NJC Prelim P3 Solutions

NJC 2018 SH2 H2 Chemistry Prelim Paper 3 Solutions: Question 1 (a)(i) Suggested solution Both octan-1-ol and iodine have simple covalent structure Strength of Td-Id in iodine due to its large elec tron cloud size is comparable to the stronger hydrogen bonding between octan-1-ol molecules. (ii) Suggested solution Graphite has a giant molecular structure with strong covalent bonds between carbon atoms to be overcome during melting. Fullerene has a simple covalent structure with weaker td-id interactions between fullerene molecules to be overcome during melting. Hence, more energy is required to melt graphite and a higher temperature is required. (b) Suggested solution Step 1: reaction 1 ;(-COCHO) Step 2: K2Cr2O7/KMnO4 with H2SO4(aq) and heat; (-COCOOH) Step 3: NaBH4 in ethanol or H2 with Nickel/Pt ; (compound B) c(i) Suggested solution PCO2 = ଴.଴ଷହ ଵ଴଴ × 101.3kPa = 0.03546kPa KH = [஼ைమ(௔௤)] ௉಴ೀమ [CO2(aq)]= 1.17 x 10-5 mol dm−3 c(ii) Suggested solution Kc = [ுమ஼ைమ(௔௤)] [஼ைమ] [H2CO3] = 1.52x 10-8 mol dm−3 (iii) Suggested solution Ka1 is larger than Ka2. H2CO3 will dissociate accordingly to eqm 1 first and [H +] from eqm 1 suppresses the dissocation of HCO3− in eqm 2. OR HCO 3− from eqm 1 is low as Ka1 is small. Hence [H+] from HCO3− is negligible. [H+] is largely from eqm 2. c(iv) Suggested solution Ka1 = [ு஼ைయష][ுశ] [ுమ஼ைయ] [H+]=1.915 x 10-6 moldm−3 pH = 5.71 d Suggested solution

Question 2 (a) Suggested solution: At anode: Co has ௢௫ ௢ (Co/Co2+) = +0.28V more positive that ௢௫ ௢ (Cu/Cu2+) = −0.34V, hence Co will be oxidised together with Cu and dissolve as Co 2+, becoming part of the electrolyte. Ag on the other hand has more negative ௢௫ ௢ (Ag/Ag+) of −0.80V, hence will not be oxidised. It will fall below anode as sludge. At cathode: Co2+ has a more negative ௥௘ௗ ௢ (Co2+/Co) = −0.28 V than ௥௘ௗ ௢ (Cu2+/Cu) = 0.34 V, hence Co2+ is not reduced at the cathode, remain dissolved as electrolyte. (b) (i) Suggested solution • mass of Ag = 0.07 g • No of moles of Co oxidised at anode = no of moles of Co(C 4H7N2O2)2 = ଴.ହହ ହ଼.ଽାଶ(ସ×ଵଶା଻ାଶ × ଵସାଶ × ଵ଺) = 0.0019038 mol Mass of Co in alloy (anode) = 0.0019038 x 58.7 = 0.112 g Mass of Cu = 1.25 – 0.07 – 0.112 = 1.068 g b(ii) % purity = 1.068 /1.25 x 100 = 85.4% b(iii) Suggested solution: Any one of the suggestion below is acceptable: (i) Mass of Ag and Co(C 4H7N2O2)2 should be heated to consistent mass to ensure all water is driven off, so that mass measured is that of dry mass of Ag and Co(C4H7N2O2)2. (ii) As mass of Ag (sludge ) is rather small, use electronic balance of higher precision in order to reduce % uncertainty in mass measurement. (iii) Repeat the experiment using fresh sa mples so that an average of the mass measurements can be taken to reduce random error. b(iv) Suggested solution:

Total no of moles of e used for electrolysis = 2.15 x 28.0 x 60 ÷ 96500 = 0.03743 No of moles of Cu expected to be discharged = 0.043523 ÷ 2 = 0.018715 Mass expected = 0.018715 x 63.5 = 1.19 g (3sf) (c) Suggested solution: Co(C4H7N2O2)2 + aq ⇌ Co2+ + 2 C4H7N2O2− −1.23 x 10−4 +1.23 x 10−4 + 2 x 1.23 x 10−4 Ksp = [Co 2+][C4H7N2O2]2 = (1.23 x 10−4)( 1.23 x 10−4 x 2)2 = 7.44 x 10−12 mol3dm−9 (d)(i) Suggested solution d(ii) Suggested solution: ି ିቤ Ionic radii of Cl− = 0.181, ionic radii of H− = 0.208nm They have the same product but LiH has a larger interionic distance than LiC l, therefore LE magnitude of LiH is smaller. (e) Suggested solution: Amt of AlCl3 = ହ ଶ଻ାଷହ.ହ ௫ ଷ) = 0.037453 mol

Amt of LiH = ହ ଻.ଽ = 0.6329 mol AlCl3 is limiting since 0.037453 mol of AlCl3 requires 0.037453 x 4 mol of LiH4 = 0.14981 mol < 0.6329 mol. No of moles of LiAlH4 formed = 0.037453 mol 0.037453 x ΔHreaction = − 1.24 x 8.4 ΔHreaction = − 1.24 x 8.4 ÷ 0.037453 = −278 kJmol−1 (f)(i) suggested solution ΣnΔHf(products) - ΣnΔHf(reactants) = −454 – 3(−152.5) = +3.5 kJmol−1 f(ii) Suggested solution ΔGo = ΔHo − TΔSo −27.7 = +3.5 − 298 x ΔSo ΔSo = +0.105 kJmol−1K−1 ΔS is positive as there is an increase of 3 moles of gas molecules after the reaction; there are more ways to distribute the molecules and their energies, increasing entropy level of the system at the end of reaction. f(iii) Suggested solution Temp at which decomposition becomes spontaneous is the cross over temperature. ΔG = 0 +3.5 = 0.105 x T T = 33.3 K Question 3 (a)(i) Suggested solution:

(ii) Suggested solution: A is formed from the primary carbocation as shown. This primary carbocation is less stable due to the positive charge not being resonance stabilised by the aromatic ring. Hence, the carbocation is formed in trace amounts, leading to trace amounts of A being formed. (iii) Suggested solution:

(b)(i) Suggested solution: In the presence of ligands, d-orbitals of Fe 3+ are split into two different energy levels with small energy gap. d-d transition, where the el ectron is promoted from a lower energy d orbital to a higher energy d orbital, is possible. Energy that corresponds to the wavelength of light in the visible region of the electromagnetic spectrum is absorbed. Co lour observed is complementary to the wavelength of visible light absorbed. (ii) Suggested solution: (iii) Suggested solution: G has a larger energy gap. Since the electronic configuration of Fe 3+ in G is in a ‘low spin’ state, energy required to overcome the energy gap in adding subsequent electrons to higher energy d-orbitals is more than that required to overcome inter-electronic repulsion when electrons paired up in the lower energy d-orbitals. (c)(i) Suggested solution: Oxidation number of C in carbon dioxide: +4 Oxidation number of C in methanol: -2

(ii) Suggested solution: Electrode 1 : CH3OH + H2O  CO2 + 6H+ + 6e− Electrode 2 : O2 + 4H+ + 4e−  2H2O Overall: 2CH3OH + 3O2  2CO2 + 4H2O (iii) Suggested solution: When [CH3OH] is decreased, oxidation of methanol becomes less favoured OR by LCP, the position of equilibrium for CH3OH + H2O ⇌ CO2 + 6H+ + 6e− shifts to the left to partially increase [CH3OH]. Thus, Eox(CH3OH/CO2) becomes less positive. A less positive Eox(CH3OH/CO2) will cause Ecell to be less positive since Ecell = E + Eox. (iv) Suggested solution: CH3OH is a liquid at room temperature and thus can be easily transported and stored than hydrogen gas OR CH3OH is less explosive than H2 gas OR CH3OH is less expensive to maintain than H2 gas Question 4 (a)(i) (ii) Suggested solution: Hydrogen bonding - lone pair

- δ+ / δ- - dotted lined to show bond (iii) Suggested solution: Leucine and Valine. Their side chains consist of hydrocarbon chains which are able to form temporary dipole- induced dipole interactions with hydrophobic groups to transport them in the blood stream. (b)(i) Suggested solution: -OOC C N+H3 H CH2CH(CH3)2 (ii) Suggested solution: Melting point Unionised form: Hydrogen bonding Zwitterionic form: Ionic bonding More energy is required to overcome the stronger ionic bonding compared to the hydrogen bonds. Hence the zwitterionic form will have a higher melting point. Solubility: Unionised form: Hydrogen bonds with water Zwitterionic form: ion-dipole interactions with water Stronger ion-dipole interactions produce more energy to overcome the hydrogen bonds between water molecules and the interactions between the solute. Hence the zwitterionic form will be more soluble in wa