TJC Prelim P2 Solutions

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 2 2008 TJC H2 Prelim P2 Answers 1 (a) For decades, scientists were puzzled by why dinosaurs suddenly became extinct 65 million years ago. In studying core samples of rock dating back to that period, scientists found an unusual high level of iridium. This possibly came from an iridium- rich asteroid that struck the Earth’s surface and the mystery to dinosaurs’ extinction was solved. Iridium exists as two naturally occurring isotopes, iridium-191 and iridium-193. For Examiner’s Use (i) Define the term relative atomic mass. · Relative atomic mass is the average mass of one atom of an element on a scale in which one atom of 12C isotope of carbon has a mass of 12 units. (ii) The relative atomic mass of iridium is 192.2. Calculate the natural abundance, in percentage, of each isotope. Let y be the percentage of iridium-191 · 191 x 100 )y100(193 100 y -´+ = 192.2 191y + 19300 – 193y = 19220 y = 40 · Iridium-191 (40%) , Iridium-193 (60%) [3] (b) Aqua regia (a mixture of 75% nitric acid and 25% hydrochloric acid by volume) is highly corrosive. Only noble metals like iridium are inert to this solution. A 5 g sample of platinum-iridium alloy required 24.6 cm3 of aqua regia for complete reaction. Platinum was completely oxidised to platinum(IV) ions by nitric acid and 0.5 g of metal was recovered. (i) Find the percentage of each metal in the alloy. · Since 0.5 g of iridium metal was recovered, Percentage of iridium = 1005 5.0 ´ = 10 % Percentage of platinum = 90%

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 3 (ii) The concentration of nitric acid used to make aqua regia is 5.0 mol dm-3. Assuming that the reaction between the sample and aqua regia is complete, construct the balanced equation for the reaction. · No. of moles of platinum = 195 5.05 - = 0.0231 mol No. of moles of nitrate = 51000 4 36.24 ´ ´ = 0.0923 mol · (1 mark working for construction of balanced equation) Pt ºººº 4e- ºººº Pt4+ No. of moles of electrons transferred = 0.0231 x 4 = 0.0924 mol No. of moles of electrons transferred per mole of nitrate = 0.0924 / 0.0923 = 1 NO3 - is reduced to NO2 NO3 - + 2H+ + e- ®®®® NO2 + H2O Pt ®®®® Pt4+ + 4e- · Pt + 4NO3 - + 8H+ ®®®® Pt4+ + 4NO2 + 4H2O [4] [Total: 7]

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 4 2 Propene readily undergoes electrophilic addition of bromine to give 1,2-dibromopropane. For Examiner’s Use (a) (i) Calculate the enthalpy change for this reaction H3CCH=CH2 + Br-Br H3C-CHBrCH2Br Bonds broken are C=C and Br-Br, bonds formed are C-C and 2 C-Br · H = Bonds broken – Bonds formed = + 610 + 193 – 350 – 2(280) · = -107 kJmol-1 of propene (ii) By referring to the electrophilic addition mechanism, sketch the reaction pathway diagram for this reaction. Label your graph, showing clearly the intermediate formed and the enthalpy change of the reaction. · first step is slow step therefore higher EA than second step and an intermediate is formed · the reaction is exothermic (labels) (iii) Bromine adds to propene readily under normal conditions whereas iodine adds only at low temperatures; the 1,2-diiodopropane that results is unstable, decomposing back to propene and iodine at room temperature. The enthalpy change for this addition reaction is – 69 kJ mol-1. Sketch the reaction pathway diagram for iodine addition on the same diagram above labeling your curve clearly. · 1 mark for curve which is less exothermic and show intermediate formed The reaction is reversible thus the addition of iodine should have a lower Ea [5] Energy Extent of reaction -107 kJmol-1 -69 kJmol-1 CH3CHICH2I CH3CH BrCH2Br + CH3 CHCH2I CH3 CH=CH2 + Br2 (or I2) + CH3 CHCH2Br

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 5 (b) 1,2-dibromopropane formed from electrophilic addition of bromine to propene undergoes a series of reactions as shown in the flow chart below. For Examiner’s Use 1,2-dibromopropane NaOH (aq) reflux AgNO3 (aq) No precipitate Acidify with HNO3 followed by addition of AgNO3 (aq) A (Cream precipitate) Evaporate to dryness Solid B Conc H2SO4, heat concentrated NH3 (aq) precipitate dissolves Mixture of Gases (i) Identify the compounds A and B · A : AgBr B : NaBr (ii) Explain why no precipitate is formed when aqueous silver nitrate was added to 1,2-dibromopropane while a cream precipitate is formed when 1,2- dibromopropane was first heated with sodium hydroxide then acidified, followed by the addition of aqueous silver nitrate. · 1,2-dibromopropane undergoes nucleophilic substitution to form 1,2- propandiol giving Br-. AgNO3 reacts with “ free” Br- to give a cream ppt, AgBr. (BrCH2CH(Br)CH3 +OH- HOCH2CH(OH)CH3 + Br- Ag+ +Br- AgBr). ····No precipitate is formed with unhydrolysed 1,2-dibromopropane as no “ free” Br- is released. (iii) Explain why the cream precipitate dissolves upon the addition of concentrated aqueous ammonia. · When NH3(aq) is added, it reacts with the silver ions in solution to from diamminesilver(I) complex ion as shown in equation (1) Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq) ------ (1) Ag+(aq) + Br- (aq) AgBr (s) ------------ (2) · This lowers the concentration of free silver ion in water causing the ionic product of the silver bromide, [Ag+][Br-], to decrease. · By LCP, the position of equilibrium in the equation (2) shifts left. In this instance, the ionic product, [Ag][Br] will decrease to a value lower than the Ksp of AgBr and the precipitate of AgBr will dissolve. (iv) Write balanced equation(s) of the reaction between solid B and concentrated H2SO4. · 2NaBr + 2H2SO4 Br2 + SO2 + 2H2O + Na2SO4 OR · NaBr + H2SO4 NaHSO4 + HBr · 2 HBr + H2SO4 Br2 + SO2 + 2H2O [9] [Total: 14]

TJC H2 Chemistry Paper 2 2008 (Answers) [Turn over 6 3 (a) Trace amounts of elements in a sample can be determined by spectroscopy.In atomic spectroscopy, the concentration of an element in a solution can be determined by its absorbance. The relationship between concentration of aluminium in a solution and its absorbance is given in the graph below. For Examiner’s Use Absorbance