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ÿþM00/420/H(2)M INTERNATIONAL BACCALAUREATE BACCALAURÉAT INTERNATIONAL BACHILLERATO INTERNACIONAL MARKSCHEME May 2000 CHEMISTRY Higher Level Paper 2 12 pages  6  M00/420/H(2)M SECTION A 1. (a) (i) Endothermic/heat absorbed/energy absorbed / increase in enthalpy / needs (a lot of) energy [1] (ii) Kc is decreased / OWTTE. [1] Since heat is removed / since equilibrium moves to left / reverse reaction favoured. [1] (iii) N2 + O2 l 2NO (16-x) (16-x) 2x [1] . . [NO]2 Kc = [1] [N2][O2] [1] [NO] = 0.065 mol dm-3 (If candidate uses 1.6 instead of (1.6 - x) , which gives an answer of 0.066, award [2] unless it is specifically stated that an approximation has been made or 1.6 x .) (b) (i) O3 : 1st order plus reasonable attempt to justify (e.g. double [O3] doubles rate). [1] NO: 1st order plus reasonable attempt to justify (e.g. triple [O3] and triple [NO], rate is × 9.) [1] (Two correct orders but no reasoning, award [1].) Rate = k [O3][NO] (accept rate expression consistent with stated orders  [1] ECF principle) (ii) 0.66 ×10-4 = k × 3×10-6 ×10-6 k = 22 ×107 [1] dm3 mol-1 s-1 [1] . [2] (Apply error carried forward (ECF) from rate expression in (i). U-1 may be applied.) (iii) Rate (experiment 4) = 2.2 ×107 × 4.5×10-6 × 7.2 ×10-6 = 7.13×10-4 [1] 3 8 OR Rate (experiment 4) = Rate (experiment 3) × × = 713×10-4 . 2 10 (Apply ECF from rate expression in (i), must be experiment 4, units not required.) 2. (a) C3H8 has higher boiling point; [1] since it has greater Mr / greater number of electrons / greater number of carbons; [1] so greater intermolecular forces / more energy needed. [1] (b) CH3CH2OH has higher boiling point; [1] hydrogen bonding between molecules; [1] so more energy needed to separate molecules / so greater intermolecular forces. [1]  7  M00/420/H(2)M 3. (a) Acidic because H+ donor and basic because H+ acceptor. [1] Suitable equation OWTTE involving water [1] (b) O2- /oxide ion (allow O-2 ) [1] (c) Electrical conductivity OR pH meter (or indicator paper) [1] Strong: good conductor OR Strong: low pH [1] [1] Weak: poor conductor OR Weak: high pH (Allow full range indicator, do not allow litmus, use judgement on other methods.) [1] Ratio of moles = 60×0.2:40×0.15 (d) (acid:alkali) = 2 : 1 Acid is in excess and reacts to form salt. Moles acid:moles salt = 1:1 [1] [H+ ][A- ] [HA] Ka = / [H+] = Ka [1] [HA] [A-] pH = 486 [1] . (Weak acid calculation based on excess acid giving a pH of 3.04 [3 max]; Weak acid calculation based on total acid giving a pH of 2.89 [2 max]; Solution based on ½ neutralisation worth [4]; Working must be shown.)  8  M00/420/H(2)M 4. (a) Reducing agent donates/loses electrons / OWTTE. [1] (b) Current flow: Al Æ Ni [1] V Al , Ni H" (both) [1] (c) [2] 2Al + 3Ni2+ ’! 3Ni + 2Al3+ (Award [1] for correct species on correct sides of equation and [1] for correct coefficients, even if equation reversed.) E =+1.43V [2] (Award [1] each for sign and value. Allow -1.43 V if equation reversed  ECF principle. If signs not appropriate but value correct, award [1]. If E values added, award [0].) (d) Seconds = 2 ¥ 3600 OR 7200 [1] Coulombs = 8 ¥ 2 ¥ 3600 OR 57600 [1] 57 600 = "F OR 0.5970 [1] 96480 57 600 = "6 [1] 96480 ¥ 6 Answer = 0.09950 = 0.10(00) [1] (Correct answer with no working, [4 max].)  9  M00/420/H(2)M SECTION B 5. (a) (If lines are shown without H atom attached, penalise once only.) H H H H C C C O H 1-propanol / propan-1-ol (I) (do not accept propanol) H H H H H H H C C C H 2-propanol / propan-2-ol (II) H O H H H H H H C O C C H methoxyethane (III) (accept ether or alkoxyalkane) H H H 3 × [1] 3 × [1] (b) (I) partially [1] oxidised [1] to CH3CH2CHO [1], propanal [1] (if state CH3CH2COOH propanoic acid instead of propanal, award [1]) (II) oxidised (if not in (I), award [1]) to CH3COCH3 [1], propanone [1] (I) or (II): orange to green [2] (c) alkanols show bands above 3000 cm-1 [1] III is the choice [1] since it has C O( O) but no  O H [1] (d) A is I [1] 3 Hs in CH3, 2 Hs in adjacent CH2, 2 Hs in next CH2 , 1 H in OH [1] B is II [1] 6 Hs in the two CH3s, 1 H in CH, 1 H in OH [1] (e) I and II [1] [1] both give CH3CH = CH2 / CH3CHCH2 CH3CH = CH2 + H2 ’! CH2CH2CH3 OR CH3CH = CH2 + HBr ’! CH3CHBrCH3 (or CH3CH2CH2Br ) OR CH3CH = CH2 + H2O ’! CH3CHOHCH3 (or CH3CH2CH2OH ) OR CH3CH = CH2 + Br2 ’! CH3CHBrCH2Br Reagents [1] Product [1] CH3 H OR CH3CH = CH2 +high pressure, high temperature/catalyst ’! C C H H n idea of polymerisation [1] content of bracket [1]  10  M00/420/H(2)M 6. (a) (i) MgO ionic [1] SiO2 covalent [1] both giant structures [1] ionic bonds strong [1] covalent bonds strong [1] PO6 simple molecular [1] 4 SO2 covalent [1] weak intermolecular forces / bonds [1] [8] (ii) Oxide Solubility Acidic/Alkaline/Neutral Magnesium Soluble Alkaline Silicon Insoluble Neutral Phosphorus Soluble Acidic Sulphur Soluble Acidic 4 correct = [3] 4 correct = [3] 3 correct = [2] 3 correct = [2] 2 correct = [1] 2 correct = [1] max [6] MgO + H2O Æ Mg(OH)2 [1] [2] PO6 + 6H2O ’! 4H3PO3 (formula of acid [1], balanced [1]) 4 [1] SO2 + H2O Æ H2SO3 (Accept suitable ionised versions, e.g. Mg2+ + 2OH- instead of Mg(OH)2 , - HSO3 + H+ instead of HSO3 .) 2 [1] (b) Ti 1s22s22p63s2 3p64s2 3d2 / [Ar] 4s23d2 Variable valency / oxidation state / OWTTE [1] removal/sharing of several electrons [1] coloured compounds [1] splitting of d orbitals, electron transitions [1] complex compounds [1] accepting of electron pairs [1] catalytic activity [1] complex formation/change of valency/can easily be oxidised or reduced [1] any three [3] plus appropriate reason [3]  11  M00/420/H(2)M [1] 7. (a) (i) "H is positive Reaction is endothermic (because products are at higher energy) [1] Bonds in reactants must be stronger than those in products (because more energy must be added than is released). [1] [1] (ii) "G is negative because reaction is spontaneous [1] [1] "S is positive Since "H is positive, "S must be positive in order to make "G negative. [1] ("G = "H - T"S ) Products must be more disordered than reactants. [1] (iii) Known volumes of reactant solutions at the same temperature are mixed and temperature is monitored. [3] Mol of limiting reactant calculated from volume and known concentration. [1] q = "T ¥ mass of solution ¥ Cp [1] [1] "H = q mol-1 of limiting reactant Use of insulated reaction vessel [1] Stir the mixture [1] Note: [6] max which must include: (a) known concentration of one volume; (b) excess or equal reacting mols of second solution; (c) temperature change; (d) q = mass ¥ specific heat capacity ¥ "T (iv) If reactants do not react completely. [1] If container is not insulated adequately, heat will be gained from surroundings. [1] Insulate container sufficiently. [1] (v) Reaction becomes more spontaneous as T is increased [1] less spontaneous as T is decreased. [1] "G T"S term will become larger as T is raised so will become more negative. [1] T"S term will become smaller as T is lowered so "G will become less [1] negative (or even positive as +"H exceeds T"S ). (Accept arguments based on Le Chatelier s principle.) [2] "Hreaction = 2(A A) bond energy + B B bond energy- 4(A B) bond energy (b) (i) [1] for correct signs [1] for correct coefficients (Number of bonds should be clear.) (ii) Tabulated bond energies are average values and may differ from those in specific compounds. [1] The best agreement is achieved when few bonds are broken / specific bond energies are used / for gaseous reactions. [1]  12  M00/420/H(2)M .. .. .. .. .. 8. (a) (i) :N N: HN:H HN:N:H H:N::N:H : : .. .. .. H H H (Penalise missing lone pair once only) [1] each [4] (ii) Bond angles in HNNH will be slightly larger than those in H2NNH2. [1] 3 sets of electrons around the N atoms in HNNH (double bond, bond to H, lone pair) will be farthest apart at about 120 but the 4 sets in H2NNH2 will adopt a tetrahedral geometry with bond angles that are slightly less than 109 / OWTTE [2] (iii) N2  sp hybridisation, NH3  sp3 hybridisation, HNNH  sp2 hybridisation [1] each [3] (iv) HNNH has two isomers H H H N N N N H [1] each [2] nonpolar polar [1] each [2] (b) (i) H C O H O 4 sigma bonds 1 pi bond [2] sigma bonds lie directly between the bonded nuclei / sigma bond strong [1] pi bonds lie above and below the line between the nuclei / pi bond weaker [1] (ii) one longer than the other [1] [1] C == O shorter/extra e- pair makes bond shorter. (iii) C O bonds of same length [1] because of delocalisation / idea of resonance. [1] - (iv) O AND Ÿ H C H C Ÿ O - [1] [1] [2] (Negative charge omitted  no penalty, electrons on O omitted  [0].) Intermediate bonding or other sensible alternative statement. [1] (Accept 1½ bonds / À electrons spread across C O bonds.)

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