2025 H1 Chem Prelim Paper answers HCI

2025 Prelim Exam H1 Chemistry For internal circulation only 1 2025 Preliminary Examination H1 Chemistry (8873) Suggested Solutions Paper 1 1 D 11 B 21 C 2 A 12 D 22 A 3 C 13 B 23 D 4 B 14 D 24 B 5 B 15 A 25 A 6 C 16 B 26 D 7 C 17 C 27 B 8 A 18 B 28 C 9 C 19 B 29 D 10 B 20 D 30 C

2025 Prelim Exam H1 Chemistry For internal circulation only 2 Paper 2 Section A 1(a)(i) E and F have the same number of protons but different number of neutrons. [1] no partial credit (ii) G has 7 electrons pairs [1] (iii) A and C [1] (iv) C and D have the same number of protons, but C has one more quantum shell than D. Hence, C has a larger shielding effect and smaller effective nuclear charge than D, resulting in weaker attraction between the nucleus and outermost electrons. [1] same no. of proton + larger shielding effect. Award 0.5 mark if only say ‘smaller effective charge’ [1] C has one more quantum shell (v) B and G are deflected to the negative plate as they are cations. [0.5] E is deflected to the positive plate as it is an anion. [0.5] 0.5 mark for ‘opposite directions’ only angle of deflection α charge mass angle of E equals angle of G as they have the same charges and masses [0.5] angle of B is smaller than angle of E (or G) as B has a greater mass [0.5] • Directions + reasons in terms of identity or charges • Relative angle + reasons in terms of charges and masses (b)(i) A large difference in electronegativity between the element and chlorine results in (transfer of electrons and) ionic bonding, while a small difference results in (sharing of electrons and) covalent bonding. [0.5] explanation for ionic bonding [0.5] explanation for covalent bonding (ii) NaCl(s) → Na+(aq) + Cl−(aq) [1] PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq) [1] state symbols not required pH of mixture = 1 [1] award 0.5 mark for correct pH of separate solutions

2025 Prelim Exam H1 Chemistry For internal circulation only 3 2(a) It states that if the conditions of a system in equilibrium are changed, the position of equilibrium will shift in a way to reduce that change. [1] no partial credit for definition (b)(i) From Fig 2.1, w hen temperature increases, yield increases which implies position of equilibrium shifted right. Hence, forward reaction is endothermic as it absorbs the additional heat to offset the increase in temperature. [1] correct shift in p.o.e. [0.5] endothermic [0.5] reasons in terms of heat (ii) From Fig 2.1, w hen pressure increases, yield increases which implies position of equilibrium shifted right. Hence, forward reaction results in a decrease in total number of gas molecules as this will offset the increase in pressure. [1] correct shift in p.o.e. [1] decrease in total no. of gas molecules (c)(i) Kc = [H2][CO] [H2O] [1] (ii) H2O(g) + C(s) H2(g) + CO(g) initial / mol 2.0 0 0 change / mol -0.3 +0.3 +0.3 eqm / mol 1.7 0.3 0.3 [1] correct eqm amount of CO and H2O Kc = 0.3 × 0.3 1.7 = 0.0529 mol dm–3 [1] ecf from eqm amount, correct s.f. 3(a)(i) Trichlorofluoromethane [1] (ii) Freon-14 [1] (b) According to VSEPR theory, to minimise electronic repulsion, electron pairs (or groups) will arrange themselves as far apart as possible. Since there are 4 bond pairs around C atom, Freon-11 has a tetrahedral shape. [0.5] correct shape [0.5] correct number of bond pairs [1] correct explanation using VSEPR (c) Freon-12 is a polar molecule. [0.5] C-F bond and C-Cl bond have different polarities (as electronegativity of C < Cl < F). Hence, the individual dipoles do not cancel off (or net dipole moment is not zero) as the molecule has a tetrahedral shape. [0.5] correct explanation in terms of bond polarities and net dipole

2025 Prelim Exam H1 Chemistry For internal circulation only 4 (d)(i) [1] use different symbols for adjacent atoms (ii) C-F bond: 485 kJ mol–1 C-Cl bond: 340 kJ mol–1 [0.5] quote both values C-Cl bond is much weaker and breaks more easily. or C-F bond is much stronger and does not break easily. [0.5] (iii) 2O3 → 3O2 [1] (iv) [1] correct diagram; deduct 0.5 mark if incomplete labels / curve did not start from origin Chlorine atoms provide an alternative reaction pathway with a lower activation energy, increasing the fraction of particles with energy greater than or equal to lower Ea. [0.5] Hence, the frequency of effective collisions increases and rate increases. [0.5] (e)(i) [1] Draw only 18 carbon atoms and at least two complete hexagonal rings (ii) The  electrons in 3D graphene can move through each layer. [1] or The delocalised electrons in 3D graphene can move through each layer. [1] (iii) lattice energy  q+× q− r++ r− F− and Cl− ions have the same charge but F− ion has a smaller radius. [0.5] Lattice energy of MgF2 has a larger magnitude (or is more exothermic) so more energy [0.5] is needed to overcome the stronger ionic bonds between Mg2+ and F− ions. [0.5] MgF2 has a higher melting point. [0.5] kinetic energy Ea No. of molecules 0 proportion of molecules with KE  Ea Ea with catalyst

2025 Prelim Exam H1 Chemistry For internal circulation only 5 4(a) Weak BrØnsted-Lowry base is a proton acceptor that dissociates partially in water. [1] no partial credit for definition (b)(i) nHCl = 0.10 × 38 1000 [0.5] = 0.0038 mol = nCH3NH2 [CH3NH2] = 0.0038 25 × 1000 = 0.152 mol dm–3 [0.5] ecf from nCH3NH2 correct units, s.f. (ii) CH3NH2 and CH3NH3+ [0.5] each correct species (iii) Since the two species are a conjugate-acid base pair, the solution is a buffer. [0.5] When HCl is added, position of equilibrium CH3NH2 + H+ ⇌ CH3NH3+ shifts right to produce CH3NH3+. [0.5] As the solution contains relatively large amounts of both species, [CH3NH2] and [CH3NH3+] remain almost constant, hence [H+] and pH remain almost constant , and the slope only changes gradually. [0.5] large amount / conc of base & conjugate acid or small amount of H+ added (WTTE) [0.5] pH remains almost constant (iv) CH3NH3+ + OH– → CH3NH2 + H2O [1] X (c)(i) Condensation [1] (ii) [1] deduct 0.5 mark if not displayed formula (d)(i) + 2 O=O → 2 O=C=O + 2 H-O-H broken formed C–C 350 4(805) C=O C–H 3(410) 4(460) O–H C=O 740 C–O 360 O–H 460 O=O 2(496) Hc = 4132 – 5060 = –928 kJ mol−1 correct sign, units, s.f. endo exo [0.5] correct BE of bonds broken [0.5] correct number of each bond [0.5] correct BE of bonds formed [0.5] correct number of each bond [1] correct ΔHc (bonds broken - bonds formed) ECF (ii) Ethanoic acid and water exist as liquid, but they are in the gaseous state in the calculation using bond energies. [1] Hence, the calculated enthalpy change is different from the value quoted.

2025 Prelim Exam H1 Chemistry For internal circulation only 6 5(a)(i) HDPE has linear polymer chains with minimal branching [0.5]. The chains can pack closely [0.5] together resulting in higher density. (ii) LDPE has highly branched [0.5] polymer chains that prevent close packing, resulting in weaker dispersion forces between the polymer chains. [1] Hence, less energy [0.5] is needed to overcome these forces during melting. (iii) The C-C and C-H bonds are strong [0.5] and non-polar [0.5] so the polymer is resistant to microbial attack. (iv) LDPE can jam the sorting machine as they are soft and flexible. [1] accept other reasons related to properties of LDPE (v) volume of pellet = 50 cm x 40 cm x 2 cm = 4000 cm3 [0.5] mass of flakes needed = 0.95 x 4000 = 3800 g [0.5] ecf number of milk bottles needed = 3800 / 70 = 54.3 = 54 [1] ecf (b)(i) PVC is water-resistant [1] (ii) –CH2CHCl– + 5 2O2 → 2CO2 + H2O + HCl [1] (iii) NaOH [1] accept other basic oxide or hydroxide e.g. Mg(OH)