2023 ASRJC-NYJC-SAJC H3 Chemistry Prelim Answer

H3 Section A A collaboration between ANDERSON SERANGOON JUNIOR COLLEGE NANYANG JUNIOR COLLEGE St. ANDREW’S JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 3 CANDIDATE NAME CLASS CHEMISTRY 9813/01 Paper 1 21 September 2023 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet I n s e r t READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer two questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 42 printed pages and 0 blank page and 1 insert.

2 H3 Chemistry 9813/01 Answer all questions in this section. 1 (a) (i) Zn  ZnO Zn + H 2O  ZnO + 2H+ + 2e– Zn + 2OH–  ZnO + H2O + 2e– (ii) [O] Zn + 2OH−  ZnO + H2O + 2e− [R] O2 + 2H2O + 4e−  4OH− Overall: 2Zn + O2  2ZnO (iii) Zn + ½ O2  ZnO E 0cell = +0.40 – (−1.25) = +1.65 V G0 = – 2 × 96500 × 1.65 = −318450 J mol−1 = −318 kJ mol−1 (3 sig fig) (iv) Since air is a mixture, the pressure of oxygen in the atmosphere is less than 1 bar. This will cause the E0(O2/OH−) to be less positive resulting in a less positive E 0cell and hence a smaller magnitude in the Gibbs free energy change. (b) CO2 + OH−  HCO3− This will decrease the concentration of OH− in the electrolyte and makes the E0(ZnO/Zn) more positive which will make the E0cell less positive. Explanation (not required in answer) ZnO + H2O + 2e– ⇌ Zn + 2OH– E0(ZnO/Zn) = −1.25 V When [OH −] decreases, position of the equilibrium will shift to the right and favor the reduction reaction. This will cause E0(ZnO/Zn) to become more positive. Since E 0cell = E 0(O2/OH–) – E 0(ZnO/Zn), a more positive E 0(ZnO/Zn) will cause the E 0cell to become less positive. OR CO2 + OH− + Zn2+  ZnCO3 + H2O The insoluble zinc carbonate formed can clog up the pores of the carbon electrode and can prevent the reduction of oxygen to hydroxide ions. (c) (i) Oxidation state of O in LiO2: −0.5 Oxidation state of O in Li2O2: –1 Oxidation state of O in Li2O: –2 (ii) O2 + Li+ + e−  LiO2 (1) 2LiO2  Li2O2 + O2 (2) LiO2 + Li+ + e−  Li2O2 (3) Li2O2 + 2Li+ + 2e−  2Li2O (4) (1) x 3 + (2) + (3) + (4) × 2 3O2 + 3Li+ + 3e−  3LiO2

3 H3 Chemistry 9813/01 [Turn Over + 2LiO2  Li2O2 + O2 + LiO2 + Li+ + e−  Li2O2 +2Li2O2 + 4Li+ + 4e−  4Li2O 2O2 + 8Li+ + 8e−  4Li2O O2 + 4Li+ + 4e−  2Li2O (iii) C C OO x x x xxO x x O x O x O 2- x x xxx x xx x x xxxx (d) (i) O OH HO O H+ O OH HO O H O H H O OH OO H OH H H HO OH OO H OH H HO OH O HO OH H O OH OO H OH H H+ (ii) Mechanism A OLiO O O e OLiO O O C O O OLiO O O O O Li+ OLiO O O OLi O Mechanism B OLiO O O O O OLiO O O C O O e OLiO O O O O Li+ OLiO O O OLi O (e) (i) Energy produced by 1 kg of Li in a Li-air cell = 32702 × 2.91 × 3.6 × 1000

4 H3 Chemistry 9813/01 = 3.426 × 108 = 3.43 × 108 J Energy produced by 1 kg of Na in a Na-air cell = 6328 × 2.33 × 3.6 × 1000 = 5.308 × 107 = 5.31 × 107 J (ii) Mass of Li required to produce 140 MJ of energy = 6 8 140 10 13.426 10   = 0.4086 kg Cost of Li-air cell to produce 140 MJ energy = 0.4086 × $68 = $27.80 Mass of Na required to produce 140 MJ of energy = 6 7 140 10 15.308 10   = 2.638 kg Cost of Na-air cell to produce 140 MJ energy = 2.638 × $2 = $5.28 Cost of Na-air cell to produce 140 MJ of energy is less than the cost of Li-air cell to produce 140 MJ energy. (shown) (iii)  For a given mass, a Li-air cell can store more energy.  Since lithium has a lower atomic mass t han sodium, a Li-air cell is lighter which improves the portability of the system.  Since Li is less reactive than Na, it is safer to store and transport. (f) Advantage:  Air (oxygen) is readily available.  Metal-air cell has high energy density. Disadvantage: Metal-air cell with aqueous electrolyte  There is additional cost incurred to remove carbon dioxide from air due to the use of selective membrane.  Carbon electrode could be oxidized and will need to be replaced.  Catalyst will be corroded over time and will need to be replaced. Metal-air cell with organic electrolyte  Organic solvent is costly.  Organic solvent is combustible.  It produces CO 2 which is a greenhouse gas.

5 H3 Chemistry 9813/01 [Turn Over 2 (a) (i) CHO CHOCHO transition state (ii) The transition state has an aromatic character due to the six delocalised  electrons. This will stabilise the transiti on state and lowers its energy level. Hence, the activation energy of the reaction is low enough for reaction to proceed without a catalyst. (iii) The cis conformation of the diene have the p orbitals of the 2 terminal carbons closer to each other such that they are able to overlap with the p orbitals of the dienophile. (iv) HOMO: LUMO: (v) The energy level of the molecular orbitals of the dienophile for the normal demand reaction will be lower than that of the inverse demand reaction. The electron-withdrawing group on the dienophile will be able to stabilise it (lower energy MO) through resonance effect where the  electrons are delocalised into the electron-withdrawing group. or inductive effect of electron-withdrawing group that helps to disperse the electron density of the C=C bond. 1 2 Energy 4 3

6 H3 Chemistry 9813/01 (vi) For normal demand, For inverse demand, 1 2 Energy 4 3 1 2 Diene Dienophil e 1 2 Energy 4 3 1 2 Diene Dienophil

7 H3 Chemistry 9813/01 [Turn Over (vii) (1) O O CH3 OCH3 (2) NH2 H3COOC NH2 H3COOC NH2 COOCH3 more stable transition state less stable transition state Note: Although the formation of this product did not follow the correct orientation as provided in the question (i.e. R 1 and R2 to be in 1,4-position), it is formed via a more stable transition state. Marks wi ll still be awarded for students who gave the 1,4-position product below. CO2CH3 NH2 CO2CH3 NH2or (b) (i) B is the major product as the nucleophile, C l, will attack at the back side of the less hindered carbon via the SN2 mechanism. Nucleophilic Substitution (SN2) B Cl Br OH CH3 H H Cl H+ O Br CH3 O Br CH3 H

8 H3 Chemistry 9813/01 (ii) Peaks 214 216 218 Ions