RVHS H2 CHEM P2 ANS RV Prelim

Answer key to Paper 2 1 (a) RvCO3 + 2HCl → RvCl2 + CO2 + H2O [1] (b) [2] (c) Preparation of saturated RvCO3 solution 1. Using a 100 cm 3 measuring cylinder , place about 100 cm 3 of deionised water (accept any volume between 80 – 250 cm 3) into a polystyrene cup. 2. Measure and record the initial temperature of the water using a thermometer. 3. Using a spatula, add solid RvCO 3 into the polystyrene cup and stir/swirl gently with the thermometer / stirrer. Continue to add more solid and stir, until some solid RvCO3 remains undissolved to ensure a saturated solution is obtained. 4. Record the lowest temperature reached on the thermometer. 5. Filter the saturated solution using a dry filter funnel and dry filter paper into a dry 250 cm 3 conical flask / beaker to ensure the saturated solution is not diluted. Justification of concentration of standard HCl solution For a 25.0 cm3 aliquot of saturated RvCO 3(aq), a reasonable HC l titre value should be 25.00cm3 (accept any value between 20−30 cm3). RvCO3 + 2HCl → RvCl2 + CO2 + H2O RvCO3 ≡ 2HCl Given that solubility of RvCO3 is approximately 0.25 mol dm−3, (250 cm3) beaker polystyrene cup thermometer lid (100 cm3) H2O + RvCO3

2 [HCl] required = = 0.500 mol dm−3 Titration of saturated RvCO3 solution against HCl 1. Pipette 25.0 cm 3 of saturated RvCO 3 solution into a 250 cm 3 conical flask. 2. Add 2-3 drops of methyl orange indicator. 3. Titrate with the standard 0.500 mol dm −3 HCl from the burette until the colour of the solution changes from yellow to orange. 4. Repeat the titration to obtain two consistent results within 0.10 cm 3 in difference. [6] (d) (i) nHCl used = = 1.00 x 10−3 WC mol RvCO3 ≡ 2HCl nRvCO3 in 25.0 cm3 of saturated solution = 1.00 x 10−3 WC = 5.00 x 10−4 WC mol solubility of RvCO3 = 5.00 x 10−4 WC = 0.0200 WC mol dm−3 (ii) Heat taken in by V cm3 solution = VCΔT = 4.18 V ΔT J nRvCO3 dissolved in V cm3 solution = 0.0200 WC = 2.00 x 10−5 WCV mol ΔHsoln = J mol−1 [3] [Total: 12]

3 2 (a) (i) initial nBr2 = = 6.70 x 10−4 mol At 900 K, Br2(g) → 2Br(g) initial nBr = 2 x 6.70 x 10−4 = 1.34 x 10−3 mol (ii) W(s) + 4Br(g) ⇌ WBr4(g) initial amount / mol − 1.34 x 10−3 0 change in amount / mol − −4y +y equilibrium amount / mol − 1.34 x 10−3 −4y y mass of gas at equilibrium = (1.34 x 10−3 −4y)(79.9) + (y)(184 + 4(79.9)) = 0.107 + 184y 0.107 + 184y = 0.003 x 50 g y = 2.34 x 10−4 mol equilibrium nWBr4 = 2.34 x 10−4 mol equilibrium nBr = 1.34 x 10−3 −4(2.34 x 10−4) = 4.04 x 10−4 mol (iii) equilibrium pWBr4 = = 3.50 x 104 Pa equilibrium pBr = = 6.04 x 104 Pa Kp at 900 K = = 2.63 x 10−15 Pa−3 (iv) When the temperature increases from 900 K to 2800 K, the Kp decreases. This implies that the equilibrium position shifts to the left to favour the backward reaction near the tungsten filament. (v) The addition of bromine gas allows the formation of WBr 4(g), which is converted back to tungsten at the filament , slowing down the thinning of the tungsten filament. [9] (b) (i) PCl5 + 4H2O → H3PO4 + 5HCl pH of resultant solution = 2

4 (ii) PCl5 is able to undergo hydrolysis in water because P atom is able to use its energetically accessible vacant 3d orbitals for dative bonding with water molecules. NCl3 does not undergo hydrolysis because N atom does not have energetically accessible vacant or bitals/ N atom is in Period 2 and not Period 3. (iii) 6PCl5(g) + P4O10(s) 10POCl3(g) Using Hess’s law, ΔHr = 6(+84.2) +2967 + (−1226) + 10(−286) = −614 kJ mol−1 (3sf) [7] [Total: 16] 10PCl3(g) + 5O2(g) 6(+84.2) 6PCl3(g) + 6Cl2(g) + P4O10(s) +2967 6PCl3(g) + 6Cl2(g) + P4(s) + 5O2(g) −1226 10(−286) ΔHr

5 3 (a) (i) Both copper and calcium have giant metallic structure . In Cu, both the 4s and 3d electrons can be contributed to the sea of delocalised valence electrons as they are very close in energy . The resulting copper ions have a smaller ionic radius than Ca2+. A larger amount of energy is required to overcome the stronger electrostatic attraction between the copper ions and the delocalised/mobile valence electrons compared to that between Ca 2+ and delocalised/mobile valence. (ii) Cu conducts electricity in the solid state due to the presence of mobile (delocalised) valence electrons that can migrate freely through the giant metallic structure when a potential difference is applied. For CuO in solid state, the Cu2+ and O 2– ions are held in fixed positions in a giant ionic lattice structure and are not mobile. [4] (b) (i) Ksp = [Cu2+][OH–]2 = (1.77  10–7)(2  1.77  10–7)2 = 2.22  10–20 mol3 dm–9 (ii) Common ion effect refers to the reduced solubility of a sparingly soluble salt (when compared to its solubility in water) in an aqueous solution containing an ion common to that salt. Let x be the solubility of Cu(OH)2 in 0.050 mol dm–3 of NaOH Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH–(aq) Initial [ ] / mol dm–3 0 0.050 Change in [ ] / mol dm–3 + x +2x Eqm [ ] / mol dm–3 x 2x + 0.050 x (2x + 0.050)2 = 2.22  10–20 Assume x is small such that 2x + 0.050 ≈ 0.05 x = 8.88  10–18 mol dm–3 << 1.77  10–7 (iii) (Pale) Blue precipitate dissolves to give a dark blue solution. Cu(OH)2 + 4NH3 + 2H2O  [Cu(NH3)4(H2O)2]2+ + 2OH– [6]

6 (c) (i) Ppt X Identity : CuI Colour : off-white/cream Solution Y Identity : I2 Colour : (reddish) brown Solid Z Identity : Cu (ii) Reducing agent [4] (d) 2Cu(s) + CO2(g) + H2O(g) + O2(g)  Cu(OH)2(s) + CuCO3(s) [1] [Total: 15]

7 4 (a) (i) Anode: 2Cl(l) → Cl2(g) + 2e Cathode: Mg2+(l) + 2e → Mg(l) (ii) Amount of Mg per day = (20  1000  1000)  24.3 = 8.23  105 mol Mg  2e Amount of electrons per day = 1.65  106 mol I = nF/t = (1.65  106  96500)  (24  60  60) = 1.84  106 A [4] (b) (i) E⦵(H2O/H2) = 0.83 V E⦵(Mg2+/Mg) = 2.38 V Electrolyis of seawater would produce hydrogen gas as water is preferentially reduced since the E⦵(H2O/H2) is more positive than E⦵(Mg2+/Mg) . (ii) The melting point of magnesium oxide is much higher than magnesium chloride. Melting magnesium oxide would require significantly more energy and thus incur higher cost. (iii) Magnesium – White Strontium/calcium – crimson/(brick) red [6]

8 (c) (i) Experiment 1: Oxygen Experiment 2: Chlorine E⦵(O2/H2O) = +1.23 V E⦵(Cl2/Cl) = +1.36 V In experiment 1 water is preferentially oxidised to oxygen gas as E⦵(O2/H2O) is more negative than E⦵(Cl2/Cl). In experiment 2 chloride anion is pre ferentially oxidised to chlorine gas as the increased concentration of choride ions shifts the position of equilibrium C l2 +2e ⇌ 2Cl to the left, making E(Cl2/Cl) more negative than E(O2/H2O). (ii) 3Cl2(g) + 6OH–(aq) → 5Cl −(aq) + ClO3 −(aq) + 3H2O(l) [5] [Total: 15]

9 5 (a) (i) (ii) + AlCl3 + (iii) [4] (b) The carbon is sp3 hybridised. The 60° bond angle makes the three - membered ring highly strained hence epoxides react in ring