YJC H1 CHEM markscheme

Yishun Junior College H1 Chemistry Preliminary Examinations 2016 Paper 1 Answers 1 C 7 C 13 B 19 B 25 C 2 C 8 B 14 D 20 D 26 A 3 B 9 B 15 A 21 C 27 D 4 B 10 D 16 C 22 B 28 D 5 D 11 B 17 D 23 A 29 B 6 C 12 A 18 A 24 D 30 C Paper 2 Answers 1 (a) (i) C7H16 + 11O2  7CO2 + 8H2O [1] (ii) Bonds broken kJ mol-1 Bonds formed kJ mol-1 6 x C−C 6(+350) 14 x C=O 14(−740) 16 x C−H 16(+410) 16 x O−H 16(−460) 11 x O=O 11(+496) Total +14116 [1] Total −17720 [1] ∆Hc = +14116 – 17720 = −3604 kJ mol−1 = - 3600 kJ mol−1 [1] [3] (b) (i) D and E [1] (ii) Geometric isomerism arises due to the restricted rotation of the C=C double bond and the presence of 2 different groups attached to each of the carbon atoms in the C=C double bond. [1] (c) F and B are structural isomers (OR have the same molecular formula). F is more branched than B and so, it has a smaller surface area of contact. [1] Hence, less energy is needed to overcome the weaker intermolecular instantaneous dipole-induced dipole interactions for F. [1] [2] (d) Reagent and conditions: KMnO4, H2SO4(aq), heat [1] Observations: Purple KMnO4 is decolourised and the gas formed gives white precipitate with limewater. [1] [2]

(e) H I conditions (a) CCl4 solvent OR room temperature OR (b) UV light anhydrous FeBr3 OR Fe catalyst type of reaction (a) addition OR (b) substitution substitution any 2 correct – [1] all 4 correct – [2] [2] Total: 12 2 (a) (i) 𝐾𝑐 = [𝐻𝐼]2 [𝐻2] [𝐼2] [1] (ii) H2 I2 2HI Initial amt / mol 4.00 x 10−3 8.00 x 10−3 0 Change in amt / mol −3.4 x 10−3 −3.4 x 10−3 +6.8 x 10−3 Eqm amt / mol 6.0 x 10−4 4.6 x 10−3 6.8 x 10−3 Eqn conc / mol 3.0 x 10−4 2.3 x 10−3 3.4 x 10−3 𝐾𝑐 = (3.4 × 10−3)2 (3.0 × 10−4)(2.3 × 10−3) = 16.8 [1] [1] [1] (b) (i) By Le Chatelier’s principle, the position of equilibrium shifts to the left, favouring the backward endothermic reaction, so as to absorb the heat added.[1] Hence, [HI] decreases while [I2] and [H2] increases, leading to the value of Kc to decrease. [1] [2] (ii) As the amount of gaseous reactants and products are equal, the position of equilibrium will not shift. Hence, the amount of HI will remain the same at the new equilibrium. [1] Total:7 3 (a) (i) 1st t1/2 (when [C6H5N2Cl] falls from 0.2 to 0.1 mol dm−3) = 63 min 2nd t1/2 (when [C6H5N2Cl] falls from 0.1 to 0.05 mol dm−3) = 63 min (+ show both t1/2 on graph) [1] Since t1/2 is constant, order of reaction wrt C6H5N2Cl = 1 [1] [2] (ii) rate equation: rate = k[C6H5N2Cl] [1] units for k: min-1 or s-1 [1] [2]

(iii) Since excess water was used in the experiment, the concentration of water is almost constant throughout the experiment. Hence, the reaction rate will not be affected by the concentration of water. [1] (b) Diagram – [1] At the higher temperature, the average energy of the reacting particles is increased. Hence the proportion of particles with energy equal to or greater than the activation energy increases significantly as seen in the shaded area in the diagram. [1] In addition, when the temperature of the reactants is increased, their average speeds increase and therefore the frequency of collisions between reacting particles increases. This increases the frequency of effective collisions and consequently the rate constant increases and rate of reaction increases. [1] [3] Total: 8 4 (a) (i) 𝑀𝑟 = 385 [1] 𝜂𝑐ℎ𝑜𝑙𝑒𝑠𝑡𝑒𝑟𝑜𝑙 = 300×10−3 385 = 7.79 × 10−4 𝑚𝑜𝑙 [1] [2] (ii) η𝐻2 = 1 2 × 7.792 × 10−4 = 3.896 × 10−4𝑚𝑜𝑙 [1] 𝑉𝐻2 = 3.896 × 10−4 × 24 = 0.00935 𝑑𝑚3 = 9.35 𝑐𝑚3 [1] [2] (b) Due to the large non-polar hydrocarbon chain, the predominant interactions between cholesterol molecules is intermolecular instantaneous dipole-induced dipole (id-id) interactions. [1] Hence, the hydrogen bonds formed between cholesterol and water molecules do not give off enough energy to overcome the stronger intermolecular id-id interactions in cholesterol and intermolecular hydrogen bonds in water. [1] [2] (c) (i) step I : H2, Ni, heat OR H2, Pt [1] step II : KMnO4 (OR K2Cr2O7), H2SO4(aq), heat under reflux [1] step III: HCN, NaOH(aq) OR NaCN(aq) 10-20 oC [1] [3] Ea Number of molecules Energy T1 T2 T1 < T2

(ii) [1] (d) (i) conditions: conc H2SO4, reflux [1] type of reaction: condensation [1] [2] (ii) O O [1] Total:13 5 (a)  MgCl2 undergoes partial hydrolysis(**) to give a slightly acidic solution. Approximate pH of resultant solution is 6.5. (*) MgCl2(s) + 6H2O (l)  [Mg(H2O)6]2+(aq) + 2Cl–(aq) (*) [Mg(H2O)6]2+(aq) + H2O(l) ⇌ [Mg(H2O)5(OH)]+(aq) + H3O+(aq) (*)  SiCl4, PCl3 & PCl5 undergo complete hydrolysis(**) Low electron density Si and P provide empty d-orbitals to accommodate lone pair of electrons from H2O. pH of solution is 1.(*) SiCl4 (l)+ 4H2O  SiO2.2H2O (s) + 4HCl (*) PCl5 (s) + H2O (cold or limited)  POCl3(aq) + 2HCl PCl5 (s) + 4H2O  H3PO4 + 5HCl (*) Each * ½ mark [5] (b) (i) Dative / Co -ordinate bond is a covalent bond in which a pair of electrons is shared between 2 atoms but ONLY ONE of them provides both electrons that make up the bond. [1] (ii) Cation : NH4+ Anion : Cl- Both correct – [1] [1]

(iii) Each ion [1] [2] (c) (i) HCl(g) [1] Addition Reaction [1] [2] (ii) Cold Concentrated H2SO4, followed by H2O, warm. [1] (iii) NH3 in ethanol/alcohol and heat in sealed tube. [1] [2] (iv)  C-Cl is a stronger bond with bond energy of 340 kJ mol-1 and while C-Br is weaker with bond energy of C-Br 280 kJ mol-1, hence easier to break. [2]  Hence the rate of reaction for CH3CH2CH2CH2Cl will be slower than that of CH3CH2CH2CH2Br. [1] [3] (d) Conditions: UV light [1] Type of reaction: Substitution [1] + Cl2  + HCl [1] [3] Total: 20 CH3 CH2Cl

6 (a) Primary alcohol and carboxylic acid [1] (b) Test: Add 2,4-DNPH to each compound separately. Observation: For ethanedial, orange ppt of hydrazone is observed. For glycolic acid, no orange ppt is observed. C C O H H O + NHNH2 O2N NO22 NH N NO2 O2N NHN O2N NO2 C C H H + 2H2O Test: Add Fehling’s solution to each compound separately and warm. Observation: For ethanedial, brick -red ppt of Cu 2O is observed. For glycolic acid, no brick red ppt is observed. CHOCHO + 4 Cu2+ + 10 OH–  (CO2–)2 + 2 Cu2O + 6 H2O Test: Add Tollens’ reagent to each compound separately and warm. Observation: For ethanedial, silver mirror is observed. For glycolic acid, no silver mirror is observed. CHOCHO + 4 [Ag(NH3)2]+ + 6 OH–  (CO2–)2 + 4 Ag + 8 NH3 + 4 H2O or CHOCHO + 4 Ag+ + 6 OH–  (CO2–)2 + 4 Ag + 4 H2O For each test, correct reagent [1]; correct observations [1]; and correct equation [1] [3m x 2 sets = 6m] [6] (c) (i) A weak Bronsted acid is one which dissociates partially in solution to donate protons, H+. [1] (ii) Ka = + 22 22 [HOCH CO ] [H ] [HOCH CO H]  [1] (iii) Since [CH2(OH)COO–] = [H+], Ka = +2 22 [H ] [HOCH CO H] 1.48 x 10–4 = +2[H ] 0.20 [H+] = 5.44 x 10–3 mol dm–3 [1] pH = –lg (5.44 x 10–3) = 2.26 [1] [2]

(iv) A buffer solution is one which is capable of