2020 HCI Prelim P2 Answers

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 HWA CHONG INSTITUTION 2020 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 2 1 (a) (i) HCl: does not decompose even on strong heating HBr: gives brown fumes of Br2(g) on strong heating HI: gives large amounts of violet fumes of I2(g) when a red-hot rod is plunged into a jar of HI gas Observation of brown & violet fumes for HBr and HI: 1m Description of trend in ease/rate/extent of decomposition: 1m 1 (a) (ii) The thermal stabilities of these halides decrease in the order HC l > HBr > H I as the H-X bond strength/bond energy decreases in the same order , and less energy is needed to break the H-X bond. [1] 1 (b) (i) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) [1] (Minus 0.5m for wrong or no state symbols, minus 0.5m if acid/base not written as a compound) 1 (b) (ii) HCl(aq) + NH3(aq)  NH4+(aq) + Cl(aq) [1] (Minus 0.5m for wrong or no state symbols, minus 0.5m if acid/base not written as a compound) Acceptable alternatives: HCl(g) + NH3(g)  NH4Cl(s) HCl(aq) + H2O(l)  H3O+(aq) + Cl (aq) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) – provided this answer was not given in (b)(i) 1 (c) (i) From the pH, [H+] = 102.10 = 7.94 × 103 mol dm3 [1] Since [H+] < [HF], HF does not dissociate fully to give H + ions and is a weak acid. [1] OR If HF is a strong acid, pH of a 0.100 mol dm3 HF = lg 0.100 = 1.00 [1] Since the actual pH of HF is higher than if it were a strong acid, HF does not dissociate fully to give H+ ions and is a weak acid. [1] OR

1 Degree of dissociation,  = (102.10)  0.100 = 0.0794 [1] Since the degree of dissociation is less than 1, HF does not dissociate fully to give H+ ions and is a weak acid. [1] 1 (c) (ii) Ka = (102.10)2  (0.100 – 102.10) = 6.85 × 104 mol dm3 [1] 1 (c) (iii) Both titration curves would show a decrease in pH as the acid is added. OR Both titration curves would show a sharp drop in pH when 12.50 cm3 of the acid is added. [1] 1 (c) (iv) 1. The pH at equivalence point is more than 7 when HF is used but equal to 7 when HCl is used [], as F is able to hydrolyse in water, but not Cl[]. 2. After equivalence point, the pH remains relatively constant/curve is more gentle or flatter when HF is used, while the pH drops more when HC l is used []. This is due to formation of a buffer solution consisting of F and excess HF after equivalence point but no buffer is formed for Cl and HCl []. 3. When 50 cm3 of the acid is added, pH of solution is lower when HCl is used [] as HCl is a stronger acid which dissociates to give higher concentration of H+ than HF of the same concentration []. 0.5m for each description [] and each corresponding explanation [] 2 (a) 26Fe: 1s22s22p63s23p63d64s2 26Fe2+: 1s22s22p63s23p63d6 26Fe3+: 1s22s22p63s23p63d5 Small jump from 6 th to 7th IE, indicating 7 th electron is removed from inner 3p subshell of Fe2+ / there are 6 electrons in the outer 3d subshell. Therefore x = 2. [1] x = 2 & attempt to explain, e.g. only mention small jump from 6 th to 7th IE but never interpret this jump [1] complete reasonable explanation 2 (b) (i) Diagram must include:  Concentration of 1 mol dm−3 for all aqueous solutions (H+, C4H4O62–, HCO2–)  1 bar pressure for CO2(g), H2(g) , 25 oC for set-up  (Platinised) platinum electrodes / graphite electrode  Salt bridge (no need to mention concentrated KNO3) & Voltmeter (symbol V accepted) with external wires [1x2] correct each half cell [1] voltmeter and salt bridge

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 2 (b) (ii) Both reactants S2O82– and C4H4O62– ions are negatively charged, will repel [1/2] each other, thus reaction is slow due to high activation energy. [1/2] 2 (b) (iii) Step 1 S2O82– + 2e– 2SO42– Eo = +2.01 V Fe3+ + e– Fe2+ Eo = +0.77 V Overall equation for step 1: S2O82– + 2 Fe2+ → 2SO42– + 2 Fe3+ [1] Eo cell = +2.01 – (+0.77) = +1.24 V > 0  reaction is spontaneous [1/2] Step 2 2CO2 + 2HCO2– + 6H+ + 6e–  C4H4O62– + 2H2O Eo = _1.02 V Fe3+ + e– Fe2+ Eo = +0.77V Overall equation for step 2: 6Fe3+ + C4H4O62– + 2H2O → 6Fe2+ + 2CO2 + 2HCO2– + 6H+ [1] Eo cell = +0.77 – (_1.02) = +1.79 V > 0  reaction is spontaneous [1/2] 2 (c) Tartaric acid has a lower pKa1. (A) (B) Conjugate based of tartaric acid (A) is more stable than that of malic acid (B). The negative charge on the carboxylate ion of A is dispersed to a larger extent [1] due to the presence of 3 electron withdrawing groups (2−OH and 1−CO2H)/ more electron withdrawing groups [1], vs only 2 electron withdrawing groups/ lesser withdrawing groups in B. OR The negative charge on the carbox ylate ion of A is dispersed to a larger extent [1] due to more electronegative atoms [1] helping to disperse the negative charge for A [No marks for no explanation] 2 (d) (i) In Step I, the π bond of the unsaturated carbon in C=O is broken and a new C−O bond is formed [1], suggesting that an addition reaction takes place. In Step III, the C−OH 2+ bond is broken and a C=O π bond is formed with an elimination of a H2O molecule [1], suggests elimination reaction takes place.

2 2 (d) (ii) [1] 3 (a) (i) PCl5(g)  PCl3(g) + Cl2(g) Initial /mol 0.2 3.0 0 Change/mol -0.05 +0.05 +0.05 Equi/mol 0.15 3.05 0.05 Total amount of gas = 0.15 + 0.05 + 3.05 = 3.25 mol [1] 3 (a) (ii) P = nRT/V = 3.25  8.31  500 / 10  103 = 1.3504 106 = 1.35 106 Pa (3 s.f.) [1] ecf from (i) 3 (a) (iii) P(PCl5) = 0.15  1.3504 106 / 3.25 = 6.2326  104 Pa P(PCl3) = 3.05  1.3504 106 / 3.25 = 1.2673  106 Pa P(Cl2) = 0.05  1.3504 106 / 3.25 = 2.0775  104 Pa [1] Kp = (1.2673  106)  (2.0775  104) / (6.2326  104) = 4.22  105 Pa (3 s.f.) [1] ecf from (i) & (ii) 3 (a) (iv) When volume of container is decreased, total pressure increases hence position of equilibrium shifts left [1] so as to favor production of fewer number of moles of gas to offset the increase in pressure [1]. 3 (a) (v) Since value of Kp is large, position of equilibrium is on the right side, implying that the reaction is spontaneous, the sign of Gr would be negative. [1] OR Given that Gr = RTlnK (and K > 1), therefore Gr must be negative. [1] 3 (a) (vi) No effect as adding a catalyst only changes the rate of both forward and backward reactions to the same extent, achieving equilibrium faster. It does not affect the amount of reactants and products produced. [1] OR No effect as Kp is only affected by temperature. [1] 3 (a) (vii) Graphs that show that PCl3 deviates more from ideality than Cl2. [1] Examples:

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 3 (a) (viii) The gases would behave less ideally at 300 K. [1] At lower temperature of 300 K, the gas particles would have lesser kinetic energy. Intermolecular forces (dispersion forces) between gas molecules would be more significant in this case resulting in less ideal behaviour. [1]

3 3 (b) (i) Isomers with unbranched structure: OH OH Pentan-2-olPentan-1-ol 0.5m each for correct structure; 0.5m each for correct name Isomers with branched structure (non-exhaustive): OH 3-Methylbutan-1-ol OH 2-Methylbutan-1-ol OH OH 3-Methylbutan-2-ol 2-Methylbutan-2-ol 0.5m each for correct structure; 0.5m each for correct name 3 (b) (ii) CH3SO3– is the weaker base. [1] 3 (b) (iii) CH3SO3– is more stable or (better leaving group ) than OH –, therefore B can undergo nucleophilic substitution more readily with CN–. [1] 4 (a) (i) Each structure [1] x 2 Step 1: Ethanolic KOH,