Paper 2 (7404/2): Organic and Physical Chemistry Mark scheme

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1 AQA Qualifications AS hemistry Paper 2 (7404/2): Organic and Physical hemistry Mark scheme 7404 Specimen paper Version 0.5

2 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments AOa 2 2 O 0.2 AO2c 2 of 4

3 MARK SEME AS hemistry Specimen paper Stage : consider the groups joined to right hand carbon of the = bond Extended response Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. onsider the atomic number of the atoms attached AOa M can be scored in stage or stage 2 has a higher atomic number than, so 2 O takes priority Stage 2: consider the groups joined to L carbon of the = bond Both groups contain atoms, so consider atoms one bond further away, ( and ) from ethyl group has higher atomic number than, ( and ) from methyl group, so ethyl takes priority Stage 3: conclusion The highest priority groups, ethyl and 2 O are on same side of the = bond so the isomer is Z Allow M5 for correct EF conclusion using either or both wrong priorities deduced in stages and 2 The rest of the IUPA name is 3-methylpent-2-en--ol 3 of 4

4 MARK SEME AS hemistry Specimen paper Moles of maleic acid = 0.0/6.0 = AND mass of organic product expected = ( ) 98.0 = 8.45 g Or moles of organic product formed = 6.53 / 98.0 = AO3 a % yield = /8.45 OR = 00 ( ) / ( ) = = 77.3% AND statement that the student was NOT correct AO3 a 4 of 4

45 g Or moles of organic product formed = 6.53 / 98.0 = 6.66 0 2 AO3 a % yield = 00 6.

5 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments O O 2 6O O 02.2 Temperature rise = 20. q = = 420 (J) AO2h Mass of alcohol burned = 0.54 g and M r alcohol = 00.0 mol of alcohol = n = 0.54/00 = AO2h eat change per mole = q/000n OR q/n = 778 kj mol OR J mol Δ = 778 kj mol OR J mol AO2h AOa M4 is for answer with negative sign for exothermic reaction Units are tied to the final answer and must match 02.3 Less negative than the reference AO3 b eat loss OR incomplete combustion OR evaporation of alcohol OR heat transferred to beaker not taken into account AO3 b 02.4 Water has a known density (of.0 g cm 3 ) AO3 2a Therefore, a volume of 50.0 cm 3 could be measured out AO3 2a 5 of 4

0054 AO2h eat change per mole = q/000n OR q/n = 778 kj mol OR 778 000 J mol Δ = 778 kj mol OR 778 000 J mol AO2h AOa M4 is for answer with negative sign for exothermic reaction Units are tied

6 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments 03. (ompounds with the) same molecular formula but different structural / displayed / skeletal formula AOa 03.2 (basic) elimination Mechanism points: orrect arrow from lone pair on :O to on adjacent to Br orrect arrow from bond to orrect arrow from Br bond to Br Structure of chosen product AOa O Br OR 3 Br 3 O of 4

03.2 (basic) elimination Mechanism points: orrect arrow from lone pair on :O to on adjacent to Br

7 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments 04. Percentage of oxygen by mass = = 54.6 AOb O % Divide by A r 2 6 = 3.4 = 4.5 = 3.4 AO2b Divide by smallest = 3.4 = 4.5 = = Nearest whole number ratio = = 3 : 3.96 : 3 Nearest integer ratio = 3 : 4 : 3 AO2b Empirical formula 3 4 O 3 Empirical formula mass = 88 = molecular formula mass Therefore, molecular formula is same as the empirical formula O 3 AO2b O O + 2O 2 AOa 7 of 4

4 3.4 Nearest whole number ratio = 3.32 3 3 = 3 : 3.

8 MARK SEME AS hemistry Specimen paper Advantage ethanol is produced at a faster rate Disadvantage more energy is used / required in the reaction AO2e AO2e 04.4 Air gets in / oxidation occurs AOa 04.5 Alcohol O absorption in different place ( cm ) from acid O absorption ( cm ) The =O in acids has an absorption at cm AO2e AO2e 8 of 4

required in the reaction AO2e AO2e 04.4 Air gets in / oxidation occurs AOa 04.

9 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments 05. UV light AOa l 4 l 3 + l 05.2 l + O 3 lo + O 2 AOa lo + O 3 l + 2O 2 AOa 05.3 M r of F 3 l = 04.5 Moles freon = / 04.5 = Number of molecules = = Molecules in 500 cm 3 = ( ) / 00 = AOb AOb AOb Allow answer in the range Answer must be given to this precision 9 of 4

5 Moles freon =.78 0 4 0 3 / 04.5 =.70 0 3 Number of molecules =.70 0 3 6.02 0 23 =.

10 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments 06. Alkenes AOa OR orrectly drawn molecule of cyclobutane or methyl cyclopropane, need not be displayed formula (or correct alkane structure with 6 carbons) Allow hexane or any other correctly named alkane with 6 carbons 06.3 Poly(but-2-ene) AOa 06.4 igh pressure AOb Allow pressure MPa Mention of catalyst loses the mark 0 of 4

11 MARK SEME AS hemistry Specimen paper This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. Level marks Level marks All stages are covered and the explanation of each stage is generally correct and virtually complete. Answer communicates the whole process coherently and shows a logical progression from stage and stage 2 (in either order) to stage 3. All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. 6 AOa 5 Indicative chemistry content Stage : consider effect of higher temperature on yield (Or vice versa for lower temperature) Le hatelier s principle predicts that equilibrium shifts to oppose any increase in temperature Exothermic reaction, so equilibrium shifts in endothermic direction / to the left So a igher T will reduce yield Stage 2: consider effect of higher temperature on rate (Or vice versa for lower temperature) Level 2 marks Level 0 0 marks Answer is mainly coherent and shows progression. Some steps in each stage may be out of order and incomplete. Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. Answer includes isolated statements but these are not presented in a logical order or show confused reasoning. Insufficient correct chemistry to gain a mark. At higher Temperature, more high energy molecules more collisions have E>Ea So rate of reaction increases/time to reach equilibrium decreases Stage 3: conclusion Industrial conditions chosen to achieve (cost-effective) balance of suitable yield at reasonable rate of 4

Answer communicates the whole process coherently and shows a logical progression from stage and stage 2 (in either order) to stage 3.

12 MARK SEME AS hemistry Specimen paper 2 Question Marking guidance Mark AO omments 07. Measured volume would be greater Level in burette falls as tap is filled before any liquid is delivered AO3 b AO3 b 07.2 Drop sizes vary AO3 b Allow percentage error for amount of oil will be large as the amount used is so small 07.3 Use a larger single volume of oil Dissolve this oil in the organic solvent Transfer to a conical flask and make up to 250 cm 3 with more solvent Titrate (25 cm 3 ) samples from the flask AO3 2b AO3 2b AO3 2b AO3 2b 2 of 4

2 Drop sizes vary AO3 b Allow percentage error for amount of oil will be large as the amount used is so small 07.

13 MARK SEME AS hemistry Specimen paper Stage Extended response calculation Mass of oil = 0.92 ( ) = 0.23 (g) Mol of oil = 0.23 / 885 = AO2h AO2h To gain 4 or 5 marks, students must show a logical progression from stage and stage 2 (in either order) to stage 3 Stage 2 Mol bromine = / 000 = AO2h Stage 3 Ratio oil : bromine : Simplest ratio = / : / = : 3 AO2h ence, 3 = bonds AO3 a M5 cannot be awarded unless working for M4 is shown 3 of 4

6 0 4 AO2h AO2h To gain 4 or 5 marks, students must show a logical progression from stage and stage 2 (in either order) to stage 3

14 MARK SEME AS hemistry Specimen paper 2 Section B In this section, each correct answer is awarded mark. Question Key AO 8 B AO2b 9 AOa 0 D AO2d 2 D AOb 3 B AOa 4 AOb 5 A AOb 6 D AOa 7 D AOa 8 AOa 9 AOa 20 B AOa 2 A AO3 2b 22 AO3 2b

Question Key AO 8 B AO2b 9 AOa 0 D AO2d 2 D AOb 3 B AOa 4

Paper 1 (7404/1): Inorganic and Physical Chemistry Mark scheme

Paper 1 (7404/1): Inorganic and Physical Chemistry Mark scheme AQA Qualifications AS Chemistry Paper (7404/): Inorganic and Physical Chemistry Mark scheme 7404 Specimen paper Version 0.6 MARK SCHEME AS Chemistry Specimen paper Section A 0. s 2 2s 2 2p 6 3s 2 3p 6