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 produced and then consumed in the reaction mechanism. In the given mechanism, \(AB_2\), \(AB_2C\), \(B_2C_2\) are produced in one step and consumed in subsequent steps.

Step2: Identify catalyst

A catalyst is a species that is consumed in an early step and regenerated in a later step. Here, \(A\) is consumed in the first step (\(A + B_2
ightarrow AB_2\)) and regenerated in the third step (\(AB_2C + C
ightarrow B_2C_2+A\)).

Step3: Determine overall equation

Add up all the steps:

$$
LATEXBLOCK0
$$

Cancel out species that appear on both sides (\(A\), \(AB_2\), \(AB_2C\), \(B_2C_2\)). The overall equation is \(2C + B_2
ightarrow D + E\).

Step4: Write rate law

The rate - determining step is the slow step (\(A + B_2
ightarrow AB_2\)). The rate law is based on the reactants of the slow step. So the rate law is \(Rate = k[A][B_2]\)

Answer:

a. Reaction intermediates: \(AB_2\), \(AB_2C\), \(B_2C_2\) (circle these)
b. Catalyst: \(A\) (box this)
c. Overall equation: \(2C + B_2
ightarrow D + E\)
d. Rate law: \(Rate = k[A][B_2]\)