Edexcel Jan 2012 Paper 4 Q21 (with explained solutions) |

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Edexcel Jan 2012 Paper 4 Q21

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21 This question is about the kinetics of the reaction between bromoethane and aqueous hydroxide ions. (a) The results of an experiment to find the initial rate of the reaction are shown in the table below. [CH3CH2Br] / mol dm(cid:237)(cid:22) 0.100 [OH(cid:237)] / mol dm(cid:237)(cid:22) 0.150 Initial rate / mol dm(cid:237)(cid:22) s(cid:237)(cid:20) 1.5410(cid:237)(cid:25) The rate equation for the reaction is rate = k[CH3CH2Br][OH(cid:237)] (i) Calculate the value of k. Give your answer to three significant figures and include units. (ii) Calculate the initial rate if the concentrations of both reactants were changed to 0.020 mol dm(cid:237)(cid:22). (b) (i) State the order of the reaction. PMT (3) (1) (1)(ii) The mechanism for this reaction can be inferred from the rate equation. Draw the transition state formed in the reaction between bromoethane and hydroxide ions. (2) *P39304A01524* 15 Turn over<br />
(c) The rate constant for the reaction between bromoethane and hydroxide ions was determined at five different temperatures. The results are shown in the table below. PMT Temperature (T) / K 1/Temperature (1/T) / K(cid:237)(cid:20) Rate constant, k (cid:21)(cid:28)(cid:22) 303 313 323 333 3.4110(cid:237)(cid:22) 3.3010(cid:237)(cid:22) (cid:22)(cid:17)(cid:20)(cid:28)(cid:3)(cid:238)(cid:3)(cid:20)(cid:19)(cid:237)(cid:22) 3.1010(cid:237)(cid:22) 5.8310(cid:237)(cid:24) 1.6710(cid:237)(cid:23) 5.2610(cid:237)(cid:23) 1.3610(cid:237)(cid:22) 3.7710(cid:237)(cid:22) (i) Complete the missing values in the table. ln k (cid:237)(cid:28)(cid:17)(cid:26)(cid:24) (cid:237)(cid:27)(cid:17)(cid:26)(cid:19) (cid:237)(cid:26)(cid:17)(cid:24)(cid:24) (cid:237)(cid:25)(cid:17)(cid:25)(cid:19) (ii) Plot a graph of ln k against 1/T. Calculate the gradient of your graph and use this to calculate the activation energy, EA. The Arrhenius equation can be expressed as E A R1T + a constant ln k(cid:3) (cid:3) (cid:3) (cid:62)(cid:42)(cid:68)(cid:86)(cid:3)(cid:70)(cid:82)(cid:81)(cid:86)(cid:87)(cid:68)(cid:81)(cid:87)(cid:15)(cid:3)(cid:53)(cid:3)(cid:32)(cid:3)(cid:27)(cid:17)(cid:22)(cid:20)(cid:3)(cid:45)(cid:3)(cid:46)(cid:237)(cid:20)(cid:3)mol(cid:237)(cid:20)] (2) (5) 16 *P39304A01624*<br />
PMT 1/T/K(cid:237)(cid:20) ln k (Total for Question 21 = 14 marks) TOTAL FOR SECTION B = 49 MARKS *P39304A01724* 17 Turn over<br />

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Question Number 21(a)(i) k = (1.54 x 106)(0.1 x 0.15) (1) Acceptable Answers Mark Reject 1.02 x 104 (= 1.0267 x 10-4) = 1.03 x 104 (1) must be to 3 SF dm3 mol1 s1 (1) Unit mark is stand alone and units can be in any order Correct answer with units but no working (3) marks Question Number 21(a)(ii) If correct unrounded answer to (a) (i) stored in Acceptable Answers Reject Mark calculator then 4.1067 x 10-8 = 4.1 x 108 (mol dm3 s1) OR If 1.0267 x 10-4 used then 4.1068 x 10-8 = 4.1 x 108 (mol dm3 s1) OR If 1.03 x 104 used then 4.12 x 10-8 = 4.1 x 108 (mol dm3 s1) IGNORE sf except 1sf IGNORE units even if incorrect TE from (a)(i) Question Number 21(b)(i) 2(nd)/second/two/(1 + 1) = 2 (order) Acceptable Answers Reject Mark 6CH0401 1201<br />
Question Number 21(b)(ii) Acceptable Answers Reject Mark 6CH0401 1201<br />
Question Number 21(c)(ii) Appropriate scale (1) Acceptable Answers Reject Mark K1 Plotted points must cover at least half of the graph paper on each axis. Points plotted correctly and straight line drawn (1) through all points Gradient = 10230500 (1) Example Ea = 10230 x 8.31(1) allow TE from incorrect gradient Ea = (+) 85.0 kJ(mol1)/(+) 85 000 J (mol-1) (1) 3 sf Ea range from 80.9 to 89.2 kJ mol-1 ALLOW TE from incorrect gradient IGNORE SF except 1 TOTAL FOR SECTION B = 49 MARKS 6CH0401 1201<br />
Mark Mark Mark Acceptable Answers Section C Question Number 22 (a)(i) Question Number 22(a)(ii) (266.9 + 186.2)310.1 (1) (+)186.2 (J mol-1 K-1) Acceptable Answers Reject Reject = + 143 (J mol-1 K-1) (1)143 scores (1) Correct answer with sign and no working scores (2) marks ALLOW TE from (i) Question Number 22(a)(iii) Yes, as reaction produces 2 molecules/moles from Acceptable Answers Reject one/more molecules/moles (1) (and) all products are gases (1) IGNORE references to volumes More moles/molecules of gas produced scores (2) OR Yes, (as the reaction is endothermic) Ssurroundings is negative (1) Since the reaction takes place/goes (spontaneously) Stotal is positive and therefore Ssystem is positive (1) ALLOW TE from (a)(ii) i.e. No, as. 6CH0401 1201<br />

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