# 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?

Jul 12, 2016

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 $\implies$, and complex reactions using $\to$.

Here is an example of a bimolecular ozone destruction mechanism.

${O}_{3} \left(g\right) + C l \left(g\right) \stackrel{{k}_{1} \text{ }}{\implies} C l O \left(g\right) + {O}_{2} \left(g\right)$ --- (elementary step 1)
$C l O \left(g\right) + O \left(g\right) \stackrel{{k}_{2} \text{ }}{\implies} {O}_{2} \left(g\right) + C l \left(g\right)$ --- (elementary step 2)
$\text{----------------------------------------}$
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 ${k}_{\text{obs""*}}$, and is written as:

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

$\text{*}$ This reaction has a catalyst: $\text{Cl}$.

Note that we do not know what the special rate constant ${k}_{\text{obs""*}}$ actually is yet, but we do know that both ${\text{O}}_{3}$ and $\text{O}$ 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:

${O}_{3} \left(g\right) + C l \left(g\right) \stackrel{{k}_{1} \text{ }}{\implies} C l O \left(g\right) + {O}_{2} \left(g\right)$ (step 1)

• step 1 is first order in ${\text{O}}_{3}$ and first order in $\text{Cl}$. This is because the mechanistic step occurs as-written. That is, if you witness this step occurring in real life for one molecule of ${\text{O}}_{3}$ and one atom of $\text{Cl}$, you would see one ${\text{O}}_{3}$ molecule colliding with one $\text{Cl}$ atom, and deduce that this is first-order in each.

$C l O \left(g\right) + O \left(g\right) \stackrel{{k}_{2} \text{ }}{\implies} {O}_{2} \left(g\right) + C l \left(g\right)$ (step 2)

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

Basically, comparing (1) and (2), the value of ${k}_{\text{obs""*}}$ is not the same as ${k}_{\text{obs}}$ 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, $r \left(t\right) \text{*} \ne r \left(t\right)$, and therefore, ${k}_{\text{obs""*" ne k_"obs}}$.

Ultimately, what we have is:

• When reaction A's ${k}_{\text{obs}}$ matches reaction B's ${k}_{\text{obs}}$, 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.