SAJC H2 CHEM P2 ANS Prelim

1 (a) 5Fe2+ + MnO4 ‒ + 8H+  5Fe3+ + Mn2+ + 4H2O [1] (b) If volume of titre is assumed to be 25 cm3: Amount of KMnO4 required = 0.5 x 0.025 = 0.0125 mol Amount of Fe2+ = 0.0125 x 5 = 0.0625 mol Mass of Fe2+ = 0.0625 x 55.8 = 3.4875 g Maximum mass of tablet = (100/80) x 3.4875 = 4.36 g Minimum mass of tablet = (100/90) x 3.4875 = 3.88 g [2] (c)  Weigh a dry and clean weighing bottle.  Add iron supplement tablet into the weighing bottle and weigh the bottle + tablet.  Tip the tablet into a small beaker and reweigh the emptied weighing bottle to determine the actual mass of tablet used.  Add excess dilute sulfuric acid to the small beaker containing the tablet.  Stir with a glass rod to dissolve the tablet.  Transfer the solution with several washings into a clean 250 cm 3 volumetric flask.  Make up to the mark with distilled water.  Stopper the volumetric flask and shake well to obtain a homogeneous solution.  Pipette 25.0 cm 3 of the iron solution prepared into a 250 cm3 conical flask.  Fill the burette the standard solution of KMnO 4.  Titrate the iron solution against KMnO 4 from the burette, with continuous swirling.  Stop when one drop of solution from the burette causes a colour change from colourless to pale pink. [6] (d) Step Expected observation Identity of cation in ppt Pour 2 cm3 of the solution into a test tube. Add NaOH(aq) dropwise until excess. Pale blue ppt formed in colourless solution. Cu 2+ Filter the mixture into a separate test tube. Add excess H 2SO4(aq) to the filtrate. White ppt formed. [½] Ba 2+ * Can identify Ba2+ first [3]

2 a) i) ii) iii) Both compounds are simple covalent. Phosphorus tribromide is polar with permanent dipole-permanent dipole while boron tribromide is non polar with induced dipole-induced dipole. More energy required to break the stronger pd-pd interactions of phosphorus tribromide so PBr3 has a higher boiling point. [5] b) PBr3(l) +3H2O(l)  H3PO3(aq) +3HBr(aq) [1] c) Energy/ kJ mol-1 BE = +250 kJmol -1 [4] d) i) Nucleophilic substitution PBr3(l) PBr3(g)  P(g) + 3/2Br2(l) P(g) + 3Br(g) P(s) + 3/2Br2(l)  0  ‐185  +39  3 x BE  3/2 x (+193) +315

ii)   iii) To form C(CH3)3Br would require the starting alcohol to be a tertiary alcohol which is not feasible for a SN2 reaction due to steric hindrance/large bulky groups/electron donating methyl groups. iv) CH 3COBr [7] e) (i) (ii) [8] Total: 25 marks

3 4 a) a) H 2O b) Fro whe Fro dou The Fro Whe The Rat c) Mec Ste rea   d) Ene                                         i) % Mol Sim n (Co n = 1 O2 +2 I- + 2H m graph 1, en [H+] chan m graph 1, ubled from 0 erefore first m graph 2, en [H+] = 1.0 erefore first te = k [H2O2 chanism A p 1 is the r ctants matc rgy       ∆H                                                e ratio mplest ratio oN4H12Cl3 ) H+  2H2O since graph nges. There when [I-] is 0.02/2s to 0 order wrt [H 0 mol dm3, order wrt [I- 2] [I-] . rate determ ches the po                              Co 25.2 0.428 1 = 233.4, O + I2 h is a straig efore zero o constant, [ .02/s. H2O2]. , [H2O2] = 2 -]. units for k: mining step ower of the       Reaction N 24.0 1.71 4 ght line grap order wrt [H+ [H2O2] is do .0 mol dm3 : mol-1dm3s- as the stoic e [H2O2] and  Pathway  H 5.1 5.1 12 ph, rate of re +]. oubled, rate 3, t1/2 is cons -1 chiometric d [I-] in the [T 1 1 2 eaction is co (gradient) i stant at ½ t coefficient rate equati Total: 11 m Cl 45.7 1.29 3 onstant is t. of the on.   marks]

molecular formula = CoN4H12Cl3 . Ligands: NH3 and Cl- ii) 0.01 mol of P contains 9.965 x 10-3 mol of free chloride ions 1mol of P contains 1 mol of free chloride ions formula of P : [Co(NH3)4Cl2]+ Cl- [6] b) i) Catalyst The reaction becomes faster when it is added as seen from the vigorous effervescences The solution turns from pink to green and back to pink colour as the catalyst is regenerated 4c) i) Violet ii) In the isolated gas phase, the d orbitals of the Co2+ is degenerate. In the complex, partially filled 3d orbitals split into two groups with a small energy gap between them. When light shines, the complex absorbs light energy from the visible light spectrum to promote electrons from the lower to the higher energy group, (d-d transition). The light not absorbed will be reflected and seen as the colour of the complex. [7] [Total: 13marks] 5 a) i) Number of chiral centers = 2 ii)

iii) H2O Chloroethane iv) Reagents and conditions: KMnO4, aq. H2SO4, heat Observations: S decolourise purple KMnO4 but purple KMnO4 remains in R *K2Cr2O7 accepted. [6] (b)                                                 J (i) It is further from the electron donating cyclopentane group which destabilise the carboxylate anion. (ii) Geometric isomer/ cis-trans 5 (b) (iii) Anion after dissociation is stabilised by intra hydrogen bonding of the carboxylate group with –OH [5] [Total: 11 marks] ~~~END~~~