SAJC Prelim P1 P2 ANS

2009 H1 CHEMISTRY PRELIMS PAPER 1 MCQ Answer Key 1. B 6. B 11. C 16. B 21. C 26. D 2. C 7. D 12. C 17. A 22. C 27. C 3. A 8. C 13. D 18. B 23. B 28. B 4. B 9. C 14. D 19. D 24. A 29. A 5. A 10. A 15. B 20 B 25. C 30 C 2009 H1 CHEMISTRY PRELIMS PAPER 2 MARKS SCHEME SECTION A: 1. (a) (i) It is the ratio of the average mass of 1 atom of the element to 12 1 the mass of an atom of carbon-12. [1] (ii) From the data booklet, Ar of Ga = 69.7 69.7 69 0.64 (1 0.64) 0.36 69.7 44.16 70.9 x x x       x = 71 Relative isotopic mass of the other Ga isotope = 71 [½] [½] (b) (i) 5 C 2O4 2- + 2MnO4 - + 16H+ → 10CO2(g) + 2Mn2+ + 14H2O [1] (ii) No. of moles of MnO4 - used = 25.60 0.021000  = 5.12 × 10-4 mol No of moles of C2O4 2- in 25.0 cm3 = 4 55.12 10 2   = 1.28 × 10-3 mol No of moles of C2O4 2- in 250 cm3 = no. of moles of H2C2O4 =0.0128 mol Mass of H 2C2O4 in 250 cm3 = 0.0128 × 90 = 1.152 g Percentage purity = 1.152 1001.27  =90.7% [½] [½] [½] [½] 2. (a) Element W belongs to Group VI. There is a large increase between 6 th and 7th ionisation energy, suggesting that the 7th electron is removed from an inner shell. [1] [1] (b) (i) WO2 or WO3 [½]

Covalent bonding [½] (ii) pH = 2 [1] 3. (a) C1: sp2 C2: sp3 (b) Carbon Shape Bond angle C1 Trigonal planar 120° [1] C2 tetrahedral 109° [1] (c) [1] 4. (a) (i) Standard enthalpy change of formation of 2-methylpropene is the enthalpy change when 1 mole of 2-methylpropene is formed from carbon and hydrogen elements under standard conditions. [1] (ii) 4 C (s) + 4 H2 (g) → (CH3)2CH=CH2 (g) [1] (b) ∆Hf° (2-methylpropene) = 4 × ∆Hc° (C) + 4 × ∆Hc° (H2) - ∆Hc° (2-methylpropene) = 4(-393) + 4(-286) – (-2520) = - 196 kJ mol-1 [1/2] for ans [1/2] for units [1] [1] 5. (a) Show 2 half-lives on graph half-life = 10 s half-lives are constant hence 1 st order [½] [½] [1] (b) (i) : Max vol = 60 cm3 ; half-life = 20s (ii): Same half life as original curve = 10 s; [½] [½] [½] [½]

max vol = 144 cm3 6. (a) KC =    Y X Z 2 =   4 . 005 . 0 08 . 0 04 . 02  [1/2] for correct substitution; [1/2] for ans KC = 0.4 [1] [1] (b) 0.05 mol dm -3 of Y was added into the system. [1] (c) show all 3 concentrations halved no change in equilibrium position [½] [½] 7. (a) A buffer solution is a solution whose pH does not change significantly when a small amount of acid or base is added to it. [2] 0.10 concentration / mol dm-3 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0 1 2 3 4 5 6 7 [Y] [X] [X] [Z] [Y] 8 9 10 time / min [Z] [Y [X Z[

Or A buffer solution resists changes in pH on addition of a small amount of acid or alkali. either phrases – both underlined terms must be present else [0]

7. (b) (i) HPO4 2- + H+ → H2PO4 - The additional H+ ions are removed by the large concentration of HPO4 2- in the buffer. Therefore the pH of the solution remains almost unchanged. [1] (ii) H2PO4 - + OH- → HPO4 2- + H2O The additional OH- ions are removed by the large concentration of H2PO4 - in the buffer. Thus, H+ changes very slightly and the pH remains almost unchanged. If only equations were written, -[1] [1] 8. (a) Reagent Organic Product heat with ammonia dissolved in ethanol in a sealed tube O CH2CH2 OC 3H H2N OH acidified potassium dichromate(VI) (aq) with distillation CCH2 O CH3 Cl OH O H O accept –OH as the oxidation of the 2 o alcohol after hydrolysis may not take place due to distillation hot aqeuous sulfuric acid CH2CH2 O CH3 Cl OH HO OH [1] per structure [3] 8. (b) The rate of reaction would increase C-Cl is a stronger bond as compared to C-Br bond and (requires [1] [½]

more energy before a reaction could take place / has a larger activation energy)[1]. Therefore the rate of reaction would be significantly faster for C-Br then C-Cl. [½] (c) butan-2-ol < butanoic acid [1] Carboxylic acids act as acids by donating their protons and forming carboxylate ions according to the following equilibrium equation: R C O O H + H2O R C O O - + H3O+ [1] Hence carboxylic acids are stronger acids than alcohols. [1] [1] 8. (d) Step 1 – HCl (aq) or H2SO4 (aq) and heat Step 2 – aq. I2, excess NaOH / KOH and heat Step 3 – LiAlH4 in dry ether Each reagent and condition [½]

SECTION B: 9. (a) (i) N2H4(l) + O2(g) N2 (g) + 2H2O(g) [1] (ii) Energy required for bond –breaking = BE(N-N) + 4x BE(N-H) + BE(O=O) = 160 + 4(390) + 496 = 2216 kJ Energy released from bond-forming = BE(N≡N) + 4x BE(O-H) = 994 + 4(460) = 2834 kJ Enthalpy change of formation of hydrazine = 2216 - 2834 =- 618 kJ mol -1 [1] [1] [1] (iii) Reaction progress Energy/ kJ N2H4(g) + O2(g) N2(g) + 2H2O(g) Ea enthalpy change of combustion Marking points:  Label both axes  Enthalpy change of formation shown.  Label Activation energy  Reactants and products label as N 2H4 and oxygen, water and steam respectively. Each point = [½] mark [2]

9. (a) (iv) Since the process is only 80% efficient, Total amount of energy needed = 6100 1080  =1.25 × 106 J No of moles of hydrazine needed = 3 6 10 618 10 25 1  . = 2.02 mol Relative molecular mass of hydrazine = 32 Mass of hydrazine needed = 2.02 × 32 = 64.6 g [1] [1] [1] 9. (b) (i) CH 3CH(OH)COOH CH3CH(OH)COO- + H+ Ka of lactic acid = ]CH(OH)COOHCH [ ] ][ CH(OH)COOCH [ 3 - 3 H [1/2] [1/2] (ii) CH 3CH(OH)COOH + NaOH  CH3CH(OH)COONa + H2O No. of moles of NaOH used = 025 01000 30 24 ..  = 6.075 × 10-4 mol No. of moles of lactic acid = 6.075 × 10-4 mol Concentration of lactic acid = 3 3 10 10 10 075 6    . = 0.0608 mol dm-3 [1] [1/2] [1/2] (iii) Phenolphthalein [1] 9. (c) (i) Reagent: Acidified KMnO4 Condition: Heat Observation: C6H5CH2CH3 will decolourise purple KMnO4 while C 6H11CH2CH3 will not. [½] [½] [1]

(ii) Reagents and conditions: Fehling’s reagent, warm Aliphatic aldehyde will give a red-brown precipitate of Cu2O but no ppt seen for the aromatic aldehyde. [1] [½] [½] (iii) CH 3CH2COOCH(CH3)2 and CH3OCOCH(CH3)2 Reagent: (i) Dilute HCl/ H2SO4 ; (ii) acidified K2Cr2O7 Condition: (i)Heat ; (ii) heat Observation: CH 3CH2COOCH(CH3)2  CH3CH2COOH + HOCH(CH3)2 CH3OCOCH(CH3)2  CH3OH + HOOCCH(CH3)2 The alcohols ini bold will undergo oxidation to further distinguish them: CH 3OH will turn orange dichromate green while HOCH(CH3)2 will not be oxidised. *Equations for the hydrolysis of esters are not required, but students should at least show the alcohols formed which undergoes oxidation in the second step. [1] [1] [1] 10. (a) (i) Comparing expts 1 and 2, when [NO] triples, rate increases by 9 times. 2nd order w.r.t. [NO