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Chemistry 9701/52 May/June 2024: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_s24_qp_529701_s24_ms_529701/52/M/J/24

Paper
Paper 5 Planning, Analysis and Evaluation
Total marks
30
Time
1 hour 15 minutes
Questions
2
Code on the paper
9701/52/M/J/24
9701_s24_ms_52.pdf

9701/52 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 115 marks

Question

Calcium carbonate, CaCO3(s), decomposes when heated, as shown. CaCO3(s) CaO(s) + CO2(g) The enthalpy change of reaction, ∆Hr , for the thermal decomposition of CaCO3(s) cannot be measured directly. Instead, a procedure involving two experiments is used. In each experiment, the enthalpy change of a different reaction is determined. The equation for the reaction in experiment 1 is shown. The enthalpy change for this reaction is ∆H1. experiment 1 CaCO3(s) + 2HCl (aq) CaCl 2(aq) + H2O(l) + CO2(g) The equation for the reaction in experiment 2 is shown. The enthalpy change for this reaction is ∆H2. experiment 2 CaO(s) + 2HCl (aq) CaCl 2(aq) + H2O(l) Experiment 1 step 1 Weigh a 0.0500 mol sample of powdered CaCO3(s). step 2 Transfer 50.00 cm3, an excess, of 2.00 mol dm–3 hydrochloric acid, HCl (aq), into a small glass beaker. step 3 Start a timer and measure the temperature of the HCl (aq) in the beaker every 30 seconds for 212 minutes. step 4 After 3 minutes add the sample of CaCO3(s) to the HCl (aq) in the beaker. Continue measuring the temperature of the reaction mixture every 30 seconds for a further 5 minutes. Experiment 2 Repeat experiment 1 using calcium oxide, CaO(s), instead of CaCO3(s). (a) Suggest why the enthalpy change of reaction for the thermal decomposition of calcium carbonate cannot be measured directly. …… [1] (b) (i) Calculate the mass, in g, of CaCO3(s) to be weighed using a two-decimal-place balance in step 1. mass of CaCO3(s) = …… g [1] (ii) Outline how a student should weigh by difference using a weighing boat in order to determine the exact mass of CaCO3(s) added to HCl (aq) in the beaker. Draw a results table, with appropriate headings, ready for the student to complete. [2] (c) Identify which piece of apparatus should be used to measure the volume of HCl (aq) in step 2 and give a reason for your choice. …… [1] (d) Without making any changes to the apparatus, suggest an instruction to be added to step 3 and step 4 to make the experiment more accurate. …… [1] (e) A student carries out experiment 1 and obtains the results given in Table 1.1. Table 1.1 time / minutes 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 temperature / °C 19.0 19.0 19.0 19.0 19.0 27.5 36.0 34.5 32.5 32.0 31.0 time / minutes 6.5 7.0 7.5 8.0 temperature / °C 29.0 28.0 26.0 25.5 (i) Plot a graph on the grid in Fig. 1.1 to show the relationship between temperature and time. Use a cross (×) to plot each data point. The points and line of best fit for the data before 3 minutes have been drawn for you. Draw a line of best fit for the data after 3 minutes that will enable you to determine the theoretical temperature increase at 3.0 minutes. 45 40 35 temperature 30 / °C 25 20 15 0 1 2 3 4 5 6 7 8 time / minutes Fig. 1.1 [2] (ii) Use your graph to determine the theoretical temperature increase at 3.0 minutes. theoretical temperature increase at 3.0 minutes = …… °C [1] (f) Suggest why the temperature measured at 3.5 minutes is lower than the temperature measured at 4.0 minutes. …… [1] (g) A student carries out experiment 2 and determines a temperature increase of 62.0 °C. The heat released by the reaction, q, is given by: q = mc∆T where m is the mass of HCl (aq). Assume that 1.00 cm3 of HCl (aq) has a mass of 1.00 g and that the specific heat capacity of the solution, c, is 4.18 J g–1 K–1. Calculate q, in J, for experiment 2 and hence determine ∆H2 in kJ mol–1. q = …… J ∆H2 = …… kJ mol–1 [2] (h) Use the energy cycle below, your answer to (g) and the information given to determine ∆Hr for the thermal decomposition of CaCO3. Enthalpy change for experiment 1, ∆H1 = –84 kJ mol–1. (If you were unable to calculate a final answer in (g), assume a value of 179 kJ mol–1. This is not the correct answer and the sign has been omitted.) CaCO3(s) + 2HCl (aq) CaO(s) + 2HCl (aq) + CO2(g) CaCl 2(aq) + H2O(l) + CO2(g) ∆Hr = …… kJ mol–1 [1] (i) Identify the main weakness of the experimental procedure and suggest one improvement to overcome this weakness. The main weakness is not the type of thermometer used. main weakness …… improvement …… [2] [Total: 15]

Mark scheme

1(a) you cannot (easily) measure temperature (change) whilst applying heat 1 1(b)(i) 5.01 g 1 1(b)(ii) M1 order of weighing: 2 1 weighing boat + CaCO3 is weighed (CaCO3 added to beaker containing HCl) 2 weighing boat is weighed after CaCO3 is added to HCl M2 table + units: /g 1 mass of boat + CaCO3 (before transfer) 2 mass of boat (+ residue of CaCO3) (after addition) mass of CaCO3 added 1(c) burette AND burette measures to nearest 0.05 cm3 1 1(d) stir the solution / mixture 1 1(e)(i) M1 10 points plotted correctly from t = 3.5 mins to t = 8 mins 2 M2 line of best fit drawn involving points from 4 to 8 minutes AND extrapolated back to 3 minutes 1(e)(ii) 38.5 – 19.0 = 19.5 °C 1 1(f) the reaction is not complete (at 3.5 minutes) 1 1(g) M1 q = 50 g x 4.18 J g–1 K–1 x 6 K = 12958 J 2 M2 ΔH2 = M1 / 0.0500 = 12 958 J / 0.0500 mol = 259 160 J mol–1 = –259 kJ mol–1 1(h) ∆Hr = ∆H1 – ∆H2 = –84 – (–259) = (+)175 (kJ mol–1) 1 1(i) M1 weakness: heat loss to the surroundings 2 M2 improvement: use a lid

Question 215 marks

Question

This question is about an experiment to investigate the effect of temperature on the equilibrium constant, K1, of the reaction shown. Fe3+(aq) + SCN–(aq) FeSCN2+(aq) The data collected is used to determine the value of the enthalpy change of the reaction. To set up the equilibrium, aqueous iron(III) nitrate, Fe(NO3)3(aq), is mixed with aqueous potassium thiocyanate, KSCN(aq). Aqueous iron thiocyanate ions, FeSCN2+(aq), have a red colour. A colorimeter is used to measure the absorbance of the reaction mixture. A calibration graph can then be used to determine the concentration of FeSCN2+(aq) in the reaction mixture. Table 2.1 shows the solutions for the experiments. Table 2.1 solution ion concentration / mol dm–3 A SCN–(aq) 0.00920 B SCN–(aq) 0.00200 C Fe3+(aq) 0.00200 D Fe3+(aq) 0.500 (a) Describe how you would prepare 100.0 cm3 of solution B from solution A. Include a calculation of the volume of solution A required for the preparation of solution B. Give the name and capacity of any key apparatus that should be used. Write your answer as a series of numbered steps. …… [3] (b) Before starting the experiment, solutions B and D are used to produce a calibration graph. Known volumes of each solution are added together and the absorbance for each mixture is recorded. The calibration graph is shown in Fig. 2.1. The concentration of solution D is much greater than the concentration of solution B in order that solution D is in excess. Suggest a reason why solution D is in excess. …… [1] 1.2 1.0 0.8 relative 0.6 absorbance 0.4 0.2 0 2 4 6 8 10 12 14 [FeSCN2+] / 10–5 mol dm–3 Fig. 2.1 (c) The following experimental procedure is used. step 1 Half-fill a large beaker with water at room temperature (25 °C). step 2 Transfer about 40 cm3 of solution B into a boiling tube and place the boiling tube in the beaker of water. step 3 Transfer 5.00 cm3 of solution C into a test-tube and place the test-tube in the beaker of water. step 4 Wait for 10 minutes. step 5 Transfer 5.00 cm3 of solution B from the boiling tube to the test-tube containing solution C. Stir the mixture in the test-tube and record the temperature of the mixture. step 6 Measure the absorbance of the mixture in the test-tube using the colorimeter. Change the temperature of the water in the beaker and repeat steps 3 to 6 for different temperatures. (i) Identify the dependent variable. …… [1] (ii) Describe how you would adjust the temperature of the water in the large beaker to obtain a temperature of 10 °C. …… [1] (d) A student obtains the results given in Table 2.2. Table 2.2 1 2 3 4 temperature / °C relative absorbance [FeSCN2+] / 10–5 mol dm–3 K1 25 0.60 55 0.42 The value of the equilibrium constant, K1, can be determined using equation 1. x equation 1 K1 = (0.0010 – x)2 x is the value of [FeSCN2+] in mol dm–3. (i) Use the calibration graph in Fig. 2.1 to complete column 3 in Table 2.2. Record values to one decimal place. [1] (ii) Use equation 1 to complete column 4 in Table 2.2. Record values to the nearest whole number. [1] (e) Another student does the same experiment for seven different temperatures, plots a graph and draws the line of best fit, as shown in Fig. 2.2. 2.2 2.1 2.0 log K1 1.9 1.8 1.7 0.0028 0.0030 0.0032 0.0034 0.0036 0.0038 1 –1 T /K Fig. 2.2 Theory predicts that the relationship between K1 and T is given by equation 2. –∆H equation 2 log K1 = + constant 2.303RT ∆H is the enthalpy change of reaction and T is the temperature in Kelvin. (i) Explain why the graph supports the relationship between K1 and T given in equation 2. …… [1] (ii) Circle the point on the graph in Fig. 2.2 that you consider to be most anomalous. There were no errors in the measurements in the experiment. A student correctly suggests that the anomaly was caused because the absorbance was lower than expected by the line of best fit. Suggest why the absorbance was lower than expected. …… [2] (iii) Determine the gradient of the line of best fit in Fig. 2.2. State the coordinates of both points you use in your calculation. These must be selected from the line of best fit. Give the gradient to three significant figures. coordinates 1 …… coordinates 2 …… gradient = …… K [2] (iv) Use the gradient calculated in (e)(iii) and equation 2 to calculate a value for the enthalpy change of reaction, ∆H. –∆H equation 2 log K1 = + constant 2.303RT (If you were unable to obtain an answer to (e)(iii), then use the value 635 K. This is not the correct answer.) ∆H = …… kJ mol–1 [2] [Total: 15] Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 12 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

2(a) M1 volume of solution A = 21.75 cm3 (to nearest 0.05 cm3) 3 M2 use burette to transfer 21.75 cm3 of solution A and add to a volumetric flask M3 top the (100 cm3 volumetric) flask up to 100 cm3 mark with distilled water 2(b) so that concentration of [Fe(SCN)2+] is known (as all solution B / SCN– reacts) 1 2(c)(i) (relative) absorbance 1 2(c)(ii) add ice (to the water in the beaker) 1 2(d)(i) 8.9 and 6.2 in column 3 1 2(d)(ii) 107 and 70 in column 4 1 2(e)(i) the graph is a straight line 1 2(e)(ii) M1 anomalous point ringed 2 M2 not all solution B was transferred / added (in step 5) (making concentration of FeSCN2+ too low) 2(e)(iii) M1 co-ordinates read and recorded correctly 2 M2 gradient calculated to three significant figures 2(e)(iv) M1 gradient / 2(e)(iii) = –∆H / 2.303R OR any correct rearrangement of this e.g. ∆H = gradient / 2(e)(iii) × –2.303 × 8.31 2 M2 correct calculations of ∆H including conversion to kJ

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Chemistry 9701/52 for May/June 2024 is out of 30 marks, with 1 hour 15 minutes to answer it, across 2 questions.

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