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OCR A Jun 2011 Paper 5 Q3
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![In aqueous solution, methanoic acid, HCOOH, reacts with bromine, Br2. HCOOH(aq) + Br2(aq) 2H+(aq) + 2Br (aq) + CO2(g) A student carried out an investigation on the rate of this reaction. The student used a large excess of methanoic acid which ensured that its concentration was effectively constant throughout. During the reaction, bromine is used up and its orange colour becomes less intense. The intensity of the bromine colour can be measured with a colorimeter to give the bromine concentration. The graph below was plotted from the experimental results. 0.010 0.009 0.008 0.007 0.006 0.005 0.004 0.003 0.002 0.001 100 200 300 500 600 700 400 time / s [Br2] / mol dm3OCR 2011<br />
In this investigation, a large excess of methanoic acid was used. Under these conditions, the reaction is effectively zero-order with respect to methanoic acid.Using the graph, determine the order of reaction with respect to bromine.Using the graph, determine the initial rate of the reaction. Calculate the rate constant, k, for the reaction between methanoic acid and bromine under these conditions. (cid:4) In your answer you should make clear how your conclusions fit with the experimental results, including working shown on the graph and units where appropriate.[9]OCR 2011 [Total: 9] Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN11_question_3-0.jpg)
![F325 Question Answer 3 Evidence of at least two half-lives measured on graph OR within text (would need evidence of two half-lives) (cid:57) Any half-life value stated in range 180220 s OR constant half-life (cid:57) 1st order (cid:57) Note: This is only correct response for order (ie no ECF). If not stated separately, this mark can be awarded from a rate equation, e.g. rate = k[Br2]1 OR rate = k[Br2] Evidence of tangent on graph drawn to line at t = 0 s (cid:57) e.g. Mark Scheme June 2011 Mark Guidance ANNOTATE ALL Q3 WITH TICKS AND CROSSES, etc MARK ON GRAPH OR IN TEXT LOOK FOR STATEMENT ON GRAPH OR WITHIN TEXT ALLOW almost constant half-life ------------------------------------------------------------------------------ Note: Response may use an alternative approach from half-life for the 1st two marks based on gradients of tangents: 1st mark would be awarded for evidence of two tangents drawn on graph 2nd mark would be awarded for stating that ratio of concentrations = ratio of rates, e.g. gradient of tangent at 0.010 mol dm3 has twice the value of gradient of tangent at 0.005 mol dm3 ---------------------------------------------------------------------------- MARK TANGENTS ON GRAPH ALLOW some leeway but tangent must coincide with part of curve that is straight (ie between [Br2] = 0.0100.009 and MUST NOT cross the curve<br />
F325 Question Answer 3 = 0.000040 OR 4.0 x 105 (cid:57) rate = 0.010 250 units: mol dm3 s1 (cid:57) (cid:57) (cid:57) rate = k[Br2] OR k = rate [Br2] k = calculated result from calculated value for rate units: s1 (cid:57) 0.010 Mark Scheme June 2011 Mark Guidance ALLOW values from 1 SF (0.00004 OR 4 x 105) up to calculator value, correctly rounded ALLOW range ~ 0.010 160 to 0.010 300 : i.e. ALLOW a calculated gradient in the range 6 x 1053 x 105 from a tangent drawn at t = 0 IF tangent is drawn on graph at a different time or incorrectly (e.g. crossing curve), then mark rate calculation by ECF using the gradient of the tangent drawn by the candidate (ie not the range above). IF no tangent is drawn ALLOW a value in the range above ONLY Credit only attempts at tangents, not just a random straight line IGNORE asign DO NOT ALLOW rate = k[Br], ie Br instead of Br2 DO NOT ALLOW just k[Br2], ie rate = OR r = must be present Calculation of k is from candidates calculated initial rate From 0.00004, k = 0.000040 0.010 Note: IF order with respect to Br2 has been shown as 2nd order, then mark this part by ECF, e.g. if Br2 shown to be 2nd order, rate = k[Br ]2 calculated value for rate k = calculated result from units: dm3 mol1 s1 OR mol Note: Units mark must correspond to the candidates stated rate equation, NOT an incorrectly rearranged k expression 0.0102 1 dm s = 4 x 103 s 1 Total<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN11_answer_3-0.jpg)
