2024 Prelims YIJC H2 Chem P2 (Ans)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUGGESTED ANSWERS CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 10 September 2024 2 hours READ THESE INSTRUCTIONS FIRST This document consists of 17 printed pages and 3 blank pages. For Examiner’s Use 1 / 9 2 / 9 3 / 16 4 / 21 5 / 20 Penalty units significant figures Overall / 75 Write your name and class in the spaces at the top of this 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. 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.

©YIJC [Turn over 2 Answer all the questions in the spaces provided. 1 (a) Calcium carbide, CaC2 is a solid used in the production of ethyne, C2H2. Calcium carbide is produced when calcium oxide reacts with carbon forming calcium carbide and carbon monoxide. (i) Write the electronic configuration of calcium in calcium carbide. Ca2+: 1s2 2s2 2p6 3s2 3p6 (ii) Write an equation, with state symbols for the formation of calcium carbide. CaO(s) + 3C(s) → CaC2(s) + 3CO(g) (iii) Predict all possible intermolecular forces which could exist between carbon monoxide molecules. Explain how these forces arise. Carbon monoxide has a simple molecular struct ure, with instantaneous dipole -induced dipole interactions and permanent dipole-permanent dipole interactions between the molecules. As electrons in orbitals are in constant random motion, there is an uneven distribution of electrons in a molecule/ distortion of the electron cloud. This separation of charges creates an instantaneous dipole in the molecule. Instantaneous dipole induce neighbouring molecules to undergo electron cloud distortion/ induce the formation of dipoles in the neighbouring molecule s and hence instantaneous dipole – induced dipole attraction occurs. Carbon monoxide being polar/ has net dipole moment/ has permanent net separation of charges due to difference in electronegativity between C and O atoms . Hence, permanent dipole-permanent dipole attraction occurs.

©YIJC [Turn over 3 (b) The German chemist Friedrich Wohler discovered that calcium carbide, CaC 2 reacted with water releasing ethyne gas, C2H2 and calcium hydroxide. The ethyne gas was burnt in miner’s lamps and headlights of early motor vehicles for the purpose of illumination. The equation for the formation of ethyne from calcium carbide is as shown. CaC2(s) + H2O(l) → Ca(OH)2(aq) + C2H2(g) An impure sample of calcium carbide of mass 0.752 g was added to 50 cm3 of water. After all the calcium carbide had reacted, 20.00 cm3 of the reaction mixture was removed and titrated against 0.250 mol dmꟷ3 of hydrochloric acid and 34.60 cm3 of hydrochloric acid was required to neutralise the sample. It can be assumed that none of the impurities reacted. Calculate the percentage purity of calcium carbide. Ca(OH)2 + 2HCl → CaCl2 + 2H2O Amount of HCl = 0.0346 × 0.250 = 0.00865 mol Amount of Ca(OH)2 in 20 cm3 = 0.00865 × ½ = 0.004325 mol Amount of Ca(OH)2 in 50 cm3 = 50/20 × 0.004325 = 0.01082 mol Amount of CaC2 = 0.01082 mol Mass of CaC2 = 0.01082 × (40.1 + 24.0) = 0.69308 g % purity = 0.69308 0.752 × 100 % = 92.2 % [Total: 9]

©YIJC [Turn over 4 2 (a) Sodium halides are ionic compounds which have the same crystal structures. Table 2.1 shows data concerning solubility of sodium halides. Table 2.1 value / kJ mol−1 enthalpy change of solution of NaCl(s) −2 enthalpy change of solution of NaI(s) +2 enthalpy change of hydration of Na+(g) −390 lattice energy of NaCl(s) −772 lattice energy of NaI(s) −694 (i) Use the data in Table 2.1, together with an appropriate energy cycle or otherwise , to calculate a value for the enthalpy change of hydration of the chloride ion. Method 1 Hhyd(Cl−) = −384 kJ mol−1 Method 2 Hsol = [Hhyd(Na+) + Hhyd(Cl−)] – lattice energy −2 = [(−390) + Hhyd(Cl−)] – (−772) Hhyd(Cl−) = −384 kJ mol−1

©YIJC [Turn over 5 (ii) By quoting appropriate data from the Data booklet , explain whether the magnitude of enthalpy change of hydration of iodide ion be larger or smaller compared to your answer in (a)(i). ΔHhyd ∝ charge size of the ion (charge density) I− which has an ionic radius of 0.216 nm while Cl− has an ionic radius of 0.181 nm. Since the ionic radius of I− is larger than the ionic radius of C l−, the charge density of I− is smaller than C l−. I− attracts water less strongly or forms weaker ion–dipole attraction between I− and water and hence the magnitude of ∆Hhyd for I− is smaller. (b) Lattice energies are not measured directly. By using Hess’ Law and Born Haber cycle, they can be obtained from experimental data. From the knowledge of the distances between cations and anions in the crystal structure, and the charge on each ion, it is possible to calculate the theoretical values for lattice energies. Table 2.2 shows the numerical values of lattice energies of sodium chloride and sodium iodide. Table 2.2 compound experimental value / kJ mol−1 theoretical value / kJ mol−1 NaCl(s) −772 −776 NaI(s) −699 −685 (i) Define, using sodium chloride as an example, what is meant by the term lattice energy. Heat evolved when one mole of the solid ionic compound, NaCl is formed from its gaseous ions, Na+ and Cl− at 298K and 1 bar. (ii) There is close agreement between experimental and theoretical values of lattice energy for sodium chloride but not sodium iodide. Suggest a reason for this. I− has a larger ionic radius than Cl− and hence I− is more readily polarised by Na+ leading to some covalent character in the ionic bond of NaI. or NaI is an ionic compound with a higher degree of covalent character because I− is more polarisable than Cl−.

©YIJC [Turn over 6 (c) The ionic radii of some Group 1 metal cations and the halides ions are given in Table 2.3. Table 2.3 ion radius, r+/ nm ion radius, r−/ nm Na+ 0.098 Cl− 0.181 K+ 0.133 Br− 0.196 Rb+ 0.148 I− 0.219 Cs+ 0.167 Complete Table 2.4 and use your results to predict the tastes of sodium bromide and potassium iodide. Table 2.4 salt NaBr KCl NaI RbCl RbBr CsCl KI (r+ + r−) taste salty salty salty + bitter bitter bitter salt NaBr KCl NaI RbCl RbBr CsCl KI (r+ + r−) 0.294 0.314 0.317 0.329 0.344 0.348 0.352 taste salty salty salty salty + bitter bitter bitter bitter [Total: 9]

©YIJC [Turn over 7 3 (a) Write equations to show the acidic or basic property of (i) phenylamine C6H5NH2 + H+ → C6H5NH3+ or C6H5NH2 + H2O ⇌ C6H5NH3+ + OH− (ii) phenol C6H5OH + OH− → C6H5O− + H2O or C6H5OH + H2O ⇌ C6H5O− + H3O+ (b) Ethylamine, CH3CH2NH2, is a weak base. CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH− (i) Using the equation given, explain why ethylamine is a BrØnsted-Lowry base. Ethylamine is a proton acceptor (and produces OH− when in solution). (ii) Identify the two different conjugate acid−base pairs in the