2nd order autocatalytic reaction

An autocatalytic reaction is one in which a product catalyses the reaction. An example is the Mn2+-catalysed oxidation of oxalic acid by potassium manganate (VII):

2MnO_4^{\; -}+16H^++5C_2O_4^{\; 2-}\rightarrow 2Mn^{2+}+10CO_2+8H_2O

The equation can be reduced to A + P → 2P, or simply, A P, with the rate law:

\frac{d[A]}{dt}=-k[A][P]\; \; \; \; \; \; \; \; 13a

Using the same logic described in a previous article, we can rewrite the rate law as:

\frac{dx}{dt}=k(a-x)(p+x)

where a = [A0] and p = [P0].

Integrating throughout, we have

\int_{0}^{x}\frac{dx}{(a-x)(p+x)}=k\int_{0}^{t}dt\; \; \; \; \; \; \; \; 14

Substituting the partial fraction expression \frac{1}{(a-x)(p+x)}=\frac{1}{a+p}\left ( \frac{1}{a-x}+\frac{1}{p+x} \right ) in eq14 and after some algebra, we have,

kt=\frac{1}{a+p}ln\frac{a(p+x)}{p(a-x)}

which is equivalent to

kt=\frac{1}{[A_0]+[P_0]}ln\frac{[A_0][P]}{[P_0][A]}

 

Question

The mechanism of the above reaction is found to include the following steps:

A+P\rightleftharpoons AP

AP\rightarrow P+P\; \; \; \; \; \; (rds)

With this in mind, derive the rate law and show that it is consistent with eq13a.

Answer

We can write the rate law as:

\frac{d[P]}{dt}=2k_{rds}[AP]=2k_{rds}K[A][P]=k[A][P]

where K is the equilibrium constant for the 1st step and k = 2krdsK. Furthermore, the overall reaction is AP, which means that

\frac{d[P]}{dt}=-\frac{d[A]}{dt}=k[A][P]

 

 

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