Auto-catalysis Mechanism

Question

I just read about the theory of auto-catalysis, and here's one thing which is sort of unclear to me:

In this type of catalysis, one of the reaction product catalyses the reaction. For example, in the oxidation of oxalic acid by potassium permanganate, Mn(II) cation is formed which is known to accelerate the reaction. So when potassium permanganate is added to a warm solution of oxalic acid in presence of dilute sulphuric acid (acid medium catalysis), initially there is a time lag before decolourisation occurs. As more permanganate is added, the decolourisation is observed to be nearly instantaneous.

Points I didn't understand:

1. How does the formation of Mn(II) cation accelerate the reaction? Could someone please help me with understanding the mechanism?

2. What is the reason for time lag in decolourisation of the solution? Also, what exactly do they mean by decolourisation here? My guess is that it's referring to the bright purple coloured permanganate solution losing its colour after getting reduced to Mn(II) cation.

3. How does the time lag reduce? Are there ways to estimate or determine the same? Of course, it'd depend on the reaction kinetics; but I'm looking for a mathematical viewpoint on how the catalyst influences reaction kinetics, except for decreasing the activation energy.

Thank you.

Answer 1

(1) There are several web sites giving details of this reaction and so this need not be copied out here.

(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B $\to$ P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).

If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;

$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$

which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,

$$ \ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)} \right| \right) =-(a_0+b_0)kt$$

and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}} $. This and also the increase of $B =a_0+b_0-x$ are plotted.

Figure Autocatalysis $\ce{A + B $\to$ P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.