2019 HCI Prelim H2 Chem P1 ANS

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 Paper 1 ANSWERS: Comments 1 D The definition for atomic mass is the ratio of the average mass of one atom of an element to one -twelfth the mass of one atom of 12C. In option D, the mass of one mole of atoms of an element has already taken into account all the isotopes and their relative abundances. 2 A The electronegativity increases across Period 3 as the effective nuclear charge increases. The electrical conductivity of the metals Na to A l increases while Si is a semiconductor and the non-metals P, S and Cl are non-conductors. The atomic radii across Period 3 shows a generally decreasing trend due to the increase in effective nuclear charge across the period. The melting po int increases from Na to A l on account of the increasing metallic bonding strength; the trend peaks at Si which has a giant covalent structure with extensive strong covalent bonding, before showing a decreasing trend for the simple covalent compounds which only have weak dispersion forces between molecules. 3 D Ice has a simple covalent structure. There is hydrogen bonding between water molecules, and covalent bonding between atoms of H and O in each molecule of water. HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS 1 2 3 4 5 6 7 8 9 10 D A D B C C A B C B 11 12 13 14 15 16 17 18 19 20 B B B A C D A B C C 21 22 23 24 25 26 27 28 29 30 D D D A C B A C D C

Iodine has a simple covalent str ucture. There are dispersion forces between iodine molecules, and covalent bonding between atoms of iodine in each molecule of iodine. Aluminium chloride has an ionic lattice structure with a large amount of covalency in the solid state at room temperature. Its structure becomes simple covalent just below its melting point resulting in dispersion forces between A l2Cl6 molecules. Hence permanent dipole interactions are not found in the solid state. Graphite has a giant covalent structure with dispersion forces holding the sheets of carbon atoms together. 4 B Species NH4+ XeF4 Shape Tetrahedral Square planar Bond Angle 109.5o 90o Species NF3 NH3 Shape Trigonal pyramidal Trigonal pyramidal Bond Angle 102o 107o F is more electronegative than N and so the electron cloud of the N–F bond is distorted towards F. This causes the electron density about N to decrease. Thus, bond pair-bond pair repulsion is weaker around the central N atom in NF 3 than in NH3. Hence, bond angle in NF3 is smaller. Species NO2+ SO2 Shape Linear Bent Bond Angle 180o <120o Species BCl3 PCl 3 Shape Trigonal planar Trigonal pyramidal Bond Angle 120o 107o 5 C Since the number of moles of gas in the three containers did not change before and after the taps are opened, the number of moles of gas found in the individual containers before and after the taps are opened should be the same. gas in first container + gas in second container + gas in third container = gas in overall container Note that there is no gas in the second container since it contains a vacuum. Thus it is not included in the calculation of the total number of moles of gas. 𝑃1𝑉1 𝑇1 + 𝑃2𝑉2 𝑇2 = 𝑃𝑓𝑖𝑛𝑎𝑙 × 𝑉𝑓𝑖𝑛𝑎𝑙 𝑇𝑓𝑖𝑛𝑎𝑙 1 × 420 293 + 2 × 75.5 273 = 𝑃𝑓𝑖𝑛𝑎𝑙 × (420 + 75.5 + 13) 298

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 Pfinal = 1.16 bar 6 C No. of moles of sodium percarbonate = 10.0 x 10–3 x 0.100 = 0.00100 1 mole of sodium percarbonate produces x moles of CO2. No. of moles of carbon dioxide = 48 ÷ 24000 = 0.00200 x = 0.00200 0.00100 = 2 No. of moles of KMnO4 = 0.0500 x 24.0 x 10–3 = 0.00120 No. of moles of H2O2 = 0.00120 x 5 ÷ 2 = 0.00300 1 mole of sodium percarbonate produces y moles of H2O2. y = 0.00300 0.00100 = 3 Ratio of y : x = 3 : 2 7 A G = H – TS In general, H & S hardly changes with temperature unless there is a change in the physical state of the reactants or products. Normally, G is dependent on the temperature as seen from the equation above. For this particular reaction, S is almost 0 regardless of the temperature since there are equal number of moles of gas on both sides of the arrow. Since S ≈ 0, therefore TS ≈ 0 and so G ≈ H. Thus the G of the reaction becomes relatively independent of temperature. Note also that the equilibrium constant, K, is always affected by temperature unless H = 0. 8 B The diagram below shows that the magnitude of the Hatom for graphite is larger than that for diamond but the magnitude of Hc for graphite is smaller than that for diamond.

In graphite, each C atom has a 2p orbital containing one electron that is not used in the formation of C –C bonds. The 2p orbitals can overlap with each other to form a two-dimensional delocalised  bonding system spreading throughout the whole sheet of atoms. Thus, the C–C bonds in graphite have partial double bond character. Thus the C –C bonds in graphite are stronger than those in diamond. This is also evident from the larger magnitude of Hatom of graphite. The rearrangement of atoms in a tetrahedral structure of carbon atoms in diamond to the trigonal planar structure of carbon atoms in graphite requires a lot of energy as many strong covalent bonds are broken. This constitutes a very high activation energy such that conversion of diamond to graphite is kinetically unfavorable despite the process having a negative G. 9 C The number of half -lives that it takes rock sample to decay to a ratio of 2:3 for thorium-234 to protactinium-234 is calculated as follows. Hence fraction of thorium- 234 in the rock sample = 2 2+3 = 2 5 1 × (1 2) 𝑛 = 2 5 n = 1.32 half-lives where 1 = amount of thorium-234 at the start, and 2 5 is the amount of thorium at the end of the time period of radioactive decay. Time taken = 1.322 x 24.1 = 31.9 hr In this question, the strategy is to recognize that more than one half-life, but less than two half -lives, have passed. So the logical answer would lie between 24.1 h and 48.2 h. 10 B

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 In order to find the correct rate equation based on the suggested reaction mechanisms, the following steps must be taken. The species found in the rate equation should only include the reactants and not any of the intermediates formed in the mechanism. It is helpful to write down th e overall chemical equation so that the reactants are not confused with any of the intermediates. The stoichiometry of each reactant in the slow step of a mechanism is reflected as the order of reaction of that reactant in the overall rate equation. 1. Overall equation: 2NO + 2H2 → N2 + 2H2O The reactants in the slow step consist of those found in the overall equation. This means that two NO molecules and one H 2 molecule are involved in the rate determining step. The rate equation should reflect this. Hence the rate equation should be rate = k[NO]2[H2]. 2. Overall equation: 2NO + O2 → 2NO2 The reactants in the slow step contain the intermediate N2O2 and reactant O2. Hence the rate law based on the slow step would look like this. rate = k[N2O2][O2] --- equation 1 However, N2O2 should not appear in the rate equation. We need to substitute [N2O2] with the concentration of reactants that produced this intermediate into equation 1, giving rate = k[NO2]2[O2]. Although the rate equation seems to indicate a termolecular reaction mechanism, the suggested mechanism shows otherwise. 3. Overall equation 2O3 → 3O2 In the slow step, one molecule of O 3 reacts with one atom of C l. Hence the overall rate equation should be rate = k[O 3][Cl]. The species C l is a homogenous catalyst since it is used up in step one and regenerated in step 2. It should appear in the rate equation as it affects the rate of reaction even if it does not appear i