DHS Prelim H2 CHEM P2 ans

© DHS 2016 9647/02 [Turn over Answer all questions in the spaces provided. 1 (a) Repulsion between two negatively charged ions causes the activation energy to be high. [1] (b) Preparation of reaction mixture 1. Using a 50.0 cm 3 measuring cylinder, add 25.0 cm 3 of peroxodisulfate ions into a 250 cm3 conical flask. 2. Using a 10.0 cm 3 measuring cylinder, add 2.0 cm 3 of aqueous copper ( II) ions into the same conical flask. 3. Using a 50.0 cm3 measuring cylinder, add 45.5 cm 3 of deionised water into the conical flask. 4. Using a 10.0 cm 3 measuring cylinder, measure out 2.5 cm 3 of ethanedioate ions. 5. Transfer the ethanedioate ions into the conical flask and insert a rubber bung into the conical f lask. The rubber bung has a delivery tube connected to rubber tubing with the other end of the tubing inserted into an inverted burette filled with water. 6. Start the stopwatch immediately and gently swirl the conical flask continuously. 7. Monitor the water level in the burette and stop the stopwatch when 40.00 cm3 of CO2 is produced. Record the time taken. 8. Repeat step 1 to 7 using the following volumes of reactants, copper ( II) ions and deionised water shown in experiment 2 to 4, at the same temperature. Experiment Volume of S2O82 / cm3 Volume of C2O42 / cm3 Volume of Cu2+ / cm3 Volume of water / cm3 1 25.0 2.5 2.0 45.5 2 25.0 5.0 2.0 43.0 3 2.5 25.0 2.0 45.5 4 5.0 25.0 2.0 43.0 Treatment of results 1. Relative rates (= 1/t) of expt 1 to 4 are calculated. Since total volume of mixture is constant in expt 1 to 4 , concentration of reactant is proportional to its volume used. 2. Compare the relative rates of expt 1 to 2 and 3 to 4, to find the order of reaction with respect to (w.r.t) C2O42 and S 2O82 respectively. If volume of C2O42 doubles and rate remains the same, it is zero order w.r.t C 2O42. If volume of C 2O42 doubles, and rate remains doubles, it is first order w.r.t C2O42. If volume of C 2O42 doubles and rate quadruples, it is second order w.r.t C2O42. The same applies for S2O82. 3. The rate law of reaction can then be found, rate = k[S2O82]n[ C2O42]m, where n and m are the orders of reaction w.r.t to S2O82 and C2O42 respectively. [8]

2 © DHS 2016 9647/02 [Turn over (c) 1. Repeat one of the experiments (e.g. expt 2) at two (at least) different temperatures, other than that for experiment 1 to 4 above. This can be done by immersing conical flasks in water bath maintained at different constant temperatures e.g. 40 °C and 60 °C. 2. Using the results in experiment 2 and that of the two further experiments, relative rate, followed by ln (relative rate) for each expt is calculated. 3. A graph of ln (relative rate) against T 1 is plotted. The gradient of the best fit line is then determined, where gradient = – R Ea . E a is given by gradient x R. [3] [Total: 12] 2 (a) (i) Half–equations Polarity Anode Pb(s) + SO42–(aq)  PbSO4(s) + 2e– – Cathode PbO2(s) + 4H+(aq) + SO42–(aq) + 2e–  PbSO4(s) + 2H2O(l) + [3] (ii) The relative density of the sulfuric acid will decrease. During discharging, as sulfuric acid is used up to produce PbSO 4, the concentration of sulfuric acid will decrease, resulting in a decrease in density. [2] (iii) Overcharging will result in electrolysis of water. Identity of other gas: Oxygen [2] (iv) E = 1.47 – (–0.13) = +1.60 V [2] (v) Lead–acid battery has a higher voltage as PbSO 4 formed is insoluble, which resulted in a lower concentration of Pb 2+ in the electrolyte, thus driving the forward reaction. [2]

3 © DHS 2016 9647/02 [Turn over (b) (i) To 1 cm 3 of halide ions, add AgNO 3(aq) dropwise. Then add aqueous NH 3 in excess. Cl– ions form white ppt with AgNO3 which is soluble in excess NH3(aq). Br– ions form cream ppt with AgNO3 which is insoluble in excess NH3(aq) [2] (ii) Ag+ (aq) + X– (aq) AgX (s), where X– is Cl– and Br– Both halide ions form insoluble ppt with Ag+. Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq) Ag+ ions form soluble diammine complex which decreases the [Ag +], resulting in a corresponding decrease in ionic product (i.e. I.P = [Ag +][X–]) for both halide ions. As Ksp value of AgCl is much larger than that of AgBr, the ionic product of AgCl will fall below its Ksp but not for AgBr. [2] [Total: 15] 3 (a) (i) Co3+: 1s22s22p63s23p63d6 [1] (ii) Suggest the identities of X and Y. X: Co(OH)3 Y: [Co(NH3)6]3+ [2] (iii) [Co(H2O)6]3+ + 3OH– Co(OH)3 + 6H2O ––––(1) When aqueous ammonia is added in excess, a ligand exchange reaction occurs. The stronger NH 3 ligands replace weaker H 2O ligands in the [Co(H2O)6]3+ ions to form [Co(NH3)6]3+ complex. [Co(H2O)6]3+(aq) + 6NH3(aq) [Co(NH3)6]3+(aq) + 6H2O(l) ––––(2) This decreases the concentration of [Co(H 2O)6]3+ in solution. By Le Chatelier’s Principle, equilibrium (1) shifts left to increase the concentration of [Co(H2O)6]3+(aq). Hence the brown precipitate of Co(OH)3 dissolves [1]. [3]

4 © DHS 2016 9647/02 [Turn over (b) [Co(H2O)6]2+ + 4Cl– [CoCl4]2– + 6H2O There is a change of co–ordination number from 6 to 4. [2] (c) (i) The partially –filled d orbitals of V 2+ are split into two groups of different energy levels by H2O ligands. When white light shines on the complex, a d electron undergoes d –d transition and is promoted to a higher energy d orbital. During the transition, the d electron absorbs light from the yellow region of the visible spectrum. The colour observed is the colour of transmitted light, which is a mixture of remaining wavelengths that have not been absorbed. [2] (ii) The reactant molecules are physically adsorbed onto the catalyst surface. This allows for formation of weak bonds between reactants and the surface catalyst, thus weakening the intramolecular bonds in the reactants and helps to catalyse the reaction. After reaction, the reactant molecules desorb from the catalyst surface. [2] (iii) 2SO2(g) + O2(g) 2SO3(g) Initial pressure /atm 0.6667x 0.3333x 0 Change in pressure /atm –0.6333x –0.3167x +0.6333x Equilibrium pressure /atm 0.0334x 0.0166x 0.6333x Let the initial total pressure be x atm. Kp = (0.6333x)2 / (0.0166x)(0.0334x)2 = 7200  x = (0.6333)2 / 7200(0.0166)(0.0334)2 = 3.01 atm [2] (iv) Rate of conversion will slow down/decrease due to poisoning of the catalyst. [1] [Total: 15] 4 (a) (i) No. of moles of HY3– required = 0.0149 x 25.55 1000 = 0.00038070 mol

5 © DHS 2016 9647/02 [Turn over No. of moles of Ca2+ present = 0.00038070 mol No. of moles of CaCO3 = 0.00038070 mol Mass of CaCO3 = 0.00038070 x [40.1 + 12 + 3(16)] = 0.0381g (3 sf) [1] (ii) ppm CaCO3 = 0.038107 50 x 106 = 762 (3 sf) The water is very hard. [2] (b) Ca2+ can be removed by heating the water sample. By Le Chatelier’s Principle, the position of equilibrium will shift right to remove the heat supplied causing a decrease in the concentration of Ca2+ ions. OR Ca2+ can be removed by heating the water sample. Heating the water sample removes CO 2 (g) hence decreasing the concentration of CO2 (aq). By Le Chatelier’s Principle, the position of equilibrium will shift right to increase the concentration of CO 2 (aq) causing a decrease in the concentration of Ca 2+ ions. [2] (c) (i) [C18H29SO3–] = √(1.2 x 10–17 / 2.5 x 10–4) = 2.1908 x 10–7 mol dm–3 =