RI Prelim P3 (Ans)

1 2020 RI H2 Chemistry Prelim Paper 3 – Suggested Solutions Section A 1(a)(i) OR 1(a)(ii) CH3OCH3 has 2 bond pairs and 2 lone pairs around O , hence it has a bent shape. Since lone pairs exert a greater repulsion than bond pairs, it has a bond angle of 105o 1(a)(iii) The permanent dipole-permanent dipole (pd-pd) intermolecular forces in CH3OCH3 is stronger than the pd−pd intermolecular forces in CF2Cl2. This is because CH3OCH3 has a greater net dipole moment than CF2Cl2. 1(b)(i) The two assumptions are:  The gas particles exert negligible attractive forces/intermolecular forces of attraction on each other.  The volume of gas particles is negligible. Other assumptions:  The gas particles are in constant random motion.  Collision between gas particles are perfectly elastic.  The average kinetic energy of gas particles is proportional to absolute temperature. 1(b)(ii) 1 p ideal gas 293 K

2 1(b)(iii) At higher temperature of 500 K, the N2O gas molecules have higher kinetic energy and are able to overcome the intermolecular attractive forces. Therefore, the intermolecular attractive forces between them are less significant and the N2O gas molecules deviate from ideal ity to a smaller extent. 1(c)(i) Kp = PCH3OCH3PCO2 (PCO)3 (PH 2 )3 units: atm−4 OR Pa−4 1(c)(ii) pV = nRT p = nRT/V Partial pressure of CH3OCH3 at equilibrium = 732  8.31  500 101325  10 = 3 atm (shown) 1(c)(iii) Let the initial partial pressure of CO be x. 3CO(g) + 3H2(g) ⇌ CH3OCH3(g) CO2(g) initial / atm x 40 – x 0 0 change / atm –0.6x = –9 –0.6x +3 +3 equilibrium/ atm 0.4 x 40 – 1.6x 3 3 Change in partial pressure of CO: 0.60 x = (3)(3) = 9 atm Equilibrium pressure of CO = 0.4x = ( 0.4 0.6) (9) = 6.0 atm 1(c)(iv) Equilibrium pressure of H2 = 28 – 6 – 3 – 3 = 16.0 atm Kp = P(CH3OCH3)P(CO2) P(CO)3P(H2)3 = (3)(3) (6)3(16)3 = 1.02  10−5 atm−4 OR 9.65  10−26 Pa−4 1 p 500 K ideal gas 293 K

3 1(c)(v) When Ar(g) is added at constant volume, although the total pressure in the vessel increases, the partial pressure of the reactants and products remains the same. Hence there is no shift in position of equilibrium (or Q still equals to K) and the partial pressure of CH3OCH3 will remain the same. 1(c)(vi) When temperature increases, position of equilibrium (1) will shift left to favour the backwards endothermic reaction. As a result less products and more reactants will be formed and the Kp value decreases. 2(a)(i) The halogen radicals formed from Halon 1211 remove/combine with the radicals in the combustion reaction to terminate the chain reaction. 2(a)(ii) Cl2(aq) + 2Br−(aq)  Br2(aq) + 2Cl−(aq) HBr(g) dissolves in water to form H+(aq) and Br−(aq). Br− can be oxidised/displaced by Cl2 (a halide can be oxidised by the halogen above it in the group) to form Br2(aq), which is orange. 2(a)(iii) BE (H–Cl) = +431 kJ mol –1 > BE (H–Br) = +366 kJ mol –1 Since H–Cl has a higher bond energy, it is harder to break the H–Cl bond and the thermal stability of HCl is greater than that of H–Br. 2(b) Cl2 + Fe → 2Cl – + Fe2+ Ecell = +1.36 – (–0.44) = +1.80 V > 0 Cl2 + Fe2+ → 2Cl – + Fe3+ Ecell = +1.36 – (+0.77) = +0.59 V > 0 Fe is oxidised by Cl2 to form Fe3+. I2 + Fe → 2I – + Fe2+ Ecell = +0.54 – (–0.44) = +0.98 V > 0 I2 + Fe2+ → 2I – + Fe3+ Ecell = +0.54 – (+0.77) = –0.23 V < 0 (not spontaneous) Fe is oxidised by I2 to form Fe2+. Since Cl2 can further oxidise Fe2+ to Fe3+ (or to a higher oxidation state) while I2 cannot, this shows that Cl2 is a stronger oxidising agent than I2. 2(c)(i) Type of reaction involving CO2: Reduction The oxidation number of C decreases from +4 in CO2 to +2 in HCOOH. 2(c)(ii) Cathode: CO2 + 2H+ + 2e–  HCOOH Anode: 2H2O  O2 + 4H+ + 4e– Ecell = Ecathode – Eanode = –0.61 – (+1.23) = –1.84 V ∆G = –nFE = – (4)(96500)(–1.84) = +710240 = +710000 J mol–1 Since Ecell < 0 / ∆G > 0, it shows that the reaction in equation 1 is not spontaneous. Thus, a power supply is required for the reaction to occur.

4 2(d)(i)  CO2 has low solubility in aqueous solution. The presence of gas diffusion electrode enables the reaction between CO2 and H+ to occur without the need to dissolve CO2 into the electrolyte. OR The presence of gas diffusion electrode prevents CO2 from dissolving in water to form carbonic acid. OR The porous gas diffusion electrode will have a larger surface area for the reaction to take place and hence increasing the rate of reaction at the cathode.  The Nafion™ membrane prevents HCOOH molecules to come into contact with the anode, thereby preventing the oxidation of HCOOH. 2(d)(ii) CO2 + 2H+ + 2e− ⇌ HCOOH E = –0.61 V A higher pressure of CO2 will cause the position of equilibrium to lie more to the right/ favour the forward reaction at the cathode. This results in a less negative E(CO2/HCOOH) value and the Ecell (which is equal to E(CO2/HCOOH) – E(O2/H2O)) will become less negative, hence requiring a lower power supply. 2(e) 2(f)(i) Reduction half-equation: MnO4 + 2H2O + 3e  MnO2 + 4OH Oxidation half-equation: HCOO + 3OH  CO32 + 2H2O + 2e 3HCOO + 2MnO4 + OH  2MnO2 + 3CO32 + 2H2O 2(f)(ii) Amount of MnO2 collected = 0.45 86.9 = 0.005178 mol Amount of HCOOˉ reacted = 3 2 x 0.0051784 mol = 0.0077675 mol Concentration of HCOOˉ reacted = 0.0077675 ÷ 30.00 1000 = 0.259 mol dm−3 V T = 298 K [H+ (aq)] = 1 mol dm–3 H2 (1 bar) Pt salt bridge e– [H+(aq)] = 1 mol dm–3 [HCOOH(aq)] = 1 mol dm–3 CO2 (1 bar) Pt e–

5 3(a)(i)  Let the volume of C6H5CH(OH)COOH (denoted by HA) and the volume of NaOH used be V dm3. Initial 0.25 molHAnV Initial 0.125 molNaOHnV   formed reacted 0.125 molNaOHAn n V   Unreacted 0.25 0.125 0.125 molHAn V V V  The resultant solution is an acidic buffer at maximum buffering capacity. pH of resultant solution = pKa = –lg (3.89  10–4) = 3.41  Initial nHA = (0.25)(10/1000) = 0.0025 mol Initial nNaOH = (0.125)(20/1000) = 0.0025 mol = nA  The resultant solution is a salt solution containing A. A(aq) + H2O(l) ⇌ HA(aq) + OH(aq) [A] = 0.0025 30 x 1000 = 0.0833 mol dm–3 Kb of A– = 10−14 / (3.89  10–4) = 2.57 x 10–11 mol dm–3                         11 6 3 6 OH A 2.57 10 0.0833 1.46 10 mol dm lg 1.46 10 5.83 bK pOH pH = 14 – 5.83 = 8.17 3(a)(ii) Initial nHA = (0.25)(10/1000) = 0.0025 mol Initial nNaOH = (0.125)(x/1000) = 0.000125x mol nA- = 0.000125x mol Unreacted nHA = (0.0025 – 0.000125x) mol pH = pKa + lg ( [A] [HA]) 3.89 = 3.41 + lg ( [A] [HA]) lg ( [A] [HA]) = 0.48 [A] [HA] = 100.48 = 3.02 0.000125x / V (0.0025 – 0.000125x)/V = 3.02 0.000125x = 3.02(0.0025 – 0.000125x) 0.000125x + 0.0003775x = 0.00755 x = 15.0 cm3

6 3(b) 3(c) The weaker intermolecular hydrogen bonds in lactic acid require a much smaller amount of heat energy to overcome than the strong ionic bonding / strong electrostatic forces of attraction between oppositely charged ends of neighbouring zwitterions in alanine. 3(d)(i) ala-met-his-asp ser-phe-ala-met asp-his his-ala-glu-thr