NJC Prelim P1 P2 ANS

NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions 8872/02 17 September 2009 2 hours For Examiner’s Use Section A 1 2 3 4 5 Section B B6 B7 B8 READ THE INSTRUCTION FIRST Write your subject class, registration number and name 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 paper clips, highlighters, glue or correction fluid. Section A Answer all questions . Section B Answer any two questions. 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. Total This question paper consists of 13 printed pages (including this page).

Section A Answer all the question s in this section in the spaces provided. 1 5.00 g of impure iron solid was dissolved in an excess of sulfuric acid and the resulting solution containing Fe2+ was made up to 250 cm 3. 10.0 cm3 of this solution was titrated against a solution of KMnO 4 of concentration 0.01 mol dm 3. It was found that 24.00 cm3 of KMnO4 was required. (a) Suggest a reason why an indicator was not required for this titration. There is a distinct colour change from purple to pale pink /pale orange at end point. [1] (b) Write a balanced redox equation for the reaction between Fe 2+ and MnO4 . 5 Fe2+(aq) + MnO4 (aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) [2] (c) A student claims that the percentage purity of iron in the sample is 33.5%. Justify if the statement made by the student is true. No. of moles of MnO 4  used = 24.00 x 103 x 0.01 = 2.4 x 104 mol No. of moles of Fe2+ in 10.0 cm3 = 5 x 2.4 x 104 = 1.2 x 103 mol No. of moles in 250 cm3 =1.2 x 103 x 10 250 = 0.03 mol Mass of Fe in sample = 0.03 x 55.8 = 1.674 g % of Fe in sample = 00 . 5 674 . 1x 100 = 33.5% Student’s claim is true. [3] [Total: 6] 2

2 The graphs below shows the rate of hydrolysis of ethyl ethanoate, CH3COOC2H5 in an acidic medium. The reaction was followed twice with different concentrations of HCl and the following results were obtained. 77.5, 0.004 155, 0.002 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0 50 100 150 200 250 (a) Using the graphs, determine the order of reaction with respect to both ethylethanoate and HCl. From the graph of [HCl] = 0.15 mol dm 3, t ½ (1) = t ½ (2) = 77.5 min, thus the reaction is 1st order with respect to ethylethanoate. To find order with respect to HCl, compare initial rates by drawing tangents at t = 0 when the graphs are the most linear. For [HCl] = 0.15 mol dm 3, rate = 0 50 006 . 0 01 . 0   = 8.00 x 105 mol dm3 min1 For [HCl] = 0.10 mol dm3, rate = 0 75 006 . 0 01 . 0   = 5.33 x 105 mol dm3 min1 Hence, when [HCl] increase 1.5 x (from 0.10 to 0.15 mol dm3), rate increases 1.5 times from 5.33 x 105 to 8.00 x 105 mol dm3 min1. (b) Deduce the rate equation and calculate the value of the rate constant, giving its units. R = k [HCl][ CH 3COOC2H5] k = ) 15 . 0 )( 01 . 0 ( 10 x 85 = 0.0533 mol1 min1 dm3 (c) On the same axis above, sketch the graph when the experiment is repeated using 0.008 mol dm3 of ethylethanoate and 0.15 mol dm3 HCl. [Total: 7] [HCl] = 0.15 mol dm3 [HCl] = 0.10 mol dm3 [ethyl ethanoate]/ mol dm3 Time/min 77.5 t ½ (1) 155 t ½ (2) 3

3 Methanol is a possible alternative to hydrocarbons as a liquid fuel. (a) Given that the enthalpy of combustion of methanol is –715 kJ mol –1, calculate the mass of methanol that should be burnt in order to boil 1000 cm 3 of water starting from an initial temperature of 20.0 °C. Assume that 50% of the heat obtained from the combustion of methanol is lost to the surroundings and that the specific heat capacity of water is 4.2 J K –1 g–1. Q = 1000 x 4.2 (100  20) = 336 kJ Heat from combustion of methanol = 336 x 50 100 = 672 kJ No. of moles of methanol = 715 672 = 0.94 mol Mass of methanol = 0.94 x (15.0 + 16.0 + 1.0) = 30.1 g [3] (b) Methanol can be produced by using a reversible reaction between carbon monoxide and hydrogen. 2H 2(g) + CO(g) CH3OH(g) (i) Given the following standard enthalpy change of combustion and energy cycle, apply Hess’s Law and calculate the enthalpy change for the production of methanol. ΔH c (H2) =  286 kJ mol1 ΔHc (CO) =  283.3 kJ mol1 2H2(g) + CO(g) CH3OH(g) +O2(g) 2H2O(l) + CO2(g) +1/2O2(g) +3/2O2(g)   Hreaction = 2H1 + H2  H3 = 2(286)  283.3 (715) = 140 kJ mol1 [2] H1 H2 H3 4

(ii) When 2.00 mol of hydrogen and 1.00 mol of carbon monoxide were mixed and heated to a high temperature in a container of volume 1.50 dm 3, 0.80 mol of methanol was formed at equilibrium. Calculate a value for the equilibrium constant, K c, for this reaction at this temperature and give its units. 2H 2(g) + CO(g) CH3OH(g) Initial no. of moles 2.0 1.0 Change 2(0.8) 0.8 +0.8 Final no. of moles 0.4 0.2 0.8 Conc/ mol dm3 5 . 1 4 . 0=0.267 5 . 1 2 . 0=0.133 5 . 1 8 . 0=0.533 Kc = ] [ ] [ ] [ 2 2 3 CO H OH CH= ) 133 . 0 ( ) 267 . 0 ( 533 . 0 2 = 56.2 mol2 dm6 [3] (iii) State and explain the effect on the position of equilibrium for the above reaction when: (I) the pressure is increased. (II) the temperature is increased. (I): By Le Chatelier’s Principle, equilibrium shifts right towards less moles of gas to partially offset the increase in pressure. (II) By Le Chatelier’s Principle, equilibrium shifts left to favour the backward endothermic reaction to use up some of the heat so as to decrease the temperature. [3] [Total: 11] 5

4 The diagra m below shows a reaction scheme starting from methylbenzene. CH3 Stage 1 CH3 Cl CH3 Cl + Stage 2 CH2Cl NaOH(aq) reflux Stage 3 C O H KCN(s) H2SO4(aq) 2,4-dinitrophenylhydrazine methylbenzene A B C (a) Draw the structures for Compounds AC in the boxes provided above. [3] C OH CNH CH2OH CNN O2N NO2 H H 6

(b) Suggest the reagents and conditions required for Stages 13. Stage Reagents and Conditions 1 Cl2. AlCl3 2 Cl2, uv light (or heat to 300oC) 3 K 2Cr2O7, H2SO4, heat with immediate distillation [3] [Total: 6] 7

5 This question is about the extraction of iron from its ore, Fe 2O3 (also known as haematite) using two different methods. The first method of extraction is by using the Blast Furnace. The second method is through a process known as the Thermite Reaction. Extraction of iron using the Blast Furnace. Ninety percent of all mining of metallic ores is for the extraction of iron. Industrially, iron is produced starting from iron ores, principally hematite, by reduction with carbon in a blast furnace at temperatures of ab out 2000 °C. The following table shows the composition of haematite. Compound Percentage Composition Fe2O3 92.0 % SiO2 6.0 % Others 2.0 % Haematite, carbon in the form of coke, and limestone are continuously fed into the top of the furnace, while a blast of heated air is forced into the furnace at the bottom. In the blast f urnace, the following reactions take place. Stage 1: The coke reacts with oxygen in the air blast to produce carbon monoxide: 2C(s) + O 2(g) → 2CO(g) Stage 2: The carbon monoxide formed then react with the iron ore to form molten iron, becoming carbon dioxide in the process: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g) The main impurity i