Reaction Kinetics (2) First Order Kinetics: The Iodine Clock Reaction

 In a previous blog I showed you how ideas about reaction kinetics can be derived from a simple consideration of a straightforward experiment like that of putting magnesium in hydrochloric acid.

Just putting an excess of magnesium in hydrochloric acid solution generated hydrogen gas and this was measured experimentally over time.

Processing the volume vs time data led us to a mathematical model of the reaction rate:

Reaction Rate =  k. [HCl]¹

This equation we called a rate equation,

k we called a rate constant,

the species in square brackets was a measure of the concentration of hydrochloric acid

the power to which this concentration term was raised we called the reaction order. 

Let’s now look at what units are involved:

If concentration has units: moles.dm^-3 and reaction rate has units: moles.dm^-3s^-1

then the units of the rate constant must be s-1

Let’s look again at the graph of the raw data collected from the experiment.

Here is another observation associated with first order kinetics.

The time taken for half the particles of hydrochloric acid to react is a constant that we call the half-life of the reaction or t _½. 

A quick look at the graph’s red line confirms this (approximately!!)

40-20 in 1.5mins

20-10 in 1.25mins

10-5 in about 1.2mins 

So the half-life of this reaction is about 1.25minutes

Though the data may be real, it is not great, I have to say.

Here are a few reasons why these half-lives decrease instead of remaining constant:

First, the magnesium might not have been in excess so it is consumed before the full data is collected, shortening the half-lives.

Second, the reaction is not carried out at constant temperature and so as the reaction proceeds the temperature increases (the reaction is very exothermic!!) and the reaction goes faster and faster until the acid runs out.

Third, the apparatus may affect the result especially if the gas is collected in gas syringe because gas syringes are notorious for being sticky and not moving smoothly as the gas is evolved. 

Let’s now look at another experimental way of getting at first order kinetics.

The reaction between potassium iodide solution and hydrogen peroxide is said to follow first order kinetics.

Here’s how to carry out an experiment to see if that assumption is true.

(Make sure before you start you are using freshly made up solutions because both hydrogen peroxide and sodium thiosulphate are known to degrade easily.)

Investigating the effect of concentration of a reactant on the reaction rate: the iodine clock reaction

Introduction

Hydrogen peroxide enters into a redox reaction with iodide ions in which the hydrogen peroxide is reduced to water and the iodide ions are oxidised to iodine.

In the presence of sodium thiosulphate the iodine is reduced back to iodide ions.  However, when all the thiosulphate has been used up the iodide produces sufficient iodine to turn starch black.  The sudden appearance of the starch-iodine complex is quite dramatic and has led this reaction to be called the “iodine clock” reaction.

Overall equation: 

H₂O₂(aq)    +    2I^-(aq)    +     2H⁺(aq)   =    2H₂O(l)    +    I₂(aq) 

Part 1:  Investigating the dependence of reaction rate on the concentration of the hydrogen peroxide solution.

You are supplied with the following solutions:   

Hydrogen peroxide    20vol

Potassium Iodide       6g/dm³

Sulphuric acid            1M

Sodium thiosulphate 7.5g/dm³

Starch 1%

You will need the following apparatus:       

100ml beaker

Dropping pipette

3 of 25ml measuring cylinders

1 of 50ml measuring cylinder

stopwatch

Place the following solutions in the 100ml beaker: 10ml potassium iodide, 10ml sodium thiosulfate solution, 10ml sulphuric acid solution and 10 drops starch.  Now add 30ml hydrogen peroxide solution and time how long it takes for the blue-black starch iodine complex suddenly to appear.  Complete the following 5 experiments topping up with distilled water in each case to keep the volume constant (see table).  In the table below also write down your value for the time and work out the reaction rate.

Reaction table

Expt #	H2O2 /ml	Water /ml	KI(aq) /ml	Na2S2O3(aq) /ml	H2SO4 /ml	Time t /s	Rate i.e. 1/t /s-1
1.	30	0	10	10	10
2.	25	5	10	10	10
3.	20	10	10	10	10
4.	15	15	10	10	10
5.	10	20	10	10	10
6.	5	25	10	10	10
7.	0	30	10	10	10

Questions

a) When you have completed the 7 experiments and noted the time it takes to give the blue/black colour, calculate the rate of each reaction.

b) Plot a graph of reaction rate against concentration of hydrogen peroxide (use Excel)

c) What can you say about the relationship between the rate and the concentration of hydrogen peroxide?

d) What is the slope of your graph? (Hint: use the options window on Excel)  This is the value of the rate constant k

e) Write out the rate equation in the form Rate = k[X].

Part 2:  Investigating the dependence of reaction rate on the concentration of potassium iodide solution.

You are supplied with the same solutions as in the previous experiment:

         Hydrogen peroxide    20vol

         Potassium Iodide       6g/dm³

Sulphuric acid            1M

         Sodium thiosulphate 7.5g/dm³

         Starch 1%

Place the following solutions in the 100ml beaker: 10ml potassium iodide, 10ml sodium thiosulfate solution, 10ml sulphuric acid solution and 10 drops starch.  Now add 25ml hydrogen peroxide solution and time how long it takes for the blue-black starch iodine complex suddenly to appear.  Complete the following 5 experiments in the same way varying the volume and hence concentration of potassium iodide solution and top up with distilled water in each case.  In the table below insert your results for the time and calculate the reaction rate

Reaction table

Expt #	H2O2 /ml	Water /ml	KI(aq) /ml	Na2S2O3(aq) /ml	H2SO4 /ml	Time t /s	Rate i.e. 1/t /s-1
1.	25	0	10	10	10
2.	25	2	8	10	10
3.	25	4	6	10	10
4.	25	6	4	10	10
5.	25	8	2	10	10
6.	25	10	0	10	10

Questions

a) When you have completed the 6 experiments and noted the time it takes to give the blue/black colour, calculate the rate of each reaction.

b) Plot a graph of reaction rate against concentration of potassium iodide (use Excel)

c) What can you say about the relationship between the rate and the concentration of potassium iodide?

d) What is the slope of your graph? (Hint: use the options window on Excel)

e) Write out the rate equation in the form Rate = k[X].

f) Write out the overall rate equation.

g) Calculate the rate constant k from your graph

h) A proposed mechanism for the reaction involves three steps:

Step 1:               H₂O₂     +   I^-    =     H₂O     +   IO^-

Step 2:               H⁺    +     IO^-    =    HIO

Step 3:               HIO    +    H⁺     +   I^-     =       I₂    +    H₂O

Which step is likely to be the rate determining step i.e. the slowest step of the mechanism?  Why?

Conclusion and analysis

What I would expect you to find is that the plot of Rate vs [H₂O₂] is a straight line.

The slope of this straight line gives you a value for the rate constant.

This allows you to agree that the kinetics are first order with respect to the [H₂O₂]. 

But what happens if you repeat the experiment keeping the [H₂O₂] constant and changing the [KI] ?

Again I would expect you to obtain a straight line suggesting that the reaction kinetics are also first order with respect to the [KI].

The implication of these results is that the overall order of the reaction is 2 and the rate equation looks like this:

Rate = k  [KI]. [H₂O₂]

So therefore the slowest step of the reaction should include both species i.e. I^- and H₂O₂ or Step 1 in the above mechanism.

Well in the next blog we’ll take this study of reaction kinetics a little further and see what zero order reactions are.