EJC 2022 GCE A-Level H1 Chemistry Paper 1 Suggested Solution

Eunoia Junior College 8873 H1 Chemistry 2022 Paper 1 Suggested Solution 1 Only A and B with 2 protons has a charge of +2 on the nucleus. The mass of a n atom or ions is due primarily to protons and neutrons in the nucleus as an electron is ~ 1 1836 the mass of a proton or neutron. B with 4 nucleons (neutrons and protons) will have a greater mass than A with 3 nucleons.  B 2 A : 22Ti: 1s2 2s2 2p6 3s2 3p6 3d2 4s2 → 22Ti2+: 1s2 2s2 2p6 3s2 3p6 3d2 B : 24Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 C ✓: 26Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 D : 29Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 → 29Cu+: 1s2 2s2 2p6 3s2 3p6 3d10  C 3 Atomic radius decreases across the period due to increasing effective nuclear charge. Excluding Group 18, the element with the smallest atomic radius in its period must be a Group 17 element. For an element to have o nly 2 electrons in its p sub -shell, the element must have a valence electronic configuration of ns2 np2, i.e. in Group 14.  A 4 As is [Ar] 3d10 4s2 4p3 (Period 4, Group 15) Sb is [Kr] 4d10 5s2 5p3 (Period 5, Group 15) Se is [Ar] 3d10 4s2 4p4 (Period 4, Group 16) Te is [Kr] 4d10 5s2 5p4 (Period 5, Group 16) There will be a significant jump betwee n the 3rd and 4th IE for a Group 15 element (A and C) since the 4 th electron is removed from a lower energy ns subshell, while the first 4 electrons are removed from the np subshell for a Group 16 element , showing a more gradual increase (B or D). A period 5 element in the same group will have lower xth IE compared to a period 4 element as the electrons removed are further away from the nucleus.  B 5 1 ✓: Ice adopts an open hexagonal lattice structure due to hydrogen bonding , resulting in ice having a lower density than water. 2 ✓: The high surface tension of water is due to the strong intermolecular hydrogen bonds between water molecules on the surface. 3 : Water can act as an acid since it can donate a H +, and as a base since it can accept a H +, exemplified by its auto-ionisation: H2O(l) + H2O(l)  H3O+(aq) + OH–(aq) 4 : The pH of pure water at 25 ºC is 7.00 due to the auto-ionisation of water: H2O(l) + H2O(l)  H3O+(aq) + OH–(aq) resulting in equal concentration of H+ and OH– of 10–7 mol dm–3 each.  B 6 The Arrhenius de finition of a base is a substance that dissociates to produce hydroxide ions in aqueous solution.  C 7 pH = –lg [H+] [H+] at pH 4 = 10–4 mol dm–3 [H+] at pH 6 = 10–6 mol dm–3 change in [H+] = 6 4 10 1 10 100 − − =  B 8 ( ) ( )2Ba OH 3 1.00 137.3 16.0 1.0 2 5.838 10 mol n − = + +  = ( ) 3 3 2 5.838 10Ba OH 0.01168 mol dm500 1000 − − == Ba(OH)2 → Ba2+ + 2OH– ( ) 3 2OH 2 Ba OH 0.02335 mol dm−−  ==  pOH lg OH 1.63 pH 14 pOH 12.37 −=− = = − =  D 9 A buffer contains a large reserv oir of a weak acid, HA, and its conjugate base, A–. The conjugate base A – is able to remove small amount of H+ added in: A– + H+ → HA The weak acid HA is able to remove small amount of OH– added in: HA + OH– → A– + H2O  B 10 A : P4O10 dissolves in water to give an acidic solution (H3PO4; pH < 7) B : Al2O3 is a solid with strong ionic bonds between Al3+ and O 2–, hence very high melting point , while P 4O10 has a simple covalent st ructure and does not conduct electricity at r.t. C : Both A lCl3 and PC l5 reacts rapidly with water, producing acidic fumes. D ✓: Both A lCl3 and PC l5 reacts rapidly with water, producing acidi c fumes . In both cases, an acidic solution results, which turns blue litmus red. 2AlCl3 + 3H2O → Al2O3 + 6HCl PCl5 + 4H2O → H3PO4 + 5HCl  D 11 When an atom loses electron(s) to form a cation, the re is a decrease in shielding, while nuclear charge remains the same. Hence effective nuc lear charge increases, resulting in a decrease in radius. When an atom gains electron(s) to form an anion, the re is a n inc rease in shielding, while nuclear charge remains the same. Hence effective nuclear charge decreases, resulting in an increase in radius.  A 12 3 Ag 0.216 deposited 2.002 10 mol107.9n −= =  3 23 21 total no. of Ag atoms deposited 2.002 10 6.02 10 1.205 10 −=    = 3 21 18 no. of Ag deposited per cm 1.205 10 150 8.03 10 = =  A 13 1 : They can have different functional groups, undergoing different chemical reactions. E.g. 2 : The compounds can be made up of different multiples of CH 2O, hence different Mr. 3 : Compounds made up of diff erent multiples of CH2O are not isomers as they have different numbers of atoms, i.e. different molecular formula 4 ✓: Consider (CH2O)n. ( ) percentage by mass of C 12.0 100%12.0 1.0 2 16.0 12 100% 40%30 n n n n = +  + =  =  D 14 4 3 CH 6.4 10 400 mol12.0 1.0 4n == + 24H CH 3 1200 molnn == 22 3 H H m 3 1200 24 dm 28 800 dm V n V=  =  =  D 15 2HI → I2 + 2H+ + 2e– Since HI reacts with HNO3 in a 2 : 1 ratio, the nitrogen in the +5 oxidation state in each molecule of HNO 3 accepts two electrons, becoming +3 in the oxide.  C 16 1 : The activation energy for reaction X is the same as the activation energy for the slowest stage, i.e. stage Y. So the magnitude should be E1. 2 ✓: The magnitude of activation energy of stage Y is the difference between the energy of P and the transition state, E1. The magnitude of the enthalpy change for stage Y is t he difference between the energy of P and intermediate Q, E3. 3 ✓: The magnitude of activation energy of stage Z is the difference between the energy of intermediate Q and the transition state, E6. The magnitude of the enthalpy change for stage Z is the difference between the energy of intermediate Q and R, E7.  C 17 Lattice energy is the energy released when one mole of a solid ionic lattice is formed from its constituent gaseous ions, all under standard states. Ca2+(g) + O2–(g) → CaO(s)  D 18 L.E. qq rr +− +−  + is always exothermic. A : 2Mg Na 2qq ++= ( ) ( )2L.E. MgC L.E. NaC ll B : 22Mg Carr ++ ( ) ( )L.E. MgO L.E. CaO

C ✓: FCrr − −l ( ) ( )L.E. KC L.E. KF l D : Na Krr ++ ( ) ( )L.E. NaBr L.E. KBr  C 19 2Br(g) + 6F(g) → 2BrF3(g) involves forming 6 Br–F bonds. Using the cycle and the information that 1 2 Br2(l) + 3 2 F2(g) → BrF3(l) fH 1300 kJ mol−=− ( ) ( ) ( ) ( ) 1 6B.E. Br–F 700 2 300 90 1210B.E. Br–F 202 kJ mol6 − − =− + +  − + −= =+−  B 20 Let rate = k[CNO–]n[NH + 4]m . Using expt 1 and 2, ( ) ( ) ( ) ( ) 0.5 2.55.80 1.16 0.5 0.5 5 5 1 nm nm m k k m = =  = Using expt 2 and 3, ( ) ( ) ( ) ( ) ( ) 1 1 0.5 0.51.16 23.20 2.0 2.5 11 120 54 n n n k k n = =  = Using expt 2 and 4, ( ) ( ) ( ) ( ) 11 11 0.5 0.51.16 X 1.0 2.0 1.16 1 X 9.28X8 k k = =  =  C 21 A ✓: Adsorption of reactants on the heterogeneous catalyst leads to formation of bonds with the catalyst and weake ning of bonds within the reactants, hence lowering Ea. B : A c atalyst speeds up a reaction by providing an alternative pathway with lower Ea. C : Reactants (alkene and hydrogen) are adsorbed on the nick el. After reaction, the products (hydrogenated alkene) must be desorbed to release active sites for other reactants to bind. D : There is no change to the enthalpy change for the overall reaction as the reactants and products are the same. The catalyst only provides a n alternative pathway with lower Ea.  A 22 The spread of energy among the particles increases with temperature. Hence curve 2 corresponds to a higher temperature than curve 1. Rate constant, k, increases with temperature. Hence curve 2 applies to the reaction occurring with the higher rate constant. Point X indicates particles with energy E1, which is lower than E2 at point Y.  D 23 At the start, t = 0, there are no PCl3(l) and Cl2(g). Hence rate of reverse reaction is 0. As the PCl5(s) is warmed, it starts to dissociates to form more PC l3 and Cl2.