ASRJC H2 Chem 2022 Prelim P4 Soln

1 ASRJC JC2 PRELIM 2021 9729/04/H2 2022 H2 Chemistry Prelim Paper 4 Marks Scheme with Markers’ Comments 1 Determination of a value for an enthalpy change of solution (a) mass of capped bottle / weighing bottle and FA 1 / g 9.514 mass of capped bottle and residual FA 1 OR mass of weighing bottle with residue OR mass of weighing bottle after transfer / g 5.519 mass of FA 1 used / transferred / g 3.995 Ti / oC 29.4 Tm / oC 21.0 T / oC 8.4  May record data in a single table or have one table for mass and one table for temperature [1]: correct headers and units. [1]: mass readings to 3 d.p. and temperature readings to 1 d.p. [1]: correctly determined maximum temperature change and mass of FA 1 used 3 Accuracy 2 Comments: The majority of students presented their data clearly. It is more accurate to describe the mass after reweighing as ‘mass of weighing bottle with residue’ than to describe as ‘mass of empty bottle’ as there might not be complete transfer of all solids into the cup. It was disappointing that some students were still recording mass to 2 d.p. instead of 3 d.p., and surprisingly some students did not realise something must be amiss when the mass of the bottle they measured was less than 1.000 g! Students shou ld have a judgement of the magnitude of the quantity they are measuring and make corrections if necessary. Students could have used the notations (Ti, T m, T) given for the various temperatures measured in their tabulation instead of giving long descripti ons. Descriptions of ‘final’ or ‘maximum’ temperature were not accepted as the temperature dropped in this experiment. Majority was able to score the accuracy marks. (b)(i) Calculate heat change using result from 1(a) heat change (q1) = mcT = (25 x 1.00) x 4.18 x (temp drop) = ________ J Ignore sign 1 (b)(ii) H2 = + {(q1) / n(NaHCO3)} x 2 = + _______ J mol-1 or kJ mol-1 with correct sign 1

(b)(iii) Tav = (25 × 31.2) + (50 × 27.6) (25+50) = 28.8 C 1 (b)(iv) heat change (q2) = mcT = (25+50) x 1.00 x 4.18 x (28.1 – 28.8) = 219.45 J n(NaHCO3) = 50 1000 × 1.00 = 0.050 mol H3 = + {(219.45 / 0.050)} x 2 = + 8778 = + 8780 J mol-1 or + 8.78 kJ mol-1 1 1 Final answer to 3 s.f. and appropriate units for (b)(i), (b)(ii), (b)(iii), (b)(iv). 1 1 Comments: (b)(i): Instead of using the information given that the density of the solution is 1.00 g cm -3, giving the mass, m, of solution as 25 g , many students made the mistake of adding the mass of the solid to 25 . Some even made the mistake of using the mass of the solid rather than the mass of the solution. A few students were still making the mistake of adding ‘273’ to temperature change. (b)(ii): Most students were unable to score this mark. Common mistakes include:  Did n ot realise that the value of {(q 1) / n(NaHCO 3)} gives the heat absorbed for only 1 mol of NaHCO 3 used, while H2 is the enthalpy change of the reaction (per balanced equation ) for 2 mol of NaHCO 3, hence the necessity to multiply by 2.  Did not recogni se the endothermic reaction (temperature dropped) and incorrectly giving a negative sign to H2.  Dividing the heat change by the Mr instead of the number of moles of solid used.  Wasting time to determine the limiting reagent when it was given in the question stem that ‘the sulfuric acid is in excess’. Students should read questions carefully to avoid unnecessary waste of time. (b)(iii): Most students correctly calculated the average initial temperature using the formula given. (b)(iv): Similar errors were made in this part as in (b)(ii). Another common error made here was in using the wrong volume of 25 or 50 cm3 instead of the total volume of reaction mixture which is 75 cm3. Some students did not make use of the Tav they have correctly calculated to find T, not understanding the purpose of the calculation in (b)(iii).

(c) + 8.78 2NaHCO3(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) + 2CO2(g) (H3) 2H1 [1(b)(ii)] (H2) 2NaHCO3(s) + H2SO4(aq) H1 = [1(b)(ii)]  (+ 8.78)] ÷ 2 = ________ J mol-1 or kJ mol-1 or H1 = [x  (+ 8.78)] ÷ 2 or H1 = [x  y] ÷ 2 [1]: correct application of Hess’ Law [1]: correct answer; awarded only if 1(b)(ii) and 1(b)(iv) applied correctly Comments: Many students attempted this part using the most straightforward method of drawing an energy cycle. However, many were not careful in balancing the species in the cycle and not multiplying H1 by 2, leading to an incorrect value for the calculated H1. Students need to revisit their concepts in Energetics especially on the definition of the various enthalpy changes and application of Hess’ Law. [Total: 14]

2 To determine the order of reaction with respect to the concentration of iodine in the iodination of propanone reaction Mark (a) Correct header with units Include for transfer time, in min and s for each entry t td/ min initial burette reading / cm3 final burette reading / cm3 Volume of FA 6 / cm3 4 min 18 s 4.0 0.00 17.40 17.40 8 min 2 s 8.0 18.00 34.20 16.20 12 min 0 s 12.0 30.00 44.90 14.90 16 min 0 s 16.0 0.00 13.70 13.70 20 min 0 s 20.0 14.00 26.40 12.40 1 Records all • volumes to 0.05 cm3 • transfer time, t, to consistent precision, i.e. nearest s • correctly calculates decimal values of td and records to 1 d.p. 1 5 sets of titration results and 1 aliquot is taken at between 3.5 min to 5 min and student choses well–spaced values of time of transfer, where the longest time ≤ 20 min 1 Comments:  Some students did not follow instructions to reflect the recording of data to the appropriate significant figures/decimal places or units.  Many wrote the units for time incorrectly as mins or sec, or did not leave t d as 1 decimal place.  The maximum time taken should not exceed 20 min and the 5 readings taken should be well-spaced from 4 to 20 min.  A number of students also forgot to include t as part of the table and thus lost marks for recording. (b)(i) Axes correct way round + correct labels + scale + units the scale must be chosen so that the y intercept would fall within the scale range; and, that the plotted points occupy at least half the graph grid in the x direction (including x = 0) and the plotted points and the y intercept together occupy at least half the graph grid in the y direction. 1 Plotting – within ±½ small square. Check all points; put ticks if correct. 1 Graph line is straight and is the best-fit line with a fair scatter of points either side of the line. 1 Accuracy 1 Comments: As the volume of FA 6 at t d = 0.00 min is required, you should clearly indicate the coordinates of the y-intercept on the graph. Since the question does not need you to find the x -intercept from the graph, it is unnecessary to include y=0. Students who did this usually ended up with a scale that does not fulfill the requirement for the plotted points to occupy at least half the graph grid in both x and y directions. Please show the scale/division for every 10 smal l boxes to avoid plotting wrongly and also for the ease of reference by the marker.

As the reaction is between I 2 and CH3COCH3, the amount of I 2 will be decreasing during the reaction, thus the graph of volume of S2O32- used against time shoul