SRJC_H1_CHEM_P2 ANS

SRJC 2009 8872/02/PRELIM/2009 [Turn Over SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 CANDIDATE NAME C L A S S CHEMISTRY 8872/02 Preliminary Examination 26 Aug 2009 Paper 2 Mark Scheme 2 hr Additional Materials: Data Booklet Answer Paper 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 work. SECTION A : Answer all questions in the space provided. SECTION B: Answer any two questions on separate answer paper. 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. For Examiner’s Use MCQ P1 /30 A1 /10 A2 /10 A3 /10 A4 /10 B5 /20 B6 /20 B7 /20 Total /110 Percentage Grade This document consists of 16 printed pages and 0 blank page

2 SRJC 2009 8872/02/PRELIM/2009 [Turn Over For Examiner’s Use 1 The industrial preparation of the polymer, poly(tetrafl uoroethene), is based on the synthesis of its monom er, tetrafluoroethene (C 2F4). Tetrafluoroethene is produced from the thermal cracking of chlorodifluoromethane (CHC lF2), as shown by the following equation: 2CHClF2(g)  C2F4(g) + 2HCl(g) (a) Use relevant data from the Data Booklet to calculate the enthalpy change of the reaction. State any assumptions that you have made. 2 C H Cl F F → CC F F F F + 2 H-Cl Let BE(C-F) be x. ∆H rxn = ∑BE(rxts) - ∑BE(pdts) = 2[BE(C-Cl) + BE(C-H) + 2BE(C-F)] – [BE(C=C) + 4BE(C-F) + 2BE(H-Cl)] = [2(340 + 410 + 2x)] – [610 + 4x + 2(431)] [1] = 1500 + 4x – 1472 – 4x = +28 kJ mol─1 [1] Assumption: The average bond energy of C –F is the same in CHClF2 and C2F4. [1] [3] (b) By using your answer to (a) and the following information, Hf(CHClF2(g)) = –485 kJ mol –1 Hf(HCl(g)) = –92 kJ mol –1 calculate the enthalpy change of formation of C2F4(g). Hrxn = ∑Hf( products) - ∑Hf(reactants) 28 = [2 x Hf (HCl) + Hf (C2F4)] – [2 x Hf(CHClF2)] [1] 28 = [2 x (-92) + Hf (C2F4)] – [2 x –485] 28 = Hf (C2F4) + 786 H f (C2F4) = -758 kJ mol–1 [1] [2] (c) (i) Explain the following phenomenon: Chlorine can react with oxygen to form C lO─, C lO2 ─, C lO3 ─ and C l2O7 but fluorine can react with oxygen to form OF2 only. Chlorine can react with oxygen to form compounds of varying oxidation states as it is able to expand its octet using the empty 3d-orbitals [0.5] since