Rates of reaction - kinetics (A2)
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Hello! I was asked by my teacher to do the typical experiment of reacting iodine with propanone. We are not using the colorimetry method but the quenching with thiosulphate(VI) one. We are asked to plot a graph of the time at which samples of the reaction mixture were added to the sodium hydrogencarbonate solution (x-axis) against the volume of thiosulphate needed to react with the remaining iodine (y-axis). I don't understand how the rate of reaction can be determined from the graph, because from my understanding, rate of reaction = concentration/time, so why is volume/time relevant? How can I determine the rate of reaction of iodine and propanone? Please, please, please help me! Thanks a million.
Igloo writes ...
The reaction between propanone and iodine:
CH3COCH3(aq) + I2(aq) ----> CH3COCH2I(aq) + HI(aq)
is catalysed by acid, so the quenching of the reaction is achieved by adding samples to an excess of aqueous sodium hydrogencarbonate (not thiosulphate, as you mention in your question). This mixture, which has now virtually stopped reacting, can be titrated with a standard solution of sodium thiosulphate to find out how much iodine remains in the mixture.
The titration reading is proportional to the concentration of iodine in the mixture, so there is no need to carry out a calculation in each case to convert these titres into iodine concentrations. Instead you can plot raw titration readings (which are effectively iodine concentrations) against time. As you will know, the shape of this curve is linked to the order of reaction, so you need to do the following:
Plot the curve, then study the plot:
If it is a straight line (the curve has a constant gradient), the reaction is zero order.
If it is a curve with a constant half-life, the reaction is first order.
If it is a curve with a steadily increasing half-life, which is inversely proportional to concentration, the reaction is second order.
I use the phrase “order of reaction”, but this needs further explanation. If in your experiment you used propanone, iodine and acid solutions of equal concentration then the order you have measured in an overall order of reaction. If, however, two of the reactants are in excess then the measured order is the one with respect to the reactant not in excess. For example, if you used propanone of concentration 1.0 mol dm-3, sulphuric acid of the same concentration, but the iodine solution was only 0.02 mol dm-3, then the concentrations of propanone and acid will remain virtually constant throughout whilst that of the iodine decreases from 0.02 mol dm-3 down to zero, and your investigation is effectively measuring the order of reaction with respect to iodine.
I am aware that your question asks about how the actual rates of reaction can be determined, but I suspect that you thought this necessary in order to calculate orders of reaction. You don’t need to, as I have explained above.
If I’ve missed the point and gone off-track, please re-submit your question to React making it clear what you want.
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updated: 14 December 2006
