OCR A Jan 2012 Paper 5 Q6 (with explained solutions) |

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OCR A Jan 2012 Paper 5 Q6

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Free energy changes can be used to predict the feasibility of processes. 16 (a) Write down the equation that links the free energy change with the enthalpy change and temperature.[1] (b) You are provided with equations for five processes. For each process, predict the sign of S. sign of S 2CO(g) + O2(g) NaCl(s) + (aq) process 2CO2(g) NaCl(aq) H2O(l) H2O(s) Mg(s) + H2SO4(aq) CuSO4(s) + 5H2O(l) MgSO4(aq) + H2(g) CuSO45H2O(s) [2] (c) Ammonia can be oxidised as shown in the equation below. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Standard entropies are given in the table below. substance S o/ J K1 mol1 NH3(g) 192 O2(g) 205 NO(g) 211 H2O(g) 189 Calculate the standard entropy change, in J K1 mol1, for this oxidation of ammonia. S o =J K1 mol1 [2]OCR 2012<br />
(d) The exothermic reaction below occurs spontaneously at low temperatures but does not occur at very high temperatures. 17 Explain why. 2SO2(g) + O2(g) 2SO3(g)[2] (e) An ore of iron contains iron(III) oxide, Fe2O3. Iron is extracted from this ore by heating with carbon. The equation below shows one of the reactions which takes place. Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g) S = +543 J K1 mol1 and H = +493 kJ mol1 Calculate the minimum temperature at which this reaction becomes feasible. Show all your working. minimum temperature =[3] [Total: 10] Turn overOCR 2012<br />

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F325 Mark Scheme January 2012 Question Expected answers 6 a b G = HTS (cid:57) process Marks Additional guidance (cid:57) (cid:57) (211 + 189)(192 + 205) incorrect sign ALLOW ECF from working line above from a single error --------------------------------------- COMMON ERRORS (+)3 (J K1 mol1)185 (J K1 mol1) ANNOTATIONS MUST BE USED DO NOT ALLOW just TS increases DO NOT ALLOW At high T, TS is greater (than H) APPROACH BASED ON TOTAL ENTROPY: With increasing temperature H/T is less negative OR H/T increases OR H/T decreases OR |H/T | decreases OR magnitude of H/T decreases (cid:51) ALLOW at high temperatures SH/T < 0 sign 2CO(g) + O2(g) (cid:127)(cid:111) 2CO2(g) NaCl(s) + (aq) (cid:127)(cid:111) NaCl(aq) H2O(l) (cid:127)(cid:111) H2O(s) Mg(s) + H2SO4(aq) (cid:127)(cid:111) MgSO4(aq) + H2(g) CuSO4(s) + 5H2O(l) (cid:127)(cid:111) CuSO45H2O(s) All 5 correct 4 correct (cid:127)(cid:111) 2 marks (cid:57)(cid:57) (cid:127)(cid:111) 1 mark (cid:57)++S = (4211 + 6189)(4192 + 5205) (cid:51) S = (+)185 (J K1 mol1) (cid:51)(cid:3) With increasing temperature TS is more negative OR TS decreases OR TS increases OR |TS| increases OR magnitude of TS increases (cid:51) At high temperature TS is more negative that H OR at high T, TS outweighs/is more significant than H OR At low temperature HTS < 0 OR At high temperature HTS > 0 (cid:51)(cid:3) 2 2 19<br />
F325 Mark Scheme January 2012 Marks Additional guidance OR S is more negative than H/T OR S outweighs/ is more significant than H/T ALLOW total entropy statement: S(total) = 0 OR S(total) >0 ALLOW 0 = 493T543 (cid:57) i.e. This mark focuses on G OR HTS being = 0 and NOT on conversion of S value into kJ K1 mol1 Mark temperature given on answer line ALLOW 3 SF up to calculator value 907.9189687 correctly rounded, e.g. 907.9, 907.92 ALLOW temperature in C: i.e. ALLOW by subtraction of 273: 635, 634.9, 634.91 oC ALLOW by subtraction of 273.15: 635, 634.8, 634.77 oC up to calculator value correctly rounded ALLOW C for C; K for K IF S has not been converted to kJ, DO NOT ALLOW 2nd mark BUTALLOW calculated answer = 493/543 = 0.91 K (calculator: 0.907918968) ALLOW 2 marks only for absence of one of the statements required for 1st marking point Question Expected answers 6 e (For feasibility,) G < 0 OR G = 0 OR 0 < HTS OR 0 = HTS OR 0 = 493T x 543/1000 (cid:57) T = H S = 908 K (cid:57) Units of temperature are required = 493 x 1000/543 (cid:57) Total 10 20<br />

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