Edexcel Jun 2011 Paper 4 Q17 (with explained solutions) |

Skip to content

Edexcel Jun 2011 Paper 4 Q17

Answers available below

17 Ammonia is manufactured using the reaction N2(g) + 3H2(g) (cid:85) 2NH3(g) (a) (i) Calculate (cid:39)S (cid:57) system for this reaction at 298 K. Give your answer in J mol1 K1 and include a sign. You will need to refer to your data booklet. [Note that the standard molar entropy values for gaseous diatomic elements are given for half a mole of molecules, and not per mole of molecules eg entropy for 1 mol of N2 is 295.8 J mol1 K1.] (ii) Using ideas about disorder, explain whether the sign of your answer to (a)(i) is as expected. PMT (2) (2)(b) At 700 K, the enthalpy change for this reaction, (cid:39)H = (cid:237)110.2 kJ mol1. (i) Calculate the entropy change of the surroundings, (cid:39)Ssurroundings, at 700 K. Include a sign and units in your answer. (2) 14 *P38482A01424*<br />
(ii) Calculate (cid:39)Ssystem for this reaction at 700 K. At this temperature the total entropy change, (cid:39)Stotal = (cid:237)78.7 J K(cid:237)1 mol(cid:237)1. Include a sign and units in your answer. (iii) What does the value of (cid:39)Stotal, which is (cid:237)78.7 J K(cid:237)1 mol(cid:237)1 at 700 K, indicate about the relative proportions of nitrogen, hydrogen and ammonia at equilibrium? PMT (1) (1)(c) A mixture of nitrogen, hydrogen and ammonia is at equilibrium at 150 atm. The partial pressures of nitrogen and ammonia in the mixture are 21 atm and 36 atm respectively. (i) Write an expression for the equilibrium constant, Kp, for the formation of ammonia, in terms of partial pressures for this reaction, and calculate its value at 700 K. Include units in your answer. (4) *P38482A01524* 15 Turn over<br />
PMT (ii) In the manufacture of ammonia, pressures of between 100 and 250 atm are used. State and explain one advantage, in terms of the yield of ammonia, of using a pressure above 100 atm. (1)*(iii) In the manufacture of ammonia, a temperature of about 700 K is used. For this exothermic reaction how does (cid:39)Ssurroundings change as temperature increases? Explain how this change affects the value of (cid:39)Stotal and the equilibrium constant as temperature increases. Hence explain the disadvantage of using a temperature higher than 700 K. (4)(iv) Suggest one advantage of using a temperature higher than 700 K. (1)(Total for Question 17 = 18 marks) 16 *P38482A01624*<br />

Show answer

Question Number 17 (a)(i) Question Number 17 (a)(ii) Mark Reject 198 Reject Just entropy decreases Mark Acceptable Answers Ssystem, = ((2 x 192.3) (2 x 95.8)(2x3x65.3)) (1) = -198.8 / -199 (J mol-1 K-1 ) Allow200 (2 SF) If units are not those in which data is given, must be correct. (1) Note check working Correct answer without working (2) Correct choice of multiples and data but wrong answer scores first mark (1) Correct value with wrong sign based on entropy of reactantsentropy of products (giving +199) (1) TE for second mark if multiples for hydrogen, nitrogen and ammonia are missed/ incorrect, but correct data used. or multiples correct and one error in data. Acceptable Answers If answer to (a)(i) is negative: Disorder decreases / order increases (as reaction goes forward) (1) Reference to order or disorder required for the mark. As number of (gas)molecules/moles/particles decreases (1) OR 4 moles of gas produces 2 moles Ignore comments on number of different types of molecule in equilibrium mixture If answer to (a)(i) is positive: Must say this is unexpected with correct reasons to score 2 marks No marks if the positive answer is expected 6CH04/01 1106<br />
Question Number 17 (b)(i) Question Number 17 (b)(ii) Question Number 17 (b)(iii) Reject Mark Acceptable Answers Ssurr = -(-110.2 x 1000)/ 700 (1) (+157.4285) = (+) 157.4 / 157 (J mol-1 K-1) OR (+) 0.1574 / 0.157 kJ mol-1 K-1 (1) Ignore sf except 1 Correct answer without working (2) Correct value with negative sign (1) Use of Ssurr = -H/T but wrong answer (1) Acceptable Answers Reject (Ssystem = Stotal - Ssurr ) =(-78.7-157.4)) = -236.1/ -236 (J mol-1 K-1 ) OR -0.2361 / -0.236 ( kJ mol-1 K-1 ) Allow -235.7 if 157 used and -238.7 if 160 used Ignore units unless value in kJ given as J or vice versa TE from (b)(i) values in kJ added to values in J Acceptable Answers Reject Reactants predominate / more nitrogen and hydrogen (than ammonia) Just Equilibrium lies to the left Just no ammonia is present. The gases are present in ratio 1:3:2 Mark Mark 6CH04/01 1106<br />
Question Number 17 (c)(i) Question Number 17 (c)(ii) Mark Mark Acceptable Answers Reject Kp = (pNH3)2/ (pN2)(pH2)3 (1) Can be written in other formats eg p2NH3 etc pH2 = (150 21 36) = 93 (atm) (1) Kp =((36)2/ (21)(93)3) = (7.6724994 x 105) =7.67 x 105 (1) Ignore sf except 1 TE on incorrect pH2 atm2 (1) TE for units on incorrect Kp expression Correct answer including units without quoting Kp expression scores 3 Square brackets in first mark No TE for value on incorrect Kp Expression Units other than atm Acceptable Answers Reject Just equilibrium moves right (Yield of ammonia is increased) because there are fewer moles / molecules (of gas) on the right OR System tries to reduce the pressure by going to the side with fewer moles/ molecules (of gas) Ignore comments about value of Kp changing Ignore comments about more collisions occurring/more molecules having energy greater than or equal to activation energy 6CH04/01 1106<br />
Question Number *17 (c)(iii) Question Number 17 (c)(iv) Reject Mark Acceptable Answers First mark At higher temperature Ssurr is less positive/ decrease/more negative (1) Second mark making Stotal more negative / less positive/decreases No TE for 2nd mark if Ssurr is said to increase. (1) Third mark (so) Kp decreases (1) Third mark depends on second mark being correct/neutral answer Fourth mark so equilibrium position further left /in endothermic direction/ in reverse direction OR lower yield of ammonia / reaction is less feasible (1) Fourth mark is a stand alone mark Acceptable Answers Reject Rate (of reaching equilibrium)is higher / faster Ignore comments about increasing numbers of successful collisions at higher temperature Mark 6CH04/01 1106<br />

Question 15 16 17 18 19

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.