JJC H2 CHEM P3 Ans

© Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 Suggested Answers to 2015 JC2 H2 Chemistry Preliminary Examination Paper 3 1 (a) (i) Hf(NH3(g)) = 1 2 (+994) + 3 2 (+436) – 3(+390) = –19.0 kJ mol1 (ii) Hr = 2(–119) – 2(–19.0) = –200 kJ mol1 (iii) Na is oxidised because O.S. of Na increases from 0 (in Na) to +1 (in NaNH2). NH3/H is reduced because O.S. of H decreases from +1 (in NH3) to 0 (in H2). (iv) (v) NH2  + CH3OH  NH3 + CH3O (b) (i) Electron-withdrawing –OH groups in ‘Tris’ makes the lone pair of electrons on N atom less available for protonation. OR Presence of intramolecular hydrogen bonding between –OH and –NH 2 groups in ‘Tris’ makes the lone pair on N atom less available for protonation. Thus, ‘Tris’ is a weaker base and hence a smaller Kb than tert-butylamine. (ii) (CH2OH)3CNH2 + HCl  (CH2OH)3CNH3 + Cl (iii) [OH] = Kb [base] [salt] 10(147.40) = (1.20  106) [base] [salt] OR pOH = pKb + lg [salt] [base] 14 – 7.40 = lg (1.20  106) + lg [salt] [base]  [base] [salt] = 0.209 Since HCl  salt, [salt] in buffer A = 100 1000  0.500 = 0.0500 mol dm3 [base] in buffer A = 0.209  0.0500 = 0.0105 mol dm3 Mass of ‘Tris’ needed = (0.0500 + 0.0105)  121 = 7.31 g (c) (i) Amide (ii) Step 1: substitution; Step 2: acid-base reaction (iii) (iv) ONH3C C

2 © Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 2 (a) A transition element is a d-block element which forms one or more stable ions with incompletely filled d-orbitals. (b) (i) A catalyst provides an alternate reaction path of lowered activation energy, Ea’. Thus, the number of particles with energy ≥ Ea’ increases. Therefore, the frequency of effective collisions increases and hence rate increases. (ii) electrophilic substitution Br2 + FeBr3  Br+ + FeBr4 – (iii) p–p orbital overlap results in the delocalisation of lone pair of electrons on N atom into the benzene ring of phenylamine, making the benzene ring highly electron rich and thus more susceptible towards electrophilic substitution. Hence, there is no need of FeBr 3 to generate stronger Br+ electrophile for the halogenation of phenylamine. (c) (i) Nickel(II) complexes are coloured because Ni( II) has d8 configuration/ partially unfilled d-orbitals so electron transition between dorbitals is possible. In an octahedral nickel( II) complex ion, the presence of ligands causes the dorbitals to split into 2 (different) energy levels. Energy gap is relatively small such that radiation from the visible light spectrum is absorbed when an electron transits from a dorbital of lower energy to a dorbital of higher energy. Hence, the colour seen is the complement of the colours absorbed. (ii) Since G < 0 for both reactions, it implies that the ligand exchange is energetically feasible for both reactions.  H2O is the weakest ligand among the three. Since G for reaction 2 is more negative than that of reaction 1, it implies that the ligand exchange in 2 is more energetically feasible.  en is a stronger ligand than NH3. EaEa' Number of particles Energy / kJ mol1 0 No. of particles with energy  Ea of uncatalysed reaction No. of particles with energy  Ea’ of catalysed reaction

3 © Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 (iii) Since the solution changes from violet to orange, it implies that ligand exchange reaction has occurred. The stronger CN  ligands displace the weaker en ligands in [Ni(en)3]2+ to form stronger dative bonds with Ni 2+ to give a more stable [Ni(CN)4]2 complex. (iv) In each reaction, six Ni O bonds are broken and six Ni N bonds are formed. Hence, H for both reactions are similar. (v) This is due to a larger increase in disorder of the system since the reaction 2 proceeds with an increase in number of particles. (vi) The chelate effect is entropy driven. Since both reactions have similar H, the more positive S in reaction 2 must be the cause for its more negative G . 3 (a) (i) CH3CO2H + 4[H]  CH3CH2OH + H2O (ii) (iii) B has 3 bond pairs and no lone pair so the shape is trigonal planar. (iv) Electron–deficient B atom of BH 3 uses its empty p orbital to accept the lone pair of electrons on O atom of RCOOH to form dative bond so that B can achieve stable octet configuration. (v) Three more electron–donating –CH 3 groups in E increases the electron density of O atoms and makes the lone pair of electrons on O atom more available for dative bond formation.  E will react faster with BH3 as compared to ethanoic acid. (vi) structure of F: structure of G: (b) (i) electrophilic addition (ii) structure of H: structure of J: Br Br structure of K:

4 © Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 (c) Amount of B2H6 formed =    0.0100 21 0 . 8 + 61 . 0 = 3.62 104 mol Since 2BF3  B2H6, amount of BF3 required = 2  (3.62  104) = 7.25  104 mol Assuming ideal gas behaviour (i.e. pV = nRT), volume of BF3 required =         4 5 7.25×10 8.31 0 + 273 0.5 1.01 10 = 3.26  105 m3 = 32.6 cm3 (d) Like SiO 2, B 2O3 has a giant covalent structure. Large amount of energy is required to overcome the strong covalent bonds between B and O atoms. B2O3 has high melting point (510 C). 4 (a) (i) 2H2O  O2 + 4H+ + 4e (ii) Amount of gas collected = 0.0104 24 = 4.33  104 mol Amount of electrons = 4  4.33  104 = 1.73 x 10—3 mol Using Q = It = nF, I  (2 x 60 x 60) = (1.73  103)(96500) I = 0.0232 A (b) E  Cl2 + 2e  2Cl +1.36 V MnO4  + 8H+ + 5e  Mn2+ + 4H2O +1.52 V E cell = (+1.52)  (+1.36) = +0.14 V Since E cell is positive, C l from HC l will also react with MnO 4  ions, the amount of MnO 4  used in the titration will not be an accurate reflection of how much Fe 2+ ions there is in the solution. (c) (i) green precipitate: Fe(OH)2 red-brown precipitate: Fe(OH)3 (ii) Amount of MnO4  used =  36.30.0250 1000 = 9.08  104 mol Since MnO4   5Fe2+, amount of Fe2+ in 0.35 g of sample = 5  9.08 x 104 = 4.54  103 mol Mass of Fe in 0.350 g of sample = 4.54  103  55.8 = 0.253 g Mass of O in 0.350 g of sample = 0.350 – 0.253 = 0.0968 g Amount of oxygen in 0.350 g of sample = 0.0968 16.0 = 6.05  103 mol Mole ratio of Fe : O = 4.54  103 : 6.05  103 = 1 : 1.33 = 3 : 4 Hence, identity of oxide is Fe3O4.

5 © Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 (d) (i) Amino acids are soluble in water because their zwitterions can form ion-dipole interactions with water molecules. (ii) At pH 7, both amino acids exist as anions and will migrate towards the positive terminal. Cysteine, having a bigger mass, will migrate slower towards the positive terminal and thus it will be closer to point M as compared to glycine. (iii) C N H O C O N H hydrogen bonds polypeptide chain (iv) (v) SO2 + H2O  H2SO3 H+ from H 2SO3 causes COO— to become COOH, disrupting the ionic interactions in the tertiary and quaternary structures of the protein. H+ from H2SO3 causes NH2 to become NH3 +, disrupting the hydrogen bonds in the tertiary and quaternary structures of the protein. Hence the protein loses its shape and undergoes denaturation. (e) cys – arg – val – tyr – ile – met – pro – phe

6 © Jurong Junior College 9647/03/ J2 PREIMINARY EXAMINATION/2015 5 (a) (i) Sodium burns with an orange/yellow flame and magnesium burns with bright white flame. (ii) Since O 2 2 has the same ionic charge but a larger ionic radius than O 2 ion, |LE(Na2O2)| is smaller than |LE(Na2O)|. (iii) Na2O(s) + H2O(l)  2NaOH(aq) MgO(s) + H2O(l)  Mg(OH)2(s) or Mg(OH)2(aq) pH of NaOH = 13 and pH of Mg(OH)2 = 9 (b) (i) 2I + S2O8 2–  I2 + 2SO4 2– (ii) Co