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Edexcel Jun 2011 Paper 4 Q16
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![PMT 16 Nitrogen(IV) oxide, NO2, is a brown gas which is a pollutant in air. It is produced in the reaction below. 2NO(g) + O2(g) (cid:111) 2NO2(g) (a) The table below shows the results of a series of experiments to measure the rate of this reaction at 298 K. Experiment number 1 2 3 Initial concentration / mol dm(cid:237)3 [NO(g)] [O2(g)] 0.0050 0.0125 0.0125 0.0100 0.0100 0.0250 Initial rate / mol dm(cid:237)3 s(cid:237)1 5.1010(cid:16)4 10.210(cid:16)4 40.810(cid:16)4 (i) State, with reasons, the order of reaction with respect to oxygen and the order of reaction with respect to nitrogen(II) oxide, NO. (2)12 *P38482A01224*<br />
(ii) Write the rate equation for the reaction. (iii) Calculate the value of the rate constant. Include units in your answer. (b) Nitrogen(IV) oxide in air reacts with carbon monoxide in car exhausts. The following two-step reaction mechanism has been suggested. 2NO2(g) (cid:111) NO(g) + NO3(g) Step 1: Slow Step 2: NO3(g) + 2CO(g) (cid:111) NO(g) + 2CO2(g) Fast (i) Write the equation for the overall reaction which takes place. (ii) The overall reaction is second order. Suggest a rate equation for this reaction, justifying your answer. PMT (1) (2) (1) (2)(Total for Question 16 = 8 marks) *P38482A01324* 13 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2011_QP_-_Unit_4_Edexcel_Chemistry_A-level_16-0.jpg)
![Question Number 16 (a)(i) Question Number 16 (a)(ii) Question Number 16 (a)(iii) Question Number 16 (b)(i) Acceptable Answers Reject O2 : first order as increasing [O2] x 2 increases rate x 2 / as rate is (directly) proportional to oxygen concentration (1) (Experiments 1 and 2 or [NO] constant) NO: second order as increasing [NO] x 2 increases rate x 4/ by 22 (1) (Experiments 2 and 3 or [O2] constant) Two correct orders with no explanation (1) only Two correct orders based on stoichiometry Acceptable Answers Reject Rate = k [O2][NO]2 Rate equation must be consistent with answer in (a)(i) Just k [O2][NO]2 i.e. no rate/R Non square brackets Reject Acceptable Answers Rate = k[O2][NO]2 TE from (i) k=((5.10 x 104 )/(0.005)(0.0125)2) = 652.8 / 653/650 OR k=((10.2 x 104)/(0.0100)(0.0125)2) = 652.8 / 653/650 OR k=((40.8 x 104 )/(0.0100)(0.025)2) = 652.8 / 653/650 (1) TE for value of k from rate equation given dm6 mol-2 s-1 (allow any order) (1) Acceptable Answers Reject NO2 + CONO + CO2 Allow multiples Equation not cancelled down eg NO3 on both sides. Mark Mark Mark Mark 6CH04/01 1106<br />
Question Number 16 (b)(ii) Mark Acceptable Answers Reject Equations involving CO to power other than zero Rate = k[NO2]2 OR Rate = k[NO2]2[CO]0 OR Rate = k[NO2]2[CO]0 [NO3]0 (1) Only molecules/reactant in slow step are (2)NO2 OR CO appears after the rate determining/slow step (and 2NO2 molecules in slow step) OR CO is not involved in rate determining / slow step OR Only the molecules in the slow step are in the rate equation OR Step 1 is slowest so determines rate equation (1) Second mark: No TE on rate equation containing incorrect species. Only allow TE if k missing in correct rate equation 6CH04/01 1106<br />](https://learnah.org/wp-content/uploads/2018/08/June_2011_MS_-_Unit_4_Edexcel_Chemistry_A-level_16-0.jpg)
