OCR A Jun 2017 Paper 1 Q25 (with explained solutions) |

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OCR A Jun 2017 Paper 1 Q25

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25 This question is about alkenes. (a) The combustion of ethene is shown in equation 25.1 below. 22 C2H4(g) + 3O2(g)2CO2(g) + 2H2O(g) H = 1318 kJ mol1 equation 25.1 (i) Explain, in terms of bond breaking and bond forming, why a reaction can be exothermic.[1] (ii) Average bond enthalpies are shown in the table. Bond OH O=O CH C=O Average bond enthalpy /kJ mol1 +464 +498 +413 +805 Calculate the average bond enthalpy of the C=C bond. Use the average bond enthalpies in the table and equation 25.1. average bond enthalpy =kJ mol1 [3]<br />
(b) An alkene D is a liquid at room temperature and pressure but can easily be vaporised. When vaporised, 0.1881 g of D produces 82.5 cm3 of gas at 101 kPa and 373 K. 23 Determine the molar mass and molecular formula of alkene D. Show all your working. molar mass =g mol1 molecular formula =[5] END OF QUESTION PAPER<br />

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H032/01 Question Answer 2 5 (a) (i) More energy is released by forming bonds than energy required when breaking bondsMark Scheme Marks (ii ) FIRST, CHECK THE ANSWER ON ANSWER LINE IF bond enthalpy = (+)612 (kJ mol1) award 3 marks IF bond enthalpy = ()316 (kJ mol1) award 2 marks Energy for bonds made ( 4C=O + 4OH ) 4805 + 4464 OR 3220 + 1856 OR 5076 (kJ)Energy for bonds broken ( 4CH + 3O=O ) 4413 + 3498 OR 1652 + 1494 OR 3146 (kJ)C=C bond enthalpy correctly calculated C=C bond enthalpy = 13183146 + 5076 = (+)612 kJ mol1Mark is for answer (b) FIRST check the molar mass on answer line MUST be derived from pV = nRT, Award 4 marks for calculation for:answer = 70OR answer that rounds to 69.9 OR 70.0 Guidance June 2017 bond breaking is endothermic AND bond making is exothermic ORA Response needs link between energy, breaking and making bonds ALLOW ALLOW within labelled energy diagram FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------- - IGNORE sign IGNORE sign ---------------------------------------- ALLOW ECF DO NOT ALLOWsign COMMON ERRORS + 2106 3248 omission of 3O=O 2 marks 1318 + 31465076 2 marks FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------- - If there is an alternative answer, check to see if there is any ECF credit possible using working 15<br />
H032/01 Question Mark Scheme June 2017 Answer Marks Guidance -------------------------------------------------------------------------- Rearranging ideal gas equation to make n subject n = pV RT Substituting all values including conversion to Pa and m3 (101103)(82.5106) n = 8.314373 n = 2.686930731032.69103 (mol)unrounded rounded to 3 SF Calculation of molar mass, M M = m n 0.1881 2.68693073103 = 70(.0) (g mol1) 0.1881 2.69103 = 69.9 (g mol1)C5H10Molecular formula of D ------------------------------------------------------------------- IF candidate has failed to derive suitable value of n, ALLOW value of M from 0.1881 AND 24000 with alkene closest to calculated value for last 2 marks See Guidance column. below 1st mark may be implicit by direct substitution of correct values below into rearranged equation. ONLY award this mark if n has been derived from correct rearranged ideal gas equation ALLOW 3 SF up to calculator value, correctly rounded NOTE: ALLOW 69.970.0 AND 70 (2 SF) Calculator from unrounded: 70.00552634 ALLOW any unambiguous structure ALLOW ECF provided that formula given is an alkene and matches M calculated from 0.1881 AND pV = nRT ------------------------------------------------------------------- 0.1881 0.1881 M = 82.5/24000 OR 3.4375103 = 54.72 OR 54.7 OR 55ALLOW 54.68 from use of 3.44103 From 54.72, ONLY ALLOW = C4H8Total 16<br />

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