NAT IONAL QUALIFICAT IONS CURRICULUM SUPPORT
Chemistry
Staff Notes for
Unit 2: Principles of Chemical
Reactions
[ADVANCED HIGHER]
Gavin Whittaker
University of Edinburgh

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Acknowledgements
This document is produced by Learning and Teaching Scotland as part of the National
Qualifications support programme for Chemistry. Grateful thanks are due to the Education
Division of the Royal Society of Chemistry, Scottish Region, for comment on the drafts.
The support of the Higher Still Development Unit and the editorial advice of Douglas
Buchanan are also acknowledged.
The author and publisher are grateful to the following for permission to reproduce
photographs: Marc Day, Lawrence Berkeley National Laboratory, USA (hydrogen fire, page
21); International Fuel Cells and Ballard Power Systems (fuel cell images, pages 54 –56).
First published 2001
Electronic version 2002
© Learning and Teaching Scotland 2001
This publication may be reproduced in whole or in part for educationa l purposes by
educational establishments in Scotland provided that no profit accrues at any stage.
ISBN 1 85955 913 1
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CONTENTS
Section 1: Thermochemistry
1
Section 2: Reaction feasibility
17
Section 3: Equilibria
29
Section 4: Fuel cells
51
Section 5: Kinetics
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Note on safety
Teachers/lecturers are advised that it is their responsibility
to take notice of employers’ regulations with regard to the
safe practices to be followed. Where necessary, prior to
using chemicals, the relevant advice may be consulted in
the Hazardous Chemicals Manual (SSERC), either in
printed form or in CD-ROM version.
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SECTION 1
Introduction
Thermochemistry is the branch of thermodynamics that deals with the
interchange of heat in chemical processes.
The science of thermodynamics itself was developed through necessity in
the nineteenth century and led directly to the steam engine. It is a
macroscopic science, in that it allows us to deal with properties of matter
without any assumptions about the nature of matter itself. Thermodynamics
examines the relationships between two physical quantities – energy and
entropy. Energy may be regarded as the capacity to do work, while entropy
may be regarded as a measure of the disorder of a system.
In the chemical context, the relationship between these properties is the
driving force behind chemical reactions and determines their feasibility.
Since energy is either released or taken in, and since the level of disorder is
changed in all chemical processes, thermodynamics enables us to predict
whether a reaction may occur or not without the need to consider structure,
bonding or reaction mechanism.
There are limitations to the practical scope of thermochemistry that should
be borne in mind. Consideration of the heat change in the course of a
reaction is only one part of the story. Although hydrogen and oxygen will
react to release a great deal of energy under the correct conditions, both
gases will coexist indefinitely without reaction. Thermodynamics te lls us
about the potential for chemical change, not the rate of chemical change –
that is the domain of chemical kinetics. Because it is such a commonly held
misconception that the potential for change depends on the release of
energy, it should also be held in the back of one’s mind that it is not energy
but entropy that is the final arbiter of chemical change.
Energy is transferred as either heat or work, which, while familiar, are
not always easily defined. One of the most useful definitions is deriv ed
from the mechanical fashion in which energy is transferred either as
heat or work. Heat is the transfer of energy as disorderly motion, as
the result of a temperature difference between the system and its
surroundings. Work is the transfer of energy as orderly motion. In
mechanical terms, work is due to energy being expended against an
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opposing force. The total work is equal to the product of the force and the
distance moved against it. Work in chemical or biological systems generally
manifests itself in only a limited number of forms. The most commonly
encountered are pressure–volume work and electrical work.
Enthalpy
The majority of chemical reactions, and almost all biochemical processes in
living organisms, are performed under constant pr essure conditions and
involve relatively small volume changes. When a process takes place under
these conditions, a useful measure of heat transfer is enthalpy, denoted by
the letter H.
The rigorous definition of enthalpy, H, is beyond this text, but a change in H
(H) may be conveniently defined as the heat given out or taken in during a
reaction (or any other process) at constant pressure. In other words, it is the
heat change under the most commonly encountered reaction conditions.
State f unctions and path f unct ions
An i mportant classification of ther modynami c properties is
whether they are regarded as state functions or path functions. If
the value of a particul ar property for a system depends solel y on
the state of the system at that ti me, then such a propert y is referred
to as a state f unction . Examples of state functions are volume,
pressure, internal ener gy and entropy. Where a propert y does not
depend on the state of the system, but on the path by which a
system in one state is changed into another stat e, then that
property is referred to as a path f unction. The distinction is
important because in performing calculations on state functions we
need take no account of how any state of interest was prepared.
This distinction may be compared to altitud e. A person at the base
of a mountain has a specific altitude, as does one at the summit .
The amount of wor k that is done in cli mbing from the base to the
summit depends on the route that is taken, and this wor k is a path
function. Thus, a person cli mbi ng from Fort William to the
summit of Ben Nevis changes their altitude (a state function) by
4000 feet whether they take the most direct route or whether they
travel via Ulan Baator, although the amount of wor k that is done (a
path function) will clearly b e different in each case!
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Properties of enthalpy
Enthalpy is a state function. In other words a system possesses a defined
value for any particular system at any specific conditions of temperature and
pressure. However, practical considerations mean t hat its absolute value in a
system cannot be known, since there would be contributions to the absolute
enthalpy from, for example, nuclear binding energies. Fortunately, changes
in enthalpy can be measured, and are denoted by the symbol H, where
H = H fina l – H init ia l .
Enthalpy changes may result from either physical processes (e.g. heat loss to
a colder body) or chemical processes (e.g. heat produced during a chemical
reaction).
An increase in the enthalpy of a system leads to an increase in its
temperature (and vice versa), and is referred to as an endothermic process.
Loss of heat from a system lowers its temperature and is referred to as an
exothermic process. The sign of H indicates whether heat is lost or gained
by the system. For an exothermic process, where heat is lost from the
system, H has a negative value. Conversely, for an endothermic process in
which heat is gained by the system, H is positive. The sign of H indicates
the direction of heat flow and should always be explicitly stated, e.g.
H = +2.4 kJ mol –1 .
For chemical reactions, the most basic relationship follows directly from the
fact that enthalpy is a state function. The enthalpy change that acco mpanies
a chemical reaction is equal to the difference between the enthalpy of the
products and that of the reactants:
H reaction = H products – H reactants
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Standard state
The enthalpy changes associated with any reaction are, to varying
degrees, dependent on temperature, pressure, and the states of the
reactants and products. In order to utilise enthalpy measurements,
therefore, it is important that the conditions under which the reaction
takes place are identical. For this reason, it is convenie nt to specify a
standard state for a substance. The standard state for a substance is
defined as being the pure, undiluted substance at 1 atmosphere pressure
and at a specified temperature. The temperature does not form part of the
definition of the standard state, but for historical reasons data are
generally quoted for 298.15K (25°C). For solutions, the definition of the
standard state of a substance is a 1 mol l –1 concentration.
Table 1
Some common enthalpy changes (note that the enthalpy change is n ot the
same as the free energy change – see page 25).
Standard enthalpy changes
Quantity
Enthalpy associated
with:
Notation
Example
Enthalpy of ionisation
Electron loss from a
species in the gas phase
Hi
Na(g)  Na+(g) + e–(g)
Enthalpy of electron
gain
Gain of an electron
Hec
½F2(g) + e–(g)  F–(g)
Enthalpy of vaporisation
Vaporisation of a
substance
Hv
H2O(l)  H2O(g)
Enthalpy of sublimation
Sublimation of a
substance
Hsub
CO2(s)  CO2(g)
Enthalpy of reaction
Any specified chemical
reaction
H
2Fe2O3(s) + 3Zn(s)  Fe(s) + 3ZnO(s)
Enthalpy of combustion
Complete oxidation of
a substance
Hc
H2(g) + ½O2(g)  H2O(g)
Enthalpy of formation
Formation of a substance
from its elements in
their standard state.
Hf
C(s) + 2H2(g)  CH4(g)
Enthalpy of solution
Dissolution of a
substance in a specified
quantity of solvent
Hsol
NaCl(s)  Na+(aq) + Cl–(aq)
Enthalpy of solvation
Solvation of gaseous ions
Hsolv
Na+(g) + Cl-(g)  Na+(aq) + Cl–(aq)
from an ionic substance
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The definition of a standard state allows us to define the standard enthalpy
change as the enthalpy change that comes about when reactants in their
standard states are converted into products in their standard states. The
enthalpy change may be the result of either a physical process, such as
melting, or a chemical process, such as oxidation. The standard enthalpy
change for a process is denoted as H  298K , with the subscript denoting the
temperature.
In order to aid concise discussion, a number of chemical and physical
processes are given specific names. Thermodynamically, there are no
differences between the processes, since they all refer to the heat change as
the components in the system convert from one form to another. The only
reason for the use of these specific terms is convenience and brevity. A
selection of the more important processes is listed in Table 1.
Measuring enthalpy changes
Enthalpy would clearly be of limited importance if we were not able to
measure enthalpy changes. Fortunately, there are a number of
experimental methods that allow us to quantify the enthalpy changes that
accompany almost all conceivable chemical reactions or physical
processes, such as melting or boiling. The methods may be broadly
categorised as direct or indirect. Direct methods, as the name suggests,
involves the direct measurement of the enthalpy associated with the
reaction or process. The most common method for this uses a
calorimeter, such as a bomb calorimeter. Indirect methods are used to
determine enthalpy changes for processes that are difficult, or practically
impossible, to measure based on Hess’s law, and involve calculating the
enthalpy change of interest indirectly from en thalpy changes that can be
accurately measured.
Direct methods
Direct methods are reserved for reactions that can be performed with a high
degree of selectivity (i.e. the reaction proceeds cleanly and completely to
products without forming by-products), with the measurements being
performed using calorimetry – literally ‘to measure heat’ (being a mongrel
word from the Latin calor, heat, and the Greek metron, measure).
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The most familiar technique in calorimetry (if not always the best
understood!) is the bomb calorimeter, in which a material is burnt in a bomb
at constant volume in a very high pressure oxygen atmosphere. The bomb is
placed in a volume of water, and the heat from the reaction increases the
temperature of the water. By knowing the te mperature change in the water
and the heat required to cause this temperature change (the calorimeter
having previously been calibrated using a known substance of an electric
heater), the heat given out from the reaction can be determined.
It is not always appreciated that the heat measured by this method does not
necessarily provide a direct value for the enthalpy change. This is because
enthalpy changes are measured at constant pressure, and this method provides
a measure of the heat given out at constant volume. The difference between
the two values arises because at constant pressure some energy has to be used
up in pushing back the atmosphere.
Sensitivity
For most people, their experience of calorimeters is of poorly reproducible
measurements of highly exothermic reactions. The level of sophistication
and accuracy in modern calorimetry is therefore somewhat surprising
(modern being a relative term – a high level of accuracy has been available
for at least 50 years!). The rapidly growing inter est in biological
reactions, for example, has led to some interesting approaches to the
calorimetry of reactions that release only small amounts of heat. The
experimental measurement of H  for these reactions is called
microcalorimetry. In the 1950s the biophysicist T H Bezinger was
largely responsible for the heat-burst calorimetry technique. Using this
method, Bezinger measured the heat evolved when 5 mmol of adenosine
triphosphate (ATP) was hydrolysed in the presence of the enzyme myosin
to be 0.36 J. So sensitive is the technique that it was necessary to
compensate for the heat evolved as protons released by the reaction
reacted with the buffer solution. The advent of modern electronics and
highly sensitive temperature sensors has made even this ex traordinary
achievement seem insensitive.
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Hess’s law
We have already described enthalpy as a state function, and it follows from
this that the absolute enthalpies associated with the reactants and products in
a reaction are independent of the process by which they were formed.
Consequently, the overall enthalpy change during the course of a reaction
(calculated as H reactants – H products) is independent of the reaction
pathway. Hess’s law of constant heat summation is a recognition of this
fact, and states that:
‘The overall enthalpy change for a reaction is equal to the sum of the
enthalpy changes for the individual steps in the reaction measured at
the same temperature.’
This law is particularly useful when measurement of a specific enthalpy
change is impractical or unfeasible. A classic example of this is the
measurement of the enthalpy change associated with the burning of carbon to
form carbon monoxide. The difficulty with this measurement is that
formation of some carbon dioxide cannot be prevented, making direct
experimental measurement of the enthalpy change impossible. However, the
enthalpy change may be measured indirectly, by considering the direct (one step) reaction as an indirect (two-step) process (Figure 1).
Figure 1
H  1 = H  2 – H  3 = –271 kJ mol –1
C + O2
CO + ½O 2
H  2 = –394 kJ mol –1
H  3 = –123 kJ mol –1
CO 2
Hess’s law indicates that the total enthalpy change by either path is identical,
in which case we may write H 1 = H 2 + H 3 , so allowing us to obtain a
value for H 1 without the need for direct measurement.
It is also possible to calculate the enthalpy by addition or subtraction of
reactions, and in this example this may be done as follows:
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1. C(s) + O 2 (g)  CO 2 (g)
H  = –394 kJ mol –1
2. CO(g) + ½O 2 (g)  CO 2 (g)
H  = –123 kJ mol –1
1–2 C(s)+ O 2 (g)+ CO 2 (g)  CO 2 (g)+ CO(g)+ ½O 2 (g)
H  = –394 –(–123) kJ mol –1
C(s) + ½O 2 (g)  CO(g)
H  = –271 kJ mol –1
This approach is effective, but can become unwieldy and confusing when
several reactions are added and subtracted. It is often easier and more
instructive to use graphical methods, but the best approach is the one that the
student finds easiest to visualise and this is a subjective decision for the
teacher and student.
The enthalpy of formation
As defined in Table 1, the enthalpy of formation for a substance is the
enthalpy associated with the formation of a substance in its standard state
from its elements in their standard states. The usefulness of this concept is
most readily appreciated when it is used in conjunction with Hess’s law.
While, in principle, it is possible to tabulate enthalpy values for all
measurable reactions, such data are limited and would be very difficult to
index and access. Tables listing the enthalpies of formation of a wide rang e
of materials, on the other hand, may be found in the chemical literature and
are more readily accessible than the enthalpy change associated with a
specific reaction. For any reaction, it is possible to construct reaction
pathways that proceed via the elemental components of both the reactants and
the products (Figure 2). The value for H reaction is readily calculated from:
H reaction = H f products – H f reactants
Hence, for the example reaction in Figure 2, the reaction enthalpy is given
by:
H reaction =
[H f(CH 3 COO CH 3 ) + H f(H 2 O)] – [H f(CH 3 COO H) + H f (CH 3 OH)]
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Figure 2
H  reaction
CH 3 COO H + CH 3 OH
CH 3 COO CH 3 + H 2 O
H  f(CH 3 COO H + CH 3 OH)
H  f(CH 3 COO CH 3 + H 2 O)
4H 2 + 1½O 2 + 3C
The enthalpy of combustion
The enthalpy of combustion is the enthalpy associated with the burning of
(usually one mole of) a substance in oxygen. This is a useful quantity, as it is
possible to usefully combine the enthalpy of combustion and Hess’s law, just
as Hess’s law is useful in combination with the enthalpy of formation. The
obvious limitation of this method is that it may only be applied to reactions
or processes that involve combustible substances. Take the example of the
reaction between methane and water, which may be used to generate
hydrogen for use as fuel in cars powered using fuel cells (see Section 4). The
energy required to accomplish this may be calculated using tabulated values
for enthalpies of combustion:
CH 4 + H 2 O  CO +3H 2
We may add sufficient oxygen to both sides of the equation to formally
combust the reactants and products (Figure 3)
Figure 3
H reaction
CH 4 + H 2 O + 2O 2
CO + 3H 2 + 2O 2
H c (CH 4 )
H c (CO + 3H 2 )
CO 2 +3H 2 O
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The overall enthalpy of reaction is unaffected by this alteration, but
H reaction may now be calculated using Hess’s law (note the change of sign
as compared to the previous expression):
H reaction = –H c products + H c reactants
From which it may be seen that:
H reaction = H c (CH 4 + 3H 2 ) – H c (CO + 3H 2 )
The advantage of this method is that enthalpies of combustion are more
readily obtained than heats of formation. The disadvantage is that it can only
be applied to reactions involving combustible substances, a restriction that
generally also excludes materials in solution.
Enthalpy of solvation
The enthalpy of solvation, H so lv , is another specific type of enthalpy change.
In this case, however, the process is an almost purely hypothetical case,
which is difficult (if not impossible) to generate experimentally, and is
almost completely impossible to measure. The enthalpy of solvation is
defined as the enthalpy change when a mole of gaseous ions from an ionic
substance is solvated in an infinite amount of solvent, for example:
Na + (g) + Cl – (g)  Na + (aq) + Cl – (aq)
Enthalpies of solvation are calculated using Hess’s law.
Enthalpy of solution
The enthalpy of solution, H so l , is the enthalpy change associated with the
dissolution (dissolving) of a substance in a solvent. It is most often quoted
for a mole of substance dissolving in an infinite amount of solvent, for
example:
NaCl(s)  Na + (aq) + Cl – (aq)
The enthalpy of solution may be determined by measurement of the enthalpy
associated with finite ratios of solute:solvent, followed by extrapolation of
these values to the value associated with an infinite quantity of solvent.
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The Born–Haber cycle
The Born–Haber cycle is a specific application of the first law of
thermodynamics – that energy may not be created or destroyed – using Hess’s
law. The ‘cycle’ – a series of reactions that form a closed path – allows the
lattice enthalpy of an ionic solid to be calculated. This is the enthalpy
associated with the direct combination of gaseous ions to form an ionic
lattice:
n
M m+ (g) + m X n– (g)  M n X m (s)
Because direct measurement of this process is generally impractical, an
indirect path is created. Using KCl as an example, the cycle illustrated in
Figure 4 is obtained.
Figure 4
The Born–Haber cycle for KCl
K+(g) + Cl(g) + e½Hd(Cl2)
K+(g) + ½Cl2(g) + eHi (K)
Hec (Cl)
K+(g) + Cl-(g)
K(g) + ½Cl2(g)
K(s) + ½Cl2(g)
Hs(K)
Hl (KCl)
Hf (KCl)
KCl(s)
If KCl is taken as the starting point, the following expression may be
generated, as enthalpy is a state function and the enthalpy change over the
complete cycle must therefore equal zero:
–H f(KCl) +H s (K) + H i(K) + ½H d (Cl 2 ) + H ec (Cl) + H l(KCl) = 0
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rearranging gives:
H l(KCl) = H f (KCl) – H s (K) – H i (K) – ½H d (Cl 2 ) – H ec (Cl)
The terms on the right-hand side of this equation may all be obtained by
direct physical or spectroscopic methods, giving a value for the lattice
enthalpy:
H l(KCl) = –431 – 89 – 419 – 124 – (–349) = –714 kJ mol –1
Bond enthalpy
In the course of a chemical reaction, some (or all) of the bonds are broken in
the reactants, and new bonds are created as the products form. Since, by
definition of a chemical process, this is essentially the only source of energy
change in the process, it should – at least in principle – be possible to
calculate this energy change. If the energy of the bonds in the reactants is
calculated and subtracted from that of the products, this should give the
energy change in the reaction.
For example, the decomposition of H 2 O 2 (g) takes place to give water and
oxygen:
2H 2 O 2 (g)  2H 2 O(g) + O 2 (g)
On the left-hand side of the equation there are 2  2 H–O bonds and
2  O–O bonds. On the right-hand side there are 2  2 H–O bonds and
1  O=O bond.
If the energy of an X–Y bond is denoted E(X–Y), then the energy change for
this reaction is given by:
{2  2  E(H–O) + E(O=O)} – {2  2  E(H–O) + 2  E(O–O)}
The problem in such calculations is that the energy of a bond between two
atoms, for example H–O, varies between different molecules. Consider the
breaking of the O–H bonds in water, for example:
H–OH(g)  •H(g) + •OH(g)
•O–H(g)  •H(g) + O : (g)
H = +492 kJ mol –1
H = +427 kJ mol –1
In some molecules, the energy required to break the bond (the bond
enthalpy) is greater than +492 kJ mol –1 , whereas in others it is less than
+427 kJ mol – 1 .
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Another common example is the stepwise breaking of C –H bonds in CH 4 :
CH 4  CH 3 +
CH 3  CH 2 +
CH 2  CH +
H
H
H
 C
H
CH
+
The H  values associated with each step of this process progressively
decrease. It is obvious that there is no single value for the O –H or C–H bond
enthalpies, and since calculation of the energy of a specific bond is not
trivial, it is convenient (if less accurate) to use average bond enthalpy
values. A selection of average bond energies is given in Table 2 for bonds
between several first row elements. Notice that the average bond enthalpy for
double and triple bonds between elements is not twice or three ti mes that of a
single bond, since the nature of the bonding in multiple bonds is quite
different from that of single bonds.
For the decomposition of hydrogen peroxide, these average values yield an
enthalpy change of:
[(2  2  463) + (2  146)] – [(2  2  463) + 497]
= (1852 + 292) – (1852 + 497)
= 2144 – 2349
= –205 kJ mol –1
This compares reasonably well with the experimental value of 196 kJ mol –1 .
The difference arises because the average bond enthalpies do n ot represent
the precise value of the bond enthalpies of the molecules. For many practical
purposes, however, the calculated value is sufficiently close to allow semi quantitative evaluation of the reaction. In this example, the calculation
demonstrates that the reaction is highly exothermic, which is clearly in
accord with experiment.
The decomposition of H 2 O 2 (aq) is slow at room temperature but may be
demonstrated more effectively with the use of a piece of liver (calf’s liver,
for example, from a butcher). The liver is chopped into small chunks and
added to a beaker of hydrogen peroxide solution. Enzymes in the liver
catalyse the exothermic decomposition reaction to water and oxygen. The
rise in temperature can be measured and used to calculate an approximate
value for the enthalpy change. The difference from the value calculated
above would be due to both experimental error and enthalpy changes
associated with changes in state.
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Table 2
Average bond enthalpies for bonds between some common elemen ts
(– indicates a single bond, = a double bond and  a triple bond)
H
436
C
412
348 –
612 =
838 
N
388
305
613
890
163
409
946
–
=
O
463
360 –
743 =
F
565
484
743 =
Cl
431
338
H
C
157
270
200
N
146 –
497 =
185
203
O
155
254
242
F
Cl

–
=

Bond enthalpies can be determined experimentally to a high level of accuracy
using spectroscopic methods that are beyond the scope of the Advanced
Higher. These techniques involve measurement of the energy required to
vibrate the bond to varying amplitudes and extrapolating this behaviour to the
point at which the vibration is infinitely large, a situation that corresponds to
complete separation of the atoms.
In the example of methane, it is possible to measure the bond enthalpy by th is
method, and since the bonds are indistinguishable, it is not surprising that
each bond is found to have the same value for the bond enthalpy. If the
bonds are broken in the stepwise fashion given above, then four different
bond enthalpies are observed. Note, however, that by Hess’s law the sum of
the four bond enthalpies from the stepwise processes must equal the sum of
four of the ‘average’ bond energies. This may be represented as shown in
Figure 5.
Figure 5
C + 4H
H4
H3
H2

CH + 3H
CH 2 + 2H

4  H average
CH 3 + H

H1
CH 4
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It is possible to determine bond enthalpies by using enthalpies of formation,
often in conjunction with bond enthalpies that have been more easily
determined, such as those of diatomic molecules (H 2 , N 2 , Cl 2 , etc.).
For example, the average O–H bond enthalpy in H 2 O may be determined
using Hess’s law as shown in Figure 6.
Figure 6
2H + O
bond enthalpy of H 2
2  O–H bond
enthalpy
H2 + O
½  bond enthalpy of O 2
H 2 + ½O 2
H f (H 2 O)
H2O
From this it is evident that:
2  O–H bond enthalpy
= H f (H 2 O) + (–½  bond enthalpy of O 2 ) + (–bond enthalpy of H 2 )
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RE AC T IO N F E AS IB I L I TY
SECTION 2
Reaction feasibility
Chaos umpire sits,
And by decision more embroils the fray
By which he reigns; next him high arbiter
Chance governs all.
John Milton, Paradise Lost, Book II, line 907.
John Milton was so close to the truth in this quote that it is probably the most
poetic version of the second law of thermodynamics in the English language.
We will come to that law shortly, but let us first set it in context.
It isn’t by accident that chaos is the result of neglect in any system, be it a
chemical system or a filing system. We are used to the idea that some things
happen spontaneously, while other things don’t. If a vase is placed on the
edge of a table, it is only a matter of time before it is knocked off, falls to the
floor and shatters into hundreds of pieces with the release of a crashing
sound. If a broken vase is placed at the foot of a table, we would never
expect sound waves to spontaneously concentrate on those pieces and for
them to jump up and reform an unblemished vase on the table. That is the
way of things in our corner of the universe at least, and it is how we
recognise the passage of time. We can generally tell when a film is being run
backwards because in our experience things do not tend to fall upwards,
steam does not tend spontaneously to enter a steam engine’s exhaust and
ripples in a pond do not tend to run into one a nother and throw out a stone
from the bottom of the pond in the process.
In short, the universe is running down by becoming increasingly disordered.
This disorder is not limited to physically tangible objects – energy is also
running down into a higher state of disorder: the ordered electrical energy
that flows into our homes is converted to light energy that is spread around
the room, and converted into low-grade heat energy as it is absorbed by the
walls and fittings in the room. This heat is little use for anything, but this
again is the way of the universe. Eventually, all the energy will be
distributed uniformly throughout the universe, rendering any chemical or
even physical events impossible.
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RE AC T IO N F E AS IB I L I TY
The driving force for change
Disorder is important because it is the single factor that determines whether
any physical process, including a chemical reaction, takes place
spontaneously. (‘Spontaneous’ is actually a technical term – see the box on
page 19 for an brief explanation of what it means.) A pr ocess will only take
place spontaneously if it increases the disorder in the universe.
It is a common misconception that reactions tend to move towards the lowest
energy level. In fact, all reactions and physical processes move towards the
state that generates the largest amount of disorder in the universe. This
disorder applies to both the spatial disorder of the molecules and the disorder
of the energy. Two examples that may be easily reproduced in the laboratory
demonstrate that the concept of increasing disorder is more important than
that of decreasing energy.
Consider first two iron blocks, one at 200°C and one at 0°C. If they are
brought together, heat will flow spontaneously from the hot block to the cold
block. Clearly, the hot block is moving to a lower energy state, but at the
same time the cold block is moving to a higher energy state. The heat, on
the other hand, is being distributed more evenly (and therefore more
randomly) throughout the blocks.
Next consider the case of hydrogen cyanide gas released from a gas cylinder
bottle into a room. At first, the hydrogen cyanide molecules are ordered, in
that they occupy a well-defined volume. After release they diffuse through
the room spontaneously, but note that there is no change i n energy when two
gases mix, so neither the hydrogen cyanide nor the air can be regarded as
moving to a low-energy state. Once again, the process is moving towards a
state of higher disorder. This example may be demonstrated by filling a
small plastic bag with mains gas and allowing this gas to diffuse through the
classroom. The smell of the gas (actually due to the tertiary butyl thiol
(CH 3 ) 3 CSH, which is added to make gas smell) rapidly pervades the room,
showing that the spread of the gas is rapid and spontaneous. Hydrogen
cyanide should not be used to demonstrate this process, even with disruptive
students.
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Spontaneous and non-spontaneous processes
Any process may be defined as being either spontaneous or non spontaneous. A spontaneous process is a process that has a natural
tendency to occur, without the need for input of work into the system.
Examples are the expansion of a gas into a vacuum, a ball rolling down a
hill or the flow of heat from a hot body to a cold one. ‘Spontaneous’ is a
formal definition and is not used here in the colloquial sense. If a
process is described as spontaneous, it does not mean that it is either fast
or random. For a spontaneous process to take place, the system must be
at a position where it is ready for change to come about without the need
for work to be done on it. A spontaneous process may be used to do
work on another system.
A non-spontaneous process is a process that does not have a natural
tendency to occur. Examples might include the compression of a gas into
a smaller volume, the raising of a weight against gravity, or the flow of
heat from a cold body to a hotter one in a refrigeration system. For a
non-spontaneous process to be brought about, energy in the form of work
must be input into a system. In the case of a ball on a hill, the
spontaneous process is for the ball to roll under the influence of gravity
to the base of the slope, releasing energy as heat in the process. The
reverse process – that the ball takes in heat from the surroundin gs and
rolls up the slope – does not occur spontaneously. Note that although the
process does not occur naturally, it is possible to effect a non spontaneous process, but work must be put into the system for this to
come about. In the example given, mechanical work must be done in
order for the ball to be raised against gravity. In any system, the reverse
of a spontaneous process must be non -spontaneous.
Entropy
Although disorder is a rather diffuse term, it is possible to quantify it in terms
of entropy, which is a thermodynamic property of a system. Entropy is
denoted by S and, like enthalpy and internal energy, it is a state function. In
other words, a specified amount of a substance at a specified temperature and
pressure will have a specific value of entropy, no matter how those conditions
were reached.
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Statistical definition of entropy
It is possible to define entropy in statistical terms, so providing an insight
into the real meaning of entropy and entropy changes. For any system,
the entropy is given by the Boltzmann equation:
S = k B ln(w)
where w is the number of possible configurations of the system and k B is
Boltzmann’s constant. This definition allows entropy to be understood
as a measure of the disorder in a system. If we take, for example, a
hypothetical crystal containing six 127 I126 I molecules, then the number of
ways in which the molecules can be arranged if the crystal is perfectly
ordered is 1. If two molecules are reversed, so increasing the disorder,
then the number of distinguishable arrangements increases to 15.
Reversing three of the molecules reduces the number of possible
arrangements to 10. If all arrangements are energetically equivalent, the
most probable situation is the one with the highest number of possible
configurations, and hence this is the most ‘disordered’. This also means
that the perfectly ordered situation, which has the lowest number of
possible configurations and lowest entropy, is also the most improbable.
The second law of thermodynamics
The second law of thermodynamics is a restatement of our previous
discussion, and states that:
‘for a spontaneous process, the total entropy change in a system and in the
system’s surroundings will increase’
In other words, a process that increases the over all entropy of the universe is
a process that is thermodynamically possible. This does not mean that more
ordered systems cannot be generated – it is obvious that a large number and
range of processes that lead to increased order take place every day in o ur
own direct experience. Ice is more ordered than water, yet we form ice in our
freezers. Similarly, we can crystallise salt crystals out of a disordered
solution. In all such cases, however, we are only looking at part of the story
– remember that it is the total entropy that must increase. Ice forms in our
freezers because heat is pumped out into the surroundings, heating them up
and increasing the disorder there. Salt crystallises out of solution as the
solution releases heat to the surroundings and as the water molecules
evaporate, the effect of both of which is to increase the entropy of the
surroundings.
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We can generalise this discussion. Clearly, it is possible in a spontaneous
process for the system to become increasingly disordered, but ev en where it
does not, then providing that the surroundings become more disordered to
compensate, the process may be spontaneous.
We can increase or decrease entropy in the surroundings by giving out heat or
taking it in to the system (usually a chemica l reaction in this context) of
interest. If a reaction, for example, is exothermic, then heat is dissipated into
the surroundings, making the motion of the molecules there more chaotic, and
the entropy of the surroundings increases (Figures 8a and 8c). I f a reaction is
endothermic, heat is taken in from the surroundings and this decreases the
random motion there (Figure 8b). The result is that the entropy in the
surroundings decreases.
Figure 8a
Fireworks, exothermic reactions in
which  S system >0 and  H<0, so
 S surroundings >0
Figure 8b
KNO 3 dissolving in water, an
endothermic process in which
 S system >0 and  H>0,
so  S surroundings <0
Figure 8c
H 2 burning in oxygen to give liquid
water, an exothermic reaction in
which  S syste m <0 and  H<<0, so
 S surroundings >>0
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The third law of thermodynamics
The third law of thermodynamics states that:
‘the entropy of a perfectly crystalline solid at the absolute zero of
temperature is zero’
For a perfectly crystalline solid, there can b e only one possible spatial
configuration of the components of the crystal, and as the material is at the
absolute zero of temperature, there are no dynamic changes in the crystal
either. In other words, the entropy of the crystal is equal to zero.
In practice, absolute zero cannot be reached and perfect crystalline solids
cannot be made, but it is still possible to apply the third law. For most
practical purposes, the entropy of materials drops to infinitesimally small
values at low temperature, so that it may conveniently be made equal to zero
at the low temperatures that can be routinely achieved in the laboratory.
This becomes important because it makes possible the measurement of
entropy changes from a reference point. (Measuring the entropy cha nges is
beyond the scope of Advanced Higher, but it may be noted that they may be
performed very easily through heat capacity by plotting C p /T against T. The
entropy change between any two temperatures is the area under the curve
between the two temperatures.) Unlike enthalpy and internal energy,
therefore, entropy has a measurable absolute value.
It is possible to plot entropy as a function of temperature for a substance
(Figure 9) and this yields a number of interesting points.
• There is a general increase in the entropy of a substance as the temperature
increases. This is not surprising, since the higher energy that is available
at higher temperatures increases the motion of the molecules in the
substance. The increased disorder that this creates translates into an
increase in entropy.
• For most substances, there are two points at which the entropy goes
through a step increase, that is it increases by a finite amount over an
infinitesimally small temperature. These two points are easily ident ified
as the melting and boiling points. At each of these points there is an
increase in the special disorder of the molecules – in other words, they are
much more free to move around above these temperatures than they are
below them. The increased disorder again translates into an increase in
entropy.
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CH EMI ST R Y
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Figure 9
The variation of entropy with temperature for chloromethane
S/JK – 1 mol – 1
gas
liquid
solid
0
0
300
T/K
Predicting entropy changes
The preceding points about entropy also allow us to inspect reactions and
predict the likely change in entropy. Consider, for example, the reaction:
CaCO 3 (s)  CaO(s) + CO 2 (g)
Here one mole of relatively ordered solid is converted into one mole of a
similarly ordered solid and one mole of highly disordered gas. The reaction
is predicted to yield an increase in entropy, and indeed S q for this reaction is
found by experiment to be +160.6 J K –1 mol –1 .
Similarly, nitrogen and hydrogen react in the Haber –Bosch process to yield
ammonia:
N 2 (g) + 3H 2 (g)  2NH 3 (g)
In this case, we have a reaction that proceeds from 4 mol of highly disordered
gas on the left-hand side of the equation to 2 mol of highly disordered gas on
the right-hand side of the equation. We predict a decrease in entropy as the
reaction takes place, a prediction that is again borne out by experiment as
S  = –198.8 J K –1 mol –1 . Although this works against the ability to perform
the reaction, this problem is overcome in the industrial Haber process by the
use of high pressures but moderate temperatures.
CH EMI ST R Y 2 3
RE AC T IO N F E AS IB I L I TY
In a third example, we take the reaction between hydrogen gas and oxygen
gas to give liquid water. In this case, we examine the reaction and see that
we proceed from 1½ mol of disordered gas to 1 mol of relatively ordered
liquid:
H 2 (g) + ½O 2 (g)  H 2 O(l)
We predict, and find, that the entropy change is negative
(S= –327 J K –1 mol –1 ). Although the entropy decreases for the system, it is
important to appreciate that the second law of thermodynamics (that entropy
always increases) refers to an isolated system. Most experimental systems
cannot be regarded as being isolated, in which case the universe effectively
becomes the isolated system. In this case, the total entropy change is the sum
of the entropy change in the system and in the surro undings, and this total
must be greater than or equal to zero to comply with the second law of
thermodynamics:
S tot al + S surroundings = S t ot al
and
S tot al 0
Another example is the entropy change as hydrogen and fluorine gases react
to generate liquid hydrogen fluoride. This is found to be –210 J K –1 mol –1 .
Although this represents a decrease in entropy, the reaction proceeds
spontaneously because the total entropy change is greater than zero. The
positive total entropy change arises because the reaction is exothermic and
the heat lost to the surroundings increases the disorder there, causing
S surroundings to be positive and of greater magnitude than S syst em .
Standard entropy change
Any non-equilibrium process leads to a change in entropy. As entropy is a
state function, entropy changes may be calculated from the standard entropies
of the initial and final states of the system:
S  process = S  final st at e – S  init ial st at e
For a chemical reaction, for example, the standard entropy of reaction is
the difference between the standard entropies of the reactants and products,
and may be calculated from:
S  reaction = S  reactants – S  products
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RE AC T IO N F E AS IB I L I TY
This expression resembles those used with other state functions, such as
enthalpy, and despite the slightly simpler form, the similarity with
expressions for enthalpy is even closer than is initially evident . In the case of
enthalpy, for example, the corresponding equation is
H  reaction = H  products – H  reactants. Here, H  f is the enthalpy of
formation of a substance from an arbitrary standard value of zero, whereas in
the expression for the entropy change, S  represents the entropy of formation
of a substance from an absolute value of zero.
Free energy
The total entropy change determines whether or not a reaction is spontaneous,
and so it is useful to have some property that quantifies the tota l entropy. A
measure of the total entropy change is the free energy. The Gibbs free
energy is used under conditions of constant pressure, and so the symbol G is
used for free energy under these conditions.
At constant pressure and temperature, change s in the free energy may be
expressed as:
G t ot al = H syst em – TS syst em
G t ot al is equal to –TS t otal (see box on page 26), and the free energy may be
regarded as a measure of the total entropy change (both in the system and the
surroundings) for a process. While a spontaneous process gives ris e to a
positive value of S tot al, however, G t ot al must be negative because of the
minus sign in G t ot al = –TS t ot al.
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RE AC T IO N F E AS IB I L I TY
Free energy and total entropy change
The total entropy change that accompanies a process is the sum of the
entropy change in the system and in the surroundings:
S tot al = S syst em + S surroundings
However, S surroundings is related to the enthalpy change in the system at
constant pressure through the relationship S surroundings = q/T, where q is the
heat added to the surroundings. If the heat lost by the system is H syst e m ,
the heat gained by the surroundings must be –H syst em , therefore S surroundings
= – H syst em/T:
S tot al = S syst em + –H s yst em /T
If this expression is multiplied by –T, it yields the relationship:
–TS t ot al = H s yst em – TS syst em
The quantity –TS t ot al is referred to as the Gibbs free energy change, G.
Since S tot al must be positive (increasing entropy) for a spontaneous
process, it follows that G syst em must be negative for a process to be
spontaneous.
An important property of the free energy is that it represents the maximum
amount of work that may be obtained from a process. This differs from the
heat that may be obtained from a process because the total entropy change
must be greater than zero. For example, in the case of a reaction for which
S syst em is negative, some heat must be lost to the surroundings and contribute
to S surroundings in order that S tot al is greater than zero. The value of the heat
that is then unavailable for conversion into work is given by TS syst em .
Free energy and spontaneity
For a spontaneous process, S of the universe is positive and G for the
system is therefore negative. The relationship G = H – TS allows
prediction of the conditions under which a reaction is spontaneous. As T
must be positive, the relationships may be summed up as in Table 3.
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Table 3
Free energy and the spontaneity of reactions
H system
S system
Spontaneous?
Spontaneity favoured by
Negative
Negative
Positive
Positive
Positive
Negative
Positive
Negative
Under all conditions
If |TS| < |H|
If |TS| > |H|
Never
All conditions
Low temperatures
High temperatures
No conditions
Temperature has a major impact on the spontaneity of some reactions, as
indicated in Table 3. For an exothermic reaction ( H<0) where S<0, |TS|
(i.e. the absolute value of TS) will be less than |H| provided that T is small,
and such a reaction will be spontaneous at lower temperatures. Conversely,
in an endothermic reaction (H>0), if S>0, |TS| will be greater than |H|
provided that T is large, and such a reaction will become spontaneous at
higher temperatures. In both cases, the temperature at which the reaction
becomes spontaneous is simply given by T= H/S.
Properties and applications of the Gibbs free energy
The Gibbs free energy can be applied in a similar manner to other state
functions, and many of the expressions that are encountered are similar in
form to those seen for enthalpy. For example, the standard reaction free
energy, G  , is the change in the Gibbs free energy that accompanies the
conversion of reactants in their standard states into products in their standard
states. It is possible to calculate the free energy of a reaction from the
standard enthalpy and energy changes for the reaction: G  = H  – TS  ,
with H  and S  being obtained either from tabulated data or from direct
measurement. We also employ the concept of the standard free energy of
formation, G  f . This is defined as the free energy that accompanies the
formation of a substance in its standard state from its elements in their
standard states. Calculation of the standard free energy change of a
reaction may be expressed as:
G  reaction = G  f products – G  f reactants
The standard reaction free energy is useful, but only deals with the
extremes of a reaction. In other words, it only compares the difference in
free energy between the pure reactants and the pure products, and does not
take any account of the reaction conditions in between. This is clearly
limited in its usefulness, since most reactions do not start and end with
pure substances. To deal with this, we consider the reaction free energy.
The reaction free energy is the change in free energy when
CH EMI ST R Y 2 7
RE AC T IO N F E AS IB I L I TY
a reaction takes place under conditions of constant composition. The
difference may be illustrated by the reaction:
cyclopropane, C 3 H 6 (g)
propene, C 3 H 6 (g)
If one mole of pure reactants react to generate one mole of products, then G  =
–41.7 kJ mol –1 . This is the standard reaction free energy change. If the free
energy change is now measured when one mole of reactants gives one mole of
products within a mixture of one million moles of cyclopropane and two million
moles of propene, it is found to be –39.9 kJ mol –1 . This reaction free energy is
given the symbol G. The difference between G and G  arises because of the
different conditions under which the two reactions take place.
Equilibria and free energy
We have already described how the universe naturally moves towards a position
of maximum disorder (i.e. maximum entropy) and that this is the driving force
for change in each individual chemical reaction or process. Achieving the most
disordered state of the system itself clearly plays an important part in
determining the position of maximum overall entropy. This is particularly true
when the value for H  (which is responsible for entropy changes in the
surroundings) is small. It follows then that in such reactions there is a strong
driving force for the system to reach the most disordered state.
When we attempt to use DG q to determine the most disordered state of the
system, we immediately encounter a limitation of this quantity. Because G 
only compares the properties at the start and finish of a reaction, it can lead
us to believe that the maximum entropy is at the point where we have either
pure reactants (if G  is positive) or pure products (if G  is negative).
When we look more closely at the system, however, it is readily appreciated
that the maximum disorder – the maximum entropy – is not obtained when it
is composed of pure products or pure reactants, but when it consists of a
mixture of the two.
It follows that an intermediate composition often represents the most
disordered state of the system, and the system will spontaneously progress to
this position. At this point, the reaction is i n a state of equilibrium (see
Section 3), where we observe a state of constant composition that is
comprised of both reactants and products.
We can represent this argument in plots of free energy against reaction
composition. When the free energy is at i ts minimum value, the total entropy
is maximised, and the system will tend to move spontaneously towards this
position.
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CH EMI ST R Y
E Q UI L IB R I A
SECTION 3
Equilibria
Reactions are almost universally written so as to imply that reactants produce
products in a one-way process. Reactions that really do behave in this way
are in the minority. The majority of reactions, such as many of the
biochemical reactions within our own bodies, are not like this. For such
reactions it is necessary to take account of the fact that the rev erse reaction is
both possible and important. An example is the dissociation of a weak acid,
such as ethanoic acid, CH 3 COO H. The forward reaction, the one that we
would normally consider to go to completion, is given by;
CH 3 COO H(l)  CH 3 COO – (aq) + H + (aq)
If the reaction could only proceed in one direction, then the ion would
become fully dissociated. It so happens that the ethanoate ion, CH 3 COO – (aq),
is capable of reacting with a proton to give ethanoic acid:
CH 3 COO – (aq) + H + (aq)  CH 3 COO H(l)
This is clearly the reverse of the first reaction, and both reactions take place
simultaneously in a sample of ethanoic acid. We indicate this by using the
equilibrium symbol,
, thus:
CH 3 COO – (aq) + H + (aq)
CH 3 COO H(l)
In a chemical system such as this, therefore, there are two reactions, one
forming products and a competing reverse reaction that re -forms reactants.
At some point, the number of forward reactions that take place in a given
time equals the number of reverse reactions that take place in that same
period. When this takes place, the concentrations of reactants and products
do not change, and the system is in a state of equilibrium. The state of
equilibrium, where the concentration of reactants and products reaches a state
of balance, should not be compared to balancing a pair of scales. It is
extremely important, and should be very strongly emphasised , that although
the concentrations of reactants and products no longer change, the reaction
does not stop but reaches a state of dynamic equilibrium, in which the rates
of forward and reverse reactions are equal.
CH EMI ST R Y
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E Q UI L IB R I A
Equilibrium at a party – an analogy
The concept of a dynamic equilibrium, as encountered in a chemical
equilibrium, can be a difficult concept for students to grasp. The idea that
a chemical system can have a constant composition, but that the reaction
has not stopped seems at first to be contradictory.
An analogy can be drawn with a group of people at a party. As the party
starts, most of the guests may start off in the living room, with none of
them in the kitchen. As the party progresses, a proportion of the guests
may decide to go into the kitchen and spend time there. From this point
on, the relative number of people in the living ro om and the kitchen
remains reasonably constant, but we can immediately recognise that the
situation is not static, as the same people do not remain in the kitchen
throughout the party. There is a flow of people between the kitchen and
living room, but if for each person that moves into the kitchen another
person moves out into the living room, then there are stable populations in
each. This is a state of dynamic equilibrium, and if the living room is the
analogue of the reactants, the kitchen can be regar ded as the analogue of
the products.
Some comments on equilibria
A major problem in chemistry, as in any subject, is trying to see the wood for
the trees, and when we consider chemical equilibria it is very easy to become
confused by the different equilibria that are encountered. It often helps to
realise that all types of equilibria are of the same general form:
chemicals in state 1
chemicals in state 2
As chemists we encounter several types of equilibria, including, for example:
reaction equilibria
H 2 (g) + I 2 (g)
2HI(g)
phase equilibria
H 2 O(s)
acid dissociation equilibria
CH 3 COO H(aq)
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CH EMI ST R Y
H 2 O(l)
H + (aq) + CH 3 COO – (aq)
E Q UI L IB R I A
autoprotolysis equilibria
2H 2 O(l)
H 3 O + (aq) + OH – (aq)
solution equilibria
AgCl(s) + H 2 O(l)
Ag + (aq) + Cl – (aq)
The only thing that distinguishes these equilibria from one another is the
manner in which we, as chemists, choose to categorise them.
The equilibrium constant, K
It is important to be able to quantify the relative concentrations of reactants
and products in an equilibrium reaction. Let us assume that we have an
equilibrium of the form:
aA + bB
cC+ dD
If the reaction is in a state of equilibrium, then we can define a quantity
known as the equilibrium constant for the forward reaction, approximately
given by:
C  D
a
b
 A  B
c
K=
d
where [A], [B], [C] and [D] are the equilibrium concentrations of the
reactants and products, and a, b, c, and d are the molar ratios of compounds
A, B, C and D respectively, as given in the chemical equilibrium.
Under any specified set of physical conditions, the value of K is constant for
a given reaction. When the system is at equilibrium, the concentrations of the
reagents must be such that when their values are placed in the equation, they
equal K.
Example
The reaction between hydrogen and iodine at a temperature of 590K is:
H 2 + I2
2HI
K=2
As before, we recognise that K is given by the expression:
 HI 
 H 2  I2 
2
K=
CH EMI ST R Y 3 1
E Q UI L IB R I A
From this, it is easy to see that the conditions for equilibrium will be met if
[HI] = 2 mol l –1 , [H 2 ] = 2 mol l –1 , and [I 2 ] = 1 mol l –1 . This set of
concentrations, however, is not the only one that represents a state of
equilibrium. The conditions for equilibrium will also be met if
[HI] = 5 mol l –1 , [H 2 ] = 5 mol l –1 , and [I 2 ] = 2.5 mol l –1 . Needless to say, there
is an infinite set of reactant and product concentrations that act as
mathematical solutions to the expression above. This illustrates the point
that, while K is constant at a given temperature, the concentrations of the
reactants and products need not be.
Some general points about K
We start with a common point of confusion. Nineteenth century
physical chemists were not noted for their originality in naming
physical constants, and the equilibrium constant for a reaction,
K (upper case), is often confused with the reaction’s rate constant,
k (lower case). It is worth noting, however, that the magnitude of
K is given by the ratio of the forward to backward rate constants,
K = k f/k b . This further emphasises the dynamic nature of the chemical
equilibrium.
Another point about the equilibrium constant is that K is dimensionless –
it is a pure number with no units. This i s a non-negotiable fact. There is,
admittedly, continuing confusion on this matter since many textbooks,
biochemistry courses and even chemistry courses attribute K with units.
This misunderstanding arises from a failure to appreciate that the use of
concentrations in the expression for K is an approximation. Because this
provokes such strong opinions in many people we include here a summary
of the relevant background to the use of activities.
The rigorous definition of K uses activities, which have no units:
a  C a  D 
c
K=
d
a  A a  B
a
b
where a(A) is the activity of compound A, etc.
The concept of activity is beyond the scope of the Advanced Higher
syllabus, but a brief explanation for its use is that concentration
does not directly influence the physical properties (such as electrical
conductivity) of a solution. While the conductivity of a 10 –4 mol l –1
solution of NaCl may reflect the presence of 10 –4 mol l –1 of noninteracting Na + and Cl – ions, a 1 mol l – 1 solution does not show 10 4
32
CH EMI ST R Y
E Q UI L IB R I A
times the same conductivity. This is because, at high concentrations, the
ions interact with one another, and the effective concentration is less than
the actual concentration. Activity may be considered to be a measure of
the effective concentration (or pressure, for gases), relative to a standard
concentration (or pressure).
For a solute, A, at low concentrations, the activity of A is approximately
equal to the concentration of A divided by the standard concentration. As
the standard concentration is 1 mol l –1 , this means that in a solution
containing 0.001 mol l –1 of A, A has an activity of 0.001 mol l –1 /1 mol l –1
which is equal to 0.001, as the units cancel out. As this is the case for all
components in the equilibrium, all the units cancel out, and so K itself has
no units. At this level, it is sufficient to say that K is calculated using
concentration, but without using the dimensions of the concentrations.
Note, however, that concentration must be measured in mol l –1 and
pressure in atmospheres, as the standard concentration and pressure is 1 in
each of these units.
Vanishing solids, liquids and solvents
The concept of chemical activity allows us to explain why pure liquids
and solids disappear from the equilibrium expression. As the activity of a
material may be regarded as the ratio of its effective concentration to its
standard concentration, and as a solid or liquid is only very slightly
different from its standard state, then it becomes clear that their activity is
1. Hence, for a solid such as silver chloride dissolving in water, we have
the equilibrium:
AgCl(aq) + H 2 O(l)
Ag + (aq) + Cl – (aq)
The full expression for the equilibrium constant is:
[Ag+ ][Cl – ]
K=
[H 2 O][AgCl]
Although we write this using the approximation of concentrations, we
recognise that AgCl is a solid, and so ‘[AgCl]’ =1. Also, although the
water contains some dissolved AgCl, it is little different from pure water,
and so ‘[H 2 O]’ = 1. Hence, K = [Ag + ][Cl – ].
The remainder of this text will use the approximation of concentration and
partial pressure.
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Changing the conditions – Le Chatelier’s principle
Le Chatelier’s principle states that:
‘When a system at equilibrium is subjected to a disturbance, the
composition adjusts to minimise the effect of this disturbance.’
Thus, when a chemical species that forms part of the equilibrium reaction is
added to the system at equilibrium, reaction occurs to remove that species.
Also, when the total pressure of a system involving gases at equilibrium is
increased, the system adjusts to reduce the total number of m oles of gas (and
hence the volume) and offset this pressure increase. Finally, when the
temperature of a system is increased, the system adjusts to take in energy and
reduce this temperature increase. This is a useful principle that allows the
effect of any perturbation on the equilibrium to be predicted.
For the general reaction aA + bB
cC + dD at equilibrium:
C  D
a
b
 A  B
c
K=
d
However, when a species on the left-hand side of the equation, e.g. A, is
added so that [A] increases, this removes t he equilibrium condition. Species
on the left-hand side of the equation (A, B) are consumed in order to produce
more on the right-hand side (C, D). This continues until a new equilibrium
position is reached and the concentrations are again related by the
equilibrium constant expression above. Note that throughout this, the value
of K remains unchanged. In contrast, if C or D is added, this again perturbs
the equilibrium, and the backward reaction is favoured over the forward
reaction. Equilibrium is again re-established with the consumption of C and
D and the production of A and B until the concentrations are related by the
equation for K above, with the value of K remaining unchanged.
In a system involving gases, the effects of pressure changes are i mportant.
For example:
N 2 (g) + 3H 2 (g)
2NH 3 (g)
The equilibrium constant for the reaction as written may be expressed in
terms of the partial pressures, p x , of each component:
Kp =
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CH EMI ST R Y
p 2 NH 3
p2N 2 p3H 2
E Q UI L IB R I A
This is closely related to the equilibrium constant expressed in terms of
concentration. Note that concentration is a measure of the number of a
given species per unit volume ( n/V). For an ideal gas, x, p x V=n x RT, and
so p x = RT(n x /V) = RT[x].
We use Le Chatelier’s principle, almost unwittingly, in organ ic
reactions. Consider the hydrolysis and formation of an ester. In
effect, we have an equilibrium between the ester and the carboxylic
acid and alcohol:
R–OH + RCOO H
alcohol
acid
R–CO–O–R + H 2 O
ester
water
To prepare an ester, we employ reagents, such as concentrated
sulphuric acid or a polar organic solvent, that stabilise the H 2 O, so
encouraging the forward reaction and the generation o f more ester. To
hydrolyse an ester we use water as a solvent which, by Le Chatelier’s
principle, leads to the dominance of the reverse reaction and the
hydrolysis of the ester to the acid and alcohol.
Increasing the overall pressure by decreasing the vo lume of the container
causes an increase in all of the partial pressures. As the equilibrium
constant involves more moles of gas on the left -hand side of the equation
than on the right -hand side, equilibrium is lost and the reaction quotient,
and the forward reaction (N 2 and H 2 reacting to form NH 3 ) takes place
until the partial pressures are again related by the equilibrium constant
expression given above.
Example
At a particular temperature the reaction is found to be at equilibrium
when the partial pressure of each component is 1 atmosphere (partial
pressures must be measured in atmospheres – see ‘some general points
about K, above):
Kp =
p 2 NH 3
= 1 2 /(1 2 1 3 ) = 1
p2N 2 p3H 2
If the volume of the container is halved, then the new pressure of each
component must be 2 atm. Placing these new values in the expression
shows that the new pressures do not represent a system at equilibrium as
they no longer equal K:
p 2 NH 3
22
4
1
Kp 
=
=
=
2
3
2 3
p N2 p H 2
2 2
32
8
CH EMI ST R Y 3 5
E Q UI L IB R I A
The values in the expression must now alter in order to give the correct value
for K. This can only be done if p 2 NH 3 increases and (p 2 N 2 p 3 H 2 ) decreases, i.e.
if the forward reaction takes place.
Looking
pressure
does not
pressure
does not
at Le Chatelier’s principle in this way explains why increasing the
by adding helium, for example, which plays no part in the reaction,
influence the position of equilibrium. In this case, while the overall
of the system increases, the partial pressures of th e reaction gases
change, and so the equilibrium is not perturbed.
Extremophiles, athletes and Le Chatelier
Figure 10
Did you ever wonder why you never see yaks or deep sea viper fish at the
zoo? The reason is that these creatures are extr emophiles – they live in
extreme conditions of high altitude and very deep seas. Amongst other
factors, these environments have low oxygen partial pressures, and this
means that the equilibrium between oxygenated and deoxygenated
haemoglobin in the lungs:
haemoglobin + O 2
oxyhaemoglobin
is shifted to the left relative to that of an animal at sea level, because of
Le Chatelier’s principle. Muscle tissue still requires the same amount of
oxygen as at sea level and to survive in these conditions these animals
must transport sufficient oxygen into the bloodstream. These animals
have responded to this need by producing larger quantities of
haemoglobin, which shifts the equilibrium back to the right and allows a
higher proportion of the available oxygen to be utilised. Unfortunately,
this also means that too much oxygen is transported into the body at sea
level, and makes life under ambient conditions very difficult for these
animals.
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CH EMI ST R Y
E Q UI L IB R I A
The human body also responds to oxygen deficienc y by releasing extra
haemoglobin, in the short term. It is for this reason that athletes undergo
training at high altitude before competitions – the extra haemoglobin in
the blood remains in place for some weeks after being at altitude. Once
again, Le Chatelier’s principle means that the equilibrium is shifted to the
right compared to that for an unacclimatised athlete, and more oxygen is
then available in the body, which allows more rapid and sustained
respiration in the muscle cells.
Catalysts
The role of a catalyst is to increase the rate of reaction at a given
temperature. There are a number of points about catalysts that should be
appreciated.
Contrary to popular myth, which runs along the lines of ‘a catalyst increases
the speed of a reaction without taking part in it’, catalysts do take part in
reactions – they have to, or how else could they have any effect on the rate of
a reaction? This myth comes about because the catalyst is regenerated at the
end of a reaction – it only appears not to have been involved.
Further to the previous point, the reason why catalysts act to increase reaction
rates is that they provide an alternative reaction pathway. In the journey from
reactants to products, there are energy barriers to be overcome. If an
alternative route with lower energy barriers becomes available through the
use of a catalyst, then it may be possible to move from reactants to products
more easily.
The catalyst provides an alternative route between reactants and products, and
does not influence the energy difference between them. As this is the case, a
catalyst cannot make possible a reaction that would not otherwise take place.
Furthermore, as the position of an equilibrium is influenced by
thermodynamics (recall that the equilibrium consta nt may be calculated from
G  = –RT ln K), where a reaction does take place and reaches a condition of
equilibrium the position of equilibrium is not altered by the presence of the
catalyst. The equilibrium condition is achieved more rapidly, however, in the
presence of a catalyst.
CH EMI ST R Y 3 7
E Q UI L IB R I A
Catalysts – ropes for molecular mountaineering
The role of a catalyst may be compared to the journey across a mountain
from one valley to another (Figure 11). In the normal course of events,
our mountaineer can only travel from A to B over the ridge, which is slow
because it requires a large gain in altitude and therefore a lot of energy.
If, on the other hand, our mountaineer has access to ropes, he can take an
alternative route up the low cliffs into a small valley betwee n the two
major valleys – an option that he does not have without ropes. This route
requires less energy, as he only has to make a modest gain in altitude and
so he can proceed much faster into the next valley to get to B.
The first route in this analogy is the equivalent of an uncatalysed reaction.
The rope acts as the catalyst in providing an alternative, and much faster,
route.
Figure 11
A
B
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CH EMI ST R Y
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Demonstrating the effect of catalysis and the influence of surface area
Hydrogen peroxide decomposes slowly, over a period of weeks at room
temperature, into water and oxygen:
H 2 O 2  H 2 O + ½O 2
This decomposition can be catalysed by a large number of catalysts.
These experiments demonstrate this and the effect of the surface area of
the catalyst.
Our bodies, and those of all living organisms, contain natural catalysts –
enzymes. One of these, catalase, can be found in potatoes. Catalase may
act as a catalyst in the decomposition of hydrogen peroxide. Place
approximately 50 cm 3 of hydrogen peroxide solution (say 30 volume
diluted to 10 volume) in a 1000 cm 3 plastic measuring cylinder, and add a
few drops of washing-up liquid. When a piece of potato is added, a
column of bubbles is formed as oxygen is released in the decomposition
process.
If the previous experiment is repeated using a piece of potato crushed with
a little sand, the column of bubbles forms, but much faster than in the
previous experiment. This increased catalytic activity is due to the higher
surface area of the potato, which is capable of interacting with more
hydrogen peroxide molecules per second than the low surface area piece.
The decomposition is also relatively fast when manganese dioxide, MnO 2 ,
is used, due to the high surface area of the catalyst.
In a final, dramatic, demonstration, the ultimate high surface area is used.
The iodide ion also catalyses hydrogen peroxide decomposition, and if an
aqueous solution of KI is added to H 2 O 2 (aq) prepared as in previous
demonstrations, every I – ion acts as a catalytic centre. The reaction now
proceeds very rapidly, and the column of soap bubbles literally flies out of
the measuring cylinder (wear goggles!)
One final point about catalysed reactions is that the reaction profile is often
incorrectly illustrated as a simple lowering of the energy barrier relative to
the uncatalysed reaction. This simplification gives an incorrect impression of
the catalytic process. Because the catalyst takes part in the reaction it forms
at least one intermediate species, and so there has to be at least one valley in
the reaction profile between the reactants and products.
CH EMI ST R Y 3 9
E Q UI L IB R I A
Phase equilibria
Partition coefficient
We started this section by recognising that all equilibria take a similar form:
chemicals in state 1
chemicals i n state 2
The partition coefficient is a measure of the distribution of a solute between
two immiscible solvents, and is simply a specific case of an equilibrium:
solute in solvent 1
solute in solvent 2
As with any equilibrium it may be described by an equilibrium constant, K,
where
K = [solute] solvent 2 /[solute] solvent 1
This is called the partition coefficient, and gives a measure of the relative
solubility of a solute in the two solvents.
Solvent extraction
Solvent extraction is the process of purifying a compound by using its
differential solubility between two solvents. This is a practical example of
using the partition coefficient. Figure 12 provides a schematic representation
of the process – in real situations better extraction is provided by repeating
the process several times.
Figure 12
Schematic representation of solvent extraction
Mixture
40
CH EMI ST R Y
E Q UI L IB R I A
Example
Suppose that a mixture of 0.1 mol each of two compounds, A and B, is
extracted between 300 cm 3 dichloromethane and 300 cm 3 water. Suppose
further that A has a partition coefficient of 0.2 between the two solvents, and
that B has a partition coefficient of 5.
For compound A, if m is the number of moles of A in water, then 0.1 –m
moles of A are dissolved in dichloromethane:
[A] wat er /[A] dichloro met hane = 0.2
(m/0.3)/{(0.1–m)/0.3} = 0.2
m/(0.1–m) = 0.2
m = 0.0167
Similarly for B, where n is the number of moles of B dissolved in the water:
[B] wat er /[B] dichloro met hane = 5
(n/0.3)/{(0.1–n)/0.3} = 5
n/(0.1–n) = 5
n = 0.08
The material in the aqueous layer, therefore, is enriched in B relative to A.
Rather than the 1:1 molar ratio that we started with, the material that is
extracted in the aqueous layer contains a 4.8:1 ratio of B:A. Using
calculations analogous to those just given it is straightforward to show that,
by repeating this process, this ratio can be increased to 23:1 and then to
110:1.
Chromatography
Although it is possible to repeatedly extract compounds between two
solvents, it is time-consuming and can require many extractions to obtain
sufficiently pure products.
Chromatography is analogous to solvent extraction, in that it is based
on the use of the partition coefficients. The difference is that one of the
phases (the mobile phase) is a solvent, in which the solute can dissolve,
and the other phase is a solid (the stationary phase) to which the solute
can adhere. At any one time there will be an equilibrium,
CH EMI ST R Y 4 1
E Q UI L IB R I A
represented by the partition coefficient, for the solute distribution between
the stationary and mobile phases. The mobile phase travels over the
stationary phase and carries each component of the sample with it, but each at
a different rate. As some solutes spend longer on the stationary phase than
others (i.e. those that have a high partition co efficient between the stationary
phase and the mobile phase), they will travel along the chromatography
column more slowly than those that spend a greater proportion of the time in
the mobile phase (i.e. those that have a small partition coefficient betwee n the
stationary phase and the mobile phase).
Shopping chromatography
The process of chromatography can be compared to a couple on a
shopping trip on a busy street. There is a mobile phase – the crowds on
the pavement who tend to draw our shoppers with them along the length
of the street – and there is a stationary phase – the shops.
Suppose that our couple have different personalities – one (A) is an avid
shopper while their more reluctant partner (B) has little interest in
shopping. Person A will therefore spend long periods in the shops, and
less time in the street – they have a large partition coefficient between the
shop and the street. On the other hand, person B will spend long periods
in the street, and less time in the shops – they have a small partition
coefficient between the shop and the street.
As our couple start off at one end of the street, A will start to spend a few
minutes in a shop, while B is drawn down the street by the crowd. The
couple are now slightly separated. Although A comes out of this first
shop and is also drawn down the street by the crowd, A will soon drop
into another shop and the process repeats. The couple become
increasingly separated, and longer streets tend to cause greater
separations.
B may occasionally spend some time in the shops, but since this amounts
to much less time than that spent by A, it is not sufficient for B to catch
up and does not prevent the couple separating.
The stationary phase can be almost any solid material, depending on the
solvent and the solute, although some are more suitable than others –
cellulose (e.g. paper), alumina and silica are among the most common. The
mobile phase can be any liquid or, as we will see later, gas.
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CH EMI ST R Y
E Q UI L IB R I A
Chromatography was developed by M S Tsvet arou nd 1906, and gains its
name from the Greek words  (chromatos graphos), ‘colour
writing’. Tsvet worked on the highly coloured chlorophylls and carotenes,
and these are still routinely used as demonstrations. A small amount of fluid,
extracted from grass or leaves, is dropped approximately 1 cm from the base
of a 10 cm strip of blotting paper. The strip is placed in a 250 cm 3 beaker
containing a 0.5 cm depth of light petroleum ether. As the liquid (the mobile
phase) moves up the paper (the stationary phase) different pigments are
carried with the solvent at different rates. The orange –yellow carotenes are
transported farthest up the paper, with the yellow xanthophylls showing only
a small movement away from the original spot. Finally, the green
chlorophylls remain almost unmoved from the original spot.
Chromatography is a huge multi-billion dollar industry – it is universally
used to separate components from a reaction. In the laboratory it can be used
as a very simple method of separating the products from a new reaction.
Thin-layer chromatography (TLC) most commonly uses cellulose, silica or
alumina on an aluminium sheet in place of blotting paper. After separation,
the sheet can be cut up to physically separate the different bands. Thereafter,
each band may be redissolved and analysed – modern analytical methods
require only milligrams of material for a good quality analysis.
Gas chromatography
Chromatography need not be performed using liquids as the mobile phase – it
is also possible to use gases. Gas chromatography (GC) w as developed from
work during the Manhatten Project, the US atomic bomb project in World
War II. In order to manufacture a bomb, it is necessary to separate 235 U from
238
U. This is accomplished by converting uranium to gaseous UF 6 and
allowing this to diffuse down extremely long packed tubes – the tube required
is several kilometres in length. The 238 U spends longer in the columns than
the 235 U, and this allows the two isotopes to be separated. Although it works
well, in this form this is an extremely inefficient and time-consuming
technique. Modern derivatives of this technique are extremely rapid and
efficient, and can separate materials that could not be separated by any other
method. By using chiral materials (i.e. a material in only one of its two
mirror image forms) in the stationary phase, it is even possible to separate
chiral molecules from their mirror images.
In GC, the stationary phase is a packed column or coated capillary that is
held in an oven, with an inert gas such as H 2 , He or N 2 as the mobile
CH EMI ST R Y 4 3
E Q UI L IB R I A
phase carrier gas. Each component of the sample is carried through the
column by the carrier gas at a different rate, and is detected as it exits the end
of the column. A range of detector types may be used, the simplest of which
merely detects the presence of a chemical in the gas stream. In this case,
chemicals appear as peaks in the signal from the detector, and are identified
by the length of time that they spend on the column (the retention time), these
having previously been ascertained using pure chemicals. An example of this
is shown in Figure 7, which displays the results for a wine, each component
having been identified by its retention time on the column. Notice the very
small amount of material that is required for this techniqu e – 1 cm 3 . Notice
also that the chemical hazard data sheets for many of the components of wine
make grim reading!
Figure 7
Gas chromatograph of a 1 cm 3 sample of wine
It is also possible to use more sophisticated detectors, a common one being
the mass spectrometer in combination with a gas chromatograph, known as
GC-MS. This was used in the analysis of materials from the Martian surface
on the Viking Mars landers in the 1970s.
GC is now an extremely sensitive technique, with femtomole sensitivity, and
finds an extremely wide range of applications: drug analysis (e.g. forensic
science), sports testing, environmental monitoring (e.g air pollutants from car
exhaust gases, pesticide residues), quality control in pharmaceuticals
manufacture, brewing, distillation etc., to name but a few.
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An innovative detector has been employed by researchers in an attempt to
identify cockroach pheromones. Pheromones are natural scents that have a
strongly attractive power for specific species. To identify the specific
chemicals that acted as cockroach pheromones, a cockroach was liquidised
and passed through a GC column. The chromatography column separated out
each of several thousand chemical components of the cockroach, and these
were passed through the column in turn. As it was only necessary to identify
the pheromone, all that was required was a sensitive detector that would
identify which of these myriad chemicals was the pheromone. Fortunately,
Nature herself had provided the best pheromone detector of all – a cockroach
antenna. Once the retention time of the pheromone had been determined by
monitoring the movement of an antenna, it was possible to isolate larger
quantities of the chemical for more detailed analysis. Once analysed, the
pheromone was chemically synthesised and used to trap cockroaches.
Acid–base equilibria
In Brønsted–Lowry theory, an acid is a proton (H + ) donor and a base is a
proton (H + ) acceptor. Examples of acids are HCl, CH 3 COOH, H 3 O + and H 2 O.
Examples of bases are NH 3 , CH 3 COO – , H 2 O and OH – . H 2 O can therefore act
as either an acid or a base. A reaction between an acid and a base ( an acid–
base reaction) involves the exchange of a proton, for example:
HCl + NH 3  Cl – + NH 4 +
acid base
These definitions apply under all conditions, bu t the most important acid–
base systems use water as a solvent. In this case, equilibria are set up in
water, which for HA (a general acid) is:
H 3 O + (aq) + A – (aq)
HA(aq) + H 2 O(l)
H 3 O + (aq) + Cl – (aq)
e.g. HCl(aq) + H 2 O(l)
and for B (a general base) is:
B(aq) + H 2 O(l)
e.g. NH 3 (aq) + H 2 O(l)
OH – (aq) + BH + (aq)
OH – (aq) + NH 4 + (aq)
In the first equilibrium the H 2 O molecule acts as a base, accepts a proton
and forms the hydronium ion, H 3 O + , which is the hydrated form of the
CH EMI ST R Y 4 5
E Q UI L IB R I A
proton in solution. In the second equilibrium the H 2 O molecule acts as an
acid and forms the hydroxide ion, OH – . The base that results from the
transfer of a proton from the acid is called the conjugate base of the acid. A –
is the conjugate base of HA and OH – is the conjugate base of H 2 O. Similarly,
the acid that results from the acceptance of a proton by the base is called the
conjugate acid of the base. This means that BH + is the conjugate acid of B
and H 3 O + is the conjugate acid of H 2 O.
If an acid is regarded as a strong acid (i.e. it readily donates a proton), then it
is always the case that its conjugate base is a weak base (i.e. it does not
readily accept a proton). An example of this is H 2 SO 4 , which is a strong acid
while its conjugate, HSO – 4 , is a weak base. Conversely, a weak acid donates
a proton to generate a strong base, an example being the weak acid, acetic
acid, CH 3 COOH, which has a very strong conjugate base, CH 3 COO – .
Note that the concept of conjugate acids and bases is only a relative one since
a species can be, for example, a base or a conjugate base, depending on the
specific equilibrium of interest. Take, for example, the ethanoate ion,
CH 3 COO – :
CH 3 COOH(aq) + H 2 O(l)
CH 3 COO – (aq) + H 2 O(l)
H 3 O + (aq) + CH 3 COO – (aq)
OH – (aq) + CH 3 COOH(aq)
In the first equilibrium, the ethanoate ion is acting as a conjugate base; in the
second, it is acting as a base. The ion is the same in both cases, it is only its
relative position in the equilibrium that determines whether or not it is
regarded as a conjugate species.
The acidity constant, Ka
It is useful to compare the relative proton donating strength of an acid, and
this is done by considering the acid dissociation equilibrium:
HA(aq) + H 2 O(l)
H 3 O + (aq) + A – (aq)
As with any other equilibrium, the position of this equilibrium may be
measured through the equilibrium constant:
K = [H 3 O + ][A – ]/[HA][H 2 O]
Strictly, we should use activity, and as the water is the solvent, its activity
is approximately equal to 1. This equilibrium constant is no different
46
CH EMI ST R Y
E Q UI L IB R I A
from any other, apart from its specificity to this particular type of equilibrium,
and the equilibrium constant is referred to as the acidity constant, K a :
K a = [H 3 O + ][A – ]/[HA]
Clearly, a strong acid will give rise to relatively high values of [H 3 O + ] and
[A – ], and so K a for a strong acid will be larger than for a weak acid.
As K a varies over such a large range (several tens of orders of magnitude), it is
convenient to use the negative of the logarithm of K a , known as pK a (cf. the pH
scale):
pK a = –log (K a )
Indicators
An indicator for use in acid–base reactions is comprised of a weak acid that
displays distinctly different colours in its protonated and deprotonated forms.
An indicator has an equilibrium of the general form:
HIn
acid colour
H+
+
In –
base colour
where In – is typically a large, complex, organic molecule. This equilibrium
has an equilibrium constant, K in , given by:
K in = [H + ][In – ]/[HIn]
This may be rearranged to:
[HIn]/[In – ] = [H + ]/K in
If the concentration of H + (aq) is high (i.e. it is an acidic solution), then HIn
will dominate the solution and the acid colour will be most apparent . If the
concentration of H + (aq) is low, then In – will dominate and the solution will
have the colour of the base species.
Clearly, if both forms are in solution, then the colour will be a mixture of the
acid and base forms. Generally, the colour will be distinctly that of the acid
form if [HIn]>10[In – ] and of the base form if [In – ]>10[HIn].
For the acidic colour:
10 = [H + ]/K i n
CH EMI ST R Y 4 7
E Q UI L IB R I A
or taking –log 10 of each side and rearranging:
pH = pK a –1
and for the basic colour:
0.1 = [H + ]/K in
and as above:
pH = pK a +1
Hence an effectively complete transition from acid colour to base colour
takes place over the range:
pH = pK in  1
From this it is easy to appreciate that the range of the indicator is determined
by its acidity constant, pK in .
Buffers
When a weak acid such as ethanoic acid is titrated against a strong base such
as sodium hydroxide, the titration curve is typical ly of the form shown in
Figure 13. The main features are a small, rapid, increase in pH as the first
drops of base are added, with only a slow pH increase thereafter. This
plateau region is maintained until the pH once again increases very rapidly as
the stoichiometric point is approached and passed through, before reducing to
a small, positive slope beyond.
A buffer is a solution that exhibits only small changes in its pH in response to
the addition of an acid or base. With this in mind, the importanc e of the
titration curve becomes evident. The section of the titration curve between
the initial rise and the rise to through the stoichiometric point has only a very
gentle slope. In other words, the solution in this portion of the titration
responds to the addition of strong base with a very small increase in pH.
Likewise, the addition of a small amount of strong acid (which is equivalent
to removing strong base) only generates a small decrease in the measured pH.
Inspection of this portion of the pH curve shows that the centre of this buffer
region is half-way between the start point and the stoichiometric point. At
this point, half of the acid has been neutralised to its salt, and half remains as
the acid. A buffer solution of this composition wi ll therefore be composed of
equal molar concentrations of acid and its salt, and will be equally capable of
buffering acids and bases.
48
CH EMI ST R Y
E Q UI L IB R I A
Figure 13
Titration curve for a weak acid and strong base
pH
12
10
v/2
v/2
8
Stoic hiometric point
6
4
Buffer region
2
V
Volume of titrant
Many reactions are highly sensitive to hydrogen ion concen tration (pH).
Buffer solutions are very important and are widely used in chemistry and
biochemistry in order that pH conditions may be maintained very specifically
during the course of a reaction or process. Moreover, animal and plant
metabolisms are also highly sensitive to pH, and much of the body’s chemical
efforts are put into maintaining the optimum pH for certain enzymes. For
example, the digestive enzyme trypsin operates most effectively in the pH
range 2–4, whereas alkaline phosphatase is most ef fective at pH 9–11.
Of all the buffered solutions in the body, blood is probably the most familiar.
Control of its pH is critically important, and for normal metabolic function
must be retained in the very narrow range between 7.35 and 7.45. There are a
large number of components to this particular buffer, but the component that
is most readily understood by students at Advanced Higher level is that
involving carbonic acid and the hydrogen carbonate salts:
H 2 O(l) + CO 2 (g)
H 2 CO 3 (aq)
H 2 CO 3 (aq)
H + (aq) + HCO – 3 (aq)
The acid component of this scheme is the carbonic acid, H 2 CO 3 , and the salt
is represented by the hydrogen carbonate, HCO – 3 . The pH is calculated
using the Henderson–Hasselbalch equation:
CH EMI ST R Y 4 9
E Q UI L IB R I A
pH
=
=
pK a +
pK a +
log 10 ([salt]/[acid])
log 10 ([HCO – 3 ]/[H 2 CO 3 ])
The pK a for carbonic acid is approximately 6.1, and the pH is held within the
required limits with a 20:1 ratio of [HCO – 3 ]:[H 2 CO 3 ]. In order to maintain
these proportions, the carbonic acid concentration is controlled via the
CO 2 /H 2 CO 3 equilibrium, with CO 2 concentration being controlled through
respiration. The HCO – 3 concentration, on the other hand, is controlled
through its excretion in the urine.
The body is perfectly capable of controlling the blood pH to the required
level under normal circumstances, but such things as asthma, pneumonia,
emphysema or smoke inhalation all reduce the ability to lose CO 2 by
exhalation. The result is an increase in H 2 CO 3 , leading to a drop in pH. This
is known as respiratory acidosis, and if unchecked can lead to coma and,
ultimately, death.
A similar effect occurs when the body undergoes physical exertion, but in this
case the pH decreases as a result of the lactic acid that is generated in the
muscles. Eventually, the capacity of the blood to buffer the lactic acid is
exhausted and the result is metabolic acidosis, which displays the same
symptoms and effects as respiratory acidosis.
The buffering capacity of the blood is tested to breaking point i n cases of
severe burns, which are often accompanied by the effects of smoke
inhalation. Burns victims often succumb not to the burns themselves, but to
the metabolic effects that result. The smoke causes impaired breathing,
leading to respiratory acidosis, while the loss of body fluids into oedema
(swelling) leads to a loss in blood volume, which reduces O 2 transport to
tissues. This in turn causes the release of lactic acid and other acidic
metabolic by-products into the bloodstream and results in meta bolic acidosis.
Furthermore, the loss of electrolytes into the oedema (a process known as
natrosis) also impacts on the body’s ability to function normally.
Burns victims are now treated using a solution known as Ringer’s lactate,
which contains Na and K salts of lactic acid. The fluid increases blood
volume and replaces electrolytes. From the perspective of buffer solutions,
the solution is interesting as the lactate ion is a strong base, which acts as the
salt to the lactic acid. This increases the blood pH and restores its buffer
capacity. Introduced carefully, Ringer’s lactate enables blood pH to be
restored to the correct range and to save the life of the burns victim. It is one
of the main reasons why burns patients have a greater chance of sur vival than
they did 50 years ago.
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SECTION 4
Fuel cells
Electrochemistry – Fuel cells
Electrochemical cells – commonly referred to as a battery, although strictly
this refers to a combination of cells – are extremely useful devices, finding a
wide range of applications in, for example, portable electrical equipment and
as back-up power for computers. In spite of their usefulness, they are
discrete devices with a finite amount of reactant and a limited energy storage
capacity. Furthermore, the electrical potential (the voltage) across the
terminals decreases with use. Because the potential decreases, the rate at
which energy may be accessed (i.e. the power) also decreases with use. A
normal zinc–carbon cell, for example, provides an initial electr ic potential of
1.5 V, but this rapidly decays over time, until the potential and power
available are too low to be of use for most purposes. The zinc battery was
invented over 100 years ago by Leclanché, and the more recent alkaline
battery (cell reaction Zn + 2MnO 2  ZnO + Mn 2 O 3 ) fares a little better,
maintaining a useable potential for slightly longer. Better still are lithium
batteries, which are capable of maintaining a more consistent power and
voltage level. All of these cells suffer from one common factor – they are
sealed units with a predetermined energy storage capacity. Rechargeable
batteries, of course, are one solution to this limitation, but the electricity must
still be generated somewhere, and in remote locations this remains a
limitation.
An electrochemical cell generates electricity by consuming reactants and
forming products, and the limitation in its storage capacity arises from the
quantity of reactants that are available. One solution to the storage capacity
problem is simply to add more reactants, but while this works, it inevitably
results in larger cells (compare, for example, an AAA -sized battery with an
A-sized battery). This solution is unsatisfactory in many cases where size is
an issue (portable electronic equipment, for example).
The other solution to the problem of capacity is to continuously feed
reactants into the cell and to remove the products from it – this is the
principle of the fuel cell. By virtue of the need to remove and
introduce material into the cells, the reactants a nd products are
CH EMI ST R Y
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FU E L C E L L S
invariably either gases or liquids. The first practical fuel cells were the result
of work by engineer Francis Bacon, in 1932, using hydrogen and oxygen as
the fuel and oxidant, and using nickel electrodes. The father of the fuel cell,
however, is generally regarded as Sir William Grove. Grove recognised that
it might be possible to reverse the reaction that produces hydrogen and
oxygen from the electrolysis of water in order to generate electricity. Even
today, the most common – and certainly the most commercially profitable –
fuel cells are based around hydrogen and oxygen, with only the electrolyte or
the electrodes being varied between devices. The hydrogen is generally
derived from the partial oxidation of fuels such as methane or methanol, and
there is active research continuing to try to make cells based on these fuels,
rather than on hydrogen.
The principle of the fuel cell is elegantly simple (Figure 14). Hydrogen is fed
in at one electrode (the positive electrode) of the fu el cell and oxygen is fed
in at the other (the negative electrode). The electrode materials are chosen
for their special characteristics in facilitating (catalysing) the cleavage of the
H–H and O=O bonds. Platinum group metals, particularly nickel or pla tinum,
are commonly used for this purpose. Interestingly, platinum is able to
catalytically cleave the H–H bond with virtually zero energy of activation,
and so the process is highly efficient. This latter point may be illustrated
using a hydrogen–oxygen mixture – the catalytic effect of adding platinum to
this mixture is so efficient that it immediately leads to rapid combustion.
Once generated, the hydrogen atoms are ionised to give a proton and an
electron (i.e. the reverse of the formal half -cell reaction: H + + e –  H).
Figure 14
Schematic representation of the processes in a hydrogen fuel cell
½O2
H2
2H+
2e-
H2O
The cell is arranged so that a proton transfer medium allows the proton to
travel from the positive to the negative electrode. The proton transfer
medium may be one of a number of substances, such as phosphoric acid,
molten salts (e.g. calcium carbonate at 1000°C) or polymeric proton exchange
membranes.
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CH EMI ST R Y
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At the negative electrode, oxygen molecules are dissociated into atoms:
O 2  2O
These oxygen atoms combine with two protons, which have travelled through
the proton transfer medium, and two electrons to give water as a waste
product, giving a half-cell reaction at the cathode:
2H + + O + 2e –  H 2 O
As with any cell, the electrons must travel around an external circui t to reach
the negative electrode, and in doing so may perform work, such as lighting a
light or running a motor.
Note one important aspect of the cell although the overall cell reaction is the
oxidation of hydrogen:
H 2 + ½O 2  H 2 O
This is not burning. Little heat is produced in this reaction, instead most of
the reaction free energy is collected as electrical energy.
When compared with conventional fossil-fuel power sources, fuel cells are
both clean and efficient. The only waste product is water, a lthough fuel cells
run using natural gas do also produce some carbon dioxide, and the efficiency
of the device means that there is less per usable energy unit than would be
created if the fuel were burned.
Variations on the cells that we have around us to day have existed for a long
time, having provided electricity on spacecraft since the 1960s (those who
have read the book Apollo 13 – or seen the film – will have encountered
numerous references to the fuel cells). Their terrestrial applications include
electricity-generating plants in remote regions – it is often more practical
to transport gas as liquid or in cylinders to an area than to build hundreds
of miles of overheard cables.
At the time of writing, one company alone has sold over 80 fuel cell
units (ONSI Corporation, USA) that are now in operation (Figure 15).
These phosphoric acid fuel cells are the size of a transport container
and generate 200 kW of electricity. They are primarily used to power
hospitals and remote hotels where centrall y generated power is expensive
and reliability is critical (the cells have to remain in operation for 95%
of the time, which is a greater display of reliability than that of a
diesel-powered generator). Furthermore, one installation in
CH EMI ST R Y 5 3
FU E L C E L L S
Connecticut consumes methane-generated landfill waste, so generating useful
power from a powerful greenhouse gas.
Figure 15
A static phosphoric acid fuel cell
Phosphoric acid-based cells are efficient, but tend to be heavy, which makes
them less than ideal for use in vehicles.
To meet this need, Ballard Power Systems in Canada has developed a
lightweight fuel cell, the proton-exchange membrane (PEM) type. The PEM
replaces the phosphoric acid of the ONSI cells. Pilot fleets of Ballard’s fuel
cell powered buses now run in Vancouver and Chicago (Figure 16).
Figure 16
Fuel cell powered bus
54
CH EMI ST R Y
FU E L C E L L S
The driving force for much of this technology is the US state and federal
governments’ drive to produce zero emission vehicles – vehicles that do not
produce nitrogen/sulphur oxides, carbon monoxide or carbon particulates.
Toyota and Daimler-Benz have exhibited PEM cell powered automobiles
(Figure 17).
Figure 17
Fuel cell powered car
These vehicles are not test vehicles that will never see production – the
imperative to fulfil the legislation means that they are becoming increasingly
visible in the US. The cars are visible but unfortunately not very audible, and
that is a disadvantage. The fuel cells are silent, and pedestrians easily
overlook silent cars – ironically, one solution is to incorporate noise
generators into these vehicles to prevent accidents!
The advantage of the PEM is its ability to miniaturise fuel cell technology.
Figure 18 shows a fuel cell that has been developed for use in a mobile
phone, along with the same technology applied to a camcorder and a laptop
computer.
CH EMI ST R Y 5 5
FU E L C E L L S
Figure 18
A miniature fuel cell for use in a mobile phone(top), and similar technology
applied to a video camera and a portable computer.
The mention of hydrogen fuels almost invariably meets the same response –
that hydrogen is dangerous. Although easily ignited, hydrogen is actually far
less dangerous than petrol – it burns with a cool flame that rapidly rises up
and away from the source and does not tend to ignite its surr oundings very
efficiently. This was illustrated during the Hindenberg disaster of 1937
(Figure 19) in which the burns that victims sustained were the result of
burning petrol from the engines, and not the hydrogen (indeed, hydrogen may
not have been responsible for the accident at all – see, for example,
http://www.ttcorp.com/nha/advocate/ad22zepp.htm).
The main difficulty with using fuel cells in passenger vehicles is in storing or
supplying the hydrogen. Practical hydrogen storage problems remain – being
a light gas, large, heavy cylinders are required to store sufficient quantities
for journeys of any great length.
One solution is to generate the hydrogen from stored methane in a steam reforming reaction over metal catalysts just prior to injection i nto the fuel
cell:
CH 4 + 2H 2 O  2H 2 + CO 2
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CH EMI ST R Y
FU E L C E L L S
Fuel cells still have problems such as cost, although with greater use that may
be dealt with through the economies of scale. One thing is certain – in a
world with increasingly expensive oil and a greater awareness of the need for
efficient energy generation, they are no longer only of interest to astronauts.
Figure 19
The Hindenburg disaster, Lakehurst, New Jersey, 1937. Erroneously cited as
the apotheosis of the dangers of hydrogen. Notice that the hydrogen, which is
burning at this point in the disaster, rises up and away from the airship. Of
the 97 people on board, 49 survived. Had this been a petrol fire, it is
doubtful that anyone would have survived, and it is also likely that more
people would have been killed on the ground.
CH EMI ST R Y 5 7
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K IN E T I CS
SECTION 5
Kinetics
Rate of a reaction
It is a common misconception that the rate of a reaction is just another way of
expressing the time required for the reaction to reach completion. There are
two problems with this – firstly, many reactions do not reach completion (see
Section 3), and secondly, rate is a rather more complex idea than this.
Interestingly, the first studies of reaction rate, performed by C F Wenzel in
1777, were attempts to investigate the time required for complete dissolution
of metals in mineral acids and fell into this misconception of rate being
equivalent to reaction time. Later investigations led to a better understanding
of rate, but we should not belittle Wenzel’s studies – remember that these
were groundbreaking investigations in their time and paved the way for our
current knowledge.
The rate of a reaction is defined in terms of the rate of loss of a reactant or
the rate of formation of a product. With few exceptions, reactions do not
progress at a constant rate, and the rate varies (generally decreasing)
throughout the reaction. The rate must not, then, be viewed in terms of the
overall rate, but in terms of the instantaneous rate – the rate of change in
concentration of a reactant or product over an infinitesimally sm all period of
time.
That rate does not mean the time for reaction may be illustrated by a
simple analogy. In a journey from the country into a city centre in a car,
we may start the journey on a major road, with a 60 mph speed limit. We
progress rapidly along this road but as we get closer to the city, we
encounter first a 50 mph speed limit, and then a 40 mph speed limit.
Assuming that we drive legally, our rate of travel – our speed – decreases
as we get closer to the city centre until we are travell ing very slowly in a
traffic jam. If the overall journey is 15 miles, and we are travelling this
distance in half an hour, then our mean rate is clearly 30 mph.
CH EMI ST R Y
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K IN E T I CS
Note two important things:
• The mean speed (analogous to the mean rate of reaction) is n ot the
same as the actual speed (the instantaneous rate of reaction) at any one
point along the journey, which has varied between 60 mph and 0 mph.
• Not only is the value for the time for the journey only very indirectly
related to the instantaneous speed (or rate of reaction), it does not even
have the same units as a rate expression, which will always have units
of (quantity time –1 ), often mol l –1 sec –1 .
If we follow the concentration of a reactant or a product in a reaction, we
may observe how that concentration varies with time. Just as the gradient of
distance against time for a car journey gives a measure of the rate of travel –
the speed – the gradient of the concentration of a reactant or product against
time gives a measure of reaction rate.
Exactly how we measure that quantity depends on the reaction itself. For a
relatively slow reaction, such as the preparation of an ester over a period of
an hour or two, we may take out a sample of material, say, every 10 minutes,
stop the reaction in that sample by rapid cooling and analyse the
concentration of the reagents by standard chemical methods. For a very fast
reaction, we do not have time to remove a sample for analysis and must
measure the concentration within the reaction as it happens. An extreme
example of this sort of study is given by the 1999 Nobel prize for chemistry,
awarded to Ahmed Zewail for his work on femtosecond spectroscopy. This
involves measurements of processes, and the rates of processes, that take
place in only 10 –15 seconds (1 femtosecond). Put very crudely, his work
involved measuring the light absorbed by chemicals as they reacted and in
this way discovering what was going on within them. To give an idea of the
timescales involved, 1 femtosecond is to one second as 1 second is to 32
million years.
The rate of a reaction is not constant for any given reaction, but varies with a
number of parameters – concentration, temperature, pressure, voltage and
current (for electrochemical reactions) are just some of the th ings that can
affect a reaction rate. We only consider the effect of concentration here.
Rate laws
Students are almost universally familiar with the idea that increasing the
concentration of a reactant is likely to increase the rate of a reaction –
indeed, it makes intuitive sense.
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Consider a reaction such as that between hydrogen and iodine:
H 2 (g) + I 2 (g)  2HI(g)
If we increase the concentration of hydrogen, we find that the rate increases
and likewise for iodine. If we perform a series of experiments, we actually
find that there is a very simple relationship between the rate and the
concentration:
rate = k[H 2 ][I2 ]
The constant, k, is known as the rate constant. It is clear from this that if, for
example, the concentration of hydrogen is doubled, then the rate doubles, and
if, for example, the concentration of iodine is halved, then the rate is also
halved.
Another example is the reaction between NO and O 2 to give NO 2 :
2NO(g) + O 2 (g)  2NO 2 (g)
Once again, we get the rather simple relation:
rate = k[NO] 2 [O 2 ]
This illustrates the concept of the order of a reaction – the order is the power
to which the concentration is raised in the rate equation. Thus the order of
reaction with respect to NO in the previous reaction is 2 ([NO] 2 ), and the
order with respect to O 2 is 1 (recall that [O 2 ] = [O 2 ] 1 ). The overall order of
the reaction is the sum of the individual orders, and in this case is 3 (= 2 + 1).
Orders need not be integral and indeed can take almost any value, which
brings us to an important point about rate equations. Unfortunately, examples
such as we have just encountered give rather a rosy picture of the rate laws,
and appear to suggest that the rate law is related to the stoi chiometric
equation. This is, however, not the case – there is no simple relationship
between the stoichiometric equation and the rate law . A rate law can only be
determined by experiment. To illustrate this point, consider the bromine
analogue of the previous reaction:
H 2 (g) + Br 2 (g)  2HBr(g)
CH EMI ST R Y 6 1
K IN E T I CS
From this stoichiometry, we might expect a simple rate law. However, if we
carry out some rate measurements on this reaction, we find that the rate is
given by the rather complex expression:
rate = k[H 2 ][Br 2 ] 3/2 /([Br 2 ] + k[HBr])
which suitably illustrates the point that rate laws can only be experimentally
determined.
The determination of rate laws
For simple cases, rate laws may be determined by carrying out only a few
experiments. As we saw with the reaction between hydrogen and iodine, if
we double the concentration of iodine, we double the rate of reaction, and
likewise for hydrogen. Assuming that we were not aware of this, we might
have performed a series of reactions, with a range of starting concentrations
of hydrogen and iodine:
Expt.
Initial concentration
of hydrogen/mol l –1
Initial
concentration of
iodine/mol l –1
1
2
3
1.0
1.0
2.0
1.0
2.0
2.0
Initial
reaction
rate/mol
l –1 s –1
1  10 –4
2  10 –4
2  10 –4
From this we see that doubling the concentration of iodine doubles the ra te of
reaction (experiments 1 and 2), so the relation must be of the form:
rate = X[I2 ]
where X is a constant that must incorporate the rate constant and some power
of the hydrogen concentration.
Likewise, we also see that doubling the concentration of hydrogen doubles the rate
of reaction (experiments 2 and 3), so the rate equation must also be of the form:
rate = Y[H 2 ]
where Y is a constant that must incorporate the rate constant and some power
of the iodine concentration.
Combining these facts, we see that the overall rate equation is of the form:
rate = k[H 2 ][I 2 ]
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K IN E T I CS
Rate laws and reaction mechanism
Not surprisingly, the rate law for a reaction originates from the reaction
mechanism. Almost any reaction can be broken down into a series of
elementary reaction steps. As an analogy, we might consider the task of
making a cup of tea. This is not a single-step process, but is comprised of a
series of more elementary steps. We take a kettle, fill it with water, plug it in
and then turn it on. We then get a cup, put in the tea and add the water when
it has boiled, etc. Eventually, we get the product – a cup of tea. A reaction,
also, is seldom comprised of a single process (compare, for example, the free
radical reaction between methane and chlorine with its stoichiometric
reaction). How quickly we get our product depends less on how quickly the
rapid parts of the reaction take place than on how slowly the slowest part of
the reaction takes place. This slowest step is known as the rate-determining
step. In the case of making a cup of tea, most of the operations – the
elementary steps – only require a few seconds, but the rate at which a series
of cups of tea can be made is generally determined by the rate at which the
water can be boiled. Boiling water is therefore the rate-determining step. A
chemical example is the addition of HCl to an unsaturated carbon –carbon
bond:
HCl + (CH 3 ) 2 C=CH 2  (CH 3 ) 2 C—CH 2 Cl
This relatively simple reaction proceeds via two elementary steps:
HCl + (CH 3 ) 2 C=CH 2
— slow 
[(CH 3 ) 2 C—CH 3 ] + + Cl –
[(CH 3 ) 2 C—CH 3 ] + + Cl –
— fast 
(CH 3 ) 2 Cl—CH 3
The rate of the slow step is the bottleneck in this process and determines the
overall rate of the reaction, just as the rate at which water can be boiled
determines the time required to make tea or the time taken to pass through a
set of traffic lights may determine the time required for a short journey.
Not surprisingly, the rate of this reaction is determined by the rate at which
HCl and (CH 3 ) 2 C=CH 2 molecules collide. This in turn is related to the
concentration of each of the reactants (just as the rate at which males and
females meet one another at a party is greater when t here are a larger
numbers of each in a given area). The result is that the rate is proportional to
the concentration of each, or put another way:
rate = k[HCl][(CH 3 ) 2 C=CH 2 ]
CH EMI ST R Y 6 3
K IN E T I CS
This is the experimentally observed rate expression. Note that:
• This expression is not derived from the stoichiometric equation, but from a
proposed mechanism.
• A reaction mechanism cannot itself be used to create a valid rate
expression – it remains the case that a rate expression is experimentally
derived. A mechanism is only valid if the rate expression that it generates
is the same as that which is observed experimentally. The experimental
evidence lends weight to support the proposed mechanism, but does not
prove it – other experimental evidence would be required for this .
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