2025 H2 Chem Prelim P2 Answers VJC

@VJC 2025 9729/02/PRELIM/2025 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Questions 29 August 2025 2 hours Additional Materials: Data Booklet ________________________________________________________________________________________________________________ READ THESE INSTRUCTIONS FIRST Write your name and CT group on this cover page. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 18 2 / 16 3 / 17 4 / 9 5 / 15 Total / 75 This document consists of 18 printed pages and 2 blank pages

2 © VJC 2025 9729/02/PRELIM/2025 Answer all the questions in the spaces provided. 1 Alkynes are a class of organic compounds with the general formula CnH2n−2. (a) (i) With the aid of a labelled diagram, explain how the orbitals overlap to form the C≡C bond in ethyne, H–C≡C–H. [3] • correct diagram • correct description for σ bond • correct description for the two π bonds (ii) Table 1.1 shows the carbon-hydrogen bond length in ethane, ethene and ethyne. Table 1.1 Molecule Carbon-hydrogen bond length / nm ethane 0.114 ethene 0.109 ethyne 0.106 Use the concept of hybridisation to explain the difference in bond length of the carbon - hydrogen bond between the molecules as shown in Table 1.1. [2] Bond length decreases in the following manner: ethane, ethene, ethyne. Hence bond strength increases in the following manner: ethane, ethene, ethyne • The sp hybridised carbon atom in ethyne has the highest percentage s character , followed by the sp2 hybridised carbon atom in ethene and lastly, sp3 hybridised carbon atom in ethane. • Hence, the extent of orbital overlap between the sp hybridised carbon atom and the s orbital of H atom is the largest, resulting in a shortest bond length. (b) Table 1.2 contains data that is relevant for this question. Table 1.2 Equation Ho / kJ mol–1 3C(s) + H2O(l) → CO(g) + C2H2(g) +401

3 © VJC 2025 9729/02/PRELIM/2025 [Turn over 2C(s) + O2(g) → 2CO(g) −221 2H2O(l) → 2H2(g) + O2(g) +572 (i) Write an equation to represent the standard enthalpy change of formation of C2H2(g). [1] • 2C(s) + H2(g) → C2H2(g) (ii) Use data from Table 1.2 to calculate the standard enthalpy change of formation of C2H2(g). [2] • correct energy cycle or working By Hess’ Law, Hof = 401 − 572/2 + 221/2 • = +225.5 kJ mol−1 (c) (i) Using the bond energies in the Data Booklet , calculate the enthalpy change of combustion of ethyne shown in equation 1.1. equation 1.1 H–CC–H(g) + 5 2O=O(g) → 2O=C=O(g) + H–O–H(g) [2] Hc = [2BE(C–H) + BE(CC) + 5 2 BE(O=O)] − [4BE(C=O) + 2BE(O–H)] • = [2(410) + (840) + 5/2(496)] − [4(805) + 2(460)] • Hc = −1240 kJ mol−1 (ii) Explain what is meant by the term entropy. [1] • Entropy is a measure of the randomness or disorder of matter and energy of a system. The higher the disorderliness, the higher is the entropy. (iii) The entropy change of combustion of ethyne is –2150 J K−1 mol−1 at 305 oC. With reference to equation 1.1, explain why the entropy change of combustion of ethyne has a negative sign. [1] • There is a decrease from 3.5 moles of gaseous reactants to 3 moles of gaseous products, hence ∆S < 0. Hfo

4 © VJC 2025 9729/02/PRELIM/2025 (d) Alkynes undergo similar reactions as alkenes. Ethyne can be reduced to ethane in a two - stage process using a transition metal as the heterogeneous catalyst as shown in Fig. 1.1. Fig. 1.1 The higher the activity of a catalyst, the more effective it is at catalysing the reaction. Fig 1.2 shows the relative activity of each catalyst against ΔHads, the enthalpy change of adsorption of hydrogen gas onto the catalyst surface in the reduction of alkyne. Fig. 1.2 (i) State the meaning of the term heterogenous catalyst. [1] • A catalyst increases rate of reaction , by providing an alternative pathway with lower activation energy, and remains chemically unchanged at the end of the reaction. A heterogenous catalyst exists in a different phase OR physical state from the reactants. (ii) State which catalyst is the most effective in the reduction of alkyne. [1] • Pd (iii) Use your knowledge of the mode of action of heterogenous catalysts , suggest an explanation for the trend observed in Fig.1.2. [2] • When ΔHads is less negative/ less exothermic, relative activity is low as the hydrogen gas is only weakly bound to the catalyst. The covalent bond in the hydrogen molecule is not weakened sufficiently for the reaction to occur . Or Reactant molecules desorb before reaction can occur. • When ΔHads is very negative/ highly exothermic, the hydrogen molecule is too strongly adsorbed so it is unable to react with ethyne/the product formed is unable to desorb from the catalyst surface. (iv) In 1952, Herbert Lindlar found that adding a thin layer of impurity, such as lead(II) oxide, to palladium catalyst reduced its activity, allowing the reaction to stop at the alkene stage rather than reducing to the alkane as shown in Fig. 1.3.

5 © VJC 2025 9729/02/PRELIM/2025 [Turn over Fig. 1.3 This is also known as “poisoning” the catalyst. By considering the shape of the molecules shown in Fig. 1.3, suggest how the addition of lead(II) oxide “poisons” the palladium catalyst. [2] • With addition of lead( II) oxide, active sites on the catalyst surface is no longer flat or active sites blocked partially • preventing the bigger ethene from adsorbing efficiently, thus decreasing activity of the catalyst. Ethyne is smaller and hence can access the active site more readily. [Total: 18] 2 An aquatic system thrives on a delicate balance based on key chemical processes. (a) Ammonia is the primary component of fish waste. When the concentration of ammonia in an aquatic system is too high, aquatic life is adversely affected. Ammonia can be removed with oxygen in the presence of nitrifying bacteria to form nitrite, NO2–, and water. (i) Write a balanced equation for the reaction of ammonia and oxygen. [1] • 2NH₃(aq) + 3O₂(aq) + 2OH–(aq) → 2NO₂⁻(aq) + 4H₂O(l) (ii) Draw the dot-and-cross diagram of nitrite ion, NO2–. [1] • (iii) Some NO3– ions may also be formed from ammonia by the action of nitrifying bacteria. Given that the shape of NO 3– ion is trigonal planar , use VSEPR theory to explain the difference in the bond angles between NO2− and NO3– ion. [2] • Nitrate ion has 3 bond pairs and 0 lone pairs of electrons about the N atom and a trigonal planar shape. Nitrite ion has 2 bond pairs and 1 lone pair of electrons about the N atom. As the lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, • hence bond angle of nitrite ion is smaller OR 118 compared to 120 for nitrate ion. (iv) The ammonia levels in a 50 dm 3 freshwater aquarium tank was investigated. It is found that the concentration of dissolved ammonia and oxygen in the tank were 0.020 mol dm−3 and 0.030 mol dm−3 respectively. Upon adding nitrifying bacter