2019 HCI Prelim H2 Chem P4 ANS

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 Paper 4 1 (a) (i)  Tabulates volumes and temperature data in (a)(i) Table should contain correct headers and units. Data headers to include volume of FA 1, volume of FA 2, TFA1, Tmax and T. [1] Volume of FA 1 / cm3 Volume of FA 2 / cm3 TFA 1 / °C Tmax / °C T / °C 10.0 40.0 32.0 36.0 4.0 20.0 30.0 32.0 39.5 7.5 25.0 25.0 32.0 41.5 9.5 30.0 20.0 32.0 41.0 9.0 35.0 15.0 32.0 38.5 6.5 40.0 10.0 32.0 36.0 4.0  Complete 6 sets of volume/temperature readings in (a)(i) Required volumes: Volume of FA 1 = 10.0 cm3, 20.0 cm3, 25.0 cm3, 30.0 cm3, 35.0 cm3 and 40.0 cm3 and use appropriate volume of FA 2 such that the total volume of reacting mixture in each set of data is 50 cm3 [1]  Records all temperature data in (a)(i) to 0.5°C, all volumes for FA 1 and FA 2 in (a)(i) to 1 d.p. [1]  Correctly calculates all T values to 1 d.p. in (a)(i) [1] (a) (ii)  Axes correct way round and correct labels and units and scale [1] Note: Scale chosen must allow for the lines to be extrapolated to cross each other. The plotted points should occupy at least half the grid in both directions.  All points are correctly plotted to within ± ½ small square. [1]  All drawn graph lines are straight best-fit lines and are extrapolated to cross each other. and there are at least three points on each side of the graphically determined Tmax [1] HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(a) (iii) Tmax and Vmax are read correctly to ± ½ small square from graph From the sample graph, Tmax = 10.4 °C Vmax = V(H2SO4) = 27.25 cm3 [1] (b) (i) H2SO4 + 2 NaOH → Na2SO4 + 2 H2O V(NaOH) = 50.0 – 27.25 = 22.75 cm3 n(NaOH) = 1.50 × (22.75/1000) = 0.03413 mol n(H2SO4) = 0.03413 / 2 = 0.01706 mol [H2SO4] = 0.01706 / (27.25/1000) = 0.626 mol dm−3 [1] (ii) Heat change = msolutioncTmax = 50.0 × 4.18 × 10.4 = 2174 J [1] (c) n(H2O) = n(NaOH) = 0.03413 mol Hneut = − 2174 / 0.03413 = − 6.37 × 104 J mol−1 = − 63.7 kJ mol−1 (Sign must be negative) [1] (d) Hneut would be less exothermic as malonic acid is a weak acid. Energy is absorbed to ionise the un-ionised weak acid. [1] (e) Suggested sources of errors and its appropriate improvements: [1] heat loss to the surrounding and use a cup lid to minimise heat exchange with the surrounding air VFA 1 and VFA 2 is to 1 d.p. as measured using less precise measuring cylinders and can be measured using more precise burettes to give 2 d.p. initial temperature of FA 2 was not accounted for and weighted initial temperature should be calculated where Tweighted initial = (Volume of FA 1 x TFA1) + (Volume of FA 2 x TFA2) Volume of FA 1 + Volume of FA 2 heat capacity of the calorimeter (Styrofoam cup) was not accounted for and heat absorbed by the styrofoam cup can be included in the calculation of heat change 2 (