Edexcel Jun 2016 (IAL) Paper 4 Q (with explained solutions) |

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Edexcel Jun 2016 (IAL) Paper 4 Q

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12 This is a question about entropy changes. (a) Consider the reaction between solid ammonium carbonate and pure ethanoic acid. The equation for this reaction is (NH4)2CO3(s) + 2CH3COOH(l) (cid:111) 2CH3COONH4(s) + H2O(l) + CO2(g) (i) State what you would observe as this reaction occurs. (1)D O N O T W R I T E I N T H I S A R E A *(ii) Predict the sign of the entropy change of the system, (cid:507)S (cid:57) system. Fully justify your answer. No calculation is required. (2)16 *P46664A01628* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(b) The rhombic allotrope of sulfur reacts with fluorine to produce sulfur hexafluoride: S(s, rhombic) + 3F2(g) (cid:111) SF6(g) (cid:507)Hf (cid:57) = 1209 kJ mol1 (i) Use the standard molar entropies on pages 2, 3 and 29 of the Data Booklet to calculate the standard molar entropy change of the system ((cid:507)S (cid:57) Include a sign and units in your answer. system) for this reaction. Note that the standard molar entropies of the elements are given per atom so that the standard molar entropy of fluorine, S (cid:57)[F2(g)] = +158.6 J mol1 K1. (ii) Use the value of the standard enthalpy change of formation ((cid:507)H f above to calculate the entropy change of surroundings ((cid:507)S (cid:57) reaction at 298 K. Include a sign and units in your answer. (cid:57) ) given surroundings) for this (2) (2) (iii) Use your answers to (b)(i) and (b)(ii) to calculate the total entropy change ((cid:507)S (cid:57) total) for the formation of one mole of sulfur hexafluoride. Include a sign and units in your answer. (1) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P46664A01728* 17 Turn over<br />
(iv) What would be the effect, if any, of an increase in temperature on the value of total calculated in (b)(iii)? Justify your answer and state any assumptions that (cid:507)S (cid:57) you have made. (3)D O N O T W R I T E I N T H I S A R E A(c) The equations for dissolving two sulfates are shown below. MgSO4(s) + aq (cid:85) Mg2+(aq) + SO4 2(aq) (cid:507)S (cid:57) total = +20 J mol1 K1 BaSO4(s) + aq (cid:85) Ba2+(aq) + SO4 2(aq) (cid:507)S (cid:57) total = 190 J mol1 K1 (i) Compare the values of the total entropy changes for dissolving these two sulfates and show that they are consistent with the trend in the solubility of Group 2 sulfates. (2)18 *P46664A01828* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(ii) The values of the total entropy change and the equilibrium constant of a reaction are related by the following equation. (cid:507)Stotal = R lnK Calculate the value of the equilibrium constant, K, for the dissolving of magnesium sulfate at 298 K. R = 8.31 J mol1 K1 (1) A E R A S I H T N I E T I R W T O N O D (Total for Question 12 = 14 marks) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P46664A01928* 19 Turn over<br />

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ALLOW Two vertical sections not at 10/20 cm3 scores (1) if M2 and M3 not awarded (Total for question 11 = 13 Marks) Reject Mark Acceptable Answers Question Number 12(a)(i) Effervescence / bubbles / fizzing IGNORE gas evolved / temperature increase Acceptable Answers Question Number *12(a)(ii) A statement that entropy is positive needs to be made once only and can be used to award M1 and M2 Penalise omission of statement that entropy is positive once only M1 Entropy (of the system) positive and solid and liquid reactants form (a solid, a liquid and) a gas ALLOW gas formed / gas is a product (1) M2 Entropy (of the system) positive and EITHER 3 moles4 moles OR more moles of products (than reactants) ALLOW molecules for moles OR More ways of distributing energy OR (1) Mark (2) Reject If entropy of system is negative / decreases scores (0) particles<br />
More ways of distributing quanta (1)<br />
Acceptable Answers Question Number 12(b)(i) S(reactants ) = (31.8) + 3(2x 158.6) =) +983.4 J mol1 K1 (1) Ssystem = (291.7 983.4 =) 691.7 J mol1 K1 / 0.6917 kJ mol1 K1 (1) Correct answer no working scores (2) If monoclinic sulfur is used (32.6) final answer = 692.5 scores (1) Acceptable Answers Question Number 12(b)(ii) Ssurroundings = ( HT)=1209000 J mol-1 (1) 298 K = (4057.04698) = + 4057 J mol1 K1 / + 4.057 kJ mol1 K1 (1) Correct answer without working scores 2 Acceptable Answers Question Number 12(b)(iii) Stotal = Ssystem + Ssurroundings Stotal = ans (b)(i) + ans (b)(ii) = 691.7 + 4057 = +3365.3 J mol1 K1 / +3.3653 kJ mol1 K1 TE on answers from (b)(i) and (b)(ii) Acceptable Answers Question Number 12(b)(iv) Marking points may be in any order Mark all 3 points independently M1: Ssurroundings becomes less positive / smaller (magnitude) / decreases (in magnitude) (because you are dividing H by a larger T) M2: (1) Reject Mark (2) Reject Mark (2) Reject Mark (1) Reject Mark (3)<br />
Ssystem / Hf /H is not (significantly) affected (by an increase in temperature) (1) M3: (So) Stotal decreases / less positive (1) Becomes negative Acceptable Answers Question Number 12(c)(i) Accept reverse arguments throughout M1 sulfates get less soluble as you descend Group 2 (1) ALLOW barium sulfate is less soluble than magnesium sulfate M2 (total) entropy / Stotal gets more negative/ less positive as you go from MgSO4 to BaSO4 ALLOW Stotal is positive for MgSO4 and negative for BaSO4 (1) IGNORE Re-stating the numerical entropy values No TE on incorrect trend Acceptable Answers Question Number 12(c)(ii) (K= inv ln (20/8.31)=) 11.098 = 11 IGNORE any units ALLOW any SF except 1SF Mark (2) Reject References to exothermic / endothermic Just decreases, gets smaller Reject Mark (1) (Total for question 12 = 14 Marks)<br />

Question 12 13 14 15

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