Chapter 22: Reaction kinetics
H2O2(aq) + I–(aq) H2O(l) + IO–(aq) In order for this catalysis to work, the standard electrode
H2O2(aq) + IO–(aq) H2O(l) + I–(aq) + O2(g) potentials for the reactions involving the catalyst must lie
between the electrode potentials involving the two reactants
The overall equation is: (Figure 22.22). The use of electrode potentials in this way
only predicts that the catalysis is possible. It does not give any
I– information about the rate of reaction.
2H2O2(aq) 2H2O(l) + O2(g) uncatalysed
reaction
Ions of transition elements are often good catalysts because
of their ability to change oxidation number.
Examples of homogeneous catalysis Enthalpy change S2O82– + 2I– catalysed
Fe3+ → Fe2+ reaction
The iodine–peroxodisulfate reaction reaction 1
Fe2+ → Fe3+ 2SO42– + I2
Peroxodisulfate (persulfate) ions, S2O82–, oxidise iodide reaction 2
ions to iodine. This reaction is very slow.
S2O82–(aq) + 2I–(aq) 2SO42–(aq) + I2(aq)
The peroxodisulfate and iodide ions both have a negative Progress of reaction
charge. In order to collide and react, these ions need Figure 22.21 Energy level profiles for the catalysed and
considerable energy to overcome the repulsive forces when uncatalysed reactions of peroxodisulfate ions with iodide ions.
like charges approach each other.
E /V
Fe3+(aq) ions catalyse this reaction. The catalysis +2.01
involves two redox reactions.
■■ Reaction 1: reduction of Fe3+ ions to Fe2+ ions by I– ions: S2O82– + 2e– 2SO42– 341
2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I2(aq)
■■ Reaction 2: oxidation of Fe2+ ions back to Fe3+ by S2O82– ions:
2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq)
In both reactions 1 and 2, positively charged iron ions +0.77 Fe3+ + e– Fe2+
react with negatively charged ions. As ions with unlike
charges are attracted to each other, these reactions are
more likely to occur than direct reaction between S2O82– +0.54 I2 + 2e– 2I–
and I– ions.
You should notice that it doesn’t matter what the order Figure 22.22 The electrode potential diagram for the
is of the two reactions. The oxidation of Fe2+ ions to Fe3+ catalysis of the reaction S2O82– + 2I– 2SO42– + I2.
by S2O82– ions could happen first: Oxides of nitrogen and acid rain
2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq) Sulfur dioxide is produced when fossil fuels containing
sulfur are burnt. When sulfur dioxide escapes into the
followed by atmosphere it contributes to acid rain. One of the steps
in the formation of acid rain is the oxidation of sulfur
2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I2(aq) dioxide to sulfur trioxide.
This reaction is catalysed by Fe3+(aq) and it is also SO2(g) + _21 O 2(g) SO3(g)
catalysed by Fe2+(aq).
This oxidation is catalysed by a wide variety of
Figure 22.21 shows an energy level profile for the mechanisms. Nitrogen(IV) oxide present in the
catalysed and the uncatalysed reactions. Note that the atmosphere from a variety of sources (see page 184) can
catalysed reaction has two energy ‘humps’ because it is a
two-stage reaction.
Cambridge International A Level Chemistry
catalyse the oxidation of sulfur dioxide. The nitrogen(IV) The stages in adsorption of hydrogen onto nickel are:
oxide is reformed by reaction with atmospheric oxygen.
■■ hydrogen gas diffuses to the surface of the nickel
SO2(g) + NO2(g) SO3(g) + NO(g) ■■ the hydrogen is physically adsorbed onto the surface –
NO + _12 O2 NO2(g) weak van der Waals’ forces link the hydrogen molecules to
the nickel
Question ■■ the hydrogen becomes chemically adsorbed onto the
15 a State which pairs of substances i to iv below might surface – this causes stronger bonds to form between the
hydrogen and the nickel
catalyse the reaction: ■■ this causes weakening of the hydrogen–hydrogen
covalent bond.
S2O82–(aq) + 2I–(aq) 2SO42–(aq) + I2(aq) Examples of heterogeneous catalysis
Explain your answer.
i Ni2+(aq) / Ni(s) E —O = –0.25 V Iron in the Haber process
ii Mn3+(aq) / Mn2+(aq) E —O = +1.49 V
iii Ce4+(aq) / Ce3+(aq) E —O = +1.70 V Particular conditions of temperature and pressure are
required to form ammonia from nitrogen and hydrogen
iv Cu2+(aq) / Cu+(aq) E —O = +0.15 V (see page 182). The reaction is catalysed by iron. The
b D escribe in terms of oxidation number change, catalyst works by allowing hydrogen and nitrogen
which species are being oxidised and which are molecules to come close together on the surface of the
being reduced in these equations: iron. They are then more likely to react. Figure 22.24
i SO2(g) + NO2(g) SO3(g) + NO(g) shows the five steps in this heterogeneous catalysis.
ii NO + _21 O2 NO2(g) 1 Diffusion: nitrogen gas and hydrogen gas diffuse to the
342 surface of the iron.
2 Adsorption: the reactant molecules are chemically adsorbed
Heterogeneous catalysis onto the surface of the iron. The bonds formed between the
reactant molecules and the iron are:
Heterogeneous catalysis often involves gaseous
molecules reacting at the surface of a solid catalyst. The
mechanism of this catalysis can be explained using the – strong enough to weaken the covalent bonds within
the nitrogen and hydrogen molecules so the atoms
theory of adsorption. Chemical adsorption (also called can react with each other
chemisorption) occurs when molecules become bonded to
atoms on the surface of a solid. Transition elements such as – weak enough to break and allow the products to
nickel are particularly good at chemisorbing hydrogen gas. leave the surface.
Figure 22.23 shows the process of adsorption of hydrogen
onto a nickel surface. 3 Reaction: the adsorbed nitrogen and hydrogen atoms
react on the surface of the iron to form ammonia.
You must be careful to distinguish between the words
adsorb and absorb. Adsorb means to bond to the surface 4 Desorption: the bonds between the ammonia and the
surface of the iron weaken and are eventually broken.
of a substance. Absorb means to move right into the 5 Diffusion: ammonia diffuses away from the surface of
substance – rather like a sponge absorbs water. the iron.
HH bonds bonds weaken HH
strengthen HH
atoms on Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni Ni
nickel surface Ni Ni Ni Ni Ni Ni
Ni Ni Ni
chemisorbed
hydrogen atoms
Figure 22.23 The adsorption of hydrogen onto a nickel surface.
Chapter 22: Reaction kinetics
H N with platinum, palladium or rhodium. These act as
H N heterogeneous catalysts. Possible steps in the catalytic
process include:
1 diffusion to ■■ adsorption of nitrogen oxides and carbon monoxide onto
the surface
the catalyst surface
surface of ■■ weakening of the covalent bonds within the nitrogen oxides
the iron
and carbon monoxide
H HN N ■■ formation of new bonds between
2 adsorption – adjacent nitrogen atoms (to form nitrogen molecules)
– carbon monoxide and oxygen atoms to form
3 reaction new bonds forming
(in several HHN N carbon dioxide
■■ desorption of nitrogen molecules and carbon dioxide
steps) H
molecules from the surface of the catalyst.
HNH
Question
4 desorption
16 a Describe in general terms what is meant by
H desorption.
HN H b N ickel acts as a catalyst for the hydrogenation of
alkenes. For example:
5 diffusion
away from CH2 CH2 + H2 Ni CH3 CH3
surface Suggest how nickel catalyses this reaction by
Figure 22.24 A possible mechanism for catalysis in the referring to the processes of adsorption, reaction 343
Haber process. on the metal surface and desorption.
c I n catalytic converters, rhodium catalyses the
Transition elements in catalytic converters reduction of nitrogen(II) oxide, NO, to nitrogen.
Draw diagrams to suggest:
In Chapter 15 (page 205) you learnt how catalytic i h ow NO is adsorbed onto the surface of the
converters convert harmful nitrogen oxides and carbon
monoxide present in the exhaust gases from car engines rhodium metal
to harmless gases. The ‘honeycomb’ structure inside ii h ow nitrogen is formed.
the catalytic converter contains small beads coated
Summary ■ The overall order of reaction is the sum of the orders
in the rate equation. For the previous example:
■ Rate of reaction is a measure of the rate at which
reactants are used up or the rate at which products overall order is m + n
are formed. The units of rate are mol dm–3 s–1.
■ The order of reaction for a given reactant can be
■ Rate of reaction is related to concentrations of determined experimentally by either:
reactants by a rate equation, which can only be – measuring the initial rate of reaction using
determined by experiment. different concentrations of a given reactant while
keeping the concentrations of all other reactants
■ The general form of the rate equation is: fixed, or
rate = k[A]m[B]n, where: – determining the change in concentration of a
– k is the rate constant specific reactant as the experiment proceeds,
– [A] and [B] are the concentrations of those the rate being calculated from tangents taken at
reactants that affect the rate of reaction several points on the graph.
– m is the order of the reaction with respect to A
and n is the order of reaction with respect to B.
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■ The order of reaction can be determined from graphs ■ The order of a reaction can be predicted from a given
of reaction rate against concentration. reaction mechanism knowing the rate-limiting step.
■ The half-life of a reaction is the time taken for the ■ Homogeneous catalysis occurs when a catalyst and
concentration of a reactant to halve. the reactants are in the same phase.
■ In a first-order reaction the half-life is independent of ■ The mechanism of homogeneous catalysis usually
the concentration(s) of the reactant(s). involves redox reactions.
■ The half-life of a first-order reaction may be used in ■ Examples of homogeneous catalysis include:
calculations to find the first-order rate constant using
the relationship – the catalytic role of atmospheric oxides of
nitrogen in the atmospheric oxidation of sulfur
t_1 = _0_.6_9_3_ dioxide
2 k
– Fe2+ or Fe3+ ions catalysing the reaction between
iodide ions and peroxodisulfate ions.
■ The rate-determining step is the slowest step in a ■ Heterogeneous catalysis occurs when a catalyst is in
reaction mechanism. The rate-determining step a different phase from the reactants.
determines the overall rate of reaction.
■ The mechanism of heterogeneous catalysis involves
■ The order of reaction with respect to a particular
the processes of adsorption, reaction
reactant shows how many molecules of that and desorption.
reactant are involved in the rate-determining step ■ Examples of heterogeneous catalysis include:
of a reaction. – the use of iron in the Haber process
■ The rate equation provides evidence to support the – the catalytic removal of oxides of nitrogen in the
suggestion of a reaction mechanism. exhaust gases from car engines.
344
End-of-chapter questions
Questions 1a, 4b and 6a all require you to plot a graph. Make sure that you do the following:
■■ the axes (e.g. 0–120 minutes and 0–0.080moldm–3) take up more than half of the graph paper you use
■■ label the axes sensibly so that plotting is easy; for example, if on your time axis, two large squares = 30
minutes, it is much easier to make plotting mistakes
■■ plot points precisely with a sharp pencil
■■ draw a smooth curve through the points, with care.
These skills are all essential and should be practised every time you plot a graph.
1 The rate of reaction between butanone and iodine is studied. In this experiment, iodine is in excess. The
concentration of butanone is measured at various time intervals. The results are shown in the table below.
Time / min 0 10 20 30 40 50 60 80 100 120
[butanone] / 0.080 0.055 0.035 0.024 0.015 0.010 0.007 0.003 0.001 0.001
mol dm–3
Chapter 22: Reaction kinetics
a Plot these data on a suitable graph. [3]
b Show from your graph that these data are consistent with the reaction being first order with respect [2]
to butanone. [2]
c Find the gradient of your graph when the butanone concentration is: [3]
[2]
■ 0.070moldm–3
■ 0.040moldm–3
■ 0.010moldm–3
d Use your answers to part c to plot a suitable graph to show rate of reaction (on the vertical axis) against
concentration (on the horizontal axis).
e Explain how the graph you plotted in part d is consistent with the reaction being first order with respect
to butanone.
Total = 12
2 The reaction
A+B+C ABC
is zero order with respect to one reactant, first order with respect to another reactant and second order [2]
with respect to another reactant.
a i Explain what is meant by the term order of reaction with respect to a given reactant.
ii Use the data in the table below to deduce the order with respect to each of the reactants, A, B and c.
Experiment [A] / mol dm–3 [B] / mol dm–3 [C] / mol dm–3 Rate / mol dm–3 s–1 345
0.00783
1 0.100 1.00 1.00 0.00802
0.00796
2 0.200 1.00 1.00 0.00008
0.00031
3 0.300 1.00 1.00 0.00073
0.00078
4 1.00 0.100 1.00 0.00158
0.00236
5 1.00 0.200 1.00
6 1.00 0.300 1.00
7 1.00 1.00 0.100
8 1.00 1.00 0.200
9 1.00 1.00 0.300
b i Write the rate equation for this reaction. [9]
ii State the overall order of the reaction. [1]
iii Calculate the value of the rate constant using experiment 6. Include the units in your answer. [1]
[3]
c Suggest a possible mechanism consistent with the rate equation you have proposed and the
chemical equation [3]
Total = 19
A+B+C ABC
cambridge international A Level chemistry
3 The rate equation for the reaction between iodine and propanone is:
rate = k[CH3COCH3][H+][I2]0
a State the order of reaction with respect to iodine. [1]
[1]
b State the overall order of reaction. [1]
c i What is meant by the term half-life? [1]
[4]
ii In an experiment a large excess of iodine is reacted with a small concentration of propanone in
the presence of H+(aq). The concentration of propanone is measured at regular time intervals.
What happens to the value of the half-life of the propanone concentration as the concentration
of propanone decreases?
d Copy the sketch graph. Plot additional points at 10-second intervals up to 50s. Join all the points with
a smooth curve.
8
[propanone] / mol dm–3 6
4
2
346
0
0 10 20 30 40 50 60
Time / s
e Explain the term rate-determining step. [2]
f Suggest a possible mechanism for the rate-determining step for the reaction between iodine [3]
and propanone.
Total = 13
4 The decomposition of hydrogen peroxide, H2O2, to oxygen and water is catalysed by manganese(IV) oxide. [2]
a Define the term catalyst.
b The results for the decomposition of a sample of hydrogen peroxide are shown in the table.
Time / min 012345678
[H2O2] / mol dm–3 1.60 1.04 0.61 0.40 0.25 0.16 0.10 0.06 0.04
i Draw a graph of concentration of hydrogen peroxide (vertical axis) against time (horizontal axis). [3]
Draw a curve of best fit.
ii Use your graph to determine the half-life of the reaction. Show your working. [2]
iii Use your graph to find the rate of reaction after 2min. [4]
c i Give the rate equation for the reaction. Explain your answer. [3]
ii Use your answer to part b, iii, to calculate the value of the rate constant, k, and give the units. [3]
iii Using your rate equation, find the rate of reaction when [H2O2] = 2moldm–3. [2]
Total = 19
Chapter 22: Reaction kinetics
5 Peroxodisulfate ions, S2O82–, react with iodide ions in aqueous solution to form iodine and sulfate ions.
S2O82–(aq) + 2I–(aq) 2SO42–(aq) + I2(aq) equation 1
The initial rates of reaction are compared by timing how long it takes to produce a particular amount of
iodine using four different initial concentrations of S2O82–. The results are shown in the table.
[S2O82–] / mol dm–3 Initial rate of reaction / s–1
0.0200 4.16 × 10–3
0.0150 3.12 × 10–3
0.0120 2.50 × 10–3
0.0080 1.70 × 10–3
a Plot a suitable graph to calculate rate of reaction. [3]
b Deduce the order of reaction with respect to peroxodisulfate ions. Explain your answer. [2]
c The reaction is first order with respect to iodide ions. Use this information and your answer to part b
[1]
to write the overall rate equation for the reaction.
d The reaction between peroxodisulfate ions and iodide ions is slow. The reaction can be speeded up
by adding a few drops of Fe3+ (aq) ions. The following reactions then take place:
2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq) equation 2
2Fe2+(aq) + S2O82–(aq) 2Fe3+(aq) + 2SO42–(aq) equation 3 347
i What type of catalysis is occurring here? Explain your answer. [2]
ii By referring to equations 1, 2 and 3 above, suggest why Fe3+(aq) ions catalyse the reaction between [4]
peroxodisulfate ions and iodide ions.
Total = 12
6 The rate of reaction between butanone and iodine is studied. In this experiment, butanone is in excess.
The concentration of iodine is measured every 10 minutes for 1 hour. The results are shown in the table.
Time / min 0 10 20 30 40 50 60
[I2] / mol dm–3 0.060 0.051 0.041 0.032 0.022 0.012 0.003
a Plot these data on a suitable graph. [3]
b Show from the graph that these data are consistent with the reaction being zero order with respect [1]
to iodine.
c The balanced chemical equation for the reaction is
CH3CH2COCH3 + I2 CH3CH2COCH2I + HI
Could this reaction occur in a single step? Explain your answer. [2]
d The rate equation for the reaction is
rate = k[CH3CH2COCH3] [4]
Explain the different meanings of the balanced chemical equation and the rate equation.
Total = 10
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7 Nitrogen oxides can be removed from the exhaust gases of a car engine by using a catalytic converter.
Many catalytic converters contain metals such as platinum and rhodium. These act as heterogeneous catalysts.
a i What is meant by the term heterogeneous catalysis? [2]
ii Explain in general terms how heterogeneous catalysts work. [4]
b Nitrogen(IV) oxide and carbon monoxide from car exhausts can react in a catalytic converter.
NO2(g) + CO(g) NO(g) + CO2(g)
The rate equation for this reaction is
rate = k[NO2]2 [2]
Suggest a two-step reaction mechanism for this reaction that is consistent with this rate equation.
c Nitrogen(IV) oxide is formed when nitrogen(II) oxide reacts with oxygen.
2NO(g) + O2(g) 2NO2(g)
The table shows the data obtained from a series of experiments to investigate the kinetics of this reaction.
Experiment [NO] / mol dm–3 [O2] / mol dm–3 Initial rate / mol dm–3 s–1
1 0.001 00 0.003 00 21.3
2 0.001 00 0.004 00 28.4
3 0.003 00 0.004 00 256
348 i Deduce the order of reaction with respect to each reactant. In each case, show your reasoning. [4]
ii Deduce the rate equation for this reaction. [1]
iii State the units of the rate constant, k, for this reaction. [1]
Total = 14
8 Bromate(V) ions react with bromide ions in acidic solution to form bromine.
BrO3–(aq) + 5Br–(aq) + 6H+(aq) 3Br2(aq) + 3H2O(l)
a Suggest two methods of following the progress of this reaction. For each method explain your answer. [4]
b The initial rates of reaction were compared using the initial concentrations of reactants shown in the table.
Experiment [BrO3–] / mol dm–3 [Br–] / mol dm–3 [H+] / mol dm–3 Relative rate of
formation of bromine
1 0.040 0.20
2 0.040 0.20 0.24 1
3 0.080 0.20 0.48 4
4 0.040 0.10 0.48 8
0.48 2
i Deduce the order of reaction with respect to each reactant. In each case, show your reasoning. [6]
ii Deduce the rate equation for this reaction. [1]
iii State the units of the rate constant, k, for this reaction. [1]
Total = 12
chapter 23: 349
entropy and Gibbs free energy
Learning outcomes ■■ calculate the entropy change for a reaction,
∆S —O, given the standard entropies of the reactants
You should be able to: and products
■■ explain that entropy, ∆S, is the measure of the ■■ define standard Gibbs free energy change of
disorder of a system and that a system becomes reaction by means of the equation
energetically more stable when it becomes
more disordered ∆G —O = ∆H —O – T∆S —O
■■ explain the difference in magnitude of entropy: ■■ state whether a reaction or process will be
– for a change in state spontaneous by using the sign ∆G —O
– for a change in temperature
– for a reaction in which there is a change in the ■■ predict the effect of temperature change on the
number of gaseous molecules spontaneity of a reaction given standard enthalpy
and entropy changes.
■■ predict whether the entropy change for a given
process is positive or negative
cambridge international a level chemistry
Introduction Figure 23.1 Potassium reacts spontaneously
with water.
about 150 years ago, many scientists thought that all chemical
reactions gave off heat to the surroundings. they thought that
all chemical reactions were exothermic. we now know this not
true. many chemical reactions and processes are endothermic.
enthalpy changes alone cannot help us predict whether or not a
reaction will occur. if we want to predict this, we need to consider
the entropy change of the reaction.
the term entropy was first given by German physicist rudolf
clausius in 1865. From experimental results, he suggested a
relationship between entropy change (∆S), the energy transferred
reversibly from the surroundings (q) and the temperature (T):
ΔS = _qT_
Introducing entropy either the methane or oxygen is completely used up. For
a reaction to be spontaneous, it does not need to happen
Entropy is a measure of the dispersal of energy at a specific rapidly. Many spontaneous reactions are slow or need an
temperature. Entropy can also be thought of as a measure input of energy to start them.
of the randomness or disorder of a system. The higher
the randomness or disorder, the greater the entropy of Entropy can also be thought of as a dispersal of energy,
350 the system. A system is the part under investigation. In either from the system to the surroundings or from
chemistry this is the chemical reaction itself, i.e. reactants the surroundings to the system. The system becomes
being converted to products. The system of magnesium energetically more stable when it becomes more disordered.
reacting with sulfuric acid in a test tube to form magnesium Chance and spontaneous change
sulfate and hydrogen releases energy to the surroundings.
The surroundings include:
Diffusion
■■ the solvent (in this case water)
■■ the air around the test tube If you spill a few drops of perfume in a closed room with
■■ the test tube itself no air draughts, the smell spreads gradually throughout
■■ anything dipping into the test tube (e.g. a thermometer). the room. The molecules in the perfume vapour, which
are responsible for the smell, move randomly in straight
Changes that tend to continue to happen naturally are lines until they collide with air molecules, other perfume
called spontaneous changes. Once started, a spontaneous molecules or with the walls of the room. After collision,
change will carry on. When a light is applied, methane the perfume molecules change direction. This process of
gas reacts with oxygen in a spontaneous reaction to form random movement and random collisions of molecules is
carbon dioxide and water. The reaction is spontaneous called diffusion. The reason molecules in a vapour diffuse
because the methane continues to burn in the oxygen until is because of the laws of chance and probability.
Ar atom
He atom before mixing after mixing
increase in entropy
Figure 23.2 The spontaneous mixing of helium atoms ( ) with argon atoms ( ).
Chapter 23: Entropy and Gibbs free energy
We can make a model to show how, during a There are eight different ways of arranging the three
spontaneous process, the entropy of the system increases. molecules between two gas jars. We can express this as:
Figure 23.2 shows a system of two flasks connected by
a stopcock. One flask contains helium and the other number of molecules
contains argon. These gases do not react.
2 × 2 × 2 = 23 = 8
When the stopcock is opened, the gas atoms move
spontaneously by diffusion. After mixing, the gases are two gas jars
mixed up and there is more disorder than before mixing.
The entropy has increased. Each of these ways is equally likely (probable). So the
chance that all the molecules will stay in gas jar A is 1 in
Diffusion and number of ways 8. Similarly, the chance that all three molecules will move
over to gas jar B is 1 in 8. The molecules diffuse because
We can show that the molecules in a vapour diffuse by there are more ways of them being spread out than
chance by thinking about the probability of finding them remaining in the same place.
at one place at any one time. Consider the simplified
model shown below. If we started with five molecules in gas jar A, the
number of ways of arranging the molecules is 25 = 32
partition different ways. If we scaled this up to the numbers of
gas molecules that we might find in a container in the
12 3 gas jar B laboratory, the number of ways of arranging the
molecules is extremely large; for example, for a million
gas jar A molecules between two gas jars it would be 21 000 000, a
number that is too large for your calculator to deal with.
The three molecules in gas jar A cannot move into gas So diffusion happens because there is an overwhelming 351
jar B. likelihood of it taking place as a result of the large number
of ways of arranging the molecules. The idea of the
In this model we assume that: ‘number of ways’ of arranging either molecules or the
energy within molecules dictates whether the changes
■■ there are only a few molecules in gas jar A that take place are the ones that are most likely to happen.
■■ there are no other particles present This applies to chemical reactions as well as to physical
■■ the molecules move randomly and change directions processes such as diffusion.
when they collide. Question
After we remove the partition, the molecules can move 1 a For this question refer back to Figure 23.3. If
randomly not only within gas jar A but also into gas jar there are four molecules in the gas jar on the left,
B. There are three molecules and two places in which they how many ways of arranging the molecules are
can be (gas jar A and gas jar B). The number of ways of there when the partition is removed?
arranging the molecules after removing the partition is
shown in Figure 23.3. b W hat is the probability of finding all four
molecules in the right-hand gas jar?
123 1 32
c Which of the following changes are likely to
1 23 23 1 be spontaneous?
2 31 31 2 i sugar dissolving in water
ii t he smell from an open bottle of aqueous
3 12 1 23
ammonia diffusing throughout a room
Figure 23.3 The eight possible arrangements of molecules iii w ater turning to ice at 10 °C
after removing the partition between the gas jars. iv ethanol vaporising at 20 °C
v w ater mixing completely with cooking oil
vi li mestone (calcium carbonate) decomposing
at room temperature
Cambridge International A Level Chemistry
Comparing entropy values
To make any comparison of entropy values fair, we must
use standard conditions. These standard conditions are the
same as those used for ΔH:
■■ a pressure of 105 Pa
■■ a temperature of 298 K (25 °C)
■■ each substance involved in the reaction is in its normal
physical state (solid, liquid or gas) at 105 Pa and 298 K.
Under these conditions and for a mole of substance, the
unit of standard molar entropy, S —O, is J K–1 mol–1. Standard
molar entropy is the entropy of one mole of substance in
its standard state. The symbol —O indicates that the entropy
is at standard conditions.
Table 23.1 shows some values for some standard
molar entropies.
Substance S —O / Substance S —O / Figure 23.4 A diamond has a very low entropy value because
diamond (s) J K–1 mol–1 methanol (l) J K–1 mol–1 it is a solid element with atoms regularly arranged. Bromine
has a high entropy value because it tends to spread out.
2.4 239.7
graphite (s) 5.7 water (l) 69.9 carbonate (solid) has a higher entropy value than mercury
(liquid).
calcium (s) 41.4 carbon 197.6 ■■ Simpler substances with fewer atoms have lower
352 lead (s) 64.8 monoxide (g) 130.6 entropy values than more complex substances with a
hydrogen (g)
calcium oxide (s) 39.7 helium (g) 126.0 greater number of atoms. For example, for calcium oxide,
calcium 92.9 ammonia (g) 192.3 CaO, S —O = 39.7 J K–1 mol–1 but for calcium carbonate,
carbonate (s) CaCO3, S —O = 92.9 J K–1 mol–1. Carbon monoxide, CO, has a
mercury (l) 76.0 oxygen (g) 205.0 lower entropy value than carbon dioxide, CO2.
bromine (l) 151.6 carbon 213.6 ■■ For similar types of substances, harder substances have
dioxide (g) a lower entropy value. For example, diamond has a lower
entropy value than graphite and calcium has a lower
entropy value than lead.
Table 23.1 Standard molar entropy values of some solids, ■■ For a given substance the entropy increases as a solid
liquids and gases. The states are shown as state symbols after
each substance. melts and then changes to a vapour (see Figure 23.5). For
example, the molar entropy of ice just below its melting
point is 48.0 J K–1 mol–1; the molar entropy for water
The values of all standard molar entropies are positive. is 69.9 J K–1 mol–1, but just above its boiling point, the
Remember that elements have positive standard molar value increases to 188.7 J K–1 mol–1. There is a gradual
entropy values. Do not get this muddled with the case of increase in entropy as the temperature of a substance is
enthalpies, where the elements in their standard states increased. Increasing the temperature of a solid makes
have entropy values of zero. The entropy values are the molecules, atoms or ions vibrate more. Increasing
compared to a theoretically perfect crystal. The Third Law the temperature of a liquid or gas increases the entropy
of Thermodynamics states that ‘All perfect crystals have because it increases the disorder of the particles. When
a substance melts or vaporises, there is a large increase
the same entropy at a temperature of absolute zero’. The in entropy because there is a very large increase in the
nearest we can get to this is a perfect diamond weighing disorder of the particles.
12 g cooled to as low a temperature as possible.
From the values in the table and other data we can When a solid changes to a liquid:
make some generalisations:
■■ the regularly arranged lattice of particles close together in
the solid changes to
■■ Gases generally have much higher entropy values than
liquids, and liquids have higher entropy values than
solids. There are exceptions to this. For example, calcium ■■ an irregular arrangement of particles, which are close
together but rotate and slide over each other in the liquid.
Chapter 23: Entropy and Gibbs free energy
Entropy / J K–1 because high values of entropy are associated with gases.
The more gas molecules, there are, the greater is the
number of ways of arranging them and the higher the
entropy. For example in the reaction:
CaCO3(s) CaO(s) + CO2(g)
boiling point there is an increase in entropy of the system because the
a gas is being produced (high entropy) but the reactant,
melting point calcium carbonate, is a solid (low entropy).
Temperature / K In the reaction
Figure 23.5 The change in entropy as a substance melts and 2N2O5(g) 4NO2(g) + O2(g)
then boils.
we should expect an increase of entropy of the system
When a liquid changes to a vapour: because there are a greater number of moles of gas
■■ the irregular arrangement of particles in the liquid which molecules in the products (5 molecules) than in the
reactants (2 molecules). In addition, there are two
are close together and rotating changes to different product molecules but only one type of reactant
■■ an irregular arrangement of particles, which are free to molecule. This also contributes to a greater disorder in
the products compared with the reactants. The system
move around rapidly because they are far apart from becomes energetically more stable when it becomes
each other. more disordered.
Question In the reaction
2 Explain the difference in the entropy of each of the
N2(g) + 3H2(g) 2NH3(g) 353
following pairs of substances in terms of their state
and structure. we should expect a decrease in the entropy of the
a Br2(l) S —O = 151.6 J K–1 mol–1 system because there is a reduction in the number of gas
molecules as the reaction proceeds. So the entropy change
and I2(s) S —O = 116.8 J K–1 mol–1 of the system is negative. The reactants, hydrogen and
b H2(g) S —O = 130.6 J K–1 mol–1 nitrogen, are more stable than the product, ammonia.
and CH4(g) S —O = 186.2 J K–1 mol–1 Question
c Hg(l) S —O = 76.00 J K–1 mol–1
3 For each of the following reactions, suggest whether
and Na(s) S —O = 51.20 J K–1 mol–1 the entropy of the reactants or the products will be
d SO2(g) S —O = 248.1 J K–1 mol–1 greater or whether it is difficult to decide. Explain
your answers.
and SO3(l) S —O = 95.60 J K–1 mol–1
a NH3(g) + HCl(g) NH4Cl(s)
Entropy changes in reactions
b S(l) + O2(g) SO2(g)
In a chemical reaction, if we compare the entropies of
the reactants and products, we can try to explain the c 2Mg(s) + CO2(g) 2MgO(s) + C(s)
magnitude of the entropy change and whether or not it
increases or decreases. We will assume that gases have d 2Li(s) + Cl2(g) 2LiCl(s)
high entropy and solids have low entropy. If there is a
change in the number of gaseous molecules in a reaction, e H2O(g) + C(s) H2(g) + CO(g)
there is likely to be a significant entropy change. This is
f 2HI(g) H2(g) + I2(g)
g 2K(s) + 2H2O(l) 2KOH(aq) + H2(g)
h MgCO3(s) MgO(s) + CO2(g)
Cambridge International A Level Chemistry
Calculating entropy changes Note that:
1 We need to take account of the stoichiometry of the
Entropy changes in exothermic and
endothermic reactions equation (as we did in calculations involving ΔH —O).
2 When looking up entropy values in tables of data, we
Energy can be transferred from the system to the
surroundings (exothermic change) or from the need to choose the data for the correct state, solid,
surroundings to the system (endothermic change). The liquid or gas.
surroundings are so large that when energy exchange
takes place there is such a small change in temperature or worked examples
pressure that we can ignore these.
1 Calculate the entropy change of the system for
■■ For an exothermic reaction, the energy released to the the reaction:
surroundings increases the number of ways of arranging the
energy. This is because the energy goes into rotation and
translation (movement from place to place) of molecules 2Ca(s) + O2(g) 2CaO(s)
in the surroundings. So there is likely to be an increase in
The standard entropy values are:
entropy and an increased probability of the chemical change S —O [Ca(s)] = 41.40 J K–1 mol–1
occurring spontaneously.
■■ For an endothermic reaction, the energy absorbed S —O [O2(g)] = 205.0 J K–1 mol–1
S —O [CaO(s)] = 39.70 J K–1 mol–1
from the surroundings decreases the number of ways of
arranging the energy. So there is likely to be a decrease
in entropy and a decreased probability of the chemical ΔS —O = S —Oproducts– S —O
change occurring spontaneously.
system reactants
= 2 × S —O [CaO(s)] – {2 × S —O [Ca(s)] + S —O [O2(g)]}
Total entropy change = 2 × 39.70 – {(2 × 41.40) + 205.0}
We can use entropy values to predict whether a chemical = 79.40 – 287.8
reaction will occur spontaneously or not. When a
354 ΔS —O = –208.4 J K–1 mol–1
system
chemical reaction takes place there is a change in entropy The negative value for the entropy change
because the reactants and products have different entropy shows that the entropy of the system has decreased.
values. The symbol for standard entropy change is ΔS —O.
We know, however, that calcium reacts
spontaneously with oxygen. So the entropy of the
The total entropy change involves both the system and the surroundings must also play a part because the total
surroundings. For the system (reactants and products) we entropy change must be positive for the reaction to
write the entropy change as ΔS —Osystem. be feasible.
For the surroundings we write the entropy change as 2 Calculate the entropy change of the system for
ΔS —Osurroundings. the reaction:
The total entropy change is given by: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
ΔS —O = ΔS —Osystem + ΔS —O The standard entropy values are:
total surroundings
If the total entropy change increases, the entropy S —O [CH4(g)] = 186.2 J K–1 mol–1
change is positive, oec.gc.uΔr Ssp—tOootanl tiasn+e4o0u Js Kly–.1W meosl–a1y. The S —O [O2(g)] = 205.0 J K–1 mol–1
reaction will then that the S —O [CO2(g)] = 213.6 J K–1 mol–1
reaction is feasible. S —O [H2O(g)] = 188.7 J K–1 mol–1
If the total entropy change decreases, the entropy
rcehaacntgioenisisntehgeantivneo,tel.igk.eΔlySt—tOootaol ciscu–r4.0 J K–1 mol–1. The ΔS —O = S —O – S —O
system products reactants
Calculating the entropy change of the = {S —O [CO2(g)] + 2 × S —O [H2O(g)]}
system – {S —O [CH4(g)] + 2 × S —O [O2(g)]}
In order to calculate the entropy change of the system we = {213.6 + (2 × 188.7)} – {186.2 + (2 × 205.0)}
use the relationship:
= 591.0 – 596.2
ΔS —O = –5.2 J K–1 mol–1
system
ΔS —O = S —O – S —O
system products reactants
Chapter 23: Entropy and Gibbs free energy
worked examples (continued) 2 The negative sign in front of ΔH e—rOenactthioanlpisypcahratnogfet.hIef the
The negative value for the entropy change shows equation and not the sign of the
enthalpy change is negative, the whole –ΔH —O/T term
that the entropy of the system has decreased becomes positive.
slightly. We know, however, that methane burns
in oxygen once it is ignited. So the entropy of the worked examples
surroundings must also play a part in the overall
entropy change. 3 Calculate the entropy change of the surroundings for
the reaction:
Question 2Ca(s) + O2(g) 2CaO(s)
ΔH —O = –1270.2 kJ mol–1
4 Calculate the standard entropy change of the system reaction
in each of the following reactions using the standard
molar entropy values given here. Step 1 Convert the enthalpy change into J mol–1 by
multiplying by 1000.
(Values for S —O in J K–1 mol–1: Cl2(g) = 165.0, –1270.2 × 1000 = –1 270 200 J mol–1
Fe(s) = 27.30, Fe2O3(s) = 87.40, H2(g) = 130.6,
H2O(l) = 69.90, H2O2(l) = 109.6, Mg(s) = 32.70, Step 2 Apply the relationship
MgO(s) = 26.90, Na(s) = 51.20, NaCl(s) = 72.10,
NH4NO3(s) = 151.1, N2O(g) = 219.7, O2(g) = 205.0) ΔS —Osurroundings = – Δ__H_—Or_eT_a c_t_io_n
a 2H2O2(l) 2H2O(l) + O2(g) = _–_( –_1_ 2_2_97_08_ 2_0_0_)
= +4262.4 J K–1 mol–1
b NH4NO3(s) N2O(g) + 2H2O(g)
c 2Mg(s) + O2(g) 2MgO(s) 4 Calculate the entropy change of the surroundings for
the reaction:
d 2Na(s) + Cl2(g) 2NaCl(s) 355
e 3Mg(s) + Fe2O3(s) 3MgO(s) + 2Fe(s) CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
ΔH —O = –890.3 kJ mol–1
reaction
Calculating the entropy change of the Step 1 Convert the enthalpy change into J mol–1 by
surroundings multiplying by 1000.
–890.3 × 1000 = –890 300 J mol–1
Many chemical reactions are accompanied by large Step 2 Apply the relationship
enthalpy changes. These enthalpy changes change
the number of ways of arranging the energy in the ΔS —O = – Δ__H_—O_reT_a_ ct_io_n_
surroundings. So, in many chemical reactions the
value of the entropy changes in the surroundings cannot surroundings
be ignored.
= –__( –_8_29_90_8 _3 _0_0_)
The entropy change of the surroundings is calculated
using the relationship: = +2987.6 J K–1 mol–1
ΔS —O = _– _ Δ_H__T—rO_e _ac_ti_o_n Question
surroundings
where 5 Calculate the entropy change of the surroundings in
each of the following reactions. Assume that the value
■■ ΔH —O is the standard enthalpy change of the reaction of ΔH does not change with temperature.
reaction
■■ T is the temperature in kelvin. At standard temperature,
this value is 298 K. a C(s) + O2(g) CO2(g) ΔH —O
reaction
= –393.5 kJ mol–1
Note: carried out at 0 °C
1 tWhehvenalupeerofof rΔmHin—rOegacctioanlciunlkatJi monosl–t1osfhinoudlΔd Sb—eOsumrrouunltdiipnlgised
b 2C(s) + N2(g) C2N2(g) ΔH —O
by 1000. This is because entropy changes are measured
in units of joules per kelvin per mole. carried out at 300 °C reaction
= +307.9 kJ mol–1
Cambridge International A Level Chemistry
Question (continued) Question
c H2(g) + F2(g) 2HF(g) ΔH —O 6 Calculate the total standard entropy change in each
of the following reactions using the standard molar
reaction entropy values given here.
= –271.1 kJ mol–1
carried out at standard temperature
d Si(s) + 2H2(g) SiH4(g) ΔH —O (Values for S —O in J K–1 mol–1: C(graphite) = 5.700,
C2N2(g) = 242.1, C3H8(g) = 269.9, CO2(g) = 213.6,
reaction H2(g) = 130.6, H2O(l) = 69.90, H2S(g) = 205.7,
= +34.30 kJ mol–1 N2(g) = 191.6, O2(g) = 205.0, P(s) = 41.10,
P4O10(s) = 228.9, S(s) = 31.80)
carried out at –3 °C
Calculating total entropy change a S(s) + H2(g) H2S(g) –20.6 kJ mol–1
ΔH —O =
The total entropy change is given by:
reaction
ΔS —O = ΔS —Osystem + ΔS —O b 2C(graphite) + N2(g) ΔH C2N2(g) +307.9 kJ mol–1
—O =
total surroundings
reaction
We can also write this as:
c 4P(s) + 5O2(g) P4O10(s)
ΔS —O = ΔS —O – ΔH —rOeaction/T ΔH —O = –2984.0 kJ mol–1
total system reaction
The total entropy change for the examples given above for d C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
the reaction of calcium with oxygen and the combustion
of methane are calculated by simply adding the entropy ΔH —O = –2219.2 kJ mol–1
reaction
change of the system to the entropy change of the
surroundings.
Entropy in equilibrium reactions
In equilibrium reactions both products and reactants are
356 worked examples present. How can the total entropy change be positive in
5 Calculate the total entropy change for the reaction: both directions? There is an additional increase in disorder
and hence an increase in entropy associated with this
2Ca(s) + O2(g) 2CaO(s) mixing. Figure 23.6 shows how the increase in entropy
changes as a reaction progresses, starting either from pure
ΔS —O = –208.4 J K–1 mol–1 reactants or pure products to reach equilibrium.
system
ΔS —O = +4262.4 J K–1 mol–1
surroundings
So:
ΔS —O = ΔS —Osystem + ΔS —O
total surroundings equilibrium
= –208.4 + 4262.4
ΔS —Ototal = +4054.0 J K–1 mol–1 Increase in entropy
6 Calculate the total entropy change for the reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
ΔS —O = –5.2 J K–1 mol–1
system
ΔS —O = +2987.6 J K–1 mol–1
surroundings
So:
ΔS —O = ΔS —Osystem + ΔS —O Pure N2O4 Pure NO2
total surroundings
= –5.2 + 2987.6
ΔS —O = +2982.4 J K–1 mol–1 Figure 23.6 The total entropy change, ΔS —Ototal, when N2O4 is
converted to an equilibrium mixture of NO2 and N2O4 and NO2
total is converted to the same equilibrium mixture.
As mixing proceeds, the rate of increasing disorder
You can see that in both of these worked examples, dfreocmreaNs2eOs a4.sAmtosroemaensdtamgoerienNthOe2rmeaoclteicounl,etshaerreaftoeromf etdhe
the large positive entropy change of the
surroundings has more than compensated for
the negative entropy change of the system. The
total entropy change is positive and the reactions
are feasible.
Chapter 23: Entropy and Gibbs free energy
forward reaction equals the rate of the backward reaction. ΔS —O = +174.8 – 131.5 J K–1 mol–1
Equilibrium has been reached. The same argument applies = +43.3 J K–1 mol–1
to the reverse reaction. At the position of equilibrium the total
total entropy change of the forward reaction equals the
total entropy change of the backward reaction, and under ΔS —O
standard conditions the overall entropy change is zero.
total
You can see that at 298 K the total entropy change
is negative, so the reaction does not occur at this
temperature. At 550 K the total entropy change is positive,
so the reaction is spontaneous at this temperature.
Entropy and temperature ■■ When the total entropy change in a reaction shows a
large increase, e.g. +200 J K–1 mol–1, the reaction can
We have seen that the entropy change of the surroundings be regarded as going to completion. It is definitely
is given by spontaneous.
ΔS —O = _– _ Δ_H__T—rO_e _ac_ti_o_n ■■ When the total entropy change shows a large decrease,
e.g. –600 J K–1 mol–1, we can deduce that there is very little
surroundings likelihood of a reaction occurring.
If we carry out reactions at temperatures above standard Question
temperature, an increase in temperature makes the
entropy change of the surroundings less negative or more
positive. If we carry out reactions at temperatures below 7 The decomposition of calcium carbonate,
standard temperature, a decrease in temperature makes CatarCoOo3m(s)temperCaatuOr(es). + CO2(g), does not take place
the entropy change of the surroundings more negative or
less positive. In both these cases we make the assumption a E xplain in terms of entropy changes why
that ΔH does not change significantly with heating the calcium carbonate to a high
—O ΔH does change slightly with temperature increases the likelihood of this
reaction taking place.
reaction
temperature. In realit y —O b I n a closed system at high temperature, the
reactants and products are in equilibrium.
reaction
temperature, but we can often disregard this change.
We can see how increasing the temperature affects 357
the ability of zinc carbonate to undergo thermal CaCO3(s) CaO(s) + CO2(g)
decomposition by comparing the entropy changes at 298 K
and 550 K. However, we have to take into account both i E xplain the meaning of the term closed
the surroundings and the system. We assume that neither system.
the standard molar entropies nor the enthalpy change of ii E xplain in terms of entropy changes what
formation change with temperature. happens when the pressure on this system
is increased.
ZnCO3(s) ZnO(s) + CO2(gΔ)H = +72.3 kJ mol–1 iii W hat is the value of the standard total
entropy change at equilibrium?
—O
reaction
For this reaction S —O = +174.8 J K–1 mol–1 (the same for
system
both temperatures).
Entropy, enthalpy changes and
Reaction at 298 K free energy
= _–_ 72_29_ 3_80_ 0_
ΔS —O For an exothermic reaction such as:
surroundings
= –242.6 J K–1 mol–1 CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
ΔS —O = +174.8 – 242.6 J K–1 mol–1 ΔH —O = –890.3 kJ mol–1
= –67.8 J K–1 mol–1
total reaction
ΔS —O the entropy change of the system is negative. But the
large negative value of the enthalpy change more than
total compensates for the negative entropy change of the system
because it causes the term –ΔH —rOeaction/T to have a high
Reaction at 550 K positive value. So the total entropy change is positive
= _–_ 75_25_ 30_0_ 0_ and the reaction, once started, is spontaneous. In highly
ΔS —O
surroundings
= –131.5 J K–1 mol–1
Cambridge International A Level Chemistry
eaxnodthneergmatiicver,etahcetieonntsh, awlphyercehtahnegveailsutehoefdΔrHivi—rOneagctifoonricselaorfge We can also write the expression without having to
the reaction. consider the entropy changes of the surroundings:
In endothermic reactions, the entropy term tends to
be more important. The term –ΔH —rOeaction/T has a negative ΔG = ΔHreaction – TΔSsystem
value. If the value orefaΔcSti—oOsynstewmiallnndoΔt Sbe—Osusrproounndtianngseaorues.both
negative, then the of is positive and large Where T is the temperature in kelvin.
However, if the value ΔS Gibbs free energy is a useful concept because it includes
—O
both enthalpy change and entropy change.
system To make any comparison of Gibbs free energy values
enough, it can compensate for the negative value of the
iΔsSsp—Osuorrnotuandninegosusso. that ΔS becomes positive. The reaction fair, we must use standard conditions. These standard
—O conditions are the same as those used for ΔH and ΔS:
■■ pressure of 105 Pa
total ■■ temperature of 298 K (25 °C)
■■ each substance involved in the reaction is in its normal
Chemists are usually interested in the system of
reactants and products rather than having to consider the physical state (solid, liquid or gas) at 105 Pa and 298 K.
energy changes with the surroundings. Fortunately for
us, there is a way in which we can take account of both The standard molar Gibbs free energy of formation is
system and surroundings in a more straightforward way. the free energy change that accompanies the formation
This involves a quantity called Gibbs free energy or, more of one mole of a compound from its elements in their
simply, free energy. It can also be called Gibbs energy or standard state.
Gibbs function, G.
The symbol for standard molar Gibbs free energy of
formation is ΔG —Of . The units are kJ mol–1.
For example:
358 Mg(s) + _12 O2(g) MgO(s) ΔG —Of = –569.4 kJ mol–1
Derivation
Gibbs free energy can easily be derived from the equation
relating total entropy to the entropy changes of system and
surroundings.
As:
ΔS —O = ΔS —O – _Δ_HT__ —O_
total system
Multiplying by –T: –TΔS —O = –TΔS —Osystem + ΔH —O
to Gibbs
total —O .
freeTehneetregrymch–TanΔgSe—Osoysftetmh e+r eΔaHct—iOoins equivalent the
Figure 23.7 Gibbs free energy is named after American system ΔG
scientist Josiah Willard Gibbs, who applied the concept of So –TΔS = ΔG and so ΔG = ΔH – TΔS —Osystem.
entropy and ‘applied energy’ changes to chemical reactions —O —O —O —O
and physical processes.
total
Gibbs free energy
Gibbs free energy and spontaneous
What is Gibbs free energy? reactions
Fpoorsiativreea. cTthioenvatolubeeosfpTonistaanlweoauyss,pΔoSsi—tOtoitvale must be
In determining whether a chemical reaction is likely to on the absolute
be spontaneous we use the quantity Gibbs free energy (kelvin) temperature scale. So applying these signs to
change, ΔG. The Gibbs free energy change is given by the relationship ΔG —O = –TΔS —tOotal, the value of ΔG must
the relationship: be negative for a reaction to be spontaneous. So, when a
spontaneous reaction occurs at constant temperature and
ΔG = –TΔStotal pressure, the Gibbs free energy decreases. If the value of
ΔG is positive, the reaction is not spontaneous.
Chapter 23: Entropy and Gibbs free energy
Applying the equation Question (continued)
ΔG —O = ΔH —O – TΔS —O system
a H2(g) + Cl2(g) 2HCl(g)
We can calculate the Gibbs free energy change for a
reaction if we know: ΔH —O = –184.6 kJ mol–1
■■ the entropy change of the system in J K–1 mol–1 r
■■ the enthalpy change of the system in J mol–1; we have to
b CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
multiply the value of the enthalpy change by 1000 because ΔH —O = –890.3 kJ mol–1
the entropy change is in joules per kelvin per mol
■■ the temperature; under standard conditions, this is 298 K. r
c 2Na(s) + O2(g) Na2O2(s) –510.9 kJ mol–1
ΔH —O =
r
d Mg(s) + Cl2(g) MgCl2(s) –641.3 kJ mol–1
ΔH —O =
worked example
r
7 Calculate the Gibbs free energy change for the e Ag2CO3(s) Ag2O(s) + CO2(g) +167.5 kJ mol–1
decomposition of zinc carbonate at 298 K. ΔH —O =
r
ZnCO3(s) ZnO(s) + CO2(g) ΔH —O = +71.0 kJ mol–1
r
(Values for S —O in J K–1 mol–1: CO2(g) = +213.6, Temperature change and reaction
ZnCO3(s) = +82.4, ZnO(s) = +43.6) spontaneity
Step 1 Convert the value of ΔH —O to J mol–1: For a reaction to be spontaneous, ΔG must be negative.
The temperature can influence the spontaneity of a
r reaction. We can deduce this by considering the Gibbs free
energy as a combination of two terms in the relationship
+71.0 × 1000 = 71 000 J mol–1
∆G = ∆Hreaction – T∆Ssystem
Step 2 Calculate ΔS —Osystem: first second
term term
Δ S —O = S —O – S —O 359
system products reactants
= S —O [ZnO(s)] + S —O [CO2(g)] – S —O [ZnCO3(s)]
= 43.6 + 213.6 – 82.4
Δ S —O = +174.8 J K–1 mol–1
system
Step 3 Calculate ΔG —O : Assuming that the value wofeΔcHanresaecetiotnhdatotehs envoat lcuheaonfge
much with temperature,
ΔG —O = ΔH —O – TΔS —O
reaction system TΔSsystem may influence the value of ΔG.
= 71 000 – 298 × (+174.8)
ΔG —O = +18 909.6 J mol–1 ■■ For an exothermic reaction, the first term (ΔHreaction) has a
= +18.9 kJ mol–1 (to 3 significant figures) negative value.
As the value of ΔG —O is positive, the reaction is not – If the value of ΔSsystem is positive, the second term
spontaneous at 298 K. (–TΔSsystem) is negative and the reaction will be
Question spontaneous because both ΔHreaction and –TΔSsystem
are negative. So ΔG is negative.
8 Calculate the standard Gibbs free energy of reaction
in each of the following using the standard molar – If the value of ΔSsystem is negative, the second term is
entropy values given. Express your answers to 3 positive. The reaction is likely to be spontaneous if the
significant figures in kJ mol–1, and in each case state
whether the reaction is spontaneous or not under temperature is low because ΔHreaction is more likely to
standard conditions. have a greater negative value than the positive value of
(Values for S —O in J K–1 mol–1: Ag2CO3(s) = 167.4, the second term. So ΔG is negative. If the temperature
Ag2O(s) = 121.3, CH4(g) = 186.2, Cl2(g) = 165, is very high, the second term may be positive enough
CO2(g) = 213.6, H2(g) = 130.6, HCl(g) = 186.8, to overcome the negative value of ΔHreaction and
H2O(l) = 69.9, Mg(s) = 37.2, MgCl2(s) = 89.6, make ΔG positive. So the reaction is less likely to be
Na(s) = 51.2, Na2O2(s) = 95.0, O2(g) = 205.0) spontaneous at a higher temperature. This mirrors
what we know about the effect of temperature on
equilibrium: for an exothermic reaction, a higher
temperature shifts the position of equilibrium in favour
of the reactants.
Cambridge International A Level Chemistry
■■ For an endothermic reaction, the first term (ΔHreaction) has Comparing Gibbs free energy values
a positive value.
Table 23.2 shows some values for some standard molar
– If the value of ΔSsystem is negative, the second term is Gibbs free energy changes of formation.
positive. The reaction will not occur because both You learnt in Chapter 6 (page 92) that the standard
terms are positive, making the value of ΔG positive.
enthalpy change of an element is zero. Similarly, the
– If the value of ΔSsystem is positive, the second term is standard Gibbs free energy change of formation of an
negative. The reaction is unlikely to be spontaneous if element is zero. Many compounds in the solid state
have high negative values of Gibbs free energy change of
the temperature is low because ΔHreaction is more likely formation. Many gases and liquids have standard Gibbs
to have a greater positive value than the negative
value of the second term. So ΔG is positive. If the
temperature is very high, the second term may be free energy change values that are negative but many
others, such as ethene, have positive values. The standard
negative enough to overcome the positive value of Gibbs free energy change of formation also depends on
ΔHreaction and make ΔG negative. So the reaction is
more likely to be spontaneous at a higher temperature. the —Osf t[aHte2.OF(ogr)]exisam– 2p2le8,.6Δ kGJ —mOf [oHl–21O. (l)] is – 237.2 kJ mol–1 but
This mirrors what we know about the effect of ΔG
temperature on equilibrium: for an endothermic
reaction, a higher temperature shifts the position of
equilibrium in favour of the products. Gibbs free energy calculations
We can see the effect of temperature on the spontaneity Gibbs free energy change of reaction
of the reaction if we rework Worked Example 7 at a
temperature of 1200 K. The standard Gibbs free energy change of reaction is
the Gibbs free energy change when the amounts of the
reactants shown in the stoichiometric equation react under
worked example standard conditions to give products. The reactants and
8 Calculate the Gibbs free energy change for the products must be in their standard states.
360 decomposition of zinc carbonate at 1200 K. The method of calculating Gibbs free energy change of
+71.0 kJ mol–1 reaction uses an energy cycle similar to the enthalpy cycles
ZnCO3(s) ZnO(s) + CO2(g) ΔH = you used to calculate the enthalpy change of reaction in
—O Chapter 6 (see Figure 23.8).
r
(Values for S —O in J K–1 mol–1: CO2(g) = +213.6, ΔG —O
reaction
ZnCO3(s) = +82.4, ZnO(s) = +43.6)
reactants products
ΔH —O = +71.0 kJ mol–1 ΔS —O = +174.8 J K–1 mol–1
r system
ΔG —O = ΔH —O – TΔS —O ΔG —O ΔG —O
reaction system 1 2
= 71 000 – 1200 × (+174.8)
= 71 000 – 209 760 elements in their
standard states
ΔG —O = –139 kJ mol–1 Figure 23.8 A free energy cycle for calculating the standard
Gibbs free energy of reaction. The dashed line shows the
indirect (two-step) route.
A s the value of ΔG —O is negative, the reaction is
spontaneous at 1200 K.
Substance ΔG —Of / kJ mol–1 Substance ΔG —Of / kJ mol–1
carbon (s) 0 water (l) –237.2
calcium (s) methanol (l) –166.4
bromine (l) 0 chlorobenzene (l) +93.6
helium (g) water (g) –228.6
calcium oxide (s) 0 ethane (g) +68.2
calcium carbonate (s) ammonia (g) –16.5
magnesium oxide (s) 0 magnesium ion, Mg2+ (aq) –454.8
zinc sulfide (s) –604.0 carbonate ion, CO32– (aq) –527.9
–1128.8
–569.4
–201.3
Table 23.2 Standard molar Gibbs free energy changes of some solids, liquids, gases and aqueous ions. The states are shown as
state symbols after each substance.
Chapter 23: Entropy and Gibbs free energy
Using the same ideas as in Hess’s law, we see that: worked examples (continued)
ΔG —O2 = ΔG —1O + ΔG —O
reaction ΔG —O = ΔG —O2 – ΔG —O
So: reaction 1
ΔG —O = ΔG —O – ΔG —O1 ΔG —O = ΔG —O [Na2CO3(s)] + ΔG —O [CO2(g)]
reaction 2 reaction f f
+ ΔG —O [H2O(l)] – 2 × ΔG —O [NaHCO3(s)]
f f
= (–1044.5) + (–394.4) + (–237.2)– {2 × (–851.0)}
Another way of writing this is: = –1676.1 – (–1702)
ΔG —O = ΔG —O – ΔG —O = +25.9 kJ
reaction products reactants The value of ΔG —O is positive. So under standard
To calculate the Gibbs free energy change of reaction from reaction
an energy cycle like this, we use the following procedure: conditions, the reaction is not spontaneous. However,
ΔG does var y with temperature. At a higher
—O
reaction
temperature the reaction is spontaneous.
■■ write the balanced equation at the top
■■ draw the cycle with the elements at the bottom 10 Calculate the standard Gibbs free energy change of the
reaction between hydrogen and oxygen.
■■ draw in all arrows making sure that they go in the
correct directions
2H2(g) + O2(g) 2H2O(l)
■■ calculate ΔG —O = ΔG —O – ΔG —O taking into account the
reaction 2 1
number of moles of reactants and products.
(The relevant Gibbs free energy value is:
ΔG —O [H2O(l)] = –237.2 kJ mol–1)
f
worked examples Note that the values of ΔG —O for both hydrogen and
f
oxygen are zero, as they are elements in their standard
states.
9 Draw a Gibbs free energy cycle to calculate the
standard Gibbs free energy change of decomposition ΔG —O = ΔG —O – ΔG —O
of sodium hydrogencarbonate.
reaction products reactants
ΔG —O 361
reaction —O [H2O(l)] – —O [H2(g)] + ΔG —O [O2(g)]}
= 2 × ΔG {2 × ΔG
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l) f f f
The relevant Gibbs free energy values are: = 2 × (–273.2) – 0 + 0
ΔG —O [NaHCO3(s)] = –851.0 kJ mol–1 ΔG —O = –546.4 kJ
f reaction
ΔG —O [Na2CO3(s)] = –1044.5 kJ mol–1 The value of ΔG —O is negative. So under standard
f reaction
conditions, the reaction is spontaneous.
ΔG —O [CO2(g)] = –394.4 kJ mol–1
f
ΔG —O [H2O(l)] = –237.2 kJ mol–1 Gibbs free energy and work
f Gibbs free energy change can be thought of as part of the
enthalpy change that is needed to do work. If we rearrange
The Gibbs free energy cycle is shown in Figure 23.9. the equation ΔG = ΔH – TΔS as ΔH = ΔG + TΔS, we can
regard the +TΔS part as being the energy unavailable to do
2 NaHCO3(s) ∆G reaction Na2CO3(s) + CO2(g) + H2O(l) work because it is involved with the disorder of the system.
The ΔG part is free energy that is available to do work, e.g.
2 ∆G f NaHCO3(s) ∆G1 ∆G f [Na2CO3(s)] driving the charge in electrochemical cells.
∆G2 +∆G f [CO2(g)]
total energy change
+∆G f [H2O(l)]
2Na(s) + 2C(graphite) + 3O2(g) + H2(g)
Figure 23.9 The free energy cycle for the ∆G T∆S
decomposition of sodium hydrogencarbonate.
The dashed line shows the two-step route.
energy available energy
to do work not available
to do work
Figure 23.10 The enthalpy change of a reaction at constant
temperature can be split into two parts.
Cambridge International A Level Chemistry
Gibbs free energy change and direction ■■ The reaction can be regarded as not being feasible
of chemical change (spontaneous) at all if the value of ΔG is high and
positive, e.g. +60 kJ mol–1. —O
Gibbs free energy of formation is a measure of the stability
reaction
of a compound. The more negative the value of ΔG —Of , the Question
greater the stability of the compound. It is unlikely to 9 Calculate the standard Gibbs free energy change
decompose. If ΔG —Of is positive, the compound is likely to be
unstable with respect to its elements. For example: of reaction in each of the following using the
standard molar values for Gibbs free energy
_ 12 H2(g) + _12 I2(s) HI(g) ΔG —Of [HI(g)] = +1.7 kJ mol–1 change given here. In each case, comment on
whether the reaction is spontaneous or not, under
The Gibbs free energy change of reaction is also a standard conditions.
measure of the feasibility of a reaction. Reactions with
negative values of ΔG are likely to be feasible
—O (Values for G —O in kJ mol–1: C3H8(g) = –23.4,
reaction
(spontaneous), whereas those with positive values are less CO2(g) = –394.4, Fe2O3(s) = –742.2, H2O(l) = –273.2,
likely to be spontaneous.
H2O2(l) = –120.4, MgO(s) = –569.4, NaCl(s) = –384.2,
NH4NO3(s) = –184.0, N2O(g) = +104.2)
■■ When a system is in chemical equilibrium and the amounts
of products and reactants balance, the value of ΔG
is zero (ΔG = 0). —O a 2H2O2(l) 2H2O(l) + O2(g)
—O
reaction
reaction
The products predominate if the value of ΔG —O has a b NH4NO3(s) N2O(g) + 2H2O(g)
■■ fairly low negative value, e.g. –10 kJ mol–1.
reaction
The reactants predominate if the value of ΔG —O has a c 2Mg(s) + O2(g) 2MgO(s)
slightly positive value, e.g. +10 kJ mol–1.
■■ reaction d C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
■■ The reaction can be regarded as complete if the value of e 3Mg(s) + Fe2O3(s) 3MgO(s) + 2Fe(s)
–60 kJ mol–1.
362 ΔG —O is high and negative, e.g.
reaction
Summary ■ Solids generally have smaller entropies than liquids,
and liquids have smaller entropies than gases.
■ Entropy (S) is related to the degree of randomness
or disorder in a system. The greater the disorder, the ■ A knowledge of the structures and states of
greater the entropy. the reactants and products helps us to make
generalisations about whether the entropy of the
■ Standard molar entropy (S —O ) is the entropy reactants or products is greater.
when the substance is in its normal state at 298 K
and 105 Pa. ■ The total entropy change in a reaction is given by
ΔStotal = ΔSsystem + ΔSsurroundings
■ In a chemical reaction, the system is the chemical
reactants and products themselves and the ■ The entropy change in the system is given by
surroundings is everything not involved in the ΔSsystem = ΔSproducts – ΔSreactants
system, e.g. the air around the reaction vessel.
■ The entropy change in the surroundings is given
■ A system becomes energetically more stable when it by ΔSsurroundings = –ΔHreaction/T, where T is the
becomes more disordered temperature in kelvin.
■ A spontaneous change is one that, once started, ■ An increase in temperature makes the entropy
tends to continue to happen. change of the surroundings less negative.
■ A spontaneous change involves an increase in ■ Standard Gibbs free energy of formation is
total entropy (ΔS is positive). If a reaction is the Gibbs free energy change when 1 mole of a
not spontaneous, there is a decrease in entropy compound is formed from its elements in their
(ΔS is negative). normal states under standard conditions.
■ The entropy increases as a substance changes state
from solid to liquid to gas.
Chapter 23: Entropy and Gibbs free energy
■ standard Gibbs free energy of reaction is the Gibbs the enthalpy change of reaction and entropy
free energy change when the amounts of reactants
shown in the stoichiometric equation react under change of the system by the relationship
standard conditions to form the products. —O —O – —O
ΔG = ΔG ΔG
reaction products reactants
■ The Gibbs free energy change of formation of an
■ Gibbs free energy is related to the enthalpy change element is zero.
of reaction and entropy change of the system by the ■ spontaneous (feasible) chemical changes involve a
decrease in Gibbs free energy (ΔG is negative).
relationship ΔG —O = ΔH —O – TΔS —O
reaction system
■ The Gibbs free energy change of a reaction can ■ Chemical reactions tend not to be spontaneous if there
is an increase in Gibbs free energy (ΔG is positive).
be calculated from Gibbs free energy changes
of formation using the relationship is related to
End-of-chapter questions
1 Graphite and diamond are both forms of carbon. Their standard molar entropies are:
Δ S —O = 5.70 J K–1 mol–1, ΔS —O = 2.40 J K–1 mol–1
graphite diamond
a i Suggest why the standard molar entropy of graphite is greater than that of diamond. [2] 363
[1]
ii Calculate the entropy change of the process Cgraphite Cdiamond at 298 K [1]
iii Explain why you would be unlikely to make diamonds from graphite at atmospheric
temperature and pressure. [4]
[1]
b The standard molar enthalpy change for Cgraphite Cdiamond is +2.00 kJ mol–1. [1]
i Calculate the total entropy change for this reaction at 25.0 ºC.
ii Explain why you would be unlikely to make diamonds from graphite at atmospheric
temperature and pressure.
c Graphite reacts with oxygen to form carbon dioxide. Would you expect the entropy of the
products to be greater or less than the entropy of the reactants? Explain your answer.
Total = 10
2 3268 kJ are required to change 1 mole of ethanol into its gaseous atoms.
C2H5OH(g) 2C(g) + 6H(g) + O(g)
a Calculate the entropy change of the surroundings during this process when it is carried out at 150 °C. [3]
b Explain why the total entropy change of this reaction is likely to be negative. [2]
c When ethanol undergoes combustion, carbon dioxide and water are formed.
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH —O = –1367 kJ mol–1
reaction
Calculate the total standard entropy change for this reaction.
(Values for S —O in J K–1 mol–1: C2H5OH(l) = 160.7, CO2 =213.6, H2O(l) = 69.90, O2(g) = 205.0) [6]
Total = 11
cambridge international a level chemistry
3 At 10 °C ice changes to water.
H2O(s) H2O(l) ΔH —O = +6.01 kJ mol–1
a Explain why the entropy change of the system is positive. [2]
[4]
b The entropy change for the system is +22.0 J K–1 mol–1. Calculate the total entropy change at 10.0 °C. [2]
c Explain why the process shown in the equation does not occur at –10.0 °C.
d At 273 K ice and water are in equilibrium.
H2O(s) H2O(l) ΔH —O = +6.01 kJ mol–1
i Calculate the entropy change of the surroundings for this process. [2]
ii What main assumption did you make when doing this calculation? [1]
iii Use your answer to part d i to deduce the value of ΔSsystem at 278 K. Explain your answer fully. [3]
e At –10.0 °C, the value of ΔSsystem is –22.0 J K–1 mol–1 and the value of ΔSsurroundings is +22.9 J K–1 mol–1.
Calculate the total entropy change in the process. Include the correct sign. [2]
Total = 16
4 Barium carbonate decomposes when heated.
BaCO3(s) BaO(s) + CO2(g)
a Use your knowledge of Hess cycles from Chapter 6 to calculate the enthalpy change of this reaction.
Express your answer to three significant figures.
ΔH —O [BaCO3(s)] = –1216.0 kJ mol–1
364 f
ΔH —O [BaO(s)] = –553.5 kJ mol–1
f
ΔH —O [CO2(g)] = –393.5 kJ mol–1 [2]
f
b Use your answer to part a and the data below to calculate ΔStotal for this reaction under standard
conditions. Express your answer to three significant figures.
ΔS —O [BaCO3(s)] = +112.1 J K–1 mol–1 [5]
ΔS —O [BaO(s)] = +70.40 J K–1 mol–1
ΔS —O [CO2(g)] = +213.6 J K–1 mol–1
c Is this reaction spontaneous (feasible) at 298 K? Explain your answer. [1]
d i Calculate the temperature at which this reaction has a total entropy value of zero. [4]
ii What main assumption did you make in your calculation in part d i? [1]
Total = 13
5 a For each of the following changes state whether the entropy of the system decreases or increases.
In each case, explain your answer in terms of the order or disorder of the particles
i NaCl(s) + aq Na+(aq) + Cl–(aq) [5]
ii H2O(g) H2O(l) [3]
b The table shows the formula, state and standard molar entropies of the first six straight-chain alkanes.
Alkane CH4(g) C2H6(g) C3H8(g) C4H10(g) C5H12(l)
ΔS —O / J K–1 mol–1 186.2 229.5 269.9 310.1 261.2
i Describe and explain the trend in the standard molar entropies of these alkanes. [4]
ii Estimate the value of the standard molar entropy of the straight-chain alkane with the formula C6H14. [1]
Total = 13
Chapter 23: Entropy and Gibbs free energy
6 a i Define standard free energy change of formation. [2]
ii Write a balanced equation to represent the standard free energy change of formation of ethane. [2]
Include state symbols in your answer.
[1]
b The standard free energy change of formation of ethane is –32.9 kJ mol–1. The standard entropy [3]
change of the system for this reaction is –173.7 J K–1 mol–1.
i State the relationship between standard free energy change of formation, standard
entropy change of the system and the enthalpy change.
ii Explain why the reaction can be regarded as spontaneous, even though the value of the standard
entropy change of the system is negative.
c Ethane undergoes combustion to form carbon dioxide and water.
C2H6(g) + 3_21O2(g) 2CO2(g) + 3H2O(l)
Calculate the free energy change of combustion for this reaction.
(Values for ΔG —O in kJ mol–1: C2H6(g) = –32.9, CO2(g) = –394.4, H2O(l) = –237.2) [3]
Total = 11
f
7 Calcium carbonate decomposes when heated to form calcium oxide and carbon dioxide.
CaCO3(s) CaO(s) + CO2(g)
a Calculate the entropy change of the system for this reaction.
(Values for S —O in J K–1 mol–1: CaCO3(s) = +92.9, CaO(s) = +39.7, CO2(g) = +213.6) [2]
b Calculate the enthalpy change for this reaction. [2] 365
[5]
(Values for ΔH —O in kJ–1 mol–1: CaCO3(s) = –1206.9, CaO(s) = –635.1, CO2(g) = –393.5) [3]
f
c Use your answers to parts a and b to calculate the Gibbs free energy change of this reaction at
298 K by a method that does not include calculating the entropy change of the surroundings.
d Explain why the reaction is not spontaneous at 298 K even though the entropy change of the
system has a positive value.
Total = 12
8 Water is formed when hydrogen burns in oxygen.
2H2(g) + O2(g) 2H2O(l) ΔH —O = –561.6 kJ–1 mol–1
r
a Calculate the entropy change of the system for this reaction.
(Values for S —O in J K–1 mol–1: H2(g) = +130.6, O2(g) = +205.0, H2O(g) = +69.9) [3]
b Use your answer to part a and the information at the start of the question to calculate a value for [5]
the Gibbs free energy change of this reaction.
c Is the reaction spontaneous at room temperature? Explain your answer. [1]
d Use your answer to part b to suggest a value for the standard Gibbs free energy of formation of H2O(l). [2]
Explain your answer.
Total = 11
Chapter 24:
transition elements
Learning outcomes
You should be able to:
■■ explain what is meant by a transition element, ■■ describe the use of cis-platin as an anticancer drug
in terms of d-block elements forming one or more ■■ explain qualitatively that ligand exchange may occur,
stable ions with incomplete d orbitals including the complexes of copper(II) ions with
■■ state the electronic configuration of a first-row water, hydroxide and ammonia
transition element and of its ions ■■ describe and explain ligand exchanges in terms of
366 ■■ contrast, qualitatively, the melting point, density, competing equilibria
atomic radius, ionic radius, first ionisation energy ■■ describe the term stability constant, Kstab, of a
complex ion as the equilibrium constant for the
and conductivity of the transition elements with
formation of the complex ion in a solvent from its
those of calcium as a typical s-block element
constituent ions or molecules
■■ describe the tendency of transition elements to have
■■ deduce expressions for the stability constant in
variable oxidation states
ligand substitution
■■ predict from a given electronic configuration, the
likely oxidation states of a transition element ■■ explain ligand exchange in terms of stability
constants and understand that a large value of Kstab
■■ describe and explain the use of Fe3+/Fe2+, is due to the formation of a stable complex ion.
MnO4–/Mn2+ and Cr2O72–/Cr3+ as examples of redox
systems (see also Chapter 19) ■■ sketch the general shape of a d orbital
■■ predict, using E —O values, the likelihood of redox ■■ describe the splitting of degenerate d orbitals into
reactions two energy levels in octahedral complexes using
the complexes of copper(II) ions with water and
■■ define the term ligand as a species that contains a
ammonia as examples
lone pair of electrons that forms a dative covalent
bond to a central metal atom/ion including ■■ explain the origin of colour in transition element
complexes resulting from the absorption of light
monodentate, bidentate and polydentate ligands
energy as an electron moves between two non-
■■ describe and explain the reactions of transition
degenerate d orbitals
elements with ligands to form complexes, including
the complexes of copper(II) and cobalt(II) ions with ■■ describe, in qualitative terms, the effects of different
water and ammonia molecules and hydroxide and ligands on the absorption of light, and hence colour,
chloride ions using the complexes of copper(II) ions with water,
hydroxide and ammonia as examples.
■■ describe the types of stereoisomerism (cis-trans and
optical isomerism) shown by complexes, including
those associated with bidentate ligands
Chapter 24: Transition elements
Introduction extract and use other, rarer transition elements. For
example, titanium is used in jet turbines because it can
Many transition elements and their compounds are withstand high temperatures and is very strong.
used in construction and engineering. Iron and copper
have been used for many centuries (Figure 24.1).
In more recent times, we have discovered how to
Figure 24.1 The transition elements iron and copper are very important in the construction industry.
What is a transition element? In atoms of the transition elements, the 4s subshell is 367
normally filled and the rest of the electrons occupy orbitals
The transition elements are found in the d block of the in the 3d subshell. However, chromium and copper
Periodic Table, between Groups 2 and 13. However, not all atoms are the exceptions. Chromium atoms have just one
d-block elements are classified as transition elements. electron in the 4s subshell. The remaining five electrons
are arranged in the 3d subshell so that each orbital is
A transition element is a d-block element that forms one occupied by one electron. Copper atoms also have just one
or more stable ions with an incomplete d subshell. electron in the 4s subshell. The remaining ten electrons are
arranged in the 3d subshell so that each orbital is filled by
We do not define Sc and Zn as transition elements. two electrons.
■■ Scandium forms only one ion (Sc3+) and this has no
Element Electronic configuration
electrons in its 3d subshell – the electronic configuration of titanium (Ti) 1s2 2s2 2p6 3s2 3p6 3d2 4s2
Sc3+ is (Ar) 3d0 4s0. vanadium (V) 1s2 2s2 2p6 3s2 3p6 3d3 4s2
■■ Zinc forms only one ion (Zn2+) and this has a complete chromium (Cr) 1s2 2s2 2p6 3s2 3p6 3d5 4s1
3d subshell – the electronic configuration of Zn2+ is (Ar) manganese (Mn) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
3d10 4s0. iron (Fe) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
In this chapter we will be looking at the transition cobalt (Co) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
elements in the first row of the d block. These are the nickel (Ni) 1s2 2s2 2p6 3s2 3p6 3d8 4s2
metals titanium (Ti) through to copper (Cu), according to copper (Cu) 1s2 2s2 2p6 3s2 3p6 3d10 4s1
the definition above.
Table 24.1 Electronic configurations of the first row of
Electronic configurations transition elements.
Atoms
Table 24.1 shows the electronic configurations of the atoms
in the first row of the transition elements.
Cambridge International A Level Chemistry
Ions Notice the partially filled d subshells (see definition of
transition element on the previous page) in the following
The transition elements are all metals. In common with examples of ions:
all metals, their atoms tend to lose electrons so they form
positively charged ions. However, each transition metal can V atom = 1s2 2s2 2p6 3s2 3p6 3d3 4s2
form more than one ion. For example, the common ions of V3+ ion = 1s2 2s2 2p6 3s2 3p6 3d2 4s0
copper are Cu+ and Cu2+. We say that the transition metals
have variable oxidation states. The resulting ions are often Fe atom = 1s2 2s2 2p6 3s2 3p6 3d6 4s2
different colours (Figure 24.2). Fe3+ ion = 1s2 2s2 2p6 3s2 3p6 3d5 4s0
Cu atom = 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Cu2+ ion = 1s2 2s2 2p6 3s2 3p6 3d9 4s0
a bcd The most common oxidation state is +2, usually formed
when the two 4s electrons are lost from the atoms. The
maximum oxidation number of the transition elements at
the start of the row involves all the 4s and 3d electrons in
the atoms. For example, vanadium’s maximum oxidation
state is +5, involving its two 4s electrons and its three
3d electrons. At the end of the row, from iron onwards,
the +2 oxidation state dominates as 3d electrons become
increasingly harder to remove as the nuclear charge
increases across the period.
368 Figure 24.2 Vanadium and its oxidation states: a a solution Question
scoolnuttaioinnincgonVOta2i+niionngsV; 3b+ ioansso;ludt iaonsoclountitoanincinogntVaOin2+iniognVs2;+cio ans.
1 a State the electronic configurations of the following
Table 24.2 shows the most common oxidation states of the atoms or ions:
first row of the transition elements.
i Ti iv Fe3+
ii Cr v Ni2+
Element Most common oxidation states iii Co vi Cu+
titanium (Ti) +3, +4 b E xplain why scandium (which forms only one ion,
vanadium (V) +2, +3, +4, +5 Sc3+) and zinc (which forms only one ion, Zn2+)
chromium (Cr) +3, +6 are not called transition elements.
manganese (Mn) +2, +4, +6, +7
iron (Fe) +2, +3 c W hy is the maximum oxidation state of
cobalt (Co) +2, +3 manganese +7?
nickel (Ni) +2
copper (Cu) +1, +2 d L ook back at the different oxidation states
of vanadium shown in Figure 24.2. State
Table 24.2 Common oxidation states of the transition the oxidation state of the vanadium in each
elements. photo a–d.
The existence of variable oxidation states means that the e Z irconium (Zr) is in the second row of transition
names of compounds containing transition elements must elements beneath titanium in the Periodic Table.
have their oxidation number included, e.g. manganese(IV) Its electronic configuration is [Kr]4d2 5s2, where [Kr]
oxide, cobalt(II) chloride. represents the electronic configuration of krypton,
the noble gas with atomic number 36.
When transition elements form ions, their atoms
lose electrons from the 4s subshell first, followed by i P redict the maximum stable oxidation state of
3d electrons. zirconium and explain your answer.
ii G ive the formula of the oxide of zirconium,
assuming zirconium exhibits the oxidation state
given in e, part i.
Chapter 24: Transition elements
The higher oxidation states of the transition elements across a period. However, the transition elements show
are found in complex ions or in compounds formed only a very small variation.
with oxygen or fluorine. Common examples are the
chromate(VI) ion, CrO42–, and the manganate(VII) ion, Comparing the transition elements with
MnO4–. an s-block element
Physical properties of the The s-block metal that lies immediately before the first row
transition elements of d-block elements in the Periodic Table is calcium (Ca),
in Group 2. When we compare the properties of calcium
The transition elements commonly have physical with the first row of transition elements we find some
properties that are typical of most metals: differences despite the fact that they are all metals. You
need to know the following comparisons:
■■ they have high melting points ■■ the melting point of calcium (839 °C) is lower than that of a
■■ they have high densities
■■ they are hard and rigid, and so are useful as construction transition element (e.g. titanium at 1660 °C)
■■ the density of calcium (1.55g cm–3) is lower than that of a
materials
■■ they are good conductors of electricity and heat. transition element (e.g. nickel at 8.90 g cm–3)
■■ the atomic radius of calcium (0.197 nm) is larger than that of
The first ionisation energy, the atomic radius and the ionic
radius of transition elements do not vary much as we go a transition element (e.g. iron at 0.116 nm)
across the first row. The data are given in Table 24.3. ■■ the ionic radius of the calcium ion, Ca2+, (0.099 nm) is larger
Element 1st ionisation Atomic Ionic than that of a transition element (e.g. Mn2+ at 0.080 nm)
energy / kJ mol−1 radius / nm radius / nm ■■ the first ionisation energy of calcium (590 kJ mol–1) is
titanium (Ti) 661 0.132 Ti2+ 0.090 lower than that of a transition element (e.g. chromium at 369
653 kJ mol–1 or cobalt at 757 kJ mol–1)
vanadium 648 0.122 V3+ 0.074 or ■■ the electrical conductivity of calcium is higher than that of a
(V) V2+ 0.090 transition element (with the exception of copper).
chromium 653 0.117 Cr3+ 0.069 or Question
(Cr) Cr2+ 0.085
2 a Explain why the 1st ionisation energy of calcium is
manganese 716 0.117 Mn2+ 0.080 lower than that of cobalt.
(Mn)
b E xplain why the density of calcium is lower than
the density of nickel.
iron (Fe) 762 0.116 Fe2+ 0.076
cobalt (Co) 757 0.116 Co2+ 0.078 Redox reactions
nickel (Ni) 736 0.115 Ni2+ 0.078 We have seen how the transition elements can exist in
copper (Cu) 745 0.117 Cu2+ 0.069 various oxidation states. When a compound of a transition
element is treated with a suitable reagent, the oxidation
Table 24.3 There are comparatively small variations in 1st state of the transition element can change. Whenever a
ionisation energy, atomic radius and ionic radius of the first- reaction involves reactants changing their oxidation states
row transition elements. the reaction is a redox reaction, involving the transfer
From previous work looking at periodic properties in of electrons. Remember that a species is reduced when
Chapter 10, we would expect the 1st ionisation energy, its oxidation state is reduced (gets lower in value). Its
atomic radius and ionic radius of positively charged ions to oxidation state is lowered when it gains electrons, and, as
vary across a period. In general, the 1st ionisation energy well as being reduced, it is acting as an oxidising agent. For
would increase as the increasing nuclear charge has a example, in the half-equation:
tighter hold on electrons filling the same main energy level
or shell, and the shielding effect stays roughly the same. F e3+(aq) + e– Fe2+(aq)
For similar reasons the atomic radius and ionic radius of
a positively charged ion would be expected to decrease p ale yellow pale green
Fe3+ has been reduced to Fe2+ by gaining one electron. In the
equation as shown Fe3+ is acting as an oxidising agent. For this
Cambridge International A Level Chemistry
reaction to happen another half-equation is needed in which ■■ A known volume (e.g. 25 cm3) of an unknown concentration
the reactant loses one or more electrons, i.e. acts as a reducing of Fe2+(aq) is placed in a conical flask.
agent. In Chapter 20 (page 286) we saw how we can use
standard electrode potential values, E —O, to predict whether or ■■ A solution of a known concentration of potassium
not such reactions should take place. manganate(VII) is put in a burette.
Another half-equation we could consider would be: ■■ The potassium manganate(VII) solution is titrated against the
solution containing Fe2+(aq) in the conical flask.
M npOurp4l–e( aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)
■■ The purple colour of the manganate(VII) ions is removed in
pale pink the reaction with Fe2+(aq). The end-point is reached when
the Fe2+(aq) ions have all reacted and the first permanent
Both half-equations are written below as they appear in purple colour appears in the conical flask.
tables of data showing standard electrode potentials:
Fe3+(aq) + e– Fe2+(aq) E —O = +0.77 V
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)
E —O = +1.52 V
Can Fe3+ ions oxidise Mn2+ tFoeM2+ nioOn4s–tioonFse,3+oriocnasn? MnO4–
ions in acid solution oxidise
The magnitude of the positive values provides a
measure of the tendency of the half-equations to
proceed to the right-hand side. The values show us that
Min nthOe4–fo(arqw)airsdmdoirreecltiikoenly, to accept electrons and proceed
changing to Mn2+(aq), than
370 Fe3+(aq) is to accept electrons and change to Fe2+(aq).
MFoxen3id+O(ias4qi–n)(a.gqTF)heie2sr+ea(afomqr)oeiroMenpnsoOtwo4e–for(afruqm)l oxidising agent than
ions are capable of
Fe3+(aq), and the top
half-equation proceeds in the reverse direction.
We can now combine the two half-cells to get the
overall reaction. Note that the sign of the Fe(III)/Fe(II)
half-cell has changed by reversing its direction. the
Fe2+/Fe3+ equation also has to be multiplied by 5 so that
the electrons on either side of the equation cancel out (this
does not affect the value of E —O).
Figure 24.3 Manganate(VII) ions can be used to determine
5Fe2+(aq) 5Fe3+(aq) + 5e– E —O = –0.77 V the percentage of Fe2+ in an iron tablet.
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) You can achieve a more accurate result for the mass of Fe2+
E —O = +1.52 V in oaxsiodliusteioitninbyautsitirnagtidoinc.hTrohmis aisteb(eVcIa)uisoencso, Cmrp2oOu7n2–d(asqs)u,ch
to
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq) as potassium dichromate(VI) can be prepared to a higher
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
degree of purity than potassium manganate(VII). In a
titration with Fe2+(aq) and dichromate(VI) we need an
E —O = +0.75 V indicator of the end-point that will be oxidised as soon as
The relatively large positive value of E —O (+0.75 V) tells the Fe2+(aq) has all reacted.
us that the reaction is likely to proceed in the forward The half-equation and value for E —O for the use of
direction as written. In fact we use this reaction to dichromate as an oxidising agent is:
calculate the amount of iron (Fe2+ ions) in a sample, such
as an iron tablet, by carrying out a titration. Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)
E —O = +1.33 V
Chapter 24: Transition elements
worked example questions
1 0.420 g of iron ore were dissolved in acid, so that all of 3 a Write two half-equations for the reactions
the iron present in the original ore was then present that take place when Fe2+(aq) is oxidised by
as Fe2+(aq). The solution obtained was titrated against dichromate(VI) ions.
0.0400 mol dm–3 KMnO4(aq). The titre was 23.50 cm3.
b C ombine the two half-equations and write
a C alculate the number of moles of MnO4– in the the equation for the oxidation of Fe2+(aq) by
titre. dichromate(VI) ions.
Use the equation: c W ork out the E —O value of the cell made when the
two half-cells in part a are connected and the
reaction in part b takes place. Explain what this
n = V × c value predicts about the likelihood of Fe2+(aq)
w here n = number of moles, V = volume of solution being oxidised by dichromate(VI) ions.
in dm3 and c = concentration
d H ow many moles of Fe2+(aq) can 1 mole of
n = 2_ 1_30_.05_00_ × 0.0400 = 0.000 940 mol dichromate(VI) oxidise?
b C alculate the number of moles of Fe2+ in the
e I n a titration, 25.0 cm3 of an solution containing
solution. Fe2+(aq) ions was completely oxidised by 15.30 cm3
The equation for the reaction in the titration is:
of 0.00100 mol dm–3 potassium dichromate(VI)
solution.
5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H2O(l) i H ow many moles of potassium dichromate(VI)
are there in 15.30 cm3 of 0.00 100 mol dm–3
number of moles of Fe2+ = 5 0 × 0.000 940 solution?
= 0.004 70 mol ii H ow many moles of Fe2+ were present in the
25.0 cm3 of solution?
c C alculate the mass of iron in the solution
(Ar of iron is 55.8). iii W hat was the concentration of the Fe2+(aq) in 371
the flask at the start of the titration?
moles of Fe = moles of Fe2+ = 0.004 70 mol
mass of Fe = n × Ar
= 0.004 70 × 55.8
= 0.262 g
d C alculate the percentage mass of iron in the 0.420 g
of iron ore.
percentage mass of iron = _00 _..24_26_02_ × 100%
= 62.4%
Ligands and complex formation H 2+
H
In the previous section on redox reactions we have learnt H
about the oxidation of Fe2+(aq) ions. When these ions H O
are in solution the Fe2+ ion is surrounded by six water H
O
O
molecules. Each water molecule bonds to the central Fe H
Fe2+ ion by forming a dative (co-ordinate) bond from the
oxygen atom into vacant orbitals on the Fe2+ ion (Figure H
24.4). The water molecules are called ligands and the O O
resulting ion is called a complex ion. Its formula is H
H
O H
written iasso[cFtea(hHe2dOra)6l].2+. The shape of a complex with six H
ligands H
Fbiegtuwreee2n4a.n4 Fe[F2e+(iHo2nOa)6n]d2+;stixhewcaotemrpmleoxleiocnufloesrm. Itedis called a
hexaaquairon(II) ion.
Cambridge International A Level Chemistry
Figure 24.5 shows the shape of complexes with four ligands. CH2 2+
H2C NH2
2– Cl 2– H2N NH2
NC CN CH2
Co
Ni CH2
NC CN Cl Co H2N NH2
a [Ni(CN)4]2– [CoCl4]2– H2C
Cl Cl NH2
b CH2
[Co(en)3]2+
Figure 24.5 The complex ion formed between a transition Figure 24.6 [Co(en)3]2+ is an example of a complex ion
metal ion and a larger ligand can only fit four ligands around containing the bidentate ligand NH2CH2CH2NH2 (abbreviated
the central ion. These are arranged in either a square planar to ‘en’).
shape (as in a [Ni(CN)4]2–) or a tetrahedral shape (as in
b [CoCl4]2–).
All ligands can donate an electron pair to a central Note in Table 24.4 that the charge on a complex is simply
transition metal ion. The co-ordination number of a the sum of the charges on the central metal ion and on
complex is the number of co-ordinate (dative) bonds to the each ligand in the complex. Some complexes will carry no
central metal ion. Some ligands can form two co-ordinate charge, e.g. Cu(OH)2(H2O)4.
bonds from each ion or molecule to the transition metal H+
ion. These are called bidentate ligands, as shown in Figure H
24.6. Most ligands, such as water and ammonia, form just H N Ag N H
372 one co-ordinate bond and are called monodentate ligands. HH
A few transition metal ions, e.g. copper(I), silver(I), Figure 24.7 The diamminesilver(I) cation has a linear
gold(I), form linear complexes with ligands. The
co-ordination number in these complexes is 2 structure.
(Figure 24.7). Table 24.4 shows some common ligands.
Name of ligand Formula Example of Co-ordination Shape of complex
complex number octahedral (see Figure 24.4)
water H2O [Fe(H2O)6]2+
ammonia NH3 [Co(NH3)6]3+ 6
chloride ion Cl– [CuCl4]2–
cyanide ion CN– [Ni(CN)4]2– 6 octahedral
4 tetrahedral (see Figure 24.5b)
4 square planar
(see Figure 24.5a)
hydroxide ion OH– [Cr(OH)6]3– 6 octahedral
thiocyanate ion SCN– [FeSCN]2+ or
[Fe(SCN)(H2O)5]2+ 6 octahedral
ethanedioate ion (abbreviated as –OOC COO– [Mn(ox)3]3– 6 octahedral
‘ox’ in the formulae of complexes)
[Co(en)3]3+
1,2-diaminoethane NH2CH2CH2NH2 6 octahedral (see Figure 24.6)
(abbreviated as ‘en’ in the formulae
of complexes)
Table 24.4 Some common ligands and their complexes.
Chapter 24: Transition elements
Question no double bond exists. In this case, the term geometric
isomerism refers to complexes with the same molecular
4 a What is the oxidation number of the transition formula but different geometrical arrangements of their
metal in each of the following complexes? atoms. Examples are cis- and trans-platin (Figure 24.8).
In cis-platin, the chlorine atoms are next to each other
i [Co(NH3)6]3+ in the square complex but in trans-platin, they are
ii [Ni(CN)4]2– opposite. The properties of these geometrical isomers
iii [Cr(OH)6]3– are slightly different.
iv [Co(en)3]3+
v Cu(OH)2(H2O)4 CI NH3 Cl NH3
b E DTA4– ions can act as ligands. A single EDTA4– Pt Pt
ion can form six co-ordinate bonds to a central CI NH3 H3N Cl
transition metal ion to form an octahedral cis-platin
complex. It is called a hexadentate ligand. Give the trans-platin
formula of such a complex formed between Ni2+
and EDTA4–. Figure 24.8 The geometrical isomers, cis-platin and
c Which ligands in Table 24.4 are bidentate? trans-platin.
Stereoisomerism in transition metal Cis-platin has been used as an anti-cancer drug. It acts by
complexes binding to sections of the DNA in cancer cells, preventing
cell division.
In Chapter 14 (pages 195–6) we learnt about two Stereoisomerism is commonly shown by octahedral
types of stereoisomerism: geometric isomerism and (six co-ordinate) complexes associated with bidentate
optical isomerism. The presence of a double bond in ligands. An example is the complex containing
1,2-dibromoethane means that two geometrical isomers nickel as the transition metal and 1,2-diaminoethane
(cis-trans isomers) are possible. Geometric isomerism (2N4.H9)2.CTHhe2CtwHo2NisoHm2)earss the bidentate ligand (Figure 373
is also possible in transition metal complexes, where are stereoisomers because the
two
different molecules are mirror images of each other and
cannot be superimposed.
a CH2 2+ CH2 2+
H2C NH2 H2N CH2
H2N NH2 CH2 H2N NH2
Ni CH2 H2C Ni
H2N NH2 H2C NH2
H2C NH2 H2N H2N CH2
CH2 CH2
b en 2+ 2+
en
Ni en en Ni
en en
Figure 24.9 The two non-superimposable optical isomers of Ni(NH2CH2CH2NH2)32+: a the full structure; b a simplified structure
with ‘en’ representing a molecule of 1,2-diaminoethane.
Cambridge International A Level Chemistry
Question Two water ligands are replaced by two hydroxide ligands
5 a Cobalt forms a complex with the simplified in the reaction:
structure: [Cu(H2O)6]2+(aq) + 2OH–(aq)
NH2 blue solution
H2N CI
Cup(OaleHblu)e2(pHrec2iOpit)a4te(s) + 2H2O(l)
CO If you now add concentrated ammonia solution, the pale
blue precipitate dissolves and we get a deep blue solution:
H2N CI Cu(OH)2(H2O)4(s) + 4NH3(aq)
the curve represents NH2 [Cu(dHee2pObl)u2e(sNoluHtio3n) 4]2+(aq) + 2H2O(l) + 2OH–(aq)
CH2 CH2
i Give the co-ordination number in this complex. The first reaction can also be achieved by adding
ii Draw the stereoisomer of this complex. concentrated ammonia solution to copper sulfate solution
iii Explain why this is a stereoisomer. drop by drop or by adding a dilute solution of ammonia.
The pale blue precipitate formed will then dissolve and
b D raw the two geometrical isomers of Ni(CN)2(Cl)22–. form the deep blue solution when excess ammonia is
Label the cis-isomer and the trans-isomer. added. The structure of [Cu(H2O)2(NH3)4]2+(aq) is shown
in Figure 24.10.
Substitution of ligands H2O 2+
The ligands in a complex can be exchanged, wholly or H3N NH3
374 partially, for other ligands. This is a type of substitution Cu
reaction. It happens if the new complex formed is
more stable than the original complex. The complexes H3N NH3
of copper(II) ions can be used to show ligand H2O
substitution reactions. Figure 24.10 The structure of [Cu(H2O)2(NH3)4]2+(aq).
Whenever we write Cu2+(aq) we are really referring to
the complex ion [Cu(H2O)6]2+. This ion gives a solution
of copper sulfate its blue colour. On adding sodium Water ligands in i[fCwue(Had2dO)c6o]n2+cecnantraatlesod be exchanged for
hydroxide solution, we see a light blue precipitate forming. chloride ligands hydrochloric acid
drop by drop. A yellow solution forms, containing the
complex ion [CuCl4]2– (Figure 24.11):
start here
[CuCl4]2− the [Cu(H2O)6]2+ Cu2+ [Cu(H2O)2(NH3)4]2+
this yellow complex forms well-known blue this dark blue complex forms
on adding concentrated HCl on adding concentrated NH3
complex with water
Figure 24.11 The equations for the changes are:
[Cu(H2O)6]2+ + 4Cl– [CuCl4]2– + 6H2O and [Cu(H2O)6]2+ + 4NH3 [Cu(H2O)2(NH3)4]2+ + 4H2O.
Chapter 24: Transition elements
[Cbluue(sHolu2tOion)6]2+(aq) + 4Cl–(aq) [CuyeCllolw4]s2o–lu(atioqn) + 6H2O(l) Stability constants
Aqueous solutions of transition element ions are hydrated.
They are complex ions with water as a ligand. Different
Aqueous cobalt(II) compounds also form complex ions. ligands form complexes with different stabilities. For
Whenever we write Co2+(aq), we are referring to the example when we add concentrated aqueous ammonia to
complex oiofnco[bCaol(t(HII2)Os)u6]lf2a+t.eTihtsispiionnkgciovleosuarn. Oaqnuaedoduisng an aqueous solution of copper(II) sulfate, the ammonia
solution ligands displace water ligands in a stepwise process.
sodium hydroxide solution, we see a blue precipitate of [Cu(H2O)6]2+ (aq) + NH3(aq)
cobalt(II) hydroxide forming, which turns red when [Cu(NH3)(H2O)5]2+(aq) + H2O(l)
warmed if the alkali is in excess. Water ligands in
[Co(H2O)6]2+ can also be exchanged for ammonia ligands [Cu(NH3)(H2O)5]2+(aq) + NH3(aq)
if we add concentrated aqueous ammonia drop by drop. [Cu(NH3)2(H2O)4]2+(aq) + H2O(l)
[Co(H2O)6]2+(aq) + 6NH3(aq[C) o(NH3)6]2+(aq) + 6H2O(l) As we increase the concentration of ammonia, this
The addition of concentrated hydrochloric acid drop by process continues until four of the water molecules are
drop to an aqueous solution of cobalt(II) ions results in replaced by ammonia to form a deep blue solution. The
the formation of a blue solution containing the tetrahedral overall process is:
complex [CoCl4]2–(aq).
[Cu(H2O)6]2+ + 4NH3(aq)
[Co(H2O)6]2+(aq) + 4Cl–(aq) [CoCl4]2–(aq) + 6H2O(l) [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Aqueous cobalt(II) ions usually form tetrahedral
complexes with monodentate anionic ligands such as Cl–, We can think of this exchange of ligands in terms of 375
SCN– and OH–. competing equilibria of the forward and backward
reactions. The position of equilibrium lies in the direction
Question of the more stable complex. In this case, the complex with
ammonia as a ligand is more stable than the complex with
6 a Blue cobalt chloride paper gets its blue colour just water as a ligand. If we dilute the complex with water,
from [CoCl4]2– ions. What is the oxidation number the position of equilibrium shifts to the left and a complex
of the cobalt in this complex ion? with more water molecules as ligands is formed.
b B lue cobalt chloride paper is used to test for The stability of the complex is expressed in terms of
water. If water is present the paper turns pink as the equilibrium constants for ligand displacement. This is
a complex forms between the cobalt ion and six called the stability constant. Usually an overall stability
water ligands (Figure 24.12). Write an equation to cTohnessttaanbti,liKtystacbo, nisstgainvetnisrtahteheerqtuhilainbrtihuemstceopnwsitsaenct ofonrsttahnets.
show the ligand substitution reaction that takes formation of the complex ion in a solvent from its constituent
place in a positive test for water. ions or molecules. The method for writing equilibrium
expressions for stability constants is similar to the one we
used for writing Kc (see page 304). So for the equilibrium
[Cu(H2O)6]2+ + 4Cl–(aq) [CuCl4]2–(aq) + 6H2O(l)
the expression is:
Kstab =_ [ [_C_u_(_H[_[C2_O_u_)C 6_]l_24 +]_2]_– [(_Ca_ql–_)(_]a_q_)]_4_
Note:
Figure 24.12 A positive test for the presence of water ■■ water does not appear in the equilibrium expression
using anhydrous cobalt chloride paper. because it is in such a large excess that its concentration is
regarded as being constant
Cambridge International A Level Chemistry
■■ the units for the stability constant are worked out in the Question
same way as for the units of Kc (see page 124). For example,
in the above case: 7 a Write expressions for the stability constants for the
following reactions:
Kstab = _ [ [_C_u_(_[H[_C_2Ou__C) _6l]4_2 ]_+2]–_ ([_aC_ql_)–_](a_q_)_]_4 _ ( m__o_l _d_m(_m–_ 3_o)_l× _d _(mm_–_o3_)l_ d_m__–3_)_4
= dm12 mol–4 i [PtCl4]2–(aq) + 2NH3(aq)
PtCl2(NH3)2(aq) + 2Cl–(aq)
Stability constants are often given on a log10 scale. When
expressed coannableogu1s0esdcatolec, othmepyahraevtehenostuabniiltist.ySotaf bainliytytwo ii [Cr(H2O)6]3+(aq) + 2Cl–(aq)
constants [Cr(H2O)4Cl2]+(aq) + 2H2O(l)
ligands. The values quoted usually give the stability of the
complex relative to the aqueous ion where the ligand is iii [Ni(H2O)6]2+(aq) + 4NH3(aq)
water. The higher the value of the stability constant, the [Ni(NH3)4(H2O)2]2+(aq) + 4H2O(l)
more stable the complex. Table 24.5 gives some values
of stability constants for various copper(II) complexes b A n iron(III) ion, Fe3+, in aqueous solution has six
relative to their aqueous ions. water molecules bonded to it as ligands.
Ligand CO2– log10 Kstab i D raw the structure of this ion.
5.6
chloride, Cl– 13.1 ii W hen thiocyanate ions, SCN–, are added to
ammonia, NH3 16.9 an aqueous solution of iron(III) ions, the
2-hydroxybenzoate solution turns red and one water molecule is
replaced by a thiocyanate ion. Use the concept
376 OH 25.0 of stability constants to explain why the
1,2-dihydroxybenzene OH reaction occurs.
OH iii D educe the formula of the ion forming the
Table 24.5 The stability constants of some complexes red solution.
of copper.
iv T he stability constant for aqueous Fe3+ ions
with SCN– as a ligand is 891 dm3 mol–1. The
stability constant for aqueous Fe3+ ions with
fluoride ions, F–, as a ligand is 2 × 105 dm3 mol–1.
A solution containing fluoride ions is added to
the red solution. Would you expect to observe
any changes? Explain your answer.
From the table you can see that, in general, complexes The colour of complexes
with bidentate ions (2-hydroxybenzoate and
1,2-dihydroxybenzene) have higher stability constants than You cannot fail to notice the striking colours of complexes
those with monodentate ligands. We can use the values of containing transition metal ions. But how do these colours
the stability constants to predict the effect of adding different arise? White light is made up of all the colours of the
ligands to complex ions. For example, the addition of excess visible spectrum. When a solution containing a transition
ammonia to tahdeacrokmbpluleexs[oCluutCioln4]2o–f(athq)e should result in the metal ion in a complex appears coloured, part of the
formation of ammonia complex visible spectrum is absorbed by the solution. However, that
because the stability constant of the ammonia complex is still doesn’t explain why part of the spectrum is absorbed
higher than that of the chloride complex. The position of by transition metal ions. To answer this question we must
equilibrium is shifted to the right in the direction of the more look in more detail at the d orbitals in the ions.
stable complex. Addition of excess 1,2-dihydroxybenzene
to the dark blue ammonia complex results in the formation The five d orbitals in an isolated transition metal atom
of a green complex with 1,2-dihydroxybenzene. This is or ion are described as degenerate, meaning they are all at
because the stability constant with the 1,2-dihydroxybenzene the same energy level (Figure 24.13).
is higher than that for ammonia:
In the presence of ligands a transition metal ion is
[Cu(NH3)4 ]2+ + 2 OH OO 2– not isolated. The co-ordinate bonding from the ligands
OH Cu causes the five d orbitals in the transition metal ion to
+ 4H++ 4NH4+ split into two sets of non-degenerate orbitals at slightly
OO different energy levels (Figure 24.14). In a complex with six
Chapter 24: Transition elements
zz z
yx y x y
z dxy
x dxz
dyz z
x yx y
dx2–y2 dz2
Figure 24.13 The degenerate d orbitals in a transition metal atom.
ligands, the ligands are arranged in an octahedral shape in a complex with ligands. The difference in the energy
around the central metal ion. The lone pairs donated by between the non-degenerate d orbitals is labelled ΔE. ΔE is
the ligands into the transition metal ion repel electrons in part of the visible spectrum of light. So, when light shines
tthheantwthoodsxe2–iny2 tahnedodthz2erortbhirteaelsdshoorbwitna in Figure 24.13 more on ethleectsrooluntaiobnsocrobnsttahinisinagmtohuenCt uo(fHen2Oer)g62y+. complex,
ls. This is because an It uses this
these d orbitals line up with the co-ordinate bonds in energy to jump into the higher of the two non-degenerate
the complex’s octahedral shape and so they are closer to energy levels. In copper complexes, the rest of the visible 377
the bonding electrons in the octahedral arrangement, spectrum that passes through the solution makes it appear
increasing repulsion between electrons. Therefore the blue in colour.
orbitals are split, with these two d orbitals at a slightly The exact energy difference (ΔE) between the non-
higher energy level than the dyz, dxz and dxy orbitals degenerate d orbitals in a transition metal ion is affected
(Figure 24.14). by many factors. One of these factors is the identity of the
A Cu2+ ion has an electronic configuration of ligands that surround the transition metal ion. As you
[Ar] 3d9. Figure 24.14 shows how the nine d electrons are have seen, saosluotluiotniocnocnotnaitnaiinnignCg uC(uN(HH32)O2()H622+Ois)2a2+liigshat blue,
distributed between the non-degenerate orbitals formed whereas a very
dx2–y 2 dz 2 excited electron
absorbs energy
3d ∆E energy absorbed ∆E
from visible spectrum
Cu2+(...3d9) dyz dxz dxy that corresponds the energy absorbed can be
degenerate to ∆E worked out by the equation:
3d orbitals
∆E = hν
splitting the where ν is the frequency
d orbitals to give of light absorbed and
non-degenerate h is Planck’s constant
orbitals
Figure 24.14 The splitting of the 3d orbitals in a Cu(H2O)62+ complex ion.
Cambridge International A Level Chemistry
deep shade of blue. The colour change arises because the questions
presence of the ammonia ligands causes the d orbitals to
split by a different amount of energy. This means that the 8 a What do we mean by degenerate atomic orbitals?
size of ΔE changes and this results in a slightly different b E xplain why an octahedral complex of a transition
amount of energy being absorbed by electrons jumping
up to the higher orbitals. Therefore a different colour is element is coloured.
absorbed from visible light, so a different colour is seen. c D raw the non-degenerate 3d orbitals in a Ni2+ ion
on a diagram similar to Figure 24.14. The electrons
should be shown in the configuration that gives
the lowest possible energy.
9 a A solution of Sc3+ ions is colourless. Suggest a
reason for this.
b A solution of Zn2+ ions is colourless. Suggest a
reason for this.
Summary
■ Each of the transition elements forms at least one ■ Some transition element complexes exist as
ion with a partially filled d orbital. They are metals geometrical (cis-trans) isomers, e.g. cis- and trans-
with similar physical and chemical properties. platin; others, especially those associated with
378 ■ When a transition element is oxidised, it loses bidentate ligands with co-ordination number 6, may
electrons from the 4s subshell first and then the 3d exist as optical isomers.
subshell to form a positively charged ion.
■ cis-platin can be used as an anti-cancer drug by
binding to DNA in cancer cells and preventing
■ Transition elements can exist in several oxidation cell division.
states.
■ Ligand exchange can be described in terms of
■ Many reactions involving transition elements are competing equilibria.
redox reactions. Some redox reactions are used in
titrations to determine concentrations. ■ The stability constant, Kstab, of a complex ion is
the equilibrium constant for the formation of the
■ A ligand is a molecule or ion with one or more lone
pairs of electrons available to donate to a transition complex ion in a solvent from its constituent ions
metal ion. or molecules.
■ Transition elements form complexes by combining ■ The higher the value of the stability constant, the
more stable is the complex ion formed.
with ligands. Ligands bond to transition metal ions
by one or more co-ordinate bonds. ■ Transition metal compounds are often coloured
because of d orbital splitting, caused by ligands.
■ Ligands can be exchanged for other ligands in a Different ligands will split the d orbitals by different
complex. This can result in a change of colour. amounts, resulting in differently coloured complexes.
Chapter 24: Transition elements
End-of-chapter questions [1]
[1]
1 Define the following terms: [2]
a transition element
b ligand
c complex ion.
Total = 4
2 Use subshell notation (1s22s22p6, etc.) to give the electronic configurations of the following: [1]
[1]
a an Fe atom [1]
b a Co2+ ion
c a Ti3+ ion.
Total = 3
3 a Give the formulae of two iron compounds in which iron has different oxidation states. Give the [4]
oxidation states of the iron in each compound. [3]
b Explain why complexes of iron compounds are coloured.
Total = 7
4 Write balanced ionic equations and describe the observations for the reactions that occur when: [2]
a sodium hydroxide solution is added to a solution containing [Cu(H2O)6]2+(aq) [3]
b excess concentrated ammonia solution is added to a solution containing [Cu(H2O)6]2+(aq).
Total = 5
379
5 The half-cell reactions given below are relevant to the questions that follow.
Cl2 + 2e– 2Cl– E —O = +1.36V
Fe3+ + e– Fe2+ E —O = +0.77V
MnO4– + 8H+ + 5e– Mn2+ + 4H2O E —O = +1.51V
SO42– + 4H+ + 2e– SO2 + 2H2O E —O = +0.17V
sInulofurdriecratcoidstaannddathrdeinsemaasdoeluittiuopntoofaKtMontaOl4v,oalustmuedeonf t2w50ecigmh3ewd iotuhtd5is.5ti6llgedofwFaetSeOr.4S.7hHe2tOh,ednistsooolkve2d5.i0t in
cm3
portions of this solution, and added 10cm3 of 2.00moldm–3 sulfuric acid to each. She then titrated these
solutions against the potassium manganate(VII) solution. The average titre was 21.2cm3.
a Using the electrode potentials, explain why she used sulfuric acid and not hydrochloric acid in her [4]
titrations. [3]
b What was the concentration of the iron(II) sulfate solution? [2]
c i Write the full ionic equation for the reaction between the manganate(VII) solution and the iron(II) [1]
[2]
sulfate.
[5]
ii How does the student know that she has reached the end-point for the reaction?
d What is the concentration of the manganate(VII) solution?
e If the student passed sulfur dioxide gas through 25.0cm3 of the manganate(VII) solution, what volume of
gas would be required to completely decolorise the manganate(VII)? (1mol of gas occupies 24.0dm3 at
room temperature and pressure)
Total = 17
Cambridge International a level Chemistry
6 Copper forms complexes with chloride ions and with ammonia.
log10Kstab for aqueous Cu2+ ions with Cl– as a ligand is 2.80.
log10Kstab for aqueous Cu2+ ions with NH3 as a ligand is 4.25.
a A few drops of hydrochloric acid are added to blue aqueous copper(II) sulfate.
i Copy and complete the equation for this reaction.
[Cu(H2O)6]2+(aq) + [CuCl4]2– (aq) + [2]
[3]
light blue yellow-green [2]
ii Describe and explain what happens when excess water is then added to the reaction mixture in part i. [2]
[1]
iii Explain what happens in terms of ligand exchange, when concentrated aqueous ammonia is added
to the reaction mixture from part ii. [4]
b Aqueous copper ions form a complex with 1,2-diaminoethane, NH2CH2CH2NH2. They also form a
complex with 1,3-diaminopropane NH2CH2CH2CH2NH2.
log10Kstab for aqueous Cu2+ ions with 1,2-diaminoethane as a ligand is 20.3
log10Kstab for aqueous Cu2+ ions with 1,3-diaminopropane as a ligand is 17.7
i What type of ligands are 1,2-diaminoethane and 1,3-diaminopropane? Explain your answer.
ii Which one of these ligands forms a more stable complex? Explain your answer.
iii 1,2-diaminoethane can be abbreviated as ‘en’.
A complex of nickel ions, Ni2+, with 1,2-diaminoethane is octahedral in shape with a co-ordination
number of 6. Draw the structures of the two stereoisomers of this complex.
380 Total = 14
Chapter 25: 381
Benzene and its compounds
Learning outcomes – hydrogenation of the benzene ring to form a
cyclohexane ring
You should be able to:
■■ describe the mechanism of electrophilic substitution
■■ interpret, name and use the general, structural, in arenes and the effect of the delocalisation of
displayed and skeletal formulae of benzene and electrons in such reactions
simple aryl compounds
■■ interpret the difference in reactivity between
■■ describe and explain the shape of, and bond angles benzene and chlorobenzene
in, benzene molecules in terms of o and π bonds
■■ predict whether halogenation will occur in the side-
■■ describe the reactions of arenes, such as benzene chain or in the benzene ring in arenes depending on
and methylbenzene, in: reaction conditions
– substitution reactions with chlorine and with
bromine ■■ apply knowledge relating to position of substitution
– nitration in the electrophilic substitution of arenes (see
– Friedel–Crafts alkylation and acylation Table 25.4).
– complete oxidation of the side-chain to give a
benzoic acid
Cambridge International A Level Chemistry Figure 25.1 This is a vanilla
orchid. Its seed pods contain
Introduction a substance called vanillin.
Its molecules are based on a
In the English language, ‘aromatic’ means benzene ring – as represented by
‘having an agreeable, sweet or spicy odour’. the hexagon with a circle inside
In chemistry, the word ‘aromatic’ was once in the structure shown below.
used to describe chemical compounds Vanillin is used to flavour foods
containing a particular structure of six such as ice cream and chocolate.
carbon atoms arranged in a ring – the Its structure is:
benzene ring. These compounds are now
called aryl compounds. However, many OH
chemicals with structures containing one or O CH3
more benzene rings do have strong odours.
C
HO
The benzene ring As analytical techniques were developed, however,
chemists found out that the benzene molecule was a
The ‘benzene ring’ is a particularly important functional planar, perfectly symmetrical molecule. Kekulé’s structure
group found in many organic compounds. A benzene would suggest three shorter double C C bonds and three
382 ring is a hexagon made of six carbon atoms bonded longer single C C bonds in the ring. This would produce
a distorted hexagonal shape, not the perfect hexagonal
together in a particular way. Benzene rings are found in arrangement of carbon atoms in benzene’s actual
many compounds that are commercially important – for molecules. Figure 25.3 shows how we represent benzene’s
example as medicines, dyes and plastics. skeletal formula.
Organic hydrocarbons containing one or more
benzene rings are called arenes. In general, compounds
of benzene are known as aryl compounds or aromatic
compounds; an example is chlorobenzene, which is one of
the halogenoarenes. Figure 25.3 The skeletal formula of benzene. It can also be
The simplest arene is benzene itself. Its formula is used in displayed formulae of aryl compounds, as on the
iCts6Hst6r.uTchtuereea. rHlyoowregvaenri,cacrohuemndis1ts86st5rFurgigelderdictho work out previous page showing the structure of vanillin.
August We can now measure actual bond lengths. This was
Kekulé seemed to have solved the problem. He proposed impossible in the 19th century when Kekulé worked.
a hexagonal ring of carbon atoms, each bonded to one Table 25.1 shows that the bond length of the carbon–
hydrogen atom. In Kekulé’s structure the hexagonal ring carbon bonds in benzene lie between the values for C C
contained three double C C bonds (Figure 25.2). single bonds and C C double bonds.
This is reflected in benzene’s name, which has the same
ending as the alkenes.
Bond Bond length / nm
H
HC C C H also shown as C C 0.154
HC C C H C C 0.134
H carbon to carbon bond in benzene 0.139
Figure 25.2 Kekulé’s structure of benzene. Table 25.1 Comparing bond lengths.
Chapter 25: Benzene and its compounds
The chemistry of benzene also suggests that the Kekulé Naming aryl compounds
structure is incorrect. If there were three C C bonds
in benzene it would undergo addition reactions like the You saw how to name aryl compounds with alkyl side-
alkenes (see page 209). However, this is not the case. For chains on page 193. Some aryl compounds have functional
example, ethene will decolorise bromine water on mixing, groups that are substituted directly into the benzene ring
but benzene needs much harsher conditions. in place of a hydrogen atom. You need to know the names
of the compounds in Table 25.2.
The actual structure of benzene can be explained by
considering the bonding in the molecule. Each carbon Skeletal formula Name
atom in the hexagonal ring is sp2 hybridised (see page 57), of aryl compound chlorobenzene
sharing:
Cl
■■ one pair of electrons with one of its neighbouring
carbon atoms NO2 nitrobenzene
OH phenol
■■ one pair of electrons with its other neighbouring
carbon atom OH 2,4,6-tribromophenol: note the 383
numbering of the carbon atoms in the
■■ one pair of electrons with a hydrogen atom. Br Br benzene ring to describe the position
These are σ (sigma) bonds; covalent bonds with the pair of each substituted group (see page
of electrons found mainly between the nuclei of the atoms 193)
bonded to each other. Each carbon atom forms three σ
bonds. That leaves one electron spare on each of the six Br
carbon atoms. Each carbon atom contributes this one
electron to a π (pi) bond (see page 57). However, the π NH2 phenylamine
bonds formed are not localised between pairs of carbon
atoms as in an alkene C C bond (see page 193). Instead,
the π bonds in benzene spread over all six carbon atoms
in the hexagonal ring. The six electrons in the π bonds are
said to be delocalised.
The π bonding in benzene is formed by the overlap
of carbon p atomic orbitals, one from each of the six
carbon atoms. To achieve maximum overlap, the benzene
molecule must be planar. The lobes of the p orbitals
overlap to form a ring of delocalised electrons above
and below the plane of the carbon atoms in the benzene
molecule. This is shown in Figure 25.4.
HH HH Table 25.2 The names of some aryl compounds. The phenyl
group can be wisrCit6tHe5nNaHs2C. 6H5; e.g. the structural formula of
phenylamine
H HH H
HH HH
overlap of p orbitals
produces a ring of
delocalised electrons
above and below the
plane of benzene’s
carbon atoms
Figure 25.4 The π bonding in benzene. The three bond angles
around each of the sp2 hybridised carbon atoms are 120°.
Cambridge International A Level Chemistry
questions We can think of the electrophile as a Br+ cation:
1 a How many electrons are involved in the π bonding Bδ+r Bδ–r FeBr3 Br+ + [FeBr4]–
system in a benzene molecule?
The Br+ cation and the ‘electron-rich’ benzene ring are
b From which type of atomic orbital do the electrons attracted to each other, as the mechanism below shows.
in part a come from? Remember that the curly arrows show the movement of a
pair of electrons.
c What do we mean by the term ‘delocalised
electrons’ in benzene?
d What is the difference between the π bonding in H Br Br
benzene and the π bonding in hex-3-ene?
2 a Draw the displayed or skeletal formula of: Br+ ⎯st⎯ag⎯e 1→ + stage 2 + HBr
i 1,3,5-tribromobenzene ⎯⎯⎯→ (+FeBr3)
[FeBr4]–
ii 1,3-dichloro-5-nitrobenzene. A similar reaction happens when chlorine gas is bubbled
b Name the molecules below: through benzene at room temperature in the presence
i OH ii Br of a catalyst such as iron(III) chloride or aluminium
CH3 Cl chloride. The products of this electrophilic substitution
are chlorobenzene and hydrogen chloride. The catalysts in
Cl these reactions, i.e. FeBr3, AlCl3 and FeCl3, are k nown as
‘halogen carriers’.
When we halogenate methylbenzene or other
Reactions of arenes alkylarenes, the halogen atom substitutes into the benzene
ring at positions 2 or 4. These positions are ‘activated’
384 Most reactions of benzene maintain the highly stable by any electron-donating groups bonded directly to the
delocalised ring of π bonding electrons intact. This occurs benzene ring (see Table 25.4 on page 390). Other
by substituting an atom, or group of atoms, for one or more examples of benzene compounds that are activated in
hydrogen atoms attached to the benzene ring. The initial these positions are phenol and phenylamine. So when
attack is usually by an electrophile, which is attracted to the we react methylbenzene with chlorine gas, using an
high electron density around the benzene ring. anhydrous aluminium chloride catalyst, two products can
Electrophilic substitution with chlorine be made:
or bromine CH3 AlCl3 CH3 CH3
+ 2Cl2 Cl + 2HCl
Benzene will react with bromine in the presence of an +
anhydrous iron(III) bromide catalyst. The catalyst can be
made in the reaction vessel by adding iron filings to the
benzene and bromine. The substitution reaction is:
Cl
anhydrous Br 2-chloromethyl- 4-chloromethyl-
+ Br2 FeBr3 catalyst + HBr benzene benzene
If excess chlorine gas is bubbled through, we can
form 1-methyl-2,4-dichlorobenzene, 1-methyl-2,6-
At first sight the electrophile that starts the attack on dichlorobenzene and 1-methyl-2,4,6-trichlorobenzene.
benzene is not obvious. The electrophile is created when (Remember that the 2 and 6 positions in substituted arenes
an iron(III) bromide molecule polarises a bromine are equivalent.)
mboonldecwulieth. TihroenB(IrI2Im) borloecmuildeefobrymdsoanadtaintigvea(lcoon-eorpdaiinraotfe) The carbon–halogen bond in halogenoarenes is
stronger than the equivalent bond in a halogenoalkane
electrons from one bromine atom into an empty 3d orbital because one of the lone pairs on the halogen atom
in the iron. This draws electrons from the other bromine overlaps slightly with the π bonding system in the benzene
atom in tthheeBerle2cmtroolpehcuillee. making it partially positive, ring. This gives the carbon–halogen bond a partial double
creating bond character.
Chapter 25: Benzene and its compounds
Notice that the methyl side-chain is not affected under the nitronium ion (or nitryl cation). This is made from a
conditions used in the reaction above. However, we learnt mixture of concentrated nitric acid and concentrated
on page 206 that chlorine will react with alkanes in the sulfuric acid:
presence of ultraviolet (UV) light or strong sunlight. This
is a free radical substitution reaction. So if the chlorine gas HNO3 + 2H2SO4 NO2+ + 2HSO4– + H3O+
is passed into boiling methylbenzene, in the presence of
UV light, the following reaction takes place: This ‘nitrating mixture’ is refluxed with benzene at about
55 °C to make nitrobenzene:
CH3 CH2Cl NO2
+ Cl2 boil + HCl + HNO3 ⎯→ + H2O
UV light
chloromethyl- The mechanism of the electrophilic substitution is:
benzene
Note that there is no substitution into the benzene ring H NO2 NO2
under these conditions. In excess chlorine, eventually all
three of the hydrogen atoms on the methyl side-chain will NO2+ stage 1 + stage 2 + H+
be replaced by chlorine atoms.
Question In stage 1 in the mechanism, the electrophile, NO2+, is
attracted to the high electron density of the π bonding
3 a Write the equation for the reaction of chlorine system in benzene. A pair of electrons from the benzene
with benzene in the presence of an iron(III) ring is donated to the Antittrhoigsepnoaitnotm, biennzNeOne2’+s, and forms
chloride catalyst. a new covalent bond. delocalised 385
ring of electrons is disrupted. There are now four π
b What do we call this type of mechanism for the bonding electrons and a positive charge spread over five
reaction in part a? carbon atoms.
However, the full delocalised ring is restored in stage 2
c Draw the mechanism of the reaction in part a, with when the C H bond breaks heterolytically. Both electrons in
Cl+ as the attacking species and using curly arrows the C H covalent bond go into nitrobenzene’s π bonding
to show the movement of electron pairs. system, and hydrogen leaves as an H+ ion. There are now
six electrons spread over the six carbon atoms, so the
d Draw the displayed formula of the ‘tri-substituted’ chemical stability of the benzene ring is retained in this
halogenoarene produced if methylbenzene is substitution reaction.
added to excess bromine at room temperature in Further nitration of the nitrobenzene produces
the presence of iron(III) bromide. 1,3-dinitrobenzene and 1,3,5-trinitrobenzene. Unlike
the electron-donating methyl group in methylbenzene
e How would the reaction in part d differ if the
methylbenzene and bromine were boiled in the
presence of UV light?
f Name the mechanism of the reaction in part e.
(which activates the 2 and 4 positions in the benzene
ring), the NO2 group is electron-withdrawing. This type
Nitration of benzene of group (which includes COOH) deactivates the 2 and
The nitration of benzene is another example of an 4 positions in the benzene ring. Therefore, when there
electrophilic substitution. Nitration refers to the is a nitro group bonded to the benzene ring, further
rineatrcotidouncttihoeneolefctthreopNhOil2e group into a molecule. In this substitution takes place at the 3 and 5 positions (see Table
is the NO2+ ion, known as the 25.4, page 390).
Cambridge International A Level Chemistry
Question The same type of reaction can also be used to introduce an
acyl group into a benzene ring. An acyl group contains an
4 a Copy and complete the two equations below, alkyl group and a carbonyl (C O) group:
which can be used to show the nitration of
methylbenzene: O CH3 O CH2CH3
C C
i C6H5CH3 + NO2+ +
+
H2SO4
ii C6H5CH3 + HNO3
iii Name the possible mono-substituted products phenylethanone phenylpropanone
in parts i and ii.
iv 1-methyl-2,4-dinitrobenzene and 1-methyl- Friedel–Crafts reactions result in the introduction of a
2,4,6-trinitrobenzene are formed on further side-chain into a benzene ring.
nitration of methylbenzene. Draw the They are also called alkylation or acylation reactions.
displayed formula of each compound.
For example:
b Benzene also undergoes electrophilic substitution
when refluxed with fuming sulfuric acid for several + CH3CH2Cl AlCl3 CH2CH3
hours. This is called sulfonation. The electrophile catalyst + HCl
is the SO3 molecule and the product formed is
benzenesulfonic acid, C6H5SO3H.
i Which atom in the SO3 molecule accepts an
electron pair in the mechanism of sulfonation?
ii Write an equation in the style of part a i ethylbenzene
for the sulfonation of benzene to form
benzenesulfonic acid. O CH3
C
386
Alkylation (or acylation) of benzene + CH3COCl AlCl3 + HCl
(Friedel–Crafts reaction) catalyst
Friedel–Crafts reactions, named after the chemists who phenylethanone
first discovered them, are a third example of electrophilic
substitution into the benzene ring. The reactions involve attack on the benzene ring by an
electrophile carrying a positive charge on a carbon atom,
Sometimes chemists need to change the structure of an i.e. a carbocation.
arene in order to make a new product. Examples include
the manufacture of detergents or the reactants needed to The electrophile is often formed by adding an
make plastics, such as poly(phenylethene) – commonly aluminium chloride catalyst to a halogenoalkane. This
known as polystyrene. They can use a Friedel–Crafts creates the carbocation electrophile:
reaction to substitute a hydrogen in the benzene ring for 1st step
an alkyl group, such as a methyl ( CH3) or an ethyl H
( C2H5) group:
H3C C Cl AlCl3 +
CH3 CH2CH3
CH3CH2 + [AlCl4]–
H
methylbenzene ethylbenzene The carbocation electrophile then attacks the
benzene ring:
2nd step H CH2CH3
C+H2CH3 [AlCl4]– +
Chapter 25: Benzene and its compounds
The aluminium chloride catalyst is regenerated in the Phenol
final step:
Phenol, C6H5OH, is a crystalline solid that melts at
3rd step CH2CH3 CH2CH3 43 °C. It is used to manufacture a wide range of products
H (Figure 25.5). Its structure is:
+ + [AlCl4]– + HCl + AlCl3 OH
Further alkylation of the benzene ring can take place as
the reaction proceeds.
Oxidation of the side-chain in arenes
The presence of the benzene ring in an alkylarene, such as
methylbenzene, can affect the characteristic reactions of
its alkyl side-chain. For example, alkanes are not usually
oxidised by a chemical oxidising agent such as potassium
manganate(VII). However in alkylarenes, the alkane side-
chain is oxidised to form a carboxylic acid. For example,
methylbenzene produces benzoic acid when refluxed with
alkaline potassium manganate(VII), and then acidified
with dilute sulfuric acid, or another strong oxidising
agent such as acidified potassium dichromate(VI):
CH3 COOH
387
+ 3[O] + H2O Figure 25.5 Araldite adhesive, compact discs and TCP®
questions antiseptic (which contains 2,4,6-trichlorophenol, as
does Dettol®) are all manufactured using phenol as a
starting material.
5 a Copy and complete the two equations below, The melting point of phenol is relatively high for an aryl
which can be used to show the alkylation and compound of its molecular mass because of hydrogen
acylation of benzene, to produce the mono- bonding between its molecules. However, the large
substituted products. non-polar benzene ring makes phenol only slightly
soluble in water, as it disrupts hydrogen bonding with
i C6H6 + CH3CH2CH2Cl + water molecules.
ii C6H6 + CH3CH2CH2CH2COCl +
Phenol is weakly acidic, losing an H+ ion from its
b i Name the mono-substituted organic product hydroxyl group:
in part a i.
C6H5OH(aq) Cp6hHen5oOxid–e(iaonq) + H+(aq)
ii What class of compound is formed in part a ii?
6 Hexylbenzene is refluxed with alkaline potassium The position of this equilibrium lies well over to the
manganate(VII) and then acidified with dilute sulfuric left-hand side. However, phenol is still a stronger acid
acid. The same experiment is carried out using hexane than water Roremanemalbcoerh:otlh. eThheigvhaelruethseofvaplKuae are shown in
and the oxidising agent. Compare what would happen Table 25.3. of pKa, the
in these experiments. weaker the acid (see page 307).
Cambridge International A Level Chemistry
Weak acid Dissociation in water 2p5K a°Cat Question
10.0
phenol C6H5OH(aq) C6H5O–(aq) + H+(aq) 7 a Place the following molecules in order of their
water 14.0 acidity, starting with the most acidic:
ethanol H2O(l) H+(aq) + OH–(aq) CH3COOH C6H5OH HCl C3H7OH H2O
16.0
C2H5OH(aq) C2H5O–(aq) + H+(aq) b Would you expect methanol to be more or less
acidic than phenol? Explain your answer.
Table 25.3 Comparing the acidity of phenol, water
and ethanol.
Phenol is more acidic than water, with ethanol being the Reactions of phenol
least acidic of the three compounds. We can explain this
by looking at the conjugate bases formed on the right-hand We can divide the reactions of phenol into those involving
side of the equations in Table 25.3. The phenoxide ion, the hydroxyl group, OH, and those involving substitution
C6H5O–(aq), has its negative charge spread over the whole into the benzene ring.
ion as one of the lone pairs on the oxygen atom overlaps
with the delocalised π bonding system in the benzene ring. Breaking of the OH bond in phenol
O– Although phenol is only slightly soluble in water, it
dissolves well in an alkaline solution. As you have just
CH3CH2 O– learnt, phenol is a weak acid so it will react with an alkali
to give a salt plus water:
OH + NaOH O–Na+ + H2O
phenoxide ion, ethoxide ion, with
388 with negative negative charge The salt formed, sodium phenoxide, is soluble in water.
charge spread over concentrated on Phenol also reacts vigorously with sodium metal, giving
the whole ion the oxygen off hydrogen gas and again forming sodium phenoxide:
This delocalisation reduces the charge density of the
negative charge on the phenoxide ion compared with
OH–(aq) or aCtt2rHac5Oted–(atoq)t.hTehpehreefnoorxeiHde+(iaoqn) ions are not 2 OH + 2Na ⎯→ 2 O–Na+ + H2
as strongly as they are to
hydroxide or ethoxide ions, making phenoxide ions less
likely to re-form the undissociated molecules. Substitution into the benzene ring
Alternatively, we can explain the greater acidity of of phenol
phenol by saying that phenol ionises to form a more
stable negative ion, so the ionisation of phenol is more Compared with benzene, phenol reacts more readily
likely. This results in the position of equilibrium in the with electrophiles. The overlap of one of the lone pairs of
phenol equation in Table 25.3 lying further to the right- electrons on the oxygen atom in the OH group with the
hand side (i.e. more molecules donating an H+ ion) than π bonding system increases the electron density of the
the other equations. benzene ring in phenol. This makes the benzene ring more
Ethanol is a weaker acid than water because of the open to attack from electron-deficient electrophiles. It
electron-donating alkyl (ethyl) group attached to the ‘activates’ the benzene ring, especially at positions 2, 4
oxygen atom in the ethoxide ion. This has the effect of and 6 (see Table 25.4, page 390).
concentrating more negative charge on this oxygen atom,
which more readily accepts an H+ ion. This explains why
the position of equilibrium lies further to the left-hand
side, favouring the undissociated ethanol molecules.
Chapter 25: Benzene and its compounds
Bromination of phenol
Phenol undergoes similar reactions to benzene,
but phenol does so under milder conditions. For
example, bromine water will not react with benzene
at room temperature. To produce bromobenzene we
need pure bromine (not a solution) and an iron(III)
bromide catalyst.
However, bromine water reacts readily with phenol,
decolorising the orange solution and forming a white
precipitate of 2,4,6-tribromophenol (see Figure 25.6):
OH OH
Br Br
+ 3HBr
+ 3Br2 →
Br Figure 25.6 Bromine water is added to aqueous phenol.
Similar reactions happen between phenol and chlorine Question 389
or iodine.
8 a Place these molecules in order of ease of nitration,
This activation of the benzene ring is also shown in with the most reactive first:
the nitration of phenol. With benzene, we need a mixture
of concentrated nitric and sulfuric acids to reflux with C6H6 C6H5CH3 C6H5COOH C6H5OH
benzene at about 55 °C for nitration to take place (see b i Write a balanced equation to show the reaction
page 385). However, the activated ring in phenol
readily undergoes nitration with dilute nitric acid at when excess chlorine is bubbled through
room temperature: phenol at room temperature.
ii How would the reaction conditions differ
OH OH OH from those in part b i if you wanted to make
dil. HNO3 NO2 chlorobenzene from benzene and chlorine?
+
NO2
If we use concentrated nitric acid we get
2,4,6-trinitrophenol formed, shown below:
OH
O2N NO2
NO2
Cambridge International A Level Chemistry
Summary
■ The benzene molecule, C6H6, is symmetrical, with a ■ When the OH group is joined directly to a benzene
planar hexagonal shape. ring, the resulting compound is called a phenol.
■ Arenes have considerable energetic stability because ■ Phenols are weakly acidic, but are more acidic
of the six delocalised π bonding electrons that lie than water and alcohols. The acidity of phenol is
above and below the plane of the benzene ring. due to delocalisation of the negative charge on the
■ The main mechanism for arene reactions is phenoxide ion into the π bonding electron system on
electrophilic substitution. This enables arenes to the benzene ring.
retain their delocalised π electrons. Hydrogen atoms ■ When reacted with sodium hydroxide, phenol forms
on the benzene ring may be replaced by a variety of
a salt (sodium phenoxide) plus water. The reaction
notithreor(ato NmOs2)ogrrgoruopusp, sa,sinwcelulldaisnaglhkyalloogreancyaltogrmosupasnd
in Friedel–Crafts reactions. of sodium metal with phenol produces sodium
■ Despite the name ending in -ene, arenes do phenoxide and hydrogen gas.
not behave like alkenes. Arenes undergo ■ The OH group enhances the reactivity of the benzene
ring towards electrophiles. The OH group is
electrophilic substitution, whereas alkenes undergo said to activate the benzene ring. For example,
electrophilic addition. bromine water is decolorised by phenol at room
temperature, producing a white precipitate of
■ Sometimes the presence of the benzene ring 2,4,6-tribromophenol.
affects the usual reactions of its side-chain, e.g. ■ Table 25.4 summarises the positions activated by
methylbenzene is oxidised by refluxing with alkaline different substituents in a benzene ring:
390 potassium manganate(VII) to form benzoic acid.
Substituent groups in the benzene ring that direct Substituent groups in the benzene ring that direct
the in-coming electrophile to attack the 2, 4 and/ the in-coming electrophile to attack the 3 and/
or 6 positions. These groups activate attack by or 5 positions. These groups de-activate attack
electrophiles (because they tend to donate electrons by electrophiles (because they tend to withdraw
into the benzene ring) electrons from the benzene ring)
NH2, NHR or NR2 NO2
OH or OR N+H3
CN
NHCOR CHO
CH2, alkyl COOH, COOR
Cl
Table 25.4 Summary of positions attacked in electrophilic substitution into substituted benzene compounds.
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Cambridge International AS and A Level Chemistry Coursebook 2nd Edition
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Cambridge International AS and A Level Chemistry Coursebook 2nd Edition
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