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?

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.