2022 EJC Prelims Paper 1 Worked Solutions

2022 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 24 H OH 25 1.0 10 2.50 10 mol1000nn+− −−= =   =  1 Ca2+ ≡ 2 H+ 2 4 Ca H 1 1.25 10 mol2nn ++ −= =    2 4 2 Ca 4 33 1.25 10CaSO Ca 50 1000 2.50 10 mol dm n V + − + −− = = = =  A 2 Large jump between 7th and 8th I.E.  7 electrons in valence shell  Group 17  C 3  C 4 pV nRT= mpV RTM= Plot of pV against T will be a straight line that passes through the origin (0 K) with gradient mR M . Since M(E) < M(F), the gradient for E will be steeper than the gradient for F.  A 5 G = H – TS 1 : H > 0, S < 0, G > 0 (non- spontaneous) at all T 2 : H < 0, S < 0, G > 0 (non- spontaneous) when HT S   3 ✓: H < 0, S > 0, G < 0 (spontaneous) at all T  D 6 A ✓: 1 mol of H 2O is formed from the reaction of an acid and a base B : lattice energyH ( )( )3NaHCO s is for the reaction Na+(g) + HCO – 3(g) → NaHCO3(s) C ✓: Being a strong acid, the HSO– 4 ion will fully dissociate in water to give H+(aq) and SO2– 4 (aq) D ✓: formationH involves forming (NH4)2SO4 from the constituent elements, S, H 2, N 2 and O 2 in their standard states  B 7 1 ✓: From the slow step, rate = k[(CH3)2C+(OH)][H2O] = k′[(CH3)2C(=O)][H3O+], k′ = kKc At low pH, [H 3O+] is high and is essentially a constant. rate = k″[(CH3)2C(=O)] 2 : rate constant, aE RTk Ae − = depends only on the Ea and T. 3 ✓: I2 is only involved in a fast step after the slow (rate determining) step. Hence rate of reaction is independent of [I2].  D 8 A : Since there is no change in the number of gaseous particles upon reaction, the total pressure in the vessel remains constant. Hence the amount of reactants and products cannot be determined. B : Slow cooling allows the equilibrium to readjust to the new t emperature. Hence, the Kc determined will be that at 20 ºC and not 500 ºC. C ✓: Rapid cooling will slow down the reaction, allowing the amount of I2 in the equilibrium mixture at 500 ºC, and hence Kc, to be determined. D : If the HI present is decomposed back to H2 and I2, the amount of reactants and product in the equilibrium mixture cannot be determined.  C 9 A : Since H < 0, position of equilibrium lies more to the left with increasing T. Hence p (g) 1K p= G es with ing T. B : npV nRT p RT CRTV=  = =   c p (g) 11 (g) KK p RT RT= = = G G C :  depends only on k1, k2 and pG. It is independent of the total surface area of the catalyst. D ✓: At high pG, 11 22 1 kk ppkk+ GG . Hence, 1 , which is independent of T.  D 10 As HCl is titrated against Na2CO3, Na2CO3 is added from the burette into HCl in the conical flask, the pH increases with the volume of titrant. HCl will completely react with any Na2CO3 added Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Hence there will be only one end-point.  A 11 Considering the species present: 1.9 8.1 5.02p +=I (zwitterion is neutral)  B 12 A : ( )sp 2 3 43 Ag COAg 2 2.53 10 4 K+− = =  B : ( ) 5 spAg AgC 1.34 10K+− = =  l C ✓: ( ) 7 spAg AgBr 7.35 10K+− = =  D : ( ) 4 sp 3Ag Ag O 1.79 10K+− = =  I  C 13 Residue: SiO2 (insoluble in H2O & H2SO4) Filtrate: acidic (gives H2 with Zn)  HCl  B 14 Reducing power (c hemical reactivity) of Group 2 metals es down the group, since  atomic radius es   attraction of nucleus for valence e–s  easier to loss valence e–s Atomic radius increases as the number of electron shells es, rendering valence electrons further from nucleus.  D 15 A : Electronegativity es with ing size down Group 17 , due to weaker attraction for shared pair of electrons. B : Oxidising power ( tendency to get reduced) es with ing size down Group 17, due to weaker attraction for incoming e–. C ✓: Polarisability of X – es with size down Group 17 as electrons are generally further from the nucleus. D : Oxidising power es (B) down Group 17, hence E ( )2XX − becomes less positive (es).  C 16 1 ✓: 2 ✓: 3 ✓:  D 17 A : Homolytic fission also occurs in the second step where the X–O bond in X–O• is broken to give X•. B : Termination steps involves the removal of X• through reaction with another radical. C ✓: X• serves as a catalyst to speed up the breakdown of O3 into O2. D : CCl2F2 → •CClF2 + •Cl instead as the C–F bond is much stronger than the C–Cl bond.  C

18 1 ✓: C=C in 2 -methylbuta-1,3-diene reacts with Br2(aq) 2 : There are no chiral elements (centres) in 2-methylbuta-1,3-diene 3 : The C=C –C=C  network is highly electron-rich and will repel approaching nucleophiles.  A 19 The two benzene rings can only be joined together via a Friedel-Crafts alkylation involving the aromatic hydrogen and a halogenoalkane:  A 20 Both Cl undergoes elimination with excess ethanolic KOH:  B 21 Halogenoarenes do not undergo nucleophilic substitution with ethanolic AgNO3. A : 1 mol of AgCl (143.4 g) B : 1 mol of AgI (234.8 g) C : 1 mol of AgCl + 1 mol of AgI (378.2 g) D : 2 mol of AgBr (375.6 g)  C 22  C 23 R gives a red ppt with Fehling’s solution  R is an aldehyde Oxidation of Q give R, an aldehyde  Q is a 1º alcohol  B 24 The carboxylic acid group reacts with alcohols (but not phenols) to give esters. CH3CHO and C 6H5COCl both reacts with the amino group.  C 25 A ✓: B : C ✓: D ✓:  B 26 A : The form of asparagine in aqueous solution depends on the pH of the solution. It exists solely in the form of the zwitterion at the isoelectric point. B ✓: The side chain, –CH2CONH2, being an amide is neutral. C : Hot dilute H 2SO4 hydrolyses the amide side chain to give NH + 4 (aq) and not NH3(g) D : All the 26 amino acids are crystalline solids at room temperature due to strong ionic bonding between the zwitterions.  B 27 [O]: Fe3+ + e–  Fe2+ E 0.77 V=+ [R]: O2 + 4H+ + 4e–  2H2O E 1.23 V=+ cellE reductionE= oxidationE− E= ( )22O H O E− ( ) ( ) 32Fe Fe 1.23 0.77 0.46 V ++ = + − + = + ( ) ( )cell reduction oxidationE E E E E= − = − TS 1 : [Fe2+] and [Fe 3+] es by the same extent. Eqm is not affected. ( )EE =S ( ) 32Fe Fe++ . No change to Ecell. 2 ✓: CN– forms complex with Fe 3+ and Fe2+. New eqm established [Fe(CN)6]3– + e–  [Fe(CN)6]4– with E 0.36 V=+ . ( )EE S ( ) 32Fe Fe++ . Hence Ecell es (more positive) 3 : [H+] es. Eqm shifts left. ( )EE T ( )22O H O Hence Ecell es (less positive)  D 28 Cathode: 2H2O + 2e– → H2 + 2OH– Anode: 2H2O → O2 + 4H+ + 4e– ( )15.0 5 60 4500 CQt= =   =I 4500 0.04663 mol96500e Qn F − = = = 22gas O H 1 1 3 4 2 4 3 0.04663 0.03497 mol4 e e en n n n n n − − −= + = + = =  =  C 29 r 3 mass mass of an atomdensity volume volume of an atom A r=   1 ✓: Ar es across the Period 4 TM, while r remain relatively constant, hence density es 2 ✓: Number of nucleons (protons and neutrons) es  mass of atom es, hence density es 3 :  in number of outer shell electrons has negligible effect on mass of the atom pn 1 or 1833 em m m   B 30 Within the tetradentate ligand, the two N atoms with 3 bonds (and 1 lone pair) are neutral, but the two nitrogen with only 2 bonds (and 2 lone pairs) are negatively charged. Hence the tetradentate ligand has a –2 charge. The remaining two ligands, CH 3OH an d , are neutral. Hence for an overall charge of +1 for the complex cation, Co must be in the +3 oxidation state.  C Answer Key Qn Ans Qn Ans Qn Ans 1 A 11 B 21 C 2 C 12 C 22 C 3 C 13 B 23 B 4 A 14 D 24 C 5 D 15 C 25 B 6 B 16 D 26 B 7 D 17 C 27 D 8 C 18 A 28 C 9 D 19 A 29 B 10 A 20 B 30 C