Edexcel Jun 2015 Paper 4 Q17 (with explained solutions) |

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Edexcel Jun 2015 Paper 4 Q17

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SECTION C Answer ALL the questions. Write your answers in the spaces provided. 17 Adipic acid, HOOC(CH2)4COOH, is a dicarboxylic acid used in the production of polymers. It can be made by the reaction of buta-1,3-diene with carbon monoxide and water. CH2CHCHCH2(g) + 2CO(g) + 2H2O(l) (cid:314) HOOC(CH2)4COOH(s) (a) (i) Use the Data Booklet to complete the table below. (2) CH2CHCHCH2(g) CO(g) H2O(l) HOOC(CH2)4COOH(s) (cid:507)H(cid:57) f / kJ mol1 S(cid:57) / J mol1 K1 +109.9 278.7 994.3 250.0 (ii) Use data from the table to calculate the standard enthalpy change, in kJ mol1, when adipic acid is formed from buta-1,3-diene, carbon monoxide and water. (iii) Use data from the table to calculate the standard entropy change of the system, in J mol1 K1, when adipic acid is formed from buta-1,3-diene, carbon monoxide and water. (2) (2) 24 *P42978A02428*<br />
(iv) Use your answers to (a)(ii) and (a)(iii) to calculate (cid:507)Ssurroundings and (cid:507)Stotal for the reaction at 298 K. (v) It was suggested that decreasing the temperature of the reaction to less than 298 K would produce a greater yield of adipic acid. Explain, in terms of the effect on (cid:507)Ssystem, (cid:507)Ssurroundings and hence (cid:507)Stotal, whether this would be the case. (3) (3)(b) Infrared spectroscopy can be used to follow the progress of reactions. During the reaction to produce adipic acid, suggest one peak which diminishes and one peak which appears. Use information from the Data Booklet to identify two such possible peaks, giving their wave numbers and the bonds involved. (2)*P42978A02528* 25 Turn over<br />
(c) Adipic acid is used as an additive in some fruit jellies. Suggest what effect the adipic acid will have on the flavour of the jelly. (1)(d) An organic compound, Q, is found to contain 49.3% carbon and 6.8% hydrogen by mass. (i) Use these data to confirm its empirical formula is C3H5O2. (3) 26 *P42978A02628*<br />
(ii) The structure of Q is shown below. H3C H2 C OH H C CH3 The table below summarises some information about parts of the nmr spectrum of compound Q. Use the Data Booklet, and your knowledge of splitting patterns, to complete the table with respect to the features of compound Q shown in bold. Feature of compound Q CH3 CH2 COOH (4) Splitting pattern Chemical shift / ppm 0.11.9 singlet (Total for Question 17 = 22 marks) TOTAL FOR SECTION C = 22 MARKS TOTAL FOR PAPER = 90 MARKS *P42978A02728* 27<br />

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Section C Correct Answer Reject Mark Question Number 17 (a)(i) Question Number 17 (a)(ii) CH2CHCHCH2 CO H2O HOOC(CH2)4COOH Hf +109.9 110.5 285.8 994.3 / kJ mol1 S / 278.7 197.6 69.9 250.0 J mol1 K-1 4 values correct (2) marks 3 / 2 values correct (1) mark 0 / 1 values correct (0) marks Correct Answer 994.3[+109.9 + (2 x 110.5) + (2 x 285.8)] (1) =311.6 (kJ mol1) (1) Allow TE from (a) NOTE If both 110.5 and 285.8 are not doubled, answer CQ = 707.9 (kJ mol1) for 1 mark Ignore SF except 1 SF Reject Mark<br />
Question Number 17 (a)(iii) Question Number 17 (a)(iv) Rejec t Mark Correct Answer 250(.0)[278.7 + (2 x 197.6) + (2 x 69.9)] (1) = 563.7 (J mol1 K1) (1) Allow TE from (a) NOTE If both 197.6 and 69.9 are not doubled, answer CQ = 296.2 (J mol1 K1) for 1 mark Ignore SF except 1 SF Reject Mark (1) Correct Answer Ssurr at 298 K = H/T =(311.6 x 1000) / 298 = (+) 1045.6 (J mol1 K1) Allow TE from (a)(ii) e.g. Ssurr = (+)2375.5(0) (J mol1 K1) scores (2) if no doubling in (a)(ii) (1) Stot = Ssurr + Ssys / Stot = 1045.6 563.7 / Stot = (+) 481.9 (J mol1 K1) Allow TE from (a)(ii) and (a)(iii) (1) Allow correct answers given in kJ mol1 K1 e.g. 0.4819 kJ mol1 K1 Ignore SF except 1 SF If candidates forget to convert H into J mol1, then Stot = 562.7 (J mol1 K1) would score (2) if correct working is included<br />
Question Number 17 (a)(v) Correct Answer (Decrease in T) 1st mark: consideration of Ssystem Ssystem is not (significantly) changed / is unchanged / remains (approximately) constant Reject Mark (1) (1) 2nd mark: consideration of Ssurr Ssurr or H/T is more positive / larger / greater COMMENT ALLOW less negative 3rd mark: consideration of Stotal (So) increases Stot / makes Stot more positive / makes Stot greater (1) NOTE IF no reference / an incorrect reference made to Ssystem, then only the 2nd and 3rd marks can be awarded NOTE If candidate states that Ssurr becomes less +ve, no M2 But if then states CQ that Stot decreases award M3 as a TE<br />
Question Number 17 (b) Correct Answer Reject Mark DIMINISHING: (Peak between) 16691645 (cm1) (due to C=C) OR (Peak between) 30953010 (cm1) (due to alkene C-H) INCREASING: (Peak between) 17251700 (cm1) (due to C=O in carboxylic acid) OR (Peaks due to alkane CH bonds at) EITHER 29622853 (cm1) OR 14851365 (cm1) ALLOW (Peak between) 33002500 (cm1) (due to OH in carboxylic acid) (1) (1) 17401720 37503200<br />
Question Number 17 (c) Correct Answer (Makes it taste) sour / sharp / tart IGNORE acidic / bitter NOTE Contradictory answers (e.g. sharp and sweeter) score (0) Reject Mark fruity sweet(er) none<br />
Correct Answer Question Number 17 (d) (i) 1st mark: (% of oxygen =) 43.9 (%) (1) 2nd mark: Amount of C = 49.3/12 = 4.1 (mol) Amount of H = 6.8/1 = 6.8 (mol) Amount of O = 43.9/16 = 2.7 (mol) Reject Mark 3rd mark: Ratio 1.5 C : 2.5 H : 1 O (3 C : 5 H : 2 O) ALLOW for 3rd mark:- Decimal values that round up to these values (e.g. 1.497 C : 2.478 H : 1 O scores the 3rd mark) (1) ALLOW Mr of C3H5O2 = 73 (g mol-1) %C = 36 x 100 = 49.3% 73 and %H = 5 x 100 = 6.8% 73 %O = 43.9% ALLOW 43.8% (1) (1) (1) (1)<br />
Question Number 17 (d) (ii) Correct Answer Reject Mark For Chemical shift column, allow any range or any single value within range Feature of compound Chemical shift / ppm Splitting pattern CH3 0.11.9 Triplet (1) Allow (splits into) three CH2 1.73(.0) Quartet (1) (1) Allow quadruplet / (splits into) four 10(.0)12(.0) (1) singlet OH (TOTAL FOR QUESTION 17 = 22 marks) Total for paper = 90 marks<br />

Question 13 14 15 16 17

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