2009 Nov (9746) Paper 1

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1 ov (9746) Paper 1 1. [09 P1 Q01 Alcohols] moles of sodium moles of gaseous hydrogen A Sugar, , has 5 groups which reacts with a(s) to give hydrogen gas. a + a + a (g) a a + 2. [09 P1 Q02 Mole] [92J P4 Q02] B 40 cm 3 2Mn 4 (aq) (aq) (aq) 2Mn 2+ (aq) (g) (l) 2 mol of mol of K mol 2 mol of KMn 4 52 mol of vol of KMn 4 ( 1000) cm cm 3 3. [09 P1 Q03 Atomic Structure] proton number electronic configuration A Since the ion is discharged at the cathode, it is a positively charged ion (a cation). Thus, the number of electrons as shown by the electronic configuration must be less than the proton number. ence, option A. 4. [09 P1 Q04 Ionisation Energies] A o To form Al 2+, 1 st I.E. + 2 nd I.E kj mol 1 To form o 2+, kj mol 1 5. [09 P1 Q05 Bonding] delocalised electrons All four ions contain double bond(s) and so, have delocalised π electrons. 6. [09 P1 Q06 Bonding] B It is a non-conductor of electricity. ry hydrogen chloride has no reaction with dry methylbenzene. ence exists as molecules in the resultant solution and so, is a non-conductor of electricity. 7. [09 P1 Q07 Gases] ( ) Pa Using ideal gas equation, pv nrt where p is in Pa, V is in m 3, and T is in K. 3 p nrt ( ) V 3 Pa [09 P1 Q08 Energetics] [03 P1 Q09] 120 o carbonate 109 o 120 o ethanoate 120 o phenoxide nitrate alcium ions have a lower enthalpy change of hydration than magnesium ions. Mg 2 4 is soluble in water but not a 2 4 suggests that a 2 4 has a lower K sp than Mg 2 4. sol hyd latt Since a 2+ is bigger (in size) than Mg 2+, it has a lower (less exothermic) hyd and smaller (less exothermic) lattice energy, latt. owever, hyd decreases much more than latt (which remains almost unchanged due to the large size of ions). ence, sol is less exothermic for a 2 4, which accounts for it being insoluble in water. 120 o

2 [09 P1 Q09 Entropy] 12. [09 P1 Q12 hemical Equilibria] S G B + B p In the purification process, pure water is separated from that contaminated with impurities (other molecules and ions) and so, there is an increase in orderliness; i.e. S < 0. Since G T S and is zero, G and S, therefore, must have opposite signs. 10. [09 P1 Q10 Entropy] A 66 J K 1 When water freezes, 6.0 kj mol 1. At the freezing point, equilibrium exists and so, G 0. Since G T S 6.0 S T J K 1 mol 1 (0+ 273) mol of 2 m Mr 54 3 mol entropy change 3 ( 22.0) 66 J K [09 P1 Q11 Electrochemical ells] Y only e 2 E o 0.00 V Zn e Zn E o 0.76 V o o o E cell E R E L (0.00 ( 0.76)) V V X: Increase in [Zn 2+ ] shifts the position of o equilibrium to the right and so, E L becomes o less negative. ence, E cell < V. Y: Increase in [ + ] shifts the position of o equilibrium to the right and so, E R becomes o more positive (i.e. E R > 0 V). o ence, E cell > V. X 2 (g) 2X(g) +ve As temperature increases, the position of equilibrium shifts to the right favouring the forward endothermic reaction so as to remove some of the extra heat. As more X 2 (g) decomposes to give X(g), pressure increases (since 1 mol of X 2 gives 2 mol of X ). ence, graph B. 13. [09 P1 Q13 Enzymes] At high ethanal concentrations all the active sites in the enzyme molecules are occupied by ethanal molecules. The flattening off of the curve shows that rate is constant, i.e. increase in [ethanal] has no effect on the rate of reaction. At this point, the enzyme is saturated with its substrate (ethanal). Each enzyme active site has a substrate bound to it, and all the enzyme molecules are continuously catalysing the conversion of substrate to product. 14. [09 P1 Q14 Periodicity] B T / K Water vapour is produced on decomposition. 2Al() 3 Al The water vapour produced when Al() 3 decomposes acts as an extinguisher and so, helps to delay the spread of flames in the event of a fire. 15. [09 P1 Q15 Transition Elements] A u atom has two more occupied electron shells than a Mg 2+ ion. u 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1 Mg 2+ 1s 2 2s 2 2p 6 ence, a u atom has two more occupied electron shells (3 rd and 4 th quantum shells) than a Mg 2+ ion.

3 [09 P1 Q16 Redox] after reaction with after reaction with potassium potassium iodide dichromate(vi) 0 2 range r turns green suggests that it is reduced to r 3+. ence, 2 2 is oxidised to 2 and so, oxidation number of oxygen after the reaction is zero. E o /V r e 2r e 0.68 r r E o cell ( ) V +0.65V (> 0 reaction is feasible). KI solution changes from colourless to brown suggests that I is oxidised to I 2. ence, 2 2 is reduced to 2 and so, oxidation number of oxygen after the reaction is 2. E o /V 2I I 2 + 2e e I I E o cell Since E o cell > 0, reaction is feasible. 19. [09 P1 Q19 Bonding] B 118 Each nitrogen atom in di-imine has one lone pair and two bond pairs. ence, the bond angle at each nitrogen atom is [09 P1 Q20 Isomerism] carvone Q * * * 2 * carvone Q ence, carvone has 1 chiral centre while Q has 3 chiral centres (marked by *). 21. [09 P1 Q21 Alkenes] o di-imine 17. [09 P1 Q17 Transition Elements] A Bromine acts as an electrophile. B o, i, u, Zn The graph shows a steady increase in 1 st I.E., which suggests that the elements are in the d- block or transition elements. From the ata Booklet, the 1 st I.E. (in kj mol 1 ) of o, i, u and Zn are 575, 736, 745 and 908 respectively which agrees with the trend shown. 18. [09 P1 Q18 Redox Potentials] 0.39 V When u 2+ (aq) and I (aq) are mixed, the brown precipitate formed is u 2 I 2 (s) in brown I 2 (aq). n adding S (aq), the brown colour disappears which suggests that I 2 is reduced to I. The white precipitate that remains is u 2 I 2. u 2+ + e u + E o V 2I I 2 + 2e E o 0.54 V 2u I 2u + + I 2 E o cell 0.39 V The reaction involves electrophilic addition of Br 2 across bond, where the δ+ end of the polarised δ+ Br Br δ acts as an electrophile. 22. [09 P1 Q22 alogen erivatives] nucleophilic substitution The reaction involves nucleophilic substitution of halogenoalkane, 2 2, with acting as the nucleophile.

4 [09 P1 Q23 alogen erivatives] [90J P1 Q25] Little or no precipitate was seen when S was boiled under reflux with ethanolic silver nitrate suggests that S does not undergo hydrolysis. ence, S is an aryl halide (option ). 24. [09 P1 Q24 Phenol] It reacts with Br 2 (aq) to incorporate up to 4 atoms of bromine in each molecule. alkene With Br 2 (aq), both alkene and phenol reacts to give the Br 2 following Br product. Br 25. [09 P1 Q25 Ketones] B ( 2 ) ketone ajulemic acid carboxylic acid phenol 3 ( 2 ) Br alcohol ketone 1 alcohol ketone cortisone alkene When cortisone is reacted with 2 /Pt, the alkene is hydrogenated while the ketone groups are reduced to the corresponding 2 alcohols. When the product is oxidised by warm acidified KMn 4, both the 1 and 2 alcohols are oxidised to carboxylic acid and ketones respectively. The final product has the structure shown and so, has 4 double bonds [09 P1 Q26 Aldehydes] A ab 4 in 3, then Q With ab 4, only the aldehyde group is reduced to a 1 alcohol, which is then oxidised by K 2 r 2 7 / + to a carboxylic acid. 27. [09 P1 Q27 arboxylic Acids] P K 2 r 2 7 / + heat Acidity: 2 2 > 2 Br 2 > > 6 5 arboxylic acids (1, 2, 4) are stronger acids than phenol because the carboxylate ion, R 2, is stabilised to a greater extent by delocalisation of the R - negative charge over the atom and both atoms. hloroethanoic acid is the strongest acid (its acidity being enhanced by the presence of an electron-withdrawing atom). Br is less electronegative than and so, bromoethanoic acid is less acidic than chloroethanoic acid. Propanoic acid is the least acidic among the three carboxylic acid due to the presence of an electron-donating ethyl group.

5 [09 P1 Q28 Amines] [84 P3 Q29] shaking the mixture with dilute aqueous acid Amine is basic and so, dissolves in aqueous acid whereas benzene is insoluble (forms an immiscible layer), which can then be separated from the aqueous layer by using a separating funnel. 29. [09 P1 Q29 Amino Acids] When insulin is heated in, it undergoes acid hydrolysis (peptide linkage breaks to give the corresponding carboxylic acid and amines). The hydrolysis products are: 30. [09 P1 Q30 Proteins] B The hydrolysis of peptide X gives 5 amino acids suggests that X has 4 peptide linkages (or 4 mol of 2 are lost in forming peptide X). M r of X (2 75) (2 105) (4 18) [09 P1 Q31 Formulae] S mole ratio : 54.5 : : : 2 molecular formula elements : ratio ,, 1 : ,, 1 : S,, S 1 : 2 ence, option A. 32. [09 P1 Q32 Solids] 1 The bond angle between nearest neighbours is smaller in diamond than in graphite. bond angle is smaller in diamond (109.5, tetrahedral) than in graphite (120, trigonal planar). bond is shorter in graphite (0.142 nm; sp 2 sp 2 overlap) than in diamond (0.154 nm; sp 3 sp 3 overlap and so, has more p-character). iamond has a giant covalent structure and all covalent bonds are of the same strength. Graphite has a layered structure. The covalent bonds within each layer are of the same strength (but weak van der Waals' forces exist between the layers). 33. [09 P1 Q33 Electrochemical ells] 1 The electrode potential of electrode 1 becomes more negative as the concentration of ethanol increases. As [ethanol] increases at electrode 1, position of equilibrium shifts to the right to remove some of the extra ethanol. xidation occurs and so, the electrode potential becomes more negative. At electrode 2, oxidation number of remains as +1. ence, hydrogen is not reduced. xygen is reduced only at electrode 2 (.. decreases from 0 in 2 to 2 in 2 ) but not at electrode 1 (.. remains as 2).

6 [09 P1 Q34 Kinetics] 1 The half-life is s. 2 The reaction is first order with respect to [ ]. The same time ( s) is taken for the relative [ ] to decrease from 10.0 to 5.0 and then, from 5.0 to 2.5. This suggests that the half-life is s. It also suggests that the reaction is first order with respect to [ ] (since half-life is constant). There is insufficient data given to deduce the overall order of reaction. 35. [09 P1 Q35 Group II] [06J P1 Q35] 1 Strontium ions are nearly the same size as calcium ions and so may easily replace them in the hydroxyapatite. Sr 2+ ions can easily replace a 2+ ions because its size (0.113 nm) is nearly the same as that of a 2+ ions (0.099 nm). Group II hydroxides become more soluble down the group. ence, Sr() 2 is more soluble than a() 2 and so, precipitates less readily. There is no metallic bonding in hydroxyapatite. 36. [09 P1 Q36 Group VII] 1 Silver(I) ions act as an oxidising agent. When Ag 3 (aq) reacts with Fe 2 (aq), a grey precipitate of Ag(s) forms which does not dissolve in 3 (aq). ence, Ag + is reduced to Ag(s) which suggests that Ag + ions act as an oxidising agent. When Ag 3 (aq) reacts with Mg 2 (aq), a white precipitate of Ag(s) forms which dissolves in 3 (aq) due to the formation of complex ion, [Ag( 3 ) 2 ] + ; i.e. 3 complexes with Ag + ion (and not with Ag metal). 2Ag 3 + Ba 2 2Ag(s) + Ba( 3 ) 2 Ag(s) [Ag( 3 ) 2 ] + (aq) [09 P1 Q37 arbonyl ompounds] The sp 2 hybridised carbon () becomes sp 3 hybridised in the product formed The sp 2 hybridised carbon () becomes sp 3 hybridised in the carbocation formed Benzene undergoes electrophilic substitution reaction. In the intermediate, the reactive carbon atom is sp 3 hybridised (tetrahedral). 38. [09 P1 Q38 arbonyl ompounds] sp 2 sp 3 sp 2 planar (sp 2 hybridised atoms) / Pt 2 / Pt + 2 sp 3 sp 3

7 [09 P1 Q39 Esters] 1 The identity of the reactants used to produce pepthidine could be determined from the products of hydrolysis of the ester. 3 ence, option A. 3 3 from carboxylic acid or acyl chloride from alcohol [09 P1 Q40 Amino Acids] 2 Equal concentrations of 3 + R 2 and 3 + R 2 are present at p pk 1. 3 There is no net charge on the amino acid at the point when the slope of the curve is at a maximum at its centre. 3 + R R R R At p pk 1, which is equivalent to the point of half-neutralisation in an acid-base titration, [ 3 + R 2 ] [ 3 + R 2 ]. The point in the titration curve when the slope is at its maximum in the centre corresponds to the isoelectric point. ence, the species present is 3 + R 2 which has no net charge. 3 + R 2 is the most common species present at the isoelectric point, which need not be at p 7. Isoelectric point ranges from p 5.5 to 6.2 depending on the nature of the amino acid. K 1 K 2 3 pepthidine [09 P1 MQ Key] Q. Key Q. Key Q. Key Q. Key 1 A A 31 A 2 B 12 B A A 14 B B B 35 6 B A B A B 9 B 19 B A B 40

8 ov (9746) Paper 2 1. [09 P2 Q01 Energetics / Solids] (a) The lattice energy of Mg is the heat energy evolved when one mole of crystalline Mg(s) is formed from its separate gaseous ions, Mg 2+ (g) and 2 (g), under standard conditions. Mg 2+ (g) + 2 (g) Mg(s) L.E. (b) q+ q (i) L.E. ( r+ + r ) ionic radii/nm:, 0.181; Br, 0.195; I, Lattice energies decrease from a to ai due to increase in ionic size of the X anions ( < Br < I ) while the charge remains the same in each case (a +, a + Br, a + I ), which results in weaker electrostatic forces of attraction between a + and X ions from a to ai. q+ q (ii) L.E. ( r+ + r ) ionic radii/nm: Mg 2+, 0.065; a +, Lattice energy of Mg is considerably larger than those of sodium halides due to the higher charge (charge on Mg 2+ 2 is double that on sodium halides, a + X ) and smaller ionic radii (Mg 2+ < a + ; 2 < X ), which results in much stronger electrostatic forces of attraction between Mg 2+ and 2 ions than that between a + and X ions. (c) AgI is not fully ionic but has considerable percentage of covalent character because I ion (0.216 nm) is much larger than F ion (0.136 nm). ence, I anion is readily polarised by the Ag + ion leading to electron density between the Ag + and I ions; i.e. electrons are incompletely transferred in forming Ag + and I ions. (d) (i) o-ordination number is the maximum number of ions that can be placed around another ion of opposite charge in a crystal lattice; i.e. the number of its nearest neighbours. 2. [09 P2 Q02 Transition Elements] (a) Metals 3 and 4 are Zn and u respectively. (b) u e u E o V e 2 E o 0.00 V Zn e Zn E o 0.76 V Al e Al E o 1.66 V Mg e Mg E o 2.38 V The solid residue is copper because E o (u 2+ /u) is positive, so u has no reaction with acids (i.e. cannot reduce + ). (c) (i) Sulfuric acid acts as an oxidising agent. (ii) S e S Zn Zn e overall equation is S Zn Zn 2+ + S (d) (i) Al() 3 (ii) Al 3+ (aq) + 3 (aq) Al() 3 (s) Al() 3 (s) + (aq) Al() 4 (aq) aluminate (e) Zn [Ar] 3d 10 4s 2 Like Group II elements, Zn has a fully filled outer s-orbital. ence, Zn shows only +2 oxidation state in its compounds; i.e. Zn forms Zn 2+ only. (f) (i) Zn( 3 ) 2 Zn (ii) Zn 2+ (0.074 nm) is larger than Mg 2+ (0.065 nm) and so, has smaller polarising power which results in the 3 anion being less polarised. ence, Zn( 3 ) 2 is relatively more stable to heat and so, decomposes at a higher temperature than Mg( 3 ) 2. (g) Zn 2+ (aq) + 2 (aq) Zn() 2 (s) Zn() [Zn( 3 ) 4 ] (ii) Ionic radii: s + > K + > a + The much larger ionic size of s + ion allows more ions to be packed around it. ence co-ordination number in s lattice is larger than those in a and K.

9 [09 P2 Q03 arboxylic Acids] (a) Benzoic acid and ethanoic acid are stronger than carbonic acid. [A stronger acid has a larger K a value.] (b) 6 5 (aq) 6 5 (aq) + + (aq) + [65 ][ ] K a [65] Since the degree of dissociation, α, is very small, [ 6 5 ] eqm [ 6 5 ] [ ] K a since [ [ ] 6 5 ] [ + ] 6 5 [ + ] K a [ 6 5 ] mol dm 3 p log 10 [ + ] log 10 ( ) 5.44 (c) 6 5 is more acidic (larger K a ) than 3 because the electron-withdrawing benzene ring weakens the bond, thereby causing phenol to dissociate readily to give + ions. In addition, the anion 6 5 is stabilised by delocalisation of the charge over the ring. 3 is less acidic (smaller K a ) because the negative charge in 3 is intensified by the electron-donating methyl group making it less stable. (d) observation: reddish-brown Br 2 (aq) decolourised, steamy fumes of Br evolved and white precipitate formed. structural formula of product: Br Br (white ppt) Br (e) (i) Benzoic acid is heated with methanol under reflux in the presence of a little conc. 2 S 4 as catalyst to give methyl benzoate. (ii) S S Ethanoic acid is first converted to ethanoyl chloride by reacting with S 2. Ethanoyl chloride is then reacted with phenol to give phenyl ethanoate. 4. [09 P2 Q04 arboxylic Acids] (a) (i) A, and [no silver mirror with Tollens' reagent not aldehyde] (ii) A and [no ppt with 2,4-dinitrophenylhydrazine not carbonyl compound ] (iii) A and [molecular formula shows Br added bond present ] (iv) A and [two a in molecular formula of product two groups present ] (v) [two a in molecular formula of product two acidic groups present ] (b) (i) Reaction (a)(v). (ii) (c) a + 2 a [ Product formed is a salt of benzene-1,4-dicarboxylic acid because the oxidation is carried out in alkaline medium.] (d) A produces produces 2 a + 2 a + [Phenol has no reaction with a 3.]

10 [09 P2 Q05 Esters] (a) (i) mole ratio : : : 1. 0 : : 5.90 : : 4 : 1 empirical formula of E is 4 4. (ii) Using the ideal gas equation, pv nrt m RT (since n m ) M r M r M r of E mrt ( ) 6 pv Let the molecular formula of E be 4n 4n n. M r of 4n 4n n n(12.0) + 4n(1.0) + n(16.0) n 2 molecular formula of E is (b) (i) structure E1 structure E2 2 dissolves in water to give carbonic acid, 2 3, which is a stronger acid than phenol but a weaker acid than both ethanoic acid and benzoic acid. ence only sodium phenoxide reacts with 2 3 to give phenol. Sodium ethanoate and sodium benzoate have no reaction with 2 3. (ii) E1 produces E2 produces a + a + + a (c) mixture from E1 produces a + mixture from E2 produces no reaction +

11 ov (9746) Paper 3 1. [09 P3 Q01 Proteins / hem Equilibria] (a) The haemoglobin molecule is a tetramer consisting of four polypeptide chains (two α- chains and two β-chains, called subunits) each with its own haem group. This is a quaternary structure of protein. In each polypeptide chain, the amino acids are bonded together by peptide linkages (covalent bonds). The sequence of amino acids in each polypeptide chain gives the primary structure of protein. The secondary structure refers to the detailed configurations of the polypeptide chains how the chains may be coiled or folded to give the α - helix or β -pleated sheets. The structures are stabilised by hydrogen bonds between the group of one amino acid unit and the group of another along the main chain. The tertiary structure refers to the overall 3- shape of the protein involving folding or coiling of the chains. It shows how protein molecules are arranged in relation to each other. The four types of R-group interactions that hold the tertiary structure in its necessary shape are hydrogen bonds between polar R- groups, ionic bonds between charged R- groups, hydrophobic interactions between nonpolar R-groups, and disulfide linkages (covalent bonds). The quaternary structure of proteins refers to the spatial arrangement of its subunits. It shows how the individually folded subunits are packed together. The quaternary structure is stabilised mainly by hydrophobic interactions between non-polar areas on the surface of the individual polypeptide chains (subunits). (b) (i) 2 2 (ii) The aggregation of b-s molecules is due to the presence of van der Waals forces (arising from induced dipoles) between the non-polar side-chain of valine residue. (iii) Molecules of normal haemoglobin do not attract each other because of the repulsion between haemoglobin chains due to the negatively charged glutamate. (c) [b( 2) 4] (i) K c 4 dm 12 mol 4 [b][ ] [b( 2) 4] (ii) K c -6 4 [b] ( ) since [b] [b( ( ) 2 ) 4 ] dm 12 mol 4 (iii) 99% of b converted to b( 2 ) 4 [ b(2) 4] 99 [b] 1 [b( 2) 4] since K c 4 [b][ 2] [ 2 ] [ 2 ] [ 2 ] mol dm 3 (d) Mb(aq) + 2 (aq) Mb 2 (aq) [Mb2] K c [Mb][2] [Mb2] -6 [Mb] ( ) [Mb ] or [Mb2 ] 7.6[Mb] [Mb] [Mb2] % of Mb [Mb] + [Mb ] 7.6[Mb] 100 [Mb] + 7.6[Mb] 88.4 % 2

12 [09 P3 Q02 Alcohols / Energetics] (a) (i) Let the oxidation no. of in 3 be x. so, x + 3(+1) + ( 2) or x 2 oxidation no. of in 3 2 (ii) A B (iii) To convert methanol to A: reagent: acidified K 2 r 2 7 (aq) condition: distil To convert methanol to B: reagent: acidified K 2 r 2 7 (aq) condition: heat under reflux To convert methanol to : reagent: acidified KMn 4 (aq) condition: heat under reflux (b) (i) heat evolved mc T ( ) J J M r of 2 5 2(12.0) + 6(1.0) mol of ethanol m mol M r c heat evolved J mol 1 mol of ethanol kj mol 1 (ii) c value calculated in (i) is much less than the true value of c because of heat lost to the surroundings and to the copper can; i.e. not all the heat from the burning ethanol are used to heat the water. (c) (i) ½ Bonds broken ( ) Bonds formed ( ) 1 (+350) 1 ( 350) 5 5(+410) 4 4( 410) 1 (+360) 1 1( 740) 1 (+460) 2 2 ( 460) ½ ½ (+496) c kj mol 1 (ii) By ess' Law, ( 182) + ( 1367) 1185 kj mol 1 (iii) c for 3 2 is expected to be less exothermic than that for 3 because more energy is needed to break the strong and bonds in 3 2 while the same amount of energy is released when bonds are formed (since same amount of 2 and 2 are formed in each case). (d) (i) ( 2 ) 3 E (ii) and E are primary alcohols, F is a secondary alcohol, and G is a tertiary alcohol. (iii) F is chiral (chiral carbon indicated with *) F G * 2 3 (iv) F contains 3 () group and so, reacts with alkaline aqueous iodine to give I 3 and (v) F (a secondary alcohol) is oxidised to give a ketone, which is a non-acidic organic product.

13 [09 P3 Q03 Periodicity / Kinetics] (a) Magnesium burns in 2 with a brilliant white flame to give the oxide, Mg (white residue). 2Mg + 2 2Mg P 4 burns in 2 with a pale bluish-green flame to give the oxides, P 4 6 or P 4 10 (white solid). P P 4 6 ; P P 4 10 Sulfur burns with a blue flame to give S 2, which is oxidised to S 3 in excess oxygen. S + 2 S 2 ; 2S S 3 (b) P 4 6 /P 4 10 (covalent oxides) dissolve readily in water to give strongly acidic solutions, which turn universal indicator red. P 4 6 (s) (l) 4 3 P 3 (aq) P 4 10 (s) (l) 4 3 P 4 (aq) S 2 /S 3 (covalent oxides) dissolve readily in water to give acidic solutions, which turn universal indicator red. S 2 (g) + 2 (l) 2 S 3 (aq) S 3 (g) + 2 (l) 2 S 4 (aq) (c) (i) a + a + 2 mol of a in 25.0 cm 3 mol of mol mol of a in 100 cm mol (ii) 2Mn Mn mol of 2 2 in 25.0 cm 3 25 mol of Mn mol mol of 2 2 in 100 cm mol (iii) a a mol of a 2 2 mol of mol mol of a from a mol of mol mol mol of a from a mol a a mol of a 2 21 mol of a mol (d) (i) rate 1. The faster the reaction, the time shorter is the time taken for the blue colour to appear. (ii) expt no. time taken /s 1 time / s Since the total volume of reaction mixture is constant (100 cm 3 ), the volume of each reagent used is proportional to its concentration. ompare expt 2 and 4: [I ] and [ + ] constant, when [ 2 2 ] doubles, the initial rate is also doubled. Reaction is first order w.r.t ompare expt 2 and 3: [ 2 2 ] and [ + ] constant, when [I 15 ] increases by times, the initial rate increases by times. Reaction is first order w.r.t. I. Let rate equation be rate k [ 2 2 ] [I ] [ + ] n From expt 1, k (15)(10)(5) n (1) From expt 2, k (5)(10)(10) n (2) (1), (2) k(15)(10)(5) k(5)(10)(10) n n n 0 Reaction is zero order w.r.t. +. (iii) rate k [ 2 2 ] [I ] units of k units of rate 2 (units of concentration) -1 s -3 2 (mol dm ) dm 6 mol 2 s 1 [ Also accepted: usual second-order units of dm 3 mol 1 s 1.] 4. [09 P3 Q04 Bonding / Amines] (a) In an ideal gas, the gas particles have negligible size/volume, negligible intermolecular forces of attraction, and the collisions of gas particles are perfectly elastic.

14 09-14 (b) (i) 3 has a much higher boiling point than 4. This is because the intermolecular hydrogen bonding in 3 is.. much stronger than the weak δ+ δ intermolecular van : der Waals' forces in 4, and so requires hydrogen bonding much more energy to overcome. (ii) When methane is liquefied, the entropy of the sample decreases ( S < 0) because 4 molecules in the liquid state display more order than in the gaseous state. (iii) G o o T S o At 33, G o 0.0 kj mol 1 o or S o T S o o / T ( ) ( ) J K 1 mol 1 (c) (i) rder of base strength: ethylamine > ammonia > phenylamine Ethylamine is a stronger base than 3 because the electron-donating ethyl group ( 3 2 ) releases electrons towards the atom, making the lone pair on the atom more available to accept a proton than that in 3. Phenylamine is a weaker base than 3 because the lone pair of electrons on the atom is delocalised over the benzene ring, and this makes it less available to accept a proton than that in 3. Besides, the charge on is more dispersed than for the smaller 4 + ion, making ion more stable. Whereas, the charge on is intensified, making ion less stable. (ii) [253 ][ ] K b [252] Since the degree of dissociation, α, is very small, [ ] eqm [ ] 0 2 [ ] K b [ ] 2 (0.100)( ) [ ] mol dm 3 p log 10 [ ] log 10 ( ) 2.10 p 14 p (iii) ( 3 ) 2 Product: ( 3 ) Br 2 (aq) Product: 2 Br Br + 3Br Br (d) (i) ucleophilic substitution. (ii) J is 1,5-dibromooctane (iii) The four isomeric alkenes are: Br Br cis isomer cis isomer trans isomer trans isomer

15 [09 P3 Q05 Group VII / Arenes] (a) 2 is a yellowish-green gas, Br 2 is a dark red liquid and I 2 is a black solid. Volatility decreases down the group (from 2 to I 2 ) due to increasing number of electrons in the molecules (as the halogen molecule becomes larger) and hence, increasing intermolecular van der Waals' forces. (b) When ion reacts with concentrated 2 S 4, only steamy fumes of is produced. 2 S 4 + (g) + S 4 When Br ion reacts with concentrated 2 S 4, steamy fumes of Br together with some orange-brown fumes of Br 2 are produced. 2 S 4 + Br Br(g) + S 4 2 S 4 + 2Br Br 2 + S Some of the Br produced is oxidised by concentrated 2 S 4 to Br 2. When I ion reacts with concentrated 2 S 4, purple vapour of I 2 are produced together with some steamy fumes of I. 2 S 4 + I I(g) + S 4 2 S 4 + 8I 4I S Most of the I produced is oxidised by concentrated 2 S 4 to I 2. (c) Phenol is nitrated more easily than methylbenzene; the nitration reaction does not require the strong electrophile 2 + produced by the reaction of 3 with 2 S 4. This is because the group in phenol has an electron-donating effect due to the interaction of the unshared pair of electrons on the atom with the delocalised π orbitals of the benzene ring, thereby making the ring more electronrich and so, is more susceptible to electrophilic attack; i.e. group activates the ring toward electrophilic substitution. (d) K 3 L 2 2 Q M P 2 With 2 and Al 3, methylbenzene undergoes electrophilic substitution to give K, which is either 2-chloro or 4-chloro methylbenzene Al [Take K as (4-chloro)methylbenzene.] With more 2 in the presence of light, K undergoes free-radical substitution to give L. 3 2 light When heated with a in ethanol, L undergoes nucleophilic substitution to give M. When heated with dilute 2 S 4, M undergoes acid hydrolysis to give a carboxylic acid. 2 2/ + reflux When heated with a(aq), L undergoes nucleophilic substitution to give P. K L M L When a mixture of P and is heated with a small amount of c. 2 S 4, esterification occurs to give an ester Q a ethanol a heat c. 2 S 4 heat 2 Q 2 L P M P