OCR A Jun 2014 Paper 5 Q2 (with explained solutions) |

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OCR A Jun 2014 Paper 5 Q2

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This question looks at different aspects of entropy. (a) Three processes are given below. For each process, state and explain whether the change would be accompanied by an increase or decrease in entropy. (i) The freezing of water. increase or decreaseexplanation[1] (ii) The reaction of calcium carbonate with hydrochloric acid. increase or decreaseexplanation[1] (iii) The formation of O3(g) from O2(g). increase or decreaseexplanation[1] (b) The enthalpy and entropy changes of a reaction both have a negative sign. Discuss how the feasibility of this reaction will change as the temperature increases.[2]OCR 2014<br />
(c) The metal tungsten is obtained on a large scale from its main ore, wolframite. Wolframite contains tungsten(VI) oxide, WO3. Tungsten is extracted from wolframite by reduction with hydrogen: WO3(s) + 3H2(g) W(s) + 3H2O(g) H = +115 kJ mol1 Standard entropies are given in the table below. Substance S / J K1 mol1 WO3(s) 76 H2(g) 131 W(s) 33 H2O(g) 189 (i) Calculate the free energy change, G, in kJ mol1, for this reaction at 25 C. Show your working. (ii) Calculate the minimum temperature, in K, at which this reaction becomes feasible. G at 25 C =kJ mol1 [2] Show your working. minimum temperature =K [2] [Total: 9] Turn overOCR 2014<br />

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F325 Question (a) (i) Answer (entropy) decreases AND (solid/ice has) less disorder/ more order/ fewer ways of arranging energy/ less freedom/ less random molecules (cid:57) (a) (ii) (entropy) increases AND (CO2) gas is formed (cid:57) Could be from equation with CO2(g) (a) (iii) entropy decreases AND 3 mol O2 form 2 mol O3 OR 3O2 (cid:111) 2O3 OR 3 mol gas form 2 mol gas (cid:57) Mark Scheme June 2014 Marks Guidance ORA decreases and reason required for mark ASSUME change is for freezing of water unless otherwise stated DO NOT ALLOW atoms are more ordered increases and reason required for mark ASSUME gas is CO2 unless otherwise stated BUT DO NOT ALLOW an incorrect gas (e.g. H2) ALLOW more gas decreases and reason required for mark For mol, ALLOW molecules ALLOW multiples, ALLOW O2 + O (cid:111) O3 Note: DO NOT ALLOW 2 mol gas forms 1 mol gas unless linked to O2 + O (cid:111) O3 IGNORE reaction forms fewer moles/molecules e.g. 1O2 (cid:111) O3; O2 + O2 (cid:111) O3<br />
F325 (b) Mark Scheme June 2014 CARE: responses involve changes of negative values -------------------------------------------------------------------------- Feasibility AND G Reaction becomes/is less feasible/not feasible AND G increases OR G becomes/is less negative/more positive OR G > 0 OR HTS > 0 OR HTS becomes/is less negative/more positive OR H > TS (cid:51) OR TS becomes/is more negative than H (cid:51) ------------------------------------------------------------------------- Effect on TS TS becomes more negative OR TS decreases OR TS becomes more positive OR TS increases OR magnitude of TS increases OR | TS | increases (cid:51) --------------------------------------------------------------------------------- FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- As alternative for less feasible ALLOW less spontaneous OR a comment that implies reaction no longer take place ALLOW for G increases G < 0 only at low T DO NOT ALLOW TS > H (comparison wrong way round) NOTE: Last statement automatically scores 2nd mark ALSO IGNORE significance IGNORE magnitude for 1st marking point ------------------- DO NOT ALLOW TS increases IGNORE significance ---------------------------------------------------------------------- APPROACH BASED ON TOTAL ENTROPY: Feasibility with increasing temperature H/T is less negative OR H/T increases OR H/T decreases OR magnitude of H/T decreases (cid:51) Reaction becomes less feasible/not feasible AND SH/T OR Stotal decreases/ less positive (cid:51) Effect on H/T<br />
F325 (c) (i) (c) (ii) Mark Scheme June 2014 DO NOT ALLOW 131 ALLOW ECF from incorrect calculated value of S ALLOW total entropy statement: S(total) = 0 OR S(total) >0 ALLOW ECF from incorrect calculated value of S from 2(c)(i) ALLOW 878 up to calculator value of 877.862595 correctly rounded FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 75.962 OR 75.96 OR 76.0 OR 76, award 2 marks -------------------------------------------------------------------- S = (33 + 3 (cid:117) 189)(76 + 3 (cid:117) 131) = (+)131 (J K1 mol1) (cid:57) G = 115(298 (cid:117) 0.131) = (+) 75.962 OR 75.96 OR 76.0 OR 76 (kJ K1 mol1) (cid:57) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 878 OR 877.9 OR 877.86, award 2 marks -------------------------------------------------------------------- (Minimum temperature when) G = 0 OR HTS = 0 OR (For feasibility) G = 0 OR G < 0 OR HTS < 0 OR T = H S T = 115 0.131 = 878 K (cid:57) (cid:57) Total<br />

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