Motivation

Complex reaction kinetics is a fairly mathematical subject in undergraduate chemistry lectures. Although kinetic rate laws for elementary reactions are easily defined and manipulated, the math quickly becomes unwieldy in more complex situations where multiple reactions might occour simultaneously or in sequence.

Although complex reactions give rise to a number of fascinating phenonema such as oscillating reactions, the beauty of the chemistry is easily lost in lectures, which can readly turn into a form of mathematical torture.

The Chameleon reaction is an example of a complex reaction where the overall reduction of permanganate ($\ce{MnO_4^-}$) to brownestone ($\ce{MnO_2}$) proceeds through manganate ($\ce{MnO_4^{2-}}$):

\[ \ce{MnO4^- -> MnO4^{2-} -> MnO2} \]

This reaction is particularly suited for class room demonstrations as the different oxidation states of manganese yield vibrant and easily distinguishable colors. The reaction will start out with the distinct purple of permanganate, transition to green whilest manganate is formed, and then become a yellow/browne suspension of manganese oxide particles. The clear visibility of the intermediate manganate provides an impressive example of a sequential reaction with intermediate.

Experimental

The recepie is simple. Two aqueous solutions should be produced and mixed while stirring:

• 2mg potassium permanganate ($\ce{KMnO4}$) in 50ml deionised water.
• 6g glucose ($\ce{C6H12O6}$) and 10g sodium hydroxide ($\ce{NaOH}$) in 750ml deionised water.

By using less hydroxide and glucose, the reaction progresses more slowly, as shown in take two.

Discussion

Reaction mechanism

The permanganate is reduced in two steps, while the sugar is oxidised. Sugar is a carbohydrate with many secondary alcohol groups (-(H)C(OH)-). These alcohol groups are easily oxides to ketone groups (-C(=O)-) as shown in the figure above. Manganese has a large number of valence states enabling allows step-wise reduction from +7 (purple) to +6 (green) to +4 (browne).

Reaction rate (simplified)

The goal is to find a set of equations that provide the concentrations of the Mn species as a function of time. Let’s define the following symbols to save some writing

Using this abstract notation, the reaction that we discuss is: \[ \ce{A ->[k_1] B ->[k_2] C} \]

Step 1: Construct system of equations

Writing the elementary reaction rates as \[ \dot r_1 = = k_1 [{\rm A}] \] \[ \dot r_2 = = k_2 [{\rm B}] \] the following system of ordinary differential equations can be constructed: \[ \begin{align} \partial_t \ce{[A]} & = - \dot r_1 = -k_1\ce{[A]}\\ \partial_t \ce{[B]} & = + \dot r_1 - \dot r_2 = +k_1\ce{[A]} - k_2\ce{[B]} \\ \partial_t \ce{[C]} & = + \dot r_2 = +k_2\ce{[B]} \end{align} \] Note that the resulting system is a system of ordinary differential equations. Time is the only independent variable. The system is complete as we have three (linearly independent) equations for three unknown concentrations.

Step2: Solve differential equations

The strategy to solve this coupled set of differential equations mirrors the sequential nature of the reaction. The first equation can be integrated straight away and gives the usual result for an elementary first order reaction:

\[ \ce{[A]} = \ce{[A]_0} \cdot \exp\{-k_1 \cdot t\} \]

where $\ce{[A]_0}$ is the initial concentration of $\ce{A}$.

After rearranging and introduction of the result for $\ce{[A]}$, the second differential equation can be written as \[ \partial_t\ce{[B]} + k_2\ce{[B]} = k_1\ce{[A]_0}\cdot\exp\{-k_1\cdot t\} \] This is an ordinary first order inhomogeneous differential equation. Calculus provides a number of ways to solve for $\ce{[B]}$. We will play a little trick that will allow us to use integration by parts and multiply through with $e^{k_2\cdot t}$: \[ \partial_t\ce{[B]}\cdot\exp\{k_2\cdot t\} + \ce{[B]}\cdot k_2\exp\{k_2\cdot t\} = k_1\ce{[A]_0}\cdot\exp\{-(k_1-k_2)\cdot t\} \] You will notice that the left-hand side now has the form \[ \dot x\cdot y + x\cdot \dot y= \frac {d}{dt} \left( xy \right) \] which allows to write \[ \begin{align} \frac{\rm d}{\rm dt}\left(\ce{[B]}\exp\{k_2t\}\right) & = k_1\ce{[A]_0}\cdot\exp\{-(k_1-k_2)\cdot t\} \\ \int {\rm d}\left(\ce{[B]}\exp\{k_2t\}\right) & = k_1\ce{[A]_0}\int {\rm dt}\,\exp\{-(k_1-k_2)\cdot t\} \\ \end{align} \] Integration on both sides gives \[ \begin{align} \ce{[B]}\exp\{k_2t\} = k_1\ce{[A]_0} \left[ \frac{ \exp\{-(k_1-k_2)\cdot t\} }{ -(k_1-k_2) } \right]_0^t\\ \ce{[B]} = \frac{k_1}{k_2-k_1}\ce{[A]_0} \frac{ \exp\{-(k_1-k_2)\cdot t\} - 1 }{ \exp\{k_2\cdot t\} } \\ \end{align} \]

\[ \ce{[B]} = \frac{k_1}{k_2-k_1}\ce{[A]_0} \left[ \exp\{-k_1\cdot t\} - \exp\{-k_2\cdot t\} \right] \\ \]

To find an expression for $\ce{[C]}$, we do not need to solve the third differential equation (although it would be easy to do so, know that we know $\ce{[B]}$). Rather, we can invoke mass conservation and write

\[ \ce{[C]} = \ce{[A]_0} - \ce{[A]} - \ce{[B]} \]

The qualitative behaviour of $\ce{[A]}$, $\ce{[B]}$, and $\ce{[C]}$ can be studied using this graph:

It plots the relative concentration normalised with $\ce{[A]_0}$ over normalised time $t\cdot k_1$. The relative rate $k_2$ of the second reaction $\ce{B->C}$ can be adjusted with the slider above the graph.