Edexcel Jan 2012 Paper 4 Q22 (with explained solutions) |

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Edexcel Jan 2012 Paper 4 Q22

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SECTION C Answer ALL the questions. Write your answers in the spaces provided. PMT 22 The hydrocarbon butane can be cracked to form propene and methane by passing it over a heated aluminium oxide catalyst at a temperature of 700 K. The equation for the reaction is C4H10(g) (cid:111) C3H6(g) + CH4(cid:11)(cid:74)(cid:12)(cid:3) (cid:3) (cid:3) (cid:3) (cid:3) (cid:3) (cid:3) (cid:3) (cid:168)H(cid:3)(cid:32)(cid:3)(cid:14)(cid:26)(cid:20)(cid:17)(cid:28)(cid:3)(cid:78)(cid:45)(cid:3)(cid:80)(cid:82)(cid:79)(cid:237)(cid:20) (a) (i) Use page 20 of the data booklet to complete the table below. Hydrocarbon S (cid:57)(cid:18)(cid:3)(cid:45)(cid:3)(cid:80)(cid:82)(cid:79)(cid:237)(cid:20)(cid:3)K(cid:237)(cid:20) C4H10(g) C3H6(g) CH4(g) +310.1 (cid:14)(cid:21)(cid:25)(cid:25)(cid:17)(cid:28) (cid:3) (cid:3) (cid:11)(cid:76)(cid:76)(cid:12)(cid:3) (cid:38)(cid:68)(cid:79)(cid:70)(cid:88)(cid:79)(cid:68)(cid:87)(cid:72)(cid:3)(cid:87)(cid:75)(cid:72)(cid:3)(cid:86)(cid:87)(cid:68)(cid:81)(cid:71)(cid:68)(cid:85)(cid:71)(cid:3)(cid:72)(cid:81)(cid:87)(cid:85)(cid:82)(cid:83)(cid:92)(cid:3)(cid:70)(cid:75)(cid:68)(cid:81)(cid:74)(cid:72)(cid:3)(cid:82)(cid:73)(cid:3)(cid:87)(cid:75)(cid:72)(cid:3)(cid:86)(cid:92)(cid:86)(cid:87)(cid:72)(cid:80)(cid:15)(cid:3)(cid:168)S (cid:57) system, for this reaction. Include a sign in your answer. (iii) Was the sign for your answer as you expected? Fully justify your answer. (1) (2) (2)18 *P39304A01824*<br />
PMT (cid:3) (cid:3) (cid:11)(cid:76)(cid:89)(cid:12)(cid:3) (cid:38)(cid:68)(cid:79)(cid:70)(cid:88)(cid:79)(cid:68)(cid:87)(cid:72)(cid:3)(cid:87)(cid:75)(cid:72)(cid:3)(cid:72)(cid:81)(cid:87)(cid:85)(cid:82)(cid:83)(cid:92)(cid:3)(cid:70)(cid:75)(cid:68)(cid:81)(cid:74)(cid:72)(cid:3)(cid:82)(cid:73)(cid:3)(cid:87)(cid:75)(cid:72)(cid:3)(cid:86)(cid:88)(cid:85)(cid:85)(cid:82)(cid:88)(cid:81)(cid:71)(cid:76)(cid:81)(cid:74)(cid:86)(cid:15)(cid:3)(cid:168)Ssurroundings, at 700 K. Include a sign and units in your answer. Use this value and your answer to (ii) to explain why butane cracks into propene and methane at this temperature. (3)(v) Calculate the minimum temperature needed for this reaction to be thermodynamically feasible. (3) *P39304A01924* 19 Turn over<br />
PMT (b) The aluminium oxide behaves as a heterogeneous catalyst. Explain both what is meant by the term heterogeneous and how, in terms of activation energy, the catalyst is able to speed up the reaction. (3)(Total for Question 22 = 14 marks) 20 *P39304A02024*<br />

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Mark Mark Mark Acceptable Answers Section C Question Number 22 (a)(i) Question Number 22(a)(ii) (266.9 + 186.2)310.1 (1) (+)186.2 (J mol-1 K-1) Acceptable Answers Reject Reject = + 143 (J mol-1 K-1) (1)143 scores (1) Correct answer with sign and no working scores (2) marks ALLOW TE from (i) Question Number 22(a)(iii) Yes, as reaction produces 2 molecules/moles from Acceptable Answers Reject one/more molecules/moles (1) (and) all products are gases (1) IGNORE references to volumes More moles/molecules of gas produced scores (2) OR Yes, (as the reaction is endothermic) Ssurroundings is negative (1) Since the reaction takes place/goes (spontaneously) Stotal is positive and therefore Ssystem is positive (1) ALLOW TE from (a)(ii) i.e. No, as. 6CH0401 1201<br />
Acceptable Answers Question Number 22(a)(iv) Ssurr = H/T (1) Mark Reject 1 or 2 sf = 71900/700 = 102.7 J K-1 mol-1/ 0.1027 kJ K-1 mol-1 (1) Correct answer and sign with no working scores (2)0.103 J K-1 mol-1 scores (1) Third mark So Stotal is positive (so reaction is feasible) (1) OR Stotal = +40.3 J K-1 mol-1 (so reaction is feasible) (1) ALLOW TE from (a)(ii) Reject Mark Question Number 22(a)(v) Stotal = 0 Acceptable Answers OR Ssurroundings = 143 (1) T = HSsurroundings OR T = () 71900()143 (1) = 502.8 (K) (1) IGNORE sf except 1sf Correct answer with no working scores (3) ALLOW 0.5028 (K) for (2) marks ALLOW502.8 (K) for (2) marks ALLOW0.5028 (K) for (1) mark ALLOW TE from (a)(ii) If the calculation is not based on Stotal = 0 then a maximum of (2) marks can be awarded if done correctly 6CH0401 1201<br />
Question Number 22(b) Acceptable Answers Reject Mark The catalyst is in a different state/phase to the reactants (1) IGNORE references to products Any two from It provides an alternative (reaction) route/mechanism/gases adsorbed on catalyst surface (1) Of lower activation energy/weakens bonds in reactants (1) Greater proportion of molecules have EEa (1) 6CH0401 1201<br />

Question 17 18 19 20 21 22 23

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