RI 2022 Y5 Promotional Examination Suggested Solutions

2022 Y5 H2 Chemistry Promotion Examinations Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C B B D D D C C A A B A A C B Question 1 (C) From the question, combustion of CS2 and H2S gives CO2 and SO 2; the hydrogen in H 2S would be converted to H2O, i.e. CS2 + H2S + O2 ⎯→ SO2 + CO2 + H2O Balancing the equation gives CS2 + H2S + 9/2O2 ⎯→ 3SO2 + 1CO2 + H2O Question 2 (B) From period 7 of the Periodic Table, Assuming the ions have a nucleon number of 267, ion no. of protons no. of e– no. of neutrons = 267 – proton no. no. of neutrons – no. of e– Rf2+ 104 104 – 2 = 102 163 61 Db3+ 105 105 – 3 = 102 162 60 Sg4+ 106 106 – 4 = 102 161 59 Bh5+ 107 107 – 5 = 102 160 58 Question 3 (B) A 3rd IE of Na : Na2+(g) ⎯→ Na3+(g) + e– [He]2s22p5 [He]2s22p4 3rd IE of Ne : Ne2+(g) ⎯→ Ne3+(g) + e– [He]2s22p4 [He]2s22p3 Incorrect. Na 2+ and Ne 2+ have the same number of electron shells. Na2+ has a higher IE due to greater nuclear charge and approximately constant shielding compared to Ne2+. B Correct. The following equations show the 4th IE of the five elements. O3+(g) ⎯→ O4+(g) + e– [He]2s22p1 [He]2s2 F3+(g) ⎯→ F4+(g) + e– [He]2s22p2 [He]2s22p1 Ne3+(g) ⎯→ Ne4+(g) + e– [He]2s22p3 [He]2s22p2 Na3+(g) ⎯→ Na4+(g) + e– [He]2s22p4 [He]2s22p3 Mg3+(g) ⎯→ Mg4+(g) + e– [He]2s22p5 [He]2s22p4 The 4th electrons were all removed from the 2p subshell. D Incorrect. Successive IE involve successive removal of electrons from the same element, hence the nuclear charge does not change. This document is copyrighted, please do not reproduce it without permission

Question 4 (D) molecule shape polarity NF3 4 regions of e– density + 1 lone pair trigonal planar pyramidal polar CH2F2 4 regions of e– density + 0 lone pair tetrahedral non-polar SO2 3 regions of e– density + 1 lone pair linear bent polar PCl5 5 regions of e– density + 0 lone pair trigonal bipyramidal non-polar (The dipole pointing up is cancelled out by the dipole pointing down. The remaining 3 dipoles lie on a trigonal plane and cancel each other out exactly) Question 5 (D) 1 Incorrect. NaHF2 is an ionic compound which dissolves in water to form aqueous ions. These ions form ion-dipole interactions with water. 2 Incorrect. structure no. of lone pairs 4 2 no. of H 2 2 average no. of hydrogen bonds 2 2 H2O2 and N 2H2 form the same average number of hydrogen bonds i.e. they have the same extensiveness of hydrogen bonding. 3 Correct. SF4 is polar which allows it to form permanent dipole – permanent dipole interactions with other polar molecules. 4 Correct. Despite the presence of the –OH groups which allows the molecules to form intermolecular hydrogen bonds, CH3(CH2)17OH has a very long alkyl chain with a very large and polarisable electron cloud, causing its predominant intermolecular force to be instantaneous dipole -induced dipole interactions. This document is copyrighted, please do not reproduce it without permission

Question 6 (D) pV = (nR)T y = (m)x For an ideal gas, the graph of pV against T gives a straight line with a positive gradient passing through the origin. Question 7 (C) 1/2H2SO4 + NaOH ⎯→ ½Na2SO4 + H2O 0.04 mol of NaOH requires 0.02 mol of H2SO4 for reaction. Hence, H2SO4 is in excess and NaOH is limiting. Heat change, q = mcT = (20 + 20)(4.2)(14) = 2352 J Enthalpy change of neutralisation = – q nwater = – q nNaOH = – 2352 0.04 = –58800 J mol–1 = –58.8 kJ mol–1 Question 8 (C) 1 Incorrect. Since Mg +(g) is at a lower energy than Mg 2+(g), Mg +(g) is more stable than Mg2+(g). 2 Since Mg2+ has a higher charge and a smaller cationic radius than Mg+ (resulting in a smaller interionic distance for MgC l2), MgC l2 has a more negative lattice energy than MgCl. 3 Since MgC l2 is the more stable form, the standard enthalpy change of formation of MgCl2 is more negative than that of MgCl. Question 9 (A) For the forward reaction to be spontaneous, ∆G < 0 i.e. ∆H –T∆S < 0. (–197) – T0(– 189 1000) < 0 T0 < 197 0.189 = 1040 K (to 3sf) As T increases, the positive –T∆S term outweighs the negative ∆H. Hence, the value of ∆Gr becomes less negative i.e. increases with increasing temperature. Question 10 (A) This is a clock reaction. • Volume of reactant ⍺ [reactant] • Rate ⍺ [I2] t (given) ⍺ volume of I2 t Calculating the rates of each experiment experiment rate 1 4 1 = 4 2 2 0.5 = 4 3 4 2 = 2 4 8 8 = 1 Hence, option 3 is incorrect. The rates of experiments 1 and 2 are equally fast. Comparing experiments 1 and 2, when volume of iodine x 2, rate is constant i.e. order of reaction wrt I2 is 0. Comparing experiments 1 and 3, when volume of propanone x 2, rate x 2 i.e. order of reaction wrt propanone is 1. energy / kJ mol–1 Mg(g) Mg+(g) + e– 1st IE = 736 Mg2+(g) + 2e – 2nd IE = 1450 At high T, a real gas behaves more ideally and approaches the ideal gas line. At low T, a real gas deviates more from ideality and deviates from the ideal gas line. The intermolecular forces of attraction become significant, causing the gas to occupy a smaller volume. At a given temperature, the value of pV of a real gas will be smaller than an ideal gas. ideal gas T / K pV This document is copyrighted, please do not reproduce it without permission

Comparing experiments 1 and 4, when volume of sulfuric acid x 4, rate x 4 i.e. order of reaction wrt sulfuric acid is 1. Note: while volume of iodine is also halved, the order of reaction wrt iodine is 0 and does not affect the rate. Rate = k[propanone][H2SO4] Hence, option 2 is incorrect. Units of rate constant = mol dm–3 s–1 (mol dm–3)( mol dm–3) = mol–1 dm3 s–1 Hence, option 1 is correct. Question 11 (B) The initial rate of this reaction is very low (near 0) – option A is incorrect. Since this is an autocatalytic reaction, as time passes, more Mn 2+, which acts as the catalyst, is formed. This causes the reaction rate to increase. The reaction rate increases until a maximum before decreasing. This is because, while there is a high [Mn 2+], the concentration of the reactants decrease, causing a decrease in the reaction rate. The graph in option D describes what happens if a heterogeneous catalyst is present, but Mn 2+ is a homogeneous catalyst. The gradient of the graph in option C is proportionate to the rate of reaction. The rate initially starts slow (near 0 gradient) and increases all the way to the end of reaction. This is unlikely as the rate would slow down (i.e. the gradient should become more gentle) when the concentration of reactants such as MnO4– are very low. Question 12 (A) This question tests students on their knowledge of the axes of the Boltzmann distribution curve. Question 13 (A) p / atm 2Z(g) ⇌ X(g) + 2Y(g) ptotal Initial 0.86 0 0 Change –2x +x +2x Eqm 0.86–2x x 2x 1.11 At eqm, 0.86–2x + x + 2x = 1.11 ⇒ x = 0.25 Degree of dissociation = amt of Z dissociated initial amt of Z = 2x 0.86 since p ⍺ n = 2(0.25) 0.86 = 0.581 Option 1 is correct. Option 2 is incorrect. Addition of a catalyst increases both the forward and backward rate, resulting in the same value of Kp i.e. the position of equilibrium does not shift with the addition of a catalyst. The equilibrium partial pressure of X does not change. Option 3 is incorrect. The inert gas is added at constant volume, causing an increase in pressure that is proportional to the amount of inert gas adde