NYJC H1 CHEM P1 P2 Ans

Nanyang Junior College J2 H1 Chemistry 2017 1 2017 J2 H1 Chemistry Prelim Answers Paper 1 Answer Key 1 B 6 A 11 A 16 D 21 A 26 C 2 C 7 B 12 B 17 A 22 D 27 B 3 D 8 C 13 C 18 C 23 D 28 A 4 D 9 B 14 A 19 A 24 C 29 B 5 C 10 B 15 B 20 D 25 D 30 C

Nanyang Junior College J2 H1 Chemistry 2017 2 Paper 2 Section A Answers 1 (a) (i) The carbon-carbon single bond in the ethane molecule consists of 1  bond. The carbon=carbon double bond in ethane molecule consists of 1  bond and 1  bond. A  bond is weaker than a  bond due to less effective overlap, hence C=C bond is less than twice of C-C bond. (ii) (iii) Ethanol and butanol differs in the size of the non-polar hydrocarbon chain. Butanol is insoluble in water even though it can form hydrogen bonds with water. Its predominantly forms dispersion forces with water due to its long, non-polar hydrocarbon chain. The energy released during formation of these dispersion forces is not enough to overcome the hydrogen bonds between water molecules and the dispersion forces between butanol. (b) (i) (ii) The delocalised  electron cloud results in stability, so the loss of this aromatic character is not energetically favored. Instead, benzene tends to undergo substitution reactions so that its  electron cloud remains intact to maintain aromatic stability. (iii) Benzene do not undergo oxidation. Methylbenzene is oxidized by heating with KMnO 4, H2SO4 (aq)

Nanyang Junior College J2 H1 Chemistry 2017 3 (c) Bond energy of C-H bond = 410 kJ mol-1 Bond energy of C-Cl bond = 340 kJ mol-1 As fluorine atom is smaller than chlorine, bond length of C-F bond is shorter than C-Cl bond. Therefore C-F bond is expected to be stronger than C-Cl bond, hence they do not break easily to form free fluorine atoms to attack the ozone layer. 2 (a) (i) CrO 42– + 8H+ + 3e  Cr3+ + 4H2O (ii) Amount of electrons gained by 0.0150 mole of CrO 42– = 0.0450 mol Amount of electrons lost by 0.0225 mole of SO2 = 0.0450 mol Therefore each mole of SO 2 lost = (0.0450/0. 0225) = 2 mol of electron Original oxidation number of sulfur = +4 New oxidation number of sulfur = +4 + 2 = +6 (iii) SO 42– (b) (i) 2- 27 2- 2 + 2 4 [Cr O ]Kc = [CrO ] [H ] (ii) Amount of PbCrO4 initiallly = 20.00/ (207.2+52.0 +16.0x4) = 0.06188 mol [CrO42–] initially = 0.06188/(100/1000) = 0.6188 ≈ 0.619mol dm-3 (iii) [CrO 42–] at equilibrium = 0.6188 / 5 = 0.1237 = 0.124 mol dm-3 [Cr2O72–] at equilibrium = 0.6188 0.1237 2  = 0.2475 ≈ 0.248 mol dm-3 (iv) Since KC = 7.55 x 1012 mol–3 dm9 2- + -6 -327 2- 2 2 12 4 [Cr O ] 0.2475[H ] = = = 1.463 x 10 mol dm [CrO ] x Kc 0.1237 x 7.55 10 x pH = 5.8 (v) By Le Chatelier’s Principle, the system will react to reduce the added amount of NaOH. Backward reaction is favoured and position of equilibrium shifts to the left. [1] The solution will appear yellow in color due to formation of aqueous CrO42–.

Nanyang Junior College J2 H1 Chemistry 2017 4 3 (a) Place a lighted splint near the mouth of the test tube. A pop sound would be heard to confirm its identity as hydrogen gas. (b) (i) From orange to green (ii) Oxidation (c) (i) X: CH3CH2CH2OH Y: CH3CH2CHO (ii) CH3CH2CHO + 2Cu2+ + 5OH  CH3CH2CO2- + Cu2O + 3H2O (iii) Test: Add Tollens’ to solution and warm. Observation: Silver mirror observed with Y. No silver mirror with X. Test: Add 2,4-DNPH to solution and warm Observation: orange precipitate formed with Y. No orange precipitate with X. 4 (a) Energy change when one mole of a substance is completely burnt in excess oxygen under standard conditions. (b) (i) Carbon: C (s) + O 2 (g) ¹ CO2 (g) Methane: CH4 (g) + 2O2 (g) ¹ CO2 (g) + 2H2O (l) (ii) 1 MJ = 106 J = 1000 kJ Since ∆H C (C) = - 394 kJ mol-1, amount of carbon need to be burned to produce 1 MJ of heat = 1000 / 394 = 2.54 mol Since ∆HC (CH4) = - 890 kJ mol-1, amount of methane need to be burned to produce 1 MJ of heat = 1000 / 890 = 1.12 mol (iii) Since efficiency of coal power station = 40% , amount of carbon need to be burned to produce 1 MJ of heat = 2.538 / 0.40 = 6.35 mol Since efficiency of natural gas power station = 51% , amount of methane need to be burned to produce 1 MJ of heat = 1.124 / 0.51 = 2.20 mol (c) Mass of C (s) need to be burn to produce 1 MJ of electrical energy = 6.35 x 12.0 = 76.2 g Mass of ash produced = 76.2 (5/95) = 4.01 g (d) (i) CO 2 is a greenhouse gas that causes global warming, leading to droughts and rising sea levels.

Nanyang Junior College J2 H1 Chemistry 2017 5 (ii) Coal is found in the solid state, which is easier to store and transport. Hence it is easier and cheaper to operate power station that burn natural gas which is harder to store and transport. (e) (i) Amount of methane need to be bur ned to produce 1 MJ of heat = 1.124 mol Volume of methane at rtp = 1.124 x 24.0 = 27.0 dm3 (ii) CO + ½ O2  CO2 ∆HC = -283 H2 + ½ O2  H2O ∆HC = -242 Since 1 mole of water-gas contain ½ mole of CO and ½ mole of H2, amount of heat energy produced by 1 mole of water-gas = ½ (283) + ½ (242) = 262.5 kJ Therefore amount of water-gas needed to produce 1 MJ of heat energy = 1000/262.5 = 3.810 mol Volume of water-gas at rtp = 3.810 x 24.0 = 91.4 dm 3 (iii) - Volume of methane needed to be burn to produce 1 MJ is lower, hence it is safer and easier to operate a power station using natural gas. Paper 2 Section B Answers 5 (a) n(MnO 4–) = 0.0200 × 27.30 / 1000 = 5.46 × 10–4 mol n(Fe2+) in 25.0 cm3 = 5.46 × 10–4 × 5 = 2.73 × 10–3 mol n(Fe 2+) in 250 cm3 = 2.73 × 10–3 × 250 / 25.0 = 2.73 × 10–2 mol mass of FeCO 3 = 2.73 × 10–2 × 115.8 = 3.161 g percentage by mass of FeCO 3 = 3.161 / 5.00 × 100% = 63.2% (b) (i) CH3CH2CH2O¯ is the least stable as the electron donating alkyl (−CH2CH2CH3) group on the CH 3CH2CH2O¯ ion increases the electron density on the oxygen atom, making it even more negative, hence destabilising the CH 3CH2CH2O¯ ion. Thus propanol is less acidic than propanoic acid. CH 3CH2COO¯ is the most stable as the p orbital of the oxygen atom overlaps with the π electron cloud of the –C=O bond and the lone pair of electrons on the oxygen atom delocalise into the –C=O. The negative charge is dispersed over the carbon atom and the two electronegative oxygen at oms, stabilising the CH 3CH2COO¯ ion. Thus propanoic acid is more acidic than propanol. (ii) It is more difficult to remove a proton from an anion.

Nanyang Junior College J2 H1 Chemistry 2017 6 (iii) A buffer solution is a solution that is able to maintain a fairly constant pH when a small amount of acid or base is added. (iv) HO2CCH2CO2Na + H+ ¹HO2CCH2CO2H + Na+ HO2CCH2CO2Na + NaOH ¹NaO2CCH2CO2Na + H2O (c) (i) Na2O (s) + H2O (l)  2NaOH (aq) pH = 13