2024 Prelims EJC H2 Chem P1 (Ans)

2024 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 mass of Fe in 0.32 g of Fe2O3 2 55.8 0.32 g 0.2238 g2 55.8 3 16.0 =  = +  0.2238% by mass of Fe in ore 0.6 37.3% = =  B 2 2Y 25.0 0.400 0.0100 mol1000n + =  = 4MnO 40.00 0.10 0.00400 mol1000n − =  = 5Y2+ ≡ 2MnO – 4 MnO – 4 is an oxidising agent, per mole, it takes in 5 moles of electrons: MnO – 4 + 8H+ + 5e–  Mn2+ + 4H2O 5Y2+ ≡ 10e–, each mole of Y 2+ gives out 2 moles of electrons.  Final oxidation of Y is +4.  A 3 nuclide charge, ratiomass, q m 24Mg+ 1 0.041724 = 48Ti3+ 3 0.062548 = 59Co2+ 2 0.033959 = 101Ru4+ 4 0.0396101= Since 48Ti3+ has the largest q m ratio amongst the four nuclides, it will experience the greatest deflection.  B 4 A: bent B: trigonal planar C: square planar D✓: octahedral  D 5 Using 1 1 2 2 1 2 21 1 2 2 1 pV p V V T ppT T V T=  =   ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 227 273 K100 cm40 kPa 100 400 cm 227 273 K 8 kPa 227 273 K400 cm20 kPa 100 400 cm 127 273 K 20 kPa p p +=   ++ = +=   ++ = K L final pressure of mixture 8 kPa 20 kPa 28 kPa pp=+ =+ = KL  A 6 1✓: Lattice energy is the energy released when 1 mole of ionic compound , KCl(s), is formed from its constituent gaseous ions, K +(g) and C l–(g). The lattice energy of potassium chloride is ∆H3. 2: The enthalpy change of solution is the energy change when 1 mole of substance in its standard state , KCl(s), is completely dissolved in a solvent to give a solution of infinite dilution, K +(aq) and C l–(aq). The enthalpy change of solution of potassium chloride is ∆H4. 3✓: Enthalpy change of formation is the energy change when 1 mole of a substance in its standard state , KCl(s), is formed from its constituent elements in their standard states , K(s) and Cl2(g). The enthalpy change of formation of solid potassium chloride is (∆H1 + ∆H2 + ∆H3).  B 7 1✓: (G1 + G2) is G for dissolving of AgCl(s) in NH 3(aq); (G3 + G4) is G for dissolving of AgBr(s) in NH3(aq). As solid AgC l is more soluble than solid AgBr in NH 3(aq), the former must be more exergonic, i.e. G is more negative. Thus, (G1 + G2) < (G3 + G4). 2✓: Eqm 2 and 4 are identical, with C l– and Br– being spectator ions: Ag+(aq) + 2NH3(aq)  [Ag(NH3)2]+(aq) Hence, G2 = G4. 3: From 2, G2 = G4. 4✓: Since solid AgCl is more soluble than solid AgBr in water, G1 must be more exergonic, i.e. more negative, than G3. Thus G1 < G3.  A 8 A: Forward reaction of reaction 1 leads to the formation of Ag atoms, causing the photochromic glass to darken. Reaction 2 and 3 together, is the backward reaction of reaction 1, catalysed by Cu + and Cu 2+. This removes Ag atoms, causing the glass to becomes transparent again. Forward reaction of reaction 1 must be much faster than the backward reaction (reaction 2 + reaction 3) when intensity of UV light is high to ensure the glass darkens. B✓: The forward reaction of reaction 1 is only favoured when intensity of (high energy) UV light is high, showing that the forward reaction is endergonic, i.e. G > 0 (non-spontaneous) C: When glass darkens in UV light, [Ag] increases while [Ag+] decreases, which will cause reaction 3 to be more spontaneous. D: The position of equilibrium of reaction 1 shifts to the right in strong UV light, leading to formation of Ag atoms which cause the photochromic glass to darken.  B 9 The fraction of m olecules with energy greater than or equal to a certain energy E is represented by the area under the graph beyond E. So, W+X+Y+Z = 1. When a catalyst is added, the activation energy is lowered. Hence, the fraction of molecules with energy greater than or equal to activation energy for the catalysed reaction will be X+Y+Z.  C 10 1: Rate is a kinetics property related to Ea, while Kc is a thermodynamic s property related to Gr. They are independent of each other. 2: The temperature of the solution does not change since boiling point of water is also at 373 K. Therefore, there is no change in the Kc value. 3✓: The presence of a catalyst will not change the Kc value.  A 11 A: As T increased, amount of A remaining at equilibrium increased. This shows that position of equilibrium shifts left, hence backward reaction is endothermic. Forward reaction is thus exothermic. B✓: Position of equilibrium shifts left with increasing T to favour endothermic reaction. Thus Kp will decrease with increasing T. C: At 650 ºC, amount of A remaining is 0.8 mol. Thus, 1.2 mol of A had decomposed. Percentage of A decomposed is 1.2 100 60%2.0= . D: The amount of A used up is 1.2 mol and hence amount of C produced is 1.2 mol at equilibrium.  B 12 23H 10 0.0100 mol dm+ − − == 3 0.0100 0.0100H after mixing 2 0.00500 mol dm VV V V V + − + = ++ = ( )pH lg H lg 0.00500 2.3+=− =− =  C 13 For a weak acid, HA,       a H A H salt HA acidK + − +        = Hence,     a acidH saltK+   C 14 CaC2O4(s)  Ca2+ (aq) + C2O 2– 4 (aq) ---- (1) H2C2O4(aq)  2H+ (aq) + C2O 2– 4 (aq) ---- (2) As pH increases, [ H+] decreases. H ence the position of eqm (2) shifts to the right , causing [C2O 2– 4 ] to increase. This will cause the position of equilibrium for eqm (1) to shift to the left, reducing the solubility of CaC2O4(s).  D 15 A: Although nuclear charge increases, shielding effect is similar due to same number of inner shell electrons, hence, effective nuclear charge increased, leading to the general decrease in atomic radius across the period. B: First ioni sation energy generally increases due to increase in effective nuclear charge, however there is a decrease from Mg to A l (due to electron being removed from 3p

instead of 3s orbital), and from P to S (due to interelectronic repulsion between paired electrons). C✓: +1 in Na2O, +2 in MgO , +3 in A l2O3, +4 in SiO2, +5 in P4O10, +6 in SO3 D: NaCl and MgC l2 are ionic chlorides with higher melting points than the covalent AlCl3, SiCl4 and PCl5.  C 16 The solubility of a salt is related to ∆Gsol = ∆Hsol – T∆Ssol where ∆Hsol = |L.E.| – |∆Hhyd| A: The charge density of Mg2+ is indeed higher such that Mg 2+ is better at organising water molecules around it, which leads to a more negative ∆Ssol. So, if ∆Ssol is more negative, then ∆Gsol is more positive for MgC2O4, which means that MgC 2O4 should be less soluble instead. B✓: Mg2+ has a higher charge density and forms stronger ion -dipole interaction and hence ∆Hhyd is more negative. This potentially contributes to a more negative ∆Hsol, which implies a more negative ∆Gsol and greater solubility. C: While it is true that Mg 2+ has a more negative L.E. due to Mg 2+ having a smaller ionic radius than Ca 2+, a more negative L.E. leads to a more positive ∆Hsol and hence lower ∆Gsol. D: While this is true, electronegativity does not affect solubitlity.  B 17 1✓: As the atomic radius es down the group, the valence orbitals used for bonding are larger and more diffused. This leads to less effective overlap of orbitals, resulting in X –X bond energy ing down the group. 2✓: The number of electrons es down the group, resulting in  polarisability of electron cloud and thus strength of the id-id interactions. Hence volatility es (higher boiling point) down the group. 3: EꝊ(X2|X–) es down the group, there is a  in tendency for X 2 to be reduced. Oxidising power of Group 17 elements es, i.e. they become weaker oxidising agents down the group.  B 18 A: Compound R has 5 chiral centres, hence 25 = 32 stereoisomers. B: Compound R and S have exactly the same connectivity of atoms. C✓: HO(CH2)13OH cannot rotate plane polarised light since it has a plane of symmetry. D: Since the line structure of compound R does not