VJC Prelim P2 QP and Ans

© VJC 2018 9729/02/PRELIM/18 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Candidates answer on the Question Paper. 11 September 2018 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. 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. 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 / 15 Total / 75 This document consists of 20 printed pages.

© VJC 2018 9729/02/PRELIM/18 2 Answer all the questions in the spaces provided. 1 (a) Bromine exists naturally as a mixture of two stable isotopes, 79Br and 81Br, in a 1:1 ratio. (i) Write down the full electronic configuration of 79Br2+. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [1] (ii) Define the term relative isotopic mass. Mass of an atom of isotope relative to 1 12 the mass of an atom of carbon– 12 isotope. [1] (b) Chlorine atom exists naturally as two isotopes, 35Cl and 37Cl, in a 3:1 ratio. When equimolar amounts of bromine and chlorine were mixed together, an interhalogen compound, BrC l, is formed. The product mixture contains four species with three different mass numbers 114, 116 and 118. 79Br35Cl is one of the four species. (i) With the help of the information given in (a), state the species that corresponds to each mass number. Hence, calculate the relative abundance for each mass number. mass number species relative abundance 114 79Br35Cl ½ × ¾ = 3 8 116 79Br37Cl and 81Br35Cl (½ × ¼) + (½ × ¾) = 1 2 118 81Br37Cl ½ × ¼ = 1 8 [3] (ii) Explain whether BrCl or Cl2 has a greater enthalpy change of vaporisation. BrC l has a greater enthalpy change of vaporisation as it has stronger instantaneous dipole–induced dipole interactions due to the greater number of electrons in the larger BrCl molecule. OR BrC l has a greater enthalpy change of vaporisation as it is polar with stronger permanent dipole–permanent dipole interactions than the instantaneous dipole–induced dipole interactions in Cl2. [1] (iii) Suggest with a reason how the first ionisation energy of 79Br is compared to 81Br. First ionisation energy of 79Br is the same as that of 81Br because they have the same number of protons. [1]

3 © VJC 2018 9729/02/PRELIM/18 [Turn over (c) Bromine reacts with an element A to form a compound with empirical formula ABr3. The percentage by mass of A in ABr3 is 4.31%. Calculate the relative atomic mass of A. Let Ar of A be y. Element A Br No. of moles/mol 4.31 y 1004.31 79.9 Mole ratio of A : Br = 1 : 3 = ૝.૜૚ ࢟ : ૚૙૙ ି૝.૜૚ ૠૢ.ૢ No. of moles of Br atoms = 3 × no. of moles of A atoms 1004.31 79.9 = 3 × 4.31 y y = 10.8 [1] (d) Bromine and fluorine react to form the pale yellow liquid, bromine trifluoride, as shown in Reaction 1. Reaction 1 Br2(l) + 3F2(g) → 2BrF3(l) Some thermochemical data are given below. Standard enthalpy change of formation of BrF3(l) / kJ mol−1 –301 Standard Gibbs free energy change of formation of BrF3(l) / kJ mol−1 –241 Standard entropy of Br2(l), So(Br2) / J mol−1 K−1 152 Standard entropy of BrF3(l), So(BrF3) / J mol−1 K−1 178 (i) The above reaction is spontaneous at 298 K even though ΔSo is negative. Explain qualitatively why ΔHo is the predominant factor that causes the reaction to be spontaneous. ΔGo = ΔHo – TΔSo < 0 at 298 K ΔHo is the predominant factor as it is exothermic (or negative) and drives the reaction. [1] (ii) Sketch a graph to show how ΔGo varies with temperature in K for Reaction 1. Label the y–intercept. [1] ΔGo / kJ mol−1 T / K ΔHo 0

© VJC 2018 9729/02/PRELIM/18 4 (iii) Given that ΔSo = 2 × So(BrF3) – [So(Br2) + 3 × So(F2)] for Reaction 1, calculate the standard entropy of F2(g), So(F2), at 298 K. ΔGo = ΔHo – TΔSo ΔSo = (ΔHo – ΔGo) / T = 2(–301 + 241) × 103 / 298 = –403 J mol−1 K−1 –403 = 2(178) – [152 + 3 So(F2)] So(F2) = 202 J mol−1 K−1 [2] (e) Similar to water, liquid BrF 3 can be used as a solvent and it undergoes minimal self–ionisation. 2H2O ⇌ H3O+ + OH− 2BrF3 ⇌ BrF2+ + BrF4− When (BrF2+)2(SnF62−) and Ag+(BrF4−) react in BrF3, an insoluble Ag2SnF6 is formed. (i) Construct an equation for the reaction between (BrF 2+)2(SnF62−) and Ag+(BrF4−). (BrF 2+)2(SnF62−) + 2Ag+(BrF4−) → 4BrF3 + Ag2SnF6 [1] (ii) State and draw the shapes of BrF 2+ and BrF4−, including lone pairs of electrons. bent square planar BrF2+ BrF4− [2] (f) One of the most readily prepared sulfur nitrides is S 4N4, which can be made by passing dry NH 3(g) into a solution of SC l2 in an organic solvent. A proposed structure of the molecule of S4N4 is shown below. S S N N N N S S

5 © VJC 2018 9729/02/PRELIM/18 [Turn over (i) Using the data given below, construct a suitable energy level diagram to calculate the S–N bond energy in S4N4. ΔHfo [S4N4(g)] = +460 kJ mol−1 ΔHato [S(s)] = +279 kJ mol−1 ΔHato [nitrogen] = +497 kJ mol−1 Bond energy of (S–S) in S4N4 = +204 kJ mol−1 By Hess’s Law, 8BE(S–N) + 2(+204) = –(+460) + 3104 BE(S–N) = +279.5 kJ mol −1 [3] (ii) The nitrogen atoms in S4N4 show their usual valency of 3. All four sulfur atoms have the same oxidation number. Add to the structure below to show which sulfur–nitrogen bonds are single bonds and which are double bonds. [1] (iii) Hence, explain why t he calculated bond energy of sulfur–nitrogen bond in S4N4 from (f)(i) is between that of a S−N bond and a S=N bond. This is due to the delocalisation of pi electrons / formation of resonance structures between the two sulfur−nitrogen bonds in S−N−S / N−S−N. [1] [Total: 20] S S N N N N S S 4S(s) + 2N2(g) S4N4(g) +460 4S(g) + 4N(g) 4(+279) + 4(+497) = +3104 8BE(S–N) + 2(+204) Energy / kJ mol−1

© VJC 2018 9729/02/PRELIM/18 6 2 (a) Dinitrogen tetraoxide, N 2O4, and nitrogen dioxide, NO2, exist in dynamic equilibrium with each other as shown below. N2O4(g) ⇌ 2NO2(g) (i) The diagram below shows the variation of the average molecular mass of the equilibrium mixture with pressure. Predict a value for y and account for the shape of the graph. y = 2(14.0) + 4(16.0) = 92.0 By Le Chatelier’s Principle, as pressure increases, the position of equilibrium would shift left to decrease the amount of gaseous molecules. Hence, more N2O4 will be produced and average Mr increases. [2] 0.0100 mol of inert N 2 with a partial pressure of 0.27 bar and 0.0500 mol of N2O4 were placed in a sealed vessel of volume 1.00 dm 3 and temperature of 50 oC. When equilibrium was established, the total pressure of all gases was 1.95 bar. (ii) With reference to the Data Booklet, calculate the average molecular mass, Mr, of the N 2O4/NO2 equilibrium mixture. Give your answer to three significant figures. Mass of N 2O4