RI 2019 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2019 A-Level H2 C hemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B C B C C A D B D D A B D D D 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B A A D D C A C B B C C B C(B) Q1 (B) 84 38 Sr has 38 protons, 46 neutrons, 38 e– ⇒ W, X and Y have 46 neutrons 84 2 38 Sr + has 38 protons, 46 neutrons, 36 e– ⇒ W, X– and Y2– have 36 e– ⇒ W has 36 e–, X has 35 e– and Y have 34 e– ⇒ W has 36 protons, X has 35 protons and Y have 34 protons Nucleon no. of W = 46 neutrons + 36 protons = 82 Nucleon no. of X = 46 neutrons + 35 protons = 81 Nucleon no. of Y = 46 neutrons + 34 protons = 80 Q2 (C) Electrons, being negatively charged, are deflected towards the (+) plate. Protons, being positively charged, are deflected towards the (–) plate. Electrons, having the same magnitude of charge but lighter than protons, have a higher charge/mass ratio, and are deflected more than protons. Q3 (B) The ionization energy (IE) data for As, Sb, Se and Te are not available in the Data Booklet. For group 16 element, ns2 np4 ns2 np3 ns2 np2 ns2np1 ns2 For group 15 element, ns2 np3 ns2 np2 ns2 np1 ns2 ns1 IE 4 of a Group 15 element involves the removal of e– from the inner ns subshell, requiring a larger than expected amount of energy compared to IE 3 which involves the removal of e – from the higher energy np subshell. Comparing the values of IE 3 and IE4 in the options, options A and C have large differences between IE3 and IE4 i.e. A and C are group 15 elements. B and D corresponds to Group 16 elements. Since Tellurium is below Selenium, the IE’s of Tellurium are lower than that of Se since IE’s decrease down the Group. Hence, B is Tellurium. Q4 (C) After forming a single bond between the two oxygen atoms, each oxygen will gain one more e – to achieve octet configuration i.e. peroxide has a 2– charge. Barium, a group 2 element, forms a cation with a 2+ charge. Q5 (C) A All bonds are sigma bonds. B CO2 contains two sigma bonds and two pi bonds. C CH3CHO contains six sigma bonds and one pi bond. D CH2CHCH3 contains eight sigma bonds and 1 pi bond. IE1 IE2 IE3 IE4 IE1 IE2 IE3 IE4

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q6 ( A) r pV nRT masspV RT M = = Since the volume, mass, and temperature for every gas are kept constant, 1 r p M∝ i.e. higher M r ⇒ lower P. Compound CH4 HCHO CH3Cl HCO2H Mr 16.0 30.0 50.5 46 Relative Pressure 1 (highest) 2 4 (lowest) 3 Q7 (D) 2NaN3(s) → 2Na(s) + 3N2(g) Amount of NaN3 reacted = 5.00 23.0 3(14.0)+ = 0.07692 mol Amount of N2 gas produced = (3/2)(0.07692) =0.1154 mol Volume of N2 gas = nRT p = 4 0.1154(8.31 )(273 30.0) 9.85 10 + × = 0.00295 m3 = 2.95 dm3 Q8 (B ) From the question, Germanium (a period 4, group 14 element) will have similar properties to Silicon (a period 3, group 14 element). Since silicon has high melting point and is a semiconductor, it can be predicted that Germanium has the same properties. Q9 (D ) From QA notes, solution Y contains Mg 2+ since, when reacted with NaOH, a white ppt is formed which is insoluble in excess NaOH. Gas Z contains the other period 3 element, which reacts with air to give water and a white solid that is insoluble in dilute acid or alkali. The oxide of phosphorus, P 4O6, is acidic and therefore reacts with alkali. The oxide of silicon, SiO 2, does not react with acids and only reacts with concentrated alkaline solutions. Note: Gas Z is actually SiH 4 and is formed from the reaction of X with HCl(aq). MgSi(s) + 4HCl(aq) → MgCl2(aq) + SiH4(g) X Y Z Q10 (D) A Electron affinity decreases down the group. B Electronegativity decreases down the group. C The chemical reactivity of the elements is determined by the valence electrons. The increase in nuclear charge down the group is outweighed by the increased shielding effect experienced by the valence electron from the nucleus. This results in weaker electrostatic attraction between nucleus and valence electron, and an increased ease of loss of their valence electrons. Hence, the increase in nuclear charge does not result in increase in reactivity. D Due to the increase in the number of electron shells down the group, the valence electrons are further away from the nucleus and experience greater shielding effect , decreasing the attraction between the nucleus and the valence electrons. Hence, down the group, less energy is required for the atoms to lose their valence electrons in a reaction. Q11 (A ) B2O3 PbO mass in 100g of solder glass / g 16 84 amount / mol 16 2(10.8) 3(16.0) 0.2299 + = 84 207.2 16.0 0.3763 + = Amount of Pb = 0.3763 mol Since 1 mol of B 2O3 contains 2 mol of B, amount of B = 2(0.2299) = 0.4598 mol. Mole ratio of Pb/B = 0.3763 0.818 0.820.4598 = ≈

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q12 (B ) A The equation for ∆H1 shows gaseous H2O being formed as the product of combustion. For the standard enthalpy change of combustion (i.e. at 298 K and 1 bar) , liquid H2O should be produced. B The equation for ∆H2 correctly shows the standard enthalpy change of formation of liquid C6H12 from the corresponding elements in their standard states i.e. C graphite(s) and H2(g). C The standard enthalpy change of atomisation of H2(g) is the energy required to form 1 mole of H(g) from H 2(g) under standard conditions i.e. ½ H2(g) → H(g). Since ∆H3 involves 6H2(g) → 12 H(g), ∆H3 = 12 x ∆Hatomisation of H2(g). D This option used the standard enthalpy of formation which involves the formation of the required substance from the corresponding elements in their standard states . The standard state of hydrogen is H 2(g) and not H(g). For the option to be correct, the reaction should read 6C graphite(s) + 6H2(g) + 9O 2(g) → 6CO2(g) + 6H2O(g). Q13 (D) 1 Incorrect. The reaction involves an increase in entropy due to the formation of a gaseous product (which is more disordered) from non- gaseous reactants (which are less disordered) i.e. ∆S > 0. Since ∆H < 0 and ∆S > 0, ∆G = ∆H – T∆S < 0 for all temperatures i.e. ∆ G is negative at 20 °C. 2 Correct. It can be deduced from the information provided that the given decomposition reaction occurs very slowly at room temperature, meaning that the reaction has very high activation energy. 3 Incorrect. See explanation of option 1. Q14 (D ) 1 Incorrect. As can be seen from the graph, if [substrate] is increased beyond x, the graph is a horizontal line. This means that the value of initial rate does not change when [substrate] is increased beyond x. 2 Incorrect. When [substrate] is increased beyond x, the initial rate does not change i.e. the reaction has become zero order with respect to the substrate. 3 Correct. When [substrate] is sufficiently high (at x and beyond), all the active sites of the enzyme are occupied. There are no active sites available to catalyse the reaction Q15 (D ) 1 Incorrect. Changing the concen tration of the reactants has no effect on the rate constant. 2 Incorrect. Increasing the concentration increases the number (or amount ) of particles having energy greater than activation energy. The pr oportion remains the same. This is because the total number of particles is greater and the number of particles with energy greater than activation energy will be proportionally greate