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Stretch and Challenge
What is Stretch and Challenge?

What it is:
 the application of knowledge and/or problem solving rather than just
recalling data
 demonstrating your breadth of knowledge (including facts learned
during your AS course) – not the narrow testing of one specification
topic
 being examined by more open-ended questions which may have no
single correct answer
 applying your scientific knowledge to new contexts you haven't met
before
 part of your normal A2 examination paper.

What it isn’t:
 additional content to learn
 lots of extra essay questions – although some Stretch and Challenge
questions may be tested by extended prose questions
 being examined by questions that are flagged: Stretch and Challenge
 only for those students aiming for an A* grade – your examination
paper contains questions that will test you on your knowledge of
chemistry and, whatever grade you are aiming for, you need to
answer all the questions to the best of your ability.
Why should I be concerned about Stretch and Challenge?

It's about learning new skills and ways of thinking
Stretch and Challenge should help improve your problem solving and thinking
skills. You will probably be required to apply your knowledge in new contexts as
well as make links across the various topics you've encountered throughout
your AS and A2 studies.

Improving your marks
The skills Stretch and Challenge help develop could mean the difference
between one grade and another, especially if you're aiming for an A or A* grade
– although these skills should be helpful in attaining whatever grade you are
aiming for.

Useful beyond A level
The Stretch and Challenge skills should also be useful if you want to carry on
with your studies in further education or want to start a career. Universities
which use entrance exams or interviews will be looking for students who can
show more than just rote learning – the thinking skills demonstrated through
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Stretch and Challenge help to prepare you for these situations. Similarly, most
businesses will be looking for people who have good problem-solving skills and
the ability to apply knowledge in different contexts – so practising such skills
during your A2 course will stand you in good stead whatever you choose to do.
How should I prepare for Stretch and Challenge questions?
Stretch and Challenge questions are just part of your examination paper, so all the
usual advice for revision applies. You need to know the content of the course as laid
out by the specification. However, to do well in Stretch and Challenge you also need
to practise thinking in different ways rather than just learning the content by rote.

Don't limit yourself to revising topics in isolation
Try to spot links between modules and use strategies such as mind maps to
connect up different topics. You could use the textbook spreads which introduce
each chapter for examples.

Read around the subject
Use different sources, for instance, newspapers, magazines (e.g. New
Scientist), websites (e.g. NewScientist.com – the link for which can be found in
on this CD). Your goal isn't to learn lots of additional information, but to be
aware of how many of the topics or ways of scientific thinking are relevant to
everyday life. It will also help you recognise that different views on the same
topic exist.

Ask your teacher for advice
This could be about the best practice questions to use for Stretch and
Challenge – including past examination questions, specimen examination
questions or other materials specific to your school. Look at the examples in this
document and try answering the questions.

Think carefully about your exam technique
Allow yourself enough time to answer all your questions and be prepared to
have a go – you’re sure to get some marks! Bear in mind you need to do as well
as you can on the whole paper.
What will Stretch and Challenge questions look like?
Stretch and Challenge questions can crop up anywhere in an A2 examination paper,
although they will not be flagged as such. However, the examples below should help
you to spot the Stretch and Challenge opportunities.

Stretch and Challenge questions can account for around 5–10% of the marks in
your examination paper, so they could make the difference between one grade
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and another.

Sometimes, a Stretch and Challenge question could be the final part of a long
question consisting of several parts, while at other times it will be a single
question.

It is quite likely that Stretch and Challenge questions will require some sort of
extended prose answer, but could equally ask you to draw a diagram, calculate
something or solve a problem.

Stretch and Challenge questions often use the command words explain,
discuss, evaluate, or suggest.
Sample questions
Below is a set of sample questions. However, the nature of Stretch and Challenge
questions means they may include material from other topics. They are not past
exam questions nor are they written by the examination board, they are designed to
give you a flavour of the type of question you may be asked.
As with most questions, even if you do not achieve full marks you can usually pick up
several marks using a careful approach. The answers have been laid out to show any
potential marking points – i.e. if the question carries six marks, the corresponding
answer will have six points to it; if the question has been allocated eight marks then
the corresponding answer will have eights points, etc. Look carefully through each
question. There are many opportunities to gain marks and don’t be put off by the
slightly more complex wording of the question.
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A2 Unit F324: Rings, polymers and analysis
Module 1 – Rings, acids and amines
Question 1
Civetone can be extracted from the anal gland of the civet cat. It was used as an
ingredient in expensive perfumes. The skeletal formula for civetone is shown in Figure 1.
Figure 1 Civetone
(a)
What is the molecular formula of civetone?
[1]
(b)
Predict three reactions for civetone.
For each reaction give the:
o reagents and conditions
o skeletal formula of the organic product.
[6]
(c)
Cyclohex-3-en-1-one contains the same functional groups as civetone. Cyclohex-3en-1-one has the carbon-carbon double bond in the Z-orientation. Suggest why the
carbon-carbon double bond cannot exist in the E-orientation.
[2]
(d)
Ammonia reacts with carbonyl compounds to form imines.
R2C=O + NH3  R2C=NH + H2O
This reaction involves a nucleophilic addition followed by an elimination. The reaction
is catalysed by the presence of hydrogen ions.
(i) Draw the displayed formula of the imine R2C=NH. Show on the formula all of the
bond angles.
[1]
(ii) Use the curly–arrow model to suggest a mechanism for the nucleophilic addition
stage of the reaction. Show any relevant dipole and lone pair.
[4]
(iii) 2-methylbutanal reacts with methylamine. Draw the organic structure of the
product from this reaction.
[2]
(iv) Explain why dimethylamine will not react in a similar way with carbonyl
compounds.
[1]
[Total: 17]
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Sample answer:
(a)
C17H30O
(b)
NaBH4/ethanol solvent gives secondary alcohol
Br2(aq)/dibromo derivative
HBr(g)/room temperature gives bromo derivative
(c)
The carbon–carbon double bond cannot exist in the E-orientation as the C–C bonds
would be as far apart from each other as possible, so the ring would be under too
much strain and the ends of the carbon chain would be too far away from each
other to form a ring.
(d)
(i)
All bond angles are 120°.
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(ii)
(iii)
CH3CH2CH(CH3)CH=NCH3
(iv)
It is impossible to have the elimination stage because there is no proton to
be lost.
Examiner’s comments:
(a) Be careful not to forget the hydrogen atoms, which are not shown in the skeletal
formula.
(b)
Stretch and Challenge questions such as this allow you to demonstrate knowledge
and understanding from different parts of the specification. Firstly, you need to
identify the functional groups present in the molecule and then decide the most
appropriate reactions to describe. Since you have the choice of which reactions to
describe, it is best to choose the most straightforward ones, such as reduction or
addition. An alternative answer could be to describe the more complex reaction with
2,4-dinitrophenylhydrazine to make 2,4-dinitrophenylhydrazone.
(c)
In a cyclohexane ring the bond angles are approximately 109 °; with an E double
bond it is impossible for the molecule to distort to this bond angle.
(d)
This illustrates another aspect of Stretch and Challenge questions –
i.e. you must use your knowledge of organic chemistry to answer questions about a
novel type of reaction. Here, you need to apply your knowledge regarding the
chemistry of carbonyl compounds to this new type of reaction. For example, using
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your answer from part (i) the bond angles will be similar to those found in carbonyl
compounds or alkenes. From part (ii), the mechanism is very similar to that for the
reaction of NaBH4 with carbonyl compounds. It also relies on your familiarity with the
nucleophilic nature of amines.
In part (ii), remember a curly arrow shows the movement of an electron pair either
from a bond or a lone pair. You must show the correct use of the curly arrow starting
from the lone pair on the nitrogen atom to attack the electron deficient carbon atom. It
is also important to include the partial charges on C and O.
The next step shows the addition of a proton to the electron-rich nitrogen atom.
There is no need to use the curly arrow model for the proton transfer step.
(iii)
(iv)
A displayed formula could also be drawn.
If you look at the mechanism in part (ii), the last stage involves loss of a
proton. Since the nitrogen atom does not have a proton to lose, the reaction
cannot happen.
Question 2
Research chemists synthesise many compounds to test as possible anticancer drugs. The
structure of one of these compounds is shown in Figure 2.
Figure 2 Compound A
(a)
The primary amine attached to the benzene ring is less basic than the other –NH2
group in the molecule. The secondary amine group is more basic than either of the
the –NH2 groups. Explain these differences in basicity.
[3]
(b)
The research chemists dissolve compound A in aqueous sodium hydroxide. They
then react the solution with aqueous bromine. Complete the structure shown in
Figure 3 to show the product formed.
[2]
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Figure 3
(c)
The research chemists add compound A to excess dilute nitric acid. Compound A
dissolves to form a solution containing ions. Complete the structure shown in Figure
3 to show the organic ion formed.
[2]
(d)
The research chemists add compound A to an excess of potassium dichromate(VI)
solution and dilute sulfuric acid. The mixture is refluxed for one hour. Complete the
structure shown in Figure 3 to show the product formed.
[1]
[Total: 8]
Sample answer:
(a)
In the –NH2 group attached to the benzene ring, the lone pair is donated into the πsystem of the benzene ring and so is less available to be donated to a proton.
Secondary amine has two electron releasing groups attached, which makes the
lone pair more available and so more basic.
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(b)
(c)
(d)
Examiner’s comments:
This is another question where you have to apply your knowledge and understanding
regarding the chemistry of several functional groups.
In part (a), it would have been better to ring the structure but the answer clearly indicates it
is the secondary amine group that is the most basic. Remember that amines are basic
because they can donate a lone pair of electrons to accept a proton.
In parts (b) and (c), you need to apply your knowledge about the reactions of phenylamine
and phenols in order to answer the question. Remember to look at the mark allocation;
even with a Stretch and Challenge question this will help you organise your answer. Since
parts (b) and (c) are worth two marks each, it is likely that there may be reactions at more
than one site in the molecule. In part (c) you could also give products that involved
electrophilic substitution. The phenol could react with dilute nitric acid to substitute
hydrogens for a nitro group. The electrophilic substitution could occur on positions 2, 4 and
6 relative to the –OH group.
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In part (d), you should recognise that acidified potassium dichromate is an oxidising agent.
Module 2 – Polymers and synthesis
Question 1
Figure 4 shows the molecular structure for the molecule atropine. Atropine is present in the
plant commonly called deadly nightshade and is toxic to humans. However, it can also act
as one of the few antidotes to nerve gases such as sarin.
Figure 4 Atropine
(a)
Deduce the empirical formula for atropine.
[1]
(b)
Calculate the percentage by mass of nitrogen present in atropine.
[1]
(c)
Atropine contains several functional groups.
(i) Describe, with the aid of an equation, three different reactions for atropine.
Each reaction should involve a different functional group.
(ii) Draw the structure of the products from each of these reactions.
[9]
(d)
Indicate with an asterisk * every chiral carbon atom that is present in a molecule of
atropine.
[1]
(e)
Synthetic atropine can be prepared by chemists.
(i) Suggest one difference between synthetic atropine and natural atropine
(ii) The last stage in one synthetic route to produce atropine involves two
compounds reacting to make an ester. Suggest the identity of the two
compounds.
[1]
[2]
[Total: 15]
Sample answer:
(a)
C17H23O3N
(b)
4.84%
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(c)
(i) and (ii) Esterification:
RCH2OH + 2[O]  RCOOH + H2O
Bromination:
C6H5R + Br2  C6H4RBr + HBr
This reaction needs the presence of a halogen carrier such as AlBr3.
o Amine will react in presence of HCl(aq) to make:
R3N + HCl  R3NH+ + Cl–
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(d)
(e)
(i)
Synthetic atropine is a mixture of optical isomers but natural atropine will be
just one optical isomer.
(ii)
Examiner’s comments:
(a) Remember that an empirical formula is the simplest mole ratio of each element in the
compound.
(b) An error-carried-forward mark would be available from an incorrect empirical
formula.
(c)
Three different reactions can be taken from the chemistry of esters, alcohols, amines
or benzene rings. In each case, you must name the functional group and the
reagents for the reaction. You must also write the correct equations and products.
Stretch and Challenge questions often allow you to demonstrate knowledge and
understanding from different parts of the specification. Again, you first need to identify
the functional groups present in the molecule and then decide the most appropriate
reactions to describe. As you can choose which reactions to describe, it is best to
choose the most straightforward ones, such as oxidation of the alcohol. You also
need to write the equation – you can use the notation rather than writing out the full
structure, e.g. RCH2OH for the alcohol reaction, R’COOR for the ester, RR’NCH3 for
the amine, and for the benzene ring just show an R-group attached to the benzene
ring.
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Other possible reactions you can have described are
o Esters reacting with hot NaOH(aq) to make:
o Alcohols reacting with Na or with PCl5
o The benzene ring reacting with HNO3/H2SO4
o Finally, remember not to describe three reactions of the same functional
group.
Question 2
Table 1 shows some amino acids and their isoelectric points.
Table 1
Amino acid
Structure
Isoelectric
point
Glycine
6.1
Glutamic acid
3.1
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Lysine
10.0
Proline
6.3
(a)
(i)
(ii)
Explain why proline is an α-amino acid.
Proline has two stereoisomers. Draw these two stereoisomers, clearly
showing the 3D relationship between the two isomers.
[1]
[2]
(b)
Draw the structure of glycine at a pH of 6.1.
[1]
(c)
Draw the structure of lysine at a pH of 6.1.
[1]
(d)
Draw the structure of glutamic acid at a pH of 6.1.
[1]
(e)
Proline is refluxed with hot sodium hydroxide. Draw the structure of the organic
product.
[1]
(f)
Proline and lysine react together to form two different dipeptides. Draw the
structures of these two dipeptides.
(g)
[2]
Glutamic acid is refluxed with excess ethanol and concentrated sulfuric acid
catalyst. Write an equation for this reaction.
[2]
[Total: 11]
Sample answer:
(a) (i) Although it is a secondary amine rather than a primary amine the amine group is
on the number 2 carbon.
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(ii)
(b)
(c)
(d)
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(e)
(f)
(g)
Examiner’s comments:
This question involves the application of your knowledge and understanding of the
chemistry of amino acids. The Stretch and Challenge aspect is that one of the amino acids
chosen is less familiar than the other three and includes a ring structure.
(a) In part (i), remember α-amino acids are 2-aminocarboxylic acids. Make certain that
you show the three-dimensional arrangement in part (ii); it is easier to show the bonds
of the ring in the plane of the paper.
(b), (c) and (d) The isoelectric point is the pH at which the amino acid exists as a
zwitterion. Above this pH the amino acid will be negatively charged and
below this pH the amino acid will be positively charged. You also need to
consider the functional groups on the side chain and how they would react
with acids or bases.
(e) This is just an acid–base reaction to make a salt.
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(f) Remember to form the peptide linkage using the amino group on carbon number 2 and
not the one on the side chain.
(g) The reagents suggest esterification, so it is the carboxyl group on the side chain that
reacts.
Module 3 – Analysis
Question 1
A research chemist analysed the percentage composition by mass of the drug, compound
A. He found the percentage composition to be 59.90% carbon, 4.44% hydrogen and
35.6% oxygen.
The research chemist also analysed the drug using three different spectroscopic
techniques.
Look at the three different spectra for compound A.
Figure 5 NMR spectrum for compound A
When D2O was added to the sample and the NMR spectrum repeated, the signal at δ =
11.2 ppm disappeared.
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Figure 6 Mass spectrum for compound A
Figure 7 IR spectrum of compound A
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Use the given information and the three different spectra from Figures 5–7 to identify, as
far as you can, compound A. Explain your reasoning carefully.
[Total: 14]
Sample answer:
(a)
Empirical formula calculation
Mole ratio: C:H:O is 4.99:4.44:2.23
2.24:2.00:1.00
Empirical formula is C9H8O4
Molecular formula is C9H8O4 as NMR spectrum indicates eight protons and m/z
from mass spectrum indicates Mr is 180.
(b)
NMR spectrum
Contains six different types of protons.
δ = 2.3 ppm indicates a –CH3 attached to a carbonyl group or benzene ring.
δ = between 7.5 to 8.2 ppm indicates a C–H attached to a benzene ring.
δ = 11.2 ppm indicates a proton in a carboxylic acid – confirmed by loss of signal in
D2O.
(c)
Mass spectrum
Molecular ion is m/z = 180
So Mr is 180.
m/z = 43 could indicate CH3CO– or C3H7 e.g. CH3CH2CH2– or CH3CHCH3 present
in molecule.
(d)
IR spectrum
Broad absorption between 2500 to 3000 cm–1 indicates O–H stretching in
carboxylic acid.
Absorptions at 1680 and 1750 cm–1 suggest C=O stretching in two carbonyl
groups.
Strong absorptions 1200 and 1300 cm–1 (possibly) indicate C–O stretching.
Presence of C–O and C=O suggest presence of an ester.
(e)
Compound A’s structure
Examiner’s comments:
Although this Stretch and Challenge question is not structured, you must make certain that
you analyse all the data you have been given and that your final suggested structure is
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consistent with all of the given information.
It is better to analyse each piece of data separately, however, you must also use any
information you have already deduced from other analysis. When you have analysed each
piece of information, you must bring all this data together when suggesting the final
structure for the compound.
It is not possible to decide the substitution pattern in the benzene ring.
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A2 Unit F325: Equilibria, energetics and elements
Module 1 – Rates, equilibrium and pH
Question 1
A student wants to determine the solubility of calcium hydroxide in water. She decides to
use a titration method.
Stage 1
A sample of 5.00 g calcium hydroxide Ca(OH)2(s) was added to a conical
flask.
Stage 2
About 100 cm3 of distilled water was added to the conical flask and the
flask with its contents was stoppered. The flask was then shaken for five
minutes and left for a further five hours.
Stage 3
The contents of the conical flask were filtered into a clean, dry conical flask.
The filtrate obtained was saturated calcium hydroxide.
Stage 4
A 25.0 cm3 sample of the saturated calcium hydroxide was titrated against
0.0750 mol dm–3 hydrochloric acid.
In Stage 2, an equilibrium between solid calcium hydroxide and its constituent aqueous
ions is established.
Ca(OH)2(s)
Ca2+(aq) + 2OH–(aq)
In Stage 4, the titre was 22.40 cm3.
(a)
Calculate the solubility in g dm–3 of the saturated calcium hydroxide.
(b)
The student decides to repeat the experiment but makes a mistake in Stage 2.
She adds about 100 cm3 of 1.00 mol dm–3 sodium hydroxide instead of water.
She finds that the solubility of the calcium hydroxide is much smaller than
before. Explain why.
[1]
(c)
The student puts 20.0 cm3 of 0.005 mol dm–3 calcium hydroxide into a conical
flask. She places a pH probe connected to a datalogger into the calcium
hydroxide. She then slowly adds 0.010 mol dm–3 hydrochloric acid using a
burette. Eventually, she adds 40.0 cm3 of hydrochloric acid. Sketch on the graph
shown in Figure 8 how the pH of the solution in the conical flask changes during
the experiment.
[4]
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Figure 8
[Total: 9]
Sample answer:
(a)
Moles of HCl = 1.68 × 10–3
Moles of Ca(OH)2 = 8.40 × 10–4
[Ca(OH)2] = 3.36 × 10–2 mol dm–3
Solubility = 2.49 g dm–3
(b)
To minimise increase in [OH–(aq], the equilibrium shifts to the left so that [Ca2+(aq)]
is lowered.
(c)
The [OH–(aq)] is 0.020 mol dm–3, therefore pOH is 2.00 and pH is 12.00.
At the end of the titration there will be excess acid. The concentration of the [H +(aq)
cannot be greater than 0.010 mol dm–3 so the pH cannot be lower than 2.00.
Two moles of HCl reacts with one mole of Ca(OH)2 but double the concentration of
acid is needed so that neutralisation occurs when 20.0 cm3 of acid is added.
So, the graph will start at pH 12.0 and will have a vertical section at 20 cm3 that
covers a range of 6 pH units and will end at just above pH 2.0.
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Examiner’s comments:
In part (a) the calculation is unstructured and this is typical of Stretch and Challenge
questions.
Part (b) is an application of le Chatelier’s principle.
In part (c) the mark allocation suggests that you need to do more than just a sketch graph.
The starting pH, finishing pH and the volume of hydrochloric acid needed to just neutralise
the aqueous calcium hydroxide need to be calculated.
Question 2
This question is about reaction kinetics of gas phase reactions.
(a)
The following data was obtained for the reaction between hydrogen and nitrogen(II)
oxide.
2H2(g) + 2NO(g)  2H2O(g) + N2(g)
Table 2
[NO(g)]/mol dm–3
[H2(g)]
/mol dm–3
initial rate of
reaction
mol dm–3 s–1
0.70
1.40
1.40
0.39
0.39
0.78
0.19
0.76
1.52
Suggest, with reasons, a mechanism for the reaction that is consistent with the data.
[4]
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(b)
Ozone is made in the stratosphere when oxygen atoms react with oxygen molecules.
The concentration of oxygen atoms in the stratosphere is estimated to be
3.0 × 10–14 mol dm–3 and the concentration of oxygen molecules,
1.3 × 10–14 mol dm–3.
O(g) + O2(g)  O3(g)
This reaction is first order with respect to both oxygen atoms and oxygen molecules.
The rate constant k for the reaction is 3.9 × 10–5 dm3 mol–1 s–1.
Calculate the number of ozone molecules made in one dm 3 of the stratosphere every
second.
[2]
[Total: 6]
Sample answer:
(a)
The reaction is first order with respect to H2 because, as concentration of H2
doubles, the rate doubles.
Reaction is second order with respect to NO because, as the concentration of NO
doubles, the rate quadruples.
Therefore, rate = k[H2][NO]2 (2)
This means that the slowest step involves one molecule of H2 reacting with two
molecules of NO.
Slow step H2 + 2NO  H2O + N2 + O
Fast step H2 + O  H2O
(b)
Rate = 3.9 × 10–5 × (3.0 × 10–14) (1.3 × 10–14) = 1.521 × 10–32 mol dm–3 s–1
Examiner’s comments:
Although parts (a) and (b) of the question are unstructured, you must make certain that
you structure your answer.
In part (a), if the question had been structured you would have had to work out the orders
of reaction and/or the rate equation and use this to decide the particles involved in the
rate-determining step. Finally, the overall equation needs to be considered to suggest a
mechanism. All of these things should be shown for a complete answer to the question.
In part (b), write down the rate equation and substitute in the values given.
Question 3
Buffer solutions minimise pH changes when small amounts of acids or alkalis are added.
(a) Calculate the pH of a buffer solution that is 0.100 mol dm–3 with respect to sodium
methanoate and 0.200 mol dm–3 with respect to methanoic acid. The Ka of
methanoic acid is 1.8 × 10–4 mol dm–3.
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[3]
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(b) What changes to the concentrations of the sodium methanoate and the methanoic
acid would you need to make to decrease the pH of the buffer by one unit?
[1]
[Total: 4]
Sample answer:
(a)
[HCOOH(aq)]eqm ≈ 0.100 mol dm–3 and [HCOO–(aq)]eqm ≈ 0.200 mol dm–3
Ka = [HCOO–(aq)]eqm[H+(aq)]
[HCOOH(aq)]eqm
+
So [H (aq)] = 9.0 × 10–5 mol dm–3
pH = 4.05
(b)
[HCOOH(aq)] = 1.00 mol dm–3 and [HCOO–(aq)] = 0.200 mol dm–3
or
[HCOOH(aq)] = 0.100 mol dm–3 and [HCOO–(aq)] = 0.0200 mol dm–3
Examiner’s comments:
Part (a) is another unstructured question typical for a Stretch and Challenge question.
Remember that the final step is to use pH = –log10[H+(aq)].
In part (b), a decrease of one unit of pH means a change of a factor of ten in the [H +].
Question 4
Dinitrogen(IV) oxide is in equilibrium with nitrogen dioxide.
N2O4(g)
2NO2(g)
0.500 mol of N2O4 was allowed to react and reach equilibrium in a 5.00 dm3 sealed
container. At equilibrium, there were 0.375 mol of NO2 present.
(a)
Calculate the numerical value for the equilibrium constant.
[3]
(b)
Explain the effect on the position of equilibrium and the equilibrium constant if the
volume of the sealed container was suddenly reduced but the temperature remained
constant.
[2]
[Total: 5]
Sample answer:
(a) If 0.375 moles of NO2 is made then 0.1875 moles of N2O4 must have reacted
So at equilibrium, N2O4 = 0.3125 moles
Therefore, [N2O4]eqm = 0.0625 mol dm–3 and [NO2]eqm = 0.0075 mol dm–3
Kc = [NO2(g)] 2eqm
[N2O4(g)] eqm
Kc = 9.00 × 10–4 mol dm–3
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(b)
Position of equilibrium moves to the left but equilibrium constant stays the same.
Examiner’s comments:
In part (a), it is easier to use moles rather than the equilibrium concentrations; however,
the equilibrium constant Kc is based on the equilibrium concentrations in mol dm–3.
In part (b), remember that the equilibrium constant will only change if the temperature
changes.
Module 2 – Energy
Question 1
Magnesium oxide is a giant ionic solid.
(a)
Use the following data in Table 3 to calculate the first ionisation energy of
magnesium.
Table 3
Process
Enthalpy change/kJ mol–1
Lattice enthalpy for magnesium oxide
–3889
Enthalpy of atomisation for oxygen
+248
Enthalpy of atomisation for magnesium
+150
Second ionisation energy of magnesium
+1450
First electron affinity for oxygen
–141
Second electron affinity for oxygen
+798
Enthalpy change of formation of magnesium
oxide
–602
[4]
(b)
(c)
Explain the difference in enthalpy change between the first and second electron
affinity of oxygen.
[2]
Magnesium nitrate decomposes when heated to form magnesium oxide, nitrogen
dioxide and oxygen.
2Mg(NO3)2(s)  2MgO(s) + 4NO2(g) + O2(g)
Use oxidation numbers to show that this reaction involves both oxidation and
reduction.
(d)
[2]
Magnesium carbonate does not decompose at room temperature but will
decompose if heated strongly with a Bunsen burner.
(i)
Using the data in Table 4, calculate the free energy change of reaction ΔG in
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kJ mol–1 for the decomposition of magnesium carbonate at 300 K (room
temperature).
MgCO3(s)  MgO(s) + CO2(g)
Table 4
Substance
Enthalpy change of formation/kJ mol–1
Entropy/J K–1 mol–1
MgCO3(s)
–1113
66
MgO(s)
–602
27
NO2(g)
+33
240
[5]
(ii)
(e)
Explain why magnesium carbonate does not decompose at room temperature.
[1]
Use the data in Table 5 to predict if lithium chloride or sodium chloride has a more
exothermic enthalpy change of solution.
Table 5
Process
Enthalpy change/kJ mol–1
Lattice enthalpy for lithium chloride
–848
Lattice enthalpy for sodium chloride
–776
Enthalpy change of hydration for chloride ion
–381
Enthalpy change of hydration for lithium ion
–499
Enthalpy change of hydration for sodium ion
–390
[2]
[Total: 16]
Sample answer:
(a)
∆Hle = –∆H1ea –∆H2ea –∆H1ie –∆H2ie –∆HatMg –∆HatO +∆Hf
so
∆1ie = –∆Hle –∆H1ea –∆H2ea –∆H2ie –∆HatMg –∆HatO +∆Hf
= +3889 +141 –798 –1450 –150 –248 –602
= +782 kJ mol–1
(b)
First electron affinity is exothermic because the electron gained is attracted by the
positive nucleus.
Second electron affinity is endothermic because the electron gained is repelled by
the overall negative charge of the O– ion.
(c)
N goes from +5 to +4, which is reduction.
O goes from –2 to 0, which is oxidation.
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(d)
(i)
(ii)
(e)
∆H = +1113 +(-603) + (+33) = +544 kJ mol–1
∆S = -66 +(27)+(240) = +201 J K–1 mol–1
∆G = ∆H − T∆S
= +544000 – (300 × +201) J K–1 mol–1
∆G = +483.7 kJ mol–1
This is not a spontaneous reaction because of the positive free energy
change.
LiCl ∆Hsoln = +848 + (–381) + (–499) = –32 kJ mol–1
NaCl ∆Hsoln = +776 + (–381) + (–390) = +5 kJ mol–1, so LiCl has a more exothermic
enthalpy change of solution.
Examiner’s comments:
For part (a), instead of using a normal Born–Haber cycle to calculate the lattice enthalpy
you will need to calculate the first ionisation energy of magnesium. Drawing a Born–Haber
cycle will help you answer the question.
The answer to part (b) requires knowledge and understanding from more than one part of
the specification. It is best to define both terms and then tackle the explanation.
In part (c), it is best to write the oxidation numbers underneath each symbol in the
equation.
The calculation in part (d) is unstructured and you must remember to ensure that the
correct units are used in the equation that links free energy with the enthalpy change and
the entropy change. Entropy data is normally given in J K–1 mol–1 so that either it must be
converted to kJ K–1 mol–1 or the enthalpy change converted to J mol–1.
Question 2
This question is about redox reactions.
(a)
The half-equation for the reduction of iodate(V) ions is shown below.
IO3–(aq) + 6H+(aq) + 6e–  I–(aq) + 3H2O(l)
Iodate(V) ions oxidise iodide ions in the presence of acid.
Construct the ionic equation for this oxidation.
(b)
[2]
Look at the standard electrode potentials shown in Table 6.
Table 6
Process
Zn2+(aq) + 2e–
Standard electrode
potential/V
Zn(s)
–0.76
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VO3–(aq) + 4H+(aq) e–
VO2+(aq) + 2H2O(l)
+1.00
VO2+ + 2H+(aq) + e– ⇌
V3+ + H2O(l)
+1.00
V3+(aq) + e–
V2+(aq)
–0.26
V2+(aq) + e–
V(s)
–1.20
Excess zinc powder is added to acidified aqueous ammonium vanadate(V), NH 4VO3.
Use the information in Table 6 to predict the final products of the reaction. Include
equations in your answer.
[5]
(c)
Aqueous ferrate(VI) ions FeO42– decompose in the presence of hydrogen ions,
forming iron(III) ions, oxygen and water. Construct the ionic equation for this reaction.
[2]
[Total: 9]
Sample answer:
(a)
2I–(aq)  I2(aq) + 2e–
So adding 6I–(aq)  3I2(aq) + 6e–
To IO3–(aq) + 6H+(aq) + 6e–  I–(aq) + 3H2O(l)
Gives IO3–(aq) + 6H+(aq) + 6e– + 6I–(aq)  I–(aq) + 3H2O(l) + 3I2(aq) + 6e–
Cancelling out this will give
IO3–(aq) + 6H+(aq) + 5I–(aq)  3I2(aq) + 3H2O(l)
(b)
2VO3–(aq) + Zn(aq) + 8H+(aq)  2VO2+(aq) + Zn2+(aq) + 4H2O(l)
2VO2+(aq) + Zn(aq) + 4H+(aq)  2V3+(aq) + Zn2+(aq) + 2H2O(l)
2V3+(aq) + Zn(aq)  2V2+(aq) + + Zn2+(aq)
The more positive the electrode potential the more likely the reduction so VO 3–(aq),
VO2+(aq) and V3+(aq) are more likely to be reduced than Zn2+(aq)
Zn2+(aq) is more likely to be reduced than V2+(aq)
So that VO3–(aq) should be reduced to V2+(aq).
(c)
2FeO42–(aq) + 10H+(aq)  5H2O(l) + 1½O2(g) + 2Fe3+(aq)
Examiner’s comments:
For part (a), you must be able to recall the equation for the reduction of iodine and, when
the two half-equations are combined, it is important to cancel out species that are included
on both the left-hand and right-hand sides of the overall equation.
Part (b) involves applying your knowledge about electrode potentials should allow you to
make the required prediction. Remember, you must show your reasoning and not just the
answer.
The equation for part (c) is very difficult to balance – start by trying to balance the charges.
One mark is available for getting the reactants and products correct, i.e.
FeO42–(aq) + H+(aq)  H2O(l) + O2(g) + Fe3+(aq)
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Module 3 – Transition elements
Question 1
A student has isolated some crystals of a compound.
The student knows that the compound has the formula (C2O4)2KxHy.zH2O.
The values of x, y and z can be determined by titration.
Stage 1
Stage 2
Stage 3
Water is added to a known mass of the compound to make 250 cm3
of solution B.
A 10.0 cm3 sample of solution B is titrated against 0.100 mol dm–3
sodium hydroxide solution.
A 25.0 cm3 sample of solution B is added to a conical flask. This
sample has 25.0 cm3 of 1 mol dm–3 sulfuric acid added to it. The
flask is heated to 60 °C and the hot contents of the flask are titrated
against 0.0200 mol dm–3 potassium manganate(VII).
In Stage 1, the compound dissolves.
In Stage 2, hydrogen ions react with hydroxide ions.
In Stage 3, ethanedioate ions are oxidised to form carbon dioxide.
5C2O42–(aq) + 16H+(aq) + 2MnO4–(aq)  2Mn2+(aq) + 10CO2(g) + 8H2O(l)
(a)
In Stage 3, the flask is heated before heating. Suggest why this is done.
[1]
(b)
In Stage 2, the titre was 9.55 cm3. Calculate the concentration of hydrogen ions in
solution B.
[2]
(c)
In Stage 1, the student used 2.017 g of the compound (C2O4)2KxHy.zH2O.
In Stage 3, the titre was 31.75 cm3.
Calculate the values of x, y and z.
[5]
[Total: 8]
Sample answer:
(a)
So the reaction is fast enough to do a titration
(b)
Moles of H+ = 2.38 × 10–3
(c)
Moles MnO4– = 6.35 × 10–4
Moles C2O42– = 2½ × moles MnO4– = 1.59 × 10–3
Each mole of the compound provides two moles of C2O42–, so moles of compound =
7.94 × 10–4.
M = mass ÷ moles = 254
y = 2.38 × 10–3 ÷ 7.94 × 10–4 = 3
x=1
z = (254 – 48 – 128 – 3 – 39) ÷18 = 2
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Examiner’s comments:
In part (a), you must make a sensible suggestion.
Part (b) is a straightforward calculation.
The calculation for part (c) is unstructured and relies on you using the relationships
between molar mass, mass and moles, and between moles, volume and concentration. It
is important that you organise your answer and indicate at each stage the quantity that you
are calculating. Error-carried-forward marks would be available, but only if it is clear how
the error had been carried forward.
Question 2
Copper reacts with excess hot, concentrated sulfuric(VI) acid to form a blue solution A, a
colourless gas B and a liquid C. The gas B contains 50.0% by mass of oxygen. When
excess aqueous ammonia was added slowly to the reaction mixture containing solution A,
an exothermic reaction initially took place forming salt D. As more aqueous ammonia was
added, a blue precipitate E was formed and finally a dark blue solution F was formed.
Concentrated hydrochloric acid was added to a small sample of solution F and a green
solution G was formed.
Identify compounds A to G showing your reasoning.
Write equations for all the reactions that take place.
[Total: 13]
Sample answer:
2H2SO4 + Cu  CuSO4 + 2H2O + SO2
A is CuSO4.
B is SO2 because, using the percentage data, the mole ratio S:O is 1.56:3.13.
C is H2O.
2NH3 + H2SO4  (NH4)2SO4
D is (NH4)2SO4 because sulfuric acid is neutralised by ammonia.
Cu2+(aq) + 2OH–(aq)  Cu(OH)2(s)
[Cu(H2O)6]2+ + 4NH3  [Cu(NH3)4(H2O)2]2+ + 4H2O
Or
Cu(OH)2 + 4NH3 + 2H2O  [Cu(NH3)4(H2O)2]2+ + 2OH–
E is Cu(OH)2 because copper(II) ions initially form a blue precipitate with aqueous
ammonia. This precipitate re-dissolves in excess aqueous ammonia to form the complex
ion F, which is [Cu(NH3)4(H2O)2]2+.
NH3 + HCl  NH4Cl
[Cu(NH3)4(H2O)2]2+ + 4Cl–  [CuCl4]2– + 4NH3 + 2H2O(l)
The excess ammonia is neutralised by hydrochloric acid and then a ligand substitution
takes place to make G, which is CuCl42–.
Examiner’s comments:
This unstructured Stretch and Challenge question focuses on the chemistry of copper
compounds. It is better to organise your answer and focus on each of the sets of reactions
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described.
The use of the percentage composition by mass enables you to work out that sulfuric acid
is reduced to sulfur dioxide.
You should remember the colours of the copper complex ions mentioned in the question
and this helps to identify the unknowns.
Question 3
Iron forms many complex ions. Table 7 shows the stability constant Kstab for two complexes
of iron(III).
Table 7
Complex
Stability
constant
Colour of
complex
[Fe(CN)6]3–
4.08 × 1052
dark blue
102
blood red
[Fe(H2O)5(SCN)]2+
9.17 ×
(a)
Write an expression for the stability constant of [Fe(H2O)5(SCN)]2+.
[1]
(b)
An aqueous solution containing [Fe(CN)6]3– and SCN– is mixed together.
Use the stability constants to predict what you would observe.
[2]
[Total: 3]
Sample answer:
(a)
Kstab =
(b)
{[Fe(H2O)5(SCN)]2+(aq)}
[Fe3+(aq)] [SCN-(aq]
The solution would stay dark blue.
The Kstab for [Fe(CN)6]3– is much larger than that for the other complex
[Fe(H2O)5(SCN)]2+, therefore the equilibrium will be entirely on the side of
[Fe(CN)6]3–.
Examiner’s comments:
In part (a), curly brackets have been used instead of square brackets to avoid confusion
with the use of square brackets in complex ions. Water is not shown in this expression
since the primary role of water is as a solvent and its concentration does not change.
Kstab is just an equilibrium constant but it allows you to compare the relative stabilities of
complex ions. In part (b), since [Fe(CN)6]3– has such a large Kstab, it is a more stable
complex than [Fe(H2O)5(SCN)]2+.
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And finally...
The three key pieces of advice to remember if you want to aim for those Stretch and
Challenge marks are:



Aim to answer all the questions you need to on the examination paper as well as
you can.
Look for questions which invite you to make links across the different but familiar
topics.
Don’t be afraid of questions which appear to be about something you haven't
studied – look hard, because it’s probably a Stretch and Challenge question which
requires you to use your existing knowledge but in a new context.
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