Edexcel Jan 2010 Paper 4 Q25 (with explained solutions) |

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Edexcel Jan 2010 Paper 4 Q25

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PMT SECTION C Answer ALL the questions. Write your answers in the spaces provided. 25 The oxidation of iron metal in the presence of oxygen is spontaneous. (a) Explain the meaning of spontaneous in a thermodynamic context. 4Fe(s) + 3O2(g)2Fe2O3(s) (1)(b) (i) Find the values of the standard molar entropies of iron and of iron(III) oxide from your data booklet. (1)(ii) The standard molar entropy at 298 K for oxygen molecules O2 is +205 J mol1 K1. Calculate the standard entropy change of the system for the reaction between iron and oxygen. Include a sign and units in your answer. (iii) The standard enthalpy change for the reaction at 25 C is 1648 kJ mol1. Calculate Ssurroundings. (2) (1) *N36289A02128* 21 Turn over<br />
PMT (iv) Use your answers to (b)(ii) and (iii) to calculate the total standard entropy change for the reaction. Include a sign and units in your answer. *(v) The reaction is thermodynamically spontaneous. Use your answers to (b)(ii), (iii) and (iv) to explain, in terms of the physical states of the substances in the reaction and the movement of the molecules in the surroundings, why this is so. (2) (3)(Total for Question 25 = 10 marks) 22 *N36289A02228*<br />

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Mark Mark Mark Acceptable Answers Reject Just the entropy is positive Reject Reject total0 total is positive / So So with or without superscript NOTE: This mark may be awarded from answer to Q25(b)(v) Accept Go is negative Acceptable Answers (+)27.3 and (+)87.4 (J mol1 K1) IGNORE incorrect units Acceptable Answers sys = (2x87.4){(4x27.3 + (3x205.0)} So (1) = 549.4 / 549(J mol1 K1) (1) Correct answer with or without correct units (2) IGNORE any wrong units Accept TE from (b)(i) NOTE: +549/+549.4 scores (1) Check working NOTE: 1st mark: for x2, x4 and x3 2nd mark: for (productsreactants), with correct arithmetic Section C Question Number 25 (a) Question Number 25 (b)(i) Question Number 25 (b)(ii)<br />
Acceptable Answers Reject Just (+)5.53 with no units OR (+)5.53 kJ mol-1 Ssurr =H T =(1648 x 103)298(.15) (J mol1 K1) = (+) 5530 (J mol1 K1) OR = (+) 5.53 kJ mol-1 K1 (1) NOTES:Correct answer, with or without working, scores (1)If 5530 (J mol1 K1) given, IGNORE any subsequent incorrect attempts to convert it to a value in kJ mol-1 K1 IGNORE s.f. except one s.f. Acceptable Answers Reject Mark Mark Just the formula: Stotal = So sys + Ssurr Stotal = (-549.4) +(+5530) = +4980.6/+ 4981 J mol1 K1 OR +4.981 kJ mol-1 K1 (1) for value (1) for correct sign and units IGNORE s.f. except one s.f. Accept TE from (b)(ii) and (b)(iii) Question Number 25 (b)(iii) Question Number 25 (b)(iv)<br />
Question Number 25 (b)(v) Mark Acceptable Answers Reject (Ssystem is negative): as loss of disorder as gas solid OR more order as gas solid OR as decrease in entropy as gas solid (1) (Ssurr is positive): (heat) energy released (increases kinetic energy and hence movement of the surrounding molecules) (1) Stotal is positive because Ssurr is (numerically) greater than Ssys OR Ssurr outweighs Ssys OR Ssurr sufficiently large so that Stotal is positive Just reaction is exothermic Stotal is negative (0) for third scoring point (1)<br />

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