VJC 2025 H1ChemistryPrelimP2 Ans

© VJC 2025 8873/02/PRELIM/25 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 ANSWERS ……………………………………………….………….. …………………………….. CHEMISTRY 8873/02 Paper 2 Structured Questions Candidates answer on the Question Paper. 29 August 2025 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 / 11 2 / 20 3 / 9 4 / 7 5 / 13 Section B 6 / 7 / 20 Total / 80 This document consists of 17 printed pages and 1 blank page.

2 © VJC 2025 8873/02/PRELIM/25 Section A Answer all questions from this section in the spaces provided. 1 (a) The element sodium can exist as a number of isotopes. Complete the Table 1.1 for two isotopes species of sodium. • 1 mark for correct cation • 1 mark for correct number of protons, neutrons and electrons • 2 marks. 1 mark for each correct electronic configuration (ecf) Isotopic species protons neutrons electrons electronic configuration Na𝟏𝟏 𝟐𝟑 11 12 11 1s22s22p63s1 N𝟏𝟏 𝟐𝟒 a+ 11 13 10 1s22s22p6 [4] (b) Sodium oxide, Na2O and sodium peroxide, Na2O2 are two compounds that can be produced when sodium metal is heated in oxygen under suitable conditions. Both Na2O and Na2O2 are ionic compounds. Each compound contains only one single anion and the charge on each anion is the same. (i) Write the formulae of the anions in Na2O and Na2O2. • O2– in K2O and O22– in K2O2 [1] (ii) Draw a ‘dot-and-cross’ diagram of the anion in Na2O2, showing the outer shell electrons only. • [1] (iii) Predict the relative magnitudes of the melting points of Na 2O and Na 2O2. Explain your answer. L.E.  𝑞+𝑞− 𝑟++𝑟− • Both O2– and O22– are doubly charged / same charge and O2– is smaller in size than O22-. • Hence, Na 2O has a larger magnitude of lattice energy than that of Na 2O2 [or stronger ionic bonds / electrostatic forces of attraction in Na2O]. Melting point of Na2O is higher than that of Na2O2. [2] (c) (i) When the Group 1 cation, M+, is passed through an electric field, it is deflected through an angle of +5.0. The angle of deflection is dependent on the relative masses and charge of the particles as shown.

3 © VJC 2025 8873/02/PRELIM/25 [Turn over Angle of deflection ∝ charge mass Given that the same electric field deflected 92Sr3+ through an angle of +22, calculate the relative atomic mass (Ar) of M. Hence suggest a possible identity of the M. Angle of deflection ∝ k |charge/mass| For Sr3+, 22 = k |3/92| • k = 675 For unknown Group 1 cation, 5.0 = 675 |1/Ar| Ar = 135 • Hence the Group 1 metal is likely to be caesium. [2] (ii) Explain why the second ionisation energy of M is more endothermic than its first ionisation energy. • The second electron is removed from an inner shell , which is closer and more strongly attracted to the nucleus. More energy is needed to remove the second electron, and second ionisation energy is more endothermic. [1] [Total:11] 2 Nitrogen is a vital element for sustaining life on Earth. Despite its high abundance in the atmosphere, the availability of reactive nitrogen compounds for biological and industrial use was historically limited. The development of the Haber process for the synthesis of ammonia, followed by the Ostwald process for the large-scale production of nitric acid from ammonia, in the early 20th century marked a significant breakthrough in expanding the availability of reactive nitrogen species for agricultural and industrial applications. (a) Ammonia is manufactured from nitrogen and hydrogen by the Haber process as shown in the equation: N2(g) + 3H2(g) 2NH3(g) Hrxn = –92 kJ mol–1 Under certain conditions , the system reache d dynamic equilibrium with the following gas concentrations. gas concentration / mol dm−3 nitrogen 1.36 hydrogen 1.84 ammonia 0.142 (i) Explain what is meant by dynamic equilibrium. • Dynamic equilibrium exists in a reversible reaction in a closed system in which the rates of the forward and backward reactions are equal, resulting in no net change in the concentrations of the reactants and products. [1] (ii) Write an expression for the equilibrium constant, Kc, for the Haber process. • Kc = [NH3]2 [N2][H2]3

4 © VJC 2025 8873/02/PRELIM/25 [1] (iii) Calculate the value of Kc given the following equilibrium concentrations and state the units of Kc. Kc = (0.142)2 (0.136)(1.84)3 • = 2.38  10−3 • for correct units mol−2 dm6 [2] (iv) Suggest a reason why the activation energy for the Haber process is high. • The activation energy of the process is high due to the high bond energy of the NN bond. [1] (v) Describe and explain the conditions required for the favourable production of ammonia in Haber process. • High pressure (200 atm) favours the production of ammonia. By Le Chatelier’s Principle, the equilibrium position will shift right to reduce the increase in pressure leading to greater amount of ammonia. • Low temperature favours the production of ammonia. By Le Chatelier’s Principle, the equilibrium position will shift right to produce more heat leading to greater amount of ammonia. However low temperature will mean slower rate of reaction . Hence temperature is kept high (450 oC) for faster rate of reaction. • Addition of catalyst (finely divided iron) is used to increase the rate of reaction but not the yield of ammonia. • Continuous removal of NH3 as it forms shifts the position of equilibrium to the right, thereby favour the production of NH3. Total 4 marks. Max 3 marks [3] (b) A large proportion of the ammonia manufactured is then used to manufacture nitric acid, which is another industrially important compound. In the chemical industry, synthesis of nitric acid (HNO3) is commonly carried out via the Ostwald process. The first stage in the process involves the catalytic oxidation of ammonia to nitric oxide (NO): 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H = –905 kJ mol–1 (i) Explain, in terms of bonds, why the enthalpy change of the reaction is exothermic. • The energy absorbed to break bonds is less than the energy released in forming bonds. [1] (ii) Using relevant bond energies in the Data Booklet and the given value of enthalpy change for the reaction, determine the N=O bond energy in nitric oxide. • ∆H = ∑B.E. (reactants) – ∑B.E. (products) –905 = 12 (B.E. N–H) + 5 (B.E O=O) – [4 (B.E. N=O) + 12 (B.E O–H]] • –905 = 12(390) + 5(496) – 4(B.E. N=O) – 12(460) –905 = 7160 – 4(B.E. N=O) – 5520 –905 = 1640 – 4(B.E. N=O) 4(B.E. N=O) = 2545 • B.E (N=O) = +63