OCR A Jun 2017 Paper 3 Q1 (with explained solutions) |

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OCR A Jun 2017 Paper 3 Q1

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Answer all the questions. 1 Within the permafrost in Arctic regions of the Earth, large amounts of methane are trapped within ice as methane hydrate, CH4xH2O. Methane makes up about 13.4% of the mass of methane hydrate. Scientists are concerned that global warming will melt the permafrost, releasing large quantities of methane into the atmosphere. (a) The HOH bond angle in ice is about 109 but about 105 in gaseous H2O. Explain why there is this difference.[3] (b) Why are scientists concerned about the release of methane into the atmosphere?[1] (c) Determine the formula of methane hydrate, CH4xH2O. In the formula, show the value of x to two decimal places. formula =[2]<br />
(d) Calculate the volume of methane, in dm3, that would be released from the melting of each 1.00 kg of methane hydrate at 101 kPa and 0 C. Give your answer to three significant figures. volume =dm3 [4] (e) Suggest why some industries are interested in the presence of methane hydrate in regions of the Earth.[1]<br />

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Answer Mark Scheme Marks Guidance June 2017 ThroughoutALLOW bonding regions for bonded pairsALLOW diagrams for communicating two bonds, two lone pairs and hydrogen bonding in iceIce IGNORE responses about open lattice/tetrahedral structure in ice Ice has hydrogen bonds/bondingH2O(g) 2 bonded pairs AND 2 lone pairsRepulsion Lone pairs repel more (than bonded pairs)It increases/causes/contributes to global warming OR CH bonds vibrate OR absorb IRFIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = CH45.74 H2O OR 5.74 award 2 marks ---------------------------------------------------------------------------- Mole ratio n(CH4) : n(H2O) = 13.4 16.0 OR 0.8375 : 4.81186.6 18.0 ALLOW more hydrogen bonding/H bonds For H2O(g),ALLOW waterIGNORE hydrogen bonding ALLOW it is a greenhouse gas/increases temp IGNORE ozone, radicals OR acid rain Working to at least 3 SF but IGNORE trailing zeroes, e.g. ALLOW 16 for 16.0 ------------------------------------------------------------------ ALLOW algebraic approach, e.g. H432A/03 Question (a) (b) (c) Formula CH45.74 H2O OR 5.74(d) FIRST CHECK THE ANSWER ON THE ANSWER LINE n(CH4) = n(CH4xH2O) 13.4 16.0 x = 5.74 16.0 + 18x 100 ALLOW ECF from incorrect mole ratio ----------------------------------------------- For 1 mark, ALLOW x with < 2 DP:x = 5.7x = 6x = 5.73x = 5.71 from 0.8375 and 4.8 from 0.84 and 4.811 from 0.84 and 4.8<br />
H432/03 Question Mark Scheme June 2017 Answer IF answer = 188 (dm3) AND use of ideal gas equation Award 4 marks for calculation ---------------------------------------------------------------------------- n(CH4) in 1 kg n(CH4) = 1103 16.0 13.4 100 = 8.375 OR 8.38 (mol)Rearranging ideal gas equation V = nRT Substitution of values into V = nRTCalculated value of n(CH4) (Use ECF)R = 8.314 OR 8.31T in K:p in Pa OR kPa 273 K 101 OR 101103 OR 1.01105 e.g. 8.3758.314273 (101103) OR 8.3758.314273 101 Marks Guidance ALLOW use of M(answer to (c) OR 119.32 Examples From n(CH45.74 H2O) 1103 119.32 = 8.38(1)188 (dm3) From n(CH45.7 H2O) 1103 118.6 = 8.43(2)189 (dm3) From n(CH46 H2O) 1103 124.0 = 8.06 (mol)181 (dm3) ---------------------------------------------------------------- IF V = is omitted, ALLOW when values are nRT substituted into rearranged ideal gas equation. Final volume in dm3 to 3 SF V = 188 (dm3)COMMON ERRORS Use of 298 K ALLOW ECF 3 marks max Example n(CH45.74 H2O) = 8.375V = 8.3758.314298 101103205 (dm3)ALLOW ECF from n(CH4) 2 marks max for n(CH4) and V in dm3 Use of 24.0 dm3 OR 22.4 dm3 24.0 dm3 22.4 dm3 13.4% (13.4/100) omitted n(CH45.74 H2O) = 8.375n(CH45.74 H2O) = 8.375n = = 62.5 (mol)1103 16 V = 8.37524.0 = 201 (dm3)V = 8.37522.4 = 188 (dm3)3 marks V = 62.58.314273 1011031400 (dm3)(e) For fuel OR energyALLOW responses linked with energy. e.g.to generate electricity<br />
Mark Scheme June 2017 Guidancefor burning/heat ALLOW (chemical) feedstock IGNORE cooking H432/03 Question Answer Marks Total 11<br />

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