EJC Prelim H2 Chemistry Paper 3 Solution

1 Section A Answer all the questions in this section. 1 Indirect calorimetry is used to estimate energy expenditure in humans by measuring the difference in oxygen concentration between inhaled and exhaled air. During aerobic respiration at 37 ºC , glucose, C 6H12O6, is oxidised to carbon dioxide and water while producing energy as shown in reaction 1 below. reaction 1 C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) HꝊ = –2818 kJ mol−1 (a) (i) It is assumed that there are negligible intermolecular forces of attraction between gas particles in an ideal gas. State two other basic assumptions of kinetic theory as applied to an ideal gas. [2] (ii) Air contains about 21% oxygen, by volume. Exhaled air contains 14% oxygen. At complete rest, a typical adult exchanges approximately 500 cm3 of air per breath at a rate of 12 times per minute at a temperature of 37 ºC at 1 atmospheric pressure. By assuming oxygen to be an ideal gas, calculate the volume of oxygen gas consumed per minute. Hence, determine amount of oxygen gas used per minute. [3] (iii) Assuming that the oxygen inhaled is used for respiration directly, use relevant information in reaction 1 and your answer in (a)(ii), calculate the approximate amount of energy released per minute. [1] EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 General Certificate of Education Advanced Level Higher 2 Chemistry Paper 3 Suggested Solutions with Marker’s comments (i) The gas particles are in constant, random motion, the volume of ideal gas particles are zero, and there are perfectly elastic collisions between gas particles.

2 (ii) volume of O2 consumed per minute = (21 − 14)/100 × 500 cm3 × 12 = 420 cm3 = 4.20 × 10−4 m3 pV = nRT (101325)(4.20 × 10−4) = n(8.31)(37 + 273) n = 0.0165 mol = 1.65 × 10−2 mol (iii) From equation 1, total of 6 mol of O2 releases 2818 kJ of energy Amt of energy released per minute = (2818 ÷ 6) × 0.0165 = 7.7495 kJ = 7.75 kJ

3 (b) Glucose, C6H12O6, is the ubiquitous source of energy for cells in the body. (i) Define the standard enthalpy change of formation of glucose, C6H12O6(s). [1] (ii) Use the following data, reaction 1 and appropriate data from the Data Booklet, to construct an energy cycle and calculate the standard enthalpy change of atomisation of C(s). Show your working. enthalpy change of formation of C6H12O6(s) = –1270 kJ mol–1 enthalpy change of combustion of H2(g) = –286 kJ mol–1 [3] (i) The energy change when 1 mole of glucose is formed from its constituent elements in their standard states. (ii) –2818 –1270 = 6H Ꝋ at(C(s)) + 6BE(O=O) –12BE(C=O in CO2) + 6H Ꝋ c (H2(g)) –2818 – 1270 = 6H Ꝋ at(C(s)) + 6(+496) –12(+805) + 6(–286) 6H Ꝋ at(C(s)) = +4312 kJ mol–1 H Ꝋ at(C(s)) = +718.67 kJ mol–1  +719 kJ mol–1 C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) –2818 kJ mol–1 6C(g) + 12O(g) + 6H2(g) + 3O2(g) 6C(s) + 6H2(g) + 9O2(g) –1270 kJ mol–1 6H Ꝋ at(C(s)) + 6BE(O=O) –12BE(C=O in CO2) 6H Ꝋ c (H2(g))

4 (c) Glucose exists in two forms, -glucose and -glucose as shown in Fig. 1.1. If a solution of - glucose is left some time, it will come into dynamic equilibrium with -glucose. Fig. 1.1 When plane-polarised light is passed through an aqueous solution of glucose, the angle of rotation of the light is dependent upon the structure of the molecule. The angles of rotations of plane -polarised light caused by the two forms of glucose solutions under identical conditions are shown in Table 1.1. Table 1.1 solution angle of rotation of plane-polarised light 1.0 mol dm–3 of -glucose +111º 1.0 mol dm–3 of -glucose +19º (i) When 1 dm3 of a freshly prepared solution of 1.0 mol dm–3 -glucose is left till equilibrium is achieved, the measured rotation is +53º. Assuming that the angle of rotation due to each glucose is directly proportional to its concentration, calculate a value for the equilibrium constant, Kc, for the conversion of -glucose into -glucose. [2] (ii) The conversion of -glucose into -glucose is catalysed by acids. State and explain the effect on the final measured rotation if the conversion is now carried out in the presence of dilute sulfuric acid. [2] (i) -glucose  -glucose initial conc / mol dm–3 1 – change in conc / mol dm–3 –y +y eqm conc / mol dm–3 1–y y At equilibrium, optical rotation due to -glucose + optical rotation due to -glucose = 53o (1–y)(111º) + (y)(19º) = 53º 92y = 58 y = 58/92 = 0.63

5 (d) Oxygen used in respiration binds to haemoglobin in red blood cells, which can undergo ligand exchange with water in tissues to regulate oxygen delivery. Both deoxyhaemoglobin and oxyhaemoglobin contain iron atoms in the +2 oxidation state. The oxygen-containing ligand is H2O in deoxyhaemoglobin, Hb, and O2 in oxyhaemoglobin, Hb(O2)4. One molecule of deoxyhaemoglobin, Hb, can bind with four molecules of oxygen, as shown in equilibrium 1. equilibrium 1 Hb(aq) + 4O2(aq)  Hb(O2)4(aq) HꝊ = –630.9 kJ mol−1 GꝊ = –160.7 kJ mol−1 (i) State Le Chatelier’s Principle. [1] (ii) During the initial stage of vigorous exercise, rapid muscle contractions generate heat that spreads through the body, raising core temperature. Using Le Chatelier’s Principle, explain how the initial increase in temperature affects equilibrium 1. [1] (iii) Suggest how the magnitude of Kc for equilibrium 1 is likely to be and explain its significance. [2] (iv) Carbon monoxide, CO, can bind to haemoglobin at the same binding site as oxygen. Explain why CO is poisonous. [2] [Total: 20] (i) Le Chatelier’s principle states that if a system in equilibrium is subjected to a change which disturbs the equilibrium, the system responds in such a way to counteract the effect of the change imposed, in order to re-establish the equilibrium of the system. Kc = [-glucose] / [-glucose] = y / (1–y) = 0.63 / (1– 0.63) = 1.71 (ii) The dilute sulfuric acid catalyst does not affect on the final measured rotation since a catalyst does not affect the equilibrium position since the rates of both forward and reverse reactions are increased to the same extent. It only enables the equilibrium (i.e. the final rotation) to be established at an earlier time.

6 (ii) The equilibrium position shifts to the left in order to favour the endothermic reaction to absorb excess heat. (iii) The magnitude of Kc is likely to be very large . Hence, the extent of this reaction is effectively complete or equilibrium position lies mostly on the right since GꝊ is highly negative implying that the forward reaction is thermodynamically spontaneous.

7 2 (a) The Williamson ether synthesis involves nucleophilic substitution between a halogenoalkane (RX) and alkoxides (RO –), the conjugate base of an alcohol. The alkoxide serves as the nucleophile in the reaction. An example of the reaction can be seen below: A solution containing CH 3ONa, is reacted separately with 1 -bromopropane and 2-bromo-2-methylpropane. (i) Predict the predominant mechanism for: I) the reaction of 1-bromopropane with CH3ONa II) the reaction of 2-bromo-2-methylpropane with CH3ONa Explain your reasoning. [3] (ii) For each mechanism, state and explain the stereochemical outcome of the nucleophilic substitution reaction. [2] (i) I) 1-bromopropane is a primary halogenoalkane. If it were to undergo substitution via SN1, the intermediate produced is a highly unstable primary carbocation. Rear-side attack by the nucleophile is relatively unhindered . Hence, 1-bromopropane will react via SN2 mechanism. II) 2-bromo-2-meth