2024 Prelims SAJC H2 Chem P2 (Ans)

1 [TURN OVER ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 3 S CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 27 August 2024 2 hours READ THESE INSTRUCTIONS FIRST Write your name and class on all the work that you hand in. 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. 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. 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. This document consists of 22 printed pages (including this cover page). For Examiner’s Use Q1 8 Q2 17 Q3 11 Q4 14 Q5 25 Total 75

2 [TURN OVER 1 Hydrogen sulfide, H2S is a toxic gas which is commonly formed as a by -product in oil and gas refining industries. The Claus process has been widely used to remove H2S. It involves two main reactions as shown. Reaction 1: 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(l) ΔHθ1 Reaction 2: 2H2S(g) + SO2(g) → 3 8S8(s) + 2H2O(l) ΔHθ2 = − 108 kJ mol−1 (a) (i) Use the data in Table 1.1 to calculate the standard enthalpy change of reaction for Reaction 1, ΔHθ1. Table 1.1 Compound ΔHθf / kJ mol−1 H2S(g) −20.2 SO2(g) −296.8 H2O(l) −285.8 [2] Ho1 = [2(–296.8) + 2(–285.8)] – (2 x – 20.2) [1] = – 1120 kJ mol–1 (to 3 sf) [1] (ii) The standard Gibbs free energy, Go2, for Reaction 2 is −89.6 kJ mol−1. Calculate the entropy change, So2, for Reaction 2 and explain the significance of its sign with respect to the process that is occurring. [2] Gro = Hro – TSro Sro = (−89.6 + 108)/(–298) = - 0.0617 kJ mol–1 K–1 = - 61.7 J mol–1 K–1 [1 with correct units] ∆S< 0 . The system becomes less disordered as t he number of gaseous particles decreases from 3 to 0. [1] (b) Besides the Claus process, organic solvents like methanol can also be used to remove H2S.

3 [TURN OVER With reference to the intermolecular forces, explain why methanol is a good solvent to remove H2S. [2] Energy released from the permanent dipole -permanent dipole interactions between H 2S and methanol molecules is sufficient to overcome the hydrogen bonds between methanol molecules and the permanent dipole-permanent dipole interactions between H2S molecules. Hence, H2S is soluble in methanol. [2] – 4 points [1] – 2, 3 points (c) One of the key advantages of the Claus process is the recovery of S 8, which is a valuable product that can be used for other purposes. S8 consists of 8 sulfur atoms bonded to each other in a crown conformation as shown in Fig. 1.1. S S S SSS S S Fig. 1.1 Use the VSEPR theory to state and explain the shape and bond angle around each S atom in S8. [2] bent 105o [1] Around each S, there are 2 bond pairs and 2 lone pairs of electrons. The electron pairs arranged as far apart as possible to minimise electrostatic repulsion and maximise stability [1] [Total: 8]

4 [TURN OVER 2 Sulfur dioxide, SO2, and sulfite, SO32–, are found in food and beverages as preservatives. Despite this, they may cause allergic reactions. Hence, the concentration of sulfur dioxide or sulfite in any food and beverages cannot exceed 10 parts per million (ppm). For this question, 1 ppm refers to 1 g of SO32– for every 1 000 000 g of food sample. (a) Magnesium oxide, aluminium oxide and sulfur dioxide are Period 3 oxides. Describe reactions that illustrate the variation in acid-base behaviour of these three oxides. Write equations for all the reactions described. [6] MgO is a basic oxide and it reacts vigorously with acids to form salt and water but no reaction with bases. [1] MgO + 2HCl → MgCl2 + H2O [1] Al2O3 is an amphoteric oxide and reacts with both acids and alkalis to form salt and water. [1] Al2O3 + 6HCl → 2AlCl3 + 3H2O Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4 [1] SO2 is an acidic oxide and reacts readily with alkalis to form salt and water, but has no reaction with acids. [1] SO2 + 2NaOH → Na2SO3 + H2O [1] (b) To determine the concentration of SO 32–, a food sample is treated to obtain the extracted solution, before being titrated against potassium iodate-iodine solution, with starch as the indicator. In the presence of strong acid, a solution of potassium iodate(V), KIO3, and potassium iodide, KI, will liberate iodine, I2. The equation to represent this is as shown. IO3− + 5I− + 6H+ → 3I2 + 3H2O SO32− which is present in the food sample will react with I2 to form iodide and sulfate ions. When SO32− has completely reacted, I2 will then combine with starch to form a blue-black complex, which is the colour seen at the end-point.

5 [TURN OVER The procedures for an experiment on SO32– analysis are as follows. Step 1 1.0 kg of food sample was treated to obtain the extracted solution. The solution was diluted to 1 dm3 in a volumetric flask. Step 2 0.010 g of KIO3 was dissolved in water. Excess KI and excess aqueous acid were added to this solution, before making up to 250 cm 3 in a volumetric flask. Step 3 25.0 cm3 of the food sample solution from Step 1 required 18.00 cm 3 of acidified potassium iodate-iodide solution to reach end-point. (i) Write an ionic equation for the reaction between SO32− and I2. [1] SO32− + H2O + I2 → SO42− + 2H+ + 2I− [1] (ii) Calculate the concentration, in mol dm–3, of SO32− in 25.0 cm3 of the food sample solution. [3] Amount of KIO3 = 0.01 39.1+126.9+(16 × 3) = 4.6728 x 10–5 mol Amount of I2 in 250 cm3 = 4.6728 x 10–5 x 3 = 1.4018 x 10-4 mol [1] [I2] = 1.4018 x 10-4 / 0.25 = 5.6074 x 10-4 mol dm–3 Amount of I2 formed = Amount of SO32– reacted = 5.6074 x 10-4 x 0.018 = 1.0093 x 10-5 mol [1] [SO32–] = 1.0093 ×10–5 25.0 1000⁄ = 4.04 x 10-4 mol dm–3 (3 s.f) [1] (iii) Calculate the concentration of SO32− in the food sample in ppm. Hence, state whether the food sample is safe for consumption. [2] No. of moles of SO32– in 1 dm3 of food sample = 0.000404 mol Mass of SO32– in food sample = 0.000404 x (32.1 + 3 x 16) = 0.0323604 g Concentration of SO32– in ppm by mass = (0.00323604 / 1000) x 106 = 32.4 ppm (to 3sf) [1] The concentration of SO32– has exceeded the safe limits of 10ppm. Hence, the food sample is unsafe for consumption. [1]

6 [TURN OVER (c) 100 cm3 of 0.106 mol dm–3 sodium sulfite, Na2SO3(aq), is added to 100 cm3 of 0.500 mol dm–3 sodium hydrogensulfite, NaHSO3(aq), to produce a buffer solution X. The pH of the resultant solution is 6.5. (Ka of HSO3–(aq) = 6.73 x 10–8 mol dm−3 at 298K) (i) Explain what is meant by the term buffer solution. [1] A buffer solution is a solution whose pH remains almost unchanged/ resists changes in pH when a small amount of H+ or OH– is added to it. [1] (ii) Write an equation to show what happens when small amounts of OH –(aq) are added to solution X. [1] HSO3–(aq) + OH–(aq) → SO32–(aq) + H2O(l) [1] No need state symbols (iii) 5 cm3 of 0.100 mol dm-3 of Ba(OH)2 was added to 75.0 cm3 of solution X. Calculate the pH of the resulting solution at 298K. [3] Amount of OH– added = 2 x 0.100 x 0.005 = 0.001 mol Total volume = 80.0 cm3 species HSO3–(aq) OH– SO32– (aq) Initial amount / mol 0.1 x 0.5 x 75/200 = 0.01875 0.001 0.1 x 0.106 x 75/200 = 0.003975 Change in amount / mol -0.001 -0.001 +0.001 Fi