NJC H2 Chem 2013 Prelim P3 Soln

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 1 1a) (i) or Shape about N: Bent (ii) Type of isomerism: cis-trans or geometric isomerism More stable isomer 1b) (i) Dynamic equilibrium refers to an equilibrium reaction when the rate of forward is equal to the rate of backward reaction. Kp = ) P)(P ( P 22 2 2 F N F N Units: Pa-1 or atm-1 (ii) pV = nRT pV = RTM m M = 65 10x500x10x12. 1 500x31. 8 x578. 0 pV mRT  =42.9 gmol-1 N2 F2 N2F2 Initial n/mol 2 1 0 Change of n/mol -x -x x Equilibrium n/mol 2-x 1-x x Total number of moles = (2-x) + (1-x) + x = 3-x 144

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 2 (Note: sum of mole fraction x Mr for each gases = average Mr) 9 .42)66(x 3 x)38(x 3 x 1)28(x 3 x 2    x = 0.576mol Mole fraction of N 2F2 = gases of moles total F N of moles2 2 = 237. 0576. 0 3 576. 0 x 3 x  (ii) Partial pressure of N 2F2 = mole fraction of N2F2 x total pressure = 0.237 x 1.12 x 105 = 26500 Pa (iii) ∆S is negative. There is a decrease in number of moles of gas in the reaction and the system becomes more ordered. (iii) ∆G=∆H–T∆S. Since ∆S is negative and T is always positive, therefor e, –T∆S is positive. In order for ∆G to be negative, ∆H must be negative and have a magnitude greater than T∆S. 1 c) A = P 2O4 P O 49% 51% Ar mass% 57. 11 .31 49  18. 30 .16 51  Simplest ratio 1 2 Empirical formula of A = PO2 [1] P2O4 + 3H2O → H3PO3 + H3PO4 A 145

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 3 1 d) (i) C H C H Cl H In chloroethene, as Cl is adjacent to the C=C bond, p-orbital of Cl can overlap with the π electron cloud of C=C bond . Hence lone pair of electrons on Cl is delocalised into the π electron cloud of C=C bond , strengthening C –Cl bond (partial double bond character). Thus C–Cl bond in chloroethene cannot be cleaved, no free Cl– ions formed, hence AgCl ppt is not produced. 1 d) (ii) Greater reactivity of the acyl chloride C –Cl bond in C6H5COCl/ the acyl chloride is more reactive because  the acyl carbon atom is bonded to 2 electronegative atoms, Cl and O which makes the carbon atom more electron-deficient/more partial positive. OR  the electron withdrawing C=O bond makes the carbon atom bonded to Cl more electron-deficient/more partial positive. Hence the δ+ C is more susceptible to nucleophilic attack by water molecules. C–Cl bond breaks readily to give Cl– ions which gives AgCl, white ppt with AgNO3(aq). 2 a) Reducing agent: H2O Relevant data required: Reduction: MnO4  + 4H+ + 3e  MnO2 + 2H2O Eo red = +1.67V Oxidation: 2H2O  O2 + 4H+ + 4e Eo oxid = 1.23V Overall 4MnO4  + 4H+ + 4OH  4MnO2 +2H2O + 3O2 + 4OH Eo cell = +0.44V i.e. 4MnO4  + 2H2O  4MnO2 + 3O2 + 4OH is a spontaneous rxn. 2 b) As SO 3 2 reacts with MnO 4  in a 1:1 ratio and SO 3 2 can be oxidised to SO 4 2, thus SO3 2  MnO 4   2e and a likely formula for the manganese species is such that Mn is in the +5 oxidation state. Relevant eqn: SO3 2 + MnO4   SO4 2 + MnO3  OR SO3 2 + MnO4  + 2OH  SO4 2 + MnO4 3 + H2O 2 c) Reaction between 2 negatively charged ions, C 2O4 2 and MnO 4 , involves high activation energy  reaction rate is slow at room temp. Warming/heating is required to hasten the reaction rate as at a higher temp, more molecules will have acquired energy greater than or equal to the activation energy. 146

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 4 2 d) (i) Plot a graph of [MnO4 ] against time. Determine 2 half-life from the graph and show that they have the same value or approximately the same (t1/2  15s). Rate = k[MnO 4 ] 2 d) (ii) Rate of consumption of C2O4 2 = 5/2 x Rate of consumption of MnO4  = 3.0 x 104 mol dm3 s1 2 d) (iii) Homogeneous catalyst 2 d) (iv)  At the beginning of the reaction, the production of CO 2 is slow and low because fruitful collision of negatively charged MnO 4 - and C 2O4 2- ions is difficult.  Once a certain amount of Mn 2+ has been formed, the production of CO 2 increases rapidly because Mn 2+ acts as the catalyst for the reaction (autocatalyst).  Vol of CO 2 remains constant after some time even though there is an adequate supply of catalyst as the reaction has reached completion. 2 e) Observations Deductions C gives yellow ppt with alkaline aq I2.  C contains either methyl carbonyl group C O CH3 or C OH CH3 H structure. C has no reaction with Tollen’s reagent.  C is not an aldehyde. C (C6H10O2)       4/ KMnO Hhot D (C4H6O3) only product  C is oxidized (vigorous oxidation/ oxidative cleavage of C=C bond).  D contains a carboxylic acid group Vol of CO2 time 147

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 5 ( increase in no. of O atoms)  C loses 2 carbon atoms as CO2 gas.  i.e. oxidative cleavage of C=C bond resulted in ethanedioic acid which is further oxidized to CO2 / C=C bond is between C2 and C3. C (C6H10O2)    4NaBH E (C6H12O2)  C is reduced.  C contains an alcohol group and a methyl carbonyl group C O CH3 which is reduced to C OH CH3 H in E. C C H C H H OH H C H H C O CH3 C O CHO C H H C O CH3 D C C H C H H OH H C H H C OH CH3 E H 3 a) (i) Cl2(g), or Cl2 in CCl4, room temperature, dark Electrophilic addition 148

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 6 Cl + Cl Cl Cl Cl + Cl Cl Cl + Cl Cl Cl Cl slow Note:  The type of mechanism reaction must be stated.  The mechanism arrows must be shown in the equation to illustrate the movement of electrons.  Lone pair on the chloride ion must be clearly shown.  Slow step and the partial charges on chlorine must be indicated in the mechanism 3 a) (ii) The geometry about the positive charge carbon atom of the carbocation from the slow step is trigonal planar. Hence, :Cl- has equal chances of attacking from both above and below the plane , to produce equal amount of the 2 enantiomers , resulting an racemic mixture whereby the optical activity of the two enantiomers cancel out each other completely. Thus, it is not optically active.  + Cl Cl + Cl 149

NJC SH2 Prelim 2013 H2 Chemistry Paper 3 Solution 7 3 a) (iii) Q R O OH C OH OH N 3 a) (iv) CHI3 3 b) Concentrated sulfuric acid acts as catalyst (Hint: limited amount of conc H2SO4). 3 c) (i) The standard enthalpy change of formation of HI(g) is the heat absorbed when 1 mole of HI(g) is formed from H 2(g) and I 2(s) (or constituent elements in their standard state) under standard conditions of 1atm at 298K. or ½ H 2(g) + ½ I2(s)  HI(g) H>0 3 c) (ii) As Hf of H-X becomes less exothermic from HCl(g) to HBr(g) to HI(g), the reaction does not occur as readily since the formation of HX is less energetically favourable. Thus, reactivity of chlorine, bromine and iodine with hydrogen decreases. Or From chlorine to bromine to iodine, the halogen exists as gas to liquid to solid. Thus, more energy is required to overcome the Van der Waals force from chlorine to bromine to iodine to form HX, resulting Hf of H-X to become more endothermic. Therefore, the reaction does not occur as readily and thus, reactivity of chlorine, bromine and iodine with hydrogen decreases. Or As H-X bond strength decreases from H-Cl to H-Br to H-I, the sta