2024 H1 Chem Prelim P2 Ans VJC

© VJC 2024 8873/02/PRELIM/24 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 ……………………………………………….………….. …………………………….. CHEMISTRY 8873/02 Paper 2 Structured Questions Candidates answer on the Question Paper. 12 September 2024 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Name and CT group in the spaces at the top of this page. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all the questions. Section B Answer one question. 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 Section A 1 / 16 2 / 25 3 / 19 Section B 4 / 5 / 20 Total / 80 This document consists of 23 printed pages.

2 © VJC 2024 8873/02/PRELIM/24 Section A Answer all questions in this section in the spaces provided. 1 (a) (i) Write down the electronic configurations of the following species. • Na :1s22s22p63s1 • Al2+ : 1s22s22p63s1 [1] (ii) The electronic configuration of copper atom is 1s22s22p63s23p63d104s1. Sketch the shape of the filled orbital of copper atom with the highest energy on the axes in Fig. 1.1 and label the orbital. • Fig. 1.1 [1] (iii) Write a balanced equation for the third ionisation energy of aluminium. • Al2+(g) → Al3+(g) + e– [1] (iv) Explain why the third ionisation energy of aluminium is greater than the first ionisation energy of sodium. • Al2+ has a greater nuclear charge than Na as it has more protons. The shielding effect is the same for both A l2+ and Na as both have the same electronic configuration (OR number of electrons). • Hence, A l2+ has a greater effective nuclear charge leading to a greater third ionisation energy than the first ionisation energy of sodium. [2] 4s

3 © VJC 2024 8873/02/PRELIM/24 [Turn over (b) Period 3 elements and their compounds show trends in their physical and chemical properties. (i) Sketch a graph to show the melting points of the first five elements in Period 3 on Fig. 1.2. Fig. 1.2 1m: Na, Mg and Al (increasing trend), 1m: Si (highest) and P (lowest) [2] (ii) Explain the variation in melting point of the first five Period 3 elements in terms of their structure and bonding. • Na, Mg and Al have giant metallic structure. Melting point increases from Na to Al due to increasing strength of metallic bonds caused by i ncreasing delocalised electrons. More energy is needed to break the stronger metallic bonds. • Si has a giant molecular structure. It has the highest melting point because a large amount of energy is needed to break the strong covalent bonds between the silicon atoms. • P4 has a simple molecular structure. It has a low melting point because a small amount of energy is needed to overcome the weak instantaneous dipole - induced dipole interactions between the molecules. Accept intermolecular forces of attractions instead of id-id. [3] (iii) An excess of cold water is added to the oxide of sodium and the chloride of silicon separately. Describe the reactions of these compounds with water. Write relevant equation for any reaction that occurs and suggest the pH of the mixture produced in each case. • Na2O dissolves in water to form sodium and oxide ions. The oxide ions react with water to form a strongly alkaline solution of pH 13 (accept pH 12–14). Heat is released as the reaction is exothermic. • Na2O + H2O → 2NaOH • SiCl4 reacts violently with water with heat released. Being covalent, it undergoes complete hydrolysis to give a strongly acidic solution of pH 2 (accept pH 1–3) • SiCl4 + 2H2O → SiO2 + 4HCl [4] • • • • •

4 © VJC 2024 8873/02/PRELIM/24 (c) Tennessine, Ts, is an unstable man -made element. It is found below astatine, At, in Group 17. The chemical properties of Ts and its compounds can be predicted. Suggest an equation for the reaction of NaTs and Br2, assuming Ts follows the same trends as the other elements in Group 17. Explain your answer. • 2NaTs + Br2 → 2NaBr + Ts2 • Br2 is a stronger oxidising agent than Ts 2. Down the group, the oxidising strength of the halogen decreases as it is harder to gain electrons due to the weakening of electrostatic forces of attraction between the nucleus and incoming electrons. [2] [Total: 16] 2 (a) The reaction of ammonia with oxygen to form nitrogen monoxide, NO, is an important industrial process to manufacture nitric acid. The equation for this reaction is shown in equilibrium 2.1. 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) ΔH = −905 kJ mol−1 equilibrium 2.1 (i) The forward reaction in equilibrium 2.1 converts NH3 into NO. Draw the energy profile for this reaction in the diagram below. Label the following parts in your diagram. • reactants • products • activation energy, Ea • enthalpy change, ΔH • correct shape of graph + showing energy level of reactants higher than products + label + reactants and products • correct labels: (1) activation energy, (2) enthalpy change with correct direction of arrows [2] energy progress of reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Ea ΔH

5 © VJC 2024 8873/02/PRELIM/24 [Turn over (ii) Write an expression for the equilibrium constant, Kc in equilibrium 2.1. • Kc = [H2O]6[NO]4 [NH3]4[O2]5 [1] (iii) Deduce the units of Kc in equilibrium 2.1. • mol dm−3 [1] (iv) The reaction in equilibrium 2.1 is carried out in a laboratory. 34.0 g of ammonia and 32.0 g of oxygen was allowed to react in a reaction chamber with a volume of 50 cm 3. At equilibrium, it was found that 20% of ammonia has reacted. Calculate a value for the equilibrium constant, Kc. 4NH3 + 5O2 ⇌ 4NO + 6H2O initial amount 34 14.0+3 × 1.0 = 2.00 32 16.0 × 2 = 1.00 0 0 change − ( 20 100 ×2.0) = − 0.400 − ( 5 4 ×0.400) = − 0.500 + 0.400 + (6 4 × 0.400) = 0.600 equilibrium amount 1.60 0.500 0.400 0.600 equilibrium concentration 1.60 50 1000⁄ = 32.0 0.500 50 1000⁄ = 10.0 0.400 50 1000⁄ = 8.00 𝟎.60 50 1000⁄ = 12.0 Kc= (12.0)6(8.00)4 (32.0)4(10.0)5 = 0.11664 = 0.117 mol dm−3 Correct concentrations for both reactants and products at equilibrium Correct value for Kc (no need to mark unit again.) [2] (v) Explain how a low temperature and a low pressure would lead to a high yield of nitrogen monoxide in equilibrium 2.1. • At low temperature, according to Le Chatelier’s principle (LCP), the position of equilibriu