Question

What role do elementary reactions play in determining reaction order with respect to each reactant? How is the mechanism related to how a rate constant for a complex reaction compares to that of a one-step reaction?

Answer 1

IMPORTANT: There is no direct correlation between stoichiometric coefficients and rate law exponents (reactant orders) in an overall (complex) reaction! If that ever occurs, it is a coincidence!

We will denote elementary reactions or reaction steps using `=>`, and complex reactions using `->`.

Here is an example of a bimolecular ozone destruction mechanism.

`O_3(g) + Cl(g) stackrel(k_1" ")(=>) ClO(g) + O_2(g)` --- (elementary step 1)
`ClO(g) + O(g) stackrel(k_2" ")(=>) O_2(g) + Cl(g)` --- (elementary step 2)
`"----------------------------------------"`
`underbrace(\mathbf(O_3(g) + O(g) stackrel(k_("obs")" ")(->) 2O_2(g)))`
`""" "" "" "^("overall reaction")`

For this, the overall rate law has the special rate constant `k_"obs""*"`, and is written as:

`\mathbf(r(t)"*" = k_"obs""*"["O"_3]["O"])`

`"*"` This reaction has a catalyst: `"Cl"`.

Note that we do not know what the special rate constant `k_"obs""*"` actually is yet, but we do know that both `"O"_3` and `"O"` are reactants in the overall reaction that do NOT:

• appear in the middle of the reaction and disappear later (intermediates)
• disappear in the middle of the reaction and reappear later (catalysts)

So, this is a valid rate law for the overall reaction.

It does not, however, reveal what the order of each reactant is, necessarily.

These are not the same:

`O_3(g) + O(g) stackrel(k_("obs""*")" ")(->) 2O_2(g)` (1)

`r(t)"*" = k_"obs""*"["O"_3]^m["O"]^n`

`m = ?`, `n = ?`

`O_3(g) + O(g) stackrel(k_("obs")" ")(=>) 2O_2(g)` (2)

`r(t) = k_"obs"["O"_3]^m["O"]^n`

`m = n = 1`.

That's what's causing you confusion, because examining the reaction mechanism of (1) based on what you have been taught, you would say:

`O_3(g) + Cl(g) stackrel(k_1" ")(=>) ClO(g) + O_2(g)` (step 1)

• step 1 is first order in `"O"_3` and first order in `"Cl"`. This is because the mechanistic step occurs as-written. That is, if you witness this step occurring in real life for one molecule of `"O"_3` and one atom of `"Cl"`, you would see one `"O"_3` molecule colliding with one `"Cl"` atom, and deduce that this is first-order in each.

`ClO(g) + O(g) stackrel(k_2" ")(=>) O_2(g) + Cl(g)` (step 2)

• step 2 is first order in `"ClO"` and first order in `"O"`. This is because the mechanistic step occurs as-written. That is, if you witness this step occurring in real life for one molecule of `"ClO"` and one atom of `"O"`, you would see one `"ClO"` molecule colliding with one `"O"` atom, and deduce that this is first-order in each.

Basically, comparing (1) and (2), the value of `k_"obs""*"` is not the same as `k_"obs"` because they are not the same mechanism. The difference is this:

• (1) is a complex reaction, known to occur in two steps.
• (2) is an elementary reaction, known to occur in one step.

The order of each reactant is able to be determined for an overall one-step reaction.

The order of each reactant is unclear for the overall two-step reaction, but each reactant's order is able to be determined for the elementary steps.

Remember that catalysts speed up rates of reaction (like in (1)!), so since (1) has a catalyst, `r(t)"*" ne r(t)`, and therefore, `k_"obs""*" ne k_"obs"`.

Ultimately, what we have is:

• When reaction A's `k_"obs"` matches reaction B's `k_"obs"`, AND the reactants in reaction A match those in reaction B, reaction A IS reaction B.
• When one or both things do not match, reaction A and B are NOT the same reaction and the reactants' orders are not necessarily the same.