TJC Prelim P2 Ans

1 2017 TJC H2 Chemistry Preliminary Exam [Turn over s CANDIDATE NAME CIVICS GROUP / CENTER NUMBER S INDEX NUMBER CHEMISTRY 9729/02 Paper 2 Structured Questions 11 September 2017 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Civics Group, centre number, index number and name on all the work you hand in. 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. 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 18 printed pages For Examiner’s Use 1 / 22 2 / 15 3 / 21 4 / 8 5 / 9 Paper 1 / 30 Paper 3 / 80 Total PRELIMINARY EXAMINATIONS HIGHER 2

2 2017 TJC H2 Chemistry Preliminary Exam [Turn over Answer all the questions 1 Silver chloride is an important photosensitive inorganic material widely used in photographic applications. It is industrially produced by mixing solutions of silver nitrate and sodium chloride. Ag+(aq) + Cl-(aq) AgCl(s) ∆Hppto = -65.7 kJ mol-1, ∆Sppto = -34.3 J mol-1 K-1 (a) (i) Suggest whether a lower or higher temperature should be used to increase the yield of silver chloride. Explain your answer. [2] • A lower temperature should be used. • By Le Chatelier’s Principle , the system will favour the forward exothermic reaction to counteract the lowered temperature. Hence, the position of equilibrium shifts to the right increasing the yield of silver chloride. (ii) Calculate ∆Go of the precipitation of AgCl. [1] • ΔGᶿ = ΔHᶿ - TΔSᶿ = -65.7 – (298)(-34.3/1000) = -55.5 kJ mol-1 (iii) In view of your answer in (a)(ii), comment on the solubility of silver chloride in water. [1] • Since ΔGᶿ is negative, precipitation is spontaneous. Hence, Ag Cl is insoluble/sparingly soluble in water. (b) (i) Lattice energies are not measured directly. The value calculated using Hess’ Law and the Born-Haber cycle is the experimental lattice energy. Using ∆H ppto of AgCl(s), the following data and relevant data from the Data Booklet, calculate the experimental lattice energy of AgCl(s). standard enthalpy change of formation of Ag+(aq) +106 kJ mol–1 standard enthalpy change of formation of Cl-(aq) –167 kJ mol–1 standard enthalpy change of atomisation of Ag(s) +285 kJ mol–1 first electron affinity of chlorine –349 kJ mol–1 [3] Ag+(g) + Cl-(g) AgCl(s) Ag(g) + Cl(g) Ag(s) + 1/2 Cl2(g) Ag+(aq) + Cl-(aq) • cycle with state symbols L.E -65.7 106 + (-167) +285 +𝟐𝟒𝟒 𝟐 +731 -349

3 2017 TJC H2 Chemistry Preliminary Exam [Turn over By Hess’ Law, • +285 + 731 + 122 – 349 + L.E = +106 – 167 – 65.7 L.E = • - 916 kJ mol-1 The theoretical value of lattice energies are calculated using the distances between the cations and anions in the crystal structure, and the charge on each ion. The table below shows the numerical values of experimental and theoretical lattice energies for sodium chloride and silver fluoride. compound experimental value/ kJ mol-1 theoretical value/ kJ mol-1 NaCl -771 -766 AgF -967 -953 (ii) It can be seen that the experimental and theoretical values for sodium chloride and silver fluoride are quite similar. However, the theoretical value for the silver chloride is significantly less exothermic than the experimental value. Suggest a reason for this. [1] • The theoretical value of lattice energy is based on the assumption that AgC l is ionic. However, there is some covalent character in AgCl due to the large size of Cl- ion which can be polarised by Ag+. (c) Explain, with the aid of equations, why silver chloride is soluble in aqueous ammonia while silver iodide does not dissolve in aqueous ammonia. [3] Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) --------------------(1) •NH3 forms a stable soluble complex [Ag(NH 3)2]+ with the free Ag + ion causing [Ag+(aq)] to decrease. Ag+(aq) + Cl-(aq) AgCl(s) ----------------(2) By Le Chatelier’s Principle, the position of equilibrium in (2) shifts left to counteract the decrease in the [Ag+(aq)]. The ionic product [Ag+][Cl-] decreases to a value lower than the Ksp of AgCl and AgCl dissolves. The Ksp of Ag I is much lower than K sp of AgCl . The ionic product is still easily exceeded even with the decrease in [Ag+(aq)] and hence it does not dissolve. • • •

4 2017 TJC H2 Chemistry Preliminary Exam [Turn over (d) (i) Most of the energy our bodies need comes from carbohydrates and fat. Starch is broken down into glucose,C6H12O6. Glucose exist mainly in cyclic forms with a small percentage in open chains. open-chain cyclic State what you would observe when glucose is added to an alkaline solution of ammoniacal silver(I) nitrate. [1] • Silver mirror is observed. Glucose is transported to the cells to react with oxygen via a series of steps to form carbon dioxide, water and energy. (ii) Write a balanced equation for the reaction of glucose with oxygen. [1] • C6H12O6 + 6O2 → 6CO2 + 6H2O (iii) Using data from the Data Booklet, calculate the amount of energy released per mole of glucose using the cyclic structure. [2] Using the cyclic structure of glucose, Bond-breaking 5 x C – C 5 x O – H 7 x C – H 7 x C – O 6 x O = O Bond-Forming 12 x C = O 12 x O – H • Energy released = +(5 x 350 + 5 x 460 + 7 x 410 + 7 x 360 + 6 x 496) – (12 x 805 + 12 x 460) = • - 2760 kJ mol-1 (iv) The literature value for the amount of energy released per mole of glucose is – 2800 kJ. Apart from bond energies are average values, suggest another reason for the difference between this value and that calculated in (d)(iii). [1] • The H calculated using bond energies applies for the reactants and products in the gaseous phase but the reaction involves liquid H2O rather than gaseous H2O. Like carbohydrates, fats are metabolised into carbon dioxide and water and when subjected to combustion in a bomb calorimeter. The reaction of tristearin, C57H110O6, a typical fat is as follows:

5 2017 TJC H2 Chemistry Preliminary Exam [Turn over C57H110O6 + 163 2 O2 → 57 CO2 + 55 H2O ∆Ho = -37760 kJ mol-1 The