RI_H1_CHEM_H1_ANS

Answers to H1 Chemistry Prelims Paper 1 1. B 6. A 11. D 16. C 21. D 26. C 2. C 7. D 12. A 17. B 22. B 27. B 3. B 8. A 13. D 18. C 23. C 28. A 4. C 9. C 14. D 19. B 24. C 29. C 5. C 10. B 15. B 20. A 25. D 30. A

Suggested Answers to H1 Chemistry Prelims Paper 2 Section A 1. (a) Dot and cross diagram for hydrazine: Shape of molecule: Trigonal pyramidal Bond angle of H − N− H: 107 o (b) N2H5 +(aq) + H 2O (l) /horizontalharpoonextender/horizontalharpoonextender⇀ /horizontalharpoonextender ↽/horizontalharpoonextender/horizontalharpoonextender /horizontalharpoonextender N2H4(aq) + H 3O+(aq) N2H5 + is able to hydrolyse in water to give acidic protons / N2H5 + is the conjugate acid of N 2H4. (c) NO 3 − (aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l) N2H4(l) → N 2(g) + 4H +(aq) + 4e − Overall balanced equation: 4NO 3 − (aq) + 4H+(aq) + 3N2H4(l) → 4NO(g) + 8H2O(l) + 3N2(g) (d) Enthalpy change of reaction = [4( − 394) + 12( − 242)] – [5( − 20) + 4(+53)] = − 4590 kJ mol − 1 (3 s.f.) (e) Diphosphine molecules experience permanent dipole-permanent dipole forces of interactions whereas hydrazine molecules experience intermolecu lar hydrogen bonding. It requires more energy to overcome hydrogen bonding as compared to permanent dipole-permanent dipole forces of attraction. (f) Structure of carbonyl compound: COCH 3 CH 2CHO Type of reaction in Step I: condensation 2. (a) (i) Group I. Largest increase in energy to remove the second o utermost electron . X could be potassium . N ×× ×× o N ×× ×× ×× ×× ×× ×× o H ×× ×× o H o o o ×× ×× o ×× ×× H H 107 o or

(ii) K2O + H 2O → 2KOH or X 2O + H 2O → 2XOH Acceptable answer: pH 12 to 14 . (b) 3. (a) ‘Standard enthalpy change of neutralisation ’ is the heat change when an amount of acid neutralises a base to form 1 mole of water (in dilute aqueous solution) at 298 K and 1 atmosphere pressure. (b) (i) Amount of Ba(OH) 2 added = 30.0 x 10 -3 x 0.40 = 0.012 mol ∴ amount of OH − added = 2 x 0.012 = 0.024 mol Amount of H + added = amount of HC l added = 40.0 x 10 -3 x 0.80 = 0.032 mol ∴ HC l is in excess. Amount of water formed from acid-base reaction = 0.024 mol Heat evolved from reaction = (30.0 + 40.0) x 4.3 x 4.0 = 1204 J Enthalpy change of neutralisation = 1204 1000 0.024 ÷− = − 50.2 kJ mol -1 (sign and correct units) (ii) Heat energy was lost during the experiment and hence, the experimental errors led to the difference in the calculated valu e from the theoretical value. OR • The density of water is used as an approximate to the density of the solution. Ionic Radius Na Mg Al Si P S C l Do not mark for Si. Electrical Conductivity Na Mg Al Si P S Cl 0

• The specific heat capacity of the water is used as an approximate to the specific heat capacity of the