YJC H1 CHEM Suggested Answers

8872 / YJC / 2017 / Preliminary Examination / Suggested Answers 1 2017 H1 Chemistry 8872 Preliminary Examinations Suggested Answers Paper 2 Section A: Structured Questions 1 (a) (i) Al(OH)3 + 3H+  Al3+ + 3H2O (or Al(OH)3 reacting with HCl) (ii) Aluminium hydroxide or calcium carbonate is insoluble in water and therefore will not increase the pH of blood. (b) (i) 2 550 0.024554 0.0246 mol22400 COn    (ii) 32 0.0246 molCaCO COnn  3 0.024554 {40.1 12.0 3(16.0)} 0.024554 100.1 2. 4565 2.46 gCaCOm         (iii) mass of CaCO3 in one tablet = 2.4565 0.491 g5  The claim is valid, as the mass if CaCO 3 is approximately the same as what was claimed by the manufacturer [Total: 7 marks] 2 (a) (i) The melting point of the elements increases from Na to Si (with Si significantly higher than that for Al), The melting point decreases drastically from Si to P and is relatively low from P to Cl (or the melting point of P to Cl is much lower than that for the Na to Si). (ii) trend: atomic radius decreases from Na to Cl explanation: nuclear charge increases and shielding effect remains constant (b) sodium: metallic bonding and giant metallic structure ‘sea’ of delocalised electrons are available to conduct electricity and so it has high electrical conductivity silicon: covalent bonding and giant molecular structure or giant covalent structure In the giant molecular structure, there are some free electrons and ‘holes’ which can be used to conduct electricity, and so silicon is a semi-conductor chlorine: covalent bonding and simple molecular structure There are no mobile electrons to conduct electricity, and so chlorine is a non-conductor (c) (i) There are 17 protons and 18 neutrons concentrated (within a very small volume) at the nucleus / centre of the atom There are 17 electrons surrounding the nucleus and moving randomly (ii) 1s2 2s2 2p6 3s2 3p5

8872 / YJC / 2017 / Preliminary Examination / Suggested Answers 2 (d) (i) (ii) There are two bond pairs and two lone pairs of electrons (allow e.c.f.) Hence the electron pairs spread themselves out as far apart as possible to minimise repulsion giving rise to the bent shape. [Total: 16 marks] 3 (a) (i) C3nH6n + 9 2 n O2  (3n)CO2 + (3n)H2O (ii) carbon (soot) (iii) benzene and methyl benzene (iv) brown Br2(aq) is decolourised CH2CH2 + Br2 + H2O  BrCH2CH2OH + HBr or CH2CH2 + Br2  BrCH2CH2Br (accept equations involving propene as well) (v) hydrogen and methane collected can be used as fuels. or benzene collected (in tube B) can be used to manufacture styrene and phenol or reduce the emission of CO2(g) to the atmosphere (accept any other reasonable answer) (b) cis–hex–2–ene trans–hex–2–ene cis–hex–3–ene trans–hex–3–ene cis–4–methylpent–2–ene trans–4–methylpent–2–ene

8872 / YJC / 2017 / Preliminary Examination / Suggested Answers 3 [Total: 8 marks] 4 (a) (i) mass of sucrose = 3110 330 10 36.3 g   (ii) Mr of sucrose = 12(12.0) 22.0 11(16.0) 342.0   number of moles of sucrose = 36.3 0.10614 0.106 mol342.0 (b) (i) Hrxno = Hfo(products) − Hfo(reactants) = 12Hfo(CO2) + 11Hfo(H2O) − Hfo(C12H22O11) = 12(−394) + 11(−286) − (−2226) = −7874 + 2226 = −5648 kJ mol−1 (or −5650 kJ mol−1 to 3 s.f.) [1] for correct equation, i.e. coefficient of the terms (for sucrose, not glucose) [1] for correct substitution of the values (regardless of correct equation or not) [1] for correct final answer with units (ii) quantity of energy = 5648 0.10614 599.48 599 kJ   (allow e.c.f.) (iii) number of ‘calories’ = 599.48 142.734.2  percentage = 142.73 100% 5.71%2500  [Total: 9 marks] Paper 2 Section B: Free Response Questions 5 (a) (i) A reversible reaction is one that can proceed in both the forward and the backward direction. (ii) 3 2 2 3 2 3 2 3 2 [ ][ ] [ ][ ] c CH CH COOCH CH H OK CH CH COOH CH CH OH (iii) 3 2 2 3 2 3 2 3 2 [ ][ ] [ ][ ] c CH CH COOCH CH H OK CH CH COOH CH CH OH 32 (1.15)(1.15)3.94 (0.18)[ ]CH CH OH  3 32 (1.15)(1.15)[ ] 1.86 mol dm (0.18)(3.94)CH CH OH  trans–3 – methylpent – 2 – ene cis–3 – methylpent – 2 – ene

8872 / YJC / 2017 / Preliminary Examination / Suggested Answers 4 (b) + +CH3CH2CH2OH CH3CH2CO2H KMnO4 or K2Cr2O7 H2SO4(aq), heat under reflux OH22[O] Observation: Purple KMnO4 is decolourised. Or Orange K2Cr2O7 turns green. +CH3CH2CHO CH3CH2CO2H KMnO4 or K2Cr2O7 H2SO4(aq), heat under reflux [O] Observation: Purple KMnO4 is decolourised. Or Orange K2Cr2O7 turns green. + +CH3CH2CN CH3CH2CO2H NH4 +H2SO4(aq), heat under reflux 2H2O + H + Observation: No visible observation. + + CH3CH2CH=CH2 CH3CH2CO2H OH25[O] + CO2 KMnO4 H2SO4(aq), heat under reflux Observation: Purple KMnO4 is decolourised. Effervescence is observed, and gas evolved forms white precipitate with limewater. (c) (i) 2K + CH3CH2COOH  2CH3CH2COO−K+ + H2 (ii) Redox reaction (accept acid-metal / base reaction) (iii) The gas will ‘pop’ with a lighted splint (iv) The effervescence / bubbling will be more vigourous As K is a bigger atom than Na (or K has a larger atomic radius than Na, or the outermost electron of K is further away from the nucleus than Na), the outermost electron of K is more loosely held by the nucleus (or the attractive force between the outermost electron and the nucleus of K is weaker) than Na,  K undergoes oxidation to form K+ more easily (d) Using HCl(aq): CH3CH2CH2 C O O CH3 CH3CH2CH2 C O OH CH3 OH + OH2 + Using NaOH(aq): CH3CH2CH2 C O O CH3 CH3CH2CH2 C O O - Na + CH3 OH + NaOH + correct structure of methylbutanoate [Total: 20 marks]

8872 / YJC / 2017 / Preliminary Examination / Suggested Answers 5 6 (a) order of reaction with respect to a given reactant is the power to which the concentration of that reactant is raised in an experimentally determined rate equation or In an experimentally determined rate equation : Rate = k [A] m m = order of reaction with respect to reactant A The half-life of a reaction, t1/2, is the time taken for the concentration of a reactant to fall to exactly half its value (b) (i) Comparing experiment 1 and 2, When [NO] is doubled, the initial rate is quadrupled  order of reaction with respect to NO is 2 Comparing experiment 2 and 3, When [H2] is doubled, the initial rate is doubled  order of reaction with respect to H2 is 1 rate = k [H2] [NO]2 (ii) Using the values from experiment 1: 3 3 3 23.0 10 (2.0 10 )(3.0 10 )k       3 5 2 6 1 3 3 2 3.0 10 1.67 10 mol dm h(2.0 10 )(3.0 10 )k       [1] for correct value, [1] for units (allow e.c.f. for both) (c) (i)- (iii) (iv) When temperature increases, the reactant particles have greater average kinetic energies, and the frequency of collisions increase As seen from the diagram, a larger fraction of the reactant particles will have kinetic energies greater than or equal to the activation energy, and so the frequency of effective collisions increases Hence, the rate constant, k, increases leading to the increase in the rate of the reaction. (d) (i) A weak acid is one that dissociates partially in water HNO2 ⇌ H+ + NO2− (ii) 2 2 [ ][ ] [] a H NOK HNO   units = mol dm−3 (iii) 3.72 4 3[ ] 10 1.91 10 mol dmH      energy fraction of particles 300 °C (c)(i) fraction of particles with E > Ea at T1 fraction of particles with E > Ea at T2 310 °C (c)(ii) Ea