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

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Edexcel Jun 2015 (IAL) Paper 4 Q17

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SECTION C Answer ALL the questions. Write your answers in the spaces provided. 17 Ethanoic acid, CH3COOH, is a carboxylic acid with many uses, including as a food additive. It can be made by the reaction of butane with oxygen. 2CH3CH2CH2CH3(g) + 5O2(g) (cid:85) 4CH3COOH(l) + 2H2O(l) (a) (i) Use the Data Booklet to complete the table below. CH3CH2CH2CH3(g) O2(g) CH3COOH(l) (cid:507)H (cid:57) f / kJ mol1 S(cid:57) / J mol1 K1 205 (3) H2O(l) (ii) Use data from your table to calculate the standard enthalpy change, in kJ mol1, for this reaction. 2CH3CH2CH2CH3(g) + 5O2(g) (cid:314) 4CH3COOH(l) + 2H2O(l) (iii) Use data from your table to calculate the standard entropy change of the system, in J mol1 K1, for the same reaction. 2CH3CH2CH2CH3(g) + 5O2(g) (cid:314) 4CH3COOH(l) + 2H2O(l) (2) (2) 24 *P45075A02428*<br />
(iv) Use your answer to (a)(ii) to calculate (cid:507)Ssurroundings and use this and your answer to (a)(iii) to calculate (cid:507)Stotal for the reaction at 298 K. (v) It was suggested that increasing the temperature of the reaction to more than 298 K would produce a greater yield of ethanoic 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. Using information from the Data Booklet, suggest one way this technique could be used to follow the progress of the reaction in (a) to produce ethanoic acid. (1)*P45075A02528* 25 Turn over<br />
(c) Ethanoic acid is the food additive E260. Suggest the role it may have when added to foodstuffs. (1)(d) An organic compound, Q, is found to contain 52.5% carbon and 7.5% hydrogen by mass. (i) Use these data to confirm its empirical formula is C7H12O4. (ii) Explain how the mass spectrum of Q could be used to confirm that its relative molecular mass is 160. (3) (1)26 *P45075A02628*<br />
(iii) The table below summarises some information about parts of the nmr spectrum of compound Q. Use the Data Booklet, and your knowledge of features in nmr spectra, to complete the table with respect to the features of compound Q shown in bold. (4) CH3 O H3C C H OH H C CH3 Feature of compound Q Chemical shift / ppm for TMS Splitting pattern Relative area below peak CH3 CH COOH 0.11.9 doublet singlet (Total for Question 17 = 23 marks) TOTAL FOR SECTION C = 23 MARKS TOTAL FOR PAPER = 90 MARKS *P45075A02728* 27<br />

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Mark Acceptable Answers Ho / kJ mol1 So / J mol1 K1 CH3CH2CH2CH3 O2 CH3CO2H H2O 126.5 484.5 285.8 310.1 205 159.8 69.9 6 values correct 3 marks 4 / 5 values correct 2 marks 2/3 values correct 1 mark 0/1 values correct 0 marks Ignore values multiplied by balancing numbers in addition to correct values eg for water 2 x 285.8 (=571.6) Question Number 17(a) (ii) Mark Acceptable Answers If answer is2256.6 /2257 (kJ mol1), award 2 marks [(2 x 285.8) + (4 x 484.5)](2 x 126.5) (1) =2256.6 /2257 (kJ mol1) (1) Allow answer converted to J mol-1 Allow TE from incorrect data in table in (a)(i) Allow (1) for cycle wrong way round eg (+) 2256.6 / (+)2257 (kJ mol1) Allow (1) for using correct values but not multiplied by balancing numbers eg -643.8 (kJ mol1) Ignore SF except 1SF Section C Question Number 17(a)(i)<br />
Question Number 17(a) (iii) Question Number 17(a) (iv) Mark Acceptable Answers If answer is 866.2 (J mol1K1), award 2 marks [(2 x 69.9) + (4 x 159.8)][(2 x 310.1) + (5 x 205)] (1) 866.2 (J mol1K1) (1) Allow answer converted to kJ mol-1K1 Allow TE from incorrect data in table in (a)(i) Allow (1) for cycle wrong way round eg (+) 866.2(J mol1K1) Allow (1) for using correct values but error(s) in balancing numbers eg -285.4 (J mol1K1) Ignore SF except 1SF Mark Acceptable Answers If answer is (+)6706.3 J mol1 K1 or (+)6.7063 kJ mol1 K1, award 3 marks Ssurr at 298 K = H/T =(2256.6 x 1000) / 298 (1) = 7572.483 (J mol1K1) Allow rounding to 3SF or more (1) Allow correct answers given in kJ mol-1 K-1 eg 7.5725 kJ mol-1 K-1 Stot = Ssurr + Ssys / Stot = 866.2 + 7572.5 / Stot = (+)6706.3 J mol1 K1 OR -0.8662 + 7.5725 / Stot = (+)6.7063 kJ mol1 K1 (1) Allow TE from (a)(ii) and (a)(iii) Ignore SF except 1SF in final answer<br />
Mark Question Number 17(a)(v) 1st mark: consideration of Ssystem Acceptable Answers Ssys is not (significantly) changed /is unchanged /remains (approximately) constant (1) 2nd mark: consideration of Ssurr (Higher temperature makes) Ssurr /-H/T is smaller / decreases / less positive Comment Allow more negative (1) No TE if Ssurr is ve in (a)(iv) 3rd mark: consideration of Stotal EITHER reduces Stot / makes Stot less positive / makes Stot closer to zero (so would not produce a greater yield) OR Stot is very large (so K is very large) so the effect of change in temperature is negligible (1) NOTE if Ssurr is ve in (iv), then allow increases Stot / makes Stot more positive / makes Stot closer to zero (so would produce a greater yield). NOTE IF no reference / an incorrect reference made to Ssystem, then only the 2nd and 3rd marks can be awarded<br />
Question Number 17(b) Question Number 17(c) Mark Mark Acceptable Answers Note: All we are looking for are the correct ranges, exactly as given below (i.e. the bonds do not have to be stated, as they follow from the correct ranges) Peak between 17251700 (cm1) (would appear due to C=O group (in alkyl carboxylic acid)) Allow peak between 33002500 (cm1) (due to OH group (in carboxylic acid)) Acceptable Answers increase sourness / sharpness of flavour OR preservative / prevents growth of microbes / prevents food decay / prevents food decomposition /kills microbes OR acidity regulator / buffer Allow improves flavouring Ignore reduce pH/ make (slightly) acidic/just flavouring<br />
Mark Question Number 17(d)(i) Working must be shown Acceptable Answers EITHER % of oxygen = 40% (1) Amount of C = 52.5/12 = 4.375 (mol) Amount of H = 7.5/1 = 7.5 (mol) Amount of O = 40/16 = 2.5 (mol) (1) Ratio 1.75 C : 3 H : 1 O7 C : 12 H : 4 O Ignore SF in mol and ratios (1) OR % of C in C7H12O4 = 84 x 100 = 52.5% (1) 160 % of H in C7H12O4 = 12 x 100 = 7.5% (1) 160 % of O in C7H12O4 = 64 x 100 = 40 % (1) 160 OR No C atoms = 52.5 x 160 = 7 (1) 100 x 12 No H atoms = 7.5 x 160 = 12 (1) 100 x 1 No O atoms = 40 x 160 = 4 (1) 100 x 16 Question Number 17(d) (ii) Acceptable Answers Reject Mark Largest/highest m/e or m/z value (is 160) OR Mass (/charge ratio) or m/e or m/z of molecular/parent ion/ C7H12O4 (=160(=Mr)) Allow last peak / peak on rhs (is at 160) Allow peak before last (is at 160 due to M+1 peak at 161) Highest peak Just there is a peak at 160<br />
CH3 CH COOH 1.83.0 (1) 1012.0 (1) septuplet / heptuplet / splits into 7 / 7 splits (1) singlet Allow heptet / septet /sevenlet and similar words that indicate 7 Total for Question 17 = 23 marks Question Number 17(d) (iii) Acceptable Answers For chemical shift column, allow any range or any single value within range and allow range in the opposite order eg 3.0-1.8 Mark Feature of compound X Chemical shift / ppm for TMS Splitting patterns 0.11.9 doublet Relativ e area below peak 3 (1)<br />

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