Chemistry Advanced Level Paper 3 (9CH0/03) 1(a)(i) suitable scale and axes labelled including units (1) all points plotted correctly (1) line of best fit (1) Plotted points use at least half the available space in both directions ±½ a square Smooth line, no dot to dot NOTE: Points at t = 000 and 6000 s should not be on the line of best fit. 3 1
1(a)(ii) 1st order with respect to SO Cl as half-life is constant Example of graph (1) values of two half-lives shown in graph in range 000 400 (s) (1) rate = k[so Cl ] (1) 3
1(a)(iii) tangent drawn at 0.00 mol dm -3 SO Cl (1) Example of calculation value of gradient (1) e.g. gradient = (-) 0.375/6400 = (-) 5.859 x 10 5 must be consistent with tangent shown on graph k = rate / 0.00 (1) value of k to or 3 SF and units (1) k = 5.859 x 10 5 / 0.00 =.9 x 10 4 s 1 /.93 x 10 4 s 1 Allow TE from incorrect rate equation / allow TE from incorrect gradient Units must be consistent with rate equation 4 1 (b) RDS Step 1 SO Cl SO Cl + Cl (1) The RDS must involve one mol of SO Cl Step SO Cl + Cl SO + Cl (1) The equations taken together must add up to the overall equation SO Cl SO + Cl e.g. SO Cl SO + Cl Cl Cl Ignore state symbols (Total for 1 = 1 marks) 3
(a) Example of calculation mass of C (1) mass of H (1) mass of oxygen and mols of C, H and O (1) ratio (1) mass of C =.83 x 1/44 = 0.77 g mass of H = 0.693 x /18 = 0.077 g mass of oxygen = 0.95 (0.77 + 0.077) = 0.103 g C H O Moles 0.77/1=0.0643 0.077/1=0.077 0.103/16=0.0064 ratio 0.0643/0.0063=10 0.077/0.00643=1 0.0064/0.0064=1 empirical and use of molar mass to determine molecular formula (1) Empirical formula = C 10 H 1 O, this has mass =148 so molecular formula = C 10 H 1 O 5 (b)(i) An answer that makes reference to the following point: contains benzene ring / arene Ignore comments about alkenes / unsaturated 1 (b)(ii) is an aldehyde or ketone / is a Both aldehyde and ketone required for first alternative 1 carbonyl compound (b)(iii) (Q is a) ketone Allow not an aldehyde if first alternative given in (b)(ii) 1 4
(c) An answer that makes reference to the following points: Award 1 mark for any structure of C 10 H 1 O which includes a benzene ring and a ketone e.g. Chemical shift (1) peak at 7.5 ppm indicates a benzene ring and peak at.55 ppm indicates H on carbon adjacent to C=O and peak at 1.5 ppm indicates alkyl hydrogens () Splitting pattern () Any from 3: complex multiplet indicates a benzene ring septuplet indicates 6 equivalent protons on adjacent carbons doublet indicates one hydrogen on adjacent carbon Area under peak () Any from 3: five H s indicate C 6 H 5 / monosubstituted benzene ring one H indicates one H on C next to carbonyl group six H s indicates six H s on two methyl groups 7 (Total for = 15 marks) 5
3(a) (i) An answer that makes reference to the following points: EITHER the solution of manganate(vii) ions (1) OR as distilled water/other solutions remaining in burette will dilute the solution of manganate(vii) ions (1) Distilled water, followed by the solution of manganate(vii) ions (1) So the distilled water will not dilute the solution of manganate(vii) ions (1) 3(a) (ii) An answer that makes reference to the following points: measure from the bottom of the meniscus (of the solution of manganate(vii) ions) (1) with eyes / head at same level (as meniscus) or ensuring the burette is vertical or use of white background to emphasise meniscus or ensuring funnel is removed from top of burette or ensuring there is no air bubble in the tip of the burette (1) 6
3(a) (iii) An explanation that makes reference to the following point: it will not change the amount / moles (of reactants) in the flask 1 3(b) H C O 4 CO + H + + e / C O 4 - CO + e and MnO 4 + 8H + + 5e Mn + + 4H O (1) 5H C O 4 + MnO 4 + 6H + Mn + + 10CO + 8H O / 5C O 4 - + MnO 4 + 16H + Mn + + 10CO + 8H O () Ignore state symbols Allow for 1 mark: 5H C O 4 + MnO 4 + 16H + Mn + + 10CO + 8H O + 10H + 3 3(c) mean =.7 /.70 (cm 3 ) (1) 1 7
3(d) Example of calculation calculates amount of MnO 4 (1) Amount of MnO 4 =.7/1000 x 0.0105 =.3835 x 10 4 mol TE on mean in (c) calculates amount of H C O 4 (in 5 cm 3 ) (1) Amount of H C O 4 (in 5 cm 3 ) =.3835 x 10 4 x.5 = 5.95875 x 10 4 mol TE on mol ratio in equation in (b) calculates amount of H C O 4 (in 50 cm 3 ) (1) calculates relative molecular mass of acid (1) Amount of H C O 4 (in 50 cm 3 ) = 5.95875 x 10 4 x 10 = 5.95875 x 10 3 mol Relative molecular mass of acid = 0.747 / 5.95875 x 10 3 = 15.4 calculates x (1) So relative mass of water of crystallisation = 15.4 90 = 35.4 35.4/18 = 1.97 so x = Do not award x = 1.97 5 8
3 (e) An answer that makes reference to the following point: more MnO 4 would show greater amount of H C O 4 (1) or so molecular mass of xh O would seem to be less, so value for x would be less (1) more MnO 4 means lower molecular mass (1) therefore mass of water of crystallisation is less so x is less than (1) (Total for 3 = 16 marks) 9
4 (a) An explanation that makes reference to the following point: production of CO (1) NaHCO 3 + HCl NaCl + CO + H O (1) Allow HCO 3 + H + CO + H O Ignore state symbols 4 (b) An answer that makes reference to the following points: drying agent (1) Allow to remove water mixture becomes transparent / clear (from cloudy) (1) Allow drying agent stops clumping together 4 (c) An explanation that makes reference to the following points: they will reduce the yield (1) as some of the -chloro--methylpropane will react with the water (and reform the alcohol) (1) Allow as some of the -chloro--methylpropane might dissolve in the water 10
4(d) breaking of CO bond in protonated alcohol (1) structure of intermediate (1) correct charge on carbon in intermediate (1) Arrow must originate from bond and finish on oxygen attack of chloride ion with lone pair (1) Arrow must originate from lone pair on Cl 4 (Total for 4 = 10 marks) 11
*5 (a) This question assesses a student s ability to show a coherent and logically structured answer with linkages and fully-sustained reasoning. Guidance on how the mark scheme should be applied: The mark for indicative content should be Marks are awarded for indicative content and for how the answer is structured and shows lines of reasoning. added to the mark for lines of reasoning. For example, an answer with five indicative marking points which is partially structured The following table shows how the marks should be awarded for indicative content. with some linkages and lines of reasoning scores 4 marks (3 marks for indicative of indicative marking points seen of marks awarded for indicative marking points content and 1 mark for partial structure and some linkages and lines of reasoning). If there are no linkages between points, the same five indicative marking points in answer would yield an overall score of 3 marks (3 6 4 marks for indicative content and no marks 5 4 3 for linkages). 3 1 1 0 0 The following table shows how the marks should be awarded for structure and lines of reasoning. Answer shows a coherent and logical structure with linkages and fully sustained lines of reasoning demonstrated throughout Answer is partially structured with some linkages and lines of reasoning Answer has no linkages between points and is unstructured of marks awarded for structure of answer and sustained line of reasoning 1 0 1
Indicative Content: technique A allows heating for a longer time period / technique B is suitable for a shorter time period technique A ensures oxidation / technique B may result in loss of reactant before oxidation technique A minimises release of flammable or harmful vapours / technique B allows release of flammable or harmful vapours, unless carried out in fume cupboard technique A involves more expensive equipment or is more complex to set up / technique B is simpler to set up or uses less expensive equipment as complete oxidation is not necessary to qualitatively show it is an alcohol that can be oxidised technique B is most appropriate as simpler / technique A is most appropriate as oxidation is guaranteed 6 13
5 (b) An explanation that makes reference to the following pairs: either no, as chloride ions could be oxidised by dichromate(vi) ions (reducing amount of oxidant / producing toxic chlorine) (1) E o values for dichromate(vi) and chloride ions being oxidised to chlorine are sufficiently close (even though E o is negative = -0.03 V) for chlorine to be formed (1) or yes, as reaction is dependent on H + ions (not anions) (1) E o value is negative (-0.03 V) (1) 5 (c) 1 (Total for 5 = 9 marks 14
6 (a) suitable extrapolation (1) Vertical at 3 minutes and smooth curve of best fit back from 10 minutes to 3 minutes T in the range of 9 33( o C) (1) 6 (b) (i) An explanation that makes reference to the following point: as reaction is not instantaneous / takes time to release heat (1) so some heat is lost as mixture heats up (1) 15
6 (b) (ii) An explanation that makes reference to the following points: A greater surface area of zinc by using more finely powdered zinc (1) Less time for heat loss because it will increase the rate of reaction (1) 6 (c) Calculation of Q using Q = mc T (1) Using T from (a) Example of calculation 5 x 4.18 x 30.5 = 3 187 (J) Allow 4. for 4.18 amount of CuSO 4 (aq) (1) energy per mol = Q/0.018 (1) H = -Q/1000 kj mol 1 (1) = 5/1000 x 0.70 = 0.018 3 187/0.018 = 177 kj mol 1 / 180 kj mol 1 / 177 000 J mol 1 / 180 000 J mol 1 Use of 4. gives 178 kj mol 1 Must have correct sign and units for 4 th mark 4 16
6 (d) (i) (17 answer to (c) / 17) x 100 Example of calculation (17 177 / 17) x 100 = 18.4% 1 6 (d) (ii) An answer that makes reference to two of the following points: If answer in (c) is less negative than -17, points one and three are valid, and for point two no lid is used (1) candidates must state less than 4.18 for SHC SHC of solution may not be 4.18 (1) energy absorbed by zinc / energy absorbed by thermometer / polystyrene cup (1) mass of solution is greater than 5 grams (1) If answer in (c) is more negative than -17, only point two can be awarded for candidates stating SHC is greater than 4.18 Do not award marks for just heat loss to surroundings 6 (d) (iii) An explanation that makes reference to the following points: T would be less, so r H / enthalpy change calculated would be less negative (1) because heat energy is transferred to greater volume of water than expected volume (1) (Total for question 6 = 15 marks) 17
7 (a) +5 1 7 (b) (i) choice of suitable reducing agent From Data Booklet of the standard electrode potentials, the right hand substance in numbers 16-4 1 7 (b) (ii) An answer that makes reference to the following point: Allow reaction may be too slow may be kinetically hindered or 1 activation energy may be too high 7 (b) (iii) An explanation that makes reference to the following points: (yellow to) green then blue (1) when the blue colour initially forms, it mixes with the yellow unreacted VO + (to turn green) (1) Ignore colours caused by contamination with reducing agent chosen 18
7 (b) (iv) An explanation that makes reference to the following point: E o value is between +0.34 V and 0.6 V justification of choice (1) correct equation for reaction (1) Allow by comparison of electrode potentials / calculation of E cell / use of anti-clockwise rule S O 3 + VO + + H + V 3+ + H O + ½S 4 O 6 or multiples 7 (c) (i) SO + ½O SO 3 Allow multiples Ignore state symbols 1 19
*7 (c) (ii) This question assesses a student s ability to show a coherent and logically structured answer with linkages and fully-sustained reasoning. Marks are awarded for indicative content and for how the answer is structured and shows lines of reasoning. The following table shows how the marks should be awarded for indicative content. of of marks indicative marking awarded for indicative points seen in marking points answer 6 4 5 4 3 3 1 1 0 0 The following table shows how the marks should be awarded for structure and lines of reasoning. Answer shows a coherent and logical structure with linkages and fully sustained lines of reasoning demonstrated throughout Answer is partially structured with some linkages and lines of reasoning Answer has no linkages between points and is unstructured of marks awarded for structure of answer and sustained line of reasoning 1 0 Guidance on how the mark scheme should be applied: The mark for indicative content should be added to the mark for lines of reasoning. For example, an answer with five indicative marking points which is partially structured with some linkages and lines of reasoning scores 4 marks (3 marks for indicative content and 1 mark for partial structure and some linkages and lines of reasoning). If there are no linkages between points, the same five indicative marking points would yield an overall score of 3 marks (3 marks for indicative content and no marks for linkages). 0
Indicative Content: Ignore comments related to collision theory V O 5 acts as a heterogeneous catalyst SO is adsorbed onto surface and reacts with the V O 5 SO 3 forms and is only weakly bonded to surface the SO 3 desorbs from the surface oxygen reacts with the VO, regenerating the V O 5 catalyst reaction follows different mechanism with lower activation energy (Total for 7 = 14 marks) 6 1
8(a) reagent for step 1 (1) KCN and HCN or KCN and H + or HCN product of step 1 (1) reagent for step (1) Allow structural or displayed formulae Allow name or formula of any mineral acid 3 8(b) (1) esterification (1)
8(c) NO + ion (1) Ignore equation to produce electrophile curly arrow from ring to electrophile (1) formula of intermediate (1) To any part of the electrophile including the plus sign The horseshoe must cover at least three carbon atoms, opening towards the carbon atom approached by the electrophile and at least part of the plus charge within the horseshoe curly arrow from CH bond to ring and formation of H + (1) (Total for 8 = 9 marks) 4 3
9(a) For Mg allow 0 or 8 electrons in outermost shell Ignore inner shell electrons 9(b)(i) amount of NaOH (1) Example of calculation 11.7/1000 x 0.100 = 0.00117 / 1.17 x 10 3 [NaOH] (1) [NaF] eq (1) K c expression (1) Value of K c (1) 0.00117/(5/1000) = 0.0468 mol dm 3 0.50 0.0468 = 0.03 mol dm 3 K c = [NaOH] / [NaF] 0.0530 (no units) 5 9 (b) (ii) An answer that makes reference to the following point: No, as temperature is likely to be different on different days and this affects value of K c Allow yes, but only if the temperature is the same 1 (Total for 9 = 8 marks) 4
10(a)(i) orange to colourless Allow brown / yellow to colourless 1 10(a)(ii) A explanation that makes reference to the following points: becomes polar / forms temporary dipole (1) Polar bromine can be shown by using a diagram as electron density in C=C bond repels bond pair (of electrons in bromine) (1) e.g. 10(a)(iii) addition (1) Example of structure: structure (1) 5
10(a)(iv) Ignore addition of square brackets and n 1 6
10(b) Raw materials (1) wood is renewable and polystyrene uses slightly more petroleum fractions Energy use (1) paper cup uses nearly 4x as much energy Water use (1) paper cup uses nearly 50x as much water Air emissions (1) paper cup produces chlorine and more sulfur dioxide both of which are toxic / acidic and polystyrene cup produces a lot of hydrocarbons Justification () Selection of either cup with two valid reasons For example: polystyrene is more sustainable, as although it uses up non-renewable raw materials, its production requires much less energy which is likely to save on fossil fuel use or paper is more sustainable, as no hydrocarbons are emitted into the atmosphere and the raw materials are renewable and use less petroleum fractions 6 (Total for 10 = 1 marks) 7