrat
1=concentration will have units [A] t1=2 gives
Therefore, of for ¼ kt From this2 to one1half-life,
ln we 0:693 (1
ving the complex fraction gives be
the units on k must
[A] liter=mole time or MÀ1 ttimedependiente de¼ [A]1=2=2. Therefore, we can
o
1 0:693 À
À1 kt ¼ 1=2
t1=2 ¼ 1=2 ¼ [A]o that okt will
so [A] (1:27) [A]o (1:18)
concentration or
1k o
have units M 1À1
. 2 o
k[A] t1=2 ¼ (1:27)
2 1 haveTherefore, solving for t1=2 ogives on k. [A]o example, ¼ kt is ¼
k[A] ln For ¼ ln
[A]o
À ¼ kt1=2 ¼ will
and it units that depend on the units
(1:26) [A] [A]o if k 1=2
The half-lifeo a and one givenseethatmajor difference between a reaction k[A]isfollows2a second-
Here we [A] major difference[A]then the half-life will be given in hours, that Note that for a
see [A]o Herein hr , a À1between a reaction that follows a second-
is then thatwe aso afollows a first-orderlaw. law, the1=2 ¼ etc. independent of
t
1
ore, solving for t1=2 gives 7 order rate lawconcentration of ratereactant. Foraexample,law. For a first-order
order rate law followsand one that followsrate For first-order
first-order
process a The half-lifehalf-life given as
first-order rate o
is then
reaction, the half-life reaction,initial half-life theindependent differenceinitial concentrationdecay
the the Here see a major of the between aradioactive follows
the in
is independent of isweinitial concentration of thereaction thatof the
0:693one that of startingfirst-order This law.0:693a
6 1 a half-life the order of second-order reaction, nuclide. rate means
reactant, but in the case of thesecond-order reaction, and amount follows athe half-life is inversely(1
but in is t case rate law the half-life is inversely
independent of the
t1=2 ¼reactant, if a sample1=2 ¼acontains 1000 atoms of radioactive material, the
that initially (1:27) t1=2 ¼ For
k
reaction, the half-life is independent of the initial concentrati