DHS H2 Chem 2012 Prelim P1 Soln

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 2 6 C Calcium and chlorine form CaC l2 rather than CaCl because DHf(CaCl2) is more exothermic than DHf(CaCl). DHf is the sum of enthalpy changes involved in a series of processes leading from the elements to the compounds: 1. Forming separate atoms from elements (DHat) 2. Forming positive (from the metal atom) or negative (from the non -metal atom) ions. (Sum of I.E. or sum of E.A.) 3. Combining the ions together in an ionic lattice held together by the attraction between oppositely charged ions (LE). For CaCl, w less energy is required to ionise the Ca atom only once, to Ca+ w only one C l atom needs to be formed from Cl2 and converted into a Cl- ion w lattice energy released by forming a 1:1 lattice of singly charged ions is less exothermic than that for CaCl2, which involves Ca2+ ions. Option C is the best answer. 7 C -100 -75 -50 -25 0 25 50 75 100 reactants products progress of reaction Ea = +50 kJ mol-1 formation of intermediate DH = -100 kJ mol-1 8 A In 1.0 mol dm –3 sulfuric acid, [H +] = 2.0 mol dm–3. To make the electrode a standard hydrogen electrode, either change the acid to a 1.0 mol dm –3 monoprotic acid (option A), or halve the concentration of sulfuric acid used. 9 B Since forward reaction is exothermic, higher temperatures will favour the backward reaction. This increases [NH 3] and [O 2] and decreases [N2] and [H 2O] at higher temperatures. Thus the two downward– sloping graphs apply to either N2 or H2O. At a higher pressure, backward reaction is favoured, and [N2] and [H2O] decreases. Thus, z > y. 10 A H2O(l) Ý H+(aq) + OH–(aq) Kw = [H+][OH–] [H+] = wK A P At 25 oC, [H+] = 1.0 × 10–7 mol dm–3 At 10 oC, [H+] = 0.5 × 10–7 mol dm–3 At 0 oC, [H+] = 0.3 × 10–7 mol dm–3 B O Ionic dissociation of water increases by a factor of 3.3 between 0 oC and 25 oC. C O Extent of hydrogen bonding cannot be deduced from the given information. D O [H+] = [OH –] at 0 oC, 10 oC and 25 oC. Thus water remains a neutral liquid at these temperatures. However pH of neutral water is no longer 7.0, but increases with decreasing temperature.

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 3 11 A Rate law is determined by slow step in proposed mechanism, i.e. Rate = k’[O3][O]. This rate law cannot be compared directly with the experimental rate equation because it contains the concentration of an intermediate, O. Thus we need to express rate law in a way that removes the intermediate O. From Step 1, ] O [ ] O ][ O [ 3 2=K Þ [O] = ] O [ ] O [ 2 3K Ass uming that Step 1 equilibrium is established quickly before O is reacted with O3 in Step 2, Rate = k’[O3][O] = ] O [ ] O [ ' 2 23K k = ] O [ ] O [ 2 23k where k = k’K 12 C [S(C2O4)2(NH3)2]– Let the oxidation number of S be x. x + 2(–2) + 2(0) = –1 Þ x = +3 S3+ : ­ ­ ­ (d3) S0 : ­ ­ ­ ­ ­ ­¯ 13 D A (Wrong) NH3 acts as a weak base and a ligand