SAJC Prelim P2 Answers

NAME Class ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Chemistry (9729) Paper 2 Structured Questions 12 September 2018 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS: Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. 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. For Examiner’s use: Question 1 2 3 4 5 Total Marks 19 20 10 7 19 75 This document consists of 20 printed pages (including this page).

2 GOx Answer all the questions. 1 Glucose oxidase (GOx) is an enzyme found in certain species of insects and fungi that catalyses the oxidation of glucose to gluconic acid. C6H12O6 + O2 + H2O C6H12O7 + H2O2 (a) Write two half-equations to show that this is a redox reaction. [2] C6H12O6 + H2O  C6H12O7 + 2H+ + 2e─ O2 + 2H+ + 2e─  H2O2 (b) The binding of reactants to GOx can be simplified with a diagram as shown below. (i) Explain why GOx can be described as a biological catalyst. [2] GOx allows certain reactants to bind specifically to it for the conversion into products. GOx speeds up the rate of reaction by providing an alternative pathway of lower activation energy. OR It is regenerated / remained chemically unchanged at the end of the reaction. (ii) Based on the diagram above, suggest why the sign of ∆S is negative for the reaction. [1] The 2 reactants and GOx must come together to form one large entity. OR There is a decrease in number of (gaseous) particles. Hence, the products are less disordered than the reactants. glucose gluconic acid glucose O2 glucose O2 GOx

3 [Please Turn Over (iii) Sketch a graph showing how the rate of this GOx-catalysed reaction varies with the concentration of glucose. [1] (c) The overall equation can be re-expressed to show the change in functional group. C 5H11O5CHO + O2 + H2O C 5H11O5COOH + H2O2 glucose gluconic acid Calculate the enthalpy change of reaction for the conversion of glucose into gluconic acid. Use relevant data from the Data Booklet. [2] Bonds broken in reactant: C─H, C=O, O=O, 2O─H Bonds formed in product: C=O, C─O, O─H + 2O─H + O─O ∆H = (410 + 496) − (360 + 460 + 150) = − 64.0 kJ mol−1 (d) Experiments were done to determine the kinetics of the reaction. In the first investigation, the following reaction mixture was prepared. initial [glucose] = 5.00 x 10−3 mol dm−3 initial [GOx] = 1.00 x 10−2 mol dm−3 initial [O2] = 1.00 x 10−2 mol dm−3 The following results in Table 1 were obtained. [glucose] rate

4 Table 1 t/s [glucose] / mol dm −3 0 5.00 x 10 −3 10 3.40 x 10 −3 20 2.50 x 10 −3 30 1.80 x 10 −3 60 6.00 x 10 −4 (i) To determine the order of reaction with respect to [glucose], use these data to plot a suitable graph on the grid below. [2]

5 [Please Turn Over Guidelines for scale: x-axis: t/s, 1 big square = 20s y-axis: [glucose] / mol dm−3, 1 big square = 1.0 X 10−3 mol dm−3 (ii) Hence, deduce the order of reaction with respect to [glucose], showing all your working and drawing clearly on your graph. [2] Show two t1/2 clearly on the graph. Since t1/2 is (approximately) constant at 20s, the reaction is first order with respect to [glucose]. In the second and third investigations, the concentrations of oxygen and GOx were changed, but the initial [glucose] was kept the same as before. The following results in Table 2 were obtained. Table 2 Investigation Initial [O 2] (mol dm−3) Initial [GOx] (mol dm−3) Initial rate (mol dm−3 s−1) 1 1.00 x 10 −2 1.00 x 10 −2 Y 2 5.00 x 10 −3 1.00 x 10 −2 2.00 x 10 −4 3 5.00 x 10 −3 2.50 x 10 −3 5.00 x 10 −5 (iii) Use your graph in (d)(i) to determine the initial rate Y, showing all your working and drawing clearly on your graph. Hence, use the information in Table 2 to determine the orders of reaction with respect to [O 2] and [GOx]. Explain your reasoning. [3] Y = (5.0 X 10 −3 – 1.0 X10−3) = 2.00 X 10−4 20 Using investigation 1 and 2, when [O 2] is halved, the initial rate remains the same. Hence, the reaction is zero order with respect to [O2].

6 Using investigation 2 and 3, when [GOx] decreases 4 times, the initial rate decreases 4 times. Hence, the reaction is first order with respect to [GOx]. (e) GOx can be used in a biosensor to convert glucose present in body fluids into gluconic acid. The amount of hydrogen peroxide produced is then reduced electrochemically to determine the amount of glucose present. A 0.1 cm 3 of blood sample from a patient was tested to diagnose if he was at risk of diabetes. The diagnosis is based on the concentration of glucose in the blood. Condition [glucose] in blood (x 10 −3 mol dm−3) Normal less than 5.6 Pre-diabetes 5.6 − 6.9 Diabetes More than 6.9 The biosensor gave a current of 1.01 mA for 1 min. (1000 mA = 1A) (i) Calculate the number of moles of hydrogen peroxide produced. Hence, diagnose the condition of the patient. [3] H 2O2 + 2H+ + 2e─  2H2O Amt of charge = 1.01 X 10−3 X 60 = 0.0606 C Amt of e─ = 0.0606 / 96500 = 6.2798 X 10−7 mol Amt of H2O2 = 6.2798 X 10−7 / 2 = 3.14 X 10−7 mol (3sf) Amt of glucose = amt of H2O2 = 3.14 X 10−7 mol [glucose] = 3.14 X 10−7 / (0.1 X 10−3) = 3.14 X 10−3 mol dm−3 The patient is normal. (ii) Before the test, the blood sample has to be treated to remove some components present. Suggest why a treated blood sample was necessary for the biosensor to give an accurate reading. [1] The blood has to be treated to remove some components that can be reduced or oxidised by the biosensor or GOx .

7 [Please Turn Over [Total: 19] 2 This question deals with carbon and silicon which are both elements in Group 14. (a) C60 and diamond are allotropes of carbon. C 60 is a simple covalent molecule while diamond is a giant covalent molecule. State the type of bonding and describe the lattice structure of solid C 60. C60 [2] Type of bonding: Within each C 60 molecule, there is strong covalent bonds between carbon atoms. Describe lattice structure: C60 exists as a regular lattice of simple covalent molecules with instantaneous dipole-induced dipole interactions between C60 molecules. (b) 0.144 g of C60 was placed in a 100 cm3 container of hydrogen gas at 20 °C and 1.00 × 105 Pa. The reaction occurred as shown in the equation. C60 (s) + xH2 (g) → C60H2x (s) When all the C60 had reacted, the pressure was found to be 2.21 × 10 4 Pa at the same temperature. (i) Calculate the amount, in moles, of C60 that reacted. [1] Amount of C60 = 0.144 / 720 = 2 × 10–4 (ii) Calculate the amount, in moles, of hydrogen gas that reacted with C60. [2] pV = nRT ∴ ∆n = (p1 – p2)V / RT ∆n = (1.00 × 105 – 2.21 × 104).100 × 10–6 / 8.31 × 293 = 0.00320 mol

8 (iii) Use your answers from (i) and (ii) to deduce the molecular formula of the hydrocarbon, C60H2x. [2] C60:H2 = 2.00 × 10–4 : 0.00320 = 1:16 X = 16 C 60H32 (c) (i) Graphite is another allotrope of carbon. State the type of hybridisation and draw the arrangement of the hybrid orbitals about each C atom. [2] sp2 (ii) Graphite is a good conductor of electricity. Explain, with