RI 2017 Y5 Promo Suggested Solutions

1 © Raffles Institution 9729/ Year 5 Promo /2017 RAFFLES INSTITUTION 2017 YEAR 5 H2 CHEMISTRY PROMOTION EXAMINATION S ECTION A 1 2 3 4 5 6 7 8 9 10 D D A D C B D B A A 11 12 13 14 15 16 C C B B D C 1 Answer: D relative molecular mass = weighted average mass of 1 molecule of the substance 1 12 x mass of 1 atom of carbon-12 S tudents need to find the option which most closely describes the above. A. Inc orrect. If only the relative isotopic mass of the most abundant isotope was considered for each atom, it will not be able to account for the mass of the different isotopes of the sam e el ement, which is required for the weighted average mass. B. Incorrect. The denominator should have been t he mass of 1 12 of one 12C atom. C. Incorrect. The denominator should have been the mass of 1 12 of 1 mole of 12C atoms. D. Correct. This option most accurately defines relative atomic mass. 2 Answer: D G2+ ion has the same number of electrons as PH3, i.e. G2+ ion has 18 electrons. So, G must have 20 electrons. As the 4s orbital will be filled before the 3d orbitals, the ground state electronic configuration of G will be 1s2 2s2 2p6 3s2 3p6 4s2. 3 Answer: A 1. Incorrect. Cu2O contains Cu+ ion. The ground state electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1. To f orm Cu+ ion, the electron will be removed from the 4s orbital. The ground state electronic configuration of Cu+ is 1s2 2s2 2p6 3s2 3p6 3d10. Cu+ ion does not have an unpaired d electron. 2. Incorrect . Zn(NO3)2 contains Zn2+ ion. The ground state electronic configuration Zn is 1s2 2s2 2p6 3s2 3p6 3d10 4s2. To f orm Zn2+ ion, the 2 electrons will be removed from the 4s orbital. The ground state electronic configuration of Zn2+ is 1s2 2s2 2p6 3s2 3p6 3d10. Zn2+ ion does not have an unpaired d electron. 3. Correct. Ti2(SO4)3 contains Ti3+ ion. The ground state electronic configuration of Ti is 1s2 2s2 2p6 3s2 3p6 3d2 4s2. To form Ti 3+, 2 electrons will be removed from the 4s orbital, followed by 1 electron from a 3d or bital. This document is copyrighted, please do not reproduce it without permission

2 © Raffles Institution 9729/ Year 5 Promo /2017 The ground state electronic configuration of Ti3+ is 1s2 2s2 2p6 3s2 3p6 3d1. Ti3+ ion has an unpaired d electron. H ence, only option 1 is correct. 4 Answer: D The actual bond length of N(1)≡N(2) (0.112 nm) in N2O is between that of N=N (0.120 nm) and N≡N (0.110 nm), which indicates that the N(1)≡N(2) bond in N2O is stronger than a N=N bond but weaker than a N≡N bond. Also, the actual bond length of N (2)→O (0.119 nm) in N 2O is between that of N –O (0.147 nm) and N=O (0.115 nm) , which indicates that the N(2) →O bond in N 2O is stronger than a N –O bond but weaker than a N=O bond. This is due to the delocalisation of the π electrons between N(1) and N(2) over the N(2)→O bond, which weakens the N(1)≡N(2) bond but strengthens the N(2)→O bond. 5 Answer: C amount of L = 0.058 46.0 = 1.2608 × 10–3 mol partial pressure of L = 1.2608 × 10−3 × 8.31 × 373 80 × 10−6 = 4.8850 × 104 Pa partial pressure of J = 101000 – 4.8850 × 104 = 5.2150 × 104 Pa mole fraction of J = 52150 101000 = 0.516 6 Answer: B A. I ncorrect. Though the radius of oxide ion is greater than that of the fluoride ion, it does not explain why the LE of MgO is more exothermic than that of NaF. B. Correct. The higher charge on Mg2+ ion compared to Na+ ion contributes to the more exothermic LE of MgO. C. Incorrect. The radius of the atom is not needed for predicting the magnitude of the LE of ionic compounds using the mathematical relation above. D. Incorrect. The difference in electronegativity is not needed for predicting the magnitude of the LE of ionic compounds using the mathematical relation above. This document is copyrighted, please do not reproduce it without permission

3 © Raffles Institution 9729/ Year 5 Promo /2017 7 Answer: D Students need to choose an option with a suitable method for monitoring the rate of the reaction. Also, in order to find the order of reaction with respect to H 2SO4, the experiment needs to be repeated with different initial [H 2SO4] each time at the same temperature to determine how the rate of the reaction changes with the initial [H2SO4]. A. Incorrect. The method for monitoring the rate of reaction is suitable as [H +] changes during the course of the reaction. However, the experiment should be repeated using a different initial [H2SO4] each time. B. Incorrect. The method for monitoring the rate of reaction is suitable as [ I2] changes during the course of the reaction. However, the experiment should be repeated using a different initial [H2SO4] each time. C. Incorrect. The method for monitoring the rate of reaction is suitable as electrical conductivity changes during the course of the reaction. However, the experiment should be repeated using a different initial [H 2SO4] each time at a constant temperature. D. Correct. The method for monitoring the rate of reaction is suitable as [ I2] and its colour intensity changes during the course of the reaction and the experiment is repeated using a different initial [H 2SO4] each time. 8 Answer: B The total volume of the reaction mixture in experiments 1 to 4 is kept constant (100 cm 3). Hence, initial [reactant] ∝ volume of reactant used. t is the time taken for the complete disappearance of the red colour of the indicator. W e are monitoring the time taken for a fixed and small amount of Br2 to be formed. Hence, initial rate ∝ 1/t. Consider experiments 1 and 2. W hen the volume of H 2SO4 is do ubled, the initial [H2SO4] doubled. With the other concentrations of reactants kept constant, the initial rate of the reaction quadrupled. Hence, the order of reaction with respect to H2SO4 is 2. Consider experiments 2 and 3. W hen the volume of KBrO 3 doubled, the initial [KBrO3] doubled. With the other concentrations of reactants kept constant, the initial rate of the reaction doubled. Hence, the order of reaction with respect to KBrO3 is 1. Using experiments 3 and 4, rate(3) rate(4) = 1 13 1 208 = k(3) k(4) 101 51 25n 50n 602 152 n = 1 Henc e, the order of reaction with respect to KBr is 1. A. Incorrect. Compared to experiment 4, the volume of each reactant and the total volume of the reaction mixture have doubled. The initial [reactants] remain the same as those in experiment 4. The rate, and hence, t, should be the same as that in experiment 4 (208 s). This document is copyrighted, please do not reproduce it without permission

4 © Raffles Institution 9729/ Year 5 Promo /2017 B. Correct. Rate = k[KBrO3][KBr][H2SO4]2. The overall order of the reaction is 4. C. Incorrect. The units of the rate constant are mol–3 dm9 s–1. D. Incorrect. Increasing the concentration of phenol added will increase the amount of Br 2 needed to react with phenol before discolouration of the red colour is observed. Hence, t will be larger. 9 Answer: A rate = k[H2O2][I–] Since [H2O2] << [I–], rate ≈ k’[H2O2], where k’ = k[I–] Since the reaction is approximated to a pseudo-first order reaction, t1/2 = ln 2 k' = ln 2 k[I–] = 80 s (reaction I) Since the [I–] in reaction II is half of that in reaction I, t1/2 = ln 2 k' = ln 2 k × 1 2 × [I–] = 80 1 2 = 160 s (reaction II) [H2O2] / mol dm–3 [I2] / mol dm–3 ratio of H2O2 to I2 no. of half-lives time taken / s 0.0200 0 - 0 0 0.0100 0.0100 1 : 1 1 160 0.0050 0.0150 1 : 3 2 320 0.0025 0.0175 1 : 7 3 480 10 Answer: A Since the initial volumes of SO2 and O2 added were equal, an equal number of moles of SO2 and O2 has been added, and the initial partial pressure must be the same. 2SO2(g) + O2(g) 2SO3(g) Initial pressure /atm x x 0 Change in pressure /atm –2y –y +2y Equilibrium pressure /atm x – 2y x – y 2y