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 greater extent than to give OH . Hence HBO3 2− is a stronger base than HPO4 2−. Strength of acids: H2PO4 − > H 2BO3 − Strength of bases: HBO3 2− > HPO 4 2− Ans: A 8. BaL2(s) ⇌ Ba2+(aq) + 2L−(aq) Given Ksp = [Ba2+(aq)][L−(aq)]2 = q mol3 dm−9 Let [L−] at equilibrium be x mol dm−3  Ksp = x 2 (x)2 = q  3x 2 = q  x = (2q) 3 1 = [L−] at equilibrium Ans: D 9. Given t1/2 = 2.0 hour After first half-life, mass of X = ½(320) = 160 mg After the second half-life, mass of X = ½(160) = 80 mg After the third half-life, mass of X = ½(80) = 40 mg Hence, drug X will still be effective after 3 half- lives (3 x 2.0 = 6.0 hour). Ans: C conjugate acid of HPO4 2− conjugate base of H2BO3 −

2012 JJ Prelim H2 Chemistry (9647) Paper 1 (9 647/01) Suggested Worked Solutions Page 3 of 11 10. Rate = k[H2O2][I], For expt 1: Rate = k’[H2O2] where k’ = k[I] The graph of [H2O2] against time is a downward sloping curve with decreasing gradient. For expt 2: Rate = k’ where k’ = k[I][H2O2] The graph of [H+] against time is a decreasing straight line with a constant gradient. Ans: B 11. EAg+|Ag = +0.80 V, EZn2+|Zn = 0.76 V, ECu2+|Cu = +0.34 V A : () E(Ag+/Ag) is more positive than E(Cu2+/Cu)  Cu has the greater tendency to be oxidised than Ag. Hence, silver is less electropositive (tendency to lose e undergoes oxidation) than copper. B : () E(Ag+/Ag) is more positive than E(Zn2+/Zn)  Zn has the greater tendency to be oxidised than Ag. Hence, zinc displaces silver from a solution containing silver ions. C : () E(Zn2+/Zn) is more negative than E(Cu2+/Cu)  Zn2+ has the lower tendency to be reduced than Cu2+. D : () E(Ag+/Ag) is more positive than E(Zn2+/Zn)  Zn has the greater tendency to be oxidised than Ag. Hence, zinc has a higher tendency than silver to form positively charged ions. Ans: A 12. Result Negative terminal (anode) Positive terminal (cathode) Voltage (V) (1) Pb x 0.35 (2) y Pb 1.10 (3) z Pb 2.60 Using (1), Pb undergoes oxidation and x undergoes reduction.  The reducing po wer of Pb is higher than that of x. Using (2) and (3), both y and z undergo oxidation and Pb undergoes reduction.  The reducing power of y and z is higher than that of Pb. Since the voltage recorded in (3) is higher than that in (2), the reducing power of z is higher than that of y.  from weakest to strongest reducing power: x, Pb, y, z Ans: D 13. A : () Reactivity of Group II elements with cold water increases down the group. {Mg reacts slowly; Ca reacts readily; Ba & Sr react vigorously.} M(s) + 2H 2O(l)  M(OH)2(aq) + H 2(g) , where M = Ca, Sr and Ba B : () Solubility of Group II hydroxides and oxides increases down the group. As a result, the pH of the resul ting hydroxide solutions increases {from pH 9 in Mg(OH) 2 to pH 12 in Ca(OH) 2 to pH 14 in Ba(OH)2}. MO(s) + H 2O(l)  M(OH)2(aq), where M = Ca, Sr and Ba MgO(s) + H 2O(l)  Mg(OH)2(aq) C : () Second ionisation energy of elements decreases from Be to Ba. This is due to an increase in ionic radii of M +(g) and the increase in shielding effect which outweighs the effect of increasing nuclear charge. D : () Down the group, thermal stability of Group II carbonates increases. MCO3(s)  MO(s) + CO 2(g) As the ionic radius of the M 2+ increases down the Group, the charge density of the M 2+ decreases. The ability of M2+ to polarise the large CO3 2–anion decreases down the group. Ans: C

2012 JJ Prelim H2 Chemistry (9647) Paper 1 (9 647/01) Suggested Worked Solutions Page 4 of 11 14. A : () Atomic radius of At is larger than Cl and AtAt bond is longer and weaker than C lCl bond and so At 2 dissociates more readily than Cl2. B : () It reacts explosively with hydrogen. X2(g)+H2(g) 2HX(g),Hr (where X = C l, Br and I ) Hr = E(XX) + E(HH) –2E(HX) As HX bond energy decreases,Hr becomes less exothermic(less energetically feasible).  Reactivity of X2 with H2 decreases down the group from Cl2 to I2. C : ( ) Down the group, E(X2/X–) becomes less positive, indicating that the o xidising power decreases down the group. Thus At2 will not be able to oxidise I to I2. D : ( ) Volatility of halogens (X 2) decreases down group VII from C l2 to I2. There is an increase in strength of van der Waals' forces between X 2 molecules as the no. of electrons per X 2 molecules increases from C l2 to At 2. Thus At2 exists as crystalline solid. Ans: D 15. From the large drop in 1 st ionisation energy from M to N, we can deduce that M is the l ast element in the 2 nd Period (Ne) and N is the first element in the 3rd Period (Na). A : () R is Si. It has the highest melting point. (Giant covalent structure, with strong SiSi covalent bonds) B : ( ) J is nitrogen and K is oxygen. Both N 3 and O2 are isoelectronic, thus as nuclear charge increases from N3 and O 2, while the shielding effect by inner shell electr ons remains relatively constant, the ionic radius of N3 (J) is larger than that of O2 (K). C : ( ) Q is aluminium. Al forms an insolub le hydroxide, which dissolves in excess dilute