# Question #9df55

Jul 22, 2015

Here's how you can approach such problems.

#### Explanation:

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

$A + B \to M + E$

$E + F \to G + R$

$J + G \to B + K$

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.

$A + \cancel{B} + \cancel{E} + F + J + \cancel{G} \to M + \cancel{E} + \cancel{G} + R + \cancel{B} + K$

This is equivalent to

$A + F + J \to M + R + K$

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

$B \to \textcolor{b l u e}{\text{catalyst}}$

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

$M \to \textcolor{\mathmr{and} a n \ge}{\text{product}}$

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

$G \to \textcolor{g r e e n}{\text{intermediate}}$

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

$F \to \textcolor{red}{\text{reactant}}$