DHS Prelim P2 Answer scheme

1  DHS 2015 9647/02 2015 DHS Year 6 H2 Chemistry Preliminary Examination (Answer Scheme) 1 Planning (P) 1 (a) (i) 1. Connect the nozzle of hydrogen gas can to the effusion hole of a 100 cm 3 well- greased syringe using rubber tubing. 2. Flush the syringe with hydrogen at least once before filling the syringe with hydrogen gas at constant temperature of 25 C. Piston should be above the start volume. 3. Detach the tubing from the effusion hole and allow hydrogen gas to effuse out. (Free movement, not pushed to start mark.) 4. Start the stopwatch when the piston passes the 80 cm 3 mark on the syringe and stop the stopwatch when the piston falls to 30 cm3 mark on the syringe. Start the stopwatch when the piston passes the 80 cm 3 mark on the syringe and stop the stopwatch after 10 minutes. 5. Repeat step 1 – 4 with another sample of hydrogen gas and obtain the average effusion time OR volume at constant temperature of 25 C. 6. Repeat step 1 – 4 with the four other gases (butane, oxygen, carbon dioxide and chlorine gas) at constant temperature of 25 C. Gas Mr Effusion time for expt 1 / s Effusion time for expt 2 / s (Average) effusion time / s Rate of effusion / cm3 s–1 Hydrogen 2.0 Butane 58.0 Oxygen 32.0 Carbon dioxide 44.0 Chlorine 71.0 OR Gas Mr Effusion volume in 10 min for expt 1/ cm3 Effusion volume in 10 min for expt 2 / cm3 Average effusion volume in 10 min/ cm3 Rate of effusion / cm3 min–1 Hydrogen 2.0 Butane 58.0 Oxygen 32.0 Carbon dioxide 44.0

2  DHS 2015 9647/02 (ii) Rate of effusion/ cm s–1 Mr 28.0 Using the results, plot a graph of rate of effusion against Mr of the gas, read off the rate of effusion of nitrogen gas, x, at its Mr of 28.0. (b) (i) Moles of Cu(NH3)4SO4 to be formed  Moles of CuSO4 required = 0.02196 mol Moles of NH 3 required  4  Moles of CuSO4 required = 0.08787 mol  Volume of CuSO4 required = 0.02196 / 0.50 = 43.9 cm3  Volume of NH3 required = 0.08787/ 2.0 = 43.9 cm3 (ii) Procedure 1. Using a 50 cm 3 measuring cylinder, transfer 50 cm3 of CuSO4 (aq) into a 150 cm3 beaker. 2. Using another 50 cm 3 measuring cylinder, gradually add 50 cm 3 of NH3 (aq) into the 50.00 cm3 of CuSO4 (aq), with stirring until all the solids have dissolved. 3. Leave the mixture to stand for 15 minutes undisturbed, so that an equilibrium can be established. 4. Place the beaker in an ice bath and slowly add (drop wise), using a dropper, about 30 cm3 of acetone to the beaker with continuous stirring. 5. Leave the solution to remain in the ice bath for about 15 minutes. 6. Filter off the crude product, using (vacuum) filtration (or Buchner apparatus), and wash it with a little cold acetone. 7. Dissolve the crude product in minimum volume of hot acetone in a small conical flask. When all crude product has dissolved completely, allow the solution to cool. 8. When recrystallisation is completed, filter off the pure product using (vacuum) filtration. Dry the crystals between filter papers then transfer them to a sample tube and leave them to dry in air. x

3  DHS 2015 9647/02 2 (a) (i) Mr of UDMH = 60; Mr of N2O4 = 92 Mole ratio of UDMH : N 2O4 is 1:2 Since total mass =244 kg = 1(60) + 2(92) Mass of UDMH used = 60 kg Moles of UDMH used = 60 000/60 = 1000 mol x = 1000 Moles of gaseous products formed = 9 x 1000 = 9000 mol pV = nRT, V = nRT/p = 9000 (8.31) (263)/600 = 3.28 x 10 4 m3 (ii) It is the energy change when 1 mole of (liquid) (CH 3)2N2H2 is formed from its constituent elements, carbon, nitrogen and hydrogen, under standard conditions, i.e.. 298 K and 1 atm pressure. (iii) G r = Hr – TSr H = G + TS = (–2107) + 298(844/1000) = –1855 kJ mol–1 Hr = Hf(products) – Hf(reactants) –1855 = 2(–394) + 4(–242) – [x + 2(9.0)] x = +81.0 kJ mol–1 (b) (i) Gradient of graph = –14194 (units: K) – R aE = –14194 Ea = –14194 x 8.31 = 117 952 J mol–1 = 118 kJ mol–1 (ii) Rate = k [NO2]2 [NO2] = 6.0/3.0 = 2.0 mol dm–3 Rate = (3.16)(2.0) 2 = 12.6 mol dm–3 s–1 (c) (i) N2H4(l) + 2H2O2(l)4 H 2O(l) + N2(g) N2H4 is oxidised. Oxidation state of N changes from –2 in N2H4 to O in N2. (ii) Anode : CH 3OH + H2OC O 2 + 6H+ + 6e– Cathode : H 2O2 + 2H+ + 2e– 2H2O

4  DHS 2015 9647/02 (iii) E cell = E red – E oxid +1.75 V = 1.77 – E oxid E oxid = +0.02 V i.e. E(CO2/CH3OH) is +0.02 V. If written as E(CH3OH/CO2) = + 0.02 V, no credit is given. 3 (a) Ca(s) + 2H 2O(l)C a ( O H ) 2(s) + H2(g) (b) (i) Ksp = [Ca2+][C2O4 2–] z = spK (ii) The oxalate anion is a conjugate base of the weak acid hydrogen oxalate. Since the extent of dissociation of hydrogen oxalate depends on the pH of the solution, the concentration of oxalate ion resulting from the dissociation of hydrogen oxalate will also depend on the pH of the solution. Therefore, the term K 2 must be taken into account when considering the solubility of oxalate. The term [H +] is to account for the pH of the solution that calcium oxalate is subjected to. (iii) The solubility of calcium oxalate will increase with decreasing pH value (can be inferred from the given equation due to the increase in [H +].) At low pH, oxalate ions will be protonated to form hydrogen oxalate ions. This will reduce the concentration of oxalate ions. By Le Chatelier’s principle, the solubility product equilibrium of calcium oxalate will shift to increase the concentration of oxalate ions. Hence, the solubility of calcium oxalate will increase. (iv) Sodium hydroxide might precipitate calc ium hydroxide which would decrease the concentration of calcium ions thereby potentially increasing the solubility of calcium oxalate at high pH. (c) (i) CH 3CH(NH2)COOH + 3O2 2 5 CO2 + 2 5 H2O + 2 1 CO(NH2)2 OR CH3CH(NH2)COOH + 6[O] 2 5 CO2 + 2 5 H2O + 2 1 CO(NH2)2 (ii) Moles of alanine in 1.00 g = 89 1 = 0.01123 mol

 DHS 2015 4 (a) (b)    (c) Am Q Q (d) Q 1. L By Hes H rxn = Hence, Volume o Ammeter Time take mount of hy = It = (5.6) = 1680 C uantity of ch = 1.92 x = 2Le 92 x 105 = ( = 6.00 x 10 CH3C s’ Law, –1577 + 6 amount of of hydrogen reading / C en ydrogen gas (5 x 60) C harge per m 105 C (2)(L)(1.60 023 CH(NH2)COO + 32 = –945 K energy evo gas Current s = 0.210 / 2 mole of hydr x 10-19) OH(s) + 3 O2 3 CO O2(g) 5 9647/02 KJ mol–1 olved = 945 24 = 0.0087 rogen gas = 2(g) O2(g) + H2O H  0.01123 75 mol = 1680 / 0.0 C O(l) + N2(g Hrxn = 10.6 kJ 00875 CO2(g) + H ) + O H2O(l) + CO O2(g) O(NH2)2(l)

6  DHS 2015 9647/02 (e) There is resistance in the circuit / Heat loss due to resistance. 5 (a) (i) On going across Period 3, the hydroxides go from basic to acidic. This systematic variation is the result of (1) increasing electronegativity of the elements, which leads to (2) the difference in the bonding between the elements and the hydroxide.  Going across Period 3, an increase in the effective nuclear charge of the elements, resulting in the increase in their electronegativity.  Difference in electronegativity between the elements and the hydroxide becomes smaller on going across the Period 3 hydroxides, resulting in the bonding between them changing from ionic to covalent.  Ionic oxides/hydroxides are basic because of the presence of OH – ions, and the covalent oxides/hydroxides are acidic due to the interactions of the partially positive elements with water, releasing H+ ion. (ii) Cl(OH) or