AQA Jun 2017 Paper 1 Q9 (with explained solutions) |

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AQA Jun 2017 Paper 1 Q9

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17There are several stages in the industrial production of methanol from methane. The first stage involves a gaseous equilibrium between the reactants (methane and steam), and some gaseous products. Figures 1 and 2 show the percentage conversion of methane into the gaseous products under different conditions at equilibrium. Figure 1 Figure 2 Deduce the optimum conditions for the industrial conversion of methane and steam into the gaseous products. Explain your deductions. [6 marks] 0 9 0 9 . 1 *17* Turn overIB/M/Jun17/7405/1<br />
18*18* IB/M/Jun17/7405/1<br />
19The equation shows the final stage in the production of methanol. CO(g) + 2H2(g)CH3OH(g) 0 9 . 2 20.1 mol of carbon monoxide and 24.2 mol of hydrogen were placed in a sealed container. An equilibrium was established at 600 K. The equilibrium mixture contained 2.16 mol of methanol. Calculate the amount, in moles, of carbon monoxide and of hydrogen in the equilibrium mixture. [2 marks] Amount of carbon monoxide Amount of hydrogen mol mol Question 9 continues on the next page *19* Turn overIB/M/Jun17/7405/1<br />
200 9 . 3 A different mixture of carbon monoxide and hydrogen was allowed to reach equilibrium at 600 K At equilibrium, the mixture contained 2.76 mol of carbon monoxide, 4.51 mol of hydrogen and 0.360 mol of methanol. The total pressure was 630 kPa Calculate a value for the equilibrium constant, Kp, for this reaction at 600 K and state its units. [6 marks] Value of Kp Units 14 *20* IB/M/Jun17/7405/1<br />

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MARK SCHEMEA-LEVEL CHEMISTRY7405/1JUNE 2017 Answers Mark Additional Comments/Guidance Indicative Chemistry content Stage 1: Deductions from graph 1a Yield increases as temperature increases (or converse) 1b After a certain temperature yield no longer increases 1c Yield decreases as pressure increases (or converse) Stage 2: Optimum temperature and explanation 2a High temperature results in high energy costs/expensive 2b (After a certain temperature) yield no longer increases therefore there is no gain in using a higher temperature 2c Optimum temperature is between 780-880oC Stage 3: Optimum pressure and explanation 3a Low pressure may be too slow 3b So compromise pressure required 3c Optimum pressure is 1000-2000kPa or moderate pressure used 30 of 35 56 marks Level 2 34 marks Level 1 12 marks Question 09.1 This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. Level 3 All stages are covered and the explanation of each stage is generally correct and virtually complete. To access Level 3, statement 3a must be considered. Answer is communicated coherently and shows a logical progression from stage 1 (including 1b) to stage 2 and stage 3 All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows progression from stage 1 to stage 2 and/or stage 3. Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. Answer includes isolated statements but these are presented in a logical order, with sensible reasoning.<br />
MARK SCHEMEA-LEVEL CHEMISTRY7405/1JUNE 2017 Level 0 0 marks Insufficient correct chemistry to gain a mark. Question Answers Mark Additional Comments/Guidance Moles of carbon monoxide 17.9 09.2 Moles of hydrogen 19.9 Allow 17.94 Allow 19.88 (CH3OH) ignore brackets p = (CO)(H2)2 Total moles of gas = (2.76 + 4.51 + 0.36) = 7.63 pp(CO) = x 630 (kPa) (= 228 (kPa)) pp(H2) = x 630 (kPa) (= 372 (kPa)) 09.3 pp(CH3OH) = x 630 (kPa) (= 29.7 (kPa)) Kp = = 9.4(1) x 10-7 or 9.4(1) x 10-13 if pp in Pa can also score M1 from this expression kPa2 or Pa-2 (if converted to 630 000) Total If Kp expression incorrect can only score M2 & M3 & M4 If CE in M2 allow ecf for M3, M4 and M6 If no total moles calculated then can only score M1 and M6 All 3 pp of CO, H2 and CH3OH = 2 marks 2 pp correct = 1 mark Allow 9.39 to 9.50 x 10-7 (kPa-2) If no marks awarded allow M6 only for kPa2 or Pa-2 1 1 14 31 of 35<br />

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