2021_VJC_H3_Chemistry_Prelims_Answer

Page 1 VICTORIA JUNIOR COLLEGE PRELIMINARY EXAMINATION 2021 CHEMISTRY (Higher 3) Answers 1 (a) (i) Gaseous particles exert negligible intermolecular forces on each other. Volume of gas particles are negligible compared to the volume of the container. [2] (ii) H2(g) ⇌ H2(aq) initial amount of H2 / mol 0 6.96 × 10–4 change in amount of H2 / mol x – x final amount of H2 / mol x 6.96 × 10–4 – x Using ideal gas equation, PH2Vgas = xRT PH2 = -----(1) = 24.5 x (PH2 is in atm) Using equation 1.1, PH2 = = x = 6.96 × 10–4 – 7.82 × 10–4 × PH2 -----(2) sub eqn (2) into (1), PH2 = 24.5 × (6.96 × 10–4 – 7.82 × 10–4 × PH2) 1.019PH2 = 1.7052 × 10–2 PH2 = 1.67 × 10–2 atm [3] (iii) Partial pressure of H 2 will increase as solubility of H 2 decreases with increasing temperature . As temperature increases, kH, which is similar to Kc, decreases. As temperature increases, more energy is available to overcome the permanent dipole -induced dipole interactions between H 2 and H 2O, allowing more gaseous H 2 molecules to form. [2] (b) (i) TCE underwent reduction as the ON of C bonded to H decreases from 0 in TCE to –2 in the product. [1]

Page 2 (ii) [2] (c) (i) To allow time for equilibrium to be established between the aqueous phase and the headspace. . [1] (ii) From graph, concentration of H2 = 5.9 ppm Partial pressure of H2 = 5.9 × 10–6 × 1 atm = 5.9 × 10–6 atm Assuming temperature is constant throughout the experiment and any increase in pressure due to H 2 is negligible (assumption is valid since 5.9 × 10–6 << 1). [3] (iii) SH2 = 5.9 × 10–6 ÷ (6.87 x 104) = √√8.59 × 10–11 (no units) [H2] dissolved = no. of moles of H2 ÷ volume of H2O = 8.59 × 10–11 ÷ (1 × 18 ÷ (0.998)) × 103 = 4.76 × 10–9 mol dm–3 Since 4.76 × 10 –9 mol dm –3 > 1.0 × 10 –9 mol dm –3, it is likely that anaerobic biodegradation was taking place in this sample. [3] 2 (a) (i) [1] (ii) Ratio of M : M+2 : M+4 = ¾ x ¾ : 2 x ¾ x ¼ : ¼ x ¼ = 9 : 6 : 1 [1] (iii) Na(s) [1]

Page 3 (iv) [3] (v) E: 6.6 and 7.1 ppm, 2 doublets, 2 protons each, aromatic protons F: 7.1 ppm, complex multiplet, 4 protons, aromatic protons [1] (vi) Chemical Shift / ppm No. of Protons Multiplicity Deduction / Structural Features 1.5 6 s -C(CH3)2 6.6 4 d Two 1,4-disubstituted benzene, aromatic protons deshielded by magnetic anisotropic effect 7.0 4 d 9.1 2 s 2 phenol protons, they are labile and disappear in presence of D2O These protons deshielded by electronegative oxygen atom and magnetic anisotropic effect of benzene G is symmetrical (due to very simple 1H NMR spectrum) G: [3] (vii)