Question

The rate equation for the reaction $2AB + B_2 \rightarrow 2AB_2$ Rate = k[AB][$B_2$] The possible mechanism consistent with this rate law is

• A. $2AB + B_2 \overset{Slow}{\longrightarrow} 2AB_2$
• B. $AB + AB \overset{Fast}{\rightleftharpoons} A_2B_2$ $A_2B_2 + B_2 \overset{Slow}{\longrightarrow} 2AB_2$
• C. $AB + B_2 \overset{Slow}{\longrightarrow} AB_3$ $AB_3 + AB \overset{Fast}{\longrightarrow} 2AB_2$
• D. $AB + B_2 \overset{Fast}{\rightleftharpoons} AB_3$ $AB_3 + AB \overset{Slow}{\longrightarrow} 2AB_2$

Answer 1

Answer: (C)

C

Answer 2

The rate law for a multi-step reaction is determined by its slowest step, known as the rate-determining step (RDS). If an intermediate is involved in the RDS, its concentration must be expressed in terms of reactants using the equilibrium of a fast preceding step.

The given rate equation is: Rate = k[AB][$B_2$]

Let's analyze each proposed mechanism:

A. $2AB + B_2 \overset{Slow}{\longrightarrow} 2AB_2$

This is a single-step reaction. If this is the slow step, the rate law would be: Rate = k[AB]$^2$[$B_2$]

This does not match the given rate law.

B. Step 1: $AB + AB \overset{Fast}{\rightleftharpoons} A_2B_2$

Step 2: $A_2B_2 + B_2 \overset{Slow}{\longrightarrow} 2AB_2$

The rate-determining step is Step 2. The rate law for this step is: Rate = k'[A$_2$B$_2$][$B_2$]

From the fast equilibrium in Step 1, the equilibrium constant $K_{eq}$ is: $K_{eq} = \frac{[A_2B_2]}{[AB]^2}$

So, $[A_2B_2] = K_{eq}[AB]^2$.

Substitute this into the rate law: Rate = k'($K_{eq}[AB]^2$)[$B_2$] = k''[AB]$^2$[$B_2$]

This does not match the given rate law.

C. Step 1: $AB + B_2 \overset{Slow}{\longrightarrow} AB_3$

Step 2: $AB_3 + AB \overset{Fast}{\longrightarrow} 2AB_2$

The rate-determining step is Step 1. The rate law for this step is directly: Rate = k'[AB][$B_2$]

This matches the given rate law (where k = k'). The intermediate $AB_3$ is formed in the slow step and consumed in the fast step, which is consistent.

D. Step 1: $AB + B_2 \overset{Fast}{\rightleftharpoons} AB_3$

Step 2: $AB_3 + AB \overset{Slow}{\longrightarrow} 2AB_2$

The rate-determining step is Step 2. The rate law for this step is: Rate = k'[AB$_3$][AB]

From the fast equilibrium in Step 1, the equilibrium constant $K_{eq}$ is: $K_{eq} = \frac{[AB_3]}{[AB][B_2]}$

So, $[AB_3] = K_{eq}[AB][B_2]$.

Substitute this into the rate law: Rate = k'($K_{eq}[AB][B_2]$)[AB] = k''[AB]$^2$[$B_2$]

This does not match the given rate law.

Therefore, only mechanism C is consistent with the given rate law.