DHS Prelim P3 Ans

Dunman High School 2017 Year 6 H2 Chemistry Preliminary Examination Paper 3 (Answer Scheme) Section A 1 (a) (i) 1. The gas particles have negligible volume compared to the volume of the container. 2. There are no intermolecular forces of attraction between gas particles. Its electron cloud size is larger hence volume of NO 2 molecules is significant compared to the volume of the gas, unlike H2. Both NO 2 and H 2 have simple molecular structure. Since NO 2 has stronger permanent dipole -permanent dipole interaction than instantane ous dipole - induced dipole interactions in H2 OR NO2 has a larger electron cloud size than H2, the electron cloud of NO 2 is more polarisable and hence stronger intermolecular forces of attraction. (ii) 𝑝V = nRT = (m/M)RT 𝑝 = (m/V)(RT/M) 𝑝 ρ = RT/M At 100 °C and very low pressure, RT/M = 110.3 M = (8.31 x 373) / 110.3 = 28.1 g mol–1 Mr = 28.1 (iii) CO (b) NO2 + SO2 → SO3 + NO NO + ½ O2 → NO2 NO2 catalyses by oxidising SO 2 to SO3, while itself is reduced to NO. NO is rapidly re-oxidised to NO2 by oxygen, regenerating the catalyst, NO2. SO3 + H2O → H2SO4 OR SO3 dissolves in the water vapour in the atmosphere/ rain to form sulfuric acid, which causes acid rain. (c) (i) Order w.r.t [O2] is zero as the graph of [O2] against time is a downward sloping straight line / the gradient of the line, i.e. rate of reaction, is constant with changing [O2]. When [SO2] = 0.8 mol dm-3, r1, rate of reaction = 0.04 0.05 88  = 1.13 x 10-4 mol dm–3 s–1 When [SO2] = 1.2 mol dm-3, r2, rate of reaction = 0.030 0.05 80  = 2.50 x 10-4 mol dm–3 s–1 r2/r1 = 2.21 ≈ 2.25 When [SO2] x 1.5 times, rate of reaction x 2.25 times, reaction is second order w.r.t [SO2].

2 © DHS 2017 9729/03 [Turn Over (ii) rate = k[SO2]2 When [SO2] = 0.8 mol dm-3, k = 1.77 x 10-4 mol-1 dm3 s-1 (k = 1.74 x 10-4 mol-1 dm3 s-1, if [SO2] = 1.2 mol dm-3) (iii) [SO2] / mol dm-3 (iv) (iii) 2 (a) For hexane / cyclohexane mixture,  Energy produced from instantaneous dipole – induced dipole (id -id) interactions between hexane and cyclohexane molecules after mixing  is just sufficient to overcome id -id interactions between hexane molecules and between cyclohexane molecules before mixing.  Thus no heat change is observed. For hexane / ethanol mixture,  Energy produced from weak id -id interactions between hexane and ethanol molecules after mixing  is insufficient to overcome stronger hydrogen bonds between ethanol molecules before mixing.  Thus heat is absorbed from the surroundings for mixing to occur. (b) (i) bonds broken bonds formed H / kJ mol–1 step 1 1 C=C 1 C–C 1 C–I (–350 – 240) + 610 = +20 step 2 1 H–Cl 1 C–H –410 + 431 = +21 (ii)  Only the free radical addition of HBr to an alkene is likely to occur since Hstep 1, Hstep 2 are all exothermic.  This reaction is however unlikely to occur if HCl and HI were used.  In each case, there is one propagation step which is endothermic — step 2 for HCl and step 1 for HI. (iii) CH3CHICH3 2 (c) (i) Q : AgI R : AgCl x ½x 1/4x 0 time t1/2, 1 t1/2, 2

3 © DHS 2017 9729/03 [Turn Over  Reaction I is the nucleophilic substitution of a halogenoalkane in which C–X bonds (where X = Cl, I) are broken to release X– ions into solution to be precipitated as AgX.  Rate of substitution depends on the strength of C–X bond. From the Data Booklet, BE(C–Cl) : 340 kJ mol–1 BE(C–I) : 240 kJ mol–1  C–I bond is weaker OR more easily broken than C –Cl bond OR less amount of energy is needed to break C–I than C–Cl bond.  Thus, time taken for I– ions to be released is shorter than that for C l– ions OR C–I bond is broken first OR AgI is precipitated first. (ii)  LiAlH4 is a nucleophilic reducing agent OR is electron-rich.  It attacks the electrophilic OR electron-deficient carbonyl C atom in carbonyl compounds, carboxylic acids and their derivatives but does not reduce electron-rich C=C bonds in alkenes. (iii) Step III: I2(aq) with NaOH(aq), heat 3 (a) (i) CH3 R O H CH3 R O + P Br BrH + CH3 R O + P Br BrH Br- + CH3 R Br + P Br Br OH Br-P Br Br Br fast slow 3 (a) (ii) Lewis acid. PBr3 accepts a lone pair of electrons from oxygen in the first step of the mechanism. (iii) For phenol, the lone pair of electron on the oxygen atom is delocalised into the benzene ring and hence less available for donation and hence, phenol will be a weaker nucleophile, resulting in a relatively slower first step. (iv) Phenol undergoes electrophilic substitution instead of addition to prevent the loss of its aromaticity (and resonance stability). (v) Chemical test 1: Add KMnO4 and dilute H2SO4, heat (in water bath) Observation for phenol: Purple KMnO4 does not decolourise Observation for 2° alcohol used in (a): Purple KMnO4 decolourises OR Chemical test 2: Add I2 and NaOH, warm/heat (in water bath) Observation for phenol: No yellow ppt forms Observation for 2° alcohol used in (a): Yellow precipitate forms OR Chemical test 3: Add neutral FeCl3(aq) Observation for phenol: Violet colouration observed δ+ δ-

4 © DHS 2017 9729/03 [Turn Over Observation for 2° alcohol used in (a): No violet colouration observed OR Chemical test 4: Add K2Cr2O7 and dilute H2SO4, heat (in a water bath) Observation for phenol: Solution remains orange Observation for 2° alcohol used in (a): Orange solutions turns green 3 (b) (i) PCl3(g) + Cl2(g) PCl5(g) Initial partial pressure / atm 2 1.5 0 Change in partial pressure / atm –x –x +x Eqm Partial pressure/ atm 2 – x 1.5 – x x Hence, (2 – x) + (1.5 – x) + x = 3.3 Partial pressure of PCl5 at eqm, x = 0.2 atm Kp = (0.2) (2−0.2)(1.5−0.2) = 0.0855 atm–1 (ii) Since Kp is much less than 1, the position of the equilibrium lies to the left (or to the reactant side). This means that the forward reaction is not likely to be spontaneous and hence, ∆G should be positive. (c) (i) AsCl5 + 4H2O → H3AsO4 + 5HCl Or AsCl5 + H2O → AsOCl3 + 2HCl AsCl5 completely hydrolyses in water to produce a strong acid, HCl, which is responsible for the very low pH. (ii) OR Bond angle: 180° 4 (a) (i) Constitutional isomers are compounds with the same molecular formula but differs in structural formula. (ii) CH3 OH CH3 OH trans cis

5 © DHS 2017 9729/03 [Turn Over (b) (i) If Cu2+ is not complexed, it will form Cu(OH)2 solid in alkaline medium and hence will not be able to react with the (aldehyde) functional group. (ii) Cu2+ is a transition metal ion while K+ is not. Cu2+ has low lying partially filled orbitals (or vacant subshell of low energy) which allows it to accept lone pair of electrons from tartrate ions to form a dative bond in complex formation. K+, however, does not low lying orbitals to form dative bonds with tartrate ligands. (ii