IODINE CLOCK COURSEWORK A LEVEL

We can now examine theoretically, the effect of changing individual concentrations on the rate of reaction of a more complicated rate expression of the form.. Collecting a gaseous product in a gas syringe or inverted burette. A plot of HI concentration versus time above was curved showing it could not be a zero order reaction with respect to the concentration of HI. A small and constant amount of sodium thiosulfate and starch solution is added to the reaction mixture. Then you would get two negative gradients one steeper than the other for the greater concentration. From experimental results you need to know how the speed of a reaction varies with respect to individual reactant concentrations. Therefore the order with respect to A is 2 or 2nd order.

You just substitute the values into the full rate expression: These examples do NOT involve graphs directly, but a ‘graphical’ section of examples has been added in section 5. Since the gradient rate changes with concentration, it cannot be a zero order reaction. The following rate data was obtained at 25 o C for the reaction: You can think the other way round i.

To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it. Orders of reaction can ONLY be determined by rate experiments The rate of reaction was is then plotted against HI concentration to test for 1st order kinetics.

iodine clock coursework a level

Some possible graphical results are shown above. We can now examine theoretically, the effect of changing individual concentrations on the rate of reaction of a more complicated rate expression of the form. CH 3 3 C—Cl concentration. The table below gives leve initial courswork for the reaction: In the zero order graph the gradient is constant as the rate is independent of concentration, so the graph is of a linear descent in concentration of reactant.

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The oxidation of iodide to iodine by potassium peroxodisulfate can be followed by a method known as the ‘ iodine clock ‘. From runs i and iikeeping [B] constant, by doubling [A], the rate is doubled, so 1st order with respect to reactant A.

This proved that the decomposition of hydrogen iodide reaction is a 2nd order reaction.

Iodine clock coursework level

For more details see section 7. It is the constancy of the half—life which proves the 1st order levek. The same argument applies if you imagine the graph inverted and you were following the depletion of a reactant. The mathematics of 1st order rate equations units.

Some rate data for the inversion of sucrose is given below. Analysing a single set of data to deduce the order of reaction. Therefore the reaction is 2nd order overalland the rate expression is The graph on the left illustrates the initial rate method for the formation of product. For latest updates see https: This proved to be a curve – compare the blue rate data curve with the black ‘best straight line’ courtesy of Excel!

Therefore the reaction is 1st order overalland the rate expression is Oodine plot of HI concentration versus time above was curved showing it could not be a zero order reaction with respect to the concentration of HI.

The rate of radioactive decay is an example of 1st order kinetics. I’ve made the numbers quite simple to follow the logic of the argument. From the point of view of coursework projects the detailed analysis described above is required, but quite often in examination questions a cock limited amount of data is given and some clear logical thinking is required.

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Iodine clock coursework level

The reacting mole ratio coursewofk 2: The numbers in bold show the factor change in concentration and its effect on the rate. Kinetics of the thermal decomposition of hydrogen iodide.

iodine clock coursework a level

Only then is it aa to derive a rate expressionwhich summarises what controls the speed of a particular reaction in terms of the relevant concentrationswhich is not necessarily all the reactants! Of course [4] to [7] could simply represent inaccurate data!

The Iodine Clock – GCSE Science – Marked by

The graph below shows what happens to a reactant with a half—life of 5 minutes. In reality the results would be not this perfect and you would calculate k for each set of results and quote the average! An individual order of reaction is the power to which the concentration cpock is raised in the rate expression.

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