Help with Kinetics 2 Step Reactions?

Consider the following two-step reaction.
NO2(g) + SO2(g) → NO(g) + SO3(g)
2 NO(g) + O2(g) → 2 NO2(g)
Identify which component is a catalyst and which is a reaction intermediate. Please Explain your answer.

The .................................... is a catalyst because it enables the reaction to move forward but is not consumed in the overall reaction.
The..................................... is a reaction intermediate because it is produced in one step of the reaction and consumed in the next.

Please try to write a few sentences of explanation. Thank you for your knowledge.

1 Answer
Feb 14, 2018

From the two-step mechanism, the overall reaction adds up to be:

#cancel("NO"_2(g)) + "SO"_2(g) → cancel("NO"(g)) + "SO"_3(g)#
#ul(cancel(2) "NO"(g) + "O"_2(g) → cancel(2) "NO"_2(g))#
#"NO"(g) + "SO"_2(g) + "O"_2(g) -> "NO"_2(g) + "SO"_3(g)#

From Hess's Law, you can add up the reaction, and the same reactant and product on both sides cancel. For the overall reaction,

#"NO"#, #"SO"_2#, and #"O"_2#

are all reactants, and

#"NO"_2# and #"SO"_3#

are products.

  • One equivalent of #"NO"# was made in step 1 and consumed in step 2 as a reactant. That makes it an intermediate. It has to be gone by the time the reaction cycle is over.

  • One equivalent of #"NO"_2# was consumed in step 1 and comes back as a product in step 2. That makes it a catalyst. It has to come back to react again in the next reaction cycle.

As I said before, this isn't a great example, because you have one #"NO"# being a reactant and one #"NO"# being an intermediate, while you have one #"NO"_2# being a product and one #"NO"_2# being a catalyst. That mixes up the roles of the substances...

How about:

#color(green)(cancel("Cl"(g))) + "O"_3(g) -> color(purple)(cancel("ClO"(g))) + "O"_2(g)#

#ul(color(purple)(cancel("ClO"(g))) + "O"_3(g) -> color(green)(cancel("Cl"(g))) + 2"O"_2(g))#

#2"O"_3(g) -> 3"O"_2(g)#

What are the catalyst and intermediate here?