NYJC H2 Chemistry P3 Answers Prelim

H2 Chemistry 9647/03 NYJC J2/14 PX [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CHEMISTRY 9647/03 Paper 3 Free Response 24 September 2014 2 hours Candidates answer Section A on the Question Paper Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. 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. Section B Answer all questions on separate answer paper. For Examiner’s Use Section A 1 2 3 4 5 Total This document consists of 20 printed pages and 0 blank page.

2 H2 Chemistry 9647/03 NYJC J2/14 PX Answer any four questions. 1 In 1939 the German chemist Gerhard Dom agk was awarded the Nobel Prize in Physiology or Medicine for the discovery of the antibacterial effects of Prontosil. It was found that sulfanilamide was the active component of Prontosil. (a) Benzene can be used as the starting reagent for the synthesis of sulfanilamide by the following route. (i) Suggest reagents and conditions for steps I and II, and the structure of compounds A and B. Step I: Reagents: Concentrated HNO 3, Concentrated H2SO4 Condition: 55 – 60 °C [1 mark for complete answer; no marks for partial answer] Step II: i) Reagents: Sn, HCl Condition: Heat under reflux ii) Reagent: NaOH(aq) Condition: Room temperature [1 mark for complete answer; no marks for partial answer] A B [2 x 1 mark] (ii) Step III involves the reaction of concentrated H 2SO4 together with SO 3 to generate the required electrophile. Write a balanced equation for the generation of the electrophile in Step III. SO3 + H2SO4 → SO3H+ + HSO4 - [1 mark] (iii) State the types of reaction in steps II and V.   Step II: Reduction Step V: Nucleophilic substitution [1 mark each]  sulfanilamide

3 H2 Chemistry 9647/03 NYJC J2/14 PX (iv) Explain why sulfanilamide is soluble in water. Sulfanilamide is soluble in water as they can form favourable hydrogen bonds with water. [1 mark] A derivative of sulfanilamide, compound C, is shown below. (v) The amino group of sulfanilamide has a p K b value of 12.0. Suggest a p Kb value for the amino group of compound C. Explain your answer. 9 < pKb < 12.0 (pKb of aniline = 9) [1 mark] Compound C has an electron-donating ethyl substituent on the amine substituent of the benzene. Hence, the electron density on the N atom is increased, making its lone pair of electrons to be more readily available to accept a proton. [1 mark] [10] (b) Describe and explain the trend in thermal decomposition of Group II nitrates, writing an equation for any reaction that occurs. [3] 2M(NO3)2(s)  2MO(s) + 4NO2(g) + O2(g) [1 mark] Down the group from Mg to Ba, cationi c radius increases but ionic charge remains the same. Therefore, charge density of Group II cations decreases, causing electron cloud of nitrate i on to be less easily polarised. [1 mark] Hence, thermal stability of Group II nitrates increases from Mg to Ba. [1 mark] (c) Ammonium nitrate is widely used as an ex plosive. It can easily decompose when ignited to give nitrogen gas, oxygen gas and steam. (i) Write a balanced equation, with stat e symbols, which represents the enthalpy change of decomposition of 1 mole of ammonium nitrate. NH4NO3(s)  N2(g) + 1/2O2(g) + 2H2O(g) [1 mark; NH4NO3 must be in 1 mole] (ii) Define the term enthalpy change of atomisation of ammonium nitrate. The enthalpy change of atomisation of ammonium nitrate is the energy required when 1 mole of ammonium nitrate is broken to form its constituent gaseous atoms. [1 mark] C sulfanilamide (pKb = 12.0)

4 H2 Chemistry 9647/03 NYJC J2/14 PX (iii) The standard enthalpy change of atomisati on of ammonium nitrate is +2928 kJ mol 1. Using suitable data from the Data Booklet , construct an energy cycle to calculate the ent halpy change of decomposition of ammonium nitrate. NH 4NO3(s)  N2(g) + 1/2O2(g) + 2H2O(g) 2N(g) + 4H(g) + 3O(g) H r = +2928 – (+994) – ½(+496) –4(+460) = –154 kJ mol –1 [3 marks; 1 mark each for correct cy cle with labelling, input of values and final answer] (iv) In a laboratory experiment, the decomposition of 1.0 g of ammonium nitrate is able to increase the temperature of 100 g of water by 4 C. Calculate the heat evolved from t he decomposition of one mole of ammonium nitrate from this experiment. Comment on the difference between the calculated values from (iii) and (iv). Qevolved = mcT = 100 x 4.18 x 4 = 1672 J For 1 mole of ammonium nitrate, Q evolved = 1672 ÷ (1.0/80) = 133.76 kJ [1 mark] The value in (c)(iv) is lower than th at in (c)(iii) as some of the heat evolved are lost to the surroundings. [1 mark] [7] [Total: 20] Hatm(NH4NO3(s)) BE(NΞN) + 1/2BE(O=O) + 4BE(O-H)

5 H2 Chemistry 9647/03 NYJC J2/14 PX 2 Phosphorus is the most abundan t element of Group V, acco unting for 0.10% of the mass of the Earth’s crust. Phosphorus exists in two common allotropic forms: white phosphorus and red phosphorus. White phosphor us contains discrete tetrahedral P4 molecules. Red phosphorus, by contrast, has a polymeric structure. (a) Compounds A and B contain phosphorus in ox idation states of 3 and +5 respectively. The relative molecular masses of A and B are 34.0 and 208.5 respectively. The molecular shape of A is the same as that of ammo nia and it burns easily in air to form phosphoric acid only. B is an off-white solid that melts at 167 oC and gives white fumes and phosphoric acid on contact with moist air. Suggest the identities and draw the shapes of compounds A and B, writing equations to illustrate all the reactions mentioned. [4] A is PH3 P H H H B is PCl5 P Cl Cl Cl Cl Cl PH3 + 2O2  H3PO4 PCl5 + 4H2O  H3PO4 + 5HCl 1 mark for suggesting both the identity of A and B correctly 1 mark for drawing both the shapes correctly 1 mark for writing each equat ion correctly; total 2 marks (b) When phosphorus burns in oxyg en, it yields either P 4O6 or P4O10 depending on the amount of oxygen present. Both P 4O6 and P 4O10 are acidic oxides, and they react with water to form aqueous solutions of phosphorous acid and phosphoric acid, respectively. Phosphorous acid is diprotic , while phosphoric acid is triprotic and their successive dissociation constants are as follows: P O HO O H H P O OO O H H H Phosphorous acid, H3PO3 Phosphoric acid, H 3PO4 Ka1 = 1.0 x 102 mol dm3 Ka1 = 7.5 x 103 mol dm3 Ka2 = 2.6 x 107 mol dm3 Ka2 = 6.2 x 108 mol dm3 Ka3 = 4.8 x 1013 mol dm3

6 H2 Chemistry 9647/03 NYJC J2/14 PX A 5.00 g sample of white phosphorus was burned in oxygen. The oxide produced was dissolved in enough water to make 250 cm3 of solution. When the solution was treated with an excess of aqueous Ca(NO 3)2, 19.4 g of a white precipitate was obtai ned. The white precipitate could be either CaHPO 3 or Ca3(PO4)2 depending on the oxide t hat was formed when white phosphorus burned in oxygen. (i) Write equations for the reactions of phos phorus with oxygen to yield either P 4O6 or P4O10 and for the subsequent reactions of both oxides with water. P4 + 3O2  P4O6 P4O6 + 6H2O  4H3PO3 P4 + 5O2  P4O10 P 4O10 + 6H