YJC H1 CHEM P1 Worked Solution

2017 JC2 H1 Chemistry 8872 Preliminary Examinations Paper 1 Worked Solutions Page 1 of 4 Paper 1 Worked Solutions 1 A 2 B 3 D 4 D 5 B 6 B 7 A 8 A 9 B 10 A 11 C 12 B 13 C 14 D 15 D 16 B 17 B 18 C 19 A 20 C 21 D 22 C 23 B 24 C 25 A 26 D 27 B 28 B 29 D 30 A 1 Bearing in mind the mole ratio of the three gases when writing the balanced equation, you should be able to conclude that 1 mole of Ca(NO3)2 requires 3 moles of C: Ca(NO3)2 + 3C  CaO + 2CO2 + CO + Z It should also be clear that all the number of O atoms on both sides of the equation are already equal (i.e. balanced), and thus Z have to be N 2 so that the number of N atoms (the only remaining element to be balanced) on both sides are equal: Ca(NO3)2 + 3C  CaO + 2CO2 + CO + N2 Answer: A 2 A: 91.1(28) 7.9(29) 1.0(30) 28.099100 rA  B: 92.2(28) 4.7(29) 3.1(30) 28.109100 rA  C: 95.0(28) 3.2(29) 1.8(30) 28.068100 rA  D: 96.3(28) 0.3(29) 3.4(30) 28.071100 rA  Answer: B 3 To get the balanced equation: H2S(aq)  S(s) + 2H+(aq) + 2e− (2) SO2(aq) + 4H+ + 4e−  S(s) + 2H2O(l) 2H2S(aq) + SO2(aq)  3S(s)+ 2H2O(l) Dividing throughout by 3, we get: 2 3 H2S(aq) + 1 3 SO2(aq)  S(s)+ 2 3 H2O(l) Answer: D 4 For 10B: number of protons = 5; number of neutrons = 10 − 5 = 5  ratio of proton : neutron = 1 : 1 A: 40Ar: number of protons = 18; number of neutrons = 40 − 18 = 22  ratio of proton : neutron = 9 : 11 B: 40K: number of protons = 19; number of neutrons = 40 − 19 = 21  ratio of proton : neutron = 19 : 21 C: 32P: number of protons = 15; number of neutrons = 32 − 15 = 17  ratio of proton : neutron = 15 : 17 D: 32S: number of protons = 16; number of neutrons = 32 − 16 = 16  ratio of proton : neutron = 1 : 1 Answer: D 5 Around the S-atom, there are 3 bond pairs and 1 lone pair of electrons (a double bond is considered one bond pair). By VSEPR theory, the four electron pairs will space themselves as far apart as possible to minimise repulsion, leading to an electronic geometry of tetrahedral, and a bond angle of 109°. However, as the lone pair-bond pair repulsion are stronger than bond pair-bond pair repulsion, the bond angle will be smaller than 109°,  approximately 107°. Answer: B 6 In order for the anion to have a square pyramidal shape, it must have 5 bond pairs and 1 lone pair of electrons. Since Sb is from group 15, it has 5 valence electrons, which are used to form normal single bond with the five F- atoms (i.e. 5 bond pairs). Hence Sb must receive two electrons from external sources, in order to have one lone pairs of electrons.  n = 2, i.e. the anion is SnF 52−. Answer: B 7 Since both butane and methane are made up of non-polar molecules, only id-id interactions exist between their respective molecules. As butane has more electrons than methane, its electron cloud is more polarisable, and so the instantaneous dipole-induced dipole interactions between its molecules are stronger, and hence it is easier to liquify butane than methane. Answer: A

2017 JC2 H1 Chemistry 8872 Preliminary Examinations Paper 1 Worked Solutions Page 2 of 4 8 .. qqLE rr     Cationic radii: Li+ < Na+ Anionic radii: O2− < S2− sodium sulfide, Na2S, has the least exothermic lattice energy while lithium oxide, Li2O, has the most exothermic lattice energy. Answer: A 9 Bond breaking: 2(C−H) +6(C−Cl) + (O=O) = 2(+410)+6(+340)+(+496) = +3356 kJ mol−1 Bond forming: 2(C=O) +4(C−Cl) + 2(H−Cl) =2(−740)+4(−340)+2(−431) =−3702 kJ mol−1 H = (+3356) + (−3702) = −346 kJ mol−1 Answer: B 10 For the first pair of acid and base, we learnt that H for a strong acid (HCl) and a strong base (NaOH) is −57.0 kJ mol−1. Since the magnitude of H for the second pair of acid and base is smaller than 57.0 kJ mol−1, a weak acid must have reacted with NaOH.  P must be ethanoic acid (option A or B). Since the magnitude of H for the third pair of acid and base is smaller than 57.0 kJ mol−1, HCl must have reacted with a weak base.  Q must be ammonia (option A or C). Since H for the fourth pair of acid and base is also − 57.0 kJ mol−1, nitric acid must have reacted with a strong base.  R must be potassium hydroxide (option A or C). Answer: A 11 Definition of a dynamic equilibrium: an equilibrium where the forward and reverse reactions are continuing at the same rate (or that the forward and reverse reaction are taking place, but the rate is not equals to zero). Answer: C 12 When pressure is reduced at constant temperature, equilibrium position will shift to the side with a larger number of moles of gases to increase pressure (option B or D). When temperature is increased, the endothermic reaction will occur to a greater extent to absorb heat (option A or B) Answer: B 13 K w = [H+] [OH−] (by definition) At 30 °C, Kw = 1.44 × 10−14 mol2 dm−6  [H+] [OH−] = 1.44 × 10−14 Since [H+] = [OH−] for pure water,  [H+]2 = 1.44 × 10−14  [H+] = 1.2 × 10−7 pH = −log [H+] = −log (1.2 × 10−7) = 6.92 (< 7) Answer: C 14 Catalyst increases the rate constant (options C and D). Catalyst lowers the activation energy (all four options). Catalyst does not alter the energy level of the reactants and the products (options B and D). Answer: D 15 When [acid] is low, reaction is first order with respect to acid  rate increases linearly as [acid] increases (options A or D). When [acid] is high, reaction is zero order with respect to acid.  rate remains the same as [acid] increases i.e. the graph approaches a verticlal line (option D). Answer: D 16 Number of protons increases from Na + to Al3+, and hence nuclear charge increases. Na+, Mg2+ and Al3+ have the same total number of electrons. As a result, the effective nuclear charge (net electrostatic force of attraction between the nucleus and valence electrons) increases from Na + to Al3+, and so ionic radii decreases. Answer: B 17 Since MgO will react with and dissolve in HCl (as a soluble salt is formed): MgO + 2HCl  MgCl2 + H2O and Si will not react with or dissolve in HCl. Hence Si can be removed from the solution by filtration.  Method 1 will work (option A or B). Since both MgO and Si have very high melting and boiling points, neither of them will vapourise on gentle heating.  Method 2 will not work (option B or D). Answer: B

2017 JC2 H1 Chemistry 8872 Preliminary Examinations Paper 1 Worked Solutions Page 3 of 4 18 The four possible isomers with molecular formula C4H8O2 are: C O OCH2CH2CH3H C O OHC CH3 CH3 H C O OCH3 CH2 CH3 C O OCH3CH2 CH3 Answer: C 19 I is an addition reaction as the first propanone molecule is added across the C=O double bond of the second propanone molecule to produce an alcohol: C O CH3 CH3 CCH3 C H H H O C OH CH3 CH3 C H H C O CH3+ II is an elimination reaction as an unsaturated alkene is formed with the elimination of a water molecule from the alcohol: C OH CH3 CH3 C H H C O CH3 CC H 3 CH3 C H C O CH3 + H2O Answer: A 20 A: Correct. Since there are 3 bond pair and 0 lone pair of electrons around each of the two C-atoms, the shape around each C-atom is trigonal planar. Hence all two C-atoms and four H-atoms lie on the same plane. B: Correct. Molecular formula of ethane is C2H4, and so the empirical formula (showing the lowest mole ratio) is CH2. C: Not correct. As the shape around each C- atom is trigonal planar, the bond angle is 120°. D: Correct. Each of the two C-atoms forms one -bond wit