2021 YIJC Prelim P2 Ans

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME CG INDEX NO H1 GROUP CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 30 August 2021 2 hours READ THESE INSTRUCTIONS FIRST This document consists of 24 printed pages and 0 blank page. For Examiner’s Use Paper 1 /30 Paper 2 1 /14 2 /6 3 /10 4 /14 5 /16 6 or 7 /20 Penalty /80 Overall (Paper 1 & 2) Percentage (%) /110 Write your name, index number and CG on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use stap les, paper clips, highlighters, and glue or correction fluid/tape. 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.

2 ©YIJC [Turn over Section A Answer all the questions in this section, in the spaces provided. 1 At 573K phosphorus( V) chloride, PC l5(g), decomposes to form phosphorus( III) chloride, PCl3(g), and chlorine, Cl2(g). A dynamic equilibrium is established as shown. PCl5(g) ⇌ PCl3(g) + Cl2 (g) numerical value of Kc = 0.625 (a) (i) Explain what is meant by the term dynamic equilibrium. [1] A dynamic equilibrium refers to a reversible reaction where the forward and reverse reactions are continuing at the same rate. (ii) Suggest and explain two changes which will increase the decomposition of PCl5. [2] Addition of PC l5 / removal of C l2 or PC l3 which will cause the position of equilibrium to shift to the right to remove the added PCl5 A smaller pressure / larger volume which will cause the equilibrium position to shift to the right to favour the formation of more gaseous particles. (iii) The experiment was repeated at 473K. The numerical value of Kc was found to be 8.3 x 10 –3. Deduce whether the decomposition reaction is exothermic or endothermic. Explain your answer. [2] - A lower temperature favours exothermic reaction (so as to release heat and decrease temperature) - At lower temperature, the value of Kc decreases, indicating that the backward reaction is favoured . Link: Since the backward reaction is e xothermic, the decomposition reaction (forward reaction) is endothermic . (iv) The first experiment at 573K was repeated but the total pressure was doubled. Predict the effect this would have on the value of Kc at 573K. Explain your answer. [No calculations are required.] [1] Kc remains unchanged, as temperature is constant and Kc is dependent on temperature only. (b) The rate of this reaction was measured at different initial concentrations of the two reagents. The table shows the results obtained. experiment [CH3CH2CHClCH3] [I] relative rate 1 0.06 0.03 3 2 0.10 0.03 5

3 ©YIJC [Turn over 3 0.06 0.05 5 4 0.08 0.04 (i) Deduce the order of reaction with respect to each of [CH3CH2CHClCH3] and [I]. Explain your reasoning. [2] M1: Using experiment 1 and 2, when [CH3CH2CHClCH3] increases by 5/3 times keeping [I] constant, rate increases by 5/3 times. Order of reaction wrt [CH3CH2CHClCH3] = 1 M2: Using experiment 1 and 3, when [I] increases by 5/3 times keeping [CH3CH2CHClCH3] constant, rate increases by 5/3 times . Order of reaction wrt [I] = 1 (ii) Write the rate equation for this reaction, stating the units of the rate constant, k. [2] rate = k[I][CH3CH2CHClCH3] units of k = dm3 mol–1 s–1 (iii) Calculate the relative rate for experiment 4. [1] 3 = k (0.06)(0.03) k = 1.6667 x 103 rate = (16667)(0.08)(0.04) = 5.33 mol dm–3 s–1 (c) (i) An important reaction of CHC l3(g) is the manufacture of CHC lF2(g), using the following reversible reaction. CHCl3(g) + 2HF(g) ⇌ CHClF2(g) + 2HCl(g) Use the data to calculate the enthalpy change of reaction, ΔHr , for the formation of CHClF2(g) as shown in the equation. compound enthalpy change of formation, ΔHf / kJ mol1 CHCl3(g) 103.2 CHClF2(g) 482.2 HF(g) 273.3 HCl(g) 92.3 [2] ∆Hr = (–482.2) + 2(–92.3) – (–103.2) – 2(–273.3) = –17.0 kJ mol1 (ii) Calculate the enthalpy change for the forward reaction in the equilibrium above using bond energy values from the Data Booklet. [2] ∆Hr = BE(reactants) – BE(products) = BE(C-H) + 3BE(C-Cl) + 2BE(H-F) – [BE(C-H) + BE(C-Cl) + 2BE(C-F) + 2BE(H-Cl)] = 410 + 3(340) + 2(562) – [410 + 340 + 2(485) + 2(431)) = –28.0 kJ mol1

4 ©YIJC [Turn over (iii) Explain why the value calculated in 1(c)(ii) is different from t hat calculated in 1(c)(i). [1] Bond energies from Data booklet are average values and thus the value for the enthalpy change of combustion is only an estimated value. [Total: 16]

5 ©YIJC [Turn over 2 An experiment was carried out to determine the percentage of iron in a sample of iron wire. A 3.35 g piece of the wire was reacted with dilute sulfuric acid, in the absence of air, so that all of the iron atoms were converted to iron( II) ions. The resulting solution was made up to 250 cm3. (a) A 25.0 cm 3 sample of this solution was acidified and titrated with 0.0250 mol dm –3 potassium dichromate(VI). The results are seen below titration number 1 2 Initial burette reading / cm3 0.00 0.50 Final burette reading / cm3 32.10 32.50 Volume of potassium dichromate used / cm3 32.10 32.00 (i) Complete the table above and find an average volume of potassium dichromate(VI) used in the titration. Leave all answers to 2dp. H ence, use the average volume of potassium dichromate to calculate the number of moles of dichromate(VI) ions used in the titration. [3] [1] for both correct values in the table, must be 2dp Average volume of K2Cr2O7 used = 32.10+32.00 2 = 32.05 cm3 Number of moles of Cr2O72 = 0.025 × 32.05/1000 = 8.0125 × 10–4 = 8.01 × 10–4 mol (ii) Using data from the Data Booklet, complete and balance the ionic equation for the reaction between the dichromate(VI) ions and the iron(II) ions. [1] Cr2O7(aq) + 6Fe2+(aq) + ...14....H+(aq)  ...2... Cr3+ (aq) + 6Fe3+(aq) + ...7..H2O(l) (iv) Calculate the mass of iron in the 3.35 g piece of wire. [3] No. of moles of Fe2+ in 25 cm3 of solution = 8.0125 × 10–4 × 6 = 4.8075 × 10–3 No. of moles of Fe2+ in 25 cm3 of solution = 4.8075 × 10–3 × 250/25.0 = 4.8075 × 10–2