CJC H2 CHEM P3 SOL Prelim

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 CHEMISTRY 9647/03 Paper 3 Free Response Thursday 27 August 2015 2 hours Additional Materials: Data Booklet Answer Paper READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] 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 any four questions. Write your answers on the answer paper provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. [Turn over 9647/03/CJC JC2 Preliminary Exam 2015 ANSWER SCHEME & EXAMINERS’ COMMENTS

2 9647/03/CJC JC2 Preliminary Exam 2015 Answer any four questions. 1 Butanone is an industrial organic solvent used in the manufacture of plastics, lacquers and varnishes. (a) Butanone reacts with iodine in the presence of an acid according to the equation: CH3CH2COCH3 + I2  CH3CH2COCH2I + HI In a series of experiments, the reaction was carried out with different concentrations of reagents, and the following initial rates were obtained: Expt. [CH3CH2COCH3] / mol dm–3 [I2] / mol dm–3 [H+] / mol dm–3 Initial rate / mol dm–3 min–1 1 0.010 0.010 0.010 8.31 x 10–8 2 0.010 0.010 0.013 10.8 x 10–8 3 0.013 0.010 0.013 14.0 x 10–8 4 0.018 0.013 0.018 26.9 x 10–8 (i) Using the data given, deduce the order of reaction with respect to each of the three reagents, clearly showing how you arrive at your answers . Hence, write a rate equation for the reaction. Let rate = k[CH3CH2COCH3]a[I 2]b[H+]c Using data from experiments 1 and 2, cba cba k k )010.0(]010.0()010.0( )013.0()010.0()010.0( 1031.8 108.10 8 8      c = 1 rate is first order with respect to H+ Using data from experiments 2 and 3, )013.0()010.0()010.0( )013.0()010.0()013.0( 108.10 100.14 8 8 ba ba k k     a = 1 rate is first order with respect to CH3CH2COCH3 Using data from experiments 3 and 4, ]013.0[]010.0][013.0[ ]018.0[]013.0][018.0[ 100.14 109.26 8 8 b b k k     b = 0 rate is zero order with respect to I2 rate = k[CH3CH2COCH3][H+]

3 9647/03/CJC JC2 Preliminary Exam 2015 (ii) Explain how the rate of reaction would change if the solution was diluted with an equal volume of inert solvent, using r to represent the original rate before dilution. Concentration of all solutions would be halved. Since rate = k[CH3CH2COCH3][H+] = r (original rate) New rate = k {[CH3CH2COCH3]/2} {[H+]/2} = r /4. Hence, the rate would be ¼ the original rate. (iii) The reaction can also be carried out in the presence of a strong base instead of an acid. State what would be observed when a strong base is used and w rite an equation for the reaction that occurs. Yellow precipitate of CHI3 observed. CH3CH2COCH3 + 4OH- + 3I2 →CH3CH2CO2 - + CHI3 + 3I- + 3H2O (OR CH3CH2COCH3 + OH - + 3I2 →CH3CH2CO2 - + CHI3 + 3HI) (Less accurate as HI further reacts with OH-) [7] (b) Elemental iodine, I2, has limited solubility in water. Its solubility is greatly increased in a solution containing iodide ion, I-, due to the formation of triiodide ion, I3 -, as shown: I2(aq) + I-(aq) ⇌ I3 -(aq) (i) Write an expression for Kc for this reaction, stating its units. Kc = ]][I[I ][I - 2 - 3 mol-1 dm3 (ii) Experiments have shown that when [ I-] = 1.4 x 10-3 mol dm-3, the concentrations of I2 and I3 - are equal. Use this information to calculate a value of Kc. Since [I2] = [I3 -], Kc = ]][I[I ][I - 2 - 3 = )10 x ](1.4[I ][I 3- 2 - 3 = )10 x (1.4 1 3- = 714 mol-1 dm3 (iii) Using your answer from (ii), calculate the concentration of iodide ions necessary for 99 % of the iodine to be dissolved. ][I ][I 2 - 3 = 1 99 Kc = ]][I[I ][I - 2 - 3 = ][I 99 - = 714 [I-] = 714 99 = 1.39 x 10-1 mol dm-3 [4]

4 9647/03/CJC JC2 Preliminary Exam 2015 (c) Explain, using a Maxwell-Boltzmann distribution curve, how the rate of a reaction would change if the temperature was increased. From the Maxwell -Boltzmann distribution of molecular energies, an increase in temperature results in more molecules having greater kinetic energy. This means that the fraction of particles with energy  Ea would increase. Thus, more molecules would have sufficient energy to overcome the energy barrier. Hence, the frequency of effective collisions increases, leading to increase in rate. [3] (d) Grignard reagents are organo -magnesium halides, commonly used in synthesis to prepare a variety of organic compounds. The carbon-magnesium bonds in Grignard reagents are highly polar and this makes it extremely useful in organic synthesis as it is able to react with other polar organic molecules to form carbon-carbon bonds. An example of the use of a Grignard reagent is the two -step reaction of CH 3CH2MgBr with butanone, CH3CH2COCH3, to form 3-methylpentan-3-ol. T2 > T1 Number of particles with energy ≥ Ea at 273k Number of particles with energy ≥ Ea at 298k Number of molecules with energy E

5 9647/03/CJC JC2 Preliminary Exam 2015 (i) Suggest the type of reaction that has taken place in step 1. Nucleophilic Addition (ii) Suggest the structural formula of the organic compound to be used with CH3CH2MgBr to form propan -1-ol, if the reaction undergoes a similar two -step process. (iii) Suggest a suitable Grignard reagent and another organic compound to be used if propan-2-ol is to be prepared using a similar two-step process. [4] (e) There are certain practical issues with the use of Grignard reagents, notably slow formation of the Grignard reagent and its susceptibility to water in air. The slow formation of the Grignard reagent may be offset by cutting the precursor magnesium into smaller pieces or scraping the sides of the metal be fore use. Using a balanced equation, explain why magnesium, which has been stored in dry ambient conditions, may not react as quickly as expected. Mg(s) + ½ O2(g)  MgO(s) Mg reacts with the oxygen in air to form magnesium oxide which acts as a protective layer and so, prevents the magnesium from reacting with the organic compounds. [2]

6 9647/03/CJC JC2 Preliminary Exam 2015 2 The direct oxidation of alcohols in a fuel cell represents potentially the most efficient method of obtaining useful energy from renewable fuel. The diagram below illustrates the functional parts of a direct -ethanol fuel cell (DEFC). The anode and cathode compartments are separated by the proton exchange membrane . The protons are transported across the proton exchange membrane to the cathode where they react with oxygen to produce water. Diagram of a DEFC (a) (i) Explain, in terms of the change in average oxidation number of the carbon atoms, whether electrode 1 or 2 acts as the anode. Electrode 1 acts as the anode as oxidation occurs since there is an increase in average oxidation state for carbon from -2 to +4. (ii) Write the half -equations for the reactions which take place at the electrodes of the DEFC, and hence an overall equation for the cell reaction. Anode: CH3CH2OH + 3H2O → 2CO2 + 12H+ + 12e- Cathode: O2 + 4H+ + 4e- → 2H2O Overall: CH3CH2OH + 3O2 → 2CO2 + 3H2O (iii) The cell is capable of producing an e.m.f. of 1.47 V. By using suitable data from the Data Booklet, calculate a