2023 EJC H2 Chemistry Prelims Paper 2 (solution with comments)

© EJC 9729/02/J2PE/23 [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2023 General Certificate of Education Advanced Level Higher 2 Paper 2 Suggested Solutions with Marker’s comments 1 (a) In the past two decades, scientists have observed a steady decrease in the pH of the oceans. This phenomenon is known as ocean acidification. (i) Ocean acidification occurs when carbon dioxide dissolves and reacts with water in the ocean to form carbonic acid, H2CO3. Write equations to illustrate how carbon dioxide decreases the pH of seawater. ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ....................... [2] (ii) In the past two centuries, the average pH of seawater has dropped from 8.2 to 8.1. Calculate the percentage increase in the concentration of H+(aq) ions in sea water. [1] (iii) The significant change in pH of seawater is unexpected as the oceans have a natural buffering system involving CO 2– 3 (aq) and HCO – 3(aq). Write an equation to show how the CO 2– 3 (aq)/HCO – 3(aq) buffer system helps resist the decrease in the pH of seawater when carbon dioxide dissolves. ................................ ................................ ................................ ....................... [1] Comments: • Generally, well done except for a small group of students who were careless in (iii) • Question analysis and references(where applicable) This is a simple application question that requires students to pick out information from the question and construct relevant equations. • Common mistakes: - For (ii), not calculating percentage increase. - For (iii), Using → instead of ⇌. - Misinterpreting a decrease in pH as increase in OH‒ in the system instead of H+. - Writing HCO – 3 + H + → H2CO3. This is incorrect as the buffer system comprises of CO 2– 3 , the conjugate base and HCO – 3(aq), the acid. To resist a drop in pH, the conjugate base will remove acid (H+) to maintain relatively constant pH. 25.9% 8.1 8.2 8.2 10 10percentage increase 100% 10 −− − −=  = CO 2– 3 (aq) + H+(aq) → HCO – 3(aq) CO2(g) → CO2(aq) CO2(aq) + H2O(l) → H2CO3(aq) H2CO3(aq)  2H+(aq) + CO 2– 3 (aq) or H2CO3(aq)  H+(aq) + HCO – 3(aq) HCO – 3(aq)  H+(aq) + CO 2– 3 (aq)

2 © EJC 9729/02/J2PE/23 (b) The main cause of ocean acidification is the rising carbon dioxide levels in the atmosphere, which is a result of the increase in the use of fossil fuels. To combat this, alternative fuels such as methanol are being explored. (i) Define the term standard enthalpy change of combustion, cH . ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ....................... [1] (ii) When 1.86 g of methanol , CH 3OH, was combusted in a bomb calorimeter, a temperature rise of 5.1 K was measured. In a separate experiment, 1.00 g of benzoic acid, C 6H5CO2H, produced a temperature rise of 3.2 K when combusted in the same bomb calorimeter under identical conditions. Given that the enthalpy change of combustion of benzoic acid is −3226 kJ mol−1, calculate the enthalpy change of combustion of methanol. OR Amount of energy produced from combustion of benzoic acid = CΔT where C is the heat capacity of the bomb calorimeter 26.44  103 = C(3.2) C = 8263.3 J K-1 It is the energy released when 1 mole of a substance is completely burnt in excess oxygen under standard conditions (at 1 bar, at a specified temperature, usually 298 K). amount of energy produced from combustion of benzoic acid sample 1.003226 7 12.0 6 1.0 2 16.0 26.44 kJ amount of energy produced from combustion of methanol sample 26.44 5.1 42.14 kJ3.2 enthal = − −   +  +  = =  = 1725 kJ mo l py change of combustion of methanol 1.8642.14 12.0 4 1.0 16.0 − = −  +  + =−

© EJC 9729/02/J2PE/23 [Turn over Since the same bomb calorimeter is used, the heat capacity will be constant. Amount of energy produced from combustion of methanol = -8263.3  5.1 = -42.14  103 J = -42.14 kJ Enthalpy change of combustion of methanol 1.8642.14 12.0 4 1.0 16.0 − = −  +  + =− 1 725 kJ mol [3] Comments: • This part was not as well done as expected. • Question analysis This part of the question involved simple handling of data given in the question. While it is not a difficult application question, students need to be clear of the concepts on calculation heat change from experimental data. • Common mistakes: - Using mass of methanol and benzoic acid as m in q = mcT. m refers to the mass of solution that the temperature change is measured from. In this question, there is no mention of such mass. Students cannot use q = mcT to calculate heat change. - Some students tried to calculate m by using specific heat capacity of water, i.e. 26.44  103 = m(4.18)(3.2)  m = 1976.9 g. Students should not assume that the temperature change is measured via water medium. - Some students tried to calculate c by 26.44  103 = (1)(c)(3.2)  c = 8263.3 J K–1. Then calculating heat released from combustion of methanol by q = (1.86)(8263.3)(5.1) This is incorrect as by substituting c , students are assuming that the specific heat capacity of benzoic acid and methanol are the same.

4 © EJC 9729/02/J2PE/23 (c) Methanol can be synthesi sed through the reaction between carbon dioxide and hydrogen, as shown in equilibrium 1 below. equilibrium 1 CO2(g) + 3H2(g)  CH3OH(l) + H2O(g) 1H = −49.5 kJ mol−1 At the same time, a side reaction, as shown in equilibrium 2, can occur. equilibrium 2 CO2(g) + H2(g)  CO(g) + H2O(g) 2H = +41.2 kJ mol−1 (i) Explain why the synthesis of methanol is usually conducted at low temperature and high pressure. ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ....................... [2] (ii) CO(g) is sometimes added to the reaction mixture at constant volume after synthesis has occurred to a certain extent. Suggest one advantage of doing so. ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ........................... ................................ ................................ ................................ ....................... [2] [Total: 12] At low temperature, the positio