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OCR A Jun 2015 Paper 5 Q3
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![This question looks at two reactions involving sulfur compounds. (a) Hydrogen reacts with carbon disulfide as shown below. 4H2(g) + CS2(g) CH4(g) + 2H2S(g) For this reaction, H = 234 kJ mol1 and S = 164 J K1 mol1. (i) Why does the reaction have a negative entropy change?[1] (ii) Standard entropies are shown in the table below. substance S o / J K1 mol1 CS2(g) 238 CH4(g) 186 H2S(g) 206 Calculate the standard entropy for H2. S o =J K1 mol1 [2] (iii) Explain, with a calculation, whether this reaction is feasible at 25 C. Show your working.[3] (iv) Explain, with a calculation, the significance of temperatures above 1154 C for this reaction.[2]OCR 2015 Turn over<br />
(b) A chemist investigated methods to improve the synthesis of sulfur trioxide from sulfur dioxide and oxygen. 2SO2(g) + O2(g) 2SO3(g) The chemist:mixed together 1.00 mol SO2 and 0.500 mol O2 with a catalyst at room temperaturecompressed the gas mixture to a volume of 250 cm3 allowed the mixture to reach equilibrium at constant temperature and without changing the total gas volume. At equilibrium, 82.0% of the SO2 had been converted into SO3. (i) Determine the concentrations of SO2, O2 and SO3 present at equilibrium and calculate Kc for this reaction. Kc =units[6]OCR 2015<br />
(ii) Explain what would happen to the pressure as the system was allowed to reach equilibrium.[1] (iii) The value of Kc for this equilibrium decreases with increasing temperature. Predict the sign of the enthalpy change for the forward reaction. State the effect on the equilibrium yield of SO3 of increasing the temperature at constant pressure. H:Effect on SO3 yield:[1] (iv) The chemist repeated the experiment at the same temperature with 1.00 mol SO2 and an excess of O2. The gas mixture was still compressed to a volume of 250 cm3. State and explain, in terms of Kc, how the equilibrium yield of SO3 would be different from the yield in the first experiment.[3] [Total: 19]OCR 2015 Turn over<br />
10 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGEOCR 2015<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN15_question_3-0.jpg)
![F325 Question 3 (a) (i) 5 mol/molecules (of gas) forms 3 mol/molecules (of gas) (cid:57) Answer (a) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer =(+)131 (J K1 mol1) , award 2 marks164 = (186 + 2206)(4S + 238) OR 4 S = 164 + (186 + 2206)238 (cid:57) S = (+)131 (J K1 mol1) (cid:57) Mark Scheme June 2015 Marks Guidance ALLOW reaction forms fewer moles/molecules IF stated, numbers of molecules MUST be correct IGNORE comments related to (cid:168)G OR disorder (even if wrong) NOTE: IF any values are omitted, DO NOT AWARD any marks. e.g. 164 may be missing ALLOW FOR 1 mark 131 wrong final sign 49 wrong sign for 164 79.5 no use of 2 524 no division by 4 38 wrong sign for 186 75 wrong sign for 206 250 wrong sign for 238 Any other number: CHECK for ECF from 1st marking point for expressions using ALL values with ONE error only e.g. one transcription error:, e.g.146 for 164 12<br />
F325 Question 3 Mark Scheme June 2015 Answer Marks Guidance (a) (iii) NOTE: DO NOT ALLOW answer to 3(a)(ii) for (cid:168)G calculation(cid:168)G calculation: 2 marks (cid:168)G = 2342980.164 (cid:57) = 185 (kJ mol1) (cid:57) IGNORE units (even if wrong) 185 subsumes 1st mark) Feasibility comment for negative (cid:168)G answer: 1 mark (Forward) reaction is feasible / spontaneous AND (cid:168)G < 0 / (cid:168)HT(cid:168)S < 0 (cid:57) (a) (iv) ((cid:168)G =) 2341427164 1000 = 0 (calculator 0.028(kJ) OR 28 (J)) (cid:57) 2nd mark only available if 1st mark has been awarded (Above 1427K/1154C), reaction is not feasible/not spontaneous(cid:57) OR 1427 K is maximum temperature that reaction happens 2 1 ALLOW (cid:168)G correctly calculated from 3 SF up to calculator value of 185.128 ALLOW working in J, ie: (cid:168)G = 234000298164 (cid:57) = 185000 (J mol1) (cid:57) ALLOW 1 mark for use of 25 OR mixture of kJ and J, e.g. (cid:168)G = 234250.164 = 229.9 (cid:168)G = 234298164 = +48638 ALLOW ECF if calculated value for (cid:168)G is +ve Then correct response for 3rd mark would be not feasible/not spontaneous AND (cid:168)G > 0 / (cid:168)HT(cid:168)S > 0 ALLOW (When (cid:168)G = 0) T = 234 0.164 For 2nd mark, IF (cid:168)G is +ve from (a)(iii) ALLOW ECF for: Above 1427 K, reaction is feasible / spontaneous OR 1427 K is minimum temperature that reaction happens IGNORE LESS feasible IGNORE comparisons of the signs of T(cid:168)S and (cid:168)H, e.g IGNORE T(cid:168)S is more negative than (cid:168)H = 1427 K OR 234000 164 = 1427 K 13<br />
F325 Question 3 (b) Mark Scheme June 2015 Answer Marks Guidance (b) (ii) (Pressure) decreases AND fewer molecules/moles (cid:57) 14 (i) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 57.6 dm3 mol1, award 6 marks IF answer = 57.6 with incorrect units, award 5 markEquilibrium amounts in mol 2 MARKS n(SO2) = 0.180 (mol) ALL 3 correct: (cid:57)(cid:57) n(O2) = 0.090 (mol) ANY 2 correct: (cid:57) n(SO3) = 0.820 (mol) Equilibrium concentrations (moles4) 1 MARK SO2 = 0.720 (mol dm3) AND O2 = 0.360 (mol dm3) AND SO3 = 3.28 (mol dm3) (cid:57) Calculation of Kc and units Kc = [SO3 ]2 OR [SO2 ]2 [O2 ] = 57.6 (cid:57) dm3 mol1 (cid:57) At least 3SF is required 3.282 (0.720)2(0.360) 3 MARKS (cid:57) FULL ANNOTATIONS NEEDED IF there is an alternative answer, check to see if there is any ECF credit possible using working belowALLOW ECF from incorrect moles of SO2, O2 AND SO2 ALL three concentrations required for this mark ALLOW ECF from incorrect concentrations NO ECF for numerical value with a square missing For Kc, ALLOW 3 significant figures up to calculator value of 57.64746228 correctly rounded For units, ALLOW mol1 dm3 DO NOT ALLOW dm3/mol ALLOW ECF from incorrect Kc expression for both calculation and units COMMON ERRORS 0.0294 from SO2 = 0.820, O2 = 0.410, SO3 = 0.180 (mol) For fewer moles, ALLOW 3 mol2 mol ALLOW more moles of reactants 3 marks + units mark<br />
(b) (iii) (cid:168)H is negative // ve AND yield of SO3 decreases (cid:57) Answer Mark Scheme June 2015 F325 Question 3 (b) (iv) IGNORE le Chatelier responses --------------------------------------------------------------------------------- Each marking point is independent Kc Kc does not change (with pressure/ concentration) (cid:57) Comparison of conc terms with more O2 [O2]/concentration of oxygen is greater OR denominator/bottom of Kc expression is greater (cid:57) QWC: yield of SO3 linked to Kc (Yield of) SO3 is greater/increases AND numerator/top of Kc expression is greater/increases (cid:57) Total 15 Marks Guidance IGNORE exothermic and endothermic FULL ANNOTATIONS NEEDED ALLOW Kc only changes with temperature IF 1st marking point has been awarded, IGNORE comments about Kc decreasing or Kc increasing and assume that this refers to how the ratio subsequently changes. i.e DO NOT CON 1st marking point. IGNORE O2 is greater/increases ALLOW (Yield of) SO3 is greater/increases AND to reach/restore Kc value (cid:57) 19<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN15_answer_3-0.jpg)
