IJC 2008 H2 Chemistry Prelim II Paper3 answers

Innova JC Preliminary Examination H2 Chemistry Paper 2 Mark Scheme © INNOVA 2008 Answer Scheme 9746/03 1(a) Ecell betw Cu & VO3 – = +1.00 – (+0.34) = +0.66V >0 (energetically feasible) E cell betw Cu & VO2+ = +0.34 – (+0.34) = 0V The reaction is in equilibrium. Hence, products and reactants are in equal proportion. Cu can reduce VO2+ to V3+. Colour changes from: VO 3 – (yellow) to VO2+ (blue) to Blue green (VO2+ + V3+) Balanced equation: 2 VO3 – + 8H+ + Cu Æ 2VO2+ + 4H2O + Cu2+ 2VO 2+ + 4H+ + Cu Æ 2V3+ + 2H2O + Cu2+ (b)(i) CuSO4(s) dissolves in water to give a blue solution containing [Cu(H2O)6]2+ ions. Cu2+ + 6H2O Æ [Cu(H2O)6]2+ (ii) With excess NH3(aq), CuSO4(aq) gives a deep blue solution due to formation of soluble complex ion, [Cu(NH3)4]2+. [Cu(H2O)6]2+ + 4NH3 Ù [Cu(NH3)4]2+ + 6H2O Blue soln deep blue complex NH3 is a stronger ligand than water, so it can replace water from [Cu(H2O)6]2+ to form [Cu(NH3)4]2+. (c)(i) CH2OH [ ]1M (ii) CH3CH2OH + OH– Æ CH3CH2O– + H2O [ ]1M (iii) The 9nucleophile for compound G is OH–. Compound H: 9nucleophile is CH3CH2O–. (iv) Test: add NaOH(aq) to each sample & heat. Then add excess HNO3(aq), followed by AgNO3(aq). Obs: Compound F: white ppt of AgCl Bromobenzene: no white ppt. (d)(i) Reagents/conditions: PBr 3, heat Or HBr, reflux. (ii) Compound K: (CH 3)2CHCN Compound L: (CH3)2CHCOOCH2CH3

2 © INNOVA 2008 Answer Scheme 9746/03 (e)(i) Discharging (using it as a galvanic cell) Eqm (1) PbO 2(s) + 4H+(aq) + 2e Ù Pb2+(aq) + 2H2O(l) +1.47V Eqm (2) Pb2+(aq) + 2e Ù Pb(s) –0.13V Overall : PbO 2(s) + 4H+(aq) + Pb(s) Æ 2Pb2+(aq) + 2H2O(l) Ecell = E , (PbO2/Pb2+) – E , (Pb2+/Pb) = +1.47 – (–0.13) = +1.60 V > 0 (energetically feasible) (ii) SO 4 2-(aq) + Pb2+(aq) Æ PbSO4(s) The precipitation of PbSO4(s) reduced the [Pb2+(aq)] which caused eqm (1) to shift to the right, Ered (PbO2/Pb2+) is more positive than +1.47 V and eqm (2) to the left. Eoxid (Pb2+/Pb) is more negative than –0.13 V. As a result, the overall E cell increases to 2.0 V. 2 (a) K c = [] 3 2 − − ⎡⎤⎣⎦ ⎡⎤⎣⎦ I I I mol-1 dm3 (b) I - (aq) + I 2 (aq) I 3 - (aq) Initial amt 0.058 0.080 0 Change –0.0375 –0.0375 +0.0375 Eqm amt 0.0205 0.0425 +0.0375 Eqm conc 0.0205/0.5 0.0425 / 0.5 0.0375/0.5 Kc = [] 3 2 − − ⎡⎤⎣⎦ ⎡⎤⎣⎦ I I I = 0.0375 0.5 0.0205 0.0425 0.5 0.5 ⎡⎤ ⎢⎥⎣⎦ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ = 21.5 mol-1 dm3 (c) (i) Amount of iodine in toluene layer = 0.0525 x 800/1000 = 0.042 mol Amount of iodine in aqueous layer = 0.0425 – 0.042 = 5 x 10-4 mol (ii) Partition coefficient = [] 2 2 [I in toluene] I i n w a t e r = ( ) ⎛⎞ ⎜⎟ ⎜⎟⎝⎠ -4 0.0525 5x10 5001000 [ ]1M = 52.5

3 © INNOVA 2008 Answer Scheme 9746/03 (iii) Iodine is Umuch more solubleU in toluene than in water. This is because iodine is a simple molecular compound which forms weak van der waals forces of attraction between the iodine molecules. It is able to form similar van der waals forces of attraction with the toluene molecule. As iodine is not able to form hydrogen bonding or ion-dipole interaction with the water molecules, the solution of iodine will involve the breaking of stronger hydrogen bonding between the water molecules, hence it will be too endothermic for the reaction to occur. (d) (i) UPressure: The pressure should be Uhigh Uto that equilibrium will shift to the UrightU to form Ufewer gaseous moleculesU to Ureduce the pressure. UTemperature: The temperature should be UlowU so that the equilibrium will shift to the UrightU to favour Uexothermic reaction, releasing the heatU to Uincrease the temperatureU. (ii) Logical answers accepted e.g.: - The products from step 1 can be easily fed into step 2 and reduce transport cost. - The heat released from the exothermic step 2 reaction can be used to supply the heat energy required for endothermic step 1. (e) (i) (ii) 3CH3OH + PI3 3CH 3I + H3PO3 UTo neutralise the acidU, H3PO3 formed in the reaction. Excess Na2CO3 will remain as suspension. CH 3I will undergo nucleophilic substitution with NaOH to form back methanol. (ii) [ ]1M Tetrahedral about phosphorus atom. A dative bond is formed between P and B atom. 12 M⎡ ⎤⎣ ⎦

4 © INNOVA 2008 Answer Scheme 9746/03 3 (a) [RCOOR']/ mol dm-3 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 5 10 15 20 25 30 35 40 45 50 Time / min Experiment 2 with [HCl] = 0.40 mol dm-3 Experiment 1 with [HCl] = 0.20 mol dm -3 (b) (i) Since the half lives for the hydrolysis of ester are relatively constant, 30 min and 31 min respectively, the reaction is first order with respect to RCOOR’ (ii) hydrochloric acid, HCl Initial rate when [HCl] is 0.20 mol dm -3 = 1.777 x 10-3 mol dm-3 min-1 Initial rate when [HCl] is 0.40 mol dm-3 = 3.556 x 10-3 mol dm-3 min-1 When concentration of HCl doubles, initial rate of reaction increase by approximately 2 times, -3 -3 3.556 10 1.777 10 × × . Hence the reaction is first order with respect to HCl. (iii) Rate = k[RCOOR’][HCl] using [H +] = 0.20 mol dm-3 k = 5.49 x 10-2 mol-1 dm3 min-1 (c) The Uhydrolysis of ester is a reversible processU, hence not 100% of the carboxylic acid is formed when the system reaches equilibrium. 30 min 31 min

5 © INNOVA 2008 Answer Scheme 9746/03 (d) • Lowers the activation energy (E a) of the reaction. Hence, more reactant particles obtain energy > E a. The frequency of effective collisions increases, hence the rate increases. • The Urate constant increasesU with arrhenius equation, k = A aE RTe − . (e) (i) CH3 + (ii) Step II Reagent: Br2 in CCl4 Condition: uv light Step III Reagent: K 2Cr2O7, Dilute H2SO4 Condition: Heat with distillation Step IV Reagent: Concentrated H 2SO4 Condition: 170oC. (iii) Compound R C O-Na+ C H H H O H C H C H O O- H (iv) UFree radical substitutionU This reaction is Unot specific U or very Udifficult to control the position of Br as it can be substituted to two different C atoms U In addition, other side products can be formed. It can give rise to a Umixture of mono-, di- or tri- substituted products U from free radical sub orU, electrophilic substitution is also possible with phenol.U (v) Test: add NaOH(aq) to each sample & heat. Then add excess HNO3(aq), followed by AgNO3(aq). Obs: Compound Q: cream ppt of AgBr Compound P: no cream ppt. (vi) C OH H Br CH=CH2 C OH Br H CH=CH2 CH3 H3C Optical isomerism

6 © INNOVA 2008 Answer Scheme 9746/03 4 (a) Hess’s law states that because enthalpy is a state function, the enthalpy change of a reaction is the same regardless of what pathway is taken to achieve the product. (b) (i) ∆Hd = 54 + 7(-394) + 2.5(-242) = - 3310 kJ mol-1 (ii) Heat lost to surrounding. (iii) The numb