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OCR A Jun 2012 Paper 5 Q2
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![2 A chemist carries out an investigation on the equilibrium system shown below. 2CO(g) + 2NO(g) 2CO2(g) + N2(g) H = 788 kJ mol1 The chemist mixes 0.46 mol of CO with 0.45 mol of NO. The mixture is left to reach equilibrium at constant temperature. The student analyses the equilibrium mixture and finds that 0.25 mol NO remains. The total volume of the equilibrium mixture is 1.0 dm3. (a) (i) Write the Kc expression for this equilibrium. (ii) What are the units of this equilibrium constant?[1] [1] (iii) Determine the value of Kc for this equilibrium mixture. Show all your working. Kc =[4]OCR 2012<br />
(iv) What does your value of Kc suggest about the position of equilibrium in this experiment?[1] (b) The chemist increases both the temperature and the pressure of the equilibrium mixture. The mixture is left to reach equilibrium again. (i) What is the effect, if any, on the value of Kc? Explain your answer.[1] (ii) Explain why it is difficult to predict what would happen to the position of equilibrium after these changes in temperature and pressure.[2] [Total: 10]OCR 2012 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_question_2-0.jpg)
![F325 Mark Scheme June 2012 Question (a) (i) (ii) Answer (Kc = ) [CO2]2 [N2] dm3 mol1 (cid:57) [CO]2 [NO]2 (cid:57) Marks Guidance Square brackets required for ALL four concentrations ALLOW mol1 dm3<br />
F325 Mark Scheme June 2012 Question Answer Marks Guidance (a) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 0.95 award 4 marks -------------------------------------------------------------------- Equilibrium amounts: n(CO) = 0.460.20 = 0.26 mol (cid:57) n(CO2) = 0.2(0) mol (cid:57) n(N2) = 0.1(0) mol (cid:57) Kc calculation Must use calculated equilibrium amounts AND 0.25 (Kc = ) 0.202 (cid:117) 0.10 0.262 (cid:117) 0.252 = 0.95 (dm mol 1 ) (cid:57) ANNOTATIONS MUST BE USED IF there is an alternative answer, apply ECF by checking working for intermediate marks ---------------------------------------------------------------------------- APPLY ECF from incorrect starting n(CO) By ECF, n(N2) = n(CO2)/2 For all parts, ALLOW numerical answers from 2 significant figures up to the calculator value Correct numerical answer with no working scores 4 marks ALLOW calculator value: 0.946745562 down to 0.95 (2SF), correctly rounded, e.g. 0.947 IGNORE units, even if incorrect --------------------------------------------------------------- Common errors 1.89 3 marks use of n(N2) = 0.2(0) mol 3 0.202 (cid:117) 0.20 0.262 (cid:117) 0.252 = 1.893491124 (dm mol 0.45 and 0.46 swapped over 1 ) (cid:57) (Kc = ) 1.29 3 marks 1.024 3 marks 0.45 used twice n(CO) = 0.450.21 = 0.24 mol (cid:57) n(CO2) = 0.21 mol (cid:57) n(N2) = 0.105 mol (cid:57) 0.212 (cid:117) 0.105 0.242 (cid:117) 0.252 = 1.28625 (dm mol (Kc = ) 1 ) (cid:57) n(CO) = 0.450.20 = 0.25 mol (cid:57) n(CO2) = 0.2(0) mol (cid:57) n(N2) = 0.1(0) mol (cid:57) 0.202 (cid:117) 0.10 0.252 (cid:117) 0.252 = 1.024 (dm mol (Kc = ) 0.46 used twice n(CO) = 0.460.21 = 0.25 mol (cid:57) n(CO2) = 0.21 mol (cid:57) n(N2) = 0.105 mol (cid:57) 0.212 (cid:117) 0.105 0.252 (cid:117) 0.252 = 1.185408 (dm mol (Kc = ) 1 ) (cid:57) 1 ) (cid:57) 1.185 3 marks<br />
F325 Mark Scheme June 2012 Question Answer (a) (iv) Mark ECF from (iii) (b) (i) (ii) Effect of T and P on equilibrium IF Kc from (iii) < 1 equilibrium to left/towards reactants OR IF Kc from (iii) > 1 equilibrium to right/towards products (cid:57) Kc has decreased AND H is negative OR (forward) reaction is exothermic (cid:57) (increased) temperature shifts equilibrium to left AND (increased) pressure shifts equilibrium to right AND fewer (gaseous) moles on right-hand side (cid:57) Overall effect on equilibrium Difficult to predict relative contributions of two opposing factors (cid:57) Marks Guidance First look at Kc value for (iii) at bottom of cut ------------------------------------------------- ALLOW favours reverse reaction For correct Kc value in (iii) of 0.95, ALSO ALLOW equilibrium position near to centre (cid:57) Statement AND reason required for mark ALLOW for reason: reverse reaction is endothermic Reason ONLY required for pressure Temperature and H had been required in (i) ALLOW ratio of (gas) moles is 4:3 ALLOW opposing effects may not be the same size ALLOW effects could cancel each other out ALLOW effects oppose one another DO NOT ALLOW just it is difficult to predict equilibrium position (in question) For the 2nd mark, we are assessing the idea that we dont know which factor is dominant Total 10<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_answer_2-0.jpg)
