VJC Prelim P2 ANS

 VJC 2015 9647/02/PRELIM/15 [Turn over 1 Victoria Junior College 2015 H2 Chemistry Prelim Exam 9647/2 Suggested Answers 1 Planning You are provided with 5 unlabelled bottles containing pure natural acids from the extracts of roasted coffee. Each bottle contains one of the following natural acids:  lactic acid,  maleic acid,  oxalacetic acid,  pyruvic acid,  quinic acid, All the acids above are soluble in water. You are also provided with any other common laboratory reagents and apparatus. (a) All natural acids contain one or more c arboxylic acid functional groups. Other than the carboxylic acid functional group, what other functional groups are also present in these natural acids? lactic acid: secondary alcohol maleic acid: alkene oxalacetic acid: ketone pyruvic acid: ketone quinic acid: Secondary alcohol and tertiary alcohol [2] (b) Two of these natural acids are colourless liquids and the rest are white crystalline solids. Using relevant chemical knowledge, identify the two liquids. Explain your choices. The two liquids are lactic acid and pyruvic acid. Explanation: These two compounds have a relatively smaller Mr and thus weaker dispersion forces between their respective molecules than the other 3 compounds (due to smaller number of electrons present). [2] OH CO2H HO HO HO CH3COCO2H HO2CCOCH2CO2H CH3CH(OH)CO2H HO2CCH=CHCO2H

 VJC 2015 9647/02/PRELIM/15 [Turn over 2 (c) Suggest a reagent that could be used to carry out a test -tube ex periment to distinguish the two liquids. Reagent: Brady’s reagent [or 2,4-dinitrophenylhydrazine] Describe what would be observed for each compound in the experiment. Observation: Pyruvic acid will give an orange ppt. but lactic acid will not. [2] (d) Outline a logical sequence of chemical tests that would enable you to identify the remaining 3 solids. You should aim for a minimum number of reactions. Your plan should include  a positive test to confirm the identity of each compound;  detailed procedure (including quantities of chemicals and conditions used);  expected observations for each compound in each test. Step 1: Prepare aqueous so lutions of each of the 3 solid samples by dissolving 1-cm depth of solid in about 5 cm depth of deionised water in each test-tube. Step 2: To 1 -cm depth of each of the 3 samples in separate test -tubes, add equal volume of Brady’s reagent. Oxalacetic acid will give an orange ppt. while the other 2 samples will not. Step 3: To 1 -cm depth of each of the 2 remaining samples in separate test-tubes, add a few drops of Br2(aq). Maleic acid will decolourise reddish -brown Br 2 while the remaining sample will not. Step 4: To 1-cm depth of the last sample in a test-tube, add a few drops of acidified KMnO4(aq). Heat the mixture in a hot water -bath. Quinic acid will turn purple KMnO4 colourless. [2]

 VJC 2015 9647/02/PRELIM/15 [Turn over 3 (e) How would you ensure the reliability of the test result for quinic acid? To ensure a colour change for KMnO 4, the acidified KMnO 4(aq) must be added slowly and dropwise. It should not be added in excess. T he mixture must also be heated to prevent incomplete reducti on to black MnO2 solid. [Or hydrochloric acid cannot be used to acidify KMnO 4 as the chloride ions can be oxidised by KMnO 4. As a result, there may be a colour change of KMnO4 due to oxidation of chloride to Cl2.] [1] (f) Suggest a safety measure that you would consider in carrying out your plan. (g) Use a hot water bath for heating instead of using a direct naked flame from the bunsen burner as most organic compounds are highly flammable. [1] Draw a set -up of the apparatus for the synthe sis of pyruvic acid from lactic acid. State the required reagents and conditions. Reagents and conditions: KMnO4(aq) / H2SO4(aq), reflux [OR K2Cr2O7(aq) / H2SO4(aq), reflux] [2] [Total: 12]

 VJC 2015 9647/02/PRELIM/15 [Turn over 4 2 Use of the Data Booklet is required for this question. (a) A 2.85 g sample of haematite iron ore, Fe2O3, was dissolved in hydrochloric acid and the solution diluted to 250 cm 3 in a standard flask. A 25.0 cm 3 of this solution was completely reduced with excess tin (II) chloride to form a solution of iron(II) ions. After t After the remaining tin(II) ions were removed with a suitable reagent, the solution of iron(II) ions was titrated against an acidified solution of 0.020 mol dm -3 potassium dichromate(VI) and required 26.40 cm3 for complete oxidation back to iron(III) ions. (i) Give the balanced equatio n for the reaction between iron (II) and dichromate(VI) ions. Cr2O7 2–(aq) + 14H+(aq) + 6Fe2+(aq)  2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) [1] (ii) Calculate the percentage of iron(III) oxide, Fe2O3, in the ore. Amount of Cr2O7 2- used = 0.0200 x = 5.28 x 10-4 mol Amount of Fe2+ present in 25.0 cm3 solution = 6 x 5.28 x 10-4 mol Total amount of Fe2+ present = 6 x 5.28 x 10-4 x = 3.168 x 10-2 mol Hence amount of Fe2O3 present = 3.168 x 10-2 x mol Mass of Fe2O3 present = 3.168 x 10-2 x x 159.6 = 2.528 g % of Fe2O3 present = x 100 = 88.7% [3] (b) (i) Suggest whether the acidified pot assium dichromate (VI) can be replaced by potassium manganate (VII) for oxidising iron (II) back to iron (III). Explain your answer. KMnO4 cannot be used as it is a stronger oxidising agent than K 2Cr2O7 (more positive E ⍬). Thus chloride ions in both hydrochloric acid and SnCl2 would also be oxidised leading to inaccurate titration results. [2]

 VJC 2015 9647/02/PRELIM/15 [Turn over 5 (ii) Explain why the excess tin (II) ions have to be removed before titr ation with potassium dichromate(VI). The excess tin( II) ions would also be oxidised to tin( IV) by Cr 2O7 2- ions hence again leading to inaccurate titration results. [1] (c) The compounds of manganese catalyse a wide variety of reactions, one instance being manganese dioxide, MnO 2, which catalyses the decomposition of hydrogen peroxide: (i) With the aid of a sketch of the Bol tzmann distribution curve, explain how the presence of a catalyst increases the rate of reaction. Correct shape of Boltzman distribution curve and axis labels When a catalyst is used in a reaction, it provides an alternative reaction pathway with lower activation energy . Hence, number of reacting particles with energy ≥ E a increases as shown by the shaded area . This leads to higher frequency of effective collisions , thereby increasing the rate of reaction. [2]

 VJC 2015 9647/02/PRELIM/15 [Turn over 6 (ii) The relationship between the rate constant and temperature is generally governed by the Arrhenius equation: k = A where k = rate constant A = Arrhenius constant Ea = activation energy in J mol-1 R = molar gas constant = = 8.31 J K-1 mol-1 T = temperature in K To determine the activation energy of the above reaction, the rate constant was determined at various temperatures and the experimental results were then processed by plotting a graph of ln k vs 1/T (K-1): Use the given information to determine a value for the activation energy of the reaction. Taking ln on both sides of the Arrhenius equation gives: ln k = ln A – (Ea / R) × (1/T) Hence in the graph of ln k vs 1/T, gradient = – Ea / R Gradient = = – 6250