2024 Prelims NYJC H2 Chem P3 (Ans)

[Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS CHEMISTRY 9729/03 Paper 3 Free Response 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. You may use an HB pencil for any diagrams or 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. Circle the question you attempted in the box below. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided 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 /20 2 /20 3 /20 4 or 5 /20 Total /80 This document consists of 34 printed pages and 1 blank page. H

2 H2 Chemistry 9729/03 NYJC J2/24 PX [Turn Over For Examiner's Use Section A Answer all questions in this section. 1 Acrylic acid is a major building block used in the manufacture of a wide range of industrial and consumer products. In 2022, the global market for acrylic acid was approximately 6.7 million tonnes valued at USD 14.1 billion. O OH acrylic acid (a) One method of synthesising acrylic acid is by reacting propene and steam with palladium (Pd) catalyst in a fuel cell as shown in Fig. 1.1 . Pd serves as both an electrode and a catalyst in the fuel cell. Propene is pumped in at the Pd electrode, while air (as a source of oxygen) is pumped in at the other electrode (platinum, Pt). The cell is operated in the gas phase at 365 K and atmospheric pressure. Electricity is generated during the reaction, making th e process economically viable. Fig. 1.1 (i) Explain fully why Pd is suitable to serve both roles in the fuel cell. [2] Pd is able to serve as an electrode as it is a good electrical conductor due to presence of delocalised valence electrons which can act as charge carriers. [1] Pd is able to serve as a (heterogeneous) catalyst as it is a transition metal which contains partially filled (4)d subshell/orbitals which can accept electrons from the reactant molecule/ are able to form weak temporary bonds with reactant molecules during adsorption. [1] Cation exchange membrane (H3PO4 in silica wool) acrylic acid propene, H2O O2 H2O Pd electrode Pt electrode External circuit Cation exchange membrane (P4O10 in silica wool (SiO2))

3 H2 Chemistry 9729/03 NYJC J2/24 PX [Turn Over For Examiner's Use (ii) Write half-equations for the reactions occurring at the cathode and the anode in the fuel cell. State the polarities of each electrode. [3] [R] Cathode (+ve): O2 + 4H+ + 4e → 2 H2O [1] [O] Anode (-ve): O OH + 2H2O + 6H+ + 6e OR CH2CHCH3 + 2H2O → CH2CHCOOH + 6H+ + 6e [1] ecf 1m for equations if mixed up cathode and anode. BOD for C3H6. [1] for correct polarity. No ecf. The cell potential of the fuel cell was measured to be +0.70 V under the operating conditions. (iii) Use your answer in (a)(ii) and the Data Booklet to determine the electrode potential of the acrylic acid/propene half-cell. [1] Ecell = E[R] – E[O] = E(O2/H2O) – E(acrylic acid/propene) +0.70 = +1.23 – E(acrylic acid/propene) E(acrylic acid/propene) = +1.23 – 0.70 = + 0.53 V [1] (iv) Calculate G for the following overall equation of the fuel cell. O OH 2 + 3O2 2 + 2H2O [2] Electrons transferred when 3 moles O2 or 2 moles propene is reacted = 12 ecf (ii) ΔG = – n F Ecell = – (12)(96500)(+0.70) = – 810600 [1] J mol–1 [1] sign and units for both (iii) and (iv) (v) Describe and explain how electrode potential of the oxygen/water half -cell will change when pH of the whole cell was increased slightly. Given electrode potential of the acrylic acid/propene half-cell will be affected similarly, deduce the overall effect on the cell potential of the fuel cell. Explain your reasoning. [2] If pH is increased, [H+] decreases, position of equilibrium of O2 + 4H+ + 4e ∏ 2 H2O shifts left to produce more H + i.e. oxidation is more favoured E(O2/H2O) becomes less positive. [1] Since overall equation doesn’t contain H+ on either side, it will not be affected by changes in [H+] hence Ecell remains unchanged. [1] OR Similarly, E(acrylic acid/propene) becomes less positive as POE of CH2CHCOOH + 6H+ + 6e ∏ CH2CHCH3 + 2H2O shifts left when [H+] decreases. Ecell = E(O2/H2O) – E(acrylic acid/propene). The effect on E(O2/H2O) and E(acrylic acid/propene) cancels out as there are equal number of H+ on either side of the overall equation.

4 H2 Chemistry 9729/03 NYJC J2/24 PX [Turn Over For Examiner's Use Accept also if student recognise equal H+ on either side of the overall equation AND CH3CHCOOH may react with OH– at higher pH hence POE of overall equation shifts right and Ecell becomes more positive. (vi) Write an equation to show the acid-base reaction taking place in the cation exchange membrane when pH of the whole cell was increased via addition of OH –(aq) ions. [1] P4O10 + 12 OH– → 4 PO43– + 6 H2O [1] Note SiO2 reacts only with conc alkali. (b) Acrylic acid can be treated with different reducing agents to form different products as shown in Fig. 1.2. O OH O OH OH no reaction propanoic acid 2-propen-1-ol NaBH 4 (aq) LiAlH4 in dry ether H2 with Ni, heat Fig. 1.2 (i) Suggest reasons to explain Fig. 1.2. You may find it helpful to discuss strength of reducing agents and how the reducing agents work in your answer. [3] NaBH4 is a weaker reducing agent than LiAlH 4 as the B–H bond is shorter hence stronger than Al –H bond hence less easily broken to form H – nucleophile OR B is more electronegative than Al hence B –H bond is less polar and H – nucleophile is formed less readily hence no reduction occur with NaBH4. [1] LiAlH4 supplies the H – nucleophile which is able to attack the electron deficient carboxylic acid carbon to reduce it to a primary alcohol. [1] H– nucleophile does not reduce alkene as C=C bonds are non-polar / they are repelled by electron rich C=C. H2 with Ni catalyst is able to reduce alkene as alkenes and hydrogen molecules can be adsorbed onto active sites of Ni catalyst for heterogeneous catalysis to take place. [1] (ii) Acrylic acid has a pKa of 4.25, while propanoic acid has a p Ka of 4.72. Explain why the pKa of acrylic acid is lower than that of propanoic acid. [1]

5 H2 Chemistry 9729/03 NYJC J2/24 PX [Turn Over For Examiner's Use In O O - , the p orbital of oxygen overlaps with both the π electron cloud/orbital of the C=O and C=C bonds. This further disperses the negative charge, stabilising the conjugate base to a greater extent, making acrylic acid a stronger acid (with a lower pKa). [1] (iii) Propanoic acid has the molecular formula C3H6O2. There are two isomeric esters with the same molecular formula as propanoic