2018 HCIS IB2 HL Chem Prelim P2 (ans)

2018 IB2 HL Chem Prelim P2 – Answers 1 2018 PRELIMINARY EXAMINATION International Baccalaureate 2 Chemistry Higher level Paper 2 1. (a) (i) Weighted average mass of its isotopes when compared to 1 12 the mass of a 12C atom. [1] (ii) Ar = 69(65)+71(35) 100 = 69.7 [1] (b) (i) 1s2 2s2 2p6 3s2 3p6 3d10 [1] (ii) Negative Decrease in number of gaseous particles, less disordered [1] [1] (iii) ΔG = ΔH – TΔS Since ΔH < 0 and ΔS < 0, For a reaction to be spontaneous, ΔG < 0. Hence, |ΔH| > |TΔS| Reaction is only spontaneous at low temperatures. [1] [1] (iv) nGa = 10.0 69.72 = 0.1434 mol nN2 = 1 2 (0.1434) = 0.07172 mol vol N2 = 0.07172(22.7) = 1.63 dm3 Absolute uncertainty = 0.2 69.72 ( 1 2 )(22.7) = 0.04 dm3 (1sf) [1] [1] [1] (c) (i) nNaOH = 14.00 1000 (0.100) = 0.0014 mol [1] (ii) nHCl reacted with NaOH = 0.0014 mol total nHCl = 200 1000 (0.300) = 0.06 mol nHCl in excess = 0.06 – 0.0014 = 0.0586 mol [1] [1] (iii) nGa2O3 = 0.0586 ( 1 6) = 0.009767 mol mGa2O3 = 0.009767 (69.72 × 2 + 16.00 × 3) = 1.83 g [1] [1]

2018 IB2 HL Chem Prelim P2 – Answers 2 2. (a) Na2O + H2O → 2NaOH NaOH releases OH− ions. pH = 13 P4O6 + 6H2O → 4H3PO3 H3PO3 releases H+ ions. pH ≈ 2 / 3 / 4 [Award 1 mark only if both pH are stated correctly.] [1] [1] [1] (b) (i) Both have giant ionic structure / ionic lattice structure, with strong ionic bonds. Charge of Mg2+ > Na+ and radius of Mg2+ < Na+ MgO has stronger ionic bonds, more energy needed to overcome. [1] [1] [1] (ii) SiO2 has giant covalent structure, and strong Si−O covalent bonds. P4O6 has simple covalent/molecular structure, and weak van der Waals’ forces / intermolecular forces. Covalent bonds are stronger than van der Waals’ forces, more energy required to overcome. [Award full 3 marks only if comparison of the strength of the bonds to be overcome is stated.] [1] [1] [1] (c) (i) Lewis structure I Lewis structure II [Award 1 mark each for each correct Lewis structure.] [Award 1 mark each for correct formal charges indicated on both S AND O atoms on each structure.] [4] (ii) Lewis structure I. All atoms have a formal charge of 0. Recap: More stable: (i) structure with greater no. of 0 FC; (ii) negative FC on electronegative atom e.g. F, O, N [1] (iii) sp3 (SiO2 is like diamond) [1] (d) (i) Energy change when 1 mol of MgO (s) is formed from its elements Mg (s) and O2 (g) under standard conditions. [Accept general definition.] [1] (ii) A : enthalpy change of atomization of Mg D : Sum of 1st and 2nd electron affinity of O (g) [1] [1] (iii) ΔH = −(−602) + (702) + (2186) + (248) + (150) = +3888 kJ mol−1 [Award 1 mark for final answer only if sign is shown.] [1] [1] (iv) 2nd IE = 2186 – 738 = +1448 kJ mol−1 After an electron is removed, the remaining electrons experience greater [1] [1]

2018 IB2 HL Chem Prelim P2 – Answers 3 nuclear attraction / more strongly attracted, More energy required to remove the 2nd electron. 3. (a) (i) When [HI] doubles, rate quadruples. Order of reaction w.r.t. HI = 2 [1] (ii) Rate = k[HI]2 [1] (iii) 0.41 = k (1.67)2 k = 0.147 Units of k = mol−1 dm3 s−1 [1] [1] (iv) energy Reaction progress [Axis must be clearly and correctly labelled. Energy of reactants lower than products. Ea of forward reaction labelled.] [Award 1 mark for correct value and sign of ΔH.] [3] (v) Catalyst provides an alternative pathway of lower activation energy. More particles have energy ≥ activation energy Frequency of effective collisions increases [Do not accept frequency of collisions. Accept number of effective collisions per unit time.] [1] [1] [1] (b) (i) Kc = [H2][I2] [HI]2 [1] (ii) [HI] initial = 0.10 2 = 0.05 mol dm−3 [HI] equilibrium = 0.564 2 = 0.0282 mol dm−3 HI H2 I2 Initial 0.05 0 0 Change −0.0218 +0.0109 +0.0109 Eqm 0.0282 0.0109 0.0109 Kc = (0.0109)(0.0109) (0.0282)2 = 0.149 No units [1] [1] [1] HI (g) H2 (g) + I2(g) +21 kJ mol−1

2018 IB2 HL Chem Prelim P2 – Answers 4 (iii) No effect on / No shifting of position of equilibrium Equal number of gaseous particles on both sides of the equation. [1] [1] 4. (a) (i) Electrophilic addition [1] (ii) H I I-+ + slow+ - I-+ I [Two step mechanism. Step 1 is the slow step. Equation must be balanced. Correct direction of arrows to indicate movement of electrons (double headed arrow) Lone pair of electrons on I− must be shown.] [3] (iii) Addition polymerization [1] (iv) C H H C H CH2 C H H C H CH2 [1] (b) (i) O O [1]

2018 IB2 HL Chem Prelim P2 – Answers 5 (ii) C I H H2C CH3 C I H CH2 H3C [Correct chiral carbon and 3D diagram. Correctly drawn mirror images.] [1] [1] (iii) Both SN1 and SN2 occurs. SN2 gives ≈ 100% of inversion of configuration, while S N1 gives ≈ 50% of each configuration. [1] [1] (iv) C−I bond is weaker than C−Br bond. [Students can quote bond length or bond energy to explain.] [1] (v) Concentrated HNO3, concentrated H2SO4, 30 oC ES K2Cr2O7, H2SO4 (aq), heat / acidified K2Cr2O7 (aq), heat oxidation [1] [1] (c) (i) Chemical shift / ppm Splitting Integration factor 1.2 Doublet 3 2.0 Singlet 1 2.7 Doublet 2 3.8 Multiplet 1 7.2 Singlet 5 [Award full marks for all correct answers. Award 1 mark for at least 2 correct answers.] [2] (ii) C Ha Ha C OH Hb C Hc Hc Hc Hb has two different neighbouring environment. Ha splits the signal of Hb into a triplet Hc splits the signal of Hb into a quartet. They overlap into multiplet. [OWTTE] [1] [1] (d) One similarity : Strong peaks at 2850 – 3090 cm−1 (C–H bond) One difference : (2−iodopropyl)benzene shows a peak at 490 – 620 cm −1 (C–I) absent in 1−phenylpropan−2−ol / [1] [1]

2018 IB2 HL Chem Prelim P2 – Answers 6 1−phenylpropan−2−ol shows a peak at 3200 – 3600 cm −1 (O–H), absent in (2−iodopropyl)benzene 5. (a) (i) Ka = [C5H7CO2 -][H+] [C5H7CO2H] [1] (ii) x2 0.22 = 7.4 × 10−4 x = 0.0128 mol dm−3 pH = 1.89 [1] (iii) A solution that resists changes in pH when small amounts of acid or base is added. [Accept H+ or OH− is added.] [1] (iv) When H+ is added, C5H7O4CO2− + H+ → C5H7O4CO2H When OH− is added, C5H7O4CO2H + OH− → C5H7O4CO2− + H2O [1] [1] (v) pH = pKa for a solution at maximum buffer capacity pH = −log10 (7.4 × 10−4) = 3.13 [1] (b) H2O H+ + OH− ΔH > 0 / endothermic Increasing temperature shifts position of equilibrium to the right to favour the forward endothermic reaction, [H+] increases, pH decreases [1] [1] 6. (a) (i) Anode : H2O (l) → ½ O2 (g) + 2OH– (aq) + 2e− Cathode : Cu2+ (aq) + 2e− → Cu (s) [1] [1] (ii) Effervescence occurs at the anode / Pink copper deposited at the cathode / Solution becomes paler blue. [Accept any two observations] [2] (iii) Current / Time [1] (iv) Cl2 (g) ½ Cl2 (g) + e− Cl− (aq) +1.36 V ½ O2 (g) + 2H+ (aq) + 2e− H2O (l) +1.23 V Increasing [Cl−], (eqm shift left to decrease [Cl–]) E < 1.36 V, falls below 1.23 V Cl− is more likely to be oxidized [1] [1] [1] (v) Cu2+ has partially filled 3d orbitals 5 degenerate 3d orbitals split into two sets of orbitals with different energy An electron is excited from lower energy d orbital to higher energy d [3]

2018 IB2 HL Chem Prelim P2 – Answers 7 orbital, Absorbing energy / light from the visible light spectrum. Colour observed is complementary to the colour absorbed. [4 required points. Deduct 1 mark for each missing point.] (b) (i) Maintain charge neutrality / Allows movement of ions between half−cells to complete the circuit [1] (ii) E = (+0.34) – (−0.45) = +0.79 V [1] (iii) From Fe elec