# Which species can't appear in the rate expression for this reaction scheme?

Question
Which species can't appear in the rate expression for this reaction scheme?
Here is the reaction scheme in steps: \begin{align} \ce{A + B~(slow) &\rightarrow C} \\ \ce{C~(fast) &\rightarrow D + B}\\ \ce{D + E~(fast) &\rightarrow F}\\ \ce{F~(fast) &\rightarrow G}\\ \end{align}

The answers are apparently F and G, but I don't see why D and E couldn't also be valid (considering the first step is the slowest). A and B will inevitably be in the rate equation since they are reactants in the rate determining step. C produces B (alongside D) and so will also be in the rate equation as a result. However, why can't we rule out D and E of being in the rate equation, considering they don't seem to have any effect on the concentrations of A and B.

• B seems to me like a catalyst, and C is an intermediate. – M.A.R. Jun 17 '15 at 15:46

Answering this question requires assuming that all steps are "elementary", e.g. that $\ce{A + B -> C}$ is an irreversible bimolecular reaction that is first-order in $[\ce{A}]$ and first-order in $[\ce{B}]$. (Obviously we are also assuming that activities can approximated by concentrations.)

With that assumption, the fact that $\ce{C}$ conversion is "fast" means that we can replace the existence of $\ce{C}$ in the scheme entirely:

\begin{align} \ce{A + B~(slow) &\rightarrow D + B} \\ \ce{D + E~(fast) &\rightarrow F}\\ \ce{F~(fast) &\rightarrow G}\\ \end{align}

Likewise if $\ce{F}$ conversion is "fast", then it can be replaced:

\begin{align} \ce{A + B~(slow) &\rightarrow D + B} \\ \ce{D + E~(fast) &\rightarrow G}\\ \end{align}

Hopefully this makes it clear why $\ce{E}$ must appear in any rate law for formation of $\ce{G}$: there is no source of it in the two reactions we are left with.

For $\ce{D}$, I agree that the situation is a bit more complex. It might appear in the rate equation, or it might not. Necessary reactants at the beginning of the reaction if we are ever to see product $\ce{G}$ are $\ce{A}$, $\ce{B}$, and $\ce{E}$, so it would make sense to eliminate $\ce{D}$ from the rate law if possible. However, that could be algebraically difficult, or perhaps $\ce{D}$ is easier to measure, etc. There is no reason $\ce{D}$ can't appear in the rate law.

Huge caveat: "fast" and "slow" are relative terms and as such without further context they are ambiguous. The problem as constructed sounds a bit sloppy. For example, is the "fastness" of $\ce{F -> G}$ faster or less fast than the fastness of $\ce{D + E -> F}$? The problem doesn't say but the simplifications I did above actually depend on these sorts of details.