1994 A-level Chemistry paper II marking scheme 1994 AL CHEM P II Page 1

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1 1994 A-level hemistry paper II marking scheme 1.(a)(i) P A V A =n A RT P B V B = n B RT P A /P B = n A /n B (V B /V A ) = 1/2(1/2) = 1/4 or P A = 1/4P B or P B = 4P A (1m) (ii) P(O 2 ) < P A Gas pressure is produced by the collision of gas molecules on wall of container. At low pressure, the attraction / intermolecular force / van der Waal's force between O 2 molecules is a dominated force, so the collision force on the wall is reduced. (b) FeNS 2+ (aq) ions absorb visible light in electromagnetic spectrum, hence FeNS 2+ (aq) can be determined by measuring the absorbance of the solution. alibrate a colorimeter by measuring the absorbance of several solutions which contained known concentrations of FeNS 2+ (aq) [1/2m + 1/2m] (In these solutions, the NS - ions are in excess, so that Fe3+ ions present can be considered as existing in the complex form, FeNS 2+ ) By mixing comparable volumes of standard Fe 3+ (aq) and NS - (aq) solutions and measuring the [FeNS 2+ (aq)] formed with the colorimeter, [Fe 3+ (aq)], [NS - (aq)] and [FeNS 2+ (aq)] at equilibrium can be found. [1/2m + 1/2m] ence, the equilibrium constant Kc = [FeNS 2+ ] / ([Fe 3+ (aq)][ns - (aq)]) can be determined. (c)(i) The graph of log [] versus t is not a straight line order 1 [1/2m] The graph of 1/[] versus t is a straight line order = 2 [1/2m] [2m for the two graphs] (ii) For second order reaction, 1994 AL EM P II Page 1

2 1/[] = 1/[] o - kt From the slope of graph, k=1.997x10-2 (mol dm -3 ) s -1 [1m for ans.; 1m for unit] (1.95 x 10-2 to 2.05 x 10-2 (mole dm -3 ) -1 s -1 ) (iii) Arrhenius Equation ln k = - E a /RT ln k = - E a /R(1/298)... (1) ln k + ln 2 = - E a /R(1/306)...(2) (2) - (1) ln 2 = E a /R(1/298-1/306) E a = 65.7 kj [1/2m for answer; 1/2m for unit] 2.(a)(i) 298 = (-285.9) - (-75.0) = kj [1/2m for answer; 1/2m for unit] (ii) 4 (g) + 2O 2 (g) 2 2 O(g) + O 2 (g) = kj -) 4(g) + 2O 2(g) 2 2O(l) + O 2(g) = kj 2 2 O(l) 2 2 O(g) = 88.6 kj vaporization ( 2 O) = kj (iii) comb. [(diamond)] = kj mol -1. comb. [(graphite)] = kj mol -1. graphite is relatively more stable allotrope onversion from diamond to graphite involves the rearrangement atoms in giant covalent network would require a high activation energy. conversion does not occur at room temperature. (b)(i) In 3 PO 4, the intermolecular force is -bond. The structure of 3 PO 4 is O / O=P-O- [1/2m] \ O (or 3 PO 4 is capable of forming 3--bonds with its neighbours 1/2m) With strong intermolecular forces, 3 PO 4 has a high viscosity [1/2m] (ii) In ice the intermolecular attraction is -bond, each 2 O molecule is tetrahedrally surrounded by 4 other 2 O molecules or (.N. = 4) [1/2m + 1/2m] i.e. an open structure [1/2m] In liquid water, the molecules are packed closer together water has a higher density than ice [1/2m] For the endothermic reaction, 2 O(s) 2 O(l) Increase in pressure, shifts the equilibrium to the right m.p. of ice decreases with increases in pressure (c)(i) [1m for labelling the two axes; 1/2m for the line drawn] Mole fraction of F = 4/(4+1) = 4/5 [1/2m] From the graph, V.P. of mixture = 1.8P o or, by calculation P E = 0.2 P o P F = 0.8(2P o ) = 1.6P o V.P. of mixture = 1.8P o (1m) 1994 AL EM P II Page 2

3 (ii) If deviation is negative P < 1.8P o If deviation is positive P > 1.8P o Explanation: In case of negative (positive) deviation, the interaction between E and F is stronger (weaker) than the average of interaction between E and E and between F and F. The molecules have a smaller (greater) chance to escape, thus resulting in a lower (higher) vapour pressure. (Max. 2m, if only one of the cases is discussed.) 3.(a)(i) From the graph When [Ag + (aq)] = 1.0 M, e.m.f. of cell = V Standard electrode potential of Ag(s) Ag + (aq) half-cell = (0.34) V = V [1/2m for answer, 1/2m for unit] (ii)(i) From the graph, when e.m.f. = V, log 10 [Ag + (aq)] = -3.2 (-3.1 to -3.3) [1/2m] i.e. [Ag + (aq)] = 6.31 x 10-4 mol dm -3 [1/2m] (7.94 x 10-4 to 5.01 x 10-4 mol dm -3 ) (II) K sp [AgBrO 3 ] = [Ag + (aq)][bro - 3 (aq)] = (6.31 x 10-4 )(0.1) = 6.31 x 10-5 mol 2 dm -6 (5.01 x 10-5 to 7.94 x 10-5 mol 2 dm -6 ) (b)(i) The dissociation of water 2 2 O(l) 3 O + (aq) + O - (aq) is an endothermic process / = +ve Increase in temperature, the equilibrium shifts to the right [1/2m] and hence [ 3 O + (aq)] increases. In pure water, [ 3 O + (aq)] at 323 K is higher than that at 298 K, p of pure water at 323 K is less than 7.0. [1/2m] (ii) Acid-base indicators are weak acids/bases. The dissociation of which can be represented by: In(aq) + 2 O(l) 3 O + (aq) + In - (aq) The colour of an indicator depends on the relative concentrations of the acidic form. In and the alkaline form In-(aq) which are of different colours. [1/2m] The dissociation constant K i for different acid-base indicators are different, [1/2m] thus they change colour over different p range. The p range of methyl orange is below 7, [1/2m] while that of phenolphthalein is above 7, [1/2m] at p 7, methyl orange shows its alkaline colour, while phenolphthalein shows its acidic colour. (c)(i) K a = [ + ][PrOO - ] / [PrOO] Assuming [ + ] = [PrOO - ], [PrOO] = 0.1 M 1.5 x 10-5 = [ + ] 2 / 0.1 [+] = x 10-3 M p = 2.91 (ii) Assuming [PrOO] = 0.05 M [PrOO - ] = 0.05 M 1.5 x 10-5 = [ + ] x 0.05/0.05 [ + ] = 1.5 x 10-5 M p = 4.82 ( > 2 d.p. or < 2 d.p. -1/2m) (iii) After addition of 1.0 x 10-3 moles solid NaO [PrOO] = x 10-3 } [PrOO - ] = x 10-3 } 1.5 x 10-5 = [ + ](0.051) / (0.049) [+] = 1.44 x 10-5 M p = 4.84 (> 2 decimal places or < 2 decimal places -1/2m) 4.(a) eating conc. l in the presence of manganese(iv) oxide [1/2m + 1/2m] MnO 2 (s) + 4l(aq) Mnl 2 (aq) + 22O(l) + l 2 (g) or add conc. l to solid KMnO 4 (1/2m + 1/2m) 10l + 2MnO l O + 2Mn 2+ (1m) Pass the l 2 produced through cold water to remove l, dry the gas by passing through conc. 2 SO 4 and collect by downward delivery [1/2m + 1/2m] 1994 AL EM P II Page 3

4 The experiment should be carried out in a fume cupboard. (b) The anomalous boiling point of F is due to extensive -bonding between F molecules [1/2m + 1/2m] (diagram showing -bonds in F) (Accept alternative answers such as in l, Br, I there are no extensive -bonds.) (c) ighest O.S. of I = +7 Example: IF 7, IO - 4 (d)(i) Nal + 2 SO 4 (l) NaSO 4 + l (ii) NaI + 2 SO 4 I + NaSO 4 8I(g) + 2 SO 4 (l) I 2 (s) + 2 S(g) O(l) or 2I(g) + 2 SO 4 (l) I 2 (s) + SO 2 (g) O(l) (1m) I, being a stronger reducing agent than l, will be oxidized by conc. 2 SO 4 to give I 2. [1/2m + 1/2m] (e)(i) N 2 N 2 : IO - 3 = 1 : 1 In the reduction of IO - 3, O.S. of I decreases from +5 (in IO - 3 ) to +1 (in Il) [1/2m] In the oxidation of N 2 N 2, 4 moles of e - should be released from each mole of N 2 N 2 [1/2m] In 1 mole of N2N2 there are 2 moles of N atoms. O.S. of N in the reaction product is 0. [1/2m] The colourless gas is probably N2 [1/2m] (ii) IO l + 4e - Il O +) N 2N 2 Il + N O IO l + N 2 N 2 Il + N O Or, IO l + 4e - Il O (1m) +) N 2N 2 Il + N O (1m) IO N 3 N l Il + N O + + (1m) 5.(a)(i) Elements with valence electrons in d-orbitals [1/2m + 1/2m] (ii) each (iii) Mn 2+ with half-filled 3d orbitals is relatively stable. Removing e - disturbs this stable configuration: relative high energy is required., Mn 2+ is not readily oxidized to Mn 3+. Fe 2+ with an electronic configuration of d 6, after removal of an e- from the outermost shell will attain the relatively stable e- configuration with half-filled 3d orbitals., the Fe 2+ ions are easily oxidized to Fe 3+. (iv) Variable oxidation states of d-block elements allow electron transfer between reactants and products by means of catalyst changing betwen two oxidation states. Any one example of transition metals or its compound as catalyst V 2 O 5 e.g. 2SO 2 + O 2 2SO 3 (example must involve redox process, accept aber process, but not accept hydrogenation of alkene.) (v) (b)(i) MnO - 4 is unstable in the presence of sunlight / undergoes photodecomposition; it will decompose to form a brown solid / MnO 2 (ii) obalt(ii) chloride, in anhydrous state is blue [1/2m] in colour. After absorption of water / forming complex with water, it changes colour to pink. [1/2m] (iii) In aqueous solution, Fe 3+ exists as aquo-complex. Addition of O - causes the formation of a brown ppt. [Fe( 2 O) 6 ] O - Fe(O) 3 (s) O(l) In the presence of excess N -, Fe 3+ forms a very stable complex [Fe(N) 6 ] 3-, concentration of free Fe 3+ ions is lowered addition of 2M NaO cannot cause [Fe 3+ ][O - ] 3 to exceed the K sp of Fe(O) AL EM P II Page 4

5 6.(a) Sodium hydrogencarbonate in the self-raising flour undergoes thermal decomposition to give O2 gas. 2NaO 3 (s) Na 2 O 3 (s) + O 2 (g) + 2 O(g) [1/2m for product; 1/2m for balanced equation] The O 2 gas produced causes the cakes to rise. (b) The small size of Li + ion and the relatively large O - 3 ions do not favour the formation of a stable crystal lattice. (accept explanation in terms of the polarization of O - 3 ion by Li + ion) The highly exothermic hydration energy of Li + ion, causes lithium hydrogencarbonate to exist in solutions only. (c) 2PbO 2 (s) 2PbO(s) + O 2 (g) [2m] (Award 1 mark for an unbalanced equation) (d)(i) Tin(II) chloride hydrolyses in water to give insoluble Sn(O)l [1/2m + 1/2m] Snl O Sn(O)l + l thus the solution appears cloudy. Addition of + shifts the equilibrium to the left, the solution becomes clear. Addition of O - forms Sn(O) 2-4 (Since Sn(O) 2 in amphoteric), thus the solution becomes clear. (ii) ydrolysis of Snl 4 involves the attack e - pairs in 2 O: on the vacant lower energy d orbitals of Sn In, there is no low energy vacant 3d orbitals for the incoming water molecule No low energy hydrolysis reaction path for l 4. (e) Reaction with acidified KMnO 4 [1m for method] Snl 2 decolorizes the purple colour of MnO 4 - while Snl 4 does not. [2m for observation] or pass 2 S(g) into aqueous solutions of the two chlorides (1m) Snl 2 gives a brown ppt. while (1m) Snl 4 gives a yellow ppt. (1m) 7.(a) Structural isomerism - occurence of more than one structure for a given molecular formula. [1/2m +1/2m] Stereoisomerism - occurence of more than one configuration (different arrangements of groups in space) for a given structural formula. [1/2m + 1/2m] Examples to illustrate the two types of isomers. [1/2m + 1/2m] (b)(i) Empirical formula of G is 2 4 O Since, relative molecular mass of G is between 80 to 100 molecular formula of G is 4 8 O 2 G is a sweet smelling liquid it is an ester Possible structures of G (any 3 of the structure shown below): (ii) Structure possesses a chiral carbon. (any 3) [2m] 1994 AL EM P II Page 5

6 l l l l l (iii) J reacts with Pl 5 to give l J carries an -O group [1/2m] Strong oxidation of J gives, J is a 1,2-disubstituted aromatic compound. Mild oxidation of J gives a ketone, J is a 2 alcohol [1/2m] Possible structures of J Iodoform test / I 2 in NaO Observation: [2m] and will gives a pale yellow ppt. while will not 8.(a) (Any chemically reasonable transformation are acceptable.) (i) (ii) [3m] (iii) 2 2 O O (b)(i) e.g. 2 Br O O 2 NaO 2 _ 2 2 O O O /Ni conc. 2 2 SO o 2 [OUT] _ [3m] N _ 3 Br N Br N + Br [3m] (ii) No, N is a weak acid, it ionizes only slightly in water. In N(aq), [N - (aq)] is low, hence rate of SN2 reaction with R-Br is low.[2m] (c)(i) 4,5-dimethylhex-1-ene 1994 AL EM P II Page 6

7 (ii) propyl butanoate (iii) 4-hydroxyhexan-3-one 9.(a)(i) P: (ii) Q: (iii) R: (iv) (v) U & V: ( ) 3 OO - Na + & ( ) 3 2 O [2m] (b)(i) There are 4 - bonds and 1 = bond is an ethene molecule. The 's in ethene are sp 2 hybridized. The - bond is formed by overlapping of a sp 2 hybrid oribtal with an atomic orbital of. The = bond consisted of a sigma bond and a pi bond. These bonds are formed by the overlapping of two unhybridized p orbitals on 's respectively. Since for sp 2 hybridization, the hybrid orbitals are arranged on a triangular plane. Ethene molecule is planar and -- bond angles are 120 or = π bond (1/2m) by overlapping of P z orbitals (1/2m) - σ bond (1/2m) by overlapping of sp 2 orbitals (1/2m) shape - planar (1/2m); triangular or the angle is 120 (1/2m) 1994 AL EM P II Page 7

8 (ii) (iii) (iv) The structure of PV can be represented by [2m] The attraction between polymer chain of PV is dipolar attraction while that between PE polymers is van der Waal's force which is weaker when compared with former. ence PV is more rigid and more durable than PE. (v) Incineration of PV produces O 2, 2 O... and also l(g) which causes more serious pollution problem than the incineration of PE (no l is produced) AL EM P II Page 8