MARK SCHEME for the October/November 2007 question paper 9701 CHEMISTRY. 9701/04 Paper 4 (Theory 2), maximum raw mark 100

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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2007 question paper 9701 CHEMISTRY 9701/04 Paper 4 (Theory 2), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

2 Page 2 Mark Scheme Syllabus Paper 1 (a) (i) K a = [H + ][RCO 2 ]/[RCO 2 H] (ii) pk a = log 10 K a or logk a or log [H + ] 2 /[RCO 2 H] NOT ln; (b) (i) acid strength increases from no. 1 to no. 3 or down the table or as Cls increase due to the electron-withdrawing effect/electronegativity of chlorine (atoms) stabilising the anion or weakening the O-H bond NOT H + more available (ii) chlorine atom is further away (from O-H) in no. 4, so has less influence (iii) either: ph = ½ (pk a log 10 [acid]) or K a = 10 pka = x 10 3 = ½ ( ) [H + ] = (K a. c) = 3.55 x 10 4 = 3.4 (allow 3.5) ph = 3.4 ecf ( for correct expression & values; for correct working) [6] (c) (i) catalyst (ii) CH 3 CH 2 CO 2 H + Cl 2 CH 2 CHClCO 2 H + HCl (iii) nucleophilic substitution NOT addition/elimination (iv) M r (CH 3 CH 2 CO 2 H) = 74 M r (CH 2 CH(NH 2 )CO 2 H) = g should give 10 x 89/74 = g percentage yield = 100 x 9.5/12.03 = 79% ecf ( for correct answer) [5] (d) + NH 3 -CH(CH 3 )-CO 2 correct atoms Allow charges on H of H 3 N, and COO but not C-O-O correct charges [Total: 15]

3 Page 3 Mark Scheme Syllabus Paper 2 (a) solubility decreases (down Group II) lattice energy decreases solvation/hydration energy (of cation) decreases but more so than does lattice energy/is not able to overcome LE H soln becomes more endothermic/positive/less exothermic [max 4] (b) identities of A and B 2 x Mg(OH) 2 + H 2 C 2 O 4 MgC 2 O 4 (aq) + 2H 2 O (A) MgC 2 O 4 (aq) + Ca(NO 3 ) 2 Mg(NO 3 ) 2 + CaC 2 O 4 (s) (B) [max 3] (c) (i) (K sp =) [Mg 2+ ][OH ] 2 units are mol 3 dm 9 ecf from K sp (ii) (call [Mg(OH) 2 (aq)] = [Mg 2+ ] = x) K sp = 2 x = 4x 3 x = 1.71 x 10 4 mol dm 3 ecf (iii) less soluble because of the common ion effect or the equilibrium Mg(OH) 2 (s) Mg 2+ (aq) + 2OH (aq) is moved to the left [5] [Total: 12]

4 Page 4 Mark Scheme Syllabus Paper 3 (a) K = 22.4/39.1 = thus ratio is: 1 Cr = 29.8/52.0 = Cl = 20.3/35.5 = O = 27.5/16.0 = or KCrClO 3 (scores 2) (b) K 2 Cr 2 O 7 + 2HCl 2KCrClO 3 + H 2 O (c) (i) redox or oxidation (ii) E o data and half equations: Cr 2 O H + + 6e 2Cr H 2 O E o = 1.33 V Cl 2 + 2e 2 Cl E o = 1.36 V overall ionic equation: Cr 2 O Cl + 14H + 2Cr Cl 2 + 7H 2 O (iii) (dilution will) lower E o for Cr 2 O 2 7 /Cr 3+ or raise E o for Cl 2 /Cl or lower [Cl ] or [H + ] will shift equilibrium in eqn to the left hand side (iv) Br 2 /Br = V, so Cr(VI) would oxidise Br (easily) [6] [Total: 9]

5 Page 5 Mark Scheme Syllabus Paper 4 (a) CCl 4 is unreactive. (The rest react (with increasing vigour)) no d-orbitals or available/low-lying empty orbitals in carbon or unable to expand octet e.g. SiCl 4 + 2H 2 O SiO 2 + 4HCl (or GeCl 4 etc) or Si(OH) 2 Cl 2 or Si(OH) 4 (allow balanced equations for partial hydrolysis) [3] (b) (i) E(Cl-Cl) = 244 kj mol 1 ; 2 E(C-Cl) = 2 x 340 = 680 kj mol 1 H = 436 (kj mol 1 ) (ii) H = = +30 (kj mol 1 ) (iii) since reaction (ii) is endothermic, the +4 oxidation state is less stable or the +2 oxidation state is more stable (down the group) [3] [Total: 6] 5 (a) 2 MnO H 2 O H + 2 Mn H 2 O + 5 O 2 o (b) E cell = = (V) (c) (i) (as KMnO 4 is added), colour changed (from purple) to colourless NOT pink or effervescence/bubbles (of O 2 ) are produced at end-point, change is to (first) pink (ii) n(mno 4 ) = 0.02 x 15/1000 = 3 x 10 4 since H 2 O 2 : MnO 4 = 5:2, n(h 2 O 2 ) = (5/2) x 3 x 10 4 = 7.5 x 10 4 in 25 cm 3 [H 2 O 2 ] = 7.5 x 10 4 x 1000/25 = 3.0 x 10 2 mol dm 3 [4] [Total: 6]

6 Page 6 Mark Scheme Syllabus Paper 6 (a) (i) O Na C is allow ONa but no covalent O-Na bond (ii) amide, ester 2 x (iii) CO 2 or H 2 CO 3 or Na 2 CO 3 CH 3 NH 2 or CH 3 NH + 3 Cl OH (iv) H 3 O + and heat >80 or OH (aq) and heat >80 [7] (b) (i) Br 2 (aq) (or other suitable solvent) (ii) dilute/aqueous HNO 3 (c) (i) OH D is NO 2 (ii) tin/fe + HCl NOT LiAlH 4 (iii) OCOCH 3 NHCOCH 3 mark each side chain separately 2 x [4] (d) (i) (allow any orientation of groups) O H 2 O Cu H 2 N N H 2 OH 2 O penalise missing H on NH 2 (ii) [Cu(NH 3 ) 4 ] 2+ or [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ NOT [Cu(NH 3 ) 6 ] 2+ (iii) ligand substitution/exchange [3] [Total: max 15]

7 Page 7 Mark Scheme Syllabus Paper 7 (a) HNO 3 + H 2 SO 4 at C (or 60 C) not dilute or (aq) (b) 2H 2 SO 4 + HNO 3 2HSO 4 + H 3 O + + NO 2 + (allow equ. with only one H 2 SO 4, giving H 2 O) (c) G is CH 3 Cl H is CO 2 H reaction I: Cl 2 + AlCl 3 /accept other halogen carriers NOT aq, nor u.v. reaction II: KMnO 4 + H + NOT HCl nor HNO 3 reaction III: KMnO 4 + H + NOT HCl nor HNO 3 reaction IV: Cl 2 + AlCl 3 /accept other halogen carriers NOT aq, nor u.v. + both I + IV both II + III [4] [Total: 7]

8 Page 8 Mark Scheme Syllabus Paper 8 (a) (i) Two interlinked spirals or chains or strands woven round each other (ii) By hydrogen bonds between bases (b) Transcription (1)DNA/RNA/nucleic acid unravels (2)strand is used as a template (3)mRNA reads the sequence on this strand/ produces complementary strand Translation (4)mRNA binds to the ribosome (5)tRNA translates the codon from mrna (6)tRNA carries amino acids to ribosome/adds a.a. to chain [max 4] (c) (i) Disruption of the secondary/tertiary/quaternary/3d structure of the protein (could be answered in terms of bonds e.g. hydrogen bonds break) (ii) The covalent/peptide bonds in the (protein) chain are too strong (d) Energy is provided by the breakdown/hydrolysis of adenosine triphosphate (ATP) ATP (+ H 2 O) ADP + P i (+ energy) or in words ATP is produced during respiration/krebs cycle/oxidation of glucose, fats or proteins/ in mitochondria/adp is recycled [3] [Total: 11]

9 Page 9 Mark Scheme Syllabus Paper 9 (a) Suitable diagram showing origin of two energy states/or description Needs to mention applied magnetic field/electron transfer negates Indication that energy difference is in the radio frequency range Indication that frequency of absorption or gap between the 2 energy states depends on the nature of nearby atoms or the chemical environment of the 1 H [3] (b) They do not damage tissues/x-rays harmful/nmr of lower energy They are not obscured by bones/skeleton They can be tuned to examine particular tissues/tumours/organs/protons [max 2] (c) (i) M : M+1 = 100/(1.1n) n = = = = 4 carbon atoms Check for 1.1 in divisor, if missing, penalise (ii) Singlet at δ 2 suggests methyl adjacent to C=O Quartet at δ 4 suggests a CH 2 - group (adjacent to a methyl group) (allow OCH 2 - ) Triplet at δ 1.2 suggests a methyl group (adjacent to a CH 2 ) G is ethyl ethanoate (or structure)/if methyl propanoate given here cannot score first marking point [5] [Total: 10]

10 Page 10 Mark Scheme Syllabus Paper 10 (a) Iron is higher in the reactivity series than copper (owtte)/allow use of E o Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq) If conversion to Fe 3+ given, E cell is 0.38 (b) It does not require investment in machinery/labour It requires little energy accept it produces little/no pollution/noise Do not accept comparison with electrolytic method [max 2] (c) The process takes a long time/requires smaller workforce (d) (i) 0.75% is 7.5 kg in every tonne of ore Hence 150,000 tonnes of ore yield or 1,125 tonnes Cu tonnes 1125 x 0.6 = 675 tonnes (accept 680) (ii) 450 x 0.17 = 76.5 tonnes (accept 77) or 1125 x 0.17 = tonnes (accept 191) this is an ecf if 675 not in (i) (e) Aluminium is too high in the reactivity series/very reactive/aluminium forms bonds with oxygen which are too strong/aluminium ore doesn t exist as sulphide /Fe unable to displace Al (f) Control the ph (greater than ph 6.0) Bioremediation/growth of special plants (to remove heavy metals) Other reasonable suggestions such as displacement by a more reactive metal/ precipitation/ion exchange [Total: 9]

MARK SCHEME for the October/November 2006 question paper 9701 CHEMISTRY. 9701/04 Paper 4 (Theory 2), maximum raw mark 60

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2006 question paper 9701 CHEMISTRY 9701/04 Paper 4 (Theory 2),