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Edexcel Jun 2014 (IAL) Paper 4 Q19
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![19 (a) A student investigated the kinetics of the reaction in which an acidified, aqueous solution of hydrogen peroxide oxidizes iodide ions to iodine. H2O2(aq) + 2H+(aq) + 2I(aq) (cid:314) 2H2O(l) + I2(aq) (i) The results of three experiments are shown in the table below. Experiment number [H2O2(aq)] / mol dm3 [H+(aq)] / mol dm3 [I(aq)] / mol dm3 Initial rate of reaction / mol dm3 s1 0.5 1.0 0.5 1.0 1.0 2.0 0.5 0.5 0.5 7.0 (cid:117) 106 1.4 (cid:117) 105 7.0 (cid:117) 106 Determine the order with respect to hydrogen peroxide and H+(aq) ions. Explain your reasoning. Order with respect to H2O2(aq) (3)Order with respect to H+(aq)14 *P42992A01428*<br />
(ii) The student carried out three more experiments in which the initial concentrations of hydrogen peroxide and H+(aq) ions were kept constant and the initial concentration of iodide ions was changed. A graph of the results is shown below. Initial rate / mol dm3 s1 [I(aq)] / mol dm3 State the order with respect to I(aq) and explain your reasoning. (2)(iii) Use your answers to (a)(i) and (a)(ii) to write the rate equation for the reaction. (iv) Use the data from the table and your answer to (a)(iii) to calculate the value for the rate constant. Include units in your answer. (1) (2) *P42992A01528* 15 Turn over<br />
(b) Nitrogen dioxide reacts with carbon monoxide. NO2(g) + CO(g) (cid:314) NO(g) + CO2(g) The reaction is second order with respect to nitrogen dioxide, and zero order with respect to carbon monoxide. (i) Suggest a possible two-step mechanism for this reaction. Write the rate determining step first. Rate determining step (2)Step 2(ii) Calculate the standard entropy change of the system, (cid:507)S(cid:57) system, for the reaction between nitrogen dioxide and carbon monoxide at 298 K. Include a sign and units in your answer. You will need to use the standard molar entropies on pages 20 and 27 of the Data Booklet. (iii) Given that (cid:507)H(cid:57) 298 for the reaction between nitrogen dioxide and carbon monoxide at 298 K is 226 kJ mol1, calculate the standard entropy change of the surroundings, (cid:507)S(cid:57) surroundings, for this reaction at 298 K. Include a sign and units in your answer. (2) (2) 16 *P42992A01628*<br />
(iv) Calculate the total entropy change, (cid:507)S(cid:57) total, for this reaction at 298 K. State what the sign of (cid:507)S(cid:57) total indicates about this reaction at 298 K. (2)(Total for Question 19 = 16 marks) *P42992A01728* 17 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2014_IAL_QP_-_Unit_4_Edexcel_Chemistry_A-level_19-0.jpg)
![Acceptable Answers Question Number 19(a)(i) First marking point - Orders Reject Mark Either/both explanations if inconsistent with order(s) stated above H2O2 first/1st order/order 1 H+ zero/0 order/order 0 both correct (1) Explanations Second marking point H2O2 - as [H2O2] doubles and [H+] and [I] stay the same/other concentrations stay the same/using experiments 1 and 2 (this could be shown on the table) and the rate doubles/working to show this ALLOW reverse argument ie as [H2O2] halves and [H+] and [I] stay the same/other concentrations stay the same/using experiments 1 and 2 (this could be shown on the table) and the rate halves/working to show this (1) Third marking point H+ - as [H+] doubles/halves and [H2O2] and [I] stay the same/other concentrations stay the same/using experiments 1 and 3 (this could be shown on the table) Notedo not penalise omission of this if it has been penalised in second marking point and the rate stays the same /working to show this OR As [H+] doubles and [H2O2] halves and [I] stays the same /using experiments 2 and 3 (this could be shown on the table) (also see note above) and rate halves due to [H2O2] change so rate does not change due to [H+]/working to show this ALLOW reverse argument (1) WCH0401 1406<br />
Acceptable Answers Question Number 19(a)(ii) First marking point First/1st order /order 1 (1) Second marking point-consequential on correct order graph shows rate is (directly) proportional to [I] OR as [I] increases, the rate increases proportionally OR as [I] doubles, the rate doubles ALLOW graph (of rate against [I]) is a straight line (through the origin) /gradient is constant ALLOW Increase in rate is constant (1) Reject Mark Any other order scores zero overall iodine/I2 Penalise once only in (ii) and (iii) Question Number 19(a) (iii) Question Number 19(a) (iv) Acceptable Answers Reject Mark rate/r/R = k[H2O2][I] ALLOW [H2O2]1[I]1 ALLOW [H+]0 [I2] ALLOW upper case K consequential on their orders from (a)(i) and (ii) Acceptable Answers Reject Mark 2.8 x 10-5 (1) dm3mol-1s-1 (1) ALLOW units in any order Note: value and units are consequential on their rate equation from (a)(iii) and must be consistent Ignore SF unless 1. Comment If zero order wrt [I], k = 1.4 x 10-5 s-1 WCH0401 1406<br />
Reject Mark NO2 or any multiple other than 2 of NO2 on LHS for first mark only Acceptable Answers Question Number 19(b)(i) (Rate determining step) Any balanced equation that has just 2NO2 on the LHS (1) (Step 2) Any balanced equation for which the two steps combine to the overall equation/double the overall equation (1) Examples of matching pairs: 2NO2N2O4 N2O4 + CONO + NO2 + CO2/ N2O4 + 2CO2NO + 2CO2 OR 2NO22NO + O2 2CO + O22CO2/CO + 1/2O2CO2 OR 2NO2N2 + 2O2 N2 + 2O2 + CONO + NO2 + CO2 OR 2NO2 2NO + 2O NO + 2O + CONO2 + CO2 OR 2NO2NO + NO3 2CO + NO32CO2 + NO ALLOW NO2 + NO2 on LHS NOTE NO2 NO + 1/2O2 then CO + 1/2O2 CO2 Does not score the first mark but scores 1 for the consequential second equation IGNORE state symbols even if incorrect WCH0401 1406<br />
Acceptable Answers Question Number 19(b)(ii) In (b)(ii) penalise incorrect/missing units but allow Reject Mark J/mol K and J/mol/K and lower case k FIRST CHECK THE FINAL ANSWER, If answer is -13.3 J mol-1 K-1 or -0.0133 kJ mol-1 K-1, award 2 marks 13.3 J mol-1 K-1 /0.0133 kJ mol-1 K-1 (sign omitted ) (1) -13.3 / -0.0133 (units omitted) (1) If none of the above answers is given First mark for correct data used (1) Second mark value, sign and units consequential on incorrect entropy value(s) used in the correct expression (So (1) IGNORE SF except 1SF system =) [210.7 + 213.6][197.6 + 240.0] Acceptable Answers Question Number 19(b)(iii) FIRST CHECK THE FINAL ANSWER, Reject Mark -0.75839 / -758.39 with no units If final answer is +0.75839 kJ mol-1 K-1 (2) +758.39 J mol-1 K-1 (2) EITHER (So surroundings =) H OR -(-226) (1) T 298 = 0.75839 kJ mol-1 K-1 (1) Second mark is for answer with correct unit OR (So surroundings =) H OR-(-226000) (1) T 298 = 758.39 J mol-1 K-1 (1) Second mark is for answer with correct unit ALLOW -0.75839 kJ mol-1 K-1 /-758.39 J mol-1 K-1 (1) IGNORE SF except 1 SF WCH0401 1406<br />
Acceptable Answers Question Number 19(b)(iv) First marking point Reject Mark total = So system + So surroundings ) (ii) and (iii) added together with different units for first mark only (So (= -13.3 + 758.39) / (-0.0133 + 0.75839) =(+)745.09 J K-1 mol-1/(+)0.74509 kJ K-1 mol-1 (1) TE on (ii) and (iii) added together with the same units IGNORE SF except 1 SF IGNORE units, even if incorrect Second marking point (sign is positive so) reaction is (thermodynamically) feasible / spontaneous (1) ALLOW feasible / will occur / reaction goes / it reacts (at 298 K) reactants thermodynamically unstable COMMENT If value for So consequential mark for (sign is negative so) reaction is not feasible / not spontaneous / will not occur / reaction / it does not take place (at 298 K) total is negative, then allow (Total for Question 19 = 16 marks) WCH0401 1406<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_IAL_MS_-_Unit_4_Edexcel_Chemistry_A-level_19-0.jpg)
