HCI Prelim P3 ANS

HWA CHONG INSTITUTION 2009 C2 Higher 2 CHEMISTRY 9746 PRELIMINARY EXAMINATION PAPER 3 Free Response Questions (ANSWERS) 1 (a) (i) -320 = 314 + 3/2(244) – 3BE(P−Cl) -32  BE(P−Cl) = +323 kJ mol –1 (ii) Shape with respect to each P: Trigonal pyramidal; Bond angle: 107 o (b) (i) Electrophilic addition Energy / kJ m ol–1 0 P (s) + 3/2 C l2 PCl3 -320 3/2 (+244) 3BE(PCl)P (g) + 3/2 Cl2 (g) +314 P (g) + 3Cl (g) PCl3 (g) +32 1

2 (ii) Rate = k[but-1-ene][HCl] (iii) BE / kJ mol–1 H−Cl 431 H−I 299 The rate of reaction will be faster. The rate determining step involves the breaking of the H-X bond. Since H-I bond is weaker than H−Cl, the rate of reaction should be faster if HI is used. (c) The product obtained will not rotate plane-polarised light, because the carbocation intermediate formed is planar. Hence in the second step, there is an equal chance of the C l– ion attacking the carbocation from either side of the plane to produce equal amounts of the two enantiomers, giving rise to a racemic mixture. The two enantiomers rotate plane-polarized light by equal amount and in opposite directions, hence cancelling each other’s optical activity. (d) (i) Both C l 2 and I 2 exist as diatomic molecules and have dispersion forces existing between molecules. I, being lower down in the group than C l, has a bigger electron cloud size and hence, stronger dispersion forces. Therefore I 2 is less volatile than Cl2. (ii) 4Cl2 (aq) + S2O3 2– (aq) + 5H2O (l)  8Cl – (aq) + 2SO4 2– (aq) + 10H+ (aq) I 2 (aq) + 2S2O3 2– (aq)  2I – (aq) + S4O6 2– (aq) E, / V Cl2 + 2e = 2Cl – +1.36 I2 + 2e = 2I – +0.54 E, Cl2/Cl– is more positive than E , I2/I–, indicating that C l2 is a stronger oxidizing agent than I2. Hence, Cl2 can oxidize S 2O3 2– to SO4 2– while I2 can only oxidize S 2O3 2– to S4O6 2–.

3 n. 2 (a) (i) When a re versible reaction reaches a state of dynamic equilibrium, reaction continues to occur but the rates of the forward and backward reactions are equal such that there is no net change in the concentrations of the reactants and products. (ii) K C CrO4 2–  2 H  2 initial [CrO 4 2–] = [CrO 4 2–] at eqm = [Cr2O7 2–] at eqm = Substituting into K C, [H +] = 1.793 x 10–6 mol dm–3 pH = -lg (1.793 x 10 –6) = 5.75 (iii) When chromate ions are precipitated, the chromate concentration will drop and the equilibrium will shift to the left. This increases the amount of H + present and lowers the pH value. (b) (i) 2CrO 4 2– + 3SO2 + 10H+  2Cr3+ + 3H2SO4 + 2H2O E, cell = + 1.16 V The overall cell potential (E cell) for the redox reaction between CrO 4 2– and SO2 is a large positive value, meaning that the reaction will go in the forward direction, i.e. chromate will be reduced to Cr3+. (ii) This could be due to the absence of water or H + ions in the atmosphere and on the painting / the amount of SO 2 in the atmosphere is low. Hence slowing down the rate of the reactio (iii) The pigment turns from yellow to green. (c) (i) No. of moles of S 2O3 2– used = (21.50/1000) x 0.20 = 4.30 x 10–3 2S2O3 2– + I 2  S4O6 2– + 2 I – Hence no. of moles of I 2 formed in 25.0 cm3 = 4.30 x 10–3 x ½ = 2.15 x 10–3 Cr 2O7 2– + 6 I – + 14H+  2Cr3+ + 3 I 2 + 7H2O  mole ratio of Cr2O7 2– : I 2 = 1 : 3  Cr2O7 2–   100 1000 8 194.2  m-30.4119 mol d 1 5 x 0.4119 0.08238 mol dmŠ3 4 5 x 0.4119 2 0.1648 mol dmŠ3 0.1648 0.08238 2 H  2 7.55 x 1012 

4 No. of moles of Cr2O7 2– reacted in 25.0 cm3 = (2.15 x 10–3) / 3 = 7.167 x 10–4 No. of moles of Cr 2O7 2– present in 100 cm3 = 200 x 7.167 x 10–4 / 25.0 = 5.733 x 10–3 [Cr 2O7 2–] in sample = 5.733 x 10–3 / (100/1000) = 0.0573 mol dm–3 (ii) B undergoes condensation with 2,4-DNPH to give a hydrazone, indicating it is an aldehyde or ketone. However, B undergoes further oxidation with dichromate to give C. Hence B is an aldehyde, and A has a primary alcohol group. C is therefore a carboxylic acid. B and C undergo oxidative cleavage with KMnO 4 to give CH 3CO2H. Loss of 2 carbon atoms. Hence B and C contains alkene group, i.e. CH3C=C group. From the molecular formula of A, A: CH3CH=CHCH2OH B : CH3CH=CHCHO C : CH3CH=CHCO2H Hydrazone: B undergoes reduction of alkene group and aldehyde group with hydrogen, while C undergoes reduction (or hydrogenation) of the alkene group only. D: CH3CH2CH2CH2OH E: CH3CH2CH2CO2H NN H C CH3CH=CH H O2N NO2

3 (a) Na Mg Al Si P Si Cl Ar M.p. /oC High melting point from Na to A l as they exist as giant metallic structures with strong metallic bonds of increasing strengths due to smaller cationic radius and increased number of delocalized electrons. Very high melting point for Si as it exists as a giant covalent structure with an extensive network of strong covalent bonds. These strong bonds require a lot of energy to break before melting can occur. Low melting point from P to Ar as they exist as simple molecular structures, consisting of discrete molecules with weak dispersions forces between the molecules. Melting point decreases from S 8 > P 4 > Cl2 > Ar because the size of the electron clouds decreases from S8 > P4 > Cl2 > Ar, such that dispersion forces are weaker. (b) Mg(OH)2 is basic, and reacts only with acids: Mg(OH)2(s) + 2HCl (aq)  MgCl2(aq) + 2H2O(l) Be(OH)2 is amphoteric, and reacts with both acids and bases: Be(OH)2(s) + 2HCl (aq)  BeCl2(aq) + 2H2O(l) Be(OH)2(s) + 2NaOH(aq)  Na2Be(OH)4(aq) (c) (i) ∆G, soln for MgSO4 = (–91.2) – 298(–0.210) = –28.5 kJ mol-1 ∆G, soln for BaSO4 = (+26.3) – 298(–0.103) = +57.0 kJ mol-1 MgSO4 is soluble but BaSO 4 is insoluble because ∆G, soln is negative only for MgSO4. (ii) The negative ∆ S, soln indicates that there is a decrease in entropy when the two compounds dissolve to form aqueous ions. This could be due to a more orderly arrangement of H2O molecules when the aqueous ions are formed. (d) (i) From pV = nRT, 32 x 103 x 150 x 10–6 = 0.080 Mr x 8.31 x 303 Mr = 42.0 S Melting point /oC 5

{or CH2CO} (ii) Trigonal planar with respect to one C and linear with respect to the other C. (iii) CH3CO2C2H5 (e) (i) A undergoes tri-iodomethane reaction  A contains CH 3CH(OH)- or CH 3CO- group. From the given molecular formula, A is CH3CH(OH)CH2CO2H. A undergoes halogenation (nucleophilic substitution) with PC l5 to form B. A contains –OH and/or –CO2H groups. B is CH3CHClCH2COCl. The –COCl group in B undergoes hydrolysis with water to give C. C is CH3CHClCH2CO2H. C undergoes nucleophilic substitution with KCN, followed by acid hydrolysis to D. D is CH3CH(CO2H) CH2CO2H. C undergoes nucleophilic substitution with NH3 to form E. E is CH3CH(NH3 +) CH2CO2 - (reject the molecular form) 6

7 4 (a) Any two properties: Fe has a higher melting/boiling point than Al. Fe has stronger metallic bonds as both the 3d and 4s electrons can be used in metallic bonding due to their proximity in energies, hence more energy is required to overcome the s