RVHS Prelim P4 Soln

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution RIVER VALLEY HIGH SCHOOL JC 2 PRELIM PRACTICAL EXAMINATION H2 CHEMISTRY 9729 PAPER 4 SUGGESTED SOLUTION 1 (a) Ksp = [Ca2+][IO3]2 (b) Results titration 1 2 final burette reading / cm3 19.60 39.20 initial burette reading/ cm3 0.00 19.60 volume of FA 2 used / cm3 19.60 19.60 (c) (i) average volume of FA 2 used = 19.60 19.60 2  =19.60 cm3 VFA 2 = 19.60 cm3 (ii) amount of S2O32 = 19.600.100 1000 = 0.00196 mol IO3  3I2  6S2O32 amount of IO3 in 25.0 cm3 of FA 1 = 0.00196  6 = 3.267  104 = 3.27  104 mol amount of IO3 in 25.0 cm3 of FA 1 = 3.27  104 mol (iii) amount of dissolved IO3 in the mixture from (b) at equilibrium = 4 100.03.267 10 25.0  = 1.307  103 mol amount of IO3(aq) in the mixture = 1.307  103 mol

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution (d) (i) amount of IO3 from KIO3 = 100.00.0100 1000 = 0.00100 mol amount of IO3 from dissolved Ca(IO3)2 = 1.307  103  0.00100 = 3.07  104 mol amount of IO3(aq) from dissolved Ca(IO3)2 = 3.07  104 mol (ii) amount of Ca2+ in mixture = 43.07 10 2  = 1.535  104 = 1.54  104 mol (to 3 s.f.) amount of Ca2+(aq) in the mixture = 1.54  104 mol (e) [Ca2+]eqm = 41.535 10 0.100  = 1.535  103 mol dm3 [IO3]eqm = 31.307 10 0.100  = 1.307  102 mol dm3 Ksp = [Ca2+][IO3]2 = (1.535  103)( 1.307  102)2 = 2.62  107 mol3 dm9 (to 3 s.f.) Ksp = 2.62  107 mol3 dm9 (f) FA 1 would be diluted/ [IO3] in FA 1 is lowered/ lower amount of IO3 in 25.0 cm3 of FA 1. This forms less I2 for reaction with S2O32. Therefore, the mean titre would be lower. (g) The student did not carry out the experiment at 25 °C. Hence, the position of equilibrium for Ca(IO3)2(s)  Ca2+(aq) + 2IO3(aq) shifted to the left. OR When the student carried out the experiment, equilibrium has not been reached before filtering in (b) step 4. Hence, the [Ca2+] and [IO3] is lower.

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution 2 (c) Results Shift 1 and timed practice Experiment VFA 5 / cm3 2HOV for solution A / cm3 VFA 6 / cm3 2HOV for solution B / cm3 t /s 1 t /s1 1 5.0 30.0 20.0 10.0 22.1 0.0425 2 10.0 25.0 20.0 10.0 11.1 0.0901 3 5.0 30.0 10.0 20.0 48.1 0.0208 Shift 2 Experiment VFA 5 / cm3 2HOV for solution A / cm3 VFA 6 / cm3 2HOV for solution B / cm3 t /s 1 t /s1 1 10.0 25.0 20.0 10.0 15.4 0.0649 2 20.0 15.0 20.0 10.0 8.0 0.125 3 10.0 25.0 10.0 20.0 28.3 0.0353 Shift 3 Experiment VFA 5 / cm3 2HOV for solution A / cm3 VFA 6 / cm3 2HOV for solution B / cm3 t /s 1 t /s1 1 10.0 25.0 20.0 7.0 11.8 0.0847 2 20.0 15.0 20.0 7.0 6.1 0.163 3 10.0 25.0 10.0 17.0 23.0 0.0435 (d) (Using results from shift 1)  IO3 (Let rate = k[IO3]n[H+]m[SO32]p. Since k, [H+] and [SO32] are constant, n 1 2 rate in experiment 1 rate in experiment 2    V V FA 5 FA 5 .) n 1 2 rate 0.0425 5.0=rate 0.0901 10.0    n = 1.08 = 1 (nearest integer)

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution OR (Concentration is directly proportional to the volume used when total volume is kept constant.) Comparing experiments 1 and 2, VFA 6 (and VFA 7) are kept constant, relative rate is doubled when VFA 5 is doubled. Therefore, the reaction is first order with respect to IO3. order of reaction with respect to IO3= 1  H+ (Since k, [IO3] and [SO32] are constant, m 1 3 rate in experiment 1 rate in experiment 3    V V FA 6 FA 6 .) m 1 3 rate 0.0425 20.0=rate 0.0208 10.0    m = 1.03 = 1 (nearest integer) OR (Concentration is directly proportional to the volume used when total volume is kept constant.) Comparing experiments 1 and 3, VFA 5 (and VFA 7) is kept constant, relative rate is halved when VFA 6 is halved. Therefore, the reaction is first order with respect to H+. order of reaction with respect to H+= 1 (e) Repeat each experiment and take average of the timings for the dark blue colour to appear. Use the average t to calculate the relative rate. OR Conduct the experiment with other VFA 5, keeping VFA 6 and VFA 7 constant. Plot the graph of relative rate ( 1 t ) against (VFA 5)n to obtain a best fit straight line. Use the best fit straight line to obtain the relative rate ( 1 t ) at various VFA 5. Then, conduct the experiment with other VFA 6, keeping VFA 5 and VFA 7 constant. Plot the graph of relative rate ( 1 t ) against (VFA 6)m to obtain a best fit straight line. Use the best fit straight line to obtain the relative rate ( 1 t ) at various VFA 6.

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution (f) (i) Shift 1, 2 and timed practice (VFA 7 = 5.0 cm3) 2 3 initial 0.0400 5.0[SO ] 75.0   = 2.667  103 mol dm3 rate of experiment 1 = 32.667 10 22.1  = 1.21  104 mol dm3 s1 rate of experiment 1 = 1.21  104 mol dm3 s1 Shift 3 (VFA 7 = 8.0 cm3) 2 3 initial 0.0400 8.0[SO ] 75.0   = 4.267  103 mol dm3 rate of experiment 1 = 34.267 10 15.4  = 2.77  104 mol dm3 s1 rate of experiment 1 = 2.77  104 mol dm3 s1 (ii) When total volume is unchanged, [SO32]  VFA 7. Since the reaction is first order with respect to SO32, when the [ SO32] is doubled, the rate of reaction is doubled. (iii) 1 t cannot be used as relative rate. t measured is the time taken for varying amount of SO 32 to react completely/ [SO32]initial to fall till 0 mol dm3 .

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution (g) (i) Plot, on the grid provided in page 13, a graph of 1ln t on the y-axis against 1 T on the x-axis. Draw a line of best-fit taking into account all your plotted points. (ii) gradient of line = 33 4.20 ( 3.125) 3.36 10 3.175 10       = 5.81  103 K gradient of line =  5.81  103 K

River Valley High School JC 2 H2 Chemistry 9729 2020 Prelim Practical Examination Suggested Solution (iii)  5.81  103 =  aE R Ea = 5.81  103  8.314 = 48.3 kJ mol1 Ea = 48.3 kJ mol1 (iv) At t = 65.0 s, 1ln 65.0   =  4.174 From graph, 1 T = 3.355  103 K1 temperature = 3 1 2733.355 10   = 25.1 C temperature = 25.1 C 3 (a) FA 8 is a black solid. FA 9 is a colourless solution. FA 8 dissolved/ react to give a green/ bluish green solution. Green/ bluish green filtrate and black residue are obtained. (b) Unless otherwise stated, use a 1 cm depth of the filtrate from (a) in separate test tubes for each of the following tests. Test Observations (i) To a portion of the filtrate from (a), add aqueous ammonia till no more change is observed. To a portion of the resultant mixture, add dilute sulfuric acid till no more change is observed. (Pale/ Light) green/blue ppt formed dissolved in excess NH 3(aq) to give a dark blue/ deep blue solution. (Pale/ Light) blue ppt formed dissolved in excess H2SO4(aq) to give a light/pale blue solution. (ii) To a portion of FA 3 in a large test tube, add a few drops of filtrate from (a), followed by 3 cm height of FA 2. Shake for a few minutes. Cream/off-white ppt formed in brown solution. Brown solution decolourised/ turned colourless when FA 2 is added leaving a cream/off-white ppt. (Cream/off-white) ppt dissolved to give a colourless solution after shaking for a few minutes.

River Valley High School JC 2 H2 Che