DHS H2 Chem 2013 Prelim P3 Soln

This question paper consists of 17 printed pages and 1 blank page. © DHS 2013 [Turn over Name: Mark Scheme Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY 9647/03 Paper 3 24 September 2013 Paper 2 2 hours Additional Materials: Data Booklet, Graph Paper & Writing Paper INSTRUCTIONS TO CANDIDATES 1 Answer any four questions. 2 Begin each question on a fresh sheet of paper. 3 At the end of the examination: • Staple or fasten all your work securely together with the Cover Sheet on top. • Hand in the question paper separately. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings and calculations. You are reminded of the need for good English and clear presentation in your answers. 534

2 © DHS 2013 9647/03 [Turn Over Answer any four questions. Begin each question on a fresh sheet of paper. 1 Dinitrogen pentoxide, N2O5, can be produced by the following reaction sequence in the car engine: I N2(g) + O2(g) → 2NO(g) ∆HӨ = +180 kJ mol-1; ∆SӨ = +24.8 J mol-1 K-1 II NO(g) + 1/2O2(g) → NO2(g) ∆HӨ = –57.0 kJ mol-1 III 2NO2(g) + 1/2O2(g) → N2O5(g) ∆HӨ = –110 kJ mol-1 (a) (i) Explain why reaction I does not take place at room temperature but occurs in car engine. ∆GӨ = ∆HӨ – T∆SӨ ∆GӨ (+172.6 kJ mol -1) is positive at 298 K, hence reaction I is not spontaneous at room temperature The high temperature in the car engine causes the T ∆SӨ to become more positive (or – T ∆SӨ to become more negative). This will lead to a more negative ∆GӨ and hence reaction I becomes spontaneous. (ii) Predict in reaction II, with reasons, the sign of ∆SӨ. The amount of gases decreases from 1 1/2 mol to 1 mol, hence there are lesser number of ways to arrange the gaseous particles in the system. ∆SӨ is negative. (iii) Nitrogen dioxide is a pollutant that is often produced in car engine. Suggest how the pollutant can be removed in the car engine. Pt 2NO2(g) → N2(g) + 2O2(g) The nitrogen dioxide is converted to nitrogen and oxygen gas in the catalytic convertor (with Pt as catalyst) in the car engine. (iv) By drawing a suitable energy cycle and using the data above, calculate the standard enthalpy change of formation of dinitrogen pentoxide. N 2(g) + 5/2O2(g) N2O5(g) 2NO(g) + 3/2O2(g) 2NO2(g) + 1/2O2(g) By Hess’ law, ∆H f Ө(N2O5) = +180 + 2(–57.0) – 110 = –44.0 kJ mol-1 +180 2(–57.0) –110 ∆Hf Ө(N2O5) 535

3 © DHS 2013 9647/03 [Turn Over (b) The rate of reaction for reaction II was investigated and the following kinetics data was obtained. Time/minutes Experiment 1, with [NO] = 0.10 mol dm-3 Experiment 2, with [NO] = 0.05 mol dm-3 [O2] / mol dm-3 [O2] / mol dm-3 0 0.0050 0.0050 5 0.0031 0.0045 10 0.0019 0.0040 15 0.0011 0.0036 20 0.0007 0.0032 25 0.0005 0.0029 30 0.0004 0.0026 (i) Explain why nitrogen monoxide is used in large excess. This is to ensure that the concentration of nitrogen monoxide remains approximately constant so that the rate of reaction will be independent of the concentration of nitrogen monoxide. (ii) Using the same axes, plot graphs of [O2] against time for the two experiments. (iii) Use your graphs to determine the order of reaction with respect to O 2 and NO, showing your workings clearly. Using the curve for experiment 1, t 1 ≈ t 2 ≈ 7 min. The half-lives of O 2 are approximately constant at 7 min, hence the reaction is first order with respect to O 2. When [NO] = 0.1 mol dm-3, initial rate = 4 x 10-4 mol dm-3 min-1 When [NO] = 0.05 mol dm-3, initial rate = 1 x 10-4 mol dm-3 min-1 When [NO] is doubled, the initial rate is quadrupled. Hence, the reaction is second order with respect to NO. (iv) State the rate equation of reaction II and hence calculate the rate constant, 0 0.001 0.002 0.003 0.004 0.005 0.006 0 1 02 03 04 0 [O2] / mol dm‐3 time / min experiment 1, [NO] = 0.10 mol dm‐3 experiment 2, [NO] = 0.05 mol dm‐3 t2 = 7mint1 = 7min 536

4 © DHS 2013 9647/03 [Turn Over including its units. rate = k[O2][NO]2 Using initial rate method: 4 x 10 -4 = k(0.005)(0.1)2 k = 8.00 mol-2 dm6 min-1 or using half-life method: rate = k’[O2] where k’ = k[NO]2 k’ = ln2 / 7 = 0.09902 min-1 k = 0.09902 / (0.1)2 = 9.90 mol-2 dm6 min-1 (v) Explain how the half-life of oxygen will be affected when the concentration of nitrogen monoxide is doubled. rate = k[O 2][NO]2 rate = k’[O2] where k’ = k[NO]2 t1/2 = ln2 / k’ = ln2 / k[NO]2 When [NO] is doubled, the half-life of O2 will be reduced by 4 times. (c) The nitrogen dioxide produced in reaction II is able to form dinitrogen tetraoxide as shown below: 2NO2(g) N2O4(g) brown colourless Explain whether the enthalpy change of dimerisation is endothermic or exothermic. Hence, predict the colour change of the reaction mixture when the temperature is increased. The forward reaction is exothermic as N-N bond is formed during dimerisation. By Le Chatelier’s principle, an increase in temperature favours the backward reaction and the position of equlibrium shifts to the left. Hence, the colour of the reaction mixture becomes darker brown. [Total: 20] 2 Transition elements, such as cobalt, copper and chromium, have different properties that can distinguish themselves from the main group element such as magnesium. With their unique properties, transition elements are capable of having variable oxidation states and forming coloured complexes. (a) What do you understand by the term transition element? A transition element is a d-block element that is capable of forming at least one ion with a partially-filled d-subshell. (b) When air is bubbled through an aqueous solution containing CoC l2, NH4Cl and NH3, and the resulting solution evaporated, crystals of a salt X can be isolated. X has the following composition by mass: Co, 25.2 %; N, 24.0 %; H, 5.1 %; Cl, 45.7 % 537

5 © DHS 2013 9647/03 [Turn Over On adding an excess of aqueous silver nitrate to an aqueous solution containing 0.01 mol of X, 1.43 g of silver chloride is precipitated. Determine the formula of the octahedral cation in X. Hence, calculate the oxidation number of the cobalt atom in X. Co N H Cl % mass 25.2 24.0 5.1 45.7 % mole 0.428 1.714 5.1 1.287 Mole ratio 1 4 12 3 n(AgCl) = 1.43 / 143.5 ≈ 0.01 mol 1 mol of complex contains 1 mol of free chloride ions. Hence, the formula of the octahedral cation is [Co(NH 3)4Cl2]+ Oxidation number of cobalt = +3 (c) Haemocyanin is a copper containing oxygen transport molecule in horseshoe crabs. Both oxygenated and deoxygenated forms of haemocyanin contain copper ions. Haemocyanin is colourle ss when deoxygenated and blue when it is exposed to oxygen. Suggest the oxidation states of copper in the oxygenated and deoxygenated forms of haemocyanin. Hence explain the difference in colour observed. Oxidation state of Cu in oxygenated state = +2 Oxidation state of Cu in deoxygenated state = +1 In the oxygenated state, as the higher energy d-orbital is empty and d-electron in the lower energy d-orbital can be promoted to a higher energy level in Cu 2+ (d9), the colour of haemocyanin is blue. In the deoxygenated state, all d orbitals are fully filled in Cu + (d10), hence d-d transition is not possible which results in colourless haemocyanin. (d) Reagents containing transition elements are commonly used in organic syntheses. A common reage