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 th