HCI Prelim C2 P3 MS

2 © Hwa Chong Institution 2017 9729 / 03 / C2 Prelim 2017 2017 C2 Prelims Paper 3 Mark Scheme 1 (a) No. of moles of NaOH = 0.100 x 26.50/1000 = 2.65 x 10-3 mol No. of moles of H2SO4 (in 25cm3) = ½ x 2.65 x 10-3 = 1.325 x 10-3 mol [1/2] No. of moles of H2SO4 (in 250cm3) = 1.35 x 10-3 x 250/25.0 [1/2] = 0.01325 mol No. of moles of H2SO4 reacted = (0.425 x 100/1000) – 0.01325 [1/2] = 0.02925 mol No. of moles of Cu3(CO3)2(OH)2 = 0.02925 x 1/3 [1/2] = 9.75 x 10-3 mol Mass of azurite = 9.75 x 10-3 x 344.5 [1/2] = 3.359 g % by mass of pure azurite in the powdered rock = 3.359/3.70 x 100% = 90.8% [1/2] (b) (i) x = 2 [1] [Cu(NH3)2]+ [1] (ii) [O] [Cu(NH3)2]+ + 2NH3 [Cu(NH3)4]2+ + e- [R] O2 + 2H2O + 4e- 4OH- Overall: 4[Cu(NH3)2]+ + O2 + 8NH3 + 2H2O  4[Cu(NH3)4]2+ + 4OH- 2 half-equations: [1/2 each] Overall equation: [1] (c) (i) I2(s) + 3Cl2(g)  2ICl3(s) +38 2(+60) 3(244) I2(g) 2ICl3(g) +151 6BE (I– Cl) 2I(g) + 6Cl(g) -162 = +38 + 151 + 3(244) – 6BE(I– Cl) – 120 BE(I– Cl) = +160.5 kJmol–1 energy cycle: [2] application of Hess’s Law: [1] correct stoichiometry coefficients: [1]

3 © Hwa Chong Institution 2017 9729 / 03 / C2 Prelim 2017 (ii) Gf = Hf - TSf -40.4 = -81 - 298Sf Sf = -0.136 kJ mol–1 K–1 = -136 J mol–1 K–1 [1] Sf is negative as there is a decrease in the number of gaseous molecules, resulting in fewer number of ways that the particles and the energy can be distributed. [1] (d) (i) Electrophilic substitution, reduction and condensation. [1 each] x 3 (ii) and CH3CO2H [1/2 each]

4 © Hwa Chong Institution 2017 9729 / 03 / C2 Prelim 2017 2 (a) (i) CnH2n+1SH + 3(n+1)/2 O2 → n CO2 + SO2 + (n+1)H2O [1] (ii) 2 2 2 2 2 SO SO CO SO H O      = 1/(n+1+n+1) = 1/(2n+2) shown [1] (iii) 2 2 SO SO total p p  1/(2n+2) = 16900 / 101325 n = 2 [1] (iv)   2 2 316900(1.65 10 ) 0.00876 mol8.31 110 273 SO SO pV x RT      = ethanethiol reacted [1] Mass of ethanethiol = 0.00876 x 62.1 = 0.544 g [1] (b) (i) As Reaction 1 is effectively complete whereas ethanol does not react with NaOH, this shows that ethanethiol is a stronger acid than ethanol. [1] (ii) Nucleophilic substitution [1] (iii) Reaction 1 was carried out to generate CH 3CH2S−, a stronger nucleophile than CH3CH2SH. [1] (iv) For bromobenzene, the lone pair of electrons on the bromine atom is delocalised into the ring. As a result, there is partial double bond character to the C-Br bond, so its bond strength is higher than a typical C-Br in a halogenoalkane and it is very difficult to break. [1] Sterically, the rear side of the C -Br bond in bromobenzene is blocked by the benzene ring. Or The pi-electron cloud of the benzene ring will repel the lone pair of electrons of the incoming nucleophile , rendering attack of the nucleophile difficult. [1] (c) (i) Base [1] (c) (ii) As the equilibrium position lies towards the products (K>1), ΔG is negative. [1] As the equilibrium constant for Reaction 1 is a very large number, as ΔG=-RTlnK, the magnitude of ΔG is very large. [1]

5 © Hwa Chong Institution 2017 9729 / 03 / C2 Prelim 2017 (iii) A: HNO3 [1] B: NO2 [1] (iv) The temperature in the furnace is 400 °C. [1] (accept any temperature between 170 °C and 630 °C) The ionic radii of Cu 2+, Pb2+ and Ba 2+ are 0.073nm, 0.120nm and 0.135nm respectively. [1/2] The charge density of the cation decreases from Cu2+ to Pb2+ to Ba2+. [1] The polarising power of the cation decreases from Cu 2+ to Pb 2+ to Ba 2+. OR the cation is less able to distort the electron cloud of the nitrate, weakening the N -O bonds within the nitrate anion to a smaller extent. [1] Hence more energy is required to decompose Pb(NO3)2 compared to Cu(NO3)2 but less compared to Ba(NO3)2. [1/2] (accept “ease of decomposition” / “decomposition temperature”) (d) (i) Reaction (A) is more likely to occur as it is easier to break a C -C bond (350 kJ mol−1) compared to a C-H bond (410 kJ mol−1) [1] (ii) Reactions (C), (D) and (E) [1] (iii) Hydrogen [1] (iv) The ●CH2CH3 radical loses a hydrogen (is oxidised) to form ethene and gains a hydrogen (is reduced) to form ethane. As it is both oxidised and reduced, this is a disproportionation. [1] OR The average oxidation number of carbon in the ●CH2CH3 radical is -2.5, but -2 in ethene and -3 in ethane. As carbon is both oxidised and reduced, this is a disproportionation. [1]

6 © Hwa Chong Institution 2017 9729 / 03 / C2 Prelim 2017 3 (a) (i) Concentration of a solid is constant. [1] (ii) (C18H29SO3Na) present in 1.00 g of detergent = 34800.1100 4.17       = 5.00  104 mol [C18H29SO3] = 5.00  104 mol dm3 [1] IP = (2.50  104)(5.00  104)2 = 6.25  1011 mol3 dm9 [1] Since IP > Ksp, a precipitate will form. [1] (iii) For the detergent to be effective, no precipitate is formed  IP < Ksp [Ca2+][C18H29SO3]2 < 1.20  10