T5-3P2 [216 marks]
1a.
[2 marks]
Markscheme
reactants and products in same phase/state;
rate of forward reaction = rate of reverse reaction;
concentrations of reactants and products remain constant / macroscopic properties remain constant;
Do not accept concentrations are equal.
Examiners report
Part (a) of this question focused on equilibrium and many candidates were able to show a good understanding of what would happen
when the conditions were changed and were able to deduce the equilibrium expression. Most could describe the properties of a
homogeneous equilibrium but some said that concentrations of reactants and products were equal at equilibrium as opposed to
constant. The candidates also could state and explain the effect of a catalyst.
1b.
[1 mark]
Markscheme
(K c) = [
[HI]2
H2][I2]
;
Examiners report
Part (a) of this question focused on equilibrium and many candidates were able to show a good understanding of what would happen
when the conditions were changed and were able to deduce the equilibrium expression. Most could describe the properties of a
homogeneous equilibrium but some said that concentrations of reactants and products were equal at equilibrium as opposed to
constant. The candidates also could state and explain the effect of a catalyst.
1c.
[2 marks]
Markscheme
no change to position of equilibrium;
no change to value of K c;
Examiners report
Part (a) of this question focused on equilibrium and many candidates were able to show a good understanding of what would happen
when the conditions were changed and were able to deduce the equilibrium expression. Most could describe the properties of a
homogeneous equilibrium but some said that concentrations of reactants and products were equal at equilibrium as opposed to
constant. The candidates also could state and explain the effect of a catalyst.
1d.
[1 mark]
Markscheme
the reaction is exothermic/heat is given out/ ΔH is negative;
Examiners report
Part (a) of this question focused on equilibrium and many candidates were able to show a good understanding of what would happen
when the conditions were changed and were able to deduce the equilibrium expression. Most could describe the properties of a
homogeneous equilibrium but some said that concentrations of reactants and products were equal at equilibrium as opposed to
constant. The candidates also could state and explain the effect of a catalyst.
1e.
[2 marks]
Markscheme
no effect (on the value of the equilibrium constant);
as it speeds up forward and reverse reaction / concentrations of reactants and products do not change / position of equilibrium does not
change / no change in yield;
Examiners report
Part (a) of this question focused on equilibrium and many candidates were able to show a good understanding of what would happen
when the conditions were changed and were able to deduce the equilibrium expression. Most could describe the properties of a
homogeneous equilibrium but some said that concentrations of reactants and products were equal at equilibrium as opposed to
constant. The candidates also could state and explain the effect of a catalyst.
1f.
[2 marks]
Markscheme
nickel / platinum / paladium;
150 − 200 °C/ heat;
Accept temperatures in this range.
Accept room temperature as an answer if platinum or palladium used.
Examiners report
Part (b) proved more problematic and relatively few could describe the necessary conditions for hydrogenation, and even fewer could
correctly state a definition of average bond enthalpy. The calculation of the bond enthalpy of propene proved difficult for many and
although some gained marks by ecf few obtained the correct answer -125. Candidates also had difficulty explaining why the process
was exothermic in terms of the relative strengths of the bonds being made and broken.
1g.
[2 marks]
Markscheme
the enthalpy change when (one mole of) the gaseous bond is broken (or formed) / X– Y(g) → X(g) + Y(g)/X(g) + Y(g) → X– Y(g);
averaged for the same bond in a number of similar compounds / OWTTE;
Examiners report
Part (b) proved more problematic and relatively few could describe the necessary conditions for hydrogenation, and even fewer could
correctly state a definition of average bond enthalpy. The calculation of the bond enthalpy of propene proved difficult for many and
although some gained marks by ecf few obtained the correct answer -125. Candidates also had difficulty explaining why the process
was exothermic in terms of the relative strengths of the bonds being made and broken.
1h.
[2 marks]
Markscheme
energy in: C=C + H–H and energy out: C–C + 2C–H;
Accept energy in C–C + 6C–H + C=C + H–H and energy out 2C–C + 8C–H.
ΔH = (612 + 436) − (347 + 826) = 1048 − 1173/ − 125 (kJ mol− 1);
Award [2] for correct final answer.
Award [1] for +125.
If old Data Booklet values then allow: ΔH = 1048−1172 = −124 (kJ mol –1)
Examiners report
Part (b) proved more problematic and relatively few could describe the necessary conditions for hydrogenation, and even fewer could
correctly state a definition of average bond enthalpy. The calculation of the bond enthalpy of propene proved difficult for many and
although some gained marks by ecf few obtained the correct answer -125. Candidates also had difficulty explaining why the process
was exothermic in terms of the relative strengths of the bonds being made and broken.
1i.
[1 mark]
Markscheme
due to the relative strength of the C–C and 2C–H bonds compared to the C=C and H–H bonds / bonds in products stronger than bonds
in reactants;
Examiners report
Part (b) proved more problematic and relatively few could describe the necessary conditions for hydrogenation, and even fewer could
correctly state a definition of average bond enthalpy. The calculation of the bond enthalpy of propene proved difficult for many and
although some gained marks by ecf few obtained the correct answer -125. Candidates also had difficulty explaining why the process
was exothermic in terms of the relative strengths of the bonds being made and broken.
1j.
[2 marks]
Markscheme
(i) addition of bromine/bromine water;
the bromine colour remains with propane and propene decolourizes the bromine / solution changes from brown to colourless;
Do not accept “clear” instead of “colourless”.
Examiners report
Part (c) was also based in organic chemistry and although most candidates could suggest bromine as a test for unsaturation, they did
not all state a correct test result.
Candidates must make sure that they state that the bromine becomes colourless and not clear. Many realised that propene polymerises
by addition polymerisation but few could successfully draw the structure of the repeating unit. Also few could suggest a reaction of
alkenes of economic importance- such as hydration to make alcohols.
1k.
[2 marks]
Markscheme
addition (polymerization);
−(−CH(CH3)−CH2−)− / −CH(CH3)CH−;
Continuation bonds necessary for mark, displayed formula or condensed structural formula can be given.
Accept if more than one repeating unit is shown.
Examiners report
Part (c) was also based in organic chemistry and although most candidates could suggest bromine as a test for unsaturation, they did
not all state a correct test result.
Candidates must make sure that they state that the bromine becomes colourless and not clear. Many realised that propene polymerises
by addition polymerisation but few could successfully draw the structure of the repeating unit. Also few could suggest a reaction of
alkenes of economic importance- such as hydration to make alcohols.
1l.
[1 mark]
Markscheme
hydrogenation (of vegetable oils) / manufacture of margarine / manufacture of ethanol / addition of water;
Accept manufacture of alcohol.
Do not accept hydrogenation of alkenes.
Examiners report
Part (c) was also based in organic chemistry and although most candidates could suggest bromine as a test for unsaturation, they did
not all state a correct test result.
Candidates must make sure that they state that the bromine becomes colourless and not clear. Many realised that propene polymerises
by addition polymerisation but few could successfully draw the structure of the repeating unit. Also few could suggest a reaction of
alkenes of economic importance- such as hydration to make alcohols.
2a.
[2 marks]
Markscheme
energy needed to break (1 mol of) a bond in a gaseous molecule/state/phase;
average calculated from a range of similar compounds / OWTTE;
Do not accept similar bonds instead of similar compounds.
M2 can be scored independently.
Examiners report
The definition of average bond enthalpy in part (a), proved challenging even though it has appeared on recent examination papers and
very few scored two marks. A good number of candidates omitted gaseous and did not state that it is the energy needed to break 1 mol
of a bond in a gaseous molecule and many did not understand that it is the average calculated from a range of similar compounds.
2b.
[4 marks]
Markscheme
(i)
Bonds breaking:
2 × (C−C) + 8 × (C−H) + 3.5 × (O=O)
= (2)(347) + (8)(413) + (3.5)(498)
= 5741(kJ mol− 1);
Bonds forming:
3 × (CO) + 8 × (O−H)
= (3)(1072) + (8)(464) = 6928 (kJ mol− 1);
Enthalpy change:
(5741 − 6928 =) − 1187 (kJ mol− 1);
Award [3] for correct final answer.
(ii)
+
+
NH+
/ammonium / N2H+
5 /hydrazinium / CH3NH3 /methylammonium / methanaminium / H2NO3 /nitrooxonium;
4
Examiners report
In Part (b) (i), the typical errors were using the incorrect bond enthalpies from the Data Booklet and using the sum of the bond
enthalpies of bond forming (products) minus bond breaking (reactants) instead of the reverse. In Part (b) (ii), instead of NH4+,
−
−
candidates identified a range of incorrect answers including NH+
3 , NF, CN , NO3 , N2 and even NaCl, although the question asked for
a nitrogen containing positive ion.
3a.
[1 mark]
Markscheme
mass spectrometry/spectroscopy / MS;
Examiners report
Candidates were for the most part, able to correctly deduce the structural features from the different spectroscopic evidence but were
not able to deduce the structure of X as the molecular formula was not considered.
3b.
[4 marks]
Markscheme
(i)
presence (or absence) of particular bonds;
Accept functional groups.
(ii)
A: O–H/hydroxyl;
B: C=C/carbon-carbon double bond;
(iii)
no C=O/carbonyl present;
Examiners report
Candidates were for the most part, able to correctly deduce the structural features from the different spectroscopic evidence but were
not able to deduce the structure of X as the molecular formula was not considered.
3c.
[4 marks]
Markscheme
(i)
protons/H’s in three different chemical environments / OWTTE;
2:1:1 ratio of protons/H’s (in these environments) / OWTTE;
Accept 4:2:2
HO–CH2–CH=CH–CH2–OH / CH2=C(CH2OH)2;
(ii)
Examiners report
Candidates were for the most part, able to correctly deduce the structural features from the different spectroscopic evidence but were
not able to deduce the structure of X as the molecular formula was not considered.
3d.
[3 marks]
Markscheme
(i)
heat =
4.18 × 20.0 × (51.3−17.8)
1000
;
= 2.80 (kJ);
(ii)
2.80
enthalpy of combustion = ( 0.421
=) − 6.65 (kJ g− 1);
Examiners report
This part was generally well answered but there were some cases where 33.5 °C was converted into Kelvin. Many candidates had
serious problems with unit conversions and gave the answer as 2800 J or 2800 kJ. Some candidates had correct value for (ii) but lost
the mark because of the omission of the negative sign.
3e.
[2 marks]
Markscheme
Name:
steroids;
Role:
(sex) hormones;
OR
Name:
phospholipids;
Role:
membranes;
Examiners report
Part (b) was well answered.
3f.
[1 mark]
Markscheme
lipids less oxidized/contain less oxygen / carbohydrates partially/more oxidized/contain more oxygen / OWTTE;
Examiners report
Very few candidates linked the fact that lipids have higher energy content due to being less oxidized.
3g.
[2 marks]
Markscheme
α: C-1 OH below plane
;
β: C-1 OH above plane
;
Examiners report
Part (a) was generally well answered with full marks awarded to more than half of the candidates. Those who did not score full marks
usually reversed OH on carbon 4. Glucose has an aldehyde functional group which can undergo oxidation and this was covered in
Topic 10. The fact that cellulose is a polymer of glucose and has beta-1,4 linkages seemed to have been overlooked by many
candidates.
3h.
[3 marks]
Markscheme
(i)
aldehyde/alkanal/CHO;
(ii)
ketone/alkanone/CO;
(iii) glucose;
Examiners report
Glucose has an aldehyde functional group which can undergo oxidation and this was covered in Topic 10.
3i.
[2 marks]
Markscheme
cellulose is (condensation) polymer of β-glucose;
(rings in cellulose) joined by β-1,4 linkages;
Examiners report
The fact that cellulose is a polymer of glucose and has beta-1,4 linkages seemed to have been overlooked by many candidates.
3j.
[1 mark]
Markscheme
micronutrients required in much smaller quantities/very small amounts/less than 0.005% of body mass;
Accept opposite statement for macronutrients.
Examiners report
Most candidates gained full marks for this question and was answered quite well.
3k.
[1 mark]
Markscheme
Name: steel and other element: carbon;
Examiners report
This question was generally answered correctly.
3l.
[2 marks]
Markscheme
atoms/ions of the alloying element are a different size/larger/smaller;
prevents the layers of atoms/ions sliding across each other;
Examiners report
Part (d) proved to be most challenging although it deals with the fundamental idea of the behaviour of alloys.
3m.
[2 marks]
Markscheme
Similarity:
both turn chemical energy into electrical energy / use chemical reactions to produce electricity/lectrical energy / OWTTE;
Difference [1 max]:
rechargeable batteries have reversible reactions but fuel cells do not;
fuel cells consume fuel but rechargeable batteries do not require (external) fuel;
rechargeable batteries can be recharged by electricity but fuel cells cannot;
Examiners report
Some candidates were able to write one similarity and one difference between fuel cells and rechargeable batteries.
3n.
[3 marks]
Markscheme
All correct [3], 4 or 5 correct [2], 2 or 3 correct [1]
Examiners report
Part (b) was very poorly answered.
3o.
[2 marks]
Markscheme
Positive electrode:
O2(g) + 4H+ (aq) + 4e− → 2H2O(l);
Negative electrode:
H2(g) → 2H+ (aq) + 2e− ;
Examiners report
None of the candidates scored full marks particularly in part (c) where it was rare to see any correct half-equations; the candidates also
overlooked the fact that the electrolyte was acidic.
3p.
[2 marks]
Markscheme
large surface area;
changes only occur on the surface / where electron transfer occurs / OWTTE;
Examiners report
Part (d) seldom had any correct answers.
3q.
[1 mark]
Markscheme
Any two for [1]
petrol/gasoline
kerosene/paraffin/aviation fuel
diesel
fuel oil/gas oil
petroleum gas/refinery gas
Examiners report
In part (a) a significant number of candidates named two fuels obtained from petroleum.
3r.
[2 marks]
Markscheme
global warming;
carbon dioxide;
OR
air pollution;
carbon monoxide / particulates / oxides of nitrogen/NO/ NO2 / VOC s;
Accept oxides of sulphur/SO 2.
OR
acid rain;
oxides of nitrogen/NO/NO2;
Accept oxides of sulphur/SO 2.
Examiners report
A significant number of candidates described the environmental problem.
3s.
[1 mark]
Markscheme
slow decomposition / not biodegradeable;
Examiners report
The non-biodegradable property of plastics was stated correctly by many candidates.
3t.
[2 marks]
Markscheme
chemically stable;
liquid crystal phase over a suitable range of temperatures;
rapid switching speed;
Examiners report
The properties of a material that made it suitable for use as a liquid crystal display demonstrated poor understanding by many
candidates.
3u.
[2 marks]
Markscheme
Advantage:
easily taken/convenient / no specialist equipment needed;
Disadvantage:
stomach acid reacts with drugs / slow effect / only small fraction of drug absorbed / vomiting / requires conscious patient / harm
digestive system/can cause stomach bleeding;
Examiners report
In general, this question was well answered.
3v.
[2 marks]
Markscheme
inhalation
parenteral / by injection / intravenous / intramuscular / subcutaneous
rectal
skin patches
eye/ear/nose drops
topical
Award [2] for all 3 correct, [1] for 2 correct, [0] for 1 correct.
Examiners report
In general, this question was well answered.
3w.
[5 marks]
Markscheme
(i)
Bacteria:
tuberculosis/TB / syphilis / cholera / salmonella / bronchitis / botulism / lyme disease / (stomach) ulcers / anthrax / diptheria / meningitis
/ MRSA / gonorrhea / chlamydia / septicaemia;
Viruses:
influenza / common cold / AIDS / herpes / rabies / small pox / polio / rubella / yellow fever / measles / mumps / encephalitis / chicken
pox / shingles / mononucleosis;
Do not accept name of an organism (such as e-coli) rather than a disease.
(ii)
bacteria larger than viruses / viruses are smaller than bacteria;
bacteria are cells / viruses comprise DNA in a protein coat;
bacteria have cell wall/nucleus/cytoplasm / viruses do not have cell components;
bacteria can reproduce without a host / viruses require host/cell for replication/reproduction;
bacteria are not always harmful/parasitic / viruses are always parasitic;
Examiners report
Many candidates stated the correct name of one bacterial and one viral disease but some had problems stating differences between
bacteria and viruses. Candidates must realize that AIDS is a viral disease not HIV.
3x.
[2 marks]
Markscheme
patient non-compliance / not completing courses / OWTTE;
overprescription;
use for animals/in animal feedstock;
Accept overuse.
Do not accept overdose.
Examiners report
In part (c), the terms “over dosage” and “over prescription” were often used interchangeably.
3y.
[2 marks]
Markscheme
becomes part of DNA of virus / alters virus DNA/genetic material / blocks enzyme (polymerase) which builds DNA;
changes the cell membrane so that it inhibits the virus entry/bonding to the cell;
prevents virus from leaving the cell (after reproduction);
prevents virus from using cell to multiply/reproduce/replicate;
Examiners report
In part (d), the method of action of anti-viral was poorly explained. It seemed candidates, at times, learned several key phrases without
clearly understanding their meaning and so used them in inappropriate context.
3z.
[3 marks]
Markscheme
shorter wavelength/UV/high energy radiation from sun passes through;
long wavelength/infrared/IR radiation from Earth’s surface (some of this radiation) is absorbed (by gas);
Do not accept “trapped” or “blocked”.
Do not accept “IR from sun”.
causes (increased) vibration in bonds;
re-radiates heat back to the Earth;
Do not accept “reflects/bounces”.
Examiners report
There seems to be a very poor understanding of the interaction of greenhouse gases with radiation although this question has
frequently appeared in the examinations. A surprising number quoted ozone depletion and the use of terms often used in the media e.g.
“trapped”, “bounces”.
3aa.
[1 mark]
Markscheme
higher concentration / more abundant/increased combustion of fossil fuels (than other anthropogenic sources);
Examiners report
In part (b), the candidates failed to state the increased combustion of fossil fuels.
3ab.
[2 marks]
Markscheme
methane/CH4;
decomposition of organic matter / animals / oil fields / gas fields / intensive farming / landfills;
OR
dinitrogen monoxide/nitrous oxide/N2O;
Do not accept NOx, NO, NO 2, nitrogen oxides.
decomposition of organic matter/fertilizers;
OR
ozone/O3;
photochemical smog;
OR
CFCs;
aerosol cans / air conditioners / solvents / foam production / refrigerants;
OR
sulfur hexafluoride/SF6;
electrical insulator;
OR
nitrogen trifluoride/NF3;
production of electronic components;
[1] for any correct gas and [1] for the corresponding source.
Examiners report
Although the question stated “other than carbon dioxide and water identify one other green house gas”, many candidates identified CO2
and H2O.
3ac.
[2 marks]
Markscheme
dissolved carbon dioxide / CO2(g) + H2O(1) ⇌ H2CO3(aq);
+
−
H2CO3(aq) ⇌ H+ (aq) + HCO−
3 (aq)/CO2(aq) + H2O(1) ⇌ H (aq) + HCO3 (aq);
Examiners report
Many candidates identified that dissolved carbon dioxide causes the rain water to be acidic but did not show with the help of equation
partial dissociation of carbonic acid.
3ad.
[5 marks]
Markscheme
(i)
internal combustion engine / high temperature combustion;
N2(g) + O2(g) → 2NO(g);
(ii)
catalytic converters / exhaust recirculation;
2NO(g) + 2CO(g) → N2(g) + 2CO2(g);
(iii)
nitric acid/ HNO3 / nitrous acid/ HNO2;
Examiners report
The source of nitrogen monoxide and the method used to control its emission appeared to be well known and many could include
appropriate chemical equations. Very few candidates could identify that nitric or nitrous acid is responsible for acidity in lakes and rivers.
Some candidates formed H2SO4, H2SO3, CH3COOH and H2CO3 from nitrogen monoxide.
3ae.
[3 marks]
Markscheme
A: protein / polypeptide/tripeptide;
B: carbohydrate / sugar / monosaccharide / glucose;
C: lipid / triglyceride / vegetable oil / fat;
Examiners report
This question was generally well answered by many candidates. Though most had a general idea of the difference between a food and
a nutrient, many did not appreciate the distinction between an “unhealthy” food and one that isn’t a nutrient.
3af.
[2 marks]
Markscheme
vitamins;
minerals;
water;
Examiners report
This question was generally well answered by many candidates. Though most had a general idea of the difference between a food and
a nutrient, many did not appreciate the distinction between an “unhealthy” food and one that isn’t a nutrient.
3ag.
[3 marks]
Markscheme
hydrolytic rancidity and oxidative rancidity;
hydrolytic rancidity involves (reaction with water) breaking ester bond / formation of a fatty acid and glycerol / OWTTE;
oxidative rancidity involves reaction of carbon-carbon double bond/C=C with oxygen / addition reaction with oxygen;
Examiners report
The majority of candidates struggled in part (a) where the difference between hydrolytic and oxidative rancidity was seldom written
correctly.
3ah.
[2 marks]
Markscheme
all (C–C) bond lengths equal / C–C bond lengths intermediate between C–C and C=C;
benzene normally undergoes substitution not addition;
thermochemically more stable than predicted / produces less heat when hydrogenated/combusted than predicted;
Examiners report
This part was generally well answered by many candidates.
4a.
[2 marks]
Markscheme
tetramethylsilane/TMS;
Any of these for second mark:
strong single peak (as there are 12 protons in identical chemical environment);
absorbs upfield/away from most other protons/H’s;
low boiling point/bp / volatile (so easily removed from sample);
not toxic / unreactive / does not interfere with sample;
Examiners report
The reference standard for NMR and the reasons for its choice were well known.
4b.
[5 marks]
Markscheme
Any three for first three marks:
both increase rate of chemical reactions;
both reduce activation energy;
both provide alternative pathways for reaction;
enzymes more specific;
enzymes have active site that substrate bonds to / “lock and key” mechanism;
enzymes much more readily denatured by changing conditions;
Both required for final two marks:
competitive inhibitors and non-competitive inhibitors;
competitive inhibitors bond to active site and non-competitive inhibitors denature/alter shape of enzyme;
Examiners report
The final section comparing enzymes and catalysts was generally well answered, though quite a few neglected to discuss the
similarities of the two.
4c.
[5 marks]
Markscheme
Addition:
double bond converted to single bond forming new bonds to other monomers / OWTTE;
poly(e)thene / polyprop(yl)ene / PVC / polystyrene / Teflon/PTFE;
Accept suitable diagram.
Condensation:
monomer contains two functional groups;
small molecule/water produced when monomers join together / OWTTE;
polyester/Terylene/Dacron / nylon/polyamide/Kevlar;
Accept suitable diagram.
Examiners report
This was the best answered question in the option with many candidates being able to outline polymerization processes and examples
of the products, though full marks were rare. The role of cross-linking and examples of polymers displaying this, was however very
limited.
4d.
[3 marks]
Markscheme
Softening polymer:
poly(e)thene / polypropylene / PVC / PET / polystyrene / Teflon/PTFE / polyester/Terylene/Dacron / nylon/polyamide;
Non-softening polymer:
phenol-urea/Bakelite / phenol-methanal/formaldehyde/Melamine;
Explanation:
rigid polymers have cross-links between polymer chains / polymers that soften do not have cross-links between polymer chains;
Examiners report
This was the best answered question in the option with many candidates being able to outline polymerization processes and examples
of the products, though full marks were rare. The role of cross-linking and examples of polymers displaying this, was however very
limited.
4e.
[2 marks]
Markscheme
long rigid/rod-shaped molecules;
polar molecules / align with same orientation;
Examiners report
A few students could identify some properties that liquid crystals must have.
4f.
[2 marks]
Markscheme
chemically stable;
liquid crystal phase over a suitable range of temperatures;
rapid switching speed;
Examiners report
Very few students were aware of the additional properties required for commercial application.
4g.
[1 mark]
Markscheme
(solution that only displays a liquid crystal state) over a range of/at certain concentrations;
Examiners report
The meaning of lyotropic was reasonably well known.
4h.
[1 mark]
Markscheme
infrared/IR;
Examiners report
This was well answered. Most students knew that the IR absorption is used for ethanol detection.
4i.
[4 marks]
Markscheme
4-membered ring/β-lactam ring;
easily broken / very reactive / highly strained ring;
binds to proteins/deactivates proteins that form cell wall / interferes with cell wall formation / prevent formation of crosslinks within cell
wall;
makes cell wall porous / allows water to pass;
causes cell to burst;
Examiners report
Most candidates knew that AIDS was a viral disease, but knowledge of a bacterial disease (rather than the name of a bacterium) was
more sketchy. The differences between bacteria and viruses and reasons for drug resistance in bacteria were generally well known.
Many candidates also gave good answers about the mode of action of penicillin, but the action of antiviral drugs could only be explained
by the best candidates.
4j.
[2 marks]
Markscheme
one enantiomer/isomer of thalidomide relieves nausea/morning sickness;
other enantiomer/isomer of thalidomide causes foetal deformities;
OR
one enantiomer/isomer of taxol is an anti-cancer drug;
other enantiomer/isomer of taxol is ineffective;
OR
cis-isomer of diamminedichloroplatinum(II)/cisplatin is an anti-cancer drug;
trans-isomer is ineffective;
Examiners report
Whilst a drug in which stereochemistry was important was well known (usually Thalidomide), the other two parts of the question were
poorly answered in very vague terms. Few seemed to realize the importance of the 3D modelling of the drug and its active site or to
have grasped the concept that combinatorial chemistry involved the random ordering of specific chemical building blocks.
4k.
[3 marks]
Markscheme
oxygen has double bond/bond order 2;
ozone has delocalized bond/bond order 1.5;
oxygen bond/bond in oxygen (molecule) is stronger/shorter / ozone bond is weaker/longer;
ozone absorbs light of lower energy/lower frequency/longer wavelength;
Accept suitable diagram.
Examiners report
Most candidates were aware of the difference in bonding between diatomic oxygen and ozone, and the effect this has on the UV
frequencies they absorb.
5a.
[2 marks]
Markscheme
Number of peaks:
1;
Splitting pattern:
singlet / it is not split;
Examiners report
The great majority scored full marks in (b).
5b.
[2 marks]
Markscheme
oxidation state of transition element/number of d electrons/charge on ion;
type/identity/charge density of ligands;
stereochemistry/shape of complex/number of ligands;
Examiners report
Many scored both marks in (a), although some blanks were seen.
5c.
[3 marks]
Markscheme
molecule colourless because energy absorbed in UV region/not absorbed in visible region;
anion pink because of greater conjugation/more alternating single and double (C=C) bonds;
anion/coloured form/more conjugated form absorbs in visible region/lower energy radiation/green light;
complementary colour seen;
Examiners report
Part (b) was generally well answered, with most showing a good understanding of the material being tested, although a few referred to
d-d electron transitions.
5d.
[3 marks]
Markscheme
at low substrate concentrations/at first rate is (directly) proportional to (substrate) concentration / OWTTE;
Do not accept only qualitative statement such as “rate increases as concentration increases”.
at high substrate concentrations/eventually rate reaches maximum/levels off/becomes constant / OWTTE;
active sites become blocked/saturated / OWTTE;
Examiners report
Although most had some idea of what sort of answer part (a) required, it was rare to find full marks being awarded – the most common
reasons were a qualitative answer for the first mark, and the absence of a reference to active sites for the third mark.
5e.
[4 marks]
Markscheme
Competitive inhibitor:
Vmax same;
inhibitor occupies active site;
Non-competitive inhibitor:
Vmax lower;
inhibitor binds elsewhere causing distortion in shape of active site / OWTTE;
In each part, explanation mark cannot be awarded without correct reference to V max.
Examiners report
In (b), the distinction between competitive and non-competitive inhibitors was well known, although a surprising number of answers
contained explanations without stating the effect on Vmax.
5f.
[3 marks]
Markscheme
(i)
sketch graph with rate and pH labels and bell-shaped curve (showing rate has maximum);
(ii)
(at higher or lower pH value of) charges on enzyme/amino acid (residues) changes;
so (shape of) active site changes / tertiary structure lost / OWTTE;
Examiners report
Most sketch graphs in (c) were sufficiently well drawn to score the mark, although many would have benefited from a scale that
indicated a narrow pH range; the explanation was generally well known.
5g.
[2 marks]
Markscheme
fuel cells produce only water / Cd and Ni are toxic (heavy metals);
fuel cells are more expensive;
fuel cells can operate continuously/do not need recharging;
fuel cells are more unwieldy/less portable/less self-contained/need supply of O2 and H2;
Accept opposite statements for NiCd cells.
Examiners report
Part (c) was better answered, with most able to gain at least one of the four possible scoring points.
5h.
[2 marks]
Markscheme
Thermotropic:
show liquid-crystal behaviour over a range of temperatures/at certain temperatures;
Lyotropic:
show liquid-crystal behaviour over a range of concentrations/at certain concentrations;
Examiners report
In (a), most understood the differences between thermotropic and lyotropic liquid crystals and scored both marks.
5i.
[3 marks]
Markscheme
polar groups/N–O/C≡N help molecules to line up in a common direction / OWTTE;
biphenyl groups/benzene rings help to make the molecule rigid/rod-shaped/linear / OWTTE;
alkyl/hydrocarbon chains help prevent too close packing/help maintain liquid-crystal state / OWTTE;
Examiners report
Answers to (b) rarely scored full marks. Some clearly knew very little about the relevant molecular features, and many of those who did,
failed to link the features with the properties.
5j.
[5 marks]
Markscheme
each pixel contains a liquid-crystal film sandwiched between two glass plates;
glass plates have fine scratches and can polarize light;
liquid-crystal molecules line up to form a twisted arrangement / twisted nematic geometry;
liquid-crystal interacts with (plane-) polarized light which is rotated 90° as it passes through the film;
when two polarizers are aligned (with the scratches) light will pass through and pixel will appear bright;
when a potential difference is applied across the film molecules align with the film losing their twisted structure;
they no longer interact with polarized light and pixel appears dark;
Apply OWTTE throughout.
Examiners report
Part (c) was sometimes left blank, and few answers scored more than one or two marks. This part tested AS C.11.2, but most answers
were vague and rambling and did not come close to matching the markscheme, which was very similar to the teacher’s notes for this
assessment statement.
5k.
[2 marks]
Markscheme
interferes with enzymes/chemicals that bacteria need to make cell walls / interferes with cell wall formation;
osmosis/osmotic pressure causes cell wall to break/burst / water enters cell causing it to burst / OWTTE;
Examiners report
Most answers scored at least one mark in (c), or came close to it – most that did not were not specific enough or failed to mention cell
walls.
5l.
[5 marks]
Markscheme
(i)
(ii)
;
ring strain / bond angles are approx 90° / should be 109° or 120° / OWTTE;
ring breaks / produces reactive amide group / OWTTE;
(so) penicillin can become bonded to enzyme/penicillinase;
(iii)
amide;
Examiners report
In (d), the β-lactam ring was usually correctly identified, as was the circled amide group. In part (d)(ii), few answers scored full marks,
although most identified the relevance of the bond angles in causing ring strain.
5m.
[2 marks]
Markscheme
caused by overprescription/overuse/overdose / not completing course of penicillin / use of antibiotics in animal feed / OWTTE;
penicillins with modified side chains must be developed/cocktail of drugs must be taken to overcome resistant bacteria / OWTTE;
Examiners report
In (e), many more scored the overprescription mark than the one for modifying the side chain.
5n.
[3 marks]
Markscheme
bonds in oxygen are double and bonds in ozone are (intermediate) between single and double;
bonds in oxygen are stronger / bonds in ozone are weaker;
oxygen absorbs higher energy UV light / ozone absorbs lower energy UV light;
Examiners report
Although part (a) was generally well answered, many candidates missed the first mark through failing to refer to bonds in both oxygen
and ozone.
6.
[1 mark]
Markscheme
B
Examiners report
[N/A]
7.
[1 mark]
Markscheme
B
Examiners report
[N/A]
8.
[1 mark]
Markscheme
A
Examiners report
[N/A]
9.
[1 mark]
Markscheme
C
Examiners report
[N/A]
10.
[1 mark]
Markscheme
A
Examiners report
[N/A]
11a.
[1 mark]
Markscheme
(1)
2
2
2
4
−−−−N(NO2)3(g) + −−CH3OH(l) → −−N2(g) + −−CO2(g) + −−H2O(l);
Examiners report
Most candidates got the correct stoichiometric coefficients for the equation in part (a).
11b.
[3 marks]
Markscheme
−
bonds broken: (6 × 305) + (3 × 158) = 1830 + 474 = 2304 (kJ mol 1);
−
bonds made: (2 × 945) + (3 × 498) = 1890 + 1494 = 3384 (kJ mol 1);
−
enthalpy change: 2304 − 3384 = −1080 (kJ mol 1);
Award [3] for correct final answer.
Award [2 max] for +1080 (kJ mol –1) .
Accept –234 kJ mol –1 which arise from students assuming that 305 kJ mol –1 refers to the strength of a single N–O bond. Students may
then take N=O from the data book value (587 kJ mol–1).
bonds broken: (3 × 305) + (3 × 587) + (3 × 158) = 915 + 1761 + 474 = 3150 (kJ mol–1)
bonds made: (2 × 945) + (3 × 498) = 1890 + 1494 = 3384 (kJ mol –1)
enthalpy change: 3150 – 3384 = –234 (kJ mol –1) .
Award [2 max] for correct calculation of the enthalpy change of reaction for the equation in part (a), which gives –2160 (kJ mol –1).
Award [1] if the final answer is not –2160 but the candidate has correctly calculated the bonds broken in trinitramide as 2304 (kJ mol –1).
Examiners report
In Part (c), the typical errors were using the incorrect bond enthalpies from the Data Booklet and using the sum of the bond enthalpies of
bond forming (products) minus bond breaking (reactants) instead of the reverse. Some candidates surprisingly used the combustion
equation from part (a) for their extensive calculations which was partially given credit.
11c.
[2 marks]
Markscheme
(N–N bond in) trinitramide is longer/nitrogen (gas) is shorter / 0.145 nm in trinitramide versus 0.110 nm in nitrogen;
trinitramide has single (N–N) bond and nitrogen (gas) has triple bond;
Examiners report
Part (d) was well answered although a number of candidates thought that nitrogen has a single or double bond instead of a triple bond
which was worrying. VSEPR theory however was exceptionally poor and most candidates demonstrated little or no understanding.
Many incorrect geometries were cited, especially trigonal planar and even linear and v-shaped! Very few candidates related the
geometry to four negative charge centres or electron domains around the central nitrogen atom.
11d.
[3 marks]
Markscheme
106°–108°;
Accept <109°.
Any two for [2 max].
4 (negative) charge centres/electron pairs/electron domains around central nitrogen;
central nitrogen has a lone/non-bonding pair;
lone/non-bonding pairs repel more than bonding pairs;
molecule will be (trigonal/triangular) pyramidal;
(negative) charge centres/electron pairs/electron domains will be tetrahedrally arranged/orientated/ have tetrahedral geometry;
Do not apply ECF.
Examiners report
[N/A]
11e.
[2 marks]
Markscheme
polar;
net dipole moment present in molecule / unsymmetrical distribution of charge / polar bonds do not cancel out / centre of negatively
charged oxygen atoms does not coincide with positively charged nitrogen atom;
Marks may also be awarded for a suitably presented diagram showing net dipole moment.
Do not accept “unsymmetrical molecule”.
For polarity, apply ECF from part (e).
Examiners report
In part (f), polarity typically involved just guess work and only few candidates could explain the reason for the polarity or gave a diagram
showing the net dipole moment which suggested poor understanding of the topic.
11f.
[3 marks]
Markscheme
burn/combust a (known) mass/volume/quantity/amount of methanol (in a spirit burner) / weigh methanol/spirit burner before and after
combustion;
use flame to heat a (known) mass/volume/quantity/amount of water;
measure the increase/rise/change in temperature (of the water);
Examiners report
Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good
understanding of calorimetry.
11g.
[3 marks]
Markscheme
calculate the heat gained by the water / calculate the heat evolved by the burning methanol / substitute in q = mcΔT ;
calculate the amount/moles of methanol / divide the mass of methanol by its molar mass;
divide the heat gained by the water by the amount/moles of methanol;
Examiners report
Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good
understanding of calorimetry.
11h.
[2 marks]
Markscheme
result would be less exothermic/less negative;
Accept “less/smaller/lower”.
heat loss / incomplete combustion;
Accept methanol is volatile/evaporates / beaker/material of calorimeter absorbs heat.
Examiners report
Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good
understanding of calorimetry.
12a.
[1 mark]
Markscheme
(1)
2
2
2
4
−−−−N(NO2)3(g) + −−CH3OH(l) → −−N2(g) + −−CO2(g) + −−H2O(l);
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12b.
[3 marks]
Markscheme
−
bonds broken: (6 × 305) + (3 × 158) = 1830 + 474 = 2304 (kJ mol 1);
−
bonds made: (2 × 945) + (3 × 498) = 1890 + 1494 = 3384 (kJ mol 1);
−1
enthalpy change: 2304 − 3384 = −1080 (kJ mol );
Award [3] for correct final answer.
Award [2 max] for +1080 (kJ mol –1).
Accept –234 kJ mol –1 which arise from students assuming that 305 kJ mol –1 refers to the strength of a single N–O bond. Students may
then take N=O from the data book value (587 kJ mol–1).
bonds broken: (3 × 305) + (3 × 587) + (3 × 158) = 915 + 1761 + 474 = 3150 (kJ mol –1)
bonds made: (2 × 945) + (3 × 498) = 1890 + 1494 = 3384(kJ mol –1)
enthalpy change: 3150 – 3384 = –234(kJ mol –1).
Award [2 max] for correct calculation of the enthalpy change of reaction for the equation in part (a), which gives –2160 (kJ mol –1).
Award [1] if the final answer is not –2160 but the candidate has correctly calculated the bonds broken in trinitramide as 2304 (kJ mol –1).
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12c.
[2 marks]
Markscheme
increase in the number of moles of gas;
gases have a greater entropy/degree of randomness (than liquids or solids);
Award [1 max] for answers stating that positive value indicates an increase in disorder/randomness.
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12d.
[3 marks]
Markscheme
ΔG = ΔH − T × ΔS;
700
= −1080 − 300 × 1000
;
−1290 (kJ mol− 1);
Award [3] for correct final answer.
Award [2 max] for incorrect conversions of units.
If no answer to part (c), using ΔH = –1000 kJ mol –1, gives –1020 (kJ mol –1).
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12e.
[2 marks]
Markscheme
no change in spontaneity / temp has no effect on spontaneity / spontaneous at all temperatures;
ΔG negative at all temperatures / exothermic/ ΔH negative and involves an increase in entropy/ ΔS positive;
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12f.
[2 marks]
Markscheme
(N–N bond in) trinitramide is longer/nitrogen (gas) is shorter / 0.145 nm in trinitramide versus 0.110 nm in nitrogen;
trinitramide has single (N–N) bond and nitrogen (gas) has triple bond;
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12g.
[3 marks]
Markscheme
106° – 108°;
Accept < 109°.
Any two for [2 max].
4 (negative) charge centres/electron pairs/electron domains around central nitrogen;
central nitrogen has a lone/non-bonding pair;
lone/non-bonding pairs repel more than bonding pairs;
molecule will be (trigonal/triangular) pyramidal;
(negative) charge centres/electron pairs/electron domains will be tetrahedrally arranged/orientated/ have tetrahedral geometry;
Do not apply ECF.
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
12h.
[2 marks]
Markscheme
polar;
net dipole moment present in molecule / unsymmetrical distribution of charge / polar bonds do not cancel out / centre of negatively
charged oxygen atoms does not coincide with positively charged nitrogen atom;
Marks may also be awarded for a suitably presented diagram showing net dipole moment.
Do not accept “unsymmetrical molecule”.
Apply ECF from part (h).
Examiners report
Most students could insert the coefficients to balance the equation provided and many recognized the benign nature of the products
formed. Though the structure of trinitramide was not given this did not seem to hinder students in calculating the required enthalpy
change. A worryingly high number of students however used bond enthalpies to calculate the enthalpy change in the part (a) equation
rather than the much simpler decomposition asked for, so to allow them to gain some credit, the mark scheme was adjusted. The
sections relating to entropy and free energy changes were generally well tackled, as was the comparative lengths of the N-N bonds.
Predicting the shape and polarity of the trinitramide molecule often proved more difficult, especially explaining the polarity of the
molecule. Explanations of the effect of external pressure on boiling point, in terms of vapour pressure, and of the effect of temperature,
in terms of kinetic theory, often lacked clarity.
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