YJC H1 CHEM ANS

YISHUN JUNIOR COLLEGE 2009 JC2 PRELIMINARY EXAMINATIONS H1 CHEMISTRY MARK SCHEME Paper1 1 2 3 4 5 6 7 8 9 10 C B B A C D C C B C 11 12 13 14 15 16 17 18 19 20 A D D C B C A A B C 21 22 23 24 25 26 27 28 29 30 D A A B D C C A A D Paper 2 Section A 1 (a) (i) Electrons [1] (ii) Electrostatic forces of attractions between the electrons and protons [1] (iii) Protons and neutrons [1] (b) (i) species protons neutrons electrons 11 12 11 8 8 8 8 10 10 [1] [1] [1] (ii) Na2O + 2HNO3 → 2NaNO3 + H2O [1] (iii) Let x % = abundance of oxygen-16 = 75% = 25% [1] (c) (i) Nitrogen: 1s2 2s2 2p3 (electrons have to be unpaired) Oxygen: 1s2 2s2 2p4 [1] [1]

(ii) 1st I.E. of N = 1400 kJ mol-1 1st I.E. of O = 1310 kJ mol-1 The 1st I.E. of O is lower than that of N. For O, electron is removed from a 2p orbital that is occupied by another electron. Thus, electron experiences inter-electronic repulsion and less energy is required to remove it. [1] [1] (iii) N3- [1] (iv) A lot of energy is required to remove all 5 electrons in nitrogen and 6 electrons in oxygen to form a stable octet. [1] 2 (a) (i) [1] (ii) bent, 104.5º [1] (b) [1] (c) Hydrogen peroxide is simultaneously oxidized and reduced in this reaction. Oxidation state of oxygen atom in hydrogen peroxide increase from -1 to 0 to oxygen and decrease from -1 to -2 in water. [1] (d) (i) 5.50 x 10-4 mol [1] (ii) 1 mol of vanadium donates 2 mol of electrons to hydrogen peroxide. Original oxidation state of vanadium is +3 [1] [1]

3 (a) (i) Na Mg Al P formula NaCl MgCl2 AlCl3 PCl3 PCl5 pH 7 6.5 2 1 [1] [1] (ii) PCl3 + 3H2O → H3PO3 + 3HCl or PCl5 + 4H2O → H3PO4 + 5HCl [1] (b) The lattice energy of MgO is more exothermic than NaCl. Mg2+ and O2- have higher charge than Na+ and Cl-. Hence the attraction between Mg2+ and O2- is stronger than Na+ and Cl-. [1] [1] (c) (i) AlF3 is an ionic compound and a lot of energy is required to overcome the strong electrostatic forces of attractions between the Al3+ and F- ions. AlCl3 is a simple covalent molecule and less energy is required to overcome the weak van der Waals forces of attraction between the AlCl3 molecules. [1] [1] (ii) [1] 4 (a) C H O % 69.8 11.6 18.6 No. of moles 5.81 11.6 1.16 5 10 1 Empirical formula = C5H10O Mr = 86 = 5 × 12.0 + 10 × 1.0 + 1 × 16.0 Molecular formula = C5H10O [1] [1] (b) (i) A contains a C=C double bond [1] (ii) A contains a –OH group / is an alcohol [1]

(iii) A has the structure [1] (c) (i) Orange precipitate [1] (ii) Secondary alcohol (do not accept alcohol) [1] (d) 1. Presence of restricted rotation about the C – C bond due to the presence of π bond; 2. No two similar groups is bonded to the C with the double bond [1] [1] (e) cis trans [2]

Section B 5 (a) [1] (b) (i) order of reaction w.r.t PABA is 1 order of reaction w.r.t ethanol is 1 order of reaction w.r.t sulfuric acid is 0 Rate equation: Rate= k[PABA][Ethanol] [1] [1] [1] [1] (ii) k = 0.0200 mol-1dm3s-1 Award if both value and unit are correct. [1] (c) Boltzmann curve – 2 marks, every mistake,-1 From the graph, increase in temperature increases the average energy of the reacting particles. Hence the proportion of particles with energy equal to orgreater than the activation energy increases significantly. Therefore the frequency of collisions between reacting particles increases. This increases the number of effective collisions and consequently the rate of reaction. [2] [1] [1] Number of particles Energy T1 T2 Ea T2>T1

(d) (i) heat evolved = mc∆T = 1 x 200 x 4.2 x 13.4 = 11256 J or 11.3 kJ [1] (ii) No of mole of water = 0.025 mol Enthalpy change of reaction = -11256/0.025 = -450 kJ mol-1 [1] [1] (e) (i) HX + H2O H3O+ + X- [H3O]+ = √ka.[HX] = √1.5 x 10-5 x 0.1 = 1.22 x 10-3 mol dm-3 pH = -log10 [H3O]+ = -log101.22 x 10-3 = 2.9 [1] [1] [1] (ii) A buffer solution is one that is able to maintain a fairly constant pH, even when relatively small amounts of acid or alkali are added to it. When small amount of H+ is added, it is removed by H+ + X-  HX When small amount of OH- is added, it is removed by OH- + HX  H3O + X- [1] [1] [1] 6 (a) (i) Heat evolved when one mole of liquid ethanol is burnt completely / in excess oxygen under standard conditions [1] (ii) ∆H = – 1411 – (– 1367) = – 44 kJ mol-1 [1] [1] (iii) Under standard conditions, water and ethanol exist as liquids. [1]

(b) (i) [2] (ii) ROH + RCOOH Ý Ester + H2O ni 1 1 0 0 nc - 0.6 - 0.6 + 0.6 + 0.6 ne 0.4 0.4 0.6 0.6 [1] [1] [1] (iii) The equilibrium will shift to the right as concentrated H2SO4 is a dehydrating agent thus increasing the yield of the ester. It also acts as a catalyst and thus increasing the rate of the reaction. [1] [1] (c) (i) PCl5 / PCl3 / SOCl2 at room temperature or HCl, heat [1] (ii) Warm the two compounds with NaOH (aq), followed by HNO3 (aq) and then AgNO3 (aq) Chloroethane: White precipitate Iodoethane: Yellow precipitate [1] [1] [1] (d) (i) C: CH3CHO D: CH3CH(OH)CN [1] [1] rate time Rf (Forward rate) Rb (Reverse rate) teqm

(ii) Reaction I: K2Cr2O7 (aq), H2SO4 (aq), distil Reaction II: HCN, trace amount of NaOH / NaCN, 10 – 20 oC Reaction III: any mineral acid, heat [1] [1] [1] 7. (a) [4] [4] (b) (i) Acids are proton donor, bases are proton acceptor [1] (ii) [H+] = 2 x 0.1 pH = - log [0.2] = 0.70 [1] [1] (iii) No of mole of water = 0.025 mol heat evolved = 0.025 x 126 = 3.15 kJ ∆T = 3.15/75 x 4.2 = 10 The final temperature of the solution is 35ºC. [1] [1] [1] [1]

(iv) [2] (v) another line showing lower Ea, same energy level of products obtained. [1] (c) CH2ClCOOH, CHCl2COOH, Cl3COOH CCl3COOH is the strongest acid as it contains the most number of electron withdrawing chlorine atoms. This is because the charge on the COO− group of CCl3COO−, is dispersed more extensively than in CHCl2COO−and CH2ClCOO− [1] [1] Energy Reaction coordinates NaOH + H2SO4 Na2SO4 + H2O Ea -126 kJ (v)