9701_w08_qp_4
A paper of Chemistry, 9701
Questions:
10
Year:
2008
Paper:
4
Variant:
0

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Use In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide ions in acidic solution. H2O2 + 2I– + 2H+ 2H2O + I2 This reaction is considered to go by the following steps. step 1 H2O2 + I– IO– + H2O step 2 IO– + H+ HOI step 3 HOI + H+ + I– I2 + H2O The general form of the rate equation is as follows. rate = k[H2O2]a[I–]b[H+]c Suggest how the appearance of the solution might change as the reaction takes place. Suggest values for the orders a, b and c in the rate equation for each of the following cases. case numerical value a b c step 1 is the slowest overall step 2 is the slowest overall step 3 is the slowest overall A study was carried out in which both [H2O2] and [H+] were kept constant at 0.05 mol dm–3, and [I–] was plotted against time. The following curve was obtained. 0.001 0.0009 0.0008 0.0007 0.0006 0.0005 0.0004 0.0003 0.0002 0.0001 120 150 180 210 240 270 300 time / s [I– ion] / mol dm–3 Examiner’s Use To gain full marks for the following answers you will need to draw relevant construction lines on the graph opposite to show your working. Draw them using a pencil and ruler. Calculate the initial rate of this reaction and state its units. rate = units Use half-life data calculated from the graph to show that the reaction is first order with respect to [I–]. Use the following data to deduce the orders with respect to [H2O2] and [H+], explaining your reasoning. [H2O2] / mol dm–3 [H+] / mol dm–3 relative rate 0.05 0.05 1.0 0.07 0.05 1.4 0.09 0.07 1.8 order with respect to [H2O2] = order with respect to [H+] = From your results, deduce which of the three steps is the slowest (rate determining) step.
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