2020 JPJC Prelim Paper 2 answers

NAME CLASS 19S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2020 CHEMISTRY 9729/02 Higher 2 Paper 2 Structured Questions 17 September 2020 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number 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. 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 Paper Question Mark 2 1 13 2 9 3 14 4 12 5 6 6 21 Penalty (delete accordingly) Lack 3sf in final ans –1 / NA Missing/wrong units in final ans –1 / NA Total 75 This document consists of 21 printed pages.

2  Jurong Pioneer Junior College 9729/02/J2 Prelim/2020 Answer all the questions in the spaces provided. 1 (a) The following graph below shows the second ionisation energies of Period 3 elements with consecutive proton number. For Examiner’s Use (i) Explain why the second ionisation energy generally increases from B to H. Across the period, the nuclear charge/ proton number increases and the radius decreases while the shielding effect is relatively constant (due to same number of inner shell electrons). Hence, increase in effective nuclear charge OR nuclear attraction on the electron to be removed increases and hence, the second IE generally increases. [2] (ii) Which of the above elements, A to H, is aluminium? Element C [1] (iii) Explain the following:  big drop in second ionisation energy from A to B  slight drop in second ionisation energy from C to D A to B: 2nd ionisation energy involves removal of 3s electron from B+ and removal of 2p electron in A+. B+ has 1 more quantum shell with electrons thus the electron to be removed from H + is further away from nucleus, hence the nuclear attraction on the electron to be removed is weaker and hence, the 2nd IE of B is much lower than that of A. C to D: C+: ns2 D+: ns2 np1 The np electron removed from D+ is further away from the nucleus and has a higher energy than the ns electron removed from C+. Thus more energy is required to remove it. [2] 0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,000 C D E F G H A B 2nd Ionisation Energy/ kJ mol–1

3  Jurong Pioneer Junior College 9729/02/J2 Prelim/2020 (b) Among the elements of Group 14, those towards the top, carbon to germanium, have very different properties from th ose at the bottom, tin and lead. For example, the melting points show a marked change after germanium. element C Si Ge Sn Pb mp / C >3550 1410 937 232 327 Carbon, silicon and germanium each form a solid with the same type of structure. (i) Explain why the melting points of these elements decrease from carbon to germanium. Size of atom increases, covalent bonds between atoms become longer and weaker (BE: C – C > Si – Si > Ge –Ge) Lesser energy required to break the decreasing strength of covalent bonds. [1] (ii) Carbon and silicon each form a tetrachloride. CC l4 has no reaction with water; SiCl4 reacts violently with water. Suggest an explanation for the inertness of CCl4 to water. CCl4 has no reaction with water because C atom has no available empty d orbitals to form dative bonds with water molecules. Si atom in SiCl4 has available low -lying empty d -orbitals, which can accept lone pair of electrons from water molecules, allowing dative bond formation between H 2O and Si before the Si –Cl bond needs to be broken. [1] (c) The chalcogens are Group 16 elements which form compounds with carbon . The properties of some of these compounds, along with CO 2, are given in Table 1 below. Table 1 compound structure dipole moment boiling point / C CO2 O=C=O 0 sublimes CS2 S=C=S 0 46 COS S=C=O 0.71 –50 COSe Se=C=O 0.73 –22 (i) Explain, in terms of structure and bonding, the difference in the boiling point of CS2 and COS. Both CS2 and COS have simple molecular or simple covalent structures. CS2 has a larger number of electrons (or larger electron cloud) to be polarised than COS. More energy is required to overcome the stronger instantaneous dipole- induced dipole interactions between CS2 molecules than the permanent dipole-dipole interactions between COS molecules. Hence, CS2 has a higher boiling point. [2]

4  Jurong Pioneer Junior College 9729/02/J2 Prelim/2020 (ii) Explain why  CO2 has no overall dipole moment.  COSe has a greater dipole moment than COS.  CO2 is linear and hence the dipole moments of C=O bonds cancel out.  C=S bond is more polar than C=Se since S is more electronegative than Se. There is smaller difference between the dipole moment of C=O and C=S than that between C=O and C=Se. [2] (d) Chalcogens also form compounds with halogens known as chalcohalides. One such compound is selenium tetrafluoride, SeF4, which is used as a fluorinating reagent in organic syntheses. Draw a ‘dot -and-cross’ diagram showing the electrons (outer shell only) in a SeF4 molecule. Use the VSEPR (valence shell electron pair repulsion) theory to predict its shape. Se F F F F x x x x xx SeF4 has 4 bond pairs, 1 lone pair, which gives a see-saw shape. [2] [Total:13] 2 (a) The key reaction during the Contact process for the manufacture of sulfuric acid is as follows. 2SO2(g) + O2(g) ⇌ 2SO3(g) A mixture of SO2 and O2 in a 2 :1 molar ratio was introduced into a sealed vessel and heated to 450K. At equilibrium it was found that the total pressure was 4.2 atm, and the mole fraction of O2 was 0.0476. For Examiner’s Use (i) Write an expression for the equilibrium constant, Kp for this reaction.       3 22 2 SO p 2 SO O p pp K  [1] (ii) Calculate the equilibrium partial pressures of O2, SO2 and SO3. 2SO2(g) O2(g) ⇌ 2SO3(g) Initial amount/ mol 2 1 0 Change amt/ mol -2x -x +2x Eqm amt /mol 2-2x 1-x 2x Ratio: 2 : 1 [2]

5  Jurong Pioneer Junior College 9729/02/J2 Prelim/2020 Let nT be the total amount of gaseous particles present. equilibrium amount of SO2 = 2 x eqm amount of O2 = 2  0.19nT = 0.38nT equilibrium amount of SO3 = nT - 0.38nT - 0.19nT =0.43T Using pA = A T n n  pT where A T n n is the mole fraction of A in a gaseous mixture, 2SO2(g) + O2(g) ⇌ 2SO3(g) Eqm pressure/ atm 0.0952(4.2) 0.0476(4.2) 0.857(4.2) 0.400 () 0.200 () 3.60() (iii) Calculate the value of Kp for the reaction at 450K. [1] (iv) Explain clearly the effect on the yield of SO3, when the volume of reaction mixture is reduced.