AJC H2 CHEM P1 detailed soln

H2 Chemistry 9647 2014 JC2 Prelim P1 Solutions 1 As dilute H 2SO4 is added to aqueous Ba(OH) 2, all Ba 2+(aq) ions present in solution will precipitate out as BaSO 4(s), a white precipitate, while the OH – is neutralised by H + to form H 2O. Hence, the number of ions present in solution decreases. H 2SO4(aq) + Ba(OH)2(aq)  BaSO4(s) + 2H2O(l) When dilute H 2SO4 is added in excess, the number of ions present in solution increases as H2SO4(aq) ionises completely in solution. H2SO4(aq)   2H+(aq) + SO4 2–(aq) A 2 Oxidation: Fe Fe2+ + 2e– Reduction: Fe3+ + e– Fe2+   Overall: Fe + 2Fe3+  3Fe2+ Fe + 2Fe 3+ 3Fe2+ initial (mol) 1 5 0 change (mol) –1 –2 +3 remaining (mol) 0 3 3 Given that reaction goes to completion, you may complete the above table using each of the options to come up with the remaining values below. n(Fe) n(Fe 3+) n(Fe 2+) A 0 0 3 B 0 1 3 C 0 3 3 D 0.5 0 4.5 Alternatively, for n(Fe2+) formed = 3 (start with the smallest whole no.) = n(Fe3+) left initial n(Fe) : n(Fe 3+) 1 (reacted) 2 (reacted) + 3 (left) 1 5 (option C) C 3 mass charge deflection of angle α angle of deflection of 241Am+ = +2 ( 241 1 ) angle of deflection of 32S– = 2)( 32 241  = –15.1 B 4 Each carbon atom in a molecule of ethene is sp2 hybridised.                                                                                                                                                                                      D 2s 2p 2s 2p excitation hybridisation sp2 2p

5 A At constant T, pV = nRT  )V 1a( p where a = nRT = constant A graph of p against V 1 is a straight line that passes through the origin. B pV = nRT pV = aT where a = nR = constant A graph of pV against T is a straight line that passes through the origin. C At constant T, pV = nRT  pV = a where a = nRT = constant A graph of pV against V is a horizontal line with y–intercept = a D At constant p, pV = nRT  V = aT where a = p nR = constant A graph of V against T is a straight line that passes through the origin. D 6 A   COO     C O OO   180  120  B B H HH B H HH H 120 109.5  C N HH N HH H 105 107  D Xe FF Xe FF FF 180 90  C 7 C CH3 CH2Cl H Cl and C CH3 ClCH2 H Cl     are a pair of optical isomers with identical phys ical properties (except their interactions with plane–polarised light) A

8 Na+(g) + H-(g) NaH(s)L.E Na(g) + H(g) 1st I.E 1st E.A Na(s) + 1/2 H2(g) A 9 [Cu(H2O)6]2+ + EDTA4– [Cu(EDTA)]2– + 6H2O A There is an increase in the number of particles (from 2 to 7) hence S for the forward reaction is positive. B The oxidation state of Cu remains unchanged at +2. C Since the coordination number of Cu in [Cu(H 2O)6]2+ and that in [Cu(EDTA)] 2– is 6, 6 coordinate (dative) bonds are broken and 6 coordinate (dative) bonds are formed during this ligand exchange reaction. D Since the coordination number of Cu in [Cu(H 2O)6]2+ and that in [Cu(EDTA)] 2– is 6, both [Cu(H2O)6]2+ and Cu(EDTA)]2– are octahedral complexes. B 10 300 150 75 37.5 3 half–lives  24 hours (1 day) 1 half–life  24 / 3 = 8 hours C 11 Eo(Ni2+/Ni) = –0.25 V Eo(Cr3+/Cr2+) = –0.41 V Eo(Mn2+/Mn) = –1.18 V Eo(Pb2+/Pb) = –0.13 V Eo(V2+/V) = –1.20 V Eo(Cr3+/Cr) = –0.74 V Eo cell = Eo reduction – Eo oxidation Eo reduction – (–0.25) = –0.13 – (–0.25) = +0.12 V When E o cell > 0, reaction is spontaneous. C 12 At anode (oxidation): 2Cl–   Cl2 + 2e– At cathode (reduction): 2H2O + 2e–  H2 + 2OH– n(Cl–) = n(NaCl) = 58.5 x 103 / (23.0 + 35.5) = 1000 mol n(Cl2) = ½ x n(Cl–) = ½ x 1000 = 500 mol mass of Cl2 = 500 x (35.5 x 2) = 35500 g = 35.5 kg n(H2) = ½ x n(Cl–) = ½ x 1000 = 500 mol mass of H2 = 500 x (1.0 x 2) = 1000 g = 1 kg n(NaOH) = n(OH–) = n(Cl–) = 1000 mol mass of NaOH = 1000 x (23.0 + 16.0 + 1.0) = 40000 g = 40 kg A Hf Hat Hat

13 A At 7 min, the concentration of the gases remains constant. This means that the system has reached dynamic equilibrium. At dynamic equilibrium, the rate of forward reaction equals rate of backward reaction with no net change in the concentration of gases. B At dynamic equilibrium at 7 min, 31- 2 2 c dm mol 41700.6 x 80 20 ][CO][C ][COC .. . l l K C At 7.5 min, the concentrations of all the gases in the equilibrium mixture increase. This is due to the decrease in volume as a result of an increase in pressure at constant temperature. After 7.5 min, the concentration of CO and Cl2 decreases, at the same time the concentration of COC l2 increases. This shows that the position of equilibrium has shifted to the right in accordance with Le Chatelier’s Principle.  (The forward reaction is accompanied by a decrease in the number of moles of gas molecules. Increasing the pressure w ill shift the position of equilibrium to the right to decrease the pressure by producing fewer gas molecules.) D At 10 min, the concentration of C l2 increases as more C l2 is added to the equilibrium mixture. According to Le Chatelier’s Principle , the position of equilibrium shifts to the right to remove some of the excess C l2. Thus, the concentration of CO and Cl2 decreases while that of COCl2 increases. B 14 CO 2(g) + H 2(g) CO(g) + H 2O(g) initial (mol) 1 1 0 0 change (mol) –x –x +x +x eqm (mol) 1 – x 1 – x x x Total no. of moles at eqm = 1 – x + 1 – x + x + x = 2 p(CO 2) = p(H2) = atm 2 x-1 atm 1 x 2 x-1  p(CO) = p(H2O) = atm 2 x atm 1 x 2 x  0.460 x 7260 x-1 x 0.726 )2 x-1( )2 x( )p(H x )p(CO O)p(H x p(CO) 2 2 22 2 p    . K      mole fraction of CO = x/2 = 0.460/2 = 0.230 A 15 NH4 + is the conjugate acid of a weak base, NH3. NH 4 + dissociates partially in water to produce H3O+ as follows: NH4 + + H2O  NH3 + H3O+ Mg(OH)2(s) then undergoes an acid–base reaction (neutralisation) with the H3O+ ions produced. Mg(OH)2(s) + 2H3O+(aq)  Mg2+(aq) + 4H2O(l) D

16 NH4 + + H2O  NH3 + H3O+ 54950 ][NH ][NH ][NH ]][10[NH10 ][NH ]O][H[NH 4 3 4 9- 374414 4 33 a . ).(         K Let V = volume (in dm3) of 0.1 mol dm–3 NH3 added, [NH3] after mixing = (V x 0.1) / (V + 0.01) [NH4 +] after mixing = 0.01 x 0.1 / (V + 0.01) = 0.001 / (V + 0.01) 333- 4 3 cm 5.50dm 10 x 5.50V 549500.001 V10 ][NH ][NH   .. A 17 Ca(NO3)2(s) CaO(s) + 2NO2(g) + ½ O2(g) 1 mol of Ca(NO 3)2 undergoes thermal decomposition to produce 2.5 mol of gaseous products. Volume of gas produced at r.t.p = 2.5 x 24 = 60 dm3 C 18 A All the Group VII elements are coloured with increasing intensity down the group. F 2 is pale yellow in colour. B Acid strength of HX increases down Group VII due to decreasing bond strength of H X. Hence, HF is the weakest acid. C Increasing atomic radius down Group VII us ually leads to weaker covalent bonds. One exception in this trend is fluorine, which has a weak F–F bond. This could be due to the lone pairs of electrons on the small fluorine atoms being so close together that they strongly repel each other, hence weakening the bond. D The melting points of Group VII elements increase down the group due to stronger instantaneous dipole–induced dipole (id–id) attraction between the X 2 molecules. Hence, F2 has the lowest m.p. C 19 The yellow solid is Ag I. The other ppt soluble in excess NH 3(aq) is AgC l. Hence, the powder consists of NaI and NaCl, which dissolve in water to produce I– and Cl– ions. D 20 dmpe is a bidentate ligand, i.e. each ligand forms 2 coordinate (dative) bonds with the central metal ion due to presence of a lone pair of electrons on each of the P atoms. The f