# How to tell whether a reaction is an elementary reaction

BrianC12
How can you tell whether a reaction is an elementary reaction without knowing the reaction mechanism? I had this question on a recent midterm and got it wrong:

A -> Product

Is this an elementary reaction?

I said yes because it has no intermediates. In retrospect there's no way I could know that unless I knew the reaction mechanism. The only other information given is that the reaction is zero order. Thanks for any help!

## Answers and Replies

That Neuron
BY checking to see if the stochiometric coefficients equal the exponents in a chemical equilibrium equation.

That Neuron
aA + bB -----> cC + dD, where all lower case letters are coefficients and all capitals are the Chemical formulae of each reactant/product. So in a chemical equilibrium equation, r=k[A]^a (^b)]/[[C]^c ([D]^d)]. If the exponents found to match the chemical reaction differ from this, the reaction is composed from two or more elementary steps. [] is notation for the concentration in molarity mass of solute/volume of solution.

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BrianC12
So that would mean this is NOT an elementary reaction because the coefficient of 1 for A doesn't match the power of 0 for the equilibrium equation? But if that's true, then doesn't that imply that any zero order reaction isn't an elementary reaction since a coefficient > 0 can obviously never equal 0.

BrianC12
Ok I'm just confused now. The second part of the question asked to write the differential rate law. I wrote -d[A]/dt = k[A]^0 = k and got full credit. If this isn't an elementary reaction then I wouldn't be able to write the rate law. But according to the TA who graded my test, there are intermediates in the reaction, meaning it can't be an elementary reaction...

Homework Helper
Gold Member
Not elementary my dear Brian :tongue:

If the reaction is zero order it cannot be an elementary reaction.

Looks like there is no other participant, just A goes to something.

In an elementary reaction all the A molecules - say they split into something - have the same chance of doing so. So the actual rate of reaction - number of molecules of A that change into something else per second - is proportional to A. I think you can write the rate law and name it.

But with zero order kinetics - that means say you put in 10 times more A, yet only as many as before react per second. Each molecule does not have the same chance of reacting as before. That strongly suggests that in order to react they need to interact with something else, e.g. a catalyst, before they can split. Not elementary.

Zero order kinetics are typical of reactions involving solid catalysts or enzymes or some free radical catalyses when you get above a certain concentration (i.e. saturation).

BrianC12
Ahh alright that makes sense, thanks. I actually checked with my friend's midterm. He wrote "this is not an elementary reaction because elementary reactions are characterized by collisions with other molecules"(that's the definition in the book) and got full credit. I don't really understand how that explains anything since first of all unimolecular elementary reactions don't involve any collisions and your explanation implies that there must be a collision, which my friend's answer contradicts. I asked him and he said he didn't actually know, he just saw a previous midterm with a similar question where the person got it wrong, so he switched the answer and wrote the definition from the book. Is that an insufficient answer or am I just missing something?

On top of that doesn't the book's definition contradict itself too? Because it states that definition and then goes on to show an example of a unimolecular reaction where an energized molecule of N2O5 dissociates into NO2 and NO3, which is just a "reaction" at high temperature that does not involve any collisions.

I hope I don't come across as arrogant for doubting everything, in my mind it just seems like a lot of the information I've been getting from professors, TA's, and the book contradict each other and I'm trying to figure out what's right and what's wrong.

Again, thanks for the help!

Homework Helper
Gold Member
I think your reasoning is OK. I don't think there is this contradiction. When I said "all the A molecules - say they split into something..." your unimolecular N2O5 dissociation is an example of that.