Therefore we can say that in this reaction, the oxygen atoms in the centre of the peroxodisulphate have been reduced, and the iodide ions have been oxidised. Therefore we can construct the half equations for this reaction: Oxidation step:
2I- I2 + 2e-
S2O82- + 2e- 2SO42-
Theory behind clock reactions
The main reaction taking place is the reaction between peroxodisulphate and iodide, as described above. The reaction produces iodine, the change in concentration of which can be measured over a given length of time. However, any colour change in this reaction is very gradual, and so although it is possible to measure this colour change, it is subject to a high degree of inaccuracy. Therefore we also add 2 other chemicals to the reaction – sodium thiosulphate (Na2S2O3) and starch. The S2O32- ion reacts with any iodine produced in the first reaction according to the following equation: 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq)
Because we add a known volume of thiosulphate to the equation, we can work out the number of moles of thiosulphate that reacts with iodine, and
therefore work out the number of moles of iodine that reacts. The starch acts as an indicator – when all of the thiosulphate has been used up, any surplus iodine produced reacts very quickly with the starch to produce a blue-black colour. Timing how long it takes for this blue-black colour to appear tells us how long it takes for the thiosulphate to become used up, therefore the time taken to produce a known number of moles of iodine. From this data, the rate of production of iodine (and therefore the rate of the initial reaction) can be calculated. Rates of reaction
Collision theory says that a reaction will occur when the particles of the reactants collide, provided they collide with a certain minimum kinetic energy. However not every collision causes a reaction. Particles in liquids and gases are constantly moving and colliding with each other but they only react when the conditions are right. When particles collide, kinetic energy is converted into potential energy and the potential energy of the reactants starts to rise. Existing bonds start to break and new bonds start to form. Only the pairs with enough combined kinetic energy on collision to overcome the energy barrier or the activation enthalpy will go on to produce products.  Activation enthalpy is the minimum amount of kinetic energy particles need to react. The particles need this energy to break the bonds to start the reaction.  Reactions with low activation enthalpies happen quite quickly whereas reactions with high activation enthalpies don’t and more energy is required. The Maxwell-Boltzmann distribution curve explains collision theory in terms of the number of molecules that have the minimum amount of activation energy. The shaded part of the graph shows that only some molecules have energy more than the activation enthalpy needed to start the reaction.
Figure 3: Maxwell-Boltzmann distribution curve http://www.webchem.net/notes/how_far/kinetics/maxwell_boltzmann.htm&docid=xBJ_P7m The effect of temperature on the rate of a reaction:
When you raise the temperature of a reaction, molecules move faster because they have more kinetic energy. This means that the molecules will collide more often and that there is a greater chance that the colliding molecules
will have energy equal to or greater than the activation enthalpy for the reaction.
Figure 4: The effect of temperature on the number of collisions with energy greater than the activation enthalpy. http://www.4college.co.uk/a/aa/rate.php Figure 4 shows the effect of raising the temperature by 10oC on the number of collisions with energy equal to or greater than the activation enthalpy. This Maxwell-Boltzmann distribution curve shows that ‘reactions go faster at higher temperatures’ because there are more colliding molecules that have energy equal to or greater than the activation enthalpy, which is needed to start the reaction.
The effect of concentration on rate:
Increasing the concentration of reactants speeds up the rate of a reaction because there are more particles in the solution and they are closer together. This means that collisions happen more often and it is more likely that the collisions will have energy equal to or greater than the activation enthalpy for the reaction to start. According to collision theory, this will increase the rate of reaction. 
Figure 5: the effect of concentration on the rate of a reaction. Source: http://www.kentchemistry.com/links/Kinetics/FactorsAffecting.htm
A catalyst is a substance that alters the rate of a reaction and remains unchanged at the end of the reaction. Catalysts usually increase the rate of a reaction. They do this by providing an alternative reaction pathway for the breaking and remaking of bonds that have a lower activation enthalpy. A decrease in activation enthalpy will mean that a higher proportion of collisions will have a combined kinetic energy that is equal to or greater than the activation enthalpy for the reaction.
Figure 6: the effect of a catalyst on the activation enthalpy of a reaction. Source: http://mychemblog-sol.blogspot.com/
There are two types of catalysts: Heterogeneous and Homogeneous. Heterogeneous catalysts are catalysts that are in a different state to the reactants. Heterogeneous catalysts work by adsorbing to the surface of the reactant and then weaken and break the bonds of the reactant. The catalyst then forms new bonds and leaves the surface of the catalyst. Homogeneous catalysts are catalysts that are in the same state as the reactants. They work by forming an intermediate with the reactants. The intermediate compound then breaks down to give the product and reform the catalyst.  Heterogeneous catalysts are more commonly used in industry but homogeneous catalysts can be more specific and controllable and can produce less waste products.
Rates of Reaction
Rate can be calculated using the equation: Rate = ∆ concentration (of reactant or product) Time
Typical rate equation:
Rate = k [A]x [B]y
Where [A] and [B] are the reactants of the experiment.
A rate equation can be constructed once the order of each of the reactants is known. The rate constant is ‘k’ in the equation. The rate constant ‘k’ is a constant for any given temperature. Raising the temperature increases the value of the rate constant, ‘k’. The rate constant can be found by rearranging the rate equation. So the equation to work out ‘k’ would be: k = Rate
Rate equation for peroxodisulphate and iodide ions:
“Rate = k ” [“S2O82-” (“aq” ) ]”x ” [“I-” (“aq” ) ]”y”
There are three common orders for a reactant. Zero, First and Second order. The order of a reaction with respect to a particular reactant tells you how the reactants concentration affects the overall rate of the reaction. You
can only find out orders from doing experiments, you cannot work them out from just the chemical equation. If you double the reactants concentration and the rate stays the same and then order with respect to that reactant is 0. If you double the reactants concentration and the rate also doubles then the order with respect to that reactant is 1. If you double the reactants concentration and the rate quadruples then the order with respect to that reactant is 2. You can use experimental data to work out the order of a reactant by seeing how the rate changes when the concentration is doubled.
Typical Graphs of orders:
Figure 7: typical graphs for orders. Source: http://sites.google.com/site/kwokthechemteacher/Home/kwokthechemteacher/chemical-kinetics/calculating-order-of-reactions
You can also work out the order of a reactant by using the half-life method. A half-life is the time taken for a reactant to halve in quantity. It is easy to work out the half-life of a reactant using a concentration against time graph.
Figure 8: Zero order. Source: http://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/Halflife.html For a reactant to be zero order, the half-life decreases as the concentration decreases.
Figure 9: half-life; first order. Source: http://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/Halflife.html So for a reactant to be first order, the length of each half-life must be constant as the concentration decreases.
Figure 10: Second order graph. Source: http://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/Halflife.html And for a reactant to be second order, the half-life needs to increase as the concentration decreases.
The link between the mechanism and orders
The rate determining step is the slowest step in the reaction. You can predict the rate determining step from the rate equation. If the reactants are in the rate equation than they are involved in the rate determining step but if they are, for example zero order and are not in the rate equation but are involved in producing the products then they won’t be involved in the rate determining step but in a second step. Reactants that in the rate determining step and in the rate equation directly affect the rate of the reaction. Catalysts can appear in the rate equation.  S2O¬82-(aq) + 2I-(aq) 2SO42-(aq) + I2(aq)Slow(reaction 1) 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq) Fast(reaction 2) I2(aq) + starch blue-black complex Slow(reaction 3)  In reaction 1, the reaction is slow because there are Thiosulphate ions present that prevent the formation of iodine until they have all been used up. The Thiosulphate in reaction 2, the fast step, quickly reacts with any iodine. Once all the Thiosulphate is used up, reaction 3 can occur when the remaining iodine reacts with starch to produce the blue-black complex. Reaction 1 is the rate determining step. According to the reactions above, there are 2 iodide ions and 1 peroxodisulphate ion involved in the rate determining step, therefore this would suggest that:
S2O82- is a FIRST order reactant
I- is a SECOND order reactant.
The rate constant is always the same for a specific reaction at a specific temperature. Rate constants vary with temperature, a higher temperature means a higher rate constant. When you increase the temperature, the concentration of the reactants do not change, but the rate of the reaction does increase so this suggests that it is the rate constant , k, that increases.  You can work out the rate constant if you are given or have worked out the: concentration of the reactants, the order of the reactants and the rate of the reaction. Since rate = k x [A]x [B]y. The units of k vary so you have to work them out.
Linked to the effect of temperature.
The fraction of molecules that have energy greater than the activation enthalpy can be calculated using the Arrhenius equation.
Figure 11: Arrhenius Equation. Source: http://www.chemguide.co.uk/physical/basicrates/arrhenius.html So the equation is:
k = Ae-Ea/RT
If we take natural logs of both sides of the equation it becomes…  ln k = ln A + (- Ea/RT )
Compare this equation to the equation of a straight line: y = mx +c. Where m is the gradient and c is the y-intercept. y = ln k
c = ln A
m = -Ea/R
x = 1/T
ln k = ln A + (- Ea/R x 1/T )
So you can plot a graph of y (ln k) against x (1/T) with the data and find the gradient of the graph. Gradient = -Ea/R
R = 8.31
So from this you can work out the Ea (activation enthalpy) of the reaction in kJ.
Skill C – References
1. Denby, D. (1996). Clock reactions. Chemistry Review. 6 (2), p14-15. (accessed via QMC http://viewpoint.qmc.ac.uk/C17/Chemistry/Chemistry%20Coursework%20A2/01%20-%20Investigation%20Starter%20Materials/Investigation%20Guidance%20Articles/Clock%20Reaction%20Investigation%20Article.pdf) 2. Textbook: Burton, G ; Holman, J ; Lazonby, J ; Pilling, G ; Waddington, D (2000). Chemical Ideas. 2nd ed. Bath: Heinemann. p155, p205-206 and p221-242. 3. Cartwright, H. (2011). Chemical safety database. Available: http://cartwright.chem.ox.ac.uk/hsci/chemicals/hsci_chemicals_list.html. Last accessed 28th Nov 2011. 5. Nuffield Advanced Chemistry. (2011). Oxidation states and redox reactions. Available: http://ww.chemistry-react.org/go/default/Faq/Faq_30389.html. Last accessed
14th Nov 2011. 6. Dr. Russell, S. (2010) Kinetics of the Persulfate – iodide clock reaction. Available: http://science.csustan.edu/CHEM1112_4/Chemical%20Kinetics%27/Persulfate_files/Clock%20Reaction%20S10.pdf. Last accessed 31st Jan 2012. 7. Clark, J. (2002). rate constants and the arrhenius equation. Available: http://www.chemguide.co.uk/physical/basicrates/arrhenius.html
Skill D – Results
Tube no.Time /sClass results
Skill E Part 1 – Analysis
Working out concentrations of tubes
Diluted concentration of tube 1 = 5/10 x 1.00 = 0.50 mol dm-3 Diluted concentration of tube 2 = 4/10 x 1.00 = 0.40 mol dm-3 Diluted concentration of tube 3 = 3/10 x 1.00 = 0.30 mol dm-3 Diluted concentration of tube 4 = 2/10 x 1.00 = 0.20 mol dm-3 Diluted concentration of tube 5 = 1/10 x 1.00 = 0.10 mol dm-3
Tube no.[KI] / mol dm-3Time /s
Calculating initial rate of reaction: (Using the Sophisticated Approach)
No. of moles of Na2S2O3 =
(Volume cm3 / 1000) x concentration =
2cm3 / 1000 x 0.005 = 1 x 10-5 moles Na2S2O3
According to the balanced symbol equation:
2S2O32- + I2 S4O62- + 2I-
Mole ratio S2O32- : I2 is 2:1
Therefore 1 x 10-5 /2 x 1 = 5 x 10-6 moles I2.
Concentration of I2 =
No of moles / (Volume cm3 / 1000) =
5 x 10-6 / (10/1000) = 5 x 10-4 mol dm-3 I2.
Rate of reaction =
Change in concentration / Time taken
Results (See Graphs 1 and 3)
Initial concentration / mol dm-3Rate of reaction / mol dm-3 s-1 0.50= 5 x 10-4 / 18 = 2.78 x 10-5
0.40= 5 x 10-4 / 27 = 1.85 x 10-5
0.30= 5 x 10-4 / 43 = 1.16 x 10-5
0.20= 5 x 10-4 / 80 = 6.25 x 10-6
0.10= 5 x 10-4 / 262 = 1.91 x 10-6
Alternative results: (See Graphs 2 and 4)
Initial concentration / mol dm-3Rate of reaction / mol dm-3 s-1 0.50= 5 x 10-4 / 21 = 2.38 x 10-5
0.40= 5 x 10-4 / 28 = 1.79 x 10-5
0.30= 5 x 10-4 / 46 = 1.09 x 10-5
0.20= 5 x 10-4 / 86 = 5.81 x 10-6
0.10= 5 x 10-4 / 191 = 2.62 x 10-6