AJC 2013 ALevel Chem P1 solns (2020 version)

1 H2 Chemistry 9647 2013 ‘A’ Level P1 Worked Solutions 1 Two H2O molecules are eliminated when three H3PO4 molecules undergo condensation. The molecular formula of triphosphoric acid produced can be deduced by construct ing a balanced chemical equation as shown: 3H3PO4 H5P3O10 + 2H2O FYI: A water molecule is eliminated between two H3PO4 molecules as shown: HO P O O OH H O P OH O OH H B 2 NO, H, D, 14 7 16 8 1 1 2 1 D3O+ H3O+ NH2– OD– no. of protons 3 + 8 = 11 3 + 8 = 11 7 + 2 = 9 8 + 1 = 9 no. of electrons 3 + 8 – 1 = 10 3 + 8 – 1 = 10 7 + 2 + 1 = 10 8 + 1 + 1 = 10 no. of neutrons 3 + 8 = 11 0 + 8 = 8 7 + 0 = 7 8 + 1 = 9 B 3 C+ N Si– P+ no. of electrons on gaining an e– 5 + 1 = 6 7 + 1 = 8 15 + 1 = 16 14 + 1 = 15 electronic config. 1s2 2s2 2p2 1s2 2s2 2p4 1s2 2s2 2p6 3s2 3p4 1s2 2s2 2p6 3s2 3p3 D 4 Largest increase from 4th to 5th I.E (4550 kJ mol–1)  5th electron is removed from the next inner quantum shell  4 electrons in outermost (valence) shell  M is a Group 4 element  M forms a chloride with formula MCl4 C 5 Solid graphite (giant covalent) conducts electricity (due to presence of delocalised e–). A 6 The behaviour of a gas is most ideal at low pressure and high temperature. At low pressure, the gas molecules are far apart. The volume of the gas molecules therefore becomes insignificant compared to the volume of container. Similarly, the intermolecular forces of attraction are negligible. At high temperature, the gas molecules move faster (higher K.E.) and the intermolecular forces of attraction become negligible. C

2 7 Hr = Hf (products) – Hf (reactants) reactants: CH3CO2Na(aq) and H2O(l) products: CH3CO2Na.3H2O(s) A 8 Temperature falls  endothermic reaction, i.e. H is positive Vigorous reaction occurs  reaction is spontaneous, i.e. G is negative Since G = H – TS –TS = G – H TS = –(G – H) S must be positive as –(G – H) is always positive. B 9 Au3+ + 3e– Au n(Au) = 6.0 / 197.0 = 0.03045 mol n(e–) = 3 x 0.03045 = 0.09135 mol Q = neF = 0.09135 x 96500 = 8815 C t = Q/I = 8815 / 0.10 = 8.8 x 104 s D 10 Ionic product of CaCO3 = [Ca2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 < Ksp(CaCO3)  no ppt Ionic product of FeCO3 = [Fe2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 > Ksp(FeCO3)  ppt forms Ionic product of MnCO3 = [Mn2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 > Ksp(MnCO3)  ppt forms  FeCO3 and MnCO3 only are precipitated. C 11 • [H+] = [OH–] at all temperatures for a neutral solution • Position of equilibrium (POE) lies furthest to the right (towards ionisation of water) at 50 C since Kw at this temperature is the largest (since the larger the K w, the greater the extent of ionisation of water) • When temperature increases, Kw increases  POE is shifted to the right; forward reaction is favoured Increase in temperature favours endothermic reaction to absorb the extra heat  forward reaction is endothermic  Options A–C are incorrect. Kw = [H+][OH–] = [H+]2 (since [H+] = [OH–] at all temperatures) [H+] = (Kw)1/2 As temperature increases, Kw increases  [H+] increases  pH decreases (since pH = –log10 [H+])  neutral pH of water at higher temperatures is <7. D

3 12 NO2(g) + SO2(g) NO(g) + SO3(g) NO(g) + ½ O2(g) NO2(g) (NO2 is regenerated at the end of the reaction) SO3(g) formed reacts with rain water to give very dilute sulfuric acid (acid rain) SO3(g) + H2O(l) H2SO4(aq) C. NO2 catalyses the formation of ozone; NO causes ozone depletion. D. N2O is a greenhouse gas. B 13 Ionisation energy generally increases across the Period.  Options A–C are incorrect. D 14 Metals are electrical conductors as their metallic lattices contain delocalised electrons which can act as mobile charge carriers . Hence, the greater the number of delocalised electrons the greater the electrical conductivity. The electrical conductivity of A l is greater than that of Mg as each Al atom contributes three valence electrons to the ‘sea’ of delocalised electrons while each Mg atom only contributes two valence electrons. The electrical conductivity of Cu (transition metal) is greater than that of Ca (s –block metal) as the 3d and 4s electrons in Cu can be delocalised for metallic bonding s ince the 3d and 4s orbitals in Cu are similar in energy level. Each Ca atom only contributes two valence electrons to the ‘sea’ of delocalised electrons. D electrical conductivity Na Mg Al Si P S Cl radius Na Mg Al Si P S Cl melting point Na Mg Al Si P S Cl

4 15 Fe2+ KMnO4 Fe3+ + Mn2+ pale green pale yellow colourless A redox reaction occurs between acidified Fe 2+(aq) and KMnO 4(aq). As KMnO 4(aq) is added until in large excess, the solution in the conical flask changes from pale green to pink , and finally purple. B 16 CH2ClCHICO2H Na, heat NaCl(s) + Na I(s) dil. HNO3 NaCl(aq) + Na I(aq) AgNO3(aq) AgCl(s) + AgI(s) white yellow conc. NH3 excess AgI(s) yellow AgCl dissolves The ppt appears more yellow as the white AgCl ppt dissolves on addition of excess conc. NH3. Note: AgI is insoluble in conc. NH3. D 17 A. H S H B. Cl Cl heterolytic fission Cl + Cl C. N H H HH D. 29Cu2+ : [Ar] 3d9 (Note: 29Cu : [Ar] 3d10 4s1) d9 : ___ ___ ___ ___ ___ D acidified Fe2+(aq) (pale green) KMnO4(aq) (purple) The ppt appears more yellow single unpaired electron

5 18 D. X can be either P or S only (Period 3). B. X is P since chloride of sulfur is not in syllabus. C. This is a correct statement if X is P. Element X is deduced to be phosphorus, P A. PCl3 + 3H2O H3PO3(aq) + 3HCl(aq) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 2NaOH(aq) + H3PO3(aq) Na2HPO3(aq) + 2H2O(l) White ppt is not produced. B. PCl3 + Cl2 PCl5 C. P is a solid at room temperature. D. P4O10 + 6H2O 4H3PO4 (phosphoric acid) A 19 C 20 C sp2 sp2 sp2 sp2 sp2 sp2

6 21 C H H C H CH3 H I C H CH3CH3 I propene 2-iodopropane I–Cl bond is polarised; Cl being more electronegative than I By pattern recognition, C H H C H CH3 C H CH3ICH2 Cl propene I Cl Alternatively, by Markovnikov’s rule: C H H C H CH3 I Cl slow C H CH3C H H I C H CH3C H H I 2o carbocation 1o carbocation Cl Cl C H CH3C H H I Cl C H CH3C H H ICl major minor 2o carbocation is more stable than 1o carbocation. C 22 HO2CC CCO2H H H HO2CC H CCO2H H OH H steam & H2SO4 electrophilic addition (hydration) hot acidified KMnO4 oxidation of 2o alcohol to ketone HO2CC CCO2H H O H oxaloacetic acidT D + – + –

7 23 A. n(C2H3OCl) = 1 / 78.5 = 0.01273 mol n(Cl–) produced from hydrolysis with NaOH(aq) = 0.01273 mol n(AgCl) produced = 0.0127 mol B. n(C6H10Cl2) = 1 / 153 = 6.536 x 10–3 mol n(Cl–) produced from hydrolysis with NaOH(aq) = 2 x 6.536 x 10–3 = 0.01307 mol n(AgCl) produced = 0.0131 mol C. Cl Cl does not undergo hydrolysis (nucleophilic substitution) with OH- D. n(C4H4O2Cl2) = 1 / 155 = 6.452 x 10–3 mol n(Cl–) produced from hydrolysis with NaOH(aq) = 2 x 6.452 x 10–3 = 0.0129 mol n(AgCl) produced = 0.0129 mol B 24 Br aq NaOH OH P Br alc. NaOH Q Br alc. NaCN CN R Order of increasing Mr: Q, P, R B

8 25 A. OH B. OH gives positive Tri–iodomethane Test (produces pale yellow ppt with alkaline aq. I2) CH3 C OH R/