Download Measuring Rates

328
Chemical models of the mechanism of a chemical reaction are useful in understanding the process, in predicting the effect of changing concentration and temperature on reaction rates, and in devising strategies
to control the chemical reaction. However, to build
those models we need experimental information
about how the actual reaction proceeds as a function
of time. The analysis of how the concentration of reProduct
actants or products changes with time can be used
to derive the order of the reaction with respect to
different species and to establish the rate law for the
Reactant
process. Measurements of the rate of the reaction at
different temperatures can be used to calculate the
activation energy for the process. All these data, toTime
gether with experimental studies about intermediate
species, are critical in making judgments about the kinetic stability of a chemical
system and in building hypotheses about reaction mechanisms.
The analysis of kinetic data demands a basic understanding of the mathematical relationships between concentration and time for chemical reactions with different rate orders. It also requires that we understand how experimental data can
be represented in different ways to elicit underlying trends. Developing these skills
is one of the central goal of this module.
Concentration
U5: MODULE 4
Measuring
Rates
THE CHALLENGE
Many Options
Once formed, amino acids may undergo a variety of chemical reactions such
as decomposition, dimerization, and racemization. Which process dominates
may vary with temperature and pressure.
•
What types of experimental measurements would you complete to determine which chemical process dominates at various conditions?
Share and discuss your ideas with one of your classmates.
This module will help you develop the type of chemical thinking used to answer questions similar to those posed in the challenge. In particular, the central
goal of Module 4 to help you understand how to derive kinetic information
from changes in the concentration of reacting species as a function of time.
Chemical Thinking
U5
329
How do we predict chemical change?
Reaction Rates
The analysis of the kinetic stability of biomolecules has been crucial in the evaluation of different theories about the origin of life. For example, chemical species
needed for the synthesis of amino acids, such as CH4 and NH3, are abundant in
hydrothermal vent regions with temperatures between 60 oC and 400 oC. How
could we determine how stable amino acids are under such conditions? From the
thermodynamic point of view, we could try to determine the DGrxn for the decomposition at various temperatures. So, for example, it is known that the
decomposition of alanine (Figure 5.14) is thermodynamically favored
under standard conditions. However, this process may take so long that
amino acids may combine to form proteins before decomposing.
Alanine Decomposition
•
of alanine into ethyl amine
and carbon dioxide.
LET’S THINK
1.0
0.8
[Ala] (mM)
The kinetics of the decarboxylation of alanine has
been explored by measuring the concentration of
the amino acid, [Ala], as a function of time, t, in
aqueous solutions at various temperatures. The
data at T = 200 oC is shown in the graph, where
the concentration is expressed in milimoles per
liter (mM) and the time is measured in days.
Figure 5.14 Decarboxylation
How would you use these data to calculate the
rate of the reaction at any given time?
Share and discuss your ideas with a classmate.
0.6
0.4
0.2
0
0
6
12
18
t (days)
24
30
36
Experimental data for the concentration as a function of time for any reacting
species involved in a chemical reaction of the type A
B can be used to calculate the average rate of reaction between any two data points using this relationship:
RATEAVERAGE = – D[A]/Dt = D[B]/Dt
Where D[X] is the difference in the concentration of species X between any two
times (D[X] = [X]2 - [X]1). The average rate of the reaction is a positive quantity
expressed in units of concentration per unit time that measures the average speed
at which reactants are being consumed or products are being produced. For example, in the case of the decarboxilation of alanine at 200 oC, the average rate of
the reaction between days 3 and 6 is:
RATEAVERAGE = – ([Ala]2 – [Ala]1)/(t2– t1) = – (0.79 – 0.89)/(6 – 3) = 0.033 mM/day
The changes in the concentration of a reacting species as a function of time
can be more accurately described by calculating the instantaneous rate of reaction,
which is a measure of the instantaneous speed at which the substance is consumed
at any given time. The instantaneous rate is a measure of the slope of the tangent
line to the concentration versus time graph at any selected time. It can be estimated using experimental results as shown in Figure 5.15.
1.0
Rateinst = - D[Ala]/Dt
0.8
[Ala] (mM)
(5.19)
0.6
0.4
D[Ala]
0.2
0
0
Dt
6
12
18
t (days)
24
30
Figure 5.15 The instantaneous rate at
a given time can be estimated by calculating the slope of the tangent line.
36
330
MODULE 4
Measuring Rates
The instantaneous rate of reaction is mathematically defined as the derivative
of the concentration as a function of time:
(5.20)
RATEINST= – d[A]/dt
As we will see in the next section, this relationship can be used to derive how the
concentration [A] changes as a function of time from experimental data.
LET’S THINK
The data tables included in this activity show the variation
of the concentration of alanine as a function of time for the
decarboxylation process at two different temperatures: 150
o
C and 200 oC.
•
Calculate the average rate of reaction between years 2
and 3, and between years 6 and 7 for T = 150 oC.
•
Discuss how you would go about estimating the instantaneous rate of the reaction at year 4.
•
How does the rate of reaction changes as the concentration of alanine in the system decreases? How would you
explain this result?
•
How does the rate of reaction changes when raising the
temperature? How would you explain this result?
Kinetic Stability
150 oC
200 oC
Time
(years)
[Ala]
(mM)
Time
(days)
[Ala]
(mM)
0
1
0
1
1
0.933
1
0.962
2
0.871
2
0.925
3
0.812
3
0.889
4
0.758
4
0.855
5
0.707
5
0.823
6
0.660
6
0.791
7
0.616
7
0.761
8
0.574
8
0.732
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
According to the kinetic model of chemical reactions, the reaction rate can be
expected to decrease as the concentration of reactants decreases and there are fewer
collisions between reacting particles. How fast the rate decreases as the concentration of reactants diminishes depends on the order of the reaction. Thus, we can
infer reaction orders by analyzing changes in concentration as a function of time.
LET’S THINK
Data Analysis
A central goal of this module is to learn how to use kinetic experimental data for chemical reactions to infer the associated rate law. Consider, for example, typical experimental data obtained in
the analysis of enzyme (E) catalyzed reactions: S + E
P at high substrate
Time
[S]
(S) concentrations ([S] >> [E]).
(min)
(mM)
•
How is the average rate of reaction changing over time?
•
What is the rate law for this reaction?
•
What is the rate order for this reaction?
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
0
0.500
5
0.413
10
0.326
15
0.239
20
0.152
Chemical Thinking
U5
How do we predict chemical change?
331
Reaction Order and Integrated Rate Laws
The derivation of rate laws from experimental measurements can be systematized
by using mathematical procedures for data analysis. Consider generic chemical
reactions governed by rate laws of the form: RATEINST = k [A]m where k is the rate
constant for the reaction and m is the reaction order. Given the definition of instantaneous rate of reaction, we can then write
(5.21) m
RATEINST = – d[A]/dt = k [A]
If we assume that m =1 (first order reaction), Eq. (5.21) indicates that the rate
at which [A] decreases over time is directly proportional to the actual value of [A].
This is the typical behavior of an exponential decay and it can be shown that [A]
decreases with time according to the following equation:
(5.22)
ln [A]
[A] = [A]o exp ( – kt )
where [A]o is the concentration of species A at t = 0 and [A] is the concentration at
time t (this relationship can also be expressed as [A]2 = [A]1 exp [– k(t2 – t1)] where
[A]1 and [A]2 are the concentrations at time t1 and t2, respectively). This expression
is known as the integrated rate law for a first order reaction.
Eq. (5.22) can be rewritten as ln [A] = ln [A]o – kt by taking the natural logarithm of both sides of the equation. The structure of this relationship suggests a
strategy to infer the reaction order and the value of the rate constant k from experimental data. The relationship has the general structure of the equation for a
straight line: y = b + mx if we make the following identifications:
Variables: y = ln [A] , x = t
t
ln [A]o
m=–k
Constants: b = ln [A]o , m = – k
Figure 5.16 Schematic representation of the variation of ln [A] as a function of time t in a first order reaction.
Thus, if a reaction is first order with respect to the measured concentration [A], a
plot of ln [A] versus time t should result in a straight line with a slope m = – k and
a y-intercept b = ln [A]o (Figure 5.16).
USEFUL TOOLS
Linearizing equations is the process of modifying an equation to generate new variables that
can be plotted to produce a straight line graph.
The mathematical equation for a straight line
has the form:
y = b + mx
where x and y are variables, while b and m are
constants. The constant b represents the y-intercept of the line and m is its slope. We can
linearize an equation if we can get it in the form
variable1 = constant1 + constant2 x variable2
If this is possible, then a plot of variable1 versus
variable2 should produce a straight line with a
y-intercept equal to constant1 and a slope given
by constant2.
To linearize an equation it is convenient to:
a) Rearrange the equation to get a variable, or
a function of, it on the left side of the equation.
This becomes the y variable.
b) Regroup the right side of the equation to create a term containing the other variable, or a
function of it. This becomes the x variable.
c) Identify the constant term that multiply the x
variable (slope) and any additive constants that
are left over (y-intercept).
Consider the equation: A = 4pr2. We can
linearize it by taking logarithms of both sides:
ln A = ln (4pr2)
or
ln A = ln (4p) + 2 ln r
A plot of ln A versus ln r will produce a straight
line with b = ln (4p) and m = 2. In this particular case, we could also simply plot A versus r2.
The graph would be a straight line with a slope
m = 4p and a y-intercept b = 0.
332
MODULE 4
Measuring Rates
Let us apply the previous ideas to explore whether the decarboxylation of amino acids is a first order reaction. Data for the process at 150
o
C is presented in the table below. In this table we also include the values of
ln [Ala] needed to build the graph ln [Ala] versus t shown in Figure 5.17.
As we can see in this figure, ln [A} is a linear function of time which is indicative of a first order reaction. The y-intercept is equal to zero given that
[A]o = 1 mM, and the slope of the line m = – k = – 0.0693 years-1. Thus, we can
express the rate law for the decarboxylation of alanine as:
Figure 5.17 Analysis of experi-
mental data for the decarcoxylation od alanine to determine
whether the reaction is first
order with respect to [Ala].
0
ln [Ala]
- 0.1
- 0.2
- 0.3
- 0.4
- 0.5
0
RATEINST = 0.0693 [Ala]
This relationship indicates that the rate of the reaction is directly proportional
to the concentration of alanine and thus it
decreases as a function of time. The order
Time
[Ala] ln [Ala]
t (years)
of the reaction is independent of tempera(years)
(mM)
2
4
6
ture but the value of the rate constant k
0
1
0
will change as T increases or decreases.
1
0.933 –0.0694
Once the rate law of the reaction is de2
0.871 –0.138
rived, it can be used to predict the instantaneous rate of the reaction at any concen3
0.812 –0.208
tration of reactant. For example, the rate
4
0.758 –0.277
of the reaction when [Ala] = 0.7 mM is
ln [A] = – 0.0693 t
5
0.707 –0.347
equal to 0.0693 x 0.7 mM/year (i.e., the
concentration [Ala] decreases at a rate of
6
0.660 –0.416
0.0485 mM per year). Given that the reaction is first order with respect to the concentration of the amino acid, the reaction rate decreases linearly with a decrease in
the value of [Ala].
Data Analysis
The rate of decarboxylation of alanine depends on temperature. The table in
this activity provides information about changes in the concentration of the
amino acid as a function of time at 200 oC.
200 oC
Time
(days)
[Ala]
(mM)
0
1
1
0.962
•
Use the data provided to verify that the reaction is first order with respect
to [Ala] at 200 oC. Click on the graphing tool to plot the data.
•
Determine the value of the rate constant k at this temperature.
2
0.925
•
Use the integrated rate law for the process to calculate the time it will
take for the concentration of alanine to decrease from 1 mM to 0.5 mM.
3
0.889
4
0.855
•
Calculate the time it will take for the concentration of alanine to decrease from 0.5 mM to 0.25 mM, and from 0.25 mM to 0.125 mM.
Based on your results, discuss how long it takes to halve the concentration of alanine. How would you explain this result?
5
0.823
6
0.791
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
CLICK TO PLAY
http://www.chem.arizona.edu/tpp/chemthink/resources/U1_M1/graph.html
LET’S THINK
Chemical Thinking
U5
How do we predict chemical change?
333
The integrated rate law for a chemical reaction expresses how the concentration of a relevant reacting species changes as a function of time. Thus, it can be
used to predict the time it will take for a reactant or product to reach a given
concentration, or to predict such concentration at a selected time. For a first order
reaction, the exponential decay of the concentration with time facilitates making
such predictions because the time required for the concentration of a reacting species to be reduced in a certain fraction (e.g., a half, a quarter) is independent of the
actual concentration of the substance.
To illustrate this property of first order reactions, let us determine the time it
takes for any concentration [A]o of the reacting species to decrease to half its value
[A] = [A]o/2. Based on the integrated rate law for this type of process as expressed
in Eq. (5.22), we have:
ln [A]o/2 = ln [A]o – kt1/2
or
where we have used the symbol t1/2 to represent the time
required to halve the concentration of the reacting species. This quantity is commonly known as the half-life of
the process. According to Eq. (5.23), the half-life for a
first order reaction only depends on the value of the rate
constant k, which is a function of temperature but is independent of concentration. This behavior is illustrated
in Figure 5.18 for the decarboxylation of alanine at 150
o
C. As we have seen, at this temperature the rate constant
k = 0.0693 years–1. The half-life for the process is then:
tration of alanine is reduced by half
every ten years (t1/2 = 10 years).
t1/2 = ln (2) / k
1.0
0.75
[Ala] (mM)
(5.23)
Figure 5.18 At 150 oC, the concen-
t1/2 = ln (2)/ k = ln (2) / 0.0693 = 10 years
1 half-life
0.5
2 half-lives
0.25
0
The half-life of a substance during a first order decomposition process is often used as an indication of its kinetic
stability.
Data Analysis 3 half-lives
0
10
20
t (years)
LET’S THINK
The graph shown in this activity depicts the concentration [R] as a function of time t at three different temperatures (300 K, 330 K, 400 K) for a reaction of the form: 2 R
P
•
•
•
•
•
•
•
Which set of data corresponds to each temperature?
What is the order of this reaction?
What is the half life of the reaction at each temperature?
Express the rate law for the reaction at each T.
Determine the reaction rate when [R] = 0.5 at 400 K.
Predict the value of [P] after 30 h at 330 K.
Predict the time at which [R] = 0.125 M at 300 K.
Clearly justify your reasoning.
30
40
334
MODULE 4
Measuring Rates
1 / [A]
m=k
t
We can apply the same ideas and mathematical procedures just discussed to
generate the integrated rate law for second order reactions. In this case, we need to
identify a function that satisfies the relationship: – d[A]/dt = k[A]2. Such a function can be expressed as:
(5.24)
1 / [A]o
Figure 5.19 Representation of the
variation of 1/ [A] as a function of
time t in a secondorder reaction.
1 / [A] = 1 / [A]o + kt
The structure of this relationship suggests that if a chemical reaction is second
order with respect to [A], a graph of experimental data representing the inverse of
the concentration 1/[A] versus time t should yield a linear plot with a y-intercept
b = 1/[A]o and a slope m = k (Figure 5.19). Once the rate constant k is known, Eq.
(5.24) can be used to predict the value of [A] at any given time or to determine the
time t it will take for this concentration to reach a certain value.
Molecules of amino acids may react with each other to form dimers. The table lists data for the
dimerization of alanine at 200 oC:
•
Use the data determine the rate order for the dimerization of alanine.
Click on the graphing tool to plot the data.
•
Determine the value of the rate constant k at this temperature.
We can use the integrated rate law for a second order reaction to develop an
expression for the half-life t1/2 for substances involved in this type of process:
•
Derive an expression for t1/2 in a second order reaction. Compare your
result with that for a first order process. Discuss differences and similarities between the two expressions and evaluate the extent to which t1/2
for a second order reaction may be a useful quantity in making judgments
about kinetic stability.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Time
(s)
[Ala]
(mM)
14
0.150
15
0.0932
16
0.0692
17
0.0552
18
0.0460
19
0.0394
CLICK TO PLAY
Activation Energy
ln k
ln A
m = – Ea/R
The type of data analysis discussed in the previous section can be used to derive the
values of the rate constant for a chemical reaction at different temperatures. According to the kinetic model of chemical reactions, changes in the values of k are
determined by Arrhenius equation as introduced in Eq. (5.17). By taking natural
1/T logarithms of both sides of this equation, this relationship can be expressed as:
(5.25)
ln k = ln A – (Ea / R)(1 / T)
Figure 5.20 Schematic representation of the variation of ln k as a function of the inverse temperature (1/T)
according to Arhenius equation.
The structure of this equation suggests that if know the values of the rate constant
k at different temperatures T, then plotting (ln k) versus (1/T) should yield a linear
plot with y-intercept b = ln A and slope m = Ea / R (Figure 5.20). Given that the
value of R is well known, the activation energy Ea can be easily derived from the
experimental data.
http://www.chem.arizona.edu/tpp/chemthink/resources/U1_M1/graph.html
LET’S THINK Second Order Reactions
Chemical Thinking
U5
How do we predict chemical change?
Data Analysis These data listed in the table correspond to the decarboxylation of
alanine at different temperatures.
•
•
Use these data to calculate the activation energy for the
process Ea as well as the value of the pre-exponential factor A.
Click on the graphing tool to plot the data.
Predict the value of the rate constant k at 623 K and calculate
the half-life of alanine at this temperature
Share and discuss your ideas with a classmate, and clearly justify
your reasoning.
LET’S THINK
Temperature
(K)
k
(years-1)
323
1.17 x 10-8
373
8.10 x 10-5
423
6.93 x 10-2
473
1.42 x 101
523
1.06 x 103
573
3.71 x 104
Many chemical reactions do not occur in a single step but involve many
elementary processes. Thus, the activation energy determined from experimental
data is likely a measure of the height of the activation barrier for the rate determining step in the reaction mechanism. It is important to point out that many
chemical reactions satisfy Arrhenius equation only in a small range of temperatures. Deviations are commonly observed when exploring reaction rates over wide
temperature intervals. In these cases, the plot of (ln k) versus (1 / T) is not linear,
which indicates that the value of the activation energy Ea depends on temperature.
The value of Ea at any given temperature can then be estimated by finding the
slope of the tangent line to the plotted curve at the corresponding value of 1/T.
Thermodynamics and Kinetics
Chemists frequently emphasize the distinction between the types of predictions
that can be made using thermodynamic data, such as DHorxn, DSorxn, and DGorxn,
and those that can be derived from kinetic data, such as k and Ea. Thermodynamic
data allow us to predict the directionality and extent of a reaction, but do not provide information about the time it will take for the process to reach chemical equilibrium. On the other hand, kinetic data can be used to predict how much time
will be needed for the concentration of reactants and products to reach certain
values, but tell us little about reaction extent. The rate of a chemical reaction can
be altered by using catalysts that change the value of the activation energy Ea but
do not have any effect on the value of DGorxn, which determines reaction extent.
Although understanding the scope of thermodynamic and kinetic data is of
central importance when working with chemical systems, it is also critical to recognize the relationship that exists between these different descriptions of chemical
processes. To illustrate such connection, let us consider a simple unimolecular
reaction of the form A
B that reaches chemical equilibrium after some time.
The rate law for the transformation of A into B (A
B) can be expressed as
RATEA-B = kf [A], where kf is the rate constant for what we will identify as the forward process. This constant is a measure of the probability of A transforming into
B at a given temperature. However, once B starts forming in the system, there
is a probability kb of B transforming back into A. The rate law for this backward
process (A
B) is given by RATEB-A = kb [B].
335
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M4/eqab.html
336
MODULE 4
Measuring Rates
CLICK TO PLAY
Figure 5.21 Click on the image to
run a simulation of an A <--> B elementary reaction. You may change
the values of kf and kb to observe
the effect on the ratio of concentrations at chemical equilibrium.
Given that the rate of the forward and backward processes are proportional
to the concentration of different species, we can expect these rates to change as
the chemical reaction proceeds. If we start the process with pure substance A in a
container, the forward RATEA-B will be faster than the backward rate RATEB-A at the
beginning of the reaction. As time passes and A is consumed and B is produced,
the forward process will slow down while the backward reaction will speed up. At
some point, the rate at which A is consumed (or B is produced) in the forward
process will be equal to the rate at which A is produced (or B is consumed) in the
backward process. Thus, the concentration of all species will remain constant and
the system will reach chemical equilibrium (Figure 5.21). From the kinetic point
of view, chemical equilibrium is defined by the equality of the reaction rates for
the forward and backward processes: RATEA-B = RATEB-A, which in this example
corresponds to the condition
kf [A] = kb [B]
or
kf / kb = [B]/[A]
This relationship implies that the ratio of concentration of products to reactants
remains constant at chemical equilibrium. We have already defined this ratio as the
equilibrium constant K for the chemical reaction, which has a value determined
by the change in standard Gibbs free energy at temperature T, DGorxn(T) (see Eq.
(5.12)). Then, according to our analysis, the ratio of the kinetic rate constants for
the forward and backward processes for an elementary reaction is directly related
to the thermodynamic equilibrium constant for the reaction as expressed in the
following equation:
(5.26)
kf / kb = K = exp ( – DGorxn(T) / RT )
LET’S THINK
Consider an elementary process of the form: A + B
Kinetics and Equilibrium
2C
• Express the rate laws for the forward and backward processes;
• Identify the relationship between the rate constants for the forward (kf ) and backward
(kb) processes and the concentrations of each species at chemical equilibrium;
• Imagine you start the reaction with equal concentrations of
each reactant [A]o = [B]o. Predict the ratio of concentrations
at equilibrium for chemical systems in which a) kf / kb = 1; b)
kf / kb = 2; c) kf / kb = 0.5.
• Use the simulation that you can launch by clicking on the image to verify your predictions.
Share your ideas with a classmate and justify your reasoning.
CLICK TO PLAY
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M4/eqab2c.html
Although most chemical reactions do not occur on a single step like the reaction A
B analyzed in the previous example, the equilibrium constant of more
complex processes can also be expressed in terms of the rate constants for the different elementary processes involved in the reaction mechanism
Chemical Thinking
FACING THE CHALLENGE
Although a variety of experiments have shown
that small organic molecules can form from the
reaction between simple inorganic molecules such
CO2, H2O, and NH3, one of the unresolved issues in moderns theories about the origin of life
is how small organic molecules then combined to
form complex polymers such as proteins, RNA,
and DNA. Analyses of reaction mixtures as those
generated in the Miller-Urey and Oró experiments
show that besides the synthesis of the basic components of common biopolymers, such as amino
acids and nucleotides, the process also leads to the
formation of chemical compounds known to hinder polymerization processes. Additionally, polypeptides (polymeric chains of amino acids) and
nucleic acids may break apart through their reaction with water (hydrolysis).
Polymeric chains made up of 20-100 monomeric units seem to be required to have any primitive catalytic and replication functions. However,
the polymerization of amino acids and nucleotides
are thermodynamically unfavorable processes.
Some scientists have proposed that the selective
adsorption of monomers onto mineral surfaces
may have promoted their polymerization, a process that has been confirmed in the laboratory.
Once the polymers were formed, they may have
detached from the surface by thermal motion or
by the action of concentrated salt solutions, a process that could have taken place in tidal regions
during evaporation or freezing of seawater.
Hydrothermal vents may have been critical
sites for the formation of short polypeptides, as
peptide bond formation becomes more thermodynamically favored at higher temperatures. Experiments carried out with glycine and other amino
acids under conditions similar to those present in
hydrothermal vents have led to the formation of
molecules with up to 6 amino acid units. However, these amino acids rapidly decompose by hydrolysis with water. To solve this problem, it has been
proposed that the development of autocatalytic
species may have helped sustain a steady concen-
How do we predict chemical change?
tration of short peptides.
Autocatalytic species are chemical substances that catalyze their own synthesis. An example
of an autocatalytic elementary step in a reaction
mechanism may take the form: A + B
2B.
CLICK TO PLAY
Oscillating chemical reactions, such as the
Belousov-Zhabotinsky reaction, involve
autocatalytic steps in their mechanism.
One may speculate that as polymerized molecules
increased in length and complexity, some of them
may have adopted configurations that facilitated
their binding to other molecules. These molecules
may have worked as primitive catalysts for different types of chemical reactions. As the variety of
polymeric combinations increased, some of these
molecules may have been able to catalyze their
own self-replication. Although these self-replicating molecules may have been scarce ar first, they
would have become increasingly more abundant
given their ability to autocatalyze their own synthesis. Several known examples of self-replicating
molecular systems have been already discovered
and studied in the laboratory.
The appearance of the first molecular species
capable of replication, catalysis, and multiplication
would have marked the origin of life and evolution
in our planet. It is judged that the first “living”
molecular entities must have had the following
properties: they could make imperfect copies of
themselves, to allow for the formation of more or
less effective “mutant” molecules; they could facilitate the replication of other molecules; and they
could survive long enough to replicate before decomposing. It is believed that the first molecules of
these type may have had a structure similar to that
of RNA in modern living organisms.
337
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M4/bz.html
Polymerization and Replication
U5
338
MODULE 4
Measuring Rates
Let’s Apply
Amino acids formed in the primitive Earth may have been involved in a variety of chemical process, such as decomposition, racemization, and dimerization. In fact, it is suspected that the formation of dimers, the first step in the polymerization process that leads to the formation of proteins,
may have helped amino acids accumulate and escape rapid decomposition at the high temperatures
that were common in our planet.
Dimerization Kinetics
Let us explore the dimerization of alanine. The dimerization process has the basic
structure: 2 A
A2 + B.
•
Click on the image below to launch the simulation of the dimerization of alanine.
The data that you can collect using this simulation can be used to determine the order of
the reaction, the values of the rate constant at different temperatures, and the activation
energy for the process.
CLICK TO PLAY
•
Run the simulation at a constant temperature and
collect data for the concentration of alanine (mM) as a
function of time (s). Graph the data using the proper
selection of variables to determine the order of the reaction and the value rate constant for the dimerization
process at the selected temperature.
•
Repeat the process at different temperatures to obtain a collection of values for
the rate constant k as a function of temperature. Use these data to determine the
activation energy Ea for the reaction.
Calculate the half-lives of a 1 M solution of alanine at 100 oC, 150 oC, and 200
o
C.
Compare all your results for the dimerization of alanine with the equivalent
results for the decomposition (decarboxilation) of this amino acid generated
through the different activities in this module. Based on these results, identify
whether there is a range of temperature in which one could expect alanine to
dimerize before undergoing decomposition.
•
•
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M4/dimer.html
ASSESS WHAT YOU KNOW
Monomers or Dimers?
Chemical Thinking
U5
How do we predict chemical change?
339
Hydrolysis Kinetics
Once an amino acid dimer is formed, it may react with water and get decomposed back into
the free amino acids: A2 + B
2 A. Processes in which chemical bonds are broken by interaction with water molecules are generically known as “hydrolysis.” The kinetic study of the
hydrolysis of amino acid dimers and polymers at high temperatures and pressures is of central
importance to understand how proteins may have developed in the primitive Earth.
•
t (h)
0
[GG] (mol/L)
0.020
10
20
30
40
50
0.0178 0.0158 0.0140 0.0125 0.0111
Use these data to determine A) the order of the hydrolysis reaction, B) the value of the
associated rate constant kh, and C) the half-life in days of the GG dimers under the given
experimental conditions.
The following tables show data for the half-lives t1/2 of the GG dimers at different temperatures
and pressures:
P = 25 atm
•
P = 265 atm
T (oC)
t1/2 (h)
T (oC)
t1/2 (h)
100
2.40 x 103
100
1.75 x 102
160
2.90 x 10
160
2.15 x 10
220
1.03
220
5.17
280
7.51 x 10-2
280
1.58
Use these data to determine the activation energy for the hydrolysis reaction at each of the
experimental pressures. Discuss how pressure affects the rate of hydrolysis.
The values for the equilibrium constant for the hydrolysis of GG dimers at 265 atm of pressure
are KC = 1.9 x 10-2 at 160 oC and KC = 4.8 x 10-2 at 220 oC.
•
•
Use the information provided to determine the rate constants kd for the dimerization of
glycine at 160 oC and 220 oC at 265 atm.
Estimate the value of the activation energy for the dimerization process, and discuss
whether GG dimers can be expected to form at the given conditions.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
The following table shows experimental data for the hydrolysis of glycine
dimers (GG) at 120 oC and 265 atm:
340
MODULE 4
Measuring Rates
Let’s Apply
Drug Degradation
ASSESS WHAT YOU KNOW
Many pharmaceutical drugs undergo chemical reactions that alter their chemical composition. The
rate of the degradation or decomposition processes depends on temperature and relative humidity.
Analyzing reaction kinetics is critical to estimate the shelf life of these pharmaceutical products.
Paracetamol
Paracetamol, or acetominophen, is a widely used over-the-counter
pain reliever and fever reducer (active component of Tylenol). Under adverse storage conditions, this substance may decompose by
reacting with water. The data in the following table shows changes
in the amount of paracetamol (p-ac) present in a 200 mg commercial tablet as a function of time at 15 oC and 75% relative humidity:
•
•
•
•
t (days)
0
50
100
150
200
250
300
p-ac (mg)
200
198.3
196.7
195.0
193.4
191.8
190.2
Calculate the average rate of decomposition between 100 - 150 days and 250-300 after
production. Sketch how the amount of paracetamol changes over time.
Use the data to determine A) the order of the decomposition reaction for paracetamol,
and B) the value of the rate constant k for the process at 15 oC and 75% humidity;
Calculate the half-life t1/2 (in days) for the decomposition of paracetamol;
The shelf life of a pharmaceutical drug is usually defined as the time it takes for 10% of
the drug to degrade (or the time it takes for the initial amount to decrease to 90%). Calculate the shelf life (expressed in months) of paracetamol at 15 oC and 75% humidity.
The table shows the values of
the rate constants for the decomposition of paracetamol
at three different temperatures
and two different relative humidities (RH):
•
T (oC)
25
37
45
k (days-1) 75% RH
1.92 x 10-4 2.17 x 10-4 2.42 x 10-4
k (days-1) 100% RH
1.69 x 10-4 1.92 x 10-4 2.12 x 10-4
Use these data to estimate the value of the activation energy Ea for the decomposition
process at different relative humidities. Compare your results and discuss how humidity
affects the rate of decomposition.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
Where From?
Where To?
How do we predict chemical change?
reactants is kinetically stable. As we have seen
in this unit, kinetic stability depends on factors
such as concentration of reactants, temperature,
and activation energy of the process. Predictions
about the rate of a chemical reaction can be simplified if we have a sense of the sequence of steps
that lead from reactants to products. Thus, the
determination and analysis of reaction mechanisms is a central activity for many chemical scientists. This work involves careful experimental
exploration of the nature of the intermediate species generated during a chemical reaction, as well
as analysis of how the concentrations of reactants
and products change as a function of time under
different conditions of temperature and pressure.
The intellectual and experimental tools discussed in this Unit are invaluable in the analysis and prediction of the properties and behavior
of reactive systems, from the hydrosphere to the
atmosphere in our planet, from single-celled to
multicellular organisms, from the combustion
engines in our cars to the batteries that power
our cell phones. In all these cases, understanding
change requires that we pay attention to the thermodynamic and kinetic factors that determine
the directionality, extent, and rate of any potential transformation.
In the following Unit of the textbook we
will seek to apply some of the ideas that we have
discussed to develop strategies to control chemical reactions. To simplify our task, we will focus
our attention on the study of chemical processes
that occur rapidly but often with limited extent.
This will allow us to discuss issues related to thermodynamic control without having to worry
about kinetic issues. We will leave for Unit 7 the
analysis of systems in which kinetic control is
also relevant. However,
before we begun a
new Unit, let us explore if “YOU ARE
READY” to apply what you have
learned to the analysis of some interesting
systems.
341
By Nina Matthews (Own work)
[Generic 2.0] via Flickr Commons
The central goal of this
unit was to introduce
the qualitative and quantitative ways of thinking
used in chemistry to
predict the directionality, extent, and rate
of chemical processes.
Thus, we analyzed how to use compositional and
structural cues to qualitatively compare the relative potential energy and entropy of reactants and
products. These types of comparisons are useful
in predicting the likelihood of a chemical process.
However, actual experimental data can be used to
quantify changes in the enthalpy and the entropy
of a system, and quantitatively predict reaction
extent based on the analysis of the sign and value
of the change in the Gibbs free energy due to the
reaction.
During our discussions, we emphasized that
the analysis of the thermodynamic stability of a
chemical system demands that we pay close attention to both energetic and entropic factors. Some
chemical processes mainly occur because random
atomic movements lead to states with lower potential energy in which strong attractive interactions among particles limit further rearrangements. Other processes happen because random
particle motions take the system to a state with
so many different configurations (high entropy)
that is unlikely that the system will revert back
to its original state. In some cases, energetic and
entropic factors act together in driving a chemical reaction. In other cases, these factors are in
competition and the final outcome depends on
the working temperature.
Although a chemical process may be favored
from the thermodynamic point of view, it may
take so long for it to happen that the mixture of
U5
342
Are you Ready?
Are You Ready?
Many of the ideas discussed in this unit can be applied to the study of the directionality, extent, and rate of
physical changes. This is, transformations in which the chemical nature of the submicroscopic species that
make up the system (e.g., atoms, ions, molecules) does not change during the process.
Phase Changes
A change of state, such as melting or boiling, is a typical example of a physical transformation. During these types of processes, energy in the form of heat
is exchanged between the system and its surroundings and the entropy of the
system may increase or decrease. Consider, for example, the changes in enthalpy
(DH) and entropy (DS) when water melts or evaporates at different temperatures under atmospheric pressure at sea level (1 atm):
H2O(s)
T
(K)
•
•
•
•
H2O(l)
DHfus
DSfus
(kJ mol-1) (J K-1 mol-1)
H2O(l)
T
(K)
H2O(g)
DHvap
DSvap
(kJ mol-1) (J K-1 mol-1)
233.15
6.18
22.7
333.15
42.3
113.7
253.15
6.10
22.3
353.15
41.5
111.3
273.15
6.01
22.0
373.15
40.7
109.0
293.15
5.92
21.7
393.15
39.8
106.8
313.15
5.84
21.4
413.15
39.0
104.7
Calculate the change in Gibbs free energy DG for each of these processes
at each of the given temperatures;
Analyze the evolution of DG as the temperature increases. What do the
DG values indicate about the directionality of the phase transitions at different temperatures?
The melting point of water is Tm = 273.15 K (0 oC) and its boiling point
is Tb = 373.15 (100 oC) at 1 atm of pressure. What is the value of DG for
each of the phase changes at the temperature of the transition? How do
you explain these results?
Based on you analysis, establish a general relationship between the DH,
DS, and T for any substance at the point in which a phase transition can
be expected to occur.
By Disdero (Own work)
[Share Alike 2.5]
via Wikimedia Commons
Share and discuss your results with a classmate, and justify your reasoning.
Chemical Thinking
U5 How do we predict chemical change?
343
Physical Transformations
Dissolution
The extent to which a solute dissolves into a solvent can be predicted by evaluating the
change in Gibbs free energy DG for the process. Consider the values for DHosoln, DSosoln, and
DGosoln for the dissolution in water of various ionic compounds under standard conditions:
Dissolution Process
3.8
42.9
–9.0
NaBr(s)
Na+(aq) + Br-(aq)
–0.6
52.4
–16.2
Na (aq) + I (aq)
–9.3
66.4
–29.1
NaNO3(s)
Na+(aq) + NO3-(aq)
20.7
88.7
-5.7
MgCl2(s)
Mg2+(aq) + 2 Cl-(aq)
–159.9
–113.6
–126.0
AlCl3(s)
Al3+(aq) + 3 Cl-(aq)
–335.5
–264.8
–256.5
CaCO3(s)
Ca (aq) + CO (aq)
–12.3
–201.7
47.7
Zn2+(aq) + S2-(aq)
94.4
–145.5
137.8
3 Ca2+(aq) + 2 PO43-(aq)
–62.4
–836.3
186.9
Ca3(PO4)2(s)
•
DGosoln
(kJ mol-1)
Na+(aq) + Cl-(aq)
ZnS(s)
•
DSosoln
(J K-1 mol-1)
NaCl(s)
NaI(s)
•
DHosoln
(kJ mol-1)
+
2+
-
23
Calculate the equilibrium constant for each of the above dissolution processes under standard conditions. Based on the data available and your own results, discuss
which compounds can be expected to be soluble or insoluble in water;
Analyze the data to evaluate how energetic and entropic factors affect the solubility
of these different compounds. For example, how does the charge of the ions seems
to affect the values of DHosoln and DSosoln? How does the size of the ions affect these
quantities? How would you explain these different effects?
How would you use the data to justify the solubility rule that suggests that most
ionic compounds in which both cations and anions are multi-charged are likely to
be insoluble in water?
Consider the following thermodynamic data for AgCl(s) and AgBr(s):
Dissolution Process
•
DHosoln
(kJ mol-1)
DSosoln
(J K-1 mol-1)
DGosoln
(kJ mol-1)
AgCl(s)
Ag+(aq) + Cl-(aq)
65.7
33.8
55.6
AgBr(s)
Ag+(aq) + Br-(aq)
84.8
47.1
70.8
How would you explain the unexpected solubility properties of these compounds?
Share and discuss your results with a classmate, and clearly justify your reasoning.
344
Are you Ready?
Physical Transformations
Clathrates
Clathrates are ice-like systems in which methane or other gases are
caged by water molecules. On freezing, water molecules commonly
arrange in a hexagonal crystal structure. In the presence of sufficient
methane and other gases, water crystallizes in a cubic lattice that traps
gas molecules. Large deposits of methane clathrate have been found on
the ocean floors of our planet. The amount of carbon present in these
hydrates is estimated to be at least twice the amount found in all fossil
fuels on Earth. The study of methane clathrates have recently gained
much attention because their potential contribution to global warming
if they were to release methane, a greenhouse gas, into the atmosphere.
The rate constants of methane clathrate formation kf have been measured at different temperatures at high pressure in the laboratory. The following table presents typical results.
•
•
Use this information to estimate the activation energy
for the formation of methane clathrates at high pressure.
The activation energy for the dissociation (destruction)
of the clathrate is close to 32 kJ/mol. Compare this value
with the activation energy for clathrate formation and
generate a hypothesis to explain the difference.
T (K)
kf
230
4.70 x 10-6
240
1.03 x 10-5
250
2.11 x 10-5
260
4.09 x 10-5
Share and discuss your results with a classmate, and clearly justify your reasoning.
Zeolites and Thermodynamics
Zeolites are microporous crystalline solids commonly used as adsorbents and catalysts. Their porous structure is generally made up of aluminum, silicon, and oxygen atoms (aluminosilicates)
and it can trap different types of cations such as Na+, K+, Mg2+, and Ca2+. Thus, these materials are
widely used in the water purification industry to eliminate undesired species. Consider the data
for the standard heat of adsorption DHoad and the standard entropy of adsorption DSoad for Cu2+,
Zn2+, and Cd2+ on a typical zeolite Z.
•
•
How would you explain the positive values of DHoad and
DSoad for the adsorption process (transfer of ions from solution to the zeolite surface)?
Calculate the standard Gibbs free energy of adsorption DGoad
for each of these ions and determine which adsorption process is more thermodynamically favored.
Ion
DHoad
DSoad
(kJ mol-1) (J mol-1 K-1)
Cu2+
19
75
Zn2+
20
77
Cd2+
23
82
Chemical Thinking
U5 How do we predict chemical change?
Are You Ready?
Zeolites and Kinetics
The following tables show values of the rate constants for adsorption kad, measured in grams of zeolite Z per
milligram of adsorbed ion per minute (g mg-1 min-1), for Cu2+, Zn2+, and Cd2+ at different temperatures:
kad
M2+ + Z
M2+-Z
kdes
•
•
•
Cu2+
Zn2+
Cd2+
T (oC)
kad
(g mg-1 min-1)
T (oC)
kad
(g mg-1 min-1)
T (oC)
kad
(g mg-1 min-1)
30
1.35 x 10-2
30
1.31 x 10-2
30
9.70 x 10-3
40
1.45 x 10-2
40
1.41 x 10-2
40
1.42 x 10-2
50
2.60 x 10-2
50
2.43 x 10-2
50
2.14 x 10-2
60
3.28 x 10-2
60
2.86 x 10-2
60
2.01 x 10-2
Use these data to estimate the activation energy of adsorption on the zeolite for each of the three ions.
Based on the available data, sketch energy diagrams to represent the adsorption of each of the ions on
the zeolite. Discuss which of the adsorption processes is more kinetically favored.
Use the information provided, together with your own results, to calculate the rate constants for desorption kdes (transfer of ions from the zeolite to aqueous solution) for each of the three ions at 30 oC.
Share and discuss your results with a classmate, and clearly justify your reasoning.
Zeolites as Catalysts
Zeolites can serve as catalysts for different chemical reactions. The associated reaction mechanisms typically
involve fast adsorption steps followed by a slow step in which the adsorbed species undergo the chemical
transformation into the product P. Consider the following two generic reaction mechanisms:
MECHANISM 1:
A + Z
AZ
Fast
MECHANISM 2: A + Z
AZ
P + Z Slow
B + Z
AZ + BZ
AZ
Fast
BZ
Fast
P + Z Slow
Let us call CA and CB the concentration of the different reacting species, CZ the concentration of empty
adsorbing sites on the zeolite, and CAZ and CBZ the concentration of zeolite sites occupied by species A and
B, respectively.
•
Derive the overall rate law for each of these mechanisms in terms of CA, CB, and CZ. Assume that the
fast steps in the mechanisms reach equilibrium with associated equilibrium constants K1 and K2.
345
346
Are you Ready?
Are You Ready?
We have already discussed that ammonia, NH3, is one of the most important industrial chemical substances
in the world. Its industrial synthesis is accomplished taking advantage of our understanding of the thermodynamic and kinetic factors that affect the chemical process.
Synthesis and Thermodynamics
Most industrial ammonia is produced by combining N2(g) and H2(g):
Synthesis Process
1/2 N2(g) + 3/2 H2(g)
•
•
•
•
NH3(g)
DHorxn
(kJ mol-1)
DSorxn
(J K-1 mol-1)
–45.9
–99.1
Justify the signs of DHorxn and DSorxn based on the analysis of the composition
and structure of reactants and products;
Calculate the standard change in Gibbs free energy DGorxn for the reaction.
Express the equilibrium constant for the process KP in terms of the pressures of
reactants and products. Calculate its value under standard conditions.
The synthesis of ammonia is thermodynamically favored at some temperatures
but not others. Discuss whether the reaction will be favored at low or high
temperatures. Estimate the temperature at which the reaction goes from being
thermodynamically favored to thermodynamically unfavored.
Share and discuss your ideas and results with a classmate, and justify your reasoning.
Thermodynamics versus Kinetics
The activation energy for the synthesis of NH3(g) from H2(g) and N2(g) in the absence
of catalyst is high, Ea = 325 kJ/mol. The rate of the reaction may be increased by carrying out the synthesis at high temperatures.
•
•
•
•
Use the data that has been provided to build an internal potential energy diagram for the synthesis of ammonia;
Determine how many times faster the reaction would be at 500 oC than at 25
o
C;
Determine how many times smaller the equilibrium constant for the synthesis
reaction KP would be at 500 oC than at 25 oC.
Evaluate the trade-offs associated with running the synthesis at high T.
Chemical Thinking
U5 How do we predict chemical change?
347
Revisiting Ammonia
The synthesis of ammonia is typically carried out at high temperatures (300 - 550
o
C) and high pressures (150 - 250 atm)
•
•
CLICK TO PLAY
Predict how an increase in pressure is
expected to affect A) the rate of the reaction, B) the amount of NH3 produced the
reaction, and C) the equilibrium constant
KP for the process;
Click on the image to launch a simulation of the synthesis reaction. Use these
interactive tool to verify your predictions.
The reaction can be speeded up by using a catalyst that reduces the activation energy
of the process. The most popular catalysts are based on iron (Fe) and ruthenium
(Ru). The rate of reaction for the catalyzed process depends on the concentration of
nitrogen gas, N2(g), in the reactor:
•
The following table shows the variation in the concentration of N2(g) as a
function of time for a reaction carried out at 500 oC and 200 atm. Use these
data to determine A) the order of the reaction with respect to N2(g) and
B) the value of the overall rate constant k for the synthesis process at the given
conditions.
t (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
[N2(g)] (mol/L)
2.00
1.68
1.42
1.19
1.01
0.846
0.710
The rate constant for the synthesis reaction in the presence of a Fe-based catalysts has
been determined at various temperatures. The following table presents the results:
T (oC)
k (s-1)
25
100
200
300
400
3.24 x 10-9 2.32 x 10-6 5.81 x 10-4 2.11 x 10-2 2.64 x 10-1
•
Use the data provided to determine the activation energy in the presence of
the Fe-based catalyst;
•
Estimate how many times faster the reaction is in the presence of catalyst
compared to the reaction rate in its absence (Ea = 325 kJ/mol) at 500 oC.
Share and discuss your ideas and results with a classmate, and justify your reasoning.
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M4/nh3.html
Catalysis
348
Are you Ready?
Revisiting Ammonia
Mechanism
The following image depicts the reaction path for the synthesis of NH3 from N2 and
H2 over the surface of a ruthenium (Ru) catalyst. The symbol * denotes an empty
site on the catalyst and X* represents an adsorbed species.
Ep
•
•
•
Describe the sequence of elementary steps that take place. Identify all the
intermediate species. Discuss how the catalyst provides a reaction path with a
lower effective activation energy. What processes decrease the internal potential energy of the reacting species?
Identify the rate-limiting step in the process. Assuming that all of the steps
that only involve adsorption or desorption of species are fast and achieve
chemical equilibrium, derive the overall rate law for the synthesis reaction in
the presence of a catalyst. Use the following notation to represent the concentration of relevant species: [N2], [H2], [*] (this symbol represents the concentration of empty adsorption sites on the catalyst);
Based on your results, discuss how doubling the concentration of [N2] and
[H2] by increasing the pressure should affect the rate of the reaction.
By ESA
Share and discuss your ideas and results with a classmate, and justify your reasoning.
Chemical Thinking
U5 How do we predict chemical change?
349
Unit 5: Image Attributions
Module 1
Most of the images in this module have been generated by the authors (Own work). Other attributions include, P282 & P283: Background
derivative from “The Origin of Life” poster: Lynn Margulis & Dorian Sagan; P284: Top right “Rust on iron” by Laitr Keiows (Own work) [Share
Alike 3.0] http://commons.wikimedia.org/wiki/File:Rust_on_iron.jpg; P285: Fig. 5.1 “Et baal” by Einar Helland Berger (Own work) [Share Alike
2.5] http://commons.wikimedia.org/wiki/File:Et_baal.jpg; P293: Center By NASA/Jenny/Mottar.
Module 2
Most of the images in this module have been generated by the authors (Own work). Other attributions include, P298: Top right “AerialViewPhotochemicalSmogMexicoCity 2” by Fidel Gonzalez (Own work) [Share Alike 3.0] http://commons.wikimedia.org/wiki/File:AerialViewPho
tochemicalSmogMexicoCity_2.jpg; P302: Fig. 5.6 “Bombenkalorimeter mit bombe” by Harbor1 (Own work) [Generic 3.0] http://commons.
wikimedia.org/wiki/File:Bombenkalorimeter_mit_bombe.jpg; P309: Center “Miller-Urey experiment-en” by GYassineMrabetTalk (Own work)
[Share Alike 3.0-2.5-2.0-1.0] http://commons.wikimedia.org/wiki/File:Miller-Urey_experiment-en.svg; P310: Bottom center By NASA; P311:
Center photograph of mural at Page Museum (La Brea Tar Pits).
Module 3
Most of the images in this module have been generated by the authors (Own work). Other attributions include, P321: Center right Derived from
“Competitive inhibition” by Jerry Crimson Mann (Own work) [Public domain] http://commons.wikimedia.org/wiki/File:Competitive_inhibition.svg; P322: Top right “Induced fit diagram” by Created by TimVickers (Own work) [Public domain] http://commons.wikimedia.org/wiki/
File:Induced_fit_diagram.svg; P323: Center “Chirality with hands” (Public domain) http://commons.wikimedia.org/wiki/File:Chirality_with_
hands.svg.
Module 4
Most of the images in this module have been generated by the authors (Own work). Other attributions include, P341: Top left Derivative from
“The Origin of Life” poster: Lynn Margulis & Dorian Sagan; Bottom right “Day 139 - IF you Run, Jog or Cycle Look below O_O” by Nina Matthews (Own work) [Generic 2.0] https://www.flickr.com/photos/21560098@N06/6722653055/in/set-72157627444093697/; P342 & P343:
Background “DeadSea3667”by Disdero (Own work) [Share Alike 2.5] http://commons.wikimedia.org/wiki/File:DeadSea3667.jpg; P348: Right
By ESA (European Space Agency).
General
Activity icons: Clip art from Microsoft Office; Molecular structures: Derived from online public software (Chemical Education Digital Library,
Jmol) and via Wikimedia Commons (Public domain images).