2022 YIJC H2 CHEM PRELIM P3 MS

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUGGESTED ANSWERS CG INDEX NO CHEMISTRY Paper 3 Free Response Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 12 September 2022 2 hours READ THESE INSTRUCTIONS FIRST This document consists of 26 printed pages and 6 blank pages For Examiner’s Use Section A 1 / 17 2 / 18 3 / 25 Section B 4 or 5 / 20 Penalty units significant figures Overall / 80 Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. 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 the questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

2 ©YIJC [Turn over Section A Answer all the questions in this section. 1 (a) Describe and explain the trend in the thermal stability of the hydrogen halides, HCl, HBr and HI. Include an equation for the thermal decomposition reaction in your answer. [3] HX(g) → ½H2(g) + ½X2(g) The thermal stabilities of the hydrogen halides decrease down the Group from HCl to HBr to HI. This is because • the size of the halogens increases from Cl to I and the valence orbital used for bonding is larger and more diffuse; • the effectiveness of the orbitals overlap decreases; • this result in the weaker covalent bond formed between the hydrogen and halogen atoms or quoting H-X bond energies; • lesser amount of energy is required to overcome the covalent bond between the hydrogen and halogen atoms. (b) Alkanes are generally considered to be unreactive compounds, showing an inertness to common reagents such as NaOH, H2SO4, and K2Cr2O7. (i) Suggest a reason why these reagents do not react with an alkane such as propane. [1] The C-H bond is non-polar. The C-H bond does not break heterolytically, only homolytically. Propane can be converted into 2-chloropropane when it reacts with chlorine in ultraviolet (UV) light. (ii) Describe the mechanism of the reaction between propane and chlorine in UV light. [3] Free radical substitution initiation propagation termination

3 ©YIJC [Turn over (c) (i) Define the term lattice energy. [1] Lattice energy is the heat evolved when 1 mole of a solid ionic compound is formed from its constituent gaseous ions. (ii) Use the data in Table 1.1, together with data from the Data Booklet, to calculate a value for the lattice energy of silver iodide, AgI(s). Show your working. Table 1.1 value / kJ mol−1 electron affinity of iodine, I(g) + e− → I−(g) −295 enthalpy change of sublimation of iodine molecules, I2(s) → I2(g) +62 standard enthalpy change of atomisation of Ag(s) +285 standard enthalpy change of formation of AgI(s) −62 [3] Lattice energy = −62 – [31 + 75.5 + (−295) + 285 + 731] = −889.5 = −890 kJ mol−1

4 ©YIJC [Turn over (d) Air bags in car inflate rapidly during an accident to protect the front passengers. The air bag contains sodium azide, NaN3, silicon dioxide, SiO2, and potassium nitrate, KNO3. On impact, three reactions take place. The sodium azide first decomposes to sodium and nitrogen. 2NaN3(s) → 2Na(s) + 3N2(g) The nitrogen formed inflates the air bag while the sodium formed reacts with potassium nitrate to form sodium oxide, potassium oxide and additional nitrogen gas, which may be used to fill the air bag. Potassium oxide and sodium oxide then react with silicon dioxide to form harmless metal silicates. K2O(s) + Na2O(s) + SiO2(s) → K2Na2SiO4(s) (i) Write an equation, with state symbols, for the reaction between sodium and potassium nitrate. [1] 10Na(s) + 2KNO3(s) → 5Na2O(s) + K2O(s) + N2(g) (ii) Calculate the mass of sodium azide needed to inflate an air bag of capacity 60 dm3 at room temperature and pressure. [2] Amount of N2 needed to fill a 60 dm3 air bag = 60 ÷ 24 = 2.50 mol 2 mol of NaN3 produces 3 mol of N2 and 2 mol of Na; 10 mol of Na produces 1 mol of N 2 in its reaction with KNO 3 which means 2 mol of Na produces 0.2 mol of N2 Hence 2 mol of NaN3 produces a total of 3.2 mol of N2. Amount of NaN3 needed to produce 2.5 mol of N2 = 2.5 3.2 × 2 = 1.5625 mol Mass of NaN3 required = 1.5625 × (23.0 + 14.0 × 3) = 102 g

5 ©YIJC [Turn over (iii) To determine the amount of sodium azide in an impure sample, the azide present is first reacted with excess iodine. 2N3− + I2 → 3N2 + 2I − The amount of unreacted iodine is then titrated with a standard solution of sodium thiosulfate. I2 + 2S2O32− → 2I− + S4O62− 0.120 g of an impure sample of sodium azide was dissolved in water. The mixture was reacted with 25.0 cm3 of 0.050 mol dm−3 of aqueous iodine. The excess iodine was found to require 23.10 cm 3 of 0.040 mol dm −3 aqueous sodium thiosulfate for reaction. Calculate the percentage purity of sodium azide in the sample. [3] I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq) Amount of S2O32− used = 0.040 × (23.10 × 10−3) = 9.240 × 10−4 mol Amount of excess I2 = 0.5 × (9.240 × 10−4) = 4.620 × 10−4 mol Initial amount of I2 used = 0.050 × 0.0250 = 1.250 × 10−3 mol Amount of I2 reacted with N3− = (1.250 × 10−3) – (4.620 × 10−4) = 7.880 × 10−4 mol Amount of NaN3 = 2 × (7.880 × 10−4) = 1.576 × 10−3 mol Mass of NaN3 = (1.576 × 10−3 ) × (23.0 + 14.0 × 3) = 0.1024 g Percentage purity of NaN3 in the sample = 0.1024 0.120 × 100% = 85.4% [Total: 17] 2 (a) Ammonia is manufactured by the following reaction. equation 2.1 N2(g) + 3H2(g) ⇌ 2NH3(g) The value of the equilibrium constant, Kp, measured at two different temperatures is shown in Table 2.1. Table 2.1 temperature / K Kp / atm−2 600 1.33 × 10−2 750 1.33 × 10−4 (i) Write the expression for the equilibrium constant, Kp, for this reaction. [1] Kp = (pNH3 )2 (pN2)(pH2)3

6 ©YIJC [Turn over (ii) A plant is designed to convert, at equilibrium, 50% of the reactants into ammonia. Assuming that the reactants are a mixture of N 2 and H2 in a 1 : 3 ratio by volume, calculate the total equilibrium pressure necessary to bring about a 50% conversion at • 600 K, and • 750 K. [2] Let the number of moles of N2 be x N2(g) + 3H2(g) ⇌ 2NH3(g) initial / mol x 3 x 0 change / mol −0.5 x −1.5 x + x equilibrium / mol 0.5 x 1.5 x x Total amount of gase