2024 JPJC Chem Prelim P3 solutions

© Jurong Pioneer Junior College [Turn Over NAME CLASS 23S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2024 CHEMISTRY 9729/03 Higher 2 Paper 3 Free Response Questions 12 September 2024 2 hours Candidates answer on the Question paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 2 3 4 or 5 Penalty (delete accordingly) Lack 3sf in final answer –1 / NA Missing/wrong units in final ans –1 / NA Bond linkages –1 / NA Total 80 This document consists of 32 printed pages.

2 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2024 Section A Answer all the questions in this section. 1 (a) One version of the Fischer-Tropsch process for manufacturing methane is the reaction of carbon monoxide with hydrogen. CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) H210 kJ mol-1 A mixture of CO and H2 in a 1:3 molar ratio was introduced into a sealed vessel and heated to 1200K. At equilibrium, 40% of the CO ha d reacted. The total pressure in the vessel was 12 atm at equilibrium. For Examiner’s Use (i) Write an expression for the equilibrium constant, Kp for this reaction. ( )  =  24 2 HO C H p 3 CO H pp pp K [1] Examiner’s comments: This was generally well answered. Common mistakes: • Excluding partial pressure of water in the expression. • Using [ ] (ii) Use your expression to calculate the value of Kp for the reaction at 1200K. Include its units. 3H2(g) + CO(g)  CH4(g) +H2O (g) Initial amount/mol 3x x − - Change/mol −3(0.4) x −0.4 x + (0.4) x + (0.4) x Equilibrium amount/mol 1.8x 0.6 x 0.4 x 0.4x Total amount at equilibrium = 1.8x + 0.6x + 0.8x = 3.2x 2HP = 121.8x×3.2x = 6.75 atm COP = 0.6x×123.2x = 2.25 atm 4CHP=2HOP = 40. x×123.2x = 1.5 atm ( ) ( )( )= 2 p 3 1.50 6.75 2.25 K = 0.00325 atm−2 [3] Examiner’s comments: This was generally well answered. Most students were able to get some marks. Answers in kPa are accepted but students are advised to leave their answers in units given in question and not to spend time converting the units.

3 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2024 [Turn Over (iii) Given that the value of Kp decreases with increasing temperature, deduce the sign for the enthalpy change, H of the forward reaction. Explain your reasoning. Since Kp decreases with increasing temperature, it implies that the position of equilibrium has shifted left to favour endothermic reaction or absorb some heat. Hence, the forward reaction is exothermic and the sign of H of the forward reaction is negative. [1] Examiner’s comments: This was generally well answered. Some of the students failed to read the question carefully and just state that the forward reaction is exothermic without giving the sign. (b) Describe and explain how the entropy of each of the following systems will change during the stated process. • 1 mol of N2(g) at 298K is added to 1 mol of CH4(g) at 298K. • 1 mol of Cl2(g) at 298 K is heated to 373 K. Assume the pressure of each gas remains at 1 atm throughout and no reaction occurs between N2 and CH4. • 1 mol of N2(g) at 298K is added to 1 mol of CH4(g) at 298K. S > 0 as there is an increase in disorder of system since there is more ways of arranging the particles OR distributing the energy among the particles due to mixing. • 1 mol of Cl2(g) at 298 K is heated to 373 K. S > 0 as there is an increase in disorder of system since an increase in temperature increases the (average) kinetic energy of gas molecules causing a broadening of the Boltzmann energy distribution (emphasise in debrief), resulting in more ways of arranging the energy quanta among the particles in the hotter system. [2] Examiner’s comments: This was generally well answered. Most students missed out “mixing” to explain for the increase in disorderliness in the first part. Students are reminded to be carefully in their answers. The correct answer should be “Increase in entropy, S” or “change in entropy, S, is positive”. (c) (i) During the monobromination of pentane, three different bromoalkanes are formed as shown in Fig. 1.1. Br Br Br + + 1-bromopentane 2-bromopentane 3-bromopentane Fig. 1.1 Predict the expected theoretical ratio in which these three products would form if the monobromination of pentane occurs randomly. Explain your answer. [2]

4 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2024 1-bromopentane: 2-bromopentane: 3-bromopentane = 6:4:2 = 3:2:1 The substitution reaction is random. There are 6 chemically equivalent H atoms that can be substituted to form 1-bromopentane, 4 chemically equivalent H atoms that can be substituted to form 2-bromopentane, 2 chemically equivalent H atoms that can be substituted to form 3-bromopentane. [2] Examiner’s comments: • This was generally well answered. (ii) When the monobromination was practically carried out in an experiment, the percentage of 1-bromopentane was obtained in the lowest proportion. Suggest an explanation for the difference between this experimental result and what you predicted in (c)(i). The intermediate formed in free radical substitution to form 1-bromopentane is a primary radical which is most electron deficient and least stable, as it has least electron-donating alkyl groups bonded to it. Thus it is less favourably formed and is the minor product. [1] Examiner’s comments: • This was generally not well answered. • Common wrong responses were “the intermediate is a primary carbocation” or “less electron donating alkyl groups to disperse the charge”. • A significant number missed out “least stable” or “electron-donating”. (d) The Wurtz reaction below shows two iodoalkanes react with sodium metal in dry ether, to form a new carbon−carbon bond, resulting in the formation of a new alkane. R−I + R’−I + 2Na → R−R’ + 2NaI Reaction of a single iodoalkane with sodium metal in dry ether via Wurtz reaction will give a good yield of symmetrical alkane product. Draw the organic product formed when (iodomethyl)benzene reacts with sodium metal in dry ether. (iodomethyl)benzene [1]

5 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2024 [Turn Over Examiner’s comments: • This was generally well answered. (e) Bromoalkanes and alkoxides react in the Williamson ether synthesis to form the ether functional group containing C-O-C, an example of which is shown below. RBr + R’O− → ROR’ + Br− alkoxide ether It is known that SN2 mechanism is usually undergone in Williamson ether synthes