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– → 2Mn