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U5: MODULE 3
Understanding
Mechanism
We have discussed how to evaluate the thermodynamic stability of chemical systems and how to predict reaction directionality and extent. However, the values
of DGorxn or the equilibrium constant for a chemical process
tell us nothing about how long it will take for the reaction
to happen. Thus, in this module we will shift our attention
to issues related to the kinetic stability of chemical systems.
The rate of a chemical reaction is influenced by a variety of
internal and external factors. The chemical composition and
structure of the species involved determine the value of the
activation energy needed to initiate the process, as well as
the configuration effectiveness of collisions between reacting
substances. Concentration, temperature, and pressure affect the aver- Enzymes increase the rate
of specific reactions.
age rate at which reacting particles collide.
To understand and predict the kinetic behavior of a chemical system, it is
important to explore how the reaction actually happens. This is, what sequence
of events, or mechanism, leads from reactants to products. Knowledge about the
mechanism of a chemical process allows us to better understand and control the
effect of variables such as concentration and temperature on the rate of reaction.
As we will see in Unit 7, it also helps us predict the likely structure of the products
of a chemical reaction.
THE CHALLENGE
Mirror Images
Molecules of a given amino acid may exist in one of two forms, commonly
called D and L isomeric forms. The D form is the mirror image of the L form.
These types of molecules can transform into each other. The conversion process involves a two-step mechanism.
•
How would you expect the rate of conversion to depend on the concentration of the amino acid and on the external temperature?
Share and discuss your ideas with one of your classmates.
This module will help you develop the type of chemical thinking that is used
to answer questions similar to those posed in the challenge. In particular, the central goal of Module 3 is to help you understand how to use information about
reaction mechanism to predict rates of reaction.
Chemical Thinking
U5
How do we predict chemical change?
315
Reaction Mechanism
There are chemical processes that are thermodynamically favored but take too
long to happen. The reactant mixture in these types of cases tends to be kinetically
stable although thermodynamically unstable. Consider, for example, the following
two potential routes for the synthesis of the amino acid glycine on the primitive
Earth:
Figure 5.10 Schematic rep-
CH2O(g) + HCN(g) + H2O(l)
2 CH4(g) + NH3(g) + 5/2 O2(g)
C2H5NO2(s)
C2H5NO2(s) + 3 H2O(l)
DG
DG
o
rxn
o
rxn
= –154 kJ
= –965 kJ
resentation of the change in
Gibbs free energy, DG, along
the two reaction paths for the
synthesis of glycine.
Figure 5.10 shows a schematic representation of the changes in the
Gibbs free energy of the reacting mixtures, DG, as each of these DG
two reactions proceeds from reactants to products. As shown in
this figure, although the second process represented above is more
thermodynamically favored than the first one, it has such a large
free energy of activation that it is likely that the first process will
2 CH4(g) + NH3(g)
occur much faster and be a more common path for the synthesis
+ 5/2 O2(g)
of glycine under standard conditions.
Analyzing chemical systems from both the thermodynamic and kinetic point of view is crucial in evaluating the actual
C2H5NO2(s)
CH2O(g) + HCN(g)
chemical stability of substances. For example, the transformation
+ H2O(l)
of diamond into graphite has a DGo = –2.9 kJ/mol, which makes
it thermodynamically favored, but the activation energy for the
C2H5NO2(s) + 3H2O(l)
process is so high (Ea = 728 kJ/mol) that the process takes millions
of years to occur without external intervention. From this perspecReaction Path
tive, diamond can be considered a chemically stable substance.
To better understand the kinetic behavior of chemical systems it is useful to
build models that describe how the reaction takes place at the submicroscopic
level. These models recognize that most chemical reactions involve a sequence of
steps in the path from reactants to products. This sequence of processes define the
mechanism of the reaction. In this mechanistic approach, the overall reaction is
Bimolecular Step
viewed as the result of multiple elementary processes, or steps, occurring simultaneously across the system. The different elementary steps describe the likely order
in which chemical bonds are broken and formed during the reaction. The most
common elementary steps involve one of the following processes:
a) Unimolecular step, corresponding to the direct transformation of a
single molecule into another;
b) Bimolecular step, involving the collision of two particles of the same or
Transition State
different type (Figure 5.11).
c) Termolecular step, involving the simultaneous collision of three particles
of the same or different types.
As a result of these types of processes, atomic rearrangements will lead to the formation of new particles with different composition and structure. Some of these
particles may suffer additional transformations in other reaction steps and will
not be present in the final products. Such chemical species are known as reaction
intermediates.
Figure 5.11 Representation of a bimolecular step involving the collision of two particles. During any type of elementary step, there is only one transition state.
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MODULE 3
Understanding Mechanism
LET’S THINK
Molecularity
In chemistry, molecularity refers to the number of colliding particles that are involved in a single
reaction step. Determining the molecularity of a reaction step thus mean to identify whether the
process is unimolecular, bimolecular, etc.
•
Determine the
molecularity of
each of the two
reaction steps
associated with
this reaction
mechanism.
A + A
B
B + C
C + E
•
Identify the intermediate species in the mechanism and build a symbolic representation of
the overall reaction that this mechanism is modeling.
•
Sketch the potential energy diagram for this reaction, showing the relative potential energy
of reactants, products, and intermediates. (Hint: Analyze the number and nature of the chemical
bonds broken and formed in each step of the mechanism).
•
Discuss how you would expect the rate of each step to change when the concentration of
the reacting species in each step (i.e., A in the first step and B in the second step) is doubled.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Concentration Effects
If we know how many particles are involved in a reaction step (molecularity), we can predict the rate of the process will vary with changes in the concentration of the reacting species. The rate of the reaction is a measure of how
the concentration of the reactants, or of the products that they form, changes
per unit time. To begin our analysis, let us consider a unimolecular process in
which a single particle A is involved. Imagine that we have nA moles of particles
of the reactant A in a given volume V; the molar concentration would then be
[A] = nA/V. For any reaction of this type, there is a probability k that a particle
will undergo a transformation in a given unit of time. The larger this probability,
the faster the rate of the reaction. The larger the concentration of A particles, the
larger the number of such particles that will be transformed per unit time. In fact,
we can expect the reaction rate to be directly proportional to [A]. Thus, the rate of
a unimolecular step can be expressed as:
(5.14)
RATE = k [A]
Unimolecular Process
The constant k in this expression is commonly called the rate constant for the
process. In this particular case, the rate constant has units of (1/time). Rates of
reaction are commonly expressed in units of concentration per unit time (e.g.,
molarity/second). Given that the rate of reaction is directly proportional to the
concentration of the reactant to the first power, we say that this is a “first order”
reaction.
Chemical Thinking
U5
317
How do we predict chemical change?
In bimolecular processes, the rate of the reaction depends on the probability
that a collision between two particles A and B leads to a chemical transformation.
This probability, k, depends on factors such as the activation energy of the reaction, the configuration effectiveness of the collision, and the temperature of the
system. The concentration of each of the colliding species will also affect the rate.
The higher the concentration of each of substance, the higher the likelihood that
their particles will collide and the faster the reaction rate. In this case, the relationship between rate of reaction and concentration can be expressed as:
(5.15)
RATE = k [A] [B]
Bimolecular Process
The reaction rate depends on the product of the concentration of each species
because the likelihood of a collision depends on the product of the probability
of finding one particle of each type in the same location. Thus, if we double the
concentration of each reactant, the reaction rate increases four times. The rate constant k for bimolecular processes has units of 1/(concentration x time).
A bimolecular process is said to be first order with respect to each reacting
species. However, overall, the process is a second order reaction. In general, the
relationship between the rate of a reaction and the concentration of each reactant
involved in the process is known as the Rate Law for the reaction. This rate law has
the following general form:
(5.16)
RATE = k [A]a [B]b [C]c
where the exponents a, b, and c are the order of the reaction with respect to each
species (i.e, a = 1, first order; a = 2, second order, etc), and the overall reaction
order is given by the sum a+b+c.
Ozone Decomposition
LET’S THINK
With the development of photosynthetic organisms on the primitive Earth, the amount of O2(g)
and O3(g) in the atmosphere increased considerably. The formation and decomposition of ozone
molecules was induced by UV radiation from the Sun. The two-step reaction mechanism for the
decomposition of O3(g) in the atmosphere can be represented as follows:
O3(g)
O2(g) + O(g)
O3(g) + O(g)
2 O2(g)
•
•
•
Determine the molecularity of each step and identify all the reaction intermediates;
Express the rate law for each step in the mechanism. Determine the rate order with respect to each species as well as the overall rate order for each step.
Write the chemical equation that represents the overall process.
The potential energy diagram for the overall reaction is shown to the right:
•
Ep
Reactants
Products
Reaction Path
Locate in the diagram the different reaction steps and the species involved in each of them.
Identify the fastest step in the decomposition process.
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MODULE 3
Understanding Mechanism
As we will discuss in Module 4, there are well established methods to determine the rate law for an overall chemical reaction. This information can be
combined with results from other experimental techniques used to identify reaction intermediates to propose a reaction mechanism. In the mechanistic model
of chemical reactions, the overall rate of the reaction is determined by the rate of
one or more of the individual steps. In particular, for reactions in which one of the
steps is considerably slower than the others, the overall rate of reaction in determined by the slowest step (Figure 5.12). The speed at which chemical species are
produced in this step sets the maximum speed at which the reaction can proceed.
The relationship between the rate law of the slowest step in a reaction mechanism (rate determining step) and the rate law of the overall reaction is direct when
the slowest step is the first step in the mechanism. Consider, for example, the following hypothetical mechanism for an overall reaction of the type A
C:
A + A
B
C
Figure 5.12 In this analogy, the
liquid passing through each filter
represents a different reaction
step. The rate at which the flask
is filled is determined by the
speed at which the liquid passes
through the slowest filter.
B + C
SLOW
(Rate Determining Step)
FAST
The rate law for the first (bimolecular) step in the mechanism can be expressed
as: RATE1 = k1[A]2, where k1 is the associated rate constant. For the second step
we have: RATE2 = k2[B]. Given that the first step is the slowest in the mechanism, the overall rate of the reaction should be equal to the rate of the first step:
RATE = k1[A]2. If the above mechanism is correct, the experimental determination of the reaction order for the overall reaction should indicate that the process
is second order with respect to the concentration of the reactant A.
The above example can also be used to illustrate the problems that we face
in deriving the rate law for an overall reaction for which the slowest process in
the mechanism is not the first step. Imagine now that the first step in our example was faster than the second step. In this case, the overall rate law should be:
RATE = k2[B]. The problem with this relationship is that it is expressed in terms
of the concentration of the intermediate species B; a concentration that we cannot
easily determine and control. To derive a more useful rate law we need to express
[B] in terms of the concentrations of reactants, or of the reactants and products.
Finding this relationship is not always straightforward but it can be more easily
accomplished when we have a mechanism in which the reaction associated with
the first step reaches chemical equilibrium at a rate much faster than the rate of the
second step. In this case we can write:
FAST
SLOW
A + A
B
B + C
C
K = [B][C]/[A]2
(Rate Determining Step)
where K is the equilibrium constant for the first process. If we are able to determine the value of this constant, the concentration of the intermediate B can be
expressed as [B] = K[A]2/[C]. Thus, the overall rate of the reaction takes the form:
RATE = k2[B] = k2K[A]2/[C] = k[A]2/[C],
where the overall rate constant k = k2K. The experimental analysis of this reaction
should show a reaction rate that is second order with respect to A but also depends
on the inverse of the concentration of product C.
Chemical Thinking
U5
Dimerization
How do we predict chemical change?
LET’S THINK
Amino acids formed on the primitive Earth somehow combined to form dimers, trimers, and
eventually polymers (proteins). Consider the overall reaction for the dimerization of alanine
(Ala):
Experimentally, it is found that the rate law for this process can be expressed as: RATE = k[Ala]2.
Several possible mechanisms have been suggested for this reaction. Schematic representations of
two of them are shown below:
Ep
Ep
Ala-Ala*
2 Ala
Reactants
Ala-Ala
+ H2O
Products
Reaction Path
•
•
•
•
2 Ala
Reactants
Ala-Ala
+ H2O
Products
Reaction Path
Write the individual steps that correspond to each mechanism;
Identify the molecularity of each step and the associated rate law;
Evaluate which of the two mechanisms would fit the experimental result for the overall rate
law for the reaction;
Propose a different reaction mechanism that involves a quick first step in which reacting
species reach chemical equilibrium, followed by a slow step. Evaluate whether this type of
mechanism would fit the experimental results for the overall rate law.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Temperature Effects
According to the kinetic model of chemical reactions, for a chemical process to
occur colliding particles must be oriented in a manner that allows reacting groups
to interact effectively. Additionally, colliding particles must have enough energy to
reach a transition state that leads to the formation of the new products. The energy
needed for colliding particles to adopt the structure of the transition state is the
activation energy, Ea, of the reaction. The larger the value of Ea, the less likely that
colliding particles will have enough energy to overcome the activation barrier and
the lower the rate of the reaction will be. Increasing the temperature of the system
leads to an increase in the average kinetic energy of the particles, increasing the
probability of favorable collisions and raising the reaction rate.
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320
MODULE 3
Understanding Mechanism
The effect of configuration effectiveness, activation energy, and temperature
on the rate of a reaction is represented in the value of the rate constant k in the
associated rate law. This constant is a measure of the probability that collisions will
lead to the formation of the transition state at a given temperature. This probability can be calculated using Arrhenius equation expressed as:
k = A exp ( –Ea / RT)
(5.17)
where the pre-exponential factor A is a measure of the frequency of collisions with
the proper configuration, and the term exp(-Ea/RT) is proportional to the fraction of colliding particles with enough energy to react at a given temperature. In
general, we can assume that the frequency factor A and the activation energy Ea
are constants for a given chemical process, independent of the value of T. Consequently, we can compare reaction rates at two different temperatures T1 and T2 by
calculating the ratio of the corresponding rate constants k1 and k2:
(5.18)
k2 / k1 = exp ( –Ea/RT2) /exp ( –Ea/RT1) = exp[–Ea/R(1/T2– 1/T1)]
LET’S THINK
First Amino Acids
The overall activation energy for the dimerization of alanine (Ala) that we studied in the previous
problem is close to 90 kJ/mol.
•
Calculate how many times faster the dimerization of alanine is at a) 100 oC, and b) 400 oC,
than at 25 oC. These temperatures represent the range of values suspected to exist in high
pressure hydrothermal vents were life may have started in our planet.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Figure 5.13 The catalyst reduces
the overall activation energy by
changing the mechanism from a
single step to a two-step process.
Ep
Changing Mechanism
It has been suggested that one crucial step in the origin of life in our planet was the
synthesis of substances that could speed up the rate of certain chemical reactions.
In chemistry, such substances are called catalysts and
they act by directly reducing the activation energy
Ea of a reaction or by changing the reaction mechaUncatalyzed
nism, which often results in lower activation barriers (Figure 5.13). Catalyst may work by the action
of intermolecular forces on the reacting particles or
by actually forming chemical bonds with some species. These interactions facilitate the formation of
a transition state that leads to the new products.
Some catalysts may be solid, but they could also
DH
Catalyzed
be species in solution. Catalysts participate in the reaction but they are not consumed during the process.
They can be separated at the end of the reaction.
Reaction Path
Chemical Thinking
U5
321
How do we predict chemical change?
Catalysts are of great importance for living organisms because many metabolic processes would take too long in their absence. The substances that catalyze
reactions in biological systems are called enzymes, which tend to be proteins with
a composition and structure that varies depending on the process in which they
participate. Most enzyme-catalyzed reaction rates are million times faster than
those of the same uncatalyzed process. As is the case with all catalysts, enzymes
participate in the chemical reactions but are not consumed in the process, nor they
change the directionality and extent of a chemical reaction. However, enzymes
differ from most chemical catalyst in their high specificity for the processes that
they help accelerate. The action and high specificity of enzymes has been explained
using different models that we will explore in the following activities.
Lock and Key Model
LET’S THINK
Enzyme (E) + Substrate (S)
Enzyme-Substrate Complex (ES)
•
•
Enzyme-Substrate Complex (ES)
Enzyme (E) + Product (P)
FAST
SLOW
Build a potential energy diagram representing the reaction path from reactants to products;
Write the rate law for each individual step in the reaction mechanism, as well as the equilibrium constant for the first step. Use these relationships to derive the overall rate law for the
reaction in terms of concentrations that can be measured experimentally.
The overall rate law that you derived in the previous step is only valid at low concentrations of
substrate. At high concentrations, the reaction rate does not
Maximum Rate (Vmax)
change much with increasing [S} as shown in the graph.
•
•
Use the lock and key model to explain the flattening of
the reaction rate at high substrate concentrations.
What is the order of the reaction under such conditions?
Catalyzed
Uncatalyzed
Share and discuss your ideas with a classmate.
By Jerry Crimson Mann (Own work)
[Public domain] via Wikimedia Commons
The process of enzyme catalysis begins with the binding of a substance, commonly called the
substrate, to the active site of the enzyme. The active site is a region of the enzyme that interacts
relatively strongly with the substrate. The binding process alters the electron distribution in the
substrate molecules and induces structural changes that facilitate their transformation. In the “Lock
and Key” model of enzyme action, the active site is assumed to have a unique geometric shape that
is complementary to the shape of
the substrate molecule. The enzyme-substrate system is thought
to be analogous to a key-lock pair
in which the lock is the enzyme
and the key is the substrate. This
model explains the high specificity of enzymes as only the correct key (substrate) can fit into the key hole (active site) of the lock
(enzyme). The reaction mechanism for enzymatic action in this model can be represented as:
MODULE 3
Understanding Mechanism
LET’S THINK
Induced Fit Model
E + S
ES
ES
EP
EP
E + P
•
•
•
FAST
SLOW
FAST
Build a potential energy diagram representing the reaction path from reactants to products;
Write the rate law for each individual step in the reaction mechanism, as well as the equilibrium constant for the first step. Use these relationships to derive the overall rate law for the
reaction in terms of concentrations that can be measured experimentally.
Compare the overall rate law predicted by the “Induced Fit Model” with that of the “Lock
and Key Model.” Discuss whether their predictions could be used to identify the best model
for enzymatic action.
Share your ideas with a classmate, and clearly justify your reasoning.
LET’S THINK
Temperature Effects
For a range of temperatures, the rate of an enzyme-catalyzed reaction increases as the temperature is
raised. A ten degree rise in temperature will increase the activity of most enzymes by 50 to 100%.
•
Consider enzymes working at body temperature (37 oC).
Estimate the range of values for the activation energy of
reactions catalyzed by these types of enzymes.
The graph to the right depicts the effect of temperature on the
reaction rate of enzyme-catalyzed processes.
•
Build an explanation for the effect of temperature on the
rate of reaction.
Temperature
By Tim Vickers (Own work)
[Public domain] via Wikimedia Commons
The “Lock and Key” model provides an inaccurate description of the behavior of most enzymes.
The model has been modified to take into account changes in the structure of the enzyme as a result
of its interaction with the
substrate. In the “Induced
Fit Model” the initial interactions between enzyme
and substrate are assumed
to be relatively weak, but
these interactions induce
conformational changes in
the enzyme that strengthen
the binding. The reaction mechanism for enzymatic action involves three steps represented by the
following set of generic chemical equations:
Reaction Rate
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Chemical Thinking
FACING THE CHALLENGE
Mirror Images
How do we know if two molecules are identical?
In principle, they should have the same chemical
composition and structure. However, sometimes
making this judgment is not so easy. Consider the
two molecules shown in the image. Are they the
same? One way to answer this
question is to explore whether
the molecules are superimposable; this is, if we can make one
lay on top of the other having
the same distribution of atoms.
If we do that with the amino
acids in the image, it turns that
the molecules are not identical.
One is the mirror image of the other, like our two
hands, and they are not superimposable. They are
said to be enantiomers or optical isomers. When
samples made up of these two types of molecules
interact with polarized light, each of the samples
rotates the polarization plane in different directions. Pairs of enantiomers are often designated
with the letters D and L to differentiate them.
Molecules that have at least one non-superimposable mirror image are said to be chiral. Molecular chirality is caused by the presence
of carbon atoms that are bonded to
four different groups. If all carbons
in a molecule are bonded two only
three or fewer different groups, the
molecule will be achiral and it will
not have enantiomers. Molecular
chirality is relevant to our discusChiral
sions in this module because many
biologically active molecules are
chiral and exist as only one type of enantiomer in
living organisms. For example, biologically active
amino acids are of type L but not D, while naturally occurring sugars are D but not L.
Chirality is of central importance for many
biological functions. Chiral molecules have most
of the same properties, but they interact differently with L and D enantiomers of other chiral substances. Thus, for example, if the active site of an
U5
How do we predict chemical change?
enzyme has chiral components, it will not interact
in the same way with the D or L isomers of a chiral
substance. This property enhances the selectivity
of the enzyme as a catalytic species.
The origin in the asymmetry of the chirality of biologically molecules (homochirality) is
not well understood. Many chemical reactions
in which chiral molecules are produced generate
the different enantiomers in the same proportion.
Additionally,
enantiomers
can transform into each other
through a chemical process
known as racemization. The
rate of interconversion varies
from substance to substance,
but for many naturally occurring compounds the rate is not
negligible. Thus, it is not clear
how living organisms developed to selectively use
and produce only one type of enantiomet.
There are several hypotheses about the origin
of homochirality of naturally occurring molecules
in our planet. For example, analysis of amino acids
found in meteorites hitting our planet reveals the
presence of preferentially one enantiomer. There
is evidence that certain type of radiation to which
these meteorites may have been exposed triggers
the formation of optical isomers. These results suggest that homochirality on Earth
may have had an extraterrestrial
Achiral
origin.
However, there are other types
of processes that result in the preferential formation of one optical
isomer over the other. Stirring a
solution as it crystallizes leads to
one chiral form almost exclusively
over the other. Similarly, boiling a
supersaturated solution perturbs the crystallization
process, leading to the formation of a single chiral
phase. Once an enantiomeric imbalance is created,
it may be amplified and propagated by chemical
processes that favor the formation of one type of
isomer over the other. Identifying the actual cause
of homochirality thus demands a careful analysis
of the mechanisms behind these types of chemical
processes.
323
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MODULE 3
Understanding Mechanism
Let’s Apply
ASSESS WHAT YOU KNOW
Racemization
Pure samples of L or D enantiomers eventually convert into a mixture of both forms. This process
is called racemization and the mixture that is formed is a racemic mixture. Racemization in vitro
or in vivo of some substances, such as chiral pharmaceutical drugs, may be a serious problem as
one enantiomer may be the one that triggers the response that we want and the other may induce
unwanted metabolic processes.
Amino Acids
The following potential energy diagram represents the racemization of the amino acid
alanine. The process is characterized by having DHorxn = 0 and DSorxn = 0.
Ep
L
L- + H+
+ H+
H+ + D-
D
H+ +
Reaction Path
•
•
•
•
Represent the reaction mechanism for the racemization of alanine as a sequence
of steps. Determine the molecularity of each step and identify all reaction intermediates.
Establish the rate law for each reaction step and derive the rate law for the overall
racemization process. Determine the order of the reaction with respect to the concentration of the amino acid. Discuss how a change in concentration would affect
the rate of racemization.
The activation energy for the first step of the racemization of alanine is close to
130 kJ/mol. Determine the temperature at which the rate of racemization would
be double than that corresponding to standard conditions.
Given the nature of this process, describe what you would expect to happen to
a pure sample of L alanine as the time passes. What would be the value of ratio
[D]/[L] when the system reaches chemical equilibrium?
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
U5
How do we predict chemical change?
325
Thalidomide
Thalidomide is a drug introduced in the 1950s to treat morning sickness in pregnant women
and aid sleep. It was sold until 1961, when it was withdrawn from the market after being found
that the drug caused birth defects. The thalidomide molecule is chiral and each enantiomer has
a different effect in the human body. While the D isomer is a sedative, the L isomer can induce
malformations (teratogenic). Unfortunately, the rate of racemization is relatively high, taking
only a few hours for a pure sample of one of the enantiomer to generate a racemic mixture inside
the body. The interconversion process can be represented in the following way:
Thalidomide-D
•
•
•
•
Thalidomide-L
Write each of the reaction steps associated with the above process. Determine the molecularity of each step and identify all reaction intermediates.
Build a potential energy diagram to represent the racemization process;
Establish the overall rate law for the process and identify the order of the reaction with
respect to the concentration of thalidomide.
Measurements of the rate constant for the racemization process at two different temperatures generate the following values: k(15 oC) = 0.37 days-1; k(37 oC) = 0.30 hours-1. Use
this information to determine the activation energy for the racemization.
The rate of the racemization reaction in aqueous solution varies depending on the concentration of hydronium ion, H3O+ (which determines the acidity of the solution). If we represent the
D-thalidomide molecule using the symbol T(D)=O, the reaction mechanism for the first half of
the racemization process can be expressed as:
T(D)=O + H3O+
T(D)=OH+ + H2O
•
•
•
T(D)=OH+ + H2O
T(D)-OH + H3O+
FAST
SLOW
Determine the molecularity of each step and identify all reaction intermediates.
Establish the rate law for each step, as well as the equilibrium constant for the first
process. Use these relationships to find overall rate law for the process and identify the
order of the reaction with respect to the relevant species. Identify the overall order of the
reaction.
The hydronium ion is said to “catalyze” the racemization process. Discuss any evidence
you have based on the analysis of the reaction mechanism to support such a claim. Build
an energy potential diagram that represents how the H3O+ ions may affect the reaction
path and speed up the process.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
Chiral Carbon
326
MODULE 3
Understanding Mechanism
Let’s Apply
ASSESS WHAT YOU KNOW
Photochemical Smog
Photochemical smog is a type of air pollution
produced when sunlight acts upon motor vehicle exhaust gases, such as nitrogen oxides (e.g.,
NO, NO2) to form harmful substances such as
ozone (O3). The substances responsible for this
phenomenon participate in a reaction cycle that
determines the concentration of pollutants at different times during the day.
Ozone Formation
Nitrogen dioxide gas, NO2(g), is formed by chemical reactions involving N2(g) and O2(g) at
the high temperatures inside internal combustion engines in our cars. In the presence of sunlight, NO2(g) reacts with O2(g) to generate O3(g) as described by the following overall reaction:
NO2(g) + O2(g)
NO(g) + O3(g)
This chemical process occurs in two steps, as illustrated in the potential energy diagram.
In the second step of the
process, the radical species O reacts with O2.
In order for the process
to occur, the collision
should involve a third
particle M that can absorb excess energy from
the collision (e.g., N2).
•
•
•
•
•
Ep
Ea = 306.5 kJ/mol
Ea = 107 kJ/mol
O + NO
O2 + M
NO + O3 + M*
NO2
Reaction Path
Write the chemical equations for each reaction step and determine the molecularity of
each of them. Identify all reaction intermediates.
Express the rate law for each reaction step and derive the rate law for the overall process.
The activation energy required to initiate the process is provided by sunlight. Calculate
the minimum frequency n of the radiation needed to start the reaction.
Estimate how many times faster this reaction is at 30 oC versus 10 oC.
Based on the information provided, evaluate whether this process can be expected to
occur to a large extent.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
U5
How do we predict chemical change?
327
Ozone formation is faster during the day due to the
action of sunlight on NO2 molecules. Thus, O3(g)
accumulates in the troposphere and its concentration peaks sometime in the afternoon. During the
night, ozone reacts with NO(g) according to the
following reaction:
NO(g) + O3(g)
PPB
Ozone Consumption
NO2(g) + O2(g)
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•
•
Hour
Build potential energy diagram that represents the transformation of reactants to
products. Indicate on the diagram the activation energy and the heat of the reaction
for this process.
Derive the rate law for this process.
Estimate how many times slower this process is at 0 oC than at 25 oC.
Share your ideas with a classmate, and clearly justify your reasoning.
Additional Reactions
The oxygen atoms generated by the decomposition of NO2 molecules
in the presence of light
are highly reactive and
interact with other substances to generate different types of pollutants.
Consider the following
sequence of reactions involving methane, CH4:
•
•
Chemical Reaction
NO2
NO + O•
O• + H2O
CH4 + OH•
CH3• + O2
CH3O2• + NO
CH3O• + O2
CH O + OH•
Rate Constant k
(25 oC)
(molecules-1 cm3 s-1)
Varies: Light Intensity
2 OH•
H2O + CH3•
CH3O2•
NO2 + CH3O•
2.0 x 10-10
6.3 x 10-15
1.8 x 10-12
7.7 x 10-12
CH2O + HO2•
1.9 x 10-15
HCO• + H O
9.0 x 10-12
2
2
Analyze this reaction
HCO
•
+
O
CO
+
HO
•
5.2 x 10-12
mechanism. Identify
2
2
reactants, products,
2 HO2•
H2O2 + O2
1.7 x 10-12
and reaction intermediates. Derive the overall chemical equation for this process.
Use the data to evaluate which step may determine the overall rate of reaction.
Share your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
This is a single-step process with an activation energy Ea = 12.5 kJ/mol and a DHrxn = –199.5 kJ/mol.