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I was studying for a PhD entrance exam and this question made me wonder about some topics in kinetics:

Consider the multi-step reaction below:

$\ce{A_2 → 2 A}$

$\ce{2A + X → Y + B}$

$\ce{B → A2 + Z}$

Which one can't be determined from the given equations:

a) The general balanced equation

False, just by adding the reactions, we have $\ce{X -> Y + Z}$

b) The identity of the catalyst

False, A2 is the catalyst: consumed in the first step and reformed in the last

c) If an intermediate is formed

False, B is an intermediate

d) The rate law for this reaction

e) None of the above

It is not stated which step is the rate-determining step. However, since $\ce{A_2}$ is a catalyst, its decomposition can't be the slow step (it doesn't make sense to have a catalyst that slows the reaction). The second step is termolecular and forms an intermediate, which makes it a strong candidate for rate-determining step. B is the decomposition of an intermediate, which should be fast considering that intermediates in reactions are usually unstable species that are quickly consumed.

My main question is: can we determine the rate law for the reaction with the given information?

Secondary questions, which could be added as separate questions, would be:

  • Can the decomposition of a catalyst be the slow step?
  • Is the formation of an intermediate always slower than its consumption? In other words, are there examples that don't follow the Steady State Approximation (considering simple chemical reactions only, not enzyme-catalyzed)?
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    $\begingroup$ I think the criteria for something to be a catalyst is that it reduces the speed of the overall reaction. So, in theory, it should be possible to have a catalyst where the decomposition of the catalyst is the slow step, as long as it is faster than the uncatalysed reaction. However, I can't think of any example at the moment. (A case which might be similar to this is the catalysis of ammonia synthesis by Fe, where the binding of nitrogen with catalyst is the slow step) $\endgroup$
    – S R Maiti
    Commented Jan 25, 2021 at 18:57
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    $\begingroup$ As for your second question, I think there can be reactions where the formation of intermediate is faster than its consumption (particularly in the case of low energy intermediates). $\endgroup$
    – S R Maiti
    Commented Jan 25, 2021 at 19:01
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    $\begingroup$ $\ce{A_{2}}$ is a catalyst if and only if the rate of the reaction is increased by using it. In other words, without it, $\ce{X->Y + Z}$ is slower. $\endgroup$
    – Zhe
    Commented Jan 25, 2021 at 21:05
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    $\begingroup$ "it doesn't make sense to have a catalyst that slows the reaction" That's true for the overall reaction. The problem is that the overall reaction is $\ce{X->Y + Z}$, so when you add $\ce{A2}$, as long as the overall reaction is faster, then $\ce{A2 -> 2A}$ can be the slow step. $\endgroup$
    – Zhe
    Commented Jan 25, 2021 at 21:08

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