OCR A Jun 2013 Paper 5 Q2 (with explained solutions) |

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OCR A Jun 2013 Paper 5 Q2

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In aqueous solution, benzenediazonium chloride, C6H5N2Cl, decomposes above 10 C. C6H5N2Cl(aq) + H2O(l) C6H5OH(aq) + N2(g) + HCl(aq) A student investigates the rate of this reaction using an excess of water at 50 C. The student takes measurements at intervals during the reaction and then plots his experimental results to give the graph shown below. [C6H5N2Cl (aq)] / mol dm3 6.0 5.0 4.0 3.0 2.0 1.0 20 40 60 80 100 120 140 160 180 time / s (a) The student uses half-life to suggest the order of reaction with respect to C6H5N2Cl. (i) What is meant by the half-life of a reaction?[1] (ii) Confirm the order of reaction with respect to C6H5N2Cl. Show your working on the graph.[2]OCR 2013<br />
(iii) What would be the effect, if any, on the half-life of this reaction of doubling the initial concentration of C6H5N2Cl ?[1] (b) The student predicts that the rate equation is: rate = k [C6H5N2Cl]. (i) Using the graph and this rate equation, determine the rate of reaction after 40 s. Show your working on the graph. (ii) Calculate the rate constant, k, for this reaction and give its units. rate = units[3] k =units[2] (c) The order of this reaction with respect to H2O is effectively zero. Explain why.[1] [Total: 10]OCR 2013 Turn over<br />

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F325 Mark Scheme June 2013 Question 2 (a) Answer (i) Time for concentration (of reactant) to fall to half original value (cid:57) (ii) At least two half-lives correctly shown on graph AND half-life stated as approx. 54 s (cid:57) 1st order has a constant half-life (cid:57) (iii) No change (cid:57) (b) (i) Tangent = 0.045 OR 4.5 x 102 (cid:57) On graph, tangent drawn to curve at t ~ 40 s (cid:57) Calculation of rate from the tangent drawn e.g. rate = 5.2 116 Units mol dm3 s1 (cid:57) Independent mark Marks 1 3 Guidance ALLOW time for concentration to fall by half DO NOT ALLOW concentration of product to fall by half ALLOW mass OR amount as alternative to concentration ALLOW time for reactant/substance/atoms to decrease by half ALLOW half-life in range 5056 s ALLOW half-life shown on graph Care: Initial concentration is ~5.8 and NOT 6.0 For constant half-life, ALLOW half lives are the same, two half-lives are 54 s, etc. ALLOW 2 tangents drawn, one at half conc of first AND evidence that gradient ( rate) halves Annotate tangent on graph Note: This mark can only be awarded from a tangent ALLOW ECF for tangent drawn at different time from 40 s ALLOW 10% of gradient of tangent drawn ALLOW 2 SF up to calculator value ALLOW trailing zeroes, e.g. 0.04 for 0.040 IGNOREsign for rate Note: IF candidate calculates rate via ln 2 method (shown in (ii), consult with TL)<br />
Answer Mark Scheme Marks F325 Question (ii) 2 (b) (cid:57) 3.45 k = answer to (b)(i) units: s1 (cid:57) Independent mark June 2013 Guidance = 0.013 From 0.045, k = 0.045 3.45 ALLOW concentration range 3.43.5 ALLOW use of unrounded calculator answer from (b)(i) even if different from answer given on (b)(i) answer line Many will keep this value in calculator for (b)(ii) ALLOW k = ln 2/t1/2 = 0.693/half life from (a)(iii) For 54 s, k = 0.693/54 = 0.013 ALLOW 2 SF up to calculator value IGNORE water does not affect the rate (c) water is in excess OR concentration of H2O is very large/does not change (cid:57) Total 10<br />

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