Question

1. examine the reaction mechanism found below:

a + b₂ → ab₂ (slow)
ab₂ + c → ab₂c (fast)
ab₂c + c → b₂c₂ + a (fast)
b₂c₂ → d + e (fast)
overall equation:

a. draw a circle around any reaction intermediates.
b. draw a box around any species that can be considered a catalyst.
c. in the box above, provide the overall equation for this process.
d. in the box below, provide the rate law based on this mechanism.
rate law:

Explanation:

Step1: Identify reaction intermediates

Reaction intermediates are species that are formed in one step and consumed in a subsequent step. In the given mechanism, \(AB_{2}\), \(AB_{2}C\), and \(B_{2}C_{2}\) are formed in one step and consumed in another.

Step2: Identify catalyst

A catalyst is a species that is consumed in an early step and regenerated in a later step. In the given mechanism, \(A\) is consumed in the first step and regenerated in the third step.

Step3: Determine overall equation

Add all the steps together and cancel out the intermediates (\(AB_{2}\), \(AB_{2}C\), \(B_{2}C_{2}\)) and the catalyst (\(A\)):

$$\begin{align*} A + B_{2}&\to AB_{2}\quad(\text{slow})\\ AB_{2}+C&\to AB_{2}C\quad(\text{fast})\\ AB_{2}C + C&\to B_{2}C_{2}+A\quad(\text{fast})\\ B_{2}C_{2}&\to D + E\quad(\text{fast})\\ \hline 2C + B_{2}&\to D + E \end{align*}$$

Step4: Determine rate law

The rate - determining step is the slow step. The rate law is based on the reactants of the slow step. The slow step is \(A + B_{2}\to AB_{2}\). So the rate law is \(rate = k[A][B_{2}]\)

Answer:

a. Reaction intermediates: \(AB_{2}\), \(AB_{2}C\), \(B_{2}C_{2}\)
b. Catalyst: \(A\)
c. Overall equation: \(2C + B_{2}\to D + E\)
d. Rate law: \(rate = k[A][B_{2}]\)