ACJC Chemistry H1 2008 Prelims P2

Examiner’s Report 2008 1 Examiner’s report H1 Prelims 2008 General Comments for Section A The examiners felt that there was visible attempt form the students to answer the questions as thoroughly as they can. However, what was most glaring was the insufficient amount of key-words and jargons provided by the students’ answers, which actually highlights that there was insufficient detailed studying of the topics. Of the 4 questions in Section A, Q2 and Q4 were generally poorly answered. 1a No. of mol of Cu2O = 0.286 ÷ (63.5+63.5+16.0) = 0.00200 I mol of C6H12O6 ≡ 1 mol of Cu2O No. of mol of C6H12O6 = 0.00200 Mr of C6H12O6 = 12.0(6) + 1.0(12) + 16.0(6) = 180.0 Mass of C6H12O6 = 0.00200 x 180.0 = 0.360g Percentage of glucose in the food sample = 0.360 / 5.00 x 100% = 7.20% Note: Answers are to show 3 significant figures to gain full marks. b 1s2 2s2 2p63s2 3p63d9 c i Isotopes are atoms of the same element with the same number of protons and different number of neutrons. ii Relative atomic mass of copper = (69.2/100 x 63) + (30.8/100 x 65) = 63.62 = 63.6 (1 dec place) d Number of protons= 29 Number of neutrons= 34 Number of electrons= 29 e i ii As angle of deflection α charge / mass, the angle of deflection for a sample of 32S2- is predicted to be twice that of 63Cu2+ as its mass is two times smaller than 63Cu2+. 63Cu2+ 32º 32S2- 16º ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 2 2a Dynamic equilibrium is reached when the rate of the forward reaction and the rate of the backward reaction is the same. Comments: • Generally well answered. b i CH3CH2COOH (l) + CH3CH2OH (l) CH3CH2COOCH2CH3 (l)+H2O(l) No marks awarded if no state symbols given ii Ethyl propanoate c CH3CH2COOH + CH3CH2OH CH3CH2COOCH2CH3+H2O Initial no. of mole 0.6 0.5 0 0 Δ in no. of mole 0.4 – 0.6 = -0.2 -0.2 +0.2 +0.2 No. of mole at eqm 0.4 0.5 – 0.2 = 0.3 0.2 0.2 Let the volume of the vessel be V dm3. Kc = [CH3CH2COOCH2CH3][H2O] [CH3CH2COOH][CH3CH2OH] = (0.2/V) (0.2/V) (0.4/V) (0.3/V) = 0.333 ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 3 d i Position of equilibrium will shift to the right hand side. By Le Chatelier’s principle, when ethanol is added, the system will favour the forward reaction so as to decrease the concentration of ethanol. Hence, the position of equilibrium will shift to the right hand side. ii The position of equilibrium shifts to the left-hand side. By Le Chetalier’s principle, increasing temperature of the system will favour the endothermic reaction to absorb the excess heat. Hence, the backward reaction is favoured as it is endothermic, and position of equilibrium shifts to left hand side. ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 4 ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 5 3 a i Aspirin is insoluble in water due to the presence of the large hydrophobic benzene ring. ii OO H O O CH3 + NaOH(aq) → OO - Na + OC 3 O H + H2O In NaOH(aq), aspirin forms an ionic salt , which is able to form ion-dipole interactions with water molecules. Hence, it is soluble in water. b i Amphoteric refers to the ability to function as either as an acid or a base. ii Reaction of Al(OH)3 with acids is typical of a metal. Al(OH)3 + 3H+ Æ Al3+ + 3 H2O Reaction of Al(OH) 3 with bases is typical of a non-metal. Al(OH)3 + OH- Æ Al(OH)4 - 3 c i Across the period, there is a general increase in the 1 st IE due to the increase in the nuclear charge and the constant shielding effect from one element to the other because electrons are added successively to the same shell. Hence the effective nuclear charge increases, the nuclear attraction on the valence electrons and more energy is needed to remove the electron. ii The 1st IE of B is lower than that of Be. 5B: 1s2 2s2 2p1 4Be: 1s2 2s2 The 2s subshell is nearer to the nucleus than the 2p subshell, hence the electrostatic attraction between the nucleus and the 2s electron is stronger than between the nucleus and the 2p electron. Hence, a lower amount of energy is required to remove the 2p electron from B than to remove the 2s electron from Be. The 1 st IE of O is lower than that of N. 8O 1s 2s 2p 7N 1s 2s 2p The mutual electronic repulsion of the paired electrons in a 2p orbital of O makes the removal of one electron from that orbital easier compared to a 2p electron of N which does not experience such repulsion. Hence less energy needed to remove the electron in O. ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 6 4a i Comparing experiment 1 and 3. [NaOH] is kept constant. Increasing the [chloromethylbenzene] by 1.5 times causes the rate of reaction to increase by 1.5 times. Therefore, reaction is first order with respect to chloromethylbenzene. Comparing experiment 1 and 2. [NaOH] doubles and [chloromethylbenzene] doubles too. But rate of reaction only doubles. Since, reaction is first order with respect to chloromethylbenzene, this implies that the reaction is independent of NaOH. Therefore, reaction is zero order with respect to NaOH. Rate = k[chloromethylbenzene] ii 0 [phenylmethanol] time t1 t1 0.5 n 0.75 n Diagram with correct labels. Reflecting constant half-life of chloromethylbenzene (since rate of formation of phenylmethanol is dictated by chloromethylbenzene) b i Strength of the covalent bond is dependent on the effectiveness of overlapping of the atomic orbitals. The smaller the atoms, the more effective the overlap, the stronger the covalent bond. Cl is the smallest atom followed by Br and then I being the largest. Between C and Cl, there will be the most effective overlapping of orbital and hence the C-Cl bond is the strongest and the C-I bond is the weakest. ii From the rate equation, the rate of formation of phenylmethanol is determined by the strength of the C – X bond. The stronger the C – X bond, more energy is required to break it and hence the faster the rate of reaction. c Rate of reaction slows down. ACJC 2008 8872/02/Aug/08 Preliminary Examination

Examiner’s Report 2008 7 Number of molecules Energy T1 Ea0 T2 = 0 oc T1 > T2 Diagram with labels Diagram showing at 0oC having very few molecules with min energy. The number of particles with the minimum kinetic energy to overcome activation energy for effective collision to occur decreases. Hence rate of reaction decreases. 5 (a) Limonene is soluble in benzene / tetrachloromethane / any non-polar solvents. Limonene is a non-polar molecule and is soluble in non-polar solvents through induced dipole-induced dipole interactions. (b) (i) Observations: 1. Reddish-brown bromine solution is decolourised. 2. White fumes of HBr observed. Structural formulae of the product: CH3 C H3 CH2Br Br Br Br (ii) Reagents/Conditions: Na metal Observations: Colourless, odourless gas evolved that extinguishes lighted splinter with a ‘po