RI 2018 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2018 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 D D D B C D C C C B C D A A A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B A C D C A B D A B B C D D Q1 (D) Particle Direction of deflection Angle of deflection proton (positively charged) towards the negatively charged plate x° ( x° < y°) electron (negatively charged) towards the positively charged plate y° Since angl e of deflection ∝ � charge mass �, and a proton is heavier than an electron, the angle of deflection for proton is smaller, i.e. x° < y°. Q2 (D) The high charge density of Al3+ causes the distortion of the electron cloud of C l– to such an extent that electron sharing becomes predominant. Hence AlCl3 and A l2Cl6 are predominantly covalent compounds i.e. they exist as discrete molecules. Al2Cl6 is formed from the dimerisation of A lCl3, in which one C l atom of each AlCl3 donates a pair of electrons to the vacant, low -lying orbital of the A l atom in the neighbouring A lCl3 to form a dative covalent/coordinate bond (represented by ' →' from the Cl donor atom to the Al acceptor atom). Q3 (D ) Since the electronegativity of C, Se and C l are all different, all three molecules contain polar covalent bonds. Molecule no. of lp & bp Shape Bond dipoles cancel? Polar? CSe2 2 bp linear yes × SeCl2 2 bp 2 lp bent no √ CCl4 4 bp tetrahedral yes × Q4 (B ) Cs+, I− and Xe are isoelectronic species (i.e. with same total number of electrons) ⇒ their outermost e− experience the same shielding effect. However, nuclear charge of Cs+ > Xe > I−. ∴E lectrostatic attraction between the nucleus and the outermost e− of Cs+ > Xe > I−. ∴ E nergy required to remove the outermost e − decreases in the order: ∆H1 > ∆H3 > ∆H2 Q5 (C ) A Incorrect. (A mole of substance is the amount of that substance which contains as many elementary entities as there are carbon atoms in 12 g of carbon-12.) Since 1 mole of compound (e.g. CO 2) contains more than 1 mole of atoms (e.g. 1 mole of C and 2 moles of O atoms), it will not contain the same number of atoms as there are atoms in 12 g of carbon-12. B Incorrect. Relative isotopic mass = mass of 1 atom of the isotope 1 12 × mass of 1 atom of carbon-12 C Correct. Relative atomic mass = (weighted) average mass of 1 atom of the element 1 12 × mass of 1 atom of carbon-12 D Incorrect. Relative molecular mass = (weighted) average mass of 1 molecule of the substance 1 12 × mass of 1 atom of carbon-12 Q6 (D ) Atomic radius decreases across a period and increases down a group. Group Period 1 13 17 4 K Br 5 Rb In ∴ Rb has the largest atomic radius

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q7 (C ) At sea level, PO2 = 1/5 × 1 = 0.2 bar In the tank, PO2 = 0.2 bar Total pressure of gas mixture = 4 bar Percentage of O2 = 0.2 / 4 × 100% = 5% Q8 (C ) A Incorrect. 2Al3+(g) + 3O2−(g) → Al2O3(s) ∆Hlattice energy(Al2O3(s)) [only 1 mol of Al2O3(s) is formed] Recall: Lattice energy (LE) of an ionic compound is the energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions at 298 K and 1 bar. B Incorrect. H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) 2∆Hneutralisation [2 mol of H2O(l) are formed] Recall: Standard enthalpy change of neutralisation (∆Hneuto) is the energy change when an acid and a base react to form one mole of water at 298 K and 1 bar. C Correct. D S8(s) + 8O2(g) → 8SO2(g) 8∆Hformation (SO2(g)) [8 mol of SO2(g) are formed] Recall: Standard enthalpy change of formation (∆Hf) of a substance is the energy change when one mole of the pure substance in a specified state is formed from its constituent elements at 298 K and 1 bar. Q9 (C ) Recall: Bond energy (BE ) of a bond is the average energy absorbed when 1 mole of the bonds are broken in the gaseous state. -890 kJmol-1 C(g) + 4H(g) + 4O(g) 2 ∆ Hvap H C H H H (g) + 2 O O (g) O C O (g) + 2 O H H (l) O C O (g) + 2 O H H (g)4 BE(C-H) + 2 BE(O=O) 2 BE(C=O) + 4 BE(O-H) By Hess’ law, 2∆Hvap + (–890) + 2 BE(C=O) + 4 BE (O-H) = 4 BE(C-H) + 2 BE(O=O) 2∆Hvap + (–890) + 2(805) + 4(460) = 4(410) + 2(496) 2∆Hvap = +72 kJ mol−1 ∆Hvap = +72 / 2 = +36 kJ mol−1 Q10 (B ) 2I− → I2 + 2e− n(I−) = n(e−) = 0.01 mol n(I2) formed from oxide = 0.006 – ½(0.01) = 0.001 mol 0.001 mol of oxide contains 2 × 0.001 mol of I Mole ratio of oxide : I (in oxide): e− = 0.001 : 2(0.001) : 0.01 = 1 : 2 : 10 = 0.5 : 1 : 5 1 mol of I (in oxide) accepts 5 mol of e− to form I2. oxidation number of I (in oxide) + 5 (–1) = 0 ∴ ox idation number of I (in oxide) = +5 Q11 (C ) A Incorrect. rate = k[NO2] Overall eqn: CO + NO2 → CO2 + NO B Incorrect. rate = k[NO2]2 Overall eqn: CO + 2NO2 → CO2 + 2NO + O C Correct. rate = k[NO2]2 Overall eqn: CO + NO2 → CO2 + NO D Incorrect. rate = k[NO2][CO] Overall eqn: CO + NO2 → CO2 + NO Q12 (D ) For constant temperature and amount (or conc) of catalase (enzyme): • At low [H2O2], not all of the catalase active sites are occupied. Rate ∝ [H2O2] and reaction is first order wrt H2O2 (substrate). • At high [H 2O2], the active sites of catalase become saturated with H2O2 (substrate). Further increase in [H2O2] will not have any effect on the reaction rate. Reaction is zero order wrt H2O2. Q13 (A ) Formula conc. of cation / mol dm−3 charge / radius A Al2(SO4)3 0.2 3 / 0.050 = 60.0 B CuSO4 0.1 2 / 0.073 = 27.4 C MgSO4 0.1 2 / 0.065 = 30.8 D Na2SO4 0.2 1 / 0.095 = 10.5 The hi gher the charge density (charge / radius), the greater the extent of hydrolysis of the cation to form H+ ions. Al2(SO4)3 forms the most acidic solution as it contains the highest concentration of the cation with the greatest charge density.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q14 (A ) H2O(l) ⇌ H+(aq) + OH–(aq) As temperature increases from 10 oC to 40 oC, • Kw increases ⇒ equilibrium position lies more to the right • greater extent of self-ionisation of water • hi gher conc of H+ and OH– (note: [H+] is still equal to [OH–]) • pH ↓ (since pH = – lg [H+]) Q15 (A ) In order to maintain the pH at about 10 (>7), an alkaline buffer consisting of a weak base and its conjugate acid must be used. A Correct. Weak base: NH3 ; Conjugate acid: NH4+ B Incorrect. Weak base: NH3 ; Strong base: NaOH [conjugate acid of NH3, i.e. NH4+, is absent] C Incorrect. Strong base: NaOH ; Weak base: CH3COO− [conjugate acid of CH3COO−, i.e. CH3COOH, is absent] D Incorrect. Only strong base, NaOH, is present Q16 (A ) Summing the first two equilibria: AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2Cl(aq) ∆G1 + ∆G2 Summing the last two equilibria: AgBr(s) + 2NH3(aq) ⇌ Ag(NH3)2Br(aq) ∆G3 + ∆G4 The more negative (or less positive) ∆G is, the more thermodynamically favourable / spontaneous the reaction. 1 Correct. Since AgCl is more soluble in NH3 than AgBr, (∆G1 + ∆G2) < (∆G3 + ∆G4) 2 Correct. In the aqueous medium, the spectator ions in both the second and fourth equations can be removed to obtain the following ionic equation: Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) ∴ ∆G2 = ∆G4 3 Incorrect. See explanation for option 2. 4 Correct. From options 1 and 2, (∆G1 + ∆G2) < (∆G3 + ∆G4) and ∆G2 = ∆G4 ∴ ∆G1 < ∆G3 [Values of Ksp of AgCl = 2.0 × 10−10 ; Ksp of AgBr = 5.0 × 10−13] Q17 (B) If a redox reaction occurs, I− must be oxidised and H2O2 must be reduced. E cell = Ecathode – Eanode = E(H2O2/H2O) – E( I2/ I−) = +1.77 – (+0.54)