JJC H2 Chem 2012 Prelim P1 Soln

2012 JJ Prelim H2 Chemistry (9647) Paper 1 (9 647/01) Suggested Worked Solutions Page 1 of 11 Suggested Worked Solutions for 2012 JC2 Prelim Paper 1 (9647/01) 1. B 11. A 21. B 31. A 2. C 12. D 22. C 32. D 3. D 13. C 23. A 33. B 4. D 14. D 24. C 34. A 5. B 15. D 25. C 35. B 6. B 16. C 26. C 36. A 7. A 17. B 27. D 37. B 8. D 18. A 28. A 38. A 9. C 19. A 29. C 39. B 10. B 20. D 30. C 40. D 1. Co3+ + e  Co2+ Co  Co2+ + 2e Overall eqn: Co + 2Co3+  3Co2+ For the resulting mixture to contain Co 3+, Co 3+ must be in excess (more than double the amount of Co). When x=1, y=3, Co + 2Co3+  3Co2+ Initial amt/mol 1 3 0 Final amt/mol 0 1 3 Ans: B 2. charge/mass angle of deflection Particle charge/mass 157Gd+ 1 157 = 0.00637 87.6Sr2+ 2 87.6 = 0.0228 Angle of deflection for 87.6Sr2+ = ( 0.0228 0.00637 x 2) = 7.2o. Ans: C 3. Ans: D I II A CH3CH2CH2F Polar molecule. Permanent dipole- permanent dipole forces of attraction between molecules. CH3CH2CHF2 Polar molecule with larger dipole moment than CH3CH2CH2F. Stronger permanent dipole-permanent dipole forces of attraction than that of I. B trans CH3CCl=CClCH3 Non-polar molecule. van der Waals’ forces of attraction between molecules. cis CH 3CCl=CClCH3 Polar molecule. Stronger permanent dipole- permanent dipole forces of attraction than weaker VDW forces of attraction between I molecules. C CH4 van der Waals’ forces of attraction between molecules H2O Stronger hydrogen bonding between molecules than the weaker VDW forces of attraction between I molecules. D CH3CH2COOH More extensive hydrogen bon ds between molecules as they exist as dimers (formation of more extensive hydrogen bonding between 2 RCO 2H molecules) than that of II. CH3CH2CH2OH hydrogen bond between molecules.

2012 JJ Prelim H2 Chemistry (9647) Paper 1 (9 647/01) Suggested Worked Solutions Page 2 of 11 4. Given, 2Y(g) + 6H(g) Y2H6(g) H = –2775 kJ mol1 Y2H6(g)  2Y(g) + 6H(g) H = +2775 kJ mol1 Hr = E(YY )+6E(YH) E(YY) = +2775 6(395) = + 405 kJ mol–1 Ans: D 5. Using the formula, H rxn=H f(products)  H f (reactants) H r = 2H f(H2O)  2H f(H2O2) = 2(–286) –2(–188) = –196 kJ mol–1 Note: H f(O2) = 0 kJ mol–1as O2 is an element. Ans: B 6. Let x be the volume of NOCl(g) that reacted. NOCl (g) ⇌ NO(g) + ½ Cl2(g) Eqm volume/cm3 100x x ½x Total volume of gases at eqm = (100+½x) cm3 % of NO at eqm = x 100 +1/ 2x x 100 % = 40 % Solving, x=50 cm3 Hence, total volume of gases at eqm =100 + ½(50) = 125 cm3 Ans: B 7. HPO4 2−(aq) + H2BO3 −(aq) ⇌ H2PO4 −(aq) + HBO 3 2−(aq) base acid H2PO4 − and H2BO3 − are acids and HBO3 2− and HPO4 2− are bases. Hence option C and D are wrong. Since the equilibrium constant < 1.0, there will be a higher [ HPO4 2−(aq)] and [H 2BO3 −(aq)] and a lower [H2PO4 −(aq)] and [HBO3 2−(aq)] at equilibrium.  Since [H2PO4 −(aq)] < [H 2BO3 −(aq)]  H2PO4 − ionises to a greater extent than H 2BO3 − to give H +. Hence H2PO4 − is a stronger acid than H2BO3 −  Since [HBO3 2−(aq)] < [HPO 4 2−(aq)]  HBO3 2− ionises to a gr