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.

  • $\begingroup$ B seems to me like a catalyst, and C is an intermediate. $\endgroup$
    – M.A.R.
    Commented Jun 17, 2015 at 15:46

1 Answer 1


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.


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