RI 2025 Chem Eqm Tutorial Answers

-1- RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2025 Tutorial 7 – Chemical Equilibria (Suggested answers for Practice Questions) Question 7 (a) Initial amt of CH3OH = 20/1000 x 0.50 = 0.01 mol Initial [CH3OH]mix = 0.01 ÷ 50/1000 = 0.200 mol dm–3 Initial amt of ethanedioic acid = 30/1000 x 0.40 = 0.012 mol Initial [ethanedioic acid]mix = 0.012 ÷ 50/1000 = 0.240 mol dm–3 (COOH)2(aq) + 2NaOH(aq) → (COONa)2(aq) + 2H2O(l) Amt of ethanedioic acid reacted with NaOH = 1/2 x (30/1000 x 0.10) = 0.00150 mol At eqm, [ethanedioic acid] = 0.0015 ÷ 10/1000 = 0.150 mol dm–3 2CH3OH(aq) + (COOH)2(aq) ⇌ (COOCH3)2(aq) + 2H2O(l) Initial conc / mol dm−3 0.200 0.240 0 – Change in conc / mol dm−3 –0.180 –0.0900 +0.0900 – Eqm conc / mol dm−3 0.0200 0.150 0.0900 – [CH3OH]eqm = 0.0200 mol dm–3 K c = [(COOCH3)2] [CH3OH]2[(COOH)2] = (0.0900) (0.0200)2(0.150) = 1500 m ol –2 dm6 (b) The titration must be done quickly to minimise the shift in the position of equilibrium during titration as the acid (COOH)2 is continually being neutralised by NaOH(aq). If the titration was done too slowly , the shift in the position of equilibrium will cause the equilibrium concentrations of all species to differ from the actual values . As a result, Kc determined would not be accurate for that temperature. Question 8 (a) (i) 42 C 4 2 [Hb(O ) ]= [O ] [Hb]K mol–4 dm12 (ii) y y 20 -4 1242 C 4 -6 4 2 [Hb(O ) ]= = = 3.00 × 10 mol dm[O ] [Hb] (7.6×10 ) ( )K (iii) Let x mol dm-3 be the initial [Hb] x x 20 -4 12 4 2 -5 -3 2 0.99 = 3.00 × 10 mol dm[O ] (0.01 ) [O ] = 2.40 × 10 mol dm

-2- (b) Let the total eqm conc of MbO2 & Mb in the mixture be x mol dm–3, fraction of MbO2 be y and fraction of Mb be 1 – y. [MbO2]eqm = xy mol dm–3 [Mb]eqm = (1 − y)x mol dm–3 6 -1 32 C -6 2 ) [MbO ] xy= = =1 × 10 mol dm[O ][Mb] (7.6×10 (1-y)xK Solving the equation, y = 0.884 Therefore, the % of MbO 2 in the Mb-MbO2 equilibrium mixture is 88.4%. Alternative method: [MbO2] [Mb][O2] = 1 ×106 Since [O2] = 7.6 ×10-6 [MbO2] [Mb] = 7.6 [MbO2] = 7.6[Mb] % MbO2 = 7.6 7.6+1 ×100% = 88.4% Question 9 Let the initial pressure of N2O4 be a & the equilibrium partial pressure of O2 be y. 2NO2(g) ⇌ 2NO(g) + O2(g) Initial p a 0 0 Change in p – 2 y +2y +y Eqm p a – 2y 2y y Total pressure at eqm = a + y Since the equilibrium total pressure is 20% greater than initial pressure, a + y = 1.2a y = 0.2a Mole fraction of O2 at equilibrium = equilibrium partial pressure of O2 / total equilibrium pressure = y / (a+y) = 0.2a / 1.2a = 0.167 Question 10 Assuming ideal gas behaviour, amt of (HCOOH)2 before reaction = (101325)(0.40 1000) (8.31)(300+273) = 8.512 x 10–3 mol am t of gases