TMJC 2025 H2 Chem P3 Answer

CANDIDATE NAME CIVICS GROUP H2 CHEMISTRY 9729/03 Paper 3 Free Response 23 September 2025 2 hours Candidates answer on the Question Paper. Additional materials: Data Booklet This document consists of 30 printed pages and 2 blank pages. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION READ THESE INSTRUCTIONS FIRST Write your name and Civics Group in the spaces at the top of the page. 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 one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Percentage Paper 1 / 30 / 15 Paper 2 / 75 / 30 Paper 3 Section A 1 / 19 2 / 20 3 / 21 Section B 4 / 20 OR 5 / 20 Paper 3 Total / 80 / 35 Paper 4 / 55 / 20 Grand Total / 100

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry Section A Answer all the questions in this section. 1 (a) Describe and explain the trend in the thermal stabilities of the Group 2 carbonates. [2] ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. (b) Calcium carbonate is used in flue -gas desulfurisation applications to remove harmful SO2 and NO2 emissions from fossil fuels burnt in power stations. (i) Draw dot-and-cross diagrams to show the bonding in the molecules of SO 2 and NO2. [2] (ii) With reference to your answer in (b)(i), explain why the bond angle in SO2 is found to be 118 while that in NO2 is 134. [2] ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... (b) (i) (ii) For SO2, as there are 2 bond pairs and 1 lone pair around the central S atom. Since lone pair–bond pair repulsion > bond pair – bond pair repulsion. the bond angle is about 118. However, in NO2, the repulsion between the unpaired electron (or lone electron) and a bond pair is weaker compared to the repulsion between a lone pair and a bond pair. This results in a larger bond angle of 134. (a) Thermal stability of Group 2 carbonates increases down the group. Down Group 2, size of cations increases, and charge density and polarising power of the cations decreases. The cations distort the CO32− / anion electron cloud, and weaken C–O bonds to a smaller extent, and hence more energy is required to decompose the CO32−.

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn over (c) Magnesium carbonate, MgCO 3, can be prepared in the laboratory by the reaction between magnesium chloride solution and aqueous sodium bicarbonate according to the equation: MgCl2(aq) + 2 NaHCO3(aq) MgCO3(s) + 2 NaCl(aq) + H2O(l) + CO2(g) Ho = + 107 kJ mol–1 So = + 360 J mol–1 K–1 (i) Explain why the reaction shows an overall positive value for So. [1] (ii) Calculate G, in kJ mol–1, for the reaction at 50 C. [1] (iii) Explain qualitatively how temperature affects the feasibility of this reaction. [2] ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ⎯⎯ → (c) (i) So is positive as there is an increase in disorder / entropy. This is due to an increase in the number of moles of gaseous particles from 0 mol to 1 mol, resulting in more ways for the particles to arrange themselves. (ii) G = Ho – TSo = +107 – (273 + 50) (0.360) = – 9.28 kJ mol–1 (iii) Both H and S are positive, and –TS is negative. At low temperatures, | H | > | TS |. Hence, G = H − TS > 0. (or G is positive) At high temperatures, |H | < | TS | (or the negative –TS outweighs the positive H. Hence, G = H − TS < 0. (or G becomes negative). Hence, reaction is non- spontaneous/ not feasible at low temperatures, but spontaneous / feasible at high temperatures.

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry (d) High purity magnesium carbonate, MgCO3, is produced industrially at high pressure via a two-step process. step 1: Mg(OH)2 + 2CO2 Mg(HCO3)2 step 2: Mg(HCO3)2 MgCO3 + CO2 + H2O When 150 kg Mg(OH) 2 was reacted with excess carbon dioxide gas, and the resultant Mg(HCO3)2 vacuum dried to remove carbon dioxide and water, 168 kg MgCO 3 was produced. Determine the percentage yield of this industrial process. [2] ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. (e) Titanium dioxide, TiO2, is a widely used white pigment employed to provide whiteness and opacity for paints, papers and toothpaste. Explain, in terms of structure and bonding, why carbon dioxide sublimes at –78 C while titanium dioxide is a crystalline solid with a melting point of 1840 C. [2] ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ……………………………………………………………………………………………...……. ……………………………………………………………………………………….................. ⎯⎯ → ⎯⎯ → (d) Amount of Mg(OH)2 reacted = 150 ×1000 58.3 = 2573 mol = Theoretical amount of MgCO3 produced Theoretical yield of MgCO3 = 2573 x 84.3 = 216900 g = 216.9 kg % yield = 168 216.9 x 100 = 77.5 % (e) CO2 has a simple molecular structure. Little energy is required to overcome the weak instantaneous dipole – induced dipole forces of attraction between CO 2 molecules, resulting in a low sublimation point. TiO2 has a giant ionic structure. A lot of energy is required to break the strong ionic bonds / electrostatic forces of attraction between Ti 4+ ions and O 2– ions / oppositely charged ions , resulting in a high melting point.

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistr