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