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 after eqm is reached = (101325)(0.60 1000) (8.31)(300+273) = 1.277 x 10–2 mol (HCOOH)2(g) ⇌ 2HCOOH(g) Initial amt / mol 8.512 x 10–3 0

-3- Change in amt / mol –x +2x Eqm amt / mol 8.512 x 10–3 – x 2x 8.512 x 10–3 – x + 2x = 1.277 x 10–2 x = 4.256 x 10–3 mol Amt of (HCOOH) 2 at eqm = 8.512 x 10–3 – 4.256 x 10–3 = 4.256 x 10–3 mol Partial pressure of (HCOOH) 2 = (4.256 x 10–3/1.277 x 10–2) x 1 atm = 0.3333 atm Partial pressure of HCOOH(g) = 1 – 0.3333 = 0.6667 atm Kp = pHCOOH 2 p(HCOOH)2 = 0.66672 0.3333 = 1.33 atm

-4- Alternative method: Under constant P and T, V α n Hence by Avogadro’s law, reacting volumes of gases are proportional to reacting amounts, and we construct ICE to track stoichiometric changes using reacting volumes. (HCOOH)2(g) ⇌ 2HCOOH(g) Initial volume / dm3 0.40 0 Eqm volume / dm3 0.40 – x 2x At eqm, total volume = 0.60 dm3 0.40 – x + 2x = 0.40 + x = 0.60 x = 0.2 Since V α n, relative eqm amount of HCOOH = 0.4 (no units) relative eqm amount of (HCOOH)2 = 0.2 (no units) At eqm, total pressure = 1 atm Partial pressure of HCOOH = 0.2 0.6 × 1 atm = 1 3 atm Partial pressure of (HCOOH)2 = 0.4 0.6 × 1 atm = 2 3 atm Kp = pHCOOH 2 p(HCOOH)2 = 2 3� 2 1 3� = 1.33 atm ** For teachers to note in discussions: Please note that this method is based on Avogadro’s law, where reacting volume is proportional to reacting amount. The ICE table is a method which tracks the stoichiometry of the reaction. “Eqm volume need not be taken as partial volume”, but rather, the result together with reacting volumes. Question 11 (a) (b) K p = 3 22 2 SO 2 SO O p pp atm–1 (c) At equilibrium, 3SOp = 4.7 atm 𝑝𝑝SO2 = 2/3 x (5.0 – 4.7) = 0.2 atm; 𝑝𝑝𝑂𝑂2 = 1/3 x (5.0 – 4.7) = 0.1 atm Substituting into expression in (b) , Kp = 4.72 (0.2)2 0.1 = 5.52 x 103 atm–1 (d) When the temperature is decreased, the position of equilibrium will shift to the right to favour the forward exothermic reaction that releases heat. More SO3 will be produced. Since PSO3 increase, and PSO2 and PO2 decrease, equilibrium constant is larger at 300 K. (e) Although the yield of SO3 is higher at 300 K, the rate of production of SO3 may be too slow at 300 K. Thus, a compromise is needed and a moderately high temperature of 800 K is preferred to ensure a reasonable rate of production and yield of SO3. Rate time forward backward 0 teqm forward backward

-5- Question 12 (a) Dynamic equilibrium refers to a state in a reversible reaction within a closed system where the forward and backward reactions are continuing at the same rate, resulting in no net change in the macroscopic properties (e.g. concentrations or partial pressur es) of the reactants and products. (b) (i) The same amount of carbon used in the form of lumps has a smaller surface area than in powder form. Hence, the rate of reaction is slower, and it will take a longer time to reach equilibrium. (ii) The position of equilibrium remain s the same as the vapour pressure and concentration of a solid is a constant at a given temperature. (iii) The numerical value of K p remains the same as Kp is only dependent on temperature. Question 13 (a) H2O(g) + CO(g) ⇌ CO2(g) + H2(g) Initial amt / mol 4 4 0 0 Change in amt / mol –x –x +x +x Equilibrium amt / mol 4 – x 4 – x x x Let the volume of the vessel be V dm3. K c = 22 2 [CO ][H ] [H O][CO] = xx() () VV 4 - x 4 - x() () VV = 9.0 ⇒ x = 3 mol Hence, equilibrium amt of CO = 4 – 3 = 1 mol (b) Amount of H2 = amount of CO2 = x = 3 mol Amount of CO = amount of H2O = 1 mol (c) (d) 0 teqm CO, H2O 4 3 2 1 CO2, H2 Amount/ mol Time

-6- (i) When temperature is lowered, equilibriu m position shifts right to favour the forward exothermic reaction, so the equilibrium amoun t of CO 2 is higher . As temperature is lowered, rate is lower. (ii) When volume of vessel is decreased, eq uilibrium position does not shift as the total amount of gases on each side of the equilibrium is equal. As initial pressure is higher, initial rate is higher. (iii) A catalyst does not change the equilibrium position, but it speeds up the reaction. The same am ount of CO 2 is obtained, but the rate at which this amount is obtained is higher. Question 14 (Answer: A) A is an incorrect statement. When solid K2Cr2O7 is added, it dissolves in the aqueous solution and [Cr2O72–(aq)] increases. The equilibrium system will counteract the increase in [Cr 2O72–(aq)] by favouring the backward reaction which removes Cr 2O72–(aq) ⇒ equilibrium position shifts to the left. B is a correct statement. Solid NaOH dissolves to give OH –(aq) which reacts with H+(aq) to form water. The decrease of [H+] causes the equilibrium position to shift left, forming more yellow CrO42–. C is a correct statement. Water is