2024 Prelims JPJC H2 Chem P1 (Ans)

Jurong Pioneer Junior College 2024 H2 Chemistry Paper 1 Worked Solutions 1 D The definition for atomic mass is the ratio of the average mass of one atom of an element to one-twelfth the mass of one atom of 12C. In option D, the mass of one mole of atoms of an element has already considered all the isotopes and their relative abundances. 2 C A Al2+ ions 1s22s22p63s1; F+ ions. 1s22s22p4 Incorrect as Al3+ has 1 outer shell electron vs F+ 6 outer shell electron. 2nd IE of F is greater than the 3rd IE of Al as 3s electron is Al2+ is higher in energy and further away from the nucleus (hence lower nuclear attraction) then the 2p electron in F+. B Incorrect as the 3rd IE involves the removal of 3rd electron from Al is from 3rd principal quantum shell as compared to the removal of 3 rd electrons from the 2 nd PQM for the remaining 4 species. ✓C Na3+ 1s22s22p4+ vs Ne2+ 1s22s22p4+ Same no of electrons but as Na 3+ has more protons than Ne, hence higher nuclear charge, thus nuclear attraction of outermost electrons in Na 3+ is higher, thus higher 4th IE for Na compared to 3rd IE for Ne . D Incorrect as the removal of electrons are removed from different PQM. 3 Due to the no of bond pairs and lone pairs of electrons in each molecule – BCl3 : 3 Bond Pairs ; PH3: 3 Bond Pairs & 1 Lone pair The shape of BCl3 and PH3 are trigonal planar and trigonal pyramidal respectively 4 D (1, 2 and 3) ✓1: Only id –id attraction can be formed between hydrocarbon molecules and H 2O molecules which is weaker than the hydrogen bonds between H2O molecules. Hence, the energy released upon forming the less favourable interaction is not sufficient to compensate the energy required to break the stronger hydrogen bonds. Hence, hydrocarbon molecules are not solvated by water and hence the two layers a re immiscible. ✓2: See (1). ✓3: The stronger hydrogen bonds between H 2O molecules pulls the molecules closer to each other and hence, the volume of water is smaller. Given similar mass, the density of water is higher and hence, it will be below the hydrocarbon layer. P H H H

5 B Under constant n and T, the ideal gas equation is simplified to pV = k (where k is a constant and k = nRT). A Rearranging pV = k such that y = 1 p and x = V ( i.e. 1 p = k’V) , a graph of y = mx is obtained (i.e. straight line passing through origin). ✓B Since pV is a constant and x = pV, a graph of x = c is obtained (i.e. vertical line) C Rearranging pV = nRT such that y = p and x =  (i.e. p = r RT M  ), a graph of y = mx is obtained (i.e. straight line passing through origin). D Rearranging pV = nRT such that y = pV T and x = p (i.e. pV T = nR = constant), a graph of y = c is obtained (i.e. horizontal line) 6 C Mol ratio of O2 = 150 500 =0.3 Hence amt of O2 = 0.3 x 1.2 = 0.36 mol Amt of N2 = 5.76 24 =0.24 mol Hence amt of Ar = 1.2 – 0.36 – 0.24 =0.6 mol Thus pAr = 0.6 1.2 × 500 =250 kPa 7 A 1 only When aqueous ammonia is added to a solution containing hexaaquairon(III) ions, [Fe(H2O)6]3+, a red-brown precipitate is formed. NH3 + H2O ⇌ NH4+ + OH- where NH3 is Brønsted-Lowry base where it accepts proton to release OH - which will then ppt with Fe3+ to form red-brown ppt, Fe(OH)3. Fe3+ + 3OH- → Fe(OH)3 Ppt does not dissolve when excess ammonia is added, indicates that there is no further reaction or no ligand exchange to form a soluble complex. 8 A Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O n(Mn+) used = 0.100  15.00 1000 = 0.0015 mol n(Cr2O72−) used = 0.0250  20.00 1000 = 0.0005 mol ratio Cr2O72− : e− : Mn+ 0.0005 6(0.0005) =0.003 0.003 0.0015 2 : 1 Hence, 1 Mn+ loses 2 electrons and the oxidation state of M changes by 2.

9 A C6H12 + 9O2 → 6CO2 + 6H2O Since 6CO2  6H2O  1C6H12, n(CO2) formed = n(H2O) formed = x mol p Q M M = 2 2 mass of H O mass of CO = ( ) ( ) 18.0 44.0 x x   = 0.41 10 C Energy absorbed by water = 250 × 4.18 × (100 – 12) × 10–3 = 91.96 kJ Energy evolved by combustion of butane = 91.96 × 100 47 = 195.6 kJ 195.6 = 2877 × m(butane) 58 m(butane) = 3.944  3.94 g 11 D H2O(g) + C(s) → H2(g) + CO(g) S > 0 as there is an increase in disorderliness from 1 to 2 mol of gas particles. ve ve ve TG H S − + −  =  −  If H <0, G < 0 at all temperatures. However, G1 becomes more negative as temperature increases; hence ve ve ve TG H S + + −  =  −  When temperature increases, −TS becomes more negative. At high enough temperatures, G < 0 since |−TS| > |H|. Reaction is spontaneous at high enough temperatures.when temperature increases 12 B With the addition of another mole of gas, the area under the graph increases. When the temperature increased, the Maxwell Boltzmann graph peak will be shifted to the right. 13 A Graph 1: The melting point increases from Na to A l as the metallic bonding is stronger. Si has the highest melting point as it is a giant covalent lattice which requires the largest amount of energy to overcome the strong network of Si -Si bond. S 8 and P 4 are simple covalent molecule. As the number of electrons for S 8 increases, the id -id increases, hence, the melting point increases. Graph 2: Across the period, nuclear charge increases while the increase in shielding effect is insignificant as electrons are added to the same shell. The effective nuclear charge increases, hence, the 1st ionisation energy increases. However, the first ionisation energy for Al is lower than expected because less energy is needed to remove the 3p electron which is further away from nucleus and experience additional shielding from 3s electrons.

14 A 1 only When a system is at a state of dynamic equilibrium, • the concentration of all reactants and products remains constant and an equilibrium mixture is obtained. • the rate of forward reaction = the rate of the backward reaction. • Equilibrium can only be achieved in a closed system, where there is no loss or gain of substances to and from the surroundings. 15 B When [maltose] is very low as compared to [enyzme], many empty active sites of the enzyme molecules available for binding so the reaction is approximately 1st order w.r.t. maltose. As [maltose] increases, more active sites of enzymes are occupied by maltose molecules so reaction is no longer 1st order w.r.t maltose At high enough [maltose], all active sites are occupied by maltose molecules (i.e. saturated) so any further increase in [CO2] will not increase the rate. Hence, the reaction becomes zero order w.r.t maltose. 16 A Ca(OH)2 (s) ⇌ Ca2+ (aq) + 2OH—(aq) … (1) A INCORRECT. When common ion Ca2+ is added, [Ca2+] increases, hence POE in (1) shifts to the left, [OH--] decreases, hence pH increases ✓B CORRECT. Increasing temperature increases solubility of solids ✓C CORRECT. When Na2O is added into the solution, NaOH is formed. When common ion OH-- is added, [OH--] increases, hence POE in (1) shifts to the left, solubility of Ca(OH)2 decreases. ✓D CORRECT. [OH-]= 10-1.7 = 0.0200 mol dm-3 [Ca2+] = 0.0100 mol dm-3 Ksp = [Ca2+][OH-]2 = 0.01 x 0.022 = 4 x 10-6 mol3 dm-9 17 D O2 + 4H+ + 4e ⇌ 2H2O Eo = +1.23V O2 + 2H2O + 4e ⇌ 4OH- Eo = +0.40V Sn4+ + 2e ⇌ Sn2+ Eo = +0.15V Co3+ + e ⇌ CO2+ Eo = +1.89V MnO4- + 8H+ + 5e ⇌ Mn2+ + 4H2O Eo = +1.52V V3+ + e ⇌ V2+ Eo = -0.26V A INCORRECT. Sn will not show any colour change even though there is a reaction between 2Sn2+ + O2 + 4H+ → 2H2O + 2Sn4+ (Eocell = +1.08V > 0) as Sn ions have no colour B INCORRECT. Eocell = +1.23 – 1.89 < 0; not energetically feasible C INCORRECT. No reaction as both will undergo reduction reactions. ✓D CORRECT Eocell = +1.23 – (-0.26) > 0; energetically feasible 4V2+ + O2 + 4H+ → 2H2O + V3+

18 B When conc. HCl is added to CuSO4 [Cu(H2O)6]2+(aq) + 4Cl –(aq) = [CuCl4]2–(aq) + 6H2O(l) pale blue solution yellow solution