Edexcel Jun 2016 Paper 4 Q13 (with explained solutions) |

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Edexcel Jun 2016 Paper 4 Q13

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13 This is a question about entropy changes. Consider the reaction between the two solids, hydrated barium hydroxide and ammonium chloride. When these substances are mixed together, a white paste is formed and the temperature decreases. An equation for this process is given below. Ba(OH)2.8H2O(s) + 2NH4Cl(s) (cid:108) 2NH3(g) + 10H2O(l) + BaCl2(s) (a) (i) Identify one hazard associated with a named substance in this reaction. (1)D O N O T W R I T E I N T H I S A R E A (ii) Use the standard molar entropies below to calculate the standard entropy change of the system ((cid:168)S(cid:57) answer. system) for this reaction at 298 K. Give a sign and units with your Compound Ba(OH)2.8H2O(s) NH4Cl(s) NH3(g) H2O(l) BaCl2(s) S(cid:57) / J mol1 K1 427 95 192 70 124 (3) 16 *P46660RA01628* 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 />
*(iii) Give two reasons why the sign of your answer to (a)(ii) is as you would expect. (2)(b) The standard enthalpy change for this reaction is (cid:168)H (cid:57) r = +162 kJ mol 1. Use this value to calculate the standard entropy change of the surroundings ((cid:168)S(cid:57) surroundings) for this reaction at 298 K. Include a sign and units in your answer. (c) Use your answers to (a)(ii) and (b) to calculate the total entropy change ((cid:168)S(cid:57) total) for this reaction. Include a sign and units in your answer. (d) What would be the effect, if any, on the value of (cid:168)S(cid:57) total from (c) of a small increase in temperature? Justify your answer and state any assumptions that you have made. (2) (1) (3)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*P46660RA01728* 17 Turn over<br />
(e) The values of total entropy change and equilibrium constant of a reaction are related by the following equation. (cid:168)Stotal = R lnK The equation for the dissolving of barium hydroxide is Ba(OH)2(s) + aq (cid:85) Ba2+(aq) + 2OH(aq) (cid:168)S(cid:57) total = 44 J mol1 K1 (i) Calculate the value of the equilibrium constant, K, for this equation at 298 K. R = 8.31 J mol 1 K1 (ii) What does the value of the equilibrium constant suggest about the solubility of barium hydroxide? Justify your answer. (1) (1) 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(iii) For the dissolving of calcium hydroxide, the value of the total entropy change is 106 J mol 1 K1 Compare the values of the total entropy changes for these two hydroxides and show that they are consistent with the trend in the solubility of Group 2 hydroxides. (2)(Total for Question 13 = 16 marks) 18 *P46660RA01828* D O N O T W R I T E I N T H I S A R E A<br />

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Mark (1) Reject References to just barium Ammonium chloride is toxic Reject Mark (3) Acceptable Answers Question Number 13(a)(i) Ammonia / barium chloride is toxic OR Ammonia / barium chloride is poisonous OR Barium hydroxide is corrosive / caustic OR Ammonia (solution) is corrosive OR Ammonium chloride is harmful / eye-irritant ALLOW Barium hydroxide is toxic / poisonous IGNORE Use of fume cupboard / gloves, etc Acceptable Answers Question Number 13(a)(ii) S(products) = ((2 x 192) + (10 x 70) + 124 =) (+)1208 (J mol-1 K-1) (1) S(reactants) = ((2 x 95) + 427 =) (+)617 (J mol-1 K-1) (1) Ssystem = (1208617 =) +591 J mol-1 K-1 Allow units in any order (1) Correct answer without working scores 3<br />
Acceptable Answers Question Number *13(a)(iii) (Positive value as expected because ) 3 moles13 moles / more moles of products (than reactants) Allow molecules for moles If numbers (of compounds) are stated, then these must be 3 and 13 COMMENT: Ignore any type of particle(s) mentioned (1) (Two) solidsa gas / a liquid (+ 1 solid) OR No gaseous reactants, but gaseous products (formed) (1) Reject Mark (2) (0) Overall if Ssystem negative or entropy decrease Reject Mark (2) (1) 543 543 Acceptable Answers Question Number 13(b) Ssurroundings = (HT) = 162 000 J mol-1 298 K = 543.6241611/544 J mol-1 K-1 Allow 0.544 kJ mol-1 K-1 (1) Correct answer without working scores 2 IGNORE sf except 1 sf Question Number 13(c) Acceptable Answers Stotal = Ssystem + Ssurroundings Stotal = ans (a)(ii) + ans (b) = +591544 = +47 J mol-1 K-1 TE on answers from (a)(ii) and (b) Reject Mark (1)<br />
Acceptable Answers Reject Mark M1: Ssurroundings becomes less negative / more positive smaller in MAGNITUDE (because you are dividing H by a larger T) IGNORE Just smaller / just decreases / just bigger / just greater M2: Ssystem / H are not (significantly) affected by a change in temperature M3: (So) Stotal increases ALLOW a TE for M3 Stotal decreases, only if incorrect M1 (i.e. Ssurroundings becomes less positive) Mark M1, M2 and M3 in any order within candidates answer (3) (1) (1) (1) Question Number 13(d)<br />
Question Number 13(e)(i) Acceptable Answers (K = e(-44/8.31)=) 0.005017/ 5.017 x 103 Ignore any units Allow any sf except 1 sf Acceptable Answers Question Number 13(e)(ii) Barium hydroxide will not be (very) soluble / will be sparingly soluble and K value suggests that the equilibrium lies to the left-hand side / reactants OR (1x1010<) K <1 so reactants predominant No TE on incorrect large value in (e)(i) Acceptable Answers Question Number 13(e)(iii) M1: Reject Mark (1) Reject Mark (1) Just K is small Reject Mark (2) Hydroxides get more soluble as you descend Group 2 M2: Stotal gets less negative / more positive as you go from Ca(OH)2 to Ba(OH)2 IGNORE Just smaller / just decreases / just bigger / just greater (1) (1) ALLOW Reverse argument No TE on calculated value more negative for Ba(OH)2 Mark M1 and M2 independently TOTAL FOR QUESTION 13 = 16 MARKS<br />

Question 12 13 14 15

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