NJC 2017 A Level Chem Solution

H2 Chemistry TYS 2017 suggested answers Paper 1 Answer Key 1 C 11 A 21 A 2 B 12 A 22 A 3 D 13 D 23 D 4 C 14 D 24 C 5 D 15 A 25 D 6 B 16 B 26 A 7 B 17 A 27 A 8 B 18 D 28 B 9 C 19 A 29 C 10 B 20 C 30 A 2017 A level Paper 1 suggested answers 1) Let % of 29Si be x, % of 30Si be (100−92.23− x) 92.23×28+𝑥×29+(100−92.23−𝑥)×30 100 = 28.10 2582.44 + 29x + 233.1−30x = 2810 x = 5.54 Ans: C 2) These are the most common isotopes for Li to Mg (Qn stated that the nucleon number is the closest whole number to its Ar) Element Li Be B C N O F Ne Na Mg Nucleon number 7 9 11 12 14 16 19 20 23 24 No. of protons 3 4 5 6 7 8 9 10 11 12 No. of neutrons 4 5 6 6 7 8 10 10 12 12 Only statements 1 and 2 are correct. Ans: B 3) C O HH For methanal, there are 3 bond pair regions and 0 lone pair region around the central C atom. It is trigonal planar with bond angle of 120°. Ans: D

4) Statement 1 is correct. More energy is required to overcome the stronger hydrogen bonding between H2O molecules as compared to the weaker instantaneous dipole-induced dipole attraction between methane molecules. Statement 2 is wrong. During boiling, only the intermolecular forces of attractions are broken. Covalent bonds are not broken. Statement 3 is wrong. Both H2O and CH4 has the same number of electrons in the molecules (10 electrons). Note that size of electron cloud affects strength of id-id. Ans: C 5) Gases liquefies under pressure as the intermolecular forces of attractions becomes more significant when the molecules are brought closer together. Hence the molecules are arranged in a more orderly arrangement in a liquid state as compared to gaseous state. Ans: D 6) Lewis base is an electron pair donor. Bronsted-Lowry acid is a proton(H+) donor. H2O can be BOTH an electron pair donor and H+ donor. Ans: B 7) Using information from the Data Booklet, atomic radius of Mg>Al>Si>P. Electronegativity increases across the period, hence electronegativity of P>Si>Al>Mg. Atomic radius of Si(0.117nm) is only slightly larger than P(0.110nm). Ans: B 8) Fifth I.E. shows the removal of electron for the following process: X4+ → X5+ + e− The sharp change in 5th I.E. from R to S shows that the 5th electron for R is removed from an inner shell as compared to S. Hence R4+ has a configuration of ns2np6. R has configuration of ns2np6(n+1)s2(n+1)p2 and is in Group 14. Q is an element from Group 13. Ans: B 9) Mg2+ has the highest charge density, and hence the strongest polarising power. It is least likely to form a stable peroxide. Ans: C

10) G = (−1.9) − 298(0.0034) = −2.9kJ mol−1. The forward reaction is spontaneous and the reverse reaction is not spontaneous. The reaction does not occur readily at room temperature and pressure even though G is negative, this implies that the rate of reaction must be very slow. Hence the forward reaction must have an extremely small rate constant. (rate constant, k, is affected by temperature and catalyst). Even though statement 2 is correct, it does not explain why the forward reaction does not occur. Ans: B 11) After 2 half-lives (60minutes), 75% of X2 will be converted to 2X atoms. X2 → 2X Initial mol 1 0 Change in mol −0.75 +1.5 Final mol 0.25 1.5 In a sealed vessel, volume of container is kept constant. 1 mol of gas exert a pressure of p. Hence, 1.75 mol of gas will exert a pressure of 1.75p. Ans: A 12) The correct rate equation for the reactions are: Reaction A: rate = k[NO]2[H2] Reaction B: rate = k[H2] Reaction C: rate = k[HBr]2[O2]1/2 Reaction D: rate = k[H2O2]2[I−] Ans: A Note: If the slow step involves a reaction intermediate, the intermediate compound is affected by concentration of reactants in the previous steps. e.g. Reaction D: rate = k[H2O2][IO−] since IO− is affected by H2O2 and I−, we can replace [IO−] with [H2O2][I−] Hence overall rate equation is rate = k[H2O2]2[I−] 13) In pure water, [H3O+] = [OH−] 𝐾𝐶= [𝐻3𝑂+]2 [𝐻2𝑂]2 [H3O+] = [H2O] × √𝐾𝐶 [H2O] = 997𝑔𝑑𝑚−3 18.0𝑔𝑚𝑜𝑙−1 = 997 18.0 𝑚𝑜𝑙𝑑𝑚−3 In 1.00dm3 of pure water, amount of H3O+ = 997 18.0× √𝐾𝐶×1 = 997 18.0 × √𝐾𝐶 mol Number of H3O+ ions = 997 18.0×√𝐾𝐶×𝐿 (where L is the Avogadro constant, 6.02 × 1023) Note : number of particles = mol × 6.02 × 1023 Ans: D

14) When G<0, the forward reaction is spontaneous, the position of equilibrium lies very much to the right, hence [H2] >> [H2O]. When G>0, the forward reaction is NOT spontaneous, the position of equilibrium lies very much to the left, hence [H2] << [H2O]. When G=0, using G=−RTlnK, Kc = 1, [H2] = [H2O] Ans: D 15) Solubility of PbCrO4 is 1.3 × 10−7 mol dm−3 (mol of solid dissolved in 1dm3 solution) PbCrO4 (s) + aq ⇌ Pb2+(aq) + CrO42− (aq) Initial / mol dm−3 − 0 0 Change / mol dm−3 −1.3 × 10−7 +1.3 × 10−7 +1.3 × 10−7 Eqm / mol dm−3 − 1.3 × 10−7 1.3 × 10−7 In a saturated solution, [Pb2+] = [CrO42−] = 1.3 × 10−7 moldm−3. Ksp = [Pb2+] [CrO42−] = (1.3×10−7)2 = 1.69×10−14 mol2dm−6. Ans: A 16) Option 1: CH3CH(NH2)CH3 and CH3C(Br)(CH3)CH3 Option 2: CH3CH2COOCH2CH3 and CH2(OH)CH(OH)CH2CH3 Option 3: CH3CH2CH2CN and CH3CH(CH3)CHO Note: Butanenitrile is 4 Carbon atom including the CN functional group. Ans: B 17) When 0.01 mol of acrylonitrile is electrolysed, amount of electron transferred = 0.01 mol At the anode, [O] 2H2O → O2 + 4H+ + 4e− Amount of O2 produced = ¼ × ne = ¼ × 0.01 = 0.0025 mol Vol of O2 produced at rtp = 0.0025 × 24 = 0.06 dm3 = 60 cm3 Ans: A

18) Molecular formula of -selinene is C15H24. It reacts with 2 mol of HBr to give C15H26Br2. There are 4 possible products, 3 of which are tertiary bromoalkanes. BrBr Br Br Br Br Br Br (Not tertiary bromoalkanes) Ans: D 19) Reaction A produces a racemic mixture as the trigonal planar carbonyl C atom is attacked by CN− nucleophile from top and bottom with equal probability. Reaction B only produces one product as the reaction is on the primary halogenolalkane that is likely to undergo SN2 mechanism. The reaction does not affect the existing chiral carbon, no new chiral carbon is formed during the reaction. The products for reaction C and D do not contain a chiral carbon. Ans: A 20) Statement 1 is correct. The electronegative F and Cl atoms withdraw electrons away from the O-H bond, weakening the bond and the O -H bond dissociates more readily to give H +. Hence chloroethanoic and fluoroethanoic acids are stronger acid than ethanoic acid. Statement 2 is wrong. The electronegative F and Cl atoms disperses the negative charge on −COO− and stabilize the carboxylate ion. Statement 3 is wrong. The electron donating methyl group increases the negative charge on −COO− and destabilize the carboxylate ion. Ans: C

21) Amide CH3CH2CONHCH2CH3 is the least basic as the lone pair electron on N is delocalised significantly to the C=O bond containing highly electronegative O atom. The lone pair electron on N atom of amide is unable to accept H+. Amide is neutral / least basic. C6H5NHCH2CH3 is less basic than (CH3CH2)2NH as the lone pair electron on N is delocalised slighly into the benzene ring. The lone pair electron on N atom is less available to accept H+ as compared to (CH3CH2)2NH. (CH3CH2)2NH is the most basic as there are two electron donating alkyl groups that increases the electron density of N atom, the lone pair electron on N atom is able to accept H+ most readily. Ans: A 22) The phenylamine and phenol group of HAA is able to react with ethanoyl chloride to give amide and ester respectively. Ans: A Note: Phenol can react directly with acyl chloride to form ester. If phenol is converted to phenoxide ion, phenoxide ion is a better nucleophile and the reaction can proceed with greater rate and yield. 23) Compound W and Y do not form a precipitate when heated with ethanolic silver nitrate. The lone