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 it is in period 3 [0.5] OR Fluorine is in period 2 . [0.5] Can only form OF 2 as it cannot expand its octet as there are no empty 3d orbitals available. [0.5] [1]

3 SRJC 2009 8872/02/PRELIM/2009 [Turn Over (ii) Predict, with reasoning, the shape of OF2. O FF There are 2 bond pairs and 2 lone pairs around O. [0.5] To minimize repulsion, the 4 electron pairs are directed to the corners of a tetrahedron. [0.5] Lone pair-lone pair repulsion > l one pair-bond pair repulsion > bond pair – bond pair repulsion. [0.5] Shape of OF2 is bent. [0.5] [2] (iii) Predict, with reasoning, the solubility of OF2 in CCl4. OF2 has a simple molecular [0.5] structure with weak intermolecular Van der Waals’ forces of attraction. [0.5] It is soluble [0.5] in CCl4 as it is able to form favourable solute-solvent interactions [0.5] with CCl4 molecules. [2] [Total: 10]

4 SRJC 2009 8872/02/PRELIM/2009 [Turn Over 2 (a) 2-bromopropane can react with sodium hy droxide in two different ways depending on conditions to give two different products A and B. CH3CHBrCH3 C3H6 C3H8O C D A B I II NaCN H+/heat (i) Draw the ‘dot-and-cross’ diagram of NaOH. Na O H + - X X X X X XX [1] or 0 (ii) Suggest and explain the polarity of 2-bromopropane. C-Br bond is polar. In the molecule, the dipole moments associated with the polar bonds and lone pair of electrons do not cancel out exactly. [0.5] Therefore 2-bromopropane is polar. [0.5] (iii) Describe the condition(s) necessary for each one of the reactions I and II. In each case, state the type of reaction undergone. Reaction I : Condition: aqueous NaOH, heat [0.5] Type of reaction: nucleophilic substitution [0.5] Reaction II: Condition: alcoholic NaOH, heat [0.5] Type of reaction: elimination [0.5] (iv) Using the above reaction scheme, dr aw the structural formulae of compounds C, D and E. [7] C: [1] D: [1] CH3CH(CN)CH3 CH3CH(COOH)CH3 E: CO C O CH3 CH3 C CH3 H HCH3 [1] conc H SO2 4, heat E C H O6 12 2

5 SRJC 2009 8872/02/PRELIM/2009 [Turn Over (b) Suggest a chemical test to distingui sh 2-chloropropane, 2-bromopropane and 2- iodopropane, stating clear ly the observations made. [3] Add aqueous sodium hydroxide and reflux [0.5], followed by addition of dilute HNO3 [0.5] and aqueous silver nitrate. [0.5] N.B. r.t.p is optional for the addition of silver nitrate. Observations: 2-chloropropane white precipitate is obtained. [0.5] 2-bromopropane cream precipitate is obtained [0.5] 2-iodopropane yellow precipitate is obtained. [0.5] [Total: 10]

6 SRJC 2009 8872/02/PRELIM/2009 [Turn Over 3(a) Propanol can be converted to propanoic acid under appropriate condition. Suggest the type(s) of hybridisation of all the carbon atoms in propanoic acid. HC 1 H H C2 H H C3 O OH C1: sp3 C2: sp3 C3: sp2 [1] / 0 (b) Propyl propanoate is synthesised indus trially mainly via the classic Fischer esterification reaction of propanol and propanoic acid which is exothermic in nature. This mixture will reach dynamic equilibrium and the ester will be produced at a yield of 65%. (i) With the aid of an appropriate rate-time graph, explain what is meant by dynamic equilibrium. Graph [1] Axes not labeled [-0.5] Fwd / bkward rate not labeled [-0.5] Labelling of t1 optional It is a state of equilibrium whereby the rate of forward reaction = rate of backward reaction [0.5] (at t1) The substances are still reacting together even though the concentration of the reactants and products remains constant [0.5] (ii) Given that the equilibrium mixture at 298 K contains the following: Propanol 0.33 mol Propanoic acid 0.33 mol Propyl Propanoate 0.66 mol Water 0.66 mol Calculate the value of Kc at 298 K for the reaction. CH3CH2COOH(l) + CH 3CH2CH2OH(l) ⇌CH3CH2CO2CH2CH2CH3(l)+H2O (l Time Rate Forward rate Backward rate t1 )

7 SRJC 2009 8872/02/PRELIM/2009 [Turn Over c =      K  32 223 2 32 322 CH CH COOCH CH CH H O CH CH COOH CH CH CH OH Kc =   2 4.00V 2 0.66 0.33V [0.5] [0.5] Volume missing [-0.5] (iii) Predict, with reasoning, the effect on the value of Kc (if any) when more propanol is added to the equilibrium mixture. CH3CH2CH2OH + CH3CH2CO2H ⇌ CH3CH2CO2CH2CH2CH3 + H2O When more propanol is added, by Le Chatelier’s Principle, equilibrium position will shift right [0.5] favoring the formation of ester and water. Kc value will remain constant. [0.5] (Kc is dependent on temperature and not concentration) (iv) Predict, with reasoning, the composition of the equilibrium mixture when the system in (b)(ii) is subjected to an increase in temperature? exo CH3CH2C