VJC Prelim P1

Detailed Ans wers for H1 Chemistry Prelim Paper 2009 1 C 6 B 11 C 16 A 21 D 26 B 2 A 7 A 12 B 17 D 22 B 27 D 3 A 8 B 13 D 18 B 23 C 28 C 4 B 9 D 14 B 19 D 24 A 29 A 5 C 10 C 15 C 20 D 25 C 30 A Page 1 of 3 1 A Correct: 1 mole of methane contains 4 moles of H atoms = 4 x 6.02 x 1023 = 2.41 x 1024 H atoms. B Correct: Ar of 12C is 12.0. C Wrong: One mole of hydrogen gas contains 2 moles of hydrogen atoms = 2 x 6.02 x 1023 atoms of hydrogen. D Correct: One mole of methane (16.0g) contains one mole of carbon (12.0g) 12.0 100%16.0 = 75% Answer: C 2 This is a disproportionation reaction where Cu 2O is reduced to Cu and is oxidised to CuSO 4 simultaneously. Hence, it can act as both an oxidising and reducing agent. H 2SO4 is not reduced or oxidised in this reaction. Answer: A 3 Let the oxidation state of Fe in [Fe(CN) 6]4- be x. x + 6(-1) = - 4 x = +2 Electronic configuration of Fe is [Ar]3d 64s2. Electronic configuration of Fe2+ is [Ar]3d6. Answer: A 4 The state of a compound depends on its bonding. CO2 gas is made up of simple molecule with weak dispersion forces (a type of van der Waals’ forces) of attraction between the molecules. SiO 2 has strong covalent bonds throughout the giant molecular structure. Therefore, more energy is required to overcome the strong covalent bonds in SiO 2 compared to the weak dispersion forces in CO 2 gas. The boiling point of SiO 2 is much higher than CO 2, therefore it is a solid at r.t. Answer: B 5 A CO2 is linear (0 lp, 2 bp); SO2 is bent (1 lp, 2 bp) B NH3 is trigonal pyramidal (1 lp, 3 bp); NO 3 - is trigonal planar (0 lp, 3 bp) C CO2 is linear (0 lp, 2 bp); I3 - is linear (3 lp, 2 bp).The linear shape of I3 - is due to 3 lp occupying the equatorial positions in a trigonal pyramid. D BCl 3 is trigonal planar (0 lp, 3 bp); PCl3 is trigonal pyramidal (1 lp, 3 bp) Answer: C 6 The standard enthalpy change of formation is when one mole of a pure compound is formed from its constituent elements in their standard states , under standard conditions of 298K and 1 atm. Answer: B 7 H c (C3H4) = Hf (product) - Hf (reactant) – 1938 = (-394 X 3) + (-286 X 2) - Hf (C3H4) Hf (C3H4) = + 184 kJ mol-1 Answer: A 8 For C2H6(g) ⇌ C2H4(g) + H2(g), ∆Hrxn = Total bond energy for reactants – Total bond energy for products = [BE(C-C) + 6BE(C-H)] – [BE(H-H) + BE(C=C) + 4BE(C-H)] = [350 + 6(410)] – [436 +610 + 4(410)] = +124 kJ mol-1 > 0 When temp increases, position of eqm shifts left to counteract the increase in temp, hence, eqm yield of ethene decreases. Rate of formation of ethene will decrease at lower temperature due to smaller frequency of collision between reactants molecules to form the products. Answer: B 9 A 2 c 2 [H ][CO]K Unit: mol dm [H O] -3 B  2 3 4 c 24 3 [Cu NH ]K  [Cu ][NH ] Unit: (mol dm-3)-4 C 2 2 c 24 [NO ]K Unit: mol dm [N O ] -3 D 32 3 2 c 33 [CH CO CH ][H O]K [CH OH][CH CO H] 2 Unit: - Answer: D 10 A Ammonia acts as a proton acceptor (a base). B Ammonia acts as a ligand to form the complex ion. C Ammonia acts as a proton donor (an acid). Acid- metal reaction gives salt and hydrogen gas. D Ammonia acts as a nucleophile or a Lewis base (electron pair donor). Answer: C 11 pH = 2.0 means that [H +] = 10-2 mol dm-3, and no. of moles of H+ = 2.0 x 10-2 Upon dilution, no. of moles of H+ remains unchanged at 2.0 x 10-2. But, new pH = 3.0 means that new [H +] = 10-3 mol dm-3 Let v = vol. of water added. Hence, new [H +] = 10-3 mol dm-3 = 2.0 x 10-2 / (2.0 + v) v = vol. of water added= 18 dm3 Answer: C 12 HCO3 2- being the acid, removes the small amt of base added. HC O3 - + OH-  CO3 2- + H2O CO3 2- being the conjugate base of HCO 3 -, removes the small amt of acid added. CO3 2- + H3O+  HCO3 - + H2O

13 Rate t 1  Rate = 1 t x When the [S2O8 2-] is halved, the initial rate is halved. Hence, the reaction is first order with respect to S2O8 2-. When the [I-] is decreases by 3 times, the initial reaction is decreased by 3 times. Hence, the reaction is first order with respect to I -. Rate = k[S2O8 2-][I-] Answer: D 14 For [Z] against time graph, a straight line means that it is zero order wrt Z since the rate of rxn remains unchanged with changing [Z]. For [X] against time graph, for [Y] = 2 mol dm -3, a constant t1/2 of 0.7 min means it is first order wrt X. Let Rate= k[X][Y] a For [Z] against time graph, when [Y] is 2x, (1/2) a = (0.02/0.08) a = 2 Hence, Rate= k[X][Y] 2, for [X] is halved and [Y] is doubled, new rate is increased by (1/2)(2)2 = 2 times. Answer: B 15 Page 2 of 3 According to Maxwell-Boltzmann distribution curve, the number of molecules with kinetic energy greater than the activation energy is greater at a higher temperature. With increased kinetic energy, the frequency of effective collisions between molecules with kinetic energy greater than the activation energy is greater at a higher temperature. Answer: C 16 X is Al since Al 2O3 has a giant ionic structure while AlCl3 has a simple molecular structure. A The melting pt of MgO is the highest (higher than that of A l2O3) among all the oxides of Period 3 elements. B X2+(g)  X 3+(g) + e, 3rd IE and Na +(g)  Na2+(g) + e, 2nd IE Since [X 2+] = [Na +] = 1s 22s22p6, 3rd IE of X = 2nd IE of Na. C A lCl3 undergoes hydrolysis in water to give rise to an acidic solution: [Al (H2O)6]3+ + H2O ⇌ [Al(H2O)6(OH)]2+ + H3O+ H 2 gas is evolved from acid-base reaction: 2H + + Mg  Mg2+ + H2 D Al2O3 + 2KOH + 3H2O  2KAl(OH)4 Answer: A 17 Outermost shell electronic configuration of M (P) is 3s 23p3 . Outermost shell electronic configuration of N (S) is 3s 23p4 . Hence, less energy is required to remove an electron from paired 3p electrons in S due to interelectronic repulsion. 1st IE of M is > 1st IE of N. For ionic radius of L (Al3+), it only has two quantum shell comparing to the three quantum shells which M (P) has, hence, L has a smaller ionic radius than M. 18 A ketone is reduced to 2o alcohol. Hydride behaves as RA. B H- is not able to eliminate hydrogen from ethane. C aldehyde is reduced to 1 o alcohol. Hydride behaves as RA. D ethanoic acid is neutralized to form ethanoate ion. Hydride behaves as a base. Answer: B 19 A 2 mono-sub products formed  1-chloro-2-methylpropane and 2-chloro- 2methylpropane. B 4 mono-sub products formed  1-chloro-3-methylpentane, 2-chloro-3- methylpentane, 3-chloro-3-methylpentane and 3- (chloromethyl)pentane C 2 mono-sub products formed  1-chloro-2,3-dimethylbutane and 2-chloro-2,3- dimethylbutane D Symmetrical 2,2-dimethylpropane yields only one mono-sub product  1-chloro-2,2- dimethylpropane. Answer: D number of molec 20 A Methylbenzene can be oxidized to benzoic acid by hot KMnO 4/H+ B methylbenzene + conc H2SO4 + Conc HNO3 will yield 2-nitomethylbenzene since -CH 3 is ortho,para directing, C It undergoes side-chain free radical substitution with Br2 to form (bromomethyl)benzene. D - CH3 is ortho, para directing, thus 3- bromomethylbenzene is cannot be made directly. Answer: D 21 A Absence of CH 3CO and CH 3CH(OH) structures no yellow ppt with alkaline I2. B Absence of aldehyde group, no silver mirror formed. C HCN(aq) with trace base react with ketone to form nitrile compound. D Answer: D 22 Since the two substituents bonded to hydroxyl- containing carbon comes from Grignard reagents, the substituents must be –CH 3 and -CH2CH3 only. 4-ethylheptan-4-ol has substituent of – CH2CH2CH3. Thus not possible to form. Answer: B 23 Which process will not give a good yield of CH3CO(CH2)3CO2H? Ea T2 ules T2 > T1 KE

A Page 3 of 3 B C CH3CH(OH)(CH2)3CH2Cl Hot KMnO4 (aq), H3O+ NaOH (alcohol) Heat CH3CH(OH)(CH2)2CH=CH2 +C O 2 CH3CO(CH2)2CO2H D Answer: