A-level CHEMISTRY (7405/1)

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1 A-level CHEMISTRY (7405/) Paper : Inorganic and Physical Chemistry Mark scheme Specimen paper

2 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk

3 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 0. 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 3 All stages are covered and the explanation of each stage is generally correct and virtually complete. 5 6 marks Level marks Level 2 marks Answer is communicated coherently and shows a logical progression from stage to stage 2 then 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. Answer is mainly coherent and shows progression from stage to stage 3. 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. 6 2 AOa 2 AO2a 2 AO2b Indicative chemistry content Stage : Electrons round P P has 5 electrons in the outside shell With 3 electrons from 3 fluorine, there are a total of 8 electrons in outside shell so 3 bond pairs, non-bond pair Stage 2: Electron pair repulsion theory Electron pairs repel as far as possible Lone pair repels more than bonding pairs Stage 3: Conclusions Therefore, tetrahedral / trigonal pyramidal shape With angle of 09(.5) decreased to 07 Level 0 0 marks Answer includes isolated statements but these are not presented in a logical order or show confused reasoning. Insufficient correct chemistry to gain a mark. 3 of 6

4 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 0.2 s 2 2s 2 2p 6 3s 2 3p 6 3d 7 AOa Allow correct numbers that are not superscripted 0.3 Too many electrons in d sub-shell / orbitals AO3 b 0.4 Tetrahedral (shape) 09.5 AO2a AO2a Allow 09 4 of 6

5 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 02. The number of protons increases (across the period) / nuclear charge increases AOa Therefore, the attraction between the nucleus and electrons increases AOa Can only score M2 if M is correct 02.2 S 8 molecules are bigger than P 4 molecules Therefore, van der Waals / dispersion / London forces between molecules are stronger in sulfur AOa AOa Allow sulfur molecules have bigger surface area and sulfur molecules have bigger M r 02.3 Sodium oxide contains O 2 ions These O 2 ions react with water forming OH ions AO2c AO2c O 2 + H 2 O 2OH scores M and M P 4 O 0 + 2OH 4PO H2 O AO2d 5 of 6

6 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN Question Marking Guidance Mark AO Comments 03. The ions in the ionic substance in the salt bridge move through the salt bridge To maintain charge balance / complete the circuit AOb AOb 03.2 F AO3 a 03.3 E Ɵ SO 2 4 / SO 2 < E Ɵ Br 2 /Br AO3 a Allow correct answer expressed in words, eg electrode potential for sulfate ions / sulfur dioxide is less than that for bromine / bromide (V) AO2d 03.5 A fuel cell converts more of the available energy from combustion of hydrogen into kinetic energy of the car / an internal combustion engine wastes more (heat) energy AO3 b 6 of 6

7 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 04. Bonds broken = 2(C=O) + 3(H H) = H H Bonds formed = 3(C H) +(C O) + 3(O H) = AOb Both required 49 = [ (H H)] [ ] 3(H H) = [ ] = 450 AOb Both required H H = 483 (kj mol ) AOb Allow 483.3(3) 04.2 Mean bond enthalpies are not the same as the actual bond enthalpies in CO 2 (and/or methanol and/or water) AOb 04.3 The carbon dioxide (produced on burning methanol) is used up in this reaction AO3 b mol of gas form 2 mol At high pressure the position of equilibrium moves to the right to lower the pressure / oppose the high pressure This increases the yield of methanol AO3 b AO3 b 04.5 Impurities (or sulfur compounds) block the active sites AOb Allow catalyst poisoned 7 of 6

8 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 04.6 Stage : moles of components in the equilibrium mixture CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) Extended response question Initial moles Eqm moles ( 0.86) = 0.4 ( ) = Stage 2: Partial pressure calculations Total moles of gas = 2.28 Partial pressures = mol fraction p total p CO2 = mol fraction p total p H2 = mol fraction p total = /2.28 = 30.7 kpa = /2.28 = 92. kpa M3 is for partial pressures of both reactants Alternative M3 = pp CO2 = pp H2 = p CH3OH = mol fraction p total = /2.28 = 88.6 kpa p H2O = mol fraction p total = /2.28 = 88.6 kpa M4 is for partial pressures of both products Alternative M4 = Stage 3: Equilibrium constant calculation K p = p CH3OH p H2O / p CO2 (p H2 ) 3 pp CH3OH = pp H2O = Hence K p = / 30.7 (92.) 3 = = AOb Answer must be to 2 significant figures Units = kpa 2 8 of 6

9 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 05. S = = 80 J K mol AOb G = H T S AOa = ( 80) 000 AOb = +45. kj mol AOb Units essential 05.2 When G = 0, H = T S therefore T = H/ S = / 80 = 272 (K) AOb AOb Mark consequentially to S in 5. 9 of 6

10 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 05.3 Diagram marks δ- δ+ Diagram of a molecule showing O H bond and two lone pairs on each oxygen AO2a Labels on diagram showing δ+ and δ- charges AO2a Allow explanation of position of δ+ and δ- charges on H and O Diagram showing δ+ hydrogen on one molecule attracted to lone pair on a second molecule AO2a Explanation mark Hydrogen bonding (the name mentioned) is a strong enough force (to hold methanol molecules together in a liquid) AO2a 0 of 6

11 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 06. Burette AO3 b Because it can deliver variable volumes AO2g 06.2 The change in ph is gradual / not rapid at the end point AO3 a An indicator would change colour over a range of volumes of sodium hydroxide AO3 a Allow indicator would not change colour rapidly / with a few drops of NaOH 06.3 [H + ] = 0 ph = K w = [H + ] [OH ] therefore [OH ] = K w / [H + ] Therefore, [OH ] = 0 4 / = (mol dm 3 ) AO2h AO2h AO2h Allow (mol dm 3 ) 06.4 At this point, [NH 3 ] = [H + ] Therefore K a = [H + ] 2 [NH + 4 ] [H + ] = = K a = ( ) 2 /2 = (mol dm 3 ) Allow (mol dm 3 ) 06.5 When [NH 3 ] = [NH 4 + ], K a = [H + ] therefore log K a = log [H + ] AO2h Answer using alternative value Therefore ph = log 0 ( ) = 9.50 AO2h M2 ph = log 0 ( ) = 8.32 Allow consequential marking based on answer from 6.4 of 6

12 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 07. Y AO3 a 07.2 X AO3 a 07.3 Jump in trend of ionisation energies after removal of fifth electron Fits with an element with 5 outer electrons (4s 2 3d 3 ) like V AO2b 07.4 Explanation: Two different colours of solution are observed AO2g Because each colour is due to vanadium in a different oxidation state AO2g 2 of 6

13 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 07.5 Extended response Stage : mole calculations in either order Moles of vanadium = /000 = 4.00 x 0 2 Moles of SO 2 = pv/rt = ( )/( ) = AO2d AO2d Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. Stage 2: moles of electrons added to NH 4 VO 3 When SO 2 (sulfur(iv) oxide) acts as a reducing agent, it is oxidised to sulfate(vi) ions so this is a two electron change AO2d Moles of electrons released when SO 2 is oxidised = = AO2d Stage 3 : conclusion But in NH 4 VO 3 vanadium is in oxidation state 5 AO2b mol vanadium has gained mol of electrons therefore mol vanadium has gained / = mol of electrons to the nearest integer, so new oxidation state is 5 =4. AO2d 3 of 6

14 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 08. An electron pair on the ligand Is donated from the ligand to the central metal ion 08.2 Blue precipitate Dissolves to give a dark blue solution [Cu(H 2 O) 6 ] NH 3 + Cu(H 2 O) 4 (OH) 2 + 2NH 4 Cu(H 2 O) 4 (OH) 2 + 4NH 3 [Cu(NH 3 ) 4 (H 2 O) 2 ] OH + 2H 2 O AOa AOa AOb AOb AO2d AO2d 08.3 [Cu(NH 3 ) 4 (H 2 O) 2 ] H 2 NCH 2 CH 2 NH 2 AO2b [Cu(H 2 NCH 2 CH 2 NH 2 ) 2 (H 2 O) 2 ] NH Cu N bonds formed have similar enthalpy / energy to Cu N bonds broken AO3 b And the same number of bonds broken and made AO3 b particles form 5 particles / disorder increases because more particles are formed / entropy change is positive AO2e Therefore, the free-energy change is negative AO2e M2 can only be awarded if M is correct 4 of 6

15 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 09. Start a clock when KCl is added to water AO3 2b Record the temperature every subsequent minute for about 5 minutes Plot a graph of temperature vs time Extrapolate back to time of mixing = 0 and determine the temperature AO3 2b AO3 2a AO3 2a Allow record the temperature at regular time intervals until some time after all the solid has dissolved for M Heat taken in = m c T = = 28.6 J AO2h Max 2 if 4.6 C used as T Moles of KCl = 5.00/74.6 = AO2h Enthalpy change per mole = +28.6/ = J mol AO2h = +6.8 (kj mol ) AOb Answer must be given to this precision 09.3 H solution = H lattice + H(hydration of calcium ions) + 2 H(hydration of chloride ions) H lattice = H solution H(hydration of calcium ions) 2 H(hydration of chloride ions) H lattice = 82.9 ( ) = (kj mol ) 09.4 Magnesium ion is smaller than the calcium ion AO2a Therefore, it attracts the chloride ion more strongly / stronger ionic bonding AO2a 5 of 6

16 MARK SCHEME A-LEVEL CHEMISTRY 7405/ - SPECIMEN 0. Q is calcium or magnesium AO3 b Mark this question independently bromide AO3 b R is aluminium AO3 b chloride AO3 b S is iron(iii) AO3 b sulfate AO3 b 0.2 Ba 2+ + SO 4 2 BaSO 4 AOa [Fe(H 2 O) 6 ] OH Fe(H 2 O) 3 (OH) 3 + 3H 2 O AOa 2[Fe(H 2 O) 6 ] CO 3 2 [Fe(H 2 O) 6 ] Cl 2Fe(H 2 O) 3 (OH) 3 + 3H 2 O + 3CO 2 [FeCl 4 ] + 6H 2 O AOa AOa 6 of 6

A-level Chemistry (7405/1)

A-level Chemistry (7405/1) A-level Chemistry (7405/) Paper : Inorganic and Physical Chemistry Specimen 205 v0.5 Session 2 hours Materials For this paper you must have: the Data Booklet, provided as an insert a ruler a calculator.