RVHS 2026 Prelim P3 Ans

2026 RVHS Preliminary Examination H2 Chemistry Paper 1 Worked Solutions Section A 1 (a) (i) Ca(NO3)2(s) → CaO(s) + 2NO2(g) + O2(g) (ii) cations Ionic radius/nm Pb2+ 0.120 Ca2+ 0.099 Zn2+ 0.074 When ionic radius of the metal cation decreases from Pb 2+ to Ca 2+ to Zn 2+, its charge density increases . As a result, the ability of the metal cation to polarise the electron cloud of the (large NO 3–) anion increases and the N−O bonds are weakened to a bigger extent . Less energy is required to decompose the metal nitrate. Hence Zn(NO 3)2 is least thermally stable and decomposes at the lowest temperature followed by Ca(NO3)2 and lastly Pb(NO3)2. (b) (i) 3Ca(s) + ½ P4(s) + 4O2(g) Ca3(PO4)2(s) 3Ca(g) 3Ca2+(g) + 2PO43– (g) −4175 = 3(+178) + 3(590 + 1150) + 2Hf(PO43–(g)) + (−10520) Hf(PO43–(g)) = +295.5 kJ mol–1 = +296 kJ mol–1 (ii) The standard enthalpy change of solution of a substance is the enthalpy change when one mole of the substance is completely dissolved to give an infinitely dilute solution under standard conditions, so that no further enthalpy change takes place on adding more solvent. OR when one mole of the substance is completely dissolved in enough solvent under standard conditions , so that no further enthalpy change takes place on adding more solvent. 3(+178) 2Hf(PO43–(g)) −10520 3(590 + 1150) −417

2 River Valley High School 9476/03/Prelims/26 (iii) Hsoln = −(−10520) + 3(−1562) + 2(−2876) = +82.0 kJ mol–1 (iv) Chemical Bonding Teacher’s copy River Valley High School 2015 Year 5 H2 Chemistry 3.3 Dot-and-cross diagrams (Lewis diagrams) for ionic bonding ‘Dot-and-cross’ diagrams involve using dots ‘•’ or crosses ‘ ’ to represent electrons. Only the valence electrons of the species are shown. NaCl: MgCl2: Checkpoint 3 Draw a ‘dot-and-cross’ diagram for Na2O. *Note: there is no need to draw a legend for dot-and-cross diagram at A-Levels as long as it is clear which electrons belong to which elements. 3.4 Factors affecting strength of ionic bonds Ionic bond strength can be deduced from the lattice energy of an ionic compound. Lattice energy is the energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions (refer to topic of Chemical Energetics). The magnitude of lattice energy is dependent on the charges of the cations and anions, as well as the ionic radii of the cations and anions. It can be described with the relationship below: Magnitude of lattice energy (LE) + + ZZ rr where, Z+ charge of the cation Z charge of the anion r + radius of the cation r radius of the anion In other words, ionic bonds formed between ions with high charge and small size (i.e high charge density) are the strongest. Correspondingly, the lattice energy of such ionic compounds will be the most exothermic. Lattice energy is exothermic because gaseous ions need to lose energy before being brought together into an ordered arrangement in the giant ionic lattice structure. Mg Cl 2+ 2 Na Cl + Inter-ionic distance, r + + r Fig. 3.3 Illustrating inter-ionic distance r+ r Ca3N2 > Ca3(PO4)2 > CaSO4 The charge and size of Ca2+ are the same for all 3 compounds. Since N 3− and PO43– have the same charge, and N3− has a smaller ionic radius/smaller anion than PO 43–, the magnitude of lattice energy is larger for Ca3N2 . SO42– has the smallest charge , so CaSO 4 has the smallest magnitude of lattice energy. (c) (i) − − =  =   +=     = 56 1 0.400 8.31 (273 70) 1.2 1.01325 10 (150 10 ) 62.5 g mol mpV RTM mRTM pV (ii) NCl3 has a larger electron cloud size/more electrons than NH 3 and hence stronger instantaneous dipole-induced dipole interactions between molecules. OR NCl3 has a bigger molecular size than NH 3. The volume of NC l3 molecules is more significant than NH3 when compared to the volume of the container. (iii) Standard entropy change is positive as there is an increase in amount/number of gaseous molecules from 0 to 4 mol and hence an increase in disorder of the system. (iv) A higher temperature will favour the removal of NCl3. ∆S > 0, −T∆S < 0 Since ∆H > 0, at higher temperatures, the −T∆S term becomes more negative/ |−T∆S| > |∆ H| and hence ∆G becomes more negative . Reaction becomes more spontaneous at a higher temperature. [Total: 16] 2 (a) H2(g) molecules are very small and can more easily diffuse/leak through gaps in gas tanks. H2(g) is flammable/ explosive and requires specific safety precautions to store. Very high pressure storage requirement results in high costs

3 River Valley High School 9476/03/Prelims/26 [Turn over Expensive and difficult to liquify due to very low boiling point of H 2(g) for storage and transport. (b) (i) Electrons flow from electrode X to Y. (ii) Anode/oxidation half: CH3OH(l) + H2O(l) → CO2(g) + 6H+ + 6e− Cathode/ reduction half: O2(g) + 4H+ + 4e− → 2H2O(l) Overall: CH3OH + 3/2O2(g) → CO2(g) + 2H2O(l) (iii) Ecell = E(O2/H2O) – E(CO2/CH3OH) (+1.21) = (+1.23) – E(CO2/CH3OH) E(CO2/CH3OH) = +0.02 V (iv) It = nF n = (4.34)(60 x 60) / 96500 = 0.1619 mol Volume of CO2 produced = 0.1619 ÷ 6 × 24 = 0.648 dm3 (v) The concentration of oxygen is higher. This will shift the position of equilibrium of O 2(g) + 4H + + 4e −  2H2O(l) to the left, causing its electrode potential value to become more positive. This in turns causes the increased/more positive electromotive force generated. (c) (i) The enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions. (ii) Bonds broken = 3 × E(CH) + E(CO) + E(OH) + 3/2 × E(OO) = 3(410) + 360 + 460 + 3/2(496) = 2794 kJ mol−1 Bonds formed = 2 × E(CO) + 4 × E(OH) = 2(805) + 4(460) = 3450 kJ mol−1 Enthalpy change = 2794 − 3450 = −656 kJ mol−1 (iii) The enthalpy change calculated from bond energies assumes that the all species exist in gaseous state, but the standard states of methanol and water are liquid. (iv) CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)

4 River Valley High School 9476/03/Prelims/26 Ecell = +1.21V G = −nFEcell = −(6)(96500)(1.21) = −701 kJ mol−1 (v) G = H − TS −701 = (−726) − (298)(S) S = −0.0839 kJ mol−1 K−1 (d) (i) Addition polymerisation (ii) Step 1: Step 2: Step 3: (iii) Let the number of repeat unit be n Formula: Mr = (2)[(1.0)+(32.1)+(3)(16.0)]+2(16.0)+n[(2)(12.0)+4(19.0)] = 13426 226.2 + 100n = 13426 n = 132

5 River Valley High School 9476/03/Prelims/26 [Turn over (iv) The termination of the polymer chain happens randomly when any two free radicals collide. Therefore, the resulting polymer chains have varying lengths and hence a range of Mr values. (v) Increase reaction time with higher concentration/ amounts of monomers. Reduce/ Limit amount of initiator. (vi) The CF and C C bond energy is strong and requires a significant amount of energy to overcome. [Total: 24] 3 (a) (i) (ii) Bond angle: 117 (b) Both hex-1-ene and 3,3 -dimethylbut-1-ene have simple covalent structures . Hex-1-ene is a straight chain isomer while 3,3 -dimethylbut-1-ene is a branched isomer. Since hex -1-ene has a larger surface area , more energy is needed to overcome the more extensive instantaneous dipole -induced dipole interactions between its molecules, giving rise to its higher boiling point. (c) J undergoes oxidation with ozone to form K and L.  J is an alkene/ has CC bond. J, K and L undergo oxidation with alkaline I2(aq).  They have –COCH3 or –CH(OH)CH3 groups. L undergoes condensation with 2,4-DNPH.  L is a ke