TJC Prelim P3 Qn & Ans

1 9729/03/TJC Prelims/2018 [Turn over CHEMISTRY 9729/03 Paper 3 Free Response 14th September 2018 2 hours Candidates answer on separate paper. Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your Centre number, index number, name and civics group on all the work you hand in. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. 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.

2 9729/03/TJC Prelims/2018 [Turn over Section A Answer all the questions in this section. 1 (a) There are several stages in the industrial production of methanol from methane. The first stage involves a gaseous equilibrium between the reactants (methane and steam), and some gaseous products. Figures 1.1 and 1.2 show the percentage conversion of methane into the gaseous products under different conditions at equilibrium. (i) Use the information from the graphs above to deduce the sign of the • enthalpy change, ΔH and • entropy change, ΔS for the first stage industrial conversion reaction of methane and steam into the gaseous products. [2] • the enthalpy change, ΔH As temperature increases from 600 0C to 880 0C, % conversion of methane increases. This indicates that the forward reaction is favoured as it absorbs energy. Hence, forward reaction is endothermic, ΔH is positive. • entropy change, ΔS As pressure increases from 1000 kPa to 5000 kPa, % conversion of methane decreases. Forward reaction is not favoured and this indicates that the number of moles of gaseous products particles is more than the gaseous reactants particles. Hence, ΔS for the forward reaction is positive. (ii) The optimum temperature for the industrial conversion of methane and steam into the gaseous products is between 780-880oC. Explain why this is so. [2] • Lower than 780 0C, rate of reaction is slow / % conversion is low, hence not cost effective. • Higher than 880 0C, results in high energy costs / expensive & also after a certain temperature, yield does not increase significantly. Therefore, there is no gain in using a higher temperature. (b) The equation shows the final stage in the production of methanol.

3 9729/03/TJC Prelims/2018 [Turn over CO(g) + 2H2(g) CH 3OH(g) 3.12 moles of carbon monoxide and 5.23 moles of hydrogen were placed in a sealed container. An equilibrium was established at 600 K and the total amount of gaseous molecules was found to be 7.63 moles. The total pressure was 630 kPa. Calculate the equilibrium constant, K p, for this reaction at 600 K and state its units. [3] Let the amount of CH3OH produced be x mol CO(g) + 2H 2(g) CH 3OH(g) Initial/mol 3.12 5.23 - Change/mol -x -2x +x Equilibrium/mol 3.12-x 5.23-2x x Total moles of gas = (3.12-x) + (5.23-2x) + x = 7.63 x = 0.36 no. of moles of CO = 2.76; no. of moles of H 2 = 4.51; no of moles of CH3OH = 0.36 P(CO) = (2.76/7.63) x 630 = 228 kPa P(H2) = (4.51/7.63) x 630 = 372 kPa P(CH3OH) = (0.36/7.63) x 630 = 29.7 kPa = 9.41 x 10—7 kPa-2 • No. of moles of CO, H2 & CH3OH • Partial pressures of CO; H2 & CH3OH • Correct substitution & value for Kp, correct units (c) Carbon monoxide also reacts with steam. CO(g) + H2O(g) CO 2(g) + H2(g) At 1100 0C, Kc = 1.00 In an experiment, 1 mole of carbon monoxide was mixed with 1 mole of steam, 2 moles of carbon dioxide and 2 moles of hydrogen. Deduce, with reasons, the direction in which the reaction will shift to reach equilibrium. [2] Or K c = 1, the concentrations of products must be equal to concentrations of the reactants at equilibrium Or Kc = 1, [CO2][H2] = [CO][H2O] at equilibrium • [CO2] & [H2] need to decrease and [CO] & [H2O] need to increase So position of equilibrium shifts to the left.

4 9729/03/TJC Prelims/2018 [Turn over (d) The compound responsible for the hot taste of chilli peppers is capsaicin. Its molecular structure can be deduced by the following reaction scheme. Capsaicin reacts with sodium metal and compounds C, D and E react with Na2CO3(aq). (i) Suggest the reagents and conditions for reactions 1 and 2. [2] • Reaction 1: Aqueous HCl or H2SO4, heat • Reaction 2: KMnO4 or K2Cr2O7 & dilute H2SO4(aq), heat under reflux (ii) Draw the structural formulae for the compounds C, D, E, F and capsaicin in the above reaction scheme. [5] • Capsaicin: [Do not accept : ]

5 9729/03/TJC Prelims/2018 [Turn over • C is • D is • E is • F is : or (e) and A B C A, B and C nitrogen containing compounds. State and explain the relative basicities of these 3 compounds. [3] Basic strength: B < C < A • A is a stronger base than B & C as the benzyl group, , being electron-releasing in nature, increases the electron density on nitrogen atom. The lone pair of electrons on N is more available for dative bonding to a proton. • The lone pair of electrons on the nitrogen atom of C can be delocalised into the benzene ring, making it less available for dative bonding to a proton. Thus C is a weaker base than A. • B is an amide which is neutral because the lone pair electrons on N is delocalised over O-C-N, i.e: (resonance structure), & hence it is not available for donation to a proton. [Total:19]

6 9729/03/TJC Prelims/2018 [Turn over 2 This question is about nitrogen compounds and their roles in organic synthesis. Azide is the anion with the formula N3-. Azide is used as a chemical preservative in hospitals and laboratories. The azide ion in sodium azide (NaN 3) can be oxidised by iodine to form nitrogen gas. To determine the amount of sodium azide in an impure sample, the azide present is first reacted with excess iodine. The amount of unreacted iodine is then titrated with standard sodium thiosulfate solution: I 2 + 2S2O32− → 2I− + S4O62− 0.120 g of an impure sample of sodium azide was dissolved in water. The mixture was reacted with 25.0 cm 3 of 0.050 mol dm −3 of aqueous iodine. The excess iodine was found to require 23.10 cm3 of 0.040 mol dm−3 aqueous sodium thiosulfate for reaction. (a) (i) Write the equation for the azide ion reacting with iodine. [1] 2N 3 - → 3N2 + 2e (x1) I2 + 2e → 2I- (x1) • 2N3- + I2 → 3N2 +