2025 Prelim P1 Teaching Solutions H2 Chem HCI

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 1 HWA CHONG INSTITUTION 2025 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION MARK SCHEME / SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 D D D C A B A A B A 11 12 13 14 15 16 17 18 19 20 B D D C B C B A C A 21 22 23 24 25 26 27 28 29 30 A A B D D C A B C B Comments 1 D The largest difference in consecutive ionisation energies occurs between the 5 th and 6th ionisation energies, implying that the 6th electron is removed from an inner quantum shell and element X has 5 valence electrons. Since element X belongs to Group 15, it is likely to form a chloride of the formula XCl5. 2 D Isotopes have the same number of protons but different number of neutrons. The number of protons can be determined from the number of electrons and charge. Once the number of protons are determined, the number of neutrons can be found from the nucleon number. E.g., in A, H has 10 electrons and a charge of −2, so H has 8 protons. Thus, H has 10 neutrons. G H no. of protons no. of neutrons no. of protons no. of neutrons A 8 10 8 10 B 18 19 16 18 C 16 20 18 20 D 20 22 20 20 In D, G and H have the same number of protons (20) but different number of neutrons (22 and 20). Note that A is incorrect as G and H have the same number of protons and neutrons. 3 D The shape of a species depends on the number of lone pairs and bond pairs of electrons around the central atom. Refer to Topic 2 Chemical Bonding Sec. 6.1. A AlCl3 PCl3 3 b.p. and 0 l.p. – trigonal planar 3 b.p. and 1 l.p. – trigonal pyramidal B BF3 NH3 3 b.p. and 0 l.p. – trigonal planar 3 b.p. and 1 l.p. – trigonal pyramidal C SF4 XeF4 4 b.p. and 1 l.p. – see saw (expanded octet) 4 b.p. and 2 l.p. – square pyramidal (expanded octet) D PH4+ BF4ˉ 4 b.p. and 0 l.p. – tetrahedral 4 b.p. and 0 l.p. – tetrahedral

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 2 4 C Using nt = ns + nb Since T is constant, PtVt = PsVs + PbVb After connection: Vt = Vs + Vb Pt = (50)(10) + (120)(30) (10+30) = 102.5 kPa = 103 kPa 5 A This question tests the definition of relative masses found in Topic 1 & 4. Option B is wrong as the relative atomic masses must take into account all the isotopes of the element, hence the term ‘average mass’ must be used. Option C is wrong as the reference should be 12C & not 1H. Option D is wrong as the reference should be to 1/12 the mass of one atom of 12C. 6 B Always check if the sum of the abundances of the isotopes add up to 100. In the case, they add up to 101.32. Ar (Ge) = ( 19.01 101.32 ×70) + ( 24.96 101.32 ×72) + ( 7.76 101.32 ×73) + ( 41.0 101.32 ×74) + ( 8.59 101.32 ×76) = 72.8 7 A Use the equation method to obtain the answer: • Flip the first equation to get the MgO on the right side like in the combustion equation. Remember to flip the sign on its H value. • Half the third equation to ensure that the H 2 from the second equation and this third equation cancels off, and you get ½ O 2 as required in the combustion equation. Remember to halve its H value. • Add up all the equations to check that you get the combustion equation as given. And then add up all H values to get the answer. MgCl2(aq) + H2O(l) → MgO(s) + 2HCl(aq) +89 Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) – 435 H2(g) + ½O2(g) → H2O(l) – 572 x ½ = – 286 Mg(s) + ½O2(g) → MgO(s) – 632 kJ mol–1 ns Vs = 10 m3 Ps = 50 kPa nb Vb = 30 m3 Pb = 120 kPa nt = ns + nb Vt = Vs + Vb Pt = ?? kPa

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 3 8 A Option 1: S increases as increasing the temperature at a fixed pressure increases the volume of the gas & increases the average kinetic energy of the gas particles and thus results in more ways to distribute the particles and energy. Option 2: S increases as doubling the number of moles of non-interacting gas particles at a fixed pressure increases the volume of the gas results in more ways to distribute the particles and energy. Option 3: S increases as doubling the number of moles of gas particles at a fixed pressure increases the volume of the gas results in more ways to distribute the particles and energy. Option4: S decreases as the number of moles of gas particles at a fixed pressure drops from 1 to 0 on the right-hand side of the equation. 9 B The rate equation can be written by inspecting the chemical equation of the slow step in the reaction mechanism. Hence it should look like: rate = k[N2O2][H2]. Rate equations should contain only reactants or product terms but not intermediates. In this case, N2O2 is an intermediate. To replace it, we see that N2O2 is related to a reactant NO by the reversible reaction which can be expressed in the form of the chemical equilibrium constant. Thus K= [N2O2] [NO]2 ; rearranging gives [N2O2] = K[NO]2. Substituting [N2O2] in rate = k[N2O2][H2] gives rate = kK[NO]2 [H2] or rate = k’[NO]2 [H2]. The units of the rate constant k’ can be worked out as follows: rate = k’[NO]2 [H2] mol dm–3 s–1 = k’ x (mol dm–3)2 x (mol dm–3) Simplifying gives k’ = mol–2 dm6 s–1 10 A The area under the curve represent the total number of molecules. Thus this is represented by P + R. Since only the T is changed, the total number of molecule, P + R does not change. Thus the expression R P + R gives the fraction of molecules that have energy greater than the activation energy of the lower temperature. Note that the steeper curve represents the lower temperature while the shallower curve is for the higher temperature. 11 B Option A is wrong as the correct temperature is 450 oC. Option C is wrong as the iron acts as an catalyst which speeds up the reaction but does not affect the yield. Option D is wrong as removing NH 3 shifts the position of the equilibrium to the right but does not affect the rate of reaction.

2025 HCI C2 H2 Chemistry Prelim Exam Teaching Solutions/Paper 1 4 12 D MnO4−(aq) + 8 H+(aq) + 5 e− → Mn2+(aq) + 4 H2O(l) SO2(g) + 2 H2O(l) → SO42−(aq) + 4 H+(aq) + 2 e− Obtain the overall equation through multiplying the first half-equation by 2 and the second half-equation by 5 in order to eliminate the electrons: 2MnO4– + 5SO2 + 2H2O → 2Mn2+ + 5SO42– + 4H+ Notice that there is a net production of H + so [H +] increases as the reaction proceeds. Since pH = −lg [H+], pH decreases over time. 13 D Titration Curve 1 = strong acid-weak base curve (equivalence pH is acidic) Titration Curve 2 = weak acid-weak base curve (no vertical portion in the curve) Option A is wrong as there are no suitable indicators for weak acid-weak base titrations. Option B is wrong as give above. Option C is wrong as Titration Curve 1 involves a strong acid as depicted by the flat smooth curve at pH between 0 & 2. A weak acid will give a sharper rise in pH before flattening out shortly after and then rising sharply again near the equivalence point (i.e. a buffer region). 14 C The decomposition of H2O2 is usually written as 2H2O2 → 2H2O + O2. In basic medium, it may be written as 2HO 2– → 2OH– + O2, where a proton is lost from both H2O2 & H2O. Re-writing it to make clear that two ions of HO 2– are reacting with each other in the decomposition process: HO2– + HO2– → 2OH– + O2. Thus HO2– → 2OH– (reduction) ➔ HO2– + H2O + 2e → 3OH– E⊖ = +0.88 V And HO2– → O2 (oxidation) ➔ HO2– + OH– → O2 + H2O + 2e E⊖ = –0.08 V Thus Ecell = + 0.96 V The semi -permeable membrane a llows ions to pass through to maintain electrical neutrality in both cells just like a salt bridge does. 15 B The % composition of the alloy is a distractor and is not needed in the calculations. The total mass of metal deposited as the cathode represents the metal that was reduced. Use of the data booklet shows that both Cu & Ni may be both oxidized at the anode while the reduction poten