MARKSCHEME SPECIMEN CHEMISTRY

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1 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M MARKSCHEME SPECIMEN CHEMISTRY Higher Level Paper 7 pages

2 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M Suject Details: Chemistry HL Paper Markscheme Mark Allocation Candidates are required to answer ALL questions. Maximum total = [95 marks].. Each row in the Question column relates to the smallest supart of the question.. The maximum mark for each question supart is indicated in the Total column. 3. Each marking point in the Answers column is shown y means of a tick () at the end of the marking point. 4. A question supart may have more marking points than the total allows. This will e indicated y max written after the mark in the Total column. The related ruric, if necessary, will e outlined in the Notes column. 5. An alternative wording is indicated in the Answers column y a slash (/). Either wording can e accepted. 6. An alternative answer is indicated in the Answers column y on the line etween the alternatives. Either answer can e accepted. 7. Words in angled rackets in the Answers column are not necessary to gain the mark. 8. Words that are underlined are essential for the mark. 9. The order of marking points does not have to e as in the Answers column, unless stated otherwise in the Notes column. 0. If the candidate s answer has the same meaning or can e clearly interpreted as eing of equivalent significance, detail and validity as that in the Answers column then award the mark. Where this point is considered to e particularly relevant in a question it is emphasized y OWTTE (or words to that effect) in the Notes column.. Rememer that many candidates are writing in a second language. Effective communication is more important than grammatical accuracy.

3 3 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M. Occasionally, a part of a question may require an answer that is required for susequent marking points. If an error is made in the first marking point then it should e penalized. However, if the incorrect answer is used correctly in susequent marking points then follow through marks should e awarded. When marking, indicate this y adding ECF (error carried forward) on the script. ECF acceptale will e displayed in the Notes column. 3. Do not penalize candidates for errors in units or significant figures, unless it is specifically referred to in the Notes column. 4. If a question specifically asks for the name of a sustance, do not award a mark for a correct formula unless directed otherwise in the Notes column, similarly, if the formula is specifically asked for, unless directed otherwise in the Notes column do not award a mark for a correct name. 5. If a question asks for an equation for a reaction, a alanced symol equation is usually expected, do not award a mark for a word equation or an unalanced equation unless directed otherwise in the Notes column. 6. Ignore missing or incorrect state symols in an equation unless directed otherwise in the Notes column.

4 4 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 3. a i (.05.5)(0.5).0 cm a ii / 0.00 mol 000 a iii / mol dm 5.00 a iv / 0.44 moldm i NaClO: for chlorine and I : 0 for iodine ii ClO since chlorine reduced/gains electrons ClO since oxidation state of chlorine changes from + to /decreases ClO since it loses oxygen / causes iodide to e oxidized iii produces chlorine gas /Cl on reaction with ClO which is toxic OWTTE iv oxidation states are not real oxidation states are just used for electron ook-keeping purposes average oxidation state of sulfur calculated to e + max ut the two sulfurs are onded differently/in different environments in thiosulfate so have different oxidation states OWTTE

5 5 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M c Valid: addition of oxygen signifies an oxidation reaction so C is oxidized loss of hydrogen signifies an oxidation reaction so C is oxidized oxidation state of C changes from 4 to +4/increases Not valid: loss of electrons might suggest formation of ionic product ut not valid since CO is covalent loss of electrons might suggest formation of ionic product ut not valid since reaction only involves neutral molecules OWTTE d i 4 [Ne]3s 3p Electrons must e given as superscript. d ii s s p 6 3s 3p 4

6 6 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M Question Answer Notes Total. a radical / unpaired electron c Change Shift Reason Increase in temperature LHS since forward exothermic reaction/ H 0 Increase in pressure RHS since fewer gaseous molecules on RHS Addition of a catalyst to the mixture No change since affects rate of forward and reverse reactions equally 3 Activation energy without catalyst Reactants Activation energy with catalyst Products correct positions of reactants and products correct profile with laels showing activation energy with and without a catalyst

7 7 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M d NO (g) HO(l) HNO 3(aq) HNO (aq) Ignore state symols. e ionic radius of nitrogen is atomic radius which is etween electrons 46pm/46 0 m which is greater than 7pm/70 m due to increased repulsion Values must e given to score mark. 3. a HCOO (aq) H O(l) OH (aq) HCOOH(aq) Equilirium sign must e given for mark. 4 K a.8 0 c 4 Kw.00 K 4 Ka d x K [OH (aq)].6 0 moldm Award [] for correct final answer of [OH (aq)]. Assumption: 0. x ~ 0. Accept any other reasonale assumption. 6 e poh log(.60 ) 5.59 ph Award [] for correct final answer. 3

8 8 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 4. a 4 curly arrow going from lone pair/negative charge on O in HO to C Do not allow curly arrow originating on H in HO curly arrow showing I leaving Accept curly arrow going from ond etween C and I to I in -iodoethane or in the transition state. Do not allow arrow originating from C to C I ond. representation of transition state showing negative charge, square rackets Do not award M3 if OH---C ond is represented. and partial onds at 80 to each other formation of organic product CH 3 CH OH and I Rate expression: rate k[oh ][CH CH I] Molecularity of RDS: imolecular 3 Inversion of configuration must e shown to score M4.

9 9 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M c S N : polar, protic solvents decrease nucleophilic reactivity due to hydrogen onding polar, protic solvents have a cage of solvent molecules surrounding anionic nucleophile resulting in increased stailization so are slower polar, aprotic solvents have no hydrogen onding so S N reactions are favoured since nucleophiles do not solvate effectively so have an enhanced/pronounced effect on nucleophilicity of anionic nucleophiles so are faster S N : polar, protic solvents favour S N reactions since the carocation intermediate is solvated y ion-dipole interactions y the polar solvent d DMF since aprotic solvent so favours S N e A is indicative of frequency of collisions and proaility that collisions have proper orientations f ( ) 4 k exp ln (.00 ).0 (8.3 98) S N implies second-order so mol dm 3 s

10 0 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 5. a only water/h O produced so non-polluting Bond reaking: ()(H H) (4)(C H) ()(C=C) ()(436) (4)(44) ()(64) 706 kj mol Bond formation: (6)(C H) ()(C C) (6)(44) ()(346) 830 kj mol kj mol Award [ max] for +4 kj mol. Award [3] for correct final answer. 3

11 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 6. a i Lines, x s or dots may e used to represent electron Lewis (electron dot) structure pairs. Charges may e included in Lewis structures of Ozone ozone ut are not required. Sulfur hexafluoride a ii Electron domain geometry Molecular geometry Award [ max] for either oth electron domain geometries correct for either oth molecular geometries correct. Ozone trigonal/triangular planar v-shaped/ent/angular Sulfur hexafluoride octahedral/square ipyramidal octahedral/square ipyramidal a iii sulfur hexafluoride/sf 6 a iv Ozone: Accept any angle greater than 5 ut less than 0 Experimental value of ond angle in O 3 is 7. and Sulfur hexafluoride: 90 (and 80 )

12 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 6. a v Doule-headed arrow not necessary for mark. Lines, x s or dots may e used to represent electron pairs. i Lewis (electron dot) structure FC of O on LHS FC of central N FC of N on RHS A 0 + B + 0 C + + Award [] for all nine FCs correct, [] for six to eight FCs correct. ii smallest FC difference for A or B, so either is preferred however B is preferred as oxygen is more electronegative than nitrogen, even though FC per se ignores electronegativity Reason required for M. OWTTE c i CH 4(g) 5O (g) CO (g) HO(g) O 3(g) c ii Ö H ( 393.5) ()( 4.8) ()( 4.3) ( 74.0) kjmol c iii standard enthalpy change of formation/ is always zero H f Ö of an element in most stale form c iv Ö S ( 3.8) ()( 88.8) ()( 37.6) ( 86) (5)( 05.0) 44.4 J K mol c v Ö Ö Ö 44.4 G H TS ( 660.8) (98) 67.8 kjmol 000 c vi spontaneous since negative G Ö

13 3 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 6. d i O has a doule ond Do not award mark for I on its own with no justification. O 3 has intermediate onds etween doule and single onds 3 O 3 has a ond order of ½ ond in O is stronger therefore I needs more energy d ii C Cl ond reaks since weakest ond Allow representation of radicals without consistent throughout. as long as hv CCl F CClF Cl Cl O3 ClO O 5 ClO O O Cl ClO O3 Cl O

14 4 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 7. a Award [] for all four correct, [] for two or three correct. I: caroxamide II: phenyl Do not allow enzene. max III: caroxyl / caroxy Do not allow caroxylic/alkanoic acid. i IV: hydroxyl : 6.6(mol).0 nc and H 9.09 : 0.649(mol) 4.0 nn and O 6.55 n : 6.49(mol) and n : 0.648(mol) 6.00 Do not allow alcohol or hydroxide. Award [ max] for correct final answer without working. 3 nc: nh: nn: no 9.5:0:: Empirical formula: C9H0NO ii C9H0NO iii (0.5)(40 0 ) 9 iv A: C H and B: C=O v O H and N H frequencies/stretches due to O H and N H occur aove 300 cm which are not present in IR of ute c i ::6

15 5 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M 7. c ii c iii CH3OCHCH 3 c iv Similarity: oth have fragment corresponding to Difference: X has fragment corresponding to X has fragment corresponding to Y has fragment corresponding to Y has fragment corresponding to ( M 5) / m/z 45 r r ( M 7) / m/z 43 ( M 43) / m/z 7 r ( M 3) / m/z 9 r ( M 9) / m/z 3 r Allow oth have same molecular ion peak/m + / oth have m/z = 60. However in practice the molecular ion peak is of low aundance and difficult to oserve for propan--ol.

16 6 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M (Question 7 continued) c v oth X and Y will exhiit hydrogen onding with water molecules diagrams showing hydrogen onding

17 7 SPEC/4/CHEMI/HP/ENG/TZ0/XX/M (Question 7 continued) 7. d i Names of complexes are not required. Complexes may e drawn without tapered onds. d ii Cisplatin Transplatin Cisplatin Transplatin Cis: polar and trans: non-polar d iii X-ray crystallography Accept NMR spectroscopy. d iv Similarity: oth involve shared pair of electrons / oth are covalent Difference: Pt N: pair of electrons comes from nitrogen / coordinate ond and N H: one electron comes from each onded atom d v London / dispersion / instantaneous induced dipole-induced dipole dipole-dipole hydrogen onding Award [] for all three correct, [] for any two correct. max

Chemistry. Higher and standard level. Specimen papers 1, 2 and 3

Chemistry. Higher and standard level. Specimen papers 1, 2 and 3 Chemistry Higher and standard level Specimen papers 1, 2 and 3 For first examinations in 2016 CONTENTS Chemistry higher level paper 1 specimen question paper Chemistry higher level paper 1 specimen markscheme