Rates of reaction - kinetics (A2)
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I am doing my coursework on the iodine clock reaction. I have found potassium iodide and potassium peroxodisulphate both to be 1st order and I have worked out the activation energy with and without the catalyst (Fe2+) ions - which was 21 kJ mol-1without and 19 kJ mol-1 with the catalyst. I worked these out by calculating the gradient of the graph log 1/time against the recipricol of temperature (1/temp K-1) then I multiplied this by the gas constant. However they seem too close together. Do I need to repeat the temperature experiments again? I used 3 repeats for each.
Also, is it possible to find the rate constant in the rate equation by multiplying the gradient of the graph to find the order for reactant A by that of reactant B? I found a site which suggested this as they were both 1st order, but I'm unsure if it would work. When I tried substituting the different rate values I got results that changed with each concentration - decreasing with the concentration.
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Ulex writes
Firstly, are your values for the activation energy too close for comfort? I think not. From the Arrhenius equation, it is the logarithm of k which depends on Ea. Consequently a small change in Ea results in a small change in ln k which gives a large change in k itself. There is an analogy here with the Richter scale of earthquake violence, which is also logarithmic. An earthquake measuring 5 on the Richter scale is not 1.25 times as violent than one measuring 4, it is ten times as violent!
The problem in using 1/t to represent rate is that it is only proportiona to the rate, it is not the rate itself, which would have to be expressed in moles per dm3 per second. This is fine when using the Arrhenius equation, but it won’t do if you are to get a meaningful value for k.
It may be that you can go back to your order determinations and work out what the real rate of reaction was at a particular reactant concentration, substitute this rate and concentration into the first order rate equation and find the apparent value of k. Then do the same for the other order determination, find another value for k and multiply the two values together to get an overall rate constant.
Having done all that – is it really worth doing? You can’t actually do anything with the rate constant when you have got it! The only thing you can do with a rate constant is to compare it with other rate constants and you can only do that if the units are the same!
It might be very much more interesting to speculate about what is going on in the reaction and how the catalyst works.
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updated: 14 April 2008
