5.2: Background

Kinetics deals with the rate at which a process occurs and chemical kinetics deals with the rates of chemical reactions. This lab will investigate the kinetics associated with the following reaction.  

\[\underbrace{2Fe^{+3}(aq)}_{\text{Ferric Iron}} + \underbrace{2I^-(aq)}_{Iodide}  \rightarrow \underbrace{2Fe^{+2}(aq)}_{\text{Ferrous Iron}} + \underbrace{I_2(aq)}_{\text{iodide}} \]

This reaction is called an iodine clock reaction because we can measure the effect of varing reagent concentrations and temperature on the time it takes for the iodine to react with starch and turn blue. The following video from Thammasat University in Thailand demonstrates this reaction. 

Video \(\PageIndex{1}\): The last 46 seconds of a video showing the reduction of Iron(III) by Iodide reaction that we will be studying, uploaded by Kanchanok Duangkhai  (https://youtu.be/azUuxhclcm0).

The rate law is
\[R = k [Fe^{+3}]^m [I^-]^n\] 
where R is the depedent variable. Although it looks like there are there are two independent variables,([Fe⁺³], [I^-]), there are really three as \(k=Ae^{-\frac{E_A}{RT}}\) and so T is also an independent variable. We will run three sets of experiments, where you vary one of the three  independent variables ([Fe⁺³], [I^-] or T) while holding the other two constant and measure the effect on the rate (dependent variable).

The Reactions

In this experiment, we will mix two solutions as in video \(\PageIndex{1}\) and measure the time for the mixture to turn blue.  Figure \(\PageIndex{1}\) shows the two reagents in each of the two reagent flasks (see table \(\PageIndex{1}\) of the experimental procedures to see the experimental concentrations).

Figure \(\PageIndex{1}\): When all the thiosulfate (S₂O₃^-2) is consumed Iodine (I₂) accumulates and forms a Dark Blue Complex,  (Copyright; Belford cc.0.0)

The chemical reactions in this equation are described below. The basis of the reaction is once the iodine reaches a sufficient concentration it reacts with the starch and the system turns blue, but this can not happen until all the thiosulfate (S₂O₃^-2) is consumed.
\[\begin{align} 2Fe^{+3}(aq) + 2I^-(aq) & \rightarrow 2Fe^{+2}(aq) + {\color{red}I_2(aq)} \; \; \; \;\; \; \;\text{Relatively Slow Step}\\  {\color{red}I_2(aq)} + 2S_2O_3^{-2}(aq) & \rightarrow 2I^-(aq) + S_4O_6^{-2}(aq) \; \; \;\ \; \text{Very Fast Step}\\  {\color{red}I_2(aq)} + Starch(aq) & \rightarrow \underbrace{\color{blue}{I_2\text{-Starch Complex}}}_{ \textcolor{blue}{(\large{DARK BLUE})}} \end{align}\]

Note how the iodine (in red) produced in the first step is consumed in both the second and third steps. That is, there are two competing reactions that are trying to consume the iodine generated in the first step.  The thiosulfate (S₂O₃^-2) of the second step reacts so fast that it is considered to be an "iodine scavenger" and as long as there is thiosulfate present the iodine never accumulates to a sufficient concentration to react with the starch of the third step and turn blue. But once the thiosulfate is consumed the iodine accumulates, reacts with the starch and the solution turns blue.

Tip \(\PageIndex{1}\)

In figure 4.2.1 we are mixing two solutions together, how does this affect the concentrations?

Tip

We start with an initial molarity of each of our reagents, but when they are mixed together we have a new volume of solution. How does this affect the final Molarity of solution?

Dilutions Review

Measuring the Rate

In section 14.1 we learned that the rate of a reaction can be expressed in terms of the consumption of any reactant or the production of any product. Lets define rate on consumption of ferric ion (-\(\frac{[Fe^{+3}]}{\Delta t}\)). We need to know how much ferric iron is consumed from the moment we mixed the solutions until they turned blue. Thiosulfate (S₂O₃^-2) reacts so fast that it is considered to be an "iodine scavenger" and as long as there is thiosulfate present the iodine never accumulates to a sufficient concentration to react with the starch and turn blue. But once the thiosulfate is consumed the iodine accumulates, reacts with the starch and the solution turns blue. So we need to relate the ferrous ion concentration to the thiosulfate concentration. If we add equations 5.2 and 5.3 (which are coupled by the iodine intermediate) we get:

\[2Fe^{+3}(aq) + 2S_2O_3^{-2}(aq) \rightarrow 2Fe^{+2}(aq) + S_4O_6^{-2}(aq)\]

If we look at the relative rates (sections 14.1.4) we see

\[ R=-\frac{1}{2} \frac{\Delta [S_2O_3^{-2}]}{\Delta t}=-\frac{1}{2}\frac{\Delta [Fe^{+3}]}{\Delta t}\]

so

\[ R= \frac{\Delta [S_2O_3^{-2}]}{\Delta t}=\frac{\Delta [Fe^{+3}]}{\Delta t}\]

and since we know the thiosulfate goes from the initial concentration to zero, we can calculate the rate of ferrous ion consumption by measuring the rate of thiosulfate consumption

Note, for this to work

\[ [S_2O_3^{-2}]<<[Fe^{+3}]
\]

and so if we know the concentration of thiosulfate ([S₂O₃^-2]) and measure the time to turn blue, we can determine the rate of reaction.

 

Before proceeding with this experiment you should review section 14.3 and section 14.5.2  on the rate law and the Arrhenius equation. The rate law for this equation is

\[R = k [Fe^{+3}]^m [I^-]^n\]

If we substitute the Arrhenius Equation (\(k=e^{-\frac{R_a}{RT}}\)) we get

\[\underbrace{R}_{\textcolor{red}{R=Rate}}=\underbrace{Ae^{-\frac{E_A}{RT}}[Fe^{+3}]^m[I^-]^n}_{\textcolor{red}{\text{R in RT is the Ideal Gas Constant}}} \]

This is a multivariable equation (section 14.4.3) with one dependent, three independent variables and five constants (m,n,R,A & E_A)

R=Rate (Dependent)
[Fe⁺³] = Ferrous ion concentration (independent)
[I^-] = Iodide concentration (independent)
T = Temperature (independent)

When we make a plot we measure the affect of the independent variable on the dependent variable.  That is, for the generic function Y=fct(X) we change X and measure the value of Y.  So we need to run three sets of experiments where we successively hold two of the variables constant.

Further reading on Experimental Design Considerations in 4.5: Differential Rate Law Lab Resources