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:  D