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 m