2025 JC2 Prelims H1 Chem Paper 2 Solutions TJC

8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN 2025 TJC H1 Chemistry Prelim Paper 2 Worked Solutions Section A 1 (a) (i) [1] W. It has the highest q r + + value Note : C is not a metallic element (ii) [1] V or Cr (iii) [✓] Giant metallic structure with strong [✓] electrostatic forces of attraction between cations and sea of delocalised electrons. [1] 2[✓] = 1m (iv) [1] Adding carbon atoms into the space between the iron atoms in the lattice will prevent the iron atoms from sliding over each other easily and hence do not go out of shape easily. (b) FeCl2 has a [✓] giant ionic structure while [✓] SiCl4 and SiI4 have simple molecular structures. Between Fe2+ and Cl- ions, there are [✓] strong ionic bonds while between SiCl 4 molecules and between SiI 4 molecules, there are [✓] weak instantaneous dipole -induced dipole attractions (id-id). Hence [✓] more energy is required to overcome the strong ionic bonds than the weak id-id attractions resulting in a high m.p. for FeCl2. SiI4 has a [✓] bigger electron cloud compared to SiCl 4 resulting in [✓] greater polarization leading to [✓] stronger id -id attractions between SiI 4 molecules. Hence [✓] more energy required to overcome the id-id attractions resulting in higher m.p. for SiI4. 3[✓] = [1]

3 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN [Turn over 2 (a) [1] The product of reduction of H 2O2 is water/oxygen which is clean / non-pollutants / environmentally friendly. Note: accept O2 (b) [✓] Sufficient energy is released from the formation of [✓] hydrogen bonds between water and H2O2 to overcome the [✓] hydrogen bonds between water molecules and hydrogen bonds between H2O2 molecules. [✓] Insufficient energy is released from the formation of [✓] instantaneous dipole - induced dipole interactions between anthraquinone and water to overcome the hydrogen bonds between water molecules. [✓] H2O2 dissolve in water / anthraquinone cannot dissolve in water. 3[✓] = [1] (c) (i) [1] Palladium catalyst is not in the same phase as the reactants. (ii) The reactant molecules are [✓] adsorbed unto the surface of the solid catalyst. This [✓] lowers the activation energy by [✓] increasing the surface concentration of reactant molecules (or: reactant molecules are brought closer together) and [✓] weakens the covalent bonds of/within the reactant molecules. [✓] Number of particles with energy greater than or equal to Ea’ increases. The [✓] frequency of effective collisions is increased. [✓] Rate constant increases and hence the rate of the reaction increases. After the reaction, [✓] desorption of product molecules occur at the catalyst surface. The product molecules eventually [✓] break free and diffuse away from the catalyst surface. Other reactant molecules may then approach the catalyst surface. 3[✓] = [1] ; Diagram = [1] (d) (i) [1] Nanoparticles are defined as particles with all dimensions between 1-100nm on the nanoscale. (ii) [1] Finely dividing the palladium into nanoparticles increases its surface area to volume ratio. Hence, there are more active sites on the catalyst surface for adsorption of reactants. (iii) [1] Due to their small size, nanoparticles can enter human body through pores on skin, breathing/respiratory system, or ingestion (by digestive system). These particles can travel around the body via the circulatory system, get deposited in body tissues/cells/organs and cause the latter to be damaged/inflamed/develop cancer/cause immune response. (iv) [1] Gold particles are larger and t he deposition of the nanoparticles decreases the chances / prevents them from being inhaled or ingested.

8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN 3 (a) (i) Either one (accept any form of structural formula): [1] correct structure ; [1] name Note: Answer must be a primary alcohol. (ii) methanol, conc. H2SO4, heat [1] (b) (i) Ka = [H+][CH3(CH2)3CO2 −] [CH3(CH2)3CO2H] [1] Reject use of round brackets (ii) pKa = –log10Ka Ka = 10–4.82 = 1.51 x 10–5 mol dm–3 [1] both value and units (iii) A buffer solution is a solution that resists changes in pH upon addition of a small amount of an acid or base. [1] Comments : must mention small amounts of acid or base (iv) CH3(CH2)3CO2– + H3O+ → CH3(CH2)3CO2H + H2O [1] Reject if reversible arrow is used or H+ used (v) Amount of KOH reacted = amount of valeric acid = 0.05 x 22 1 000 = 0.00110 mol [1] Amount of valeric acid in 250 cm3 buffer = 0.00110 x 10 = 0.0110 mol [1] (vi) Initial amount of valeric acid = amount in 250 cm3 buffer + amount reacted with NaOH = 0.0110 + 0.1 x 100 1 000 = 0.0210 mol [1] Initial molar concentration of valeric acid = 0.0210 0.150 = 0.140 mol dm–3 [1] Note : the initial amount of valeric acid would include the unreacted acid (from (v)) and the amount of acid that reacted with NaOH to form the salt. (c) Heat released = mcT = (100 + 100) x 4.18 x 5.5 = 4598 J [1] Since valeric acid is the limiting reagent Amount of H2O formed = amount of valeric acid reacted = 0.9 x 100 1 000 = 0.09 mol Hn = – 4598 0.09 = –51.1 kJ mol–1 [1] with sign and units

5 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN [Turn over 4 (a) (i) PVAc is a thermoplastic as there are no cross links  between the polymer chains. The weak permanent dipole-permanent dipole interactions  (OR instantaneous dipole induced dipole interactions ) between polymer chains are easily overcome hence it can be remolded / soften when heated . Two  [1] (ii) PVAc will not have a weak absorption peak from 1635 to 1690 cm–1 as it does not contain C=C bonds. [1] Note : Need to quote IR absorption values and specify which functional gp is present. (iii) Step 1: ethanolic NaOH, heat [1] V: CH3CH2OH [1] (iv) C2H4 + CH3COOH + ½O2 → CH3CO2CHCH2 + H2O [1] (v) CH3CHO [1] (b) acid: CH3CHClCOOH and conjugate base: CH3CHClCOO– [1] base: NH3 and conjugate acid: NH4+ [1] [1] each both pairs correctly identified

8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN 5 (a) [1] accept allene Comments : Answer must be a non-cyclic branched HC with only 5 C atoms. Also, there must be 2 C=C based on the reaction with Br2. (b) [1] Systematic name: buta-1,3-diene (c) (i) Cis-trans isomerism. [1] There is restricted rotation about the C=C bond and each carbon of the C=C bond is bonded to two different groups. [1] Comments : Must clearly state that the 2 different groups are attached to each of the C of the double bond. (ii) Structure of polymer Y (trans isomer), need to show extended bonds [1]. Diagram must show trigonal planar around C=C. (iii) polymer Z (cis isomer) [1] (d) Carbon-carbon bonds are non-polar / too strong / not attacked by nucleophiles / cannot by hydrolysed. [1 with explanation] (note: the focus is on C -C bonds not C -H bonds as need to break these bonds in order to degrade the polymer) (e) (i) [1] with n indicated and bonds extended Comment: Not repeat unit and must be open-ended (ii) Reactions with a high percentage atom economy are more efficient and less wasteful of resources / less pollution. [1] Accept: maximises the use of raw materials in the process into useful products, saves resources

7 DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 8873 / TJC Prelims / 2025 DO NOT WRITE IN THIS MARGIN [Turn over (iii) The percentage atom economy of the reaction to form PET is lower than that to form polymer X as H2O / another molecule is also formed during the condensation or reaction while all the monomers are used to form polymer X. [1] Comment: insufficient to just mention a