ASRJC 2021 J2Prelims H2Chem P4 soln

2021 H2 Chemistry Paper 4 Suggested Solution Qn Teaching points Marks 1(a) mass of capped bottle and FA 1 / g 7.378 mass of capped bottle and residual FA 1 / g 3.488 mass of FA 1 used / g 3.490 initial temperature / oC 28.8 lowest / minimum temperature reached / oC 24.8 decrease in temperature / maximum change in temperature / oC 4.0 • May record data in a single table or have one table for mass and one table for temperature; Tabulation may be vertical or horizontal; lines are not essential • For “temperature drop” allow “temperature change” sign not essential as it will be accounted for later in the sign for the enthalpy change. [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 marks compare student’s and teacher’s [2] (b)(i) Calculate heat change using result from 1(a) Heat change (q1) = mcDT = (50 x 1.00) x 4.18 x (temp drop) = ________ J [1] (b)(ii) Determine value of DHsol(KHCO3) with correct sign DHsol(KHCO3) = + (q1) / n(KHCO3) = + _______ J mol-1 [1] (b)(iii) Calculate correctly initial Tav = 28.6 oC [1] (b)(iv) Heat change (q2) = mcDT = (25+50) x 1.00 x 4.18 x (28.6 – 28.2) = 125.4 J DHr(KHCO3(aq)) = + (125.4) / (3.450/100.1) = +3640 J mol-1 Final answer to 3 s.f. or 4 s.f. and appropriate units for (b)(i), (b)(ii), (b)(iii) and (b)(iv). [1] [1] [1] T m D

(c) 2 x [1(b)(iv)] 2KHCO3(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l) + 2CO2(g) 2 x [1(b)(ii)] DHr(KHCO3(s)) 2KHCO3(s) + H2SO4(aq) 2 x [1(b)(ii)] + 2 x [1(b)(iv)] = DHr(KHCO3(s)) = ________ J mol-1 [1]: correct application of Hess’ Law [1]: correct answer; awarded only if (b)(ii) and (b)(iv) are correct and applied correctly (ignore units) [2]

Qn Teaching points Marks 2(a) Preliminary Calculations - Calculate the mass of MgSO4 to use for the experiment. - Assuming that 100 cm3 of water was used in the experiment and a temperature change of 5 oC is measured and no heat loss to surroundings, nsalt x 78.9 x 103 = 100 x 4.3 x 5 Þ nsalt = 0.02724 mol minimum mass of MgSO4 to use = 0.02724 x 120.4 = 3.28 g Given the solubility of MgSO4 at 20 oC = 0.292 mol per 100 cm3 maximum mass that can dissolve in 100 cm3 of water = 0.292 x 120.4 = 35.2 g Hence a mass of about 10 g of MgSO4 can be used for the experiment. (10 g of MgSO4 is a suitable mass as it can be easily measure and from the above preliminary calculations, we know that this mass chosen will be able to give a temperature rise of about 5°C and will completely dissolv