PJC H1 CHEM P2 ANS

PIONEER JUNIOR COLLEGE 2015 JC2 PRELIMINARY EXAMINATION HIGHER 1 CHEMISTRY 8872/02 Paper 2 15 September 2015 Candidates answer Section A on the Question Paper Additional Materials: Data Booklet Writing Paper 2 hours READ THESE INSTRUCTIONS FIRST Write your name, index no and CT group on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Section A Answer all questions. Section B Answer any two questions on separate writing paper. Begin each question in a fresh sheet of writing paper. 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. This document consists of 16 printed pages Name: Index No.: CT Group: 14 FOR EXAMINER’S USE Section A Section B 1 / 10 5 / 20 2 / 9 6 / 20 3 / 9 7 / 20 4 / 12 Penalty s.f. / units TOTAL / 80

2 Section A (40 marks) Answer all questions in the spaces provided. 1 The diagram below shows the reactions of an aqueous salt, A. orange solution B brown solution C colourless solution colourless solution cream precipitate yellow precipitate colourless A(aq) AgNO3(aq)AgNO3(aq) A has the following composition by mass: K, 41.1%; S, 33.7%; O, 25.2%. The relative formula mass, Mr, of A is 190.4. One formula unit of A contains only one type of anion. (a) Determine the formula of the salt A. Let the mass of salt A be 100g. K S O Mass / g 41.1 33.7 25.2 Amount / mol 1.0512 1.0498 1.575 Ratio 1.0013 1.000 1.500 Simplest Ratio 2 2 3 Empirical formula of salt is K2S2O3. Let formula of salt be (K2S2O3)n. [2(39.1) + 2(32.1) + 3(16.0)]n = 190.4 n = 1 [2] Formula of salt is K2S2O3.

3 (b) Suggest the identities of the cream precipitate and the yellow precipitate. Cream precipitate: AgBr Yellow precipitate: AgI [1] (c) Suggest the identities of B and C. B: Br2 C: I2 [1] (d) With reference to the Data Booklet, show, by means of two ionic equations, how the yellow precipitate is formed from C. 2S2O3 2-(aq) + I2(aq)  2I-(aq) + S4O6 2-(aq) I-(aq) + Ag+(aq)  AgI(s) [2] (e) For the reaction of A and B, a titration was carried out to determine if the sulfur - containing product is S4O6 2-or SO4 2-. It was found that 25.0 cm 3 of 0.200 mol dm -3 of B required 10.0 cm3 of 0.125 mol dm-3 of A for complete reaction. Use the data to determine the final oxidation state of sulfur in the product. Hence, write a balanced equation for the reaction of A and B. 25.0 x 0.200 mol of Br2 ≡ 10.0 x 0.125 mol of S2O3 2- 4 mol of Br2 ≡ 1 mol of S2O3 2- 4 mol of Br2 ≡ 8 mol of e- ≡ 1 mol of S2O3 2- Each S atom will have a gain in O.S. by +4. Final oxidation state of S = +2 + 4 = +6 4Br2 + S2O3 2- + 5H2O  8Br - + 2SO4 2- + 10H+ [3] [Total: 9]

4 2 For a reaction X → Y, the following experimental results were obtained. Time / s 0 20 40 60 80 100 120 140 ∞ [Y] / mol dm-3 0 0.12 0.20 0.26 0.30 0.33 0.35 0.37 0.40 When the reaction is complete, the concentration of Y is 0.40 mol dm-3. (a) Plot the graph of [ Y] ag ainst time on the grid below to find the half -life of the reaction. Hence, determine the order of reaction with respect to X. Half-life of the reaction = 40 s Since the two half-lives are constant, hence the order of reaction with respect to X is one. [4] (b) Determine the initial rate of the reaction. From the graph, gradient at t = 0 is the initial rate of reaction = 0.20 / 32 = 0.00625 mol dm-3 s-1 [1] (c) Hence, c alculate the rate constant, stating its units , given that the initial concentration of X is 0.40 mol dm-3. 0.00625 = k(0.40) k = 0.0156 s-1 [1]

5 (d) With the aid of a sketch of the Boltzmann Distribution, explain why an addition of catalyst increases the rate of reaction. A catalyst  provides an alternative reaction pathway with lower activation energy  increases the number of reacting particles with energy  Ea  increases the number of effective collisions per unit time hence increase the rate of reaction [3] [Total: 9] 3 (a) Describe what you would see when sodium burns in oxygen and write the balanced equation, with state symbols, for the reaction described. Sodium b urns very vigorously in O 2 with a yellow flame , leaving behind a white residue of Na2O. 4Na(s) + O2(g) → 2Na2O(s) [2] (b) Describe what you would see when the residue from (a) is dissolved in water containing Universal Indicator solution. Write balanced equation, with state symbols, for the reaction described. Na2O(s) + H2O(l) → 2NaOH(aq) The white solid dissolves. Universal Indicator changes from green to purple / violet. [2]

6 (c) Each of the following oxides can react with sodium hydroxide and/or hydrochloric acid. For each oxide, write a balanced equation for its reaction with either hydrochloric acid or sodium hydroxide.  magnesium oxide MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)  aluminium oxide Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) Or Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq)  phosphorus(V) oxide P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) [3] (d) Explain, in terms of structure and bondi ng, why aluminium oxide melts at 2070 ºC while aluminium chloride sublimes at 178 ºC. Al2O3 has a giant ionic lattice structure whereas A lCl3 has a simple covalent structure. The strong electrostatic forces of attraction / ionic bonds between Al3+ & O2- ions require a large amount of energy to be weakened. The weak van der Waals’ ( instantaneous dipole ) forces between molecules of A lCl3 require only a small amount of energy to overcome. Hence, Al2O3 has a much higher melting point than AlCl3. [3] [Total: 10]

7 4 (a) Consider the following reaction scheme, starting from ethanol. CH3CH2OH CH3CHO C OH CO2H H CH3 CH3CH(OH)CN C OH H H C H H HO2C Isomer E Isomer D Compound A Compound B Compound C excess concentrated H2SO4, 170 oC Br2(aq) hot ethanolic KCN hot H2SO4(aq) step I step II step III (i) Suggest reagents and conditions for steps I, II and III. step I: K2Cr2O7(aq), H2SO4(aq), distill step II: HCN, trace NaCN