NYJC Prelim P1 P2 P3 P4 Ans

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 1 2017 H2 Chemistry Prelim Answers Paper 1 Answer Key 1 B 6 D 11 A 16 D 21 D 26 C 2 D 7 B 12 C 17 A 22 B 27 A 3 C 8 A 13 C 18 B 23 A 28 A 4 C 9 A 14 D 19 D 24 C 29 A 5 C 10 A 15 D 20 B 25 B 30 D

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 2 Paper 2 Answers 1 (a) (i) Shape of ammonium ion: tetrahedral Shape of nitrate ion: trigonal planar (ii) ΔHsol = -(-Lattice Energies) + (-ΔHhyd) = |L.E.| - | ΔHhyd | |L.E.| ∝ ||qq rr   | ΔHhyd | ∝ ||q r Anionic radius of nitrate ion is larger than chloride, therefore the decrease in | ΔH hyd | of nitrate ion is larg er than the decrease in |L.E.|. ΔHsol is expected to be more endothermic. (b) (i) Assuming a temperature change of 5 oC and no heat loss to surroundings, n(salt) x 15 000 = 100 x 4.3 x 5 n(salt) = 0.1433 mol minimum mass = 0.1433 x 53.5 = 7.67 g (ii) 4.0 ΔT

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 3 (iii) ΔGo = ΔHo – TΔSo ΔGo is negative since reaction is spontaneous and ΔH o is positive since reaction is endothermic. Therefore sign of ΔS o is positive as there are more ways to arrange the particles when the solid dissolves in water. (c) (i) Since Zn is oxidised, Zn(s) → Zn2+(aq) + 2 e− Overall equation: Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + ZnCl2(aq) + 2NH3(aq) + H2O(l) (ii) E°cell = (+0.5) – (-0.76) = +1.26V (iii) ΔGo = - nFEo = - 2 x 96500 x 1.26 = - 243 000 J mol-1 = - 243 kJ mol-1 The sign of ΔGo is negative and hence the reaction is spontaneous. 2 (a) (b) (i) Quenching is required to stop or slow down the reaction so as to achieve a more accurate titre value at that time or to find the concentration at that instance. Quenching agent: large volume of cold water / add large volume of acid to remove the NH3 (in this question). (ii) Cl2 + 2I−  2Cl− + I2 (iii) I2 + 2S2O32−  2I− + S4O62− (iv) Starch indicator is added when the solution turns pale yellow. The end -point can be recognised when one drop of sodium thiosulfate added cause the dark blue solution to permanently turn colourless.

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 4 (c) (i) (ii) [2-iodobutane] = 0.20 / 2 = 0.10 mol dm-3 n(2-iodobutane) : n(I−) : n(S2O32−) = 1 : 1 : 1 n(2-iodobutane) = n(S2O32−) = 10 1000 × 0.10 = 0.001000 mol V(S2O32−) required when all 2 -iodobutane reacted = 0.001000 0.0250 = 0.04000 dm3 = 40.00 cm3 When volume of sodium thiosulfate increases from 0 to 20 cm3, 1st t1/2 = 9.5 min . When volume of sodium thiosulfate increases from 20 to 30 cm3 2nd t1/2 = 9.5 min. Since the 1st t1/2 is approximately equal to 2nd t1/2, it is 1st order with respect to 2-iodobutane. (iii) Since the rate for 2.00 mol dm−3 of ethanolic ammonia reaction is half the rate for 4.00 mol dm −3 of ethanolic ammonia reaction , it is 1st order with respect to ethanolic ammonia. (iv) rate = k [2-iodobutane][ethanolic ammonia] 30.0 × × × × × × × × 30 40 0 10 20 Volume of sodium thiosulfate used / cm3 time / min 1st t1/2 = 9.5 min 2nd t1/2 = 9.5 min 3rd t1/2 = 9.5 min 0 5.0 10.0 15.0 20.0 25.0 30.0 × ×

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 5 (v) Nucleophilic Substitution (SN2) (vi) After mixing equal volume of 0.20 mol dm −3 2-iodobutane and 4.00 mol dm−3 ethanolic ammonia, [2 -iodobutane] = 0.10 mol dm −3 and [ethanolic ammonia] = 2.00 mol dm−3. Since [ethanolic ammonia] is in large excess, rate = k’ [2-iodobutane], where k’ = k [ethanolic ammonia] t1/2 = ln 2 k′ = ln 2 k [ethanolic ammonia] 9.5 = ln 2 k (2.00) k = 0.0365 [1] mol−1 dm3 min−1 (vii) I. 2-iodobutane is replaced with 2-chlorobutane. BE(C−Cl) = 340 kJ mol−1 BE(C−I) = 240 kJ mol−1 As more energy is required to overcome the stronger C−C l bond, the rate of reaction decreases. II. ethanolic ammonia is replaced with ethanolic ethylamine. Electron donating ethyl group increases the electron density around the nitrogen atom of ethylamine, making the lone pair of electrons more available (or make the nitrogen more nucleophilic) to attack the electrophilic carbon. Hence, rate of reaction increases.

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 6 Energy 3 (a) From Cr to Co,  number of protons increases, nuclear charge increases  additional electron is added to the penultimate 3d subshell.  Hence, screening/shielding effect also increases as presence of the 3d orbital shields the 4s electrons from the nuclear attraction.  The effective nuclear charge experience by the outer 4s electrons increases only very gradually.  Energy required to remove the 4s electron is relatively invariant. (b) (i) Ksp = [Cr3+][OH-]3 6.3 ×10–31 = (0.010/2)[OH-]3 [OH-] = 5.013 x 10-10 mol dm-3 (ii) Ksp = [Fe3+][OH-]3 4 × 10–38 = [Fe3+](5.013 x 10-10)3 [Fe3+] = 3.174 x 10-10 mol dm-3 << 0.005 mol dm-3 hence effective (iii) To the solution of Cr3+ and Fe3+ ions, add sodium hydroxide until excess. Cr(OH)3 ppt forms and is soluble in excess. Cr3+ + OH-  Cr(OH)3 Cr(OH)3 + 3OH-  [Cr(OH)6]3- Fe(OH)3 ppt is insoluble in excess sodium hydroxide. Fe3+ + OH-  Fe(OH)3 Filter the mixture and Fe(OH)3 is the residue and [Cr(OH)6]3- is the filtrate. (c) (i) The electrons in orbitals that lie along the same axes as the ligands experiences greater repulsion, hence the energy is raised. (ii) In the presence of ligands, the 3d orbitals split into 2 groups with an energy gap. When visible light passes through the iron complex, the violet wavelength of light corresponding to the energy gap is absorbed by the 3d electron in the lower energy level. This electron is promoted to a vacant 3d orbital at the higher energy level. The complementary colour, corresponding to unabsorbed wavelengths is observed.

Nanyang Junior College 2017 H2 Chemistry Prelim Answers 7 4 (a) At low pressure, when volume increases, pressure of chloromethane falls more than that of ideal gas as permanent dipole -permanent dipole interaction between CH3Cl molecules hold the particles closer together, hence they strike the walls of the container with less force, resulting in lower pressure. OR At low pressure, volume of chloromethane gas is lower/decreases more than ideal gas for a given pressure as permanent dipole -permanent dipole interaction between molecules is significant and the molecules are attracted closer to each other. (b) (i) Electrophilic Substitution. (ii) A: CH2 CH CH3CH3 C O CH3 [1] B: CH2 CH CH3CH3 C CH2 CN or CH2 CH CH3CH3 C CH3 COOH OH (iii) Reagents and conditions Step I anhydrous AlCl3,(CH3)2CHCH2Cl, room temperature Step IV Al2O3, heat at 350 oC [1] (OR conc H2SO , 170 oC less preferred as hydrolysis of nitrile may occur) Any dilute acid, heat under reflux Step VI H2, Ni catalyst, heat OR H2