Edexcel Jan 2012 Paper 3 Q3 (with explained solutions) |

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Edexcel Jan 2012 Paper 3 Q3

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PMT 3 Magnesium carbonate reacts with dilute nitric acid as shown in the equation below. MgCO3(s) + 2HNO3(aq) (cid:111) Mg(NO3)2(aq) + CO2(g) + H2O(l) The enthalpy change for this reaction can be determined as follows: Procedure 1. Weigh 3.50 g of finely powdered magnesium carbonate. 2. Transfer 50.0 cm3 of 2.00 mol dm3 nitric acid into a polystyrene cup and record the temperature of the acid. 3. Add the magnesium carbonate to the nitric acid. 4. Stir the mixture and record the maximum temperature reached. Results Temperature of nitric acid before addition of magnesium carbonate Final temperature of solution 21.0 (cid:113)C 29.7 (cid:113)C (a) Explain why the magnesium carbonate used in this experiment should be finely powdered rather than in lumps. (1)(b) (i) Calculate the number of moles of magnesium carbonate in 3.50 g. [Assume the molar mass of magnesium carbonate is 84 g mol1.] (ii) The volume of dilute nitric acid used contained 0.100 mol of HNO3. Suggest why this amount is suitable. (1) (1)*P39305A0616*<br />
(c) (i) Calculate the heat energy transferred, in joules, in this reaction between magnesium carbonate and nitric acid. Use the expression energy = mass ofspecific heat capacity transferred (J) solution of solution rise temperature[Assume the mass of solution is 50 g and its specific heat capacity is 4.18 J g1 (cid:113)C1.] (ii) Calculate the enthalpy change, (cid:39)H, for the reaction of one mole of magnesium carbonate with nitric acid. Your answer should be in units of kJ mol1, expressed to three significant figures, and include a sign. PMT (1) (2) (d) (i) The nitric acid for this experiment could be measured using either a pipette or a measuring cylinder. Give one practical advantage of using each piece of apparatus. (2) PipetteMeasuring cylinder*P39305A0716* Turn over<br />
PMT (1) (2) (ii) The total error in measuring the mass of the magnesium carbonate was 0.01 g. Calculate the percentage error in the weighing. (e) State and explain the effect, if any, on the calculated enthalpy change, (cid:39)H, if (i) a copper calorimeter were used instead of the polystyrene cup.(ii) 3.50 g of damp magnesium carbonate were used. (2)(f) The experiment was repeated with a change in the procedure. The temperature of the dilute nitric acid was measured every minute for three minutes. After exactly three and a half minutes, the magnesium carbonate was added and the mixture was stirred. The temperature of the mixture was then taken every minute for another six minutes. A graph of the temperature readings against time using this procedure is shown opposite. *P39305A0816*<br />
Temperature / (cid:113)C 32 30 28 26 24 22 20 6 4 Time / minutes 10 (i) Use this graph to calculate the maximum temperature change. Show your working on the graph. Maximum temperature change =(cid:113)C (ii) Why does this method give a more accurate result than the original procedure? PMT (2) (1)(Total for Question 3 = 16 marks) *P39305A0916* Turn over<br />

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Question Number 3(a) Question Number 3(b)(i) Question Number 3(b)(ii) Question Number 3(c)(i) Mark Mark Mark Correct Answer Reject So that there is less (time for) cooling ALLOW so that reaction is fast/to increase the rate of reaction OR using lumps will be slow To ensure all magnesium carbonate reacts Just To make surface area larger Correct Answer Reject 0.04, 0.041, 0.0416 - 2 (mol) (3.50/84) = 0.041667 = 0.0417/ 0.042/4.2 x 10 ACCEPT 3.50/84.3 = 0.041518 = 0.0415/0.042 (mol) Mark FINAL answer IGNORE sf except 1 sf IGNORE units Correct Answer Reject To ensure all magnesium carbonate reacts/ it is in excess/moles of acid is more than twice the number of moles of magnesium carbonate ALLOW To ensure all magnesium carbonate dissolves Statement that acid is in excess even if calculation not totally correct To ensure all the acid reacts Calculation of number of moles without indication that acid is in excess To ensure there is enough/adequate acid There are more moles of acid than carbonate Correct Answer Reject Mark (50 x 4.18 x 8.7) = 1818.3 = 1820/ 1818/1800 (J) accept 1.8183 = 1.82 kJ IGNORE sf except 1 sf IGNORE signs 2000 (J) 53.5 x 4.18 x 8.7 = 1945.58 (1) 6CH0701 1201<br />
Question Number 3(c)(ii) Correct Answer Reject Mark First mark: Answer in (c)(i)/number of moles (1) Second mark: Correct value with negative sign and 3sf (1) Correct answer with no working scores (2) = -43.6 (kJ mol-1) ACCEPT answers from -43.3 up to -45.5 which may arise from using different roundings, but check they come from values in (c)(i) and a suitable number of moles. 1818.3 1818.3 1818.3 1818 1820 1818.3 1818 1820 1818.3 1818 1820 1818.3 1818 1820 1820 1820 0.042 0.04 0.041667 0.041667 0.041667 0.0417 0.0417 0.0417 0.041518 0.041518 0.041518 0.0415 0.0415 0.04 0.0417 0.0415 43.3 45.5 43.6 43.6 43.7 43.6 43.6 43.6 43.8 43.8 43.8 43.8 43.8 45.5 43.6 43.9 Answer to c(i) divided by one (for one mole of magnesium carbonate) scores second mark only, if sign and sf correct and in kJ mol-1 6CH0701 1201<br />
Question Number 3(d)(i) Question Number 3(d)(ii) 3(e)(i) Question Number 3(e)(ii) Correct Answer Reject Pipette : more accurate/more precise/lower % error (1) Measuring cylinder : faster/easier/more convenient (1) Cost Safer Measures a range of volumes/ larger volumes Easier to clean More suitable Correct Answer Reject Mark Mark (0.01 x 100 ) 3.50 = ()0.28571 = () 0.286/ ()0.29/ ()0.3 % IGNORE sf Correct Answer (calculated) enthalpy change less negative/less exothermic (1) allow (value) less or smaller or decreases (Temp rise will be lower) due to heat loss/more heat absorbed by calorimeter (1) Mark independently 0.285, 0.28 Reject Mark Correct Answer Reject (calculated) enthalpy change less negative/less exothermic (1) allow (value) less or smaller or decreases Less MgCO3 to react (so temp rise will be lower) (1) Mark independently Water will not affect the reaction Mark 6CH0701 1201<br />
Question Number 3(f)(i) Correct Answer Reject Mark Diagonal line from 3 or 3.5 Extrapolated horizontal line, extrapolated (best fit) cooling line and vertical line at 3.5 minutes (1) 9.3(cid:176) C0.2 (1) Mark independently Question Number 3(f)(ii) Corrects for (loss of heat during) Correct Answer cooling/checks original solution is at equilibrium with surroundings Reject Mark There is no heat loss The original method included heat loss to the surroundings IGNORE reference to anomalous readings 6CH0701 1201<br />

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