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 solution. (c) (i) (ii) Enzymes are proteins which act as biological cataly sts which speed up the reaction by providing an alternative pathway that lowers the activation energy of the reaction such that more molecules now have energy greater than or equal to Ea’ (catalysed reaction). 4. (a) The C=C double bond restricts free rotation about t he double bond. There are 2 different adjacent groups attached to each of the doubly bonded carbon. (b) (i) Step I – reduction (ii) CH 2 C NH 2 H C OH H CN (iii) HCN is a weak acid which dissociates partially: HCN(aq) /horizontalharpoonextender/horizontalharpoonextender/horizontalharpoonextender ⇀ ↽ /horizontalharpoonextender/horizontalharpoonextender /horizontalharpoonextender H+(aq) + CN − (aq) Adding a trace amount of base will remove H + and shifts the position of the equilibrium to the right to increase the concen tration of CN − . The rate increases with increasing concentration of CN − . Ea =79 kJ mol -1 ∆ H = - 98 kJ mol -1 H2O + ½ O 2 H2O2 Reaction Coordinate Energy

5. (a) CH 3 CH 2CH 2Br CH 3 CH 3 CH 3 COCH 3 CH=CH 2 CH(OH)CH 3 (b) CH 3 CH 2CH 2Br CH 3 CH 2CH 2 ethanolic NaOH, heat 1) Cold conc. H 2SO 4 2) Warm with H 2O Do not award marks if KMnO 4 is used. K2Cr 2O7, H2SO 4, heat FeBr 3 Or any other suitable catalyst + HBr

Section B 6. (a) (i) Lattice energy of NaC l = ∆ H1 - ∆ H2 - ∆ H3 - ∆ H4 - ∆ H5 = − 776 – (109) – (244) – 494 – ( − 364) = − 1260 kJ mol − 1 (ii) ∆ H1 – Enthalpy change of formation of NaC l(s) ∆ H4 – First ionisation energy of Na(g) (iii) Lattice energy ∝ q q r r + − + − × + The lattice energy of MgC l2 will be more exothermic compared to that of NaC l. This is because the Mg 2+ ion has a higher charge and smaller ionic radius/ higher charge density compared to Na + ion. (b) Empirical fomula of P is C 4H9Br, since Mr is 136.9, the molecular fomula is C4H9Br. Explanation: • P reacts via substitution with aq. sodium hydroxide to give Q. Q is likely to be an alcohol. • Q on oxidation gives R, which forms orange ppt with 2,4- dinitrophenylhydrazine via condensation. R is a carbonyl compound. • Positive iodoform test with Q and R. Q has the structure –CH(OH)(CH 3) while R has the structure : C CH 3 O • P undergoes elimination with hot ethanolic NaOH to give V, W and X. V, W and X are likely to be alkenes. Structure of P: CH 3CH 2 CH CH 3 Br Structure of Q: CH 3CH 2 CH CH 3 OH

Structure of R: CH 3CH 2 C CH 3 O Structure of V, W and X: C C H CH 2CH 3 H H C C CH 3 H H CH 3 C C CH 3 HH CH 3 (c) (i) Test: Add acidified KMnO 4 to both compounds and heat. Bubble any gas evolved into Ca(OH) 2(aq). Observations: CH 2CH 3 will produce CO 2 gas which gives a white ppt with Ca(OH) 2(aq). (ii) Test: Add aqueous alkaline iodine to both compounds and heat. Observations: O O CH 3 will give a yellow ppt with aqueous alkaline iodine. 7. (a) (i) Weak acids dissociate partially in water. (ii) 3 2 3 [HCO ( )][H ( )] [H CO ( )] a aq aq K aq − + = ; mol dm − 3 (iii) Addition of H + : HCO 3 – + H +  → H 2CO 3 Addition of OH – : H2CO 3 + OH –  → HCO 3 – + H 2O Cis & Trans Isomers

With a large reservoir of H 2CO 3 and HCO 3 – present, since the amount of H+ and OH – added is small, there will be insignificant change s in [H 2CO 3] and [HCO 3 – ]. Hence pH is maintained. (iv) 3 2 3 [HCO ( )][H ( )] [H CO ( )] a aq aq K aq − + = = 7.94 × 10 -7 - 3 2 3 [HCO ] =19.95 [H CO ] (or ratio 20:1) (v) Yes. Kc changes with temperature and an increase in temper ature will cause the position of the equilibrium to shift, therefore changing the concentrations of the ratio of hydrogen carbonate i on and carbonic acid in plasma. (b) (i) The melting points from Na to A l are high due to their giant metallic structures and melting involves overcoming strong m etallic bonds. The melting point increases from Na to A l as the strength of the metallic bond increases from Na to A l due to an increasing number of mobile valence electrons used in bonding. Si has the highest melting point due to its giant m olecular structure. A large amount of energy is required to break the str ong covalent bonds [1] in Si. P 4, S 8, Cl 2 and Ar have lower melting points since they are si mple molecules with weak van der Waals’ forces between molecules. The strength of the van der Waals’ forces increases wit h the size of the electron cloud in the order of Ar, C l2, P 4, S 8. (ii) Ga 2O3 has a giant ionic structure. Ga 2O3 is amphoteric. Ga 2O3(s) + 6HC l(aq) → 2GaC l3(aq) + 3H 2O( l) Ga 2O3(s) + 2NaOH(aq) + 3H 2O( l) → 2NaGa(OH) 4(aq) (c) No. of mol of C l atoms in MC l3 = No. of mol of AgC l formed = 33.00 1000 ×