55. For an autocatalytic second order reaction R → P, the rate law is [where v is rate of the reaction and k is the rate constant]

(A) v = k[R]

(B) v = k[R][P]

(C) v = k[R]²

(D) v = k[P]²

Autocatalytic Second Order Reaction Rate Law – Complete Explanation with Reaction Mechanism

Unlike ordinary reactions where catalysts are added externally, an autocatalytic reaction generates its own catalyst during the course of the reaction. As the concentration of product increases, the reaction rate also increases, producing the characteristic S-shaped kinetic curve. Understanding this behavior is essential for predicting the correct rate law.

Understanding an Autocatalytic Reaction

An autocatalytic reaction is one in which one of the products acts as a catalyst for the same reaction. Initially, the reaction proceeds very slowly because almost no product is present. As the reaction progresses, more product is formed, and this newly formed product accelerates the conversion of reactant into additional product.

Since both the reactant and the product influence the reaction rate, the rate depends simultaneously on the concentration of R and P.

Rate Law for an Autocatalytic Second-Order Reaction

For the reaction

R → P

where the product acts as the catalyst, the rate law is

v = k[R][P]

The reaction is first order with respect to the reactant and first order with respect to the product.

Therefore, the overall order of the reaction becomes

1 + 1 = 2

This is why the reaction is called an autocatalytic second-order reaction.

Why Product Appears in the Rate Law

In ordinary chemical reactions, products usually do not influence the reaction rate. However, in an autocatalytic reaction, the product itself behaves as the catalyst. Consequently, increasing the concentration of product increases the reaction rate.

This unique feature distinguishes autocatalytic reactions from conventional first-order and second-order reactions.

Initially, when the concentration of product is very small, the reaction proceeds slowly. As product accumulates, the rate increases rapidly. Near completion, the reactant concentration becomes low, causing the rate to decrease again. This produces the characteristic sigmoid (S-shaped) kinetic profile of autocatalytic reactions.

Explanation of Every Option

Option (A): v = k[R]

This option is incorrect because it represents a simple first-order reaction. In this equation, the rate depends only on the concentration of reactant, while the catalytic role of the product is completely ignored. Therefore, it cannot describe an autocatalytic reaction.

Option (B): v = k[R][P]

This is the correct answer. The reactant supplies the reacting molecules, while the product functions as the catalyst. Consequently, both concentrations appear in the rate equation. The reaction is first order with respect to reactant and first order with respect to product, giving an overall reaction order of two.

Option (C): v = k[R]²

This option is incorrect because it describes a conventional second-order reaction involving two reactant molecules. It does not account for the catalytic action of the product, which is the defining feature of autocatalysis.

Option (D): v = k[P]²

This option is also incorrect. Although the product participates in catalysis, the reaction cannot occur without the reactant. A rate law depending only on the product concentration is chemically unrealistic because the reactant concentration must influence the reaction rate.

Concept Behind Autocatalysis

Autocatalytic reactions differ from ordinary catalytic reactions because the catalyst is not added externally. Instead, it is generated during the reaction itself. As soon as a small amount of product is formed, it catalyzes the conversion of additional reactant into more product. This positive feedback causes the reaction rate to accelerate before eventually slowing as the reactant is consumed.

This behavior explains why autocatalytic reactions often exhibit induction periods followed by rapid reaction rates and finally a gradual decrease as the reaction approaches completion.

Final Answer

Correct Option: (B)

Rate Law: v = k[R][P]