RI 2024 Chem Eqm Tutorial Answers (Practice Questions) for uploading

-1- RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2024 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%. 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 O 2 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 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

-3- 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. 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 pressures) 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 H 2 = amount of CO2 = x = 3 mol Amount of CO = amount of H2O = 1 mol Rate time forward backward 0 teqm forward backward

-4- (c) (d) (i) When temperature i s 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 v olume 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 t he 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 (A nswer: 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 a solvent in this equilibrium and hence present in large excess. This means that the [H 2O] is approximately constant and hence, excluded from the Kc expression. 0 teqm CO, H2O 4 3 2 1 CO2, H2 Amount/ mol Time Amount of CO2/ mol Time Original CO2 curve (i) At lower temperature (ii) Decreased volume (iii) With catalyst teqm

-5- D is a correct statement. Diluting the solution with water will cause the concentration of all ions in solution to decrease. The equilibrium system will counteract this by favouring the backward reaction which produces more ions, increasing the concentration of ions in solution. The solution thus turns more yellow due to formation of more CrO42–. Question 15 The mixture is kept within a movable syringe, so internal pressure is equal to external pressure at equilibrium. Change #1 is correct. When the plunger is pulled, the total volume of the system increases. As the number of moles of gases remains constant, all the concentrations of gases are simultaneously decreased, accounting for the sudden dip in concentration of H2. The system restores equilibrium by favouring the backward reaction which produces more gaseous particles. Thus, [H2] gradually increases until the new equilibrium