2024 Prelims VJC H2 Chem P3 (Ans)

© VJC 2024 9729/03/PRELIM/24 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/03 Paper 3 Free Response Candidates answer on the Question Paper Additional Materials: Data Booklet 16 September 2024 2 hours READ THESE INSTRUCTIONS FIRST Write your name and CT group 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 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 Section A 1 / 23 2 / 19 3 / 18 Section B 4 / 20 OR 5 / 20 Total / 80 This document consists of 30 printed pages and 0 blank page.

2 © VJC 2024 9729/03/Prelim/24 Section A Answer all the questions in this section. 1 (a) Chlorine is bubbled through 100 cm3 of hot 4.0 mol dm–3 sodium hydroxide until the reaction is complete. 6NaOH(aq) + xCl2(aq) → yNaCl(aq) + zNaClO3(aq) + 3H2O(l) (i) State the type of reaction that occurs. Explain your answer in terms of changes in oxidation numbers. [1] • Disproportionation reaction as Cl2 is reduced and oxidised simultaneously. Cl2 is reduced to Cl–, oxidation number of Cl decreases from 0 to –1. Cl2 is oxidised to ClO3–, oxidation number of Cl increases from 0 to +5. (ii) Determine the values of x, y and z. [1] • x = 3 y = 5 z = 1 (iii) Determine the concentration of Na+(aq), in mol dm–3, after the reaction. [1] • Concentration of Na+(aq) after the reaction = 4.0 mol dm–3 (Na+ does not participate in the reaction thus its concentration remains the same. ) (b) Compound A is an ether with molecular formula C 4H10O. When A is heated in a sealed container, an equilibrium mixture is produced. C4H10O(g) ⇌ C2H6(g) + CO(g) + CH4(g) H = –7.00 kJ mol–1 Table 1.1 shows the activation energy, Ea, for the reaction in the presence and absence of I2. Table 1.1 Ea (with I2) / kJ mol–1 Ea (without I2) / kJ mol–1 143 224 (i) State the role of I2 in this reaction and explain what effect it has on the value of Kc. [1] • I2 is a catalyst in the reaction and it has no effect on the value of Kc as it does not affect the position of equilibrium. (ii) Complete the energy profile diagram for this reaction in Fig. 1.1. Include labels to show the enthalpy change and the activation energy data in Table 1.1.

3 © VJC 2024 9729/03/Prelim/24 [Turn over Fig. 1.1 [2] 1m: 2 curves and exothermic 1m: all correct labels include state symbols (iii) Suggest the effect of increasing the pressure on the position of equilibrium. [1] • When the pressure is increased, by Le Chatelier’s Principle, the position of equilibrium will shift to the left to decrease total number of moles of gas. (c) Potassium chloride, KCl, and magnesium chloride, MgCl2, are both ionic solids. The following data can be used to answer some parts of this question. Table 1.2 standard enthalpy change value / kJ mol–1 standard enthalpy change of solution, Hosol, of KCl +15 lattice energy, Holatt, of KCl(s) –701 standard enthalpy change of hydration, Hohyd, of K+ –322 reaction progress energy / kJ mol–1 C4H10O(g) reaction progress energy / kJ mol–1 C4H10O(g) C2H6(g) + CO(g) + CH4(g) –7 224 143

4 © VJC 2024 9729/03/Prelim/24 standard enthalpy change of hydration, Hohyd, of Cl– –364 standard enthalpy change of solution, Hosol, of MgCl2 –155 lattice energy, Holatt, of MgCl2(s) –2493 (i) Define the term entropy and state the effect on the entropy of the chemical system for the following reaction. Explain your answer. [2] K+(g) + Cl–(g) → KCl(s) • • Entropy is a measure of the degree of disorder of a system, indicated by the physical arrangement (physical chaos) of particles and spread of energy (thermal chaos) in the particles of the system. Entropy decreases as the number of moles of gaseous particles decrease (from 2 mol to 0 mol) and there is a decrease in number of ways of arranging fewer gaseous particles. (ii) Potassium chloride dissolves readily in water at 25°C. By considering the enthalpy change and Gibbs free energy change, state and explain the sign of the standard entropy change for the dissolution of potassium chloride. [1] • Since ∆Go < 0, and Ho > 0, from ∆Go = Ho – TSo, magnitude of Ho < magnitude of 298So (T = 298 K under standard conditions). Thus So > 0 under standard conditions. (iii) Define enthalpy change of hydration. [1] • The energy change when one mole of gaseous ions is dissolved in a large amount of water. (iv) Complete the energy cycle involving the enthalpy change of solution (Hsol), lattice energy ( Hlatt) of magnesium chloride, and the enthalpy changes of hydration (Hhyd). Label the enthalpy changes in your diagram. State symbols should be used. [2] MgCl2(s)

5 © VJC 2024 9729/03/Prelim/24 [Turn over 1m: correct Mg2+(aq) + 2Cl–(aq) and Mg2+(g) + 2Cl–(g) 1m: correct directions of arrows with labels (v) Hence, calculate the enthalpy change of hydration of magnesium ions, Mg2+. Show your working. [1] • Hsol = Hhyd – Hlatt –155 = Hhyd(Mg2+) + 2(–364) – (–2493) Hhyd(Mg2+) = –1920 kJ mol–1 (vi) Explain why the lattice energy of MgC l2 is more exothermic than the lattice energy of KCl. [1] • Lattice energy  q+q− r++ r− Mg2+ has a greater charge and smaller size than K+. Hence, there is greater attraction (OR stronger ionic bonds) between Mg 2+ and Cl– ions, leading to more exothermic lattice energy in MgCl2. (vii) Molten magnesium chloride is electrolysed for 15.0 minutes by a constant current. At the cathode, 4.75 ⨯ 1022 magnesium atoms are produced. Calculate the value of the current used. [2] • • Mg2+ + 2e– → Mg Amount of e– used = 2( 4.75 ⨯ 1022 6.02 ⨯ 1023) = 0.158 mol Q = 0.158 ⨯ 96500 = I ⨯ 15.0 ⨯ 60 I = 16.9 A (d) An electrochemical cell consisting of an Fe3+/Fe2+ half-cell and a Cl2/Cl– half-cell is set up. The cell reaction for the electrochemical cell is shown below. Cl2 + 2Fe2+ → 2Cl– + 2Fe3+ In this experiment, the Fe 2+ concentration is 0.15 mol dm –3. Concentrations of all other species remain at their standard values. The Nernst equation is shown below. E = Eo + 0.059 n lg [oxidised species] [reduced species] MgCl2(s) Hsol Mg2+(aq) + 2Cl–(aq) OR MgCl2(aq) Hlatt Hhyd Mg2+(g) + 2Cl–(g)

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