Chapter 10: Catalysis and Catalytic Reactors

Frequently Asked Questions (FAQs)

In this chapter we have an irreversible reaction: T + H₂ → M + B
and we have a table that says

All the intermediate steps show reversible reactions so why do we call the reaction irreversible?
Because the overall equilibrium constant for the reaction is very large.
It seems like sometimes you add species on the denominator when they are weakly adsorbed. Can you define weakly adsorbed species?
It means that you know it has to be on the surface for the reaction to occur, but the value of K_AP_A is small with respect to 1.0, i.e. K_AP_A<<1.
What other applications does chemical vapor deposition have?
Making diamonds.
Please tell us an industrial reaction where the catalyst is completely regenerated.
Catalytic cracking and most reactions where coking is the decay mechanism.
Is it possible for one step to be rate limiting under one set of conditions (concentrations, temperature, etc.) and another step to be rate limiting under other conditions? If so, how would the rate law be modeled?
Yes, you can do this with the same as we do in a computer program using if/then statements.
What methods are used to prevent catalyst decay?
Remove poisoning agents such as sulfur and lead, and operating at temperatures below 40% of the melting temperature of the catalyst.
It says "catalyst support becomes soft and flows, resulting in pore closure." What does soft and flow have to do with pore closure? All I understand is that flow can cause pore closure, but what does "soft" mean?
It means it becomes like soft butter and flows to close off the pores.
Can you ever have a catalytic reaction where the reaction is a decomposition reaction instead of just an isomerization?
Yes.
If so, are all of the products bound to the active site after reaction?
No, some of the products could go directly into the gas phase.
I've heard of octane ratings that are greater than 100, how can this be?
The fuels have a compression rate at which the knock intensity is exceeded that is greater than that for normal octane
Why is it that for more than 75% of heterogeneous reactions that aren't diffusion limited is it that surface reaction is the rate liming step?
The adsorption steps are usually rapid. At high temperatures the rate limiting step is usually diffusion rather than surface reaction.
In a home problem we're supposed to find the a resulting form the maximal profit. Shouldn't the maximal profit occur when the catalyst is completely used up at the end of the reactor. Obviously if the activity, a, is >0 at the exit, then the catalyst isn't being used at it's optimum. However, if the catalyst is 0 before the end of the reactor, shouldn't that be a sign of inefficiency?
No. It costs money to pass the catalyst through the reactor.