GE Edexcel GE hemistry (8080, 9080) 6244/01 Summer 2005 Mark Scheme (Results) Edexcel GE hemistry (8080, 9080)
IGNE state symbols in all equations 1. (a) (i) (Magnesium oxide is) ionic / electrovalent (Sulphur dioxide is) covalent NOT giant / dative (ii) MgO + 2 O Mg(O) 2 MgO + 2 O Mg 2+ + 2O ontains/produces O - ions This mark is dependent on an O in the product of the equation SO 2 + 2 O 2 SO 3 SO 2 + 2 2 O 3 O + + SO 3 SO 2 + 2 O + + SO 3 SO 2 + 2 O 2 + + SO 3 2 QW* ontains/produces + / 3 O + If 2 SO 4 formed in equation, can score last mark (iii) (Silicon dioxide) giant covalent/ giant atomic/ giant molecular/ macromolecular Strong/ covalent bonds (have to be) broken (for reaction) Reference to Si=O or van der Waals forces scores (0) for this mark (4 marks) (so reactants are) kinetically stable / activation energy too high (for reaction) / not enough energy released in bond formation to overcome energy required in bond breaking IGNE any references to reaction mechanism (b) (i) [Al( 2 O) 6 ] 3+ + 2 O / [Al( 2 O) 5 O] 2+ + 3 O + AEPT one more deprotonation of aluminium ions i.e.[al( 2 O) 4 (O) 2 ] + (3 marks) (ii) Sil 4 + 2 2 O SiO 2 + 4l AEPT products SiO 2.x 2 O / Si(O) 4 in a balanced equation (c) (i) (The trend is) increasing stability of the +2 state relative to the +4 state (or instability of +4 etc) i.e. answer must be a comparison of oxidation states (ii) (Tin(II) chloride will) reduce / be oxidised to Sn(IV) (Fe 3+ goes) to Fe 2+ orrect equation e.g. 2Fe 3+ +Sn 2+ 2Fe 2+ +Sn 4+ scores both marks i.e. species balancing. Do NOT penalise an unbalanced equation if 1 st two marks are awarded (Lead(II) chloride has) no reaction (3 marks) Total 15 marks 1
2. (a) (i) (The first electron affinity) is the energy/ enthalpy/ heat change when 1 electron is added to each atom in 1 mole - must not imply endothermic process e.g. energy required of gaseous atoms energy change per mole (for) A(g) + e A (g) (ii) orrect labelling of a(s) to a 2+ (g) (+193, +590, +1150) orrect labelling of I 2 (s) to 2I - (g) (+214, 2 x EA) orrect labelling of lattice energy and f of ai 2 (s) (-2074, -534) Labelling can be done with symbols, words or numbers (iii) Mark consequentially on their labels in (ii) f = a of calcium + 1 st IE calcium + 2 nd IE calcium + 2 x a iodine + 2 x EA I(g) + LE ai 2 (s) -534 = +193+590+1150+2x107+ 2 EA + (-2074) EA = ½(2074-534-193-590-1150-214) = -303.5 / -304 (kj mol -1 ) Other possible answers: One EA and +107on cycle gives EA = -500 (2) One EA and +214 on cycle, but 2EA shown in working gives EA = -303.5 / -304 (2) One EA and +214 on cycle but EA shown in working gives EA = -607 (3 marks) 2
(b) QW* (i) Potassium ion / K + larger than a 2+ Must not refer to atoms K + smaller charge than a 2+ Must not refer to atoms, but AN say potassium has a smaller charge ( than calcium ) harge density of K + is less than charge density for a 2+ without explanation is worth out of these 1 st two marks Less attraction between (K + and I ) ions NOT just weaker bonds AEPT reverse argument IGNE references to extent of covalency (3 marks) (ii) Potassium ion / K + less polarising (than a 2+ ) KI (close to) 100 % ionic / no covalent character ai 2 partially/ significantly covalent orrect description of anion polarisation in ai 2 NOT just distortion of anion (3 marks) Total 13 marks 3
3. (a) Ethylmagnesium bromide or formula, or any other halide NOT 2 5 BrMg, Dry ether/ ethoxyethane Followed by hydrolysis/ acid/ water Grignard reagent/ named reagent with incorrect alkyl group scores (0) for reagent but can score both condition marks. If halogenoalkane given as reagent, can score 1 st mark if Mg included under conditions. (b) (i) Observation effervescence/ bubbles/ fizzing NOT gas evolved (3 marks) (ii) 2 2 5 OO + Na 2 O 3 2 2 5 OONa + O 2 + 2 O Observation steamy/ misty/ white fumes NOT smoke 2 5 OO + Pl 5 2 5 Ol + POl 3 + l (c) Reagents potassium dichromate ((VI)) / K 2 r 2 O 7, sulphuric acid/ 2 SO 4 / hydrochloric acid/ l but conseq. on an oxidising agent ALLOW acidified potassium dichromate / + and r 2 O 7 2- (2) ALLOW acidified dichromate ions (2) Acidified dichromate (without ion) scores just AEPT Potassium manganate(vii) / potassium permanganate/ KMnO 4 / Tollens * / Fehling s* Acidified /alkaline*/neutral * need to acidify to liberate free acid for 2 nd mark (d) (i) Reagent (any one of) N N or KN KN N ondition (to match) and KN (buffered between) p between 6 and 9 + acid / + NOT excess + base / O NOT excess (2) Type of reaction Nucleophilic addition - both words needed (ii) Reagent ondition (any one of) (to match) ydrogen Pt / Ni / Pd (catalyst) IGNE ref to temp. Sodium (in) ethanol Lithium aluminium hydride dry ether/ ethoxyethane Sodium borohydride (in) aqueous/ water/ ethanol/ methanol Type of reaction Reduction AEPT redox / hydrogenation (not addition) AEPT nucleophilic addition if metal hydrides used (3 marks) (2) (3 marks) 4
(e) (i) O N 3 O O 3 ALLOW O N 3 If O represented as O, penalise once only (ii) O + - N 3 l + N 3 l - N 3 l O + N 3 + O N 3 NOT N 3 l O + If show all bonds in N 3, + charge must be shown on N atom ie + N 5
(f) Optical NOT stereo 2 N 2 2 N 2 O 2 3 3 2 O 2 N 2 ALLOW - 2 5 for - 2 3 MUST show the two as object and mirror image e.g. are acceptable (2) but NOT must not be bonded to in O group Near-miss molecule plus mirror image The two solid lines in 3D structure must not be at 180 (3 marks) Total 21 marks 6
4. (a) (i) K p = p(o 2 ) allow without brackets, IGNE p[ ] (ii) 1.48 (atm) Penalise wrong unit Answer is consequential on (a)(i) e.g. 1 must have atm -1 1.48 (b) (i) K p = p(l 2 ) x p(no) 2 allow without brackets, penalise [ ] (p(nol)) 2 (ii) 2NOl 2NO + l 2 Start 1 0 0-0.22 +0.22 +0.11 eq moles 0.78 0.22 0.11 total moles of gas 1.11 mole fractions above values 1.11 0.7027 0.1982 0.09910 partial pressure / atm above values x 5.00 3.51 0.991 0.495 K p = 0.495 atm x (0.991 atm) 2 (3.51 atm) 2 = 0.0395/ 0.0394 atm range of answers 0.0408/0.041 0.039/0.0392 NOT 0.04 AEPT 2 S.F orrect answer plus some recognisable working (5) Marks are for processes Equilibrium moles Dividing by total moles Multiplying by total pressure Substituting equilibrium values into expression for K p alculating the value of K p with correct consequential unit. (iii) As the reaction is endothermic stand alone the value of K p will increase (as the temperature is increased) consequential on 1 st answer (if exothermic (0) then K p decreases ) For effect on K p mark, must have addressed whether reaction is endothermic or exothermic (5 marks) (iv) Because (as the value of K p goes up), the value of pl 2 x (pno) 2 /(pnol) 2 (the quotient) must also go up and so the position of equilibrium moves to the right stand alone But mark consequentially on change in K in (iii) If position of equilibrium moves to right so K p increases (max 1) IGNE references to Le hatelier s Principle Total 12 marks 7
5. (a) 3 OO labelled as base and linked to 3 OO 2 + labelled (conjugate) acid 2 SO 4 labelled acid and linked to SO 4 - labelled (conjugate) base If acids and bases correct but not clearly or correctly linked 1 (out of 2) Just link but no identification of acids and bases (0) (b) (i) (p) more than 7/ 8-9 Indicator: phenolphthalein ALLOW thymolphthalein thymol blue (mark consequentially on p) Mark consequentially on p but if p7 do not allow either methyl orange or phenolphthalein QW* (ii) As O - / base removes + ions / neut is per mole of 2 O produced / + + O - = 2 O the equilibrium shifts to the right and so all the ethanoic acid reacts (not just 1% of it) Endothermic (O) bond breaking is compensated for by exothermic hydration of ions for 3 OO + 2 O 3 OO - + 3 O + = +2 kj mol -1 / almost zero / very small neut [ 3 OO] = +2 + neut [l] the same (for both acids) neut is per mole of 2 O produced (heat) energy required for full dissociation (of weak acid) so neut slightly less exothermic (for weak acid) (iii) [ + ] 2 = K a [ 3 OO] = 1.74x10-5 x 0.140 = 2.44x10-6 [ + ] = 0.00156 (mol dm -3) p = 2.81 consequential on [ + ] but not p>7 AEPT 2.80/2.8 (answers to 1 or 2 dp) (3 marks) The assumptions are two from: [ + ] = [ 3 OO - ] - this mark can be earned from working / negligible [ + ] from ionisation of water (iv) [ 3 OO] = 0.140 [ + ] 0.140 (mol dm -3 ) / ionisation of acid negligible solution at 25 o 1.74 x 10-5 = [ + ] [salt] [acid] max (2) (4 marks) [ + ] = 1.74 x 10-5 x 0.070 = 1.22 x 10-5 0.100 p = 4.91 / 4.9 / 4.92 NOT 5 Max 2 if 0.140 / 0.200 is used (3 marks) Total 14 marks TOTAL F PAPER: 75 MARKS 8