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OCR A AS Specimen Paper 2 Q3
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![3 Nitrogen can be reacted with hydrogen in the presence of a catalyst to make ammonia in the Haber process. N2(g) + 3H2(g) 2NH3(g) H = 92 kJ mol1 (a) Describe and explain the effect of increasing the pressure on the rate of this reaction.(b) A mixture of N2 and H2 was left to react until it reached equilibrium. The equilibrium mixture had [2] . . A higher temperature: . . [3]OCR 2014 H032/02 the following composition: (i) Calculate a value for Kc for this equilibrium. 1.20 mol dm3 2.00 mol dm3 0.877 mol dm3 N2 H2 NH3 Specimen mixture. Use of a catalyst: (ii) Explain how the following changes would affect the amount of NH3 present in the equilibrium Kc =dm6 mol2 [3]<br />
(c) 1.00 tonne of ammonia from the Haber process is reacted with carbon dioxide to prepare the fertiliser urea, NH2CONH2. 2NH3(g) + CO2(g)NH2CONH2(s) + H2O(l) 1.35 tonnes of urea are formed. Calculate the percentage yield of urea. Show all your working. Specimen yield =% [3]OCR 2014 H032/02 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/Specimen_QP_-_Paper_2_OCR_Chemistry_A_AS-Level_3-0.jpg)
![H032/02 Question 3 (a) (b) Answer (Increase in pressure) increases the rate AND because molecules are closer together so there are more collisions per unit time (cid:57) (cid:57) (i) Expression: Kc = [NH3]2 / [H2]3[N2] (cid:57) Calculation: = (0.877)2/(2.00)3(1.20) (cid:57) = 0.0801 (cid:57) (dm6 mol2) (ii) Catalyst: No effect, it only changes the rate of reaction (cid:57) Higher temperature: Forward reaction is exothermic (cid:57) so position of equilibrium moves to the left and there will be less NH3 (cid:57) Mark Scheme June 20XX Marks Guidance ALLOW more particles per unit volume NOT molecules move faster or have more energy Square brackets required ALLOW from 1 sig fig up to calculator display Correct answer alone scores all marks Specimen 10 P M T<br />
H032/02 Question (c) Mark Scheme June 20XX Answer Marks Guidance FIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = 76.5 (%) award 3 marks n(NH3) = (1106) / 17 = 5.88104 (58824) (mol) AND Theoretical yield: n(NH2CONH2) = 5.88104 / 2 = 2.94104 (29412) (mol) (cid:57) Actual yield: n(NH2CONH2) = 1.35106 / 60 = 2.25104 (22500) (mol) (cid:57) % yield = (2.94104 / 2.25104)100% = 76.5(%) (cid:57) If there is an alternative answer, check to see if there is any ECF credit possible using working below ALLOW up to full calculator display For 2nd and 3rd marks, ALLOW calculation in mass. Theoretical mass yield: m(NH2CONH2) = 605.88104 / 2 = 1.764 tonne (cid:57) % yield = (1.35 / 1.764)100 = 76.5% (cid:57) ALLOW 76% (2 sig figs) up to calculator answer correctly rounded from previous values ALLOW ECF from calculated actual and theoretical yields Specimen Total 11 11 P M T<br />](https://learnah.org/wp-content/uploads/2018/08/Specimen_MS_-_Paper_2_OCR_Chemistry_A_AS-Level_3-0.jpg)
