Edexcel Jun 2014 Paper 4 Q9 (with explained solutions) |

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Edexcel Jun 2014 Paper 4 Q9

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 9 This question is about magnesium chloride, MgCl2. It can be formed by burning magnesium in chlorine. Mg(s) + Cl2(g) (cid:111) MgCl2(s) (cid:507)S(cid:57) surroundings = +2152 J mol1 K1 Remember to include a sign and units in your answers to the calculations in this question. (a) (i) The standard molar entropy at 298 K for 1 mol chlorine molecules, Cl2, is +165 J mol1 K1. Use this, and appropriate values from your Data Booklet, to calculate the standard entropy change of the system, (cid:507)S(cid:57) system, for this reaction. *(ii) Explain fully why the sign for the standard entropy change of the system, (cid:507)S(cid:57) system, is as you would expect. (2) (2)(b) Calculate the total entropy change, (cid:507)S(cid:57) your answer to three significant figures. total, in J mol1K1, for this reaction, giving (2) *P42972A0924* Turn over<br />
(2) (1) (2) (c) Use the standard entropy change of the surroundings, (cid:507)S(cid:57) standard enthalpy change, (cid:507)H (cid:57), in kJ mol1, for the reaction at 298 K. surroundings, to calculate the (d) 0.0300 mol of magnesium chloride, prepared by burning magnesium in chlorine, is added to 51.5 cm3 of water. 50.0 cm3 of 1.00 mol dm3 solution is formed, and the temperature rise, (cid:507)T, is 22.5C. (i) Calculate the energy transferred in joules for this process using: Energy transferred in joules = volume of solution (cid:113) 4.2 (cid:113) (cid:507)T (ii) Calculate the enthalpy change of solution, (cid:507)Hsolution, of magnesium chloride in kJ mol1. 10 *P42972A01024*<br />
*(iii) The enthalpy change of hydration of Mg2+(g) is 1920 kJ mol1. Use this, your value from (d)(ii), and the experimental lattice energy from your Data Booklet, to calculate the enthalpy change of hydration of Cl(g). (3) Mg2+(g) + 2Cl(g) MgCl2(s) Mg2+(aq) + 2Cl(aq) AnswerkJ mol1 (iv) Draw a diagram to represent a hydrated chloride ion. (v) Suggest why the addition of anhydrous magnesium chloride to water results in an increase in temperature and a decrease in volume. (1) (2) Temperature increasesVolume decreases(Total for Question 9 = 17 marks) *P42972A01124* 11 Turn over<br />

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Section B Question Number 9 (a)(i) Acceptable Answers +89.6[+32.7 + 165] (1) = 108.1 J mol1 K1/ J K1 mol1 Value, sign and units (1) Ignore SF except one Internal TE for recognisable numbers allowed, for example: Ho Halving So [Cl2] (82.5-25.6) Correct answer with no working (2) +/no sign 108.1 J mol1 K1/ J K1 mol1 (1) at magnesium chloride (147.7 -223.1) Reject Mark 6CH0401 1406<br />
Acceptable Answers Reject Question Number 9 (a)(ii) Mark Energy (Positive) Answer is as expected (0) 6CH0401 1406 (The sign is negative because) Any two from:(A solid and) a gas reacting to form a solid. OR (Entropy decreases because) a gas reacting to form a solid.There are fewer ways of arranging particles in a solid than a gas or vice- versa. OR Decrease in disorder as solid more ordered than gas or vice versaTwo mol(es) of reactant forming one mole of product. (Ignore two molecules form one molecule) OR Number of mol(es)/molecules decreases OR Fewer/less mol(es) of products than reactants<br />
Question Number 9 (b) system total = So surroundings + So Correct Answer So OR = +2152 + (-108.1) = (+)2043.9 Value 2043.9 / 2044 (1) = (+)2040 (J mol1 K-1) 3SF This mark conditional on correct value or correct TE value from (a)(i) (1) Accept TE from (a)(i), for example, -223.1+1928.9+1930 -25.6+2126.4+2130 Correct answer (2040, etc) with or without working scores 2 Mark Reject 6CH0401 1406<br />
Question Number 9 (c) surroundings x 298 Correct Answer1 So surroundings = - Ho 298 Ho =So OR Ho- = 2152 x 298 (1) = 641.296 = 641.3 (kJ mol1) (1) ALLOW = 641.3 x103 J mol1 Note 1. -640.1338 = -640.1 (if 2040/answer to part (b) used to recalculate entropy change of surroundings first.) (2) 2. Ho = +641.3 (kJ mol1) (1) 3. Ho = - So surroundings (0) 298 Ignore SF except one Reject Mark 6CH0401 1406<br />
Reject Mark Reject Mark Question Number 9 (d)(i) Correct Answer 50 x 4.2 x 22.5 = 4725 (J) Ignore sign ALLOW 4.725 kJ Ignore SF except one Correct Answer Question Number 9 (d)(ii) There are two legitimate answers to this part. If both methods have been used, you must send the item to review under mark scheme ()47250.0300 = 157.5 (kJ mol-1) /157500 J mol1 OR ()47250.0500 = /94.5 (kJ mol-1) /94500 J mol1 ALLOW TE answer (d)(i)0.0300/0.0500 Ignore SF except one Value (1) Sign (1) The mark for the negative sign is awarded for their calculation even if value is wrong, providing any energy divided by moles or energy multiplied by 1/number of moles calculation has been done. 6CH0401 1406<br />
Reject Mark Question Number 9 (d)(iii) There are two correct answers: Correct Answer Using 0.03 gives the answer of -381.75 kJ mol-1 Using 0.05 gives the answer of -350.25 kJ mol-1 Both these answers score full marks with or without correct working. First mark Appreciation of Hesss Law either in words, numbers, symbols or on the diagram For example, Hsolution + Lattice energy = Hhydration Mg2+ + (2)Hhydration Cl (1) Second mark 2 Hhydration Cl = 2526157.5(1920) = 763.5 OR 2 Hhydration Cl = 252694.5(1920) = 700.5 ALLOW Any number or group of numbers minus (-1920) (1) Third mark Hhydration Cl =381.75 (kJ mol1) OR Hhydration Cl =350.25 (kJ mol1) Any number, wherever it has come from, 6CH0401 1406<br />
divided by two can score this mark, provided that the sign is consistent. (1) Ignore SF except one Use of lattice energy2326 gives 281.75/250.25 scores (2) ALLOW TE from (d)(ii) 6CH0401 1406<br />
Question Number 9 (d)(iv) Correct Answer OROne/several water molecule(s) all correctly orientated.H/ hydrogen (one or two hydrogens from each water molecule) towards chloride ion with negative charge either on chlorine or on the whole hydrated ion. ALLOWA minus sign with a ring around it for the ClBonds shown by lines/broken lines/dotted lines/wedges Reject Mark Cl-.H2O H/ H/ H Cl / Cl (with no charge) 6CH0401 1406<br />
Correct Answer Both marks may be awarded in either part. First mark (Temperature increases) because the reaction/process/dissolving/hydration of ions is exothermic. OR Strong(er) forces between the + H and Cl OR Strong(er) forces between theO and Mg2+ OR Strong(er) ion-dipole forces OR Formation of bonds releases energy OR Strong(er) bonds formed OR Enthalpy of hydration is greater than lattice energy (1) Second mark (Volume decreases so) shorter bonds between ion and water molecules ALLOW Water molecules more tightly arranged/pack better/occupy less space OR Water molecules more ordered/ clustered (around the ions). Reject Mark The breaking of the lattice is exothermic. Ions more tightly arranged Ions more ordered (1) Total for Question 9 = 17 marks Question Number 9(d)(v) 6CH0401 1406<br />

Question 8 9 10 11 12 18 19 20 21 22

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