iodine clock experiment order of reaction

There are two fundamentally different approaches to this - you can either investigate what happens to the initial rate of the reaction as you change concentrations, or you can follow a particular reaction all the way through, and process the results from that single reaction. We will look at these two approaches separately. Don't expect full practical details.

This is going to be a very long page. I wouldn't really recommend that you try to read it all in one go.

The simplest initial rate experiments involve measuring the time taken for some easily recognisable event to happen very early on in a reaction.

This could include the time taken for, say, 5 cm of gas to be produced. Or it could be the time taken for a small measurable amount of precipitate to be formed. Or you could measure the time taken for some dramatic colour change to occur. We will look at examples of all these below.

You then change the concentration of one of the components of the reaction, keeping everything else constant - the concentrations of other reactants, the total volume of the solution and the temperature and so on. Then you find the time taken for the same event to take place with that new concentration.

This is repeated for a range of concentrations of the substance you are interested in. You would need to cover a reasonably wide range of concentrations, taking perhaps 5 or so different concentrations varying from the original one down to half of it or less.

Obviously, you could then repeat the process by changing something else - the concentration of a different substance, or the temperature, for example.

We will take a simple example of an initial rate experiment where you have a gas being produced. This could be a reaction between a metal and an acid, for example, or the catalytic decomposition of hydrogen peroxide.

If you plotted the volume of gas given off against time, you would probably get the first graph below.

A measure of the rate of the reaction at any point is found by measuring the slope of the graph. The steeper the slope, the faster the rate. Since we are interested in the initial rate, we would need the slope at the very beginning.

If you then look at the second graph, enlarging the very beginning of the first curve, you will see that it is approximately a straight line at that point. That is only a reasonable approximation if you are considering a very early stage in the reaction. The further into the reaction you go, the more the graph will start to curve.

Measuring the slope of a straight line is very easy. The slope in this case is simply V/t.

Now suppose you did the experiment again with a different (lower) concentration of the reagent. Again, we will measure the time taken for the same volume of gas to be given off, and so we are still just looking at the very beginning of the reaction:

The initial rates (in terms of volume of gas produced per second) are:

Now suppose you didn't actually know what the volume V was.

Suppose, for example, that instead of measuring the time taken to collect 5 cm 3 of gas, you just collected the gas up to a mark which you had made on the side of a test tube. Does it matter?

If you are simply wanting to compare initial rates, then it doesn't matter. If you look at the expressions in the table above, you should recognise that the initial rate is inversely proportional to the time taken. In symbols:

In experiments of this sort, you often just use 1/t as a measure of the initial rate without any further calculations.

You can then plot 1/t as a measure of rate against the varying concentrations of the reactant you are investigating. If the reaction is first order with respect to that substance, then you would get a straight line. That's because in a first order reaction, the rate is proportional to the concentration.

If you get a curve, then it isn't first order. It might be second order - but it could equally well have some sort of fractional order like 1.5 or 1.78.

The best way around this is to plot what is known as a "log graph". The maths of this might not be familiar to you, but you may find that you are asked to do this as a part of a practical exam or practical exercise. If it is an exam, you would probably be given help as to how to go about it.

The maths goes like this:

If you have a reaction involving A, with an order of n with respect to A, the rate equation says:

rate = k [A] n

If you take the log of each side of the equation, you get:

log(rate) = log k + n log[A]

If you plotted log(rate) agains log[A], this second equation would plot as a straight line with slope n. If you measure the slope of this line, you get the order of the reaction.

So you would convert all the values you had for rate into log(rate). Convert all the values for [A] into log[A], and then plot the graph. This should be a straight line. If it isn't, then you have done something wrong! Measure the slope to find the order, n.

Note:   Don't worry if you don't understand logs (logarithms), or how I got from the first equation to the second one! I suspect that in the unlikely event of you needing it in an exam at this level, it would be given to you.

All you need to do is find the log button on your calculator and use it to convert your numbers. Practise to start with by trying to find log 2. It should give a value of 0.3010(etc). You probably have to enter 2 and then press the log button, but on some calculators it might be the other way around. If you do it the wrong way around, you will just get an error message.

Some sample reactions

The catalytic decomposition of hydrogen peroxide

This is a simple example of measuring the initial rate of a reaction producing a gas.

A simple set-up to do this might be:

The reason for the weighing bottle containing the catalyst is to prevent introducing errors at the beginning of the experiment. Since this is the part of the reaction you are most interested in, introducing errors here would be stupid!

You have to find a way of adding the catalyst to the hydrogen peroxide solution without changing the volume of gas collected. If you added it to the flask using a spatula, and then quickly put the bung in, you might lose some gas before you got the bung in. Alternatively, as you pushed the bung in, you might force some air into the measuring cylinder. Either way, it makes your results meaningless.

To start the reaction, you just need to shake the flask so that the weighing bottle falls over, and then continue shaking to make sure the catalyst mixes evenly with the solution.

You could also use a special flask with a divided bottom, with the catalyst in one side, and the hydrogen peroxide solution in the other. They are easy to mix by tipping the flask.

If you use a 10 cm 3 measuring cylinder, initially full of water, you can reasonably accurately record the time taken to collect a small fixed volume of gas.

You could, of course, use a small gas syringe instead.

If you were looking at the effect of the concentration of hydrogen peroxide on the rate, then you would have to change its concentration, but keep everything else constant.

The temperature would have to be kept constant, so would the total volume of the solution and the mass of manganese(IV) oxide. You would also have to be sure that the manganese(IV) oxide used always came from the same bottle so that its state of division was always the same.

You could, of course, use much the same apparatus to find out what happened if you varied the temperature, or the mass of the catalyst, or the state of division of the catalyst.

The thiosulphate-acid reaction

If you add dilute hydrochloric acid to sodium thiosulphate solution, you get the slow formation of a pale yellow precipitate of sulphur.

There is a very simple, but very effective, way of measuring the time taken for a small fixed amount of precipitate to form. Stand the flask on a piece of paper with a cross drawn on it, and then look down through the solution until the cross disappears.

So . . . you put a known volume of sodium thiosulphate solution in a flask. Then you add a small known volume of dilute hydrochloric acid, start timing, swirl the flask to mix everything up, and stand it on the paper with the cross on. Time how long it takes for the cross to disappear.

Then repeat using a smaller volume of sodium thiosulphate, but topped up to the same original volume with water. Everything else should be exactly as before.

If you started with, say, 50 cm 3 of sodium thiosulphate solution, you would repeat the experiment with perhaps, 40, 30, 20, 15 and 10 cm 3 - each time made up to a total of 50 cm 3 with water.

The actual concentration of the sodium thiosulphate doesn't have to be known. In each case, you could record its relative concentration. The solution with 40 cm 3 of sodium thiosulphate solution plus 10 cm 3 of water has a concentration which is 80% of the original one, for example. The one with 10 cm 3 of sodium thiosulphate solution plus 40 cm 3 of water has a concentration which is 20% of the original one.

When you came to plotting a rate against concentration graph, as we looked at further up the page, you would plot 1/t as a measure of the rate, and volume of sodium thiosulphate solution as a measure of concentration. Alternatively, you could plot relative concentrations - from, say, 20% to 100%. It doesn't actually matter - the shape of the graph will be identical.

You could also look at the effect of temperature on this reaction, by warming the sodium thiosulphate solution before you added the acid. Take the temperature after adding the acid, though, because the cold acid will cool the solution slightly.

This time you would change the temperature between experiments, but keep everything else constant. To get reasonable times, you would have to use a diluted version of your sodium thiosulphate solution. Using the full strength solution hot will produce enough precipitate to hide the cross almost instantly.

Iodine clock reactions

There are several reactions which go under the name "iodine clock". They are all reactions which give iodine as one of the products. This is the simplest of them, but only because it involves the most familiar reagents.

The reaction we are looking at is the oxidation of iodide ions by hydrogen peroxide under acidic conditions.

The iodine is formed first as a pale yellow solution darkening to orange and then dark red, before dark grey solid iodine is precipitated.

There is a very clever way of picking out a when a particular very small amount of iodine has been formed.

Iodine reacts with starch solution to give a very deep blue solution. If you added some starch solution to the reaction above, as soon as the first trace of iodine was formed, the solution would turn blue. That doesn't actually help!

However, iodine also reacts with sodium thiosulphate solution.

If you add a very small amount of sodium thiosulphate solution to your reaction mixture (including the starch solution), it will react with the iodine that is initially produced, and so the iodine won't affect the starch, and you won't get any blue colour.

However, when that small amount of sodium thiosulphate has been used up, there is nothing to stop the next lot of iodine produced from reacting with the starch. The mixture suddenly goes blue.

Note:   There is a neat piece of video on YouTube showing an iodine clock reaction (not necessarily the one I am talking about here, but it doesn't matter).

It shows three reactions side by side:

The right-hand one is done at room temperature.

The left-hand one is also done at room temperature, but at three times the concentration of one of the reagents.

The central one is colder than room temperature, and presumably at the same concentration as the right-hand one.

The blue colours appear in exactly the order you would predict.

In our example, you could obviously look at the effect of changing the hydrogen peroxide concentration, or the iodide ion concentration, or the hydrogen ion concentration - each time, of course, keeping everything else constant.

That would let you find the orders with respect to everything taking part in the reaction.

Following the course of a single reaction

Rather than doing a whole set of initial rate experiments, you can also get information about orders of reaction by following a particular reaction from start to finish.

There are two different ways you can do this. You can take samples of the mixture at intervals and do titrations to find out how the concentration of one of the reagents is changing. Or (and this is much easier!) you can measure some physical property of the reaction which changes as the reaction continues - for example, the volume of gas produced.

We need to look at these two different approaches separately.

Sampling the reaction mixture

Bromoethane reacts with sodium hydroxide solution as follows:

During the course of the reaction, both bromoethane and sodium hydroxide will get used up. However, it is relatively easy to measure the concentration of the sodium hydroxide at any one time by doing a titration with some standard acid - for example, with hydrochloric acid of a known concentration.

You start with known concentrations of sodium hydroxide and bromoethane, and usually it makes sense to have them both the same. Because the reaction is 1:1, if the concentrations start the same as each other, they will stay the same as each other all through the reaction.

So all you need to do is to take samples using a pipette at regular intervals during the reaction, and titrate them with standard hydrochloric acid in the presence of a suitable indicator.

That is a lot easier said than done!

The problem is that the reaction will still be going on in the time it takes for you to do the titration. And, of course, you only get one attempt at the titration. By the time you take another sample, the concentration of everything will have changed!

There are two ways around this.

You can slow the reaction down by diluting it, adding your sample to a larger volume of cold water before you do the titration. Then do the titration as quickly as possible. That's most effective if you are doing your reaction at a temperature above room temperature. Cooling it as well as diluting it will slow it down even more.

But if possible (and it is possible in the case we are talking about) it is better to stop the reaction completely before you do the titration.

In this case, you can stop it by adding the sample to a known volume (chosen to be an excess) of standard hydrochloric acid. That will use up all the sodium hydroxide in the mixture so that the reaction stops.

Now you would titrate the resulting solution with standard sodium hydroxide solution, so that you can find out how much hydrochloric acid is left over in the mixture.

That lets you calculate how much was used up, and so how much sodium hydroxide must have been present in the original reaction mixture.

This sort of technique is known as a back titration . These calculations can be quite confusing to do without some guidance. If you are interested, you will find back titrations discussed on pages 72-75 of my chemistry calculations book .

Processing the results

You will end up with a set of values for concentration of (in this example) sodium hydroxide against time. The concentrations of the bromoethane are, of course, the same as these if you started with the same concentrations of each reagent.

You can plot these values to give a concentration-time graph which will look something like this:

Now it all gets pretty tedious!

You need to find the rates of reaction at a number of points on the graph, and you do this by drawing tangents to the graph, and measuring their slopes.

You would then draw up a simple table of rate against concentration.

The quickest way to go on from here is to plot a log graph as described further up the page. You would convert all your rates into log(rate), and all the concentrations into log(concentration). Then plot log(rate) against log(concentration).

The slope of the graph gives you the order of reaction.

In our example of the reaction between bromoethane and sodium hydroxide solution, the order would turn out to be 2.

Notice that this is the overall order of the reaction - not just the order with respect to the reagent whose concentration you were measuring. The rate of reaction was falling because the concentrations of both of the reactants were falling.

Note:   This all takes ages to do - not just the practical which would probably take at least an hour, but all the graph drawing, and processing the results from the graphs. There is no obvious way this could be asked in any normal practical or written exam at this level.

I can see that it is just possible that you might be asked in principle how you would do it, but actually doing it could only reasonably be a part of a coursework exercise.

Following the course of the reaction using a physical property

An example where a gas is given off

A familiar example of this is the catalytic decomposition of hydrogen peroxide that we have already looked at above as an example of an initial rate experiment.

This time, you would measure the oxygen given off using a gas syringe, recording the volume of oxygen collected at regular intervals.

So the practical side of this experiment is straightforward, but the calculation isn't.

The problem is that you are measuring the volume of product, whereas to find an order of reaction you have to be working in terms of the concentration of the reactants - in this case, hydrogen peroxide.

That means that you will have to work out the concentration of hydrogen peroxide remaining in the solution for each volume of oxygen you record. To do this, you have to be happy with calculations involving the ideal gas law, and also basic mole calculations.

Having got a table of concentrations against time, you will then process them in exactly the same way as I described above. Plot the graph, draw tangents to find rates at various concentrations, and then plot a log graph to find the order.

Note:   It seems to me fairly unlikely that you could ever be asked to do this in an exam situation. And, at this level, you would almost certainly be given some guidance with the calculations

In a practical exam, few schools could provide a class set of the very expensive gas syringes accurate enough to produce meaningful results, and the time taken to process the results would be far greater than was available in any normal exam. That is equally true of a theory paper.

If you know that you can follow the course of a reaction which produces a gas using this method, that is probably all you will need. But check your syllabus, and past papers and mark schemes.

Colorimetry

In any reaction involving a coloured substance (either reacting or being produced), you can follow the course of the reaction using a colorimeter.

All of this is contained in one fairly small box.

The colour of the light can be changed by selecting a particular coloured filter (or using some more sophisticated device like a diffraction grating). The colour is chosen so that it is the frequency of light which is absorbed by the sample.

Taking copper(II) sulphate solution as a familiar example, you would choose to use a red filter, because copper(II) sulphate solution absorbs red light. The more concentrated the solution is, the more of the red light it will absorb.

Note:   For an explanation of why absorbing red light makes copper(II) sulphate solution blue, see the first part of the page about the colours of complex metal ions . You don't need to read about the origin of the colour for now.

A commonly quoted example of the use of colorimetry in rates of reaction is the reaction between propanone and iodine in the presence of an acid catalyst.

The solution of iodine in propanone starts off brown, and then fades through orange to yellow to colourless as the iodine is used up.

A colorimeter lets you measure the amount of light which is absorbed as it passes through a solution - recorded as the absorbance of the solution.

It is common to plot a calibration curve for a colorimeter by making up solutions of the coloured substance of known concentration and then measuring the absorbance of each under the same conditions as you will do the experiment. You then plot a graph of absorbance against concentration to give your calibration curve.

During your rate of reaction experiment, you read the absorbance from the meter at regular intervals, and then use your calibration curve to convert those values into concentrations.

Then you are faced with the same graphical methods as before.

Note:   In truth, these days, you are more likely to plug your colorimeter into a computer with the right software to do it all for you!

Two other methods

pH measurements

If you have a reaction in which hydrogen ions are reacting or being produced, in principle you should be able to follow changes in their concentration using a pH meter.

You may be aware that pH is a measure of hydrogen ion concentration, and it isn't difficult to calculate an actual hydrogen ion concentration from a pH.

However, if you are measuring pH over a fairly narrow range of hydrogen ion concentrations, the pH doesn't change all that much.

For example, the pH of a solution containing 0.2 mol dm -3 H + has a pH of 0.70. By the time that the concentration has fallen to 0.1 mol dm -3 , the pH has only increased to 1.00.

Whether it is feasible to use a pH meter obviously depends on how accurate it is. If the pH meter only recorded to 0.1 pH units, your results aren't going to be very good.

Conductivity measurements

The electrical conductivity of a liquid depends on the number of ions present, and the nature of the ions. For example, we looked at this reaction much further up the page:

During the course of the reaction, as hydrogen ions and iodide ions get used up, the conductivity of the mixture will fall.

Note:   I am not giving any more detail on this, because conductivity measurements aren't a part of any of the syllabuses that I am tracking. CIE expect you to know that it is possible to use conductivity measurements to follow the course of a reaction involving changes in the ions present, but not how you would actually carry out the experiments or process the results.

Where would you like to go now?

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© Jim Clark 2011 (modified October 2013)

 Free A-level and IB Chemistry Notes, Resources and Tutorials 

A2-Level Rates of Reaction

• Rates of Reaction
• Orders of Reaction
• The Rate Equation
• Rate and Mechanisms
• Clock Reactions

Video   Tutorial  Iodine Clock Reaction

Quick Notes  Clock Reactions

• The length of time taken to form a small amount of product is measured.
• A colour change or observation is used to show when this small amount of product has formed.
• Measuring the initial rate of a reaction is difficult because the concentrations of the reactants are constantly changing, meaning the rate of reaction is also changing.

• The thiosulfate ions are instantly converting any iodine molecules formed into iodide ions.
• Iodine molecules will no longer be reacted back to iodide ions by the thiosulfate ions and they can react with the starch (causing colour change).
• This is used to find how long it took to form a specific amount of iodine and give the initial rate of reaction.

Full Notes  Clock Reactions

The initial rate of a reaction refers to how fast a reaction is happening at the very start of the reaction . The rate of a reaction changes as the reaction proceeds, due to changes in the concentrations of the reactants ( see Rates of Reaction ).

To find a rate of a reaction, the change in concentration of a reactant (or product) needs to be measured against time. Finding changes of a concentration during the initial stages of a reaction can be very difficult because the concentration is changing rapidly as the reaction proceeds.

One effective method is to time how long it takes to produce a small amount of product compared to the starting amount of reactants . This means that, although the amount of the reactants is changing, the change is very small compared to the concentration overall – meaning the change to the rate is minimal, enabling us to study the initial rate of the reaction.

In order to determine when a small amount of product has been made, we need a visual indication of some kind – usually an indicator (although sometimes precipitate formation can be used). The problem is indicators can be very sensitive and will often change colour as soon as a product is formed – this is no good as we won’t know exactly how much product has been formed to cause the colour change. The colour change would also happen so fast, we would never be able to time it accurately.

To overcome this, another substance is added to the mixture that reacts with the product from the main reaction. This means that no colour change will occur until all of this substance is used up . By adding a known amount of this ‘new reactant’, we can determine how much product must have been formed from the first reaction by the time a colour change occurs.

This process is called a clock reaction. Clock reactions can be confusing to A-level students, but their idea is actually very simple.

Iodine Clock Reaction

A common example of a clock reaction at A-level Chemistry is the iodine clock reaction.

The basic reaction involves hydrogen peroxide and potassium iodide (in the presence of an acid catalyst).

The hydrogen peroxide oxides the iodide ions and iodine (I 2 ) is formed.

Using a simple starch indicator, we can see when iodine has been formed (starch turns dark blue (blackish) in the presence of iodine ).

However, as we are trying to determine the initial rate of the reaction we also need to time how long it takes to form a specific amount of iodine.

To do this, a known amount of sodium thiosulfate is added. Thiosulfate ions react with iodine to form iodide ions.

Now, every time a molecule of iodine is made in the first reaction, it is instantly converted back into iodide ions by the thiosulfate ions. As long as the thiosulfate ions are reacting with the iodine formed, there will be no colour change to the mixture (as there is no iodine to turn the starch indicator dark blue).

As soon as the thiosulfate in the mixture is used up, however, the iodine formed by reaction one stays as iodine. This means the solution turns dark blue (due to the starch indicator).

The time it takes the solution to change colour is determined by the amount of thiosulfate ions there are at the start. The higher the amount of thiosulfate ions, the longer it will take for the solution to change colour. As we are trying to find the initial rate of reaction, we want the amount of thiosulfate ions to be very small compared to the amount of hydrogen peroxide and iodide ions in the mixture (see above) .

If, for example, one mole of thiosulfate ions is present in the mixture when the hydrogen peroxide and iodide ions are mixed, and it takes 30 seconds for the solution to change colour, this means it has taken 30 seconds to produce 0.5 moles of iodine (the reacting ratio of thiosulfate ions to iodine is 2:1).

We now have a ‘rate of reaction’, as we know how long it took to produce a set amount (0.5 moles) of iodine.

By changing the concentration of hydrogen peroxide or potassium iodide, but keeping the amount of thiosulfate ions the same, we can see how the rate of reaction changes as we change the concentrations of each reactant. This is now just the same as determining a rate equation from reaction data ( see Rate Equation ).

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Iodine clock reaction demonstration

In association with Nuffield Foundation

• Five out of five

Introduce your students to rates of reaction and kinetics with the iodine ‘clock’ reaction. Mix a solution of hydrogen peroxide with potassium iodide, starch and sodium thiosulfate to see a colourless solution suddenly turn dark blue

This demonstration can be used at secondary level as an introduction to some of the ideas about kinetics. It can be used to stimulate discussion about what factors affect the rate of reaction. It also makes a useful starting-point for a student investigation.

Watch a video of this demonstration

This experiment is featured in our Get set … demonstrate video . A version of the experiment for students is used in our Rates of reaction video, 16-18 years , along with supporting resources , including illustrated technician notes , questions, worksheets and more.

The procedure described below outlines this experiment as a demonstration, best done on a large scale for the greatest visual impact. The demonstration itself takes less than one minute. For a student investigation, the quantities required would be smaller and they would have to measure the quanities quite accurately with, for example, disposable plastic syringes.

• Eye protection
• Balance (1 or 2 decimal places)
• Beaker (1 dm 3 )
• Beaker (250 cm 3 )
• Volumetric flasks (1 dm 3 ) x2
• Measuring cylinder (50 cm 3 )
• Measuring cylinders (100 cm 3 ) x2
• Stirring rod or magnetic stirrer and follower (optional)
• Stopwatch/timer
• Deionised or distilled water, 2 dm 3 .
• Soluble starch, 0.2 g
• Anhydrous sodium ethanoate (sodium acetate), 4.1 g
• Potassium iodide, 50 g
• Sodium thiosulfate–5–water, 9.4 g
• Glacial (concentrated) ethanoic acid (CORROSIVE), 30 cm 3  (note 2)
• Hydrogen peroxide solution, 20 volume (IRRITANT), 500 cm 3   (note 2)

Chemical notes

• Solutions A and B should be made up before the demonstration. The solutions will keep overnight, but best results are obtained if the solutions are made up on the day. Sodium thiosulfate will react with acids to give sulfur dioxide and a precipitate of sulfur, hence the sodium thiosulfate and ethanoic acid are separated in solutions A and B respectively.
• If you have access to 1 M dilute ethanoic acid, use 500 cm 3  of this to make solution B. Mix 500 cm 3  of 20 volume hydrogen peroxide with 500 cm 3  of 1M ethanoic acid. 

Health, safety and technical notes

• Read our standard health and safety guidance
• Wear eye protection.
• Anhydrous sodium ethanoate, CH 3 CO 2 Na(s) – see CLEAPSS Hazcard HC038a .
• Potassium iodide, KI(s) – see CLEAPSS Hazcard HC047a .
• Sodium thiosulfate–5–water, Na 2 S 2 O 3 .5H 2 O(s) – see CLEAPSS Hazcard HC095a .
• Ethanoic acid, CH 3 CO 2 H(l), (CORROSIVE) – see CLEAPSS Hazcard HC038a and CLEAPSS Recipe Book RB039. Handling glacial ethanoic acid requires care and should be done in a fume cupboard using gloves and eye protection.
• Hydrogen peroxide, H 2 O 2 (aq) (IRRITANT) – see CLEAPSS Hazcard HC050 and CLEAPSS Recipe Book RB045. 

Solution A is made up as follows

• Make a paste of 0.2 g of soluble starch with a few drops of water in a beaker. Pour onto this approximately 100 cm 3 of boiling water and stir.
• Pour the resulting solution into a 1 dm 3 beaker and dilute to around 800 cm 3 .
• Add 4.1 g of sodium ethanoate, 50 g of potassium iodide and 9.4 g of sodium thiosulfate. Stir until all the solids have dissolved and allow to cool to room temperature.
• Pour the mixture into a 1 dm 3 volumetric flask and make up to 1 dm 3 with water.

Solution B is made up as follows

• In a 1 dm 3 volumetric flask mix 500 cm 3 of 20 volume hydrogen peroxide with 30 cm 3 of glacial ethanoic acid and dilute to 1 dm 3 with water. 

The demonstration

• Measure 100 cm 3 of solution A and 100 cm 3 of solution B in separate 100 cm 3 measuring cylinders.
• Both solutions are colourless although solution A will be slightly cloudy.
• Pour both solutions simultaneously into a 250 cm 3 beaker to mix. Ensure thorough mixing with a stirring rod or magnetic stirrer.
• After about 20 seconds at room temperature the mixture will suddenly turn dark blue. The appearance of the blue colour may be timed – an assistant or a student can start and stop the timer.

Additional notes

• Hydrogen peroxide is capable of oxidising thiosulfate ions to tetrathionate ions but the reaction is too slow to affect this demonstration.
• The ethanoic acid/sodium ethanoate is added to buffer the pH.
• The acid will react slowly with sodium thiosulfate and produce a cloudy suspension of sulfur and release sulfur dioxide which is TOXIC - see CLEAPSS Hazcard HC097. To avoid this, the acid and sodium thiosulfate are separated in solutions A and B.
• The CLEAPSS Guide L195 ‘Safer chemicals, safer reactions’ gives useful information about how to carry out rates experiments involving acid/thiosulfate mixtures more safely.

Teaching notes

A white background will help so that the impact of the sudden and spectacular colour change is not lost. Scaling up the volumes of solution that are mixed may help in a large room. There is no warning of when the blue colour is about to appear.

It may help understanding if the students are already familiar with the reactions of starch and iodine, and iodine and sodium thiosulfate, so it may be worth demonstrating these beforehand.

The basic reaction is:

H 2 O 2 (aq) + 2I – (aq) + 2H + (aq) → I 2 (aq) + 2H 2 O(l)

(For more advanced discussions or investigations – this reaction is the rate determining step and is first order with respect to both H 2 O 2 and I – .)

As soon as the iodine is formed, it reacts with the thiosulfate to form tetrathionate ions and recycles the iodide ions by the fast reaction:

2S 2 O 3 2– (aq) + I 2 (aq) → S 4 O 6 2– (aq) + 2 I – (aq)

As soon as all the thiosulfate is used up, free iodine (or, strictly, I 3 - ions) remains in solution and reacts with the starch to form the familiar blue-black complex.

The time for the blue colour to appear can be adjusted by varying the amount of thiosulfate in solution A so a ‘clock’ of any desired time interval can be produced.

More resources

Add context and inspire your learners with our short career videos showing how chemistry is making a difference .

Additional information

This is a resource from the  Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry.

Practical Chemistry activities accompany  Practical Physics  and  Practical Biology . 

© Nuffield Foundation and the Royal Society of Chemistry

• 14-16 years
• 16-18 years
• Demonstrations
• Physical chemistry
• Rates of reaction

Specification

• 4.3.1 use simple rate equations in the form: rate = k[A]ˣ[B]ʸ(where x and y are 0, 1 or 2);
• 4.3.4 deduce, from a concentration-time or a rate-concentration graph, the rate of reaction and/or the order with respect to a reactant.
• Rate of reaction.
• Nature of reactants.
• 2. Develop and use models to describe the nature of matter; demonstrate how they provide a simple way to to account for the conservation of mass, changes of state, physical change, chemical change, mixtures, and their separation.
• 4. Classify substances as elements, compounds, mixtures, metals, non-metals, solids, liquids, gases and solutions.
• 7. Investigate the effect of a number of variables on the rate of chemical reactions including the production of common gases and biochemical reactions.
• 9. Consider chemical reactions in terms of energy, using the terms exothermic, endothermic and activation energy, and use simple energy profile diagrams to illustrate energy changes.
• C5 Investigation the effect of surface area, concentration and temperature on the rate of a chemical reaction
• 7.1b observing a colour change (in the reaction between sodium thiosulfate and hydrochloric acid)
• 8 Investigation the effect of surface area, concentration and temperature on the rate of a chemical reaction
• 5 Investigate how changes in concentration affect the rates of reactions by a method involving measuring the volume of a gas produced and a method involving a change in colour or turbidity. This should be an investigation involving developing a hypothesi…
• RP19 Investigation of how changes in concentration affect the rates of reactions by a method involving measuring the volume of a gas produced and a method involving a change in colour or turbidity. This should be an investigation involving developing…
• 11 Investigate how changes in concentration affect the rates of reactions by a method involving measuring the volume of a gas produced and a method involving a change in colour or turbidity. This should be an investigation involving developing…
• AT a: Use appropriate apparatus to record a range of measurements (to include mass, time, volume of liquids and gases, temperature).
• AT e: Use volumetric flask, including accurate technique for making up a standard solution.
• AT l: Measure rates of reaction by at least two different methods, for example: an initial rate method such as a clock reaction, a continuous monitoring method.
• Use initial concentration–time data to deduce the initial rate of a reaction.
• 4 i. understand experiments that can be used to investigate reaction rates by: an initial-rate method, carrying out separate experiments where different initial concentrations of one reagent are used
• 13a and 13b. Following the rate of the iodine-propanone reaction by a titrimetric method and investigating a ‘clock reaction’ (Harcourt-Esson, iodine clock)
• an initial rate method such as a clock reaction
• l) measurement of rates of reaction by at least two different methods, for example: an initial rate method such as a clock reaction; a continuous monitoring method.
• h) the techniques and procedures used to investigate reaction rates by the initial rates method and by continuous monitoring, including use of colorimetry
• PAG.10 Rates of reaction – initial rates method
• Collision theory can be used to explain the effects of the following on reaction rates: concentration; pressure; surface area (particle size); temperature; collision geometry.
• (b) the effect of changes in temperature, concentration (pressure) and surface area on rate of reaction
• (h) measurement of reaction rate by gas collection and precipitation methods and by an ‘iodine clock’ reaction

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IODINE CLOCK REACTION

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Lab Handout & Procedure

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Complete Lab Guide

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GoogleSheet Data & Analysis

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Introduction

Chemical kinetics is the branch of chemistry that is concerned with the mechanisms and rates of chemical reactions.  

The mechanism of a chemical reaction is a description of what happens to each molecule at a very detailed level—which bonds are broken, which new bonds are formed, and how the three-dimensional shapes of the chemicals change during the course of the reaction. The rate of the reaction is a measure of its speed. The rate of a chemical reaction can be measured by how quickly the reactants disappear, or by how quickly the products are generated. 

The iodine clock reaction is a favorite demonstration in chemistry classes because it has an element of drama. Two clear solutions are mixed, producing a new clear solution. Then, after a period of several seconds, the solution turns dark blue. A demonstration of this reaction is shown in the video below. 

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Iodine Clock Reaction

Warning : never let solution B stand in open beaker for over an hour. Test existing stock solutions. If demonstration does not work, discard solution B.

Chemicals and Solutions

• Iodine clock solution A
• Iodine clock solution B

Solution Preparation

0.02 M KIO 3  

Solution B 

4g soluble starch, 0.2g sodium metabisulfite ( Na ₂ S ₂ O ₅), 5mL 1M sulfuric acid in 1L solution

• Electronic stop clock
• Ice bath in large refrigerator dish
• 100 mL cylinders
• 400 mL beakers
• Magnetic stirrer and stirring bars for both beakers
• Into cylinders, put 100 mL of solution A.
• Into beakers, put 100 mL of solution B.
• Put one cylinder and beaker into ice bath at least 15 minutes before expected use.
• Mix together and observe the color change.
• ALTERNATIVE: to illustrate effect of concentration, put differing amounts of solution A in cylinders and top off with DI water, mixing each cylinder with an equal volume of solution B (e.g. 50 mL solution A with 50 mL DI water mixed with 100 mL solution B). Measure rate of reaction with a stop clock.

A simplified explanation of the reaction is as follows:

I- reacts with IO ₃ - to form I ₂ .

\( \ce{ 5HI_{(aq)} + HIO3_{(aq)} -> 3I2 + 3H2O } \)

I ₂ is immediately consumed by reaction with HSO ₃ -.

\( \ce{ I2 + HSO3- + H2O -> 2I- + SO42-_{(aq)} + 3H+_{(aq)} } \)

When HSO ₃ - has been consumed, I ₂ accumulates. I ₂ + starch forms a blue colored starch-I ₂ complex.

When you dilute solution A (0.02 M KIO ₃ ) in half, it takes twice as long to form the blue starch - I ₂ complex. When solution A is diluted to 1/4 the concentration, it takes four times as long to form the blue starch-I ₂ complex. The reaction is much slower at colder temperatures.

The solutions can be poured down the drain. (The bisulfite and iodate are consumed and the iodine is complexed with starch so there is no oxidizer hazard). Demo generates 800 mL of 0.2% starch iodine complex aq.

Summerlin and Ealy, Chemical Demonstrations , pp. 75-76.

Clock II (Oscillating Clock)

30% hydrogen peroxide is very reactive.

Solution #1

36 mL of 30% H ₂ O ₂ to 100 mL (make fresh)

Solution #2

43 g  KIO ₃, 4.3 mL conc. sulfuric acid, in 1L solution

Solution #3

• 15.6 g malonic acid, 3 g of MnSO ₄ in 970 mL of water.
• Stir in 30 mL of a 1% starch solution.
• magnetic stirrer and bar
• Three 100 mL graduated cylinders
• 400 mL beaker
• Measure out 100 mL of each solution into graduated cylinders.
• With stirring, quickly add each solution to the 400 mL beaker. The solution will oscillate between colorless, amber and dark blue.
• Clock will oscillate for about 5 minutes typically.

Hint : When no stir bar is used, regions of the solution will change first.

The oscillations are due to the shifting concentrations of I ₂ and I - . The amber color is due to the presence of I ₂ . When I - is present, it reacts with I ₂ and starch to produce a dark blue complex. This color fades as iodine is consumed.

A very simplified explanation of this reaction is:

\( \ce{ $\underset{\text{gold}}{\ce{ 2HIO3 + 5H2O2 -> I2 + 5O2 + 6H2O }}$ } \)

\( \ce{ I2 + CH2(COOH)2 -> ICH2(COOH)2 + H+ + I- } \)

\( \ce{ $\underset{\text{dark blue}}{\ce{ I2 + I- + starch -> starch-iodine complex }}$ } \)

\( \ce{ $\underset{\text{colorless}}{\ce{ I2 + 5H2O2 -> 2HIO3 + 4H2O }}$ } \)

The hydrogen peroxide and iodate are consumed and the iodine is complexed with starch so there is no oxidizer hazard. Therefore the solution can be rinsed down the drain. Demo generates 300 mL of 2% starch iodine complex aq.

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Iodine Clock Reaction

Try an at home version of this experiment using a few things you may have in your bathroom medicine cabinet. In may ways this experiment feels almost like magic. Two colorless liquids are mixed together and after a few moments the mixture turns a dark blue color. There are actually a couple of simple chemical reactions going on at the same time to make this “clock reaction” occur. This version of the classic “iodine clock reaction” uses safe household chemicals most people have on hand at home.

What you need:

• distilled water (tap water will work OK as well)
• a couple plastic cups
• 1000 mg vitamin C tablets
• tincture of iodine (2%)
• hydrogen peroxide (3%)
• liquid laundry starch

What to do:

• Make a vitamin C solution by crushing a 1000 mg vitamin C tablet and dissolving it in 2 oz of water. Label this as “vitamin C stock solution”.
• Combine 1 tsp of the vitamin C stock solution with 1 tsp of iodine and 2 oz of water. Label this “solution A”.
• Prepare “solution B” by adding 2 oz of water to 3 tsp of hydrogen peroxide and 1/2 tsp of liquid starch solution.
• Pour solution A into solution B, and pour the resulting solution back into the empty cup to mix them thoroughly. Keep pouring the liquid back and fourth between the cups.

What’s going on?

There are actually two chemical reactions going on at the same time when you combine the solutions. During these reactions two forms of iodine created – the elemental form and the ion form.

In Reaction # 1 iodide ions react with hydrogen peroxide to produce iodine element which is blue in the presence of starch. BUT, before that can actually happen, the Vitamin C quickly reacts and consumes the elemental iodine.

The net result, at least for part of the time is that the solution remains colorless with excess of iodide ions being present. Now after a short time as the reactions keep proceeding in this fashion, the Vitamin C gets gradually used up. Once the Vitamin C is used up, the solution turns blue, because now the iodine element and starch are present.

Safety Precautions

Be careful when working with the iodine – it stains, and it stains really well. Be very careful not to spill any of the solution.

Waste Disposal

Dig deeper into the science behind clock reactions in  this paper  from the Journal of Chemical Education.

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7 Kinetics of the Iodine Clock Reaction

Determine the rate law and activation energy of an iodine clock reaction.

Learning Outcomes

• Calculate the rate of reaction given experimental data.
• Use the method of initial rates to determine the rate law and rate constant for a reaction.
• Graphically determine the activation energy of a reaction using the Arrhenius equation.

Introduction

In this experiment, you will study the rate of the reaction

proceeds as a function of reactant concentration and as a function of temperature.

General Principles Associated

Defining the rate of reaction.

The rate of a reaction is related to the change irange of soln concentration (in terms of molarity) of reactant/product present per unit time. That is to say, for the reaction

the rate of reaction is given by

• When you are dealing with the change in concentration of a reactant, a negative sign is placed in front. This ensures that the rate of reaction is positive .
• In order to ensure that the rate of reaction has the same numerical value regardless of the reactant/product being studied, the rate of change of concentration of a species is divided by its coefficient in the expression for the rate of a reaction

For the reaction

the rate of reaction can (equally) be expressed in terms of the change of concentration of N 2 O 4 or NO 2 as

The dependence of reaction rates on concentration can be summarized by rate laws , which take the formy

Rate laws cannot be found simply by reading off the balanced chemical equation. The only way to find rate laws is by conducting experiments, for which there are several methods. In this experiment, you will use the method of initial rates to determine the rate law for the iodine clock reaction.y

Effect of Temperature on Rates of Reaction

In accordance with collision theory, there are two reasons why reactions occur at a faster rate at higher temperatures:

• Molecules move at a faster speed.
• A greater proportion of molecules have sufficient energy to react when they collide.

In reality, the second effect is the more important one. As a result, the temperature dependence of the rate constant can be given by the Arrhenius equation :

The Iodine Clock Reaction

In this experiment, you will study the iodine clock reaction between iodate and iodide under acidic conditions:

and looking at the stoichiometric coefficients of the iodine clock reaction,

and putting all of this together, we can show that the reaction rate is

You can therefore, by varying the initial concentrations and temperatures in each of the tubes, determine the rate law.  From this, you can determine the rate constants for the reaction at different temperatures, and hence determine the activation energy for this reaction.

Experimental Procedure

In this experiment, you will use Dr. Gary Bertrand’s virtual lab setup to obtain simulated data associated with this reaction.  Click on the Experiment tab to proceed.

You can vary independently the concentrations of each of the reactants as well as the temperature.  The initial concentrations at the instant of mixing (before any reactions occur) will appear on the right hand panel of the reaction system.  It is incumbent upon you to select a number of different set of solution concentrations and conditions to make the experiment work for you.

Determination of the Rate Law

For this part of the experiment, you will set up each of the concentrations at a particular temperature that works for you.  Record the initial concentrations and conditions upon mixing, and then select “mix the solutions” to make the lab bench setup appear.

• Wait and watch carefully until the instant the solution starts showing purple.  Hit “stop” to stop the stopclock.
• Hit “reset” to go back to the state just before mixing so you can repeat the experiment.  For each set of conditions, please be sure to repeat this at least three times.

To prepare another mixture, go to “prepare solution”.

Determination of the Activation Energy

To determine the activation energy, you will need to have a set of conditions from which to vary the temperature.

Repeat the above experiment for one set of concentrations of solutions, but be sure to do it for every temperature.

• Sometimes, you can see concentrations of products or very rarely concentrations of intermediates that exist in this reaction. For most purposes, however, intermediates will not likely appear in the rate law. ↵

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