HCI H2 Chem Prelim P3 Ans

HWA CHONG INSTITUTION 2008 C2 H2 CHEMISTRY 9746 PRELIMINARY EXAMINATION PAPER 3 FREE RESPONSE (SUGGESTED ANSWERS) 1 (a) (i) Dynamic equilibrium refers to a state of balance in a reversible reaction or process in which the rates of the forward and reverse processes are equal. (ii) OH] H [C H] CO [CH O] [H ] H C CO [CHK 5 2 2 3 2 5 2 2 3c = CH3CO2H + C2H5OH CH3CO2C2H5 + H2O = Initial mol: 0.5 1.0 0 0 Eqm mol: 0.5 – 0.42 1.0 – 0.42 = 0.58 0.42 0.42 = 0.08 Kc = V 0.58 V 0.08 V 0.42 V 0.42 × × (where V = total volume of mixture) = 3.80 (iii) NaOH neutralises the ethanoic acid. [CH3CO2H] decreases. By Le Chatelier’s Principle, position of equilibrium shifts to the left. (iv) CH3COCl (b) (i) Ka = H] CO [CH ] [H ] CO [CH 2 3 2 3 + − No. of moles of KOH used = 0.61000 30 × = 0.0180 No. of moles of CH3CO2H = No. of moles of KOH used = 0.0180 [CH3CO2H] in vinegar sample = 25/1000 0.018 = 0.720 mol dm–3 pH of vinegar sample = 2.44, ∴ [H+] = 10–2.44 = 3.63 × 10–3 mol dm–3 ∴ Ka = 3 2 3 10 3.63 0.72 ) 10 (3.63 − − × − × = 1.84 × 10–5 mol dm–3 (shown) HCI 2008 1

(ii) Solution A contains unreacted CH3CO2H and its salt, CH3CO2 –K+. It is a buffer. When a small amount of acid is added, CH3COO– + H+ → CH3CO2H. When a small amount of base is added, CH3CO2H + OH– → CH3CO2 – + H2O. The small amount of H+ or OH– added is removed. Hence, the solution resists changes in pH. (iii) No. of moles of KOH added = 0.61000 10 × = 6.00 × 10–3 No. of moles of CH3CO2H remaining = 0.0180 – 6.00 × 10–3 = 0.0120 Ka = H] CO [CH ] [H ] CO [CH 2 3 2 3 + − ⇒ [H+] = ] CO [CH H] CO [CH K 2 3 2 3 a − = 2 2 5 V 0.006 V 0.01210 1.84× ×− (where V2 = volume of buffer) [H+] = 3.63 × 10–3 mol dm–3 pH = –lg (3.63 × 10–3) = 4.43 (c) CH 3CH2OH CH 2=CH2 CH 2ClCH2Cl excess conc. H HO 2CCH2CH2CO2H dilute H (d) Warm each compound with a solution containing AgNO3 in excess NH3(aq) (Tollens’ reagent). Only P gives silver mirror. OR Warm each compound with complex copper( II) ions in alkaline solution (Fehling’s solution). Only P gives reddish-brown precipitate. OR Heat each compound with dilute H 2SO4 and K2Cr2O7(aq). Only P turns the orange solution green. 2SO4 heat 2SO4 170 C o in CCCl2 l4 (in the dark) KCN in ethanol heat under refluxCH2CH2 CNCN HCI 2008 2

2 (a) (i) 1s2 2s2 2p6 3s2 3p6 3d5 (ii) Step 1 Relevant data: Mn 3+ + e– = Mn2+ E, = +1.49 V MnO4 – + 8H+ + 5e– = Mn2+ + H2O E, = +1.52 V Overall equation for step 1: MnO4 – + 8H+ + 4Mn2+ → 5Mn3+ + 4H2O E, cell = +1.52 – 1.49 = +0.03 V > 0. ∴ reaction is feasible Step 2 Relevant data: 2CO2 + 2e– = C2O4 2– E, = –0.49 V Mn3+ + e– = Mn2+ E, = +1.49 V Overall equation for step 2: 2Mn 3+ + C2O4 2– → 2Mn2+ + 2CO2 E, cell = +1.49 – (–0.49) = +1.98 V > 0. ∴ reaction is feasible (b) (i) A complex ion is formed by a central metal ion or atom, datively-bonded by surrounding ligands, which can be ions or molecules. (ii) [Mn(C2O4)3]3– (iii) In the presence of ligands, d-orbitals (of the transition metal ion) split into 2 energy levels with an energy gap. Electrons are able to be promoted from a lower energy d-orbital to the higher energy one by absorbing energy from the visible spectrum. We observe the complementary colour / the wavelengths that are transmitted. For Group II metal ions, electronic transition may take place from the ground state to the excited state by absorbing energy outside the visible region, thus compounds are not coloured. (c) (i) Solid dissolved in dilute H2SO4 Withdraw a sample of the solution Titrate with the KMnO4 solution from a burette until the solution (in conical flask) turns from yellow to orange Add in a catalyst (eg, MnSO4) or heat the solution, then start the titration Repeat titration, noting the volume of KMnO4 each time, and stop when values are consistent (to within ±0.10 cm3) Balanced equation: 24H+ + 3MnO4 – + 5FeC2O4 → 3Mn2+ + 5Fe3+ + 10CO2 + 12H2O HCI 2008 3

(ii) No. of moles of KMnO4 = 310 2.05 0.11000 20.50 −× = ×mol No. of moles of FeC2O4 = 33 10 3.42 53 10 2.05 −− × = ×× mol Mass of FeC2O4 in sample = 0.491 g ∴ percentage purity = 81.8 % 3 (a) (i) Oxidation half-equations to be considered at anode: E, ox / V (1) 2F – = F 2 + 2e– –2.87 (2) 2H 2O = O 2 + 4H+ + 4e– –1.23 Suppose concentrated NaF is used. Higher concentration of F– causes position of redox equilibrium (1) to shift to the right, Eox(F– / F2) increases. However, difference between the E, ox values is too large. E, ox(H2O / O2) will still be higher than E, ox(F– / F2). Hence, H2O is still oxidised in preference instead of F–. O2 not F2 will be obtained. (ii) Anode : 2C l–(aq) → Cl2(g) + 2e– Cathode : 2H 2O(l) + 2e– → H2(g) + 2OH–(aq) (iii) No. of moles of Cl2 = 35.5 2 10 16 × × = 14084.5 No. of moles of electrons passed = 14084.5 × 2 = 28169 Q = 96500 × 28169 = 2.718 × 109 C I = 60 60 24 10 2.7189 × × × = 3.15 × 104 A (b) (i) 3 bond pairs, 2 lone pairs around central atom ∴ T-shape molecule Cl F F F (ii) 6H2O(l) + 4ClF3(g) → 3O2(g) + 2Cl2(g) + 12HF(g) (c) (i) Both HF and HI are simple molecules. HF forms intermolecular hydrogen bonding and that is stronger than the van der Waals’ forces (OR dispersion forces) for HI. Therefore, more energy needed for boiling HF. (ii) The reaction between NaCl(s) and concentrated H2SO4 is: NaCl(s) + H2SO4(l) → HCl(g) + NaHSO4(s) For NaI(s) and conc. H2SO4: HCI 2008 4

NaI(s) + H2SO4(l) → HI(g) + NaHSO4(s) 2HI(g) + H2SO4(l) → I2(g) + SO2(g) + 2H2O(l) OR 8H I(g) + H2SO4(l) → 4I2(g) + H2S(g) + 4H2O(l) Concentrated H2SO4 is able to oxidise HI to I2, but not strong enough an oxidising agent to oxidise HCl. (OR explain in terms of HI being a better reducing agent than HCl.) HI is so easily oxidised that very little of it remains. (iii) The hot glass rod causes HI to decompose: 2HI(g) → H2(g) + I2(g) The violet vapour is I2. Reaction involves breaking of H–I bond. H–Br bond is stronger than H–I bond; evidence: bond energy of +366 & +299 kJ mol–1 respectively OR justify in terms of size of Br vs I atom. Heat provided by hot glass rod unable to decompose HBr. 4 (a) Boltzmann diagram, correct labels for axes, appropriate shading of areas. When temperature increases, there is an increase in the fraction of reactant particles with kinetic energy larger than or equal activation energy EA (refer to shaded areas). Thus, frequency of effective collision increases, reaction rate increases. (b) (i) Order of reaction with respect to a reactant is the power to which its concentration is raised in the rate equation. The orders of reaction must be found experimentally. Rate constant (k) is the proportionality constant in the rate equation. It is a constant at a given temperature. (ii) Rate = k [CH3COCH3] [H+] (iii) 0.05 0.05 10 1.25 ] [H ] COCH[CH Ratek 6 3 3 × ×== − + = 5.00 × 10–4 mol–1 dm3 s–1 (iv) Catalyst (v) linear graph [I2] 0.002 Time/s Logical justification in terms of relative concentrations of reagents & simplification of rate equation HCI 2008 5

I2 in NaOH(aq), warm (c) (i) A: C CHCH 2 CO2H OH O CH3 OR HO2C CHCH 2 C OH CH3 O (ii) D: C CHCH 2 C O Cl Cl O Cl white fumes of HCl(g) (iii) condensation phenol in NaOH(aq) F & G: (iv) OH and HO2C CHCH 2 CO2H Cl 5 (a) (i) A Bronsted-Lowry base is a proton acce