RI 2009 A-Level H2 Chemistry Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q1 O C HC HC HC HC HC OH OH OH OH OH H H + 5Na O C HC HC HC HC HC O-Na+ O-Na+ O-Na+ O-Na+ O-Na+ H H + 2.5H2 Ans : A Q2 Amt of C2O42 = 2 × 103 mol since 1 mol of KHC2O4.H2C2O4 gives 2 mol of C2O42 Amt of MnO4 = -32.0 10 25   = 8 × 104 mol Conc of MnO4 = 4 33 3 8.0 10 1000 / 0.020 mol cm dm moldm     = 40cm3 Ans: B Q3 Proton No Electronic Configurn Ion/ Atom A 11 2.8 Na+ The only cation, that could be reduced at cathode Na+ + e  Na B 16 2.8.8. S2 C 17 2.8.8. Cl D 18 2.8.8. Ar Ans: A Q4 From Data Booklet, Metal 1st I.E. 2nd I.E 1st+2nd I.E. A Co 757 1640 2397 B Cr 653 1590 2243 C Cu 745 1960 2705 D Ni 736 1750 2486 Al 577 1820 2397 Ans:A Q5 C O- O- O CO3 2- N O O- O NO3 - C C O- O CH3CO2 - H HH O- C6H5O- 109o All have delocalized electrons. Ans: C Q6 In the absence of water, HCl(g) remained as an undissociated simple molecule, and not dissociated into H+ and Cl ions, thus it does not have acidic properties. Ans: B Q7 PV = nRT P = nRT V P = 3 1 1 33 (1.6 10 )(8.31 )(273 273) (3.0 10 ) mol JK mol K m       Pa Ans: D Q8 Ans: C (A) CaC 2O4 should have lower solubility product than MgC2O4 as it is less soluble and they are both similar salts. (B) CaC 2O4 should have numerically lower lattice energy than MgC2O4 as Ca2+ is larger than Mg2+ & .. qqLE rr    (C) Hsoln = Hhyd  L.E. Generally, the more exothermic or more negative Hsoln is, the more soluble the salt. hyd qH r    and Ca2+ is larger than Mg2+, thus Ca2+ has lower Hhyd than Mg2+ considering that both ions have same charge. (D) Electropositive values are not related to solubility. Suggested Solutions to N2009 H2 Chemistry Paper 1 (9746/01) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A B A A C B D C B A C B C B D C B D B D 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 A D D D B A C C C B A D D B D D A B D C Detailed Working for Suggested Solutions to N2009 H2 Chemistry Paper 1 (9746/01)

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q9 G= H  TS H  0  G=  TS  S & G have opposite signs Since pure water is separated from that contaminated with ions (and molecules), with the incre ased ordering of state, ΔS must be negative G is positive (non-spontaneous). Ans: B Q10 Freezing at melting is a reversible process as both solid and liquid co-exist at dynamic equilibrium, G=0. H is negative since heat is evolved. G= H  TS = 0 G = 3154( )( 6.0 10 )18.0 mol Jmol   (273K)(S)= 0 S =  66 J K1 Ans: A Q11 Cathode: 2H+ +2e  H2 (reduction) Anode: Zn  Zn2+ +2e (oxidation) Statement X is wrong as [Zn2+] > 1 moldm-3 favours reduction at anode Ecell < Ecell. Stat ement Y is correct as [H+] > 1 moldm-3 favours reduction at cathode, Ecell > Ecell. Ans: C Q12 PV = nRT P = nR TV Gradient of the graph is directly proportional to amount of gaseous particles. Amount of gaseous particles increases as dissociation of X2 is favoured at higher temperature since bond dissociation is endothermic. Gradient of graph should be doubled when X2 is dissociated into X at higher temperature. Ans: B Q13 At lower [ethanal], active sites of enzymes are not saturated and available, thus reaction is 1st-order with respect to ethanal. At higher [ethanal], active sites are saturated and reaction is zero order with respect to ethanal. Ans: C Q14 Water vapour would be produced on the decomposition of amphoteric aluminium hydroxide. The heat absorbed by the decomposition reaction acts as a retardant by delaying ignition of the associated substance. The water released would also act as an extinguisher (dilutes any combustible gases and inhibits oxygen from aiding the combustion). Ans: (B) Q15 (A) Cu atom (Z=29) has more electrons than Mg(Z=12) atom. (B) Cu2+:1s22s22p63s23p63d9 Mg2+:1s22s22p6 (C) Cu2+:1s22s22p63s23p63d9 Mg: 1s22s22p63s2 Same number of quantum shells (D) Cu: 1s 22s22p63s23p63d104s1 Mg2+:1s22s22p6 Cu has 2 more electron shells (quantum shells: n=3 & 4) than Mg2+. Ans: D Q16 3H2O2 + Cr2O72 + 8H+  2Cr3+ + 7H2O + 3O2 (orange) (green) (0) H2O2 + 2I + 2H+  H2O + I 2 (colourless) (2) (brown) Ans: C Q17 From Data Booklet, (A) B C N O 1st IE/kJmol-1 799 1090 1400 1310 (B) Co Ni Cu Zn 1st IE/kJmol-1 757 736 745 908 (C) Mg Ca Sr Ba 1st IE/kJmol-1 736 590 548 508 (D) Si P S Cl 1st IE/kJmol-1 786 1060 1000 1260 Ans: B Q18 Cu2+(aq) + 4I(aq)  2CuI(s) + I 2(aq) White ppt brown solution 2S2O32 + I2  S4O62 + 2I  Ecell = +0.15V  (+0.54V) = 0.39V Ans: D Q19 N N H H There is 1 lone pair and 2 bonding pairs about N atom. Bonding pair-lone pair repulsion is stronger than bonding pair-bonding pair repulsion, thus the bond angle is approximately 118. Ans: B 

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q20 C CH3 CC HO H2C CH2 C HC CH2H3C carvone + 3H2 HO CH3 H3C CHH3C * * * * Ans: D Q21 Bromine is an electrophile that is added onto the C=C bond. Ans: A Q22 O- Cl Cl is the nucleophile that replaces Cl in ClCH2CO2H, thus reaction is nucleophilic substitution. Ans: D Q23 Halogen compound S, a halogenoarene, cannot undergo nucleophilic substitution with ethanol due to partial double bond character in CCl bond. Thus, it unable to release free Cl ions that can be precipitated with free Ag+ ions. Ans: D Q24 (A) It reacts with HNO 3(aq) to form a substitution product (shown below), not an addition product. OH3C CH3 CO2H OH CH3 CH3 (CH2)5CH3 NO2 OH3C CH3 CO2H OH CH3 CH3 (CH2)5CH3 NO2 OR (B) It will only form a single ester(shown below), not a diester upon heating with ethanol with conc H2SO4. OH3C CH3 CO2CH2CH3 OH CH3 CH3 (CH2)5CH3 (C) K 2Cr2O7 cannot oxidise the C=C bond or the alkyl side chain of the aromatic ring,thus no colour change is expected. (D) The product formed is shown below. Note that 4 Br atoms are incorporated. OH3C CH3 CO2H OH CH3 CH3 (CH2)5CH3 Br Br Br Br Ans: D Q25 H3C OH CCH2OHO CH3 O Cortisone H2/Pt H3C OH CCH2OHO CH3 O H+/KMnO4 warming H3C OH CCO CH3 O O OH O HO O Ans : B Q26 NaBH4 can only reduce aldehydes and ketones. H2C CHCH 2OH P H2C CHCO 2H Q Ans : A

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q27 Phenol is the least acidic of the 4 compound. Carboxylic acids are stronger acid. The negative charge is delocalised between 2 oxygen atoms in the anions of the other 3 carboxylic acids. Bromoethanoic acid and chloroethanoic acid are more acidic than propanoic acid due to presence of electron-withdrawing halogen atoms which further de-intensify the negative charge on the anions of the carboxylic acids. Chlorine is a stronger electron-withdrawing group than bromine as it is more electronegative, thus chloroethanoic acid is the most acidic of the 4 compounds. Ans: C Q28 Amine is basic and dissolves in acid to form an ionic salt that is soluble (miscible) in the aqueous layer. Benzene has no reaction with aqueous acids, thus remains immiscible with the aqueous layer. Ans: C Q29 OH CH2 -CONHCHCONHCHCONHCHCONHCHCONH- CH2CH CH2OH CH2CONH2 CH3H3C 6 moldm-3 HCl Heat for pro