Edexcel Jan 2011 Paper 4 Q17 (with explained solutions) |

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Edexcel Jan 2011 Paper 4 Q17

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 17 Hydrogen can be manufactured by reacting methane with steam, as shown in the equation below. CH4(g) + H2O(g) (cid:85) CO(g) + 3H2(g) (cid:39)H (cid:57) 298 = +206.1 kJ mol1 Use these values: the standard entropy of 1 mol of H2(g) is (2 (cid:117) 65.3) = 130.6 J mol1 K1 the standard entropy of 1 mol of H2O(g) is 188.7 J mol1 K1 You will also need to refer to the data booklet in the calculations which follow. (a) Calculate the standard entropy change of the system, (cid:39)S (cid:57) system, for this reaction at 298 K. PMT (2) (2) (b) Calculate the standard entropy change of the surroundings, (cid:39)S (cid:57) surroundings , for this reaction at 298 K. Include a sign and units in your answer. (c) Calculate the total entropy change, (cid:39)S (cid:57) total, for this reaction at 298 K. Explain why this value shows that the reaction is not spontaneous at this temperature. (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . *N37949A0924* Turn over<br />
PMT (d) The composition of an equilibrium mixture produced at 2.0 atmospheres pressure and at a much higher temperature is shown below. CH4(g) + H2O(g) (cid:85) CO(g) + 3H2(g) Amount in equilibrium 0.80 0.80 1.20 3.60 mixture / mol *(i) Write the expression for the equilibrium constant, Kp, of the reaction and calculate its value. Include units in your answer. (6) (1) (ii) The total entropy change in J mol1 K1 is related to the equilibrium constant by the equation (cid:39)S (cid:57) total = R lnKp or (cid:39)S (cid:57) total = 2.3R logKp Calculate the total entropy change at the temperature of the reaction. [R = 8.31 J mol1 K1] 10 *N37949A01024*<br />
(iii) Calculate the temperature at which this equilibrium is reached using your total. Assume that (cid:39)H is still +206.1 kJ mol1 and that system = +225 J K1 mol1. (This is not the same as the value for (cid:39)S (cid:57) answer to (ii) for (cid:39)S (cid:57) (cid:39)S (cid:57) calculated in (a) which is at 298 K.) system *(e) Use the magnitude and signs of the entropy changes to explain the effect of a temperature increase on the equilibrium constant of this endothermic reaction. PMT (2) (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (Total for Question 17 = 17 marks) *N37949A01124* 11 Turn over<br />

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$Section B Question Number 17 (a) Question Number 17 (b) Acceptable Answers D Ssystem = (3x2x65.3 +197.6)(186.2 + 188.7) Correct data for CH4 and CO (186.2 and 197.6) (1) = (+) 214.5 / 215 (J mol-1 K-1) / (+) 0.2145 / 0.215 kJ (mol-1 K-1) (1) Units must be shown if data has been converted to kJ Full marks (2) for correct answer without working Ignore sf except 1 Answer of -214.5 scores (1) Answer of +18.6 if entropy of H not doubled scores (1) Answer of 46.7 if entropy of H2 not tripled scores (1) ALLOW TE in second mark for minor error in data e.g. writing 63.5 instead of 65.3. No TE if data used is not entropy of compounds. Acceptable Answers (D Ssurroundings) = -D H T Expression or use of expression, -206.1x(1000) (1) 298 = -691.6 J (mol-1 K-1 )/ -0.6916 kJ (mol-1 K-1 ) (1) Ignore sf except 1 Mark Reject 214 0.214 Reject Mark 6CH04/01 1101<br /> Question Number 17 (c) Mark Acceptable Answers Reject D Stotal = (214.5+(-691.6)) = -477.1 (J mol-1 K-1) / - 0.4771 (kJ mol-1 K-1) (1) ALLOW TE for answer to (a) plus answer to (b). If 214.5 is added to -0.69 no TE unless -0.69 is specified to be in joules. Ignore sf except 1 Negative / less than zero (so not spontaneous) / would be positive if spontaneous. (1) ALLOW feasible for spontaneous. If answer to calculation is positive, accept comment that it would be expected to be negative if not spontaneous Addition of value in J to specified value in kJ Comments on kinetic stability 6CH04/01 1101<br /> Question Number *17 (d) (i) Mark Acceptable Answers Reject Square brackets TE for Kp expression with addition, not multiplication Kp = (pH2)3x(pCO) (pCH4)(pH2O) (1) 4 Correct partial pressures (3) CH4 H2O CO H2 pp 0.25 0.25 0.375 1.125 ALLOW partial pressures as fractions Kp = (1.125)3 x (0.375) = 8.54 atm2 (0.25)(0.25) value of Kp (1) unit (1) (Stand alone mark) Correct calculation without working scores the 5 calculation marks. TE from Kp expression if inverted Ignore sf except 1 If any partial pressures are incorrect: Calculating total number of moles (6.4) (1) Calculating mole fractions (0.125, 0.125, 0.1875, 0.5625 if total number of moles is correct) (1) Multiplying mole fractions by total pressure (x 2 atm) (1) value of Kp (1) unit (1) (stand alone mark) ALLOW TE in value of Kp only from incorrect partial pressures, not using values in question as not using equilibrium moles If treated as a Kc calculation following Kp expression : Kp expression (1) units atm2 (1) Max. mark (2) 6CH04/01 1101<br /> Mark Mark Question Number 17 (d) (ii) Question Number 17 (d) (iii) Reject Reject Acceptable Answers D Stotal = (8.31 ln 8.54) = (+)17.8 (J mol-1 K-1) Accept any value that rounds to 17.8 TE from value in (i) Kp value of 87.48 (obtained by treating calculation in (i) as Kc) gives D Stotal = 37.16 / 37.12 Acceptable Answers 17.8 = 225 - 206.1 x 1000 (1) T T = (206.1 x 1000) = 995 / 990 (K) (1) 207.2 Correct answer with no working shown scores 2 Correct method with wrong answer or missing 103 scores 1 TE from (ii) Kp value of 87.48 gives T = 1097 OR If D Stotal is taken as zero 0 = 225 - 206.1 x 1000 (1) T T = 916K (1) Kp value of 87.48 gives T = 916 Ignore sf except 1 6CH04/01 1101<br /> Question Number *17 (e) Mark Acceptable Answers Reject D Ssurroundings / -D H T becomes less negative making D Stotal more positive (as T increases) OR D Ssurroundings / -D H T becomes less negative making D Stotal greater (as T increases) OR (magnitude of) D Ssurroundings becomes less / lower making D Stotal more positive / greater (as T increases) (1) Because D Stotal increases equilibrium constant increases (1) OR value of D Stotal at new temperature is more than at 298K (1) (must be clear that the two D Stotal values at the different temperatures have been considered) Because D Stotal increases equilibrium constant increases (1) Le Chatelier statements without reference to entropy changes Just as temperature increases D Stotal increases 6CH04/01 1101<br />$

Question 17 18 19 20 21

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