Question #9df55

1 Answer
Jul 22, 2015

Answer:

Here's how you can approach such problems.

Explanation:

First, list all the three reactions that are a part of the three-step mechanism.

#A + B -> M + E#

#E + F -> G + R#

#J + G -> B + K#

To write the overall reaction, simply add these three chemical equations together. The net reaction will always be the sum of the elementary steps given to you.

#A + cancel(B) + cancel(E) + F + J + cancel(G) -> M + cancel(E) + cancel(G) + R + cancel(B) + K#

This is equivalent to

#A + F + J -> M + R + K#

Now take a look at the four species given to you, #B#, #M#, #G#, and #F#.

Notice that #B# starts off on the reactants' side and ends up unchanged on the products' side. This is characteristic of a catalyst, which is a chemical species that simply speeds up the reaction without being consumed in the overall process. So I'd say that

#B -> color(blue)("catalyst")#

On to #M#, which is produced by the first step of the reaction and is left unchanged throughout the ensuing two steps. This means that #M# is actually a product of the overall reaction, so I'd say

#M -> color(orange)("product")#

Now take a look at #G#. This compound is produced by the second step of the reaction and is immediately consumed in the third step.

This tells you that #G# is a very reactive compound in the context of this reaction, which, in addition to the fact that it's produced and consumed by the reaction, makes #G# match the characteristics of an intermediate species. I'd say that

#G -> color(green)("intermediate")#

FInally, compound #F#. Notice that this compound is being added to the reaction in the second step, in which it reacts with #E#, another intermediate, to form #G# and #R#, which is a product of the overall reaction.

This means that #F# is actually a reactant, so

#F -> color(red)("reactant")#