2024 Prelims RVHS H2 Chem P3 (Ans)

River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination [Turn over River Valley High School 2024 JC 2 H2 Chemistry 9729 Prelim Paper 3 Suggested Solutions 1 (a) (i) Dynamic equilibrium refers to a reversible process at equilibrium in a closed system, in which the rate of forward reaction is equal to the rate of backward reaction, where there is no net change in concentration of reactants and products. (ii) 𝐾𝑐 = [𝐻2][𝐼2] [𝐻𝐼]2 (iii) Initial amount of HI = 15.4 = 0.1204 mol127.9 [HI]initial = 30.1204 = 0.06020 mol dm2.00 − [HI]change = 30.06020 0.34 = 0.02047 mol dm − 2HI(g) = H2(g) + I2(g) []initial/ mol dm−3 0.06020 0 0 []change/ mol dm−3 − 0.02047 + 0.010235 + 0.010235 []eqm/ mol dm−3 0.03973 0.010235 0.010235 ( ) ( ) 2 c 2 0.010235 K = = 0.0664 (to 3 s.f.) 0.03973

2 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination (iv) The KC value will increase. [✓] Position of equilibrium will shift right [✓] to favour the forward endothermic reaction [✓] to absorb the some of the heat. [✓]. (Hence the concentration of hydrogen and iodine will increase while the concentration of hydrogen iodide will decrease.) (b) Benzylamine is the most basic because the electron-donating benzyl group (-CH2C6H5) makes the lone pair of electrons on the N atom the most available for donation/protonation. Phenylamine is the less basic than benzylamine because the lone pair of electrons on the N atom delocalised into the the benzene ring due to the overlap of the p-orbital of the N atom with the -electron cloud of benzene. 3-aminobenzonitrile is the least basic because the presence of the additional electron withdrawing nitrile (CN) increases delocalisation of the lone pair of electrons on the N atom into the benz ene ring. This causes the lone pair of electrons to be least available for donation/protonation.

3 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination [Turn over (c) (i) (ii) Electrophilic substitution Fe + 3/2 Cl2 → FeCl3 FeCl3 + Cl2 → FeCl4− + Cl+ [✓]

4 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination (d) (i) Hc = mc T n − Hc = (0.600 1000)(4.18)(40.0 25.0) 3.0078.0 −− Hc = 1978.12 = 978 kJ mol (to 3 s.f.)−−− (ii) Copper is a good conductor of heat and will - maximise heat transfer to the water. - allow more efficient transfer of heat. Any logical answer that points to maximising transfer of heat to water Copper will allow the calorimeter to withstand high pressure of the oxygen pump in.

5 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination [Turn over (iii) The apparent value obtained is less exothermic. [✓] The layer of black solid is soot/ carbon [✓] which suggest that incomplete combustion [✓] of benzene took place. 2 (a) (i) Na2O(s) reacts vigorously with water to give a strongly alkaline solution of pH = 12 to 14. Na2O(s) + H2O(l) → 2NaOH(aq) P4O10(s) reacts vigorously with water to give an acidic solution of pH = 2. P4O10(s) + 6H2O(l) → 4H3PO4(aq) OR P4O6(s) reacts vigorously with water to give an acidic solution of pH = 2. P4O6(s) + 6H2O(l) → 4H3PO3(aq)

6 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination (ii) Divide the powder into two portions. To one portion, add H2SO4(aq). To the other portion, add (concentrated) NaOH(aq). If the powder reacts with acid only, it is MgO. MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l) If the powder reacts with both acid and alkali, it is Al2O3. Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l) Al2O3(s) + 2OH−(aq) + 3H2O(l) → 2[Al(OH)4]−(aq) If the powder reacts with alkali only, it is SiO2. SiO2(s) + 2OH−(aq) → SiO32−(aq) + H2O(l) Marker’s comments: • Many candidates tried to distinguish the oxides by dissolving them in water, not realising that it would be ineffective since MgO is sparingly soluble while Al2O3 and SiO2 are insoluble. • Some candidates tried to use litmus or universal indicator to distinguish the oxides. • A significant number of candidates wrongly assumed that the oxides would dissociate into aqueous ions and a cation test could be performed with dropwise/excess NaOH(aq). (b) (i) Lattice energy is the enthalpy change when one mole of MgSiF6 solid is formed from gaseous Mg2+ and SiF 62– ions under standard conditions. OR Mg2+(g) + SiF62–(g) → MgSiF6(s) H⦵ = lattice energy of MgSiF6 Marker’s comments: • A significant number of candidates failed to define lattice energy in the context of magnesium silicofluoride. • The phrase “under standard conditions” was commonly missing.

7 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination [Turn over (ii) By Hess’ Law, −2580 = +148 + 456 + 2(+158) + 736 + 1450 − 4(+565) + (−639) + LE(MgSiF6(s)) LE(MgSiF6) = −2787 = −2790 kJ mol–1 Marker’s comments: • Common errors include revers ing the enthalpy change for bond energy, and wrong multiplying factors for enthalpy changes. • Some candidates left out or gave wrong state symbols for chemical species.

8 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination (c) (i) Between 300 K and 400 K, ½Ti(s) + Cl2(g) → ½TiCl4(l) S < 0 due to a decrease in disorder as the amount of gaseous molecules decreases from 1 to 0 mol . Since H and S are both negative, −TS becomes more positive with increasing temperature . Hence the graph for reaction 1 becomes more positive between 300 K and 400 K. Between 400 K and 2000 K, ½Ti(s) + Cl2(g) → ½TiCl4(g) S  0 / is less negative/ has a smaller magnitude as the amount of gaseous molecules decreases from 1 to ½ mol , so −TS remains relatively constant with increasing temperature . Hence the graph for reaction 1 becomes almost independent of temperature from 400 K to 2000 K. Marker’s comments: • This part proved to be a challenge. • Most candidates did not appreciate how S varied when TiCl4 underwent a phase change, and did not further explain in terms of amount of gaseous molecules. • Most candidates could not relate −TS to the shape of graph. • Common misconception: Entropy of TiC l4 increased as it boiled, and remained constant beyond boiling point. (ii) To prevent the hot titanium extracted from oxidising back to TiO2. Or To prevent the hot magnesium from getting oxidised to MgO. Marker’s comments: • Many candidates gave an ambiguous answer, pointing out that argon was an inert gas due to its stable octet configuration and would not interfere in the reaction. (iii) ½TiCl4 + Mg → ½Ti + MgCl2 Marker’s comments: • Some candidates failed to deduce that they should combine the equations for reaction 1 and 2. These candidates often included irrelevant species like Cl2 and TiCl2.

9 River Valley High School 9729/03/PRELIMS/24 2024 Preliminary Examination [Turn over (iv) ½Ti + Cl2 → ½TiCl4 G at 1110 K = −350 kJ mol−1 Mg + Cl2 → MgCl2 G at 1110 K = −460 kJ mol−1 G = −460 − (−350) = −110 kJ mol−1 Note: G will be doubled if the stoichiometric coefficient in previous part is doubled. Marker’s comments: • Some candidates failed to deduce that they should combine the equations for reaction 1 and 2. • Some candidates misread G values from the graph. • Among the candidates who managed to read off the values correctly, it was common for them to forget to double G for the reaction when their stoichiometric coefficients in (c)(i ii) were doubled. (v) 1870 K (when T > 1870 K, G > 0) Marker’s comments: • Some candidates either left this part blank or misread the value from the graph axis.

10 River Valley High School 9729/03/PRELIMS/24 2024 P