VJC 2022 H2 Chem Prelim P3 ans

1 © VJC 2022 9729/03/PRELIM/22 [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 19 September 2022 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 a soft 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 / 18 2 / 22 3 / 20 Section B 4 OR 5 / 20 Total / 80 This document consists of 25 printed pages.

2 © VJC 2022 9729/03/PRELIM/22 Section A Answer all the questions in this section. 1 (a) (i) Under what conditions of temperature and pressure would you expect the behaviour of a real gas to be most like that of an ideal gas? [1] • High temperature and low pressure (ii) Barium ethanedioate, BaC 2O4, decomposes on heating to produce a mixture of two different gases, A and B, and an oxide only. Neither gas A nor gas B is an ideal gas. They have the following boiling points. gas boiling point / oC A –191.5 B –78.5 The graph below shows the variation of pV RT with pressure, p, for 1 mol each of gas A and gas B at constant temperature. Identify the graph that corresponds to gas A and explain your choice. [2] • Graph I corresponds to gas A. • Gas A has lower boiling point than B, hence, it has weaker intermolecular forces of attraction leading to less deviation from ideality. (iii) Free volume, V, refers to the volume of space between gas molecules. For an ideal gas, the free volume is essentially the same as the volume of the container. This can be calculated using the ideal gas equation, pV = nRT. The pressure of a 72 g gaseous sample containing gas A and gas B in a container of volume 400 cm3 is measured to be 3.36 ⨯ 107 Pa at 527 oC. Using the ideal gas equation, calculate the free volume of this gaseous sample in cm 3. Assume the gaseous sample has an average Mr = 36. [2] n = 72 / 36 = 2.00 mol pV = nRT • V = 2.00 x 8.31 x (527+273) 3.36 x 107 = 3.96 ⨯ 10–4 m3 • = 396 cm3 ideal gas p 1 I II pV RT

3 © VJC 2022 9729/03/PRELIM/22 [Turn over (iv) Explain why the volume you have calculated in (a)(iii) differs from that of the volume of the container. [1] The gaseous sample behaves non –ideally and the volume calculated in (a)(iii) is smaller than the volume of the container. • Under high pressure, the volume of the gaseous molecules is not negligible / significant. Hence, the actual free volume between the gas molecules is smaller. (v) An impure sample of barium ethanedioate, BaC 2O4, of mass 0.500 g, is added to 50.0 cm 3 of 0.0200 mol dm –3 acidified MnO4–(aq) and heated. A redox reaction takes place and all BaC 2O4 are reacted. The resulting solution is titrated with Fe 2+(aq). The end-point is reached when 30.40 cm3 of 0.0500 mol dm–3 Fe2+(aq) has been added. C2O42– ⇌ 2CO2 + 2e– MnO4– + 8H+ + 5e– ⇌ Mn2+ + 4H2O Fe2+ ⇌ Fe3+ + e– Calculate the percentage by mass of BaC2O4 in the 0.500 g impure sample. Show your working. [Mr: BaC2O4, 225.3] [4] • Initial total amount of MnO4– = 0.0200 ⨯ 50.0  10–3 = 1.00 ⨯ 10–3 mol Amount of Fe2+ used = 0.0500 ⨯ 30.40 ⨯ 10–3 = 1.52 ⨯ 10–3 mol 5Fe2+ ≡ MnO4– • Amount of MnO4– unreacted with C2O42– = 1.52 ⨯ 10–3 / 5 = 3.04 ⨯ 10–4 mol Amount of MnO4– reacted with C2O42– = 1.00 ⨯ 10–3 – 3.04 ⨯ 10–4 = 6.96 ⨯ 10–4 mol • Amount of C2O42– reacted = 6.96 ⨯ 10–4 ⨯ 5/2 = 1.74 ⨯ 10–3 mol • Mass of BaC2O4 reacted = 1.74 ⨯ 10–3 ⨯ 225.3 = 0.392 g Percentage by mass of BaC2O4 = 0.392 0.500 ⨯ 100% = 78.4% (b) The elements of Group 14 can form monoxides and dioxides. The monoxides are un stable and will disproportionate into the ir element and dioxide. The equations for the disproportionation reactions are given in Table 1.1, together with some thermodynamic data for the reactions. Table 1.1 disproportionation equation So / J mol–1 K–1 Ho / kJ mol–1 Go / kJ mol–1 2CO(g) → C(s) + CO2(g) –176 –173 –120 2SiO(g) → Si(s) + SiO2(s) –363 –712 –603 2GeO(s) → Ge(s) + GeO2(s) –13.6 –127 –123 2SnO(s) → Sn(s) + SnO2(s) –9.20 –9.10 –6.36 2PbO(s) → Pb(s) + PbO2(s) –4.00 +157 +158

4 © VJC 2022 9729/03/PRELIM/22 (i) Explain why the entropy change for the disproportionation of SiO(g) is much more negative than that for CO(g). [1] • The disproportionation of SiO(g) leads to a larger decrease in the amount of gas than that for the disproportionation of CO(g) . Thus, there are less ways of arrangement, leading to a greater decrease in disorder liness and hence a much more negative entropy change. (ii) Explain why the entropy change for the disproportionation of PbO(s) is close to zero. [1] • The disproportionation of PbO(s) only involves the same amount of solid reactants and solid products . Thus, there is very little change in disorderliness and hence a close to zero entropy change. (iii) Use data from Table 1.1 to deduce the temperature above which the disproportionation of CO(g) becomes unfavourable. [1] When disproportionation of CO(g) becomes unfavourable, ∆G > 0 ∆H – T∆S > 0 –173 – T(–176 ⨯ 10–3) > 0 T > 983 K • When T is greater than 983 K , the disproportionation of CO(g) becomes unfavourable. (iv) Explain why CO(g) does not spontaneously disproportionate at room temperature. [2] CO contains strong CO bond (or triple bond) which requires a lot of energy to be broken. The activation energy is high and the reaction is kinetically not feasible at room temperature. (v) Carbon monoxide, CO, is a gas at room temperature and pressure. It contains a coordinate bond. Explain what is meant by a coordinate bond. [1] • A coordinate bond is a covalent bond in which both electrons come from only one of the atoms in the bond. (vi) Dicarbon monoxide, C2O, is extremely reactive and is not encountered in everyday life. It is found in dust clouds in space and analysis has shown that the central atom is carbon with no unpaired electrons while the other carbon atom has a lone pair of electrons. Draw the structure of dicarbon monoxide, stating its shape and bond angle. [2] • C=C=O • Shape: linear, bond angle: 180o [Total: 18]

5 © VJC 2022 9729/03/PRELIM/22 [Turn over 2 (a) Alanine, CH3CH(NH2)CO2H, is an amino acid that is used to make proteins. Its pKa values are 2.34 and 9.87. (i) Sketch a graph to show how the pH of the solution would change during the